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Basic Pharmacokinetics Michael C. Makoid, Ph.D. Professor of Pharmaceutical Sciences Creighton University School of Pharmacy and Allied Health Sciences, Omaha, Nebraska
Phillip J. Vuchetich Pharm.D. Candidate Creighton University School of Pharmacy and Allied Health Sciences, Omaha, Nebraska
Umesh V. Banakar, Ph.D. Professor of Pharmaceutical Sciences St. Louis College of Pharmacy St. Louis, Missouri
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Copyright 1996-1999 The Virtual University Press All rights reserved.
ISBN 0-000-000000-0
ABCDEFGHIJ-DO-89
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Acknowledgement When I first started teaching, I had the good fortune to work with another new Ph.D., John Cobby. We struggled through our first five years on the otherside of the podium together and learned many of the tenents upon which this book is based, not content but process. First and formost, it was his belief that students are bright, enthusiastic and hardworking. We should tell them what to do and get out of their way. We both prepared extensive handouts complete with even more extensive practice problems so that the student could experience the scientific method as a detective might solve a murder mystery. The idea was to make learning pharmaceutical science interesting and fun. Through the years, as the methods became more refined, student perceptions and performance improved dramatically. John ultimately abandoned academe to go to work in the “real world” of industry, clearly their gain and our loss. I approached him some years ago to co-author this text. He declined believing himself to be too far removed from the cutting edge of this discipline. That may be true (I doubt it!), but what can not be argued is that he was a major contributor to this book in his philosophy and class notes. Over the years, the explainations were rewritten and revised. Many new problems were added and some were suplanted. These teaching aides have evolved but their origins are clear. Using the industry standard regarding authorship, which defines an author as one whose contributions significantly alters the content of the paper, Dr. Cobby is an author of this book.
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CHAPTER 1
Introduction Basic Pharmacokinetics 1-2 Course Objectives: 1-3 Course Arrangement: 1-3 Learn the tools; get the pharmacokinetic parameters from patient information. 1-3 Learn the modifications of the pharmacokinetic parameters which result from illness. 1-5 Apply the tools; use the pharmacokinetic parameters to develop dosage regimens. 1-5 Apply the tools in specialized drug classes. 1-5
Exams
1-7
Library Assignment in Pharmacokinetics 1-7
Blooms taxonomy for the Hierarchy of Educational Objectives Course Contract 1-10 Computers in the course 1-14 Survival Kit 1-15
1-9
Things for your Survival Kit! 1-15 What you will gain: (your goals) 1-16
Tentative Schedule
1-17
Study Group 1: Learn the tools - obtain pharmacokinetic parameters from data. 1-17 Study group 2: Learn how the parameters are modified. 1-19 Study Group 3: Apply the tools in compromised patients. 1-20 Study Group 4: Apply the tools in special cases. 1-20
Competency Statements Related To Pharmacokinetics
1-22
Specific Competency Statements addressed in this course
1-22
Pharmacokinetic Symbolism
1-25
Amount Terms (unit: mass) 1-25 Concentration terms (units mass/volume) 1-26 Volume Terms (unit: volume) 1-26 Time Terms (unit: time) 1-27 Rate Constant Terms (unit: reciprocal time (*), mass/time (**) 1-27 Clearance Terms (units: volume/time) 1-28 Rate Terms (units: mass/time (*), mass/time, volume (**), volume/time (***) 1-28 Other Terms 1-29 Subscripts 1-30 Superscripts 1-30
First Lesson in Pharmacokinetics
CHAPTER 2
1-32
Mathematics Review Concepts of Mathematics 2-2 Mathematical Preparation 2-3 Zero and Infinity 2-3 Expressing Large and Small Numbers 2-3 Significant Figures 2-4 Rules of Indices 2-4 Logarithms 2-6 Natural Logarithms 2-6 Negative Logarithms 2-9
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Using Logarithmic and Anti-logarithmic Tables 2-10 Dimensions 2-11 Dimensional Analysis 2-12
Calculus
2-14
Differential Calculus 2-14 Non-linear Graphs 2-14 Slope of Non-linear Graph 2-15 Value of the Slope 2-15 Differentiation from First Principles Rule of Differentiation 2-17 Three Other Derivatives 2-17 A Seeming Anomaly 2-18 Integral Calculus 2-19 Rule of Integration 2-19 The Constant of Integration 2-20 The Exception to the Rule 2-20 A Useful Integral 2-20 Example Calculations 2-21
Graphs
2-16
2-24
Graphical Conventions 2-25 Straight Line Graphs 2-26 The Slope of a Linear Graph (m) 2-28 Linear Regression: Obtaining the slope of the line Parallel lines 2-31 Graphical Extrapolations 2-32 Significance of the Straight Line 2-32 Graphical Honesty 2-32 Axes with Unequal Scales 2-33 Graphs of Logarithmic Functions 2-34 Semilogarithmic Coordinates 2-34 Log - Log Coordinates 2-38 Pitfalls of Graphing: Poor Technique 2-38 Graphical analysis 2-40
Pharmacokinetic Modeling
2-29
2-44
Making a Model 2-45 One Compartment Open Model 2-47
The LaPlace Transform
2-48
Table of LaPlace Transforms 2-49 Symbolism 2-49 Conventions used in drawing pharmacokinetic schema. 2-50 Steps for Integration Using the LaPlace Transform 2-53 Example Integration Using the LaPlace Transform 2-54 Second Example Integration Using the LaPlace Transform 2-56 Third Example Integration Using the LaPlace Transform 2-57 Conclusions 2-58 Table of LaPlace Transforms 2-60 LaPlace Transform Problems 2-62 LaPlace Transform Solutions 2-63
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CHAPTER 3
Pharmacological Response Pharmacological Response
3-2
The Hyperbolic Response Equation 3-2 Interrelationships between concentration, time and response 3-3
Change in Response with Time
3-4
One-Compartment Open Model: Intravenous Bolus Injection 3-4 One-Compartment Open Model: Oral Administration 3-4 Duration of Effective Pharmacological Response 3-4 Pharmacokinetic Parameters from Response Data 3-5 “Delayed” Response 3-5 Response of active metabolite: 3-6
Therapeutic Drug Monitoring
3-7
Therapeutic monitoring: Why do we Care?
Problems Answers: Answers: Answers:
CHAPTER 4
3-9
3-11 Oxpranolol 3-18 Minoxidil 3-21 Propranolol 3-23
I.V. Bolus Dosing I.V. Bolus dosing of Parent compound
4-2
Plasma 4-2 Iv bolus, parent compound, plasma Problems Urine 4-47
Metabolite
4-7
4-51
Plasma 4-51 Urine 4-56
CHAPTER 5
I.V. Infusion Parent compound
5-2
Plasma 5-2
Problems
5-10
CHAPTER 6
Biopharmaceutical Factors
CHAPTER 7
Oral Dosing Oral dosing
7-3
Valid equations: ( oral dosing, plasma) 7-3 Utilization 7-3
CHAPTER 8
Bioavailability, Bioequivalence, and Drug Selection Bioavailability, Bioequivalence and Drug Product Selection
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-6
Relative and Absolute Bioavailability 8-2 Factors Influencing Bioavailability 8-4 Methods of Assessing Bioavailability 8-11 Study Design 8-15 In-vitro Dissolution and Bioavailability 8-15 In-vitro / in-vivo correlation studies- 8-18
Bioequivalence
8-20
Bioequivalence Regulations 8-23 Study Design 8-26 Assessment of bioequivalence 8-29 Controversies and Concerns in Bioequivalence 8-31 Generic Drugs and Product Selection 8-35 The Orange Book 8-38 Therapeutic equivalence 8-38 Therapeutic equivalence evaluation codes- 8-39
Drug Product Selection
8-44
Considerations in selecting a manufacturer 8-45 Special Cases 8-50
Summary
8-54
Questions 8-55 Answers to Questions
8-57
Bioavailibility Equations Problems 8-60 Solutions 8-80
8-58
“Caffeine” on page 61 8-80 “Cefetamet Pivoxil” on page 62 8-83 “Cefixime” on page 63 8-86 “Ceftibuten” on page 64 8-87 “Cimetidine” on page 65 8-88 “Diurnal Variability in Theophylline Bioavailability” on page 66 8-89 “cis-5-Fluoro-1-[2-Hydroxymethyl-1,3-Oxathiolan-5-yl] Cytosine (FTC)” on page 67 8-90 “Hydromorphone” on page 68 8-91 “Isosorbide Dinitrate” on page 69 8-92 “Ketanserin” on page 70 8-93 “Methotrexate” on page 71 8-94 “Moclobemide” on page 72 8-95 “Nalbuphine” on page 73 8-96 “Nefazodone” on page 74 8-97 “Ondansetron” on page 75 8-98 “Omeprazole” on page 76 8-99 “Paroxetine” on page 77 8-100 “Ranitidine” on page 78 8-101 “Sulpiride” on page 79 8-102
References
CHAPTER 9
8-103
Clearance Equations 9-2 Definitions and Terms
9-3
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Measurement of Creatinine Clearance Model Correlations 9-5
9-4
Renal Clearance 9-5 Systemic Clearance and Metabolic Clearance Use in Pharmacokinetic Equations 9-6
Physiological Factors Affecting Clearance
9-5
9-7
Intrinsic Clearance 9-7 Extraction Ratio (E) 9-8
Hepatic Function and Clearance
9-10
Alterations in Hepatic blood Flow 9-10 Alterations in Hepatic Intrinsic Clearance Tabulated or Graphical Alterations 9-11
9-10
Renal Function and Clearance 9-12 General Equations for Changes in Clearance
9-14
Plasma/Blood ratio 9-14 Half life and elimination rate constant in relationship to clearance 9-16 Effects of alterations in protein binding on clearance 9-16
Problems
CHAPTER 10
9-17
Dosage Regimen (Healthy, Aged, and Diseased Patients) Therapeutic Drug Monitoring
10-2
Therapeutic Range 10-2 Therapeutic monitoring: Why do we Care? Steady State 10-5
10-4
Diseases - Dosing the Compromised Patient Problems 10-12 Answers 10-36
CHAPTER 11
10-10
Multicompartment Modeling Executive Summary 11-2 Equations 11-3 PHARMCOKINETICS: MAMMILLARY MODELS Begin 11-90 Problems 11-91
11-4
Two-compartment Model Equations 11-119 Answers 11-120
CHAPTER 12
Protein Binding
CHAPTER 13
Nonlinear (Michaelis-Menton) Kinetics Problems 13-2 Nonlinear Equations
13-14
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Answers 13-14
CHAPTER 14
Practice Exams: Section 1 Nifedipine - Section 1
3
Nifedipine Data 3 Nifedipine Questions 4 Nifedipine Solutions 6
Enalapril - Section 1 11 enalapril data 11 Enalapril Questions 13 enalapril Solutions 14
Ciprofloxacin Section 1 19 Ciprofloxacin data 19 Ciprofloxacin Questions 20 Ciprofloxacin Solutions 21
Methylphenidate Section 1
23
methylphenidate data 23 MethylPhenidate Questions: 24 methylPhenidate Solutions 25
Adinazolam - Section 1
27
ADINAZOLAM DATA 27 Adinazolam Questions 28 Adinazolam Solutions 30
Labetalol - Section 1
32
labetalol data 32 labetalol Questions 33 labetalol Solutions 34
Zidovudine Section 1
36
zidovudine data 36 Zidovudine Questions 38 Zidovuldine Solutions 40
Fosinopril Section 1
42
fosinopril data 42 Fosinopril Questions 44 Fosinopril Solutions 46
Omeprazole
49
Omeprazole Data 49 Omeprazole Questions 50 Omeprazole solutions 53
EXP3312, an Experimental Drug
57
EXP3312 DATA 57 EXP3312 & M1 Questions: 59
Graph Paper
CHAPTER 15
62
Practice Exams: Exam 2 Nifedipine: Exam 2
2
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Nifedipine Questions: 3
Valproate: Exam 2
6
Valproate Questions 7 Valproate Solutions 10
Methyl phenidate
11
Methyl phenidate Questions: 12 Methyl phenidate Solutions: 13
Verapamil
15
Verapamil Questions 16 Verapamil Solutions 17
Hydromorphone hydrochloride
19
Hydromorphone hydrochloride Questions 20
Fosinopril Sodium
23
Fosinopril Questions 24 Fosinopril Sodium Solutions
Remoxipride
26
28
Remoxipride Questions 29 Remoxipride Solutions 31
Naproxen
33
Naproxen Questions 34 Naproxen Solutions 36 38
CHAPTER 16
Exam 3 Pharmacokinetics Final Exam
16-2
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CHAPTER 1
Introduction
Author: Michael Makoid and John Cobby Reviewer: Phillip Vuchetich
OBJECTIVES At the completion of this chapter, the successful student shall be able to: 1.
define pharmacokinetics
2.
state the overall objectives of the course
3.
state the major themes of the course
4.
state the course organizational structure with respect to study sections
5.
state the objectives of each study section
6.
state the examination structure and objectives
7.
state student performance expectations
8.
state the schedule and timeline
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Introduction
1.1 Basic Pharmacokinetics What is pharmacokinetics?
Pharmacokinetics is the mathematics of the time course of Absorption, Distribution, Metabolism, and Excretion (ADME) of drugs in the body. The biological, physiological, and physicochemical factors which influence the transfer processes of drugs in the body also influence the rate and extent of ADME of those drugs in the body. In many cases, pharmacological action, as well as toxicological action, is related to plasma concentration of drugs. Consequently, through the study of pharmacokinetics, the pharmacist will be able to individualize therapy for the patient.
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Introduction
1.2 Course Objectives: The Roman numerals refer to the cognitive complexity as described in Bloom's Taxonomy of Educational Objectives described elsewhere in this document. At the completion of this course, the successful student will be able to: What will you be able to do?
• Calculate (III) patient and drug specific pharmacokinetic parameters from patient data, • Predict (calculate - III) the changes in relevant pharmacokinetic parameters in the patient with selected diseases,
• Utilize the above parameters to individualize patient therapy (devise a dosage regimen - V), • Communicate his/her therapy recommendations to another competent health professional (write a consult - V).
1.2.1
COURSE ARRANGEMENT: Two courses are described below. The first, a two credit (Creighton University required) and the second, a three credit (CU optional) version. The two credit course will consist of major themes one through three and exams one and two, while the three credit course will add theme four and exam three. The four major themes are entitled:
How is the course arranged?
• Learn the tools; get the pharmacokinetic parameters from patient information. • Learn the modifications of the pharmacokinetic parameters which result from illness. • Apply the tools; use the pharmacokinetic parameters to predict patient response and develop dosage regimens for the normal as well as for the compromised patient.
• Apply the tools in specialized drug classes.
Each major theme of the course is further broken down into study sections, each with their own set of general objectives as shown below:
1.2.2
LEARN THE TOOLS; GET THE PHARMACOKINETIC PARAMETERS FROM PATIENT INFORMATION. A. Basic Mathematical skills objectives:
What will I be required to be able to do? How will examination questions be written for this material?
1. 2. 3. 4.
Given a data set containing a pair of variables, the student will properly construct (III) various graphs of the data. Given various graphical representations of data, the student will calculate (III) the slope and intercept by hand as well as using linear regression. The student shall be able to interpret (V) the meaning of the slope and intercept for the various types of data sets. The student shall demonstrate (III) the proper procedures of mathematical and algebraic manipulations.
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5.
The student shall demonstrate (III) the proper calculus procedures of integration and differentiation. 6. The student shall demonstrate (III) the proper use of computers in graphical simulations and problem solving. 7. Given the assumptions for the model, the student will construct (III) models of the ADME processes using Laplace Transforms. 8. The student shall develop (V) integrated equations associated with the above models. 9. The student shall generate a pharmacokinetic model based on given information. 10. The student shall interpret a given model mathematically. 11. The student shall predict changes in the final result based on changes in variables throughout the model.
B. Pharmacological Response objectives: 1.
2.
Given patient data of the following types, the student will be able to properly construct (III) a graph and compute (III) the slope: response (R) vs. concentration (C), response (R) vs. time (T), concentration (C) vs. time (T) Given any two of the above data sets, the student will be able to compute (III) the slope of the third.
C. IV one compartment model, plasma and urine objectives: 1.
2.
Given patient drug concentration and/or amount vs. time profiles, the student will calculate (III) the relevant pharmacokinetic parameters available ( V d , K, k m , k r , AUC , Clearance, MRT) from IV data. Given the appropriate pharmacokinetic parameters, the student shall simulate (III) I.V. bolus/infusion dosing for parent compounds Plasma concentration vs. time profile analysis
3. 4. 5.
Rate vs. time profile analysis Given patient specific pharmacokinetic parameters, the student shall provide professional communication regarding IV bolus/infusion information The student shall utilize computer aided instruction and simulation Given patient drug concentration and/or amount vs. time profiles, the student will calculate (III) the relevant metabolite (active vs. inactive) pharmacokinetic parameters available ( V d , K, k m , k r , AUC , Clearance, MRT) from IV data.
D. Biopharmaceutical factor objectives: the student shall be able to discuss: 1. 2. 3. 4. 5.
physiology and mechanisms of absorption effects of diffusion, cardiac output / blood perfusion, physical properties of the drug and body on distribution biotransformation, first pass effect, and clearance renal, biliary, mammary, salivary, other forms of excretion. the effects of physiological changes with age, sex, and disease on the absorption, distribution, metabolism, and excretion of a drug.
E. Oral one compartment model objectives: 1.
Given patient drug concentration and/or amount vs. Time profiles, the student will calculate (III) the relevant pharmacokinetic parameters ( V d , K, k m , k r , k a , AUC , Clearance, MRT, MAT) available from oral data.
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Introduction
F. Bioavailability objectives: 1.
Given sufficient data to compare an oral product with another oral product or an IV product, the student will estimate (III) the bioavailability (compare AUCs) and judge (VI) professional acceptance of the product with regard to bioequivalence (evaluate (VI) AUC, T p and ( C p )max ).
2.
1.2.3
The student will write (V) a professional consult using the above calculations.
LEARN THE MODIFICATIONS OF THE PHARMACOKINETIC PARAMETERS WHICH RESULT FROM ILLNESS. G. Clearance objectives: 1.
1.2.4
Given patient information regarding organ function, the student will calculate (III) changes in clearance and other pharmacokinetic parameters inherent in compromised patients.
APPLY THE TOOLS; USE THE PHARMACOKINETIC PARAMETERS TO DEVELOP DOSAGE REGIMENS. H. Dosage regimens objectives: 1. 2. 3. 4.
1.2.5
Given population average patient data, the student will devise (V) dosage regimens which will maintain plasma concentrations of drug within the therapeutic range. Given specific patient information, the patient will justify (VI) dosage regimen recommendations. Given patient information regarding organ function, the student will devise (V) and justify (VI) dosage regimen recommendations for the compromised patient. The student will write (V) a professional consult using the above calculations
APPLY THE TOOLS IN SPECIALIZED DRUG CLASSES. I. Two Compartment Model objectives: 1.
Given patient Concentration and/or Amount of Drug vs. Time, profiles the student will calculate (III) the relevant pharmacokinetic parameters( V d1 , Alpha, A 1 , Beta, B 1 , k 10 , k 12 , k 21 , AUC , Clearance, compartmental amount ratios) available from two com-
2. 3. 4. 5.
partment data. Given population average patient data, the student will devise (V) a dosage regimen which will maintain plasma concentrations of drug within the therapeutic range. Given specific patient information, the patient will justify (VI) the optimal dosage regimen. Given patient information regarding organ function, the student will devise (V) and justify (VI) the optimal dosage regimen for the compromised patient. The student will write (V) a professional consult using the above calculations.
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Introduction
J. Protein Binding objective: 1. 2. 3. 4.
Given population average patient data, the student will devise (V) dosage regimens which will maintain plasma concentrations of unbound drug within the therapeutic range. Given specific patient information, the patient will justify (VI) the optimal dosage regimen. Given patient information regarding organ function, the student will devise (V) and justify (VI) dosage regimens for the compromised patient. The student will write (V) a professional consult using the above calculations.
K. Non-linear kinetics objective: 1. 2. 3. 4.
Given population average patient data, the student will devise (V) a dosage regimen which will maintain plasma concentrations of drug within the therapeutic range. Given specific patient information, the patient will justify (VI) the optimal dosage regimen. Given patient information regarding organ function, the student will devise (V) and justify (VI) dosage regimens for the compromised patient. The student will write (V) a professional consult using the above calculations.
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Introduction
1.3 Exams How are the exams made?
1.3.1
Exams will consist of problems which will be linked directly back to an objective (above) and a library assignment in which you will be asked to evaluate a research article with the tools available to you by the time of the exam as discussed below.
LIBRARY ASSIGNMENT IN PHARMACOKINETICS L. Library Assignment Objectives
What do I have to do in the library?
1. 2.
Given a suitable primary research article in the area of pharmacokinetics, the student shall calculate the pharmacokinetic parameters from the data using the tools learned in class. The student shall communicate in writing the results of such calculations with suitable commentary regarding differences and interpretations.
Format of the “paper”: How should the paper look?
1.
Tell me what type of paper you have chosen to evaluate:
The problem sets show what data you need for each of these. First Exam What content should I look for in the paper and what is its relative worth?
IV Bolus Parent compound
10 pts
IV Bolus Parent metabolite
12.5 pts
IV Infusion
12.5 pts
Pharmacological Response
15 pts
Second Exam Oral Dosing / Bioavailability
10 pts
Third Exam (2 credit course) Multiple dosing
10 pts
Clearance and disease
12.5 pts
Dosage Regimen
15 pts
Third Exam (3 credit course)
2.
Two compartment model
10 pts
Protein Binding and Disease
12.5 pts
Non-linear kinetics and disease
15 pts
Include a Xerox copy of the entire paper. I need to evaluate it, too.
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Introduction
3. Enlarge the graph by successive Xeroxes so that you can accurately evaluate the data. 4.
Do analysis of data by hand and by PKAnalyst, or Scientist.
5.
Compare your work with the author’s (short paragraph).
6. Comment on any differences of parameter calculation or interpretation. See objectives above (Paragraph). 7. Write an exam question to obtain pharmacokinetic parameters. You know from the first exam what they should look like. Why do I need to do this library assignment?
Each of the above sections is designed to bring the student an understanding of the information and the processes necessary to operate as a competent professional in the area of pharmacokinetic evaluation and consulting. Consequently, the course will evolve from a quantitative, manipulative mathematics course to a course which stresses communication skills. Consults will be graded not only on content (the proper dosage regimen for the patient) but also grammar, punctuation, spelling, organization and neatness. You may have the best medical information in the world, but if it is poorly executed, it will be ignored.
Can I cram the night before?
This course will probably be one of the more rigorous ones that you will have experienced in your college career to date. It will be one of the first ones which attempt to show some clinical relevance. The course can be successfully completed with your current skills and background. It is not difficult IF (and that is a big IF) taken slowly, in small bites. Its just like eating an elephant - you can't do it all in one sitting. Some of you may try to get it all the night before the exam, regardless of my admonitions and those of your upper-class friends (ask them!). In many cases, that has been more than sufficient to get A's and B's on exams in previous courses. Past experience tells many of you that you can do it. I suggest that the requirements and expectations of a professional school are considerably more than your undergraduate experience and it most likely will not work in many courses which require assimilation of the information presented, as is expected in a professional program.
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Introduction
1.4 Blooms taxonomy for the Hierarchy of Educational Objectives Blooms taxonomy for the Hierarchy of Educational Objectives describes the expectations of a course in increasing order of complexity as: What is cramming good for? Lowest level of cognitive skills.
I. To Know: means to memorize (recognize, recall) (Many college courses require only this level of cognitive effort, hence the extensive experience with “multiple guess” exams). II. To Comprehend: means to translate; to be able to put information into your own words. (Essay exams routinely call for this level of effort on the part of the student).
This is where we begin.
III. To Apply: means to be able to use knowledge, rules and principles in an unfamiliar situation. (This is the lowest level of skill necessary to function at a technician level).
This is where we need to be in school.
IV. To Analyze: means to be able to critically examine a body of knowledge and to be able to identify the relationships. (This is where a B.S. graduate should operate. Education obviates the need for teachers.)
This is where we need to be as graduates.
V. To Synthesize: means to put together information, not necessarily previously so organized, in order to get a new piece of information. (This is the beginning level of professional judgment). VI. To Evaluate: means to be able to judge the worth of an idea, form hypotheses and do problem solving, research, invent new knowledge. (This is the doctoral level of participation in the area).
Can’t I just do it the same way that I have always studied?
A professional routinely operates at level IV and V with occasional forays into level VI. This is where you will operate in this course and in most subsequent courses in the professional curriculum. You will note that each of the objectives for the course contains specific action words followed by the level in the taxonomy at which you will operate. These are the standard descriptive terms for use in instructional objectives. You will be asked to do critical thinking, not simply recite or recognize the right answer. Problems challenge thinking skills and demand the synthesis of material into concepts. To facilitate this transition we both must work very hard.
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Introduction
1.5 Course Contract I Will: What will the teacher do in this class? Act as a facilitator.
1. Provide individualized learning methods: Some students learn by hearing and others by seeing (auditory or visual learners). I have designed the course to accommodate both types of learners. In class, I will provide you with executive summaries of what you read. I will provide group leaders with detailed reviews of materials for which they are responsible. I will tell you what I'm going to say, say it, and then tell you what I said. I will also attempt to write it out and draw appropriate graphs, charts and pictures as well as appropriate visual aids in class and with the homework problems. I will provide you with ample examples of the types of manipulations that you will be expected to do. I will provide you with ample problem sets so that you may practice those manipulations. I will provide you with computer simulations so that you may see these manipulations in action and begin to get a feel for the numbers and their magnitude. Feedback and interaction is encouraged. If I am not meeting your perceived needs, you must tell me. Some students might feel too intimidated to ask questions. To obviate this problem, you will elect a group team leader, an ombudsman, whose job it will be to carry your questions, concerns, and comments to me. It is your job and his responsibility to see that the group interaction facilitates the learning process. This is not to prevent you coming to see me but offered as another avenue of communication.
Will I learn anything relevant in this course?
2. Provide clinical relevance to the practice of pharmacy. This will be stressed at all times. I will also relate real clinical experiences; virtually all of the problems come from real patients. Some educators believe examples must fit the theory exactly. This gives the student a false set of reality parameters. Consequently, when “the data does not fall on the line” the student rejects relevant information. You will become familiar with real data, and the problems associated with real data.
How will I know how I’m doing?
3. Give adequate feedback: Evaluation of your performance will be available to you at all times. A running evaluation, updated weekly will be on my door for your review. You may check any thing with me at any time. I expect that you will see me outside of class time either individually if you need help or in supervised review sessions. You must see me for assistance if your performance is unsatisfactory.
What will the teacher be doing? Engaging you in an active learning process.
4. Teach: As an operational definition this means: clarifying what you read, demonstrating how and why things work as they do, and unifying the material attempting to generate the A - HA! syndrome. The correlate of teach from the student view is learn. Neither is a passive process. I can not open your head and pour the knowledge in. A saying in education is: “Knowledge maketh a bloody entrance”. You must expend the effort necessary for you to learn.
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What must I do in this active learning process? You MUST participate in class and in your assigned groups!!!
5. Facilitate Learning. You received objectives (above) and a summary for each study section (chapters in this text), of exactly what is expected of you with examples in the problem sets at the end of each chapter. We will have ample time during class to field questions generated by the correlated reading and problem sets, as well as homework assignments. I will not be duplicating any book's efforts. Student participation in class is required. You will answer (as well as ask) questions, do problems in class. You will sound things out and get feedback from me and your fellow colleagues. Remember - the class is to help you learn. It is not the sole means of learning, nor am I the source of all knowledge. Its’ only reason for being is to help you organize and summarize what you learn. It has a relatively simple plan with multiple examples. From these examples you will develop concepts which will obviate the need for memorizing individual facts (or actually me entirely). I will assist you in the formation of these concepts. It is patently obvious that I can not give you every possible example of every type of question that you will be asked during your professional career. For one thing I don't know what questions you will be asked nor problems you will encounter. Going from the specific to the general forms concepts which will allow you to go from the general to the specific, even if you have never been there before. The total medical knowledge is now doubling at a rate of every 4 years. I can not teach you the content necessary to operate 5 years in the future, let alone 40. You must learn to learn. Hence, if you plan to become a competent professional, you must operate at least in Bloom's level V.
How do I get in touch with the teacher?
6. Be available: I do not have office hours. I believe them to be restrictive from your view point. What I do have is a schedule prepared two weeks in advance of when I am NOT available. You may set an appointment, at least a 1/2 day in advance to guarantee that I see it, any other time. Of course, appointments are not necessary if I'm in my office, but you take the chance of my not being there or someone else being there ahead of you if you do not sign up. You may contact me by e-mail:
[email protected], or by phone: 402-280-2952.
How can I tell the teacher how things are going?
7. Be responsive: After each study section, you will be asked to provide me with a one minute summary of the topic consisting answering the following questions: a.
What was the main thrust of the study section?
b.
What was clear about the study section? What was done well?
c.
What was unclear? How could it be done better?
This will provide me with a running monitor of my effectiveness as well as a framework of what to stress and what to change.
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Do I have any say in the examination questions?
After each exam, in addition to working out the problems, we will decide whether any individual question was not covered by the objectives. Note: Not that it was tough, not that you got it wrong, not that it didn't allow you to tell me what you knew, but did I tell you that I was going to ask you to do it? (Was it covered by the objectives?)
How are the exams graded?
8) Evaluate your performance fairly and honestly: Quite simply, I'm going to tell you what I expect that you will do. I will show you how to do it. I will provide you with practice in doing it. I will provide you with an exam which tests your ability to do it. The exams, as well as the whole course, will use real data and/or pharmacokinetic parameters for real drugs in real patient settings, much like the state board exams (and hopefully real life). Like both of these situations, all answers are interconnected. What that means is, if you improperly calculate a parameter which is needed to make another calculation which is used to make a third, etc. ALL are wrong. Conversely, if you can't get a particular calculation by one method or equation, try another. That's simply the way it is. You probably wouldn't get much sympathy if you calculated a dosage regimen properly based on a wrong elimination rate constant and ended up killing your patient. You Will:
What do I have to do? How much work is really expected?
1) Come prepared to participate in class. This is your full time job. If you are working full time, it is usually 40 to 60 hours per week. If you go to college 15 to 18 credits and prepare/study 2 hours for each credit, you work 45 to 55 hour per week - you have a full time job. Your commitment is the 45 to 50 hour week not just the contact hours and a night for each exam. This specifically means for each 1 hour class, I expect no less than 2 hours of preparation on your part. Each of you will be assigned to a study group. You will work the problems together and teach each other both in and out of class. We will have group discussion of class as well as group problem solving. It will be your responsibility that every member of your group be adequately prepared to answer for the group during recitation. There will be a grade for group participation. Part of your grade will be based on quizzes, how well your group performed both on the material for which you were responsible as well as overall and your peer evaluation.
Do I have to read the text?
2) Read the text. When you read, read critically. Do you understand each idea? Place a <+> (in pencil) in the left side of each paragraph after you read and understand it. If you don't get it, place a and come prepared to ask about it in class.
Why do I have to do the problem sets?
3) Work the problems. Check the answers. These come from old exams, so they are the type that you are likely to see. Work them in your study groups so that everyone can see your thought processes. Bring them to class if you can not do them or come and see me privately. Be prepared to show me your attempts at solving the problem. I will show you how to get started and give direction to your thought. I will not work the problem for you. You would not learn if I did it for you. It is crucial that you work the problems. Each has a specific objective. Over-
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all, they contribute to your gaining facility in the processes that a pharmacokineticist must know how to do. Can I just coast through?
4) Do not delude yourself with respect to your performance. If you received a grade that was less than satisfactory for you, do not simply console yourself by saying “I knew the stuff, I just made a little error.” Can you get it right consistently? That's when you know the stuff. That is not a laudable goal. That's what a professional does. There have been several students in the past that “knew that stuff” right up till the time that they had to repeat the course (and sometimes beyond).
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1.6 Computers in the course Can I get through with just paper and pencil?
Computer literacy is necessary in this field. Consequently computers will pervade the course. The homework problems (above) are to done both by hand and checked with the computer. This will help your understanding of pharmacokinetics in general and that homework objective in particular. Computers are natural adjunctive tools in the teaching of pharmacokinetics. The are able to simulate the concentration vs. time profiles and do difficult repetitive calculations which allow the student to get a broad view of the processes involved.
What are the programs that I will be using?
Programs that are currently being used in the course are The Scientist and PKAnalyst both from MicroMath Scientific Software, P.O. Box 71550, Salt Lake City, UT 84171-0550; or http://www.micromath.com. A working student version of the software is available free for the downloading for your own work at home. A full working version is on the Pharmacy Server. In addition to the above course objectives, there are specific objectives for the use of computers in the course. They are:
What will I be expected to do with the computers?
1. Simulation. The student will construct a graph of the drug time course using classical pharmacokinetics. The student will demonstrate effects of changes in pharmacokinetic parameters on the ADME processes and correlated pharmacological / therapeutic response. 2. The student shall statistically evaluate models with regard to fit of data using both linear and nonlinear regression analysis. The student will calculate pharmacokinetic parameters which best describe the processes of ADME.
How can I use the computers in the homework and library questions?
These objectives will be met in a variety of ways. Clearly, the most direct method is the solution of the problem sets by computer. First, I expect that you would do the problem by hand, complete with graphs and other supporting calculations followed by computer simulation and data analysis. Just how close did you come to the best fit? Next, a portion of each exam will be a library exercise in which you will find and evaluate a published article according to the principles that you learned in class utilizing the computer facilities. How close did you come the authors numbers? Do you, in fact, even agree with the authors? You will prepare a short consult in which you describe the patient and what the authors did along with your support (or non-support) of the authors conclusions.
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1.7 Survival Kit 1.7.1
THINGS FOR YOUR SURVIVAL KIT!
What do I need to buy RIGHT NOW?
1. You will need a good calculator - One with have it ASAP. You will use it in class.
e
x
and
ln x
functions. You must
2. You will also need 2 cycle semilog paper and a clear straight edge ruler for use in class. These are available in the book store or at an office supply store. 3. You will need access to a computer (486 DX or higher). Micromath has made available a student version of the programs for a nominal fee (This software is pre-loaded in the Criss computer lab; you may purchase a copy for home use) 4. You will need a 3” D three-ring binder for collecting and maintaining all the pages in this book as well as your class notes. What do I need to do in and out of class?
Work in your study groups. You never learn it so well as when you teach it to someone else. Everyone benefits from a well run prepared study group. You are not in competition with your fellow classmates. If everyone earns an “A”, then everyone will receive one.
How can I organize this material?
Organize and label your study notes. This is basic survival. This is one strategy that I find works well. I recommend it highly. Good study notes are formatted on loose-leaf in a three ring binder. The individual pages have a line drawn down about 1/3 the way in. The notes are taken on the right (2/3) of the page, while labels go in the left. The labels on the left are often written as questions, which are answered in the text on the right. Loose leaf binders allow for the incorporation of reading summaries as well as relevant problems and homework to be organized with a divider all in one place. You should write intelligently, with proper punctuation and spelling as if you were preparing a consult for a physician. Organization is the key. Remember: you may have all the information in the world at your fingertips; be able to solve the most difficult therapy problem and no one will listen to you if you can't communicate intelligently. Chapters in the book will be organized as above.
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1.7.2
WHAT YOU WILL GAIN: (YOUR GOALS)
How is the course to be graded?
1. At the lowest level, a decent grade for a significant course. Specific grades will be earned by attaining the following averages: • • • • • • •
A 90 and above B+ 85 to 89 B 80 to 84 C+ 75 to 79 C 70 to 74 D 60 to 69 F 59 and below
The number of exams and point distribution will be determined in class. 2. At the next higher level, I will guarantee that if you comprehend this material at level V, you will have no trouble passing any state board anywhere with regard to pharmacokinetics. 3. You will gain a useful skill that will make you an integral part of the health care team. Do I really need a teacher to learn?
4 You will learn to learn. There is an old proverb which goes: “Give a man a fish and you feed him for a day. Teach a man to fish and you feed him for a lifetime.” The B.S. Degree is designed to eliminate teachers. An educated man is one who has learned to how to learn, not one who memorized a page in a book. That is what you need to be a professional. The total medical knowledge is doubling at a rate of every 3-4 years. That means that you will be out of date shortly after graduation (if not before) if you simply memorized content and don't learn to learn and continue to learn throughout your career.
What about cheating?
One last piece of information: Neither you nor I will not tolerate any academic misconduct. Anyone caught will minimally receive an “F” for their efforts and I will recommend dismissal from the program. The profession has no room for unprofessional behavior. I will prosecute.
