E XAMKRACKERS
MeAT
VERBAL REASONING & MATHEMATICAL TECHNIQUES 7 TH EDITION
O St}TE PUBLISHING
Acknowledgements Although I am the author, the hard work and expertise of many individuals contributed to this book. The idea of writing in two voices, a science voice and an MeAT voice, was the creative brainchild of my imaginative friend Jordan Zaretsky. I would like to thank David Orsay for his help with the verbal passages. I wish to thank my wife, Silvia, for her support during the difficult times in the past and those that lie ahead. Finally, I wish to thank my daughter Julianna Orsay for helping out whenever possible. '
Copyright © 2007 Examkrackers. Inc
TABLE OF CONTENTS INTRODUCTION: INTRODUCTION TO MCAT INCLUDING MCAT MATH •....•..........•...•......•..••.•.. 1 intro.1
The Layout of the MCAT .................................................................... ...... .. .............................. 1
intro.2 The Writing Sample .................................................................. .. ..................... .. ....... .. .............. 3 intro.3 How to Approach the Science Passages .......... ........ ... ............ .... .... .......... .. ............. ............... .4 intro.4 MCAT Math ................................................................ ...................... .. ............. .. ....... ................ 5 intro.5 Rounding ........................ ........ ......... .................. .. ...... .. ...... .. ..... ........ ... ...... .. .... ........ ................. 5 intro.6 Scientific Notation ............................................................................................ ........................ 9 intro.7
Multiplication and Division ................................ ...... .. ............................... ........ ..................... 10
intro.S Proportions .............................................................. ........ ........ .. ...... .. .................................... 13 intro.9 Graphs .................... ........ ......... ........ ........ .. ....... ................ .................... ............. ... ...... ............ 16 LECTURE
1:
STRATEGY AND TACTICS ................................................................................ 19
1.1
The Layout of the Verbal Reasoning Section ........ ......... ........ ........ ................ ....................... 19
1.2 Other Verbal Strategies ............................................... ................................ ............... ............ 19 1.3 Take Our Advice .............................................................. ........ ........ ........ ......... ..................... 20 1.4 Expected Improvement ............................................................................ ......... .. ................... 20 1.5 The ExamKrackers Approach to MCAT Verbal Reasoning ............ ......... .......... .. ...... ............. 20 1.6 Tactics .......................................................................................... ................................ .......... 23 LECTURE
2:
ANSWERING THE QUESTIONS ........................................................................ 27
2.1
Tools to Find the Answer ...................................................................................................... 27
2.2 Answer Choices ...................................................................................................................... 34 2.3
Identifying the Correct Answer ........................................... ..................... .......... .......... .......... 35
2.4 Simplification of the Question and Answer Choices ............... ........................... .. .................. 36 2.5 Marking Your Test to Improve Your Score ........................................ ................. ........... ......... .44 2.6 When to Bubble .................................................. ............. ................... ....... .. ........................ .44 LECTURE
3:
THE MAIN IDEA ..........................................................................................45
3.1
The Main Idea ........................................................................................................................45
3.2 Constructing the Main Idea ...................... ........ ... ..... ........ ........ ........ ......................... ............ .45 3.3 Confidence .............................................. ......... .. ..... ........ ........ ........ ........ ......... ......... ........... .46 3.4 Know Your Author ............................................... .............. .. ..... ......... .. ................................... 46 3.5
Ignore the Details and See the Big Picture ...... ....... .. ........................ .. ...... ........... .......... ........ 46
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LECTURE 4:
How TO
STUDY FOR THE VERBAL REASONING SECTION .................................. 53
30-MINUTE IN-CLASS EXAMS ........................................................................................55 In-Class Exam for Lecture 1 ......... ........ ........ ......... ....... ......... ....... ..•...... " ..... .................. ......... 55 In-Class Exam for Lecture 2 ......... ........ ......... ........ ................ ................. ...... .. ........ ......... ........ 63 In-Class Exam for Lecture 3 .........•................•...... ................. ........ ................. ........ ................. 71
ANSWERS
&
EXPLANATIONS TO IN-CLASS EXAMS ............................................................ 79 Answers and Scaled Score Conversion for In-Class Exams .................................................. 80 Explanations to In-Class Exam for Lecture 1 ....................................... ........................... ........ 81 Explanations to In-Class Exam for Lecture 2 ................ ......................... ................................. 90 Explanations to In-Class Exam for Lecture 3 ................ ..•....... .• ..... .••......•.......•. ........ .............. 99
Copyright © 2007 Exarn krackers, Inc.
Introduction to MCAT Including MCAT Math i.1
MeAT Format
At the time this book is published, MCAT administration will be completely computerized. Although many students bemoaned this change, the computerized MCAT is actually a much better option for you! The many advantages of this format are listed in the table below. In addition to making the actual test computer-based, AAMC has also made several administrative changes to decrease test duration, increase the number of available test days, and speed up score reporting.
Advantages of Computerized Format
• It is easier to retest with 20 more test adnunistration
Environment
Time
Registering
• The test is 30% shorter
•
The computerized test day is about half as long due to
dates less administrative require• You can no~etest up to 3 ments times a year • New afternoon sessions are • You can monitor YOUT own great for those w ho struggle breaks within the given time in the morning hours limit • Scores are reported twice as • Weekday administration allows you to not have to fast (now 30 days) ruin a weekend
• The test is 30% shorter • TIle test taking environment is controlled for climate and sound
• The testing groups are • •
•
smaller Ergonomic chairs Noise reduction headsets available Lockers and locks provided for personal belongings
Even better news is that the computerized version refl ects the same topics, uses the same scoring system, and costs the same as the paper version. Also, you can still review and make changes w ithin each section. So for anyone that has previously taken the MCAT, or already started studying, the test should still be familiar. Similarly, strategies and tools for successfully navigating the MCAT will remain the same.
In addition to the above, AAMC is working to reduce score reporting to 14 days. They are also investigating teclmology which may allow you to make notations directly on the computer screen. Note that this comp uterized test is NOT currently a computerized adaptive test (CAT) like the GRE, meaning that everyone gets the same test questions for any given v ersion of the MeAT. However, this is open to change in the future.
,
2
VERBAL REASONING
&
MATHEMATICAL TECHNIQUES
The MCAT consists of four sections: 1.
Physical Sciences
2.
Verbal Reasoning
3.
Writing Sample
4.
Biological Sciences
Physica l Sciences This section covers topics from undergraduate physics and inorganic chemistry. Passages average approximately 200 words in length and are often accompanied by one or more charts, diagrams, or tables. Generally there are 6-10 questions following each passage, as well as 3 sets of stand-alone multiple-choice questions, for a total of 52 questions. The top score on the Physical Sciences Section is a 15.
Verbal Reasoning The Verbal Reasoning Section has been shortened (to everyone's pleasure) from 85 to 60 minutes. It now consists of only 40 (previously 60) multiple-choice questions with answer choices A through D. There are 9 passages followed by 4 to 10 questions each. Passages average approximately 600 words in length. There is a wide variety of passage topics ranging from economics and an thropology to poetic analysis, and most intentionally soporific. The top score on the Verbal Reas.oning Section is also 15.
Writing Sample The Writing Sample Section consists of two 30 minute periods, without any break in between. For each essay, the test-taker is given a general statement to analyze in
a standard cookie-cutter fashion. This section is scored on an alphabetic scale from
J to T, with T being the highest score. This scale translates to a score of 1-6 on each essay resulting in a combined score of a 2-12 represented by J through T.
Bio logical Sciences The Biological Sciences Section covers science topics from a wide range of undergraduate biology topics, organic chemistry and genetics. The set up of this section is exactly the same as the Physical Sciences section, as is the scori ng. ""~,
Test Section Tutorial ) I
Physical Sciences Break Verbal Reasoning Break Writing Sample Break Biological Sciences
Questions
Time Allotted
(Optional)
10 minutes
52
70 minutes
(Optional)
10 minutes
40
60 minutes
(Optional)
10 minutes
2
60 minutes
(Optional)
10 minutes
52
70 minutes
Time/Question - 1.35 minutes (81s) 1.5 minutes (90s) 30 minutes/essay -1.35 minutes (81s)
10 minutes
Survey
Total Content Time
4 hours, 20 minutes
Total Test Time
4 hours, 50 minutes
Total Appointment Time
5 hours, 10 m inutes
Copyright © 2007 Examkrackers, Inc.
INTRODUCTION TO THE MCAT INCLUDING MCAT MATH . 3
i.2
The Wri t ing Sample
Please Note: The section that follows includes material from the MCAT Practice Test !II. These materials are reprinted with permission of the Association of American Medical Colleges (AAMC). In the U.S., your writing sample score is unlikely to affect w hether or not you gain admittance to medical schooL Curren tly, medical schools do not give this section m uch weight in their decision making process . Medical schools do no t see your actual writing sample. They only see your score. The writing sample functions to wear you down for the Biological Sciences Section.
111e writing sample is more of an exercise in following directions than it is a test of your ability to write. You should no t attempt to be creative on the writing sample or try to make your reader reflect deeply. Instead, follow the simple three s tep process given below. Two sample statements are given w ith each step followed by an example of how you r essay should appear for that statement. A similar set of d irections is always given with each statement. Don't waste time reading the d irections on the real MCAT. The d irections can be summarized into the following three step process: 1.
Explain the statement as thoroughly as possible using an example to clarify. Statement: An understanding of the past is necessary for solving the problems of the present. Paraphrase: History is an integral part of tile learning process. By studying the past, we can analyze repercussions of certaiu behavior and action patterns. Statem ent: No matter holV oppressive a government r violent revollttion is
never justified. Paraphrase: The fa miliar idiom "He who lives by the sword shall die by the sword", is echoed in any statement that condemns violence. It is fI very simple principle based on a very logical argument. Violence invites more of the same. If a government is overthrown by violent means, then there is a precedent set and there is nothing stopping others froln doing the same again. Do not: begin your essay with the statement "so and so" lIIeans that.. 2.
Give a specific example contradicting the statement. Statement: An understanding of the past is necessary for solving the problems of the present. Example: On the other hand, some problems exist today that are totally independent of any historical event. The current 'sslle of AIDS .. Statement: No matter how oppressive a government, violent revolution is never justified. Example: However, there can be times when extreme action becOllles necessary. It was the violence of the Russian revolution that brought an end to the suffering of the masses during WWI. D o n ot: Llse controversial topics as examples, sllch as abortion or contemporary
political issues. 3.
Give a guideline that anyone might qse_to determine when the statement is true and when it is false. Statement: An understanding of the past is necessan) for solving the problems of the present. G uidelin e: When then is the past crllcial to our lInderstanding of the current events? It is important only, and especially, when it relates to the present situntion ..
Copyright © 2007 Examkrackers, Inc.
4
VERBAL REASON ING
&
MATHEMATICAL TECHNIQUES
Statement: No matter how oppressive a government, violent revolution is never justified. Guideline: Whether or not violent r<cualution is justified depends upon whether some form of oppression is lifted from the masses. Spend the first 5 minutes of the essay writing an outline of these three steps. Write 2 pages. Be sure to finish your essay. The outline should help you do this. Above all, write neatly. Use proper grammar correctly. Don't misspell words. Don't use words if you are not certain of the meaning. Historical eXaInples are much bet~ ter than personal examples; "Martin Luther King said ... " is a much better example than "My mother always said ... " To think of examples, think of wars or famous people. Feel free to paraphrase liberally: "Socrates once said that he was the smartest man because he understood how little he really knew." This is not an accurate quote; it is an acceptable paraphrase. Socrates said something like this, and this is close enough.
i.3
How to Approach Science Passages
The following guidelines should be followed when working an MeAT science passage: 1.
Read th e passage first. Regardless of your level of science ability, you should read the passage. Passages often give special conditions tha t you would have no reason to suspect without reading and which can invalidate an otherwise correct answer.
2.
R ead quickly; do not try to master the information given in the pass age. Passages are full of information both useful and irrelevant to the adjoining questions. Do not waste time by attempting to gain complete understanding of the passage.
3.
Quickly check tables, g raphs, and charts. Do not spend time studying tables, graphs, and charts. Often, no questions will be asked concerning their content. Instead, quickly check headings, titles, axes, and obvious trends.
4.
When multiple hypothes es or experiments are posited, make note of obvious contrasts in the m argin alongside the respective paragraphs. Making note in the margin w ill accomplish two things. First, it will distinguish firmly in your mind each of the hypotheses or exp·eriments. (At least one question w ill require such discernment.) Second, by labeling them you prevent confusion and thus obviate rereading (and avoid wasting precious time).
5.
Pay close attention to detail in the questions. The key to a question is often found in a single word, such as "net force" or "constant velocity".
6.
Read answer choices immedi ately, before doing calculations. Answer choices give infornlation. Often a question that appears La require exlensive calculations can be solved by intuition or estimation due to limited reasonable answer choices. Sometimes answer choices can be eliminated for having the wrong units, being nonsensical, or other reasons.
7.
Fill in your answer grid question by question as you go. This is the best way to a void bubbling errors. This method avoids time w asted trying to find your place. The posited reason for doing differently is that you can relax your brain while you transfer your answers. Try it. It's not relaxing. In fact, if you do relax, you are likely to make errors.
8.
If time is a factor for you, skip the questions and/or passages that you
find d ifficult. If you usually do not finish this section, then make sure Copyright © 2007 Examknckers, Inc.
INTRODUCTION TO THE MeAT INCLUDING MeAT MATH . 5
that you at least answer all of the easy questions. In other words, guess at the difficult questions and come back to them if you have time. Be sure to make time to answer all of the free-standing questions. The free-standing questions are usually easier than those based on passages. By the time you ha ve finished this course, you should not need to skip any questions.
i.4
MeAT Math
MeAT math will not test your math skills beyond the contents 01 this book. The MCAT does require knowledge of the follow ing up to a second year high school algebra level: ratios; proportions; square roo ts; exponents and logarithms; scientific notation; quadratic and simultaneous equations; graphs. In addition, the MCAT tests: vector addition, subtraction; basic trigonometry; very basic probabilities. The MCAT does not test dot product, cross product or calculus. Calculators are neither allowed on the MCAT, nor would they be helpful. From this moment until MCAT d ay, you should do all math problems in your head whenever possible. Do not use a calculator, and use your pencil as seldom as possible, w hen you do any math. If you find yourself doing a lot of calculations on the MCAT, it's a good indication
that you are doing something wrong. As a rule of thwnb, spend no more than 3 minutes on any MCAT physics question. Once you have spent 3 minutes on a question with no resolution, yo u should stop what you're doing and read the question again for a simple answer. If you don' t see a simple answer, yo u should make your best guess and move to the next question.
i.S
Rounding
Exact numbers are rarely useful on the MCAT. In order to save time and avoid errors when making calculations on the test, use round numbers. For instance, the gravitationa l constant g should be rounded up to 10 mis' for the purpose of calculations, even when instructed by the MCAT to do otherwise. Calculations like 23.4 x 9.8 should be thought of as "something less than 23.4 x 10, which equ als something less than 234 or less than 2.34 x 10 2 " Thus if you see a question requiring the above calculations followed by these answer choices:
A. B. C. D.
1.24 x 1.8 1 x 2.28 X 2.35 X
la' 10' 10' 10'
An swer is something less than 23.4 x IO ~ 234.
Wrong way
Right way
answer choice C is the closest answer urder 2.34 x 10' , and C should be chosen qUickly without resorting to complica ted calculations. Rarely will there be two possible answer choices close enough to prevent a correct selection after rounding. If two answer choices on the MCAT are so close that you lind you have to write down the math, it' s probably because you've made a mistake. If yo u find yourself in that situation, look again at the question for a simple solution. If you don't see it, g uess and go on. Copyright © 2007 Examkrackers, Inc.
6
VERBAL REASONING
&
MATHEMATICAL TECHNIQUES
It is helpful to remain aware of the direction in which you have rounded. In the above example, since answer choice D is closer to 234 than answer choice C, you may have been tempted to choose it. However, a quick check on the direction of rowlding would confirm that 9.8 was roWlded upward so the answer should be less than 234. Again, assuming the above calculations were necessary to arrive a t the an~ sw er, an answer choice w hich would prevent the use of roWld ing, like 2.32 x 10' for instance, simply would not appear as an answer ch oice on a real MCAT. It would not appear for the very reason that such an answer ch oice wo uld force the test taker to spend time making complicated calculations, and those aren ' t the skills the MCAT is d esigned to test. If a series of calculations is used where rounding is perfo rmed a t each step, the rounding errors can be compounded and the resulting answer can be useless. For instance, we may be required to take the above exa mple and further divide "23.4 x 9.8" b y 4.4. We might roWld 4.4 do wn to 4, and div ide 240 by 4 to get 60; how ever, each of our ro Wldings w ould ha ve increased our result compounding the error. Instead, it is be tter to roWld 4.4 up to 5, dividing 235 by 5 to get 47. This is closer to the exact answer of 52.11 82. In an attempt to increase the accuracy of mul~ tiple estima tions, try to compensate for upward rounding with downward rounding in the same calculations. Don't hurt yourself with complicated calculations!
N otice, in the example, that w hen we increase the d enOlni..nator, we are d ecreasing the entire term . Fo r instance: 625 = 26.042
625 = 25
24 25 Rounding 24 up to 25 results in a decrease in the overall term. When rounding squares remember that you are really rounding twice. (2.2)' is re~ ally 2.2 x 2.2, so wh en we say that the answer is som ething greate r than 4 we n eed to keep in mind tha t it is significantly greater because we have rounded dow n tw ice. One way to increase your accuracy is to roWld just one of the 2.2s, leaving you wi th some thing grea ter than 4.4. This is much closer to the exact answer of 4.84. Another strategy for roWld ing an exponential term is to remember tha t diffic ult~to~ solve exponential terms must lie behveen two easy-ta-sol ve exponential terms. Thus 2.2' is between 2' and 3' , closer to 2'. This strategy is especially helpful for square roots. The square root of 21 must be between the square root of 16 and the square root of 25. Thus, the MCAT square root of 21 must be between 5 and 4 or about 4.6.
.fi5 = 5 ..fil =? ,116 = 4 For more comp lica ted roots, recall that any root is si mply a fractional exponent. For instance, the square root of 9 is the same as 9 ' J'. This m eans tha t the fourth root of 4 is 411'. This is the same as (4 ' /2)'/2 or,fi . We can combine these techniques to solve even more comp licated roots:
,
l{i7 = 3
4 3 =V4' =ifl6 = ?=2.51
'18 =2 It's worth your time to m em orize .f2 ~ 1.4 and .J3
~ 1.7. Copyright © 2007 Examkrackers, Inc.
INTRODUCTION TO THE MCAT INCLUDING MCAT MATH . 7
The MCAT is likely to give yo u any values that you need for trigonometric functions; however, since MCAT typically uses common angles, it is a good idea to be familiar w ith trigonometric values for common angles. Use the paradigm below to remember the values of common angles. Notice that the sine values are the reverse of the cosine values. Also notice tha t the numbers under the radical are 0, 1, 2, 3 and 4 from top to bottom for the sine function and bottom to top for the cosine function, and all are divided by 2.
e
sine
cosine
0°
ro2
14
30°
.J1
-!3
2
2
45°
.J2
.J2
2
2
60°
-!3 - -
.J1 --
90"
14
ro
2 2
2
2 2
Less practiced test takers may perceive a rounding strategy as risky. On the contrary, the test makers actually design their answers with a rounding strategy in mind. Complica ted numbers can be intimidating to anyone not comfortable with a rounding strategy.
Copyright © 2007 Examkrackers, Inc.
8 .
VERBAL REASONING
&
MATH EMATICAL TECHNIQUES
Questions
Solve the following problems by rounding. Do not lise n pencil or n enlculator.
1.
5.4 x 7.1 x 3.2
4.6'
A. B. C.
D. 2.
2.2 3.8 5.8 7.9
.,/360 x 9.8 6.2
A. B. C.
D.
9.6 13.2 17.3 20.2
(F2) X23 50
3.
A. B. C.
D.
0.12 0.49 0.65 J.J
4.
A. B. C.
D. 5.
II 39 86 450
~2X9.8 ' 49
A. B. C. D.
0.3 0.8 1.2 2
'66,,6'r 51 laM5ue pexa
a4~
'pallOJ
SJ a
's
'(;08<;'6 511aMSUU pexa
al{~
'pallOJ
sf V
'r
',86[<; 511aMsue pexa
a4~
'paUOJ
sf :::J
'1
SJaMSU\f
Copyright © 2007 Examkrackers, Inc.
INTRODUCTION TO TH E MCAT INCLUDING MCAT MATH . 9
i.6
Scientific Notation
One important math skill tested rigorously by the MeAT is your ability to use scientific notation. In order to maximize your MeAT score, you must be familiar with the techniques and shortcuts of scientific notation. Although it may not seem so, scientific notation was designed to make math easier, and it does. You should practice the following techniques until you come to view scientific notation as a valuable ally. This manual will define the terms in scientific notation as follows:
exponential term~ exponent
~XI0 4 mantissa
base
Magnitude: You should try to gain a feel for the exponential aspect of scientific notation. 10-<3 is much greater than 10-12 . It is 10,000 times greater! Thus, when comparing one solution whose concentration of particles is 3.2 x 10-11 mol/L with a second solution whose concentration of particles is 4.1 x 10-9 mol/L, you should visualize the second solution as hundreds of times more concentrated than the first. Pay special attention to magnitudes when adding. For example try solving: 3.74x10-15
+ 6.43 X 10-3 On the MeAT, the answer is simply 6.43 x 10-3 . This is true because 6.43 x 10-3 is so much greater than 3.74 x 10-1' that 3.74 x 10-15 is negligible. Thus you can round off the answer to 6.43 x 10-3 • After all, the exact answer is 0.00643000000000374. Try solving: 5.32 X 10-4
x 1.12 X 10-13 The MeAT answer is something greater than 5.3 x 10-17 • We cannot ignore the. smaller number in this case because we are multiplying. In addition or subtraction, a number 100 times smaller or more can be considered negligible. This is not true in multiplication or division. The fastest way to add or subtract numbers in scientific notation is to make the exponents match. For instance: 2.76x10 1
+ 6.91 x 10' The MeAT answer is something less than 7.2 x 105 • To get this answer quickly we match the exponents and rewrite the equation as follows: 2.76 x 10'
+ 69.1x10' This is similar to the algebraic equation: 2.76y
+ 69.1y Copyrigh t © 2007 Examkrackers, Inc.
10
V ERBAL REASONI NG
&
MATHEMATICAL T ECH NIQUES
where y = 10'. We simply add the coefficients of y. Rounding, we have 3y + 69y = 72y. Thus 72 x 10' , or 7.2 x 10' is the answer. When rearranging 6.91 x 10' to 69.1 x 10', we simply multiply by 10/ 10 (a form of 1). In other words, we divide 72 by 10 and multiply 10' by 10.
xlO
~X~=69.1 XI04 .,. 10 A useful mnemonic fo r remembering which way to move the decimal point when we add or subtract from the exponent is to use the acronym LARS,
i.7
Multiplication and Division
When multiplying similar bases with exponents add t.he exponents; when dividing, subtract the exponents. 10' x 10' = 109 10'/10-6 = 10 10 When multiplying or dividing with scientific notation, we deal with the exponential terms and m antissa separately, regardless of the nllmbet of terms. For instance: (3.2 x 10') x (4.9 x 10.... ) (2.8 x 10- 7 )
should be rearranged to: 3x5 10' xlO-s -x _
3
10 '
giving us an MeAT answer of something greater than 5 x 10' . (The exact answer, 5.6 x 10', is greater than our estimate because we decreased one term in the numerator by more than we increased the other, which would result in a low estimate, and because we increased the term in the denominator, w hich also results in a low estimate.) When taking a term written in scientific notation to some power (such as squaring or cubing it), we also deal with the decimal and exponent separately. The MeAT answer to: (3.1
X
10')2
14
is something greater than 9 x 10 • Recall that when taking an exponential term to a power, we multiply the exponents. The first step in taking the square root of a term in scientific notation is to make the exponent even. Then we take the square root of the mantissa and exponential term separa tely. .JS.l x 10' Make the exponent even. .JS1x10'
Copyrig ht © 2007 Exam krackers, Inc.
INTRODUCTION TO THE MCAT INCLUDING MCAT MATH • 11
Take the square root of the mantissa and exponential term separately.
.,J8i x M = 9 xl02 Notice how much more efficient this method is. What is the square root of 49,000? Most students start thinking about 700, or 70, or something with a 7 in it. By using the scientific notation method, we quickly see that there is no 7 involved at all. .J49,000 x.J4.9xl0 4 =2.1xl02
Try finding the square root of 300 and the square root of 200.
Copyright © 2007 Examkrackers, Inc.
12 .
VERBAL REASONING
&
MATHEMATICAL TECHNIQUES
Questions Solve the following problems without a calculator. Try not to use a pencil.
1.
2.3 X la' x 5.2 x 10-5
A. B. C.
.D. 2. (2.5
A. B. C. .D. 3. [(1.1 A. B.
c.' D.
1.2 X 10-1 2.8 3.1 x 10 5.6 X 102 10-7 x 3.7
X
1.3 5.1 4.2 1.3 X
X X
X X
X
X
10'
I!
1010- 10 10' 10 15
10-4) + (8.9
1.1 X 1.4 X 1.8 X 2.0 X
10-<5) + 4.2
11
X
10-5)]1/2
10-2 10-2 10-' 10-'
4. 112(3.4 X 102 ) (2.9 X 108)' A. 1.5 X 10 18
B. C. D.
3.1x1018 1.4 X 1019 3.1 X 10 19
1.6 X 10-19 x 15
5.
36 2 A. B. C.
D.
1.9 X 10-21 2.3 X 10-17 1.2 X 10-9 3.2 X 10-9
',-OT
x
LOU'T
S! JaMSUB pBXa al.J.L 'jJaJJOJ S! \I
"£
'jUBOy)lril!S\If alB SJaqumu J"l[IO al[L 'paJJOJ S!;) 'r
SJ&MSUV
Copyright © 2007 Examkrackers, Inc.
INTRODUCTION TO THE MCAT INCLUDING MCAT MATH . 13
i.8
Proportions
On the MeAT, proportional relationships between variables can often be used to circumvent lengthy calculations or, in some cases, the MeAT question simply asks the test taker to identify the relationship directly. When the MeAT asks for the change in one variable due to the change in another, they are making the assumption that all other variables remain constant. In the equation F = rna, we see that if we double F while holding rn constant, a doubles. If we triple F, a triples. The' same relationship holds for m and F. This type of relationship is called a direct proportion.
x2
I=ml
x2
F and a are directly proportional. Notice that if we change the equation to F = rna + b, the directly proportional relationships are destroyed. Now if we double F while holding all variables besides a constant, a increases, but does not double. In order for variables to be directly proportional to each other, they must both be in the numerator or denominator when they are on opposite sides of the equation, or one must be in the numerator while the Other is in the denominator when they are on the same side of the equation. In addition, all sums or differences in the equation must be contained in parentheses and multiplied by the rest of the equation. No variables within the sums or differences will be directly proportional to any other variable. If we examine the relationship between m and a, in F :::: rna, we see that when F is held constant and rn is doubled, a is reduced by a factor of 2. This type of relationship is called an inverse proportion. Again the relationship is destroyed if we add b to one side of the equation. In order for variables to be inversely proportional to each other, they must both be in the numerator or denominator when they are on the same side of the equation, or one must be in the numerator while the other is in the denominator when they are on opposite sides of the equation. In addition, all sums or differences in the equation must be contained in parentheses and multiplied by the rest of the equation. No variables within the sums or differences will be directly proportional to any other variable.
F=/zl x2
x2
m and a are inversely proportional. If we examine a more complicated equation, the same rules apply. However, we have to take care when dealing with exponents. One method to solve an equation using proportions is as follows. Suppose we are given the following equation: APnr
4
Q= 8TJL This is Poiseuille's Law. The volume flow rate Q of a real fluid through a horizontal pipe is equal to the product of the change in pressure t.P,1t, and the radius of the pipe to the fourth power divided by 8 times the viscosity TJ and the length L of the pipe.
r"
Water (TJ = 1.80 X 10-3 Pa s) flows through a pipe with a 14.0 cm radius at 2.00 Lis. An engineer wishes to increase the length of the pipe from 10.0 m to 40.0 m withCopyright © 2007 Examkrackers, Inc.
14
VERBAL REASONING
&
MATHEMATICAL TECHNIQUES
Water (11 = 1.80 x 10-3 Pa s) flows through a pipe with a 14.0 em radius at 2.00 L/s. An engineer wishes to increase the length of the pipe from 10.0 m to 40.0 m without changing the flow rate or the pressure difference. What radius must the pipe have?
A. B. C.
D.
12.1 em 14.0 em 19.8 em 28.0 em
Answer: The only way to answer this question is with proportions. Most of the information is given to distract you. Notice that the difference in pressure between the ends of the pipe is not even given and the flow rate would have to be converted to m 3 /s. To answer this question using proportions, multiply L by 4 and r by x. Now pullout the 4 and the x. We know from that, by definition, Q = M"r' / 811L; thus, x' / 4 must equal 1. Solve for x, and this is the change inthe radius. The radius must be increased by a factor of about 1.4. 14 x 1.4 = 19.6. The new radius is approxirnately19.6 crn. The closest answer is C.
Q
M,,(xr)' 811( 4L)
x=.[i
Copyright © 2007 Examkrackers, Inc.
INTRODUCTION TO THE MCAT INCLUDING MCAT MATH • 15
Questions: 1. The coefficient of surface tension is given by the equation y = (F - mg)l(2L), where F is the net force necessary to pull a submerged wire of weight mg aud length L through the surface of the fluid in question. The force required to remove a submerged wire from water was measured and recorded. If an equal force is required to remove a separate submerged wire with the same mass but twice the length from fluid x, what is the coefficient of surface tension for fluid x. (Yw",,, = 0.073 mN/m)
A. B. C. D.
0.018 0.037 0.073 0.146
mN/m mN/m mN/m mN/m
2. A solid sphere rotating about a central axis has a moment of inertia
4. The kinetic energy E of an object is given by E = Ih mv' where m is the object's mass and v is the ve1ocity,of the object. If the velocity of an object decreases by a factor of 2 what will happen its kinetic energy? A. B. C. D.
Kinetic energy Kinetic energy Kinetic energy Kinetic energy
will increase by a factor of 2. will increase by a factor of 4. will decrease by a factor of 2. will decrease' by a factor of 4.
5. Elastic potential energy in a spring is directly proportional to the square ofthe displacement of one end of the spring from its rest position while the other end remains fixed. If the elastic potential energy in the spring is 100 J when it is compressed tq half its rest length, what is its energy when it is compressed to one fourth its rest length. A. B. C. D.
50J 150 J 200 J 225 J
I =~ MR2 3
where R is the radius of the sphere and M is its mass. Although Callisto, a moon of Jupiter, is approximately the same size as the planet Mercury, Mercury is 3 times as dense. How do their moments of inertia compare? A. B. C. D.
The moment of inertia for Mercury is 9 times greater thau for Callisto. The moment of inertia for Mercury is 3 times greater than for Callisto. The moment of inertia for Mercury is equal to the moment of inertia for Callisto. The moment of inertia for Callisto is 3 times greater than for Mercury.
3. The force of gravity on an any object due to earth is given by the equation F = G(moM/?) where G is the gravitational constant, M is the mass of the earth, mf> is the mass of the object and r is the distance between the center of mass of the earth and the center of mass of the object. If a rocket weighs 3.6 x 106 N at the surface of the earth what is the force on the rocket due to gravity when the rocket has reached an altitude of 1.2 x 104 km? (G = 6.67 X 10-11 Nm'/kg', radius of the earth = 6370 km, mass of the earth = 5.98 x 1024 kg)
A.
1.2 x 10sN
B. C. D.
4.3 x 105 N 4.8 X 10' N 9.6 X 10' N
.
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2007 Examkrackers, Inc.
16
VERBAL REASONING
&
MATHEMATICAL TECHN IQUES
i.9
Graphs
The MCAT requirE," that you recognize the graphical relationship between two variables in certain types of equations. The three graphs below are the most commonly used. You should memorize them. The first is a directly proportional relationship; the second is an exponential relationship; and the third is an inversely proportional relationship.
t
t
y
y
..
x y ~
Note: n is greater than zero for the graph of y ~ nx, and n is greater than one for the othertwo gra phs.
y
I must have been multiplied by a negative constant.
x
..
x-
nx
y~
IIX'
Notice that, if we add apositive constant b to the right side, the graph is simply raised vertically by an amount b. If we subtract a positive constant b from the right side, the graphs are shifted downwarQs.
t
t
As long as the value of n is within the given parameters, the general shape of the graph will not change.
y
y
y
y
b
b-l-- -
b
x
..
y ~ nx + b
x
•
..
x y ~ 1/,,' + b
When graphs are unitless, multiplying the right side of an equation by a positive constant w ill not change the shape of the graph. If one side of the equation is negative, or multiplied by a negative constant, the graph is reflected across the x axis. Whenever the M CAT asks you to identify the graphical relationship between two variables you should assume that all other variables in the equation are constants unless told otherwise. Next, manipulate the equation into one of the above forms (with or without the added constant b, and choose the corresponding graph. If you are unsure of a graphical relationship, plug in 1 for all variables except the variables in the question and then plug in 0, 1, and 2 for x and solve for y. Plot your results and look for the general corresponding shape.
Copyright @ 2007 Examkrackers, Inc.
INTRODUCTION TO THE M CAT INCLUDING M CAT M ATH . 17
3. Which of the following graphs shows the relationship between frequency and wavelength, of electromagnetic radiation through a vacuum? (c = VA)
Questions
1. The height of an object dropped from a building in the absence of air resistance is gi ven by the equation h = ho + Vof + 1/2 gr, where ho and v are the initial height and velocity respectively and g is - 10 m/s'- If Y, is zero which graph best represents the relationship betwee n h and I?
C.
A.
t
Q
;>
C.
A.
t
t
..,
j~
1----
D.
B.
A____
D.
B.
..:::
t ____
;>
A----
h,
ho
/
t t ;>
A____ A- - - -
h"
'!<>
1 ----
1----
4. Which of the following gra phs best describes the magnitude of the eleclrostatic fo rce F = k( qq)/r created by an object with negative charge on an object with a positive charge as the distance r between them changes?
C.
A. 2. Whi ch of the following graphs best describes the magnitude of th e force (F) on a spring obeying Hooke'; law (F - ktu) as it is compressed to dxOliiX?
C.
A.
t tJ.x
r ____
t
....
.... ~
tJ.x
Llxmax
D.
B.
1
1
=
D.
B.
----
~mat
1
1 r ____
.... Ax
----
Ax mu
Copy righ t © 2007 Examk rackers. Inc.
tJ.x
----
Llx'mu
r ____
r----
':)UI 's)a:p e)~w ex 3 LOOl' @ +116!lAd o J
Answers
1. A is correct. Since v, is zero we have h = h, + l/2gf. Since g is in the opposite direction to h, and ho is a constant we can rew rite this equation as h =_1/2 gl' + h, where g = 10. This is the same form as y = X'. The negative sign flips the graph vertically. In addition a constant has been added to the right side so the graph intercepts the y axis at h,. 2. A is correct. The question asks for magnitude. Thus the negative sign is ignored and the equation has the form y = nx. 3. D is correct. Manipulation of thls formula produces v =ciA.. Which is in the form of y = Ilx. 4. D is correct. The form of this equation is y = IIX'. The negative can be ignored because the question asks for magnitude. 5. D is correct. The equation has the form y = nx wh ere n is l i t.
_ _M
.-M
~
~ '(1
·s
_ _M ' .
.-M
~ ':J
'V
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8~
Strategy and Tactics
1.1
The Layout of the Verbal Reasoning Section
The Verbal Reasoning Section of the MeAT is composed of nine passages, averaging 600 words per passage. Generally, a passage discusses an area from the humanities, social sciences, or natural sciences. Six to ten multiple-choice questions follow each passage for a total of 60 questions. Answers to these questions do not require information beyond the text of the passage. The test taker has 85 minutes to complete the entire section.
1.2
Other Verbal Strategies
Dogma about the Verbal Section is abundant and free, and that's an accurate reflection of its value. There are many cock-a-mamie verbal strategies touted by various prep companies, academics, and well-wishers. We strongly suggest that you ignore them. Some test prep companies design their verbal strategy to be marketable (to make money) as opposed to being efficient (raise your score); the idea being that unique and strange will be easier to sell than commonplace and practicaL Desperate techniques such as note taking and skimming are prime examples. Some colleges offer classes designed specifically to improve reading comprehension in the MeAT Verbal Section. Typically, such classes resemble English 101 and are all but useless at improving your score. They are often taught by well-meaning humanities professors who have never even seen a real MeAT verbal section. Being a humanities professor does not qualify you as an expert at the MeAT Verbal Section. The emphasis in such classes is usually on detailed analysis of what you read rather than how to eliminate wrong answers and find correct answers. Improvements are predictably miserable. There are those who will tell you that a strong performance on the verbal section requires speed-reading techniques. This is not true. Most speed-reading techniques actually prove to be an impediment to score improvements by shifting focus from comprehension to reading technique. It is unlikely that you will improve both your speed and comprehension in a matter of weeks. As you will soon see, speed is not the key to a good MeAT verbal score. Finishing the Verbal Section is within the grasp of everyone, if they follow the advice posited by this book. A favorite myth of MeAT shtdents is that copious amounts of reading will improve scores on the Verbal Section. This myth originated years ago when one prep company having insufficient verbal practice materials suggested to their students to "read a lot" rather than use the other cOlnpanies materials. The Inyth has perpetuated itself ever since. "Reading a lot" is probably the least efficient method of
20
VERBAI~ REASONING
&
MATHEMATiCAL TECHNIQUES
improving your verbal score. If you intend to take the MCAT four or five years hence, you should begin "reading a lot". If you want to do well on the verbal this year, use the strategies that follow.
1.3
Take Our Advice
Most smart students listen to advice, then pick and choose the suggestions that they find reasonable while disregarding the rest. This is not the most efficient approach for preparing to take the MCAT Verbal Section. In fact, it is quite counter productive. Please abandon all your old ideas about verbal and follow our advice to the letter. Don't listen to your friends and f?mily. They are not experts at teaching students how to score well on the MCAT Verbal Reasoning Section. We are.
1.4
Expected Improvement
Taking the MCAT verbal section is an art. (Not exactly what a science major wants to heaL) Like any art form, improvement comes gradually with lots of practice. Imagine attending a class in portraiture taught by a great artist. You wouldn't expect to become a Raphael after your first lesson, but you would expect to improve after weeks of coaching. The verbal section is the same way. Follow OUf directions to the letter, and with practice you will see dramatic improvements over time.
1.5
The EK Approach to MeAT Verbal Reasoning
We shall examine the verbal section on two levels: strategic and tactical. The strategic point of view will encompass the general approach to the section as a whole. The tactical point of view will explain exactly what to do, passage by passage, and question by question.
Strat e gy There are four aspects to strategy: 1.
Energy
2.
FocllS
3.
Confidence
4.
Timing
Energy Pull your chair close to the table. Sit up straight. Place your feet flat on the floor, and be alert. This may seem to be obvious advice to some, but it is rarely followed. Test-takers often look for the most comfortable position to read the passage. Do you really believe that you do your best thinking when you're relaxed? Webster's Dictionary gives the definition of relaxed as "freed from or lacking in precision or stringency." Is this how you want to be on your MeAT? Your cerebral cortex is most active when your sympathetic nervous system is in high gear, so don't deactivate it by relaxing. Your posture makes a difference to your score. One strategy of the test writers is to wear you down with the verbal section before you begin the biology section. You must mentally prepare yourself for the tremendous amount of energy necessary for a strong performance on the verbal section. Like an intellectual athlete, you must train yourself to concentrate for long periods 'of time. You must improve your reading comprehension stamina. Practice! Practice! Practice ! always under timed conditions. And always give 100% effort when you practice. If you give less than 100% when you practice, you will be teaching yourself to relax when you take the verbal section, and you will be lowering your score. It is more productive to watch TV than to practice with less than complete effort. If Copyright @ 2007 Exarnkrackers. Inc.
LECTURE
you are not mentally worn after finishing three or mo re verbal passages, then you have not tried hard enough, and you have trained yourself to do it incorrectly; you have lowered your score. Even when yo u are only prac ticing, sit up straight in your chair and attack each p assage.
Focus The verbal section is made up of nine passages with both interesting and boring topics. It is sometimes difficult to switch gears from " the migration patterns of the Alaskan tit-mouse" to "economic theories of the post-Soviet Union. " In other w ords, sOIlJetimes you ma y be reading one passage w hile thinking ab9ut the prior passage. You must lea rn to focus your attention on the task at hand. We will d iscuss methods to increase your focus when we discuss tactics. During the real MCAT, it is not unlikely tha t unexpected interruptions occur. People get physically ill, nervous students brea the heavily, air conditioners break down, and ligh ts go out. Your score w ill not be adjusted for unwelcome interruptions, and excuses will not get you into med school, so learn to focus and ignore distractions.
Confidence There are two aspects to confidence on the Verbal Section: 1) b e confiden t of yo ur score and 2) be arrogant when you read . Imagine taking a m ultiple choice exam and narrow ing 50% of the questions d own to just two answer choices, and then guessing. On a physics exam, this would almost certainly indicate a very low grade. Yet, this exact situation describes a stellar performance on the Verbal Section of the MCAT. Everyone of whom we know that has earned a perfect score on the Verbal Section (in cluding many of our own students) has guessed on a large portion of the answers. The test writers are aware that mos t students can predict their grade on science exaln s based up on their perfonnance, and that guessing makes science majors extremely uncomfortable. The Verbal Section is the most dissatisfy ing in term s of perceived performance. You should realize that even tlle best test takers finish tlle Verbal Section with some frustration and insecurity concerning their performance. A perceived dissatisfac tory performance early in the testing day is likely to reflect poorly in scores on the Biology Section. You should ass ume that yo u have guessed correctly on every answer of the verbal section and get psyched to ace the Biological Sciences Secti on. The second aspect of confidence concerns how you read the passage. Read the passages as if you were a Harvard p rofessor grading high school essays. Read critica lly. If you are confu sed while reading the passage, assume that it is the passage writer, and not you, who is at fault. If you find a contradiction in the reasoning of the argument, trust your reasoning ability that you are correct. The questions w ill focus on the author 's argument an d you must be confident of the strong and weak points. In order to identify the strong and weak points, yo u must read with confidence, even arrogance.
Tim ing If you want a 10 or better on the Verbal Section, yo u must read every passage and attempt to answer every question. If yo u want to go to med ical school, you should attempt to score 10 on the Verbal Section. Therefore, read every passage in the order giv en, and attemp t every question.
Skippin g around in the Verbal Section to find the easiest passages is a marketable stra tegy for a prep compan y but an obvious waste of time for you. It is a bad idea that makes a lot of money for some p rep companies because it's an easy trick to sell. 'Cherry pi cki ng' is an unfortunate carry over from SAT strategy where it works beCopyright © 2007 Examkrackers, Inc-
1:
STRATEGY AND TACTICS .
21
22 .
VERBAL REASONING
&
MATHEMATICAL TECHNIQUES
cause the questions are prea rranged in order of difficulty. On the MCAT, some passages are difficult to read, but have easy questions; some passages are easy to read, but have difficult questions. Some passages start out difficult and finish easy. You have no way of knowing if a passage is easy or difficult until you have read the entire passage and attempted all the questions, so 'cherry picking' lowers your score. If you begin reading a passag~ and are asking yourself "Shall I continue, or shall I move on to the next p assage? Is this passage easy or difficult?", then you are not reading with confidence; you are not concentrating on what the author is saying; and yo u are wasting valuable time. Your energy and focus should be on doing well on each passage, not on trying to decide which passage to do first.
•
Out of Order
Hmmm. Let's see. I must knock down all nine blocks. Is it faster and more efficient to Imock them down in order, or is it faster to decide which one is heaviest and then run back and forth and knock them down out of order" In Order Check your 't imer only once during the Verbal Section. Constantly checking your timer is distracting an9- not very useful since some passages take longer to finish tha n others. Instead, check your timer only once, and after you have finished the 'fifth ,passage. You should have about 40 minutes left. A well-practiced test taker will develop a sense of timing acute enough to obviate looking at a timer at all. Don't spend too much time with the difficult questions. G uess at the difficult questions and Dlove on. Guessing is very difficult for science students, who are accustomed to being certain of the answer on an exam or getting the answer w rong. Test writers are aware of this, and use it to their advantage. You should learn to give up on difficult questions so that you have more time on easier questions. Accurate guessing on difficult questions is one of the keys to finishing the exam and getting a perfect score. To accurately guess, you must learn to use all your tools fo r answering the questions. We wi ll discuss this w hen we discuss tactics. Many test-takers are able to guess on difficult questions during a practice exam , but when it comes to the real MeAT, they wa nt to be certain of the answers. This meticulous approach has cost such students dearly on their scaled score. Learn to guess at difficult Questions so you have time to
Finish the entire section with two minutes to spare, no more, no less. If you have more than hvo Ininutes to spare, you missed questions on wh ich you could have spent more time. Finishing the exam early and returning to d ifficult questions is not a good strategy. The stress of exam taking actually makes you more perspicacious while yo u take the exam. When you finish an exam, even if you intend to go back and check your wo rk, you typically breathe a sigh of relief. Upon doing so, you lose yo ur perspicacity. The best strategy is to use your time efficiently during yo ur first and only pass through the exam.
;:m~wp.r
the easy questions.
Some people have difficulty finishing the exam. These people often think that they ca n't finish because they read too slowly. This is not the case. In tactics, we will discuss how fini shing the exam does not depend upon reading speed and that anyone can finish the exam.
1.6
Tactics
Although, at first glance, it may not appear so, the following techniques are designed to increase your pace and efficiency. Tactics is where many students begin to pick and choose a verbal method that they think best suits their own personality. Please don't do this. Follow these steps exactly for each passage and afte r much Copyright © 2007 Examkrackers, !nc.
LECTURE 1 : STRATEGY AND TACTICS . 23
practice your verbal score will move to a ten or above. •
Take a five second break
•
Read every word
•
Construct a main idea
•
Use all four tools to answer the questions: 1.
going back;
2.
the main idea;
3.
the question stems; and
4.
the answer choices.
•
Th e Five Second Break If you were to observe a room full of MCAT tak~rs jus t after the sentence "You may break the seal on your test booklet and begin," you would see a room full of peop le frantically tear open their test booklets,. read for 20 to 30 seconds, pause, and then begin rereading from the beginning. Why? Because as they race through the first passage, they are thinking about what is happening to them (''I'm taking the real MCATI Oh my God l "), and not thinking about what . they are reading. They need a moment to become acc ustomed to the idea that the MeAT has actually begun. They need a moment to focus. However, they don' t need 20 to ,30 seconds' They take so much time because they are trying to do two thil)gs afonce; calm themselves down and understand the passage. They end up accomplishing neither. nus loss of concentration may also occur at the beginning of each new-passage, when the test-taker ma y still be struggling with thoughts of the previous passage w~ie reading the new passage.
,
If you continued to observe the test-takers, you wo uld see them in the midst of a passage suddenly stop everything, lift up their head, stretch, yawn, or crack their knuckles. This is their beleaguered mind forcing them to take a break. No one has an attention span 85 minutes long. If you don' t allow yourself a break, your mind will take one. How man y tirnes have you been reading a passage when sudden ly you realize, you \veren ' t concentrating? You're forced to start the passage over. More time is wasted.
There is a simple method to preven t all this lost time. Instead of taking breaks at random, incon venient moments, p lan your breaks. Before each passage, including the first passage, take five seconds to focus your thoughts. Remind yo urself to forget the las t passage and all other thoughts not related to the task at hand. Remind yourself to sit up straight, concentrate, and focus. For these five seconds, look away from the page, stretch your muscles and prepare to give your full attention to the next passage. Then begin and don't break your concentration w1til you have finished answering all the questions to that passage. The five second break is like a little pep-talk before each passage. Unfortunately, most students w ill not take the five second break. Understand one thing. All students w ill take breaks d uring the verbal section. Most will take them without realizing it, and most w ill take them at inopportune moments. If your goa l is to get the highest verbal score of w hich you are capable, you should take the five second break a t planned intervals.
Copyright © 2007 Examkrackers, Inc.
24
VERBAL REASONI NG
&
MATHEMATICAL T ECHNIQUES
Reading the Passage Most test takers have difficul ty finishing the verbal section in the 85 nUnutes allowed. Many fi nish as few as six passages. Strangely enough, an y premed without a read ing disorder is capable of reading 5400 words and taking a one hour nap in 85 minutes. A very slow read er can easily read every word of a 600 word passage in 3 n1inutes. Try it! It's true! This leaves 58 minutes to answer the questions, or nearly one minu te per questiqn to answer the ques tions. In other words, over two thirds of your time is spent answering questions on the MCAT Verbal Section, and less than one third is spent reading. If you read TWICE as fas t as you do now, you would have about 70 seconds, instead of 60 seconds, to answer each question. So increasing you r reading speed has very little effect on your verbal score. If you're not finishing now, yo u won't fi:;1ish by reading faster.
•
85 minutes Tin1e spen t reading
Time spent answering questions
So why do so many test-takers fail to finish the verbal section? The answer is "because they spend too much tin1e hunting for the answer in the passage, and end up reading the passage many times over. " We'll talk mo re about "going back" to the passage when we discuss w here to find the answer choice. For now, just believe us that you can read every word in the verbal section and easily finish the exam, so you should. Improving your efficiency at answering questions will be more profitable than increasing your reading speed and allow you more time to read the passage . If you increase your read ing speed by 10%, a strong improvement, you will only gain 2 minutes and 12 seconds on the entire exam. Spread over 60 questions, this allows you an additional 2.2 seconds per question. Not too fru itfuL If you increase your efficiency at answering quesUons by 10%, a rather simple task as you will soon see, you gain 5 minutes and 48 seconds. This is almost enough ti me to read two additional passages!
Have you ever tri ed skimming through a novel, not rea ding every word? Try it and see how much you understand. If you don't usually understand much when you skim, then why wo uld you skim on the most in1portan t test of your life; especially w hen doing so won' t give you much more time to answer the questions? Don't skim. H ave you ever mapped o ut a novel by writing a brief synopsis alongside each paragraph as you read? Try it. We think you will fa ll asleep from boredom. You will understand less of what yo u read, not more. Passages are intended to be read in their entirety as a single work presenting one overriding theme. MCAT expects you to understand this theme. The details w ithin this theme are far less important. Don't distract yourself by writing in the margins. The people that write the MCAT know that most of us are scientists. They know that we like to find the exact answer to things. Give us a mysterious powder and let us analyze it, and we will tell you exactly what it is. Show us the exact words in a passage as an answer choice and we will probably choose it. Don't fall for this trap. The Verbal Section tests yo ur ability to detect and understand ambiguities and gray areas, not details. Rely heavily on your main idea and give little weight to details. Tf you are highly certain of all your answers on the Verbal Section, then you probably have fallen for all its traps. Mastering this section is as much an art as a science. With practice, you will develop a 'feel' for a good MCAT answer. Learn to use this 'feel' to help you move faster t11Tough the Verbal Section. If yo u teach yourself not to expect the concrete certainty that you get with science questions, you will become more comfortable w ith the Verbal Section and your score will increase. The biggest mistake you can make on the verbal section is to consciously attempt to remember what yo u are reading. The vast majority of the questions will not concern the details of the passage and will not be answerable by searching the passage for facts. Most questions are about the main idea of the passage. The main idea will not be found in a list of details. In order to learn the main idea, the passage as a whole Copyright © 2007 Examkrackers, Inc.
LECTURE
1:
STRATEGY AND TACTICS .
25
must be understood. Read the passage the way you would read an interesting novel; concentrate on the main idea, not the detail. An often posited tactic is to read the questions first; don't do it! You will not remember even one question while you read the passage,. much less the 6 to 10 questions that accompany every passage. Tn fact, a short term memory can contain
5 items; that may be why the passages are followed by six or more questions. Not only that, reading the questions first w ill force you to read for deta il and you will never learn the main idea. You w ill probably end up rereading the passage many times but never straight through. This results .in a tremendous waste of time. Don't circle or underline words. This is another marketing teclu1ique that has sold well but is counterproductive. It is ve ry unlikely that und erlining or circling a sentence or a word will assist you in answering any ques tions. Ha ve
you ever answered a question correctly because you underlined or circled something in the passage? Underlining and Circling w ords forces you to concentrate on detail; fine for the SAT, not good for the MeAT. When you underline or circle something, you are reading it at least twice. This interrupts the flow of the passage. It distracts you from the main idea. This is an old SAT trick, inappropriately applied to MeAT. Some of the Verbal topics w ill fascinate you and some will bore you . The rnallenge will be to forget the ones that fascinate you as soon as yo u move to the next passage, and to pay close attention to the ones that bore you. Train yourself to become excited and interested in any and every passage topic. This will increase your comprehension. However, don't become so engrossed in a passage
that you slow your pace. Don't use fancy speed reading techniques where you search for meaningful words or try to read entire phrases in one thought. This will only distract you from your goal. Read the way you normally read. Your reading speed is unlikely to mange significantly in 10 weeks, and your reading speed is not the problem anywa y. Finishing the entire section depends upon how long you spend on the questions, not how long it takes you to read the passages. You also cannot assume that the pa ssages are written well enough so that you can read just the first and last sentence of earn paragraph. They are sometimes barely intelligible when yo u read every word. You must read every word, read quickly and concentrate. Read each passage like you are listening to a friend tell you a very interesting story. Allow the details (names, dates, times) to slip in one ear and out the other, while you wait with baited breath for the main point. The funny thing about this type of reading is that, when you practice it, you can't help but remember most of the details. Even if you were to forget some of the details, it only takes about 5 seconds to find a name, number, or key word in a 600 word passage. Thus, when yo u rW1 into a ra re question about a detail that yo u've forgotten, it is easy to find the answer. Another convenient aspect of this type of reading is that you are trying to accomplish exactly w hat the verbal section w ill be testing: your ability to pick out the main idea. The best thing about thjs type of reading is t1"t you have practiced it every day of your life. This is the way that you read novels, newspapers and magazines. Read the passages the way tha t you read best; read for the main idea. When you read, ask yo urself, "What is the autl10f trying to say? What is his paint? Is he in favor of idea A or against jt? If this author were sitting righ t in front of me, wou ld he want to discuss idea Aor is his real interest in idea B?" Creating an image
of the author in your mind wiII help you understand him. Use your life experiences to stereotype th e author. This will help you make quick, intuitive decisions abo ut how the author might answer each MeAT question about his passage. Make Copyright © 2007 Examkrackers, Inc.
When I create a great soup, you do not taste the salt, and each spice separately. You must experience the whole soup as a single , wonderful consomme' .
26
VERBAL REASONING
&
MATHEMATICAL TECHNIQUES
careful m ental note of anything th e author says tha t may not fit your stereotype. Use the stereotype to help guide your intuition on th e questions.
The Main Idea When you h ave finish ed reading a passage, ta ke twenty secon ds an d constr uct a main idea in the form of one or two complete sentences. Verbal Reasoning Leclure 3 w illcover h ow to construct a main idea. On a timed MeAT, writing the main idea requires too much time, so yo u should spend 20 second s men tally contemp lating the m ain idea b efore you b egin the questions. After yo u have completed an entire timed exam, scored yourself, and taken a break, it is a good idea to go back to each passage and write the m ain idea for practice.
A nswering t he Questions Answering the questions w ill be covered thoroughly in Verbal Reasoning Lecture 2. For now, attempt to answer the questions based upon the main idea and not the details.
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Answering the Questions
2.1
Tools to Find the Answer
For most students, the Verbal Reasonffig Section is literally a test of their ability to comprehend what they have read. Such students read a question and choose the correct answer based upon what was said in the passage. If they do not arrive at an answer, they eliminate answer choices based upon what was said in the passage. If they still don't arrive at an answer, they search the passage for relevant information that they may have missed or don't recall. If they still don't arrive at a single answer choice, which is likely to be about 50% of the time with this method, they repeat the process until they give up and make a random guess in frustration. This method uses only about 50% of the information provided by the test. When you consider that a portion of the questions on a lnultiple choice test will be answered correctly by luck, it's no coincidence that the national mean score on the MeAT is attainable by answering only about 61% of the questions correctly. When you can't identify an answer, 'thinking harder' (whatever that means) is not an effective solution. Nor is an effective solution to search the passage until the answer jumps out at you. However, both use up your precious time. In addition to your understanding of the passage, there are four tools that you should use to help you answer the questions. These four tools go beyond your understanding of the passage. They force you to consider additional information presented to you in the question stems and answer choices that is often overlooked or otherwise noticed only on a subconscious level. The four tools are: 1.
going back;
2.
the main idea;
3.
the question stems; and
4.
the answer choices.
Going Back 'Going back' refers to actually rereading parts of the passage to search for an answer. 'Going back' should be used only when: 1.
you are regularly finishing an exam on time;
2.
you know what you're looking for; and
3.
you know where you can find the answer.
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VERBAL REASON ING
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MATHEMATICAL TECHNIQUES
" Going back' is the most time consuming and least useful of the four tools. Unfortunately, it is the tool most often relied upon by inexperienced test takers. It is true that forgotten details can be found by rereading parts of the passage. However, lllost questions require an understanding of the main idea, not your memory of details. The main idea cannot be found by rereading parts of the passage. 'Word-far-word' and other traps have been set for the unwary test taker looking for the Ifeel-good' answer. The 'feel-good' answer is an answer where a section of the passage seems to unequivocally answer the question so that the test taker feels good when choosing it. This is often a trap. Remember, the Verbal Section is ambiguous and a simple clear ans~er is rarely the correct answer. You should learn to use 'going back' as seldom as possible. Most of the time, you should force yourself to choose the best answer without going back to the passage. This is a difficult lesson to accept, but it is extremely important to achieving your top score. Going back to the passage for anything but a detail will take large amounts of your testing time, and allow the test writers to skew your concept of the main idea by directing you toward specific parts of the passage. If you are unable to finish the test in the time given, it is because you are overusing the 'going back' tool. If you are not finishing, you should not go back at all until you can regularly finish. Questions sometimes refer to line numbers in the passage. Don't assume that you must go back to the given line number. Usually these types of questions should be answered without going back to the given line numbers. Often times the most helpful part of the passage ih answering the question is nowhere near the lines mentioned. If you do go back, you may have to begin reading 5 lines or more above the actual reference in order to place the reference in the correct context. Your number one goa I should be to finish the Verbal Section. Difficult questions are worth no more than easy questions. Don't sacrifice five easy questions by spending a long time answering a single difficult question. If you usually finish the Verbal section with time to spare, you can ~go back~ to the passage more often; if you don.' t usually finish the Verbal section, you shou ld stop going back to the passage until you begin finishing within the allotted time on a regular basis. "Going back" is a useful tooL Just use it wisely.
M ain Idea The main idea is the most powerful tool for answering MeAT verbal questions. We will 'discuss the main idea in Verbal Reasoning Lecture 3.
Q uestion St em s "The section that follows includes material from the MeAT Practice Test l / Practice Items. These materials are reprinted with permiSSion of the AssoCiation
of
Colleges (AAMC) .
American
. The question stems hold as much information as the passage. Read them and see how much you can learn about the passage from just the question stems. 1.
The author of the passage believes that the fiction written by the current generation of authors:
2.
The overall point made by the passage's comparison of movies to fiction is that:
3.
According to the passage, John Gardner concedes that preliminary good advice to a beginning writer might be, "Write as if you were a movie camera." The word concedes here suggests that:
4.
The fact that the author rereads Under the Volcano because it has been made into a movie is ironic because it:
Medical
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.~
LECTURE
5.
The passage suggests that a reader who is not bored by a line-by-line description of a room most lik:ly: .
6.
The passage suggests that if a contemporary writer were to write a novel of great forcefulness, this novel would most likely:
7.
The passage places the blame for contemporary writers' loss of readers on the:
Ask yourself some questions about the author. What does he/she do for a living? How does he/she dress? What does he/she like to eat? How does he/she vote? How old is he/she? Is he/she a he or she? Look closely at each question stem and see what kind of information you get from it. Why are certain adjectives used? Who is John Gardner? What can I learn about the passage from these question stems? Now, in the space below, write down everything that you can think of that is revealed about the passage from each stem. Include an answer to each of the seven question stems. (Warning: If you read on without writing the answers, you will miss an important opportunity to improve your verbal skills. Once you read on, the effect of the ex~r cise will be ruined.)
1.
2.
3.
4.
5.
6.
7.
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2:
ANSWERING THE QUESTIONS .
29
30 .
VERBAL REASONING
&
MATHEMATICAL TECHNIQUES
Information that can be gained from the seven previous question stems: 1.
From the first question stem, we immediately know that the passage was about the writing of fiction. The word' current' suggests a comparison between authors of fiction from the past and the present.
2.
From the second question stem we learn that there is also a comparison between movies and fiction. We also know that this was central to the authorfs point. Movies are a 'currene phenomenon. Hrnmm. What is the significance of this?
3.
In question stem three, you need to wonder "Who is John Gardner?" You know he is not the author of the passage because on the MeAT, you never know the name of the author. Thus, a named identity will be someone wj.lom the author used either to support his point or as an example of someone who has a bad idea. You should decide which. Now, even if the question didn't ask this, you should have asked yourself about the word 'concedes'. When you concede, you give in. So 'concedes' here indicates that Mr. Gardner is giving in to a point when he says "Write as if you were a movie camera." Mr. Gardner's argument must be that writing (or fiction) is not good when it's like the movies, but it is okay to write like a movie camera when you are a beginning writer. Notice how hesitant the wording is. Beginning is stressed by the use of both words 'preliminary' and 'beginning', and the word 'might' is also used.
At this point, you should begin forming a feeling of what this passage was about: movies versus fiction; current fiction versus past fiction; someone implying that movies don't make for good fiction. The author believes something about current fiction and makes a point about fiction and movies. Three question stems with no passage and not even answer choices to the questions, and we can already get a sense of the passage. The remaining question stems will confirm what the passage is abou!. 4.
The fourth question stem indicates that a movie makes the author read a book. The question states that this is ironic. That means the actual result is incongruous with the expected result. Apparently, according to the author's argument, watching a movie should not make him read the book. Thus, part of the author's argument in the passage must be that movies make people less interested in reading. It is also reasonable to assume from this that the author used John Gardner in stem #3 to support his argument, so the author probably believes that fiction written like a movie is not good fiction. Extrapolating further from the comparison of movies to fiction and the stated dichotomy between current and past fiction, the author is probably arguing that current fiction is not as good as old fiction.
5.
The fifth passage compares the phrase 'line-by-line description' with the idea of boredom. It is a simple logical jump to equate 'line-by-line description' with past fiction as opposed to current fiction or movies. From our conclusions thus far about the author's argument, it would be logical to conclude that someone who is NOT bored by 'line-by-line descriptions' would NOT be bored by past fiction, but would, in fact, appreciate it as the author obviously does.
6. 'Question stem six reinforces OUI conclusion about the author's argument. The 'If' indicates that 'contemporary writers' do not 'write novels of great forcefulness'. Instead, they must be writing novels that resemble movies. The only question is 'What would a novel of great forcefulness' do? Answering this question is as simple as seating the author in front of you and asking him. The amazing thing is that we already have a Copyright © 2007 Examkrackers, Inc.
LECTUR E
.
stereotypical idea of this author just by reading six question stems! This guy is a college English pr?fessor fed up with the quick fix satisfaction offered by movies. He would love a novel of great forcefulness. Does he think that we would appreciate it? Be careful here. He appreciates the novel because he truly believes that the novel itself is great, and not because he thinks he is great or better than everyone else. The answer is yes, he thinks that we would appreciate a novel of great forcefulness as well. 7.
This last question stem answers the previous question. The seventh question stem says that current fiction is losing readers. It asks for the explanation. Of course, the author's whole point is to explain why current fiction is losing readership. It is because it is like movies and not forceful like past fiction.
What should be revealing and even shocking to you is that we can accurately answer every question on this actual AAMC passage without reading the passage. In fact, we can accurately answer every question without reading the passage OR the answer choices. Did you realize that there was this much information in the question stems alone? Have you ,been using this information to answer the questions on the MCAT? If you haven't, you are capable of scoring many points higher on the MCAT Verbal Section. You can't expect to always be able to answer questions without the passage or the answer choices, but you can expect to gain much information about the passage from the question stems.
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2:
ANSWERING THE QUESTIONS .
31
32
VERBAL REASONING
&
MATHEMATICAL TECHNIQUES
Compare your answers with the actual answer choices below. HThe section that follows includes material from the MeAT Practice Test l/Practice Items. These materials are reprinted with permission of the Association of American Medical Colleges (AAMC).
4.
1.
A. B.
C. D.
lacks the significance of fiction written by previous generations. is, as a whole, no better and no worse than fiction written by previous generations. brilliantly meets the particular needs of contemporary readers. is written by authors who show great confidence in their roles as writers.
I.
seems to go against the overall point of the passage concerning fiction and film. II. implies that the film version was a box-office failure. III. hints that the author was dissatisfied with the novel.
,
A. B. C. D.
I only I! only III only I! and III only
2.
A. B.
C. D.
contemporary authors have strengthened their fiction by the application of cinematic techniques. the film of Under the Volcano is bound to be more popular than the novel. great fiction provides a richness of language and feeling that is difficult to re-create in film. contemporary authors would be well advised to become screenwriters.
5.
A. B.
C. D.
6. I. confuse and anger lovers of great literature. II. exist in stark contrast to the typical contemporary novel. III. win back some of the readers contemporary writers have lost.
3. I.
Gardner'·s approach to writing has been influenced by the competing medium of film. II. Gardner must have written screenplays at one point in his life. III. Gardner dislikes the medium of film.
A. B. C. D.
I only I! only I and I! only II and III only
prefers the quick fix of the movies. would be bored by a single shot of a room in a film. has no tolerance for movies. displays the attitude demanded by good fiction.
A. B. C. D.
I only I! only I and I! only I! and III only
7. I. competition presented by movies. II. writers themselves. III. ignorance of the public.
A. B. C. D.
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I only I! only I and I! only I, I! and III
LECTURE
Answers to the Questions QUESTION
1
Choice A, the answer to question #1 is exactly what we expected: past fiction is better than current fiction. Notice that we can simplify the choices to: A.
Current fiction is not as good as past fiction. Current fiction is equal to p ast fiction. Current fiction is good. Current fiction is good.
B.
e. D.
Simplifying the question and the answer choices can make the correct answer easier to find. We'll discuss simplification later in this lecture. The main idea is all you need to answer this question. QUESTION
2
Choice C, the answer to question #2 is also exactly what We expected. The choices can be rephrased to: A. B. C. D.
Movies have been good for fiction. Movies are more likeable than fiction. Movies aren't as good as good fiction. Authors of fiction should make movies.
When we put these questions to our author, the choice is
QUESTION
2:
ANSWERING THE QUESTIONS •
33
4
The answer to question #4 is exactly what we expected. The choices can be rephrased to: I. Seeing the movie shouldn't have made the author read the book. II. The movie flopped. III. The author didn' t like the book. Only choice I addresses the 'irony' suggested in the question. Only choice I pertains to the main idea . The answer isA. QUESTION
5
Choice D, the answer to question #5 is exactly what we expected. The choices can be rephrased to: A. B.
e. D.
.If you' re patient, you'll prefer the fast pace of movies. If you're patient, you won't like waiting for action. If you're patient, you won't have the patience for the fast pace of movies. If you're patient, you'll like the careful pace of good fiction.
Choices A, B, and C seem to be self contradictory.
obvious. QUESTION QUESTION
3
The answers to' question #3 are not w hat we expected. We expected a more sophisticated question pertaining to the use of the word 'concedes'. Although the question told us much about the passage, the answer choices match a much simpler question than we anticipated, "Who is John Gardner?" The choices can be rephrased as: I. John Gardner has been influenced by movies. II. John Gardner wrote movies. III. John Gardner dislikes movies. Clearly John Gardner has been influenced by movies if he is suggesting that writing like a movie might be good advice for a beginning writer. From the answer choices, we can see that if I is correct, then III is likely to be incorrect. If Gardner dislikes movies, it would be unlikely that he . would be influenced by them. II is incorrect because Gardner would not have to have written movies in order to be influenced by them. Even if III were an option, and even if Gardner is like the author, liking good fiction _ more than movies isn't the same as disliking movies. Choice A is correct.
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•
6
Choice D, the answer to question #6 is exactly w hat we expected. Remember that 'a novel of great forcefulness' describes past fiction to our author, and our author would expect u s to like past fiction. This describes choices II and III. QUESTION
7
\
Choice C, the answer to question #7 is exactly what we expected . Choice I restates ti).e main idea that movies have, hurt fiction. Certainly, our author is criticizing current authors, so choice II is also true. Choice III is not true based upon our idea that the author w ould expect us to like a forceful novel. The answer here is C.
34 . VERBAL REASON ING
&
MATHEMATICAL TECHNIQUES
2.2
Answer Choices
Each MeAT question has four possible answer choices. One of these will be the correct answer and the other three we ,,~ call distracters. Typically, when a verbal question is written, the correct answer choice is written first and then distracters are created. Because the correct answer is written to answer a specific question and a
distracter is written to confuse, the two can often be distinguished without even referencing the question. In other words, w ith practice, a good test-taker can sometimes distinguish the correct answer among the distracters without even read-
ing the question or the passage. This is a difficult skill to acquire and is gained only through sufficient practice. Begin by leaming to recognize typical distracter types. Among other things, effective distracters may be: a statement that displays a subtle misunderstanding of the main idea; a statement that uses the same or similar words as in the passage but is taken out of context; a true statement that does not answer the question; a statement that answers more than the question asks; a statement that relies upon information commonly considered true but not given in the passage. In order to help you recognize distracters, we have artificially created five categories of suspected distracters. It is unlikely, but not impossible that the correct answer choice might also fan into one of these categories. Thus, you must use this tool as a
guide to assist you in finding the correct answer, and not as an absolute test. •
Round-About: a distracter that moves arow1d the question but does not directly answer it
•
Beyond: a distracter whose validity relies upon information not supplied by (or information beyond) the passage
•
Contrary: a distracter that is contrary to the main idea
•
Simpleton: a distracter that is very sin1ple and / or easily verifiable from the passage
•
Unintelligible: a distracter that you don't W1derstand
The Round-About Round-about distracters simply don't answer tl,e question as asked. They may be true statements. They may even concur with the passage, but they just don't offer a direct answer to the question. A Round-about is the answer yo u expect from a politician on a Sunday moming talk show; a lot of convincing words are spoken but nothing is really sa id.
Beyonds Often times, a distracter will supply information beyond that given in the question and passage without substantiating its veracity. These distracters are called beyonds. When you read a beyond, you typically find yourself wondering something like "This answer sounds good, but this passage was on the economics of the post Soviet Union, I don't remember anything about the Russian revolution." Beyonds can also play upon current events. A passage on AIDS may have a question with an answer choice about cloning. Cloning may be a hot topic in the news, but if it wasn't mentioned in the passage or in the question, you should be very suspicious of it being in an answer choice. Don' t confuse a beyond with an answer choice that directly asks you to aSSW11e information as true.
Copyright © 2007 Examkrackers, Inc.
LECTURE
2:
ANSWERING THE QUESTIONS . 35
Contraries A contrary distracter contradicts the main idea. If the question is not an EXCEPT, NOT or LEAST, the answer choice is extremely unlikely to contradict the main idea. Most answer choices support the main idea in one fonn
Of
another.
Simpletons If the correct answers on the Verbal Section were simple, direct, and straigh t forward, then everyone would do well. Instead, the correct answers are vague, ambiguous, and sometimes debatable. This means that an answer choice that is easily verifiable from a reading of the passage is highly suspect and often incorrect. These answer choices are called simpletons. Simpletons are not always the wrong answer choice, but yo u should be highly suspicious when you see one.
Typical of simpletons is extreme wording like always and never. Here's a manufactured exa mple of a simpleton: 13. In mid-afternoon in December in Montana , the author believes that the color of the sky most closely resembles:
B.
cotton balls floating on a blue sea.
If this w ere the answer, everyone would choose it. This is unlikely to be the correct answer.
Unintelligibles Unintelligibles are answer choices that you don't understand. Whether it's a vocabulary word, or a concept, avoid answer choices that you don't understand. These are likely to be traps. Strangely enough, many test takers are likely to choose an answer that confuses them. This is apparently because the MCAT is a difficult test so students expect to be confused. Test writers sometimes purposely use distracters with obscure vocabulary or incomprehensible diction in order to appeal to the test taker who finds comfort in being confused. As a general rule, don't choose an answer that you don't w1derstand unless you can positively eliminate all other choices. Be confident, not confused.
2.3
Identifying the Correct Answer
Besides identifying distracters, you should become familiar with the look and feel of a typical correct answer choice. Typical correct answer choices contain softeners. Softeners are words that make the answer true tmder more circumstances, s uch as most likely, seemed, had a tendency to, etc. An answer choice with a softener is not necessarily correct; it is just more likely tu be co rrect.
2.4
Simplification of the Question and Answer Choices
It is often helpful to simplify the question and answer choices in terms of the main idea. For instance, reexamining the questions and answer choices from our original seven AAMC question stems we have a passage with the following main idea:
"Great fiction provides a richness of language and feeling that is difficult to recreate in film. Contemporary authors emulating fihrn have lost this richness and their a udience with it. "
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"I always never pick the simpleton answer!~
36
V ERBAL REASON ING
&
MATHEMATICAL TECH NIQ UES
This is a nice complete main idea but can be difficult to understand all at once. It is helpful to simplify it as follows: There is past fiction, current fiction, and movies. •
Past fiction is good;
•
Current fiction is bad;
•
Current fiction is like movies.
When analyzing the questions and answer choices, resta te them in terms of these ideas, keeping in mind that this is a simplification. For instance, a reference to 'a great, forceful novel' or 'a line-by-Iine description' can be replaced by 'past fiction'. 'The passage suggests' can be replaced by 'the author thinks'. Tills is much like using the concept of an ideal gas to approximate the behavior of a real gas and then adding the characteristics of a real gas for the detailed work.
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L ECTURE
2:
ANSWE RING THE Q UESTIONS .
37
Compare the following restatements with the original seven AAMC questions:
Restatement 1. The author believes current ficti on is: A. B.
e. D.
not as good as past fiction. equal to past fiction. good. good.
Original Question 1. The author of the passage believes that the fiction written by the current generation of authors: A. B. C. D.
2. The au thor compares movies to fiction in order to show that: A. B. C. D.
movies have been good for fiction. movies are more likeable than fiction. movies aren ' t as good as good fiction. authors of fiction should make movies.
2. The overall point made by the passage's comparison of movies to fiction is that: A. B. C. D.
3. John Gardner says, "Write like the movies," therefore: I. be has been influenced by movies. II. he wrote movies. m. he dislikes movies.
lacks the significance of fictio n written by previous generations. is, as a whole, no better an d no worse than fiction written by previous generations. brillia ntly meets the particular needs of contemporary readers. is written by authors who show great confidence in their roles as writers.
contemporary authors have strengthened their fiction by the application of cinematic techniques. the film of U oder the Volcano is bound to be more popular than the nove l. great fictio n provides a richness of language and feeling that is difficult to re-create in film. contemporary authors would be well advised to become screenwriters.
3. According to the passage, John Gardner concedes that preliminary good advice to a beginning writer might be, "Write as if you were a movie camera." The word concedes here suggests that:
I. Gardner's approach to writing has been influenced II.
m.
4. The author sees a movie that causes him to read a book, this:
I. II.
m.
weakens his argument. means the moyie was bad. means the author didn't like the book.
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by the competing medium of film. Gardner must have written screenplays at one point in his life. Gardner dislikes the medium of film.
4. The fact that the author rereads Under the Volcano because it has been made ioto a movie is ironic because it: seems to go against the overall point of the passage concerning fiction and film. II. implies that the film version was a box·office fail· ure. UI. hints that the author was dissatisfied with the novel. I.
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5. The author says that if you like past fiction: A. B. C. D.
you'll like movies. you'll be bored by past fiction. you won'r like movies. you'll like past fiction.
5. The passage suggests that a reader who is not bored by a line·by·line description of a room most likely: A. B. C. D.
6. If a new novel were like old fiction: I. people who like old fiction wouldn 't li ke the novel.· II. the novel would not be like current fiction. III. people would like to read it.
7. No one reads current fiction because:
I. movies are as good. II. currenl fiction writers write bad fiction . III. people are ignorant.
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prefers the quick fix of the movies. would be bored by a single shot of a room in a film. has no tolerance for movies. displays the attitude demanded by good fiction .
6. The passage suggests that if a contemporary writer were to write a novel of great forcefulness, this novel would most likely: I. confuse and anger lovers of great literature. II. exist in stark contrast to the typical contemporary novel. 01. win back some of the readers contemporary writers have lost.
7. The passage places the blame for contemporary writers' loss of readers on the: I. competition presented by movies. II. writers themselves. III. ignorance of the public.
L ECTURE
2.5
Summary
You have four tools for finding the correct answer (going back, main idea, question stems, and answer choices). In order to get your best MCAT score, you should use all of them. Your fourth tool is the most difficult to master. When evaluating the answer choices for distracters, keep in mind that there are no absolutes, just suspects. When necessary, restate complicated questions using the simplified concep ts from the main idea.
STOP! (DO NOT LOOK AT THE FOLLOWING QUESTIONS UNTIL CLASS. IF YOU WILL NOT BE ATTENDING CLASS, GIVE YOURSELF 30 MINUTES TO COMPLETE THE FOLLOWING SET OF QUESTIONS.) **Th e section that follows includes material from the MeAT Practice Test 1/Practice Items. Th ese materia ls are reprinted with permission of the Association of American Medical Colleges (AAMC) .
Copyright © 2007 Examkrackers, Inc.
2: A NSWERING
TH E Q UESTIONS .
39
40
VERBAL REASONING
&
MATHEMATICAL TECHNIQUES
The following questions come from three passages. Each page represents a different passage. The passages have been removed to force you to pay attention to the questions and the answer choices. Try to answer the questions, and compare your scaled score to your normal practice MeAT score.
Passage 1 (Questions 1-7)
1. According to the passage, an image is a versatile tool that: A. B. C. D.
is always visual, never abstract. can be either abstract or visual. is always abstract, never visual. is neither visual nor abstract.
5. Which of the following findings would most weaken the claim that the use of symbolic imagery is unique to humans? A. B.
C. D.
2. An experiment found that dogs can remember a new signal for only five minutes, whereas six-year-oId children can remember the same signal much longer. Based on the information in the passage, this finding is probably explained by the fact that: A.
B. C. D.
a human being possesses a larger store of symbolic images than a dog possesses. the human brain evolved more quickly than the brain of a dog. the children were probably much older than the dogs. most dogs are color-blind.
6. It has been said that language does not merely describe reality but actually helps to bring reality into existence. Which of the points made in the passage would best support this claim? A. B.
C. D.
3. In order to defend poets from the charge that they were liars, Sidney noted that "a maker must imagine things that are not" (line 38). Sidney's point is that: A. B. C. D.
a true poet must possess a powerful imagination. in order to create something, one must first imagine. poets are the most creative people in our society. imagination is not a gift unique to poets, but is possessed by all creative people.
Chimpanzees are capable of learning at least some sign language. Certain species of birds are able to migrate great distances by instinct alone. Human beings have larger frontal lobes than do other animals. Some animals have brains that are larger than human brains.
To imagine means to make images and move them about in one's head. The tool that puts the human mind ahead of the animal's is imagery. There is no specific center for language in the brain of any animal except the human being. Images play out events that are not present, thereby guarding the past and creating the future.
7. According to the author, the most important images are: A. B. C. D.
words. poetic images. images of the past. images of the future.
4. In the context of the passage, the statement "if thereby we die a thousand deaths, that is the price we pay for living a thousand lives" (lines 52-54) is most likely meant to suggest that: A. B. C.
D.
we must guard against using our imaginations toward destructive ends. although imagination sometimes causes pain, its positive aspects outweigh its negative ones. it is possible to be too imaginative for one's own good. without imagination, the uniquely human awareness of death would not exist.
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GO ON TO THE NEXT PAGE.
L ECTURE
Passage 2 (Questions 8-16)
8. Why is the San Luis Valley site being investigated urgently? A. B. C. D.
Artifacts are few in number. Artifacts are being eroded by the wind. Bison bones are few in number. Excessive rainfall is damaging the site.
9. According to the passage, which of the following activities was common to each band of Folsom Indians? A. B. C. D.
Cultivating a number of different crops Eating a wide variety of wild game Interacting with other bands Making tools out of nearby rocks
10. The passage suggests that the presence of bum an remains, tools, and animal bones at a single location means that: A. B. C. D.
bison and other animals migrated from one place to another. communal tasks were performed at the site. erosion has not yet occurred at the site. extensive interactions occurred among bands of Paleoindians.
11. Assume that a new Folsom hunter site has just been discovered in northern Texas. On the basis of the infonnation contained in the passage, this site would most likely contain all of the following EXCEPT: A. B. C. D.
clusters of bones and tools. human bones. remains of hearths. tools made of Colorado flint.
2: A NSWERING
THE Q UESTIONS . 41
13. According to the passage, bands of Paleoindians did not trade between one another. What is the evidence for this statement? A. B. C. D.
Tools of a band came only from local sources. Tool shapes were unique to each band. Food sources were unique to each band. Each band had its unique language and customs.
14. Given the information contained in the passage, if a large number of deer bones were discovered at the San Luis Valley site, the most likely explanation for tbeir presence would be that the deer: A. B. C. D.
accidentally died at the scene. competed with bison for food. migrated from another region. served as food for the Indians.
15. Which of the fo llowing discoveries would most strengthen the hypothesis that Folsom hunters killed the bison they ate? A.
B. C. D.
Bone breaks consistent with the shapes of the Folsom hunters' pointed tools No evidence of an alternative animal food source Bison bones at a Folsom site Similar accumulation of bison bones at many Folsom sites
16. If the Paleoindians had eaten small game sucb as rabbits instead of large game, the finding of small animal skeletons and individual tools with many edges at the same sites would LEAST support the conclusion that: A. B. C. D.
certain tools had many uses. small animals made up the people's main diet. the animals were killed at tbe site. tools were used to prepare the animals for use.
12. If a Folsom hunter site containing tools made of petrified wood were discovered in Iowa, where there is little petrified wood, this discovery would weaken which of the fonowing conclusions made in the passage?
I. Paleoindians hunted bison. TI. Folsom hunters did not travel great distances. TIl. There was little trading among bands of I'olsom hunters. A. B. C. D.
I only III only I and II only Il and TIl only
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GO ON TO THE NEXT PAGE.
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VERBAL REASONING
&
MATHEMATICAL TECHNIQUES
Passage 3 (Questions 17-23)
21. According to the author, most historians of science do NOT believe that:
17. The example concerning Galileo (lines 23-31) best supports the author's c1aim that:
A. B. C. D.
science" and society usually coexist hannoniously.
science works in an unpredictable manner. cultural bias limits scientific theorizing. scientific fact occasionally forces a change in cultural assumptions.
18. Based on the passage, a scientific claim has the best chance of being free from cultural influence when the claim has:
A. B. C. D.
much supporting evidence and much social impact. little supporting evidence and little social impact. much supporting evidence and little social impact. little supporting evidence and much social impact.
19. The author mentions the abandonment of eugenics in America and Hitler's use of arguments for sterilization and racial purification primarily to support the claim that: A. B. C. D.
science is often misused. science is impartial. scientific attitudes are sometimes affected by social movements. science should avoid involvement in social issues.
20. The author. believes that the view that science "inexorable march toward truth'" (lines 66-67) is: A. B. C. D.
one of the myths of science. supported by good evidence. clearly proven by the case of Galileo. accepted by most historians of science.
Copyright © 2007 Examkrackers, Inc.
IS
A. B. C. D.
scientific facts lead to effective theories. most theories are developed by straightforward induction from facts. objectivity is a worthwhile goal in scientific investigation. facts are influenced by cultural assumptions.
22. When the author states that "science cannot escape its curious dialectic" (line 54), he is emphasizing science's: A. B. C. D.
dilemma between truth and mere theories. interrelationship with social factors. quest for truth. imprecise methodology.
23. According to the author, one reason that scientists have a difficult time escaping cultural assumptions is that scientists often: A. B. C. D.
formulate hypotheses that can only result in the verification of accepted beliefs. project their research findings onto society. attribute too much significance to scientific data as opposed to social belief. base theories on too much data.
an STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
STOP.
LECTURE
Don't look at the answers yet. Question 1: If we look at this question and use common sense, we know that an image can be both abstract and visual. The word "versatile" in the question also helps us find the answer. Question 2: Ask yourself "Why might a child remember a signal longer than a dog?" B, C, and 0 don't seem like reasonable answers. For answer Bf what does it mean for a htunan's brain to evolve more quickly? This answer is somewhat unintelligible. Answer C compares the age of a human child with a dog in terms of memory as if they were equivalent. This doesn't seem to be reasonable. At best, it calls for outside information about a dog's ability to remember based upon its age. For answer choice D, the question doesn't say anything about vision. Where does color-blind come in. This is a beyond. Question 3: Notice that the question asks what is meant by the quote. For this type of question, just match the answer to the quote. Answer B is a paraphrase of the quote. Sidney himself is superfluous information. Question 4: This is the same type of question as the last. Match the answer to the quote. Notice answer choice D. This is for those who want to see things in black and white, and take the quote very literally. It also does not match the quote. Question 5: What would weaken the claim that the use of symbolic imagery is unique to humans? An example of a non-human using symbolic imagery. A is correct. Question 6: Here we are asked to interpret a paraphrase. Just match the paraphrase to the answer choice. "bringing reality into existence" is the same as "creating the future". Question 7: This is difficult to answer without the passage. However, look at the other questions. Ask yourself, "What is the main idea of this passage?" It is certainly about Images, symbols, and language. Which answer fits most closely? Notice that the word image is in all the answers except the correct one. This makes choices B C and D Simpletons. A is correct. ' , Question 8: The word "urgently" helps to narrow down the choices to Band D. It is difficult to choose between these two without reading the passage. Question 9: The question appears to be impossible to answer without reading the passage; however, we will find that it is easy to answer after we answer the other questions. We'll come back to it.
2:
ANSWERING THE QUESTIONS •
43
special about location and Folsom hunter sites. Looking at the answer choices, "Colorado" also stands out. What is special about Colorado? 0 seems like a pretty good answer. There is certainly no reason to choose any other answer. But then again, there doesn't seem to be any reason not to either. We'll come back to this one too. Question 12: Here it is again; location! This question gIVes us another clue. Apparently the author believes that tools are made from material found near a site because if tools were found that were not made from material near the site, this :",ould somehow weaken the author's argument. Here, It makes sense that if the author argued for either II or III, both would be weakened. There doesn't seem to be any reason why choice I would be weakened. B or 0 must be the answer. But wait, if the author thought that Folsom hunters did travel great distances, then they might have carried tools with them, and III would not be weakened. Therefore, 0 must be the answer. Question 9 revisited: Now we know that the answer to question 9 must be D.
Question 10 revisited: Now we know that 0 must be wrong. Notice the word "single" emphasizing the importance that the tools appear together. Choice Balsa addresses this togetherness with the word" communal". A and C don't seem to have a mechanism which would explain them. The correct answer is B. Question 11 revisited: Clearly, if tools must be made from nearby materials, "tools made from Colorado flint" would not be found at a site in "northern Texas". 0 is correct. Question 13: This question even tells you part of the answer to question 12. Answer A just confirms what we've already discovered. A is correct. Question 14: This question can be answered using the other queslions as background. The passage is about the Folsom hunters. Only 0 incorporates this into its answer. The main idea is always the best choice. Question 15: The question asks for something that would prove (show) that the hunters killed the bison that they ate. In other words, the question wants something that would show that they didn't scavenge the bison after finding them dead, but they actually killed them. Choice A shows that the bison were killed with tools made by the hunters. Choice A is also the only choice that seems consistent with a main idea that apparently has to do with tools.
Question 10: This is also a difficult question to answer before we answer the others. We'll come back to it.
Question 16: Notice that the question emphasizes small versus large. The only answer choice that addresses this emphasis is choice C. If the game were small, the hunters could have carried their prey to the site.
Question 11: The word "Texas" should stand out here. What is special about Texas? There must be something
Question 17: If you know that Galileo was forced by the church to recant his theories, you know that A is wrong.
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VERBAL REASON ING
& MATHEMATI CAL TECHNIQUES
This information also seems to suggest CorD. We'll corne back to this one. Question 18: This is just common sense. Now we know what the p~ssage is about.
Question 19: Certainly B is wrong. The problem with both A and D is that the example includes a 'good thing' about science; the abandonment of eugenics, and a bad
thing about science, Hitler's use of it. (Eugenics is the creation of the perfect race through breeding.) A and D suggest that both are bad, so they are the wrong answer choices. C is correct.
Question 21: The main idea is about the relationship between social events and science. Clearly B is correct. B is also very MCAT-like in its non-absolutism. Question 22: B is the main idea; it must be correct. A and Dare beyonds. C is not MCAT-like.
Question 23: 'c and D don't make sense; they seem to contradict the logic of the question. A answers the question better than B because B has science affecting society as opposed to the way the question has society affecting science.
verbal thought: "science is an inexorable march toward
Question 17 revisited: We were stuck between C and D. Unfortunately, we're still stuck. We'll have to just guess, or read the passage. We still get at least an MCAT score of 12.
truth." Verbal is ambiguous, not absolute. This is extremely unlikely to be the_belief of an author in the verbal section. A is clearly the correct answer because it is so MCAT-like.
The correct answers are: 1. B, 2. A, 3. B, 4. B, 5. A, 6. D, 7. A, 8. D, 9. D, 10. B, 11. D, 12. D, 13. A, 14. D, 15. A, 16. C, 17. C, 18. C, 19. C, 20. A, 21. B, 22. B, 23 A.
Question 20: This is a perfect example of a non-MCAT
This exercise is not to convince you not to read the passage. You should always read the passage. It should show you that there is a large amount of information in the questions and answer choices. If you scored higher without reading the passage, then you probably haven't been taking advantage of the wealth of information in the questions and ans,ver choices.
2.5
Marking Your Test to Improve Your Score
As you review the possible answer choices to a question, you should mark the letters of each answer choice with one of four symbols meaning: 1. absolutely incorrect; 2. probably incorrect; no idea; and 3. possibly correct. A diagonal line through a letter indicates that you have dismissed that answer as absolutely incorrect. A small x to the left of the letter indicates that you have a feeling that the answer is wrong, but you are not certain. A small horizontal line to the left of the letter indicates that you have no inclination as to whether the answer 'choice is right or wrong. A small circle to the left of the answer choice indicates that you like the answer but are not certain that it is correct. Below is an example: ~ Absolutely incorrect )( B. Probably incOITect - C, No idca o D. Possibly correct
This marking system saves time by helping you keep track of your thoughts concerning each answer choice. Recording your in1pressions next to each letter also helps guide your instincts providing for more accurate guesses.
2.6
When to Bubble
VVhen you decide upon an answer, circle the correct letter and bubble in the answer on your test immediately. DO NOT wait till you have answered several questions and then bubble in several answer choices at a time. This is the most common way that bubbling errors occur. Bubbling in answer choices is NOT a productive way to spend your 5 second break. Bubble as you go, one answer at a time. This is the fastest, most accurate and efficient method.
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The Main Idea
3.1
The Main Idea
The main idea is a sUlnmary of the passage in one or hvo sentences. It should reflect the author's opinion (if presented or implied), and it should emphasize minor topics to the same extent as they are emphasized in the passage. It is not a list of topics discussed in the passage nor an outline of those topics. It is a statement about the passage topics, and includes the author's opinion.
In one form or another, 90% of the Verbal Section questions will concern the main idea. Notice that the main idea cannot be found by going back to the passage and searching for details. You must concentrate on the main idea while you read the entire passage. If you read for detail, if you try to remember what you have read rather than process what you are reading, you will have to guess at 90% of the questions. It is important to have a clear concept of the main idea before reading any ques-
tions. MCAT Verbal section questions are designed to take your inchoate thoughts concerning the passage and subtly redirect them away from the tme main idea. Each successive question embellishes on insidious pseudo-themes steering unwary followers into an abyss from which there is no return. Like a faithful paladin, your clearly stated main idea unmasks these impostors and leads you toward the holy grail of Verbal Section perfection. Writing the main idea on paper is an ilnportant step to-ward iInproving your ability to find the main idea; however, it requires too much time while taking the exam. Instead, the a few days after taking a practice exam, go back to each passage and write out the main idea. While taking the exaln, make a 20 second pause after reading a passage, and construct the main idea in your head.
Most students resist writing out their main idea until they are halfway through the course and the materials. At this point they begin to realize how ilnportant the main idea is. Unfortunately, they must start from scratch and begin writing out th~ main idea with only four weeks until the MCAT. Don't do this. Start now by going back to used passages and writing out the main idea. It's very painful at first, but it will get easier, and it will dramatically improve you r score.
3.2
Constructing the Main Idea
A good main idea can be formed as follows:
1.
After reading the passage, write down the main topics. Each topic should be from one to four words.
46 .
VERBAL REASONING
&
MATHEMATICAL TEC HNIQUES
2.
From these topics, choo~e the most important ones two or three at a time, and write a short phrase relating them to each other and the passage.
3.
Now connect the phrases into one or two sentences which still concern the most important topics but incorporate the other topics as well. Be sure to include the author's opinion if it was given or implied. Try to emphasize €aGh topic to the same extent to which it was emphasized in the passage. This is your main idea. Over time, you will be able to construct the main idea in your head. /
3.3
Confidence
Often on the MCAT, passages seem incomprehensible. Don't get bent! Re,member, most questions are answered correctly by 60% or more of test-takers, and only two or three are answered incorrectly by less than 40%, so no group of questions will be that difficult. Have the confidence to keep reading. Don't reread a line or paragraph over and over until you master it. If a line or paragraph is incomprehensible to you, then it is probably incomprehensible to everyone else, and understanding it will not help your score. Instead, continue reading until you get to something that you do understand. Just get the general sense of what the author is trying to say. Chances are good that this will be enough to answer all the questions. Remember, after you read the passage you have four tools beyond your understanding of the passage to help you answer the questions.
3.4
\
Know Your Author
You must become familiar with the author. Who is he or she? Is the author young or old; rich or poor; male or female; conservative or liberal? Do you love or hate this author? Take a guess. Create a picture of the author in your mind. Use YOlli
prejudices to stereotype the author. Your harsh judgment of the author is everything to understanding what he is trying to say. The better you understand the author, the easier the questions will be. Read with emotion and judge harshly. Now that you know the author intimately, when you get to a question, ask yourself "If this author were right here in front of me, how would he answer this question?" The way that the author would answer the question, is the correct answer.
3.5
Ignore the Details and See the Big Picture
There is no reason to remember the details of a passage. They can be found in seconds, and are rarely important to answering a question. Instead, focus on the big picture. Ask yourself "What is the author trying to say to me? What's his beef?" The author's 'beef' will be the main idea, and the key to answering 90% of the questions.
STOP! DO NOT LOOK AT THE FOLLOWING PASSAGE ANb,QUESiiONS UNTIL CLASS. IF YOU WILl NOT BE ATTENDING CLASS, READ THE PASSAGE IN THREE MINUTES AND ANSWER THE QUESTIONS WHICH FOLLOW. HThe section that follows includes material from the MeAT Practice Test 1/Practice Items. These materials are reprinted with permission of the Association of American Medical Colleges (AAMC). Copyright © 2007 Examkrackers, Inc.
..
LECTURE
Passage
5
10 /
15
20
It is rougbly a century since European alt began to experience its first significant defections from the standards of painting and sc ulpture that we inherit from the early Renaissance: Looking back now across a long succession of innovative movements and stylistic revolutions, most of us have little trouble recognizing that such aesthetic orthodoxies of the past as the representative convention, exact anatomy ruld optical perspective, the casement-window canvas, along with the repertory of materials and subject matters we associate with the Old Masters- that all this makes up not "art" itself in any absolute sense, but something like a school of art, one great tradition among many. We acknowledge tbe excellence which a Raphael or Rembrandt could achieve withi n the canons of that school ; but we have grown accustomed to the idea that there are other aesthetic visions of equal validity. Indeed. innovation in the arts has become a convention in its own right with us, a "tradition of the new;' to such a degree that there are critics to whom it seems to be intolerable that any two painters should paint alike. We demand radical originality, and often confuse it with quality.
Yet what a jolt it was to our great-grandparen ts to see the certainties of the academic tradition melt away before their eyes. How distressing, especially for the academi25 cians, who were the guardians of a classic heritage embody ing time-honored techniques and standards whose perfection bad been the labor of genius. Suddenly they found art as they understood it being rejected by upstarls who were unwilling to let a single premise of the inherited 30 wisdom stand unchallenged, or so it seemed. Now, with a little hindsight, it is not difficult to discern, continui ties where OUI predecesso rs saw only ruthless disj unctions, To see, as well, that the artistic revolutionaries of the past were, at their best, only opening our minds to a more global 35 conception of art which demanded a deeper experie nce of light, color, a nd form. Through their work, too, the art of our time has done much to salvage the values of the primitive and childlike, the dream, the immediate emotional response, the life of fantasy, and the transcendent symbol. 40
]n aliT own day, much the same sort of rurning point has been reached in the history of science. It is ::ts if the
aesthet ic ground pioneered by the artists now unfolds before us as a new ontological awareness. We are at a moment when the reality to which scientists address them45 selves comes mOre and more to be recognized as but one segment of a far broader spectrum. Science, for so long regarded as our single valid picture of the world, now emerges as, also, a school: a schooL of conscious-ness, beside which alternative realities take their place. 50
There are, so far, only fragile and scattered beginnines of this perception. They are still the ~ubterranean history of our time. How far they will carry toward liberat-
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3:
TH E MAIN IDEA • 47
ing us from the orthodox world view of the technocratic establishment is still doubtful. These days, many gestures 55 of rebellion are subtly denatured, adjusted, and converted into oaths of allegiance. In our society at large, little beyond submerged unease challe nges the lingering authority of science and technique, that dull ache at the bottom of the soul we refer to when we speak (usually too glibly) of an 60 "age of anxiety," an "age of longing."
48
VERBA L REASONING
&
MATHEMATICAL TECHN IQUES
Answer the following questions without going back to the passage. If you don't know the answer, guess.
YOU MAY NOT LOOK AT THE PASSAGE! •
Is the author male or female?
•
Does the author have long or short hair?
•
How old is the author?
•
"Vhat political party is the author a member of?
•
Would the author prefer a wild party ur a night at the upera?
•
Do you think you would like the author?
•
What does the author do for a living?
These are the types of questions that you should be able to answer with prejudice if you have read the passage the way you should. If you can answer these questions, you have compared the author to people of your past and categorized the author accordingly. This means that you have a better understanding of who the author is, and how he would answer the MeAT questions about his own passage. The previous questions were asked to make you realize how you should be trying to understand the author. You should not be asking yourself these questions on a real MeAT. Here are some questions that you should ask yourself on a real MeAT:
YO U MAY NOT LOOK AT THE PASSAGE! •
If the author were sitting in front of you, would he or she want to discuss science or art?
•
What emotion, if any, is the author feeling?
•
Is the author a scientist?
•
Is the author conservative, liberal, or somewhere in the middle?
The answers to these questions are unequivocal. This author is discussing science, not art. Art is used as a lengthy, nearly incomprehensible introduction to make a point about science. The author doesn't even begin discussing the main idea until the beginning of the third paragraph. "In our own day, much the same sort of turning point has been reached in the history of science." When you read this, you should have been startled. You should have been thinking "Where did science come from? I thought we were talking about some esoteric art history crap that I really wasn't understanding." This one sentence should have said to you Ahaa' That other stuff was appetizer, now the author is going to discuss his real interest." Notice that it is at the beginning of the third paragraph that the writing actually becomes intelligible. In other words, the second hvo paragraphs are much easier to read. This is because the author is interested in this topic and knows what he wants to say. The art stuff was a poorly written introduction and the author had not thought it through with any clarity. If you spent lots of time rereading the first two paragraphs, trying to master them, you wasted your time. The author didn't even master them; how could you? U
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LECTURE 3: THE MAIN IDEA .
The author is frustrated and possibly even bitter. He is so angry, that he is n ameca lling. For instance, he calls the scientific community "the technocratic establishment". The tone of the p assage is like that of a whining child. He blames scientists for being too con servative and thus creating "an age of anxiety", as if the anxiety of most people wo uld be relieved if scientists were less practical. In the last sentence, he even blames u s, his reader, for not taking his issue more seriously. The a uthor is positively paranoid. Notice tha t his adversaries move against him "subtly" as if tryin g to hide their evil intentions. They take "oaths of allegian ce" like some NAZI cult. This is way overdone when you consider that the guy's only complaint is that science isn' t liberal enough in its approach. The author is certainly not a scientist. First of all, he writes like a poet not a scientist: "orthodox world view of the technocratic establishment", "subterranean history of our time", "gestures of rebellion subtly denatured". Secondly, his whole point is that he is upset with scientists. (An entire separate a rgument can be made tha t his point results from his misunderstanding of how science progresses.) And finally, he talks like a member of some pyramid cult, not a scientist: "alternative rea lities" and "ontological awareness". This author probably flunked high school physics and just can't get over it. The author is cer tainly liberal, or anti-establishment. H e talks abou t "liberating us" and "rebellion" among other things. Now, with this understandin g of the author, answer the questions on the next page.
YOU MAY NOT LOOK AT THE PASSAGE!
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49
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&
MATHEMATICAL
TECHNIOU ES
1. The author believes that in "the subterranean history of our time" (line 5 1-52) we find the beginnings of a: A.
B. C. D.
renewaJ of allegiance to traditional values. redefinition of art. redefinition of scie nce. single va lid picture of the world.
6. The claim that the unease mentioned in line 57 is "submerged" most directly illustrates the idea that: A. B. C.
2. The author compares the idea that: A.
B.
C.
D.
art
and science mainly in support of
the conven tions of science, like those of art, are now
beginning to be recognized as but one segment of a far broader spectrum. aesthetic orthodoxies of the past, unlike scientific olthodoxies of the present, make up only one traditi on among many. artistic ac;. well as scie ntific revolutionaries open our minds to a more global conception of art. artists of the past have provided inspiration to the scientists of the present
3. The two kinds of art discussed in the passage are the: A. B. C. D.
aesthetic and the innovative. dull and the shocking. traditional and the innovative. representative and the traditional.
D.
7. Based on the information in the passage, the author would most likely claim that someone who did NOT agree with his view of, ~cience was:
A. JI. C. D.
A. B. C.
D.
technocratic establishment is opposed to scientific inquiry. traditional perception of science is identical to the world view of the technocratic establishment. cu rrent perceptions of science are identical to those of art. technocratic establishment has the same world view as the artistic revolutionaries of the past.
5. Whic h of the following concepts does the author illustrate with specific examples? A. B. C. D.
Scientific Scientific Aesthetic Aesthetic
in novatio ns of th~ pff~sent innovations of the past innovations of the present orthodoxies of the past
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dishGnest. conformist. rebellious. -imaginative.
8. Hased ,on infonnation in the passage, which of the following opinions could most reasonably be ascribed to the author? A. B.
e. 4. The author's statement "How far [new perceptions of science] will carry toward liberating us from the orthodox world view of the technocratic establishment is still doubtful" (lines 52-54) assumes that the:
our great-grandparents were jolted by the collapse of academic certainty. we have grown accustomed to the notion that there is more than one valid aesthetic vision. so far, new perceptions of science are only fragile and scattered. the authority of science is rapidly being eroded.
D.
It is misguided to rebel against scientific authority. The world views of other disciplines may have something valuable to teach the scientific community. Art that rebels against established traditions cannot be taken seriously. The main cause of modern anxiety and longing is our rash embrace of new scientific and artis tic theones.
me author's views as presented in the passage would most likely mean acknow ledging that:
9. Adopting A.
B. C. D.
it is not a good idea to accept traditional beliefs simply because they are traditional. we must return to established artistic and scientific values. the future is bleak for today's artists and scientists. the scientific community has given us little of benefit.
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
STOP.
LECTURE
Don't worry about the correct answers yet.
YOU MAY NOT LOOK AT THE PASSAGE YET! The first thing to notice is that only qu estion 5 requires a ny information from the first two paragraphs, and question 5 was a question about detail, not concept. This is because the first two paragraphs are n ot abo ut the main idea. The second thing to notice is that none of the questions require us to go back to the passage, even though some refer u s to specific line numbers. All but question 5 are answerable directly from the main idea. Question 5 is a detailed question, but before you run back to the passage to fin4 the anpwer, look a t the possibilities. The chances are that you remembered RaphaeLand Rembrandt from the first paragraph. These are specific examples of "aesthetic orthodox ies of the past". No tice tha t many of the questions can be rephrased to say "The author thinks _ _ _ _ ." This is typical of an MCAT passage, and that's w h y you must "know your author". Question 1: Forget about the quote for a m oment. Simplify the question to say "The author tl1inks tha t we find the begirmings of a:" Answer C is the main idea. Certainly the author would disagree with A, B, and D. Question 2: liThe author thinks: " that scien ce is Uke art, and that conventions of both are but p art of a larger spedrum. B says science is not like art; the opposite of what the a uthor thinks. C says that scientific revolutionaries are changing science; the a uthor is fru strated because this is not really h appening . D says scientists of the present a re opening their minds to new ideas; the author complains that they are not. Question 3: The main idea of the passage contains the theme of traditional vs. inno va ti ve. Question 4: Ignore the quotes w1til you need them. Without the quotes, the questions says "The author's statement assumes that the:" In oth er words, "The author thinks ." C and D are exactly opposite to w ha t the author thinks. Answer A plays a common game on the MCAT. They take the author 's view too far. They want you to think "the author d oesn' t like the scientists; therefore, he thinks the scientists can' t even do science." Even this author wo uldn't go that far. A is incorrect. Answer B requires you to realize that the " technocra tic establishment" is conservative. Question 6: A nswer D is out because it disagrees with the main idea, and C is the only an swer that supports the main idea. However, this question is best answered b y comparing the anSwer choices with the question. The question asks for an example of "submerged W1ease". "jolted" in answer choice A certainly doesn't describe s ubme rged unease. "grow n accustomed" in answer choice B certainly does not describe submerged unease. Answer choice C could describe submerged W1ease, and it does describe the main idea. It is the best an swer. Question 7: The author is rebellious and imaginative. If yo u disagree w ith h im, he thinks yo u are a conformist, which, by the way, is wo rse than dishonest as far as he's concerned . Question 8: "The a uthor thinks ." The whole p oint of the intra is to say that the scientific community should learn fram the diScipline of art.
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3:
THE MAIN IDEA . 51
52 .
VERBAL REASONING
&
MATHEMATICAL TECHNIQUES
Question 9: "The author thinks ." The author is a rebel. He thinks you should always question authority. Notice choice D is another example of taking things too far. No sane individual could argue that science has provided little benefit. Answer choice C would be incorrect even if it had not included 'artists'. It would have been too extreme.
NOW YOU MAY LOOK AT THE PASSAGE. Hopefully, we have demonstrated the power of knowing the author and understanding the main idea. Remember to use all four of your tools, and, most importantly, read and answer questions with confidence. , If you have problems, go back to the basics of this manual. Figure out what part of our strategy and tactics you aren't using, and use it.
(
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How to Study for the Verbal Reasoning Section 4.1
How to Study for the Verbal Reasoning Sectio n
One might think that studying the correct answers to many verbal questions in order to discover why they are correct would be a helpful exercise toward improving your score. On the contrary, it's probably a waste of yo ur precious time. It is rarely useful to go back to old tests and learn the logic used to explain why the correct answers are correct. By doing so, you may learn something about the topic of the passage, but you do not learn what you can do differently next time to improve your score. Since most explanations justify answers by pointing to a specific place in the passage that is claimed to support or even prove the correct answer, such practices can even lower your MeAT score by giving you the false impression that answers can be found in a specific place in the p assage. Most MeAT answers require an lmderstanding of the passage as a whole and cannot be proven correct by reading from one place in the passage. In most verbal materials, explanations tend to be too brief and not particularly insightful. It is doubtful that reading them w ill increase your reading comprehension skills. One method for increasing your reading comprehension skills is to join a book club or organize a reading group, and discuss things that yo u have read. This is not particularly p rac tical for most premeds, and even this idea is only eifective if there are strong, insightful readers in your club or group. The often posited advice of reading lots of magazine and newspaper articles on yo ur own is a Significantly less effective method for improving your reading comprehension skills. At the very least, you should be spending your reading time doing verbal passages followed by questions and not just articles without questions. The most effective method of study to improve your MeAT verbal score is to do the following: 1.
Take a verbal test under strict timed conditions and score yourself.
2.
Take a break from verba l for at least one day.
3.
Take the set of questions for the first passage in the verbal exam that you recently finished and examine the questions a nd each answer choice as if you had never read the passage, as was done in Lecture 2 of this book. If this step takes you less than 30 minutes per passage, then do it again because yo u missed quite a bit.
4.
Repeat step 3 for each passage.
5.
Take a break from verbal for at least one d ay.
54 .
V ERBAL REASON ING
&
MATHEMATICAL TECHNIQUES
6.
Carefully read the first passage in the same verbal test, and write out a precisely worded main idea in one or two complete sentences being certain that your main idea expresses the author 's opinion or stance on the issues.
7.
Match your main idea to each question and all the answer choices and see what insights you gain into answering MCAT questions as was done in Lecture 3 of this book.
8.
Repeat steps 6 and 7 for each passage.
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STOP! DO NOT LOOK AT THESE EXAMS UNTIL CLASS.
30-MINUTE IN-CLASS EXAM FOR LECTURE 1
.
-'
55
Passage I (Questions 1-7)
Some Marxist theorists have speculated that Marx would tar both Kant and Hume as "bourgeois" philosophers. Answering why is an extremely difficult question .... It may 55 lie in this sentence from Marx's Manuscripts of 1844: "The interests of the capitalist and those of the workers are therefore, one and the same, assert the bourgeois economists." ... In the directly preceding passage, Marx explained how capitalists try to mask the class struggle and exploitation 60 inherent in capitalist production, by monopolizing cultural institutions to establish hegemonic control over the working class. This is why, he explains, workers are actually conditioned to be grateful to the factory owner for allowing them to produce goods, only to have these taken away and sold. 65 Based on these writings, Marx would probably see any system. that sought out a universal theory of morality as ignoring the opposing economic classes in society, and easily adapted to give the workers a false perception of the unitary interest of them and their oppressors.
Philosophers Immanuel Kant and David Hume both spent their professional careers searching for a universal principle of morality. Considering that they began their searches with seemingly irreconcilable ideas of where to 5 look, the similarity in the moral systems th.ey constructed is surprising .... Hume decided at the outset that a moral system must be practical, and maintained that, since reason is only useful for disinterested comparison, and since only sentiment 10 (emotion) is capable of stirring people to action, the practical study of morality should be concerned with sentiment. Hume begins with the assumption that whether something is judged moral or praiseworthy depends on the circumstances. He says, "What each man feels within himself is 15 the standard of sentiment." . By contrast, Kant begins by assuming that, while some time should be d~voted to studying practical morality ("ethics"), it is also valuable to have an absolute system of morality based solely on reason, to be called "meta~ 20 physics". Thatforms the core of his laborious exploration of pure logic, called Grounding for the Metaphysics of Morals.
}, The word tar (line 53) is used in the sense of:
A. B,
C.
D. For both authors, t1\e problem of subjectivity threatened to prevent unbiased analysis of morality. So both 25 invented systems of "moral feedback," in which the philosophic actor tries to imagine the results of his actions as if someone else were performing them. Still, the final evaluation of the action's worth must, in the end, be subjective. Kant's ultimate standard of morality is the "categorical 30 imperative". It is phrased, "[To follow your] duty, act as if the physical act of your action were to become a universal law of nature." This means that, before one does anything, one shou1d forget one's own motives for a moment, and ask if he would want everyone to do as he does. If the answer is 35 no, then his subjective desire is different from his objective assessment, and the action is contrary to duty.
asphalt. suggest. label. stick.
2. According to the passage, Marx would have disagreed with Kant and Hume over which of the following ideas? A. B. C. D.
What each man feels within himself is the standard of sentiment. There is unifonnity in people's reason or emotion. The interests of the capitalists and the workers are one and the same. Morality is dependent upon the class struggle.
Hume says that moral actions are those that create agreeable sentiments in others, as well as in oneself. You can therefore judge what kind of sentiments your actions 40 may cause others to feel by imagining someone else performing that action, and thinking about what kind of sentiments it would inspire in you. To make both of these constructions possible, there is a notion of some kind of unifonnity in people's reason or emotion in both works; 45 Kant's reason for using pure logic was preCisely to bypass empirical differences between people and individual ' circumstances, and ... he has the belief that people using only logic must inevitably reach the same conclusions about the morality of their own actions. Hume also admits 50 that "the notion of morals implies some sentiment common to 011 mankind." ... Copyright © 2007 Exarnkrackers, Inc.
56
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6. Based upon passage information, Kant's system of " moral feedback" (line 25) d iffered from Hume's in that it might result in a situation wherein:
3. Assume that a universal principle of morality can be proven to exist. Which of the following hypotheses does . this assumption suggest?
A.
The author is correct; despite their genesis. it is not - surprising that Kant and Hume constructed similar systems. B. The author is correct; Marx, Hume, and Kant all constructed similar systems. C. The author is incorrect; Marx, Hume, and Kant di,d not all construct similar systems. D. The auth or is incorrect; despite their genesis, it is not swpri sing that Kant and Hume constructed similar systems. A.
B. C.
D,
7. According to the passage, Hume's ideas evolved to the point where he:
4. Accord ing to the author. in creating his moral system, Hume equated: A. B. C. D.
A.
circumstances with disinterested comparison. circumstances with absolutism. prac ti ~ality with stirring people to action. pract icality with reason.
B. C. D.
5. Based upon passage information. Marx most li ke ly believed that: , A. B. C. D.
one realized that his action might be 'right' so long as it didn' t become a universal law of nature. one realized that a universal law of nature was unnecessary in determining duty. one.-realized that his actio n might be 'wrong' even though it created agreeable sentiments in others, as well as in oneself. one realized that his action might be illogical if sentiment was not further considered.
realized tb at reason was an inseparable part of a uni versal system of morality. was considering the sentiments of others as well a~ himself. chose to essentially agree with Kant on a universal system of morality. decided that sentiment without action was a necessary component of his morality system.
there is no universal theory of morality. philosophers were part of the bourgeois. the workers could be easily fooled . Kant' s book supported capitalist exploitation.
"
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Passage II (Questions 8-14)
Both of these theologians wou ld hold that uprisings generally occur for the wrong reasons (i.e. worldly ambitions), because no tyrant can keep a true Christian from sal55 vation, which should be all Christians ' only concern in life. For Luther, salvation lies in faith emanating from personal understanding of Biblical teachings, and "it is impossible that anyone should have the gospel kept from him ... for it is a public teaching that moves freely." In Augustine's under60 standing, salvation is granted through the mercy of God, who sends hardship and death to test men's faith. The true Augustinian Christian would maintain his faith through any ordeal, and even if his body perishes, God will save him for his conviction.
In the Bible, when Jesus Christ was queried by skeptical IsraeHtes on how C hristians could continue to pay taxes to support the (pagan) Roman governors of Israel, he counseled; uRender unto Caesar what is Caesar's; render 5 unto God what is God's". The basic meaning of this guidance was that Christians may practice their faith while coexisting with the secular government. ... Later Christian theorists followed this example, often urging peaceful coexistence even with goyernmen~ which violated every pre10 cept oJ..Christian teachings .... The early Catholic bishop (St.) Augustine of Hippo and the Protestant dissident leader Martin Luther both advocated submission to the rule of tyrants, employing the analogy of a dual city or government, where earthly rule is 15 often oppressive, yet is balanced by the assurance of brotherly love in the kingdom of heaven....
8. Which of the following stateme nts, if true, would most directly challenge the ·principles of Martin Luther? A.
20
25
30
35
According to Luther, true Christians should be wi lling to suffer persecution, without seeking to resist it by the antiChristian methods of taking up arms in violent revolt, or seeking redress in the courts of the unbelievers. Luther even took the analogy so far as to suggest that it is the Christians themselves who most benefit from harsh sec ul ar 1aws, which protect them from exploitation and persecution by false Christians and heathen. In fact, when commenting on popular "Christian" uprisings, Luther lays more blame on self-righteous Christian rebels (such as various German peasant rebels of the age) than upon their oppressors. teaching that, regardless of their rulers ' faults, rebels immediately cease to be Christians up~m taking up arms. and incur further displeasure from God by blasphemously arrogating His name and sCriptures for an un-Christian cause: Luther claims that this is furthermore exacerbated because what the rebels often want is material benefit rather than religious freedom .... According to Luther, it is the role of God alone to punish rulers, and rebels' usurpation of that authority for themselves adds another sin to the list of charges against them; they effectively revolt against God's justice.
B.
e. D.
9. Some theologians believe that kill ing and violence are acceptable when used in self-defense. An appropriate clarification of the passage would be the stipulation that: A.
both Luther and Augustine would bave disagreed . with this belief. B. both Lutber and Augustine would have agreed with this belief. e. only Augustine might have agreed with this belief. D. only Luther might have agreed with this belief.
10_ If the information in lines 38-55 is correct, one could most reasonably conclude that, compared to Luther, Augustine was:
St. Augustine would have agreed. His Confessions and other theological writings maintained that God gran ts 40 earthly rule to·C hri stians and pagans alike, but that His inscrutable wi ll is always just. Tn the case of a revolt, Luther claimed, both sides inevitably incur divine punishment, since "God hates both tyrants and rebels; therefore He se ts them on each other". Luther's vision of God raising up the 45 peasa nts to punish their tyrants perfectly matches Augustine's frequent portrayal of the Germanic " barbarians" who finally sacked Rome as the brutal instruments of God, sent to crush the corrupt Romans' arroganc e. Augustine never considers the individual fate of these living 50 tools, but Luther maintains that God sends the devil to stir them up with lies, and afterwards they go to eternal torment.
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The Bible's Old Testament refers to a period before the birth of Jesus. The Bible alone contains only a small part of what Jesus intended for his followers. The German "barbarians" who sacked Rome had been previously converted. Augustine's understanding of salvation granted through the mercy of Christ was flawed.
A. R C. D.
58
much more reasonably inclined. more prepared to 'define God 's will. less eager to send people to eternal torment. less willing to announce God's final judgment on those who had sinned.
GO ON TO THE NEXT PAGE.
13. The author's primary purpose in the passage is apparently:
11. The author's attitude toward the tbeories of Augustine and Luther in the passage is most accurately described as:
A. B. .
C. D.
A.
disapproving. mistrustful. neutral. favorable.
to clarify the differences between the ways in which the early Catholics and Protestants dealt with persecution.
B.
to justify the persecution of early Christians by sec-
C.
to consider the similarities between the ways in
ular governments.
12. What is tbe meaning of the phrase; "The true AlI&ustinian Christian would maintain his faith through any ordeal, and even if his body perishes, God will save him for his conviction" (lines 61-64)? A. B. C. D.
which the early Catholics and Protestants dealt with persecution.
D.
This Christian would end up in heaven because of his beliefs. God would save this Christian for judgment at the end of the Christian's ordeal. God would save this Christian from his ordeal and judge him. It was not necessary for this Christian to die for him
14. What is the most serious apparent weakness of the information described? A.
LO be convicted.
B.
C.
D.
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to question the passive practices of the early Catholics and Protestants when faced with persecution.
59
While implying that Christians' may coexist with a secular government, it differentiates between Catholics and Protestants. While implying representation of Augustine and Luther, its conclusions are based primarily on information according to Luther. While implying representation of all Christian theorists, only Augustine and Luther are mentioned. While implying agreement between Augustine and Luther, their attitudes were clearly opposed.
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Passage III (Questions 15-21)
50
... She stayed there for three years, in relative isolation. It was during that time that she was dubbed Typhoid Mary. Mallon despised the moniker and protested all her life that she was healthy aud could not be a disease carrier. As she told a newspaper, "I have never had typhoid in my 55 life and have always been healthy. Why should I be banished like a leper and compelled to li ve in solitary confinement ... ?"
Of all the bizarre and melancholy fates that could befall au otherwise ordinary person, Mary Mallon's has to be among the most sad aud peculiar.... Like millions before and si nce, she came to this country from Ireland, seeking a 5 !>;etter life. Never " lried" in any sense, i~ead, slie was forced by public health officials to live for a total of 26 years on a tiny island in the East River, isolated from and shunned by her fellow humans. And, while she was not the only one of her kind, her name became synonymous with 10 disease and death. She was Typhoid Mary, and her story really begins on Long Island.
After a short period of freedom in which Mallon failed to comply with the health inspector's requirements, she was 60 eventually sent back to North Brother Island, where she lived the rest of her life, alone in a one-room cottage. In 1938 when she died, a newspaper noted tbere were 237 other typhoid carriers li ving under city health department observation. But she was the only one kept isolated for 65 years, a result as much of prejudice toward the Irish aud noncompliant women as of a public health threat.
In the summer of 1906, Mallon, was working as a cook for a wealthy New York banker, Charles Henry Warren, and his family. The Warrens had rented a spacious 15 house in Oyster Bay, "in a desirable part of the village," for the summer. From August 27 to September 3, six of the II people in the house came down with typhoid fever, including Mrs. Warren, two daughters, two maids and a gardener. Two investigators were unable to find contaminated water 20 or food to explain the outbreak. Worried they wouldn't be able to rent the house unless they figured out the source of the disease, the owners, in the winter of 1906, hired George Soper, a sanitary engineer.
15, The author probably mentions tl13t Mallon was " never 'lried' in any sense" (line 5) in order: A. B.
Soper soon dismissed " soft clams" and other potential 25 contaminants as the cause and began to focus aD the family. He later wrote, " It was found that !he family had changed cooks about three weeks before the typhoid epidemic broke out ... She remained with the family only a short time, leaving about three weeks after the outbreak occurred ... [and] 30 seemed to be in perfect health." Soper became convinced that this woman was a healthy carrier of the disease, and, in so doing. was the first to identify a healthy typhoid carrier in the United States. Although his deduction was undoubtedly brilliant, his handling of Mallon was not.
C,
D,
to demonstrate the power of the wealthy at that time. to provide a comparison with people who have actually committed a crime. to illustrate . the persistence of Soper's investigations. to support the claim that she deserved at least a hearing.
16. According to the passage, the first two investigators were unable to find the cause of the outbreak (lines 19-20). The information presented on typhoid makes which of the following explanations most plausible? A.
35
Soper tracked Mallon down to a home on Park Avenue in Manhattan where she was a cook. Appearing without warning, Soper told her she was spreading death and disease through her cooking aud that he wanted samples of her feces, urine, and blood for tests. In a later description, 40 Soper wrote, " It did not take Mary long to react to this suggestion. She seized a carving fork and advanced in my direction. I passed rapidly down the narrow hall through the tall iron gate."
B,
C. D.
They focused too closely on the "soft clams" that Soper later discredited. Typhoid is not really passed through contaminated food or water. They never considered that typhoid could be carried by a healthy person. By this time, Mallon was no longer e mployed by the Warren family.
Convinced by Soper's information, Ibe New York City 45 health inspector in March 1907, carried Mallon off, screaming and kicking, to a hospital, where her feces did indeed show high concentrations of typhoid bacilli. She was moved to an isolation cottage on the grounds of the Riverside Hospital, between the Bronx and Rikers Islaud.
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17. The author's argument that Mallon's isolation was "a result as much of prejudice ... as of a public health threat" (lines 64-65) is most weakened by which idea in the passage?
A. B. C. D.
20. According to passage information, Mallon worked for the Warren family for approximately: A. B. C. D.
Mallon's primary occupation was as a cook. Mallon did not believe that she was a carrier of the disease. Mallon would not abide by the health inspector's requirements. Mallon actually was the source of the typhoid outbreak in the Warren home.
21. The contention that in "1938 when [Mallon] died, ... there were 237 other typhoid carriers living under city health department observation. But she was the only one kept isolated for years" (lines 61-65), can most justifiably be interpreted as support for the idea that:
18. Which of the following statements is the most reasonable conclusion that can be drawn from the author's description of the typhoid outbreak in the house at Oyster Bay? A. B.
C. D.
two weeks. three weeks. five weeks. six weeks.
A. B.
The Warren family did not hire Soper. The two investigators were hired by the Warrens. The Warren family hired Soper. The owners were anxious to sell the house.
C. D.
Mallon was unfairly treated by the city health department. Mallon's isolation might have stemmed from the health department's early ignorance of the disease. The "other" 237 typhoid cartiers were all kept isolated at one time or another. The "other" 237 typhoid cartiers were much like Mallon.
19. Passage information indicates that which of the following statements must be true? A. B.
C. D.
Mallon had probably not infected anyone prior to the Warren family. Mallon was almost certainly not washing her hands prior to preparing the Warren's meals. Being labeled 'Typhoid Mary' by the press was the primary reason for her confinement. The health inspector was doubtless prejudiced toward the Irish. .
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STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
61
STOP.
3D-MINUTE IN-CLASS EXAM FOR LECTURE 2
63
Passage I (Questions 22-28)
fully, then count out loud slowly, while enunciating each nnmber clearly, aiming for a count of 60.
Polling research shows !hat the ideal speaking voice should be clear and intelligible, of moderate volume and pace, and inflected to suggest !he emotions ·exRressed. To suggest credibility, the voice's tone should be pitched as 5 low as is naturally possible... A low pitch is' desirable in both genders, since it is popularly associated with trilthtelling. However, an artificially lowered voice can sacrifice intelligibility, which is irritating to listeners. Thus, speakers should experiment to find their optimal level, which will be 10 at their lowest intelligible pitch .... To deepen the pitch, speakers should make an extra-deep inhalation before speaking, and then exhale fully as they speak.
55
Loudness, or volume, is di stinct from pitch, though the remedy for overly soft-spoken people is similar. They can manage to speak louder by first inhaling more deepl y, which allows them added lungpower to project their sentences. Alternately, they can pause more often, say :fewer 60 words in every breath, and thu s leaving more air power for each. There are those who speak too softly not because of improper breathing, but due to psychological factors: they may be shy and not wish to be obtrusive, or may not hear that their words are too soft to be intelligible at a distance. 65 ... For them, one useful exerci se is to recognize the Jive ' . basic volume levels (whisper, hushed, conversation, loud, and yelling) by practicing speaking a word in each of these modes ....
HabilS to be avoided (because !hey irritate most listeners) include monotony, mumbling, grating, pretension, 15 high-pitched whining, and brea!hiness . ... Among the best practitioners ' of mainstream vocal "propriety" are famous news anchors, like Walter Cronkite, Dan Rather, and Jane Pauley, for whom vocal image is a key component of their job success. However, everyone who communicates should 20 be aware that their voice is a critical component of their audience's perceptions of them, comprising abollt 38% of the overall impression imparted by their presentation. (By comparison, appearance accounts for about 50% of the speaker's impact, and the quality of content accounts for a 25 mere 6%.) .. '
22. According to the passage informati on, which of the following would be most likely, if a person, who was talking to you, attempted ,to make their voice sound unusually low? A. B.
C. D.
For those not born with naturally pleasant voices, or worse still, those with naturally unpleasant ones, speech training can be invaluable in improving impressions. No voice teacher is necessarily required, since practice alone 30 can ' produce significant improvements. However, some special equipment is needed. Because talking always causes cranial resonance. which distorts the speaker's hearing, DO one can hear what his voice really sounds like to a listener. Thus, some sort of tape recorder or o!her feedback is a 35 virtual necessity.
You might think that they were lying. They could be irritated with you. They might well sound monotonous. You could find them difficul t to understand.
23. The author most likely believes that one of the main purposes of speaking, during a face-to-face meeting, should be to: A. B. C. D.
convey a favorable impression. effeEtively transmit your ideas. gain leverage. communicate as naturally as possible.
Speaking begins with breathing, since speech is just exhaled air that sets the vocal cords to resonating. If ins ufficient air is inhaled before speaking, the words formed must necessarily be strained and breathless . ... 40 Diaphragmatic breathing results in a deeper voice than . upper-lung breathing. People tend to stick out their cheslS and inhale shaUowly with the upper lungs, resulting in a high-pitched voice, which must also be rapid to avoid running out of breath before the sentence ends. In diaphrag45 matic breathing, the lower abdomen moves out, inflating the bottom two-thirds of the lungs full y, but the shoulders do not rise. To practice switching from shallow breathing to deep, diaphragmatic inhalations, it helps to lie on the floor and breath e naturally, since diaphragmatic breathin g is 50 natural when prone .... Increasing lung capacity will also deepen the voice and permit longer sentences without paus-' ing .... One method to increase lung capacity is to inbale Copyrig ht © 2007 Examkracke rs, Inc.
64
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24. The author provides a list of "habits to be avoided" (lines J 3·15). Whicb of the babits wou Jd the suggestions in this passage not help a speaker to curb? A.
monotony
B. C. D.
breathiness
27. Which of the following assertions is most clearly a thesis presented by tile author? A. B.
pretension
hi gh-pitched whining
C. D.
25. The term ideal speaking voice (line 1) refers implicitly to
a voice that is: A.
B. C. D.
the most pleasant to listen to. the most persuasive. the least irritating. the most natural.
Speakers can gain by improving his or her speaki ng voices. The tone of tbe ideal speaking voice should be pitched as low as possible. What you are saying is more important tban bow you are saying it. Emotional inflections can be an irritating aspect of a speaker's voice.
28. Tbe ideas discussed in this passage would likely be of most use to: A. B. C. D.
26. Passage information indicates that a person speaking in a high· pitched voice might be doing all of the following
A doctor A journalist A radio show personality A television commentator
EXCEPT: I. U.
m.
Breathing with iheir upper lungs Breathing deeply I.ying
A. B. C. D.
I only II only III only I and III only
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Passage II (Questions 29-35)
Peruvian territory", their mission is both to keep guerrillas and drugs out of Peru .... Though understandable, this has 55 in turn, pushed Columbia to respond in kind with more Columbian border troops facing Peru. This brinksmanship seriously depletes resources from these needy countries , which might be bener spent elsewhere.
A great deal of international conflict arises from border disputes. Tbroughout history. particularly along borders which have been "artificially" defined, rather than utilizing more-natural pre-existing cultural and' geograpbi5 cal demarcations, there has been a constant ebb and flow as nations have sought to consolidate their borders and their security. However. with ever-increasing economic disparity between many bordering countries, these conflicts have changed and now center more around issues of immigra10 tion. . .. Such situ ati ons are prevalent today in countries such as New Zealand, the Colombian-Peruvian border, and the U.S.'s Mexican border. These instances ... exemplify the problems caused by such disputes.
Traditionally, these two countries might have been 60 attempting to secure their borders from invading countries, or even seeking to expand their own territories and acquire additional resources. However, Ecuador and Peru are protecting their borders from rogue drug traffickers and guerillas, not Colombia's government. .. , Neither side is 65 attempting to acquire new territory. but rather to secure and protect that which they already hold ....
Presently, New Zealand's conflict stems from illegal 15 immigration into its tenitory, mostly from the Chinese island-province Pujian .... Fujian is situated on China's southern coast, near Taiwan. Many Fujianese immigrants use New Zealand, because of its location, as a steppingstone to their final goal, the U.S. Their transpon is usually 20 a smuggling boat's hold, where living conditions are inadequate and sometimes dangerous, with insufficient food, sanitation, and ventilation. Within the past year, U.s. officials found tbree Chinese immigrants in a smuggling boat's sealed cargo container, dead from suffocation .... Recently, 25 New Zealand attempted to deal with these aliens hy enacting new immigration laws, which hasten the process required (0 depon them.
29. The author's discussion of "push" and "pull" factors (line 29) most accurately implies that: A. B. C.
D.
"pull" factors compel someone to leave, "push" factors induce someone to come. "pull" factors induce someone to come, "push" factors also induce someone to come. "push" factors require someone to leave, "pull" factors also compel someone to leave. "push" factors compel someone to leave, "pull" factors induce someone to come.
while while while
while
30. Given the information in the passage, if "'anificially' defined" borders (line 3) were eliminated throughout the world, which of the following outcomes would most likely occur?
The reasons for the Chinese immigrant's journey stems from both "push" and "pull" factors relative to the 30 countries of origin and destination. For example, the Fujianese feel compelled ("pushed") to leave because of the area's low standard of living. The pocr wages, bad housing, and lack of political freedom can also be seen as "pull" factors, due to the idea that the Fujianese understand that 35 life would be better in other countries. The U.S. and New Zealand offer much higher wages, a better standard of living, and political freedom ... These push and pull factors are powerful incentives .... What keeps New Zealand from experiencing an even more profound illegal immigration 40 problem is that the immigrants often do not settle there.
A. B. C. D.
People would naturally immigrate to areas with higher standards of living. Nations would encounter less traditional border strife. Nations would require greater border security measures. People wou ld live more harmoniously.
The border issue between Colombia and its neighbors is another illustration of international conflict. Colombia lies along a corridor from South to Central America. This region has historically been politically uns.table, partially 45 due to regional narcotics trafficking, and the wars this engenders. Colombia, itself. is notorious for its export of drugs, especially cocaine. This reputation forces neighboring countries to strengthen patrols over adjoining borders. Recently, Peru deployed additional soldiers to its border 50 with Colombia. Although Peruvian President Fujimori denied any diplomatic problems and stated his troops were there "to guarantee the sovereignty and integrity of Copyright © 2007 Examkracke rs. Inc.
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31. Which of the following assertions does the author support with an example? A. B. C. D.
34. The author implies that which of the following is not one of the reasons that Peruvian President Fujimori deployed soldiers to its borders with Columbia?
Transportation methods used by illegal immigrants are sometimes dangerous. Peru and Columbia are seeking to expand their owl! territories. New Zealand has enacted laws that hasten deportation proceedings. The mission of the Peruvian troops is to keep guerillas and drugs out of Peru.
I. Fujimori is attempting to keep drugs out of his country. n. Fujimori fears that Columbia is seeking to expand its territories. III. Fujimori is probably concerned that Columbia wants to acquire additional resources. A. B. C. D.
32. The passage as a whole suggests that in order for a nation to slow the exodus of its inhabitants to other countries, it must:
A. B. C. D.
35. It seems likely that New Zealand may be suffering less from immigration issues than the United States for which of the following reasons:
become more attractive to those who are leaving. abandon the traditional methods of guarding borders. respond in some way'to the conflicts arising from border disputes. answer the challenges set forth by adjoining coun-
I. The u.s. offers higher wages than New Zealand. II. New laws, enacted in New Zealand, a1low faster deportation proceedings. III. Immigrants often do not settle in New Zealand.
tries.
A. B.
33. If the passage information is correct. what inference is justified by the fact that virtually no immigration from West Berlin to adjoining East Berlin occurred, over the 40 years before the period described? A. B. C.
D.
J only II only III only II and III only
C. D.
II only III only II and III only J, II, and III
Crossing the heavily guarded borders between West and East Berlin was very dangerous. It was understood that life would be better in East Berlin. The inhabitants of both 'Berlins' were happy to remain where they were.
The economic conditions of West Berlin were much more favorable than those of East Berlin.
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Passage III (Questions 36-42)
T?rt law, gradually realizing the difficulty of proving negligence, has moved towards allowing recovery with ever less proof of negligence.... Potentially, the most promis55 ing development in tort (personal injury) law has been the advent of strict liability, which waives plaintiffs' need to prove the defendant's carelessness in certain instances where the carelessness is- obvious, or could -have resulted from no factor other than negligence. Unfortunately, the 60 application of strict liability is severely constrained by legal doctrine, which limits its application _ to a small range of "unusually hazardous activities".
Perhaps the greatest problem with the law of personal injury is its uncertainty about its own purpose-does it exist to compensate victims fully, or to deter careless \yrol1gdoers fully? It must choose, because these two aims are 5 mutually exclusive: tort awards cannot fully compensate and correctly deter, as long as there are administrative costs involved in obtaining an award. Assume a plaintiff's lawyer charges a 30% contingency fee upon winning a case (or the equivalent flat fee). If the plaintiff is awarded 100% of the 10 damages suffered, she only receives comp~nsation for 70% of her injuries. If she is paid in full, then the defendant is paying 130% of the actual harm caused, and is overdeterred.
Sometimes, the' criteria for imposing strict liability seem arbitrary.... For example, the law permits strict liabil65 ity only when the expected darnage-a product of risk and probable harm-is high .... Yet it is equally appropriate when the damage is slight. Consider the same tanker spilling toxic chemicals along a 200-mile stretch of farmland. Imagine the total cost of decontamination is 70 $1,000,000, but the costs are borne by 150 small farmers. In this scenario, the total damage is high, but comes to only $6,667 per plaintiff. Since just proving negligence may cost more, few will actually sue. The same applies for small harms; there is neither compensation nor deterrence, 75 because plaintiffs bear the loss, and defendants effectively have no incentive to prevent small harms ....
In reality, compensation tends toward ina_dequacy, and 15 not just because of administrative costs .. There is no compensation unless the plaintiff proves - "negligence", meaning a person may cause any amount of harm, but be excused from paying because she acted "reasol1ably" rather than carelessly. The hurdle of proving negligence also tends 20 toward inadequate deterrence, because even negligent injurers escape liability if plaintiffs cannot collect" convincing proof of negligence.
On the other hand, in a few cases, both compensation and deterrence are exorbitant, especially when a single jury 25 award tries to be both. Consider the following permutation on actual events. A tanker passing through a residential neighborhood leaks acrylonitrile, destroying several homes and poisoning one. Angry residents sue the company for designing its tanker cars negligently. At trial, the company's 30 counsel-a good economist, but a poor lawyer-admits safer tankers were available, but the cost is prohibitive. After extensive cost-benefit analyses, he says, the company found it cheaper just to pay victims for their losses, as it now offers to do. Sound fair? The company would be lucky 35 to escape punitive damages! Remember, these have been applied where judges deemed that even full compensation is inadequate deterrence; generally when the plaintiff's conduct is seen as malicious. However, they may also be awarded when there is "a conscious and deliberate disre40 gard of others."... In this case, the company's cost-benefit analysis is economically "correct" and justifiable: if the new tanker costs more than it saves, it is inefficient. On the other hand, how many juries--or even judges-will see this very analysis as anything but a cold 45 and calculating balancing of profits against the costs of human life? If (arbitrary) punitive damages are granted, plaintiffs emerge overcompensated, and defendants payout of proportion to harm. Since repeated punitive awards are allowed, the company may be forced to buy the expensive 50 _tankers. This is unfortunate, because it results in a waste of resources.
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36. What is the author's response to the standard story about the woman who spills hot McDonald's coffee in her lap, sues and gets several million dollars? A. B,
C. D.
68
This story does not reflect that compensation is usually insufficient. This story is a good example of just the right amount of compensation. This story does not reflect that deterrence is costly. In this story, the woman was malicious.
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40. Suppose that a study found that police agencies routinely set aside large amounts of money in their yearly budgets, which they expect to payout in lawsuits against their agency. Which of the following statements is an assumption of the author about the effects of lawsuit awards that would b'e called into question?
37. Which of the following assertions is the most effective argument against the author's opinion that personal injury law cannot satisfactorily compensate and deter "as long as there are administrative costs involved in obtaining an award" (lines 1-7)? A. B. C. D.
These administrative costs are inconsequential. , Attorneys are a necessary part of the judicial system and should be compensated for their work. The administrative costs should be added to the compensation received by the plaintiff. The administrative costs should be subtracted from the compensation received by the plaintiff.
. Simply proving negligence can be a very costly process. B. Many people will not sue because the process is too costly. C. If a plaintiff receives full compensation and administrative costs, the defendant is over-deterred. D. Depending upon the size of the award, a defendant police agency might not be deterred at all. A.
38. The passage indicates that its author would NOT agree with which of the following statements?
A. B.
C.
D.
41. Which of the following conclusions can justifiably be drawn from the experience of the tanker company's counsel mentioned in the passage?
Tanker companies are a good example of defendants who are under-deterred. Negligence on the part of the defendant is generally not difficult for the plaintiff to prove. The costs associated with suing and defending against suits can be tremendous. In many situations, over-deterrence results in primarily economic ramifications.
A. B.
C. D.
39. Assume that since the 9-11 terrorist attacks on the World Trade Center (WTC) buildings. all lawsuits have been settled by the WTC insurance companies, who have now mandated that they will no longer insure any building in the world that is over five stories tall. The author's comments suggest that this situation could reasonably be interpreted as evidence that: A. B. C. D.
the the the the
Good economists make for poor attorneys. Costs should never be considered prohibitive where safety is concerned. Toxic materials should not be shipped fhrough residential neighborhoods. Honesty is not always the best policy for an attorney.
42. The author argues that, "Potentially, the most promising development in tort (personal injury) law has been the advent of strict liability" (lines 54-56). These beliefs imply that: A.
insurance companies were over-deterred. insurance companies were under-deterred. plaintiffs were under-compensated. plaintiffs were overcompensated.
B. C. D.
the use of strict liability has become increasingly popular for defendants. the uses of strict liability should remain limited in scope. the author approves of waiving the requirement for proof, where carelessness is evident. the author approves of compensation where carelessness is evident.
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69
STOP.
3D-MINUTE IN-CLASS EXAM FOR LECTURE 3
71
Passage I (Questions 43-49)
coulg also argue that certain activities were labeled male, and· Chris molded her interests to include them, so she, would/be considered "male." It is virtually impossible to . 55 unravel which· came first, the labeling of her favored activities as "male" or her interest in them. Almost certainly, there is a -complex interaction between the two.
For most people, gender and gender identity go handin-hand. In other words, a female acts like a "woman". The fact that Gender Identity Disorder exists in the DSM IV [the official handbook of psychiatri,c disordersl as a diagnosis is 5 an admission on the part of psychologists that our society has clearly defined gender roles. These contribute to what it is generally considered "normal". Implied in the name of the disorder is an incongru~ty bet.ween assigned sexual organs and gender identity.
Remaining, however, is the question of treatment for Chris. Although she is currently a well-adjusted individual, 60 she acknowledges a de~ire to change her sexual assignment ,to match her gender. One might ask if this is really necessary, since she is already well adjusted. Part of what we strive for as psychologically healthy individuals is an acceptance of ourselves in a "natural" state. It is often the 65 ,case that, through psychotherapy, one learns that one may not necessarily have to change oneself as much as one's perception of self.... The effect of a physical sex reassignment operation on Chris' happiness cannot be foretold with 'complete certainty.
l a T h e most comprehensive case study of the disorder i.8 the recently published case history of "Chri§," a female patient who identified more closely with males, and stated a wish to become a male. In response, Chris adopted masculine mannerisms such as wearing men's clothes, 15 deepening her voice, sporting a masculine h,aircut, playing sports against male teams, etc. Chris believed she was born the wrong'sex; she was a man inside a woman's body. The incongruity is undisputed; she did everything that was within her means to have her peers recognize'her as a man.
In conclusion, it is interesting to note that Chris' desire to have the anatomy of a man is considered part of a "disorder," while a small-breasted woman's desire for breast implants would usually be construed as a desire to increase her femininity and not be labeled as such. Perhaps quis is 75 somewhere on a male-female continuum and, like the woman who desires the breast augmentation, is pushing herself toward the closer end of the spectrum into our neatly constructed gender dichotomy.
70
20
It is important to stress when looking for causal factors for Chris' disorder that Chris was in no way confused about her identity. In some ways, this makes Chris an ideal subject for studying the causes of the disorder. She is described as a relatively well-adjusted individual who did 25 well in school, was relatively well liked among her peers, and seemed capable of creating long-term intimate relationships with others. Because the disorder is clearly present (Chris is aware of it herself), basic defects in personality (i.e. maladjustment, inability to make friends) can be more 30 or less ruled out as causes of her uneasiness with her assigned sex. One must then ponder the age-old question of society as the cause.
43. The passage suggests that its author would probably disagree with which of the following statements? A.
It is possible that Chris participated in "male" activ-
B.
It is possible that Chris naturally participated in
C. D.
"male" activities. Chris was not confused about her identity. Most cultures have clearly defined gender roles.
ities in order to be considered male. The notion of gender identity is so heavily dependent on societal norms that, in this case, many psychologists 35 may believe society is the culprit. Similar to the ideas advanced in labeling theory, the mere labeling of some behaviors as "masculine" and others as "feminine" may have created the criteria for this disorder to be labeled as deviant or abnormal. In another culture, where the labeling 40 is different, would Chris have even felt the need to identify herself as distinctly male? Along the same lines of thinking, would the incongruity even be considered a "disorder" in another culture? Evidence varies regarding the two sides of the issue. 45 Upbringing and/or some biochemical abnormalities could account for the etiology of the disorder, indicating that it is not simply due to societal labels. Indeed, it may seem convincing to argue that Chris strongly identified with males very early in her life, and that this was reflected in 50 her interest in "male" activities (e.g. sports) and unease in using girls' bathrooms and playing on girls' teams. Yet one Copyright © 2007 Examkrackers, Inc.
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48. The author admits that, "The effect of a physical sex reassignment operation on Chris' happiness cannot be foretold with complete certainty" (lines 67-69). The author most likely believes that:
44. Implicit in the passage is the assumption that:
L
one should be )mppy in one's "natural" state. one can be well-adjusted, yet unhappy with one's "naturar state. III. one's perception of self is most important.
n.
A, B.
C. D.
I.
it is just as likely that psychotherapy would help Chris to change her perception of self. II. in our society, in the body of a woman, Chris will not be happy. III. a "sex reass ignment operation" would make Chris happier.
I only II only III only I and HI only
A. B. C. D.
45. The author of the passage would be most likely to agree with which of the following ideas expressed by other psychologists?
A.
B.
C.
D.
A DSM TV 'disorder' may not actually be a disorder at aU. The DSM TV is a poor descriptor abnoffilal behavior and desires since it is easily influe nced by societal norms. Some DSM TV 'disorders' are simply an attempt to characterize socially abnormal behavior and desires. Behavior and desires must fall within the parameters of the DSM IV to be considered normal by society.
of
I only II only III only II and III only
49. The author's attitude toward "our" societal norms is most accurately described as:
A. B. C. D.
favorable. neutral. distrustful. disapproving.
46. The author hints that the fact that Chris is well-adjusted indicates that her "uneasiness with her assigned sex" (JineS 30-31): A. B. C. D.
is a problem which should be overcome through psychiatry. . is due to the culture she li ves in. ca n be overcome through surgery. is a basic personality defect.
47. Suppose it is discovered that prescnptlOn medication allows Chris to become somewhat more comfo~b le with her "natural" state. Does this discovery support the author's argument? A. B. C. D.
Yes; it confIrms it. No: it does not affect it. No: it weakens it. No; it disproves it.
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Passage II (Questions 50-56)
giving people more information, they have a wider basis from which to make decisions and form opinions. No one 55 can argue that giving additional and different perspectives is inherently bad. Yet, Postman does have a valid point that the information is useless. unless it is thoroughly examined. Simply put, the challenge is not necessarily to reduce the amount of information available, but to allow it to be better 60 tailored to individual recipients' needs. Thus, we as a society must consciously attempt to not only obtain information, but to analyze it Jor meaning and relate this back to our own values ....
Human values-these 'ends', which are generally viewed as "desirable"-are not static. They change according to certain factors surrounding the people who create them. One of the more significant of these factors is tech5 nology. Due to this, people must continually question the reasons for the use of technology in their own surroundings; in their "information ecologies". Just as "freedom of speech" is not clearly defined, the purpose of some technology is ambiguous. To decide 10 whether a certain technology is a good fit for a certain "ecology," one must use critical analysis. [Analysts] Nardi and G'Day create a three-pronged format for analyzing technology's worth .... These methods include working from core values, paying attention, and asking strategic 15 questions.
50. The author of the passage would be most likely to agree with which of the following ideas expressed by other technology theorists? A.
Making sure that technology fits with a person's or group's "core values"-its essential function/purpose-is the first step in examining its use. If technology doesn't assist in promoting these values, then it cannot be consid20 ered useful. For example, a for-profit business' core value is profit maximization; any technology which does not increase revenues or reduce costs should be viewed with suspicion .... Paying attention keeps people from taking a technology for granted. If the technology is taken for 25 granted, the actual purpose for using it can become clouded. Finally, a~king strategic and open-ended questions fosters a discussion about all uses for the technology. Asking who, what, why, where, and how questions is the best method in analyzing the use for technology in an information ecology. 30 If, by examining the situation, it becomes evident that the existing technology is not useful, then the person or grop-p that expects to use it should have direct input into finding a replacement. This ensures that technology will be used for the right purposes.
B. C. D.
Soliciting further ideas and diverse ideas is not always wrong. Obtaining additional and differing perspectives is always beneficial. For information to be of value, it must challenge our human values. A great deal of the information which we receive is useless because it does not pertain to us.
51. The passage argument suggests that information recipients might benefit from: I. a more rapid flow of infonnation. II. thoroughly examining their information. III. limiting non-pertinent infonnation. A. B.
C. D.
I only II only III only II and III only
Technology may force people away from their own values. This idea, presented by theorists Neil Postman and Jacques Ellul, creates a disturbing image of a world where technology itself becomes the center of a society without values. One example is "information glut", a term coined by 40 Postman. He states that the rapid flow of information allowed by automation and Internetlemail provides vast amounts of data to many users quickly and simultaneously, but doesn't allow humans to understand its meaning fast enough to keep up. He goes on to say that much of the 45 information we receive does not pertain to us, so it is at best an irrelevant distraction rather than the useful news it was intended as. This may create a problem when, for example, images of violence in the Middle East do not affect us because they are "here today, gone tomorrow" "and are too 50 far away to influence us. However, this view is somewhat one-sided. It is argued that too much information is bad, yet it can also be argued that more information is good. By 35
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52. In describing Neil Postman's "infomnation glut", the author uses the example of "images of violence in the Middle East [that] do not affect us" (lines 48). The author's point is that: A. B. C. D.
55. According to the passage, one drawback of technology is ,- that it can:
the Middle East is at best an irrelevant disrraction. the Middle East is poorly understood by us. these images should affect us. these images should not be so easily accessible.
B. C.
D.
supersede all other val ues. be used to destroy human values.
D.
subtly change people's values. become an irrelevant distraction.
56. Suppose it could be established that technology is most efficient when it performs its function unobtrusively; without being noticed. The author of the passage would be most likely to respond to this information by:
53. If the following statements are true, which would most weaken the argument of the author? A.
A. B. C.
The degree to which technology determines human values is questionable. Information must be analyzed for its relevance to
A.
suggesting that this determination ratifies his thesis.
B.
proposing that we mu st still analyze the information for meaning.
human values.
C.
Human values are uncbanging. Human values are culturally dependent.
D.
asserting that efficiency is usually degraded when a technology is taken for granted. explaining that we must still remain· aware of the technology and its intended pnrpose.
54. The author argues, "more information is good" (line 52). Unlike Neil Postman, the author does not consider-which of the following to be a factor that might limit the usefulness of information?
A. B. C. D.
Space Distance Frequency Time
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Passage III (Questions 57-63)
50
In Scene II, Faust is transformed into a younger man, who courts the young woman of his dreams, a commoner named Margaret. At this point, the play devolves into stock characters and slapstick. Faust and Margaret sing very forgettable arias about the supremacy of feeling over 55 reason, a theme which is not really congruent with the Faust myth. The shallowness of the libretto's throwaway lyrics is compounded by Margaret's emotionless singing.
From its very beginning, ... the New York City Opera production of "Mephistopheles" deserves high marks for visual excellence. It begins with an audiovisual show featuring stars, religions images projected onto swirling 5 mist, and very, very loud brass winds, intended for drama and only slightly corny.
Those who read the book know the next scene as the Witch's Sabbath on .Walpurgis Night, though the libretto 60 itself offers little explanation for the abrupt change of scene. Mephistopheles now appears as the leader of hedonistic sinners, ca11ing them with the aria "Come on, onward, onward". Again, the portrayal of his character is less evil than rebellious and hedonic, he recognizes the power of 65 mankind's pursuit of earthly pleasures, singing, "Here is the world, round and empty" while holding the globe. It is unclear whether it is his effort, or human nature itself, which is responsible for sin. At one point, he laments mankind's cruelty and cunning, concluding, "How I laugh 70 when I think what's in store for them!!Dance on; the world is lost." ...
The scene then shifts to Hell, where a naked, disheveled Mephisto, singing from his broken throne, sarcastically apologizes for not being up to Heavenly stan10 dards of singing, providing the proof that harmony is still a longer way off in some places than in others. In the director's vision, the characterization of Mephisto is akin to Milton's rebellious, but somewhat sympathetic anti-hero, a dissident angel who dares to fight a vastly superior power to 15 preserve his vision of the world. Accordingly, this production features a Mephisto who is flippant but clever, blasphemous but thought provoking, and possessed of both sympathy and contempt for human weaknesses. He stt:ikes a balance between his boldness and his cowering before 20 Heaven. A chorus presents the essential plot of Faust, reduced from its several incarnations. When the angels point out the mortal Dr. Faust, as God's incorruptible servant on earth, the devil Mephisto promises to tum him from God through 25 temptation. Thus, this version presents temptation as essentially a wager, or struggle, between God and the Devil (which, at one time, was a remarkably blasphemous notion, as it contradicted the dogma that God is all-powerful over evil). As the divine host departs, Mephisto regains his 30 mocking manner, singing, "It's nice to see the Eternal Father talking with the Devil-in such a human way!"
57. Assume that several others who had attended the same opera were interviewed. If they remarked that Margaret sang with tremendous passion, these remarks would weaken the passage assertion that:
A. B. C. D.
Mephisto tempts Faust in the middle of a country fair thronged with revelers, which is meant to symbolize the worldly pleasures. This symbolism hearkens back to the 35 ancient morality play Vanity Fair, which also featured a bazaar extolling sins. Mephisto, garbed in virtue as a gray robed beggar monk, finally announces himself to be Mephisto. In a good aria, much of which is delivered while dancing or rolling on the ground, the devil again introduces 40 himself sympathetically as God's constructive critic, one who "thinks of evil! but always achieves the good," singing menacingly of how he wages an eternal dissent against God:
45
Mephisto had rendered her irresistible to Faust. the lyrics which she sang were "throwaway". the young woman of Faust's dreams sang without emotion. the young woman of Faust's dreams was a commoner.
"Light has usurped my power, seized my scepter in rebellion; I hurl forth this single syllable-NO!" In this version of Faust, it is Faust who seizes the devil's bargain: Mephisto must furnish him with a single moment so lovely it deserves to last forever.
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58. On the basis of tbe passage, it is reasonable to conclude that: A. B.
C. D.
62. According to the passage, through what primary means is the fundamental plot transmitted to the audience?
Faust lost his soul to the devil. the "country fair thronged with revelers" was not in the book. the operatic interpretation differed from the book. the author did not enjoy the performance.
A. B. C. D.
63. Regarding the devil's bargain with Faust, the passage strongly implies that:
59. According to the passage, the author felt that the New York City Opera production of "Mephistopheles":
A. B. C. D.
A.
was plagued with a poor characterization of
B. C. D.
Mephisto. suffered from noticeable weaknesses beginning in Scene II. was enhanced by Dr. Faust's singing. could have been improved in Scene III.
60. According to the passage, the author seems to have most enjoyed:
A.
B. C. D.
Via visual imagery Via Faust's musings Via the chorus Via Mephisto
it is Faust who got the better deal. it is the devil who got the better deal. in other versions, the bargain was with Margaret. in other versions, it is Faust who does the bargaining.
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the music of. the opera. the singing ofthe opera. the plot of the opera. the images of the opera.
61. Which of the following does the author suggest was a component of the original "Faust myth" (lines 55-56)? I. Reason triumphing over feeling II. A more evil Mephisto III. A more powerful God
A. B. C. D.
I only II only III only II and III only
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ANSWERS & EXPLANATIONS FOR
30-MINUTE IN-CLASS . EXAMINATIONS
79
80 .
VERBAL REASONING
&
MATHEMATICAL TECHNIQU ES
ANSWERS TO THE IN-CLASS EXAMS Exam 1
MeAT VERBAL REASONING AND MATH
Exam 2
Exam 3
e
22. 0
43. 0
2. B
23. A
44. B
3. 0
24.
I.
e
45.
e
Raw Score
Estimated Scaled Score
23
15
22
14
21
13
19- 20
12
e
18
11
25. A
46. B
16-17
10
5. A
26. B
47. B
15
9
e
27. A
48. 0
13-14
4.
6.
-
8
7. B
28.
e
49. 0
12
7
8. B
29. 0
50. A
10-11
6
9. A
30. B ·
51. B
9
5
10. 0
31. A
52.
e
32. A
53.
12. A
33. 0
54. 0
e
34. B
55.
11.
13.
e
e e
35.
15. B
36. A
57.
e e
37. B
58.
38. B
59. B
18. A
39. A
60. 0
19. B
40.
e
61. A
20. 0
41. 0
62. C
21. B
42.
e
63. 0
17.
~
4
e
14. B
16.
7-8
56. 0
e e
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EXPLANATIONS FOR THE 30- MINUTE IN-CLASS EXAMS . 81
EXPLANATIONS FOR 30-MINUTE IN-CLASS: EXAM 1 Passage I (Questions 1-7) 1.
The word tar (line 53) is used in the sense of: "Some Marxist theorists have speculated that Marx would tar both Kant and Hume as "bourgeois" philoso-· phers" (lines 52-53). '
2.
A.
asphalt. : WRONG: This is not the intended sense, or mearung, of the word. "Tar" is n ot used literally.
B.
suggest. WRONG: This is not the intended sense, or meaning, of the word. This is not strong enough.•
C.
label. CORRECT: This is the intended sense, or meaning, of the word. The reference harkens to a literal " tarring and feathering" of individuals for the purpose of ri
D.
stick. WRONG: This is not the intended sense, or meaning, of the word. This word is not specific enough.
According to the passage, Marx would have d isagreed with Kant and Hume over which of the following ideas? The correct answer requires the embodiment of an idea which is l. shared by Kant and Hume, and 2. that Marx would have disagreed with. . A.
What each man feels within himself is the standard of sentiment. WRONG: This is not an idea that Kant and Hume share. This is Hume's idea only (lines 14-15).
B.
There is uniformity in people's reason or emotion. CORRECT: First, this is an idea that Kant and Hume share; "To make both of [Kant's and Hume's) constructions pOSSible, there is a notion of some kind of uniformity in people's reas.o n or emotion in both works" (lines 43-44). Second, according to the passage, this is clearly an idea with which Marx would have disagreed. "Based on these writings, Marx would probably see any system thatsought out a universal theory of morality as ignoring the apposing economic classes in society ... " (lines 65-67).
c. . The interests of the capitalists and the workers are one and the same. WRONG: This is not clearly an idea that Kant and Hume share. It is outside the scope of information in the passage. D.
3.
Morality is dependent upon the class struggle. WRONG: This is not an idea from the paSsage.
Assume that a universal principle of morality can be proven to exist. Which of the following hypotheses d oes this assumption suggest? A.
The author is correct; despite their genesis, it is not surprising that Kant and Hume constructed similar systems. WRONG: This hypothesis is not suggested by the assumption in the question. "Considering that they began their searches with seemingly irreconcilable ideas of where to look, the similarity in the moral systems they constructed is surprising" (lines 3-6).
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VERBAL REASONING
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MATHEMATICAL TECHN IQUES
B.
The author is correct; Marx, Hume, and Kant all constructed similar systems. WRONG: This hypothesis is not suggested by the assumption in the question. The author never suggests that all three of these people did construct similar systems. ~
C.
The author is incorrect; Marx, Hume, and Kant did not all construct similar systems. WRONG: This hypothesis is not suggested by the assumption in the question. The author never suggests that all three of these people did construct similar systems.
D.
The author is incorrect; despite their genesis, it is not surprising that Kant and Hume constructed similar systems. CORRECT: This hypothesis is suggested by the assumption in the question. "Considering that they began their searches with seemingly irreconcilable ideas of wh~re to look, the similarity in the moral systems they constructed is surprising" (lines 3-6). It is not so surprising once given the assumption that a universal principle of morality can be proven to exist. Hume and Kant simply arrived at this "Truth" by different paths.
According to the author, in creating his moral system, Hume equated: A.
circumstances with disinterested comparison. WRONG: There is no support for this answer in the passage.
B.
circumstances with absolutism. WRONG: The two are diametrically opposed in the passage and not equated by Hume.
C.
practicality with stirring people to action. CORRECT: "Hume decided at the outset that a moral system must be practical, ... and since only sentiment (emotion) is capable of stirring people to action, the practical study of morality should be concerned with sentiment" (lines 7-9).
D.
practicality with reason. WRONG: Hume was not much impressed with reason. He focused on sentiment.
Based upon passage information, Marx most likely believed that: A.
there is no universal theory of morality. CORRECT: This is Marx's most likely belief. "Marx would probably see any system that sought out a universal theory of morality as ignoring the opposing economic classes in society ... [which he felt were involved in a class struggle)" (lines 65-67).
B.
philosophers were part of the bourgeois. WRONG: Based upon passage information, this is not Marx's most likely belief. This is a vast generalization without any real support. We have no way of knowing what Marx thought about philosophers in general.
C.
the wo rkers could be easily fooled. WRONG: Based upon passage information, this is not Marx's most likely belief. This is a vast generalization without any real support.
D.
Kant's book supported capitalist exploitation. WRONG: Based upon passage information, this is not Marx's most likely belief. There is no passage information that would tell us whether Kant's book had ever been read by Marx, whether it was written before Marx's time, or written well afterwards. This is outside the scope of the available passage information.
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ExPLANATIONS FOR THE 3 D- M INUTE IN-C LASS ExAMS . 83
Based upon passage information, Kant's system of "moral feedback" (line 25) differed from Hume's in that it might result in a situation wherein: A.
one realized that his action might be 'right' so long as it didn't become a universal law of nature. WRONG: Hume did not consider "universallaws of nature". Further, and more importantly, the idea of a '~universallaw of nature" (lines 31-32) was all important to Kant.
B.
one realized that a universa l law of nature was unnecessary in d etermining duty. WRONG: It was Hume who did not consider "universallaws of nature". Further, and m ore importantly, the idea of a "universallaw of nature" (lines 31-32) was all important to Kant.
C.
one realized that his action might be 'wrong' ev.en though it created agreeable sentiments in others, as well as in oneself. CORRECT: Unlike Hume, Kant's system relied upon reason and logic [think Vulcan]; "before one does anything, one should forget one's own motives for a moment, and ask if he would want everyone to do as he does. If the answer is no, then his subjective desire is different from his objective assessment, and the action is contrary to duty" (lines 32-36). Sentiment was not a factor for Kant. A "subjective desire" that might, in the short term please himself and others, might in the long term not be best established as a "universallaw of nature" (lines 31-32).
D.
one realized that his action might be illogical if sentiment was not further considered. WRONG: Sentiment was not a factor for Kant.
According to the passage, Hume's ideas evolved to the point where he: Note that the correct answer requires that you know and understand where Hume's ideas stood when he began formulating them. "Hume's ideas evolved ... ". You must consider the paragraph beginning, "Hume decided at the outset ... " (lines 7-15). A.
realized that reason was an inseparable part of a universal system of morality. WRONG: Hume did not think much of reason (lines 7-9), but based his ideas upon sentiment (emotion ).
B.
was conSidering the sentiments of others as well as himself. CORRECT: As indicated by the question, Hume's ideas evolved. TIley changed from the time he first began formulating them to the end. At first, Hume says "What each man feels within himself is the standard of sentiment" (lines 14-15); others are not considered . Later, "Hurne says that moral actions are those that create agreeable sentiments in others, as well as in oneself" (lines 37-38).
C.
chose to essentially agree with Kant on a universal system of morality WRONG : This is not correct. Even in the end, the two differed in their reliance on reason / logic as opposed to sentiment/emotion.
D.
decided that sentiment without action was a necessary component of his morality system. WRONG: Hume chose sentiment specifically because he felt that it was only sentiment which elicited action. " Hume decided at the outset that a mora l system must be practical, ... and since only sentiment (emotion) is capable of stirring people to action ... " (lines 7-10).
Passage 1\ (Questions 8-14) 8.
Which of the following statements, if true, would most directly challenge the principles of Martin Luther? A.
The Bible's Old Testament refers to a period before the birth of Jesus. WRONG: This does not most directly challenge the principles of Martin Luther. 1his answer is way outside of the scope of the passage.
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VERBAL REASONING
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MATHEMATICA L T ECHNIQUES
B.
The Bible alone contains only a small part of what Jesus intended for his followers. CORRECT: This most directly challenges the principles of Martin Luther. This is within the scope of the passage and relates directly to information provided in the passage on the beliefs of Martin Luther; "For Luther, salvation lies in faith emanating from personal understanding of Biblical teachings" (lines 56-57).
C.
The German "barbarians" who sacked Rome had been previously converted. WRONG: This does not most directly challenge the principles of Martin Luther. Whether they had been converted or not is irrelevant. Either way, t)1ey would have been considered "living tools" of God by both Luther and Augustine.
D.
Augustine's understanding of salvation granted through the mercy of Christ was flawed. WRONG: This does not most directly challenge the principles of Martln Luther. It has little bearing on the principles of Martin Luthe r. If anything, this information would strengthen Luther' s ideas and principles.
Some theologians believe that killing and violence are acceptable when used in self-Defense. An appropriate clarification of the passage would be the stipulation that: A.
both Luther and Augustine would have disagreed with this belief. CORRECT: This stipulation would be an appropriate clarification. "Both of these theologians would hold that. .. salvation, ... should be all Christians' only concern in life" (lines 52-55). There is no evidence in the passage that either theologian made exceptions to their explanations of Christ's teachings wherein "killing and violence" are acceptable.
B.
both Luther and Augustine would have agreed with this belief. WRONG: This stipulation would not be an appropriate clarification. There is no evidence in the passage that either theologian made exceptions to their explanations of Christ's teachings wherein "killing and violence" are acceptable.
C.
only Augustine might have agreed with this belief. WRONG: This stipulation would not be an a ppropriate clarification. " In Augustine's understandlng, [God] sends hardship and d eath to test men's faith. The true Augustinian Christian would maintain his faith [and presumably his passivity] through any ordeal, and even if his body perishes, God will save him for his conviction" (lines 59-64).
D.
only Luther might have agreed with this belief. WRONG: This stipulation would not be an appropriate clarification. See above. ;
10.
If the information in lines 38-55 is correct, one could most reasonably conclude that, compared to Luther,
Augustine was: A.
much more reasonably inclined. WRONG: This is not a reasonable conclusion. This is a fuzzy value judgment based solely upon personal opinion. Unless the passage itself provided direct evidence regarding the "reasonableness" of someone's inclinations or ideas, this type of answer would be a poor choice.
B.
more prepared to define God's will. WRONG: This is not a reasonable conclusion. There is no evidence in support of this. What does "more prepared" mean? This answer approaches the vagueness of Answer A.
C.
less eager to send people to eternal torment. WRONG: This is nota reasonable conclusion. This answer is attractive. Particularly given the last sentence of the quo te, "Augustine never considers the individual fate of these living tools, but Luther maintains that ... they go to eternal torment" (lines 49-51). However, neither of the theologians is "sending people to eternal torment". According to the passage and the information it is God who decides who is 'sent' to hell, not Lu ther or Augustine.
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EXPlANATIONS FO R THE 3D-M INUTE IN-C lASS EXAMS . 85
less willing to announce God's final judgment on those who had sinned . CORRECT: This is the only reasonable conclusion based on the information. "Augustine never considers the individual fate of these [barbarians], b u t Luther maintains that ... they go to eternal torment" (lines 49-51). This answer choice more correctly d efines the role that Luther or Augustine m ight p lay as theologians in relation to God . The theologians can only" announce" what God has decided; if they think they know. They themselves d o not have the power to "send" anyone to heaven or hell.
The au thor's attitude toward the theories of Augustine and Luther in the passage is most accurately described as: This type of question must usually be gleaned from the overall impression given by the author as the passage is read. There is no 'going back' to the passage. The reader must ask if there w ere any derogatory, sarcastic, praiseworthy, or other type of information or words used that p rovide clues to the author's attitude. If the passage did not try to persuade or argue, then it is probably neutral.
12.
A.
d isapp roving. WRO NG: This is not the most accurate description of the author 's attitude.
B.
m istrustful. WRONG : This is not the m ost accurate d escription of the author's attitude.
C.
neutral. CORRECT: lhis is the most accurate description of the author's attitude. This type of passage is not persuasive, but purely informa tive.
D.
favorable. WRONG: This is not the most accurate d escription of the author 's attitude.
What is the meaning of the phrase; "The true Augustinian Christian would maintain his faith through any ordeal, and even if his body perishes, God will save him for his conviction" (lines 61-64)? Unfortunately, passages and sentences are not always provided in their clearest form . The tortured syntax of this sentence begs for clarification. Particularly w here the word "conviction" is used in its less common form; 'belief'. We are used to thinking of a prosecuting attorney seeking a criminal's 'conviction'. A.
This Christian would end up in heaven because of his beliefs. CORRECT: This is the meaning of the phrase. Restated in other words, 'The true Ch ristian wquld maintain his faith through an y ordeal, and even if he phYSically died, God w ill save him because of his belief.
B.
God would save this Christian for judgment at the end of the Christian's ordeal. WRONG: This is not the meaning of the phrase. This answer mistakenly equates "for his conviction" w ith 'for his judgment'.
C.
God would save this Christian from his ordeal and judge him. WRONG: This is not the meaning of the phrase. First, this Christian w ill not be saved from his ordeal. It is clear that God is not going to save this Christian from his 'earthly' ordeal. "[E]ven if his body perishes ... " means the Christian will probably die. Secondly, this answer m istakenly equates "for his conviction" with 'for his judgmen t'.
D.
It w as not necessary for this Christian to die for him to be convicted. WRONG: This is not the meaning of the phrase. First, this Christian w ill not be saved from his ordeal. It is clear that God is not going to save this Christian from his 'earthly' ordeal. "[E]ven if his body perishes . .. " means the Christian will probably die. Secondly, though this answer is not completely clear, it seems to m istakenly equate "for his conviction" w ith "finding him guilty".
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86 .
VERBAL REASONING
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MATHEMATICAL TECHN IQUES
The author's primary purpose in the passage is apparently: These questions are less value judgments on the word "primary" than they are accuracy questions. On these types of questions the "primary purpose" usually means the answer choice which accurately restates passage information. Generally~ three are inaccurate and one is accurate. However, if two answer choices are accurate, one will clearly be more all encompassing (primary) than the other. A.
to clarify the differences between the ways in which the early Catholics and Protestants dealt with persecution.
WRONG: This is not the author's primary purpose. There seem to be no differences in the passage regarding the way in which the early Catholics and Protestants dealt with persecution. The only differences were in their understandings of how to achieve salvation and pronouncing God's judgment on others.
14.
B.
to justify the persecution of early Christians by secular governments. WRONG: This is not the author's primary purpose. The only justification of persecution in the passage comes indirectly from Augustine's writings "that His inscrutable will is always just" (lines 40-41), and that "God ... sends hardship and death to test men's faith" (lines 60-61).
e.
to consider the similarities between the ways in which the early Catholics and Protestants dealt with persecution. CORRECT: This is the author's primary purpose. A topic sentence in the first paragraph is, "Later Christian theorists followed this example, often urging peaceful coexistence even with governments which violated every precept of Christian teachings" (Jines 7-10). The author then goes on to support this idea with Augustine and Luther by considering their "similarities". Though Luther's ideas predominate in the passage, "S!. Augustine would have agreed" (line 38), and "Both of these theologians wOllld hold that ... " (line 52), etc.
D.
to question the passive practices of the early Catholics and Protestants when faced with persecution. WRONG: This is not the author's primary purpose. The author is not questioning the practices, but explaining the reasoning behind them in a rather objective manner.
What is the most serious apparent weakness of the information described? These types of questions are not necessarily the value judgments that they might first seem to be. At least two of the answer choices will inaccurately restate conclusions or passage information. Of the remaining two, one will obviously weaken the passage, while the other may even strengthen it. A.
While implying that Christians may coexist with a secular government. it differentiates between Catholics and Protestants. WRONG: This is not the most serious apparent weakness of the information described. What little differentiation there is strengthens the passage by supporting the idea that "Later Christian theorists ... " (i.e. supporting the topic sentence).
B.
While implying representation of Augustine and Luther, its conclusions are based primarily on information according to Luther. CORRECT: This is the most serious apparent weakness of the information described. The author chooses the Catholic pillar St. Augustine and the Protestant representative Martin Luther tu stand for later "Christian theorists". However, the second paragraph merely paraphrases the two of them. The large third paragraph is all Luther. Finally at line 38, "St. Augustine would have agreed" but by the next sentence we are back to Luther. The passage does not even come close to equally representing the two Christian theorists and is therefore weakened when attempting to use both men's theories in support of the topic.
e.
While implying representation of all Christian theorists, only Augustine and Luther are mentioned. WRONG: This is not the most serious apparent weakness of the information described. If you felt that there were other Christian theorists from that time you would be bringing in outside information. For instance, Flmdamentalists, Mormons, and other sects did not exist at that time, for one thing.
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ANSWERS
D.
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EXPLANATIONS FOR THE 30-M INUTE IN-C LASS EXAMS •
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While implying agreement between Augustine and Luther, their. attitudes were clearly opposed. WRONG: This is not the most serious apparent weakness of the information described. This is inaccurate and there is no support for this answer.
Passage III (Questions 15-21)
15.
16.
The author probably mentions that Mallon was "never ' tried' in any sense" (line 5) in order: A.
to demonstrate the power of the wealthy at that time. WRONG: This is not the probable reason for t1\e author's mentioning. Though the privilege of the wealthy is an underlying theme of the passage, there is no evidence that the "wealthy", nor their power, had anything to do with Mallon's confinement.
B.
to provide a comparison with people who have actually committed a crime. CORRECf: This is the most probable reason for the author's mentioning. This answer supports the fact that the author obviously feels that Mallon's treatment was unjust. This is evidenced in the first and last sentences of the passage: "Of all the bizarre and melancholy fates that could befall an otherwise ordinary person, Mary Mallon's has to be among the most sad and peculiar.", and "But she was the only one kept isolated for years, a result as much of prejudice toward the Irish and noncompliant women as of a public health threat.". This is the best answer.
C.
to illustrate the persistence of Soper 's investigations. WRONG: This is not the p robable reason for the author's mentioning. There is no evidence that Soper's investigations were 'persistent' in the first place.
D.
to support the claim that she deserved at least a hearing. WRONG: This is not the probable reason for the author's mentiOning. There is no that she "deserved at least a hearing".
~'claim"
by the author
According to the passage, the first two investigators were unable to fmd the cause of the outbreak (lines 19-20). The infolmation presented on typhoid makes which of the following explanations most plausible? A.
They focused too closely on the "soft clams" that Soper later discredited. WRONG: This explanation is not the most plausible. There is no evidence or information that the investigators focused on "soft clams" at all.
B.
Typhoid is not really passed through contaminated food or water. WRONG: This explanation is not the most plausible. It seems that typhoid is passed through contaminated food or water.
C.
They never considered that typhoid could be carried by a healthy person. CORRECf: This explanation is the most plausible. We know that Soper "was the first to identify a healthy typhoid carrier in the United States" (lines 32-33). Also, Soper 's investigation and discovery occurred after the two investigators canle to the Warren house. Therefore, this answer is the most likely.
D.
By this time, Mallon was no longer employed by the Warren family. WRONG: This explanation is not the most plausible. There is no timeframe given regarding the "two investigators", thus this is purely speculative. Further, since, Soper was the first person to identify a healthy typhoid carrier in the United States, it is urilikely that the two investigators would have suspected Mallon had she been even directly interviewed by them ..
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88 .
VERBAL REASONING
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The author's argument that Mallon's isolation was "a result as much of prejudice ... as of a public health threat" (lines 64-65) is most weakened by which idea in the passage? N ote that the correct answer must satisfy two criterions. It must 1). be an idea in the passage, and 2). weaken the author 's argument. A.
Mallon's primary occupation was as a cook. WRONG: The author 's argument is not weakened by this idea, to the extent that Answer C weakens it. First, though it does seem that this answer is promoted in the passage, it is an assumption. If Answer C were not available this might be a good choice. However, it is not the fact that Mallon was a cook which doomed her, but it was the fact that she "failed to comply with the health inspector's requirements" (lines 58-59), which presumably would have precluded her fmm preparing meals. C is the better answer.
B.
Mallon did not believe that she w as a carrier of the disease. WRONG: The author's argument is not w eakened by this idea, to the extent that Answer C weakens it. If Answer C were not available this might be a good choice. However, similarly to Answer A, it is not the fact that Mallon did not believe that she was a carrier which doomed her, but it was the fact that she "failed to comply w ith the health inspector 's requirements" (lines 58-59). C is the better answer.
C.
Mallon would not abide by the health inspector's requirements. CORRECT: The author's argument is weakened by this idea. Ultimate, this is the explicitly provided reason for Mallon's reincarceration. "After a short period of freedom in which Malloll failed to comply with the health inspector's requirements, she was eventually sent back to North Brother Island, where she lived the rest of her life, alone in a one-room cottage" (lines 58-Q1).
D.
Mallon actually was the source of the typhoid outbreak in the Warren home. WRONG: The author's argwnent is not w eakened by this idea, to the extent that Answer C weakens it. Had MaHon complied with the "health inspector's requirements" (line 59), she might well not have been reincarcerated.
18.
19.
Which of the following statements is the most reasonable conclusion that can be drawn from the author's description of the typhoid outbreak in the house at Oyster Bay? A.
The Warren family did not hire Soper. CORRECT: This is a reasonable conclusion. "Worried they wouldn' t be able to rent the house unless they figured out the source of the disease, the owners .. . hired George Soper ... " (lines 20-21).
B.
The two investigators were hired by the Warrens. WRONG: This is not a reasonable conclusion. It is not certain who it was w ho hired the two investigators. Why not the owners, or the city health department, for instance?
C.
The Warren family hired Soper. WRONG: This is not a reasonable conclusion. "Worried they wouldn't be able to rent the house unless tlley figured out the source of the disease, the owners ... hired George Soper ... " (lines 20-23) .
D.
The owners were anxious to sell the house. WRONG: This is not a reasonable conclusion. The owners were worried that they wouldn' t be able to rent the house. "Worried they wouldn' t be able to rent the house unless they figured out the source of the disease, the owners .. . hired George Soper ... " (lines 20-23).
Passage information indicates that which of the following statements must be true? A.
Mallon had probably not infected anyone prior to the Warren family. WRONG: There is no information that this must be true.
B.
Mallon was almost certainly not was hing her hands prior to preparing the Warren's meals. CORRECT: We know that Mallon's feces "did indeed show high concentrations of typhoid bacilli" (lines 46-47). Therefore, there are only a few unlikely alternatives to the idea that the bacilli were being transCopyright © 2007 Exa mkrackers, Inc.
ANSWERS
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EXPLANATIONS FOR THE 30-MiNUTE IN-CLASS EXAMS .
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ferred from the feces to Mallon's hands, to the food she prepared, and then to those becorrUng sick. None of the grotesque alternatives is really plausible.
20.
21.
C.
Being labeled 'Typhoid Mary' by the press was the primary reason for her confinement. WRONG: There is no information that this must be true.
D.
The health inspector was doubtless prejudiced toward the Irish. WRONG: There is no information that this must be true.
According to passage information, Mallon worked for the Warren family for approximately: A.
two weeks. WRONG: See explanation for Answer D.
B.
three weeks. WRONG: See explanation for Answer D.
C.
five weeks. WRONG: See explanation for Answer D.
D.
six weeks. CORRECT: According to Soper's own words describing Mallon's tenure in Oyster Bay, which are quoted in the passage, "It was found that the family had changed cooks about three weeks before the typhoid epidemic broke out ... She remained with the family only a short time, leaving about three weeks after the outbreak occurred" (lines 26-29).
The contention that in "1938 when [Mallon] died, ... there were 237 other typhoid carriers living under city health department observation. But she was the only one kept isolated for years" (lines 61-65), can most justifiably be interpreted as support for the idea that: A.
Mallon was unfairly treated by the city health department. WRONG: This idea is not justifiably supported by the contention. "Fairness" requires a comparative analysis of some sort. Mallon was the "first" of her kind to be identified. The contention regards 1938, decades after Mallon's discovery. Her treatment cannot be compared to the 237 other typhoid carriers.
B.
Mallon's isolation might have stemmed from the health department's early ignorance of the disease. CORRECT: This idea is justifiably supported by the contention. This answer might be gleaned through process of elimination also. However, the key" early ignorance" is a giveaway given that the contention regards 1938, decades after Mallon's discovery and isolation.
C.
The" other 237 typhOid carriers were all kept isolated at one time or another. WRONG: This idea is not justifiably supported by the contention. Though this is tempting because the contention provides that" she was the only one kept isolated for years", this is not a strong interpretation of the contention. It is entirely possible that at least one of the 237 "other" carriers was never isolated at all, negating this answer.
D.
The "other" 237 typhoid carriers were much like Mallon. WRONG: This idea is not justifiably supported by the contention. This is very weak. For one thing, the contention regards 1938, decades after Mallon's discovery. Further, it can be presumed, based upon the passage, that those who were free had at least agreed and cooperated in not cooking and serving food; unlike Mallon.
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VERBAL REASONING
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MATHEMATICAL T ECHNIQUES
ANSWERS & EXPLANATIONS FOR 30-MINUTE IN-CLASS EXAM 2 Passage I (Questions 22-28) 22.
23.
24.
According to the passage information, which of the following would be most likely, if a person, who was talking to you, attempted to make their voice sound unusually low? A.
You might think that they were lying. WRONG: This would not be most likely. The key is "unusually low". Though a speaker who was lying, might be aware that in order to successfully lie, or "suggest credibility, the voice's tone should be pitched as low as is naturally possible" (lines 4-5), this is a much more 'tortured' answer choice. In other words, in contrast to the straightforward aspect of Answer D, this answer requires several assumptions that are outside the scope of the question.
B.
They could be irritated with you. WRONG: This would not be most likely. First, it is lack of "intelligibility" that is irritating. Second, it is the listener who is irritated, not the speaker.
C.
They might well sound monotonous. WRONG: This would not be most likely. There is no passage link between monotony and pitch. Monotony has to do with inflection, not pitch.
D.
You could find them difficult to understand. CORRECT: This would be most likely. The key is "unusually low". "[A]n artificially lowered voice can sacrifice intelligibility" (lines 7-8).
The author most likely believes that one of the main purposes of speaking, during a face-to-face meeting, should be to: A.
convey a favorable impression. CORRECT: This is not the most likely belief of the author. Almost the entire passage is about conveying favorable impressions, not ideas or concepts. The author admits that "the quality of content accounts for a mere 6%" (lines 24-25) of the overall impression given by the speaker.
B.
effectively transmit your ideas. WRONG: This is not the most likely belief of the author. There is no support for this answer within the passage. The author admits that "the quality of content [i.e. ideas] accounts for a mere 6%" (lines 24-25) of the overall impression given by the speaker.
C.
gain leverage. WRONG: This is not the most likely belief of the author. There is no support for this answer within the passage.
D.
communicate as naturally as possible. WRONG: This is not the most likely belief of the author. For instance, the author would not be a proponent of communicating" as naturally as possible" if you naturally were 1. shy, 2. spoke with a high-pitched whine, 3. were 'breathy', etc.
The author provides a list of "habits to be avoided" (lines 13-15). Which of the habits would the suggestions in this passage not help a speaker to curb? Notice that, of the four answer choices, Answer C. "pretension" differs in 'kind' from the other choices. In other words, if you had a list of only these four choices with a new question asking, 'Which one of the following items doesn't belong with the others?', you would still choose "pretension".
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A.
monotony WRONG: There is information within the passage which would help a speaker to curb "monotony". Speech should be "inflected to suggest the emotions expressed" (line 3).
B.
breathiness WRONG: There is information within the passage which would help a speaker to curb "breathiness". See lines 36-54.
C.
pretension CORRECT: There is no information within the passage which would help a speaker to curb "pretension". Though we may make some assumptions, the passage does not convey how "pretension" is conveyed. Since the symptoms of 'pretension' are not defined, there is little in the way of help which can be gleaned from the passage for the person who suffers from 'pretension'.
D.
high-pitched whining WRONG: There is information within the passage which would help a speaker to curb "high-pitched whining". "To deepen the pitch, speakers should make an extra-deep inhalation before speaking, and then exhale fully as they speak" (lines 10-12).
The term ideal speaking voice (line 1) refers implicitly to a voice that is: A.
the most pleasant to listen to. CORRECT: This is not the most strongly implied answer. The author gives us our first clue when describing a voice which might "sacrifice intelligibility, which is irritating to listeners. Thus, speakers should experiment to find their optimal level, which will be at their lowest intelligible pitch" (lines 7-10). At this point, we might assume that, thus, the "optimal level" or ideal speaking voice is the "least irritating" (Answer C.). However, overall, the implication of the passage is seeking to be the best, not (forgive the tortured syntax) the "least" worst of the worst.
B.
the most persuasive. WRONG: This is not the most strongly implied answer. This has to do with' content'. We can surmise that since "the quality of content accounts for a mere 6%" (lines 24-25) of the overall impression given by the speaker "most persuasive" is not a characteristic of the ideal speaking voice. f
26.
C.
the least irritating. WRONG: This is not implied. The author seeks the best, not (forgive the tortured syntax) the "least" worst of the worst.
D.
the most natural. WRONG: This is not implied. For instance, the author would not be a proponent of the idea speaking voice being the "most natural" if your "most natural" speaking voice was 1. shy sounding, 2. a high-pitched whine, 3. or 'breathy', etc.
Passage information indicates that a person speaking in a high-pitched voice might be doing all of the followingEXCEPT: I. Breathing with their upper lungs
WRONG: This is not an exception. A person speaking in a high-pitched voice might be "breathing with their upper lungs". "Diaphragmatic breathing results in a deeper voice than upper-lung breathing" (lines 40-41). II. Breathing deeply CORRECT: This is an exception. A person speaking in a high-pitched voice would probably not be "breathing deeply". For instance, "To deepen the pitch, speakers should make an extra-deep inhalation before speaking, and then exhale fully as they speak" (lines 10-12).
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III. Lying WRONG: This is not an exception. A person speaking in a high-pitched voice might be "lying". "To suggest credibility, the voice's tone should be pitched as low as is naturally possible ... A low pitch is desirable in both genders, since it is popularly associated with truth-telling" (lines 3-7).
27.
28.
A.
Ionly
B.
II only CORRECT: See above answer explanations.
C.
III only
D.
I and III only
Which of the following assertions is most clearly a thesis presented by the author? A.
Speakers can gain by improving his or her speaking voices. CORRECT: This is clearly a thesis presented by the author. It is presented through the negative aspects associated with having a 'poor' speaking voice, such as "irritating" the listener. It is presented through the examples of "famous news anchors" who had great speaking voices. And, it is presented through statistics within the passage; "everyone who communicates should be aware that their voice is a critical component of their audience's perceptions of them, comprising about 38% of the overall impression imparted by their presentation" (lines 19-22).
B.
The tone of the ideal speaking voice should be pitched as low as possible. WRONG: This is not clearly a thesis presented by the author, and inaccurately paraphrases the passage. Speakers are admonished to find their "lowest intelligible pitch" (line 10), and to pitch their voices as "low as it naturally possible" (line 5), else they will risk irritating their listeners.
C.
What you are saying is more important than how you are saying it. WRONG: This is not clearly a thesis presented by the author. It is an antithesis. "[T]he quality of content accounts for a mere 6%" (lines 24-25) of the audience's perception of the speaker.
D.
Emotional inflections can be an irritating aspect of a speaker's voice. WRONG: This is not clearly a thesis presented by the author. The ideal speaking voice" should be ... inflected to suggest the emotions expressed" (lines 2-3).
The ideas discussed in this passage would likely be of most use to: A.
Adoctor WRONG: This is not most likely.
B.
A journalist WRONG: This is not most likely. There is no reason to believe that the ideal speaking voice is of more particular use to a journalist who writes for a living, than to anyone else.
C.
A radio show personality CORRECT: This is most likely. The passage is almost solely on the "ideal speaking voice". Unlike the "famous [television] news anchors" used by the author as examples, 'radio show personalities' would not require coaching and advice on their visual appearance.
D.
A television commentator
WRONG: This is not most likely. The passage is almost solely on the "ideal speaking voice". Despite the author's use of "famous [television] news anchors" which might make this answer seem attractive, clearly, television relies at least somewhat if not just as much, on visual components and appearance, as it does on the speaking voice. However, beyond mentioning that" appearance accounts for about 50% of the speaker's impact" (lines 23-24), there is no further discussion of these ideas within the passage. Nor is there any advice given regarding improving the "appearance". Thus, this passage would not be as com-
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pletely useful to a 'television commentator' as it would to a 'radio show personality' to whom his speak~is entire persona.
ing voice is
Passage II (Questions 29-35)
29.
The author's discussion of "push" and "pull" fadors (line 30) most accurately implies that: A.
"pull" factors compel someone to leave, while "push" factors induce someone to come. WRONG: This is not the most accurate implication.
B.
"pull" factors induce someone to come, while "push" factors also induce someone to come.
WRONG: This is not the most accurate implication. C.
"push" factors require someone to leave, while "pull n factors also compel someone to leave.
WRONG: This is not the most accurate implication. D.
II
push" factors compel someone to leave, while "pull" factors induce someone to come.
CORRECT: This is the most accurate implication. "For example, the Fujianese feel compelled ("pushed") to leave because of the area's low standard of living. The poor wages, bad housing, and lack of political freedom can also be seen as "pull" factors, due to the idea that the Fujianese understand that life would be better in other countries" (lines 31-35). 30.
Given the information in the passage, if '''artificially' defined" borders (lines 3) were eliminated throughout the world, which of the following outcomes would most likely occur? A.
People would naturally immigrate to areas with higher standards of living. WRONG: This outcome is not the most likely to occur. Though the author does provide that people do tend to immigrate to areas with higher standards of living, there is no particular reason to believe that elimination of artificially defined borders would increase or decrease this tendency, without more information from the question.
B.
Nations would encounter less traditional border strife. CORRECT: This outcome is the most likely to occur. The key here is "traditional" border strife. In contrast to the immigration problems predominantly focused upon in the passage, this type of strife is characterized by nation/states lIattempting to secure their borders from invading countries, or even seeking to expand their own territories and acquire additional resources" (lines 60-63). Why? Because they would be separated by nahlral geographical" boundaries such as wide rivers, mountain ranges, and oceans_ Or, cultural boundaries. One can presume that these are boundaries where people sharing the same cultural characteristics, such as language, religious beliefs, etc., have chosen to live together_ II
31.
C.
Nations would require greater border security measures. WRONG: This outcome is not the most likely to occur. Natural "geographical" boundaries such as wide rivers, mountain ranges, and ocean would be separating peoples who shared the same cultures.
D.
People would live more harmoniously. WRONG: This outcome is not the most likely to occur. This is possible. However, this answer is vague when compared with Answer B.
Which of the following assertions does the author support with an example? A.
Transportation methods used by illegal immigrants are sometimes dangerous. CORRECT: This assertion is supported with an example. "Within the past year, U.s. officials found tluee Chinese immigrants in a smuggling boat's sealed cargo container, dead from suffocation" (lines 22-24).
B.
Peru and Columbia are seeking to expand their own territories. WRONG: This assertion is not supported with an example. Presumably because it is not a passage assertion or an assertion of the author's!
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C.
New Zealand has enacted laws iliat hasten deportation proceedings. WRONG: This assertion is not supported wiili an example.
D.
The mission of the Peruvian troops is to keep guerillas and drugs out of Peru. WRONG: This assertion is not supported wiili an example.
The passage as a whole suggests that in order for a nation to slow the exodus of its inhabitants to other countries, it must: A.
become more attractive to those who are leaving.
CORRECT: This is suggested by the passage as a whole. The information regarding the "push" and "pull" factors suggests that poor conditions in the native country 'push' the immigrant, while higher standards in an adjoining country 'pull' the immigrant. Presumably, by becoming "more attractive to those who are leaving" a nation might slow the leaving of its inhabitants. Though "more attractive" might seem vague, it encompasses a variety of options and changes such as higher pay, better standards of living, better political system, more political freedom, etc., that make this the best answer.
33.
B.
abandon the traditional methods of guarding borders. WRONG: This is not suggested by ilie passage as a whole. We don't even really know what ilie "traditional methods of guarding borders" refers to. What was it. How are borders guarded now? IITraditional" refers to what occurred along borders which countries shared, not the methods by which iliey were guarded.
C.
respond in some way to ilie conflicts arising from border disputes. WRONG: This is not suggested by the passage as a whole. This really has nothing to do with the inhabitants leaving or staying.
D.
answer ilie challenges set forth by adjoining countries. WRONG: This is not suggested by ilie passage as a whole. This really has nothing to do wiili the inhabitants leaving or staying. One might infer iliat by "challenges", this answer refers to "economic challenges" and rising to meet the standards of ilie more attractive neighboring country. However, iliis is a stretch. "Challenges" might just as well mean challenges to war, or sporting challenges.
If ilie passage information is correct, what inference is justified by the fact that virtually no immigration from West Berlin to adjoining East Berlin occurred, over ilie 40 years before the period described? A.
Crossing ilie heavily guarded borders between West and East Berlin was very dangerous. WRONG: This inference is not justified by ilie new "fact." There is no passage information to support the premise iliat more 'heavily guarded borders' would be effective in an effort to prevent or slow immigration. Or, iliat a "very dangerous" situation would prevent immigration. On the contrary, the example of the suffocated Chinese immigrants indicates iliat danger would not stop a motivated person from attempting to immigrate.
B.
It was understood iliat life would be better in East Berlin. WRONG: This inference is not justified by the new "fact." Just as the Fujianese migrated to countries and areas where the cunuitioIlS were lIlUre favurablt:~, if this answer were accurale, lhe question would provide iliat 'a great deal of immigration has taken place from West Berlin to East Berlin.'
C.
The inhabitants of both 'Berlins' were happy to remain where they were. WRONG: This inference is not justified by the new "fact." The use of the word 'happy' is simplistic and should alert you that this is probably not ilie best choice. The passage heavily incorporates economics as causal "push" and "pull" factors. Furilier, you assume iliat iliere is no immigration from East Berlin to West Berlin. Notice iliat iliis possibility is left open in ilie question'S factual premise. This is not the best answer.
D.
The economic conditions of West Berlin were much more favorable than those of East Berlin. CORRECT: This inference is justified by the new "fact." Again, it is certainly in the realm of possibility Copyright © 2007 Examkrackers, Inc.
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that there is immigration taking place from East Berlin to West Berlin. But whether this is true or not, this answer incorporates passage economic ideas in advancing a reason why there would be "virtually no immigration from West Berlin to East Berlin." 34.
The author implies that which of the following is not one of the reasons that Peruvian President Fujimori deployed soldiers to its borders with Columbia? The question requires an implication of the author's. Note the absence of an implication. Notice that this type of question differs dramatically from one reading: "The author implies that Peruvian President Fujimori deployed soldiers to its borders with Columbia for all of the following reasons EXCEPT:" I. Fujimori is attempting to keep drugs out of his country.
WRONG: The author implies that this is one of the reasons the soldiers were deployed. II. Fujimori fears that Columbia is seeking to expand its territories. CORRECT: The author implies that this is not one of the reasons the soldiers were deployed. "Traditionally. these two countries might have been attempting to secure their borders from invading countries, or even seeking to expand their own territories and acquire additional resources. However, Ecuador and Peru are protecting their borders from rogue drug traffickers and guerillas, not Colombia's government" (lines 59-64).
III. Fujimori is probably concerned that Columbia wants to acquire additional resources. WRONG: This is not an implication of the author's. The question requires an implication of the author's, not the absence of an implication. This is not an 'exception' type of question. Notice that this type of question differs dramatically from one reading: "The author implies that Peruvian President Fujimori deployed soldiers to its borders with Columbia for all of the following reasons EXCEPT:"
35.
A.
I only
B.
II only CORRECT: See above answer explanations.
C.
III only
D.
II and III only
It seems likely that New Zealand may be suffering less from immigration issues than the United States for
which of the following reasons: I. The U.S. offers higher wages than New Zealand.
WRONG: There is no support for this answer in the passage. II. New laws, enacted in New Zealand, allow faster deportation proceedings. CORRECT: This answer choice "seems likely". "Recently, New Zealand attempted to deal with these aliens by enacting new immigration laws, which hasten the process required to deport them" (lines 24-27). III. Immigrants often do not settle in New Zealand.
CORRECT: This idea not only "seems likely", it is provided in the passage as a reason. "What keeps New Zealand from experiencing an even more profound illegal immigration problem is that the immigrants often do not settle there" (lines 38-40). A.
II only
B.
III only
C.
II and III only CORRECT: See above answer explanations.
D.
I, II, and III
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Passage III (Questions 36-42) 36.
What is the author's response to the standard story about the woman who spills hot McDonald's coffee in her lap, sues and gets several million dollars? Since this story is not actually in the passage, the author is not actually responding to it. However, corollaries can be drawn from within the passage. The author's response can be predicted. This type of question is very similar to others found on the MCAT such as, "The author would argue that ... ", "On the basis of the passage, it is reasonable to conclude that ... ", and "The author would be most likely to respond by ... ".
37.
A.
This story does not reflect that compensation is usually insufficient. CORRECT: This is the most likely response to the "standard story". Remember that the author emphasizes that, "In reality, compensation tends toward inadequacy" (line 14). He gives several reasons, one of which is the "hurdle" of proving negligence. Thus, the author would not believe that the woman's case was indicative of most other torts.
B.
This story is a good example of just the right amount of compensation. WRONG: This story is not a good example of anything that the author promotes. We certainly have no way of knowing if the compensation was "just the right amount".
C.
This story does not reflect that deterrence is costly. WRONG: This is way outside of the scope of the passage. Deterrence is costly? The only information we have regarding the cost of deterrence is for the defendant. Finally, the question has to do with the woman's compensation, not deterrence.
D.
In this story, the woman was malicious. WRONG: This makes little sense given the abbreviated information in the question. It has nothing to do with the passage.
Which of the following assertions is the most effective argument against the author's opinion that personal injury law cannot satisfactorily compensate and deter as long as there are administrative costs involved in obtaining an award" (lines 1-7)? JI
A.
These administrative costs are inconsequential. WRONG: This is not the most effective argument against the author's opinion. First, one must realize that the author has essentially defined 'administrative costs' as 'attorney's fees' (lines 6-13). Then, one must ask if it is reasonable to argue that attorney's should be compensated for their work. It is reasonable to argue that anyone should be compensated for their work (we are not saying how much or how little), and thus, yes, it is reasonable that attorney's should be compensated for their work. This answer is not reasonable. Further, if the costs are around "30%" of the awards, that is probably not considered by anyone to be "in_ consequential".
B.
Attorneys are a necessary part of the judicial system and should be compensated for their work. CORRECT: This is the most effective argument against the author's opinion. First, one must realize that the author has essentially defined 'administrative costs' as 'attorney's fees' (lines 6-13). Then, one must ask if it is reasonable to argue that attorney's should be compensated for their work. It is reasonable to argue that anyone should be compensated for their work (we are not saying how much or how little), and thus, yes, it is reasonable that attorney's should be compensated for their work. Finally, unlike some of the other answer choices, this argument is one that the author has not responded to in the passage. This is the best answer.
C.
The administrative costs should be added to the compensation received by the plaintiff. WRONG: This is not the most effective argument against the author's opinion. The author responds to this idea in the passage. In the example, "If she is paid in full [i.e. administrative costs are added to the full compensation], then the defendant is paying 130% of the actual harm caused, and is over-deterred" (lines 11-13). This cannot be the best argument because it is answered already in the passage.
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& EXPlANATIONS FOR THE 30- M INUTE IN-C lASS ExAMS • 97
The administrative costs should be subtracted from the compensation received by the plaintiff. WRONG: This is not the most effective argument against the author's opinion. The author responds to this idea in the passage. In the example, "If the plaintiff is awarded 100% of the damages suffered, she only receives compensation for 70% of her injuries" (lines 9-11). This cannot be the best argument because it is answered already in the passage.
The passage indicates that its author would NOT agree w ith which of the following statements? A.
Tanker companies are a good example of defendants who are under-deterred. WRONG: There is no basis from which to judge whether the author would agree or not agree with this answer. The story of the tanker company is a "permutation on actual events"; it is a fabrica tion.
B.
Negligence on the part of the defendant is generally not difficult for the plaintiff to prove. CORRECf: 111e author would not agree with this statement. The author argues in several ways the "hurdle" (line 19) of proving negligence. Besides that fact that 'hurdle' connotes difficulty in surmounting, the author also tellsus tha t "compensation tends toward inadequacy" (line 14) specifically because negligence is so hard to prove.
C.
The costs associated with suing and defending against suits can be tremendous. WRONG: There is no basis from which to judge whether the author would agree or not agree with this answer. The only dollar amounts provided are in the last paragraph. The idea of "tremendous" costs is relative and opinions between Bill Gates and a homeless person might vary.
D.
In nlan y situations, over-deterrence results in primarily economic ramifications.
WRONG: The author would probably agree. Over-deterrence "results in a waste of resources" (lines 50-51). 39.
Assume tha t since the 9-11 terrorist attacks on the World Trade Center (WTC) buildings, all lawsuits have been settled by the WTC insurance companies, who have now mandated that they w ill no longer insure any building in the world Ulat is over five stories tall. The author's comments suggest that this situation could reasonably be interpreted as evidence that: A.
the insurance companies were over-deterred.
CORRECT: A reasonable interpretation of the situation, based upon the author's information in the passage could lead one to this conclusion. This idea parallels the 'tanker trial' example offered by the author. Here, after settling the law suits from the WTC torts, the insurance companies have mandated a very extreme policy by anyone's estimation. Perhaps the WTC people were at fault which resulted in huge payments to the plaintiffs. Nevertheless, it seems that the WTC insurance companies were, in fact, "over-deterred", resulting in "a waste of resources" (lines 50-51). B.
the insurance companies were under-deterred . WRONG: This is not a reasonable interpretation of the situation, based upon the author's information in the passage. "Under-deterrence" results in a defendants "effectively hav[ing] no incentinve" to change or prevent future h arms. Here, after settling the lawsuits from the WTC torts, the insurance companies have mandated a very extreme policy by anyone's estimation.
C.
the plaintiffs were under-compensated. WRONG: This is not a reasonable interpretation of the situation, based upon the author's information in the passage. This is possible, but very unlikely. There might have been so many plaintiffs that their rather small compensatory damage awards simply overwhelmed the insurance companies. However, it is not probable. It is not even as probable as Answer D, which is also not the best answer. Review the explanations for Answer D and Answer A.
D.
the plaintiffs were overcompensated. WRONG: This is not a reasonable interpretation of.the situation, based upon the author's information in the passage. This is possible. With a poor understanding of the au thor's concepts from the passage, this answer might seem to go hand-in-hand with Answer A; it might seem just as "correct". lt is not. We have no way of knowing if the p laintiffs were overcompensated. Every one of the plaintiffs might have been
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making billions of dollars a year, yet rendered helpless quadriplegics requiring full-time 24-hour care which was not completely compensated by the lawsuits. There might have been so many plaintiffs that their rather small compensatory damage awards simply overwhelmed the insurance companies. We have no way of knowing based upon the information provided. Consider that, as mentioned, it is conceivable that over-deterrence and under-compensation occur simultaneously, to the benefit of no one. This is not the best answer. 40.
Suppose that a study found that police agencies routinely set aside large amounts of money in their yearly budgets, which they expect to payout in lawsuits against their agency. Which of the following statements is an assumption of the author about the effects of lawsuit awards that would be called into question? The correct answer must satisfy two criteria. It must 1. be an assumption of the author about the effects of lawsuit awards, and 2. be called into question by the given supposition. A.
Simply proving negligence can be a very costly process. WRONG: This statement is not a clear assumption of the author, and is not clearly called into question by the supposition.
B.
Many people will not sue because the process is too costly. WRONG: This is an implication of the author's since the idea was brought up in an example (lines 72-73), though the word "many" makes it highly questionable. Further, the supposition clearly states that the 'set aside' money is not for defending against lawsuits, but in order to "pay o ut" the awards .
41.
C.
If a plaintiff receives full compensation and administrative costs, the defendant is over-deterred. CORRECT: This statement is very clearly an assumption of the author's (lines 11-13). Further, this statement is called into question by the supposition. If the police agency "routinely" sets aside/budgets the money it will have to payout in awards, then it seems that this is just a business-as-usual approach. The author 's definition of over-deterrence results in a "change". There is an "incentive to prevent [harm]".
D.
Depending upon the size of the award, a defendant police agency might not be deterred at all. WRONG: This is not clearly an assumption of the author, since it sp ecifies police agencies, though it does somewhat parallel the idea that in situations where there is no award"defendants effectively have no incentive to prevent small harms" (lines 75-76). However, this statement is actually supported by the supposition . It is certainly not questioned by it.
Which of the following conclusions can justifiably be drawn from the experience of the tanker company's counsel mentioned in the passage? A.
Good economists make for poor attorneys. WRONG: This conclusion cannot be justifia bly drawn from the experience of the counsel (i.e. attorney). This answer is attractive because of the author's reference to "the company's counsel - a good economist, but a poor lawyer" (lines 29-30). However, there is no indication that being a good economist made this person a poor attorney.
B.
Costs should never be considered prohibitive where safety is concerned. WRONG: This conclusion cannot be justifiably drawn from the experience of the counsel (i.e. attorney). This is outside the scope of the question. The question does not ask about the tanker incident in general, but the tanker company's counsel specifically.
C.
Toxic materials should not be shipped through residential neighborhoods. WRONG: This conclusion cannot be justifiably drawn from the experience of the counsel (i.e. attorney). This is outside the scope of the question. The question does not ask about the tanker incident in general, but the tanker company's counsel specifically.
D.
Honesty is not always the best policy for an attorney. CORRECT: This conclusion can be justifiably drawn from the experience of the counsel (i.e. attorney). The author has already informed the reader that the tanker lawyer is a "poor lawyer". Thus, his actions would Copyrigh t © 2007 Examkrackers , Inc.
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not be an example for others to follow. The lawyer"admits safer tankers were available, but the cost is prohibitive. After extensive cost-benefit analyses, he says, the company found it cheaper just to pay victims for their losses, as it now offers to do" (lines 30-34). The author emphatically (!) announces that the company would be lucky to "escape punitive damages" (line 35). Apparently, the attorney was simply being honest, which (at least from the standpoint of his clients) was probably not the best idea. 42.
The author argues that, "Potentially, the most promising development in tort (personal injury) law has been the advent of strict liability" (lines 54-56). These beliefs imply that: Note that you must understand what "strict liability" is. A.
the use of strict liability has become increasingly popular for defendants. WRONG: This is not implied by the beliefs in the author's argument.
B.
the uses of strict liability should remain limited in scope. WRONG: This is not implied by the beliefs in the author's argument. According to the passage, strict liability is limited in scope. The author believes that, "Unfortunately, the application of strict liability is severely constrained by legal doctrine, which limits its application to a small range of 'unusually hazardous activities'" (lines 59-63).
C.
the author approves of waiving the requirement for proof, where carelessness is evident. CORRECT: This answer is implied by the beliefs in the author's argument. "Potentially, the most promising development in tort (personal injury) law has been the advent of strict liability, which waives plaintiffs' need to prove the defendant's carelessness in certain instances where the carelessness is obvious, or could have resulted from no factor other than negligence" (lines 54-59).
D.
the author approves of compensation where carelessness is evident. WRONG: This is not implied by the beliefs in the author's argument. This answer may be accurate information according to the passage. However, that differs from specifically answering the question. You cannot simply choose an answer that provides accurate information to the question. It must be accurate and the most responsive answer to the question.
ANSWERS & EXPLANATIONS FOR 30-MINUTE IN-CLASS EXAM 3 Passage I (Questions 43-49) 43.
The passage suggests that its author would probably disagree with which of the following statements? A.
It is possible that Chris participated in "male" activities in order to be considered male.
WRONG: The author would not disagree with this statement. From paragraph (lines 45-59) the author discusses the "two sides of the issue"; this statement being one of the sides. "Almost certainly, there is a complex interaction between the two" (lines 56-57). B.
It is possible that Chris naturally participated in "male" activities.
WRONG: The author would not disagree with this statement. From paragraph (lines 45-59) the author discusses the "two sides of the issue"; this statement being one of the sides. "Almost certainly, there is a complex interaction between the two" (lines 56-57). C.
Chris was not confused about her identity. WRONG: The author would not disagree with this statement. "Chris was in no way confused about her identity" (lines 21-22).
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D.
44.
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MATHEMATICAL TECHNIQUES
Most cultures have clearly defined gender roles. CORRECT: The author would disagree with this statement. The word, "most", in this statement renders it disagreeable. The author alludes to other cultures where Chris might not have felt the need "to identify herself as distinctly male" (lines 40-41). The culture/society wherein Chris lives is clearly defined as "our" (line 5 and 77) culture/society, implying that there are others. There is simply no way of quantifying this statement. This answer choice can also be arrived upon through process of elimination since the other answers the author would clearly not disagree with.
Implicit in the passage is the assumption that: I. one should be happy in one's "natural" state.
WRONG: This is not implied. Though we may strive for" an acceptance of ourselves in a 'natural' state" (line 64), and psychotherapy may help us to arrive at this point, there is not value-judgment-type "should" or "should not" implication in the passage from which to choose this answer. The author seems to have no problem with the idea that Chris would want a sex change operation. II. one can be well-adjusted, yet unhappy with one's "natural" state.
CORRECT: This is clearly implied. We know from the passage that Chris "is described as a relatively welladjusted individual" (line 24). Additionally, we know that Chris felt "she was a man in a woman's body" (line 17). "Unhappiness" can be used to describe this state because of the information at lines 67-69 "The effect of a physical sex reassignment operation on Chris' happiness cannot be foretold with complete certainty". Apparently, Chris "happiness" was an issue. III. one's perception of self is most important.
WRONG: This is not implied. This assumption is defined by the quote from the passage that "through psychotherapy, one learns that one may not necessarily have to change oneself as much as one's perception of self' (lines 65-67). There is no indication or implication in the passage that this is "most important", or more important than changing the "natural state" to fit our perception of self. A.
I only
B.
II only CORRECT: See above answer explanations.
45.
C.
III only
D.
I and III only
The author of the passage would be most likely to agree with which of the following ideas expressed by other psychologists? A.
A DSM IV 'disorder' may not actually be a disorder at all. WRONG: The author would not most likely agree with this idea. This idea is perhaps a second best choice. But it is much more vague than Answer C. It is clear from the passage that the DSM is a reflection of societal norms. In this society, in "our" society, even the author admits that Gender Identity Disorder is a disorder.
B.
The DSM IV is a poor descriptor of abnormal behavior and desires since it is easily influenced by societal norms. WRONG: The author would not most likely agree with this idea. Though it seems that the DSM is influenced by societal norms, there is no indication that it is "easily" influenced, or that overall it is a u poDr descriptor" of abnormal behavior. It may actually be a very accurate descriptor of behavior that society has determined is abnormal.
C.
Some DSM IV 'disorders' are simply an attempt to characterize socially abnormal behavior and desires. CORRECT: The author would most likely agree with· this idea. The author argues that in another culture, Chris' disorder might not even be a disorder. "The fact that Gender Identity Disorder exists in the DSM IV [the official handbook of psychiatric disorders] as a diagnosis is an admission on the part of psychologists Copyright © 2007 Examkrackers, Inc.
ANSWERS
&
EXPLANATIONS FOR THE 30-MINUTE IN-CLASS EXAMS • 101
that our society has clearly defined gender roles. These contribute to what it is generally considered "normal"" (lines 2-7). In this society, in "our" society, even the author admits that Chris' is considered to have a disorder.
D.
46.
47.
Behavior and desires must fall within the parameters of the DSM IV to be considered normal by society. WRONG: The author would not most likely agree with this idea. First, behavior falling "within" the parameters of the DSM IV is considered abnormal, not normal. Second, it is societal norms which the author believes have determined the DSM parameters. Not the other way around.
The author hints that the fact that Christ is well-adjusted indicates that her "uneasiness with her assigned sex" (lines 30-31): A.
is a problem which should be overcome through psychiatry. WRONG: This is not hinted at by the author. This is too strong. The author admits that the treatment of Chris is still a "question" (line 58). There is no implication that she" should" overcome her problem through psychiatry. "Psychologically healthy individuals" accept themselves in a "natural" state (lines 63-64). However, the author seems to believe that the "natural" state may just as easily be altered, as one's perception of self.
B.
is due to the culture she lives in. CORRECT: This is hinted at by the author. Paragraph 33-43, in its entirety, strongly argues this answer; beginning with, "in this case, many psychologists may believe society is the culprit" (lines 34-35).
C.
can be overcome through surgery. WRONG: This is not hinted at by the author. This is too strong. The author admits that the treatment of Chris is still a "question" (line 58). There is no implication that she "can" overcome her problem through surgery.
D.
is a basic personality defect. WRONG: This is not hinted at by the author. There is no support for this answer in the passage.
Suppose it is discovered that prescription medication allows Chris to become somewhat more comfortable with her "natural" state. Does this discovery support the author's argument?
What is the author's argument? This passage is not a completely objective representation of a psychological case study. His main argument is that "society is the culprit". "One must ... ponder the age-old question of society as the cause" (lines 31-32). A.
Yes; it confirms it. WRONG: This discovery does not support the author's argument. For this to be a correct answer choice, the author would have to be arguing that there is a psychological or neurological "causal factor". However, the author finds that Chris is "well-adjusted" and dismisses this idea and posits that, "One must ... ponder the age-old question of society as the cause" (lines 31-32).
B.
No; it does not affect it. CORRECT: This discovery supports the author's argument. This passage is not a completely objective representation of a psychological case study. His argument is that "society is the culprit". "One must ... ponder the age-old question of society as the cause" (lines 31-32). He implies in paragraph 33-43 that in another culture, Chris might not even be suffering from a disorder. Her unhappiness can be linked to society's unwillingness to accept her. Therefore, even if drugs help her become "somewhat" more comfortable with her "natural" state, the author's argument is not affected by this discovery.
C.
No; it weakens it.
WRONG: This discovery does not support the author's argument. The author's argument that "society is the culprit", is not weakened by an unspecified dosage and type of "prescription medicine" rendering Chris only "somewhat more comfortable". The supposition in the question is vague enough to assume that she was on a very strong dosage of a very powerful psychotropic drug. This is not the best answer. Copyright © 2007 Examkrackers, Inc
102 .
VERBAL REASONING
D.
48.
&
MATHEMATICAL TECHNIQUES
No; it disproves it. WRONG: This discovery does not support the author's argument. This answer and Answer A are the poorest of the four choices. See the explanations to Answer B.
The author admits that, "The effect of a physical sex reassignment operation on Chris' happiness cannot be foretold with complete certainty" (lines 67-69). The author most likely believes that: I. it is just as likely that psychotherapy would help Chris to change her perception of self.
WRONG: This is not what the author most likely believes. The author believes that Chris is "well-adjusted" and that "society is the culprit" (line 35). The author's nod to psychotherapy is lukewarm at best and provided only in the spirit of seeming to be objective (lines 64-67). II. in our society, in the body of a woman, Chris will not be happy.
CORRECT: The author most likely believes this. This passage is not a completely objective representation of a psychological case study. His argument is that "society is the culprit". "One must ... ponder the ageold question of society as the cause" (lines 31-32). He implies in paragraph 33-43 that in another culture, Chris might not even be suffering from a disorder. Her unhappiness can be linked to society's unwillingness to accept her acting as a man in the body of a woman. III. *a "sex reassignment operation" would make Chris happier.
CORRECT: The author most likely believes this. This answer is tantamount to Answer II. See the above explanation.
49.
A.
I only
B.
II only
C.
III only
D.
II and III only CORRECT: See the above answer explanations.
The author's attitude toward "our" societal norms is most accurately described as:
A.
favorable. WRONG: This is not an accurate description.
B.
neutraL WRONG: This is not an accurate description.
c.
distrustfuL WRONG: This is not an accurate description because it is not strong enough. Further, "trust" or "distrust" are poor descriptors for the author's attitude in the passage. See the explanation for Answer D.
D.
disapproving. CORRECT: This passage is not a completely objective representation of a psychological case study. The author's argument is that "society is the culprit" (line 35). "One must ... ponder the age-old question of society as the cause" (lines 31-32). He implies in paragraph 33-34 that in another culture, Chris might not even be suffering from a disorder. Her unhappiness can be linked to society's unwillingness to accept her acting as a man in the body of a woman. "Our" societal norms are described disparagingly as a uneatly constructed gender dichotomy" (lines 77-78).
Copyright © 2007 Examkrackers, Inc.
ANSWERS & EXPLANATIONS FOR THE 30-MINUTE IN-CLASS EXAMS . 103
Passage II (Questions 50-56) 50.
51.
The author of the passage would be most likely to agree with which of the following ideas expressed by other technology theorists? A.
' Soliciting further ideas and diverse ideas is not always wrong. CORRECT: The author would be most likely to agree with this idea. This answer is restating the rather stilted statement from the passage, "No one can argue that giving additional and different perspectives is inherently bad" (lines 54-56).
B.
Obtaining additional and differing perspectives is always beneficial. WRONG: The author would not be likely to agree with this idea. The word "always" in this idea is extreme. This idea is certainly not the same as the rather stilted statement from the passage, "No one can argue that giving additional and different perspectives is inherently bad" (lines 54-56). At most the author offers that "it can also be argued that more information is good" (line 52). However, conspicuous by its absence is the necessary"always" good.
C.
For information to be of value, it must challenge our human values. WRONG: The author would not be likely to agree with this idea. This is an antithesis. According to Nardi and OUay, whom the author seem s to agree with, "ti technology doesn't assist in promoting ... [our human] values, then it cannot be considered useful" (lines 18-20).
D.
A great deal of the information which we receive is useless because it does not pertain to us. WRONG: The author would not be likely to agree with this idea. He certainly would not be as likely to agree with this idea as the idea given in Answer A. Answer 0 is actually an idea of Postman's, not the author's; "[Postman] goes on to say that much of the information we receive does not pertain to us" (lines 44-45). First, the author does not agree with some of what Postman posits (lines 50-56). Further, even Postman does not say that this renders the information "useless". He refers to it as "at best an irrelevant distraction" (lines 45-46).
The passage argument suggests that information recipients might benefit from: I. a more rapid flow of information.
WRONG: This is not suggested. Though the author does not completely agree with everything that Postman says, even the author seems to think that recipients are receiving all the information that they can handle at this point.
II. thoroughly examining their information. CORRECT: This is suggested. The author argues that "information is useless unless it is thoroughly examined" (line 57) and "we as a society must consciously attempt to not only obtain information, but to analyze it for meaning" (lines 60-63). There are no counterpoints offered in the passage to this argument.
Ill. limiting non-pertinent information. WRONG: This is not suggested. The idea is only implied as counterpoint to the main passage arguments. Only Postman, whom the author does not completely agree with, says thal "much of the information we receive does not pertain to us [and that] this may create a problem ... " (lines 44-47). Yet, the reader is left only to assume that Postman would go on to 'limit non-pertinent information'; this is not stated. However, the author then argues (lines 50-56) that Postman's view is too extreme. The author sees Postman as wanting to "reduce the amount of information available" (lines 58-59) and is against this. A.
I only
B.
II only CORRECT: See above answer explanations.
C.
III only
D.
II and III only
Copyright © 2007 Examkrackers, Inc.
104 .
52.
53.
54.
VERBAL REASO NING
&
MATHEMATICAL T ECHNIQUES
In describing Neil Postman's "information glut", the author uses the example of "images of violence in the Middle East [that] do not affecl us" (line 48). The author's point is that: A.
the Middle East is at best an irrelevant distraction. WRONG: This is not the author's point. Perhaps Postman would suggest this. However, the author, who d?"s not want to "reduce the amount of information available" (lines 58-59), would not agree.
B.
the Middle East is poorly understood by us. WRONG: This is not the author 's point, and neither is)t suggested in the passage.
e.
these images should affect us. CORRECT: This is the author's point. The author prefaces his example of the Middle East categorizing it as a "problem" (line 47) that w e are not affected by these images. .
D.
these images should not be so easily accessible. WRONG: This is not the author 's point. The author would not limit "access" to information, w hich is tantamount to limiting the amount of information available. Instead, he would better tailor the information "to individual recipient's needs" (line 60).
If the following statements are true, which would most weaken the argument of the au thor? A.
The degree to which technology determines human values is questionable. WRONG: This statement would not 'm ost weaken the argument of the author. This answer / statement admits that technology determines human values. It is only a question of the "degree" of change. To paraphrase the author: Human values "are not static", but change according to certain factors, one of the most significant being technology. Based solely upon passage information and this statement it is difficult to determine if there is any disagreement at all.
B.