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Introduction
1.8 Tentative Schedule 1.8.1
STUDY GROUP 1: LEARN THE TOOLS - OBTAIN PHARMACOKINETIC PARAMETERS FROM DATA. A: Introduction 1. Texts 2. Literature 3. Grading Policy 4. Course Philosophy B: Math review 1. Numbers and exponents 2. Graphs and reaction order 3. Calculus 4. Laplace transform 5. Computer Introduction 6. Computer simulation and problem sets C: Pharmacokinetic modeling 1. What a model is and what it isn't. 2. Why we model 3. Philosophy of modeling D: Pharmacological Response 1. Michaelis - Menton Mass balance equation 2. Interrelationships between Concentration, time and response. E: I.V. Bolus dosing 1. Parent compound I. Plasma a. Plasma concentration vs. time profile analysis b. Computer aided instruction and simulation c. Problem sets d. Professional communication. II Urine a. Amount vs. time profile analysis b. Rate vs. time profile analysis c. Computer aided instruction and simulation d. Problem sets e. Professional communication. 2. Metabolite
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I. Plasma a. Plasma concentration vs. time profile analysis b. Computer aided instruction and simulation c. Problem sets d. Professional communication. II. Urine a. Rate vs. time profile analysis b. Computer aided instruction and simulation c. Problem sets d. Professional communication. F: I.V. infusion 1. Parent compound I. Plasma a. Plasma concentration vs. time profile analysis b. Computer aided instruction and simulation c. Problem sets d. Professional communication. II. Urine a. Amount vs. time profile analysis b. Rate vs. time profile analysis c. Computer aided instruction and simulation d. Problem sets e. Professional communication. 1. Metabolite I. Plasma a. Plasma concentration vs. time profile analysis b. Computer aided instruction and simulation c. Problem sets d. Professional communication. II. Urine a. Amount vs. time profile analysis b. Rate vs. time profile analysis c. Computer aided instruction and simulation d. Problem sets e. Professional communication. End of Material for first exam (six weeks for semester, two weeks for summer session)
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1.8.2
STUDY GROUP 2: LEARN HOW THE PARAMETERS ARE MODIFIED. G: Biopharmaceutical factors 1. Absorption I. Physiology II. Mechanisms III. Physiological changes with age, sex, disease 2. Distribution I. Diffusion II. Cardiac output / blood perfusion III. Physical properties of the drug IV. Physical properties of the body V. Physiological changes with age, sex, disease 3. Metabolism I. Biotransformation methods II. First pass effect III. Clearance IV. Physiological changes with age, sex, disease 4. Excretion I. Renal II. Biliary III. Mammary IV. Salivary V. Misc. VI. Physiological changes with age, sex, disease H: Oral dosing 1. Parent compound I. Plasma a. Plasma concentration vs. time profile analysis b. Computer aided instruction and simulation c. Problem sets d. Professional communication. II. Urine a. Amount vs. time profile analysis b. Rate vs. time profile analysis c. Computer aided instruction and simulation d. Problem sets e. Professional communication. 2. Metabolite I. Plasma a. Plasma concentration vs. time profile analysis b. Computer aided instruction and simulation
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c. Problem sets d. Professional communication. II. Urine a. Amount vs. time profile analysis b. Rate vs. time profile analysis c. Computer aided instruction and simulation d. Problem sets e. Professional communication. I: Bioavailability, Bioequivalence, Drug product selection 1. Relative and Absolute Bioavailability 2. Factors Influencing Bioavailability 3. Methods of Assessing Bioavailability I. in vivo II. in vitro III. Correlation 4. Bioequivalence 5. Bioavailability 6. Drug Product Selection
1.8.3
STUDY GROUP 3: APPLY THE TOOLS IN COMPROMISED PATIENTS. J: Dosage regimen (Healthy, aged and diseased patients) 1. Multiple dose kinetics 2. Optimization of dosage regimen 3. Computer aided instruction 4. Computer simulation and problem sets 5. Computer aided consultation 6. Professional consultation process End of Material for two credit course (six weeks semester - 2 weeks summer)
1.8.4
STUDY GROUP 4: APPLY THE TOOLS IN SPECIAL CASES. K: Multicompartment Modeling 1. Parent compound plasma vs. time profile analysis 2. Multiple dose considerations 3. Computer aided instruction 4. Computer simulation and problem sets 5. Computer aided consultation 6. Professional consultation process L: Protein Binding (healthy, aged and diseased patients) 1. Mass balance considerations / drug interactions 2. Effects of protein binding on pharmacokinetic parameters
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3. Computer aided instruction 4. Computer simulation and problem sets 5. Computer aided consultation 6. Professional consultation process M: Non - linear (Michaelis - Menton) kinetics 1. Computer aided instruction 2. Computer simulation and problem sets 3. Computer aided consultation 4. Professional consultation process End of material for three credit course (4 weeks semester, 2 weeks summer)
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1.9 Competency Statements Related To Pharmacokinetics The profession of pharmacy has determined that there are minimum, entry level abilities necessary for a pharmacist. These form the Standards of Practice for the profession of pharmacy, as written by The National Association of Boards of Pharmacy (who make the NAPLEX, coincidentally). It is important to note that these abilities are not thought up by some faculty member who sits in his ivory tower saying what he thinks is important. These are what pharmacists do. They have been promulgated as competency statements They are also the basis for the state board exams as well as the basis far your coursework while in the School of Pharmacy. They are broken down into three general areas: Area 1: Manage Drug Therapy to Optimize Patient Outcomes (Approximately 50% of Test) Area 2: Assure the Safe and Accurate Preparation and Dispensing of Medications (Approximately 25% of Test) Area 3: Provide Drug Information and Promote Public Health (Approximately 25% of Test) For a complete listing of competency statements please refer to the National Association of Boards of Pharmacy’s web site, www.nabp.net
1.9.1
SPECIFIC COMPETENCY STATEMENTS ADDRESSED IN THIS COURSE Area 1: Manage Drug Therapy to Optimize Patient Outcomes (Approximately 50% of Test) 1.1.0 Evaluate the patient and/or patient information to determine the presence of a disease or medical condition, to determine the need for treatment and/or referral, and to identify patient-specific factors that affect health, pharmacotherapy, and/or disease management. 1.1.1 Identify and/or use instruments and techniques related to patient assessment and diagnosis. 1.1.2 Identify and define the terminology, signs, and symptoms associated with diseases and medical conditions.
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1.1.4 Identify patient factors, biosocial factors, and concurrent drug therapy that are relevant to the maintenance of wellness and the prevention or treatment of a disease or medical condition. 1.2.0 Assure the appropriateness of the patient's specific pharmacotherapeutic agents, dosing regimens, dosage forms, routes of administration, and delivery systems. 1.2.1 Identify drug products by their generic, trade, and/or common names. 1.2.3 Evaluate drug therapy for the presence of pharmacotherapeutic duplications and interactions. 1.2.5 Identify physicochemical properties of drug substances that affect their solubility, pharmacokinetics, pharmacologic actions, and stability. 1.2.6 Interpret and apply pharmacokinetic principles to calculate and determine appropriate drug dosing regimens. 1.2.7 Interpret and apply biopharmaceutic principles and the pharmaceutical characteristics of drug dosage forms and delivery systems, to assure bioavailability and enhance patient compliance. 1.3.0 Monitor the patient and/or patient information and manage the drug regimen to promote health and assure safe and effective pharmacotherapy. 1.3.2 Evaluate patient information to determine the safety and effectiveness of pharmacotherapy. 1.3.5 Identify and remedy interactions or contraindications with diagnostic or monitoring tests or procedures. Area 2: Assure the Safe and Accurate Preparation and Dispensing of Medications (Approximately 25% of Test) 2.1.0 Perform calculations required to compound, dispense, and administer medication. 2.1.1 Calculate the quantity of medication to be compounded or dispensed; reduce and enlarge formulation quantities and calculate the quantity or ingredients needed to compound the proper amount of the preparation. 2.1.3 Calculate the rate of drug administration.
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2.1.4 Calculate or convert drug concentrations, ratio strengths, and/or extent of ionization. 2.2.0 Select and dispense medications. 2.2.3 Interpret and apply pharmacokinetic parameters and quality assurance data to determine equivalence among manufactured drug products, and identify products for which documented evidence of inequivalence exists. 2.2.5 Identify and describe the use of equipment and apparatus required to administer medications. Area 3: Provide Drug Information and Promote Public Health (Approximately 25% of Test) 3.1.0 Access, evaluate, and apply information to promote optimal health care. 3.1.1 Identify the typical content and organization of specific sources of drug and health information. 3.1.2 Interpret and evaluate data presented in textual, tabular, or graphic form. 3.1.3 Evaluate the suitability, accuracy, and reliability of information from reference sources by explaining and evaluating the adequacy of experimental design and by applying and evaluating statistical tests and parameters.
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1.10 Pharmacokinetic Symbolism Pharmacokinetics was developed in several locations simultaneously. Because of this, the symbols used in the literature are not consistent. Provided each symbol is rigorously defined prior to use, this inconsistency should not prove an insurmountable difficulty when assessing the literature. In this book, the symbolism below will be generally used, though, as an illustration of the variety, some deviation may be anticipated on occasions.
1.10.1
AMOUNT TERMS (UNIT: MASS) ARE
amount remaining to be eliminated (excreted)
D
dose (or maintenance dose)
DL
loading dose
Xa
amount of drug remaining to be absorbed at any time
X
amount of unchanged drug in body at any time
Xm
amount of metabolite in body at any time
Xu
cumulative amount of unchanged drug excreted into urine up to any time
X mu
cumulative amount of metabolite excreted into urine up to any time
X max maximum amount of unchanged drug in body X min
minimum amount of unchanged drug in body
X
average amount of unchanged drug in body (also Laplace transform)
X eff minimum amount of unchanged drug in body necessary for pharmacological response
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Introduction
1.10.2
CONCENTRATION TERMS (UNITS MASS/VOLUME) Cb
concentration of drug in blood at any time
Cp
concentration of drug in plasma at any time
Cm
Concentration of metabolite in plasma (or blood) at any time
( C p )ss “average” steady-state concentration of drug in plasma during a dosing interval ( C p )max maximum concentration of drug in plasma ( C p )min minimum concentration of drug in plasma Cp
average concentration of drug in plasma
KA
dissociation constant of drug-protein complex
KM
Michaelis-Menton rate constant
KR
dissociation constant of drug-receptor complex
MEC minimum effective concentration of drug or metabolite MTC minimum toxic concentration of drug or metabolite
1.10.3
VOLUME TERMS (UNIT: VOLUME) Vd
apparent volume of unchanged drug distribution in compartment
Vm
apparent volume of metabolite distribution in compartment
Vw
physiological volume of plasma water
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Introduction
1.10.4
1.10.5
TIME TERMS (UNIT: TIME) t
time since administration of dose
T
duration of zero-order input
t'
time since cessation of zero-order input
t0
lag time
t
mean time during sampling interval
t½
elimination half-life (“biological half-life”)
t 0.5
time for 50% removal
t max
time when maximum amount or concentration occurs
t dur
duration of effective pharmacological response
τ
dosing interval (greek theta)
b
time variable used in association with zero-order input
RATE CONSTANT TERMS (UNIT: RECIPROCAL TIME (*), MASS/TIME (**) K, k e ,K d, K i apparent first-order rate constant for elimination, Summation of all the ways the drug is eliminated (*) ka
apparent first-order rate constant for absorption (*)
k u, k r apparent first-order rate constant for urinary (renal) excretion of unchanged drug (*) km
apparent first-order rate constant for metabolism of unchanged drug (*)
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Introduction
1.10.6
1.10.7
k mu
apparent first-order rate constant for excretion of metabolite (*)
k ij
apparent first-order transfer rate constant (*)
k0
zero-order input rate constant (**)
Q
zero-order infusion rate constant (**)
R
rate constant for decline in pharmacological effect (usual units:%/time)
α
hybrid first-order rate constant (*) (greek alpha)
β
hybrid first-order rate constant (*) (greek beta)
CLEARANCE TERMS (UNITS: VOLUME/TIME) Cl
total body clearance (TBC)
Cl r
renal clearance (RC)
Cl m
metabolic clearance (MC)
Cl cr
creatinine clearance
Cl H
hepatic clearance (HC)
RATE TERMS (UNITS: MASS/TIME (*), MASS/TIME, VOLUME (**), VOLUME/TIME (***) dX ------dt
instantaneous rate of change of amount of unchanged drug (*)
X ---t
measured rate of change of amount of unchanged drug (*)
RH
rate of plasma flow through the liver (***)
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Introduction
1.10.8
Rr
rate of plasma flow through the kidney (***)
VM
theoretical maximum rate of a process (**)
OTHER TERMS AUC area under the plasma concentration-time curve (units: time * mass/volume) AUMC area under the first moment of the plasma concentration-time curve (units: 2
time ⋅ mass ⁄ volume
)
MRT Mean Residence Time (units:
time )
MAT Mean Absorption Time (units:
time )
MDT Mean Dissolution Time (units:
time )
E
intensity of pharmacological effect
EH
steady-state hepatic extraction ratio
Er
steady-state renal extraction ratio
E max maximum intensity of pharmacological effect F
fraction of administered dose ultimately absorbed
FRE
fraction remaining to be eliminated (excreted)
H
hematocrit (fractional volume of erythrocytes in whole blood)
N
number of elimination half-lives in a dosing interval
R
accumulation factor
b
intercept
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1.10.9
f
fraction of drug that is free (unbound)
f ss
fraction of steady-state
m
slope (sometimes specifically for log dose-response plot)
n
number of doses
s
Laplace operator
[ ]
indicates molar concentration
SUBSCRIPTS 0
at time zero
∞
at time infinity
ss
during steady-state
t
at time t
T
at time T
n
following dose n
diff
difference between extrapolated and observed
int
intrinsic
i
index (i.e., 1,2,3)
j
index (i.e., 1,2,3)
1.10.10 SUPERSCRIPTS x
extrapolated
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Introduction
x°
last measured value
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Introduction
1.11 First Lesson in Pharmacokinetics It should be intuitively obvious to the most casual observer that the relative bioavailability of 2 simultaneous I.V. bolus doses of a drug is equal to the following: (EQ 1-1) 1 –1 –1 1 2 2 1 δ ln lim ( [ x ] – [ x ] )! + --- + ( sin q ) + ( cos q ) = δ δ → ∞
∞
cosh p 1 – ( tanh p )
2
∑ ------------------------------------------------n 2
(EQ 1-2)
n=0
given that 100% bioavailability of a single I.V. bolus dose is equal to 1, and both doses contain an equal mass of active drug. For the struggling pharmacokinetics student, we would like to show the veracity of this statement. Of course, it is obvious that; the reverse of the transpose is equal to the transpose of the inverse in matrix theory. i.e.: 1 –1
[x ]
–1 1
= [x ]
(EQ 1-3)
Also, it should be obvious that: 0! = 1
(EQ 1-4)
Consequently, 1 –1
([x ]
–1 1
– [ x ] )! = 1
(EQ 1-5)
which means that: 2 2 1 δ ln lim 1 + --- + ( sin q ) + ( cos q ) = δ δ → ∞
∞
2
cosh p 1 – ( tanh p ) ∑ ------------------------------------------------n 2 n=0
(EQ 1-6)
By definition, δ e = 1 + 1--- δ
(EQ 1-7)
and 1 = cosh p 1 – ( tanh p )
2
(EQ 1-8)
Thus:
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Introduction
2
∞
∑ ----n-
2
ln e + ( sin q ) + ( cos q ) =
1
(EQ 1-9)
n = 02
Also, ∞
2 =
∑ ----n1
(EQ 1-10)
n = 02
and 1 = ln e
(EQ 1-11)
and 2
1 = ( sin q ) + ( cos q )
2
(EQ 1-12)
So, as we observed in equation 1, 1+1 = 2
(EQ 1-13)
under the stated conditions, two I.V. bolus doses given simultaneously will have twice as much available drug as a single I.V. bolus dose. You will agree, however, equation 1-1 is obvious and therefore is more easily understood by a pharmacokineticist!
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CHAPTER 2
Mathematics Review
Author: Michael Makoid, Phillip Vuchetich and John Cobby Reviewer: Phillip Vuchetich
BASIC MATHEMATICAL SKILLS OBJECTIVES 1.
Given a data set containing a pair of variables, the student will properly construct (III) various graphs of the data.
2.
Given various graphical representations of data, the student will calculate (III) the slope and intercept by hand as well as using linear regression.
3.
The student shall be able to interpret (V) the meaning of the slope and intercept for the various types of data sets.
4.
The student shall demonstrate (III) the proper procedures of mathematical and algebraic manipulations.
5.
The student shall demonstrate (III) the proper calculus procedures of integration and differentiation.
6.
The student shall demonstrate (III) the proper use of computers in graphical simulations and problem solving.
7.
Given information regarding the drug and the pharmacokinetic assumptions for the model, the student will construct (III) models and develop (V) equations of the ADME processes using LaPlace Transforms.
8.
The student will interpret (IV) a given model mathematically.
9.
The student will predict (IV) changes in the final result based on changes in variables throughout the model.
10.
The student will correlate (V) the graphs of the data with the equations and models so generated.
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2.1 Concepts of Mathematics Pharmacokinetcs is a challenging field involving the application of mathematical concepts to real situations involving the absorbtion, distribution, metabolism and excretion of drugs in the body. In order to be successful with pharmacokinetics, a certain amount of mathematical knowledge is essential. This is just a review. Look it over. You should be able to do all of these manipulations.
This chapter is meant to review the concepts in mathematics essential for understanding kinetics. These concepts are generally taught in other mathematical courses from algebra through calculus. For this reason, this chapter is presented as a review rather than new material. For a more thorough discussion of any particular concept, refer to a college algebra or calculus text. Included in this section are discussions of algebraic concepts, integration/differentiation, graphical analysis, linear regression, non-linear regression and the LaPlace transform. The Scientist and PKAnalyst are the computer programs used in this course.
Something new LaPlace transforms. Useful tool.
A critical concept introduced in this chapter is the LaPlace transform. The LaPlace transform is used to quickly solve (integrate) ordinary, linear differential equations. The Scientist by Micromath Scientific Software, Inc.1 is available for working with the LaPlace transform for problems throughout the book.
1. MicroMath Scientific Software, Inc., P.O. Box 21550, Salt Lake City, UT 84121-0550, http://www.micromath.com
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2.2 Mathematical Preparation 2.2.1
ZERO AND INFINITY Any number multiplied by zero equals zero. Any number multiplied by infinity ( ∞ ) equals infinity. Any number divided by zero is mathematically undefined. Any number divided by infinity is mathematically undefined.
2.2.2
EXPRESSING LARGE AND SMALL NUMBERS Large or small numbers can be expressed in a more compact way using indices.
How Does Scientific Notation Work?
316000 becomes 3.16 × 10
Examples:
5
–3
0.00708 becomes 7.08 ×10 In general a number takes the form: A × 10
n
Where A is a value between 1 and 10, and n is a positive or negative integer The value of the integer n is the number of places that the decimal point must be moved to place it immediately to the right of the first non-zero digit. If the decimal point has to be moved to its left then n is a positive integer; if to its right, n is a negative integer. Because this notation (sometimes referred to as “Scientific Notation”) uses indices, mathematical operations performed on numbers expressed in this way are subject to all the rules of indices; for these rules see Section 2.2.4. A shorthand notation (AEn) may be used, especially in scientific papers. This may n
be interpreted as A × 10 , as in the following example: 4
2.28E4 = 2.28 ×10 = 22800
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2.2.3
SIGNIFICANT FIGURES A significant figure is any digit used to represent a magnitude or quantity in the place in which it stands. The digit may be zero (0) or any digit between 1 and 9. For example: TABLE 2-1. Significant
Value
Figures
Significant Figures
Number of Significant Figures
(a)
572
2,5,7
3
(b)
37.10
0,1,3,7
4
(c)
10.65 x 104
0,1,6,5
4
(d)
0.693
3,6,9
3
(e)
0.0025
2,5
2
How do I determine the number of significant figures?
Examples (c) to (e) illustrate the exceptions to the above general rule. The value 10 raised to any power, as in example (c), does not contain any significant figures; hence in the example the four significant figures arise only from the 10.65. If one or more zeros immediately follow a decimal point, as in example (e), these zeros simply serve to locate the decimal point and are therefore not significant figures. The use of a single zero preceding the decimal point, as in examples (d) and (e), is a commendable practice which also serves to locate the decimal point; this zero is therefore not a significant figure.
What do significant figures mean?
Significant figures are used to indicate the precision of a value. For instance, a value recorded to three significant figures (e.g., 0.0602) implies that one can reliably predict the value to 1 part in 999. This means that values of 0.0601, 0.0602, and 0.0603 are measurably different. If these three values cannot be distinguished, they should all be recorded to only two significant figures (0.060), a precision of 1 part in 99. After performing calculations, always “round off” your result to the number of significant figures that fairly represent its precision. Stating the result to more significant figures than you can justify is misleading, at the very least!
2.2.4
RULES OF INDICES
What is an index?
An index is the power to which a number is raised. n
Example: A where A is a number, which may be positive or negative, and n is the index, which may be positive or negative.
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Sometimes n is referred to as the exponent, giving rise to the general term, “Rules of Exponents”. There are three general rules which apply when indices are used. (a) Multiplication n
m
n
m
A ×A = A A ×B
n+m
n n+m = A --- × B B
(b) Division n
A - = An – m -----m A n
A - = A n × Bn – m ------- B m B (c) Raising to a Power n m
(A )
= A
nm
There are three noteworthy relationships involving indices: (i) Negative Index A
–n
1- As n tends to infinity ( n → ∞ ) then A – n → 0 . = ----n A
(ii) Fractional Index 1 --n
A =
n
A
(iii) Zero Index 0
A = 1
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2.2.5
LOGARITHMS
What is a logarithm?
Some bodily processes, such as the glomerular filtration of drugs by the kidney, are logarithmic in nature. Logarithms are simply a way of succinctly expressing a number in scientific notation. In general terms, if a number (A) is given by A = 10
n
then log ( A ) = n where ‘log’ signifies a logarithm to the base 10, and n is the value of the logarithm of (A). 5
Example: 713000 becomes 7.13 × 10 , and 7.13 = 10
0.85
, thus 713000 becomes 10
0.85
5
× 10 = 10
( 5 + 0.85 )
= 10
5.85
and log ( 713000 ) = 5.85 Logarithms to the base 10 are known as Common Logarithms. The transformation of a number (A) to its logarithm (n) is usually made from tables, or on a scientific calculator; the reverse transformation of a logarithm to a number is made using anti-logarithmic tables, or on a calculator. What is the characteristic? the mantissa?
2.2.6
The number before the decimal point is called the characteristic and tells the placement of the decimal point (to the right if positive and to the left if negative). The number after the decimal is the mantissa and is the logarithm of the string of numbers discounting the decimal place.
NATURAL LOGARITHMS
What is a natural logarithm?
Instead of using 10 as a basis for logarithms, a natural base (e) is used. This natural base is a fundamental property of any process, such as the glomerular filtration of a drug, which proceeds at a rate controlled by the quantity of material yet to undergo the process, such as drug in the blood. To eight significant figures, the value of the transcendental function, e, is e = 2.7182818 ....
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Strictly speaking, ∞
e = 1+
∑ ---x! 1
x=1
Where x is an integer ranging from 1 to infinity ( ∞ ) , ∞
∑
denotes the summation from x = 1 to x = ∞ , and
x=1
! is the factorial (e.g., 6! = 6x5x4x3x2x1= 720) n
In general terms, if a number (A) is given by A = e , then by definition, ln ( A ) = n Where, ‘ln’ signifies the natural logarithm to the base e , and n is the value of the natural logarithm of A . Natural logarithms are sometimes known as Hyperbolic or Naperian Logarithms; again tables are available and scientific calculators can do this automatically. The anti-logarithm of a natural logarithm may be found from exponential tables, which n
give the value of e for various values of n. How are natural logarithms ln x and common logarithms log x related?
Common and natural logarithms are related as follows: ln ( A ) = 2.303 × log ( A ) , and log ( A ) = 0.4343 × ln ( A ) Because logarithms are, in reality, indices of either 10 or e , their use and manipulation follow the rules of indices (See Section 2.2.4). (a) Multiplication: n
m
n
To multiply N × M , where N = e and M = e ; NM = e × e
m
= e
n+m
.
By definition,
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ln ( NM ) = n + m ; but n = ln ( N ) and m = ln ( M ) , hence ln ( NM ) = ln ( N ) + ln ( M ) Thus, to multiply two numbers (N and M) we take the natural logarithms of each, add them together, and then take the anti-logarithm (the exponent, in this case) of the sum. (b) Division N- = ln ( N ) – ln ( M ) ln ---M (c) Number Raised to a Power m
ln ( N ) = m × ln ( N ) There are three noteworthy relationships involving logarithms: (i) Number Raised to a Negative Power ln ( N
–m
1- ) = – m × ln ( N ) = m × ln --N
As m tends to infinity ( m → ∞ ) , then ln ( N
–m
) → –∞
(ii) Number Raised to a Fractional Power
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1
ln ( m
--m 1 N ) = ln N = ---- × ln ( N ) m
(iii) Logarithm of Unity ln ( 1 ) = log ( 1 ) = 0
2.2.7
NEGATIVE LOGARITHMS The number 0.00713 may be expressed as: –3
7.13 ×10 , or 10 10
0.85
–3
× 10 , or
– 2.15
.
Hence, log ( 0.00713 ) = – 2.15 , which is the result generated by most calculators. However, another representation of a negative logarithm (generally used by referencing a log table): log (0.00713) = 3.85 The 3 prior to the decimal point is known as the characteristic of the logarithm; it can be negative (as in this case) or positive, but is never found in logarithmic tables. The .85 following the decimal point is known as the mantissa of the logarithm; it is always positive, and is found in logarithmic tables. In fact 3 is a symbolic way of writing minus 3 (-3) for the characteristic. In every case the algebraic sum of the characteristic and the mantissa gives the correct value for the logarithm. Example: log (0.00713) = 3.85 Add -3 and 0.85 Result is -2.15, which is the value of log ( 0.00713 )
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The reason for this symbolism is that only positive mantissa can be read from antilogarithmic tables, and hence a positive mantissa must be the end result of any logarithmic manipulations. Note that while there are negative logarithms (when N < 1), they do not indicate that number itself is negative; the sign of a number (e.g., N) is determined only by inspection following the taking of anti-logarithms.
2.2.8
USING LOGARITHMIC AND ANTI-LOGARITHMIC TABLES Though the preferred method to using logarithms is with a calculator or computer, the understanding of how the number is being manipulated may be important in understanding the use of logarithms. (See the end of this chapter for Logarithm tables). (a) Find the log of (62.54) 1
62.54 = 6.254 ×10
Look up the mantissa for 6254 in a table of logarithms: it is 7962. 1
Hence, 6.254 ×10 = 10
0.7962
1
× 10 = 10
1.7962
and log ( 62.54 ) = 1.7962
(b) Find the log of (0.00329) –3
0.00329 = 3.29 ×10
The mantissa for 329 is 5172 Hence, log(0.00329) = 3.5172. Note that in both examples the value of the characteristic is the integer power to which 10 is raised when the number is written in scientific notation. How do I multiply using logarithms?
(c) Multiply 62.54 by 0.00329 log (62.54) = 1.7972 log (0.00329) = 3.5172 log (62.54 + log (0.00329) = 1.7962+3.5172 = 1.7962-3+ 0.5172=-0.6866 0.6866=1.3134
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(d) We wish to find anti-log (1.3134) Look up the anti-log for the 0.3134 (mantissa) in a table: it is 2058. –1
Antilog (1.3134) = 2.058 ×10
Hence, antilog (1.3134) = 0.2058 How do I divide using logarithms?
log (62.54) - log (0.00329) = 1.796 - 3.5172=1.796 +3 - 0.5172=4.2788 antilog 4.2788 = 19002
2.2.9
DIMENSIONS
What is a unit?
There are three fundamental dimensions which are used in various combinations to express the properties of matter. Each of these dimensions has been assigned a definite basic unit, which acts as a reference standard. TABLE 2-2 Dimensions
Dimension
Dimensional Symbol
Unit
Unit Symbol
Length
L
meter
m
Mass
M
gram
g
Time
T
second
sec
How are units made bigger and smaller?
In the metric system, which emerged from the French Revolution around 1799, there are various prefixes which precede the basic units and any derived units. The prefixes indicate the factor by which the unit is multiplied. When the index of the factor is positive the prefixes are Greek and have hard, consonant sounds. In contrast, when the index is negative, the prefixes are Latin and have soft, liquid sounds. (see Table 2-3).
How big is big?
Examples: An average adult male patient is assumed to have a mass of 70 kilograms (70 kg). An average adult male patient is assumed to have a height of 180 centimeters (180 cm). A newly minted nickel has a mass of 5.000 g. Doses of drugs are in the mg (10-3 g) range (occasionally g) never Kg (103 g) or larger. Students have told me that the dose that they have calculated for their patient is 108 g (converting to common system - ~ 100 tons). I doubt it. Get familiar with this system. Note that the plural of Kg or cm is Kg or cm; do not add an “s”. In pharmacy there are two derived units which are commonly used, even though they are related to basic units. The Liter (L) is the volume measurement and is a cube 10 cm on a side (1L = (10cm)3 = 1000 cm3 ) while the concentration measurement and has the units of Mass per Volume.
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Why should I use units?
Whenever the magnitude of a measured property is stated, it is imperative to state the units of the measurement. Numbers are useless by themselves. Example: The procainamide concentration range is 4-8 µ g/ml; stating the range without units may lead to a potentially lethal error in which procainamide is administered in a sufficient dose to attain a range of 4-8 mg/ml, which is 1000 times too large and would give rise to cardiac arrest.
TABLE 2-3 Scale
Name
Symbol
exa-
E
peta-
P
tera-
T
giga-
G
mega-
M
kilo-
k
hecto-
h
deca-
da
of Metric system and SI Multiplication Factor 10 10 10
Name
Symbol
18
deci-
d
15
centi-
c
12
milli-
m
9
micro-
µ
6
nano-
n
3
pico-
p
2
femto-
f
1
atto-
a
10 10 10 10 10
Multiplication Factor 10 10 10 10 10 10 10 10
–1 –2 –3 –6 –9
–12 –15 –18
TABLE 2-4
2.2.10
Dimension
Dimensional Symbol
Volume
V
liter
l
Concentration
C
grams/liter
g/l
Unit
Unit Symbol
DIMENSIONAL ANALYSIS
How are units useful?
It is a general rule that the net dimensions (and units) on the two sides of any equation should be equal. If this is not so, the equation is necessarily meaningless. Consider the following equation which defines the average concentration of a drug FDin blood after many repeated doses, ( C b )∞ = ---------VKτ
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Where: •
F is the fraction of the administered dose ultimately absorbed (Dimensions: none),
• D is the mass of the repeated dose (Dimension: M), • V is the apparent volume of distribution of the drug (Dimension: V = L ) • K is the apparent first-order rate constant for drug elimination (Dimension: T
–1
),
• and τ is the dosing interval (Dimension: T )
Writing the dimensions relating to the properties of the right-hand side of the equation gives: M M ------------------------ = ----–1 V V⋅T ⋅T Thus ( C b )∞ has the dimensions of M ----- , which are correctly those of concentration. V Sometimes dimensional analysis can assist an investigator in proposing equations which relate several properties one with the other. If the units cancel, and you end up with the correct unit of measure, you probably did it right. If you obtain units that do not make sense, it’s guaranteed sure that you did it wrong.
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2.3 Calculus What is Calculus?
Calculus concerns either the rate of change of one property with another (differential calculus), such as the rate of change of drug concentrations in the blood with time since administration, or the summation of infinitesimally small changes (integral calculus), such as the summation of changing drug concentrations to yield an assessment of bioavailability. In this discussion a few general concepts will be provided, and it is suggested an understanding of graphical methods should precede this discussion.
2.3.1
DIFFERENTIAL CALCULUS
2.3.2
NON-LINEAR GRAPHS Consider the following relationship: y = x TABLE 2-5 x,
y sample data
x
0
1
2
3
4
y
0
1
8
27
64
3
As can be seen from the graph (Figure 2-1), a non-linear plot is produced, as expected. FIGURE 0-1.
y=x3
70 60 50 40 30 20 10 0 1
2
3
4
(Question: How could the above data be modified to give a linear graph?)
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2.3.3
SLOPE OF NON-LINEAR GRAPH As with a linear graph, y2 – y1 ∆y ---------------- = -----x2 – x1 ∆x Where ∆y is the incremental change in y and ∆x is the incremental change in x But, as can be seen (Figure 1), the slope is not constant over the range of the graph; it increases as x increases. The slope is a measure of the change in y for a given change in x. It may then be stated that: “the rate of change of y with respect to x varies with the value of x.”
2.3.4
VALUE OF THE SLOPE 3
We need to find the value of the slope of the line y = x when x = 2 (See Figure 1). Hence, we may choose incremental changes in x which are located around x ≈ 2. FIGURE 0-2. ∆y / ∆x
when x ≈ 2
x1
x2
∆x
y1
y2
∆y
∆y -----∆x
0
4
4
0
64
64
16.000
1
3
2
1
27
26
13.000
1.5
2.5
1.0
3.375
15.625
12.250
12.250
1.8
2.2
0.4
5.832
10.648
4.816
12.040
1.9
2.1
0.2
6.859
9.261
2.042
12.010
1.95
2.05
0.1
7.415
8.615
1.200
12.003
As may be seen, the value of the slope
∆ -----y- ∆x
tends towards a value of 12.000 as the
magnitude of the incremental change in x becomes smaller around the chosen value of 2.0. Were the chosen incremental changes in x infinitesimally small, the true value of the slope (i.e., 12.000) would have appeared in the final column of the above table.
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Calculus deals with infinitesimally small changes. When the value of ∆x is infinitesimally small it is written dx and is known as the derivative of x. Hence, dy ------ = f ( x ) dx Where dy/dx is the derivative of y with respect to x and f ( x ) indicates some function of x.
2.3.5
DIFFERENTIATION FROM FIRST PRINCIPLES Differentiation is the process whereby the derivative of y with respect to x is found. Thus the value of dy/dx, in this case, is calculated. (a) Considering again the original expression: y = x
3
(b) Let the value of y increase to y + dy because x increases to x + dx . Hence, y + dy = ( x + dx )
3
(EQ 1-14)
Multiplying out: 3
2
2
y + dy = x + 3x ( dx) + 3x ( dx ) + ( dx )
3
(EQ 1-15)
(c) The change in y is obtained by subtraction of the original expression from the last expression. (i.e., Eq. 2 - Eq. 1) 2
2
dy = 3x ( dx ) + 3x ( dx ) + ( dx )
3
(EQ 1-16)
Dividing throughout to obtain the derivative, 2 2 dy ------ = 3x + 3x ( dx ) + ( dx ) dx
When dx is infinitesimally small, its magnitude tends to zero ( dx → 0 ) . The limiting value of this tendency must be dx = 0 . At this limit,
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2 dy ------ = 3x dx
(EQ 1-17)
2
Hence the derivative of y with respect to x at any value of x is given by 3x . (d) In section 2.3.4 we saw how the true value of the slope (i.e., dy/dx) would be 12.0 when x = 2 . This is confirmed by substituting in Equation 1-16. 2 2 dy ------ = 3x = 3 ( 2 ) = 12 dx
2.3.6
RULE OF DIFFERENTIATION Although the rate of change of one value with respect to another may be calculated as above, there is a general rule for obtaining a derivative. Let x be the independent variable value, y be the dependent variable value, A be a constant, and n be an exponential power. The general rule is: If y = Ax
n
then n–1 dy ------ = nAx dx
The Rules of Indices may need to be used to obtain expressions in the form y = Ax
n
(e.g., if y =
2.3.7
5
x)
THREE OTHER DERIVATIVES 0
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dy Hence, ------ = 0 dx Thus the derivative of a constant is always zero. (b) Accept that if y = ln ( x ) dy 1 then ------ = --- . dx x This derivative is important when considering apparent first-order processes, of which many bodily processes (e.g., excretion of drugs) are examples. (c) Accept that if y = Be Ax dy base then ------ = ABe dx
Ax
where B and A are constants, and e is the natural
This derivative will be useful in pharmacokinetics for finding the maximum and minimum concentrations of drug in the blood following oral dosing.
2.3.8
A SEEMING ANOMALY Consider the following two expressions: n
(a) If y = Ax , then n–1 dy ------ = nAx dx n
(b) If y = Ax + A , n–1 n–1 dy + 0 = nAx then ------ = nAx dx
Both of the original expressions, although different, have the same derivative. This fact is recognized later when dealing with integral calculus.
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2.3.9
INTEGRAL CALCULUS Generally integral calculus is the reverse of differential calculus. As such it is used to sum all the infinitesimally small units (dy) into the whole value (y). Thus,
∫ dy 2.3.10
= y , where
∫ is the symbol for integration.
RULE OF INTEGRATION The derivative expression may be written: n dy ------ = Ax , or dx n
dy = Ax ⋅ dx To integrate, y =
∫ dy
=
∫ Ax
n
n
dx = A ∫ x dx
A general rule states: n+1
n Ax - + A A ∫ x dx = --------------n+1
Where A is the constant of integration However, there is one exception - the rule is not applicable if n = – 1 2 Example: If dy ------ = 3x (See section 2.3.5), dx 2+1
then y = 3x --------------- + A , and 2+1 3
y = x +A
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2.3.11
THE CONSTANT OF INTEGRATION There has to be a constant in the final integrated expression because of the seeming 3
3
anomaly referred to in section 2.3.8. As mentioned, both y = x and y = x + A 2 will give, on differentiation, dy ------ = 3x . dx So whether or not a constant is present and, if so, what is its value, can only be decided by other knowledge of the expression. Normally this other knowledge takes the form of knowing the value of y when x = 0 . In the case of our graphical example we know that when x = 0 , then y = 0 . The integrated expression for this particular case is: 3
y = x + A , therefore 3
0 = 0 + A , thus A = 0 In some examples, such as first-order reaction rate kinetics, the value of A is not zero.
2.3.12
THE EXCEPTION TO THE RULE It occurs when n = – 1 –1 1 y = A ∫ x dx = A ∫ --- dx x
Upon integration, y = A ⋅ ln ( x ) + A This is the reverse of the derivative stated in section 2.3.10 (b).
2.3.13
A USEFUL INTEGRAL Accept that if, Ax dy ------ = Be dx
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then, Ax
Be + A y = ----------A This integral will be useful for equations which define the bioavailability of a drug product.
2.3.14
EXAMPLE CALCULATIONS (a) Consider, 2
c = 3t ( t – 2 ) + 5 Where c is the drug concentration in a dissolution fluid at time t . Then, multiplying out, 3
2
c = 3t – 6t + 5 The rate of dissolution at time t is 2 dc ------ = 9t – 12t dt
So at any time, the rate may be calculated. 2 dc (b) Consider, ------ = 3t ( t – 4 ) = 9t – 12t dt
Then rearranging, 2
dc = 9t ⋅ dt – 12t ⋅ dt The integral of c is: c =
∫ dc
3
2
= 3t + A – 6t + B
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where B is a second constant. Adding the two constants together, 3
2
c = 3t = 6t + D where D = A + B We know, from previous work, that when t = 0 , then c = 5 Substituting 5 = D , the final expression becomes: 2
2
c = 3t + 6t + 5 Which is the initial expression in example (a) above. (c) Following administration of a drug as an intravenous injection, – dC p ------------- = KC p dt Where C p is the plasma concentration of a drug at time t K is the apparent first-order rate constant of elimination. Rearranging, 1- ⋅ dC – K ⋅ dt = ----p Cp – Kt = – K ∫ dt =
1
- ⋅ dC p ∫ ----Cp
This integral is the exception to the rule (see section 2.3.12). – Kt = ln ( C p ) + A We know that when t = 0 , C p = ( C p ) 0 .
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Substituting, 0 = ln ( C p )0 + A Or, A = – ln ( C p )0 Hence – Kt = ln ( C p ) – ln ( C p ) 0 or, ln ( C p ) = ln ( C p ) 0 – Kt or, Cp = ( C p )0 ⋅ e
– Kt
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2.4 Graphs Why do we graph?
We would like to agonize the chaotic world around us so that we can predict (see into the future) and retrodict (see into the past) what will happen or has happened. Our recorded observations are collectively known as data. We make a theory about what we think is happening and that theory is expressed in an equation. That determines our paradigm of how we see the world. This paradigm is expressed as a graph. The language of science is mathematics and graphs are its pictures.
TABLE 2-6
English
What is a graph?
Science
Observations
Data
Theory
Equations
Paradigms (pictures)
Graphs
A graph is simply a visual representation showing how one variable changes with alteration of another variable. The simplest way to represent this relationship between variables is to draw a picture. This pictorializing also is the simplest way for the human mind to correlate, remember, interpolate and extrapolate perfect data. An additional advantage is it enables the experimenter to average out small deviations in experimental results (non-perfect, real data) from perfect data. For example: TABLE 2-7 Perfect
vs. Real data
Perfect
Real
-3
-5
-4.6
-2
-3
-3.4
-1
-1
-0.6
0
+1
+0.8
+1
+3
+3.4
+2
+5
+4.4
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FIGURE 0-3.
Plot of Perfect vs. Real data
6
4
y
2
0
-2
-4
-6 -3
-2
-1
0
1
2
x
Simply looking at the columns x and y (real) it might be difficult to see the relationship between the two variables. But looking at the graph, the relationship becomes apparent. Thus, the graph is a great aid to clear thinking. For every graph relating variables, there is an equation and, conversely for every equation there is a graph. The plotting of graphs is comparatively simple. The reverse process of finding an equation to fit a graph drawn from experimental data is more difficult, except in the case of straight lines.