Information must be analyzed for its relevance to human values. WRONG: This statement would not most weaken the arg ument of the author. This is the thesis of the passage.
e.
Human values are unchanging. CORRECT: This statement would most weaken the argument of the author. The author argues that "Human values ... are not static [i.e. unchanging]" (lines 1-2). His thesis rests upon this idea since he considers technology to be a "significant" factor affecting human values.
D.
Human values are culturally dependent. WRONG: This statement would not most weaken the argument of the author. It would matter little to the author's arguments whether they were culturally dependent or not.
The author argues, "more information is good" (line 52). Unlike Neil Postman, the author does not consider which of the following to be a factor that might limit the usefulness of information? The correct answer must be one that Postman considers a "limiting factor", but the author does not. A.
Space WRONG: This is not implied as a limiting factor by Postman, and is not alluded to by the au thor.
B.
Distance WRONG: This is not implied as a limiting factor by Postman, but is offered as an example of Postman's 'information glut' idea by t"eauthor. Providing this example of Postman's ideas, the author writes "for example, images of violence in the Middle East do not affect us because they are ... too far away to influence us" (lines 47-50). Further, the author never responds to this idea of distance as a limiting factor.
e.
Frequency WRONG: This is not implied as a limiting factor by Postman, and is not alluded to by the author. "Frequency" cannot be assumed to be the same as 'quantity', or 'amount'. Copyright © 2007 Exam krackers , Inc.
A NSWERS
D.
55.
&
EXPLANATIONS FOR THE 30- M INUTE IN -C LASS ExAMS '
10S
Tune CORRECT: This is implied as a limiting factor by Postman, and is alluded to by the author, w ho does not seem to agree that it might limit the usefulness of information, Postman "states that the rapid flow of inform ation allowed by automation and Internet/email provides vast amounts of data to m any users quickly and simultaneously, but doesn't allow humans to understand its meaning fas t enough to keep up" (lines 4044), In other w ords, they don 't have time to process the information, However, the author apparently doesn' t think that this is a problem, He seems to believe that people generally have all the time in the w orld, "By giving people more information [which admittedly takes more time for them to process it), they have a wider basis from which to make decisions and form opinions, No one can argue that giving additional and different perspectives is inherently bad [except that we usually don 't have the time]" (lines 52-56) ,
According to the passage, one drawback of technology is that it can: A,
supersede all o ther values, WRONG: This is not a d rawback of technology, First, technology is not implied to be a "value", The "other" in this answer clearly implies that technology is a value. Though the passage provides that, "Technology may force people away from their own values [which] creates a disturbing image of a world where teclmology itself becomes the center of a society without values" (lines 35-39), This idea still does not make technology a value,
B.
be used to destroy human values, WRONG: This is not a drawback of technology. Though the p assage p rovides that, "Technology may force people away from their own values [which] creates a disturbing image of a world w here technology itself becomes the center of a society without values" (lines 35-39), there is no inference that "the technology can be used to destroy human values", The two do not have the same meaning,
C.
subtly change people's values, CORRECT: This is a drawback of technology, Notice the use of the 'softener ' "subtly" which makes this answer even more palatable, We know that technology is a "significant factor" that can change people's values (lines 1-5), "Technology may force people away from their own values" (lines 35-38), This is a drawback illustrated by Ellul's "disturbing image" (lines 37-39), The passage p rovides that technology should fit with the values and not vice versa, One way to ensure this is the "three-pronged format" posited by N ardi and O'Day.
D.
become an irrelevant distraction.
WRONG: This is not a drawback of technology, It is "information" that can become an "irrelevant distraction rather tllan the useful news it was intended as" (lines 46-47), 56.
Suppose it could be established that technology is most efficient when it performs its function unobtrusively; without being noticed, The au thor of the passage w ould be most likely to respond to this information by: This is a classic MCAT question, Consider your best answer choices those that would allow the author to either support or, in the wors t case, resurrect or rehabilitate his main arguments and thesis, No author would be likely to admit he was completely wrong or abandon his thesis. A.
su ggesting that this determination ratifies his thesis, WRONG: The author would not be most likely to respond in this way. The supposition in the question is somewhat in opposition to his tilesis, The author agrees with Nardi and O'Day tilat, " If the technology is taken for granted, the actual purpose for u sing it can become clouded " (lines 24-25), TIle author would be required to do some explaining to resurrect his ideas.
B.
proposing that we must stiU analyze tile information for meaning, WRONG: The author would not be most likely to resp ond in this way, Though the entire second half of the passage is about informa tion, this is tangen tial to the overall thesis on technology, It begins with the example of 'information glut' and snow balls, However, this answer is not really responsive to the supposition, Tnformation technology would be only one small aspect of technology,
Copyright © 2007 Exam kracker.s, Inc
106
VERBAL REASO NING
&
MATHEMATICAL T ECHNIQUES
C.
asserting that efficiency is usually degraded when a technology is taken for granted. WRONG: The author would not be most likely to respond in this way. The supposition in the question is somewhat in opposition to his thesis. The author agrees with Nardi and O'Day that, " If the technology is taken for granted, the actual purpose for using it can become clouded" (lines 24-25). However, it is a leap of logic to assume that what the author and these analysts specifically mean is that "efficiency is usually degraded" .
D.
explaining that we must still remain aware of the technology and its intended purpose. CORRECf: The author would be most likely to respond in this way. The supposition in the question is somewhat in opposition to his thesis. The author agrees with Nardi and O'Day that, "If the technology is taken for granted, the actual purpose for using it can become clouded" (lines 24-25). Therefore, by responding in the fashion of Answer D, the author has responded to the seemingly opposing-supposition without losing ground in his own arguments.
Passage III (Questions 57-63) <
57.
58.
Assume that several others who had attended the same opera were interviewed. If they remarked that Margaret sang with tremendous passion, these remarks would weaken the passage assertion that: A.
Mephisto had rendered her irresistible to Faust. WRONG: This answer is not a IIpassage assertion", which is a requirement for a correct answer.
B.
the lyrics which she sang were "throwaway". WRONG: This answer is a "passage assertion". However, it is not weakened by the assumption in the question. The remarks have to do with the "tremendous passion" of Margaret's singing, not the "lyrics". This answer is not as responsive to the question as Answer C.
C.
the young woman of Faust's dreams sang without emotion. CORRECf: This answer is a "passage assertion". First, the young woman of Faust's dreams is a "commoner named Margare t" (lines 51-52). Second, the passage asserts and alludes to "Margaret's emotionless singing" (line 57).
D.
the young woman of Faust's dreams was a commoner. WRONG: This answer is a "passage assertion". The young wonlan of Faust's dreams is a "commoner named Margaret" (lines 51-52). However, the remarks do not w eaken this assertion because they have no relation to it.
On the basis of the passage, it is reasonable to conclude that: A.
Faust lost his soul to the devil. WRONG: This is not a reasonable conclusion. There is no sure way of knowing this from the passage.
B.
the " country fair thronged with revelers" was not in the book. WRONG: This is not a reasonable conclusion. There is no sure way of knowing this from the passage.
C.
the operatic interpretation differed from the book. CORRECT: This is a reasonable conclusion. There are several instances which allude to and support this conclusion: "In the director's vision ... " (lines 11-12), and "A chorus presents the essential plot of Faust, reduced from its several incarnations ... " (lines 21-22), and" ... this version presents ... " (line 25 al1d line 48), and "a theme which is not really congruent with the Faust myth .. . " (lines 55-56).
D.
the author did not enjoy the performance. WRONG: This is not a reasonable conclusion. This is certainly arguable. Though the author indicates that at Scene II the opera was not particularly to his liking, he begins the passage by asserting that "the New York City Opera production of "Mephistopheles" deserves high marks for visual excellence" (lines 1-3).
Copyright © 2007 Exa mkrackers. Inc.
ANSWERS
59.
60.
&
EXPLANATIONS FOR THE 30-MlNUTE IN-CLASS EXAMS • 107
According to the passage, the author felt that the New York City Opera production of "Mephistopheles": A.
was plagued with a poor characterization of Mephisto. WRONG: This is not supported by passage information. The author actually seems to have liked Mephisto's characterization.
B.
suffered from noticeable weaknesses beginning in Scene II. CORRECT: "In Scene II, ... the play devolves into ... very forgettable arias [and] ... a theme which is not really congruent with the Faust myth" (lines 50-56).
C.
was enhanced by Dr. Faust's singing. WRONG: This is not supported by passage information. There are no allusions to Faust's singing.
D.
could have been improved in Scene III. WRONG: This is not supported by passage information. There are no allusions to Scene III.
According to the passage, the author seems to have most enjoyed: A.
the music of the opera. WRONG: This is not supported by passage information. It is not clear if the author really enjoyed the music. He said that the music was" only slightly corny" (line 6). He certainly did not like the music as much as the "visual excellence" of the production.
B.
the singing of the opera. WRONG: This is not supported by passage information. The author cared little for Margaret's" emotionless singing" for instance. f
61.
C.
the plot of the opera. WRONG: This is not supported by passage information. All this we know from the passage, is that this operatic version differed from the book and other versions which the author was familiar with. Further, at Scene II, and at the "Walpurgis night" scene, the author seems not have cared for the way in which the opera was presented.
D.
the images of the opera. CORRECT: " ... the New York City Opera production of "Mephistopheles" deserves high marks for visual excellence" (lines 1-3).
Which of the following does the author suggest was a component of the original "Faust myth" (lines 55-56)? I. *Reason triumphing over feeling CORRECT: This is clearly suggested. From the passage we know that in the opera "Faust and Margaret sing very forgettable arias about the supremacy of feeling over reason, a theme which is not really congruent with the Faust myth" (lines 53-56). II. A more evil Mephisto WRONG: This is not clearly suggested. The second paragraph, for one, describes the character of Mephisto in the operatic production, and does give the impression that he has been portrayed/ characterized differently in other versions. However, the author gives no indication that a more evil Mephisto" "was a component of the original 'Faust myth"'. There is not enough specificity in the passage information for one to draw this conclusion. /I
III. A more powerful God WRONG: This is not clearly suggested. This answer seems attractive because of the passage statement, "Thus, this version presents temptation as essentially a wager, or struggle, between God and the Devil (which, at one time, was a remarkably blasphemous notion, as it contradicted the dogma that God is allpowerful over evil)" (lines 25-29). However, it is not at all clear that the phrase "at one time" refers to the "Faust myth" or about people's attitudes and belief in God in general. This is not the best answer. Copyright © 2007 Examkrackcrs, Inc.
108 .
62.
63.
VERBAL REASONING
&
MATHEMATICAL TECHNIQUES
A.
I only CORRECT: See the above answer explanations.
B.
II only
C.
III only
D.
II and III only
According to the passage, through what primary means is the fundamental plot transmitted to the audience? A.
Via visual imagery WRONG: This is not described as a means of transmitting the 'fundamental' plot to the audience.
B.
Via Faust's musings WRONG: This is not described as a means of transmitting the 'fundamental' plot to the audience.
C.
Via the chorus CORRECT: This is specifically described as a means of transmitting the 'fundamental' plot to the audience. "A chorus presents the essential plot of Faust, reduced from its several incarnations" (lines 21-22).
D.
Via Mephisto WRONG: This is not described as a means of transmitting the 'fundamental' plot to the audience.
Regarding the devil's bargain with Faust, the·passage strongly implies that: A.
it is Faust who got the better deal. WRONG: This is not implied.
B.
it is the devil who got the better deal. WRONG: This is not implied.
C.
in other versions, the bargain was with Margaret. WRONG: This is not implied.
D.
in other versions, it is Faust who does the bargaining. CORRECT: This is implied. "In this version of Faust, it is Faust who seizes the devil's bargain ... " (lines 47-48).
Copyright © 2007 Examkrackers, Inc.
EXAMKRACKERS
MeAT
BIOLOGY 7 TH EDITION
'.,."'" ~
O~SQItt '\
Acknowledgements Although I am the author, the hard work and expertise of many individuals contributed to this book. The idea of writing in two voices, a science voice and an MeAT voice, was the creative brainchild of my imaginative friend Jordan Zaretsky. Jordan and others also wrote many of the questions in the lectures. I would like to thank Jerry Johnson for lending his science talent and exceptional science skills to this project. I would like to thank many years worth of ExarnKrackers students for doggedly questioning every explanation, every sentence, every diagram, and every punctuation mark in the b ook, and for providing the creative inspiration that helped me find new ways to approach and teach inorganic chemistry. Finally, I wish to thank my wife, Silvia, for her support during the difficult times in the past and those that lie ahead.
Copy righ t (l) 200'7 Examkrackers, Inc.
READ THIS SECTION FIRST! This manual contains all the biology tested on the MeAT and more. It contains more biology than is tested on the MeAT because a deeper understanding of basic scientific principles is often gained through more advanced study. In addition, the MeAT often p resents passages with imposing topics that may intimidate the test-taker. Although the questions don't require knowledge of these topics, some familiarity will increase the confidence of the test-taker. In order to answer questions quickly and efficiently, it is vital that the test-taker understand what is, and is
not, tested directly by the MeAT. To assist the test-taker in gaining this knowledge, this manual will use the following conventions. Any term or concept which is tested directly by the MeAT will be written in bold and brown. To ensure a p erfect score on the MeAT, you should thoroughly understand all terms and concepts that are in bold and brown in this manual. Sometimes it is not necessary to memorize the name of a concept, but it is necessary to understand the concept itself. These concepts will also be in bold and brown. It is important to note that the converse of the above is not true: just because a topic is not in bold and brown, does not mean that it is not important. Any formula that must be memorized will be written in
large, red, bold type.
If a topic is discussed purely as background knowledge, it will be written in italics. If a topic is written in ital-
ics, it is not likely to be required knowledge for the MeAT but may be discussed in an MeAT passage. Do not ignore items in italics, but recognize them as less important than otl1er items. Answers to questions that directly test knowledge of italicized topics are likely to be found in an MeAT passage. Text written in orange is me, Salty the Kracker. r wiH remind you what is and is not an absolute mllst for MCAT. I wi ll help YOll develop your MCAT intuition. In addition, I will offer mnemonics, simple methods of viewing a complex concept, and occasionally some comic fcHef. Don't ignore me, even if you think 1 am not funny, because l11y comedy is des igned to help you understand Jnd remenlber. If you ~hink I an1 funny, tell the boss. I could usc a raise.
Each chapter in this manual should be read three times: twice before the class lecture, and once immediately following the lecture. During the first reading, you should not write in the book. Instead, read purely for enjoyment. During the second reading, you should both highlight and take notes in the margins. The third reading should be slow and thorough. The 24 questions in each lecture should be worked during the second reading before coming to class. The inclass exams in the back of the book are to be done in class after the lecture. Do not look at them before class. Warning: Just attending the class will not raise your score. You must do the work. N ot attending class will obstruct dramatic score increases. If you have Audio Osmosis, then listen to the appropriate lecture before and after you read a lecture. If you are studying independently, read the lecture twice before doing the in-class exam and then once after doing the in-class exam. If you have Audio Osmosis, listen to Audio Osmosis before taking the in-class exam and then as many times as necessary after taking the exam.
A scaled score conversion chart is provided on the answer page. This is not meant to be an accurate representation of your MeAT score. Do not become demoralized by a poor performance on these exams; they are not accurate reflections of your p erformance on the real MeAT. The thirty minute exams have been designed to educate. They are similar to an MeAT but with most of the easy questions removed. We believe tha t you can answer most of the easy questions without too much help from us, so the best way to raise your score is Copyright © 2007 Examkrackers, Inc
to focus on the more difficult questions. This method is one of the reasons for the rapid and celebrated suc. cess of the Examkrackers prep course and products. If you find yourself struggling with the science or just needing more practice materials, use the Examkrackers 1001 Questions series. These books are designed specifically to teach the science. If you are already scoring lOs or better, these books are not for you.
You should take advantage of the bulletin board: at www.examkrackers.com. The bulletin board allows you to discuss any question in the book with an MCAT expert at Examkrackers. All discussions are kept on file so you have a bank of discussions to which you can refer to any question in this book. Although we are very careful to be accurate, errata is an occupational hazard of any science book, especially those that are updated regularly as is this one. We maintain that our books have fewer errata than any other prep book. Most of the time what students are certain are errata is the student's error and not an error in the book. So that you can be certain, any errata in this book will be listed as it is discovered at www.examkrackers.com on the bulletin board. Check this site initially and periodically. If you discover what you believe to be errata, please post it on this board and we will verify it promptly. We understand that this system calls attention to the very few errata that may be in our books, but we feel that this is.the best system to ensure that you have accurate information for your exam. Again, we stress that we have fewer errata than any other prep book on the market. The difference is that we provide a public list of our errata for your benefit. Study diligently; trust this book to guide you; and you will reach your MCAT goals.
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TABLE OF CONTENTS
LECTURE
1:
MOLECULAR BIOLOGY; CELLULAR RESPiRATION ... . . . .... . .. .. .. ... .. ... ..... .. ... . ............ 1
1.1
Introduction ............................................................................................... ......... .. ..... ............... 1
1.2 Water ......... ......................... ..... ...... .. ................. ................................................ .......... .. ...... ...... 1 1.3 Lipids ............................... ................ ................. ........ ........ ......... ... ..... ....... .. ....... ......... .............. 2 1.4 Proteins ................................ .. .................. ................ ........ ........ ..... ..... ...... ,.... ..... ..................... .4 1.5 Ca rbohydrates ........ ........ ......... ......................... ... ...................... ............... .... ... ... ......... ............. 6 1.6 Nucleotides ............ ..... ............ ............. ... ...................................... ............ ........ .......... ............. 8 1.7 Minerals ........................................... ................. .. ....... ...... .. .. ...... ... .... ......... .. ...... .. ..................... 9 1.8 Enzymes .......................................... .... .................................... ................ .... ...... ... ................... 11 1.9 Enzyme Inhibition ........... ........ ................. .. ......... ............. ................. ......... .. ...... .......... .......... 12 1.10 Enzyme Reg ulation ...... ... ................ ..... .... .. ............. ... ....... ...... ......... ......... .. .. ......................... 13 1.11
Enzyme Classification ............ ... ....................... ..... .... ................................. ... ...... ........ ....... .... 15
1.12 Cellula r Metabolism ...... .... ............................................................ .. ........................... ........... 17 1.13 Glycolysis ................................. ...... .. ......... ..................... ,........ .......... ..... .. ........... ........ ... ......... 17 1.14 Fermentation ........................... ................. ....... ... ............................................ .... .................... 18 1.15 Aerobic Respiration .......... .. ...... ......... ...... .. ........ ................ ...... ...... .. .. ......... ............................ 21 1.16 Krebs Cycle ............................. ........ ........ ..... ...... ............... ............. .. .. .. .. ... .... ......................... 21 1.17 El ectron Transport Chain ............ .... ..... .... ... ....... ...... ......... :........ ................. ... ......................... 22 LECTURE
2:
GENES ...... •.. . ...... . .... .•. . .•.. . .••. . ..••..... . ...........•.. . .••. . . .•. . ..•. . .. . .•.•..• .•• . •.••.••..•..• 25
2.1 The Gene ......... ........ .. ............... ................ .... .............. ............... .. ....... .......... ........... ...... ..... .. .. 25 2.2 DNA. ................................................. ......... .. ........ ....... ...... ... ...... .. ....... ... .............. ......... .... ..... .25 2.3 Replication ..... ........... .......... ... ......... .... ... ..... .................... ..... .......... ... ..... ........... .................... 27 2.4 RNA .......... .............. ......... .................. ....... ......... ........ ........ ......... ......... ....... ........ ........ ............ 28 2. 5 Tran scription ................ ... ..... ... ....... ........... ......... ........ ........ ............... .... ........ ......................... 29 2.6 Post-transcriptional Processing .. ... ... ................. ................................ ......... .. ...... ... ...... ........... 31 2. 7 DNA Technology ........ ....... .. .............................. ......... ........ .... .... ........ ......... ........ ....... ............ 32 2.8 The Genetic Code .................... ..... ....... ........................ ........ ...... ........ ....... ........ .......... ........... 38 2.9 Tra nslation ...................... ..... ..... ...... .. ... .... ........ ................. ..................... .. .......... ....... ... .. .... ..... 39 2.10 Mutations ........ ... ......... ............ ......................................... ................... ................................ .. ..41 2.11
Cancer .................................. .. ..... ............. ............. ..... .................. ..... ... ....... ..... ... ............... .... 43
2.12 Chromosomes ..... .... ................ ... ...... ... ......................... ............. ... ..... ........ ... .......... ,.............. .45 2.13 Cell Life Cycle ......... ... ...... ... ........ .. :..... ...... .... ............................................. ... ........ ................ .46 2.14 Mitosis .............................. ........ ........ .... ..... ... ..... ............................ .... .......... ... ... ... ...... ........... .47 2.15 Meiosis ........... ........ ..... .... ........ ................. ................................ ......... ...... .. ... ..... ................... .49
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LECTURE
3:
MICROBIOLOGY .; ........................................................................................53
3.1
Viruses ........................................................... ................ ...................•........•.......................... 53 I
3.2 Defense Against Viral Infection ........ ,.............. ....................................................................... 55 3.3
Prokaryotes ....................................................................... ............................... ...................... 58
3.4 The Structure of Bacteria ................ ................................................................................. ....... 59 3.5
Membranes ................................................................................ ............... ...................... ....... 61
3.6 Membrane Transport .......................................................................... .................................... 62 3.7
Bacterial Envelope ..................................................... ................. ... ............. ............................ 64
3.8
Bacterial Reproduction ........... ........ ................... ,...... ............................................................. 67
3.9 Endospores ....................................................... ........ ........................ ................. .. .................. 68 3.10 Fungi ........................................................................... ....... ....... .. .................................... ....... 70 3.11
Fungal Reproduction and Life Cycle ...................................................................................... 70
4:
THE EUKARYOTIC CELL; THE NERVOUS SYSTEM ............................................ 73
4.1
The Nucleus .......................................................................................................................... 73
LECTURE
4.2 The Membrane ......................... ................ .. ....................... ........ ............................................ 74 4.3 Cellular Filaments ..................... ........ ........................................................................ ............. 76 4.4 Cellular Junctions ............................................................... .................................. ................. 78 4.5
Mitochondria ................................................................. ....... .. ..... ................ ........................... 79
4.6 The Extracellular Membrane .................................................................................................. 79 4.7 Organization in Multicellular Eukaryotes ...... .. ..................... ........ .. ....... ........ .. ........................ 80 4.8 Intercellular Communication ............................... ................. ........................... .. ........ .. ........... 82 4.9 Paracrine System ................................................................... .. ............................................... 82 4.10 Nervous System ............................ ........ .. ....................... .. .............. .. ...................... ................. 82 4.11
The Action Potential ...................................................................... ............... .. ...... ................. 85
4.12 The Synapse .................................. ...... ............ .... .. ................................................................ 87 4.13 Support Cells .......................................................................................................................... 89 4.14 The Structure of the Nervous System .......... .. ...... .. .................................. .............................. 92 4.15 The Central Nervous System ........................................................................... ......... .. ............ 95 4.16 Sensory Receptors .................................................................................... ........ ...................... 96 4.17 The Eye ................................. ................................................................................................. 96 4.18 The Ear ............................................ ......... .... ................ ..................................... .................... 98 4.19 The Nose and Mouth .......... .. .............. .. ....... .. ................................. ............... ........ .. ............. 98 LECTURE
5:
THE ENDOCRINE SYSTEM .......................................................................... 101
S.1
Hormone Chemistry ..................... ........ .......................................................... .. ....... .. .......... 101
5.2 Negative Feedback .............................................................................................................. 103 5.3 Specific Hormones and Their Functions ................. ... .... .... ........ ...... .. ..... ......... .. ....... .. ......... 105
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5.4 Anterior Pituitary ~........................................ .......................................................... ............... 105 5.5 Posterior Pituitary ............. ................. ........ .•....... ................ ...................," ..... .......... .............. 106 5.6 Andrenal Cortex ........•........•...... .••...... ................ ......... ........ ........ ........ ......... .........••............ 107 5.7 Thyroid ....................................................... ................. ........ ........ ........ ......... ....................... 108 5.8 Pancreas (Islets of Langerhans) ........... ........................................ .........•...... ......... .•.............. 108 5.9 Parathyroid .......................................................... ................. ....... ........................................ 109 5.10 Reproduction ................................................................ ................. ....................................... 111 5.11
The Male Reproductive System ...........................................•. ..... ........ .• .......•.............. ........ 111
5.12 The Female Reproductive System ........•................•............... ................. ........ ........ .............. 113 5.13 Fertilization and Embryology .............................................. ...........................••..... .............. 115 LECTURE
6:
THE DIGESTIVE SYSTEM; THE EXCRETORY SYSTEM ........................................ 119
6.1
Anatomy ...............................................................................................•.............................. 119
6.2 The Mouth and Esophagus ......... .................................................................... ..................... 120 6.3 The Stomach ................................................ .................................................. ........ .............. 121 6.4 The Small Intestine ............ ................. ........ .•....... ........ .•...... ........ .........•....... ......... .............. 122 6.5 The Pancreas ................................................. ................. ....... .................................. ............. 123 6.6 The Large Intestine .............................................................................................................. 124 6.7 Gastrointestinal Hormones Involved in Digestion ...................... ................. .•....... .............. 125 6.8 Absorption and Storage ....................................................... ......... ....... .......... .. ................... 127 6.9 Carbohydrates ...............................••....... ,.............................. ......................... ...................... 127 6.10 Proteins ........................................................... ............... .. ............ .. ...... .... ........ .... .. ............. 129 6.11
Fats ............................... ................ .. .............. .... ..... .. .......... ...... .. ...... .. ........ .............. ............. 130
6.12 The Liver ..................... ................................... ... .............. ............... ........ .. ............................ 131 6.13 The Kidney .......................................................................................................................... 134 6.14 The Juxtaglomerular Apparatus ............ ... ............ .. ............... .. .......... .............. .. ......... .. ....... 135 LECTURE
7:
THE CARDIOVASCULAR SYSTEM; THE RESPIRATORY SYSTEM •..•.......•..••.•.••••••• 139
7.1
Cardiovascular Anatomy .................................................................................... .................. 139
7.2 The Respiratory System ........................................................................................................ 145 7.3 The Chemistry of Gas Exchange ............. .......................... ................ ................................... 146 7.4 The Lymphatic System .................................................................................... ........ ............ 150 7.5 The Blood ................................................................ ........ ........ ....... .. .............. .......... ........... 151 7.6 The Immune System .... ......... ........ .. ...... .................. ........ .. ...... .. ....... .. ..... .. ........ .... ...... ......... 152
LECTURE
7.7
Blood Types ........................................... ................. ........ .................................................... 154
8:
MUSCLE, BONE, AND SKIN ........................................................................ 157
8.1
Muscle ..................................................... ......... ....................... ................................. ............ 157
8.2 Skeletal Muscle .................................................................... ..... .. ......... ... ..... ........ .... .... ........ 157 CopyriSJht
(i;J
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8.3
Physiology of Skeletal Muscle Contraction .................. .......................................... .............. 158 /
8.4 A Motor Unit ......................................................... .............................. .. ................ ............... 160 8.5 Skeletal Muscle Type .................. .... ................. ........ ....................................... .. ....... .. ........... 161 8.6 Cardiac Muscle ............................. .......................... ................ ........ ....... .......... ..... .. .. ........... 163 8.7 Smooth Muscle ................ .. ....... ................ ............. .... ................. ................ ......................... 164 8.8 Bone .................................................................................................................................... 166 8.9 Bone Function in Mineral Homeostasis ....... ................ ........................ .... ... .......................... 166 8.10 Bone Types and Structures .................................................................................... .............. 167 8.11
Cartilage .................................................................................................................. ............ 168
8.12 Joints ................. ................ ................. ............................................ ..... ................. ................ 168 8.13 Skin ............................................ ........ ................. ........ .... ................ ..................................... 169
9:
POPULATIONS .......................................................................................... 171
9.1
Mendelian Concepts .......... ........ ......... ........ ................................ ......... .......... ........ .............. 171
LECTURE
9.2 Evolution ...................................... ............................... .......... .. .... ......... .... ..... ........ .... ........... 175 9.3 Symbiosis ............................................ .. ....... ........ .. ....... ................ .. ..... .. ......... .. ...... .. ............ 176 9.4 Hardy-Weinberg Equilibrium ................................................ .... .................................. .......... 177 9.5 Origin of Life ........................................ .. .............. ................................................................ 179 9.6 Chordate Features .............................. ........ ......... ........ ......... ....... ... ............. ......................... 179 30-MINUTE IN-CLASS EXAMS ........................................................................................ 183 In-Class Exam for Lecture 1 ........ .. ............... .. ........ .. .... .. ........ ....... .. ....... .. ...... .. ..................... 183 In-Class Exam for Lecture 2 ..... : ............ .. ..... .......... .. ............. ................. .. .............. .. ............. 189 In-Class Exam for Lecture 3 .......................................................................... ........ ................ 195 In-Class Exam for Lecture 4 ................ .......... ....... .......... ....... ................................ ................ 201 In-Class Exam for Lecture 5 ......................... .................................................. .. ..... .... ............ 207 In-Class Exam for Lecture 6 ................. ................. ................................................................ 213 In-Class Exam for Lecture 7 .......... ................. .. ..... .. .......................... ............. ....................... 219 In-Class Exam for Lecture 8 .................. ........ ......... ............................................................... 225 In-Class Exam for Lecture 9 .................................................................... .............................. 231 ANSWERS
&
EXPLANATIONS TO IN-CLASS EXAMS ............................................................ 237 Answers and Scaled Score Conversion for In-Class Exams ................................................ 238 Explanations to In-Class Exam for Lecture 1 ........................................ .. ......... .. ................... 239 Explanations to In-Class Exam for Lecture 2 ................ ............ .... ........................................ 240 Explanations to In-Class Exam for Lecture 3 ................. ........ ........ ........ ........ ....................... 242 Explanations to In-Class Exam for Lecture 4 .......................................................... .. ............ 244 Explanations to In-Class Exam for Lecture 5 ........................ .. .............. .. .............. ................ 245
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Explanations to In-Class Exam for Lecture 6 .................... .. ,...... ......... ........ ... ....................... 247 Explanations to In-Class Exam for Lecture 7 ............. ........ ........ ......... ........ ........ .................. 249 Explanations to In-Class Exam for Lecture 8 .............•.......•.......•........•.......•...... .•........•........ 250 Explanations to In-Class Exam for Lecture 9 ..................................................... ................... 252
ANSWERS
&
EXPLANATIONS TO QUESTIONS IN THE LECTURES .•.•••.•.••..•••..•..•....•....•••....•... 255 Answers to Questions in the Lectures ...................................... ............................................ 256 Explanations to Questions in Lecture 1 .............•........•........•.......•.........•.........••.......•......... 257 Explanations to Questions in Lecture 2 ........................................... ......... .......... ................ 258 Explanations to Questions in Lecture 3 ................................................................ ......•....... 259 Explanations to Questions in Lecture 4 .......... ........ ........ .•.............. .•........•........ .•.......•....... 260 Explanations to Questions in Lecture 5 ............•..... ..........•...... ....... ..•................ ................. 262 Explanations to Questions in Lecture 6 .............................................................................. 263 Explanations to Questions in Lecture 7 .................................... ................ .......... ................ 265 Explanations to Questions in Lecture 8 .......... .................. ............... ..........••............... ........ 266 Explanations to Questions in Lecture 9 .......... .................................................................... 268
INDEX ••...••.•.•••..••••.••••...•••...••..••...••...••......••....•....•....•...••...••...••••....•...••..•..•..•..•.•••....• 271
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BIOLOGICAL SCIENCES DIRECTIONS: Most questions in the Biological Sciences test are organized into groups, each preceded by a descriptive passage. After studying the passage, select the one best answer to each question in the group. Some questions are not based on a descriptive passage and are also independent of each other. You must also select the one best answer to these questions. If you are not certain of an answer, eliminate the alternatives that you know to be incorrect and then select an answer from the remaining alternatives. Indicate your selection by blackening the corresponding oval on your answer document. A periodic table is provided for your use. You may consult it whenever you wish.
PERIODIC TABLE OF THE ELEMENTS -
I
2
I
He 4.0 10
0 I
4
.i
Be
B
C
N
0
F
Ne
9 1
9.0 12
10.8 13
12.0 14
14.0 15
16.0 16
19.0 17
20.2 18
5
a Mg
6
7
8
9
AI
Si
P
S
CI
Ar
27.0 31
28.1 32
31.0 33
32.1 34
35.5 35
39.9 36
.0 9
24.3 20
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
.1 7
40.1 38
45.0 39
47.9 40
50.9 41
52.0 42
54.9 43
55.8 44
58.9 45
58.7 46
63.5 47
65.4 48
69.7 49
72.6 50
74.9 51
79.0 52
79.9 53
83.8 54
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
,,
~
21
22
23
24
25
26
27
28
29
30
b
Sr
V
Zr
Nb Mo
Tc
.5 5
87.6 56
88.9 57
91.2 72
92.9 73
95.9 74
(98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 75 76 77 78 79 81 82 80 83 84 85 86
Ba La*
Hf
Ta
W
5
Re
Os
Ir
Pt
Au
Hg
TI
Pb
Bi
Po
At
Rn
1.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222) 7 104 105 106 107 108 109 88 89
r
Ra Aco Unq Unp Unh Uns Uno Une
!3) 226.0 227.0 (261) (262) (263) (262) (265) (267)
*
58
59
60
63
64
65
66
67
68
70
71
Ce
Pr
Nd Pm Sm Eu
61
62
Gd
Tb
Dy
Ho
Er Tm Vb
69
Lu
140.1 140.9 144.2 (145) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0 91 92 93 94 95 96 97 98 99 100 101 102 103 , 90
Th
Pa
U
Np
Pu Am Cm Bk
Cf
Es
Fm Md No
Lr
232.0 (231) 238.0 (237) (244) (243) (247) (247) (251) (252) (257) (258) (259) (260)
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Molecular Biology; Cellular Respiration 1.1
Introduction
This lecture discusses the structure and b,;lsic functions of the major chemical components of living cells and their surroundings. Although most of the details of this biochemistry are not required on the MeAT, this knowledge does create a strong base from which to understand the rest of the manual.
Most biological molecules can be classified as lipids, proteins, carbohydrates or nucleotide derivatives. Each of these types of molecules possesses a carbon skeleton. Together with water and minerals, they form living cells and their environment.
1.2
Water
Water is the solvent in which the chemical reactions of living cells take place . 70 to 80 percent of a cell's mass is due to water (Fig. 1-2). Water is a small polar molecule that can hydrogen bond. Most compounds as light as water would exist as a gas at typical cell temperatures. The ability of water to hydrogen bond allows it to maintain its liquid state in the cellular environn1ent. Hydrogen bonding also provides strong cohesive forces between water molecules. These cohesive forces "squeeze" hydrophobic (Greekhydros:water, phobos:fear) molecules away from water, and cause them to aggregate. Hydrophilic (Greekphilos:love) molecules dissolve easily in water because their negatively charged ends attract the positively charged h ydrogens 6f water, and their positively charged ends attract the negatively charged oxygen of water (Fig. 1-1). Thus, water molecules surround (solvate) a hydrophilic molecule separating it from the group.
~;,
L-i:-(:
L
t' .
II
'\.....0../ '
\Vater
Figure 1-1 The solvent properties of wa te r Besides acting as a solvent water often acts as a reactant or product. Most Inacromolecules of living cells are broken apart via hydrolysis (Greeklysis:separation), and are formed via dehydration synthesis. (Hydrolysis and dehydration are discussed in Biology Lecture 7, and in the Organic Chemistry manual.)
2
MeAT
B,OLOGY
Lipid _________ RJ'l A ~_- Carbohydrate
b ;;;;;;i.......[:C-=--:::____-__ -
Inorganic ~ ________~- DNA
H,o
Other organic
Figure 1-2 Relative mass of the molecular components of a living cell
1.3
Lipids
A lipid is any biological molecule that has low solubility in water and high solubility in nonpolar organic solvents. Because they are hydrophobic, tllPy rna ke excellent barriers separating aqueous envirorunents. Six major groups of lipids are: fatty acids; triacylglycerols; phospholipids; glycolipids; steroids; and terpenes (Fig 1-3). Besides being lipids themselves, fatty acids are the building blocks for most, but not all, complex lipids. They are long chains of carbons truncated at one end by a carboxylic acid. There is usually an even number of carbons, with the llldximum number of carbons in humans being 24. Fatty acids can be saturated or unsaturated. Saturate d fatty acids possess only single carbon-carbon bonds. Unsaturated fatty acids contain one or more carbon-carbon doub le bonds. Oxidation of fatty acids liberates large amounts of chemical energy for a cell. Most fa ts reach the cell in the form of fatty acids, and not as triacylglycerols. Triacylglycerols, p h ospholipids, and glycolipids are sometimes referred to as fatty acids. Triacylglycerols (Latin:tri:three), commonly called triglycerides or simply fats and oils, are constructed from a three carbon backbone called glycerol, which is attached to three fatty acids. Their function in a cell is to store energy. They may a lso fu n ction to provide thermal insulation and padding to an organism. Adipocytes (Latin: adips:fat, Greek:kytos:cell), also called fat cells, are specialized cells whose cytoplasm contains almost nothing but triglycerides. Phospholipids are built from a glycel'ol backbone as well, but a polar phosphate group replaces one of the fatty acids. The phosphate group lies on the opposite side of the glycerol from the fa tty acids making the phospholipid polar at the phosphate end and nonpolar a t the fa tty acid end. This condition is called amphipathic (Latin:ambo:both) and makes phospholip ids especially well suited as the major component of membranes.
Glycolipids (Greek:glucus:sweet) are similar to phospholipids, except that glycolipids .have one or more carbohydrates attached to the three-carbon glycerol backbone instead of the phosphate group. Glycolipids are also amphipathic. They are found in abundance in the membranes of myelinated cells composing the human nervous system. Steroids are four ringed structures. They include some hormones, vitarnin D, and cholesterol, an impor tant membrane cOlnponent.
Tnpenes are a sixth class of lipids which include vitamin A, a vitamin important for vision .
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LECTURE 1 : MOLECULAR BIOLOGY; CELLULAR RESPIRATION
3
Another class of lipids (not shown in Figure 1-3 but often listed as a fatty acid) is the 20 carbon eicosal1oids (eikosi:Greektwen ty). Eicosanoids include prostaglandins, thromboxanes, and leukotrienes. Eicosanoids are released from cell membranes as local hormones that regulate, among other things, blood pressure, body temperature, and smooth muscle contraction. (See Paracrine System in Lecture 4 for more on local hormones.) Aspirin is a commonly used inhibitor of the synthesis of prostaglandins.
leI!
~-t'. 1:1
Cit .
tlJ, l1'~-CIJ, Cholesterol Phosphatidylcholine
Phospholipids
ell, H CH
H
I
Galactocerebroside
Steroids
Glycolipids
ct!, H
'
H,
'~j , '''1'\/'''1'\/'\0'"
H
CH. l 1 H
H
H
H
,.
Vitamin A,
Terpenes
Triacylglycerols
Fatty acids
Lipids Figure 1-3 Since lipids are insoluble in aqueous solution, they are transported in the blood via lipoproteins. A lipoprotein contains a lipid core surrounded by phospholipids and apoproteins (apoproteins are discussed below). Thus the lipoprotein is able to dissolve lipids in its hydrophobic core, and then move freely through the aqueous solution due to its h ydrophilic shell. Lipoproteins are classified by their density. The greater the ratio of lipid to protein, the lower the density. The major classes of lipoproteins in humans are chylomicrons, very low density lipoproteins (VLDL), low density lipoproteins (LDL), and high density lipoproteins (HDL). (For more on lipoproteins, see Biology Lecture 6-11 - Fats)
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2007 Exarnkrackers, Inc.
Know these major fu nctions of lipids: phospholipids serve as a structu ral compo nent of membranes; triacylglycerols store metabolic energy, provide thermal insulatioll and padding; steroids regulate metabolic activities; and some fatty acids (eicosa noids) even
serve as local 11Ormones.
4
MeAT
BIOLOGY
1.4
The amino acid stuctures shown here are artificial. Amino acids in solution will always ca rry one or more cha rges. The position and nature of the charges will depend upon the pH of the solution.
Proteins
Proteins are built from a ch ain of anpno acids linked together by peptide bonds (Peptide bonds are disc ussed in Organic Chemistry Lecture 4). Thus proteins are sometimes referred to as polypeptides (Greek:polys:man y). Nearly all proteins in a ll sp ecies a re built from the sam e 20 a-am ino acids. They a re called alpha amino acids because the am.ine is attached ~the carbon in the a lpha position to the carbonyl. In humans, ten of the amino acid s are essential . Tn other words the body cannot manufacture these 10, so they must be ingested directly. Each amino acid in a polypeptide chain is referred to as a resjd.!le; very smail pplypep tid es are sometimes referred to as p eptides. Th e amino acids typically d iffer from each other only in their side chains, often d esignated as the R group. The side chain is also attached to the a -carbon. Digested proteins reach the cells of the human body as single amino acids. 111e 20 amino acids are shown in Figure 1-4.
Nonpolar R Groups
CH,
I
I-I:N- C- COOH H
H,N -
Glycine
C H
COOH
~ /
H,N -
Phenylalanine
H,N -
CI-!
COOH
H,N -
CH
COOH
H,N -
C
CH ,
CH
COOH
Isoleucine
Leucine
S
NH
COOH
Tryptophan
I I CI-!, I
CH,
I
I
CH
I '
H
TH ,
TH,
C - COOH H ""
CH
CH,
Valine
C= CI-!
CH,
CH
CH
Alanine
¥ J-I.N -
I
TH,
CH ,
CH, H
c\,J/ 0 -1,
I-!,N -
C H
(/ COOH
HN-CI-!
Methionine
COOH
Proline
Polar R Groups CH,
OH
J-~N
I C H, I
CI-!
H- C COOH
Serine
SI-!
I
OI-!
I
H,N - C- COOH H
Threonine
Acidic R Groups
CH, H,N -
CH
COOI-!
H,N -
Cysteine
Chl
COOI-!
H,N
Tyrosine
C H
Asparagine
HN ~
I .
T T H
COOH
H
C-
H
NHl
NH
,
I
I I-!,
CI-I
H
CH,
CH,
Glutamic Acid
fiN _ .. C -- COOH
COcm
Aspartic Acid
G lutamine
"C I
,H,
CH,
.
H
Basic R Groups NH.
H,N -
COOH
,
I '
I .
.
H
Lysine
H,N -
C H
CooH
Arginine
C H
COOI-!
Histidine
The 20 Common Amino Acids Figure 1-4 Copyright
2007 EXull1krackers, Inc.
LECTURE
1: MOl FCUI AR
BIOLOGY; CELLULAR RESPIRATION
.
5
The number and sequence of amino acids in a polypeptide is called the primary structure. Once the primary structure is formed, the single chain can twist into an a-helix, or lie along side itself and form a ~-pleated sheet. With ~-pleated sheets, the connecting segments of the two strands of the sheet can lie in the same direction (parallel) or in opposite directions (al/liparallel). Both a -helices and p-pleated sheets are reinforced by h ydrogen bonds behveen the carbonyl oxygen and the h ydrogen on the amino group. A single protein usually contains both structures at various locations along its chain . The a,helix and the p-pleated sheets are the secondary structure and contribute to the conforllIation of the protein. All proteins have a p rim ary structure and most have a secondary structure. Larger proteins (globular, fibrous/structural, etc.: .) can have a tertiary and quaternary structure. The tertiary structure refers to the three dimensional shape formed when the peptide chain curls and folds. Five forces create the tertiary strllcture: 1.
covalent disulfide bonds between two cysteine an1ino acids on djfferent parts of the chain;
2.
electrostatic (ionic) interactions mostly between acidic and basic side cha ins;
3.
hydrogen bonds;
4.
van der Waals forces;
5.
hydrophobic side chains pushed away from water (toward center of protein).
In addi tion to these forces, the alnino acid pro line induces turns in th e polypeptide that will disrupt both ah elix and p-pleated sheet formation. When two or more polypeptide chains bind together, they form the quaternary structure of the protein. The sam e fi ve forces at work in the tertiary structure can also act to form the quaternary structure. (Note: The structure of each polypeptide subunit within the qua ternary structure is not typically identical as shown in Figure 1-5.)
.'
P- Plea ted Sheet
R~ ~~C" .
I
When the conformation is disrupted, the protei n is said to be denatured. A denatured protein ha s lost most of its secondary, tertiary, and quaternar y structure. Some denaturing agents and the forces that they disrupt are given in Table 1-1. Very often, once the denaturing age nt is remo ved , the protein will spontaneous ly refold to its original conformation. This suggests that the amino acid sequence plays a key role in the conformation of a protein.
Secondary Structure
N
,
c r-;." c~
I
(0)
N
IR
-
e
a-Helix
Secondary Structure
Several Polypeptides bonded togethe r
Quaternary Structure Twisted Polypeptide
Tertiary Structure
Protein Structure
. As alluded above, there are two types of Figure 1-5 proteins: "lobular and slruclural. There are more types of globular proteins than types of strllctllra l proteins. Globular proteins function as enzymes (i.e. pepsin), honnones (i.e. insulin), membrane pumps and channels (i.e. Na+/ K+ pump and voltage gated sodium charIDe!s), membrane Copyright (c) 2007 Examkrackers, Inc.
6
MeAT BiOlOGY
Denaturing Agents Forces Disrupted hydrogen bonds urea salt or change in pH electrostatic bonds mercaptoethanol disulfide bonds hydrophobic forces organic solvents all fo rces heat Table 1-1 Proteins are important. Understand the different structures, 1 0 , 2°, 310 , and 4 0
,
and the bonding involved. Know what denaturation means. The rest is just good background knowledge to help you read MeAT passages. You don't have to memorize the structures of each amino
acid, but recognize the basic structure of a generic amino acid. Although nucleic
receptors (i.e. nicotinic receptors on a post-synaptic neuron), intercellular and intracellular transport and storage (i.e. hemoglobin and myoglobin), osmotic regulators (i.e. albumin), in the immune response (i.e. antibodies), and more.
Structural proteins are made from long polYlners. They maintain and add strength to cellular and matrix structure. Collagen, a structural protein made from a unique type of helix, is the most abundant protein in the body. Collagen fibers add great strength to, among others, skin, tendons, ligaments, and bone. Microtubules, which make up eukaryotic flagella and cilia, are made from globular tubulin, which polymerizes under the right conditions to become a structural protein.
Glycoproteins are proteins w ith carbohydrate groups attached. These are a component of cellular plasma melnbranes. Proteoglycans are also a mixture of proteins and carbohydrates, but they generally consist of more than 50% carbohydrates. Proteoglycans are the nlajor component of extracellular matrix as discussed in Biology Lecture 4 - Cellular Matrix.
acids, some lipids, and even some
carbohydrates contain nitrogen, when you see nitrogen on the MeAT, think protein .
Cytochromes (Greekkytos:cell, chroma:color or pigment) are proteins which require a prosthetic (nonproteinaceous) hellle (Greek:haima:blood) group in order to function. Cytochromes get their name from the color that they add to the cell. Examples of cytochromes are hemoglobin and the cytochromes of the electron transport chain in the inner-melnbrane of mitochondria. Proteins containing nonproteinaceous components are called conj1lgated proteins.
1.5
Know the basic structure of a carbohydrate . You should also be familiar with the different polysaccllalides and where they come from . For glucose, remember that animals eat the alpha linkages, but only bacteria break the beta linkages.
Carbohydrates
As implied by the name, carbohydrates (also called sugars or saccharides) are made from carbon and wa ter. They have the empirical formula C(H,O). Five and six carbon carbohydrates (pentoses and hexoses) are the most comlnon in nature. The six carbon carbohydrate called g luco se (Greekglucus:sweet) is the most commonly occurring six carbon carbohydrate. Glucose normally accounts for 80% of the carbohydrates absorbed by humans. Essentially all digested carbohydrates reaching body cells have been converted to glucose by the liver or enterocytes. Glucose exists in aqueous solution in an unequal equilibriuln heavily favoring the ring form over the chain form (see Fig. 1-6). The ring form has two anomer s . In a-glucose the hydroxyl group on the anomeric carbon (carbon number one) and the methoxy group (carbon number six) are on oppqsite sides of the carbon ring. In ~-glucose the hydroxyl group and the methoxy group are on the same side of the carbon ring. The cell can oxidize glucose transferring its chemka l energy to a more readily useable form, ATP. If the cell has sufficient ATP, glucose is polymerized to the polysaccharide, g lycog en or converted to fat. As shown in Figure 1-6, glycogen is a branched glucose polymer with alpha linkages. Glycogen is found in all animal cells, but especially large amounts are found in muscle and liver cells. The liver regulates the blood glucose level, so liver cells are one of the few cell types capable of reforming glucose from glycogen and releasing it back into the blood stream. Only certain epithelial cells in the digestive tract and the proximal tubule of the kidney are capable of absorbing glucose against a concentration gradient. This is done via a secondary active transport lnechanisln down the concentration gradient of sodium. All other cells absorb glucose via facilitated diffusion. Insulin increases the rate of facilitated diffusion for glucose and other monosaccharides. In the absence of insulin, only neural and hepatic cells are capable of absorbing sufficient amounts of glucose via the facilitated transport system. Plants form starch and cellulos e from glucose. Starch comes in two forms: amylose and amylopectin. Amylose is an isomer of cellulose that may be branched or unbranched and has the same alpha linkages as glycogen. Amylol pectin resembles glycogen but has a different branching strucCopyri ght © 2007 Examki-ackers, Inc.
LECTUF~ E 1 : MOLECULf-,\f;: BIOLOGY; CELLULAR RESPIRATION
ture. Cellulose has beta linkages. Animals have the enzymes to digest the alpha linkages of starch and glycogen but not the beta linkages of cellulose. Some animals such as cows and termites have bacteria in their digestive systelTIS that release an enzyme to digest the beta linkages in cellulose.
13: Glucose
g}Glucose
QD,QQQQ Starch
Glycogen
Cellulose
t
0;-(1- 4) linkage
0;-(1 - 4) linkage
~- (1-
4) linkage
Glucose and Glucose Polymers Figure 1-6
Copyright (9 2007 ExamkrJckers, Inc.
. 7
8
MeAT
BIOLOGY
1.6
Nucleotides
Nucleotides are composed of three components:
You shou ld know that a nucleotide consists of three parts: the pentose sugar, the phosphate group , and the nitrogenous base, We'll hear more about nuc!eotides when we discuss genetics in the next lecture.
1.
a five carbon sugar;
2.
a n itrogenous base;
3,
a phosphate group,
The most common nitrogenous bases in nucleotides are adenine, guanine, cytosine, thymine, and uracil, Nucleotides form polymers to create the u ucleic a cids, DNA and RNA , In nucleic acids, nucleotides are joined together by phos p hodiester bonds between the phosphate group of one nucleotide and the 3'" carbon of the pentase of the other nucleotide forming long strands. By convention, a strand of nucleotides is written as a list of their nitrogenous bases. A nucleotide attached to the number 3 carbon (3') of its neighbor, follows that neighbor in the list. In other words, nucleotides are written 5'->3', In typical DNA, two strands are joined by h ydrogen bonds to make the structure called a d ouble helix , Adenine and thymine form two hydrogen bonds, while cytosine and guanine fornl three. By convention, DNA is written so that the top strand runs 5'->3' and the bottom runs 3'->5', In typical RNA there is only one strand and no helix is formed; also uracil replaces thymine, (More will be said about nucleic acids in Biology Lecture 2,)
O·
O·
P--- O-
P-
!
' O-
II
o
Phosphate group
O·
!
I
O-
P=
I
I
0
\
Nitrogenous . base NH,
O
<~-r~I JN
.
C ~ 'H O~ ~~ ;
H
H :
Pentose HO
H
H
OH
\ \1
I
Nucleoside \
v
;
Nucleotide
ATP Figure 1-7 Other important nucleotides include ATP (adenosine triphosphate: Fig, 1-7), the source of readily available energy for the cell; also cyclic AMP, an important component in many second messenger systems; NADH and FADH21 the coenzylnes involved in the Krebs cycle.
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LECTU RE 1: MOLECULAR BIOLOGY; C ELLULAR RESPIRATION .
1.7
M in era ls
Minerals are the dissolved inorganic ions inside and outside the cell. By creating electrochemical gradients across membranes, they assist in the transport of substan ces en tering and exiting the cell. They can combine and solidify to give strength to a matrix, such as hydroxyapatite in bone. Minerals also act as cofa ctors (discussed later in this lecture) assis tin g enzylue or protein function. For instan ce, iron is a m inera I fo und in hellle, the prosthetic group of cytochromes.
This completes a n overview of the ITI8jor chemicals ac ting in the body. Your e mphasis should be on understanding, not memorizing. Th,1t doesn't mean thilt you should breeze through this section . A la rge portion of the biology section on the MeAT is reading comprehension . Since it is impossible to know the exact topic of the passages, YO ll can't just memorize facts and do well. Instead, it is helpful to become fam il i'H with the terms and d ia lect in w hich the passages a re w ritten . Try to see the big picture a nd ho\<\' each little piece fits into the big picture. Be certain thot you thoroughly understand a l\ the terms bolded and brown! Your goal by the time you finish this book should be to a ttain a general overview of how the body functions on each level of organization (Le. molecular, cc lllllal~ ti~suef organ, organ systenl , organism), and to know the specific contribution of ,my bold and brown term or concept.
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9
6. Metabolism of carbohydrate and fat spare protein tissue. All of the following are tnte of fats EXCEPT:
Questions .1 . through 8 .are NOT based on a -descriptive ,passage.
A. ' 8.
C.
1. The most common catabolic reaction in the hum an body is: A.
B.' C. D.
, D.
dehydration. hydrolysis. :condensat,ion. elimination.
A. 8. C. D.
EXCEPT: B. C. D.
deoxyribose sugars polypeptide bonds phosphodiester bonds nitrogenous bases
thymine a double helix an additional hydroxyl group hydrogen bonds
8. Like cellulose, chitin is a polysaccharide that cannot be digested by animals. Chitin differs from cellulose by possessing an acetyl-amino group at the second carbon. What molecule is a reactant in the breaking of the ~- 1,4glycoside linkages of cellulose and chitin?
3. Whi ch of th e following is a carbohydrate polymer that is stored. in plants and digestible by animals?
A. B. C. D.
,
7. Which of the followin g is found in the RNA but not the DNA of a living cell?
2. A molecu le of DNA contains all of the following A.
Fats may be used in cell structure. Fats may be used as hormones. Palli are a more efficient form of en¢~y storage than proteins. Fat') are a less efficient form of energy storage than carbohydrates
A~
starch glycogen cellulose glucose
B,'
C. D.
water oxygen a-lA-glucosidase ~- I ,4-glucosidase .
4. Excess ive amounts of nitrogen are found in the urine of an indi vidual who has experienced a period of extended fasting. This is most likely due to: A.
B. C. D.
glycogenolysis in the liver. the breakdown of body proteins. lipolysis in ad ipose tissue. a tumpr on the posterior" pituitary causing excessive ADH secretion . .
5. prolip'e
i~
not technically an a-amino acid. Due to the ring
sttUcture Of pro'Iine, it cannot conform to the geometry of 'the a -heli x. and creates a bend in the polypeptide chain. This phenomenon assists in the creation of what level of pr(lte iA stru(;t~re?
,
A. ' . primary B. s~c('mdary C. tertiary D. qu aternary
10
STOP.
LECTURE 1 : MOLECULAR BIOLOGY; CELLULAR RESPIRATION .
-1,.8
Enzymes
Virtually a ll biological reactions are governed by enzymes. Although there are a few nucleic acids that act as enzymes, typically enzymes are globular proteins. The function of any enzyme is to act as a catalyst, lowering the energy of activation for a biological reaction and increasing the rate of that reaction. "Enzymes increase reaction rates by magnitudes of as much as thousqnds of trillions. This is a much greater increase th ~ n typical lab catalysts. Such extreme control over reaction rates gives enzymes the ability to pick and choose which reactions w ill or will not occur inside a cell. Enzy mes, like any catalysts, are not consumed nor permanently altered by the reactions which they catalyze. Only a small a mount of catalyst is required for any reaction. Like any catalyst, enzymes do not alter the equilibrium of a reaction. (See Chemistry Lecture 2 for more on equilibrium and catalysts.)
11
,
,
'
The reactan t or reactants upon w hich an enzym e works are called the substrates. Substrates are generally J maller than the enzyme. The position on the enzyme to where the substrate binds, usually with numerous noncovalent bonds, is called the active site. The enzyme bound to the substrate is called the enzyme-substrate complex. Normall y, enzymes are designed to work onl y on a sp ecific substrate or group of closely related substrates. This is called enzyme specificity. The lock and key theory is an example of enzyme specificity. In this' ¢e~ry, the active site of the e nzy me has a specific shape like a lock that only fits a specific substrate, the key. The lock and key model ex plei ns so me but not all enzymes. In a second theory called the indnced fit mod el, the shape of both the enzyme and the substrate are altered upon binding. Besides increasing specificity, the alteration actuall y helps the reaction to proceed. In reactions with more than one substrate, the enzyme may also orient the substrates relati ve to each other, creating optimal cond itions for a reaction to take place.
~ ",: I. . . . . . . . . . . . .: :."'. .~-:
Enzy mes exhibit satnration kinetics; as the relative concentration of substrate increases, the rate of the reaction also increases, but to a lesser and lesse r degree until a m aximum ra te (Vllln.\,) has been K", achieved (Fig. } .. 8). This occurs because as m ore substrate is added, Substrate Concentration individual substrates must begin to wait in line for a free enzyme. Thus, VIIIIlX is proportional to enzyme concentration. TlIrnover number Figure 1 .. 8 EnzymatiC reactions is the number of subs trate molecules one enzyme ac tive site 'c an concompared at different enzyme concentrations vert to product in a given unit of time when an enzyme solution is saturated wHh substrate. Related to V is the Michaelis cons tant (K/IJ. Don't fret over Vm~x and Kill' they're not on Kill is the substrate concentration at which the reaction rate is equal to 1/2Vlllllx . Unlike the MCAT. If you are able to understand V!Hrlx' Kill does not vary when the e~lzynle concentration is changed. Under certain them, however, they provide a deeper conditions, Kill is th erefore a good indi cator-of an en zym e's affinity for-its substrate. understanding of enzyme kinetics, which Temperature and pH also affect en zymatic reactions. At first, as the temperature in .. is on . the MeAT. We'll hear more about creases, the reaction ra te goes up, but at some point, the en zyme dena tures and the Vm;n and Kmlater, but remember, the stuff in italics is not tested on the MCAT rate of the reaction drops off precipitously. For enzymes in the human body, the op" timal temperature is most often around 37" C. Enzymes also function within specific directly. pH ranges. The optimal pH varies depending upon the enzyme. Pepsin, active in the stomach, pre fers a p H below 2, w hile trypsin, active in the small intestine, works best at a pH between 6 and 7. IIIIl X
In order to reach their optimal activity, many enzymes require a non-protein component called a cofactor (Latin: co .. :with or together). Cofactors can be coenzyme~ or me tal ions. Coenzymes are divid ed into two types: cosubstmtes and prosthetic grolips. Both types are organic molecules. Cosllbstrates reversibly bind to a specific enzym e, a nd transfer some chemical group to another substrate. The cosubstrate is Copyri~Jht \~ ?007 Examkr<1ckers, Inc.
12
MeAT BIOLOGY
Just know that some enzymes need cofactors to function, and that cofactors are either minerals or coenzymes. Also remem ber that many coenzymes are vitamins or their derivatives.
then reverted to its original form by another enzymatic reaction. This reversion to original fo rm is w hat dis tinguishes a cosubs trate from normal s ubstra tes. ATP is an example of a cosubs trate type of coenzyme. Prosthetic gro ups, on the other ha nd, remain cova lently bound to the enzyme throu ghout the reaction, and , like the enzy me, e me rge from the reaction unchanged. Coenzymes are often vitamins or vitamin derivatives. (Vitamins are essential [caru10 t be produced by the body] organic molecules.) As mentioned before, hellle is a prosthetic group. Herne binds with catalase in peroxisom es to degrade h ydrogen p erox ide. Metal ions are the second type of cofactor. Me tal ions can act alone or with a pros thetic group. Typical Inetal ions that fun ction as co[aclors in the human bouy are iron, copper, manganese, magnesium, calciuln, and zinc. An enzYlne without its cofactor is called an apoenzyme (Greek:apo-:away from) and is completely nonfunctional. An enzyme with its cofacto r is called a holoenzyme (Greek:holos:whole, entire, complete).
reaction ra te
The important relationships between enzymes and their environment are represented by these three graphs. Memorize these graphs and understand them.
J\ -
-
LIl
reaction ra te /
reaction rate
~
(
--
-
pH
1.9
~
te mpe ra tu re
s ubstrate co n centration
Enzyme Inhibition
Enzylne activity ca n be inhibited. Enzyme inhibitors ca n be classified according to three different mechanisms: irreversible inhibitors, competitive inhibitors, and noncompetitive inhibitors (Fig . 1-9). Agents w hich bind cova len tl y to enzymes and disrupt their function a re irreversible inhibitors. A few irreversible inhibitors bind n oncovalentl y. Irreversible inhibitors tend to be highly toxic. Penicillin is a n irre versibl e inhibitor tha t bind s to a b acte ria l enzym e that assists in the manulacturing of peptidoglycan cell walls. Competitive inhibitors compete w ith the subs trate by binding reversibly w ith noncovalent bonds to the active site. Since, typicall y, they bind directly to the active site for only a fraction of a second, they block the subs trate from binding during that time. Of course, th e reverse is also true; if the substrate binds first, it blocks the inhibitor from binding. Thus, competitive inhibitors raise the a pparent Kill but do not change V nmx ' In other words, in the presen ce of a competitive inhibitor, the rate of the reaction can be increased to the original, uninhibited V by increasing the concentration of the s ubstrate. Overcoming inhibition by increasing substrate concentration is the classic indication of a compe titive inhibitor. Competitive inhibitors often resemble the substrate. Sulfanilamide is an antibiotic which competitively inhibits a bacterial enzyme that manufactures fo li c acid. This leads to the death of bacterial cells, w hich require the en zyme, but d oes not harm humans, wllo require folic acid but don't u sc the SuIDe enzym a tic pathway to manufacture it. Noncompetitive inhibitors b ind noncov alently to an enzy me at a spot other than tl1e active site and change the con forma tion of the en zyme. No nc'o mpetitive inhibitors do not prevent the s ubstrate from binding, and they bind jus t as readily to enzym es that have a subs tra te as to those that don' t. Noncompe titive inhibitors do not resemble the substra te, so commonly act on m ore than one enzym e. Unlike competitive inhibitors, they Caru10t be overcome by excess substrate, and they lower V They do not, however, lo wer the enzyme affinity fo r the s ubstrate, so Kill remains the Saine. nHlI'
IIIIII'
•
Copyright
~.:'>
2007 EXilmkl·ackers, Inc.
LECTURE
Enzyme-Substrate Complex
Competitive Inhibition
1:
MOLECULAR BIOLOGY; CELLULAR RESPiRATION
NoncOlnpetitive Inhibitor Enzyme-Substrate Complex
Noncompetitive Inhibition
•
i
~
r JV~~~i~~;e
u::~ Enzyme
Irreversible Inhibition
Enzyme Inhibition Figure 1-9
1.10 Enzyme Regu lation Enzymes control which reactions take place within a cell, so the cell must regulate enzyme activity. EnzYlnes are regulated by four primary means: 1.
Proteolytic cleavage (irreversible covalent modification)-Many enzymes are released into their environment in an inactive form called a zymogen or proenzyme (Greek:pro:before). When specific peptide bonds on zymogens are cleaved, the zymogens become irreversibly activated. Activation of zymogens may be instigated by other enzymes, or by a change in environment. For instance, pepsinogen (notice the "-ogen" at the end indicating zymogen status) is the zymogen of pepsin and is activated by low pH.
2.
Reversible covalent modification- Some enzymes are activated or deactivated by phosphorylation or the addition of some other modifier such as AMP. The removal of the modifier is almost always accomplished by hydrolysis. Phosphorylation typically occurs in the presence of a protein kinase.
3.
Control proteins-Control proteins are protein subun its that associate with certain enzymes to activate or inhibit their activity. Calmodulin or G-pl'Oteins are typical examples of control proteins.
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.
13
14
MeAT B,OLOGY
4.
Hold onl Don't go overbo ard here. You don't have to memorize the names of each type of inhibition. You j ust have to understand the concepts.-Nor will you be asked to distinguish allosteric inhibition from noncompetitive in hibition . Many good bio texts don't distinguish them, and the MCAT certainly won '\. However, you must understand negative feedback . Negative feedback will be tested on the MCAT. Sometlling else. Other proteins as well as enzymes undergo the types of regulation described above. Hemoglobin is arr example of a protein that is not an enzyme . but exhibits severa! of these characteristics.
Allosteric interactions-Allosteric regulation is the modification of the enzyme con6guration resulting from the binding of an activator or inhibitor at a specific binding site on the enzyme. (Allosteric regulation is discussed below.)
Normally, an enzyme governs just one reaction in a series of reactions. If one of the products downstream in a reaction series comes back and inhibits the enzymatic activity in an early reaction, this phenomenon is called negative feedback or feedback inhibition. Negative feedback prov ides a shut down mechanism for a series of enzYlnatic reactions w hen that series has produced a sufficient amount of product. Most enzymes work w ith in some type of negative feedback cycle. Positive feedback also occurs, where the product returns to activate the enzyme. Positive feedback mechanisms occur less often than negative feedback.
•
• III
(-)
E~ C+D - G
A+B -
Negative feedback
Feedback inhibitors do not resemble the substrates 6f the enzymes that they inhibit. Instead, they bind to the enzyme and cause a conformational change, This is called allosteric regulation (Greekallos:different or othel; stereos:solid). There exist both , allosteric inhibitors and allosteric activators, All allosteric inhibitors and activators are not necessarily noncompetitive inhibitors, because many alter Kill without affecting V"',, . Allosteric enzymes do not exhibit typical kinetics because they normally have several binding sites for different inhibitors, activators, and even substrates. At low substrate concentrations, small increases in substrate concentration increase enzyme efficiency as well as reaction rate. The first substrate changes the shape of the enzyme allowing other substrates to bind m ore easily. This phenomenon is called positive cooperativity. Negative cooperativity occurs as well. It is cooperativity in the presence of the allosteric inhibitor 2,3BPG that gives the oxygen dissociation curve of hemoglobin its siglnoidal shape.
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LECTURE 1 : MOLECULAR B ,OLOGY; CELLUUAR RESPIRATION .
15
1 .11 Enzyme Classification Enzymes are named according to the reactions that they catalyze. Very often, the suffix" -ase" is simply added to the end of the substrate upon which the enzyme acts. For instance, acetylcholinesterase acts upon the ester group in acetylcholine.
Enzymes are classified into six categories: 1.
oxidoreductases
2.
tmnsferases;
3.
hydrolases;
4.
lyases;
5.
isomerases;
6.
ligases.
The only distinction between classifications that might be of interest to an MCAT taker is between Iyases and ligases. A lyase that catalyzes addition of one substrate to a double bond of a second substrate is sometimes called a synthase. ATP synthase is an example of a lyase. A ligase also governs an addition reaction, but requires energy from ATP or some other nucleotide. Ligases are sometimes called synthetases.
Kinases and phosphatases may also come up on the MCAT. A kinase is an enzyme which phosphorylates something, while a phosphatase is an enzyme which dephosphorylates something. Often times a kinase phosphorylates another enzyme in order to activate or deactivate it. Hexokinase is the enzyme which phosphorylates glucose as soon as it enters a cell.
Copyright © 2007 Examkrackers, Inc.
Look for the "·ase" ending. Often a seemingly complicated question asking about a complicated chemical will depend upon the simple fact that tile chemical is an enzyme. and the on ly clue is the" -ase" at the end of the name . Once you recognize that the chemical is an enzyme, you know that it contains nitrogen, and it is subject to denaturation .
You don't have to memorize the six categories of enzymes.
13. One mechanism of enzyme inhibition is to inhibit an enzyme without blocking the active site, but by altering the shape of the enzyme molecule. This mechanism is called:
Questions 9 through 16 are NOT based on a descriptive passage.
A. B.
9. Enzymes are required by all living things because enzymes:
C.
A. raise the free energy of chemical reactions. B. 'properly orient reactants and lower act~vation energy. C. increase the temperature of reacting molecules. D. increase the number of reacting molecules.
D.
14. The continued production of progesterone caused by the release of RCG from the growing embryo is an example of:
A. B. C. D.
10. All of the following must change the rate of an enzymecatalyzed reaction EXCEPT: A. B. C. D.
changing the pH. lowering the temperature. decreasing the concentration of substrate. adding a noncompetitive inhibitor.
11. Since an increase in temperature increases the reaction rate, why isn't the elevation of temperature a method nonnally used to accelerate enzyme-catalyzed reactions?
B. C. D.
B. C.
Raising the temperature causes the reaction to occur too quickly. Raising the temperature does not sufficiently surmount the activation energy barrier. Heat changes the configuration of proteins. Heat does not increase the probability of molecular collision.
D.
A. B. C. D.
I. It often acts by inhibiting enzyme activity. II. )t works to prevent a build up of excess nutrients. III. 1t only acts through enzymes.
C.
D.
increase their function in the small intestine due increased hydrogen ion concentration. decrease their function in the small intestine due increased hydrogen ion concentration. increase their function in the small intestine due decreased hydrogen ion concentration. decrease their function in the small intestine due decreased hydrogen ion concentration.
to to to to
16. The rate of a reaction slows when the reaction is exposed to a competitive inhibitor. Which of the following might overcome the effects of the inhibitor?
12. Which of the following is (are) true concerning feedback inhibition?
A. B.
positive feedback negative feedback feedback inhibition feedback enhancement
15. Peptidases that function in the stomach most likely: A.
A.
competitive inhibition. noncompetitive inhibition. feedback inhibition. positive inhibition.
decreasing enzyme concentration increasing temperature increasing substrate concentration The effects of competitive inhibition cannot be overcome.
I only II only I and II only I, II. and III
Copyright © 2007 Examkrackers. ,Inc
16
STOP.
LECTURE 1 : MOLECULAR B iOlO\iY; CELLULAR RESPIRATION . 17
1.12 Cellular Metabolism Metabolism is all cellular chemical reactions. It consists of an'abolism (Greek:ana:up, ballein:to throw), molecular synthesis, and catabolism (Greek:kata:down), molecular degradation. There are three basic stages of metabolism: 1) lyIacromolecules (polysaccharides, proteins, and lipids) are broken down into their constituent parts (monosaccharides, amino acids, and fatty acids and glycerol) releasing little or no energy; 2) Constit.uent parts are oxidized to acetyl CoA, pyruvate or other metabolites forming some ATP and reduced coenzymes (NADH and FADH,) in a process that does not directly utilize oxygen; 3) If oxygen is available and the cell is capable of using oxygen, these metabolites go into the citric acid cycle and oxidative phosphoFylation to form large amounts of energy (more NADH, FADH" or ATP); otherwise the coenzyme NAD' and other bypro ducts are either recycled or expelled as waste. The second and third stages, the energy acquiring stag;s, are called respiration. If oxygen is used, the respiration is aerobic; if oxygen is not used, the respiration is anaerobic.
1.13 Glycolysis Anaerobic respiration (Latin: an-:not or without, aer:ail~ _ respirare:to breath) is respiration in which oxygen is not required. Glycolysis is the first stage of anaerobic and aerobic respiration. Glycolysis (Figure 1-10) is the series of reactions that breaks a 6-carbon glucose molecule into two 3-carbon molecules of pyruvate (pyruvate is just the conjugate base of pyruvic acid). Other important products from glycolysis are two molecules of ATP each from ADP, inorganic phosphate and water, and two molecules of NADH each from the reduction of NAD' . All living cells and organisms are capable of breaking down glucose to pyru.. -vate; the most common chemical pathway for this is glycolysis. Glycolysis will operate in both the presence and absence of oxygen; it neither requires oxygen, nor is poisoned by it. The reactions of. glycolysis occur in the cytosol (fluid portion) of living cells. The first step of glycolysis (See Figure 1-10) occurs upon the entry of glucose into any human cell. Hexokinase phosphorylates gluc.ose to glucose 6-phosphate with a phosphate group from ATP. (The liver and pancreas use an isozyme [an enzyme with the same function] of hexokinase called glucokinase.) Under normal cellular conditions the phosphorylation of glucose is irreversible, and assists the facilitated diffusion mechanism which transports glucose into the cell. (The liver, which must make glucose from glycogen and export it, possesses a special enzyme, glucose 6phosphatase, which dephosphorylates glucose 6-phosphate to reform glucose. Glucose 6-phosphatase is also found in kidney cells.) Phosphorylated molecules cannot diffuse through the membrane. Although this is the first step in glycolysis, the process does not necessarily continue. Glucose 6-phosphate may be converted to glucose 1-phosphate and then to glycogen. If glucose 6-phosphate follows the glycolytic pathway, it goes to fructose 6-phosphate in the second step of glycolysis. In the third step of glycolysis, a second phosphate group is added at the expense of one more ATP. This step is irreversible and commits the molecule to the glycolytic pathway. Now, the six carbon fructose 1,6-bisphosphate is broken into Glyceraldehyde 3-phosphate (PGAL) and dihydroxyacetone phosphate. Dihydroxyacetone phosphate is subsequently converted to PGAL. Up to this point"no energy has been captured from the breakdown of glucose, but twoATPs have been spent. Next each 3-carbon molecule is phosphorylated while reducing one NAD' to NADH. The resulting 3-carbon molecules each transfer one of their phosphate groups to an ADP to form one ATP each in substrate level phosphorylation. (Substrate level phosphorylation is the formation of ATP from ADP and inorganic phosphate using the energy released from the decay of high' energy phosphorylated compounds as opposed to using the energy from diffusion.) The remaining 3-carbon molecules go through three more steps before donating their phosphate group to ADP to yield
Copyright © 2007 Examk ({)ckers, In c
18
MeAT BiOLOGY
ATP and pyruvate. All together, 2 ATPs are spent and 4 ATPs are produced. The two pyruvate molecules and the two NADH molecules that are left, are still relatively high energy molecules. The products of carbohydrate digestion in the alilnentary tract are approximately 80% glucose, and 20% fructose and galactose. Fructose and galactose are monosaccharides. Much of the fructose and galactose ingested by humans is converted into glucose in the liver enterocytes; however, fructose can enter glycolysis as fructose 6phosphate or glyceraldehyde 3-phospiJate, and galactose can be converted to glucose 6-phosphate to enter glycolysis. Simple table sugar is a disaccharide made from glucose and fructose. Lactose is a d isaccharide found in milk, and is broken do-won into glucose and galactose in the small intestine. 95% of the monosaccharides in the blood are glucose.
1.14 Fermentation Fermentation is anaerobic respiration. It includes the process of glycolysis, the reduction of pyruvate to ethanol or lactic acid, and the oxidation of the NADH back to the NAD+. Yeast and some microorganisms produce e thanol, while hUlnan lTIUSde cells and other m icroorganisms produce lactic acid. Fermentation takes place w h en a cell or organism is either unable to assimilate the energy from NADH and pyruvate, or has no oxygen available to do so. In fe rmentation, the N AD+is restored for use in its role in glycolysis as a coenzYlne, and the lactic acid or ethanol is expelled from the cell along with carbon dioxide as a waste product.
Notice that glycolysis has two stages: a six carbon stage and a three carbon stage. The six carbon stage expends two Arps to phosphorylate the molecule; kind of!ike "priming a pump", The three carbon stage synthesizes two ATP with each three carbon molecule. Also recognize vyruvate and NADH (the names, not the structures), but don't "worry too much about the names of the other chemicals. Just recognize then1 as part of glycolysis. Knmv the products of glycolysis, especially the net production of 2 AT1's. Of course, all these reactions are governed by enzymes. A typical i\lCAT question will ask what happens ""vhen a certain enzyme is inhibited by a poison, The poison V\lill create a build up of reactants and a dramatic reduction of products at the reaction that the enzynlc governs. Understand that fermentation recycles, NADH back to N AD-'-.
Copyr"i ght @ 7007 EX21'Y1i
LECTURE 1 : MOLECULAR BiOlOGY; CELLULAR RESPIRATION
Glucose'
vF-:~H
· ······· ~di:-rbH
~
ATP
H
OH
I.... Phos pha te
ADP--i
CH,O
Glucose 6-phosphate .
1
4
Phosphate
Fructose 6-phosphate· .
I
. .. ' I
H
H
I
HO
I
~
ATP
•
Ol-I'~O CH,OH I I . HO
I
OH
I
]-1
..... Phosp hatc
ADP---i
OH'~ ~ 0
~
? ~T-
HO- C C
I • CHD
•
Fructose l,6-bisphosphate .. . . . ... . .
Dihydroxyacetone phosphate
r" Phosphate
I
. .... H
Glyceraldehyde 3-phosphate
H I
HO I
HO
H
OH
9H?7
... . . ... H- C- C - C- O -----.... Phosphate 1 1 H H
0 - ........ Phos phate
?H?7
1
H
H
l ,3-Bisp hosphoglycerate· .
Phosphate
ADP ~
- o-C- C - C- O -----oolllo Phosph ate 1 1 H
ATP--i 3-Phosphoglycerate ........ . ...... .
-
H
OHO I-I
I
1 '
o-C- C - C- O -----oOOll" Phosphate 1 1
H H
1 2-Phosphoglycerate .. . .. .
ADP
1 ---i
ATP
.-./~
Phosphoenolpyruv ate" .
-
o
H H
I 1 1
o-C- C- C- OH
Phosphate
1
1
- 0
H
? C= C~
o c
1 Phosphate
- 0
~
- ?
1 H
Pyruv ate··· ···· · ·· · · · ··· · ····· o- C-~-T-I-l
o
Glycolysis Fig u re 1-10 Copyright © 2007 Examkrackers, Inc.
H
.
19
20
MeAT BIOLOGY
Intermembrane ATP Space
Cytosol Fatty
acid .----~ .<
Acyl CoA
Glycolysis t -€.ADH
Pyruvate ADH FAD FADH, '--_~~iN A DH
Acetyl CoA
~~
(
NAD+
~DH
/
Pyruvate~ / , ) - - Acetyl CoA
co,~
:::: 17""~rebS U,"\ H'+----+-
~AD+ = Q e FADH b
H'
_
:[e . Cycle 'J-":.'D'
H I-
2
CC\
FAD ~
, 4H++O,
FADH,
~
Succ~
FAD
2H 0
ADP
2
/::;;:;\=~
ATP ADP + iP
7"-
ADH
CO~ ' a-ket0'~,;;~ate
) S~ccinyl CoPl J~.GDP
ADH
ATP
H'
ATP /
ATP synthas~;::
-~
Mitochondrial Matrix
The Krebs Cycle Figure 1-11 Copyright © 2007 Examkrackers, Inc
LECTURE
1:
MOLECUlAR BIOLOGY; CELLUlAR RESPIRArION .
1.15 Aerobic Respiration Aerobic respiration requires oxygen (Fig . 1-11). If oxygen is present in a cell that is capable of aerobic respiration, the products of glycolysis (p yruvate and NADH) will move into the matrix of a mitochondrion. The outer m embrane of a mitochondrion is permeable to small molecules, and both p yruvate and NADH pass via faci litated diffusion through a large membrane protein called porill . The inner mitochondrial membrane, howeve r, is less permeable. Although pyruvate moves into the matrix via fa cilitated diffusion, each N ADH (depending upon the m echanism used for transport) mayor may not require the hyd rolysis of ATP. Once inside the Inatrix, pyruvate is converted to acetyl CoA in a reaction that produces NADH a nd CO2 ,
1.16 Krebs Cycle Acetyl CoA is a coenzy me which transfers two carbons (two carbons from pyruvate) to the 4-carbon oxaloacetic acid to begin the Krebs cycle (also called the citric acid cycle). Each turn of the Krebs cycle produces 1 ATP, 3 NADH, and 1 FADH,. The process of ATP production in the Krebs cycle is called substrate-level phosphorylation. DurIng the cycle, two carbons are lost as CO" and oxaloacetic acid is reproduced to begin the cycle ove r again.
[PfoteiI"iSJ
! Polysaccharides!
j
j
[Amino acidS]
[Simple ~ -Ig;;JiJ -a-r' s glycerol tt'tt'tJ'
I
~
~
PGAl
j
I
~
i~ ~
p:p:p:
Pyruvic acid ~----
~
-.< ,-.< --<
)
. ~
Acetyl CoA
j
( ;:;) cycle
~
Digestion Metabolism of Proteins and Fats , Figure 1-12
Triglycerides can also be catabolized fo r ATP. Fatty acids are converted into acyl CoA along the outer m embrane of the mitochondrion and e ndoplasmic reticulum at the expense of 1 ATP. They are then brought into the m a tri x, and two carbons at a tiIne are d eaved from the acyl eoA to make acetyl eoA. This reaction also produces FADH, and NADH for every two carbons taken from the original fatty acid. Acetyl eoA then enters into the Krebs cycle as usual. The glyce rol backbone is con ver ted to PGAL. Amino acids arc dea minated in the live r. The deaminated product is either chemica lly converted to pyru vic acid or acetyl CoA, or it may enter the Krebs cycle a t various stages depending upon which amino acid was d eamin ated.
Copyright © 2007 Examkrackers, Inc.
21
22
MCAT B,OLOGY
1.17 Electron Transport Chain
You shou ld know that aerob ic respiration produces about 36 net ATPs (includes glycolysis) . You should also know that 1 NADH brings back 2 to 3 ATPs a nd that 1 FADH, brings back ab out 2 ATPs. Know how many NA DHs, FA DH,s, and ATPs are produced in each turn of the Krebs cycle , and that one glucose produces two turns. Don't worry too much about the fatty acids and a mino acids. Just real ize that they can be c,atabolized for energy via th e Krebs cyc le.
The electron transport chain (ETC) is a series of proteins, including cytochromes with heme, in the inner membrane of the mitochondrion. The first protein complex in the series oxidizes NADH by accepting its h igh energy electrons. Electrons are then passed down the protein series and ultimately accepted by oxygen to form water. As electrons are passed along, p rotons are pumped into the intermernbrane space for each NADH. This establishes a proton gradient called th e proton-motive force wh ich propels protons through ATP synthase to manufacture ATP. Production of ATP in this fashion is called oxidative phosphorylation. From 2 to 3 ATPs are manufactured for each NADH. FADH2 w orks in a similar fashion to NADH, except FADH, reduces a pro tein further along in the ETC series, and thus only produces about 2 ATPs.
Notice that the <:nvirol1JTIcnt in the intcrmembrnne space has a low,er pH than the matrix . You don't have to memorize the names of all the chen1icals; ho\-\,cver, you should recognize theI)) as being in the Krebs Cycle. You should pay special attention to acety l CoA and ATP synthase. Also know the difference between oxidative phosphoryla tion and substrate level phosphorylation . You should definitely know the products and reactan ts for respiration : Glucose + 0,
¢
CO, + H,o
(This re
This is a combustion reaction. FinallYf be sure th~1t you remember that the final electron acceptor is oxygen. This isvvhy oxygen is necessary for aerobic respiration.
Copyright
It)
2007 Exarnkrackers, Inc.
22. Glycolysis takes place in the cytoplasm of an animal cell. Which of the following is NOT a product or reactant in glycolysis?
Questions 17 through 24 are NOT based on a descriptive passage.
A. B.
17. As electrons are passed from one protein complex to another, the final electron acceptor of the electrontransport chain is: A. B.
C. D.
ATP H,o NADH
B. Co' D.
0,
A. B. C.
O2
D.
OATP 2ATP 4ATP 8ATP
24. Heart and liver cells can produce more ATP for each molecule of glucose than other cells in the body. This is most likely results of:
cytosol. mitochondrial matrix. inner mitochondrial membrane. intermembrane space.
A. B.
19. As electrons move within the electron transport chain, each intermediate carrier molecule is: A.
D.
23. What is the net ATP production from fermentation?
18. In a human renal cortical cell, the Krebs cycle oq::urs in the: A. B. C. D.
C.
glucose pyruvate ATP
C.
oxidized by the preceding molecules and reduced by the following molecule. reduced by the preceding molecule and oxidized by the following molecule. reduced by both the preceding and the following molecules. oxidized by both the preceding and the following molecules.
D.
a more efficient ATP synthase on the outer mitochondrial membrane. an additional turn of the Kreb's cycle for each glucose molecule. a more efficient mechanism for moving NADH produced in glycolysis into the mitochondrial matrix. production of additional NADH by the citric acid cycle.
20. In aerobic respiration, the energy from the oxidation of NADH: A. B.
C. D.
directly synthesizes ATP. passively diffuses protons from the intermembrane space into the matrix. establishes a proton gradient between the intermembrane space and the mitochondrial matrix. pumps protons through ATP synthase.
21. Which of the following processes occurs under both aerobic and anaerobic conditions? A. B.
C. D.
Fermentation Krebs cycle Glycolysis Oxidative phosphorylation
Copyright © 2007 Exanlkrackers, Inc.
23
STOP.
Genes
2.1
The Gene
A gene is a series of DNA nucleotides that generally codes for the production of a single polyp eptide or mRN A, rRNA, or tRNA. Eukaryotes have more than on e cOPl' of some gen es, w hile prokaryote. have only one copy of each gene. Genes a re often referred to as unique sequence DNA; while regions of non-coding DNA found only in eukaryo tes are called repetitive sequence DNA . Even in e ukaryotes, uniquesequence DNA domin a tes. Eukaryotic genes that are being actively transcribed by a cell are associated with regions of DNA ca lled euchrOlna tin, w hile genes not bejn g ac tively transcribed a re associated with tightly packed regions of DNA called heterocllromatin. Repetitive sequence DNA is found mainly in heterochromatin .
Generally speaking: One gene; one polype ptide . One exception is postranscriptional processing RNA.
There are between 26,000 and 38,000 gen es in the human genome. The entire DNA sequence of an organism is called the genome. Only a little over 1% of a huma n genome actilally codes for protein. Variation of the nucleotide sequence am ong humans is small; human DNA differs from individual to individual at approximately 1 nucleotide out of every 1200 or about 0.08%. The variatio n between humans and chilnpanzees is abo ut 2%.
A small variation in a genome can make a big difference.
The Central Dogma of gene exp ression is that DNA is transcribed to RNA, w hich is transla ted to amino acids fo rming a protein . All living organisms use th is sa me method to express thei r genes.
DNA ¢ RNA ¢ Protein
NH,
2.2
0
H.~ I ~o
Adenine (A)
Thymine (f)
)
t il'NII !
!
N H,
0
N
NH
~NH
60 !
Guanine (G)
Cytosine (C)
Purines
Pyrimidines
Figure 2-1
DNA
DNA (deoxyribonucleic acid) is a polymer of nucleotides. DNA nucleotides differ from each other only in their nitrogenous b ase. Four nitrogenous bases exist in DNA: adenine (A); guani ne (G); cytosine (C); thymine (T) (show n in Figure 2-1 ). Adenine and guanine are two ring stnlctures ca lled purines, while cytosine and thymine are single ring structures ca ll ed pyrimidines. (See' Figure 2-1.) The DNA n ucleotide w ith the base adenine is ca lled adenosine phosphate; however, it is COlnmon to refer to the nucleotides by their base name only. Each nucleotide is bound to the next by a phosphodiester bond between the third carbon of one deoxyribose and the
26
MeAT
BIOLOGY
phosphate backbone of a single strand of DNA w ith a 5 ' .... 3 ' directionality. The 5' and 3' indicate the carbon numbers on the sugar (See Figure 2-2). The end 3' carbon is attached to an -OH group and the end 5' carbon is at tached to a phosphate group. In a living organism, two DNA strands lie side by side in opposite 3· .... 5' directions (antipara Ile l) bound together by hydrogen bonds between nitrogenous bases to form a d o u ble stranded structure. This hydrogen bonding is commonly referred to as base-pairing. The length of a DNA strand is measured in base-pairs (bp) . Under normal circum stances, the hydrogen bonds form only between specific purine-pyrimidine pairs; adenine forms 2 hydrogen bonds with thymine, and guanine forms 3 hydrogen bonds with cytosine. Therefore, in order for two Strands to bind together, their bases must match up in the correct order. Two strands that match in 'such a fashion are called complementary strands. When complementary strands bind together, they curl into a double helix. (See Figure 2-2.) The double helix contains two distinct grooves called the major groove and the min or groove . Each groov e spirals once around the doubJe helix for .every ten base-pairs. The diameter of the double helix is about 2 nanometers or 13 times the
3' end
5'end
o I
-o- p- o I
\
($)CH V
O
,
alphabe tical order
I
<~_
_ ...
I- A C
K HbT:; '1bood" I
3
DNA Double Helix Figure 2-2
Remember that DNA is a polymer of nucleotides and each nucleotide made up of three parts : the phosphate group, the 5-carbon sugar, and the nitrogenous base. Know the names of the purines (adenine and guanine) and the pyrimidines (cytosine and thymine). A good \-vay to remember this is : "pyrirnidine" contains a "y," and so do "cytosine" and IIthymine." Uradl, a nitrogenous base in RNA, is also a pyrimidine. This is easy to remember since thynline, the base it replaces, is a pyri.m idine os well. Know the pairings (AT, Ge) and the nllmber of II-bonds between each pair (two in AT, three in CG; use IlA2T, C3G" to remen1ber, or write them the way I have done here.)
Copyright © 2007 Examkrackers, Inc.
LECTURE
2.3
2:
GENES •
27
Replication
One time in each life cycle, a cell replicates its DNA. DNA replication is semiconservative. This means tha t when a new double strand is created, it contains one stra nd from the original DNA, and one newly synthesized strand. The process of DNA replication (Fig. 2-3) is governed by a group of proteins called a replisome. Replication does not begin a t the end of a chromosome, but toward the middle at a site called the origin of replication. A single eukaryotic chromosome contains multiple origins on each chromosome, while replication in prokaryotes usually takes place for a Single origin on the circular chromosome. From the origin, two rep Ii somes proceed in opposite directions along the chromosome making replication a bidirectional process. The point w here a replisome is attached to th e chromosome is called the replication fork. Each chromosome of eukaryotic DNA is replicated in many discrete segments called replication units or replieons.
Replication proceeds in both directions from an origin. Each direction produces a leading and a lagging strand.
DNA polymerase lD complex
/5'
Helicase . "• .--/1
JI/V/ Y/ V/ W,3'
3'
VJJI/'dJY/YJY/VlJ!J~
5'
3'
Fork movement , 5
~Leading strand Lagging strand
3,--1/' , - - , 5' _
Parental strands
pnmase~
5'
J'
]'
5'
5' 3' , 5'
.
~)
.\ /
J
\
5'
SSE tetrarner
-
~
7
SSE tetra mer
:>
DNA polymerase III complex
tW1o-\ 3' / 'tY.1r
Okazaki fragment
5'
"II' 0,...wfI',W '. .\:.)r(' (I' W(r,\ 5'
'--' \.VJ ...v
f{ \
,W . .
3'
A Prokaryotic Replisome Figure 2-3 As pa rt of the replisom e, DNA heliease unwinds the double helix separating the two strands. DNA polymerase, the enzyme that builds the new DNA strand, cannot initiate a strand from two nucleotid es, but can only add nucleotides to an existing strand. Primase, an RNA polymerase, creates an RNA primer app roximately 10 ribonucleotides long to initiate the strand. DNA polymerase adds deoxynucieotides to the primer and moves along each DNA strand creating a new complementary s trand. DNA polymerase reads the parental strand in the 3' -+ 5' direction, creating the new complementary s trand in the 5' -J 3' direct.ion. (By cunvention, the nucleotide sequence in DNA is written 5' .... 3' as well. This direction is sometimes referred to as downstream and the 3' .... 5' direction as upstream.) Each nucleotide added to the new strand requires the removal of a pyrophosphate group (two p hosphates bonded together) from a deoxyn ucieotide triphosph ate. Some of the energy derived from the hydrolysis of the pyrophosphate is used to drive replication . For instance, the DNA nucleotide containing adenine is m ade by cleaving the second phosphate bond in ,the deoxy-version of ATP. (The ' deoxy-version of ATP' just means tha t the hydroxyl group on the 2' ca rbon has been replaced with a h ydrogen.)
Copyright © 2007 Examkrackers, Inc.
DNA polymerase requires an RNA primer to get started .
Here's a mnemonic to help you remember tile direction of replication:
DNA is complicated to understand, so reading it is like paddling upstream, from 3' to 5 '. However, once it's read, synthesizing DNA is downstream, from 5' to 3 '.
28
MeAT BiOLOGY
Replication has five stepsi 1. Helicase unzips the double helix; 2. RNA Polymerase builds a primer; 3. DNA Polymerase assembles the leading and lagging strands; 4. the primers are removed; 5. Okazaki fragments are joined.
Since DNA polymerase reads in only one direction, one strand of DNA is looped arolU1d the replisOlne giving it the same orientation as the other. The single stran9 in the loop is prevented from folding back onto itself by the SSB tetmmer proteins (also called helix destabilizer proteins). As is shown in Figure 2-3, the polymerization of the new strand is continuously interrupted and restarted with a ne,v primer. This
interrupted strand is called the lagging strand; the continuous new strand is called the leading strand. The lagging strand is made from a series of disconnected strands called Okazaki fragments . Okazaki fragments are about 100 to 200 nucleotides long in eukaryotes and about 1000 to 2000 nucleotides long in prokaryotes. DNA ligase (Latini Iigareito fasten or bind) moves along the lagging strand and ties the Okazaki fragments together to complete the polymer. Since the fonnation of one strand is continuous and the other fragmented, the process of replication is said to be semidiscontin u o us. ' Besides being a polymerase, one of the subunits in DNA polymerase is an exonucle-
ase (it removes nucleotides from the strand). This enzyme autOlnatically proofreads each new strand, and makes repairs when it discovers any n1islnatched nucleotides, such as thymine matched with guanine. DNA replication in eukaryotes is extremely accurate. Only one base in 109_1011 is incorrectly incorporated.
In order to complete the copy of an entire genome, replication must be fast. The DNA polymerase shown in Figure 2-3 moves at over 500 nucleotides per second. DNA polymerase in humans moves much more slowly at around 50 nucleotides per second. However, multiple origins of replication allow the over 6 billion base pairs that make up the 46 human chromosomes to be replicated quite quickly. Replication in a human cell requires about 8 hours. DNA replication is fast and accurate.
The ends of eukaryotic chromosomal DNA possess telomeres. Telomeres are repeated six nucleotide units from 100 to 1,000 units long that protect the chromosomes from being eroded through repeated rounds of replication. Telomerase catalyzes the lengthening of telomeres.
2.4
RNA
Although there are some differences, replication in eukaryotes and prokaryotes is very similar. Unless otherwise specified, the process described above is accurate for both. You should be able to list the differences between DNA and RNA. DNA is made from deoxyribose i RNA is made from ribose. DNA is double stranded; RNA is single stranded. DNA has thyminei RNA has uracil. DNA is produced by repl ication i RNA is produced by transcription. In an imals, DNA is on ly in the nucleus and mitochondrial matrix; RNA is also in the cytosol. There IS one major type of DNA; the re are three major types of RNA.
RNA (ribonucleic acid) is identical to DNA in structure except thalr 1) carbon number 2 on the pentose is not "deoxygenated" (it has a hydroxyl group attached); 2) RNA is single stran ded; and 3) RNA contains the pyrimidine uracil (shown in Figure 2-4) instead of thymine. Unlike DNA, RNA can move through the nuclear pores and is not confined to the nucleus. RNA exists in three forms: mRNA; rRNA; and tRNA mRNA (messenger RNA) delivers the DNA code for amino acids to the cytosol where the proteins are manufactured. rRNA (ribosomal RNA) combines with proteins to form ribosomes, the cellular complexes that direct the synthesis of proteins. rRNA is synthesized in the nucleolus . tRNA (trans fer RNA) collects amino acids in the cytosol, and transfers them to the riboson1es for incorporation into a protein. Notice the Uracil (U) similarity between uracil and thymine. This is a common cause Figure 2-4 of mutations in DNA
Copyright @ 2007 Examkrackers, inc.
LECTURE
2.5
2 : GENES . 29
Transcription
All RNA is manufactured from a DNA template in a process called transcription. Since DNA cannot leave the nucleus or the mitochondrial matrix eukaryotic transcription must take place only in these two places. The beginning of transcription is called initiation. In initiation, a group of proteins called initiation factors finds a promoter on the DNA strand, and assembles a transcription initiation complexl which includes RNA polymerase. Prokaryotes have one type of RNA polymerase, whereas eukaryotes have three: one for each type of RNA. A promoter is a sequence of DNA nucleotides that designates a beginning point for transcription. The promoter in prokaryotes is located at the beginning of the gene (said to be upstream) . The transcription start point is part of the promoter. The first base-pair located at the transcription start point is designated +1; base-pairs located before the start point such as those in the promoter are designated by negative numbers. The most commonly found nucleotide sequence of a promoter recognized by the RNA polymera?e of a given species is called the consensus sequence. Variation from the consensus sequence causes RNA polymerase to bond less tightly and less often to a g iven promoter, which leads to those genes being transcribed less frequently. f
Transcription requ ires a promoter, wherea s replication requires a primer. A promoter is a spot on the DNA that teils RNA polymerase where to begin
transcription . A primer is a short piece of RNA that jump starts replication .
Gene Promoter
~-\>I"fI\H '
L£l
I
Q& IV
f1
,:.,
~ .;s
'--\ " .....})
RNA polymerase
-[I}'\..-
Consensus sequence / Start
RNA polymerase scans the DNA molecule for the promot
/
~~r;,\
~\
.
IVrrV
,£
lfJ!T.
~igna l
(,, )
Q . '- I"~ ~~I~W;W~ I ' I),/b
~-----------------\..1 )".,
RNA polymerase recognizes the promoter.
Transcription bubble
A transcription bubble is formed and elongation begins.
Transcription Figure 2-5
Copyright @ 200 7 [xamkrackers, Inc
30
MeAT B,OLOGY
Only the template strand of the DNA double helix is transcribed . The coding strand resembles the universal code sequence of RNA.
After binding to the promoter, RNA polymerase unzips the DNA double helix creating a transcription bubble. Next the complex switches to elongation mode. In elongation, RNA polymerase transcribes only one strand of the DNA nucleotide sequence into a complementary RNA nucleotide sequence. Only one strand in a molecule of double stranded DNA is transcribed. This strand is called the template strand or (-) antisense strand. The other strand, called the coding strand or (+) sense strand protects its partner against degradation. Like DNA polymerase, RNA polymerase moves along the DNA strand in the 3'->5' direction building the new RNA strand in the 5'->3'. Transcription proceeds ten times more slowly than DNA replication. In addition, RNA polymerase does not contam a proofreading mechanism, and the rate of errors for transcription is higher than for replication. (Errors in RNA are not called mutations.) Since the errors are created in RNA, they are not transmitted to progeny. Most genes are transcribed many times in a cell life cycle, so the problems arising from errors in transcription are not generally harmful. The end of transcription is called terminatio n , and requires a special termination sequence and special proteins to dissociate RNA polymerase from DNA.
Most genetic regulation occurs at transcription when regu latory proteins bind DNA and activate or inhibit its transcription . In other words, the amount of a given type of protein within a cell is likely to be related to how much of its mRNA is transcri bed. One reason for thiS is that mRNA has a short half-life in the cytosol, so soon after its transcription is over, th e mRNA is degraded and its protein is no longer translated. A second reason is that many proteins can be transcribed from a single mRNA, so there is an amplifying effect.
Replication makes no distinction between genes. Instead, genes are activated or deactiva ted at the leve l of transcription. For all ce11s, most regulation of gene expression occurs at the level of transcription via proteins called activators and re pressors. Activators and repressors bind to DNA close to the promoter, and either activate or repress the activity of RNA polymerase. Activators and repressors are often allosterically regulated by small molecules such as cAMP. Promoter I
I
V/v/v/vn " " T.,):" /yV/ V/ V/ V/ V/ V/ V/'..V/ V/ V/V/ V/',,V/\f/ W y Activator binding site
Repressor binding site
Prokaryotic Gene ,
Figure 2-6
The primary function of gene regulation in prokaryotes is to respond to the environmental changes. Changes in gene activity are a response to the concentration of specific nutrients in and around the ce~l. In contrast, lack of change or homeostasis of the intracellular and extracellular compartments is the hallmark of multicellular organisms. The primary function of gene regulation in multicellular organisms is to control the intra- and extracellular environments of the body.
Prokaryotic mRNA typically includes several genes in a single transcript (polycistronic), whereas eukaryotic mRNA includes only one gene per transcript (monocistronic). The genetic unit usually consisting of the operator, promoter, and genes that contribute to a single prokaryotic mRNA is called the operon. A commonly used exmnple of an operon is' the lac operon. The lac operon codes for enzymes that allow E. coli to hnport and metabolize lactose when glucose is not present in sufficient quantit.ies. Low glucose levels lead to high cAMP levels. cAMP binds to and activates a catabolite activator protein (CAP) . The activated CAP protein binds to a CAP site located adjacent and upstream to the promoter on the lac operon. The promoter is now activated allowing the formation of an initiation complex and the subsequent transcription and translation of three proteins. A second regulatory site on the lac operon, called the operator, is located adjacent and downstream to the promoter. The operator provides a binding sHe for a lac repressor protein. The lac repressor protein is inactivated by the presence of lactose in the cell. The lac repressor protein will bind to the operator unless lactose binds to the lac repressor protein. Copyright © 2007 Examkrackers, Inc.
LECTURE 2: GENES . 31
The binding of the lac repressor to the operator in the absence of lactose prevents the transcription of the lac genes. Lactose, then, can induce the transcription of the lac operon only when glucose is not present. The promoter and gene for the Inc repressor is located adjacent and upstream to the CAP binding site. Promoter for lac operon
Gene for repressor protein
CAP binding Promoter \ fo r J gene .
site
\
Gene for
/
~-galactosidase
I
\ P, \
~ 3'
\
transa\~ etYlaSe
y
A
...-------* )' "' Active
5' ~ mRNA
Gene for permease Gene for
~z
,....,J.-..,-:.;--,~
\
I.~
Operator
An operon is a sequence of Bacterial DNA containing an operator, a promqter, and related genes. The genes of an operon are transcribed on one mRNA. Genes outside the operon may code for activators and repressors .
repressor
C
protein---------..
Allolactose (inducer)
_ _ _ _ _", ,... _ _ _ _ _
J/I. ~
~
Inactive _ _ _ _ _repressor ....:.._ _ ___,
J
Coding' region
Regulat(;;y region
\~-----------------------, ,~----------------------------------------I Lac operon
Lac operon Figure 2-7 Gene regulation in eukaryotes is more complicated involving the interaction of ll1any genes. Thus n10re room is required than is available near the prOlnoter. Enhancers are regulatory proteins commonly used by eukaryotes. Their function is siluilar to activators and repressors but they act at a much greater distance from the prOlTIoter.
2.6
Post-transcriptional Processing
Post~transcriptional processing of RNA occurs in both eukaryotic and prokaryotic cells. In prokaryotes, rRNA and tRNA go through posttranscriptional processing, but almost all mRNA is directly translated to protein. In eukaryotes, each type of RN A undergoes posttranscriptional processing and posttranscriptional processing allows for additional gene regulation.
TIle initial mRNA nucleotide sequence arrived at through transcription is called the primary transcript (also called pre-mRNA, or heterogeneous nuclear RNA lhnRNAJ). The primary transcript is processed in three ways: 1) addition of nucleotides; 2) deletion of nucleotides; 3) modification of nitrogenous bases. Even before the eukaryotic mRNA is completely transcribed, its 5' end is capped in a process using GTP. The 5 ' cap serves yS an attachment site in protein synthesis and as a protection against degradation by exonucleases. The 3' end is polyadenylated with a poly A tail, also to protect it from exonucleases.
The primary transcript is much longer than the mRNA that will be translated into a protein. Before lea ving the nucleus, the primary transcript is cleaved into introns and exons. Enzyme-RNA complexes called small nuclear ribonucleoproteins (snRNPs "snUfps") recognize nucleotides sequences at the ends of the introns. Several snRNPs associate with proteins to fonu a complex called a spliceosome. Inside the spliceosome, the introns are looped bringing the exons together. The introns are then excised by the spliceosomes and' the exons are spliced together to Copyright @ 200 7 EXZlm kraci
You don't need to memorize anything about the lac operon . Just understand how it works.
32
MCAT
BIO LOGY
Remember that intrans remain in th e nucleus, and exons exit the nucleu s to
be tran slated . Most of a typica l gene con sists of introns removed by snRNPs in the nucleus.
form the single mRNA strand that ultimately codes for a polypeptide. Introns do not code for protein and are degraded within the nucleus. The exons of some gen~s may be spliced together in different order allowing them to code for different polypeptides. Although there are only 27,000 to 38,000 genes in the human genome, there are about 120,000 proteins made possible by differential splicing of exons. GenerallYf intron sequences are much longer than exon sequences . Intron5 represent about 24% of the genome, while exons represent about 1.1 %. The average number of exons per gene is seven. The sequences of DNA that code for intron5 and exons are also called intron5 and exons. f
G 5'
3' G '~_ _~v,.._ _-,J '~ _ _~\I~_ _."J
Cap
'~_ __
"" I~ _ _-,J
Exons
'~---", r~---') \~---, ,..-
Poly-A tail
Introns Figure 2-8
2.7
To Denature DNA means to separate the two strands of the double helix.
DNA prefers to be double stranded and wi ll look for a compl ementary partner.
DNA Technology
Y\lhen heated or immersed in high concentration salt solution or high pH solution, the hydrogen -bonds cOlmecting the two strands in a double stranded DNA molecule are disrupted, and the strands separate; the DNA molecule is said to be d e n a ture d or melted. The temperature needed to separate DNA strands is called the melting temperatllre (T,,,!. Since guanine and cytosine make 3 hydrogen bonds while thymine and aden ine make only two, DNA with more G-C base pairs has a greater Tm. Heating to 95°C (just below the boiling point of water) is generally sufficient to denahue any DNA sequence. Denatur~d DNA is less viscous, denser, and more able to absorb UV light. Separated strands will spontaneously associate with their original partner or any other complementary nucleotide sequence. Thus, the following double stranded combinations can be formed through nucle ic acid hybridization: DNA-DNA, DNA-RNA, and RNA-RNA. Hybridization teclmiques enable scientists to identify nucleotide sequences by binding a known sequence with an unknown sequence. One method bacteria use to defend themselves from viruses is to cut the viral DNA into fragments with Restriction enzytnes. The bacteria protect their own DNA from these enzymes by methylation (adding a -CH3)' Methylation is usually, but not always, associated with inactivated genes. Restriction enzymes (also called restriction endonucleases) digest (cut) nucleic acid only at certain nucleotide sequences along the chain (see Figure 2-9). Such a sequence is called a restrictioll site or recognition sequence. Typically, a restriction site will be a palindromic sequence four to six nucleotides long. (Palindromic means that it reads the same backwards as forwards). Most restriction endonucleases cleave the DNA strand unevenly, leaving complementary single stranded ends. These ends can reconnect through Copyright © 2007 Examkrackers, Inc.
LECTURE
2: GEN ES
.
33
hybrid iza tion and are termed sticky el1ds. Once paired, the phosphodiester bonds of the fragments can be joined by DNA ligase. There are hundreds of restriction endonucleases know n, each attacking a different res triction site. A given sample of DNA is likely to contain a recognition sequence for an y restriction endonuclease. Two DNA fragmen ts cleaved by the same endonuclease can be joined together regard less of the origin of the DNA. Such DNA is called recombinant DNA; it has been a rtificially recombined.
Endonuclease
3'
5'
Endonuclease Figure 2-9 Recombinant DNA can be made long enough for bacteria to replicate and then placed within the bacteria u sing a vector, typica lly a plasmid or sometimes an infec tive virus. The bacteria can then be grown in large quantity forming a clone of cells con taining the vector with the recombinant DNA fragment. The clon~s can be saved sepa rately producing a clone library. Beca use no t all bac teria take up the vector and not all vectors take up the DNA fragInent, a library may contain some clon es that do not contain vectors or contain vectors that do not contain the recombinant DNA fragment. By including in the original vector a gene for resistance to a certai.n antibiotic and the lacZ gene, which enables the bacteria to metabolize the sligar X-gal, libraries can later be screened for the appropriate clones. When an antibiotic is added to a library of clones, clones wi tho ut resistance will be eliminated. Clones w ithout resis tance mus t not have taken up the vector. In order to screen out clones that contain the original vector and not the DNA fragment, an endonuclease with a recogni tion site that cuts the IneZ gene in two should be used to p lace the DNA fragment into the vector. Since the endon uclease cleaves the IneZ gene, the IneZ gene will not w ork when the DNA fragment is placed in the vector. Clones with an active IneZ gene turn blue in the presence of X-gal. Clones with the cleaved form of the gene do not turn blue. Clones w ith the DNA fragment then will not turn blue when placed on a medium with X-gal. Copyright © 2007 Exarnkrackers, Inc.
To make a DNA library, take your DNA fragment, use a vector to insert it into a bacterium , and reproduce that bacterium like crazy. Now you have a clone of bacteria with your DNA fragment.
34 ' MeAT BIOLOGY
The cloning process isn't perfect, so some bacteria in a library don't have the vector and some vectors don't have the DNA fragment To screen out these undesirable elements, you can include the lael gene and an antibiotic resistant gene when you originally prepare your clone, Also when preparing your clone, use an endonuclease that wi ll insert your DNA fragment into the middle of the lacZ gene and inactive it.
o
fl11lp R
(ampicillin
Iacz gene (lactose ~ breakdown)
resistance gene)
Plasmid (vector) and Human DNA are isolated,
/
Restriction site
Bacterial Plasmi
l
e
(
Human DNA fragment is inserted into plasmid: ~
Human DNA containing gene of interest
/
~
"'-"
, Sticky -- _____ ends ~-\
HumanONA~
4)
Human DNA fragments and plasmids are mixed, Some p lasmids join w ith the gene of interest.)
are digested with the same
restriction enzyme . •
DNA ligase is added,
/ Human DNA fragment containing gene of interest
Recombinan t plasmid
-
- Nonfunctional IneZ gene
ECOliceu {~
.,
Cells are cloned:
Clone (bacterial only)
Bacterial clone carrying many copies of human gene of interest
Plasmid is placed into
laeZ bacteria by transformation.
•
Bacteria is plated onto Inedium with ampicillin and x-gaL Only clones containing the recombinant plasmid are able to grow in the presence of ampicillin, Colonies that also contain the DNA fragment are w hite because their IneZ gene is nonfunctional preventing the break down of x-gaL
. , Identify clone carrying gene of interest.
Figur e 2-10 Copyright (9 2007 Exumkrackers, Inc.
LECTURE
2: GENES
.
35
Part of the screening process involves actually finding the desired DNA sequence from a library. One technique to find a particular gene in a library is hybridization. The radioactively labeled complementary sequence of the desired DNA fragment (called a probe) is used to search the library. The radiolabeled clones are identified by laying them over photographic film which they expose and non-radiolabeled clones do not.
Eukaryotic DNA contains introns. Since bacteria have no mechanism for removing introns, it is useful to clone DNA with no introns. In order to do this, the mRNA produced by the DNA is reverse transcribed using reverse transcriptase. The DNA product is called complementary DNA or cDNA. Adding DNA polymerase to cDNA produces a double strand of the desired DN A fragment.
A much faster method of "cloning" called polymerase chain reaction (peR) has been developed using a specialized polymerase enzyme found in a species of bacterium adapted to life in nearly boiling waters. In FCR, the double strand of DNA to be "cloned" or (speaking more precisely) amplified is placed in a mixture with many copies of two DNA primers, one for each strand. The mixture is heated to 95' C to denature the DNA. When the mixture is cooled to 60'C, the primers hybridize (or anneal) to their complementary ends of the DNA strands. Next, the heat resistant polymerase is added with a supply of nudeotides, and the mixture is heated to 72' C to activate the polymerase. The polymerase amplifies the complementary strands doubling the amount of DNA. The procedure can be repeated many times without adding more polymerase because it is heat resistant. The result is an exponential increase in the amount of DNA. Starting with a single fragment, 20 cycles produces over one million copies (220 ). What used to require days with recombinant DNA techniques, can now be done in hours with FCR. Another advantage of FCR is that minute DNA samples can be amplified. PCR requires that the base sequence flanking the ends of the DNA fragment be known, so that the complementary primers can be chosen.
5·,
peR is a fast way to "clone " DNA.
( 5' 5'~
{JI"
' 3' { (,
.'
( ) f .' • 3'
cDNA is just DNA reverse ttanscribed from mRNA. The great th ing about cDNA is that it lacks th e introlls that would normally be found in eukaryotic DNA.
'\
-",""~
3
3' _
J
f
3·
5'
S\J\J\J\J\J\J'J\ 3' 3 ' ~#
3'
Target DNA is denatured and mixed with many complementary primers.
5' ............
5'
5'~
3'
\J\J\J\J\J\J'\A5· 3 - •.,~ \J\J'i Primers hybridize with DNA fragments
V/\'!'..'5· 3·
Specialized polymerase replicates DNA fragments
peR
3' ~
5' ,
\
Figure 2-11 Folymerase
Target DNA
3'
\
5'
RNA Frimers
Southern blotting is a teclmique used to identify target fragments of known DNA sequence in a large population of DNA. In a southern blot, the DNA to be identified is cleaved into restriction fragments. The fraglnents are resolved (separated) according to size by gel electrophoresis. Large fragments move more slowly through the gel than small fragments. Next, the gel is made alkaline to denature the DNA fragments. A membrane, such as a sheet of nitrocellulose, is used to blot the gel which transfers the resolved single stranded DN A fragments onto the membrane. A radio-labeled probe with a nucleotide sequence complementary to the target fragCopyright © 2007 Examkrackers, Inc
A Southern blot identifies specific sequences of DNA by nucleic acid hybridization , and a Northern blot uses the same techniques to identify specific sequences of RNA.
36
MeAT
B,OLOGY
The recipe for a Southern blot is: 1. Chop , ment is added tQ the membrane. The probe hybridizes with and marks the target up some DNA; 2. Use an electric field to fragmenLThe membrane is exp'osed to radiographic film which reveals the locali on spread out pieces accord ing to size; 3. of the probe and the target fragment. Blot it onto a membrane; 4. Add a radioactive probe made from DNA or A Northern blot is just like a Southern blot, but it identifies RNA fragment?, not RNA' and; 5. visualIZe ' with radiographic DNA fragments. film . . If Western blot sh ows up on the MCAT, it is likely to be described thoroughly in a passage. Just recognize that this is the one that detects a protein with antibodies.
A Western blot can detect a particular protein in a mbyture of proteins. First a mixture of proteins are resolved by size using electrophoresis. Next they are blotted onto a nitrocellulose membrane. An antibody (the primary antibody) specific to the protein in question is then added and binds to that protein. Next, a secondary antibody-enzyme conjugate is added. The secondary antibody recognize,s and binds to the prilnary antibody and marks it with the ,e nzyme for subsequent visualization. The ' reaction catalyzed by the enzyme attached to the secondary antibody can produce a colored, fluorescent or radioactive reaction product which can be visualized or detected with xray film.
Restriction fragment length polymorphisms (RFLP) analysis identifies individuals as opposed to identifying specific genes. The DNA of different individuals possesses different restriction sites and varying di-stances between restriction sites. The population of humans is polymorphic for their restriction sites. After fragmenting the DNA sample with endonucleases, a band pattern unique to an individual is revealed on radiographic film via South~rn blotting techniques. RFLPs (pronounced "rillips") are the DNA fingerprints used to identify criminals in court cases. The genome of one human differs from the genome of another at about one nucleotide every 1000. These differences have been called single nucleotide polymorphisms (SNPs).
in
,,
Copyr-ight}C;l 2007 Exam krackers , Inc.
29. The gene for triose phosphate isomerase 'from maize (a corn plant) spans over 3400 base pairs of DNA and contains eight introns and nine exons, Which of the following would most likely represent the number of nuc1eotides found in the mature mRNA after posttranscriptional processing?
Questions 25 through 32 are NOT based on a descriptive passage.
25. Which of the following is always true concerning the base composition of DNA?
A. B. C. D.
In each single strand, the number of adenine residues equals the number thymine residues. B. In each single strand. the number of adenine residues equals the number of guanine residues. C. In a molecule of double stranded DNA. the ratio of adenine residues to thymine residues equals the ratio of cytosine residues to guanine residues. D. In a molecule of double stranded DNA. the number of adenine residues plus thymine residues equals the -number of cytosine residues plus guanine residues. A.
30. Complementary strands of DNA are held together by: A. B. C. D.
B. C. D.
A.
the primary transcript was cut as it crossed the nuclear membrane. normally mUltiple copies of the mRNA are produced and spliced. introns in the primary. transcript are excised. several expressed regions of the primary transcript have equal numbers of base pairs,
B.
C. D.
Transcription from DNA in the cytoplasm followed by post transcriptional processing on the ribosome Transcription from DNA in the nucleus followed by post transcriptional processing in the nucleus Tnmslation from DNA in the nucleus followed by post-transcriptional processing in the nucleus Translation from DNA in the cytoplasm followed by post-transcriptional processing on the ribosome
32. Tn Southern blotting, DNA fragments are separated based upon size during electrophoresis. Which of the following is true of this process?
27. In PCR amplification, a primer is hybridized to the end of a DNA fragment and acts as the initiation site of replication for a specialized DNA polymerase. The DNA fragment to be amplified is shown below. Assuming that the primer attaches exactly to the end of the fragment, which of the following is most likely the primer? (Note: The N stands for any nucleotide.)
A. B.
C.
5'-ATGNNNNNNNNNNNNNGCT-3' DNA fragment A. B. C. D.
phosphodiester bonds covalent bonds hydrophobic interactions hydrogen bonds
31. Eukaryotic mRNA production occurs in the following sequence:
26. An mRNA molecule being translated at the rough endoplasmic reticulum is typically shorter than the gene from which it was transcribed because: A.
1050 3400 6800 13.600
D.
Y-GCT-3' Y-TAC-3' Y-TCG-3' Y-AGC-3'
Positively charged DNA fragments move toward the cathode. Positively charged DNA fragments move toward the anode, Negatively charged DNA fragments move toward ihe'cathode. Negatively charged DNA fragments move toward the anode.
28. Which of the following is NOT true concerning DNA replication? A. B. C. D.
DNA ligase links the Okazaki fragments Helicase unwinds the DNA double helix. Only the sense strand is replicated. DNA strands are synthesized in the 5' to 3' direction,
Copyright © 2007 Examkrackers, Inc.
37
STOP.
38
MCAT
B IOLOGY
2.8
A C G T
A AA AC AG AT
C CA CC CG CT
G GA GC GG GT
T TA TC TG IT
Table 2-1
Memorize the start codon, AUG , and the stop codons 1JAA, UAG , and UGA. It is necessary to unde rsta nd that a single codon (such as GUC) always codes for only one amino acid, in this case valine ;
but that there are other codons (GUU, GUA, GUG) that also code for valine . The first part of this means that the code is unambiguous , and the second part means that the code is degenerative. Fina ll y, you must understand probabilities. For instance, you must be able to figure out how many possible codons exist As discussed above, four possible nucleotides can be placed in each of 3 positions giving 43 ~ 64, Try til is: A polypeptide contains 100 amino acids. How many possible am ino acid sequences are there for th is polypeptide? Answer: 20 possible amino acids (you should know that) and 100 positions gives 20"10 possible sequences.
The Genetic Code
mRNA nucleotides are strung together to form a genetic code which translates the' DNA nucleotide sequence into an amino acid sequence and ultimately into a protein. There are four different nucleotides in RNA that together must form an unambiguous code for the 20 common amino acids. The number of possible combinations of a row of two nucleotidesr where each nucI~otide might contain anyone \ of the four nitrogenous bases, is 4' = 16, not enough to code for 20 amino acids (see Table 2-1). Therefore, the code must be a combination of any three nucleotides, However, any three nucleotides gives 43 = 64 possible combinations. These are more possibilities than there are amino acids. Thus more than one series of three nucleotides may code for any amino acid; the code is degenerative , But any single series of three nucleotides will code for one and only one amino acid; the code is unambiguous , In addition, the code is almos t universal; nearly every living organism uses the same code. First Position 5' end
t
I
Second position
U C IA IG
Third POSItIon 3' end
t
Phe Phe Leu Leu
Ser Ser Ser Ser
Tyr Tyr STOP STOP
Cys Cys STOP Trp
U C
C
Leu Leu Leu Leu
Pro Pro Pro Pro
His His Gin Gin
Arg Arg Arg Arg
U C
A
He lie He Met
Thr Thr Thr Thr
Asn Asn Lys Lys
Ser Ser Arg Arg
V
U c
--
G
A G
I
A G
C
A G -
Val Val Val Val
I I
Ala , Ala Ala Ala
Asp Asp Glu Glu
Gly Gly Gly Gly
V C A G
Table 2-2 Three consecutive nucleotides on a strand of mRNA represent a codon. All but three possible codons code for amino acids. The relnail1ing codons, VAA, UGA, and VAG, are stop codons (also called t e rmination codons ), Stop codons signal an end to protein synthesis, The start codon, AVG, also acts as a codon for the amino acid methionine.
The genetic code is given in Table 2-2, Be certain that you can read this table, For instance, the codons for lYSine (Lys) are AAA and AAG, By convention, a sequence of RNA nucleotides is written 5' ... 3',
Copyrig ht © 2007 Exall1kra cke rs, Inc.
LECTURE
2.9
2:
GENES •
39
Translation
Translation (Fig. 2-12) is the process of protein synthesis directed by mRNA. Each of the th ree major types of RNA plays a unique role in translation. mRNA is the template which carries the genetic code from the nucleus to the cytosol in the form of codons. tRNA contains a set of nucleotid es that is complementary to the codon, called the anticodon. tRNA sequesters the amino acid that correspond s to its anticodon. rRNA with protein makes up the ribosome, which provides the site for translation to take place. rRNA ac ti vely participates in the translation process. The ribosome is composed of a small subunit and a large subunit mad e from rRNA and many separate proteins. The ribosome and its subunits are Ineasured in terms of sedimentntioll coefficients given i.n Svedberg units (5). The sedimentation coefficient gives the speed of a particle in a centrifuge, and is proportional to mass, and rela ted to shape and density. P rokaryotic ribosom es are s malle r than eukaryotic ribosomes. Prokaryotic ribosomes are made from a 305 and a 50S subunit and have a combined sedimen tary coefficient of 70S. Eukaryotic ribosomes are made of 405 and 605 subunits and have a combined sedimentary coefficient of 80S. The complex structure of ribosomes requires a special organelle called the nucleolus in which to manufacture the m. (Prokaryotes do not possess a nucleolus, but synthesis of prokaryotic ribosomes is similar to that of eukaryotic ribosomes.) Although the ribosome is assembled in the nucleolus, the small anei large subunits are exported separately to the cytoplasm. After posttranscriptional processing in a eukaryote, m RN A leaves the nucleus through the nuclear pores and enters the cytosol. With the help of initiation fnctors (proteins), The 5' end attaches to the small subunit of a ribosome. A tRNA possessing the 5'-CAU-3' a nticodon sequesters the amino acid methionine and settles in at the P site (peptidyl site). This is the signal for the large subunit to join and fo rm the initiation complex. This process is termed initiation.
"' ~
mRNA
tRNA
"
with amino acid
A site tRNA without Polypeptide
amino aci d
NH,
Translation Figure 2-12 Now elongation of the polypeptide begins. A tRNA with its corresponding amino acid attaches to the A site (am inoacyl site) at the expense of two CTPs. The C-/er/IIinus (ca rbox yl end) of methionine attaches to the N-termil1us (amine end) of the amino acid at the A site in a dehydration reaction catal yzed by peptidyl trollsferose, an activity possessed by the ribosome. In an elongation step ca lied translocation, Copyright It) 2007 Exumkri::lckers, Inc
(
Notice that the sedimentation coefficients don't add up: 40 + 60 . 80.
40
MeAT BIOLOGY
the ribosome shifts 3 nucleotides along the mRNA toward the 3' end. The tRNA that carried methionine moves to the E site where it can exit the ribosome. The tRNA carrying the nascent (newly formed) dipeptide moves to the P site, clearing the A site for the next tRNA. Translocation requires the expenditure of another GTP. The elongation process is repea ted until a stop codon reaches the P site. Know the process of translation: initiation : elongation and; termination . Know the role of each type of RNA.
Translation ends when a stop codon is reached-in a step called tennination. When a stop (or nonsel1se) codon reaches the A site, proteins known as release factors bind to the A site allowing a water molecule to add to the end of the polypeptide chain. The polypeptide is freed from the tRNA and ribosome, and the ribosome breaks up into its subunits to be used again for another round of protein synthesis later. Even as the polypeptide is being translated, it begins folding. The amino acid sequence determines the folding conformation and the folding process is assisted by proteins called chaperones. Tn post-translational modifications, sugars, lipids, or phosphate groups may be ad ded to amino acids. The polypep tide may be cleaved in one or more places. Separate polypeptides may join to form the quaternary structure of a protein. Translation may take place on a free floating ribosome in the cytosol producing proteins that function in the cytosol, or a ribosome may attach itself to the rough ER during transla tion ·and inject proteins into the ER lumen. Proteins injected into the ER lumen are destined to become membrane bound proteins of the nuclear envelope, ER, GogH, lysosomes, plasma membrane, or to be secreted from the cell. Free
1 Polypeptide synthesis begins on a free floating ribosome in the cytosol.
2 Signal peptide is recogn ized by the SRP.
3 The SRP ca rries the entire riboso me complex to a receptor protein on the ER.
4 The protein grows across the membrane where it is either released into the lumen or remains p
Signa l. recognitiOl particle (SRP) SRI' receptor protein ""-
CYTOSOL
Signal peptide t t.,,"'" removed ...~
,
"
.•
5 The signa l peptide is usually removed by an enzyme .
6 The protein undergoes post-transla tiona 1 modiiications.
E], /
membrane
S _ _ Protein
ER LUMEN
Polypeptide Synthesis Figure 2-13 Copyright @ 2007 Examkrackers, Inc.
LECTURE
floating ribosomes are identical in stru cture to ribosomes tha t attach to the ER. The growing polypeptide itself may or may not cause the ribosome to a ttadl to the ER depending upon the polypeptide. A 20 amino acid sequence called a signal peptide n ear the fron t of the poly p e ptid e is recognized by protein-RNA signal-recognition particle (SRP) that carries the e ntire ribosome comp lex to a receptor p ro tein on the ER. There the protein grows across the membrane w here it is either released into the lumen or remains p artia lly a ttached to the ER. The signal peptide is usua lly removed by an enzyme. Signal p ep tides may also be attached to polypeptides to target them to mitochondria, the nucleus, or other organelles.
2:
GENES . 41
Translation begins on a free fioatlOg ribosome . A signal peptide at the beginning of the translated polypeptide may direct the ribosome to attach to the ER, in which case the polypeptide is injected into the lumen. Polypeptides injected into the lumen may be secreted from the cell via the Golgi or may remain partially attached to the membrane.
2.10 Mutations Any a lte ration in the genome that is not genetic recombina tion is called a mutation. (The genome is the total complement of DNA.) Mu tations may occur at the chromosom al level, or the nucleotide level. A gene mutation is the alteration in the sequence of DNA nudeotides in a single gene. A chromosomal mutation occurs w h en the structure of a chromosome is ch an ged. In nlulticellular o rganisms; a mutation in a somatic cell (Greek: som a: bod y) is called a somatic mlltation . A som atic muta tion of a single cell ma y have very little effect on an orga nism w ith millions of cens. A mu ta tion in a germ cell, from which all oth er cells arise, can be very serious for the offspring. Only about one out of every million gametes will carry a mutation for a given gene.
Mutations can be spontaneous (occurring due to random errors in the natural p rocess of replica tion and genetic recombination) or induced (occurring due to physical o r ch emical age nts called mutagens). The effects on the cell are the same in either case. A mutagen is any physical or chemical agen t that increases the frequency of mutation above the frequen cy of spontaneous mutations. If a mutation changes a single base-pair of nudeotides in a double strand of DNA, that mutation is called a point mutation. One type of point mutation ca lled base-pair substitution mutation results w hen one base-pair is replaced by another. A base pair substitution involving a complete switch from the ad enine-thymine base-pa ir to the guanine-cytosine base-pair or the opp osite is also called a transition mutation. A base-p air su bstitution tha t involves a reversal of the sa me base-p airs is called a transversion 111utation . A missense mutation is a base-pair mu tation that occurs in the amino acid coding sequence of a gene. A m issense mutation may or m ay not alter the a mino acid sequence of a protein, and an a lteration of a single amino acid mayor ma y not have serious effects on the function of the protein. (Sickle cell anemia, for instan ce, is a disease cau sed by a sing le am ino ac id difference in he moglobin.) If there is no change in prote in function, th e muta tion is caned' a ne/./ tral mutation, and if th e amino acid is not ch anged, it is ca lled a silent mutation. Even a silent mutation may be significant becau se it ma y ch an ge the rate of transcription.
a
A second type of point mutation, an insertion or deletion of a base-pair, may result in a frameshift mutation. A frameshift mutation results w h en the deletions or insertions' occur in multiples other than 3. This is because the genetic code is read in groups of three nucleotides, and the en tire seq uence after the muta tion will be shifted so that the three base sequences w ill be grou ped incorrectl y. For instance, if a single T nucleotide were inser ted into the series: AAA IGGG Ieee I AAA, so that it reads AAT I AGG IGee IeAA I A, each 3-nucleotid e sequence dow nstream from the mutation would be changed because the entire series w ould be sh ifted one to the righ t. On the other h and, if th ree T nucleotides were inserted randomly, the d ownstream sequ ence w ould not be shifted, and onl y one or a few 3-nucleotide sequences would be ch an ged. AAT I T AG IGTG Ieee IAAA This is a nonframeshift mutation. Frameshift mutations often result in a comple tely nonfunctional protein, w hereas nonframeshift mutations may still result in a partially or even comple tely ac tive protein.
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Mutations are rare. Mutations in somatic cells are not passed to offspring; mutations in germ cells are.
42
MeAT BIOLOGY
If a base-pair substitution or an insertion or deletion mutation creates a stop codon, a nonsense mutation results. Nonsense mutations are usually very serious for the cell because they prevent the translation of a functional protein entirely. . Structural changes may occur to a chromosome in the form of deletions, duplications, translocations, and inversions. Chromosomal deletions occur when a portion of the ChrOlTIOSOn1€ breaks off, or when aportion of the chromosome is lost during homologous recombination and/or crossing over events. Duplications occur when a DNA fragment breaks free of one chromosome and incorporates into a holTIologous chromosOlne. Deletion or duplication can occur with entire chromosomes (aneuploidy) or even entire sets of chromosomes (polyploidy). Down syndrome is the result of aneuploidy where there are three copies of chromosome 21.
When a segment of DNA from one chromosome is inserted into another chromoSOlne, the resulhng mutation is called a translocation. In inversion the orientation of a section of DNA is reversed on a chromosome. Translocation and inversion can be caused by transposition. Transposition takes place in both prokaryotic and eukaryotic cells. The DNA segments called transposable elements or transposons can excise themselves from a chromosome and reinsert themselves at another location. Transposons can contain one gene, several genes, or just a control element. A transposon within a chromosome will be flanked by identical nucleotide sequences. A portion of the flanking sequence is part of the transposon. When moving, the transposon may excise itself from the chromosome and move; it may copy itself and move, or copy itself and stay, moving the copy. Transposition is one mechanism by which a somatic cell of a multicellular organism can alter its genetic makeup without meiosis.
Flanking sequences -~~-
.........
~
DNA
Trans
-
............
-
-~~~
080n
DNA
DNA
Transposon Figure 2-14 Copyright © 2007 Examkrackers, Inc.
LECTURE
A mutation can be a forward mutation or a backward mutation. These terms refer to an already mutated organism that is mutated again. The mutation can be forward, tending to change the organism even more from its original state, or backward, tending to revert the organism back to its original state. The original stat~ is called the wild type. For example, you may be working in the lab with bacteria that normally produce histidine, and you mutate a sample so that the bacteria in that sample no longer produce histidine. Now you have the wild type, hi5+, and the mutants, his-. If you back mutated the his-, you would produce the wild type, his+. If you forward mutated the his-, they may lose the ability to produce some other aInino acid in addition to histidine. Here's a diagram that I use to keep my mutations straight:
Mutations Nucleotide . / Replacem7
/
Base Pair Mutations No Stop /
COdOY
Missense
/
Stop
Nucleotide
Insertions and Deletions /
~OdO/
Nonsense
"
~dition or Remov~l
Non-MUltipley /MUltiples of 3 /
Frameshift
"{ 3
NonFrameshift (Base pair mutations are nonframeshift mutations as well)
2.11 Cancer Cancer is the unrestrained and uncontrolled growth of cells. Normal cells divide 20 to 50 times before they stop dividing and die, but cancer cells continue to grow and divide indefinitely. A mass of cancer cells is called a tumor. A tumor is benign if it is localized in a small lump. When an individual has a tumor invasive enough to impair function of an organ, the tumor is said to be malignant and the individual is said to have cancer. Cancer cells may separate from the tumor and enter the bodies circulatory systems and establish tumors in other parts of the body. This process is called metastasis. Certain genes that stimulate normal growth in hUlnan cells are called proto-oncogenes . Proto-oncogenes can be converted to oncogenes, genes that cause cancer, by mutagens su ch as UV radiation, chemicals, or sin1ply by randon1 mutations. Mutagens that can cause cancer are called carcinogens.
Copyrig ht (c) 2007 EX(-Jrnkrackcrs, Inc.
2: GENES
•
43
Questions 33 through 40 are NOT based on a descriptive passage.
38. One difference between prokaryotic and eukaryotic translation is: A.
33. If each of the following mRNA nucleotide sequences contains three codons, which one contains a start codon? A. B. C. D.
B.
3'-AGGCCGUAG-5' 3'-GUACCGAAC-Y 5'-AAUGCGGAC-3' 5'-UAGGAUCCC-3'
C. D.
34. Translation in a eukaryotic cell is associated with each of the following organelles or locations EXCEPT: A. B. C. D.
the the the the
39. During translation the growing polypeptide can be found attached to a tRNA at which site on the ribosome?
mitochondrial matrix. cytosol. nucleus. rough endoplasmic reticulum.
A. B. C. D.
35. Which of the following is true concerning the genetic code? A. B. C. D.
the E site the P site theA site the Z site
40. During translation, a signal peptide is synthesized and attaches to an SRP complex in order to:
There are more amino acids than codons. Any change in the nucleotide sequence of a codon must result in a new amino, acid. The genetic code varies from species to species. There are 64 codons.
A. B. C. D.
36. The large subunit of an 80S ribosome is made from: A. B. C. D.
eukaryotic ribosomes are larger containing more subunits. prokaryotic translation may occur simultaneously with transcription while eukaryotic translation cannot. prokaryotes don't contain supra molecular complexes such as ribosomes. prokaryotic DNA is circular so does not require a termination sequence.
inactivate the new protein. activate the new protein. prevent the ribosome from attaching to the endoplasmic reticulum. direct the ribosome to attach to the endoplasmic reticulum.
rRNA only. protein only. rRNA and protein only. rRNA and protein bound by a phospholipid bilayer.
37. A tRNA molecule attaches to histidine. The anticodon on the tRNA is 5'-AUG-3 '. Which of the following nucleotide sequences in an mRNA molecule might contain the codon for histidine? A. B. C. D.
3'-GCUAGGCCU-5' 3'-GGTACCTAC-5' 5'-CATTCTTAC-3' 5'-UCAUGGAUC-3'
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44
STOP.
LECTURE
2 : GENES
.
45
2.12 Chromosomes If a double strand of all the DNA in a single human cell were stretched out straight, . it would measure around 5 ft. Since the nucleus is much smaller than this~ the sections of DNA that are not in use are wrapped tightly around globular proteins called, histones. Eight his tones wrapped in DNA form a nudeosome .. Nucleosomes, in turn, wrap into coils called solenoids, which wrap into supereoils. The entire DN A/protein complex (includ.i ng a very small amount of RNA) is called chromatin (Greek: chroma: color) (See Figure 2-15). By mass, chfClmatin is about one third DNA, two thirds protein, and a small amount of RNA. Chromatin received its name because it absorbs basic dyes due to the large basic amino acid content in histones. The basicity of his tones gives them a net positive charge at the normal pH of the cell.
In animals. DNA is found on ly in the nucieus and the mitochondria .
Coiling within Chromoson1es
Supercoil within chromosome
Chromatin
Chromatin fiber
Chromatin Figure 2-15
Chromatin condensed in the manner described above is called heterochromatin (Greek: heteros: other). Some chromatin, called constitutive heterochromatin, is permanently coiled. When transcribed, chromatin must be uncoiled. Chromatin that can be uncoiled and transcribed is called euchromatin (Greek: eu: well or properly). Euchromatin is only coiled during nuclear division. Copyrig ht © 2007 Examkrackors, Inc.
46
MeAT BIOLOGY
The tricky thing about chromosomes is "How many are there?" In the nucleus of human cells, there are 46 chromosomes before replication, and 46 chromosomes after replication. The duplicates can be referred to separately as sister chromatids. Diploid means that the cell has homologous pairs. It's that simple . Diploid - homologues
Inside the nucleus of a human somatic cell, there are 46 double stranded DNA molecules. The chromatin associated with each one of these molecules is called a chromosome (Greek: chroma: color, soma: body). Each chromosome contains hund reds or thousands of genes. In human cells, each chromosome possesses a partner that codes for the same traits as itself. Two such chromosomes are called homologues (G reek: homologein: to agree with, homo: same, logia: collection). Although the traits are the sa me, the actual genes may be different. For instance, the trait may be eye color, but the eye color gene on one chromosome may code for blue eyes while the other codes for brown eyes. Any cell that contains homologous pairs is sa id to be diploid (Greek: di-: twice) . Any cell that does not contain homologues is sa id to be haploid (Greek: haploos: single or simple).
2.13 Cell Life Cycle Every cell has a life cycle (Fig. 2-16) that begins with the birth of the cell and ends with the dea th or division of the cell. The li fe cycle of a somatic cell of a multicellular organism can be divided into four stages: the first grow th phase (G,); synthesis (5); the second growth phase (G,); mitosis or meiosis (M); and cytokinesis (C). G" 5, and G, collecti vely are called interphase. G, checkpoint
M 'Checkpoint
G, checkpoint
Cell Life Cycle Figure 2-16 In G" the cell has just split. and begins to grow in size producing new organelles and proteins. Regions of heterochromatin have been unwound and decondensed into euch romatin. RNA synthesis and protein synthesis are very active. The cell must reach a certain size, and synthesize sufficient protein in order to continue to the next stage. Cell grow th is assessed at the G, checkpoint near the end of G,. If conditions are favorable for division, the cell enters the 5 phase, otherwise the cell enters the Go phase. The main factor in triggering the begirming of 5 is cell size based upon the ratio of cytoplasm to DNA. G, is normally, but not always, the longest stage. In G o' is a nongro wing s tate d istinct from interphase. The Go phase allow s for the
differences in length of the cell cyd e. In humans, enterocytes of the intestine divide more than twice per da y, while liver cells sp end a grea t deal of time in Go di viding less"than once per yea r. Ma ture neurons and muscle cells remain in Go permanently. In S, the cell devotes most of its energy to replicating DNA. Organelles and proteins
are produced more slowly. In this stage, each chromosome is exactly duplicated, but, by convention, the cell is still considered to have the same number of chromosomes, only now, each chromosome is made of two identical sister chromatids. Copyright @ 2007 Examkrackers, Inc.
LECTURE
2: GENES . 47
In G" the cell prepares to divide. Cellular organelles continue to duplicate. RNA and protein (especially tubulin for microtubules) are actively synthesized. G, typically occupies 10-20% of the cell life cycle. Near the end of G2 is the G, checkpoint. The G, checkpoint checks for mitosis promoting factor (MPF). When the level ofMPF is high enough, mitosis is triggered. There is an M checkpoint during mitosis that triggers the start of G, .
2.14 Mitosis Mitosis (Greek: mitos: cell) is nuclear division without genetic change. Mitosis has four stages: prophase; metaphase; anaphase and; telophase (Fig. 2-17). These stages in turn are also divided, but this is beyond the MCAT. Mitosis varies among eukaryotes. (For instance, fungi don't have centrioles and never lose their nuclear Inelnbranes.) The following stages describe mitosis in a typical animal cell. Prophase is characterized by the condensation of chrOluatin into chromosomes. Centrioles move to opposite ends of the cell. First the nucleolus and then the nucleus disappear. The spindle apparatus begins to form consisting of aster (microtubules radiating from the centrioles), kinetochore microtubules growing from the centromeres (a group of proteins located toward the center of the chromosome), and spindle microtubnles connecting the two centrioles. (The kinetochore is a structure of protein and DNA located at the centromere of the jOined chromtids of each chromosome.)
In metaphase (Greek: meta: between) chromosomes align along the equator of the cell. Anaphase begins when sister chromatids split at their attaching centromeres, and move toward opposite ends of the cell. This split is termed disjunction. Cytokinesis, the actual separation of the ceJJuJar cytoplasm due to constriction of microfilmnents about the center of the cell, may commence toward the end of this phase.
Know the basic conditions that define each phase of mitosIs. It is especially important to realize that mitosis results in genetically identical daughter cells.
Remember mitosis by the acronym : In telophase (Greek: teleios: complete) the nuclear membrane reforms followed by the reformation of the nucleolus. ChrOlnosomes decondense and become difficult to see under the light microscope. Cytokinesis continues.
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(<;)
2007 Examkrackcrs, In c.
PMAT.
48 . MeAT
BIOLQ(JY
-Interphase
Metaphase
Mitosis Figure 2-17 Anaphase
Interphase
Interphase
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2007 Examkrackers, Inc.
LECTURE
2: GENES
. 49
2.15 Meiosis Meiosis (Fig. 2-18) is a double nuclear division which produces four haploid gametes (also called germ cells). In humans, only the spermatogonium and the oogonium undergo meiosis. All other cells are somatic cells and undergo mitosis only. After replication occurs in the S phase of interphase, the cell is called a primary spermatocyte or primary oocyte. In the human female, replication takes place before birth, and the life cycle of all germ cells are arrested at the primary oocyte stage until puberty. Just before ovulation, a primary oocyte undergoes the first meiotic division to become a secondary oocyte. The secondary oocyte is released upon ovulation, and the penetration of the secondary oocyte by the sperm stimulates anaphase II of the second meiotic division in the oocyte. Meiosis is tvvo rounds of division called meiosis I and meiosis II. Meiosis I proceeds similar!y to mitosis with the following differences: In prophase I homologous chromosomes line up along side each other, matching there genes exactly. At this time, they mayor exchange sequences of DNA nuc1eotides in a process called crossing over. Genetic recombination in eukaryotes occurs during crossing over. Since each duplicated chromosome in prophase I appear as an 'x', the side by side homologues exhibit a total of four chromatids, and are called tetrads (Greek: tetras: four). If crossing over does occur, the two chromosomes are "zipped" along each other where nucleotides are exchanged, and form what is called the synaptonemal complex. Under the light microscope, a synaptonemal complex appears as ' a single point where the two chromosomes are attached creating an 'x' shape called a chiasma (Greek: chiasmata: cross). Genes located close together on a chromosome are more likely to cross over together, and are said to be linked.
-®> ------------------------------@ .
46 ~ ~ ---------Chromosomes Spermatogonium 46
~~
-
t ______________________________ @t
__________ @
Chromosomes
Primary Spermatocyte
23
OogonIum
Primary OOfyte
Meiosis I
~ _____ ' - - ______ ~_ (Re~_~c_t~on_D~~~si~_~)__ \t@ ,
Chromosomes
I!
Secondary
! ....
Chromosom es
~ __
Meiosis II
@ ____ ® __ @ ____ ®
46
sprmatts
Chromosomes
!
? ? ?? Spermatogenesis
Meiosis Figure 2-18
\ \ ifI ifI Second
II)
Polar Bodies
Zygote
Oogenesis
Spermatozoa
Copyright © 2007 Exarnkrackers, Inc.
____ @
First \ ,\{olar Body
Secondary Oocyte
SP\ermatoc/y.tes\
23
~
Meiosis is like mitosis except that In meiosis there are two rounds, the daughter cells are haplOid, and genetic recombination occurs. You must know the names of the cells at tile different stages and whether or not those cells are haploid or diploid. Recognize that, under the light microscope , metaphase in mitosis would appear like metaphase II in meiosis and not like metaphase !.
50
MeAT
BIOLOGY
In metaphase I the homologues remain attached, and move to the metaphase plate. Rather than single chromosomes aligned a long the plate as in mitosis, tetrads align in meiosis.
Anaphase I sepa rates the h omologues from their partners. In telophase I, a nuclear membrane mayor may not reform, and cytokinesis may or may not occur. In humans the nuclear membrane doe~ reform and cytokinesis does occur. If cytokinesis occurs, the new cells are h aploid with 23 replicated chromosomes, and are ca lied secondary spermatocytes or secondary oocytes. In the case of the fema le, one of the oocytes, called the first polar body, is much smaller, and degenerates. This occurs in order to conserve cytoplasln, which is contributed only by the ovum. The first polar body mayor may not go through meiosis II producing two polar bodies. These four phases together are called meiosis I . Meiosis I is reduction division.
Meiosis II proceed s w ith prophase II, metaphase II, anaphase II, and telophase.II. appearing under the light microscope much like normal mitosis. Th e final products a re haplOid g!,metes each with 23 ch romosomes. Tn the case of the sperma tocyte, fo ur sperm cells are formed. In the case of the oocyte, a single ovum is formed. Tn the female, telophase II produces on e gamete and a second polar body. If during anaphase T or II the centromere of any chromosome does not split, this is called nondisjunction. As a result of primary nondisjunction (nondisjunction in anaphase J), one of the cells will have two extra chromatids (a comple te extra duomosome) and the other will be missing a chromosome. The extra chromosome will typically line up along the metaphase plate and behave normally in meiosis II. Nondisjunction in anaphase II will result in one cell having one extra chromatid and one cell lacking one chromatid. Nondisjunction can a lso occur in mitosis but the ramifkatio1l5 are less severe since the genetic information in the new cells is not passed on to every cell in the body.
!
Copyright ((.') 2007 Examkrackers, Inc.
Questions 41 through 48 are NOT based on a descriptive passage.
46. Which of the following characterizes mitotic prophase? A.
.
B. C. D.
41. How many chromosomes does a human primary spermatocyte contain? A. B. C. D.
23 46 92 184
47. All of the following might describe events occurring in prophase I of meiosis EXCEPT: A. B. C. D.
42. In which of the following life cycle phases does translation, transcription, and replication take place? A. B. C. D.
GI S G2 M
tetrad formation spindle apparatus formation chromosomal migration genetic recombina,tion
48. When a human female is born. the development of her oocytes is arrested in: (\. B~
43. A scientist monitors the nucleotide sequence of the third chromosome as a cell undergoes normal meiosis. What is the earliest point in meiosis at which the scientist can deduce with certainty the nucleotide sequence of the third chromosome of each gamete?
A. B. C. D.
chromosomal alignment along the equator of the cell separation of sister chrolh_atids centriole migration to the cell poles cytokinesis
C. D.
prophase of mitosis prophase I of meiosis prophase II of meiosis interphase
prophase I metaphase I prophase II telophase II
44. Which of the following is a process undergone by germ cells only? A. B. C. D.
meiosis mitosis interphase cytokinesis
45. Which of the following represents a germ cell in metaphase l?
A.
c.
B.
D.
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51
STOP.
Microbiology
3. 1
Viruses
Viruses are tiny infectious agents, much smaller than bacteria. They are comparable in size to large proteins (Some viruses are larger and some are much smaller.) In its most basic form, a virus consists of a protein coat, called a capsid, and from one to several hundred genes in the form of DNA or RNA iflside the capsid. No virus contains both DNA and RNA. Most animal viruses, some plant viruses, and very few bacterial viruses surround themselves with a lipid-rich envelope either borrowed from the membrane of their host cell or syntheSized in the host cell cytoplasm. The envelope typically contains some virus-specific proteins. A Inature virus outside the host cell is called a virion. All organisms experience viral infections .
Although there is debate as to the vitality of viruses, viruses are not currently classified as living organisms; they do not belong to any of the taxonomical kingdoms of organisms. Viruses differ from living organisms in the following ways. Although viruses can reproduce through a process involving the transfer of genetic information, th ey always require the host cell's reproductive machinery in order to do so. Viruses do not metabolize organic nutrients. Instead they use the ATP made available by the host cell. Unlike Jiving organisms, viruses in their active form are not separated from their external environment by some type of barrier such as a cell wall or membrane. All living organisms possess both DNA and RNA; viruses possess eith er DNA or RNA, but never both. Viruses can be crystallized without losing their ability to infect.
Attachment
Tai1 contraction
Figure 3-2
Tail Fiber
Bacteriophage
Penetration
and Injection
Adsorption and Injection
./
"
A v iral infection begins when a v irus adsorbs to a specific 'chem ica l receptor site on the host. The host is the cell that is being infected. The chemical receptor is usually a specific glycoprotein on the host cell membr8nf'. The virus cannot infect the cell if
Landing
Capsid or H ead (Contains N ucleic Acid)
Figure 3-1
54 . MeAT
BIOLOGY
the specific receptor is not available. Next, the nucleic acid of the virus penetrates into the cell . In a bacteriophage (Greek: phagein: to eat), a virus that in fects bacteria (Fig. 3-1), the nucleic acid is normally injected thro ugh the tail after viral enzymes have digested a hole in the cell wall (Fig. 3-2). (Notice that this indicates tha t some v iruses a lso include enzymes within their capsids.) Most viruses th at infect eukaryotes are engulfed by an endocytotic process. Once ins ide the cell, there are two possible paths: a lysogenic infection, or a lytic infection (Fig. 3-3). In a lytic (Greek: lysis: separation) infection, the virus commandeers the cell's reproductive machinery and begins reproducing new viruses. There is a brief period before the first fu lly formed virion appears. This period is called the eclipse period. The cell may fill w ith new viruses until it lyses or bursts, or it may release the new viruses one at a time in a reverse endocytotic process. The period from infection to lysis is called the latent period. The latent period encompasses the eclipse period. A virus following a lytic cycle is called a virulent virus.
Uninfected Cell Virions O
) {'®~+-
AdSO~O
LYTlC CYCLE
Virus to Cell Wan
Viral Nucleic Injected into
AC~ C ce:1~
0
o
Active Virus
)
Reduction ( to Provirus
Viral DNA Integrated into Chromosome
Assembly of New Viruses
Induction of Provirus to Active Virus
Il :SOGENI~ Il U U
)
CYCLE
\. ffJ .1 Reproduction of Lysogenic Bacteria
Viral Life Cycles Figure 3-3 Cupyright
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LECTURE
3: M,CROBIOLOGY
Tn a lysogenic infectioll the viral DNA is incorporated into the host genome, or, if the virus is an RNA virus and it possesses the enzyme reverse transcriptase, DNA is actually reverse-transcribed from RNA and then incorporated into the host cell genome. When the host cell replicates its DNA, the viral DNA is replicated as well. A virus in a lysogenic cycle is called a temperate virus. A host cell infected with a telnperate virus may show no symptoms of infection. VVhile the v iral DNArelnains incorporated in the host DNA, the virus is said to be dormant or latent,t. and is called a provirus (a prophage [Greek: pro: before, phagein: eatl if the host cell is a bacterium). The dormant virus may become active w hen the host cell is under some type of stress. Ultraviolet light or carcinogens also Inay activate the v irus. When the virus becomes active, it becomes v irulent. f
There arc many of types of viruses. One way to classify them is b y the type of nucleic acid that they possess. A virus with unenveloped plus-strand RNA is responsible for the common cold. (Therefore, not all animal viruses are enveloped.) The "plus-strand" indicates that proteins can be directly translated from the RNA . Enveloped plus-strand RNA viruses include retroviruses such as the virus that causes AIDS (Fig. 3-4). A retrovirus carries the enzyme reverse transcriptase in order to create DNA from its RNA. The DNA is then inReverse corporated into the genome of the host cell. Transcri ptase Minus-strand RNA viruses include measles, rabies, and the flu. Minus-strand RNA is the complement to mRNA and must b e transcribed to p lus-RNA before RNA being translated. There are even double stranded RNA viruses, and single and double stranded DNA viruses.
Viroids are a related form of infectious agent. Viroids are small rings of naked RNA without capsids. Viroids only infect plants. There also exists naked proteins called prions that cause infections in animals. Prions are capable of reproducing themselves, apparently w ithout DNA or RNA.
HIV Figure 3-4
3.2
Defense Against Vira l Infection
The human body fights viral infections with antibodies, w hich bind to a' viral protein, and with cytotoxic T cells, which destroy infected cells. Although the envelope is borrowed from the host cell, spike proteins encoded from the viral nucleic acids protrude from the envelope. These proteins bind to receptors on a new host cell causing the virus to be infectious. However, it is also the spike proteins that human antibodies recognize when fighting the infection. Since RNA polymerase does not contain a proofreading mechanism, changes in the spike proteins are COlumon in RNA viruses. When the spike proteins change, the antibodies fail to recognize them, and the virus may avoid detection until new antibodies are formed. A vaccine can be either an injection of antibodies or an injection of a nonpathogenic v irus with the same capsid or env elope. The later allows the host ilumune system to create its own antibodies. Vaccines against rapidly luutating viruses arc generally not very effective. Another difficulty of fighting viral infections is that more than one animal may act as a carrier population. Even if all viral i.nfections of a certain type \-vere eliminated in humans, the virus may continue to thrive in another animal, thus maintaining the ability to reinfect the human population. For instance, ducks carry Copyright l{;) 2007 Exarnkr0ckcrs, Inc
)
.
55
56
MeAT BIOLOGY
You should know the structure of a virus: capsid , nucleic acid, and lipid-rich
protein envelope for some viruses: tail,
the flu virus, apparently without any adverse symptoms. One of the reasons that the fight against small pox was so successful was because the virus can only infect humans.
base plate, and tail fibers for most bacteriophages. Viruses are very small.
Remember, a bacterium is the size of a mitochondrion, and hundreds of viruses may fit within a bacterium. Be able to distinguish between the lysogenic and lytic life cycles : "lysogenic" is a longer word and a longer cycle. You don't need to know about any specific types of viruses .
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54. Which of the following describes a lysogenic cell?
Questions 49 through 56 are NOT based on a descriptive passage.
A. B. C. D.
49. Which of the following events does NOT playa role in the life cycle of a typical retrovirus? A. B. C. D.
Viral DNA is injected into the host cell. Viral DNA is integrated into the host genome. The gene for reverse transcriptase is transcribed and the mRNA is translated inside the host cell. Viral DNA incorporated into the host genome may be replicated along with the host DNA.
55. A bacteriophage is easily recognizable due to: A. B. C. D.
50. A mature virus outside the host cell is called a virion. A virion may contain all of the following EXCEPT: A. B.
C. D.
A cell that harbors an inactive virus in its genome. A cell that has developed immunity from viral infection. Any cell infected with a virus. A cell that is about to lyse as a result of viral infection.
a lysogenic life cycle a protein capsid circular nucleic acids a tail and fibers
56. Which of the following would never be found in the capsid of a virion?
a capsid. an envelope made from a phospholipid bilayer. core proteins. both RNA and DNA.
A. B. C. D.
single stranded DNA double stranded RNA ribosomes reverse transcriptase
51. Prior to infecting a bacterium, a bacteriophage must: A. B. C. . D.
reproduce, making copies of the phage chromosome. integrate its genome into the bacterial chromosome. penetrate the bacterial cell wall completely. attach to a receptor on the bacterial cell membrane ..
52. Most viruses that infect animals: A. B.
C. D.
enter the host cell via endocytosis. do not require a receptor protein to recognize the host cell. leave their capsid outside the host cell. can reproduce independently of a host cell.
53. Viruses most closely resemble:
A. B. C. D.
facultative anaerobes. aerobes. saprophytes. parasites.
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3.3
Don't get caught up in all the minute differences between Archaea and Bacteria. Just know that there is a distinction and that,although both Archaea and Bacteria are prokaryotes, Archaea have sim ilarities to eukaryotes.
MCAT expects you to know two aspects of the classification system : 1) energy source and ; 2) carbon source . Autotrophs and heterotrophs differ in their source of carbon : autotrophs use CO 2 and heterotrophsuse organic matter. 'Photo' and 'chemo' refer to where the organism derives its energy: 'photo' from light and 'chemo' from chemicals. Only prokaryotes can acquire energy from an inorganic source other
than light.
Prokaryotes
Prokaryotes do not have a membrane bound nucleus. They are split into two domains called Bacteria and Archaea. Archaea have as much in COfnmon with eukaryotes as they do with bacteria. They are typically fou nd in the extreme environments such as salty lakes and boiling hot springs. Unlike bacteria, the cell walls of archaea are not made from peptidoglycan. Most known prokaryotes are members of the domain Bacteria (Greek: bakterion: small rod). The introduction of the two domains makes the kingdom Monera obsolete. The kingdom manera was the kingdom containing all prokaryotes. In order to grow, all organisms require the ability to acquire carbon, energYf and electrons (usually from hydrogen). Organisms can be classified according to the sources from which they gather these commodities. A carbon source can be organic or inorganic. Most carbon sources also contribute oxygen and hydrogen. CO2 is a unique inorganic carbon source because it has no hydrogens. To some degree, all microorganisms are capable of fixing CO, (reducing it and using the carbon to create organic molecules usually through a process called the Calvin cycle). However, the reduction of CO, is energy expensive and most microorganislns cannot use it exclusively as their carbon source. Autotrophs (Greek: autotrophos: supplying one's own food, aut-:sell, trephein:to nourish) are organisms that are capable of u s ing CO 2 as their sale source of carbon. Heterotrophs (Greek: heteros:different or other) use preformed, organic molecules as their source of carbon. Typically these organic molecules come from other organisms both living and dead, but it is believed that at the dawn of life they formed spontaneously in the environment of primitive earth. All organisms acquire energy from one of two sources:
1.
light; or
2.
oxidation of organic or inorganic Inatter.
Organisms that use light as their energy source are called phototrophs; those that use oxidation of organic or inorganic matter are called chemotrophs. Electrons or hydrogens can be acquired from inorganic matter by fithotrophs, or organic matter by orgnnotroplrs. All organisms can be classified as one of each of the three types. For instance, a flesh eating bacteriUln is a chemoorganotrophic heterotroph. Most heterotrophs also use organic n1atter as their energy source, making them smne type of chemotrophic heterotrophs. Bacteria are found in all classifications. Some bacteria are capable of fixing nitrogen. Atmospheric nitrogen is abundant, but in a s trongly bound form that is useless to plants. Nitrogen fixation is the process by which N2 is converted to ammonia. Most plants are unable to Llse amn10nia, and must wait for other bacteria to further process the nitrogen in a process called nitrification. Nitrification is a tvvo step process that creates nitrates, which are useful to plants, from 31nn10nia. Nitrifica tion requires two genera of chemoautotrophic prokaryotes. The relevant reactions are shown below.
NH/ + 11/ , 0, .... N0 2- + H 20 + 2H+ N0 2- + 1/,02 .... N03Chelnoautotrophy is an inefficient mechanism for acquiring energy, so chemoautotrophs require large amounts of substrate. This means that chemoautotrophs have a large envirmunental impact, which is reflected in processes like nitrification. All known chemoautotrophs are prokaryotes. Copyrighl
(Cj
2007 Exan,kracke:-s, Inc.
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3:
MICROBIOLOGY ,
59
Structure of Prokaryotes
The typical structure of prokaryotes (Fig. 3-5) is simpler than that of eukaryotes. The most basic distinction between eukaryotes and prokaryotes is that prokaryotes don't have a nucleus, and eukaryotes always have at least one nucleus. Instead of a nucleus, prokaryotes usually have a Single, circular double stranded molecule of DNA. This molecule is twisted into supercoils and is associated with histones in Archaea and with proteins that are different from histones in Bacteria. The DNA, RNA and protein complex in prokaryotes forms a structure visible under the light microscope called a nucleoid (also called the chromatin body, nuclear region, or nuclear body). The nucleoid is not enclosed by a membrane.
Recognize that the name of the bacteria often reveal s the shape . Lil,e: spiro plasma; staphylococcus or; pneumococcus.
There are two major shapes of bacteria: cocci (round) and bacilli (rod shaped). There are many other shapes, including helical. H elically shaped bacteria are called spirilla, if they are rigid. Otherwise they are called spirochetes. Certain species of spirochetes may have given rise to eukaryotic flagella through a symbiotic relationship. Prokaryotes have no complex, membrane-bound organelles. All living organisms contain both DNA and RNA, so prokaryotes have RNA. Since they translate proteins, prokaryotes have ribosomes. Prokaryotic ribosomes are smaller than eukaryotic ribosomes. They are made from a 50S and a 30S suburnt to form a 70S ribosome.
Prokaryotes lack a nucleus. In fact, they have no complex, membrane-bound organelles at all. The key words are 'complex' and 'membrane-bound '. They have organelles: ribosomes, nucleoid , mesosomes etc ... ; just not complex, mem brane-bound organelles.
Inclusion body Ribosome
Plasmid
Nucleoid with Chromosome Mesosome
Bacterium Figure 3-5
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B iOLOGY
A prokaryote mayor may not contain a mesosome. Mesosomes are invagina tions of the plasma membrane. They may be in the shape of tubules, lamellae, or vesicles. Under the light microscope mesosomes may appear as bubbles inside the bacterium. Their exact fW1ction is unknown, but they m ay be involved in cell wall formation during cellular division. Prokaryotes also have inclusion bodies. Inclusion bodies are granules of organic or inorganic matter that may be visible under a light microscope. Inclu sion bodies may or may not be bow1d by a single layer membrane .
....1
Fatty Acid
""
Fatty Acid
o~
U
d C)
Phosphorylated alcoh ol
H0/
Palmitic acid
,.,-~-o-~J~)-~-H I
r
H-C-H
H
I
I;-l T-J .
I + I I NI 1'
0
\" "H
H-C-O-C-(uc~C-(~JH I II I I I I I 0
II
?-?-O-r o-1'-H
H-
H
HH
0-
7H
H
/ Oleic acid
7H
H
H- C~ H
I
H
\
. Phosphatidylethan olamine
I ~,
I
Polar &
----->
Nonp olar & Hydrophobic
HYd~\h~~ C
Phospholipid Figure 3-6
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LECTURE 3 : M,CROB,OLOGY • 61
3.5
Membranes
The cytosol of nearly all prokaryotes is surrounded by a phospholipid bilayer called the plasma membrane. (The membranes of archaea differ in their lipid structure. This is unlikely to be on the MeAT.) The phospholipid (Fig. 3-6) is composed of a phosphate group, two fatty acid chains. and a glycerol backbone. 111e phospholipid is often drawn as a balloon with two strings. The balloon portion represents the phosphate group. and the strings represent the fatty acids. The phosphate group is polar, while the fatty acid chains are nonpolar, making the molecule amphipathic (having both a polar and a nonpolar portion). When placed in aqueous solution, amphipath ic molecules spontaneously aggregate, turning their polar ends toward the solution, and their nonpolar ends toward each other. The resulting spherical structure is ca lled a miceIle (Fig. 3-7). If enough phospholipids exist, and the solution is subjected to ultrasonic vibrations, liposoJ/les may form. A liposome is a vesicle surround ed and filled by aqueous solution. It contains a lipid bilayer like that of a plasma membrane. The inner and outer layers of a membrane are referred to as leaflets. As well as phospholipids, the plasma membrane contains other types of lipid s s uch as glycolipids. Unlike eukaryotic membranes, prokaryotic plasma membranes usually d o not contain s teroids such as cholesterol. Instead, some bacterial membranes contain steroid-like molecules called hopanoids (see Figure 3-8). Different lipid types are arranged asymmetrically between the lea flets . For instance, glycolipids are found on the outer lea flet only.
Micelle Figure 3-7 Also embedded within the plasma membrane arc proteins. Most of the functional aspects of Inembranes are due to their proteins. MClnbrane proteins act as transporters. receptors. attachment sites. and enzym es. Amphipathic proteins that traverse the membrane from the inside of the cell to the outside are called integral or intrinsic proteins. Peripheral or extrinsic proteins are situated entirely on the surfaces of the membrane. They are ionically bonded to integral proteins or the polar group of a lipid. Both integral and peripheral proteins may contain carbohydrate chains making them glycoproteins. The carbohydrate portion of membrane glycoproteins always protrudes toward the outside of the cell . Lipoproteins (sometimes called lipid anchored proteins) also exist in some plasma membranes with their lipid p ortions embedd ed in the membra ne and their protein portions at the surface s. Membrane proteins are distributed asymmetricall y throughout the membra ne and between the leaflets. Neither proteins nor lipids flip easily from one leaflet to the other. Since the forces holding the entire membrane together are intermolecular, the membrane is fluid; its parts can move laterally but cannot separate. The model of the membrane as just described is known as the fluid mosaic model. A mosaic is a picture made by placing many small. colored pieces side by side. The mosaic aspect of the membrane is reflected in the asymmetrical layo ut of its proteins. In eukaryo tic membranes. cholesterol moderates membrane fluidity. In the prokaryotic plasma membrane. hopanoids prObably reduce the fluidity of the membrane. Copyright © 2007 Examkruckcrs, Inc.
Steroid OHO/-{ OUOI'!
Hopanoid Figure 3-8
You should be familiar with the fluid mosaic model of a membrane. Most prokaryotic membranes differ only slightly from eukaryotic membranes.
62
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B,OLOGY
3.6
Membrane Transport
A membrane is not only a barrier between two aqueoLls solutions of different composition; it actually creates the difference in the compositions of the solutions. At normal temperatures for living organisms, all molecules move rapidly in random directions frequently col1iding with one another. This random movement is called Brownian motion. Brownian motion creates in compounds the tendency to mix completely with each other over time. If two compounds, X and Y, are placed on opposite sides of the same container, the net movement of X will be toward Y. This movement is called diffusion. For molecules without an electric charge, diffusion occurs in the direction of lower concentration. In Figure 3-9, X diffuses in the direction of lower concentration of X, and Y toward lower concentration of Y. A gradual . change in concentration of a compound over a distance is called a chemical concentration gradient. The chemical concentration gradient is a series of vectors pointing in the direcbon of lower concentrati0f\. For molecules with a charge, there is also an electrical gradient pointing in the direction that a positively charged particle will tend to move. The two gradients can be added to form a single electrochemical gradient for 'a specific (,ompound. The electrochemical gradient for compound X points in the direction that particle X will tend to move. (Note: There are other factors affecting the direction of diffusion, including heat, and pressure. In this text, we shall assume these factors to be included in the electrochemical gradient. In strict terms, diffusion occurs in the direction of decreasing free energy, or in the strictest terms, in the direction of increasing universal entropy.)
If compounds X and Yare separated by an impermeable membrane, diftusion is stopped. However, if the molecules of X can wiggle their way across the membrane, then diffusion is only slowed. Since the membrane slows the diffusion of X, but does not stop it, the membrane is semipermeable to compound X.
fb
w
lSI <$ft> IXl
<$
~ ~
fJJ
lSI
IXl
4
~
lSIft>w
• •
W
Diff;'sion
1Xl - - - - -- · Concentration gradient of
[X]
Figure 3-9 Figure 3-11 shows examples of melnbrane transport. Natural membranes are semipermeable to most compounds, but there are degrees of semipermeability. There are two aspects of a compouod that affect its semipermeability: size and polarity. The larger the molecule, the less permeable the membrane to that molecule. Toward polar molecules with a molecular weight greater than 100, a natural membrane is generally impermeable without some type of assistance. The greater the polarity of a molecule (or if the molecule has a charge), the less permeable the membrane to that molecule. Very large lipid soluble (nonpolar) molecules like steroid hormones can moye right through the membrane. When conSidering permeability, it is important to consider both size and polarity. For example, water is larger than a Copyright (C) 2007 Exarnkrackers, Inc
LEcrURE
3:
MICROBIOLOGY .
sodium ion, but water is polar, while the sodium ion possesses a complete charge. Therefore, a natural membrane is more permeable to water than to sodium; in this case, the charge difference outweighs the size difference. A naturallnembrane is, in fact, highl y permeable to water. However, if the membrane were made only of a phospholipid bilayer, and did not contain proteins, the rate of diffusion for water would be very slow. Most of the diffusion of polar or charged molecules across a n atural membrane takes place through inciden tal holes (sometimes called leakage channels) created by the irregular shapes of integral proteins. The function of these proteins is not to aid in diffusion. This is merely an ll1Cidental contribution.
Glycoprotein
Carbohydrate Integral Protein
Glycolipid
f1
Phospholipid Bilayer
Cell Membrane Figure 3-10
The diffusion described above, w here molecules move through leakage channels across the membrane due to randOn11TIotion, is called passive diffusion. As mentioned previously, some molecules are too large or too charged to passively diffuse, yet they are needed for the survival of the celL To assis t these 11101ecules ill moving across the nlembrane, specific proteins are embedded into the Inenlbrane. These proteins, called transport or carrier proteins, are designed to facilitate the diffusion of specific molecules across the membrane. There are several mcchanislns used by transport proteins in facilitated diffusion, but, in order for the passage to be calied facilitated diffusion, diffusion must occur down the electro-chemical gradient of all species involved. Most, but not all, human cells rely on facilitated diffusion for their glucose supply. Facili ta ted diffusion is said to make !he membrane selectively permeable because it is able to select behveen molecules of similar size and charge.
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MeAT BiOLOGY
Passive Difw sion
Diffusion
.. Passive
\
&>
Transport Protein Facilitated Passive Diffusion Diffusion
0
Active Transport The Na'/K' Pump
Diffusion There are some important concepts here
6
that requ ire understanding as well as memorization. The membrane stuff is
worth a second read through . Basically, passive diffusion through the membrane depends on lipid solubility (are you nonpolar enough to slide right through the phospholipid bilayer?) and size (can you fit through the cracks around the integral proteins?). If you're big and polar, you must rely upon facil itated diffusion ; a helper protein to open up a space designed just for you . To move against your electrochemical gradient, it doesn 't matter if you are large or small, polar or nonpolar, you need active transport. Only active tran sport can move something against its concentration gradient.
Glucose
..
Water
0
0
K'
Na'
eX? Steroid
• ..,;
... . ~
ATP
Membrane Transport Figure 3-11 A living organism must be able to concentrate some nutrients against their electrochemical gradients. Of course, diffusion will not do this. Movement of a compound against its electrochemical gradient requires active transport, Active transport requires expenditure of energy. Active transport can be accomplished by the direct expenditure at ATP to acquire or expel a molecule against its electrochemical gradient. It can also be accomplished indirectly by using ATP to create an electrochemical gradient, and then using the energy of the electrochemical gradient to acquire or expel a molecule. The latter method is called secondary active transport.
3,7
Bacterial Envelope
The bacterial plasma membrane and everything inside it is called the protoplast. A protoplast is not a complete bacterium. Surrounding the protoplast is the bacterial envelope (Fig. 3-12). The component of the envelope, adjacent to the plasma membrane is the cell wall. (Archaea possess cell walls with a different chemical composition than will be described here. This will not be on the MeAT unless it is explained in a passage.) One of the functions of the cell wall is to prevent the protoplast from bursting. Most bacteria are hypertonic (Greek: hyperi above) La their environment. This means that the aqueous solution of their cytosol contains more particles than the aqueous solution surrounding them. Compare isotonic------(Greek: i50: same) where the cytosol contains the same amount of particles and hypotonic (Greek: hypoi below) where th e cytosol contains less particles. When there are more particles on one side of a barrier than the other, the particles want to move down their electrochemical gradient to the other side of the barrier. If the particles are prevented from crossing the barrier, water will try to cross in the opposite direction. (This is actually water moving down its electrochemical gradient.) The cell wall is strong and able to withstand high pressure. As the cell fills with water and the hydrostatic pressure builds, it eventually equ;ls the osmotic pressure, and the filling stops (see Physics Lecture 5 for more on Copyright © 2007 Exarnkrackers, Inc.
LECTURE
3: M ,CROB,OLOGY . 65
hydrostatic and osmotic pressure). Water continues to move in and out of the cell very rapidly, but an equilibrium is reached. If the cell wall is removed, the plasma membrane cannot withstand the pressure, and the bacterium will burst. The cell wall is made of peptidoglycan (also called murein). (Archaea do not have peptidoglycan cell walls.) Peptidoglycan is a series of disaccharide polymer chains with amino acids, three of which are not found in proteins. These chains are connected by their amino acids, or crosslinked by an interbridge of more amino acids. The chains are continuous, forming a single molecular sac around the bacteriuln. Peptidoglycan is more elastic than cellulose. (Cellulose is the component of plant cell walls. [see Biology Lecture 1] ). It is also porous, so it allows large molecules to pass through. Many antibiotics s11ch as penicillin attack the amino acid crosslinks of peptidoglycan. Lysozyme, an enzyme produced naturally by humans, attacks the disaccharide linkage in peptidoglycan. In both cases the cell wall is disrupted, and the cell lyses killing the bacterium. One method of classification of bacteria is according to the type of cell wall that they possess. A staining technique, called gram staining, used to prepare bacteria for viewing under the light microscope, stains two major cell wall types differently. The first type is called gram-positive bacteria because its thick peptidoglycan cell wall prevents the gram stain from leaking out. These cells show up as purple when stained with this process. Gram-positive bacteria have a cell wall that is approximately four times thicker than the plasma membrane. The space between the plasma membrane and the cell wall is called the periplnsmic space. The periplasmic space contains many proteins that help the bacteria acquire nutrition, such as hydrolytic enzymes. Gram-negative bacteria appear pink when gram stained. Their thin peptidoglycan cell wall allows most of the gram stain to be washed off. The peptidoglycan of gram-negative bacteria is slightly different from that of gram-positive. Outside the cell wall, gram-negative bacteria have a phospholipid bilayer, This second membrane is more permeable than the first, even allowing molecules the size of glucose to pass right through. It is similar in structure to the plasma membrane, but also possesses lipopolysaccharides. TI,e polysaccharide is a long chain of carbohydrates which protrudes outward from the cell. These polysaccharide chains can form a protective barrier from antibodies and many antibiotics. A lipoprotein in the outer membrane called Braun's lipoprotein points inward toward the cell wall and attaches covalently to the peptidoglycan. In gram-negative bacteria the periplnsmic space is the space between the two membranes. (Different species ofArchae a may stain positive or negative.) . Many bacteria are wrapped in either a capsule or a slime layer. Both capsules and slime layers are usually made of polysaccharide. Slime layers are easily washed off, while capsules are not. A capsule can protect the bacterium from phagocytosis, desiccation, SOlne viruses, and some components of the immune response of an infected host. Some gram-negative bacteria possess fimbrine or pili (not to be confused with the sex pilus discussed below). Fimbriae are short tentacles, usually numbering in the thousands, that can attach a bacterium to a solid surface. They are not involved in cell motility.
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You should have some idea about the differences between gram positive and gram negative bacteria, but most of the details on the structure of bacteria are just good background information .
66
MeAT
B,OLOGY
Disaccharide Unit
Peptidoglycan
Peptidoglycan
~
Capsule
Plasma
Cytosol
I
Cytosol
Gram Positive Envelope
Peri plasmic Liposac~haride Space Layer
Gram Negative Envelope
Bacterial Envelope Figure 3-12
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200l Exarlik',-lCkers, Inc.
LECTURE
3: M:CROS:OLOGY . 67
Bacterial flagella are long, hollow, rigid, helical cylinders made from a globular protein ,called flagellin; these should not be confused with eukaryotic flagella, which composed of microtubule,. Theyrotate counterclockwise (from the point of view of looking at the cell from the outside) to propel the bacterium in a single direction. When they are rotated clockwise, the bacterium tumbles. The tumbling acts to change the orientation of the bacterium allowing it to move forward in a new direction. The flagellum is propelled using the energy from a proton gradient rather than by ATP. Some bacteria can move via a gliding motion that has not yet been explained. Spirochetes, the flexible, helica l shaped bacteria, can move through viscous fluids by flexing and spinning.
3.8
Bacteria l Reproduction
Sexual reproduction is one method of recombining the genetic infonnation between individuals of the same species to produce a genetically different individual. Sexual reproduction requires meiosis. Bacteria do not undergo Ineiosis or mitosis, and cannot reproduce sexually. However, they have three alternative fonns of genetic recombination~ conjugation, transformation, and transduction. They are also capable of undergoing a type of cell division called binary fission (Latin: fissus:split). Binary fission is a type of asexual reproduction. In binary fission, the circular DNA is replicated in a process silnilar to replication in eukaryotes. (See Biology Lecture 2 for the replication process.) Two DNA polymerases begin at the same point on the circle (origin of replication) and move in opposite directions making complementary single strands that combine with their template strands to form two complete DNA double stranded circles. The cell then divides, leaving one circular chromosome in each daughter cell. The two daughter cells are genetically identical.
The first method of genetic recombination, conjugation (Fig. 3-13), requires that one of the bacterium have a plasmid w ith the gene that codes for the sex pilus. Plasmids are small circles of DNA that exist and replicate independently of the bacterial chromosome. If the plasmid can integrate into the chromosome it is also called an episome. Plasmids are not essential to a bacterium which carries them. Not all bacteria with plasmids can conjugate. In order for a bacterium to initiate conjugation, it luust contain a conjugative plasmid. Conjugative plasmids possess the gene for the sex pilus. The sex pilus is a hollow, protein tube that connects two bacteria to allow the passage of DNA. The passage of DNA is always from the cell containing the conjugative plasmid to the cell that does not. The plasmid replicates differently than the circular chromosome. One s trand is nicked, and one end of this strand begins to separate from its complement as its replacement is rep licated. The loose strand is then replicated and fed through the pilus. There are two important plasmids that may be mentioned on the MeAT: the F plasmid and the R plasmid. The F plasmid is called the fertility factor 01' F factor. It was the first plasmid to be described. A bacterium with the F factor is called F+, one without the F factor is called F-. The F plasmid can be in the form of an episome, and if the pilus is made while the F factor is integrated into the chromosome, some or all of the rest of the chromosome may be repUcated and transferred. The R plasmid donates resistance to certain antibiotics. It is also a conjugative plasmid. ft was once common practice to prescribe multiple antibiotics for patients to take at one time. Such conditions promote conjugation of different R plasrnids providing different resistances to antibiotics to produce a super-bacterium that contains many antibiotic resistances on one or more R plasmids. Some R plasn1ids are readily transferred between species further promoting resistance and causing serious health problems for humans.
F+
F-
F Plasmid
o o o o
o
o o ©
o o Conjugation Figure 3-13
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MeAT B,OLOGY
Transformation is the process by which bacteria may incorporate DNA from their external environment into their genome. DNA may be added to the external environment in the lab, or it may occur due to lyses of other bacteria. The typical experimental procedure which demonstrates transformation is when heat-killed virulent bacteria are mixed with harmless living bacteria. The living bacteria receive the genes of the heat-killed bacteria through transformation, and become virulent. Sometimes, the capsid of a bacteriophage will mistakenly encapsulate a DNA fragment of the host cell. When these virions infect a new bacteriul=!l, they inject harmless bacterial DNA fragments instead of virulent viral DNA fragments. This type of genetic recombination is called transduction. The virus that med iates transduction is called the vector. Transduction can be mediated artificially in the lab.
3.9
Endospores
Some gram-positive bacteria can form endosporcs (Greek: endon: within, speirein:to seed or sow) that can lie dormant for hundreds of years. Endospores are resistant to heat, ultraviolet radiation, chemical disinfectants, and desiccation. Endospores can survive in boiling water for over an hour. Endospore formation is usually triggered by a lack of nutrients. In endospore formation, the bacterium divides within its cell wall. One side then engulfs the other. The chemistry of the cell wall of the engulfed bacterium changes slightly to form the cortex of the endospore. Several protein layers lie over the cortex to form the resistant structure called the spore cont. A delicate covering, called the exosporium, sometimes surrounds the spore coat. The outer cell then lyses, releaSing the dormant endospore. The endospore must be activated before it can germinate and grow. Activation usually involves heating. Germination is triggered by nutrients.
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61. The exponential growth that occurs in E. coli at 37°C following inoculation of a sterile, nutrient-rich solution results from:
Questions 57 through 64 are NOT based on a descriptive passage.
A. B. C. D.
57. Which of the fol1owing structures are found in prokaryotes? I. A cell wall containing peptidoglycan. II. A phism~ membrane lacking cholesterol. III. Ribosomes A. B. C.
D.
62. A staphylococcus infection is most likely caused by an organism that is:
I only II only I and II only I. II, and III
A. B. C. D.
58. DNA from phage resistant bacteria is extracted and placed on agar with phage-sensitive E. coli. After incubation it is determined that these E. coli are now also resistant to phage attack. The most likely mechanism for their acquisition of resistance is: A. B. C. D.
C. D.
rod-shaped spherical a rigid helix a non-rigid helix
63. Which of the following is a mechanism for reproduction in Bacteria?
A. B.
C. D.
transduction. sexual reproduction. transformation. conjugation.
transduction conjugation binary fission transformation
64. Penicillin interferes with peptidoglycan formation. Penicillin most likely inhibits bacterial growth by disrupting the production of:
59. Bacteriophages are parasites that infect bacterial cells in order to carry on their life function. When a phage transfers bacterial DNA from one host to another this process is called:
A. B.
conjugation. transformation. sporulation. binary fission.
A. B. C. D.
transformation. transduction. conjugation. transmission.
bacterial plasma membranes prokaryotic cell walls the bacterial nucleus bacterial ribosomes
60. The lipopolysaccharide layer outside the peptidoglycan cell wall of a gram negative bacterium: A. B. C.
D.
absorbs and holds gram stain. protects the bacterium against certain antibiotics. does not contain phospholipids. allows the bacterium to attach to solid objects.
69
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MCAT
B iOLOGY
3.10 Fungi
•
The fungi are far too diverse and complex to justify memorizing a lot of specifics. The MeAT will not require know ledge of specifics beyond what is in bold and underlined above. Just remember the basics about fungi. Fungi are eukaryotic heterotrophs and spend most of their lives in the haploid state . They can reproduce sexually or asexua lly; remember when and why.
Fungi represent a distinct kingdom of organisms with tremendous diversity. Three divisions exist within this kingdom: Zyg01uycotn, Ascol1lycotn, and Basidiornycota (Greek: mykes: mushroom). (Oo111ycota, which are slime n101ds and water Inolds, are not true fungi, and are in the protista kingdom.) Like plants, fungi are separated into divisions not phyla. All fungi are eukaryotic heterotrophs that obtain their food by absorption rather than by ingestioni they secrete their digestive enzymes outside their bodies and then absorb the products of digestion. Although most fungi are considered saprophytic (Greek saprosi rotten or decayed), many fungi do not distinguish betw.een living and dead matter, and thus can be potent pathogens (disease causing). (Saprophytic means to live off dead organic matter.) Most fungi possess cell walls, called septa (Latini saeptumrfence or wall), made of the polysaccharide, chitin. Chitin is more resistant to microbial attack than is cellulose. IN s the same substance of which the exoskeleton of arthropods is made. (Arthropods are insects and crustaceans.) Septa are usually perforated to allow exchange of cytoplasm between cells, called cytoplasmic streaming. Cytoplasmic streaming allows for very rapid growth. One division of fungi, zygomycota, possesses no cell walls at all, except in their sexual structures. vVith the exception of yeasts, fungj are multicellular. (Yeasts are unicellular fungi.) A fungal cell may contain one or more nuclei. The nuclei in a single cell mayor may not be identical. Fungi lack centrioles, and mitosis in fungi takes place entirely within the nucleus; the nuclear envelope never breaks down. In their growth state, fungi consist of a tangled mass (called a myceliu m ) of multiply branched thread-like structures, called hyph ae.
3.11 Fungal Reproduction and Life Cycle
H ffi Gametes
~ Zygote
Fungal Reproduction Figure 3-14
Like most organisms, fungi alternate between haploid and diploid stages in their life cycle; however, the h aploid stage predominates, and is their growth stage. Hyphae are haploid. Some hyphae may form reproductive structures called spomngiophores in Zygomycota, col1diophores in Ascomycot17. These structures release haploid spores that give rise to new mycelia in asexual reproduction. Basidiol11ycotn, which rarely reproduce asexually, produce bnsidiospores via sexual reproduction. (Note that spore formation is not always via asexual reproduction.) Spores are borne by air currents, water, or animals to locations suitable for new mycelial growth. Yeasts rarely produce sexually by producing spores. More often in yeasts, asexual reproduction occurs by budding (also called cell fission ), in which a smaller cell pinches off from the single parent cell. When sexual reproduction occurs it is between hyphae from two mycelia of different mating types + and - (fig. 3-14). These two hyphae grow towards one another eventually touching and forming a conjugation bridge. The tip of each hypha forms a complete septum in all divisions of fungi, and becomes a gamete-producing cell, called a gametangium. In Zygomycota, the gametangia remain attached to the parent hyphae and the nuclei fuse with one another to produce a diploid zygote, called a zygospore. The zygospore separates from the parent hyphae and usually enters a dormant phase. When actjvated by the appropriate environmental conditions, the zygospore undergoes meiosis to produce haploid cells, one of which ilnmediately grows a short sporangiophore to asexually reproduce many spores. Except for the zygospore, all cells of Zygomycota are haploid. Sexual reproduction in Ascomycota and Basidiomycota is similar, but slightly more complicated. The important thing to understand about fungal reproduction is that asexual reproduction nonnally occurs "vhen conditions are good; sexual reproduction normally occurs when conditions are tough. This is because if conditions are good for the parent, they will be good for asexually reproduced offspring that are exactJllike the parent, but if conditions are bad for the parent, they may not be bad for sexually reproduced offspring that are different from the parent.
Copyright
(~l
2007 [xcHYlkrackers, Inc.
70. Which of th e following is the best explanati on for why fungicides tend to cause more side effects in humans than do bacte ri al antibioti cs?
Questions 65 through 72 are NOT based on a descriptive passage.
A. 65. Fungi are classified as a di stinc t kingdom because: A. B. C. D.
B. C.
they don't undergo mitosis. they reproduce asex ually. sexual reproduction involves the union of differe nt mating types. plus and minus strains. they have characteristics that are both plant-like and animal-like.
D.
71. The Kingdom of Fungi is divided into:
A. B.
66. Which of the following is not a result of sexual reproduction in fun gi? A. B. C. D.
C. D.
Hyphae of + and - mycelia meet and fuse. Fertilization prod uces a dipl oid state. dipl oid cell undergoes meiosis. Cell division lengthens mycelia.
D.
phyla divisio ns orders species
72. Which of the following is true of Fungi? A. II. C. D.
67. Fungi are considered saprophytic because: A. B. C.
Chitin is more difficult to break down than peptidoglycan. Fungus doesn' t respond to penicillin. Fungal cells are more similar to human cells than are bacterial cells. Fungus is a topical infection.
Fun gi prey upon only dead organi c matter. Fungi digest their food outside their bodies. Fungi undergo meiosis during asexual reproductio n. All fungi are obligate aerobes.
they reproduce by budding. they reproduce by sporulation. they absorb chemicals through a mass of tiny threads. they acquire energy from the break down of the dead re mains of living organisms.
68. All of the following statements are true concernin g most fungi EXCEPT: A. B. C. D.
They have cell walls made of chitin. They are autotrophs. Their growth stage is composed of filaments COntaining many nuclei. Their life cycle alternates between a haploid and dipl oid stage.
69. What selective advantage is offe red by the haploid state of fungi? A. B. C. D.
The haploid state can reproduce more quickly than the diploid state under favo rable conditions. The haploid state is more genetically diverse. The haploid state produces a large number of celk The haploid state requires less energy to sustain itself.
Copyright @ 2007 EXilmkrackers, Inc.
71
STOP.
The Eukaryotic Cell; Th~ Nervous System 4.1
The Nucleus
The major feature distingu ishing eukaryotic (Greek eu:well, karyos: kemel) cells from prokaryotic cells is the nucleus of the eukaryo te (Fig. 4-1). The nucleus contains all of the DNA in an animal cell (except for a small amount in the mitochondria) . The aqueous 'soup' inside the nucleus is ca lled the nucleoplasm. The nucleus is Smooth cndoplasm.ic
Flagellum
reticulum
N ucleLis Nucleolus
Rough endoplasmic reticulum
Ribosome
(free floating)
Centrioles Centrosome Lysosome
Peroxisome ----
Goigi Complex
I
Eukaryotic Cell Figure 4-1
/ Mitochondrion
Only eukaryotes have nuclei. If you remember that DNA cannot leave the nucleus, then you will remember that transcription must take place in the nucleus. RNA leaves the nucleus through
nuclear pores.
74
MeAT B,OLOGY
wrapped in a double phospholipid bilayer called the nuclear envelope or membrane. The nuclear envelope is perforated with large holes called nuclear pores. RNA can exit the nucleus through the nuclear pores, but DNA callilO!. Within the nucleus is an area called the nucleolus where rRNA is transcribed and the subunits of the ribosomes are assembled. The nucleolus is not separated from the nucleus by a membrane.
4.2
The Membrane
Besides transport across the membrane (discussed in Biology Lecture 3), cells can acquire substances from the extracellular environment through endocytosis (Fig. 4-2). There are several types of endocytosis: phagocytosis (Greek: phagein: to eat), pinocytosis (Greek: pinein: to drink), and receptor mediated endocytosis. In phagocytosis, the cell membra"e protrudes outward to envelope and engulf particulate matter. Only a few specialized cells are capable of phagocytosis. The impetus for phagocytosis is the binding of proteins on the particulate matter to protein receptors on the phagocytotic cell. In humans, antibodies or complement proteins (discussed in Biology Lecture 8) bind to particles and stimulate receptor proteins on rnacrophages and neutrophils to initiate phagocytosis. Once the particulate matter is engulfed, the membrane bound body is called a phagosome.
Phagocytosis
. •: Pinocytosis Receptor mediated endocytosis
Endocytosis Figure 4-2 In pinocytosis, extracellular fluid is engulfed by small invaginations of the cell membrane. This process is performed by most cells, and in a random fash ion; it is nonselective.
On the MeAT, you will probably not have to distinguish between the diffelent types of endocytosis, but you shou ld understand the basIc concept, and be aware that multiple methods for particles to gain access to the interior of
Receptor mediated endocytosis refers to specific uptake of macromolecules such as hormones and nutrients. In this process, the ligand binds to a receptor protein on the cell membrane, and is then moved to a clathrin coated pit. Clathrin is a protein that forms a polymer adding structure to the underside of the coated pi!. The coated pit invaginates to fonn a coated vesicle. One way that this process differs from phagocytosis is that its purpose is to absorb the ligands, whereas the ligands in phagocytosis exist only to act as signals to initiate phagocytosis of other particles.
a cell. Exocytosis is simply the reverse of endocytosis. Th e structure of the phospolipid bilayer of the eukaryotic membrane is similar to the prokaryotic plasma membrane discussed in Biology Lecture 3. However, in the e ukaryotic cell, the membrane invaginates and separates to form individual, membrane bound compartments and organelles. The eukaryotic cell contains a thick Copyright © 2007 Exarnkrackers, Inc.
LECTURe
4:
THF EUKARYOTIC CELL; THE NERVOUS SYSTEM •
maze of membranous wa lls called the endoplasmic reticulum (ER) separating the cytosol (the aqueous solution inside the cell) from the ER lumen or cisternal space (the "extracellular fluid" side of the ER). In many places the ER is contiguous w ith the cell membrane and the nuclear membrane. The ER lumen is contiguo us in
places with the space between the double bilayer of the nuclear envelope. ER near the nucleus has many ribosomes attached to it on the cytosolic side, givi ng it a granular appearance, hence the name granular or rough ER. Translation on
the rough ER propels pro teins into the ER lumen as they are created. These pro teins are tagged with a sigllnl seqllellee of amino acids and sometimes glycosylnted (carbohydrate chains are added) . The newl y synthesized proteins are moved through the lumen towa rd the Golgi apparatus or Golgi- complex. The Golg i apparatus is a series of fl attened, membrane bound sacs. Small trallsport vesicles bud off from the ER and carry the proteins across the cytosol to the Golgi. The Golgi organizes and concentrates the proteins as they are shuttled by tra nsport vesicles progressively o utward from one compartment or cisterna of the Golgi to the next. Proteins are distinguished based upon their signal sequence and ca rbohydrate chains. Th ose proteins no t possessing a Signal sequence are packaged into secretory vesicles and
expelled from the cell in a process ca lled blllkflow. The Golgi may change proteins chemically by glycosylntioll or by removing amino acids. Some polysaccharide formation also takes place within the Golgi. The end-product of the Golg i is a vesicle fu ll of proteins. These protein filled vesicles may either be expelled from the cell as secretory vesicles, released from the Golgi to mature into lysosomes, or transported to other p arts of the cell such as the mitochondria or even back to the ER.
Secretory vesicles (sometimes called zymogen grnnules) ma y contain enzymes, growth fa ctors, or extracellular matrix components . Secretory ves icles release the ir contents thro ugh exocytosis. Since exocytosis incorpora tes vesicle membranes into
the cell membrane, secretory vesicles also act as the vehicle with which to supply the cell membrane w ith its integral proteins and lipids, and as the mechanism for membrane expansion. In the rev erse process, endocytotic vesicles made at the cell
membrane are shuttled back to the Golgi for recycling of the cell membrane. Secretory vesicles are continuousl y released by most cells in a process called cOllstitlltive secretion. Some specia lized cells ca n release'secretory vesicles in response to a
certain chemical or electrical stimulus. This is called reglliated secretioll. Some proteins are activated within the secretory vesicles. For instance, proinsulin is cleaved to insulin only after the sec retory vesicle buds off the Golgi.
Lysosomes contain ncid hydrolnses (h ydrol ytic enzymes that function best in an acid environment) s uch as proteases, liposes, Iwe/eases and glycosidases. Together, these e nzymes are capable of breaking dow n every major type o f n1acromolecule w ithin th e
cell. Lysosomes generally have an interior pH of 5. They fuse with end ocyto tic vesicles (the vesicles form ed by phagocytosis and pinocytosis), and digest their contents. Any material no t degraded by the lysosome is ejected from the cell through exocytosis. Lysosomes also take up and degrade cytosolic proteins in an endocy to tic process. Under certain conditions lysosomes w ill rupture and release
their contents into the cytosol killing the cell in a process called nlltolvsis. Auto lysis is useful in the formati on of certain organs and tissues, like in the destruction of the
tissue between the digits of a human fetus in order to form fingers. Endoplasm ic reticulum, w hich lacks ribosomes, is ca lled agranular or smooth en-
doplasmic reticulum. Rough ER tends to resemble flattened sacs, whereas smooth ER tends to be tubular. Smooth ER p lays several important roles in the cell. Smooth ER contains glucose 6-pilosphntnse, the enzyme used in the liver, the intestina l epithelia l cells, and renal tubule epithelia l cells, to hydrolyze glucose 6-phosphate to g lucose, an important s tep in the produc tion of glu cose from glycoge n. Triglycerides are produced in the smooth ER and stored in fat droplets. Adipocytes are cells containing predominately fat drople ts. Such cells are important in energy Copyright to 2007 hamkrackers, Inc.
75
76
MeAT B,OlOGY
storage and body temperature regulation. The smooth ER and the cytosol share in the role of cholesterol formation and its subsequent conversion to various steroids. Most of the phospholipids in the cell m em brane are originally synthesized in the smooth ER. The phospholipids are all synthesized on the cytosol side of the membrane and then some are flipped to the other side by proteins called phospholipid translocators located exclusively in the smooth ER. Finally, smooth ER oxidizes foreign substances, detoxifying drugs, pesticides, toxins, and pollutants.
~
~~
>..,.
Peroxisomes are vesicles in the cytosol. They grow by incorporating lipids and proteins from the cytosol. .R ather than budding off membranes like lysosomes from the golgi, peroxisomes self-replicate. They are involved in the production and breakdown of h ydrogen peroxide. Peroxisomes inactivate toxic substances _... ..~~ such as alcohol, regulate oxygen concentration, playa role in the synDelivery for a \ thesis and the breakdown of lipi ds, and in the metabolism of nitrogenous bases and carbohydrates.
~
Mr. Membrane. ~/-------' _~~__ _ _
4"'!"·
-----
,~
Ther..,€ is a lot of bac~g,round information here that is. not required by th e MCAT. For the MCAf you should know the followmg:
1. Although there (Ire many compartments in a celt it CiJn be di\'ided into two sides: the cytosol and the ER IUlnen, In order to reach the cytosot a substllncc
must cross a membrane via passive or facilitated diffll~ion, or active transport, but it CiJll reach the EH. lumen viiJ endocytosis without ever transporting across a membrane.
Golgi Salty
2. Rough ER has rib osomes attached to its cytosol side, and it synthesizes virtually all p rote ins not used in the cytosol. Proteins synthesi7:ed on the ro ugh ER iJre p ushed in to the ER lumen and sent to the Go\gi. 3. The Golgi modifies and packages protejns for use in other parts of the cell and outside the cell. 4. Lysosomes con tain hydrolytic enzymes that digest substances taken in by endocytosis. l.ysosomes come from the Golgi . 5. Smooth . ER is the site of lipid synthesis including steroids. The smooth ER also helps to detOX ify sonle d r ugs.
4.3
You must know the difference between microtubules and microfilaments . Microtubu les are larger and are involved in flagella and cilia con struction, and the spindle apparatus. In humans, cilia are found only in the fa llopian tubes and th e respiratory tract. Microfilaments squeeze the membrane together in phagocytosis and cytokinesis. They are also the contractile force in microvi lli and muscle.
Cellular Filaments
The structure and motility of a cell is detennined by a network of filaments known as the cytoskeleton. The cytoskeleton ancho rs some membrane proteins iJnd other cellular cOlnponents, n10ves components within the cell, and moves the cell itself. Two major types of filaments in the cytoskeleton are microtubules (Fig. 4-3) and microfilam ents. Micro tubules are larger than microfilaments. They are rigid hollow tubes made fro m a protein called tubulin. Although tubulin is a globular protein, under certain cellular conditions it polymerizes into long straight filaments. Thirteen of these filaments lie alongside each other to form the tube. The sp.i ra l appearance is due to the two types of tubulin, a and ~, used in the synthesis. The mitotic spindle (sec Biology Lecture 2) is made from microtubules.
Flagella and cilia are specialized structures also made from micro tubules. The major portion of each flagellum and dlitun, called the axoneme, contains nine pairs of microtubules forming a circle around two lone microtubules in an arrangement known as 9+2 . Cross bridges made frOln a protein called dynein connect each of the outer pairs of microtubules to their neighbor. The cross bridges cause the microtubule pairs to slide along their neighbors creating a whip action in cilia causing fluid to move laterally, or a wiggle action in flagella causing fluid to move directly away from the cell.
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2007 Examkrackc(s, Inc
LECTURE 4: THE EUKARYOTIC CELL; THE NERVOUS SYSTEM . 77
00~TUbU1in
C)
-
Microtubule pair
~
\
'
Sperm
Dynein cross bridge
9+2
Flagellum or cilium
Microtubule
Structure of Flagella and Cilia Figure 4-3 Microtubules have a + and - end. The - end attaches to a microtubule-organizing center (MTOC) in the cell. A microtubule grows away from an MTOC at its + end. The major MTOC in animal cells is the centrosome. The centrioles function in the production of flagella and cilia, but are not necessary for microtubule production. Microfilalnents are slnaller than microtubules. The polymerized protein actin fonns a Inajar con1ponent of microfilaments. Microfilaments produce the contracting force in muscle (discussed in Biology Lecture 8) as well as being active in cytoplasmic streaming (respo nsible for amoeba-like movement), phagocytosis, and microvilli movement.
Copyright 0 2007 EX(lmkr.xkcrs, Inc.
Don 't confuse eukaryotic flagella with prokaryotic flagella . Eukaryotic flagella are made from a 9 + 2 microtubule configuration; a prokaryotic flagellum is a thin strand of a single protein called flagellin . Eukaryotic fl agella undergo a whip-like action , while prokaryotic flagella rotate .
78 . MeAT BIOLOGY
4.4
You can think of tight junctions like the plastic rings around a six pack of beer. The beer cans are the cells. The plastic rings hold the cans together and provide a water tight barrier around them . Although the beer cans are impermeable to water, real cells mayor may not be impermeable.
Cellular Junctions
There are th ree types of junctions or attachments tha t connect anima l cells: tight junctions; desmosomes; and gap junctions (Fig. 4-4). Each junction performs a differen t function. Tight junctions form a w atertight sea l from cell to cell that can block water, ions, and other molecules from m oving around and past cells. Tissue held together by tight junctions may act as a complete fluid barrier. Epithelial tissue in organs like the bladder, the intestines, and the kidney are held together by tight junctions in order to prevent waste materials from seeping around the cells and into the body. Since proteins have some freedom to move laterally abo ut the cell membrane, tight juncti ons also act as a barrier to protein movement between the apiml and the basalaternl surface of a cell. (The part of a cell facing the lumen of a cavity is called the ap ica l surface. The opposite side of a cell is called the basolateral surface. )
Tight Junction
Gap Junction Remember the three types of cellular junctions. Tight junctions act as a fluid barrier around cells. Desmosomes are like spot-welds holding cells together. . Gap junctions are tunnels between celis allowing for the exchange of small molecules.
Cellular Junctions Figure 4-4
Desmosomes join two cells at a single point. They attach directly to the cytoskeleton of each cell. Desmosomes do not prevent fluid from circulating around all s id es of a cell. Desmosomes are found in tissues that norma lly experience a lot of stress, like Salt¥'s skin or intestina l epitheliunl. Desmdsome _ ~~ Desmosomes often accompany tight junctions.
--M~
Gap junctions are small tunnels cOIU1ecting cells. They allow small molecules and ions to move between ce lls. Gap junctions in cardiac muscle provide for the spread of the ac tion potential from cell to cell. Copyright © 2007 Exarnkracke rs, lnc.
LECTURE 4 : T HE EUKARYOTIC CELL; THE NERVOUS SYSTEM . 79
4.5
Mitochondria
Mitochondria (Fig. 4-5) are the powerhouses of the euka ryo tic cell. We have already seen that the Krebs cycle takes p lace inside the mitochondria. According to the endosymbiont theory, mitochondria may have evolved from a symbiotic relationship between ancient prokaryotes and euka ryotes. Like prokaryotes, mitochondria have their own circular DNA that replicates independently from the eukaryotic cell. This DNA contains no his tones or nucleosomes. Most animals have a few dozen to several hundred molecules of circular DNA in each mitochondrion. The genes in the mitochondrial DNA code for mitochondrial RNA that is distinct from the RNA in the rest of the cell. Thus mitochondria have their own ribosomes with a sediment coefficient of 55-60S in humans. However, most proteins used by mitochondria are coded for by nuclear DNA, not mitochondrial DNA. Antibiotics that block translation by prokaryotic ribosomes but not eukaryotic ribosomes, also block translati on by mitochondrial ribosomes. Interestingly, some of the codons in mitochondria differ from the codons in the rest of the cell, presenting an exception to the universal genetic code! Mitochondrial DNA is passed maternally (from the mother) even in organisms whose male gamete contributes to the cytoplasm. Outer membrance
\
Matrix Crista
Mitochondrion Figure 4-5
Mitochondria are surrounded by two phospholipid bilayers. The inner membrane in vag inates to form cristae. Tt is the inner membrane that holds the e lectron transport cha in . Between the inner and outer membrane is the intermembrane space.
4.6
The Extracel lular Matrix
Most cells in multicellular organisms form groups of s imilar celis, or cells that work together for a common peupose, called tisslle. In some tissues, cells called fibroblasts secrete fibro us proteins such as elastin and collagen that form a molecular network
that holds tissue cells in place called an extmcefllllnr mntrix. Different tissues form dramatically different matrices. The matrix can constHute most o f the tissue as in bone, where a few cells are interspersed in a Jarge matrix, or the matrix may be only a sma ll part of the tissue. The consistency of the matrix may be liquid as in blood, or solid as in bone. Copyright © 2007 Extlmk rnc kers, Inc.
Memorize the parts of the mitochondrion and its purpose. Relate the parts to respiration discussed in Biology Lecture 1.
80
MeAT
BIOlOGY
An extracellular matrix may provide structural support, help to determine the cell shape and motility, and affect cell growth, Three classes of molecules make up animal cell matrices:
L
glycosaminoglycnns and proteoglyeans;
2.
structural proteins;
3,
adhesive proteins,
Glycosaminoglycans are polysaccharides that typically have proteoglycans attached, They make up over 900;:, of the matrix by mass. This first class of molecules provides pliability to the matrix, Structural proteins provide the matrix with strength. The most common extracellular matrix structural protein in the body is collagen, Collagen is the structural protein that gives cartilage and bone their tensile strength, Adhesive proteins help individual cells within a tissue to stick together, Don 't wOrlY about memorizing anything about the extracellular matrix, Just know that it's the stuff that surrounds the cell and that it is formed by the cell itsell,
You may see basal lamina (also called basement membrane) in an MCAT passage. The basal lamina is a thin sheet of matrix material that separates epithelial cells from sllpport tissue, (Epithelial cells separate the outside environment from the inside of the body. Support tissue is composed of the cells adjacent to the epithelial cells on the inside of the body) Basal lamina is also found around nerves, and muscle and fat cells, Basal lamina typically acts as a sieve type barrier, selectively allowing the passage of some molecules but not others, Many animal cells contain a carbohydrate region analogous to the plant cell wall or bacterial cell wall, called the glycocalyx, The glycocalyx separates the cell membrane from the extracellular matrix; however, a part of the glycocalyx is made from the same material as the matrix, Thus, the glycocalyx is often difficult to identify The glycocalyx may be involved in cell-cell recognition, adhesion, cell surface protection, and permeability
4.7
Org;;lnization
In
Multicellular Eukaryotes
In multicellular eukaryotes, groups of cells work together, each type of cell performing a unique function that contributes to the specialized function of the group, These groups of cells are called tissues, Cells in the same tissue usually have similar embryology; they arise from the same embryonic germ layer, There are four basic types of tissue in animals: epithelial tissue, muscle tissue, connective tissue, and nervous tissue, Epithelial tissue separates free body surfaces from their surroundings, Simple Epithelium is one layer thick, while stratified epithelium is two or more layers thick Simple epithelium includes endothelium lining the various vessels of the body including the heart. Connective tissue is characterized by an extensive ~atrix. Examples include: blood, lymph, bone, cartilage, and COlU1ective tissue proper making up tendons and ligan1ents. Muscle and nervous tissue will be discussed later. Differenl Lissue lypes work logether to fOI11) organs. For exanlple, the stOll1ach is an organ with an outer layer made fronl epitbelial tissue and connective tissue, a second layer of muscle tissue, and an innermost layer of epithelial tissue.
Organs that work together to perform a common function are called systems, The remainder of this manual (except for lecture 9) is devoted to the study of biological systems in the human body Many of the details are not required knowledge for the MCAT Howeve]; if you keep in mind the holistic concept (the body is not simply many disconnected parts but an entire organism with systems that work in conjunction with each other), you will attain a stronger recall of the details and a greater lU1derstanding of material. Copyright @ 200? Exarnkrackers, Inc
78. Which of the following is not a membrane bound organelle?
Questions 73 through 80 are NOT based on a descriptive passage.
A. B. C. D,
73. All of the following are composed of microtubules EXCEPT: A. B. C. D.
the tail of a sperm cell the spindle apparatus the cilia of the fallopian tubes the flagella of bacteria
79. One function of the liver is to detoxify alcohol taken into the body. The organelle within the liver cell that most directly affects this process is: A. B.
74. Which of the following is true concerning the nucleolus? A. B. C. D.
C. D.
It is bound by a phospholipid membrane. It disappears during prophase. It is the site of translation of ribosomal RNA. It is found in most bacteria.
A. B. C.
active transport. facilitated transport. passive transport. osmosis.
C. D.
the the the the
smooth endoplasmic reticulum nucleus Golgi apparatusrough endoplasmic reticulum
80. When a primary lysosome fuses with a food vesicle to become a secondary lysosome:
75. In some specialized cells, glucose is transported against its concentration gradient via an integral protein using the energy of the sodium ion electrochemical gradient. If no ATP is used for this transport. it is most likely: A. B.
The golgi body The nucleus The smooth endoplasmic reticulum The ribosome
D.
its pH drops via active pumping of protons ipto its interior. its pH drops via active pumping of protons out of its interior. its pH rises via active pumping of protons into its interior. its pH rises via active pumping of protons out of its interior.
76. Which of the following cells would be expected to contain the most smooth endoplasmic reticulum? A. B.
a liver cell. an islet cell from the pancreas a mature sperm a zygote
C. D.
77. Which of the following statements tight junctions?
IS
true concerning
I. They connect adjacent cells. II. They may form a barrier to extracellular fluids. III. They have the greatest strength of all cellular adheSIOns. A. B. C. D.
Copyright
I only I and II only II and III only T, II, and III
(i)
2007 Examkrackers, Inc.
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STOP.
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MeAT B,o,oGY
4.8
Intercel lular Communication
In multicellular organisms, cells must be able to communicate with each other so that the organism can function as a single unit. Communication is accOlnplished chemically via three types of molecules: 1) neurotransmitters; 2) local mediators; 3) hormones. These methods of cOlnmunication are governed by the nervous system, the paracrine system, and the endocrine systelTI respectively. There are several distinctions between the methods of communication, the major distinction being the distance traveled by the mediator. Neurotransmitters travel over very short intercellular gaps; local mediators function in the iInmediate area around the cell from which they were released; hormones travel throughout the organism via the blood stream. For the MeAT, you should focus on the distinctions between neurotransmitter and hormonal mediated comlnunication. Neurotransmitters are released by neurons. Neuronal communication tends to be rapid, direct, and specific. Hormonal communication, on the other hand, tends to be slower, spread through out the
body, and affect many cells and tissues in many different ways .
4 .9
Specific knowledge of the paracrine system will not be tested by the MeAT. However, you should be aware of the existence of this intermediate
Paracrine System
Local mediators are released by a variety of cells into the interstitial fluid (fluid behveen the cells) and act on neighboring cells a few millimeters away. Local mediators lnay be proteins, other amino acid derivatives, or even fatty acids. Prostaglandins are fatty acid derivative local mediators. Prostaglandins affect smooth muscle contraction, platelet aggregation, inflamlnation and other reactions. Aspirin inhibits prostaglandin synthesis and thus is an anti-inflammatory. Growth factors and lymphokines, discussed later, are other exmnples of locallnediators.
communication system.
The remainder of this lecture and the next are devoted mainly tu the nervous system and the endocrine system . Besides mernorizing the details of each system, keep in mind that they represent two different methods of cellular comlTIunication . After rl'
4.10 Nervous System The nervous system allows for rapid and direct communication behveen specific parts of the body resulting in changes in muscular contractions or glandular secretions. Included within the nervous system are the brain, spinal cord, nerves and neural support cells, and certain sense organs such as the eye, and the ear. The functional unit of the nervous system is the neuron (Fig. 4-7). A neuron is a highly specialized cell capable of transmitting an electrical signal from one cell to another via electrical or chenlical nleans. TIle neuron is so higl11y specialized that it has lost the capacity to divide. In addition, it depends almost entirely upon glucose for its chemical energy. Although the neuron uses facilitated transport to move glucose from the blood into its cytosol, unlike most other cells, the neuron is not dependent upon insulin for this transport. The neuron depends heavily on the efficiency of aerobic respiration. However, it has low stores of glycogen and oxygen, and must rely on blood to supply sufficient levels of these nutrients. Neurons in different parts of the body have a different appearance, but all neurons have a basic anatOlny consisting of many dendrites, a Single cell body, and usually one axon with many small branches.
CopyriS3ht
(9
2007 Examkrackers, Inc.:.
LECTURE 4 : THE EUKARYOTIC CELL; THE NERVOUS SYSTEM .
;
'.
:w ..Il..
Q
J
[] []
1
JL
[J ...£L
IT
u
u
. Unipolar (sensory only)
. Bipolar (Retina, inner ear, olfactory area of the brain)
Multipolar (most neurons of the brain)
Possible Neural Cell Structures Figure 4-6
The dendrites receive a signal to be transmitted. Typically, the cytosol of the cell body is highly conductive and any elec trical stimulus creates a disturbance in the electric field tha t is transferred immediately to the axon hillock. If the stimulus is great enough, the axon hillock generates an action potential in all directions, including down the axon. The membrane of the cell body usually does not contain enough ion channels to sustain an action potential. The axon, however, carries the action potential to a synapse, which passes the signal to another cell.
CoPyri9ht © 2007 Exarnkracke:"s, Inc.
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MeAT B,OLOGY
Nucleus
Axon hillock Mitochondrion
- - Node of Ranvier
~
You must know the basic anatomy of a neuron. Remember that a signal travels from the dendrites to the axon hillock,
Layers of
~yelin
sheath
where an action potential is generated and moves down the axon to the synapse. Neurons do not depend upon
insul in to obtain glucose.
Axon terminal
Myelinated Neuron Figure 4-7
Copyri gh t G) 2007 [xamkrackers, inc.
LeCTURE
4 : THE EUKARYOT:C CELl; THE NERVOUS SYSTEM
4. 11 The Action Potentia l The action potential is a djsturbance in the electric field across t11e membrane of a neuron. To understand the action potential, we first lTIUst understand the resting potential (Fig. 4-8). The resting potential is established mainly by an equilibrium between passive diffusion of ions across the membrane and the Na+ jK+ pUlnp. The Na+ /K+ pump moves three positively charged sodium ions out of the cell while bringing two positively charged potassium ions into the cell. This action increases the positive charge along the membrane just outside the cell relative to the charge along the membrane on the inside of the cell. As the electrochemical gradient of Na+ becOlnes greater, the force pushing the N a+ back into the cell also increases. The rate at which Na+ passively diffuses back into the cell increases until it equals the rate at which it is being pumped out of the cell. The same thing happens for potassium. When all rates reach equilibrium, the inside of the membrane has a negative potential difference (voltage) compared to the outside. This potential difference is called the resting potential. Although other ions are involved, Na+ and K+ are the major players in establishing the res ting potential. Na'
Interstitial fluid
Passive diffusion (incidental leakage through membrane proteins)
•
+++++++ 111111111
Electric field
Cytosol Passive diffusion directly through membrane
ATP
The Na+IK' Pump
Resting Potential Figure 4-8 The n1embrane of a neuron also contains integrallTIen1brane proteins called voltage gated sodium channels. These proteins change configuration when the voltage across the membrane is disturbed. Specifically, they allow Na+ to flow through the 111embrane for a fraction of a second as they change configuration. As Na' flows into the cell, the voltage changes s till fur ther, causing more sodium channels to change configuration, allowing still more sodium to flow into the cell in a positive feedback mechanislTI. Since the Na+ concentration moves toward equilibrium, and the K+ concentration remains higher inside the cell, the membrane potential actually reverses polarity so that it is positive on the inside and negative on the outside. This process is called depolarization. The neuronalluembrane also contains voltage gated potassium channels. The potassium channels are less sensitive to voltage change so they take longer to open. By the time they begin to open, most of the sodium channels are closing. Now K' flows out of the cell making the inside more negative in a process called repolarization. The potassium channels are so slow to close that for a fraction of a second, the inside membrane becomes even more negative than the resting potentia1. This portion of the process is called hyperpolarization. Passive diffusion returns the membrane to its resting potential. The entire process just described is called the action potential. Throughout the action potential, the Na ' /K+ pump keeps working.
.
. . '. ADP
85
86
MeAT
BIOLOGY
You must understand the dynamics of the membrane potential. Use the Na' /K' pump to help you remember that the inside of the membrane is negative with respect to the outside. Different cells have different action potentials. If you understand the principle, you should be able to understand any action potential. Remember that an action potential originates at the axon hillock. This section is im portant for the MeAl If you don't thoroughly understand it, then you should reread it.
The action potential occurs at a point on a membrane and propaga tes along th at melnbrane by depolarizing the section of membrane irnn1ediately adjacent to it. In Figure 4-9, the protein channels marked 1 are about to recieve the action potential while the protein channels marked 5 have already recei ved the action potential. Therefore, the action potential is traveling from right to left along the membrane; the synapse would be to the left of the portion of the membrane shown. The voltage as a function of time at any given point on the Inelnbrane is given by the wave shown . The entire action potential as measured a t one point on the membrane of a neuron takes p lace in a fraction of a lTIillisecond.
I
PotasSiUlTI channels begin to open.
+50mV
I I
Relative refractory period
I
I I I I I
I
I
I I I
om V
- - - - - - - - - - - - - - - - - - - - - - - -I
I
------
- ~- - - --- - - - - ------ - ' -- -- 1' -
~c
v. <\;~
-70mV
I I I
~
..?
c
~
(/,;~
<;)
I
~
~~
Threshold stimulus' . . . . - . - ..
- J - _ •••••• •• - - _ ••••• 1
~. I
~
Absolute refractory period
I I I
I
Sod iurn channels begin to open.
i Hyperpolarization I
! K'
Na' 0
0
in
K'
Na' 0
0
IHlilu
CD O K'
@
@ O K'
0 N a'
0 Na'
C:O Membrane is at rest. Sodium and potassium channels are closed_
CD
Sodium channels open and the cell depolarizes_ @ Potassium channels open as sodium channels begin to inactivate, C'!:) Sodiulu channels are inactivated. Open potassium channels repolarize the membrane. ® Potassium channels close and the melubrane equilibrates to its resting potential.
Action Potential Figure 4-9
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(~::;
2007 Exarnk rJckcrs , Inc.
LECTURE
4 : THE
EUKARYC'
'c
CELL; THE
NERVOUS SyS rEM
.
An ac tion poten tial is all-or-nothing; the m embran e comple tely depolarizes or no action potentia l is generated. In order to create an action potentiat the stimulus to the m embrane must be greater than the threshold stimulus. Any stimulus greater than the threshold stimul us creates the same size ac tion poten tial. If the threshold stimulus is reached, but is reached very slowly, an action potential still may not occur. This is called nccommodaUon. Once an action potential has begun, there is a sh ort pe riod of time called the absollite refractory period in w hi ch no stimulus w ill create another ac tion potential. The relative refrnctory period gives the time during which only an abnorma ll y large stimulus w ill create an action potential. Other cells, such as skeleta l and cardiac muscle cells also conduct action potentials. Although these action potentials are slightly different in duration, shape, and even the types of ions, they work on the same princip les.
4.12 The Synapse Neural impulses are transmitted from one cell to another chemically or electrically via a synapse. The transmission of the signal from one cell to another is the slowest part of the process of nervous system cellular communicati on, yet it occurs in a fraction of a second. Electrical synapses are uncommon. They are composed of gap junctions between cells. Ca rdiac muscle, visceral sn100th muscle, and a very few neurons in the central nervous system contain electrical synapses. Since they don't involve diffusion of chemica ls, they transmit signals Inuch faster than chemical syna pses and in both directions. A more common syn apse, a chemical synapse (Fig 4-10) (called a motor end p late when conn.ecting a neuron to arnuscle), is unidirectional. In a chemical synapse, sm all vesicles filled with neurotransmitter rest just inside the presynaptic membrane. The membrane near the synapse contains an un usually large number of Ca" voltage gated channels. When an action potential arrives at a synapse,. these Presynaptic chatu1els are activated allowing Ca 2 to flow into the neuron cell [n a mechanism not completely lmderstood , the sudden influx of calcium ions causes some of the neurotransmitter vesicles to be released throu gh a n exocy totic process into the synaptic cleft. Th e n eurotransmitter difCa lcium fuses across the synaptic cleft v ia Enzyme to channel Brownian motion (the random motion of ca tabolize neurotran smitter the molec ul es). The post~ynaptic membrane contains neurotransmitter receptor proteins. When the neurotransmitter attac h es to the receptor protein s, the postsynaptic me mbrane becomes Inore permeable to ions. Ions move across the postsyn aptic membrane through p ro teins called iOllopllOl'es, completing the transfer of the ne ural impu lse. In this way, the impulse Postsynaptic is not attenuated by e lectrical resistance as it neuron moves from one cell to the next. If a cell is fired too often (hundreds of times per second for several minutes) it will not be abl e to replenish its supp ly of neurotransmitter vessicles, and the resu lt is fatigue (the impulse will not pass to the postsynap ti c neuron). Figure 4-10 -t-
Synapse
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lC)
2007 Exarnkrackcrs, Inc
Neurotransmi tter vesicles
Synaptic cleft
Receptor protein
87
88
MeAT BIOLOGY
The neurotransmitter attaches to its receptor for only a fraction of a second? and is released back into the synaptic cleft. If the neurotransmitter remains in the synap-
tic cleft, the postsynaptic cell may be stin1ulated over and over. There are several mechanisms by wh ich the cell deals w ith this problem. The neurotransmitter may be destroyed by an enzyme in the matrix of the synaptic cleft and its parts recycled by the presynaptic cell. It may be directly absorbed by the presynaptic cell via active transport. The neurotransmitter may also diffuse out of the synaptic cleft. Ncurotra nsmi tter Recep tor protein
0"
Membrane enzyme
Intracellul aJ enzyme activation
Gene transcription ATP or
G-protein Figure 4-11
G~li CGMP
Over 50 types of neurotransmitters ha ve been iden tified . Different neurotransmitters are characteristic of different parts of the nervous system (i.e . certain neurotransmitters are found in certain areas of the nervous system). A single synapse usually releases only one type of neurotransmitter and is designed either to inhibit or to excite, but not both. A single synapse cannot change from inhibitory to excitatory, or vice versa. On the other hand, some new'otransmitters arc capable of inhibition or excitation depending upon the type of receptor in the postsynaptic membrane. Acetylcholine, a common neurotransmitter, has an inhibitory effect on the heart, but an excitatory effect on the visceral smooth muscle of the intestines. Receptors may be ion channels themselves, which are opened w hen their respective neurotransmitter attaches,. or they may act via a second messenger system acti-
vating another molecule inside lIle cell to make changes. For prolonged change, such as that involved U1 memory, the second messenger system is preferred. G-profe;ns (Fig . 4-11 ) common ly initi atf' sf'cond messenger systelns. A G-protein is attached to the receptor protein along the insid e of the postsynaptic membrane. When the receptor is stimulated by a neurotransmitter, part of the G-protein, called the a-sllbLlIlit, breaks free. The a -subunit may: 1.
ac ti va te separate specific ion chaIu1els;
2.
activate a second messenger (i.e. cyclic AMP or cyclic GMP);
3.
ac tivate intracellular enzymes;
4.
activate gene transcription.
Copyright © 2007 EX
LECTURE 4 : THE EUKARYOTIC CELL; THE NERVOUS SYSTEM . 89
A single neuron may make a few to as many as 200,000 synapses (Fig. 4-12). Most synapses contact dendrites, but some may directly contact other cell bodies, other axons, or even other synapses. The firing of one or more of these synapses creates a change in the neuron cell potential. Th is change in the cell potential is called either the excitatory postsynaptic potential (EPSP) or the inhibitory postsynaptic potential (IPSP). Normally, 40-80 synapses must fire simultaneously on the same neuron in order for an EPSP to create an action potential within that neuron.
The chemical synapse IS the important synapse for the MeAT. Understand that it is the slowest step in the transfer of a nervous signal. and that it can only transfer a signal in one direction . Also recogn ize what a second messenger system is.
Axon
Motor Neuron with Synaptic Terminals Figure 4-12
4.13 Support Cells Besides neurons, nervous tissue contains many support cells called glial cells or neuroglia. In fact, in the human brain, glial cells typically outnumber neurons 10 to J. Neuroglia are capable of cellular division, and, in the case of traun1atic injury to the brain, it is the neuroglia that multiply to fiJI any space created in the central nervous system.
There are six types of glial cells: microglia; ependymal cells; satellite cells; astrocytes; oligodendrocytes; and neuroleml11ocytes or SchWalm cells. Microglia arise from white blood cells called monocytes. They phagocytize microbes and cellular debris in the central nervous system. Ependymal cells are epithelial cells that line the space containing the cerebrospinal fluid . Ependymal cells use cilia to circulate the cerebrospinal fluid. Satellite cells support ganglia (groups of cell bodies in the peripheral nervous system) . Astrocytes are star-shaped neuroglia in the central nervous system that give physical support to neurons, and help maintain the mineral and nutrient balance in the interstitial space. Oligodendrocytes wrap many times around axons in the central nervous system creating electrically insulating sheaths called m yelin. In the peripheral nervous system, myelin is produced by Schwann cells . Myelin increases the rate at ·w hich an axon can transmit signals. To Copyright @ 2007 Exarnkrackers, Inc.
For support celi s, you should understand how myelin increases the speed With which the action potential moves down the axon. Memorizing the names of support cells is not necessary. but know the general functions of support cells. Only vertebrates have myelinated axons.
90
MeAT
B,OLOGY
the naked eye, myelinated axons appear white wll ile the neuronal cell bodies appear gray. Hence the name white matter and gray matter. Tiny gaps between myelin are called nodes of Ranvier. When an action potential is generated dovvn a Inyelinated axon, the action potential jumps from one node of Ranvi er to the next as quickly as the disturbance moves through the electric field between them. Th is is called saltatory conduction (Latin snltus:a jump).
Schwann cell
Satellite cell
.'.....
~
Axon
Microglia
i
Cilia
Neuron cell body of a ganglion
Ependymal cell
Neuron
As trocy te NeUfon .",,_
Oligodendrocyte
Neuroglia Figure 4-13 ,
\
85. White matter in the brain and spinal cord appears white because:
Questions 81 through 88 are NOT based on a descriptive passage.
A. B. C. D.
81. Which of the following gives the normal direction of signal transmission in a neuron? A. B. C.
D.
from the trom the from the from the
axon to the cell body to the dendrites dendrites to the cell body to the axon cell body to the axon and dendrites dendrites to the axon to the cell body
86. The jumping of an action potential from one node of Ranvier to the next is known as: A. B. C. D.
82. Novocain is a local anesthetic used by many dentists. Novocain most likely inhibits the action potential of a neuron by: A. B. C. D.
stimulating calcium voltage gated channels at the synapse. increasing chloride ion efflux during an action potential. uncoiling Schwann celIs wrapped around an axon. blocking sodium voltage gated channels.
B. C. D.
become more positive because potassium ion centration would increase inside the neuron. become more positive because potassium ion centration would increase outside the neumn. become more negati-.:e because potassium ion centration would increase inside the neuron. become more negative because potassium ion centration ~ould increase outside the neuron.
Brownian motion saltatory conduction a threshold stimulus an all-Of-nothing response
87. What is the ratio of sodium ions to potassium ions transferred by the Na+/K+ pump out of and into the cell? A. B. C. D.
83. A cell membrane is normally slightly passively permeable to potassum ions. If a neuronal membrane were to become suddenly impermeable to potassium ions but retain an active Na+/K+-ATPase, the neurons resting potential would: A.
it contains large amounts of myelinated axons. it does not contain any myelinated axons. it is composed primarily of cell bodies. it contains a high concentration of white blood cells to protect the central nervous system from infection.
2 sodium ions 3 sodium ions 3 sodium ions 2 sodium ions
in; 3 potassium ions out in; 2 potassium ions out out; 2 potassium ions in in; 3 potassium ions in
88. Which of the following is found in vertebrates but NOT in invertebrates? A. B.
conC. conD.
a dorsal. hollow nerve chord mylenation to increase the speed of nervous impulse transmission along the axon axons through which the nervous impulse is conducted Na+/K+-pump
concon-
84. If an acetylcholinesterase inhibitor were administered into a cholinergic synapse, what would happen to the activity of the postsynaptic neuron? A. B.
C. D.
It wOl}ld decrease, because acetylcholine would be degraded morc rapidly than normal. It would decrease, because acetylcholinesterase would bind to postsynaptic membrane receptors ' less strongly. It would increase, because acetylcholine would be produced more rapidly than normal. It would increase, because acetylcholine would be degraded more slowly than normal.
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91
STOP.
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MeAT BiOLOGY
4.14 Th e Structure of the Nervous System Neurons may perform one of three functions. 1.
Sensory (afferent) neurons receive signals from a receptor cell that interacts with its environment. The sensory neuron then transfers this signal to other neurons. 99(Yo of sensory input is discarded by the brain.
2.
Interneurons transfer signals from neuron to neuron. 90% of neurons in the human body are interneurons.
3.
Motor (efferent) neurons carry signals to a muscle or gland called the effector. Sensory neurons are located dorsally (toward the back) from the spinal cord, while motor neurons are located ventrally (toward the front or abdomen).
Figure 4-14 shows a simple reflex arc using all three types of neurons. Some reflex arcs do not require an interneuron. Neuron processes (axons and dendrites) are typically bundled together to form nerves (called tmcts in the eNS as discussed below).
Sensory Neuron Interneuron
Recep tor
o
Motor Neuron
Simple Reflex Arc of the Somatic Nervous System Figure 4-14
The nervous system has two major divisions: the central nervous system (eNS) and the peripheral nervous system (PNS) . The eNS consists of the interneurons and support tissue within the brain and the spinal cord. The function of the eNS is to integrate nervous signals betvveen sensory and motor neurons. For the MCAT, think of the CNS as the brain and spinal cord, and the PN S as everything else.
The eNS is connected to the peripheral parts of the body by the PNS. Parts of the PNS, such as the cmnial nerves and the spinal nerves, project into the brain and spinal cord. The PNS handles the sensory and motor functions of the nervous system. The PNS can be further divided into the somatic nervous system and autonomic nervous system (ANS) . The somatic nervous system is designed primarily to respond to the external environment. It contains sensory and motor functions. Its motor neurons innervate only skeletal muscle. The cell bodies of somatic Illatar Copyrighl (07007 E=xDrn krackers, Inc
LECTURE 4 : THE EUKARYOT" CELL; THE NtRVOIIS S'YFM . 93
neurons are located in the ventral horns of the spinal cord. These neurons synapse directly on their effectors and use acetylcholine for their neurotransmitter. The motor functions of the somatic nervous system can be consciously controlled and are considered voluntary The sensory neuron cell bodies are located in the dorsal rool gal1gliol1.
The sensory portion of the ANS receives signals primarily from the viscera (the organs inside the ventral body cavity). The motor portion of the ANS then conducts these signals to smooth muscle, cardiac muscle, and glands. The function of the ANS is generally involuntary The motor portion of the ANS is divided ioto two systems: sympathetic and parasympathetic (Fig. 4-15). Most internal organs ar innervated by both with the two systems working antagonistically. The sympathetic ANS deals with "fight or flight" responses. For instance, its action on the heart would be to increase beat rate and stroke volume; it works to constrict blood vessels around the digestive and excretory systems io order to increase blood flow around skeletal muscles. Parasympathetic action, on the other hand, generally works toward the opposite goal, to "rest and digest". Parasympathetic activity slows the heart rate and increases digestive and excretory activity. Sympathetic signals origina te in neurons whose cell bodies are found in the spinal cord, while parasympathetic signals origioate in neurons whose cell bodies can be
Don't make the mistake of thinking that sympathetic stimulates and parasympathetic inhibits; this is INCORRECT. Their actions depend upon the organ that they are innervating. For instance , parasympathetic neurons inhibit the heart but stimulate the digestive system . Use "fight or flight" and "rest or digest" to make predictions.
found in both the brai n and spioa! cord. (A group of cell bodies located in the eNS is called a nne/eus; if located outside the eNS, it is called a gallglioll.) These neurons extend out from the spina! cord to synapse with neurons whose cell bodies are located outside the eNS. The former neurons arc called preganglionic nellrons; the later are called pastgallgliollic l1ellrOI1S. The cell bodies of sympathetic postganglionic neurons lie far from their effectors generally withio the pamverlebml gal1glion, which runs parallel to the sp inal cord, or within the prevertebral ganglia in the abdomen. The cell bodies of the parasympathetic postganglionic neurons lie io ganglia inside or ncar their effectors. With few exceptions, the neurotransmitter used by all preganglionic neurons io the ANS and by postganglionic neurons in the pa-rasympathetic system is acetylcholine; the postganglionic neurons of the sympathetic nervolls system use either epinephrine or norepinephrine (also called adrenaline and noradrenaline). Receptors for acetylcholine are called cholinergic receptors. There are two types of cholinergic receptors: nicotinic and III11 Scnril1ic. Generall y, nico tinic receptors are found on the
postsynaptic cells of the synapse between ANS preganglionic and postganglionic neurons and on s keletal muscle membranes at the neuromuscular junction. Muscarinic receptors are
found on the effectors of the parasympathetic nervolls system. The receptors for epinephrine and norepioeplu-ioc are called adrellelsic.
Adrenergic CholmNpc
..
:
.
rhe autonomic nervous system is involuntary and innervates ca rdiJc and smoo th muscle, and some glands; the somatic nervous system innen'i:ltes skeletal mu~clp . Autonomic pathways arc controlled mainly by the hypoth(llamLis. You
must
be
f
\-vith
the
(' lpyriyht
© 2007 Examkrackers, Inc.
~\~----\\"'-. ----
ffector
}-~-"",,---< Utcctor
0;-,-·----< Parasympathetic
rwurotranSfl1itters,
<1cetylcholine, epinephrine, and norepinephrine. You should relate acetylcholine to the SOnlC1tic and pUr
~
Spinal
cord
Motor Pathways
ffector
94
MeAT BIOLOGY
Sympathetic
Parasympathetic
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Postganglionic N
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,
,
:;'
\
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The Autonomic Nervous System Figure 4·15
Copyright
(t~
2007 Exarnkrackers, Inc.
LECTUR E 4 : THE EUKARYOTIC CELL; THF NEfNOUS SYSTEM
4.15 The Central Nervous System The CNS consists of some of the spinal cord, the lower brain, and all of the higher brain. Although the spinal cord acts mainly as a conduit for nerves to reach the brain, it does possess lilnited integrating functions such as walking reflexes, leg stiffening, and limb withdrawal from pain. (See Figure 4-14 for the reflex arc.) The lower brain consists of the medulla, pons, 11'1CSencephalon, hypothalamus, t h alamus, cerebellum, and basal ganglia . It integrates subconscious activities such as the respiratory system, arterial pressure, salivation, emotions, and reaction to pain and pleasure.
Cerebrum (Cerebral cortex)
The higher brain or cortical brain consists of the cerebrum or cerebral cortex. The cerebral cortex is incapable of functioning without the lower brain. It acts to store memories and process thoughts.
Cerebellum
Medulla
The Brain Figure 4-16
Central Nervous System
/~ Spinal Cord
Peripheral Nervous System
/~ Autonomic
Somatic·
/~
Sensory
/~
Motor Sensory
Motor
/~
Sympathetic
Neural Salty
Parasympathetic
You have to know the structure of tile nervous system . especia ll y the difference between sympathetic and parasympathetic. Sympathetic is for fight or flight responses. It activates your heart and major skeletal muscles. dilates your pupils for night hunting, redirects blood from your digestive and excretory systems. etc ... Parasympathetic is to rest and digest. It deactivates all of the above and activates your intestines and excretory system. Caveat: Keep in
mind that fight or flight is a memorizing technique and does not always make the correct prediction . The only way to be sure of the effects of sympathetic or parasympathetic is to memorize them. This is Impractical and unnecessary for the MeAT.
Copyriqh t (g 2007 Exa mk rc)cKe rs, Inc.
95
96
MeAT
BIOLOGY
4.16 Sensory Receptors Although the somatic sensory neurons transfer signals from the external environment to the brain, they are incapable of distinguishing between different types of stimuli, and are not designed to be the initial receptors of such signals. Instead, the body contains 5 types of sensory receptors: Very little knowledge concerning the sensory rece ptors is required for the
MeAT. However, you very basic anatomy Also rea lize that transduce physical signa ls.
shou ld know some of the eye and ear. sensory receptors stimulus to neural
1.
rnechanoreceptors for touch;
2.
thennoreceptors for temperature change;
3.
nociceptors for pain;
4.
electromagnetic receptors for light; and
5.
chemoreceptors for taste, smell and blood chemistry.
Each receptor responds strongly to its own type of stimulus and weakly or not at all to other types of stimuli. Each type of receptor has its own neural pathway and terluination point in the central nervous system which resu lts in the varlous sensations.
4.17 The Eye For the MeAT, you should know the basic anatomy of the eye (Fig. 4-17), and understand the function of a few of its parts. A good way to remember this is to follow the path of light as it enters the eye. The eye is more likely to show up on a physics passage. You Sllould understand the lens of the eye as a converging lens, and understand that flattening the eye by relaxing the cilia ry muscles makes the lens less powertul. Making the lens less powertul moves the focal point away from the lens.
Light reflects off an object in the external environment and first strikes the eye on the cornea. (The light first strikes a very thin, protective layer known as the corneal epithelium) The cornea is nonvascular and made largely from collagen. It is clear with a refractive index of about 1.4, which means that the most bending of light actually occurs at the interface of the air and the cornea and not at the lens. From the cornea, the light enters the anterior cavity, which is filled w ith aqueous humor. Aqueous humor is fanned by the ciliary processes and leaks out the ca/1al of Schlemm. Blockage of the canal of Schlemm increases intraocular pressure resulting in one form of glaucoma and possibly blindness. From the anterior cavity, light enters the lens . The lens would have a spherical shape, but stiff suspensory ligaments tug on it and tend to flatten it. These ligaments are connected to the ciliary mus cle . The Ciliary muscle circles the lens. When the Ciliary muscle contracts, the opening of the circle decreases allowing the lens to become more like a sphere and bringing its focal point closer to the lens; when the muscle relaxes, the lens flattens increasing the focal distance. The elasticity of the lens declines with age making it difficult to focus on nearby objects as one gets older. The eye system just described focuses light through the gel-like vitreous humor and onto the retina. Since the eye acts as a converging lens, and the object is outside the focal distance, the image on the retina is real and inverted. (See Physics Lecture 8 for more on lenses.) The retin a covers the inside of the back (distal portion) of the eye. It contains light sensitive cells called rods and cones. These cells are named for their cDaracteristic shapes. The tips of these cells contain light sensitive photochemicals called pigments that go through a chemical change when one of their electrons is struck by a single photon. The p igment in rod cells is called rhodopsin. Rhodopsin is made of a protein bound to a prosthetic group called retinal which is derived from vitamin A. The photon isomerizes retinal causing the membrane of the rod cell to become less permeable to sodium ions and hyperpolarize. The hyperpolarization is transduced into a neural action potential and the signal is sent to the brain. Copyright rt-;J 2007 Examkrackers, Inc.
LECTURE 4 : THE EUKARYOTIC CELL; THE NERVOUS SYSTEM • 97
Rods sen se all photons w ith wavelengths in the visible sp ectrum (390 nm to 700 nm). Thus rod s crum ot distinguish colors. The re are tlu'ee types of con es, each wi th a d ifferent p igmen t that is stimulated by a slightl y d ifferen t spectrum of wavelengths. Thus cones distinguish colors. Vitamin A is a precursor to a ll the pigmen ts in rod s and cones.
Remember that cones distinguish colors and rods don't.
The foven is a small p oint on the retina con tainin g mostly cones. The fo vea marks the point on the retina where vision is most 3Cllte. One oth er feature of the eye with which you shou ld be fa miliar is the iris (Greek: irid:colored circl e). The iris is the colored portion of the eye that creates the opening ca lled the pupil. The iris is made from circular an d radial muscles. In a dark en virorunent, the sympathetic ner vous system contracts the iris d ilating the pupil and a llowing more light to e nte r Anterio r cav ity the eye. In a bright en vironment, the (agueous h umor) p a rasympa the tic n ervo us system contracts the circular muscles of th e iri s constricti ng the pupil and screening out Lightlight.
Vitreou s cbamber (vitreous bumor)
Cornea Optic nerve Light
Dark
Na+ channels - -'
Na+ channe ls closed
open
In acti ve ~
rhodopsin
Sclera
The Eye Figure 4-17
R
Act ive rhodopsin
-
(J0
J
Glutamate rclcllScd
~., ...
Bipolarccll either ~ . 0:;,' inhibited or excited (depending on glutamate receptors)
:J
Rod cell depolarized
R rod cel l C cone cell B bipolar cell
Rod cell hyperpolarized optic nerve fibe rs
LIGHT
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G ga nglion cell H horizontal cell A amacrine cel!
98
MeAT BIOLOGY
4.18 The Ear Like the eye, you should know the basic parts of the ear (Fig. 4-18) and the function s of th ese p arts. The ear is divided ioto three parts: 1.
the outer ear;
2.
the middle ear and;
3.
the inner ear.
Following the path of a sound wave through the ear can be helpful in rememberiog the parts. The allricle or pil/na is the s kin and cartilage flap that is commonly ca lled the ear. The auricle functions to direct the sound wave into the extenrn! auditory calla/. The external auditory canal carries the wave to the tympanic membrane or eardrum. The tympanic membra ne begins the middle ear. The ear is also more likely to sh ow up on a physics passage concerning waves or the mechanics of the middle ear. Just be sure to memorize the few parts that are underlined. Also know that the Cochlea detects sound, while the semici rcular canals detect orientation and movement of the head.
The middle ear con tains the three small bones: 1. the malleus; 2. the incus and; 3. the stapes. These three small bones ac t as a lever system transla ting the wave to the oval wil/dow. Like any lever system, these bones change the combination of force and displacem ent from the inforce to the outforce. The displacement is actually lessened, wh ich crea tes an increase in force. In addition, the oval window is smaller than the tympanic membrane, acting to increase the pressure. (See Physics Lecture 4 for more on machines and mechanical advantage.) This increase in force is necessary beca use the wave is being transferred from the ai r in the outer ear to a more resistant fluid (the perilymph) within the inner ear. The wave in the imle r ear moves through the scala vestibuli of the cochlea to the center of the spira l, and then spirals b ack out along the scala tylllpani to the round windolV. As the wave moves through the coch lea, the alternating increase and decrease in pressure moves the vestibular membrane in and out. This movement is detected by the hair cells of the organ of Corti and transduced into ne ural signals, which are sent' to the braio. The hair cells do not actually con tain h air, bu t contain ios tead a specia lized m icrovilli called stereocilia, w hich d etect movement. Also in the inner ea r are th e semicircular canals. The semicircuiar canals are responSible for ba lance. Each canal con tains fluid and hair cells. When the body moves or the head position changes w ith respect to gravity, the momentum of the fluid is changed impacting on the hair cells, and the body senses motion. The canals are oriented a t righ t an gles to each other, in order to d etect movement in all directions.
4.19 The Nose and Mouth The sen ses of smell and tas te are called olfactory and gustatory, respec tive ly. These sen ses iovolve chemoreceptors. Different chemoreceptors sense d ifferent chemicals. There are only fo ur pri.mary taste sensations: 1.
bitte r;
2.
sour;
3.
sa lty and;
4.
sweet.
All taste sensatio ns are cOlnbinations of these four.
lPyri9ht
'2007 EX(lIn~ lCkers, Inc.
LFCTURF 4: THF EUKt\RYOTIC CEi.l ; THE NERVO US SYSTEM
Middle ear Outer ear
Inner ear Semicircular Canal
[MalleuS ! Incus
Vestibulocochlear Nerve
I
Oval window
Cochlea
,j ~~"
~
,
Stape~
'~~.
~"
[
~J
:
i Pinna
R. ound window
Eardrum (Tympanic Membrane)
~
~
External Auditory Canal cochlear canal vestibular canal
cochlear nerve
tympanic canal cochlear canal tectorial
Cross-section of Cochlea cochlear nerve
basilar rn"rnlbra~ Hair cell
Organ of Corti
Structure of the Ear Figure 4-18
. 99
93. Reflex arcs:
Questions 89 through 96 are NOT based on a descriptive passage.
A. B. C.
89. Which of the following activities is controlled by the cerebellum? A. B. C.
D.
D.
Involuntary breathing movements Fine muscular movements during a dance routine Contraction of the thigh muscles during the kneejerk reflex Absorption of nutrients across the microvilli of the small intestine
94. Which of the following structures is NOT part of the central nervous system?
A. B.
C.
90. If an acetylcholine antagonist were administered generally into a person. all of the following would be affected EXCEPT: A. B. C. D.
D.
a parasympathetic effector the medulla the hypothalamus the cerebral c0l1ex
95. Which paJ1 of the brain controls higher-level thought processes?
the neuroeffector synapse in the sympathetic nervous system. the neuroeffector synapse in the parasympathetic nervous system the neuromuscular junction in the somatic nervous system. the ganglionic synapse in the sympathetic nervous system.
A. B. C. D.
the thalamus the cerebellum the cerebrum the medulla
96. A cook touches a hot stove and involuntarily withdraws his hand before he feels pain. Which of the following would not be involved in the stimUlus-response pathway described?
91. Which of the following occurs as a result of parasympathetic stimulation? A. B. C. D.
involve motor neurons exiting the spinal cord dorsally. require fine control by the cerebral cortex. always occur independently of the central nervous system. often involve inhibition as well as excitation of muscle groups.
A. B. C. D.
Vasodilation of the arteries leading to the kidneys Increased rate of heart contraction. Piloerection of the hair cells of the skin. Contraction of the abdominal muscles during exercise.
a a a a
neuron in the cerebellum neuron in the spinal cord motor neuron sensory neuron
92. Pressure waves in the air are converted to neural signals at the: A. B. C. D.
retina tympanic membrane cochlea semicircular canals
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100
STOP.
The Endocrine System
5.1
Hormone Chemistry
The neurotransmitters and local mediators discussed in Biology Lecture 4 are often referred to as local hormones. General hormones are the hormones released by the endocrine system. They are referred to as 'general' because they are released into the body fluids, often the blood, and may affect many cell types in a tissue, and multiple tissues in the body. Although the following section concentrates on general hormones, the chemistry described is accurate for local hormones as well .
The endocrine glands differ from exocrine glands in the following manner. Exocrine glands release enzymes to the external environment through ducts. Exocrine glands include sudoriferous (sweat), sebaceous (oil), mucous, and digestive glands. Endocrine glands release hormones directly into body fluids. For instance, the pancreas acts as both an exocrine gland, releasing digestive enzymes through the pancreatic duct, and an endocrine gland releasing insulin and glucagon directly into the blood. (See Biology Lecture 6 for more or; the pancreas.) The effects of the endocrine system tend to be slower, less direct, and longer lasting than those of the nervous system. Endocrine hormones may take anywhere from seconds to days to produce their effects. They do not move directly to their target tissue, but are released into the general circulation. All hormones act by binding to proteins called receptors. Each receptor is highly specific for its hormone. One method of hormone regulation occurs by the reduction or increase of these receptors in the presence of high or low concentrations of the hormone. Some hormones have receptors on virtually all cells, while other hormones have receptors only on specific tissues. Very low concentrations of hormones in the blood have significant effects on the body. In general, the effects of the endocrine system are to alter metabolic activities, regulate growth and development, and gUide reproduction. The endocrine system works in conjunction with the nervous system. Many endocrine glands are stimulated by neurons to secrete their hormones. Hormones exist in three basic chemistry types:
1.
peptide hormones;
2.
steroid hormones and;
3.
tyrosine derivatives.
Always remember that hormones need a receptor, either on the membrane or inside the cell. Also, when comparing the endocrine system with the nervous system remember that the endocrine system is slow, indirect, and long lasting.
102
MCATBloloGY Peptide hormones are derived from p eptides. They may be large or small, and often include carbohydrate portions. All peptide hormones are manufactured in the rough ER, typically as a preprohormone that is larger than the active hormone. The preprohormone is cleaved in the ER lumen to become a prohormone, and transported to the Golgi apparatus. In the Golgi, the prohormone is cleaved and sometimes modified w ith carbohydrates to its final form . The Golgi packages the hormone into secretory vesicles" and. . upon stimulation by anothe r hormone or a nervous signal. . the cell releases the vesicles via exocytosis. Since they are peptide derivatives, peptide hormones are water soluble, and thus move freely through the blood, but have"difficulty diffusing through the cell membrane of the effector. (The effector is the target cell of the hormone; the cell that the hormone is meant to affec!.) Instead of diffusing through the membrane, peptide hormones attach to a membrane-bound receptor. Once bound by a hormone, the receptor may act in several ways. The receptor may itself act as an ion channel increasing membrane permeability to a specific ion, or the receptor may activate or deactiva te other intrinsic membrane proteins also acting as ion channels. Another effect of the hormone binding to the receptor may be to activate an intracellular second messenger such as cAMP, cGMP, or calmodulin. These chemicals are called second messengers because the hormone is the original, or first messenger, to the cell. The second messenger activates or deactivates enzymes and / or ion channels and often creates a 'cascade' of chemical reactions that amplifies the effect of the hormone. A cascade is one way that a small concentration of hormone can have a Significant effect .
• Th e peptide hormones that yo u must know for the MeAT are: You should know which hormones are steroids, which are tyrosines, and which are peptides. Then you should know where and how each of these types of hormones reacts. This isn't really so tough. First. steroid hormones come on ly from the adrenal cortex, the gonads, or the placenta. Secon d, tyrosines are the thyroid horm ones and the catecholamines (the adrenal medulla hormones). The rest of the hormones disc uss~ d in this book are peptide hormones. Since steroids are lipids, they diffuse through the membrane and act in th e nucleus. Since peptides are proteins, they ca n't diffuse through the membrane, so they bind to receptors on the membrane and act through a second messenger. The tyrosines are split: thyroids diffuse illio Ihe nucleos and catecholamines act on receptors al the membrane.
1.
the anterior pituitary hormones: FSH, LH, ACTH, hGH, TSH, Pro1actin;...
2.
the posterior pituitary hormones: ADD and oxytocin,-
3.
the parathyroid hormone PTH,-
4.
the' pancreatic hormones: glucagon and insulin.
The specifics of these hormones w ill be discussed later in this lecture. Steroid hormones are derived from and are often chemically similar to cholesterol. They are formed in a series of steps taking place mainly in the smooth endoplasmic reticulum and the mitochondria. Since they are lipids, steroids typically require a protein transport molecule in order to dissolve into the blood stream. (Usually, a fraction of the steroid concentration is bound to a transport molecule and a fraction is freeform in the blood.) Being lipid soluble, steroids diffuse through the cell membrane of their effector. Once inside the cell, they combine with a receptor in the cytosol. The receptor transports the steroid into the nucleus, and the steroid acts at the transcription level. Thus, the typical effect of a steroid hormone is to increase certain membrane or cellular proteins within the effector. The important steroid hormones for the MCAT are: 1. the glucocorticoids and mineral corticoids of the adrenal cortex: cortisol and aldosterone; 2. the gonadal hormones: estrogen, progesterone, testosterone. (Estrogen and progesterone are also produced by the placenta.) The specifics of these steroids will be discussed later in this lecture. The tyrosine derivatives are: the thyroid hormones: T 3 (triiodothyronine contains 3 iodine atoms) and T, (thyroxine contains 4 iodine atoms) and; the Copyright © 2007 Examkrackers, Inc.
LECTURE
catecholamines fonned in the adrenal medulla: epinephrine and norepinephrine. All tyrosine derivative hormones are formed by enzymes in the cytosol or on the rough ER. Thyroid hormones are lipid soluble and must be carried in the blood by plasma protein carriers. They are slowly released to their target tissues and bind to receptors inside the nucleus. Their high affinity to their binding proteins in the plasma and in the nucleus create a latent period in their response and increase the duration of the effect of thyroid hormones. Thyroid hormones increase the transcription of large numbers of genes in nearly all cells of the body. Epinephrine and norepinephrine are water soluble and dissolve in the blood. They bind to receptors on the target tissue and act mainly through the second messenger cAMP. The specifics of the tyrosine derivative hormones will be discussed later in this lecture.
5.2
Negative Feedback
Endocrine glands tend to over secrete their hormones. Typically, some aspect of their effect on the target tissue will inhibit this secretion. This is an example of negative feedback (discussed in Biology Lecture 1). An important aspect to understand about negative feedback in endocrine glands is that the control point of the feedback is the conduct of the effector, not the concentration of hormone. In other words, the gland lags behind the effector. For instance, high insulin levels do not typically create low blood glucose. Instead, high insulin levels are caused by high blood glucose, and low blood glucose would cause high blood glucagon levels. So if an MeAT question indicates that a patient has high blood glucose and asks whether high levels or insulin or high levels of glucagon would be expected, the correct answer is the hormone that is responding to the condition, not creating it; or insulin. There will be a negative feedback question on the MCAT. See if you get the idea . If ADB holds water in the body decreasing urine output and increasing blond pressure, does d person with high blood pressure (holding water) have a high ADH blood level or a low ADB blood level? If you said high ADI-l, thel1 you reasoned that the ADH created the high blood pressure. WRONG! If you said low ADH, then you reasoned that the ADH output responded to the body. CORRECT'
Hm·v about another? A secondary effect of aldosterone is to increase blood pressure. Would expected aldosteronL' levels be hjgh or low in a person w ith low hlood pressure?
The nnswer: Since flldostcronc inrrcflscs hlood pn.'ssufC, and the body tries to bring pressure back to normal, the adrenJI cortex should n-:-lt'Clse more a ldosterone into the blood. Expected aldosterone levels would be higher than normal.
blood
•
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5:
THE ENDOCRINE SYSTEM .
103
101. Which of the following is true of all steroids?
Questions 97 through 104 are NOT based on a descriptive passage.
A. B.
97. Aldosterone exerts its effects on target cells by: A. B. C. D.
C.
binding to a receptor at the cell surface, setting off a second-messenger cascade. diffusing into adrenal cortical cells. where it influences transcription of certain DNA sequences. flowing across the synapse, where it binds and initiates an action potential. entering into target cells, where it increases the rate of production of sodium-potassium pump proteins.
D.
102. The pancreas is a unique organ because it has both exocrine and endocrine function. The exocrine function of the pancreas releases: A. B. C. D.
98. A patient develops an abdominal tumor resulting in the secretion of large quantities of aldosterone into the bloodstream. Which of the following will most likely occur? A. B. C. D.
Levels of renin secreted by the kidney will increase. Levels of oxytocin secreted by the pituitary will increase. Levels of aldosterone secreted by the adrenal cortex will decrease. Levels of aldosterone secreted by the tumor will decrease.
They They They They
digestive enzymes straight into the blood. digestive enzymes through a duct. hormones straight into the blood hormones through a duct
103. Most steroid hormones regulate enzymatic activity at the level of:
A. B. C.
D.
replication transcription translation the reaction
104. Which of the following side-effects might be experienced by a patient who is administered a dose of thyroxine?
99. Which of the following is true for all endocrine hormones? A. B. C. D.
The target cells of any steroid include every cell in the body. Steroids bind to receptor proteins on the membrane of their target cells. Steroids are synthesized on the rough endoplasmic reticulum. Steroids are lipid soluble.
A. B. C. D.
act through a second messenger system. bind to a protein receptor. dissolve in the blood. are derived from a protein precursor.
An increase in endogenous TSH production A decrease in endogenous TSH production An increase in endogenous thyroxine production A decrease in endogenous parathyroid hormone production
100. All of the following act as second messengers for hormones EXCEPT: A. B. C. D.
cyclic AMP calmodulin acetylcholine cyclic GMP
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104
STOP.
LECTURE
5.3
Specific Hormones and Their Functions
Memorization of several major hormones, their glands, and their target tissues is re-
quired for the MCAT. As a memory aid, you should group hormones according to -the gland that secretes them. A given gland produces one of either steroids, peptides, or tyrosine derivatives, but not two. (The adrenal glands are really two glands. The cortex produces steroids; the medulla 'produces catecholamines. The thyroid is a true exception. The thyroid secretes T3 and T4, which are tyrosine derivatives, and calcitonin, which is a peptide.) We will start by discussing the hormones of the anterior pituitary.
5.4
Anterior Pituitary
The anterior pituitary (Fig. 5-1) (also called the adenohypophysis) is located in the brain beneath the hypothalamus. The hypothalamus controls the release of the anterior pituitary hormones with releasing and inhibitory hormones of its own. These releasing and inhibitory hormones are carried to the capillary bed of the anterior pituitary by small blood vessels. The release of the releasing and inhibitory hormones is, in tum, controlled by nervous signals throughout the nervous system. The anterior pituitary releases six major hormones and several minor hormones. All
of these are peptide hormones. For the MCAT you should be familiar with the six major hormones, their target tissues, and their ftmctions. The hormones are: l.
human growth hormone (hGH);
2.
adrenocorticotropin (ACTH);
3.
thyroid-stimulating hormone (I SH);
4.
follicle-stimulating hormone (FSH);
5.
leutinizing hormone (LH) and;
6.
prolactin.
hG H Human growth hormone (hGH) (also called somatotropin), a peptide, stimulates growth in almost all cells of the body. All other hormones of the anterior pituitary .have specific target tissues. hGH stimulates growth by increasing episodes of mito,sis, increasing cell size, increasing the rate of protein synthesis, mobilizing fat stores, increasing the use of fatty acids for energy, and decreasing the use of glucose. T he effect on proteins by hGH is accomplished by increasing amino acid transport across the cell membrane, increasing translation and transcription, and decreasing the breakdown of protein and amino acids.
ACTH Adrenocorticotropic hormone (ACTH), a peptide, stimulates the adrenal cortex to release glucocorticoids via the second messenger system using cAMP. Release of ACTH is stimulated by many types of stress. Glucocorticoids are stress hormones.
(See below for the effects of the adrenal cortical hormones.)
TSH Thyroid-stimulating hormone (TSH) (also called thyrotropin), a peptide, stimulates the thyroid to release T, and T, via the second messenger system using cAMP. Among other effects on the thyroid, TSH increases thyroid cell size, number, and the rate of secretion ofT, and T,. It is important to note that T3 and T, concentrations 11ave a negative feedback effect on TSH release, both at the anterior pituitary and the hypothalamus. (See below for effects of T3 and T,.)
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5: THE
ENDOCR INE SYSTEM
.
10S
106
MeAT BIOLOGY
FSH and LH (These pep tides are discussed in this lecture under reproduction.)
Prolactin Prolactin, a peptide, promotes lactation (milk production) by the breasts. The reason that milk is not normally produced before birth is due to the inhibitory effects of milk production by progesterone and estrogen. Although the hypothalamus has a stimulatory effect on the release of all other anterior pituitary hormones, it mainly inhibits the release of prolactin. The act of suckling, which stimulates the hypothalamus to stimulate the anterior pituitary to release prolactin, inhibits the menstrual cycle. It is not known whether or not this is directly due to prolactin. The milk production effect of prolactin should be distinguished from the milk ejection effect of oxytocin.
5.5
Posterior Pituitary
The posterior pituitary is also called the neurohypophysis because it is composed mainly of support tissue for nerve endings extending from the hypothalamus. The hormones oxytocin and ADH are synthesized in the neural cell bodies of the hypothalamus, and transported down axons to the posterior pituitary where they are released into the blood. Both oxytocin and ADH are small polypeptides.
Oxytocin Oxytocin is a small peptide hormone that increases uterine contractions during pregnancy and ca uses milk to be ejected from the breasts.
ADH Antidiuretic hormone (ADH) (also called vasopressin) is a small peptide hormone which ca llses the collecting ducts of the kidney to become permeable to w ater reducing the amou nt of urine and concentrating the urine. Since fluid is reabsorbed, ADH also incr.eases blood pressure. Coffee and beer are ADH blockers that increase urine volume.
Hypothalamus
Anterior pituitary l.FSH 2. J.l-{
3. ACTH 4.hGH 5. TSH 6. Prolactin
Pituitary Figure 5-1
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LECTURE 5 : THE EN DOCRINE SYSTEM'
5.6
107
Andrenal Cortex
The adrenal glands (Fig. 5-2) are located on top of the kidneys. They are generally separated into the adrenal cortex and the adrenal medulla. The adrenal cortex is the outside portion of the gland. The cortex secretes only steroid hormones. There are two types of steroids secreted by the cortex: mineral corticoids and glucocorticoids. (The cortex also secretes a small amount of sex hormones, significant in the female but not the male.) Mineral corticoids affect the electrolyte balance in the blood stream; glucocorticoids increase blood glucose concentration and have an even greater effect on fat and protein metabolism. About 30 corticoids have been isolated from the cortex, but the major mineral corticoid is aldosterone, and the Inajar glucocorticoid is cortisol.
Adrenal cortex
Aldosterone
1. Aldosterone 2. Cortisol
Aldosterone, a steroid, is a mineral corticoid that acts in the distal convoluted hlbule and the collecting duct to increase Na· and Cl- reabsorption and K+ and H+ secretion. It creates a net gain in particles ll1 the plasma, which results in an eventual increase in blood pressure. Aldosterone has the same effect, but to a lesser extent on the sweat glands, salivary glands, and intestines.
Adrenal medulla 1. Epinephrine 2. Norepinephrine
For the MCAT, the main effect of aldosterone is the Na· reabsorption and K · secretion in the collecting tubule of the kidney. The increase in blood pressure is a secondary effect.
Cortisol Cortisol, a steroid, is a glucocorticoid that increases blood glucose levels by stimulating gluconeogenesis in the liver. (Gluconeogenesis is the creation of glucose and glycogen, mainly in the liver, from amino acids, glycerol, and/or lactic acid.) Cortisol also degrades adipose tissue to fatty acids to be used for cellular energy. In addition, cortisol causes a moderate decrease in the use of glucose by the cells. Cortisol causes the degradation of nonhepatic proteins, a decrease of nonhepatic amino acids and a corresponding increase in liv~r and plasma proteins and amino acids. Cortisol is a stress hormone. The benefit of excess cortisol under stressful situations is not fully understood. One explanation may include anti-inflammatory properties possessed by cortisol. Cortisol also diminishes the capacity of the immune system to fight infection.
Catecholamines The catecholamines are the tyrosine derivatives synthesized in the adrenal medulla: epinephrine and norepinephrine (also called adrenaline and noradrenaliue). The effects of epinephrine and norepinephrine on the target tissues are similar to their effects in the sympathetic nervous system but they last much longer. Epinephrine and norepinephrine are vasoconstrictors (they constrict blood vessels) of most ineternal organs and skin, but are vasodilators of skeletal muscle (they increase blood flow); this is consistent with the 'fight-or-flight' response of these hormones. Because of their 'fight or flight' response, the catecholarnines are also considered stress hormones.
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Figure 5-2
108
MCAT B,OLOGY
5.7
Thyroid
The thyroid hormones are triiodothyronine (13)' thyroxine (T4), and calcitonin. The thyroid (Fig 5-3) is located along the trachea just in front of the larynx.
T3 and T4 T3 and T4 are very similar in effect, and no distinction will be made on an MeAT qu estion unless it is thoroughly explained in a passage. T3 contains three iodine atoms, and T4 contains four. Both hormones are lipid soluble tyrosine derivatives that diffuse through the lipid bilayer and act in the nucleus of the cells of their effector. Their general effect is to increase the basal metabolic rate (the resting metabolic rate). Thyroid hormone secretion is regulated by TSH.
Calcitonin Calcitonin is a large peptide hormone released by the thyroid gland. Calcitonin slightly decreases blood calcium by decreasing osteoclast activity and number. Calcium levels can be effectively controlled in humans in the absence of calcitonin.
5.8
Pancreas (Islets of Langerhans)
The pancreas (Fig. 5-3) acts as both an endocrine and an exocrine gland. For the MCAT, the two important endocrine hormones released into the blood by the pancreas are the peptide hormones insulin and glucagon. Somatostatin, not likely to be seen on the MCAT, is released by the o.cells of the pancreas. Somatostatin inhibits both insulin and glucagon. The role of somatostatin may be to extend the period of time over which nutrients are absorbed.
Insulin Insulin, a peptide hormone, is released by the ~cells of the pancreas. It is associated with energy abundance in the form of high energy nutrients in the blood. Insulin is released when blood levels of carbohydrates or proteins are high. It affects carbohydrate, fat, and protein metabolism. In the presence of insulin, carbohydrates are stored as glycogen in the liver and muscles, fat is s tored in adipOse tissue, and amino acids are taken up by the cells of the body and made into proteins. The net effect of insulin is to lower blood glucose levels. Insulin binds to a membrane receptor beghming a cascade of reactions inside the cell. Except for neurons in the brain and a few other cells which are not affected by insulin, the cells of the body become highly permeable to glucose upon the binding of insulin. The insulin receptor itself is not a carrier for glucose. The permeability of the
Parathyroid 1. PTH
Thyroid
~
1. T 3r T4
2. Calcitonin
Pancreas 1. Insulin 2. Glucagon
Figure 5-3
membrane to amino acids is also increased. In addition, intracellular metabolic enzymes are activated and, much more slowly, even translation and transcription rates are affected.
Glucagon Glucagon, a peptide hormone, is released by the a-cells of the pancreas. Ihe effects of glucagon are nearly opposite to those of insulin. Glucagon stimulates glycogenolysis (the breakdown of glycogen), and gluconeogenesis in the liver. It acts via the second messenger system of cAMP. In higher concentrations, glucagon breaks down adipose tissue ,yin creasing the fatty acid level in the blood. Ihe net effect of glucagons is to raise blood glucose levels.
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LECTURE 5: THE ENDOC RINE SYSTEM .
5.9
Pa rathyroid
There are four small parathyroid glands (Fig. 5-3) attached to the back of the thyroid. The parathyroid glands release parathyroid honnone.
PTH Parathyroid honnone (PTH), a peptide, increases blood calcium. It increases osteocyte ab sorption of calcium and phosphate from the bone and stimulates proliferation of osteoclasts. PTH increases renal calcium reabsorption and renal phosphate excretion. It increases calcium and phosphate uptake from the gut by increasing renal production of the steroid, 1,25 dihydroxycholecnlciferol (DOHCC ), derived from vitamin D. PTH secretion is regulated by the calcium ion plasma concentration, and the parathyroid glands shrink or grow accordingly.
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109
109. Osteoporosis is an absolute decrease in bone tissue mass, especially trabecular bone. All of the following might be contributory factors to the disease EXCEPT:
Questions 105 through 112 are NOT based on a descriptive passage.
A. 105. Sympathetic stimulation results in responses most similar to release of which of the following hormones? A. B. C. D.
B. C. D.
Insulin Acetylcholine Epinephrine Aldosterone
110. All of the following hormones are produced by the anterior pituitary EXCEPT: A. B. C. D.
106. When compared with the actions of the nervous system. those of the endocrine system are: A.
increased sensitivity to endogenous parathyroid hormone defective intestinal calcium absorption menopause abnormally high blood levels of calcitonin
quicker in responding to changes, and longer-Iast-
thyroxine growth hormone prolactin leutinizing hormone
mg.
B. C. D.
quicker in responding to changes, and shorter-lasting. slower in responding to changes, and longer-lasting. slower in responding to changes, and shorter-last-
111. Which of the following hormonal and physiologic effects of stress would NOT be expected in a marathoner in the last mile of a marathon? A. B. C. D.
ing.
107. Insulin shock occurs when a patient with diabetes selfadministers too much insulin. Typical symptoms are extreme nervousness, trembling, sweating, and ultimately loss of consciousness. The physiological effects of insulin shock most likely include: A. B. C. D.
112. Parathyroid hormone is an important hormone in the control of blood calcium ion levels. Parathyroid hormone directly impacts:
a pronounced increase in gluconeogenesis by the liver
I. bone density II. renal calcium reabsorption III. blood calcium concentration
a rise in blood fatty acid levels leading to atherosclerosis a dramatic rise in blood pressure dangerously low blood glucose levels
A. B. C. D.
108. Vasopressin, a hormone involved in water balance, is produced in the: A. B. C. D.
Increased glucagon secretion Increased heart rate Decreased ACTH secretion Decreased blood flow to the small intestine
I only I and II only I and III only I, II and III
hypothalamus. posterior pituitary. anterior pituitary. kidney.
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110
STOP.
,
.
LECTURE 5 : THE E NDOCRINE SYSTEM .
5.10 Reproduction Except for FSH, LH, HCG, and inhibin, which are peptides, the reproductive hormones discussed below are steroids released from the testes, ovaries and placenta.
5.11 The Male Reproductive System You should know the basic anatomy of the male and female reproductive systems (Fig. 5-4 and 5-9). The male gonads are called the testes. Production of sperm (Fig. 5-5 and 5-6) occurs in the seminiferous tubules of the testes. Spermatogonia located in the seminiferous tubules arise from epithelial tissue to become spermatocytes, spermatids, and then spermatozoa. Serloli cells stimulated by FSH surround and nurture the spermatocyte and spermatids. Leydig celis, located in the interstitium between the tubules, release testosterone when stimulated by LH. Testosterone is the primary androgen (male sex hormone), and stimulates the germ cells to become sp erm. Testosterone is also responsible for the development of secondary sex characteristics such as pubic hair, enlargement of the larynx, and growth of the penis and seminal vesicles. While testosterone helps to initiate the growth spurt at puberty, it also stimulates closure of the epiphyses of the long bones, ending growth in stature. Sertoli cells secrete inhibin, a peptide hormone (actually a glycoprotein) which acts on the pituitary gland to inhibit FSH secretion.
Vas Deferens
Urethra Penis Prostate Gland Cowper's Gland Ep ididymis Testis
Male Anatomy Figure 5-4
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Scrotum
11 1
112
MeAT BIOLOGY
The spermatid has the characteristics of a typical cell. However, as it becomes a spermatozoon it loses its cytopla sm and forms the head, midpiece, and tail shown in Figure 5- 6. The head is composed of the nuclear material and an acrosome. The acrosome contains lysosome-like enzymes for penetrating the egg during fertilization. The midpiece contains many mitochondria to provide energy for movement of the tail. Only the nuclear portion of the sperm enters the egg. Once freed into the tubule lumen, the spermatozoon is carried to the epididymus to mature. Upon ejaculation, spermatozoa are propelled through the vas deferens into the urethra and out of the penis. Semen is the complete mixture of spermatozoa and fluid that leaves the penis upon ejaculation. Semen is composed of fluid from the s eminal vesicles, the prostate, and the bulbourethral glands (also called Cowper's glands). Spermatozoa become activated for fertilization in a process called capacitation, which takes place in the vagina.
Tubule lumen
Tail
____-Spermatozoon
~_
Sertoli cell
- - Sertoli cell nucleus
Spermatozoon Figure 5-6
~ Spermatogonium Basement
(!:r-- Leydig cell Cross-section of a Portion of a Seminiferous Tubule Figure 5-5 Copyright (6) 2007 Examkracke rs, Inc.
LECTURE
5:
THE ENDOCRINE SYSTEM •
5.12 The Femal e Reproductive System Oogenesis begins in the ovaries of the fetus. All the eggs of the female are arrested as primary oocytes at birth. At puberty, FSH stimulates the growth of granulosa cells around the primary oocyte (Fig. 5-7). The granulosa cells secrete a viscous substance around the egg called the zona pelludda. The structure at this stage is called a p,-il11ary follicle. Next, theca cells differen tiate fro m the interstitial tissue and grow around the follicle to form a secondary follicle. Upon stimulation by LH, theca cells secrete androgen, which is converted to estradiol (a type of estrogen) by the granulosa cells in the presence of FSH and secreted into the blood. The estradiol is a steroid hormone that prepares the uterine wall for pregnancy. The follicle grows and bulges from the ovary. Typically, estradiol inhibits LH secretion by the anterior pituitary. However, just before ovulation (the bursting of the follicle), the estradiol level rises rapidly, actually causing a dramatic increase in LH secretion. This increase is called the luteal surge. The luteal surge results from a positive feedback loop of rising estrogen levels which increase LH levels, which increase estrogen. The luteal surge causes the follicle to burst, releasing the egg (now a secondary oocyte) into the body cavity. The egg is swept into the Fallopian (uterine) tube or oviduct by the fimbriae. The remaining portion of the follicle is left behind to become the corpus luteum. The corpus luteum secretes estradiol and progesterone throughout p regnancy, or, in the case of no pregnancy, for about 2 weeks lmtiJ the corpus luteum degrades into the corpus albicans.
o •
Corpus luteum degenerates to corpus albicans if no fertilization of egg
0
Primary oocyte ~ within follicle
Corpus luteum develops from
..... ® .", ~.
remnants of
follicle
~
- -. ,
r
- ..;
~ ,~,0~: .. '.
i;
.(
0
':
~.~!:~ e
t
~ .:
', .... ,~~-
'"
.~~J
-',;,
'~ _ !:I: ~ -:...~... .:J- :~-'
Ovary eSecondary (Graffian) follicle
( Secondary oocyte with
Follicle ruptures, releaSing secondary oocyte
corona radiata
Ovulation Figure 5-7 Copyright © 2007 Examkrackers, Inc.
Growing follicle consists of theca cells surro unding granulosa cells which surround the zona pellucida ------and the oocyte
113
114
MeAT BIOLOGY
The cycle just described repeats itself approximately every 28 days after puberty unless pregnancy occurs. This cycle is called the menstrual cycle (Fig. 5-8). With each menstrual cycle, several primordial oocytes may begin the process, but, normally, only one completes the development to ovulation. The cycle is divided into three phases: 1..
the follicular phase, which begins with the development of the follicle and ends at ovulation;
2.
the luteal phase, which begins with ovulation and ends w ith the degeneration of the corpus luteum into the corpus albicans;
3.
flow, which is the shedding of the uterine lining lasting approximately 5 days.
·
--..-!-._ _' \....._ _ _ _-"":ir'
FSH_ _
·
LH~ --~----L-H-s-ur~~--------------~----__
progeste~r~o;.n;.;,e;;..._..;.------y ~~ Ovu!a tion
"'-......
_"'-'--
~
:~
Corpus Luteum
Mature Follicle
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 1
Menstruation : Proliferative: Secretory Phase . Phase Ovulation
2
3
4
5
enstruation
The Menstrual Cycle Figure 5-8
Copyright © 2007 Examk rackers, Inc.
LECTURE
5.13 Fertilization and Embryology Once in the Fallopian tube, the egg is swept toward the uterus by cilia. Fertilization normally takes place in the Fallopian tubes. The enzymes of the acrosome in the sperm are released upon contact with the egg, and d igest a path for the sperm through the granulosa cells and the zona pellucida. The cell membranes of the sperm head and the oocyte fuse upon contact, and the sperm nucleus enters the cytoplasm of the oocyte. The entry of the sperm causes the cortical renction, which prevents other sperms from fertilizing the same egg. Now the oocyte goes through the second meiotic division to become an ovum and releases a second polar body. Fertilization occurs when the nuclei of the ovum and sperm fuse to form the zygote.
Cleavage begins while the zygote is still in the Fallopian tube. The zygote goes through many cycles of mitosis; when the zygote is comprised of eight or more cells, it is called a morula. The embryo at this stage does not grow during cleavage. The first eight cells formed by cleavage are equi valent in size and shape and are said to be totipotent, meaning that they have the potential to express any of their genes. Anyone of these eight cells at this stage could produce a complete individual. The cells of the morula continue to divide for four days forming a hollow ball filled with fluid . This fluid filled ball is called a blastocyst. It is the blastocyst that lodges in the uterus in a process called implantation on about the 5'" to 7'" day after ovulation. The blastocyst is made up of embryonic stem cells that each have the ability to develop into most of the types of cells in the hu ma n bod y. Upon implantation, the female is said to be pregnant. Upon implantation, the egg begins secreting a peptide hormone called human chorionic gonadotropin (HCG) . HCG prevents the degeneration of the corpus luteum, and maintains its secretion of estrogen and p rogesterone. HCG in the blood and urine of the mother is the first ouh" ard sign of pregnancy. A placenta is formed
Fallopian Tube Ovary
Cervix
_ - - - Uterus Urinary Bladder Rectum Urethra
Clitoris Vagina
Female Anatomy Figure 5-9
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5: T HE
ENDOCRINE SYSTEM • 115
116
MeAT B,OLOGY
from the tissue of the egg and the mother, and takes over the job of hormone secretion. The placenta reaches full development by the end of the first trimester, and begins secreting its own estrogen and progesterone while lowering its secretion of
HCG. As the embryo develops past the eight cell stage, the cells become different from each other due to cell-cell interactions. This process where a cell becomes committed to a specialized developmental path is called determination. Cells become determined to give rise to a particular tissue early on. The specializaton that occurs at the end of the development forming a specialized tissue cell is called differentiation. The fate of a cell is typically determined early on, but that same cell usually doesn't differentiate into a specialized tissue cell until much later at the end of the developmental process. Recent research has shown that the fate of even a fully differentiated cell can be altered given the proper conditions.
The formation of the gastrula occurs in the second week after fertilization in a process called gastrulation. Cells begin to slowly move about the embryo for the first time. In mammals, a primitive streak is formed, which is analogous to the blastopore in aquatic vertebrates. Cells destined to become mesoderm migrate into the primitive streak. During gastrulation, the three primary germ layers are formed: 1.
the ectoderm;
2.
the mesoderm;
3.
the endoderm.
Although there is no absolute rule for memorizing which tissues arise from which germ layer, for the MCAT certain guidelines can be followed. The ectodermal cells develop into the outer coverings of the body, such as the outer layers of skin, nails, and tooth enamel, and into the cells of the nervous system and sense organs. The endodermal cells develop into the lining of the digestive tract, and into much of the liver and pancreas. The mesoderm is the stuff that lies between the inner and outer covering of the body, the muscle, bone, and the rest. (WARNING : These are just guidelines, not absolute rules.) In the third week, the gastrula develops into a neurula in a process called neurulation. In neurulation, the notochord (made from mesoderm) induces the overlying ectoderm to thicken and form the neural plnte. The notochord eventually degenerates, while a neural tube forms from the neural plate to become the spinal cord, brain, and most of the nervous system. For the MeAT you must know that induction occurs when one cell type affects the direction of differentiation of another cell type. Part of normal cell development is programmed cell death or apoptosis. Apoptosis is essential for development of the nervous system, operation of the immune system, and destruction of tissue between fingers and toes to create normal hands and ff'pt in humans. Damaged cells .m ay lUldergo apoptosis as well. Failure to do so may result in cancer. Apoptosis is a complicated process in humans, but it is baSically regulated by protein activity as opposed to regulation at the transcription or translation level. The proteins involved in apoptosis are present but inactive in a normal healthy cell. In mammals, mitochondria play an important role in apoptosis.
Copyright © 2007 Examkrackers, Inc.
LECTURE 5: THE ENDOCRINE SYSTEM . 117
I
Gland
l-
I
Anterior pituitary
---
Hormone
I I -+
-
I
Effect
I
hGH
Growth of nearly all cells
ACTH
Stimulates adrenal cortex
FSH
Growth of follicles in female; Sperm production in male Causes ovulation; stimulates estrogen and testosterone secretion -Stimulates release of T, and T, in the thyroid
LH
-----------.-
--
I
-
- -
TSH -
Posterior pituitary Adrenal cortex
Prolactin
Promotes milk production
Oxytocin
Milk ejection and uterine contraction
ADH
wa ter absorption by the kidney; increase blood pressure
Aldosterone
Reduces Na' excretion; increases K' excretion; raises blood pressure
Cortisol
Increases blood levels of carbohydrates, proteins, and fats
-
.-
-----------_._---- -
--.- -.--.---------------.--
Adrenal medulla
-
Epinephrine _. - - -..-
--- .-.-----.-.-.~---
-
--I
Simulates sympathetic actions -
Simulates sympathetic actions
Norepinephrine
-
Thyroid .
Calcitonin
Parathyroid
Raises blood calcium
-
-
Insulin
Pancreas
- - - - - - - -----.-
I
PH
-----
-
Increases basal metabolic rate Lowers blood calcium
T" T, -
...
-
""---'-'---
--
Glucagon
Promotes entry of glucose into celis, decreasing gl ucose blood level -- -,------- _.'._----Increases gluconeogenesis, increasing glucose blood leyel
I
I
-
Ovaries -
-
-
Growth of mother sex organs; causes LH surge.
Estrogens -
-
-
Progesterone
Prepares and maintains uterus for pregnancy
Testes
Testosterone
Secondary sex characteristics; closing of epiphysea l plates
Placenta
HCG
---- . ----- ---'-- . ~-~ -~
Copyright © 2007 Examkrackers, Inc.
-
Estrogens
-Stimu lates corpus luteum to grow and release estrogen and progesterone Growlh of mol her SP\. orgdll"i; Gltlses U I sllrg('.
Progesterone
f'rep,lt'(·"
____ L __
and ll1
for prq.;n<1!1()'
l
, I I
117. The inner linings of the Fallopian tubes are covered with a layer of cilia. The purpose of this layer is to:
Questions 113 through 120 are NOT based on a descriptive passage.
A. B.
113. A drug that causes increased secretion of testosterone from the interstitial cells of a physically mature male would most likely: A. B. C. D.
C.
cause tbe testes to descend prematurely. delay the onset of puberty. cause enhanced secondary sex characteristics. decrease core body temperature.
D.
118. Which of the following endocrine glands produce testosterone? A. B. C. D.
114. During the female menstrual cycle, increasing levels of estrogen cause: A. B.
C. D.
a positive feedback response, stimulating LH secretion by the anterior pituitary. a positive feedback response, stimulating FS H secretion by the anterior pituitary. a negative feedback response, stimulating a sloughing-off of the uterine lining. a negative feedback response, stimulating decreased
A. B.
C.
115. The function of the epididymus is to:
B. C. D.
store.: spenn until they are released during ejacu lation. produce and secrete testosterone. cond uct the ovum from the ovary into the uterus. secrete FSH and LH to begin the menstrual cycle.
D.
B. C. D.
A. B. C.
thickening of the endometrial lining in preparation for implantation of the zygote. increased secretion of LH, leading to the lu teal surge and ovu1ation. degeneration of the corpus luteum in the OVilly. increased secretion of estrogen in the follicl e, leading to the flow pbase of the menstrual cycle.
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The solid ball of cells produced during cleavage i, caBed a morula. The size of the embryo remains constant throughout the cell divisions of cleavage. Cell division occurs in one portion of the egg in meroblastic cleavage. Daughter cells are genetically identical to parent cells.
120. The heart, bone and skeletal muscle most likely arise from ~hich of the following primary germ layers?
116. Decreasing progesterone levels during tbe luteal pbase of the menstrual cycle are associated with: A.
The anterior pituitary The pancreas The adrenal cortex The adrenal medulla
119. Which of the following does NOT describe cleavage in human embryos?
progesterone secretion by the anterior pituitary.
A.
remove particulate matter that becomes trappe~ in the mucus layer covering the Fallopian tubes. maintain a layer of warm air close to the inner lining, protecting the ovum from temperature changes occurring in the extema1 environment. kill incoming sperm, thus preventing fertilization facilitate movement of the ovum towards the uterus.
D.
118
The ectoderm The endoderm The gastrula The mesoderm
STOP.
The Digestive System; The Excretory System 6.1
Anatomy
Digestion is the break down of ingested foods before they are absorbed into the body. The major reaction involved in the digestion of all macromolecules is hydrolysis. You should kno w the basic anatomy of the digestive tract (Fig. 6-1), w hich goes as follows: mouth; esophagus; stomach; small intestine (dllodenum, ileum, jejUlIlIm); large intestine (ascending c%n, transverse c%n, descending c%n, sigmoid colon); rectum and; anus.
Esophagus Liver
Stomach Gall bladder
Pancreas
J ?uodenum
l
lLl~~~:~:--tf:~==~~::])
Large ..... Intestine
Small Intestine
. .'
Digestive Tract Figure 6-1
120
MeAT
B,OLOGY
6 .2
Remember that digestion begins with carbohydrates in the mouth via a amylase, and that there is no digestion in the esophagus. Also understand peristalsis, the wave-like motion of smooth muscle that moves food through the digestive tract.
Gastrin
The Mouth a nd Esophag us
Digestion begins in the mouth with a-amylase contained in saliva. Starch is the major carbohydrate in the human diet. a-amylase begins breaking down the long straight chains of starch into polysaccharides. Chewing also increases the surface area of food, which enables more enzymes to act on the food at anyone time. Chewed food forms a clu mp in the mouth called a bolus. The bolus is pushed into the esophagus by sw allowing, and then moved down the esophagus via peristaltic action. (Technically, swallowing includes the movement of the bolus from the esophagus into the stomach, and is composed of a voluntary and involluntary stage.) Peristaltic action is a wave motion, similar to squeezing a tube of toothpaste at the bottom and sliding your fingers toward the top to expel the toothpaste. The peristaltic movement is performed by smooth muscle. Saliva acts to lubricate the food help ing it to move down the esophagus. No digestion occurs in the esophagus.
G cell
Mucous cell
secreting gastrin
containing rough ER and Golgi to make mucus
HCl Pepsinogen
\
Parietal cell Chief cell
with the many mitochondria needed to produce sufficient energy to establish a proton gradient
synthesizing pepsinogen on rough ER
Gastric Gland Cell Types Figure 6-2 Copyright © 2007 Examkrackers, Inc.
LECTURE
6.3
6: THE
D ,GEST,VE SYSTEM ; T HE EXCRETORY SYSTEM •
The Stomach
The bolus moves into the stomach through the lower esophageal sphincter (or cardiac sphincter) . (A sphincter is a ring of muscle that is normally contracted so that there is no opening at its center.) The stomach is a very flexible pouch that both mixes and stores food, reducing it to a semifluid mass called chyme. The stomach contains exocrine glands (two types that are very similar) whose gastric pits are shown in Figure 6-2. Another important function of the stomach is to begin protein digestion w ith the enzyme pepsin. The low pH of the stomach assists this process by denaturing the proteins. A fu ll stomach has a pH of 2. The low pH also helps to kill ingested bacteria. There are four major cell types in the stomach (Fig. 6-2 and 6-3): 1.
mucous cells;
2.
chief (peptic) cells;
3.
parietal (oxyntic) cells and;
4.
G cells. Parietal cell
Gastric pits
G cell Mucous cells
cell
Th ree muscle layers in the wall of the stomach
(M" 'a 4-_4__ a~~0i5~iB=f=Wi~~~~~4? A Sectional View of the Stomach Figure 6-\
"'. \ Copyrigh t © 2007 Exam krackcrs, Inc
\
"
,
121
122
MeAT BIOLOGY
There are different types of mucous cells, but all of them perform fhe same basic function, secreting mucus. The mucous cells line the stomach wall and fhe necks of the exocrine glands. Mucus, composed mainly of a sticky glycoprotein and electrolytes, lubricates the stomach wall so that food can slide along its surface without causing damage, and mucus protects the epithelial lining from fhe acidic environment of the stomach. Some mucous cells also secrete a small amount of pepsinogen.
Chief cells are found deep in the exocrine glands. They secrete pepsinogen, the zymogen precursor to pepsin. Pepsinogen is activated to pepsin by the low pH in fhe stomach. Once activated, pepsin begins protein digestion.
For the MeAT, be fam iliar with the different cell types , and especially be aware that protein digestion begins in
the stomach. No absorption occurs in the stomach .
Parietal cells are also found in the exocrine glands of the stomach. Parietal cells secrete hydrochloric acid (HCI), which diffuses to the lumen. The exact method used by the parietal cells to manufacture HCI has not been agreed upon, but the amount of energy necessary to produce the concentrated acid is great. Carbon dioxide is involv ed in the process, making carbonic acid inside the cell. The hydrogen from fhe carbonic acid is expelled to the lumen side of the cell, while the bicarbonate ion is expelled to the interstitial fluid side. The net result is to lower the pH of the stomach and raise fhe pH of the blood. Parietal cells also secrete intrinsic factor, which helps the ileum absorb Bu ' G cells sem,te gastrin into the interstitium. The gastrin, a large peptide hormone, is absorbed into the blood and stimulates parietal cells to secrete He!. The major hormones that affeel the secretion of the stomach juices are acetylcholine, gastrin, and histamine. Acetylcholine increases the secretion of all cell types. Gastrin and histamine lnainly increase Hel secretion.
6.4
The Small Intestine
About 90% of digestion and absorption occurs in fhe small intestine. In a living human the small intestine is about 3 m in length. (In a cadaver the length increases to about 6 m due to loss of smooth muscle tone.) The small intestine is divided into three parts. From smallest to largest they are the duodenum, jejunum, and ileum. Most of digestion occurs in the duodenum, and most of the absorption occurs in the jejunum and ileum. The wall of the small intestine is similar to fhe wall of the stomach except fhat the outermost layer contains finger-like projections called villi (Fig. 6-4). The villi increase the surface area of the intestinal wall allowing for greater digestion and absorption. Within each villus are a capillary network and a lymph vessel, called a lacteal. Nutrients absorbed through the wall of the small intestine pass into the capillary network and the lacteal. On the apical (lumen side) surface of the cells of each villus (cells called enterocytes) are much smaller finger-like projections called microvilli. The microvilli increase the surface area of the intestinal wall still further. Under a light microscope the microvilli appear as a fuzzy covering. This fuzzy covering is called the brush border. The brush border contains Inembrane bound digestive enzymes, such as the carbohydrate-digesting enzymes; dextrinase; maltase; sucrase and; lactose; protein-digesting enzymes called peptidases; and nucleotide-digesting enzymes called nucleosidases. Some of the epithelial cells are goblet cells that secrete mucus to lubricate the intestine and help protect the brush border from mechanical and chemical damage. Dead cells regularly slough off into the lumen of the intestine and are replaced by new cells.
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LECllJRE
6 : THE
DIGESTIVE SYSTEM; THE EXCRETORY SYSTEM • 123
Located deep between the villi are the intestinal exocrine glands, the crypts of Lieberkuhn. These glands secrete an intestinal juice with a pH of 7.6 and lysozyme. Lysozyme helps to regulate the bacteria w ithin the intestine.
Intestinal Lumen Microvilli
Lacteal
Small Intestine Figure 6-4
Afferent Blood flow
Efferent Blood flow
Villus 6.5
The Pancreas
The sem ifluid chyme is squeezed out of the stomach through the pyloric sphincter and into the duodenum. The fluid inside the duodenum has a pH of 6 due mainly to bicarbonate ion secreted by the pancreas. The pancreas also acts as an exocrine gland, releasing enzymes from the acinar cells through the pancreatic duct into the duodenum. The major enzymes released by the p ancreas are t.ypsin, chy motrypsin, pancreatic amylase, lipase, ribonuclease" and deoxyribonuclease. All enzymes are released as zymogens. Trypsin is activated by the enzyme enterokinase located in the brush border. Activated trypsin then activates the other enzymes.
Trypsin and chymotrypsin degrade proteins into small polypeptides. Another pancreatic enzyme, carboxypolypeptidase, cleaves amino acids from the sides of these p ep tides. Most proteins reach the brush border as small polypeptides. Here they are reduced to amino acids, dipeptides, and tripeptides before they are absorbed into the enterocytes. Enzymes within the enteracytes (the cells of the brush border) reduce the dipeptides and tripeptides to amino acids.
Copyright © 2007 ExamkrClckers, Inc.
The small intestine is where the action is
in digestion and absorption .
You must know the pancreatic enzymes
tryps in . chymotrypsin, amylase , and lipase, and know their functions in the small intestine.
124
MCAT
BIOLOGY
Like salivary amylase, pancreatic amylase hydrolyzes polysaccharides to disaccharides and trisaccharides; however, pancreatic amylase is much more powerfuL Pancreatic amylase degrades nearly all the carbohydrates from the chyme into small glucose polymers, The brush border enzymes finish degrading these polymers to their respective monosaccharides before they are absorbed. Understand that bile is necessary to increase the surface area of fat, but that it does not digest the fat. In other words, bile physically separates fat molecules, but does not break them down cllemically,
Lipase degrades fat, specifically triglycerides, However, since the intestinal fluid is an aqueous solution, the fat clumps together, reducing its surface area, This problem is solved by the addition of bile, Bile is produced in the liver and stored in the gall bladder, The gall bladder releases bile through the cystic duct, which empties into the common bile duct shared with the liver. The common bile duct empties into the pancreatic duct before connecting to the duodenum at the ampulla of Vater, Bile emulsifies the fat, which means it breaks it up into small particles without changing it chemically This increases the surface area of the fat, allowing the lipase to degrade it into mainly fatty acids and monoglycerides, These products are shuttled to the brush border in bile micelles, and then absorbed by the enterocytes, Bile also contains bilirubin, an end product of hemoglobin degradation, Much of the bile is reabsorbed by the small intestine and transported back to the liver, Chyme is moved through the intestines by peristalsis , A second type of intestinal motion, segmentation, mixes the chyme with the digestive juices.
6.6 Whenever you get a large intestine question on the MeAT, think water reabsorption. Profuse water loss in tIle form of diarrhea ofteo results when there is a problem with the large intestine. You should be aware that there is a mutualistic symbiosis between humans and bacteria in the large intestine. Bacteriilget our leftovers; we get certain vitamins from them .
the Large Intestine
The large intestine, or colon, has four parts: l.
ascending colon;
2.
transverse colon;
3.
descending colon and;
4.
sigmoid colon.
The major functions of the large intestine are water absorption and electrolyte absorption. When this function fails, diarrhea results. The large intestine also contains the bacteria E. coli . The bacteria produce vitamin K, B12' thiamin, and riboflavin. Healthy feces are composed of 75% water. The remaining solid mass is 30% dead bacteria, 10-20% fat (mainly from bacteria and sloughed enterocytes), 10-20% inorganic matter, 2-3% protein, and 30% roughage (i.e. cellulose) and undigested matter (i.e. sloughed cells).
Copyright © 2007 Examkrackers. In c.
LECTURE 6 : THE D ,GEST,VE SYSTEM; THE EXCRETORY SYSTEM . 125
6. 7
Gastro intestina l Hormones Involved in Digestion
Secretin, cholecystokinin, and gastric inhibitOl"y peptide are local peptide hormones secreted by the small intestine after a mea l. Each of these hormones increases blood insulin levels especially in the presence of glucose. Gastric inhibitory peptide is released in response to fat and protein digestates in the d uodenum, and to a lesser extent, in response to carbohydrates. It has a mild effect in decreasing the m otor activity of the stomach. Hydrochloric acid in the duodenum causes secretin release. Secretin stimulates sodium bicarbonate secretion by the p ancreas. Food in the upper duodenum, especially fat digestates, causes the release of cholecystokinin. Cholecystokinin causes gallbladder contraction and pancreatic enzyme secretion. It also decreases the motility of the stomach allowing the duodenum more time to digest fat. You do n't have to re member
thCSl'
gastrointt'stinal hormones although they may
dppl'dr in () passage. Instead, undcrstilnd the idl'ilS of digl'Stiol1. The body cats to gain energy in the form of food . Tht:' digL'sti\"(~ systCIll brctlks down the food sn it call be absorbed in to the body. One problem is that the food may mon:' too fast through the digestive tract ('mel. corne out unaigested . The stomach stores food, and releases Sill <1 II amounts ,1t a time to be dig<..'stcd and absorbed by the intestine. This 'WdY, the body C.111 take in (cat) a large alllount of food at d single tinlc, ilnd ti1ke ., long tinw to d igest it. Olll' of the jobs of the g.lstro intestinal horrnones just described is tn help regulate this process.
We have looked at d igestion, tIl(' break down of fond. l\'ext, dbsnrption, the assi milation of the by-products of digestion.
Copyright © 2007 Examkrackers. Inc.
WI..'
\'\'ill look at
125. One function of the large intestine is:
Questions 121 through 128 are NOT based on a descriptive passage.
A. B. C. D.
121. As chyme is passed from the stomach. to the small intestine, the catalytic activity of pepsin: A. B.
C. D.
B. C. D.
absorb water secrete excess water digest fat secrete urea
126. Salivary a-amylase begins the digestion of:
increases because pepsin works synergistically with trypsin. increases because pepsin is activated from its zymogen form. decreases in response to the change in pH. decreases because pepsin is digested by pancreatic amylase in the small intestine.
A. B. C. D.
lipids nucleic acids proteins carbohydrates
127. All of the following enzymes are part of pancreatic exocrine function EXCEPT:
122. Which of the following is the best explanation for why pancreatic enzymes are secreted in zymogeJ1 form? A.
to to to to
A. B. C. D.
A delay in digestion is required in order for bile to increase the surface area chyme. Enzymes are most active in zymogen form. Zymogens will not digest bile in the pancreatic duct. Pancreatic cells are not as easily replaced as intestinal epithelium.
bile chymotrypsin pancreatic amylase lipase
128. In humans, most chemical digestion of food occurs in the: A. B. C. D.
mouth stomach duodenum ileum
123. Omeprazole is used to treat duodenal ulcers that result from gastric acid hypersecretion. Omeprazole blocks the secretion of HCl from the parietal cells of the stomach. Which of the following is LEAST likely to occur in a patient taking omeprazole? A. B. C. D.
an increase in microbial activity in the stomach a decrease in the activity of pepsin an increase in stomach pH a decrease in carbohydrate digestion in the stomach
124. Which of the following reaction types is common to the digestion of all macronutrients? A. B. C. D.
hydrolysis reduction glycolysis phosphorylation
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126
STOP.
LECTURE
6.8
6:
THE DIGESTIVE SYSTEM; THE EXCRETORY SYSTEM
Absorption and Storage
The function of the entire digestive tract described in the previous section is to convert ingested food into basic nutrients that the small intestine is able to absorb. Once absorbed into the enterocytes, nutrients are processed and carried to the individual cells for use. The following section describes the process of absorption and the postabsorpti ve fates of the major nutrients: carbohydrates; proteins; and fats. This section is provided mainly as background knowledge; very little of this information will be tested directly on the MCAT.
6.9
Carbohydrates
By far the major carbohydrates in a human diet are sucrose, lactose, and starch. Cellulose (the polysaccharide making up the cell wa ll of plants) cannot be digested by hum ans, and is considered roughage. Sucrose and lactose are disaccharides made from glucose and fructose, and from glucose and galactose, respectively. Starch is a straigh t chain of glucose molecules. Typically, 80% of the end product of carbohydrate digestion is glucose. 95% of the carbohydrates in the blood are glucose. Ca rbohyd rate absorption is shown in Figure 6-5. Glucose is absorbed by a secondary acti ve transport ill_e chanisrn down the concentra tion gradient of sodium. Sodium is actively pumped out of the enterocyte on the basolateral side. The resulting low sodi um concentration inside the enterocyte drags sodium from the intestinal lu men into the cell through a transport protein, but only after glucose has also attached itself to the protein. Thus glucose is dragged into the enterocyte by sodium. As the concentration of glucose inside the cell builds, it moves out of the cell on the basolateral side via facilitated transport. At high concentrations of lumenal glucose, glucose builds up in the para cellular space and raises the osmotic pressure there. The aqueous solution of the lumen is dragged into the paracellular space pulling g lucose along with it. Glucose is absorbed by this second method only when present in high concentrations. Galactose follows a similar absorp tion path to glucose. Fructose is absorbed via facilitated diffusion, and much of it is conve rted to glucose while inside th e enterocyte. All carboh ydra tes are absorbed into the bloodstream an d carried by the portal vein to the liver. One of the jobs of the liver is to maintain a fairly constant blood glucose level (90 mg/ dl between meals to 140 mg/ dl after a meal). The liver absorbs the carbohydrates and converts nearly all the galactose and frllctose into glucose, and then into glycogen for storage. The formation of glycogen is called glycogenesis. When the blood glucose level decreases, glycogenolysis takes place in the liver, and glucose is returned to the blood. In all cells except enterocytes and the cells of the renal tubule, glucose is transported from high concentra tion to low concentration via fac ilitated diffusion. Nearly all cells are capable of producing and storing some glycogen; however, only muscle cells and especially liver cells store large amounts. When the cells have reached their saturation point with glycogen, carbohydrates are converted to fatty acids and then triglycerides in a process requiring a small amollnt of energy.
Copyright © 2007 Examkrackers, Inc .
.
127
128
MeAT B,OLOGY
Intestinal lumen large c c" polysaccharides ",,,,o small;$I , to ~"'~ polysaccharides","P,,,l small to ~ _,"'0;;;Y polysacchandes gI ucose, 'C c fructose, and galactose
glucose (only when concentrations are high)
To the
glucose galactose fructose
Carbohydrate Digestion and Absorption Figure 6-5
YOLl can ignore rnost of the details here and concentrate on the big picture of carbohydrntc digestion, absorption, and metabolism. Rcbte this information to glycolysis and the Krebs cycle in Biology Lecture 1 for iJ complete pidur~. Notice that most of the glucosl' is stored for later liSC. When the glycogen stores arc [ult the glucose is cOl1verted to fat, a long-term form of energy storage. The conversion of glucose to fat takes place in the liver and fat cells ilnd is stored in tht, fat Cl'IlS. Kt'l'P in mind the role of the liver in processing carbohydrates. WewiH talk more about this later.
Co pyr igh t (c) 2007 Examkrackers, Inc.
LECT URE
6: THE
DIGESTIVE SYSTEM ; THE EXCRETORY SYSTEM .
129
6.1 0 Proteins Protein digestion results in amino acids, dipeptides, and tripeptides (Fig. 6-6). Absorption of many of these products occurs via a cotransport mechanism down the concentration gradient of sodium, similar to the mechanism used by glucose. A few amino acids are transported by facilitated diffusion. Because the chemistry of amino acids varies greatly, each transport mechanism is specific to a few amino acids or polypeptides. Nearly all polypeptides absorbed into an enterocyte are hydrolyzed to their amino acid constituents by enzymes within the enterocytes. From the enterocytes, amino acids are absorbed directly into the blood and then quickly taken up by all cells of the body, especially the liver. Transport into the cells may be facilitated or active, but is never passive, since amino acids are too large and polar to diffuse through the membrane. The cells immediately create proteins from the amino acids so that the intracellular amino acid concentration remains low . However, most proteins are easily broken down and returned to the blood when needed. When the cells reach their upper limit for protein storage, amino acids can be burned for energy (see Biology Lecture 1) or converted to fat for storage. The energy that can be gained from burning protein is about 4 Calories per gram of protein. This can be compared to carbohydrates, w hich produce about 4.5 Calories per gram, and fat, which produces about 9 Calories per gram. Anunonia, a nitrogen containing compound, is a by-product of gluconeogenesis from proteins. Nearly all ammonia is converted to urea by the liver and then excreted in the urine by the kidney.
Stomach lumen proteins
Intestinal lumen 'If
~-$
small pep tides
to 4;'" polypeptides
to
~o
0'>
/
.\'>
""
--0'",::, e,~
smaller pep tides
~ \~ePtideS
'00" ~0'"
q
and
to
--&-".,," ,,~maller polypeptides ~
and
"'-". G amino acids,
dipeptides, tripeptides amino acids
~~_ __
Protein Digestion and Absorption Figure 6-6
Copyright (g 2007 Exam krackers, In c.
Again, concentrate on tile big picture, It may be ilelpful to remember tilat virtually all dietary protein is completely broken down to its amino acids before being absorbed into the blood. In fact, any protein that is not broken down completely may cause allergic reactions. Also, wilen you think of proteins, think
nitrogen .
130
MCAT
B IOLOGY
6.11 Fats Most of d ietary fat consists of triglycerides, which are broken down to monoglycerides, and fatty acids, before they are shuttled to the brush border by bile micelles, and diffuse through the enterocyte membrane (Fig. 6-7). After delivering their cargo, the micelles sh uttle back to the chyme to pick up more fat digestates. Micelles also carry other fat d igestates such as small amounts of h ydrolyzed phospholipids and cholesterol, which also diffuse through the enterocyte membrane. Once inside the enterocyte, monoglycerides, and fatly acids, are turned back into triglycerides at the smooth endoplasmic reticulum. The newly synthesized triglycerides aggregate within the smooth endoplasmic reticular lumen along with some cholesterol and phospholipids. These amphipathic molecules orient themselves like a micelle with their charged ends pointing outward toward the aqueous solution of the lumen. Apoproteins attach to the outside of these globules. (See Biology Lecture 1 fo~ more on apopro teins.) The globules move to the Golgi apparatus and are released from the cell into the interstitial fluid via exocytosis. Most of these globules, now called chylomicrons, move into the lacteals of the lymph system (shown in Figure 6-4). 80-90% of ingested fat that is absorbed by this process, moves through the lymph system, and is emptied into the large veins 01 the neck at the thoracic duct. Small amou nts of more water soluble fatty acids (short chain fatty acids) are absorbed directly into the blood of the villi. The chylomicron concentration in the blood peaks about 1-2 bours after a meal, but falls rapidly (chylomicrons bave a half life of about 1 hour) as the fat digestates are absorbed into the cells of the body. Tbe major absorption of fat occurs in the liver and adipose tiss ue. Chylomicrons stick to the side of capillary walls wbere lipoprotein lipase bydrolyzes the triglycerides, the products of which immed iately diffuse into the fat and liver cells. Inside the fat and liver cells, the triglycerides are reconstituted at the smooth endoplasmic reticulum. Thus, tbe first stop for most of the digested fat is the liver. From adipose tissue, most fatty acids are transported in tbe form of free fatty acid, which combines immediately in the blood witb albumin. A single albumin molecule typically carries 3 fatty acid molecules, but is capable of carrying up to 30. Although lipoproteins are a hot topic, they are not a required topic for the MCAT. Once again, look at the big picture here. Keep in mind that fat is insoluble in water, so typically requires a carrier (i.e. a lipoprotein, or albumin). For the MCAT, you should associate fat with effiCient long-term energy storage; lots of calories (energy) with little weight.
Take a look at Orgo Lecture 4 and relate the chemistry to this biology section.
Between m eals (called the postabsorptive state) 95% of lipid s in the plasma are in the form of lipoproteins. Lipoproteins look like small chylomicrons, or, more precisely, chylomicrons are large lipoproteins. Besides chylomicrons, there are four differen t types of lipoproteins: 1.
very low-denSity lipoproteins;
2.
intermediate-density lipoproteins;
3.
low-density lipoproteins and;
4.
high-density lipoproteins.
All are made from triglycerides, cholesterol, phospholipids, and protein. As the density increases, first the amount of triglycerides decrease, and then the amount of cholesterol and phospholipids decrease. Thus, very low-density lipoproteins have a lot of triglycerides, and high-density lipoproteins have very few triglycerides. Most lipoproteins are made in the li ver. Very-low density li poproteins transport triglycerides from the liver to aclipose tissue. Intermediate and low-density lipoproteins transport cholesterol and phospholipids to the cells of the body. The function of high-denSity lipoproteins is less well understood. Hardening of the arteries seems to be induced by the lower density lipoproteins, but impeded by high-density lipoproteins. Copyright © 2007 Exarnkrackers, Inc
LECTURE 6 : THE DIGESTIVE SYSTE M; THE EXCRETORY SYSTEM .
Intestinal lumen cholesterol, "e .. phospholipids \)" trIglycerIdes and triglycerid~s Wi'"se to emulsified into micelles ~ monoglycerides, ---' fatty acids
Micelle
shuttle to brush border
Micelle
Lacteal monoglycerides, fatty acids
monoglycerides, fa tty acids to triglycerides
~
Fat Digestion and Absorption
lymph fluid to the thoracic duct
figure 6-7
6.12 The Liver The liver is positioned to receive blood from the capillary beds of the intestines, stomach, spleen, and pancreas via the hepatic portal vein. This blood is 'worked upon' by the liver. A second blood supply, used to oxygenate the liver, is received through the hepatic artery. All blood received by the liver moves through large flattened spaces called the hepatic sinusoids and collects in the hepatic vein, which leads to the vena cava.
Copyright © 2007 EX<:lmkrockecs, Inc.
1 31
132
MCAT
BIOLOGY
The liver has the following interrelated functions (Fig. 6-8).
ProUnombin and fibnnogen are two important clotting factors . Albumin is the major osmoregulatory protein in the blood . Globulins are a group of proteins that include antibod ies. Antibodies, however, are not made in the liver. They are made by plasma cells.
•
Blood storage: the liver can expand to act as a blood reservoir for the body.
•
Blood filtration: Kupfer cells phagocytize bacteria picked up from the intestines.
'.
Carbohydrate metabolism: The liver maintains normal blood glucose levels through gluconeogenesis (the production of glycogen and glucose from noncarbohydrate precursors), glycogenesis, and storage of glycogen.
•
Fat metabolism: The liver synthesizes bile from cholesterol and converts carbohydrates and proteins into fat. lt oxidizes fatty acids for energy, and forms most lipoproteins.
•
Protein metabolism: The liver deaminates amino acids, forms urea from ammonia in the blood, synthesizes plasma proteins such as fibrinogen, prothrombin, albumin, and most globulins, and synthesizes nonessential amino acids.
•
Detoxification; Detoxified chemicals are excreted by the liver as part of bile or polarized so they may be excreted by the kidney.
•
Erythrocyte destruction: Kupfer cells also destroy irregular erythrocytes, but most irregular erythrocytes are destroyed by the spleen.
•
Vitamin storage: The liver stores vitamins such as vitamins A, D, and BJ2.The liver also stores iron combining it w ith the protein apoferritin to form of ferritin.
When the liver mobilizes fat for energy, it produces acids called ketone bodies. This often results in a condition called ketosis or acidosis. For the MCAT, you should know that when the liver mobilizes fat or protein for energy, the blood acidity increases.
o
irregular blood cells to gallbladder
'.
----~-------bile formation Oq,
glycogenolysis
~ o
~
glycogen
,--t;lucose ~'-----~
lipoproteins
o
~cogenesis
-------------gluconeo enesis
~
r ~
O
conversion of galactose and fructose to glucose 0
~
to cells of the body
,
"
fat
detoxificatio
~----0~---drugs
~
plasma proteins
Functions of the Liver Figure 6-8 Copyright © 2007 EX<:lmkr()ckers. In c.
134. Most of the glycogen in the human body is stored in the liver and the skeletal muscles. Which of the following hormones inhibits glycogenolysis?
Questions 129 through 136 are NOT based on a descriptive passage.
A. B. C. D.
129. A stomach ulcer may increase the acidity of the stomach. The stomach cells most affected by a stomach ulcer are: A. B.
e. D.
goblet cells. parietal cells. chief cells. G cells.
135. Free fatty acids do not dissolve in the blood, so they must be traosported within the body bound to protein carriers. The most likely explaoation for this is: A.
130. Which of the following occurs mainly in the liver? A. B.
e. D.
fat storage protein degradation glycolysis gluconeogenesis
B. C. D.
131. Dietary fat consists mostly of neutral fats called triglycerides. Most digestive products of fat: A. B.
C. D.
Cortisol Insulin Glucagon Aldosterone
Blood is an aqueous solution and only hydrophobic compounds are easily dissolved. Blood is ao aqueous solution and only hydrophilic compounds are easily dissolved. Blood serum contains chylomicrons which do not bind to fatty acids. Blood serum is lipid based and the polar region of a fatty acid will not be dissolved.
136. Essential amino acids must be ingested because they cannot be synthesized by the body. In what form are these amino acids likely to enter the blood stream?
enter intestinal epithelial cells as chylomicrons. are absorbed directly into the capillaries of the intestines. are degraded to fatty acids by the smooth endoplasmic reticulum of enterocytes. enter the lymph system before entering the blood stream.
A. B.
e. D.
single amino acids dipeptides polypeptides proteins
132. Which of the following is not true concerning the digestive products of dietary protein?
A. B. C. D.
They are used to synthesize essential amino acids in the liver. Some of the products are absorbed into the intestines by facilitated diffusion. Energy is required for the intestinal absorption of at least some of these products. Deamination of these products in the liver leads to urea in the blood.
133. Cholera is ao intestinal infection that can lead to severe diarrhea causing profuse secretion of water and electrolytes. A glucose-electrolyte solution may be administered orally to patients suffering from cholera. What is the most likely reason for mixing glucose with the electrolyte solution? A. B. C. D.
When digested, glucose increases the strength ofthe patient. The absorption of glucose increases the uptake of electrolytes. Glucose is an electrolyte. Glucose stimulates secretion of the pancreatic enzyme, amylase.
Copyright © 2007 Examkrackers, Inc.
133
STOP.
134
MeAT B!OLOGY
6.13 The Kidney The functipn of the kidney is! 1.
to excrete w aste products r such as urea, uric acid, ammonia, and phos-
phate; 2.
to maintain homeostasis of the body fluid volulue and solute COll1position and;
3.
to help control plasma pH.
There are two kidneys. Each kidney is a fist-sized organ made up of an outer cortex and an inner medulla. Urine is created by the kidney and emptied into the renal pelvis. The renal pelvis is emptied by the ureter, which carries urine to the bladder. The bladder is drained by the urethra. The functional unit of the kidney is the nephron (Fig. 6-9 and 6-10). Blood flows into the first capillary bed of the nephron called the g lomerulus . Together, Bowman's capsule and the glomerulus make lip the renal corpuscle . Hydrostatic pressure forces some plasma through fenestrations of the glomerular endothelium and into Bowman's capsule. Like a sieve, the fenestrations screen
out blood cells and large proteins from entering Bowman's capsule. The fluid that finds its way into Bowman's capsule is called filtrate or primary urine. Filtrate moves from Bowman's capsule to the proxima:! tubule. The proximal hlbule is where most reabsorption takes place. Secondary active transport proteins in the apical membranes of the proximal tubule cells are responsible for the reabsorption of nearly all glucose, most proteins, and other solutes. These transport proteins can be"" transport proteins is come saturated. The concentration of a solute that saturatesJts called the transport maximum. Once a solute has reached itS"!iiansport maximum, any more solute is washed into the urine. Some solutes that are n'b t actively reabsorbed are reabsorbed by passive or facilitated diffusion. Water is reabsorbed into the renal interstitium of the proximal tubules across relatively permeable tight junctions due to the favorable oemotic gradient. Drugs, toxins, and other solutes are secreted into the filtrate by the cells of the proximal tubule. Hydrogen ions are secreted through an antiport system with sodium, which is driven by the sodium concentration gradient. This antiport sys-
tem is similar to the transport system of glucose w ith sodium, except the proton crosses the membrane in the opposite direction to sodium. Urk acid, bile pigments, antibiotics and other drugs are also secreted into the proximal tubule.
The net result of the proximal tubule is to reduce the amount of filtrate in the nephron while changing the solute composition without changing the osmolarity.
From the proximal tubule, the filtrate flows into the loop of Henle. The loop of Henle nips into th p mpol11la. Thp funr:tion of the loop of Henle is to increase the solute concentration, and thus the osmotic pressure, of the medulla. As filtrate descends into the medulla, water passively diffuses out of the loop of Henle and into the medulla. The descending loop of Henle has low permeability to salt, so filtrate osmolarity goes up. As the filtrate rises out of the medulla, salt diffuses out of the ascending loop of Henle, passively at first, then actively. The ascending loop of Henle is nearly impermeable to water. A second capillary bed, called the vasa recta, surrounds the loop of Henle and helps to maintain the concentration of the medulla.
COpYi"i ght @ 2007 Exarnkrackers, Inc. '
LECTURE 6 : THE D,GEST,VE SYSTEM; THE EXCRETORY SYSTEM . 135
The distal tubule reabsorbs N a' , an d Ca2• w hile secreting K+, H +, and HCO; . Aldos terone acts on the d is tal tubule cells to increase sodium and potassium mem' brane trans port prote ins. The net effect of th e dis tal tubule is to lower the filtrate osmolarity. At the end of the distal tubule, called the col/ecting tllbllie (Th e collecting tubule is a por tion of the d istal tubule, not to be confused w ith the collecting duct), A DH ac ts to increase the permeability of the cells to water. Therefore, in the pres, ence of ADH, wa ter flows from the tubule, concentrating the filtrate. The distal tubule empties into the collecting duct. The collecting duct carries the filtrate into the highly osmotic medulla. The collecting duct is impermeable to wate r, but is also sensitive to ADH. In the presence of ADH, the collecting duct becomes p ermeable to water allowing it to passively diffu se into the medulla, concentra ting the urine. Many collecting ducts line up side by side in the medulla to lnake the renal pyramids. The collecting d u cts lead to a rena l calyx, which empties into the renal pelvis.
6.14 The Juxtaglomerular Apparatus The juxtaglomerular apparatus monito rs fil trate p ressure in the distal tubu le. Specialized cells, called gran ular cells, in the juxtag lomerula r apparatus secrete the enzym e renin. Renin initiates a regulatory cascade producing angiotensin I, II, and III, w hich ultimate ly stimulates the adre n a l cortex to secrete aldosterone. Aldos terone ac ts on the distal tubule, stimulating the formation of membrane proteins that a bsorb sodium and secrete potassium.
Always Digging Holes
There are many details about the kidney that you must be able to recall for the MeAT. You should know the function of each section of the nephron: filtration occurs in the renal corpuscle; reabsorption and secretion mostly in the proxima l tubule ; the loop of Henle concentrates solute in the medulla; the distal tubule empties into the collecting duct the collecting duct concentrates the urine. Understand that the amount of filtrate is related to the hydrostatic pressure of the glomerulus. You should know that the descending loop of Henle is permeable to water, and that the ascending loop of Henle is impermeable to water and actively transports sodium into the kidney. Don't lose sight of Trl'e big picture; the function of the kidney, is homeostasis.
ADH is always digging holes in the collecting duct
-.
Copyriyht © 2001 Exnrnkmckers , Inc .
136 . MeAT B,OLOGY
Juxtaglomerular Apparatus
Glomerulus
Distal Convoluted Tubule
Proximal Convoluted Tubule
Bowman's Capsule
Cortex Thick Ascending Limb of the Loop of Henle
- M;dulla
Renal Medulla Renal Cortex
Vasa Recta ,-rn:-l-t-:~=-----
Npn hron
in
Collecting Duct
Loop of Henle
Kidney
The Nephron Figure 6-9
Copyright © 2007 Examkrackers, Inc.
b'
~.
"';;@ N
Proximal tubule
Bowman's capsule
8
Amino Acids
"
~
3
"
~
\in
Filtratio n --+-""""
i'
5"
o
Loop of Henle
Distal tubule
Secretion
~ -I
G'~~c:
H:O~ \
.\ I 300
---
100
\;!)
Collecting duct
~'1
Reabsorption
AmIno
Na'
Collecting tubule
only
Wi;:A DH
100
3~~
with
ADH
increased by aldosterone
cH,o
..Cor.tex..
Medulla
Na t in creased by aldosterone
:;;-
~
Concentration of Urine Figure 6-10
'"m
'"-i
ICH,o
H,O /
I
m
oc; m
Na
J
Na'
1/
600 or
___' ~-----
1200 with
ADH
Na'
Xso I or
1200 I with
H , o /./11 only with AOl-l
ADH
~ ~
~ m
.?: -i I
m
~
Urea only with AOH
n
'" ~ ~
'{'
~
?:
~
W
.....
141. Which of the following correctly orders the structures through which urine Ilows as it leaves the body?
Questions 137 through 144 are NOT based on a descriptive passage.
A. B. C. D.
137. Bowman's capsule assists in clearing urea from the blood by: A.
B. C. D.
actively transporting urea into the filtrate using ATP-driven pumps. exchanging urea for glucose in an ~tiport mechanism. allowing urea to diffuse into the filtrate under filtration pressure. converting urea to amino acids.
142. How are the blood levels of vasopressin and aldosterone in a dehydrated individual likely to compare with those of a healthy individual? A. B.
C.
138. Tests reveal the presence of glucose in a patient's urine. This is an indication that: A. B.
C. D.
D.
glucose transporters in the loop of Henle are not functioning properly. the patient is healthy. as glucose normally appears in the urine. the proximal tubule is over-secreting glucose. glucose influx into the filtrate is occurring faster than it can be reabsorbed.
B.
e. D.
A. B.
C.
increase the amount of filtrate that reaches the loop of Henle. increase the surface area available for the absorption. slow the rate of at which the filtrate moves through the nephron. move the filtrate through the nephron with cilia like action.
D.
D.
Copyri~J ht (f~)
A. B.
The patient's blood pressure would increase. Platelets would be found in the urine. The amount of filtrate entering Bowman's capsule would increase. Sodium reabsorption by the distal tubule would decrease.
2007 EX<Jrnkrackers, Inc.
these fenestrations would constrict resulting in decreased urinary output. Filtrate volume would be expected to be larger due to increased fluid pressure. Filtrate volume would be expected to be smaller due to increased fluid pressure. Urinary output will most likely be diminished due to increased solute concentration.
144. Long loops of Henle on juxtamedullary nephrons allow for greater concentration of urine. For an individual with highly concentrated urine, filtrate entering the loop of Henle is likely to be:
140. If a patient were administered a dmg that selectively bound and inactivated renin, which of the following would most likely result? A. B. C.
Vasopressin and aldosterone levels are likely to be lower in a dehydrated individual. Vasopressin and aldosterone levels are likely to be higher in a dehydrated individual. Vasopressin levels are likely to be higher while aldosterone levels are likely to be lower in a dehydrated individual. Vasopressin levels are likely to be lower while aldosterone levels are likely to be higher in a dehydrated individual.
143. An afferent arteriole in a glomerular tuft contains microscopic fenestrations which increase fluid flow. In a hypertensive patient (a patient with high blood pressure):
139. The epithelial cells of the proximal convoluted tubule contain a brush border similar to the bmsh border of the small intestine. The most likely function of the bmsh border in the proximal convoluted tubule is to: A.
urethra, urinary bladder, ureter, collecting duct collecting duct, urinary bladder, urethra, ureter collecting duct, ureter, urinary bladder, urethra ureter, collecting duct, urethra, urinary bladder
C. D.
138
more concentrated than filtrate exiting the loop Henle. less concentrated than filtrate exiting the loop of Henle. more voluminous than filtrate exiting the loop of Henle. less voluminous than filtrate exiting the loop of Henle.
STOP.
The Cardiovascular System; The Respiratory System 7.1
Ca rdiovascular Anatomy
111e cardiovascular system consists o f the heart, blood, and blood vessels (Fig. 7-1). For the MeAT, you must be able to trace the ci rculatory path of the blood . Beginning with the left ventricle, blood is pumped through the aorta. From the aorta, bra nch many smaller arteries, whitch themselves branch into still smalier arterioles, which branch into still smaller capillaries. Blood from the capillaries is collected into venules, which themselves collect into larger veins, which collect again into the superior and inferior vena cava. The vena ca va emp ty into the
Su perior -~
vena cava Pulmonary
va lve
--------Left atrium
Right atrium
Aortic valve
Tricuspid va lve
Mitral valve Le ft ventricle
In ferior vena cava
Trunk and Lower Extrem ity
The Heart Figure 7-1
140
MeAT BIOLOGY
Don 't be concerned with the narn es of th e different valves of the heart, or the specifi c narn es of any srn aller blood ve sse ls. Instead, concentrate on function. The left ventricle contracts with the most force to propel the blood through the system ic ci rcul alion .
right atrium of the heart. This first half of the circulation as just described is called the systemic circulation. From the right atrium, blood is squeezed into the right ventricle. The right ventricle pumps blood through the pulmonary arteries, to arterioles, to the capillaries of the lungs. From the capillaries of the lungs, blood collects in venules, then in veins, and finally in the pulmonary veins leading to the heart. (True capillaries branch off arterioles, and do not represent the only route between an arteriole and venule.) The pulmonary veins empty into the left atrium, which fills the left ventricle. This second half of the circulation is called the pulmonary circulation. Since there are no openings for the blood to leave the vessels, the entire system is said to be a closed circulatory system.
The heart itself is a large muscle. Unlike skeletal muscle, it is not attached to bone. Instead, its fibers form a net and the net contracts upon itself squ ~ezing blood into the arteries. Systole occurs when the ventricles contract; diastole occurs during relaxation of the entire heart and then contraction of the atria .
.21
.20
Transmission of the Cardiac Impulse Shmvn in fractions of a second Figure 7·2
The on ly nerve that you must know, is the vagus , wtl icll is parasympathetic, innervating the heart and the digestive system . The vagus nerve slows the rate of heart con tractions and increases digestive activity in the intestines. Know the role and Incation of the Purkinje fibers.
The blood is propelled by the hydrostatic pressure created by the contraction of the heart. The rate of these contractions is controlled by the autonomic nervous system, but the autollOlnic nervous system does not initiate the contractions. The heart contracts automatically, paced by a group of specialized cardiac muscle cells called the sinoatrial node (SA node) located in the right atrium. The SA node is autarhythmic (contracts by itself at regular intervals), spreading its contractions to the surrounding cardiac muscles via electrical synapses made from gap junctions. The pace of the SA node is faster than normal heartbeats but the parasympathetic vagus nerve innervates the SA node, slowing the contractions. The action potential generated by the SA node spreads around both atria causing them to contract, and, at the same time, spreads to the atrioventricular node (AV node) located in the interatrial septa (the wall of cardiac muscle between the atria). The AV node is slower to contract, creating a delay which allows the atria to finish their contraction, and to squeeze their contents into the ventricles before the ventricles begin to contract. From the AV node, the action potential moves down conductive fibers called the bundle of His. The bundle of His is located in the wall separating the ventricles. The action potential branches out through the ventricular walls via conductive fibers called Purkinje fibers , From the Purkinje fibers, the action potential is spread through gap junctions from one cardiac muscle to the next. The Purkinje fibers in the ventricles allow for a more unified, and stronger, conh'action. Copyright @ 2007 Examkrackers, Inc.
LECTURE
7:
THE CARDIOVASCULAR SYSTEM; THE RESP IRATORY SYSTEM'
Arteries are elastic, and stretch as they fill with blood. When the ventricles finish their contraction, the stretched arteries recoil, keeping the blood moving more smoothly. Arteries are wrapped in smooth muscle that is typically innervated by the sympathetic nervous system. Epinephrine is a powerful vasoconstrictor causing arteries to narrow. Larger arteries have less smooth muscle per volume than medium size arteries, and are less affected by sympathetic innervation. Medium sized arteries, on the other hand, constrict enough under sympathetic stimulation to reroute blood. Arterioles are very small. They are wrapped by smooth muscle. Constriction and dilation of arterioles can be used to regulate blood pressure as well as rerouting blood. Smooth muscle fiber
Arteriole
Intercellular cleft ~'"'-_-./
_ _ _.~~/7 y
Pinocytotic vesicle
\
\/ ,- - - Capillary - nucleuscell
~\ Venule
\ - Basement ) membrane
,
~
Capillary lumen Fenestrated pore
Cross-section of a Capillary
~
To heart
Capillaries Figure 7-3 Capillaries are microscopic blood vessels (Fig. 7-3). Capillary walls are only one cell thick, and the diameter of a capillary is roughly equal to that of a single red blood cell. Nutrient and gas exchange with any tissue other than vascular tissue takes place only across capillary walls, and not across arterioles or venules. There are four methods for materials to cross capillary walls: 1.
pinocytosis;
2.
diffusion or transport through capillary cell membranes;
3.
movement through pores in the cells called fenestrations;
4.
movement through the space between the cells.
Capillaries are found close to all cells of the body. As blood flows into a capillary (Fig. 7-4), hydrostatic pressure is greater than osmotic pressure, and net fluid flow is out of the capillary, and into the interstitium. Although osmotic pressure remains relatively constant throughout the capillary, hydrostatic pressure drops from the arteriole end to the venule end. Thus, osmotic pressure overcomes hydrostatic pressure near the venule end of a capillary, and net fluid flow is into the capillary and out of the interstitium. The net result of fluid exchange by the capillaries is a 10% loss of fluid to the interstitium.
Copyright
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2007 Examkrackers, Inc.
141
142
MeAT BIOLOGY
~
Hydrostatic pressure
10% to Lymph
forces . ...... __ .. ~~t Flmd Flow /
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....
Net Fluid Flow
p~essure
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B
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... --------____
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....".,_~.,jI~ HydTOstatic .. ~ pressure Blood flow forces
Capillary
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Fluid Exchange in the Capillaries Figure 7-4
Venules and veins are similar in structure to arterioles and arteries. The lumen is larger than the lumen of cOlnparable arteries and veins contain a far greater volume of blood. (Fig 7-5) Veins, venules, and venus sinuses in the systeluic circulation hold about 64% of the blood in a body at rest, and act as a reservoir for blood. Arteries, arterioles, and capillaries in the systemic circulation contain about 20% of the blood. f
Pulmonary circulation 9%
Arteries 13% Heart 7%
Arterioles and capillaries 7%
Veins, venules, and venus sinuses 64%
The cross-sectional area of the veins is about four tin1€S that of the arteries. The total cross-sectional area of the capillaries is far greater than the cross-sectional area of the arteries or veins. Since the blood volume flow rate is approximately constant, the blood velocity is inversely proportional to the cross-sectional area. Therefore, the blood moves the slowest through the capillaries. Although Bernoulli's equation tells us that pressure is inversely related to cross-sectional area, it is evident from Figure 7-7 that this is not the case in the blood vessels. The blood is not an ideal flow, and you should memorize Figure 7-7 for the MCAT. The pumping force of the heart is the major contributor to pressure in the b lood vessels. To compensate for the lower pressure, veins have a valve system that prevents back flow of blood. Contraction of skeletal muscle helps blood move through veins; however, the Inajor propulsive force moving blood through tJle veins is the plunping of the heart. An artery carries blood away from the heart; a vein carries blood toward the heart. Don ' t confuse oxygenated blood \-vith the definition for arteries. Tl1e pulJnonary arteries contain the most deoxygenated blood in the body.
Blood Volume Figure 7-5
Copyright <9J 2007 Examkr-a cKers, Inc.
LECTURE
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Velocity: A single artery IS much bigger than a capillary, but there are far more capillaries than arteries. The total cross-sectional area of all those capillaries put together is much greater than the cross sectional area of a smgle aorta or a few arteries. Blood flow fo llows the Continuity Equation , Q ~ Av, rea sonably we ll , so velocity is greatest in the arteries where crosssectional area is smallest, and velocity is lowest where crosssectional area is the greatest.
149. Which chambers of Ihe heart pump oxygenated blood?
Questions 145 through 152 are NOT based on a descriptive passage.
A. B. C,
D,
145. The atrioventricular node: A. B. C. D.
is a parasympathetic ganglion located in the right atrium of the heart. conducts an action potential from the vagus nerve to the heart. sets the rhythm of cardiac contractions. delays the contraction of the ventricles of the heart.
150, Hypovolemic shock represents a set of symptoms that mocur when a patient's blood volume falls abruptly. Hypovolemic shock is most likely to occur during: A. B, C. D.
146. Cardiac output, which is the product of the heart rate and the stroke volume (the amount of blood pumped per contraction by either the left or the right ventricle) would most likely be: A. B. C. D,
C. D.
arterial bleeding venous bleeding low oxygen intake excess sodium consumption
151. The capillary network comprises the greatest cross sectional area of blood vessels in the body with the highest resistance to blood flow. In a healthy individual, the highest blood pressure would most likely be found in:
greater if measured using the stroke volume of the left ventricle. greater if measured using the stroke volume of the right ventricle. . the same regardless of which stroke volume is used. dependent on the viscosity of the blood.
A. B. C. D.
the the the the
aorta vena cavae systemic capillaries pulmonary capillaries
152: Gas exchange between the blood and tissues occurs:
147, Which of the following is responsible for the spread of the cardiac action potential from one cardiac muscle cell to the next? A. B.
The right and left atria The right and left ventricles The right atria and the left ventricle The left atria and the left ventricle
A, B.
C.
gap junctions desmosomes tight junctions acetylcholine
D,
throughout the circulatory system. in the arteries, arterioles and capillaries. in the systemic arteries only. in the capillaries only.
148. In the congenital heart defect known as patent ductus arteriosus, the ductus arteriosus, which connects the aorta and the pulmonary arteries during fetal development, fails to close at birth. This will likely lead to all of the following EXCEPT:
A. B,
C, D,
equal, or increased, oxygen concentration in the blood that reaches the systemic tissues. increased oxygen concentration in the blood that reaches the lungs. increased work load imposed on the left ventricle. increased work load imposed on the right ventricle.
Copyright © 2007 Examkrackers, Inc.
144
STOP.
LECTURE
7.2
7:
THE CARDIOVASCULAR SYSTEM; THE RESPIRATORY SYSTEM' 145
The Respiratory System
The respiratory system provides a path fo r gas exchange between the external environment and the blood (Fig. 7-8). Air enters through the nose, moves through the p harynx, laryn x, trachea, bronchi, bronchioles, and into the alveoli where oxygen is exchanged for carbon dioxide w ith the blood . Inspiration occurs w hen the medulla oblongata of the midbrain signals the diaphragm to contract. The diaphragm is skeletal muscle, and innervated by the phrenic nerve. When relaxed, the diaphragm is dome-shaped. It fl attens upon contrac tion, exp anding the chest cavity and creating negative gauge press ure. (Gauge p ress ure is measu red relative to loca l atmospheric conditions. See Physics Lecture 5.) intercostal muscles (rib muscles) also help to expand the chest cavity. Atmospheric pressure forces air into the lungs. Upon relaxation of the d iaphragm, the chest cavity shrinks (aided by different intercostal muscles and abdominal muscles), and the elasticity of the lungs along with the increased pressure in the chest cavity forces air out of the bod y. The nasal cavity is the space inside the nose. It filters, moistens, and warms incoming air. Coarse hair at the front of the cavity traps large dust particles. Mucus secreted by goblet cells traps smaller dust particles and moistens the air. Capillaries within the nasal cavity warm the air. Cilia mo ves the mucus and dust back towa rd the pharynx, so that it may be removed by spitting or swallowing. The pharynx (or throat) functions as a passageway for food and air. The larynx is the voice box. It sits behind the epiglottis, which is the cartilaginOUS member that prevents food from entering the trachea during swallowing. When nongaseous material en ters the larynx, a coughing reflex is triggered forcing the material back out. The larynx contains the vocal cords. The trachea (or windpipe) lies in front of the esophagus. It is composed of ringed ca rtilage covered by ciliated mucous cells. Like the nasal cavil)', the m ucus and cilia in the trachea collect dust and usher it toward the pharynx. Before entering the lungs the trachea splits into the right and left bronchi. Each bronchus branches man y more times to become tiny bronchioles. Bronchioles terminate in grape-like clusters called alveolar sacs composed of tiny alveoli. From each alveolus, oxygen d iffuses into a capillary where it is picked up by red blood cells. The red blood cells release carbon dioxide, which d iffuses into the alveolus, and is expelled upon exhala tion.
Know the basic anatomy given here. Also. understand that the job of the respirato ry system is to deliver oxygen to the blood and expel carbon dioxide. Part of the respiratory tract functions to prepare the air by warming, moistening, and cleaning. Know that since microtubules are found in cilia . and ciliated cells are found in the respiratory tract (and the Fallopian tubes and ependymal cells of the spinal cord), a problem in microtubule production might result in a problem in breathing (or fertility or circulation of cerebrospinal fluid) .
Capillary Nasal ca vity Pharynx Erythrocyte
-+---l'l Larynx-..--J Trachea -y-~--~
Nucleus
Bronchus -r----r-.-..I.
Air
Lung
Alveolar cell
Alveolus
Diaphragm
Respiratory System Figure 7-8 Copyright © 2007 Examkrackers, Inc.
Macrophage
146
MeAT
BIOLOGY
7.3
The Chemistry of Gas Exchan ge
Typ ically, the air we inspire is 79% nitrogen and 21% oxygen, w ith negligible am ounts of o ther trace g ases. Exhaled air is 79% nitrogen, 16% oxygen, and 5% car-
bon d ioxide and trace gases. Inside the lungs, the partial pressu re of oxygen is approximately 110 mm Hg, and carbon dioxide is approximately 40 mm Hg. Under these pressures, oxygen diffuses into the capillaries, and carbon dioxide diffuses into the alveoli. 98% of the oxygen in the blood binds rapidly and reversibly with the protein hemoglobin insid e the eryth rocytes forming oxyhemoglobin. Hemoglobin is composed of four polypeptide subuni ts, each with a single he111e cofactor. The heme cofactor is an o rganic molecule with an atom of iron at its center. Each of the four iron atoms in hemoglobin can bind w ith one 0 , molecule. When one O 2 molecule binds w ith an iro n atom in helnoglobin, oxygena tion of the other heme groups is ac-
celerated. Sim ilarl y, release of an 0 , molecule by any of the hem e gro up s, accelerates release by the others. Th is phenomenon is ca lled cooperntivity. As 0 2 pressure increases, the 0 2 saturation of hemoglobin increases sigmoidally. The OXylJeIlfOglobill (HbO,) dissociatiol1 curve (Fig. 7-9) shows the percent of hemoglobin tha t is bOWld w ith oxygen at various partial pressures o f oxygen. In the arteries
of a normal person breathing room air, the oxyge n sa turation is 97%. The flat portion of the curve in this region shows that small fluctuations in oxygen pressure have li ttl e effect.
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The oxygen saturation of hemoglobin also depends upon carbon dioxide pressure, pH, and temperature of the blood. The oxygen dissociation curve is shifted to the [[ght by an increase in carbon dioxide pressure, hydrogen ion concentraCopyright i&:) 2007 [xal1lkrackers, Inc
LcC"URr 7: THE CARDIOVASCULAR 5YS EM; THe RESPIRATORY SYSTC.M •
tion, or temperature. A shift to the right ind icates a lowering of hemoglobin's affin ity for oxygen. The shift due to pH ch ange is called th e Bohr shift. 2,3-DPG, a chemical fo und in red blood cells, also shifts the curve to the right. Carbon monoxide h as m ore than 200 times grea ter affinity for he moglob in than does oxygen but shifts the curve to th e left. In cases of carbon m on oxide po isoning, pure oxygen can b e adminis tered to displace the CO from h em oglobin. Oxygen pressure is typically 40'm1'1 H g in body tissues. As the blood moves throug h the sys temic capillaries, oxygen diffuses to th e tissues, and carbon dioxide diffuses to the blood. Carbon dioxide is carried by the blood in three forms: 1.
in ph ysical solution;
2.
as bica rbo nate ion and;
3.
in carbamino compounds (combined w ith hemoglobin and other proteins).
147
You don't need to memorize too many of the details of gas exchange; however, you must understand the mechanisms. You should know the effects that temperature. pH, and carbon dioxide pressure have on hemoglobin. Be sure that you can read a dissociation curve. and predict shifts.
Ten tin1es as m uch is carried as bicarbonate than as eith er of the other forms . The bica rbonate ion forma tion is gov erned by the enzyme carbonic anhydrase in the revers ible reacti on:
Beca use ca rbonic anhydrase is inside the red blood cell and n ot in the pla sma, w hen carb on d iox id e is absorbed in the lungs, bicarbon ate io n diffuses into the cell. To bala nce the electros tatic forces, chlorine m oves o u t of th e cell in a phenomenon called the chloride shift (Fig. 7-10). HCo,-
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CLUVC,
w hich rela tes blood content of car-
bon dioxide with carbon dioxide pressure. The greater the pressure of carbon d ioxide, the greater the blood content o f ca rbon dioxide. H owever, when hemog lobin becom es satura ted with oxygen, its capacity to ho ld ca rbon dioxide is reduced . This is ca lled the Haldane eff ect. The Haldane effec t fac ilitates the transfer of carbon dioxide fro m blood to lungs, and from ti ssues to blood. Reduced h emoglobin (Hb) (hem oglob in w ithout oxygen) acts as a blood buffer b y acce pting protons. It is the g rea ter ca pacity of reduced hemoglobin to form ca rbamino hemoglobin that explains the Haldan e effect. Copyright (\. 200"1 EX
Understand the connection between rate of breathing and carbon dioxide. pH. and oxygen levels in the blood. For instance. in the case of acidosis (too much acid in the blood), the body compensates by increasing the breathing rate thereby expelling carbon dioxide and raising the pH of the blood.
148
MCAT
B,OLOGY
The rate of breathing is affected by cel1tral chemoreceptors located in the medulla , and peripheral chemoreceptors located in the carotid arteries and aorta. Central and peripheral chemoreceptors monitor carbon dioxide concentration in the blood and increase breathing when levels get too high. Oxygen concentration and pH are monitored mainly by peripheral chemoreceptors. What about a ll that nitrogen? What effect does nitrogen have on the body? Remember your chemistry. Nitrogen is extremely stable due to its s trong triple bond. Thus, nitrogen diffuses into the blood, but doesn 't react with the chemicals in the blood . However, people that go diving must be carefu l. As the pressure increases w ith depth, more nitrogen diffuses into the blood . "Vhen divers con~e back up, the pressure decreases and the gas vollune increases. If they don't a I lo\-" e nough
time for the nitrogen to d iffuse Ollt of the blood and into the lungs, the nitrogen w ill form bubbles. Among other problems, these bubbles may occlude (block) \'essds causing decompression s ickness also known as 'the bends'.
Copyright @ 2007 Examkrackers, Inc.
157. Carbon dioxide partial pressure:
Questions 153 through 160 are NOT based on a descriptive passage.
A. B.
153. Alkalosis is increased blood pH resulting in a leftward shift of the oxy hemoglobin dissociation curve. Which of the following might cause alkalosis? A. B. C. D.
C. D.
hypoventilation hyperventilation breathing into a paper bag adrenal steroid insufficiency
158. At high altitude, water vapor pressure in the lungs remains the same and carbon diqxide pressure falls slightly. Oxygen pressure falls. The body of a person remaining at high altitudes for days, weeks, and even years will acclimatize. All of the following changes assist the body in coping with low oxygen EXCEPT:
154. Which of the following would most likely occur in the presence of a carbonic anhydrase inhibitor? A. B. C. D.
The blood pH would increase. The carbamino hemoglobin concentration inside erythrocytes would decrease. The rate of gas exchange in the lungs would decrease. The oxy hemoglobin concentration inside erythrocytes would increase.
A. B. C. D.
C.
D.
Blood pH will decrease in the active tissues. Less oxygen will be delivered to the tissues due to increased cardiac contractions resulting in increased blood velocity. Capillm'ies surrounding contracting skeletal muscles will constrict to allow increased freedom of movement. The respiratory system will deliver less nitrogen to the blood.
A. B. C. D.
B.
C.
D.
smooth muscle spasms of the bronchioles. cartilaginous constriction of the trachea. edema in the alveoli. skeletal muscle spasms in the thorax.
160. Sustained heavy exercise results in all of the following changes to blood chemistry except: A. B. C. D.
156. An athlete can engage in blood doping by having blood drawn several weeks before an event, removing the blood cells, and having them reinjected into her body a few days before an athletic activity, Blood doping is most likely an advantage to athletes because: A.
increased red blood cells. decreased vascularity of the tissues. increased pulmonary ventil ation . increased diffusing capacity of the lungs.
159. In an asthma attack, a patient suffers from diffic ulty breathing due to constricted air passages. The major causative agent is a mixture of leukotrienes called slow reacring substance of anaphylaxis. During an asthma attack, slow reacting substance of anaphylaxis most likely causes:
155. Which of the following will most likely occur during heavy exercise? A. B.
increases in the blood as it travels from the systemic venules to the inferior vena cava. increases in the blood as it travels from the pulmonary arteries to the pulmonary veins. is greater in the blood in the systemic capillary beds than in the alveoli of the lungs. . is greater in the blood in the systemic capillary beds th ~Ul in the systemic tissues.
lowered pH raised CO 2 tension increased temperature decreased carboxyhemoglobin
the increased concentrati6n of immune cells in the blood after reinjection can decrease the chances of becoming ill just before the competition, the inc reased red blood cell count in the blood after reinjection can facilitate greater gas exchange with the tissues. the increased blood volume after reinjection can ensure that lhe athlete maintains adequate hydration during the event. the decreased red blood cell count of the blood in the weeks before the competition can facilitate training by decreasing the viscosity of the blood,
Copyright © 2007 Exarnkrackers, Inc
149
STOP.
150
MeAT
B,OLOGY
7.4
The Lymphatic System
The lymphatic system collects excess interstitial fluid and returns it to the blood. Proteins and large particles that cannot be taken up by the capillaries, are removed by the lymph system. The pa thway to the blood takes the excess fluid through lymph nodes, which are well pre pared to elicit an immune response if necessary. Thus, the lymph system recycles the interstitial fluid and monitors the blood for infection. In addition, the lymph system reroutes low solubl e fat digestates around the small capillaries of the intestine and into the large veins of the neck. Most tissues are drained by lymphatic channels. A notable excep tion is the central nervous system. The lymph system is an open syst em . In other words, fluid enters at one end and leaves at the other. Lymph capillaries are like tiny fingers protruding into the tissues. To enter the lymph system, interstitial fluid flows between overlapping endothelial cells (Fig. 7-11). Large particles lite rall y push their way between the cells into the ly mph. The cells overlap in such a fash ion tha t, once inside, large particles cannot push the ir way out. Typically, interstitial fluid pressure is slightly negati ve. (Of course, we mean gauge pressure. See PhYSic Lecture 5.) As the interstitial pressure rises toward zero, lymph flow increases. Factors that affect interstitial pressure incl ude: blood pressure; plasma osmotic pressure; interstitial osmotic pressure (e.g. from proteins, infection response, etc.); permeability of capillaries. Like veins, lymph vessels are constructed with intermittent valves, which allow fluid to flow in only one direction. Fluid is propelled through these valves in two ways. First, sm ooth muscle in the walls of larger lymph vessels contracts when stretched. Second, the lymph vessels may be squeezed by adjacent skeletal muscles, body movements, arterial pulsations, and compression from objects outside the body. Lymph flow in an ac tive individual is considerably greater than in an individual at rest.
Force
One-way valve Lymph cell
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Lymph Vessel Figure 7-11 The lymph system empties into large vei ns at the thoracic dllct and the right lym phatic dllct. Lymph from the right arm and head enters the blood through the right lymp hati c duct. The rest of the body is drained by the thoracic duct. Throughout the lymphatic system are many lymph nodes, containing large quantities of lymp hocytes. Copyright CO 2007 Examkrackers, Inc.
LECTURE
7,5
7:
THE CARDiOVASCULAR SYSTEM; THE RFSPiRATORY SYSTEM .
151
The Blood
The blood is connective tissue. Like any cOlmective tissue, it contains cells and a matrix. Blood regulates the extracellular environment of the body by transporting nutrients, waste products, hormones, and even heat. Blood also protects the body from injury and foreign invaders. When a blood sample is placed in a centrifuge (Fig 7-12), it separates into three parts: 1. the plasma; 2. the buffy coat (white blood cells); 3. red blood cells. The percentage by volume of red blood cells is called the hematocrit. Hematocrit is normally 35-50%, and is greater in men than women. Plasma contains the matrix of the blood, which includes water, ions, urea, alnmonia, proteins, and other organic and inorganic compounds. Important proteins contained in the plasma are albumin, immunoglobulins, and clotting factors. Albumins transport fatty acids and steroids, as well as acting to regulate the osmotic pressure of the blood. Immunoglobulins (also called antibodies) are discussed below. Plasma in which the clotting protein fibrinogen has been removed is called serum. Alb Ulnin, fibrinogen, and nl0st other plasma proteins are formed in the liver. Galnma globulins that constitute antibodies are made in the lymph tissue. An important function of plasma proteins is to act as a source of amino acids for tissue protein replacenlent.
Erythrocytes (red blood cells) are like bags of hemoglobin. They have no organelles, not even a nucleus, w hich means they do not reproduce nor undergo mitosis. They are disk-shaped vesicles whose main function is to transport 0 2 and CO 2 , "Squeezing through capillaries wears out their plasma membranes in about 120 days. Most worn out red blood cells burst as they squeeze through channels in the spleen or, to a lesser extent, in the liver.
Plasma -
Buffy coat
~C I ~._.--./
Red blood cells
~
Blood Composition
Leukocytes (white blood cells) do contain organelles, but do not
Figure 7-12
contain hemoglobin. They function to protect the body from foreign invaders. All blood cells differentiate from the same type of precursor, a stem cell residing in the bone marrow. Erythrocytes lose their nucleus while still in the marrow. After entering the blood stream as reticulocytes, they lose the rest of their organelles within 1 or 2 days. Leukocyte formation is more complex due to the many different types. The granular leukocytes are neutrophils, eosinophils, and basophils . With respect to dyeing techniques, neutrophils are neutral to acidic and basic dyes, eosinophils stain in acid dyes, and basophils stain in basic dyes. Generally, granulocytes remain in the blood only 4 to 8 hours before they are deposited in the tissues, where they live for 4 to 5 days. Agranular leukocytes include monocytes, lymphocy tes, and megakaryocytes. Once deposited in the tissues, monocytes become macrophages and may live for months to years. Lyrnphocytes may also live for years.
Platelets are sm all portions of membrane-bound cytoplasm torn from megakaryocytes. Megnkaryocytes remain mainly in the bone marrow. Platelets are similar to tiny cells w ithout a nucleus. They contain actin and myosin, residuals of the Golgi and the ER, mitochondria, and are capable of making protaglandins and some important enzymes. Its membrane is designed to avoid adherence to healthy endothelium w hile adhering to injured endothelium. When platelets come into conact with injured endothelium, they become sticky and begin to swell releasing various chelnicals and activating other platelets. The platelets stick to the endothelium and to each other forming a loose platelet pillg. Healthy individuals have many platelets in their blood. Platelets The platelet has a half-life of 8-12 days in the blood. Copyriqht @ 2007 Exarnkr3ckc;rs, Inc
Know tllat the job of erythrocytes is to deliver oxygen and remove carbon dioxide.
Notice that granulocytes live a very short time, whereas agranulocytes-other white blood cells-live a very long time. This is because granulocytes function no"specifically against all infective agents, whereas most agranulocytes work against specific agents of infection. Thus, agranulocytes need to hang around in case tile same infective agent returns; granulocytes multiply quickly against any infection . and then die once the infection is gone.
152 . MCAT B,OLOGY
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Just know that the coagulation process involves many factors starting wth platelets and including the plasma proteins prothrombin and fibrin.
Coagulation occurs in three steps: 1) A dozen or so coagu lation factors form a comlex called protrombin activator. 2) Protrombin activator catalyzes the conversion of prothrombin, a plasma protein, into thrombin . 3) Thrombin is an enzyme that governs the polymerization of the plasma protein fibrinogen to fibrin threads that attach to the platelets and form a tight plug. This blood cot fo rmation (or coagulation) begins to appear in seconds in small injuries and 1 to 2 minutes in larger injuries. The following is the leukocyte composition in the blood: Neutrophils 62% Lymphocytes
30%
Monocytes
5.3%
Eosinophils
2.3%
Basophils
0.4%
7.6
Th e Imm une System
The human body protects itself from infectious microbes and toxins in two ways: innate immunity and acquired immunity. Innate ilnrnunity involves a generalized protection from most intruding organisms and toxins. Acquired immunity is protection
against specific organisms or toxins. Acquired immunity develops after the body is first attacked. IImate immunity incl udes: 1.
the skin as a barrier to organisms and toxins;
2.
stomach acid and digestive enzymes to destroy in gested organisms and toxins;
3.
phagocytotic cells; and
4.
chem ica ls in the blood.
Injury to tissue results in inflammation, which includes dilation of blood vessels, increased permeability of capillaries, swelling of tissue ceils, and migration of granCopyright © 2007 Exarnkrackers, Inc.
LECTURE
7:
THE CARDIOVASCULAR SYSTEM; THE RESPIRATORY SYSTEM .
ulocytes and macrophages to the inflamed area. Histamine, prostaglandins, and lymphokines are just some of the causative agents of inflammation that are released by the tissues. Part of the effect of inflammation is to 'wall-off' the effected tissue and local lymph vessels from the rest of the body, impeding the spread of the infection.
Infectious agents that are able to pass through the skin or the digestive defenses and enter the body are first attacked by local macrophages. These phagocytotic giants can engulf as many as 100 bacteria. Neutrophils are next on the scene. Most neutrophils are stored ill the bone marrow until needed, but some are found circulating in the blood or in the tissues. Neutrophils move toward infected or injured areas, drawn by chemicals (a process called chemotaxis) released from damaged tissue or by the infectious agents themselves. To enter the tissues, neutrophils slip between endothelial cells of the capillary walls, using an amaeboid-like process called diapedesis. A single neutrophil can phagocytize from 5 to 20 bacteria.
Monocytes circulate in the blood until they, too, move into the tissues by diapedesis. Once inside the tissues, monocytes mature to become macrophages.
When the neutrophils and macrophages engulf necrotic tissue and bacteria, they die. These dead leukocytes, along with tissue fluid and necrotic tissue, make up what is known as pus.
Eosinophils work mainly against parasitic infections. Basophils release many of the chemicals of the inflammation reaction. There are two types of acquired immunity: humoral or B-cell immunity; ceIl-mediated or T-cell immunity. Humoral immunity is promoted by B lymphocytes. B lymphocytes differentiate and mature in the bone marrow and the liver. Each B lymphocyte is capable of making a single type of antibody or (immunoglobulin), which it displays on its membrane. An antibody recognizes a foreign particle, called an antigen. The portion of the antibody that binds to an antigen is highly specific for that antigen. The portion of the antigen that binds to the antibody is called an antigenic determinant. An antigenic determinant that is removed from an antigen is called a hapten. Haptens can only stimulate an in1ffiune response if the individual has been previously exposed to the full antigen. Macrophages present the antigenic determinants of engulfed microbes on their surfaces. If the B lymphocyte antibody contacts a matching antigen (presented by a macrophage), the B lymphocyte, assisted by a helper T cell, differentiates into plasma cells and memory B cells. Plasma cells begin synthesizing free antibodies, and releasing them into the blood. Free antibodies may attach their base to mast cells. When an antibody whose base is bound to a mast cell also binds to an antigen, the n13st cell releases histamine and other chemicals. When other free antibodies contact the specific antigen, they bind to it. Once bound, the antibodies may begin a cascade of reactions involving blood proteins (called complement) that cause the antigen bearing cell to be perforated. The antibodies may mark the antigen for phagocytosis by macrophages and natural killer cells. The antibodies may cause the antigenic substances to agglutinate or even precipitate, or, in the case of a toxin, the antibodies may block its chemically active portion. The first time the immune system is exposed to an antigen is known as the primary response. The primary immune response requires 20 days to reach its full potential. Memory B cells proliferate, and remain in the body. In the case of re-infection, each of these cells can be called upon to synthesize antibodies, resulting in a faster acting and more potent affect called the secondary response. The secondary response requires approximately 5 days to reach its full potential. Humoral immunity is effective against bacteria, fungi, parasitic protozoans, viruses, and blood toxins.
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153
154
MCAT
Blo'-OGY
We shall drive back
Bone marrow stem cell
I
I
Erythrocyte
I
Monocyte
~\ §~
the invaders.
Leukocyte
I
I
Megakaryocyte
Granulocyte
l~
Neutrophil
Platelets
I Lymphocyte
I
Mast c;::ell
Eosinophil Basophil
Macrophage Natu'ral Killers
B-Iymphocyte
T-Iymphocyte
Plasma
Killer
Memory
Helper Memory Suppressor
Cell-mediated immunity involves T-Iyrnphocytes. T-Iymphocytes mature in the thymus. Similar to B lymphocytes, T lymphocytes have an antibody-like protein at their surface that recognizes antigens. However, T-lymphocytes never make free an-
tibodies. In the thymus, T-lymphocytes are tested against self-antigens (antigens expressed by normal cells of the body). If the T-lymphocyte binds to a self-antigen, that T lymphocyte is destroyed. If it does not, it is released to lod ge in lymphoid tissue or circulate between the blood and the lymph fluid. T lymphocytes that are not destroyed differentiate into helper T cells, memory T cells, suppressor T cells, and killer T cells (also called cytotoxic T cells). As discussed above, T helper cells assist in activating B lymphocytes as well as killer and suppressor T cells. Helper T cells are the cells attacked by HIY. Memory T cells have a similar function to Memory B cells. Suppressor T cells playa negative feedback role in the immune system. Killer T cells bind to the antigen-carrying cell and release per/orin, a protein which punctures the antigen-carrying cell. Killer T cells can attack many cells beca use they do not phagocytize their victims . Killer T cells are responsible for fighting some forms of cancer, and for attacking transplanted tissue. ,1 -'
Cell-mediated immunity is effective against infected cells. Let's imagine a bacterial infection . First we have inflamtnation . Macrophages, then ncutrophils, engulf the bacteria. Interstitial fluid is flushed into the lymphatic system vvhere lymphocytes wa it in the lymph nodes. Macrophagcs process and present
the bacterial antigens to B lymphocytes. With the help of Helper T cells, G lymphocytes differentiate into rnemory cel ls and plasma cells. The memory cells arc preparation in the event that the ~all1c bacteria evcr attack again (the secondary response). The plasma cells produce antibodies, which are released into the blood to attnck the bacteria . You must know that '-1 single ant ibody is specific for a single antigen, and that a single 13 lymphocyte produces only one antibody type.
Copyright
2007 Examkrackers, Inc.
LECTURE
7.7
7:
THE CARDIOVASCULAR SYSTEM; THE RESPIRATORY SYSTEM . 155
Blood Types
Blood types are identified by the A and B surface antigens. For instance, type A blood means that the red blood cell membrane has A antigens and does not have B antigens. Of course, if the erythrocytes have A antigens, the immune system does not make A antibodies. Type 0 blood has neither A nor B an tigens, and makes both A and B antibodies. Thus, a blood d ono r may donate blood only to an ind ividua l that does not make antibodies against the donor blood. Figure 7- 13 shows a '+' sign when blood agglutinates (is rejected), and a '-' sign when no-agglutination occurs. Notice that an individual with type 0 blood may donate to anyone Call minuses in the donor column), and an individual with ty pe AB blood may receive from anyone Call minuses in the recipient column). The genes which produce the A and B antigens are co-dominant. Thus, an individual having type A or B blood may be heterozygous or homozygous. An individ ual with type 0 blood has two recessivk alleles.
~
:::
..QJ. . 0.. ..<.J .. QJ
~
Donor A
B
AB
0
A
-
+
AB
-
-
+ + -
-
B
+ -
-
0
+
+
+
-
Blood type
Genotype
A
("I A or l'i
B
1"1" or I"i
AB
1"1"
0
ii
Blood Types Figure 7-13
Rh factors are surface proteins on red blood cells first identified in Rhesus monkeys. Individuals having genotypes that code for nonfunctional products of the Rh gene are said to bc Rh-negative. All others are Rh-positive. Transfusion reactions involving the Rh factor, if they occur at all, are usually mild. Rh factor is more of a concern during the pregnancy of an Rh-negative mother with an Rb-positive fetus. For the first pregnancy, the mother is not exposed to fetal blood until giving birth and problems are rare. Upon exposure, the mo ther develops an immune response against the Rh-positive blood. In a second pregnancy, the second fetus that is Rh-positive may be allaeked by the antibodies of the mother, which are small enough to pass the placen tal barrier. The p roblem is life threa tening, and treatment usually involves complete replacement of the fetal blood with Rh-negative blood for the first few weeks of life.
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to 2007
Ex;)mkr3ckcr~. Inc.
166. Which of the followin g is true concerning type B negative blood?
Questions 161 through 168 are NOT based on a descriptive passage.
A. B.
161. Anemia (decreased red blood cell counl) can be caused by over activity of which of the following organs?
C.
A . . Thymus B.· Thyroid C. Spleen D. Lymph nodes
D.
167. Which of the following would you not expect to find in a lymph node?
162. "Swollen glands" are often observed in the neck of a person with a cold. The most likel y explanation for this is that: A. B. C. D.
A. B. C. D.
blood pools in the neck in an attempt to keep it warm. lymph nodes swell as white blood cells proliferate within them to fight the infection. the infection sets off an inflammatory response in the neck, causing fluid to be drained from the area. fever causes a general expansion of the tissues of the head and neck.
B lymphocytes proteins discarded by tissue cells invading bacteria old erythrocytes
168. An individual exposed to a pathogen for the first time will ex hibit an innate immune response involving: A. B. C. D.
163. Humoral immunity involves the action of: A. B. C. D.
Type B negalive blood will make antibodies that attack type A. antigens but not type B antigens. Type B negative blood will make antibodies that attack type B antigens but not type A antigens. Type B negative blood will make antibodies that attad type 0 antigens only. Type B negative blood will make antibodies that attack both type A and type 0 an tigens.
B lymphocytes T lymphocytes granulocytes An individual exposed to a pathogen for the first time must acquire immunity before it can respond.
cytotoxic T lymphocytes. stomach acid. pancreatic enzymes. immunoglobulins.
164. Antibodies function by: A. B. C.
phagocylizing invading antigens. adhering to circulating plasma cells and marking them for destruction by phagocytizing cells. preventing the production of stem cells in h/,)e bone
D.
attach ing to antigens via their variable portions.
marrow.
165. Lymphatic vessels absorb fluid from the interstitial spaces and carry it to the: A. B. C. D.
kidneys, where it is excreted. large intestine, where it is absorbed and returned to the bloodstream. lungs, where the fluid is vaporized and exhaled. lymphatic ducts, which return it Lo the circulation.
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'"
156
STOP.
Muscle, Bone, and Skin
8.1
Muscl e
There are three types of muscle tissue: L
skeletal muscle;
Know the three types of muscle and their four possible functions.
2. cardiac muscle; 3.
smooth muscle.
An y muscle tissue gen e rates a force only by contracting its cells. The mechanism s by which muscle cells contract differ be tween the three types of tissue, and are described below. M uscle contraction h as four possible functions: L
body movement;
2.
stab ilization of body position;
3. m ovement of substances through the body; 4.
8.2
generating heat to maintain bod y temperature.
Skeletal Muscle
Skeletal muscle is voluntary muscle tissue. It can be consciously controlled. Skele tal muscle COlulects one bone to anoth er. The muscle d oes n ot attach directly to the bone, but instead is attached v ia a tendon. (A tendon connects muscle to bone; a ligament comlects bone to bone.) Typically, a muscle stretches across a joint. The muscle origin is on the larger bon e, which rem ains relatively station ary, and its insertion is on the smaller bon e, w hich mo ves rela ti ve to the larger bon e up on contraction of the muscle. Muscles work in groups. The agonist (the muscle responsible for the m ovement) co.o tT:' <=Icts, w hli p a s pcond musc.Je, th e antagonist, stretches. When the antagonist contracts, the bone moves in the opposite direction, stretching the agonist. An example of antago nistic muscles is th e upper arm m uscles, the biceps and the triceps. In addi tion to antagonistic muscles, there are usu ally synergistic muscles. Synergistic muscles assis t the agonist by stabilizing the origin bone or by positioning the insertion bone during the m ovement. In this way, skeletal oluscle allows for m ovement and posture.
Contraction of skeletal muscle m ay squeeze blood and lymph vessels aiding circulation.
A muscle uses leverage by applying a force to a bone at its insertion point and rotating the bone in some fashion about the joint. This is a likely MeAT topic because It applies the phYSics concept of leverage to a biological system . It may seem strange, but most lever systems of the body typically act to increase the required force of a muscle contraction. In
other words, a greater force than mg is required to lift a mass 111. This is done in
order to reduce the bu lk of the body and increase the ra nge of movement. If the muscle has a shorter 'lever arm, it is
closer to the body, and thus creates less bulk.
158
MeAT BIOLOGY
Contraction of skeletal muscle produces large amounts of hea t. Shivering, controlled by the hypo thalamus upon stimulation by receptors in the skin and spinal cord, is the rapid contraction of skeletal muscle to warm the bod y
8.3 It'S not important to memorize the parts of a sarcomere, but it is necessary to understand the components of muscle contraction. In other words, given the labeled diagram, you should be able to te ll that the H zone and I band get smaller, while the A band does not change size.
Physiology of Skeletal Muscle Contraction
The smallest functional unit of skeletal muscle is the sarcomere (Fig. 8-1 ). A sarcomere is composed of many strands of h¥o protein filaments, the thick and the thin filament, laid side by side to form a cylindrical segment. Sarcomeres are positioned end to end to form a myofibril. Each myofibril is surrounded by the specialized endoplasmic reticulum of the muscle cell ca lred the sarcoplasmic reticulum. The lumen of the sarcoplasmic reticulum is fill ed with Ca2+ ions for reasons th at sh all become clear shortly. Lod ged between th e m yofibrils are mitochondria and many nuclei. Skeletal muscle is multinucleate. A modified memb rane called the sarcolemma wraps several myofibrils together to form a muscle cell or muscle fiber. Many muscle fibers are further bound into a fas ciclIllls, and many fa sciculae m ake up a single muscle. The thick filament of a sarcomere is made of the protein myosin. Several long myosin molecules wrap around each other to form one thick filament. Globular heads protrude along both ends of the thick filament. The thin filament is composed mainly of a polymer of the globular protein actin. Attached to the actin are the proteins troponin and tropomyosin.
Recognize the importance of ca lcium in muscle contraction, and be fami liar with the functions of the sarcoplasmic reticulum and the Hubules.
Myosin and ac tin work together sliding alongSide each other to create the contractile force of skeletal muscle. Each myosin head crawls along the actin in a 5 stage cycle (Fig. 8-2). First, tropomyosin covers an acti ve site on the actin preventing the myos in head from binding. The myosin head remains cocked in a high-energy position with a phosphate and ADP group attached . Second, in the presence of Ca' + ions, troponin pu lls the tropomyosin back, exp osing the active site, allowing the m yosin head to bind to the actin. Third, the myos in head expels a phosphate and ADP and bends into a low energy position, dragging the ac tin along with it. This is called the p ower stroke beca use it causes the shortening of the sarcomere and the muscle contraction. In the fo urth stage, ATP attaches to the myosin head . This releases the m yosin head from the active site, which is covered immediately by tropom yosin. Fifth, ATP splits to inorganic phosp hate and ADP causing the myosin head to cock into the high-energy position. This cycle is repeated many times to form a contraction. A muscle contraction begins w ith an action potential. A neuron attaches to a muscle cell forming a neuromuscular synapse. The ac tion p otential of the neuron releases acetylcholine into the synaptic cleft. The acetylcholine activates ion channels in the sa rcolemma of the muscle cell creating an ac tion potential. The action potential moves deep into the muscle cell via small turmels in the membrane called T-tubules. T-tubules allow for a uniform contrac tion of the m uscle by allowing the action potential to spread through the muscle cell more rapidly. The action potential is tran sferred to the sarcoplasmic re ticulum, w hi ch s udd e nl y bpcomes permeable to Ca 2+ ions. The Ca2 +ions begin the 5 stage cycle described above. At the end of each cycle, Ca 2+ is actively pumped back into the sarcoplasmic reticulum.
Copyright © 2007 Examkrackers, Inc.
LECTURE
~ Muscle
8:
MUSCLE, BONE, AND SKIN .
Muscle Fasciculus
~--r;::.:;;;:::
Muscle Fiber
Muscle Fiber
Nucleus Capillary
A band
I
Z line
I
I
~ mm:
ff4f4f( 1mm
· · ·
. . .
HIHH
Hzone
Z line 1
~
Actin thin filament
'>----------"'~
~
n:n
Jband
Structure of Skeletal Muscle Figure 8-1
Copyright © 2007 Examkr
\ Myosin thick filament
159
160
MeAT BIOLOGY
8.4
A Motor Unit
The muscle fibers of a single muscle do n ot all contract a t once. Instead, from 2 to 2000 fibers spread through out the muscle are innervated by a single neuron. The neuron and the m uscle fibers that it innervates are called a motor IInit. Motor units are independent of each other. The force of a con tracting muscle depends u pon the number and size of the acti ve motor units, and the frequency of action potentials in each neuron of the motor unit. Typically, smaller motor units are th e first to be activa ted, and larger motor units are recruited as needed. This results in a smooth increase in the force genera ted by the muscle. Another important point concerning motor units is that muscles requiring intrica te movements, like those in the finger, have smaller lnotor units, whereas muscles requiring greater force, su ch as those in the back, have larger mo tor units.
C"
0 Troponin ~(ffQ;~~~ Actin Filament
Tropomyosin
/
Myosin
cD-=~'
\
fir
ADP
Troponin-Tropomyosin
/
-:Ii
/
Complex Blocks Actin-M yosin Interaction
Movement of Troponin - Tropomyosin Allows Actin-Myosin interaction
Physiology of Skeletal Muscle Contraction Figure 8-2
Copyright © 2007 Examkrackers, Inc.
LECTURE
8.5
8:
MUSCLE, BONE, AND SKIN .
161
Skeleta l Muscle Type
There are three types of skeletal muscle fibers: 1) slow oxidative (type 1) fibers; 2) fast oxidative (type 1I A) fibers; andfost glycolytic (type 1I B) f ibers. Type I or slow-twitch m uscle fibers are red from large amounts of myoglobin. Myoglobin is an oxygen storing protein similar to hemoglobin, but having only one protein subunit. Type I fibers also contain large amoun ts of mitochond ria. They split ATP at a slow rate. As a result, they are slow to fatigue, but also have a slow contrac tion velocity. Type 11 A or fast-twitch A fibers are also red, but they split ATP at a high rate. Type II A fibers contract rapidly. Type II A fibers are resistant to fatigue, but not as resistant as type I fibers. Type II B or fast-twitch B fibers have a low m yoglobin content, appear white under the light microscope, and contract very rapidly. They contain large amounts of glycogen.
Myoglobin stores oxygen inside muscle cells. A molecule of myoglobin looks like one subunit of hemoglobin . It is capable of staling only one molecule of oxygen .
Most muscles in the body have a mixture of fiber types. The ra tio of the mixture depends upon the contraction requirements of the muscle and upon the genetics of the individual. Large amounts of type I fibers are found in the postural muscles. Large amounts of type II A fibers are found in the upper legs. Large amounts of type II B fibers are found in the upper arms. Adult human skeletal muscle does not generally undergo mitosis to create new muscle cells (hyperplasia). Instead, a number of changes occur over time when the muscles are exposed to forceful, repetitive contractions. These changes include: the diameter of the muscle fibers increases, the nwnber of sarcomeres and mitochondria increases, and sa rcorneres lengthen. This increase in muscle cell diameter and change in muscle conformation is called hypertrophy.
Copyright © 2007 Examkrackers, Inc.
Like many ceil types, human muscle cells are so specialized that they have lost the ability to undergo mitosis. Only in rare cases does one muscle cell split to form two cells.
173. When one of a pa ir of antagon istic mu scles contracts ,
Questio ns 169 t hrough 176 are NOT based on a descripti ve passage.
what usually happens to the other muscle to produce moveme nt? A.
169. During a muscular contraction: A. B. C. D.
B. C. D.
both th e thi n and th ick filaments con tract. the thin fIl ament contracts, bu t th e thi ck ti lament doc s not. the thick fila me nt contracts, but the thin fila ment does not. neither the thi n no r the thick filament contract.
174. When unde rgoin g physica l exercise, hea lthy adult skeletal mu scle is li kely to respond wi th an increase in all of the followi ng except:
170. Irreversib le seq ues tering of calc ium in the sarcoplasmic
A. B.
retic ulum wou ld most likely: A. 8.
C. D.
re sult jn perm anent contractio n of the mu scle fibers, simi lar to what is seen in ri gor mo rtis. create a sharp increase in bone density as calci um is reso rbed fro m bo nes to replace the sequeste red cal-
A. B. C. D.
preve nt myosin from bind ing to actin. depoJymeri ze actin fil a me nts in the sarcomere.
171. Sh ive rin g increases body tem'perature by:
A. B.
C. D.
glycolys is the Citric Ac id Cycle mitosis prote in prod uction
175. T he biceps muscle is connected lO the radius bone by:
cium. C. D.
Tt acts synergisticall y by contrac ting to stabil ize the moving bone. It relaxes to all ow moveme nt. It con tracts in an isometric action. Its inse rtion sli des down the bone to all ow a larger ran ge of movement.
serving as a warn in g that body temperature is too low, promp ting the person to seek warmer locatio ns. caus ing bones to ru b together, c reati ng heat through fri ctio n. increas ing the ac tiv ity of muscles. conv inc ing the hypothalamus that body tem peratu re is higher than it actu all y is.
biceps tendon an nul ar li gamen t of the rad ius articular cartilage th e triceps muscle
176. Skeletal mu scle contraction may ass ist in all of the follow in g EXCEPT: A. B. C. D.
movement of flu id through the body body temperature regu latio n posture peristalsis
172. M uscles cause move me nt at joints by: A. B.
C. D.
Copyri~Jht
inci tin g neuro ns to inili ate an e lectrical " twitch" in tendons. increasi ng in length , thereby pushin g the muscle's o ri g in and inserti o n farther apart. fi lling wi th blood, thereby expandin g and inc reasin g the d istan ce between the ends of a muscle. dec reas in g in le ngth, thereby brin g ing the m uscle's orig in and inSCl1io n closer to~cth cr.
© 2007 EX
162
STOP.
LECTURE
8.6
8: MUSCLE,
BONE, AND SKIN .
Cardiac Muscle
The human heart is composed mainly of cardiac muscle (Fig. 8-3) . Like skeletal Inuscle, cardiac muscle is striated, which m,eans that it is composed of sarcomeres. However, each cardiac muscle cell contains only one nucleus, and is separated from its neighbor by an intercalated disc. The intercalated discs contain gap jtffiCHons w hich allow an action potential to spread from one cardiac cell to th e next via electrical synapses. The mitochondria of cardiac muscle are larger and more numerous. Skeletal muscle connects bone to bone via tendons; cardiac muscle, on the other hand, is not connected to bone. Instead, cardjac muscle forms a net w hich contracts in upon itself like a squeezing fi st. Cardiac muscle is invohintary. Like . skeletal muscle, cardiac muscle grows by h ypertropJ:y. The action poten tial of cardiac muscle exhibits" plateau after depolarization. The plateau is created by slow voltage-gated calcium channels which allow calcium to en ter and hold the inside of the membrane at a positive potential elifference. The plateau lengthens the time of contraction.
Mitochondrion Transverse Tubule Sarcoplasmic Reticulum
Sarcolemma
+-- Sarcomere ----+
Cardiac Muscle Tissue Figure 8-3 Copyright Ij;) 2007 Examkrackers, Inc
163
164 . MCAT B,OLOGY
8.7
Smooth Muscle
Smooth muscle is mainly involuntary, so it is innervated by the autonomic nervous system (Fig. 8-4). Like cardiac muscle, smooth muscle cells contain only one nudeus. Smooth muscles also contain thick and thin filaments, but they are not organized into sa rcomeres. In addition, smooth muscle cells contain intermediate filaments, which are attached to dense bodies spread throughout the cell. The thick and thin filaments are attached to the intermediate fil aments, and, when they contract, they ca use the intermediate filaments to pull the dense bodies together. Upon contraction, the smooth muscle cell shrinks leng th-wis" There are two types of smooth muscle: 1. single-unit and 2. multiunit. Single unit smooth muscle, also called visceral, is the most common. Single-unit smooth muscle cells are connected by gap junctions spreading the ac tion potential from a single neuron thro ugh a large group of cells, and allowing the cells to contract as a single unit. Single-unit smooth muscle is found in small arteries and veins, the stomach, intestines, uterus, and urinary bladder. Each multiunit smooth muscle fiber is attached directly to a neuron. A group of multiunit smooth muscle fibers can contract independently of other muscle fibers in the same location. Multiunit smooth muscle is found in the large arteries, bronchioles, pili muscles attached to hair follicles, and the iris. In addition to responding to neural stimulus, smooth muscle also contracts or re-
la xes in the presence of hormones, or to changes in pH, O 2 and CO2 levels, temperature, and ion concentrations.
Muscle fiber Autonomic neuron
Autonomic neuron
Muscle fiber
•
Dense bodies
Intermediate filaments
Contraction
Multiunit Smooth Muscle Single-unit Smooth Muscle
Smooth Muscle Figure 8-4 Copyright © 2007 Examkrackers, Inc.
182. Cardiac muscle is excited by:
Questions 177 through 184 are NOT based on a descriptive passage.
A. B. C. D.
177. The function of gap junctions in the intercalated discs of cardiac muscle is to: A. B. C. D.
183. When left alone, certain specialized cardiac muscle cells have the capacity for self-excitation. The SA node is a collection of such cells. TIle SA node is innervated by the vagus nerve. The freq uency of self excitation of the cardiac cells of the SA Node is likely to be:
anchor the muscle fibers together. insure that an action potential is spread to all fibers in the muscle network. control blood flow by selectively opening and closing capillaries. release calcium into the sarcoplasmic reticulum.
A. B.
178. Which of the following muscular actions is controlled by the autonomic nervous system? A.
n. C. D.
C.
the knee-jerk reflex conduction of cardiac muscle action potential from cell to cell peristalsis of the ga'trointestinal tract contraction of the diaphragm
D.
B. C. D.
slower than a normal hearbeat because excitation by the vagus nerve decreases the heart rate. slower than a normal hearbeat because excitation by the vagus nerve increases the heart rate. faster than a normal hearbeat because excitation by the vagus nerve decreases the heart rate. faster than a normal hearbeat because exdtation by the vagus nerve increases the heart rate.
184. In extreme cold, just before the onset of frostbite. sudden vasodilation occurs manifested in flush skin. This vasodilation is most likely the result of:
179. During an action potential, a cardiac muscle cell remai ns depolarized much longer than a neuron. This is most likely to: A.
parasympathetic nervous excitation. constriction of T-tubules. increased cytosolic sodium concentration . increased cytosolic calcium concentration.
A. B.
prevent the initiation of another action potential during contraction of the heart. ensure that adjacent cardiac muscle cells will contract at different times. keep the neuron from firing twice in rapid succession. allow sodium voltage-gated channels to remain open long enough for all sodium to exit the cell.
C. D.
paralysis of smooth muscle in the vascular walls. paralysis of skeletal muscle surrounding the vascular walls. sudden tachycardia with a resultant increase in blood pressure. blood shunting due to smooth muscle sphincters.
180. All of the following are true concerning smooth muscle
EXCEPT: A. B. C. D.
Smooth muscle contractions are longer and slower than skeletal muscle contractions. A chemical change in the environment around smooth muscle may create 'a contraction. Smooth muscle does not require calcium to contract. Smooth muscle is usually involuntary.
181. Which of the following muscles is under voluntary conlrol? A. B. C. D.
the diaphragm the heart the smooth muscle of the large intestines the iris
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165
STOP.
166
MCAT
BIO LOGY
8.8
Bone
Bone is living tissue. Its functions are support of soft tissue, protection of internal organs, assistance in movement of the body, mineral storage, blood cell production, and energy storage in the form of adipose cells in bone marrow. Bone tissue contains four types of cells surrounded by an extensive matrix.
1.
Osteopragenitor (or osteogenic) cells differentiate into osteoblasts.
2.
Osteoblasts secrete collagen and organic compounds upon which bone is formed. Osteoblasts are incapable of mitosis.. As osteoblasts release matrix materials around themselves, they become enveloped by the matrix and differentiate into osteocytes.
3.
Osteocytes are also incapable of mitosis. Osteocytes exchange nutrien ts and waste materials with the blood.
4.
Osteoclasts resorb bone matrix, releasing minerals back into the blood. Osteoclasts are believed to develop from the white blood cells called monocytes.
A typical long bone (Fig. 8-5) h as a long shaft, called the diaphysis, and two ends, each end composed of a metapilysis and epiphysis. A sheet of cartilage in the metaphysis, called the epiphyseal plate, is where long bones grow in length. Spongy bone contains red bone marrow, the site of hemopoiesis or red blood cell development Compact bone surrounds the medullary cavity, which holds yellow bone marrow. Yellow bone marrow contains adipose cells for fat storage. Compact bone is highly organized. In a continuous remodeling process, osteoclasts burrow tunnels, called Haversian (central) canals, through compact bone. The osteoclasts are followed by osteoblasts, which la y down a new matrix onto the tunnel walls form ing concentric rings called lamellae. Osteocytes trapped between the lamellae exchange nutrients v ia canaliculi. Haversian canals contain blood and lymph vessels, and are connected by crossing canals called Volkmann's canals. The entire system of lamellae and Haversian canalis called an osteon (Haversian system).
8.9
Bone Function in Mineral Homeostasis
Calcium salts are only slightly soluble, so most calcium in the blood is not in the form of free calcium ions, but is bound mainly by proteins and, to a much lesser extent, by phosphates (HPO/-) and other anions. It is the concentration of free calcium ions (Ca",) in the blood tha t is important phYSiologically. Too much Ca'+ results in m embranes becoming hypo-excitable producing lethargy. fatigue, and memory loss; too little produces cramps and convulsions. Most of the Ca 2+ in the body is stored in the bone matrix as hydroxyapatite [Ca lO (P04MOH),j. Collagen fibers lie along the lines of tensile force of the bone, giving the bone great tensile strength. Hydroxyapatite crystals lie alongside collagen fibers, and give bone greater compressive s trength than the best reinforced concrete. Some of the body's Ca2+ exists in bone in the form of slightly soluble calcium salts such as CaHP04. It is these salts that buffer the plasma Ca2• levels. Thus bone acts as a storage site for Ca2 -t and HPO/-.
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LECTURE 8: MUSCLE, BONE, AND SKIN .
167
8. 10 Bone Types and Structures Most bones fan into one of four types: l. IOllg; 2. sltort; 3. flat; or 4. irregular. Long bones have a shaft that is curved for strength. They are composed of compact and spongy bone. Leg, arm, finger and toe bones are long bones. Short bones are cuboidal. They are the ankle and wrist bones. Flat bones are made from spongy bone surrounded by compact bone. TI1ey provide large areas for muscle attachment, and organ protection. The skull, sternum, ribs and shoulder blades are flat bone. Irregular bone has an irregular shape and variable amounts of compact and spongy bone.
Know the functions of osteoblasts , osteocytes, and osteoclasts.
i{elllcrnber that bone is not just for support, protection, and movement. Bone also stores calcium and phosphate, helping to maintain a consistent concentration of these ions in the blood. Bone stores energy in the form of fat. And bone is the site of blood cell formation.
~:=:::::-,,/ Articular Cartilage
Spongy Bone (contains red bone marrow)
Osteocyte
Compact (dense) Bone _ __Medullary (marrow) Cavity (contains yellow bone marrow in adult)
Artery
Lamellae
Spongy Bone Trabeculae
/1Blood Vessels
Blood Vessels
in Haversian Canal
Articular Cartilage
Bone Structure Figure 8-5 Copyrig ht (9) 2007 Exmnkracke rs, Inc.
\
Volkmann's Canal with Blood Vessels
168 . MeAT
BIOLOGY
8.11 Cartilage Cartilage is flexible, resilient connective tissue. It is composed primarily of collagen, and has great tensile strength. Cartilage contains no blood vessels or nerves except in its outside membrane called the pericholldrilll11. There are three types of cartilage: 1. hyaline; 2. fibrocartilage; and 3. elastic. Hyaline ca rtilage is the most common. H yaline ca rtilage reduces friction and absorbs shock in joints.
8.12 Joints Joints can be classified by structure into three types: 1.
Fibrous joints occur between two bones held closely and tightly together by fibrou s tissue permitting little or no movement. Skull bones form fibrous joints with each other, and the teeth form fibrous joints with the mandible.
2.
Cartilaginous joints also allow little or no movement. They occur between two bones tightly connected by cartilage, such as the ribs and the sternum, or the pubic symphysis in the hip bone.
3.
Synovial joints (Fig. 8-6) are not bound directl y by the intervening cartilage. Instead, they are separated by a capsule fill ed with synovial fluid . Synovial fluid provides lubrication and nourishment to the cartilage. In addition, the synovial fluid contains phagocytotic cells that remove microbes and particles which result from wear and tear from joint movement. Synovial joints allow for a wide range of movement.
Synovial fluid
Articular cartilage
Synovial Joint Figure 8-6
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LECTURE 8: MUSCLE, BONE, AND SKIN •
8.1 3 Skin The skin (Fig. 8-7) is an organ, which means that it is a group of tissues working together to perform a specific fW1ction. Some important functions of the skin are: 1. Thennoregulation; The skin helps to regulate the bod y temperature.
Blood conducts heat from the core of the body to skin. Some of this heat can be dissipated by the endothermic evaporation of sweat, but most is dissipated by radiation. Of course, radiation is only effecti ve if the body is higher than room temperature. Blood can also be shunted away from the capillaries of the skin to reduce heat loss, keeping the body warm. Hairs can be erected (piloerection) via sympathetic stimulation trapping insulating air next to the skin. Skin has both warmth and cold receptors. 2. Protection; The skin is a physical barrier to abrasion, bacteria, dehydration, m any chemicals, and ultra violet radiation. 3. Environmental sensory input; The skin gathers information from the environment by sensing tempera ture, pressure, pain, and touch. 4. Excretion; Water and salts are excreted through the skin. This water loss occurs by diffusion through the skin and is independent of sweating. Adults lose one quarter to one half liter of water per da y via this type of insensible flflid loss. Burning of the skin can increase this type of wa ter loss dramatically. 5. Immunity; Besides being a p hysical barrier to bacteria, specialized cells of the epidermis are components of the immune system. 6. Blood reservoir; Vessels in the dermis hold up to 10% of the blood of a resting adult. 7. Vitamin D synthesis; Ultra violet radiation activates a molecule in the skin that is a precursor to vitamin D. The activated molecule is modified by enzymes in the liver and kidneys to produce vitamin D. The skin has two principal parts: 1) the epidermis and 2) the dermis. Beneath the skin is a subcutaneous tissue called the sflperfic ial fascia or hypodermis. The fat of this subcutaneous layer is an important hea t insulator for the bod y. The fat helps maintain normal core bod y temperatures on cold days while the skin approaches the temperature of the environment.
The epiderm is is avascular (no blood vessels) epithelial tissue. It consists of fo ur major cell types: 1) 90 % of the epidermis is composed of Keratinocytes, which produce the protein keratin that helps waterproof the skin. 2) Melanocytes transfer /IIelanin (skin pi gment) to keratinocytes. 3) Langerhal1s cells interact with the the helper T-cells of the immune system. 4) Merkel cells attach to sensory neurons and function in the sensa tion of touch. There are five strata or layers of the epidermis. The deepest layer contains Merkel cells and stem cells. The stem cells continually divide to pro.d uee keratinocytes and other cells. Keratinocytes are pushed to the top layer. As they rise, they accumulate keratin and die, losing their cytoplasm, nucleus, and other organelles. When the cells reach the outermost layer of skin, they Slough off the body. The process of keratinization from birth of a cell to sloughing off takes two to four weeks. The outermost layer of epidermis consists of 25 to 30 layers of flat, dead cells. Exposure to friction o r pressure stimulates the epid ermis to thicken froming a callus.
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169
17 0
MeAT B,OLOGY
The dermis is connective tissue derived from mesodermal cells. The dermis is embedded by blood vessels, nerves, glands, and hair follicles. Collagen and elastic fibers in the dermis provide skin with strength, extensibility, and elasticity. The dermis is thick in the palms and soles. The skin, hair, n ails, glands, and some nerve endings make up the integumentary system. Hair, nails, and some glands are derivatives of embryonic epidermis. Hair is a column of kera tinized cells held tightly together. As n ew cells are added to its base, the hair grows. Most hairs aTe associated with a sebaceous (oil) gland that empties oil directly into the follicle and onto the skin. When contracted, smooth muscle (arrector pili), also associated w ith each hair, stands hair up pointing it perpendicular to the skin. Nails are also keratinized cells. Sudoriferous (sweat) glands are found in the skin separate from hair follicles. Cerumif10us glands produce a wax-like material found iri the ears.
Hair shaft Sweat pore
]
Epidermis
Corpuscle of touch (Meissner's corpuscle) Sebaceous (oil) gland ---t1f--:;?!f'--::--~~ Arrector pili muscle ------->tt--'<;----'cT-''<'<±:-t'''--~''_/t--''-__If'-'
Dermis
Sensory nerve - - - - - - -\.\ Hair follicle - - - - ---"1 Lameliated (Pacinian) corpuscle Sudoriferous ----~~,,--TI_j'-+__f+--~k:i~q (sweat) gland Vein - - -- - - - - -r --"'T' Artery -------~~~~~h "!' Adipose tissue --------:::,;,L.::~t""'-
Subcutaneous layer
...
Skin
Figure 8-7
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190. All of the following are found in compact bone EXCEPT:
Questions 185 through 192 are NOT based on a descriptive passage.
185. The production of which of the foll owing cells would most likely be increased by parathyroid hormone'! A.
osteoprogenitor
B.
osteocyte
C.
osteoclast
D.
osteoblast
A.
reduce friction between bone ends.
keep bone cells adequately hydrated. occupy space until the bones of the joint complete their growth. maintain a rig id connection between two flat bones.
D.
A.
muscle bone
D.
haversian canals
C. D.
canaliculi Volkman n's canals
is the mineral porti on of bone. Hydroxyapatite contains all of the following elements
except:
A. B. C. D.
calcium sulfur phosphate hydrogen
192. The spongy bone of the hips is most important in: A. B. C. D.
187. Surgical cutting of which of the following tissues would result in the LEAST amount of pain? B. C.
yellow marrow
B.
191. Hydroxyapatite
186. In a synovial j oint. the purpose of the synovial fluid is to: B. C.
A.
red blood storage red blood cell synthesis fat storage lymph tl uid production
cartilage skin
188. In a synovial jo int, the connective tissue holding the bones lOgelher are called:
A.
ligaments
B.
tendons
C.
muscles
D.
osseous tissue
189. All of the following are functions of bone EXCEPT: A. B. C. D.
mineral storage structural support . blood temperature regulation fat storage
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171
STOP.
Populations
9.1
Mendelian Concepts
Gregor Mendel was a 19'" century monk, who performed hybridization experiments with pea plants. The difference between Mendel and those who had come before him was that Mendel quantitated his results; he counted and recorded his findings. Mendel found that when he crossed purple flowered plants with white flowered plants, the 'first filial, or F, generation, produced purple flowers. He called the purple trait dominant, and the white trait recessive. Mendel examined seven traits in alt and each trait proved to have dominant and recessive alterna-
pp
x
pp
p
p
Cross pollinated
r-----_________~____----------~
Pp
Pp
Pp
p l pp l pp l p Pp Pp
Pp Self-pollinated
p
~_ _ _---~~
r - -_ _ _ _ _ _ _
p
p Ipp l pp l p Pp pp pp
Pp
Pp
Mendelian Ratio 3:1 Figure 9-1
pp
174
MeAT
B,OLOGY
tives. When Me nd el self-pollina ted the F, generation plants, the F2 genera tion exp ressed both the d ominan t and recessive traits in a 3 to 1 ratio, now referred to as the Mendelian ratio (Fig. 9-1). When the F, generation was self-pollina ted , 33% of th e domin an ts p rod uced only dominan ts, an d the rest of the d ominants produced the Mendelian ra tio. The w hi te flowered plants prod uced only w hi te flowered plants. Thus, h a lf of the F, generation expressed the d om ina nt tra it w ith the recessive tr ait latent.
To test his model, Mendel performed a
test cross, where he crossed the heterozygous F, generation (purple) with homozygous recessive parent (white). Since there were white offspring resulting from this cross of a purple Fj plant and a white parent plant, Mendel proved the F, generation was heterozygous.
Th e expression of a trait is the phenotype, and an individ ua l's gen etic lllake up is the genotype. In Fig ure 9-1 the phenotyp es are p urple and white; the genotyp es are PP, Pp, and pp . The p hen otype is expressed th rou gh the ·action of enzymes a nd other structural p ro teins. Wh ich are encoded b y gen es. In complete dominance, exhibited by the flowers in Mendel's experim ent, fo r nn y on e trait, a d iploid individ ual will h ave two ChrOlTIOSOm es each conta ini ng a sep ara te gene that codes for tha t sp ecific trait. These two chrom osomes are h omologous by definition. Their corresp o n din g gen es ar e locate d a t th e s am e locus or p os iti o n on respective chromosomes. Each gen e con trib utes an allele to the gen otype. H owever, only one allele, the dom inant a llele, is exp ressed. U both alleles arc dominant, th en the dom in ant pheno type is exp ressed; if both alleles a re recessive, then the recessive phenotyp e is expressed. An individual w ith a genoty pe ha ving two dominan t or two recessive alle les is said to be homozygous for that trait. A n individual w ith a genotype h aving one d ominant and one recessive a llele is said to be heterozygous for th e tra it, an d is called a hybrid. Men del's First La w of H ered ity, the Law of Segregation, sta tes that alleles segrega te indep en d ently of each other w hen forming gametes . Any game te is equally likely to posses any allele. Also, the phenotypic express ion of the alleles is n ot a blen d o f the two, but an expression of the d ominant allele (the p rinciple of complete dominance) .
Mendel self-pollinated hi' pea plants. Mating relatives is called inbreeding, and does not change the frequency of alleles, but does increase the number of homozygous individuals Within a population, Outbreeding, or outcrossing, IS mating of nonrclatives which produces hybrids or 11eterozygJtes.
When a he terozygo us individual exhibits a p henotyp e that is in te rmediate beh-veen its homozygous counterparts, the alleles a re referred to as partial, or illcolllplete dOIllinall ts. Alleles showing pa rtia l d ominance are represented w ith the sam e capital letter, and disting uished with a prime or superscript. For ins tance, a cross betw'een red flowered sweet peas an d wh ite flow ered sweet peas may p roduce pink fl ow ers. The genotype for the pink flowered individual would be ex pressed as either CC or C' Cwo If the heterozygote exhibits both phen otyp es, th e a lleles are [OdOlllillallt . H uman blood type alleles are cod ominant becau se a heterozygote exhibits A and B antigens on th e b lood cell membranes. Figure 9-1 shows a Punnett square for predicting genotypic ratios of offsprin g. The gen otypes of all possible gametes of each paren t are d isp layed in the first column and first row resp ectively. The alleles are then combined in the correspon d ing boxes to show the p ossible geno types of th e offspring. Since, accord in g to the law of segrega tion, each gametic genotype is equally likely, each offspring genotype is also eq ually li kely. Mendel's Second Law of He red ity, the Law of Independent Assortment, states tha t genes located on different ch ro mosomes assort independentl y of each o ther. In other w ord s, genes tha t cod e for different tra its (such as pea shap e and pea color), when loca ted on d ifferen t chrom osomes, do n ot affect each o ther d ming gamete form a tion. If tw o gen es are located on the same chromosom e, th e likelihood that they w ill remain together d urin g gamete fo rmation is indirectly p roportional to the distance sep arating them. Thus, the closer th ey are on the chromosom e, the more likely they will remain together. In Figure 9-2, we use a P unnett square to p redict the phenotypiC ratio of a dihybrid cross. 'W' is the allele fo r a round pea shape, which is do min an t, and 'w' is the allele fo r wrinkled p ea shap e, which is recessive. 'G' is the allele for yellow color, w hich is d om inant, and 'g' is the allele for green
,
Copyright © 2007 Exarnkracke rs, )nc
LECTURE
WG
Wg
wG
9: POPULATIONS
•
175
wg -
WG
0000 C) 0 e () C) ~9 ® -@ C)
WWGG WWGg WwGG WwGg Wg
WWgG WWgg wG
wg
WwgG
Wwgg --=-
wWGG wWGg wwGG
wwGg
wWGg
wwgg
wWgg
wwgG
9 yellow, round 3 yellow, wrinkled 3 green, round 1 green, wrinkled
•
Dihybrid Cross WwGgx WwGg Hgure 9-2
color, which is recessive. We make the assumption that the genes for pea shape and pea color are on separate chromosomes, and will assort independently of each other. Notice the phenotypic ratio of a dihybrid cross, 9:3:3:1 . The chromosomes of males and females differ. In humans, the 23'd pair of chromosom es es tablishes the sex of the ind iv idual, and each p arhler is called a sex chromosome. One of the 23'" chromosomes of a male is abbreviated. Instead of appearing as two XS in a karyotype (a map of the chromosomes), the chromosome · pair appears as an X and a Y. All other chromosomes appear as two Xs. When a gene is fo und on the sex chromosome it is called sex-linked. Generally, the Y chromosome does not carry the allele for the sex-linked trait; thus, the allele that is carried by the X chromosome in the male is expressed whether it is dominant or recessive. Since the fema le has two X chromosomes, her geno type is found through the normal rules of dominance. Howevel~ in Illost sOlnatic cells, one of the X ChrOlTIOSOmes wi ll condense, and most of its genes wi ll become inactive. The tiny dark object formed is called a Barr body. Barr bodies are form ed at random, so the active allele is split about evenly among the cells. Nevertheless, in most cases, the recessive phenotype is only displayed in homozygous recessive individuals. Thus, the female may carry a recessive trait on her 23'" pair of chromosomes w ithout expressing it. If she does, she is said to be a carrier " XH for the tra it. Such a recessive trait has a strong chance of being expressed .~ in her ma le offspring regardless of the genotype of her mate. V
Normal male
Xl<
y
~
Hemophilia is a sex-linked disease. The Punnett square shown in Figure 9-3 shows a cross between a female carrier for hemoph il ia and a healthy male. Since there are two possible phenotypes fOT the males, and one is the recessive phenotype, the male offspring from such a pairing have a 1 in 2 chance of ha ving the disease.
Male with 1--=--1 hemophilia
-;;" E Xh X"Xh Xhy ~
Sex-linked Traits Figure 9-3
CopYrigh t © 2007 Exarnkrackers. Inc.
,
male with the disease a nd a fe male that is not diseased, but carries the trait, produce two girls. What is the probability that neither girl carries a recessive allele?
Questions 193 through 200 are NOT based on a descriptive passage.
A. B. C. D.
193. Color-blindness is a sex-linked recessive trait. A woman who is a carrier for the trait has two boys with a colorblind man. What is the probability that both boys are color-blind?
A. B. C. D.
0% 25% 50% 66%
198. The parents of a dihybrid cross:
0% 25% 50% 100%
A; B. C.
D.
194. In the pedigree below, the darkened figures indicate an individual with hemophHia, a sex- linked recessive disease. The genotype of the female marked A is:
are genetic opposi tes at the genes of interest. are genetically identical at the genes of interest. are genetic opposites at one. gene and genetically identical at the other. have no genetic relationship.
199. Sex-linked traits in me n usually result due to genes located: A. B.
C. D.
on both chromosomes of a pair of homologous chromosomes. on one chromosome from a pair of homologous chromosomes on both chromosomes of a pair of nonhomologous chromosomes on one chromosome from a pair of nonhomologous chromosomes
200. Colorblindness is a sex-linked recessive trait. A woman is horn colorblind. What can be said with cenainty?
A.
XHX"
B. C. D.
X"X" X"X" X"Y
A. B. C. D.
Her father and mother are colorblind. Her mother and daughte r are colorblind. Her father and son are colorblind. Her father is colorblind, but'her son mayor may not
be.
195. .What fraction of offspring are likely to display both dominant phenotypes in a dihybrid cross?
A.
B. C. D.
1116 3116 9116
15116
196. In a dihybrid cross, what frac tion of the offspring are likel y to be dihybrids? A.
B. C. D.
1/16 1/4 112 9116
197. Sickle ceU anemia is an autosomal recessive disease. A Copyright © 2007 Examkrackers, Inc.
176
STOP.
LECTUR E
9.2
9:
POPULATIONS
. 177
Evolution
The gene pool is the total of all alleles in a population. Evolution is a change in the gene pool. Figure 9-4 shows the eye color alleles for a small population. Even if the ratio of blue eyed to brown eyed individuals temporarily changes, as long as the gene pool remains 30% b alleles and 70% B alleles, the population has not evolved.
BB
" B
BB B B
Bb
B
b
B B B
b B B
b
b
B
H
b B
B
B B
B
b
B B
B
30% b 70% B
Population
Gene pool
Gene Pool of a Population Figure 9-4
Most taxonomical c1assificatiOl'l systems are based upon genetic similarity. The classification system for animals that you must know for the MeAT contains ever more specific groupings in the following order: Kingdom, Phylum, Class, Order, Family, Genus, Species. (Plants and fungi use divisions instead of· phyla.) Within each gro1)P are many subgroups, which are unimportant for the MCAT (except for the subphylum Vertebrata, which is in the phylum Chordata). Since organisms within the same group have similar genetic structures, they probably share similar phylogenies (evol u tionary histories). For instance, all mammals belong to the class Mamm alia and the phylum Chordata; thus, all mammals probably share a common ancestor that they do not share with birds, which are also in the phylum Chordata, but in the class Aves. The taxonomy is changin g, and for the MCAT you may want to be awa re of the new super kingdoms called domains. There are three domains: Bacteria, Archaea, and Eukarya. This basically puts the kingdoms of Protista, Fungi, Plantae, and Animalia in):o the domain Eukarya. It makes the kingdom Monera obsolete dividing it into the domains of Bacteria and Archaea. The two domains of Bacteria and Archaea are divided into several kingdoms each. Archaea is more closely related to Eukarya than is Bacteria.
When naming an organism, the genus and species name are given in order. Typically, they are both written in italics, and th e genus is capitalized while the species is not. Copyright © 2007 Examkrackers, Inc.
You shou ld be aware that ontogeny reapitu lates phylogeny. In other words, the cou rse of development of an organism from embryo to adu lt refiects its evolutionary history. For instance, the human fetus has pharyngeal pouches reflecting a gilled ancestor.
178
MeAT B,OLOGY
Species is loosely limited to, but not inclusive of, all organisms that can reproduce fertile offspring with each other. In other words, if two organisms can reproduce fertile offspring, they might be the same species; if their gametes are incompatible, they are definitely not the same species. Another guideline for species (but still an imperfect guideline) is all organisms which normally reproduce selectively fit offspring in the wild. Oganisms of different species may be prevented from producing fit offspring by such things as geographic isolation (separated by geography), habitat isolation (live in the same location but have different habitats), seasonal isolation (mate in different seasons), mechanical isolation (physically impossible to mate), gametic isolation (gametes are incompatible), developmental isolation (fertilized embryo developes improperly), hybrid inviability or sterility (hybrid malformed), selective hybrid elimination (hybrid is less fit), and behavioral isolation (different mating rituals).
In order to survive, the members of the same species will exploit their environment in a unique manner not shared by any other species. The way in which a species exploits its environment is called its niche. No two species can occupy the same niche indefinitely. The theory of survival of the fittest predicts that one species will exploit the environment more efficiently, eventually leading to the extinction of the other with the same niche. The definition of the "fittest" organism in this theory is the organism which can best survive to reproduce offspring which will, in turn reproduce offspring and so on generation after generation. This definition may include living beyond rep rod uction in order to proVide a better chance for offspring to reproduce. In fact, there are two opposing reproductive strategies: r-selection and K-selection. r-selection involves producing large numbers of offspring that mature rapidly with little or no parental care. r-strategists generally have a high brood Inortality rate. Their population growth curves are exponential. r-strategists are generally found in unpredictable, rapidly changing environments affected by density independent factors such as floods, or drastic temperature change. K-selection is the other side of the spectrum. K-selection involves small brood size with slow maturing offspring and strong parental care. K-strategists tend to have a sigmoidal growth curve which levels off at the carrying capacity. (The K comes from an equation variable representing carrying capacity.) The carrying capacity is the maximum number of organisms that an environment can maintain. The carrying capacity is a density dependent factor. Most organisms have reproductive strategies somewhere between K- and r-selections.
Speciation is the process by which new species are formed. When gene flow ceases between two sections of a population, speciation begins. Factors which bring about speciation include geographic, seasonal, and behavioral isolation. Adaptive radia.. tion occurs when several separate species arise from a single ancestral species, such as the 14 species of Galapagos finches that all evolved from one ancestor. A species may face a crisis so severe as to cause a shift in the allelic frequencies of the survivors of the crisis. This is called an evolutionary bottleneck. Divergent evolution exists when two or lnore species evolving from the same group maintain a similar structure from the common ancestor (called a homologolls structure). However, hvo species may independently evolve similar structures in conve rgent evolution. Such similar structures are said to be analogous or homoplastic. An example of homoplasticity is the wings evolved by bats and birds; the two do not share a common ancestor from which they received their wings.
Some phenotypic forms vary gradually within a species, such as height. There are short people, tall people and every height in between short and tall. Other forms are distinct, like flower color, either red or white, or chicken plumage, either barred or non-barred. The occurence of distinct forms is called polymorphism .
•
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LECTURE
9.3
9:
POPULAT:ONS
Symbiosis
A symbiosis is a relationship between two species. The relationship can be beneficial for both, called mutualism; beneficial for one and not affect the other, called commensalism; beneficial for one and detrimental to the other, called parasitism. There is even a symbiosis called enslavement where one species enslaves another.
9.4
In the early part of the 20 th century, there was some question as to how less frequent alleles might be maintained in the population. Hardy and Weinberg came up with the explanation simultaneously. They showed statistically that there should be no change in the gene pool of a sexually reproducing population possessing the five following conditions:
large population;
2.
mutational equilibrium;
3.
immigration or emigration must not change the gene pool;
4.
random mating; and
5.
no selection for the fittest organism.
A population with these five characteristics is considered to be in Hardy-Weinberg equilibrium. No real population ever possesses these characteristics completely. Small populations are subject to genetic drift where one allele may be permanently lost due to the death of all members having that allele. Genetic drift is not caused by selective pressure, so its results are randOlTI in evolutionary terms. Mutational equilibrium means that the rate of forward mutations exactly equals the rate of back mutations. This rarely occurSLIl real populations; however, ill the short term, mutations are seldom a major factor in changing allelic frequencies. Any imn1igration or en1igration lTIUst not change the gene pool. This condition may occur in some isolated populations and is not typically a major factor in genetic change. The last two conditions probably do not occur in natural populations and are the lTIOst influential m echanisms of evolution. The binon1ial theorem:
p2 + 2pq + q2 predicts the genotype frequency of a gene with only two alleles in a population in Hardy-Weinberg equilibrium. Imagine that 'A' is the dominant allele and 'a' is the recessive allele, and they are the only alleles for a specific gene. Now imagine that SO% of the alleles are 'A' . This means that SO% of the gametes will be 'A' and 20% will be 'a'. The probability that two 'A's come together is simply 0.S2 = 0.64. The probability that two :a's come together is 0.22 = 0.04. Any remaining zygotes will be heterozygous, leaving 32% heterozygotes. (2 x 0.8 x 0.2 = 0.32) Using the formula, we represent' A' as p and' a' as q. Since there are only two alleles, p + q = 1 .
Copyright It) 2007 Exarnkrach>rs, Inc
~~
commensalism
~
Q
parasitism
~
@
Figure 9-5 Symbiosis
Hardy-Weinberg Equilibrium
1.
mutualism
•
. 179
206. The wolf, or Canis lupus, is a member of the family Canidae. Which of the following is most likely to be true?
Questions 201 through 208 are NOT based on a descriptive passage.
A. B.
201. If a certain gene possesses only two alleles. and the dominant allele represents 90% of the gene pool. how many individuals display the recessive phenotype. A. B. C.
D.
C.
D.
0% 1% 18% 10%
207. If, in a very large population, a certain gene possesses only two alleles and 36% of the population is homozygous dominant, what percentage of the population are heterozygotes? '
202. Which of the following would least likely disrupt the Hardy-Weinberg equilibrium? A. B. C.
D.
A.
emigration of part of a population a predator that selectively takes the old and sick a massive flood killing 15% of a large homogeneous population exposure of the entire population to intense radiation
B.
C. D.
B. C. D.
16% 24% 36% 48%
208. Although human behavior ensures the success of each new generation of corn, selective breeding by humans has genetically altered corn so that it could not survive ~n the wild without human intervention. Corn population is controlled, and most of the corn seeds are eaten or become spoiled. The relationship between humans and corn is best described as:
203. Which of the following is most likely an example of two organisms in the same species?
A;
There are more living organisms classified as Canidae than as Canis. There are more living organisms classified as lupus than as Canis. An organism may be classified as Canis but not as Canidae. An organism may be classified as lupus, but not as Canis.
A cabbage in Georgia and a cabbage in Missouri that mate and produce fertile offspring only in years of unusual weather patterns. Two fruit: flies on the same Hawaiian island with very different courtship dances. Two South American frogs that mate in different seasons. Two migratory birds that nest: on different: islands off the coast of England.
A. B. C. D.
commensalism because humans benefit and com is neither benefited nor harmed. commensalism because there is no true benefit to either species. parasitism because humans benefit and corn is harmed. mutualism because both species benefit.
204. All of the following factors would most likely favor an r-selection reproductive strategy over a K-selection strategy EXCEPT: A. B. C. D.
intense seasonal droughts a short growing season limited space large scale commercial predation by humans
205. If two species are members of the same order, they must also be members of the same:
A. B. C. D.
habitat family class biome
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180
STOP.
LECTURE
9.5
Origin of Life
The universe is 12 to 15 billion years old. According to the Big Bang Theory, The universe began as a tiny spec of highly concentrated mass and exploded outward. In the early moments, only hydrogen gas existed. As the universe cooled, helium was able to form. The explosion was irregular, and gravitational forces created clumping of the mass. Heavier elements, and solar systems formed from these clumps of mass.
Our solar system is approximately 4.6 billion years old. The earth itself is about 4.5 billion years old; however, due to the volatile nature of early earth, there are no rocks on earth older than 3.9 billion years old. Early earth probably had an atmosphere made mainly from nitrogen and hydrogen gas, and very little oxygen gas. One theory holds that the atmosphere contained clouds of H,S, NH" and CH4 creating a reducing environment. From this environment, the formation of carbon based molecules that we associate with life required
little energy to form. Experiments attempting to recreate the atmosphere of early earth have resulted in the autosynthesis of molecules such as urea, amino acids, and even adenine. The Urey-Miller experiment was one of the early experiments to make such an attempt. The first cells are thought to have evolved from coacervates, lipid or protein bilayer bubbles. Coacervates spontaneously form and grow from fat molecules suspended in water. Organisms may have initially assimilated carbon from methane and carbon dioxide in the early atmosphere. The earliest organisms were probably heterotrophs subsisting on preformed organic compounds in their immediate surroundings. Fossils of these organisms have
been dated at 3.6 billion years old. As preformed compounds became scarce, some of these organisms developed chemosynthetic autotrophy followed by photosynthetic autotrophy. . Around 2.3 billion years ago, the ancestors of cyanobacteria evolved. They were able to use sunlight and water to reduce carbon dioxide. These were the first oxygen producing, photosynthetic bacteria. The atmosphere began to fill with oxygen. Eukaryotes evolved about 1.5 billion years ago, and did not develop into multicellular organisms until several million years later.
9.6
Chordate Features
Chordata is the phylum containing humans. Chordata does not mean backbone. All chordates have bilateral symmetry. They are deuterostomes, meaning their anus develops from or near the blastopore. (Compare protostomes, where the mouth develops from or near the blastopore.) Chordates have a coelom (a body cavity within mesodermal tissue). At some stage of their development they possess a notochord (an embryonic axial support, not the back bone), pharyngeal slits, a dorsal, hollow nerve cord, and a tail. Members from the subphylum Vertebrata have their notochord replaced by a segmented cartilage or bone structure. They have a distinct brain enclosed in a skull. Most chordates are vertebrates. Vertebrata is composed of two classes of jawless fish (Agnatha), the cartilaginOUS fish, bony fish, amphibians, reptiles, birds, and mamCopyright © 2007 Examkrackers, Inc.
9:
POPULATIONS
•
181
182
MeAT
BIOLOGY
mals. The agnatha arose first and seperatel y from the rest about 470 million years ago. Amphibians arose from bony fish. Reptiles arose from amphibians about 300 million years ago. Birds and mammals arose from reptiles. Mammals arose from reptiles about 220 million years ago. This last section contains b its of science trivia. Although it is probable that the MeAT will ask something from this section, it will be only one question or it will be explained in a passage. Keep this in mind, when you decide how much time you . want to spend memorizing the details of this section. You have now reviewed all the science tested by the MeAT. I suggest that you go . back and review all of the tests that you have taken to this poin t. When you are done, you shou ld pick your weakest area and master it. Then go to your next weak· est area, and so on.
Next week is Zen Week. This is an importan t week of mental preparation. Be sure to attend. If you'/re not in the class, see the website at www.examkrackrs.com fo r information on Zen Week. See you.
•
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214. Humans are members of the order:
Questions 209 through 216 are NOT based on a descriptive passage.
A. B.
209. Which of the following was the earliest to evolve on earth?
e.
Primata
D.
Homididae
215. What do the results of the Urey-Miller experiment
A. B. C.
protists
A.
D.
fish
B.
plants prokaryotes
demonstrate?
C.
210. All of the following are characteristics of members of the phylum Chordata at some point in their life cycle EXCEPT: A. B. C. D.
D.
a tail a notochord a backbone gills
A. B. C. D.
Chordata?
A. D.
the existence of life on earth that small biological molecules cannot be synthesized from inorganic material that li fe may have evolved from inorganic precursors that humans have evolved from photosynthetic cyanobacteria
216. If the first living organisms on earth were heterotrophs, where did they get their energy?
211. Which of the following are not members of the phylum
B. C.
Vertebrata Chordata
from from from from
eating each other eating naturally formed organic molecules the su n eating dead organisms.
tunicates apes
birds ants
212. Which of the following was probably not necessary for the origin of life on earth?
A.
H,O
B.
hydrogen
e.
0,
D.
carbon
213. Which of the followi ng is the phylum to which Homo sapiens belong?
A. B. C.
D.
Mammalia Chordata Vertebrata Homo
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183
STOP.
STOP! . DO NOT LOOK AT THESE EXAMS UNTIL CLASS.
30-MINUTE IN-CLASS EXAM FOR LECTURE 1
'.
185
Passage I (Questions 1-7)
3. Which of the following is the best explanation for why attempts at predicti~g protein configuration based upon amino acid sequence have been unsuccessful?
The three dimensional shape of a protein is ultimately determined by its amino acid sequence. The folding pattern itself is a sequential and cooperative process where initial folds assist in aligning the protein properly for subsequent folds. For many smaller proteins, the amino acid sequence alone can direct protein configuration, but for other proteins assistance in the folding process is necessary. Two types of proteins may assist in the folding of a polypeptide chain: enzymes which catalyze steps in the folding process, and proteins which stabilize partially folded intermediates. Proteins in the latter group are called chaperones.
A.
B.
C.
D.
An example of an enzyme which cat;:tlyzes the folding process is protein disulfide isomerase. This enzyme assists in the creation of disulfide bonds. The enzyme is not specific for any pmilcular disulfide bond in a given chain. Instead, it simply increases the rate of formation of all disulfide combinations, and the most stable disulfide formations predominate.
4. Chaperones assist in the formation of a protein's: A.
B. C. D.
Chaperones also assist in protein folding. As the protein folds, chaperones bind to properly folded sections and stabilize them. Chaperone synthesis can be induced by application of heat or other types of stress, and they are sometimes referred to as heat shock proteins or stress proteins.
A.
B.
1. Which of the following statements concerning the function of protein disulfide isomerase in the formation of proteins is tme?
B. C. D.
primary structure. secondary structure. tertiary structure. quaternary structure.
5. Natural selection has resulted in increased chaperone synthesis in the presence of elevated temperatures. How might increased chaperone production in the presence of heat be advantageous to a cell?
Although the amino acid sequence determines the configuration of a protein, attempts at predicting protein configuration based upon amino acid sequence have been unsuccessful.
A.
It is impossible to know the amino acid sequence of a protein without knowing the DNA nucleotide, sequence. Enzymes and chaperones help to determine the three dimensional shape of a protein. The three dimensional shape of a protein is based upon hydrogen and disulfide bonding between amino acids, and the number of possible combinations of bonding amino acids makes prediction difficult. The amino acid sequence of the same protein may vary slightly from one sample to the next.
Protein disuffide isomerase increases only the rate at which disulfide bonds are formed. Protein disulfide isomerase increases only the rate at which disulfide bonds are broken. Protein disulfide isomerase increases both the rate at which disulfide bonds are formed and broken. Protein disulfide isomerase increases the rate at which disulfide bonds are formed and decreases the rate at which disulfide bonds are broken.
C.
D.
Heat destabilizes intermolecular bonds making protein configuration more difficult to achieve. Chaperones counteract this by stabilizing the partially folded intermediates. Increased temperatures increase reaction rates creating an excess of fully formed proteins. Chaperones stabilize the partially folded intermediates and slow the process. Increased chaperone production requires energy. This energy is acquired from the kinetic energy of molecules and thus cools the cell. Elevated temperatures result in increased cel1ular activity requiring more proteins. Chaperones Increase the rate of polypeptide formation.
1 2, According to the passage, the folding pattern of a protein is determined by the protein's: A. B. C. D.
primary structure secondary structure tertiary structure quaternary structure
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Passage II (Questions 8-14)
6. Which of the following bonds in a protein is likely to be LEAST stable in tbe presence of heat? A. B. C. D.
a disulfide bond a hyd rogen bond a polypeptide bond the double bond of a carbonyl
In polyacrylamide gel electrophoresis (PAGE), electrically charged proteins are dragged by an electric field through the pores of a highly cross-linked gel matrix at different rates depending upon their size and charge. In a SDS-PAGE, proteins are separated by size only. Since different proteins have different native charges, the protein mixture is first dissolved in SDS (sodium dodecyl sulfate) solution. SDS anions disrupt the noncovalent bonds of the proteins
7. Protein disulfide isomerase most likely: A.
lowers the activation energy of the fonnation of cys-
B.
tine. raises th e activation energy of the formation of cys-
C. D.
and a')Sociate with the peptide chain s, approximately one mole-
cule of SDS for every two residues of a typical protein. The resu1ting net negative charge on the protei n is normally much greater th an the charge on the native protein. Mercaptoethanol is usually added in the presence of heat to reduce disulfide bonds and complete the denaturization process. This solution is then applied to a porous gel and an electric field is applied. The rate of movement through the gel is inversely proportional to the logarithm of the molecular weight of the protein.
tine. lowers the activation energy of the formation of proline. raises the activatio n energy ~f the formatio n of pro-
line.
The proteins are then stai ned with a dye such as coomassie blue. Lines are formed at differe nt points along the gel corresponding to the molecular weight of the proteins. A second type of electrophoresis, caned isoelectricfocusing, distinguishes proteins based upon their isoelectric points. A permanent pH gradient is established within a polyacrylamide gel by applying an electric field to polyacrylamide polymers with different pIs. When the native proteins are applied to this gel in the absence of SDS , each protein moves until it reaches its pl.
8. SDS PAGE would be least effective in distinguishing between the masses of different: A. B. C. D.
carbohydrate-ric h glycopro teins. acidic proteins. polar proteins. enzymes.
9. What is the purpose of cooma"ie blue? A. B. C. D.
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187
Coomassie blue increases the separation of the proteins by increasing their mass. Coomassie blue increases the separation of the proteins by increasing their charge. Coomassie blue allows the results of electrophoresis to be visualized. Coomassie blue stops further m{)vement of the protcins by increasing their mass.
GO ON TO THE NEXT PAGE.
Passage III (Questions 15-21)
10. Which of the following proteins would move tbe most slowly through the ge l in SDS PAGE? A. B. C. D.
a large protein a small protein a protein with a high native charge a protein with a low native charge
Gl ycolysis is the metabolic breakdown of glucose into the readily useable form of chemical energy, ATP. For some human cells, such as neurons and erythrocytes, glucose is the qnly source of chemical energy available under typical circu'm· stances.
11. Electrophoresis is also used to analyze nucleic acids. [n
Glucose
electrophoresis of nucleic acids. SDS is unnecessary because: A.
ATP~
Hexokinase
ADP
nucleic acids already contain negatively charged phosphate groups in proportion to their size. nucleic acids are already large enough to separate appreciably on their own.
Glucose 6-phosp hate
C.
nucleic acids don't have a quaternary or tertiary
Fructose 6-phosphate
D.
structure to disrupt. ' nucleic acids do not contain hydrogen bonds.
ATP~
B.
L Phosphohexose isomerase Phosphorructokinase (PFK)
ADP 12. SDS is a detergent that does not cleave covalent bonds.
Fructose 1.6-bisphosphate
Which protein structure cannot be dismpted by SDS ?
A. B. C. D.
} Aldolase primary secondary tertiary quaternary
Dihydroxyacetone ~ Glyceraldehyde PhosPhat/':~ 3-phosphate Triose phosphate Isomerase
13. Which of the following would most likely occur if a mllltisllbunit protein were subjected to the electrophoresis techniques used in SDS PAGE? A. B.
C. D.
The protein would remain intact and separate from other prote ins according to its native charge. The prote in would re main intact and separate from other prote ins according to its size. Each subunit would separate independendy. according to its native charge. Each subunit would separate independently, according to its size.
Phosphoglycerate kinase .
1
2-Phosphoglycerate
LEnolase Phosphenolpyruvate
ADP~ Pyruvate ATP
SDS PAGE is likely to be an expensive process. SDS PAGE can not easily distinguish between protei ns of similar molecu1ar weight. Any proteins used in SDS PAGE are denatured. The native charge on a protein does not always cor· respond to i L~ size.
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1.3~~~~hO SPhoglycerate ATP
mixture?
C. D.
3.phosphate dehydrogenase
NADH
3-Pbosphoglycerate Phosphoglycerate mutase
14. Whi ch of the following is most likely a limitation to SDS PAGE in identifying different proteins within a protein A. B.
NAD;i Glyceraldehyde
kinase
Pym vate
Figure 1 Glycolysis Each reaction in the glycolytic pathway is governed by an enzyme. Glucose is phosphorylated as it enters the cell in .an irreversible reacti on with hexokinase. It is not until the reacti on governed by phosphofructokinase (PFK), however. that the molecule is committed to the glycolytic pathway. The PFK reaction is called the committed step. PFK activity is inhibited when cellular energy is plentiful. and stimulated when energy is low.
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19. Why is the PFK reaction. and not the hexokinase reaction, the cOIrunitted step?
Glycolysis can be interrupted by poisons that interfere with glycolytic enzyme activity. Arsenate. a derivative of arsenic. is a deadly poison that acts as a substrate for glyceraldehyde 3phosphate dehydrogenase.
A. B.
C. 15. The net result of aerobic respiration can be summarized most accurately as: A. B. C.
D.
the the the the
D.
oxidation of glucose. reduction of glucose. elimination of glucose. lysis of glucose.
20. From the information in the passage, which of the following might he an allosteric activator of PFK?
16. The process of the synthesis of ATP.in the glycolytic reac-
A. B. C. D.
tion governed by phosplwglycerate kinase is called: A. B. C. D.
oxidative phosphorylation. substrate-level phosphorylation. exergonic phosphate transfer. electron transport.
C. D.
Glucose + Glucose + 2ATP + 2 Glucose + Glucose +
citrate insulin ADP ATP
21. The action of arsenate on glyceraldehyde 3-phosphale dehydrogenase is best describe as:
17. Which of the following gives the net reaction for glycolysis? A. B.
The hexoki nase reaction is irreversible, but the PFK reaction is not. The PFK reaction requires the hydrolysis of ATP, but the hexoklnase.reaction does not. Glucose 6-phosphale is a higher energy molecule than fructose 1,6-bisphosphate. Glucose 6-phosphate may be converted into glycogen in some circumstances, but fructose 1,6-bisphosphate has only one possible chemical fate in the cell.
A. B. C. D.
4 ADP .... pyruvate + 2 ATP 2 ADP + 2 P; + 2 NAD+ .... 2 pyruvate + NADH O2 .... CO2 + H20 + 2 ATP 0 2 .... 2 pyruvate + 2 ATP + 2 NADH
competitive inhibition. noncompetitive inhibition. allosteric inhibition . negative feedback.
18. Which of tbe following would most likely occur inside a cell in the presence of arsenic? A. B.
C. D.
The concentration of glyceraldehyde 3-phosphate dehydrogenase would decrease. The concentration of glyceraldehyde 3-phosphate would increase. The concentration of aldolase would increase. The co ncentration of 1.3-bisphosp hoglycerate would increase.
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23. The substrate concentration in a reaction which is governed by an enzyme is slowly increased to high levels. As the substrate concentration increases the reaction rate:
Questions 22 through 23 are NOT based on a descriptive passage.
A. B. C. D.
22. Cofactors are best described as: A. B. C.
D.
nonprotein substances required for all enzyme activity. small, nonprotein, organic molecules. metal ions or coenzymes that activate an enzyme by binding tightly to it. catalysts.
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continues to increase indefinitely. continues to decrease indefinitely. increases at first, then levels off. does not change.
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
190
STOP.
30-MINUTE IN-CLASS EXAM FOR LECTURE 2
J
191
Passage I (Questions 24-28)
26, As presented in the passage, the theory that Pelomyxa represents an early stage in the evolution of eukaryotic cells would be most weakened by which of the following?
The giant amoeba-like Pelomyxa palustris, the only member of the phylum Caryoblastea, exhibits one of the most primitive forms of cell division in eukaryotes; it does not undergo mitosis. Instead, the nucleus simply splits into two daughter nuclei. Like most other protists, the nucleus of the Pelomyxa is bound by a nuclear envelope. During cellular division, the chromosomes of the Pelomyxa double in number and assort randomly to the daughter cells, Multiple copies of each chromosome ensure that each daughter cell maintains the necessary amount of genetic material to specify the organism. Pelomyxa has no centrioles or mitochondria; however, it does contain two bacterial symbionts which may function similarly to mitochondria. Pelomyxa may represent an early stage in the evolution of eukaryotic cells,
A. B.
C. D,
A protist containing mitochondria and undergoing mitosis, proved to be an ancestor to Pelomyxa. Diatoms were found to have evolved from Pelomyxa. New evidence placed dinoflagellates in the kingdom plantae. A second species belonging to the phylum Caryoblastea is discovered and found to undergo mitosis.
27, The separation of duplicate chromosomes of dinoflagellates most closely resembles which two phases in mitosis: A, B, C. D.
Unicellular protists from the phylum Pyrrhophyta, commonly called dinoflagellates, undergo a form of mitosis where the nuclear membrane remains intact. Microtubules extending through the nuclear membrane attach to chromosomes. The nuclear membrane grows between attached chromosomes separating them and creating two daughter nuclei, The dinoflagellate then divides with each daughter cell accepting a nucleus. Most dinoflagellates are photosynthetic, and most are protected by cellulose plates.
prophase and metaphase metaphase and anaphase anaphase and telophase anaphase and cytokinesis
28, The bacterial symbionts in Pelomyxa most likely: A, B. C.
Mitosis in diatoms from the phylum chrysophyta is similar to that in dinoflagellates, though slightly more advanced.
D.
parasitically infect the Pelomyxa. provide energy for the Pelomyxa from the metabolism of absorbed nutrients. reproduce independently from the Pelomyxa through a primitive mitosis. function in lipid synthesis.
24. Which of the following is true according to the passage? A.
B. C. D.
Eukaryotes that lack centrioles cannot undergo mitosis. Some prokaryotes undergo mitosis, while others do not. Not all eukaryotes undergo mitosis. All protists lack mitochondria.
25, During mitosis, the nuclear envelope of a mammalian cell: A. B. C. D.
disintegrates during replication of the chromosomes. disintegrates while crossing over is taking place. disintegrates while the chromosomes condense. remains intact.
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I
Passage II (Questions 29-34)
29. Which of the following would most strongly indicate that the poly-A tail is added after transcription and not during transcription?
For many genes in eukaryotes, the mRNA initially transcribed is not translated in its entirety. Instead, it is processed in a series of steps occurring within the nucleus. The initial unprocessed mRNA is called pre-mRNA. Just after transcription begins, the 5' end of the pre-mRNA is capped with GTP forming a 5'-5' triphosphate linkage. The cap protects the mRNA from exonucleases acting on its 5 " end, and acts as an attachment site for ribosomes. At the 3' end, the polyadenylation signal, AAUAA, signals the cleavage of the pre-mRNA 10 to 20 nucleotides downstream. Severa] adenosine residues are added to the 3 ' end to form a poly-A tail. The poly-A tail wraps tightly around an RNA-protein complex protecting the 3 ' end from degradation.
./
A.
B. C. D.
30. Small nuclear ribonucleoproteins (snRNPs) catalyze the splicing reaction in the post-transcriptional processing of RNA. Based upon the information in the passage, in which of the following locations would snRNPs most likely be found?
As much as 90% of the pre-mRNA may be removed from the nucleotide sequence as introns. The remaining exons are spliced together in a single mRNA strand, and leave the nucleus to be translated 1n the cytosol or on the rough endoplasmic reticulum. intron
f
,---L-
5' .J-
A. B. C. D.
exon
the cytosol of a prokaryotic cell the cytosol of a eukaryotic cell the lumen of tbe endoplasmic reticulum the nucleus of the eukaryotic cell
,--L-,
i
[ I---iJr---;l lc---, ----1 ___ J,---- .....j J
I
"1
LJ
31. The loops in Figure 2 represent:
pre-mRNA
A. B. C. D.
Poly-A tail ;--.
5' ICap-l:~
Evidence of an enzyme in the nucleus that catalyzes the synthesis 0f a sequence of multiple adenosine phosphate molecules A sequence of multiple thymine nucleotides following each gene in eukaryote DNA A sequence of mUltiple uracil nucleotides following each gene in eukaryote mRNA A sequence of multiple adenosine nucleotides following each gene in eukaryote DNA
;
Ii,
MaturemRNA Figure 1 Post-transcriptional processing of mRNA
DNA introns DNA exons RNA introns RNA exons
32. In R looping the displaced DNA strand contains the complementary nucleotide sequence for: A. B. C. D.
In R looping, a technique used to identify introns, a fully processed mRNA strand is hybridized with its double stranded DNA counterpan under conditions which fa vor formation of hybrids between DNA and RNA strands. In this process, the RNA strand displaces one DNA strand and binds to the other DNA strand along complementary sequences. The results can be visualized through electron microscopy and are shown in Figure 2.
the entire DNA strand shown in Figure 2. the entire RNA strand shown in Figure 2. the DNA introns only. the RNA introns only.
33. The intron sequences of identical genes in closely related species are often very different. Whi ch of the following is most strongly suggested by this evidence? A.
In prokaryotes, mRNA is normally translated without modi fication.
B.
....
C. D,
Identical genes in closely related species may code for different proteins. Changes in the amino acid sequence of a protein do not necessarily change protein function. Intron sequences are heavily characterized by selective pressure. Selective pressure has little or no role in the development of intron sequences .
..•. Figure 2 R looping
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Passage III (Questions 35-42)
34. Which of the following represents the S' end of eukaryotic mRNA?
A.
C. C H,
0
1
;Guanine I I
.~¥ . ;t;;;""'HA 6H •
I phosphate
IG~aninel
0
H"j
)., H
H~H
OH group I
I phosphate
group I
Iphosph~te group I
Iphosph~te group] phosph~te group
I ,
CH 1
An alteration in cellular DNA other than genetic recombination is called a mutation. Mutations can occur in both germ and somatic cells. Germ line mutations are transmitted to offspring and can result in genetic diseases. Somatic mutations lead to neoplastic diseases, which are responsible for 20% of all deaths in industrialized countries. The basal mutation rate is the natural rate of change in the nucleotide sequence in the absence of environmental mutagens. Such mutations result mainly from errors in DNA replication. A common error during replication is a tautomeric shift. Thymine, in its keto form, pairs with adenine; however, in its enol form, it prefers guanine. Adenine usually exists in an amino fonn that prefers thymine, and, rarely, in an imino fonn that pairs with cytosine. Tautomeric shifts result in mismatched base pairs.
Iphosph~te group I
I
I
~H, I
0
mRNAchain
]Adenine]
~H,
i
•
H~
]Thymine]
~
OH
! !
0
I
~nRNA chain
B.
D.
CH
0] Guanine]
~
I
.
0
,
!
~ , I
• H
mRNA chain
mRNAchainl
~H
H'
H
\
phosphate group!
I
I
I
I
: Adenine I
'
Environmental mutagens such as chemicals and radiation can increase the mutation rate. Ionizing radiation, such as gamma rays and x-rays, excites electrons creating radicals inside the cell that react with DNA. Ultraviolet radiation from normal sunlight cannot penetrate the skin nor make radicals; however, it can form pyrimidine dimers from adjacent pyrimidines. Chemicals, such as deaminating agents, can convert adenine, guanine, and cytosine into hypoxanthine, xanthine, and uracil, respectively, which lead to mismatched base pairs and thus errors during replication. Though deamination of adenine and guanine are rare, deamination of cytosine is fairly common.
I
phosphate group I , , I phosphate group j
phosphate group!
,
]Guanine]
i
phosphate group i
,
0 '
~ 6H OH
phosph~te group I
CH
CH
,
i
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0
]Cytosine]
~
Because many mutations occur initially to only one strand of a DNA double helix, they can be repaired. In a healthy cell there are specialized enzyme systems which monitor and repair DNA. However, once the DNA is replicated, these enzymes can no longer recognize the mutation and repair becomes very unlikely. For errors occurring during replication there are enzymes called glycosylases that can remove mismatches and small insertions and deletions. Glycosylases, which also work on the deaminated bases. must be capable of identifying mismatched base pairs and distinguishing between the old and new DNA strand.
194
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35. It has been hypothesized that in ancestral organisms DNA contained uracil and not thymine. Why might DNA that contains thymine have a natural selective advantage over DNA illat contains uracil? A.
C.
,
A. B.
DNA that contains uracil would be unable to utilize glycosylases that repair mutations caused by the deamination of cytosine. Uracil does not undergo a tautomeric shift and thymine does. DNA that contains uracil cannot form a double helix. Uracil is a purine, and thymine is a pyrimidine.
B.
.
39. Which of the following pairs of nitrogenous bases might form a dimer in DNA when ex posed to UV mdiation?
D.
C. D.
40. 5-bromouracil resembles thymine enough to become incorporated into DNA during replication. Once incorporated, however, it rearra nges to resemble cytosine. If 5-bromouracil were present during the replication of the sense strand shown below, which of the following might be the sense strand formed in the following replication?
36. Radiati on therapy is used to treat some form s of cancer by damaging DNA and thus killing the rapidly reproducing cancerous cells. Why might radiation treatment have a greater effect on cancer cells than normal cells? A.
5'-GGCGTACG-3' A. B. C. D.
Normal ceUs have more time between S phases to re pai r damaged DNA. Damaged DNA is more reactive to rad iation. Cancer cells have lost the ability to repair damaged DNA. The affect of radiation is the same, but there are more cancer cells than normal cells.
B. C. D.
A. B. C. D.
Tumors may produce neoplastic diseases. Neoplastic diseases are hereditary. Neoplastic diseases begin with the mutation of a normal cell. Exposure to chemical mutagens may lead to a neoplastic disease.
D.
A.
B. C. D.
Mutations do not occur in the absence of radiation or chemical mutagens. Industrial countries have a higher basal mutation rate than non-industrial countries. Hypoxanthine is an example of a chemical mutagen. Mutations occurring in rapidly reproducing cells are more likely to become a permanent part of the cell genome.
Prokaryotes do not possess a ligase enzyme to join separate DNA molecules. Prokaryotic DNA is single stranded. Eukaryotes have matching pairs of chromosomes to act as a template for repair. Eukaryotes contain more DNA making the consequences of a break less severe.
B. C. D.
Co pyright
a frameshift mutation a base pair insertion a base pair substitution a chromosomal aberrat ion
42. According to the information in the passage, which of the following is true conceming mutations?
38. Ionizing radiation can create double stranded breaks in DNA. Eukaryotes are able to repair some of these breaks, but prokruyotes are not. Which of the following gives th e most likely explanation for this difference? A.
S'-GGCGCACG-3' 5'-GGCGATCG-3' 3'-CCGCAGGC-5 ' 5'-GGCGTGCG-3 '
41. According to the passage, in the absence of glyeosylases, a tautomeric shifi would most likely result in which of the following mutations?
37. From the information in the passage, which of the following is LEAST likely to be true concerning neoplastic diseases?
A. B. C.
thymine - thymine thymine - adenine thymine - guanine uracil - uraci l
(~';)
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./
45. A primary spermatocyte is:
Questions 43 through 46 are NOT based on a descriptive passage.
A. B. C. D.
43. DNA replication occurs during: A.
prophase. metaphase. telophase. interphase.
B. C.
D.
46. Crossing over occurs in: A. B. C. D.
44. Turner's syndrome occurs due to nondisjunction at the sex chromosome resulting in an individual with one X and no Y chromosome. Color-blindness is a sex-linked recessive trait. A -color-blind man marries a healthy woman. They have two children both with Turner's syndrome. One of the children is color-blind. Wbich of the following is true?
mitosis, prophase. meiosis, prophase I. meiosis, prophase II. interphase.
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
Nondisjunction occurred in the father for both children. Nondisjunction occurred in the mother for both children. Nondisjunction occurred in the mother for the color-blind child and in the father for the child with normal vision. Nondisjunction occurred in the father for the colorblind child and in the mother for the child with normal vision.
A.
B. C.
D.
Copyright
haploid and contains 23 chromosomes. haploid and contains 46 chromosomes. diploid and contains 23 chromosomes. diploid and contains 46 chromosomes.
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47. Certain eating utensils are treated with a sanitizer. After 3 minutes the number of microbes is reduced from 6.25 x 10 12 to 2.5 X 10". According to the passage, 3 minutes more exposure to the sanitizer would reduce the number of microbes to:
Passage I (Questions 47-53) Disease-causing microbial agents are called pathogens. Microbial growth is affected by temperature, 02' pH, osmotic activity, and radiation. Physical methods of pathogen control include chemical, heat, filtration, ultraviolet radiation, and ionizing radiation. Moist heat is generally more effective than dry heat.
1.0 5.0 6.0 1.9
A.
B. C.
D.
Sterilization is the removal or destruction of all living cells, viable spores, viruses, and viroids. Sometimes it is only deemed necessary to kill or inhibit pathogens. This is called disinfection. Sanitation reduces the number of microbes to levels considered safe by public health standards.
Substrate
C. D.
Phosphate buffer
z value
12 seconds 2.4 minutes 12 minutes 2.0 x lOll minutes
B.
(0C)
C. peifringens Culture media
D 12I = 0.20
10
D9f)= 3-5
6-8
Salmonella
Chicken a la king
Dwl = 0.40
5.0
S. aureus
Chicken a la king
DW)= 5.4
5.5
S. aureus
Turkey stuffing
Dw= 15
6.8
S. aureus
0.5% NaCl
D6fj = 2.0-2.5
5.6
C.
50. A nutrient rich agar is seeded with a few bacteria, which quickly become a thriving colony. Which of the following most accurately depicts the exponential portion of the population growth on a logarithmic scale?
A.
"'lL/-~
c.
/
~
,I
""
I
.5 j
Table 1 D and z values for some food-borne pathogens
time
time
Like population growth, population death is exponential. However, as the popUlation reduces to very low levels, the propOliion of resistant strains increases slowing the overall rate of death.
B.
D.
~L-----dJ
-3 / ""I o I
The food processing industry relies on D and z values for guidelines in controlling contamination. For instance, after being canned, food must be heated sufficiently to destroy any endospores of Clostridium botulinum, the bacteria responsible for botulism. C. botulinum is an obligatory anaerobic, gram positive bacterium found in soil and aquatic sediments. When food containing endospores of C. botulinum is stored, the endospores may germinate and produce a deadly neurotoxin. Although the disease is fatal to 113 of untreated patients, it can be effectively treated with an injection of antibodies produced by horses.
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49. If federal regulations require that canned food be heated at 121°C long enough to reduce a colony of C. botulinum in a phosphate buffer from lOll bacteria to 1 bacterium, how long must canned food be heated?
D.
C. botulinum
X
0.24 seconds 2.4 seconds 24 seconds 40 minutes
A.
B.
A.
Bacteria
X
48. From Table I, the D value for Salmonella in Chicken a la king at 70'C is:
The decimal reductibn time (D) or D value is the time required to kill 90% of microorganisms or spores in a sample at a specified temperature. Environmental factors may affect D values. The subscript on the D value indicates the temperature at which it applies. Increasing the temperature decreases the D value. The z value is the increase in temperature necessary to reduce a D value by a factor of 10. Table I shows D and z values for some food-borne pathogens.
D value (0C) in minutes
1010 10 '0 10 '0 10"
X
/
~I
time
198
11// time
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51. According to the passage, which of the following is true concerning a surface that has been disinfected? A. B.
C. D.
Passage II (Questions 54-58) Even within mature cells, reshuffling of genetic material can take place. The vehicles by which nucleotides move from one position on a chromosome to another, or from one chromosome to another, are collectively termed transposable genetic elements (TGE). TOEs in prokaryotes include insertion sequence (IS) elements, and transposons.
No living microorganisms exist on the surface. Some living microbes remain on the surface, but all or most microbes capable of producing disease have been destroyed or reduced. The surface has been cleansed of bacteria, but not necessarily viruses or viroids. All pathogens on the surface are destroyed or removed, while all nonpathogens on the surface remain alive.
IS elements are segments of double stranded DNA. Each single strand of DNA in an IS element possesses a nucleotide sequence at one end that is complementary to the reverse sequence of nucleotides at its other end. These end sequences are termed inverted repeat (IR) sequences. Between the IR sequences is a tran,<,posase gene which codes for an enzyme that allows the IS element to integrate into the chromosome. IS elements do not contain entire genes other than the transposase gene. Insertion of an IS element into a gene activates or deactivates that gene, depending upon the location of the insertion and the orientation of the IS element.
52. Which of the following statements is best supported by the data in Table I? A. B. C. D.
A bacterium's resistance to heat is directly related to its z value. C. botulinum is more likely to contaminate commercially processed food than S. aureus. A bacterium's resistance to heat may vary depending upon its environment. In chicken a Ia king, Salmonella is more resistant to heat than S. aureus.
The TOEs known as transposons exist mainly as a series of complete genes sandwiched between two opposite oriented IS elements. As well as existing in prokaryotes, transposons are the main type of TGEs found in eukaryotes. They are similar to retroviruses. Both IS elements and transposons are found in bacterial plasmids.
53. Which of the following statements does NOT contradict the information in the passage concerning C. botulinum? A. B.
C. D.
C botulinum thrives in the presence or absence of oxygen. C. botulinum has a lipid bilayer outside its peptidoglycan cell wall. A glass containing C botulinum may be disinfected by immersion in boiling water. Animals have no natural defense against the neurotoxin produced by C botulinum.
A TOE may move as a complete entity, called conservative transposition, or it may move a duplicate copy of itself, called replicative transposition.
54. One strand of an IS element begins with the nucleotide sequence 5'-ACTGTIAAG-3'. The same strand must end with the nucleotide sequence: A. B. C. D.
5'-GAATIGTCA-3' 5'-TGACAATIC-3' 5'-CTIAACAGT-3' 5'-ACTGTIAAG-3'
55. When the passage states that transposons are similar to retroviruses, to what aspect of the life cycle of a retrovirus is the passage most likely referring? A. B. C. D.