2.4.1
GRAPHICAL CONVENTIONS
How are graphs made?
Certain conventions have been adopted to make the process of rendering a data set to a graphical representation extremely simple. The ‘y’ variable, known as the dependent variable, is depicted on the vertical axis (ordinate); and the ‘x’ variable, known as the independent variable, is depicted on the horizontal axis (abscissa). It is said that ‘y’ varies with respect to ‘x’ and not ‘x’ varies with ‘y’. A decision as to which of the two related variables is dependent can only be made be considering the nature of the experiment. To illustrate, the plasma concentration of a drug given by IV bolus depends on time. Time does not depend on the plasma concentration. Consequently, plasma concentration would be depicted on the ‘y’ axis and time on the ‘x’ axis.
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Any point in the defined space of the graph has a unique set of coordinates: 1) the ‘x’ value which is the distance along the ‘x’ axis out from the ‘y’ axis and always comes first; and 2) the ‘y’ value which is the distance, along the ‘y’ axis up or down from the ‘x’ axis, and always comes second. Several points are shown in Figure 2. For example, (0,1) is on the line and (1,0) is not. The intersection of x and y axis is the origin with the coordinates of (0,0). In two dimensional spaces, the graph is divided into 4 quadrants from (0,0), numbered with Roman numerals from I through IV. It should be readily apparent that the coordinates for all points within a particular quadrant are of the same sign type i.e., TABLE 2-8 Quadrants
on a cartesian graph
Quadrant II (-x, +y)
Quadrant I (+x, +y)
Quadrant III (-x, -y)
Quadrant IV (+x, -y)
A line (or curve) on a graph is made up of an infinite number of points, each of which has coordinates that satisfy a given equation. For example, each point on the line in Figure 2 is such at its coordinates fit the equation y = 2x + 1 . That is for any value of x (the independent variable), multiplying the x value by 2 and adding 1 results in the y value (the dependent variable).
2.4.2
STRAIGHT LINE GRAPHS
What is a straight line?
A graph is a straight line (linear) only if the equation from which it is derived has the form y = mx + b Where: y = dependent variable x = independent variable m = slope of the straight line =
∆y -----∆x
b = the y intercept (when x = 0) or if the equation can be “linearized”, e.g., y′ = b′e
mx′
is not linear
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However. ln y′ = ln b′ + mx′ is of the same general form as: y = b + mx and consequently a plot of ln y′ (the dependent variable) versus x′ (the independent variable) will yield a straight line with a slope of m and an intercept of ln b′ . Expressions of any other form are non linear. For example: An expression relating the plasma concentration of a drug ( C p ) over time ( t ) . C p = C p0 e
– Kt
this relationship put in linear perspective yields: ln C p = ln C p0 – Kt , which is in the form y = b + mx The graphs that yield a straight line are the ones with the ordinate being ln C p0 , and the abscissa being t . Any other combination of functions of C p and t will be non-linear, e.g., C p versus t C p versus ln t ln C p versus ln t The appropriate use of a natural logarithm in this case serves to produce linearity. However, the use of logarithms does not automatically straighten a curved line in all examples. Some relationships between two variables can never be resolved into a single straight line, e.g., y = k 0 + k 1 x
( n – m)
+ k2 x
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(n – m + 1)
+ … + ( k n )x
x
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where n ≥ 2 ;n = m + 1 or K a FD –K t Kt C p = ------------------------ ⋅ (e – e a ) V ( Ka – K ) (It is possible to resolve this equation into the summation of two linear graphs which will be shown subsequently.)
2.4.3
THE SLOPE OF A LINEAR GRAPH (M)
What is the slope of a straight line?
From the equation a prediction may be made as to whether the slope is positive or negative. In the previous example, the slope is negative, i.e: m = – K TABLE 2-9 Sample
data of caffeine elimination
t (min)
µg- C p ------ mL
ln C p
12
3.75
1.322
40
2.80
1.030
65
2.12
0.751
90
1.55
0.438
125
1.23
0.207
173
0.72
-0.329
The differences in both the y-values and the x-values may be measured graphically to obtain the value of the slope, m. Then knowing the value of m, the value of K may be found.
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FIGURE 0-4.
Plasma Concentration
( C p ) of caffeine over time
Caffeine Concentration (
g/mL) u
101
2.4.4
100
10-1 0
50
100
150
200
Time (min)
LINEAR REGRESSION: OBTAINING THE SLOPE OF THE LINE The equation for a straight line is: y = m⋅x+b Where y is the dependent variable x is the independent variable m is the slope of the line b is the intercept of the line The equation for the slope of the line using linear regression is: ( Σ( x ) ⋅ Σ ( y ) ) – ( n ⋅ Σ ( x ⋅ y )) m = -------------------------------------------------------------------2 2 [ Σ ( x ) ] – ( n ⋅ Σ( x ) ) And the intercept is b = y – (m ⋅ x)
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TABLE 2-10
Linear Regression for data in table 2-9 2
X⋅Y
X
Y
X
12
1.322
144
15.864
40
1.030
1600
41.2
65
0.751
4225
48.815
90
0.438
8100
39.42
125
0.207
15625
25.875
173
-0.329
29929
-56.917
ΣX = 505
ΣY = 3.239
ΣX = 59623
2
ΣXY = 114.257
2
( ΣX ) = 255025 Σx x = ------ = 4.167 n
Σy y = ------ = 0.5398 n
Using the data from table 2-10 in the equation for the slope of the line · ( 505 ⋅ 3.239 ) – ( 6 ⋅ 114.257 ) m = --------------------------------------------------------------------- = – 0.01014 255025 – ( 6 ⋅ 59623 )
and the intercept would be b = ln C . In oder to find the b
Cpo = e = e
1.4229
0.5398 – ( – 0.01014 ⋅ 4.167 ) = 1.4229 . C p0 ,
the anti-ln of
b
Note that this is must be taken. i.e.
= 4.15
It is important to realize that you may not simply take any two data pairs in the data set to get the slope. In the above data, if we simply took two successive data pairs from the six data pairs in the set, this would result in five different slopes ( ∆x ⁄ ∆y ) ranging from -0.0066 to -0.0125 as shown in table 2-11. Clearly, this is unacceptable. Even to guess, you must plot the data, eyeball the best fit line by placing your clear straight edge through the points so that it is as close to the data as possible and look to make sure that there are an equal number of points above the line as below. Then take the data pairs from the line, not the data set.
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TABLE 2-11 Sample
2.4.5
slope data from figure 0-4
Time (x)
ln Conc. (y)
∆x
∆y
∆y -----∆x
12
1.322
-28
0.292
-0.0104
40
1.030
-25
0.28
-0.0112
65
0.751
-25
0.312
-0.0125
90
0.438
-35
0.231
-0.0066
125
0.207
-48
0.536
-0.0112
173
-0.329
PARALLEL LINES Two straight lines are parallel if they have the same slope. Calculating for the intercept of a linear graph (b): (a) Not knowing the value of m; The graph may be extrapolated, or calculations performed, at the situation where t = 0 . In this case b = ln C p0 . (b) Knowing the value of m; There are two ways: for any point on the graph: y 1 = mx 1 + b b = y 1 – mx 1 Hence, b may be calculated from a knowledge of y 1 and x 1 . Secondly, the graph may be extrapolated or calculations performed, at the situation where t = 0 . In this case, b = ln C p
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2.4.6
GRAPHICAL EXTRAPOLATIONS
How far can I predict?
It is dangerous to extrapolate on non-linear graphs, and it is unwise to extrapolate too far on linear graphs. Most often extrapolation is used to find the value of y at a selected value of x. If the size of the graph does not permit physical extrapolation to the desired value, the required result may be obtained by calculation. The values of m and b must be found as shown above. Then: y' = mx' + b , where x' is the selected value of x, and y' is the new calculated value for y.
2.4.7
SIGNIFICANCE OF THE STRAIGHT LINE The more closely the experimental points fit the best line, and the higher the number of points, the more significant is the relationship between y and x. As you may expect, statistical parameters may be calculated to indicate the significance.
What good is a straight line?
By using all the experimental data points, calculations may be made to find the optimum values of the slope m, and the intercept, b. From these values the correlation coefficient (r).and the t-value may be obtained to indicate the significance. Exact details of the theory are available in any statistical book, and the calculations may most easily be performed by a computer using The Scientist or PKAnalyst in this course. The advantage of computer calculation is that it gives the one and only best fit to the points, and eliminates subjective fitting of a line to the data.
2.4.8
GRAPHICAL HONESTY
How many points are needed?
Any graph drawn from 2 points is scientifically invalid. Preferably, straight-line graphs should have at least 3 - 5 points, and non-linear graphs a few points more.
Can I discard points that don’t fit?
As a graph is a visual representation which enables the experimenter to average out the small deviations in results from the “perfect” result, no one result can be unjustifiably ignored when the best fitting line is drawn. Still, an “errant” point may be justifiably ignored if there were unusual experimental circumstances which may have caused the deviation. Thus it is not justifiable to omit a point solely because it “does not fit”.
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2.4.9
AXES WITH UNEQUAL SCALES In mathematical studies, the scales of the x and y are almost always equal but very often in plotting chemical relations the two factors are so very different in magnitude that this can not be done. Consequently, it must be borne in mind that the relationship between the variables is given by the scales assigned to the abscissa and ordinate rather than the number of squares counted out from the origin. FIGURE 0-5. y = 0.1 x
0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 0
2
4
6
8
10
0
2
4
6
8
10
10 8 6 4 2 0
For example (shown in Figure 0-5), these two parabolic curves represent the same equation the only difference is the scales are different along the y axis. Frequently it is not convenient to have the origin of the graph coincide with the lower left hand corner of the coordinate paper. Full utilization of the paper with suitable intervals is the one criteria for deciding how to plot a curve from the experimental data. For example, the curve below (Figure 0-6) is poorly planned, where the following (Figure 0-7) is a better way of representing the gas law PV = nRT
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FIGURE 0-6.
Poorly presented graph
50 40 30 20 10 0 0
FIGURE 0-7.
4
8
12
16
20
Well arranged graph
25 20 15 10 5 0 1
2
3
4
5
6
7
8
9 10
Pressure (atm)
2.4.10
GRAPHS OF LOGARITHMIC FUNCTIONS 2
Previously variables were raised to constant powers; as y = x . In this section x
constants are raised to variable powers; as y = 2 . Equations of this kind in which the exponent is a variable are called (naturally) exponential equations. The most x
important exponential equation is where e is plotted against x .
2.4.11
SEMILOGARITHMIC COORDINATES Exponential or logarithmic equations are very common in physical chemical phenomenon. One of the best ways of determining whether or not a given set of phenomenon can be expressed by a logarithmic or exponential equation is to plot the logarithm of one property against another property. Frequently a straight line is obtained and its equation can be readily found. For example:
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In the following table the plasma concentration ( C p ) of the immunosuppressant cyclosporine was measured after a single dose (4mg/kg) as a function of time. TABLE 2-12 Plasma
concentration of cyclosporine Concentration
Time (hours) 0.25
1900
.75
1500
1.5
1300
4
900
6
600
8
390
ng -----ml
D’mello et al., Res. Comm. Chem. Path. Pharm. 1989: 64 (3):441-446
These can be illustrated in three different ways (Figures 0-8, 0-9, 0-10), A. Concentration vs. time directly B. Log concentration vs. time directly C. Log concentration vs. time with concentration plotted directly on to logarithmic scale of ordinates. FIGURE 0-8.
Concentration (ng/ml) vs. time (hr)
2000 1800 1600 1400 1200 1000 800 600 400 200 0
1
2
3
4
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6
7
8
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FIGURE 0-9.
Log Concentration vs. time
3.200 3.100 3.000 2.900 2.800 2.700 2.600 2.500 2.400 2.300
Log Concentration - Time Curve
0
FIGURE 0-10.
1
2
3
4
5
6
7
8
5
6
7
8
Log concentration (on log scale) vs. time
10000
1000
100 0
1
2
3
4
Graphing is much easier because the graph paper itself takes the place of a logarithmic table, as shown in Figure 1-10. Only the mantissa is designated by the graph paper. Scaling of the ordinate for the characteristic is necessary. The general equation y = Be ax can be expressed as a straight line by basic laws of indices. ax
ln y = ln B + ln ( e ) → ln y = ln B + ax ln y = ax + ln B
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One axis is printed with logarithmic spacing, and the other with arithmetic spacing. It is used when a graph must be plotted as in the example (Figure 1-4) y = log [ C p ] and x = t . In this example, the vertical logarithmic axis is labelled “Plasma concentration of cyclosporine” and the values plotted are the ordinary values of [ C p ] . Thus, there is no need to use logarithmic tables, because the logarithmic spacing is responsible for obtaining a straight line. Two problems may occur when graphing on a logarithmic mantissa: a) there are not enough cycles to incorporate all the data b) obtaining the value of the slope is difficult. In this instance the slope is given by: ln [ C p ] 2 – ln [ C p ] 1 y2 – y1 m = --------------- = -------------------------------------------x2 – x1 t2 – t1 Hence, before calculating the value of m, the two selected values of [ Cp ] 1 and [ Cp ] 2 must be converted, using a calculator, to ln [ Cp ] 1 and ln [ C p ] 2 in order to satisfy the equation. The same problem may arise in obtaining the intercept value, b. The two problems may be avoided by plotting the same data on ordinary paper, in which case the vertical axis is labelled “log plasma concentration”. However, in this instance the ordinary values of [ C p ] must be converted to ln [ Cp ] prior to plotting. It is the ln [ Cp ] values which are then plotted. The calculation of the slope is direct in this case, as the values of read from the graph.
y1
and
y2
may be
Hence, one must consider the relative merits of semilogarithmic and ordinary paper before deciding which to use when a log plot is called for. In the case of semilog graphs the slope may be found in a slightly different manner, i.e., taking any convenient point on the line ( y 1 ) we usually take the as the second point, ( y 2 ) one half of ( y 1 ) . Thus, y1 1 - ln ------------------- ln -------- ln y 1 – ln ( ( 1 ⁄ 2 )y 1 ) ( 1 ⁄ 2 )y 1 1 ⁄ 2 ln 2 - = 0.693 = ------------------------------= --------------------- = ------------------------m = ----------------------------------------------t 1 – t2 t1 – t2 –t1 ⁄ 2 t 1 – t2 t 1 – t2 Basic Pharmacokinetics
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(in which case, t 2 – t 1 is called the half-life t½ ). Since because
2.4.12
0.693 m = ------------- = – k –t1 ⁄ 2
and
t1 < t2 ,
then
t1 – t2 = –t1 ⁄ 2
0.693 k = ------------- . t1 ⁄ 2
LOG - LOG COORDINATES a
Functions of the type y = Bx give straight lines when plotted with logarithms along both axis. i.e., equation in logarithmic form is: log y = log B + a log x or log y = a log x + log b which is in the form y = mx + b This is directly applicable to parabolic and hyperbolic equations previously discussed (see Figure 1-5).
2.4.13
PITFALLS OF GRAPHING: POOR TECHNIQUE The utility of these procedures requires proper graphing techniques. The picture that we draw can cause formation of conceptualizations and correlations of the data that are inconsistent with the real world based simply on a bad picture. Consequently the picture must be properly executed. The most common error is improper axes labelling. On a single axis of rectilinear coordinate paper (standard graph paper), a similar distance between two points corresponds to a similar difference between 2 numbers. Thus,
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FIGURE 0-11.
Graphing using standard number spacing
40 30 20 10 0 0
FIGURE 0-12.
5
10
15
20
25
30
Nonstandard (incorrect) graph
40 30 20 10 1 0 2
0
5
10
20
30
Obviously, the distance (Time) on the graph 12 between 0 and 2 hours should not be the same as the distance between 10 and 20 hours. It is, and therefore Figure 012 is wrong. Similarly, the use of similar paper may result in some confusion. With logarithms the mantissa for any string of numbers, differing only by decimal point placement, is the same. What differentiates one number from another, in this case, is the characteristic. Thus, TABLE 2-13 Logarithmic
graphing
Number
Mantissa
Characteristic
Log
234
.3692
2
2.3692
23.4
.3692
1
1.3692
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TABLE 2-13 Logarithmic
graphing
Number
Mantissa
Characteristic
Log
2.34
.3692
0
0.3692
0.234
.3692
-1
1.3692
The paper automatically determines the relationship between strings of numbers (mantissa) by the logarithmic differences between the numbers on the axis within a cycle. The student must determine the order of magnitude (characteristic) to be relegated to each cycle. FIGURE 0-13.
Logarithmic mantissa Logarithmic Plot 103 234
102
Y axis (units)
23.4
101 2.34
100 0.234 10-1 1.0
1.5
2.0
2.5
3.0
3.5
4.0
X axis (units)
Thus, we see, in Figure 0-13, the cycle on the semilog paper to relate to orders of magnitude (e.g., 1, 10, 100, 1000, etc.) and consequently the characteristic of the exponent. The third common problem is labelling the log axis as log “y”. This is improper. It is obvious from the spacing on the paper that this function is logarithmic, and thus the axis is simply labelled “y”. There are almost as many different errors as there are students and it is impossible to list them all. These few examples should alert you to possible problems.
2.4.14
GRAPHICAL ANALYSIS We will look at several different types of plots of data:
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FIGURE 0-14.
0
Straight line going down on semi-log paper
1
2
3
4
5
Find the slope by taking any two values on the Y axis such that the smaller value is one half of the larger. The time that it takes to go from the larger to the smaller is the half-life. Dividing 0.693 by the half-life yields the rate constant. Extrapolating the line back to t = 0 yields the intercept.
FIGURE 0-15.
0
Curved line which plateaus on semi-log paper.
2
4
6
8
10
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FIGURE 0-16.
0
Curved line which goes up and then straight down on semi-log paper.
5
10
15
20
25
Find the terminal slope by taking any two values on the Y axis such that the smaller value is one half of the larger. The time that it takes to go from the larger to the smaller is the half-life. Dividing 0.693 by the half-life yields the rate constant. Plot type one is reasonably easily evaluated. There are 2 important things that can be obtained: Slope and Intercept. However, the slope and intercept have different meanings dependent on the data set type plotted. The slope is the summation of all the ways that the drug is eliminated, -K.
TABLE 2-14 Plot
type 1
Data Type
Y axis
X axis
Slope
Intercept
IV Bolus Parent
Drug Conc. parent compound
Time
-K
dose C p0 = ----------Vd
IV Bolus Parent
dXu ---------- urine rate of excretion dt parent compound
Time (mid)
-K
Kr ⋅ X 0
IV Bolus Parent
Xu ∞ – Xu Cumulative urine
Time
-K
kr -------------K ⋅ X0
data
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Plot type two is not usually evaluated in its present form as only the plateau value can be obtained easily. But again it has different meanings dependent on the data plotted. TABLE 2-15 Plot
Type 2 Y axis
X axis
Plateau Value
IV Bolus Parent
Data Type
Xu Cumulative urine data parent compound
Time
kr Xu ∞ = -------------K ⋅ X0
IV Infusion Parent
Drug concentration parent compound
Time
Q - = ---Q ( C p )ss = ----------K⋅V cl
Usually urine data of this type (parent compound - IV bolus) is replotted and evaluated as plot 1 (above). Infusion data can be replotted using the same techniques, but usually is not. Plot type 3 must be stripped of the second rate constant from the early time points, thus: There are 3 things that can be obtained from the plot: the terminal slope (the smaller rate constant), the slope of the stripped line (the larger rate constant) and the intercept. The rate constants obtained from a caternary chain (drug moving from one box to another in sequence in compartmental modeling) are the summation of all the ways that the drug is eliminated from the previous compartment and all the ways the drug is eliminated from the compartment under consideration. See LaPlace Transforms for further discussion. Again, dependent on the data set type being plotted they will have different values.
TABLE 2-16 Plot
Data Type
Type 3 Y axis
X axis
S1
S2
Intercept
IV Bolus Parent
Metabolite conc.
Time
-Ksmall
-Klarge
km ⋅ X 0 ------------------------------------------------------( K l arg e – K small ) ⋅ V dm
IV Bolus Parent
dXmu --------------- excretion dt rate of metabolite into urine
Time (mid)
-Ksmall
-Klarge
k mu ⋅ k m ⋅ X 0 -----------------------------------K l arg e – K small
Oral
Drug conc.
Time
-Ksmall
-Klarge
k a ⋅ fX 0 --------------------------------------------------( K l arg e – K small ) ⋅ V d
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2.5 Pharmacokinetic Modeling It has been observed that, after the administration of a drug, the concentration of the drug in the body appear to be able to be described by exponential equations. Thus, it appears that, even though the processes by which the drug is absorbed. distributed, metabolized and excreted (ADME) may be very complex, the kinetics (math) which mimics these processes is made up of relatively simple first order processes and is called first order pharmacokinetics. A second observation is that the resulting concentration is proportional to dose. When this is true, the kinetics is called linear. When this math is applied to the safe and effective therapeutic management of an individual patient, it is called clinical pharmacokinetics. Thus, in clinical pharmacokinetics, we monitor plasma concentrations of drugs and suggest dosage regimens which will keep the concentration of drug within the desired therapeutic range. Pharmacodynamics refers to the relationship between the drug concentration at the receptor and the intensity of pharmacological (or toxicological) response. It is important to realize that we want to control the pharmacological response. We do that indirectly by controlling the plasma concentration. In order for this to work, we assume kinetic homogeneity, which is that there is a predictable relationship between drug concentration in the plasma (which we can measure) and drug concentration at the receptor site (which we can not measure). This assumption is the basis for all clinical therapeutics. Models are simply mathematical constructs (pictures) which seem to explain the relationship of concentration with time (equations) when drugs are given to a person (or an animal). These models are useful to predict the time course of drugs in the body and to allow us to maintain drug concentration in the therapeutic range (optimize therapy). The simplest model is the one used to explain the observations. We model to summarize data, to predict what would happen to the patient given a dosage regimen, to conceptualize what might be happening in disease states and to compare products. In every case, the observations come first and the explanation next. Given that a data set fits a model, the model can be used to answer several different types of questions about the drug and how the patient handles the drug (its disposition), for example: if the drug were to be given by an oral dose, how much is absorbed and how fast? Are there things which might affect the absorption, such as food or excipients in the dosage form itself. What would happen if the drug were to be given on a multiple dose regimen? What if we increased the dose? etc. You should be able to: • be facile in the use of the equations. You should be able to graphically manipulate data sets and extract pharmacokinetic parameters, applying the appropriate equations or variations of them.
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• define all new words used in this section. e g.: Succinctly define, stating rigorously the meaning of any symbols used and the dimensions of measurement.
• compare and contrast new concepts used in this section. e. g.: rate and rate constant, zero and first order kinetics, bolus and infusion methods, excretion and elimination, the assumptions made in pharmacokinetic models with physiological reality. Why can these assumptions be made?
• pictorially represent any two variables (graph) one vs. the other. e.g. for each of the following pairs of variables (ordinate against abscissa), draw a graph illustrating the qualitative profile of their relationship. Where appropriate, indicate the nature of important slopes, intercepts, and values. Unless you specifically indicate on your plot that semi-log paper is being considered (write “S-L”), it will be assumed that rectilinear paper is being considered. Graphs are for a drug given by IV Bolus where applicable.
2.5.1
MAKING A MODEL
X
ka
The differential equations used result from the model which is our conceptualization of what is happening to the drug in the body.
The box (compartment) is the area of interest. We want to find out how the mass of drug, X , changes with time in that compartment, the rate, and how the rates change with time, the differential equations. How do we make a differential equation?
The picture that we build is made up of building blocks, consisting of the arrow and what the arrow touches. The arrow demonstrates how quickly the mass of drug, X , declines. The arrow times the box that the arrow touches = the rate. Rates can go in, i.e. arrows pointing to a box mean drug is going in (+ rate). Rates can go out, i.e. arrows pointing away means drug is going out (- rate). Rate = rate constant (arrow) times mass of drug (box). So the arrow and box really is a pictorial representation of a rate where the rate is the rate constant on the top of the arrow times what the tail of the arrow touches. Again, the rate constant, k , tells the magnitude of the rate,
k⋅X.
Consider the following simple chain:
X1
k12
X2
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k23
X3
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The building blocks are k 12 ⋅ X1 and k 23 ⋅ X2 . Every arrow that touches the compartment of interest becomes part of the differential equation. If the arrow goes to the box, it’s positive; if it goes away from the box, it’s negative. To find dX 1 ⁄ dt (the rate of change of X 1 with time), we simply add up all of the rates which affect X1 (all of the arrows that touch X1 ) dX 1 --------- = – k 12 ⋅ X 1 dt and thus: dX 2 --------- = k 12 ⋅ X 1 – k 23 ⋅ X2 dt dX 3 --------- = k 23 ⋅ X 2 dt (Note: the first subscript of the rate constant and the subscript of the box from which it originates are the same.) You should be able to develop the series of interdependent differential equations which would result from any model. The integration of those equations by use of the Laplace Table is done by transforming each piece of the equation into the Laplace domain (looking it up on the table and substituting). The algebra performed solves for the time dependent variable: put everything except the variable (including the operator, s) on the right side and put the variable on the left. Find the resulting relationship on the left side of the table. The corresponding equation on the right side of the table in the integrated form. You should be able to integrate any differential equation developed from any model (within reason) that we can conceptualize. (Note: Each subsequent variable is dependent on the ones that precedes it. In fact, the solutions to the preceding variables are substituted into the differential to remove all but one of the time dependent functions - the one that we are currently attempting to solve.)
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2.5.2
ONE COMPARTMENT OPEN MODEL A simplified picture (mathematical construct) of the way the body handles drug is one where the body can be conceived to be a rapidly stirred beaker of water (a single compartment). We put the drug in and the rate at which the drug goes away is proportional to how much is present (first order). Thus the assumptions are: • • • •
Body homogeneous (one compartment) Distribution instantaneous Concentration proportional to dose (linear) Rate of elimination proportional to how much is there. (First order)
It is important to note that we know some of these assumptions are not true. It is of little consequence, as the data acts as if these were true for many drugs. The visual image which is useful is one of a single box and a single arrow going out of the box depicting one compartment with linear kinetics. The dose is placed in the box and is eliminated by first order processes. In many cases, more complicated models (more boxes) are necessary to mathematically mimic the observed plasma versus time profile when one or more of these assumptions are not accurate. For example, the two compartment (or multi-compartment) model results when the body is assumed to not be homogeneous and distribution is not instantaneous.
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2.6 The LaPlace Transform Why do we need to know the LaPlace transform?
One of the important facets of biopharmaceutics is a familiarity with the principles of pharmacokinetics. This latter discipline describes the study of the dynamic processes by which the body “handles” or “disposes of” an administered drug. These processes (absorption, distribution, metabolism, and excretion (ADME) are dynamic in that they represent the time-dependent changes occurring to the drug. Thus, in pharmacokinetics the time course of these changes, which overall describe the fate of the administered drug, is described mathematically. If the mathematical principles are understood, it is then possible to use pharmacokinetics in clinical practice, such as the design of rational dosage regimens (T.S. Foster and D.U.A. Bourne, Amer. J. Hosp. Pharm., 34, 70-75 (1977). Understanding (Bloom’s level 5) is not simply memorizing (Bloom’s level 1) nor calculating using a memorized equation (Bloom’s level 3). The authors believe that the proper conceptualizing of the process and the subsequent derivation of the appropriate equations will lead to an understanding of the mathematical principles, and thus, a better, more optimal dosing regimen. Since a mathematical description of the time-dependent ADME processes is required, it becomes necessary to deal with their corresponding rate equations. Inevitably this will involve calculus (mainly integral calculus). However, the LaPlace Transform provides a method whereby calculus can be performed with minimal trauma. If a conscientious effort to learn the method is made and applied, a potentially serious obstacle (the fear of calculus) to the understanding and appreciation of biopharmaceutics will be removed. Indeed, many students will find they no longer fear integration and are thus free to comprehend the principles underlying pharmacokinetics, which, after all, is the primary aim. So, the LaPlace Transform is a tool which is of great assistance in pharmacokinetics; its utility and importance should not be lightly disregarded.
The LaPlace Transform: What Is It?
There is, of course, a theoretical background to the LaPlace Transform. However, it can be used without recourse to a complete theoretical discussion, though appropriate pharmaceutical use of the method is found in the following references: M. Mayersohn and M. Gibaldi, Amer. J. Pharm. Ed., 34, 608-614 (1970). M. Gibaldi and D. Perrier, “Pharmacokinetics”, Marcel Dekker, pp. 267-272 (1975). Basically, the LaPlace Transform is used to solve (integrate) ordinary, linear differential equations. In pharmacokinetics such equations are zero and first-order rate equations in which the independent variable is time. For instance, if a differential equation describing the rate of change of the mass of drug in the body with time is
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integrated, the final equation will describe the mass of drug actually in the body at any time. The procedure used is to replace the Independent variable (time) by a function containing the LaPlace Operator, whose symbol is “s”. In doing so we have replaced the time domain by a complex domain. This is analogous to replacing a number by its logarithm. Once in the complex domain, the transformed function may be manipulated by regular algebraic methods. Then the final expression in the complex domain is replaced by its equivalent in the time domain, yielding the integrated equation. This ultimate process is analogous to taking an antilogarithm.
2.6.1
TABLE OF LAPLACE TRANSFORMS
A table of useful LaPlace transforms is given in Section 2.7. Page 2-56.
The replacement of expressions in one domain by their equivalents in another is accomplished by reference to tables. One column shows time domain expressions, stated as f ( t ) , and second column shows the corresponding complex domain expressions, stated as the LaPlace Transform. Note that “ f ( t ) ” simply means – at
“some function of time”. For example, when f ( t ) is Be , then the LaPlace Transform is B ⁄ ( s + a ) , where “B” is a constant and “a” is a rate constant For example, when the LaPlace Transform is
2.6.2
A⁄s
2
, then f ( t ) is At .
SYMBOLISM For simplicity in writing transformed rate expressions (and to distinguish them from untransformed (time domain) expressions), the following symbolism will be employed: “a bar will be placed over the dependent variable which is being transformed”. Example: If X is the mass of unchanged drug in the body at any time, then X is the LaPlace Transform of this mass.
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2.6.3
CONVENTIONS USED IN DRAWING PHARMACOKINETIC SCHEMA. When drugs enter the body, they will encounter many different fates. It is important to set up the possible fates of the drug by creating a well thought out flow chart or scheme in order to follow all the events that are occurring in the body as described by the pharmacokinetic description of the drug. For example, a drug may be excreted unchanged or may undergo hepatic metabolism to yield active or inactive metabolites. All of these components are part of pharmacokinetics, which by definition, includes ADME (the Absorption, Distribution, Metabolism and Excretion of drugs), and must be considered. This flow chart becomes the backbone or the framework upon which to build the equations which describe the pharmacokinetics of the drug. The differential equations result as a direct consequence of the flow chart. Using Laplace transforms, the integration of these differential equations are simplified and provide the pharmacokineticist to (easily?) keep track of all of the variables in the equation. If the drug scheme or flow chart is set up incorrectly, this would have a definite negative impact or the expected equations (as well as the answers and your grade). Below are two examples of how to construct a flow chart. Note that not all drugs follow the same flow chart and it is quite possible that you will need only to use a portion of these examples when construction your own. In general, schema are relatively consistent in the placement of the compartments in relationship to one another. You might consider, for example a drug, given by IV bolus, which is metabolized and both the metabolite and the parent compound are excreted unchanged as shown below: Feces
Body
Urine
Dose kf
Parent Compound Xf
ku X
Xu
km kmf Metabolite
Xmf
kmu Xm
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Using the pharmacokinetic symbolism from chapter one, the compartments are named and placed: metabolites below (or above the plane of the parent compound): compounds going into the urine, to the right; and compounds going into the feces, to the left of the compounds in the body. The rate constants connecting the compartments also follow the symbolism from chapter one. In the above flow chart, K1, the summation of all the ways that X is removed from the body, is ku + kf + km while K2, the summation of all the ways that Xm is removed from the body, is kmu +kmf. Only those compartments are used which correspond to the drug’s pharmacokinetic description, thus when a drug is given by IV bolus and is 100% metabolized with the metabolite being 100% excreted into the urine the model would look like this: Dose X km kmu Xm
Xmu
Thus in this flow chart, K1, the summation of all the ways that X is removed from the body, is km while K2, the summation of all the ways that Xm is removed from the body, is kmu.
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Drugs sometimes are metabolized to two (or more) different metabolites. In the first case, the drug is metabolized by two separate pathways resulting in this flow chart: Xmf1
kmf1
Xm1
Dose
Xf
kf
kmu1
Xmu1
km1 ku
X
Xu
km2 Xmf2
kmf2
kmu2
Xm2
Xmu2
In this flow chart, K1, the summation of all the ways that X is removed from the body, is ku + kf + km1 + km2 while K2, the summation of all the ways that Xm1 is removed from the body, is kmu1 +kmf1 and K3, the summation of all the ways that Xm2 is removed from the body, is kmu2 + kmf2. While in a second case, the drug is metabolized and the metabolite is further metabolized resulting in this flow chart: Dose Xf
kf
ku
X
Xu
km1 Xmf1
kmf1
Xm1
kmu1
Xmu1
km2 Xmf2
kmf2
Xm2
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kmu2
Xmu2
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In this flow chart, K1, the summation of all the ways that X is removed from the body, is ku + kf + km1 while K2, the summation of all the ways that Xm1 is removed from the body, is kmu1 +kmf1+ km2 and K3, the summation of all the ways that Xm2 is removed from the body is kmf2 + kmu2. Both of these flow charts result in very different end equations, so it is imperative that the flow charts accurately reflect the fate of the drug.
2.6.4
STEPS FOR INTEGRATION USING THE LAPLACE TRANSFORM • Draw the model, connect the boxes with the arrows depicting where the drug goes. • The building blocks of the differential rate equations are the arrows and what the tail touches. • Write the differential rate equation for the box in question. The box is on the left side of the equal sign and the building blocks are on the other. If the arrow goes away from the box, the building block is negative, if it is going towards the box, the building block is positive.
• Take the LaPlace Transform of each side of the differential rate equation, using the table where necessary.
• Algebraically manipulate the transformed equation until an equation having only one transformed dependent variable on the left-hand side is obtained.
• Convert the transformed expression back to the time domain, using the table where necessary to yield the Integrated equation.
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2.6.5
EXAMPLE INTEGRATION USING THE LAPLACE TRANSFORM Following an intravenous injection of a drug (bolus dose), its excretion may be represented by the following pharmacokinetic scheme:
(Scheme I)
X
ku
Xu
Where X is the mass of unchanged drug in the body at any time. X u is the cumulative mass of unchanged drug in the urine up to any time, and k u is the apparent first-order rate constant for excretion of unchanged drug. Consider how the body excretes a drug a. The building block is the arrow and what it touches. This first box (compartment) of interest is [ X ] . The arrow ( k u ) is going out, therefore, the rate is going out and is negative, thus dX ------- = –k u X dt
(EQ 1-18)
The negative sign indicates loss from the body. Taking the LaPlace Transform of each side of equation 1-18: sX – X0 = – k u X
(EQ 1-19)
Note that because the independent variable (time) did not appear on the right-hand side of equation 1-18, neither did the LaPlace Operator, s, appear there in equation 1-19. All that was necessary was to transform the dependent variable ( X ) into X . Hence, the table was only required for transforming the left-hand side of equation 1-19. Manipulating the transformed equation: 1. Get only one variable which changes with time 2. Get
X
(X)
on the left and everything else on the right.
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sX + k u X = X0 X ( s + k u ) = X0 X0 X = ------------s + ku LetX 0 = A
Letk u = a
(EQ 1-20)
A X = ---------------(s + a)
(EQ 1-21)
Note that X is the only transformed dependent variable and is on the left-hand side of equation 1-20. Converting back to the time domain:
X = X0e
–ku t
(EQ 1-22)
A Note that the right-hand side of equation 1-22 was analogous to ---------------- in the (s + a) table, because X0 is a constant (the initial dose administered). The left-hand side of equation 1-22 could be converted back without the table. The final expression is the familiar first-order integrated expression.
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2.6.6
SECOND EXAMPLE INTEGRATION USING THE LAPLACE TRANSFORM Look at Scheme I again. Consider how the drug goes from the body into the urine. The next box of interest is Xu . The arrow is coming in, therefore the rate is coming in and is positive. thus, dX u --------- = k u X dt
(EQ 1-23)
(b) Taking the LaPlace Transform of each side of equation 1-23: sX u – ( X u ) 0 = k u X
(EQ 1-24)
But, at zero time, the cumulative mass of unchanged drug in the urine was zero: that is ( Xu ) 0 = 0 . sX u = k u X
(EQ 1-25)
(c) Manipulating the transformed equation: ku X Xu = -------s
(EQ 1-26)
Note that there are two transformed dependent variables. One of them ( X ) can be replaced by reference to equation 1-20. ku X0 Xu = -------------------s(s + ku) Let ( k u X0 ) = A
Let ( k u ) = a
(EQ 1-27)
A X = -----------------s(s + a)
(EQ 1-28)
(d) Converting back to the time domain: –k t
kuX0 ⋅ ( 1 – e u ) X u = ---------------------------------------ku
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Where k u X 0 and k u are analogous to “A” and “a” respectively in the table. Simplifying, Xu = X0 ( 1 – e
2.6.7
– kut
)
(EQ 1-29)
THIRD EXAMPLE INTEGRATION USING THE LAPLACE TRANSFORM During the intravenous infusion of a drug, its excretion may be represented by the following pharmacokinetic scheme: (Scheme II)
Infusion
Q
ku
X
Xu
Where Q is the zero-order infusion rate constant (the drug is entering the body at a constant rate and the rate of change of the mass of drug in the body is governed by the drug entering the body by infusion and the drug leaving the body by excretion). The drug entering the body does so at a constant (zero-order) rate. dX ------- = Q – k u X dt
(EQ 1-30)
(b) Taking the LaPlace Transform of each side of equation 1-30: Q sX – X 0 = ---- – k u X s Note that because Q is a rate, and is therefore a function of the independent variable (time), its transformation yields the LaPlace Operator. In this case, Q was analogous to “A” in the table. But, at zero time, the mass of unchanged drug in the body was zero: that is, X0 = 0 Q- – k X sX = --u s
(EQ 1-31)
(c) Manipulating the transformed equation: Q X = -------------------s ( s + ku )
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(EQ 1-32)
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Let ( Q = A )
A X = ------------------s( s + a)
Let ( k u = a )
(EQ 1-33)
(d) Converting back to the time domain: –kut
Q(1 – e ) X = ----------------------------ku
2.6.8
(EQ 1-34)
CONCLUSIONS The final integrations (Eqs. 24, 24, and 28) are not the ultimate goal of pharmacokinetics. From them come the concepts of: 1.
(a) elimination half-life
1.
(b) apparent volume of drug distribution
2.
(c) plateau drug concentrations
These, and other concepts arising from still other equations, are clinically useful. Once the method of LaPlace Transforms is mastered, it becomes easy to derive equations given only the required pharmacokinetic scheme. Under such circumstances, it no longer becomes necessary to remember a multitude of equations, many of which, though very similar, differ markedly in perhaps one minute detail. As with any new technique, practice is required for its mastery. In this case, mastery will banish the “calculus blues.” It is also possible to see certain patterns which begin to emerge from the derivation of the equations. For example, for a drug given by IV bolus the equation is monoexponential, with the exponent being K1, summation of all the ways that the drug is removed form the body. A graph of the data (Cp v T on semi-log paper) results in a straight line the slope of which is K1, always. If the drug is entirely metabolized K1 = km. If the drug is entirely excreted unchanged into the urine, K1 = ku. If the drug is metabolized and excreted unchanged into the urine, K1 = km +ku. thus K1 can have different meanings for different drugs, depending on how the body removes the drug. Following the drug given by IV bolus a second example of a pattern would be that of the data of the metabolite of the drug. From the LaPlace, the equation for the plasma concentration of the metabolite of the drug has in it K1 and K2, the summation of all the ways that the metabolite is removed from the body, always. K2 would have different meanings depending on how the metabolite is removed from the body.
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After several years teaching, I was fortunate to have a resident rotate through our pharmacokinetic site. She had come with a strong Pharmacokinetcs background and during our initial meeting, she had told me that she had a copy of John Wagner’s new textbook on pharmacokinetcs. She was excited that, finally, there was a compilation of all the equations used in pharmacokinetics in one place. “There are over 500 equations in the new book and I know every one,” she said. “I’m not sure which one goes with which situation, though.” OOPS! Throughout this text and on each exam, each equation is derived from first principles using scientific method, modeling and LaPlace Transforms in the hopes that memorization will be minimized and thought (and consequently proper interpretation) would be maximized.
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2.6.9
TABLE OF LAPLACE TRANSFORMS
TABLE 2-17 Table
A, B
are constants
a, b, c
are rate constants ( a ≠ b ≠ c )
s
is the LaPlace Operator
x
is a variable, dependent on time ( t )
m
is a power constant
of LaPlace Transforms Time Function, F ( t )
LaPlace Transform, f ( s )
A
A --s
At
A ---2 s
m
A ( m! -) -------------m+1 s
– at
A ----------s+a
m – at
A -------------------------m+1 (s + a)
At Ae
At e
– at
A ------------------s (s + a )
A(1 – e ) -------------------------a – at
A( 1 – e ) At ----- – -------------------------2 a a
A --------------------2 s (s + a)
– at
As – B-----------------s (s + a )
Ae
– at
B( 1 – e ) – -------------------------a – at
– e - – Bt A + B --- 1-------------------- a a a
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As – B --------------------2 s (s + a)
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TABLE 2-17 Table
of LaPlace Transforms Time Function, F ( t ) – bt
LaPlace Transform, f ( s )
– at
A -------------------------------(s + a)(s + b)
A( e – e ) --------------------------------(a – b) A – e - – 1----------------– e - 1------------------------------- (a – b) b a
A -----------------------------------s( s + a )( s + b )
1 – e – bt 1 – e – at At A ------ – ---------------- ------------------ – ------------------ ab ( a – b ) b2 a2
A -------------------------------------2 s ( s + a) ( s + b)
1 [ ( B + Aa )e – at – ( B + Ab )e – bt ] ---------------( a – b)
As – B -------------------------------(s + a)(s + b)
( 1 – e – bt ) ( 1 – e – at ) – bt – at 1 ---------------- A ( e – e ) – B ---------------------- – ---------------------- (a – b) b a
As – B -----------------------------------s( s + a )( s + b )
( 1 – e – bt ) ( 1 – e – at ) 1 B B Bt ---------------A + -----------------------– A + -- ---------------------- – -----(a – b) b b a a ab
As – B -------------------------------------2 s ( s + a) ( s + b)
– bt
– ct
– at
– bt
– at
e e e A -------------------------------- + --------------------------------+ --------------------------------(a – c)(b – c) (a – b )(c – b) (b – a )(c – a) – ct
– bt
– at
(1 – e ) (1 – e ) (1 – e ) A ------------------------------------ + ------------------------------------- + ------------------------------------c( a – c )( b – c ) b( a – b)( c – b ) a( b – a )( c – a)
A ------------------------------------------------(s + a)(s + b)(s + c) A ---------------------------------------------------s( s + a )( s + b )( s + c )
(1 – e ) (1 – e ) (1 – e ) At- – A -----------------------------------------+ --------------------------------------- + --------------------------------------2 2 2 bc c ( a – c) ( b – c ) b ( a – b) ( c – b) a ( b – a )( c – a)
A ------------------------------------------------------2 s ( s + a) ( s + b ) ( s + c )
dX ------dt
sX – X 0
– ct
– bt
– at
(m + 1 )
A ----m s
At ------------------(m + 1)
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2.6.10
LAPLACE TRANSFORM PROBLEMS By means of the LaPlace Transform, find the equation for: 1.
The amount of drug in the body when the drug is given by IV Bolus (assume no metabolism).
2.
The amount of drug in the urine when the drug is given by IV Bolus (assume no metabolism).
3.
The amount of metabolite in the body when the drug is given by IV Bolus (assume no parent drug excretion)
4.
The amount of metabolite of the drug in the urine when the drug is given by IV Bolus (assume no parent drug excretion)
5.
The amount of metabolite of the drug in the urine when the drug is given by IV Bolus (assume both parent drug and metabolite excretion)
6.
The amount of drug in the body when the drug is given by IV infusion (assume no metabolism).
7.
The amount of drug in the urine when the drug is given by IV infusion (assume no metabolism).
8.
The amount of metabolite in the body when the drug is given by IV infusion (assume no parent drug excretion).
9.
The amount of metabolite in the urine when the drug is given by IV infusion (assume no parent drug excretion).
10.
The Rate of excretion of the metabolite into the urine for a drug given by IV bolus when km+ku=kmu.
11.
The amount of the principle metabolite (Xm1) when the drug is eliminated by several pathways (Xu, Xm1,Xm2,Xm3,etc)
12.
X- , in the body when the drug is given orally by a delivery system The concentration of drug, ----Vd
which is zero order. What is the concentration at equilibrium ( T ∞ ). 13.
The amount of metabolite of a drug in the body when the drug is given by IV Bolus and concomitant IV infusion.
14.
Disopyramide (D) is a cardiac antiarrythmic drug indicated for the suppression and prevention of ectopic premature ventricular arrythmias and ventricular tachycardia. It appears that disopyramide is metabolized by a single pathway to mono-dealkylated disopyramide (MND). In a recent study, the pharmacokinetics of disopyramide were attempted to be elucidated by means of a radioactive tracer. Since both D and MND would be labeled by the tracer, any equations showing the time course of the label would show both the D and MND. By means of the laPlace transform, find the equation for the rate of appearance of tracer into the urine if the drug were given by IV Bolus.
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2.6.11
LAPLACE TRANSFORM SOLUTIONS
1. The amount of drug in the body when the drug is given by IV bolus (assume no metabolism).
Dose X
ku
Xu
X = Xo At time zero, all of the IV bolus is in the compartment. Here K1 = ku
dX ------- = – k u X dt sX – X0 = –kuX sX + kuX = X o X ( s + k u )= Xo Xo X = ------------s + ku – kut A Let ( X0 ) = A , k u = a ,X = ---------------X = Xo e (s + a)
2. the amount of a drug in the urine when the drug is given by IV bolus (assume the drug is NOT metabolized
Dose X
ku
Xu
Again K1 = ku
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dX --------u- = k u X dt
sX – Xo = – k m X sX + k m X = Xo
sX u – Xuo = k u X
X ( s + k m ) = Xo
sX u = k u X
Xo X = -------------s + km
X o ku sX u = -----------------(s + ku)
A Let ( X 0 ) = A , k m = a , X = ---------------(s + a)
Xo ku X u = -------------------s(s + ku)
X = Xo e
A Let ( X0 k u ) = A , k u = a , X u = -----------------s( s + a )
– kmt
dX m ---------- = k m X – k mu X m dt
– kut
ku X o ( 1 – e ) X u = -----------------------------------ku Xu= Xo ( 1 – e
– kut
sX m – Xm0 = k m X – kmu X m km X 0 sX m = -------------– k mu Xm s + km
)
3. the amount of metabolite of a drug in the body when the drug is given by IV bolus (assume no parent drug excretion).
km ⋅ Xo X m = ----------------------------------------( s + k mu ) ( s + k m ) Let ( k m X 0 ) = A , k m = a = K1 , k mu = b = K2 A X m = --------------------------------(s + a)(s + b )
Dose
( km ⋅ X o ) –k t –k t X m = ------------------------⋅ ( e mu – e m ) or ( k m – k mu )
X
( km ⋅ Xo ) – K 2t – K 1t X m = ------------------------⋅ (e –e ) in general terms. ( K1 – K2 )
km Xm
kmu
Xmu
NOTE: We could also:
Let ( k m X 0 ) = A , k m = b , k mu = a Here K1 = km and K2 = kmu
and then
dx ------ = – k m X dt
( km ⋅ X o ) –k t –k t X m = ------------------------⋅ ( e m – e mu ) or ( k mu – k m )
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( km ⋅ Xo ) – K 1t – K2t ⋅ (e –e ) X m = ------------------------( K2 – K1 )
dX mu ------------- = k mu Xm dt
Both of those equations are identical.
sX mu – Xom = k mu X m
4. the amount of metabolite of a drug in the urine when the drug is given by IV bolus (assume the parent compound is not excreted).
Remember at time zero,
X om = 0
Here, again, K1 = km and K2 = kmu
sX mu = k mu X m
Dose
( k mu ) ( k m X o ) sX mu = ----------------------------------------( s + k mu ) ( s + k m ) ( k mu ) ( k m ) ( X o ) X mu = -------------------------------------------s ( s + kmu ) ( s + k m )
X
k mu k m Xo 1 – e – kmt 1 – e – kmut X mu = -------------------- --------------------- – ----------------------- k mu – k m k m k mu
km Xm
kmu
Xmu 5. the amount of metabolite of a drug excreted in the urine when both the parent and metabolite are excreted.
dXm ----------- = k m X – k mu X m dt
Here K1 = km + ku and K2 = kmu
sX m – Xom = k m X – kmu X m
Dose sX m + k mu Xm = k m X X s + km
X
o substitute previously solved X = --------------
Xm
km X o X m = ----------------------------------------( s + k mu ) ( s + k m ) – kmt
– kmut
Xu
kmu
Xmu
km
km Xo X m ( s + k mu ) = -------------s + km
km Xo ( e –e ) X m = -------------------------------------------------( k mu – k m )
ku
dX ------- = – k u X – k m X dt sX – Xo = – k u X – k m X
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sX + k u X + k m X = X o X ( s + ku + km ) = Xo
Q
X
ku
Xu
( k u + k m ) = K1 Xo X = --------------s + K1 X = Xo e
– K1t
dXm ------------ = k m X – k mu Xm = k m X – K2Xm dt sXm – X om = k m X – K2Xm
At time zero, all the drug is still in the IV bag, therefore there is no drug in the body. X0 = 0 Here K1 = ku
dX ------- = Q – k u X dt ---- – K u X sX – X0 = Q s
sX m + K2Xm = k m X
Q sX + k u X = --s
km Xo X m = ----------------------------------------( s + K2 ) ( s + K1 )
Q X = -------------------s ( s + ku )
dX mu ------------ = k mu Xm dt
Q- ( 1 – e –k u t ) – K1t Q X = ---) or X = ------- ( 1 – e ku K1
sX mu – Xom = k mu X m = K2Xm sX mu = K2X m ( k mu ) ( kmX o ) sX mu = ----------------------------------------( s + K2 ) ( s + K1 ) ( k mu ) ( k m ) ( Xo ) X mu = -------------------------------------------s ( s + K2 ) ( s + K1 ) k mu k m X o 1 – e – K1t 1 – e – K2t X mu = --------------------- --------------------- – --------------------K2 – K1 K1 K2
7. the amount of drug in the urine when the drug is given by infusion (assume the drug is NOT metabolized).
Q
X
ku
Xu
Here K1 = ku
dX ------- = Q – k u X dt 6. The amount of drug in the body from a drug given by IV infusion (assume no metabolism).
Q- – k X sX – Xo = --u s QsX + k u X = --s
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QX ⋅ (s + k u ) = --s
dX ------- = Q – k m X dt
Q X = ------------------------s ⋅ (s + ku )
Q- – k X sX – Xo = --m s
dX --------u- = k u X dt
sX + k m X = Q ---s
sX u – Xo = k u X
QX ( s + k m ) = --s
sX u = k u X
Q X = --------------------s ( s + km )
ku Q sX u = ------------------------s ⋅ (s + ku)
– kmt
(1 – e ) X = Q -------------------------km
ku Q X u = --------------------------2 s ⋅ ( s + ku )
dX m ---------- = k m X – k mu X m dt
– e – kut k u Qt X u = ----------– Qk u 1------------------- ku k2
sX m – Xo = k m X – k mu Xm
u
– kut
– K1t
(1 – e ) (1 – e ) X u = Qt – Q ------------------------- or Xu = Qt – Q -------------------------ku K1
8. the amount of metabolite of a drug in the body from a drug given by IV infusion (assume no parent drug excretion) Here K1 = km and K2 = kmu
Q
sX m = k m X – k mu Xm km Q X m = -------------------------------------------s ( s + k mu ) ( s + k m ) A X m = ---------------(a – b)
– bt
– at
– e 1 – e - 1---------------- b - – ----------------a
1 – e – k m t 1 – e – k mut km Q X m = ------------------------ -------------------- – --------------------- or ( k mu – k m ) k m k mu – K1t – K2t kmQ – e - – 1-------------------– e - 1-------------------X m = ------------------------( K2 – K1 ) K1 K2
X km Xm
kmu
Xmu
9. the amount of metabolite of a drug in the urine from a drug given by IV infusion (assume that the parent compound is not excreted). Here K1 = km and K2 = kmu
.
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sX mu – Xomu = k mu X m substitute X m
Q
k mu k m Q X mu = ---------------------------------------------2 s ( s + k m ) ( s + k mu )
X
k mu k m Qt k mu k m Q ( 1 – e –kmut ) ( 1 – e – kmt ) X mu = --------------------- – --------------------- ----------------------------- – -------------------------- 2 2 k m ⋅ k mu k m – k m u k k
km
mu
Xm
kmu
Xmu
or
m
k mu k m Qt k mu k m Q ( 1 – e –K2t ) ( 1 – e – K1t ) X mu = --------------------- – -------------------- -------------------------- – -------------------------- 2 2 K1 ⋅ K2 K1 – K2 K2 K1
dX ------- = Q – k m X dt Q sX – Xo = ---- – k m X s sX + k m X = Q ---s X ( s + km ) = Q ---s Q X = --------------------s ( s + km )
10. the rate of excretion of the metabolite into the urine for a drug given by IV bolus when
k m + k u = k mu In this case, K1 = ku +km and K2 = kmu and thus K1 = K2. This is not normal but could happen. The problem arises when we get to the LaPlace that assumes the rate constants are different (i.e. a ≠ b ) because for this special case a = b .
dX m ---------= k m X – k mu X m dt
Dose
sX m – Xo = k m X – k mu Xm X
sX m + k mu X m = k m X km Q X m ( s + k mu ) = --------------------s ( s + km )
dX mu ------------ = k mu Xm dt
Xu
kmu
Xmu
km Xm
km Q X m = -------------------------------------------s ( s + k m ) ( s + k mu ) k m Q 1 – e – kmut 1 – e – kmt X m = -------------------- ----------------------- – --------------------- k m – k mu k mu k m
ku
dX
. ------- = – k u X – k m X dt
sX – Xo = – k u X – k m X sX + k u X + k m X = X o
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the principal metabolite ( X m 1 ) when the drug is
X ( s + ku + km ) = Xo
11.
Xo X = -----------------------------( s + k u + km )
In this case K1 = km1 + km2 + ku, K2 = kmu1 and K3 = kmu2
cleared by several pathways ( X u, X m 1, X m 2 )
K1 = ( k u + k m ) k mu = K2
Xm1
Xo X = -------------------( s + K1 ) X = Xo e
Dose
kmu1
Xmu1
km1
– K1t
ku
X
Xu
dX m ---------= ( k m X – k mu X m ) dt km2
sX m – Xom = ( k m X – K2X m )
Xm2
sX m + K2X m = k m X km X o X m ( s + K2 ) = -----------s+K km Xo X m = ----------------------------------------( s + K2 ) ( s + K1 )
X m = k m Xo te
Xmu2
dX ------- = – k u X – k m1 X – k m2 X dt (remember- K2 = K1)
km Xo X m = ----------------------------------------( s + K1 ) ( s + K1 ) km Xo X m = ---------------------2 ( s + K1 )
kmu2
sX – Xo = – k u X – k m1 X – k m2 X sX + k u X + k m1 X + k m2 X = X o X ( s + k u + k m1 + k m2 ) = Xo Let K1 = ku + km1 + km2
(kmX0 = A)
– K1t
dX mu ------------ = k mu Xm dt dX mu – K1t ------------ = k mu k m Xo te dt
Xo X = -------------------( s + K1 ) K1 = ( k u + k m1 + k m2 ) and K2 = k mu1 dX m1 ------------- = k m1 X – K2X m1 dt sX m1 – Xm1o = k m1 X – K2X m1 X m1o = 0 k m1 Xo sX m1 + K2X m1 = -------------------( s + K1 )
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k m1 Xo X m1 ( s + K2 ) = -------------------( s + K1 )
Xo= 0 ka Q sX + K1X = -------------------s ( s + ka )
k m1 X o X m1 = ----------------------------------------( s + K1 ) ( s + K2 )
ka Q X ( s + K1 ) = -------------------s ( s + ka )
k m1 Xo – K1t – K2t (e –e ) X m1 = -------------------K2 – K1
ka Q X = -----------------------------------------s ( s + k a ) ( s + K1 ) 12. the concentration of drug X ⁄ Vd in the body when the drug is given orally by a delivery system which is zero order. What is the concentration in the body at equilibrium ( t ∞ ) Here K1 = ku
Q
ka
Xa
X
ku
Xu
k a Q ( 1 – e –K1t ) ( 1 – e – kat ) X = ---------------------- --------------------------- – ------------------------- ka K1 ( k a – K1 ) ka Q ( 1 – e –K1t ) ( 1 – e – kat ) C = ----------------------------- --------------------------- – ------------------------- ( k a – K1 )Vd K1 ka k Q
1 1 a If t= ∞ , then e –kt = 0 , thus C = ----------------------------- ------- – ----- ( k a – K1 )Vd K1
ka
Q simplified yields: C = --------------
dX --------a- = Q – k a X a dt Q sX a – Xao = ---- – k a Xa s
K1Vd
13 the metabolite of a drug in the body Xm given by IV infusion and concomitant IV bolus dose. Infusion: Here K1 = km and K2 = kmu
X a0 = 0
Dose
sX a + k a Xa = ( Q ⁄ s ) X a ( s + ka ) = ( Q ⁄ s ) Q X a = -------------------s(s + ka)
Q
X km
– kat
Q( 1 – e ) X a = ----------------------------ka
Xm
kmu
Xmu
dX ------- = k a X a – K1X dt sX – Xo = k a Xa – k1X
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dX ------- = Q – K1X dt
dX m ---------- = k m X – K2X m dt
sX – Xo = Q ---- – K1X s
sX m – Xo = k m X – K2X m
sX + K1X = Q ---s
sX m + K2X m = k m X
X ( s + K1 ) = Q ---s Q X = ----------------------s ( s + K1 ) dX m ---------= k m X – K2X m dt
km X o X m ( s + K2 ) = --------------s + K1 km Xo X m = ----------------------------------------( s + K1 ) ( s + K2 ) km Xo – K2t – K 1t X m = ------------------------(e –e ) ( K1 – K2 ) Thus,
sX m – Xo = k m X – K2Xm sX m + K2X m = k m X km Q X m ( s + K2 ) = ----------------------s ( s + K1 ) km Q X m = -------------------------------------------s ( s + K1 ) ( s + K2 ) k m Q ( 1 – e – K1t ) ( 1 – e – K2t ) X m = ------------------------- -------------------------- – -------------------------- ( K2 – K1 ) K1 K2
km X o – K2t – K 1t (e –e ) + below X m = ------------------------( K1 – K2 ) k m Q ( 1 – e –K2t ) ( 1 – e –K1t ) -------------------- -------------------------- – -------------------------- K2 K1 K1 – K2 14. By means of the LaPlace transform, find the equation for the rate of appearance of the tracer in the urine if the drug were given by IV bolus. Here K1 = ku + km and K2 = kmu
IV Bolus:
dX ------- = – K1X dt
Dose
sX – Xo = – K1X
X
ku
Xu
kmu
Xmu
sX + K1X = X o km
X ( s + K1 ) = Xo
Xm
Xo X = --------------s + K1
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dX ------- = – k u X – k m X dt sX – Xo = – k u X – k m X sX + k u X + k m X = X o X ( s + ku + km ) = Xo ( k u + k m ) = K1 Xo X = -------------------( s + K1 ) X = Xo e
– K1t
dX – K1t --------u- = k m X = k u ( Xo e ) dt dX m ---------= k m X – K2X m dt sX m – Xo = k m X – k2Xm k m Xo sX m + K2X m = -------------------( s + K1 ) km X o X m ( s + K2 ) = --------------s + K1 km Xo X m = ----------------------------------------( s + K2 ) ( s + K1 ) km Xo – K1t – K2t X m = ------------------------{e –e } ( K2 – K1 ) k mu k m Xo – K1t – K2t dX mu ------------ = k mu Xm = ------------------------{e –e } dt ( K2 – K1 ) k mu k m X o dX u dX mu – K1t { e – K1t – e – K2t } --------- + ------------- = k u ( X o e ) + ------------------------ ( K2 – K1 ) dt dt
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CHAPTER 3
Pharmacological Response
Author: Michael Makoid and John Cobby Reviewer: Phillip Vuchetich
OBJECTIVES After completing this chapter, the student will be able to: 1.
Given patient data of the following types, the student will be able to properly construct (III) a graph and compute (III) the slope using linear regression: response (R) vs. concentration (C), response (R) vs. time (T), concentration (C) vs. time (T)
2.
Given any two of the above data sets, the student will be able to compute (III) the slope of the third by linear regression.
3.
Give response vs. time and response versus concentration data, the student will be able to compute (III) the terminal (elimination) rate constant and half life of the drug.
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3-1
Pharmacological Response
3.1 Pharmacological Response Drug must get into blood and blood is in contact with receptor.
3.1.1
One theory (A.J. Clarke) on the mechanism of action of drugs is the occupation theory. It suggests that the intensity of a pharmacological response (E) is proportional to the concentration of a reversible drug-receptor complex
THE HYPERBOLIC RESPONSE EQUATION A mathematical description of the occupation theory, assuming complete and instantaneous drug distribution, yields [ D ]E max E = ---------------------KR + [D ]
(EQ 1-35)
where E is the intensity of the pharmacological response, Emax is the maximum attainable value of E , [ D ] is the molar concentration of free drug at the active complex and K R is the dissociation constant of the drug-receptor complex.
PKAnalyst Plot 1.0
0.8
0.6
E
If
0.4
E
is plotted against
[D]
a hyperbolic curve will result; the asymptote will be
E max .
0.2
0.0 0.0
0.8
1.6
2.4
3.2
4.0
D
a. If linear pharmacokinetics hold, the molar concentration of free drug at the active site is proportional to the plasma concentration of the drug once equilibrium has been established. Hence, a plot of E against Cp will also be hyperbolic. b. Because the mass of drug in the body is hyperbolic.
X = V ⋅ Cp ,
a plot of
E
against
X
will be
c. For a series of doses the value of X at the same given time after dosing is proportional to the dose (D). Thus, a plot of E against D will also be hyperbolic at a specific time. d. Any hyperbolic curve, if plotted on reverse semilogarithmic paper (i.e., abscissa is logarithmic), has a sigmoid shape. If we plot E against Cp (of X , or D ) in this manner, the plot is virtually linear in the range E ⁄ Emax = 0.2 → 0.8 ; and if this is the clinical range of responses, linear equations may be written. For example,
1.0
Response
0.8 0.6 0.4 0.2 0.0 10 -810-710 -610-510 -410 -310-210 -1
E = m ⋅ ln x + b
Conc.
where
m
is the slope
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3-2
Pharmacological Response
Plot of Response vs. Ln(C) is a straight line in the middle (if you squint), but only between 20% and 80% maximum response
3.1.2
This example equation shows that, in the clinical range, the intensity of a pharmacological response is proportional to the logarithm of the administered dose, providing response is measured at a consistent time after dosing. The proportionality constant (slope, m ) is a function of the affinity of the drug for the receptor. In fact, equation yields a log-dose response plot. Note that doubling the dose does not double the response.
INTERRELATIONSHIPS BETWEEN CONCENTRATION, TIME AND RESPONSE Pharmacological Response (R), Concentration (C), and Time (t) are interrelated. The response and concentration relationship is studied in pharmacology. The concentration and time relationship is studied in pharmacokinetics. The response and time relationship is applied in therapeutics.
Remember: Use only the data between 20% and 80% of maximum response for the straight part of both response vs. Ln(c) and response vs. t.
You should know what the various graphical relationships look like. Response vs. natural log of concentration is sigmoidal. (S shaped). We are interested in the middle almost straight part. The slope is dR ⁄ d ln c . Response vs. time is a straight line. The slope is
dR ⁄ dt .
Natural log of concentration vs. time (drug given by IV bolus) is a straight line. The slope is d ln c ⁄ dt . You should be able to obtain the slope of each of these relationships from data sets. You should be able to obtain the third slope’s relationship given the other two (or data sets with which to get the other two). dR dR d ln c ------- = ----------- ⋅ ----------dt d ln c dt dR dR ⁄ dt ----------- = -------------------d ln c d ln c ⁄ dt
NOTE: Only between 20% and 80% of maximum response!!!!!!
(EQ 1-36)
(EQ 1-37)
d ln c dR ⁄ dt (EQ 1-38) ----------- = ---------------------dt dR ⁄ d ln c You should be able to apply the equation y = mx + b to each of the above relationships. Given the slope (or having obtained the slope) and two of the three variables (y, x, b), you should be able to find the third.
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Pharmacological Response
3.2 Change in Response with Time 3.2.1
ONE-COMPARTMENT OPEN MODEL: INTRAVENOUS BOLUS INJECTION X = X0 e
– Kt
= De
– Kt
(EQ 1-39)
or Ln ( X ) = Ln ( D ) – Kt
(EQ 1-40)
Substituting twice from eq. once at time t and once at zero time E –b = E 0–b ------------------------- – Kt → E = E0 – Rt m m
(EQ 1-41)
Hence a plot of the intensity of the pharmacological response at any time ( E ) against time declines linearly. The slope is – R = ( – K ⋅ m ) and the intercept is E0 (the initial intensity).
3.2.2
ONE-COMPARTMENT OPEN MODEL: ORAL ADMINISTRATION
Response follows plasma profile.
3.2.3
Because E is proportional to ln x at any time, a plot of E against t will be analogous to a plot of ln x against t . Hence E will rise at first and then decline with time. When t is large, the terminal slope will be – R .
DURATION OF EFFECTIVE PHARMACOLOGICAL RESPONSE ( t dur )
Duration of action is related to how long plasma concentration is above Minimum Effective Concentration.
Once equilibrium has been established, there is a minimum plasma concentration below which no pharmacological response is seen; this concentration is ( C p ) eff or MEC . For an intravenous bolus injection, the time to reach ( C p ) eff is t dur . ( C p )eff = ( C p ) 0 e
– Kt dur
multiplying by the volume of distribution we obtain ln ( X eff ) = ln ( D ) – Kt dur
(EQ 1-42)
Rearranging,
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t dur
D ln -------X eff = ------------------K
(EQ 1-43)
The duration of effective pharmacological response is proportional to the (natural) logarithm of the dose. A second rearangement of equation 1-42 results in : ln ( Dose ) ln ( Xeff ) t dur = ----------------------- – -----------------K K
(EQ 1-44)
Thus a plot of duration of action vs ln dose would result in a straight line with a slope of 1/K and an x intercept of
3.2.4
ln ( X eff ) . – ------------------K
PHARMACOKINETIC PARAMETERS FROM RESPONSE DATA
How can I get the elimination rate constant from pharmacological data? Use this “cookbook.”
The measurement of pharmacological effect provides a non-invasive means of obtaining the value of t 1 ⁄ 2 (but not V ).
Remember: Use only the data between 20% and 80% of maximum Response for both of these plots.
b. Find the slope
3.2.5
a. Obtain a log dose-response plot (Eq. 1-37). The response must always be measured at the same time after administering the dose. (m)
of this plot.
c. Obtain a response against time plot for a single dose (Eq. 1-36). d. Find the terminal slope e. Calculate
R K = ---- . m
f. Calculate
0.693 t 1 ⁄ 2 = ------------- K
( –R )
of this plot.
.
“DELAYED” RESPONSE
Two compartment model - biophase is in second compartment.
If a drug does not distribute instantaneously to all the body tissues (including the active site), the pharmacological response will not always parallel the drug concentrations in the plasma. In such a situation the response may parallel the mass of drug presumed to be in a second compartment ( X 2 ) , and hence seem “delayed”. Eventually, however, once equilibrium is attained, the response will parallel plasma concentrations. In such a case, E is proportional to ln X 2 .
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Thus a plot of E against X1 (or E against Cp ) will show a hysteresis loop with time, most noticeably during an intravenous infusion.
3.2.6
RESPONSE OF ACTIVE METABOLITE:
Parent compound (inactive) yields active daughter compound.
In the case of an inactive prodrug yielding an active metabolite, the response curves will mirror the active metabolite plasma profile (assuming the biophase is the plasma) and not the prodrug plasma profile.
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3.3 Therapeutic Drug Monitoring Part of Pharmaceutical Care!
The pharmacokinetics of a drug determine the blood concentration achieved from a prescribed dosing regimen. During multiple drug dosing, the blood concentration will reflect the drug concentration at the receptor site; and it is the receptor site concentration that determines the intensity of the drug’s effect. Therefore, in order to predict a patient’s response to a drug regimen, both the pharmacokinetics and pharmacological response characteristics of the drug must be understood. Pharmacological response is closely related to drug concentration at the site of action. We can measure plasma concentration and assume that the site of action is in rapid equilibrium with the plasma since we usually do not measure drug concentration in the tissue or at the receptor site. This assumption is called “kinetic homogeneity” and is the basis for clinical pharmacokinetics.
Need to keep plasma concentration in the therapeutic range to optimize therapy.
There exists a fundamental relationship between drug pharmacokinetics and pharmacologic response. The relationship between response and ln-concentration is sigmoidal. A threshold concentration of drug must be attained before any response is elicited at all. Therapy is achieved when the desired effect is attained because the required concentration has been reached. That concentration would set the lower limit of utility of the drug, and is called the Minimum Effective Concentration (MEC). Most drugs are not “clean”, that is exhibit only the desired therapeutic response. They may also exhibit undesired side effects, sometimes called toxic effects at a higher, (hopefully a lot higher), concentration. At some concentration, these toxic side effects become become intolerable/and or dangerous to the patient.. That concentration, or one below it, would set the upper limit of utility for the drug and is called the Maximum Therapeutic Concentration or Minimum Toxic Concentration (MTC). Patient studies have generated upper (MTC) and lower (MEC) plasma concentration ranges that are deemed safe and effective in treating specific disease states. These concentrations are known as the “therapeutic range” for the drug (Table 2-18). When digoxin is administered at a fixed dosage to numerous subjects, the blood concentrations achieved vary greatly. Clinically, digoxin concentrations below 0.8 ng ⁄ ml will elicit a subtherapeutic effect. Alternatively, when the digoxin concentration exceeds 2.0 ng ⁄ ml side effects occur (nausea and vomiting, abdominal pain, visual disturbances). Drugs like digoxin possess a narrow therapeutic index because the concentrations that may produce toxic effects are close to those
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Pharmacological Response
required for therapeutic effects. The importance of considering both pharmacokinetics and pharmacodynamics is clear. TABLE 2-18 Average
therapeutic drug concentration
DRUG
RANGE
digoxin
0.8-2.0 ng ⁄ ml
gentamicin
2-10 µg ⁄ ml l
lidocaine
1-4 µg ⁄ ml
lithium
0.4-1.4 mEq ⁄ L
phenytoin
10-20 µg ⁄ ml
phenobarbitol
10-30 µg ⁄ ml
procainamide
4-8 µg ⁄ ml
quinidine
3-6 µg ⁄ ml
theophylline
10-20 µg ⁄ ml
Note that drug concentrations may be expressed by a variety of units. Pharmacokinetic factors that cause variability in plasma drug concentration are: • • • • •
drug-drug interactions patient disease state physiological states such as age, weight, sex drug absorption variation differences in the ability of a patient to metabolize and eliminate the drug
If we were to give an identical dose of drug to a large group of patients and then measure the highest plasma drug concentration we would see that due to individual variability, the resulting plasma drug concentrations differ. This variability can be attributed to factors influencing drug absorption, distribution, metabolism, and excretion. Therefore, drug dosage regimens must take into account any disease altering state or physiological difference in the individual. Therapeutic drug monitoring optimizes a patient’s drug therapy by determining plasma drug concentrations to ensure the rapid and safe drug level in the therapeutic range. Two components make up the process of therapeutic drug monitoring:
• Assays for determination of the drug concentration in plasma • Interpretation and application of the resulting concentration data to develop a safe and effective drug regimen.
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The major potential advantages of therapeutic drug monitoring are the maximization of therapeutic drug benefits and the minimization of toxic drug effects. The formulation of drug therapy regimens by therapeutic drug monitoring involves a process for reaching dosage decisions.
3.3.1
THERAPEUTIC MONITORING: WHY DO WE CARE? The usefulness of a drug’s concentration vs. time profile is based on the observation that for many drugs there is a relationship between plasma concentration and therapeutic response. There is a drug concentration below which the drug is ineffective, the Minimum Effective Concentration (MEC), and above which the drug has untoward effects, the Minimum Toxic Concentration (MTC). That defines the range in which we must attempt to keep the drug concentration (Therapeutic Range). The data in Table 2-18 are population averages. Most people respond to drug concentrations in these ranges. There is always the possibility that the range will be different in an individual patient. For every pharmacokinetic parameter that we measure, there is a population average and a range. This is normal and is called biological variation. People are different. In addition to biological variation there is always error in the laboratory assays that we use to measure the parameters and error in the time we take the sample. Even with these errors, in many cases, the therapy is better when we attempt to monitor the patient’s plasma concentration to optimize therapy than if we don’t. This is called therapeutic monitoring. If done properly, the plasma concentrations are rapidly attained and maintained within the therapeutic range throughout the course of therapy. This is not to say all drugs should be monitored. Some drugs have a such a wide therapeutic range or little to no toxic effects that the concentrations matter very little. Therapeutic monitoring is useful when: • • • •
a correlation exists between response and concentration, the drug has a narrow therapeutic range, the pharmacological response is not easily assessed, and there is a wide inter-subject range in plasma concentrations for a given dose.
In this era of DRGs, where reimbursement is no longer tied to cost, therapeutic monitoring of key drugs can be economically beneficial to an institution. A recent study (DeStache 1990) showed a significant difference with regard to length of stay in the hospital between the patients on gentamicin who were monitored (and their dosage regulated as a consequence) vs. those who were not. With DRGs the hospital was reimbursed a flat fee irrespective of the number of days the patient stayed in the hospital. If the number of days cost less than what the DRG paid, the
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Pharmacological Response
hospital makes money. If the days cost more than the hospital loses money. This study showed that if all patients in the hospital who were on gentamicin were monitored, the hospital would save $4,000,000. That’s right FOUR MILLION per year. I would say that would pay my salary, with a little left over, and that is only one drug! The process of therapeutic monitoring takes effort.
• • • • •
First the MD must order the blood assays. Second, someone (nurse, med tech, you) must take the blood. Someone (lab tech, you) must assay the drug concentration in the blood. You must interpret the data. You must communicate your interpretation and your recommendations for dosage regimen change to the MD. This will allow for informed dosage decisions.
• You must follow through to ensure proper changes have been made. • You must continue the process throughout therapy. Therapeutic drug monitoring, in many cases, will be part of your practice. It can be very rewarding.
Thus, if we have determined the therapeutic range, we could use pharmacokinetics to determine the optimum dosage regimen to maintain the patient’s plasma concentration within that range. Selected References
1. Nagashima, R., O’Reilly, RA., and Levy, G, Kinetics of pharmacologic effects in man: the anticoagulant action of warfarin. Clin. Pharm. Therap, 10 22-35 (1969). 2. Wagner, J.G, Relations between drug concentration and response. J. Mond. Pharm., 4, 279-310 (1971). 3. Gibaldi M. and Levy, G. Dose-dependent decline of pharmacologic effects of drugs with linear pharmacokinetics characteristics. J.Pharm.Sci, 61, 567-569 (1972). 4. Brunner, L., Imhof, P., and Jack, D. Relation between plasma concentrations and cardiovascular effects of oral oxprenolol in man. Europ. J. Clin. Pharmacol., 8, 3-9 (1975). 5. Galeazzi, R.L., Benet, L.Z., and Sheiner, L.B. Relationship between the pharmacokinetics and pharmacodynamics of procainamide. Clin. Pharm. Therap., 20, 67-681 (1976). 6. Joubert, P., et al. Correlation between electrocardiographic changes, serum digoxin, and total body digoxin content. Clin. Pharm. Therap., 20, 676-681 (1976). 7. Amery, A., et al. Relationship between blood level of atenolol and pharmacologic effect. Clin. Pharm. Therap., 21, 691-699 (1977).
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3.4 Problems What to do.---> We want to get pharmacokinetic data (elimination rate constant) from pharmacological response data (Response vs concontration and Response vs time graphs) . Response vs Time Graph 1. Plot Response vs Time on Cartesian (regular) Graph Paper. 2. Use Response data between 20 and 80 percent of maximum (Pick the straight part) to do the linear regression on. (Rule of thumb: Connect first and last data point with a straight line. If all the points fall on one side of the line, its not straight!
1.0
Response
0.8
0.6
3. Find the slope of the straight line, dR ------- , (eyeball the rise over the run or use linear regression as dT
0.4
0.2
0.0 10
10
10 Time
10
10
10
required). Important: you must determine the best fit line through all of the points that you will use. Eyeball method: Get the line as close to the points as possible placing as many points above the line as below the line. Take two points on the line (not data points) to calculate the change in Y over the change in X.
Response vs Ln(Concentration) Graph 1. Turn semi-log paper on its side so that the numbers are on the top.
0
0.6 0.4 0.2 0.0 10 -810-710 -610-510 -410-310 -210 -1
Response
Response
10010
10
0.8
10 10
1
10010
1.0
1010
Conc.