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reverse transcription of viral RNA proliferation of multiple copies of viral genome leading to the lyses of the host cell capsid formation the procedure by which the viral genome integrates into the host genome
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Passage III (Questions 59-65)
56. Which of the following is not typically associated with genetic recombination in prokaryotes?
A. B.
C. D.
A phage (pronounced lambda phage) is a double stranded DNA bacteriophage exhibiting both a lytic and a lysogenic Iifecycle. The virion injects a single molecule of double stranded DNA containing 40 genes into the host cell. The two ends of the DNA molecule are single strands of DNA that are reverse complements of each other. Once inside the cell. the host cell DNA ligase links the single stranded ends together forming a small circle of DNA. The virus may now enter a lytic phase, or it may insert itself between the galactose and biotin operons. Insertion into the host cell genome req uires the viral enzyme integrase. Integrase is translated soon after infec ti on of the host cell.
transduction TOEs binary fission Iransformation
57. Which of the following mechanisms of genetic recomb ination between prokaryotes involves plasmids? A. B.
C. D.
transduction conjugation meiosis binary fission
At the same time that integrase is translated, a protein called
58. Bacterium A is able to live on a histidine deficie nt
A repressor is also translated. A repressor prevents the rranscription of all A phage genes except its ow n. For as long as the concentration of A repressor is maintained above a critical
medium. Bacterium B is not. After initiating conjugation with bacterium B. Bacterium A is unable to live on a histidine defic ie nt medium. Which of the following statements is most hkely true concerning the conjugation? A.
B.
C.
D.
limit. the virus remains lysogenic. This can last hundreds of thousands of cell generations. Damaged DNA activates a cellular protease that degrades A repressor. When the concentration of A. repressor falls below the critical limit, the gene for excisionase is transcribed. Excisionase cuts the viral DNA from the host cell chromosome. The virus then switches from the lysogenic to the lytic pathway which ultimately results in the lysis of the host cell.
Conservative transposition occurred duri ng co njugation where bacterium A transferred the his+ gene to bacterium B. Replicative transposition occurred during conjugation where bacterium A transferred the his+ gene to bacterium B. Conservative transposition occurred during conjugation where bacterium B transferred the his+ gene to bacterium A . Replicative transposition occurred during conjugation where bacterium B transferred the his+ gene to bacterium A.
59. While integrated into th e host cell DNA a A phage is
called a: A. B.
C. D.
virion. prophage. c hromosome. plasmid.
60. A phage most likely integrates into the host cell DNA in the: A. B. C. D.
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host cell nucleus. host cell mitochondria. lumen of the host cell endoplasmic reticulum. cytoplasm of the host cell.
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64. Which of the following is a likely host cell for a A phage?
61. Which of the following would most likely lead to lysis of a host cell infected with A phage in the lysogenic stage?
A. B. C. D.
infection by a second A phage binary fission exposure to ultraviolet radiation mitosis
C. D.
an E. coli bacterium
C. D.
a paramecium
T-Iymphocyte a neuron
65. Which of the following is never fo und inside the capsid of a virus?
62. Which of the following must be true in order for infection with A phage to take place? A. B.
A. B.
A. B. C. D.
The host cell must be diseased or weakened. The host cell membrane or wall must contain a specific receptor protein. Some type of mutagen must be present. Viral DNA must penetrate the host cell nuclear
single stranded RNA only double stranded RNA only single stranded DNA on ly RNA and DNA
membrane.
63. Which of the following describes in the correct chronological order the event. of an infection with A phage? A.
B.
C.
D.
viral DNA is injected into the host cell-viral DNA is translated and replicated-the capsid is formedthe host cell lyses releasing hundreds of viral progeny. viral DNA is injected into the host cell-viral DNA is transcribed-viral RNA is translated and reverse transcribed-the capsid is formed- the host cell lyses releasing hundreds of viral progeny. viral DNA is injected into the host cell- viral DNA is transcribed and replicated- the capsid is formed-the host cell lyses releasing hundreds of viral progeny. viral DNA is injected into the nucleus- viral DNA is tran sc ribed and replicated-the capsid is fo rmed-the host cell lyses releasing hundreds of viral progeny.
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68. The proteins and glycoproteins which make up the capsid, envelope and spikes of a virus determine the infective properties of that virus. All of the following are
Questions 66 through 69 are NOT based on a descriptive passage.
true concerning viruses EXCEPT:
A.
66. Bacteria contain all of the following EXCEPT: A. B. C. D.
membrane bound organelles double stranded DNA ribosomes circular DNA
B. C.
D. 67. Which of the following statements are true concerning yeast?
69. Which of the following is found either in prokaryotes or
I. Yeasts are eukaryotic. II. Yeasts are unicellular. III. Yeasts are facultative anaerobes.
A. B. C. D.
Some complex viruses replicate without the synthetic machinery of the host cell. Animal viruses enter their host via endocytosis. A bacteriophage sheds its protein coat outside the host cell and injects its nucleic acids through the host cell wall. A latent bacteriophage consisting only of a DNA fragment is called a prophage.
eukaryotes, but not in both?
I only I and II only I and III only I. II. and III
A.
a cell wall
B. C. D.
ribosomes
RNA centrioles
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
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30-MINUTE IN-CLASS EXAM FOR LECTURE 4
203
Passage I (Questions 70-76)
72. Which of the following cell types most likely contain adrenergic receptors (receptors that respond to epinephrine)?
Smooth muscle and visceral organs of the body are innervated by the autonomic nervous system, which is controlled mainly by centers within the spinal cord, brain stem, and hypothalamus. The two branches of the autonomic nervous system are the sympathetic and parasympathetic. Most visceral organs are innervated by both branches.
A. B. C. D.
73. Which of the following is not an autonomic response to temperature change?
The sympathetic nervous system is composed of motor pathways consisting of two neurons, the preganglionic and the postganglionic. Preganglionic nerve fibers exit the spinal cord between segments Tl and L2. The neuronal cell bodies of the postganglionic neurons are mainly contained in the sympathetic paravertebral chain ganglion located on either side of the spinal cord. One exception to this rule is the nerves innervating the adrenal medulla, which synapse directly onto the chromaffin cells. Chromaffin cells are themselves modified postganglionic neurons.
A. B.
C. D.
A. B. C. D.
A. B. C. D.
70. From the information in the passage, it can be presumed that chromaffin cells secrete:
the the the the
somatic nervous system. sympathetic nervous system. parasympathetic nervous system. endocrine system.
76. Amphetamines cause epinephrine to be released from the ends of associated neurons. Which of the following is most likely NOT a symptom of amphetamine usage?
cortisol epinephrine ACTH renin
A. B. C. D.
Which of the following is not innervated by the autonomic nervous system?
increased heart rate elevated blood glucose levels constricted pupils increased basal metabolism
the arteries of the heart the iris musculature of the eye the diaphragm sweat glands
A. B. C. D.
Copyright
sympathetic parasympathetic both sympathetic and parasympathetic somatic
75. Upon anival to a high altitude environment, individuals may experience symptoms of nausea and vertigo. These symptoms generally subside after a few days exposure to the environment. The system most likely responsible for acclimatizing the body in this case is:
The delivery of neurotransmitter to effector organs by the autonomic nervous system is less precise than in neuromuscular junctions of the somatic nervous system.
7},
piloerection shivering constriction of cutaneous vessels sweating
74. Which of the following nervous systems is responsible for the simple reflex arc?
The parasympathetic nervous system is also composed of motor pathways consisting of two neurons; however, the bodies of the postganglionic cells of the parasympathetic system are located on or near the effector organs. Most parasympathetic nerves exit the central nervous system through cranial nerves, the vagus nerve containing 75 percent of all parasympathetic nerve fibers.
A. B. C. D.
chromaffin cells sympathetic postganglionic neurons parasympathetic postganglionic neurons cardiac muscle cells
(~)
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79. Compared to a healthy individual, an individual who ingests large amounts of phenobarbital over an extended period of time will most likely:
Passage 11 (Questions 77-83)
In a healthy cell, smooth endoplasmic reticulum (ER) performs several functions including carbohydrate metabolism, lipid synthesis, and oxidation of foreign substances such as drugs, pesticides, toxins and pollutants.
A. B. C. D.
The liver maintains a relatively stable level of glucose in the blood via its glycogen stores. An increase of cyclicAMP activates protein kinase A, which leads to the formation of glucose I-phosphate from glycogen. Glucose I -phosphate is converted to glucose 6-phosphate, which still cannot diffuse through the cell membrane. Glucose 6-phosphatase, associated with smooth ER, hydrolyzes glucose 6-phosphate to glucose, which is then transported from the cell into the blood stream.
80. A given type of phospholipid may exist in different concentrations on either side of the same membranous structure. Which of the following enzymes most likely contributes to such an asymmetric arrangement? A. B. C. D.
The smooth ER synthesizes several classes of lipids, including triacylglycerols which are stored in the ER lumen, cholesterol and its steroid honnone derivatives, and phospholipids for incorporation into the various membranous cell structures. Phospholipids are synthesized only on the cytosol side of the ER. They are then selectively flipped to the other side by phospholipid translocators.
A. B. C. D.
A. B. C.
D.
A. B.
C.
C.
Copycigh t
reduced oxidized removed inverted
83. Some cells, called adipocytes, specialize in the storage of triacylglycerols synthesized by the smooth ER. The primary function of adipocytes is to:
skeletal muscle adrenal cortex intestinal epithelium cardiac muscle
78. Where are many of the enzymes necessary for phospholipid synthesis likely to be located?
D.
smooth ER rough ER Golgi apparatus. cellular membrane.
82. In the final step of the reactions governed by mixed-function oxidases, oxygen is converted to water. In this step. the iron atom in P-4S0 is most likely:
77. Which of the following tissues would be expected to have especially well developed smooth ER?
A. B.
glucose 6-phosphatases phospholipid translocators cytochrome P-4S0 protein kinase A
81. The primary structure of mixed function oxidases is most likely synthesized at the:
Most detoxification reactions in the smooth ER involve oxidation. Such reactions usually involve the conversion of hydrophobic compounds into hydrophilic compounds, and are governed by a system of enzymes called mixed-function oxidases. Cytochrome P-450, a group of iron-containing integral membrane proteins, is a central component of one mixed-function oxidase system. Mixed-function oxidases also govern the oxidation of steroids and fatty acids. Ingestion of the depressant phenobarbital triggers an increase in smooth ER and mixed-function oxidases but not in other ER enzymes.
A. B. C. D.
detoxify dangerous drugs more slowly. be more able to maintain steady blood glucose levels. degrade phenobarbital more quickly. be more responsive to therapeutically useful drugs such as antibiotics.
D.
maintain chemical homeostasis of the body. filter and remove toxins. provide for cholesterol synthesis. serve as a reservoir of stored energy.
ER lumen cytosol lysosome Golgi complex
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84. Influx of Ca2+ ions contribute most to which of the numbered sections from the cardiac action potential in Figure
Passage III (Ouestions 84-90)
27
Tbe action potenti al of cardiac muscle differs from the action poten ti al in skeletal muscle in two important ways. First, depolarization in skeletal muscle is created by the opening of fast Na+ voltage-gated channels; depolarization of cardiac muscle is effected by both fast Na+ voltage-gated cbannels and slow Ca2 ""- Na+ voltage-gated channels. Fast Na+ voltage-channels exhibit three stages: closed; open; and inactivated. Upo n an increase in membrane potential they open for a fraction of a second allowing Na+ ions to rush into the cell, and then immediately beco me inactivated. Slow Ca'+-Na+ voltage-gated channels open more slowly and close more slowly allowing both Na+ a nd Ca2+ to enter the cell. The second major difference between skeletal and cardiac action potentials is that, upon depolarization. the cardiac muscle membrane becomes highly impenneable to K+. As soon as Ca2+-Na+ voltage-gated channels are closed, the me mbrane suddenly becomes very permeable to K+.
A. B. C. D.
A. B. C. D.
A.
o
2
C. D.
A. B. C. D.
5
4
3
4
Section Section Section Section
number 2 number 4 number 4 number 4
in Figure 2 in Figure 2 in Figure I in Figure 2
will lengthen. will lengtben. will shorten. will shorte n.
87. Wbicb of the following contributes most to section number 3 of the action potential sbown in Figure I ?
---.---_-L.
3
2
86. In a lab experiment a student placed a beating frog beart in to saline solution. Whicb of the following is true if the student adds acetylcholine to the same solution?
B.
~
4
85. In Figure 1 the numbered section of the skeletal muscle action potential which best represents depolarization is:
30
A
2 3
time (msec) Figure 1 a skeletal muscle acti on potential
Na+ ions are diffusing into tbe cell. K+ ions are diffusing into the cell. K+ ions are diffusing out of the cell. Ca'+ ions are diffusin g out of the cell.
88. Between action potentials, a potential difference called the resting potential exists across the neuron cell membrane. All of the followin g help to establisb the resting potential of a neuronal membrane EXCEPT:
o
2oo
tOO
A. B. C. D.
3oo
selective penneability of the cell membra ne. the Na+/K+ pump. the Na+ voltage gated channels. the electrochemical gradient of mUltiple ions.
lime (msec) Figure 2 a cardiac muscle action potential
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89. Which of the following cell types most likely contain Na+ voltage gated cbannels? A. B. C. D.
Questions 91 through 92 are NOT based on a descriptive passage.
an epithelial cell from the proximal tubule of a nephron a pari etal cell from the lining of the stomach an a·ce ll from the islet of Langerhans in the pan· creas a muscle fiber from the gastrocnemius
91. All of the following are true concerning a typical motor neuron EXCEPT: A. B.
90. According to Figure I. at 2 msec after the action potential begins, Na+ voltage gated channels are most likely:
A. B. C. D.
C.
open closed inactivated activated
D.
The K+ concentration is greater inside the cell than outside the cell. K+ voltage-gated channels are more sensitive than Na+ voltage-gated channels to a change in membrane potential. Cl- concentrations contribute to the membrane resting potential. The action pctential begins at the axon hillock.
92. In saltatory conduction: A.
an action pctential jumps along a myelinated axon from one node of Ranvier to the next.
B.
C. D.
an action potential moves rapidly along the membrane of a Schwann cell, which is wrapped tightly around an axon. an action potential jumps from the synapse of one neuron to the next. ions jump from one node of Ranvier to the next alon g the axon.
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
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STOP.
I
(, L
3D-MINUTE IN-CLASS EXAM FOR LECTURE 5
r
209
Passage I (Questions 93-100)
phate from ATP to select protein amino acid residues. The biological effects of cyclic AMP are mediated by changes in protein phosphorylation.
General hormones which circulate in the blood can be divided into 3 categories: I) Steroid hormones, synthesized in the smooth endoplasmic reticulum of endocrine glands and secreted from the cell via exocytosis, bind to proteins for transport to their target tissue. They then diffuse through the target cell membrane and normally bind to a large receptor protein in the cytosol, which in turn carries them into the nucleus where they exert their effect directly at the trauscriptional level. 2) Peptide hormones bind to membrane bound receptors and act via a second messenger system. 3) Tyrosine derivatives are further divided into thyroid hormones and the catecholamines, epinephrine and norepinephrine. Thyroid hormones diffuse through the cell membrane and bind to receptors in the nucleus. They also act directly at the transcription level. The catecholamines act at the cell membrane through a second messenger system.
In 1957, Earl Sutherland found that liver homogenates incubated with either epinephrine or glucagon would stimulate the activity of glycogen phosphorylase, an enzyme which governs the conversion of glycogen to glucose-I-phosphate. However, if the membranes present in the homogenate were removed by centrifugation. glycogen phosphorylase could no longer be activated by epinephrine or glucagon, but could still be activated by the addition of cyclic AMP.
93. Cortisol most likely binds to a receptor protein: A.
B. C. D.
94. Phosphodiesterase breaks down cyclic AMP to AMP. Caffeine, a drug abundant in coffee, suppresses the activity of phosphodiesterase. According to the information in the passage. which of the following would most likely be found in a blood sample of someone who has recently drunk large amounts of coffee?
~hormone
..r-receptor ,II ;';':
ii,l,l
" ;':Ii
; I! ' "
I,; II
iIY~' •...
.' , ... / a,
....
/
GTP
1" sumu atIOn
adenylyl cyclase
A.
B. C. D.
~
G protein
ATP:
i
bound to the cell membrane. in the cytosol. on the nuclear membrane. just outside the cell.
cAMP I
;GDP
high ADH levels low insulin levels high glucose levels low glucose levels
activati0y' 95. A certain mutant tumor cell line has normal epinephrine receptors and normal adenylyl cyclase; however, it fails to increase cAMP in the presence of epinephrine. The most likely explanation for this is:
Protein Kinase A
~hOSPh~rYlati~
A.
./,
VariOllS
B.
proteins
Figure 1
C.
One second messenger system, shown in Figure I, works as follows: The activated membrane bound hormone receptor activates a protein inside the cell, called a G-protein. Once activated, the G-protein exchanges GDP for GTP, which causes a portion of the G-protein to dissociate. Depending upon the type of G-protein, the dissociated portion may stimulate or inhibit adenylyl cyclase, another membrane bound protein. Inhibitory O-proteins are calJed OJ-proteins; stimulating 0proteins are called Os-proteins. After activating adenylyl cyclase, the dissociated portion of the O-protein must hydrolyze GTP to GDP in order to become inactive and recombine with its other portion. Adenylyl cyclase converts ATP to cyclic AMP. Cyclic AMP activates protein kinase A. Kinases are a family of enzymes that catalyze the transfer of y-phos-
D.
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The Os-protein in the ce11 is either missing or malfunctioning. The Gi-protein in the cell is either missing or malfunctioning. Epinephrine diffuses directly into the cell to act on protein kinase A. The mutation results in a change in the structure of protein kinase A.
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96. Glucagon works via a G-protein system. Which of the following best explains why glucagon stimulates glycogen breakdown in liver cells, but stimulates lipid break down in fat cells. A. B. C. D.
99. Vibrio cholerae, the bacterium responsible for Cholera, releases an enterotoxin which acts on Os-proteins of the intestinal mucosa inhibiting their OTPase activity. Which of the following is most likely to occur inside an intestinal mucosal cell of an individual infected with cholera?
Fat cells contain a Gs-protein while liver cells contain a G;-protein. Liver cells don't contain adenylyl cyclase. The two cell types contain a different set of proteins phosphorylated by protein kinase A. Glycogen and lipid breakdown are not governed by cyclic AMP levels.
A. B. C. D.
100. Which of the following is the most likely effect of aldosterone?
97. Which of the following was most likely not part of the membrane presence removed by Sutherland during his experiment? A. B. C. D.
decreased activation of protein kinase A increase electrolyte concentration decreased rate of hormones binding to membrane bound receptors increased concentration of cyclic AMP
A. B. C. D.
glucagon and epinephrine receptor proteins adenylyl cyclase cyclic AMP G-protein
activation of ion channels via binding to membranebound protein channels increase of cyclic AMP activation of protein kinase A increased production of membrane bound protein.
98. Epinephrine binds to several types of receptors called adrenergic receptors. Heart muscle cells contain ~1adrenergic receptors, where as the smooth muscle cells of the gut contain many ~-adrenergic receptors. Which of the following is most likely true concerning these two types of receptors? A. B. C. D.
~I-adrenergic receptors activate Os-proteins while (X2-adrenergic receptors activate OJ-proteins. ~l-adrenergic receptors activate OJ-proteins while {X2-adrenergic receptors activate Os -proteins. Both ~l-adrenergic receptors and ct2-adrenergic receptors activate OJ-proteins. Both ~l-adrenergic receptors and ct2-adrenergic receptors activate Os-proteins.
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Passage II (Questions 101-106)
103. Syncytiotrophoblast cells in the placenta are responsible for manufacture and re lease of HCG. Which of the following statements most accurately describes this process?
Huma n chorionic gonadotropin (}ICG) is secreted by the placenta)t can be detected in maternal plasma or urine within 9 days of co nception, which is shortly after the blas/ocyst implants in the uterine wall. Maternal blood levels of HCG increase expone ntially for the first 10 to 12 weeks of pregn ancy and decline to a stable plateau for the remainder of the pregnancy.
A, B.
C. D.
HCG sti mul ates the corpus luteum to secrete estrogens and progesterone until the placenta assumes the synthesis of these steroids. Duri ng this time the corpus luteum grows to approx imately twi ce its ini tial size. After 13 to 17 weeks, the corpus luteum involutes.
104. The corpus luteum mentioned in the passage is: A.
B. C. D. HCG acts 0 11 a G-protein-coupled receptor on the target cell membrane. LH acts on the same receptor and FSH ac ts on a very simila r receptor. Once activated, the G protein stimulates an increase in cyclicAMP, which activates protein kinase A.
C. D.
A.
B. C. D.
LH FSH estrogen progesterone
C. D.
HCG LH
106, Which of the following hormones most likely acts as the substrate molecule for synthesis of cortisol and aldosterone in the fetal adrenal gland ?
A, B. C.
FSH HCG estrogen progesterone
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progesterone
B. , estrogen
102. Whic h of the following hormones is most responsible fo r preventing the sloughing off of the uterine wall during preg nancy? A,
a gland developed in the fe tal ovary during the third trimester of pregnancy. a group of neuronal cells in the hypothalamus.
105, Early pregnancy tests use an antibody to bind to a hormone in the urine. Pregnancy is indicated when binding occurs. To which hormone does the antibody most likely bind?
101. HCG acts most like which of the following hormones?
B.
a permanent functional part of any healthy ovary. the remainder of the follicle which produced the ovum.
HCG acts to stimula te the testes of the male fetus to secre te testosterone. It is thi s testosterone that accounts for the male sex organs.
A.
HCG is manufactured in the rough ER and modified in the Golgi apparatus before secretion. HCG is manufac tured by the smooth ER and diffused into the blood stream. HCG is manufactured in the nucleus and transported via a protein carrier in to the blood. HCG is manufactured at the membrane of the syncytiotrophobast and re leased into the plasma.
D.
212
HCG estrogen
LH FSH
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108. Which of the following two cell types possess the most similar phylogeny (developmental history)"
Passage III (Questions 107-114)
A, B. C. D.
Neurons of the hypothalamus secrete gonadotropin-releasing hormone (GnRH), a IO-amino acid peptide which stimulates the anterior pituitary to release the gonadotropins LH and FSH. LH secretion keeps close pace with the pulsatile release of GnRH, whereas FSH changes more gradually in response to long term changes in GnRH levels. Both FSH and LH act by changing intracellular cyclicAMP levels. In women, GnRH secretion occurs monthly guiding the menstrual cycle. In men, GnRH secretion occurs in bursts throughout each day to maintain a relatively steady blood level of the hormone.
109. Androgens are sometimes taken by athletes to improve performance. Which of the following may be a side effect of taking large quantities of androgens? A. B. C.
Production of spermatozoa occurs in the seminiferous tubules of the testes where a single Serfo/i cell envelopes and nurtures the developing spermatozoa. FSH stimulates Sertoli cells, which, in turn, secret inhibin that acts at the pituitary level to inhibit production of FSH independently of LH. LH stimulates the Leydig ceLls located between the seminiferous tubules to secrete testosterone, which acts on the hypothalamus to inhibit GnRH. Testosterone stimulates spermatogenesis. Both FSH and LH are required for spermatogenesis.
D.
A,
GnRH FSH LH D . . testosterone B, C,
tIt. Some post-menopausal women suffer from osteoporosis, a lowering in density of the bones. Administration of estrogens is an effective treatment for this disease. This demonstrates that estrogen most likely inhibits the activityof: A. B. C. D.
t07. The follicle in its earliest stages is called a primordial follicle. The primordial follicle most likely contains a:
osteoblasts osteoc1asts osteocytes hemopoietic stem cells
112. Which of the following are re leased into the fallopian tube during ovulation?
haploid, primary oocyte. diploid, primary oocyte. haploid, secondary oocyte. diploid, secondary oocyte.
I. the follicle II, the secondary oocyte III. some granulosa cells A, B,
C. D,
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infertility due to decreased endogenous testosterone production infertility due to decreased secretion of GnRH increased fertility due to decreased endogenous testosterone production increased fertility due to increased Sertoli cell activity
110, Which of the following hormones would most likely be found in the nucleus of a somatic cell of a pregnant woman?
Oogenesis occurs in the ovaries of the female. The oocyte develops surrounded by theca and granulosa cells. The entire structure is called a follicle. As a follicle develops, granulosa cells secrete a viscous glycoprotein layer, called the zona pellucida, which sun-ounds the oocyte. The granulosa cells remain in contact with the oocyte via thin strands of cytoplasm. Theca cells differentiate from interstitial cells and [onn a thin layer surrounding the granulosa cells. The follicle does not require FSH to reach this stage. When stimulated by LH, theca cells supply granulosa cells with androgen, which is then converted to estradiol and secreted into the blood along with inhibin. Like testosterone, estradiol inhibits GnRH secretion from the hypothalamus.
A. B. C. D.
theca cells and Sertoli cells Leydig cells and granulosa cells Leydig cells and theca cells granulosa cells and oocytes
213
I only II only II and III only I, II, and Tll
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113. Which of the following hormones is most closely associated with ovulation?
A. B. C. D.
Question 115 is NOT based on a descriptive passage.
estrogen FSH LH GnRH
115. A competitive inhibitor of TSH binding to TSH receptors on the thyroid would lead to a rise in the blood levels of which of the following:
114. A vaccine that stimulates the body to produce antibodies against a hormone has been suggested as a long term male contraceptive. In order to insure that the vaccine has no adverse effects on androgen production, which hormone should be targeted?
A. B.
C. D.
A. B. C.
D.
FSH LH GnRH testosterone
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TSH Thyroxine PTH epinephrine
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30-MINUTE IN-CLASS EXAM FOR LECTURE 6
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Passage I (Questions 116-122)
117. Most dietary fat first enters the blood stream: A.
Dietary fat is mmnly composed of triglycerides and smaller amounts of cholesterol and phospholipids. Lingual lipase secreted in the mouth digests a very small portion of fat while in the mouth and small intestine. but mainly in the stomach. Enterocytes in the small intestine also release tiny amounts of e nteric lipase. However, the most important e nzy me for the digestion of fats is pancreatic lipase.
B.
C. D.
from tion. from from from
the right lymphatic duct into arterial circulathe thoracic duct into venous circulation. the small intestine into capillary circul ation. the intestinal enterocyte into the lacteal.
liS. Most tests for serum cholesterol levels do not distinguish between HDLs and LDLs. Life insurance rates generally increase with increasing serum cholesterol levels. Which of ti,e following supports the claim that it is important to determine whether HDLs or LDLs are responsible for high serum cholesterol when evaluating a patient, risk for coronary heart disease?
Since fats are not soluble in the aqueous solution in the small intesti ne, fat digestion would be very ineffi cient were it not for bile salts a nd lecithin, which increase the surface area upon which lipase can act. In addition, bile forms miceUes w ith the fatty acid and monoglyceride products of e nzymatic hydrolysis of tri glycerides, and carries these micelles to the brush border of the intestine where they are absorbed by an enterocyte.
A. B.
Once inside the enterocyte, the fatty acids are taken up by the smooth endoplasmic reticulum and new tri glycerides are formed. These triglycerides combine with cholesterol and phospholipids to form new globules called c hylomicrons that are secreted through exocytosis to the basolateral side of the e nterocyte. From there, the chylomicrons move to the lacteal in the intestinal villus.
C. D.
The risk from coronary heart disease doubles from an HDL level of 60mg/lOOml to 30mgllOOml. The incidence of coronary heart disease rises in linear fashion with the level of serum cholesterol. The optimal serum cholesterol for a middle aged man is probably 200mg/IOOml or less. VLDL is the mmn source of plasma LDL.
119. Lingual lipase most likely functions best at a pH of:
Chylomicrons are just one member of the li poprotein families which transport lipids through the blood. The other me mbers are: very low density lipoproteins (VLDL); low density lipoprote ins (LDL); and high density lipoproteins (HOL). VLDLs are degraded by lipases to LDL. LDLs account fo r approximately 60-75% of plasma c holesterol and their levels are directly related to cardiovascular risk. HDLs, on the other hand, account for only 20-25% of plasma cholesterol and HDL plasma levels are inversely related to cardiovascular risk. HDL levels are positi vely correlated with exercise and moderate alcohol intake a nd inversely related to smoking, Obesity and use of progestin-containi ng contraceptives.
A.
2
B.
5
C. D.
7 9
120. Pancreatic enzymes are released from the pancreas in an inactive form called a zymogen. Which of the following activates the zymogen form of pancreatic lipase.
A. B. C.
D.
bile salts trypsin low pH high pH
116. Bile allows lipases to work more efficiently by increasing the surface area of fa t. If fat globules are assumed to be spherical, then each time bile decreases the diameter of all the fat globules in the small intestin e by a factor of two, the surface area of the fat is increased by a factor of: A. B. C.
D.
2
4 8 16
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Passage II (Questions 123-130)
121. The process by which bile increases the surface area of dietary fat is caUed: A. B. C. D.
The rate at which a solute is excreted in the urine is given by the product of the co ncentration (U) of the solute in the urine and the urine flow rate (V). Dividing tbis number by the blood plasma concentration (P) of the solute, gives the minimum volu me of blood plasma necessary to supply the solute. This number is called the renal clearance (C).
lipolysis adi polysis malabsorption em ulsification
122. A pati ent that has had his pancreas surgically re moved would most likely need to supplement his diet with all of the following enzymes EXCEPT: A. B.
C. D.
C= U x V P
lactase lipase amy lase proteases
Because almost no solute is completely filtered from the blood in a si ngle pass through the renal corpuscles, the renal clearance does not represent an actual volume of plasma. However, the organic dye. p-aminohippuric acid (PAH), is not only filtered but also secreted by the kidney. As a resul~ 90% of PAH that e nters the kidney is excreted. Thus, the clearance of PAH is an approximation o f the rellal plasma flow (RPF) to within 10%. Due to secretion, resorption, and metabolism in the nephron, the clearance of a solute may not be equal to its rate of filtration by the glomerulus. Inulin, a nonmetabolizable polysaccharide, is neither secreted nor resorbed. Thus the clearance of inulin is equal to the glomerular filtration rate (GFR). In a healthy adult, the GFR for both kidneys is approximately 125 mllmin.
I
The filtration fraction (FF) is the fractio n of the plasma that filters through the glomerular membrane. The filtration fraction is given by the equation:
FF = GFR RPF
123. Which of the following must be true of a solute that has a renal clearance greater than the GFR? A, B. C,
D.
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The solute is being resorbed by the nephron. The solute is being secreted by the nephron. The plasma concentration of the solute must be greater than the plasma concentration of inulin. The solute has exceeded its transport maximum.
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128. If a patient has a PAH clearance of 625 mllmin, and an inulin clearance of 125 ml/min. then approximately what percent of plasma is li ltered by the patient's kidneys each minute?
124. Since inulin is neither secreted nor resorbed, what must also be true about inulin, if its clearance is to accurately represent the GFR? A. B. C. D.
Inulin lowers the renal blood flow. Inulin raises the filtration rate. The fil tered concentration of in ulin is exactly equal to its concentration in the plasma. 100% of inulin is liltered in a single pass through the kidney.
A.
B. C.
D.
129. The blood concentration of creatinine, a naturally occurring metabolite of skeletal muscle, tends to remain constant. Creatinine flows freely into Bowman's capsule, and only negligible amounts of creatinine are secreted or absorbed. A patient with a normal GFR has a urine output of 1 ml/min with a creatinine concentration of2.5 mg/ml. What is the patient's plasma concentration of creatinine?
125. Inulin is most likely:
A. B. C. D.
smaller than an amino acid. the same size as albumin. the size of a red blood cell. larger than glucose.
A. B. C. D.
126. Which of the following would LEAST affect the renal clearance of a solute? A. B. C. D.
size and charge plasma concentration glomerular hydrostatic pressure the concentration of PAH
B. C. D.
A. B. C. D.
0 ml/mi n 60 mllmi n 125 mllmin 145 m1!min
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2 rug/ looml 5 mg/ looml 2 mg/rul 5 mg/ml
130. If an individual has a hematocrit of 50%, and a PAH clearance of 650 ml/min, what is his renal blood flow (RBF)?
127. Based upon the information in the passage. the renal clearance of glucose in a healthy adult is most like ly:
A.
0% 0.2 % 20% 50 %
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325 mllmin ·650 mllmin 1300 mllmin 2600 ml/min
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Passage //I (Questions 131-136)
131. Amino acids are reabsorbed in the proximal tubule by a secondary acti ve transport mechanism down the concentration gradient of sodium. A high protein diet would most likely lead to:
Like other circulatory syste ms, renal perfusion is autoregulated. The juxtaglomerular apparatus (lOA) assists in regul ating the volume and pressure of the renal tubule and the glomerular arterioles. The lOA connects the arterioles of the glomerul us with the distal convoluted tubule of the nephron via specialized smooth muscle cells, called juxtaglomerular cells. Renin, a proteolytic enzyme, is synthesized and stored in these cells. Renin release is stimulated by low plasma sodi um, low blood pressure, and sympathetic stimulation via ~-adrenergic receptors on the juxtaglomerular cells.
A. B.
C. D.
Once released into the blood, renin acts on angiotensinogen, a plasma protein, to fOlm angiotensin 1. Angiotensin I, an inactive decapeptide, is cleaved in the lungs to form angiotensin 1I. Angiotensin II stimulates the release of aldosterone, causes systemic vasoconstriction, enhances neurotransmitter release from sympathetic nerve endings. stimulates ADH release, and acts in the brain to cause thirst. Angiotensin n preferentially constricts the efferent arterioles of the glomerulus.
a decrease in glomerular filtration rate and rena] blood flow. an increase in glomerular filtration rate and renal blood flow. a decrease in glomerular filtration rate and an increase in renal blood flow. an increase in glomerular filtration rate and a decrease in renal blood flow.
132. Renal artery stenosis (partial blockage of the renal artery) leads to activation of the renin-angiotensi n cascade resulting in: A.
B. C. D.
Distal tubule
high renal blood pressure onl y low renal blood pressure only high systemic blood pressure low systemic blood pressure
133. Which of the following is least likely to be affected by angiotensin II?
Macula
A. B. C. D.
the the the the
adrenal cortex thyroid posterior pituitary autonomic nervous system
cells 134. The role of renin in the conversion of angiotensinogen to angiotensin I is most likely to:
arteriole
A. B.
Bowman' s Capsule
C. D.
act as a cofactor. bind to angiotensinogen making it soluble in the aqueous solution of the blood. lower the energy of activation of the reaction. add to angiotensinogen in a hydrolytic reaction.
Figure 1 A renal corpuscle Specialized epithelial cells of the distal tubule, called macula densa, are contiguous with the juxtaglomerular cells. The presence of low sodium chloride concentration causes the macula densa to signal the juxtaglomerular cells to release renin into the blood, and to lower the resistance of the afferent anerioles. A low resistance in the afferent arterioles increases the glomeru lar filtrdtion rdte. When functioning properly, the lOA operates to maintain the glomerular tilhation rate in the face of large fluctuations in arterial pressure.
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135. The renin-angiotensin cascade increases all of the following EXCEPT: A. B. C. D.
Questions 137 through 138 are NOT based on a descriptive passage.
urine volume blood pressure Na+ reabsorption K+ excretion
137. The ability to produce a concentrated urine is primarily based on the presence of functional kidney nephrons. The most important structure involved in concentrating urine within the nephron is the:
136. Angiotensin II most likely: A. B. C. D.
A. B.
diffuses into the nucleus of the target cell. attaches to a receptor on the membrane of the target cell and acts through a second messenger system. acts on the cells of distal tubule increasing transcription of sodium transport proteins. creates a chemical reaction in the plasma that increases the permeability of the collecting duct.
e. D.
glomerulus. proximal convoluted tubule. loop of Henle. Bowman's capsule.
138. Which of the following statements about digestion is NOT true? A. B. C. D.
Carbohydrate metabolism begins in the mouth. Most dietary protein is absorbed into the body in the stomach. The large intestine is a major source of water reabsorption. The liver produces bile which is stored in the gallbladder.
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3D-MINUTE IN-CLASS EXAM FOR LECTURE 7
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140. Which of the following is a cell that a nude mouse would be unable to produce?
Passage I (Questions 139-145)
A. B. C. D.
The immune system has several methods of selecting and destroying particles that are foreign to the body. One such mechanism works as follows: Large phagocytotic cells called macrophages engulf some antigens and process them internally, ultimately producing antigen fragments that protrude from the outer surface of the membrane of the macrophage. The antigen fragments are held to the membrane by MHC class II molecules. A special type of T-cell, called a helper T-cell, recognizes and binds to the MHC class II-antigen complex. The helper T-cell produces protein local mediators called interleukins which stimulate the T-cell to divide. A newly formed helper T-cell activates a B-cell capable of producing antibodies that specifically bind to the antigen. The B-cell begins cell division into plasma cells and memory B cells. Memory B cells resemble unstimulated B cells and do not secrete antibodies. Plasma cells produce antibodies that bind to the antigen. Once bound to the antigen, the antibody may initiate a chemical chain reaction which results in lysis of the antigen carrying cell Of, if the antigen is a chemical such as a poison, the antibody may simply inactivate it. In addition, cells with antibodies bound to their surfaces may be engulfed by macrophages or punctured by natural killer cells.
141. The experiment demonstrated that: A. B. C. D.
A. B.
C. D.
by T-cells. by B-cells only after expoproduce antibodies. produce T-cells.
via a second messenger by binding to a membrane bound protein receptor. by diffusing through the membrane of the helper Tcell and binding to a receptor in the cytosol. at the transcriptional level by binding directly to nuclear DNA. via the nervous system.
143. All of the following arise from the same stem cells in the bone marrow EXCEPT: A. B. C.
D.
helper T-cells B lymphocytes erythrocytes osteoblasts
144. Which of the following is a foreign particle capable of provoking an immune response? A. B. C. D.
139. Which of the following is most likely to create the immune system response mediated by MHC class II m01ecules?
antigen antibody interleukin histamine
145. The function of the memory B-cell is most likely to:
a bacterial infection a cell infected by a virus a tumor a foreign tissue graft
A. B. C. D.
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antibodies are produced antibodies are produced sure to T-cells. nude mice are unable to nude mice are unable to
142. Illterieukills most likely act:
In an effort to discover which cell type creates antibodies, a scientist performed the following experiment. A nude mouse, which lacks a thymus and cannot form antibodies, was injected with healthy lymphocytes trom the thymus of a donor mouse. The host mouse was then injected with an antigen. Antibodies were produced in the host mouse. Lymphocyte samples were then removed from the spleen of the host mouse and the host cells or the donor cells were selectively destroyed from separate samples. The scientist found that the host cell samples were still able to produce antibodies against the antigen while the donor cell samples were not.
A. B. C. D.
an antibody a T-cell a B-cell a macrophage
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remain as an immune system reserve against different antigens. attract killer T-cells to the infected area. magnify the immune response by releasing antibodies into the blood stream. allow for rapid production of antibodies in the case of reinfection with the same antigen.
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147. Permanent immunity may be imparted to an individual by a vaccine containing:
Passage" (Questions 146-152)
A. B. C. D.
The function of plasma cells in the immune system is to secrete specifi c proteins called antibodies, also caned imm unoglobulins. which bind to antigens a nd mark them for destruction. One plasma cell can only make antibodies that are specific for a single epitope for a single a nti gen.
antibodies antigens plas ma cells white blood cells
148. The light chain is represented by which of the followi ng labeled segments in Figure I?
Antibodies exist mainly in blood plasma but are also present in tears, milk, saliva, and respiratory and intestinal tract secretions.
A. B. C.
Treatment of a nti bodies with the protease "papain" yields three fragmen ts: two Fab fragments and one Fe fragment. The Fc fragment is nearl y ide ntical in all antibodies. Mercaptoethallo1 cleaves disulfide bonds. Treatment of antibodies with mercaptoethanol results in two light chain polypeptides (25.000 daltons) and two heavy chain polypeptides (50.000 daltons). Both the light and heavy chains possess constant, variable, and hypervariable regions in their amino acid structures. The light chain contains no part of the Fc region.
D.
A. B. C. a nd D C. D. E. and F C and D E and F
149. Which of the following labeled segments from the antibody in Figure I most likely attaches to an antigen?
A.
A
B. C.
D and F D but not F
D.
Cand D
Hypervari
150. Which of the following structures in the plasma cell most likely produces antibodies?
Variable ~/ regions
A. B. C. D.
B
ribosomes in the cytosol rough e ndoplasmic reticulum smooth e ndoplasmic retic ulum cellular membrane
151. Which of the fo llowing statements concerning the immune system is NOT true?
Constant egions
A.
B. A
C. D.
Figure 1 An antibody
146. What level of protein structure is di srupted by mercap-
Plasma cells which provide immu nity against one disease may also pro vide immunity against a closely related disease. Two a ntibodies from the same plasma cell must bind to the same antigen type. Plasma cells ari se from T lymphocytes. Memory B cells help the immune system to respond to the same antigen more quickly during a secondary imm une response.
toethanol? A. B. C. D.
152. Which of the following labeled segments from Figure I is part of the F, region of an antibody?
primary secondary tertiary on ly tertiary and quaternary
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A. B. C. D.
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AandB A, B. C. and D A. B. C. and E C. D. E. and F
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Passage III (Questions 153-159)
153. Which of the following most li kely acted as the evolutionary selective pressure which led to the decreased affinity of he moglobin and myoglobin for CO as compared to the isolated heme group?
Oxygen in erythrocytes is stored by hemoglobin. In adults. hemoglobin consists of four polypeptide chai ns. two alpha (X) and two beta (~). each held together by noncovalent interactions. The interior of each folded chain consists almost en tirel y of nonpolar residues. Each polypeptide chain contains a heme group with a single oxygen binding site. The heme group is a nonpolypeptide unit with an iron atom at its center. The organic portion of the heme, protoporphyrin, binds the iron atom at its center with four nitrogen atoms, leaving the iron atom with a +2 or +3 oxidation state and capable of making two more bonds. Carbon monoxide is a byproduct of the break down of the heme group. An isolated heme group binds to CO 25.000 times as strongly as it binds to 0 2' However, the binding affin ity of hemoglobin and myoglobin for CO is only about 200 times as great as for 0 2'
A. B. C.
D.
The emergence of industri al societies which significantly increased the level of environmental CO. The products of aerobic respiration. The endogenous production of CO due to the breakdown of heme. Carbonic acid in the blood.
154. Hemoglobin and myoglobin are:
A. B. C. D.
The oxygen CalTief in skeletal muscle tissue is myoglobi n. Myoglobin is very similar to hemoglobin. except that it consists of only a single polypeptide chain. Myoglobin does not show a decreased affinity for 0, over a broad range of pH nor in the presence of CO,. Both decreased pH and increased CO, enhance the release of 0, in hemoglobin. Hemoglobin has a lower affinity for oxygen than does myoglobin. partially due to BPG. a chemical in red blood cells. BPG affects the characteristic sigmoidal oxygen dissociation curve of hemoglobin.
similar in their primary and tertiary structure. similar in their quaternary structure but differ in their primary structure. similar in their tertiary structure but differ in their primary structure. similar in their primary structure but differ in their quaternary structure.
155. Which of the following gives the oxygen dissociation curves for hemoglobin H and myoglob in M?
A.
~~ ~
H
""
Po,
~
Although the polypeptide chain of myoglobin is identical to the (X chain of hemoglobin only at 24 of 141 amino acid positions . the three dimensional shap es of myoglobin and hemoglobin ex. chains are very similar. Inter-species comparison of the three dimensional shape of hemoglobin reveals this same similarity. A further comparison of the amino acid sequences in the he moglobin of different species revea ls that nine positions are the same in nearly all known species. Severd.1 of these invaria nt residues affect the oxygen binding site.
i~ ""
224
i~
"" D.
B.
Po,
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c.
Po,
!1l2 M
~
""
Po,
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156. increased concentration of lactic ac id due to active skeletal muscle most likely results in which of the following? A. B. C.
!
I
D.
Questions 160 through 161 are NOT based on a descriptive passage.
increased off· loading of 0 , by both he moglobin and myoglobin . increased off-loading of 0 , by hemoglobin but not myoglobin. decreased off-loading of 0 , by both hemoglobin and myoglobin. decreased off-loading of 0, by hemoglobin but not myoglobin.
160. Which of the following is NOT tme concerning the lymphatic system? A. B.
\
C.
157. In a healthy human body. where is BPG likely to have its greatest effect? A. B. C. D.
in in in in
D.
Ihe al veoli the capillaries of the lungs the muscle cells of the heart the capillaries of skeletal muscle
161. Which of the following is a cell that does NOT contain a nucleus? A. B. C. D.
158. The teltiary structure of human myoglobin most likely: A. B. C. D.
represents a fundamental design for oxygen carriers in nature. arose relatively late in human evolution. varies among individuals of the same species. allows for more than one heme group.
erythrocyte platelet macrophage B lymphocyte
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159. Which of the following is LEAST likely to increase the breathing rate of an individual? A. B. C. D.
The Lymphatic system removes large particles and excess fluid from the interstitial spaces. The lymphatic system is a closed circulatory system. The lymphatic system contains lymphocytes that function in the bodies immune system. Most fatty acids in the diet are absorbed by the lymphatic sys tem before enteri ng the blood stream.
increased pH in the blood low oxygen levels in the blood increased muscle activity increased CO 2 levels
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30-MINUTE IN-CLASS EXAM FOR LECTURE 8
227
For active muscle cells the delivery of oxygen to the cell is far too slow to maintain sufficient energy levels, thus many muscle cells contain large amounts of myoglobin, whi ch along with the cytochromes in Ihe mitochondrial membrane impart a red hue to the muscle.
Passage I (Questions 162-169) The functional unit of a ske letal muscle cell is the sarcomere. The protein polymers actin and myosin lie lengthwise along a sarcomere creating the various regions shown in Figure 1. Z Line
I Band HZone
162, Which of the following proteins is an ATPase?
I
A. B. C. D.
actin myosin phosphocreatine calsequestrin
163. What is the fun ction of ca/seqllestrin? A. I A Band
B.
I
Sarcomere
C, D.
Figure 1 A sarcomere
to lower the free Ca2+ ion concentration inside the lumen of the sarcoplasmic reticulum to raise the free Ca2f ion concentration inside the lumen of the sarcoplasmic reticulum to make Ca2• for release into the cy tosol to pump Ca2+ into the sarcoplas mic reti culum
164, Two types of skeletal muscle are named after their characteristic color, red muscle and white muscle. Which of the following statements is most likely true concerning these muscle types?
A group of muscle cells within a muscle may be innervated by a single neuron making up a motor unit. The neuron carries an action potentia] to each muscle cell in the motor unit. The
action potential is delivered deep into each muscle cell via tubul ar invaginations in the sarcolemma or cell membrane called T-tubules. The chan ge in membrane potential is transferred to the sarcoplasmic reticulum, which causes it to become penneable to Ca2+ ions and to release its large stores of calcium into the cytosol. Once in the cytosol, th e Ca2 + ions cause a
A. B. C.
conformational change in the protein lrOponin, which in rum acts upon a second protein, tropomyosin, exposing an act inmyosin binding site. Upon binding. the actin and myosin slide past eac h other shortening the sarcomere and creatin g a muscular co ntraction. ATP the n binds to myosin releasing it from ac tin. Myosin immediately hydrolyzes ATP using the energy to return to its ready position. The Ca2+ ions are removed from the cytosol by extremely efficient calcium pumps via active transport. Once the Ca2+ has been sequestered back into the lumen of the sarcoplasmi c reticulum, the Ca2• ions are bound by
D.
Both muscle types contain large amounts of myoglobin and mitochondria. White muscle contains more mitochondria than red muscle. White muscle is capable of longer periods of contraction than red muscle. Red muscle is capable of longer periods of contraction than white muscle.
165, Which of the followi ng concentrations c hanges the most within the cytosol during the contraction of a muscle cell? A. B. C. D.
calsequestrin.
ATP Myosin Actin Ca2+
Although the direct source of energy for muscle contraction comes from ATP, the ATP concentration in actively contracting muscle remains virtually constant. In addition. it has been shown that inhibitors of glycolysis and cellular respiration have no effect on ATP levels in actively contracting muscle over the short term. Instead, phosphocreatine donates its phosphate group to ADP in a reaction catalyzed by creatine kinase . Phosphocreatine levels are replenished via ATP from glycolysis and cellular respiration.
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166. If a creatine kinase inhibitor is administered to an active muscle cell, which of the following would most likely occur? A. B. C.
D,
Passage II (Questions 170-175)
Connective tissues secrete large molecules that make up their extracellular matrix. Specialized cells called fibroblasts play the major role in the formation of the matrix. Three types of molecules characterize a matrix: proteoglycans; structural proteins; and adhesive proteins. Proteoglycans contain hydrated protein and carbohydrate chains, and can be very large. They typically create a gelatinous structure between the cells. Structural proteins add strength and flexibility to the matrix, and the adhesive proteins hold the cells together.
ATP concentrations would diminish while muscle contractions continued. Phosphocreatine concentrations would diminish while muscle contractions continued. ATP concentrations would remain constant while the percent saturation of myoglobin with oxygen would diminish. Cellular respiration and glycolysis would increase to maintain a constant ATP concentration.
The main component of both bone and cartilage is the matrix. The cellular element of cartilage is called a chondrocyte. In their immature form, they secrete the matrix which includes collagen, the most abundant protein in the body. Three polypeptide chains wrap around each other to form the triple helix which gives collagen much of its strength and flexibility. Cartilage is not innervated and has no blood supply.
167, Muscle cell T-tubules function to: A. B. C. D,
create an action potential within a muscle cell. receive the action potential from the presynaptic neuron. deliver the action potential directly to the sarcomere. supply Ca2+ to the cytosol during an action potentiaL
The human fetus has a cartilaginous endoskeleton which is gradually replaced with bone before and after birth until adulthood. Bone forms within and around the periphery of small cartilaginous replicas of adult bones. Bone fonning osteoblasts differentiate from fibroblasts of the perichondrium. In long bone formation, periosteum ossifies around the shaft or diaphysis of the bone. Chondrocytes enlarge within the diaphysis and then break down leaving a honeycombed cartilage. Calcium deposits form and vascular connective tissue invades the area. Ossification of the inner portion of the bone proceeds from the center toward each end. Osteoclasts differentiate from certain phagocytotic blood cells and begin to burrow through older bone. The osteoblasts line up around the periphery of the newly formed tunnels and deposit concentric layers of bone. Vascular connective tissue and nerves move into these tunnels which are called osteons. In long bones, most osteons form along the length of the bone, and the columnar shape of the osteons gives these bones tremendous strength.
168, All of the following are true concerning skeletal muscle cells EXCEPT: A. B. C. D.
SkeletaJ muscle cells contain more than one nucleus. Human skeletal muscle cells continue normal cell division via mitosis throughout adult life. Skeletal muscle cells contain more than one sarcomere. During muscle contraction, only the H band and the I band change length.
169. Which of the following is most likely NOT true concerning the uptake of Ca1+ ions from the cytosol during muscle contraction? A. B. C. D.
It requires ATP.
The mechanism involves an integral protein of the sarcolemma. It occurs against the concentration gradient of Ca2+. It is rapid and efficient.
170. Each concentric layer of bone formed by osteoblasts in an osteon is called a: A. B. C. D.
lacuna lamella collagen fibril diaphysis
171. Osteoblasts which become trapped in small spaces within the bone and mature are called:
A. B. C. D.
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229
fibroblasts. chondrocytes. osteoclasts. osteocytes.
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Passage'" (Questions 176-182)
172. Which of the following matrix components best describes collagen?
A. B. C. D.
The strength of muscle contraction is directly related to the cross-sectional area of the muscle. with a maximal force of 3 to 4 kg/cm'. The rate at which a muscle can perform work, or the muscle power, varies over time and is given in Table I.
strnctural protein proteoglycan adhesive protei n glycosaminoglycan
173. Which of the following hormones most likely stimulates osteoclasts? A. B. C. D.
parathyroid hormone calcitonin epinephrine prostaglandin
D.
B.
C. D.
First 10 seconds
7000
Next 1.5 minutes
4000
Next 30 minutes
1700
Energy for muscle contraction is derived directly from ATP and ADP. However, a muscle cell's original store of ATP is used up in less than 4 seconds by maximum muscle activity. Three systems work to maintain a nearly constant level of ATP in a muscle cell during muscle activity: the phosphagen system; the g/ycogen-laclic acid system; and the aerobic syslellt. Phosphocreatine contains a higher energy phosphate bond than even ATP and is used to replenish the ATP stores from ADP and AMP. This is the phosphagen system and it can sus tain peak muscular activity for abo ut 10 seconds.
Bone is connective tissue. Bone is innervated and has a blood supply. Yellow bone marrow stores triglycerides as a source of energy for the body. Bone is the only nonl ivi ng tissue in the body.
175. Which of the following cells most likely arises from the same stem cell in the bone marrow as an erythrocyte?
A.
Power (kg mlmin)
Table 1 Muscle power variance over time
174. All of the following statements are true concerning bone EXCEPT: A. B. C.
Time
The glycogen-lactic acid system is relied upon for muscular activ ity lasting beyond 10 seconds but not more than 1.6 minutes. This system produces ATP from glycolysis.
osteoblast fibroblast osteoclast chondrocyte
Aerobic metabolism of glucose, fatty acids and amino acids can sustain muscular activity for as long as the supply of nutrients lasts. The recovery of muscle after exercise involves replacement of oxygen and glycogen. Before exercise, the body contai ns approximately 2.5 liters of oxygen in the lungs, he moglobin, myoglobin, and the body fluids. This oxygen is used up in approximately I minute of heavy exercise. [n a resting adult having just finished 4 minutes of heavy exercise, the oxygen uptake is increased dramatically at first and then levels back down to normal over a I hour period. The extra oxygen taken in is called the oxygen debt, and is about 11.5 liters. Glycogen stores are replenished in 2 days for individuals on a high carbohydrate die~ but can take several more days for those on a high fat or high protein diet.
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180. Fast twitch muscle fi bers contract with more force than slow twitch muscle fibers. However, slow twitch muscle fibers accou nt for most of the workload in an endurance event. What organelle is most likely more abundant in slow twitch muscle fibers than fast twitch muscle fibers?
176. Based upon the information in Table I, wbich of the following are the most likel y rates of molar production of ATP for the phosphagen system, the glycogen-lactic acid system, and the aerobic system respectively? A. B. C. D.
1.7 Mlmin, 4 Mlmin, 7 Mlmin 2 Mlmin, 2 Mlmin, 2 Mlmin 4 Mlmin, 3 Mlmin, 2.5 Mlmin 4 Mlmin, 2.5 Mlmin, I Mlmin
A. B.
C. D.
177. Why is the oxygen debt greater than the amount of oxygen stored in the body?
A. B.
C.
D.
181. [n which of the following sports do athletes most likel y rely primarily upon the phosphagen system for muscle contraction?
Exercise increases the hemoglobin content of the blood so that it can store more oxygen. Heavy breathing after exercise takes in more oxygen. In addition to replenishing the stored oxygen, oxygen is used to reconstitute the phosphagen and lactic acid systems. In addi tion to replenishing the stored oxygen, oxygen is used to reconstitute the phosphagen, lactic acid system, and aerobic systems.
A.
B. C.
D.
C. D.
tennis boxing diving cross-country skiing
182. Which of the followi ng is depleted first in tbe body of an athlete performing max imal exercise?
A. B. C.
178. According to the information in the palOsage, in order for an individual on a high carbohyd rate diet to perform at his peak in an athletic event, he should not engage in heavy exercise for at least: A. B.
mitochondria smooth endoplasm ic reticulum free floating ribosomes lysosomes
D.
phosphocreatine ATP glycogen glucose
1.6 minutes before the event. 30 minutes before the event. 48 hours before the event. 10 days before the event.
179. Most of the lactic acid produced by muscle activ ity is reconver[ed into glucose via the Cori cycle. Which organ plays the major role in the reconstitution of lactic acid to glucose? A. B. C. D.
kidney liver spleen muscle tissue
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184. Which of the following statements concerning bone is false?
Questions 183 through 184 are NOT based on a descriptive passage.
A. B.
183. All of the following are true concerning the musculoskeletal system EXCEPT: A. B. C. D.
C.
Skeletal muscles function by pulling one bone toward another. Tendons connect muscle to muscle. Ligaments connect bone to bone. The biceps works antagonistically to the triceps.
D.
Bone functions as a mineral reservoir for calcium and phosphorous. Bone contains cells which differentiate into red and white blood cells. Bone acts as a thermostat for temperature control of the body. Bone provides a framework by which muscles can move the body.
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232
STOP.
30-MINUTE IN-CLASS EXAM FOR LECTURE 9
233
Passage I (Questions 185-190)
186. Which of the following could be true co ncerning a population that has li ved in a stable environment for hundreds of years and whose individuals have an average life expectancy of 20 years?
Two schedules summarize the most important demographic information of a closed population: the survivorship schedule and tbe fertility schedule.
A. B.
The survivorship schedule records the probability that an individual survive to a particular age, and is shown in Figure 1. There are three basic forms of survivorship curves in nature: type I; type II; and type III. Most species in nature exhibit a type III survivorship curve. 1.0 -.:----~
,9-;;.-. ~-
~;:
01 .
C. D.
type I
187. The net reproductive rate is most likely based only on females and not on males because:
type II
o:E
A. B.
'" .> - .J:J
t: 8
~ 0.
The number of individuals that are 13 years old doubles every 100 years. If most of the individuals in the population are over 18 years old, then the population is aging. Most individuals in the population live at least forty years. If 60% of the population is 10 years old, then, in two years, 60% of the population will be 12 years old.
0.01
c.
type III 0.001 -..L---===~
Age of individual
D.
males reproduce faster than females . females live longer than males. the maximum rate at which a female gives birth is not changed by the number of males in a population. early population scientists designated this as the conventional stand ard.
Figure 1 Survivorship Schedule The fertility schedule records the average number of daughters that will be produced by all the females in the population at each particular age. The net reproductive rate is the average number of females produced in the lifetime of a single female. The net reprod uctive rate can be obtained from the survivorship and fertility schedules as the sum of the yearly products of the probability of survivorship and the average number of females born to a single female during life expectancy.
188. Which form of survivorship curve would most likely be exhibited by a pure r strategist?
Any population reproducing in a constant environment (other than species breeding synchronously at a single age) will attain a stable age distribution. In such a population, the proportion of individuals at a given age wil1 remain constant.
189. Assuming a zero growth rate, (he survivorship curve that most closely represents the population of humans in the U.S. during the late 20,h century is:
A. B. C. D.
A. B. C.
185. The annual ad ult mortality of white storks is approximately 2 1 percent regardless of age. What form of survivorship curve is exhibited by white storks?
A. B. C. D.
D.
type I type jj type III In order to predict the survi vorship curve, tbe fertility schedule must be known.
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type I type II type III r strategists do not exhibit typical survivorship curves.
type I type II type III The survivorship curve for a zero growth rate population would be flat.
190. Which of the following is the LEAST impol1ant characteristic for the survival of a species living in a short-lived, unpredictable habitat? A. B. C. D.
234
large brood size rapid development early reproduction efficient utilization of available resources
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194. A human male may inherit all of Ule following EXCEPT:
Passage II (Questions 191-196)
Isogenic mice are mice that have been inbred until they have nearly identical genotypes. Tn 1955, Eichwald and Silmser transplanted ski n between isogenic mice. Histocompatibility is the acceptance of tissue grafts. A Y-Iinked gene located at the H-Y locus produces the H-Y antigen. Tissue grafts in isogenic mice may be rejected due to the H-Y antigen. In Eichwald and Silmser ' s experi ment, only female mice rejected the skin grafts, and then only from male mice.
A.
his mother's mother's X c hromosome.
B. C.
his mother's father's X chromosome.
D.
his father's father's Y chromosome. his father's mother's X chro mosome.
195. Colorblindness is a sex linked recessive trait. What are the possible genotypes of the mother of a colorblind female, if the mother's father was not colorblind? A. B. C. D.
In 1984, McLaren bred mice in which the H-Y gene segregated separately from maleness. However, males which lacked the H-Y antigen had a defect in spermatogenesis. Then in 1987, Page mapped a Y-linked gene that coded for a factor important in the development of male genitalia, called the testis-determining factor (Tdf), to a different region on the Y chromosome than that of the H -Y gene.
homozygous recessive only homozygous dominant only heterozygous only homozygous recessive or heterozygous
196. Isogenic mice are produced by breeding only siblings for many generations. Which of the following is true concerning the production of a population of isogenic mice?
Not all animals have an X- Y chromosome makeup like that found in mammals. In birds, butterflies, moths; and some fish. the homozygous partner is male. while the heterozygous partner is female.
A. B. C.
The Hardy-Weinberg law is not violated. It is an example of speciation. Isogenic mice are more likely to have autosomal recessive diseases than are normal mice.
D.
Isogenic mice have fewer chromosomes than normal mice.
191. In chickens, nonbarred plumage is a sex linked recessive trait. What is the probability that a barred female and a nonbarred male produce a barred chic as their first female offspring?
A.
0
B.
25% 50% 100%
C. D.
192. Which of the following mice do not produce H- Y antibodies? A.
isogenic males of the Eichwald and Silmser experiment
B. C. D.
isogenic females of the Eichwald and Silmser experiment males of the McLaren experiment that lack the H-Y gene. wild female mice.
193. Which of tbe foUowing is NOT true concerning the H-Y gene and the gene for Tdf? A. B. C. D.
They obey the Mendelian Law of Segregation. They obey the Mendelian Law of Independent Assol1ment. They both code for proteins. They are normally found in males but not in females.
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Passage III (Questions 197-202)
199. If the red flower trait is dominant and the white flower trait is recessive in garden peas, what is the expected ratio of red flowered plants. pink flowered plants, and white flowered plants when step 3 is performed?
Gregor Mendel, an Austrian monk, performed the first quantitative studies of inheritance. Using the garden pea, Me ndel performed a series of hybridization experiments with lines differing in seven traits. His experiments followed these three basic steps:
A. B.
C. D.
I. Pea plants were allowed to self-fertilize for several generations and only plants which faithfully reproduced their original traits in each generation were used.
A.
3. He allowed the hybrids to self-fertilize for several
B.
generations.
e.
These steps had been done before by other scientists with the exception that, this time, Mendel counted and recorded the results.
D.
By chance, each of the traits that Mendel studied followed the Law of Independent Assortment. Each trait also exhibited complete dominance.
C.
D.
A. B. C. D.
Alleles for the same trait would not have separated independently of each other. Alleles for different traits would not have separated independently of each other. Homologous chromosomes would exhibit alleles coding for different traits at the same loci. Phenotypes of a single trait would resemble both parental phenotypes on the same individual.
the Law of Segregation the Law of Independent Assortment the Hardy-Weinherg Principle pleiotropic separation
202. A dihybrid cross is made between individuals with the genotype BbFf. What is the ratio of tbe following genotypes BBFF. BBff. bbFF, bbff in the progeny?
A. B.
198. What was accomplished in step I of Mendel's experiment? A. B. C. D.
The offspring of two pea plants with wrinkled peas has smooth peas. The offspring of two pea plants with smooth peas has wrinkled peas. The offspring of a pea plant with smooUl peas and a pea plant with wrinkled peas has smooth peas. The offspring of a pea plant with smooth pea. and a pea plant with wrinkled peas has wrinkled peas.
201. Which of the following explains why a cross hetween a plant with green pea pods and a plant with yellow pea pods results in offspring of plants with either yellow or green pea pods. and does not result in plants with yellowish green pea pods?
197. If the traits that Mendel had studied had not followed the Law of Independent Assortment, how might Mendel's findings have changed?
B.
I I I I
200. If the wrinkled pea trait is dominant to the smooth pea trait, which of the following is not possible?
2. Mendel then cross-fertilized plants having different traits.
A.
9,3, 3,0, I, I, 1,2.
C.
9,3,3, 1 4,3,2,1 1,3,3, 1
D.
I, I. I, I
Pure homozygous populations were created.
Pure heterozygous populations were created. Any sick plants were naturally selected against. Recessive alleles were removed from the population.
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206. Sickle cell anemia is an autosomal recessive trait. The ability of homozygous recessives to survive and reproduce is greatly reduced. Which of the following must be true in order for the sickle cell gene to remain in the gene pool indefinitely?
Questions 203 through 207 are NOT based on a descriptive passage.
203. All of the following habitats would probably favor an r strategist over a K strategist EXCEPT: A. B. C. D.
A.
the weedy cover of new clearings in forests the mud surfaces of new river bars the bottoms of nutrient-rich rain pools a cave wall
B. C. D.
204. Healthy organisms living in the wild can typically mate and reproduce fertile offspring if they are members of the same: A. B. C. D.
207. All of the following are examples of density dependent factors affecting population growth EXCEPT:
specIes genus family phylum
A. B. C. D.
205. Colorblindness is a sex-linked recessive trait. If a colorblind man and a woman that is a carrier for the trait have two girls aud two boys, what is the probability that at least one of the girls will be colorblind? A. B. C. D.
competition for resources. catastrophic weather predation disease
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0% 50 % 75 % 100 %
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Homozygous dominates must have increased fitness over heterozygotes. Heterozygotes aud homozygous dominates must have equal fitness. Heterozygotes have increased fitness over homozygous dominates. Heterozygotes must also have decreased fitness.
237
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ANSWERS & EXPLANATIONS FOR
30-MINUTE IN-CLASS EXAMINATIONS
239
240
MeAT
BIOLOGY
ANSWERS TO THE 30-MINUTE IN-CLASS EXAMS Lecture 1 Lecture 2 Lecture 3 Lecture 4 Lecture 5 Lecture 6 Lecture 7 Lecture 8 Lecture 9
e
24.
e e
47. A
70. B
2. A
25.
48. A
71.
e e
26. A
49. B
5. A
93. B
116. A
139. A
162. B
185. B
e
117. B
140. B
163. A
186. A
72. 0
95. A
118. A
141. B
164. 0
187.
e
73. B
96.
119. A
142. A
165. 0
188.
28. B
51. B
74. 0
97.
e e
120. B
143. 0
166. A
189. A
6. B
29. A
52.
75. 0
98. A
121. 0
144. A
167.
e
190. 0
7. A
30. 0
53.
e
99. 0
122. A
145. 0
168. B
191. A
8. A
31. A
54.
e e e
77. B
100. 0
123. B
146. 0
169. B
192. A
e
32. A
55. 0
78. B
101. A
124.
e
147. B
170. B
193. B
10. A
33. 0
56.
e
102. 0
125. 0
148. 0
171. 0
194. 0
11. A
34. A
57. B
80. B
103. A
126. 0
149. B
172. A
195.
12. A
35. A
58. A
81. B
104. B
127. A
150. B
173. A
196.
13. 0
36. A
59. B
82. B
105.
e
174. 0
197. B
14. B
37. B
60. 0
83. 0
106. B
129. A
e
198. A
15. A
38.
e
84. B
107. B
130.
176. 0
199. B
16. B
39. A
62. B
85. A
108.
40. 0
63.
e
86. B
109. B
132.
e e
200. B
17. B
e e e
18. B
41.
e
64. A
87.
110. 0
133. B
156. B
179. B
202. 0
19. 0
42. 0
65. 0
88.
e e
111. B
134.
e
157. 0
180. A
203. 0
e
43. 0
66. A
89. 0
112.
135. A
158. A
181.
e
204. A
e
67. 0
90.
e
113.
136. B
159. A
182. A
205.
45. 0
68. A
91. B
114. A
137.
e
160. B
183. B
206.
46. B
69. 0
92. A
115. A
138. B
161 A
184.
1.
3. 4.
9.
20.
21. A 22. 23.
e e
27.
44.
e
e
50.
61.
e
76.
79.
e
94.
e
e
e e
128.
e
151.
152. A
e
153.
131. B
154.
e
155.
175.
177. 178.
e
e e
e e
201. A
e e
207. B
BIOLOGICAL SCIENCES
Raw Score
Estimated Scaled Score
23
15
22
14
21
13
19-20
12
18
11
17
10
15-16
9
14
8
12-13
7
11
6
10
5
8-9
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EXPlANATION S FOR TH E 30- M INUTE IN-ClASS EXAM INATIONS .
241
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 1 Passage I 1.
C is correct. Protein disulfide isomerase is an en zyme. The function of any enzyme is to lower the activation energy of both the forward and reverse reactions. Enzym es cannot alter the equilibrium constant of a reaction; they can only increase the ra te at which a reaction proceed s towards equilibrium.
2.
A is correct. The first line of the passage says that folding is dependent upon amino acid sequence, which is the primary structure. The folding pattern itself is the tertiary structure. This question may seem ambiguous, but, especially with the directions "according to the passage", the best answer is A.
3.
C is correct. Process of elimination is the best technique for this question. To help clarify a question, it is sometimes helpful to restate the question as a statement with "because", and add the answer choices. For example, "Attempts at predicting protein configuration based upon amino acid sequence have been unsuccessful because." Even if A were true, it does not answer the question. B is false because enzymes are catalysts. They do not determine the product; they only increase the rate of the reaction. 0 is generally false except in rare cases of neutral mutation. C is a direct explanation to the question and happens to be true.
4.
C is correct. The folding of a peptide chain is called the tertiary structure of a protein. The passage states that chaperones assist in the "folding process". This is the best answer.
5.
A is correct. Choice A describes exactly the cole of chaperones as explained in the pas:sage. A cell with more chaperones can synthesize proteins more easily, giving it a selective advantage. B is incorrect. Chaperones don' t slow the process of protein folding. C is wrong for a number of reasons. For one, the energy for protein synthesis comes from ATP not heat. For 0, chaperones don' t affect the rate of polypeptide synthesis; they affect polypeptide folding.
6.
B is correct. Why does a protein denature in the presence of heat? The hydrogen bonds and other non-covalent interactions in the secondary and tertiary structure are disrupted. Another way to answer this question is to notice that all the bonds are covalent except hydrogen bonds. Although hydrogen bonds are the strongest type of inte nnolecular bond, they are much weaker than covalent bonds, and will be disrupted before covalent bonds when heat is applied.
7.
A is correct. This question requires that you know that disulfide bond formation occurs in the synthesis of cystine, not proline, and that you know that an enzyme lowers the energy of activation of a reaction. This question is pushing the envelope in how much detail is required by MCAT. In other words, this may be too detailed for MCAT. However, notice that even in this question, the amino acids chosen were ones with which you are likely to be familiar.
Passage II 8.
A is correct. This question calls for some speculation. We must assume that carbohydrates d o not react with SOS in the same way that proteins do. The passage explains that SOS gives proteins a charge "in approximate proportion to their size." It then says that the rate of m ovement is related to the molecular weight. The carbohydrates of a glycoprotein increase the mass without reacting to the 50S. Thus, they disrupt the relationship between mass and movement. Process of elimination can help. Acidic proteins are polar, so C can't be correct unless B is also correct. Enzymes include acidic proteins and polar proteins, so 0 can't be correct unless B and C are correct.
9.
C is correct. According to the passage, coomassie blue is a dye added to the proteins after they have separated. Thus, it does not affect the results of the experiment. C is the only choice that does not affect the results.
10.
A is correct. The passage states that SOS separates proteins based upon size only. The native charge is small compared to the SOS. The force on a protein is proportional to the charge, which is proportional to the size. So why wouldn't all proteins move at the same rate? The answer is because large proteins have difficulty fitting through the pores of the gel and thus move more slowly.
11.
A is correct. The passage explains that SOS gives proteins a charge "in approximate proportion to their size." Only A offers an explanation why this is unnecessary with nucleic acids. C and 0 are simply false anyway.
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242 . MeAT BIOLOGY 12. 13.
A is correct. Primary structure is the order of amino acids, which is determined by covalent peptide bonds. The question says that SDS cannot disrupt covalent bonds. D is correct. Because SDS breaks noncovalent bonds, and mercaptoethanol breaks disulfide bonds, the quaternary structure of a protein is disrupted. Once disrupted, the sublll1its separate according to their size as
explained in the passage. 14.
B is correct. Since SDS PAGE identifies proteins based upon size, two proteins with similar molecular weight would not be easily distinguished using SDS PAGE. C is true but is not a limitation in identifying the different proteins in a mixture. D is the very reason that SDS is used; to make charge correspond to size.
Passage III ~
15.
A is correct. Aerobic respiration is the oxidation of glucose: glucose + oxygen also combustion.
carbon dioxide + water. It is
16.
B is correct. Substrate-level phosphorylation is when an energy-rich intermediate transfers its phosphate group to ADP, forming ATP, without requiring oxygen. Oxidative phosphorylation is the production of ATP via the electron transport chain and ATP synthase.
17.
B is correct. C and D are wrong because glycolysis is independent of oxygen. A has only one pyruvate, no NADH, and isn't balanced.
18.
B is correct. The passage says that arsenate acts as a substrate for Glyceraldehyde 3-phosphate dehydrogenase. Thus, arsenic would prevent the reaction of glyceraldehyde 3-phosphate to 1,3-bisphosphoglycerate leading to a build up of the former.
19.
D is correct. D directly explains the word committed. A contradicts the passage. B contradicts the diagram. C is not true and does not answer the question; reactions in chemical pathways do not have to create progressively lower energy molecules.
20.
C is correct. The passage states that PFK activity is stimulated when cellular energy is low, and inhibited when energy is high. Low cellular energy corresponds to high ADP concentration because ATP has been used. ATP is, itself, an allosteric inhibitor of PFK, and an indicator that cellular energy is high, so D is wrong. (By the way, the concentration of ATP is held relatively constant within a cell, and is usually much higher than the concentration of ADP.) Glucagon is a peptide hormone that doesn't enter the cell, and therefore could not allosterically inhibit PFK, an enzyme of glycolysis, which takes place in the cytosol. Thus, B is wrong. Citrate is a intermediate in the Krebs cycle and actually acts as an allosteric inhibitor of PFK. This makes sense, because if there is an abundance of citrate, then there must be ample energy being produced in the cell. Thus, A is wrong.
21.
A is correct. Arsenate competes for the active site; this is the definition of competitive inhibition. Arsenate would have to be a product of the glycolytic pathway in order for its action to be considered as negative feedback.
Stand Alones 22.
C is correct. A is wrong because not all enzymes require cofactors. B is wrong because not all cofactors are organic. D is wrong because cofactors aren't catalysts.
23.
C is correct. The graph of reaction rate versus substrate concentration for enzymes is shown in Figure 1-8; enzyme activity is saturable.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 2 Passage I 24.
C is correct. According to the passage, the Pelomyxa is a eukaryote and does not undergo mitosis. A is not true and not implied in the passage. Fungi lack centrioles and undergo mitosis. B is false. Prokaryotes do not undergo mitosis. They undergo binary fission. D is not true, nor does the passage imply it. Copyright @ 2007 Examkrackers, Inc.
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EXPLANATIONS FOR THE 30- MINUTE IN-CLASS EXAMI NATIONS •
243
25.
C is correct. In prophase, the nuclear membrane begins to disintegrate and chromosomes condense. Replication occurs during the S phase so A is wrong. There is no crossing over in mitosis so B is wrong. D is incorrect as well.
26.
A is correct. The claim that Pe/omyxn is a primitive eukaryote is based upon its apparently primitive method of nuclear division, and the fact that it has not acquired certain organelles such as mitochondria. Choice A would indicate that Pelomyxa underwent mitosis and contained mitochondria at one point and lost the ability and organelles through evolution. This would indicate that Pelomyxa was not a primitive eukaryote. B is wrong because it only tells us that Pelomyxa is phylogenetically older than diatoms. C is wrong because it is irrelevant. D is wrong because two species in the same phylum don't necessarily have similar phylogenic age.
27.
C is correct. Anaphase is the separation of chromosomes. The formation of two daughter nuclei marks telophase. These two phases most closely resemble the separation of chromosomes and formation of the nuclear membrane between chromosomes in dinoflagellates.
28.
B is correct. The passage states that the symbionts function like mitochondria. Mitochondria provide energy in the form of ATP from the metabolism of nutrients.
Passage II 29.
A is correct. Choice A is tl,e only choice that indicates that the poly-A tail is not coded for by DNA but is synthesized separately. It offers a mechanism by which synthesiS may take place other than transcription. B seems to indicate that the poly-A tail is transcribed from DNA at the end of the gene. Even if C and D made sense, they would indicate that the poly-A tail was transcribed from a gene.
30.
D is correct. The question is really asking, "Where does post-transcriptional processing take place?" The obvious answer is in the nucleus. Some (very little) post-transcriptional processing of mRNA does take place within prokaryotes, but snRNPs are not found in prokaryotes. Post-transcriptional processing of rRNA and tRNA commonly takes place in prokaryotes.
31.
A is correct. Figure 2 is an electron micrograph of R looping. The passage explains that R looping is a technique where DNA and mature RNA are hybridized. Mature RNA has no introns. Therefore, the loops are parts of the DNA that correspond to the removed sections of the RNA; the loops are DNA introns.
32.
A is correct. The DNA strand that is displaced is the template for the DNA strand shown.
33.
0 is correct. The question says that introns change while the genes that encompass them remain intact. TItis suggests that the intron plays little or no role in the phenotype. The phenotype is affected by selective pressure. Thus, selective pressure has no apparent mechanism by which to affect introns.
34.
A is correct. The passage states that the 5' end is capped forming a 5' -5' triphosphate linkage. You must understand that the 5' refers to the carbon number on the ribose. You don' t have to know how the carbons are numbered, just that the same carbon on each ribose must be attached to the triphosphate group. TItis leaves only A or C. C has thymine, so it cannot be RNA. Note: Sometimes, carbon 2 on the first two nucleotides after guanosine has an O-methyl group instead of a hydroxyl group.
Passage III 35.
A is correct. The passage says that glycosylases recognize uracil in DNA as a product of deamination of cytosine. After recognizing uracil, the glycosylases change it to cytosine. If DNA naturally contained uracil, glycosylases would have no way to distinguish the good uracil from the uracil that is produced by the deamination of cytosine. B indicates an advantage of haVing uracil, not a disadvantage. Who knows if C is true or not? If it is, it is knowledge that is beyond that required by the MCAT, and, therefore, cannot be the correct answer. D is a false statement.
36.
A is correct. Only answer choice A gives a logical reason. The passage says that much DNA damage is not permanent if repaired before replication. The time between S phases (replication) is shorter for rapidly reproducing cells. Cancer cells reproduce rapidly; thus, cancer cells cannot repair the damage as thoroughly as normal cells. B is not true, and we are given no reason to believe that it is. C is not true, and we are given no reason to believe that it is. D is very unlikely; if tumor cells outnumbered normal cells, the patient would have to be dead or near death. Anyway, there is no strong scientific reason that we would know if D is true or not, whereas A is supported by the passage and our knowledge of the cellular life cycle.
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37.
B is correct. The passage states that neoplastic diseases arise from mutations in somatic cells. Thus, they cannot be passed on to offspring. This is not the same thing as saying that susceptibility to neoplastic diseases cannot be inherited. For instance, breast cancer is a neoplastic disease. Susceptibility to breast cancer is inheritable; breast cancer is not.
38.
C is correct. C is the only choice that answers the question and is true. A and B are false statements; prokaryotes must possess a ligase to connect DNA molecules because they replicate DNA in a very similar fashion to the way eukaryotes replicate; they make Okazaki fragments. Prokaryotic DNA is double stranded. D does not answer the question of why eukaryotes can repair the breaks.
39.
A is correct. We are looking for two pyrimidines that are both found in DNA. Only A meets this criterion. B and C do not contain two pyrimidines, and D must be RNA.
40.
D is correct. This is tricky. The sense strand might code for an antisense strand where bromouracil substitutes for T. The next replication, the new antisense strand with bromouracil would code for a guanine at that spot. Thus the new sense strand would be identical to the original except that guanine would replace adenine. (Sense and antisense are ambiguous terms, but their ambiguity is irrelevant to the answer to this question.) original sense strand = S'-GGCGTACG-3' new antisense strand in presence of Br = 3' -CCGCAJiGC-S' new sense strand = S'-GGCGTGCG-3' Since there is no answer choice where bromouracil might be present in the second replication, we don't worry about that pOSSibility.
41.
C is correct. The passage states that a tautomeric shift causes a mismatch base pair. Although this may lead to a deletion upon repair by glycosylases, in their absence, the resulting mutation is a substitution. Choice A requires a deletion or insertion. Choice D is the result of an improper number of chromosomes. There is no mechanism by which a tautomeric shift would result in more or fewer chromosomes.
42.
D is correct. D is a summary of the last paragraph. Rapidly reproducing cells have less time to repair DNA between replications. A is wrong because there is the basal mutation rate which the rate of mutations in the absence of mutagens. B is wrong because the environmental mutagens simply add to the basal mutation rate, they don't change it. C is wrong because Hypoxanthine results from a mutagen; it is not a mutagen itself.
Stand Alones 43.
D is correct. DNA replication takes place during the S stage of interphase. You should memorize the stages of the cell life cycle.
44.
C is correct. Both children are female, because the genotype of Turner's syndrome, as per the question, must be XO. The 0 came from the parent within whom nondisjunction occurred; the X came from the other parent. The mom only has healthy Xs to pass on, the father has only one colorblind X to pass on. Thus the colorblind child got her X from Dad and her 0 from Mom, and the healthy child got her X from Mom and her 0 from Dad.
45.
D is correct. The primary spermatocyte is the spermatogonium just after DNA replication. Humans never have more than 46 chromosomes. The primary spermatocyte has 92 chromatids but only 46 chromosomes.
46.
B is correct. Memorize this for the MCAT.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 3 Passage I 47.
A is correct. The passage says that the death rate is exponential. This means that it decreases by the same fraction at even time intervals. In the first three minutes the population dropped to 1/2S of its original size. In the . second three minutes it should do this again. 2.5 x 1011 /2S = 1.0 x 1010.
48.
A is correct. The z value for Salmonella is soc. This means that for each increase of SoC, the D value is reduced by a factor of 10. Table 1 reports that the D value for Salmonella at 60°C is 0.4 minutes or 24 seconds. 70' is 10 Copyright © 2007 Examkrackers, Inc.
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degrees more than 60°. 10 degrees is two z values for Salmonella. Thus, we divide 24 seconds by 10 twice. We get 0.24 seconds. 49.
B is correct. In order to go from 1012 to 1 (1 = 10°), we divide by 10 twelve times. This is 12 D values (a 90% reduction is the same as dividing by 10). One D value is 0.2 minutes, so 12 D values is 2.4 minutes. While answer D may be appealing mathematically, it is logically ridiculous. It is equivalent to 380,500 years. These regulations would be a little stiff.
50.
C is correct. Exponential growth on a logarithmic scale is a straight line. Get used to reading log scales. They may be on the MCAT.
51.
B is correct. The passage states that disinfection is the killing or inhibition of pathogens. Pathogens are disease causing microbes (stated in the passage). Disinfection methods (including the controlling methods discussed in the passage) are likely to destroy some nonpathogens as well as pathogens.
52.
C is correct. The three D values for S. aureus vary according to substrate. This supports statement C. A is not true, nor is there sufficient evidence in the table to support it as strongly as C. D is false, and contradicts Table 1. There doesn't appear to be any evidence for B.
53.
C is correct. C. botulinum can be killed by boiling water. However, it takes a long time. Table 1 predicts a D value of 20 minutes for C. botulinum at 101 0c. A is not true because C. botulinum is an obligate anaerobe. B is not true because C. botulinum is a gram positive bacterium. 0 is false because, according to the passage, horses make antibodies to C. botulinum toxin.
Passage II 54.
C is correct. The passage says that the ends of the strands are reverse compliments of each other. The complement sequence is 5'-TGACAATIC-3'. The reverse of this is 5'-CTIAACAGT-3'.
55.
D is correct. D is the only choice that is consistent with both viruses and transposons. A, B, and C are all unique to viruses. It should be clear from the passage that transposons are not viruses, they are a form of TGE, which is a vehicle for genetic shuffling.
56.
C is correct. Binary fission typically produces identical daughter cells. It is not a method of genetic recombination in prokaryotes.
57.
B is correct. Conjugation is genetic recombination between prokaryotic individuals, which typically involves the F plasmid or some other plasmid. D is not genetic recombination. C does not take place in prokaryotes. 'A: does not typically involve plasmids.
58.
A is correct. Conjugation is a one way transfer of genetic information from the initiator or F+ to the receiver or F-. The passage says that replicative transposition is when a copy of the gene is transferred. In this case, bacterium A lost the ability to live on histidine lacking medium. Thus the gene must have been completely removed and transferred; conservative transposition.
Passage III 59.
B is correct. A virion is the inert form of a virus that exists outside the host cell. A prophage is the name used to describe the virus while it is incorporated into the host cell DNA.
60.
D is correct. The passage states that A phage is a bacteriophage. Bacteria don't have nuclei, mitochondria, or endoplasmic reticulum. Their DNA is in the cytoplasm.
61.
C is correct. You should know that UV light causes DNA damage. The passage states that DNA damage leads to the lytic cycle. A is wrong because the genes of a second A phage would be repressed by the A repressor of the first A phage. B is wrong because the lysogenic stage can last hundreds of thousands of generations. D is wrong because bacteria don't undergo mitosis.
62.
B is correct. All viruses need a specific protein to bind in order to attach to and infect a host cell. The host cell does not have to be weakened. Mutagens are an entirely separate topic from viral infections. Bacteria don't have nuclei.
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63.
C is correct. A is wrong because DNA is not translated. B is wrong because only some RNA viruses contain reverse transcriptase and lambda phage isn't one of them. D is wrong because bacteria have no nuclei and viruses are never injected into a nucleus.
64.
A is correct. The passage states that lambda phage is a bacteriophage. This means it attacks bacteria.
65.
D is correct. No virus has both DNA and RNA.
Stand Alones 66.
A is correct. This question should be easy. Bacteria have no complex membrane bound organelles. The MCAT is likely to leave out the 'complex' part.
67.
D is correct. Yeast is probably the only member of the Fungi kingdom about which you must know anything specific.
68.
A is correct. All viruses require the host cell in order to replicate.
69.
D is correct. Only eukaryotes have centrioles. Prokaryotes have peptidoglycan cell walls, and some eukaryotes have cellulose or chitin walls. Of course, both have RNA and ribosomes, so that they can synthesize protein. All living organisms have both DNA and RNA. (Viruses are not living.) By the way, since ribosomes are made from RNA, if you chose B, then C would also have to be true.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 4 Passage I 70.
B is correct. Chromaffin cells are modified postganglionic sympathetic neurons, so they most likely secrete norepinephrine and epinephrine.
71.
C is correct. The diaphragm is skeletal muscle and is not innervated by the autonomic nervous system. You may be able to eliminate A, B, and D from memory. By the way, sympathetic activity does not innervate skeletal muscle, but its effects increase glycogenolysis in skeletal muscle.
72.
D is correct. Cardiac cells are innervated by both autonomic nervous systems. Sympathetic neurons release epinephrine onto cardiac muscle cells. A and B would be expected to have similar receptors because chromaffin cells are modified postganglionic sympathetic neurons. All autonomic preganglionic neurons release acetylcholine onto postganglionic neurons.
73.
B is correct. You should know that A, C, and D are autonomic responses. A and C are autonomic because they have to do with smooth muscle. Sweating is controlled by both the sympathetic and the parasympathetic as well. Shivering is skeletal muscle contraction, which you should know is not controlled by the autonomic nervous system.
74.
D is correct. The somatic nervous system governs skeletal muscles, which are involved in the simple reflex arc.
75.
D is correct. The nervous system acts directly and immediately (in seconds or less), while hormones are indirect and require more time.
76.
C is correct. Epinephrine is a sympathetic neurotransmitter. The sympathetic nervous system dilates the pupils (so you can hunt in the dark [memory aid]).
Passage II 77.
B is corred. From the passage, we know that one function of the smooth ER is hormonal synthesis. Thus, B is the best answer.
78.
B is correct. The passage states that all phospholipid synthesis takes place on the cytosol side of the smooth ER.
79.
C is correct. The passage states that ingestion of phenobarbital leads to greater production of smooth ER and mixed-function oxidases but not other enzymes. More mixed-function oxidases means more efficiency in de-
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grading phenobarbital. This is why people that take sleeping pills must take increasingly greater doses and are less responsive to antibiotics. 80.
B is correct. The passage states that phospholipid translocators flip phospholipids from one side of the membrane to the other.
81.
B is correct. Proteins are synthesized on the rough ER or on ribosomes in the cytosol.
82.
B is correct. Since oxygen is reduced to water, the iron must be oxidized.
83.
D is correct. Fat is stored energy.
Passage III 84.
B is correct. Although the calcium channels begin opening immediately, they are slow to open and slow to close. The result is the extended plateau of section 2. The sodium channels close very quickly and are the major contributors to depolarization or section 1.
85.
A is correct. Depolarization is the initial influx of sodium ions.
86.
B is correct. Acetylcholine is the neurotransmitter used by the parasympathetic nervous system, particularly the vagus nerve innervating the heart. The acetylcholine binds to muscarinic receptors, which stimulate the opening of K+ channels and thus inhibit depolarization. You should know that the time between heartbeats is increased by acetylcholine. Thus, section 4 lengthens.
87.
C is correct. Near the end of the action potential potassium channels are slow to close while potassium ions exit the cell. Calcium ions are being pumped back into the sarcoplasmic reticulum. Sodium channels are closed.
88.
C is correct. Na+ voltage gated channels contrib ute to the action potential but not to the resting potential. They are only open during an action potential. If you removed them, the resting potential would not be affected.
89.
D is correct. Only cells that experience action potentials contain Na' voltage gated channels. Muscle cells and neurons are two examples. Since the passage is about action potentials in muscle, this question should be deducible even without this information.
90.
C is correct. You should know that Na' channels are shut at the end of section 1 and the beginning of section 2 of the graph in Figure 1. From the passage, you should deduce that they are inactivated not closed .
Stand Alones 91.
B is correct. You should know that Na' channels are more sensitive than K' channels. That's why they open first.
92.
A is correct. Choice A describes saltatory conduction.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 5 Passage I 93.
B is correct. Cortisol is a steroid. The passage states, and you should know, that steroids diffuse through the membrane and bind to a receptor in the cytosol, where they are carried to the nucleus.
94.
C is correct. Caffeine inhibits phosphodiesterase, which leads to an increase in cyclic AMP. This much is derived from the question. Cyclic AMP activates phosphorylase, resulting in breakdown of glycogen to glucose. This is from the last sentence in the passage.
95.
A is correct. From Figure 1, the only missing link in the chain reaction from hormone to cyclic AMP as explained in the passage is the G,-protein. G,protein does not activate adenylyl cyclase; it inhibits, so B must be incorrect. C is incorrect as epinephrine binds to plasma .m embrane receptors. Nowhere in the passage is D discussed; protein kinase A is downstream of the receptOt; G,protein, and adenylyl cyclase.
96.
C is correct. C describes a mechanism by which the same hormone can stimulate different processes in different cells through increasing cyclic AMP levels. A is wrong because liver cells must contain a G,-protein, since
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248
MCAT B,OLOGY addition of cyclic AMP activates phosphorylase. B is wrong because from the passage we know that liver cells are capable of production of cyclic AMP. D is wrong because the p assage says that glucagon works through cyclic AMP levels.
97.
C is correct. Cyclic AMP is in the cytosol. Look at Figure 1. Cyclic AMP could not be a part of the membrane presence because it is not part of the membrane. It must be the right answer.
98.
A is correct. Epinephrine is a sympathetic hormone and stimulates the heart while inhibiting smooth muscle of the gut. The answer choices only give the option of a mechanism differing in G,-proteins and Grproteins. The answer choice must coincide with epinephrine stimulating heart muscle and inhibiting smooth muscle.
99.
D is correct. Since the G,-protein can't hydrolyze GTP, it can't turn off. This leads to increased cAMP. A is a bad answer because the passage states that the G-protein must hydrolyze GTP in order to be inactivated. You have no way of knowing about electrolyte concentration so B must be wrong. C is wrong because nothing in the passage mentions any connection between G-proteins and hormone binding to receptors on the outside of the cell.
100.
D is correct. Aldosterone is a steroid . Its effect is at the level of transcription. It increases protein production.
Passage II 101.
A is correct. These two hormones even use the same receptor protein as stated in the passage.
102.
D is correct. Progesterone prepares the uterus for pregnancy.
103.
A is correct. The passage states that HCG uses a membrane bound receptor; therefore, HCG is a peptide hormone. It is really a glycoprotein like LH and FSH.
104.
B is correct. The remaining part of the follicle after it bursts to release the egg is the corpus luteum. The corpus luteum secretes estradiol and progesterone until it degrades into the corpus albicans. You should know this for the MCAT. .
105.
C is correct. Since HCG is produced by the placenta, it is only found when there is a pregnancy. All other hormones normally occur during the menstrual cycle.
106.
B is correct. Only estrogen is a steroid like all the cortical hormones, so only estrogen could act as a substrate.
Passage III 107.
B is correct. Know your mitosis. You should also know that at birth the oocytes are arrested in prophase of meiosis I until puberty.
108.
C is correct. These two cells arise from the interstitial ceUs and secret steroids. They are phylogenetically related.
109.
B is correct. Androgens are male hormones like testosterone. From the passage, testosterone inhibits GnRH. GnRH stimulates FSH and LH which are required for gamete production. Choice A is wrong because, although there would be a decrease in testosterone production, the exogenous production more than makes up for this decrease. Otherwise, there w ould be no point in taking androgens. C and D are wrong for the reasons that B is correct.
110.
D is correct. The only steroid choice. Steroids act in the nucleus. Peptides don't enter the cell. Woman produce some testosterone from the adrenal cortex.
111.
B is correct. Osteoclasts breakdown bone, osteoblasts build bone. This will be covered in Lecture 8. .
112.
C is correct. The follicle bursts on ovulation and remains behind as the corpus luteum. The secondary oocyte and corona radiata are released into the body cavity and swept into the fallopian tubes by fimbriae. The corona radiata consists of the zona pellucid a and some granulosa cells.
113.
C is correct. The LH surge causes ovulation. Yes, this question seems to have more than one answer. Couldn't estrogen also be true? However, read the question closely. It says MOST. LH alone will cause ovulation; estrogen alone will not.
114.
A is correct. FSH blockage would prevent spermatogenesis by interfering wi th Sertoli cells and would not interfere with Leydig cells, which produce androgens. The others would affect testosterone production.
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Stand Alones 115.
A is correct. All hormones work through negative feedback. The negative feedback begins when the effector is overproducing. The effector of TSH is the thyroid. In this case, the effector would be under producing. TSH production would increase to try to correct this.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 6 Passage I 116.
A is correct. The volume of a sphere is 4/ 3 It,", and the surface area of a sphere is 41t,". These equations tell us that while the surface area of a sphere is proportional to the square of the radius, the volume is proportional to the cube of the radius. Since 4/3 It and 4 It in these equations are just constants, and we do not want to know the exact values of surface area and volume, we can ignore them. For the oritinal fat globule, we will arbitrarily assume its radius = 2; using this radius the volume is 8 while the surface area is 4 (ignoring constanls). The problem states that the bile is reducing the diameter (and thus the radius) by a factor of 2. Decreasing the radius by a fetor of 2 (new radius = 1) makes the volume = 1, and surface area = 1. Notice that the volume decreased from 8 to 1, while the surface area dedreased from 4 to 1. Now you must realize that no fat was lost in this process, so a decrease in volume by a facor of 8 while maintaining the same amount of fat means that the Single fat droplet must be divided into 8 smaller fat droplets (each with a volume = 1 and surface area = 1). 8 new fat globules, each with a surface area 4 times as small as the original add up to a total cumulative surface area of 8 (versus the original surface area of 4). The surface area increased by a factor of 2.
(1 ,--
-_ /
globule
117.
B is correct. You should narrow this down to A or B from the passage. The thoracic duct delivers lymph to the venous circulation from the lower part of the body and the left arm. This question requires that you know either that the thoracic duct delivers the fat to the blood, or that the lymphatic ducts empty their contents into the veins, not the arteries. This question is on the trivial side for an MCAT question, but it is not impossible that they would ask it.
118.
A is correct. Only A shows that serum cholesterol level alone might not indicate a health risk to the patient. The HDLs might ca use a high serum cholesterol but indicate a healthy patient.
119.
A is correct. From the passage, lingual lipase works in the stomach .
120.
B is correct. You should know that trypsin is activated by enterokinase and then activates the other pancreatic enzymes. C and D must be wrong because the duodenum is at a pH of 6.
121.
D is correct. You should know the word emulsification.
122.
A is correct. You should know the major pancreatic enzymes, trypsin (works on proteins), chymotrypsin (works on proteins), lipase (works on lipids), and amylase (breaks starch into disaccharides). Disaccharides are broken down by intestinal enzymes. Lactase breaks down lactose, a disaccharide in milk.
Passage II This passage offers an opportunity to see some topics that are not required by the MCAT but are likely to be used in a passage. Glomerular filtration rate (GFR) is the rate at which filtrate enters Bowman's capsule. The rate of filtration is affected by: 1) the oneotie pressure difference between the blood and Bowman's capsule (oncotic pressure is osmotic pressure that tends to move fluid back to the blood); 2) the hydrostatic pressure difference between the blood and Bowman's capsule (overpowers the oncotic pressure and moves fluid into Bowman's capsule); and 3) the rate of blood flow (the faster, the greater the GFR). Conventional thinking (MCAT) says that the change in hydrostatic pressure is the regulator of GFR. If we multiply the plasma concentration of solute z (PJ times the GFR, this should tell us the rate at which the solute is excreted. If there is no resorption or secretion, this number should equal the solute concentration in the urine (V,) times the urine volume (V). GFR x P, = V, x V. If we take into account, secretion, resorption, and change Copyng ht © 2007 Examkrackers, Inc.
250 . MeAT B,OLOGY
in volume of total filtrate, we can change GFR in the equation to C,. C, x P, = U, x V. This is the first equation in the passage. Another way to look at clearance is as a type of comparison between the concentration in the urine to the concentration in the plasma. C, = (U,/PJ x V. RBF is the rate at which blood flows through the glomerulae of the kidney. RPF is the rate at which plasma flows through the glomerulae of the kidney. Still another way to look at clearance is as the volume of plasma that must be filtered in order to produce the solute by filtrate alone (i.e. without secretion or resorption). 123.
B is correct. The GFR is the volume of plasma filtered each minute. The renal clearance of a substance is the minimum volume of plasma needed to be filtered in order to produce that much substance by filtration alone. If clearance is greater than filtration, then the urine must be receiving an additional supply of the substance from some other source. Secretion is the only answer.
124.
C is correct. To accurately measure GFR you need a substance that is cleared from the plasma solely by glomerular filtration in the absence of any complicating factors that might cause a difference in the concentration of the substance in the filtrate vs. its concentration in the plasma. These substances should not be reabsorbed or secreted by the tubules. It should also not be destroyed, synthesized or stored by the kidneys and pass through the glomerular filtration membrane unhindered. In summary, the substance must not be affected by the nephron (except to be filtered) filtrate should be equal to its concentration in the plasma barring water resorption by the tubules. Choices A and B describe a situation where inulin is affecting the phYSiology of the kidney, so they must be incorrect. 0 is incorrect because the passage states that almost no solute is completely filtered in one pass through the renal corpuscles, yet it is possible to measure GFR.
125.
0 is correct. The passage states that inulin is a polysaccharide, so it must be larger than glucose. You should know that a red blood cell is too large to be filtered, so inulin must be smaller. Albumin is just small enough to be filtered but its negative charge prevents filtration. (This knowledge concerning albumin is not required for the MCAT or this question since you know that 0 must be true of any polysaccharide.)
126.
D is correct. The renal clearance must be affected by the ability of a substance to filter into Bowman's capsule. This is dependent upon size and charge. The renal clearance is also affected by the filtration rate, which is dependent upon the hydrostatic pressure difference between the glomerulus and Bowman's capsule, and the oncotic pressure difference. PAH is a solute used to measure the plasma flow, and should not significantly affect the clearance of another solute or else its accuracy would be diminished .
127.
A is correct. You should know that glucose is completely resorbed in the proximal tubule of a healthy adult.
US.
C is correct. A close reading of the question reveals that it is asking for the filtered fraction. This is equal to GFR/ RPF. The clearance of inulin is the GFR, and the clearance of PAH is the RPF. This equals 125/ 625 = 1/ 5
=0.2 =20%
129.
A is correct. You must see from the passage that a normal GFR is 125 ml / min. Then you must recognize that creatinine is just like inulin, since neither is absorbed nor secreted. Thus, the clearance of creatinine is equal to the GFR. Now use the equation: GFR=C=U x V P
• ••
130.
Ux V p = GFR
=
2.5 mg/ml x 1 mlfmin 125 mllmin
C is correct. A hematocrit level of 50% means that 50% of the blood by volume is red blood cells and the other 50% is plasma. The renal blood flow, then, is twice the renal plasma flow. The rena l plasma flow is equal to the PAH clearance.
Passage III
131.
B is correct. Because amino acids and sodium are reabsorbed together in the proximal tubule, less sodium reaches the macula densa cells in the distal tubule. Renin is released, leading to decreased resistance in the afferent arterioles, increased resistance in the efferent arterials, and increased renal blood flow. The increased renal blood flow increases the GFR.
132.
C is correct. The renin-angiotensin system causes increased systemic blood pressure as per the passage. The . renal blood pressure does not increase because blood flow to the kidneys is impeded by the stenosis. Copyright © 2007 Examkrackers, Inc.
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133.
B is correct. Aldosterone comes from the adrenal cortex; ADH is from the posterior pituitary; and the sympathetic nervous system is part of the autonomic nervous system. The thyroid is not mentioned.
134.
C is correct. The passage states that renin is an enzyme. An enzyme is a catalyst. A catalyst lowers the energy of activation of a reaction without being permanently altered.
135.
A is correct. Urine volume is reduced by ADH secretion among other things. Aldosterone causes B, C, and D.
136.
B is correct. Angiotensin II is a peptide (from the passage). Peptide hormones act via second messenger.
Stand Alones 137.
C is correct. Urine is concentrated in the collecting ducts, but the loop of Henle plays the major role in allow ing that to happen by establishing a concentration graclient between tl1e collecting duct and the medulla.
138.
B is correct. The small intestine is the major site for absorption of nutrients.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 7 Passage I 139.
A is correct. Since antiboclies are produced, this is an example of humoral in1mtmity. Humoral in1munity is directed against an exogenous antigen (one found outside the cell) such as fungi, bacteria, viruses, protozoans, and toxins. Cell mediated in1mtmity (T-cells) works against infected cells, cancerous cells, skin grafts, and tissue transplants.
140.
B is correct. A nude mouse lacks a thymus. T-cells require a thymus for maturation. An antibody is not a cell, so A is wrong. The MCAT may be misleading in this fashion, so read the question closely.
141.
B is correct. The experin1ent begins with the premise that nude mice do not produce antibodies or T cells. They do produce B-cells. Since antibodies are produced only after exposure to the donor T cells, either the T-cells or the B-cells are producing antibodies. Since the B-cells continue to produce antibodies after the T cells are removed, B-cells produce antibodies only after exposure to helper T-cells.
142.
A is correct. Interleukins are protein hormones, so they act through a second messenger system at the membrane surface.
143.
D is correct. Yo u should know that all blood cells arise from the same stem cells in the bone marrow.
144.
A is correct. A foreign particle capable of provoking an immune responseis the definition of an antigen. Choices B, C, and D are all molecules being produced by the body (recognized as self) and should not illicit an immune response.
145.
D is correct. Each B-cell is capable of reproducing only one type of antibody specific to one an tigen. When many memory cells are produced, the body will start with many more B-cells that make antibodies against the same antigen, allowing for a faster response in case of a second infection.
Passage (( 146.
D is correct. Quaternary structure consists of the joining of separate polypeptide chains. From the Figure 1, you should see that the quaternary structure is disrupted when disulfide bonds are broken. You should know that disulfide bonds are also involved in tertiary structure.
147.
B is correct. If antigens are released within a healthy individual, that individual will make memory B cens for a secondary ll1mune response, often times making the individual immune to infection.
148.
D is correct. The light chain is the smaller polypeptide. If we break apart the antibody shown in Figure 1 at its disulfide bonds, the passage says that we are left with two heavy and two light chains. Since the light chain contains no part of the F, region, E and F must mark the light chain. The passage says that the F, region is constant from antibody to antibody; therefore, it must contain at least, A and B. Since the light chain contains no part of the F, region, E cannot be in tl1e F, region.
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252
MCAT
BIOLOGY
149.
B is correct. The variable and hypervariable regions are responsible for antigen binding. Different antibodies have different variable and hypervariable regions that make each antibody specific for particular antigens.
150.
B is correct. Proteins for secretion are produced at the rough endoplasmic reticulum.
151.
C is correct. Plasma cells arise from B-cells, not T-cells. A is a true statement. Remember how the vaccine for chickenpox was found through cowpox. This does not contradict the passage. The pathogens or disease carriers may be different but carry similar antigens.
152.
A is correct. From the passage, the Fe region is constant and not part of the light chain. (See the answer to question 148.)
Passage III 153.
C is correct. We want to know why hemoglobin evolved to have less affinity for CO than the heme group has for CO (as explained in the passage). This would only happen if there was some advantage to not binding with CO, and there would only be an advantage if CO were present. So what explains the presence of CO in the cell? Since the passage tells us that the break down of heme produces CO, this is a logical source. Thus the production of CO by the breakdown of heme. favored the selection of a form of ox,ygen carrier that had less affinity for CO. For choice D, there is no simple mechanism that an MCAT test-taker should know that would result in CO from carbonic acid in the blood. Choice A is a far too recent event to account for such broad evolutionary change, and, of course, hemoglobin existed before the Industrial Revolution. For B there is no CO involved.
154.
C is correct. They differ in their amino acid sequence (primary structure) but have a similar three-dimensional shape (tertiary). Hemoglobin is made from four polypeptides while myoglobin is made from one (quaternary structure).
155.
C is correct. The passage states that M has a greater affinity for O 2, Thus, at a given pressure the M saturation will always be higher. It also states that BPG (in red blood cells) affects the characteristic sigmoidal oxygen dissociation curve for hemoglobin. The M curve should not be sigmoidal because it is not exposed to BPG.
156.
B is correct. Lactic acid lowers pH. The passage says that only hemoglobin responds to low pH.
157.
D is correct. The passage states that BPG is found in red blood cells. A and C are not red blood cells, so they must be wrong. As per the passage, BPG affects the characteristic sigmoidal oxygen dissociation curve for hemoglobin. Thus at low pressures of O 2 the BPG has the greatest effect. Low levels of O 2 would result in the capillaries of skeletal muscle not the lungs.
158.
A is correct. Since the passage states that the tertiary structure (or three dimensional shape of the polypeptide) does not vary significantly between species it is likely that this shape is fundamental to oxygen transport.
159.
A is correct. High pH means low H' concentration and slower breathing so that one doesn't lose too much CO2 (an acid).
Stand Alones 160.
B is correct. The lymphatic system is an open system, meaning something goes in one end and out the other. A closed system is like the blood, where fluid doesn't exit or enter the system.
161.
A is correct. Platelets do not contain a nucleus and they are not cells. Human erythrocytes contain no organelles.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 8 Passage I 162.
B is correct. Myosin hydrolyzes ATP to ADP.
163.
A is correct. From the passage you know that Ca" is bound to calsequestrin in the sarcoplasmic reticular lumen. This lowers the concentration inside the lumen, which weakens the gradient that the calcium pumps must work against.
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ANSWERS
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EXPlAN.~TIONS FOR THE 30-MINUTE IN-CLASS EXAM INATIONS •
253
164.
D is correct. From the passage, we know that the red color comes from myoglobin and cytochromes within the mitochondria. These supply ATP to muscle. Muscle with only small amounts of these, such as white muscle, is capable of only short periods of contraction.
165.
D is correct. The passage states that ATP levels remain constant. Ca 2+ concentrations change to create muscle contraction.
166.
A is correct. From the passage, the function of creatine kinase is to allow phosphocreatine to give or receive a phosphate group to or from ATP. This is what maintains ATP levels during muscle contraction. The passage states that cellular respiration is not fast enough to maintain ATP levels.
167.
C is correct. T-tubules are invaginations of the sarcolemma which deliver the action potential directly to the sarcoplasmic reticulum along the center of each sarcomere.
168.
B is correct. This question is a little bit trivial for the MCAT, but is still within the realm of possibility. You should be aware that some cell types in the human body, such as muscle cells and neurons, are so specialized that they have lost the ability to undergo mitosis.
169.
B is correct. The mechanism involves an integral protein of the sarcoplasmic reticulum not the sarcolemma. The passage says that uptake is active, so it requires ATP and can occur against the concentration gradient of Ca2+.
Passage II 170.
B is correct. This knowledge may be too trivial to be required on the MCAT. Only one or two questions will be on any given MCAT that require this much detail.
171.
D is correct. As osteoblasts release matrix materials around themselves, they become enveloped by LI,.e matrix and differentiate into osteocytes. This knowledge is required by the MCAT.
172.
A is correct. The passage mentions strength and flexibility when talking about structural proteins and collagen. This is a hint. The passage also says that collagen is a protein; another hint. You should have some idea that collagen is a structural protein.
173.
A is correct. Parathyroid increases blood calcium by breaking down bone via stimulation of osteoclasts.
174.
0 is correct. Bone is living tissue containing vascular connective tissue (blood), and nerves.
175.
C is correct. This question is reading comprehension. The passage says that osteoclasts are differentiated from phagocytotic blood cells. All blood cells differentiate from the same precursor.
Passage III 176.
D is correct. These are in the same ratio as the table. Since ATP concentration remains relatively constant within the cell, it makes sense that ATP production rate would mirror muscle power for these systems.
177.
C is correct. The body must reconstitute the phosphagen system, which requires ATP to make phosphocreatine. Oxygen is required to make ATP. This is the very heavy breathing following exercise. The heavier than normal breathing that follows for approximately 1 hour takes in oxygen to convert most of the lactic acid produced by the glycogen-lactic acid system back into glucose. (This is done principally in the liver.) Choice A is ridiculous because the time frame is to short to make new hemoglobin, and the increase from 2.5 to 11.5 is way too high for the change to be due to hemoglobin. B is not an explanation of why. D is wrong because the oxygen used during aerobic respiration is breathed not stored; the passage states that the aerobic contractions can continue indefinitely or as long as nutrients last. Nutrients are not replenished with oxygen, but with food.
178.
C is correct. The passage states that glycogen is replenished in 2 days for an individual on a high carbohydrate diet.
179.
B is correct. The liver makes glucose from lactic acid.
180.
A is correct. Slow twitch must rely upon aerobic respiration.
181.
C is correct. Only C is an event that lasts for a period of 10 seconds or less.
182.
A is correct. ATP levels remain nearly constant. Phosphocreatine is the first thing used to maintain those levels.
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254
MeAT
BIOLOGY
Stand Alones 183.
B is correct. Tendons connect muscle to bone. Muscle pulls on bone not other muscles.
184.
C is correct. Bone does not act as a thermostat for the body.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 9 Passage I 185.
B is correct. The best way to answer this is to imagine a simpler example. Imagine 50% died each year. (This is simply the half-life curve. You may already know that a half-life curve is a straight line on a semilog plot.) Now start with a population of 100, record the results for 5 years, and plot them. The amount of adults left alive each year would be: 50,25,12.5,6.25,3.125. The probability of an individual living to a certain age is simply the number of individuals living to that age divided by total born. This leaves us with 0.5, 0.25, 0.125, 0.0624, and 0.03125. Plotting this on the graph in Figure 1 gives: 10
0.1-
0.001
Age of organism
186.
A is correct. As the answer explanation explains, the last paragraph is key; according to the last paragraph describes that the population in a constant environment attains a stable age distribution (a bell curve) where the proportion of individuals at a particular age remains constant (achieves a steady state). This does not mean that the population is not growing or shrinking; only that the relative number of each age group remains constant. The only way for a particular age class to maintain itself with respect to other age groups is for the birth and death rate to be equal. Choice B does not describe this; instead it describes a situation where the birth rate has plummeted and the population ages due to no great influx of newborns. Remember that the last paragraph said that the proportion of individuals at a particular age remains constant; that means that if 60% of the population is 10 years old now, then 60% of the population should be 10 years old two years from now. Choice 0 violates this because choice 0 assigns the same proportion of individuals changing in a two year span of time. Choice C is wrong because if more than half the individuals live to age 40, the average life expectancy must be over 20. A is true only if individuals of all different ages increases proportionally over the same time, and thus could be true, as the question asks.
187.
C is correct. Imagine one female human. In 90 months she could not have more than 10 babies regardless of how many males were in the population. Now imagine one male human. In 90 months, the number of babies he could produce depends only upon the number of females.
188.
C is correct. Remember that an r strategist has many babies to ensure the survival of only a few. Thus, many die early on, and type III is the most likely curve.
189.
A is correct. The population growth rate doesn't change the survivorship curve. Humans in the pected to live long lives with an increasing expectancy of death later in life.
190.
D is correct. A short-lived, unpredictable habitat is best suited for an r strategist. 0 describes a K strategist.
u.s.
are ex-
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ANSWERS
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EXPLANATIONS FOR THE 30-MlNUTE IN-CLASS EXAMINATIONS •
255
Passage II 191.
A is correct. The passage states that male birds are the homozygous partner in the sex chromosomes meaning that they are XX. Thus:
XB XbXb
X"Y remale
male
parents
Xb Xb
Y
X"x b Xby X"x b Xby barred male
: unbalTO'd f~rnal~
harred male
unb",....,d female
192.
A is correct. Those that produce the H-Y antigen will not make the H-Y antibodies. These are the same individuals that have the H-Y gene.
193.
B is correct. The law of independent assortment does not work for genes on the same chromosome. It says that alleles will separate independently (in a random fashion) during meiosis. The law of segregation says that homologous chromosomes will segregate during meiosis.
194.
D is correct. D cannot happen because the father cannot pass an X to his son. The Y must come from the father.
195.
C is correct. If the mother's father was not colorblind, then he gave her one dominant X. If the girl is colorblind, then she inherited a recessive X from her mom. Thus the mother is heterozygous.
196.
C is correct. Since siblings are more likely to have recessive alleles at the same locus, recessive diseases are more likely.
Passage III 197.
B is correct. The Law of Independent Assortment states that alleles for different traits will sort independently of each other. This is only true if the traits exist on separate chromosomes.
198.
A is correct. Any heterozygotes would have produced both phenotypes when self fertilized. Bb x Bb. When they produced the phenotype opposite to their own, they were removed from the population. Eventually, only homozygotes remained.
199.
B is correct. This is the Mendelian ratio. BB x bb
200.
B is correct. bb x bb cannot produce anything but bb. A wrinkle phenotype might be Bb or BB.
201.
A is correct. The law of segregation states that homozygous alleles segregate independently and do not blend, but show complete dominance.
202.
D is correct. You can do a Punnell square to figure this out, but why would one genotype be more likely than another? By the way, the phenotypic ratio is 9, 3, 3, 1.
Stand Alones 203.
D is correct. We are looking for a habitat that does not change very often. One in which a species would have to make efficient use of the resources in order to survive. r strategists reproduce quickly to take advantage of rapidly changing, short-lived habitats. In longer lived habitats, K strategists gain the advantage. The first three answer choices are short-lived habitats.
204.
A is correct. Members of the same wild species can typically reproduce fertile offspring. There are many exceptions to this rule, but the MeAT tests for the rules, not the exceptions. A wolf and a dog are different species. They can breed and produce fertile offspring. There are road cabbage, fish, tobacco plants, and millions of other examples. The rule is "separate species don't normally reproduce fertile offspring in the wild due to a variety of factors. f!
205.
C is corn,d. There is a 50% probability for each girl to be colorblind. This means that there is a 25% probability that both will be colorblind and 25% that neither will be colorblind, leaving 50% that either one or the other might. The boys are irrelevant. We want all situations where at least one is colorblind. 50 + 25 = 75.
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256 . MeAT B,OLOGY
206.
C is correct. If the gene is actively selected against in one form, it must be actively selected for in another form or it will be eliminated from the population.
207.
B is correct. Density independent factors alter birth, death, or migration rates without having the effects influenced by population density. A storm destroying one half of an island affects all species in the same proportions regardless of density. This is density independent.
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I ,
ANSWERS & EXPLANATIONS FOR
QUESTIONS IN THE LECTURES
257
258
MeAT BIOLOGY
ANSWERS TO LECTURE QUESTONS Lecture 1 Lecture 2 Lecture 3 Lecture 4 Lectu re 5 Lecture 6 Lecture 7 Lectu re 8 Lectu re 9 1. B
25. C
49. A
73. 0
97. 0
121. C
145. 0
2. B
26. C
50. 0
74. B
98.
e
122. 0
146.
e
170.
3. A
27. 0
51. 0
75. A
99. B
123. 0
4. B
28.
e
52. A
76. A
100.
e
e
29. A
53. 0
77. B
6. 0
30. 0
54. A
e
31. B
8. A
32. 0
56.
9. B
33. B
5.
7.
10. C
34.
e
169. 0
193. B 194. A
147. A
171.
e e
124. A
148. A
172. 0
196. B
101. 0
125. A
149. 0
173. B
197. A
78. 0
102. B
126. 0
150. A
174.
e
198. B
55. 0
79. A
103. B
127. A
151. A
175. A
199. 0
e
80. A
104. B
128.
e
152. 0
176. 0
200.
57. 0
81. B
105.
129. B
153. B
177. B
201 . B
e
82. 0
106.
e e
130. 0
154.
58.
e
178.
e
e e
35. 0
59. B
83. A
107. 0
131. 0
155. A
179. A
e
60. B
84. 0
108. A
132. A
156. B
180.
13. B
37. 0
61. 0
85. A
109. D
133. B
157.
14. A
38. B
62. B
86. B
110. A
15. D
39. B
63.
e
40. D
64. B
17. D
41. B
18. B
11.
195.
202.
e
e e
203. 0
e
204.
e
181. A
205.
134. B
158. B
182. D
206. A
e
135. B
159. A
183.
e
207. D
88. B
112. D
136. A
160. D
184. A
208. D
65. D
89. B
113.
e
209. B
42. B
66. D
90. A
114. A
138. D
162. B
186. A
19. B
43. A
67. D
91. A
115. A
139. B
163. 0
187.
20. C
44. A
68. B
92.
e
140. D
164. D
188. A
e
45. D
69. A
93. D
117. D
e
165. D
189.
e
94. A
118.
142. B
166. A
190. A
214.
e
119.
e e
143. B
167. D
191. B
215.
120. D
144. A
168.
e
192. B
216. B
12.
16.
21.
36.
e
e
71. B
95.
48. B
72.B
96. A
46.
23. B
47.
e
87.
e e
22. 0
24.
e
70.
111.
116.
e
137.
141.
e
161
e
185.
e e
210.
e e
e
211. D 212.
e
213. B
e e
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ANSWERS
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EXPLANATIONS FOR QUESTIONS IN THE L ECTURES •
259
EXPLANATIONS TO QUESTIONS IN LECTURE 1 1.
B is correct. Answers A and C are anabolic reactions. Hydrolysis is used to break down triglycerides, proteins, carbohydrates, and is involved in nucleotide catabolism.
2.
B is correct. DNA is a nucleotide polymer. A nucleotide is a ribose sugar, a phosphate group, and a nitrogenous base. The nucleotides in DNA are held together by phosphodiester bonds.
3.
A is correct. Plants store carbohydrates as starch. Animals store carbohydrates as glycogen. Glucose is not a polymer. Cellulose is found in plant cell walls and is not digestible by humans.
4.
B is correct. This question simply requires that you recognize that protein is the only major nutrient containing nitrogen. <
5.
C is correct. The bending of the polypeptide chain is the tertiary structure of a protein.
6.
D is correct. Fats are a more efficient form of energy storage than carbohydrates and proteins. The phospholipid bilayer membrane is a fatty component of cell structure. Fats such as prostaglandins behave as hormones.
7.
C is correct. DNA is double stranded with A. C, G, and T, while RNA is single stranded with uracil (U) replacing T. The 'D' in DNA stands for deoxy- meaning that DNA is lacks a hydroxyl group possessed by RNA at its second pentose carbon atom.
8.
A is correct. Both alpha and beta linkages in polysaccharides are hydrolyzed by adding water. The question asks for a reactant; choices C and 0 are enzyme catalysts which are never reactants.
9.
B is correct. Enzymes function by binding the substrates on their surfaces and in the correct orientation to lower the energy of activation. The change in free energy for the reaction, L'l.G, is the difference in energy between reactant and products and is not changed by enzymes.
10.
C is correct. Decreasing the temperature always decreases the rate of any reaction. Lowering the concentration of a substrate will only lower the rate of an enzym'ltic reaction if the enzyme is not saturated. Adding a noncompetitive inhibitor will definitely lower the rate of a reaction because it lowers V m'" Changing the pH will increase or decrease the rate of an enzymatic reaction depending upon the optimal pH.
11.
C is correct. A high temperature would denature the enzyme.
12.
C is correct. Feedback inhibition works by inhibiting enzyme activity and to prevent the build up and waste of excess nutrients. Nonenzymatic feedback mechanisms also exist, like the action potential in the neuron.
13.
B is correct. Noncompetitive inhibition changes the configuration of the enzyme.
14.
A is correct. The implanted, fertilized embryo produces human chorionic gonadotropin, which stimulates the corpus luteum to continue producing progesterone in a positive feedback mechanism; feedback enhancement is made up.
15.
D is correct. Peptidases that function in the stomach work at a low pH. Once chyme enters the small intestine, it encounters an alkaline environment, meaning high pH or low hydrogen ion concentration.
16.
C is correct. A competitive inhibitor may be overcome by increasing the concentration of substrate.
17.
D is correct. Oxygen accepts the. electrons (along with protons) to form water.
18.
B is correct. The Krebs cycle occurs within the mitochondrial matrix in all eukaryotic cells.
19. B is corred. As electrons flow, the carriers pass along one or two electrons, and are reduced (gain electrons) then oxidized (l.ose electrons) until the last carrier donates electrons to oxygen. 20.
C is correct. The electrons from NADH drive protons .outward across the inner mitochondrial membrane.
21.
.c is correct. Glycolysis occurs in aerobic and anaerobic respiration.
22.
D is correct. ATP is a product. Two ATPs !'nter the reaction to "prime the pump", and four ATPs are produced. Gluc.ose is a reactant and pyruvate is a product. Oxygen plays no role in glycolysis.
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(c;)
2007 Examkrackers, Inc.
260
MCAT B,OLOGY
23.
B is correct. The process of fermentation includes glycolysis, which produces two ATPs.
24.
C is correct. This answer can most easily be found 'by process of elimination. Choice A is incorrect because ATP synthase is on the inner mitochondrial membrane. Choice Band D are p oor answers because they mention the specific processes, Glycolysis and Krebs cycle, of which you should be familiar. A change in these processes would indicate a completely different process. Choice C, on the other hand, refers to membrane transport in a more general way allowing for the possibility that a specific mechanism of transport may differ in heart and liver cells.
EXPLANATIONS TO QUESTIONS IN LECTURE 2 25.
C is correct. Since A always binds with T and G always binds w ith C, both the ratio of A f T and the ratio of Gf C equal one. .
26.
C is correct. The introns (intervening sequences) are removed during posttranscriptional modification.
27.
D is correct. This question requires no knowledge of PCR. It requires only that you know that a DNA polymerase replicates from 5' to 3', and that you know the complementary bases. (Complementary bases will be covered in this the next section of this lecture. Since DNA is replicated from 5' to 3', the primer must match be the complement of the 3' end of the DNA fragment. In other words, the DNA polymerase can only read from 3' to 5', so it must start at the 3' end of the DNA fragment. The complement of the 3' end of the DNA fragment is answer ch oice D. (By the way, the primer does not h ave to start exactly at the end of a DNA fragment in PCR, and a primer is longer than 3 nucleotides.)
28.
C is correct. DNA replication is semiconservative, which means that both strands are replicated, and each old strand is combined with a new strand.
29.
A is correct. Introns are removed from the primary trancript during posttranscriptional processing. The number of nucleotides in the mature mRNA would have to be less than the number of base pairs of the gene.
30.
D is correct. Note that the question asks about complementary strands, w hich are the two strands in a double strand of DNA.
31.
B is correct. You should know that mRNA leaves the nucleus in its finished form and that the process of RNA production is called transcription.
32.
D is correct. Electrophoresis uses an electrolytic cell w ith a positively charged anode and negatively charged cathode. The phosphate group of the DNA fragment gives it a negative charge tllat is attracted to the positively charged anode.
33.
B is correct. The start codon is AUG. mRNA is translated 5'->3'. (Note: You did not need to know that to answer this question correctly.) We are looking for AUG 5' .... 3'. Only A, Band C have an AUG sequence. However, if there are three codons in C, they must be: AAU, GCG, and GAe. The three in A must be: GAU, GCC, and GGA.
34.
C is correct. Translation does not take place within the nucleus.
35.
D is correct. There are 4' possible different codons. There are more codons than amino acids (used in proteins). This m eans that any amino acid could have several codons. The genetic code is evolutionarily very old, and almost universal. Only a few species use a slightly different genetic code.
36.
C is correct. The ribosome is made in the nucleolus from rRNA and protein. It does not have a membrane.
37.
D is correct. The complementary sequence to 5'-AUG-3' is 5'-CAU-3'. Onl y D contains this sequence in any order. Remember, thymine is only found in DNA, no t RNA so Band C must be wrong.
38.
B is correct. Only choice B is both true and concerns translation. Ribosomes contain two subunits for both eukaryotes and prokaryotes, so choice A is incorrect. Prokaryo tes do contain ribosomes, so choice C is wrong. Translation does not concern DNA, so choice 0 is incorrect.
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AN SWERS
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EXPLANATIONS FOR QUESTIONS IN THE LECTU RES .
261
39.
B is correct. The P stands for peptidyl site, where the growing peptide chain attaches to the tRNA.
40.
0 is correct. Signal pep tides attach to SRPs to direct the ribosome to attach to a membrane such as the endoplasmic reticulum. The signal peptide is usually removed during translation.
41.
B is correct. A primary spermatocyte has finished the 5 stage of interphase bu t not the first meiotic division. Thus, it has 46 chromosomes.
42.
B is correct. Replication takes place only during the synthesis phase.
43.
A is correct. In normal meiosis, the only change in the nucleotide sequence of the third chromosome will occur d uring crossing over. Crossing over occurs in p rophase I.
44.
A is correct. Only germ cells undergo meiosis.
45.
0 is correct. In metaphase I we see tetrads.
46.
C is correct. Centrioles migrate in p rophase of mitosis. Chromosomes align in metaphase; centromeres split in anaphase; cytokinesis usually occurs during telophase.
47.
C is correct. In prophase I, a tetrad will form .and genetic recombination will occur; a spindle apparatus will always fo rm; BUT chromosomal migration describes anaphase.
48.
B is correct. The life cycle of all oocytes is arrested at the primary oocyte stage until puberty.
EXPLANATIONS TO QUESTIONS IN LECTURE 3 49.
A is correct. A retrovirus contains RNA which is reverse transcribed to DNA and then incorporated into the host cell genome.
50.
D is correct. A virus cannot con tain both DNA and RNA. Many viruses contain proteins.
51.
D is correct. The first step in the infection of a host is attachment of the phage tail to a specific receptor on the host cell membrane. The capsid on the bacteriophage does not enter the host cell.
52.
A is correct. Animal viruses attach by recognizing a receptor protein and entering through endocytosis.
53.
0 is correct. Viruses are not living and do n ot carry out any type of respiration. They require no nutrients, using energy acquired by their host cell. They cannot reproduce inside nonliving organic matter. Viruses reproduce at the expense of a host. Thus, they most closely resemble parasites.
54.
A is correct. In the lysogenic cycle of viral infection is a cell that harbors inactive viral DNA in its genome.
55.
D is correct. A bacteriophage has a tail and fibers.
56.
C is correct. Viruses do not carry ribosomes.
57.
D is correct. Prokaryo tes have a cell wall that contains peptidoglycan, ribosomes, and a plasma membrane w itho ut cholesterol.
58.
C is correct. Because DN A is acquired directly from the medium, this is transformation. Transduction is the transfer of DNA via a virus. There is no sexual reproduction in bacteria. Conjugation occurs between two bacteria.
59.
B is correct. Transduction is the transfer of DNA via a virus.
60.
B is corred. You should arrive at this answer by process of elimination . You should know that gram negative bacteria do not retain gram stain, so 'A' is wrong. You should know that the membrane is made from phospholipids, so C is wrong. (Remember, archaebacteria are not on the MCAT.) D is wrong because fimbriae allow a bacterium to hold to solid objects. Finally, you should know that a bacterium with an outer membrane is gram negative and protected against certain antibiotics such as penicillin.
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262 . MeAT
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61.
D is correct. The exponential growth in bacteria is due to binary fission-asexual reproduction.
62.
B is correct. Bacilli are rod-shaped; 'spirilli are rigid helixes; spirochetes are not rigid; AND cocci are round.
63.
C is correct. Although this is a simple question, it is a reminder that transduction, transformation, and conjugation are not methods of reproduction in bacteria. They are methods of genetic recombination, which is associated with sexual reproduction in eukaryotics, but is not necessarily associated with reproduction in prokaryotes.
64.
B is correct. Bacterial plasma membranes are a phospholipid bilayer. Bacteria do not have a nucleus. Ribosomes do not contain peptidoglycans. Bacterial cell walls are made from peptidoglycan.
65.
D is correct. Fungi are unique because they are immotile and have a cell wall (like most plants), but are heterotrophic and not photosynthetic (like most animals).
66.
D is correct. Hyphae are haplOid and lengthen through mitosis.
67.
D is correct. Fungi are saprophytic. Saprophytes are organisms that break down the dead remains of living organisms.
68.
B is correct. Fungi are heterotrophs, not autotrophs,
69.
A is correct. HaplOid spores can form and spread faster and more efficiently than diplOid zygotes because they don't undergo meiosis.
70.
C is correct. Because fungus is more like human cells, drugs that attack fungi are more likely to affect human cells.
71.
B is correct. Similar to the plant kingdom the fungi kingdom is divided into divisions.
72.
B is correct. Fungi are exodigesters. They put enzymes into their food while it is outside their bodies and then absorb the nutrients. Although dead matter is more susceptible to fungal attack, fungi may attack living or dead matter. Meiosis is associated with sexual reproduction. Yeast is an example of a facultative anaerobe.
EXPLANATIONS TO QUESTIONS IN LECTURE 4 73.
D is correct. The flagella of bacteria are made from the protein flagellin.
74.
B is correct. The nucleolus is the site of rRNA transcription not translation. It is not membrane bound and should not be confused with the nucleoid of prokaryotes.
75.
A is correct. Anytime a compound moves against its electrochemical gradient across a membrane, it is active transport. The sodium electrochemical gradient was established by the expenditure of AT!', making this secondary active transport.
76.
A is correct. We are looking for the cell that is most active in detoxification, one of the jobs of smooth ER. That would be the liver.
77.
B is correct. I and II are true, but desmosomes are anchored to the cytoskeleton and are stronger than tight junctions.
78.
D is correct. Ribosomes are made of RNA and protein. They do not have a phospholipid bilayer.
79.
A is correct. The nucleus runs the cell and makes nucleic acids; the Golgi body packages materials for transpDrt. The rough endoplasmic reticulum makes proteins for use outside the cell. The smooth endoplasmic reticulum helps to detoxify alcohol in the liver.
80.
A is correcl. The hydrolytic enzymes of.lysosomes are activated by a low pH achieved by pumping protons into the interior.
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EXPlANATIONS FOR QUESTIO NS IN THE LEC TU RES .
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81.
B is correct. A signal is typically transmitted to the dendrites to the cell body and then down the axon; however, synapses are found all along the neuron and a signal may begin anywhere on the neuron. Although an action potential moves in all directions along an axon, the cell body and dendrites do not normally contain enough sodium channels to conduct the action potential for any length.
82.
D is correct. This question is testing your knowledge of an action potential. The major ions involved in the action potential are sodium and potassium. Blocking sodium channels is the only way given that would block an action potential.
83.
A is correct. The sodium/potassium pump moves potassium inside the membrane. Potassium is positively charged making the inside of the membrane more positive. The resting potential is measured with respect to the inside.
84.
D is correct. Acetylcholinesterase is an enzyme that degrades acetylcholine. You should gather this from the name. If this enzyme is inhibited, then acetylcholine will not be catabolized as quickly, and it will bind and release repeatedly with postsynaptic receptors.
8S.
A is correct. White matter is composed primarily of myelinated axons.
86.
B is correct. This is a knowledge based question. You should know this term.
87.
C is correct. A negative potential is created inside the cell so excess positive charge is pumped out of the cell.
88.
B is correct. Invertebrates do not have myelinated axons to accelerate nervous impulse transmission. Instead, they rely upon increased size. Invertebrate is a subphylum of Chordate which is characterized by a dorsal nerve chord at some point in their development.
89.
B is correct. The cerebellum controls finely coordinated muscular movements, such as those that occur during a dance routine. Involuntary breathing movements are controlled by the medulla oblongata. The knee-jerk reflex is governed by the spinal cord.
90.
A is correct. Every type of synapse in the peripheral nervous system uses acetylcholine as its neurotransmitter except the second (the neuroeffector) synapse in the sympathetic nervous system. You may not have known what a ne'uroeffector synapse was, but you should have been able to reason that it is an end·organ synapse. An effector is an organ or a muscle, something that responds to neural innervation by making something happen in the body.
91.
A is correct. Parasympathetic stimulation results in "rest and digest" responses, or responses that are not involved in immediate survival or stress. B is a sympathetic response, as is C. 0 is mediated by skeletal muscles, w hich do not receive autonomic innervation.
92.
C is the right answer. Pressure waves, or sound, are converted to neural Signals by hair cells in the organ of Corti in the cochlea.
93.
Dis correct. In order to prevent conflicting contractions by antagonistic muscle groups, reflexes will often cause one muscle group to contract while it sends an inhibitory signal to its antagonistic muscle group. Motor neurons exit ventrally from the spinal cord, not dorsally, so A is out. Reflex arcs (at least somatic ones) are usually confined to the spinal cord; they do not require fine control by the cerebral cortex. This eliminates B. Reflex arcs may be integrated by an interneuron in the spinal cord . C is out as well.
94.
A is correc!. The central nervous system is comprised of the brain and spinal cord. An effector is organ or tissue affected by a nervous impulse.
9S.
C is correct. This is a knowledge based question. The cerebrum is also called the cerebral cortex.
96.
A is correct. The question describes a simple reflex arc which does not involve neurons in the brain.
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264 . MCAT BIOLOGY
EXPLANATIONS TO QUESTIONS IN LECTURE 5 97.
D is correct. Aldosterone, as we can tell by its name, is a steroid (any hormone whose name ends in "sterone" or something similar is a steroid). This allows us to eliminate choice A, because steroid hormones do not need cell membrane receptors or second-messenger systems. They simply diffuse across the cell membrane. We can elioOOate B because the adrenal cortex is aldosterone's source, not its target tissue. C describes events at a synapse. Aldosterone actually exerts its effect by doing what D says, increasing the production of sodiumpotassium pump proteins.
98.
C is correct. This is a negative feedback question. If another source of aldosterone exists in the body besides the adrenal cortex, negative feedback (through the renin-angiotensin system and increased blood pressure) would suppress the level of aldosterone secreted by the adrenal cortex. A is out because the levels of renin in the blood would decrease, not increase; aldosterone release would increase blood pressure, and renin is released in response to low blood pressure. OxytOCin plays no role in blood pressure (vasopressin does) and would not be affected by this tumor. Now, to choose between C and D: we know that aldosterone from the adrenal cortex would respond to negative feedback, but we aren' t sure whether the tumor would. At this point, we know enough to go with C. If we're sure C is a correct response, that must mean D is an incorrect response, and we can eliminate it. In fact, this is a good choice because normally, hormone-secreting tumors will not respond to negative feedback.
99.
B is correct. All hormones bind to a protein receptor, whether at the cell membrane, in the cytoplasm, or in the nucleus of the cell. Steroids and thyroxine require a transport protein to dissolve in the aqueous solution of the blood. Steroids are derived from cholesterol, not protein precursors.
100.
C is correct. Acetylcholine acts through a second messenger system, and is not a second messenger itself.
101.
D is correct. Steroids are lipid soluble. Different steroids may have different target cells. For instance, estrogens are very selective while testosterone affects every, or nearly every, cell in the body. Steroids act at the transcription level in the nucleus, and are synthesized by the smooth endoplasmic reticulum.
102.
B is correct. Exocrine function refers to enzyme delivery through a duct.
103.
B is correct. Steroids act at the level of transcription by regulating the amount of mRNA transcribed.
104.
B is correct. You should know that T, and T, (thyroxine) production are controlled by a negative feedback mechanism involving TSH (thyroid stimulating hormone) from the anterior pituitary; parathyroid hormone production is not be affected by thyroxine levels.
105.
C is correct. Epinephrine release leads to "fight or flight" responses, as does sympathetic stimulation. A is out because insulin causes cells to take up glucose. It is not involved in '"fight or flight" responses. B is out because acetylcholine is a neurotransnutter; it has few, if any, known hormonal actions. 0 is out because aldosterone is · involved in sodium reabsorption by the kidney; it has no role in "fight or fli ght" responses.
106.
C is correct. The nervous and endocrine systems are, in general, the two systems that respond to changes in the environment. In general, the endocrine system's responses are slower to occur but last longer.
107.
D is correct. The only important thing to recognize from the question is that high insulin levels exist. Then go to the basics; insulin decreases blood glucose.
108.
A is correct. lhis is an important distinction to be made. The hormones of the posterior pituitary are synthesized in the bodies of neurons in the hypothalamus, and transported down the axons of these nerves to the posterior pitUitary.
109.
D is correct. Calcitonin builds bone mass. Menopause contributes to osteoporosis by reducing estrogen levels leading to diminished osteoblastic activity. You are not reqUired to know this, and may have had difficulty in eliminating this answer. Instead, you should answer this question by realizing that D was the exception.
110.
A is correct. Thyroxine (T,) is produced by the thyroid gland.
111.
C is correct. Glucagon increases blood sugar, a good thing if you are running a marathon. An increased heart rate and sympathetic blood shunting might similarly be expected in someone who had just run 25 miles.
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112.
D is correct. Parathyroid honnone stimulates osteoclast (bone resorbtion) activity. It also works in the kidney to slow calcium lost in urine. It controls blood calcium levels via these two mechanisms.
113.
C is correct. A looks good (testosterone does stimulate the testes to descend) until you notice that we're dealing with a physically mature male. The testes nonnally descend during late fetal development. B is out because increased testosterone would cause puberty to occur early, and would not change the timing of puberty if it's already happened (we are, after all, dealing with a physically mature male). D is out because we don't know of any direct mechanism by which testosterone increases body temperature.
114.
A is correct. Increased secretion of estrogen sets off the luteal surge, which involves increased secretion of LH and leads to ovulation.
115.
A is correct. The epididymus is where the sperm goes to mature and be stored until ejaculation. Testosterone is secreted by the seminiferous tubules.
116.
C is correct. Decreased progesterone secretion results from the degeneration of the corpus luteum, which occurs because fertilization of the egg and implantation didn't happen. A is out because thickening of the endometrial lining occurs while estrogen and progesterone levels are high, not while progesterone secretion is decreasing. B is out because increased estrogen secretion causes the luteal surge, and because the luteal surge occurs earlier in the cycle. D is out because, while the flow phase does follow decreased progesterone secretion, it does not occur as a result of increased estrogen secretion.
117.
D is correct. The layer of cilia along the inner lining of the Fallopian tubes serves to help the egg cell move towards the uterus, where it will implant if it has been fertilized. (Fertilization usually happens in the Fallopian tubes.) A describes what the ciliary lining in the respiratory tract does. B may sound good, but the Fallopian tubes are far enough away from the extema l environment that protection from its temperature fluctuations is not an issue. C would seem to gum up the whole "continuation of the species" plan. It's not a good answer.
118.
C is correct. The adrenal cortex makes many other steroid based hormones, as well as testosterone.
119.
C is correct. Mammalian eggs undergo holoblastic cleavage where division occurs throughout the whole egg. At first glance, this question appears to ask for somewhat obscure knowledge about meroblastic cleavage. However, you should be able to eliminate A, B, and D quite easily as being part of human embryonic cleavage, so it is unnecessary to know meroblastic or holoblastic cleavage.
120.
D is correct. Generally, the inner lining of the respiratory and digestive tracts, and associated organs, come from the endodenn. The skin, hair, nails, eyes and central nervous system come from ectoderm. Everything else comes from the mesodenn. The gastrula is not a germ layer.
EXPLANATIONS TO QUESTIONS IN LECTURE 6 121.
C is correct. PepSin, whose optimum pH is around 2.0, denatures in the environment of the small intestine, whose pH is between 6 and 7. A is wrong because pepsin isn't working at all in the small intestine; it won't be working synergistically with trypsin. B is wrong because pepsinogen is activated in the stomach by low pH. D is wrong because pepsin is a catalyst, which makes it a protein; amylase digests starch.
122.
D is correct. The best answer is D because only D is both true and reveals a benefit for enzymes to be inactive while in the pancreas. A may seem logical but would only apply to lipase. B is false. C is true but is not an adequate explanation.
123.
D is correct. The stomach doesn' t digest carbohydrates. If stomach acid secretion is obstructed then: A) fewer bacteria will be killed in the stomach; B) less pepsinogen will be activated; C) the pH will rise.
124.
A is correct. All macronutrients are digested through hydrolysis, or the breaking of bonds by adding water.
125.
A is correct. The large intestine absorbs water. Fat is digested in the small intestine. Urea is secreted by the kidney.
126.
D is correct. Amylases digest sugars; lipases digest fats; and proteases digest proteins.
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127.
A is correct. Pancreatic exocrine function includes enzymes made in the pancreas and secreted through a duct. Bile is not an enzyme. It is made in the liver and stored in the gallbladder.
128.
C is correct. Most chemical digestion occurs in the first part of the small intestine, the duodenum.
129.
B is correct. Parietal cells secrete HCl. Goblet cells secrete mucus. Chief cells secrete pepsin. G cells secrete gastrin into the blood.
130.
0 is correct. Gluconeogenesis is the production of glycogen from noncarbohydrate precursors. This function is performed mainly in the liver. Glycolysis can be performed by any cell. Fat storage takes place in adipocytes. Protein degradation occurs in all cells.
131.
0 is correct. Most fat digestates enter the lymph as chylomicrons via lacteals. Smooth endoplasmic reticulum synthesizes triglycerides. .
132.
A is correct. 'Essential' means that the body cannot synthesize them. Nonessential amino acids are synthesized by the liver. Amino acids are absorbed by facilitated and active transport. Urea is the end product of amino acid deamination in the liver.
133.
B is correct. Glucose is absorbed in a symport mechanism with sodium. Sodium is an electrolyte. The absorption of glucose increases the absorption of sodium. Glucose is not an electrolyte. Glucose does not stimulate the secretion of amylase.
134.
B is correct. Insulin decreases blood sugar levels in several ways. One of the ways is by inhibiting glycogenolysis.
135.
B is correct. This is a knowledge based question. You should know that blood is an aqueous solution.
136
A is correct. Macromolecules are broken down into their basic nutrients.
137.
C is correct. The only process available for the removal of wastes by the Bowman's capsule is diffusion, aided by the hydrostatic pressure of the blood.
138.
0 is correct. Glucose is normally completely reabsorbed from the filtrate and thus does not appear in the urine. When glucose does appear in the urine, the glucose transporters in the PCT (not in the loop of Henle, as stated in answer choice A) are unable to reabsorb all of the glucose from the filtrate. C is wrong because the proximal tubule does not secrete glucose.
139.
B is correct. The purpose of the brush border is to increase the surface area available to reabsorb solutes from the filtrate. The brush border is made from villi, not cilia, and so has little or no bearing on the direction or rate of fluid movement.
140.
D is correct. Renin secretion catalyzes the conversion of angiotensin I to angiotensin II, which increases the secretion of aldosterone. If renin is blocked, then aldosterone cannot cause increased synthesis of sodium absorbing proteins, and sodium absorption decreases. Without renin secretion, production of angiotensin II would decrease. Blood pressure would decrease, not increase; A is wrong. Platelets are irrelevant; B is wrong.
141.
C is correct. This is the correct order of structures.
142.
B is correct. Vasopressin is antidiuretic hormone increasing water retention. ADH levels will be raised in response to dehydration. Aldosterone is a mineral corticoid released by the adrenal cortex in response to low blood pressure. In a severely dehydrated person, blood volume would be low, likely resulting in diminished blood pressure. Aldosterone levels will rise in response to the low blood pressure.
143.
B is correct. High blood pressure would result in more fluid being forced into Bowman's capsule.
144.
A is correct. The loop of Henle concentrates the medulla via a net loss of solute to the medulla. This process is critical to the function of other parts of the nephron; a medulla with a high concentration of solute allows for the passive absorption of water from the filtrate in other areas of the nephron.
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267
EXPLANATIONS TO QUESTIONS IN LECTURE 7 145.
0 is correct. TIle atrioventricular node, which sits at the junction between the atria and the ventricles, pauses for a fraction of a second before passing an impulse to the ventricles.
146.
C is correct. Stroke volume must be the same for both ventricles. If it weren't, we'd have a never-ending backlog of blood in one or the other circulations, ending with the faster circulation running dry. To keep the whole system running smoothly, both halves of the circulation must pump the same quantity of blood with each stroke.
147.
A is correct. The cardiac action potential is spread from one cardiac muscle cell to the next via ion movement through gap junctions.
148.
A is correct. Less oxygenated blood will reach the systemic system because some oxygenated blood will be shunted from the aorta to the lower pressure p ulmonary arteries. The pulmonary circulation will carry blood that is more oxygenated than normal. since highly oxygenated blood from the aorta is mixing with deoxygenated blood on its way to the lungs. The entire heart will pump harder in order to compensate by with more blood to th e tissues.
149.
D is correct. Oxygenated blood returning from the lungs feeds into the heart at the left atrium. From there, it flows into the left ventricle, which pumps it to the systemic circulation. The right atria and the right ventricle pump deoxygenated blood to the lungs.
150.
A is correct. Blood loss is likely to be more rapid during arterial bleeding due the greater blood pressure in the arteries.
151.
A is correct. Don't let the not-so-subtle physics reference fool you. Bernoulli's equation, which would indicate a greater pressure at the greater cross-sectional area, doesn't work here. You should memorize that blood pressure in a human is greatest in the aorta and drops until the blood gets back to the heart.
152.
D is correct. The capillaries are one cell thick and blood moves slowly to allow for efficient oxygen exchange.
153.
B is correct. Hyperventilation results in loss of CO" leading to lower concentrations of carbonic acid in the blood, and an increase in pH. Hypoventilation would result in the reverse. Breathing into a bag would increase the CO2 content of the air and lead to acidosis. Excess aldosterone may lead to metabolic alkalosis due to hydrogen ion exchange in the kidney.
154.
C is correct. Carbonic anhydrase is a catalyst. Catalysts increase the rate of a reaction . If the catalyst is inhibited, the rate decreases. Since the reaction moves in one direction in the lungs, and the opposite direction in the tissues, answer choice A is ambiguous. Unless one believes that a carbonic anhydrase inhibitor will affect transcription or degradation of hemoglobin, choices B and D are equivalen t. If one were true, the other should also be true.
155.
A is correct. Cellular respiration produces carbon dioxide, which, in turn, lowers blood pH. During heavy exercise, capillaries dilate in order to deliver more oxygen to the active tissues. Nitrogen is irrelevant to respiration.
156.
B is correct. The increased hemoglobin concentration in the blood after reinjection increases the blood's ability to deliver oxygen to the tissues, often a limiting factor in endurance competitions. A is wrong because, while this may happen, it iH not the prinlary benefit of blood doping. B iH a much better answer. C is wrong because, since we are only reinjecting blood cells, we are not increasing the body's hydration status. D is wrong because, while blood is less viscous with fewer red blood cells, we are removing whole blood, not just red blood cells, so we won't be changing the blood 's viSCOSity.
157.
C is correct. Carbon dioxide is produced in the tissues. It is transported by the blood to the lungs, where it is expelled by diffusing into the alveoli. Since the concentration gradient carries CO2 into the capillaries from the tissues and from the blood into the alveoli, we can reason that there is a higher concentration of CO, in the tissues than in the alveoli. A is wrong because blood CO, concentration will not change in the veins; it has nowhere to go. B is wrong because CO2 is expelled into the lungs at the pulmonary capillaries, so CO2 that was present in the pulmonary arteries (before the capillaries) will largely be gone in the pulmonary veins (after the capillaries). D describes the opposite of the concentration gradient that actually exists for CO, between the systemic tissues and the systemic capillaries.
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268 . MeAT
BIOLOGY
158.
B is correct. The person would need increased vascularity to deliver more blood to the tissues because the blood would carry less oxygen.
159.
A is correct. Constricted air passages is the clue. The bronchioles are surrounded by smooth muscle and small enough to constrict. Cartilage does not constrict, muscle does. The skeletal muscle in the thorax does not constrict the air passages. The alveoli are not part of the air passages.
160.
D is correct. Heavy exercise manifests increased carbon dioxide production that leads to increased carboxyhemoglobin.
161.
C is correct. Hemolytic anemia can result from abnormalities of the red bloods cells tha t make them fragile and more susceptible to rupture when they are processed by the spleen. You should know for the test that the spleen destroys old, worn out red blood cells.
162.
B is correct. "Swollen glands" are actually swollen lymph nodes that are bulging with immune cells gearing up to fight the invasion. A is simply not what's going on here. C is out because an inflammatory response would draw fluid into the inflamed area, not drain it away. D is also just a goofy answer.
163.
D is correct. Immunoglobulins, or antibodies, are involved in the humoral immune system, or the B cell system. A is out because cytotoxic T cells work in cell mediated immunity. B is out because stomach acid plays a role in the nonspecific innate immunity; humoral immunity is specific and acquired. C is irrelevant.
164.
D is correct. Antibodies bind to antigens through interactions between the antibody's variable region and the an tigen. Antibodies do not phagocytize anything, so A is out. Antibodies are produced by plasma cells, they don't normally bind to them; B is out. Plasma cells are derived from stem cells in the bone marrow. Antibodies will not usually prevent their production. C is out.
165.
D is correct. Fluid that is picked up by the lymphatic tissues is returned to the circulation at the right and left lymphatic ducts, which feed into veins in the upper portion of the chest.
166.
A is correct. Type B negative blood carries B antigens and not the Rh factor. It does not carry A antigens. There are no 0 antigens. Thus type B negative blood makes antibodies that will only attach A antigens and Rh antigens.
167.
D is correct. Old erythrocytes are destroyed in the spleen and the liver.
168.
C is correct. The innate immune response does not involve humoral immunity (B-cell) or cell mediated immunity (T-cell). The innate immune system responds to any and every foreign invader with the white blood cells called granulocytes as well as with inflammation and other actions.
EXPLANATIONS TO QUESTIONS IN LECTURE 8 169.
D is correct. Neither actin (the thin filament) nor myosin (the thick filament) changes its length during a muscular contraction; instead the proportion of actin and myosin overlap increases.
170.
C is correct. Permanent sequestering of calcium in the sarcoplasmic reticulum would prevent calcium from binding to troponin, which is what causes the conformational change that moves tropomyosin away from the myosin binding sites on actin. Choice A would occur if calcium were present and ATP were not. Loss of ATP would prevent the myosin from releasing from actin. B is incorrect because a resorption of calciunl from bone would result in a decrease in bone density, not an increase. D is incorrect because loss of calcium would not cause depolymerization of actin filaments. If this actually occurred, it would pose a serio us problem every time calcium was re-sequestered into the SR after the completion of a contraction.
171.
C is correct. Shivering results from the increase of muscle tone. When muscle tone increases beyond a certain critical point, it creates the familiar indiscriminate muscle activity typical of shivering. While shivering may serve as a warning, it does more than that, and that is not its primary purpose; A is out. B is simply not the case, nor is D.
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172.
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EXPLANATiONS FOR QUESTiONS iN THE LECTURES
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269
D is correct. Muscles cause movement by contracting (eliminating B and C), bringing origin and insertion closer together, usually by moving the insertion. Neurons cause contractions in muscles, not tendons, and the
neural signals are not initiated by the muscle. This eliminates A. 173.
B is correct. Antagonistic muscles move bones in opposite directions relative to a joint. In order to produce movement, one must relax while the other contracts.
174.
C is correct. All muscles will need more energy and protein if they are being used rigorously, but in humans, mature skeletal muscle cells do not divide.
175.
A is correct. MCAT doesn't test anatomy. This question is asking if you know that tendons connect muscle to bone. Ligaments connect bone-ta-bone. Tendons are not cartilage.
176.
D is correct. Peristalsis is a function of smooth muscle only. Shivering is an example of temperature regulation by skeletal muscle. Skeletal muscle may assist in venous blood movement and lymph fluid movement.
177.
B is correct. Gap junctions allow for the spread of the action potential throughout the heart.
178.
C is correct. Peristalsis is a smooth-muscle activity, and smooth muscle is innervated by the autonomic nerv-
ous system. The skeletal muscles are innervated by the somatic nervous system, so interruption of the autonomic nervous system would not affect the knee-jerk reflex or the diaphragm. An action potential in cardiac muscle is conducted from cell to cell by gap junctions. 179.
A is correct. 111e heart requires long steady contractions in order to pump blood. We know we want adjacent heart cells to contract at the same time; B is out. We're not concerned about a neuron here; C is out. Sodium flows into the cell, not out. D is out as well.
180.
C is correct. Smooth muscle contains thick and thin filaments, so it requires calcium to contract.
181.
A is correct. This question asks you to recognize that smooth muscle and cardiac muscle are involuntary, and then to recognize muscle types of the different structures. Of the muscles listed, only the diaphragm is skeletal muscle.
182.
D is correct. All muscles contract in response to increased cytosolic calcium concentration.
183.
C is correct. You should know that the vagus nerve is a parasympathetic nerve. Heart rate is set by the SA node, and the SA node is innervated by the parasympathetic vagus nerve. The pace of the SA node is faster than the normal heart beats, but the parasympathetic vagus nerve tonically slows the contractions of the heart to its resting pace. Without tonic inhibition from the vagus, the heart would normally beat at 100 - 120 beats per minute.
184.
A is correct. Thc word dilation here is the give away. Smooth muscle must be relaxing in order to dilate the vessels.
185.
C is correct. Parathyroid hormone increases blood calcium by increasing osteocyte activity, and increasing osteoclast number.
186.
A is correct. Synovial fluid acts as a lubricant, decreasing friction between the ends of the bones as they move. Bone cells receive circulation to keep them adequately hydrated; they do not need synovial fluid for this purpose. B is out. Synovial fluid persists in adults, after bones have stopped growing, so we know C is wrong. Synovial fluid is found in synovial joints, which allow movement. Its purpose is not to prevent movement. D is out.
187.
C is correct. Of the tissues listed, only cartilage does not contain nerves.
188.
A is correct. Ligaments attach bone to bone. Tendons attach bone to muscle.
189.
C is correct. Bone does not regulate the body or blood temperature. It does store calcium and phosphate, support and protect the body, produce blood cells, and store fat.
190.
A is correct. Yellow bone marrow is usually found in the medullary cavity of long bones.
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270 . MeAT
BIOLOGY
191.
B is correct. Hydroxyapatite is made up of calcium and phosphate in a compound that includes hydroxyl groups as well. You should know that bone acts as a storage place for phosphate and calcium. The hydroxyl you can get from the name.
192.
B is correct. Spongy bone contains the blood stem cells important for blood cell synthesis. Red blood cell storage is the job of the liver and spleen. Fat storage is in long bones.
EXPLANATIONS TO QUESTIONS IN LECTURE 9 193.
B is correct. According to the Punnell square, each time they have a boy, there is a 50% chance that he will be color-blind. The chance that both boys are colorblind is the chance of this happening twice, or 0.5'. Colorblind male
.... OJ
a
XC
y
xcx
XCy
r---r-._=.~---l.
OJ
~
Xc
X'
X'X' /X'y
OJ
"/
...
- .. /
Colorblind male
___/~
One of two possibilities or 50%
194.
A is correct. D is a male. Since the mother had the disease, the mother must have been homozygous recessive. The father could not have been a carrier. Thus, female A received an Xh from her mother and an XH from her father. Since she is only a carrier, it is possible that she passed on only her good X and that all her children are healthy.
195.
C is correct. From the dihybrid cross Punnell square we have the phenotype ratio 9:3:3:1. Thus, 9 of 16 individuals display both dominant phenotypes.
196.
B is correct. A dihybrid is heterozygous for both traits. See the dihybrid Punnell square in this chapter and count the dihybrid offspring.
197.
A is correct. Both girls must carry the trait because they receive their father's recessive chromosome.
198.
B is correct. A dihybrid cross is when two individuals that are hybrids at two different genes are crossed. i.e. AaBb x AaBb
199.
D is correct. Sex-linked traits occur on the X or Y chromosomes. The X and Y chromosomes are not homologous.
200.
C is correct. The woman has two recessive Xs. Her mother gave her one X and her father gave her one X. Her father only had one X to start with, so it must have been recessive and he must have been colorblind. Her son will receive a recessive X from her and a Y from his father so he will be colorblind.
201.
B is correct. First you must recognize that 10% of the gene pool is represented by the recessive allele. Then you must realize that only the homozygous recessives display the recessive phenotype. Now use the binomial theorem to see that homozygous recessives are represented by 0.12 of the population.
202.
C is correct. Catastrophic events will not cause significant genetic drift to a large homogeneous (well-mixed) population. Emigration, selection, and mutation all affect the HW equilibrium.
203.
D is correct. Only D does not represent a type of isolation (geographic, seasonal, behavioral) which leads to speciation. The birds are migratory, and thus are not geographically isolated. This makes D the best answer choice.
204.
C is correct. r-strategy is more efficient when the prevailing conditions are governed by density independent factors such as harsh environment, short seasons, etc. Commercial predation methods have a similar effect. rstrategists are able to beller withstand massive predation from man or others, because they produce so many Copyright
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EXPLANi'ifiONS FOR QUESTIONS IN THE LECTURES •
271
more offspring than they need to continue the species. K-strategists fare better when the prevailing conditions are governed by density dependent factors such as limited resources. K-strategists can better exploit limited resources by specializing. 205.
C is correct. A single class encompasses several orders; two different families can belong to the same order.
206.
A is correct. Remember: Darn King Phillip Came Over For Good Soup. Domain, kingdom, phylum, class, order, family, genus, species. The epithet Canus lupus indicates that the wolf is genus Canus and species lupus. Family is a broader category than either genus or species, so any organism that is of the species lupus, must also be Canus and canidae.
207.
D is correct. This is an application of the binomial theorem under Hardy Weinberg Equilibrium. p' + 2pq + q' ~ 1 and p + q ~ 1. So, p' ~ 0.36. P ~ 0.6. q ~ 0.4. 2pq ~ 0.48.
208.
D is correct. Corn depends upon Humans for its survival as a species. Humans ensure its survival. In return, humans are provided with food. Both species benefit from the relationship.
209.
B is correct. Prokaryotes arose at least 3.6 billion years ago.
210.
C is correct. Vertebrates, a subphylum of Chordata, contain backbones.
211.
D is correct. Ants are not chordates. Tunicates, sponge-like creatures that attach to the sea floor, are actually chordates.
212.
C is correct. Life originated in an atmosphere with little or no oxygen.
213.
B is correct. The taxonomy of Homo sapiens is Domain: Eukarya; Kingdom: Animalia; Phylum: Chordata; Subphylum: Vertabrata; Class: Mammalia; Order: Primata; Family: Homididae; Genus: Homo; Species: Sapiens.
214.
C is correct. You need to memorize that the taxonomy of humans is animalia, chordata, mammalian, primata, homididae, Homo sapiens.
215.
C is correct. Urey and Miller demonstrated that organic molecules may be created from inorganic molecules under the primordial earth conditions. Urey-Miller did not prove the existence of life on earth, (you do), nor did they prove that humans have evolved from bacteria, photosynthetic or not.
216.
B is correct. If you answered A, you would have to explain what the first living organism ate because he had no one else to eat. Even when there were millions of living organisms, they could not survive off each other because one organism would have to eat many others, and there just wouldn't be enough to go around initially.
Copyright © 2007 Exarnkrackers, Inc.
r
INDEX .
273
INDEX 2,3-DPG 147 5' cap 31
A A site 27, 39-40, 44 a tail 31, 57, 179, 181, 191 accommodation 87 acetyl CoA 17, 21-22 Acetylcholine 15, 88, 91, 93, 100, 104, 110, 122, 144, 158, 204 acidosis 132, 147 ACTH 102, 105, 110, 202 actin 77, 158, 162, 226 action potential 78, 83-87, 89, 91, 96, 104, 140, 144, 158, 163-165,204-205,226-227 activators 14, 30-31 active site 11-12, 16, 158, 162 active transport 6, 21, 63-64, 76, 81, 88, 127, 134,217, 226 Adaptive radiation 176 Adenine 8, 25-27, 32, 37, 41,179,192-193 Adipocytes 2, 75, 203 adrenal cortex 102-105, 107, 118, 135, 203, 217 adrenaline 93, 107 adrenergic 93, 202, 209, 217 Aerobic respiration 17, 21-23, 82, 187, 222 agglutinate 153 agonist 157 Agranular leukocytes 151 albumin 6, 130, 151, 216 aldosterone 102-104, 107, 110, 133-135, 138, 209-210, 217 alimentary tract 18 allele 172-174, 177-178 Allosteric regulation 14 alveolar sacs 145 alveoli 145-146, 149, 223 amino acids 4-5, 17, 21-22, 25, 28, 38, 40, 42, 44, 65, 75, 105, 107-108, 123, 129, 132-133, 138, 151, 179, 184, 217, 228 amphipathic 2, 61, 130 amylase 120, 123-124, 126, 133, 215 anaerobic respiration 17-18 Anaphase 47, 49-50, 190 androgen 111, 113, 211-212 aneuploidy 42 angiotensin 135, 217-218 Animalia 175 Copyright
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2007 Exarnkrackors, Inc.
ANS 92-93 antagonist 100, 157 antibiotics 65, 67, 69, 71, 79, 134, 203 antibodies 6, 55, 65, 74, 132, 151, 153-156, 196, 212, 220221,233 antibody 153-154, 210, 220-221 anticodon 39, 44 antiport 134, 138 antisense 30 anus 119, 179 aorta 139, 144, 148 apoproteins 3, 130 aqueous humor 96 archaebacteria 58, 61, 64-65 arteries 100, 130, 139-142, 144, 146, 148-149, 164, 202 arterioles 139-142,144,217 Ascomycota 70 aster 47 astrocytes 89 ATP 6,8,11,15,17-18,21-23,27,53,63-64,67,81,138, 158,161,186-187,208,226-229 atria 140, 144 autonomic nervous system 92-93,140,164-165,202,217 Autotrophs 58, 71 axon 82-84, 86, 88-89, 91, 205 axon hillock 83-84, 86, 205 axoneme 76
B B-cell immunity 153 B12 122, 124 bacilli 59 bacteria 6-7, 32-35, 43, 53, 58-61, 64-65, 67-69, 81, 121, 123-124,132,152-154,156,175,179,196-197,200 bacteriophage 53, 57, 68, 198, 200 Barr body 173 basal lamina 80 base-pair 29, 41-42 Basidiomycota 70 basophils 151, 153 behavioral isolation 176 bicarbonate 122-123, 125, 147 bicarbonate ion 122-123, 147 Big Bang Theory 179 bile 124, 126, 130, 132, 134, 214-215, 218
274
MeAT BIOLOGY
binary fission 67, 69, 198-199 blastocyst 115, 210 Blood types 154 Bohr shift 147 Bone 6, 9, 79-80, 109-110, 116, 118, 140, 151-153, 156-157, 159,t161-163, 166-169, 179,220,227-228,230 Bowman's capsule 134, 138, 216, 218 brain 82, 89, 91-93, 95-96, 98, 100, 105, 108, 116, 179, 202, 217 bronchi 145 bronchioles 145, 149, 164 Brownian motion 62, 87, 91 brush border 122-124, 130, 138, 214 budding 70-71, 76 bulbourethral glands 112 bulk flow 75 bundle of His 140
c C-terminus 39 Calcium 12, 87, 91, 108-110, 158, 162-163, 165-167, 169, 226-227,230 Calmodulin 13, 102, 104 Calvin cycle 58 cAMP 30, 102-103, 105, 108, 208 canal of Schlemm 96 canaliculi 166, 169 Cancer 43, 153, 193 capillaries 133, 139-142, 144-147, 149-152, 165,223 capillary 105, 122, 130-131, 134, 141, 144-145, 149, 152, 214 capsid 53, 55-57, 68, 197, 199-200 capsule 65, 134, 138, 168, 216, 218 carbamino hemoglobin 147, 149 carbohydrates 1-2, 6, 10, 65, 76, 102, 108, 120, 124-129, . 132 carbonic anhydrase 147, 149 carboxypolypeptidase 123 carcinogens 43, 55 cardiac muscle 78, 87, 93, 140, 144, 157, 163-165, 202-204 carrier 23, 56, 63, 108, 130, 173-174, 210, 222, 235
carrier population 56 Cartilage 80, 98, 145, 162, 166, 168-169, 179,227 Cartilaginous joints 168 catalase 12 catalyst 11, 23 cDNA35 cell membrane 53, 57, 74-76, 78, 80, 91, 102, 105, 154, 199,203-204,208,210,226
cell membranes 3, 115, 141, 172 cell-mediated 153 cellulose 6-7, 10, 65, 70, 124, 127, 190 Centrioles 47, 70, 77, 190, 200 centromeres 47 centrosome 77
cerebellum 95, 100 cerebral cortex 95, 100 cerebrum 95, 100 chaperones 40, 184 chemoreceptors 96, 98, 148 chemotaxis 152 chemotrophs 58 chiasma 49 chief (peptic) cells 121 chitin 10, 70-71 chloride shift 147 cholecystokinin 125 cholesterol 2, 61, 69, 76, 102, 130, 132, 203, 214 cholinergic receptors 93 Chordata 175, 179, 181 chromatids 46-47, 49-51 chromatin 45-47, 59 chylomicrons 3, 130, 133, 214 chyme 121, 123-124, 126, 130 chymotrypsin 123, 126 cilia 6, 76-77, 81, 89, 115, 118, 138, 145 ciliary muscle 96 cisternal space 75 citric acid cycle 17, 21, 23, 162 Class 3, 80, 175, 178, 183, 189, 195, 201, 207, 213, 219-220, 225,231 clathrin 74 Cleavage 13, 115, 118, 191 clone 33-35 CNS 92-93, 95 co-dominant 154 coacervates 179 cocci 59 cochlea 98, 100 codominant 172 codon 38-40, 42, 44 coelom 179 coenzymes 8, 11-12, 188 cofactor 11-12, 217 Collagen 6, 79-80, 96, 166, 168, 227-228 collecting duct 107, 135, 138, 218 collecting tubule 107, 135 Copyright © 2007 Examkrackers, Inc.
INDEX .
conunensalism 176, 178 Compact bone 166-167, 169 Competitive inhibitors 12 complemen t 41, 55, 67, 74, 153 complement proteins 74 complementary DNA 35 complementary strand 27 complete dominance 172, 234 cones 96-97 conjugated proteins 6 conjugation 67, 69-70, 198 connective tissue 80, 151, 168-169, 227-228 consensus sequence 29 convergent evolution 176 cooperativity 14, 146 cornea 96 corpus albicans 113-114 corpus luteum 113-115, 118, 210 cortex 68, 95, 100, 102-105, 107, 118, 134-135, 203, 217 cortical reaction 115 cortisol 102, 107, 133, 202, 208, 210 Cowper's glands 112 cristae 79 crossing over 49, 190, 194 crypts of Lieberkuhn 123 Cyanide 12 Cytochromes 6, 9, 22, 226 Cytokinesis 46-47, 50-51, 76, 190 cytoplasmic streaming 70, 77 cytosine 8, 25-26, 32, 37, 41, 192-193 cytoskeleton 76, 78 cytosol 17, 23, 28, 30, 39-40, 44, 61, 64-65, 75-76, 82-83, 102, 191, 203, 208, 220-221, 226-227 cytotoxic T cells 55, 153
D degenerative 38 dehydration 2, 10, 39 denatured 5, 32 dendrites 82-84, 88, 91-92 dense bodies 164 density dependent factor 176 density independent factors 176 deoxyribonuclease 123 depolarization 85, 163, 204 Desmosomes 78, 144 determined 69, 76, 116, 184 deuterostomes 179 dextrinase 122 Copyright «.;> 2007 Exa mkrackers, inc.
275
diapedesis 152 diaphragm 145, 165, 202 diaphysis 166, 227 diastole 140 differentiate 113, 116, 151, 153-154, 166, 211, 227, 230 diffusion 6, 17, 21, 62-64, 76, 85, 87, 127, 129,133-134, 141 dihybrid cross 172, 174, 234 dihydroxycholecalciferoll09 diploid 46, 50, 70-71, 172,194,211 disjunction 47 distal tubule 134-135, 138, 217-218 Divergent evolution 176 DNA 8, 10, 25-38,41-42,44-46,49,53,55,57,59,67-69, 73-74,79, 104, 184,191-193 DNA ligase 28, 33, 37, 198 DNA polymerase 27-28, 30, 35, 37 DOHCC 109 domains 58, 175 dormant 55, 68, 70 dorsal, hollow nerve cord 179 double helix 8, 10, 26-28, 30, 32, 37, 192-193 dynein 76
E E site 40, 44 E. coli 30, 69, 124, 199 ectoderm 116, 118 effector 92,100,102-103,108,202 eicosanoids 3 electrical synapses 87, 140, 163 electromagnetic receptors 96 electron transport chain 6, 22-23, 79 elements 34, 42,169,179,197 Endocrine 82, 101, 103-104, 108-110, 113, 118, t129, t167, 202,208 endocytosis 57, 74, 76, 200 endocytotic 53-54, 75 endoderm 116, 118 endosymbiont theory 79 Enhancers 31 enterocytes 6, 18,46,122-124,127,129,133,214 enterokinase 123 envelope 41, 53, 55-57, 64, 70, 74-75, 190, 200 enzyme specificity 11 enzyme-substrate complex 11 enzymes 5-6,11-16,18,28,30,32,53,61,65, 70,75-76, 88,101-102,104,108,112,115,120,122-124,126,129, 152,156,172,184-185,192,203,208,214-215
276
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BIOLOGY
eosinophils 151, 153 ependymal cells 89, 145 epididymus 112, 118 epiglottis 145 epinephrine 93,102-103,107,110,141,202,208-209,212, 228 epiphysis 166 epithelial cells 6, 75, 80, 89, 122, 133, 138, 217 EPSP 88-89 equilibrium 6, 11, 65, 85, 177-178 ER 40-41, 75-76, 102, 203, 210 erythrocytes 132, 146, 149, 151, 154, 156, 186, 220, 222 esophagus 119-120, 145 estrogen 102, 106, 113, 115, 118, 210-212 eubacteria 58 euchromatin 45-46 Eukarya 175 Eukaryotes 25, 28-29, 31, 45, 47, 49, 53, 59, 67, 73, 79-80, 179, 190-191, 193, 197, 200 Evolution 175-177, 190, 223 Exocytosis 74-75,102,130,208,214 exons 31-32, 37, 191 exonuclease 28 exonucleases 31, 191 exsporium 68 external auditory canal 98
F F factor. See also fertility factor facilitated diffusion 6,17,21,63-64,76,127,129,133-134 FADH2 8, 21-22 Fallopian (uterine) tube 113 Family 175, 178, 208, 235 fasciculus 158 fats 2-3,10,127,130,133,214 fatty acids 2-3, 17, 21-22, 61, 82, 105, 107, 124, 127, 130, 132-133,151,203,214,223,228 feedback inhibition 14, 16 fenestrations 134, 138, 141 Fermentation 18, 23 fertility factor 67 fibrinogen 151 fibrocartilage 168 Fibrous joints 168 filaments 71, 76, 158, 162, 164 fimbriae 65, 113 fixing nitrogen 58 Flagella 6, 59, 67, 76-77, 81
flagellin 67, 77 fluid mosaic model 61 frameshift 41-42,193 fructose 17-18, 127, 187 FSH 102, 105-106, 111, 113, 118, 210-212 Fungi 47, 70-71, 153, 175
G G cells 121-122, 133 G-proteins 13,88,208 galactose 18, 127, 198 gametangium 70 gametes 41, 49-51, 172, 176-177 Gap junctions 78, 87, 140, 144, 163-165 gastric inhibitory peptide 125 gastrin 122 gastrula 116, 118 gastrulation 116 gene 25, 30-35, 37, 41-42, 46, 57, 67, 88, 155, 172-178, 191, 197-198,233,235 gene pool 175, 177-178, 235 genetic code 38-39, 41, 44, 79 genetic drift 177 Genetic recombination 41, 49-51, 67-68, 192, 198 genome 25, 28, 32, 36, 41-42, 55, 57, 67, 193, 197-198 genotype 172-174, 177,234 Genus 175, 235 geographical isolation 176 germ cells 41, 49, 51, 111 germinate 68, 196 glaucoma 96 Globular proteins 5, 11,45 glomerulus 134-135, 215, 217-218 glucagon 101-103, 108, 110, 133, 208-209 glucocorticoids 102, 105, 107 gluconeogenesis 107-108, 110, 129, 132-133 glucose 6, 10, 15, 17-18, 22-23, 30-31, 63, 65, 75, 81-82, 84, 103,105,107-108,110,124-125,127-129,132-134,138, 186-187,202-203,208,216,228-229 glycerol 2, 17, 21, 61, 107 glycocalyx 80 glycogen 6-7,10,17,75,82,107-108,127-128,132-133, 161,187,203,208-209,228-229 glycogenesis 127, 132 Glycolysis 17-18, 21-23, 126, 128, 133, 162, 186-187, 226228 Glycoproteins 6, 61, 185, 200 goblet cells 122, 133, 145
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2007 Exarnkrackers, Inc.
INDEX '
Golgi 41, 75-76, 81, 102, 130, 203, 210 gonads 102, 111 gradient 6, 22-23, 62-65, 67, 81, 85, 127, 129, 134, 185, 204, 217, 227 gram staining 65 granular 75, 135, 151 granular cells 135 granular leukocytes 151 granulosa cells 113, 115,211 gray matter 89 guanine 8, 25-26, 28, 32, 37, 41, 192-193 gustatory 98
H hair cells 98, 100 Haldane effect 147 haplOid 46, 49-50, 70-71, 194, 211 hapten 153 Hardy-Weinberg 177-178, 233-234 Haversian (central) canals 166 HCG 16, 111, 115, 210 heart 23, 88, 93, 95,100,110,118,139-140,142,144,163, 165,202,204,209,214,223 helicase 27-28, 37 helix 5, 8, 10, 26-28, 30, 32, 37, 69, 192-193,227 helix destabilizer proteins 28 helper T cell 153 helper T cells 153-154 hematocrit 151, 216 h eme 6, 9, 12,22, 146, 222-223 hemoglobin 6, 14, 41,124,146-147,149,151,161, 222223, 228-229 hepatic cells 6 hepatic portal vein 131 hepatic vein 131 heterochromatin 25, 45 Heterotrophs 58, 70, 179, 181 heterozygote 172 heterozygous 154, 172, 177, 233-234 H exokinase 15, 17, 186-187 hGH 102, 105 histamine 122, 152-153, 220 histones 45, 59, 79 holoenzyme 12 homeostasis 30, 134-135, 166, 203 homologues 46, 49-50 homozygous 154, 172-173, 178, 233-235 hopanoids 61
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277
hormones 2-3, 5, 10, 62, 74, 82, 101-108, 110-111, 122, 125, 133, 151, 164,208-212,228 host 53, 55, 57, 68-69,197-200,220 human chorionic gonadotropin 115, 210 humoral 153, 156 hyaline 168 hybrid 172 hydrogen bond 1, 185 hydrolysis 1-2, 10, l3, 21, 27, 119, 126, 187, 214 Hydrophilic 1, 3, l33, 203 hydrophobic 1-3, 5, 37, 133, 203 Hydrostatic pressure 65, 134-135, 140-141, 216 hydroxyapatite 9, 166, 169 h yperplasia 161 hyperpolarization 85, 96 hypertonic 64 hypertrophy 161, 163 hyphae 70-7l hypothalamus 95, 100, 105-106, 110, 162, 202, 210-211 hypotonic 65
immunoglobulins 151, 156, 221 implantation 115, 118 inclusion bodies 60 incus 98 inflammation 82,152-154 inhibin 111, 211 inhibitors 12, 14, 226 initiation 29-30, 37, 39-40, 165 initiation complex 29-30, 39 initiation factors 29, 39 innate immunity 152 insulin 5-6, 75, 82, 84, 101-103, 108, 110, 125, 133, 187, 208 intercalated disc 163 Intercostal muscles 145 intermediate 23, 82, 130, 164, 172 interphase 46, 49, 51, 194 interstitial fluid 82, 122, 130, 150, 154 introns 31-32, 35, 37, 191 inversions 42 ionophores 87 IPSP 88 isomerases 15
isotonic 64
278 . Me AT
B IOLOGY
J juxtaglomerular apparatus 135, 217
K K-selection 176, 178 karyotype 173 ketone bodies 132 ketosis 132 kidney 6,17,78,104,106-107,110,129,132,134-135, 215216, 218, 229 killer T cells 153 kinase 13, 15, 187, 203, 208-210,226-227 Kingdom 58, 70-71, 175, 190 Krebs cycle 8, 21-23, 79, 128 Kupfer cells 132
L lac operon 30-31 lactase 122, 215 lacteal 122, 214 Lactose 18, 30-31, 127 lagging strand 27-28 lamellae 60, 166 large intestine 119, 124, 126, 156, 218 larynx 108, 111, 145 latent 54-55, 103, 172, 200 Law of Independent Assortment 172, 233-234 Law of Segregation 172, 233-234 leading strand 28 left ventricle 139-140, 144 lens 96-97 Leukocytes 151, 153 leukotrien es 3, 149 LH 102, 105-106, 111, 113, 118, 210-212 ligament 157, 162 ligases 15 lipase 123-124, 126, 130, 214-215 lipid anchored proteins 61 lipopolysaccharides 65 lipoproteins 3,61,130,132, 214 lithotrophs 58 liver 6, 10, 17-18, 21, 23,46,75,81,107-108, 110, 116, 124, 127-133,151,153,203,208-209,218, 229 local media tors 82, 101,220 locus 172, 233 loop of H enle 134-135, 138,218 luteal surge 113, 118 lyase 15
lymph 80, 122, 130, 133, 150-154, 156, 166, 169 lymphatic system 150, 154, 223 Lymphocytes 150-151, 153-154, 156, 220-221,223 Iymphokines 82, 152 lysogenic 53, 55-57, 198-199 lysosomes 41, 75-76, 229 lysozyme 65, 123 lytic 53-54, 56, 198
M macrophages 74, 151-154, 220 malignant 43 malleus 98 maltase 122 Mammalia 175, 181 mast cells 153 matrix 6, 9, 21-23, 28-29, 44, 75, 79-80, 87, 151, 166, 185, 227-228 mechanoreceptors 96 medulla 95, 100, 102, 105, 107, 118, 134-135, 145, 148,202 megakaryocytes 151 Meiosis 42, 46, 49-51,67,70-71, 194, 198 melting 32 memory B cells 153, 220-221 memory T cells 153 Mendel's First Law of H eredity 172 Mendel's Second Law of H eredity 172 Mendelian ratio 171-172 menstrual cycle 106, 114, 118, 211 mesoderm 116, 118 mesosome 60 metaphase 47, 49-51, 190, 194 metaphysis 166 methylation 32 micelle 61, 130 microfilaments 47, 76-77 microglia 89 microtubule-organizing center 77 Microtubules 6,47, 76-77, 81, 145, 190 microvilli 76-77, 98, 100, 122 mineral corticoids 102, 107 Minerals 1, 9, 12, 166 missense 41 Mitochondria 6, 41, 45, 73, 75, 79, 102, 112, 158, 161, 163, 190,198,226,229 Mitosis 46-47, 49-51, 67, 70-71, 105, 115, 151, 161-162, 166,190,194,199,227 monera 58, 175
Co pyrig ht © 2007 Examkrackers, Inc.
INDEX '
monocistronic 30 Monocytes 89, 151-152, 166 monosaccharides 6, 17-18, 124 morula 115, 118 motor unit 160, 226 mouth 98, 119-120, 126, 179, 214, 218 mRNA 28, 30-32, 35, 37-40, 44, 55, 57, 191-192 mucous cells 121-122, 145 multiunit smooth muscle 164 murein 65 muscarinic 93 muscle 3, 6, 18, 46, 76-78, 80, 82, 87-88, 92-93, 96, 100, 116,118,120-122,127, 140-142,144-145,149-150,157165, 167, 169,202-205, 209, 216-217, 222-223, 226-230 mutation 41-43, 192-193, 208 mutualism 176, 178 mycelium 70 myelin 89 myofibril 158 myoglobin 6, 161, 222-223, 226-228 myosin 158, 162, 226
N N-terminus 39 NADH 8, 17-18, 21-23,187 negative feedback 14, 16, 103, 105, 118, 153, 187 nephron 134-135, 138,205,215, 217-218 neural plate 116 neural tube 116 neurolemmocytes 89 neuromuscular synapse 158 neuron 6,82,84-86,88-89,91-93, 100, 158, 160,164-165, 199,204-205,226-227 neurotransmitter 82, 87-88, 93, 202, 217 neurotransmitters 82, 88, 93, 101 neurula 116 neurulation 116 neutrophils 74,151-154 niche 176 nicotinic 6, 93 nitrocellulose 35 nociceptors 96 nodes of Ranvier 89 Noncompetitive inhibitors 12, 14 nondisjunction 50, 194 nonsense 40, 42 noradrenaline 93, 107 norepinephrine 93, 102-103, 107,208
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279
Northern blot 36 notochord 116, 179, 181 nuclear envelope 41,70,74-75,190 nuclear pores 28, 39, 73-74 nucleic acids 6, 8, 11, 55, 57, 126, 186, 200 nucleoid 59-60 nucleolus 28, 39, 47, 74, 81 nucleosidases 122 nucleosome 45 nucleotide derivatives 1 nucleus 28-29, 31-32, 37, 39, 41, 44-47, 58-60, 69-70, 7375,81,93,102-103, 108,115, 151,163-164, 190-191, 198-199,208, 210-211, 218, 223, 227
o oblongata 145 oils 2 Okazaki fragments 28, 37 olfactory 98 oligodendrocytes 89 oncogenes 43 oocyte 49-51, 113, 115, 211 oogonium 49 Oomycota 70 operator 30-31 operon 30-31 Order 6, 11, 15,26, 28, 32-33, 35, 44, 46, 50, 53, 55, 58, 63, 67,69,75-76,78,87,89, 93,98,102,126, 158, 175-176, 178,181, 199,208,212,229,232,235 organ of Corti 98 organotrophs 58 origin of replication 27 osmotic pressure 65, 127, 134, 141, 150-151 osteon 166, 227 oval window 98 oviduct 113 ovulation 49,113-115,118,211-212 ovum 50, 115, 118, 210-211 oxidative phosphorylation 22-23, 187 oxidoreductases 15 oxyhemoglobin 146
p P site 39-40, 44 pancreas 17, 81, 101, 104, 108, 116, 118, 123, 125, 131, 205, 214-215 pancreatic 101-102, 123-126, 133, 156,214 paracrine system 3, 82 parasitism 176, 178
280
MeAT
B IOLOGY
parasympa thetic 93, 95, 97, 100, 140, 144, 165, 202 parathyroid hormone 102, 104, 109-110, 169,228 parietal (oxyntic) cells 121 PCR35,37 pepsin 5,11,13,121-122,126 pepsinogen 13, 122 peptide 4-5, 13, 40-41, 44, 101-102, 105-106, 108-109,111, 115,122,125,185,208,211 peptide bonds 4, 13 peptidoglycan 12,65,69,71,197 perichondrium 168,227 perilymph 98 periplasmic space 65 peristalsis 120, 124, 162, 165 peristaltic action 120 Peroxisomes 12, 76 phagocytosis 65, 74-77, 153 phagocytotic cells 152, 168, 220 phagosome 74 pharyngeal slits 179 pharynx 145 phenotype 172-173, 178 phosphate 2, 8, 17-18, 25-27, 37, 40, 61, 75, 109, 134, 158, 167,169, 186-187,191,196,203,208,226,228 phosphodiester bonds 8, 10, 33, 37 Phospholipids 2-3,61,69,76, 130,203,214 phototrophs 58 Phylum 175, 179, 181, 190,235 pili 65,164 pinocytosis 74-75, 141 placenta 102, 111, 115, 210 Plantae 175, 190 Plasma 41, 60-61, 64-65, 69, 74, 103, 107, 109, 130, 132, 134,147,150-151,153-154,156,166,210,214-218,220221 plasma cells 153-154, 156,220-221 plasmid 33, 67, 198 Platelets 138, 151 pleated sheet 5 PNS92 polar body SO, 115 poly A tail 31 polycistronic 30 polymerase chain reaction 35 polypeptides 4, 32, 40-41, 106, 123, 129, 133, 221 polyploidy 42 Positive feedback 14, 16, 85, 113, 118 Post-transcriptional processing 31, 37, 191 postganglionic neurons 93, 202
primary follicle 113 primary oocyte 49, 113, 211 primary response 153 primary spermatocyte 49, 51, 194 primary structure 5, 184, 203, 222 primary transcript 31, 37 Primase 27 primer 27-29, 37 prions 55 probe 35-36 proenzyme 13 progesterone 16, 102, 106, 113, 115, 118, 210 prokaryotes 25, 27-31, 39, 44, 58-60, 69, 79, 181, 190-191, 193, 197-198,200 Prolactin 102, 105-106, 110 proline 5, 10, 185 prophage 55, 198,200 Prophase 47, 49-51, 81, 190, 194 prostate 112 prosthetic 6, 9, 11-12 proteins 1, 4-6, 10-11, 13-14, 16-17, 22, 25, 27-32, 39-41, 45,47,53,55,57,59, 61, 63-65, 74-76, 78-80, 82, 85, 8788,101-105,107-108,121,123,126-127,129,132-135, 147,150-151,153,155-156,158,166, 184-186,191,200, 203, 208-209,218, 221,226-227, 233 Proteoglycans 6, 80, 227 Protista 175 Proto-oncogenes 43 protoplast 64 protostomes 179 provirus 55 proximal tubule 6, 134-135, 138, 205, 217 PTH 102, 109, 212 Punnett square 172-173 pupil 97 Purkinje fibers 140 pus 153 pyrimidines 25-26, 192 pyrophosphate 27 pyruvate 17-18, 21, 23, 187
Q quaternary structure 5, 40, 184, 222
R r-selection 176, 178 receptor 41, 53, 57, 74, 87-88, 92, 96, 101-102, 104, 108, 199,208-210,218,220 receptor mediated endocytOSiS 74 Copyright © 2007 Examkracke rs , Inc.
IN DEX'
recognition sequence 32-33 recombinant DNA 33, 35 rectum 119 red bone marrow 166 reduction division 50 renal corpuscle 134-135 renal pelvis 134-l35 renal pyramids 135 repetitive sequence DNA 25 replication 27-30, 37, 41, 46, 49, 51, 67, 104, 190, 192-194 replication fork 27 replication units 27 replicons 27 replisome 27-28 repolarization 85 repressors 30-31 residue 4 respiration 1, 3, 5, 7, 9,11,13,15,17-19,21-23,79,82, 187,222, 226-227 resting p otential 85, 204-205 restriction endonucleases 32-33 Restriction enzymes 32 Restriction fragment length polymorphisms 36 restriction site 32-33 reticulocytes 151 retina 96-97, 100 retroviruses 55, 197 reverse transcriptase 35, 55, 57
RFLP36 Rh factors 155 riboflavin 124 ribonuclease 123 ribosomes 28, 39, 41, 44, 57, 59-60, 69, 74-76, 79, 191, 200, 221,229 right lymphatic duct 150, 214 RNA 8, 10, 25-32, 36, 38-41, 45-47, 53, 55, 57, 59, 73-74, 79, 81, 191, 197, 199-200 RNA polymerase 27-30, 55 RNA primer 27 rods 96 round window 98 rRNA 25, 28, 31, 39, 44, 74
S saprophytic 70-71 sarcolemma 158,226-227 sarcomere 158, 162, 226-227 sarcoplasmic reticulum 158, 162, 165, 226
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281
sa tellite cells 89 saturation kinetics 11 scala vestibuli 98 seasonal isolation 176 sebaceous 101 secondary follicle 113 secondary response 153-154 secondary structure 5, 184 Secretin 125 Secretory vesicles 75, 102 sedimentation coefficients 39 Semen 112 semicircular canals 98, 100 semiconservative 27 semidiscontinuous 28 seminal vesicles 111-112 seminiferous tubules 111, 211 septa 70, 140 serum 133, 151, 214 sex chromosome 173, 194 sex-linked disease 173 side chains 4-5 signal peptide 41, 44 signal-recognition particle 41 Single nucleotide polymorphisms 36 Single unit smooth muscle 164 sinoatrial node (SA node 140 sinusoids 131 skeletal muscle 92-93,118,140,142,145,149,157-158, 161-163,165,203-204,216, 222-223,226-227 skin 6, 78, 80, 98, 100, 116, 152, 165, 169, 192, 233 slime layer 65 small intestine 11, 16, 18, 100, 110, 119, 122-127, 138, 214 smooth muscle 3, 82, 87-88, 93, 120, 122, 141, 149-150, 157, 164-165, 202, 209, 217 SNPs 36 sodium 5-6, 62-63, 81, 85, 91, 96, 104, 125, 127, 129, 134135, 138, 144, 165, 185, 217-218 solenoids 45 somatic nervous system 92-93, 100, 202 somatostatin 108 Southern blotting 35-37 Speciation 176, 233 Species 4, 29, 35, 44, 59, 63, 67, 71, 175-176, 178, 190-191, 222-223,232,235 sperm 49-50, 81, 111-112, 115, 118 Spermatogonia 111 spermatogonium 49 spike proteins 55
282 . MeAT
B IO LOGY
spindle apparatus 47, 51, 76, 81 spirilla 59 spirochetes 59, 67 sporangiophores 70 spore coa t 68 spores 70, 196 SRP 41, 44 SSB tetramer proteins 28 stapes 98 stem cell 151, 228 stem cells 115, 156, 211, 220 Steroids 2-3, 61, 76, 102, 104-105, 107, 111, 151, 203, 210 stomach 11, 16, 80, 119-123, 125-126, 131, 133, 152, 156, 164, 205, 214, 218 Structural proteins 5-6, 80, 227 substrate level phosphorylation 17, 22 Substrates 11, 14 sucrase 122 sudoriferous 101 supercoils 45, 59 suppressor T cells 153 survival of the fittest 176 Svedberg uni ts 39 symbiosis 124, 176 sympathetic 93, 95, 97, 100, 107, 110, 141, 202, 217 synapse 83-84, 86-89, 91, 93, 100, 104, 158, 202, 205 synaptic cleft 87-88, 158 synaptonemal complex 49 synergistic 157 synergistic muscles 157 synovial fluid 168-169 Synovial joints 168 synthase 15, 22-23 synthetases 15 Systole 140
T T-cell immunity 153 T-lymphocytes 153 T-tubules 158, 165, 226-227 target cell 102, 208, 210, 218 taxonomical classification 175 telophase 47, 50-51, 190, 194 temperate virus 55 tendon 157, 162 termination 30, 38, 40, 44, 96 termination sequence 30, 44 tertiary structure 5, 184, 186, 222-223
testes 111, 118, 210-211 testosterone 102, 111, 118, 210-212 tetrads 49 thalamus 95, 100 thermoreceptors 96 thiamin 124 thick 65, 74, 141, 158, 162, 164 thick filament 158, 162 thin filament 158, 162 thoracic duct 130, 150,214 threshold stimulus 87, 91 thromboxanes 3 thymine 8,10, 25-26, 28,32,37, 41,191-193 thyroid 102-103, 105, 108-109, 156, 208, 212, 217 thyroxine 102, 104, 108, 110, 212 light junctions 78, 81, 134, 144 tissue 9-10, 75, 78-80, 89, 92, 101, 103, 106-108, 110-111, 113, 115-116, 130, 141,151-153, 156-157, 166,168-169, 179, 208, 220, 222, 227-229, 233 trachea 108, 145, 149 traits 46,171-172,174,234 transcription 28-31, 37, 41, 44, 51, 73, 88, 102-105, 108, 191, 197-198, 208, 218 transduction 67-69, 198 transferases 15 transformation 67-69, 198 transition 41 Translation 30,37,39-42,44,51,75, 79, 81, 104-105, 108 translocation 39-40, 42 translocations 42 transport maximum 134, 215 transposable elements 42 transposons 42, 197 transversion 41 Triacylglycerols 2-3, 203 triglycerides 2, 21, 75, 124, 127, 130, 133, 214, 228 triiodothyronine 102, 108 tRNA 25, 28,31,39-40,44 tropomyosin 158, 226 troponin 158, 226 trypsin 11, 123, 126, 214 TSH 102, 104-105, 108,212 tubulin 6, 47, 76 tumor 10,43, 104, 208,220 tympanic membrane 98, 100 tyrosine 101-103, 105, 107-108, 208
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INDEX ·
u unique-sequence DNA 25 uracil 8, 26, 28, 191-193 urea 126, 129, 132-134, 138, 151, 179 urethra 112, 134, 138 Urey-MiJler experiment 179, 181
v vaccine 55, 212, 221 vagus nerve 140, 144, 165, 202 vas deferens 112 vasa recta 134 vector 33-34, 68 veins 130, 139-140, 142, 149-150, 164 vena cava 131, 139, 149 venules 139-142, 149 Vertebrata 175, 179, 181 v illi 122-123, 130 virion 53, 57, 198 Viroids 55, 196-197 virus 33, 53-57, 68, 198-200, 220 visceral 87-88, 164, 202 vitamin D 2, 109 vitamin K 124 vocal cords 145 Volkmann's canals 166, 169 voltage gated sodium channels 5, 85
w white matter 89, 91 wild type 43
y yellow bone marrow 166, 228
z zona pellucida 113, 115,211 Zygomycota 70 zygote 70, 81, 115, 118 zymogen 13, 75, 122, 126, 214
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283
AN U NEDITED STUDENT REVIEW .
An Unedited Student Review of This Book The following review of this book was written by Teri R-. from New York. Teri scored a 43 out of 45 possible points on the MCAT. She is currently attending UCSF medical school, one of the most selective medical schools in the country.
I
"The Examkrackers MCAT books are the best MCAT prep materials I've seen-and I looked at many before deciding. The worst part about studying for the MCAT is figuring out what you need to cover and getting the material organized. These books do all that for you so that you can spend your time learning. The books are well and carefully written, with great diagrams and really useful mnemonic tricks, so you don 't waste time trying to figure out what the book is saying. They are concise enough that you can get through all of the subjects without cramming unnecessary details, and they really give you a strategy for the exam. The study questions in each section cover all the important concepts, and let you check your learning after each section. Alternating between reading and answering questions in MCAT format really helps make the material stick, and means there are no surprises on the day of the exam-the exam format seems really familiar and this helps enormously with the anxiety. Basically, these books make it clear what you need to do to be completely prepared for the MCAT and deliver it to you in a straightforward and easy-to-follow form. The mass of material you could study is overwhelming, so I decided to trust these books - I used nothing but the Examkrackers books in all subjects and got a, 13-15 on Verbal, a 14 on Physical Sciences, and a 14 on Biological Sciences. Thanks to Jonathan Orsay and Examkrackers, I was admitted to all of my top-choice schools (Columbia, Cornell, Stanford, and UCSF). I will always be grateful. I could not recommend the Examkrackers books more strongly. Please con tact me if you have any questions. " Sincerely, Teri R-
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MeAT BIOLOGY .
About the Author Jonathan Orsay is uniquely qualified to write an MCAT preparation book. He graduated on the Dean's list with a B.A. in History from Columbia University. While considering medical school, he sat for the real MCAT three times from 1989 to 1996. He·scored in the 90 percentiles on all sections before becoming an MCAT instructor. He has lectured in MCAT test preparation for thousands of hours and across the country for every MCAT administration since August 1994. He has taught premeds from such prestigious Universities as Harvard and Columbia. He was the editor of one of the best selling MCAT prep books in 1996 and again in 1997. Orsay is currently the Director of MCAT for Examkrackers. He has written and published the following books and audio products in MCAT preparation: "Examkrackers MCAT Physics"; "Examkrackers MCAT Chemistry"; "Examkrackers MCAT Organic Chemistry"; "Examkrackers MCAT Biology"; "Examkrackers MCAT Verbal Reasoning & Math"; "Examkrackers 1001 questions in MCAT Physics", "Examkrackers MCAT Audio Osmosis with Jordan and Jon".
About the Editor Dr. Jerry Johnson earned his B.s. in biology in 1999 and his Ph.D. in Biochemistry in 2003 from the University of Houston. As a student he conducted biochemistry workshops for the University and conducted research on mechanisms of mitochondrial reactive oxygen species production and oxidative phosphorylation. As a post-doctoral and NEI fellow at the University of Houston, College of Optometry Dr. Johnson continued his biochemistry workshops, conducted research on retinal toxicology and development and accepted an adjunct professorship with the University of Houston-Downtown campus to teach Cell Biology. Beginning in the fall of 2005, Dr. Johnson joined the faculty of the University of Houston-Downtown as a full-time Assistant Professor of Biology and Biochemistry. In add ition to his teaching of biochemistry and cell biology, Dr. Johnson continues to conduct research focusing on mitochondrial physiology and enzymology, reactive oxygen species production and mechanism s of retinal development, toxicology and aging.
Copyright
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Examkrackers, Inc.
EXAMKRACKERS
MeAT
INORGANIC CHEMISTRY 7TH EDITION
Copyright ©2007 by Examkrackers, Inc., All rights reserved under International and Pan-American Copyright Conventions. Published in the United States of America by Osote Publishing, N ew Jersey. ISBN 978-1-893858-47-3 (Volume 4) ISBN 978-1-893858-49-7 (5 Volume Set) 7th Edition To purchase additional copies of this book or the rest of the 5 volume set, call 1-888-572-2536 or fax orders to 1-859-255-0066. examkrackers.com osote.com audioosmosis.com
Inside layout design: Saucy Enterprizes (www.saucyenterprizes.com) Cover design: Scott Wolfe, Visible Theory (626) 795-1885 Inside cover design consultant: Fenwick Design Inc. (212) 246-9722; (201) 944-4337 Illustrations by ExamKrackers staff.
Copyright © 2007 Examkrackers, Inc.
Acknowledgements Although I am the author, the hard work and expertise of many individuals contributed to this book. The idea of writing in two voices, a science voice and an MCAT voice, was the creative brainchild of my imaginative friend Jordan Zaretsky. I would like to thank Scott Calvin for lending his exceptional science talent and pedagogic skills to this project. I would like to thank Paul Foglino for his contribution to the lecture questions. I also must thank five years worth of ExamKrackers students for doggedly questioning every explanation, every sentence, every diagram, and every punctuation mark in the book, and for providing the creative inspiration that helped me find new ways to approach and teach inorganic chemistry. Finally, I wish to thank my wife, Silvia, for her support during the difficult times in the past and those that lie ahead.
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READ THIS SECTION FIRST! This manual contains all the inorganic ch emistry tested on the MeAT and m ore. It contains more chemistry than is tested on the MeAT because a deeper understanding of basic scientific principles is often gained tluough more advanced study. In addition, the MeAT often presents passages with imposing topics that may intimidate the test-taker. Although the questions don't require knowledge of these topics, some familiarity will increase the confidence of the test-taker. In order to answer questions quickly and efficiently, it is vital that the test-taker understand what is, and is not, tested directly by the MeAT. To assist the test-taker in gaining this knowledge, this manual will u se the following convention s. Any term or concept which is tested directly by the MeAT will be written in bold and brown. To ensure a perfect score on the MeAT. you should thoroughly understand all terms and concepts that are in bold and brown in this manual. Sometimes it is not necessary to memorize the n ame of a concept. but it is necessary to tmderstand the concept itself. These concepts will also be in bold and brow n. It is important to note that the converse of the above is not true: just because a topic is not in bold and brown, does not mean that it is not important. Any formula that must be memorized will be written in
large, red, bold type.
If a topic is discussed purely as background knowledge, it will be written in italics. If a topic is written in italics, it is not likely to be required know ledge for the MeAT but may be discu ssed in an MeAT passage. Do not ignore items in italics, but recognize them as less important than other items. Answers to questions that directly test knowledge of italicized topics are likely to be found in an MeAT p assage.
Text written in orange is me, Salty the Kracker. I will remind you what is and is not an absolute must for MCAT. 1 will help you develop your MCAT intuition. In addition, 1 will oHer mnemonics, simple me thods of vjewing a complex concept, and occasionally some comic
relief. Don't ignore me, even if you think 1 am not ftumy, because my comedy is designed to help you understand and remember. If you think I am funny, tell the boss. 1 could use a raise. Each chap ter in this manual should be read three times: twice before the class lecture, and once immediately following the lecture. During the first reading, you should not write in the book. Instead, read purely for enjoyment. During the second reading, you should both highlight and take notes in the margins. The third reading sh ould be slow and thorough. The 24 question s in each lecture should be worked during the second reading before coming to class. The inclass exams in the back of the book are to be done in class after the lecture. Do not look at them before class. Warning: Just attending the class will not raise your score. You m ust do the work. Not attending class will obstruct dramatic score increases. If you have Audio Osmosis, then listen to the appropriate lecture before and after you read a lecture. If you are studying independently, read the lecture twice before doing the in-class exam and then once after doing the in-class exam. If you have Audio Osmosis, listen to Audio Osmosis before taking the in-class exam and then as many times as necessary after taking the exam.
A scaled score conversion chart is provided on the answer page. This is not meant to be an accurate representati on of your MeAT score. Do not become demoralized by a poor performance on these exams; they are not accurate reflections of your performance on the real MeAT. The tllirty minute exams have been designed to educate. They are similar to an MeAT but with most of the easy questions removed. We believe that you can answer most of the easy questions without too much help from us, so the best way to raise your score is to Copyright © 2007 Examkrackers, Inc.
focus on the more difficult questions. This me thod is one of the reasons for the rapid an d celebra ted success of the Exarnkrackers prep course and products. If you find yourself struggling with the scien ce or just needing more practi ce materials, use the Exam krackers 1001 Qu estions series. These books are designed specifically to teach the science. If you are already scoring lOs or better, these books are not for you. Yo u sh ould take advantage of the bulletin board at www.examkrackers.com. The bulletin board allows you to discuss any ques tion in the book with an MCAT exp ert at Examkrackers. All discussion s are kept on file so you have a bank of discussions to which you can refer to any question in this book. Although we are very careful to be accurate, erra ta is an occupational hazard of any science book, especially those that are updated regularly as is this one. We maintain that our .,ooks have fewer erra ta than any other prep book. Most of the time what students are certain are errata is the ,tudent's error and not an error in the book. So that you can be certain, any errata in this book w ill be listed a, it is discovered at www.examkrackerS.com on the bulletin board. Check this si te ini tially and p eriodically. If 'lOU discover what you believe to be errata, please p ost it on this board and we will verify it promptly. We und 'rstand that this system calls attention to the very few erra ta that may b e in our books, but we feel that this is 'he best system to en sure that you have accurate information for your exam. Again, we stress that we have j ,w er erra ta than any other prep book on th e market. The difference is that we provide a public list of our err, ta for your benefit. Study diligently; trust this book to guide you; and you w ill reach your MCAT (-oals.
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TABLE OF CONTENTS
LECTURE
1: 1.1
ATOMS, MOLECULES, AND QUANTUM MECHANICS .......................................... 1 Atoms ..... .......... ....... ......... ....... ....... ..................... ...... ............. ................. .. ..................... ........ 1
1.2 Elements .......................................................................................................... ........................ 2 1.3 The Periodic Table ................................................... ................................................................ 3 1.4 Characteristics Within Groups ..................................... ............................................................. 4 1.5 Ions ...................................................................................................................................... 6 1.6 SI Units and Prefixes ............................................... ........ ...................................... ..... .. ......... 10 1.7
Molecules ....................................................... ......... .............................................................. 10
1.8 Nam ing Inorganic Compounds ...........
.. ......................................................................... 12
1.9 Chemical Reactions and Equations ....................................................................................... 12 1.10 Chemical yield ................................................. ... .................................................................... 13 1.11
Fundam enta l Reaction Types ............................................................................................. 13
1.1 2 Reacti on Symbols ............................................ ......... ...... ......... ........ ........... .............. ............. 14 1.13 Bonding in Solids ........... .. ........... ... .. ....... .. ........ ......... .............. .. ........................ ................... 14 1.14 Quantum Mechanics .................................................................................. .............. ......... .. 16 1.1 5 Quantum Numbers .................................................. ......... .. .......................... .... .. .................. 16 1.16 The Heisenberg Uncertainty Principle .................................................................. .... ... .. ......... 17 1.17 Energy Level of Electrons ..................................................................................................... 17 LECTURE
2:
GASES, KINETICS, AND CHEMICAL EQUILIBRIUM .......................................... .. 23
2.1
Gases ........................................................................................ ...... .................................... 23
2.2
Kinetic Molecu lar Theory ........................... ......... ..... ................. .. ............... .. ....................... 24
2.3
Real Gases ........................................ .................................................................................... 27
2.4 Chemical Kinetics ........ ......... ...... ..... .................. .. ...... ............ ....... .... ........ ............... ......... .... 29
2.5 The Coll ision Theory ......................................... ................................................. .................. 29 2.6 Equations for Reaction Rates ........................................................................ ..................... 30 2.7
Determining the Rate Law by Experiment ...... .. ..................................... ..... ......... ........ ......... 31
2.8
Recognizing Reaction Orders ................................................................................................ 32
2.9 Rates of Reversible Rea ctions . .
.. ........... .......... .... ........... ........ ...... ............................... 33
2.10 Catalysis .............................................................................................................................. 34 2.11
Effects of Solvent on Rate ........ ... ...................... .. ............................... ................................. 36
2.12 Equilib rium .......................................................................................................................... 38 2.13 The Partial Pressure Eq uilibrium Constant .......................... ...............
.. ..................... 39
2.14 The Reaction Quotient .................................... ..... .................... ...... .. ........................... .. ....... 39 2.15 Le Chatelier's Principle .......... ... ...... .. ...... ... ........................... ................... .......... ......... .......... 40
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LECTURE
3:
THERMODYNAMICS .............•....•....•. ....•..............•........... ...•............. ......... .. 45
3.1
Thermodynamics
..................... ......................................... .45
3.2 State Functions. ...... .. ...... ...... ........... ...... .... ....... ....... . ....... ........ ......... ........ ..................... .... .45 3.3
Heat ............ .......... ... .... ... ... .... ... .. ...... ......... ................. ............... ............ .... .......... .......... .. ..... .46
3.4 Work ...................... .
........... ............ ........ ............ ...... . ......... ... ........... .. ........................... .48
3.5 The First Law of Thermodynamics ...................................... ..... ........ ............ ..... .... ...... ........ .49 3.6 Heat Eng ines .................................................. .... ................ ... ................... .. ... ... .................... 49 3.7 Thermodynam ic Functions .................................... .... ......... ......... ... .. ........ ......... ........ ........... 53 3.8 Internal Energy ................................ .. ...................................................................................53 3.9 Temperature ........ .................. .............. .......................... ........................................... ............ 54 3.10 Pressure .......... .. ............ ....... .. .......... ......... ......... .. ....... .. ... .. ............. ............ ..... .. ............ ....... 55 3.11
Enthalpy .. .. ....... .. .................................................. ...................... .... ............. .. .............. .. ........ 55
3.12 Entropy ................................... ................ ................... ........... ........ .. ......... ................. ....... .59 3. 13 Gibbs Free Energy ........ ....... ............... .... .... ..................... ........ .... ..................... ................... 61 LECTURE
4:
SOLUTIONS ........................ ... ................................... ........ . .......... . ............. 65
4. 1 Solutions ............... .. ............ .......................................................... ............ .. ............... .. ..... ... 65 4.2 Colloids ......... .... ........... ........ ... .. ............... .... .... .. .................................... .......... ..... ....... .. ....... 66 4.3 More So lutions ................................................ ........ ............... ........................ .. ................. 66 4.4 Units of Concentration ..................................... ..................................... ..... .. ......................... 6 7 4.5 Solution Formation ................. ... ..... .......... .. ........... .. ................ ...... ... .. ........ .. ..... ................ .... 70 4.6 Vapo r Pressure ..... ............... ..... .. ...... .............................................................. ....................... 71 4.7 So lubility .................................................... .. .. ....................... .. ............................................... 75 4.8 Solubility Guidelines ........................................ .... .... ... ................................ .. ..... .. ................. 76 4.9 Solubi lity Factors ........... .................. .. ...... ................... ......... .. ...... .. ..... ........................ .. .. ........ 77 LECTURE
5:
HEAT CAPACITY, PHASE CHANGE, AND COLLIGATIVE PROPERTIES .............. ...... 79
5. 1 Phases ......... .... .............. ....... ..................................................... ...... ................. .. ................. 79 5.2 Heat Capacity ................ .. ....... ........... ...... ......... ..... ... ........ ........ .................. ....... .. .................. 80 5.3 Calorimeters .............................................................................................. .......................... 81 5.4 Phase Changes ........... .... ......... .. .......... .... .. ...... .. ............... ........... ...... ............ ........... ...... ...... 83 5.5
Ph ase Diagrams ................................ ......... ............. .... .. ...................... .... ..... .. ...... .............. ..... 84
5.6 Colligative Properties ............ .. ......... ........ ............. .... .. .............. ........ ..... ........... .......... .... ...... 88 LECTURE
6:
ACIDS AND BASES ......................................................................................93
6.1
Defin itions .................................................................. ................ .... ................................ ........ 93
6.2 How Molecular Structu re Affects Acid Strength. 6.3
Hydrid es .. ......... ............. .. ..............
............... ........................ 96
.. ........................ ................................................... 97
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6.4 Equilibrium Constants for Acid-Base Reaction~ .................. .................................. ................ 99 6.5 Finding the pH ............................................................. .......•............. ................. ................ 100 6.6 Salts .................................. ......... ................. ......................... ................ ................................ 101 6.7 Titrations ............................................................. .. ............................................... ............... 103 6.8 More Titrations and Buffered Solutions .............................................................................. 103 6.9 Indicators and the End Point ....................... ................ .. ....... .......................... .............. ........ 105 6.10 Polyprotic Titrations ..................... .................................................................... ........ ............. 106
LECTURE
7:
ELECTROCHEMISTRY .................................................................................. 109
7.1
Oxidation-Reduction ........... ................................. ...................................................... .......... 109
7.2 Oxidation States .......................................................... ............................................ .. .......... 109 7.3 Oxidation-Reduction Titrations ............ ................................................................................ 111 7.4 Potentials .......................................................................................................... .................... 113 7.5
Balancing Redox Reactions ........ ................. ......................... ... ...... ....................... ................ 114
7.6 The Galvanic Cell ........................................................ ........ ......... .......... ............... .............. 114 7.7
IUPAC Conventions ........................................................................................ .. .................. 116
7.8 Free Energy and Chemical Energy .............................................. ........ ........... .. ...... ..... : ....... 117 7.9 MoreCells .................................................................. ......................... ............................... 120
30-MINUTE IN-CLASS EXAMS ...................................................................................... 125 In-Class Exam for Lecture 1 ......... ......... ... ..... ......... ........ ......... .............................................. 125 In-Class Exam for Lecture 2 .......... ........ ........................ ........................................................ 131 In-Class Exam for Lecture 3 ............................................................ ........ ......... ... ..... ............. 137 In-Class Exam for Lecture 4 ........... ....................................................................................... 143 In-Class Exam for Lecture 5 .................................................... ................................ ... ........... 149 In-Class Exam for Lecture 6 .......... ................. ......... .............................................................. 155 In-Class Exam for Lecture 7 .................................................... .............................................. 161
ANSWERS
&
EXPLANATIONS TO IN-CLASS EXAMS .......................................................... 167 Answers and Scaled Score Conversion for In-Class Exams .......... ........ ............... .. ............ 168 Explanations to In-Class Exam for Lecture 1 ......................................... .. ................ ............. 169 Explanations to In-Class Exam for Lecture 2 .......... ................ ................. ......... .................... 170 Explanations to In-Class Exam for Lecture 3 ........................................... .. ............. .. ........... 172 Explanations to In-Class Exam for Lecture 4 ................. ......................................... .............. 174 Explanations to In-Class Exam for Lecture 5 ......... ....... ................. ....................................... 176 Explanations to In-Class Exam for Lecture 6 ........ .................. .. ..... ......... .............................. 178 Explanations to In-Class Exam for Lecture 7 ................. ......................... .. ..... ....................... 180
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ANSWERS
& EXPLANATIONS
TO QUESTIONS IN THE LECTURES ........................... '" ...... , ... 183
An swers to Questions in th e Lectures .... ... ..... ... ........... .... .... ... ..... .. .................... .... ............ .. 184 Exp lanation s to Questions in Lectu re 1 .................... ........ ................. ...... .... ......... .............. 185 Explanations to Questions in Lecture 2 ..... ....... ... ............. .. ....... .... ...... ............................... 186 Explanations to Questions in Lecture 3 ............ ....... ........ ... ..... ... .............. .......................... 188 Expl anation s to Q uesti ons in Lecture 4 .................... ................ .......................................... 190 Exp lan ations to Questions in Lecture 5 .... .. ............. ... ......... ................... ........... ..... ............ 192 Explan ations to Questions in Lecture 6 .... .. ..... .............................................. .. ..... ...... .... .... 194 Explanations to Questions in Lecture 7 .... .... ... ....... .. ... .... .... ...... .. ........ .. ...... ....... ................ 195
INDEX ........................................................................................................................ 199
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PHYSICAL SCIENCES
DIRECTIONS. Most questions in the Physical Sciences test are organized into groups, each preceded by a descriptiv:e passage. After studying the passage, select the one best answer to each question in the group. Some questions are not based on a descriptive passage and are also independent of each other. You must also select the one best answer to these questions. If you are not certain of an answer, eliminate the alternatives that you know to be incorrect and then select an answer from the re maining alternatives. Indicate your selection by blackening the corresponding oval on your answer document. A periodic table is provided for your use. You may consu lt it whenever you wish.
PERIODIC TABLE OF THE ELEMENTS ~-
1
2
H
He
1.0 3
~-.----,
.. ,
4
'
5
6
Be
I
B
C
~·~1 ~~_c Na Mg
10.8 13
12.0 14
AI
Si
23.0
27.0 28.1 31 , 32
Li
19 -
24.3 20 : 21
K . Ca . Sc
1
3~~1 1 4~~1
4: 0 9
~ Sr V ~ ~~~ !lZ.6 . 889
I 22
23
In
4il
24
~
~
l'0l?c ~
Ba
La*
Hf · Ta
l
Ra
~ !0 14.0 16.0
! ;
-.
15
16
F
Ne
19.0 17
20.2 18
P
S
CI
3 1.0 33
32.1 34
35.5 35
25
26
27
28
29
30
Fe
Co
Ni
Cu
Zn
i Ga
Ge
As
Se
Br
5~8 i ~~9
58.7 46
63.5 47
65.4 48
69.7 49
72.6 50
74.9
79.0 52
79.9 53
Ru
Pd
Ag
Cd
In
Sn
Te
I
~
Rh
,
/~51
Sb
4.0 10
9
Mn - 5::
~
ZrNb
e
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9 1.2 92.9 72 I D
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I
V
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Ar 39.9 36 ;
Kr 83.8 .
-
54
Xe
95.9 __(~81 1101 .1 : 102:9 106.4 107.9 11 2.4 114.8 11 8.7 121 .8 127.6 126.9 , 13 1.3 1 78 79 80 81 82 M ~ ' N n 83 84 851 1 86
W
Re
Os
Ir
Pt
Au
Hg
TI
Bi . Po , At
Pb
i
Rn
i
9 137.3 138.9178.5 ' 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (2091 (2101 (2221 87 88 89 104 105 106 i 107 108 109
Fr
Ac' Unq Unp i Unh i Uns Uno : Une
.(2.231 ..226.0 _227,Q,J2611 (262)~(?621 (2651 (2671 I
58 . 59
*
Ce
I
I
60 . 61
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Nd
Pa
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i
62
Pm
I Sm
Np
l
I
63
64
65
Eu
Gd
Tb . Dy
IAm
Cm
Bk
66
-.
I 67 i I Ho ,
68
69
70
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Tm
Vb
Es
Fm
Md
No
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71
Lu
. 140.1 ' 140.9 144.2 i (1451 ! 150.4 152.0 157.3 158.9 162.5 . 164.9 167.3 168.9 173.0 175. 0 92 r 93 . 94 95 90 91 96 97 100 101 102 . 103 . 98 l 99
!
Th
I
Pu
Cf
Lr
1232 .0 (2311 238.0 (237) (244) (243) (247) (2471 (251) (2521 (257) (258) (259) (260)1
Copyrig ht @ 2007 Exarnkrackers, Inc.
Atoms, Molecules and Qu antum Mechan ics 1.1
Atoms
All mass consists of tiny particles ca lled atoms. Each atom is composed of a nucleus slliTounded by one or more electrons. The radius of a nucleus is on the order of 10--1 angstroms (A). One angstrom is 10-10 m . The nucleus contains protons and neutrons , collectively called nIlcleolls, held together by the strong nuclear!oree. (More precisely, the strong nuclear force holds together the three quarks that make up each nucleon, and it is the 'spill over' from the strong nuclear force that holds together the nucleons.) Protons and neutrons are approximately equal in size and mass. (Neutrons are very slightly heavieL) Protons have a positive charge and neutrons are electrkall y neutral.
Atomic radius (reduced approximately 500 times relative to the nucleons)
\
Electron (magnified 1,000 times relative to the nucleons)
Neutron SlU'rounding the nucleus at a distance of about 1 to 3 A are electrons. The mass of one electron is over 1800 times smaller than the mass of a nucleon. Electrons and protons have opposite charges of equal magnitude. Although for convenience we often think of the charge on an elech'on as 1- and the charge on a proton as 1+, we should remember that this charge is in electron units 'e' called the electronic charge. A charge of 1 e is equal '0 1.6 x 10-19 coulombs. An atom itself is electrically neutral; it contains the sam e nU lnber of protons as electrons.
Particle
Charge
Mass (amll)
Proton
Positive (1 +)
1.0073
Neutron
Neutral
1.0087
Of course, you want to know the charges on the particles, but don't memolize the masses in the table shown. Instead, recognize the dispality in size between electrons and nucleons. Also, notice that protons and neutrons have nearly the
Electron
Negative (1- )
5.5 x 10-1
same mass, about one amu.
Since the nucleons are so small compared to the size of the. ato m, the atom itself is composed mostly of empty space. If an atom were the size of a modern football stad ium , it w ould have a nucleus th e size o f a marble.
2
MCAT INORGANIC CHEM ISTRY
1.2
Elements
Any single atom must be one of just over 100 elements. Elemen ts are the building blocks of all compounds and carmot be decomposed in to simpler substances by chemical means. Any element can be displayed as follows:
Mass Number A = protons + neutrons
Elemental Symbol
Atomic Number Z = proton s Notice that 'f{ stands for mass number and NOT atomic number.
The number of protons identifies which element. For a given element, the number of neutrons identifies which isotope.
w here A is the mass number or n umber of protons plus neutrons, and Z is th e atomic number or number of protons. The atonlic n i.tmbe r is the identity number of any element. If we know the atomic number, then we know the element. Th is is not true of the mass number or the number of electrons. Any element may have any nutnber of neutrons or electrons, but only one lllffilber of protons.
Two or more atoms of the same element that contain different nU111hers of neutrons are said to be isotopes. An atom of a specific isotope is called a Iluelide. Isotopes have similar chemical properties. Hydrogen has three important isotopes: 'H (protiu m), 2H (deuterium), and 'H (tritium) . 99.98% of naturally occurring hydrogen is protium. Examp les of three isotopes for carbon are: 12C, I3C, and 14C.
Each of carbon's isotopes contains 6 protons. 6 protons define carbon. 12C (carbon12) contains 6 neu trons, 13C (carbon-13) conta ins 7 neu trons, and "c (carbon-l4) contains 8 neutrons.
Think of an amu as approximately the mass of one proton or one neutron. Biochemists cali an amu a dalton.
Although the mass number is a good approximatio n of the mass of an atom, it is not exact. The atomic weight or molar mass (MM or M) of an atom is given in atomic mass units (abbreviated amu or with the less commonly used 51 abbreviation u ). The atomic weight of an element is actually a mass (or i.n some books a ratio) and not a weight. An amu is defined by carbon-12. By definition, one atom of 12C has an atomic weig ht of 12 amu. All other atomic weights are measured against this standard. Since carbon natura lly occurs as a mixture of its isotopes, the a tomic weight of carbon is listed as the weighted average of its isotopes or 12.011 amu. (This is very close to 12 amu because almost 99% of ca rbon occurs in nature as 12C.) 12C also defines a mole. A mole (or Avogadro'S number, 6.022 X 1023 ) is the number of carbon atoms in 12 grams of 12c. Keep ing in mind the relationship between an ilmu and grnms cnn be useful: 6.022
If you can't find moles from molecular or atomic weight, then you better get crackin'. This is basic stuff, but you need to have it down cold!
X
1023 amu = 1 g ram
We can read atomic weights from the periodic table as either amu or g/ mo!. If we are given the amount of an elemen t or compound in grams, we can d ivid e by
the atomic or molecular weight to find the number of moles in that sample.
moles
grams atomic or molecular weight
= --....:::...---:----
l..'ipyrigtlt
OC7 f
0
.el
:11(,"
LECTURE
1.3
1:
ATOMS, MOLECULES, AND QUANTUM MECHANICS .
3
The Periodic Table , ,
The periodic table lists the elements from left to right in the order of their atomic num bers. Each horizontal row is called a period. n,e vertical columns are called groups or families . There are at least two methods used to num ber the gro ups. The newer method is to number them I through 18 from left to right. An older method , which is still used, is to separate the g roups into sections A and B. These sections are then numbered with Roman numerals as shown below.
y. 01
TIle periodic table below divides the elements in to three sections: 1) nonmetals on the right (dark orange); 2) metals on the left (light orange); and 3) metalloids along the yellow-shad ed diagonal separating the m etals from the nonmetals.' Metals are large atoms that tend to lose electrons to form positive ions or form positive oxidation states. To emphasize their loose hold on their electrons and the fluid-like n ature of their valence electrons, me tals are often d escribed as atom s in a sea of electrons. The easy movemen t of electrons w ithin metals g ives them th e ir metallic character. Metallic character includes ductility (easily stretched ), m alleability (easily hammered into thin strips), thermal and electrical conductivity, and a characteristic luster. Me tal atoms easily slide past each other allowing metals to be hammered into thin sh eets or draw n into w ires. Electrons move easily from one metal atom to the next transferring energy or charge in the form of hea t or electricity. All metals but mercury exist as solids at room temperature. Metals typically fo rm ionic oxides such as BaD. (BeD is one exception that is not ionic.)
Know the characteristics of metals: lustrous, ductile, malleable, thermally and electrically conductive.
Nonmetals have diverse appearances and chemical behaviors. Generally speaking, nonmetals have lowe r melting p oints than metals. They form negative ions. Molecular substances a re typically made from only nonmetals. Nonmetals form covalen t oxides such as SiD, or CO,.
1
2
3
4
6
5
7
IA -,
8
9 10 11 U
3
IIA
,
6.9
Be ,9.0
11
12
Li
Na Mg
•
N
10.8
12.0 H
14.0 15
13
VIBVIIB~ IS 24 25 27 2. 28 29
llIB IVB VB
'0
21
22
23
K
Ca
Sc
Ti
V
39,] 37
40.1
45.0
47.9
50.9
;::I
38
39
.0
41
l5
Rb
Sr
85.5
87.6
Y 88.9
91.2
55
5.
57
72
Cs
Ba
La t
·. HI
To,
W
87
88
89
104
105
10.
107
Fr
Ra
o
......,
vum
24.3
0..
C
5
•
19
I
B
Transition Metals
23.0
Zr
7
9
10
Ne
16.0 1.
19.0
20.2
17
18
p
S
Cl
Ar 39.9
lIB
AI
S;
27.0
28.1
31.0
32.1
35.5
30
31
32
33
34
35
3.
Co
N;
Cu
Zn
Ga G.
As
So
Br
Kr
55.8
58.9 45
58.7
63.5
65.4
69.7
72.6
74 .9
79,O
79.9
83.S
44
47
48
49
50
51
52
75
76
77
78
79
80
81
82
83
Re
Os
Ir
Pt
Au
108
109
Nb Mo
Ac: 0nq
Alkali Metal!
(261)
95.9 74
54.9 43
4.
53 54 [ Tc Ru Rh Pd Ag Cd In Sn Sb Te Xe (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121 .8 127.6 126.9 131.3
o
Unp Unh Uns Uno Une (262)
(263)
(262) (265)
(267)
r
•
Alkaline Earth Metals 58
59
Co
Pr
.0
61
.2
Nd Pm Sm
63
Eu '
84
85
8.
Po
At
Rn
(209)
(21U)
(222)
t
M etals
r
Halogens
Metalloids
'Lanthanides :
~I
F
Pb B; l'lg TI 132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (223) 226.0 227.0
16 17 18
Cr Mn Fe 52.0 42
92.9 73
~
VIllA 1'2'" He llIA IVA VA VIA VIIA 4.0
Periods--+
H 1.0
D M
Noble Gases
Nonmetals 6.
65
Gd Tb
66
67
68
70
71
Dy
Ho
Er Tm Yb
.9
Lu
140.1 140.9 144 .2 (l4S) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0 101 102 103 90 92 94 9. 97 98 99 100 91 93 95
IActinides
Th 232.0
Copyrigllt~.
2007
t.xJrpk Kkl.'r~,
~.
Pa
U
(231) 238.0
Np Pu Am Cm Bk
CI
Es
Fm Md No
(237) (244) (243) (247) (247) (25]) (252) (257) (258) (259)
Lr (260)
You need to know the names of th,~ following groups: alkali metals, alkaline earth metals. halogens, and noble gases.
4
MCAT INORGANIC CHEMISTRY
Metalloids have some characteristics that resemble metals and some that resemble norurnetals. You should also recognize the names of the following four gro ups: alkali metals; alkaline ea rth metals; halogens; and noble (or rare) gases. The section A groups are known as the representative or main-group elements and the section B groups are called the transition metals. Elements in the same family on the periodic table tend to have similar chemical properties. For example, they tend to make the same number of bonds, and exist as similarly charged ions.
1.4
Characteristics Within Groups
Hydrogen is unique and its chemical and physical characteristics do not conform well to any family. It is a nonmetal. Under most conditions, it is a colorless, odorless, diatomic gas. Elements in the same family have similar chemical properties. Hydrogen is an exception to the rule.
As pure substances, Group 1A or alkali metals are soft metallic solids with low densities and low melting points. They easily form 1+ cations. They are highly reactive, reacting w ith tTIost nonmetals to form ionic compounds. Alkali metals react w ith h ydrogen to form hyd rides such as NaH. Alkali metals react exothermically w ith water to produce the metal hydroxide and hyd rogen gas. In nature, alkali metals exist only in compounds.
You can remember that alkali metals are Group 1 and alkaline earth metals are Group 2 because alkali comes before alkaline in alphabetical order.
Group 2A or alkaline earth metals are harder, more dense, and melt at higher temperatures than alkali metals. They form 2+ cations. TIley are less reactive than alkali metals. Heavier alkaline earth metals are more reactive than lighter alkaline earth metals. All the 4A elements can form four covalent bonds wi th nomnetals. All but carbon can form two additional bonds with Lewis bases. Of the 4A elements, only carbon forms strong pi bonds to make strong double and even triple bonds. Group SA elements can form 3 covalent bonds . In addition, all SA elements except nitrogen can form five covalent bonds by using their d orbitals. These elements can further bond with a Lewis base to form a sixth covalent bond. Nitrogen forms strong pi bonds to make double and triple bonds. Phosphorous can form only weak pi bonds to make double bonds. The other SA elemen ts cannot make pi bonds. Nitrogen can also form four covalent bonds by donating its lone pair of electrons to form a bond.
This is a lot of detail! Knowing some of details of the characteristics of each group on the periodic table as explained here will be helpful , but not required by the MeAT. Don't be too concerned with ali this detail.
Group 6A elements are called the ellalcogens. Oxygen and sulfur are the important chalcogens for the MCAT. Oxygen is the second most electronegative element. Oxygen is divalent and can form strong pi bonds to make double bonds. In nature, oxygen exists as 0 , (dioxygen) and 0 , (ozone). Oxygen typically reacts with metals to form metal oxides. Alkali metals form peroxides (Na,O,) and sllper oxides (KO,) with oxygen. The most common form of pure sulfur is the yellow solid 58' Metal sulfides, such as Na,S, are the most common form of sulfur found in nature. 5ulfur can form two, three, four, or even six bonds. It has the ability to pi bond making strong double bonds. The radioactively stable Group 7A elements (called halogens) are fluorine, chlorine, bromine, and iodine. Halogens are highly reactive. Fluorine and chlorine are diatomic gases at room telnperature; bromine, a diatomic liquid; iodine, a diatomic solid. Halogens like to ga in electrons. However, halogens other than fluorine can . take on oxidation states as high as +7 when bonding to hi gh ly electronegative Copyright © 2007 Examkrackers, Inc.
LECTURE 1 : ATOMS, MOLECULES, AND QUANTUM MECHANICS . 5
a toms like oxygen. When in compounds, fluorine always h as an oxidation state of - 1. This means tha t fluorine makes onl y one bond, while the other halogens can make more than on e bond. Hydrogen combines with all the halogens to form gaseo us hydrogen halides. The h ydrogen halides are soluble in water forming the hydrohalic acids. H alogens react w ith metals to form ionic h a lides. Noble gases are nonreactive. They are some times called the inert gases. Only the noble gas elements a re normally found in nature as isola ted a toms. They are all gases at room temperature. The elements that tend to exist as diatomic molecules are h ydrogen, oxygen, nitrogen, and the h alogens. Typically, w hen these elements are discussed , it is assumed that they are in their diatomic form unless otherwise stated.ln other words, the statement "Nitrogen is.nOlueactive. " refers to N 2 and not N . Notice that the size of an atom h as a significan t effect on its chemistry. If we exam ine the smallest element in a Group, we can sometim es see devia tions in its behavior due to its size. Small atoms ha ve less room to stabilize charge by spreading it out. This makes them bond more stron gly to water resulting in greater h eats of hydration. Becau se beryllium in its ionic form is n ot large e nough to stabilize its charge, it forms a covalent oxide, whereas o ther alkaline earth me tals make io nk oxides. This m eans that BeO is amphoteric whereas other alka line ea rth metal oxides are basic. Sm all atoms don' t ha ve d orbitals ava ilable to them for bond formation. Atom s w ithout d orbitals caru10t form more than four bonds. Oxygen typically forms two bonds, w hile the larger sulfur can form up to six. On the other hand, the p orbitals on atoms that are too big don't overlap significantly, so large atoms can't easily form pi bonds. The second period elements carbon, nitrogen, and oxygen are small enough to form strong pi bonds w hile their larger third row family members form onl y weak pi bonds, if they form pi bonds at all. Strong pi bond ~
p orbitals
~
~ Weak pi bond
Sigma bonds Small atoms make strong pi bonds due to overlap of p orbitals.
Copyright © 2007 Examkruckers, Inc.
Large atoms are unable to make strong pi bonds.
120°
120° 90°
Large atoms have d orbitals allowing fo r more than 4 bonds.
6 . MeAT
INORGANIC C HEMISTRY
1.5
Ions
When an element has more or fewer electrons than protons, it becomes an ion. Positive ions are called cations; negative ions, anions. The representative elements make ions by forming the closest noble gas electron config uration. (Electron config urations are discussed later in this Lecture.) Metals form cations; norune tals form anions. When the transition metals form ions, they lose electrons from their 5 subshell first and then from their d subshell. (Subshells w ill be discussed later in this lecture.) Below are some common ions formed by metals. Easy here! You don't have to memorize me charge on every cation made by transition metals. This is background knowledge. But, you should be able to predict the charge based upon two things: 1. Atoms lose electrons from the highest energy shell first. In transition metals, this means that electrons are lost from the s subshell first, and then from the d subshell.
2. Ions are looking for symm etry. Representative elements form noble gas electron configurations wh en they make ions. Transition metals try to 'even-out' their d orbitals, so each orbital has th e same number of electrons.
-
VIll A .---
IA
lilA IVA VA VIA VIl A
I1A
VIllA IIlB IVB VB VIBV II B~
Cr'
Mn" Fe"
Fe'
Co"
Ni~'
m
liB
~~;.
Zn~'
Ag'
r e'
AI'
Cd"
Sn2•
Au' H g,,. Au' Hg 2•
Pb"
Bi}<
These are not all the p ossible ions; they are just some of the more common ions. Notice that Group IB n1akes 1+ ions. There are six 3+ ions Cr3.;., Fe}+, Au 3-'-, A P+, and Be-t-. The rest are 2+ ions.
Cations are significantly smaller than their neutral atom counterparts. For instance, sodium's outermost electron is located by itself on an outer shell. When this outer electron is removed, the sodium cation is Significa ntly smaller because the remaining electrons are located in inner shells . The reverse is true for anions. The additional electrons are added to an outer shell making the anion much larger than its neutral atom counter part. Isoelectronic ions (ions w ith the saIn e number of electrons) tend to get smaller w ith increasing atomic nUluber because m ore protons pull inward on the same number of electrons. The sizes of the oxygen, f1uorme, sodium
and magnesium ions reflect this trend.
Ionic compounds are usually made from both metals and nomnetals. Coulomb's law: F = kq,q,/r', describes the electrostatic forces holding an electron to its nucleus. The distance between the electron and the nucleus is r . For q, we might plug in the positive charge of the nucleus, Z, and for q" the charge on the given electron. This wou ld wo rk fine for hydrogen, where the lone' electron feels 100% of the positive charge on the nucleus. However, in helium tl,e first electron shields some
Copyright © 2007 Examkrackers, Inc.
LECTURE
1:
ATOMS, MOLECULES, AND QUANTUM MECHANICS •
7
of the nuclear charge from the second electron, so that the second electron doesn' t
feel the entire nuclea r charge, Z. The amou nt of charge felt by the second electron is ca lled the effective nuclear charge (Z,,,) . You can see in the graph below that 2,ft for the outermost electron of helium is not 2, even though there are two protons in the helium nucleus. The Z,f( is the nuclear charge Z minus the average number of electrons between the nucleus and the electron in question. The Z",ffl not 2/ should be plugged in for q in Coulomb's law to find the force on the outermost electron. Note that the force is a function of both q (Z,.) and r (the distance from the nucleus).
and
When considering the effect of
No Shielding 4
/' :' :/ :' :' :' :' .' :'
3 Z,ff 2 (eV)
1
.'
/-------- ------ ----------._------ ------------------------ -
1
2
3
4
5
678
H He Li Be B C N 0
Perfect ...Shielding
9 10 11 F NeNa
000 The grap h above shows Z,ft values (gi ven in electron volts) for the h ighest energy electron in each element through sodium. Notice the drop in Z,(f going from helium to lithium. This is because the last electron added to make lithium is ad ded to an outer shell making the shielding effect strong. To form beryllium, an electron is added to the same shell and the shielding effect is no t as grea t. To form boron, an electron is added to the 2p subshell, a higher energy subshell, and shielding is stronger aga in. (We'll discuss subshells later in this lecture.) Going from nitrogen to oxygen, the next e lectron m ust share an orbital with one of the three p orbitals resulting in sbme shielding and a reduction in 2 df. Moving from neon to sodium, the next electron is added to an entirely new shell, the 3s subshell. This causes a strong reduction in Z cffl but notice that the outermost electron in sodium still experiences a higher Z ,ff than the outermost electron of the element immediately above it on the periodiC table, lithium. For now, think of Z,ff as increasing going from left to right and from top to bottom on the periodic table.
,
With Z d f in mind, we can make general predic tions about the elemen ts based upon their position in the periodic table. The tota lities of these predictions are called the periodic trends. Since the effective nuclear c11a rge increases when moving from left to right, each additional electron is pulled more strongly toward the nucleus. This results in a smaller atomic radius. Of course, with each add ed shell th e atom g rows larger. Th us, atomic rad ius also increases from the top of the periodic table to the bottom.
When an electron is more strongly a ttached to the nucleus, more energy is required to detach it. The energy necessary to detach an electron from a nucleus is called ionization energy. The energy necessary to detach an electron from a neutral atom is Copyright © 2007 Examkrackers, Inc.
Z\lrr,
consider the strength of Z'1f and the distance from the nucleus. The force pulling the electron inward goes up with Z" but goes down with distance according to Coulomb's law: F ~ Kqq/i'.
8
MCAT INORGANIC CHEMISTRY
, called the first ionization energy. (By definition, the atom being ionized is gaseous.) The energy for the removal of a second electron from the same atom is called the second ionization energy, and so on. The second ioniza tion energy is alw ays much grea ter than the first because when one electron is removed, the effective nuclear charge on the other electrons increases. Ionization energy generally increases along the periodic table from left to right and from bottom to top . This trend is exp lained by Z,ff' Z,ff increases w hen moving across a period to the right. making it tougher to knock off an electron. Although Z,ff also increases when moving down the p eriodic table, the d istance of the electron from the nucleus increases as well, thus decreasing the electric field at the point of the electron. The decreased electric field has less strength to hold the electron to the atom. Electronegativity is the tendency of an atom to attract an electron in a bond that it shares with another atom . The most commonly used measuremen t of electronegativity is the Pauling senle, which ranges from a value of 0.79 for cesium to a value of 4.0 for fluorine. Electronegativity also tends to increase from left to right and bottom to top on the period ic table, and is related to Z,ff in a similar fashion to ionization energy. Electronegativity values are undefined for the noble gases. Electron affinity is the willingness of an atom to accept an additional electron. More precisely, it is the energy released when an electron is added to a gaseous atom. Electron affinity tends to increase on the period ic table from left to right and from bottom to top, and is related to Z,{f' (Warning: Many books use the exothermic value for electron affinity, which is the negative of the energy released . We can state this as follows: Electron affinity is more exothermic to the right and up on the periodic table.) The noble gases do not follow this trend. Electron affinity values for the noble gases are endothermic. The final important periodic trend, metallic character, ten ds to increase from right to left and top to bottom. An easy way to remember the 5 periodic trends is as follows: if it begins with an 'E', as shown here, then it increases going to the right and up: if it doesn't begin with an 'E', then it increases in the opposite direction . Be careful! This mnemonic requires that you think of 'ionization energy' as 'energy of ionization' so that it begins with an 'E', z"" is not considered a periodic trend for this mnemonic, Keep in mind that the trends are just trends, and are violated frequently,
Energy of Ionization Electron Affinity Electronegativity
==~~~~-------E
Atomic Radius Metallic Character
The noble gases do not follow the trends for electronegativity or for electron affinity. Notice that I have moved hydrogen to a more appropriate position.
Copyright © 2007 Examkrackers, Inc.
7. In 1869 both Mendeleev and Meyer. working separately. published nearly identical classification schemes for the elements that were the forerunners of the modern periodic table. Although scientists of that time had no knowledge of atomic numbers, both schemes ordered the elements in nearly correct order from lowest to highest atomic number. Which of the following is the most likely explanation?
Questions 1 through 8 are NOT based on a descriptive passage.
1. Which of the following increases with increasing atomic number within a family on the periodic table? A. B. C. D.
electronegativity electron affinity atomic radius ionization energy
A.
Both scientists noticed similar patterns in chemical and physical behaviors among the elements. B. Atomic number generally increases with atomic weight and the scien'tist knew the atomic weights of the elements. C. ,The chemical identity was predictable from the number of valence electrons which was known at the time. D. Although the number of protons for each element was not known, the number of neutrons was.
2. Which of the following molecules has the greatest dipole moment? A. B.
H2
C.
HF
D.
HBr
O2
8. Which of the following could be a stable molecular 3. How many carbon atoms exist in 12 amu of 12C? A. B.
C. D.
structure?
A. 6.02x 10'-' 7.22xlO24
Se=Se
4. Silicon has a silvery luster at rdom temperature. Silicon is brittle, and does not conduct heat or electricity well. Based on its position in the periodic table, silicon is most likely a: A. B. C. D.
C.
12
B.
D.
Cl
I
nonmetal
CI-F-CI
metalloid metal
I
Cl
chaJcogen
5. Which of the following most likely represents the correct order of ion size from greatest to smallest? A. B. C. D.
0 2-, F, Na+, Mg2+ Mg2+, Na+, F, 0 2Na+, Mg2+, 0 2-, PMg2+, Na+, 0 2-, P-
6. A natural sample of carbon contains 99% of 12c. How many moles of 12C are likely to be found in a 48.5 gram sample of carbon obtained from nature?
A. B.
4
C.
12 49.5
D.
Copyright © 2007 Exarnkrackers, Inc.
9
STOP.
10
MCAT INORGANIC CHEMISTRY
1,6
SI Units and Prefixes
By international agreement, S1 units are used for scientific measurements. '5I Units' stands for Systeme International d'Unites, SI units predominate on the MCAT
There are seven base units in the 51 system. The seven are listed in the table below: Physical Quantity
Name of Unit
Mass Lengfh Tilne Electric current Temperature Luminous intensity Amount of substance
Kilogram Meter Second Ampere Kelvin Candela Mole
Abbreviation kg m s A K cd mol
Other SI units can be derived from these seven, such as a newton: 1 N ~ 1 kg m S- 2, There are other units still commonly in use that you may also see on the MCAT, such as atm or torr for pressure. All such units will have an S1 counterpart that you should know. We will point this out as we corne across new units. The 51 system also employs standard prefixes for each unit These prefixes are commonly seen on fhe MCAT The table below lists these standard prefixes:
1,7 STOP! You were going to skip this page ; weren 't you? Memorize the 51 units and the prefixes, They will help you in all the
sciences.
Prefix
Abbreviation
Meaning
MegaKiloDeciCentiMilliMicroNanoPicoFernto-
M k d c m
10' 10' 10-'
10-2 10-;
fL
10 ~
n p f
10-" 10-12 10-15
Molecules
Atoms can be held together by bonds, In one type of bond, two electrons are shared by two nuclei, This is called a covalent bond, The negatively charged electrons are pulled toward both positively charged nuclei by electrostatic forces. This 'tug of war' between the nuclei for the electrons holds the atoms together, If the nuclei come too close to each ofher, the positively charged nuclei repel each other. These repulsive and attractive forces achieve n balance to create a bond. The diagram COlTIpares the internuclear distance between two llydrogen atoms to their electrostatic potential energy level as a system, The bond length is defined as fhe point where fhe energy level is the lowest Two atoms will only form a bond if they can lower fheir overall energy level by doing so, Nature tends to seek the lowest energy state, If we separate the atoms by an infinite distance, the forces between them, and thus the energy, go to zero. The energy necessary to achieve a complete separation is given by the vertical distance on the graph between the bond length and zero. This is called the bond dissociation energy or bond energy, (Bond dissociation en, ergy and bond energy are very closely related,)
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LECTURo 1: ATOMS, MOLECULES, AND QUANTUM MECHANICS .
A s ubs tance mad e from two or more elements in definite proportions is called a compound. In all pure compounds, the relative number o f atoms of on e element to another can be represented by a ratio of whole numbers. This ratio is called the empirical formula. In molecular compou nds, groups of Bond length atoms form repeated, separate and a Internuclear distance ~ distinct units called molecules. 1.n mol ecular compo unds, the exact number o f elemental atoms in each molecule can be represented by a molecular formula . The empirical formula for glucose is CH,O. The molecular formula is C 6H 120 6•
t
From the empirical formula and the atomic weight of each atom, we can find the percent composition of a compound by mass. To do this, multiply an atom 's atomic weight by the number of atoms it contributes to the empirical formula. Divide your result by the weight of all the atoms in the empirical formula. This gives you the mass fraction of the compound represented by the atom. Now multiply this fraction by 100, and yo u have the p ercent composition by mass.
11
Notice from the graph that energy is always required to break a bond . Conversely. no energy is ever released by breaking a bond. (Energy from AT? is released when the new bonds of ADP and iP are formed, and not when the ATP bonds are broken .)
The percent mass of carbon in glucose (empirical formula = CHP) is found as follows:
To find the empirical formula from the percent mass
molecular weight of carbon
12
molecular weight of CH,O
11
= 0.'1.
Glucose is 40(X) carbon by mass.
compos ition, you aSSUll"le that you have a 100 gram
sample. Now the percent translates direc tly to grams. When you divide the grams by atomic weight, yo u get moles. Now divide by the greatest cornman factor. This is the number of atoms represented by each element in the empirical formul a. In order to find the molecular formula, we would need more information.
If we arc asked to find the empirical formula of a compound thi'lt is 6(~,o hydrogen and 94% oxygen by mass, we do the following: Fronl 110l gram sa "lple:
/
6 H h vdrogen . "-
.
These mllst be whole numbers
61Tloles
J g/mol is
,J.
one to one ratio
5.9 moles
The empirical formula is HO
Copyright © 2007 Examkrackers, Inc.
This stuff should be easy for you. You have to know it backwards and forwards . It will definitely be on the MeAT. and it should be a couple of fast and easy points. Practice this until you know it well.
12
MCAT INORGANIC CHEMISTRY
1,8
Naming Inorganic Compounds
There is very little tested concerning the naming of inorganic compounds. However, it is a good idea to be able to identify compounds when they are referred to. Ionic compounds are named after their cation and anion. If the cation is metal and capable of having different charges, for example, copper can take on a charge of 1+ or 2+, then its name is followed by a Roman numeral in parentheses, as in copper(I) ion or copper(II) ion. An older method for naming cations that can take on different charges is to add -ic to the ending of the cation witl1 the greater positive charge and -ous to the cation with the smaller charge, as in cupric (Cu") and cuprous (Cu') ions. If the cation is made from a nonmetal, the cation name ends in -iUID, such as amlnonium(NH:).
This nomenclature stuff is boring. Just memorize this stuff once and for all and get it over with . Don't get too involved in the myriad little ru les in nomenclature. Keep it simple because the MeAT sllre wi ll.
Monatomic anions and simple polyatOlnic anions are given the suffix -ide, such as hydride ion (H-) or hydroxide ion (OH-). Polyatomic anions with multiple oxygens end with the suffix - ite or -ate depending upon the relative number of oxygens. The more oxygenated species will use the -ate suffix, such as nitrite ion (N02-) versus nitrate ion (N03- ). If there are more possibilities, the prefixes hypo- and per- are used to indicate fewest and most oxygens respectively, such as the hypochlorite (CIO-), chlorite (ClO,-), chlorate (Cl03- ), and perchlorate (Cl04-) ions. If an oxyanion has a hydrogen, the word hydrogen is added as in hydrogen carbonate ion (HC0 3-). The old name would have been bicarbonate ion. To name an ionic compound, just put the cation name in front of the anion name as in barium sulfate (Ba504 ) or sodium hydride (NaH). Acids are named based on their anions. If the name of the anion ends in -ide, the acid name starts with hydro- and ends in -ic, as in hydrosulfuric acid (H,S). If the acid is an oxyacid, the ending -ic is used for -the species with more oxygens and - ous for the species with fewer oxygens, as in sulfuric acid (H2S0-t) and sulfurous acid (H,SOJ)'
For binary molecular compounds (compounds with only two elements), the name begins with the name of the element that is farthest to the left and lowest in the periodic table. The name of the second element is given the suffix -ide and a Greek number prefix is used on the first element if necessary (e.g., dinitrogen tetroxide, N,04)'
1.9
Chemical Reactions and Equations
When a compound undergoes a reaction and maintains its molecular structure and thus its identity, the reaction is called a physical reaction. Melting, evaporation, dissolution, and rotation of polarized light are some examples of physical reactions. When a compound LU1dergoes a reaction and changes its molecular structure to form a new compound, the reaction is called a chemical reaction. Combustion, Inetathesis, and redox are examples of chelnical reactions. Chemical reactions can be represented by a chemical equation with the molecular formulae of the reactants on the left and the products on the right.
CH, + 20, --) CO, + 2H,O An unbalanced equation is a wrong answer unless specifically asked for. By the way, know the difference between a physical and chemical reaction .
Notice that there is a conservation of atoms from the left to the right side of the equation. In other words, there is the same number of oxygen, hydrogen, and carbon atoms on the right as on the left. This means the equation is balanced. On the MCAT, if the answer is given in eguation form, the correct answer will be a balanced equation unless . specifically indicated to the contrary. Copyright © 2007 Exarnkrackers, Inc.
LECTURE
In the previous equation, 0 , is preceded by a coeffi cient of two. A coefficient of one is assumed for all molecules not preceded by a coefficient. Methane, then, has a coefficient of one. These coefficients indica te the relative number of molec ules. They represen t the number of single molecules, moles of molecules, dozens of m olecules or any other quantity. They do not represent the mass, the number of grams, or kilograms.
1:
ATOMS, MOLE CULES, AND Q UANTUM MECHANICS .
13
CH 4 + 20, ---7 CO, + 2H,O Starting amount:
4
6
0
0
Amount used up:
3
6
0
0
0
3
Ending amount: 1 To say that a reaction runs to completion means that it moves to the right until the supply of at least one of the reactants is depleted. (Reactions often don't run to completion because they reach equilibrium first.) As indicated by the equation above, two moles of oxygen (0,) are needed to burn one mole of methane (CH,). If we were to react four moles of methane with six moles of oxygen, and the reaction ran to completion, we would be left with 1 mole of methane. This is because, from the two to one ratio in the equation, six moles of oxygen are only enough to burn three moles of methane. Since we would run out of oxygen first, oxygen is our limiting reagent. Notice that the limiting reagent is not necessarily the reactan t of which there is the least; it is the reactant that would be completely used up if the reaction were to run to completion. Also from the balanced equation, the one to one ratio of methane to carbon dioxide and the two to one ratio of methane to water, teUs us that burning three moles of methane produces three moles of carbon dioxide and six moles of water.
If you don't understand where the
numbers came from in the table, take another look. You should understand this for the MCAT.
1.1 0 Chemical Yield The amount of product produced when a reaction runs to completion is called the theoretical yield. The amoun t of ac tual product after a real experiment is the actual yield. As mentioned above, reactions often don't run to completion, an d sometimes there are competing reactions that reduce the actual yield. Actual yield divided by the theoretical yield, times 100, gives the percent yield.
Actual yield . . 1 'ld x 100 = Percent YIeld Theorehca yel
MCAT would probably give you the equation for the percent yield . Just understand the concept.
1.11 Fundamental Reaction Types Reactions can be categorized into types. The following lists four reaction types using hypothetical elements or molecules A, B, C, and D. Combination: A + B ~ C Decompositi.Qn: C
~
A+B
Single Displacement: A + BC
~
Double Displacement: AB + CD
B + AC
~
(also called single replacement)
AD + CB (also called double replacement or meta thesis)
Some important reaction types not shown here are redox, combustion, BronstedLowry acid-base, and Lewis acid-base. We will cover these types later in this book. Reaction types are not mutually exclusive, so one reaction can fall into more than one type.
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Nothing to do here but memonze these reaction types in case MCAT asks one question requirtng you to identify which type. It won't be worth more than one question on an MCAT.
14
MCAT INORGANIC CHEMISTRY
1.12 Reaction Symbols The symbol /:1' usually means "change in" but' t:J,' above or below a reaction arrow indicates that heat is added. When a chemical is written above the reaction arrow, it is often a ca talyst. Two arrows pointing in opposite directions (like this ';;=') indicate a reaction that can reach equilibrium. If one arrow is longer than the other, the equilibrium favors the side to which the long arrow points. A single arrow pointing in both directions (like this 'n ') indicates resonance structures. Square brackets [ 1 around an atom, molecule, or ion indicate concentration. The naught symbol 10' indicates standard state conditions. J
1.13 Bonding in Solids Solids can be crystalline or amorphous. A crystal has a sharp melting point and a characteristic shape w ith a well ordered structlU€ of repeating lmits w hich can be atoms, molecules or ions. A crystal is classified as ionic, network covalent, metallic, or moleclllar depending upon the nature of the chemical bonding and the intermolecular forces in the crystal. Ionic crysta ls consist of oppos.itely charged ions held together by electrostatic forces. Salts are ionic crystals. Metallic crystals are single metal atoms bonded together by delocalized electrons. These delocalized electrons allow metallic crystals to efficiently conduct heat and electricity. They also make metallic crystals malleable and ductile. Network covalent crystals consist of an infinite network of atoms held together by polar and nonpolar bonds. Diamond and crystal SiO, are, common examples of network covalent crystals. It is not possible to identify individual molecules in ionic, metallic, and network covalent crystals. Molecular crystals are composed of individual molecules held together by intermolecular bonds. lee is an example of a molecular crys tal. . / The MCAT does not directly test your knowledge of the structure of solids beyond ionic and molecular solids; however, it is good to at least be aware that atoms can form substlnces in many ways. Molecu lar solids are actually less common than other types of solids. There has been an MCAT passage on this topic.
An amo rphous solid has no characteristic shape and melts over a temperature range. Glass (SiO,) is an amorph ous solid usuaUy with some impurities added to lower the melting point. Some substances are capable of forming both crystalline and amorphous solids.
Polymers are solids with repeated structural units. They can be crystalline or amorphous. Generally, rapid cooling of liquid polymers res ults in amorphous solids and slow cooling results in crysta lline solids. There are many polymers found in living systems. Examples of biopolymers are DNA, glycogen, and protein.
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14. Na me the following co mpound: C u(ClO,),.
Questions 9 through 16 are NOT based on a desc riptive passage.
A.
B. C. D.
9. What is the e mpirical formula o f a neutral co mpound containing 58.6% oxygen, 39% sulfur, 2.4% hydrogen by mass ?
A, B,
e. D,
copper(l) chlorate copper(lI) perchl orite copper(ll ) chl orate cop per(ll) perchlorate
15. Polyeth ylene is a flexible pl astic w ith many industrial uses. The syn thesis of polyethyle ne is a radical reacti on thaLfoll ows the equati on:
,
HSO HSO,H,SO, H,SO,
nC H, = CH, ...., (CH,C H,)" Polyeth yle ne is a(n):
10. Silica is a network solid of silicon a nd oxygen atoms. The em pirical fo rmul a fo r silica is SiO;:. In silica, to how many oxygen atoms is each sil icon bonded?
A.
B. C. D.
A,
e.
2 3
D,
4
S,
16. Wh at type of reacti on is shown below? 2M g(s) + 0 , ...., 2MgO(s)
A.
11, What is the percen t by mass of carbon in CO, ?
A, B,
e. D,
mol ecul ar solid . polymer. ionic solid. network solid.
B. C. D.
12% 27% 33 % 44%
combination single replaceme nt metathesis deco mposition
12. Sulfur dioxide oxidizes in the presence of 0 2 gas as per the reacti on: 2S0,(g) + O,(g) ...., 2S0 ,(g)
Approxim ately how many grams of sulfur tri ox ide are produced by the complete oxidati on of I mole of sulfur dioxide?
A,
I o"
B,
2g
C.
80 g 160 g
D.
'.
13. Whe n gaseo us ammo ni a is passed ove r solid copper(TT)ox ide at high te mperalures, nitrogen gas is formed. 2NH ,(g) + 3CuO(s) ...., N ,(X) + 3Cu(s) + 3H,o(g)
What is the limiting reagent when 34 gram s of ammon ia form 26 grams of nitrogen in a reaction th at run s to co mpleLion? A. B.
e.
D.
NH) CuO N, Cu
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15
STOP.
16
M C AT INORGANIC CHEMISTRY
1-14 Quantum Mechanics The MCAT requires a small amoun t of know led ge concerning quantum mechanics. Everything that you' ll need to know is in this Lecture. Quantum mechanics basically says that elem entary pa rticles can only gain or lose energy and certain other quantities in discrete units. Th.is is analogous to walking up stairs as op posed to walking up a ramp . If each stairstep is one foot, you can only raise or lower yourself by increments of one foot on the stairs, w hile on the ramp you can move along the ra mp until you are raised one half foo t or any other fraction of a foot. The discrete units of energy are so sm all that the gain or loss of one unit is insignificant on a macroscopic scale. Thus quantum mech anic effects are generally only important w h en dealing w ith elementary particles.
1. 15 Quantum Numbers A set of four quantum numbers is the address or 10 number for an electron in a given atom. No two electrons in the same atom can have the same four quantum numbers. The first quantum number is the principal quantum number, n . The principal quantum number designates the shell leve!. The larger the p rincipal quantum number, the greater the size and energy of the electron orbita!. For the representative elem en ts the principal quantum number for electrons in the outer most shell is given b y the period in the periodic table. The principal quantum number for the transitio~ metals lags one shell beh.ind the period, and for the lanthanides and actinides lags two shells beh.ind the period. Valence electrons, the electrons which contribute most to an element's ch emical properties, are located in the outermost she ll of an atom. Typically, but not alw ays, on ly electrons from the s and p subshells are considered valence electrons.
Is
25
35 z
t
2p,
2p,
The second quantum number is the azimuthal quantum number, I. The azimuthal quantum number deSignates the subshell. These are the orbital shapes w ith w h.ich we a re familiar su ch as s, p, d, and f lU = 0, we are in the s subshell; if Q = 1, we are in. the p subshell; and so on. For each new shell, there exists an additional subsh ell with the azimuthal qua ntum number 1 = n 1. Each subsh ell h as a peculia r shape to its orbitals. The shapes are based on probability functions of the position of the electron. There is a 90% chance of finding the electron somewhere inside a given shape. You should recognize the sha pes of the orbitals in the sand p subsh ells.
The .third quantum number is the magnetic quantum number, m,. The magnetic quantum number designates the precise orbital of a given subshell. Eac h subshell will have orbitals with magne tic quantum numbers fro m -I to +1. Thus for the first shell w ith 11 = 1, and e = 0, there is only one possible orbital, and its magnetic qua ntum number is O. For the third shell w ith II =3, and Q= 2, there are 5 p ossible orbitals w hich have the m agnetic quantum numbers of - 2, - 1, 0, +1, and +2. The fourth quantum number is the electron spin quantum number, m,. The e1ec-" tron spin quantum number can h ave values of _'/2 or +'/2. An y orbital can hold up to two electrons and no more. lf two electrons occupy the same orbital, they have the same first three quantum numbers. The Pauli exclusion principle says that no .
Copyright © 2007 Exarnkrackers, Inc.
LECTURE 1 : ATOMS, MOLECULES, AND QUANTUM MECHANICS .
two electrons in the same atom can have the same four quantum numbers. Because two electrons in the same orbital have identical P t, 2nd and 3 rd quantum numbers, they mus t have opposite electron spin quantum numbers. f
The number of total orbitals within a shell is equal to n'. Solving for the number of orbitals for each shell gives 1,4,9,16 ... Since there are two electrons in each orbital, the number of elements in the periods of the periodic table is 2, 8, 18, and 32. Number
Character
Value
Symbol
1st
Shell
II
II
2nd
subshell
r
from zero to n - 1
3rd
orb ital
tll I
4th
Spin
1lI5
"2 or-"2
1
1.16 The Heisenberg Uncerta inty Princip le The Heisenberg Uncertainty Principle arises from the dual nature (wave-particle) of ma tter. It states that there exists an inherent uncertainty in the p roduct of the position of a particle and its mom entum, and that this uncertainty is on the order of Planck's constant. !;.x!;.p
EB h
Here's the story with the Heisenberg uncertainty prJndplc: the mort' we know about
the momentum of any particle, the less we can know about the position. The amount of uncert(]inty is very small; on the order of Planck's constant (6.63 x lO-:\.l J s) . There arc other quant ities besides position and momentum to which the uncertainty principle applies, but position and momentum is the pair that you are likely to need to know for the MeAT.
1.17 Energy Level of Electrons The Aufbau principle states that with each new proton added to create a new element, a new electron is added as well. Nature typically prefers a lower energy state. All oth er things being equal, the lower the energy level of a system, the more stable the system. Thus, electrons look for an available orbital with the lowest energy state whenever they add to an atom. The orbital w ith the lowest energy will be contained in the subshell with th e lowest energy. To see w hy the energy level rises as the electrons move further from the nucleus, we must consider the attractive force between the negatively charged electrons and the positi vply ('h<'lrfjPo n llcll?l1 S. RI?(,<'Il1SI? the force is attractive, we must do work to separate them; we apply a force o ver a distance. \Alork is the transfer of energy into or out of a system. In this case, our system is the electron and the nucleus. We are doing work on the system, so we are transferring energy in to the system. This energy shows up as increased electrostatic potential energy. Like the en ergy between bonding atoms, the energy between the electron and the nucleus increases from a negative to zero as the electron moves to an infinite distance away from the nucleus.
Copyright © 2007 Examkrackers. Inc.
Let's f ummarize: The first quantu m number is the shell. It corresponds roughly to the energy level of the electrons within that shell. The second quantu m number is the subshell. It gives the shape. You need to recognize the shape of sand p orbitals. The third quantum number is the specific orbital withi n a subshell. The fourth quantum number distinguishes between two electrons in the same or-
between r and - I 1
17
bital; one is a +1/2 and the other is a - 1/2.
18
M C AT INORGAN IC CHEM ISTRY
If, for a given atom, we list the shells and the subshells in order from lowest to highest energy level, and we add a superscript to shm,v the number of electrons in each subshell, we have the electron configuratio n of that atom. (Electron configurations do not have to be from lowest to highest energy subshells, but they usually are.) Electron configura tions for several atoms are given below:
--..,;..;;.---....
Na => Is' 2s' 2p6 3s' Ar => Is' 2s' 2p' 35' 3p 6 Fe => I s' 252 2p6 3s 2 3p' 4s 2 3d' Br => 152 2s' 2p6 35' 3p6 45 2 3d'O4 p5
~ abbreviated electron configuration can be written by using the configuration of the next smallest noble gas as follows:
A simple trick to find the re lative energies of the subshells is to use th is table. The chart grows like stair-steps. An arrow is drawn down ea ch diagonal as shown . If we follow the arrows as they go down the step s, th ey show us the ord er of increas ing energy for the subshells. Notice that the energy levels are not
Na => [Ne]3s 1 Ar => [Ar] Fe => [Ar] 4s2 3d" (sometimes written [Ar] 3d 6 4s2) Br => [Ar] 45' 3d w 4p'
eu => [Ar] 45 ' 3d w
exactly in numeri cal orde r. For example,
Above are the electron configurations for atoms whose electrons are al1 at their lowest energy levels. This is called the gro und state. Electron configurations can also be given for ions and atoms with excited electrons:
the 4s subshell is at a lower energy level than the 3d .
Na+ => 152 282 2 p6 or [Ne]
Fe' + => [Ar] 3d 5 Br- => [Ar] 4s' 3d W 4 p6 or [Kr]
Be w ith an excited electron => 1s2 2S1 2pl Be certain that the total number of electrons in y our electron conflguration equa ls the total number of electrons in the a ton1. Notice that fo r the ions of the representative elements, elec tron configuration resembles that of a noble gas. The electron configurations of the transition metal ions are not the same as the nearest noble gas. As noted be fore, for transition meta ls, ions are formed by losing electrons from the subshel! w itll the highest principle quantum number first. Something else to know about transition metals is that their electron configurations aren' t ahvays that easy to predict. For ins tance, in the fourth period, the electron configurations of Cr and Cu have only one electron in the 4s orbital. This is because the 45 and 3d orbitals of these atoms a re degenerate (at the same energy level) . You don ' t have to memorize the electron configurJtion of each transition metaL Just be aware that they don't always foll ow the table given above, due to degenerate orbitals.
Also, notice tha t an electron Ca il momentarily (for a lUntter of microseconds) absorb energ y and jump to a h igher energy level creating an atOIn in an excited state.
Copyright © 2007 Examkrackers, Inc.
LECTURE 1: ATOMS , MOLECULES, AND QUANTUM M ECHANICS '
Like charges repel each otheL If we considered the energy of two particles with like charges, we would find that as the particles near each other, the mutual repulsion creates an increase in potential energy. This is the case when electrons approach each other. It explains why only two electrons can fit into one orbital. It also helps explain Hund's rule: electrons will not fill any orbital in the same subshell until all orbitals in that subshell contain at least one electron, and the unpaired electrons w ill have parallel spins. Hund's rule can be fep resented graphically as shown in the chart to the right. Electrons are represented by vertical arrows. Upward arrows represent electrons with positive spin, and downward arrows represent electrons with negative spin. When going from boron to carbon, the added electron has a choice of sharing the 2p, orbital or taking the 2py orbital for itself. Hund's rule says that the electron prefers to have its own orbital when such an orbital is available at the same energy level. Max Planck is con sidered the father of quantum mechanics. Planck's quantum theory demonstrates that electromagnetic energy is quantized (Le . it comes only in discrete units related to the wave frequency). In other words, if we transfer energy from one point to another via an electromagnetic wave, and we "wish to increase the amount of energy that we are transferring w ithout changing the frequencYf we can only change the energy in discrete increments given by:
M=hf 34
where h is Planck's constant (6.6 x 10-
t. Hydrogen
C, Helium
r'
Beryllium
H 15
2s
1~
2s
15
Boron
-,,
,
J s). ~ Nitrogen
.
<"", <,
;,
Oxygen
When an electron fa lls from a higher energy rung to a lower energy rung, energy is released from the atom in the form of a photon. The photon must have a frequency which corresponds to the change in energy of the electron as per tlE ~ hf Of course the reverse is also true. If a photon collides with an electron, it can only bump that electron to another energy level rung and not between energy level rungs. If the photon doesn't have enough energy to bump the electron to the next rung, the electron will not move from its present rung and the photon will be reflec ted away.
2p, 2p: 2p,
2p, 2py 2p,
1
H H 2s 15
""'1
The energy levels of electrons are quantized as well. The possible energy levels of an electron can be represented as an energy ladder. Each energy level is ana logous to a rung on a ladder or the spheres shown in the diagram on the next page. The electrons may occupy any rung, but do not occupy the space between rungs because this space represents forbidden energy levels.
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2s
15
Carbon
Einstein showed that if we considered light as a particle phenomenon with each particle as a photon, the energy of a single photon is given by the same equation: E pho'on = hf Neils Bohr applied the quantized energy theory to create an electron ladder model for hydrogen with each rung representing an allowed energy level for the electron. His model explained the line spectra for hydrogen but failed for atoms with more than one electron. Louis de Broglie then showed that electrons and other moving masses exhibit wave characteristics that follow the equation:
1
2p, 2p, 2p,
2p, 2p, 2p,
1
1~
H 2s
2p, 2p, 2p,
1.l Is
1~
2p, 2py 2p,
15
2s
H 1~ 2s 15 H H 15 2s
1 1
Is
1 1 1
2p, 2py 2p,
H1 1
2p, 2p, 2p,
19
20
MCAT Inorgan ic Chemistry
•
Photon released when electron faUs to lower energy shen Electron bumped up to its excited state after absorbing a photon
Q Nucleus
With the photoelectric effect Einstein demonstrated this one-to-one, photon to electron collision. He showed that the one-to-one collision proved that light was made up of particles. Einstein's reasoning went as follows: Light shining on a metal can cause the emission of electrons (sometimes called photoelectrons in the photoelectric effect). Since the energy of a wave is proportional to its intensity, we would expect that when the intensity of light shining on a metal is increased by increa~ing the nu,m ber of p h otons, the kinetic energy of an emitted electron would increase accordingly. This is not the case. Instead, the kinetic energy of the electrons increases only when intensity is increased by increasing the frequency of each photon. If the frequency is low enough, no electrons at all will be emitted regardless of the number of photons. This demonstrates that the electrons must be ejected by one-to-one photon-electron collisions and not by the combined energies of h'Vo or lTIOre photons. It also shows that if a single photon does not have sufficient energy, no electron w ill be emitted. The minimum amount of energy required to eject an electron is called the work function, <1>, of the metal. The kinetic energy of the ejected electron is given by the energy of the photon minus the work function (K.E. ~ Izf- <1».
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23. Hund's mle says that unpaired electrons in the same subshell:
Questions 17 through 24 are NOT based on a descriptive passage.
A. B.
C.
17. Whi ch of th e following species has an unpaired electron in its ground-state electronic configuration?
A. B. C. D.
D.
Ne Ca' Na-t 0 '-
24. Aluminum has only one oxidation state, while chromium has several. Whic h o f th e following is the best explanation for thi s diffe rence? A.
18. What is the electron co ntiguratio n of chromium? A. B. C. D.
have opposite spi ns. have parallel spins. occupy the same orbital. cannot exist.
B.
[Ar] 3,f' [Ar] 4s' 3d' fAr] 4s' 3d' I ArJ4s' 4p4
C. D.
Electron s in the d orbitals of Cr mayor used to form bonds. Electrons in the p orbitals of Cr mayor used to form bonds. Electrons in the d orbitals of Al may or used to form bonds. Electrons in the p orbitals o f AI may or used to fo rm bonds.
may not be may not be may not be may nol be
19. If the position of an electron is known with 100% cert ainty, which of the following cannot be determ ined for the sa me electron? A. B. C. '-D.
mass velocity charge spin quantum number
20. When an electron moves from a 2p LO a 3s orbital, the atom containing that electron :
A. B. C. D.
becomes a new isotope. becomes a new element. absorbs energy. releases energy.
21. Compared to a n electron with a pri ncipal qu antum number of 1, an electron w ith a principal quantum number of 2 will have:
A. B. C.
a hi gher energy.
a lower energy.
D.
a positive spin.
a negative sp in.
22. Which of the following best explain s wh y sulfur can make more bonds than oxygen ? A. B. C. D.
Sulfur is more electronegati ve than oxygen. Oxygen is more electron egative than sulfur. Sul fur has 3d orbitals not available to oxyge n. Sul fur has fewer valence,electrons.
Copyright © 2007 Examkrackers, Inc.
21
STOP.
Gases Kinetics and l
1
Chem ica I Equ iIibri um 2.1
Gases
A typical real gas is a loose collection of weakly attracted atoms or molecules moving rapidly in random directions. In a gas, the volume of the molecules accounts for about 0.1 percent of the total volume occupied by the gas. By comparison, molecules in a liquid account for about 70 percent of the total volume occupied by the liquid. O°C and 1 atm is called standard temperature and pressure (STP) . At STP, the average distance between gas molecules is about 35 A. This is small on a macroscopic scale, but typicaHy amounts to over a dozen molecular diameters on the microscopic scale. For instance, an oxygen molecule is about 2.5 A from end to end. Gas molecules move at tremendous speeds. At STP, the average speed of oxygen molecules is about 1,078 mph (481 m/s). The mean free path is the distance traveled by a gas molecule between collisions. The mean free path of oxygen at STP is about 1600 A. about 1 ten thousandth of a millimeter. Moving such small distances at such high speed results in one oxygen molecule making about 2,500,000,000 collisions with other lTIolecules each second. This explains why some chemical reactions can appear to occur instantaneously.
If a gas is a rruxture of compounds, then unlike liquids, the mixture will be homogeneous regardless of polarity differences. For instance, liquid gasoline and liquid weter don't mix because gasoline is nonpolar while water is polar; however, water and gasolme vapors form a~"t~. ,.... homogeneous mIXture. .~ t- ~y.1 ... h , ~~.
,~
C \. -:\ .. ,
S;",,-5.W
%:'.
Although polanty dIfferences do not cause ,', .~.;, '~:. gases to separate, when temperatures are low enough gravity causes denser gases to settle be- ~ . . i\~!- t;,~ . neath less dense gases. Cold CO2 gas from a fire extinguisher is heavier than air and smothers a fire by settling over the fire and displacing the air upward. Hot air rises because it is less dense than cold air.
ft. . }: .
Unlike liquids, all gases are miscible with each other; they mix regardless of
polarity differences. However, given time and low temperatures, heavier gases tend to settle below lighter gases.
24
M C AT INORGA NIC C HEM ISTRY
2,2
Kinetic Mo lecular Theory
To better understand the complex behavior of gases, scientists have theorized a model of an ideal gas, This model is called the kinetic molecular theory, In the kinetic molecular theory, an ideal gas lacks certain real gas characteristics. Ideal gas has the following four characteristics not shared by a real gas: Ratller than memorizing Charles', Boyle's, and Avogadro's laws, you should have a good understanding of the ideal gas law. The equations associated with these laws have been left out purposely, so that you won 't waste time memonzing them, The equations create confusion in the sense that they are true only within certain limitations. For instance, Charles' law requires constant pressure;
PV ~ oRT will solve any problem that the other three laws might solve, By the way, you won't need to memorize the name of each law either,
1.
Gas molecules have zero volume;
2.
Gas molecules exert no forces other than repulsive forces due to collisions;
3.
Gas molecules make completely elastic collisions;
4.
The average kinetic energy of gas molecules is directly proportional to the temperature of the gas.
Ideal gas obeys the ideal gas law :
pv = nRT where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles of gas, T is the temperature in Kelvin, and R is the universal gas constant (0.08206 L atm K-1 mol-lor 8.314 J K-1 mol-' ). Special cases of the ideal gas law include: Charles' law : The volume of a gas is proportional to temperature at constant pressure; Boyle's law: The volume of a gas is inversely proportional to pressure at constant temperature and; Avogadro's law: The volume of a gas is proportional to the number of moles at constant temperature and pressure.
Notice that there are four variables needed to define the state of a gas: P, V, n, and T. This can lead to confusion. For instance, it is possible to cool a gas by increasing the volume, even though the equation PV = nRT indicates that temperature is directly proportional to volume. When a gas expands in a container, the gas does work on its surrounding by pushing outward on the walls of the container. The force on the walls times the distance that the walls expand is the work done by the gas, This work represents a transfer of energy from the gas to the surroundings. The energy comes from the kinetic energy of the gas molecules, Both the temperature and the pressure are related to the kinetic energy of the molecules. The pressure is related to the kinetic energy per volume, and the temperature is related to the average kinetic energy per mole. When the gas expands, the pressure decreases due to both the loss in kinetic energy and the increase in volume. The temperature decreases due to the loss in kinetic energy via work done. So, if the volume of a gas "vere doubled, and the kinetic energy remained the same, the pressure would be reduced by a factor of two. But if no heat were added, the kinetic energy would be reduced. If the kinetic energy were reduced as the volume doubles, the pressure would be reduced by more than a factor of two. Thus the-temperature would also decrease to preserve the equality in PV = nRT. Notice that the ideal gas law does not change for different gases behaving ideally. (Of course not, it's written for an ideal gas.) This means that all gases (behaving ideally) will have the same volume, if they have the same temperature, pressure, and number of molecules. At STP one mole of any gas (behaving ideally) will occupy the standard molar volume of 22.4 liters. As well as pure gases, we can apply the kinetic molecular theory to mixtures of gases~ In a mixture of gases, each gas contributes to the pressure in the same proportion as it contributes to the number of molecules of the gas. This makes sense, given the kinetic molecular theory because molecules have no volume, no interactive ,forces other than collisions, and kinetic energy is conserved when they collide. Thus, each gas in a mixture essentially behaves as if it were in its container alone. The Co pyr ight © 2007 Examkrackers, Inc.
LECTURE
x2
x2
IV=J RT G
"
ill
'" Cl
"
"
.
"
..
. . .
.,
II
"
(I
2: GASES,
KINETICS , AND CHEMICAL EQUILIBRIUM •
x2
II =
nRT
+2
You must recognize the standard molar volume and
4 atm 11.2 L 2 moles GOG
the ideal gas law. For instance, 2 moles of gas at GOG occupying 11.2 liters will have a pressure of 4 atm. To get this result, we start with the standard molar volume, 22.4 L, at STP. First, we double the number of moles, so, according to the ideal gas law, the pressure doubles, Second, we halve the volume , so pressure doubles again.
. .. .... ... .. . " .. . . . . " " "" "
understand what it means. Learn to use this volume with
III
..
.,
1 atm 22 .4 L 1 mole GOC (273 K)
II
.
'" ""
'" '"" . ., " 2 atm 22.4 L 2 moles GOC
25
. .
....
...
G
amount of pressure contributed by any gas in a gaseous mixture is called the partial pressure of that gas. The partial pressure of a particular gas is the total pressure of the gaseous mixture times the mole fraction of the particular gas. The equation for the partial pressure is:
where P, is the partial pressure of gas 'a', and X, is the mole fraction of gas 'n', (The mole fraction is the number of moles of gas 'a' divided by the total number of moles of gas in the sample.)
Dalton's law states that the total pressure exerted by a gaseous mixture is the sum of the partial pressures of each of its gases.
From the ideal gas law, we can derive the following equation relating average translational kinetic energy and the temperature of a gas:
where the average translational kinetic energy is found from the root-mean-square (nns) velocity. (rms velocity is the square root of the average of the squares of the molecular velocities_ :rms velocity is slieht1y eTPatpf than thp avprage speed.) K.E. = 3/2 RT is valid for any fluid system, including liquids. Notice that the kinetic energy derived from this equation is the average kinetic energy for a mole of gas molecules and not the energy of every, or maybe even any, of the molecules. A gas molecule chosen at random may have almost any kinetic energy associated with it. Since the temperature dictates the average kinetic energy of the molecules in a gas, the gas molecules of each gas in any gaseous mixture must have the same average kinetic energy. For instance, the air we breathe is made up of approximately 21 % O 2 and 79% N2 by number. The molecules of 02 and N , in a sample of air have the same average kinetic energy. Howeve:(, since 02 and N 2 have different masses, their molecules have different rms velocities. By setting their kinetic energies equal to each Copyrig ht © 2007 Examk racke rs, Inc.
Dalton's law is a good way to understand an ideal gas. Each gas behaves like it is in the container by itself so all the partial pressures added together equal the total pressure,
26
MCAT INORGANIC CHEMISTRY
In a sample of gas, the kinetic energy of the molecules will vary from molecule to molecule, but there will be an average kinetic energy of the molecules that is proportional to the temperature and independent of the type of gas,
other, we can. derive a relationship between their rms velocities. This relationship, which gives the ratio of the rms velocities of two gases in a homogeneous mixture, is called Graham's law :
Notice that the subscripts are reversed from one side of the equation to the other. Graham's law also tells us that the average speed of the molecules of a pure gas is inversely proportional to the square root of the mass of the gas molecuJes.
Effusion
Interes tingly, Graham's law gives information about the rates of two types of gaseous spreading: effusion and diffusion. Effusion is the spreading of a gas from high pressure to very low pressure through a 'pinhole'. (A 'pinhole' is defined as an opening much smaller than the average distance between the gas molecules.) Because molecules of a gas with higher rms ve10city will experience more collisions with the walls of a container, the rate at which molecules from such a gas find the pinhole and go through is likely to be greater. In fact, Graham's law predicts the comparative rates of effusion for two gases at the same temperature. The ratio of the rates of effusion of two gases is equal to the inverse of the ratio of the square roots of their molecular weights and equal to the ratio of their rms velocities.
effusion rate 1 effusion rate 2
=
Diffusion is the spreading of one gas into another gas or into empty space . The ratio of the diffusion rates of two gases (acting ideally) is approximated by Graham's law. The diffusion rate is much slower than the rms velocity of the molecules because gas molecules collide with each other and with molecules of other gases as they diffuse. For example, if we wet two cotton balls, one wifh aqueous NH3 and fhe other with aqueous HCl, and place them into opposite ends of a glass tube, gaseous NH3 and HCl will diffuse toward each other through the air inside the tube. Where they meet, they will react to form NH,Cl, which will precipitate as a white solid. Graham's law accurately predicts that NB, will travel1.S times further than Hel. However, any particular molecule is likely to follow a very crooked path similar to those shown in the diagram. Recall that the mean free path of a gas 'molecule is on the order of a few hundred nanometers.
Cotton ball NH,
Cotton ball
NH,Cl
_--I",~l1:n
Diffusion rate of NH, Diffusion rate of BCl
__ JM r;-;-
HC ,
~MN"
_
-
r
6 .5 = 1.5 17
Copyright © 2007 Examkrackers, Inc
LECTURE
2.3
2:
G ASES, KINETICS, AND C HEMICAL EQUILIBRIUM .
27
Real Gases
Real gases devia te from ideal behavior w h en their molecules are close togeth er. When molecules are close together, the volume of the molecules become significant compa red to the volume around the m olecules. Also, as can be seen by Coulomb's law (F = kqq I r), w hen m olecules are close together, the electrostatic forces increase and become significant. High pressure pushes gas molecules toge ther causing devia tions from ideal behavior. Low temperatures cause gas molecules to settle close together also resulting in deviations from id eal b ehavior. Gases generally deviate from ideal behavior at pressures above ten a t.m ospher~s and temperatures near their boiling points. Substances that we typically think of as gases closely approximate ideal behavior.
,
. . . • " . • . • • . '" • .. ".. • • ,,~ \Ii
Ideal Behavior
Once aga in, be careful about the confusion with PV = nRT. It seems like increasing pressure
also
increases
temperature, so how can ideal behavior deviations occur with either an increase in
pressure or a decrease in temperature? The answer is that to create deviabons in ideal behavior, we just want to move the molecules closer together; in other words, decrease the volume. We
Call
decreClse
the volume by squeezing the molecules together with high pressure, or by lowering the temperature and letting the molecules settle close together. From PV = nRT, we see that volume decreases with either increasing pressure or decreasing
temperature .
Nonideal Behavior
Yo u should be aware of h ow real gases d evia te from ideal beh avior. Van der Waals equation: [P '+- a (Il/ V)' ]( V - lIb) = IlRT]
PV
RT
Positive dev iation due mainly to molecular volu
N, CH,
H,
approximates the real pressure and real volume of a gas~ where nand b are constan ts for specific gases. The variable b is a measure of the actual volume occupied by a mole of gas. The va riable a reflects the strength of intermolecular a ttraction s. The values of a and b generally increase w ith the molecular m ass and molecular complexity of a gas .
You d o not n eed to know van der Waals equation for the MeAT It is more importan t tha t you have a quaEtative W1dersta nding of real gas deviations from ideal behavior. Firs t, sin ce m olecules of a real gas do ha ve volum e, their vol ume must be added to the ideal volum e. Thus:
V r•• ] >
Videa]
where Video' is calculated from P V = nRT. Second, molecules in a real gas d o exhibit forces on each other, and those fo rces are a ttractivc wh en th e molecules are fa r apa rt. In a gas, repulSive fo rces are only significant during mo lecular collisions or near collisions. Since the predominant interm olecular forces in a gas are attrac tive, gas molecules are p ulled inw ard towa rd the center of the gas, and slow b efore colliding with contai ner w alls. H aving been slightly slowed, they strike the container w all with less force tha n predicted by the kinetic m olecular theory. Thus a real gas exerts less pressu re than predicted by the ideal gas law.
w here
P ideal
is calculated from
Copyright © 2007 h
Py = nRT.
1.0 ~==.L..--:".L----;;
p
Native deviation ue mainly to attractive intermolecular forces .
From PV = nRT, we expect PVjRT to equal one for one mole of ideal ges at any temperature and pressure. Since volume deviates positively from ideal behavior and pressure deviates negatively, if PVj RT is greater than one for one mole of gas, tllen the deviation due to molecular volume must be greater than the deviation due to the intermolecular forces. If PVjRT is less than one for one mole of gas, then the deviation due to intermolecular forces must be greate r than the deviation due to molecular volume.
30. Equal molar quantities of oxygen and hydrogen gas were placed in container A under high pressure. A small portion of the mixture was allowed to effuse for a very short time into the vacuum in container B. Which of the follow ing is true concerning partial pressures of the gases at the end of the experiment?
Questions 25 through 32 are NOT based on a descriptive passage.
25. A 13 gram gaseous sample of an unknown hydrocarbon occupies a volume of 11.2 L at STP. What is the hydrocarbon?
A. B.
c.
D.
A.
CH
C2H..
B.
C2H1
C3 H3 C.
26. If the density of a gas is given as p which of the following expressions represents the molecular weight of the gas?
A. B.
c. D.
D.
PplRT pRTIP nNTlPp Pp/NT
The partial pressure of hydrogen in container A is approximately four times as great as the partial pressure of oxygen in container A. The partial pressure of oxygen in container A is approximately four times as great as the partial pressure of hydrogen in container A . The partial pressure of hydrogen in container B is approximately four t.imes as great as the partial pressure of oxygen in container B. The partial pressure of oxygen in container B is approximately four times as great as the partial pressure of hydrogen in container B.
31. HC! gas and NH3 gas form NH 4C! precipitate according to the following equation:
27. Ammon ia bums in air to form nitrogen dioxide and water. 4NH3(g) + 70,(g) --> 4NO,(g) + 6H,o(l)
When colton balls are moistened with the aqueous solutions of the respective gases and inserted into either end of a glass tube, the gases diffuse toward the middle of the glass tube to form the precipitate. If a 10 em glass tube is used, at what distance x will the precipitate form?
If 8 moles of NH J are reacted with 14 moles of 0 2 in a rigid container with an initial pressure of 11 atm, what is the partial pressure of N0 2 in the container when the reaction runs to completion? (Assume constant temperature.)
A. B. C.
D.
4 atm 6 atm 11 atm 12 atm
~x --, :l1li
/[:\
28. At moderately high pressures, the PVIRT ratio for one mole of methane gas is less than one. The most likely reason for th is is:
A. B. C.
D.
Cotton ball withNH,
Methane gas behaves ideally at moderate pressures. The temperature must be very low. At such pressures, molecular volume causes a greater deviation to ideal behavior than intermolecular forces for methane. At such pressures, intermolecular forces cause a greater deviation to ideal behavior than molecular volLime for methane.
C. D.
4.0 em
A. B.
decreases. increases. remains constant. The temperature change depends upon the amount of force used .
Copyright © 2007 Exam krackers, Inc
1.5 em
2.5 em
~~
7 NH,CI precipitate
If\, \
Cotton ball with HCI
3.0 em
32. At STP, olle li ler o[ which of lhe following gases contains the most molecules?
29. A force is applied to a container of gas reducing its volume hy half. The temperature of the gas: A. B. C. D.
A. B.
10 em
H2
C.
He N2
D.
Each gas contains the same number of molecules at
STP.
28
STOP.
LECTU RE
2.4
2:
GASES, KINETICS, AND C HEMICAL EQU ILIBR IUM .
Chemical Kinetics
Chemical kinetics is the study of reaction mechanisms and rates. As of yet, there are no unifying principles of kinetics, which means kinetics is a complicated field with many opposing theories as to how reactions proceed. Additionally, the mathematics of kinetics is complicated and well beyond the scope of MCAT. MCAT will address kinetics only in its simplest form. Keep in mind that kinetics deals with the rate of a reaction typically as it moves toward equilibrium, while thermodynamics deals with the balance of reactants and products after they have achieved equilibrium. Kinetics tells us how fast equilibrium is achieved, while thermodynamics tells us what equilibrium looks like. The two disciplines are intricately related, but they should not be confused.
2.5
The Col lision Theory
The collision model of reactions provides an enlightening method for visualizing chemical reactions. In order for a chemical reaction to occur, the reacting molecules must collide. However, for most reactions, the.rate of a given reaction is found to be much lower than the frequency of collisions. This indicates that most collisions do not result in a reaction. There are two requirements for a given collision to create new molecules in a reaction. First, the relative kinetic energies of the colliding molecules must reach a threshold energy called the activation energy. The relative kinetic energy refers to the kinetic energy due to relative velocity only. In other words, velocity in a direchan away from another molecule decreases the relative kinetic energy of a collision. Second, the colliding molecules must have the proper spatial orientation.
Reaction
Result
collision
No Reaction
Result
Copyright © 2007 Exomkracke rs, Inc.
29
30
MCAT INORGAN IC C HEMISTRY
The product of the collision frequency z, the fraction of collisions having the effective spatial orientations p (called the steric jactor), and the fraction of collisions having sufficient relative energy e-Ca;RT (V\There Ea is the activation energy) gives the rate constant k of a reaction. This relationship is called the Arrhenius eqllatioll l k :::: zpe-EllfRT Does the rate of an exothermic reaction
increase with temperature? You bet it does. Remember, do not confuse kinetics with thermodynamics. Increasing the rate is not necessari ly a statement about the equilibrium. It simply means that equilibrium is achieved more quickly.
Reaction at lower tempe rature / Reaction at / higher
(often written as k:::: Ae-E1J!RT). The value of the rate constant depends upon pressure, catalysts, and temperature. Pressure dependence is typically negligible. Catalysts will be discussed later in this Lecture. The temperature dependence is seen in the Arrhenius equation. The rate constant generally doubles to triples with each increase of lO'C. Since the effect of temperature on E, is negligible for most reactions, the fraction of collisions that have at least the activation energy (e-E'iRT) increases with the temperature. This, in turn, indicates that the rate constant k increases with increasing temperature for nearly all reactions. As demonstrated by the rate law (discussed later in this lecture), the rate constant is directly proportional to the rate of a reactioll. The rate of a reaction increases with temperature mainly bec'.luse more collisions with sufficient relative kinetic energy occur each second. The temperature dependence of rate is demonstrated by the graph to the left which compares two san1ples of identical gaseous mixtures reacting at different temperatures. The area under any section of the curve represents the relative number of collisions in that energy range. Notice the area to the right of the activation energy is greater at higher temperature. At higher temperature there are more collisions with enough energy to create a reaction. Notice that the energy of activation does not change with telnperature. (Actually, E, is temperature dependent, but for most reactions the dependence is extremely small, and, for the MCAT, E, should be considered independent of temperature.)
tcmpcl:<:!turc
E, Energy of collisions
2 .6
Equations for Reaction Rates
For MCAT purposes we will consider only reactions occurring in gases and ideally dilute liquids at constant temperature. The rate of a reaction tells us how quickly the concentration of a reactant or product is changing. Rates are Inost often given in terms of molarity per second (mol V I S-I). Factors affecting the rate of a reaction are temperature, pressure, and concentration of certain substances in the reacting system; however, pressure effects on reaction rates are usually small enough to be ignored. The rate of a reaction can be viewed in terms of the change in concentration of any one of the reacting participants. Consider the following elementary reaction where the lower case letters are the stoichiometric coefficients of the balanced equation:
aA + bB --; cC + dD In an elementary reaction , the coefficient tells you how many molecules participate in a reaction producing coilision .
An elementary reaction is. a reaction that occurs in a single step. The stoichiometric coefficients of an elementary equation give the moleculnrity of the reaction. The molecularity is the number of molecules colliding at one time to make a reaction. There are three possible molecularities: unimolecular, bin101ecular, and termolecular. Since the reaction above is elementary, its Inolecularity is given by a + b. Chemical equations often represent multistep reactions called complex or composite reactions. There is no way to distinguish an elementary reaction from a complex reaction by inspection of the chemical equation. On the MCAT, the only way to know if a reaction is elementary is if you are told that it is elementary.
Copy right © 2007 Examk racke(s, Inc.
LECTURE
2: GASES,
KINETICS, AND CHEMICAL EQUILIBRIUM
The average reaction rate for any brief time interval t during the above reaction is: _~ ;:;[B] b
rate
.
31
Don't confuse the rate constant with the rate of the reaction . They are proportional. but they are NOT identical.
1
means "change in". The negatjve signs indicate that the reactant concentrations are decreasing as the reaction nl0ves forward. Although the rate equation above is strictly correct only for an elementary reaction! it is a good approximation for a lnldtistep reaction if the concentration of any internlediates is kept low. Intermediates are species that are products of one reaction and reactants of a later reaction in a reaction chain. The concentration of intennediates is often very low because they are often unstable and react as quickly as they are formed.
1/1
Chelnical reactions are reversible; as the products are formed, products begin to react to form the reactants. This reverse reaction complicates the study of kinetics. For the time being! we will consider only the forward reactions. If we consider only the forward reaction, we can write a rate law for the reaction above as: ratejorward
TIle important thing to know here is the form of the rate law.
= k}Al a[Bl ~
where kf is the rate constant for the forward reaction. a and ~ are the order of each respective reactant and the sum CJ. + ~ is the overall order of the reaction. If the reaction is elementary, ex, = a and ~ = b.
Determining the Rate Law by Experiment
2.7
Both the order of the reactants and the value of the rate constant must be determined through experiment. Finding the rate law on the MCAT is a relatively simple matter. Consider the hypothetical reaction: 2A+ B + C ->20
In this case, we will assume that no reverse reaction occurs. VVe are given the following table with experimental data:
Trial
Measured Initial Initial Initial Concentration Concentration Con centration Initi al Rate (mollL s) of B (molfL) of C (moIlL) of A (mollL)
1
0.1
0.1
0.1
S.O x 10-4
2
0.2
0.1
0.1
1.6 x 10-3
3
0.2
0.2
0.1
6.4 x 10-3
4
0.1
0.1
0.2
8.0 x 10-4
We can find the order of each reactant by comparing the rates between two Trials in "vhich only the concentration of one of the reactants is changed. For instance, comparing Trial 1 to Trial 2, the initial concentration of A is doubled and the concentrations of Band C remain the same. The reaction rate also doubles. Thus the rate of this reaction is directly proportional to the concentration of A. InJhe rate law, [AJ receives an exponent of 1, and the reaction is considered first order with respect to reactant A. Comparing Trials 2 and 3 we find that when only the concentration of B is doubled, the reaction rate is quadrupled. This indicates that the rate is proportional to the square of the concentration of B. [BJ receives a 2 for its exponent in the rate law and the reaction is second order with respect to B. Comparing Trials 1 and
Copyright C9 2007 Ex.]rnkrackers . Inc.
For MeAT, you -must be able to derive the rate law from a table of trials as done in this section . When given a rate law, you must be able to predict what changing the concentration of a reactant will do to the rate .
32
MCAT INORGANIC CHEMISTRY
The Rate Law will be determined by experiment!
4, the concentration of C is doubled, but there is no resulting change in the rate. The rate is independent of the concentration of C, so [C] receives an exponent of zero in the rate law. The reaction is zero order with respect to C. The complete rate law for our hypothetical reaction is:
also written:
By adding the exponents we find that the reaction overall is third order. Notice that the coefficients in the balanced chemical equation were not used to figure out the order.
Once we have derived the rate law from the experimental data, we can plug in the rate and concentrations from any of the experiments into the rate law, and solve for the rate constant k. Notice that the rate may be increased by increasing the concentration of the reactants. If we consider the collision model, this makes sense. The greater the concentration of a species, the more likely are collisions.
2.8
Recognizing Reaction Orders
The different orders of reactions have recognizably different characteristics. Keeping in mind that we are assuming that there is no reverse reaction taking place, we can compare graphs of the different orders of reaction for a single reactant. Plotting [A] with respect to time t for a zeroth order reaction results in a straight line with a slope -kt In a first order reaction of the form:
A --> products where there is no reverse reaction, [A] decreases exponentially. A graph of a first order reaction comparing In[A] with time t gives a straight line with a slope of -kt The same first order reaction has a constant half life that is independent of the concentration of A. The JTIolecularity of a first order reaction implies that no collision takes place since there is only one molecule reacting. However, current theory requires a collision with any other molecule, boosting the reactant to a higher energy state resulting in a reaction. Although this is a two step process, it exhibits first order kinetics (except under low pressure). This is referred to as pseudo-first order ki-
netics. For a graph of an irreversible second order reaction with a single reactant of the form: 2A --> products
rate
=
kJAl'
plotting 1/[A] gives a straight line with a slope of kt The half life of this type of second order reaction is dependent upon the concentration of A. It has the interesting characteristic that each consecutive half-life is twice as long as the last. For instance, the time required to reduce the concentration of A from lOO% to 50% is half as long as the time required to reduce the concentration from 50% to 25%. The second order reaction of the form: A + B --> products
!
rate
=
kJAHBl
does not reveal the same graph, and does not have an easily predictable half life. Copyright © 2007 Examkrackers, In c.
LECTURE
2:
GASES, KINETICS, AND CHEMICAL EQU ILIBRIUM .
33
For a graph of an irreversible third order reaction with a single reactant of the form: 3A
--7
rate = kJAl'
products
plotting 1/2[Al' gives a straight line with a slope of kf' To avoid the complications of reverse reactions, the technique of initial rates is often employed. In the initial mOlnents of a reaction starting with all reactants and no products, the rate of the reverse reaction is zero. The rate law in section 2-7 was determined with initial rates.
l~ I ~ I ~
::;.
2.9
:?
,
~
if
c
I-
t-
Zero Order
First Order
"",0 e
.sIr : S\O e
t -
t -
Second Order
Third Order
Rates of Reversible Reactions
Any complex reaction can be separated into elementary steps. The rate of the slowest elementary step determines the rate of the overall reaction and is called the rate determining step. If the slow step is the first step, the rate law can be derived directly from this step and no other. If the slow step is other than the first step, the slow step is still the rate determining step, but steps prior to the slow step will contribute to the rate law. Steps after the slow step will make no contribution to the rate law. Consider the following reaction: N0 2 (g) + CO(g)
--7
NO(g) + CO,(g)
This reaction has two elementary steps: 1. 2.
N02(g) + N02(g) N03 (g) + CO(g)
--7
--7
N03(g) + NO(g)
NO, (g) + CO2 (g)
,1,,,,.
"Cf'
fa st s t ep
Notice that if we add these two equations together, we arrive at the original equation. Elementary steps must add to give the complex reaction. Since the first step is the slow step, the rate law for the overall reaction is given by this step and is:
rate = k, [N0 21' We know that the exponent for [NO, l is 2 because we derived the rate law from an elementary equation, and in the elementary eguation two NO, molecules collide to create a reaction. Automatically using the coefficient of the balanced equation for the exponent in the rate law works only if the equation is elementary. Don't forget, the rate law above assumes negligible contribution from the reverse reaction, and it also assum,e s a sufficient concentration of CO for the fast step to occur. When the first step of a reaction series is the fast step, things can be a little tricky. If the first reaction is the fast reaction, the rate of the overall reaction is still equal to the rate of the slowest step. However, now one of the products of the fast step is a reactant in the slow step. Such a species is called an intermediate. The concentration of the intermediate is tricky to predict. An intermediate is usually not stable. If we Copyright © 2007 Examkrackers, Inc.
Remember
that
determines the rate .
the
slow
step
34
MCAT INORGANIC CHEMISTRY
assume that the fast reaction reaches equilibrium very quicklYf the concentration of the intennediate remains at its equilibrium concentration. We can use the equilibrium concentration of the intermediate in predicting the slow step. For instance: 2N O(g) + Br 2 (g) --> 2N OBr(g) This reaction has two elementary steps: 1.
NO(g) + Br2 (g) --> NOBr,(g)
2.
NOBr,(g) + NO(g) --> 2NOBr(g)
fast step
The rate law for this reaction is:
rate = k,[NOBr,][N O] Keep It Simple: When a fast step precedes the slow step, the slow step still determines the rate, but the concentration of one or more of the reactants in the slow step will be determined by the fast step. In such a case , assume that the fast step reaches and maintains equilibrium throughout the reaction, and use the equilibrium
However, the concentration of NOByz depends upon the first step. If we assume that the first step reaches equilibrium very quickly, the concentration of NOBr, can be written in terms of the equilibrium constant K, for step 1, [NOBr,] = KJNO][Br,]. Alternatively, since the first step is considered to be in equilibrium, we can set the forward reaction rate, k1[NO][Br,1, equal to the reverse rate, k_1[NOBr,1, and solve for [NOBr,]. [NOBr,] = k/k_l [NO][Br,]. The resulting rate law is:
concentration of any intermediates.
I111s method is called the equilibrium approximation, which assumes that all steps prior to the rate limiting step are in equilibrium. The equilibrium approximation requires that the slow step be significantly slower than the fast steps. Since k,k,/ k_l is a constant, it is usually rep1aced by a single constant, kobse rvoo .
rate = k, k,lk_l [N Ol'[Br2 ]
If there is not a step that is significantly slower than the others, we can use the steady state approximation. In the steady state approximation, the concentration of the intermediate is considered to be small and hardly changing. This leads to the same result as the equilibrium approximation.
2.10 Cata lysis
Cat.l1)'ze d reaction
Jn ll'J"I1wd iille
RI?
A catalyst creates a new reaction pathway which typically includes an intermediate .
A catalyst is a substance that increases the rate of a reaction without being consumed or permanently altered. Catalysts are capable of enhancing product selectivities and reducing energy consumption. A catalyst may Im,ver the activat.ion energy, Ea, or increase the steric factor (p from the Arrhenius equation). Most catalysts work by lowering the activation energy. The reaction rate depends exponentially on the activation energy. \I\Ihen the activation energy is lowered, more collisions have sufficient relative kinetic energy to create a reaction. This leads to more reactions and an increase in the overall reaction rate. This effect is shm·vn in the energy vs. reaction coordinate diagram at the top of the next page. A catalyst works by providing an alternative reaction mechanism that cOlnpetes with the uncatalyzed mechanism. A catalyst cailllot alter the equilibrium constant of a reaction, so it must increase the rate of both the forward and the reverse reaction. Although a catalyst cannot change the equilibrium constant, it can, in some cases, change the composition of tlw mixture at equilibrium; however, the effect is usually very small and should be ignored for t11e MCAT. For the MCAT, catalysts do not change the equilibrium composition. A catalyst can be either heterogeneous or homogeneous. A heterogeneous catalyst is in a different phase than the reactants and products. Heterogeneous catalysts are usually solids while the reactants and products are liquids or gases. A reactant may physically adsorb (via van der Waals forces) or, more oiten, chemically adsorb Copyright © 2007 Examkrackers, Inc.
LECTURE 2 : GASES, KINETICS, AND CHEMICAL EQUILIBRIUM • 35
(usually via covalent bonds) to the surface of a solid catalyst. (Adsorption is the binding of molecules to a surface as opposed to absorption which refers to the uptake of molecules into an interior.) Molecules bind to a metal surface because, unlike metal atoms in the interior, Inetal atoms at the surface have unfulfilled valence requirements. The binding is almost aJways exothermic and the rate of catalysis depends upon the strength of the bond between the reactant and the catalyst. If bonds are too weak, not enough adsorption occurs; if bond s are too strong, too m uch energy is req uired to remove the reactant. Depending on the reaction, reactant molecules may bind to any of the atoms at the surface, or they may bind only to surface imperfections. Once adsorbed, molecules migrate from one adsorption site to the next. The more binding that occurs, the greater the reaction rate. Thus, reaction rates can be enhanced by increaSing the surface area of a catalyst. A homogeneous catalyst is in the same phase as the reactants and products, usually in the gas or liquid phase. Aqueous acid or base solutions often act as homogeneous catalysts. Some reactions exhibit autocatalysis by generating the catalyst as a product. The acid catalyzed hydrolysis of an ester is an examp le of autocatalysis, where the carboxylic acid product acts as a catalyst to the
Reaction Coordinate
With catalyst Without
reaction.
En In the lab, concentrations of catalysts are usually small compared to the concentration of the reactants and products. Tn such cases, increasing the concentration of the catalyst increases the rate of the reaction. If the concentration of the catalyst is large compared to the reac tants and products, the rate changes little or not at all with the catalyst concentration. Since catalysts alter reaction mechanisms, reactions with ca tal ys ts require separate rate constants. Remember, a catalyst doesn' t preve nt the original reaction from proceeding, so the total ra te is given by the sum of the rates for both reactions. For instance, a first order uncatalyzed reaction ma y follow the ra te law : rate = kolA]
When the same reac tion is catalyzed by acid, the new rate law wo uld be: rate = kolA] + k".[H'][A]
TypicaJly, the rate of the original reaction is negligible compared to the catalyzed rate.
Almost every chem ical reac tion in the h uman body is quickened by a protein catalyst ca lled an enzyme. Enzymes are far more effective than inorganiC catalysts. The number of reactions o,ccurring at one active site on one enzynle is typically 1,000 per second and can be tens of thousands of times greater for the fastest enzymatic reactions. This n umber is called the turnover I1I.l111ber.
2.1 1 Effects of Solvent on Rate Roughl y speaking, liquid molecules make around 100 times more collisions per second than gas molecules because liquid molecules are much closer to each other. However, most of tl,e collisions in a liquid are w ith the solvent resulting in no reaction. The rate constant in a liquid is a functi on of the solvent as well as the temperature. The reactant in a liquid is solvated. These solvent-reactant bonds m ust be broken before a reaction can take place. In addition, the bonds may stabilize an intermediate. The degree of solvation affects k. Solvents can electrically insulate reactants reducing the electrostatic forces between them. The dielectric of the solvent affects k. Solvent viscosity can affect k as per the 'cage effect' described below. Copyright © 2001 Examkrackers, Inc .
(catalyzed)
E~ (uncatalyzed )
Energy of collisions
Remember that a catalyst increases the rate of a reaction by lowering the activation energy of that reaction. A catalyst doesn't change equilibrium and is not used up.
36
MCAT INORGANIC CHEMISTRY
Reactants in a liquid can be trapped in a cage of solvent molecules. They rattle around in such a cage at tremendous rates making hundreds of collisions before squeezing between solvent molecules and into a new solvent cage. If they are trapped in the cage with another reactant, many of their collisions are with the other reactant and a reaction is likely to occur, but if there is not another reactant in the solvent cage, they cannot react until they escape the cage. The net result is that reactants in a liquid make an approximately equal number of collisions with other reactants as they would in a gas with equal concentrations; collisions in a liquid occur at about the same rate as in a gas.
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37. As tem perat ure is increased in an exothermi c gaseous reaclion, all of the foll owin g increase EXCEPT:
Questions 33 through 40 are NOT based on a descriptive passage.
A.
B.
C.
33. Which of the fol lowi ng changes to a reac tion wil! always increase the rate constant for that reaction? A. B. C. D.
D.
decreas ing the temperature increasing the temperature increas in g the concentration of the reactants increasing the concentration of the catalyst
reac Li on rate. rate constant. activation energy. rms molecular velocity.
38. A contain er holds a pure sa mp le of compoun d X shown at t = 0 min . Co mpo und X undergoes a first order reaction to form com pound Y. The same comainer is show n at t :::: 15 min . (Compound X is s hown as white circles and co mpound Y is shown as black circles.)
a
34. All of the fo llowing may be true concernin g catalysts and th e reaction which they catalyze EXCEPT: A. B. C. D.
o
15
C9
Cata lysts are not used up by tll C reaction. Catal ysts lower the energy of activation. Catalyst'i increase the ra te of the reverse I~acti o n . Catal ysts shift th e reaction equ ilibrium to the right. If the rate of the reverse reacti on is neg li gible, which of the follow ing mig ht represe nt the contai ner at 1 :::: 30 min?
35. The table below shows 3 trials where the ini ti al rate wa s measured ror the reac ti on:
c.
A.
30
2A + B --> C
CD
Whi ch of the foll owing express ions is the correct rate law for the reaction?
~... .
Molarity of A Molarity of B
T
1 2
3 A. B.
C. D.
rate = rate = rate = rate
0.05
0.1
0.1
0.05
••" 0 •
o •• ~.
Initial Rate
0.05
0.05
·::~:o.:
5 x 10-3
B.
30
CD
3
5 x 10-
D.
::..... ...J':: ..... . . ...:..
1 x 10-'
0.1f AJ [AI
30
CD I:: .·..-.1 I-: .>;1
...,..:.J.: ....
~
IA][B]
=rAf[B]
39. Th e conversion of c is-2-bu tene {Q tran s-2-butene takes place in a reaction that is fi rst-order with respect to cis-lbute ne. The conversio n was obser ved in two separate trial s und er identical condit ions except th at in the second trial, the co ncentrati o n of cis-2-butene was doubl ed. In the second trial:
36. The reacti on below proceeds via th e two step mec hani sm as shown.
Step I:
2NO, + F, --> 2NO, F NO, + F, --> N0 2F + F
Step 2:
NO, + F --> NO,F
Overall Reaction:
A. B. C. D.
X is the rate of step 1, ami Y is the rate of step 2. If step 1 is
the the the the
reaction rate will be halved. reactio n rate wi ll be doubled. rate co nstant will be halved. rate constan t wil l be do ub led .
much slower than step 2, then the rate of the overall reaction can be rep resented by:
A.
X
B. C. D.
X+ Y X- Y
40. When a rad ioactive isotope undergoes nuclear decay, the conce ntrat ion of the isoto pe decreases ex ponenti a ll y with a constant half-life. It can be determined from th is that radioacti ve decay is a:
Y
Copyriqht © 2007 Exam krackers. Inc.
A. B. C. D.
37
zeroth order reaction. first ordcr reactio n. seco nd ordcr reactio n. third order reaction .
STOP.
38
MCAT INORGANIC CHEMISTIN
2.1 2 Equilibrium Notice that by the rate definition:
rate
1
illAI
a
t
= - ---
the rate at equilibrium is zero. Understand that zero is the net reaction rate, but there
is a forward and a reverse reaction rate at equilibnum . Equilibrium is a dynami c
process.
Chemical reactions are reversible. As reactants are converted to products, the concentration of the reactants decreases, and the concentration of the products increases. Since rates are related to concentrations, the rate of the forward reaction begins to slow, and the rate of the reverse reaction quickens as a reaction proceeds. Eventually, the hvo rates become equa1. This condition, where the forward reaction rate equals the reverse reaction rate, is called chemical equilibrium. At chemical equilibrium, there is no change in the concentration of the products or reactants. Equilibrium will be reached from either direction, beginning with predominantly reactants or predominantly products. Equilibrium is the point of greatest entropy. Consider the hypothetical first order elementary reaction: A-->B
~
A---)B
~i~ ",W"d <= .9 reaction
1-1
~
Equilibrium
Forwar~/ate == Reverse rate
;::::::I /
II
B---)A Reverse reaction
Notice from the diagram on the left that the forward rate changes faster than the reverse rate as the reaction proceeds. This indicates that the forward rate constant is greater than the reverse rate constant. The rate law for the forward and reverse reactions are rate = kJAJ and rate = k,[BJ respectively. The rate of the forward reaction is directly proportional to [AJ. From the diagram, the rate of the forward reaction is reduced by more than half, so [AJ is also reduced by more than half. For every molecule of A lost, a molecule of B is created, so at equilibrium [BJ must be greater than [AJ. Setting the rates equal, kJAJ = k,[B], we see that kfmust be greater than k, if [BJ is greater than [AJ at equilibrium.
Time For a homogeneous reaction, where all species are in the same phase, there will always be some of each species present at equilibrium; however, in some cases, the rate constant for the forward reaction is so much greater than the rate constant for the reverse reaction that for all practical purposes, the reaction runs to completion. Alternatively, if a product is continually removed as the reaction proceeds (perhaps in the form of a gas leaving an aqueous solution), the reaction can nffi to cOlTIpletion. Consider the hypothetical elementary reaction: aA + bB --> cC + dO. Since the reac- . tion is elementary, we can use the stoichiometric coefficients for the forward and reverse rate laws: ratejol"ward ::;
kJAY[B j1'
ratcrevcrsc ::;
kJCY[D]d
(WARNING: For the MCAT, never use the coefficients as the exponents in the rate law. We do it here because we are told that the reaction is elementary.) Since equilibrium occurs when these two rates are equal, then, for equilibriull1 conditions only, we set them equal to each other as shown below: kj [A] "[B ]" = k, [enD]"
With (] little olgebrak manipulation
WE'
have:
k(
[enD]"
k,
[AnB]'
Since both k's are constants, we can replace them ,,"lith a new constant called the equilibrium constant K. (2nd WARNING: This simple relationship between K equilibrium and k rate is only true for elementary equations.) The relationship between a chemical equation and the equilibrium constant is called the law of m ass action and is written as follows:
Copyright © 2007 Exa rn krackers, lnc.
LECTURE
[A]a[B] b
KINETICS, AND C HEMICAL EQUILIBRIUM .
39
Productscoefficients
[CJ<[D]d
K =
2 : GASES,
=
Reactantscoefficients
The value of K h as no d imensions beca use the concentration s are ac tually approxima tions for a dimensionless quanti ty called an activity. The law of mass action is good for all chemical equations, including non-elementary equations. In other wo rds, fo r equilibrium constants, use the ch emical equation coefficients as the exp on ents of the con centrations regardless of molecula rit )'. N otice tha t the equilibrium constant is a capital K and the rate constant is represented by lowercase k. Also notice that the equilibrium cons tant fo r the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. This is true regardless of wh ether or not the reaction is elementary. Following this same line of reasoning w ill d emons trate that the equilibrium cons tant fo r a series of reac tions is equal to the p roduct of the equilibrilun constants for each of its elementar y s tep s. Since the ra te constant depends upon temperature, the eq uilibrium constan t must also dep end upon temperature.
The equ ilibrium constant depends upon temperature only. Don't confu se the equilibrium constant with equilibrium.
The concentration of a pure liquid or a pure solid is usually given a value of one for the equilibrium expression. Concentra tions are only approxilnation s for activities. Acti vities for pure solids and pure liquids are like 'effective' m ole fra ctions. The m ole fraction of a pure solid or liquid is one. Although solvents are not actually pure, they are usu ally considered ideally d ilute on the MCAT, which means that the ir mole fra ction is one. The activity then of a pure solid or liquid is approxim ately one. Be aware tha t pure solids or liquids can s till participate in th e equilibrium. When they do, they must be present in order for equilibrium to exis t.
Don't use solids or pure liquids such as water in the law of mass action.
2.13 The Partial Pressure Equilibrium Constant For reactions w ith m ore than one p a thway, and for reactions with more than one step, the principle for detailed balance states that, a t equilibrium, the for w ard and reverse reaction ra tes for each step must be equ al, and an y two or more single reactions or series o f reactions resulting in the same products from identical reacta nts must have the same eq uilibrium constant for a give n temperature. The equilibrium constant d oes not depend upon whether or not other s ubstances are present.
For reactions involving gases, the equilibrimn constant can be w ritten in terms of p artial pressures. The concentration equilibrium con s tant and the p artial pressure equilibrium constan t d o not have the same value, but they are related by the equation: K P = K , (Rn&"
w here Kp is the partia l pressure equilibrium constant and 11 is th e sum of the coeffi cients of the products minus the sum of the coefficients of the reactants. This equation is n ot required for the MCAT, bu t you must be a ble to work with p artia l pressure equilibrium constants.
2.14 The Reaction Quotient The equilibrium cons tant describes only equilibrium condition s. For reactions not a t equilibriuln, a sin1ilar equation gives inforn1ation about the reaction:
Prod uctsCOefficients
Q=
Reactan tscoefficients
w here Q is called the reaction quotient. O f course, Q is not a constant; it can h ave a ny positive va lue . Copyright © 2007 Examkrackers, Inc.
Use the reaction quotient Q to predict the direction in which a reaction will proceed .
40
MCAT INORGANIC CHEMISTRY
Since reactions always move toward equilibrium, Q will always change toward K. Thus we can compare Q and K for a reaction at any given moment, an d learn in which direction the reaction will proceed. •
If Q is equal to K, then the reaction is at equilibrium.
•
If Q is greater than K, then the ratio of the concentration of products to the concentration of reactants, as given by the reaction quotient equation above, is greater than when at equilibrium, and the reaction increases reactants and decreases products. In other words, the reverse reaction rate will be greater than the forward rate. This is sometimes called a leftward shift in the equilibrium. Of course, the equilibrium constant does not change during this type of equilibrium shift.
•
If Q is less than K, then the ratio of the concentration of products to the concentration of reactants, as given by the reaction quotient equa tion above, is less than when at equilibriun1, and the reaction increases products and decreases reactants. In other words, the forward reaction rate will be greater than the reverse rate. This is son1etimes called a rightward shift.
2-15 Le Chatelier's Pri nciple There is a general rule called Le Chatelier's principle that can often be applied to systems at equilibrium. Le Chatelier's principle states that w hen a system at equilibrium is stressed, the system will shift in a direction that will reduce that stress. There are three types of stress that usually obey Le Chatelier's principle: 1) addition or removal of a product or reactant; 2) changing the pressure of the system; 3) heating or cooling the system. The Haber Process is an all gas reaction commonly used on the MCAT to test Le Chatelier's principle. The Haber Process is an exothennic reaction, so it creates heat. For Le Chatelier purposes, we can think of heat as a product in the Haber Process: N 2(g) + 3H2(g) ->
2NH3l~)
+ Heat
Imagine a rigid container with N" H" and N H 3 gas at equilibrium. If we add N, gas to our system, the systeln attempts to compensate for the increased concentration of nitrogen by reducing the partial pressure of N2 with the forward reaction. The forward reaction uses up H2 as well, reducing its partial pressure. N H3 and heat are created by the forward reaction. If we raise the temperature by adding heat, the reaction is pushed to the left. The concentrations of N 2 and H 2 are increased, while the concentration of NH J is decreased.
If the size of the container is reduced a t constant temperature, total pressure increases. Since there are four gas molecules on the left side of the reaction and only 2 on the right, the equilibrium shifts to the right producing heat and raising the NH3 concentration. Tnterestingly, a similar effect is found w hen a solution is concentrated or diluted. If one side contains more moles than the o ther, the equilibrium shifts to the side with fewer moles when the solution is concentrated. Warnin g: Le Chiitelier's Principle does not always predict the correct shift. Notable exceptions are solvation reactions, and pressure increase due to the addition of a nonreactive gas. The solubility of salts generally increase with increasing telnperature, even when the reaction is exothermic. This is largely due to the significant entropy increase that occurs with dissolution. The entropy factor becomes more important as the temperature increases. An example of a pressure increase where Copyright © 2007 Examkrackers, Inc
LECTURE
2: GASES,
KINETICS, AND CHEMICAL EQUILIBRIUM •
equilibrium is n ot affected is the addition of He to the Haber Process. If we add He gas to our container of N" H." and NH" the total pressure increases, but there is n o shift in equilibrium. This can be seen by examining the partial pressure equilibrium constan t. Ad ding H e to a rigid container, does not change the p artial p ressures of the other gases, so the equilibrium does not shift.
Equilibrium
'"
"'"
-c
" -:.
0
.8
"
...
.~
~
C
'" " U <.J
H, NH, N,
0
'"
E -;;; ·c ~w @
"
@
'3
'3
E ~
"-'" -"e'" "
.8
,e ~
~
z" ~ 0
;;:
'"
'"
<.J
U" 0
§ '~
~
~
'"
NH~
u
0
N,
U
'"
H,
N, NH,
Time
0-
0-
E
"-:.'" -""...
.8
~
" :§ ...'"'" 'C
~
W
·3
0-
'"
""..c:
O!l
:E
Copyright © 2007 Examkracke rs, Inc.
<
E ~
'C .0
'"'OJ
'"
NH,
<.J
" 0
U
H,
N,
H,
Time
~
0-
0-
Time
'C
E 'C
Time
41
44. Nitrous oxide is prepared by the thermal decomposition of ammonium nitrate.
Questions 41 through 48 are NOT based on a descriptive passage.
The equilibrium constant for this reaction is:
41. As temperature is increased, the equilibrium of a gaseous reaction will always: A.
B.
C. D.
A.
shin to the right. shift to the left. remain constant. The answer cannot be determined from the information given.
B. C. D.
[NH,NO,]/[ N,G][ H 2 0]' [NP][ HPl'/[NH,NO,]
c.
[NPH H,GJ'
D.
[NP][HP]
45. Which of the following is true concerning a reaction that begins with only reactants and moves to equilibrium?
42. All of the following are true concerning a reaction at equilibrium EXCEPT:
A.
B.
A. B.
The rate of the forward reaction equals the ratc of the reverse reaction. There is no change in the concentrations of both the products and the reactants. The activation energy has reached zero. The Gibbs free energy has reached a minimum.
C.
D.
43. Nitric acid is proctuc.ect commercially by oxidation in the Oswald process. The first step of this process is shown
The rate of the forward and reverse reactions decreases until equilibrium is reached. Th e rate of the forward and reverse reactions increases until equilibrium is reached. The rate of the for\vard reaction decreases, and the rate of the reverse reaction increases until equilibrium is reached. The rate of the forward reaction increases, and the rate of the reverse reaction decreases until equilibrium is reached.
below. 4NH,(g) + SO,(g)
~
4NO(g) + 6H,G(g)
A container holds 4 moles of gaseous ammonia, 5 moles of gaseous oxygen, 4 moles of gaseous nitric oxide, and 6 moles of water vapor at equilibrium. \Vhich of the following would be true if the conlainer were allowed to expand at constant temperature? A. B. C. D.
Initi ally during the expansion the forward reaction rate would be greater than the reverse reaction rate. The equilibriUm would shift to the left. The partial pressur~of oxygen would increase. The pressure insi de the container would increase.
Copyright © 2007 Exa mkracke rs , Inc.
42
GO ON TO THE NEXT PAGE.
Questio ns 46 th rough 48 refer to the diagram be low.
47. What is the equi libri um express ion for the reaction?
Calci um carbonate deco mposes to calcium ox ide and carbon dioxi de gas via the following reversible reacrion:
A.
K ~ [CO,]
B.
K ~ [CaOIlCO, i
c.
K ~
D.
K ~
[CaO][CO, i rCaCO,] rCO,] [CaCO,]
.~- ---- -
48. The part ial press ure equili brium cons tant fo r the decompos ition of CaC0 3 is KI" If Beaker II is removed, under which of the follow in g co nd iti ons woul d eq uili briu m NOT be ac hieved?
CaCO,(s) ;;= CaO(s) + CO,(g) Beaker T co ntai ns pure CaC0 3 and Beaker 11 conta ins pure CaO. T he co ntai ner is sealed.
A. B. C. D.
46. Whi ch of the followi ng statements could be true abou t the sys tem upo n achievi ng equ ilibrium ?
A. B.
C. D.
Beaker I is empty. The number of calci um atoms in Bea ker II has decreased. The number o f calcium atoms In Beaker 11 has increased. Beaker 11 contains a mixture of CaCO:; and CaO.
Copyright © 2007 Examkrackers, Inc.
43
Kp is less than the parti al pressure of CO 2 _ K" is greater than the partial pressure of CO z_ Kp is equal to the parti al pressure of CO 2 Equilibriu m coul d not be achieved under any conditions because soli d CaO is required to achieve equili brium_
STOP.
Thermodynamics
3.1 Thermodynamics is the study of energy and its relationship to macroscopic properties of chemical systems. 111ermodynamic functions are based on probabilities and are valid only for systems composed of a large number of molecules. In other words, with few exceptions, the rules of thennodynamics govern complex systems containing many parts, and they cannot be applied to specific microscopic phenomena. f
Thermodynamic problems divide the universe into a system and its surroundings. The system is the macroscopic body under study, and the surroundings are everything else. There are three systems: open, closed, and isolated. System definitions are based upon mass and energy exchange with the surroundings. Open systems exchange both mass and energy with their surroundings; closed systems exchange energy but not mass; and isolated systems do not exchange energy or mass.
System
r
..., ~
... '" /
/
Mass
3 .2
OUt
"
~
~o,--\ 0,,' " v o,,"
\-. .__J_._ ~
System Open Closed I solated
State Functions
A state is the physical condition of a system described by a specific set of thermodynamic property values. There are two types of properties that describe the macroscopic state of a system: 1) extensive and; 2) intensive. Extensive properties are proportional to the size of the system; intensive properties are independent of the size of the system. If you combine two identical systems and a property is the same
L'.E
L'.M
Yes Yes No
Yes No No
46
MeAT
INORGANIC CHEMISTRY
Extensive properties change with aroount; intensive properties do not. You need to know what a state function is. State functions are pathway independent. State properties describe the state of a systero. In other words, the change in a state property going from one state to another is the same regard less of the process via which the system was changed. Three state properties, one being extensive, describe the state of a system unambiguously.
for both the single system and the combined system, that property is intensive. If a property doubles when the systems are combined, the property is extensive. If you divide one extensive property by another, the result is an intensive property. Volume V and number of moles n are examples of extensive properties. Pressure P and temperature T are examples of intensive properties. The macroscopic state of anyone-component fluid system in equilibrium can be described by just three properties, of which at least one is extensive. All other properties of the state of the same system are necessarily specified by the chosen three properties. For instance, if for a single component gas in equilibriulTI, pressure, temperature, and volume are known, all other properties which describe the state of that gas (such as number of moles, internal energy, enthalpy, entropy, and Gibbs energy) must have a specific single value. Since the state of a system can be described exactly by specific properties, it is not necessary to knoV\T how the state was fonned or what reaction pathway brought a state into being. Such properties that describe the state of a system are called state fun ctions. Properties that do not describe the state of a system, but depend upon the pathway used to achieve any state, are called path functions. Work and heat are examples of path functions.
3.3
Heat
There are only h"o ways to transfer energy between systems: h eat q and work w. Heat is the natural transfer of energy from a warmer body to a cooler body. Any energy transfer that is not heat is work. Heat is movement of energy via conduction, convection, or radiation always from hot to cold .
The temperature difference in thermal conductivity is like the pressure difference in fluids or the potential difference in electricity. The rate of heat flow is like
Heat has three forms: conduction, convection, and radiation. Conduction is thermal energy transfer via molecular co11isions. Conduction requires direct physical contact. In conduction, higher energy molecules of one system transfer some of their energy to the lower energy molecules of the other system via molecular collisions. Heat can also be conducted through a single object. An object's ability to conduct heat is called its thermal conductivity k. The thermal conductivity of an object depends upon its composition and, to a much lesser extent, its temperature. A slab of a given substance with face area A, length L, and thermal conductivity k will conduct heat Q from a hot body at temperature T" to a cold body at temperature T,. in an amount of time t.
volume flow rate or current. Notice the
similarity in these equations for the conduction of heat, ftuids, and electricity. LlT
I R
LlP
~
QR
V
~
i R
You can use this similarity to help you
understand heat flow, fluid flow, and electron flow. For instance, in all cases, thicker conduits allow for greater flow,
longer conduits impede flow, and flow rate depends upon the difference in a property of the reservoirs at either end of the conduit.
Qlt is the rate of heat flow or heat current I. The resistance to heat flow R can be written as LI(kA). A little algebra results in f1T = IR, an equation similar in form to Ohm's law in electricity. Similar to the rate of fluid flow in an ideal fluid (see Physics Lecture 6) or to the flow of electric current through resistors in series (see Physics Copyright © 2007 Exarnkrackers, Inc.
LECTURE
3:
THERMODYNAM!CS
•
47
Lecture 7), in a steady state system the rate of heat flow is constant across any number of slabs betvveen two heat reservoirs. In other words, if a series of slabs were lined up end to end between hot and cold reservoirs, the rate of heat flow, Q/t, would be the same in all slabs even if they each had different lengths, thicknesses, and different thermal conductivities. Conservation of energy explains why. If the rate of energy transfer were not steady across all slabs, the slab that conducted heat the fastest would become cold. A cold slab could not conduct h eat to warmer slabs because heat moves from hot to cold. It also follows that since the rate of energy transfer is the same for £ach slab, the order in which we place the slabs does not affect the overall conductivity. Finally, since the rate of heat flow is the same through each slab, a higher conductivity results in a lower temperature difference across any slab of a given length. Let's see how well you understand thermal conductivity . If Thave a heavy blanket and a light blanket, which blanket should J place on top in order to stay warmer? The answer is: the orde r of the blankets will not make any difference sLllce it will not change the rate of conduction . Convection is thennal energy transfer via fluid lnovements. In convection, differences in pressure or density drive warm fluid in the direction of cooler fluid. For instance, on a warm sunny day at the beach the air above the land heats up (hand warm ing faster than the air above the water. As the air the glClss) above the land warms, it becomes less dense 3 types of Heat and rises carrying its thermal energy with it. Ocean and air currents are common examples of convection. Radiation is thermal energy transfer via electromagnetic waves. When metal is heated, it glows red, orange-yellow, white, and finally blue-white. The hot metal radiates visible electromagnetic waves. But even before the metal begins to glow, it radiates electromagnetic waves at a frequency too low to be visible to the human eye. Tn fact, all objects with a temperature above 0 K radiate heat. The rate at which an object radiates electromagnetic radiation (its power P) depends upon its temperature and surface area, and is given by the SteJan-Baltzman law:
P = GEAr where A is the surface area of the object, T is the temperature of the object in kelvins, is the Stefan-Boltzman constant (5.67 x 10-8 W m-2 K-4), and E is the emissivity of the object's surface, which h as a value between 0 and 1. If we substitute the temperature of the environment for th e temperature of the object, we find the rate at which the object absorbs radiant heat from its environment. The net rate of heat transfer will always be from hot to cold, and is given by: G
P = ci8A!T,4- T,4) where the Tc is the temperature of the environment and To is the ten1perature of the object. Newton's law of cooling states that the rate of cooling of a body is approximately proportional to the temperature difference between the body and its environment.
Notice that an object that radiates heat fa ster also absorbs heat faster. This mean s that an object that is a more
When radiation strikes an opaque surface, only a fraction is absorbed. The remainder is reflected. The fraction absorbed is indicated by the emissivity E of the surface. The emissivity depends upon surface composition. As stated above, the emissivity
with its environment more quickly. With thi s in mind, is it better to paint your house bla ck or white?
efficient radiator comes to equilibrium
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48
MCAT INORGANIC CHEMISTRY
Answer: White is better In Summer, your house is cooler than the environment and white refi ects away the heat. In Winter, your house is- warmer than the environment and white radiates away less heat.
of any surface is between 0 and 1. An object with an emissivity of 1 is called a blackbody radiator and is possible only in theory. All other objects reflect as well as absorb and radiate. Dark colors tend to radiate and absorb better than light colors, \-\Thich tend to reflect. Radiation is the only type of heat that transfers through a vacuum.
3.4
Work
From OUf discussion of heat, we know that work is any energy transfer that is not heat. In Physics Lecture 3 we defined work as an energy transfer due to a force. In physICS problems, work typically changes the motion or position of a body. However, now we want to look at work as it applies to a chemical system at rest. For our purposes, a system at rest may change its size or shape, but does not translate or change its position. A systeln at rest with no gravitational potential energy and no kinetic energy, may still be able to do PV work. Imagine a cylinder full of gas compressed by a piston. If we place two blocks of mass m on the top of the piston and allow the gas pressure to lift the masses to a height 11, the system has done work on the mass in the amount of the gravitational energy change 2mgll. Thus negative work has been done on the system. This work is called PV work because, at constant pressure, it is equal to the product of the pressure and the change in the volume (P" V). From Newton's second law F:::: rna, we know that if the masses are lifted at constant velocity, the force on the masses is constant and equal to 2mg. From the definition of pressure P = F/A, we know that if the force remained constant, the pressure also remained constant. Constant pressure conditions allow us to calculate the work done using:
w = P L1 V (constant pressure) If pressure is not constant, then calculus is required to fi nd the amount of PVwork done. In such a case, the MeAT wi ll not require you to calculate the work.
If the volume rema ins constant, no PV work is done at all.
The start point and end point of the expansion could also have been achieved by removing one mass, allowing the piston to rise, increasing the pressure to 2mg/A while holding the piston steady, and then replacing the second mass. This would have been a different pathway to achieve the same result. Since work is a path function, a different pathway results in a different amount of work; in this case the work would have been mgh. Finally, the start and end point of the expansion could have been achieved by changing the pressure as the piston rose. In this case, PV work is done, but it does not equal P"V. To calculate the work in this case, you would need to use calculus. If we examine a pressure vs. volume graph for each case, the work done is given by the area under the curve. Notice that the area is different for each case.
An MeAT problem asking about PV work will likely specify constant pressure . However, you should understand th ese graphs and understand that PV work takes place when a gas expands against a force regardless of whether or not th e pressure is constant.
2.mg
A
-~
,.~ I
2mgh
.
P ~ P ~ mgh
v
v
~)
v
Copyright © 2007 Examkrackers, Inc.
LECTURE 3: THERMODYNAMiCS .
3.5
The First Law of Thermodynamics
The First Law of Thennodynamics states that energy of the system and surroundings is always conserved. Thus an y energy change to a system must equal the heat flow into the system plus the work done on the system.
M=q+w Warning: We have chosen the convention where work on the system is positive. It is possible that an MeAT passage could define work done by the system as positive, in which case the formula would be (>'E ~ q - w.
3.6
Heat Engines
To simplify things by removing significant gravitational forces, let's turn our cylinder and piston on its side. The gas pushes against the piston which is now held by an outside force that we can controL By heating the gas, we can allow it to expand w hile maintaining a constant telnperature. For a heat source, we can use a large hot body which we will call a hot reservoir. Since temperature is the kinetic energy per mole of gas molecules, and we are holding temperature constant, the total energy of the gas does not change as it expands. But the energy that we added as heat must be going somewhere. It is changed completely into PV work done by the force against the piston. (By the way, the force cannot be constant if the temperature is constant because at constant temperature, the pressure must go down as the gas expands [PV ~ I1RT].) This experiment seems to su ggest that heat can be changed completely into work. But there is a problem. In the real world, we will run out of cylinder and we will have to push the piston back to its start point to begin again. When we do so, we must do work on the gas to compress it. In fact, compressing the gas as is will raise its temperature, requiring more work to compress it th an was gained by expanding it. So we must cool the gas first. Heat flows from hot to cold, so we cannot cool the gas by allowing heat to flow to the hot reservoir. Instead, we must have a cold reservoir into w hich we dump the heat energy. Once the gas is sufficiently cooled, it can be compressed to its original state w ith less work than was gained. The expansion
Sta rt point
erature.f~
~Force
Tern .. p.. . mcrease .
.
~
:q
..' \
Cold
'f\
.
p Work done by the gas
v Copyright © 2007 EX
~
""-
Work done"-,,_--. on the gas
'?q~.
~~
0<;> Net work ---:0 Q;;,
.. .. "il017 . . .. - . .
rlTemperature decrease
, -----------~.
49
SO
MCAT INORGANIC CHEMISTRY
The heat engine stuff is given here in order to help you understand the relationsh ip between heat and work. If it is on the MeAT, it will be explained in a passage. However, don't just ignore it. It is a possible pa ssage topi c and a good way to learn to understand heat and work. At the very least, know the second law of therrnodynarnics in terrns of heat and work: Heat cannot be cornpletely converted to worked In a cycli cal process.
process can begin again and, as can be seen by the graph, we have turned some, but not all, of the heat into work Our cylinder and piston is called a heat engine.
This is a demonstration of the second law of thermodynami cs which says "Heat cannot be changed completely into work in a cyclical process." A machine that converts heat to work is called a heat engine. A heat engine can be diagrammed as shown. Notice that, via conservation of energy, the heat entering
the engine q" must equal the net work done by the engine the engine q,. (q" ~ w + qcl
10
plus the heat leaving
We can reverse a heat engine to create a refrigerator. If we change the directions of
the arrows in the diagram, the inside of the refrigerator is represented by the cold reservoir. Notice that a refrigerator requires work, and that the heat that it generates is greater than the heat that it removes from the cold reservoir. The second law
of thermodynamics requires this result.
Cold Reservoir If we put a running refrigerator in a small closet, close the closet door and open the refrige rator door what will happen to the temperatore of the closet?
Refrigerator
The most effective cyclical conversion of heat into work is produced by the hypothetical Carnot engine and depends upon the temperature difference of the two reservoirs. The further apart the temperatures of the tvvo reservoirs the more effective the conversion. The fraction of heat that can be converted to work with a Carnot engine is called the efficiency e, and is given by: f
Answer: The closet will warm up becaose q" is greater than q,.
where TN and Tc are the temperatures of the hot and cold reservoirs respectively.
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52. A box sliding down an incline increases in temperature due to friction. The name for this type of heat is:
Questions 49 through 56 are NOT based on a descriptive passage,
A. B. C. D.
49. Which of the following is true concerning an alf conditioner that sits inside a thermally sealed room and draws energy from an outside power source?
53. Which of the following gas properties is needed to calculate the work done by an expanding gas?
It requires more energy to cool the room than if part of the air conditioner were outside the room. It will require more ti me to cool the room than if part of the air cond itioner were outside the room. It will require less energy to cool the room than if part of the air conditioner were outside the room. It cannot cool the room on a permanent basis.
A. B.
C. D.
I. The initial and final pressures II. The initial and final volumes III. The path followed during the expansion,
A. B. C. D.
50. Three blocks made from the same insulating material are placed between hot and cold reservoirs as shown below.
Block Block y X
A
Cold
B
C
I only Il only I and Il only J, n, and III
54. The heating bill for a homeowner is directly proportional to the rate at which heat is conducted out of the house and into the surroundings. The average temperature inside and outside of a house is measured on different months and recorded in Table 1.
Block Z
Hot
convection conduction radiation The energy transfer here is due to work and not to heaL
D Month
Which of the following must be true? The temperature difference between points A and B is less than the temperature difference between points C and D. II. The rate of heat flow through Block X is greater than the rate of heat flowing through Block Z. III. Switching the positions of Block X and Block Z would decrease the rate of heat tlow. I.
A.
Tonly
B.
flI only I and III only T, n, and III
C.
D.
Temperature outside CC)
Temperature inside (0C)
8
22 25 20 26
Nov Dec
5
Jan Feb
3 13
For which month would the homeowner expect to have the largest heating bill?
A. B. C.
D.
November December January February
51. Immediately upon bringing a hot piece of metal into a room , the heat is felt from 5 meters away. The type of heat transfer is probably:
A.
convection transduction radiation conduction
B. C.
D.
Copyright
(t)
2007 Exarnkrackcrs , lnc.
51
GO ON TO THE NEXT PAGE
56. Under the best possible condi tions, a stea m engine will
55. A ri gid container of co nstant volume is used to store compressed gas. When gas is pumped into the co ntainer, the pressure of the gas inside the contai ner is increased and the temperature of the co nta iner also increases. Wh ich statemenl is tru e of th e work done on the container? A.
B.
C. D.
have an efficiency of sl ightly more than 20 percent. A normal steam engine has an efficiency of about 10 percent. This means that for a normal steam engi ne: A.
B.
The work is equal to the increase in the pressure inside the container. The work is equal to the increase in th e temperature inside the co ntainer. The work is eq ual to the sum of the pressu re and temperature increases. There is no work done 00 the container.
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C. D.
52
10 percent of the input energy contr ibu tes to lhe work done by the engine. 90 percent of the input energy contributes to the work done by the engine. the iotern aj temperature will increase by 10 percent during operation. the internal temperature will increase by 90 percent during operation .
STOP.
LECTURE
3.7
3:
THERMODYNAMICS .
53
Thermodynamic Functions
To understand thermodynamics, you mus t be familiar with seven state functions:
3.8
1.
internal energy
U
2.
temperature
T
3.
pressure
P
4.
volume
V
5.
en thalpy
H
6.
entropy
5
7.
Gibbs energy
G
Internal Energy
Sin ce thermodynamics is mainly ~oncern ed with chemical energy, most problems will not deal with macroscopic mechanical energies. Instead, problem s will be concerned with in ternal energy. Internal ene rgy is the collecti ve energy of molecules measured on a microscopic scale. This energy includes v ibrational energy, rota-
tional energy, translational energy, electronic energy, intermolecular potential energy, and rest mass energy. Internal en ergy does not include mechanical energy. In other words, internal energy is all the possible forms of energy imaginable on a molecular scale. Vibrational energJ} is created by the atoms vibrating within a molecu le. Rotational energJ} is molecular movement wh ere the spatial orien tation of the body changes, while the center of mass remains fixed and each point within a m olecule remains fixed relative to all other points. Translational energy is the movement
of the center of mass of a molecule. Electronic energJ} is the potential electrical energy created by the attractions between the electrons and their respective nuclei. The intermolecuLar potentiaL energy is the energy created by the intermolecular forces between molecular dipoles. Rest mass energy is the energy predicted by Einstein's fa mous equation E ~ 111e'. The sum of these en ergies for a group of molec ules is called the internal energy.
Rota tional
Translational
Intermolecular potential
Vibra tiona 1
Electronic
Rest mass
Internal Energy Types jf we have a closed system at rest with no electric or magnetic field s, the only en ergy change will be in interna l energy, and the first law of thermodynamics can be rewritten as: ~U = q + 'lV . For a reaction within such a system invol v ing no change in volume, there is no work of any kind and the change in in terna l energy is equal to the heat: LlU ~ q.
Copyright © 2007
ExalTlk.ral.kt::'I~,
IIiL
The MCAT doesn't generally ask about internal energy directly. The stUdy of internal energy is beyond the MCAT. However, in order to understand and use conservation of energy you must consider internal energy.
The MeAT may refer to internal energy as 'heat energy', 'thermal energy', or even 'heat', Heat energy and thermal energy are really the vibrational, rotational, and translaMnal parts of internal energy. They are called thermal energy because they affect temperature. Heat is a transfer of energy, and using it as another name for internal energy can create confusion. Unfortunately, this is a common mistake.
54
MCAT INORGANIC CHEMISTRY
Tidbits of info that may help your understanding but won't be tested by MCAT.
Internal energy is a state function. For an ideal gas, any state function can be expressed as a function of temperature and volume only. For an ideaL gas, internal energy is independent of volume and is a function of tempera ture only.
3.9 The zeroth law of thermodynamics Just states that temperature exists. It's called the zeroth law because after the first, second, and third laws were already establ ished it was realized that they depended upon a law that established the existence of temperature .
50 lot
40
V/L
30 20 10
,
~
,/
V
V
;.~
Temperature
The zeroth Jaw of therlllodynamics states "Tw o systems in thermal equilibrium with a third system are in thermal equilibrium with each other." The zeroth law declares that two bodies in thermal equilibrium share a thermodynamic property, and that this thermodynamic property must be a state function . The thermodynamic property described by the zeroth law is temperature . There are several methods used to define temperature. One definition is based upon the volume of an idea l gas. For an ideal gas the volume vs. temperature graph is exactly linear for any given pressure. Although all real gases become liquids at low temperatures, if we extrapolate back along the volume vs. temperature line, the lines for all pressures intersect at the same point on the temperature axis. We can define the temperature of this point as 0 K or -273°C. To establish the size of a unit of temperature, we can arbitrarily choose the freeZing point and boiling point of water along the 1 atm line, and label those points we and 1000 e respectively. This and other definitions of temperature do not give a satisfying intuitive notion for what temperature really is. For an intuitive feel of temperature, we need to examine the motion of the molecules. When we look at internal energy, we see that the translational, rotational, and vibrational energies describe the energies of molecular motion. The sum of these energies is called thermal energy. An y increase in thermal energy increases temperature.
For a fluid , temperature is directly proportional to the translational kinetic energy of its molecules. Translational motion can be divided into -300 -200 - 100 o 100 200 300 three degrees off reedom or modes: 1) along the x axis; 2) along the y axis; T jOC and 3) along the z axis. The equipartiliol1 theory states that for a normal Volume vs. Temperature system each mode of motion will have the same average energy and for one mole of N , gas tha t the energy will be equal to l kT, where T is the temperature, and k extrapolated back to zero volume is the Boltzmann constant (1.38 x 10-23 J K-'). The Boltzmann constant k is related to the ideal gas constant R by Avagadro's number N ,,: R ~ Think of temperature as a measurement N Ak. Since there are three modes of kinetic energy each averaging l kT joules, the avof how fast the molecules are moving or erage kinetic energy of a single molecule in any fluid is given by: vibrating. When substances get hot, it is because their molecules move faster. K.E..,. =
o
,
, :::
5 tm
MCAT won't test the equiparlltion theo ry. Just know the formula : K.E. =
;kT
This is a variation of an equation from Lecture 2. The equipartition theorem is derived from classical physics and breaks down when quantum effects are significant. Quantum effects are greater for rotation than for translation, and greater still for vibration. TI1C moleculcs of a solid do not translate or rotate; they vibrate. For solids at high temperatures where quantum effects are less important, temperature is proportional to the average kinetic energy of the vib ration of molecules about their equilibrium position. For solids at low temperatures, there is deviation from this rule due to quantum effects. The greater the random translational energy of gas molecules per mole, the greater the temperature. So ideally, temperature can be thought of as the thermal energy per molecule or mole of m olecules. Recall that when we divide one extensive property by another, we get an intensive property. Energy and num ber of moles are extensive properties. This makes temperature an intensive property.
Copyright © 2007 Examkrackers , Inc.
LECTURE
3: THERMODYNAM ICS
.
55
The MeAT will use two measurement systems for temperature: degrees Celsius and Kelvin. Celsius is just the centigrade system with a new name. At 1 atm, water freezes at O°C and boils at 100°C. The lowest possible temperature is called absolute zero, and is approximately -273°C. To find approximate Kelvin from degrees Celsius, simply add 273. An increase of 1°C is equivalent to an increase of 1 K.
When in doubt. use kelvin . In chemistry, you are always safe using the Kelvin scale because the Kelvin scale is absolute .
Virtually all physcial properties change with temperature.
Good to know for the MCAT!
3.1 0 Pressure Loosely speaking, pressure of an ideal gas is the random translational kinetic energy per volume. Pressure is an intensive state function. Pressure and volume are discussed in depth in Chemistry Lecture 2 and Physics Lecture 5.
The greater the random translational kinetic energy of gas molecules per volume, the greater the pressure
3.11 Enthalpy Two systems at rest may have the same amount of internal energy, and, if they are at different pressures, they have different capacities to perform PV work. (This is one demonstration of why "energy is the capacity to perform work" is a poor definition of energy. See Physics Lecture 3.) Enthalpy is a man-made property that accounts for this extra capacity to do PV work. Unlike functions such as pressure, volume, and temperature, enthalpy is not a measure of some intuitive property. Enthalpy is defined as an equation rather than as a description of a property. Enthalpy H is defined as: H'" U + PV From OUf inexact but intuitive concept of pressure as the random translational kinetic energy per unit volume, we see that enthalpy actually counts random translational kinetic energy twice! Although enthalpy is measured in units of energy (joules), enthalpy itself is not conserved like energy. Enthalpy of the universe does not remain constant.
Enthalpy cannot be intuited . Just memorize the equation .
Since U, P, and V are state functions, enthalpy is also a state function. Like internal energy, for an ideal gas enthalpy depends only on temperature. Enthalpy is an extensive property. As with many state functions, we are interested in the change in enthalpy and not its absolute value. For the MCAT, we will be interested in the change in enthalpy under constant pressure conditions only:
Mi = AU + PAV
(canst. P)
There are no absolute enthalpy values, so scientists have assigned enthalpy values to compounds based upon their standard state. Standard state should not be confused with STP. Standard state is a somewhat complicated concept that varies with phase and other factors and even has differnet values based upon which convention you choose to follow. As usual, we can simplify things greatly for the MCAT. Recall that a 'state' is described by a specific set of thermodynamic property values. For a pure solid or liquid, the standard state is the reference form of a substance at any chosen temperature T and a pressure of 1 bar (about 750 torr or exactly 105 pascals). The reference form is usually the form that is most stable at 1 bar and T. For a pure gas there is an additional requirement that the gas behave like an ideal gas. An element in its standard state at 25°C is arbitrarily assigned an enthalpy value of 0 J/ mol. From this value we can assign enthalpy values to compounds based upon the change in enthalpy when they are formed from raw elements in their standard states at 25'C. Such enthalpy values for compounds are called stan-
Copy ri ght © 2007 Examkrackers, In c.
Don't confuse standard state and STP. An element in its standard state at 25°C is arbitrarily assigned an enthalpy value of 0 J/mol.
56
M C AT INORGANIC CHEMISTRY
dard enthalpies of formation. The standard enthalpy of formation IlH of is the change in enthalpy for a reaction that creates one mole of that compo und from its raw elements in their standard state. The symbol '0' (called na ught) indicates stand a rd state cond itions . Standa rd enthalpies of fo rm a tion can be fo und b y experiment, and are available in books. An examp le of the en thalpy of formation of water is: H ,(g) + fiO,(g) --> H ,O(l) IlWf
=- 285.8 kJ / mol
For reactions involving no change in pressure, the change in enthalpy is equal to the heat: M:l Basically, if gas is not part of the reaction, enthalpy change is equal to heat which , in the abse nce of work, is equal to a change in energy.
=
q
(canst. P, closed system at rest, PV work only)
Man y liquid and solid chemical reactions performed in the lab take place at constant pressure (1 atm) and nearly constan t volume. For these reactions (any reactions involving only solid s and liquids at mod erate pressures)!l.U Ell !l.H. Since, in many reactions in the lab, enthalpy approximates heat, the change in enthalpy from reactants to products is often referred to as the heat of reaction. M:le
Hess' law says that when you add reactions, you can add their enthalpies. Hess' law works because enthalpy is a state function.
Exothermic reactions (-!JH) release heat making the reaction system hot; endothermic reactions (+LlH) absoro heat making the reaction system cold.
f
products -
Allf e reactants
Since enthalp y is a state function, the change in enthalpy when converting one group of compounds to another is not dependent upon what reaction or even series of reactions take p lace. The change in enthalpy d epends only on the iden tities and thermodynamic states of the initial and final compounds. Thus, in any reaction, the steps taken to get from reactants to products do not affect the total change in enthalpy. Hess' law states "The swn of the enthalpy changes for each step is equal to the total enthalp y change regardless of the path chosen." For example: N, + 0 , --> 2NO 2NO + 0 , --> 2N02
!J.H = 180 kJ !J.H = - 112 kJ
N, + 20 , --> 2NO,
!J.H = 68 kJ
step 1 + step 2 complete reaction
Hess's law also inclicates that a forward reaction has exactly the opposite change in enthalpy as the reverse. If the enthalpy change is positive, the reaction is said to be endothermic; if it is negative, it is said to be exothermic. If we consider a reaction where the change in enthalpy is equal to the heat (a constant p ressure reaction ), then an exothermic reaction prod uces heat flow to the surroundings, while an endothermic reaction produces heat flow to the system.
Reactants
Products
Reaction Progress
- M:l e
reaction -
We can see this grap hically if we compare the p rogress of a reaction wi th the energy of the molecules. (Due to the close relationship between internal energy and enthalp y, the term energy is used loosely for these types of gra phs. You m ay see the y axis labeled as enthalpy, Gibbs free energy, o r simply energy.) You can see from the graph that if the reaction progress is reve rsed, the enthalp y change is exactl y reversed . Notice that there is an initial increase in energy regardless of which direction the reaction moves. This increase in energy is called the activation energy (the sa me ac tiva tion energy as in Chemistry Lecture 2). The p ea k of this energy hill represents the m olecules in a transition state where the old bonds are breaking and new bonds are fo rming. The transition state occurs during the reCopyright © 2007 Examkrackers, Inc.
LECTURE
ac tion collision. Do not confuse the transition state with intennediates, which are the products of the first step in a two step reac tion. A two step reaction has two humps as shown on page 34. The intermediates exist in the trough between the two humps.
Reaction Progress
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3:
THERMODYNAMICS •
57
Notice, on an energy diagram, how a
catalyst lowers the activation energy. The activation energy for both the forward and the reverse reactions is lowered . Although the relative amount by which the activation energies are lowered is different, if we used the Arrhenius equation (Chemistry Lecture 2) to find the new rate constants, we would find that the rate constants are raised by the same relative amounts; thus, equilibrium is unaffected by a catalyst. For the MCAT you must remember that a catalyst affects the rate and not the equilibrium . Notice that a catalyst does not affect the enthalpy change either.
61. T he heats of combustion for graphi te and diamond are as follows:
Questions 57 through 64 are NOT based on a descriptive passage.
57. What is the en thalpy change in the following reaction? Compound
;',}{ 0
CHig)
- 75 kllmol
CO 2(g)
-394 kllmol
H,O(l)
- 286 kJ/mo!
Cgmphi,JS) + O,(g) --> CO,(g) I1H
~
- 394 kJ
C"pmcpis) + 02(g) --> CO,(g ) !:ili
~
- 396 kJ
Diamond spontaneously changes to graphite. What is the change in enthalpy accompanying the conversion of two moles of diamond to graphite?
f
A. B. C.
D.
-790 kJ -4 kl 2 kl 4 kJ
CHig) + 20 2(g) --> CO 2(g) + 2H,o(l) A. B.
C. D.
62. The standard enthalpy of formation for Jiquid water is:
- 755 kJ - 891 kJ - !04 ! kJ 89 ! kJ
H,(g) + fiO,C<;) --> HP(I) 11H"1~ -285.8 kJ/mo!
Which of the following could be the standard enthalpy of formation for water vapor?
58. W hich of the following properties of a gaseous system affect its enthalpy?
I. II. III.
A.
B. C. D.
A. B. C. D.
pressure vo lume internal energy
63. For a particular reversible reaction, the forward process is exothermic and the reverse process is endothermic. \Vhich of the following statements must be true about this reaction?
TTl only I and II only II and III only r, IT. and TTl
A. 59. A catal yst will change all of the following EXCEPT:
A. B. C. D.
B.
enthalpy activation energy rate of the forward reaction rate of the reverse reaction
C. D.
60. In an exothermic reaction, which of the following will most likeJy increase the ratio of the forward rate to the reverse rate? A. B. C. D.
The forward reaction will be spontaneous under standard conditions. The reverse reaction will be spontaneous under standard conditions. The activation energy will be greater for the forward react ion than for the reverse reaction. The activation energy will be greater fo r the reverse reaction than for the forward reaction.
64. Sulfur dioxide reacts with oxygen in a reversible reaction to form sulfur trioxide as shown.
2S0,(g) + O,(g) '" 2S0 3(g)
adding thermal energy to the sys tem removi ng thermal energy from the system using a catalyst lowering the activation energy
;',}{o ~
-200 kl
If the temperature at which the reaction takes place increased, which of the following will take place? A. B. C. D.
Copyright © 2007 Examkr2ckers, Inc.
-480.7 kllmol - 285.8 kl - 241.8 kllmo! +224.6 klima!
58
IS
The rates of both the forward and reverse reactions will increase. Only the rate of the forward reaction will increase. Only the rate of the reverse reaction will incre'ase. The rates of neither the forward nor reverse reactions will inc rease.
STOP.
LECTURE
3:
THERMODYNAMICS •
59
3.12 Entropy If you have studied entropy before, you have probably heard the following exam ple: "Over time, a clean room w ill tend to get dirty. This is entropy at work." Entropy is nature's tendency toward disorder. A better definition of entropy incorporates the concept of probability. Entropy S is nature's tendency to create the most probable situation that can occur within a system. For instance, imagine four iden-
tical jumping beans that bounce randomly back and forth between two containers. If we label each bean A, B, C, and D respectively, we will find that the most likely s ituation is to have two beans in each container. The least likely situation is to have
all four beans in either of the containers. For exampl e, if we choose the left container, there is only one way for all four beans to be in the left container, but tb ere are 6 poss ible ways that two beans can be in each container. Two bean s in each conta iner is six times more likely than fo ur beans in the left container. Since the two-bean container situation is more likely, it has grea ter entropy.
If we replace the four jumping beans w ith millions of molecules moving randomly back and forth between two glass spheres connected by a glass tube, you should be able to see how the odds against having all the m olecules in one sphere become astronomical. The odds are so POOf, in fact, tha t the second law of thermodynamics s tates that it will never happen without some outsjde intervention, namely work. The second law of thermodynamics states that the entropy of an isolated system
The second law of thermodynamics is a funny law because it could be violated, but it is so unlikely to be violated that we say it won·t be violated no matter what.
w ill never decrease.
We can appl y the second law of thermodynamics to any type of system, if we recall that the surroundings of any system include everything tha t is not in the system. Thus, the system and the surroundings toge ther make up the entire universe. The universe itself is an isolated system. Therefore, the sum of the entropy changes of any system and its surroundings equals the entropy change of the universe, w hich must be equal to or g reater than zero.
!:.Ssystelll + !:.S s"rro"ndings = !:.S IlIIiversc ~ 0 So the entropy of a system can decrease, only if, at tbe same time, the entropy of the surroundi ngs increases by a greater or equal magnitude. Entropy is a state function. This means that the entropy change of a forward reaction is equal to the negative entropy change of a reverse reaction. This is true for the entropy change of the system, the surroundings, and the universe. Since the change in entropy of the universe must be pos itive or zero for all reactions, only a reaction w ith
zero universal entropy change can be reversible. Only ideal reactions create zero change in the entropy of the universe, so only ideal reactions aTe reversible from a thermodynamic definHion of reversibi1 ity. On a macroscopic scale, all real reactions are irreversible. Recall fro m the beginning of this lecture that we said that thermodynamics properties are valid only for macroscopic systems with many particles. On a microscopic scale, all rea l chemical reactions are reversible. The terms reversibLe and irre versjble are m ost often used in a relative sense where their meaning is am-
blgUOllS and not as often used in the molecular or thermodynamic senses where there mea nings are exact. Ln the relative sense, if the activation ene rgy of a chemical reaction is sufficiently high for the reverse di rection, tbe probability that it will occur may be suffi ciently low for chemists to call the reaction irreversible. The MCAT is Inost likely to use the te rms reversible and irreversible in their relative sense. An intuitive way to view entropy is as nature's effort to spread energy evenly between systems. N ature likes to lower energy of a system ·w hen it is high relati ve to the energy of the surroundings, but tbat means that nature likes to ra ise energy of a system whe n it is low relative to the ene rgy of the surroundings. A wa rm objec t Copyright © 2007 Examkrackers. Inc.
Six possible situatio ns with two beans in one pan.
Only one possible situation with fO lir beans in the left pan.
60 . MCAT
INORGANIC
CHEMISTRY
Entropy, and not energy, dictates the direction of a reaction.
w ill lose energy to its surroundings w hen placed in a cool room, but the same object will gain en ergy when p laced in a hot room. This means that it is entropy and not energy that dri ves reac tions in a given direction. The second law of thermodynamics tells us that entropy of the uni verse is the driving force that dicta tes whether or not a reaction w ill proceed. A reaction can be unfavorable in terms of enthalpy, or even energy, and still proceed, but a reaction must increase the entropy of the universe in order to proceed.
Reactions at equilibrium have achieved maximum universal entropy.
Eq uilibrium is the point in a reaction where the universe has achieved maximum entropy. A thermodynamically reversible reaction is one that remains infinitely close to a state of equilibrium at all times. (Such conditions are called qllasis!a!ic.) A quasistatic reaction moves infinitely slowly with infinitely small changes being made to the system, each one establishing a new equilibrium.
Entropy increases with number, volume, and temperature.
Entropy is an extensive property. (It increases w ith amount of substance.) All other factors being equal, entropy increases with number, volume, and temperature. On the MeAT, if a reaction increases the number of gaseous molecules, then that reaction ha s positive entropy (for the reaction system , not n ecessa ril y for the surroundings or the uni verse). The greater the temperature of a substan ce the greater its entropy. The third law of thermodynamics assigns by convention a zero entropy value to any pure substance (either an elem ent or a compound) at absolute zero and in internal equilibrium. At absolute zero, atoms have very little motion. Absolute zero temperature is tmattainable. The lmits for entropy are J/K. Entrop y change is defined mathematically by the infinitesima l change in heat dq ~, per kelvin in a reversible process:
Since entropy is a state func tion, the change in entropy for any process w ill be the same as the change in en tropy for a reversible process between the same two states. Imagine what happens when you slide a box across the floor. The kinetic energy of the box is diSSipa ted into internal en ergy of the box and floor via molecular collisions collectively called friction. O f course, energy is conserved; the increase in internal energy equals the initia l kinetic energy. Now imagine the reverse reaction: the molecules of the floor and the box happen to be moving in a coordinated fash ion so as to collide and make the stationary box sudden ly start movLook out, Crouton! ing. The internal energy of the molecu les becomes kinetic energy. Why Reverse friction could doesn't this happen in real life? Energy is still conserved in the restart t hese boxes moving verse reaction, so there is no violation of the first law of and crush you like a grape. thennodynamics. The on ly reason boxes don't spontaneously start sliding across the floor is due to the decrease in entropy \·v hkh would But there 's no need to fear! accompany such a reaction.
Captain Entropy is here.
Copyright © 2007 Examkrackers, Inc.
LECTURE
3:
THERMO DYNAM ICS . 61
3.13 Gibbs Free Energy Recall that equilibrium is achieved by maximizing the entropy of the universe. We can restate this relationship in terms of only the system by using the equation: ~Ssurmundings :::: dqrc\j T. We know that at constant pressure the change in enthalpy is equal to the heat; thus, ~S~urrounding~ :::: ~surroundingJT. Also, since ilHsurroundings :::: -IlHsyskm under these conditions, we have L'lSslIl'mundings :::: -~systcnjT. Therefore, in a closed system capable of doing only PV work, and at constant temperature and pressure, equilibrium is achieved by maximizing universal entropy via the equation:
If we multiply through by - T. and substitute t.G for -t.Son;,.""T, we have the important MeAT equation for G ibbs free en ergy G :
AG = Ml - TAS All variables in this equation refer to the system and not the surroundings. This equation is good only for constant temperature reactions, and loses some significance jf pressure is not held constant. Before \ve multiplied the !151.miverse equation by -T, we wanted to maximize .1.Sunivcrsc in order to achieve equilibrium. Since we multiplied by a negative to arrive at the Gibbs function, we must minimize the t.G in the Gibbs function in order to achieve equilibrium. Thus equilibrium is achieved when the change in the Gibbs free energy is zero (constant temperature and pressure, PV work only, and a reversible process). A reaction where the change in the Gibbs energy is negative indicates an increase in i1Sun ivnsc' and such a reaction is said to occur spontaneously. This definition of spontaneity is derived from the Gibbs function, and requires constant temperature and pressure, PV work only, and a reversible process. The true definition of a spontaneous reaction requires only that L1Sunivcrsc be positive under any conditions. For the MCAT however, a negative LlG from the Gibbs function is good enough for s pontaneity. Gibbs energy is an extensive property and a state function. It is not conserved in the sense of the conservation of energy law. An isolated system can change its Gibbs energy. The Gibbs energy represents the maximum non-PV work available from a reaction, hence the name 'free energy'. Contracting muscles, transmitting nerves, and batteries are some examples of things that do only non-PV work, making Gibbs energy a useful quantity when analyzing these systems. You must know the Gibbs function, and, most importantly, that a nega tive t.G indicates a spontaneous reactioll . Realize that the Gibbs ftmction deals with the change in enthalpy and entropy of a system . Ifa reaction produces a positive change in enthalpy and a negative change in entropy, the reaction can never be spontaneous. Conversely, if a reaction produces a negative chi'l11gc in entl1alpy and • a positive change in entropy, it must be spontaneous. lf 6G = 6H - T 6S the signs of both enthalpy and entropy are the same for a reaction, the spontC'meity of the reaction will depend + + upon temperature. A higher tempera+ ture will favor the direction favored +/by entropy . Remember, these changes Jre changes in the system + +/- + and not the surrowldings.
•
\
I
Copyright © 2007 Examkrackers, Inc.
Important MeAT equation I
A negative LlG indicates a spontaneous
reation .
62
MCAT INORGANIC CHEM!STRY
To finish this lecture, let's summarize the ](nvs of thermodynamics. It's not so important to memorize the laws, as it is to understand them and keep them in mind. The first and second laws are the important olles for the MeAT. Law: 1\'\10 bodies in thermal equilibrium with the same system are in thenual equilibrium with each other. (1n other vvords, temperature exists and is a state function.) oth
rt Law: The energy of an isolated systelTI is conserved for any reaction . 2 nd Law: The entropy of the universe increases for any reaction.
3rd Law: A perfect crystal at zero kelvin is aSSigned an entropy value of zero. All other substances and all tenlperatures have a positive entropy value. (Zero kelvin is unattainable.)
Copyright (C) 2007 Examkrackers, Inc.
70. The reaction below shows the condensation of water.
Questions 65 through 72 are NOT based on a descriptive passage,
H,o(g) --7 H,o(l)
Which of the following will be positive for the water at 25°C and 1 atm? 65. Which of the following describes a reaction that is always spontaneous'? A. B. C. D.
increasing enthalpy and increasing entropy decreasing enthalpy and decreasing entropy increasing enthalpy and decreasing entropy decreasing enthalpy and increasing entropy
B. C. D.
B.
C.
D.
C.
/,C
D.
None of the above,
/,S
AgCl(s)
--7
Ag+(aq) + Cr(aq)
During the course of the reaction above, both entropy and enthalpy are increased. If the reaction is not spontaneous at a given temperature and pressure, what car. be done to make the reaction occur spontaneously?
The entropy of a system will always increase in a spontaneous reaction. Entropy is a measure of disorder. The entropy change of a forward reaction is exactly opposite to the entropy of the reverse reaction. Entropy increases with temperature.
A. B. C. D.
67. Which of the following is a violation of the law of conservation of energy? A.
t,H
71.
66. \Vhich of the following statements about entropy is false? A.
A. B.
Increase the temperature. Decrease the temperature. Increase the pressure. Decrease the pressure.
72. The normal boiling point of benzene (C 6H 6) is 80,1 0c. If the partial pressure of benzene gas is I atm, which of the following is true of the reaction shown below at 80.1 DC?
Heat can be changed completely to work in cyclical process. A system undergoing a reaction with constant enthalpy experiences a temperature change. After sliding to a stop, a box with initial kinetic energy K has only thermal energy in an amount less than K. A bond is broken and energy is released.
A. B. C. D.
1'1S is negative 1'15 is zero 1'1G is negati ve 1'1G is zero
68. All of the following are examples of processes which increase system entropy EXCEPT: A. B. C. D.
the expanding universe aerobic respiration melting ice building a bridge
69, Which of the following statements is most likely true concerning the reaction: 2A(g) + B(g) --7 2C(g) + D(s)
A. B. C. D.
System entropy is decreasing. System entropy is increasing. The reaction is spontaneous. The reaction is nonspontaneous.
Copyright © 2007 Examk rackers. Inc
63
S T OP.
Solutions
4.1
Solutions
A solution is a hOInogeneous mixture of two or more compounds in a single phase, such as solid, liquid, or gas. The MeAT w ill probably test your knowledge of liquid solutions only. However, you should be aware that solu tions are possible in other phases as well. Brass is an example of a solid solution of zinc and copper. Generally, in a solution w ith two compounds, the compound of which there is more is called the solvent, and the compound of which there is less is called the solute. Sometimes, when neither compound predominates, both compounds are referred to as solvents. Although a compound's behavior does depend upon the molecules around it, the label of 'solvent' or 'solute' does not indicate a particular behavior.
There are ideal solutions, ideally dilute soilltions, and nonideal solutions. Ideal solutions are solutions made from compounds that have similar p roperties. In other words, the compounds can be interchanged within the solution without changing the spatial arrangement of the molecules or the intermolecular attractions. Benzene in toluene is an example of a nearly ideal solution because both compounds have similar bonding properties and similar size. In an ideally dilute solution, the solute molecules are completely separated by solvent molecules so that they have no interaction with each other. Nonideal solutions violate both of these conditions. On the MCAT, you can assume that you are dealing with an idea lly dilute solution unless otherwise indicated; however, you should not automatically assume that an MCAT solution is ideal. +
+
Solvent usually indicates the compound that predominates in a solution. The concepts of an ideal and an ideally dilute solution are not tested directly on the MeAT. They are mentioned here in order to deepen your understanding of solutions and to help explain some apparent paradoxes which result when they are not considered . In ideally dilute soluUons, the mole fracUon
,
(
•
..
~
+
+
+
., " (
(
,
'. +
+
.. + .. +
'"
+
66
MCAT INORGANIC CHEMISTRY
4.2
A colloid is like a solution, only the solute pa rticles are larger. The colloid particles are usually too small to be extracted by filtration but usua lly large enough or charged enough to be separated by a sem ipermeable
Colloids
Particles larger than small molecules may form mixtures with solvents. If gravity does not cause these particles to settle out of the mixture over time, the mixture is called a colloidal syste111, or colloid. (The term 'colloid' can also refer to the colloidal particles.) Colloidal particles are larger than solute particles, and can even be single large molecules such as hemoglobin. A colloidal system can be any combination of phases (except gas in gas). Some examples of colloidal systems are an aerosol (liquid or solid particles in a gas like fog or smoke), afoa111 (gas particles in a liquid like whipped cream), an e111ulsion (liquid particles in a liquid or solid like milk or butter), or a sol (solid particles in a liquid like paint).
membrane.
Unlike a true solution, colloidal suspensions will scatter light, a phenomenon known as the Tyndall effect. The beam of light in a smoke filled theatre is visible due to the Tyndall effect. Colloidal particles may be attracted (lyophilic) or repelled (lyophobic) by their dispersion 111ediu111. (The dispersion medium in a colloid is analagous to the solvent in a solution.) Lyophobic colloids form when amphipathic or charged particles adsorb to the surface of the colloidal particles stabilizing them in the dispersion medium. Protein in water is an example of a lyophilic colloid; emulsyfied fat in water is an example of a lyophobic colloid. Colloidal particles are usually too small to be extracted by filtration; however, heating a colloid or adding an electrolyte may cause the particles to coagulate. The larger particles produced by coagulation will settle out or can be extracted by filtration. Colloidal systems can also be separated by a semipermeable membrane, a process called dialysis.
4.3
More Solutions
When a solute is mixed with a solvent, it is said to dissolve. The general rule for dissolution is 'like dissolves like'. This rule refers to the polarity of the solute and solvent. Highly polar molecules are held together by strong intermolecular bonds Like dissolves like; polar solvents formed by the attraction between their partially charged ends. Nonpolar molecules dissolve polar solutes; nonpolar solvents are held together by weak intermolecular bonds resulting from instantaneous didissolve nonpolar solutes. pole moments. These forces are called London dispersion forces . A polar solute interacts strongly with a polar solvent by tearing the solv ent-solvent bonds apart and forming solvent-solute bonds. A nonpolar solute does not have enough charge separation to interact effectively with a polar solvent, and thus cannot intersperse itself within the solvent. A nonpolar solute can, Come on buddy! I however, tear apart the weak bonds of a nonWe're stickin' you can join our club. polar solvent. The bonds of a polar solute are together, cracker boy. too strong to be broken by the weak forces of a nonpolar solvent.
Nonpolar Nerd's Club
Nonpolar Salty
The Polar Muscle Club
Ionic compounds are dissolved by polar substances. When ionic compounds dissolve, they break apart into their respective cations and anions and are surrounded by the oppositely charged ends of the polar solvent. This process is called solvation. Water acts as a good solvent for ionic substances. The water molecules surround the individual ions pointing their positive hydrogens at the anions and their negative oxygens toward the
Copyright © 2007 Examkrackers, Inc.
LECTURE 4 : SOLUTIONS • 67
cations. When several water molecules attach to one side of an ionic compoW1d, they are able to overcome the strong ionic bond, and break apart the compound. The molecules then surround the ion. In water this process is called hydration. Something that is hydrated is said to be in an aqueous phase. The number of water molecules n eeded to surround an ion varies according to the size and charge of the ion. This number is called the hydration number. The hydration number is commonly 4 or 6. For the MCAT you should be aware of common names, formulae, and charges for the polyatomic ions listed on the right. VVhen ions form in aqueous solution, the solution is able to conduct electricity. A compound which forms ions in aqueous solution is called Hydration shell an electrolyte. Strong electrolytes create solutions which conduct electricity well and contain many ions. Weak electrolytes are compounds which form few ions in solution.
4.4
Water is a poor conductor of electricity un less it conta ins electrol ytes. Electrolytes are compo unds that form
ions in aqueous solution .
Name
Formula
nitrite
NO,-
nitrate
NO,-
sulfite sulfate
sot sot
hypochlorite
CIa-
chlorite
ClO,-
chlorate
ClO,-
perchlorate
ClO,-
carbonate
CO,'-
bicarbonate
HCO,-
phosphate
pot
Units of Concentration
There are several ways to measure the concentration of a solution; fi ve of which you should know for the MCAT: molarity (M), molality (m), mole fraction (X), mass percentage and parts per million (ppm). Molarity is the moles of the compound divided by the volume of the solution. Molarity generally has units of mol/L. Molality is moles of solute divided by kilograms of solvent. Molality generally has units of mol/kg and is usually used in formulae for colligative properties. The mole fraction is the moles of a compound d ivided by the total moles of all species in solution. Since it is a ratio, mole fraction has no W1its. Mass percentage is 100 times the ratio of the mass of the solute to the total mass of the solution. Parts per million is 106 times th e ratio of the mass of the solute to the total mass of the solution.
M=
moles of solute volume of solution moles of solute kilograms of solvent
m=
x=
moles of solute total moles of all solutes and solvent
mass
%
ppm=
=
mass of solute x 100 total mass of solution
mass of solute total mass of solution
X
106
Notice that parts per mi llion is NOT number of solute molecules per million molecules. It is the mass of the solute
per mass of solution times one million . Copyright © 2007 Examkrackers, Inc.
68
MCAT INORGANIC CHEMISTRY
Remember that solution concentrations are always given in terms of the form of the solute before dissolution. For instance, when 1 mole of NaCl is added to 1 liter of water, it is approxhnately a 1 molar solution and NOT a 2 molar solution even though each NaCl dissociates into two ions.
Normality measures the number of equivalents per liter of solution . The definition of an equivalent will depend upon the type of reaction taking place in the solution. The only time normality is likely to appear on the MeAT is with an acid-base reaction. In an acid-base reaction an equivalent is defined as the mass of acid or base that can furnish or accept one mole of protons. For instance, a 1 molar solution of H 2S04 would be called a 2 normal solution because it can donate 2 protons for each H 2S04 ,
Copyright © 2007 Exarnkrackers, Inc.
78. A solution contains 19 grams of MgClz in 0.5 liters of di stilled water. If MgCI 2 totally di ssociates, what is the co ncentrati on of chl oride ions in the solution?
Qu estions 73 through 80 are NOT based on a descriptive passage.
A. B. C. D.
73. What is the appro ximate molarity of a NaC I solution with a specific grav ity of 1.006?
A. B. C.
D.
0.05M O.06M 0.1 M 0.2M
79. A student has 0.8 li ters of a 3 molar HCl solu tion. How many liters of distill ed water must she mix with the 3 mo lar solution in order to create a I molar HCl solution?
A. B.
74. Which of the following substances is least solu ble in water?
A.
B. C. D.
C. D.
NH, NaCI HS04-
D.
0.8 1.6 2.4 3.2
L L L L
80. All of the foll owing substances are strong elec troly tes EXCEPT:
CC!4
A.
75. Which of the following solutions is the most concentrated? (Assume I L of water has a mass of I kg.) A. B. C.
0.1 M 0.2M O.4M O. 8M
B. C.
D.
I MNaCI 1 III NaCI A aqu eous solution with a NaCl mol e fraction of 0.01 55 grams of NaCI mixed w1th one li ter of water.
HN0 3 CO, NaCI KOH
76. The air we breathe is approximately 2 1% 0 , an d 79% N,. 1f th e partial pressure of ni troge n in air is 600 torr, then all of the following are true EXCEPT: A. B. C.
The mole fraction of nitrogen in air is 0.79 . The mass of nitrogen in a 22.4 L sampl e of air is 22. 1 grams at O°C. The partial pressure of oxygen is approx imately 160 torr.
D.
For every 21 grams of oxygen in an air sa mple, there are 79 grams of njtrogen.
77. A polar solute is poured in to a container wi th a nonpolar solvent. \Vhich of the fo llowi ng statements best ex plains the reaction? A. B. C.
D.
The stron g dipoles of the polar molec ules separate the weak bonds between the nonpolar molecules. The dipoles of the polar molecules are too weak to break the bonds between the nonpolar molecules. The instantaneous dipo les of the nonpol ar molecules are too weak to separate the bonds between the pol ar mo lec ules. The instantaneous dipoles of the nonpolar molecules separare the bonds between the polar molecules.
Copyright © 2007 Examkrackers, Inc.
69
STOP..
70
MCAT INORGANIC CHEMISTRY
4.5
Solution Formation
11,e formation of a solution is a physical reaction. It involves three steps: Step 1: the breaking of the intermolecular bonds between solute molecules; Step 2: the breaking 01 the intermolecular bonds between solvent molecules; Step 3: the lormation 01 intermolecular bonds between the solvent and the solute molecules. Energy is required in order to break a bond. Recallirom Chemistry Lecture 3 that at constant pressure the enthalpy change of a reaction equals the heat: Mi = q, and that for condensed phases not at high pressure (lor instance the formation 01 most MCAT solutions) enthalpy change approximately equals internal energy change: Mi 6:) flU. For solution chemistry we shall use these approximations. Thus the heat of solution is given by: Aliso!
= Llif 1
+ !l.H2 + AH3
Since energy is required to break a bond, the lirst two steps in dissolution are endothermic while the third step is exothermic.
Solute Step 1 Endothermic step. Solute-solute bond are broken.
Solvent
Solution Step 3 Exothermic step. Solvent-solute bonds are formed.
Step 2 Endothermic step. Solvent-solvent bonds are broken.
-!!.H,
+!!.H, (
You must recognize that breaking a bond always requ ires energy input. Since enthalpy and heat are equal at constant pressure, a solution with a negative enthalpy will give off heat when it forms. Thus, a solution that gives off heat when it forms is creating stronger bonds within the solution.
)
If the overall reaction releases energy (is exothermic), the new intermolecular bonds are more stable th an the old, and, in general, the intermolecular attractions w ithin the solution are stronger than the intermolecular attractions within the pure sub-
stances. (Remember, less energy in the system usually means a more stable system.) If the overall reaction absorbs energy (is endothermic), the reverse is true. Using the approximations mentioned above, the overall change in energy of the reaction is equal to the change in enthalpy and is called the heat of solution Mi"" . A negative heat 01 solution results in s tronger intermolecular bonds, while a positive h eat of solution results in weaker intermolecular bonds. (Some books combine steps 2 and 3 of solution formation for aqueous solutions calling the sum of their enthalpy changes the heat of hydration.) Copyright © 2007 Examkrackers, Inc.
LECTURE 4: SOLUTIONS •
Since the combined mixture is more disordered than the separated p ure substances, most of the time, the formation of a solution involves an increase in entropy. In fact, positional entropy always increases in the formation of a solution, so, on the MeAT, solution formation has positive entropy.
4.6
71
When solutions form , entropy increases.
Vapor Pressure
Imagine a pure liquid in a vacuum-sealed container. If we w ere to examine the space inside the container, above the liquid, we would find tha t it is no t really a vacuum. Instead it would contain va por molecules from the liquid. The liquid molecules are h eld in the liquid by intermolecular bonds. However, they contain a certain amount of kinetic energy, which depends upon the te mperature. Some of the liquid molecules at the surface contain enough kinetic energy to break the intermolecular bonds that hold them in the liquid. These molecules la unch themselves into the open space above the liquid. As the space fills with m olecules, some of the molecules crash b ack into the liquid. When the rate of molecules leaving the liquid equals the rate of molecules entering the liquid, equilibrium has been established. A t this point, the pressure created by the molecules in the open space is called the vapor pressure of the liquid.
Equilibrium between the liquid and gas phases of a compound occurs when the molecules move from liquid to gas as quickl;y as they move from gas to liquid. The vapor pressure necessary to bring the liquid and gas phases of a compound to equilibrium is called the vapor pressure of the compound.
(;.;;
0
Less vapor
Q
More vapor
}
Solution
Since vapor pressure is rela ted to the kinetic energy of the molecules, vapor pressure is a function of temperature. A derivation of the Clausius-Clapeyron equation relates vapor pressure and temperature to the heat of vaporization: Mi"p In(P,) = --R -
(1) T
+C
where Mi"p is the h eat of vaporization, and C is a constant specific to the comp ound. Vaporization is an endothermic process, so the equation indicates tha t vapor pressure increases w ith temperature. When vapor pressure equals local atmospheric pressure, a compound boils. Solids also have a vapor presssure. The melting point is the temperature at which the vapor pressures of the solid is equal to the vapor pressure of the liqUid. Above the m elting point the liquid vapor pressure is greater than that of the solid; below the m elting point the liquid vapor pressure is less than that of the solid. When a nonvolatile solute (a solute with no vapor pressure) is added to a liquid, some of those solute molecules will reach the surface of the solution, and reduce the amount of surface area available for the liquid molecules. Since the solute molecules Copyright © 2007 Examkrackers, Inc.
Vapor pressure temperature.
increases
with
Boiling occurs when the vapor pressure
of a liq uid equals the atmospheric pressure . Melting occurs when the vapor pressure of the solid phase equals the vapor pressure of the liquid phase.
72 .
MCAT INORGANIC CHEMISTRY
Raoul!,s law for nonvolatile solutes: If 97% of the solution is solvent, then the vapor pressure will be 97% of the vapor pressure of the pure solvent. Raoul!'s law for volatile solutes: If 97% of the solution IS solvent, then the vapor pressure will be 97% of the vapor pressure of the pure solvent PLUS 3% of the vapor pressure of the pure solute.
don't break free of the solution but do take up surface area, the nu mber of molecules breaking free from the liquid is decreased while the surface area of the solution and the volume of open space above the solution remain the same. From the ideal gas law, PV = nRT, we know tha t a decrease in n at constant volume and tempera ture is proportional to a decrease in P. The vapor pressure of the solution P, is given by Raoult's law, and is proportional to the mole fraction of the liquid a and the vapor pressure of th e p ure liquid P,.
If the solute is a volatile solute (a solute with a vapor pressure), the situation is a little more complicated. A vola tile solute will also compete for the surface area of a liquid. H owever, some of the molecules of a volatile solu te will escape from solution and contribute to the vapor p ressure. If the solution is an ideal solution (sol ute and solvent h ave similar properties), the partial pressures contributed by the solvent and solute can be found by applying Raoult's law separately. The sum of the partial pressures gives the total vapor pressure of the solution, and we arrive at a modified form of Raoul!'s law:
w h ere each xP term represents the partial pressure contributed by the respective solven t, and p. represents the total vapor pressure.
Negative heats of solution form stronger bonds and lower vapor pressure; positive heats of solution form weaker bonds and raise vapor pressure.
But this is not the entire story. As we saw with heats of solution, if the solution is not ideal, the intermolecular forces between m olecules will be changed. Either less energy or more energy will be reqUired for m olecules to break the intermolecular bonds and leave the surface of the solution. This means that the vapor pressure of a n onideal solution will d eviate from the predictions made by Raoul!'s law. We can make a general pred iction of the d irection of the deviation based upon hea ts of solution. If the heat of solution is negative, stronger bonds are formed , fewer molecules are able to break free from the surface and there will be a negative deviation of the vapor pressure from Raoul!'s law. The opposite will occur for a positive hea t of solution.
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LECTURE 4: SOLUTIONS . 73
The deviation of vapor pressure from Raoul!'s law can be represented graphically by comparing the mole fractions of solvents with their vapor pressures. Graph 1 below shows only the p artial pressure of the solvent as its mole fraction increases. As pred icted by Raoul!,s law, the relationship is linear. Graph 2 shows the vapor pressure of an ideal solution and the individual partial p ressures of each solvent. N otice that the p artial pressures add at every point to equal the total p ressure. This must be true for any solution. Graph 3 and 4 show the deviation s of nonideal solutions. The straight lines are the Raoult's law predictions and the curved lines are the actual pressures. N otice that the partial pressures still add at every point to equal the to tal pressure. N otice also that a positive heat of solution leads to an increase in vapor pressure, and a negative heat of solution, to a decrease in vapor pressure.
In order to really understand this section you must have a thorough understanding of many of the physics and chemistry concepts that we've studied so far (i.e. bond energy, thermodynamics, pressure, and solutions). I suggest that you re-read this section and be sure that you thoroughly understand the concepts.
+f..H /---...
. --... . .................... . --...
P...
"
IlJ H
;::l
--...--... --...,
CfJ CfJ
1
;:J
IlJ
IlJ
P,
P,
6
X,, -
,
1 0
0
X,l - 1 - Xb
Graph 2
Graph 3
- - - Psolution ----------. p. p.lfti~l l
P bPilTti
Copyright © 2007 Examk rackers, Inc.
"""-
H
1 - Xb Graph 1
IlJ CfJ CfJ
H
o
""
.......
P...
1 0
o I
X, -
-
Xb Graph 4
i 0
85. Wh en two volatile solven ts are mixed, the vapor press ure drops below the vapor pressure of either solvent in its pure form. What else can be predicted about the solution of these sol vents?
Questions 81 th rough 88 are N OT based on a descriptive passag e.
A. B.
81. NaCI di ssolves sponta neo usly in water. Based upon the fo llowing reaction: NaC I(s) -> Na+(g) + C1-(g) MI = 786 kJ/ mol
C. D.
the heat of hydrati on for NaCI must be: A. B. C. D.
negative with a magnitude less than 786. negative with a magnitude greater than 786. positi ve wi th a magnitude greater than 786. Nothing can be determined about th e heat of hydration without more info rmation.
86. A solution composed of ethanol and methanol can be thought of as ideal. At room te mperature, the vapor pressure of eth anol is 45 mmHg and the vapor press ure of methanol is 95 mm Hg. Which of the followi ng will be true regarding the vapor press ure of a solu ti on containing onl y ethanol and methanol?
82. Whi ch of the foll owing indicates an exothermi c heat of solutio n?
A. B. C. D.
A. B.
Heat is evolved. The final sol ution is acidi c. A preci pitate is formed. The reaction is sponta neo us.
C. D.
83. W hen two pure li quids, A and B , are mi xed, the
B. C. D.
The intermolecular bond strength in al least one of the liquids is less th an the intermolecular bond strength between A and B in solution. The reaction is exothermi c. The vapor press ure of the solution is less th an both the vapor press ure of pure A and pure B. The rm s velocity of the molecul es increases when the solutio n is for med.
A. B. C. D.
B. C. D.
A.
increasing the-surface area of [he liquid by pouring it into a wider container increasing the kinet ic energy of the molecules of the liqu id decreasing the temperature of the liquid adding a nonvolatile sol ute
Copyright © 2007 Examkr
60% 50% 40% 20%
benzene benzene be nzene benzene
a nd and and and
40% toluene 50% tolue ne 60% toluene 80% to luene
88. 'When solute A is added to solvent B, heat is released. Whi ch of the following must be true of the solvation process?
84. Which of the follow ing wil l increase the vapor pressure of a liqui d? A.
It will be less than 45 mmHg. [t will be greater than 45 mmHg and less than 95 mmHg. It wi ll be greater than 95 mmHg and less than 140 mmH g. It will be greater than 140 mmHg.
87. Benzene and tolue ne co mbine to fo rm an ideal solution. At 800C. vapor pressure of pure benzene is 800 mmH g and the vapor pressure of pure toluene is 300 mmHg. If the vapor pressure of the solution is 400 mm Hg, what are the mole fractions of benze ne and toluene?
te mperature of the solut ion increases. All of the following must be true EXCEPT: A.
The solution is ideal. The mole frac tio n of the more volatile solvent is greater than the mole fraction of the less volatile solvent. The heat of solutio n is exothermic. The heat of solution is endothermic.
B. C. D.
74
The bonds broken in solute A must be stronger th an the bonds broken in solvent B. The bonds broken in solute A must be weaker th an the bonds broken in solvent B. The bonds for med in the solution must be stronger than the bonds broke n in solute A and solvent B. The bonds fo rmed in the solution must be weaker than the bonds broken in solute A and solve nt B.
STOP.
LECTURE 4: SOLUTIONS . 75
4.7
So lubility
Solubility is a solu te's tendency to dissolve in a solvent. On the MeAT, the solute w ill us ually be a salt, and the solvent will most often be wa ter. Dissolution of a salt is reversible on a molecular scale; dissolved molecules of the salt reattach to the surface of the salt crys ta l. For a dissolv ing sait, the reverse reaction, calle d precipitation, takes place initially at a slower rate than dissolution. As the salt dissolves and the con centration of dissol ved salt builds, the rate of dissolution and precipitation equilibrate. When the rate of dissolution and precipitation are equal, the solution is said to be saturated; the concentration of d issolved salt has reached a maxin1U1n in a saturated solution. Just like any other reac tion the equilibrium establish ed a t the saturation point is dynamic; the concentrations of products and reactants remain constant, but the forward and reverse reactions continue at the saIn e rate. f
The equilibrium of a solvation reaction has its own equilibrium constant called the
solubility product K ,p' Use K,p the same way you wo uld use any other equilibrium constant. Remember that solids and pure liquids h ave an approx.ima te mole fraction of on e and can be excluded from the equilibrium expression. Thus, solids are left out of the solubility product expression as in the example of the K,p for barium h ydroxide shown b elow. Ba(OH)2(s) '" Ba'+(aq) + 20W(aq) K
'P
Use K,p like any other equilibrium constant to create an equilibrium expression. Set the K,pequal to products over reactants raised to the power of their coefficients in the balanced equation . As always, leave out pure solids and liquids.
= [Ba2+][OH-l'
Solubility and the solubility product are not the sa me thin g. The solubility of a substance in a given solvent is found from the solubility product. The solubility is the number of moles of solute per liter of a solution tha t can be dissolved in a given solvent. Solubility d epends upon the common ions in the solution. The solubility constan t is independent of the common ions, and can be found in a reference book.
Solubility and the solubility product are not the same thing. Solubility product or K" is a constant found in a book. Solubility is the maximum number of moles of the solute that can dissolve in solution .
For most salts, crystallization is exothermic.
We can w rite an equation for the solvation of BaF, in water as follows: BaF,(s) '" Ba'+(aq) + 2r (aq) The solubili ty product for BaF, is: K ,p = [Ba' +][rl'
If we look in a book, we find that the K,., for BaP, has a value of 2.4 x 10-3 at 25"C. Like any equilibrium constant, the K" i~ ' unit!ess. From the K,p we can find the solubility o f BaP, in any solution at 25"C. For instan ce, to find the solubility of BaF, in one liter of water, we simply saturate one liter of water with BaP2• The solubility is the m ax imum number of moles per liter that can dissolve in the solution. We call the solubility 'x', since it is unknown. If x moles per liter of BaF, dissolve, then there w ill be x 1110ies per liter of Ha z+ in solution and wice as nlany, or 2x moles per liter, of r . We plug these values irito the K,,>equa tion a nd solve. 2.4 x 10-5
x
EI)
= (x)(2x) '
1.8 X 10-'
1.8 X 10-' mol/ L is the solubility of BaF, in one liter o f water at 25°C. What would happen if we added 1 mole of P- to OllT solution in the form of NaP? The solubili ty of BaP, would change. The NaP would completely dissociate forming 1 mole of P- and 1 mole of Na ' . The Na' ions are not in the equilibrium
Copyright © 2007 Examkrackers, )nc.
The solubility product changes only with temperature. The solubility depends upon the temperature and the ions in solution.
76
MCAT INORGA NIC CHEMISTRY
xBaF,
~
t
A common ion added to a saturated solution will shift the equilibrium increasing precipitate. It does not affect the K",.
expression and (ideally) would have no effect on the equilibrium. Because they have no effect, the Na· ions are called spectator ions. The I NaF P- ions, however, do affect the equilibrium. Their disturbance of the equilibrium is called the common ion effect because it involves an ion common to an ion in the equilibrium expression. By Le Chat§Jier's principle, the addition of a common ion will push the equilibrium in a direction which tends to reduce the concentra tion of that ion. In this case, the equilibrium w ill move to the left, and the solubility of BaF, will be reduced. To find out by exactly how much the solubility will be reduced, we go back to the equilibrium expression. One key to solvin g solubility problems is realizing that the order in which you mix the solution is irrelevant, so you should mix them in the order tha t is most convenient to you. In this case it is easiest to add the NaF first, since it completely dissoci ates. Now we add BaF, to a solution of 1 liter of water and 1 mole of P-. Again, x moles will dissolve leaving x moles of Ba' · . But this time, since there is already 1 mole of P-, 2x + 1 moles of P- will be in solution at equilibrium. 2.4 X 10-5 = (x)(2x + I)'
A common ion added to a solution that
is not saturated will NOT shift the equilibrium, because in an unsaturated solution, there is no equi librium to shift.
Now h ere's a trick to simplify the math. We know that the equilibrium is shifting to the left, so x will be smaller than our earlier calculations of 1.8 x 10-'. Even 2, will be much smaller than 1. Thus, 2x + 1 is going to be very close to 1. Therefore, we drop the 2. and solve: 2.4 x 10-5 $ (x)(I)'
x $ 2.4 X 10-' Just to be sure tha t we were correct in our estimation of 2x, we plug our estimated val ue of x into the term that we dele ted (2.), and we see if it is truly much smaller than the term to which we added it (in this case 1). 2x = 4.8
X
10-5 « 1
Our assumption was valid. Thus our new solubility of BaF, is 2.4 x 10-5 mol/L.
4.8
Solubility Guidelines
Co mpounds with water solubilities of less than 0.01 mol V l are generally consid ered insoluble. MCAT w ill not require you to m emorize the solubilities of different compounds. Nevertheless, h ere are a few solubility guidelines for compounds in water:
It is very unlikely that an MeAT question would require that you know these solubilities. However, knowing them will make solution chemistry easier.
Nearly all ionic compounds containing nitrate (NO,-), ammonium (NH; ), and alkali metals (Li· , N a·, K· ... ) are solLlble. Ionic compounds con taining halogens (Cl-, Br-, r-) are solLlble, EXCEPT for silver, mercury, and lead compounds (Ag·, Hg,'·, Pb' "). Sulfate compounds (50 / -) are soluble, EXCEPT for mercury, lead, and the heavier alkaline metals (Hg/"'", Pb2 t , Ca2+, Sr2+ Ba 2+), I
Compounds containing the heavier alkaline metals (Ca'· , Sr" , Ba'· ) are solLlble when paired with sulfides (5' -) and hydroxides (OW). Carbonates, phospha tes, sulfides, and hydroxides (CO,", PO,}-, 5'-, OH-) are genera lly insoillble other than in the cases mentioned above.
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LECTURE 4 : SOLUTIONS .
4.9
77
Solubility Factors
Pressure and temperature affect solubilities. Pressure on liquids and solids has little eflect, but pressure on a gas increases its solubility. For an ideally dilute solution, the increase in pressure of gas a over a solution is directly proportional to the solubility of gas a, if the gas does not react with, or dissociate in, the solvent. This relationship is given by Henry's law:
where C is the solubility of the gas a (typically in moles per liter), k" is Henry's law constant, which varies with each solute-solvent pair, and P v is the vapor partial pressure of gas a above the solution. Strangely, Henry's law can also be written as: Po=
x.k.,
where X. is the mole fraction of a in solution, p. is the vapor partial pressure of gas a, and k" is Henry's law constant. Although both equations show that the concentration of a gas in solution is proportional to the vapor partial pressure of the gas above the solution, the Henry's law constant in the second equation has a different value than the Henry's law constant in the first equation. If we compare the second equation with Raoult's law (P, = x.,P,), th ey appear to conflict unless P, has the same value as k" . In fa ct, they do N OT agree. Both are approximations. Rault's law is most accurate when looking a t the vapor partial pressure of a solvent with high concentration. Henry's law is more accurate when looking at the va por partial pressure of a volatile solute where the solute has a low concentration. In other words, in. an ideally dilute solution, the solvent obeys Raault's law and tlte solute obeys Henry's law. One way to relnenlber this is w hen the solvent concen tration is high, each solvent molecule is surrounded by other solvent molecules, so it behaves more like a pure solvent. Thus the solvent vapor partial pressure is proportional to its vapor pressure as a pure liquid; Raoult's law. When the volatile solute concentration is low, each molecule is surrounded by solvent molecules crea ting a deviation from the behavior 01 the pure volatile solute. Thus its vapor partial pressure is not proportional to its pressure as a pure substance (Raoult's law doesn't work in this case.), b ut is proportional to some constant; Henry's law.
As shown by Raul!'s law and Henry's law, the partial vapor pressure of a solution component is always proportional to its mole fraction . If the component predominates as the solvent. Raul!'s law says that the partial vapor pressure is proportional to the pure vapor pressure. If the component represents a tiny amount of solution, Henry's law says that the vapor partial pressure is proportional to Henry's law constant.
The most important thing to remember about Henry's law is that it demonstrates that the solubility of a gas is proportional to its vapor partial pressure . We can remember this by thinking of a can of soda . Whe n we open the can and release the pressure, the solubility of the gas decreases causing some gas to rise out of the solution and create the familiar hiss and foam .
Le Chatalier's principle, when applied to solutions, should be used with caution. Because heat energy is a product of a reaction w ith a negative heat of solution, Le Chatalier's principle pred icts that a temperature increase will push such a reaction to the left decreasing the solubility of the solute. However, entropy increase is large in solution formation. From the equation llG =Mi - TllS, we see that a temperature increase emphasizes the llS term tending to result in a more negative llG and thus a more spontaneous rea ction. Due to the large increase in entropy.. the wa ter solubility of many solids increases with increaSing temperature regard less of the enth alpy change. To be absolutely certain, the change in solubili ty d ue to temperature must be found by experiment, but the solubility of most salts increases with tempera ture.
The solubility of gases, on the other hand, typically decreases with increasing temperature. You can remember this by understanding why hot waste wa ter from factories that is dumped into streams is hazardous to aquatic life. The hot water ha s a double effect. First, it holds less oxygen than cold water. Second, it floats on the cold wa ter and seals it off from the oxygen in the air above. O ther factors that affect the solubility 01 a gas are its size, and reactivity with the solvent. Heavier, larger gases experience greater va n der Waals forces and tend to be more soluble. Gases that chemically react with a solvent have greater solubility.
Copyright © 2007 Examkrackers, Inc.
As the temperature increases, the solubility of salts generally increases. Gases behave in the oppOSite fashion . As tem perature increases, the solubililty of gases decreases. The can of soda is useful here as well. If we place a can of soda on the stove, the gas escapes the solution and expands in the can causing it to explode. (This is not the only reason that the can explodes, but it is a good memory aid.)
94. \Vhich of the following expressions represents the solubility product for Cu(OH), ?
Questions 89 through 96 are NOT based on a descriptive passage.
A. B.
c.
89. When a solution is saturated: A. B. C. D.
D.
the solvent changes to solute, and the solute changes to solvent at an equal rate. the vapor pressure of the solution is equal to atmospheric press ure. the concentration of solvent is at a maximum. the concentration of solvent is at a minimum.
95. If the solubility of PbCl, is equal to x. which of the following expressions will be equal to the solubility product for PbCI,?
90. The addition of a strong base to a saturated solution of Ca(OH), would: A. B. C. D.
decrease the number of OH- ions in solution. increase the number of Ca2+ ions in solution. cause Ca(OH), to precipitate. decrease the pH.
A. B. C. D.
A. B.
4x3
C. D.
x'
2x'
r
96. A beaker contains a saturated solution of CaP2 (K.,p = 4 x 10- 11). There are some Na+ ions in the solution.lfNaF is added to the beaker, which of the following will occur? A. B. C. D.
91. Na2 S04 dissociates completely in water. From the information given in the table below, if N''-2S 0 4 were added to a solution containing equal concentrations of aqueous Ca2+, Ag+, Pb2+, and 8a2+ ions, which of the follow ing solids would precipitate first? Compound
K,p = [Cu' +J[OW]' K w = [Cu' +],[OW] Kw = [Cu2+J'[owl' K", = [Cu"][OW]
The concentration of Na+ will decrease. The concentration of Ca2+ \\fill decrease. The concentration of P- will decrease. All concentrations will remain constant.
K,p
CaSO,
6. 1 x 10-5
Ag 2S0 4
1.2 x 10
PbS0 4
1.3 x la '
BaS04
1.5 x 10 '
5
CaS04 Ag, S04 PbS04 BaS04
92. A sealed container holds gaseous oxygen and liquid water. Which of the following would increase the amount of oxygen dissolved in the water? A. B. C. D.
expanding the size of the container adding an inert gas to the container decreasing the temperature of the container. shaking the container
93. The K,p of Bae0 3 is 1.6x 10-'-1 . How many moles of barium carbonate can be dissolved in 3 liters of water? A. B. C. D.
4 X 10-5 moles 6.9 x 10-5 moles 1.2 x 10-4 moles 2.1 x 10-4 moles
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78
STOP.
Heat Capacity, Phase Change, and Colligative Properties 5.1
Phases
If aU the intensive macroscopic properties of a system are constant, that system is said to be homogeneous. Any part of a system that is homogeneous is called a phase. A phase is uniform th roughout with respect to chemical composition and physical state. Some examples of different phases are crystalline solid, amorphous solid, aqueous, pure liquid, and vapor. A system may have a number of solid and liquid phases, but it will usually ha ve only one gaseous phase. Pure substances have only one gaseous phase and usually have only one liquid phase. Most of the time, yo u can think of phases as solid, liquid, and gas. Just be aware that this is not the technical definition. And w hen we discuss things like solutions, remember that pure water is a different phase than aqueous Na ' ions. Another common example is rho mbic sulfur and monoclinic sulfur; these are hvo different solid phases of the same element.
t
80
MCAT INORGAN IC CHEMISTRY
5.2
Heat Capacity
Phase changes arise through changes in the manner in which internal energy is distributed over molecules and space. In other words, a phase change may result when the energy of each molecule is decreased or increased, or 'when the space around each molecule is reduced or enlarged. Such changes are accomplished via heat or work. In order to understand phase changes then, we must understand how substances react to heat and PV work. If, while being heated, no PV work is done by a system at rest, nearly ali the heat energy goes into increasing the temperature. When the system is allowed
Heat capacity C is a measure of the energy change needed to change the temperature of a substance. The heat capacity is defined as:
to expand at constant pressure, some
energy leaves the system as work and the temperature increase is diminished .
Thus constant pressure heat capacities are greater than constant volume heat capacities.
Don't let the name 'heat capacity' fool you. Recall that heat is a process of energy transfer, and caml0t be stored. Heat capacity was given its name before heat was fully understood. 'Internal energy capacity' would be a better, but not perfect, name. Not perfect because we can change the temperature of a substance at constant internal energy by changing only the volume. Likewise, in an isothermal expansion of a gas, we can expand a gas at constant temperature by adding heat during the expansion. There are two heat capacities for any substance: a constant volume heat capacity Cv and a constant pressure heat capacity Cpo If we recall the first law of thermodynamics for a system at rest, t.U = q + w, and remember the relationship between temperature and internal energy, we can understand why the same substance can have different responses to the same amount of energy change. For instance, if the volume of a system is held constant, then the system can do no PV work; all energy change must be in the form of heat. This means that none of the energy going into the system can escape as work done by the system. Most of the energy must contribute to a temperature change. On the other hand, when pressure is held constant and the substance is allowed to expand, some of the energy can leave the system as PV work done on the surroundings as the volume changes. Thus, at constant pressure, a substance can absorb energy with less change in temperature by expelling SOn1€ of the energy to the surroundings as work. Cp.is greater than C v -
Just think about the heat capacity of a substance as the amount of energy a substance can absorb per unit of temperature change. Don't worry too much about the difference between constant pressure and constant volume heat capacities. The MeAT might even ignore this fact completely. Use units to help you solve heat capacity problems. For instance, if a heat capacity is given in cal if' 'C', then you know thalto find the heat (measured in calones) you simply multiply by grams and degrees Celsius. This gives you the equation q ~ mcL1T. Most of the time you don't have to know the formula, if you look at the units. Also with heat capacity problems, follow the energy flow, remembering that energy is always conserved : L1E - q + w.
q Cv = !:iT
cOTIs!antvolu rne
For a solid or a liquid, both of which experience very little change in volume, there is a lnore important reason why constant volun1€ and pressure heat capacities differ. The intermolecular forces of a solid or liquid are much stronger than those of a gas. Small changes in the intermolecular distances of noncompressible phases result in large changes in intermolecular potential energy. Intermolecular potential energy does not affect temperature, and thus heat is absorbed at constant pressure with less change in temperature than when heat is absorbed at constant volume. Again: C p is greater than C V ' Heat capacity is always positive on the MCAT; the temperature will always increase when energy is added to a substance at constant volume or pressure. In the real world, heat capacity also changes with temperature; the amount of energy that a substance can absorb per change in telnperature varies with the temperature. However, unless otherwise indicated, for the MeAT, assume that heat capacity does not change with temperature.
Copyright © 2007 Exarnkrackers, Inc.
LECTURE
5:
HEAT CAPACITY, PHASE CHANGE, AND COLLIGATIVE PROPERTIES .
81
Sometimes the MCAT w ill give you the heat capacity of an entire system. For instance, a thermometer ma y be made from several substances each with its own hea t capacity. The thermometer may be immersed in a bath of oil. The oil has its own heat capacity. On the MCAT, the heat capacity of the thermometer-oil system may be precalculated and given in units of energy divided by units of temperature: i.e. J/ K or cal/ °e. For such a situation, we w ould use the following equation:
q = CtlT Sometimes the MCAT will give a specific heat capacity c. Specific means divided by mass, so the specific heat capacity is simply the hea t capacity per unit mass. A specific heat usually has units of J kg-l K-1 or cal g-l °C-1 When a specific heat is given, use the following equation:
q
=
mctlT
The 'm' in this equation is for mass, not molality. This equation is easy to remember because it looks like q ~ MCAT. Notice that the sym bol for specific heat is usually a lower case 'c' while the symbol for heat capacity is usually an upper case 'C .
By the way, don't be surprised if you see molar heat capacity or something Similar. Heat capacities can be given per mole, per volume, per gram, or per whatever. Just use the equation q " mc.!>T and rely on the units of c to find the units of m. For instance, if c is given as the molar heat capacity, m would be in moles.
For the MCAT you must know that water has a specific heat of 1 cal g-' 0C-' . This was once the definition of a calorie. Cwater
5.3
= 1 cal
g - l 0 C-
1
Calorimeters
A calorimeter is a device which measures energy change. There are both constant pressure and constant volume calorime ters. A coffee cup calorimeter is an exan1pIe of a constant pressure calorimeter beca use it measures energy change at atmospheric p ressure. In a coffee cup calorimeter, two coffee cups are used to insulate the solution. A stirrer maintains equal distribution of energy throughout, and a thermometer m easures the change in tem perat ure. Obviously, a coffee cup calorimeter cannot contain expanding gases. Reactions that take place inside a coffee cu p calorimeter occur at the constant pressure of the local atmosphere. A coffee cup calorimeter is used to meas).!re heats of reaction. (Recall that at constant pressure q =1lH.) Fo r instance, if we mix HCl and NaOH in a coffee cup calorimeter, the net ion ic reaction is:
Usmg the specific heat of water/ the mass of wa ter, and the Ineasured change In temperature, we ca n solve for q ill the equahon. q = mc.!>T. Smce q ~ IlH at constant pressure, w e have the heat of reaction.
[1
l
A bomb calorimeter measures energy change at constant volume. A bomb calorimeter tells us the internal energy change in a reac tion. (Recall that at cons tant volu me q = .c.tI.) In a bomb calorimeter, a steel container full of reac tan ts is placed inside another rigid, thermally insulated container. VVhen the reaction occurSj heat is transferred to the surrounding w ater (shown in the d iagra m). Using the known heat capacity of the container and the equation: q ~ C.!>T, we can deduce the heat of the reaction, and thus the internal energy change in the reaction.
Coffee Cup Calorimeter Copyright © 2007 ExamkrJckors, Inc.
Bomb Calorimeter
Reactants
101. In a free adiabatic expansion, a real gas is allowed to spread to twice its original volume with no energy transfer from the surroundings. All of the following are true concerning this process EXCEPT:
Questions 97 through 104 are NOT based on a descriptive passage.
A. B.
97. 20 grams of NaCI is poured into a coffee cup calorimeter containing 250 ml of water. If the temperature inside the calorimeter drops 1°C by the time the NaCI is totally dissolved, what is the heat of solution for Nael and water? (specific heat of water is 4.18 Jig 'C.)
A.
C. D.
-3 kJ/mol Questions 102 through 104 refer to the table below, which lists several common metals and their specific heats.
B. C.
-1 kJ/mol I kJ/mol D. 3 kJ/mol 98. Using a bomb calorimeter, the change in energy for the combustion of one mole of octane is calculated to be -5.5 x 103 kJ. Which of the following is true concerning this process?
A. B. C. D.
Since no work is done, the change in energy is equal to the heat. Since there is no work, the change in energy is equaJ to the enthalpy. Since work is done, the change in energy is equal to the heat. The work done can be added to the change in energy to find the enthalpy.
A. C. D.
99. Which of the following are true statements?
I.
The heat capacity of a substance is the amount of heat that substance can hold per unit of temperature. II. The specific heat for a single substance is the same for all phases of that substance. III. When heat is added to a tluid. its temperature will change less if it is allowed to expand.
D.
C. D.
A. B.
I only TTl only I and III only I. II. and III
C. D.
Fe
0.44
Au
0.13
AI
0.90
Cu
0.39
Iron Gold Aluminum Copper
2 kJ 4 kJ 12 kJ 18 kJ
104. When a sample of aluminum of unknown mass was subjected to 1.8 kJ of heat. the temperature of the aluminum sample increased from 26°C to 31°C. What was the mass of the sample? A. B.
Substance A has larger molecules than substance B. Substance B has a lower boiling point than substance A. At the same temperature, the molecules of substance B move faster than those of substance A. Substance A has more methods of absorbing energy than substance B.
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Specific Heat c (J/g-OC)
103. In an experiment. it was found that 6 kJ of heat were required to raise the temperature of a sample of copper by 15°C. If the experiment was repeated with a gold sample of the same mass, how much heat would be required achieve the same temperature change?
100. Substance A has a greater heat capacity than substance B. Which of the following is most likely true concerning substances A and B? A. B.
Metal
102. If samples of equal mass of all of the metals listed are subjected to the same heat source, which metal would be expected to show the LEAST change in temperature?
B.
A. B. C.
No work is done. Increased potential energy between molecules results in decreased kinetic energy and the gas cools. Entropy increases. The gas loses heat.
C. D.
82
200 400 600 800
g g g g
STOP.
LECTURE
5.4
5:
HEAT CAPACITY, PHASE CHANGE, AND COLLIGATIVE PROP ERTIES .
83
Phase Changes
To understand the phase change process, we shall examine H 20 at constant pressure of 1 atm. If we start with ice at _lOoe and begin heating uniformly at a constant rate, initially, the energy going into the ice increases the vibration of its molecules and raises its temperature. When the ice reaches aoc, the temperature stops increasing. Energy now goes into breaking and weakening hydrogen bonds. This results in a phase change; the ice becomes liquid water. When all of the ice has changed to water, the temperature begins to rise again; the heat goes into increased movement of the molecules. When the water reaches 100oe, the temperature stops rising. The energy once again goes into breaking hydrogen bonds. This process results in a second phase change: liquid water to steam. Once all the hydrogen bonds are broken, the heat increases the speed of the molecules and the temperature rises again. This simplified explanation of phase change is diagrammed below in a heating curve.
140 slope =
120-
1 /lI e
10080604020 -
!1H,,, ""
o -20
heat! q Notice that at ooe and 1000e the heat stops changing the temperature until the phase change is complete. At these temperatures, the heat capacity is infinite. These points are called the nonnal melting point and nonnal boiling point for water. The word 'nonnal' indicates a constant pressure of 1 atm. Since the pressure is constant, heat equals the enthalpy change (q = ilH). The enthalpy change associated with melting is called the heat of fusion; the enthalpy change associated with boiling, the heat of vaporization. Since enthalpy change is a state function, exactly the same amount of heat absorbed during melting is released during freezing. This is also true for vaporization and condensation, and sublimation and deposition.
The flat line segments of the heating curve represent a phase change.
The slope of the heating curve, where not zero, is proportional to the inverse of the specific heat. Since the mass of a substance does not change with phase change, the slope is dependent on the specific heat. Notice that each phase of a substance has a unique slope, and therefore a unique specific heat.
Each phase of a substance has its own specific heat.
Copyright © 2007 Examkrackers, Inc.
You need to know the names of the types of phase changes: melting-freezing; va porization-condensation; su blimationdepOSition.
84
MCAT
INORGANIC
CHEMiSTRY
Evaporation occurs when the partial pressure above a liquid is less than the liquid's vapor pressure, but the atmospheric pressure is greater than the
vapor pressure. Under these conditions, the liquid evaporates rather than boils.
n,e heating curve shows that melting and boiling are endothermic processes; heat is added. You should also know that melting and boiling normally increase volume and lnolecular movemenC and therefore result in increased system entropy, so entropy and enthalpy are both positive for the processes of melting and vaporizing, and both negative for the processes of freezing and condensing. From the equation, L'>G = Mi - TL'>S, we see that when enthalpy and entropy have the same sign, temperature will dictate in which direction the reaction will move. So phase changes at constant pressure are governed by temperature.
5-5
Phase Diagrams
Pressure and temperature are two important intensive properties that help determine the phase of a substance. A phase diagram indicates the phases of a substance at different pressures and temperatures. Each section of a phase diagram represents a different phase. The lines marking the boundaries of each section represent temperatures and pressures where the corresponding phases are in equilibrium with each other. Like other equilibriums in chemistry, this equilibrium is a dynamic equilibrium. For instance, when water and steam are in equilibrium, water molecules are escaping from the liquid phase at the same rate that they are returning. Notice that there is only one point where a substance can exist in equilibrium as a solid, liquid, and gas. This point is called the triple point. There is also a temperature above which a substance carmot be liquefied regardless of the pressure applied. This temperature is called the critical temperature. The pressure required to produce liquefaction while the substance is at the critical temperature is called the critical pressure. Together, the critical temperature and critical pressure define the critical point. Fluid beyond the critical point has characteristics of both gas and liquid, and is called supercritical fluid.
critical point
gas
temperature
solid
gas
temperature
COlnparing the phase diagrams for water and carbon dioxide, we notice some interesting things. Even if it were not labeled, we could approximate the location of the 1 atm mark for either diagram. We know that at atmospheric pressure, water exists in all three phases at different temperatures. Thus, we know that the 1 atmosphere mark must be above the triple point. Since carbon dioxide (dry ice) sublimes (changes from solid to gas) at one atmosphere, we know that the triple point must be above the 1 attn mark. Copyright © 2007 Examkrackers, Inc.
LECTURE
5:
HEAT CAPACITY, PHASE CHANGE, AND COLLIGATIVE PROPERTIES .
Compare the equilibrium line separating the liquid and solid phases on each diagram shown. For waterf the line has a negative slope; for carbon dioxide, a positive slope. Most phase diagrams resemble carbon dioxide in this respect. The negative slope of water explains why ice floats. Since volume decreases with increasing pressure, as we move upward on the phase diagram from ice to liquid water, the volume occupied by H 20 .must be decreasing and thus the water must be increasing in density. Therefore, water must be denser than ice. The reason for this is that the crystal structure formed by ice requires lllore space than the random arrangement of water molecules.
For phase changes you must know where the energy goes. It enters the substance as heat or PV work, but what then? During a phase change, it breaks bonds and doesn't change the tempera ture. When the phase is NOT changing~ energy increases 11101ecular movement, which increases the temperature. Th ink about this: for a single sample of a substance, P, V, II, and T are interrelated in such a \vay that if you kno\'\r three of them, you can derive the other. This means that a phase diagram ca n also be given as a comparison between volume and pressure, or volume and tenlperature. What would that look like? See the problems on the next page for the answer.
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85
108. The diagram below compares the densi ty of water in the liquid phase with its vapor phase.
Questions 105 through 112 are NOT based on a descriptive passage.
105. What is the total heat needed to change 1 gram of water from _10°C to 110°C at 1 atm? (DHfusion = 80 caI/g, i1Hv
ril.mion = 540 cal/g, specific heat of ice and steam are 0.5 calJg "C) A.
- 730 cal
B.
-630 cal
C. D.
630 cal 730 cal
liquid
pi (glcm·')
vapor
~ TI "C
106. When heat energy is added evenly throughout a block of ice at O°C and I atm, all of the foJlowing are true EXCEPT: A.
The temperature remains constant until all the ice is
B.
The added energy increases the kinetic energy of the molecules. Entropy increases. Hydrogen bonds are broken.
What is the critical temperature of water?
melted.
C. D.
A. B. C.
O"C 135"C 374°C
D.
506'C
109. In graph (a) below. isotherms for water are plotted agajnst pressure and volume. Graph (b) is a phase diagram of water with pressure vs. temperature.
107. Below is a phase diagram for carbon dioxide. Platm
10"'
,-
218
Platm 10'
"'" \
// I
10'
85
liquid,
\
\,'
15
vapor 374"C 300··C \ - 200·'C
'i-.- - - - - - \ \
!
liquid + vapor (a)
- 50
0
50 II "C
v
100 P
T
s
B
What is the critical temperature for carbon dioxide? A. B. C. D.
A
- 57°C O°C 31°C 103°C
R Q
C
(b)
T
The area inside the dashed line on graph (a) is represented on graph (b) by: A. B.
C. D.
Copyright © 2007 EXdmkrackers, Inc.
86
the line between points Rand S. the area B. the area C. parts of both area B ancl C.
GO TO THE NEXT PAGE.
110. A solid 78 gram sample of ben ze ne (C(,H6) was gradually heated until it was mel ted completely_ The heating curve for the sample is show n below.
111. At atmospheric pressure, the temperature of a pot of
boiling water remains at 100°C, when heat is added. The best explanation as to why the added energy does not rai se the temperature is that: A. B.
C.
o "-----fo 3.5
I
D.
14.4 Heat (kJ)
What is the heat of fusion of benzene?
A. B. C. D.
at the boiling point, the large heat capacity of water allows water to absorb the added energy. the hydrogen bonds of water are strong enough to absorb the added energy w ithou t breaking. as the water becomes steam, the added energy becomes ki netic energy of the gas molecules. the added energy is used to break bonds between water molecules.
112. A student has a block of an unknown solid in the laboratory. Wh ich of the following will most likely melt the block?
3.5 kllmole 10.9 kllmole 14.4 kl/mole 17.9 kl/mole
J. Heating the solid at constant pressure II. Compressing the solid at constant temperature III. Accelerati ng the soli d to high speeds to increase its kinetic energy.
A. B. C.
D.
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87
I only I and II only I and 1lI only I, Il, and III
STOP.
88
MCAT INORGANIC CHEM ISTRY
5.6
Colligative Properties
Some properties in chemistry depend solely on the number of particles, irrespective of the type of particle. Such properties are called colligative. There are four colligative properties of solutions: vapor pressure, boiling point, freezing point, and osmotic pressure. Colligative properties depend upon number, not kind.
A substance boils when its vapor
pressure equals the local atmospheric pressure.
When using the nonvolatile solute equations, be sure to consider the
number of particles after dissociation .
For boi ling point and freezing point calculations molality is used instead of molarity because molality doesn't change with temperature while molarity does.
Boling point elevation with Boiling _ ;; , point
addition of nonvolatile
solute
Melting pomt
_I' V
Freezing point depression with addilion of nonvolatile solute
Tn Chemistry Lecture 4, we saw that the addition of a nonvolatile solute will lower the vapor pressure of the solution in direct proportion to the number of particles added, as per Raoult's law. TI,e vapor pressure has an important relationship to the normal boiling point. When the vapor pressure of a solution reaches the local atmospheric pressure, boiling occurs. Thus, the boiling point of a substance is also changed by the addition of a solute. The addition of a nonvolatile solute lowers the the vapor pressure and elevates the boiling point. The equation for the boiling point elevation of an ideally dilute solution due to the addition of a nonvolatile solute is:
where kb is a specific constant of the substance being boiled, m is the molality of the solution, and i, called the van't Hoff factor, is the number of particles into which a single solute particle will dissociate when added to solution. The van't Hoff factor has two possible values: the expected value and the observed value. For an ionic compound, the expected vallie of the van't Hoff factor is the number of ions created upon complete dissociation. For instance, the expected value of i for NaCI is 2, and for MgCl2 is 3. These values are for an ideally dilute solution. It turns out that, in a non ideal solution consisting of ions, there is ion pairing. Ion pairing is the n10mentary aggregation of hvo or more ions into a single particle. Ion pairing is not the solute incompletely dissolving; ion pairs are still in the aqueous phase. Ion pairs occur due to the strong attraction between positive and negative ions. The observed value of the van't Hoff factor will take into account ion pairing. Ion pairing increases with solution concentration, and decreases with increasing temperature. In a dilute solution, the observed value will be only slightly less than the expected value. On the MCAT, use the expected value wliess otherwise instructed. You camlot apply the boiling point elevation equation to volatile solutes. As shown in Chemistry Lecture 4, a volatile solute can actually decrease the boiling point by increasing the vapor pressure. If you know the heat of solution, you can make qualitative predictions about the boiling point change when a volatile solute is added. For instance, since you know that an endothermic heat of solution indicates weaker bonds, which lead to higher vapor pressure, you can predict that the boiling point will go down. Melting point also changes when a solute is added, but it is not related to the vapor pressure. Instead, it is a factor of crystallization. Impurities (the solute) interr upt the crystal lattice and lower the freezing point. Freezing point depression for an ideally dilute solution is given by the equation:
/';,T = kfmi Again, the constant kf is specific for the substance being frozen. Be careful with freezing point depression. If you add a liquid solute, the impurities will initially lower the melting point; however, as the mole fraction of the solute increases, you will come to a point where the solvent becomes the impurity preventing the solute from freeZing. At this point, additional solute acts to reduce the impurities creating a more pure solute, and the freeZing point of the solution will rise as solute is added. Copyright © 2007 Examkrackers, Inc.
LECTURE
5:
HEAT CAPACITY, PHASE CHANGE, AND COLLIGATIVE PROPERTIES .
The fourth colligative property is osmotic pressure . Osmotic pressure is a measure of the tendency of water (or some other solvent) to Inove into a solution via osn1osis. To demonstrate osmotic pressure, we divide a pure liquid by a membrane that is permeable to the liquid but not to the solute. We then add solute to one side. Due to entropy, nature wants to make both sides equally dilute. Since the solute cannot pass through the barrier to equalize the concentrations, the pure liquid begins to move to the solution side. As it does so, the solution level rises and the pressure increases. Eventually a balance behveen the forces of entropy and pressure is achieved. The extra pressure on the solution side is called osmotic pressure. Osmotic pressure IT is given by:
11 = iMRT
89
Osmotic pressure is a funny th ing. It is not really pressure at all. For instance , the total pressure at the bottom of a 10 meter swimming pool filled with pure wate r is about 1 atm . The osmotic pressure is zero. If you dump a couple of wheel barrows full of salt into the pool ,
the
osmotic
pressure
increases
significantly, but the total pressure barely changes.
Osmotic pressure IS only relevant when comparing one solution with another.
where M is the molarity of the solution. Related to osmotic pressure is osmotic potential. Osmotic potential is a partial measure of a system's free energy. Pure water is arbitrarily aSSigned an osmotic potential of zero. When a solute is added, the osmotic potential becomes negative. At constant temperature and pressure, water flows from higher osmotic potential to lower osnl0tic potential. Water potential, another related term, is similar to osmotic potential but takes into account telnperature and pressure. Water potential is essentially the same as free energy. When water and the solution in the diagram below have come to equilibrium, points A and B have the same water potential, but the osmotic potential of point B is less than that of point A.
Students often think about osmotic
pressure as the pressure pulling into a solution, and hydrostatic pressure as the pressure pushing out of a solution . Although this is technically incorrect
because pressure is a scalar and has no direction , thinking about osmotic pressure in this way may give you some intuition about it.
FB= pghB
'"
.- .'
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Solute particles are too large to move through the pores in the melnbrane
117. 500 ml of an aqueous solution having a mass of 503 grams and containing 20 grams of an unknown protein was placed into a bulb and lowered into pure water as shown. A membrane permeable to water but not to the solute separated the solution from the water.
Questions 113 through 120 are NOT based on a descriptive passage.
113. Which of the following aqueous solutions will have the lowest boiling point?
A. B.
c.
D.
0.5 M glucose I M glucose 0.5 M NaCI O.6M NaCI
t h
114. An object experiences a greater buoyant force in seawater than in fresh water. The most likely reason for this is:
A.
Seawater has greater osmotic pressure making the
B.
Fresh water has greater osmotic pressure making
pressure difference greater at different depths. The height of the column of solution was found to be '11'. Which of the following statements is true concerning this procedure?
the pressure difference greater at different depths.
C. D.
Seawater has greater density. Fresh water has greater density.
A. 115. Glycol (C2 H(,0z) is the main component in antifreeze. What mass of glycol must be added to 10 liters of water to prevent freezing down to - 18.6°C? (The molal freezing point depression constant for water is 1.86°C kg/mol.) A. B.
C. D.
B.
C.
3.1 kg 6.2 kg 10 kg 12.4 kg
D.
B. C.
in the solution. A large value for h indicates a high osmotic pressure in the pure water. A large value for h indicates that the protein has a low molecular weight. A large value for Ii indicates that the protein has a high molecular weight.
118. Calcium chloride is sometimes sprinkled on winter sidewalks to melt snow and ice. If 333 grams of calcium chloride is dissolved completely in 1.00 kg of water, what will be the freezing point of the solution? (The molal freezing point depression constanr for water is I.S6°C kg/mol )
116. A student holds a beaker of pure liquid A in one hand and pure liquid B in the other. Liquid A has a higher boiling point than liquid B. When the student pours a small amount of liquid B into liquid A , the temperature of the solution Increases. Which of the following statements is true?
A.
A large value for h indicates a low osmotic pressure
A. B.
The boiling point of the solution is lower than either pure liquid A or B. The boiling point of the solution IS higher than either pure liquid A or B. The freezing poi~t of the solution IS higher than
-5.58°C
C.
- 9.30"C -11.7"C
D.
-16.7°C
either pure liquid A or B. D.
The vapor pressure of the solution is higher than pure liquid B.
Copyright © 2007 Examkrackers, Inc.
90
GO ON TO THE NEXT PAGE.
119. A popular experiment uses freezing point depression to find the molar mass of an unknown solute. If a known mass of an unknown non-polar solute is placed into a known mass of a known non-polar solvent and the freezing point depression is measured, which of the follo wing expressions will be equal to the molar mass of the unknown solute?
A.
B.
120. A student prepared two solutions in separate flasks. Solution A consisted of 0.1 mole of sodium fluoride in I liter of water. Solution B consisted of 0.1 mole of potassium chloride in I liter of water. The student then heated both flasks and meas ured the boiling point of each solution. Which of the sol utions would be expected to have the hi gher boiling point?
(k f Xgrams of sOlute)
A.
(llTXkg of sOlvent)
B.
(kf Xkg of sOlvent)
c.
volatile than sod ium fluoride.
(ll TXgrams of solute
D.
c. D.
Solution A, because sodium fluoride has a lower molar mass than potassium chloride. Solution B, because potassium chloride is less Solution A, because potassium chloride will not dissociate completely in water. Both solutio ns will have the same boiling point.
(.1.TXgrams of solule
(kf Xkg of sOlvent) (t.T Xkg of sOlvent) (k f Xgrams of sOlute)
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91
STOP.
Acids and Bases
6.1
Definitions
There are three d efinitions of an acid that you must know for the MeAT: Arrhenius, Bronsted-Lowry, and Lewis. These definitions are given here in the order in which they were created. An Arrhenius acid is anything that produces hydrogen ions in aqueous solution, and an Arrhenius base is anything that produces hydroxide ions in aqueous solution. This definition covers only aqueous solutions. Bronsted and Lowry redefined acids as anything that donates a proton, and bases as anything that accepts a proton. Finally, the Lewis definition is the most general, defining an acid as anything that accepts a pair of electrons, and a base as any thing that d onates a pair of electrons. The Lewis definition includes all the acids and bases in the Bronsted-Lowry and more. Lewis acids include molecules that have an incomplete octet of electrons around the cen tral atom, like AlCl3 and BF, . They also include all simple cations except the alkali and the heavier alkaline earth metal cations. The smaller the cation and the higher the charge, the stronger the acid strength. Fe" is a common example of a Lewis acid. Molecules that are acidic only in the Lewis sense are not generally called acids unless they are referred to explicitly as Lewis acids.
OHbase One measure of the hydrogen ion concentration is called the pH, where p(x) is a fLmction in which, g iven any x, p(x) =- log(x). If we measure the h ydrogen ion concentration in moles per liter ([H+] the brackets always indica te concentration), pH is g iven by:
pH
= - log[W]
Notice that in the Bronsted definition . the acid 'donates', and in the Lewis definition the acid 'accepts'.
Although you mu st memorize the definitions, it is usually convenient to think of an acid as H' and a base as OH . In fact, aqueou s solutions always contain both H' and OW. An aqueou s solution containing a greater concentration of H' than OW is acidic, while an aqueous solution containing a greater concentration of OW than H' is basic. An aqueous solution with equal amounts of H" and OH is neutral.
94
MCAT INORGANIC CHEMISTRY
You must understand some very basic ideas about logarithms for the MeAT. pH uses the base 10 logarithm . The base 10 logarithm is used to solve a problem like: 10'
=
3.16
The answer is: x - log(3 .16). Luckily on the MeAT we don't have to do calculations; instead we estimate. Since 10° equals 1. and 10' equals 10, in the problem above x must be between a and 1. The answer is: x - 0.5. Applying this to acids, if we have a hydrogen ion concentration of 10 ' , the log of 10-' is -3, and the negative log of 10-' IS positive 3. Thus the pH is 3. If we have a hydrogen ion concentration of a little more than 10 ' , say 4 x 10 ' , then the solution is a little more acidic and the pH is slightly lower than 3: say 2.4. Notice that 4 x 10 3 is not as large a number as 10 ' , so the pH is lower than 3 but not quite 2. On the MeAT you must be able to estimate pH values as shown in this paragrapll. The second and last thing you should know about logarithms is: 10g(AB) - 10g(A) + 10g(B)
This is easily verifiable: log(10' ) - 2; 10g(10' ) ~ 3; log(10' x 10' ) - 5. One more thing; many reactions in living cells involve the transfer of a proton . The rate of such reactions depends upon the concentration of H' ions orthe pH .
The sca le for pH generally runs from 0 to 14, but since any fI+concentration is possible, any pH valu e is possible. A t 25°C, a pH of 7 is neutra l; a lower pH is acidic and a higher pH is basic. Each point on the pH scale corresponds to a tenfold difference in h ydrogen ion concentration. An acid w ith a p H of 2 prod u ces 10 times as m any h ydrogen ions as an acid w ith a pH o f 3, and 100 times as many h ydrogen ions as an acid w ith a pH of 4.
From the d efinitions o f an acid, it must be clear that, if there is an acid in a reaction, there must also be a base; yo u can't have a proton d onated w ithout som e thing to accept it. We can write a h ypothetical acid-b ase reaction in aqueous solution as follows:
HA + H 2 0
A - + H,O+
--7
Here, HA is the acid, and, s ince wa ter accepts the proton, water is the base. If we look at the reverse reaction, the h ydronium ion donates a proton to A-, making the h y dronium ion th e acid and A- the base. To a vo id confu s ion, we refer to th e reactants as the acid and base, and the products as th e conjugate acid and conjugate base. Thus, in every reaction the acid h as its conjuga te base, a n d th e base has its conju gate acid. Deciding w hi ch fo rm is the conjuga te simply d e pends upon in w hich directi on you h appen to be v iew ing th e reaction. In other 'vvords, it is correct to say either: "HA is the conjugate add of base A-"; or "A is the conjugate base of acid HA." You must be able to identify conjugates on the MCAT.
You should also know that the stronger the acid, the we
t
weak acid strong conjugate
""
t
/
Ui
:;:
weak acid y\'cak conjugate
co '" £
'"
s trong acid weak conjuga te
--- ---
OJ]
;:l
g
L -________________________
Acid Strength
~
--+-
Some substances act as either an acid or a base, d epending upon their environm ent. They are called ampboteric. Water is a good example. In the reaction above, water acts as a base accepting a p roton. Water can also act like an acid by donating a proton.
Iog(lO') AcidS taste sour or talt; bases taste bitter. Bases are slippery when wet.
= log(lO) =
~
1
~
Iog(3.16) = .5
-1 Y
Iog(lOO) = log(l)
0
(
Copynght
~
2007 Examkrackers, Inc.
LECTURE
6: ACIDS
AND BAS ES .
95
For the MCAT, you need to recognize th e strong acids and bases in Table 6-1.
Strong Bases
Strong Acids hydroiodic ac id
HI
sodium hydroxide
NaOH
h ydrobromic acid
HBr
potassium h ydroxide
KOH
h ydrochloric ac id
HC!
amide ion
NH,-
nitric acid
HNO,
hydride ion
H-
perch!oric acid
HCIO,
calcimn hydroxide
Ca(OH),
chloric acid
HClO,
sodium oxide
Na,O
sulfuric acid
H ,SO,
calcium oxide
CaO
NaOH
1-1 13
NH,
12 11
HCO,-
10 9
F
wa ter H ,CO,
R
7
6 5 4 3
H C,H,O, HSO,-
2
Some acids can donate more than one proton. These ac ids are called polyprotic acids. An acid tha t can donate just two protons can be called a diprotic acid as well as a polyprotic acid. The second proton donated by a polyprotic acid is usu ally so weak that its effect on the pH is neglig ibl e. On the MCAT the second proton ca n almost always be ignored. (The rule 01 thumb is that if the K, values differ by more than 10' , th e second proton can be ignored.) For instance, the second proton from H ,SO, is a s trong acid; yet, except with dilute concentrations (concentrations less than 1 M), it h as a negligible effect on the pH of H 2SO4 solution. This is because H 2S04 is so much s h·onger than HSO,-. N otice here that the p ercent dissociation of an acid decreases with acidity, This means that ac id s dissociate less in more concentrated soluti ons. It does not mean that concentrated solutions are less acidic.
Acid dissociation decreases with acid concentration b ut aci d s trength increases wi th ac id HF conccnh'ation. Imagine the following : I have] 00 HCI acid molecules in wa ter and 50 d issociate, so that I have 50rX) dissociation and 50 hydrogen ions. If I "All values are for have ]000 add molecules in the same amount of 1M sol uti ons wa ter, now only -:l00 d issociate so that I have -:lO(l,() dissociation a nd 400 hydrogen ions. More hydrogen ions in the same anlount of water means more acid s trength . Notice that this mea ns that increasing the concentratio n of a wea k acid by a factor of ten docs NOT result in a ten fold increase in hydroge n ion concentration . 1
°
Cop yrig ht © 200 7 Examkracke rs, Inc.
You should recognize a hydronium ion H,O-. The hydronium ion is simply a hydrated proton. For MCAT acid and base reactions , a hydronium ion and a
proton are the same th ing.
By the way, when we say "strong acid ·' in inorganic chemistry, we mean an acid
that is stronger than H,O-. A strong base is stronger than OH . With bases, we often call something as strong as OH , like NaO H, a strong base. For MCAT purposes, we assu me that a strong acid or base com pletely dissociates in water.
96
. MCAT INORGANIC CHEMISTRY
6.2
How Molecular Structure Affects Acid Strength
There are three factors in molecular structure that determine w hether or not a molecule containing a hydrogen will release its h ydrogen into solution, and thus act as an acid: 1) the strength of the bond holding the h ydrogen to the molecule; 2) the p olarity of the bond; and 3) the stability of the conjugate base. If we examine the C- H bond in metha ne, which has extremely low acidity, it is nearly the sam e streng th as the H-Cl bond in hyd rochloric acid. However, the H-Cl bond is much more polar, and therefore the proton is nl0re easil y removed in aqueous solution. HCl is m ore acidic than m ethane. On the other hand, a comparison of the bond streng ths and polarities of the hydrogen halides shows that, although the H-F bond is the m ost polar, it is also the strongest bond. In addition, the sm all size of the fluorine ion concentrates the negative charge and ad ds to its instability. In this case, the bond strength and conjuga te instability outweigh the polarity, and HF is the weakest of the hydrogen halide acids.
H- F ..
•
H-Cl
H - Br
H-I
increasing polarity - - - increasing bond strength - - - - - - - increasing acidity ...""u,. ",lutmn • ~q
In a series of oxyacids, more Qxygens means a stronger acid.
Keeping conjugate stabili ty in mind, if we examine the oxyacids, we see that the electronegative oxygens draw electrons to one side of the bond, increasing polarity. The oxygens in the conjugate of an oxyacid ca n share the nega ti ve charge spreading it over a larger area and stabilizing the conjugate base. In similar oxyacids, the molecule w ith the most oxygens ma kes the strongest acid. Another way to look at this phenomenon is that the acid ity increases with the oxidation number of the central atom .
(
H - O - Cl
H - O - Cl- O
H- O
o
Cl
H
'0 hypochlorous acid
chlorous acid
chloric acid
increasing acid strength
oI
O - Cl - O I
o perchloric acid I
Copyright © 2007 Examkrackers, Inc.
LECTURE 6: AC IDS AND BASES . 97
6.3
Hydrides
Binary compounds (compounds with only two elements) containing hydrogen are called hydrides. Hydrides can be basic, acidic, or neutral. On the periodic table, the basic hydrides are to the left, and the acidic hydrides are to the right. For instance, NaH is basic; H 2S is acidic. Following this trend, metal hydrides are either basic or neutral, while !1orunetal hydrides are acidic or neutra1. (Ammonia, NH:v is an exception to this rule.) The acidity of nonmetal hydrides tends to increase going down the periodic table. H 20 < H,S < H 2Se < H ,Te -
Group 4A
SA
6A
6A
r
:§ Period 2
Period 3
u
CH,
NH,
H,G
HF
...:
Neither acidic nor basic
Weakly basic
--------
Weakly acidic
.S
SiH,
PH ,
Neither acidic nor basic
Weakly basic
bI)
H,S
HCl
Weakly acidic Strongly acidic
Increasing Acidity
Copyright © 2007 Examkrackers, in c.
'"
...'uOJ"
-'"
IV
126. A student prepared two acid soluti o ns. Solution A has a hydrogen ion concentrati o n of 6.0 x 10 5 mo le L- 1. Solution B has a hydrogen ion concentratio n of I x 10-1 mole L- 1• The pH of solution A d irrers fro m th at of solution B by:
Questions 121 through 128 are NOT based on a descriptive passage.
A.
121. Ammo nia reac ts with water La form the ammoniulll io n and hydroxide ion.
NH, + H 2 0
~
NH: +
B.
C.
D.
OH~
According to the Bronsted-Lowry detinition of acids and bases , what is the co nju gate acid of ammoni a?
A. B.
N H,
C.
OW W
D.
127. In the reaction below, amm onia and boron trifluoride combine when a coordinate coval ent bond is formed between nitroge n and boron. In thi s reaction , ammonia acts as a:
NH ~+
NH, + BF,
A. B.
122. By defin iti o n, a Lew is base: A.
B. C. D.
C. D.
do nates a proto n. accepts a proton. donates a pair of electrons. accepts a pa ir of electrons.
B.
NaH + H,G ~ Na+ + OH~ + H, Reacli on 2
NH/ p-
D.
Br~
Which of the follow ing is tru e? A.
124. Which of the fo llo wing is amphoteric? A.
an amino acid
B.
H2 SO,
C.
NaOH HF
D.
H3N BF,
N H/ + H 20 ~ NH3 + HJO+ Reac ti o n 1
CJ ~
c.
~
Lewis acid Lew is base Bronsted-Lowryac id Brondted-Low ry base
128. Two chemical reactions in volvin g water are shown below.
123. Whi ch of the fo ll owing is the strongest base in aqu eous solution ?
A.
1.3 2. 8 3.7 5.0
B. C. D.
Water acts as a base in Reac tion 1 and an acid in Reaction 2. \Vater ac ts as an ac id in Reac ri on I and a base in Reacti on 2. Water acts as a base in both reactio ns. Water acts as ne ither an acid nor a base.
125. Th e additi o n of an electron withdraw ing group to the alpha carbon of a carboxy li c acid will:
o
I
Ii
i"
\OH
- C- C A. B. C. O.
increase the ac idity of the proton by making O-H bo nd mo re polar. increase the acid ity of the proton by making O- H bo nd stron ger. decrease the ac idity of the proto n by makin g O- H bo nd more po lar. decrease th e ac id ity of the proton by stab ili z ing conju ga te base.
CopYright (£)20UI I:xJrnkr<Jckers, Inc.
the the the th e
98
STOP.
LECTURE 6: ACDS AND BASES . 99
6.4
Equilibrium Constants for Acid-Base Reactions
Pure wa ter reacts with itself to form h ydronium and h ydroxide ions as follows: H 2 0 + H,O
~
Hp' + OH-
This is ca lled the autoionization of water. K", is the equilibrium constant for this reaction. K.., = [W][ OWl
(For convenience, we have substituted H ' for H 30 '.) At 25°C the equilibrium of this reaction lies far to the left:
In a n eutral aqueous solutions at 25°C, the H ' concentra tion and the OH- concentration are equal at 10-' mol L- 1 The pH of the solution is found by taking the negative log of the hydrogen ion concentration , wh ich is: -log[lO-'] = 7. An acid or base added to an aqueous solution will change the concentrations of both H ' and OH-, but K" w ill remain 10-14 at 25°C. For example, in a solution with a pH of 2, th e ion concentrations will be: [H' ] = 10-2 mol e l and [OH-] = 10-12 mol e l . Usin g the p(x) function and the rule: 10g(AB) = 10g(A) + 10g(B), we can put this relationship into a simple equati on: pH + pOH = pK",
For an aqueous solution at 25°C: pH + pOH = 14
An acid will have its own equilibrium constant in water, called the acid dissociation constant K •. If we use our hypothetical acid-base reaction: HA + H 20 ~ H,O' + A-, then the acid dissociation constant for the acid HA is:
Corresponding to every K" th ere is a K •. The K" is the equilibrium constant for the reaction of the conjugate base with water. For the conjugate base K, the reactio n is:
K + H 20
~
OH- + HA
and the K, is: K _ [OW][HA]
, - "----' [A"--]--=' Notice that the reaction for K, is th e reaction of the conjugate base and water, and it is not the reverse of the reaction for K,. No tice also that the product of the two constants is Kuo'
K, K, =
Using the p(x) function and the rule: 10g(AB) = log(A) + 10g(B), this formula can also be written as:
pK, + pK, = 14 Copyright © 2007 Examkrackers, Inc.
It may seem like there are a lot of equations to memorize here, but it is
really very simple. First, all equil ibrium constants are derived from the law of mass action . They are all products over reactants, where pure solids and liquids are given a concentration of one. Once you know how to find one K, you should know how to find any K. The subscript on the constant is supposed to make things less complicated , not more complicated. Second, memorize that: K" ~ 10 25 °C.
14
at
Third. remember the log rule, 10g(AB) 10g(A) + 10g(B), and you can derive any of the equations. Notice that the larger the K, and the smaller the pK" the stronger the acid. A K, greater than 1 or a pK, less than zero indicates a strong acid. The same is true of the K, and pK, of a base.