1
Concentration
1
10
0
1
What we are attempting to do is get the logarithm part of the paper on the x axis and have the numbers get bigger as you go from left to right. 2. Plot concentration on the x axis and response on the y. 3. Find the slope of the line plotted this way by the rise over the run method. Run is change in ln(C). If you take any two concentrations such that C2 = 2*C1 then the run is (ln(C2) - ln(C1)). Using rules of logs, when two logs are subtracted, the numbers are devided, thus: = ln(C2/C1). If C2 = 2*C1 then ln(C2/C1) = ln(2) = 0.693. Basic Pharmacokinetics
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Rise = change in Response. Take the difference of the two responses coresponding to the concentrations picked. (R2-R1). R 2 – R1 4. The slope of the line is m = rise -------- = -----------------run
0.693
Ln(Concentration) vs Time Graph (Pharmacokinetic Data) If you have concentration vs Time data: 1. Plot Concentration vs time on semi-log paper (Y axis is concentration this time)
Concentration
10 100
1010
110 0
1
Time
2. Find the slope as before, using semi log paper (Remeber the log is on the Y axis this time, so you find two concentrations such that c2 = 2*c1 and put it in the rise this time. Thus the slop of the 0.693 = 0.693 line is m = rise -------- = -------------------------- = – k run
t2 – t1
–t1 -2
If you have pharmacological response data: 1. Divide the slope of the Response vs Time graph by the slope of the Response vs ln(C) graph: dR ------slope of r vs t dT dln(C) ------------------------------------------ = --------------- = --------------- = m = – k slope of r vs ln(c) dR dT --------------dln(C)
Both methods should be equivalent. Additional problems are available in chapter 14, practice exams.
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Pharmacological Response
Oxpranolol Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 0 - 1)
AHFS 00:00.00 GPI: 0000000000
Brunner et al, Europ. J. Clin. Pharmacol., 8, 3-9 (1975).
In humans, the pharmacological response to oxpranolol (a beta blocker) is a decrease in beats per minute (bpm) compared to placebo during physical exercise. The following approximate mean data is from 7 healthy volunteers: beats per minute (bpm) altered with time (t) after oral administration of three doses (D). TABLE 2-19
Response vs Concentration
Response vs time
BPM
Dose (mg)
BPM
Time (hr)
10
40
17.6
1
13.5
60
13.9
2
16
80
10.2
3
19
120
6.6
4
21
160
TABLE 2-20 Oxpranolol
plasma concentration following 160 mg IV dose
Time (min)
C p ng ------ ml
30
699
60
622
120
413
150
292
240
152
360
60
480
24
1. Calculate the half life ( t 1 ⁄ 2 ) of oxpranolol from the pharmacological response table. 2. Plot plasma concentration data on Cartesian graph paper directly as well as transforming Cp into ln C p . 3. Plot plasma concentration data on semilog paper. Use linear regression to find the rate constant of elimination of oxpranolol. 4. Calculate the half life obtained from the concentration data and compare it with the half life calculation based on the pharmacological response.
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Minoxidil Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 0 - 2)
AHFS 00:00.00 GPI: 0000000000
Shen et al. Clin. Pcol. Ther 17:593-8 (1975)
Minoxidil is a potent antihypertensive which lowers the mean arterial blood pressure (MAP) in certain patients. PROBLEM TABLE 0 - 2. Minoxidil
Initial decrease in MAP ( mmHg )
Dose ( mg )
17
2.5
40
5.0
53
7.5
63
10.0
76
15
PROBLEM TABLE 0 - 2. Minoxidil
25 mg I.V. Bolus yielded: Decrease in MAP ( mmHg )
Time ( hr )
75
20
66
30
56
40
48
50
From the preceding information, determine the following:
dR d ln C
1. Graph and find ------------ (slope of (R)esponse vs. ln(C)oncentration graph).
dR dt
2. Graph and find ------- (slope of (R)esponse vs. (T)ime graph).
dR-----dt 3. Find the ln(C)oncentration vs. (T)ime slope : ------------- : Note that your slope m = – K . If you are having problems dR -----------d ln C understanding this, refer to Sections 2.4.2 -2.4.4. K is the elimination rate constant.
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------------- . 4. Calculate t 1 ⁄ 2 = 0.693 K
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Pharmacological Response
Propranolol Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 0 - 3)
AHFS 00:00.00 GPI: 0000000000
Citation?
Beta blockers can be considered first line drugs of choice in the treatment of hypertension in certain patients. The following data was obtained regarding Propranolol used to treat hypertension in a group of patients. PROBLEM TABLE 0 - 3. Propranolol
Fall in Systolic BP (mmHg)
Cp
20
50
16
40
11
30
5
20
PROBLEM TABLE 0 - 3. Propranolol
I.V. Bolus dose of Propranolol Fall in Systolic BP (mmHg)
Time (hr)
24
1
20
2
19
3
9
6
From the preceding information, determine the following:
dR d ln C
1. Graph and find ------------ (slope of (R)esponse vs. ln(C)oncentration graph).
dR dt
2. Graph and find ------- (slope of (R)esponse vs. (T)ime graph).
dR ------dt 3. Find the ln(C)oncentration vs. (T)ime slope : ------------- : Note that your slope m = – K . If you are having problems dR -----------d ln C understanding this, refer to Sections 2.4.2 -2.4.4.
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Pharmacological Response
------------- . 4. Calculate t 1 ⁄ 2 = 0.693 K
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Pharmacological Response
3.4.1
ANSWERS:
OXPRANOLOL
1. Calculate the half life ( t 1 ⁄ 2 ) of oxpranolol from the pharmacological response table
Oxpranolol
Response (BPM)
22 20 18 16 14 12 10 1
10
3
2
10
10
Dose (mg)
TABLE 3.
X ln(Dose)
Dose
Y Response
X
3.689
40
10
13.61
36.89
4.094
60
13.5
16.76
55.27
4.382
80
16
19.20
70.11
4.787
120
19
22.92
90.96
5.075
160
21
25.75
106.58
ΣY = 79.5
ΣX = 98.25
ΣX = 22.03
2
2
X⋅Y
ΣXY = 359.82
2
( ΣX ) = 485.23
Σ ( x ) ⋅ Σ ( y ) ) – ( n ⋅ Σ ( x ⋅ y ) )m = (-------------------------------------------------------------------2 2 [Σ(x) ] – ( n ⋅ Σ (x ))
)
Σy y = ------ = 15.9 n
)
ΣX X = --------- = 4.41 n
Slope of the line from linear regression. Chapter 2.4.4
22.03 ⋅ 79.5 ) – ( 5 ⋅ 359.82 ) m = (------------------------------------------------------------------- = 7.93 485.32 – ( 5 ⋅ 98.25 )
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dR = 7.93 ----------the slope is equal to the linear regression of the change in response vs. ln concentration. d ln c
OXPRANOLOL 18
Response (BPM)
16 14 12 10 8 6 1.0
1.5
2.0
2.5
3.0
3.5
4.0
Time (hr)
R –R T1 – T 2
16 – 10 1.45 – 3.07
dR dt
1 2 - = --------------------------- = – 3.71 therefore, ------- = – 3.71 . The slope of this plot is m = -----------------
dR ------dt = d----------ln c = – k = –------------3.71 = – 0.4678hr – 1 ----------dR dt 7.93 ----------d ln c
ln 2 = -------------------------0.693 - = 1.48hr t 1 ⁄ 2 = -------half life (89 min). –1 k 0.4678hr
2. Plot plasma concentration data on Cartesian graph paper directly as well as transforming Cp into ln C p .
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Pharmacological Response
Plasma concentration vs. Time
Oxpranolol
800 3
10
Concentration (ng/mL)
640
Concentration (ng/ml)
480
320
160
2
10
1
10
0
0 0
100
200
300
400
100
500
200
300
400
500
Time (min)
Time (min)
3. Plot plasma concentration data on semilog paper. Use linear regression to find the rate constant of elimination of oxpranolol. Using linear regression, as described above, the elimination rate constant is approximately 0.007797 min-1 * (60 min/hr) = 0.4678 hr-1 4. Calculate the half life obtained from the concentration data and compare it with the half life calculation based on the pharmacological response.
0.693 t 1 ⁄ 2 = ------------------- = 90min = 1.5 hrs compared to 1.48 hours (89 min) from the pharmacological response 0.00763 method.
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Pharmacological Response
3.4.2
ANSWERS:
MINOXIDIL
NOTE: Answers will vary depending on whether linear regression is calculated via calculator or using the formula such as observed in Problem 1. Either method can be used. However, if you use the formula, you should be within 10% of the calculated answer. A word of caution: if you choose to do linear regression via calculator make sure you have valid data. This cannot be assured until you have graphed all the data points given. Many a student has incorrectly calculated parameters because he/she falsely assumes that all the points are valid. Blindly choosing data points for linear regression will only lead to error. Every problem in this manual has been derived from actual journal articles and will therefore be “real” data. This real-world data is inexact. dR (slope of (R)esponse vs. ln(C)oncentration graph). 1. Graph and find -----------mHg)
d ln C
R vs Ln(C) 80 60 40 20 0 10
10
10
Dose (mg)
dR = 32.96 -----------d ln C
dR dt
2. Graph and find ------- (slope of (R)esponse vs. (T)ime graph).
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Pharmacological Response
mmH
R vs T 75 70 65 60 55 50 45 20
25
30
35
40
45
50
Time (hr)
dR ------- = –0.91 dt dR-----dt 3. Find the ln(C)oncentration vs. (T)ime slope : ------------- : Note that your slope m = – K . dR -----------d ln C K = 0.028hr
–1
–------------0.91 = – ( 0.028 ) 32.96
4. Calculate t 1 ⁄ 2 = 0.693 ------------- . K
0.693 - = 24.75hr t 1 ⁄ 2 = ----------------------– 1 0.028hr
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Pharmacological Response
3.4.3
ANSWERS:
PROPRANOLOL
dR (slope of (R)esponse vs. ln(C)oncentration graph). 1. Graph and find -----------d ln C
R vs T 20 15 10 5 0 10
10
dR = 16.36 -----------d ln C dR dt
2. Graph and find ------- (slope of (R)esponse vs. (T)ime graph).
R vs T 25 20 15 10 5 0
1
2
3
4
5
6
dR ------- = –2.93 dt
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Pharmacological Response
dR-----dt 3. Find the ln(C)oncentration vs. (T)ime slope : ------------- : Note that your slope m = – K . dR -----------d ln C K = 0.179hr
–1
–------------2.93 = – 0.179 16.36
------------- . 4. Calculate t 1 ⁄ 2 = 0.693 K
0.693 - = 3.87hr t 1 ⁄ 2 = ---------------------- – 1 0.179hr
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CHAPTER 4
I.V. Bolus Dosing
Author: Michael Makoid and John Cobby Reviewer: Phillip Vuchetich
OBJECTIVES For an IV one compartment model plasma and urine: 1.
Given patient drug and/or metabolite concentration, amount, and/or rate vs. time profiles, the student will calculate (III) the relevant pharmacokinetic parameters available from IV plasma, urine or other excreta data: e.g.
V d, K, k m, k r, AUC, AUMC, CL, MRT, t 1 ⁄ 2 2.
The student will provide professional communication regarding the pharmacokinetic parameters obtained to patients and other health professionals.
3.
The student will be able to utilize computer programs for simulations and data analysis.
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I.V. Bolus Dosing
4.1 I.V. Bolus dosing of Parent compound 4.1.1
PLASMA
Valid equations: (Obtained from the LaPlace transforms derived from the appropriate models derived from the pharmacokinetic descriptions of the drug)
ln C p = – K1 ⋅ t + ln C p 0
(EQ 4-1)
ln X = – K1 ⋅ t + ln X0
(EQ 4-2)
C p = C p0 e
( ∞)
AUC =
∫ Cp dt
– K1t
(EQ 4-3)
DC p 0 = ----Vd
(EQ 4-4)
0.693 t ½ = ------------K
(EQ 4-5)
( Cp n + Cp n + 1 ) Cp last ⋅ ∆t + -------------= Σ ------------------------------------2 K1
(EQ 4-6)
0 ( ∞)
AUMC =
∫
t
t ⋅ C p dt =
Cp last
( t last ⋅ Cp last )
(EQ 4-7)
K1
0
0
Utilization: Can you determine the slope and intercept from a graph? Plot the data in table 4 -1.on semi-log graph paper. Extrapolate the line back to time = 0 to get Cp0. Find the half life. Calculate the elimination rate constant.
( t n ⋅ Cp n ) + ( t n + 1 ⋅ Cp n + 1 )
- ⋅ ∆t + --------------- + ---------------------------------∑ ------------------------------------------------------------------ 2 2 K1
MRT = AUMC -----------------AUC
(EQ 4-8)
Cl = K ⋅ Vd
(EQ 4-9)
• You should be able to plot a data set Concentration vs. time on semilog yielding a straight line with slope = – K1 and an intercept of C p0 .
TABLE 4-1.
Nifedipine 25 mg IV bolus
Time (hr)
Cp (mcg/L)
2
139
4
65.6
6
31.1
8
14.6
FIGURE 4-1.
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4-2
I.V. Bolus Dosing
Does your Graph look like this?
FIGURE 4-1.
Nifedipine IV Bolus (25 mg IV Bolus)
100
Concentration (ng/mL)
Concentration (mic/L)
10 3
Cp0 = 295 mic/L -K1 = -0.375 hr -1
10 2
50
1.85 hr 10 1 0
2
4
6
8
Time (hours)
Time (hr)
• You should be able to determine K1. A plot of the data in TABLE 4-1. results in FIGURE 41. Remember from high school algebra, the slope of any straight line is the rise over the run, dy ------ , In the case of semi-log graphs dy is the difference in the logarithms of the concentrations. dx
Thus, using the rules of logarithms, when two logs are subtracted, the numbers themselves are divided. i.e. ln ( C1 ) – ln ( C2 ) = ln C1 ------- . Thus if we are judicious in the concentrations that we C2 take, we can set the rise to a constant number. So, if we take any two concentrations such that one concentration is half of the other (In FIGURE 4-1. above, we took 100 and 50), the time it takes for the concentration to halve is the half life (in the graph above, 1.85 hr). Then –1 0.693 0.693 K1 = ------------- = ---------------- = 0.375 hr t½ 1.85hr
• You should be able to determine V d :. To do this, extrapolate the line to t = 0 . The value of Cp when t = 0 is C p0 (in the graph above, C p0 = 295 mic --------- which is equal to D ⁄ V d for an IV bolus L
dose only. Dose Dose 25mg 1000mic Thus, Cp 0 = ------------ , V d = ------------- = ------------------ ⋅ --------------------- = 85L Vd
Cp 0
295mic -----------------L
mg
The volume of distribution is a mathematical construct. It is merely the proportionality constant between two knowns - the C p0 which results from a given D 0 . It is, however, useful because it is patient specific and therefore can be used to predict how the patient will treat a subsequent dose of the same drug. You should be able to obtain the volume of distribution from graphical analysis of the data. Pay attention to the units! Make sure that they are consistent on both sides of the equation. NOTE: the volume of distribution is not necessarily any physiological space. For example the approximate volume of distribution of digoxin is about 600 L If that were a physiological space and I were all water, that would mean that I would weigh about 1320 pounds. I'm a little overweight (I prefer to think that I'm underheight), but REALLY!
• Given any three of the variables of the IV bolus equation, either by direct information (the volume of distribution is such and such) or by graphical data analysis, you should be able to find the fourth.
• You should be able to calculate Area Under the Curve (AUC) from IV Bolus data (Time vs. Cp). From the above data in TABLE 4-1. the AUC is calculated using (EQ 4-6):
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I.V. Bolus Dosing
(∞)
AUC =
∫ Cp dt 0
Cp n + Cp n + 1 Cp l = Σ --------------------------------+ -------- which in this case is: K1 ∆t
Cp 1 + Cp 2 Cp 2 + Cp 3 Cp 3 + Cp last Cp last Cp o + Cp 1 ⋅ ∆t last + --------------Σ ------------------------- ⋅ ∆t 1 + ------------------------- ⋅ ∆t 2 + ------------------------- ⋅ ∆t 3 + ------------------------------ K1 2 2 2 2 295 + 139 139 + 65.6 65.6 + 31.1 31.1 + 14.6 14.6 mcg Σ ------------------------ ⋅ 2 + ------------------------- ⋅ 2 + --------------------------- ⋅ 2 + --------------------------- ⋅ 2 + ------------- ----------hr or 2 2 2 2 0.375 L mcg mcg Σ { 434 + 204.6 + 96.7 + 45.7 + 38.9 } ----------hr = 819.9----------hr . In tabular format, the AUC calculation L L
is shown in TABLE 4-2.
TABLE 4-2 AUC t t–1
AUC
AUC
t 0
TIME
Cp
0
295
2
139
434.0
434.0
4
65.6
204.6
638.6
6
31.1
96.7
735.3
8
14.6
45.7
781.0
∞
0
38.9
819.9
The AUC of a plot of plasma concentration vs. time, in linear pharmacokinetics, is a number which is proportional to the dose of the drug which gets into systemic circulation. The proportionality constant, as before, is the volume of distribution. It is useful as a tool to compare the amount of drug obtained by the body from different routes of administration or from the same route of administration by dosage forms made by different manufacturers (calculate bioavailability in subsequent discussions). The AUC of a plot of Rate of Excretion of a drug vs. time, in linear pharmacokinetics, is the mass of drug excreted into the urine, directly.
• You should be able to calculate the AUMC from IV Bolus data (Time vs. Cp). The equation for AUMC is equation 4-7: ( ∞)
AUMC =
∫
t
t ⋅ C p dt =
0
( t n ⋅ Cp n ) + ( t n + 1 ⋅ Cp n + 1 )
Cp last
( t last ⋅ Cp last )
- ⋅ ∆t + --------------- + ---------------------------------- which in the ∑ ------------------------------------------------------------------ 2 2 K1 K1 0
data given in TABLE 4-1. is: T0 ⋅ C po + T1 ⋅ C p1 T1 ⋅ C p1 + T2 ⋅ C p2 T 2 ⋅ C p 2 + T3 ⋅ C p3 Σ ---------------------------------------------- ⋅ ∆t 1 + ---------------------------------------------- ⋅ ∆t 2 + ---------------------------------------------- ⋅ ∆t 3 + 2 2 2 T 3 ⋅ C p 3 + T last ⋅ C p last T last ⋅ C p last Cp last - + --------------- and thus, --------------------------------------------------------- ⋅ ∆t last + -----------------------------2 2 K1 K1
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I.V. Bolus Dosing
⋅ 295 + 2 ⋅ 139 2 ⋅ 65.6 + 6 ⋅ 31.1 ⋅ 2 mcg ⋅ 139 + 4 ⋅ 65.6 ⋅ 2 + 4-----------------------------------------⋅ 2 + 2---------------------------------------Σ 0------------------------------------------------hr + 2 L 2 2 6 ⋅ 31.1 + 8 ⋅ 14.6 8 ⋅ 14.6 14.6 mcg 2 ------------------------------------------ ⋅ 2 + ------------------ + ---------------2- ----------hr or 2 0.375 0.375 L
mcg 2 Σ { 278 + 540.4 + 449 + 303.4 + 311.47 + 103.82 } = 1986.1----------hr L
Thus in tabular format the AUMC for data given in TABLE 4-1. is TABLE 4-3 below. TABLE 4-3 AUMC
Cp*T
AUMC t
AUMC
TIME
Cp
0
295
0
2
139
278
278.0
278.0
4
65.6
262.4
540.4
818.4
6
31.1
186.6
449.0
1267.4
8
14.6
116.8
303.4
1570.8
∞
0
0
415.3
1986.1
t 0
The AUMC is the Area Under the first Moment Curve. A plot of T*Cp vs. T is the first moment curve. The time function buried in this plot, the Mean Residence Time (MRT), can be extracted using equation 4-8 below. It is the geometric mean time that the molecules of drug stay in the body. It has utility in the fact that, as drug moves from the dosage form into solution in the gut, from solution in the gut into the body, and from the body out, each process is cumulatively additive. That means if we can physically separate each of these processes in turn, we can calculate the MRT of each process. The MRT of each process is the the inverse of the rate constant for that process.
• You should be able to calculate MRT from IV Bolus data (Time vs. Cp) using equation 4-8 AUMC 1986.1 MRT = ------------------ = ---------------- = 2.42 AUC 819.9
Since there is only the process of elimination (no release of the drug from the dosage form, no absorption), the MRT is the inverse of the elimination rate constant, K. Thus MRT = 1/K. Flow Chart 2-1
IV Bolus
X
K
MRT(IV) = 1/K
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I.V. Bolus Dosing
Suppose the drug were given in a solution. Then the drug would have to be absorbed and then eliminated. Since the MRTs are additive, the MRT of the oral solution would be made up of the MRTs of the two processes, thus: Flow Chart 2-2
Oral Solution Ka Xa
K X
MRT(os) = MAT(os)+MRT(IV) MRT(os) =
1/Ka
+ 1/K
Consequently, if a drug has to be released from a dosage form for the drug to get into solution which is subsequently absorbed, a tablet for example, the MRT of the tablet will consist of the MRT(IV) and the MAT(os) and the Mean Dissolution Time (MDT), thus: Flow Chart 2-3
Tablet
Kd
Ka
Xd
Xa
K X
MRT(tab) = MDT
+ MAT(os)
+ MRT(IV)
MRT(tab) =
+
+
1/Kd
1/Ka
1/K
MRT(tab) = MAT(tab)
+ MRT(IV)
MRT(tab) = 1/Ka (apparent)
+
1/K
Normally, we don’t have information from the oral solution, just IV and tablet. So in that case the information obtained about absorption from the tablet is bundled together into an apparent absorption rate constant consisting of both dissolution and absorption. It should be apparent that this is a reasonably easily utilized and powerful tool used to obtain pharmacokinetic parameters.
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I.V. Bolus Dosing
4.1.2
IV BOLUS, PARENT COMPOUND, PLASMA PROBLEMS
Equations used in this section:
1.
K1 = –slope from equation 4-3
2.
ln 2 equation 4-5 t 1 ⁄ 2 = -------K1
3.
1 AUMC MRT = ------- ( estimate ) MRT = ------------------ equation 4-8 K1 AUC
4.
Cp 0 = the y-intercept of the line from equation 4-3
5.
Cp Estimate for AUC = AUC = ---------0 which is K1 (∞)
AUC =
∫
∞
∫0 Cp dt
( Cp n + Cp n + 1 ) Cp last Cp dt = Σ ------------------------------------( ∆t ) + ------------- 2 K1
0
Trapezoidal rule applied to equation 4-6 6.
Estimate for AUMC = AUMC = AUC ⋅ MRT from equation 4-8 ( ∞)
AUMC
∫
t
Cp dt =
0
( t n ⋅ Cp n ) + ( t n + 1 ⋅ Cp n + 1 )
Cp last
( t last ⋅ Cp last )
- ⋅ ∆t n + --------------- + ---------------------------------∑ ------------------------------------------------------------------ 2 K1 2 0
K1
from equation 4-7 7.
V d = Dose ------------- from equation 4-4 Cp 0
8.
Cp 0 Dose Dose Cl = K1 ⋅ V d = -----------⋅ ------------- = ------------AUC Cp 0 AUC
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4-7
I.V. Bolus Dosing
Acyclovir Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 4 - 1)
AHFS 12:34.56 Antivirals GPI: 0000000000 Antivirals
De Miranda and Burnette, “Metabolic Fate and Pharmacokinetics of the Acyclovir Prodrug Valaciclovir in Cynomolgus Monkeys”, Drug Metabolism and Disposition (1994): 55-59.
Acyclovir is an antiviral drug used in the treatment of herpes simplex, varicella zoster, and in suppressive therapy. In this study, three male cynomolgus monkeys were each given a 10 mg ⁄ kg intravenous dose. The monkeys weighed an average of 3.35 kg each. Blood samples were collected and the following data was obtained: PROBLEM TABLE 4 - 1.
Acyclovir Serum concentration
Time (hours)
( µg ⁄ mL )
0.167
26.0
0.300
23.0
0.500
19.0
0.75
16.0
1.0
12.0
1.5
7.0
2.0
5.0
From the data presented in the Preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p )0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
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4-8
I.V. Bolus Dosing
(Problem 4 - 1) Acyclovir:
2
CONCENTRATION (MIC/ML)
10
1
10
100 0.0
0.5
1.0
1.5
2.0
TIME (HR) –1
1.
k = 0.93hr
2.
t ½ = 0.75hr .
3.
MRT = 1.08hr .
4.
( C p )0 = 30.4ug ⁄ mL .
5.
AUC = 32.75ug ⁄ mL ⋅ hr .
6.
AUMC = 35.2ug ⁄ mL ⋅ hr .
7.
V d = 1.1L
8.
Cl = 1.02L ⁄ hr .
2
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4-9
I.V. Bolus Dosing
Aluminum
(Problem 4 - 2)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Xu, Pai, and Melethil, "Kinetics of Aluminum in Rats. II: Dose-Dependent Urinary and Biliary Excretion", Journal of Pharmaceutical Sciences, Oct 1991, p 946 - 951.
A study by Xu, Pai, and Melethil establishes the pharmacokinetics of Aluminum in Rats. In this study, four rats with an average weight of 375g, were given an IV bolus dose of aluminum (1 mg/kg). Blood samples were taken at various intervals and the following data was obtained: PROBLEM TABLE 4 - 2.
Aluminum
Time (hours)
ngSerum concentration, ------mL
0.4
19000
0.6
18000
1.4
15000
1.6
14500
2.3
12500
3.0
10500
4.0
8500
5.0
6500
6.0
5000
8.0
3250
10.0
2000
12.0
1250
From the data presented in the preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p )0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
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4-10
I.V. Bolus Dosing
(Problem 4 - 2) Aluminum:
5
CONCENTRATION (NG/ML)
10
4
10
3
10
0
2
4
6
8
10
12
TIME (HR)
–1
1.
k = 0.234hr
2.
t ½ = 3hr .
3.
MRT = 4.3hr .
4.
( C p )0 = 21000ng ⁄ mL .
5.
AUC = 89285ng ⁄ mL ⋅ hr .
6.
AUMC = 383926ng ⁄ mL ⋅ hr .
7.
V d = 17.86mL
8.
Cl = 4.18mL ⁄ hr .
2
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4-11
I.V. Bolus Dosing
Amgen
(Problem 4 - 3)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Salmonson, Danielson, and Wikstrom, "The pharmacokinetics of recombinant human erythropoetin after intravenous and subcutaneous administration to healthy subjects", Br. F. clin. Pharmac. (1990), p 709- 713. Amgen (r-Epo) is a form of recombinant erythropoetin. Erythropoetin is a hormone that is produced in the kidneys and used in the production of red blood cells. The kidneys of patients who have end-stage renal failure cannot produce erythropoetin; therefore, r-Epo is being investigated for use in these patients in order to treat the anemia that results from the lack of erythropoetin. In a study by Salmonson et al, six healthy volunteers were used to demonstrate that both IV and subcutaneous administration of erythropoetin have similar effects in the treatment of anemia due to chronic renal failure. The six volunteers were each given a 50 U/kg intravenous dose of Amgen. The average weight of the six volunteers was 79 kg. Blood samples were drawn at various times and the data obtained is summarized below: PROBLEM TABLE 4 - 3.
Amgen
Time (hours)
--------Serum concentration, mU mL
2
700
4
600
6
400
8
300
12
150
24
40
From the data presented in the preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p )0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
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4-12
I.V. Bolus Dosing
(Problem 4 - 3) Amgen:
Con (mU/mL)
CONCENTRATION (MU/ML)
103
102
101 0
5
10
15
20
25
TIME (HR)
–1
1.
k = 0.134hr
2.
t ½ = 5.2hr .
3.
MRT = 7.46hr .
4.
( C p )0 = 900mU ⁄ mL .
5.
AUC = 6945mU ⁄ mL ⋅ hr .
6.
AUMC = 49600 mU ⁄ mL ⋅ hr .
7.
V d = 4.44L
8.
Cl = 0.6L ⁄ hr .
2
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4-13
I.V. Bolus Dosing
Atrial Naturetic Peptide (ANP) Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 4 - 4)
AHFS 00:00.00 GPI: 0000000000
Brier and Harding, "Pharmacokinetics and Pharmacodynamics of Atrial Naturetic Peptide after Bolus and Infusion Administration in the Isolated Perfused Rat Kidney", The Journal of Pharmacology and Experimental Therapeutics (1989), p 372 - 377.
A study by Brier and Harding a dose of 45 ng was given by IV bolus to rats. Samples of blood were taken at various intervals throughout the length of the study and the following data was obtained: PROBLEM TABLE 4 - 4.
Atrial Naturetic Peptide (ANP)
Time (minutes)
pg Serum concentration, -------mL
3
380
10
280
20
170
30
130
40
100
50
70
60
50
From the data presented in the preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p )0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
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4-14
I.V. Bolus Dosing
(PG/ML) Con CONCENTRATION (pg/mL)
(Problem 4 - 4) Atrial Naturetic Peptide (ANP):
ANP
103
102
101 0 10 20 30 40 50 60
Time (min) –1
1.
k = 0.0345min
2.
t ½ = 20.09min .
3.
MRT = 28.95min .
4.
( C p )0 = 386.6pg ⁄ mL .
5.
AUC = 11206.4pg ⁄ mL ⋅ min .
6.
AUMC = 324425.4pg ⁄ mL ⋅ min .
7.
V d = 116.4mL
8.
Cl = 4.02mL ⁄ min .
2
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4-15
I.V. Bolus Dosing
Aztreonam
(Problem 4 - 5)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Cuzzolim et al., "Pharmacokinetics and Renal Tolerance of Aztreonam in Premature Infants", Antimicrobial Agents and Chemotherapy (Sept. 1991), p. 1726 - 1928.
Aztreonam is a monolactam structure which is active against aerobic, gram-negative bacilli. The pharmacokinetic parameters of Aztreonam were established in a study presented in by Cuzzolim et al in which Aztreonam (100 mg/ kg) was administered intravenously to 30 premature infants over 3 minutes every 12 hours. The group of neonates had an average weight of 1639.6g. The following set of data was obtained: PROBLEM TABLE 4 - 5.
Aztreonam
Time (minutes)
µgSerum concentration, ------mL
1
40.50
2
34.99
3
29.99
4
23.88
5
22.20
6
19.44
7
16.55
8
14.99
From the data presented in the preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p )0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
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4-16
I.V. Bolus Dosing
Aztreonam
Con (ug/mL) CONCENTRATION (UG/ML)
(Problem 4 - 5) Aztreonam:
10 2
10 1 0
2
4
TIME (MIN)
6
8
–1
1.
k = 0.144min
2.
t ½ = 4.81min .
3.
MRT = 6.94min .
4.
( C p )0 = 45.75ug ⁄ mL .
5.
AUC = 317.7ug ⁄ mL ⋅ min .
6.
AUMC = 2204.8ug ⁄ mL ⋅ min .
7.
V d = 3.58L
8.
Cl = 0.516L ⁄ min .
2
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4-17
I.V. Bolus Dosing
Recombinant Bovine Placental Lactogen Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 4 - 6)
AHFS 00:00.00 GPI: 0000000000
Byatt, et. al., "Serum half-life and in-vivo actions of recombinant bovine placental lactogen in the dairy cow", Journal of Endocrinology (1992), p. 185 - 193.
Bovine placental lactogen (bPL) is a hormone similar to growth hormone and prolactin. It binds to both prolactin and growth hormone receptors in the rabbit and stimulates lactogenesis in the rabbit. In a study by Byatt, et. al., four cows (2 pregnant and 2 nonpregnant) were given IV bolus injections of 4 mg and the following data was obtained: PROBLEM TABLE 4 - 6.
Recombinant Bovine Placental Lactogen
Time (minutes)
-----Serum concentration µg L
3.8
117
6.8
72
12.0
43
16.0
27
20.0
18
From the data presented in the preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p )0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
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4-18
I.V. Bolus Dosing
(Problem 4 - 6) Recombinant Bovine Placental Lactogen:
(MIC/L) ConCONCENTRATION (ug/L)
103
102
101 0
5
10
15
20
Time (min) –1
1.
k = 0.113min
2.
t ½ = 6.13min .
3.
MRT = 8.85min .
4.
( C p )0 = 167.8ug ⁄ L .
5.
AUC = 1484.9ug ⁄ L ⋅ min .
6.
AUMC = 13141.1ug ⁄ L ⋅ min .
7.
V d = 23.84L
8.
Cl = 2.69L ⁄ min .
2
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4-19
I.V. Bolus Dosing
Caffeine
(Problem 4 - 7)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Dorrbecker et. al., "Caffeine and Paraxanthine Pharmacokinetics in the Rabbit: Concentration and Product Inhibition Effects.", Journal of Pharmacokinetics and Biopharmaceutics (1987), p.117 - 131.
This study examines the pharmacokinetics of caffeine in the rabbit. In this study type I New Zealand White rabbits were given an 8 mg intravenous dose of caffeine. Blood samples were taken and the following data was obtained: PROBLEM TABLE 4 - 7.
Caffeine
Time (minutes)
µg Serum concentration -------mL
12
3.75
40
2.80
65
2.12
90
1.55
125
1.23
173
0.72
243
0.37
From the data presented in the preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p )0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
4-20
I.V. Bolus Dosing
(MIC/ML) ConCONCENTRATION (ug/L)
(Problem 4 - 7) Caffeine:
101 100 10-1 0
50
100 150 200 250
Time (min) –1
1.
k = 0.00997min
2.
t ½ = 69.51min .
3.
MRT = 100.3min .
4.
( C p )0 = 4.105ug ⁄ mL .
5.
AUC = 411.7ug ⁄ mL ⋅ min .
6.
AUMC = 41293.5ug ⁄ mL ⋅ min .
7.
V d = 1.95L
8.
Cl = 19.44mL ⁄ min .
2
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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4-21
I.V. Bolus Dosing
Ceftazidime
(Problem 4 - 8)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Demotes-Mainard, et. al., "Pharmacokinetics of Intravenous and Intraperitoneal Ceftazidime in Chronic Ambulatory Peritoneal Dyialysis", Journal of Clinical Pharmacology (1993), p. 475 - 479.
Ceftazidime is a third generation cephalosporin which is administered parenterally. In this study, eight patients with chronic renal failure were each given 1 g of ceftazidime intravenously. Both blood samples were taken the data obtained from the study is summarized in the following table:
PROBLEM TABLE 4 - 8.
Ceftazidime
Time (hours)
mg Serum concentration ------L
1
50
2
45
4
38
24
21
36
14
48
11
60
8
72
4
From the data presented in the preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p ) 0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
4-22
I.V. Bolus Dosing
CONCENTRATION (MG/L) Con (mg/L)
(Problem 4 - 8) Ceftazidime:
10 2 10 1 10 0 0
20
40
60
80
Time (hours) –1
1.
k = 0.0324hr
2.
t ½ = 21.39hr .
3.
MRT = 30.86hr .
4.
( C p )0 = 47.57mg ⁄ L .
5.
AUC = 1468.2mg ⁄ L ⋅ hr .
6.
AUMC = 45308.6mg ⁄ L ⋅ hr .
7.
V d = 21.02L
8.
Cl = 0.681L ⁄ hr .
2
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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4-23
I.V. Bolus Dosing
Ciprofloxacin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 4 - 9)
AHFS 00:00.00 GPI: 0000000000
Lettieri, et. al., "Pharmacokinetic Profiles of Ciprofloxacin after Single Intravenous and Oral Doses", Antimicrobial Agents and Chemotherapy (May 1992), p. 993 -996.
Ciprofloxacin is a fluoroquinolone antibiotic which is used in the treatment of infections of the urinary tract, lower respiratory tract, skin, bone, and joint. In this study, twelve healthy, male volunteers were each given 300 mg intravenous doses of Ciprofloxacin. Blood and urine samples were collected at various times throughout the day and the following data was collected: PROBLEM TABLE 4 - 9.
Ciprofloxacin
Time (hours)
mg Serum concentration ------L
2
1.20
3
0.85
4
0.70
6
0.50
8
0.35
10
0.25
From the data presented in the preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p )0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
4-24
I.V. Bolus Dosing
CONCENTRATION (MG/L) Con (mg/L)
(Problem 4 - 9) Ciprofloxacin:
101 100 10-1 0
2
4
6
8
10
Time (hours) –1
1.
k = 0.1875hr
2.
t ½ = 3.7hr .
3.
MRT = 5.33hr .
4.
( C p )0 = 1.57mg ⁄ L .
5.
AUC = 8.395mg ⁄ L ⋅ hr .
6.
AUMC = 44.74mg ⁄ L ⋅ hr .
7.
V d = 190.6L
8.
Cl = 35.74L ⁄ hr .
2
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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4-25
I.V. Bolus Dosing
The effect of Probenecid on Diprophylline (DPP) Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 4 - 10)
AHFS 00:00.00 GPI: 0000000000
Nadai et al, "Pharmacokinetics and the Effect of Probenecid on the Renal Excretion Mechanism of Diprophylline", Journal of Pharmaceutical Sciences (Oct 1992), p. 1024 - 1027.
Diprophylline is used as a bronchodilator. A study by Nadai et al was designed to determine whether or not coadministration of Diprophylline with Probenecid affected the pharmacokinetic parameters of Diprophylline. In this study, male rats (average weight: 300 g) were given 60 mg/kg of Diprophylline intravenously and a 3 mg/kg loading dose of Probenecid followed by a continuous infusion of 0.217 mg/min/kg of Probenecid. The following set of data was obtained for Diprophylline (DPP):
PROBLEM TABLE 4 - 10.
The effect of Probenecid on Diprophylline (DPP)
Time (minutes)
µg Serum concentration -------mL
16
40.00
31
27.00
60
13.00
91
6.50
122
3.50
From the data presented in the preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p )0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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4-26
I.V. Bolus Dosing
CONCENTRATION (MIC/ML) Con (ug/mL)
(Problem 4 - 10) The effect of probenecid on diprophylline (DPP):
102
10 1
100 0
20
40
60
80
100
Time (min) –1
1.
k = 0.023min
2.
t ½ = 30.13min .
3.
MRT = 43.48min .
4.
( Cp ) 0 =
5.
AUC = 2396.96ug ⁄ mL ⋅ min .
6.
AUMC = 104219.8ug ⁄ mL ⋅ min .
7.
V d = 326.5mL
8.
Cl = 7.5mL ⁄ min .
55.13ug ⁄ mL .
2
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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4-27
I.V. Bolus Dosing
Epoetin
(Problem 4 - 11)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
MacDougall et. al., "Clinical Pharmacokinetics of Epoetin (Recombinant Human Erythropoetin", Clinical Pharmacokinetics (1991), p 99 - 110.
Epoetin is recombinant human erythropoetin. Erythropoetin is a hormone that is produced in the kidneys and used in the production of red blood cells. The kidneys of patients who have end-stage renal failure cannot produce erythropoetin; therefore, Epoetin is used in these patients to treat the anemia that results from the lack of erythropoetin. Epoetin has also been used in the treatment of anemias resulting from AIDS. malignant disease, prematurity, rheumatoid arthritis, sickle-cell anemia, and myelosplastic syndrome. In a study by Macdougall et al, eight patients who were on peritoneal dialysis (CAPD) were given an IV bolus dose of 120 U/kg which decayed monoexponentially from a peak of 3959 U/L to 558 U/L at 24 hours. The following data was obtained: PROBLEM TABLE 4 - 11.