100 . MCAT INORGANIC C HEMI STRY
Finding the pH
6.5
Very strong acids and bases will dissociate almost completely. This means that the HA or BOH concen tration (for the acid and base respectively) will be nearly zero. Since division by zero is impossible~ for such acids and bases, there is no Ka or Kb" Surprisingly, this fact makes it easier to find the pH of strong acid and strong base solutions. Since the entire concentration of acid or base is assumed to dissociate, the concentration of H+or OH- is the same as the original concentration of acid or base. For instance, a 0.01 molar solution of HCl will have 0.01 mol V i of H+ ions. Since 0.01 = 10-', and - log(lO-') = 2, the pH of the solution will be 2. Likewise, in a 0.01 molar solution of NaOH, we will have 0.01 mol L- I of OH- ions. (Be careful here!) The pOH will equal 2 so the pH will equal 12. You can avoid a mistake here by remembering that an acid has a pH below 7 and a base has a pH above 7. Weak acids and bases can be a little trickier. Doing a sample problem is the best way to learn. For example, in order to find the pH of a 0.01 molar solution of HeN, we do the following: 1.
Set up the equilibrium equation: K = [W][eN ] = 6.2 x 10-10
" 2.
[HeN ]
If we add 0.01 moles of HeN to one liter of pure wa ter, then 'x' amount of that HCN will dissociate. Thus, we will have 'x' mol L-I of H+ ions and 'x' mol V i of eN- ions. The concentration of undissociated HCN w ill be whatever is left, or '0.01 - x' . Plugging these va lues into the equation above, we have: [x][x] = 6.2 X 10-10 [0.01 - x]
3.
If we solve for x, we have a quadratic equation. Forget it! You don't need this for the MeAT. We make an assumption that x is less than 5% of 0.01, and we will check it when w e are done. Throwing out the x in the denominator, we have: [x][x] ? 6.2
X
10- 111
[0.01] Thus, x is approximately 2.5 x 10-". This is much smaller than 0.01, so our assumption was valid. 'x' is the concentration of H+ ions. The pH of the solution is between 5 and 6. This is close enough for the MeAT. [- log(2.5 x 10-<) = 5.6] Just to make sure, we ask ourselves, "Is 5.6 a reasonable pH for a dilute weak acid ?" The answer is yes. For a weak base, the process is the same, except that we use K ill and we arrive at the pOH. Subtract the pOH from 14 to find the pH. This step is often forgotten. If we ask ourselves, "Is tb is pH reasonable for a weak base?", we won' t forget this step.
Copyright © 2007 Examkrackers, Inc.
LECTURE
6.6
6:
ACIDS AND BASES • 101
Salts
Salts are ionic compounds that dissociate in water. Often, when salts dissociate, they create acidic or basic conditions. The pH of a salt solution can be predicted qualitati vely by comparing the conjugates of the respecti ve ions. Just keep in mind that strong acids have weak conjugate bases and strong bases have weak conjugate acid s.
Na+ and Cl- are the conjugates of N aOH and HCl respectively, so, as a salt, NaCl produces a neutral solution. NH4N03 is composed of the conjugates of the base NH, and the strong acid HNO, respectively. Thus, NH/ is acidic and N03- is neutral. As a sa lt, NH, N03 is weakly acidic.
Copyright © 2007 Examkracke rs, Inc.
When considering salts, remember, all cations, except those of the alkali metals and the heavier alkaline earth metals (Ca" , Sr" , and 8a" ), act as weak Lewis acids in aqueous solutions.
134. The acid dissociation constant for HBrO is 2 x 10-9 . What is the base dissociation constant for BrO-?
Questions 129 through 136 are NOT based on a descriptive passage.
A. B. C. D.
129. Which of th e following is the K. for the conjugate base of carbonic acid?
A.
[HlCO, ] [H][HCO,-]
B.
[OW][HCO,-] [H,CO, ]
c.
[W][H,CO, ] [HCO,]
D.
[OW][H,CO,] [HCO, J
5x Sx SX 5X
10-5 10-<> 10-7 10-8
135. A solution of soapy water has a pH of 10. What is the hydroxide ion concentration?
A. B.
C. D.
10- 10 M 10-7 M 10-4 M 10-1 M
136. When solid sodium acetate, NaC, H, O, is added to pure water. the pH of the solution will:
130. An aqueous solution of 0.1 M HBr has a pH of:
A.
0
B. C. D.
I 2 14
A. B. C. D.
decrease because Na+ acts as an acid. increase because Na· acts as a base. decrease because C2H)02- acts as an acid. increase because C 2H)0 2- acts as a base.
131. Carbonic acid has a K, of 4.3 x 10-7 What is the pH when I mole of NaHC0 3 is dissolved in 1 liter of water?
A.
3.2
B.
3.8
C. D.
10.2 12.5
132. Stomach acid has a pH of approximately 2. Sour milk has a pH of 6. Stomach acid is:
A. B. C. D.
3 times as acidic as sour milk. 4 times as acidic as sour milk. 100 times as acidic as sour milk: 10,000 times as acidic as sour milk.
133. Which of the following salts is the most basic?
A. B.
C. D.
Nal NaNO, NH4Cl
KF
Copyright © 2007 Exam krackers, Inc.
102
STOP.
LECTURE
6.7
6:
ACIDS ANC BASES
Titrations
A titration is the drop-by-drop mixing of an acid and a base. Titrations are performed in order to find the concentration of some unknown by comparing it with the concentration of the titrant. The chan ging pH of the unknown as the acidic or basic titrant is added is represented g raphically as a sigmoidal curve. To the right is the titration curve of a strong acid titra ted by a strong base. Notice the portion of the graph that most nearly app roximates a vertical line. The midpoint of this line is called the equivalence point or the stoichiometric point. The equivalence point for a monoprotic acid is the point in the titration when there are equal equivalents of acid and base in solution. (An equivalent is the mass of acid or base necessary to produce or consume one mole of protons.) For instance, since there is a one to one correspondence between Hel with NaOH, the equi valence point for a titration of HCl w ith NaOH will be reached w hen the sa me number of moles of HCl and N aOH exist in solution . This is not n ecessari ly when they are at equal volumes. If the concentrations differ (and they probably will) the equivalence point will not be where the volumes are equal. For equally strong acid-base titrations, the equivalence point will usually be at pH 7. (Warning! For a d iprotic acid whose conjugate base is a strong acid, like I-I,SO" this is no t the case.) The graph to the right is fo r the titration of a strong acid with a strong base. In other words, we are slowly adding base to an acid. This is clea r because we start with a very low pH and finish with a very high pH. For a titration of a strong base with a strong acid, we would Simply invert the graph.
6.8
More Titrations and Buffered Solutions
The titration of a weak acid with a strong base looks slightly d ifferent than the curve above, and is shown below. The equiva lence point is also not as predictable. Of course, if the base is stronger than the acid, the equivalence point will be above 7, and if the acid is stronger than the base, the equivalence point will be below 7.
pH
t
End point ge
Equivalence point
rar
100% A -
7 pH = pK,
- - - - - -~ - -, ..,..0--
Half equivalence point 50% A -
50% HA
Volume of base
Copyright © 2007 Examkrackers, Inc.
pH
7-- - - - - - - ---
Volume of base
.
103
104
MCAT INORGANIC CHEM ISTRY
Notice the half equivalence point. This is probably more likely be tested by the MCAT than the equivalence point. The half equivalence point is the point where exactly one half of the acid has been neutralized by the base. In other words, the concentration of the acid is equal to the concentration of its conjugate base. Notice that the half equivalence point occurs at the midpoint of the section of the graph that most represents a horizontal line. This is the spot where we could add the largest amount of base or acid with the least amount of change in pH. Such a solution is considered to be buffered. The half equivalence point shows the point in the titration where the solution is the most well buffered. The Henderson-Hasselbalch equation is simply a form of the equilibrium expression for Ka:
Notice also that, at the half equivalence point, the pH of the solution is equal to the pK, of the acid. This is predicted by the Henderson-Hasselbalch equation:
[A-l
KiJ =
pH = pKa + log --=---=-[HAl
[W][Kj [HAj Recall that log(l)
K ~ [Wj [A- ]
, using the log rule: - log(K,) .
~ .
~ \'---'I'
~
[A-] - log[H+] - Jog - [HA]
[K] [HA]
pK, ~ pH - log --
-
~.
[HAl
.
-x
There is no need to memorize it, since it is so easy and so quick to derive it.
~
0; thus when [K]
~
[HAL pH
~
pK,.
If we were to make a buffer solution, we would start with an acid whose pK, is closest to the pH at which we want to buffer OUf solution. Next we would mix equal amounts of that acid with its conjugate base. We would want the concentration of our buffer solution to greatly exceed the concentration of outside acid or base affecting our solution. So, a buffer solution is made from egual and copious amounts of a weak acid and its conjugate base. It appears from the Henderson-Hasselbalch equation that we
could add an infinite amount of water to a buffered solution with no change in pH. Of course, this is ridiculous. Will adding Lake Tahoe to a beaker of buffered solution change the pH of that solution? The Henderson-Hasselbalch equation in the form above does not allow for ion pairing. ([on pairing is when oppositely charged ions in solution bond momentarily to form a single particle.) Water will generally act like a base in acidic solution and an acid in basic solution. If you add a base or y\Tater to an acidic, buffered solution, it is clear from the titration cunre that the pH will increase. It just won't increase as rapidly as other solutions less well buffered. However, a question on the MCAT is more likely to consider the ideal circumstance where adding a small anlount of water to an ideally dilute, buffered solution will have no effect on the pH.
Warning! You cannot typically use the Henderson-Hasselbalch equation to find the pH at the equivalence point. Instead, you must use the Kb of the conjugate base. You can find the Kb from the K, and the Kw' The concentration of the conjugate base at the equivalence point is equal to the number of moles of acid divided by the volume of acid plus the volume of base used to titrate. Don't forget to consider the volume of base used to titrate. Unless the base has no volume, the concentration of the conjugate at the equivalence point will not be equal to the original concentration of the acid. The pH at the equivalence point involves much more calculation than the pH at the half equivalence point. For this reason, it is more likely that the MeAT will ask about the pH at the half equivalence point.
Copyright © 2007 Examkrackers, Inc.
LECTURE 6: ACIDS AND BASES • 105
Finding the pH at the equivalence point is a good exercise, but you won't have to do it on the MCAT. Here are the steps:
Usc K., and K,,to find the K,
Sct up the K" equilibrium expression . 10KIlHA]
fA I Solve for the O H concentration, and find the pOH. Subtract the pOH from H to find the pH . 14 - pOH = pH
6.9
Indicators and the End Point
To find the equivalence point, a chemical ca lled an indicator is used. (A pH meter can also be used.) The indicator is usually a weak acid whose conjugate base is a diffe rent color. We can desig nate an indicator as HIn, where m- represents the conjugate base. In o rder for the human eye to detect a color change, the new fo rm of the indicator must reach 1/ 10 the concentra tion of the original form . For examp le, if we titrate an acid with a base, we add a small amount of indicator to our acid. (We add only a small a mount because we do n' t want the indicator to affect the pH.) At the initial low pH, the HIn form of the indicator predominates . As we titrate, and the pH increases, the In- form of the indicator also increases. When the In- concentra tion reaches 1/10 of the Hln concentration, a color change ca n be detected by the human eye. If we titrate a base with an acid, the process works in reverse. Thus, the pH of the color change d epends upon the direction of the titration. The pH values of the tw o points of color change give the range of an ind icator. An indicator's range can be predicted by using the H enderson-Hasselbalch equation as follows:
You don't need to memorize this stuff about indicators, but it's useful to understand.
[In-] pH = pK, + log-[HIn] lower range of color change ==>
pH = pI(, + log 110 ==> pH = pK, - 1
upper range of colo r change ==>
10 pH = pK, + 10g T ==> pH = pK, + 1
The point where the indicator changes color is called the endpoint. Do not confuse the eqUivalence point with the end po int. We usually choose an indicator whose range will co ver the equiv alence poin t. You can also monito r the p H with a pH meter. A pH meter is a concentration cell comparing the voltage difference between different concentra tions of H' . (See Chemistry Lecture 7 for concentration cells.) Copyright © 2007 Examkmckcrs, Inc .
By the way, you can remember that the end point is where the indicator changes color by spelling indicator as: Endica!or
106
MCAT INORGANIC CHEMISTRY
Since we establ ished that the Henderson-Hasselbalch equation is not useful to find the pH of the equivalence point, how can it be useful to find an indicator range that will include the
pH
equ ivalence point? The answer is that we are using the in the indicator concentrations Henderson-Hasselbalch equation, and
the
indicator
never
reaches
Buffer point
its
t
equ ivalence point in the titration . The indicator ions do not approach zero concentration near the color change range .
Experiment held in this pH range by buffered solution
6.10 Polyprotic Titrations Titrations of polyprotic acids will have more than one equivalence point and n10re than one half equivalence point. For the MeAT, assume that the first proton completely dissociates before the second proton begins to dissociate. (This assumption is only acceptable if the second proton is a much weaker acid than the first, which is usually the case.) Thus we have a titration curve like the one shown below.
Titration of the diprotic acid H,A with a strong base
pH 2nd equivalence point 100% A'7
.---- -='--- - -1st equiv alence point 100% HApH = pK.,, -
2nd half equivalence point 50% A'50% HA-
- - - _. 1st half equivalence point 50% HA50% H ,A Volume of base
Copyright © 2007 Examkrackers, Inc.
139. If the ex pec ted eq ui valence point for a ti trat io n is at a pH of 8.2, which of the following wo uld be the best indicator for the titration?
Questions 137 through 144 are NOT based on a descriptive passage.
137. The titration curve below represents the ti tration of:
pH
A. B. C.
D. A. B. C. D.
a a a a
stron g acid with a weak base. strong base with a weak acid. weak ac id with a stron g base. weak base wi th a stron g acid.
Indicator
K,
phenolphthalein
1.0 x 10-8
brolllthYlllol blue
7.9 x 10-'
methyl orange
3.2 x 10....
methyl violet
IA x lO
J
phenolphthalei n bromthymol blue methyl orange methyl violet
140. On the titration curve of the H 2C0 3 pictured below, at which of the following points is the concentration of
HC0 3- the greatest?
138. The foll owing is a list of acid dissociation constants for 4 D
acids.
pH
c
K, Ac id I
I.2 x l O '
Ac id 2
8.3 x 10 '
Aci d 3
3.3x I0 '
Acid 4
6.1 x 10-'
B A
A.
B.
Which ac id sho uld be used to manufacture a buffer at a pH of6.1? A.
B. C.
D.
Acid Acid Acid Acid
C.
D.
I 2 3 4
A
B C D
141. Which of the fo ll ow ing is the equivalence point when the weak acid, aceti c acid, is titrated with NaOH?
A.
107
4.3
B.
7
C.
8.7 14
D.
Copyright © 2007 Exam krackers. Inc.
point point point point
GO ON TO THE NEXT PAGE.
144. The acid di ssociation constant for HC 6H 70 6 is 8.0 X 10-5 . If a solution contains equal concentrations of HC(,H 70 6 and C(jH 7 0 ('-' what will be the pH of the sol ution?
142. A buffered solut ion has a pH that cannot readily be changed. A buffered solution wil1 be produced by mixing equal vo lumes of: A. B.
C. D.
1M 1M 1M 1M
A.
HC1 and I M NaC,Hp, HC1 and I M NaOH HC,HP, and I M NaC, Hp, HC,HP , and I M NaOH
B. C.
D.
3.0 4.1 5.3 9.0
143. All of the following statements regarding HC0 3- are true EXCEPT: A. B.
C. D.
HC0 3HCO]HCO,' HCO;
can act as a Bronstcd Lowry acid. can act as a Lewis basc. is amphotelic. is a polyprotic acid.
Copyright 0 2007 Examkrvck,;rs, Inc
108
STOP.
Electrochemistry
7.1
Oxidation-Reduction
In an oxidation-reduction reaction (called a redox reaction for short), electrons are transferred from one atom to another. The atom that loses electrons is oxidized; the atOln that gains electrons is reduced.
7.2
Oxidation States
In order to keep track of the electrons in a redox reaction, you must memorize the oxidation states of certain atoms. Oxidation states are the possible charge values that an atom may hold within a molecule. In many cases, these charges don't truly exist; it is sinlply a system to follow the electrons of a redox reaction. Even though they do not represent actual charges, the oxidation states must add lip to the charge on the molecule or ion. For instance, the sum of the oxidation states of the atoms in a neutral 11101ecule lTIUst equal zero. The oxidation states that you lnust n1emorize for the MeAT are given in Table 7-1. When a conflict arises, the rule occupying the higher position on the table is given priority.
r
I
I
110
MCAT INORGANIC C HEMISTRY
Oxidation State
For the MeAT, you probably won't need to memorize any other oxidation states than those in Table 7-1.
Atom
0
Atoms in their elemental form
-1
Fluorine
+1
Hydrogen (except when bonded to a metal: then - 1.)
-2
Oxygen (except when it is in a peroxide like H 2O,) Table 7-1
In general, when in a compound, elements in the following groups have the oxidation states listed in the table below. It is helpful to know Table 7-2 but not crucial for the MCAT.
Oxidation State +1
Group 1 elements (alkali metals)
+2
Group 2 elements (alkaline earth metals)
-3
Group 15 elements (nitrogen family)
-2
Group 16 elements (oxygen family)
-1
Group 17 elements (halogens)
To help keep oxidation and reduction straight, just remem ber:
GERrrrrr ...
Table 7-2 The idea is simple; a general guideline for oxidation states is the atom's variance from a This is definit ely noble gas configu!ati o n. However, if all not in my contract . atom s had permanent oxidation states, no redox reactions cou ld take place. The oxidation states in table 7-2 are to be used only as a general guideline. When the two tables conflict, the first table is given priority. For example, the oxidation state of nitrogen in NO; is +5 because the -2 on the oxygens have priority and dictate the oxidation state on nitrogen. (Don't forget that the oxidation states for NO; must add up to the 1charge on the molecule.) The transition metals change oxidation states according to the atoms with which they are bonded. Although each transition metal has on ly certain oxidation states that it can atta in, the MCAT will not require that you InemoriZ€ these.
L ose
G ain
Electrons
the Lion says
O xidotion
Group on Periodic Table
The foll owing is an example of a redox reaction:
Electrons
2H z + 0, --> 2H,O
H ere oxygen and hydrogen begin in their elemental form, and thus have an oxidation state of zero. Once the water molecule is formed, hydrogen 's oxidation state is +1, and oxygen's is -2. In this case, we say that hydrogen has been oxidized; hydrogen has lost electrons; its oxidation sta te has increased from 0 to +1. Oxygen, on the other hand, has been reduced; it has gained electrons; its oxidation state has been reduced from 0 tc -2. Whenever tilere is oxidation, there must also be reduction.
R eduction
Copyright
2007 EXilmktockers,
LECTURE 7: ELECTROCHEMISTRY •
Since in any redox reaction one atom is oxidized and an other atom is reduced, there is a reducing agent (also called the reductant) and an oxidizing agent (also called the oxidant). Because the reducing agent is giving electrons to an atom, an atom in the reducing agent must be giving up some of its own electrons. Since an atom in the reducing agent giv es up electro ns, an atom in the reducing agent is oxidized. The reverse is true for the oxid izing agent. Thus, the reducing agent is the compound containing the a tom being oxid ized, and the oxidizing agent is the compOLU1d containin g the a tom being reduced. For example, in the following reaction, methane is the reducing agent and dioxygen is the oxidizing agent.
111
and get outta the country fast .
Reducing Agent Salty
Carbon goes from --4 to +4 .
/~CO, + 2H, 0
C H, + 20, -
~i
Oxygen goes from 0 to - 2. Notice that the reducing agents and oxidizing agents are compounds, not atoms. In a redox reaction, th e atom is oxidized or reduced; the compound is the oxid ant or reductant. In the reaction: Cd(s) + N iO,(s) + 2H,G(/) ---> Cd(OH),(s) + Ni(OH),(s) Ni is reduced. N iO, is the oxidizing agent.
7.3
Oxidation-Reduction Titrations
In order to find the molarity of a red ucing agent, a sample can be titrated wi th a stro ng oxid izing agen t. For instance, if we want to know the molarity of 5n 2+ ions in a solution, we can titrate it with a known concentration of the strong oxidizing agent Ce-t+. 5n2+ ions oxidize to 5n4+, while Ce-i+ reduces to Ce31' . Since only one electron is required to reduce Ce4 +, and two electrons are given up to oxidize 5n 2+, two Ce-l1' ions are reduced for every 5n 21' ion oxidized. Thus, we know that the number of moles of Ceh required to reach the equivalence point is twice the number of m oles of Sn 2• in solution. Instead of measuring the pH, we measure the voltage compared to a standard solution.
Copyright © 2007 Examkrackers, Inc.
Knowledge of oxidation-reduction titrations is not required for the MeAT. However, it is pOSSible that there will be a passage which explains them . They are included here just so you won't be shocked if you see one in a passage.
150. The process below takes place in acidic solution.
Questions 145 through 152 are NOT based on a descriptive passage.
In this process, the oxidation state of nitrogen is: A. B. C. D.
145. \tVhat is the oxidation state of sulfur in HSO..-?
A.
-2
B.
+3 +6
C. D.
+7
151. Which of the following statements j's true about the reaction below?
146. Which of the following statements is true concerning the reaction:
HNO) + NaHCO) --> NaNO, + H,CO) A. B. C. D.
2AI,o) + 3C --> 4AI + 3CO, A. B. C. D.
reduced from +2 to +3. oxidized from +2 to +3. reduced from +3 to +5. oxidized from +3 to +5.
Both aluminum and carbon are reduced. Both aluminum and carbon are oxidized. Aluminum is reduced and carbon is oxidized. Carbon is reduced and aluminum is oxidized.
Nitrogen is reduced and oxygen is oxidized. Oxygen is reduced and carbon is oxidized. Hydrogen is reduced and sodium is oxidized. No oxidation or reduction takes pJace. CI 2 + 2Br- --> Br, + 2Ct
152. 147. What is the reducing agent in the following reaction:
In the reaction shown above, A. B. C. D.
2HCI + Zn --> ZnCl, + H,
A.
Zn
B.
Zn?+
C.
H'
D.
Ct
Cl 2 is Cl 2 is CI 2 is el? is
the the the the
oxidizing agent and Br- is oxidized. oxidizing agent and Br- is reduced. reducing agent and Br- is oxidized. reducing agent and Br- is reduced.
148. The first step in producing pure lead from galena (PbS) is as follows: 2PbS(s) + 30,(g) --> 2PbO(s) + 2S0 2 (g) All of the following are true concerning this reaction EXCEPT: A. B. C. D.
Both lead and sulfur are oxidized. Oxygen is the oxidizing agent. Lead sulfide is the reducing agent. Lead is neither oxidized nor reduced.
149. All of the following are always true concerning oxidation-reduction reactions EXCEPT: A. B. C. D.
An atom in the reducing age nt is always oxidized. If reduction takes place. so must oxidation. An atom in the oxidizing agent gains electrons. Jf an atom of the reductant loses two electrons. an atom of the oxidant gains two electrons.
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112
STOP.
LECTURE
7.4
7:
ELECTROCHEMISTRY .
113
Pot entials
Since in a red ox reaction electrons are transferred, and since e lectrons have ch arge, there is an electric potential E associated w ith an y redox reaction . The poten tials for the oxidati on comp on en t and reduction compone nt of a reac tion can be approximated se parately based upon a standard hydrogen electrode (SH E) discussed later in this lectu re. Each component is called a half reaction. O f course, no half reacti o n will occur by itself; an y reduction ha lf reaction m us t be acco mpanied by an oxida tion h alf reaction. There is o nly one possible po tentia l fo r an y given h alf reacti on. Since the reverse of a reduction ha lf reac tion is an oxida tion half reaction, it wo uld be redunda nt to lis t potentials fo r bo th th e oxida tion an d red uction h a lf reac ti o n s . Therefore, ha lf reac tion p o te ntia ls are u s u a ll y li s te d a s reduction potentials. To find the oxidation p otential for th e reverse h alf reacti o n, the sign of the reduction potential is reversed . Be low is a lis t of some COIlUTIon reductio n potentials.
Notice that, except for nickel, the metals used to make coins have negative oxidation potentials. In other words, unlike most metals, platinum, gold, silver, mercury, and copper do not oxidize (or dissolve) spontaneously under standard conditions in the presence of aqueous H' .
Standard Reduction Potentials al 25°C Polential EO
Half reaction A u 3+(aq) + 3e-
--7
A u (s )
O ,(g) + 4H +(aq) + 4eP t2+(aq) + 2e-
--7
--7
1.50 H,o(l )
Pt(s)
1.23 1.2
Ag'+ (aq) + 2e-
--7
Ag(s )
0.80
H g'+(aq) + 2e-
--7
H g( l )
0.80
C u +(aq) + e-
--7
C u' +(aq) + 2e-
C u ts)
--7
0.52
C u ts )
0.34
2H+(aq) + 2e- --7 H ,(g)
0.00
Fe3+(aq) + 3e-
--7
Fe(s )
- 0.036
N i'+(aq) + 2e-
--7
N i(s)
-D.23
Fe' +(aq) + 2e-
--7
Fe(s)
-0.44
Zn'+(aq) + 2e-
--7
Zn(s)
-0 .76
H ,o(l) + 2e-
--7
H , (g) + 20H- (aq)
- 0.83
Table 7-3 Recall from physics tha t electric po tentia l has no absolu te value. Thus the va lues in the table above are assign ed based upon the arbitrary assignment of a zero value to the ha lf reaction that occu rs at a stan da rd hyd rogen electrode:
EO= 0.00 V Th is is the only red uctio n poten tial tha t yo u n eed to m em orize. An exalnple of an oxida ti on potentia l take n from the table above wou ld be: Ag(s) --) AgZ.(aq) + 2e-
E" = --{).80 V
If we wish to find the potential of the fo llowing ionic reaction: 2Au'+ + 3C u --) 3Cu'+ + 2Au
Copyright © 7.007 EXrtmkrackers, lnc.
Also notice that Table 7-3 gives us the reduction potential for Ag" (aq) and the oxidation potential for Ag(s) . (Warning: The table does not give us the oxidation potential for Ag" .) The strongest oxidizing agent is shown on the upper left hand side of a reduction table. The strongest reducing agent is shown on the lower right hand side of a reduction table . Notice that water is both a poor oxidizing agent and a poor reducing agent. Finally, notice that the second half reaction in Table 7-3 is part of the final recll:uulI ill aerobic respiration where
oxygen accepts electrons to form water. Predictably, this reaction has a high positive potential.
114
MCAT INORGANIC CHEM ISTRY
we can separate the reaction into its h-vo half reactions and add the half reachon potentials: 2(Au 3 , + 3e- ~ Au) 3(Cu ~ Cu2- + 2e-)
EO ~
150V
E" ~ -0.34 V ~ 1,16 V
Warning: Since reduction potentials are intensive properties, '\,v e do not multiply the half reaction potential by the number of times it occurs,
7,5
Balancing Redox Reactions
Balancing redox reactions can be tricky. When you have trouble, follow the steps below to balance a redox reaction that occurs in acidic solution.
All th is effort to balance a redox reaction will get you, at most, one point on the MCAT. MCAT just doesn't require the balancing of redox reactions very often, so spend your time accordingly,
Divide the reaction into its corresponding half reactions.
2.
Balance the elements other than Hand O.
3.
Add H 20 to one side until the 0 atoms are balanced.
4.
Add H' to one side until the H atoms are balanced.
5.
Add e- to one side until the charge is balanced.
6.
Multiply each half reaction by an integer so that an equal number of electrons are transferred in each reaction.
7.
Add the two half reactions and simplify.
For redox reactions occurring in basic solution, follow the same steps, then neutralize the H ' ions by adding the same number of OH- ions to both sides of the reaction.
7.6
More sim ply put, a galvan ic cell tllrns chemica l energy into electrical energy.
1.
The Ga lvanic Cell
If two distinct electrically conducting chemical phases are placed in contact, and one charged species from one phase cannot freely flow to the other phase, a tiny amount of charge difference may result. This tiny charge difference crea tes an electric potential between the phases (typically one or two volts). By offering an alternative path for electron flow, a galvanic cell (also called a voltaic cell) uses the electric potential between such phases to generate a current of electrons from one phase to another in a conversion of chemical energy to electrical energy. A galvanic cell is made of a multiphase series of components with no component occurring in more than one phase. All phases must conduct electricity, but at least one phase must be impermeable to electrons. Otherwise, electrons would move freely throughout the circuit and come to a quick eqUilibrium. The phase that is impermeable to electrons is an ionic conductor carrying the current in the form of ions. The ionic conducting phase is usually an electrolyte solution in the form of a salt bridge. The components of a simple galvanic cell can be symbolized by the letter T-E-J-E'-T', where T represents the tenninals (electronic conductors such as metal wires), E represents the electrodes (also electronic conductors), and J the ionic conductor (often the salt bridge). When the cell is formed, the emf is the electric potential difference between T and r. A simple galvanic cell has two electrodes: the anode and the cathode . The anode is marked with a negative sign and the cathode is marked with a positive sign. The oxidation half reaction takes place at the anode, and the reduction half reaction takes place at the cathode. Depending upon the text, electrodes may refer to only a strip of metal or both a strip of metal and the electrolyte solution in which it is submerged. The strip of metal and solution together may also be called a half cell. Copyright © 2007 EXOlmkracke rs, Inc.
LECTURE 7: ELECCROCHEM ISTRY •
115
Only potential differences between chemically identical forms of matter are easily measurable, so the two terminals of a galvanic cell must be made of the same material. The cell potential E, also called the electromotive force (emf), is the potential difference between the terminals when they are not connected. Connecting the terminals reduces the potential difference due to internal resistance within the galvanic cell. The drop in the emf increases as the current increases. The current from one terminal to the other through the load (or resistance) flows in the direction opposite the electron flow. Since electrons in the anode have higher potential energy than those in the cathode, electrons flow through the load from the anode to the cathode.
Remember: RED CAT; AN OX: reduction cathode; anode oxidation .
The standard state cell potential is simply the sum of the standard state potentials of the corresponding half reactions. The cell potential for a galvanic cell is always positive; a galvanic cell always has chemical energy that can be converted to work. The real cell potential depends upon the half reactions, the concentrations of the reactants and products, and the tenlperature.
Electrons are negatively charged, so they are attracted to the positive cathode and repelied by the negative anode in a galvanic cell.
Below is an example of a simple galvanic cell with the standard hydrogen electrode. Hydrogen gas is bubbled over the platinum plate. The platinum acts as a catalyst in the production of H ' ions. The half reaction is shown. The platinum plate carries an electron through the wire to the silver strip. Ag' accepts the electron converting it to solid silver and allowing a chloride ion to solvate into the aqueous solution.
--1-
~
_---1~
H,(g)-E=::!~~+-
____~ Red Cat
An Ox
+
0 0 0
0
0
0
0
HCl
Anode H,(g)
~
2H\aq) + 'le-
solution
Cathode
AgCl(s)
+e .... Ag(s) + 0
(aq)
Galvanic Cell with standard hydrogen electrode (SHE) Since, by convention, the oxidation potential of hydrogen is zero, the cell potential of any electrode used in conjunction with the SHE will be exactly equal to the reduction potential of the half reaction occurring at the other electrode. Thus, some half reaction reduction potentials can be measured using the SHE.
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116
MCAT INORGANIC CHEMISTRY
Notice that there is no salt bridge in the SHE Galvanic Cell. Both electrodes are in contact with the same solution so no salt bridge is necessary. When a cell contains two different solutions, a liquid jU l1ction is required to separate the solutions. Because ions can move across a liquid junction, any liquid junction creates an additional small potential difference that affects the potential of the galvanic cell. A salt bridge is a type of liquid junction that minimizes this potential difference. Typically a salt bridge is made from an aqueous solution of KCI. The salt bridge allows ionic conduction between solutions without creating a strong extra potential within the galvanic cell. It is able to minimize the potential because the K+ ions move toward the cathode at about the same rate that the CI- move toward the anode. Below is an example of a simple galvanic cell that requires a salt bridge. Without the salt bridge, the solutions in the cell below would mix providing a low resistance path for electrons to move from Zn(s) to Cu 2>(aq) effectively short circuiting the cell, and leaving it with a cell potential of zero.
2e
salt bridge
t
An Ox -
You should sketch a couple of your own galvanic cells so that you know how they are made . Notice that the concentrations are 1 M. This represents standard conditions and allows the use of the values from the reduction half reaction table to calculate the cell potential.
+ Red Cat
.
Anode
\
~ . t _) SO,'
-----7
Cathode
~~y »~ ---c:--
~1 M Cu"
Zn
C
Cu' ·
Zn2+ + 2e-
Galvanic Cell Even in a galvaniC cell 'w ith a salt bridge, there is some leakage of ions across the liquid junction, which causes the battery to lose its chemical potential over time. Commercial cells use an insoluble salt to prevent this from happening.
7.7
IUPAC Conventions
Galvanic cells can be represented by a cell diagram. Each phase is listed from left to right begilming with the terminal attached to the anode and ending with the terminal attached to the cathode. The terminals are often omitted because they are always the same material and do not take part in the reaction. A vertical line is placed between phases. A double vertical line is used to indicate a salt bridge. A dotted vertical line indicates a boundary between two miscible liquids, and species in the san1e phase are separated by a comma.
Copyright © 2007 Examkrackers, Inc.
LECTURE 7 : ELECTROCHEMISTRY .
Pt' (s) IZn(s) IZn" (nq) IICU"(aq) ICu(s) IPt(s)
Cen Diagram The standard state emf ca n be found from the cell dia gram by subtracting the potential of the reduction half reaction on the left (the reaction at the anode) from the potential of the reduction half reaction on the right (the reaction at the cathode).
7.8
117
It is unlikely that MCAT will require that you know the IUPAC conventions for a galvanic cell diagram . You can remember that the cathode is on the right because reduction and nght both begin with an ',. Don't spend too much time here.
Free Energy and Chemical Energy
A positive cell poten tial indicates a spontaneous reaction as shown b y the following equation:
~G
=- nFE
max
w here 11 is the number of moles of electrons that are transferred in the balanced redox reaction, and F is Faraday's constant, w hich is the charge on one mole of e lectrons (96,486 C mol-I). This equation says that the free energy represents the product of the total charge /I F times the voltage E. Recall from PhYSics Lecture 7 that the product of charge and voltage equa.ls work (w = qV). Since this is electrical work, it is not the result of a change in pressure or volume, so it represents nonPV work. Recall from Chemistry Lecture 3 that the change in Gibbs free energy represents the maximum amount of nonP V work ava ilable from a reaction at constant temperature and pressure. A negative i\.G indicates that the work is being done ~ the system and no t on the system. A positive cell potential indicates a nega tive .6.G and a spon taneous reaction. When all the conditions are standard, we can write the equation above using the symbol as follows (the 'max' is part of tha definition of LlG and is assumed):
10 '
i\.G ° = -I1 FEo LlGo can be found in books, but w ha t about LlG for non-sta nd ard state conditions? There are an infi.nite number of possible combinations of concentrations of reactants and products and tenlperatures with which we could start a reaction. How can we predict the maximum available work from these combinations? In order to make predictions about reactions that d o not occur at standard state, we must use the following equation, which relates LlG with LlGo:
~G
=
~G o
+ RT In(Q)
where Q is the reaction quotient discussed in Chemistry Lecture 2, and 'In( )' is the natural IogaritiuTI . You may see this equation written using a base 10 logarithn1 as: LlG = LlGo + 2.3RT log( Q)
This is based upon the crude approxinla tion: 2.310g(x)
V In(x).
There is no need to confuse nG and i\.Go. nGo is a specific nG with specifically described parameters called standard cond itions. N otice that if we use only one molar concentrations for Q, then Q = 1, and RTln(Q) = 0, leaving us w ith nG = LlGo This is what we would expect for a reaction at standard conditions. (Remember, standard conditions don't ac tuall y indicate a particular temperature; YO ll can have standard conditions at an y temperature. Stan dard conditions are usually assumed to be 298 K.) Recall from Chemistry Lecture 3 that at equilibrium, there is no available free energy with which to d o work; LlG = 0 by d efinition. Thus, if we have equilibrium conditions, we can p lug in a value of 0 for LlG, and rewrite "LlG = LlGo + RT In(Q)" as: ~G o CopYright © 2007 Examkmckcrs, Inc.
= - RT In(K)
Memorizing this equation isn't nearly as
important as understanding what it says about the relationship between LlG, LlGo, Q and K.
118
MeAT
INORGANIC CHEMISTRY
In this equation, both K and "'GO vary with temperature. Whenever you specify a new temperature, you must look up a new .6.Go for that temperature. Notice that since this equation uses the natural log of the equilibrium constant, a value of 1 for K will result in a value of 0 for "'GO. For the MeAT you should understand the relationship between K and "'Go. This is pretty tricky stuff, but it's worth wracking your brains and killing some time on it now, rather than on the MCAl Reread this section and make sure that you understand the relationship between K, Q, "'GO , "'G , and T.
if K=l then LlGo = 0 if K>l then LlGo < 0 if K 0 Warning: This relationship does NOT say that if a reaction has an equilibrium constant that is greater than one, then the reaction is always spontaneous. That doesn't make any sense, since the spontaneity of a reaction depends upon starting concentrations of products and reactants. It does say that if a reaction has an equilibrium constant that is greater than one, the reaction is spontaneous at standard state (starting molar concentrations of exactly 1 M) and the prescribed temperature. The galvanic cell pictured on page 116 was drawn with standard conditions of 1 M concentrations. That's great for the instant that the concentrations are all one molar, but what about for the rest of the time? How can we find the potential when the concentrations aren't one molar? If we take the equation:
You do not have to know the Nernst equation for the MCAl However, you should understand how the Nemst equation expresses the relationship between chemica! concentrations and potentia! difference. For instance, the Nernst equation could be used to express the resting potential across the membrane of a neuron . Such a situation would be similar to a concentration cell as discussed in tile next section .
L1G = "'Go + RT InC Q) and substitute -nFE for L1G, and - nFP for L1G o , and then divide by -nF, we get:
E = P _ RT In(Q) nF
This is the Nernst equation. At 298 K, and in base 10 logarithm form, the Nemst equation is:
The Nernst equation allows us to plug in nonstandard concentrations to create Q and find the cell potentiaL
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157. A negative cell potential ind ica tes whi ch of the following:
Qu estions 153 through 160 are NOT ba sed on a descripti ve passage.
A. B. C.
153. Whi ch of the following statements about a galvan ic cell is false? A. B.
C. D.
D.
If E" = 0, a reaction may still be spontaneous dependin g upon the chemical concent rat ions. A galva nic cell with a positi ve potenti al ca n perform work. Redu cti on takes place at the cathode. A salt bridge balances the charge by all ow ing positi ve ions to move to the anode.
Questions 158 throu g h 160 are based on the information below. A gal vanic cell uses the react ion between solid rin and aqueous copper io ns to produce elec trical power. S n(s) + 2Cu'+(aq)
154. The val ues of all of the following are reversed when a reaction is reversed EXCEPT: A. B. C. D.
enthalpy Gibbs energy the rate constant reacti on potential
C. D.
A. B.
C. D.
The reaction is at equilibrium. At 298 K and I M concentrations of products and reactants the equilibrium constant equ als one. D.G is also zero. The reacti on is spontaneous at temperatu res greater th an 298 K.
+ 2Cu(s)
Sn' +(aq) + 2e- -7 Sn(s)
£0 = -D. 14 V
Cu'+(aq ) + 2e- -7 Cu rs)
E"= 0.15V
Half Reaction
A. B. C. D.
EO(V)
Ag2-t- + e- ---} Ag+
1.99
Fe3+ + e- ----7
F e Z.;.
0 .77
Cu
0.34
-----7
2H+ + 2e- -7 H,
0.0 1 V 0.16 V 0.29 V
0.44 V
159. The reaction in rhe cell is all owed to proceed. As the reacti on in the cell progresses, the cell pote nti al will:
156. The fo llowing is a table of half reactions:
Cu 2+ + 2e-
-7 Sn" (aq)
158. The standard state cell potential for th is reaction is:
155. Which of the following is true for a reaction, if t>Go29' = O? (The 298 subsc ript indicates a temperature of 298 K .) A. B.
Both half reactions are nonspontaneous. The reduct ion half reaction potenti al is greater than the oxidation half reac tion potential. The oxidation half reacti on potential is greater than the reduction half reacti on potential. The cell is electrolytic.
decrease as the concent rat ion of Sn 2+ increases. dec rease as the concentrati on of Sn2+ decreases. increase as (he concent ration of Sn 2+ increases. increase as the conce ntrat ion of Sn 1+ decreases.
160. Whi ch of the following is true of the ti n/copper cell reacti on? A. B. C. D.
K > 1 and K > 1 and K < 1 and K < 1 and
t>G' > 0 l!.Go < 0 t>Go > 0 t>Go < 0
0.00
Fe2.;- + 2e-
----7
Fe
-D.44
Zn z" + 2e-
----7
Zn
-D.76
The strongest red ucing agent show n in the table is:
A. B. C. D.
Zn Zn2T Ag+
Agh
Copyright © 2007 Examkracke rs, In c.
119
STOP. .
120
MCAT INORGANIC CHEMISTRY
7,9 The concentration cell is just a type of galvanic cell. It is never at standard conditions, so the Nernst equation is required to solve for the cell potential.
More Cells
A concentration cell is a limited form of a galvanic cell with a reduction half reaction taking place in one half cell and the exact reverse of that half reaction taking place in the other half cell.
salt bridge
+
An Ox
Red Cat
'tF:
,+/ -
~)
/)_
Anode
~ur'~ 0.01 M Fe'+
Fe --" Fe'+ + 2e-
0.1 M Fe'+
Cathode
Fe' ++ 2e---" Fe
Concentration Cell For cells , you should learn to diagrarn a galvanic cell by yourself. Once you can do that, the other cells can be created frorn the galvanic cell. Remember that galvanic cells have a positive cell potential ; electrolytic cells have a negative potential. Galvanic cells are spontaneous; electrolytic are forced by an outside power source, 'Electrochemical' cell can mean either 'galvanic' or 'electrolytic' cell. For any and all cells , rernember 'Red Cat, An Ox'. This translates to Reduction at the Cathode, and Oxidation at the Anode.
Of course, when we add the two half reactions we get: £0 = O. If the concentrations were equal on both sides, the concentration cell potential would be zero. You can use the Nemst equation to find the potential for a concentration celL (If you need the Nemst equation, the MCAT will give it to you.) It is much more likely that the MCAT will ask you a qualitative question like "In which direction will current flow in the concentration cell?" In this case, we must think about nature's tendency for balance; nature wants to create the greatest entropy. The more concentrated side will try to become less concentrated, and electrons will flow accordingly. To use the Nernst equation to find the potential of a concentration cell at 25' C, we must realize that Fe 2 -1- is both a product and a reactant. Thusf we simply substitute for Q the mtio of the Fe 2+ conccntrations on either sidc. For the case above we have:
n = 2 because 2 electrons are used each time the reaction occurSf and EOequals zero. Concentration cells tend to have small potentials.
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LECTURE
7:
ELECTROCHEMISTRY
.
121
If we hook up a power so urce across the resis tance of a galvanic cell, and force the cell to run backwa rds, we have crea ted another type of cell, the electrolytic cell. An y electroly tic cell on the MCAT will have a negative emf. In the electrolytic cell, the cathode is marked negative and the anode is marked positive. Reduction still takes place at the ca thode and oxidatio n at the anode.
~
,_---11Power ,..--_ Source
Red Cat
Cu
Cathode Cu" + 2e- ~ Cu
Electrolytic Cell Electroly tic cells are used in indus try for m etal plating, and for purifying m e tals. For instance, pure sodium can be collected through e lectrolysis of sodium chloride solutio n in a Downs cell. The half rea ctio ns are as follows:
N a+ + e- --> Na
EO= - 2.71 V
2CI- --> 2(- + C I,
£0 =
The assignment of positive and negative to electrodes in galvanic and electrolytic celis is based upon perspective. Galvanic celis are used to provide energy to an external load, so the electrodes are labeled so that negative electrons will flow toward the positive electrode. Ele"ctiOi1s flow from the load to the cathode, so the cathode is labeled positive in the galvanic cell. The focus of electrolytiC cells is within the cell itself. For instance, electrophoresis uses an
electrolytiC cell. It is important that negatively charged am ino acids within the electrolytic cell flow toward the positive electrode, so the anode is labeled positive in the electrolytic cell.
-1.36 V
Notice that this reaction will not run in aqueous solution beca use, from Table 7-3, we s ee that water h as a less negative red ucti on potential than sodium. In fact, this indica tes that solid sodium w ill oxidize spontaneously in water.
"'
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164. According to the Nernst eq uati on:
Qu esti o ns 161 throug h 168 are NOT based on a descripti ve passage .
(x)
- [og E -- EO-0.06 11
if a concentration cell has a po ten tial of 0.12 V, and a concentration of 0.1 M Ag+ at the anode, what is the concentration of Ag+ at the cathode?
161. A galvan ic cell is prepared with solutions of Mg2+ a nd AI.3-1- ions separated by a salt bridge. A potentiometer reads the di ffe re nce across the electrodes to be 1.05 Vo lts. T he fo llowing standard reduction pote nti als at 25°C appl y:
A. B.
C. AI'+ + 3e- --> AI
- 1.66
Mg2+ + 2e-
- 2.37
-7
D.
E O(V)
Ha lf Reaction
Mg
IO- J M 10- 1 M 1M 10M
165. A spoon is plated with silver in an electrolytic process where the half reaction at the cathode is: Ag+(aq) + e- --> Ag(s)
Whi ch of the fol lowing stateme nts is true concernin g the
galva nic cell at 25°C? A. B. C.
Magnesium is redu ced at the cathode. T he conce ntrations of ions are I M . T he reac ti on is spontaneous.
D.
For every al uminum atom reduced, an equal number
162. Whi c h o f th e following is true for an electrolyti c cell ? Reducti on takes place at the anode. T he reaction is spontaneous. electrons fl ow to th e cathode. A n e lectrolyti c cell requires a salt bridge.
B. C.
D.
A.
107.8 ifF
B.
107.8~
c.
107.8 ..!..-
D.
107.8 iF
F
tF t
166. The charge on I mole of e lectrons is given by Faraday's cons lam, 96,500 C/moie. A ga lvani c cell was operated continuously fo r 5 minutes and 0.01 moles of electrons were passed through the wire. Tf the vo ltage rema ined constant for the e nt ire ti me, what is the current generated by the cell?
163. A conce nlralion cell co ntai ns 0 .5 M aqueo us Ag+ on one side and 0. 1 M aq ueo us Ag' on the other. All of thc fo llowing are true EXC EPT: A.
E." = 0.8 V
If the c urrent i is he ld constant for I seconds, whi ch of the fo llowing ex pressions gives the mass of silver deposited on the spoon? (F is Faraday's co nstanl.)
o f magnesium atoms are oxidi zed.
A. B. C. D.
Ij
Electro ns wi ll move from the less co ncent rated side to the mo rc co ncentrated side. Elec tro ns wiII move fro m the anode to the cathode. As the cell potenti al moves toward zero, the co ncentrati ons of both sides will tend to even out.
A. B.
3A 5A
C. D.
6A 9A
t.C > 0
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167. The red uction potentjal for two hal f reaction s are given below. 2H,O(l) + 2e- --> H, (g) + 20W(aq) Na\aq) + e- --> Na(s)
E" = -0.8 V
EO= - 2.7 V
Aqueous sodium cannot be reduced to solid sodium in an electrolytic cell because: A. B. C. D.
aqueous sodium wi ll combine with aqueous hydroxide to form solid sodium hydroxide. liquid water is more easily reduced than aqueous sodium. aqueous sodium must be ox idi zed to form solid sodium. hydrogen gas is more eas il y oxidized than solid sodium.
168. The reaction above takes place in a galvanic cell. Which of the following is true? 2Ag2+ + Fe -? 2Ag"- + Fe2+ A. B. C. D.
E" = 2.4 V
Ag2+ is reduced at the a node. Ag2+ is oxidized at the anode. Ag2+ is reduced at the cathode. Ag2+ is ox idi zed at the cathode.
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123
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3D-MINUTE IN-CLASS EXAM FOR LECTURE 1
125
2, What type of bonding holds together the compound MgCI,?
Passage I (Questions 1-7)
A.
There are five types of interactions within and between molecules. Intramolecular interactions include covalent and ionic bonds. Intermolecular interactions include van def Waals's forces, dipole-dipole, and hydrogen bonds. Table 1 lists the typical energies for these interactions.
B, C. D,
3. Why do the boiling points of the noble gases increase as the period increases?
Typical Energy kJ mol-'
Interaction
A.
Van der Waals's
0.1- 5
Dipole-dipole
5 -20
B.
Hydrogen bond
5 -50
C.
Ionic bond
400 - 500
Covalent bond
150 - 900
covalent ionic hydrogen van def Waals's
D.
The bonds are stronger because larger atoms are more polarizable as period increases. The bonds are weaker because larger atoms are more polarizable as period increases. The bonds are stronger because larger atoms are less polarizable as period increases. The bonds are weaker because larger atoms are less polarizable as period increases.
Table 1 Energies of interactions 4. The atomic radius of Ne is: The boiling point of a substance increases with the strength of its intermolecular bonds. Figure I shows the boiling points of hydrides for some main-group elements and of the noble gases.
350
g ~
.S
200
.S
150
'§ o:l
HF
250
HC Siij
C
50 0
A. B. C.
H!>e
H~
100
Kr
1.5
2
Xe
D.
2.5
3
3.5
4
4.5
6. What type of intermolecular bonding occurs in gaseous CH4 ?
5
A,
Period
B. Figure 1 Boiling points of some main-group hydrides and the noble gases.
C. D,
1. Why does the boiling point of H,o and HF deviate from the trend in Figure 1? A. B. C. D,
Vaporization requires more energy than melting. Vaporization requires less energy than melting . Transition from solid to liquid involves other factors such as crystalline lattice structures. Vaporization is easier to measure.
Ar
Ne
He
greater than Ar less than Ar the same as AI cannot be determined
5. Why are boiling points a better indication of intemlolecular bonding than melting points?
Hp
300
0
~ bI)
A, B. C. D.
7. Why is a dipole-dipole interaction stronger than a van der Waals's interaction?
F and 0 both occur in the second period. The size of H 2 0 and HF are small relative to the other molecules H,o and HF are less polarizable. H,o and HF can hydrogen bond.
A. B.
C. D.
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covalent ionic hydrogen van der Waals's
126
Dipole-dipole is an electrostatic interaction. Van cler Waals's interactions rely on temporarily induced dipoles. Van der Waals's interactions require a Jarge sllrface area. Dipole-dipole interactions only occur with ionically bonded compounds.
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Passage II (Questions 8-15)
9. Which of the following elements is the most chemically similar to N a?
Figure 1 shows atomic radius as a function of atomic number for the first three periods of the periodic table. Within a period the atomic radius decreases as the atomic number increases, but the atomic radius increases as the period increases.
'. ••
2
~
•
1.5
~
w
" '"
•
'i3 H
U
S 0
~
0.5
"'"
f-
•
•
••••• •
o
• • ••
•
••
10
5
•
C.
C
D.
Cs
15
A.
Li
B. C.
Ne Rb
D.
Br
11. Why is the electronegativity scale adjusted to fluorine?
Period 1 Period 2 Period 3
A. B. C. D.
20
Atomic number
A.
Electronegativity of an atom also follows a trend in the periodic table. Electronegativity for any element (X) is based upon the difference (t.) between the actual bond energy of a bond between element X and hydrogen and the expected bond energy of the same bond:
B. C.
(H-Xtxpect~d bond energy
D.
where the expected bond energy is given by: H-Hbond energy H - Xexp bond energy
=
13. Why does the atomic radius follow the trends observed in Figure I?
2
A. B.
c. D.
8. If Se has an atomic radius of 1.16 A, what is the predicted atomic radius of As?
A.
1.05 1.15 1.25 1.97
C. D.
Electron affinity becomes more exothermic as atomic number increases in a period and a group_ Electron affinity becomes less exothermic as atomic number increases in a period and a group. Electron affinity becomes more exothermic as atomic number increases in a period and 1ess exothermic as atomic number increases in a group. Electron affinity becomes less exothermic as atomic number increases in a period and more exothermic as atomic number increases in a group.
+ X-Xoond energy
Pauling electronegativity values are assigned to each element based upon its L'!. value with respect to fluorine. Fiourine is arbitrarily given a value of 4.0. The electronegativity of elements may be used to predict the type of bonding found in a molecu1e. Large differences in electronegativities of atoms in a bond result in an ionic bond.
B.
The researcher who discovered electronegativity was working with fluorine. Fluorine has the smallest atomic radius. Fluorine has the smaliest electronegativity. Fluorine has the greatest electronegativity.
12. How does electron affinity change with atomic number?
Figure 1 Atomic radius as a function of atomic number
L1 ::::: (H-X)actual bond energy -
H Mg
10. Which element has the largest atomic radius?
I
0
A. B.
As the atomic number increases, the nuclear charge Increases. As the atomic number increases, the nuclear charge decreases. As the atomic number increases, the nuclear charge increases and as the period increases the number of electron shells increases. As the atomic number increases, the nuclear charge increases and as the period increases the number of electron shells decreases.
A A A A
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Passage III (Questions 16-21)
14. Hydrogen has a Pauling electronegativity of 2.1. What is the value of d for hydrogen?
A.
0
B. C.
1.0 2. 1
D.
4.0
The empi rical formula of a hydrocarbon can be determined using an instrument similar to the one shown in Figure 1. A sample hydrocarbon is combusted. The absorption chambers absorb all the water and carbon dioxide from the reaction. CaCl 2 can absorb both water and CO 2 • The masses of the chambers before and after the reaction are compared to find the moles of carbon and hydrogen in the sample.
15. What type of intramolecular bonding is found in a CO molecule?
A. B. C.
D.
covalent ionic hydrogen van der Waals's
Heaters
\~~
-
Chamber I
0, - ICO,& HO -
Desiccant forHO absorptfon
Chamber 2
cO,NaOH + CaCi . for COcabsorption
Figure 1 A combustion train For example: propane can be combusted in the apparatus as follows:
In an experiment using the combustion train, a gaseous fuel used in welding (containing only C and H) is reacted with 0 2. The mass of the absorbers in Chamber I increases by 0.9 grams and the mass of the absorbers in chamber 2 increases by 4.4 grams. The density of the welding gas is 1.1 g L- 1 at 25°C and atmospheric pressure. At the same conditions, 0 2 has a density of 1.3 g L- 1 •
16. What is the empirical formula of the welding gas?
A. B. C. D.
CHO CH
C2H2 C)HS
17. \Vhat is the molecular we ight of the gas?
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128
A.
13 g/mol
B.
26 g/mol
C. D.
32 g/mol 60 g/mol
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18. A compound has an empirical formula of CH,G. Using osmotic pressure, the molecular weight is determined to be 120 g/mol. What is the molecular formula for this
Questions 22 through 23 are NOT based on a descriptive passage.
compound?
A. B. C. D.
CH,G C4 HP3 C3HP3 C4 HP4
22. What is the electron configuration of a chloride ion? A.
B. C. D.
19. If I mole of C3Hs is reacted with 2.5 moles of O 2 , how many moles of H2 0 will be produced? A. B. C. D.
I 2 3 4
23. According to the Heisenberg uncertainty principle, which of the fonowing pairs of properties of an electron cannot be known with certainty at the same time?
mole of H,G moles of H,G moles of H,G moles of H,G
A. B. C. D.
20. What would happen if the order of the chambers in the combustion train were reversed? A. B. C.
D.
[Ne] 3s' 3p' [Ne] 3s2 3p 6 [Ne] 3s2 3dJO 3p 5 [Ar] 3s' 3p 6
The amount of CO 2 calculated would be higher than the actual amount produced. The amount of CO, calculated would be lower than the actual amount produced. The amount of H,G calculated would be higher than the actual amount produced. Nothing, the experiment would still give the same results.
charge and velocity spin and subshell average radius and energy level momentum and position
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
21. Why is it necessary to react the 02 in excess when using a combustion train?
A. B.
C. D.
In addition to the combustion reaction, the O 2 is used as a carrier gas. In addition to the combustion reaction, the 02 is used as a source of energy to propel the non-spontaneous reaction. 02 needs to be the limiting reagent in order for the calculations to be correct. The sample needs to be the limiting reagent in order for the calculations to be con-ect.
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3D-MINUTE IN-CLASS EXAM FOR LECTURE 2
131
24. For an ideal gas, which of the following is most likely the correct graph?
Passage I (Questions 24-29)
A.
Over the years, many anempts have been made to find an equation which represents the behavior of non-ideal gases. Althougb none of the equations are completely accurate. they do allow an investigation of some of the macroscopic properties of real gases. The most commonly used of these is the Van der Waals equation:
2.5500K
2.(}
300 K 1.5-
Z 200K
I.(}
0.5 0.0
where P is the absolute pressure, n is the number of moles, V is the volume. T is the absolute temperature, R is the gas constant (0.08206 L atm/mol K) , and a and b are co nstants determined experimentally for each gas studied. Table 1 gives tbe values of a and b for some common gases:
B. 2.5 2.0
a
b
(atm V/rnol)
(Umol)
Ar
1.4
0.032
HCI
3.7
0.041
0.5
Cl,
6.4
0.054
0.0
H,
0.25
0.027
NH3
4.3
0.037
0,
1.4
0.032
Gas
Molarity
1.5
Z
All Temps
1.0
MOlarity C. 2.5 500 K
2.0
300K
Table 1 Van der Waals Constants for Various gases 1.5 Z
To belp quantify the deviation of a real gas from ideality, a compression factor Z has been defined by Z = P VlllRT. Figure I shows how tbe compression factor for ammonia depends on pressure at several different temperatures.
0.5 0.0
2.5 500K
2.0
200K
1.0
Molarity
D.
300 K
2.5
200 K
2.0
500 K
1.5
300 K
1.0
200K
1.5 Z 1.0 0.5 0.0
Z Molarity
0.5 0.0
Figure 1 Compression factors for ammonia
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Mo larity
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Passage 11 (Questions 30·37)
25. For an ideal gas, what can be said about the constants a andb? A. B. C. D.
They They They They
are both zero. are both positive and equal to each other. depend on the temperature of the gas. must be determined experimentally.
In 1889, SvanteArrhenius proposed that the rate constant for a given reaction is given by the formula:
k=Ae
26. Based on the information in the passage, under which of the following conditions does ammonia behave most ideally? A. B.
C. D.
B.
C.
D.
Low temperatures and low pressures Low temperatures and high pressures High temperatures and low pressures High temperatures and high pressures
In heterogeneous catalysis, the catalyst is in a different phase from the reactants and products. For example, a solid may catalyze a fluid-phase reaction. Such a catalysis involves the following steps: 1. A reactant molecule diffuses through the liquid to the surface of the catalyst.
The average kinetic energy of the molecules depends only on the temperature of the gas. At constant volume in a sealed container, the pressure of the gas is directly proportional to its temperature. At constant temperature in a sealed container, the volume of the gas is directly proportional to its pressure. The intermolecular attractions between the gas molecules are negligible.
2. The reactant molecule bonds to the catalyst (adsorption).
3. Adsorbed molecules bond with each other or with a molecule which collides with the adsorbed molecules.
4. The product leaves the catalyst. In homogeneous catalysis, the catalyst is in the same phase as the reactants and products. Acids often act by this mechanism.
28. Which of the following demonstrates nonideal behavior of a gas? A. B. C. D.
30. Which of the following is true of a catalyzed reaction?
Some of the molecules move more rapidly than others. Condensation occurs at low temperatures. The gas exerts a force on the walls of its container. The average speed of the molecules in the gas is proportional to the square root of the absolute temperature.
A. B. C.
D.
29. Why must absolute temperature be used in the Van der Waals equation? A. B. C. D.
A catalyst may be the limiting reagent. At equilibrium, more products are produced when a catalyst is present. The catalyzed reaction pathway has a lower energy of activation than the uncatalyzed reacton pathway. The rate of the reverse reacton will be slower for the catalyzed reaction.
31. Consider the following mechanism:
CI + 0 3 --> CIO + 02
Because the Van der Waals equation is a nonrelativistic equation. Because it is impossible to have a negative absolute temperature. Because ratios of temperatures on other scales, such as the Celsius scale, are meaningless. Because international convention requires it.
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RT
where Eo is the activation energy for the reaction, R is the gas constant (8.314 J/mol K), T is the absolute temperature, and A is a factor, which depends on factors such as molecular size. Catalysts change the reaction pathway, which may result in a change in E", A, or both.
27. Which of the following statements is NOT true for an ideal gas?
A.
-~
°
CIO + --> CI + 0, In this mechanism, what is the catalyst?
A. B. C. D.
133
CI 03
CIO 02
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36. H2 can be added to ethylene in the presence of a heterogeneous catalyst such as solid platinum. What might account for the initial attraction between the hydrogen molecules and the solid platinum?
32. As the temperature of a reaction increases, which of the following always occurs? A. B. C. D.
The rate constant increases. The rate constant decreases. The activation energy increases. The activation energy decreases.
A. B. C. D.
33. Consider a reversible reaction. If the activation energy for the forward reaction is lowered by a catalyst, what can be said about the activation energy for the reverse reaction? A. B. C. D.
bydrogen bonding metallic bonding van der Waals attraction the plasma continuum effect
37. If the solid line in the graph below represents the reaction profile for an uncatalyzed reaction, which line might represent the reaction profile for the catalyzed reaction?
It is also lowered. It is raised. It is unaffected by the catalyst. The effect of the catalyst on the reverse reaction cannot be predicted without more information.
34. Suppose a reaction is acid-catalyzed by a solution of pH 3.0. What can be said about the pH of the resulting solution? A. B. C.
D.
It will be greater than 3.0 because the acid is consumed. It will be equal to 3.0 because the acid is regenerated. It will be equal to 3.0 because catalysts have no effect on equilibrium. It cannot be predicted without information on the acidity of the reactants and products.
35. The rate of a reaction may depend on which of the following? I. II. III. IV.
Concentrations of the reactants Concentration of a catalyst Surface area of a heterogeneous catalyst Temperature
A. B. C. D.
I only IV only I and IV only I, II, III, and IV
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A.
A
B. C.
B
D.
D
C
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Passage'" (Ouestions 38-44)
40. The following table gives the relative concentrations and rates found using the method described in the passage. What is the rate law for Reaction I ?
Peroxydisulfate (persulfate) ion reacts with the iodide anion according to Reaction 1. S, o ,2- (aq) + 31- (aq) ;= 2S0,'- (aq) + 1,- (aq) Reaction 1 The amount of 1,- formed can be determined by adding a known amount of S',o,2- and allowing it to react according to Reaction 2.
1
2
3
WI, (M)
0.060
0 .030
0.030
[S,O,"'I, (M)
0.030
0030
0.015
Rate, (Mlsec)
2S, O,'- (aq ) + 1,- (aq) ;= S. O,'- (aq) + 31- (aq) Reaction 2
If starch is also added, any excess I)- will react to form a blue-black I, complex. The formation of this complex indicates th e completion of Reaction 2. The rate of Reaction I can be determined by the following equation where t is the elapsed time from the addition o f the last component to the formation of the blue-black starch, r, complex.
A.
kWI'[S,ot ]'
B. C. D.
kW]'[S,o,2-]'
6.0 x 10-0 3.0 x 10"
1.5
X
10-'
kU-]' [S,ot ] kW][S,Ot ]
41. The rate expression for the reaction of H2o with Br2 is:
rate = k[H,J[Br,]. A.
! [s,o;-]
B.
rate = ~2,-,-_ _~ t
C. Equation 1
D.
The rate is flist order with respect to H" and first order overall. The rate is first order with respect to H2, and second order overall. The rate is second order with respect to H" and first order overall. The rate is second order with respect to H" and second order overall.
38. Why can Equation J be used to measure the rate of Reaction I ? A.
B.
C. D.
42. A student is performing the kinetic study described in the passage and forgets to add starch. What will be the result
Reaction 2 must be mucb faster tl13n Reaction I, thus the rate in Equation I is the rate of formation of 1,-. Rea~tion 2 must be much slower than Reaction 1, thus the rate in Equation I is the rate of formation ofI,-. Reactions I and 2 must occur at the same rate, thus the rate in Equation 1 is the rate of formation ofI,-. Equation 1 can be derived directl y from the rate laws of Reactions I and 2.
of the experiment? A.
B. C.
D. 39. What would happen to the time and the rate in Equation I , if the temperature were reduced? A. B. C. D.
The rates of both reactions as measured by the student will increase because th e starch slows the reactions. The rate of Reaction 1 as measured by the student will decrease because starch speeds up the reaction. The rate of Reaction I as measured by the student will stay the same because starch has no effect on the rate. The rate of Reaction I as measured by the student will not be able to be deterntined by the method described in the passage.
Time would increase and rate would decrease. Time woul d decrease and rate would increase.
Time would increase and rate would remain unchanged. Time wouJ d remain the same and rate wou1d increase.
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43. What would happen to the time and the rate in Equation 1, if more $,0,'- were added, and all other conditions re mained the same?
A. B. C. D.
Questions 45 through 46 are NOT based on a descriptive passage.
TIme would increase and rate would decrease. Time would decrease and rate would increase. Time would increase and rate would remain unchanged. Time would remain the same and rate would in-
45. Which of the following are true concerning any reaction at equilibrium?
I. The concentration of products is equal to the concentration of reactants. II. The rate of change in the concentration of the products is equal to the rate of change in the concentration of reactants. III. The rate constant of the forward reaction is equal to the rate constant of the reverse reaction.
crease. 44. What would happen to the time and the rate in Equation 1, if a catalyst is added to Reaction 1, and all other conditions remain the same?
A. B. C. D.
Time would increase and rate would decrease. Time would decrease and rate would increase. Time would increase and rate would remain unchanged. Time would remain the same and rate would increase.
A. B. C.
D.
II only [ and II only II and III only I, II, and III
46. Equal concentrations of hydrogen and oxygen gas are placed on side I of the container shown below. Side 2 contains a vacuum. A small pin hole exists in the barrier separating side I and side 2. Which of the following statements is true?
0 , and H, gas
vacuum
~
pin ho le
side I
A. B. C. D.
side 2
The partial pressure of oxygen on side I will increase. The partial pressure of hydrogen on side I will increase. The mole fraction of oxygen on side 1 will increase. The mole fraction of hydrogen on side I will increase.
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
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3D-MINUTE IN-CLASS EXAM FOR LECTURE 3
137
50. Which of the fo llowing explains the error in Che mist E's argument?
Passage I (Questions 47-52) Nickel is purified by the Mond process, which relies on the equilibrium:
A.
Ni(s) + 4CO(g) '" Ni(CO).{g)
B.
M!"
=-1 60.8 kJ, !!.S' =-409.S JK-
1
at 2S' C
C.
Reaction 1 The Mond process D. Two chemists analyze the equilibrium.
Chemist A
Higher temperatures favor a reaction that decreases entropy. The entropy change of a reaction is the entropy change of the universe. A positive e ntropy of reaction does not necessarily indicate an entropy increase for the universe. The reacti on will only run in the reverse direction until entropy is maximi zed for the reaction system.
51. What is the change in Gibbs free energy for Reaction I at standard state and 2S' C?
Chemist A argues that Reaction 1 will be spontaneous in the forward direction because th e product is more stable than the reactants. Furthennore, if the temperature is raised, the reaction will run in reverse because it is an exothermic reaction.
A. B. C. D.
- 38 kJ 38 kJ -ISO.S kJ 150.5 kJ
Chemist B 52. Consider the reaction below:
Chemist B argues that Reaction I will be spontaneous in the reverse direction because the entropy is higher for the reactants than for the products. Furthermore, if the temperature is raised, the spontaneity of the reverse reaction will increase.
C(s) + O2 (g) ..., CO, (g)
Mf' is - 393.S1 kJ and !!.S' is 2.86 JK- 1 at 2S' C If solid carbon is exposed to l atm of oxygen and 1 atm of carbon dioxide gas at room temperature, w ill carbon dioxide gas form spontaneously?
47. The Mf', for NiCs) is: A. B. C. D.
0 kJl mol
A.
-1 60.8 kJlmol 160.8 kJlmol 409.S kJlmol
B. C.
48. Which of the following is a logical conclusion of Chemist A 's argument? A. B.
C. D.
D.
No, because the enthalpy of formation for CO, is negative. No, because the change in Gibbs energy is negative. Yes. because the enthalpy of formation for CO, is negati ve. Yes, because the change in Gibbs energy is negative.
The enthalpy change of a reaction is an indicator of the relative stab ili ty of the reactants and products. The entropy change of a reaction is an indicator of the relative stability of the reactants and products. At hi gher temperatures, the reactants of Reacti on 1 will be more stable than the products. R eactions do not necessarily need to increase the entropy of the universe in order to be spontaneous.
49. What are thc most efficient conditions for purifyi ng nickel when using the Mond process? A.
B. C. D.
low pressure and low temperature low pressure and high te mperature high pressure and low temperature high pressure and high temperature
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Passage II (Questions 53-58)
53. If both Engines I and 2 are operating at the same time as shown in Figure I, and the rate of heat energy being removed from each reservoir is equal to the rate of heat energy being added then: .
A heat engine converts heat energy to work via a cyclical process which necessarily results in some of the heat energy being transferred from a higher temperature heat reservoir to a lower temperature heat reservoir. A heat engine obeys the First Law of Thermodynamics. The efficiency e of a heat engine is the fraction of heat energy input converted to useful work and is given by:
A.
B. C.
D.
W Q"
e=-
W'> W. W'< W. Engine I is a Carnot engine. The efficiency of Engine 2 is greater than the efficiency of Engine I.
54. Which of the following is true of the engines in Figure 1 if Engine I is not a Carnot eDgine and the work from Engine I is used to run Engine 2?
where W is the work done by the heat engine on the surroundings and Qh is the heat removed from the higher temperature reservoir. Work on the surroundings can also be represented by:
A. B. C. D.
where Qc is the heat energy expelled into the cold reservoir.
Q, < Q, 'and Qh < Qh '. Q, > Q,' and Qh < Q, '. Q, > Q, ' and Qh > Qh'· The work from Engine I cannot be used to run Engine 2.
55. Assume Engine 2 is running in the opposite direction as shown in Figure I. Which of the following changes to Engine 2 will increase its efficiency as a heat engine? A. B. C.
D.
Figure 1 A schematic diaram of two heat engines operating between the same two heat reservoirs. (Not drawn to scale.)
56. Which of the following would allow a Carnot engine to operate at 100% efficiency where e = 1. A. B. C.
The Carnol theorem states: No engine working between two heat reservoirs can be more efficient than a reversible engine working between those same reservoirs. Such a reversible engine is called a Carnot engine. This theorem can be proven with the Second Law of Thermodynnmics, which states that for any process other than a reversibJe process the entropy change of me universe is greater than zero, and fo r a reversible process the entropy change of the universe is zero. Thus a Carnot engine obeys the following equation:
D.
The work on the surroundings fiust be very large. The engine must be reversible. The hot reservoir must be at the same temperature as the cold reservoir. The cold reservoir must be at a temperature of absolute zero.
57. What is the minimum power required for a heat engine to lift a 80 kg mass 5 m in 20 s if it releases 1000 J of heat energy from its exhaust each second? A.
B. C. D.
Figure 1 shows a schematic representation of two heat engines working between the same two reservoirs. Engine 1 is absorbing heat energy from the hot reservoir and doing work while emitting heat energy into the cold reservoir. Engine 2 is a Carnet engine and is being run backwards removing heat energy from the cold reservoir and rejecting heat energy into the hot reservoir.
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increasing Qh ' decreasing Q,' decreasing the temperature difference between the reservoirs increasing the temperature difference between the reservoirs
139
200W 500W 1200 W 3000W
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Passage III (Questions 59-65)
58. A certain Carnot engine requires 18 kg of water in the form of steam as its working substance. When 5 x 10' J of heat energy are added at a constant temperature of 400 K the gas expauds to 4 m3 What is the approximate pressure of the gas after the initial expansion? (The ideal gas constant is R = 8.3 14 JIK mol) A. B. C.
D.
As shown in Figure 1, Reaction 1 is thermodynamically favored because the products are at a lower energy state than the reactants. However, at low temperatures this reaction will be too slow to be observed because the reactant molecules do not have enough energy to form the activated complex.
8.3 x 10' Pa 8.3 x 10' Pa 1.3 x 10" Pa 1.3 x 10' Pa
NO, + CO '" NO + CO, Reaction 1
O~ N --O--CO
~,
»
I'!'
~
~132kr \\ NO,+ CO
\ 226 kJ E,
\\ ~+CO,
Reaction Coordinate
Figure 1 Energy diagram for Reaction I. As shown in Figure 2, Reaction 2 is not thermo-dynamically favored, but it has a smaller reaction barrier. The reactants require less energy to form the activated co mplex.
H, + Br '" H + HBr Reaction 2
H --Br--Br
H+HBr 61 kJ
H,+Br Reaction Coordinate
Figure 2 Energy diagram for Reaction 2
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59. What does the symbol H--Br--Br represent in Figure 2? A. B. C. D.
64. What is M for Reaction 2 as written? A, B.
an intermediate a transition state a reactant of the second step of the reaction a product of the first step of the reaction
C. D,
60. What is Ea for Reaction 2 as written?
A.
B. C. D.
65. For the reaction,
15 kJ/mole 61 kJ/mole
Br
~+HBr
76 kJ/mole 132 kJ/mole
C. D.
+~Br
2
the ratio of products changes with temperatnre. At -80 c C, 80% of 1 and 20% of 2 form. At 40"C, 15% of I and 85% of 2 form. Assuming the relative stability of products vs. reactants does not change significantly with the change in temperature, which product is the kinetically favored one? A. 1 B. 2 C. both I and 2 D. neither 1 nor 2
NO, + CO -> NO + CO2 NO + CO, -> NO, + CO H2 + Br -> H + HBr H + HBr -> H, + Br
62. Which reaction is most thermodynamically favored? A. B.
~ 1
61. Which reaction is most kinetically favored?
A. B. C. D.
15 kJ/mole 61 kJ/mole 76 kJ/mole -76 kJ/mole
NO, + CO -> NO + CO, NO + CO, -> NO, + CO H, + Br -> H + HBr H + HBr -> H, + Br
63. If a catalyst were added to Reaction I. what would happen? A. B. C. D.
E j would be less than 132 kJ but E2 would remain unchanged. E j would be less than 132 kJ and E2 would be less than 226 kJ. E2 would be less than 226 but E j would remain unchanged. A catalyst doesn't affect the thermodynamic properties of the reactants and products. E, and E2 would remain unchanged but the reaction rate would increase.
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68. A metal rod is in thermal contact with two heat reservoirs both at constant temperature. one at 100 K and the other at 200 K. The rod conducts 1000 J of heat from the warmer to the colder reservoir. If no energy is exchanged with the surroundings. what is the total change of entropy?
Questions 66 through 69 are NOT based on a descriptive passage.
66. An iron skillet is laid on a hot stove. After a few minutes the handle gets hot. The method of heat transfer described
A. B. C. D.
IS:
A. B.
c. D.
convection. conduction. radiation. translation.
69. Two ideal gases, A and B, are at the same temperature, volume and pressure. Gas A is reversibly expanded at constant temperature to a volume V. Gas B is allowed to expand into an evacuated chamber until it also has a total volume V, but without exchanging heat with its surroundings. Which of the following most accurately describes the two gases?
67. A man straightens up his room. His action does not violate the second law of thermodynamics because: A. B. C.
D.
-5 J/K 0 J/K 5 J/K 10 JIK
the entropy of his room increased. energy of the universe was conserved. the entropy increase by the breakdown of nutrients in his body is greater than the entropy decrease by the straightening of his room. his action does violate the second law of thermodynamics.
A. B. C. D.
Gas A has a higher temperature and enthalpy than gas B. Gas A has a higher temperature but a lower enthalpy than gas B. Gas B has a higher temperature and enthalpy than gasA. Gas A and B have equal temperatures and enthalpies.
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142
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3D-MINUTE IN-CLASS EXAM FOR LECTURE 4
143
70. The unkn own solution contains which of the following ions:
Passage I (Questions 70-75)
A.
The following tests were carried out on samples of an unknown solution in order to identify any presence of nitrate and nitrite ions.
B. C. D.
nitrite but not nitrate both nitrite and nitrate nitrate but not nitrite neither nitrite nor nitrate
Experiment 1 71. In Experiment I , why is BaCl, added before the sulfamic acid crystals?
In acidic solution nitrites react with sulfamic acid (HNH,SO,) according to the following reaction:
A. B.
NO,- + NH,SO; -+ N,(g) + SO,'- + H,o
C. D.
Barium sulfate is insoluble while barium sulfamate is solu· ble. In a test tube, BaCl, is added to a few drops of a sample of the unknown solution. No precipitate is formed. A few crystals of sulfam;c acid are then mixed into the sample.
to acidify the solution to remove any sulfate ions existing prior to the reaction with nitrite to prevent sulfamic acid from reacting with BaCl, BaC1, is needed to react with nitrite and form the precipitate.
72. If the unknown solution contained nitrite what would be the expected result of Experiment I ? A.
No visible reaction occurs.
Experiment 2
B.
Active metals such as aluminum and zinc in alkaline solution reduce nitrate to ammonia. Nitrite ion will also fonn ammonia under these conditions. Devarda's alloy (50% Cu, 45% AI, 5% Zn) gives the following reaction with nitrate ion:
C.
D.
Al + NO,· -> AI(OH); + NH, (g)
A precipitate would be formed before the addition of sulfamic acid. Bubbles and precipitate would be observed after the addition of sulfamic acid. Bubbles but no precipitate would be observed after the addition of sulfamic acid. No visible reaction wou ld be observed after the addition of sulfamic acid.
73. If carbonate ion is present in the unknown , whi ch of the fo Uowing reactions might interfere with Experiment I ?
Several drops of the unknown sol ution are mixed with an equal amount of 6 M NaOH and placed into a dry test lUbe. Care is taken not to wet the walls of the lUbe. Devarda 's alloy is then added and a loose cotton plug pushed one third of the way down the tube. The tube is warmed briefly in a water batb and removed. A bent strip of red litmus with a moistened fold is then placed at the top of the tube as shone in Figure 1.
A. B. C. D.
2W + CO,l- -+ CO,(g) + H,O Ba'+ + 20W + CO, -> BaCO(s) + H,O NH; + CO,'· -+ NH, + HCO; H, CO, -+ CO(g) + O,(g) + H,(g )
74. When the solution in Experiment 2 is warmed in the water bath, all of the following may be true EXCEPT:
The moistened section of the litmus turns blue.
A. B. C. D.
--Litmus paper
The equilibrium constant K of the reaction changes. The rate constant k of the reaction changes. The rate of the reaction decreases. The rate of the reaction increases.
Cotton filter
75. Which of the following represents the reaction taking place in the litmus paper in Experiment 2? A. B. C. D.
Figure 1
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'44
NO,- + H,O -> OW + HNO, HNO, + H,O -+ H, O+ + NO; NH; + H, O -> H,o+ + NH, NH, + H,O -> NH; + OH-
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Passage II (Questions 76-82)
78. The graph of 2(SJ312 versus [WI for the data shown in Table I would most closely resemble which of the following?
A.
When Ca(I03)2 dissolves in a solution containing H+ the following two reactions occur. Ca(IO,), '" Ca'+ + 2I0 3Reaction 1
c.
~LL ~~ [H]
[H]
H+ + 103- '" HI0 3 B.
Reaction 2 HIO, is a weak acid. The K'P for Ca(IO,), and the K, for HI03 can be detennined from the solubility (S) of Ca(IO,), for solutions of varying [H+]. The solubility is related to the initial hydrogen ion concentration [WI by the following equation:
~~ ~lv ~
"
[H]
[H]
79. After filtering out excess solid, a student adds HCI to Solution I in Table I. He then adds a small amount of CaS04 , which dissolves completely. Which of the following also occurs in the new solution?
A student prepared four saturated solutions by mixing Ca(IO,), with a strong acid. Excess solid was filtered off. The student found the S for each solution with constant ionic strength, using iodometric titrations. The resulting data are shown in Table I. Solution [H'l (mol/l) 1.0 x 10-7 I 2 2.5 X 10-1 3 5.0 X 10-1 4 I
D.
A. B. C.
D.
S (molll) 5.4 X ]G-3 9.9 X 10-3 1.4 X 10-' 2.0x 10-'
Some Ca(103), precipitates when the CaSO, is added. Undissociated HI03 increases when the HCI is added. Aqueous 10; decreases when CaS04 is added. Aqueous Ca2+ decreases when CaS04 is added.
80. According to Table I, what is the value of Ca(I03 ), ?
A. B. C.
D.
Table 1 Solubility data for Ca(l03)'
1.0 X 6.4 X 5.4 X l.l X
K;p
for
10-14 10-7 10-' 10-1
81. If Ca(OH)2 is added to the Solution 3 in Table I: 76. The K,p for Ca(I0 3), and the K, for HIO,. respectively are:
A.
[Ca'+][IO,-I' and
B.
[Ca"][IO,-]' and [Ca(IO,),]
[H+][IO;]
[Ca"][IO,-], and
[H+][IO,-]
C.
D.
ICa7+][IO,-]' and
A. B. C. D.
[Hl[IO,-] [HIO,]
[HIO,]
82. How will the addition of HI03 affect Solution 2 from Table I? A. B.
[HIO,i [Hl[IO, ]
C. D.
77. As [WI increases, the solubility of Ca(IO,),:
A. B.
C. D.
increases and Ksp increases. decreases and K..'p decreases. increases and K,p does not change. does not change and Ksp increases.
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the concentration of H+ will increase. the concentration of HIO) will increase. the concentration of 103- will increase Ca(IO,), will precipitate.
145
The lower pH will shift Reaction 2 to the right. The increased hydrogen ion concentration will dissolve more Ca(IO,),. The common ion effect will shift Reaction 1 to the left. The lower pH will balance out the common ion effect and the equilibrium will not change.
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Passage III (Questions 83-89)
Many households soften hard water using ion exchange resins. These resins replace the Ca2+and M g2+ ions with smaller cations such as Na'" and H ....
Many carbonate minerals are found in the earth's crust. As a result, the waters of several lakes. rivers, and even oceans are in contact with these minerals. CaC03 is the primary co mponen t of limestone and marble, while dolomite (CaMg(CO, ),) and magnesite (MgCO,) are mi nerals found in other rock formations.
83. A scientist determines how hard the tap water is in the laboratory, using an EDTA ti tratio n. If the pipes in the building are old and some rust dissolves into the tap water. how will the results of the test change?
A.
Limestone lines many of the river and lake beds resulting in contamination of the fresh water supply with Ca2• and Mg" ·. The amount of these minerals present in water can be measured in parts per million (ppm). The "hardness" of water is determined by the ppm of Ca'· and Mg'· present. Hard water is the cause of many problems in the home. Scale buildup in pipes, on pots and pans, and in washing machines are just a few of the problems.
B.
C. D.
The hardness of water can be measured by titrating a sample of water with the ligand ethylenediamine tetraacetic acid (EDTA) and the indicator eriochrome black T. This ligand forms a coordination complex with metal cations (M) in a aneta-one stochiometry in the following association reaction:
84. When EDTA reacts with a metal ion to form a complex ion, EDTA is acting as a(n): A. B. C. D.
EDTA + M -> EDTA-M The structure of EDTA is shown in Figure I. It has six binding sites to form a very stable complex ion with most metal IOns.
O-
°II
H,
C-
C
-O-C -
II
H,
"N - C - C -~
C/ H,
H,
H,
°
°II
A. B. C. D.
C - O-
86. A 25 mL sample of hard water is titrated with a 0.00 1 M solution of EDTA, and the endpoint of the titration is reached at 50 mL of EDTA added. What is the concentration of Ca2+ and Mg 2... ions in solution?
Figure 1 T he structure of EDTA The association constants for EDTA with several metal ions are listed in Table 1.
A. B. C.
K~~
Hg' ·
6 X 10 21
Mg2T
5 X 10'
Ca2+
[t will increase. It will decrease. It will remain the same. The change in solubility cannot be determined.
II
°
Metal
oxidizing agen t. reducing agent. Lewis base. Lewis acid.
85. Salt water contains a high concentration of cr ions. These ions form complexes (CaCI· ) with Ca'·. How will the solubility of limestone change in ocean water compared to fresh water?
c--c - 0 -
"C H,
The results will not change because the EDTA titration only works with Ca2+ and Mg2+. The titration will not be able to be carried out because the tap water will be colored. The tap water will appear to have less Ca2• and Mg'· present. The tap water will appear to have more Ca' · and Mg2-t- present.
D.
0.0005 M 0.001 M
0.002M O.OO6M
10
AI"
5 x 10 I X 1016
Fe"
2 x 10 "
Fe" Cu 2+
6 X 10 18
I x 10"
Table 1 Association constants for EDTA with metal ions
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87. The EDTA titrations are carried out at a pH of 10. Why is it necessary to buffer the pH at 1O? A. B. C. D.
Questions 90 through 92 are NOT based on a descriptive passage.
A low pH will cause metal hydroxides to form. CaC0 3 requires a high pH in order to dissolve. The indicator requires pH 10 to change color. The coordinating atoms must be deprotonated in order to bond with the metal ion.
90. The vapor pressure of pure water at 25°C is approximately 23.8 torr. Which of the following is the vapor pressure of pure water at 95°C? A. B. C. D.
88. Why does replacing the cations found in hard water with Na+ or H+ soften the water (i.e., reduce the unwanted residue produced by hard water)? A.
B. C. D.
The smaller cations do not form insoluble mineral deposits. Twice as many smaller ions are necessary to react with soaps and other ligands. No minerals contain Na and H. H is found in water so there is no addition of new atoms.
91. Benzene and toluene form a nearly ideal solution. If the vapor pressure for benzene and toluene at 25°C is 94 mm Hg and 29 mm Hg respectively, what is the approximate vapor pressure of a solution made from 25% benzene and 75% toluene at the same temperature? A.
B. C.
89. 9 ppm is equivalent to an aqueous concentration of approximately 5 x 10-4 moIIL. If a water saruple were reduced from 18 ppm Mg2+to 9 ppm Mg'+ by the addition of EDTA, according to Table I what would be the concentration of the remaining unbound EDTA? A. B. C. D.
2X 5x Ix 5x
10 torr 23.8 torr 633.9 torr 800 torr
D.
29 mmHg 45 mmHg 94mmHg 123 mmHg
92. When volatile solvents A and B are mixed in equal proportions heat is given off to the surroundings. If pure A has a higher boiling point than pure B, which of the following could NOT be true?
10-9 moIIL 10-4 moIIL 10-] mollL 10-3 moIIL
A. B. C. D.
The boiling point of the mixture is less than pure A. The boiling point of the mixture is less than pure B. The vapor pressure of the mixture is less than pure A. The vapor pressure of the mixture is less than pure B.
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147
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3D-MINUTE IN-CLASS EXAM FOR LECTURE 5
149
93. Why was NaCl added to the ice bath?
Passage I (Questions 93-100)
A. Two students pedarmed the following experiment to calculate the molecular weight of an unknown substance. The apparatus shown in Figure 1 was used.
B. C.
clamp
/
D.
94. Which salt is the most efficient per gram at lowering the freezing point of water?
/tirrer
thermometer:
clarn~
cork
To lower the freezing point of the water and cool the cyclohexane solution more quickly. To lower the freezing point of the cyclohexane solution below the freezing point of the water. To lower the freezing point of the water below the freezing point of the cyclohexane solution. To raise the freezing point of the water.
~'!t!1" 22 x l7Smm
======~~~~~~~ttube
A.
Ba(OH),
B.
MgS04
C. D.
NaCl CaCl,
95. The purpose of the copper stirrer is: A. B.
crushed ice and NaCl solution
C. D.
Figure 1 Freezing Point Apparatus Both students placed 10.00 mL of cyclohexane into the test tube at room temperature. Next 0.500 gram of an unknown solid was dissolved in the cyclohexane. The test tube and contents were lowered into the ice bath, which was maintained at a temperature of -S.O°C by adjusting the relative amounts of NaCI, ice, and water. The students monitored the temperature of the cyclohexane mixtures by taking readings from the thermometer at 30 seconds intervals. The freezing point of the solution for a given trial is the temperature maintained for four consecutive readings. The experiment was repeated two more times by warming the cyc10hexane to room temperature then freezing it again.
96. According to the results in Table 1, which student had the unknown with the greatest molecular weight? (Assume no dissociation of the unknown solids occurs.) A. B. C. D.
A.
Time, seconds
B.
Studentl 22.0
60
90
6.0
4.0 -3.S -3.S -3.5 -3.S -3.5
Student 2 22.0 12.0
6.0
120
0.6
ISO
0.6
180 210
0.6
0.6
Student 1 Student 2 The molecular weights were the same. It canno:- be detennined based on the given information.
97. Student 1 recorded a lower freezing point for the experiment because:
The results obtained by the students are recorded in Table 1.
30
to ensure that the solid stays in solution. to create heat to offset the chilling effect of the ice bath. to ensure that the solution temperature remains homogenous. to allow the student to see when crystals begin to form.
240
C.
0.4 D.
Table 1 Solution temperature (0C) with time
the concentration of particles in Student l 's crushed ice solution was lower than in Student 2's crushed ice solution. the concentration of particles in Student l's crushed ice solution was greater than in Student 2's crushed ice solution. the concentration of particles in Student 1's cyclohexane solution was lower than in Student 2's cyclohexane solution. the concentration of particles in Student l's cyc1ohexane solution was greater than in Student 2's cyclohexane solution.
(The Kf for cyclohexane is 20.2°C kg/mol, the freezing point is 6.6°C, and the density is 0.78 g/mL. The Kf for water is 1.86°C kg/mol.)
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Passage II (Questions 101-106)
98. In order to ca\Culate the molecular weight of their unknown solid the students probably used all of the following data EXCEPT: A. B. C.
D.
A series of experiments are perlormed using the calorimeter shown in Figure 1.
the mass of the unknown solid added to the solution. the time required for the cyc1hexane solution to freeze.. the temperature at the freezing point of the cyc1oexane solution. the volume of the cyC\ohexane solution.
99. A professor must choose the unknown from the following solutes. Which of the following would be the most appropriate for the experiment in the passage? A. B. C.
NaCI Mg(OH), CH,oH
D.
C,Jf, Figure 1 Coffee Cup Calorimeter
100. Why does the temperature in Student 2's experiment begin to drop after 210 seconds? A. B. C. D.
A volume of 0.5 M NaOH is placed near the calorimeter, which contains an equal volume of 0.5 M HC!. The temperatures of both solutions are monitored until they equilibrate to room temperature.
Student 2 used too much ice in the ice bath. Student 2's cyC\ohexane solution was completely frozen at 210 seconds. Student 2 stopped stirling the solution at 180 seconds. Student 2 stopped dissolving the unknown solid.
The NaOH solution is added to the HC] solution through the funnel. The temperature is recorded every 30 seconds for 5 minutes. The experiment is repeated three times with three different volumes of Hel and NaOH. The results of one of these experiments are shown in the graph in Figure 2. The data for all experiments are recorded in Table 1.
~ ~ This temperature is the average of the
I
o
solution temperatures before mixing.
Time in minutes
Figure 2 Temperature change of solution over time
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105. Which trial would be expected to result in the greatest heat of solution per mole of reactants?
Final Volume Initial ofHCL and Temperature Temperature 'C 'C Trial NaOH (ml) 22.0 25.3 I 30 2 40 20.0 23.3 24.3 3 50 21.0
A. B.
C. D.
106. If the solutions in the experiment began at room temperature, which of the following explains the heat transfer between the calorimeter and its surroundings for the experiment shown in Figure 2?
Table 1
101. What reaction is taking place in the calorimeter to cause the temperature change?
A. B. C. D.
A.
W + OH- --"> H20 Na+ + Cl- --"> NaCl NaCI --"> Na+ + CINa+ + le- ---7 Na
B.
C.
102. The reaction in the calorimeter is an: A. B. C. D.
2 3 They should all be the same.
D.
endothermic reaction. exothermic reaction. oxidation reaction. isothermic reaction.
Initially heat is transferred from the surroundings to the calorimeter, and then heat is transferred from the calorimeter to the surroundings. Initially heat is transferred from the calorimeter to the surroundings, and then heat is transferred from the surroundings to the calorimeter. Heat is transferred from the surroundings to the calorimeter throughout the experiment. Heat is transferred from the calorimeter to the surroundings throughout the experiment.
103. Assuming that the heat capacity of the solution is the same as the heat capacity of water, what is the enthalpy change for the reaction in Trial 2 as recorded in Table I? (The heat capacity for water is 1.0 cal 'C- 1 mL- 1)
A. B. C. D.
-132 cal 132 cal -330 cal 330 cal
104. If 0.5 M NH,OH (a weaker base) were used instead of NaOH, how would this affect the results of the experiment? A.
B. C. D.
The temperature change would be greater because more energy is required to dissociate NH4 0H. The temperature change would be less because more energy is required to dissociate NH40H. The temperature change would be greater because less energy is required to dissociate NH4 0H. It would not change thc results bccause both bases are ionic compounds and the energy required to separate equal charges is always the same.
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Passage III (Questions 107-113)
108. According to Figure I, at - 78°C and I aim CO, will: A. B. C.
Phase diagrams show the changes in phase of a mmeriai as a function of temperature and pressure. Student A prepared a phase diagram for CO 2 , After observing the phase diagram, he concluded that raising the pressure isothermally promotes a substance to change from a gas to a liquid to a solid as demonstrated by the dashed line in Figure I.
B A
D.
109. The temperature and pressure above which the gas and liquid phases of a substance can not be distinguished is called the: A. B. C. D.
t
C
't--- D I! ,
A.
D. -78
C. D.
Temperature (0C) Figure 1 Phase diagram of CO,
B
CO, is a vapor. CO, is a liquid. CO, is in both liquid and vapor phase. the vapor and liquid phases of CO, cannot be distinguished.
111. According to Figure 2, as the pressure increases the melting point of H,o?
Student B chose to make a phase diagram of H 2 0. She observed that raising the pressure isothermally promotes a substance to convert from vapor to solid then to liquid as indicated by the dashed line in Figure 2.
A
critical point triple point boiling point super poi nt
110. At temperatures and pressures greater than point C in Figure 1:
I
E
exist as a liquid. exist in equilibrium as a gas and liquid. exist in eqUilibrium as a gas and solid. exist in equilibrium as a liquid and solid.
A. B. C. D.
increases decreases does not change increases than decreases
112. The normal boiling point for 0 , is 90.2 K. Which of the
'"C
following could be the triple point for O, ?
A. B. C. D. 100 Temperature (OC)
1.14 mmHg and 54.4 K 1.14 mmHg and 154.6 K 800 mmHg and 54.4 K 37,800 mmHg and 154.6 K
113. Describe the phase change for H,o as the pressure is raised at 100°C.
Figure 2 Phase diagram of H,0 A.
B. C. D.
107. Which of the followin g explains the discrepancy between the observations of the two students? A. B. C. D.
sublimation vaporization condensation melting
Water expands when going from liquid to solid, where as CO2 contracts. CO, expands when going from liquid to solid, where as water contracts. The two chemists observed the phase changes at different temperatures. CO, is a gas at room temperature, while H,O is a liquid.
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115. On his honeymoon the chemist, Joule, took with him a
Questions 114 through 115 are NOT based on a descriptive passage.
long thermometer with which to measure the temperature
difference between fhe waters at the top and fhe bottom of Niagra Falls. If fhe height of fhe falls is 60 meters and fhe specific heat of water is approximately 4200 J kg- 1 K- 1 ,
114. During a solid to liquid phase change, energy is: A. B. C. D.
what is the expected temperature difference?
absorbed by bond breakage. released by bond breakage. absorbed by increased kinetic energy of the liquid molecules. released by increased kinetic energy of fhe liquid molecules.
A. B. C. D.
117 K 7K 70K 700K
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154
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30-MINUTE IN-CLASS EXAM FOR LECTURE 6
155
117. How do the titrations in Trials I and 3 compare?
Passage I (Questions 116-121)
A. The solubility of Ca(OH), (Reaction 1) can be determined by titrating the saturated solution containing no precipitate against a standardized HCI solution and determining [OH-].
B.
Ca(OH), "" Ca'+ + 20H-
C.
Reaction 1
D.
Once [OH-] is determined, the solubility (S) of Ca(OH), is calculated using the following equation:
The pH of the equivalence points are the same, but more Hel is required to reach the equivalence point in Trial 3. The pH ofthe equivalence point in Trial 1 is higher, and less Hel is required to reach it. The pH of the equivalence point in Trial 3 is higher, and less Hel is required to reach it. The pH of the equivalence point in Trial 3 is higher, and more Hel is required to reach it.
118. The K,p for Reaction 1 in the presence of NaOH is:
S=~[OW] 2 Ca(OH)2 Equation 1 where [OH-]Ca(OHh is the concentration of hydroxide ion due only to Ca(OH),. The solubilities of Ca(OH)2 in a variety of solutions of varying [OH-] concentrations were determined by the above method, but the calculation of S had to be altered slightly due to the presence of additional hydroxide ions.
A.
[Ca2+][OWl'c'(OH)'
B.
C.
[Ca2+][OH-]\"", [Ca2+] [20W]2 '(OH),
D.
[Ca2+] [20H-]\"",
C
119. What is the pH of the solution in Trial 3 before the titration?
A. B.
S-~{[ow] -[ow] solvent } 2 total
C. D.
l.l 7.0 9.3 12.7
Equation 2 120, Which indicator would be best for the titration in this experiment?
The results of the experiment are summarized in Table 1. Trial
Solntion
Solubility
A. B.
1
H,o
0.0199 M
2
0.01793 M NaOH
O.OlOOM
3
0.03614 M NaOH
0.0047 M
4
0.07119 MNaOH
0.0015 M
C. D.
Phenolphthalein: (acid color is colorless, base is red, and the transition pH is 8.0 - 9.6). Thymolphthalein: (acid color is colorless, base is blue, and the transition pH is 8.3 - 10.5). Bromocresol purple: (acid color is yellow, base is purple, and the transition pH is 5.2 - 6.8). Neutral red: (acid color is red, base color is yellow, and the transition pH is 6.8 - 8.0).
Table 1 Solubility data for Ca(OH),
116. How does the solubility of Ca(OH)2 change as the [OH-] in the solvent increases?
A. B.
C. D.
It decreases because the increase Reaction I toward the left. It decreases because the increase in with the acid titration. It increases because the increase Reaction 1 toward the left. It increases because the increase in with the acid titration.
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in OH- shifts OH- interferes
in OH- shifts OH- interferes
156
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Passage 11 (Questions 122-127)
121. If a pH meter were placed into the titration beaker, what would be the resulting curve for Trial I?
A.
The reaction for the autoionization of water is shown below: 14t=-_~
2H,G
'&7
--7
H,o+ + OW
The equilibrium constant (K.,..J is temperature dependent. Table 1 lists the value of K.,." at several temperatures. Temperature ("C) Volume of Titrant Added
B.
Kw
0
0.114 x 10
14
10
0.292 x 10
14
20
0.681 x 10
14
25
1.01 x 10
30
1.47
40
2.92 x 10
50
5.47 x 10
14
60
9.61 x 10
14
X
14
10-14 14
Volume of Titrant Added Table 1 Equilibrium constants for water at different temperatures C.
14
Water has a leveling effect on acids. Any acid stronger than H30+ appears to have the same behavior in aqueous solution. For example, 1 M HCI and 1 M HCI04 have the same concentration of H30+ even though in anhydrous acetic acid, HCI04 is a stronger acid.
'&7
I Volume of Titrant Added
122. What is the pH of H,G at 400C? A.
D.
B. C. D.
14
7.5 7.0 6.7 6.0
'&7 123. At 10°C, the concentration of OH- in I M HCI is approximately: A. B. C.
Volume of Titrant Added
D.
1X IX 3X 1X
10-7 M 10- 14 M 10-15 M 10- 15 M
124. As temperature increases, the pH of pure water: A. B. C. D.
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mcreases. decreases. becomes less than the pOH. becomes greater than the pOH.
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Passage III (Questions 128-134)
125. What is the conjugate base of H 2S04 ? A.
B. C.
D.
Hp OHHS04SO,'-
Acid rain results when S03(g). produced by the industrial burning of fuel, dissolves in the moist atmosphere. SO/g) + Hp(l)
126. Why can the relative strength ofHCl and HCl04 be determined in acetic acid but not in water? A. B. C.
D.
B.
because acetic acid is a weaker acid than H30+ because acetic acid is a stronger acid than H30+ because acetic acid is a weaker Bronsted-Lowry base than HP because acetic acid is a stronger Bronsted-Lowry base than Hp
Another pollutant which dissolves in water vapor and reacts to form acid rain is SO,(g). This gas forms H 2 S0 3 (aq) which can be oxidized to H 2S04 , (The pK, values are 1.81 for H,S03 and 6.91 for HS03-.) The table below gives the color changes of many acid base indicators used to test the pH of water.
[OH-][HP+]
[OH- ][HP+] [HPJ'
C.
D.
H,S04(aq)
The rain formed from the condensation of this acidic water is an environmental hazard destroying trees and killing the fish in some lakes. (The pH of the water varies depending upon the level of pollution in the area. The pK, values are about -2 for H2S04 and 1.92 for HS04 -.)
127. The equation for Kw at SO'C is: A.
--7
[OH ][Hp+] [H 2 O] [Hp+]
Indicator
Color Change
pH of color change
Malachite green
yellow to green
0.2 - 1.8
Thymol blue
red to yellow
1.2 - 2.8
Methyl orange
red to yellow
3.2 - 4.4
Methyl red
red to yellow
4.8 - 6.0
Phenolphthalein
clear to red
8.2 - 10.0
Alizarin yellow
yellow to red
10.1 - 12.0
Table 1
128. A sample of rainwater tested with methyl orange results in a yellow color, and the addition of methyl red to a fresh sample of the same water results in a red COIOf. What is the pH of the sample? A. B. C. D.
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158
between between between between
1.2 and 1.8 3.2 and 4.4 4.4 and 4.8 4.8 and 6.0
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129. If there is no oxidant present in the air and the same number of moles of SO, and S0, are dissolved, which gas would produce acid rain witb a lower pH? A. B. C. D.
S0, S0, SO, SO,
134. A sample of rainwater polluted witb S0, is titrated with NaOH. Which of the following most resembles the shape of titration CUNe.
A.
because H,SO, has a higher pK. than H,S04' because HSO,- has a higher pK. than HSO,-. because H2S04 has a lower pK. than H,SO, . because HS04- has a lower pK. than HSO,-.
14
:a7
130. What is the pK, for HSO,-?
A.
0
B. C. D.
6.91 7.09 12.19
Volume of Titrant Added
B.
14
131. H,S04 is a stronger acid than:
:a7
I. H,0 ll. H,o' Ill. H,SO,
A.
m only
B.
[ and II only I and III only I, n, and m
C. D.
1 Volume of Titrant Added C.
132. What is the oxidation state of sulfur in H2S0 4 and H,SO, respectively? A. B.
C. D.
14
:a7
+6,+4 -1-4,+6 --{}, -4 -4 , --{}
Volume of Tilrant Added 133. What is tbe pH of a 5.0 x 10-8 M aqueous solution of H2 S04 at room temperature?
A. B. C. D.
D.
8.3 7.3 6.8 6.0
Volume of Titrant Added
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138. NH3 has a Kb of 1.8 x 10-'. Which of the following has a K" of 5.6 x 1O- 10?
Questions 135 through 138 are NOT based on a descriptive passage.
A.
B. C.
135. Which of the following is the strongest base?
A. B. C.
D.
D.
CIOCIO,CI03ClO.STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
136. A weak acid is titrated with a strong base. When the concentration of the conjugate base is equal to the concentration of the acid, the titration is at the:
A. B. C. D.
NH3 NH: NH; H+
stoichiometric point. equivalence point. half equivalence point. end point.
137. A buffer solution is created using acetic acid and its conjugate base. If the ratio of acetic acid to its conjugate base is 10 to I. what is the approximate pH of the solution? (The K" of acetic acid is 1.8 x 10-')
A. B.
C. D.
3.7 4.7 5.7 7.0
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160
STOP.
30-MINUTE IN-CLASS EXAM FOR LECTURE 7
161
141, What is the volume of dichromate added to the titration at the equivalence point?
Passage I (Questions 139-144) When Fe2+ is titrated with dichromate (Cr Z0 72-) according to Reaction I, the titration curve similar to the one shown in Figure 1 results. The curve was generated by measuring the potential difference between the reaction solution and a standard solution after each addition of a known volume and concentration of dichromate. The potential is measured using a voltmeter attached to an orp electrode.
eo =
A. B.
C. D,
142, Which of the following expressions gives the concentration of Fe2+ in the unknown solution in terms of the volume of dichromate added Vc"o,~, the molarity of dichromate Mcr,o,>-, and the originai volume of Fe2+ solution VFe ?
1.33 V
Reaction 1 700
----
I I
---- 1-
~
I I
+I
500 400
..
I I I
200 -5
1··
I
0
,
i
I, ___ .1
I
i
I,
10
15
20
-- --or ··1I
300
A.
...
600
--j"-
B,
I... I I I I
C.
I I 25
30
D.
35
mL of K,Cr,07
Figure 1. Titration Curve for Reaction 1
Cr20:;
Cr20 7
VFe
(6Vop; )(Mccp;) (Vc r:P7,)( M erp:;,.) 6VFe (2 Ve'P; )( Me,,oi ) VFc
143, If the formal potential of the solution is not lowered, which of the following will be the result of the titration when DAS is used as the indicator?
Chromium (TIl) is green in color, and dichromate solution is orange. The orange is not intense enough to be used as an indicatoI. Instead, the endpoint of the titration can be indicated by the redox indicator diphenylamine sulfonic acid (DAS), which changes from colorless to violet when oxidized. The color change observed during the titration is from green to violet, After all the Fe z+ ions have been oxidized, the dichromate ion oxidizes DAS_, However, the formal potential of the Fe2+ solution must be lowered in order to match the endpoint with the equivalence point of the titration. This is accomplished by the addition of H,S04 and H 3P04 immediately before titration.
A. B,
C, D,
The solution will tum violet before the equivalence point is reached. The solution will turn violet after the equivalence point is reached. Dichromate ion will not be able to oxidize DAS. Dichromate ion will reduce DAS to its colorless fonn.
144. Which of the following statements is true at the equivalence point of the titration in the passage?
139. Which of the following is true concerning Reaction I?
A.
Cr is oxidized and CrZ0 72- is the oxidizing agent. Cr is reduced and Cr:2072- is the reducing agent. Cr is reduced and CrzO/- is the oxidizing agent. Cr is oxidized and Cr2 0 72- is the reducing agent.
B.
A. B. C. D.
(v . )(M , ) VF c
...... 1.
,
13 mL 15 mL 25 mL 30 mL
C. D.
Each iron ion has lost one electron and chromium ion has gained three electrons. Each iron ion has lost one electron and chromium ion has gained six electrons. Each iron ion has gained one electron and chromium ion has lost three electrons. Each iron ion has gained one electron and chromium ion has lost six electrons.
each each each each
140, What is the oxidation state of Cr in Cr,O/-?
A.
+12
B.
+7 +6
C. D,
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+3
(g
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Passage II (Questions 145-151)
147. What is the oxidizing agent in the nickel cadmium battery during discharge?
Rechargeable batteries have become an esse ntial part of OUf
A. B. C. D.
environmentally conscientious society. The nickel-cadmium cell battery is a rechargeable battery used in small electronic devices. The half reactions that take place in the nickelc admium battery during discharge are:
Cd Cd(OH), NiO, 2Ni(OH),
148. Which of the following is true concerning the nickelcadmium battery w hen it is recharging?
Cd(OH),(s) + 2e- --> Cd(s) + 20HE' = -OAV Half Reaction 1
A. B.
2NiO,(s) + H,o + 2e- --> 2Ni(OH),(s) + 20HE' = 0.5 V
D.
C.
The cell The celi The cell The celi
is is is is
a a a a
nonspontaneous electrol ytic ceiL nonspontaneous galvanic ceiL spontaneous electrolytic cell. spontaneous galvanic cell.
Half Reaction 2
149. In order to recharge the ni ckel-cadmium battery back to standard conditions, what is the minimum voltage that
Other types of rechargeable batteries currently being developed are those using sodium or lithium metal as the anode and sulfur as the cathode. These batteries must operate at high temperatures because the metals must be in the liquid state, but they provide a high energy density, which means the batteries will be very light weight
must be applied across its electrodes?
145. The reaction taking place at the anode when the nickel-
A.
C. D.
C. D.
0.9V L8V
sodium in the spontaneous direction?
Cd(s) + 20W --> Cd(OH ),(s) + 2e2NiO,(s) + H 20 + 2e- --> 2Ni(OH),(s) + 20W Cd(OH ),(s) + 2e- --> Cd(s) + 20W 2Ni(OH),(s) + 20W --> 2NiO, (s) + H 20 + 2e-
A.
Na+ + e-....., Na
B.
Na ~ Na+ + eNa ....., Na 2+ + 2eNa+ + OW --> NaOH
C.
D.
151 . The nickel-cadmium battery is used to power a light bulb. The current in the light bulb flows:
146. When the nickel-cadmium battery is recharging, what is the reaction at the anode? A. B.
0.1 V 0.2 V
150. In a sodium-sulfur battery, what is the half reaction for
cadmium batteries are discharging is:
B. C. D.
A. B.
Cd(s) + 20H- --> Cd(OH)2(s) + 2e.2NiO, Cs) + H,o + 2e- --> 2Ni(OH),(s) + 20W Cd(OH),(s) + 2e- --> CdCs) + 20W 2Ni(OH),(s) + 20W --> 2NiO,(s) + H,o + 2e-
A.
B.
C.
D.
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in the same direction as the flow of electrons, from the side with Half Reaction I to the side with Half Reaction 2. in the same direction as tbe flow of electrons, from the side with Half Reactio n 2 to the side with Half Reactio n 1. in the opposite direction to the flow of electrons, from the side with H alf Reaction I to the side with Half Reaction 2. in the opposite direction to the flow of electrons, from the side with Half Re.action 7. to the ~ide with Half Reaction L
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153. What would be the approximate potential difference measured by a pH meter if the test solution had a pH of 2 and the refere nce solution had a pH of 4?
Passage III (Questions 152-158) A pH meter is a concentration ceU which measures the potential difference between a reference solution and a test solution and reports the difference in terms of pH. In a simplified version of a pH me ter the half reactions are:
A. B. C. D.
H2 + 2H 20 -7 2H30+ + 2e2H,G+ + 2e- -7 H, + 2H, O
154. The potential difference measured by a pH meter is directly proportional to:
The potential difference between the two solutions is derived from the Nemst equation as follows:
A. B. C. D.
where E is given in volts. This equation can be rewri tte n in terms of the pH of the solutions as follows:
A. B.
C. D.
A.
C.
152. What is the reaction quotient (Q) in the Nem st equation for the simple pH meter ?
B.
c.
0 1 2 8
156. How would the potential difference registered by a pH meter change for a given test solution if th e hydrogen ion concentration of the reference solution were increased by a factor of IO?
B.
D.
r
[ H }O+
the difference in the hydrogen ion concentrations of the test and reference solution. the difference in the pH of the test and reference solution. the pH of the test solution. the hydrogen ion concentration of the test solution.
155. If the reference solution of a pH meter were I M HCI, and the potential difference measured by the meter were 59 mY, what would be the pH of the test solution?
Because it is inconvenient to bubble H, gas through a solution, a more sophisticated pH meter is used in standard laboratory practice. Dilute hydrochloric acid is used as the reference solution. The test solution is in contact with a thin glass membrane in wh ich a silver wire coated with silver chloride is imbedded. This glass membrane is dipped into the test solution and the potential difference between the solutions is measured and interpreted by a computer, which displays the pH of the test solution. The same equation holds for both pH meters.
A.
-118 mV -59 mV 59 mV 118mV
(e.~( S()1uLion
The pote ntial difference wou.ld increase by 59 m V. The potential differe nce would decrease by 59 m V. The potential difference would increase by a factor of 10. The potential differe nce wo uld decrease by a factor of 10.
[H,] [H,] [ H,o+ ]2 ,~, ~!"tio"[H2 ]
.
, efel c lICC SOIUUOl1
. [H30 +]2refereoccwlulion[H, ]Ir:st solullon
D.
[H JO+Y lesl SOIUlioD
r
[ H 30+
refeIence solution
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164
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157. A galvanic cell is prepared by connecting two half cells with a salt bridge and a wire. One cell has a Cu electrode and I M CuSO,. and the other has a Cu electrode and 2 M CuSO,. Which direction will the current flow through the wire? A. B. C. D.
Questions 159 through 161 are NOT based on a descriptive passage.
159. Consider the reduction potential:
toward the 1M CuSO, solution. toward the 2M CuSO, solution. current will not flow because the half reactions are the same for both sides. current will not flow because both half cells have Cu electrodes.
Zn2+ + 2e- --> Zn(s)
B. C. D.
~
0.76 V.
When solid Zinc is added to aqueous HCI, under standard conditions, does a reaction take place? A. B.
158. Which of the following is true for an acid-base concentration cell such as the one used by the pH meter? A.
EO =
C.
Current always flows toward the more acidic solution. Current always flows toward the more basic solution. Current always flows toward the more neutral solution. Current always flows away from the more neutral solution.
D.
No, because the oxidation potential for Cl- is positive. No, because the reduction potential for Ct is negative. Yes, because the reduction potential for H+ is positive. Yes, because the reduction potential for H+ is zero.
160. Chemicals are mixed in a redox reaction and allowed to come to equilibrium. Which of the following must be true concerning the solution at equilibrium?
A. B. C.
D.
K=l !J.G" = 0 E=O !J.Go =!J.G
161. At 298 K all reactants and products in a certain oxidationreduction reaction are in aqueous phase at initial concentrations of I M. If the total potential for the reaction is E = 20 mY, which of the following must be true? A. B. C. D.
K= I = 20 mV !J.G is positive K< 1 E0 298
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
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165
STOP.
ANSWERS & EXPLANATIONS FOR
30-MINUTE IN-CLASS EXAMINATIONS
167
168
MCAT INORGANIC CHEMISTRY
ANSWERS FOR THE 30-MINUTE IN-CLASS EXAMS Lecture 1 1. 0
Lecture 2 24.
Lecture 3
Lecture 4
Lecture 5
B
47.
A
70.
e
93.
e e e
2.
B
25. A
48.
e
71. B
94.
3.
A
26.
49.
B
72. B
95.
4.
B
27.
e e
50.
e
73. A
96. B
5.
e
28.
B
51.
A
74.
6. 0
29.
52.
0
75. 0
7. B
30.
e e
53.
76.
A
8.
e
31.
A
54.
e e
77.
9.
0
32.
A
55.
10.
e
33.
A
11. 0
e
Lecture 6
Lecture 7
116. A
139.
117. A
140.
118. B
141.
e e e
0
142. B
0
120. 0
143. A
98. B
121. A
144. A 145. A
97.
119.
0
122.
e
100. B
123.
e e
0
78. B
101. A
124.
B
147.
56.
0
79. B
102. B
125.
148. A
34. 0
57.
e
80. B
103. A
126.
e e
127. A
150. B
e e
151. 0
99.
146. 0
149.
e e
e e
35.
0
58.
A
81.
0
104.
36.
e
59. B
82.
e
105. 0
128.
14. A
37.
B
60.
e
83.
0
106. 0
129.
15. A
38.
A
61. 0
84.
e
107. A
130. 0
153. A
16. B
39. A
62. A
85. A
0
154. B
B
40. 0
63. A
86.
12. 13.
17.
108.
B
e
131.
152.
0
e
109. A
132. A
155. B
18. 0
41.
B
64.
B
87. 0
110. 0
133.
e
156. A
B
42.
0
65.
A
88.
A
111. B
134. 0
157. A
20. A
43.
e
66.
B
89.
A
112. A
135. A
158. B
21.
0
44.
B
67.
90.
e
113.
e
159. 0
22.
B
45.
A
68.
e e
91. B
114. A
137. A
160.
23.
0
46.
e
69.
0
92. B
115. A
138. B
161. B
19.
e
136.
e
PHYSICAL SCIENCES Raw Score
Estimated Scaled Score
23 22
15 14
21 19-20 18 16-17 15
13
13-14 12 lO- l1 9 7--8
12
11
10 9 8
.
7
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ANSWERS
&
EXPLANATIONS FOR THE 30-M INUTE IN-CLASS EXAMINATIONS . 169
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 1 Passage I 1.
D is correct. If you get a boiling point question on the MCAT, look for hydrogen bonding. It increases the strength of intermolecular attractions. Stronger intermolecular attractions leads to higher boiling point.
2.
B is correct. You should recognize this compound as ionic because alkaline earth m etals like to form ionic compounds with halogens.
3.
A is correct. In order to explain an increase in boiling pOint, we have to look for a reason that intermolecular bond strength would increase. The intermolecular bonds in n oble gases are totally due to van d er Waals forces. H the atoms are more polarizable, instantaneous dipoles can have greater strength. Larger atoms are more polarizable because the electrons can get farther from the nucleus and create a larger d ipole moment.
4.
B is correct. This is a periodic trend. Radius increases going down an d to the left on the periodic table.
5.
C is correct. Crystallization depends upon molecular symmetry as well as intermolecular bonding. Boiling point is strongly dependent upon intermolecular bond strength.
6.
D is correct. Methane is n onpolar, so its only intermolecular bonding is through van der Waals forces.
7.
B is correct. All intermolecular bonding is v ia electrostatic forces. The dipoles in van der Waals forces are temporal whereas d ipole-dipole interactions m ay be due to permanent dipoles.
Passage" 8.
C is correct. 'As' is just to the left of 'Se' on the periodic table. Therefore, its radius should be slightly larger than Se.
9.
D is correct. Elements in the same family tend to be chemically similar. Hydrogen is an exception.
10_ C is correct. Atomic radius is a periodic trend increasing down and to the left. 11.
D is correct. Only D is a true statement. A is knowledge that would not be required by the MCAT.
12.
C is correct. Electron affinity is a periodic trend increasing (becoming more exothermic) to the right and up.
13.
C is correct. The answer we are looking for must explain shielding. With each new period, a new shell is added which shields the new electrons from the greater nuclear charge.
14.
A is correct. If you substitute H for X in the equation for d in the passage, you can only arrive at zero.
15.
A is correct. C and 0 are close together in electronegativity and will form a covalent bond.
Passage III 16.
B is correct. Only water is caught in chamber l. The change in mass of chamber 1, 0.9 grams, is all water. 0.9 grams of w ater divided by 18 glmol gives 0.05 mole of water. All the h ydrogen came fro m the sample, and all the oxygen came from the excess oxygen. For every m ole of wa ter, there are 2 m oles of hydrogens, so there is 0.05 x 2 = 0.1 mole of hydrogen in the sample. Doing the same with the carbon dioxide caught in chamber 2 we h ave: 4.4/44.2 =0.1 of CO" or 0.1 mole of carbon from the sample. This is a 1:1 ratio. The empirical formula is CH.
17.
B is corree!. Th e molarity of 0, is equal to the molarity of the welding gas or any other ideal gas a t the same temperature and pressure. Density divided by molecular weight is m ola rity. Therefore, we can set the ratios of d e nsity to molecular weight for oxygen and the welding gas equal to each other. We get: 1.3/ 32 = 1.1/ M .W.
18.
Dis corree!. CH,O has a molecular weight of 30 g/mo\. Thus, we must multiply this by 4 to get 120. So, for the molecular formula we need four times as many atoms of each element fro m the empirical formula.
19.
B is correct. 0, is the limiting reagent. Only 0.5 mole of propane can react, producing 2 moles of water.
20.
A is correct. The p assage says tha t CaCl, absorbs water. Thus if chamber 2 were in front of chamber 1, it would weigh more because it would absorb both water and carbon dioxide. The amount of carbon dioxid e is calculated from the weight of chamber 2, so the calculated value would be too high.
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170
MCAT
21.
INO RGANIC CHEM ISTRY
0 is correct. AU of the welding gas must be reacted because the mass of the original sample is divided by the moles of carbon and hydrogen to find the molecular weight. If all the gas were not reacted, the calculated molecular w eight would be too large. Adding excess oxygen ensures that all of the welding gas reacts.
Stand Alones 22.
B is correct. Chlorine takes on an additional electron to become an ion.
23.
0 is correct. This is the Heisenberg uncertainty principle.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 2 Passage I 24.
B is correct. The definition of Z = PV/(nRT) is always 1 for an ideal gas.
25.
A is correct. If a and b are both zero, the van der Waals equation becomes PV = nRT, the ideal gas law.
26.
C is correct. You can figure this out from the passage, but it's a lot easier to faU back on your previous knowledge: gases behave most ideally at high temperature and low pressures.
27.
C is correct. Volume is inversely proportional to pressure. A: K.E. assumptions underlying the derivation of the ideal gas law.
28.
B is correct. Condensation is due to intermolecular attractions, which are neglected for ideal gases. For 0, start with K. E. = 3/2 kT. Then 1/2 mv' = 3/2 kT, so v is proportional to the square root of T.
29.
C is correct. Equations involving products or ratios of temperature are meaningless if the zero of the temperature scale is not absolute zero. A an d B are true statements, but they don't explain why absolute temperature must be used.
= 3/ 2 kT. B:
PV
= nRT, 0 : This is one of the
Passage II 30.
C is correct. A catalyst acts to lower the activation energy of a reaction, so the catalyzed reactioin will have a lower activation energy than the uncatalyzed reaction.
31.
A is correct. A catalyst is neither produced nor consumed in a reaction and does not appear in the net reaction. The net reetion for the mechanism shown is:
0 3 +0 ->202 The catalyst, Cl, does not appear. ClO aslo does not appear, but it is produced and consume in the recaction making it an intermediate. 32.
A is correct. See the Arrhenius equation:
k = zpe - fA/lIT You should also memorize the fuct that temperature always increases the rate of a reaction. Even in the case of biologically catalyzed reactions, heat increases the reaction rate until the enzyme is denatured. Once the enzyme is denatured, although the reaction rate slows, the reaction takes a new pathway, and is no longer the same reaction.
33.
A is correct. If a catalyst only affected the rate in one direction, the equilibrium would be affected: A catalyst doesn't change the equilibrium. This can also be seen from a reaction profile diagram as shown in question 37.
34.
0 is correct. The catalyst is not necessarily the only factor influencing pH.
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AN SWERS & EXPLANATIONS FO R THE 30-Ml NUTE IN- CLASS EXAMINATION S '
35.
171
D is correct. Choice I is seen from the standard form of the rate law: rate = k[A )[B). For choice n, imagine the saturation kinetics exhibited by enzyme catalysts:
Substrate Concentration
This should make it clear that the ratio of concentrations of the catalyst and the substrates affect the rate of a reaction. This ratio can be changed by changing the concentration of the catalyst. Thus the concentration of a catalyst can affect the rate of a reaction. For choice Ill, a heterogeneous catalyst is one that is not in the same phase as the reactants. Increasing the surface area of a heterogeneous catalyst is like increasing the concentration. The reaction is affected for the same reasons as in choice II. The reason that a heterogeneous catalyst is typically in the form of metal shavings as opposed to a solid metal bar is to increase surface area. Choice IV you should know from the Arrhenius equation: k = zpe - E'Mf . 36.
C is correct. The MCAT sometimes uses the phrase "van der Waals" forces as a synonym for London Dispersion Forces. A more modern meaning is as a synonym for intermolecular forces. In either case, this is a correct answer. Hydrogen bonding requires a hydrogen atom bonded to a nitrogen, fluorine, or oxygen. D is from an episode of 5tar Trek.
37.
B is correct. Only the activation energy is changed by a catalyst. The initial and final states are not affected!
Passage III In this experiment, Reaction 2 uses up 1,- as it is form ed . When ail the 5,0 ,'- is used up in Reaction 2, the 1,- reacts with
the starch to turn black. The black color signals the experimenter that all the 5,0 ,'- is used up. The experimenter now knows that half as much 1,- was used up in the same time, and can calculate the rate for Reaction 1. This depends upon Reaction 2 being the fastest reaction . 38.
A is correct. If we look at Reactions 1 and 2 as two steps of a single reaction, we know that the rate of the slow step is equal to the rate of the overall reaction. Equation 1 measures the time necessary for a specifiC nmnber of moles of 1,- to be used by Reaction 2. (Notice that the rate of change of '/,[5,0 ,' -] will be equal to the rate of change of [13-]) If Reaction 2 were not the fast step, then Equation 1 would not measure the rate of Reaction 1 accurately. 5ince Reaction 2 is the fast step, the time t required to use up 1f,[5,O,'-] is equal to the time needed to produce [1,-]. The [[,-I concentration produced divided by the time necessary to produce it is the rate of Reaction 1. Equation 1 is not derivable from the rate laws of Reactions 1 and 2.
39.
A is correct. A temperature decrease reduces rate and makes the reaction take longer.
40.
D is correct. The rate law is found by comparing the rate change from one trial to the next when the concentration of only one reaction is changed. Comparing trials 1 and 2, when the concentration of 1- is reduced by a factor of two, the rate is also reduced by a factor of two. This indicates a fjr-st order reaction w ith respect to r-. D is the only possible answer.
41.
B is correct. The exponents in the fate law indicate the order of the reaction w ith respect to each concentration.
42.
D is correct. The starch is used to measure the rate of Reaction 1, and does not affect the rate. Although C is true, it does not answer the question as w ell as D.
43.
C is correct. Equation 1 gives the rate of Reaction 1. 5,0 / - is not part of Reaction 1 and its concentration does not change the rate. If rate doesn't change, then, according to Equation 1, t must increase with 5,0,'-.
44.
B is correct. A catalyst increases the rate of a reaction.
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172 . MCAT INORGANIC CHEMISTRY
Stand Alones 45.
A is correct. In a reaction at equilibrium, the rate of change in the concentrations of both products and reactants is zero. This does not mean that the concentrations of reactants and products are equal, nor that the rate constants are equal.
46.
C is correct. Some of both gases will effuse from side 1 to side 2. This means that the partial pressures of both gases will decrease. (Remember, partial pressure is the pressure of the gas as if it were alone in the container. Thus if we reduce the number of moles of a gas at constant volume and temperature, we reduce its partial pressure.) Since hydrogen will diffuse more rapidly than oxygen, the mole fraction of oxygen will increase.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 3 Passage I 47.
A is correct. Nickel is an element and is a solid in its natural state at 298 K. Thus, the enthalpy of formation of solid nickel at 298 K is zero.
48.
C is correct. Chemist A chooses the direction of the reaction based upon chemical stability, and then says that the direction will change at higher temperatures. This is tantamount to saying that the stability will switch at higher temperatures, which, by the way, is also correct. D is, of course, a false statement. The entropy shown for Reaction 1 is the entropy of the reaction. In other words, it is the entropy of the system, and not the entropy of the universe. Chemist A's statement is correct, and does not contradict the second law of thermodynamics. A and B are contradicted because Chemist A says the direction is temperature dependent.
49.
B is correct. Use Le Chatelier's principle. Read the first sentence of the passage carefully, and notice that to purify nickel, the reaction must move to the left. There are four gas molecules on the left side of the reaction and only one on the right. Pressure pushes the reaction to the right. The reaction is exothermic when moving to the right, so high temperature pushes the reaction to the left.
50.
C is correct. The reaction is the system, and everything outside the reaction makes up the surroundings. The entropy change given in the passage refers to the system not the universe. The second law of thermodynamics says that a reaction is spontaneous when the entropy of the universe is positive. The entropy of the system may be positive or negative. A and B are false statements. D is a false statement as well. The reaction runs until the entropy of the universe is maximized.
51.
A is correct. Use "'G =Mi - T"'S. Don't forget to convert J/K to kJ IK.
52.
D is correct. Spontaneity is dictated by Gibbs energy. 1 almos. is standard state for a gas, so "'G = "'G'. When Gibbs energy is negative, a reaction is spontaneous. If enthalpy change is negative and entropy change is positive, then Gibbs energy change must be negative. You can use "'GO = Mio - T"'SO. Check this as follows: If the partial pressures are 1, then the reaction quotient Q is 1, and the log of the reaction quotient is zero. From the equation "'G = "'Go + RTlnQ we see that "'G = "'GO. The reaction is spontaneous.
Passage II Note: A heat engine obeys the first law of thermodynamics. It must expel the same amount of energy as it takes in.
I
-
- - - -
Heat
'\
{E' t+:~:' Energy in equals energy out. Copyright © 2007 Examkrackers, Inc.
A NSWERS
&
EXPLANATI ONS FOR THE 30-MlNUTE IN-CLASS EXAMINATIONS . 173
53.
C is correct. If Qh' = Q h and Q,' = Q, then W' = W. Thus the efficiencies of the engines must be equal. Since only a Carno t engine can be as efficient as another Carnot engine, Engine 1 must be a Carnot engine.
54.
C is correct. If the all the work done by Engine 1 is done on Engine 2, the net work is zero. Since Engine 1 is not a Carnot engine, the entire process is not reversible. The result of any nonreversible process where no work is done must be that heat energy is transferred from the hot reservoir to the cold reservoir. Engine 1 has a lower e than Engine 2 and thus requires more heat energy to create as much work. According to conservation of energy, this extra heat energy input must be matched by extra heat energy output.
55.
D is correct. Engine 2 is a Carnot engine, and, as the passage states, it has the highest possible efficiency of any engine working between the existing heat reservoirs. Thus only a change in the heat reservoirs will increase its efficiency. The answer must therefore be C or D. For greatest efficiency we want to remove the most heat energy pOSSible from the hot reservoir and expel the least amount possible to the cold reservoir thus getting the most work with the least amount of wasted energy. Removing heat energy from the hot reservoir decreases its entropy, while adding heat energy to the cold reservoir increases its entropy. As the temperature of the hot reservoir increases, removing heat energy has less effect on the change in entropy, so more heat energy can be removed . The reverse is true for the cold reservoir. Since, in a Carnot engine, the change in entropy must be zero, the extra heat energy removed from the hot reservoir must be added to work. TIle engine becomes more efficient. Thus maximizing the temperature difference increases efficiency. This can be derived from the equation in the passage (the long method) as follows. Considering magnitudes only we have:
substituting T, for Q, we have:
Th
Qh
T e = l -.....L.
Th
56.
D is correct. From the derivation for efficiency in the previous explanation, if T, is zero, e = 1. However, it is probably easier to eliminate the other answer choices insead . Choice A is wrong because by the equation for efficiency given in the passage, e = WI Qh' a large W by itself won't give e = 1. Choice B is wrong because as per the passage all Carnot engines are reversible, but all Carnot engines are not 100% efficient. Choice C is w rong because from the entropy formula, if T, and Th were at the same temperature, Q, and Q, would also have to be at the same temperature, which gives W = Q, - Q, = 0, and an efficiency of zero. 100% efficiency is an imposSibility.
57.
C is correct. The exhaustis wasted energy. Qh = W + Q,. P = QJt = Wit + Q,l t = (nIgh)1 20 s + 1000 Jls + 1000 = 1200 J.
58.
A is correct. PV = nRT = (m/M.W.)RT. P = 1000 X 8.314 X 400/4 = 8.314 X 10' Pa. By the way, if the gas did not behave ideally, the real pressure would be lower. There is no answer lower than A, and the gas does behave very nearly ideally because it is at high temperature.
= 4000 120
Passage III 59.
B is correct. The transition state corresponds to the top of the energy curve.
60.
C is correct. The energy of activation is given by the vertical displacement from the reactants to the top of the energy curve.
61.
D is correct. The smallest energy of activation is the most kinetically favored.
62.
A is correct. The largest drop in energy is the most thermodynamically favored.
63.
A is correct. A catalyst lowers the energy of activation but does not change the energy difference between the reactants and products.
64.
B is correct. The change in energy is energy of products minus reactants.
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174 . MCAT INORGAN IC CHEMISTRY
65.
A is correct. The kinetically favored product is the one with a lower energy of activation. The difference in fheir equilibrium is due to conflicting thermodynamics and kinetics. At a low temperature (T,), fhe thermodynamically favored product does not have enough energy to reach the activated complex, so no reaction occurs. The kinetically favored reaction does reach fhe activated state and a reaction can occur. At the high temperature (T,) bofh reactions occur but fhe reverse of the fhermodynamically favored occurs only with a relatively lower probability. Thus the thermodynamically favored reaction predominates. This is not always true but is a concept of which you must be aware. T,~ '
I
T,~
(
,~
/V' / /
E
thermnlly
_ __
less
~,,",.lIy
kine' ally more vored
\
~
favo~~ _______~ __
therma lly more tavored
-- -
\ \,
" :::..-
Stand Atones 66.
B is correct. Transfer by contact is conduction.
67.
C is correct. The second law of thermodynamics says that entropy of the universe increases for any process. By straightening up his room, the man increased fhe order in his room, and thus decreased its entropy. In order for the entropy of fhe universe to have increased, fhere must be a larger increase in entropy of fhe surroundings. Only C provides an explanation for this.
68.
C is correct. The entropy of the system is equal to change in entropy of the two reservoirs. L;.S = Q/T for each reservoir. The change in entropy of fhe first reservoir is negative because heat energy is leaving fhe system (-1000/200 = - 5), and fhe change in entropy of fhe second reservoir is positive because heat energy is entering fhe system (1000/ 100 = 10). The sum of fhe two entropy changes is +5. You should have at least narrowed down the possibilities to C and D because the change in entropy for any isolated system must be positive for any irreversible process.
69.
D is correct. The temperature of Cas A remains constant because the question says so. Temperature is kinetic energy (due to random motion) per mole. Cas B does no work and doesn't exchange heat so its energy doesn' t change; it has fhe same kinetic energy (due to random motion) per mole as when it began. Thus, its temperature doesn't change eifher. Enfhalpy is PV + U. P and V are the same for bofh gases because they are at fhe same temp, volume and therefore pressure (PV = nRT). U doesn't change for Cas A because any energy removed is replaced to keep the temperature the same. U doesn't change for Cas B because no energy is exchanged wifh fhe surroundings for Cas B.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 4 Passage I 70.
C is correct. The first experiment shows fhat no nitrite was in fhe solution. Had there been nitri te, nitrogen bubbles would have formed as per fhe reaction, and then barium sulfate would have preCipitated upon fhe addition of barium chloride. The second experiment demonstrates fhat nitrates exist. 111e water in the moistened litmus paper reacled with ammonia gas to make OH- ions turning fhe paper blue. Ammonia gas resulted from the reaction of nitrate wifh Devarda's alloy.
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ANSWERS
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EXPLANATIONS FOR THE 30-MINUTE IN-CLASS EXAMINATIONS .
175
71.
B is correct. Any sulfate ions that exist in solutin before sulfamic acid is added must be removed. They are removed by the addition of BaCl,. Choice A is incorrect because BaCl2 does not acidify the solution. Choice C is incorrect because BaC~ doesn't react with sulfamic acid, it reacts with sulfate. Choice D is incorrect because nitrite doesn't form a precipitate with BaCl" sulfate does.
72.
B is correct. The precipitate would result when the sulfate ion from the reaction reacts with the barium ion to form barium sulfate. The bubbles would be created by the formation of nitrogen gas.
73.
A is correct. Choice A produces gas bubbles which could be confused with nitrogen gas bubbles. The BaCl2 actually prevents this from happening by forming a precipitate with carbonate before the sulfamic acid-nitrite reaction. Choice B is incorrect because the solution in Experiment 1 is acidic not basic, and because the precipitate would come out before the sulfamic acid-nitrite reaction. Choice C is incorrect because there is no ammonium in solution. Choice D is not carbonate ion, and although carbonic acid is formed by carbonate ion in aqueous solution, you should know that it breaks down to water and carbon dioxide, not carbon monoxide, hydrogen, and oxygen. A reaction you should know: ~C03 -7 CO,(g) + H,o
74.
C is correct. Change in temperature can change the rate constant and the equilibrium constant but it can only increase the rate of the reaction.
75.
D is correct. The litmus paper is turned blue when the basic ammonia gas from the reaction in the experiment reacts with the water in the paper.
Passage II 76.
A is correct. This is definitional: products over reactants excluding pure liquids and solids.
77.
C is correct. K" is a constant; solubility of Ca(103)2 is not. By Reaction 2, as acidity increases, 103- ions are used up, pulling Reaction 1 to the right. Or just look at Table 1. If you want to see why you can't just use Le Chatelier's principle on Reaction 2, simply add the two equations together to get: Ca(103)2 + H+
-7
Ca2+ + 103- + HI03
Now when you add H+ to this equation, it moves to the right, dissolving Ca(103),·
= mx + b form of the equation. This is the equation of a line. The b in
78.
B is correct. You should recognize the y this case is not zero, but is (K,p)'
79.
B is correct. You must realize that the new solution is no longer saturated. These means that Reaction 1 is not in equilibrium. No precipitate exists. New Ca2+ ions do not immediately create precipitate because the solution is not saturated. There is no leftward shift because there is no equilibrium. Thus A, C and D are wrong. lodic acid is in eqUilibrium however. Increasing H+ ions shifts Reaction 2 to the right, creating more HI03. This is, of course, why Solution 1 is no longer saturated after adding the acid.
80.
B is correct. The easiest way to find the K,p is to plug the value of 5 for Solution 1 into the equation. Notice that the [H+] value in Solution 1 is for neutral water. It is so small that the second term in the solubility equation becomes negligible. The equation becomes 2S 3f2 = K 'P 1/2 .
Squaring both sides gives: 4S 3 = K"
=> 4 X (5.4 X 10-3)3 = K", The only answer that is even close is choice B at 10-7 •
81.
D is correct. The net equation is Ca(I03),(s) + H +(aq) -7 Ca 2+(aq) + I0 3-(aq) + Hl0 3(aq). CaOH2 in aqueous solution will produce both OH- and Ca'+. The OH- will reduce the H+ in solution. Since the solution is saturated (in equilibrium), Le Chatelier's principle predicts that both of these changes will shift the reaction to the left producing Ca(103),(s).
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176 .
82.
MCAT INORGAN IC CHEM ISTRY
C is correct. Reaction 2 will shift left via Le Chatelier 's principle. The resulting increase in 10,- will shift Reaction 1 to the left due to the common ion effect, creating precipitate in the already saturated solution. The H+ ion concentration in the formula for solubility is from hydrogen ions in solution before iodic acid is added. If a different acid were added (like HCl), the H+ ion concentration would move Reaction 2 to the right and thus Reaction 1 to the right.
Passage III A ligand is an ion or neutral molecule that can donate a pair of electrons to form a coorrunate covalent bond with a metal ion. EOTA is a chelating agent (a ligand that makes more than one bond to a single metal ion). It wraps around its metal ion like a claw. Chele (XT)"T) means claw in Greek. 83.
D is correct. The passage states that EOTA reacts with other metal ions. If more EOTA is used up, the scientist will assume that it is being used up by calcium and magnesium ions. This will result in an overestimation of these ions.
84.
C is correct. EDTA is donating a pair of electrons in a coordinate covalent bond, so it is a Lewis Base.
85.
A is correct. TIlis is LeChatelier's Principle. The chlorine ion will remove some of the Ca2+ pulling the reaction to the right.
86.
C is correct. The passage states tha t there is a one-to-one stOichiometry between EOTA and its metal ion. (50 mL)(O.OOl moUL) = (25 mL)x. x = 0.002 maU L
87.
D is correct. D is the best explanation. Under high pH conditions, protons are stripped from the carboxylic acids allOWing the ligand to bond to the cation. Indicators change color over a range so C is wrong. Calcium carbonate dissolves in an acid solution so B is wrong. A is irrelevant.
88.
A is correct. You should know that Na+ is very soluble, and H+ does not form mineral deposits.
89.
A is correct. The association constant from Table 1 is 5 x 10". The association reaction is: Mg'+ + EDTA --; EOTA - Mg. The K.=
= 5 X 10' = [EDTA - Mg] / [EDTAJ[Mg'+]
Since half the magnesium is bound, [EOTA-Mg] is 9 ppm, which is 5 x l Q4. The remaining half of [Mg'+] is 9 ppm, which is also 5 x lQ4. Plugging these into the equilibrium expression leaves the remaining concentration [EOTA] = 1/(5 X 108) or 2 X 10-9
Stand Alones 90.
C is correct. Water boils at 100°C and atmospheric pressure, 760 torr. Boiling point is where vapor pressure meets atmospheric pressure. Thus water vapor pressure must be below 760 at 95°C. But it must also rise w ith increasing temperature, so it must be above 23.8 torr.
91.
B is correct. No calculations are required since the vapor pressure would be somewhere between the vapor pressures of the pure liquids. The solution follows Raoult's law.
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AN SWER S
92.
&
EXPLANATIONS FOR THE 30-M INUTE IN- CLASS EXAMINATIONS • 177
B is correct. Since the reaction was exothermic, the vapor pressure deviated negatively from Raoulfs law. Depending upon the ratios of the liquids in solution, the vapor pressure could be lower than either or just lower than B. (A has a higher boiling point thus a lower vapor pressure.) The boiling point must have gone up from B because the vapor pressure went down from B.
Pressure of pure B""";" '. ,-.. __
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'.'~_~;
Solution pressure is
of pure A
T~~;~~~~ ;~:t 1sU~~:~ s. . . . :::;::~:;/:·. ·' ..... greater than pure B. a 1
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1
0
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 5 Passage I
93.
C is correct. The salt lowers the freezing point of water. This is necessary in order to insure that the water can bring the unknown solution to its freezing point.
94.
C is correct. Even if the others completely dissociate, NaCl still releases more p ar ticles per gram than the others. (2 particles/58 grams) (A = 3/172, B = 2/120, and D = 3/115)
95.
C is correct. The copper stirrer acts to evenly distribute the heat throughout the solution by convection.
96.
B is correct. Since the freezing point depression was lower for Student 1, there must have b een more particles for the same amount of mass. Thus Student 1 had an unknown with lower molecular weight.
97.
0 is correct. A greater concentration of partieles lowers freezing point more. The freezing point data was collected and recorded for the cyclohexane solution, not the crushed ice solution.
98.
B is correct. Tune is not a factor in the calculation of the molecuar weight of the unknown solid. The molecular weight of the unknown solid can be calculated as follows:
m = (grams",lu.,/M.W.)/(volwne"'lvon' x densitY",lven. x kg /gram) plugging into !!.T
=K;n and rearranging we have:
M.W. = (Kf ) (grams",lu,,) / ( !!.T) (volume",I",u. x densitY",lven' X kg/gram») Kf = 20.2, (grams oolul,) = 0.5, !!.T = (6.6 - -3.5) = 10.1, volume",I",n. = 10.0, densitY"'I""" = 0.78, kg / gram = 1/1000
Thus M.W. = (Kf ) (grams.,lu ..) / (!!.T) (volume"'I"~' 99. 100.
X
density.,I"",' X kg/gram))
D is corred. 0 is the only nonvolatile, nonpolar solute that is soluble in cyelohexane. B is correct. The only explanation is B. As long as some solution remains liquid, the energy removed by the ice bath creates bonds forming a solid. As soon as the entire solution is frozen, the energy removed from solution lowers the temperature.
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178 . MCAT INORGANIC CHEMISTRY
Passage II 101.
A is correct. The acid and base are totally dissociated to begin with. This reaction takes high energy molecules and makes a low energy molecule, releasing heat.
102.
B is correct. Heat is released.
103.
A is correct. 40 mL negative.
104.
B is correct. The ammonium nitrate would require energy to dissociate before releasing energy to form water.
105.
D is correct. Heat per mole is an intensive property.
106.
D is correct. The temperature of the calorimeter is higher than the surroundings throughout the experiment. Heat always moves from hot to cold.
X
1 caire mL x -3.3'e = -132 cal. Since heat is released, we already know the answer is
Passage III 107.
A is correct. The negative slope on the phase diagram demonstrates that water expands when freezing.
108.
C is correct. The line between A and E represents equilibrium of gas and solid.
109.
A is correct. Point e in Figure 1 is the critical point, which is the temperature and pressure above which the gas and liquid phases cannot be distinguished.
110.
D is correct. Point e in Figure 1 is the critical point, which is the temperature and pressure above which the gas and liquid phases cannot be distinguished.
111.
B is correct. The negative slope between the solid and liquid phases of water in Figure 2 represents melting point at different temperatures and pressures. As pressure increases, the temperature decreases moving along the line.
112.
A is correct. The normal boiling point is the boiling point at local atmospheric conditions (1 atm). To have a normal boiling point, the triple point must be at a pressure below 1 atm. In that case, the temperature of the triple point will be below the temperature of the normal boiling point.
1 atm
r .
Normal boiling point
Triple poirft ;
90.2K 113.
C is correct. See the graph.
Stand Alones 114.
A is correct. During a phase change, temperature, and thus molecular kinetic energy, is constant. Breaking bonds always absorbs energy. Ice cools things when it melts.
115.
A is correct. The potential energy of the water at the top of the falls becomes kinetic energy as it drops, and then thermal energy at the bottom of the falls. Thus mgh = Q = mc!iT, or !iT = gh/c.
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ANSW ERS
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EXPLANATIONS FOR THE 30-MINUTE IN-CLASS EXAMINATIONS'
179
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 6 Passage I 116.
A is correct. This is the common ion effect in Reaction 1.
117.
A is correct. For all trials, at the equivalence point there are Ca'" ions, Na+ and Cl- ions. (Trial 1 has no Na" ions, but these don't affect pH anyway.) The Ca'" started as a saturated solution. As OH- ions are removed from solution by the acid, there is no precipitation of Ca(OH)2' When all the OH- ions are neutralized by the acid, the pH is 7 for all trials. The equivalence point is 7 for all trials. For trials with NaOH, the pH begins higher. More HCl is required to neutralize this extra base. The curves are exactly the same, but the trial one curve starts at a lower pH and requires less HCl.
118.
B is correct. This is just the normal
119.
D is correct. pOH = -log[OH-] In this case, the significant OH- ion contribution comes completely from NaOH which dissociates completely. 1 > pOH > 2 The pH is 14 - pOH = between 12 and 13.
120.
D is correct. The Ca(OH)2 solution begins as basic, and when the Ca(OH), is totally dissolved, it should be neutral. See answer to question #117.
121.
A is correct. This is the titration of a strong base with a strong acid. See question #117.
Kep'
Passage II 122.
C is correct. Since the Kw is higher at 40'C, the hydrogen ion concentration will also be higher. Thus the pH will decrease. Note that the hydrogen ion concentration would have to be 10 times higher in order for the pH to be higher.
123.
C is correct. Set ~ equal to 1 mole of hydrogen ions times the OH concentration.
124.
B is correct. As T increases, the hydrogen ion concentration increases.
125.
C is correct. A proton is lost to form the conjugate base of a Bronsted acid.
126.
C is correct. The leveling effect in water occurs because water readily accepts all protons from both acids. The equilibrium in water is so far to the right for both reactions that no comparison can be made. Although acetic acid accepts protons from both HCl and HClO" it does not do so as readily as water (it is a weaker proton acceptor or Bronsted-Lowry base). Thus, an equilibrium is established for both reactions, and the equilibriums can be compared.
127.
A is correct. Definitional.
Passage III 128.
C is correct. The pH must be where both indicators can have the proper color.
129.
C is correct. The lower pK, of sulfuric acid demonstrates that it is a stronger acid than sulfurous acid. The second proton is not the major contributor to the acid strength, so D is wrong.
130.
D is correct. The Kb is found by adding the base to water. If the pK, is known, the pKb of the conjugate base can be found by subtracting the pK, from 14. HS03- is the conjugate base of H,S03' The pK, of H,S03 is given as 1.81. 14 - 1.81 = 12.19.
131.
D is correct. The lowest pK, is the strongest acid. You should know that sulfuric acid is stronger than the other acids.
132.
A is correct. Minus eight for the oxygens, plus two for the hydrogens leaves minus six which must be counter balanced. Minus six for the oxygens, plus two for the hydrogens leaves minus four which must be counter balanced.
133.
C is correct. We could never raise the pH by adding an acid. Water is the main contributor of H". To find the pH, we add the 5 x 10-8 ions contributed by H 2S04 to the 1 x 10-7 ions contributed by water. This leaves 1.5 x 10-7 H".
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180
134.
MCAT INORGANIC C HEMISTRY
D is correct. This is the titration of a diprotic acid with a strong base.
Stand Alones 135.
A is correct. Perchloric acid is the strongest acid, thus it has the weakest conjugate base. In oxy acids, the more oxygens, the greater the acid strength.
136.
C is correct. This is the definition of the half equivalence point.
137.
A is correct. Use the Henderson-Hasselbalch equation. pH = pK, + 10g(A-/ HA) => pH = -log(1.8 x 10-5) + log(1/10) => pH = 4.7 - 1 = 3.7
138.
B is correct. The conjugate acid has the K, that equals Kj Kb ·
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 7 Passage I 139.
C is correct. Cr20 ," is reduced to Cr'·. Although this doesn't look like reduction from the charges, Cr in the dichromate has an original oxidation state of +6.
140.
C is correct. Oxygen is - 2. There are 7 oxygens which make - 14. The 2- charge on the ion takes care of 2 of the 14 negatives. The 2 chromiums must take care of the other 12. That's +6 for each chromium.
141.
C is correct. This is simply reading the graph.
142.
B is correct. For each mole of dichromate that is reduced, 6 moles of Fe are oxidized. The top portion of B gives the number of moles of dichromate reduced times six, which is the number of moles of Fe oxidized. (The equivalence point is where all the iron has been oxidized.) We divide this by the original volume of Fe solution and get the molarity.
143.
A is correct. The equivalence point is defined as the point when the Fe2• is completely oxidized. Changing the formal potential won't change that definition. The Endpoint is when the indicator changes color. This should be at the equivalence point, but it doesn't have to be. The passage says that the formal potential is lowered so that the equivalence point and endpoint will coincide. This indicates that the indicator will change color at a low potential. If the formal potential isn't lowered, the indicator will change color early (still at the low potential) before the equivalence point is reached. The passage states that the indicator changes from colorless to violet; choice A. Dichromate ion will still oxidize DAS, but it will do so before oxidizing Fe2. , so choice C is incorrect. Dichromate ion is an oxidizing agent and will not reduce anything, so choice D is incorrect.
144.
A is correct. Each iron goes from a +2 to a +3 oxidation state by losing one electron. Each chromium goes from a +6 to a +3 oxidation state by gaining 3 electrons.
Passage II 145.
A is correct. The half reactions must be rearranged in such a fashion so as the total voltage is positive (meaning the battery is discharging). This requires reversing the top half reaction. When reversed, this reaction becomes oxidation, which tak~s pl"ce at the anode.
146.
D is correct. When we recharge the battery, the reactions are both reversed from the positions in question one.
147.
C is correct. Ni is being reduced, so Ni02 is the oxidizing agent. The compound w ith many oxygens is often the oxidizing agent.
148.
A is correct. The cell has a negative potential and is forced to run in the nonspontaneous direction .
149.
C is correct. The two half reaction potentials must be added after they have been rearranged to represent the galvaniC cell. This means that the first half reaction is reversed . If this potential is applied, the cell can be recharged back to this potential w hich is the standard potentiaL
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EXPLANATIONS FOR THE 30- M INUTE IN-CLASS EXAMINATIONS • 181
150.
B is correct. The passage tells us that sod ium is at the anode so it is oxidized. Sodium is n ot norm ally oxidized to a +2.
151.
D is correct. Crnrent moves opposite to electrons. Since electrons flow from Half Reaction 1 to 2, current flows from Half Reaction 2 to 1.
Passage III 152.
D is correct. The reaction quotien t is in the same form as the equilibrium constant. Pure solids and liquids Should not be used in the law of mass action to solve for the equilibrium cons tant.
153.
A is correct. Plug the numbers into the Nemst equation. (102)' is 10' . The log of 10' is 4. Thus the potential is negative and the voltage is twice 0.0592 V.
154.
B is correct. See the last equation in the passage.
155.
B is correct. See the last equation in the passage. A 1 M solution of HCI h as a p H of O.
156.
A is correct. An increase of H + by a facto r of 10 is a decrease in pH of 1. E = 0.059 (pH.." crease in pHrer.~ of 1 amounts to an increase in E of 0.059 V or 59 m V.
157.
A is correct. The current w ill try to even the ch arges in the solutions. Since we h ave m ore positive charge on the concentrated side, the cturent moves to the less concentrated side.
158.
B is correct. Again, current flows toward the less positive side, which is the basic side, which h as less H + ions.
pHw.,~".)
Each d e-
Stand Alones 159.
D is correct. You should know this one reduction potential which is: 2H++ 2e- -> H, to the oxidation of solid zinc, the poten tial is positive, which means spontaneous.
160.
C is correct. A t equilibrium, there can be no potential; neither direction of the reaction is favored.
161.
B is correct. The products and reactants are at stan dard state, and therefore their potential defines the standard p otential P . A is wrong because they are not at equilibrium when Q = 1. C is wron g because th e potential is p ositive. D is wrong because Q is at 1 and Q will m ove toward K. The reaction is spon taneous from h ere so prod ucts will increase, and Q w ill increase. Therefore, K must be grea ter than 1.
181
E = O. When this is added
ANSWERS & EXPLANATIONS FOR
QUESTIONS IN THE LECTURES
183
184
MCAT INORGANIC CHEMISTRY
ANSWERS FOR THE LECTURE QUESTIONS Lecture 1
Lecture 2
Lecture 3
Lecture 4
e
Lecture 7
Lecture 5
Lecture 6
97. 0
121. B
145.
e e
146.
e e
e e
e
49. 0
73.
26. B
50. A
74. 0
98.
A
122.
3. A
27. A
51.
e
75. A
99. B
123.
4. B
28. 0
52. 0
76. 0
100. 0
124. A
148.
5. A
29. B
53. 0
77.
e
101. 0
125. A
149. 0
6. B
30.
e
54. B
78. 0
102.
B
150. 0
7. B
31.
0
55.
0
79. B
103. A
127. B
151. 0
8. B
32. 0
56.
A
80. B
104. B
128. A
152. A
B
57.
B
81. 0
105. 0
129. 0
153. 0
106. B
130.
B
154.
e
131.
e
155.
B
132. 0
156. A 157. 0
1. 2.
9.
e
25.
33.
10. 0
34. 0
58. 0
82. A
11.
B
35. A
59.
83.
12.
e
36. A
e e
13. B
37.
0
38.
14.
e
126.
e
107.
60. B
84. B
108.
e e
61. B
85.
e
109.
A
133. 0
A
147. A A
62.
e
86. B
110.
B
134.
B
158.
111. 0
135.
e
159. A
A
136.
0
160. B
15. B
39. B
63.
0
87. 0
16.
A
40. B
64.
A
88.
17.
B
41. 0
65.
0
89. 0
A
90.
e
112.
113. A
137. 0
161.
e
138. B
162.
e
e e
18. B
42.
e
66.
19. B
43.
A
67. 0
91. 0
115. B
139. A
163. 0
e
44.
68. 0
92.
140. B
164. 0
45.
69. A
93.
e e
116. B
21. B
e e
165. B
70. 0
e e
A
20.
22.
e
46. 0
23.
B
47.
A
71.
24.
A
48.
A
72. 0
e
114.
e
141.
94. A
118. 0
142.
95. A
119. A
143. 0
167. B
B
120. 0
144. B
168.
96.
117.
166. A
e
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ExPLANATIONS FOR QUESTIONS IN TH E LECTURES •
185
ANSWERS & EXPLANATIONS FOR LECTURE QUESTIONS 1.
C is correct. A family or group is the name for any vertical column on the periodic table. Of the choices given, only atomic radius increases going down a column. Although electron affinity is a possible choice depending upon the definition used, atomic radius is an unambiguous choice.
2.
C is correct. The dipole moment w ill be greatest for the atoms with greatest difference in electronegativity. Based upon periodic trends, Hand F will have the greatest dipole m oment.
3.
A is correct. By definition there are 12 amu in one atom of 12c.
4.
B is correct. Metals are lustrous, but they are also malleable and good conductors of electricity and heat. Silicon is positioned along the diagonal of elements in the periodic table sometimes referred to as metalloids.
5.
A is correct. This is an isoelectronic series, which means that the number of electrons on each ion is the same. In an isoelectronic series of ions, the n uclear charge increases w ith increasing atomic number and draws the electrons inward with greater force. The ion with fewest protons produces the weakest attractive force on the electrons and thus has the largest size.
6.
B is correct. Don't do any complicated calculations. This is the type of problem that everyone will get right, but many will spend too much time trying to be exact. First assume that 100% of the sample is 12c. Now use the formula: moles = grams/molecular weight. This is very close to 4. The 1% thatis not 12C is insignificant.
7.
B is correct. We are looking for an answer that would allow for the prediction of the order of atomic number. If atomic number increases w ith atomic weight, the scientists could have made accurate predictions.
8.
B is correct. Sulfur can form fo ur bonds. In choice A, Cl has the wrong number of electrons. In choice C, Se is too large to form stable pi bonds, so it can' t double bond. In choice D, fluorine cannot m ake more than one bond. This question may require a little too much d etailed knowledge to be on the MCAT.
9.
C is correct. We start by assuming a 100 gram sample. By dividing grams by molecular weight, we obtain moles. 58.6/16 = 3.6, 2.4/ 1 = 2.4, 39/ 32 = 1.2. Now we divide thro ugh by the lowest number of moles: 3.6/ 1.2 = 3; 2.4 / 1.2 = 2; 1.2/1.2 = 1. This gives you the molar ratio of each element. Just to reduce the necessary calculations, the question tells you that it is a neutral compound. Nevertheless, MCAT questions with this much calculation occasionally come up, but they are few and far between. Maybe three on one entire exam.
10.
D is correct. Silicon is too large to form pi bonds like carbon does. In order to complete its octet, it must make four bonds. It makes one bond with each of four oxygens. Each oxygen bonds with two silicon atoms.
11.
B is correct. C has 12 g / mol and 0 has 16 g / mol. The total weight of CO, is 44 g / mol. Carbon's weight divided by the total weight is 12/ 44= 0.27. We multiply by 100 to get 27%.
12.
C is correct. When one mole of sulfur dioxide is oxidized, one mole of sulfur trioxide is produced. O ne mole of sulfur trioxide has a mass of 80 g.
13.
B is correct. Normally, 34 grams of ammonia (2 moles) could make 28 grams of nitrogen (1 m ole), but here, only 26 grams were made. In a reaction that runs to completion, this must be due to lack of CuO.
14.
D is correct. Each perchlorate ion has a 1- charge giving the copper ion a 2+ charge.
15.
B is correct. This is an tmusuallooking reaction because it is a polymerization. Here it is drawn in a more representative form showing the repeated unit of the polymer as the product:
nCH,= CH, -
ft-t1 H H}n
16.
A is correct. This reaction has the form A + B --7 C. This is the form of a combination reaction.
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186 . MCAT INORGANIC CHEMISTRY
17.
B is correct. The quickest way to see this is by realizing that atoms like to form ions with electron configurations similar to the nearest noble gas. Of course a noble gas does not have any unpaired electrons. You should recognize that Ca likes to form a 2+ ion not a 1+ ion.
18.
B is correct. This question borders on requiring too much specific knowledge for the MCAT. The knowledge that the 4s and 3d orbitals are at the same energy level for the first row transition metals is probably beyond the MCAT. Rather than memorize specific exceptions to the Aufbau principle, answer this question by eliminating that A, C, and D must be wrong. A is wrong because there is no reason to skip the s subshell entirely. C is wrong because it contains the wrong number of electrons. D is wrong because we should be in the 3d subshell, not the 4d subshell. You may be able to see that, by Hund's rule, each electron would rather take its own orbital than share an orbital at the same energy level with another electron. Thus for Chromium, electrons fill the orbitals like this:
Chromium looks like this:
[Ar]
not like this:
[Ar]
1 45
11111
3d 3d 3d 3d 3d
1 1 11 H 3d 3d 3d 3d 45
3d
Copper is the only other first row transition metal that breaks the Aufbau principle. Its electron configuration is [Ar]4s' 3dlO • 19.
B is correct. According to the Heisenberg uncertainty principle, both the position and the momentum mv of an electron cannot be known with absolute certainty at the same time. Since we know the mass of an electron, the uncertainty must lie in the velocity.
20.
C is correct. The atom must absorb energy in order for one of its electrons to move to a higher energy level orbital.
21.
B is correct. The principle quantum number (n) represents the energy level of the electron. The lowest energy shell is n ~ 1. As n increases, the shells move farther from the nucleus and energy increases.
22.
C is correct. Because sulfur is larger than oxygen, sulfur has 3d subshells available that allow electrons to form bonds and break the octet rule of the Lewis structure.
23.
B is correct. Hund's rule says that electrons added to the same subshell will occupy empty orbitals first and the unpaired electrons will have parallel spins.
24.
A is correct. Since chromium forms more than one oxidation state and aluminum forms only one, chromium requires the variability in number of bonds formed. This means choices C and D are out. Chromium has electrons in the orbitals of the 2p subshell, but these are core electrons and not used for making bonds. Chromium has 6 valence electrons, 5 of which are in the orbitals of the 3d subshell.
Lecture 2 25.
C is correct. You should recognize that 1 mole of gas occupies 22.4 liters at STP, so there is 0.5 moles of gas in the sample. 13 g/O.5 mol ~ 26 g/mol.
26.
B is correct. Since density (p) is mass (m) divided by volume (V), and mass is moles (n) times molecular weight (MW), we have (nMW)/V ~ p. After some algebra we have: MW ~ (pV)/n. From the ideal gas law we know that V/n ~ RT/p. Substituting we have answer B.
27.
A is correct. The number of moles of gas is extra information. If the container began at 11 atm then each gas is contributing a pressure in accordance with its stoichiometric coefficient. When the reaction runs to completion, the only gas in the container is nitrogen dioxide, so the partial pressure of nitrogen dioxide is the total pressure. The volume of the container remains constant, so the pressure is in accordance with the stoichiometric coefficient of nitrogen dioxide.
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EXPlANATIONS FOR QUESTIONS IN THE LECTURES .
187
28.
0 is correct. An ideal gas has a PV / RT equal to one. Real volume is greater than predicted by the ideal gas law, and real pressure is less than predicted by the ideal gas law. Volume d eviations are due to the volume of the molecules, and pressure deviations are due to the intermolecular forces. Thus, a negative deviation in this ratio would indicate that the intermolecular forces are having a greater affect on the nonideal behavior than the volume of the molecules. (see the graph on page 27)
29.
B is correct. The force does work on the gas, which means that the inte rnal energy of the gas is increased. Since the internal energy of the gas is increased, and the number of moles remains the same, the temperature, which is average kinetic energy p er mole, also increases.
30.
C is correct. From Graham's law we know that the effusion rate for hydrogen is four times that of oxygen.
H2effusion rate 02{!lfusionr"~
Since we don't know how many moles of gas were initially in container A, nor how many moles effused out, we don't know the ratio of hydrogen to oxygen. However, since we know that four times as many moles of hydrogen effused from container A into B, we know that container B contains four times as many moles of hydrogen. We can neglect any effusion in the reverse direction since the question says a "very short time". 31.
0 is correct. The diffusion rate for NH3 is 1.5 times that of HCl. If HCl diffuses 4 cm, NH3 will diffuse 6 cm. 4 cm + 6 em = 10 cm. diffusion rate of NH, diffusion rate of HCl
=
JM,;c, JMN",
=
F6.5
= 1.5
17
32.
0 is correct. At STP, equal volumes of any gas behaving ideally contain the same number or moles.
33.
B is correct. Changing the concentration of the reactants will not change the rate constant. Increasing the concentration of a catalyst will only increase the rate of the reaction if the supply of catalyst is so small that the reactants are w aiting for a catalyst. Most of the time on the MCAT, YOll can assume that the supply of catalyst is large enough so that a change in concentration will not change the reaction rate. (See Biology Lecture 2 for the graph relating reaction rate to enzyme catalysts.) Increasing the amount of catalyst never increases the rate constant. Increasing the temperature will always increase the rate constant, and the rate of the reaction. If the reaction is catalyzed by an enzyme, the enzyme may denature, slowing the reaction; however, the reaction withau t the enzyme is considered a different reaction.
34.
0 is correct. Catalysts do not directly affect the equilibrium of a reaction. Catalysts d o increase the rate of the reverse reaction as well as the forward reaction.
35.
A is correct. When the concentration of B is doubled, the rate doesn' t change. When the concentration of A is doubled, the rate doubles. The reaction is first order overall, and first order with respect to A. By choosing a trial and plugging the values into the rate law, we find that the rate constant has a value of 0.1.
36.
A is correct. The slow step determines the rate of a reaction.
37.
C is correct. Exotherrnicity concerns the thermodynamics of the reaction, and not the rate. You can ignore it. The energy of activation is the energy required for a collision of properly oriented molecules to produce a reaction. This d oes not change with temperature.
38.
C is correct. A first order reaction has a constant half life. In the first 15 minutes, 16 out of 33 white dots (com· pound X) turned black, so 15 minutes represents approximately one half life. In the next 15 minutes, the second half life, half of the remaining 17 white dots should turn black. This represents choice C where there are 9 white dots left. Once you identify that 15 minutes is the half life, you should be able to eliminate answer A because there is no change and answer choice Band D because there are very few dots left. Even if you didn't know that a first order reaction has a constant half life, you should know that the reaction will be proportional to the concentration of white dots. In choice Band 0 , the rate of the reaction hasn't changed in the second 15 minutes even though the concentration of white d ots has been reduced after the first 15 minutes, so this can 't be right.
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MCAT INORGANIC CHEMISTRY
39.
B is correct. For a first order reaction the reaction rate is directly proportional to the concentration of reactant, according to the equation, rate ~ k[AJ. So if the concentration of cis-2-butene is doubled, the reaction rate will also double. The rate constant is not affected by changes in the concentrations of reactants.
40.
B is correct. The concentration of reactants decreases exponentially in a first order reaction. Another way of saying that is that the graph of In(reactants) will be linear.
41.
0 is correct. Equilibrium will probably shift with temperature. The direction is dictated by thermodynamics. We need more information.
42.
C is correct. The activation energy is dictated by the reaction itself and doesn't change during the reaction. We will see later that the Gibbs free energy is at a minimum when a reaction is at equilibrium.
43.
A is correct. By Le Chatelier's principle the equilibrium would shift to the right causing an increase in the forward reaction.
44.
C is correct. The equilibrium constant is products over reactants with the coefficients as exponents. However, reactants and products in pure liquid and solid phases generally have an exponent of zero, so they are not included in the equilibrium expression.
45.
C is correct. Initially there are no products, so the reverse reaction begins at zero. As the reactants are used up, the forward reaction slows down. Equilibrium is the point where the rates equalize.
46.
0 is correct. A must be false because some of both solids must be present in order for equilibrium to exist. B and C are false because, as part of a solid molecule, calcium atoms have no way of leaving their respective beakers (other than a negligible amount of vapor pressure from their respective solids). 0 is true because when CaO in Beaker II combines with CO2 gas in the container, CaC03 is formed in Beaker II. An equilibrium is achieved in both Beakers.
47.
A is correct. Pure solids are not included in the equilibrium expression.
48.
A is correct. "Kp is equal to the partial pressure of CO2 '' is the equilibrium expression for this reaction. If 1<" were less than the partial pressure of CO" the reaction would want to go to the left, but there would be no CaO to react to form CaCO,. Regarding choice 0, although solid CaO is required to achieve equilibrium, solid CaO could be formed with the decomposition of CaC03.
Lecture 3 49.
0 is correct. The second law of thermodynamics states that a heat engine cannot have 100% efficiency in converting heat to work in a cyclical process. An air conditioner is a heat engine running backwards. Thus an air conditioner must expel more heat than it takes in when it runs perpetually. A specially made air conditioner could initially cool the room, but to cool the room permanently, it must expel the heat to a heat reservoir.
50.
A is correct. 1. The temperature difference is directly proportional to the distance between two points of the same material.
51.
II. The rate of heat flow is constant throughout the blocks, or else heat would build up at the point of slowest flow. III. Since heat flow rate is constant, changing the order of the blocks won't change the rate of heat flow. C is correct. There is no type of heat transfer called transduction. Conduction through the air would take a very long time and be very inefficient. Convection would require some type of air current or breeze. Radiation is as fast as light, and is the correct explanation.
52.
0 is correct. Unless the box and the incline are at different temperatures, there can be no heat. Energy transfer due to friction is work.
53.
0 is correct. Work is not a state function, thus we must know the path in order to calculate it. Copyright © 2007 Examkrackers, Inc.
ANSWERS
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EXPLANATIONS FOR QU ESTIO NS IN THE LECTURES •
189
54.
B is correct. The rate at which heat is conducted is directly proportional to the difference in temperatures between the hot and cold reservoirs. In December, the difference is 25 -5 ~ 20 degrees. That's the largest difference on the table.
55.
D is correct. Since the container is at rest and h as constant volume, no work is done.
56.
A is conect. The efficiency of a thennodynamic process describes what percent of the input energy is converted into work. No thennodynamic process can be 100 percent efficient.
57.
B is eonect. To find the enthalpy of the reaction we use the following formula: ,Mi0reaction -- AHf oproducts
-
Mio f reactants
The table gives these enthalpies. Don't forget that enthalpy is an extensive process, so quantity matters. We must multiply the enthalpies by the number of moles formed for each molecule. The enthalpy of formation of 0 , is zero, like that of any other molecule in its elemental form at 298 K. 58.
D is correct. The definition of enthalpy is: H s U + PV
59.
A is correct. A catalyst affects the kinetics of a reaction and not the thermodynamics.
60.
B is eonect. Altering the ratio of the rates of a reaction w ill change the equilibrium. Removing thermal energy from an exothennic reaction will probably push it forward according to Le Chatelier's principle, since heat is a product. Answer C and D concern catalysts and will not change the ratio of the forward and reverse reaction.
61.
B is conect. This is Hess's law. We reverse the equation for graphite, so that graphite is a product. In doing so, we must also reverse the sign of the enthalpy. Now we add the two equations and their enthalpies. Don't forget that we must multiply by two for the two moles. Enthalpy is an extensive property.
62.
C is correct. Condensation must occur to form liquid water. Condensation is an exothermic process, so the form alion of liquid water should be more exothermic than the formation of water vapor. The standard enthalpy of formation of water vapor will not be an endothermic process, so D is wrong.
63.
D is correct. The reaction coordinate diagram below shows the energy of activation for an endothermic reactio n is greater than for an exothermic reaction.
-
exothermicendothermic
64.
A is correct. Increasing the temperature increases the energy available to both the forward and the reverse reactions, enabling both to more easily overcome the activation energy. Just so you know, because the reverse reaction is endothermic, its rate will increase more. That's what causes the increase in the concentration of reactants predicted by LeChatelier's law.
65.
D is correct. According to the equation ~G = m - T~S, to guarantee that a reaction is spontaneous, enthalpy of the system must decrease and entropy of the system must increase.
66.
A is correct. The entropy of the universe will increase in a spontaneous reaction. The entropy of a system may or may not increase.
67.
D is correct. Energy is always required to break a bond.
68.
D is correct. The process of building a bridge is an ordering process.
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MCAT INORGANIC CHEMISTRY
69.
A is correct. Since the number of moles of gas is decreasing with the forward reaction, positional entropy is decreasing. This almost always means that overall system entropy is decreasing. Since the MCAT doesn't distinguish between positional entropy and any other kind of entropy, you can always view a reaction with decreasing number of gas particles as decreasing in entropy and vice versa.
70.
D is correct. Bonds are formed when water condenses, so energy is released and m is negative. The water molecules become less random, so IlS is negative. Condensation occurs spontaneously at 25°C (room temperature), so llG is negative. Notice that you can answer this question without being given any numbers.
71.
A is correct. llG ~ !lH - TllS. For a spontaneous reaction, llG must be negative. As T is increased, the negative part of the equation increases in magnitude. If T is increased enough, eventually llG will switch from positive to negative. Changing the pressure will have no effect on a nongaseous reaction that takes place in a solution.
72.
D is correct. At the boiling point, benzene is in equilibrium between the liquid and gas phases. At equilibrium, ~G for a reaction is equal to zero. 5 is positive for the reaction shown because gases are more random than liquids.
Lecture 4 73.
C is correct. One liter of water weighs 1 kg; one liter of this solution weighs 1.006 kilograms. If we assume that the volume of water changes very little when NaCl is added, then about 0.006 kg, or 6 g, of NaCl are in each liter of solution. The molecular weight of NaCl is 58.6. 6 grams is about 0.1 moles. (By the way, even if the salt increased the volume of 1 liter of solution by 10 cubic centimeters, the molarity would still be slightly greater than 0.099 M. So this is a good approximation. Remember, for dilute solutions, the volume of the solute is negligible.)
74.
D is correct. Remember that like dissolves like. Water is polar, and will dissolve polar and ionic substances. A, B, C are ions, ionic compounds, or capable of hydrogen bonding. Carbon tetrachloride is a nonpolar molecule.
75.
A is correct. For all practical purposes, choices A and B are the same. However, since the question asks you to compare them, a one molar solution is 1 mole of NaCl in slightly less than a liter of water. This is because the NaCl requires some volume. A one molal solution is one mole in one full liter of water. (This question assumes that a liter of water has a mass of 1 kg. This is true at 1 atm. and approximately 3°e. Water at 1 atm. is at its most dense state at a temperature of slightly over 3°e.) There are 55.5 moles of water in a liter (grams/molecular weight ~ moles). 1/100 ~ 0.01 and 1/50 ~ 0.02. Thus a solution with a mole fraction of 0.01 is closer to a 0.5 molar solution than a 1 molar solution. The last answer choice is less than one mole of NaCl in one liter of water.
76.
D is correct. You should know that 1 atm is equal to 760 torr. Since the partial pressure of nitrogen is 600, the mole fraction of nitrogen is 0.79. This means that the percentages given are by particle and not by mass. D would be true if the percentages were based on mass. If you chose B, you need to go back to Lecture 3 and review standard molar volume.
77.
C is correct. No solution is formed, so either B or C must be correct. B is not true.
78.
D is correct. First calculate the number of moles of MgCI2 moles
~
grams/MW
~
19 g/95 g/mol
~
0.2 mol
0.2 moles of MgCl2 will dissociate to produce 0.4 moles of Cl- ions. [Cn 79.
~
moles/liters
~
0.4 mol/O.5 L ~ 0.8 M
B is correct. First find the number of moles of HCI. moles
~
(mol/L)(L)
~
(3 moI/L)(0.8 L)
~
2.4 moles
Now find the number of liters needed to make the solution 1 molar L ~ mol/mol/L ~ 2.4 mol/l M
~
2.4 L
Now be careful. You already have 0.8 liters of solution, so in order to get 2.4 L, you have to add 1.6 L of water.
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ANSWERS
&
EXPLANATIONS FOR QUESTIONS IN THE LECTURES •
191
80.
B is correct. A strong electrolyte is a substance that dissociates completely in water to form ions, which can fhen conduct electricity. Carbon dioxide does not dissociate to form ions so it is not an electrolyte.
81.
0 is correct. The change in entropy is positive in solution formation and Gibbs free energy is negative in a spontaneous reaction. From L'>.G = !>.H,o! - TL'>.S we see that the heat of solution may be either positive or negative in this case. The heat of hydration is fhe separation of water molecules (which requires energy) and fhe formation of bonds between fhe ions and water molecules (which releases energy). Thus, fhe value of fhe heat of hydration could be eifher positive or negative. The actual heat of hydration is -783 kJ Imol making the heat of solution +3 kJ Imol.
82.
A is correct. At constant pressure, change in enfhalpy is equal to heat.
83.
C is correct. The vapor pressure of solution might be lower than just one of the pure substances but not the ofher. You can see this from the graph below. pressure of pure 'B'
-L'>.H
84.
B is correct. You should notice fhat B and C are opposites, so one of fhem must be the answer. Molecules break free of the surface of a liquid and add to fhe vapor pressure when they have sufficient kinetic energy to break fhe intermolecular bonds.
85.
C is correct. The solution had to deviate from Raoult's law and fherefore could not be ideal. Since it deviated negatively from Raoult's law, fhe heat of solution is exothermic.
86.
B is correct. In an ideal solution, fhe vapor pressure will be somewhere in between fhe vapor pressures of the solute and fhe solvent, depending on their relative mole fractions.
87.
0 is correct. Try testing the answer choices to see which one is right. (0.2)(800 mmHg) + (0.8)(300 mmHg) = 400 mmHg.
88.
C is correct. If heat is released, fhe solvation process must be exofhermic. The breaking of bonds is an endothermic process and the forming of bonds is an exothermic process, so in order for the process to be exofhermic overall, the bonds formed must be stronger than the bonds fhat were broken.
89.
0 is correct. Think in terms of mole fraction. The concentration of solvent is at a minimum when fhe concentration of solute is at a maximum.
90.
C is correct. This is the common ion effect (very important for the MCAT).
91.
0 is correct. We can compare the solubilities in one liter of water. For the compounds that dissociate into two parts, the smallest K,p will be the least soluble and first to precipitate. This is BaSO,. We don't have to compare BaS04 with AgzSO, because Ag,S04 dissociates into fhree particles. This means fhat if fheir K,ps Were equal, then Ag,S04 would be more soluble fhan BaS04. However, fhe K,p for BaSO, is much lower, so we know for sure that it is less soluble.
92.
C is correct. Gases become more soluble under greater pressure and lower temperatures. The pressure must be the partial pressure of the soluble gas. Adding an inert gas would not change fhe partial pressure of oxygen in this example. Shaking fhe can is adding energy, and is similar to heating fhe can. Think about shaking a can of soda.
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192 . MCAT INORGANIC CHEMISTRY
93.
C is correct. The solubility of BaCO, in 3 liters of water is found from the equilibrium expression:
K,p = [Ba" ][CO,'-] 1.6 x 10-" = [xlix] x=4x lO-5
This is the saturated concentration in mol/L. We multiply this by 3 liters to get the total number of moles. 94.
A is correct. The solubility product is created by multiplying the concentrations of the products of the solvation wrule turning the coefficients into exponents.
95.
A is correct. The K,p expression is as follows. K,p = [Pb" ](CI-l'
For every PbCI, in solution, there is 1 Pb" and 2 Ct . So [Pb" ] = x and [CI-] K,p
96.
= 2x.
= (x)(2x)' = (x)(4x') = 4r'
B is correct. NaF is very soluble, so w hen it is added to the solution, it will introd uce more Na' and P- ions. The introduction of extra P- ions will shift the CaF, equilibrium toward solid CaF" w ruch removes Ca" ions from the solution.
lecture 5 97.
D is correct. First figure out the heat evolved by the reaction using q = mct;.T =>
q = 250 grams x 4.18 J/g 'c x 1 'c
= 1050 joules
Next divide by moles of NaCI (20 grams is about 1/3 of a mole). This gives you 3150 joules, which is equal to 3 kJ. Since the temperature went down, the reaction is endothermic with positive enthalpy. Notice all the rounding. Thls problem should have been d one with very little math. 98.
A is correct. Remember, IlE = w + q. There is no work done because there is no change in volume in a bomb calorime ter. Thus, the total change in energy is heat. Heat is not enthalpy. H eat equals enthalpy at' constant pressure. The pressure is not constant in a bomb calorimeter.
99.
B is correct. I is false because objects cannot contain heat, and because the same amount of the same substance can have the same amount of energy and be at different temperatures. Nevertheless, this is a treading the MCAT edge of required knowledge. Don' t feel too bad if you chose C. II is false. Different phases will have different specific heats. III is true.
100.
D is correct. A, B, and C are false. Temperature is proportional to kinetic energy not just velOCity, so more mass per molecule does not make a difference. Boiling point does not make sense; substance A might be water and substance B ice. Answer C mistakenly relies upon speed and not kinetic energy for temperature. D is the correc t choice by process of elimination. The more ways that a substance has to absorb energy, the more heat it can absorb with the least change in temperature.
101.
D is correct. No energy transfer takes place, so there is no heat or work.
102.
C is correct. Aluminum has the largest value for specific heat, which means that it can absorb the most energy while showing the smallest temperature change.
103.
A is correct. Since the specific heat for Au is one-third as large as the specific heat for Cu, one-trurd as much heat will be required to get the same temperature change.
104.
B is correct. Use the equation q = mct;.T. The change in temperature is 31 - 26 = 5. Don't forget to convert 1.8 kJ into 1800 J.
m = q/ct;.T = (1800)/(0.90)(5) = 400 g.
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ANSWERS
105.
&
EXPLANATIONS FOR QU ESTIONS IN THE LECTURES •
193
D is correct. We can solve this problem by summing the q's on the heat curve. The heat is positive because heat is added to the system. 140 IZO
~too t·~·~··~·~··~·~··~·~··~·~·~··~·~··~·~··~·~·~========:1 ]
8()
G
It '"
§ ·W zo o
heatl q
q=mci.A.T=l x .5 x lO=
~"= mC= 1
Mf..
x
5 \\
80 = 80::J
q = mcW"",bT= 1 x 1 x 100 = 100 Mfv.pOO~tioo = mC = 1 x 540 = 540 q = mc,,,j1T= 1 x .5 x 10 =---.2
....- - -
730 106.
B is correct. The added energy goes into breaking bonds, and as is demonstrated by the heat curve above, the temperature remains constant until all the ice is melted. Entropy increases moving to the right on the heat curve.
107.
C is correct. This is just a phase diagram with pressure on a log scale. There are many ways to manipulate the phase diagram. Don't be intimidated . Try to compare it to what you know.
108.
C is correct. Above the critical point, liquid and vapor water have the same density. The critical temperature w ill be the highest temperature on the graph w here the two lines meet.
109.
A is correct. The area in the dashed line is the point where water is changing phase. Like along line RS, in the dashed line area water and steam exist in equilibrium.
110.
B is correct. The heat of fusion is the amount of heat that must be added to convert one mole of a substance completely from solid to liquid. Benzene has a molecular mass of 78, so the sample contains 1 mole. The flat line on the heating curve represents the heat being added while the phase changes from solid to liquid, so the heat of fusion can be found by measuring the length of the flat line. So 14.4 kJ - 3.5 kJ = 10.9 kJ.
111.
D is correct. At the boiling point, any added energy is used to break intermolecular bonds and not to increase kinetic energy, so while the water is boiling, there is no temperature increase.
112.
A is correct. Heating the solid will raise its temperature which will eventually melt it. Compressing the solid will raise the pressure on the solid which will most likely keep it a solid. A few substances like water will melt under pressure, but for most solids, pressure changes a liquid to a solid. It is the random kinetic energy of the molecules of a solid and not the uniform translational motion kinetic energy of the solid that increases its temp erature and would make it melt.
113.
A is correct. Boiling p oint elevation is a colligative property. The more particles the higher the boiling point. NaCl dissociates so that the normality is twice the molarity. Thus, the least number of particles will be in 0.5 M glucose solution.
114.
C is correct. The osmotic pressure will not create a difference in the buoyant force . The equation for buoyant force (Fb = PVg) does not include osmotic pressure. Seawater has greater density because salts are heavier than water, and the salt added does not create an appreciable difference in volume.
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194 . MCAT INORGANIC CHEMISTRY
115.
B is correct. You must recognize from the fonnula that glycerol does not dissociate; it is not ionic. Then use ISm, which gives you a molality of 10. Molality is moles of solute divided by kg of solvent. Assume that 1 liter of water has a mass of 1 kg. Thus 100 moles of glycerol are required. Glycerol has a molecular weight of 62 g/ mol. 6200 g = 6.2 kg.
flT =
116.
B is correct. The reaction is exothermic because the temperature increased. An exothermic reaction makes stronger bonds. Stronger bonds lower vapor pressure. A lower vapor pressure means more energy is needed to raise the vapor pressure to equal atmospheric pressure. Thus, a lower vapor pressure means a higher boiling point.
117.
C is correct. The question has a lot of extra information to mislead you. A high value for h indicates a high osmotic pressure in the solution. From the formula for osmotic pressure, IT = iMRT, we know that a high osmotic pressure corresponds to a high molarity. A high molarity means many particles per gram of protein placed into the solution. Thus a high osmotic pressure means a low molecular weight.
118.
D is correct. The freezing point depression formula is flT = kmi. We know that k is 1.86. The molar mass of CoCl, is Ill, so there are 3 moles of salt in 1 kg of water and m is 3. CaCI, dissociates into 3 particles, so i is 3. flT = (1.86)(3)(3) =16.7°C
119.
A is correct. The freezing point expression is flT = kmi. The solute is non-polar, so i is 1 and we can leave it out. Now let's remember the definition of molality.
m = (moles of solute) /(kg of solvent) but (moles of solute) = (grams of solute) / (molar mass of solute) Substituting, we get flT = (k)(grams of solute) / (molar mass of solute)(kg of solvent) If you solve for the molar mass, you get choice A. 120. D is correct. Boiling point is a colligative property, which means that it depends only on the number of particles in solution, not on their specific properties. Both NaF and KCl dissociate completely into 2 particles each, so they will have the same effect on the boiling point of water.
Lecture 6 121.
B is correct. The conjugate acid is the molecule after it accepts a proton.
122.
C is correct. By definition, a Lewis base donates a pair of electrons.
123.
C is correct. NH; is an acid. The strongest base is the conjugate of the weakest acid.
124.
A is correct. An amino acid can act as an acid or a base depending upon the pH. Although the conjugate base of sulfuric acid is amphoteric, sulfuric acid cannot accept a proton and is not amphoteric.
125.
A is correct. The electron withdrawing group will further polarize the O-H bond, and polarization increases acidity in aqueous solution.
126.
B is correct. The pH of solution B is 7. The pH of solution A is between 4 and 5. The difference in pH must be between 7 - 5 = 2 and 7 - 4 = 3. Choice B is the only one in that range.
127.
B is correct. In a coordinate covalent bond, one atom donates an electron pair to share with another atom. In this case, ammonia has the unbonded pair to donate to boron, so ammonia is the Lewis base and boron is the Lewis acid.
128.
A is correct. In Reaction 1, water accepts a proton to become H 30', so it is acting as a Bronsted-Lowry base. In Reaction 2, water gives up a proton to become OH-, so it acts as a Bronsted-Lowry acid.
129.
D is correct. Kb is the reaction of the conjugate base w ith water.
130.
B is correct. HBr dissociates completely, so the concentration of H+ ions will be equal to the concentration of solution. The -log(O.I) = 1. Copyright © 2007 Examkrackers, Inc .
ANSWERS
131.
C is correct. The Kb for NaHC03 is Kj K, = _
X
&
EXPLANATIONS FOR QUESTIONS IN THE LECTURES .
195
10-'. We can set up the equilibrium expression:
[OW][H,CO,l [HCO,1 0.25x10·' ~
Thus, the pOH
~
[xll>:L /
This x is insignificant.
[I@)
between 3 and 4. Subtracting from 14, the pH
~
between 10 and 11.
132.
0 is correct. Each unit of pH is a tenfold increase of acidity.
133.
0 is correct. You should recognize P- as the conjugate base of a weak acid. Choices A and B are conjugates of strong acids, and thus weaker bases. Choice C is an acid.
134.
B is correct. BrO- is the conjugate base of HBrO, so you can find the base dissociation constant Kb by dividing. 1 x 10-14 /2 X 10-9• You get 0.5 x 10-5, which is the same as 5 x 10-".
135.
C is correct. If the pH is 10, the pOH must be 4. If the pOH is 4, then the hydroxide ion concentration must be 10-4 M.
136.
0 is correct. Acetate ion, CzH,Oz-, is the conjugate base of a weak acid, so it will act as a base in solution. Sodium ion, Na+, is the conjugate acid of a strong base, so it is neutral in solution.
137.
0 is correct. The pH starts basic so a base is being titrated. It ends very acidic so a strong acid is titrating.
138.
B is correct. A buffer is made from equal amounts of an acid and its conjugate. The buffer works best when the pH ~ pK,. -log(8.3 x 10-') ~ between 6 and 7. 8.3 is close to ten, making the pK, closer to 6.
139.
A is correct. An indicator generally changes color within plus or minus one pH point of its pK,.
140.
B is correct. The concentration of the conjugate base of the first acid is the greatest at the first equivalence point.
141.
C is correct. The equivalence point of a titration of a weak acid with a strong base will always be greater than 7. It is the same as adding the conjugate base of the acid to pure water. 14 is way too basic. Pure 1 M NaOH has apHof14.
142.
C is correct. A buffered solution is formed when equal amounts of a weak acid and its conjugate base are present in a solution. Acetic acid is a weak acid and the acetate ion is its conjugate.
143.
0 is correct. HCO,- can act as a Bronsted Lowry acid and give up a hydrogen ion to become CO,z-. It can also act as a Lewis base and donate an electron pair to a hydrogen ion to become H 2CO,. It is amphoteric because it can act as an acid or base. It is not polyprotic because it has only one hydrogen.
144.
B is correct. Prom the Henderson-Hasselbalch equation, you can see that when a weak acid and its conjugate base are present in a solution in equal amounts, the pH will be equal to the pK,. If you take the negative logarithm of 8.0 x 10-5, i.t will he hetween 4 and S. That's cholee B.
Lecture 7 145.
C is correct. Each oxygen has an oxidation state of -2, and hydrogen has an oxidation state of +1. In order for the ion to have a 1- charge, the sulfur must have a +6 oxidation state. (Notice that oxidation states are given as +n, and actual charges are given as n+.)
146.
C is correct. Aluminum begins as +3 and ends as 0, while carbon begins as
Copyright © 2007 Examkrackers, Inc.
a and ends as +4.
196
MCAT INORGANIC CHEMISTRY
147.
A is correct. The Zn is oxidized from an oxidation state of 0 to +2. Thus, it is the reducing agent.
148.
A is correct. Both A and D cannot be true, so the answer must be A or D. The trickiest part of this problem is to know that lead is comfortable at +2 and sulfur, being in the oxygen family, is comfortable at - 2; thus these are their oxidation states when they are together. But when they are with oxygen, the - 2 of the oxygen rules.
149.
D is correct. An example of where this is false is: 2HCI + Zn ...., ZnCL, + H, Here each atom of the reducing agent, zinc, loses two electrons, and the h ydrogen atom of the oxidizing agent, H CI, gains one electron. Of course, there must be two hydrogens for each zinc.
150.
D is correct. The two oxygens in NO,· have a total oxidation number of -4, so nitrogen must have an oxidation number of +3 to get a total of -Ion the polyatomic ion .. The three oxygens in NO,· have a total oxidation number of - 6, so nitrogen must have an oxidation number of +5 to get a total of - Ion the polyatomic ion. Since the oxidation state is increasing from +3 to +5, electrons are being lost and oxidation is taking place.
151.
D is correct. None of the oxidation states are changed during the course of this acid-base neutralization reaction, so no redox takes place.
152.
A is correct. Don 't forget, the oxidation states for CI and Br in CL, and Br, are zero by definitic!l because the two elem ents are in their llllcombined states. Chlorine gains electrons, so it is reduced. Since it is reduced, it is the oxidizing agent. Bromine loses electrons, so it is oxidized.
153.
D is correct. Positive ions move across the salt bridge to the cathode. You can remember this because the salt bridge is used to balance the charges. Since negative electrons move to the cathode, positive ions must balance the charge by moving to the cathode.
154.
C is correct. The forward and reverse reaction rates are only equal at equilibrium, and their rate constants are rarely equal.
155.
B is correct. This question requires knowledge of the equation: Lleo = -RT In(K). This equation is a statement about the relation ship between Lleo and K at a specific temperature. If Lleo = 0, then K = 1. The standard state for an aqueous solution is 1 M concentrations.
156.
A is correct. The strongest reducing agent is the one most easily oxidized; thus we must reverse the equations and the signs of the potentials.
157.
D is correct. Although a both a Galvanic cell and an electroly tic cell can have a positive potential, only an electrolytic cell can have a negative potential.
158.
C is correct. The potential given are reduction potentials. Since copper is reduced, we can use its potential (0.15 V) as written. Tin is oxidized, so we have to change the sign before we calculate. The total potential is 0.15 V + 0.14 V = 0.29 V. Notice that we ignore the coefficients when we do cell potential calculations.
159.
A is correct. You can use the Nemst equation here, or you can just think about LeChatelier's law. As the concentration of products increases, the reaction will become less spontaneous. The less spontaneous the reaction, the lower the reaction potential.
160.
B is correct. A galvanic cell generates power via a spontaneous reaction, so LlGo must be less than zero. From the expression Lleo = -RTln(K) , you can figure out that if Lleo is negative, then K must be greater than 1.
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A NSWERS
161.
&
E XPLANATIONS FOR Q UESTIONS IN THE LECTURES .
197
C is correct. Reactions in galvanic cells are always spontaneous. To find the reaction for this cell we must flip the more negative half reaction. Now we have a spontaneous cell. AI'+ + 3e- ~ AI Mg ~ Mg2• + 2e-
-1.66 2.37 0.71
We also have to multiply the aluminum reaction by 2 and the magnesium reaction by 3. Notice, however, that we do n o t multiply their potentials. 2Al'+ + 3Mg ~ 3Mg'+ + 2AI
EO= 0.71 V
Since the potential for this cell does not equal this, the conditions must not be standard. 162.
C is correct. Reduction always takes p lace at the cath ode in any cell. This means that the cath ode gains electrons.
163.
D is correct. A concentration cell is a special type of galvanic cell. It is always spontaneous. The concentrations in the cell even out at equilibrium.
164.
D is correct. In this cell the cathode has the greater concentration because electrons flow toward it to reduce the number of cations. Also in a concentration cell EO = 0, since the reduction h alf reaction is simply the reverse of the oxida tion half reaction. n = 1 because only one electron is transferred in each reaction. x/y must be a fraction so that the log will be negative and E will be positive. Thus we have:
E = -{0.06/ 1)log(0.1/ y) = 0.12
y = 10 so that x/y = 10-'. 165.
B is correct. Use units to solve the problem. We want to go from current to grams. Current is Cis. F is coulombs per mol of electrons. For every mol of electrons there is one mol of silver. The molecular weight of silver is 107.8 g/mol
Cis x s
= C => C x maliC = mol => mol x grams/mol = grams so: i x t x (1 / F) x 107.8 = grams
166.
A is correct. Remember, I = Q/t. The charge Q is (96,500)(0.01) = 965 C. The time t in seconds is (5)(60) = 300. So Q/t = 965/300 = 3. The answer is rounded to 3 because there is only 1 significant digit in some of the numbers used in the calculation.
167.
B is correct. If a voltage is applied to a solution containing Na+ and H 20 , the H 20 w ill take the electrons first because it has a larger reduction potentiaL As long as there is H,O present, the aqueous sodium will not react.
168.
C is correct. Ag" gains an electron to become Ag+, so it is reduced (LEO says GER). Reduction takes place at the cathode (ANOX/REDCAn.
Copyright © 2007 Examkrackers, Inc.
INDEX'
199
INDEX A acid dissociation constant 99,102, 108 acids 5,12,93,94-96,98,100,101,106, 107,121 actinides 16 activation energy 29,30,34,35,37,42,56,57,58,59 activity 39 actual yield 13 aerosol 66 alkali metals 3,4, 76, 101 alkaline earth metals 3,4,5, 101 ammonium 12, 42, 76, 98 amorphous 14, 79 amphipathic 66 angstrom 1 angstroms 1 anions 6, 12, 66 anode 114, 115,116,117,119,120,121,122,123 aqueous phase 67, 88 Arrhenius 30, 34, 57, 93 Arrhenius acid 93 Arrhenius equation 30, 34, 57 atmospheric pressure 71,78, 81, 84, 87, 88 atom 1, 2,3, 5-19,21,23,35,43,53,60,93,96, 109-112, 114, 122 atomic mass units (amu) 2 atomic number 2, 6, 9 atomic radius 7, 9 atomic weight 2,9, 11 ATP 11 Aufbau principle 17 autocatalysis 35 autoionization of water 99 Avogadro's law 24 Avogadro's n umber 2 azimuthal quantum number 16
B balancing redox reactions 114 bases 4, 93,94,95,98, 100,101 beryllium 5, 7 bimolecular 30 blackbody radiator 47 Bohr, Neils 19 boiling point 54,63,82, 83, 87, 88, 90, 91 boiling point elevation 88 Copyright © 2007 Examkracke rs, Inc.
bomb calorimeter 81, 82 bond dissociation energy 11 bond energy 11, 73 bond length 10 bonds 4,5,10,11,14,21,35,36,57,66,69,70-72,74, 83,85-88 boron 7,19,98 Boyle's law 24 Broglie, Louis de 19 bromine 4 Bronsted and Lowry 93 Bronsted-Lowry 13, 93,98 buffer solution 104
c cage effect 36 calorimeter 81, 82 carbon 2,4,5,9,12,13, 15,19,43,84,85,86,98,112 carbon dioxide 13,43,84-86 carbonates 76 Carnot engine 50 catalyst 14,30,34,35,37, 57, 58, 115 cathode 114-117, 119, 120-123 cations 4, 6, 12, 66, 93, 101 cell diagram 116, 117 cell potential 115--120, 122 celsius 55, 80 centigrade 55 cesium 8 chalcogens 4 Charles' law 24 chemical equilibrium 38 chlorine 4 coagulate 66 coffee cup calorimeter 81, 82 cold reservoir 49,50 colligative 67, 79, 88, 89 collision model 29,32 colloid 66 colloidal suspensions 66 colloidal system 66 combination 13, 15, 66 combustion 12, 13, 58, 82 common ion effect 76 concentration cell 105, 118, 120, 122
200
MCAT INORGANIC CHEMISTRY
condensation 63, 83 conduction 46,47,51, 116 conjugate acid 94, 98 conjugate base 94, 96, 98, 99, 102, 103, 104, 105 conservation of energy 47,50,53, 61, 63 convection 46,47,51 copper 12, 15,65,82, 113, 119 Coulomb's law 6,7,27 covalent 3,4,5,10,14,35,98 covalent bond 4, 10, 98 critical point 84 critical pressure 84 critical temperature 84, 86 crystal 14,62,75,85,88 crystallization 75,88
D d orbitals 4,5,6,21 d subshell 6 Dalton's law 25 decomposition 13, 15, 42, 43 degenerate 18 degenerate orbitals 18 degrees of freedom 54 delocalized electrons 14 deposition 83 deuterium 2 deviations from ideal behavior 27 dialysis 66 dia tomie 4, 5 diffusion 26 diprotic acid 95, 103 dispersion medium 66 dissolution 12,40,66,68, 70, 75 DNA 14 double bonds 4. See also pi bonds double displacement 13 double replacement 13 Downs cell 121
E Eo
30,34. See also activation energy
effective nuclear charge 7, 8 efficiency 50,52 effusion 26 Einstein 19,20,53 electric potential 113,114 electrical work 117
electrochemical cell 120 electrodes 114,116,121,122 electrolysis 121 electrolyte 66, 67, 114, 115 electrolytic cell 120, 121, 122, 123 electromagnetic waves 47 electromotive force 115 electron 1,6,7,8, 9, 16, 17, 18, 19, 20, 21, 46, 98, 111, 114,115 electron affinity 8,9 electron configuration 6, 18, 21 electron spin quantum number 16 electron units 1 electronegative 4,5,21,96 electronic charge 1 electronic energy 53 electrons 1-4,6--10,14,16-21,53,93,96,98,109, 110-117,120-122 electrostatic forces 6,10,14,27,36 element 2,4,5-7, 9, 11, 12, 16, 17, 21, 55, 60, 79 elementary reaction 30,31,38 elements 2-7,9,11-13,16--18,56,97,110,114 emf 114,115,117,120 emissivity 47 empirical formula 11, 15 emulsion 66 endothermic 8,56,58, 70, 71, 74, 84, 88 endpoint 105 energy ladder 19 enthalpy 46,53,55-58,60,61,63,70,71,77,82-84,119 entropy 38,40,46,53,59-63,71,77,82,84,86,89,120 equilibrium 13,14,29,30,34,35,37-43,46,47,54,57, 60-62,71,75,76,84,85,89,99,100,104,114,118,119 equilibrium approximation 34 equilibrium constant 34,38-43, 75, 99, 118, 119 equipartition theory 54 equivalence point 103-107, 111 equivalents 68, 103 evaporation 12, 84 exothermic 8,30,35,37,40,56,58,70,74,75 extensive properties 45,46,54
F families 3 First Law of Thermodynamics 49,53,60,80 fluorine 4-6, 8, 96 foam 66,77 free energy 42,56,61,89,117,118
Copyright cg 2007 Examkrackers, Inc.
INDEX'
freezing 54, 83, 84, 88, 90, 91 freezing point 54, 88, 90, 91 freezing point depression 88,90,91
201
hydrogen 2,4-8, 10, 12, 15, 19, 28, 83, 86, 87, 93-99, 110, 112, 113, 115, 123 hydrogen halide acids 96 hydroxide 4,12,75,93,98,99,102,123
G galvanic cell 114-123 Gibbs energy 46,53, 61, 119 Gibbs free energy 42,56, 61, 117 glucose 11, 90 glycogen 14 Graham's law 26 ground state 18 groups 3,4, 11, 110
H H
35,48,53,55,81,90,93,94,96, 98, 99, 100, 105, 112, 113, 114, 115. See also hydrogen H, 28,40,41,56,98, 112, 113, 123. See also hydrogen Haber Process 40, 41 half cell 115, 120 half equivalence point 104, 106 half life 32, 33 half reaction 113-115,117,119,120,122 halides 5, 96 halogens 3-5,76 HCl 26,28,69,81,96,100,101,103,108,112 heat 3,9,14,24,40,46-51,53,56,60,61,63,70-74,77, 79-88 heat capacity 79-83,87 heat engine 50 heat of fusion 83, 87 heat of hydration 70,74 heat of reaction 56,81 heat of solution 70,72-74,77,82,88 heat of vaporization 71, 83 heating curve 83, 84, 87 heats of hydration 5 heats of reaction 81 Heisenberg Uncertainty Principle 17 helium 6,7 Henry's law 77 Henry's law constant 77 Hess'law 56 hot reservoir 49 Hund's rule 19,21 hydration 5, 67, 70, 74 hydration number 67 hydrides 4, 97 Copyright IQ 2007 Examkrackers, Inc.
ideal gas 24,25,27,54,55,72 ideal gas law 24,25,27, 72 ideal solutions 65 ideally dilute solutions 65 indicator 105-107 inert gases 5 intensive properties 45,46,84,114 intermediates 31,34,57 intermolecular potential energy 53, 80 internal energy 46,53-56,58, 60, 70, 80, 81 iodine 4 ion 6,9,12,14,67,76,88,93-96,98,99,102,104,109, 111,115 ion pairing 88, 104 ionic compounds 4, 6, 12, 66, 76, 101 ionic oxides 3, 5 ionization energy 7,8,9 ions 3,4,6,12,14,18,66-69,75,76,78,79,88,93-95, 99-101,104,106,111,114-116,119,122 irreversible 32, 33, 59 isotopes 2
K kinetic molecular theory
24,27
L lanthanides 16 law of mass action 38,39,99 Le Chatelier's principle 40 lead 24, 76, 88, 112 Lewis 4, 13, 93, 98, 101, 108 Lewis acids 93, 101 Lewis bases 4 limiting reagent 13, 15 liquid junction 116 lithium 7 London dispersion forces 66 lyophilic 66 lyophobic 66
202
MCAT INORGANIC CHEMISTRY
M magnesium 6, 122 magnetic quantum number 16 main-group elements 4 mass 1,2, 11, 13, 15, 21, 26, 27,38,39,45, 48, 53, 67-69, 81, 82, 90, 91, 99, 103, 122 mass number 2 mass percentage 67 mean free path 23, 26 melting 3,4,12, 14, 63, 7l, 83, 84, 88 melting point 14, 71, 83, 88 melting points 3, 4 mercury 3, 76, 113 metal 3-5,9,12, 14, 18, 20, 35, 47, 51, 82, 93, 97, 110, 114, 115, 121 metal plating 121 metallic character 3, 8 metalloid 9 metalloids 3, 4 metals 3--{j, 16, 18, 76, 82, 101, 110, 113, 121 metathesis 12, 13, 15 methane 13, 28, 96, 111 modes 54 molality 67, 81, 88 molar mass 2, 91 molarity 30,37, 67, 69, 88, 89, 111 mole 2,13,15,24, 25, 27,28,39,49,54,56,65, 67--{j9, 72-77, 81, 82, 87, 88, 91, 98, 102, 103, 117, 122 mole fraction 25, 39, 65, 67, 69, 72, 73, 74, 75, 77, 88 molecular formula 11 molecular weight 2, 28, 90 molecularity 30, 32, 39 molecules 5,9, 10, 11, 13, 14, 23-30, 33, 35, 36, 40, 45, 46,49,53-57,59,60,65--{j7,69- 72, 74, 75, 77,80, 82-87, 93, 95
noble gas 5, 6, 18, 110 nonideal solutions 65, 73 nonmetals 3, 4, 6 nonvolatile solute 71,74,88 normal boiling point 83,88 normal melting point 83 normality 68 nuclear charge 6, 7, 8 nucleon 1 nucleons 1 nucleus 1,6-8, 17, 18 nuclide 2
o orbital 6,7,16-19,21 order 1,3,4,9, 11,17,18,26,29,31-33,35,37-39, 47, 50,53, 60, 61, 65, 69, 70, 73, 76, 80, 93, 100, 103, lOS, 109, 111, 117 osmosis 89 osmotic potential 89 osmotic pressure 88- 90 oxidant 111, 112 oxidation 3,5, IS, 21, 42, 96, 109-115, 119, 120 oxidation half reaction 113, 114, 119 oxidation potential 113, 115 oxidation state 3, 5, 21, 109-112 oxidation-reduction reaction 109 oxides 3,4,5 oxidizing agent 111, 112, 113 oxyacid 12, 96 oxygen 4-7, 12, 13, 15,21, 23, 28, 42, 58, 69, 77, 78, 110, 112, 113 ozone 4. See also oxygen
p p orbitals
N neon 7 NernSt equation 118, 120, 122 network 14, 15 neutron 2 neutrons 1,2,9 Newton's law of cooling 47 Newton's second law 48 NH3 15, 26, 28, 40, 41, 69, 97, 98, 101 nitrate 12,42, 76 nitrite 12 nitrogen 4,5,7, 15,28,40, 69, 98, 110, 112
5, 7, 17, 21 partial pressure 25,28,39-43,69,72,73,77,84 parts per million 67 path functions 46 Pauli exclusion principle 16 Pauling scale 8 percent yield 13 period 3,5, 8, 16, 18 periodic table 2-4, 7, 8, 9, 12, 16, 17, 97 periodic trends 7, 8 peroxides 4 pH meter 105 phase 35,38, 55,65,67,79,80, 83-86,88,114,116 phase changes 80,83-85 Copyright © 2007 Examkrackers, Inc.
(
1
I I'
INDEX'
phase diagram 84--1l6 phosphates 76 phosphorous 4 photoelectric effect 20 photoelectrons 20 photon 19, 20 pi bonds 4,5 Planck's constant 17, 19 Planck's quantum theory 19 Planck, Max 19 platinum 113, 115 polymers 14 polyprotic acids 95, 106 precipitation 75 principal quantum number 16 principle for detailed balance 39 protein 14, 35, 66,90 protium 2 protons 1,2,6,7,9,68,95, 103 purifying metals 121 PV work 48,49,55,56,61,80, 85
Q quadratic equation 100 quantum m echanics 16, 19 quarks 1
R radiation 46-48, 51 range of an indicator 105 Raoult's la w 72,73, 88 rare gases. See noble gas rate constant 30-32,36--39, 119 rate determining step 33 rate law 30- 35,37,38 reaction coordinate 34 reaction quotient 39, 40, 117 redox 12, 13, 109- 111, 113, 114, 117 redox reaction 109- 111. 113, 114, 117 reducing agent 111- 113, 119 reductant 111, 112 reduction 7, 109-ll5, 117, 119, 120-123 reduction half reaction 113--115, 117, 119,120 reduction potentials 113--115, 122 representative elements 6, 16, 18 representative 4, 6, 16, 18 resonance 14 rest mass energy 53 Copyright © 2007 Exam krackers, Inc.
203
reversible 31,33,38,43, 58-
s salt bridge 114,116,119,122 Salts 14,40,75,77,101,102 saturated solution 75, 76, 78 second law of thermodynamics 50,59, 60 SfIE 69, 113, 115, 116 shells 6, 16, 18 smelding 7 SI units 10 silver 76, 113, 115, 122 single displacement 13 single replacement. See Single Displacement sodium 6, 7, 12, 91, 102, 112, 121, 123 sol 66 solubility 40, 75, 76, 77, 78 solubility guidelines 76 solubility product 75, 78 solute 65-72, 74, 75, 77, 78, 88- 91 solvation 36,40, 66, 74, 75 solvent 36, 65-
204 . MCAT INORGANIC CH EMI STRY
sulfate 12, 76 sulfides 4, 76 sulfur 4,5, 15, 21, 58, 79, 112 sulfuric acid 12 supercritical fluid 84 surroundings 24,45, 49, 51, 56, 59, 60, 61, 80, 82 system 10,17,18,25, 30,40,43,45-49, 53-56,58-63, 66, 70, 79-81, 84, 89, 109, 117
T temperature 3-5,9, 14,23-28,30, 36, 37, 39-42, 46, 47, 49-55,58,60-63,71,72,74,75,77,78,80-89,90,115, 117-119 terminals 114--116 termolecular 30 theoretical yield 13 thermal conductivity 46, 47 thermodynamic functions 45,53 third law of thermodynamics 60 titrant 103 titration 103-107 titration curve 103, 104, 106, 107 transition m etals 4,6, 16, 18, 110 transition state 57 translational energy 53, 54 triple bonds 4. See also pi bonds triple point 84 tritium
v valence electrons 3, 9, 16, 21 Van der Waals equation 27 van der Waals forces 35, 77 van't Hoff factor 88 vapor presssure 71 vaporization 71, 83 vibrational energy 53 volatile solute 72, 77, 88 voltaic cell 114 volume 23-25, 27,28,46,48,52- 56,58,60, 67, 72, 80-82,84--86,104,117
W water 4,5, 13, 23, 28, 39, 42, 47, 54, 55, 56, 58, 63, 66-69, 74--79,81-87,89-91,94,95,98- 102,104,110, 113, 121, 123 water potential 89 work 6,17, 18,20,24,34,39,46, 48,49, 50-53,55, 56, 59,61,63,80, 82, 85,115,117- 119 work function 20
z Z 30,51,54. See nuclear charge Z'ff 7,8. also See effective nuclear charge zeroth law of thermodynamics
54
2
turnover number 35 Tyndall effect 66
u unimolecular 30 universal gas constan t
24
Copyright © 2007 Examkraders, Inc
· .'
EXAM, ,t(RACt(ERS ~7'-
EXAMKRACKERS
MeAT
7 TH EDITION
Acknowledgements Although I am the author, the hard work and expertise of many individuals contributed to this book. The idea of writing in two voices, a science voice and an MCAT voice, was the creative brainchild of my imaginative friend Jordan Zaretsky. I would like to thank Scott Calvin for lending his exceptional science talent and pedagogic skills to this project. I also must thank five years worth of ExamKrackers students for doggedly questioning every explanation, every sentence, every diagram, and every punctuation mark in the book, and for providing the creative inspiration that helped me find new ways to approach and teach physics. Finally, I wish to thank my wife, Silvia, for her support during the difficult times in the past and those that lie ahead.
Copyright (f) 2007 EXClrnkrackers, Inc.
READ THIS SECTION FIRST! This manual contains all the physics tested on the MeAT and more. It contains more physics than is tested on th e MeAT because a deeper understanding of basic scientific principles is often gained through more advanced study. In addition, the MeAT often presents passages with imposing topics tha t may intimidate the test-taker. Although the questions don't reqwre knowledge of these topics, som e fami liarity will increase the confidence of the test-taker. In order to answer questions quickly and efficiently, it is vital that the test-taker understand what is, and is not, tested directly by the MeAT. To assist the test-taker in gaining this knowledge, this manual will use the following conventions. Any term or concept w hich is tested directly by the MeAT will be written in bold and brown. To ensure a perfect score on the MeAT, you should thoroughly understand all terms and concepts that are in bold and brown in this manual. Sometimes it is not necessary to m emorize the name of a concept, but it is n ecessary to understand the concept itself. These concepts will also be in bold and brown. It is important to note that the converse of the above is not true: just because a topic is not in bold and brown, does not mean tha tit is n ot important. Any form ula that must be memorized will be written in
large, red, bold type.
If a topic is discu ssed purely as background knowledge, it will be written in italics. If a topic is written in italics, it is not likely to be required knowledge for the MeAT but may be discussed in an MeAT passage. Do not ignore ite ms in italics, but recognize th em as less important than other item s. Answers to questions that directly test knowledge of italicized topics are likely to b e found in an MeAT passage. Text written in orange is me, Stllty thl' Kracker. 1 will remind you what is and is not an absolute must for MCAT. I wil l help you develop your MCAT intuition . In addition, 1 will offer mnemoni cs, simple method s of viL'wing a com p lex concept, and occasionally some comic relief. Don't ignore me, eve n if yo u think 1 am not funn y, bt' C<:ltlSC m y comedy is designed to help YOll understand and remember. If yo u think J a m funny, tell the boss. I cou ld usc a ra ise.
Each chapter in this manual should be read three times: twice before the class lecture, and once immediately following the lecture. During the first reading, you should not write in the book. Instead, read purely for enjoyment. During the second reading, you should both highlight and take notes in the margins. The third reading sho uld be slow and thorough. The 24 qLlestions in each lecture should be worked during the second reading before coming to class. The inclass exams in the back of the book are to be done in class after the lecture. Do not look at them before class. Warning: Just attending the class will not raise your score. You must do the work. Not attending class will obstruct dramatic score increases. If you h ave Audio Osmosis, then listen to the appropriate lecture before and after yo u read a lecture. If YO ll are studying independently, read the lecture twice before doing the in-class exam and then once after doing the in-class exam. If you have Audio Osmosis, listen to Audio Osmosis before taking the in-class exam and then as many times as necessary a fter taking the exam . A scaled score conversion chart is provided on the answer page. This is not meant to be an accurate representation of your MeAT score. Do not become demoralized by a poor performance on these exams; they are not accurate reflections of your performance on the real MeAT. The thirty minute exam s have been designed to ed ucate. They are similar to an MeAT but with most of the easy questions removed. We believe that you can answer most of the easy qu estions without too much help from us, so the best way to raise your score is Copyright © 2007 Examkrackers, Inc.
to focus on the more difficult questions. This method is one of the reasons for the rapid and celebrated success of the Examkrackers prep course and products. If you find yourself struggling with the science or just needing more practice materials, use the Examkrackers 1001 Questions series. These books are designed specifically to teach the science. If you are already scoring lOs or better, these books are not for you.
You should take advantage of the bulletin board at www.examkrackers.com. The bulletin board allows you to discuss any question in the book with an MCAT expert at Examkrackers. All discussions are kept on file so you have a bank of discussions to which you can refer to any question in this book. Although we are very careful to be accurate, errata is an occupational hazard of any science book, especially those that are updated regularly as is this one. We maintain that our books have fewer errata than any other prep book. Most of the time what students are certain are errata is the student's error and not an error in the book. So that you can be certain, any errata in this book will be listed as it is discovered at www.examkrackerS.com on the bulletin board. Check this site initially and periodically. If you discover what you believe to be errata, please post it on this board and we will verify it promptly. We understand that this system calls attention to the very few errata that may be in our books, but we feel that this is the best system to ensure that you have accurate information for your exam. Again, we stress that we have fewer errata than any other prep book on the market. The difference is that we provide a public list of our errata for your benefit. Study diligently; trust this book to guide you; and you will reach your MCAT goals.
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2007 EXvll1krcKkcrs, Inc.
TABLE OF CONTENTS
LECTURE
1:
TRANSLATIONAL MOTION ....................................... .. ..... .. .. .... .. ................. ..... 1
1.1
Solvin g a Physics Problem .............. ... .................. .... .. ..... .... .............................. ........................ 1
1.2 Vectors and Scalars ................................................. ............... .... ........ ............. ..... ..... ............... 2 1.3 Adding and Subtracting Vectors ............ ....... ................................. ..... ..................... ................ 2 1.4 Multiplying Vectors .............. ....... .. .. .............. ................. ......... ........ .. ...................... ................. 2 1.5 Compon ent Vectors .............. .. .. ; .......... ................................. ............... ..... .............................. 3 1.6 Distance-Displacement, Speed-Velocity, Acceleration ................................. .. ........................ .4 1.7 Uniformly Accelerated Motion and Linear Motion .................................................................. 7 1.8 Graphs of Linear Motion .... .... .................. ............................... ........ ... .... ... ...... .. ........... ...... ...... 7 1.9 Proj ectile Motion ....... .............. .. ............. .. ............. .... .. ...... ...... .. .............. ....... .......... .............. 13 1.10 A ir Resist ance ............ ........... .. ...... .................................................... ....................... .. ............ 14 1.11
Equation Summary ...... ......... .. ...... .. ...... ................. ...... .. .. .............. ... ...... .. ...... ......... .............. 15
2:
FORCE ................... ........ . . .. ......... ............................ . ............. . ................... 17
2.1
Mass and Weight.. ........ ....... .... .............. ................ .... .... .. ....................................................... 17
LECTURE
2.2 Center of Mass .......... .... ............. ............ ...... ................ .... .......... .. .. ..... .... ..................... ......... 18 2.3 The Nature of Force .................................................................... ........ ............ ...................... 18 2.4 Newton's Laws ........ .................................. ..... ........ .. ........ ....... .... .... ........ ... ...... ........ .. ............. 20 2.5 The Law of Universal Gravitation .......... ................................................ ............ .. .......... .. ...... 23 2.6 Inclined Planes ................................................... ...... ................ .. ...... ... ...... .. .... ... ...... .. .......... .. .25 2.7 Circular Motion and Centripetal Force ......... ..... ................................................... .... ............. 27 2.8 Friction .... .................... ..... ... ..................... .. ....... ..... .. ...... ... ...... .. ...... .. ........ .................... .. .. ...... 31 2.9 Tension ................. .. ......................... ..... ..... ............ ................ ..... ... ................. ... ...... .......... .... 32 2.10 Hooke's Law .............. ............. ......................... ... ..... .... ............... ...... ................ ....... ........ ........ 32 2.11
Equation Summary .......... .. .. ....... .......................... .................................. ....................... ........ 34
3:
EQUILIBRIUM, TORQUE, AND ENERGY ................................................ ............ 37
3.1
Equilibrium .................. .. .... .. ............... ................... . ;... ................... ........................... ... .......... 37
LECTURE
3.2 Syste ms Not in Equ ilibri um ...... ...... ........ .... ................. .. .......... .. ............ ................................. 38 3.3 To rque .............. .... ........... .... .......................... .... ..... ................. .. .............. ............... : .. ... ..... ... .. 40 3.4 Energy ......... .. .. ...... .......... .... ..... ..... ......... .. .............. ... ............. ........ .. ........ ......... .... ..... .. .... ..... .44 3. S Systems ......... .. .............. .. ...... ........ ... ..... ... ..... ......... ...... .......... .. .............. ..... ..... ... .......... ....... .. .4S 3.6 Work ............................... ............... .......................... ...... ... ..... .. ...... .......... ............... .. ...... ...... .45 3.7 Conservative and Nonconservative Forces ............ ........ ....... ......... .... .................................... 46 3.8 Work and Friction ....................................... ........... ........ ...... ..... ........................... ..... .... ........ .47 3.9 Exa mples of Work ........ .... ......... ............................. ........ ........ ...... ..... ............................. ....... .49 ~:qpyri 9ht
ii.'J 2007 EX<:lrn kr;) ckcrs, Inc.
3.10 Summary of Work ....... ........ .. ....... ........ ......... ............ .................... .. ...... .. ...... .......... ............... 50 3. 11
Power .... ......... ...................... .. .. ................................ ...... ..... ........... .. ...................... ......... ... .....50
3.12 Eq uation Summary .............. .. ................ ....... .. .. ........ ..... ............... .. .... ... ........ ... ...... ... .... ........ 51 LECTURE
4:
MOMENTUM, MACHINES, AND RADIOACTIVE DECAY ...... .. ... ............. .. .. .......... 53
4.1
Momentum ... ............................................................. .... ........................................................ 53
4.2 Co llisions ........... ... ... ... ... .... ... .......... .. ......... .... ......... ....... .. ...... .. ....... ........ .................. ... ........... 53 4.3 Reverse Collisions ..... ... ........... .... ......... ...... .... ........ ....... .. ... .... .. ........... ...... ..... ...... .. ................ 55 4.4 Intuition about Coll isions ............. ........ ......... ........ .... .... .... ............. .... ... ........................ ........ 56 4.5 Impulse .. .... .... .... ................................... ............... .. ..... ...... ..... ...... ...... ............ ........... ............. 57 4.6 Machines .......... ..................... ........ ......... .. ...... ...... .................. ........ ... ...... ........ ... .................... 60 4.7 The Ramp ....... ............................... .. ...... .. ...... .. ..... ..... ..................... ........ ................... ........... .60 4.8 The Lever ........ ....................... ... ...... ...................... ... ... ... .......... .. ...... ......... .. ................ .. ... ...... .61 4.9 The Pulley ... ... .. ....... .......................................... ...... .. ............................................................ .62 4. 10 Radioactive Decay ................. ............. ................ .................... .................. .. ............................ 65 4.1 1 Half-Life ... ... .. .... ................ .... ..... ......... ............ ...... ........... ............................................ .... ... .. .. 65 4.12 Types of Radioact ive Decay .................. .. ........... ... ......... .... ..................... .. ..... .............. .. ....... 66 4.13 M ass Defect ........... ............... ....... ............ ... .. ..... ................................. ...... .... ........ ...... ............ 67 4.14 Fission and Fusion ........ ......... ........... ............ .. ........... .. .... .......... ............. .. ................ .............. 68 4.15 Equation Summ ary ............ ................................. ..... .. ............................... ........ ........ .... ... ... ... 69 LECTURE
5:
FLUIDS AND SOLIDS ... . .... . ......... . .......................... ... ..... .. .......... . . ..... ... .. ....... 73
5. 1 Fl uids ............. .............................................................. ...... ............... .. ............................ ... ..... 73 5.2 Density ......... .................... ....... .. .. ............................... .. ...... .. ............... :.............. .. ......... .......... 74 5.3 Pressure ................... .......................................................... ......... .. .............. .. ...... .. ...... .. ......... .75 5.4 Fluids at Rest ............ .. ............. ......... .. ....................... .............. ... ......... ...................................76 5.5 Hydra ulic Lift ..................... .. .... ........... .. ........ .. ................ ... ........ ......... ................ .............. ... ... 78 5.6 Arch imedes' Principle ................. ..... .... .... ........ ... ...... ....... ..... ...... ....... .. ...... .. ............... ........... 78 5.7
Fluids in Motion ..... ........... ............. ............ ................. ........ .... .............. ..... ............. ... ... ... ....... 83
5.8 Idea l Fluid .......... .. ............ .. .... ...... ......... .... ...... ... ... ..... ..... ..... ..... ......... .... ................. .... .... ....... 83 5.9 Non-id eal Fluids (Rea l Fluids) ........................................ ........., .................... ........ .... ...... ......... 86 5.10 A Melhmj for Greater Understanding of Flu id Flow ...... ............ .... ..... ................. .................. 87 5. 11
Su rface Tension .......................................... .. .. .. ......... .... ...... .. .. ....... ............................ .. ..... .... .88
5.12 Solids .. ..... ...... ................ ..... .. .... .. .. ....................... .. ...................................... ...........................91 5. 13 Thermal Expansion ........................................ .... ... ......................... ............. ............... : ... ... ..... 93 5. 14 Equation Summary ........... .... ............ ......................... .... .............. .. ..... .......................... .. ...... .94
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LECTURE
6:
WAVES •• ... .• ... ••.. .••. . .• •. . .•.. .. ... ..• •. . .. •.. ..•........• . ...• •. . .••....... •... .. . . . ... •. ... .......... •97
6. 1 Wave Charact eri stics .......................................... .... ...... ......................... .. ...... ........ ................. 97 6.2 Superpositio n. Phase. and Interference ................................ ...... .. .......... .............. .. ............ 103 6.3 Simple Harmon ic Motion .......................... ................ ........ ................. ........... .. ..... .. ............. 108 6.4 T he Doppler Effect ................... .... .......................... .. .. ................ .... ........ ....... ............... .... ... 11 0
LECTURE
6.5
Eq uat ion Summa ry .................... .... ........ ............... .. .. .......................... ............... .. ................ 11 2
7:
ELECTRICITY AND MAGNETISM .......................................... ........................ 115
7. 1 Electric Charge .. ...... .. ............................ .. .. ......... .... ... ................................................. ......... 115 7.2 M ovement of Cha rge ............. .... .. ...... .. ...... ......... .... .. ...... .... ..... ... ............ ........... ...... ........ .. . 123 7.3 Ci rcuits ..... ... .... ...................................................... .. ....................... .. ...... .. ...... .. ...... ........ .... ... 123 7.4 Power .... .. .. .. ... ... ........................... .. .............. ... ........ .............. .. ...... ... ............................ ........ . 129 7.5 AC Current .......... ............. .... ..... ..................... ..... .. .............. .. ......... .......................... .... ....... 132 7.6 Magnetism .. ................. .. ...... ... ............. ................ .... ............. .. ...... ......... ... ...................... ... ... 132 7.7 Equ ation Summary ..................... .. ............... .. .. ...... .................................. ...... .. ........... ...... ... 136 LECTURE
8:
LIGHT AND OPTICS .................................................................................. 139
8. 1 Light ....................... ............. .......... ........ ........................ ....... ..... ... ..... .............. .. .. ... ............... 139 8.2 Imag es ................ .. ........... ............. ... ..... ....... ... ....... .. ............................. .... .... .... ..... ............... 145 8.3
M irrors and Len ses ................................... .. ... ...... .. .. ........................................ ........... .. ... .... 146
8.4 A System for Mirrors and Lenses ....... ......... .... .... ......................... .. .................. ................... 152 8.5 Two-lens Syst em s ........................................ .. .... .. ..... .................... .. ............ .. .. ............ .. .. .. .. . 155 8.6 Using the Diag rams on the Faci ng Page ..... .. ................................................ ...... .. ............. 155 8.7 Equat ion Summary ........................................ ...... .. ...... .................... ............. .. ...... .. ......... .. .. 156 30-MINUTE IN-CLASS EXAMS . ..... .. . .. .. . . . ... .. ... .. .................... ... . . ....... ................ .. ...... . ... 159 In-C lass Exam fo r Lecture 1 .................... .... .. .. ..... .... .... .... ........... .. ................ .... .. ......... ......... 159 In-Class Exam for Lecture 2 .... .. .... ................ ......... .. ...... .. ... .... .. ..... .... ... ... ..... ......... ............... 165 In-Class Exam for Lectu re 3 .... ... .. .................. ........ ........ ........ .... .... .. ... .. .... ............. ............... 171 In-C lass Exam fo r Lectu re 4 .......... .. ......... .. ... ........ ................. ....... ........ .......... ............. ......... 177 •
In-Class Exam forLecture 5 .......... ........ ........ ..... ....... .. ... .... .... .. ........................................ .... . 185 In-Class Exam fo r Lect ure 6 .......... .... .... .... .. ..... .... .. .... .... .................... ..... ... .......... .. ..... ...... .... 191 In-C lass Exam fo r Lect ure 7 ...... .. ..................... .... .. .. .. ... .... ..... .................................. .. ........... . 197 In-Class Exam fo r Lecture 8 ........................... .. .... .......... ... ........ ........ ...................... ;...... .. ..... 205
ANSWERS
&
EXPLANATIONS TO IN-CLASS EXAMS .. ................... .. ........ .... ....................... 213 Answers and Scaled Score Conversion for In-C lass Exa m s ............ ........ ............................ 21 4 Explanations to In -C lass Exam for Lecture 1 ............................... ... ...................................... 215 Explanations to In -Cl ass Exa m for Lecture 2 ......... .. ...... ........ .... .. .. ........ .. ............. ............... 216
COPyri9ht I[,') 2007 Ex a mkr.3c k er~" Inc.
Explanations to In-Class Exam for Lecture 3 ........................................................................ 218 Explanations to In-Class Exam for Lecture 4 ............. ........ ........ ......... ................. ..•.............. 220 Explanations to In-Class Exam for Lecture 5 ...... .,....... ....................... ........ .......................... 222 Explanations to In-Class Exam for Lecture 6 .............. ........ .. .............. ........ ........... ............... 224 Explanations to In-Class Exam for Lecture 7 .................... .. ....................... ................... ........ 226 Explanations to In-Class Exam for Lecture 8 ...... ........................................... ...... .. ............... 229
ANSWERS
&
EXPLANATIONS TO QUESTIONS IN THE LECTURES ........................................ 231 Answers to Questions in the Lectures ............................................... .. ................................. 232 Explanations to Questions in Lecture 1 .... .. ........................................................................ 233 Explanations to Questions in Lecture 2 ........................... .. ................................................. 235 Explanations to Questions in Lecture 3 .......... ........ ........ .......... .. ........................................ 237 Explanations to Questions in Lecture 4 .................................... ........ .................. .. .............. 239 Explanations to Questions in Lecture 5 .......... ............... .......... ........ ......... .......................... 240 Explanations to Questions in Lecture 6 .......... ... ...... ...... .... ............................. .. .................. 242 Explanations to Questions in Lecture 7 .......... ........ ................ ................. ...... .. .......... ......... 243 Explanations to Questions in Lecture 8 .............................................. .............. .................. 245
INDEX ........................................................................................................................249
PHYSICAL SCIENCES DIRECTIONS. Most questions in the Physical Sciences test are organized into groups, each preceded by a descriptive
passage. After studying the passage, select the one best answer to each question in the group. Some questions are not based on a descriptive passage and are also independent of each other. You must also select the one best answer to these questions. If you are not certain of an answer, eliminate the alternatives that you know to be incorrect and then select an answer from the remaining alternatives. Indicate your selection by blackening the corresponding oval on your answer document. A periodic table is provided for your use. You may consult it whenever you wish.
PERIODIC TABLE OF THE ELEMENTS 1
H
I, , ,
1.0 3
4
Li
Be
6.9
9.0
11
12
567
T
Na . Mg 23.0 ·24.3
8
9
4.0_ 10
B C ~~:i8 -li~O AI Si
N 0 16.0 1~~o 16 p. S
19.0 I 20.2 , 18 17
27.0
31.0
35.5 , 39.9
32.1
F
He
CI
I
Ne Ar
I
28.1
1
32
33
34
35
36
Ge
As
Se
Br
Kr
72.6
83.8
19
20
21
22
23
24
25
26
27
28
29
30
31
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
39.1
40.1
45.0
47.9
50.9
52.0
54.9
55.8
58.9
65.4 ----
------'.
74.9
79.0
79.9
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
V
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
85.5
87.6
88.9
91.2
92.9
95.9
198)
55
56
57
72
73
74
75
Cs
Ba
La'
Hf
Ta
W
Re
.5.~L 63.5
_-
69.7
-----~
101.1 102.9 ....... 106.4 112.4 114.8 118.7 121.8 127.6 126.9 131.3 ....... - ...••....••• ...... _107.9 .. t-· 76 77 78 79 80 81 82 83 84 85 86 "
Os
Ir
Pt
Au
Hg
-~-
TI
Pb
Bi
Po
At
Rn
I
132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) 1222 )1 ---87 104 105 88 89 106 107 108 109 --~.
Fr
Ra
Ac Unq Unp Unh Uns Uno Une o
1223) 226.0 i 227.0 1261) 1262) 1263) 1262) 1265) 1267)
----
5
.,
C:
I
59
60
Pr
Nd
61
62
Pm Sm
63
64
65
66
67
68
Eu
Gd
Tb
Dy
Ho
Er
69,1 70 Tm Vb
71
Lu
14 0.1 (140.9 144.2 1145) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 ! 173.0 175.0 9 o I 91 92 93 94 95 96 97 98 99 103 100 ( 101 1102
1
~
Th
' Pa
U
Np
Pu Am Cm Bk
Cf
Es
Fm! Md
No
Lr ,I
23 2.0! 1231) 238.0 1237) (244) (243) 1247) (247) 1251) 1252) (257) 1(258) (259) 1260)
Copyright @ 2007 Exarnkrackers, Inc.
Translational Motion
1.1
Solving a Physics Problem
Whether they realize it or not, any good physics student has a system to solve physics problems. Some problems are so trivial that the entire system is done in the mind in a fraction of a second. Other times, each step is given careful and deliberate consideration. The following is my system that you should use to solve every single physics problem on the MCAT. For easy problems you will be able to do the entire system in your head in seconds or less, but, the moment you feel any hesitation, you should begin writing with your pencil.
Salty's Own Never Fail 5-Step-System For Solving MeAT Physics Problems: Step 1: Be Confident. Don't be intimidated by any MCAT question. Remember the MCAT only tests basic physics. After reading this manual, you will know all the physics necessary to handle any MCAT problem. Step 2: Draw a well-labeled diagram. A good diagram takes the question out of the 'MCAT environment', and puts it on your terms. Also, the act of drawing a diagram allows you to think about the problem in different ways. Step 3: Narrow your focus to only the system of bodies in which you're interested. This may be the most obvious step in physics but it is the one most often for~otten. You must learn to concentrate upon only the body or bodies about which the question asks, and ignore all extraneous infonnation. Step 4: Find a formula that uses the variables in your diagram. Write down several formulas, and then eliminate until you find the useful one. Actually write your formulas out on the test booklet. It doesn't take much time and it increases accuracy. Step 5: Plug in values and calculate the answer. Note: This last step is often unnecessary on the MCAT.
1) Confidence 2) Diagram
3) System 4) Formula
5) Plug-n -Chug
2 . MeAT PHYSICS
1.2 A vector has magnitude and direction; a scalar has magnitude only. Changing the magnitude or direction of a vector creates a new vector.
Vectors and Scalars
Appreciating the difference between vectors and scalars will help you solve MCAT physics problems. A scalar is a p hysical qu antity that has magnitude but no direction. A vector is a pbys.i.cal quantity with both magnitude and direction. A vector can be represented by an arrow. The d irection of the arrow reveals the direction of .the vector; the length of the arrow reveals the magnitude of the vector.
1.3
Adding and Subtracting Vectors
In order to add vectors, p lace the h ead of the fi rs t vec tor to the tail of the second vector, an d d raw an arrow from the tail of the fi rst to the head of the second. The resulting arrow is the vector sum of the other two vectors. Notice that the magnitude of the sum of wo vectors must be smaller than, or equal to the sum of their magnitu des, and greater th an , or equal to, the difference of their magnitudes. In other words, the sum of two velocity vectors that are 10 m/s and 7 m/s will be greater than o r equal to a velocity vector of 3 mis, but smaller than or equal to a velocity vector of 17 m/s. To subtract vectors place the heads of the nyo vectors together and draw an arrow from the tail of the first to the tail of the second, or add the negative of the vector to be subtracted. The new vector is tl1e vector difference between the two vectors.
head
/
tail
P + (- Q )
1.4
Multiplying Vectors
Vectors cannot be added to, nor subtracted from, scalars, or vice versa. However, vectors can be multiplied or divided by scalars. When a vector is multiplied or divided by a scalar the direction of the original vector is retained but the magnitude changes in proportion to the scalar.
Tedmically, one vector cannot be multiplied by another. Instead fhere is something called dot product and cross product. Although neifher product is required by the MCAT, you will be required to predict fhe results when certain vector quantities are multiplied togefher. When multiplying two vectors, first check to see if the resulting physical quantity is a scalar or vector. If a vector, fhen the vector must point perpendicularly to both of the original two vectors, and the magnitude of the new vector is the product of the magnitude of fhe original vectors times the sine of the angle Copyright if) 2007 Examkrackers, Inc.
LECTURE 1: TRANSLATIONAL MOTION '
between them. (Vpnxioct = Vj VzsinS). There will always be two possible directions that are perpendicular to both of the original two vectors. Right Hand Rule is used to decide between these two directions. Although the MeAT Student Manual lists the Right Hand Rule as being tested by the MeAT, it is unlikely that it will be. If the product of the two vectors is a scalar, the magnitude of the scalar is equal to the product of the magnitudes of the two vectors times the cosine of the angle between them. (Spn"'"" = VjVzcosS). Since there are only a few instances on the MeAT that require multiplication of vectors, students often prefer to memorize each case separately rather than memorize the above rules. This manual will cover all possible MeAT occurrences of vector multiplication on a case by case basis.
3
~",{/ I
Ill '
is a scalar III :::: 2
A vector times a scalar is a vector. The product of two vectors may be either a scalar or a vector. For instance, if we multiply vectors A and B as shown and the product is a vector, it will point into or out of the page depending upon Right Hand Rule. For the MeAT you just need to know that the vector will point perpendicularly to both A and B. The magnitude of the product vector will be ABsin8. If the product is a scalar, it will have a value equal to ABcosS.
1.5
Component Vectors
Any vector can be divided into hvo perpendicular component vectors whose vector sum is equal to the original vector. This is often convenient, since vectors acting perpendicularly to each other sometimes don't affect each other, or affect each other only in a limited fashion. We shall examine this more closely in p rojectile motion, circu1ar motion, and oth er areas. In addition, any vector has an in finite number of possible component vectors, offering great versatility in solving vector p roblems.
Resultant vector The lengths of the component vectors are foun d throu gh simple trigonometry such as the Pythagorean Theorem and SOH CAR TOA.
Copyright (0 2007 i:::xamkrClckers , Inc
The diagram on the left shows three possible pairs of components for the resultant vector. Each component is perpendicular to its partner and sums with its partner to equal the resultant. Each of the infinite number of points on the semi-circle represents a pair of possible component vectors.
4 . MCAT
P HYSICS
Any vector can be replaced by component vectors. Component vectors are always at right angles to each other and thei(sum is equal to the vector being replaced.
c
H
B
o
A
A
Pythagorean Theorem: A 2+ B2~ C 2
SOH CA H TO A
By the way, SOH CAH TOA is a little bit slow for the MCAT. You shou ld memorize the (allowing:
4
2='1
5
12
0 = HsinS A = HcosS
The MCAT will provide the values of sine and cosine when needed (and , more often , wilen not needed).
tane~ O / A
e
e As long as we 're thinking about the Pythagorean theorem, we might as well remember one of the most common triangles used on the MCAT: the 3-4 -5 triangle, and a less common cousin : the 5-1213 triangle .
sine ~ O / H case ~ A / H
PythagoSaltoras
1.6 Distance-Displacement, Speed-Velocity, Acceleration Distance and dispJacement are scalar and vector counterparts, as are speed and velocity. In other words, displacement is distance with the added dimension of direction, and velocity is speed with the added dimension of direction. The definitions of speed and velocity are given by the following formulae: spee d
~
distance . time
. ve I oClty
~
displacement
--"--c.---
hme
.. ..... . x
• 0
....
•••• ..
'"
y
.. ...... ;:
.
Point B
Point A
z
0
"
If a man walks from Point A to Point B, his distance traveled can be Ineasured by the number of steps that he takes. His displacement is his position relative to his starting point or his net distance. If Point A and Point Bare 10 meters apart, the man's displacement is 10 meters to the right; however, the distance that he has traveled is unknown, because he may have taken path X, Y, or Z. If the entire trip took 100 seconds, the Inan's average velocity is his displacelnent divided by the time or Copyri ght
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2007 Examkr,Kkcrs, Inc
LrCTURE
1: TRANSLATIONAL
0.1 m / s to the ri gh t. Notice that the average velocity is indepen dent of the path chosen. The man's average vertical velocity during the trip was zero. Since the man's distance is unknown, his speed or instantaneous velocity at any moment during the trip is unknow n. Eve n if the man took path Y, he may ha ve covered the first half of the trip in 99 second s and the second half in 1 second, thus, no t maintaining a consta nt velocity or speed. Acceleration is a vec tor, and is defined as the rate of change in velocity.
.
change in velocity
aceeIeratlOll = - - -"-- - ---'time Any change in velocity, in either magni tud e or direction, is accele ration . This means that a parlicle mus t accele ra te in order to cha nge the direction of its motion . 1\ n object tra \"eling at 10 m l s north one moment a nd 10 m / s ('(1s t the next moment has nccelerated even though it is mov ing i'l t the san1e speed. Th is a lso mea ns tha t a par ticle mov ing at constant ve locity has no acceleration . You have a natura l intuition about velOcity, but not about ncceil;'rdtion. For ins tnnce, we nil know whnt it feels like to move a t a velocity of 55 mil es/ h o Ul~ but wha t does it fecllike to accelerate at 55 Inites/ ho ur 2? Are we throw n to the back of ou r st'a ts, or do \.I,/e become impatient waiting to reach a good s peed . We can lmdcrs ta nd 55 rni les / hour 2 as a chcmge in veloc ity of 55 miles / hour every ho ur. In othe r wo rds, :-.ti:l rting from zero by the side of the highway, it would ta kp us one hour to reach (I ve loci ty of 55 miles/ ho ur, and s ti ll a no ther ho ur to reach 11 0 miles/hour. No w you know w htlt it fcd s like to accelerate a t 55 miles/ hour::! . G nL' more point \lbo ut acceleration: Ve locity and accek'ra lion d o NOT h
the sa me direction . A pa rticle can be moving to the Jeft "" hil e Dccelerating to the right, or moving up w hi le accelera ting d own . For ins ta nce, a ba ll thru w n upwnrd s is accelerating d o w n w ards even w hil t' moving upwa rd s. In fact, it is even (lcceic rntin g the mome nt it reaches its max imum height whe re its velocity is zero.
O utta the w ay Kracke r boyl
Copyright
(i~i
2007 f)(<.'lrnkrackels, In(
Sorry, no can do . Science experiment in progress.
MOTION · 5
6. An automobile that was moving forward on a highway pulled over onto the exit ramp and slowed to a stop. While the automobile was slowing down, which of the following could be true?
Questions 1 through 8 are NOT based on a descriptive passage.
A.
1. A weather balloon travels upward for 6 kIn while the wind blows it 10 kIn north and 8 km east. Approximately what is its final displacement from its initial position? A.
B. C. D.
B.
7km 10km 14 km 20km
C.
negative.
D.
2. Which of the following gives the average velocity of an
The velocity and acceleration had the same sign. either positive or negative.
7. All of the following describe the magnitude and direction of a vector EXCEPT:
athlete running on a circular track with a circumference of liz km. if that athlete runs I km in 4 minutes?
A. B. C. D.
The velocity was positive and the acceleration was positive. The velocity was negative and the acceleration was negative. The velocity was positive and the acceleration was
A.
B. C.
0 mls 2 mls 4.2 mls 16.8 mls
D.
10 m/s West 10 mls in a circle 20 m to the left 20 m straight up
8. An elephant runs at a speed of 36 kmlhour. Based on this 3. A man entered a cave and walked 100 m north. He then made a sharp turn 150° to the west and walked 87 m straight ahead. How far is the man from where he entered the cave? (Note: sin 30° = 0.50; cos 30° = 0.87.) A. B. C.
D.
information, how far can the elephant run in 10 seconds? A. B. C. D.
25m 50m 100m ISO m
10m 50m 100 m 200m
4. The earth moves around the sun at approximately 30 mls. Is the earth accelerating?
A. B. C. D.
No, because acceleration is a vector. No, because the net displacement is zero. Yes, because the speed is not constant. Yes, because the velocity is not constant.
5. An airliner flies from Chicago to New York. Due to the shape of the earth; the airliner must follow a curved trajectory. How does the curved trajectory of the airliner affect its final displacement for this trip? A.
The displacement is less than it would be if the air-
B.
The displacement is greater than it would be if the airliner flew in a straight line to New York. The displacement is the same as it would be if the airliner flew in a straight line to New York. The final displacement of the airliner is zero.
liner flew in a straight line tu New Yurko
C. D.
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6
STOP.
LECTURE 1: TRANSLATION,\L MOTION'
1.7
7
Uniformly Accelerated Mot ion and Linear Motion
Uniformly accelerated motion is motion with constant acceleration. Since acceleration is a vector, constant acceleration means that both direction and magnitude of acceleration must remain constant. A particle in unifonnly accelerated motion will accelerate at a constant rate regardless of the path traveled by the particle. The most common example of uniformly accelerated motion on the MeAT is a projectile. However, before we examine projectile motion we will examine the rules for the simpler case of uniformly accelerated motion along a straight line. For a particle in uniformly accelerated motion on a linear path, there are four basic variables that will describe its motion completely: displacement (x), velocity (v ), acceleration (a ), and time (I). The first th ree of these are vectors and the last one is a scalar. The values for these .variables can be found tluough three basic equations. These equations can be derived with calculus, but it is far better for you to memorize them. We will refer to these equations as the linear motion equations. However, remember that constant acceleration is required for all of them. 111e equations are:
v = vo +at
v2 =
V o 2+
2ax
In order to use these equations, there must be constant acceleration and linear motion. When choosing which equation to use, pick the one for which you know the value of all but one of the variables. The velocities above are instantaneous velocities, or velocities at a given moment in time. Another concept that is useful on the MeAT is average velocity. Average velocity in a uniformly accelerated motion problem is ,given by:
1.8
G raphs of Linear Mot ion
Most graphs of linear motion will be plotted as displacement, velocity, or acceleration, versus time. For these graphs, you should know the significance of the slope, the line, and the area under each curve. On a displacement versus time graph, the slope at any point is the instantaneous velocity at that time. An upward slope indicates positive velocity; a downward slope, negative velocity (velocity in the reverse direction). A straight line indicates that the slope is constant and, thus, the velocity is constant as welL A curved line represents a changing slope, which indicates a changing velocity and thus acceleration. (Acceleration is the rate of change in velocity.) The area beneath the curve has no meaning for a displacement versus time graph.
Copyright © 2007 Examkrackers, Inc.
The equations on this page require
constant acceleration.
8 . MeAT PHYSICS
Sometimes students look at these graphs and imagine that the particle follows a path that somehow resembles the line. This is incorrect. On the displacement versus time graph, if the slope is positive, the particle is moving in tile positive direction (let's say to the right) . If the slope is negative, the particle is moving in the opposite direction. Notice also that the graph tells us nothing about any perpendicular motion (up or down, in or out) that the particle may have.
30
20 10
o
10
20
30
-10 -20 -30 Time (s)
If the graph above describes the position of a particle confined to a horizontal line, and we arbitrarily choose the righhvard direction as positive, then between zero and 20 seconds the particle's velocity is 1 m/ s to the right. Between 20 and 40 seconds the particle is stationary at a position 20 meters to the right of its initial position. At 50 seconds the particle is back where it started but moving at 2 m/s to the left. Also at 50 seconds it has traveled a distance of 40 meters, yet its displacement is zero. At 60 seconds the particle changes directions and begins accelerating to the right. The average velocity of the particle after 100 seconds is 20 m/lOO sec or 0.2 m/s to the right. If you just read through this very quickly, go back now and examine the graph at each step, and try to derive the values for yourseJf. The linear motion equations can be used on any of the straight-line sections of the displacement graph, because acceleration for those sections is a constant zero. On a velocity versus time graph, the slope at any point is the instantaneous acceleration at that time. An upward slope indicates positive acceleration; a downward slope, negative acceleration. Negative acceleration is not necessarily slowing dOy\Tn. It is acceleration in the reverse direction, which means slowing down if the velocity is already in the positive direction, or speeding up if the velocity is already in the negative direction. A straight line indicates that the slope is constant and, thus, the acceleration is constant as well. A curved line represents a changing slope, which indicates a changing acceleration. The area beneath the curve can represent distance or displacement. If we label all the area between the curve and zero velocity as positive, the area represents distance. If we label the area below zero velOCity as negative, the total area represents displacement.
Copyright © 2007 Exarnkrackers, Inc.
LECTURE 1 : TRANSLAT:ONAL MOTION ' 9
30 20 ~
"'
"-
10
g
.c
'0
0
~
- 10
0
30
40
- 20 - 30 Time (5)
If, aga in, we assume that the above graph is of a particle moving horizontally with the positive direction to the right, we see that the particle began with a velocity of 10 m/s. The particle is constantly decelerating for the first 20 seconds. At 10 seconds, the particle has traveled 50 meters to the right (not zero meters). At 20 seconds the particle has a total displacement of zero from its starting point but it has traveled a distance of 100 meters. Between 20 and 40 seconds the particle has no acceleration and is moving at a constant velocity to the left. At 80 seconds the particle begins decelerating to the right or accelerating to the left (either is correct, think about it) . This deceleration is not constant, however, as is indicated by the curved line. At 100 seconds the particle has a positive, nonzero displacement. Calculating the displacement would require subtracting the area under the x-axis and above the curve from the area above the x-axis and under the curve. To find the total distance traveled would require adding these areas, because distance is not dependent upon direction. Although, when the curve is below the x-axis, the displacement is negative, the entire area between the curve and the x-axis represents positive distance traveled. (Distance has no direction.)
Copyri ~Jht
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There are two Important graphs likely to be on the MeAT: displacement vs. time (d/t graph), and velocity vs. time (v/t graph) . On the dlt graph, the slope is velocity, and the area is meaningless. On the vlt graph, the slope is acceleration, and the area is displacement. Study these graphs (and the paragraphs beneath them) until you are comfortable with all their aspects. The MeAT is very fond of questions that require you to interpret graphs.
10·
PHYSICS
Want a fast, easy way to solve projectile motion problems without using the equations? Use a v /t graph as follows. Draw a line and label the left end with the initial velocity and the right end with the final velocity. If the acceleration is constant, this line represents the line on a v /t graph. The exact middle of the line is always the average velocity. Since the displacement is the average velocity times the time, you know the displacement. If you don't know the time, it is the change in velocity divided by the acceleration, or the difference between the two ends of your line divided by the rate of change in velocity. Remember to always think of acceleration as how fast velocity is changing. It's not as complicated as it sounds. Watch. What is the distance traveled by a particle that srarts at 30 m/s and accelerates to 50 m/s in four sefonds? What is the acceleration?
50 mls
/
\
30 mls
50 mls
~I'
1) draw ancllabel your line -----' 2) find the average velocity exactly at the middle
30 m/s 3) average velocity times time is 160 m.
4) the acceleration is 50 minus 30 divided by four = 5 m/s 2
An object is dropped from a plane and falls for 5 seconds. }'low far does it fall? ~
J) Vertical velocity for a projectile changes by 10 111/s each second so final velocity is 50 m/s. 2)
l.
o m/s
I
draw and labe1 your line ________.....
50 m/s
3) find the average velocity exactly at the 4)
avcragevclocity times time is 125 nt.
middh~
~!S
o ll1/s
/
© 2007 Examkrackers, Inc.
11. Which of the following is th~ most probable description of the motion of the object depicted by the graph below'!
Questions 9 through 16 are NOT based on a descriptive passage.
c
9. Which of the following graphs best represents a particle with constant velocity?
A.
g 01-------
'il ;>
C. E
s" " ..':l U
'" ~
'i3
E
~
a
u
time
ro
'0. 'i3 ~
A. time
time
B.
B. C.
D.
"ro
"0
D.
0
.",
".g
~
~
'il u
'il u
12. A car accelerates at a constant rate from 0 to 25 mls over a distance of 25 m. Approximately how long does it take the car to reach the velocity of 25 mls?
u
u
ro
" time
time
A. B.
10. The graph below represents a particle moving along a straight line. What is the total distance traveled by the particle from t = 0 to t = 10 seconds'!
C. D.
20 10 \
"-
A. B.
0 10 20
"-
c.
/
D.
D.
-12.5m/52 -25 m/s2 -50 m/52 -100 m/s2
1/ 5
Ib
15
14. The graph below shows the displacement of a particle
time (s)
B. C.
Is 2s 4s 8s
13. A particle moving in a straight line slows down at a constant rate from 50 mls to 25 mls in 2 seconds. What is the acceleration of the particle?
30
A.
A person on a bike accelerating in a straight line, and then decelerating. A baseball thrown by a pitcher and hit by a batter. A planet in orbit. One swing on a pendulum.
over time.
-
Om 50m 100m 200m
time
The particle exhibits increasing: I. displacement II. velocity III. acceleration A. B. C. D.
Copyright (fJ 2007 EXilrnkrilckcrs, Inc.
11
I only II only I and II only I and III only
GO ON TO THE NEXT PAGE.
16. The graph below represents a particle movin g in a straight line. When t = 0, the displacement of the particle is O.
15. A driver moving at a consram speed of 20 m1s sees an accident up ahead and hits the brakes. If the car decelerates at a constant rate of -5 mls2, how far does the car go before it comes to a stop?
A. B.
C. D.
~
:s.2 ~
10m 20 m 40m 100m
u 0
~
15 10
5
0
0
5
IO
15
time (8) All of the following statements are true about the particle EXCEPT: A. B. C.
D.
The particle has a total di splace ment of 100 m. The particle moves wi th constant acceleration from o to 5 seconds. The particle moves with constant velocity between 5 and 10 seconds. The particle is moving backwards between 10 and 15 seconds.
"
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12
STOP.
LECTURE 1: TRANSI ATiONAL MOTION '
1.9
Projectile Motion
Projectile: a body projected by an external force and continuing in motion by its own inertia. Because projectile motion is not linear motion , we emmot apply the linear Inotion equations directly. However, we can separate projectile motion into perpendicular components and analyze it as two distinct linear motion problems. Separate the motion into vertical and hori zontal components. For the vertical mo-
ti on, acceleration is constant and due to gravity (10 m / s' ). For ideal situations with no air resistance (as on most MeAT problems) the horizon tal acceleration is a constant zero.
/
.. _--- -
..••...•..•... _
'.
".
",~'
, ...
1
g = -10m/s'
II
, /'
o
I~
cos0
sinO cose
-0
cosO
Tn, =
0
0, = , fi x
=
0
=a Oy = , sine n, = -10 mIs'
XOy
cosO
- sinO
In the diagram above a projectile experiencing no air resis tance is launched with a velocity v at an angle 8. In order to describe this moti on with the linear motion equations, we must separate the Illat.ion into perpendicular components . For con-
venience we choose the horizontal and vertical directions. Using SOH CAH TOA we find that the initial vertical velocity is always vsin8 and the horizontal velocity remains constant at vcos9. Notice tha t there is no acceleration in the horizontal direction, and therefore no change in horizontal velocity thro ug ho ut the flight. Notice that the vertica l accelera tion throughout the flight remains constant at g. At its peak height (h), the projectile has no vertical velocity but is still accelerating downwards at 10 m /s'. The peak height of the projectile can be found from the equation:
Copyright @ 2007 EX<Jrnkrnckcrs, Inc.
13
14 · MeAT PHYSICS
where g is positive 10 m /s'. By substitu ting vsine for v.sine, this equation gives the final velocity v of a projectile when dropped from a height II. It can be derived from special cases of the third linear motion equation and, by doing so, you can see why positive g is used. Notice that the path of a projectile not experiencing a ir resistance is independent of the mass of the projectile. In a vacuum, a bowling ball will follow the same path as a ping p ong ball if their initial velocities are the same. You should also know that vertical velocity alone d icta tes the time of flight for a projectile. The ra nge (horizontal distance) is the horizontal velocity times the time in fli ght; thus the range is dictated by both horizonta l and vertical velocities. Finally, a projectile exhibits symmetry: its path upward is the mirror image to its path downwa rd. This means tha t for a projectile over a flat plane time is the same for both halves of the fligh t, and initial speed is equal to fina l speed.
Projectil e S.ll ty
Und('r~t,1I1d p rojeclile Illo tion qualitati\'e1y and llll<1ntilativcly. In o ther words, don't just rely Oil the equations. Stop right now and contemplate projectile motioll . [ven though the p rojecti le m()\'l'~ bo th lip dnd down in the S~ll1lt' flight its acceleration is const<1nt. EVL'1l whik- the projecti le b motionit''"is ell the instant it rC'<1Ches its pellk, i.lCc('kralioll is still S . How can
in the definition of (lcceleratiol1 .
Usc the symml'try of projectile motion to help you sol\'e problt'llls. For inst,uKC', if will ,1 lways cqlhlll.cro, making
we usc only the second half of the trip, vcrtic,ll caleu lations easier,
i",
Undl'rst,lI1d lhdt vcrtiGl! velocity dictatl'~ l iml' of fl ight. If tvV() prnji.>ctiles ICa\·c thc ('{] rth w i th L1w ~\, l11e \'L'rlica l H ' loci ty, they wi II i<md d t the Sc1 I11l' ti me, reg
And, of ( ourse, " lv\'ays rCI1Wmbl'f th ilt, ill the
n:'si~t.lncc,
mass dops
not dffcct projeclilt· 11101iol1.
1. 10 Air Resistance Air resistru1ce is created when a projectile collides with air mo]ecules. Air resistance is a type of friction, We will discuss friction in Lecture 2. Air resistance takes away energy and slows a projectile. For the MeAT, you need to understand air resistance only qualitati vely. In other words, you don't need to memorize or be familiar w ith any formulae. Factors that change air resistance are: speed, surface area , and shape.
Large surface area increases air resistance because it allows for more collisions with air molecules. Shape also affects air resistance. Stre~mlined objects with smooth surfaces experience less air resistance than irregularly shaped, rough objects. Gene rally speaking, the higher the velocity, the greater the air resistance.
Mass changes the effect of air resistance, but does not change air resistance.
Mass doesn' t change the force of air resistance, but it does d1ange the path of the projectile experiencing the air resistance. Since the force of air resistance remains constant for an y mass, then, from F :::: rna, we see an in verse relationship between mass and acceleration; acceleration must decrease as n1ass increases. This acceleration is not g; it is onl y the deceleration due to air resistance. Thus, larger masses experience less deceleration due to air resistance beca use they are less affected by the same force of air resistance. Copynght (9 2007 EX<1mkrc1Ckers, Inc.
LEClUR" 1:
1'0 lIn(krst~nd hm,v air resistance affects a projectile, compare a bowling balJ (a l11a"ivc prnjectUl') w ith a volley ba ll (a less massi\c projectile'). Propel both of them do\-vn a bowling alley ,I t bowling pins. The bowling pins represent th___' air molL'cu les thtlt creak' ilir resistill1ce. Both experience the snml' resistance, but the \'o IlL-y bellI is deflected to the sidt' while the bO'vvling ball J11O\ 'l'S through the pins li ke they're not there. Air resist.ulCl' has k-ss effect on a more mast:;i\'c object.
1. 11 Equation Summary
v=
%
d = distance v = speed
~ v=
i1
a=
t
i1 = displacement 'Ii = velocity
t= time
Ii = acceleration
t = time
.x = x, + v"t +tat'
x = displacement h = height v;; velocity a = acceleration
v = v, + at a must be constant.
v' = v: + 2ax v avg =
v + v, 2
v= j 2gh
Copyright It) 2001 Cxarnkrackers, Inc.
}
Vo
must be zero.
TRANSLAfiONAL MOTION'
15
21. A projectile is launched at an angle of 30° to the horizontal and with a velocity of 100 mls. How high will the projectile be at its maximum height?
Questio ns 17 through 24 are NOT based on a descriptive passage.
A. B. C. D.
17. If an apple that is dropped from an altitude of 100 m reaches an altitude of 80 m after falling for t = 2 seconds. what alti tude will it be at in t = 4 seconds? A. B.
C. D.
60m 40m 20 m Om
22. Two balls are dropped from a tall tower. The balls are the same size, but Ball X has greater mass than Ball Y. When both balls have reached terminal velocity, which of the following is true? A. B.
18. Two skydivers are playing catch with a ball while they are falling through the air. Ignoring air resistance, in which direction should one skydiver throw the ball relative to the other if the one wants the other to catch it? A.
B. C. D.
C. D.
above the other since the ball will faU faster above the other since the ball will faU more slowly below the other since the ball will fall more slowly directly at the other since there is no air resistance
A. B.
maintain that horizontal velocity when it jumps, how high must it jump in order to clear a horizqntal distance of 20
C. D.
m..) 5m
20. Jg nori ng air re.sistance, if the initial height of a body in free fall is inc reased by a factor of 4, the final velocity whe n it hits the ground will increase by a factor of:
2
B.
4
C. D.
It depends upon the value of the initial heigh t. The velocity will remain the same.
CoPyri9f1t' ~r,J 2007 EX
3 sec 4 sec 5 sec 6 sec
24. A golfer hits a ball with an initial speed of 30 mls at an angle ofAO" to the horizontal. If the ball is in the air for 6 seconds, which of the following expressions will be equal to the horizontal distance traveled by the ball? (Ignore the effect'!; of air resistance.)
10m 20m 45 m
A.
The force of air resistance on either ball is zero. Ball X has greater velocity. The Ball X has greater acceleration. The acceleration of both balls is 9.8 mls'.
23, A hiker throws a rock horizontally off a cliff that is 4Q meters above the water below. If the speed of the rock is 30 mis, how long does it take for the rock to hit the water? (ignore air resista nce, g = 10 mls')
19. If an antelope is running at a speed of 10 mis, and ca n
A. B. C. D.
100m 125 m 250 m 500m
A. B,
C. D.
16
(15)(6)' (cos 40°) m (30)(6)(cos 40°) m (l5)(6)' (sin 40°) m (30)(6)(sin 40°) m
STOP.
Force
2.1
Mass and Weight
Whether moving or at rest, all objects tend to remain in their present state of motion. This tendency of an object to remain in its present state of motion is called inertia. Mass is the quantitative llleasure of an object's inertia. An object's 1113SS tells us how llluch that object will resist a change in its motion. On the MeAT mass is measured in kilograms (kg). H ere in space 1" am v irtually weightless. Yet, my mass is the same as it is on earth,' Regardless
of where I go, my change.
m{lSS
does not
Weight is the gravitational force an object experiences when near a much larger body such as the earth. On the MeAT, weight is measured in newtons (N). An object's weight at the surface of the earth is given by the product of its mass and the gravitational constant g. Thus, the weight of any object at the surface of the earth is 'mg'. Weight and mass are proportional to each other, but they are not the same physical quality.
18
MeAT PHYSICS
2 .2
Center of Mass
When solving mechanics problems it is often convenient to consider an object as a single particle with its mass concentrated at a single point. This can be done without error as long as the point chosen is the center of mass and all forces move through the chosen point. The center of mass of a system is the single point at w hich, for the purposes of a simple mechanics problem, all the mass of that system can be considered to be concentrated. More precisely, the center of mass is the point through which a single force may be applied in any direction causing all points on the system to accelerate equally. For the MeAT it is important that you be able to locate the approximate center of mass of a system. If a system is w1iformly dense, then its center of mass coincides with its geometric center. If the system is not uni-
formly dense, then its center of mass is located toward the denser side from its geometric center. For example, a cube made of 0\1e half lead and the other half Styrofoam would have a geOlnetric center at its middle; however, its center of mass would be located toward the lead side of that geometric center. Notice that the center of mass of an object does not have to be located wi thin that object. For example, a ring with uniform density has a center of mass located at its circumscribed center, a point where there is no mass. The center of gravity of an object is the single point at which, for the purposes of a simple mechanics problem, the force of gravity can be applied to the entire mass. If gravity is the only force, the result will be the same regardless of the orientation of that mass. The center of mass and the center of gravity will always coincide.
7 · Center of mass
The center of mass is the point where, if you could hang your system by a string, your object would be perfectly balanced in any orientation. But center of mass isn't limited to systems with only one object. A system with any number of objects also has a center of mass. For instance, if the spheres in the diagram above were of uniform density, the center of mass would be in the position shown. If these spheres were planets, and if a spaceship were far away, they would appear as a single small dot. The ship would be affected by their gravitational force as if their entire mass were concentrated at the center of mass of the system.
2.3
The N ature of Force
There are only four forces in nature. They are: 1.
the strong nuclear force;
2.
the weak nuclear force;
3.
gravitational force;
4.
electromagnetic force.
Copyright © 2007 Examkrackers, In c.
LECTURE 2 : FORCE . 19
The firs t two are not o n th e MCAT. Thus, all forces on the MCAT w ill be gravitational or electromagnetic. This would make iden tifying fo rces very easy except for one problem . Some electromagnetic forces are difficult to identify. For instance, if a person pushes a book w ith his finger, this force is actuall y electromagnetic.
The electrostatic repulsion between the atoms in the person 's finger and th e a toms in the book create a fo rce that ,ve n aturally think of as being crea ted by contac t. Since it is d ifficu lt to think of s uch contact forces as electro m agnetic, we w ill label all su ch fo rces as 'contact fo rces' ins tead of electromagnetic. Thus, for an y MCAT problem, there are only three possible forces: 1.
g rav itational;
2.
electrom agn etic; a.nd
3.
contact.
Only g ravita tion al and electromagnetic fo rces act a dis tance. These forces are easy to identify. G ravity is u su ally just mg. Electrom agnetic fo rces require a charged object or a m agnet. In order for any other fo rce to be acting o n a system, something must be making v isible contact with the system. Contact fo rces mus t act in at least one of two directions: 1.
perpendicular to a surface; and l or
2.
parallel to a surface.
(An exception is ten sion, which is a contact force that can act in an y direction away from the Object. Tension w ill be discussed la ter in this Lecture.) The perpendicular force is also called the normal force . The parallel force requires friction. Both the normal force and friction w ill be discu ssed later in this Lecture. Let's go back to my 5-step-system. The lhird s tep, choosing a system, is \vhat I want to address here. Choosing: a system is \'l~ry important. A sy~ tL'm (11 n be' any mass or group of ma:-,ses. Define y our :-,ystem
Cd rL'(UlIy.
rn o ther "vords, Il1
tain (h ..1t you knO\v ''\'hat is your sys h.·m ilnd 'what is not. When \<\forking your problem, consider o nly lhe forces that act directly on you r system . Ask yourself " I:-, then-' g l"il\'ity iJcting on my systcm?" Then label it. Ask yoursl'l f " Is m y syst('1Tl ch<:uged or is it (I magnet? " Then lc.l bl' I the e lectromagnetic forces, if any. Finally, look for Jllylhing: to uching your system, and label tl1<' contact forces crealed by those o bjccb. Aftcr this, YOll know that you h<1\,l' included all llw possible fon:l's, and C
20
MeAT
PIIYSICS
For instnl1ce, if we afe in terested in the movement of the box in the diagram below, vve should consider only the forces ilcting o n the box . The top diagram contains all kinJ s of fo rce vecto rs and is nea rl y useless. The black vectors in the bottom di
T;
T,
-f,
2.4
Newton's Laws
Newton's First Law is the law of inertia: an object in a state of rest or in a state of motion will tend to remain in that state unless it is acted upon by a net force.
Newton's Second Law tells us quantitatively that when an object is acted upon by a net force, the change in that object's state of motion will be inversely proportional to the mass (/11) of the object and directly proportional to the net force (F) acting upon the object. In formula, Newton's second law is:
F=ma Newton's Third Law states that, for every action, there exists an equal and opposite reaction. This simply means that when object A applies a force to object B, object A experiences a force of cquallnagnitude but in the opposite direction. Newton's third law forces never act on the same system. H ow c;)n the horse accelerate the cart? No matter how hard the horse pulls, the cart pulls back just as ha rd . How can it possibly move?
Copyright (0 2007 EXdmk racke rs, Inc.
LECTURE
Force on the cart
due to the horse
nl ' \ ·l'r
b
Force on the horse due to the cart
For the <1nsw(~r, choose your systerrl to be the cart on ly ilnd tlll'n dr,-l\v only the forces acting on the C
Copyright «(;) 2007 EX3rnkrackors, Inc.
\york,
2:
FORC E • 2 1
29. If F is the force of air resisrance on an object with mass m moving at a constant velocity, which of the followi ng best describes the acceleration of the object when the force of air resistance is reduced by a factor of 4?
Questions 25 th ro ugh 32 are NOT based on a descriptive passage.
25. An astronaut on the moo n applies a lOa N horizo ntal force to a 10 kg mass at rest on a table. At what rate does the mass accelerate'! (Note: The gravitational constant at the moon's surface is 1.6 mls'. Ignore friction.)
A.
B. C. D.
5 mis' 8 mis' 10 mis' 16 mis'
C. D.
C.
D.
C. D.
-Flm 2
I
C.
I - Flm
D.
3 - Flm 4
4
@
@@ 1
1
1
1
I 2 345
1
1
1
1
6 7 8 9
The center of mass of the system is located at point:
remains constant. decreases. increases. 15 zero.
A.
3
B. C.
4 5
D.
6
31. An airplane's propellers exert a force on the plane of 2500 N to the east. Wind resistance of 500 N acts to the west. If the weight of the plane is 40,000 N, what is the acceleration of the plane?
ON ION lOON 200N
A. B. C. D.
28. A SO kg skydiver and a 100 kg skydiver open their parachutes and reach a co nstant veloc ity. The net force on the larger skydiver is: . A. B.
B.
+ 1
27. A 10 kg mass is in free fall with no air resistance. In order to slow the mass at a rate equal to the magnitude of g, an upward force must be applied with magnitude:
A. B.
Flm
30. The system below consists of three sphe res of equal mass m.
26. A bottle rocket is launched into the air. The black powder, which propels it, burns leaving an exhaust trail mai nly consisting of CO, gas. If the force propelling the rocket is constant, the rate of change in its velocity: (Note: ignore air resistance) A. B.
A.
0.5 mis' to the east 0.5 m/s 2 to the west 0.05 mis' to the east 0.05 mls2 to the west
32. An automobile with a mass of 3000 kg is traveling down a straight flat road at a constant speed of 20 mls. The coefficient of friction between the tires and the road is 0.5. The net force acting on the automobile is:
equal to the net fo rce on the smaller skyd iver. twice as great as the net force on the smaller skydiver. four times as great as the net force on the smaller skydiver. half as great as the net force on the smaller skydiver.
A. B. C. D.
22
ON 30.000N 60,000 N 90,000 N
STOP.
LE
2.5
The Law of Universal Gravitation
Newton's Law of Universal Gravitation sta tes that every mass in the universe exerts an attractive fo rce on every other mass in the universe, and that the force is proportional to both of the masses 111\ and 111 2 and inversely proportional to the square of the distance r between their centers of mass. No tice th at the distance is from th e center of one mass to the cen ter of the other, and not the distance betwee n their surfaces. The formula represen ting the law of gravitation is g iven as follows:
where G is 6.67 X la- II m' kg-l 5-'. This formu la gives the magnitude of the force but not the d irection . The d irection is from the center of mass of one object to the center of mass of the other. According to New ton's third law, both masses exp erience a force of the sanle magnitude. Since this is true, the earth pulls yo u towa rd its center with a force equa I to your weight, and you, in turn, pull the earth toward your cen ter of mass w ith a fo rce also equal to yo ur w eigl1t. Wllen we use the gravitation al acceleration constant g, we conside r the force that the object exerts on the earth as negligible and assume the earth to be stationary. Of course, due to the large difference in Inass, this is a very good assumption . H owever, if we examined the two bodies below, and were asked to find how fast they wo uld accelera te toward each o ther, we would bave to app ly Newton 's second law to each mass, and then add tbe magn itudes of th eir accelera tions.
M ,, =2M,
F.,
F,
M,
IF" = F" :. 12 n., = n,l
• ", boundary A
In other w ords, suppose that the gravita tional force on object A ca used it to accelerate 10 mis' in the direction of object B. Assuming object B is half as massive as object A, although the gravitational force on B is of equal magnitude, object B accelerates at 20 m /s'- These values represent the separate accelera tions of the objects, but the two bodies are accelerating toward each other at a faster rate. To find out how fast the bodies are accelerating toward each other, we mu st add the magnitudes of their ind ividual accelera tion s fo r a val ue of 30 m /s'- Tn other words, object B is accelerating rela ti ve to object A at 30 mi s', but it is accelerating rela tive to a stationary boundary A at on ly 20 m /s'-
Copyright It) 2007 EX<Jrllkrd<.:kets, Inc.
- ?
F
c
•
23
24
MeAT PIIYSICS
Why do thing~ float when in orbit arollnd the earth? H(1\'l' they gone out of reach of carth's ~radt)'? What is th<.· rC<1ch of earth's graYity? By' N('wton's I' been out of rl'.1(:h of t'arth's g'\l\ it)'? Yet, \ye know that thing:-. fhmtt'd Oil Sky I ,
,, , , '
,,,
,, ,
,, I
,,
",
,
.,........ --------
-- ... "
...
,
... ... ... '....
S~Y Lab
'!.-';"§,c. !:Y7
,
Ld
/
I
,,, I
I I
I I I
Copyright
rr;)
2001 lxamkrackers, inc.
LEe- JRF
2.6
Inclined Planes
The inclined plane is a specific topic often tested by the MCAT. There are certain basic characteristics that exist for all inclined planes. Once we understand these chars cteristics, all inclined plane proble ms become trivial. In th e simplest, idea l case (no friction a nd nothing a ttached to the block), the only forces acting on a block on an inclined p lane are gravity p ushing straight dow nward, and the inclined plane pushing back. The force of the inclined plane pushing back against the gravita tional force is ca lled the normal force (F") . The normal fo rce is always perpe nd icular to the surface that applies it. Your diagram of a block on a frictionless inclined plane should look like the diagram on the right. Since gravity and the normal force are the on ly forces acting on the b lock, their sum is ca lled the net force . It is the n et force that should be p lugged in to New ton's second la w to find the accele ration of yo ur system . Notice fron1 the p icture below that vector addition of grav ity and the n ormal force creates a ri gh t triangle. Notice also that this triangle is similar to the triangle of the incli ned plane. Similar triangles h ave equa l corresponding angles. By SOH CA 1-1 TOA we find that the resulta nt vector has a magnitude of IIIgsin9. TI, US the force due to gravity and the normal force of a n inclined plane is always equal to mgsin9 fo r any inclined plane a nd points d irectly along th e plane.
r
Reme mber that II1gsin9 is the vector sum of the weight and the n onna l force. You may not label your system with both nzgsin8 and weight or the normal force, since this would be redundant.
m
Rc s ulta~
III
Vector ad dition of F" +rng
"
~lIlgsine '- -~
___ __ --"1
F"
e You may also notice from the diag ram above that, by the rules of SOH CAI-I TOA, the normal force is "tways equal to mgcos9.
Jpy:iqht, 2007 F. ,1nlkrw(kcrs, Inc
2:
FORCE .
25
26
MeAT
PIIYSICS
\Nhl'neYCf you Sl't' an inclined plane, think mgsinq. This is .,Iwa)',-, the net force down any inclilwd planc dUL' to gra\'ity and tht' norm.,l force. Li"l'wisl', mgcosB is Z1lways til{' normal force. Thesl' formu l;Js work rq!;:ardlcss of tht, a ng le of the plane, F\'l'11 clirn. ~d slIrf.lCl's Cdll PL' thought of as an infinite number of inclined planes,
e
You can rl'll11'll1bl'r l11~sil10
tIll' 111'1SS ~,lidl" do\\'nthl indint,
bCldU'-'C
Be careful: Thl' normal force for
d 111.1S-'; l11o\'ing: dllWIl.,
cur\'(,d
slIrf.1Cl' h.1S
t\\'o jobs:
1. a portion (mgcosO) counters some grll\itatiol1al forces, and 2. tlw rest (111\" / r) must create the centripeldl lKCl'it..'rdtion to ch.lI1gt' the direction of the \'e1ocih"" T he norlTI
By the W
f
m
90<
1
0 -9
'lJ
~
~.
0< 0 <9
m
-. 0°
0 =0
Copyright (0 2007 Examkri.'lckcrs, Inc
L
2.7
u·
). F
. 27
Circular Motion and Centripetal Force
Angular velocity (00) and angillar accelemtion (a) are not tested on the MCAT. We sha ll address them briefly for completeness. Angular velocity is a measure of the speed at wh ich an object spins and is given by the equation: Ol= v/ r w here r is the radiu s at w hich the velocity v is measured. Ang ular velocity is given in radians per second. The following formula converts angular velocity into frequency (j):
f = 00/ 2" The frequency is the number of full rotations per second.
The equations are given only for clarification and are not on the MCAT. We will refe r back to these equations frorn time to time to improve our understanding of certain concepts that are on the MeAT. Okay, the stuff below is what you really need to know about circular rnotion on the MCAT.
Angular acceleration is simply the ra te of change in angular velocity and is given by:
a = air
v2 aC = - r mv 2 FC = - r
Circular motion on the MeAT deals w ith concepts in translationa l lTIotion applied to objects spuming o r moving in circles. Examine the diagram above. An object moving in a circle at a cons tan t speed has an instantaneous velocity v at any given moment. The magnitude of th is velocity remains constant but the direction continually changes at a constant rate. The rate of this change in velocity is acceleration. So, the object's speed never changes, yet the object is continually accelerating. This type of acceleration is called centripetal acceleration (a,) . Cen tripetal acceleration always points toward the center of the circle that is circumscribed by the motion. Since, when an object moves in a circle, the direction of centripetal acceleration is constantly changing, only the magn itude of centripetal accelera tion is constant. By Newton's second law w e know that an acceleration must be accomparlied by a net force. The net force in this case is called centripetal force. Of course, centripetal force always points toward the center as well. The fo rmulae for centripe ta l force and centripe tal acceleration are s],own in the diagrmTI above.
The thing to remember about centripeta l force is that it must be at least one of the three forces: gravity, electromagnetic, or contact forces. Whenever centripetal force exists, there is a lways some other fo rce responsible for it. Most cen tripetal fo rce problems on the MeAT can be solved by equating tl1e centripetal fo rce with the responsible force. For instance, the gra vitational force of the earth causes the moon to move around it. In this case we simply set the centripetal force equation equal to the Newton's gravitational force equation. We do not add the two forces together. These Copyright © 2007 Fxamkrackers. Inc.
An easy way to remember that centripetal force must be created by another force is by thinking about a car on ice . A car on ice cannot turn or drive in a circie because there is nothing to create the centripetal force. What would normally cause the centripetal force that allows a car to turn on pavernent? That is our next topic.
28
MeAT
PHYSICS
b y a string is ano ther exa mple. Here, the centripetal force is created by the tension
in the string. Whenever there is a centripetal force, there will always be a force causing it. But first, bock to "Why do things float when in orbit <.,bout- the t'arth?" The i.mSWl'r is lhe,l they don't. Somcthing: ill orbit about tlw l\1rth is clctuillh cOllst
ground, and ,-lhvay~ by the S,lI11l' d ist
Copyright
(i
2001 FxtHllkrnc (crs, In,
36. Which of the following is true of the magnitudes of velocity and acceleration, as the ball rolls down the slope as shown? Note: Please ignore any centripetal accelera-
Questions 33 through 40 are NOT based on a descriptive passage.
tion.
33. If M is the mass of the earth, m is the mass of the moon, and d is the distance between their centers, which of the following gives the instantaneous velocity of the moon as it orbits the earth? (The universal gravitational constant is given by G.) A.
JGdM
B.
r~m
C.
A. B.
J~m
C.
D.
D. JGM md
A.
'/o
2
A. B. C. D.
4
Is 2s 4s 8s
38. An object of mass m resting on the surface of the earth experiences a force equal to its weight mg, where g is the acceleration due to gravity_ If M is the mass of the earth, G is the universal gravitation constant, and R is the radius of the emth, which of the following expressions is equal to g?
earth will: decrease by a factor of 2. remain the same. increase by a factor of 2. increase by a factor of 4.
A.
B. C. D.
Copyri ~~h t ,:t')
decreases and the acceleration
8
35. If the radius of the orbit of a satellite orbiting the earth is reduced by a factor of 2, the gravitational force on the A. B. C. D.
and the acceleration increase. and the acceleration decrease. increases and the acceleration
37. A box starts from rest and slides 40 m down a frictionless inclined plane. The total vertical displacement of the box is 20 m. How long does it take for the block to reach the end of the plane?
34. The owner of a warehouse asks an engineer to design a ramp which will reduce the force necessary to lift boxes to the top of a 1/2 m step. If there is only room enough for a 4 m ramp, what is the maximum factor by which the lifting force could be reduced.
B. C. D.
The velocity The velocity The velocity decreases. The velocity mcreases.
2007 Exarnkrackcrs. Inc.
29
GMm R' GM R' GMm R GM -R
GO ON TO THE NEXT PAGE.
40. A box rests on an incline. ·Which of the foll owing
39. A jogger is running on a circular track with a rad ius of 30 meters. If the jogger completes one trip around the track in 63 seconds. what is her average speed?
A. B. C.
D.
describes the forces on the box as the ang le of inclination is increased?
A.
0 mls I mls 2 mls 3 m/s
B.
force perpendicular to the ramp also increases.
C.
D.
Copyrig hr @ 2007 Exarn kracke rs, Inc
The force parallel to the ramp increases and the force perpendicular to the ramp decreases. The force parallel to lhe ramp increases and the
30
The fo rce parallel to lhe ramp decreases and the force perpendicular to the ramp also decreases. The force parallel to the ramp and the force perpendicular to the ramp remain constant.
STOP.
LECHJRf
2.8
2:
FORCE .
31
Friction
Any object tha t contac ts your system ma y apply forces in two directions: 1.
the normal force is always perpendicular to the con tact surface;
2.
a frictional fo rce is always parallel to the contact surface.
Friction is ca used by the attractive molecular forces between contiguous surfaces. Since the forces are attractive, friction opposes the relative Inction between contiguous surfaces. There are two important types of surface-to-surfa ce friction on the MeAT: static and kinetic frict ion. Static friction if, ) is the force opposing motion when two contiguous surfaces are not moving relative to each other. If yo u lay a block on an inclined plane and the block does not slide down the plane, it is the static frictional force that p revents it from sliding. Kinetic friction if.) is the force resisting motion once the two contiguous surfaces are sliding relative to each other. A block of wood sliding dow n an inclined plane moves more slow ly than a block of ice sliding d own the same plane, because the kinetic frictional force is greater on the wooden b lock. For any two surfaces, there are two coefficients of friction (J.!. and J.!.) ,which represent the fractions of the normal force tha t will eq ual the static and kinetic frictional forces. Thus the formulae for static friction and kinetic fri ction are respectively:
and
Since friction is usually a frac tion of the normal force, the coeffi cients of friction genera lly have a value less than one. In addition, ~, is g reater than 11k' Imagine pushing a heavy object. Once the object is moving, it is usually easier to push. This is due to ~, being grea ter than ).l ,. When faced with a friction problem on the MeAT, first decide if your system is moving relative to the surface creating the friction . If it is, then use kinetic friction. If not, then use static fri ction. If yo u d on't know, then calcu late the cOlnponent of the net force (excluding friction ) on your system that is parallel to the surface creating the friction. Next, compare that calculated net force to ~,F" . If the calculated net force is smaller than ~,F", then your system is probably' not sliding along the surface; static friction holds it in place. In this case, Sllice there is no acceleration, the static fri ction is equ al and opposite to the calculated net force. No tice that the static fri cti on could never be greater than the calculated net force. If the calculated net force is greater than ~l", then yo ur system must be sliding along the surface. In the later case, ignnrp. th p. sta ti c friction ann sllhtrrlct tlw mngn.itude of the kinetic frictional force frOJn the calculated ne t force to arri ve at a new net force tha t includes friction. There a re other types of fric tion, such as drag (i.e. air resista nce), which is fluid resistance to an object's motion through that fluid , a nd viscosity, which is a fluid 's resistance to mo tion through itself. On the MeAT these other types of friction will be d ea lt with onl y qualitatively or else a form ula w ill be provid ed for plug-n-chug calculations. We will discuss the qualitative effects of drag and viscosity in Physics Lecture 6. >I- We Si'ly "probably" because if the surfaces were already slid ing relative to one another, the calcu lated net force could still be less than ~l ln. Remember ~lk is usually less than f.!s'
Copyright ((,) 2007 FXi'lmkr,x:kpr<;, Inc
Friction does NOT oppose motion; it opposes relative motion. Always draw a fnctional force vector pointing in tile direction tllat would prevent surfaces from slid ing past eacll other. For instance, the frictional force on the front tires of an accelerating front wheel-drive car points in tile direction of motion of tile car because the force prevents the tires from sliding backwards on the road.
32
MeAT
PHYSICS
2.9 Tension is actually beyond the scope of the MCAT. However, they test it in exactly the way explained above. When you see tension on the MCAT, Just replace the rope with a force vector acting on your system. To see where the confusion may result take a look at the mass hanging from the
Tension
For the MeAT, think of tension as a force acting through a flex ible object w ith no m ass, such as a stri ng or rope. (We shall refer to all these objects as simply ropes.) Tension is equa l th roughout a rope as long as there is no friction ac ting on the rope. At any poi nt in a rope there is a tension force pulling in equal and opposite d irections. We onl y usc the force p ulling awa y from OUf system. Tension requires an equal force at both ends of the rope, and the tension in the rope is equal to only one of the forces, not both . This is tricky, but remembe r, the rope has no mass. Thus, if a net force we re a pplied to only one end of a rope, it would accelerate at an infinite rate.
string be low. We know that the tension in
the rope is mg. But is it really mg, or do we need to add the fo rce of the ceiling pulling up on the rope with a force mg and the box pulling down on the rope with a force mg? And if we add these forces do we get zero or 2mg?
LT j mg
The answer is that the tension in this case is simply mg.
m
~ mg 2.10 Hooke's Law One more ilnportant force on the MeAT is the force due to a cOlnpressed or stretched object followi ng Hooke's law. When de form ed , solid s tend to 'remember ' the ir shap e a nd reform to it. Hooke's law describes the force applied by most objects against a defo rming force . This force is directl y proportional to the amount of deforma tion o r, Juore p recisely, the change in position (Lit) . Hooke's law is given by the followin g equation:
F = -kth The most common MCAT questions concern ing Hooke's law deal with springs. 'k' is often referred to as the 'spring constan1'. The negative sign in the formula can usually be ignored for tile MCAT.
where k is a cons tant unique to a given object. The nega ti ve sign indicates that the force is in the opposite direction of the displacement. Most solids follow Hooke's law to some extent. All solids violate Hooke's law at some limit of displacement, unique to that object. The point of viola tion is called the yield poillf. When an object is defo rmed beyond its yield point, it loses son1e of its 'memory' and will not regain its original shape. At so me grea ter displacement, the object will reach a fmc tllre poillt and break.
l.c')yriqh
L)OJ E 3lnkl3.crs tn(
LECTURE
2:
FORCE
'
33
i 111g
·----1-·-----.. .
!II . .ILt
r6.~y
..... . 111
~ 111g On the MeAT, Hooke's law is most often applied to springs. The force F is reany the tension in the spring and ~x is the change fTmTI its rest position. For instance: the spring shown above has a spring constant Ie ~ mg / I1x.
Assurnlllg that my head follows Hooke's law, tile force that it produces against til is vice is equal to the change in its thick-
:11 :
i+' t.x / · ill
F
ness, l\X , times some constant, k'; 3It y' which is specific to my head . The change In the thickness of rny head is negative because I'rn getting thinner. If I were being stretched, the change in my thickness would be positive and the force I create would be in the other direction, Accord ing to Newton's third law, the vice appl ies an equal but opposite force against rne. That's the one that hurts,
F = - k"",t.x
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34
MeAT
PH YSICS
2.1 1 Equation Summary Newton's Second Law F = ma
The net force applied to the center of mass of a system always . equals the mass of the system times its acceleration.
Gravity
m1 m2 r
F=G - . ,-
The force of gravity is proportional to the mass of each body and inversely proportional to the square of the distance between their centers of gravity. G is a universal constant.
w
Inclined Planes: F=mgsin8 F" = mgcos8
The sum of the normal force and the force of gravity is mgsin8. The normal force is mgcos8.
Circular Motion An object moving in a circle at constant speed v experiences a centripetal acceleration that is proportional to the square of its speed and inversely proportional to the radius of the circle which it circumscribes. Some force F, must be applied to an object in order to give that object a centripetal acceleration.
Fiction / ,":; fl,Fn / k =flk F"
Hooke's Law F = - kru:
Contiguous surfaces may exert equal and opposite forces against each other parallel to their contiguous surfaces. If the surfaces do not slide relative to each other, this force is static friction. If the surfaces slide relative to each other, this force is kinetic friction.
When deformed, objects obeying Hooke's Law will exert a force proportional to their deformity. k is a constant uniqu e to the object.
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(~>
200/ Exarnkrackers, Inc.
44. On a particular stretch of wet pavement, the kinetic coefficient of friction ~ for a patiicular car with mass m is 0.08. If the car is moving at a velocity v, and suddenly locks its wheels and slides to a stop, which of the following expressions gives the distance that it will slide?
Questions 41 through 48 are NOT based on a descriptive passage.
41. If the rear wheels of the truck pictured below drive the truck forward, then the frictional force on the rear tires due to the road is:
A.
L mgf1
,
B.
v
2mgf1
,
_v_
C. 2gf1
A B A. B. C. D.
D.
kinetic and in the direction of A. kinetic and in the direction of B. static and in the direction of A. static and in the direction of B.
45. In order to test the strength of a rope, one end is tied to a large tree and the other end is hitched to a team of 2 horses. The horses pull as hard as they can, but cannot break the rope. If the rope is untied from the tree and attached to another team of 2 horses with equal strength, and the two teams pull in opposite directions, the tension in the rope will:
42. If a rope capable of withstanding 900 newtons of tension is attached to a wall as shown, what is the maximum force that can be applied in the direction of F before the rope will break?
A.
300 N
B.
450 N 900 N 1800 N
C.
D.
43. In many harbors, old automobile tires are hung along the sides of wooden docks to cushion them from the impact of docking boats. The tires deform in accordance with Hooke's law. As a boat is brought to a stop by gently colliding with the tires, the rate of deceleration of the boat: A. B. C. D.
A. B. C. D.
is constant until the boat stops. decreases until the boat stops. increases until the boat stops. increases and then decreases before the boat stops.
Copyright (i') 2007 Exarnkrackers, Inc.
decrease by a factor of 2. remain the same. increase by a factor of 2. increase by a factor of 4.
46. A child on a sled is sliding down a hill covered with snow. The combined mass of the child and sled is m, the angle of inclination of the hill is 8, and the coefficient of kinetic friction between the snow and the sled runners is ~. Which of the following expressions gives the frictional force on the sled?
35
A. B. C.
f.l111gcos8 f.l111gsin8 f.l111g
D.
mg
GO ON TO THE NEXT PAGE.
47. The diagram below shows two different masses hung from identical Hooke's law springs. The Hooke's law constant k for the springs is equal to:
48. In a very tall building. an elevator with weight W moves quickly upward at a constant speed. The entire weight of the elevator is supported by a single cable. The tension in the cable is: A. B. C. D.
greater than W less than W equal to W dependent on the speed of the elevator.
1.5 kg
A. B.
c. D.
2 N/cm 5 N/cm 10 N/cm 20 N/cm
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36
STOP.
Equilibrium, Torque and Energy 3.1
Equilibrium
Equilibrium is a fancy word for no translational (straight line) or angular (rotati o nal) acceleration. Stated another way, a system is in equilibrium if the translational velocity of its cen ter of m ass and angular velocities of all its parts are constant (i.e. it is m oving and rota ting at a constant velocity). If all velocities are zero, then the syste m is in static equilibrium. If any velocities are n onzero, but all velocities are also constant, then the system is in dynamic equilibrium. Remelnber, equi librium does not m ean ITIotionless; it means cons tant velocity.
force
due to gravity Normal Force
Force
Force
of air resistance
Static Equilibrium Salty
Dynamic Equilibrium Salty
Velocity is a constant zero.
Velocity is constant and not zero.
For
38
MeAT
PHYSICS
For all systems in equilibrium, the sum of all the forces acting on the system equals zero. In other words, the net force acting on a system in equilibrium is zero. A reliable and much Simpler method of viewing systems in translational equilibrium on the MCAT is as follows: The sum of the magnitudes of the upward forces eqllals the sum of the magnitudes of the downward forces, and the sum of the m agnitudes of the rightward forces eqllals the sum of the m ag nitlldes of the leftward forces. This method allows you to use only positive numbers for all your forces; it is no longer necessary to decide if g is positi ve or negative 10. g is al ways positive wi th this method . This is not the method that you learned in physics class, but it is faster and more intuitive for simple proble ms. More importantl y, it is the best method for the MeAT. Thus the formulae that yo u must know for a system in equilibrium a re: Fupward
= Fdownward
Frightward
3.2
=
F leftward
Systems Not in Equilibrium
If a syste m is not in equilibrium, it s imply m eans that the center of n1ass is acceler-
ating tra nslationally or its pa rts a re accelerating rotationally. The MeAT does not test angular acceleration, so a system not in equilibrium on the MCAT must be exhib itiJ1g translational accele ration . For a system not in equilibrium, the sunl of the forces equals the mass of the system times its acceleration or LF = 111a. On the MeAT there is a fas ter and more effective way to solve these problems. When faced with any system not in equilibrium, follow these two steps: 1. write the equations as if the system were in
equilibrium; 2. before solvi ng, add '/1/n' to the side with less force. Again, this method makes all numbers positive. Just after I jump from a plane, 1 [1m not in equilibrium . The force upon mC' due to gravity is greater lhan the force due to it ir re~istZlIlCt>. In order to find my acceleration, I put nIl up,-,vard forces on one side Force of gravity
of th e equation nnd ,111 downward forces on the other. Since I am not in equilibrium, thl' two sides arc not equal. f"p\\.ud
Now
r must
0 F"" Wll"",·, 1
decide which side h(b grl'cller force.
Si nce 1 am accelerating: downward, the down"vilrd
forces mllst be greater. In order to b.dancc the tvvo s id es of my cqU<1tiOll, Tmust add ' m,l' to the wenker side. Force of air resistance Now the two sidl':-i an.' equal and I
Cclll :-'OIV l'
for ac-
celeration.
Copyright ~;\ 2007 lXCllllkrackers, Inc.
53. The arrows shown below represent all the force vectors that are applied to a single point. Which of the following could NOT be true of the point? (Note: sinl50° ~ 0.5;
Questions 49 through 56 are NOT based on a descriptive passage.
cos 150 0
A.
ON 750 N 1500 N No amount of tension in the rope could make it perfectly straight.
2F
A. B. C. D.
50. A rescue helicopter lifts a 50 kg rock climber by a rope from a cliff face. The rock climber is accelerated vertical1y at 5 m/g 2 . What is the tension in the rope? A. B.
e.
D.
A. B,
C, D.
•
B.
C. D,
e.
ON 200 N 400 N 600 N
D.
52. A skydiver jumping from a plane will accelerate up to a maximum velocity and no greater. This constant velocity is known as terminal velocity. Upon reaching terminal velocity, the net force on the skydiver is: A. B.
e.
D,
zero and the skydiver is in equilibrium. zero and the skydiver is not in equilibrium. equal to the weight of the skydiver and the skydiver is in equilibrium. equal to the weight of the skydiver and the skydiver is not in equilibrium.
© 2007 Exarnkrackers, Inc.
northeast northwest southeast southwest
A car starts from rest and reaches a speed of 80 kmlhr after 15 seconds. A bucket is lowered from a rooftop at a constant speed of 2 m/s. A skater glides along the ice, gradually slowing from 10 mls to 5 m/s. The pendulum of a clock moves back and forth at a constant frequency of 0.5 cycles per second.
56, A child pushes a block across the floor with a constant force of 5 N. The block moves in a straight line and its speed increases from 0.2 mls to 0.6 m/s. Which of the following must be true? A.
B. C. D.
Copyri:~h t
moving at a constant velocity. not moving. accelerating at a constant rate. not accelerating.
55. Which of the following describes a situation requiring no net force? A,
A.
The point is The point is The point is The point is
54. There are 3 forces acting on an object. Two of the forces are of equal magnitude. One of these forces pulls the object to the north and one pulls to the east. If the object undergoes no acceleration, then in which direction must the third force be pulling?
350 N 500 N 750 N 1500N
51. The pulley shown below is old and rusted. When the 50 kg mass is allowed to drop, the friction in the pulley creates a constant 200 N force upward. What is the tension in the rope?
B.
-
2 )
49. A circus tightrope walker wishes to make his rope as straight as possible when he walks across it. If the tightrope walker has a mass of 75 kg. and the rope is ISO m long, how much tension must be in the rope in order to make it perfectly straight?
B. C. D.
~ .fl
39
The force applied by the child is greater than the force of kinetic friction between the block and the floor. The force applied by the child is less than the force of kinetic friction between the block and the floor. The force applied by the child is greater than the force due to the weight of the block. The force applied by the child is less than the force due to the weight of the block.
STOP.
40
MeAT
PHYSICS
3.3
Torgue
Torque (~) is a twisting force (MCAT definition) . Although torque is a vector, the MCAT allows you to think of torque as being clockwise or counter-clockwise. Torque is the vector product of both a force vector F and a position vector r. Since this is vector multiplication and the result is a vectoT, the magnitude of the resultant vector Inust include the sine of the angle between the original two vectors (see
Physics Lecture 1). The magnitude of torque is given by the following equation: ~
where e is the angle between the force and the position v ectors. In this equation, the position vector is the distance from the point of rotation to the point of application of the force. The point of rota-
/.-,
r~
• Rotation point
"
\ r
F
J
= Fr sine
~
l
tion is any fixed point of your choosing. It is convenient to choose the position vector to be from the point of rotation to the point where the force acts at 90°. Such a position vector is called a lever arm (I) . vVhen the lever arm is used, the equation for torque becomes:
't =
Fl
Compare r and I in the diagram to the left. Any prob lem on the MCAT involving torque, will be a statics problem. Therefore, use the follow ing three formulas in the order given to solve any MCAT torque problem:
F rightwa rd
'tclockwise
=
=
F leftward
'tcounter-cIockw ise
In the picture to the right, what is the distance from the left end of a massless board from which I need to hang the we ight in order to establ ish equilibrium?
Copyright
@
200; Lxarnkrackers, Inc.
LECTURE
Lt.·t's aSSUl1le th
3:
EQUILIBRIUM, TORO"E AND ENcRG'
.
41
L
d
x
m);
r
t
.. mg Fur
~I =F ,~ .. ·n
Now I go to my third l'quation . But I need to know the torques on my system. In order to find the torques, I must choose a point of rota tion . I can choose any point that 1 wdnt, but, s inct.:' I nccd all the forcl's to act at l)O degrcc.'s to their Ic\'cr ,UIllS, I will choose a point Oil the board . Let's say that I usC' thc left end of lhc board as Illy point of rotation. Now .1 dr~l\Y my ciock\.vis(' and counter-clockwise torques.
" = Tx ~"" '" ...... Point of rotation ..
'\ \.
--~~
" = mgd !l- ~===
~~\~ (; I ;' (" ,
(
\0
~
"
" = m,gL ~ ......... '"
c..opyr\Jht ~) 2007 EX
<;,
Inc
P
.
,,'
.,'
To do this, I hold my system motionless only at the point of rotation, and push it around that point in the direction of the force . Finally I set the clockwise and counter-clockwise torques equal to each other. Now I have two equations and two variables. I simply plug and chug.
59. If al1 of the forces below have equal magnitude, which one creates the most torque?
Questions 57 through 64 are NOT based on a descriptive passage.
A.
c.
B.
D.
57. A telephone pole stands as shown be low. Line A is 4 m off the ground and line B is 3 m off the ground. The tensions in line A and line B are 200 Nand 400 N respectively. What is the net torq ue on the pole?
,line A
F
IU
°
line B U
4m 3m
A. B. C.
D.
60. A one meter board with uniform density, hangs in static equilibrium from a rope with tension T. A weight hangs from the left end of the board as shown. What is the mass of the board?
ONm 400Nm 800Nm 2000 Nm
0.2 m
58. A sign hangs by a rope attached at 3~" to the middle of its upper edge. It rests against a frictionless wall. If the weight of the sign were doubled. what would happen to the tension in the string? (Note: sin 30° = 0.5; cos 30° = 0.87)
T
3 kg
A. B. C. D.
I 2 3 4
kg kg kg kg
61. Boards X and Y are both massless and 4 m in length. A 4 N force F is applied to board Y as shown. Board X is held stationary_ The two boards are nai led togelher at I m from the left e nd of board Y. If the boards do not move, what is the static frictional force between th e nail and board X?
SIgn
-
1m
A. B. C. D.
It It It It
would would would would
remain the same. increase by a factor of 1.5. increase by a factor of 2. increase by a factor of 4.
[X A. B. C.
D.
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42
4N 8N 12 N 16 N
[y
ii
:! -I
U
~
GO ON TO THE NEXT PAGE .
62. A person pushes on a door and it swings open. Where should the force be applied in order to make the door swing open as quickly as possible? A. B. C. D.
64. A carpenter who is havi ng a difficult time loosening a screw puts away his screwdriver and chooses another
with a handle with a larger diameter. He does this because:
On th e edge of the door nearest the hinges. At the center of the door. On th e edge farthest from the hinges. A force anywhere on the door will have the same effect.
A. B.
decreasing force decreases torque.
C. D.
increasing lever arm increases torque. dec rea<; ing lever arm decreases torque.
increasing force increases torque.
63. A student with a mass of 40 kg sits on the end of a seesaw with a total length of 10 meters as shown in the picture.
l
i
54
t
3
i
2
Iii 2 3 4
i
5
How far to ri ght of the center of the seesaw should a student with a mass of 50 kg sit to achieve the best balance? A. B. C. D.
1m 2m 4m 5m
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43
STOP.
44
MCAT
PHYSICS
3.4
Energy
Look inside most basic physics textbooks and yo u w ill find the sta tement "Energy is the capacity to do work." This statement, w hich is inaccurate, is an a ttempt to define energy. It is inaccu ra te beca use a system can ha ve energy and s ti ll have no capacity to do work. There is no satisfactory defin ition of energy. Energy is a manmade concept designed to assist us in understanding our uni verse, The best way to understand energy is to work physics problems. For now, thin k of energy as you ha ve always thought about it. For instance, you have an intuitive idea of what is meant by the statement "He is full of energy today." Use tha t intuition about en ergy when you work physics problems. The units o f energy used on the MeAT are the joule 0) for macroscopic systems and, for microscopic systems, the electron-volt (eVl. One joule is one kg m 2 /s'. Energy is a sca lar. Thus energy u sually provides the most convenien t method by w hich to solve mecha nics problem s. Wh enever you h ave a mechan ics problem on the MCAT, always check first to see if you ca n solve it using conserva tion of energy, w h ich we sha II discuss below. Energy can be d ivid ed into m echanical and nonmechanical energies. Mechanical energy is the kinetic en ergy and potential energy of macroscopic systems. A macroscopic system is a system that yo u can examine without a microscope. Kinetic energy (K) is the en ergy of motion. An y m oving mass has a kinetic energy given b y the equ ation:
1
K = - mv 2 2
Potential energy (II) is the energy of position. All potential energies a re position dependen t. The re are several types of potential energy. The most important types on the MeAT are gravita tiona l potential energy (Us.) a nd elastic poten tia l energy (U,). (Electrical poten tia l energy w ill be d iscussed in Physics Lecture 7.) Gravitational potential energy (Us) is th e energy due to the force of gravity. Gravitationa l potential energy between any two nlasses is given by U A = - GII/ 11112/ f, where G is the universal gravitational constant, 111 1 and 111 2 arc the two masses, and r is the distance between their centers of gravity. The negative sign indicates that energy decreases as the distance b etween objects that are attracted to each other decreases. A limited form of this equation, more useful on the MeAT, gives the gravitational potentia l energy of an object near the earth's surface. This form ula is:
w h ere In is the mass of the object, g is the free- fall acceleration at the surfacf' of thp earth, and II is the height of the object or systen1 above some arbi trary point. Elastic potential energy (U.> is the en ergy due to the resistive force app lied by a deformed object. The clastic p oten tial ene rgy fo r objects following Hooke's law is given by th e formula: 2 u e = !k/tt 2
w here k is the Hooke's law constant for the object, and LlX is the displacement of the object from its relaxed position. ( .)ryright ( ?C07 FX,lln ,rack ,rs, Inc
LECTURE
3.5
3:
EGUllI['RIUM,
stant. lonSCn",ltion or l'IWI)';! d oes n ot SdY lhdt a certdin type of l'llcrgy (i .C'. kinetic or
potl'llti.1i) mllst be consl'ITl'd; it statl's lh
sy~t(' m
onlv ,-lS \'1/01''' o r hl\lt. (Work is d isCHSSl'd next. Hl',li
is disclissed in Clll'mistr) LectU[l' J)
Wo rk
There are only two types of energy transfer: wo rk and heat. Work (W) is the transfer of energy via a fo rce. Heat is the transfe r of energy by natu ral flow from a warmer body to a colder body. (See Ch emistry Lecture 3 for mo re on hea!.) Thus a ll work is energy transfer, but all ene rgy transfer is not work. By 'transfer', we mean tran sfer from the system to the sur roundi ngs or vice versa. Therefore, the amount of work done will depend upon w hat we choose for our system. [This Lecture will not consider pressure-vol ume work (PV work). PV work is discussed in Chemistry Lecture 3. ) Work is
0
sco lor and is measured in uni ts of energy (joules).
The work done by a ny fo rce other than fricti on is: W
=
Fdcos9(for all forces except frictiun)
w he re F is the force on some system, d is the d isp lacement of the system, and 8 is the angle between F a nd d. This equa tion gives the e ne rgy tra nsferred into a system due to a force. The force may be one of many forces acting on the sys tem or it m ay be the n et force.
Frictional forces are an exception to the equation above beca use frictional forces change in ternal ene rgy as well as mechanica l energy. (Internal energy is the energy o f ind ivid ual molecules. Unfortunately, MCAT wi ll probably call this 'heat energy' o r, worse, just 'h eat'. Inte rn al energy is discussed in Chemistry Lecture 3.) J.f the total energy transfer is due to forces and none to hea t, the work done on an object is also given b y: W
= AI<. + fl.U +
AEi (no heat)
If there is neither h eat nor friction: W
C )pvrighl
ANI) ENERGY
45
Systems
Before we ca n ta lk about energy transfer, we need to h ave some understanding of systems. A systelll is a ny d efined area that we choose to consider separately from the rest of the universe. The rest of the uni verse is called the surroundings. Togethe r, mass and energy define the three basic systems in physics: the open systern, w he re energy and mass are exch anged w ith the surroundings; the closed systelll , w here energy is exchanged with the surroUJldi ngs but mass is no t: the isolnted system, w here neither energy no r mass is exchanged w ith th e surround ings. By defin ition, a lthOLlgh the form of energy in an isolated system m ay cllange, the energy of an isola ted system is conserved. Thus the Law of Conservation of Energy states that, since the uni verse is a n isolated system, the energy of the universe relnains con-
3.6
TOROL'e
I
;7007
Ex"mkr,1Ck(~IS.
= AI<. + fl.U(no friction, no heat) 1m
Some MeAT questions will be solvable by vectors or with conservation of energy. It will be much faster to solve them using conservation of energy. so always try to solve a mechanrcs problem first bv using conservation of energy.
46
MCAT
PHYSICS
where K is kinetic energy, U is any potential energy, and E; is internal energy. This makes sense in terms of conservation of energy. Since work and hea t are the only two types of energy transfer, when there is no heat, work must be responsible for any net energy change, and therefore must equal the sum of all energy changes. If there is no fri ction, all energy change is in the form of mechanical energy. In any physi cs textbook you will also see: W=L'.K
This is the Work-Energy Theorem. It is only true when all energy transfer results only in a change to kinetic energy. Tn other words, it is a very limited case of the previous equations, and is not very useful for the MCAT. The :,i mplest way to undcrstnnd "'lurk is to rcrnember th e first law of thcrmodyn,1I11ics: Fncrgy is ~l lways consl~rve d, Of
M:= W+q whL'rc
q is heat ~lI1d l1E is the tot<:11 change in energy of d closed system. This simply
Silys tlli.lt lhere arl' only h,vo \V.:l)'s thi.lt energy G111 lcd\'C or entcr i.l system: \vork dnd hl'd t.
Now, if you want to kilO"" if work is donc, do the following: Define your system . H your system is the ~i.lme tcmperature d S its surroundings then there Ci:ln be no hC'dt. Any energy ch.1ngc to sllch a system must bl' accomplished through work . Slim th e clli.ll1gC in enl"rgy ('md YO LI hd vc the vvork done on the sYSh'tll . if your system is not the 5<1111(' temperature as the surroundings, then heat must be considered and YOLI ha n ..' d therl11 odynilll1ics prob lem. IC;J\·eat: Challge ill /('I/lpcmlll/,c i~ 1101 111(' ~mll(, fllillS a.~
11m/.)
3 .7
Conservative and Nonconservative Forces
Conservative forces are called conservative because the mechanical energy is conserved within the system. If a force acts on a system as the system moves from point A to point B and back, and the total work done by the force is zero, the force is a conservative fo rce. Thus/ the net work done by any conservative force on an object moving around any closed pa th is zero. A second way to recognize a conservative force is that the energy change is the same regardless of the path taken by the system. It is a necessary but not a sufficient condition that conservative forces be fu nctions
of position only. In other words, the strength of a conservative force is dependent solely upon its position. For instance, the conserva ti ve force of gravity upon an object is dependen t upon its position within a g ravitationa l field; the conservative Hooke's law force is dependent upon the position of the sp ring or object creating it. The work done against conservative forces is conserved In potential energy; the work done against non conservative forces is not conserved.
Conservative forces have potential energ ies associated with them. Conservative forces do not change the mechanical energy of a system. Thus the Law of Conservation of Mechanical Energy states that when only conservati ve forces are acting, the sum of the mechanical energies remains constant:
Written another way:
Copyright (9 ;?007 Exalllkr(lcbrs, Inc.
LECTURE
3:
EOUIl.l8RIUM, TOROUE AND ENERGY
47
Warning: If a question asks, "How much work is done b y gravity?" (or any other conservative force), the question itself implies that gravity is not part of the system. There are three m e thods to answer such a question: 1) use Fdcose; 2) simply calculate the change in t.Ug; or 3) use: W ~ t.K + t.U + t.E, but do not include gravitational potential energy in your calculation of t.U. Technically speaking, a conservative fo rce doesn 't do work because energy is never lost nor gained by the system. Con sen'ati ve forces do not change the temperature or the internal en ergy of an object to w hich they are applied. Gravitational fo rces, Hooke's law forces, and electric and magnetic field forces, are the conservative forces that you're likely to see on the MCAT. Nonconservative forces are forces that change the mechanical energy of a system w hen they do work. Examples of nonco nservative forces are kinetic frictional forces and the pushing and pu lling forces applied by animals. For instance, if a human lifts an object from rest to a height 'il', the total mechanical energy of the object has changed. On the other h and, if an object were propelled by its kinetic energy to a height 'h', its total mechanical energy would remain constan t. Except for frictional forces, the work d one by all nonconservative forces equals the change in the mechanical energy of the systems upon which they are applied. This result is described by the equation: W = tl.K +
.6.U(nunCOnSC TV
No tice that this is the same equation as given for one of the d efiniti ons of work. This is because conservative forces don't do work. Compare this equation to the equation for the change in mech anical energy when only conservative forces act.
3.8
Work and Frict ion
Kinetic frictional forces in crease the internal energy of the systems to w hich they are applied. Thus, the entire amount of work done by such a force does not go into changing the mech an ica l e nergy. When YOll rub your hands together to warm them, YOll are doing work, via kinetic friction , which increases their interna l energy. There is no heat because your two hands are at an equa l temperature throughout. In order to find the work done by a kinetic frictional force, we must consider the internal energy. Imagine a box sliding to a stop alon g a tabletop. Kinetic friction h as done negative work on th e box; the force decreases the kinetic energy of the box. The mechanical energy change of the box is given by :
However, the box increased its internal ene rgy, so its net energy loss is not Ikd cose. Since the box and table are at the sam e tempera ture, there is no heat in this problem; all en ergy ch an ge is due to work. ThllS the wor k done by friction is not I kd case. The energy change of the box (the work done by friction) is the change in its kinetic energy, which is n egative, and the change in its internal energy, which is positive. The work done on the box is W ~ t.K + t.Ll + t.E,. The kinetic energy of the box became internal energy of the box and inte rna I energy of the table.
Copyright
if:)
2007 Examkrackers, Inc.
This is a tough topiC. A famous physicist once wrote "There are no nooconservative forces," meaning that. of the four possible forces in nature, all are conservative. However, on a macroscopic scale , mechanical energy is changed when certain forces are acting. These forces we call non conservative. It is possible that the MCAT might ask you to identify conservative and non conservative forces. But the most important thing to understand is how they affect work. If you already understand work, and can do most MCAT problems involving work. then it may be best not to worry too much about conservative and nonconservative forces. The work done against conservative forces is conserved in potential energy; the work done against nonconservative forces is not conserved.
48
MeAT PHYSiCS
SIi ding
j, ~ v~
----"-+
ION 10 m/s
Eifinal ofOO., = 70 J E i in ilial <>f t.bk
=
0J
Stopped E i inilialofbo,
KiniLi,,1
Kf,"ol
= =
lJ2
=
0J
m/ = 100 J
1S mv
2
=
0J d
"
Wdonc by frictiQ n = nE1O!O I for . ;Iher system = 30 J J¥doneonbox = 11K + Miofbox = -30 J
W"':me on table =
llEi uf lablc
=
30 J
Total energy of the isolated system of the table and box is conserved. The energy transfer into the closed system of the box is equal to the work done on the box. Notice that work done by friction can only be found if the change in internal energy is known.
Work and Friction
Wo rk done by friction is a sub tle point and is explained Ilere on ly because, witllout it, tile law of
conservation of energy is violated and some basic concepts, such as heat and work, are obscured . The MeAT wi ll probably not add ress tllese subtleties.
CorVliqht
(iJ
7()O! EXC!lnl;rackcrs, inc
LECTURE
3.9
3:
Ec.. JILI!;RllJM, TORQUE IIN[, ENERG'
Examples of Work
If we look at the diagram below, we have the following:
i-'---p-o-s-~-ti-0l-1-1--. . !
-
- - - - - J.._ -_p-o-sM -lt-lo- n- 2-....J.
d
A fo rce F acts on mass M along a frictionless surface resulting in a displacement d. It is important to reali ze that the force is ac ting through the entire displacement. The mass moves from position 1 to position 2. Many concepts can be appreciated by this simple display. First, since a force is applied resulting in a transfer of energy from the applicator of the force (whomever or w hatever that might be) to the mass, work is done. The vertical component of the force was apparently too small to move the mass off the horizontal line. TIms the vertical d isplacement is zero, and the vertical force component does no w ork. Gra vity and the norma l force are 90° to the displacement and also d o no work. The ho rizontal component of the force, however, moves the mass a displacemen t of d. To find the work done by the force, we w ould use W = Fdcos60". (Notice that Fcos60" is the horizontal component of the force .) The mass does not change height, so there is no change in potential energy, U. Thus, the work done goes completely into changing kinetic energy. The change in kinetic energy is equal to the wo rk.
Consid er the physical m anifesta tions of work in the example above. In other w ords, since work is a transfer o f energy, what are th e physical changes to a mass as a result of this energy transfer? To test yourself, illlagine the same force acting on the box at an angle of 30". How would this affect the work done on the box? Would one fo rce do more work than the other? If a d ifferent amount of wo rk is done in each case, then we should be able to see this d ifference in physical quantities. What wo uld be the physical manifesta tions of the difference in work done?
A1'---_M~ position 1
M
position 2
Ii Sin ce the force applied at 3~'' has a grea ter horizontal component, it does more work. This grea ter work would manifest itself in greater acceleration throughout the d isplacement, greater velocity at the end of the d isplacement, and less time required to achieve the displacement.
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(c)
2007 EX<JmkrclCk(>r<;, Inc
.
49
50
MeAT
PHYSICS
3.10 Summary of Work When faced \,vith ~l problt.:'ln involving work, follow my 5 step systL'm gi\,L'11
J.
Fdcos8;
2.
,1U;
3.
everything but .lU .
3.11 Power Power (P) is the rate of energy transfer. The unit of power is the watt (W), which is equivalent to J/ s. Do not con fuse the unit W w ith the concept W for work. Power is given by the following equation:
p = flE t where I is the time duri ng which energy is transferred and I'lE is the energy change of the system, which equals work W pills heat q. A more narrow definition of power, but one tha t is often used, is the rate at which a force does work: p= W t
The i.nstan taneous power due to a force is:
p = Fvcos8 where e is the angle between F and v. We ca n see by these equations that power is a scalar.
Copyright © 2007 Examkrackers, Inc.
LECTURE
3.12 Equation Summary
Equilibrium (no acceleration)
F upward = F downward F rightW:trd = F leftward
Non-equilibrium (acceleration)
F "p,~,d =F d"m"..d ± ma F rightward = F left".,1rd ± ma Add ma to the weaker side.
"I
Torque 't
=Fl
Energy K= ! mv'
U. =mgh U, =
Work W = Fdcose (C);cq* friction) W = f'..K + f'..U + f'..E , (00",,1)
Power _ f'..E P- t
P=Fvcos8
Copyright (f;) 2001
EX~Hn k rac k er s,
Inc.
!kt.x'
3:
EQUILIBRIUM, TORQUE' AND ENERGY . 51
69. A 100 N force is applied as shown to a 10 kg object for 2 seconds. If the object is initially at rest, what is its final velocity? (ignore friction: sin 30" ~ 0.5; cos 30" ~ 0.87)
Questions 65 through 72 are NOT based on a descriptive passage.
~_"_-~·L-_l_O_k_g
65. A meteor with a mass of I kg moving at 20 kmls collides
with Jupiter's atmosphere. The meteor penetrates 100 km into the atmosphere and disintegrates. What is the average force on the meteor once it enters Jupiter's atmosphere? (Note: ignore gravity)
70. A large rock is tied to a rubber band and dropped straight down. As the rock falls, the rubber band gradually stretches, eventually bringing the rock to a stop. Which of the following energy transfers is taking place in this process? A. B. C. D.
less gravitational potential energy than blocks at the middle or blocks near the top of the stack. less gravitational potential energy than blocks at the middle and the same gravitational energy as blocks near the top of the stack. the same gravitational potential energy as all other blocks. more gravitational potential energy than blocks at the middle or blocks near the top of the stack.
C. D.
I m/s 17.4m/s 34.8 m/s
D.
66. If I kg blocks were stacked one upon the other starting at the sUlface of the earth and continuing forever into space, the blocks near the bottom of the stack would have:
B.
8.7 mls
C.
A.2xIO'N B.4xlO'N C.8xIO'N D. 2 x 10' N
A.
A. B.
A. B. C. D.
3,600 J 6,000 J 3,600,000 J 6,000,000 J
72. A winch is used to lift heavy objects to the top of building under construction. A winch with a power of 50 kW was replaced with a new winch with a power of 100 kW. Which of the following statements about the new winch is NOT true? A. B.
rise rise rise rise
Kinetic to gravitational potential to elastic potential Kinetic to elastic potential to gravitational potential Gravitational potential to elastic potential to kinetic Gravitational potential to kinetic to elastic potential
71. Energy consumption in the home is generally measured in units of kilowatt hours. A kilowatt hour is equal to:
67. Objects A and B are placed on the spring as shown. Object A has twice as much mass as object B. If the spring is depressed and released, propelling the objects into the air, object A will:
A. B. C. D.
__L -__
one fourth as high as object B. half as high as object B. to the same height as object B. twice as high as object B.
C. D.
The new winch can do twice as much work in the same time as the old winch. The new winch takes twice as much time to do the same work as the old winch. The new winch can raise objects with twice as much mass at the same speed as the old winch. The new winch can raise objects with the same mass at twice the speed of the old winch.
68. A spring powered dart-gun fires a dart I m vertically into the air. In order for the dart to go 4 m, the spring would have to be depressed: A. B. C. D.
Copyrigh t
2 times 3 times 4 times 8 times
(i)
the distance. the distance. the distance. the distance.
2007 bclrnkracbo%, Inc.
52
STOP.
Momentum, Machines, and Radioactive Decay 4.1
Momentum
A baseball, when thrown by a major league pitcher, has approximately the same energy as a bowling ball thrown by a professiona l bowler. Why, then , can the baseb a ll be knocked out of the park w ith a swing of a bat, w hen the same swing would only deflect the motion of the bowling ba ll? The answer is momentum (P) . 111e momen tum of the bow ling ball is much greater. Momen tunl is a measure of a m ov ing object's tendency to continue alon g its present path. By increasing either an object's velocity or its mass, and thus its momentum, it becomes more difficult to change its path. MomentuIl1 is given by the equation:
p = mv The units of m omentum are kg m / s.Momentum is close ly related to ine rtia (Physics Lecture 2). The re are two important points to know about momentum fo r the MeAT. The first is that in an isolated system momentum is always conserved. This law is as invi o lab le as the law of conservation of en ergy. The second im portan t point is tha t momentum is a vector. When we put these two points togethe r, we find that the initial momentum of the center of mass o f a n isola ted system is alw ays equal to its final nlomentum in both nlagnitude and direction. In other words, the nl0nlentum of the center of m ass of an isola ted system is cons tant in direction and magnitude.
4.2
Co ll isions
A collision occurs in the following nlanner: Two bodies come into contact and are m omentarily deform ed while doin g so. Franl Hooke's law we k now that the force gene rated b y the d e fo rmed bodies is proportiona l to the d egree of the deformity. If the bodies follow Hooke's law p erfectly, the force is conserva ti ve and the all energy is transferred (via W = Fdcos8) back to tl,e motion of the bodies. If the bodies d o not fo llow Hooke's law perfectly, some or a ll of the energy is dissipated as inte rna l energy. The two types of collisions just d escribed arc called clastic collisions and inelas tic collisions, res pectively.
Elastic collisions are collisions w here the mechanical e ne rgy is conserved. In an c lastic collision, no energy is d issipated to internal energy. When very small hard objects w ith no inte rnal parts collide, the e ne rgy has no place to diSSipate. This is a n e las tic collision. Atom ic collisions approXimately follo w this model. Two magn ets may slide into each other's magnetic field and repulse each other w ithou t ever tou ching. Their kinetic energies would be conserved momenta rily in their magnetic
Momentum of an isolated system is always conserved.
54
vii ~AT P,IY"" ,
100 J kmetic energy 50 J internal energy
25 J kinetic ene rgy 75 J in ternal ene rgy
20 kg m!
,~~~=E~=~~==5~::=j~=~~====:::::::::=~3~0~J~kineti c
70 J inte rnal energy energy
OJ k ineti c energy 50 J internal energy
Inelastic Collision
30 kg mls
Momentum is conse rved. Kine tic energy is los t to internal energy.
fields. This represents a perfectly elastic collision. A rubbe r ball dropped from 1 meter bOlU1ces off a hard s urface and returns to the same height. This is another elastic collision. In all these collisions, only conservative forces are at work resulting in conservation of mechanical en ergy.
Elastic collision problems are easy to solve because the sum of the Inechanical energies before the collision is equal to the sum of the mechanical energies after the collision.
Since energy is a scalar, even the d irections of the collid ing objects a re irrelevant. When solving a coliision problem on the MeAT, be sure that your chosen system in step 3 is an isolated system, Repeat step 2, drawing two diagrams of your system, The first should be a diagram of the system immediately before the collision and the second a diagram immediately after the coliision , Then, for elastic coliisions, set the total mechanical energy in the first diagram equal to the tota l mechanical energy in the second diagram . For inelastic collisions, set the initial and final momentums equal.
Inelastic collisions occu r w hen th e collidi ng objects lose some of their mechanical energy to interna l energy. Any collision that is not elastic is inelastic. Stated another way, if an y mecha nical en ergy is lost, the collision is inelastic. A completely inelastic collision occurs when the collidin g objects stick together upon collision. Since lnech anical energy is not completely con served in any irtelastic collision, we must use conservation of momen tum to solve inelastic collision problems. The fo n nula fo r solving inelastic collision problems is simple enough: Pinitial
= Pfinal
The in itia l mome ntum of an isolated syslem equals th e filial fTlurn enturn of an isolated system. However, because momentum is a vcctor, we must pay close attention to its direction. For ins tance in the inelastic collisio n d iagram on the previous page, we see that by adding the momentun1 vectors of the fina l system, we a rri ve at a 20 kg 01 /S vec tor po inting down ward, the sa me as the initia l mome ntUlTI. In multidimen sional system s, the vector nature of momentum may req uire several equations. For ins tancc, iJ1 a 2 dimensional system, if we ha ve momentum in both the x and y directions, one equation is required for each direction. Thus a 2 din1en sional collision may require th e following equations: P (:r) in itia!
=
P{xlfinal
Lrc
-LJRf
4:
MOMEN ut~,
M,
The x and y components are fo und by using the cosine and sine of the angles as shown in the example below. Since this involves lengthy calculations, the MCAT probably will not test it in this manner. Instead , the MCAT is likely to ask only about the mOln entum in the x direction or only about the momentum in the y direction.
Ju st after the collision
Just before the collision
{ n
.
b
" " " " """""' X{j ~ e".
11,
P"
p
"
\ n :I
b
X
p11,sine11, + pb,sin0b, = pII, sin~J ' + p/', sin(-),J,
4.3
Reverse Co ll isions
Imagine wa tching a short film of two objects colliding and sticking together. This is a completely inelastic co lJjsion. The final momentum equals the initial momentum. If the objects stop when they collide, the final mome ntu m is zero, so the initia I momentUln must have been zero as well. Now play the film again, only this time run it backwards. The two objects start together and suddenly burst apart. This is a reverse collision. Of course, just rmming the film backwa rds didn' t change the momentulllS. The final and the initial momentlllnS must still be zero. This is true of any explosion or radioactive decay where the pieces start from rest. Notice that the vector nature of 1110mentum dictates that, in a two-piece explosion, the two pieces mus t separate in exactly opposite directions.
1111
+ 111 2
Now imagine a cat standing on a board on top of a frozen, frictionless lake. The momentum of the cat and the board together is zero. If the cat tries to jump to the right, the board must be pushed to the left in order to conserve momentmn.
Copyright
lC)
2007 Examkrackers, Inc.
l1\Jl" AND RAC '11\.(
·,1f)1
-
55
56
MeAT PHYSICS
4.4
Intuition about Col lisions
Collisions w ill be fully elastic, partially elastic, or fully inelastic. If you study the table and the graph below, you will be able to make simple predictions concerning such collisions. Knowi ng this table and graph is by no means crucial for the MCAT, but it will provid e yo u with a deeper understanding concerning collisions. Don't try to m emorize the table!
~ Equations
1'111
< In!
In ] =1'n1
m 1 >1n2
Elastic v, VI
v,
n1 1- 1112
Do= m 1 + 1'112 -V,,
Partially Elastic
0
O
Z'~
Do=
v,
~ 111 1 + 111 2
_ U_, _
V,, - V 2
=
Inelastic
v, _ 1Il2
7111
O < 'l'~
- v" < II ,
v! =v"
0< III < 2
v" < {1~ < 2v"
0< VI < v"
""
V,
-V ,,- VI
V combined
=
!.!.!. I
1112
V .:..!'
2
< p,
711 1
In } +
11;
v. = ~ 2
< 'lJ,.
< u < 2v
'
v, =
o < VC < '2v"
0 « 1, < tJ"
""2
.s.
"
V
2. < v,. < v"
The table above is based upon a collision between a mass 1111 moving at V o and a stationary Inass 11l2' The velocities V I and v 2 represent the respective velocities of the masses after the collision. The velocity v , represents the velocity of the combined mass after a fully inelastic collision.
... ... V ( luIlY ;"'·',l>i;,·
The graph to the left represents the information in the table. If you understa nd the g raph, it is an excellent guide to answering collision questio ns qualitatively. If yo u don' t unders tand the graph, don' t swea t it. It won't be on the MCAT. An example using the graph ahove: A partially inclastic collision occurs when an object moving with
Copyrig ht «~ 2007 hzunkrac kers, Inc.
LECTURE 4 : MOMENTUM, MACHINES, AND RADIOACTIVE DECAY . 57
velocity Vo collides with a statio nary object of eq ua l mass. We examine the y axis of the graph beca use the masses are equal along the y ax is. We see that the stationary object must have a velocity in the darkly shaded region along the y axis, so it must ha ve a fina l velocity less than Vo but greate r than vo/2. The first object must have a velocity in its original direction, greater than zero bu t less than vo/ 2. The graph also shows that in order for the first object to bounce backwards, it must have a mass less than the object it strikes.
4.5
Impu lse
Impulse (]) is equa l to the change in momentum.
J =!1p If you examine an y collision, you w ill notice that, if the materials approximately fol -
low Hooke's law, the force d uring the time of contac t is not constant. The average force on either collid ing body can be found from the equation:
To find the average force from the change in momentlun simply put the two impulse equations together to make:
I1mv =
Favg
Ilt
Impulse shows us lhCll if the time over which tIll'
fOI"Cl'
acts is increased, the
Si:lmc
changc' in v(']ocity Gll1 be achieved with a JO\,vcr force. For instill1ce, air bags on an automobile don't
cJhlngc the momentum, but they increase the time over \vhich the
co llision occurs, (ll1d thus dene£lsl::' the forn> on llll' driver. Thl.:' graphs below show three separ
30
cLlch Li me, the nlass re-
20
mains C(lllst,mt, and the initii-l1
10
the
SLI me
and fin(ll velocities ,lft:' the
30
2
~
S (lIllC
for each graph. However, the way thnt the final velocity is
(lch il'vcd is vl' ry dirfe renl in c"ch graph. If tl'" particle were em l'gg, w hich graph represcnts the conditions undl'[" w hich lhe I.'gg w Ollld ht· II'.'Ist likel y to break?
0 -10
J3
10 20 30 40 50 60
20 10 0 -10
time (s)
10 20 30 40 50 60 time (s)
30
~~II o
10 20 30 40 50 60
-10
time (s)
II Copyright © 2007 Examkrackcrs, Inc.
?mv
76. The chemical potential e nergy in gasoline is co nverted to kinetic e nergy in cars. If a car accelenttes from zero to 60 kmlh , compared to the energy necessary to increa.o;;e the velocity of the car from zero to 30 km/h, the energy necessary 10 increase the velocity of the car from 30 to 60 kmlh is:
Questions 73 th roug h 80 are NOT based on a descriptive passage.
73. A rocket with a mass of 7.2 x 10" kg starts from rest in outer space and fires its thrusters until it is mo vi ng with a veloc ity of lOOmis. What was the average force on lhe rockel due to the thrusters? A.
B. C. D.
A. B.
C. D.
7.2x IO' N 7.2 x JO'N 7.2 X 10" N The average force cannot be determined with the information given.
77. A 3 kg cat sitting on a 1.5 kg piece of cardboard on a frozen lake wants to jump to shore wi th out touching the ice. If there is no friction between th e cardboard and the ice, when the cat jumps, th e cardboard will move in the opposite d irectio n with a ve locity:
74. A boy is sliding down a long icy hill on his sled. In order ( 0 decrease his mass and increase his velocity, he drops
A. B. C. D.
his heavy winter coat "Iod heavy boots from (he sled while he is moving. Wi ll his strategy work? A. B.
C. D.
half as great. the same. twice as great. three times as great.
No, because he loses the potential energy of the ob· jects that he leaves behind. No, beca use although his kineti c energy increases, his mo mentum decreases. Yes, because although his kinetic energy decreases, his momentum increases. Yes, because although his momentum decreases, his kinetic energy decreases.
half as great as the eat's velocity. equal to the cal's velocity. twice as great as the cat's velocity. four times as great as the cat's velocity.
78. A block of mass m t slides across a frictionless surface with speed V j and collides with a stationary block of mass nI 2" The blocks stick together after the collision and move away with a speed v2" Which of the following statements is (are) true about the blocks?
I. II.
75. Ball A mov ing at 12 mls collides elastically with ball B as shown. If both baBs have the same mass, what is the tina l veloc ity of ball A'I (Note: sin 60° = 0 .87; cos 60° = 0.5)
+ m 2)v2
m l VI = (ml I
)
I
, -1I1,V,= - (m , + m~)v') 2 2 - -
III.
v, ~ . ,
A. B.
I o nly II onl y I and II onl y I, II and IT[
C. D.
B
A. B. C. D.
3 lOis 6 m/s
9 mls 12 m/s
58
GO ON TO THE NEXT PAGE.
79. Two I kg carl. wi th spring bumpers undergo a collision on a frictionless track as shown in the before and after pictures below. 4 m/s
4 mls
-----7
~
80. A trapeze artist who accidentall y falls builds up a great deal of momentum before he is brought safely to rest by a safety net. The sa fety ne t serves to:
A. B.
~
[{jffD
C.
Before
D.
4 mls
4 mls
~
-----7
increase the force of the collision by decreasing the colli sion time.
decrease the force of the colli sion by decreasing the collision time. increase the force of the co lli sion by increasing the collision time. decrease the force of the co llision by increasing the collision time .
~~ After
The total momentum of the system i ~ equal to:
A. B.
0 kg-m/sec before the collision and 0 kg-m/sec after the collision. -4 kg-mlsec before the collision and 4 kg-m/sec after the co llision.
C. D.
- 8 kg-mlsec before the collision and 8 kg-mlsec after the colli sion. 8 kg-mlsec before the collision and 0 kg-mlsec atier the collision.
Copyrili:J ht (Q) ?007 Fxarnkri1cke rs, In c
59
STOP.
60
MeAT PHYSICS
4.6
Machines
Now that we've covered the more simple topics in classica l mechani cs, let's examine one of the ways these topics ma y appear on the MCAT; let's examine machines. Machines are mechanica l devices that reduce force w hen doing work. Every time that you see a mach ine on the MCAT, remind yourself that ideal machines reduce force but don' t change work. (Nonideal machines lll.CreaSe work because they increase internal energy through friction.) Remembering that ideal machin es don't change work can make some otherwise difficult MCAT problems fast and simple. In this lecture we will examine the ramp, lever, and pulley. In Physics Lecture 5 we will examine one more simpl e machine called a h ydra ulic lift.
Ramp
Lever
mgL, = FL, mg L, = F L
h
h
L, mg fulcrum
sinO
=h
W = Fd
d
F = mgsinO F = mg
h d
W = mg
F=T
~d
.. .........
2T= mg T=
.. W = mgh
... F =
4.7
mg
2
mg 2
The Ramp
A ramp is simply an inclined p lane (see Physics Lecture 3). If we examine the work necessary to lift a mass III to a tabletop of height II , we find that it is the force mg times the dis tance II, or mgh. By building a frictionless ramp, we can achieve the same resu lt w ith a reduced fo rce. To push tile mass up the incline plane, we must onl y overcome the force that is pushing the m ass dow n the plane, w hich is mgsinS. Since the sine of any angle is a fraction, we know that this force is only a fraction of mg and thus reduced by the machine. To prove that the work is still the same we can multiply the force times the distance. From SOH CAH TOA, we know that the distance along the ramp is the opposite, or fr, divided by siJ1e. Thus, W = l1lgsine x fr lsina. This reduces to W = mglr, the same work as wi thout the machine. From this it beCOlTICS clear that the fraction by w hich we want to reduce the force must be the Coryriqhl (c: 2007 Examkrackcrs, Inc
LECTURE
4:
MOMENTUM, MACHINeS , I ... [\ ]D RADIOACTiVE DECAY
<
61
same as the fraction by which we increase the length of the ramp. In other words, if we want to reduce the force to 1/ 2 mg, we must make a ramp with length 21t. This is the same as saying that, when work is held constant in W:::: Fd, force and distance are inversely proportional to each other. The lever is based on the principle of torque. Again, let's examine lifting a lTIass m to a height h. Like the ramp, the lever simply allows LlS to increase the distance through which our force acts. Since wc want to move the lTIaSS at a constant velocity, we want to establish a dynamic equilibrium. This lTIeans that the clockwise torques must be equal to the counter-clockvvise torques. Torque is fo rce times lever arm. So, by dOLlbling the length of the lever arm, we reduce the force necessary by a factor of hvo. We can do this by placing our fulcrum twice as far from our force as from om mass. By the diagram below, we can see that the cmve traveled by the mass to reach height h, is only half as long as the curve traveled by the force-bearing end of the lever. Once again, the force is inversely proportional to the distance, and the work is the salTIC with or without the machine. (Notice that as soon as the lever begins to move, the lever arm shortens. However, as long as both gravity and the force point downward, the lever anTIS remain in the same proportions.)
Although the machines in the diagram to the left appear to be pulleys. they are actually modified levers. They work on the principle of torque. In each, the force F is acting on a greater lever arm than mg.Thus the force necessary to lift mg is reduced . Of course, tile work remains the same. Notice that the tension is not the same throughout these ropes as rt is througllOut the ropes of a true pulley.
Rotation point /
/
m
m F
Modified Levers
Copy(i~-Jht
(() 2007 Exarnkrdckers, Inc .
F
62
MeAT
PHYCICS
4.9
The Pulley
A pulley acts on the same principle as the ra mp and lever; it allows force to act over a greater d istance and th us d o the same am oun t of work with less force. The key to tmderstanding a pulley is reme mbering that tension throug hout a massless rope attached to a frictionJess, massless pu lley is constant. In other words, in the diagram below, the tension T is the same at every point in the rope. Now, in our S-step-systern, let's choose our system to be pulley n um ber 1. We choose pulley number one, beca use pulley number 1 w ill move exactly as the mass moves. (If you have trouble visualizing this, imagine that the rope attaching pulley number 1 to the mass is a solid, inflexible bar. Jt wo n' t change the problem.) If you first chose the mass as your system, you would not get the problem wrong, you would sim ply arrive at the conclusion that the rope connecting the lTIaSS to pu lley number 1 has a tension of mg. Then you wou ld be forced to choose a new system. Eventually, you would have to find a system on which your unknow n, T, was acting directly. This system is pulley number 1. Like the lever, we w ish to create a dynamic equilibriulTI where our mass has a constant velocity up ward. To do this, we want the upward forces to equal the downward fo rces. The down ward force is mg. The upward forces are the hvo tens io ns in the rope attached to the pulley. The tension throughout a rope in an ideal pulley is the same at every point, so the two tensions here llUlst be equal. Setting upward forces equal to d ownward forces gives us IIlg = 2T, or T= '/' IIlg. Of course the work necessary to lift the mass to the table has not changed. So, since the force is halved, how is the force applied over twice the distance? If we look closely at the pulley system and imagine tha t the m ass is raised one meter, we see that in order fo r the rope to lilt evenly, one meter must come off of both sides of the p ulley rope. Since it is all one rope, this alTIOunts to pulling the rope a distance of two meters where the fo rce F is applied. Thus we have reduced the force by hvo and increased the distance over wh ich it acts by two as well. Again, when work is held constant fo rce and distance are in versely proportional.
Pulley #2
" T T
~
--/
T
Pulley #1
~
mass
Copyright «) 2007 Exalllkrackers, inc.
83. A crate is to be lifted to a height of 3 meters wi th the assistance of an inclined plane. If the inclined plane is a non-ideal mac hine, which of (he fo llowing state me nts is most likely true?
Questi ons 81 through 88 are NOT based on a descriptive passage.
A.
81. The frictio nless pulley system below reduces the force necessary to lift any mass by a factor of 3. How much
B.
power is requ ired to lift a 30 kg object 2 me ters in 60 seconds using thi s pulley system?
C. D.
The non-ideal incl ined plane increases the force required and dec reases the work. The non- ideal inclined plane decreases the fo rce required and increases th e work. The non-i deal incli ned plane increases the force and the work required. The non-ideal inclined plane decreases the force and the work required.
F 84. A girl riding her bicycle up a steep hi ll decides to save energy by zigzaggi ng rather th an riding straight up. Ignoring fri ction, her slrategy wi ll: A.
A. B.
4W lOW
B.
C.
24 W
D.
120W
C. D.
82. An eccentric pulley can be used on a compo und bow to increase the velocity of an arrow. The pulleys piVOl around the dots as shown. Below is a compound bow in two positi ons. The tension at point A compared to point B is most likely:
85. An inventor designs a machine that he claims will lift a 30 kg object with the application of only a 25 N force. If the inventor is correct, what is the shortest possible distance through which the force must be applied for each meter that the object is raised?
A.
5m
B. C.
8m
D.
Position I A. B. C. D.
req ui re the same amount of e ne rgy but less force o n the pedals. req ui re the same a mou nt of e nergy and the same amoun t of force on the pedals. require less energy and less force on the pedals. require less energy and more force on the pedals.
12 m 15 m
Position 2
less in both position 1 and position 2. less in position I and greater in position 2. greater in both position 1 and position 2. greater in positio n 1 and less in position 2.
63
STOP.
87. The mechanical advantage for a machine is defined as the output force divided by the input force. Since the output force is typically greater than the input fo rce, this value is normal1y greater than one. For an ideal machine, what would be another way of representing the mechanical advantage?
86. The pulley system show n below operates as a modified lever. Pulley A and pulley B turn together so when a person pulls on rope A the mass attached to rope B will be lifted. Which of the following changes to the system will reduce the force needed to lift the mass? Pulley A
A. B.
Pulley B
C.
•
D.
(output distance)/(input distance) (input distance)/(outp", distance) (output distance) (input distance) (input distance) + (output distan ce)
Rope A Rope B
88. A wheelchair access ramp is to be designed so that 1000 N can be lifted to a height of 1 meter through the application of 50 N of force. The length of the ramp must be at least:
A. B. C. D.
In
A. B. C. D.
[ncrease Increase Increase Increase
the the the the
length of rope A. length of rope B. diameter of pulley A. diameter of pulley B.
Copyright © 2007 Examkrackers, Inc.
64
5m 10m 20m 100m
STOP.
LECTURE 4: MOMENTUM, MACHINr-S, AND RADiOACTIVE D ECAY .
4.10 Radioactive Decay Radioactive decay concerns atoms that spontaneously break apart. All atoms other than hydrogen are subject to som,e type of spontaneous decay. However, the rate at which decay occurs varies dramatically. Atoms with a high decay rate are said to be radioactive. Of the 2000 known nuclides (a toms and their iso topes), only 266 are stable. Atomic stability can be reasonably predicted by an atom's neutron to proton ratio. No atoms with more than 83 protons are considered stable. In smaller atoms, a stable neutron to proton ratio is 1:1. As atoms get heavier, they require a larger number of neutrons for stability and the ratio increases to as Inuch as 1.5:1.
Particle alpha beta positron gamma
Symbol
a'2 -l) or.ll) or.~e .. ~ Ol· ..~C
y:
4.11 Half-Life There is no way to predict how long a single atom will take to spontaneously decay. However, since atOlns are small, we are usually concerned with millions of them at a time. Thus we can apply the rules of probability and make predictions concerning large groups of atoms. Any substance (a large group of identical atoms) has a predictable rate of decay. This predictable rate of decay is usually given in terms of a half-life. A half-life is the length of time necessary for one half of a given amount of a substance to decay. Tn an}' half-lifeprobleln, there are 4, and only 4, possible variabl\:.~s. They c1fE' the initial dTllount of subst.mce, thE' final i.nnount of substance, tlw number of hdlf-lives (uften given as ,1 time period, in which case you simply' divide b~/ the kngth of a h
your fingers.
I (
_ / - - -.. f--'\.,,--")
:::Thcrc goes 01115; , (, 11<1 I" .,,) 1- I'I rl' . ..J
"C
-t7&J "~-;.\...nee ~ ~t®®1 .
Wi%
,~
-
(Kracker s ) ~ ~ not included ~ ~
Copyright © 2007 EX2mkr<'lck""rs. Inc
I
65
66
MeAT
PHYSICS
4.12 Types of Radioactive Decay There are five types of radioactive decay on the MCAT: alpha decay, beta decay, positron emission, electron capture, and gamma ray production. (Positron emission and electron capture are actually types of beta decay.) If you remember how each particle is written and on which side of the equation it belongs, solving a decay problem on the MCAT becomes a very simple math problem. Simply be sure that the sum of the atomic numbers and the sum of the mass nUlnbers on the left side of the equation equal the sum of the atomic numbers and the sum of the mass numbers on the right side, and look up the proper clements in the periodic table.
Alpha decay (or a-decay) is probably the easiest to remember. An alpha particle is a helium nucleus . Thus, it contains 2 protons and 2 neutrons. In alpha decay, an alpha particle is lost. An example of alpha decay is: 218U -- 'a + <12 2
23~Th 90
Beta decay (~-decay) is the expulsion of an electron. (Some books include positron elnission as a type of beta decay.) A beta particle is an electron or positron. (A positron is like an electron with a positive charge. ) Notice that beta decay is not the destruction of an electron; instead, it is the creation of an electron and a proton from a neutron, and the expulslon of the newly created electron. An example of beta decay is:
A neutrino (not shown) is also emitted during beta decay. A neutrino is a virtually massless particle. A neutrino is typically represented with the Greek letter nu (v) . Positron emission is the elnission of a positron when a proton beconles a neutron. In positron emission, a proton is transformed into a neutron and a positron is emitted . An example of positron emission is:
Electron capture is the capture of an electron along with the merging of that electron with a proton to create a neutron. In electron capture, a proton is destroyed and a neutron is created. An exanlple of electron capture is:
A gamma ray is a high frequency photon . It has no charge and does not change the identity of the atom from which it is given off. Gamma ray enlission often accompanies the other decay types. An example of gamma ray elnission is when an electron and positron collide:
Tllis is a matter-antimatter collision called annihilation. Mass is destroyed releasing energy in the form of gmnma rays.
Copyright ;) 2007 EAa:l1krackers, Inc
L!::Cl lJRf: 4: MO~'lcNTlJiVl ! MACHINE';, /\ND RN)10/,CTIVF D[CAY
4.13 Mass Defect The matter-antimatter collision between an electron and positron brings up an interesting question. Did the energy exist before the collision occurred? Is this a violation of the conservation of energy? Einstein had the answer with:
This equation gives the rest mass energy of an object. For the MeAT just think of rest mass energy as latent energy within the mass of an object. It will only appear on the MeAT if 111ass is created or destroyed. Otherwise, never even think about rest 111ass energy. If mass is created or destroyed, always use E:::: me 2 to find the answer, where m represents the amount of mass created or destroyed and c is the speed of light (3x10 8 m/ s). The forces holding the nucleons (protons and neutrons) together are the result of a change in the rest mass energy of the individual nucleons. In otber words, if we measured the 111ass of the nucleons before fonning the nucleus of an atom, and then measured the ll1ass of the nucleus, there would be a discrepancy; the nucleus would have less ll1ass than the sum of the masses of its individual parts. The difference in the masses is called the mass defect. To flnd the binding energy holding the nucleons together, plug the mass defect into E = mc' .
• proton proton neutron neutron 1.0073 amll + 1.0073 amll + 1.0087 amu + 1.0087 amu
+
•
•
electron 0.0005 amll
• +
electron 0.0005 amll
4.033 -4.003 0.030 amll
mass d efect
Helium 4.003
Cupyr iqh l
200 / EXi.JfTlkrJckc r::i, 11](
4.033 amu
.
67
68
MeAT
PIIYSICS
4.14 Fission and Fusion Fusion is the combi ning of two nudei to form a single heavier nucleus. Fission is the splitting of a single nucleus to form two lighter nuclei. How can large amounts of energy be released in both processes? The energy comes from the mass defect. If we think of the binding energy as a bond, we know that energy must be added in order to break a bond (including ATP bonds). Thus, when we make a bond in fusion, we can see from where the energy can come; energy is always released when a bond is formed. The energy comes from the bonds between the nucleons in the new nucleus. These new bonds are stronger and more stable than those of the nucleus that was just divided. Thus more energy was released in the formation of the stronger bonds than was absorbed in the breaking of the wea ker bonds.
8 7 Fission >, bIJ
...
6
OJ
5
'"0:
4
0:
3
0:
bIJ
'B ~
!
Fusion
2
1
0
I
I
100
200 Mass Number
The most stab le nucei have the strongest binding energy. Both fission and fusion produce more stable nuclei. This graph is an approximation.
Copyright
i(j
2007 ExamkrC1ckers, Inc.
LECTURE
4 .15 Equation Summary
Momentum p=mv
Inelastic collisions p ,nilial
=
P fillal
Elastic collisions Pi"iti~l = K inilial
P fin
= Kfina1
Impulse F.,gi!. t = i!.mv
Rest mass energy E=mc'
Copyright © 2007 EXilfnkrackers, Inc.
4:
MOMENTUM, MACHINES, "NO RAOIOAC IVE DECAY • 69
93. Whic h of th e fo llowing graphs best represents the
Questions 89 through 96 are NOT based on a descriptive passage.
radioactive decay of 23iSU?
A. 23!1
89. T he half-life of substance X is 45 years, a nd it decomposes to substance Y. A sample fro m a meteorite was taken which contained 1.5% of X and 13.5% ofY by mass. If substance Y is not normall y fo und on a meteorite, what is the approxi mate age of the meteori te?
A. B. C. D.
C.
\
U
"'U
time D.
90. When 224Ra undergoes alpha decay an alpha particle is emitted at 1.0 x 10' m/s. What is the velocity of the other particle? 1.6 1. 8 1. 8 5.4
x x x x
time
B.
45 years 100 years 140 years 270 years
A, B. C, D.
~ time
time
94. A diagram show ing the changing mass of an unstable isotope undergoing radioactive decay over time is shown below. What is the half-life of the isotope?
104 10' 10" 10"
70
91. In nuclear fission, a urani um nuc1eus combines with a
:§
neutron, becomes unstable, and splits into Ce and Zr plus two neutrons. The change in the mass of the interacting parts is 0.211 amu. How much energy is released in this reaction? (Note: c' = 93 1.5 MeV/amu)
A. B. C,
D.
""-
50
...
40
~
~
0
98 MeV 130 MeV 157 Me V 197 MeV
60
~ ~
30
'"
~ 20
10
0
92. 2J(,PO undergoes two alpha decays and two beta decays to
0
form:
A,
'""1'1
B. C,
214Ra
D.
212Pb 20Hpb
Copyngh t (1::) 2007 Examkrac.kprs, Inc.
70
A. B,
2.5 hours
C. D,
4.0 hours 6.0 hours
2
4 3 5 Ttme (hrs)
6
7
8
1.5 hours
GO ON TO THE NEXT PAGE.
95. Which of the sequences below could describe the decay process from Bi·210 to Pb-206?
A. B. C. D.
96. The mass number of an atom unde rgo ing radioactive decay w1ll remain unchanged in all of the processes below EXCEPT:
alpha, beta beta, beta alpha, alpha beta, beta, beta
COPyri9ht © 2007 Exarnkrackers, In c.
A. B. C. D.
71
alpha decay. beta decay. electron capture. gamma emission.
STOP.
Fluids and Solids
5.1
Fluids
Mos t substances can be classified as either a solid or a fluid . The molecules of a solid are held in p lace by molecular bonds that can permanently resist a force from any direction. A fluid is a liquid or gas. Unlike a solid, any existing molecular bonds in a fluid are constantly breaking and reforming due to the high kinetic energy of the molecules. Since the lTIolecules of a fluid are no t arranged with any order or structure, but move about in random directions relative to each other, a fluid has only ten1poral (impermanent) resistance to forces that are not perpendicular to its surface. However, since flu id molecules require romn to move, collectively they can create a pern1anent force outward frOlTI within the fluid. This outward force allows a fluid to permanently withstand forces perpendicular to its surface. In other words, the only permanent force that a resting fluid can exert is one nonnal to its surface. TI1US, a fluid is pushed and molded until its surface matches the shape of its container exactly. When the fluid comes to rest, it experiences only the normal force from the surface of its container and the force of gravity. (A liquid takes on a flat upper surface so that tlle gravitational force is also perpendicular. In a gas, gravity has an insignificant effect on the path of an individua l molecule due to the high average velocity of the molecules, and a gas will fill an enclosed container.)
7 -;/
~Aui-
\\
\ ~.:( , \
\:
~-
. .>':f/--,4/
//
~
Molecules of a solid bond strongly and vibrate in a fixed position.
fr \
~~
-==1-
~
U'
fr ~
fr // L"" ~/
I"
~(
~
~fr ' l
tC '
!
(t -
Molecules of a fluid bond weakly and rotate, spin, and move past each other.
74
MCAT
PHvSICS
In other words, a fluid conforms to the shape of its container. The ocean, a fluid, will withstand the weight of a motionless battleship forever by conforming its surface to that of the battleship's so that all forces are normal to its surface. However, the ship can move through the water propelled by a much smaller force than its own weight. This is because the net force from the moving ship is not perpendicular to the surface of the water and thus the water provides on ly temporal resistance. (The forces shown in the diagram
b>
are the forces on the water
mg
due to the ship. and not meant to represent the forces on the ship.)
r
F,co,,','oo
1\Fnet
In particle mechan.ics we discussed mass and energy. These properties were useful because we knew exactly how much substance with w hich we were dealing. Externall y, we could view the entire object and measure these properties. Properties such as these, w hich are con cerned with quantity, a re called extensive properties. Ex tensive properties change w ith the quantity of a substance. In fluid mechanics, we often d o n' t know how much of a fluid with w hich we are d ealing, thus we cannot Ineasure its mass or energy. In order to ana lyze s uch fluids we use illtensive properties or p roperties that are concern ed w ith the intrinsic nature of a substance. Intensive properties do not change w ith the quan tity of a su bstance. TI,e two intensive propcrties that are analogous to mass and energy are, respectively, density and pressure.
5.2
Density
Density (p) is the 'heaviness' of a fluid; it is how mu ch mass it contains in a specified vol Lllne (V). The formula for density is:
p = mlV The S.1. units of dc nsity nrc kg/ m' . No ticc that ch anging the amount of a given substance w ill not change the d ensity of that subs tance. Compression of a fluid ch anges its volum e without changing its mass, and the re fo re w ill change density. Since gases compress more easily than liquids, the density of a gas is easily ch anged while tha t of a liqu id is not. Unless otherw ise indicated, for the MCAT, assume that ail liquids and solids are totall y incompressible, and thus h ave constant density. In reality, gases a re far more compressible than liqUids, and liqUid s are far more compreSSible than solids. Gases on the MCAT change their volume (and thus the ir d ensity) as per the ideal gas law: PV = uRT. We have a strong intuition about the concept of m ass because we use it everyday; however, few of LI S h ave a s trong sense of density. For instance, yo u can appreciate Copyright (c) 2007 FXilrnkrackers, Inc.
LtC URE
5:
FI UIDS AND SOLIDS •
75
how it feels to lift a 13 kg m ass (about 29 pounds), but ca n you lift a bucket full of m ercury, which has a density of about 13,600 kg / m' ? Most of us have no idea. We just don' t know d ensity well enough. [n order to make density a more intuitive concept, specific gravity was created. The specific gravity (S.G.) of a substance is the density of that substance (P'''b'''"~) compared to the density of wa ter (Pw,,",) .
S.G. =
P substanc.!P water
Notice tha t a specific gravity of less than one indicates a substa nce lighter than water; a specific gravity of one indicates a substance equally as heavy as wa te r; a specific gravity grea ter than one indicates a substance heavier than wa ter. Since we all have an intuitive feel for the heaviness of water, we can relate this to a substance, if we know its specifi c gravity. The specific gravity of m ercury is 13.6, so lifting one bucket of mercury would be equivalent to lifting 13.6 buckets of water.
Mercury has a specific gravity of 13.6, therefore 1 bucke t of mercury has the same mass as 13.6 buckets of water For the MeAT you should memorize the density of water in the fo llowing two forms:
p,,,,,,
~ 1000
P Wdler ;;:
5.3
k g/m 3
1 gmJcm
3
Pressure
Recall froll1 Lecture 4 our discussion about impulse. hnpulse ,is the change in momentum or the force of a co lli sion mu ltiplied by the duration of the collision (F(;t) . Since the millions of molecules in a fluid are moving rapidly in random directions, some will collide with an object submerged in that fluid . In any given time t, such a su bmerged object will experience millions of collisions. If we measure the magnitude of the inlpulse 01 each collision and divide it by the time over which the collisions occur, we arrive at the average magnitude of force created by the collisions. Since the molecules are moving in randOlTI directions a nd at random speeds, no Single direction or speed will be more likely than any other, and the lorce on one sid e of the object w ill be exactly countered by the force on its other sid e. (We w ill ignore grav ity lor the moment and we will also assume that, if the flui d is moving, the object is movin g at the same velocity as the flui d.) If we take the average magni tude 01 the force, and divide it by the area over w hich the collisions a re taking p lace (the surface a rea of the object), we have the pressure experienced by the object. This is the fluid pressure. FILlid pressure results from the impulse of molecular collisions. It is the ave rage of the magnitud es of the change in momentum of th ese Copyright rg 2007 Examkrackcrs, Inc.
/
/ Fluid Molecule
~- .
~!
- .. .-• 0
Object
"
\
0-
" •
0
- 0
,;
•
0
V
V
76
MeAT PHYSICS
collisions divided by the ti me duration of the colli sions and the area over which these collisions occur. Pressure (P) is defined as force per unit area (A).
P=FlA The 5.1. unit of pressure is the Pascal (Pa) . Pressure is a scalar; it has no direction. Pressure exists in a fluid whether or not an object is immersed in that flu id . Another way to th ink of fluid pressure is as a measure of the kinetic energy due to the random velociti es of molecules within a fluid distributed over the fluid volume. The tmits of pressure are equivalent to energy per unit vo lume. So pressure can be thought of as a type of 'stored' energy per unit volume. Atmospheric pressure
Less than atmospheric pressure
5.4
Fluids at Rest
A fluid at rest is one that is experiencing forces only pe rpendicular to its surface. A t any given depth, the pressure is equal to the weight of the fluid above a disk w ith area A d ivided by the area of the d isk. N otice from the d iagram that the pressure is independent of the area chosen. For a flu id at rest w ith uniform densHy in a sealed container, pressure P is given by:
P=pgy where p is the density of the fluid, g the gravitational consta nt, and y is the depth of the fluid.
I • - - - - - - - - ] -'- - - - - - - , - • - -
- - - - - - -
~-
-r
- '-.- - < - - -'- - - -
Mg
>
-~ l' ----IIlg
p= Mg
A
Copyright ((; 2007
IIIg a
=
pgy
EXill11kl',3(: ker~ , Inc.
LrCTURE
y'l y'l y, "j
p.
t i
5:
FLUIDS AND SOLIDS
PS y·, p,gy,
p,gy, PlgYI
p ~ p,gy , + p,gy, + P,g!h + PS y,
Since fluid pressure is simply weight divid ed by area, additiona l fluid s on top of the first fluid simply add their weight toward the total pressure. The total pressure can be found b y summi ng the pressures due to each fluid as sho wn in the d iagram on the righ t. Air is a fluid. If we op en OUf sealed con tainer and expose it to the atmosphere, we must add atmosphe ri c pressure to any point in our fluid . ln any fluid open to the atmosphere, the pressure can be found from P = pg1j + Pa tll1()s.• (Note: Jf yo u are using meters and kilogralns, you must measure th e atmospheric pressure in pascals (Pa). PatmosPh@riC
= 1 01,000 Pa)
You can think of the .ltmospherl' ilS <1 sea of ,lif. /\s you 1ll0\' l' doser to the top of this Sl'a, its dep th (y) decreases. Nl'iJ f the top, you h.1\'(' rt:'\,\rer molecules dbove Yl)lI, w hi ch mt~ans less "veigh! nnd 10w(-'r prl'SSllrL'. tliologists often t"lk about nega tive prl'ssure Cl'l'<:ltl'd in your chl'st when YOLI Slick in air. If you th ink of the molt'cutH coll ision model of pn'ssLlI'C' (Chemistry Leclu re 2), you "vill reali/p lhcll I1cgd li \'c presslIn:' is impossible. It would indicate less thelll zero collisions; clll dbsurdity. The lll'gi1tin' pressure tl"lilt biologisb re fer to in the chest ca\'ily is gauge pressure. Gduge prC:isure is a measure of the pressure compared t{l local atmospheric pn.:-ssun.'. In other words, I()C(l I ~lllllnspheric pressure is (Hbilrilrily giYen i:l va lue of zero. So when a biologist says there is negative pressurc' ins ide yo ur chest, th ere is still pressure in your chest; it b just- less pressure thlln ntmospheric pressure. The higher pressure of the atmosphere pushes air into your lungs. The same thing happens when YOLl 'suck' fluid through a st ra',,-'. You create a partial vacllum inside the straw. But a vncuum doesn't ren ll y 'Slick' clnything into it. The atmospheric p ressure pushes down 011 the flui d outside the straw pushing up the fluid ins ide the straw. Without
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(g
2007 Examkr'Kkcrs, Inc.
Air prc:;slIrc up hl..'l"l' i ~ lo\\" making it tough to brc,lth.
77
78
MeAT
PHYSICS
By the way, pressure measured relative to a vacuum as zero is called absolute pressure. To find absolute pressure from gauge pressu re, Just add atmospheri c pressure.
Since fluid pressure is a function of depth, the shape of the container does not affect it. The pressure everywhere at a given depth in the same resting fluid will be constant.
Just as each block in a stack of blocks must bear the weight of all the blocks above it, each point in an enclosed fluid must bear any increase in pressure. This is called
Pas cal's principle . Pascal's principle states that pressure applied anywhere to an enclosed incompressible fluid will be distributed undiminished throughout that fluid. Notice that Pascal's principle does not apply to a gas because a gas is compressible.
5.5
Hydraulic Lift
The hydraulic lift is a simple machine that works via Pascal's principle. A force on piston 1 acts to apply a pressure on the incompressible fluid. This pressure is transferred undiminished to piston 2. Since piston 2 has a greater area than piston 1, the force on piston 2 is proportionally greater. However, recall t11at an ideallnachine
does not change work. Thus, the distance through which the force is applied is proportionally less. F1d1 ~ F2d2; FJ A1 ~ F2/ A 2·
Piston 1
5.6
Piston 2
Archimedes' Principle
In the 3'd century B.C., the king of Syracuse was given a crown and told that it was solid gold. Archimedes was given the task of proving or disproving that the crown was solid gold. Archimedes knew the density of gold and decided to find the density of the crown. To find the volume of the crown, hf' SUhmf'f3t"cl it into i1 hlJ( KPt of water and measured the amount of water displaced. H e then weighed the crown
and divided its mass by its volume. The density of the crown was not the density of gold; Archimedes had proven that the crown was a fake . This story reminds us that an object submerged in a fluid displaces a volume of fluid equal to its own volume. Before the object is submerged, the upward force on the fluid that it will displace must equal the weight of that fluid (Fb,,"ve,,,, ~ mgw .","). Once the object is submerged, the net upward force remains, but the fluid is gone, replaced by the object. Thus the upward force acts on the submerged object. This force is called the buoyant force. Archimedes' principle says that the buoyant force
Copyright © 2007 Examkr3ckcrs, Inc.
LECTURE 5: FLUIDS AND SOLIDS . 79
(F,,) is an upw ard fo rce acting on a submerged object, and is equal to the weight of the flLlid displaced by the submerged object. The buoyant force is given by:
I \ A 1\1\
IV
v v
1\
VI
L______ J
F t>UOl'".m,
Fb =
P fluid Vg
where V is the volume of the fluid disp laced . The buoyant force is alw ays equal to the weigh t of the flu id d isplaced. In the diagram below the buoyan t force is Fb"oy,'"
,,-----------,, \ } \...... _/
~--~-)~ 1 .
I---- ,,- ----1, \\" ~//
I11g,,,,,.,
Fraction submerged
= P lluid
A floatil1g object d isp laces an amount of fluid equal to its own weight. The fl oatil1g equatiol1 says that the submerged fraction of a floating object is equal to th e ratio of the density of the object to the density of the fluid in which it is floating. If the object is floating in water, this ratio is the specific gravity of the floating object
Copyright © 2007 Exarnkr
80
MCAT
PIIYSICS
Make sure that you understand that. since the buoyant force is due to the 'difference' in pressure, the buoyant force does not change with depth .
Fb = pVg
v = A~h
A
. h, ~h
: . Fb = pgA~h
F
Ab = pg~h
. h,
Anothl'[ W"pl'ricnccs greater A fully submerged object displaces its volume in fluid; a floating object displaces its weight in fluid.
pressure th{lll the upper surface. This pressure difference multiplied by the upper or lower surf<.1CL' ,In:d is equal to the buoY<:lnt forcC' .
Center of buoyancy
Center of
foam
I11g
If we consider an object to be a single particle, the buoyant force ac ts at the center of buoyancy. The center of buoyancy is the point where th e center of mass would be, if the object had a wlifonn density. If the object is not uniformly dense, the center of mass and the center of buoyan cy will not coincide. This could crea te a torque on the object and cause it to spin as shown in the dia gram below. Cente r of buoyancy explains why a fishing bobber al wa ys floats upright.
Center of buoyancy
Center of mass "
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99. A brick with a dens ity of 1.4 x If}' kglm' is placed on top of a piece of Styrofoam floating on water. If one half the
Questions 97 through 104 are NOT based on a descriptive passage .
volume of the Styrofoam sinks below the water, what is
the ratio of the volume of the Styrofoam compared to the volume of the brick? (Assume the Styrofoam is massless.) 97. Mercury has specific gravity of 13.6. The column of
A.
mercury in me barometer below has a height h = 76 em. If a similar barometer were made with water, w hat wo uld
be the approximate height h of the column of water?
0.7
B.
1.4
C.
2.8 5.6
D.
100. A helium balloon will rise into the atmosphere until: A.
The temperature of the helium inside the balloon is equal to the temperature of the ai r outside the bal-
B.
The mass of the helium inside the balloon is equal to the mass of the air outside the balloon. The weight of the balloon is equal to the force of the upward air current. The density of the helium in the balloon is equal to the density of the air surround ing the balloon.
loon.
C.
A.
B. C. D.
5.6 em 76 em 154 em 1034 em
D.
101. A child's bathtub toy has a density of 0.45 glc m' . What
fraction of the toy floats above the water?
98. Two identical discs sit at the bottom of a 3 m pool of water whose surface is exposed to atmospheric pressure.
A.
The first disc acts as a plug to seal the drain as shown. The
B.
second disc covers a container containing nearl y a perfect vacuum. If each disc has an area of I m2, what is the approximate difference in the force necessary to open the containers? (Note: I atm = 10 I ,300 Pa)
C. D.
5% 45% 55% 95%
102. The diagram below shows a hydraulic lift. A force is applied at side I and an output force is generated at side 2. Which of the following is true'!
almos.
F, F,
1-~--1 CiJ A, armas. A. B. C. D.
There is no difference. 3000 N 101,300 N HYI ,300 N
A. B. C.
The force at side I is greater than the force at side 2. The force at side I is less than the force at side 2. The pressure at side 1 is greater than the force at side 2.
D.
The press ure at side
is less than [he pressure at
side 2.
Copyright ttl 2007 EXarrlkliK.kefs, Inc.
81
STOP.
103. The pressure at the bottom of a cylindrical tube filled with water was measured to be 5000 Pa. If the water in the tube were replaced with ethyl alcohol, what would be the new pressure at the bottom of the tube? (The density of ethyl alcohol is 0 .8 g/cm3.) A. B.
C. D.
104. Three containers are filled with water to a depth of I meter. At the bottom of which container is the pressure
the greatest?
~
4000 Pa 4800 Pa 5000 Pa 6250 Pa
---B
A. B. C. D.
~- ~:~ \ - -- - ! \,..----~-j
"-----------C
Container A Container B Container C The pressure is the same at the bottom of all the containers .
Copyright @ 200'7 Exa rnkra<.:kcrs, Inc.
82
STOP.
LE.CTURE: 5: FLUIDS AND SOLIDS .
5.7
Fluids in Motion
The molecules of a moving fluid can be thought of as having two types of motion: 1.
the random translational motion tha t contributes to fluid pressure as in a fluid at rest and;
2.
a uniform translational motion shared equally by all the molecules at a given locatjon in a fluid.
The uniform translational motion is the motion of the fluid as a whole. This motion does no t contribute to fluid pressure. If we recall our molecular model of fluid pressure, an object ll10Ving along with the fluid w ill not experience additional collisions due to this uniform translational motion. Thus, it will not experience any additional pressure. In fact, the energy from the two types of motion can be converted back and forth; SOlne of the random translational motion can be converted to uniform translationallTIotion and vice versa. For instance, if we remove a portion of the wa11 of a container holding a fluid at rest, the fluid will move through the opening. This happens because the molecules moving in the direction of the opening do not collide with anything, but instead continue in their present direction. Some of the random motion has changed to uniforn11TIotion. Since there are fewer collisions in the fluid moving through the opening, there is less pressure. We w ill come back to this point later in this lecture.
1-
~ " I t/ ... '>0+ ;f ~... • t
\.!~
...... ...
... <1-"''''
••
.-- t.. ----.~ ...- \.- _ ... ...-7--...... "
Molecules with some random motion and some uniform motion .
/
" .........
" ,;' :-.. t... ~ '1'
""
Molecules lTIoving in random directions in a fl u id at rest.
5.8
Ideal Fluid
Because moving fluids are very complicated, it is useful to create a hypothetical fluid which lacks certain characteristics of real fluids. This hypothetical fluid is called ideal fluid . Ideal fl u id differs from rea I fluids in the following four ways: 1.
Ideal fluid has no viscosity. Viscosity is a measure of a fluid's temporal resistance to forces not perpendicular to its surface. (More precisely, viscosity is the rate of shear stress divided by the rate of strain.) For the MeAT think of a fluid's viscosity as its tendency to resist flow. For example, syrup ha s greater viscosity than water. A closely related concept to viscosity is drag. Drag is force, similar to friction, created by viscosity and pressure due to motion. Drag always opposes the motion of an object through a fluid.
2.
Ideal fluid is incompressible; it has uniform density. This is the same assumption that we make for any liquid on the MeAT unless otherwise indicated, but not for gasses.
Copyright (0 2007 E::xarnkrackers , Inc
83
84
MeAT
PHYSICS
3.
Ideal fluid lacks turbulence; it experiences steady (or laminar) flow. Steady flow means that all fluid flowing through any fixed point will have the same velocity. Remelnberr velocity specifies magnitude and direction. Turbulence means that, at any fixed point in the fluid, the velocity may vary with time.
4.
Ideal fluids experience irrotationnl flow. This means that any object moving with the ideal fluid will not rotate about its axis as it flows, but will continue to point in one direction regardless of the direction of flow. The MCAT is not likely to touch this one.
No ideal fluid actually exists. However/we can use ideal fluid to make crude predictions about real fluids. We do this by imagining how ideal fluid would behave in a given situation, and then considering how the above characteristics would affect this behavior. On the MeAT, all liquids are jdeal, unless othenvlse indicated.
Since ideal fluids are incompressible, their volume remains constant. The volunl€ of a fluid moving through a section of pipe is given by the cross-sectional area (A) of the pipe times the distance (d) of the pipe section. If this same volume of fluid moves completely through this pipe section in a given time (f), the rate (Q) at which volume passes through the pipe is Ad / t. Since the fluid moves a distance d in time t, its velocity is v = d / t. Putting these two equations together we get the continuity equation:
Q=Av where Q is called the Ivolume flow rate'. Flow can be given in terms of mass as well. For the mass flow rate (I) multiply the volume flow rate by denSity:
1 = pQ = pAv In an ideal fluid, flow rate is constant. Notice from these equations that area is inversely proportional to velocity; the narrower the pipe, the greater the velocity.
A second important equation that you must memorize for the MCAT is Bernoulli's equation:
Where K is n constnnt specific to n fluid in a given situCltion of flow, and P, h, and v refcr to the pressure, height, and velocity of the fluid at any given point. (Warning; II is not the same as yin P = pgy. h is the distance above some arbitrary point; 1f is the distance beneath the surface.) Bernoulli's equation states that, given one continuous ideal flow, the sum of its three terms is a constant at any point in the fluid. If we look closely at Bernoulli's equation, we see that it is actually a restatement of conservation of energy. Notice that if we multiply any of the terms by volume, we get units of energy. In fact, the second term gives the gravitational potential energy per unit volume (mgh/V). The third term gives the kinetic energy from the uniform translational motion of the molecules per unit volume mv')/V). The first term, pressure, is the energy per volulne from the random motion of the molecules. Because energy is conserved in ideal fluid flow, the total energy must relnain con-
«'h
Copyright
(~)
2007 Exanlkrackcrs, Inc
LECTURE
stant; thus, the sum of the three terms is constant throughout the fluid. This is an easy method for remembering Bernoulli's equation. This also aids our understanding of the terms. For instance, the h in the secon d term is similar to the h in gravitational potential energy; the zero value for h can be chosen arbitrarily. Like the h term in gravitational potential energy, it is measured from bottom to top. Notice this is the opposite direction of measurement for the y term in hydrostatic pressure: p ~ pgy. Also, recall the equation that predicts the velocity of a body in free fall when all of its potential energy is converted to kinetic energy. You may see that Bernoulli's equation predicts the same result for a fluid. If a spigot attached to a tank of fluid is opened, and we choose h ~ 0 to be the point of the spigot, the velocity of the fluid coming from the spigot can be derived from Bernoulli's equation as:
5:
FLUIDS AND
SOLIDS . 85
A helpful analogy might be to think about a swarm of bees. Imagine that the swarm represents a fluid with each bee as a fluid molecule . Bee stings represent
pressure-causing collisions. Now if I
stand still in the swarm, its gonna hurt. The bees can swarm around and sting
me at their leisure. I·m gonna get stung a lot; this is analogous to lots of molecular coliisions and high pressure . But if I run , each bee must use some of its swarming
energy to keep up with me . Therefore. they can't swarm as much, and I won't get stung as often. Less stings means less pressure. The same is true with mo lecules in a fluid; uniform translational kinetic energy is achieved by borrowing energy from the random translational kinetic energy, thus pressure goes down . ..
~
Streamlines
Copyright (i~ 2007 Examkrackers, In c.
"0 UC;hf.
~ ..
High velocity
. u
"Ouch!";.
From Bernoulli's equation we can derive an important (and possibly counter-"" ''-'- :>--/ intuitive) notion about the relationship between pressure and velocity in ideal fluid flm-y. As velocity increases, pressure decreases. The concept of streamlines was created to assist in the visualization of an ideal fluid. A streamline is a path followed by a hypothetical fluid particle. This particle follows only the uniform translational motion of the moving fluid. The magnitude and direction can change from one point to the next, but its velocity at any fixed point will remain the same. The velocity of the particle at any point along a streamline is tangent to the curve made by the streamline. The magnitude of the velocity is inversely related to the distance between stream lines; the closer the streamlines, the greater the velocity. Streamlines can never intersect, since this would indicate two possible velocities for the same fixed point. A group of streamlines three-dimensionally encompass a tube of flow.
~
86
MeAT PHYSICS
5.9
Non-ideal Fluids (Real Fluids)
All real fluids are non-ideal. The MeAT only requires that you understand nonideal fluids qualitatively. This means that you must predict the general deviation to ideal fluid when we add the first three of the four lacking characteristics. (Irrotational flow will not be on the MeAT.)
Drag and viscosity are like friction and always act to impede flow. Increasing viscosity increases drag. Drag occurs at the fluid-object interface and is a force working against flow. As we move away from the fluid-object interface the effect of drag lessens. In a real fluid flowing through a pipe, the greatest velocity would be at the center of the pipe, the spot furthest from the fluid-object interface. You can remember this general rule about drag by recalling the dusty blades of a well-used fan. The dust on the fan blade remains despite the high speed of the fan. This is because the air immediately adjacent to the fan moves extremely slowly or not at all due to drag. Because the sides of the pipe create the greatest drag, the longer the pipe, the greater the amount of fluid-object interface, and the greater is the resistance to flow. Also notice that if the radius of a pipe is reduced by a factor of 2, the fluid volume (V ~ nr'd) is reduced by a factor of 4, but the surface area is only cut in half (A ~ 2nrd). Thus, the more narrow the pipe, the greater the effect of drag. Warning: a non-ideal fluid does not behave in an opposite manner to an
ideal fluid . Narrowing a pipe increases the velocity of an ideal fluid . It will probably increase the velocity of a nonideal fluid as well. However, with a nonideal fluid you must also consider drag, which impedes flow. Thus, if you narrow the pipe in a non-ideal fluid, velocity will probably increase, but not as much as if there were no drag.
Notice from the tube of flow that fluid does not necessarily move from high pressure to low pressure. Pressure has no direction. The driving force behind the direction of fluid flow is the fluid's tendency to find its greatest entropy. For the MeAT, just use common sense. Ask yourself in which direction you would expect the fluid to flow. If all other things are equal, fluid will move from high pressure to low pressure. In a horizontal pipe of constant cross-sectional area, fluid will flow from high pressure to low pressure according to the following equation:
tlP
~
QR
w here R is the resistance to flow. In Physics Lecture 7, we will discuss the similarity of this equation wi th Ohm's law for voltage and the analogy of pressure to voltage. The volume flow rate for real fluid in a horizontal pipe with constant crosssectional area can also be givcn in terms of pressure, viscosity (11), pipe length (L), and pipe radius (r):
This equation is known as Poiscuillc's Law. It is given here because it is a common equation and you should recognize that it is concerned with real fluids, not ideal fluids.
5.10 A Method for Greater Understanding of Fl uid Flow As stated earlier, Bernoulli's equation describes conservation of energy within an ideal fluid. If each term in Bernoulli's equation is divided by the specific weight pg of the fluid, the units of each new term become meters. Each new tenn is referred to as a 'head' . pgh becomes h, and is called the elevation head. 11, pv' becomes 1I, v' / g, and is called the velocity heod. P becomes P /(pg), and is called the pressure heod. We shall refer to the original terms by the heads, but, for simplicity, we shall not divide by the specific weight. If we examine the figure below, a hypothetical fluid particle at the top of the tank is stationary. Its energy is completely contained as gravitational potential energy. The height of the fluid at this point i;; its elevation head. The zero point is arbitrary, so we will choose the floor. The velocity and pressure heads are zero because the particle is at zero velocity and zero gauge pressure. Si.nce energy is conserved in ideal Copyright © 2007 EXarnkrackers, Inc.
LECTURE
5:
F LUIDS AND SOLIDS .
fluid flow, the sum of the three heads (the piezometric head) always measures to this same value. An energy line (EL) can be drawn horizontally at this level. The piezometer tube measures the piezometric head. A static pressure tap measures the pressure head and the elevation head, but not the velocity head. The hydraulic gradient line (HGL) can be drawn along the top of the static pressure taps. With this knowledge, you can use the continuity equation (Q ~ Av) to better understand fluid flow. The difference between the piezometric tube and the static pressure tap is the velocity head. The top of the elevation head is at the center of mass of the moving fluid. In the diagram below, the displacement from where the elevation head ends to where the velocity head begins is the pressure head. If that displacement is downward, there is negative guage pressure. Notke if a static pressure tap were placed at a position where there is negative guage pressure, atmospheric pressure would push air into the fluid.
Static pressure tap
Piezometer tube
, I
Ii
pgh
PiS"
P + pgh + In a real fluid, the energy line drops as the fluid progresses.
5.1 1 Surface Tension You should understand surface tension qualitatively for the MeAT. Any formula will be provided with a passage. Although more dense than water, a tiny needle can be made to float on the surface of water. The force supporting the needle is not the buoyant force; no water is displaced. The force supporting the needle is created by surface tension. Surface tension is the intensity of the intennolecular forces per unit length. Much like a spring, when the molecules at the surface of the water are pushed downward by the weight of the needle, the intermolecular bonds of the water are stretched, and pull upward. Surface tension is also responsible for the formation of water droplets. The intennolecular forces pull inward tending to minimize the surface area by creating a more spherical shape. (A sphere has the least surface area per volume of any shape.) Since surface tension is a function of the intermolecular forces, it is dependent upon the temperature of the fluid (the higher the temperature, the weaker the surface tension) and upon the fluid with which it is interfacing. Copyright (g 2007 EX2 rnkrackers . Inc.
'2 pv ~ K
87
88
MeAT
PHYSICS
Glass tubing
Water
Mercury
Related to surface tension is the phenomenon of capillary action, where a fluid may be pulled up a thin tube. For capillary action, recognize that there are two types of forces acting: the intermolecular forces responsible for surface tension (cohesive forces); and the forces between the molecules of the tube and the fluid molecules (adhesive forces). If the cohesive forces are stronger, a convex meniscus is formed and the fluid is pulled downward by the vertical component of the surface tension. If the adhesive forces are stronger, a concave Ineniscus is formed and the fluid is pulled upward by the vertical component of the surface teoston. In the diagram above, the adhesive forces between water and glass are stronger than the cohesive forces between water molecules, so the water is pulled upward. In the other tube, the adhesive forces between mercury and glass are weaker than the cohesive forces between mercury molecules, so the mercury is pulled downward.
Copyright cg) 2007 Examkrackers, inc.
lOS. All of the following would increase the volume flow rate of a fluid bei ng pumped through a pipe EXCEPT:
Questions 105 t hrough 112 are NOT based on a descriptive passage.
A. B. C.
105. An ideal fluid with pressure P flows through a horizontal pipe with rad ius r. [f the radius of the pipe is increased by
D.
a factor of 2. which of the following most likely gives the new pressure?
A.
P
B. C. D.
4P 16P The new pressure cannot be determined without more infonnation.
increasing the pressure difference between the ends of the pipe. decreasing the fluid viscosity. increasing the pipe radius. inc rea.ing the length of the pipe.
109. A spigot is to be placed on a water tank below the surface of the water. Which of the following gives the distance of the spigot below the surface h compared to the velocity with which the water will run through the spigot? A.
106. If the container pic tured below is filled with an ideal fluid. which point in the tluid most likely has the greatest pressure?
:~ v
B.
C.
:t=v
110. Two drops of equal volume of different substances were
A. B. C. D.
placed on the same flat surface. A side view of drop A and drop B is shown below.
A B C D
Drop A
Compared to drop B, drop A has:
107. Water in moist soi l rises through capillary action. The intermol ecular forces between water molecules are: A.
B.
C. D.
A.
weaker than the intermolecular forces between water a nd soil molecuJes. equal to the intermolecular forces between water and soil molecules. stronger than the intermolecular forces between water and soil molecules. The comparative strength between the intelIDolecular forces cannot be determined with the information given.
Copyright @ 2007 Examkrackers, Inc.
DropB
B.
C. D.
89
stronger intermolecular forces and lesser surface te nsion. stronger intermolecular forces and greater surface tension. weaker intermolecular forces and Jesser surface tension. weaker in termolecular forces and greater surface tension.
GO ON TO THE NEXT PAGE.
112. A spigot was opened at the bottom of a barrel full of water and the water was allowed to run th rough the spigot until the barrel was empty. Which of the following describes the speed of the water flowing through the spigot as the barrel emptied?
Ill . The diagram below shows a cross-sectional view of a cylindrical pipe of varying diameter.
\'---- --
A. B. C, D.
If an ideal fluid is fl owing th rough the pipe, all of the following statements are true EXCEPT: A.
B. C.
D.
Always decreasing Al ways increasing Constant Decreasing, then increasing.
The cross-sectional area is greater at point A than at point B. The pressure is lower at point B than at point A. The volume flow rate is greater at point A than at point B. The flow speed is greater at point B than at point A.
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90
STOP.
LeCTURE
5.12 Solids In Chemistry Lecture 4, we d iscussed the d ifferent structures of a solid. In general, atoms or molecules tend to be held together rigid ly. However, all solids are, to some extent, el.astic. In other words, they can change their dimensions by stretching or compressing, but not breaking, these rigid bonds. To discuss the elasticity of solids we must Lmderstand two concepts: s tress and strain.
Stress is the force applied to an object d ivided by the area over which the force is applied. It has the same units as pressure but by convention we use N / m' and not Fa in order to distinguish it from pressure.
Stress = FlA Strain is the fractional change in an object's shape. Strain is a ratio of change in dimension compared to original dimension, and has no units.
lldimension . Stram = - __- : - : --'--original d imension Stress is what is done to an object, and strain is how the object responds. Like force and displacement in Hooke's law, stress and strain are proportional to each other. This proportionality can be given as a ratio known as the modulus of elasticity. .. M o d UIus 0 f e IaStiC1ty
stress straIn
= - -.-
Up to some maximum stress, the modulus of elasticity is a consta nt for a specific substance. This constant is arrived at th rough experiment. Like Hooke's law, the maximum stress point is called the yield point. An object strained to its yield point will regain its shape once the stress is removed . Up to this point, the ra tio of stress to strain is described accurately by the modulus of elasticity. Beyond the yield point an object will rema in intact, but w ill not regain its o riginal shape. The stress-strain ratio is not described accu rately by the modulus when the sh'ess exceeds the yield point. When a stress is applied that is significantly greater than the yield point, the object w ill break. This is called the fmctllre point.
Fracture point
Yield point w w
~
ifJ
Behaves / according to modulus
Strain
A typical stress vs. strain curve
Copyright © 2007 Examkrackcrs, Inc .
5:
FLUIDS AND SOLIDS .
91
92
MeAT PHYSICS
There are three separate moduli that you should know for the MeAT: 1. Young's modulus (E) for tensile stress; the shear modulus (G) for shear stress; and
2.
3. the bulk modulus (B) for compression and expansion. All moduli work in both directions. For instance, Young's modulus works for tensile cOlnpressive forces and tensile stretching forces. Also, in each case, the force used in the equation is not the sum of the fo rces, which would be zero, but the magnitude of one of the forces.
Xo
F
,): -----,':+!!.h , ,
A
f - ----- -(
",j
Ll.x
F
./
,,
,,, ,, ,, ,, ,,
":'F
'-
E=
Il/l
,, ,, , , ~r' ,
-L Ilx 110
"0
:
I:
'
L:---l~/
F G=
,
:
:
L -L
~~'~~-- -----~~ -'1)-
,
!!.p
B=
2'.V Vo
The trickicsl thing ilbout moduli on the MeAT is proportion';:liity problems. For installCl.', if the
from the Introdu ctory Lecture of th is course.
t Bulk
Tensile Salty
Shear Salty
Salty
E
G
B
Copyright © 2007 EX.:lmkrackers, Inc.
LECTURE 5: FLUIDS AND SOLIDS . 93
5.1 3 Th erm al Expansion Solids typically expand w hen heated. As their molecules absorb energy, their vibrations require more room. Expansion is ty pically considered in one dimension (linear expansion ) or three dimensions (volume expansion). The formula for linear expansion is: t.L = LCf.l'.T
where L is the original length of the object, t.T is the change in temperature, and a. is a constant unique to the particular substance. N otice that the change in temperatu re is proportional to the change in length. Don't memorize the therma l expansion equations unless you just have absolutely nothing to do and you plan to be on Jeopardy someday.
The formula for volume expansion is: t.V = V~t.T
where V is the original volume and ~ is a constant unique to the particular substance. It is a simp le exercise to show that ~ = 30.. ~~----
~:':~that ~
Copyright © 2007 FXClmkr;)ckcr"i, Inc.
MC. ATC:U~
other is fina lly payingoffl
C:> --.. ~ -
JEOPARDY
----
-----
~
-----
I
94
MeAT
PHYS ICS
5.14 Equation Summary
Fluids at rest p=
S.G. =
V
P= ~
Psubstance
P=pgy
P water
Fbuoyant
= p Vg
Fluids in motion Q = Av K=P +
1Pvz + pgh
v = J2gh
Solids stress modulus of elasticity = strain
Copyright (fJ 2007 Ex<)mkr<:1ckers, Inc.
Questions 118 through 119 are based on the table of Young's moduli shown below.
Questions 113 through 120 are NOT based on a descriptive passage.
Substance 113. If a solid will buckle under pressure greater than 12 atm. and that solid has a specific gravity of 4, what is the maximum height of a circular column made from the solid that can be built at the earth's surface? (Note: I atm = 101,000 Pa.) A. B. C. D.
4m 12m 24m 30m
A. D.
A. B. C. D.
0% 33% 300% 900%
D.
A.
B. C.
0.001% 0.1% 1% 10%
D.
Aluminum
7.0 x lOw
Magnesium
4.1 x lO lll
Lead
1.5 x lO lll
Glass
6.0 x 1010
Copper Aluminum Lead Glass
0.25% 0.50% 1.0% 4.0%
120. The bulk modulus for a substance would be most important to a researcher who is testing material that will be:
8
116. A single steel column is to support a mass of 1.5 x 10 kg. If the yield strength for steel is 2.5 x 10' N/m 2 and safety
A. B. C. D.
regulations require the column to withstand five times the
weight it presently holds, what should be the approximate cross-sectional area of the base of the column?
A. B. C. D.
1.0 x 1011
119. A glass rod is SUbjected to a stress and undergoes a fractional change in length of 1.0%. If a lead rod is subjected to the same stress, it will undergo a fractional change in length of:
115. The Young's modulus for bone is 9 x 109 N/m'. What is the percent change in length of a tibia with a crosssectional area of 6 cm 2 , if it experiences a compressive force of 5.4 x 103 N? A. B. C.
Copper
118. If all of the substances listed are subjected to the same stress, which one will undergo the smallest fractional change in length?
114. Which of the following gives the percent change to the Young's Modulus for a substance, when its cross-sectional area is increased by a factor of 3? B. C.
Young's modulus (N/m')
0.6 m' 3 m' 6m 2
used in high tension cables. submerged deep in the ocean. subjected to high temperatures. transported at great speeds.
30 m'
117. The sole of a cerlain tennis shue has a shear modulus of 4 x 107 If the height of the sale is doubled, the strain will: A. B. C. D.
decrease by a factor of two. remain the same. increase by a factor of two. increase by a factor of four.
Copyright ZD 1'007 Exarnkrackcl-s, Inc.
95
STOP.
Waves
6.1
Wave Characteristics
A wave is the transfer of momentum and energy from one point to another. There are three types of waves: mechanical; electromagnetic; and matter. Although man y of the concepts discussed early in this lecture are applicable to electromagnetic and lIlatler waves, these waves ha ve some special features that will be d iscussed toward the end of this lecture and in Physics Lecture 8. Mechanical waves obey the laws of classical physics and require some medium through w hich to tra vel. The medium, if it is perfectly elastic, is momentarily displaced by a wave and then returned to its original position. Such a medium is called nandispersive because a wave m,aintains its shape and does not disperse as it travels. Nondispersive waves can be considered ideal waves. On the MeAT assume all media to be nondispersive unless otherwise indicated. Mechanical waves can be further separated into transverse and longitudinal waves. A transverse wave is one in which the medium is displaced perpendicularly to the direction of wave propagation, such as ~aves on a string. A longitudinal wave (also called a sound wave) is one in which the medium is displaced parallel to the direction of w ave propagation, such as a sound w ave in air.
·-;'tl~lli_.'!IIIII!lmIX_IIII!mlll!)_DI!IIIIII!III)_!IIIIII!!11 11)1
Longitudinal Wave
Transverse wav;>
98
MeAT PHYSICS
Simple transverse and simple longitudinal waves can be represented mathematically by the sine function. For a transverse wave, the sine fun ction represents vertical displacement of the medium w ith respect to time or displacement of the wave. For a longitudinal wave, a phase-shifted sine function represents either the change in pressure or the horizonta l displacement of the medium with respect to the time or d isplacement of the wave. If we exa mine this mathematical representation of a wave, w hen the x-axis is dis-
placement of the wave, the wavelength (A) is measured from any point in the wave to the point where the wave begins to repeat itself. For a simple sine function, the wavelength can be measured from trough to trough, or peak to peak. For any other function, a wavelength is measured frOlTI any poi nt, to the next point where the function begins to repeat itself. Wavelength has units of meters. The frequency (f) of a wave is the n umber of wavelengths that pass a fixed point in one second. Frequency is measured in hertz (Hz), or cycles per second. It is often written simply as l /s. The product of w avelength and freguency is veloci ty.
V=f ').. The recip rocal of frequency is called the period (1) . The period is the number of seconds required for one wavelength to pass a fixed point. When the x-axis is time, the period is from any point on the wave function, to the next point where the function begins to repeat itself.
T=~
f
The amplitude A of a wave is its maxim um displacement from zero. Amplitude is always positive.
f=
).
y
r----------~--------~,
number of waves that pass th is point in one second
\
Direction of propagation
+-----~-----~----------_t----------~~----------~~--------~~~-----.x
~
T ~ the number of seconds \..~------v---------/ ).
reqUired for one w ave Lo pass Liti s poi nt
Fo r transverse waves, the y axis is the displacement of the med ium perpendicu lar to the directi on of propagation. For longitudinal waves, the y axis may be either pressure or displacement of the medium para llel to the direction of propagation.
Copyright :-. 2007
b~lI11krackers.
Inc
LECTURE
Be aware that the wave func tion can be plotted against either disp lacement or time. For a sinusoidal wave moving along a string in the x d irection, the transverse displacement y is given by:
y(x, t) = A sin(kx - mt) where k is the angular wave l1umber, and 0) is the angular frequency (see Physics Lecture 2). This equation will not be useful on the MeAT, but understanding it helps to nnderstand waves.
6:
WAVES .
99
Hold on! Too many formu las! Don't memorize all of these formulas ; just the ones given in the equation summary at the end of each Lecture. Instead of memorizing, look at the relationships each formula implies.
Although the velocity of a wave is always given by the product of the wavelength and frequency, the velocity is dictated by the medium th rough which the wave travels. A change in frequency or wavelength does not change the velocity of a wave in a given l10ndispersive tnedium. Nor does the velocity of the wave source affect the velocity of the wave itself. In other words, the sound waves made by a speeding jet travel at the same speed as the sound waves made by a crawling turtle. Only the medium affects the velocity. Two aspects of the medium affect the velocity; 1.
the mediwn's resistance to change in shape (or elasticity); and
2.
the Inedium's resistance to change in lnation (or inertia).
For instance, the velocity of a wave traveling along a perfectly elas tic string is: V=
It;
~ (Here, T is tension not period.)
where T is the tension in the string (not the period), and J.l is the mass per unit length of the string. From this equation, we see tha t we change the medium without changing the string, but by increasing the tension on the sa me string. The velocity of a sound wave is given by: V=
~p (smClll p ressure variations only) = ItP(isothermal gases only)
where B is the bulk modulus of the medium and P is the pressure of the gas. It is no t important to memorize these equations, but it is important to be able to predict how a change in mediwn might change the wave velocity. For instance, if the tension is equal, a wave will travel faster on a lighter string than a heavier one.
For instance, notice that heavier mediums tend to slow waves down, while stiffer mediums tend to speed waves up.
However, notice that, when comparing sound waves in water and air, w e calulot predict the relative velocities w ithout actual va lues. Since water is heavier than air,
order to pass through it, the inertia of the medium (its resistance to motion) tends to slow it down . On the other hand, the greater the elasticity of the medium, the faster it snaps back to position moving the wave along. The elastic component stores potential energy; the inertial component stores kinetic energy. Wave
it should slow the sonnd waves. But actually, water more than makes up for its higher density with a much greater bulk modulus, and sound waves tra vel significantly faster in water. You should also be aware tha t fnr a gas, the velocity increases with temperatUre acco rding to the equation:
v=J
~T(Here,
velocity T is
temperature.)
where y is a constant for a specific gas (about 1.4 for air) w hich compensates for tempera ture changes during contractions, R is the uni versal gas constant, and M is the molecular mass. This indicates that the random velocity of the gas molecules is a limiting factor for the velocity of a sound wave. The greater the temperature, the greater the random velocity, the greater the sound wave velocity. In fact, the velocity of a sound wave through a gas is on the order of magnitude of (but slightly less than) the random velocity of its molecules. Copyright © 2007 Exarnkracke rs, Inc.
Since a wave must move the medium in
IS
a constant in a nondispersive
mediu m; it IS independent of frequency, wavelength, and amplitude.
100 . MCAT
PHYSICS
The velocity of sound waves in a gas is limited by the average speed of the molecules with in that gas. Thus, sound waves move more quickly through hot gases than through cold gases.
SlIIfnce waves, such as waves on the surface of water, have some special properties. They are neither completely transverse nor completely longitudinal. If the MCAT were to test your knowledge of surface waves, it would probably give you a forInula in a passage. Surface waves are also called gravity waves, beca use gravity acts as the e lasti c component. Because gravity acts as the elastic component, just like a
projectile, the mass (or density) of the liquid does not change the rate at which a surface wave rises and falls. Thus, the velocity of wave propagation is not changed by the density of the liquid. Tn shallow liquid, where the depth y is much smaller than the wavelength 'A., the velocity of a surface wave is given approximately by:
Notice that the velocity increases as the depth increases. Although not shown by this approximation, the velocity also increases slightly with amplitude. In deep liquid, a dispersive mediuln, the velocity increases w ith the wavelength as follows:
(y» 'A.) Know that intensity increases with the square of the amplitude and the square of the frequency for all waves.
Power is the ra te at which a wave transfers energy. Power in w aves is typically discussed in terms of intensity (1) . It has units of W 1m' Intensity of a sound wave is given by: I = 1/,puJ A ' v where p is the density of the medium, (J) is the angular frequency, A is the amplitude, and v is the wave velocity. For a wave on a string, simply replace p with /1. Recall from Physics Lecture 2 that (J) = 2ITf, so intensity is proportional to the square of the frequency. Dependence of intensity (or power) on the square of frequency and square of amplitude is true for all types of waves. Notice that frequency and amplitude depend upon the wave source, w hile density and velocity are factors of the medium.
Intensity is a useful way to d iscuss the rate of energy transfer of waves because a wave may travel in severa) directions at o nce. For instance, if you snap your fingers in the air, some of the energy is h'ans ferred away [rom your fiIlgers in the form of a sound wave moving in all directions. Although this energy remains constant, it is mov ing away from your fjnge rs and spreading out over the surface area of an ever enlarging sphere. The increase in area means that the intensity of the so und is de-
creasing. H ere, the intensity at any given radius r is the power divided by the surface area of a sphere and given by:
This is not a useful equation to memorize becallse it is applicable only under very limited circumstances. Copyright © 2001 Exarnkrackers, Inc.
LECTURE
Althou gh intensity fo r sound w aves is a measure of energy rate tran sfer per area, humans do not perceive intensity on a linear scale. For in sta nce, the sound waves created by the rus tling of leaves are about 10 times more intense than those created by normal breathing; yet, we don't p erceive them as 10 times louder. In order to compensate for this and to make intensity more intuitive, an artificial scale for in ... tensity level (I» has been created, based upon a logarithm ic scale of intensities. The units of this scale are decibels (dB) . The relationship between ~ and I is given by: ~=
1Olog -
I
I" where Io is the thresh old intensity of human hearing (th e lowest intensity audible by the typical human). All you need to understand about decibels on the MeAT is that, if the intensity increases bv ZI factor of 10, the decibels increase bv the addition of 10 decibels. In other
word~, an increase in intensity from 30 W / n~2 to 3000W I m~ is equivalent to
an increase of 20 decibels; I added 2 ze ros to intensity, so I acid 20 decibels to the decibl'i level. If I had added 3 zeros to the intensity, I would have added 30 decibels to the decibel le\'el, and so on .
Copyright © 2007 EXdlllkroc.ker:;, Inc.
1'.'
x10 xl 0' xlO' xlO'
I'.~
= = =
+10 +20 +30 +40
6:
WAVES . 101
125. Sound waves are an example of:
Questions 121 through 128 are NOT based on a descriptive passage.
A. B.
121. If an ocean wave hits a patiicular beach once every four seconds, and the wave peaks are 12 meters apart, with what velocity are the waves coming into shore?
A. B.
c. D.
C. D.
3 mls 4 mls 12 mls 48 m/s
126. When the frequency of a sound wave is increased, which of the following will decrease? I. Wavelength II. Period III. Amplitude
122. Waves generally travel faster in solids than in gases because: A. B.
C. D.
The density of solids is generally greater than the density of gases. The density of gases is generally greater than the density of solids. Solids are less compressible than gases. Gases are less compressible than solids.
A. B. C. D.
2 5
C.
10
D.
50
A. B. C. D.
124. The sound level of the chirping made by a bird at a distance of 5 meters is measured at 30 dB. When the same bird is 50 meters away the sound level is measured at 10 dB. How many times greater is the amplitude of the sound wave at 5 meters away compared to 50 meters away? A. B. C. D.
Copyright
750 m 1500 m 3000 m 4500m
128. If the intensity of a sound is doubled. the decibel level will increase by: A. B. C. D.
3 times greater. 10 times greater. 20 times greater. 100 times greater.
2007 Exurnkrackcr-s, Inc.
I only III only I and II only I and III only
127. A ship uses a depth finder to discover the depth of water beneath it at any time. The depth finder operates by sending a sound wave towards the bottom of the ocean and measuring the time it takes for the wave to be reflected off the bottom and return to the ship. At a certain point, it takes I second for the wave to return. What is the depth at that point? (The speed of sound in water is 1500 mls)
123. One end of a string is shaken each second sending a wave with an amplitude of 10 em toward the other end. The string is 5 meters long. and the wavelength of each wave is 50 em. How many waves reach the other end of the string in each 10 second interval?
A. B.
longitudinal waves because the medium moves perpendicularly to the propagation of the wave. longitudinal waves because the medium moves parallel to the propagation of the wave. transverse waves because the medium moves perpendicularly to the propagation of the wave. transverse waves because the medium moves parallel to the propagation of the wave.
102
less than IOdB. exactly 10 dB. more than 10 dB. exactly 20 dB.
STOP.
LEe' JR[
6.2
Superposition, Phase, and Int erference
The phase of a w ave relates to its waveleng th, frequency, and pl ace and time of origin. In a nondispe rsive mediuln, the pha se is constant and g iven by:
kx - wi = the wave phase The angu lar values in this equation are beyond the MCAT. Don' t memorize this equation. It is sufficient for the MCAT to think of phase as a ho rizontal shift of a wave on a Cartesian graph as shown below. Each waveleng th represents 360' . So half a waveleng th represents 180°. Two waves that are the sa me wavelength, and begin at the same point, are said to be in phase with each othe r. Two waves that are the same waveleng th but travel different distances to arrive at the same point, will be out of phase if that dis tance is not some multiple of the wavelength. The angle by which two waves differ is called their plinse COllstnnt. Two or more waves can occupy the same space. When thi s happens, if the waves are transverse, their displacements add at each point along the wave to form a new w ave. This supe rposition of waves is called interference . lnterference can be constructive or destructi ve. Constructive interference occurs when the sum of the displacements results in a greater displacement. Destructive interference occurs when the sum of the displacements results in a smaller displacement. After passing through each other, waves that interfere w ill revert to their original shape, una ffected by the interference.
Two wa ves in phase
Two waves 180" out of phase
Constructive lnterference
Destructive Interference
Any w aveform, no matter how irregular, can be created by supe rposition of a sufficient number of sine waves w ith the correct anlplitudes and wavelengths. A special case of superposition of w aves is the phenomenon known as beats. Beats occur when tw o waves with slightly different frequ encies arc superimposed. At some points they will be nearly in phase and experience constructive interference. At other points they will be out of phase and experience destructive interference. These points will alternate with a frequency equal to the difference between the frequencies of the o riginal two waves. This di fference is called the beat frequ ency. IBeat
Copyright © 2007 EXcHTlkrackcrs, Inc
=
Ifl - 1211
6: WAVES . 103
104
MeAT PHYSICS
Notice the lines below. They occur at slightly different frequencies. Your eye perceives equally spaced light and dark spots, which correspond to the beat frequency.
A good way to remember the beat frequency is to th ink about tuning a piano . It is impossible to tune a plano perfectly. The piano tuner li stens to the beat frequency of the tuning fork and the piano note. In order to bring the piano into perfect tune , the beat frequency must be zero. The tuner wou ld have to wait forever wh ile the beats get farther and farther apart until they are infinitely separated . By the way, the beat
frequency is an alternating increase and decrease in the intensity of the noise. What the tuner actua lly hears is called the pit ch . The frequency creating the pitch wou ld be an average of the frequencies from the piano and the tuning fork . Pitch corre lates with frequency; a high note has high pitch and high frequency.
When a wave reac11es an interface between two medial some or all of the energy and mOlnentum will reflect back into the first medium. Any energy and lllOluenturn not reflected ",.rill refract into the second mediuIll. Any refracting wave will continue in the same orientation with the same frequency but with a smaller amplitude and a different wavelength. The orientation of the reflected wave will depend upon the relative density of the two n1edia. When the wave reflects off a denser Inediuffi, the wave is inverted (Its phase is shifted 180') . When the wave reflects off a less dense medium, it is reflected upright (No phase shift occurs) .
Upright reflection
Important: When a wave transfers from one medium to the next, the wavelength changes and the frequency remains the
same.
Inverted reflection
The reflection of a wave is most easil y visualized by examining a wave pulse. A wave pulse is a single wavelength. If you imagine a string attached to a thread, and then ilnagine sending a wave pulse down the string, some of the energy would continue into the thread (the light medium) as an upright wave pulse, and the rest of the energy would reflect back as an uprigl1t wave pulse. Now imagine the same string attached to a heavy rope. The reflected wave is inverted.
Copyright @ 2007 ExarnkrZlckers , Inc.
LECTURE
Now let's look at two sine waves with the same wavelength traveling in opposite directions on the same perfectly elastic string. As shown by the diagram, when they pass through each other something interesting happens. The point where they collid e is never displaced. It doesn't move at all. This point is represented by the black dot in the d ia gram and is called a node. Notice also that only the points intersected by the two vert.ical lines experience n13ximum constructi ve interference. These points are called antinodes. Now imagine two endless rows of sine waves traveling in opposite directions on the same string. The s tring would hold perfectly still at the nodes and move violently up and down at the antinodes. This condition is know n as a standing wave.
L
,
A, What would happen if we locked both ends of the string at zero displacemen t by tying them to a wall and then genera ted a row of sine waves on the string? Let's assume that, at the string-wall interface, the entire wave is reflected back to the string and no energy is refracted into the walL In this s itua tion, since two nodes, one at each wall, are already specified, only certain wavelengths wou ld create a standing wave. All other wavelengths would create very sma ll, irregular oscillations of the string. A list of the wavelengths from largest to smallest of the possible standing waves for a given situation is called a harmonic series. The harmonics are I1tUTICopynght V 2007 Exarnkrackers . Inc
6:
WAVES .
105
106 . MeAT
P~YSICS
bered from longest to shortest wavelength. The longest waveleng th, called the first harmonic (A.,) or fundamental wavelength, is created w ith the fewest number of nodes, two. This means that the distance from one wa ll to the other is half a wavelength. The second harmonic ("-.,) is created by ad ding ano the r node. This makes the wavelength of the second harmonic equal to the distance between the walls. Each successive harmonic is created by adding a node. The equation for the harmonic series where each end is tied down as a node or where each end is loose creating an antinode is: L=
n~"
(n= l , 2, 3, ... )
where L is the di sta nce between the two ends of the string and II is the number of the harmonic. This equation is the same for longitudinal waves such as sound. A pipe closed or open a t both ends with sound waves inside will follow this equation. If only one end of the stri ng is tied down, or only one end of a pipe is open, the untied or open end is an anti node. For a string tied at only one end, or a pipe open at only one end, the equation changes to: L = 111.." 4
(11 = 1, 3, 5, .. )
Notice when one end is an antinode, the even numbered hannonics are missing.
The standing waves d escribed abo ve cause the string to resonate or vibrate at its natural frequency or resonant frequency. Since velocity is constant for a given mediun1, the resonant frequency can be found for any given harmonic frOln the equa tion v = fA. All m echanical structures have na tural frequencies at whi ch they resonate. If an outside driving fo rce is applied to a structure at the resonant frequency, the structure w ill experience n1axitnum vibration velocities and maximum displacement amplitudes. The cond ition where the natural frequency and the driving frequency are equal is also called resonance. An examination of the resonating string d iscussed above will reveal the d riving force to be the reflected wave. This demonstrates that both definitions of resonance are the same. In a non-ideal situation, energy is lost to some damping effect at the resonant frequency, and must be replaced by some outside driving force at the same frequency. A m aximum d isplacem ent is produced resulting in a sta nding wave.
Copyright (c) 2007 ExamkrDckers, inc.
133. If a guitar string is 0.5 m long. what is the wavelength of
Questions 129 through 136 are NOT based on a descriptive passage.
its third harmonic?
129. How many wavelengths are shown between the dotted
A.
0.25 m
B.
0.33 m
e.
0.5 m 1m
D.
lines in the wave fo rm below?
134. Two violinists are play ing together. slightl y out of tune. If one violinist produces a frequency of 883 Hz and the other produces a frequency of 879 Hz, beats would be heard with a freq uency of: A.
A. B.
C. D.
B.
e.
2 3 4
D.
2Hz. 4 Hz. 881 Hz. 1762 Hz.
135. A vibratin g strin g has consecutive harmonics at
wavelengths of 2.0 m and 4.0 m. What is the length of the
130. When two waves are superimposed, th e resulting wave can be found by sum mi ng their: A. B.
C. D.
string?
A.
frequencies
B. C.
periods wavelengths displacements
D.
136. Waves A and B, pietured below. mayor may not be in phase. If wave A and wave Bare supeli mposed, the range of possible amplitudes for the resultin g wave will be:
131. Tn order for two sound waves to have an audible beat frequency, the two waves must be:
A. B. C. D.
1.0 m 2.0m 4.0m 8.0 m
in phase. out of phase. close in frequency_
I
of the same wavelength.
Wave A 6em
132. All of the following statements are true about a resonating strin g EXCEPT: A. B.
C. D.
10 cm
waveB ~
A reso nating string forms a standing wave. TIle wavelength of a re..~on a ting string mu st coincide with o ne of its harmonics. Some spots on a resonating string will not move at all. If left al one, the amplitude o f a wave on a resonating string will grow infinitely large.
3em A. B.
e. D.
Copyrigh t 1) 2007 Examkrackers, Ine.
107
from tram from from
0 cm to 3 cm. 0 em to 9 em. 3 cm to 6 em. 3 cm to 9 cm.
STOP.
108 . MeAT PHYSICS
6.3
Simple Harmonic Motion
Any motion that repea ts itself is called periodic or harmollic molion. If we stand directly in front of someone who is steadily peddling a stationary bicycle, and we watch the peddles, they appear to move straight up and down. They move faster in the middle of the motion and slow down at the top and bottom. If we recorded this up and down motion on paper as the paper was pulled to our right, we would draw a perfect sine wave. This type of motion is a specific type of harmonic motion. It is called simple harmonic motion, which means that it is a sinusoidal function in time. Objects in simple harmonic motion exhibit similar properties of which you must be aware for the MeAT.
The function that gives the displacement x of an object in simple harmonic motion with respect to time is a simplified form of the eq uation for a wave moving along a string. It is: x(t) = A cos(wt + <\1)
where A is the maximum displacement, W is the angular frequency (w = 2rtj), and is the phase constant that depends on what displacement x we call I = O. (When x = A at 1=0, then = 0.) You will never need this equation on the MeAT. By taking the first and second derivatives of this equation we get the velocity and acceleration of the object in simple harmonic motion. Although these equations are not required for the MeAT, by combining the d isplacement and the acceleration equations we arrive at:
This simple relationship identifies two hnportant properties of simple harmonic motion; the acceleration is d irectly proportional to the displacement, but opposite in sign, and the acceleration and displacement are related by the square of the frequency. If we multiply both sides of [a(t) = -w' X(I) ] by mass III, we get Hooke's law' F = - 11luix, and we ca n see that the spring constant k is 11/0)2, Therefore, since a spring follows Hooke's la w, a mass bounc ing on the end of a massless spring exhibits simple harmonic motion. If we lay the mass and spring on a frictionless horizontal table, stretch the spring, and release it, we get the same motion, and we can exam-
ine it without the complication of gravity. When the sprIng is fully stretched or fully compressed, the restoring force is at a maximum. Since the force is at a maximum, the acceleration is also at a maximum. Since displacement is at a maximum, the elastic potential energy, '/, ki', is also at a max imum. At this point the velOCity is zero because the mass is reversing directions. Since the velocity is zero, the kinetic energy, Ih IIW 2, is also zero. When the n1ass is crossing the equilibrium point of the spring (the length of the spring at rest), the net force on the mass is zero because ill: is zero (F = - k6x). The potential energy is zero for the same reason. The velocity, however, is at a maximum, which means that kine tic energy is also at a maximum . Of course, no force means that acceleration is zero. So we see that another characteri,stic of most systenls in simple harmonic Inotion is an oscillation between kinetic energy and potential energy. No energy is lost to the surroundings. Copyright ~ 1 2007 Examkrackers, Inc.
LECTURE
I L'l.x
F = -kL'l.x
v=O a=O
v=max
a=max
v=O ~ " '.-::
..... . Total energy remains constant.
.
K .E. := 1'2
Since k = mol, and by:
(0
6 : WAVES . 109
/Ill'
= 2rrf, the period of the motion for the mass on a spring is given
Notice that the inertial factor is on top and the elastic factor is on the bottom. Since period is inversely proportional to velocity, this is
e
what "\ve would expect from a wave phenomenon.
Another apparatus that simulates silnple harmonic motion is a pendulum swinging at a small angle. (A 5 degree angle equals approximately 0.1 % deviation from simple harmonic motion. For the MeAT, assume simple harmonic Ination on all pendulums unless otherwise indicated.) Just like the mass on a spring, the pendulum exchanges energy fonus between potential and kinetic. A pendululTI has total gravitational potential at the top of its swing and total kinetic at the bottom.
T ension
= mgcos8 + m
rv'
L
T=2nfj Simple hannonic motion can take 111any forms and is easily disgllisf:'d on tht'M(, AT For eXilmp!e: til(' nrhit of a plilnf't vif'wpd fmm the s ide; a tetherball spinning around a pole viewed from the s ide; and electrons oscillating back and forth in AC current arc all forms of simple harmonic lllOtioll .
Usc Hooke's lilW to l'"lclp you remember that the
~1Cccleration
of any system in s imple harmonic motion is proportional to the displacement of that system . Also, remember that it is proportional to the square of the frequency. Recognize that for most systems in simple harmonic motion, en ergy oscillates between kineti c and one or more forms of potential. Copy ri g ht
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2007 Exarn kracke rs, Inc.
---7-
..
Center of gravity
mg mgcose
e
110
MeAT
PHYSICS
Finally, I can't res ist giving: this mnemonic. To remember the anguldr frequencies for a mass on
[1
=
_"
(tl-
m
\-\'dck em
&
-
, L
wiggle
Even though these formulas w ill not be on the MeAT, they ([In help YOLI remember things like the period of d pendulum is independent of the mass.
6.4
The Doppler Effect
The Doppler effect results because waves are unaffected by the speed of the source which produces them. If the SOU Ice luoves relative to the receiver of the waves (h ere called 'the observer '), each wave w ill travel a different distance in order to reach the observer, and therefore the observe r w ill not receive them at the same frequency as they were emitted. For instance, if the source were to move toward the observer as fast as the waves, all the waves would arrive at the observer a t the same time; obviously this would n ot be the frequency with which they were emitted. If we a re talking about sound waves, w here pitch changes with frequency, the observer would not hear the same pitch that the source emitted. With light, the observer would not observe the sa me color that was emitted.
If needed, the formula for the Doppler Effect for mechanical waves will be given to you QIt the MeAT as:
This formula is un li kely to be useful on the MeAT. The difficult aspec t of this forlTIllla is to understand wh en to use the plus sign and when to use the minus sign . The easy way to solve this problem is to follow th ese steps: 1) assume that the observer is not moving; 2) if the source is moving toward the object, label that direction nega ti ve and use the minus sign for V,; 3) now check the direction that the observer is moving. if the d irection is the same~ use the same sign for v~ if not~ use the opposite sign for v". Th is system is possible because velocity is a vec tor and once you label a direction positive for one vector it must be so for a ll vectors. For all waves, the Doppler Effect can be approximated by:
N
V -- - -
j,
C
and
V
C
(c is not neccessarily the speed of light) You should always use these formLLle for any Doppler MeAT problem, and not the more complicated formu la above. These formulae app roximate the Doppler Effect wh en the relati ve velOCity V of the source and the observer are much smaller than the wave velocity c. f'1f and f'1).. are the change to the source frequency f and the source wavelength A.
N otice that you must make a qualita tive judgement as to the d irection of the change in frequency and wavelen gth. This is simple: when the relative velocity brings the source and observer closer, observed frequency goes up and observed wavelength goes down; in the opposite case, the opposite is true. If the objects are getting farther apart as time passes, subtract N from h or add f'1A from A,; if they are approaching each other as time passes, add c..f to f , o r subtract f'1A from A,. Copyright
2007 EXilmkrackers, Inc.
LECTURE
6: WAVES
. 111
TIle relative velocity v is sometimes difficult to gra sp. It is simply the net speed at which the source and object are approaching each other. For objects moving in the same direction, subtract their individual speeds; for objects lTIoving in opposite directions, add their indiv idual speeds.
Same direction
-
+---
6 mis
) n ,:; -~-
..
11'
----~-----
---+ 6 m/s
Same direction
--.
Opposite directions
). Ill ,S
+--(, m/s
Opposite directions
5 m /~
6 m/s
-
11 I,n
<.;
---+
1n the d iagram below, the train's velocity tends to make me hear the vvh istle with a higher pitch. Notke that the wave peaks to the right of the trclin arc closer than those to the left. The movement of the train pushes the peaks closer together. By running to the right, I tend to sprcdd the wave peaks back out again as they strike I1W . I'm not as fast as the tra in , so the ne t result is that I hear the whistle (It a higher frequency.
Since the tra in is closing in on Sa lty, he hears the train's whistle at a higher frequency and a higher pitch. This is evident fro m the closer wavelengths.
Be sure to remember that, for light, when tile source and observer are approaching each other, the wavelength shortens creating a blue shift. When they separate, a red shift is created .
Also remember that for objects moving ill tile same direction at the same speed, there is no Doppler Effect; the relative velocity is zero , so the change in frequ ency is zero.
Copyrig ht ::J 2007 Exomk rJckcrs, Inc.
112
MeAT PHYSICS
6.5
Equation Summary
Waves
v = fA. T=
f
Sound
n". (n =n". (n
P= 10 log-I
L-
4
=
135) , , , ...
f ...,= If, -f, l
L
2
=
1, 23 , , ... )
10
The Doppler Effect
N _ "Q f, - c
/',." ;::- cv
Copyright
(1.2\
2007 Examkrar:kers, Inc
141. A piano creates a musical note when a metal wire stretched between two fixed ends is struck by a hammer, creating a standing wave. As the force with which the hammer strikes the string is increased, the amplitude of the string's motion is increased. Whi ch of the following properties of the wave on the string will remain the same as the force of the ha mmer is increased?
Questions 137 through 144 are NOT based on a descriptive passage.
137. If the mass on the bob of a pendulum is increased by a factor of 3, the period of the pendulum's motion will: A. B. C. D.
I. frequency
be increased by a factor of 2. remain the same. be decreased by a factor of 2. be decreased by a factor of 4.
II. wavelength III. velocity
A. B.
C.
138. Which of the following would most accuratel y de monstrate the kine tic energy of a pendulum?
D.
I only I and II only II and III onl y r. II, and III
c.
A.
142. As a pendulum pictured below swings through point A.
K.E.
time
which of the following is at a maximum?
time
D. K.E.
A.
o·L----
B.
time
C. D.
tangential acceleration displaceme nt from rest kinetic e nergy gravitational potential energy
139. If the amplitude of a sine wave is doubled. the intensity: A. B. C. D.
remains the same. increases by a factor of 2. increases by a factor of 4. increase< by a factor of 16.
143. A clock uses the motion of a pendulum to keep time. If the clock were placed at a height several thousand kilometers above the earth's surface, it would run:
140. Which of the following factors by itself will increase the
A. B. C.
frequency at which an observer hears a so und emanati ng from a source?
D.
A. B.
C. D.
A wind blows from the source to th e observer. The source and the observer move away from each other at the same speed. The source and the observer move in the same direction at the same speed. The source moves away from the observer more slowly than the observer moves toward the source.
Copyright © 2007 Examkrackcrs, Inc.
faster than it would on the surface of the earth. slower than it would on the surface of me earth. at the sa me speed that it wou ld at the surface of the earth. at a speed that can't be determined from its speed at the surface of the e311h.
144. All of the following are examples of harmonic motion EXCEPT: A. B. C. D.
113
a a a a
pendulum moving back and forth skydiver fa lling through the atmosphere. car moving around a circular track. string vibrating on a musical instrument.
STOP.
Electricity and Magnetism
7.1
Electric Charge
Like energy, charge is an entity that defies definition. Yet, all of us have an intuitive idea about what it is; we've all experienced a shock from static electricity, for instance. Charge is intrinsic to the nature of SOllle subatomic particles; it is part of their identity. Most of us are aware that there is positive c11arge and negative charge. The 'positive' and 'negative' signify nothing more than that these charges are opposite to each other. Instead of positive and negative, they could have been called up and down, black and white, or even had their names reversed to 'negative' and 'positive'. It is an accident of science that electrons were labeled negative and not positive, and, as a result of this accident, current runs in the opposite direction of electrons. Charge (q) is given in units of coulombs (C) .
Just as there is a universal law of conservation of energy, there is a Universal Law of Conservation of Charge. The universe has no net charge. In the majority of situations (and for the MCAT), net charge is created by separating electrons from protons. If we were to put all the positive and negative charges in the universe together, they would cancel each other out, right down to the last electron and proton. Thus, anytime a negative charge is created, a positive charge is created, and vice versa.
Charge is quantized. This means that any charge must be at least as large as a certain smallest possible unit. The smallest possible unit: of charge lS one electron unit (e = 1.6 x Hr l9 C), the charge on one electron or one proton.
Opposite charges attract each other; like charges repel each other, The formula describing the magnitude of the force of the repulsion or attraction between two charged objects is called Coulomb's law, and is analogous to the formula for gravitational force:
w here k is the Coulomb constant (k = 8.988 x 109 NIm 2 /C' ), q represents the respective charges, and r is the distance between the centers of charge.
116
MeAT
PHYS1CS
qA= 2qll MA ~2M "
I FAI ~ I F, I
•
/1
2Ia A I~ l a ,, 1
•
,
a,
boundary A
This diagram assumes that the force due to gravity is negligible compared to the facre due to charge. This is often a safe assumption.
Notice that the diagram above is the same as that used for gravity in Lecture 2. We want to emphasize the similarity between these two forces. Both gravitational and electric forces change inversely with the square of the distance between the centers of mass or charge. One major difference is tl1at gravitational forces are always attractive while electrical forces may be either attractive or repulsive. Also notice that the force due to gravity in the diagram above is ignored. Coulomb's fo rces are usually of a far greater magnitude than gravitational forces, and, unless the masses are very large, gravitational forces are negligible. In defining Coulomb's law, we u sed the phrase 'center of charge'. Similar to center of mass, the center of charge is a point from which the charge generated by an object or system of objects can be considered to originate. For example, the charges on a hollow, positively charged sphere made from conducting material will repel each other so that they move as far apart as possible. This results in the positive charge spreading uniformly along the outer surface of the sphere. Due to the symmetry, the center of charge exists at the center of the sphere, even though there is no actual charge at the center of the sphere. The electrostatic force on a charged object placed outside the sphere can be found using Coulomb's Lm,v, where r is the distance between the object and the center of the sphere.
C-i-)
••
r
.'
Notice the san1e is true for gravity. The similarities between gravity and electricity stem from the fact that both mass and charge create fields. A field is a man-made concept designed to explain action Copyright @ 2007 Exam krackers, Inc.
LeCTURE
at a distance. Recall that on the MCAT, any force that acts on your system mLlst be physically contiguous to it, except for the forces of gravity, electricity. or magnetism. This is because these forces are created by fields and can act at a distance.
Any field can be represented by lines of force . Lines of force point in the direction of the field (positive to negative for electric fields, towards the mass creating the field for gravitational fields). The relative distance between lines indicates the strength of the field; the closer the lines, the stronger the field . Lines of force can never intersect, as this would indicate a field pointing in tvvo different directions from the same location, an impossibility. The lines of force for a single positive point charge are shown in the diagram above. Examine the lines of force for the field created by the positively charged, hollow sphere. Notice that the inside of the sphere has no electric field. A negatively charged sphere would produce the same result. This is because the lines of force must begin on a positive charge and end on a negative charge. This is an impossibility for lines entering the spl1ere. Thus there can be no lines of force inside a uniformly charged sphere. Again, the same is true for a gravitational field.
An electric field is defined as the electrostatic force per unit charge. The symbol for any electric field is E. E is a vector pointing in the direction of the field and has units of N I C or Vim. For a point charge, the electric field is found by dividing Coulomb's la w by q giving:
E =k~
" Copyright (g 2007 Fxarnkrackers, Inc
7:
EI_ ECTRiClTi hNO M/\GNETISrvl .
1 17
118
MeAT PHYSiCS
The electric field for a system of point charges is found by summing the fields due to each charge. Remember that E is a vector and you must use vector addition when summing fields. The symbol for the gravitationaJ fieJd near the surface of the earth is the familiar g. When we wish to discuss the gravitational force on any object at the surface of the earth, we normally use 'mg' and not IF:::: Gmm(r-'. This is because we are very familiar with the gravitational field near the surface of the earth, and thus we have created a shorthand method for describing the gravitational force for any mass; that method is the mass times the field, mg. Similarly, the force on a charge (q) in an electric field (E) is:
F=Eq To find the potential energy of a mass in the earth's gravitational field relative to some other position, we multiply this force times the displacement in the direction opposite the field, mgh. Similarly, the potential energy (U) of a charge in an electric field is the force times the displacement (d):
U = Eqd where d is measured from a zero point of our own choosing, similar to h in gravitational potential energy. Since this particular electric potential energy is dependent upon position, it is also a type of potential energy. If the electric field is created by a point charge, we can derive the electric potential energy from Coulomb's law:
u = k q]q2 r Notice that according to this formula, electric potential is zero for particles separated by an infinite distance. Since energy is a state function, its value can be arbitrarily assigned, and it is given a zero value in this case by convention.
Field
1
/~ss (
~~~s
or
. charge
or charge
W = II/gil or
liard
W ='1£11
Recalling our study of gravity, if we wanted to create a function for the work required to move any given mass along any frictionless path near the surface of the earth, what would this function be? I~ other words, we are looking for a function in trinsic to the gravitational field, which is independent of any mass. What function would give us the 'potential' of the field in terms of work gained or lost per unit mass? The answer is gh, the field times the displacement in the direction opposite the field. If we multip lied any mass times gh, we would have the work done by the field in moving that mass. This is called the potential of the field . In electricity, poCopyright @ 2007 Examkrackors, Inc
LECTURE
7:
ELECTRICITY AND MAGNETISM
•
119
tential has a special name, voltage. Voltage (V) is the potential for work by an electric field in moving any charge from one point to another.
V=Ed Voltage is given in units of volts (V), and is a scalar. You should also recognize voltage in units of J/ c. The voltage due to a point charge is:
N
V = k'll r Since voltage is a scalar, when finding the voltage due to a group of point charges, the voltages due to each individual charge can be summed directly. Notice from the diagram on the oppostie page that, like the work done by gravity, the work done by an electrostatic field is independent of the path. This is because both fields are conservative; they both conserve mechanical energy. As we will see when we discuss magnetisffi/ this is not true of all electric fields.
N C orY 'lll
U = kq~~2
Within an electric field, movement perpendicular to the field does not result in a change in potential, just as a mass moving along the surface of the earth does not experience a change in its gravitational potential. In any electric field we can define a surface normal to the field that describes a set of points all with the same potential. Examples of such surfaces are shown as dashed lines in the diagram below. TIley are called equipotential surfaces . All points on an equipotential surface are at the same voltage. An equipotential surface can be drawn at any point in the field.
Also shown in the diagram are the field lines of an electric dipole. An electric dipole is created by two opposite charges with equal magnitude. An electric dipole Copyright IT] 2007 Exarnkrac kers, Inc
J
Y or J/C When working electricity problems be sure to know what type of electric field you 're working with , The formulas in th e box above refer to an electric field created by a point charge . Point charges create electric fields that change with r If your electric field is constant, like the field inside a capacitor. you should NOT use ti,e formulas In the box above, If your field is created by a point. you should be careful about using formulas other than those in the box above,
120
MeAT PHYSICS
momen t (p = qd) is a vector w hose magnitude is the charge q on one of the charges times the distance d between the charges. In physics, this vector points in the opposite direction to the electric field, from the negative charge to the positive charge. In chemistry the vector points from positive to negative. At large distances the electric field of a dipole varies by 1/ r'. A dipole p laced in an electric field will tend to align itself along the field in the opposite orientation to the field. Thus a dipole not perfectly aligned with an external electric field will have a potential energy of U = - pEcDs8 where 8 is the angle between the d ipole and the electric field.
Copyright
(C)
2007 Exarni
148. If the distance between a point charge and an infinitely large charged plate is increased by a factor of 2-, the new force on the point charge will:
Questions 145 through 152 are NOT based on a descriptive passage.
A. B. C. D.
145. Two charged metal plates are placed one meter apart creating a constant elec tric field between them. A one coulomb charged particle is placed in the space between them. The particle experiences a force of 100 newtons due to the electric field. What is the potential difference between the plates? A.
IV
B. C. D.
lOY 100V 1000V
149. A positively charged particle starts at rest 25 cm from a second positively charged particle which is held stationary throughout the experiment. The first particle is released and accelerates directly away from the second particle. When the first particle has moved 25 cm, it has reached a ve locity of 10 mls. What is the maximum velocity that the first parlicle will reach? A. B. C. D.
146. How much work is required to move a positively charged pm"ticle along the 15 em path show n, if the electric field E is IONIC and the charge on the particle is 8 C? (Note: ignore gravity)
~ ....... ---------------.--~
~--.-.---.- ...
Oem
A. B. C.
D.
---.--
.'
10 m!s 14 m!s 20 m!s Since the first particle will never escape the electric field of the second particle, it will never stop accelerating, and will reach an infini te velocity.
150. The electric field for two point charges A and B is shown below. Which of the following is true?
I /
Scm
•
10 cm
0.8 J 8J 12 J 1200 J
A. B. C. D.
147. If the distance between two point charges is increased by a factor of 3, the new force on either charge will be: A. B. C. D.
decrease by a factor of 4. decrease by a factor of 2. re main the sa me. increase by a hlctor of 2.
Both charges are positive. Both charges are negative. The charges have opposite charges. The charges can not be determined.
decreased by a factor of 9. decreased by a factor of 3. remain the same. increased by a factor of 3.
Copyright © 2007 Exarnkrackers, Inc.
121
STOP.
152. When -10 C of charge are moved from point A to point B in the diagram below, 90 J of work is done.
151. Two particles are held in equilibrium by the gravitational and electrostatic forces between them. Particle A has mass Ina and charge q a" Particle B has mass mb and charge q". The distance between (he charges is d. Which of the following changes wi ll cause the charges to accelerate
+ - -+. A + - -+ + --+. + ~ + - - -.
towards one another?
A.
B. C. D.
is doubled and m b is doubled. is doubled and m/, is halved. q, is doubled and qb is doubled. d is doubled. mo
l1lu
B
1111
l cm The voltage between point A and point B is:
A. B. C. D.
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Ex,~tYl krJc k ers,
Inc.
122
0.9V 9V 90V 900V
STOP.
LFCTLJRE
7.2
ELECTRICITY AND MAGNETISM
.
123
Movement of Charge
When charge moves along an object (usually in the form of electrons), that object is sa id to be conducting electricity. At the same time that it is conducting electricity, the object also resists the movement of charge. All substances conduct electricity to some extent, and all substances resist movement of charge to some extent (superconductors excluded). Substances resist and conduct charge to different degrees. However, it· turns ou t that the vast majority of substances either conduct charge very well or very poorly. Thus, we can safely classify most substances as conductors or resistors. Good conductors, such as metals, allow elec trons to flow relatively freely Poor conductors (good resistors) hold electrons tightly in pl ace. Poor conductors are represented by, among other substances, network solids such as diamoild and glass.
1st conductor
J
e eO e-
e e" e
Negatively charged object 2nd conductor Since electrons flow easily along conductors, we can charge a conductor by induction. If we move a negatively charged object close to an electrically insulated conductor, the electrons on that conductor will be repelled to its opposite side. If we then touch a second conductor to the first, tbe electrons will move still further from the charged object by moving onto the second conductor. Once the second conductor is removed, the first conductor is left with fewer electrons than protons, and thus has an induced positive charge. Moving charge is ca lled current. Current is given in amps (A), or C! s. Current is a scalar, but we describe its flow to be in the direction of the movement of positive charge. Unfortunately, Ben Franklin designated electrons to be negative without realizing it. Because of tb is, current, which is usually created by flOWing electrons, is in tile opp osite direction to the flow of electrons. The flow of electrons resembles the flow of a fl uid. Like molecules in a moving fluid, electrons move very fas t in random directions, while there is a much slower uniform translational movement (called drift speed) opposite the direction of the current.
7.3
7:
Circuits
A circuit is a cyclical pathway for moving marge. As we learned earlier, all substances resist the flow of charge. The quantitative measure of this property is called resistivity (p) . The quantitative measure of an object of a particu lar shape and size to resist the flow of charge is ca lled its resistance (R) and is measured in ohms (0) . If an object is made from a homogeneous conductor, the resistance of the object when a voltage is applied unifonnly to its ends is related to the resistivity of the material that it is made from by: L R=p -
A
This form ula dem onstrates that if the length of a wire is doubled or its cross-sectional area is cut in ha lf, its resistance is also doubled. This is similar to what we Copyright © 2007 Fxamkrackers, Inc .
Notice that, since charge moves easily through a conductor, if an object is made of a conducting material, it can only hold excess charge on its surface. The excess charges try to move away from each other, and they end up only on the surface. To visualize this, imagine 10 people with terrible body odor in a round room. Obviously they want to get as far away from each other as possible. They move to the walls to maximize thei r distance from everyone else. Like the uniformly charged sphere, the electric field in side a uniform ly charged
conductor is zero .
124
MeAT
PHYSICS
would expect for fluid flowing through a pipe. Many useful analogies can be made between fluid flow and electron flow. The prod uct of the resistance (R) and the current (i) gives the voltage.
V = iR This is known as Ohm's law. Ohm's law is a very useful form ul a for analyzing circuits. This law also reveals another useful analogy between fluids and electricity. Recall from Lecture 5 that the change in pressure in a real fluid moving through a horizontal pipe w ith constant diameter is given by the product of the volume flow ra te and the resistance (!;'P = QR). Beca use we are probably more intuitive about fluids, it is often helpfu l to think of current as flow through a constant diameter pipe, and voltage as the difference in height between points in the pipe. (More precisely, voltage is analogous to gl1.) If we grasp this analogy, it makes a useful aid in remembering Kirchoff's two rules.
Kirchoff's first rule states that the amOlmt of current fl owing into any node must be the same amount that flows out. A node is any intersection of wires. If we imagine cur ren t as fluid, it becomes obvious that the rate at which fluid flows into an intersection must match the ra te at which fluid flows out. Otherwise, a pipe would burs t. Kirchoff's second rule states that the voltage arolmd any pa th in a circuit must SUln to zero. If we imagine vol tage as the height difference between two points, this rule states the obvious that the height of the starting point does not change when we go around some path (regard less of the path) and end up back where we started.
A battery adds energy to a circllit by increasing the voltage from one point to another. In Ollr ana logy to fluids, a battery pumps the fluid to a greater height. Batteries are rated with an electromotive force (EMF) . EMF is not a force at all, bu t is simply a fancy word for voltage. In fact, it is a mis take. EMF was named before scientists really understood voltage. Real batteries have internal resistance. Most of the time on the MeAT, there will be no internal resistance. Always assume that there is no internal resistance wllcss othen'Vise ind ica ted. To account for internal resistance, simply redraw the battery and place behind it, or in front of it, a resistor the size of the internal resistance.
Copyright (,\:) 2007 Exan,krackers, Inc
LECTURE 7: ELECTRICI ry AND MAGNETISM .
Here is an example o f lUY analogous fluid ci rcuit. Each spinning wheel represents resistance
[IS
it res ists the movement of the fluid . Technicc1l1y, voltage
should only drop across resistance as per O hm's law, so in
. my fluid circllit, the height of the tluid should only change when the fluid goes through a resistor. A battery works to put energy back into the circuit. To t1nalyze this circuit, we
add the voltage drop across each n:~!"islor, i.l lld set it equal to the volt.1g:c of the battery. This would
be
annlogous to the
\.vork done on each wheel se t
equal to the work done by me. (Morc preCisely, we
\Vould d i, ide the work by lhe
ITIclS!-'
to arrive at the po-
tential;
however,
unnCCCSS[l rily
this
complicates
the analogy.) A capacitor is used to temporarily store energy in a circuit. It stores it in the form of separated charge. [n a parallel plate capacitor, two plates ll1ade from condu ctive Inaterial are separated by a very small distance. On a charged capacitor, one pl ate ho lds positive charge, and the other plate holds the exact same amount of negative charge. This separation of charge creates an electric field that is constant everywhere beh'Veen the plates. The electric field is given by:
Notice that this K is not the Coulomb's constan t. This is the dielectric K, which we will discuss be low. Q is the charge on eithe r plate. The E" term is derived from Coulomb's constant k. It is related to k by:
By defin ition, capacitance is the ability to store charge per unit voltage. In other words, something with a high capacity can store a lot of charge at low voltage.
In a parallel plate capacitor, since the charge sits on the surface of the plates, the taller and wider the face of each plate, the more charge each plate will be able to store. Recall that charge sits on the surface of a charged conductor. In a charged capacitor, the charge sits on only the inside fa ce of each plate. Therefore, the thickness Copyri~Jht (£l
2007 ExnmkrDckcls, Inc.
125
126
MCAT
PHYSICS
of the plates of a capacitor w ill not in crease their ability to s tore ch arge. Recall also that voltage is d efined by d istance (V = Ed). Thus, the farther the p la tes are separated, the g reater th e voltage, and the low er the cap acitance. The phYSical makeup of a p ar allel p la te cap acitor in term s of p late area (A) and separa tion dis tance (d) is given by: Irs easy to see where the formula for energy stored by a capacitor comes from. If we imagine a capacitor with no charge . on it. the voltage across th is capacitor must be zero. For each unit of charge that we add, the voltage increases proportionally (0 ~ CV) . If we graph this, we get a straight line. The area under this line is the product of charge times voltage. or energy. The area is a triangle, which is '/l base times height or V, QV.
C =K AEo
d A capacitor 's job is to store energy (generally for quick use in the future). The energy (U) s tored in any shape capacitor is given by:
30
?
20 10 0 -10
--.
10 20 30 40 SO 60
u=
U = !QV 2
L or
e
e
voltage (V )
Q
V
or
1 Q2 U = --
2 C
If you know anyone of these equations, the others can be derived from Q = CV . In ollr fluid circuit, .1 capacitor \\'Guld look like this:
1 %b
r
C
.3
'0
>
Fluid c
As the fluid comps to the fork at pipe Band C, some fluid would move in each direction . As the tluid capacitor fills, the fluid flow through pipe C would eventually come to i:l stop and all the fluid \·vould move through pipe B. In order to 111aintain tilt' fluid c(1pacitor at hL~ighl h, tluid flow through pipe A would have to have kinetic energy equal to the grc1Yitational potential of the fluid capacitor. This rl'sults in the cqulltion v = 12gh . The flu id G~pacitor now stores Cl1l'rgy for thl' circuit. 1f flnw Copyright © 2007 EXi)mkrackers, Inc.
LFCTURE
through pipe A is suddenly blocked , the capacitor would empty w ith an initial \'Clocity of v = 12gh . You s hould recognize the s h'lpe of the \'ultage vs. time g r(lph~ for ch
The dielectric constant, K, refers to the substance between the plates of a capacitor. The substan ce between the plates must be an insulator, otherwise it would conduc t electrons from one p late to the o ther, not allowing any buildup of charge. A dielectric acts to resist the creation of an electric field, and th us allow the capacitor to store more cha rge (to have greater capacitance). Usually a dielectric contains dipoles oriented in random directions. Recall that a dipole in an external electric field has potential en ergy depending upon its orientation. When the electric field begins to build up between the plates of a capacitor, the dipoles are rotated to point in the direction of the electric field (from a phYSics n ot a chemistry sense). This rota tion requ ires energy in the form of work done on the dielectric. The work is conserved in th e field, thus the capacitor is able to store more energy. Another way to look at it is from the standpoint that each d ielectric creates its own electric field that reduces the overall electric field within the capacitor. The more charge required to build an electric field , the more en ergy stored within a capacitor. The dielectric constant of a vacuum is defined to be unity (one). Air is very close to one, and all o ther dielectric cons tan ts increase from there.
In order to analyze a circuit you must recognize the symbols for a battery, a capacitor, and a resistor.
capacitor
battery
Lines connecting compon ents sh ould be considered completely non-resistive wires.
-4
-4 Resistors ill series
Copyright © 2007
Ex~m krac k Ns .
Inc
ELECTRICITY AND M AGNETISM .
gh
I
d ischnrging
ti Illl'
+ Work is done
One oth er effect of a dielectric is to limit the value of the possible voltage across the plates. A t some maximum voltage, the dielectric will break d own and conduct electricity. This value of a dielectric is called the dielectric strength. If d ielectric strength ap pears on the MeAT, it will be explained in a passage.
resistor
7:
Resistors in parallel
on the dielectric and energy is stored
in the dielectric.
1 27
128
MCAT
PHYSICS
You also lTIUst be able to recognize when these components are in parallel and when they are in series. This has nothing to do with their orientation in space; parallel components are not always pointing in the same direction. Components lined up in a row, like train cars, are in series. More precisely, any two components not separated by a node are in series. Single components in alternate paths connecting the same nodes are in parallel. When resistors are in series, their total resistance (effective resista nce, Rt'ff) is the sum of their resistances.
(Resistors in series) When they are in parallel, their effective resistance can be arrived at through the followin g equatjon:
1
1
Reff
Rl
1
- - = - +- - +
Rz ...
(Resistors in parallel)
Capacitors are exactly opposite. In parallel their capacitance sums directly to give an effective capacitance:
Ceff = C1 + C2 + ...
(Capacitors in parallel)
in series they follow the equa tion below:
1
- -
Ceff
1
1
C1
C2
= -- + -
- + . ..
(Capacitors in series)
To solve any circuit on the MCAT, we must simplify it as shown below. We begin by replacing components in parallel and series with their corresponding effective comp onents. \Alc continue this process until we have our simplified circuit; one of each clement. Next we use Olun's law to find the missing quantity. For more complicated circuits we would ha ve to use Kirchoff's rules. The solutions to more complica ted circuits will not be required on the MeAT.