Epoetin
Time (hours)
U Serum concentration ---L
0.0
4000
0.5
3800
1.0
3600
2.0
3300
3.0
3000
4.0
2550
5.0
2350
6.0
2150
7.0
1900
From the data presented in the preceding table and assuming that the patient weighs 65 kg, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p ) 0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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4-28
I.V. Bolus Dosing
Con (U/L) CONCENTRATION (U/L)
(Problem 4 - 11) Epoetin:
104
103 0
1
2
3
4
5
6
7
Time (hours) –1
1.
k = 0.107 hr
2.
t½ = 6.5 hr .
3.
MRT = 9.38 hr .
4.
( Cp) 0
5.
Units ⋅ hr AUC = 37775-----------------------. L
6.
⋅ hr AUMC = 354697Units -------------------------- . L
7.
V d = 1.9 L
8.
L Cl = 0.2065 ----. hr
= 4023 Units/L .
2
Basic Pharmacokinetics
REV. 99.4.25
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4-29
I.V. Bolus Dosing
Famotidine
(Problem 4 - 12)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Kraus, et. al., "Famotidine--Pharmacokinetic Properties and Suppression of Acid Secretion in Pediatric Patients Following Cardiac Surgery", Clinical Pharmacokinetics (1990), p 77 - 80.
Famotidine is a histamine H2-receptor antagonist. The study by Kraus, et. al., focuses on the kinetics of famotidine in children. In the study, ten children with normal kidney function and a body weight ranging from 14 - 25 kg, were each given a single intravenous 0.3 mg/kg dose of famotidine. Blood and urine samples were taken providing the following data: PROBLEM TABLE 4 - 12.
Famotidine
Time (hours)
g Serum concentration µ-----L
0.33
300
0.50
250
1.00
225
4.00
125
8.00
70
12.00
40
16.00
15
From the data presented in the preceding table, determine the following assuming that the patient weighs 17.2 kg: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p ) 0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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4-30
I.V. Bolus Dosing
(Problem 4 - 12) Famotidine:
(MIC/L) ConCONCENTRATION (ug/mL)
10 3
10 2
10 1 0
5
10
15
20
Time (hours) –1
1.
k = 0.17 hr
2.
t ½ = 3.9 hr .
3.
MRT = 5.7 hr .
4.
( Cp ) 0 =
5.
g ⋅ hrAUC = 1600 µ---------------. L
6.
g ⋅ hr AUMC = 9000µ-----------------. L
7.
V d = 18 L
8.
Cl = 3.2L .
g 285µ-----. L
2
Basic Pharmacokinetics
REV. 99.4.25
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4-31
I.V. Bolus Dosing
Ganciclovir Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 4 - 13)
AHFS 00:00.00 GPI: 0000000000
Trang, et. al., "Linear single-dose pharmacokinetics of ganciclovir in newborns with congenital cytomegalovirus infections", Clinical Pharmacology and Therapeutics (1993), p. 15 - 21.
Ganciclovir (mw: 255.23) is used against the human herpes viruses, cytomegalovirus retinitis, and cytomegalovirus infections of the gastrointestinal tract. In this study, twenty-seven newborns with cytomegalovirus disease were given 4 mg/kg of ganciclovir intravenously over one hour. Blood samples were taken and the data obtained is summarized in the following table:
PROBLEM TABLE 4 - 13.
Ganciclovir
Time (hours)
Serum concentration
1.50
4.50
2.00
4.00
3.00
3.06
4.00
2.40
6.00
1.45
8.00
0.87
From the data presented in the preceding table and assuming the patient weighs 3.6 kg, determine the following : 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p ) 0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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4-32
I.V. Bolus Dosing
(Problem 4 - 13) Ganciclovir:
CONCENTRATION (MICMOLE/L)
10
10
10 0
2
4
6
8
TIME (HR) –1
1.
k = 0.288hr
2.
t ½ = 2.4hr .
3.
MRT = 3.5hr .
4.
( Cp ) 0 =
5.
mole ⋅ hr AUC = 80 µ-------------------------. mL
6.
mole ⋅ hr AUMC = 280µ---------------------------. mL
mole23 µ--------------. mL
2
7.
8.
1000µ g4 mg ------- ⋅ 3.6kg ⋅ -----------------kg mg - = 2.45L V d = Dose ------------- = -----------------------------------------------------------Cp 0 23 µ---------------mole ⋅ 255.23 --------------µg L µ mole L . Cl = 0.7 ----hr
Basic Pharmacokinetics
REV. 99.4.25
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4-33
I.V. Bolus Dosing
Imipenem
(Problem 4 - 14)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Heikkila, Renkonen, and Erkkola, "Pharmacokinetics and Transplacental Passage of Imipenem During Pregnancy", Antimicrobial Agents and Chemotherapy (Dec. 1992), p 2652 - 2655.
Imipenem is a beta-lactam antibiotic which is used in combination with cilastin and is active against a broad spectrum of bacteria. The pharmacokinetics of Imipenem in pregnant women is established in this study. Twenty women (six of which were non-pregnant controls) were given a single intravenous dose of 500 mg of imipenem-cilastin (1:1). Blood samples were taken at various intervals and the data obtained is summarized in the following table:
PROBLEM TABLE 4 - 14.
Imipenem
Time (minutes)
mg Serum concentration ------L
10
27.00
15
23.50
30
15.50
45
9.50
60
6.50
From the data presented in the preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p ) 0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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4-34
I.V. Bolus Dosing
(Problem 4 - 14) Imipenem: 2
CONCENTRATION (MG/L)
10
10
10
1
0
0
10
20
30
40
50
60
TIME (MIN) –1
1.
k = 0.029 min
2.
t ½ = 24 min .
3.
MRT = 34.5 min .
4.
( Cp) 0=
5.
⋅ min AUC = 1250mg --------------------. L
6.
⋅ min AUMC = 43125mg ----------------------- . L
36.2 mg ------- . L
2
7.
8.
500mg = 13.8L Vd = Dose ------------- = -----------------Cp 0 36.2mg ------L L Cl = 0.4 --------. min
Basic Pharmacokinetics
REV. 99.4.25
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4-35
I.V. Bolus Dosing
Methylprednisolone Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 4 - 15)
AHFS 00:00.00 GPI: 0000000000
Patel, et. al., "Pharmacokinetics of High Dose Methylprednisolone and Use in Hematological Malignancies", Hematological Oncology (1993), p. 89 - 96.
Methylprednisolone is a corticosteriod that has been used in combination chemotherapy for the treatment of hematological malignancy, myeloma, and acute lymphoblastic leukemia. In a study by Patel et. al., eight patients were given 1.5 gram intravenous doses of methylprednisolone from which the following data was obtained:
PROBLEM TABLE 4 - 15.
Methylprednisolone
Time (hours)
µg Serum concentration ------mL
0.5
19.29
1.0
17.56
1.8
15.10
4.0
9.98
5.8
7.10
8.0
4.70
12.0
2.21
18.0
0.71
24.0
0.23
From the data presented in the preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p ) 0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
4-36
I.V. Bolus Dosing
(Problem 4 - 15) Methylprednisolone:
CONCENTRATION (MIC/ML)
102 101
Con (ug/mL)
100 10-1 0
5
10
15
20
25
Time (hours) –1
1.
k = 0.188 hr
2.
t ½ = 3.69hr .
3.
MRT = 5.3hr .
4.
( Cp) 0=
5.
g ⋅ hrAUC = 112.5 µ---------------. mL
6.
g ⋅ hr AUMC = 598.4µ-----------------. mL
7.
V d = 71L
8.
L Cl = 13.3 ----. hr
µg 21.2 -------. mL
2
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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4-37
I.V. Bolus Dosing
Omeprazole Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 4 - 16)
AHFS 00:00.00 GPI: 0000000000
Anderson, et. al., "Pharmacokinetics of [14C] Omeprazole in Patients with Liver Cirrhosis", Clinical Pharmacokinetics (1993), p. 71 - 78.
Omeprazole (mw: 345.42) is a gastric proton-pump inhibitor which decreases gastric acid secretion. It is effective in the treatment of ulcers and esophageal reflux. In normal patients 80% of the omeprazole dose is excreted as metabolites in the urine and the remainder is excreted in the feces. In the study by Anderson, et. al., eight patients with liver cirrhosis were given 20 mg, IV bolus doses of omeprazole. The patients had a mean body weight of 70 kg. Both blood were taken at various intervals throughout the study and the following data was obtained:
PROBLEM TABLE 4 - 16.
Omeprazole
Time (hours)
ρ mole Serum concentration ---------------mL
0.75
3.49
1.00
3.25
2.00
2.46
3.00
1.86
4.00
1.40
5.00
1.06
6.00
0.80
7.00
0.61
8.00
0.46
10.00
0.26
12.00
0.15
From the data presented in the preceding table, determine the following : 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p ) 0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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4-38
I.V. Bolus Dosing
Omeprazole
CONCENTRATION (PICOMOLE/ML) Con (umol/mL)
(Problem 4 - 16) Omeprazole:
101 100 10 -1 0
2
4
6
8
10 12
Time (hours)
–1
1.
k = 0.280hr
2.
t ½ = 2.5hr .
3.
MRT = 3.57hr .
4.
( Cp ) 0 =
5.
mole ⋅ hr AUC = 15.4 ρ-------------------------. mL
6.
mole ⋅ hr AUMC = 55 ρ---------------------------. mL
mole 4.3 ρ---------------. mL
2
7.
8.
Dose- = -----------------------------------------------------------------------------------------------------------20mg Vd = -----------= 13465L Cp 0 4.3 --------------ρ mole mmole 345.42mg 1000mL - ⋅ ------------------------- ⋅ ------------------------ ⋅ -------------------mL 109 ρ mole mmole L L Cl = 3.9 ----. hr
Basic Pharmacokinetics
REV. 99.4.25
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4-39
I.V. Bolus Dosing
Pentachlorophenol Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 4 - 17)
AHFS 00:00.00 GPI: 0000000000
Reigner, Rigod, and Tozer, "Absorption, Bioavailability, and Serum Protein Binding of Pentachlorophenol in the B6C3F1 Mouse", Pharmaceutical Research (1992), p 1053 - 1057.
Pentachlorophenol (PCP) is a general biocide. That is, it is an insecticide, fungicide, bactericide, herbicide, algaecide, and molluskicide, that is used as a wood preservative. Extensive exposure to PCP can be fatal. In a study by Reigner et al, six mice (average weight: 27 g) were given 15 mg/kg of PCP by intravenous bolus. Blood samples were taken at various intervals from which the following data was obtained: PROBLEM TABLE 4 - 17.
Pentachlorophenol
Time (hours)
µg Serum concentration ------mL
0.083
38.00
4.000
22.00
8.000
14.00
12.000
7.90
24.000
1.30
28.000
0.75
32.000
0.60
36.000
0.40
From the data presented in the preceding table, determine the following : 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p ) 0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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4-40
I.V. Bolus Dosing
CONCENTRATION (MIC/ML)
(Problem 4 - 17) Pentachlorphenol:
102 101
Con (ug/mL)
100 10-1 0
10
20
30
40
Time (hours) –1
1.
k = 0.134 hr
2.
t ½ = 5.2hr .
3.
MRT = 7.5hr .
4.
µg- . ( C p ) 0 = 35.6 ------mL
5.
µg ⋅ hr AUC = 281 ----------------- . mL
6.
⋅ hr - . AUMC = 2100µg -----------------mL
7.
V d = 11.4mL
8.
Cl = 1.5 ml ------ . hr
2
Basic Pharmacokinetics
REV. 99.4.25
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4-41
I.V. Bolus Dosing
9-(2-phophonylmethoxyethyl) adenine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 4 - 18)
AHFS 00:00.00 GPI: 0000000000
Naesens, Balzarini, and Clercq, "Pharmacokinetics in Mice of the Anti-Retrovirus Agent 9-(2-phophonylmethoxyethyl) adenine", Drug Metabolism and Disposition (1992), p. 747- 752.
9-(2-phophonylmethoxyethyl) adenine (PEMA) is an anti-retrovirus (anti-HIV) agent. The pharmacokinetics of PEMA in mice were established in a study by . In this study there were three different PEMA doses given: 25 mg/kg, 100 mg/kg, and 500 mg/kg. Each of these doses was injected intravenously into male mice. The data obtained from study using the 25 mg/kg dose is summarized in the following table:
PROBLEM TABLE 4 - 18.
9-(2-phophonylmethoxyethyl) adenine
Time (minutes)
µg Serum concentration -------mL
2.0
90.3
2.9
83.9
5.6
67.3
8.9
51.5
10.5
45.2
13.5
35.4
15.0
31.3
20.0
20.9
24.0
15.1
59.6
0.9
From the data presented in the preceding table, determine the following. (Assume that the mouse weighs 200g.) 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p )0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
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(MIC/ML) ConCONCENTRATION (ug/mL)
(Problem 4 - 18) Pema:
102 101 100 10 -1 0
10 20 30 40 50 60
Time (min) –1
1.
k = 0.08min
2.
t ½ = 8.67min .
3.
MRT = 12.5min .
4.
µg( C p ) 0 = 105 ------. mL
5.
⋅ hr- . AUC = 1300 µg ---------------mL
6.
⋅ hr . AUMC = 16250µg ------------------mL
7.
V d = 47.6ml
8.
mL Cl = 3.8 --------. min
2
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Thioperamide Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 4 - 19)
AHFS 00:00.00 GPI: 0000000000
Sakurai, et. al., "The Disposition of Thioperamide, a Histamine H3-Antagonist, in Rats", J. Pharm. Pharmacol. (1994), p. 209 212.
Thioperamide is a histamine (H3) receptor-antagonist. In a study by Sakurai et al, rats were given 10 mg/kg intravenous injections of Thioperamide. The following data was obtained from the study: PROBLEM TABLE 4 - 19.
Thioperamide
Time (minutes)
µg Serum concentration -------mL
3.7
3.1
7.5
2.8
13
2.4
45
1.1
60
0.74
120
0.16
From the data presented in the preceding table, determine the following: 1.
Find the elimination rate constant, k .
2.
Find the half life, t ½ .
3.
Find MRT .
4.
Find ( C p )0 .
5.
Find the Area Under the Curve, AUC .
6.
Find the area under the first moment curve, AUMC .
7.
Find the volume of distribution, V d
8.
Find the clearance, Cl .
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(MIC/ML) ConCONCENTRATION (ug/mL)
(Problem 4 - 19) thioperamide:
101
100
10-1 0
20
40
60
80 100 120
Time (min) –1
1.
k = 0.0254min
2.
t½ = 27.3min .
3.
MRT = 39.4min .
4.
µg ( C p )0 = 3.39 -------. mL
5.
⋅ minAUC = 133.5 µg -------------------. mL
6.
⋅ min AUMC = 5256µg ---------------------. mL
7.
10 mg ------Dose kg LVd = ------------- = ------------------------------------------------------------------- = 2.95 ----Cp 0 3.39 ------µg- ⋅ -----------------mg - ⋅ 1000mL kg -------------------mL 1000µg L
8.
Cl = 0.0254min
2
–1
L- ⋅ -----------------1000ml = 75 ------------------mL ⋅ 2.95 ----. kg L min ⋅ kg
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Cocaine
(Problem 4 - 20)
Khan,vM. et. al. “Determination of pharmacokinetics of cocaine in sheep by liquid chromatography” J. Pharm. Sci. 76:1 (39-43) Jan 1987
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4.1.3
URINE From the Laplace Transform of a drug given by IV bolus we find that : ku ( – K1 ⋅ t ) X u = ------ ⋅ X0 ⋅ ( 1 – e ) K1
(EQ 4-10)
where Xu is the cumulative amount of drug in the urine at time t. Rearranging, we get: k – K1t ( X u ) ∞ – X u = ------u- ⋅ X0 ⋅ e K1
(EQ 4-11)
ku
-⋅X where the amount of drug that shows up in the urine at infinite time, ( Xu ) ∞ = -----K1 0 . Thus a plot of ( Xu )∞ – X u vs. time on semi-log paper would result in a straight line with a slope of -K1 and an intercept of ( Xu )∞ .. and we can get ku from the intercept
and the slope. Rearranging the intercept equation, we get
( X u )∞ k u = K 1 ⋅ -------------X0
This
method of obtaining pharmacokinetic parameters is known as the Amount Remaining to be Excreted (ARE) method. TABLE 4-4 Enalapril
urinary excretion data from 5 mg IV Bolus
Time (hr)
Cumulative Enalapril in urine (mg)
1
0.41
0.59
2
0.65
0.35
3
0.80
0.20
4
0.88
0.12
6
0.96
0.04
∞
1.0
------
X
∞ u
– X u mg
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Utilizations: A.R.E. Method
FIGURE 4-2.
10
Cumulative Enalapril in urine
0
Xu(inf) - Xu
0.2 0.1 -1
10
1.3 hr
half life
-2
10
0
1
2
3
4
5
6
Hours
• You should be able to transform a data set containing amount of drug in the urine vs. time into cumulative amount of drug in the urine vs. time and plot the ARE. (Amount Remaining to be Excreted ->
{ ( Xu )∞ – Xu ( cum ) } vs. time on semi-log yielding a straight line with a slope of
– K1 = – 0.533 hr
–1
and an intercept of
ku ⋅ X0 ( X u ) ∞ = --------------= 1.0 mg K1
• You should be able to determine the elimination rate constant, K1, from cumulative urinary excretion data. (Calculate the slope of the graph on SL paper.)
• You should be able to determine the excretion rate constant, ku, from cumulation urinary excretion data. (Divide the intercept of the graph by X0 and multiply by K1. ( X u )∞ –1 1.0 mg –1 k u = K 1 ⋅ -------------= 0.53 hr ⋅ ----------------- = 0.106 hr ) X0 5.0 mg
• You should be able to determine k m . K = k u + k m • You should be able to calculate percent metabolized or excreted from a data set. Thus, Percent metabolized =
km ku ------ ⋅ 100 and percent excreted unchanged = ------ ⋅ 100 assuming K1 K1
K = k u + km A second method is to plot the rate at which the drug shows up in the urine over time. Again, using the LaPlace transforms, we find that:
Utilization: Rate of excretion method
dX u – K1 t – K1 t --------- = k u ⋅ X0 ⋅ e = R0 ⋅ e (EQ 4-12) dt Thus, a plot of the rate of excretion vs. time results in a straight line on semi-log paper with a slope of -K1 and an intercept, R0 , of kuX0 . Rearranging the intercept
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equation yields rate
dXu dt
R k u = -----0- . X0
In real data, we don’t have the instantaneous excretion
, but the average excretion rate,
∆X u ---------- , ∆t
over a much larger interval. What
that means to our calculations is that over the interval of data collection, the total amount of drug collected divided by the total time interval is the average rate. In the beginning of the interval the rate was faster than at the end of the interval. So the average rate must have occurred in the middle of the interval. Thus equation 412 which is the instantaneous rate can be rewritten to ∆X u – K1 t mid – K1 t mid ---------- = k u ⋅ X0 ⋅ e = R0 ⋅ e ∆t TABLE 4-5 Enalapril
(EQ 4-13)
Urinary Rate Data ∆X u ---------∆t
Interval (hr)
t(mid)
∆t
Enalapril in urine ∆X u ,(mg)
0-1
0.5
1
0.41
0.41
1-2
1.5
1
0.24
0.24
2-3
2.5
1
missed sample
?
3-4
3.5
1
0.08
0.08
4-6
5
2
0.08
0.04
• You should be able to transform a data set containing amount of drug in the urine vs. time inter∆X
val into Average Rate, ---------u- , vs.
t ,(t mid the time of the midpoint of the interval), on semilog
∆t
yielding a straight line with a slope of
– K1 and an intercept of k u ⋅ X 0 . as shown below.
-1 0
Urinary Excretion Rate (mg/hr)
10
R0 = 0.53 mg/hr
-2 -1
10
1.3 hr
half life -2
10
0
1
2
3
4
5
T (Mid)
• You should be able to determine k u extrapolate the line to t = 0 . The value of Rate (at
t = 0 ), R0, = k r ⋅ X0 = 0.53 ( mg ⁄ hr ) which when divided by X 0 .is kr.
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R –1 0.53mg/hr - = 0.106hr Thus, -----0- = -----------------------X0
5mg
• You should be able to determine k m . K = k u + k m • You should be able to calculate percent metabolized or excreted from a data set.
The rate equation is superior clinically because the ARE method requires collection of all of the urine which is usually only possible when you have a catheterized patient while the Rate Method does not. (People don’t urinate on command, and your data could be in the toilet, literally.) An additional advantage of the rate equations is that the
AUC
∞ 0
has the units of
mass, which gives the total amount of drug excreted into the urine directly. Thus: AUC
∞
R mg/hr = 1 mg = -------0 = 0.53 -------------------------–1 K1 0 0.53 hr
AN INTERESTING OBSERVATION: If you look at the LaPlace Transform of the rate equation for any terminal compartment, you would see that the resulting equation is that of the previous compartment times the rate constant through which the drug entered the terminal compartment. Thus, the rate of drug showing up in the urine (terminal compartment) is: dX u –K1 t – K1 t --------- = k u ⋅ X 0 ⋅ e = R0 ⋅ e dt
where ku is the rate constant through which the drug entered the urine and – K1 t dX ------- = X 0 ⋅ e dt
is the equation of the previous compartment.
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4.2 Metabolite 4.2.1
PLASMA Remember, the LaPlace Transform of the metabolite data yielded ( km ⋅ Xo ) ( km ⋅ Xo ) – K 2t – K 1t – K 1t – K2t X m = ------------------------⋅ (e –e ) or Xm = ------------------------⋅ (e –e ) depending ( K1 – K2 ) ( K2 – K1 ) on which rate constant that we arbitrarily assigned to be K1, the summation of all the ways that the drug is removed from the body and K2, the summation of all the ways that the metabolite is removed from the body. When we begin to manipulate the data, we know that we have a curve with two different exponents in it. (If they were the same, the equation would be different.) We don’t know which is bigger, K1 or K2, but we can rewrite the equation to simply reflect Klarge and Ksmall, knowing that one is K1 and the other is K2 but not which is which. If we, then, devided both sides of the equation by Vdm, the volume of distribution of the metabolite, we would get :
km X0 – ( Ksmall ⋅ t ) – ( K l arg e ⋅ t ) - e –e C pm = ------------------------------------- -------- K l arg e – K small V dm Utilization: Curve Stripping
(EQ 4-14)
• You should be able to plot a data set of plasma concentration of metabolite vs. time on semi-log paper yielding a bi-exponential curve.
e
– Kt
→ 0 as t → ∞ . And e – K l arg e t
– k l arg e t
– K small t
time, t, e . In fact «e time, t, the equation becomes :
e
→ 0 faster than e – K l arg e t
– k small t
→ 0 . So, at some long
is small enough to be ignored. Thus at long
km X0 – ( Ksmall ⋅ t ) C pm = ---------------------------------- --------(e ) K –K V l arg e
small
(EQ 4-15)
dm
So that the plot of the terminal portion of the graph would yield a straight line with a slope of -Ksmall and an intercept of I =
km X0 --------------------------------- --------K V –K l arg e
small
dm
• You should be able to obtain the slope of the terminal portion of the curve, the negative of which would be the smaller of the two rate constants, K small , (either the summation of all the ways that the drug is eliminated, eliminated,
K1 , or the summation of all the ways that the metabolite is
K2 ).
• Subtracting the two previous equations yields
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km X 0 – ( K big ⋅ t ) C pm – C pm = ---------------------------------- --------(e ) K l arg e – K small Vdm
(EQ 4-16)
which is a straight line on semi-log paper with a slope of -kbig and an intercept of km X0 --------I = -----------------------------------. Note: we can get the larger of the two rate constants from this K V –K l arg e
small
dm
method. TABLE 4-6
Drug
Metabolite
(1)
(2)
(3)
(4)
Time (hr)
Cp (mcg/L)
Cpm1 (mcg/L)
Cpm
(5)
Cpm
– Cpm
0
0
181.2
181.2
0.5
24.7
175
150.3
1
44.4
168.9
124.5
2
139
71.8
157.5
85.7
4
65.6
96.5
136.9
40.4
6
31.1
100
119
19
8
14.6
94.7
12
76.5
24
34
In the above data Cp vs. Time is the plasma profile of the drug from Table 4-1 on page 2 and Cpm1 vs. Time is the plasma profile of the metabolite. A plot of Cp vs. Time yielded a straight line with a slope,(-K1) of -0.375 hr-1,
0.693 = 0.375 hr –1 and and intercept of 295 K1 = ---------------------–1 1.85 hr
mic/L,
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Figure 4-1 on page 3 (column 2 vs. 1 in Table 4-6 on page 52)
10 3
Concentration (mic/L)
Cpo = 295 mic/L
100
Concentration (ng/mL)
10 2
50
1.85 hr 10 1 0
2
4
6
8
Time (hours)
Time (hr)
while a plot of Cpm1 vs. Time( Figure 4-3 on page 53) yields a biexponential plot with a terminal slope of 0.07 hr-1 , k small = 0.693 ------------- and extrapolating the terminal line back to time = 0 10 hr
yields 181 mic/L. FIGURE 4-3.
Nifedipine Metabolite (column 3 vs. 1 in Table 4-6 on page 52)
L)
Nifedipine IV bolus - Metabolite
Concentration (mic/L)
103
mic Cpm0 = 181 --------L
80 102
40 10 hr 101 0
4
8
12
16
20
24
Time (hours)
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• You should be able to feather (curve strip) the other rate constant out of the data by plotting the difference between the extrapolated (to t = 0 ) terminal line (column 4 vs. 1 in Table 4-6 on page 52) and the observed data (at early times) (column 3 vs. 1 in Table 4-6 on page 52) yielding a straight line with the slope of the line equal to the negative of the other (larger) rate constant (column 5 vs. 1 in Table 4-6 on page 52). First you would fill in the Cpm column (column 4 in Table 4-6 on page 52) by computing Cpm for various values of time i.e Cpm = Cpm0 ⋅ e
– ks m a l l t
where – k small is the terminal slope of the
graph. Then Cpm – Cpm (column 5 in Table 4-6 on page 52) would be column 4 - column 3. Then a plot of Cpm – Cpm vs. time (column 5 vs. 1 in Table 4-6 on page 52) is shown below. FIGURE 4-4.
Curve strip of Nifedipine Metabolite data
3
10
Intercept Column 5
2
100 102
1
1.85 hr
50
Half life 1
10
0
1
2
3
4
5
6
Time (hr)
In this case, the slope of the stripped line line is -0.375 hr-1 and the intercept is 0.181.2 mic/L. The slope of -0.375 hr-1 should not be surprising as the plot of the data in Figure 4-3 on page 53 resulted in a terminal slope of -.07 hr-1 . Since the data set yielded a bi-exponential plot, separating out the exponents could only yield K1 (0.375 hr-1) or K2 as determined by our Laplace Transform information. Thus, the terminal slope could be either -K1 or -K2. Since it was obviously not -K1, it had to be -K2. Thus the other rate constant obtained by stripping has to be K1. You can determine which slope is which rate constant if you have any data regarding intact drug (i e. either plasma or urine time profiles of intact drug) as the slope of any of those profiles is always
– K1 .
• You should be able to determine V dm if you have any urine data regarding intact drug (i.e. urine time profiles of intact drug) as the intercept of those profiles allow for the solution of k m . Thus the intercept, I, of the extrapolated line of equation 4-14 could be rearranged to contain only one unknown variable, V dm
–1 mic 0.375hr ⋅ 25mg ⋅ 1000 ---------------------km ⋅ X0 mg = ----------------------------------------------- = -------------------------------------------------------------------------- = 170 L . –1 ( K l arg e – K small ) ⋅ I mic ( 0.375 – 0.07 ) hr ⋅ 181.2--------L
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Utilization: MRT Calculations
• You should be able to determine the rate constants using MRT calculations. In a caternary chain, each compartment contributes its MRT to the overall MRT of the drug, thus: Flow Chart 2-4
IV Bolus
X
K1
MRT(IV) = 1/K1 Suppose the drug were given by IV bolus. Then the drug would have to be metabolized and the metabolite eliminated. Since the MRTs are additive, the overall MRT of the metabolite would be made up of the MRTs of the two processes, thus: Flow Chart 2-5
Metabolite km
kmu
X
Xm
MRT(met) = MRT(elim)+MRT(IV) MRT(met) =
1/K2
+ 1/K1
Thus, using the data from Table 4-3 on page 5 the MRT(IV)Trap is AUMC 1986.1 AUMC 2100 MRT = ------------------ = ---------------- = 2.42 hr or about MRT = ------------------ = ------------ = 2.67 hr using calculus. AUC 819.9 AUC 787
And using the data from columns 1 and 3 from Table 4-6 on page 52 the MRT(met) using calcuAUMC- = 36000 lus is MRT = ------------------------------- = 17 hr. AUC
2116
MRT(elim) = MRT(met) - MRT(IV) = 17 hr - 2.67 hr = 14.33 hr = 1/K2. Thus K2 = 0.07 hr-1.
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4.2.2
URINE Valid equations: dX mu k mu ⋅ k m X0 –K t –K t ------------- = --------------------------------------- ⋅ { e small – e l arg e } dt ( K l arg e – K small )
Utilization:
(EQ 4-17)
as in the previous urinary rate equation, clinically we work with the average rate over a definite interval which results in rewriting equation 4-17 as: ∆Xmu k mu ⋅ k m X0 –K t –K t ------------= --------------------------------------- ⋅ { e small mid – e l arg e mid } ∆t ( K l arg e – K small )
(EQ 4-18)
• You should be able to plot a data set of rate of metabolite excreted vs. time (mid) on semi-log paper yielding a bi-exponential curve.
• You should be able to obtain the slope of the terminal portion of the curve, the negative of which would be the smaller of the two rate constants (either
K1 or K2 ).
• You should be able to feather (curve strip) the other rate constant out of the data by plotting the difference between the extrapolated (to t = 0 ) terminal line and the observed data (at early times) yielding a straight line with the slope of the line equal to the negative of the other (larger) rate constant (either
K1 or K2 ).
• You should be able to utilize MRT calculations to obtain K1 and K2 . • You should be able to determine which slope is which rate constant if you have any data regarding intact drug (i.e. either plasma or urine time profiles of intact drug) as the slope of any of those profiles is always
– K1 .
By this time, it should be apparent that data which fits the same shape curve (mono-exponential, bi-exponential, etc.) are treated the same way. When the curves are evaluated, the slopes and intercepts are obtained in the same manner. The only difference is what those slopes and intercepts represent. These representations come from the equations which come from the LaPlace Transforms which come from our picture of the pharmacokinetic description of the drug. Please refer back to the section on graphical analysis in the Chapter 1, Math review for a interpretation of slopes and intercepts of the various graphs. Temporarily, please refer to exam section 1, chapter 14 for problems for this section (until problems can be generated) as well as additional problems for the previous sections.
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CHAPTER 5
I.V. Infusion
Author: Michael Makoid and John Cobby Reviewer: Phillip Vuchetich
OBJECTIVES 1.
Given patient drug concentration and/or amount vs. time profiles, the student will calculate (III) the relevant pharmacokinetic parameters available ( V d , K, k m , k r , AUC ,
Clearance, MRT) from IV infusion data.
2.
I.V. Infusion dosing for parent compounds
3.
Plasma concentration vs. time profile analysis
4.
Rate vs. time profile analysis
5.
Professional communication of IV Infusion information
6.
Computer aided instruction and simulation
7.
Metabolite (active vs. inactive)
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I.V. Infusion
5.1 Parent compound 5.1.1
PLASMA
Valid equations: – Kt Q C p = -------------( 1 – e ) or K ⋅ Vd
(EQ 5-1)
– Kt Dose C p = --------------------------------------( 1 – e ) K ⋅ Vd ⋅ T infusion
(EQ 5-2)
at any time during the infusion
Q ( C p )ss = -------------K ⋅ Vd
(EQ 5-3)
at steady state (t is long)
C p = C p( term ) ⋅ e
– Kt
(EQ 5-4)
after termination of infusion
Where C p is the plasma concentration Dose is the infusion rate shown in equation 5-1 and equation 5-2. Q = ------------------T infusion – Kt infusion Q ) is the plasma concentration when the C p( term ) = -------------- ( 1 – e K ⋅ Vd infusion is stopped.
Rewriting equation 5-4 to an equation which may be used by a computer results in: Q - ( e – K ⋅ T∗ – e – K ⋅ T ) Cp = ----------K⋅V where
T∗ = ( T – T i v )
and
T∗ = 0
(EQ 5-5)
for ( T > T iv )
for ( T < T iv )
.
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I.V. Infusion
Using The Scientist@‘s Unit function makes the change in T∗ straight forward. In The Scientist@, Unit(+) = 1 and Unit(-) = 0, so defining T∗ = ( T – T iv ) ⋅ UNIT ( T – Ti v ) meets these needs. This equation is utilized in The Scientist@’s companion product PKAnalyst@ also by MicroMath. Since the route of administration is an infusion and we would know how much we gave (Dose), how fast we gave it (Q), and over how long the infusion lasted (Tiv), the only other variables in the equation are K and Vd. PKAnalyst asks for Tiv and yields DoverV ( Dose ------------- ) and K as parameters resulting from Vd non-linear regression analysis. Dividing Dose by Dose ------------- yields Vd. Vd
Utilization:
You should be able to determine the infusion rate necessary to obtain a desired plasma concentration. Rearranging equation 5-3 results in: K ⋅ V d ⋅ ( C p )ss = Q
(EQ 5-6)
You should be able to determine how long it would take to get to a desired plasma concentration. Using equation 5-1 and equation 5-3, it looks like it will take forever to get exactly to steady state because in order for Q = -------------Q ( 1 – e – Kt ) , e – Kt → 0 which occurs when t = ∞ . So, ( C p )ss = -------------K ⋅ Vd K ⋅ Vd how close is close enough? If ( C p ) = 0.95 ⋅ ( C p )ss , that’s good enough in most people’s estimation. So in order to find out how long it will take we use equation 5-1, setting ( C p ) = 0.95 ⋅ ( C p )ss and solve for time. Thus: Q ( 1 – e – Kt ) which results in ( C p ) = 0.95 ⋅ ( C p )ss = -------------K ⋅ Vd 0.95 = ( 1 – e
– Kt
0.95 – 1 = – e
)
– Kt
ln ( 0.05 ) = – Kt – 2.996 = – Kt Basic Pharmacokinetics
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5-3
I.V. Infusion
–---------------2.996 = t –K 2.996 -------------t 1--- = 4.32t 1--- = t , 0.693 2 2
(EQ 5-7)
or about 4.32 half lives to get to 95% of steady state. Generalizing, then, the number of half-lives it takes to get to steady-state is equal to the logarithm of the inverse of how close is close (in this case, 5% or 0.05 = 20) devided by the logarithm of two. Changing infusion rates:
Occasionally, it is necessary to change infusion rates to stabilize the patient. If a patient were started on an infusion rate, Q1, and then at some subsequent time, T>T*, the infusion rate was changed to Q2, the equation for the concentration after the change would be: – ( K ⋅ T∗ ) – ( k ⋅ ( T – T∗ ) ) Q1 Q2 - ⋅ ( 1 – e – ( k ⋅ ( T – T∗ ) ) ) )⋅e + ----------Cp = ------------ ⋅ ( 1 – e K⋅V K⋅V
(EQ 5-8)
Assuming equilibrium was reached at infusion rate Q1, we could simplify equation 5-8 by setting T = 0 at the time of the rate change (thus we would be interested in the time after the change) resulting in: –k ⋅ T –k ⋅ T Q1 Q2 Cp = -----------⋅e + ------------ ⋅ ( 1 – e ) K⋅V K⋅V
(EQ 5-9)
Under these conditions, it would be useful to determine the time to reach the new equilibrium. As before, within 5% is close enough. Thus if we are coming down Q2- and if we were (lowering the Cp, i.e. Q2 < Q1), we would want Cp = 1.05 ----------K⋅V Q2 . Taking going up (raising the Cp, i.e. Q2 > Q1), we would want Cp = 0.95 ----------K⋅V the first condition we find: –k ⋅ T Q2 Q1 Q2 - ⋅ ( 1 – e – k ⋅ T ) + ----------Cp = 1.05 ------------ = ------------ ⋅ e K⋅V K⋅V K⋅V
(EQ 5-10)
Rearranging and solving for T results in: ⋅ Q2 ln 0.05 ---------------------Q1 – Q2 T = ---------------------------------–K
(EQ 5-11)
Similarly, under the second condition, we would find: Basic Pharmacokinetics
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5-4
I.V. Infusion
0.05 ⋅ Q2 ln –------------------------Q1 – Q2 T = ------------------------------------–K
(EQ 5-12)
Combining equation 5-11 and equation 5-12 and rearranging results in: Q1 – Q2- ⋅ 20 ln ----------------------Q1 – Q2- ⋅ 20 ln ---------------------- Q2 Q2 T = ---------------------------------------------- = ---------------------------------------------- ⋅ t 1 ⁄ 2 K 0.693
(EQ 5-13)
Thus it is the absolute value of the difference of the two rates and the elimination rate constant which determine the length of time needed to establish a new equilibrium. Under the conditions of Q1 = 0 , that is no previous infusion, and the difference is maximal equation 5-13 simplifies to equation 5-7. Under the conditions of Q1 = Q2 , the equation is undifined and has no utility (as well as makes no sense, because the equation was designed to be used when there was a change in rate.) However, lim T = 0 , thus no change results in zero time to get to the new Q2 → Q1
equilibium. Similar to equation 5-7 as before, the generalization for the number of half-lives it takes to obtain the new steady-state is the logarithm of (the fractional difference of the rates (or the steady-state concentrations) times the inverse of how close is close) devided by the logarithm of two. As pharmacokinetic equations are additive, you should be able to determine a loading dose (by I.V. bolus, for example) and a maintenance dose (infusion rate) for a patient to extablish an equilibrium. If, for example, you want to give a loading dose followed by an IV infusion, the generalization for the number of halflives it takes to obtain the new steady-state is the logarithm of (the fractional difference of the concentrations, Cp0 and Cpss, times the inverse of how close is close) devided by the logarithm of two.
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5-5
I.V. Infusion
Discussion:
Example: Using population average pharmacokinetic parameters to make professional judgements.
IV infusion is a controlled way to get drug into your patient. Using patient population average pharmacokinetic parameters (K, Vd) available in the drug monographs, you are able to make a professional judgement about: 1.
the plasma concentration that you would like to achieve (from therapeutic range) and the time in which you would like to get there.
2.
the infusion rate necessary to get to the target concentration, and
3.
the time necessary to to get there.