2Q
6Ve
,.-_........ _....... .
,t20 ~ 2Q
) Equiva lent circu it
\, -- --- -- --,. ------_."
6VD'
2Q
10
, ,,
:
, ' ,,
' - - - - - - - - T - - - - - - - - --
Series resistors Parallel resistors .2 .•\
Solution for curren t
Equi valent 1A
CIrCUI t
I
-J,
6V -D3l2 Simplified circu it
Solution for vo ltage
Copyright
© 2007 EXi.1rnkrackers, Inc
LECTURE
7.4
Power
Electrical power is the Same quality as mechanical power, just as electric energy is the same quality as mechanical energy. In other words, power is power regardless of what source produced it. Power is often used on the MCAT to integrate electricity and mechanics into the same questi on. For ins tance, you may be given the voltage and current of a machine and be asked how quickly it can lift a mass to a certain height. In such a problem you woul d set the electrical power equa l to mechanical power. The equations for electric power are:
P = iV and P = i 2R and P = V 2/R If yo u remember anyone of these, the others can be derived from Ohm's law by plugging in for V, i, or R. As curren t goes through a resistor, heat is gene rated. The rate at which heat is generated is the power diSSipated. The second and th ird equation shown are appli cable on ly to energy dissipated as heat by a resistor. This is unlikely to cause any confusion on the MCAT.
Copyright © 20Ql Examkrackers, Inc.
7:
ELECTRICITY AND MAGNE ISM .
129
156. The resistors in the circuils below each represe nt a lighl bulh. If all Ihree circui ts use the same size ballery, which circuit w ill produce the mosl lighl?
Questions 153 through 160 are NOT based on a descriptive passage.
DO
153. Whal is nel force on the dipole inside the capacitor if the plates are separated by I em?
I
II
B III
2Q
16V
A.
ON
B. C. D.
4N
A. B. C. D.
2Q
I only II only III only I, II, and 1lI will produce Ihe same amount of light.
157, If all the resistors in the circuit pictured below have equal resistances, and the current flowing into resistor A is 4 amps, what is the current flowing into resistor F?
8N 16N
154. Each resistor in the circuit below has a resistance of 2 Q. The battery is a 12 volt battery. What is the current across resistor B? A
A
B
c
o
E
F
B
A.
I A
B. C. D.
2A 3A 4A
A.
c
B. C. D.
16A
158. What is the energy required to operate a 60 W light bulb for 1 minute? A. B. C. D.
155. Which of the following changes to a parallel plate capac itor would not increase its capacitance? A. B. C. D.
2A A 8A
IJ 60J 360 J 3600 J
decreas ing Ihe distance between the plates increas ing the area of Ihe plates increasing the dielectric constant increasing the voltage across the plates
Copyright © 2007 Ex,lm krackers, Inc.
130
GO ON TO THE NEXT PAGE.
160. All of the following expressions are equal to an Ohm EXCEPT:
159. The circuit shown below has three resistors connected in parallel to a battery.
A. B. When an additional resistor, R4 is added to the circuit: A. B.
C. D.
C.
The voltage produced by the battery will be increased. The voltage produced by the battery will be decreased. The current produced by the battery will be decreased. The power produced by the batter will be increased.
Copyright @ 2007 Examkrackers, Inc.
D.
131
Vsec C
W
A' A V V' W
STOP.
132
Me AT
PHYSi CS
7.5
AC Current
Up to now, we have considered only direct current (dc current) , where the net movement of electrons is in one direction around the circuit. Since movement of the electrons creates power regardless of direction, electrons do not have to be driven in one direction. Alternating current (ac cUITent) is created by oscillating electrons back and forth in simple harmonic motion. This is the current that is cOffilllonly used in honle outlets in the U.s. Since it is simple harmonic Ination, the voltage or the current can be described by a sine wave. Maximum current occurs when the electrons are at maXimUlTI velocity. The important thing to know about ac current on the MeAT is that the 11laximUlll voltage or current is given by:
The 'rms' stands for root Inean square. In other words, it is the square root of the average (lI1ean) of the squares. To find the 'rms' value of something, you square all the terms, take the average, and then take the square root. The reason that rms values are used for sine wave functions is that the average value of a sine wave is zero. The electric company wants to charge you for the electricity that you have used. The average ac current is zero. Since the current is only at a Inaximunl periodically, they can' t charge you for the maximum either. Thus they charge you for the rms value. It doesn't hurt to know that the 'rms' voltage in the U.s. is typically 120 volts. This
corresponds to a 170 volt maximum.
7.6
M agnetism
Magnetic fields have many similarities to electric fields. Magnetic field strength is measured in units of tesla, T . Like the positive and the negative of electricity, a magnetic field comes with north and south poles, where like poles repel and opposite poles attract. Unlike electric charges, magnetic poles have never been found to exist separately; one pole always accompanies the other. Similar to the electric field, the magnetic field can be represented by lines of force. The lines of force in a magnetic field point from the north pole to the south pole. Interestingly. the Earth's magnetic field points from the geographic South Pole to the geographic North Pole. Thus, what is called the magnetic north pole is rea lly the south pole of the Earth's magnetic dipole. (The actual magnetic poles are also 11,SO away from the geographic poles and constantly shifting very slowly.)
A magnet
The Earth
A changing electric field creates a magnetic field. A stationary charge does not create a magnetic field; a moving charge does. Current is moving charge. Thus, current creates a magnetic field. The magnetic field B created at a displacement r by a current i moving through a small section of wire with length L is given by: Copyright © 2007 [xarnkrilckcrs, Inc
LECTURE
B = h iLsin6 4n: r2 where ~o is a constant, called the permeability constant and 8 is the angle between the direction of the current and the displacement r. This equation (a form of the Biot-Savart Law) should not be memorized for the MeAT. It should be noted. however, that, like electric fields, magnetic field strength follows the inverse square law.
If we apply the Biot-Savart Law to a very long, straight wire the equation becomes: B= )lj 2rrr
Notice that for a very long straight wire the ll1agnetic field varies inversely with the displacement and not by its square. In other words, the magnetic field in this case varies with r, not fl. The direction of the magnetic field due to a current carrying wire is predicted by right ha11d rule. If. using the right hand. we place our thumb in the direction of the current and grab the wire, the direction in which our fingers wrap around the wire is the direction of the magnetic field.
A charge moving through a magnetic field experiences a force. The force (F) on a charge (q) moving with velocity (v) through a magnetic field (B) is:
F = qvBsin9 where e is the angle between the magnetic field and the velocity of the charge. You must know that the force is directed perpendicularly to both the velocity and the magnetic field . This leaves only two possible directions for the force. Right hand rule also predicts which of these two directions is correct. The MeAT is un likely to ask you to use the right hand rule. but it will require that you understand that the force is perpendicular to both the velocity and the magnetic field. To find the force with right hand rule. using your right hand, again point your thumb in the direction of the movjng positive charge, and point your fingers in the direction of the magnetic field. Your palm will point in the direction of the force. For a negative charge moving in the same direction, the direction of the force is reversed. In the example to the right, a positive charge moves to the left through a magnetic field pointing downward. Right hand mle predicts that the force will be diN rected out of the page.
,
Since this force is always perpendicular to the v elocity. it does no work (W = Fdcos8). It changes the direction, but never the magnitude of the velocity. Thus, this force always acts as centripetal force and can be set equal to mv 2 It to find the radius of curvature of the path of the particl e. Copyright (0 2007 EXdmkrackcrs, Inc.
[]
•
-;;;A+
7:
ELECTR:CITY AND MAGNETISM •
133
134
MeAT
PHYSICS 7
~
!
+2
x
B Directed in to the page
q
Path of
B .,(1(11::::
particle
,,' r
lll-
e ~ 'U : Si1l8:::: 1 1 [1
The force on a current carrying wire placed in a magnetic field is: F:::: iLBsin8, where L is the length of the wire within the field and
e is the angle between the wire and
the magnetic field. The direction of the force can be found using the right hand rule, and will be perpendicular to both the wire and the magnetic field.
A changing magnetic field creates an electric field, However, unlike the electric field created by a stationary charge, this field is non-conservative. The mechanical energy creating the electric field is not conserved, but is disSipated as heat in the charged object. Thus, electric potential has no meaning for electric fields induced by changing magnetic fields. Imagine a loop of wire pulled out of a magnetic field. As the magnetic field around the wire changes, an electric field is created and a current develops in the wire. The current created in the wire as it moves out of the external magnetic field creates its own magnetic fieJd. A force is required to remove the loop at a constant velocity. The work done by this force is not conserved, but, instead, creates thermal energy in the loop.
~
~ I
l'
~
J B
~
,' ~
~"
M
,
+.--
2
J
Faraday's law B
This effect is simply stated in Faraday'S law of induction, which says that a changing magnetic flux (~B/ M) induces an emf (E). The magnetic flux is the number of magnetic field lines running through the loop shown above. Since the number of these lines changes as the loop is removed from the magnetic field, an electric field and a current are produced inside the wire. By the way, the electric field is induced even if the loop is not there.
'\
/
~
..- l'
..
51
N
.' -'"
Copyright
(c)
2007 Ex;)mkrackers, Inc.
LECTURE
Lmz's law states that the induced current w ill create a magnetic fi eld opposing the
ind ucing magnetic field. Imagine a magnet moved toward a loop of wire. The magnetic flux through the loop changes, inducing a current in the wire. The current in the wire creates a magnetic field that opposes the magnetic field created by the magnet. The energy used to move the magnet becomes thermal energy in the ring. Since an induced electric field is not dependent upon the presence of a loop, sma ll eddies of current result when a cond uctor is moved through a magnetic field . Imagine a pendulum made from a conducting material swinging into and out of a magnetic field. Current eddies develop in the pendulum due to its electrons swinging through the magne tic field. The resistivity of the p endulum absorbs energy as internal energy (heat energy on the MeAT), thus changing the kinetic energy of the pendulum into internal energy. The swinging pendulum will stop swinging more quickly due to the magnetic field.
7:
ELECTRICITY AND MAGNETISM .
135
Notice that in each example of an induced electric field, mechanical energy IS transferred to internal energy. This is because the forces due to the induced
electric fields are non conservative. Hold on! Don't get bent here. Most of this information on magnetIsm is background trivia that. if it shows up at all. is likely to be explained in a passage. Concentrate on the basics. Here's what you must know :
1. A magnetic field is generated by a moving charge. and 2. a moving charge experiences force when moving through a magnetic field . From number 2 remember how to find the circular path of a charged particle moving through a magnetic field. Like poles repel; opposites attract. The only formula that you need to remember about magnetism for the MCAT is:
Ed dy B
Copyright © 2007 Examkrackers, Inc.
currents
F= qvB Don 't forget that the direction of the force is perpendicular to both the velocity of the charge and the direction of the magnetic field ,
136 . MeAT
PHYSICS
7.7
Equation Summary
Constant electric fields
Electric fields due to a point charge F= k
q,?,
U = k q,q,
qr :
V-k!ll - r
r
E= k
r
F = Eq
U=qEd
V=Ed
U=Vq
Resistors P = iV
R... =R,+R, + ... V=iR
-1 + -1 +
R,
R,
(Resistors in series)
P=i'R
...
(Resistors in
p"r~Il.. I)
P=
V' R
Capacitors 1
c=
~
1
-C, + C,+ ...
c,.. = C, + C, + ...
Alternating current
(Cap.1.citors ill ~ri{:s)
U=Y2 QV
Q'
(Capacitors in parallel)
U= Y2 C U= V2CV '
Magnetism F= qvBsine
,
Copyright ©J 2007 Examkrackers, Inc
164. The magnetic field created by a long straight current carrymg WIre:
Questions 161 through 168 are NOT based on a descriptive passage.
A.
B.
161. If the AC current delivered to a home by the electric company is delivered at 120 V rms' what is the maximum voltage across an outlet? A. B. C. D.
C. D.
86V 120V
nov nov
165. A charged oil drop is allowed to fall through the electric field created by the plates as shown. In order to give the oil drop a straight trajectory, a magnetic field should be established with field lines pointing:
162. The north pole of the earth's magnetic field is at the geographic south pole. A compass is a small magnet whose north pole end is drawn in the approximate direction of: A. ......
B. C. D.
the geographic magnetic field. the geographic magnetic field. the geographic magnetic field. the geographic magnetic field.
charged particle
.t/'
south pole along the lines of the
B. C. D.
...
+
north pole along the lines of the south pole against the lines of the A. B. C. D.
north pole against the lines of the
163. A charged particle moves horizontally through a magnetic tleld which points directly upward. The force on the particle due to the magnetic field is: A.
decreases in strength proportionally with the distance from the wire. decreases in strength with the square of the distance from the wire. increases in strength proportionally with the distance from the wire. increases in strength with the square of the distance from the wire.
166. A positively charged particle is moving through a magnetic field of strength B as shown below.
perpendicular to the magnetic field and parallel to the velocity of the particle. parallel to the magnetic field and perpendicular to the velocity of the particle. parallel to the magnetic field and parallel to the velocity of the particle. perpendicular to the magnetic field and perpendicular to the velocity of the particle.
B
The force experience by the particle due to the magnetic field is: A.
B. C. D.
Copyright :[J 2007 Exarnkrackcrs, Inc.
left to right. right to left. out of the page. into the page.
137
to the right. to the left. into the page. equal to zero.
GO ON TO THE NEXT PAGE.
167. A stationary loop of wire is placed in a magnetic field directed into the page as shown below.
x
x
x
x
x
x
x
x
x
x
x
x
168. A particle of mass m is fired into a magnetic field of strength B at a speed v. The particle travels in a circular path inside the field with a radius r. Which of the following expressions gives the magnitude of the charge on the particle?
A.
x
x
x
x
x
x
x
x
x
x
x
x
B.
x
x
x
x
x
C.
x
B
,
The current in the loop of wire is:
A.
C. D.
Copyriqht
D.
clockwise if the magnitude of the magnetic field is decreasing. clockwise if the magnitude of the magnetic field is increasing. clockwise if the magnitude of the magnetic field is increasing or decreasing. No current will flow through the loop if the magnetic field is increasing or decreasing.
B.
(C)
2007 [xarnkrackers, Inc.
vB mr mv Br mr v'B
138
mv-
Br
STOP.
Light and Optics
8.1
Light
In Lecture 7 we learned that a changing electric field creates a magnetic field and vice versa. An electromagnetic wave is the traveling oscillation of an electric and a magnetic fi eld. The fields are perpendicula r to each other and the direction of propagation is perpendicu lar to both fields. An electromagnetic wave is a transverse wave.
Electric Field
~ • • tft-· "J Electromagnet ic Wave
The speed (c) at which an electromagnetic wave propagates through free space is constant and is always equal to the ratio of the magnitudes of the electric field and the magnetic field: E c= B Although the electric field is much larger when compared in SI units, the energies of the two fi elds are exactly equal. The above equation is useless for the MCAT, It is given here to remind you of the nature of electromagnetic radiation. It is interesting to note that all electromagnetic waves are generated by the acceler-
ation of electric charge . If a charge oscillates with frequency f, it radiates energy in the form of electromagnetic radiation at the same frequency. The ra te and the direction in which an electromagnetic wave is transporting e ne rgy per unit area is described by a vector S, called the Poynting vector. The Poynting vector is always perpendicular to both E and B, and has a magnitude of EBsinS. Electromagnetic radiation exists in all wavelengths. Light is a tiny sliver from the electromagnetic spectrum. For the MCAT you should memorize tha t visible light includes a ll wavelengths from 390 H 10" m to 700 H 10" m, You should also know that the shorter wavelengths correspond to violet light and the longer wavelengths to red light. Just beyond the visible spectrum is ultraviolet (beyond violet) light on the smaller wavelength side, and infrared (beyond red) on the longer wavelength side.
140
MeAT
PHYSICS
Just remember, Roy G. Biv invented the rainbow. O.K., not really, but Roy G. Biv is an acronym for the order of the colors in the visible spectrum (Red, Orange, Yellow, Green , Blue, Indigo, Violet). You can remember that wavelengths toward violet light have more energy because ultraviolet light has so much energy that it gives you sunburn .
Notice that each waveleng th has a corresponding frequency. 111is is because the speed of ligh t in a vacuunl is constant, which means tha t we can derive frequency f from wavelength 'A. From our wave equa tion, v = fA, we have:
c=/').. Light is slower whe n propagating through a m edium. The speed of light propagating through some medium is found using a constant for that medium, called the index of refraction (n) . The index of refraction compares the speed c of light in a vacuum to the speed v of light in a particular medium. C
n =-
V
Wa velength (m) lifl~WWI~WI~Wl if l rlif~~ I ~lrlif~I~I~I~~I~ I ~ I ~ l~ I
I
I
I
I
I
long waves I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
infrared I UltraViOlet
radio waves I
I
I
I
I
I
I
I
I
I
I
I
I
gamma rays
x-rays I
I
I
I
I
10' 10' 10' 10' 10' 10' 10' 10' 10' 10" 10" 10" 10" 10" 1015 1016 10" lOIS 10" 10ID 10" 10" 10" 10" Frequency (Hz) The visible spectrum
ROYG.BIV
Since nothing exceeds the speed of light in a vacuum, all media have a refractive index greater than one. The greater the index of refraction for a medium, the slower light moves through that particular medium. Typically used on the MeAT are the indices of refraction for water and glass: 1.3 and J.5 respectively. It h elps to be familiar with these. Light is made up of photons. Each photon represents an electromagnetic wave. If we examine only the electric fields of these electrOlnagnetic waves, in typical visible light emanating from a point source (called isotropic light), the fields are oriented in random directions. If we use a device to screen Ollt all photons not having an electric field in one particular direction, the resulting light with all electric fields oriented in the same d irection is ca lled plane-polarized light. When isotropic light is polarized, it loses one half of its intensity, since it loses from its electric field a ll components in one direction and keeps all componen ts perpendicular to that d irection. Light has a dual nature. It acts like both a wave and a particle. The propagation properties of light can be described with wave theory, while the energy transformation properties of light are usually best described by particle theory. Neither wave nor particle theory alone explains the phenomenon of light. Copyright
~
2007 Examkrackers, Inc.
LECTURE
8:
LIGHT AND OPTICS'
141
We can approximate light as a ray mov ing in a straight line, and represen t it as an arrow. Th is is ca lled geometrical optics. Like any other wave, w hen light meets an interface between two meclia, some of its en e rgy re flects and some may also refract. The angles made by a light ray when it reflects o r refracts are measured from a line normal to the interface. The angle at which the lig ht ray strikes the interface is called the angle of incidence. The angle at which it reflects is called the angle of reflection. The angle at which it refracts is ca lled the angle of refraction. The angle of incidence is eq ual to the angle of reflection . Yo u can remember this because the collision of photons against the interface is completely elastic; the photons lose no kinetic en ergy.
The angle 01 relraction is given by Snell's law:
air
glass
11=1
11 = 1.5
where the subscripts 1 and 2 specify the respective interfacing m edia. N otice that in Snell s law, the a ngle of incidence and refraction are not specified ; it makes no difference if lig ht is m ovin g from medium 1 to m edium 2 or from medilun 2 to med iu m 1.
The path that light travels between any two points is the shortest possible path for light in terms of time. This should help you decide which way light will bend at any interrace. Imagine that I must rescue a fair maiden in a swimming
pool. She and I are both several yards from the edge of the pool. I must approximate at what point I should enter the pool in order to reach herthe fastest. I could either find a pen and paper and ca lculate Snell's law based upon my ve locity on land and in the pool, or I could just guess that since I am faster on land, I should travel farther on land. Just like light, I am looking for the shortest path in terms of time and I will bend my path in the same direction at the Interrace .
S,llty in ,} spced()
.... ">
Copyright (Q) 2007 Examkrackers, lnc.
142
MeAT
PHYSICS
Another simp le way to choose the direction that light will bend is to imagine a pair of wheels on an axle . The wheels move fastest where light moves fastest and they always straddle the light ray. When the wheels hit the interface, the first wheel to make contact will either speed up, if light would speed up ('n ' decreases), or slow down if light would slow down ('n ' increases). Since the wheels hit the interface at different
(~
I~). .
c ':,:)
(::, Th is \-"heel . / '-.....J .. . slows dOl,vn first.
glass
times , the axle will turn as the wheels
move at different speeds.The direction in which tile axle turns is the direction in which light will bend .
n=l
aIr
n
= 1.5
When light cros ses into a new medium, the frequency remains the same a n d the wavelength changes , If the medium's index of refraction is higher, the wavelengths become shorter; if the index is lower, then the wavelengths become longer. The energy of a single photon is given by:
E =
An interesting feature about Snell's law is that, at first glance, it seems to violate the conservation
of
energy. Since
the
frequency of the light wave does not change from one medium to the next, both the reflected light and the refracted light must have the same energy. This appears to be twice the energy with wh ich the light started . The ttlck is that the light still has the same energy per photon . However, some of the photons have reflected and some have refracted . Thus, the sum of the intensities of the refracted and reflected beam equals the mtensity of the incident beam . Energy is conserved.
hf (E is energy not e lectric field.)
where II is Planck's constant (discussed in Chemistry Lecture 6). This equation shows that higher frequencies, such as violet and blue light, have more energy than lower frequencies. (Warning: Do not be mislead by this equation: This equation gives the energy per photon. It turns out that if we double the frequency, we also double the n umber of photons increasing the intensity by a factor of four as expected .) When light is COIning frOIn a lnedium with a higher index of refraction, the angle of incidence can be so great as to cause total internal reflection. In other words, if the angle of incidence is large enough, the entire amount of photons will be reflected at the angle of reflection, and none will refract. This angle is called the critical angle. The critical angle is derived from Snell's law by recognizing that the angle of refraction is 90" and that sin90" = 1:
air
8rdr_
glass
11 =
1.5
The concept of the critical angle is used in fiber optics, where a beam of light is trapped inside a glass tube and signals are sent using the energy of the beam.
LECTURE
All types of waves refract. The index of refraction for any substance varies slightly with frequency. Longer wavelengths (lower frequencies) move faster through a medium than shorter wavelengths (higher frequ encies), and therefore bend less d ramatically at the media interface. As a result, white light, which is made up of all the frequenCies in the visible spectrum, is split by a prism in a phenomenon known as chromatic dispersion. Diffraction is another type of wave-bend ing phenomenon. All types of waves diffract. When a wave moves through a small opening, it bends around the COlners of the opening. This is called diffraction. Significant diffraction occurs only when the size of the opening is on the order of the wavelength or smaller. The smaller the opening and the larger the wavelength, the greater the bending of the wave. Diffraction is a limiting factor for geometrical optics. If we attempt to create a ray of light by blocking out all the light from a single source except for a small hole, the light corning through tl1e single hole experiences diffraction and spreads out, frustrating our efforts to create the ray. The smaller we make the hole, the greater the spreading of the light.
Diffraction of light can result in an image of light and dark bands or in dispersion and the creation of colors. Both phenomena depend upon destructive and constructive interference when waves with altered paths meet as shown below. Constructive
Waves diffract as they move through the openings. When the diffracted waves meet interference forms bright and dark bands.
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173. A piece of glass shaped as shown, with a refractive index of 1.S allows light to pass through it striking point B. In order to make the light strike point A, the piece of glass should be:
Questions 169 through 176 are NOT based on a descriptive passage.
light
169. A ray of light strikes a flat window as shown. Which ray most closely approximates the path of light as it exits the window?
so urc~
n \J" I
lens
Light.
source
1\
•A •B
J
r
I
A B
C
Note: diagr.iJl1 nor drawn to scale
A. B.
Window
A.
A
B.
B
C.
C
D.
Some light will follow all three paths.
e. D.
174. If a light on a dimmer switch is gradually turned down, it will generally show a red glow at the moment before it is turned off. This is because red light:
170. Compared to humans, bees perceive a slightly higher frequency of electromagnetic waves. Based on only this information, to wl1ich of the following flower colors is a bee mOfe likely to be attracted?
A. B.
A. B.
green
C. D.
yellow blue
raised. lowered. made thicker from top to bottom. made thinner from top to bottom.
red
e. D.
moves more slowly through air than light of any other color. moves more quickly through air than light of any other color. has more energy than light of any other color. has less energy than light of any other color.
175. The index of refraction of glass is I.S. How long does it take for light to pass through a plate of glass that is 1 em thick?
171. The Coma Cluster is a galaxy approximately 2.7 H 10" km away from earth. How many years does it take for light from the Coma Cluster to reach earth? (Note: light travels at approximately 3 H 10' m/s)
A. B.
lO" 1 I I I I 1000 A. 2.7x - 1 - x 3XTIJ' x60 x 60 x 24 x 36S x - 1-
C.
D.
2.7xlO" 1 1 1 I 1000 B. 1 - x 3x IO, x 60 x24x36Sx-l-
10'" sec 10- 11 sec 2 H 10-< sec 2 H 10- 11 sec SH S H
176. All of the foll owing are examples of wave diffraction EXCEPT:
2.7xld' 3x l0' 1 1 1 1 1000 C. --I- x - - x x 60 x 24 x x -l36S I 60
A. B. C. D.
A light wave bends when passing from air to water. Music is audible arou nd a cumer from the source. The shadow cast by statue is blurred at the edges. Ripples in water become semicircul ar after passing through a small space.
172. All of the following are indicative of the wave nature and not the particle nature of light EXCEPT: A.
B. C.
D.
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diffraction interference dispersion reflection
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144
STOP.
LECTURE
8.2
Images
Mirrors reflect light; lenses refract light. In both cases light rays are bent. TI1e mind may use reason and visual cues such as size and parallax to compensate, but the eye by itself carulot detect whether or not light rays have been bent. Without other visual cues, the mind assumes that light travels in a straight line. As a result, the lnind
traces straight back along the path of the light rays entering the eye and perceives an image. To the person in the diagram below, the fish appears to be where the image is formed because the person's eyes callilot detect the bending of the light.
Image
Fish
An image lllay or Inay not exist. A virtual image does not actually exist outside the mind of the observer; no light rays elnallate from a virtual image. If a sheet of white
paper is placed at the position of a virtual image, no image will appear on the paper. A real image exists separately from the observer. Rays of light actually intersect and then emanate from the point of intersection to form a real image. If a sheet of white paper is placed at the position of a real image, the inlage will appear on the paper. Your reflection in a flat mirror is an exalnple of a virtual image. Your reflection ap-
pears to be behind the mirror, but if you go behind the mirror and look for it, you won't find it. On a warm day, light from the sky enters the hot air just above the pavelnent, refracts, and shines into a driver's eyes, fonning an image that may appear like water. This image is called a mirage. If you went to the pavement at the position of the mirage, you would not find the image of the sky. The mirage is a virtual ima'ge. The image of the fish in the diagram above is also a virtual image.
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8:
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146
MeAT PHYSICS
8.3
Mi rrors and Lenses
There are two types of mirrors: convex and concave. There are also two types of lenses: diverging (concave) and converging (convex). You should recognize both names for e.ach lens type, but think 01 them as diverging and converging because a diverging lens acts like a convex mirror and a converging lens acts like a concave mirror.
observer
observer Concave mirror
Convex mirror
Converging (convex) lens
Diverging (concave) lens
Generally the light from the object originates from 50111C other source and reflec ts off the object. Hm,vever, to avoid confusion, when \vorking v/ith mirrors or lenses, al"w ays aSSLlme that light originates from the object. H e re 's a little trick to help identify a converg ing lens. lust rpmcmber the three C's: A thick center converges light.
Thi cker center converges If the center of a lens is thicker than its ends, it ,,,rill converge light, regardless of its sh(lpe or \vhich direction light moves through the lens. If the center is thinnel~ it 'vvill djverge light.
Oi\ l't bll1g
cen ters
Copyright © 2007 Examkrackers, Inc.
LECTURE
A small enough section of any curve can be extended to form a perfect circle. The radius of curvature for that small section of the curve is the radius of the extended circle. The diagranl below shows the radii of curvature for hvo sections of a curved line. Notice that a smaller radius of curvature indicates a sharper curve. The straighter the line, the larger the radius of curvature. A straight line has an infinitely large radius of curvature.
(r~ ~.·..· •.•.
( ~ ...
/
~'\
Radius of curvature for this curve is r1
..... .. ..... .. .
Radius of curvature for this curve is Y2 ----~
Although the mirrors in this lecture are called spherical mirrors, only a parabolically curved surface will focus all parallel lines to a single focal point. In other words, the equations for the rest of the chapter are only approxinlations for spherical mirrors, and they require that the rays of light are at small angles. For the same reasons, spherical lenses produce flawed images in a phenomenon called spherical aberration. This can lead to confusion if you are drawing ray-diagrams to find an image. Light from horizontal rays is reflected by concave mirrors (or refracted by converging lenses) to focus on a single point called the focal point. For convex mirrors and diverging lenses, horizontal rays of light are reflected and refracted ouhvard from a single point called the focal point. The focal point of convex mirrors and diverging lenses is found by tracing back along the reflected or refracted ·rays.
The focal point for any nlirror or lens is separated from the Inirror or lens by the focal length. The focal length (fmicmc) for a mirror is related to the radius of curva ture (r) as follows: fmirror =
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2007 Exal!lkrackers, Inc.
l/zf
8:
LIGHT AND OPTICS·
147
148
MeAT
PHYS ICS
The focal length for a lens is affected by the refractive index of the lens (n,) and the refractive index of the substance surrounding the lens (n 2 ). (Usually the substance surrounding the lens is air, 11 = 1.) The focal length of a lens (1;",,) is given by the lens maker's equation :
'
..
"
..
object observer
.
'
.. .
'
. '
. ' .
112 '
..
This equation has several forms, and is quite complicated to apply. Tile MeAT would give this equation to you with an explanation if needed . Just remember that tile focal point for a lens is affected by the refractive indices of the lens and the medium that the lens is in . It is also affected by the radii of curvature 01 both sides of the lens.
In the example above, f 2 is negative. Notice that this equation indicates that a lens immersed io a fluid with an equal index of refraction will not bend light. In other words, when 111 :::: 1121 a lens will not refract light.
You should be aware that since the index of refraction varies for different frequencies, the focal point of a lens also varies w ith frequency resulting in chromatic aberration. This is an entirely separate phenomenon frOln spherical aberration mentioned earlier in this lecture. A lens has something called power. This power is not the same as the power in mechanics. The power of a lens is measured in diopters, which has equ iva lent units of m ·l . 111e power of a lens is simply the inverse of the focal length.
The rest of the equations in this lecture apply equally to both mirrors and lenses.
Ray-diagrams are a useful tool in understanding Inirrors and lenses. They help locate the position of an image. However, they are not useful on the MeAT since they are time consuming and inacc urate. Later, Salty w ill offer a preferable alternative to ray-diagrams for solving optics problems on the MeAT, but for now, you should learn to draw them. When drawing a ray diagram for a mirror or lens, imagine the object emitting three photons from a single point. In the example below, this point is the tip of the candle flame. Each photon takes a different path. For simplicity, lenses will be considered infinitely thin, so refraction occurs at the center of the lens. The first photon (1") moves parallel to the ground, strikes the mirror or lens and reflects or refracts so that its path can be traced back through the focal point. By definition, all rays of light parallel to the ground will reflect or refract through the focal point or so that their paths can be traced back through the focal point. The second photon (2°) moves through, directly away from, or directly toward, a focal point and reflects or refracts parallel to the ground. The third photon (3") strikes the Copyright © 2007 EXClmkrackers, Inc.
LECTURE 8 : LIGHT AND OP1Ics' 149
mirror or lens at the exact middle and reflects back and at an angle equal to the angle of incidence, Of, in the case of a lens, moves straight through without being affected. (This is possible because the lens is considered to be infinitely thin and both sides of the lens are parallel at the middle). We follow the paths of the photons until they meet. Where the photons meet is where the image of the tip of the candle is formed. If the photons diverge, we trace their paths backwards frOlTI the mirror or lens to where they would meet.
o
~~~~;;:n(. ~. .
•
,
--"..
• 3°
Convex mirror
Concave mirror
~_ _ _2=-0_ _ _,_._~__~~=~/_;;;;~;{~
o
-
_/.
\f··
U :.
". 3°
Diverging lens
Converging lens
Ray-diagrams Since light always travels the path that takes the least amount of time, although all our intersecting photons in the ray diagram travel different distances, all rays take the same amount of time for their trip. This can be explained as follows: Because light travels more slowly while in the lens, the ray that travels the farthest, travels the shortest distance through the lens. (In very strange cases where the lens has a lower index of refraction, the opposite is true.) A mirror or a lens may magnify an iInage. The lateral magnification m is the ratio of the size of the image to the size of the object. For simplicity, we can compare only the height of the image hi with the height of the object h". The magnification is also equal to the negative of the ratio of the distance of the image di and distance of the object d" from the mirror or lens. The negative sign indicates that, if both distances are positive, the image is inverted.
Copyright (g 2007 Exarnkrac kers, Inc.
,::...:::
150
MeAT
PHYSICS
m = _ _d. l
h
=
_
do
1
ho
Another way to measure magnification is angular magnification. The closest an object can be to an individual while that individual can still focus clearly on the object is called that individual's near point. The angle occupied by the object 8"p when at the near point, compared to the angle occupied by an image 8i of the object when in front of a lens is called the angular magnification mo:
/ ,'-;;-"\
b~ . /\,///_ fr -~"8,'Lu..-Lll -f~K i
d~ ..:~,~ ~
Object in front of lens
Object at near point
Do not confuse angular magnification with lateral magnification. Note that in the diagram above, the lines are drawn to illustrate the angles, not the path of light as in a Ray-diagram.
For any mirror or lens, distance of the image is related to the focal length and the distance of the object by the following equation:
111
-= - + -
f
do
dj
This equation is called the thin lens equation; however, it applies to mirrors as well. It is called the thin lens equation because it assumes that the lens is so thin that the light bends only once when passing through the lens (instead of at both interfaces). Objects, images, focal points, and radii of curvature are all given positive or negative values based upon their position relative to the mirror or lens. The difficult aspect of optics problems is identifying when a variable is positive or negative. The next section is devoted to a system for solving optics problems.
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181. Which of the following is not a possible path for a light ray through a glass lens?
Questions 177 through 184 are NOT based on a descriptive passage.
A,
c,
B.
D.
177. The focal distance on a mirror cut from a glass sphere with a radius of 10 em is: A.
B. C. D.
2.5 ern Scm 10 ern 20 em
178. When an object is 10 em from a certain converging lens, the image is magnified by a factor of 1.5. What is the distance of the image? A. B.
C. D.
3.3 em 6.6 em 10 em IS em
182. The image seen in a flat bathroom mirror is a: A. B. C. D.
179. A glass magnifying-lens is submerged in water to view an underwater object. Compared to viewing the object with the magnifying-lens out of water, this will:
real image that appears behind the mirror. real image that appear in front of the mirror. virtual image that appears behind the mirror. virtual image that appears in front of the mirror.
183, An increase in which of the following lens properties will
A.
increase the magnification.
increase the power of a lens?
B. C. D.
decrease the magnification. not change the magnification. The magnifying glass will not work at all under water.
I. Index of refraction II. Focal length III. Radius of curvature on one side of the lens. A. B,
180. Which of the following statements is (are) true?
C. D.
I. Virtual images can be projected onto a screen. II. Real images can never be seen. III. Real images can only be created by converging lenses and concave mirrors in a single lens or single mirror system. A. B.
III only I and II only
C.
I and III only I, II, and III
D.
Copyright if) 2007 Exarnkracker-s. Inc
I only
I and II only II and III only I, II, and III
184. A concave milTor has a focal length of 4 em. What is its radius of curvature? A. B,
C. D,
151
2em 4em 8em 16 em
STOP.
152
MeAT
PHYSIC;
8.4
A System for Mirrors and Lenses
tvlirrors and lenses may seem tricky, but, luckily, I have a system. The tough part of mirrors and lenses is deciding what's negative and \vhat's positive. After tl1at, it's just plug dnd chug vvith only three equations to memorize. My system \vith only three rules and one exception, finds the positives and negatives. 1.
Begin by dra\ving your mirror or lens and an eye on the side 011 vvhich the observer \vill stand. 1. ,"vill drmv all four possibilities on the next page. No';,\' cOllles the first of three rules: UI (Eye) am positive that real is inverted." You must mClTIorize this sentence. On the sIde v\rhich the eye is drmvl1, vaite 'positive, real, inverted'. In1agcs and focal points on this side \\iill always be positive, reat and inverted. Images and focal points 011 the other side 'A/iI! be always be negative, virtual, and upright. No exceptions.
2.
Of course, everyone knows that you must stand in front of a mirror to sec anything, so the front is the side that 1 (eye) am 011. Lenses are just the opposite, but you can also remember that a camera is a lens! and 1 (eye) st~lnd behind a camera to view an object. Label the front and back of your ll1irror or lens. Nov\' cornes the second rule: Objects are always positive when they are in front of a Jens Of a fllirror and always negative when they are behind a lens or a mirror. For single lens systelns or single mirror S)lste111S, the object must be placed in front, so the object must be positive.
3.
Rule nlllnber three states: As long as the object is in front, convex mirrors and diverging lenses lllake negative, virtual, upright images. As long as the objt--'ct is in front concave mirrors and converging lenses make positive, real, inverted images EXCEPT when the object is within the focal distance, in which case they make negative, virtual, upright images.
Tn the ca::;e of a double lens or mirror systen1, silnply find the image for the first lens or mirror, and LIse that image as the object of the second lens or mirror. Ca1::l:JJ1;. For a two lens or two mirror systen1, you must be careful 'with rule number 3 because the in1age of the first ]ens or mirror may be behind the second lens or 111irror. This results in a negative object distance for the second lens or mirror. Now we can label f, d(), and d i positive or negative depending upon \vhich side the~y afe on. For a convex nlirror and a diverging lens, f is ahvays negative. For a concave 111irror and a converging lens, f is always positive. :tvIenlorize the three formulas on the bottom of the next page. Any other formula for optics \Nill be provided on the I'v1CAT, including the lens maker's equation. That's it. 1 have drawn one lInage for each mirror or lens to shov'v' you hm,v the light rays are traced. f-!Ll\VeVer, if you are tracing light rays on the rvtCAT, you are \\'Clsling your time.
Copyrigh t @ 2007 EXilrnkrackers, Inc.
LECTURE
8:
L IGHT AND OPlIcs·
An image on this side is
An image on this side is
+ real inverted
virtual upright *A concave mirror makes a positive, rea 1 inverted image.
4
eye
BACK
FRONT
A convex mirror always makes a negative, virtua l upright image .
1t
eye
mirrors lenses ob·ect o
eye
pseudo focal poin t object
BACK
• focal point
eye
*A converging lens makes a positive, real inverted image.
FRONT A diverging lens always makes a negative, virtual upright image.
*If the object moves within the focal distance (the yellow area on this chart), a concave mirror and a converging lens will make a negative, virtual upright image. 1 P=T1=1 - +d; d"
f milT"r ~
Copyright
= l...r 2
153
154
MeAT PHYSICS co
co
.-1 ~
!'ositive "'ilgnificJtion mel';;;;;:;;
:
~\ nttk1luprighl"nat."'cltd ""!;" t""dISf'L",ement
, . ";
,
........
. . .e .........-1'I
~
i
'L1\
-'-4
~ .-.. ..
'!
:' ..,. ... "
................ ..
•••••
......
1-0-------t-------------------.-.-. ~ .~~:: -------------- 1 __ __________ ___ . -:'... ~:~: -------- --- ----------- j1 _~ n.-.~ :~:::;;;;. n~ ~.-.-.~~~~, '~. ' nn~~n---f I _______ __
'-'-'-'- r --- ___ _______________
______
c
Q
•• ···•··
e
",C
- co
t:: 0=
Ncgnh\'c 1Th'~nlficalio" indicat",, ~ real. in,wtl-d image at a
2f :J
~f
.
....
Of - If - 2' 'J '.I
2f :I
pooHj,·cdi
----
~f - Itn
If
'J
n______ ___
- 2f :J
-co
Disp lacement from a concave ll1irror
r~~ ......:.:::.::.";l~"·.. .
Displacement from a converging mirror
. .. . . . ~.~. . . . _:_~;o:fO':': '~:'................~.
_~------ - ----!-----:-- --l\n -'J',_,_, I ••;;::::::--------- -~ ----=--------------- ---r-ni __ ~~:~~,." .• ;;;;::.~, ,
2f
If
Of -
'"
-2f
-co
2f
l
!
l.j
!If
Of - If - 2f
..... ...... -co
Displacement from a convex mjrror
in focal lengths
m=
Displacement from a di verging mirror in focal lengths
1
- - - O bject ............ Image
d,
m = I --
I
_~I
actual size and
position of image (virtual upright)
1
." ........... .
...........
... ........
(-
first horizontal line
second hori zon tal line actual size and position of image (real inverted)
position of object (1,5 focal lengths)
Copyright ,) 2007 f.xi:lrnkrackers, Inc.
LECTURE
8.5
Two-lens Systems
H andle a two lens system one lens at a time. Use the image of the first mirror or lens as the object of the second mirror or lens. Sometimes the image from the first mirror or lens is formed behind the second mirror or lens. The object distance for the second mirror or lens is negative in this case. In a single lens system, an object cannot be behind a Inirror or lens, nor can it have a negati ve distance. The lateral magnjfication of a two lens system is the product of the lateral magnifica tion of each lens:
M =
In , m2 •
Two lenses in contact w ith each other ha ve an effective power equal to the sum of their individual p owers:
8.6
Using the Diagrams on the Facing Page
Microscopes and telescopes can be built from two lens systems as well as from mirro rs. The MCAT w ill g ive all necessary fo rmulae concerning these apparatus. The diagrams on the facing page are meant to increase your familiarity with lenses and mirrors. Understanding them is not a requirement in ord er to do w ell on the MCAT. The solid line in each diagram represents the magnification when the object is placed at a given distance. The dotted line represents the m agnification when the image is formed at a given distance. Since the magnification for the image and object must match, you can find the corresponding positions of an image and an object as follows: draw a hori zontal line an ywhere on the graph and the object w ill be where your horizontal line crosses the solid line, while the image will be where your hor izontal line crosses the dotted line. The magnification is the y-value of your horizontal line. The bottom diagram gives two examples. Notice that a positive magnification indicates a virtuaJ, upri ght image and a negati ve magnification indicates a real, inverted image. Notice that the diverging lens and convex mirror on ly produce smaller images for all object positions in front of the mirror or lens. In other w ords, in a single lens Of single mirror systenl, a diverging lens and a convex mirror can on ly produce images slnaller than the object. No tice that for all mirrors and lenses, the magnification of the image is directl y propo rtional to the image distance from the focal point. For any lens or mirrof, if the image is two focal leng ths from the focal point, the image is m agnified by hvo; if it is three focal lengths away, it is magnified by three; if it is half of a focal length awa y, it is magnified by one half (or reduced); and so on. Notice that converging lenses and concave mirrors produce smaller images while the object is outside 2 focal lengths, but la rger images when the object is within 2 focal leng ths.
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(t)
2007 EX(s, Inc.
8:
LIGHT AND OPTICS ' 155
156 . MeAT PHYSICS
8.7
Equation Summary
Electromagnetic radiation
n,sin8, = n,sin8,
Mirrors and lenses 1
fmirror= "'2 r m=
Copyright (c) 2007 Exarnkrackcrs, Inc.
190. An object is placed at the focal point of a converging lens. The image will appear:
Questions 185 through 192 are NOT based on a descriptive passage.
A. B. C.
185. An object stands 4 cm in front of a converging lens. If the lens has a focal distance of 1 eID, where is the image formed?
A. B. C.
D.
D.
191. Light from the moon passes through a converging lens on the surface of the earth. If the lens has a focal length of 20 cm, at what distance from the lens will the image appear?
0.75 cm in front of the lens 0.75 cm behind the lens I cm behind the lens 1.33 cm behind of the lens
A. B. C. D.
186. An invet1ed image is created 5 m in front of a mirrof. Which of the following could be true about the mirror and the object?
A. B. C.
D.
IO cm 20 cm 40cm at infinity
192. The diagram below shows an object placed in front of an unknown optical device and the image produced.
The mirror is convex with less than a 5 m focal distance. The mirror is concave with less than a 5 m focal distance. The min"or is convex with more than a 5 m focal distance. The mirror is concave with more than a 5 m focal distance.
-----C5bIect----------------------:r-----Optical device
187. A 1 em candle stands 4 em in front of a concave mirror with a 2 em focal distance. The image is: A. B. C. D.
on the surface of the lens. at the focal point. at a distance of twice the focal length. not at all.
Image
The optical device is a:
inverted and 1 em tall. inverted and 2 cm tail. upright and 1 em tall. upright and 2 cm tall.
A.
B. C. D.
convex mirror. concave mirror.
convergmg lens. diverging lens.
188. The focal distance of a lens is -3 m. The lens is a: A. B. C. D.
113 diopter converging lens. - 1/3 diopter converging lens. 1/3 diopter diverging lens. -113 diopter diverging lens.
189. A !ens is manufactured in such a way as to allow the object and the image to be at the same distance from the lens. If the lens is not fiat. the only way this could be true is if the lens were: A. B.
C. D.
a diverging lens with the object at the focal distance. a diverging lens with the object at twice the focal distance. a converging lens with the object at the focal distance. a converging lens with the object at twice the focal distance.
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157
STOP.
STOP! DO NOT LOOK AT THESE EXAMS UNTIL CLASS.
3D-MINUTE IN-CLASS EXAM FOR LECTURE 1
159
Passage I (Questions 1-6)
3. Which of the following is true of a projectile in a vacuum when it reaches its maximum height?
In 1939 Emanuel Zacchini, a circus entertainer, had himself shot from a cannon over three 18 m tall Ferris wheels. He reached a velocity of 27 m1s and sailed to a height of nearly 24 m landing safely in a net on the other side of the Ferris wheels. The cannon muzzle and the net were 3 m above the ground. Zacchini's initial trajectory was at an angle of 53° above the horizontal.
A. B.
C. D.
Zacchini had two concerns about his ±light. First, he could not be completely certain of the effects of air resistance and air currents. Second, the force on him while inside the cannon was so great that he would momentarily lose consciousness during the stunt. The second problem he solved by training himself to wake quickly,
Both its kinetic and potential energies are at a maxImum. Both its kinetic and potential energies are at a minunum. Its kinetic energy is at a maximum and its potential energy is at a minimum. Its potential energy is at a maximum and its kinetic energy is at a minimum.
4. What is the vertical component of Zacchini's velocity when he exits the cannon?
A, B, C,
Projectile motion near the surface of the earth can be approximated by the following three equations:
D,
16.2 m1s 21.6 m/s 23.8 m/s 27.0 m/s
5, At which of the following points during the stunt is Zacchini's acceleration the greatest? A, B. C. D.
where x is displacement, v is either the horizontal or vertical velocity, t is time in flight, and a is either 0 or equal to the gravitational constant g which can be approximated at 10 mfs2. The subscript denotes initial values. These equations do not take into account the effects of the medium through which the projectile moves. (Note: sin 53° = 0.8, cos 53° = 0.6)
6. Ignoring the effects of air resistance, Zacchini would have flown farthest if his initial trajectory had been: A, B, C,
}, Using the above equations and adjusting the net for the effects of air resistance, Zacchini should place the center of the net at a displacement: A. B. C. D.
While he is still inside the muzzle of the cannon. The moment he exits from the muzzle. The moment before he lands in the net. Acceleration is constant throughout the stunt.
D,
300 45° 53" 60°
exactly equal to x. exactly equal to xu. greater than x. less than x.
2, From the infonnation in the passage, which of the following factors most likely plays the greatest role in Zacchini's loss of consciousness during the flight? A, B, C,
D,
velocity height acceleration momentum
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9. Which of the following accurately describes Tom's fall from the board compared to Jim's fall?
Passage II (Questions 7-13)
Two boys, Tom and Jim, are at the local pool playing a game. They take turns dropping from the diving board and throwing each other a ball. Sometimes the boy dropping from the board throws the ball to the other who is waiting at the edge of the pool. Sometimes the one at the edge of the pool throws the ball to the one dropping from the board. They always drop straight down from the end of the board, and never jump upward.
A. B.
e. D.
The diving board is 10 m above the surface of the water and 10m from the edge of the pool as shown in Figure I. Tom has a mass of 60 kg and Jim has a mass of 50 kg. The ball has a mass of 1 kg.
10. Tom throws the ball horizontally the moment he leaves the board. If Jim lets the ball hit the ground, Tom will be in the air:
Tom is bigger than Jim and is able to throw the ball faster. Tom throws the ball with an initial velocity of 10 mis, whereas Jim throws the ball with an initial velocity of 8 m/s.
A. B.
e.
D.
r
Jim leaves the board, and Jim catches the ball at the moment he hits the water, approximately what is the maximum height achieved by the ball?
e.
D.
Figure 1
A. B. C. D.
7. If Tom throws the ball at the instant Jim leaves the board. in order to hit Jim, Tom should aim:
D.
slightly below Jim. directly at Jim. slightly above Jim. Where Tom should aim will depend upon how fast Tom throws the ball.
4m 10m 14m 20m
13. If the boys use a 2 kg ball instead of the I kg ball, and Tom wants the ball to follow the same projectile path, Tom must throw the ball with an initial velocity: A. B. C. D.
8. From the moment he leaves the board, approximately how long will it take Tom to hit the water?
A. B. C. D.
am 2.5 m 5m 10m
12. If Tom throws the ball horizontally the moment he leaves the board, approximately how far from the edge of the pool must Jim stand in order to catch it?
As the boys play the game. they vary the distance that they stand from the edge of the pool. (Note: the gravitational constant g = 10 m1s 2• Ignore air resistance unless otherwise indicated.)
e.
liS as long as the ball. the same amount of time as the ball. 5 times as long as the ball. 25 times as long as the ball.
11. If Tom throws the ball to Jim releasing it at the moment
A. B.
A. B.
Tom's velocity will change faster and Tom will hit the water with greater velocity. Tom's velocity will change faster and Tom will hit the water with less velocity. Jim's velocity will change faster and Jim will hit the water with greater velocity. Both Tom and Jim will hit the water with the same velocity.
half as great as the I kg ball. the same as the I kg ball. twice as great as the I kg ball. four times as great as the I kg ball.
Is 1.4
s
2s 4s
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Passage III (Questions 14-19)
17. The projectile would reach its maximum height when fired from which of the following angles?
Students conduct an experiment to study projectile motion. A projectile is launched from a spring-loaded gun. The gun launches the projectile from a hill and with the same speed each time. The gun is aimed so that the initial velocity of the projectile has an angle e from the horizontal. The angle e is increased by 15" each time the projectile is launched. The horizontal displacement d traveled by the projectile as well as the time t spent in flight is measured and recorded. The results are shown in Table 1.
A. B. C. D.
30° 45° 60 0 90 0
18. Which of the following graphs most accurately represents the relationship between the horizontal displacement of the projectile and the angle e?
A. angle
e
displacement d
timet
0"
14.0 m
1.4 s
15°
16.4 m
1.7 s
30°
17.3 m
2.0 s
45°
16.2m
2.3 s
60°
12.6m
2.5 s
75°
6.9m
2.7 s
o
1m
5m
C.
10m 14m
D.
C.
D.
e
90
D. d
o
e
90
e
19. Which of the following statements is true concerning the flights of the projectile in the experiment?
15. Which of the following statements is true concerning the projectile in the experiment?
B.
o
90
d
A.
A.
e
B.
14. Approximately how high above the ground is the springloaded gun held when it releases the projectile?
B.
d
d
Table 1 Horizontal displacement and time of flight for a projectile shot from a spring-loaded gun
A.
C.
B. C.
The longer the projectile remained in the air, the greater was its horizontal displacement. The higher the projectile went, the greater was its horizontal displacement. The speed of the projectile was greatest just after it left the spring-loaded gun. The speed of the projectile was greatest just before hitting the ground.
D.
At its maximum height. the speed of the projectile was zero for every flight. All projectiles reached maximum acceleration just before hitting the ground. The speed of the projectile changed at a constant rate throughout the experiment. The distance traveled through the air by the projectile was smallest when launched at 0° from the horizontaL
16. Each time the projectile is launched, it leaves the springloaded gun with an initial speed of:
B. C.
1 mls 5 mls 10 mls
D.
20 mls
A.
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22. All of the follow ing will affect the time of flight for a projectile experiencing no air resistance EXCEPT:
Questions 20 throug h 23 are NOT based on a descriptive passage.
I. the mass of the projectile II. the initial horizontal velocity of the projectile III. the initial vertical velocity of the projectile
20. A man takes two strides each second. TIle same man walks at a rate of I mls. How long are his strides?
A.
1/4
B.
Ihm
C.
1m 2m
D.
A. B. . C. D.
m
23. A ball is rolled down a I m ramp placed at an angle of 30' to the horizontal. Tbe same ball is rolled down a I m ramp placed vertically. Which of the following statements is true? (Note: sin 30° = 0.5, cos 30° = 0.87)
21. The moon has no atmosphere, and has less gravity than earth. How will the patb of a golf ball struck on the earth differ from one struck on the moon? A. B. C.
D.
[ onl y III only [ and II only [ and 1Il only
A.
Both the earth's atmosphere and gravity will act to lengthen the projectile path of the ball. Both the earth's atmosphere and gravity will act to shorten the projectile path of the balJ. The earth's atmosphere will act to shorten the path of the ball but its gravity will act to lengthen the path. The earth's atmos phere will act to lengthen the path of the ball but its gravity will act to shorten the path.
B.
C.
D.
The ball required the same amount of time for both trips. The ball had the same displacement at the end of both trips. The ball accelerated at the same rate for both trips. The ball reacbed approximately L.4 times the speed on the second trip as it did on the fi rst ttip.
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
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163
STOP.
3D-MINUTE IN-CLASS EXAM FOR LECTURE 2
165
26, The centripetal force on the vehicle is:
Passage I (Questions 24-30)
A, B.
Statistically speaking, traveling on U.S. highways is more dangerous than airplane travel. At the high speeds achieved by vehicles on the highway, turns must be very gradual. As a safety precaution, highway turns are banked toward the inside. Federal guidelines specify highway curve speed limits based upon the angle of the bank, the average coefficient of friction between a vehicle and the pavement, and the radius of curvature of the turn. The radius of curvature of a turn is the radius of a circle that would be circumscribed by the vehicle if the vehicle were to complete a full circle.
e.
D.
in the in the in the in the
direction direction direction direction
of B. opposite to B. of C. opposite to C.
27. If the vehicle in Figure 1 were stationary, the net force on the vehicle would be:
A.
zero.
B.
in the direction of A. in the direction of C. in the direction of D.
C. D.
A 28, If the speed of the vehicle were doubled, the centripetal force required to turn the vehicle would:
A. B.
e.
D.
decrease by a factor of 2. remain the same. increase by a factor of 2. increase by a factor of 4.
D
8
29, If the bank angle 8 were increased to 90° and the vehicle did not fall, the !fictional force on the vehicle would be:
Figure 1
A, R
The diagram above shows a vehicle on a highway curve moving in a direction out of the page and turning to the driver's left.
C. D,
24, Which of the following statements is most likely false concerning the federal guidelines on highway curve speed limits? A. B.
e. D.
less than the weight of the vehicle. equal to the weight of the vehicle. greater than the weight of the vehicle. The laws of physics dictate that the vehicle must fall if the bank angle is increased to 90 0 •
30, Which of the following WOI>. 1 require the bank angles which currently exist on highways to be increased? A.
The speed limit increases with the radius of curvature. The bank angle 8 increases with the radius of cur-
B.
vature.
C. D.
A greater bank angle 8 allows for a greater speed limit. A smaller radius of curvature leads to a greater bank angle.
The average mass of vehicles on the highways increases. The average mass of the vehicles on the highways decreases. Curve speed limits are increased. Curve speed limits are decreased.
25. If the vehicle in Figure 1 is moving very fast, but not slipping off the bank, the !fictional force on the vehicle is most likely: A. B. C. D.
static and in the direction of vector C. static and in the direction of vector B. kinetic and in the direction of vector C. kinetic and in the opposite direction of vector C.
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Passage II (Questions 31-37)
33. From the astronaut's point of view, along what trajectory should he send his projectile if he wants it to reach the garbage receptacle?
A student imagines an astronaut sitting in his space vehicle as it orbits an unknown planet sometime in the future. His space vehicle is pressurized so he is able to remove his helmet. The astronaut has a plastic bag of juice. He opens the bag and squeezes the juice out into the air in front of himself. The juice does not pour into his lap but remains as an amorphous blob wiggling in the air in front of him. He puts his mouth to the juice and slurps it in completely. Finished with the juice, he gently tosses the empty bag toward a garbage receptacle. Then the astronaut removes his pen from his pocket and finds that it floats perfectly in the air.
A. B. C. D.
a perfect parabolic path that ends exactly at the receptacle a parabolic path that ends at the receptacle but is adjusted for air resistance a perfect parabolic path that ends just above the receptacle a straight line to the receptacle
34. How many hours will the astronaut's message take to reach earth?
He looks over at his grandfather clock that runs on a pendulum system, and notices that it is time to radio earth. He radios Earth, which is 3 x 10 13 Ian away, that he is done with breakfast. Earth replies that he must change the radius of his orbit. Earth asks him to decrease his present velocity by a factor of four, which he does. Soon afterwards he discovers that he is following the same orbital trajectory as a small moon.
A. B. C. D.
35. When the astronaut changes his velocity, the radius of his orbit:
The circular motion of the craft is governed by the equation below: r
=
A, B. C. D.
GM v2
where v is the speed of the space ship, r is the orbital radius, M is the mass of the Earth (S.98xlO24 kg), and G is the gravitational constant (6.67xlO- 11 m3/kg_s- 2)
and the moon when they are in the same orbit? (Assume that neither is using a propulsion system to maintain its orbit.) A. B. C. D.
hoth must be at the same speed. both must have the sarne mass. both must have the sarne mass and speed. must have different masses.
real pendulum on the clock in orbit would:
It will orbit in a circle forever.
A.
It will gradually spiral inward. It will gradually spiral outward. It will break from the orbit to travel in a straight line.
B.
32, If the plastic bag misses the garbage receptacle, from the astronaut's point of view, will it continue straight along its present path at a constant velocity?
D.
They They They They
37. The passage comes from the imagination of a student. A
C.
A. B. C.
increases by a factor of 16. decreases by a factor of 2. decreases by a factor of 4. decreases by a factor of 16.
36. Which of the following is true concerning the spacecraft
31. The space ship experiences centripetal acceleration while orbiting the planet. According to Newton's laws of motion, if the spaceship encounters no resisting force in the course of its circular orbit, what will be its future path? A. B. C. D.
3 X 10 13 X 3 X 108 X 60 x 60 (3 x 10 16 )/(3 x 108) x 1/60 x 1/60 (3 x 10 16 )/(3 x 108 ) x 60 x 60 (3 x 108)/(3 x 10 16) x 1/60 x 1/60
D.
swing more slowly than it would if it were on the planet below. swing more swiftly than it would if it were on the planet below. swing at the same rate as it would if it were on the planet below. not swing on the orbiting spacecraft.
Yes, because space is a vacuum. Yes, because the net force on the bag is zero. No, because the gravity of the unknown planet will change the trajectory of the bag. No, because the air molecules in the spacecraft will create air resistance and slow it down.
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40. Which of the following expressions gives the approxi, mate speed of the earth moving through its orbit?
Passage III (Questions 38-42) The- earth does not move around the sun in a perfect circle. Not only is the path very slightly elliptical, but the moon creates a wobble in the orbit. It is actuaIIy the center of gravity of the earth-moon system, called the barycenter, that follows the smooth elliptical path around the sun, Nevertheless, the earth's orbit is so nearly a circle that it can be treated as such for most calculations. The average distance between the earth and the sun is called an astronomical unit, AU.
Table 1 gives the mass, radius, and orbital radius of the sun, earth, and moon. The universal gravitational constant is: G = 6.67 X 10'" N m'/kg'
696,000
Earth
5.97 x 1024
6,378
149,600,000
Moon
7.5 x 1022
1,738
384,400
B.
G x 1.9x10'3< 5.9x10 " 1.5xlO"
C.
G x 1.9x1O'" 1.5x1O"
J
l.5x1O" G x 1.9xlO"
41. How much would a 100 kg man weigh on the moon? A. B. C. D.
Radius (km) .orbital radius (km)
1.9 x 1030
Sun
G x 1.5xl(j' 1.9x I 0'"
D.
The moon moves once around the earth every 27,3 days. The moon does not rotate relative to the earth, so the same side of the moon is always facing earth.
Mass (kg)
A.
17N lOON l70N 1000N
-
42. If F is the gravitational force created on the moon by the earth, which of the following expressions is equal to the gravitational force created on the earth by the moon?
A.
Table 1 Astronomical statistics
F
24 B. (5.97xlO )
X
F
(7.5xI0")
38. A light second is the distance that light moves in one second. How many light seconds is the moon from the earth? A. B. C. D.
22 C. (7.5xlO ) 24x F
(5.97xlO
D.
5.8 x 10,6 light seconds 1.2 x 10'3 light seconds 5.8 x 10'3 light seconds 1.28 light seconds
)
(1.738)' x F (5.97xI024)(7.5xI022 )
39. A lunar day is defined as the time that elapses from sunrise to the following sunrise on the moon at a given location. How long is one lunar day? A. B. C. D.
12 earth hours 24 earth hours 27.3 earth days one earth year
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45. The earth spins on its axis fl atte ning its spherical shape from pole to pole and bowing out at the equator. An object is placed on a scale at the equator. How does the centrifugal force and the distance from the center of gravity affect the weight Cas measured by the scale) at the equator?
Questions 43 through 46 are NOT based on a descriptive passage.
43. In a 'tug of war' two groups of men pull in opposite directions on either end of a rope. Each group applies 2000 N of force. What is the tension in the rope? A. B. C. D.
A.
ON 1000N 2000N 4000N
B.
C.
44. A 50 kg box is moved across the floor at a constant velocity of 5 m/s. The coefficient of friction between the box and the floor is 0.1. What is the net force on the box? A.
ON
B. C. D.
SON 250N 2500 N
D.
Both the increased distance and the centrifugal force act to decrease the weight of the object. The increased distance tends to decrease the weight of the object while the centrifugal force tends to increase its weight. The increased distance tends to increase the weight of the object while the centrifugal force tends to decrease its weight. The increased distance tends to decrease the weight of the object and the centrifugal force does not affect the weight of the object.
46. A 5 kg mass hangs from a spring distending it 10 cm from its restin g point. What is the spring constant k of the spring? A. B. C.
50N/m 100 N /m 250 N/m
D.
500 N/m
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
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3D-MINUTE IN-CLASS EXAM FOR LECTURE 3
171
Puncho is shown in Figure I riding his unicycle while juggling and then riding his unicycle on a tight rope.
Passage I (Questions 47-51) Puncho was a circus clown whose act consisted of juggling five O.S kg balls while riding a unicycle across a tightrope.
o o
Puncho has a mass of SO kg.
o
10m
PoieB
Pole A
Puncho the Clown Figure 1
50. Puncho throws each ball S m into the air. He throws one ball every half second with a velocity of 10 m/s. At any moment when all the balls are in the air, how much greater is their total energy than when aU the balls are at rest?
47. If the radius of the wheel on Puncho's unicycle is 0.2S m, aud Puncho is riding with a velocity of 10 mis, how many revolutions does the wheel make each second ? A. B. C. D.
rrl20 revolutions rrllO revolutions 20/" revolutions 2OTI: revolutions
A.
B. C. D.
48. If the angle 6 in Figure I is 60°, and Puncho is 2 m from Pole A, what is the net torque on Pole A? (Note: sin 60° 0.87, cos 60° O.S)
=
A. B. C.
D.
=
51. Why is it easier for Puncho to balance on his unicycle if he carries a lon g heavy pole centered horizontally at his chest?
ONm
250 N m 2SOO N m SOOON m
A. B. C.
49. If the distancc between Pole A and Pole B in Figure I is doubled and th e angle e remains the same, the tension in the tightrope will: A.
B. C. D.
SO] 125 ] 2S0 J SOO ]
D.
The pole decreases his rotational ineltia. The pole increases his rotational inertia. The weight of the pole increases the frictional force between the unicycle and the tightrope. The weight of the pole increases his momentum when he isn't moving.
decrease by a factor of 2. remain the same. increase by a factor of 2. increase by a factor of 4.
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Passage II (Questions 52-58)
A student perfonned two experiments to investigate the nature of tension. Experiment 1
Trial
m (kg)
T(N)
I
0.10
0.50
2
0.15
0.60
3
0.20
0.67
4
0.30
0.75
Table 2
The student used the apparatus shown in Figure 1 to measure the tension in a string when different masses (M) were hung from its end. The experimental results showing how the tension changed with M are given in Table 1.
52. If the smooth table in Experiment 2 is frictionless, during the experiment the mass m is: A. B. C. D.
in static equilibrium. in dynamic equilibrium. initially not in equilibrium but may achieve equilibrium if the string and table are long enough. not in equilibrium and does not achieve equilibrium during the experiment.
53. If the smooth table in Experiment 2 is frictionless, what is the maximum tension that can be achieved? Figure 1
Trial 1 2
M(kg)
T(N)
0.10 0.15
1.0 1.5
3 4
0.20 0.25
2.0 2.5
A. B. C. D.
0.1 N IN ION As long as the string does not break, there is no limit to the tension that can be achieved.
54. What is the net force exerted on mass M in Experiment 1 Trial 3?
A.
Table 1
B. C.
Experiment 2
D. The student attached one end of a string to a O.lkg mass resting on a smooth table. The student attached the other end of the string over a pulley to a hanging mass (m). The apparatus is shown in Figure 2. The hanging mass was allowed to fall and the tension in the string as it fell was measured. Several different hanging masses were used and the results recorded in Table 2. (Note: Assume massless pulleys for all questions unless otherwise indicated.)
ON 0.1 N IN ION
55. If the string in Experiment 2, Trial 1, were cut, and mass m were allowed to faIl freely, what would be the tension in the string?
A. B. C.
D.
ON 0.1 N 10 N ION
Figure 2
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Passage III (Questions 59-66)
56. Which of the di agrams below most accurately represents the speed of the block in Experiment 2 after the block begins to slide? (Note: Assume that the tension and friction forces are constant.)
A.
In the early part of World War I (WW I) the science of aeronautics was in its infancy. Primitive propeller planes reached top horizontal speeds of 30 m/s. If a plane d idn ' t come apan, greater speeds were possible in a di ve. Initiall y airpl anes were used only for reconnaissance. Enemy pilots would often salute each other with a fri endl y wave as th ey passed in the sky. However, as the war progressed, pilots began throwing bricks and other objects at each other. The tirst bombs were literally dropped from the pilots hands as he flew. In order to hit his target, a pilot would have to take into acco un t his own velocity, wind velocity, and air resistan ce.
C.
"0
"0
""0-
"0
~
~
time
B.
time
D.
time
Modern warplanes fire jet-propelled missiles. Such missiles take air in through the fronl, heat and compress it, and force it out the back along with combusted fuel. Nevel1heless, the same factors for dropping a projectile must also be considered when aimi ng a modern missile. (NOle: Ignore air resistance unless otherwise indicated.)
time
57. How does fri cti on between the 0.1 kg mass and the table in Experiment 2 affect the results of the experiment? A. B. C. D.
59. If a WW I pilot nying hori zontally at top speed dropped a 2 kg bomb from an altitude of 300 m, what would be the kinetic energy of the bomb just before hitting the ground?
The tension in the string is decreased by fricti on. The tension in the string is increased by fri ction . Friction decreases as mass In increases. Friction increases as mass m increases.
A. B.
C. D.
58. Based on th e res ults in Table 2. if another tri al were atte mpted in Experime nt 2 using a mass m of 0.4 kg, what would be the approximate tension in the string?
A. B. C. D.
Copyright
60. If a modern jet-propelled missile with a mass of 300 kg is designed to move vert icall y upward at J200 mIs, how mu ch power must be deli vered by the propulsion system?
0.80 N 0.83 N 1.3 N 4.0 N
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900 J 1800 J 6000 J 6900 J
A. B. C. D.
174
3.6 x 10' W 3.6 X 10' W 2.2 X IO IO W 2.2 X 10" W
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65. A WW I pilot increases his altitude at an angle of 30" to the horizontal and a velocity of 20 mls. If he takes a 2 kg bomb with him, starting from rest on the ground, how much work has been done on the bomb when the plane reaches an altitude of 200 m? (Note: ignore air resistance. sin 30" = 0.5)
61. If a WW I pilot dropped a 2 kg bomb from an altitude of 300 m, which of the following would result in the greatest kinetic energy for the bomb just before it hit the ground? (sin 30" = 0.5) A. B.
e. D.
The pilot releases the bomb while flying straight up at a velocity of 10 mls. The pilot releases the bomb while climbing at an angle 30° above the horizontal at a velocity of 20
A.
B. C.
m/s.
D.
The pilot releases the bomb while flying straight down at a velocity of 20 mls. The pilot releases the bomb while flying horizontally at a velocity of 25 m/s.
66. A WW I pilot flying north at top speed and an altitude of 180 m wishes to drop a bomb on a trench. At how many meters before he is over the trench should he drop the bomb? (Ignore air resistance.)
62. The gravitational potential energy of WW I propeller planes increased with altitude. Since energy is always conserved, from where did this energy most likely come? A. B. C. D.
2000 J 2400 J 4000 J 4400 J
A. B. C. D.
kinetic energy achieved on the runway kinetic energy of air molecules lifting the plane chemical potential energy from the airplane's fuel kinetic energy of the wind
60 m 120 rn 180 m 360 m
63. When a WW I airplane went into a dive, it might reach a constant terminal velocity due to air resistance. Which of the following is true concerning a diving plane that has reached terminal velocity?
A. B. C. D.
The net force on the plane is zero. The plane is accelerating at 10 mls2. The plane is in static equilibrium. The air resistance is equal to the force of the propulsive produced by the planes engines.
64. While pulling out of a dive, a pilot's apparent weight:
A. B. C. D.
lllcreases decreases remains the same The pilot is weightless while pulling out of a dive.
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69. The earth is approximately 80 times more massive than the moon. The average distance between the earth and the moon is just less than 400,000 km. If the radius of the earth is 6370 km, the center of gravity of the earth-moon system is located:
Questions 67 through 69 are NOT based on a descriptive passage.
67. A rocket is launched from earth to explore our solar system and beyond. As the rocket moves out of the earth's atmosphere and into deep space, the gravitational constant g decreases and approaches zero, and the gravitational potential energy of the rocket: A. B. C. D.
A. B. C. D.
also decreases and approaches zero. continually increases. remains constant. increases at first and then decreases and approaches zero.
at the center of the earth. just beneath the earth's surface. just above the earth's surface. exactly between the earth and the moon.
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
68. From the right end of a massless, meter stick hangs a 5 kg mass. 20 em from its left end the meter stick is attached to the ceiling by a string. What downward force F should be applied to the left end of the meter stick to balance it horizontally and in rotational equilibrium?
-
20 ern
F
5 kg
A. B.
C. D.
20N 50N lOON 200N
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3D-MINUTE IN-CLASS EXAM FOR LECTURE 4
177
Collision 4
Passage I (Ouestions 70-76)
Block A with mass 2 kg slides towards Block B with unknown speed and Block B with mass I kg slides towards Block A with unknown speed. After the collision, the two blocks stick together at rest.
Four examples are shown below to demonstrate the properties of collisions. All four examples take place on a frictionless horizo ntal smface at room temperature.
Collision I Block A, with mass 2 kg slides towards Block B with a speed of 4 mls. Block B is at rest and its mass is unknown. After the collision, Block A remains at rest and Block B moves forward at angle x relative to Block A's original motion with a speed of 4 mls.
before
after
Example 4
70. Which of the following statements must be true regarding angles x and y in Collision I and Collision 2?
®
@---before
after
A.
x~
B. C. D.
x ~ y~90°
2y 2x ~ Y
x=y=O°
Example 1 71. What is the speed of the combined blocks after Collision 2?
Collision 2 Block A with mass I kg slides towards Block B with a speed of 4 m/s. Block B is at rest and has a mass of I kg. After the collision, the two blocks stick together and move forward at an angle y relative to Block A's original motion at an unknown speed.
A. B.
I mls 2 mls
C. D.
4 mls 8 mls
72. What is the mass of Block B in Collision I?
A.
@----
B.
®
before
C. D.
0.5 kg I kg 2kg 4kg
after 73. After Collision 4, what has happened to the kinetic energy initially present in the motion of the blocks?
Example 2 Collision 3
A.
B.
Block A with mass 3 kg is initially at rest. An explosion breaks the blocks into three pieces, each with mass I kg. The three pieces each move away at unknown speeds at the angles
c. D.
It remains unchanged. It is converted into elastic potential energy. It is converted into gravitational potential energy. It is co nverted into heat energy.
shown.
74. Which of the following must be true about Collision 3?
A. B. C.
v2 = vJ v2 cos(w)
~
D.
v,sin(w)
~
VI
= v2
+ v) v,cos(z) v,sin(z)
before
Example 3
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Passage II (Questions 77-82)
75. Which of the collisions is perfectly elastic? Collision Collision Collision Collision
A. B. C. D.
1 2 3 4
Two small nuclides can join together via nuclear forces to release energy_ This process is called fusion. However, nuclear forces are only effective at relatively close range, and, in order for this reaction to occur, the particles must overcome the electrostatic repulsion between their positively charged nuclei. These repulsive forces make up the Coulomb barrier. For two protons, the height of the Coulomb barrier is about 400 ke V.
76. If v A is the speed of Block A and VB is the speed of Block B before Collision 4. which of the following must be true? A. B.
VA
C.
2VA
One way that successful collisions can occur is through high temperatures. This is called thermonuclear fusion. Temperature in thermonuclear studies are reported in terms of the most probable kinetic energy K of the interacting particles via the relation:
VA=V B
= 2VE = VB 4VA = VB
D.
K~kT
where k is Boltzmann constant (8.62 x 10-5 eV/K) and T is the temperature in kelvins. Using this method, the temperature at the core of the sun is 1.3 ke V; room temperature is approximately 0.03 eV; and the peak temperature for particles to overcome the Coulomb barrier is 400 ke V.
2xllH+1H~2H++e+ v (Q ~ 0.42 MeV) Reaction 1
Reaction 2 2xl'H +'H--7'He + y(Q
~ 5.49 MeV)
Reaction 3
I'He +'He--7'He +'H + H'(Q ~ 12.86 MeV) Reaction 4 e-:::: electron e+ :::: positron v:::: neutrino y:::: gamma particle Q :::: energy released
Figure 1 Thermonuclear fusion at the sun's core Thermonuclear fusion occurs in the sun via the reactions given in Figure 1. Fusion occurs despite the low temperature for two reasons: (1) some particles move at much greater speeds than the most probable speed; (2) some barrier tunneling can occur at temperatures lower than 400 keY. (Note: An electron carries a charge of 1.6 x 10- 19 C.)
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77. The heat from thermonuclear fusion inside the sun's core comes from: A. B. C. D.
81. In nuclear fission one high-mass nuclide is split into two middle-mass nuclides and energy is released. The nucleons (protons and neutrons) in both nuclides are held together by the nuclear binding energy Q = /y,mc'. The binding energy per nucleon of the high-mass nuclide:
kinetic energy chemical energy mass energy electrostatic potential energy
A. B.
78. What is the approximate temperature of the sun's core in kelvins?
A. B. C. D.
C.
4.6 x 10' K 1.5 X 104 K 4.6 X 106 K 1.5 X 10' K
D,
82. Which of the following is a requirement in order for energy to be released in a fusion reaction?
79. According to Figure I, the net products of thermonuclear fusion in the sun's core are: A, B. C,
D.
A.
helium, hydrogen, neutrinos, and energy. helium, neutrinos, and energy. hydrogen, neutrinos, and energy. helium, hydrogen, and neutrinos.
B. C. D.
80. If an electron and a positron have the same mass, then, according to Reaction 2 in Figure 1 what is the approximate mass of a positron?
A, B. C.
D.
is greater than the binding energy per nucleon of the middle-mass nuclides. is less than the binding energy per nucleon of the middle-mass nuclides. is equal to the binding energy per nucleon of the middle-mass nuclides. may be either greater or less than the binding energy per nucleon of the middle-mass nuclides.
The resulting nucleus must be at a higher energy state than the colliding nuclides. The nuclides must collide at a temperature of 400 ke V or greater. The number of nucleons in the fusing nuclides must be large. The number of nucleons in the fusing nuclides must be small.
10-37 kg 10-)0 kg 10-25 kg 10- 15 kg
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Passage III (Questions 83-88)
Mammalian bones have evolved divergently to meet the requirements of different mammals. Figures I and 2 show the lever system in the forelimbs of two different mammals. Each lever system al10ws for a ratio of out-force to in-force and velocity of limb movement tbat best suits its respective user. Swift running mammals take advantage of third order lever systems to reduce bulky limbs and extend 1imb movements. Large muscles can be kept close to the body requiring less energy expenditure on unnecessary movements; short contractions can be translated into long strides. In order to further maximize velocity the mass of the proximal portion of the limb has been reduced in these swift running mammals.
An important function of many mammalian bones is to act as a lever ann, transmitting an in-force to an oUl-force via a cente r of rotation or fulcrum . Three orders of lever arms exist: first order where the fulcrum separates the in-force and the outforce; second order where the in-force and out-force are on the same side but the out force is nearest to the fulcrum; and third order where both forces are also on the same side but the inforce is nearest th e fulcrum. In mammalian bone lever systems the in-force is supplied by a muscle, one end of which is attached to the bone at the point where the in-force is applied and the other end anchored to a separate bone closer to the body.
FUlcrum ~ ,
:;;~
.
~/ 0 - ,,
,?Lne o f Muscle Action :,' (i n-force) In-lever Arm
I
In-lever Arm Out-lever Arm
85. If the out-lever arm in Figure I is I m and the in-lever arm is 10 cm, and the mammal appl ies an in-force of 10 N, what will be the approximate ou t-force?
first order second order third order ]t can not be determined from the figure.
A. B.
e. D.
84. Ass uming the figures are drawn to scale, if the same inforce is applied to each lever system, which lever system will have the greatest out-force?
D.
The answer cannot be determined from the information given.
IN ION 40N 100 N
86. The animal in Figure 2 is well adapted for rapid digging. If we assume ideal conditions for the lever syste~ in Figure 2, compared to the in-force supplied by the muscle, The out-force must: A. B.
tems.
e. D.
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I I I I
Figure 2
83. What type of lever is shown in Figure 2?
The lever system in Figure I. The lever system in Figure 2. The out-force would be the same in both lever sys-
,,{ine of Muscle Action (in-force)
'--"~-'!"" Force exerted on Ground (o ut-force)
Figure 1
A. B. C.
,
Out-lever Arm:
Force exerted on Ground (out-force)
A. B. C. D.
I
"
181
do less work. do more work. be less than the in-force. be greater th an the in-force.
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87. According to the passage, which of the following conditions would most likely make the animal in Figure I a faster runner? A.
B. C.
D.
Questions 89 through 92 are NOT based on a descriptive passage.
increasing the length of the in-lever arm and decreasing length of the out-lever arm decreasing the length of the in-lever arm and increasing length of the out-lever arm increasing both the length of tbe in-lever arm and the out-lever arm decreasing both the length of the in-lever arm and the out-lever arm
89. The rate at which the thyroid gland absorbs iodine can be measured using the radio nuclide 1281. The half-life of 128r is 25 min. A patient is administered 800 ~g of 1281. If no 128 1 is absorbed by the thyroid, approximately how much will remain in the patient's blood after 2 hours? A. B.
C.
88. Which position has the greatest in-lever arm?
A.
D.
c. 90.
B.
O~g
27 ~g 55 ~g 800 ~g
D.
undergoes two alpha decays and four beta decays to become:
218pO
A.
1JOBi
B. C.
2iOPb
D.
21OpO
226Ra
91. A pulley system is attached to a massless board as shown below. The board pivots only at the pivot point. A 10 kg mass M sits exactly in the middle of the board. F
e
•"'-.....
Pivot point
If the angle 8 is 30", what is the force F necessary to lift the 10 kg mass? (Note: sin30° ~ 0.5)
A. B. C. D.
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182
12.5 N 25N SON lOON
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92. If the two objects shown below collide and remain together without spinning. what will be their final velocity? (sin 600 = 0.87; cos 60" = 0.5)
10 kg
~--
.....--...-...
9=60" 10 mls
& A. B. C. D.
IOmls •
5 mls 5.8 mls 8.7 m/s 10 mls
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
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183
STOP.
3D-MINUTE IN-CLASS EXAM FOR LECTURE 5
185
Passage I (Questions 93-98)
94. A bimetal strip consisting of brass and steel is welded together lengthwise as shown.
Because heating a solid increases the vibrational energy of its
molecules resulting in an increase in space required by each molecule, a typical solid will expand when heated. For such solids, the fractional change in their length per unit temperature
is given by a coefficient of linear expansion, which is specific for each solid. Although the change in length varies slightly with temperature for any solid, The following equation gives a good approximation.
As the temperature increases: A.
B.
where L is the original length of the solid, !!.L is the change in length, a is the coefficient of linear expansion for the particular substance, and !1T is the change in temperature.
C. D.
Of course, any solid that increases in length when heated will also expand in both other directions. The change in volume V for such a solid is given by:
A.
where ~ is the coefficient of volume expansion which is exactly 3 times the coefficient of linear expansion. The equation for the change in volume can be applied to most liquids as well as solids. Water, however, has a maximum density at 4°C.
expand, bending bending
51
Aluminum
23
Brass
19
Steel
11
Glass (ordinary)
9
D.
1.9 %
96. Which of the following substances contracts the most when cooled? A.
(lO-6rC)
Ice (at O°C)
C.
0.0019 % 0.019 % 0.19 %
B.
A list of the coefficients of linear expansion for some common substances is provided in Table 1. (l
expand,
95. If a piece of brass is slowly heated from 25°C to 35°C, by approximately what percent will its length be increased?
Ll.V = Vf3Ll.T
Substance
the brass will contract and the steel will bending the rod to the left. the steel will contract and the brass will bending the rod to the right. the brass will expand faster than the steel, the rod to the right. the steel will expand faster than the brass, the rod to the left.
B. C. D.
aluminum brass steel glass
97. A bottle is half filled with water at 4°C and sealed shut. The bottle is placed on a scale and put into a freezer. As the water nears O°C, the water level in the bottle: A. B. C.
Table 1
D.
falls and the reading on the scale remains constant. rises and the reading on the scale remains constant. rises and the reading on the scale decreases. rises and the reading on the scale increases.
93. What is the coefficient of volume expansion for glass? A. B. C. D.
9 X 1O·18;oC 27 X 1O·18/oC 9 X 1O~6;oC 2.7 X 1O~5;oC
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Passage II (Ouestions 99-105)
98. Which of the following is true concerning an aluminum buoy that floats in a lake all year round? (Note: The volume coefficient of expansion for water is 2 x IO-'I"C.)
A.
B.
C.
The pipe sbown in Figure I holds a fluid with a specific gravity of 5.0. The top of the pipe at end A is sealed so tbat only a negligible amount of vapor pressure exists above the fluid surface. A narrow flexible section extends as shown from end A and is sealed at end D. Both ends of the pipe ean be opened so that fluid flows from point A to D.
The buoy floats higher in the winter because the density of water changes more than the de nsity of aluminum. The buoy floats higher in the winter because the density of water changes less than the density of aluminum. The buoy floats lower in the winter because the density of water changes more than the density of alu-
______________ _________ ______ Jil_,pl
minum. D.
7em ---- --------------------------
The buoy floats lower in the winter because the density of water changes less than the density of aluminum.
4cm
1 Figure 1 Pipe with unknown fluid The points A, B, C, and D and the surface of the liquid are measured from an arbitrary point as shown. Assume that the unknown fluid behaves ideally unless otherwise indicated.
99. When both ends are sealed shut, the pressure is the greatest at point:
A.
A
B.
B
C.
C
D.
D
100. When both ends are open, the flow rate is the greatest at point:
A. B.
A B
C.
D
D.
The flow rate is the same at all points.
101. What is the pressure at poin t C when the pipe is closed and the fluid is at rest?
A. B.
2000 Pa 3000 Pa
C.
2000 Pa + I atm 3000 Pa + I atm
D.
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104. If both ends of the pipe were opened, all of the following would decrease significantly at point B as the unknown fluid drained from the pipe EXCEPT:
102. What is the approximate velocity of the fluid at point D when the pipe is opened at both ends?
A. B. C. D.
0.9 mls 1.1 mls 1.4 mls 2.0 mls
A.
B. C.
D. 103. A 2 kg object submerged in the unknown fluid has an apparent loss of mass of 0.5 kg. What is the specific gravity of the object?
lOS. The pipe is closed at both ends and the fluid is at rest. Compared to the pressure at point A, the pressure at point Cis:
A. B.
1.25
A.
C.
5
D.
20
B. C. D.
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volume flow rate fluid veloci ty fluid density fluid pressure
188
half as great. twice as great. 'h as great. 7/4 as great.
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Passage 1/1 (Questions 106-111)
107. The air very close to the surface of the ball is dragged along by a golf ball's spinning motion so that it moves at the sanae speed as the surface of the ball. If a golf ball in flight spins with a frequency of 60 Hz, what is the approximate speed w of the air at its surface?
The flight of a golf ball does oat strictly follow the noles of projectile motio n. The reason for this deviation is that the golf ball experiences a force called "lift" FL' The lift force is directly prop0l1ional to the difference in pressure above and below the ball caused the ball's rotation during its flight. Lift can be roughly explained using Bernoulli's theorem.
A. B. C. D.
I mls 4 mls 8 mls 20 mls
108. Which of the following changes would NOT serve to increase the lift force FL exerted on a golf ball in flight?
where M' is the pressure difference, p is the density of the air surrounding th e golf ball (p = 1.2 kg/ml), and v 2 and v, are the effecti ve airspeeds above and below the ball.
A. B.
C.
As the golf ball flies tlrrough the air, air moves past the ball at speed II. But as the ball spins, it drags some air along its surface. If the surface of the ball is moving at speed w, the n the effective airspeed above tbe ball is u + W and the efrective airspeed below the ball is u - w.
D.
Weather conditions cause an in crea~ e in the density of air. A golf ball with a lower density is used. The golf ball is struck harder, causing it to move with greater speed. The golf ball is struck with an angled club, causing it to spin more rapidly.
109. When a golf ball like the one descri bed in the passage lands in a lake, which of the fo llowi ng will be true? A.
B. Golf ball
C. D.
110. If a golf ball in flight spins in the direction opposite the one shown in Figure l. the ball will experience:
Golf ball ma nufacturers are continually experime nting with different surface patterns to improve lift properties. The mass of a ty pical golf ball is 4S grams and the di amete r is 4.3 cm. The volume of a golf ball is 42 cm'.
A. B. C.
106, Ass uming the spin on a golf ball has no effect on its horizontal acceleration, how does the flight of a ball undergoing the lift force compare to the flight of a ball that experiences no lift? A. B.
C. D.
D.
a downward force because the pressure will be
greater below the ball . a downward force because the pressure will be greater above the ball. an upward force because the pressure will be greater below the ball. an upward force because the pressure will be greater above the ball.
111. Which of the followin g e x rr~~ s i o n .~ is e'lmll to the difference between the effective airspeeds above and below a golf ball while it is in flight?
The ball that experiences lift will go hi gher, but not as far hori zontally. The ball that ex perie nces lift will go higber and farther horizontally. The ball that experiences lift will not go as high but will travel farther horizontally. The ball that experiences lift will not go as high or as far horizontally.
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The ball will sink. The ball will float with 96% of its volume submerged. The ball will fl oat with 93 % of its volume submerged. The ball will float with 87% of its volume submerged .
189
A. B.
II
C. D.
2u 2w
w
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114. A water tower is filled with water to a depth of 15 m. If a leak forms 10m above the base of the tower, what will be the velocity of the water as it escapes through the leak?
Questions 112 through 115 are NOT based on a descriptive passage.
A. B. C. D.
112.A 5 liter container weighing 2 kilograms is thrown into a lake. What percentage of the container will float above the water? (l L = 1 dm 3)
A. B. C.
D.
10% 40% 60% 90%
115. What is the approximate absolute pressure 5 m below the surface of a lake that is 20 meters deep?
A. B. C. D.
113. A brick sits on a massless piece of Styrofoam floating in a large bucket of water. If the Styrofoam is removed and the brick is allowed to sink to the bottom:
A. B.
e. D.
the water level will remain the same. the water level will fall. the water level will rise. the density of the brick must be known in order to predict the rise or fall of the water level.
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10 mls 14 mls 17 mls 20 mls
50,000 Pa 150,000 Pa 200,000 Pa 300,000 Pa
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
190
STOP.
3D-MINUTE IN-CLASS EXAM FOR LECTURE 6
191
117. A similar tunnel was proposed to be built from Boston to Washington DC. Based upon Figure I compared to the travel time for the trip from Boston to Manhattan, the travel time for the trip from Boston to Washington DC would be:
Passage I (Questions 116-122)
Due to the large volume of traffic between Manhattan and Boston, a group of engineers proposed a tunnel that wou ld allow a train to carry passengers along a perfectly straight path between the two cities. The engineers reasoned that the train could operate without consuming any energy because the force of gravity would pull the train down the first half of the tunnel. As the train accelerated during the lirst half of the journey. it would acquire exactly enough momentum to carry through the second half of the journey against the gravitational force. If friction is neglected and the graviational constant g is assumed to be constant throughout the trip, a one-way trip would be onehalf of a cycle of simple harmonic motion similar to the swinging of a pendulum from one side to the other. (The radius of the earth is 6370 km. g = 10 m/s2 • Assume ideal condirions unless otherwise instructed.)
A. B. C. D.
greater because the distance is greater. the same because the distance is greater but the train would go faster. shorter because the train would go faster. shorter because the trip would be downhill all the way.
118. Given ideal conditions, which of the following expressions most closely represents the time necessary for a trip from Boston to Manhattan?
Hartford Boston
Manhattan
B.
C.
r
~. ,,
D.
119. Which of the following diagrams shows the change in the speed of a train as it travels on a one-way trip from Boston to Manhattan?
A.
C.
B.
D.
Figure 1 Tunnel train (not drawn to scale)
116. Which of the following equations gives the acceleration of the train at the beginning of the trip? A. B.
g sinS g cosS
C.
Gm" ,,,!.:'
D.
g
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Passage II (Questions 123-127)
120. Which of the following best describes the net force and the acceleration on the train during the trip? A. B. C.
D.
The net force and acceleration are zero when the train is directly beneath Hartford. The net force and acceleration are at a maximum when the train is directly beneath Hartford. The net force is at a minimum when the train is directly beneath Hartford but the acceleration is constant throughout the trip. The net force and acceleration are constant throughout the trip.
Because bats are nocturnal hunters they rely upon sound waves to locate their prey. A horseshoe bat emits ultrasonic waves from its nostrils that reflect off its prey and return to the bat. When a horseshoe bat detects flying prey, it adjusts the frequency of the waves until the frequency of the waves rebounding off the prey is 83 kHz, the frequency at which the bat hears best. From the difference in the frequencies, the bat can judge the position of its prey and capture it. The frequency at which the moth receives and reflects the waves emitted by the bat is given by the Doppler effect equation:
121. Because the train goes downhill, then uphill, the shape of the track must be: A.
B. C,
D.
fm fb
a smooth constant curve along the entire length. a straight track along the entire length. a straight track with one bend exactly beneath Hartford where the train turns uphill. a successive series of straight track and curved track.
where f; is the frequency of the waves emitted by the bat, fm is the frequency at which the waves reflect off the moth, Vb is the velocity of the bat, vm is the velocity of the moth. The sign conventions are chosen in accordance with the Doppler effect.
122. The engineers assumed that acceleration of gravity would remain constant during the trip. If we consider that the force of gravity gets smaller as the train nears the center of the earth, how will this affect the trip? A, B.
C. D,
Certain moths can avoid being captured by bats by either flying directly away from the ultrasonic waves, or clicking to create a jamming frequency and confuse the bat.
The trip will require more energy than calculated by the engineers. The trip will require less energy than calculated by the engineers. The trip will require more time than calculated by the engineers. The trip will require less time than calculated by the engmeers.
A horseshoe bat flies at approximately 10 mls. Assume that the moth flies at 5 mls. The velocity of an ultrasonic wave in air is 340 mls.
123. By flying directly away from the ultrasonic waves, the moth most likely avoids capture because: A. B.
C.
D.
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340±v m 340±v b
193
the sound waves reflect away from the bat. the frequency of the reflected waves is decreased so that it approaches the frequency of the emitted waves and the bat may not detect the moth. the frequency of the reflected waves is increased so that it approaches the frequency of the emitted waves and the bat may not detect the moth. the frequency of the reflected waves is increased so that it separates from the frequency of the emitted waves and the bat may not detect the moth.
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126. As the humidity of air is increased, there is less time between the moment when a bat sends a signal and the moment when the bat receives the signal from its prey. This is most likely because the addition of water vapor to
124. If the bat and moth fly directly toward each other, and the bat sends ultrasonic waves at 66 kHz, at what frequency do the waves reflect off the moth?
A. B. C. D.
63 kHz 65 kHz 67 kHz 69 kHz
aIr:
A.
B. 125. Which of the following will decrease the frequency of the waves detected by the bat?
C.
I. The moth flies toward the bat. II. The bat flies toward the moth. III. The moth flies away from the bat.
A.
B. C. D.
D.
127. Which wavelength does the horseshoe bat hear best?
[only III only I and II only II and III only
A. B.
C. D.
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increases the speed of sound in air by decreasing the density of the air. increases the speed of sound in air by increasing the density of the air. decreases the speed of sound in air by decreasing the density of the air. decreases the speed of sound in air by increasing the density of the air.
194
2x 10-3 m 4 x 10-3 m
2m 4m
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131. A piano note is compared to a tuning fork vibrating at 440 Hz. Three beats per second are discerned by the piano tuner. When the tension in the string is increased slightly, the beat frequency increases. What was the initial frequency of the piano wire?
Passage III (Questions 128-134) A piano creates sound by gently striking a taught wire with a soft hammer when a key on the piano is pressed. All piano wires in a given piano are approximately the same length. However, each wire is tied down at two points, the bridge and the agraffe. The length of the wire between the the bridge and the agraffe is called the speaking length. The speaking length is the part of the wire that resonates. The point of the wire struck by the hammer is displaced perpendicularly to the wire's length. A standing wave described by Equation I is generated by the hammer strike, where v is the velocity, T is the tension in the wire, and ~ is the mass per unit length of the wire.
A. B. C. D.
A piano wire with a 90 cm speaking length resonates at a frequency of 360 Hz. What is the wavelength of the reSUlting sound wave? A. B. C.
Equation 1 Velocity of a wave on a piano wire
D.
Different notes are created by using wires of different lengths, and masses. Most piano strings are actually three parallel wires; however, some lower notes are made by two or even a single wire.
A. B. C. D.
transverse wave. longitudinal wave. standing wave. a harmonic wave.
134. If, when the hammer strikes the piano wire, the displace· ment of the wire increases, which of the following properties of the wave on the wire also increases?
128. A piano wire with a speaking length of 120 cm is dis·
A. B. C. D.
placed 0.5 cm when struck by the piano hammer. What is the length of the first harmonic resonating through the wire?
C. D.
0.94m 1.06 m 4m 40m
133. The wave on a piano wire is NOT an example of a:
Tuning a piano involves adjustment of the tension in the wires until just the right pitch is achieved. Correct pitch is achieved by listening to the beat frequency between the piano and a precalibrated tuning fork.
B.
Hz Hz Hz Hz
132. Sound waves move through air at approximately 340 mls.
v=~
A.
434 437 443 446
the frequency the wavelength the amplitude the velocity
60 em 120 em 180 em 240cm
129. A piano with which of the following properties would deliver a note with the lowest pitch? A. B. C. D.
100 cm 120 cm 100 em 120 cm
speaking speaking speaking speaking
length; 800 N length; 800 N length; 700 N length; 700 N
tension; tension; tension; tension;
130. The following are characteristics of a wave on a piano wire. Doubling which one will have the LEAST effect on the intensity of the sound produced?
A.
I!
B. C. D.
period speaking length amplitude
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195
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138. Identical wine glasses A and B contain water. If gently struck with a spoon the water and the glass will vibrate at the same frequency. Which glass will ring at the higher pitch when struck?
Questions 135 through 138 are NOT based on a descriptive passage.
135. A 5 kg mass bounces in simple harmonic motion at the end of a spring. At which point is the acceleration of the mass the greatest? A. B.
C. D.
!I A
When the spring is fully compressed and when the spring is fully extended. When the spring is at its rest length. When the spring is halfway between its rest length and its fully extended or compressed length. The acceleration is constant.
A. B.
c. 136. Two waves are traveling toward each other on the same string. If wave A has an amplitude of 3 em and a wavelength of 10 em. and wave B has an amplitude and wavelength twice that of wave A. what will be the maximum displacement of the string when the waves interfere with each other? A. B. C. D.
D.
Ocm 3 cm 6cm 9cm
B
glass B because more air in the glass will create resonance at a longer wavelength glass B because less water in the glass results in less inertia which allows the glass to vibrate at a higher frequency glass A because less air in the glass will create resonance at a shorter wavelength glass A because more water in the glass lowers the frequency at which the glass can vibrate
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
137. The period T of a pendulum is given by:
where L is the length and g is the free fall acceleration near the surface of the earth. In order to increase the period of a pendulum by 42%. the length must be: A. B. C. D.
decreased by 51 % decreased by 20% increased by 42% increased by 102%
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196
STOP.
3D-MINUTE IN-CLASS EXAM FOR LECTURE 7
197
In a similar phenomenon electrons and protons above the atmosphere are reflected back and forth between the north and south poles of the earth's magnetic field forming the two Van Allen belts high above the atmosphere. Occasionally a solar flare shoots additional electrons and protons into the Van Allen belts creating an electric field at the point where the electrons normally reflect. The electric field drives the electrons along the earth's magnetic field lines into the atmosphere where they collide with the electrons of air molecules forcing them to a higher energy level. The energized electrons quickly drop back to their lower energy level emitting photons. Oxygen atoms emit green light and nitrogen atoms emit pink light. This light fonns the curtain of lights in the sky known in the northern hemisphere as the aurora borealis and in the southern hemisphere as the aurora australis.
Passage I (Questions 139-145)
If the velocity of a charged particle moving through a uniform magnetic field has a component parallel to that field the particle will move in a helical path as shown in Figure I. The net force F on the particle can be described by the equat10n:
F= qvBsin8 where q is the charge on the particle, v is the velocity of the charge, B is the magnetic field strength, and e is the angle between the velocity and the magnetic field. The pitch p of the helix is the distance between adjacent turns. The radius r of the helix is determined by the component of velocity of the charge perpendicular to the magnetic field and is independent of the component of the velocity parallel to the magnetic field.
- - - - magnetic field - - path of charged particle W
B
x
>-<====t======
Tp 1-
Figure 2 Path of a charged particle trapped in a changing magnetic field
139. If the magnetic field in Figure 2 represents a portion of the Earth's magnetic field, at which labelled point would you expect to find the aurora borealis?
Figure 1 Path of a charged particle moving through a uniform magnetic field
A. B.
W X
C.
Y
D.
Z
140. Which of the following most accurately describes the change in the pitch and period of the helical path traveled by a negatively charged particle moving through a magnetic field that is gradually strengthening?
A nonuniform magnetic field that is stronger at its ends than at its middle can trap a charged particle by reflecting it back and forth between its ends. Such a situation is shown in Figure 2.
A. B. C. D.
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198
The pitch increases and the period decreases. The pitch increases and the period increases. The pitch decreases and the period remains the same. Both the pitch and the period decrease.
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144. Which of the following most accurately describes the work done by the Earth's magnetic field on an electron
141. The diagram below shows five paths travelled by charged
particles through a uniform magnetic field. If path X is the path of a proton, which path would most closely represent the path of an electron moving with the same speed?
trapped in the Van Allen belts? A.
No work is done because the force is always perpendicular to the motion of the electron.
B.
Work is done only by the component of the field that is parallel to the velocity of the electron. Work is done only by the component of the magnetic field that is perpendicular to the velocity of the
C.
D.
electron. The work done is equal to the force on the electron times the distance traveled by the electron.
145. In order for the VanAllen belts to form, the magnetic field created by the earth must be:
A. B. C. D.
A. B.
weakest near the North Pole. weakest near the South Pole.
c.
weakest around the equator.
D.
the same strength throughout.
A B
C D
142. If the following particles move at the same speed and
travel a helical path through a magnetic field which one is likely to experience the greatest net force?
A.
a proton with a pitch p
B.
an electron with a pitch 2p an alpha particle with a pitch p an alpha particle with a pitch 2p
C. D.
143. The electrons and protons caught in the Van Allen belts reflect from one pole to the other. Looking up from the North pole, electrons rotate counterclockwise. Protons rotate:
A.
B. C.
counterclockwise both when looking up from the North Pole or when looking up from the South Pole. clockwise both when looking up from the North Pole or when looking up from the South Pole. clockwise when looking up from the North Pole but counterclockwise when looking up from the South
Pole. D.
counterclockwise when looking up from the North
Pole but clockwise when looking up from the South Pole.
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atmosphere is approximately 400,000 V. Air is a poor conductor and thus the average current is only about 10-[ 2 Alm2 .
Passage II (Questions 146-151) Under normal conditions an electric field exists in the air near the surface of the Earth with an average sttength of approximately 100 Vim. Since the human body is a relatively good conductor of electricity it remains at the same pote ntial as the ground and the electric field in the air adjusts around the body accordingly as shown in Figure 1.
A lightning strike occurs when the bottom of a cloud has a negative electric charge that is greater than the negative charge below it on the ground. This temporari ly reverses the electric lield between the ground and the cloud, allowing electrons to flow from the cloud to the ground. The current in a lightning strike is about 10,000 amperes and a typical strike will deliver a charge of 20 coulombs.
The atmosphere is about 50,000 m deep and the total potential difference between the ground and the top of the
+300 V +200V----+lOOV
OV Figure 1 Potential grad ient near the surface of the earth 146. The electric potential of the ground was measured at the top of a hill. If the height of the hill is 20 meters above the level of the surrounding plain. what will be the measured voltage?
A.
OV
B. C. D.
5V 200V 2000V
149. Which of the following best describes the electric field vectors above a flat plain at the Earth's surface? A. B. C. D.
150. If the total electric current reaching the Earth 's surface is nearly constant at 1800 A, approximately how much electrical energy is diss ipated each second by the atmosphere?
147. 147. What is the approximate duration of the lightning strike described in the passage? A. B. C. D.
A.
2 seconds 0.5 seconds 2 X 10-3 seconds 5 x 10-< seconds
B. C. D.
148. What is the average resistance of the atmosphere'!
4 x 10-7 J 4XlO5 J
7.2
X
1.3
X
10' J 10" J
151. Which of the following describes the direction of current flow during a lightning strike?
A.
0 Q(m' )
B.
4 Q(m' )
A.
C. D.
4 x la' Q(m' ) 4 X 10 17 Q (m' )
B. C. D.
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perpendicular to the ground and pointing upward perpendicular to the ground and pointing downward parallel to the ground and pointing north parallel to the ground and pointing south
200
From a cloud at high potential to the ground lower potenlial. From a cloud at low potential to the ground higher potential. From the ground at high potential to a cloud lower potential. From the ground at low potential to a cloud higher potential.
at at at at
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Passage III (Questions 152-157)
152. In Figure I, the force on the moving elec tron due to the magnetic field is in which direction?
In 1879 Edwin Hall demonstrated that conducting electrons in a copper wire are deflected by a magnetic field. This phenomenon is now known as the Hall effect. The Hall effect al so predicts the charge of the conduction carriers inside the copper wire.
A. B.
C. D.
153. Which of the following will be true if the direction of the magnetic field in Figure 1 is reversed and the system is allowed to establish equilibrium?
To demonstrate the Hall effect a copper strip carrying a current i is placed in a magnetic field with field strength B directed into the page as shown in Figure 1. The magnetic field and the average velocity of the electrons v result in a force whi ch pushes the electrons to one side of the copper stri p. This in turn creates an electric field E inside the strip which pushes the electrons in the opposite direction. An equilibrium quickly builds up to where the force on the electrons due to E and B are equal and in opposite directions. The Hall potential difference V is proportional to the strip width d and can be measured with a voltmeter.
A. B. C. D.
v?
A. B.
n = (Bi)/(Vle)
where e is the charge on one electron and I is the thickness of the strip and equals the cro ss-sectional area divided by the strip width ; I = Ald.
X
X
X
X
X
X
X
X
X
X
X
C. D.
A.
C. D.
X
X
X
X
X
X
X
X
X
X
The electric field E is equal to the magr. etic field B. The electric fie ld E is in the opposite direction of the electric field in Figure 1. The magnetic field B is in the opposite direction of the magnetic field in Figure I. The Hall potential difference wi ll be zero.
155. If the copper strip is held stationary, the net force on the copper strip is: B.
X BX
The electric field E will be reversed. The electron velocity v will be reversed. The Hall potential difference will be zero. The magnetic field in Figure I cannot be reversed because the current creates a mag netic field directed into the page_
154. Which of the followi ng is true when the copper strip in Figure 1 is moved in the direction opposite to the e1ctrons with a velocity greater than the average electron velocity
When the electric and magnetic forces are in balance the number density of charge carriers n (number of electrons per unit volume) is given by the following equation:
x
to the left to the right into the page toward the top of the page
zero. directed into the page. directed to the ri ght. directed to the left.
I _ d __ _
Figure 1 Electrons moving through a copper strip in a magnetic field
The average velocity of the electrons can also be measured using the Hall effect. If the copper strip is moved in a direction opposite to the electron velocity, the Hall potential difference will be zero when the strip velocity equals the electron veloc· ity. (e = 1.60 x 10-" C)
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156. If an electron were to break free from the surface of the copper strip, which of the following would most accu· rately represent its path from the moment it breaks free?
Questions 158 thr oug h 161 are NOT based on a descriptive passage.
c.
A.
158. The capacitor shown below is fully charged. Both resis·
/ B.
tors have a 2.0: resistance. When the switch is opened, what is the initial current through resistor A ?
D.
o
o ,-
157. Which of the following describes the Hall potential dif· ference where R is the effective resistance of the copper strip?
A. B.
C. D.
A. B. C. D.
evB iR evR
\ '-;-,- , -- - - , ..
2A 3A 6A 12A
159. What is the charge on the capacitor in the circuit below
Ed
after the circuit has been on a long time?
2 r.l
12 V
2r.l A. B. C. D.
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202
1.2 X 2.5 X 6.0 X 1.7 X
10-' C 10-" C 10-<> C 10-7 C
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160. Charges A, B, C, and 0 are charged particles forming a square as shown. A and 0 have a charge of +2 C; Band C have a charge of -4 C. If q is a particle with a charge of + I sitting directly in the middle of the square, what is the net force on q due to the other particles?
161. A particle possessing a charge of +2 C and a mass of I g is exposed to an electric field with strength 5 N/C. How far will the particle move in 10 seconds?
A.lxlO' m B.2 xlO'm C. 5 x lo'm D. I x 10' m
A_ ----"B
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK
c A.
ON
B. C. D.
IN 1.4 N 4N
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YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
D
203
STOP.
3D-MINUTE IN-CLASS EXAM FOR LECTURE 8
205
Passage I (Questions 162-167)
Experiment 2: Double Slit In Experiment 2 another slit also of width W is made in the diaphragm at a di stance s from the frrst slit and the diaphragm is recentered as shown in Figure 3. s is large compared to the wavelength of li ght A. Again the physic ist shines a monochromatic beam of light of frequency f an the diaphragm. This tim e a different fringe pattern forms. The results of Experiment 2 are shown in Figure 4.
A physicist performed the following experiments to investigate the wave nature of light. Experiment I : Single Slit Monochromatic light with frequency f is incident on a diaphragm in which there is a single slit of width W. The li ght shining through the slit falls on a detector located a distance D to the right of the slit. D is much greater than W. The detector measures the energy delivered by the scattered wave as a function of the distance x. The entire apparatus is shown in Figure 1.
f
I
W
x
(-I
" x=o
W :t
'f
w--
I_I<-'- - - - D - - - - - > I
:t
4----D
-----+1
Figure 3 Double slit apparatus (not drawn to scale)
",'"
Figure 1 Single slit apparatus (not drawn to scale)
\l\-JV''''-__~ S
The tight falling on the detector reveals a series of dark and li ght fringes. These results are shown in Figu re 2.
•
2'A1W
sm
•
Figure 4 Intensity vs . s sine for Experiment 2
The intensity striking near the center of the diaphragm in Experiment 2 is given by the following equation:
'-~
-3/c
-2/c -/c
o
2/c
I, = 4/, cos'(,h'l')
___ wsin9
3/c
where 'I' is the phase difference between rays of light from the different slit') the moment they strike the detector at the point in question.
Figure 2 Intensity vs. w sine for Experiment I The intensity at any given point near the cen ter of the detector is a function of the phase difference «> between the rays at the top and ballom of the slit the moment they strike the same point on the detector. The intensity is given by the following equation:
1, = 1m (sin '12$)/'''$ where 1m is the intensity of light striking the detector directly
across from the slit. Copyright
(~';)
2007 Exam kracke rs, Inc
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162. What wave phenomenon or phenomena are demonstrated by the two experiments? A. B. C. D.
166. If the same light source is used for both experiments, the brightest fringe in Experiment 2 will have an intensity:
Experiment 1 demonstrates only interference and Experiment 2 only diffraction. Experiment I demonstrates only diffraction and Experiment 2 only interference. Both experiments demonstrate interference and diffraction. Neither experiment demonstrates interference or diffraction.
A. B. C. D.
167. If the amplitude of the light wave in Figure 1 were doubled, which of the following would be true?
163. In Experiment 1 which of the following equations represents the time necessary for light to travel from the slit to any point on the detector? (c = speed of light)
A. B. C. D.
half as great as the brightest fringe in Experiment 1. equal to the brightest fringe in Experiment 1. twice as great as the brightest fringe in Experiment 1. four times as great as the brightest fringe in Experiment I.
A. B. C.
xlcsin8 xle DlcsinS Die
D.
I~ would increase by a factor of 2. Is would increase by a factor of 4. The distance between the peaks in Figure 2 would increase by a factor of 2. The distance between the peaks in Figure 2 would decrease by a factor of 2.
164. [f light were purely a particle phenomenon which of the following would best describe the results the physicist could expect from either experiment? A. B. C. D.
The The The The
graph graph graph graph
in in in in
Figure Figure Figure Figure
2 4 2 4
would have would have would have would have
more peaks. more peaks. 1 peak. 1 peak.
165. If white light were used in Experiment 1 the outermost fringes of light on the detector would appear: A. B. C. D.
blue red white green
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Passage II (Questions 168-174)
168. A certain compound microscope magnifies an image 1200 times. If the eyepiece is replaced with a lens with twice the power, the image will be magnified by a factor of:
The compound microscope (Figure 1) uses two convex lenses in order to magnify small objects at short distances. The lens nearest the object is called the objective; the lens nearest the observer is called the ocular or eyepiece. The distance between the two lenses minus the sum of the magnitudes of their focal lengths is called the tube length L.
A. B. C.
D.
600 1800 2400 4800
169. The image of the object in Figure I created by the objective is: eyepIece
A. B. C. D.
objective
virtual and inverted virtual and upright real and upright real and inverted
170. lfthe eyepiece on a compound microscope has a power of 25 diopters, what is the focal length of the eyepiece"
A. B. C.
D.
Figure 1 A compound microscope If a small object is placed just outside the focal point of the objective, an enlarged image is formed just inside the focal point of the eyepiece. The lateral magnification mo of this image is given by the equation:
171. A 2 mm object is magnified 500 times by a compound microscope. The magnitudes of the focal lengths of the eyepiece and the objective are 1 cm and 0.5 em respectively. What is the distance between the two lenses when the object is in focus?
L m =-o
A. B. C.
f,
D.
where Io is the focal length of the objective.
is given by the equation:
A.
,
8.5 cm 10.0 cm 1I.5cm 15.0 cm
172. What would happen if the object were placed just inside the focal point of the objective?
The eyepiece acts as a simple magnifier on the image formed by the objective. The angular magnification of the eyepiece Me
M
0.25 cm -D.25 cm 4cm --4 cm
=_ 25cm I.
B.
where J; is the focal length of the eyepiece, and 25 cm is the closest point to the eye for which a sharp image may be formed; this distance is called the near point or the distance of most distinct vision.
C. D.
The objective would form a virtual, upright image of the object on the object side of the lens. The objective would form a virtual, upright image of the object on the side of the lens opposite to the object. The objective would form a real, inverted image of the object beyond the eyepiece. The objective would form a real, inverted image of the object behind the object.
The total magnification of the microscope is given by the product of the lateral magnification of the objective and the angular magnification of the eyepiece.
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Passage III (Questions 175· 180)
173. The word 'READ' is placed under the mi croscopic in th e upright position. Which of the following represents the word when viewed through the microscope?
A. B.
READ
C. D.
OVen l OA.35I
A refracting telescope can be made from two convex lenses separated by a tube length s. The focal points of the two lenses must coincide in order for the telescope to focus on distant objects. Although Figure I shows two simple lenses, in prac· tice, each lens is usually a co mpound le ns. The lens nearer the observer is called the eyepiece, and the other lens is called the objective.
IffiVD
174. Which of the following describes how a compound microscope magnifies an image? A.
Light refl ects off an object and diffracts througb the
B.
Light di sperses off an object and reflects through tbe lenses. Light reflects off an object and refracts through the le nses. Light refracts off an object and re flects through the
eyepiece
\ __ F,:... : F..:... . --+~rl-\
lenses.
C. D.
p
,
Figure 1 A telescope
lenses.
The angular magnification of a te lescope is given by the following equation:
where f is the focal distance of tbe respective lenses. Since a telescope is used to view di stant objects, it requires light-gathering power. Light·gatheri ng power determ ines how bright the image wi ll be. a nd is increased by increasing the diameter of the objective. Resolving power, the ability to di stingui sh between two distant points. is also important for a telescope. Refracting telescopes may have two flaws, spherical and chromatic aberration. Spherical aberration res ults in a blurred image because true spherical lenses do not focus all parallel light rays exactly on the focal poin!. This can be fixed by using parabolic lenses. Chromatic aberration OCC urs because different wavelengths refract at different angles.
175. Which of the following must be truc in ordcr for a refract· tog telescope to magnify distant images? A. B.
C. D.
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209
The focal length of the eyepiece must be greater than the focal le ngth of the objective. The focal length of the objective must be greater than the focal length of the eyepiece. Only the eyepiece must be a co mpound lens. Both the eyepiece and the objective must be compound lenses.
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176. Where will the objective fonn the image of a very distant object? A. B. C. D.
180. If the space between the lenses in the telescope in Figure I were filled with water, which of the following changes to the tube length would bring the image back into focus. (index of refraction for glass = 1.5, index of refraction for water = 1.3, index of refraction for air = 1)
in front of the objective behind the objective but in front of the focal point of the objective at the focal point of the objective behind the eyepiece
A.
B. 177. If all lenses shown below are made from the same type of glass, which lens will have the greatest positive power in diopters?
A.
B.
o K
C. D.
C.
increasing the tube length to compensate for the longer focal lengths of the lenses in water increasing the tube length to compensate for the increased power of each lens decreasing the tube length to compensate for the decrease in focal. lengths of the lenses in water decreasing the tube length to compensate for the decreased power of each lens
D.
178. Which of the following helps to explain chromatic aberration in a refracting telescope? A.
B.
C.
D.
When light enters the lens. the frequency is lowered in a greater proportion for blue light than for red light. When light enters the lens. the wavelength is shortened in a greater proportion for red light than for blue light. When light enters the lens, the wavelength is lengthened in a greater proportion for blue light than for red light. When light enters the lens, the wavelength is shortened in a greater proportion for blue light than for red light.
179. If the eyepiece of the telescope shown in Figure I has a power of 100 diopters. and the magnification of the telescope is -9, what is the focal length of the objective?
A. B. C.
D.
9cm -9 em 10m -10 m
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184. On a hot day. a driver on the highway may see a mirage that appears as a puddle of water far ahead on the hot pavement. The mirage is actually a virtual image of the sky caused by light rays bending as shown below. Which of the following helps to explain the formation of the mirage?
Questions 181 through 184 are NOT based on a descriptive passage.
181. Although waves in the open ocean propagate in all directions, waves washing into any shore usually move nearly perpendicular to the shore. Which of the following best explains the reason for this phenomenon?
A. B. C. D.
Light ray
~ warmer air
The shallow water decreases the speed of the waves causing them to refract. The shal10w water increases the speed of the waves causing them to refract. The shal10w water decreases the speed of the waves causing them to diffract. The shallow water increases the speed of the waves causing them to diffract.
I
A. B. C. D.
IDJrag e
Hot pavement Hot pavement is an efficient reflector. The thin layer of wanner air acts as an aperture through which light diffracts. As air warms the speed of light decreases. As air warms the speed of light increases.
182. A mirror has a radius of curvature of 8 em and makes a real inverted image 20 em from its surface. Where is the object? A.
B.
C. D.
STOP. IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK. YOU MAY GO BACK TO ANY QUESTION IN THIS TEST BOOKLET.
4cm 5 cm IOcm 20cm
183. A 3 cm object is placed 15 em in front of a convex mirror. The image forms 5 cm behind the mirror. How big is the image?
A. B.
C. D.
I em 3 cm 5cm 9cm
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211
STOP.
ANSWERS & EXPLANATIONS FOR
30-MINUTE IN-CLASS EXAMINATIONS
213
214
MeAT PHYSICS
ANSWERS FOR THE 30-MINUTE IN-CLASS EXAMS l ecture 1
lecture 2
lecture 3
lecture 4
lecture 5
lecture 6
lecture 7
lecture 8
1. D
24. B
47.
e
70. D
93. D
116. A
139. D
162.
e
25. A
48. A
71. B
94.
e
117. B
140. D
163. A
3. D
26. A
49. B
72.
e
95. B
118. B
141. A
164. C
4. B
27. A
50. B
73. D
96. A
119. A
142.
e
165. B
5. A
28. D
51. B
74. D
97. B
120. A
143. C
166. D
6. B
29. B
52. D
75. A
98. A
121. B
144. A
167. B
7. B
30.
e
53. B
76.
99. D
122.
e
145. C
168.
8. B
31. A
54. A
77.
e e
100. D
123. B
146. A
169. D
9. D
32. D
55. A
78. D
101. B
124. D
147.
e
170.
10. B
33. D
56. D
79. B
102.
e
125. B
148. D
171.
11. B
34. B
57. B
80. B
103. D
126. A
149. B
172. A
12. A
35. A
58. A
81. B
104.
e
127. B
150.
13. B
36. A
59. D
82. D
105. B
128. D
151.
e
37. D
60. B
83.
e
106. B
129. D
152. B
175. B
15. D
38. D
61. D
84. B
107.
e
130. A
153. A
176. C
e
85. A
108. B
131. C
154. B
177. A
2.
14.
16.
e
39.
e
62.
e e
173. 174.
e
e e e e e
17. D
40. C
63. A
86. C
109. A
132. A
155. A
178. D
18. A
41.
e
64. A
87. B
110. B
133. B
156. B
179. A
19. D
42. A
65. D
88. A
111. D
134.
e
157. D
180. A
20. B
43. C
66. C
89. B
112. C
135. A
158. B
181. A
21. B
44. A
67. B
90. D
113. B
136. D
159.
e
182. B
e
45. A
68. D
91. B
114. A
137. D
160. A
183. A
23. D
46. D
69. B
92. B
115. B
138. B
161.
e
184. D
22.
MCAT PHYSICS Raw Score
Estimated Scaled Score
23
15
22
14
21
13
19-20
12
18 16- 17 15
11
13- 14
8
12
7
10- 11
6
9
5
7-8
4
10
9
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EXPLANATIONS FOR THE 30-MINUTE IN-CLASS EXAMINATIONS'
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EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 1 Passage I 1.
0 is correct. The value x in the equation gives the distance the projectile would travel in the absence of air resistance. Like all frictional forces, air resistance creates a force in the opposite direction of motion, so Zacchini should expect to fly a shorter distance than x.
2.
C is correct. The passage says that it is the "propulsion" that causes the loss of consciousness. In other words, the force on Zacchini causes him to accelerate at a rate so great that he loses consciousness.
3.
0 is correct. Total energy is conserved throughout the flight. At maximum height gravitational potential energy mgh is maximized, so kinetic energy 1/2 mv2 is minimized.
4.
B is correct. The vertical component is vsin9 or 27 x 0.8.
5.
A is correct. While in the cannon the acceleration is so great that it makes him unconscious. While in the cannon muzzle, Zacchini goes from zero velocity to maximum velocity. This is the greatest change in velocity in the shortest distance or time. Acceleration is rate of change in velocity.
6.
B is correct. For a projectile without air resistance, the range is maximized at 45°. This is a fact that you should memorize for the MCAT. Subtracting or adding to the angle of trajectory from 45° by equal amounts results in equal ranges. For instance, 30° and 60° are both 15° from 45° and result in the same range.
Passage II 7.
B is correct. Without air resistance, both Jim and the ball will accelerate downward at the same rate, 10 m/52, This is the same situation as if Tom and Jim were two astronauts in a spaceship orbiting earth, An orbiting spaceship and everything inside it fall toward earth at the same rate. Everything in the spaceship appears to float from the perspective of the two astronauts, Tom and Jim. In this case, it is obvious that Tom should throw the ball directly at Jim. In the diagram shown on the left, the ball moves with a constant horizontal velocity. and both Jim and the ball accelerate downward at the same rate. Notice that the ball moves along a straight line Witll respect to Jim, but moves in a parabolic path with respect to Tom. At every moment, the ball appears to Jim as if it were coming straight toward him along the dotted line. This effect is independent of the speed of the ball, and independent of where Jim catches the ball.
!J.,
ft·
-----~r
<':-:.
t
Tom 1f&1 ..... . "It, .... '" ..... i:J'.
'.,
'. ",
'., ,,',
t·.· Jim
8.
B is correct. The equation is x = l/'gf. x equals 10 m. You should probably know without calculating that an object starting from rest falls 5 m in one second and 20 meters in two seconds.
9.
D is correct. The velocity of a projectile experiencing no air resistance is independent of its mass,
'.
10.
B is correct. Only vertical velocity dictates the time of flight for a projectile. When Jim throws the ball from the board, both Jim and the ball have an initial vertical velocity of zero. Again, the velocity of a projectile experiencing no air resistance is independent of its mass.
11.
B is correct. Look at the diagram for question 7. Jim will travel a vertical distance of 10 111 starting from zero m/ s and constantly accelerating. The ball will start with an initial vertical velocity upward, slow to zero, and then reverse direction to arrive at the same point and at the same time as Jim. Thus the ball could not possibly travel as far as Jim. 10 m and 5 m are out. The ball definitely rises above 0 m. Thus B is correct. Another way to figure this out is to recognize that Jim's trip requires 1.4 seconds (see question 8). This means that the ball must reach its peak at 0.7 seconds due to the symmetry of projectile motion. From the formula x = l/zgt' we have approximately 2.5 m.
12.
A is correct. Since Tom's fall lasts 1.4 seconds, the ball's flight also lasts 1.4 seconds. (See question 8. Tom's flight time is the same as Jim's. Mass is irrelevant; they have the same initial vertical velocity.) The distance traveled by the ball is its horizontal velocity (10 m/s from the passage) times its flight time. This is 14 m. Notice that the edge of the pool is already 10 m from the board, so the answer is four meters.
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13.
MCAT
PHYSICS
B is correct. Mass is irrelevant to the path of a projectile experiencing no air resistance. Now, in order to achieve that velocity, Tom must use more force on the 2 kg ball, but that's a different question.
Passage III 14.
C is correct. This can be deduced from any trial, but the easiest is from the 0" trial. Since the vertical velocity is zero in this trial, the height is found from the equation x ~ '/,gt'. This gives x ~ 10 m.
15.
D is correct. Since the projectile starts from a height of 10 meters, it reaches a maximum velocity just before it strikes the ground: potential energy turns to kinetic energy. Table 1 shows us that A is wrong. B is wrong because if we fire the projectile straight up, it will have zero horizontal displacement but maximum height.
16.
C is correct. Again, the first trial is the easiest to examine. From the first trial, we have d ~
~
vt, which gives us v
10 m/s.
17.
D is correct. This should be intuitive. v sin fl ~ ~2gh . Sin 90" is 1, so h is maximized at 90 degrees.
18.
A is correct. This is the only graph that isn't zero when 8 is zero.
19.
D is correct. The distance through the air is not the horizontal displacement. This is best solved by process of elimination: A is wrong because only vertical velocity is zero at maximum height. B is wrong because acceleration was constant for all projectiles. C is wrong because only the velocity changed at a constant rate; as the projectile climbed, speed decreased; as it fell, speed increased.
Stand Alones 2 strides/s ~ 'hm/stride
20.
B is correct. Use units. 1 m/s
21.
B is correct. The atmosphere will create air resistance, shortening the path of the ball, and the gravity will reduce the time of flight, also shortening the path of the ball.
22.
C is correct. Only the vertical velocity affects the time in flight of a projectile experiencing no air resistance.
23.
D is correct. The velocity is proportional to the square root of the initial height of a ball rolling down an inclined plane: v' ~ 2ax or v ~ ~2gh . The ball falls twice as far in the second trial. Thus the ball has 2 times the velocity. Eliminate A by taking the example to extremes. If the ball falls straight down, it will obviously take less time to travel one meter than if the ball rolls down at a very slight angle. Displacement is a vector; direction matters, so B is wrong. C is wrong because the acceleration down an inclined plane is gsin8. You also should have known C is wrong because if A is wrong, speed is distance over time, so C would be wrong as well.
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EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 2 Passage I 24.
B is correct. Band D are opposites so one must be the false statement. The larger the radius of curvature, the straighter the curve. A straight line would require no bank angle. (Radius of curvature is discussed in Physics Lecture 8)
25.
A is correct. Friction is always parallel to the surfaces that create it, and it always opposes sliding motion between the two surfaces. In this case, the car is going very fast and, if the bank were made of ice (had no friction), the car would slide up the bank. The friction is static because the vehicle has no motion relative to the bank in the direction of friction (i.e., the tires do not slide along the pavement).
26.
A is correct. If we were to draw a circle circumscribing the path of the car, arrow B would point to the center of that circle. The centripetal force always points to the center of the circle.
27.
A is correct. A stationary object experiences no net force. We will discuss this more in Physics Lecture 3.
28.
D is correct. Centripetal force is given by F, ~ mv'lr. Doubling v requires that F quadruples.
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29.
B is correct. Since there is no vertical acceleration, the net vertical force would have to be zero (F = mal. The only vertical force downward would be the weight of the vehicle, so the frictional force would have to be equal . and opposite. The normal force is the centripetal force of the wall on the vehicle. The car must be moving fast enough for the frictional force to be equal to gravity. So f, = 1l,N = Il,mv' I r. ll,mv'lf = mg.
30.
C is correct. The mass of the vehicles will not affect the required bank angle because the centripetal force, the frictional force and the force down the incline due to gravity are all proportional to mass. A greater speed limit on the curves would require a greater bank angle in order to keep the cars from sliding off the road.
Passage II 31.
A is correct. As long as there are no resisting forces like friction, the centripetal force will always have the same magnitude and will always be perpendicular to the spaceship's motion, so the speed of the spaceship will never change. The ship will continue to travel with the motion dictated by the equation in the passage unless another force causes it to change.
32.
0 is correct. The passage says the ship is pressurized, thus there are air molecules in the ship. The passage also says that there is air in the cabin. The man has his helmet off and is still alive so there must be air in the ship. Air molecules create air resistance, which will slow any projectile. The gravity of the planet acts equally on the ship, the astronaut, and the projectile, SO relative to the astronaut, the bag will move along a straight line.
33.
0 is correct. The bag will move in a straight line relative to the astronaut. (See question 32.)
34.
B is correct. t = dlv. Convert seconds to hours. You must know that radio w aves are electromagnetic, and electromagnetic waves move at the speed of light or 3 x la' ml S.
35.
A is correct. Centripetal force equals gravitational force. GmMlr' = mv'lr. Thus v' is inversely proportional to r.
36.
A is correct. Their orbits are related only to their velocities because both gravitational and centripetal force are proportional to mass. (See the equation in question 35.)
37.
0 is correct. A pendulum is propelled by gravity. Since the clock is in free fall the entire apparatus is propelled by gravity and the pendulum would not fall faster than the apparatus from which it is hanging.
Passage III 38.
0 is correct. Light travels 3 x 10" m each second, so a light second is equal to 3 x 10' m. The distance in light seconds is the orbital radius of the moon (found in Table 1), converted to meters (384,400,000 m), divided by meters per light second (3 x 10" m l light second). Using scientific notation, this comes out to something slightly grea ter than 1. .
39.
C is correct. The passage states that the same side of the moon always faces the earth and that the moon moves around the earth once every 27.3 days. Thus the moon revolves once on its axis every 27.3 days. Yes, you probably would be expected to know what defines a day on a given planet.
40.
C is correct. These are not lengthy calculations, so use your pencil! To solve this problem w e set the gravitational force of the sun on the earth equal to the centripetal force. F = G M8unm~~rth 2
r u';';I.1
solving for v we get: v = JGrM-
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MeAT
PHYSiCS
41.
C is correct. The quickest way to solve this problem is by proportions. The radius of the earth is a little less than 4 times as large as the radius of the moon. According to F ~ Gmmlr', this would lead to an increase in relative gravity on the moon by a factor of something less than 16. From Table 1 we see that the earth is about 80 times more massive than the moon. This would lead to a decrease in relative gravity on the moon by a factor of 80. Multiplying the man's weight on the earth (1000 N) by 16 and dividing by 80 we get: 16,000/80 ~ 200 N. This is easily closest to 170. Don't be afraid to round off. OR, you can solve this problem the long way: From Table 1, plug into F ~ Gmmlr2for the moon and solve.
42.
A is correct. This is Newton's 3rd law: For every force there is an EQUAL and opposite force.
Stand Alones 43.
C is correct. Just as if we suspended a 200 kg mass from the rope, the ceiling would pull upward with a force of 2000 newtons and the mass would pull downward with 2000 newtons, but the tension in the rope would still be mg or 2000 newtons.
44.
A is correct. Since there is no acceleration, the net force is zero.
45.
A is correct. From the F ~ Gmmlr2 equation, we know that the weight is reduced by increasing r. The flattened shape of the earth makes the equator further from the center of gravity of the earth. Secondly, if there were no gravity, objects along the equator would be thrown away from the earth by the centrifugal force. More precisely, they would continue straight in the direction of their present velocity iostead of turning with the earth's surface. Thus, some of the force of gravity is used up as centripetal force to turn the direction of the velocity of the objects. This results in a decrease in the weight of the object.
46.
D is correct. F ~ -kx and F ~ mg. Thus mg ~ -kx. Ignore the negative which only indicates that the force is io the opposite direction of the displacement. 5 x 10 ~ O.lk.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 3 Passage I 47.
C is correct. Each revolution moves Puncho 211:r meters, which equals 0.511: m/ revolution. If we examioe the units, we see that dividiog the velocity 10 m/s by 0.511: m/rev gives us 20/11: revolutions/second. The question asks for revolutions io one second, so the answer is 20/11: revolutions.
48.
A is correct. Since the pole is stationary, the net torque must be zero Newton meters.
49.
B is correct. As shown by the equations below, the aogle is related to the mass of Puncho and the tension in the rope, and is not related to the length of rope.
8 T]cosA] mg
50.
r .cos8 2 mg
B is correct. Sioce gravity is the only force acting on a ball while it is io the air, and because gravity is a conservative force, while in the air, the total energy of each ball remains constant. Each ball has an energy equal to the ioitial kinetic energy given to it by Puncho, Ij,mv2 ~ 25 J. Five balls gives a total energy of 125 J. Or each ball has an energy equal to the maximum potential energy which is P.E. ~ mgh ~ 0.5 x 10 x 5 ~ 25 J. Five balls gives a total energy of 125 J.
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EXPlANATIONS FOR f HE 3D-MI NUTE IN-CLASS EXAM INATIONS •
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B is correct. When Puncho is on his unicycle, he doesn't want to move laterally. Inertia is the tendency for an object to resist change in its present state of motion. Mass is a measure of inertia, so a massive pole increases inertia. As an object's mass is spread from its center of gravity, more torque is required to rotate that object; its rotational inertia is increased.
Passage" 52.
D is correct. There is an initial net force, and all the forces involved are constant, so the acceleration is constant. Acceleration means no equilibrium . The mass does not achieve equilibrium regardless of how long it falls; it always accelerates.
53.
B is correct. Think of the extreme case. If the mass m is much, much larger than 0.1 kg, the greatest acceleration is still only equal to g = 10 m/s2 In order to accelerate 0.1 kg at the rate of g, we would only need 1 N of tension in the string. (F = rna)
54.
A is correct. The mass in Experiment 1 is in static equilibrium; there is no net force.
55.
A is correct. Mass m would faU at a rate of g = 10 m / s'. Any tension in the string would s low this rate and mass m would no longer be in free fall. The tens ion in the string must be zero.
56.
D is correct. The forces on the block are constant, so, from F = 1IIa, so is the acceleration. The block does accelerate because tension had to overcome static friction to move the block initially. Kinetic friction is weaker than static friction, so the tension must be greater than kinetic friction.
57.
B is correct. Friction is a force opposing motion, which is in the opposite direction of tension, increasing the tension. Changing mass m does not change the friction. The friction is given by f = IlN, where N is the normal force which in this case is the weight of the 0.1 kg block.
58.
A is correct. You should use Table 2 to figure this out quickly. You can see from the table that there is not a linear relationship between mass 111 and tension T. From trial 1 to 3 the mass was doubled increasing the tension by 0.17 N; from trial 2 to 4 the mass was doubled increasing the tension by 0.15 N. Thus, you would expect that if we double the mass of trial 3 we would get an increase in tension of less than 0.15 N . A is the only answer that sa tisfies this criterion. Tnal I 2 3 4
m
T 0. 17 N increase O. 15 N increase [ncrease should be less than 0.15 N so T should be less than 0.83.
The long way to figure this out is as follows: Assuming no friction, first look at the 0.1 kg mass. T = 1IIa => T = O.la. For the other block mg = T + rna => 0.4(10) = T + O.4a. Putting both equations together we get 0.4(10) = T + 0.4(T /0.1) => 4 = 5T => T = 4/5 = O.S.
Passage III 59.
D is correct. Once the bomb is in the air, gravity is the only force acting upon it. Since gravity is a conservative force, total mechanical energy is conserved. Just before the bomb hits the ground, all the energy would be kinetic. The initial potential energy of the bomb is P.E. = rngh = 2 x 10 x 300 = 6000 J. The initial kinetic energy of the bomb is KE. = '12 m'; = '/, x 2 x 30' = 900 J. The total energy is the sum of these two.
60.
B is correct. The power required is P = Fv. Where the force is the weight of the rocket mg, and v is the velocity of the rocket. Thus, 1200 x 300 x 10 = 3.6 x 10' W.
61.
D is correct. Energy is a scalar and is conserved. The total initial energy will equal the final energy in every case. Also, in every case gravitational potential energy w ill be completely converted to kinetic energy. Since energy is a sca lar, the direction of the plane does not affect its initial energy. (Notice that in question 59 we were not concerned with the direction in which the plane was flying.) The initial potential energy is the same in each case. Thus the initial energy and the final energy are greatest where the initial velocity is the greatest.
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MeAT PHYSICS
62.
C is correct. The plane is not a glider. It receives its energy from fuel.
63.
A is correct. Terminal velocity is constant velocity, so the net force must be zero. The forces acting on the plane are its propulsive force, the air resistance, and gravity. The sum of these forces is zero.
64.
A is correct. The apparent weight of the pilot is the reading on a scale if he were sitting on the scale during his dive. When he pulls out of the dive, he is decelerating. This means that the scale underneath him would have to push up w ith m ore force than his weight. His apparent weight increases.
65.
D is correct. Don't use vectors to solve this problem. The work done is equal to the change in potential and kinetic energy. W =6.K.E. + ME. W ='f, m,r + mgh ='f, x 2 x 400 + 2 x 10 x 200 = 400 + 4000.
66.
C is correct. The time for the bomb to reach the ground is found from the equation x = 'hgf. t = 6 seconds. The horizontal velocity of the bomb is the same as the plane, 30 m/s. In 6 seconds, the bomb will move 180 meters horizontally.
Stand Alones 67.
B is correct. Energy is required to separate attracting bodies. The rocket is attracted to earth by gravity. Gravity is a conservative force so the added energy goes into potential energy.
68.
D is correct. The left end is 4 times closer to the balancing point than the right end, and thus has a lever arm four times smaller. A lever arm four times smaller requires a force four times greater or 4mg.
69.
B is correct. Finding the center of gravity can be done like a torque problem. To find the center of gravity; we find the balancing point. The lever arm for the moon must be 80 times longer than the lever arm for the earth because the earth is 80 times heavier. We should divide the distance from the earth to the moon into 81 equal parts, but it is easier to use 80. 400,000180 = 5,000. Thus 400,000/81 < 5,000. The center of gravity is less than 5,000 km from the center of the earth, which is just beneath the surface of the earth.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 4 Passage I 70.
D is correct. In order for momentum to be conserved, the horizontal and vertical components mus t be equal before and after the collision. There is no vertical momentum before either Collision 1 or Collision 2, so there can be none afterwards. In order to have zero vertical momentum, angles x and y m ust be equal to zero.
71.
B is correct. In order for momentum to be conserved, The momentum m,v, (before) must equal (m, + m2 )v (after). So, (1 kg)(4 m / s) = (2 kg)(v). Thus, v = 2 m /s.
72.
C is correct. To conserve momentum, mAYA (before) must be equal to /n.v. (after). Therefore (2 kg)(4 m /s) = m(4
m /s) . So In = 2 kg. 73.
D is correct. There is no motion after the collision, so there is no kinetic energy. The blocks are permanently deformed and all of the collisions take place on a horizontal surface, so there is no elastic or gravitational potential energy. So the kinetic energy must be converted into heat energy during the collision.
74.
D is correct. The vertical momentum before the explosion is zero, so it must be zero after the explosion. The momentum going up the page is (1 kg)v,sin(w), and the momentum down the page is (1 kg)v3sin(z). These two must be equal if they are to add up to zero. The other particle moves horizontnliy, so it has no effect on the vertical m omentum.
75.
A is correct. In order for a collision to be perfectly elastic, mechanical energy must be conserved. Mechanical energy is not conserved in explosions or in collisions where the colliding objects stick togefher, so only Collision 1 can be perfectly elastic.
76.
C is correct. 'The two blocks come to a stop after the collision, so the ir momenta must have been equal and opposite before the collision. Therefore (2 kg)v A (I kg)v•. So 2vA = VB.
=
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EXPLANATIONS FOR THE 30-MINUTE IN-CLASS EXAMINATIONS .
221
Passage II 77.
C is correct. Energy from a fusion reaction comes from rest mass energy: E = mc2
78.
0 is correct. The equation is given in the passage. The temperature at the sun's core is given as 1.3 keY. We divide this by Boltzmann's constant to get D. Don't forget the kilo in keY.
79.
B is correct. The net reaction found by adding all the reactions together is: 4 J H + 2e-
80.
B is correct. The energy for the reaction came from the annihilation of an electron and a positron. This energy equals me', where m represents the mass of both the electron and positron. But first we must convert to joules by multiplying 1.02 MeV (don't forget that Mega = 106) times coulombs/electron (given at the end of the passage). This is the energy in E = me'. We divide energy by the speed of light squared to get the mass of the electron and the positron. We divide by two for the mass of the positron. The answers are given to the nearest magnitude of 10, so, as usualy, round your numbers and do calculations quickly: 106 x 10-19 = m x 9 x 1016 => m = 10-13 /10 17 = 10-30. The true mass of a proton or electron is 9.1 x 10-31 kg.
81.
B is correct. If energy is released then stronger bonds must be formed. As stated in the question, the strength of the bonds comes completely from the binding energy.
82.
0 is correct. Choice A is a violation of the conservation of energy. B is contradicted twice in the passage. C is contradicted in the passage and you should know that large nuclei undergo fission and small nuclei undergo fusion.
--+
'He + 6y + 2v
Passage III 83.
C is correct. Both forces act on the same side of the fulcrum and the in-force is nearer to the fulcrum. From the passage, this is a third order lever system. The passage also states that running mammals take advantage of a third order lever system. Although Figure 1 shows a running mammal, and Figure 2 a digging mammal, the two lever systems are the same.
84.
B is correct. The lever in Figure 2 has a greater in-lever arm to out-lever arm ratio and thus applies greater leverage and more force.
85.
A is correct. The out-lever arm is 10 times greater than the in-lever arm, so the force applied by the out-lever arm must be 10 times smaller. As stated in the passage, the advantage to this system is speed not strength.
86.
C is correct. The same formula as the previous question applies. The shorter lever arm requires the greater force. Work is not changed by an ideal machine.
87.
B is correct. Decreasing the in-lever arm while increasing the out-lever arm creates greater relative velocity between the fulcrum and the point of out-force application. The animal in figure 2 is a digger and the animal in Figure 1 is a runner; notice the lever arm proportions. The passage also explains this. The passage tells us that swift runners take advantage of third order lever systems. In other words, swift rurmers take advantage of having the in-force closer to the fulcrum and the out-force further away.
88.
A is correct. The elbow is the fulcrum. The distance from the elbow to the point where the force due to the muscle is 90° is greatest in A. This is the in-lever arm.
Stand Alones 89.
B is correct. 2 hours is approximately 5 half-lives. 25 = 32. 800/32 = 25. But you should just count on your fingers.
90.
0 is correct. Two alpha decays indicate a loss of 8 in mass number and four in atomic number. 4 beta decays
indicates an increase of four in atomic number. The element remains the same and the mass number goes to 210.
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91.
MCAT
PHYSICS
B is correct. The angle is irrelevant. Both the lever and the pulley system reduce the force necessary by a factor of 2 each. The total reduction is by a factor of 4.
Pivot point
(~~ T
= F/2 = Mg!4
~r 92.
B is correct. The horizontal momentum of the two rock system equals zero so there will be no horizontal velocity. The vertical momentum is 87 kg m/ s. We divide this by 15 kg to get 5.8 m/ s.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 5 Passage I 93.
D is correct. The passage states that the coefficient of volume expansion is 3 times a. Use the value of a from the table. Don't forget that the table gives a in magnitudes of 10-<>.
94.
C is correct. Steel has a lower ex than brass and increases more slowly to temperature change.
95.
B is correct. Plug in 19 x 10-6 for a, 10 for !1T, and solve for !1L/L. !1L/L is the fractional change in length. Multiply this times 100 to get the percent change in length.
96.
A is correct. Aluminum has the highest a, and thus has the greatest change in length or volume per change in temperature.
97.
B is correct. Since the density of water is greatest at 4°C, water expands when cooled below 4 0c. The weight does not change with expansion or contraction. There are the same number of water molecules; only the space between them has changed.
98.
A is correct. The easy way to solve this problem is to take things to the extremes: Since the aluminum has a much smaller volume coefficient of expansion, imagine that the aluminum doesn't change volume at all. The buoy must displace enough water to equal its mass. The mass of the buoy doesn't change with temperature. So in our extreme example only the volume of water is changing with temperature. When the water gets warm, it expands. The same mass of water fills more volume. The buoy sinks in the summer, in order to displace enough water to equal its weight. Buoy
water in winter
water in
winter
summer
summer
In the summer water is less dense and more must be displaced in order to equal the weight of the buoy.
The more difficult way to solve this problem is as follows: PbuoyVbuoyg:::: PwatcrVwa ter displacedg => Pwilter displacedg :::: Pbuoyg(Vbuoy/Vwater displ
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EXPLANATIONS FOR THE 30-MINUTE IN-CLASS EXAMINATIONS .
223
Passage II 99.
D is correct. This is a fluid at rest. The pressure is greatest where the depth is the greatest. P measured from the surface.
= pgy; where y is
100.
D is correct. In an ideal fluid, flow rate is the same at all points.
101.
B is correct. P = pgy. P is 5 times that of water, thus p is 5000 kg/ m'- Measure y from the surface of the fluid: 6 em. P = 5000 x 10 x 0.06 = 3000 Pa. The pipe is sealed shut so there is no atmospheric pressure.
102.
C is correct. Choose ho = 0 to be point D, so h = 0.1 m, and we have v = v = ~2gh = 1.4 m/s. Important: Notice that the velocity at C and B will also be 1.4 m/ s because Q = Av. We can only use v = ~2gh at point D because the pressure is the same at point D as at the surface of the fluid.
103.
D is correct. The volume of fluid displaced by the object is equal to the volume of the object. The mass of the fluid displaced by the object is equal to the apparent loss of mass of the object (weight = 5 N). Since the specific gravity of the fluid is 5, the same volume of water would weigh 1 N. Thus, the object weighs 20 times more than water giving it a specific gravity of 20. water
object
\
't 1N
20N
104.
C is correct. As y decreases, the pressure (P = pgy) decreases, velocity (v = h = Y in this case) decreases, and flow rate decreases (Q = Av [since v decreases, Q decreases]). The density does not decrease. At first glance, this appears to violate the rule that Q is constant everywhere in an ideal fluid. The reason that it doesn't violate this rule is because the rule says Q is constant everywhere in space, not in time. In other words, Q can change with time, but not with position; at any given moment Q is constant in any given cross-section of an ideal fluid.
105.
B is correct. Point C is twice the depth as point A, so the pressure is twice as great. (P = pgy) This question is really just asking "Do you measure y from the top, or from the bottom?" Of course, you measure y from the top.
Passage III 106.
B is correct. Since lift acts against gravity, the ball will undergo less downward acceleration, so it will go higher and stay in the air longer. The horizontal distance is given by the horizonatl velocity times the time in the air. Since time in the air increases, it will also go farther horizontally.
107.
Cis corree!. The speed of a point on the surface of a rotating object can be found by multiplying the frequency of rotation by the circumference. Don't forget to convert centimeters to meters. So w = (3.14)(0.043 m)(60 Hz) = 8.1 m/ s. 60 Hz has one significant digit, so the answer can have only one significant digit. The answer is 8 m/ s.
108.
B is correct. The density of the ball has nothing to do with the force exerted on it. A less dense ball might experience greater acceleration for a given force, but it will not change the actual lift force. All of the other choices will increase the pressure difference as shown in the equation in the passage.
109.
A is correct. The density of the ball is (45 g)/(42 cm3). Since the density is greater than 1 g/cm3 (the density fa water), the specifiC gravity is greater than 1 and the ball will sink in water.
110.
B is correct. If the spin is reversed, the relative airspeed will be decreased above the ball, causing the pressure above the ball to be greater than the pressure below.
111.
D is correct. Subtract the airspeed below the ball u - w from the airspeed above the ball u + w to get the difference 2w.
Copyright {CJ 2007 Exarnkrackers, Inc.
224
MeAT
PHYSICS
Stand Alones 112.
C is correct. The density of the container is 2 kg/ 5 X 10-3 m' = 400 kg/m'. The specific gravity is 0.4 which means that 60% floats above the water. Or even simpler, fhe bottle weighs 2 kg, while 5 liters of water would weight 5 kg, so the specific gravity of fhe bottle is 2/5 = 0.4.
113.
B is correct. When fhe brick is on fhe Styrofoam, the water that is displaced is equal to the weight of the brick. When fhe brick is submerged, fhe water fhat is displaced is equal to less than the brick. More water is displaced by fhe floating brick-Styrofoam combination and thus the water level falls when fhe brick is submerged.
-
Weight of weight volume of = of water displaced brick.
- --.--
t--~~-,,;,·~· 1······ \-:-:-=, ,,,",,cc""n '·f~~~!;J. \
~~
Volume of water displaced
volume of brick
114.
A is correct. The leak is 5 m below fhe surface. We use v = ~2gh where h = 5 m.
115.
B is correct. 10 meters of water produces approximately 1 atm or 105 Pa of pressure (P = pgh). 5 m produces half of fhat. Add atmospheric pressure (105 Pal to get 150,000 Pa.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 6 Passage I 116.
A is correct. As shown in the diagram below, this is an inclined plane. The acceleration of any object down an inclined plane wifh no friction is g sinB.
r
Copyright@ 2007 EXGimkrilCkers, Inc.
ANSW ERS
117.
&
EXPLANATIONS FOR THE 30-MINUTE IN-CLASS EXAMINATIONS •
225
B is correct. The motion duplicates a pendulum with a length equal to the radius of the Earth. One period is equal to a round trip on the train. The equation for the period of a pendulum is T ~ 2" (Lig). Notice that the period is independent of the distance that the pendulum swings. Since the periods are equal, the top speed must be greater for the longer trip.
Washi
Manhattan
Boston
118.
B is correct. The passage says that the motion is similar to a pendulum. The period of a pendulum is given by T ~ 2" (Lig). Here, L is the same as r, and we only want half the period because it is a one way trip. See the diagram above.
119.
A is correct. Simple harmonic motion can be graphed as a sine curve. Since a trip from Boston to Manhattan is half a cycle, Choice A is correct answer.
120.
A is correct. At the midpoint in the trip, the train is on a flat plane perpendicular to the radius of the earth. The forces on the train are gravity acting straight toward the center of the earth, and the normal force acting straight away from the center of the earth. Since the train is obviously not accelerating in either or these directions, these two forces are equal, and the net force is zero. Newton's second law, F = rna, tells us that the acceleration must also be zero.
121.
B is correct. See Figure 1. The track is straight. It is our perception of uphill and downhill that is the problem. We perceive downhill as any vector with some component in the direction toward the center of the earth, and uphill as any vector with some component in the direction away from the center of the earth. Because we are so small compared to the earth, we are accustomed to thinking of the direction of the center of the earth as being constant. This is not the case. In the example of the tunnel train, it is the direction toward the center of the earth that changes and not the direction of the track.
122.
C is correct. Gravity decreases as we move toward the center of the earth from the surface, so acceleration decreases, and maximum velocity will be less. The trip will require more time.
Passage II 123.
B is correct. Only B gives the correct prediction of change in frequency when the moth flies away from the bat. The Doppler effect predicts that when the source moves away from the observer, the observed frequency goes down. In this case, the bat is observing the frequency reflected off the moth.
124.
D is correct. The waves reflect off the moth at the same frequency that it receives them. This is the Doppler effect. Don't use the equation in the passage. To find the frequency at which the bat receives the waves use Nil ~ vic where I ~ 66 kHz, v ~ 15 m/s (the relative velocity), and c ~ 340 m/s. Since they are moving toward each other, the frequency will increase. Thus, to find the frequency at which the waves reflect off the moth we add Nto 66 kHz.
125.
B is correct. Only movement that separates the pair will decrease the frequency.
126.
A is correct. If the signal takes less time, then it must be going faster. The speed of sound in a medium increases with decreasing density. You may think that humid air is heavier, but £his is incorrect. There is a decrease in density as water vapor is added to air that occurs because the molecular mass of water (18 g/ mol) is less than that of nitrogen (28 g/ mol) or oxygen (32 g/ mol) gases, the main constituents of air.
127.
B is correct. This is just v hertz.
Copyright It) 2007 Examkrackcrs, Inc.
~
AI The numbers from the passage are: I ~ 83 kHz; v ~ 340 m/ s. Don't forget the kilo-
226
MCAT PIIYSICS
Passage III 128.
D is correct. The length of the first harmonic is the longest possible standing wavelength, which is simply twice the length between the fastened ends of the wire.
~ ..
..'
129.
D is correct. Pitch correlates with frequency; fA ~ v. We can set this equation equal to the one in the passage to see that increased tension raises frequency, and increased length lowers frequency. (From question 128 we know that speaking length L is proportional to wavelength A..)
130.
A is correct. Intensity is given by I ~ fi110)2A'v. Doubling 11 increases intensity by a factor of only 2. Since w ~ 2ITf ~ 2IT/T, doubling T would decrease intensity by a factor of 4. The speaking length is directly proportional to the fundamental wavelength. From v ~ fA, doubling the speaking length doubles A. and reduces fby a factor of 2, reducing intensity by a factor of 4. Doubling amplitude increases intensity by a factor of 4. Thus, changing 11 has the least effect.
131.
C is correct. 3 Hz is the beat frequency. The beat frequency is the difference between the frequency of the tuning fork and the frequency of the piano, so we know that the original frequency must be 3 Hz away from 440. We just don't know which direction. Tightening the string increased the beat frequency. This means that tightening the string moved us away from the tuning fork. Tightening the string increases the frequency, so when we increase the frequency we are moving away from 440 Hz. The original frequency of the piano note must be 443 Hz.
132.
A is correct. This is just v
133.
B is correct. The piano wire moves up and down, while the wave moves along the string; the medium is moving perpendicular to the propagation of the wave. This is a transverse, not a longitudinal, wave.
134.
C is correct. This is the definition of amplitude. The velocity is dictated by the medium. The wavelength is dictated by the speaking length. The frequency is dictated by the velocity and wavelength. How hard or far you strike the string only affects the amplitude.
~
V The speaking length is extraneous information.
Stand Alones 135.
A is correct. When the spring is fully compressed or fully extended is when the Hooke's law forces are the greatest, and this indicates that the acceleration is the greatest. F ~ rna.
136.
D is correct. The maximum amplitude will result from constructive interference. This is the sum of the amplitudes.
137.
D is correct. The period is increasing by 1.42 or about the square root of 2. From T ~ 2IT (L/ g) we see that the square of the period is proportional to the length. L increases by a factor of 2, or a 100% increase.
138.
B is correct. Only B and C allow for a higher frequency, which would explain a higher pitch. Heavier objects vibrate more slowly. The water in the glass increases the inertia of the system and creates a lower frequency.
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 7 Passage I 139.
D is correct. The Earth's magnetic field points from the geographic South to the geographic North. From the passage, the aurora borealis is in the north.
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ANSWERS
&
EXPLANATIONS FOR THE 30-MiNUTE IN-CLASS EXAMINATIONS •
227
140.
D is correct. The easy way to answer this question is to look at Figures 1 and 2. From the passage and Figure 1 we see that pitch is the distance between the spirals. From Figure 2, we see that pitch is decreasing as the spirals move toward the stronger magnetic field. (The magnetic field lines are the horizontal lines. Closer field lines indicate a stronger field.) This leaves answer choice CorD. The period is the length of time necessary for the particle to make one rotation. You should know that the force on a moving particle due to a magnetic field is Rerpendicular to the velocity of the charge. This means that a magnetic field can do no work on a moving charge, which, in tum, means that it cannot change its kinetic energy or its speed. If the speed is the same, and from Figure 2 we see that the spirals are getting smaller, then the length of time to make each spiral is also decreasing. Thus, the period is decreasing.
141.
A is correct. The electron has a negative charge, so it will be turned by the magnetic field in the opposite direction of the positively charged proton. The electron has less mass than the proton, so it will tum in a smaller circle than the proton. You can use the equation for centripetal force: F = mv2 1r. When m increases, r must also increase.
142.
C is correct. From F = qvB we know that charge increases force, so the alpha particle will experience the greatest force. Only the velocity perpendicular to the magnetic field will create a force on a charged particle. Any velocity in the direction of the pitch does not increase the force. Thus the smaller the pitch, the greater the force.
143.
C is correct. The protons rotate in the opposite direction to the electrons
because they have an opposite charge. As per Figure 2, the particles continue rotating in the same direction when they reflect from one pole to the other. However, the perspective from one pole to another. Imagine a two headed coin. The noses on either side point in opposite directions. When the coin spins, if we view the front side and the head appears to tilt downward, the head on the opposite side would appear to tilt upward. If the coin continued spinning in the same direction as it followed the magnetic
field lines from the North Pole to the South Pole, we would see the opposite side of the coin. From our point of view at the south pole, the opposite side would appear to be spinning counter-clockwise. 144.
A is correct. The force on any moving charged particle due to a magnetic field can only be perpendicular to the movement of the particle, and can therefore do no work. This is why the speed of the electron doesn't change.
145.
C is correct. According to Figure 2, in order for the particles to bounce from pole to pole, the magnetic field at the poles must be stronger than at the equator.
Passage II
146.
A is correct. Use the same logic as that used in the passage to explain why the human body remains at the same potential as the ground. The ground is a conductor, so it will be at the same potential at any height.
147.
C is correct. Use the definition of current. A = C/sec. So sec = CIA = 20/10,000 = 2 x 10-3 seconds.
148.
D is correct. V
149.
= lR : From the passage: 400,000 = 10-12 X
R.
B is correct. An electric field vector points from positive to negative potential. Figure 1 shows the electric field gets more positive as we move upward.
150.
C is correct. This question concerns the rate of energy transfer or power. P because the question asks for energy.
151.
C is correct. Current always flows from high potential to low potential, so Band 0 are wrong. During a lightning strike, electrons flow from the cloud to the ground, so current must flow in the opposite direction, from the ground to the cloud.
Copyright (g 2007 Exarnkrackers, Inc.
= IV.
The answer is given in joules
228
MeAT PHYSICS
Passage III 152.
B is correct. The passage states that the electric field force is in opposition to the magnetic force. Electrons want to move against an electric field. The electric field pushes the electron to the left. You can also use the right hand rule: Point your thumb in the direction of the current, down the page (opposite to movement of electrons), point your fingers in the direction of the magnetic field , and push in the direction of yo ur palm. Your palm should be facing to the right. This is the force on the electrons due to the magnetic field.
153.
A is correct. The passage states that at equilibrium "the force on the electrons due to E and B are equal and in opposite directions." Thus the direction of B creates the direction of E and reversing B will reverse E. B and D concern the current which can be adjusted independently of the magnetic field. A Hall potential difference will be established regardless of the direction of current, so C is wrong.
154.
B is correct. Moving the strip in the direction of the current is the same as moving the strip in the opposite direction to the electrons. If the strip is moved at the same speed as the electrons, the electrons are stationary with respect to the magnetic field and there is no Hall Effect. (This is explained in the passage.) When the strip is moved faster than the electrons, the relative velOCity of the electrons w ith respect to the magnetic field is reversed. Thus the force on the electrons due to the magnetic field is also reversed. The Hall Effect is established in the opposite direction from the original.
155.
A is correct. If the copper strip is held stationary then the sum of the forces or the net force must be zero. No acceleration; no net force.
156.
B is correct. The electric field E created by the magnetic field pushing the electrons to the r ight demonstrates the direction of force on the electrons due to the magnetic field B. Also, right hand rule applies for the same result. Since the force is constant and perpendicular to the velocity, the magnitude of the velOCity cannot change but the direction changes at a constant rate. The particle follows a circular path.
157.
0 is correct. The potential difference is the electric field times the distance. Choice B is the potential difference along the length of the strip, not the Hall effect, which is the potential difference across the width of the strip.
158.
B is correct. The voltage across the capacitor is 6 volts. IR
"V
Ij
[
"> 2U
~
'oF
=
V so 3 amps = 6 V I 2 Q. See the diagram below.
.i
l:;xi G ·: s implified
Vultugt: when the capacitor is fully charged
Cu rre ll! when the cnpnc it oJ"
is fully ch:lrged
Current the moment the ,witch i, opene d
159.
C is correct. The charge is equal to the voltage times the capacitance. See the diagram above for the solution of the circui t.
160.
A is correct. According to Coulomb's law (F = kqq/r), A and D apply equal and opposite forces on q, and so do Cand B.
161.
C is correct. First set the force equal to mass times acceleration Eq = mao Then use the uniform accelerated motion equation x = '/, a!'
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ANSWERS
&
EXPLANA110NS FOR THE 30-MINUTE IN-CLASS EXAMINATIONS .
229
EXPLANATIONS TO IN-CLASS EXAM FOR LECTURE 8 Passage I 162.
C is correct. Diffraction is demonstrated by the light waves bending as they move through the opening. This happens in both experiments. Interference is demonstrated in both experiments where the path length traveled by light rays differs creating alternating high and low intensity bands due to constructive and destructive interference.
163.
A is correct. From Soh Cah Toa, x/sinS is equal to r. From distance divided by velocity equals time, r/c is the time light takes to travel to any given point on the detector.
164.
C is correct. There would be no diffraction. In experiment 1, all the light would go directly to the detector at the level of the slit to form 1 peak on the graph. In experiment 2, two peaks would be formed.
165.
B is correct. Longer wavelengths are diffracted the most. Red has the longest wavelength.
166.
D is correct. The equations show that Id max is 4 times greater than I,.
167.
B is correct. Intensity is proportional to energy, and energy is proportional to the square of the amplitude.
Passage" 168.
C is correct. From the passage we know that mome = M'o',l and me = -25cm/f,.. Thus, the focal length is inversely proportional to the magnification. Since power is inversely proportional to the focal length, doubling the power doubles the magnification.
169.
D is correct. The object is outside the focal length of a converging lens and thus creates a real inverted image.
170.
C is correct. The focal length of a converging lens is positive. A diopter is the reciprocal of the focal length in meters. Thus, 1/25 equals 0.04 meters.
171.
C is correct. Just plug and chug. lengths.
172.
A is correct. An object inside the focal point of a converging lens makes a virtual upright image on the same side as the object.
173.
C is correct. The objective inverts the first image. This first image is within the focal point of the eyepiece so the eyepiece creates a virtual image WITHOUT changing the orientation. The final image is an inverted image of the object. An inverted image is inverted up and down, and left and right. When you push the slide left, it looks to be moving to the right under the microscope. When you push the slide up, it looks to be moving down. So, left is right and up is down on an inverted image. Answer choice C has these characteristics. Choice C is the inverted image.
174.
C is correct. Go back to lecture 8 for an explanation of refraction and reflection.
MMOI
= mom, = 500 x 0.01 x 0.005/ 0.25 = L. L = 10 cm. Then add the focal
Passage '" 175.
B is correct. From the equation in the passage we can see that this is true.
176.
C is correct. Although this is an approximation, it is thf' bpst anSWf'T. T.ir;ht rays from distant objects will be approximately parallel and thus converge on the focal point. Since the rays aren't truly parallel, they would form slightly behind the focal point but this is not an answer choice.
177.
A is correct. A thicker center converges more. Only converging lenses have positive power.
178.
D is correct. When light enters the lens, velocity slows, frequency remains constant, and wavelength must shorten. This effect is greater on blue light than red light. This is called chromatic dispersion.
179.
A is correct. P = l/focallength. This means the focal length of the eyepiece is 0.01 m. The focal length of the objective must be nine times greater, or 9 cm. It is positive by the equation or because converging lenses have positive focal lengths.
Copyright @ 2007 EXilmkrackers, Inc.
230
180.
MeAT
PHYSICS
A is correct. Light bends less moving from glass to water and thus the focal lengths of the lenses would lengthen. The tube length must be increased so that the two new focal points will coincide as per the passage.
Stand Alones 181.
A is correct. The waves are not moving through an aperture, so they are refracted, not diffracted. Since the waves are moving nearly perpendicular to the shore they must be turning toward the normal, which means their speed must be decreasing. Waves refract toward the normal when a new medium slows their progress. The new medium is water that is more shallow.
182.
B is correct. The mirror is concave since it makes a real inverted image. The focal length is positive, and half the radius of curvature. From the thin lens equation (1/f = 1/ d i + 1/ do) we get do = 5 cm.
183.
A is correct. Magnification =
184.
D is correct. The mirage is created by refraction of light. The light enters the warmer air and speeds up causing it to turn upward. The diagram below shows how the lower portion of the wave speeds up as it enters the warmer air causing the wave to turn.
-dJ do
warm air
///t(/ (
, \ '-..,
~
-wi:i;Oa"r~m~eiirfaOcl"r~-~+7JL-fI-T\-'\G'c-'~-'-o---o-'
mirage
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ANSWERS & EXPLANATIONS FOR
QUESTIONS IN THE LECTURES
231
232
MeAT
PHYSICS
ANSWERS TO THE LECTURE QUESTIONS Lecture 1
Lecture 2
e e
Lecture 3
Lecture 4
Lecture 5
49. D
73. D
97. D
e
74. A
98. 99.
Lecture 6
e
25.
2. A
26.
3. B
27. D
51. B
75. B
4. D
28. A
52. A
76. D
e e
29. D
53.
30. B
54. D
78. A
102. B
126.
7. B
31. A
55. B
79. A
e e e
32. A
56. A
33. A 34. D
58.
11. B 12. B
36.
13. A
37.
14. A
38. B
62.
e
39. D
63.
16. D
e
1.
5. 6.
8. 9. 10.
15.
17.
50.
e e
121. A 122. 123.
e e
Lecture 7
Lecture 8
e
169. A
146. B
170. D
147. A
171. A
e
172. D
145.
100. D
124. B
148.
e
125. B
149. B
173.
e
150. A
174. D
103. A
127. A
151. A
175. B
80. D
104. D
128. A
152. B
176. A
57. A
81. B
105. D
129. B
153. A
177. B
e
82. B
106. B
130. D
154. B
178. D
35. D
59. A
83. B
107. A
131.
e
155. D
179. B
e e
60. B
84. A
108. D
132. D
156. A
180. A
61. D
85.
109. D
133. B
157.
e
181. B
86.
110. B
134. B
158. D
182.
e
135. B
159. D
183. A
40. A
64.
e e e
e e
112. A
136. D
160.
41. D
65. A
89.
e e
113. D
137. B
161.
e e e
66. A
90. B
114. A
138.
e
91. D
68. A
,
e
77.
e
87. B 88.
101.
111.
e e
184.
e
e e
185. D
162. B
186. B
115. B
139.
e e
163. D
187. A
92. D
116. D
140. D
164. A
188. D
e
93. B
117. B
141. D
165.
e
189. D
46. A
70. D
94. B
118. A
142.
e
166. D
190. D
23. A
47. B
71.
e
95. A
119. D
143. B
167. A
191. B
24. B
48.
72. B
96. A
120. B
144. B
168. B
192.
18. D
42.
19. A
43.
20. A
44.
21. B
45. B
69.
22. B
e
67.
e
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AN SWERS
&
EXPLANATIONS FOR Q UESTIONS IN THE LECTURES •
233
EXPLANATIONS TO QUESTIONS IN LECTURE 1 1.
C is correct. The balloon travels in three perpendicular directions. These can be considered three displacement vectors. The total displacement is the vector sum of the three. If you notice, two of the vectors have lengths of 8 and 6, multiples of 4 and 3 respectively. These are the components of a 3-4-5 triangle. Thus, the displacement from the tail of the 6 km vector to the head of the 8 km vector is 10 kilometers. This 10 km vector is perpendicular to the other 10 km vector. Using the Pythagorean theorem on the two 10 km vectors gives a total displacement of approximately 14 km. 8
10
J6' + 10' + 82 =net displacement 2.
3.
A is correct. Since the runner is on a circular track of? km, the runner will end up where hel she began. This is zero displacement and zero velocity.
B is correct. The man is making a 30-60-90 triangle. Where he turns represents the 30° angle in the triangle. The d istance back to the entrance is half the hypotenuse or 100 sin30°. You can either recognize the proportions of
a 30-60-90 triangle Of, if you have a lot of extra time on your hands while taking the MCAT, use the law of cosines: A' = B' + C 2 - 2BCcos(a). 4.
D is correct. Acceleration is the rate of change of velocity. Since velocity is a vector it specifies direction. The direction of the earth's motion is constantly changing.
5.
C is correct. Since the beginning and starting points are the same, the displacement is the same.
6.
C is correct. If the car is slowing down, the velocity and acceleration must have opposite signs. Since the car is moving forward, it's safe to assume that the velocity is positive.
7.
B is correct. The direction for a vector must specify a straight line at a specific point. You couldn' t draw an arrow to represent uin a circle."
8.
C is correct. You need to convert your units. 1hr )=10m/s (36km)(lOOOm)( 1 hr 1 km 3600 sec In 10 seconds, the elephant can run 100 m .
9.
C is correct. Velocity is the slope on a d I t graph. Constant velocity requires only a straight line on a dl t graph. Any acceleration represents a change in velocity and so can not represent constant velocity.
10.
C is correct. Since we are looking for distance and not displacement, we add up the total area between the line and the x-axis. We could also use our linear motion equations since there is constant acceleration.
11.
B is correct. The graph shows an object moving in one direction at a constant velocity and suddenly changing d irections. There is no gradual acceleration. The baseball is th&only object that suddenly changes direction. A is wrong because the description describes gradual acceleration, and there is no gradual acceleration in the graph. C is constantly changing velocity. D is a grad ual change in velocity.
Copyrig ht Ii;) 2007 Exa mkrackers, Inc.
234
12.
MeAT
PHYSICS
B is correct. This problem may be tricky because the question only implies a necessary variable. That variable is initial velocity. The initial velocity is zero. The average velocity of any constantly accelerating object that starts with zero velocity is the final velocity divided by two. This from v'"g = (v + v o )/2. Then the average velocity times time equals displacement, v,"gt = x. Thus, 25/12.5 = 2. Or, by Salty's method:
r
25 mls mlS
omls
Average velocity times time equals distance: 12.5 x 2 = 25. 13.
A is correct. This is a plug and chug problem. The correct formula is v = Vo + at. Which results in 25 = 50 + a2. Thus a = 12.5. You also could reason that if it had taken only one second to slow from 50 to 25 mls then the acceleration would have been -25 m/s'. But it took more than one second so the acceleration must be less.
14.
A is correct. The graph clearly shows that displacement increases with time. Since the displacement graph is a straight line, the particle must be moving at constant velocity, so neither velocity nor acceleration are increasing.
15.
C is correct. Acceleration is 5 ml s' so the velocity is reduced by 5 ml s each second. Starting from 20 requires 4 seconds. Average velocity is between 20 and 0 which is 10 ml s. 10 ml s for 4 seconds gives 40 meters. Or use the equation that doesn't include time.
v/ = vo' + 2ax The car comes to a stop, so vI:::::: O. If we plug in Vo = 20 and a = -5, we'll get x = 40 m.
16.
17.
D is correct. In order for the particle to move backwards, the velocity graph would have to dip below zero. Between 10 and 15 seconds, the particle is slowing down, but not going backwards.
C is correct. The best way to answer this question is to plug 4 seconds into the linear motion equations
5
= 50
+
vJ + '/wt 2 This results in 8 = '/,at2 The distance traveled by the apple is 80 m so it reaches 20 m in altitude. You could solve this problem using proportions. The square of the time is proportional to the distance. If we double the time, we multiply the distance by 4. 18.
D is correct. Since both the ball and the skydivers are accelerating at the same rate, each skydiver should aim for the other's chest.
19.
A is correct, You should use Salty's system in every single physics problem. However, sometimes the problem is easy enough for you to imagine the diagram in your head. This problem probably requires actually drawing a diagram. Once you have your diagram, you may notice that the antelope needs to be in the air for 2 seconds at 10 mls to clear 20 meters. So now you have t. You know that the upward trip equals the downward trip, so the downward trip lasts one second. You also know that the downward trip starts at zero because a projectile at its peak has zero velOCity. The question is now simplified to "How far does a free falling body starting from rest travel in one second". At the end of one second the body is traveling at 10 ml s, thus its average velocity is 5 m/s. The ubject travels 5 meters. Alternatively, once you have 2 seconds, you can plug 2 seconds into s = So + vot + 'I, at', with sand 8 0 as zero. Doing the math gives you the initial velOcity of 10 m/s. The velocity at the top is zero. Using v2 = v,/ + 2ax gives you x = 5.
20.
A is correct. This is a proportions question. The correct equation is vosin9 = (2gh). Remember, due to the symmetry of projectile motion, the velocity in this equation can be initial or final depending upon the direction of motion. Multiplying the height by four only doubles the velocity.
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21.
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EXPLANATIONS FOR Q UESTIONS IN THE LECTURES
.
235
B is correct. You can use vosinB = (2gh). The sine of 30° is 11,. Thus the vertical velocity is 50 m / s. 50' is 2500. Divide this by g = 10, and by 2 gives 125. Practice doing problems like this in your h ead to save time, build your confidence, and most of all, to sharpen your skills. The initial vertical velocity is 50 m / s; the final at max height is zero. This is a change in velocity of 50 which takes 5 seconds at 10 m / s'. Draw your line and multiply the av-
erage velocity by 5 seconds.
50 m/s
~m/S omls
22.
B is correct. At terminal velocity, acceleration is zero. The force of air resistance counters gravity exactly so the force is equal to the weight for both balls. Ball X requires more collisions with air molecules to compensate for the larger force of gravity. More collisions means greater air resistance.
23.
A is correct. The horizontal speed has no effect on the length of time that a projectile is in the air, so you don't need it here. Because the initial vertical speed is zero, you can use the equation below.
x = (1/2)gf with x = 40 and g = 10
t =..J8 = 2.8
,-----
-
Since there is only one significant figure in the numbers in the problem we round the answer up to 3. 24.
B is correct. The horizontal distance traveled for a projectile is given by vlcose. In this case, v £'30, t
e =40°.
=
6, and
EXPLANATIONS TO QUESTIONS IN LECTURE 2 25.
C is correct. This is a straight forward F = rna plug-n-chug problem. The moon is thrown in to confuse you. The gravitational force of the moon acts perpendicularly to the horizontal force and is countered by a normal force. It has no effect on the motion.
26.
C is correct. Since the mass of the rocket is decreasing, and the force remains constant, the rate of change in velocity (acceleration) must be increasing.
27.
D is correct. The downward force is IIIg = 100 N. The first 100 N upward counters this to give a net force of zero and thus a constant velocity. We want a net force of mg = 100 N upwards. This requires adding 100 more newtons for a total of 200 N.
28.
A is correct. Since both skydivers are at constant velocity, they must both experience a net force of zero.
29.
D is correct. Since the projectile is at constant velocity when the force of air resistance is F, The force propelling the projectile must have a magnitude F as well. W11en the air resistance is reduced by a fa ctor of 4, the net force must be F - '/.F = 3/. F. Thus 3/.F = rna.
30.
B is correct. Because the masses are aU on a line, you can just average the distance of the masses from the origin to get the central point. The average distance is (2 + 3 + 7)/3 = 4.
31.
A is correct. The net force on the plane is 2500 N - 500 N = 2000 N to the east. The mass of the p lane will be the weight divided by g. That's 40,000/ 10 =4000 mg. Acceleration is F / 111 =2000/ 4000 =0.5 m i s' to the east.
32.
A is correct. If the car is moving in a straight line at a constant speed, then it is not accelerating. From Newton's second law, you know that if there is no acceleration, then there is no net force. Friction is irrelevant.
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MCAT PHYSICS
33.
A is correct. This question requires setting the gravitational force between the earth and the moon equal to the centripetal force.
34.
0 is correct. The work done will remain the same. W; Fd. The distance is increased from ,/, to 4. Tms is an increase by a factor of 8. Since work remains constant, force must decrease by a factor of 8.
35.
0 is correct. This question uses misdirection by introducing the concept of circular motion. Gravitational force is not affected by movement. The gravitational force is inversely proportional to the square of the radius.
36.
C is correct. The acceleration is gsinS; however, S is constantly decreasing and with it the sine of S. Thus, the acceleration is decreasing. Since there is some acceleration throughout the drop, the velocity is increasing.
37.
C is correct. The force down the incline is mgsinS. The acceleration down the incline is gsin8. The sine of 8 is opposite over hypotenuse wmch is 20/ 40 ; ? Thus the acceleration is '/' g or 5 m / s2. If we plug tms into our linear motion equation we have x = ' j,C/2g)t2. x is the length of the incline. Thus 40 = ' j,(' j,g)f. 16; t'. t = 4 seconds.
38.
B is correct. Set the weight equal to Newton's law of universal gravitation. GMm
mg = ~' m cancels, so g =
GM
J2l
39.
0 is correct. v; d/ I. Once around a circular track is the same as the circumference, so d = 21<'. So V (6.28)(30) / 63 ; (0.1)(30) = 3 m / s.
= 21
40.
A is correct. The force parallel to the ramp is the same as the net force, mgsine. As e increases, sine increases and so does the net force. The force perpendicular to the ramp is the same as the normal force, mgcose. As e increases, cose decreases, and so does the normal force.
41.
0 is correct. If we look at the point on the tire that makes contact with the road, that point does not move relative to the road or else the tires would spin in place. Since there is no relative movement, the friction must be static. The force of friction is in the direction opposite to the way the tires are trying to slide against the road. This is the force that accelerates the vemcle. So the only way that the truck can move forward is if the force on the tires is in tms direction.
42.
C is correct. Tension in a static system is defined by the force in one direction. The rope will also experience a force from the right, but that does not double the tension. That force is necessary to make the tension equal to 9OON.
43.
C is correct. The force changes wifh the displacement of the tires. The greater the displacement, fhe greater the force, the greater the magnitude of acceleration as per the formula F = k!J.x; rna .
44.
C is correct. Since the frictional force is constant, this is a linear motion problem with constant acceleration. The normal force is 1I1g, so the frictional force is mg~. The acceleration is just g~. Using if ; v o' + Zax and plugging in g~ for the acceleration, gives us answer C.
45.
B is correct. The tension could only be as great as the force at one end . Thus, the tension could not be greater than the force applied by the first team.
46.
A is correct. On an inclined plane F" ; mgcosfJ. So the force of friction is equal to
47.
B is correct. The difference in mass between the two situations is 0.5 kg, so the difference in force is 5 N. The difference in displacement is 1 cm. These are the numbers we plug into Hooke's law. k = Fix = (5 N) / (1 cm) = 5 N / cm. By the way, we can do this because we are really subtracting one equation from the other.
~mgcosfJ.
F'5 ; - k!J.x' 5 -F, = - k!J.x,
F L5 - F, ; - k!J.X' 5 - -k!J.x, ; > MI.5 = - k(!J.x LS - - !J.x,) Copyright It> 2007 Exam krackers, Inc.
ANSWERS
48.
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EXPLANATIONS FOR QUESTIONS IN THE LECTURES .
237
C is correct. If the elevator is moving at constant speed, then there is no acceleration and no net force, so the tension in the cable must exactly balance the weight of the elevator.
EXPLANATIONS TO QUESTIONS IN LECTURE 3 49.
D is correct. When the tightrope walker stands in the middle of the rope, he is in static equilibrium. The vertical and the horizontal net force must equal zero. The force downward is the weight of the tightrope walker, 750 N. The force upward must also equal 750 N. The upward force must come from the vertical component of the tension in the rope. If the rope is perfectly straight, there is no vertical component.
50.
C is correct. If we begin by examining the problem as a static equilibrium problem, the tension in the rope with no acceleration would be 500 N. If we want to pull the climber upward, we must increase the tension by ma, which is 250 N.
51.
B is correct. The force upwards is defined by the question to be 200 N. The force on either end of a massless rope must be the same. The tension in the rope must be 200 N. This question answers itself and then attempts to confuse the issue by discussing a rusted pulley.
52.
A is correct. The skydiver has a constant velocity so the net force must be zero, and by definition, this means the skydiver is in dynamic equilibrium.
53.
C is correct. You don't have to look at the diagram for this problem. C is the only answer that could be false and still have all the rest be true. Form a triangle with the vectors to prove to yourself that they sum to zero.
54.
D is correct. The third force must have equal components pulling to the south and west to counter the other two forces. Since the two components are equal, the third force will be directed exactly to the southwest.
55.
B is correct. The bucket is moving at constant speed, so there is no acceleration. If there is no acceleration, then the system is in dynamic equilibrium. All of the other situations described include accelerated motion.
56.
A is correct. The block is accelerating across the floor so the force applied by the child must be greater than any force that is resisting the motion. The weight of the block acts vertically, so it is not directly involved in the horizontal motion. It is true that the frictional force is likely to be less than the weight, but that doesn't indicate whether the force applied by the child is greater or less than the weight.
57.
A is correct. The pole is not rotating, so tl1e net torque must be zero.
58.
C is correct. The sign is in static equilibrium. Since the wall is frictionless, there is no torque in this problem. The vertical component of the force must be equal to the weight of the sign. The vertical component is equal to the tension times sin30 mg = Tsin30". If we double mg, we double the tension. Q
59.
A is correct. The longest lever arm is on A. The entire wrench is the lever arm on A.
60.
B is correct. This is a straight forward torque problem. If we choose our point of rotation to be the point on the board where the string attaches to the board, and we understand that the weight of the board acts at the center of gravity which is the center of the board in this case, the counter-clockwise torque is 3 kg x 0.2 m. The clockwise torque is the weight of the board times 0.3 m. Setting these equal, we have the mass of the board at 2 kg.
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238 61.
MeAT
PHYSICS
0 is correct. The net torque must be zero. If we choose the rotation point to be the end of board Y, then
1 m x F = 4 m x 4 N. A little tip for torque problems: Find a force that is neither known nor asked for, and choose the point where this force acts to be your rotational point.
16
3
I
I
4
12
62.
C is correct. If you push on the edge farthest from the hinges, you'll have the greatest lever arm, which will give you the greatest torque.
63.
C is correct. For the best balance, the torques should be equivalent. (400 N)(5 m) = (500 N)(x). So x = 4 m
64.
C is correct. Increasing the diameter of the screwdriver handle increases the lever arm for the force applied by his grip. Increased lever arm gives more torque to tum the screw.
65.
66.
A is correct. The frictional force applied by Jupiter's atmosphere times the distance along which it is applied equals the change in mechanical energy of the meteor. If we ignore the gravitational force of Jupiter, the change in the mechanical energy is a loss of all kinetic energy which equals 1/, mv2. We set this equal to force times distance and solve for force. A is correct. The potential energy increases as you go up the stack. This may seem obvious if you are thinking about Imgh'; however, you would be thinking about this problem incorrectly because g does not remain constant. If you use Newton's law of gravitation to solve for g at any altitude and then plugged it into 'nzgh'(g =
Gml r2, r = h), you would get the reverse answer that potential energy decreases to zero. A fast and easy way to understand this problem is to think of work. Your system is the stack of blocks. Each time you add a block, you do work on the system. In other words, you transfer energy to your system. 67.
C is correct. A simple technique for solving this and many other physics problems is to "take the examples to extremes". Here the examples are reasonably close in mass. What if object A were one million times as massive
as object B? In other words, imagine that object A is a piano and object B is a dime. Now we place them on a spring and propel the piano one inch into the air. Will the dime be propelled one million inches into the air at the same time? Of course not. Thus we know that mass is not proportional to the height. Since all the answers are given as such, only C can be correct. We could also look at this problem and ask ourselves, "When do the masses become projectiles? They must become projectiles the moment the spring stops pushing. From F = ma, we know that this is the moment the spring stops accelerating. This means that the masses become projectiles at the same moment and with the same velocity. As we learned in Lecture 1, mass is irrelevant to projectile paths. 68.
69.
A is correct. This is a proportions problem. We set the initial elastic potential energy equal to the final energy, 11, kx 2 = mgh. and we see that the square of the displacement of the spring is proportional to the height, x' = h. Thus, to increase the height by a factor of four, we must increase the displacement of the spring by a factor of 2. C is correct. The horizontal component of the force is 87 N. To find the acceleration we use F = rna. a = 8.7 ml S2 Vo + at and arrive at v = 17.4 m/s.
We use v::::
70.
0 is correct. The rock starts out with gravitational potential energy. As it falls, it loses gravitational potential energy and gains kinetic energy. As the rubber band stretches, the rock slows to a stop and kinetic energy is transferred to the elastic potential energy of the rubber band.
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«J 2007 Examkrackers, Illc.
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71.
72.
C is correct. 1 kW - hou{ 36oo seC ) 1 hour
B is correct. P of this.
~
~
3600 kW - sec
~
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EXPLANATIONS FOR Q UESTIONS IN THE LECTU RES •
239
3,600, 000 W - sec
A Watt is a J/sec, so a W-sec
~
O/ sec)(sec) = Joule
So 3,6000,000 W-sec
~
3,600,000 J
Wit. If the power is increased, more work can be done in less time. Choice B says the reverse
EXPLANATIONS TO QUESTIONS IN LECTURE 4 ~
73.
D is correct. Impulse is equal to change in momentum, F!J.t For t. We do not have enough information to solve for F.
74.
A is correct. This is easily visualized if we take the example to extremes. If the boy were somehow able to make himself nearly weightless, would he travel at some extremely high velocity? This is actually similar to a projectile problem . The boy receives his kinetic energy from his initial potential energy. When he drops his coat and boots, he is leaving that potential energy unused at the point where he dropped them.
75.
B is correct. The initial and final m omentums must be equal. The initial vertical momentum is mvcos60°. Since ball B has no vertical momentum, all the vertical momentum remains in ball A. Ball A has no horizontal momentum after the collision, so all of its momentum is represented by mvcos60'.
76.
D is correct. To answer this question, we simply compare the total energy of the car at zero, 30, and 60 km/h. All the energy is kinetic energy. K ~ II, mv' . Without doing too much math we can see that if you double the velocity you increase energy by four times. The question asks for the difference in the energies, 4 - 1 ~ 3.
77.
C is correct. The initial momentum of the cat-cardboard system is zero, so the final momentum must be zero. Since the cat has twice the mass as the cardboard, the cardboard will have twice the velocity.
78.
A is correct. Choice I represents conservation of momentum, which is always true. Choice II is conservation of kinetic energy. In an inelastic collision, kinetic energy is not conserved. As for Choice III, if the masses are different before and after the collision and momentum is conserved, then the velocities before and after must be different.
79.
A is correct. Momentum is a vector quantity, so the two vectors with the sam e magnitude and opposite directions will add up to zero both before and after the collision.
80.
D is correct. From the impulse equation Ft ~ mv. The trapeze artist is brought gradually to a stop in the safety net, so the change in momentum takes place over' a longer time than if the person hit the floor. The increase in time means that less force is required to achieve the same change in momentum, wruch makes the fall less dangerous.
81.
B is correct. The pulley does not change the work, so it does not change the rate at which work is donp. Thp power is workl time or mgh I t.
82.
B is correct. The eccentric pulley does not work on the principles of a normal pulley but, instead, works on the same prinCiple as a lever. ~ lever arm fo r the string at point A in position 1 is greater than that for point B in position one. In position 2, the reverse is true. The lever is stationary, so the sum of the torques must equal zero, or the clockw ise torque equals the counter-clockwise torque. Where the lever arm is greater, the ten sion force m us t be less.
83.
B is correct. Machines are used because they d ecrease the force required to perform a task. An ideal machine requires the same work as would be done without the machine, but a non-ideal machine requires more work because frictional forces must be overcome.
Copyrig ht
«, 2007 EXClm krackers, Inc.
L'.mv. We know the change in momentum but not
240
MeAT
PHYSICS
84.
A is correct. By zigzagging, she is basically using a ramp. Her path is not as steep, but it is longer. The work remains the same, but the force is lessened.
85.
C is correct. The work cannot be decreased by a machine. Therefore, Fd
86.
C is correct. For a lever FIll ~ F212 • The radii of the two pulleys act as the lever arms for the system, so increasing the diameter of pulley A will decrease the force required to pull rope A. Changing the lengths of the ropes will have no effect on the machine.
87.
B is correct. The formula for machines is FinpuAnput ;::O; Foutputdoutputf where d is the distance over which the forces act. If mechanical advantage is equal to (output force)/(input force) , then it must also be equal to (input distance)/(output distance).
88.
C is correct. For an inclined plane, Fd
89.
C is correct. Since Y is not normally found in the meteorite, we assume that all the Y came from the decomposition of X. The total amount of X originally, then, must have been 15%. The amount left is 10% of 15%, or 1.5%. We count on our fingers to find the half-lives. 50%, 25%, 12.5%. A little more than 3 half-lives. 3 times 45 is 135 plus a little more is 140.
90.
B is correct. This is a reverse collision. The initial momentum is zero; so, the final momentum must be zero. The momentum of the alpha particle is approximately mv ~ 4 x 10'. The momentum of the other particle, Rn-220 must be equal in magnitude and opposite in direction. Thus, the velocity of Rn equals (4 x 10')/220. This quickly rounds to 4 x 10'/2.2 X 102 whichis slightly less than 2 x 105 or choice B.
91.
D is correct. This is a simple exercise in plug-n-chug. Whenever there i~ some lllysterious lllissing Illass, E :::: mc2 •
92.
D is correct. First count the change in protons in order to discover the identity of the final atom. In this case, each alpha decay results in the loss of 2 protons, and each beta decay results in the gain of one proton. (Remember, create a negative, create a positive.) Thus, we have a net loss of 2 protons, and we know that our atom is Pb. Now we track the change in the mass number. Each alpha decay loses 4 mass units and each beta results in no change in the mass units. (Beta decay is an exchange of a proton for a neutron.) This means a total loss of 8 mass units for a new mass number of 208.
93.
B is correct. This is a half-life curve, for equal units of time the amount divides by 2.
94.
B is correct. The half life is the amount of time it takes for half of the isotope to decay. The graph decreases from 60 g to 30 grams in 2.5 hours, so the half life must be 2.5 hours.
95.
· correct. A IS
96.
A is correct. In alpha decay, a particle identical to a helium nucleus is released, so the mass number will change by 4. None of the other processes will change the mass number.
21OB'
83
ZU6Pb +20:+ 4 Oil -1 fJ
1---7 82
~
mgh, so (50 N)d
~
~
mgh. d ~ 300/25
~
12.
(1000 N)(1 m) and d ~ 20 m.
.
EXPLANATIONS TO QUESTIONS IN LECTURE 5 97.
D is correct. Atmospheric pressure supports the column of fluid. The pressure at the bottom of the column must be equal to atmospheric pressure. The pressure is equal to pgh. If P is decreased by a factor of 13.6, the height must be increased by the same factor. Notice that, given the choices, there is no need to do the math. Every other answer is less than 10 times as tall.
98.
C is correct. The only difference between the two discs is what they are covering. Ignore everything else. The first disc has atmospheric pressure pushing upward; the second disc does not. This is the difference between the forces necessary to lift them.
99.
C is correct. The brick displaces a volume of water equal to its own weight. Since its denSity is 1400 kg/m3, its density is 1.4 times that of water, and it must displace a volume of water 1.4 times its own volume. Since this
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EXPLANATIONS FOR QUESTIONS IN THE LECTURES .
241
is only 1/2 the volume of the Styrofoam, the full volume of the Styrofoam must be 2.8 times larger than the volume of the brick. 100.
0 is correct. The balloon rises because the buoyant force is greater than the weight. When these forces are equal, the balloon will stop rising. Thus the balloon stops rising when: p,j,Vg = Phelj=Vg. The volumes are always equal because the balloon is always fully submerged in the atmosphere. Another way to look at this problem is to see that the balloon is fully submerged in the fluid atmosphere. We want the balloon to float, not rise or sink, so we use the floating equation: Fraction submerged = Pobje'/ Pllujd' The entire balloon is submerged, so the fraction submerged is equal to one.
101.
C is correct. If you have forgotten the floating equation, the quickest way to do this problem is to take the example to the extremes. If the specific gravity of the toy were 0.999, the toy would be almost the same weight as water and, of course, only a very small part would float above the water; 0.001/1 = 0.1%. The specific gravity must be how much is under the water. Now we look at the example in the question. 45% must be under water so 55% must be above. To solve this problem mathematically, we set the buoyant force equal to the weight of the toy, Pwatcr Vsubmcrgcd fraction of the toyg
=
Ptoy V toyg·
We end up with the ratio: Vsubmcrged fraction of thctoy/Vtoy
=
Ptoy/ Pwater
The right side of this equation is the specific gravity, and the left side is the fraction of the toy submerged. To find the fraction of the toy above water, subtract the submerged fraction from 1. 102.
B is correct. The pressure on both sides is the same. Force is equal to the product of pressure and area, so the force will be larger on the side with the greater area.
103.
A is correct. The formula for fluid pressure is P = pgh. If the density is changed, the pressure will change by the same ratio. Since the specific gravity of ethyl alcohol is 0.8, the pressure will decrease by a factor of 0.8.
104.
0 is correct. Pressure depends only on depth and density, not on the shape of the container.
105.
0 is correct. The cross sectional area A is increased by a factor of 4 when r is doubled: A = 1tr'. Since Q remains constant, the velocity decreases by a factor of 4, Q = Av. From Bernoulli's equation, we see K = P + 1/, pv'- We know that the 1/2 pv' term decreases by a factor of 16, however, we don't know the amount and thus we don't know by how much P increases.
106.
B is correct. The fluid at A, C, and 0 is at atmospheric pressure. The fluid at B is at atmospheric pressure plus pgh.
107.
A is correct. Since the molecules of water are more attracted to the soil than to other water molecules, the shape of the meniscus is concave. The surface tension creates a net force which pulls the water upward.
108.
0 is correct. Increasing the length increases resistance to flow, M = QR. All other factors mentioned decrease the resistance to flow. Notice that this question is concerned with a real fluid. For an ideal flow, there is no viscosity, and the radius and length of the pipe do not affect flow rate of an ideal fluid.
109.
0 is correct. The equation for velocity of fluid from a spigot is derived from Bernoulli's equation. The relationship is 2gh = v', and h is proportional to v 2, which is reflected in the graph in answer choice D.
110.
B is correct. The drop with stronger intermolecular forces will have greater surface tension, which will cause it to bead up more.
111.
C is correct. For ideal flow, volume flow rate is constant at all points, so the volume flow rate will be equal at points A and B.
112.
A is correct. The equation governing the speed is v = ~2gh . As h decreases, so does v.
113.
0 is correct. The pressure at the bottom of the column is given by pgh. Setting this equal to maximum pressure we get 12 x 10' = 4000 x 10 x h.
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242
MeAT
PHYSICS
114.
A is correct. The Young's modulus for any substance is a constant.
115.
B is correct. Using the formula for Young's modulus we have: strain = (5.4 x 103 16 x 104 )/9 x 109 Thus strain equals
116.
10~3
This is 0.1 %.
D is correct. The weight per unit area cannot exceed one fifth of the yield strength. Convert 1.5x108 kg to 1.5x109 N. Divide the yield strength by 5 and set this equal to the weight per unit area.
(O.5xHY' N/m 2 ) = (l.5xlO° N)I A A = (1.5x109 N)/(0.5x10 8 N/m')
= 30 m'
117.
B is correct. The strain must remain the same. The deformity of the shoe will double with the height vecause F1A does not change. This keeps the strain (~I ho ) the same.
118.
A is correct. Copper has the largest value for Young's modulus, so it will undergo the least strain for a given stress. glass~
119.
D is correct. Young's modulus for lead is one-fourth the modulus for be four times the change for glass.
So the fractional change in lead will
120.
B is correct. The bulk modulus describes a substance's resistance to pressure applied from all sides, which is the same as the stress encountered under water.
EXPLANATIONS TO QUESTIONS IN LECTURE 6 121.
A is correct. The period of each wave is 4 seconds so the frequency is v = fA.
122.
C is correct. The formula for this problem is v = (Pip). Although densities of solids are usually greater than the density of gases, this would make waves move more slowly. Thus, the answer must be that solids are less compressible than gasses. This means that they have a higher bulk modulus
1/4
Hz. The wavelength is 12 meters.
p.
123.
C is correct. The frequency of the waves being sent is equal to the frequency of waves being received. Everything else is irrelevant information. Every second one wave is sent. In 10 seconds, 10 waves are sent.
124.
B is correct. Sound level (P) is related to intensity (1) by P = 10 log (I/Io )' so a change of 20 dB in sound level means that I is 100 times greater at 5 meters. The intensity of a sound wave is proportional to the square of the amplitude, so to change the intensity by a factor of 100 requires a change in amplitue by a factor of only 10. This question requires you to consider two relationships. Although you should know both relationships for MCAT, that's probably one step more than a real MCAT question would require.
125.
B is correct. By
126.
C is correct. Frequency is related to both wavelength and period by inverse relationships. There is no direct mathematical relationship between amplitude and frequency.
127.
A is correct. If it takes 1 second for the wave to go to the bottom and back, it must take 0.5 seconds for the wave to reach the bottom. We know that vi = x, and x is the depth in this case. So, (1500 ml s)(O.5 sec) = 750 m.
128.
A is correct. A 10 dB increase means that the intensity is increasing by a factor of 10, so if the intensity is only doubled, there will be less than a 10 dB increase (actually, it's about a 3 dB increase).
129.
B is correct. The pattern repeats twice between the dotted lines.
130.
0 is correct. This is the definition of interference. The displacements of two superimposed waves are summed all along the wave.
131.
C is correct. An audible beat frequency requires that the two sound waves have close frequencies.
definition~
(See the first page of this lecture)
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ANSWERS
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EXPLANATIONS FOR QUESTIONS IN THE LECTUPES .
243
132.
D is correct. When energy is added to a structure at one of its natural frequencies, the amplitude reaches some maximum value. At this point, damping effects create an energy loss at the same rate at which energy is absorbed, and the total energy of the structure is constant. A standing wave is produced.
133.
B is correct. The formula for the third harmonic closed on both sides is L = 3'-/2. Here, L is 0.5 m. Thus '- = 0.33 m.
134.
B is correct. When two slightly different frequencies are sounded at the same time, they will create beats with a frequency equal to the difference. So 883 Hz - 879 Hz = 4 Hz.
135.
B is correct. The first and second harmonics are the only consecutive harmonics that have a ratio of 1 to 2. The second harmonic is the length of the string. Alternatively, lise the harmonic series formula, L = nAil, to find n. 2
n(4.0) =(n + 1)(2.0) From this you can find that n = 1. Now plug n = 1 back in to find L. L = (1)(4.0) 2
2
=
2.0.
2
= 9). If there is perfect
136.
D is correct. If there is perfect constructive interference, the amplitudes will add (6 + 3 destructive interference, they will subtract (6 - 3 = 3).
137.
B is correct. The period of a pendulum is not related to the mass of the bob. It is similar to projectile motion in this respect. It is possible to think of a pendulum like a 'guided' body in free fall.
138.
C is correct. As shown earlier in the lecture, kinetic energy of a pendulum can be described by a sine wave with energy fluctuating between zero and some maximum.
139.
C is correct. The square of the amplitude is proportional to the intensity.
140.
D is correct. For the frequency to increase, the relative velocity must move the source and observer toward each other. Be careful. The relative velocity does not dictate at what frequency the sound is heard. A wind can also change the frequency by changing the velocity of the sound, but, even with a wind, the source and the observer must have a relative velocity.
141.
D is correct. The velocity of the wave on a string is a function of the properties of the string, so it will remain constant. The wavelength of a standing wave is determined by the length of the string, so it will remain constant. If the velocity and wavelength are constant, then frequency is constant too.
142.
C is correct. The pendulum is at its greatest speed at point A, so it will be at it's greatest kinetic energy. All of the other quantities are at zero at this point.
143.
B is correct. Remember, frequency of a pendulum is related to the square root of gil. High above the earth, the acceleration due to gravity will decrease. If g decreases, the frequency will decrease. If the frequency of the timekeeper decreases, the clock will slow down.
144.
B is correct. Harmonic motion is motion that is repeated over and over again. There is no repeated action in the fall of a skydiver.
EXPLANATIONS TO QUESTIONS IN LECTURE 7 145.
C is correct. This is a units question. 100 NIC is equivalent to 100 V 1m. The one coulomb experiences 100 Newtons of force. This is a measure of the strength of the electric field: 100 N/C Another way to say 100 NIC is 100 V 1m. The plates are one meter apart, so they must have a 100 volt potential difference.
146.
B is correct. The forces are conservative so if we tum the picture 90°, this is just like gravity, mgh; the vertical distance h, and not the horizontal distance, is what matters. Similarly, in the question, only the distance against the electric field matters. The work required is the force times the distance parallel to the field or Eqd.
147.
A is correct. The force is given by Coulomb's law, F = k qqlr. The electrostatic force changes with the square of the distance between the centers of charge.
Copyright
(£!
2007 Exarnkrackers, )nc.
244
MeAT
PHYSICS
148.
C is correct. The electric field above an infinitely large electric plate remains constant with distance. You can visualize this by imagining the electric field lines. The lines are perpendicular to the plate and have nowhere to spread. By bending in one direction or another, they would increase their distance from one line, only to decrease their distance from another line. Since the lines would remain at an equal distance from one another, the electric field would remain constant.
149.
B is correct. This problem is about energy. The system has a total electric potential energy of U = kqqlr. Remember, the forces acting are conservative so mechanical energy is conserved. Thus, as the first particle is propelled away from the second, electric potential energy is converted to kinetic energy. When the first particle moves 25 em, it has doubled its distance of separation. From U = kqq I r, we know that the first particle has lost half of its potential energy to kinetic energy when r is doubled.
When the first particle is infinitely far from the second particle, it will have lost the rest of the electric potential energy to kinetic. In other words, it will have twice the kinetic that it had at 25 cm. We know from K.E. = 1/2 mv2 that if we multiply the K.E. by 2, we must multiply the velocity by the square root of 2 or approximately 1.4.1.4 times 10 equals 14 m/s.
150.
A is correct. The field lines are directed away from both charges, so by definition they are both positively charged.
151.
A is correct. Doubling both masses will increase the attractive gravitational force. Choice C is wrong because doubling both charges will increase the repulsive electrical force. Choices Band D will not change the forces at all.
152.
B is correct. Electrostatic forces are conservative, so the work done by a force against them is conserved in potential energy. A volt is a joulel coulomb, so you can get voltage by dividing work by charge. (90 JlI (10 C) = 9 J/C=9V.
153.
A is correct. The electric field inside a capacitor is constant. By definition, a dipole has equal but opposite charges on either end. The force on each end of the dipole is Eq and in opposite directions. The net force is zero.
154.
B is correct. The effective resistance is 3 0. The voltage divided by the effective resistance gives 4 amps coming out of the battery. The 4 amps split evenly at the node before A and B; 2 amps through each resistor.
155.
D is correct. Increasing the voltage across the plates would increase the amount of charge on the capacitor but not the capacitance of the capacitor. Capacitance is defined by C = QIV.
156.
A is correct. The energy for the light comes from the battery. The rate at which the energy is released is the power. P = i'R. Since the voltage remains constant, the change in the current will produce the greatest change in the power. Where more light bulbs are attached, the resistance goes up and the current goes down; thus the power goes down and less light is produced.
157.
C is correct. This is Kirchoff's first rule: current flowing into a node must also flow out. Since the resistors have equal resistances, the current is the same in both parallel branches. Thus 4 amps flow into the node from both branches. Therefore 8 amps must flow out of the node.
158.
D is correct. A Watt is a joule/sec. So you can get Joules by multiplying power and time. Don't forget to convert time to seconds. So (60 W)(60 sec) = 3600 J.
159.
D is correct. Adding a resistor in parallel decreased the overall resistance, which will increase the current and the power. The voltage of the battery is not affected by changes in the circuit.
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ANSWERS
&
EXPLANATIONS FOR QUESTIONS IN THE LECTURES
•
245
= IR, with I replaced by C/sec. Choice B comes from P = I'R. Choice 0
160.
C is correct. Choice A comes from V comes from P = V' / R.
161.
C is correct. The maximum voltage is given by Vm ,,,
162.
B is correct. A magnetic field runs from magnetic north to magnetic south. The north pole of a compass needle points to the south pole of the earth's magnetic field, which is near the geographic North Pole.
163.
D is correct. You must remember that the force is perpendicular to both the velocity and the magnetic field. You may recall from Lecture 1 that since velocity and the magnetic field are both vectors, and their product, force, is a vector, the product will always be perpendicular to the other two vectors.
164.
A is correct. We can either memorize the Bio-Savart law, B = floi/2rrr, or visualize how the energy of the field is spread from the wire. As we move away from the wire, the energy spreads out over a cylinder surrounding the wire. The circumference of this cylinder increases directly with the radius. In other words, if we double the radius/ or distance from the wire, there is twice as much room over which to spread the energy.
165.
C is correct. The electric field between the plates will push a positive charge to the right, so the magnetic field must push the drop to the left to counter the force. The right hand rule shows that a magnetic field corning out of the page will push a faIling positively charged drop to the left.
166.
0 is correct. A charged particle moving parallel to magnetic field lines experiences no force. Remember, F = qvBsin8. If 8 is zero, then F is zero.
167.
A is correct. Lenz's law says that current will flow in a loop of wire to oppose the changes in magnetic field inside the loop. If the field strength is decreasing, then current will flow to try to increase field strength. Using
=..fi ~",.,.
the right hand rule, you can see that a clockwise current will create a magnetic field inside the loop. 168.
B is correct. Since the path is circular, the particle must be traveling at 90 degrees to the magnetic field (otherwise the path would be helical). The centripetal force is equal to the magnetic force. So, qvB = mv' Jr. If you solve for q, you get mv / Br.
EXPLANATIONS TO QUESTIONS IN LECTURE 8 169.
A is correct. The ray will turn toward the normal as it enters the glass and away from the normal as it exits the glass. Light source
A B C Window
170. 171.
D is correct. Blue has the highest frequency of the given choices. Remember, ROY G. BIV. A is correct. We are given 'km'; we want 'years'. We work with the units as follows: km x slm x minis x
hrs/min x days/hrs x yrs/ days x m/km. Everything cancels but years. 172.
D is correct. Reflection is indicative of either wave or particle theory.
173.
C is correct. Light passes through the piece of glass as shown below. Only the thicker glass directs the light towardA.
Copyright (c) 2007 Examkrackers, Inc
246
MeAT
PHYSICS
Ii htsource
g
li ht,so_U",rc,e=c:l
Ii.~."~.'~ .
ass. AB
AB
AB
AB
AB
174.
0 is correct. As the light is dimmed, less and less energy is available to the light bulb until, at the last moment, there is only enough energy to produce red light.
175.
B is correct. We can find the speed of light through glass by using the index of refraction. n = clv, or v = cln= (3 x 108 )/1.5 = (2 x 108). Once we know the speed of light in glass, we can use x = vi. Rearrange the equation to solve for I and change cm into meters. t = (1 x 10-')/(2 x 108) = 0.5 X 10-10 = 5 X 10-11
176.
A is correct. Choice A describes refraction, not diffraction. Roughly speaking, diffraction occurs when waves bend around corners.
177.
B is correct. Focal distance is equal to one half the radius of curvature,
178.
0 is correct. m
179.
B is correct. According to the lens maker's equation, as the refractive indices of the lens and the surrounding medium approach one another, the lens will lose its effect. However, you don't need the lens maker's equation to visualize this. If we use our technique of taking examples to their extremes, we can imagine a lens made out of water. When we use the lens in the air, it acts as a magnifying glass; when we use it in water, it doesn't work. If the water were at a slightly different temperature to change the index only slightly, the water lens would bend light only slightly under water.
180.
A is correct. Virtual images are called 'virtual' because they are not really there, so they cannot be projected on to a screen. Real images can be seen if they are focused into the eye or projected onto a screen. Diverging lenses and convex mirrors, by themselves, can only create virtual images.
181.
B is correct. The light ray is bending in the wrong direction. This is a diverging lens, it should diverge parallel light rays.
182.
C is correct. The light rays that bounce off a flat mirror do not intersect in front of the mirror, so to create an image, you follow them back to their implied source behind the mirror. That's where the virtual image appears.
183.
A is correct. An increase in the index of refraction of a lens will increase the bending of the light rays, which will increase the power of the lens. Alternatively, you can look at the lens maker's equation to see that increasing n, will increase 1 I J. lncreasing the radius of curvature of one side will flatten the lens and reduce the amount that it refracts light, thus decreasing its power. Since P = 11f, an increase in fwill decrease the power.
184.
Cis correct.f = r12. So iffis 4, then r is 8.
f = '/, r.
= dJ do.
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\
ANSWERS
&
EXPLANATIONS FOR QUESTIONS IN THE LECTURES •
247
185.
D is correct. The thill lens equation is 1/1 = 1/4 + 1/ di . The object distance is always positive. This results in a value of 4/3 for di . Since this value is positive, the image is behind the lens where the 'eye' is. Remember, I (eye) am positive that real is inverted.
186.
B is correct. This one is tricky. First of all, a convex mITror can't make an inverted image so A and C are out. Secondly, The focal point on a concave mirror is positive. Thus, from the thin lens equation, 1/f = 1/5 + 1/ di • Since the image and the focal distance are positive, the focal distance must be less than 5 (which is the same as saying l/fmustbe greater than 1/5).
187.
A is correct. The object is outside the focal distance of a converging mirror; the image will be positive, real, and inverted. The thill lens equation gives 1/z= 1/4+ 1/ di • The image distance is four, so the magnification is negative 1. The negative means that the image is inverted.
188.
D is correct. The focal distance is negative so the lens is diverging and the power is 1/f.
189.
D is correct. See question 187. If the lens is diverging then the image and the object cannot be at the same distance, 1/f = 1/ d i + 1/ dc ' Since di is negative, the focal distance would have to be infinite; a flat lens.
190.
D is correct. When an object is placed at the focal point of a lens, the rays will emerge parallel on both sides of the lens. If the rays never intersect, then no image is formed. If you use the thin lens equation, you'll get an image distance of infinity.
111
-=-+-
f
191.
f =
B is correct. If an object is placed at a very large distance from a lens, we can think of it as being placed at infinity. For an object placed an infinite distance from a lens, the image will appear at the focal point.
1
1
1
f
=
f
-=-+192.
C is correct. A converging lens is the only one of the choices that produces an inverted image on the side opposite the object.
Copyright © 2007 Examkrackers. Inc
INDEX
.
249
INDEX A absolute pressure 78 adhesive forces 88
alpha decay 66, 70-71 alternating current 132 amplitude 98-100, 102, 104, 107, 113 angle of incidence 141-142, 149 angle of reflection 141-142 angle of refraction 141-142 angular magnification 150 angular wave number 99 annihilation 66 antinodes 105
B battery 124-125, 127, 130-131 beat frequency 103-104, 107 Bernoulli's equation 84-85, 87 beta decay 66, 71 bulk modulus 92, 95, 99 buoyant force 79-80, 88
c capacitor 119, 125-127, 130 capillary action 88-89 center of buoyancy 80 center of charge 116 center of gravity 18 center of mass 18, 22-23, 37-38, 53, 80, 87, 116 centripetal acceleration 26-27 centripetal force 26-28, 133 charge 66, 115-123, 125-127, 132-135, 138-139 chromatic aberration 148 chromatic dispersion 143 circular motion 3, 24, 27 coefficients of friction 31 cohesive forces 88 component vectors 3-4 concave mirror 146, 152, 157 conductors 123 conservative force 46-47, 50 constructive interference 103, 105, 143 contact forces 19-20, 27 continuity equation 84, 87 converging 146-147, 151-152, 155, 157 Copyright Cc) 2007 Exarnkrackers. Inc.
converging lens 146, 151-152, 157 convex mirror 146,152,155,157 Coulomb's law 115-118 critical angle 142-143 cross product 2 current 81, 109, 115, 123-124, 129-135, 137-138 current eddies 135
D decibels 101 density 18, 42, 74-75, 77-84, 99-100, 102, 104 destructive interference 103 dielectric constant 127, 130 direct current 132 displacement 4-12, 29, 32, 44-45, 49, 91, 98-99, 103, 105106,108-109,113,118,132-133 distance 4-5, 8-11, 14, 16, 19, 23, 28-29, 35, 40-41, 44, 52, 60-64,78,84-85,89, 103, 106, 110, 115-118, 120-123, 125-126,130,137,149-152,155,157 Doppler Effect 11 0-111 dot product 2 drift speed 123 dual nature 140 dynamic equilibrium 37, 61-62
E effective resistance 128 elastic collisions 53-54 elastic potential energy 44, 108 electric dipole 119 electric dipole moment 119 electric field 117-121, 123, 125, 127, 132, 134-135, 137, 139-140,142 electromagnetic force 18 electromagnetic wave 139-140 electromotive force 124 electron capture 66, 71 elevation head 87 energy line 87 equilibrium 37-42, 45, 47, 49, 51, t55, 61-62, t75, t79, 108, t117, 122 equipotential surfaces 119
F Faraday's law of induction 134 field 46-47, 53, 102, 116-121, 123, 125, 127, 132-135, 137140, 142
250 . MCAT
P HYSICS
fission 68, 70 fluid 31, 73-80, 83-90, 123-127, 148 fluid pressure 76-78, 83 fluids at Rest 76 fluids in Motion 83 flux 134 focal length 147-148, 150-151, 155, 157 focal point 147-148, 155, 157 fracture point 32, 91 frequency 27, 39, 66, 98-100, 102-104, 106-111, 113, 139140, 142-144,148 friction 14, 19, 22, 25, 31-32, 35, 39, 45-48, 50, 52, 58, 60, 63,83,86 fundamental wavelength 106 fusion 68
G gamma ray 66 gauge pressure 77-78, 87 geometrical optics 141, 143 gravitational force 17-18, 23, 25, 27, 29, 73, 115, 118 gravitational potential energy 44, 47, 52, 84-85, 87, 113, 118 gravity 13, 18-19, 23-25, 27-29, 38, 44, 46-47, 49, 52, 61, 73,75-76,79,81,95, 100,108, 116-119,121 gravity waves 100
H half-life 65, 70 harmonic series 105-106 heat 45-48, 50, 129, 134-135 Hooke's law 32-33, 35-36, 44, 46-47, 53, 57, 91, 108-109 hydraulic gradient line 87 h ydraulic lift 60, 78, 81
ideal fluid 83-87, 89-90 impulse 57, 75-76 inclined plane 25-26, 29, 31, 60 index of refraction 140,142-144,148-149,151 induction 123, 134 inelastic collisions 53-54 inertia 13, 17, 20,53, 99 infrared 139 intensity 88,100-102,104,113,140,142 intenSity level 101-102 internal energy 45, 47, 50, 53-54, 60, 135 irrotational flow 84, 86 isotropic light 140
K kinetic energy 44-47, 49, 58, 73, 76, 84-85, 99, 108, 113, 127, 135, 141 kinetic friction 31, 35, 39, 47 Kirchoff's first rule 124 Kirchoff's second rule 124
L laminar 84 lateral magnification 149-150, 155 Law of Conservation of Energy 45, 48, 53, 115 lens maker's equation 148, 152 lenses 145-148, 151-152, 155 Lenz's law 134 lever 40-41, 43, 60-62,.64 light 67, 104, 110-111,130, 139-153, 155, 157 lines of force 117, 132 longitudinal wave 97-98
M Machines 53, 55, 57, 60-61, 63, 65, 67, 69 magnetic field 47, 53,132-135, 137-139 magnetic flux l34 m ass 14-20, 22-23, 26, 29, 32-33, 35, 37-39, 41-45, 49, 5253,56-58,60-64,66-68, 70-71, 74-75, 78, 80-81, 84, 87, 95,99-100, 108-110,113,116-119,121-122,125,129, 138 mechanical energy 44-47, 53-54, 119, 129, 134-135 mirrors 145-148, 150-152, 155 modulus of elasticity 91 momentum 53-55, 57-59, 61, 63, 65, 67, 69, 75-76, 97, 104
N near point 150 Newton's First Law 20 Newton's Law of Universal Gravitation 23 Newton's Second Law 20, 23, 25, 27 Newton's Third Law 20, 23, 33 node 105-106, 124, 128 nonconservative forces 46-47 nondispersive 97, 99, 103 nondispers ive medium 99, 103 normal force 19, 25-26, 31, 49, 73
o Ohm's law 86, 124-125, 128-129
Copyright © 20 07 Examkrackers, Inc.
INDEX '
p parallax 145 Pascal 76, 78 Pascal's principle 78 pendulum 11, 39,109-110,113, 135 period 65, 98-99,102,109-110, 113 phase 98, 103-104, 107-108 phase constant 103, 108 piezometer tube 87 piezometric head 87 positron emission 66 potential energy 44-47, 49, 52, 58, 84-85, 87, 99, 108, 113, 118, 120, 127 power 50, 52, 63, 100, 129, 131-132, 148, 151, 155 poynting vector 139 pressure head 87 projectile 3, 7,10, 13-16,100 projectile Motion 3, 7, 10, 13-14 pulley 39, 60-64 Pythagorean Theorem 3-4
R radioactive decay 53, 55, 57, 61, 63, 65-67, 69-71 radius of curvature 133, 147, 151 ramp 6, 29-30, 60-62, 64 ray-diagrams 147-148 real image 145, 151 resistance 13-16, 22, 31, 33, 38, 73-74, 83, 86, 99, 123-125, 128, 130
resistivity 123, 135 resistors 123, 128, 130-131 resonant frequency 106 rest mass energy 67 right Hand Rule 3, 133-134
s scalar 2-4,7,44-45,50,54,76,119,123 shear modulus 92, 95 simple harmonic m otion 108-109, 132 sinusoidal function 108 sinusoidal wave 99 sound 97, 99-102, 106-107, 110, 113 specific gravity 75, 79, 81, 95 speed 4-6, 12, 14, 16,22,27, 30, 36,39, 52,58, 67, 76,86, 90, 99-100, 102,110-111,113,123, 138-140,142 spherical aberration 147-148 spherical mirrors 147 standing wave 105-107 static equilibrium 37,41-42 Copyrigh t © 2007 Examkrackers, Inc .
static friction 31 static pressure tap 87 strain 83, 91, 95 streamlines 85 stress 83, 91-92, 95 strong nuclear force 18 surface tension 88-89 surface waves 100
T tension 19, 28, 32-33, 35-36, 39, 42, 61-63, 88-89, 95, 99 the point of rotation 40-41 thin lens equation 150 torque 37, 40-43, 45, 47, 49, 51, t55, 61, t 75, t79-80, t117 total internal reflection 142 transverse wave 97-98, 139 tube of flow 85-86
u ultraviolet 139-140 universal Law of Conservation of Charge 115
v vector 2-7, 25, 31-32, 40, 53-55, 110, 117-118, 120, 139 velocity 2, 4-14, 16,22, 26-29,35,37,39,49,52-53,56-58, 61-63,70, 73, 76,84-87, 89,98-100, 102, 106, 108-lll, 113, 121, 127, 132-135, 137, 141
velocity head 87 virtual image 145, 151 viscosity 31, 83, 86, 89 voltage 86, 119, 122-127, 129-132, 137 volume flow rate 84, 86, 89-90, 124
W water 16, 74-75, 78-79, 81-83, 88-90, 99-100,102,140,144145, 151
wave 97-100, 102-111, 113, 132, 139-144 wave pulse 104 wavelength 98-100, 102-107, 110-111, 113,139-140, 142143
weak nuclear force 18 weight 17, 20, 22-23, 25, 29, 36, 39-40, 42, 74, 76-81, 8788, 95
work 26, 44-50, 52, 54, 58, 60-62, 77-78, 92, 118-119, 121122, 125, 127, 133-134, 151
y yield point 32, 91 Young's modulus 92, 95
A N UNEDITED STUDENT REVIEW .
An Unedited Student Review of This Book The following review of this book was written by Teri R- . from New York. Teri scored a 43 out of 45 possible points on the MCAT. She is currently attending UCSF medical school, one of the most selective medical schools in the country.
"The Examkrackers MCAT books are the best MCAT prep materials I've seen-and 1 looked at many before deciding. The worst part about studying for the MCAT is figuring ou t what you need to cover and getting the material organized. These books do all that for you so that you can spend your time learning. The books are well and carefully written, with great diagrams and really useful mnemonic tricks, so you don't waste time trying to figure out what the book is saying. They are concise enough that you can get through all of the subjects without cramming unnecessary details, and they really give you a strategy for the exam. The study questions in each section cover all the important concepts, and let you check your learning after each section. Alternating between reading and answering questions in MCAT format really helps make the material stick, and means there are no surprises on the day of the exam-the exam format seems really fa miliar and this helps enormously with the anxiety. Basically, these books make it clear what you need to do to be completely prepared for the MCAT and deliver it to you in a straightforward and easy-to-follow form. The mass of material you could study is overwhelming, so I decided to trust these books - I used nothing but the Examkrackers books in all subjects and got a 13-15 on Verbal, a 14 on Physical Sciences, and a 14 on Biological Sciences. Thanks to Jonathan Orsay and Examkrackers, I was admitted to all of my top-choice schools (Col umbia, Cornell, Stanford, and UCSF). I will always be grateful. I could not recommend the Examkrackers books more strongly. Please contact me if you have any questions." Sincerely, Teri R-
Copyright © 2007 Examkrackers. Inc.
MeAT PHYSICS .
About the Author Jonathan Orsay is uniquely qualified to write an MCAT preparation book. He graduated on the Dean's list with a B.A. in History from Columbia University. While considering m edical school, he sat for the real MCAT three times from 1989 to 1996. He scored in the 90 percentiles on all sections before becoming an MCAT instructor. He has lectured in MCAT test preparation for thousands of hours and across the country for every MCAT administration since August 1994. He has taught premeds from such prestigious Universities as Harvard and Columbia. He was the editor of one of the best selling MCAT prep books in 1996 and again in 1997. Orsay is currently the Director of MCAT for Examkrackers. He has written and published the following books and audio products in MCAT preparation: "Examkrackers MCAT Physics"; "Examkrackers MCAT Chemistry"; "Examkrackers MCAT Organic Chemistry"; "Examkrackers MCAT Biology"; "Examkrackers MCAT Verbal Reasoning & Math"; "Examkrackers 1001 questions in MCAT Physics", "Examkrackers MCAT Audio Osmosis with Jordan and Jon".
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@
2007 Exarnkracko(s, Inc.