As an example, theophylline is a bronchodilator used in asthma with a therapeutic range of 10 to 20 mg/L, a volume of distribution of 0.45 (0.3 - 0.7) L/kg and a half life of about 8 (6 - 13) hours for a non-smoking adult. Your patient weighs 200 pounds and meets these these criteria. The physician decides to maintain him at 15 mg/L. What do you do? Using population average parameters for K and Vd, equation 5-6 results in: L- ⋅ ------------kg - ⋅ 200 lb ⋅ 15 mg 0.693 = 53.2mg - ⋅ 0.45 ----------- . ----------- ------8 hr kg 2.2 lb L hr For an eight hour IV infusion, you would need 53.2 mg ------- ⋅ 8 hr = 425 mg of theophylline. hr IV Theophylline comes as aminophylline which is theophylline compound containing 85% theophylline and 15% ethylenediamine. So in order to get 425 mg of theophylline we have to give 100 mg aminophylline 425 mg theophylline ⋅ ------------------------------------------------------ = 500 mg aminophylline . So we 85 mg theophylline prepare our IV infusion using Aminophylline U.S.P. for injection (500 mg aminophylline in 20 mL) by placing the contents of the ampule in 1000 mL of D5W and calculate the drip rate using an adult IV administration set which regulates the drip to 10 gtts/mL. Thus the drip rate is: 1020 ml- ⋅ ---------------hr - ⋅ 10 gtts- ∼ 21 -------gtts- ∼ -------------7 gtts--------------------------------8 hr 60 min ml min 20 sec
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5-6
I.V. Infusion
How long to get to steady state?
After setting up the infusion, the doctor asks, “How long to steady state? Using equation 5-7, our patient who has an eight hour half life, will take about 4.32 ⋅ 8 hr = 34.6 hr to get to 95% of steady state. The patient doesn’t want relief in a day and a half. He needs to breathe NOW. What would you suggest?
Infusion takes too long. How do we get relief now? IV Bolus stat.
It might be possible to give him an IV Bolus dose stat which would get him to Dose( C p )ss right away. This is done by converting ( C p )ss = -----------to Vd V d ⋅ ( C p ) ss = Dose . L- ⋅ ------------kg - ⋅ 200 lb ⋅ 15 mg 0.45 ----------- = 613.6 mg Theophylline kg 2.2 lb L Converting to aminophylline yields: mg aminophylline ∼ 725 mg aminophylline . Thus, 613.6 mg Theophylline ⋅ 100 -----------------------------------------------------85 mg theophylline if we gave a 725 mg IV bolus dose of aminophylline followed by a concomitant IV infusion of 500 mg aminophylline over 8 hours, our patient should get to steady state right away and stay there.
Some protocols require starting with faster infusion, then changing to a slower one to get to steady state faster.
Sometimes the physician might want to just increase the infusion rate (say double it for a short time, 2Q) to get to the target concentration faster and then just back the infusion down. If that is the protocol, the question becomes, “ How long do you run the infusion in at the faster rate?” Thus: – Kt – Kt Q 2Q ( C p )ss = -------------- = -------------- ( 1 – e ) which yields 1 = 2 ( 1 – e ) and so K ⋅ Vd K ⋅ Vd – Kt – Kt 1 1 --- = 1 – e . Thus --- – 1 = – e . Taking the ln of both sides ln ( 0.5 ) = –Kt 2 2
ln ( 0.5 ) = 0.693 ------------------------------ t 1--- = t 1--- = t or that it will take one half-life to get to the target –K 0.693 2 2 plasma concentration (which is the Cpss obtained by the infusion rate of 1Q) if you run the infusion at a faster rate, 2Q. So for your patient, you might suggest an infusion of 1000 mg over 8 hours (2Q for one half life) to get to steady state quickly and then back off to 500 mg over 8 hours for the second 8 hours.
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5-7
I.V. Infusion
Clearance: New pharmacokinetic parameter
Clearance ( Cl = K ⋅ Vd ) is a pharmacokinetic parameter which relates the fraction of the volume of distribution which is cleared of the drug per unit time. The volume of distribution is a mathematical construct which relates two knowns, the Dose of the drug and the resultant Concentration. In linear kinetics, the Dose is proportional to the Concentration. C ∝ D . The units of concentration are Mass - while the units of dose are Mass. So the units of the proportionality con------------------Volume stant must be volume in order for the equation to balance. Thus, the volume of distribution is a hypothetical volume and not necessarily a real volume or physiological space. Consequently, clearance is the hypothetical volume of fluid from which the drug is irreversibly removed per unit time. So equation 5-3 can be rewritten: Q( C p )ss = ----Cl
How do we calculate Clearance from IV infusion data?
(EQ 5-14)
and equation 5-14 can be rewritten to: Q Cl = -------------( C p )ss
(EQ 5-15)
Thus, assuming steady state, the clearance can be calculated by dividing the infusion rate by the resultant steady state plasma concentration. How do we separate K and Vd out of Clearance?
Graphing equation 5-4 which relates the decline in plasma concentration after cessation of the infusion, the resultant slope of the line yields - K, the elimination rate constant. Dividing the elimination rate constant, - K, obtained by equation 5-4 into the clearance obtained by equation 5-15 results in the other necessary pharmacokinetic parameter, Vd.
How can we utilize the rate of change of plasma concentration to determine the pharmacokinetic parameters, K and Vd?
From our original model d X = Q – (K ⋅ X ) dt
(EQ 5-16)
X- . Thus , V ⋅ C p = X . Rewriting equation 5-16 yields: and Cp = ----d Vd
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5-8
I.V. Infusion
dCp Q- – K ⋅ C and rearranging and incorporating equation 5-1 yields = ----p dt Vd dCp Q- – K ⋅ -------------Q ( 1 – e – Kt ) which can be simplified to = ---- dt Vd K ⋅ Vd dCp Q- – ----Q- + ----Q- e – Kt or = ----dt V d V d Vd dC p Q- e – Kt = ----dt Vd
(EQ 5-17)
dCp Cp vs. t ( actually, ∇ ----------- vs. tmid exactly like we did in urinary dt ∇t rate graphs) of the ascending portion of the plasma profile would result in a Q- . straight line with a slope of -K and an intercept of ----Vd Thus a plot of
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5-9
I.V. Infusion
5.2 Problems Equations needed for solving the problems: 1.
k from the slope of the terminal portion of the graph of Cp vs. T
2.
0.693 t 1 ⁄ 2 = -----------k
3.
Volume of distribution from Cp = -------------- ( 1 – e
4.
Clearance Cl = K ⋅ V d
5.
You wish to maintain a plasma concentration of Cpss.
Q K ⋅ Vd
a.
– Kt
)
Calculate the infusion rate necessary to maintain Cpss. Q = Cp ss ⋅ K ⋅ V d
b.
Suggest a loading dose which would give you Cpss immediately.
Dose loading = Cp ss ⋅ V d c.
How long will it take to reach steady state?
T 95 = 4.32 ⋅ T 1 ⁄ 2 d.
Find the plasma concentration if the infusion is discontinued at time = Tdc hours.
Q ( 1 – e – ( K ⋅ Tdc ) ) . Cp dc = -------------K ⋅ Vd e.
Find the plasma concentration Tpost hours after infusion is discontinued at time = Tdc hours.
Cp post = Cp dc ⋅ e
– ( K ⋅ T post )
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5-10
I.V. Infusion
Acyclovir
(Problem 5 - 1)
Problem Submitted By: Maya Leicht
AHFS 08:18.00 Antivirals
Problem Reviewed By: Vicki Long
GPI: 1200001000 Antivirals
Laskin, O., "Clinical pharmacokinetics of acyclovir", Clinical Pharmacokinetics (1983), p. 187 - 201.
Acyclovir (225.21 g/Mole) is an antiviral drug used in the treatment of herpes simplex, varicella zoster, and in suppressive therapy. In this study, patients were given varying doses of acyclovir over one hour by infusion. Acyclovir distributes uniformly into the plasma and tissues such that the plasma concentration is representative of tissue concentration. Acyclovir is 30% metabolized and 70% renally excreted. The following data was obtained from an intravenous infusion dose of 2.5 mg/kg over one hour where the patient weighed 70 kg. PROBLEM TABLE 5 - 1. Acyclovir
Plasma concentration Time (hours) 0
0
0.25
7
0.5
12
0.75
17
1
20
2
10
3
5
5
1
umol - -----------L
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Volume of distribution
4.
Clearance
5.
You wish to maintain a plasma concentration of 25 umol ⁄ L . a.
Calculate the infusion rate necessary to maintain a plasma concentration of 25 umol ⁄ L
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 5 hours.
e.
Find the plasma concentration 2 hours after infusion is discontinued at time = 5 hours.
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5-11
I.V. Infusion
“Acyclovir” on page 11
Concentration
102
101
100 0
1
2
3
4
5
Time 1.
k = 0.751 hr-1 (from slope of graph).
2.
t 1 ⁄ 2 = 0.923 hr (from slope of graph).
3.
Volume of distribution = 26.2 L
4.
Clearance = 19.67 l/hr
5.
You wish to maintain a plasma concentration of 25 umol ⁄ L . a.
Calculate the infusion rate necessary to maintain a plasma concentration of 25 umol ⁄ L = 111 mg/hr
b.
Suggest a loading dose for the patient which would give you Cpss immediately. 148 mg
c.
How long will it take to reach steady state? 4 hr
d.
Find the plasma concentration if the infusion is discontinued at time = 5 hours. = 25 umol ⁄ L
e.
Find the plasma concentration 2 hours after infusion is discontinued at time = 5 hours. = 5.6 umol ⁄ L
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5-12
I.V. Infusion
Aminophylline
(Problem 5 - 2)
Problem Submitted By: Maya Leicht
AHFS 12:12.00 Sympathomimetics
Problem Reviewed By: Vicki Long
GPI: 4430001000 Xanthine Sympathomimetic
Gilman, T., et al., "Estimation of theophylline clearance during intravenous aminophylline infusions", Journal of Pharmaceutical Sciences (May 1985), p. 508 - 514.
Aminophylline is used in the treatment of bronchospasm. In this study, aminophylline was given by intravenous infusion to patients with a mean weight of 75.7 kg. The doses given were chosen to maintain a between 10 -20 mg/L based on desirable body weight. The doses were given at a rate of 0.5 mg/kg/hour (Theophylline) for 84 hr. The following set of data was collected. PROBLEM TABLE 5 - 2. Aminophylline
Plasma concentration Time (hours) 0
mg ------L
0.
6
5
12
8
24
11
30
11.6
36
12.0
48
12.4
54
12.5
66
12.6
72
12.8
84
12.8
88
9
92
6.4
96
4.6
100
3.2
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Volume of distribution
4.
Clearance
5.
6.
You wish to maintain a plasma concentration of 15 mg/L in your patient.
a.
Calculate the infusion rate necessary to maintain a plasma concentration of 15 mg/L.
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 5 hours.
e.
Find the plasma concentration 2 hours after infusion is discontinued at time = 5 hours.
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5-13
I.V. Infusion
“Aminophylline” on page 13
CONCENTRATION
102
101
0
10
0
20
40
60
80
100
Time 1.
k = 0.085 hr-1
2.
t 1 ⁄ 2 = 8.15 hr
3.
Vd = 35.3 L
4.
Cl = 3 L/hr
5a.
Q = 45 mg/hr
5b.
D L = 530 mg
5c.
t
5d.
C p = 5.2 mg/L
5e.
C p = 4.4 mg/L
ss = 35 hr 95%
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5-14
I.V. Infusion
Carmustine
(Problem 5 - 3)
Problem Submitted By: Maya Leicht
AHFS 10:00.00 Antineoplastics
Problem Reviewed By: Vicki Long
GPI: 2110201000 Antineoplastics, Nitrosoureas
Henner, W., et al., "Pharmacokinetics and immediate effects of high-dose carmustine in man", Cancer Treatment Reports vol.70 (1986), p. 877 - 880.
Carmustine (BCNU) is an antineoplastic agent with a molecular weight of 214.04 g. 2
In this study a 70 kg, 1.8 M2 patient was given 600 mg ⁄ m by intravenous infusion over 2 hours. The following data was obtained. PROBLEM TABLE 5 - 3. Carmustine
Plasma concentration Time (minutes) 15
.3
30
.5
60
.7
90
.75
120
.8
135
.5
142.5
.4
150
.3
mg ------L
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4
Cl
5.
A patient with a BSA of1.8 M is to be given BCNU by IV infusion. You wish to maintain a plasma
2
concentration of 2 uM . Determine the following: a.
Calculate the infusion rate necessary to maintain a plasma concentration of 2 uM .
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 10 minutes.
e.
Find the plasma concentration 1 hour after infusion is discontinued at time = 10 minutes.
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5-15
I.V. Infusion
“Carmustine” on page 15
0
CONCENTRATION
10
-1
10
0
50
100
150
TIME
1.
k = 0.031 min-1
2.
t 1 ⁄ 2 = 22 min
3.
Vd = 198 L/M2
4
Cl = 6.15 L/M2/hr
5.
A patient with a BSA of1.8 M is to be given BCNU by IV infusion. You wish to maintain a plasma
2
concentration of 2 uM . Determine the following: a.
Calculate the infusion rate necessary to maintain a plasma concentration of 2 uM .
214µg- ⋅ -----------------mg - ⋅ -----------198L ⋅ 1.8M 2 ⋅ 0.031min– 1 = 4.73mg Q = Cp ss ⋅ V d ⋅ K = 2µmole ------------------- ⋅ -------------------------------- ∼ 285mg ----------------L µmole 1000µg M2 min hr b.
Suggest
a
loading
dose
for
the
patient
which
would
give
you
Cpss
immediately.
214µg- ⋅ -----------------mg - ⋅ -----------198L ⋅ 1.8M 2 = 150mg Dose = Cp ss ⋅ Vd = 2µmole ------------------- ⋅ --------------L µmole 1000µg M 2 c.
How long will it take to reach steady state? 4.32 * T 1/2 = 97 min.
d.
Find the plasma concentration if the infusion is discontinued at time = 10 min. = 0.1197 mg/L
e.
Find the plasma concentration 1 hour after infusion is discontinued at time = 10 min. = 0.017 mg/L
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5-16
I.V. Infusion
Cefotaxime
(Problem 5 - 4)
Problem Submitted By: Maya Leicht
AHFS 08:12.06 Cephalosporins
Problem Reviewed By: Vicki Long
GPI: 0230007510 Cephalosporins - 3rd Generation
Kearns, G., Young, R., and Jacobs, R., "Cefotaxime dosage in infants and children--pharmacokinetic and clinical rationale for an extended dosage interval", Clinical Pharmacokinetics (1992), p. 284 - 297.
Cefotaxime is a third generation cephalosporin which is widely used as an antimicrobial in neonates, infants, and children. In this study, infants and children were given a 50 mg/kg dose of cefotaxime intravenously over 0.25 hour. The following data was collected: PROBLEM TABLE 5 - 4. Cefotaxime
Plasma concentration Time (hours) 0.00
0
0.05
35
0.10
70
0.20
140
0.35
155
0.60
130
0.85
110
1.20
80
1.30
75
2.00
45
2.40
35
3.40
15
4.50
8
6.50
1.7
mg ------L
From this data, assuming that the patient weighs 30 kg, determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
5.
You wish to maintain a plasma concentration of 80 mg/L. Determine the following: a.
Calculate the infusion rate necessary to maintain a plasma concentration of 80mg/L
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 0.25 hours.
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5-17
I.V. Infusion
e.
Find the plasma concentration 2 hours after infusion is discontinued at time = 0.25 hours.
“Cefotaxime” on page 17
CONCENTRATION
103
102
101
100 0
1
2
3
4
5
6
7
TIME
1.
k = 0.733 hr-1
2.
t 1 ⁄ 2 = 0.945 hr
3.
Vd = 0.276 L/kg
4.
Cl = 0.202 L/kg/hr
4a.
Q = 16.2 mg/kg/hr
4b.
D L = 22.1 mg/kg
4c.
t
4d.
C p = 13.35 mg/L
4e.
C p = 3.09 mg/L
ss = 4.1 hr 95%
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5-18
I.V. Infusion
Ganciclovir
(Problem 5 - 5)
Problem Submitted By: Maya Leicht
AHFS 08:18.00 Antivirals
Problem Reviewed By: Vicki Long
GPI: 1200002010 Antivirals
Trang, J., et al., "Linear single-dose pharmacokinetics of ganciclovir in newborns with congenital cytomegalovirus infections", Clinical Pharmacology and Therapeutics (1993), p. 15 - 21.
Ganciclovir is used against the human herpes viruses, cytomegalovirus retinitis, and cytomegalovirus infections of the gastrointestinal tract. In this study, twenty-seven newborns with cytomegalovirus disease were given 4 mg/kg of ganciclovir intravenously over one hour. Blood samples were taken and the data obtained is summarized in the following table: PROBLEM TABLE 5 - 5. Ganciclovir
Plasma concentration --------
ug mL
Time (hours) 0.5
3.10
1.5
4.50
2.0
3.80
3.0
2.90
4.0
2.30
6.0
1.50
8.0
0.88
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
4.
A patient is to be given ganciclovir by IV infusion to an infant weighing 6.1 kg. You wish to maintain a plasma concentration of 5.5 mcg/mL. Determine the following: a.
Calculate the infusion rate necessary to maintain a plasma concentration of 5.5mcg/mL.
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 1 hour.
e.
Find the plasma concentration 2 hours after infusion is discontinued at time = 1 hour.
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5-19
I.V. Infusion
“Ganciclovir” on page 19
CONCENTRATION
10 1
10 0
10-1 0
2
4
6
8
Time
1.
k = 0.255 hr-1
2.
t 1 ⁄ 2 = 2.72 hr
3.
Vd = 0.687 L/kg
4.
Cl = 0.175 L/kg/hr
5a.
mg - ⋅ -----------------1000ml ⋅ 0.687L Q = Cp ss ⋅ V d ⋅ K = 5.5µg -------------- ⋅ ------------------ ⋅ 6.1kg ⋅ 0.255 ------------- = 5.9mg -------------- ---------------ml 1000µg L kg hr hr
5b.
mg - ⋅ -----------------1000ml ⋅ 0.687L D L = Cp ss ⋅ Vd = 5.5µg -------------- ⋅ ---------------------------------- ⋅ 6.1kg = 23mg kg ml 1000µg L
5c.
T
ss = 4.32 ⋅ t1 ⁄ 2 = 11.75hr 95% 5.9mg --------------– Kt – K ⋅ 1hr Q hr = -------------- ( 1 – e ) = ----------------------------------------------------- (1 – e ) = 1.24mg -----------------K ⋅ Vd 0.255 L ------------- ⋅ 0.687L ----------------- ⋅ 6.1kg hr kg
5d.
C p term
5e.
C p = C p term ⋅ e
– K ⋅ 2hr
= 1.24mg ------------------ ⋅ 0.6 = 0.74mg -----------------L L
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5-20
I.V. Infusion
Gentamicin
(Problem 5 - 6)
Problem Submitted By: Maya Leicht
AHFS 00:00.00
Problem Reviewed By: Vicki Long
GPI: 0700002010 Aminoglycosodes
Kaojarern, S., et al., "Dosing regimen of gentamicin during intermittent peritoneal dialysis", Journal of Clinical Pharmacology (1989), p. 140 - 143.
Gentamicin is an aminoglycoside antibiotic which is frequently used in the treatment of gram-negative bacilli infections. Since it has a low therapeutic index, it is important to determine proper dosage regimens. In this study, patients on peritoneal dialysis received a 30 minute intravenous infusion of 80 mg gentamicin in 100 mL of 5% dextrose in water. The following data was collected: PROBLEM TABLE 5 - 6. Gentamicin
Plasma concentration Time (hours) 0.50
5.68
1.50
5.15
3.70
4.80
7.35
3.99
11.30
3.35
24.00
2.02
ug ------ mL-
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
5.
A patient is to be given gentamicin by IV infusion. You wish to maintain a plasma concentration of 5.2 ug ⁄ mL . Determine the following: a.
Calculate the infusion rate necessary to maintain a plasma concentration of 5.2 ug ⁄ mL
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 0.5 hours.
e.
Find the plasma concentration 2 hours after infusion is discontinued at time = 0.5 hours.
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
5-21
I.V. Infusion
“Gentamicin” on page 21
CONCENTRATION
101
100 0
5
10
15
20
25
Time
1.
k = 0.0431 hr-1
2.
t 1 ⁄ 2 = 16.1 hr
3.
V d = 14.5 L
4.
Cl =0.625 L/hr
5a.
Q = 3.25 mg/hr
5b.
D L = 75 mg
5c.
t
5d.
C p = 0.11 mg/L
5e.
C p = 0.10 mg/L
ss = 69.6 hr 95%
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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5-22
I.V. Infusion
Human Monoclonal Anti-lipid A antibody (HA-1A) Problem Submitted By: Maya Leicht
AHFS 24:06.00 Antilepemics
Problem Reviewed By: Vicki Long
GPI: 3900000000 Antihyperlipidemic
(Problem 5 - 7)
Fisher, C., et al., "Initial evaluation of human monoclonal anti-lipid A antibody (HA-1A) in patients with sepsis syndrome", Critical Care Medicine (1990), Vol.18, No. 12, p. 1311 - 1315.
HA-1A is an immunoglobulin antibody. In this study, patients received a 250 mg intravenous infusion of HA-1A over 15 minutes. Serum levels were measured before and after infusion and the following data was collected: PROBLEM TABLE 5 - 7. Human
Monoclonal Anti-lipid A antibody (HA-1A)
Plasma concentration Time (hours) 0.00
0
0.75
80
1.00
75
2.00
74
5.00
65
15.00
50
25.00
40
48.00
21
72.00
10
ug- ------ mL
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
5. A patient is to be given HA-1A by IV infusion. You wish to maintain a plasma concentration of Determine the following:
100 µg/mL.
a.
Calculate the infusion rate necessary to maintain a plasma concentration of 100 ug ⁄ mL .
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 1 hour.
e.
Find the plasma concentration 3 hours after infusion is discontinued at time = 1 hour.
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
5-23
I.V. Infusion
“Human Monoclonal Anti-lipid A antibody (HA-1A)” on page 23
CONCENTRATION
102
101 0
20
40
60
80
Time 1.
k = 0.0282 hr-1
2.
t 1 ⁄ 2 = 24.4 hr
3.
V d = 3.2 L
4.
Cl = 0.09 L/hr
5a.
Q = 9 mg/hr
5b.
D L = 320 mg
5c.
t
5d.
C p = 2.78 mg/L
5e.
C p = 2.56 mg/L
ss = 105 hr 95%
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
5-24
I.V. Infusion
Ifosfamide
(Problem 5 - 8)
Problem Submitted By: Maya Leicht
AHFS 00:00.00
Problem Reviewed By: Vicki Long
GPI: 0000000000
Lewis, L., "The pharmacokinetics of ifosfamide given as short and long intravenous infusions in cancer patients", British Journal of Clinical Pharmacology Vol. 31 (1991), p. 77 - 82.
Ifosfamide is an agent which has shown some pharmacological response in the treatment of cancer. In this study, a 5
g⁄m
2
2
dose of ifosfamide was infused over 30 minutes. The median BSA for the subjects was 1.8 m . The
following data was obtained: PROBLEM TABLE 5 - 8. Ifosfamide
Plasma concentration Time (hours) 0
0.0
0.5
285.0
1
260.0
2
220.0
4
160.0
6
112.0
8
80.0
10
60.0
24
5
ug ------ mL-
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
5. A patient is to be given ifosfamide by IV infusion. The patient has a BSA 1.8 M2. You wish to maintain a plasma concentration of 336 µg/mL. Determine the following: a.
Calculate the infusion rate necessary to maintain a plasma concentration of 336 ug ⁄ mL
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 20 min.
e.
Find the plasma concentration 2 hours after infusion is discontinued at time = 20min.
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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5-25
I.V. Infusion
“Ifosfamide” on page 25
CONCENTRATION
103
102
101
100 0
5
10
15
20
25
Time 1.
k = 0.1716 hr-1
2.
t 1 ⁄ 2 = 4.04 hr
3.
V d = 16.6 L/M2
4.
Cl = 2.85 L/hr/M2
5a.
Q = 1.725 g/hr
5b.
D L = 10 g
5c.
t
5d.
C p = 18.7 mg/L
5e.
C p = 13.25 mg/L
ss = 17.5 95%
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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5-26
I.V. Infusion
Isosorbide 5-mononitrate Problem Submitted By: Maya Leicht
AHFS 00:00.00
Problem Reviewed By: Vicki Long
GPI: 0000000000
(Problem 5 - 9)
Major, R., et al., "Isosorbide 5-mononitrate kinetics" (1983), p. 653- 660.
Isosorbide 5-mononitrate (5-ISMN) is a metabolite of isosorbide dinitrate. In this study, the kinetics of isosorbide 5mononitrate were looked at in 12 healthy patients after an intravenous infusion of 20 mg at 8 mg/hour for 2.5 hours. This drug follows one-compartment, open model kinetics. The following data was collected: PROBLEM TABLE 5 - 9. Isosorbide
5-mononitrate
Time (hours)
Plasma concentration (ng/mL)
0.25
40
0.50
91
0.75
141
1.00
181
1.50
239
2.00
305
2.50
351
3.00
335
3.50
303
4.50
257
5.50
216
7.50
162
9.50
117
11.50
77
14.50
47
18.50
24
26.50
7
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
5.
A patient is to be given 5-ISMN by IV infusion. You wish to maintain a plasma concentration of 300 ng/mL. If the volume of distribution of 5-ISMN is 44.5, determine the following: a.
Calculate the infusion rate necessary to maintain a plasma concentration of 300 ng/mL.
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 1 hour.
e.
Find the plasma concentration 2 hours after infusion is discontinued at time = 1 hour.
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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5-27
I.V. Infusion
“Isosorbide 5-mononitrate” on page 27
CONCENTRATION
103
102
101
100 0
5
10
15
20
25
30
Time 1.
k = 0.168 hr-1
2.
t 1 ⁄ 2 = 4.125 hr
3.
V d = 44.6 L
4.
Cl = 7.5 L/hr
5a.
Q = 2.25 mg/hr
5b.
D L = 13.4 mg
5c.
t
5d.
C p = 46.4 ng/mL
5e.
C p = 33.2 ng/mL
ss = 17.8 hr 95%
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
5-28
I.V. Infusion
Moclobemide
(Problem 5 - 10)
Problem Submitted By: Maya Leicht
AHFS 00:00.00
Problem Reviewed By: Vicki Long
GPI: 0000000000
Schoerlin, M., et al., "Disposition kinetics of moclobemide a new MAO-A inhibitor, in subjects with impaired renal function", Journal of Clinical Pharmacology Vol 30 (1990), p. 272 - 284.
Moclobemide is reversibly inhibits the A-isozyme of the monoamine oxidase enzyme system. In this study, twelve patients received a 96.7 mg dose as an intravenous infusion over 20 minutes. Blood samples were obtained during the infusion and after the infusion was ended and the following data was obtained: PROBLEM TABLE 5 - 10. Moclobemide
Time (hours)
Plasma concentration (mg/L)
0.0
0.000
0.2
0.6
0.4
1
0.7
0.85
0.9
0.750
1.2
0.70
1.6
0.60
1.9
0.50
2.4
0.40
3.4
0.25
4.5
0.15
5.5
0.10
6.4
0.070
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
5.
A patient is to be given moclobemide by IV infusion. You wish to maintain a plasma concentration of 1mg/L. Determine the following: a.
Calculate the infusion rate necessary to maintain a plasma concentration of 1mg/L.
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 15 min.
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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5-29
I.V. Infusion
e.
Find the plasma concentration 3 hours after infusion is discontinued at time = 15 mins.
“Moclobemide” on page 29
CONCENTRATION
10 0
10-1
10-2 0
1
1.
k = 0.44 hr-1
2.
t 1 ⁄ 2 = 1.6 hr.
3.
V d = 90.4 L
4.
Cl = 39.8 L/hr
5a.
Q = 40 mg/hr
5b.
D L = 90 mg
5c.
t
5d.
C p = 0.1 mg/L
5e.
C p = 0.028 mg/L
2
3
Time
4
5
6
7
ss = 6.8 hr 95%
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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5-30
I.V. Infusion
Obidoxime
(Problem 5 - 11)
Problem Submitted By: Maya Leicht
AHFS 00:00.00
Problem Reviewed By: Vicki Long
GPI: 0000000000
Bentur, Y., et al., "Pharmacokinetics of obidoxime in organophosphate poisoning associated with renal failure", Clinical Toxicology (1993), Vol. 31, p. 315 - 322.
Obidoxime is an agent which is used as an antidote in organophosphate poisoning. In this study, the pharmacokinetics of obidoxime were studied in a 20 year old patient who attempted to commit suicide by ingesting Tamaron (60% methamidophos, an organophosphate, in ethylene glycol monethyl ether). She was given 4 mg/kg Obidoxime by intravenous infusion over 10 minutes and the following data was collected: PROBLEM TABLE 5 - 11. Obidoxime
Time (minutes)
Plasma concentration
5
9
10
18
15
17
30
16
45
15
60
14
90
12
120
11
150
9.3
180
8
240
6.1
300
4.6
µg ⁄ mL
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
5.
A patient is to be given obidoxime by IV infusion. The patient has a body weight of 60 kg. You wish to maintain a plasma concentration of 10 µg/mL. Determine the following: a.
Calculate the infusion rate necessary to maintain a plasma concentration of 10 ug ⁄ mL .
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 30 minutes.
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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5-31
I.V. Infusion
e.
Find the plasma concentration 1 hour after infusion is discontinued at time = 30 minutes.
“Obidoxime” on page 31
CONCENTRATION
102
101
100 0
50
100
150
200
250
300
Time 1.
k = 0.00463 min-1
2.
t 1 ⁄ 2 = 150 min
3.
V d = 0.22L/kg
4.
Cl = 1 mL/min
5a.
Q = 0.61 mg/min
5b.
D L = 132 mg
5c.
t
5d.
C p = 1.3 mg/L
5e.
C p = 0.98 mg/L
ss = 10.8 hr 95%
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
5-32
I.V. Infusion
Perindoprilat
(Problem 5 - 12)
Problem Submitted By: Maya Leicht
AHFS 00:00.00
Problem Reviewed By: Vicki Long
GPI: 0000000000
Macfadyen, R., Lees, K., and Reid, J., "Studies with low dose intravenous diacid ACE inhibitor (perindoprilat) infusions in normotensive male volunteers", Journal of Pharmaceutical Sciences (1991), p. 115 - 121.
Perindoprilat and other ACE inhibitors are used in the management of hypertension and chronic congestive heart failure. In this study, a 1 mg dose was infused over a one hour period. The following data was collected: PROBLEM TABLE 5 - 12. Perindoprilat
Time (minutes)
Plasma concentration
5
4.0
10
9.0
20
16.0
30
24.0
40
30.0
50
36.0
60
42.0
65
40.0
70
38.0
80
35.0
90
32.0
100
29.0
110
27.0
120
24.0
ng ⁄ mL
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
5.
A patient is to be given perindoprilat by IV infusion. You wish to maintain a plasma concentration of 30 ng/ml. Determine the following: a.
Calculate the infusion rate necessary to maintain a plasma concentration of 30 ng/mL.
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 5 hours.
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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5-33
I.V. Infusion
e.
Find the plasma concentration 2 hours after infusion is discontinued at time = 5 hours.
“Perindoprilat” on page 33
CONCENTRATION
102
101
100 0
20
40
60
80
100
120
Time 1.
k = 0.0087 min-1
2.
t 1 ⁄ 2 = 79.6 min
3.
V d = 18.9 L
4.
Cl =164 mL/min
5a.
Q = 5 µg/min
5b.
D L = 0.57 mg
5c.
t
5d.
C p = 27.8 ng/mL
5e.
C p = 9.8 ng/mL
ss = 5.73 hr 95%
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
5-34
I.V. Infusion
Sulfonamides
(Problem 5 - 13)
Problem Submitted By: Maya Leicht
AHFS 00:00.00
Problem Reviewed By: Vicki Long
GPI: 0000000000
Boddy, A., Edwards, P., and Rowland, M., "Binding of sulfonamides to carbonic anhydrase: influence on distribution within blood and on pharmacokinetics", Pharmaceutical Research (1989), p. 203- 209
This study looks at the affinity of sulfonamides for carbonic anhydrase. Doses of 8 micromoles/kg were administered via the jugular vein cannula in approximately 0.5 mL of PEG 400 over 5 minutes at a constant rate. Samples were collected during the infusion period and for 30 minutes afterward. The following set of data was collected: PROBLEM TABLE 5 - 13.
Sulfonamides
Time (minutes)
Plasma concentration
2.0
17.0
4.0
31.0
5.0
37.0
7.5
32.0
9.0
28.0
12.0
22.5
15.0
18.0
18.0
14.0
23.0
11.0
30.0
6.5
35.0
4.5
( µM )
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
5.
A 70-kg patient is to be given a sulfonamide by IV infusion. You wish to maintain a plasma concentration of 30 µM. Determine the following: a.
Calculate the infusion rate which would be necessary to maintain the plasma concentration of 30 µM.
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 4 hours.
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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5-35
I.V. Infusion
e.
Find the plasma concentration 30 minutes after stopping infusion at time = 4 hours.
“Sulfonamides” on page 35
CONCENTRATION
102
101
100 0
5
10
15
20
25
30
35
Time 1.
k = 0.0705 min-1
2.
t 1 ⁄ 2 = 9.8 min
3.
V d = 0.18 L/kg
4.
Cl = 12.7 mL/min/kg
5a.
Q = 26.9 µmole/min
5b.
D L = 380 µmole
5c.
t
5d.
C p = 30 µmole/L
5e.
C p = 3.6 µmole/L
ss = 42 min 95%
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
5-36
I.V. Infusion
Terodiline
(Problem 5 - 14)
Problem Submitted By: Maya Leicht
AHFS 00:00.00
Problem Reviewed By: Vicki Long
GPI: 0000000000
Hallen, B. ,et al., "Bioavailability and disposition of terodiline in man", Journal of Pharmaceutical Sciences (1994), p. 1241 1246.
Terodiline is an agent which works as an anticholinergic and a calcium antagonist. It is used to treat incontinence. It is metabolized into p-Hydroxyterodiline, which is further metabolized to 3,4-dihydroxyterodiline. The parent drug and all of its metabolites are excreted into the urine as well as the feces. A patient is given 12.5 mg of Terodiline by IV infusion at a rate of 1 mL/ minute for 5 minutes. The following data is collected: PROBLEM TABLE 5 - 14.
Terodiline Time (hours)
Plasma concentration
25.000
31
50.000
23
75.000
15
100.000
12
125.000
8
150.000
6
175.000
4
200.000
3
225.000
2
µg ⁄ L
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
5.
A patient is to be given terodiline by IV infusion. You wish to maintain a plasma concentration of 40 mcg/L. Determine the following: a.
Calculate the infusion rate necessary to maintain the plasma concentration of40 mcg/L.
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 5 hours.
e.
Find the plasma concentration 2 hours after stopping infusion if the infusion ended at time = 5 hours.
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
5-37
I.V. Infusion
“Terodiline” on page 37
CONCENTRATION
102
101
100 0
50
100
150
200
250
Time 1.
k = 0.0136 hr-1
2.
t 1 ⁄ 2 = 50.9 hr
3.
V d = 283 L
4.
Cl =3.85 L/hr
5a.
Q = 0.154 mg / hr
5b.
D L = 11.32 mg
5c.
t
5d.
C p = 2.63 µg/L
5e.
C p = 2.56 µg/L
ss = 220 hr 95%
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
5-38
I.V. Infusion
Tinidazole
(Problem 5 - 15)
Problem Submitted By: Maya Leicht
AHFS 00:00.00
Problem Reviewed By: Vicki Long
GPI: 0000000000
Robson, R., Bailey, R., and Sharman, J., "Tinidazole pharmacokinetics in severe renal failure", Clinical Pharmacokinetics (1984), p. 88 - 94.
Tinidazole is an antimicrobial similar to metronidazole which is used in the treatment of trichomoniasis, giardiasis, amoebiasis, and anaerobic infections. This study focuses on the pharmacokinetics of tinidazole in patients suffering from severe renal failure. Twelve patients received 800 mg of tinidazole dissolved in 400 mL of dextrose monohydrate solution as an intravenous infusion at a rate of 60 mg/min. Blood samples were taken and the following data was obtained: PROBLEM TABLE 5 - 15.
Tinidazole Time (hours)
Plasma concentration (mg/L)
1
14.9
3
13.1
6
11.2
12
8.9
24
5.1
48
2.1
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
5.
A patient is to be given tinidazole by IV infusion. Determine the following: a.
Calculate the infusion rate necessary to maintain the plasma concentration of 25 mg/L.
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 1 hour.
e.
Find the plasma concentration 2 hours after stopping infusion if the infusion was stopped at time = 1 hour.
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
5-39
I.V. Infusion
“Tinidazole” on page 39
CONCENTRATION
102
101
100 0
10
1.
k = 0.04136 hr-1
2.
t 1 ⁄ 2 = 16.75 hr
3.
V d = 54.7 L
4.
Cl = 2.26 L/hr
5a.
Q = 56.6 mg/hr
5b.
D L = 1.37 g
5c.
t
5d.
C p = 1 mg/L
5e.
C p = 0.93 mg/L
20
Time
30
40
50
ss = 72.4 hr 95%
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
5-40
I.V. Infusion
Tobramycin
(Problem 5 - 16)
Problem Submitted By: Maya Leicht
AHFS 00:00.00
Problem Reviewed By: Vicki Long
GPI: 0000000000
Cooney, G., et al., "Absolute bioavailability and absorption characteristics of aerosolized tobramycin in adults with cystic fibrosis", Journal of Clinical Pharmacology Vol. 34, (1994), p. 255- 259.
Most persons with cystic fibrosis (CF) become colonized with Pseudomonas aeruginosa in their bronchial secretions within their second decade of life. These patients require frequent treatment with potent anti-pseudomonal antibiotics such as Tobramycin. In this study, an intravenous infusion of 2.5 mg/kg tobramycin was given over 35 minutes. The following data was collected: PROBLEM TABLE 5 - 16.
Tobramycin
Plasma concentration Time (minutes) 35
8.00
60
6.00
90
4.50
150
2.50
270
0.75
mg ------L
From this data determine the following: 1.
k
2.
t1 ⁄ 2
3.
Vd
4.
Cl
5.
A patient is to be given tobramycin by IV infusion. The patient has a body weight of 70 kg. You wish to maintain a plasma concentration of 10 mg/L. Determine the following: a.
Calculate the infusion rate necessary to maintain the plasma concentration of 10 mg/mL.
b.
Suggest a loading dose for the patient which would give you Cpss immediately.
c.
How long will it take to reach steady state?
d.
Find the plasma concentration if the infusion is discontinued at time = 30 minutes.
e.
Find the plasma concentration 1 hour after stopping infusion if the infusion wasstopped at time = 30 minutes.
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
5-41
I.V. Infusion
“Tobramycin” on page 41
CONCENTRATION
101
100
10-1 0
50
100
150
200
250
300
Time 1.
k = 0.01 min-1
2.
t 1 ⁄ 2 = 69.3 min
3.
V d = 0.269 L/kg
4.
Cl = 2.7 mL/min
5a.
Q = 0.027mg/kg/min = 1.62 mg/kg/hr
5b.
D L = 2.7 mg/kg
5c.
t
5d.
C p = 2.6 mg/L
5e.
C p = 1.43 mg/L
ss = 300 min = 5 hr 95%
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5-42
CHAPTER 6
Biopharmaceutical Factors
Author: Michael Makoid Reviewer: Phillip Vuchetich
OBJECTIVES After successfully completing this chapter, the student shall understand: 1.
Physiology and machanisms of absorbtion
2.
Effects of diffusion, cardiac output / blood perfusion, physical properties of the drug and body on distribution
3.
Biotransformation, first pass effect, and clearance
4.
Renal, biliary, mammary, salivary, other forms of excretion.
5.
identify the effects of physiological changes with age, sex, and disease on the absorption, distribution, metabolism, and excretion of a drug.
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Oral Dosing
Author: Michael Makoid and John Cobby Reviewer: Phillip Vuchetich
OBJECTIVES After successfully completing this chapter, the student shall be able to 1.
Given patient drug concentration and/or amount vs. Time profiles, the student will calculate (III) the relevant pharmacokinetic parameters ( V d , K, k m , k r , k a , AUC , Clearance, MRT, MAT) available from oral data.
2.
Given patient drug concentration and/or amount vs. Time profiles, the student will calculate (III) the K from the terminal portion of the curve.
3.
Given patient drug concentration and/or amount vs. Time profiles, the student will calculate (III) the k a from either the curve stripping Moment techniques.
4.
Given patient drug concentration and/or amount vs. Time profiles, the student will calculate (III) the Absolute Bioavailability from comparing IV and oral (or some other process which involves absorption) data.
5.
Given patient drug concentration and/or amount vs. Time profiles, the student will calculate (III) the Comparative Bioavailability from comparing the generic to the inovator product.
6.
Given patient drug concentration and/or amount vs. Time profiles, the student will qualitatively evaluate (IV) bioequivalence as determined by rate of absorption (peak time) and extent of absorption (Area Under the Curve - AUC, and ( Cp ) max ).
7.
Given patient drug concentration and/or amount vs. Time profiles, the student will evaluate (IV) bioequivalence data.
8.
Given patient drug concentration and/or amount vs. Time profiles, the student will lucidly discuss (IV) bioequivalence and recommend (V) to another competant professional if s/he believes products to be equivalent.
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9.
The student shall be able to properly use vocabulary relative to bioequivalence.
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7.1 Oral dosing 7.1.1
VALID EQUATIONS: ( ORAL DOSING, PLASMA)
where
CB
ka –k t – Kt C p = fD ------ ⋅ -------------- ⋅ (e – e a ) Vd k a – K
(EQ 5-18)
AUC ( oral ) ⁄ Dose ( oral ) f = -------------------------------------------------------------AUC ( iv ) ⁄ Dose ( iv )
(EQ 5-19)
AUC ( generic ) ⁄ Dose ( generic ) CB = ------------------------------------------------------------------------------------AUC ( inovator ) ⁄ Dose ( inovator )
(EQ 5-20)
ln ( k a ⁄ K ) t p = ----------------------( ka – K )
(EQ 5-21)
X -----a- = K ⋅ AUC ∞ – ( C p + K ⋅ AUC t ) v
(EQ 5-22)
= the comparative bioavailability
f = the absolute bioavailabilty; the fraction of dose which ultimately reaches systemic circulation (which is made up of the fraction of the dose which is absorbed times the fraction which gets past the liver (first pass effect)) ka
7.1.2
= absorption rate constant.
UTILIZATION
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Ampicillin
(Problem 5 - 17)
The following information is available for ampicillin: 90% is excreted unchanged and a 250 mg IV bolus dose yields an AUC of 11 mic/mL*hr. The following blood level profile has been reported for two brands of ampicillin which were given as 500 mg oral capsules.
TABLE 4-7
Time (hr)
µg MEAN SERUM LEVEL -------mL LEDERLE
BRISTOL
0.5
0.37
0.38
1.0
1.97
1.91
1.5
2.83
2.49
2.0
3.15
3.11
3.0
2.73
2.79
4.0
1.86
1.95
6.0
0.43
0.49
Find the following:. a.
k for both products.
b.
k a for both products.
c.
k u for both products.
d.
AUC for both products.
e.
f for both products.
f.
t max for both products.
g.
Cl
h.
Vd
i
Cp 0 for a 250 mg IV dose.
j.
Cp max
k.
Are these two products bioequivalent? Why or why not?
l.
What infusion rate would be necessary to maintain a serum
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Oral Dosing
plasma concentration of 2mcg/mL
The data was plotted as above with the best fit line drawn. From the graph the following parameters were derived: TABLE 4-8 Comparison
of Ampicillin Lederle
Bristol
–1
K ( hr )
0.688
0.635
Ratio (L/B)
K a ( hr )
–1
0.858
0.831
T max ( hr )
1.74
1.8
0.97
( C p ) max ( µg ⁄ mL )
3
2.9
1.03
AUC (trapaziodal)
11.4
11.6
0.98
2) In a clinical study (DiSanto & DeSante, JPS 64:100,1975) prednisone was administered to 22 adult healthy volunteres (average weight 64.5 kg) either as one 50 mg tablet (product A) or as ten 5 mg tablets (product B). The following data was observed: Time (hours)
Concentration (mic/100ml plasma) A
B
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0.5
40.8
57.3
1
70.0
77.1
2
79.5
82.3
3
80.7
69.4
4
68.6
60.6
6
49.4
48.0
8
35.0
33.7
12
15.3
17.4
24
2.1
3.0
Find ka's for both products. Calculate peak time and Cp max and AUC for both products. Can you conclude that these products are bioequivalent ? (Reasons should include discussion of rate and extent of absorption)
Answer:
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Oral Dosing
Product A Ka (hr^-1)
1.19
Tmax (hr)
Product B
Ratio (A/B)
1.8 2
1.52
1.31
Cmax (mcg/100mL)83.2
82.8
1.00
AUC (trapazoidal)676.52
688.81
0.976
Can you conclude that these products are bioequivalent ? No, Time to peak (Tmax) is outside guidelines. 3) Wilkenstein et al.(Gastroenterology 74:360,1978) tested 12 normal healthy volunteers in a four way crossover design of four dosage forms containing 300 mg of cimetadine. The following data was obtained:
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A.U.C. (mic/ml x hr)------
A
B
5.2
5.4
C
D
% recovered in urine intact77.177.147.149.0 Peak serum conc.(mic/ml)------ 1.53 1.44 Onset (hr)
0
Duration (hr)
4.5
Time to peak (hr)0
0.34
0.65
4.0
4.2
4.4
1.0
2.0
A = IV bolus B = IM inj. C = Oral Liq. D = Oral Tab.
The plasma concentration - time profile for product A is as follows:
time(hrs)
(ug/ml)
time(hrs)
(ug/ml)
1
1.79
6
0.45
2
1.36
12
0.08
4
0.78
a} Using linear regression, find K & Cp0. b} What is the absolute bioavailability (f) of the liquid. c} How does that correlate with % recovered intact in the urine? d} Would you consider the oral forms bioequivalent? Why/Why not?
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f} What infusion rate would you suggest to maintain a plasma concentration of 0.75 mic/ml ? g} How long would it take that infusion rate to attain a therapeutic plasma concentration of 0.5 mic/ml ? Answer:
IV Bolus Parameters: Cp max2.4 mic/mL AUC 8.5 K
0.283 hr^-1
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a} Using linear regression, find K & Cp0. (graph)
b} What is the absolute bioavailability (f) of the liquid. 5/2/8.5 = 0.61
c} How does that correlate with % recovered intact in the urine? Very well. Only 61% (f) of liquid gets in and you would expect only 77% of that to show up in the urine because only 77% of the IV dose shows up in the urine (.61*.77=.47).
d} Would you consider the oral forms bioequivalent? (No) Why/Why not? Ratio of peak times ouside guidelines.
e} What infusion rate would you suggest to maintain a plasma concentration of 0.75 mic/ml ? Q = Cpss * K * V = 0.75 mg/L * 0.283 hr^-1 * 125 L = 26.54 mg/hr
f} How long would it take that infusion to attain a therapeutic plasma concentration of 0.5 mic/ml ? Cp = Q/(K*V)(1-exp(-K*T) = 0.5 = 26.54/(0.283*125)*(1-exp(-0.283*T)) --> 3.9 hr 4) LYSERGIC ACID DIETHYLAMIDE (LSD) was given to human volunteers at the dose of 150 mic orally. (Impregnated blotter dosage form.) The following data was obtained:
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Oral Dosing
Time
Cp (ng/ml)
Time 2.0
Cp (ng/ml)
0.25
1.75
4.6
0.5
2.9
3.0
4.1
0.75
3.7
4.0
3.3
1.0
4.2
6.0
2.1
1.5
4.6
8.0
1.4
a) Find ka
b) An IV dose of 100 mic resulted in an AUC of 20.4 ng/ml*hr. Find f.
c) The volunteers ability to concentrate as measured by their ability to do standard tasks was also monitored. (100% control means no drug interference.) The following data was obtained: Cp (ng/ml)
% Control
Cp (ng/ml)
5.5
33
1.5
65
4.1
40
1.1
80
2.9
52
% Control
If 100 mic dose were given by IV bolus, how long would it be before the volunteer would regain 80% of his control? Answwer:
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Oral Dosing
Evaluation of the graph of Concentration vs. time yields:
Cpmax
4.63 ng/mL
T max
1.7 hr
AUC (trap)30.07 K
0.225 hr^-1
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Oral Dosing
m (-K)
-0.225 hr^-1
Ka
1.22 hr^-1
f (AUCoral/Doseoral)/(AUCiv/Doseiv) = .98
Evaluation of the graph of response vs ln(concentration) yields: dR/dln(c) = 27.86
Multiplying dR/dln(c) * dln(c)/dt (m of the previous graph) yields dR/dt = 27.86 * -0.225 = 6.26%/hr
100 mic dose IV yields Cp0 of (Cp0 =AUC * K = 20.4 * 0.225) 4.59ng/mL.
The response of a 100 mic dose is (R = 27.86*ln(4.59)+19.9) 62.3%
Response = Response at t=0 - dR/dt * t 20% = 62.3% - 6.26%/hr * t hours T
= 6.76 hours
5. The following data was collected from a double blind cross over study between 500 mg dose of cloxacillin made by Bristol (Tegopen@) and a generic product which you might want to put in your store. Time
(Conc. mic/ml)
Time
TEGOPEN GENERIC
(Conc. mic/ml) TEGOPEN GENERIC
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Oral Dosing
0.25
.41
0.1
1.5
6.93
7.75
0.5
8.56
6.39
2
4.95
5.16
0.75
11.97
11.44
1
11.28
11.42
9.57
9.64
1.25
3 4
2.19 1.48
2.29 1.30
Calculate the comparative bioavailability. Would you consider these products bioequivalent? Why/Why not? Answer:
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Oral Dosing
Evaluation of the above graphs yields: Tegopen
Cpmax (mic/mL)10.8
GenericRatio (G/T)
9.94
0.92
T max (hr)
0.74
0.89
1.20
AUC (trap)
21.7
21.06
0.97
K (hr^-1)
0.72
0.8
ka (hr^-1)
4.3
2.69
Actual evaluation of ka and peak time is dificult because of the pucity of data at early time points however all relavent parameters meet guidlines. 7. The F.D.A. reported the following data submitted to be consideration regarding the equivalence of Mylan Pharmaceuticals' Tetracycline with that of Lederle and an intervenous bolus dose. (Dose 250 mg). Time(hrs) Conc.(mcg/ml) Time(hrs) Lederle Mylan I.V.
Conc.(mcg/ml)
Lederle Mylan I.V.
0.5
0.55 0.20 5.2
4
2.70 2.60 2.9
1
1.80 1.35 4.8
6
2.20 1.80 2.1
1.5
2.11 1.75 4.4
9
1.35 1.25 1.26
2
2.35 2.10 4.0
12
0.83 0.74 0.76
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Oral Dosing
3
2.65 2.25 3.4
15
0.50 0.45 0.46
Would you consider Mylan to be bioequivalent to the Lederle product ? Calculate the absolute bioavailability of Lederle Tetracycline.(.77) f) Calculate the volume of distribution of tetracycline. (44.3 L) g) Tetracycline has a pKa of 9.7. Tetracyclines tend to localize in the dentin and enamel of developing teeth causing hypoplasia and permanent discoloration of teeth. Would you recomend tetracyline for a 110 pound lactating mother ? Support your argument with the dose of the child. (Child's weight 11 lbs. and he eats 2 oz of milk every 2 hours. Mom's average plasma concentration is maintained at 3 mic/ml by taking 250 qid. pH of the milk is 6.1, pH of blood is 7.4) Answer:
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Oral Dosing
Pharmacokinetic parameters: Lederle Cpmax (mic/mL)2.75 2.42 Tmax (hr)
3.04
AUC
Mylan
IV
5.65 3.08
26.4
0 23.3
k (hr^-1)
0.165
0.161
Ka (hr^-1)
0.684
0.729
31.4 0.167
Ratio of bioequivalence parameters (Cpmax, Tmax and AUC) are all within guidelines. So, the would be considered bioequivalent.
Absolute bioavailability f (= (AUCoral/DOSEoral)/(AUCiv/DOSEiv) = (26.4/ 250)/(31.4/250) is 0.84.
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Volume of Distribution (Dose/Cp0 = 250 mg/ 5.65 mg/mL) is 44.2 L
The ratio of milk to blood is about 200.
r(m/b) = (10^(pKa-pH) + 1)milk / (10^(pKa-pH) + 1)blood = (10^(9.7-6.1)+1)/(10^(7.4-6.1)+1) = 10^3.6/10^1.3 = 10^2.3 = 200
Dose the kid gets is mom's plasma concentration * Ratio(M/b) * volume of milk / day = 3 mic/mL * 200 * 60cc * 12 feedings = 432 mg.day Mom gets 1000 mg/day
Ratio of dose on a mg/kg basis (kid/mom) = (432/5)/1000/50) = 4.32 - Kid's getting more than mom. Fifty miligrams of ketameperidine was given by IV bolus. The following urinary profile was obtained for the only metabolite N-methyl-ketameperidine:
Collection period (hr)
Mean urinary excretion rate (mg/hr)
0.0 - 0.5
2.26
0.5 - 1.5
5.83
1.5 - 2.5
5.43
2.5 - 3.5
4.60
3.5 - 5.0
2.36
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Oral Dosing
5.0 - 7.0
1.47
7.0 -10.0
0.96
10.0 -18.0
0.44
Calculate K, km and ku. What Percent of ketameperidine was metabolized?
Answer: With only one data point in the early time points, the larger rate constant is in question. The terminal slope is assumed to be K. The AUC will yield the amount of ketameperidine which was metabolized (dXmu/dt * t = Xmu).
K (hours^-1) 0.216 AUC (mg)30.3
30.3 mg showed up as metabolite = 60.6% of 50 mg dose.
km = 60.6% * K = 0.131 hours^-1
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Oral Dosing
kr = K - km = 0.085 hours^-1 Aminophylline consists of THEOPHYLLINE (85% W/W) & Ethylene diamine (15% W/W) THEOPHYLLINE is the active compound measured in blood. THEOPHYLLINE has a volume of distribution of 0.45 l/kg. THEOPHYLLINE is 10% excreted unchanged and 90% metabolized to inactive metabolites. THEOPHYLLINE has a therapeutic range between 20 and 10 mg/l. AUC FROM 0 to infinity for THEOPHYLLINE (given as 400 mg AMINPHYLLINE) is 120 mg/l x hr. The average plasma concentration of THEOPHYLLINE given as 400 mg of AMINOPHYLLINE is as follows: time
conc. time
conc.
(hrs)
(mg/L) (hrs)
(mg/L)
0.5
7.24
4.0
8.06
1.0
9.56
6.0
6.89
2.0
10.00 8.0
5.57
3.0
8.84
4.53
10.0
Find f, K, ka, Vd,total body clearance. Find the infusion rate necessary to maintain a plasma concentration of 15 mg/l.
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Oral Dosing
Answer: AUC (mg/L)*hr117.8 K (hr^-1)
0.096
ka (hr^-1)
2.11
f = (AUCoral/DOSEoral)/(AUCiv/DOSEiv) = = (117.8 / 400 )/(120 / 400 ) = 0.98
Vd AUC * K = Cp0iv 120 * 0.096 = 11.52 mg/L
Vd = Dose/Cp0 = (400mg*0.85)/11.52 = 29.5 L
TBC = K * Vd = 0.096/hr * 26.5L = 2.83 L/hr
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Oral Dosing
Infusion rate = Q = Cpss * TBC = 15 mg/L * 2.83 L/hr = 42.45 mg/hr Theophylline = 42.45/.85 = 50 mg/hr Aminophylline
Abbott labs has provided the following data conserning their ORETIC tablets (hydrochlorthiazide tablets U.S.P.) Dose given was 50 mg.
time
conc. time
conc.
(hrs)
(mg/L) (hrs)
(mg/L)
0.5
0.05
3.0
0.31
1.0
0.21
4.5
0.23
1.5
0.27
6.0
0.18
2.0
0.31
8.0
0.12
a Find K, ka, Cmax,
Answer:
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Oral Dosing
The data is plotted both without (first figure) and with (second figure) a lag-time which is associated with the release of the drug from the delivery system. Note that the addition of the lag-time improves the fit.
The parameters obtained from each fit are:
WithoutWith
Cpmax (mg/L)0.22
0.31
Tmax (hr)
2.28
3.45
AUC (mg/L*hr)2.2
2.26
K (hr^-1)
0.216
0.201
ka (hr^-1)
0.380
1.10
t lag (hr)
0.0
0.393
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Oral Dosing
It takes the tablet about 20 minutes to release the drug!
Wilkenstein et al.(Gastroenterology 74:360,1978) tested 12 normal healthy volunteers in a four way crossover design of four dosage forms containing 300 mg of cimetadine. The following data was obtained: A B C D AUC(mic/ml x hr)
---
recovered in urine intact77.177.1
---
5.2
54.9
55.8
Peak serum conc.(mic/ml)--- --Onset (hr)
0
Duration (hr)
4.5
Time to peak (hr)
1.53
1.44
4.6
0
5.4
0.34
0.65
4.2
4.4
1.0
2.0
A = IV Bolus B=IM injection C = Oral liquid D= Oral tablet
The plasma concentration vs. time profile for product A is as follows: time (hrs)
conc.(ug/ml)
1
1.79
2
1.36
4
0.78
6
0.45
12
0.08
a} find K, Cp0. Both can be found from the graph. K = .283/hr Cp0 = 2.36 mic/ml
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b} What is the absolute bioavailability (f) of the liquid. 5.2/8.5 = 0.61
c} How does that correlate with % recovered intact in the urine? Very well. Only 61% (f) of liquid gets in and you would expect only 77% of that to show up in the urine because only 77% of the IV dose shows up in the urine (0.61 * .77 = .47).
d} How can you explain the variation in % recovered intact in the urine?
e} Would you consider the oral forms bioequivalent ? Why/Why not? No. The ratio of peak times is outside the guidelines.
f} What infusion rate would you suggest to maintain a plasma concentration of 0.75 mic/ml? Q = Cpss * K * V = 0.75 mg/L * 0.283/hr * 125L = 26.54 mg/hr
g} How long would it take that infusion rate to attain a therapeutic plasma concentration of 0.5 mic/ml ? Cp = Q/(K * V)(1-exp(-K*T) = 0.5 = 26.54/(0.283 *125)*(1-exp(-0.283 * T)) > 3.9 hr Roxane labs of Columbus, Ohio offers the following data for your review of their Quinidine Sulfate tablets (Dose 200 mg). It is compared against the reference standard by Ely Lilly and company at the same dose.
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Oral Dosing
Time (hours)Concentration (mcg/ml) Roxane
Lilly
1
.42
.58
2
.73
.77
3
.71
.74
4
.61
.66
6
.45
.52
8
.32
.34
12
.20
.22
a) Calculate the comparative bioavailability. b) Would you consider Roxane Quinidine Sulfate to be bioequivalent to the Lilly product ?
Answers
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Oral Dosing
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Oral Dosing
Roxane labsEli Lilly w/o
w
w/o
w
Rw/o(R/L)Rw(R/L)
Cpmax (mcg/mL)0.650.740.740.76 0.88
0.97
AUC (mcg/mL*hr)6.086.236.756.84 0.90
0.91
Tmax (hr)
2.69
2.05
2.33
2.10
T lag (hr)
0.0
0.70
0.0
0.36
1.15
0.98
Yes. Ratios are within guidelines. Shand et al. offers the following data for propranolol :
Answers:
Time Concentration (ng/ml) (hours) 10 mg I.V. 80 mg oral 0.5
--
50
1
--
77
1.5
--
100
2
29
100
3
24
90
4
18
78
5
15
59
6
11
45
7
9
32
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Oral Dosing
a) find ka b) Calculate the absolute bioavailability of propranolol. c) Calculate TBC
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Oral Dosing
IV data
Oral Data w/o
w
AUC (ng/mL*hr)201.3562.8 540 Cpmax (ng/mL) 47.7 97.8
99.7
Tmax (hr)
0
2.0
K (hr^-1)
0.239
0.324 0.421
ka (hr^-1)
---
0.715 0.548
2.1
T lag
0.0
0.02
Absolute bioabailability = (AUCoral/DOSEoral)/(AUCiv/DOSEiv) = (562.8/80) (540 / 80)
/(201.3/10) = 0.35 or using lag time data /(210.3/10) = 0.335
TBC = Dose / AUC = 10,000 mic/ 201.3 mic/L*hr
= 50 L/hr or
0.35*80,000mic /562.8 mic/L*hr = 50 L/hr Niazi et al. offers the following data for meperadine : Meperidine : is 95% metabolized has an absolute bioavailability of 0.4 has a hepatic plasma extraction ratio of 0.6 has a volume of distribution of 100 L. has a half life of 3.5 hours.
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a) Calculate TBC TBC = K * V = (0.198/hr)(100L) = 19.8 L/hr
b) Calculate the intrinsic hepatic plasma clearance of meperidine. 19.8 L/hr * .95 = 18.8 L/hr
c) Calculate the effect on total body clearance in a patient with viral hepititis (FI = 0.3). Clh*/Clh = (.3)(1)/1 + .6(.3 - 1) = .3/.58 = .517 (.517)(18.8) = 9.72 TBC = 1 + 9.72 = 10.72
d) Calculate the effect on total body clearance in a patient with stenosis (FR = 0.3). Clr*/Clr = (1)(.3)/.3 + .6(1 - .3) = .3/.72 = .417 TBC = 18.8 + .417 = 19.22
e) Comment on which patient might need modification in therapy and why. The patient with viral hepatitis would need modification in therapy. Because of the decrease in TBC, we can see that the drug is staying the body much longer than normal, therefore the dosage regimen should be decreased. Chlorthalidone is used to treat high blood pressure. The following information is offered regarding a generic and a brand name chlorthalidone 50 mg tablet: Time Conc. (mcg/ml)
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Oral Dosing
(hours)
Hygroton@Generic
.5
0.14
0.15
1
0.51
0.64
2
1.23
1.67
3
1.94
2.48
4
2.20
2.91
6
2.64
3.49
8
2.86
3.52
12
3.43
3.82
24
3.22
3.38
48
2.45
2.74
72
1.53
1.91
96
1.20
1.40
120
0.76
0.77
Pharmacokinetic parameters Cpmax (mg) 3.73
4.62
Time to peak (hr) 13.810.8 AUC (0 to Inf)293
336
Xu inf (mg)18.3
22.1
Ka (hr^-1)0.168
0.253
Ke (hr^-1)0.019
0.019
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Oral Dosing
Average mean83.1
84.5
blood presure
a) Calculate the comparative bioavailability. (336/50mg)/(293/50mg) = 1.15 b) Would you consider the generic product to be bioequivalent to the USV (Hygroton@) product? Prepare a short statement that you would tell a patient regarding why you would or would not make a generic substitution for this drug. No. The maximum concentration the generic is too much greater than that of the brand name product. They are not considered to be bioequivalent.
R(G/H) Cpmax (mg) 1.23
outside
Time to peak (hr)0.78outside AUC (0 to inf)115
ok
Buspirone is a new anxiolytic agent that has been found to be effective for the treatment of generalized anxiety disorder at a mean dose of approximately 20 mg/ day orally in divided doses. Buspirone is metabolized almost entirely. Less than 0.1% is found intact in the urine. The following data has been presented by Gammans (Am J Med:80(supp 3b),41-51;1986): Time (hours)Concentration (ng/ml) (hours)
1 mg I.V.20 mg oral
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Oral Dosing
0.25
--
1.07
0.50
4.33
1.76
1.0
3.75
2.45
2
2.80
2.51
3
2.10
2.05
4
1.57
1.60
6
0.8
0.91
a) find ka b) Find Oral Peak Time and Oral Cmax. c) Calculate the absolute bioavailability of buspirone.
answer:
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Oral Dosing
IV
Oral
Cpmax (ng/mL) 5.0 2.6 AUC (0 to inf)17.4
13.9
Tmax (hr)
0
1.5
K (hr^-1)
0.290 0.289
ka (hr^-1)
1.3
Absolute bioavailability, f, = (AUCoral/DOSEoral)/(AUCiv/DOSEiv) = ( 13.9 / 20)
/( 17.4/ 1
)
= 0.04 Valproate is a carboxylic acid anticonvulsant. Its activity may be related, at least in part, to increase concentrations of the neurotransmitter inhibitor gamma aminobutyric acid in the brain. It is used alone or in combination with other anticonvulsants. in the prophylactic management of petit mal. It appears to be almost entirely cleared by liver function with negligible amounts excreted into the urine unchanged. It comes as soft gelatin capsules of 250 mg and enteric coated tablets 250 and 500 mg as well as oral syrup of 250 mg / 5 cc. Two different formulations of Valproate (250 mg) were prepared by Abbott and compared. The data is as follows:
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Oral Dosing
Time(Hr.)Formulation BFormulation A 0.5
3.4
AUC = 287 mg/L * hr
1.0
6.0
Ka = 0.7 hr^-1
1.5
7.9
Ke = 0.065 hr^-1
2.0
9.3
2.5
10.3
3.0
10.9
4.0
11.6
6.0
11.4
8.0
10.5
12.0
8.3
18.0
5.7
24.0
3.8
1) find ka for formulation B. 2) Five hundred mg of valproate was administered by IV bolus. The AUC for that route was 574 mg/L * hr. Calculate f for formulation A. Calculate Cp0 for the IV dose. 3) Find Peak Time and Cmax for formulation A. 4) Calculate the comparative bioavailability of formulation B. 5) Would you consider formulation B to be bioequivalent to Formulation A ? Prepare a short statement in which you would substantiate that stand that you might
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need to respond to another health professional who asked you to stock that formulation for his patients. 6) Calculate the Total Body Clearance (TBC) of valproate.
Answers: Formulation B
R(A/B)
AUC = 243.3 mg/L * hr 1.18 Cpmax = 11.7 mg/L1.12 Tp max = 4.70 hr0.79 ka
= 0.493 hr^-1
K
= 0.0655
Tmax(A) = ln(ka/K)/(ka-K)= 3.75 hr cpmax = (ka/(ka-k))*(fX0/Vd)*(exp(-k*tmax)-exp(-ka*tmax) 13.3 mg/L
Absolute bioavailability, f,=(AUCoral/DOSEoral)/(AUCiv/DOSEiv) = (287/250)/(574/500) = 1.0
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Comparative bioavailability =(AUCb/DOSEb)/(AUCa/DOSEa) = 243.3/287 = 0.85 TBC = Dose / AUC = 500 mg / 574 mg /L * hr The following data was made available by Lederle Labs regarding its generic Procainamide HCl. (Dose 250mg). Procainamide is a base (pka =9.1). As the hydrochloride salt it is 87% Procainamide.
Time (hrs)Conc.(mcg/ml) Procainamide Base Lederle
Squibb
0.33
0.68
0.26
0.5
0.82
0.67
0.66
1.17
0.93
1
1.23
1.12
1.45
1.33
1.31
1.19
1.35
2
1.39
1.12
1.18
3
0.93
0.96
0.95
4
0.74
0.74
0.77
6
0.51
0.51
0.51
8
0.32
0.30
0.33
12
0.11
0.09
0.14
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I.V.
7-38
Oral Dosing
a) find ka of the Squibb product b) ka of the Lederle product. c) Calculate the comparative bioavailability. d) Would you consider Lederle to be bioequivalent to the Squibb product ? e) Calculate the absolute bioavailability of Lederle Procainamide. f) Calculate the volume of distribution of procainamide. g) Would you recommend your patient breast feed her newborn? Prepare a short consult for her physician. Support your argument with the dose of the child. (Child's weight 11 lbs. and he eats 2 oz of milk every 2 hours. Mom's average plasma concentration is maintained at 4 mic/ml from a 1 g dose ever 6 hours. pH of the milk is 6.3, pH of blood is 7.4) Procainamide is cleared about 60% by liver and 40% by kidney function. 20 % of cardiac output (70 ml/min/kg) goes to liver, 25% goes to the kidney. Mom's weight is 130 lb. Assuming her plasma vs time profile to be similar to the Lederle product (i.e. pharmacokinetic parameters obtained from this information can be used): h) Calculate Total body clearance i) Calculate the intrinsic hepatic plasma clearance of procainamide. j) Calculate the effect on her total body clearance if she were to contract viral hepatitis which effect liver function (FI = 0.4). Prepare a short consult for her physician as to whether you would recommend a change in therapy. d) Calculate the effect on her total body clearance stenosis of the liver (FR = 0.4). Prepare a short consult for her physician as to whether you would recommend a change in therapy. Answers:
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Oral Dosing
IV
LederleSquibb R(L/S)
AUC (0 to inf)8.577.46.8
1.09
Cpmax
1.8
1.28
1.25
1.02
Tmax
0
1.43
1.45
0.99
K
0.212 0.247 0.256
ka
---
1.51
1.93
t lag
0
0
0.24
Absolute bioavailability, f,=(AUCoral/DOSEoral)/(AUCiv/DOSEiv) = 7.4
/ 8.57
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Oral Dosing
= 0.86
Vd = Dose/Cp0 = 0.87*250mg/1.8mg/L= 120.8 L
Ratio of milk to blood = (10^(9.1-6.3)+1)/(10^(9.1-7.4)+1)= 12.4 Kid's dose = 4 mic/mL * 12.4 * 60 mL/feeding * 12 feedings/day * 1 mg/1000 mic = 36 mg/day Ratio of kid's daily dose/# to Mother's daily dose/# = (36mg/11#)/(1000mg*4/ 130#) = 0.42. The kid gets about half of the mother's dose!
Nifedipine (Procardia @) is a calcium channel blocker which specifically inhibits potential-dependent channels not receptor-operated channels, preventing calcium influx of cardiac and vascular smooth muscle (coronary, cerebral). Calcium channel blockers reduce myocardial contractility and A-V node conduction by reducing the slow inward calcium current. They are indicated in angina, cardiac dysrhythmias, and hypertension among others. Nifedipine appears to be metabolized entirely into an inactive metabolite, an acid and subsequently further metabolized to a lactone. Both the acid and the lactone are excreted into the urine and the feces.
Echizen and Eichelbaum (Clin Pkin 1986; 11:425-49) and Kleinbloesem et al (Clin Pcol Therap 1986; 40: 21-8) Reviewed the pharmacokinetics of Nifedipine. While the drug is not routinely given by IV bolus and does not strictly conform to a one compartment model, lets treat the data as if those problems can be ignored. The following data is offered for evaluation: 25mg IV
10 mg oral tablet Formula AFormula B
Time Cp
Cp
Cp
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Oral Dosing
(hr.)
(mic/l)
(mic/l)
0.5
29.3
33.1
1
42.1
43.7
1.5
45.7
43.7
44.4
39.8
36.2
25.5
2
(mic/l)
139
3 4
65.6
27
20.7
6
31.1
13.6
9.9
8
14.6
6.5
4.7
1.5
1.0
12
a} Find ka's of the two products. b} Calculate peak time and Cp max for both products. d} Can you conclude that these products are bioequivalent ? (you must support you argument) e) Calculate the absolute bioavailability of product A. f} What infusion rate would you suggest to maintain a plasma concentration of 30 mic/L ? Answers:
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Oral Dosing
IV
A
B
R(A/B)
Cpmax (mic/L)294.25 45.7 44.01.04 Tmax (hr) 0
1.57
1.18 1.33
AUC(0 to inf)785219.7182.7 1.20 ka (hr^-1)---
1.0
1.6
K (hr^-1) 0.375 0.374 0.375
No,Tmax is outside the guidelines.
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Oral Dosing
Absolute bioavailability, f,=(AUCoral/DOSEoral)/(AUCiv/DOSEiv) =(219.7/10)/(785/25) =0.7
Q = Cpss * K * V = 0.955 mg/hr Tetracycline HCl has a pKa of 9.7. Tetracyclines tend to localize in the dentin and enamel of developing teeth causing hypoplasia and permanent discoloration of teeth. Would you recommend tetracycline for a lactating mother ? Support your argument with the dose of the child. (Child's weight 11 lbs. and he eats 2 oz of milk every 2 hours. Mom's average plasma concentration is maintained at 4 mic/ ml she is taking 250 mg T.I.D. ( Milk pH = 6.1, Blood pH = 7.4)
Tm/Tb = 109.7 - 6.1/109.7 - 7.4 = 20/1 The concentrarion of tetracycline in the mother's milk is 80 mic/ml The child takes in 720 ml of milk per day 80 mic/ml * 720 ml = 57600 mic = 57.6 mg 57.6mg/5kg = 11.52mg/kg = dose that the child is getting from the mother's milk.
I would not recomend tetracycline for a lactating mother. The dose that a nursing child gets from the milk too high.
Oxazepam (acid, pKa 11.5) is an anxyolytic sedative with the usual adult dose 10 mg 3 times daily. If the circulating plasma concentration of oxazapam were 20
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mic/ml for nursing 120 lb mother, would her 9 lb infant be getting a comparable mg per kg daily dose if he consumes 2 oz of his mothers milk every 2 hours. Prepare a short consult for her physician in which you might (or might not) recommend the patient stop breast feeding while she is on this medication. Include appropriate calculations.
Om/Ob = 1011.5 - 6.1/1011.5 - 7.4 = 20/1 The concentration of the mother's milk would then be 400 mic/ml 400 mic/ml * 720 ml = 288000 mic given to baby = 288mg 288mg / 4.1kg = 70 mg/kg = dose/kg given to baby This dose is much greater then that given to the mother. The mother should discontinue breast feeding while taking Oxazepam. Bioequivalence studies are sometimes done within the same company to check if the tablets of the same drug, but different strengths (with the strength normalized) could be considered equivalent (i.e. could two 5 mg tablets be considered equal to one 10 mg tablet). While not strictly kosher (products are not pharmaceutical equivalents because of different strengths), it is done. Here is the results of such a study in which Zomax 100 and 200 mg tablets were compared. (Yes, I know that Zomax was removed from the market after a short life of only 6 months.)
Zomax 100 mg tablet 200 mg tablet 50 mg IV bolus Time Conc AUC
Conc AUC
Conc.
(hr)
(mg/L) (0->t)
(mg/L) (0->t)
(mg/L)
0.25
1.41
0.18
4.03
0.50
0.5
1.98
0.60
5.13
1.65
0.75
2.15
1.12
5.18
2.94
1
2.12
1.65
4.89
4.20
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1.14
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Oral Dosing
2
1.56
3.49
3.37
8.33
0.764
3
1.05
4.80
2.26
11.14
0.512
4
0.707 5.67
1.51
13.03
0.343
6
0.318 6.70
0.68
15.22
0.154
8
0.143 7.16
0.306 16.20
0.069
1) What is the elimination rate constant for zomax (hr) ? A) 0.2 B) 0.3 *C) 0.4 D) 0.5 E) 0.6
2) What is the volume of distribution of zomax given by IV bolus (L) ? A) 43.85 B) 33.3 *C) 29.4 D) 25.9 E) 0.034
AUC = D/(Vd * K) Vd = D/(AUC * K) = 50mg/(4.25 * 0.4) = 29.4 L
3) What is the volume of distribution of zomax given by 100 mg oral tablet ? A) 43.85 *B) 33.3 C) 29.4 D) 25.9 E) 0.034
Vd = 100mg/(7.48 * 0.4) = 33.4 L
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Oral Dosing
4) What is the AUC(0->infinity_ for the IV bolus dose ? A) 2.68 B) 2.85 C) 3.55 D) 4.08 *E) 4.25
5) What is the AUC(0->infinity) for the 100 mg tablet ? A) 7.16 *B) 7.5 C) 16.20 D) 17
E) 37.38
6) What is the absolute bioavailability of the 100 mg tablet ? A) 0.84 *B) 0.88 C) 1
D) 1.14 E) 1.19
(7.48/100)/(4.25/50) = 0.88
7) What is the AUC(0->infinity) for the 200 mg tablet ? A) 7.16 B) 7.5 C) 16.20 *D) 17
E) 73.98
8) What is the absolute bioavailability of the 200 mg tablet ? A) 0.84 B) 0.88 *C) 1
D) 1.14 E) 1.19
(16.9/200)/(4.25/50) = 1
9) What is K * AUC (0->infinity) for the 100 mg tablet (mic/ml) ? A) 2.9 *B) 3.0 C) 6.5 D) 6.8 E) 14.95 7.48 * 0.4 = 2.99
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10) What is the absorption rate constant for the 100 mg tablet ? A) 1.7 B) 2.2 *C) 2.6 D) 3.2 E) 3.7
11) What is the intercept of the extrapolated line for the 200 mg tablet ? A) 3.5 B) 4.1 C) 5.6 D) 6.1 *E) 7.6
12) What is the absorption rate constant for the 200 mg tablet ? A) 1.7 B) 2.2 C) 2.6 D) 3.2 *E) 4.01
13) What is the Tmax for the 100 mg tablet ? A) 0.5
B) 0.67 C) 0.75 *D) 0.85 E) 0.95
14) What is the Tmax for the 200 mg tablet ? A) 0.5
*B) 0.67 C) 0.75 D) 0.85 E) 0.95
15) Would you consider these two tablets bioequivalent (given normalization for dose) (consider all ratios to be the 100 mg / 200 mg parameter normalized as to dose where applicable)? A) Yes B) No, because the ratio of the ka's is 0.70 C) No, because the ratio of the AUCs is 0.44
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D) No, because the ratio of the Cmaxs is 0.41 *E) No, because the ratio of the Tmaxs is 1.27
16) What infusion rate would you recommend to maintain an average plasma concentration of 1 mic/ml ? A) 17.5 B) 13.3 *C) 11.8 D) 10.4 E) 9.0
Vd = D/Cp0 = 50mg/1.7mg/L = 29.4 Q = Cpss * K * V = 1mg/L * 0.4/hr * 29.4L = 11.8
17) What would be the concentration (mg/L) 2 hrs after discontinuing the infusion assuming you reached steady state ? A) 0.67 B) 0.55 *C) 0.45 D) 0.37 E) 0.30
Cpss = Cp0 * e-Kt = 1mg/L * e(-0.4 * 2) = 0.44 A 110 pound mother breast feeds her 11 pound infant while on morphine sulfate (base, pKa = 9.85). Mother's average circulating plasma levels are 0.5 ug/ml following a 10 mg IV dose q4h. (pH Milk = 6.1, pH blood = 7.4)
18) What is the Ratio of morphine concentration in the milk as compared to the blood ? A) 0.05 B) 0.5
C) 1
D) 2
*E) 20
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Oral Dosing
Mm/Mb = 10(9.85 - 6.1)/10(9.85 - 7.4) = 20
19) How much (mg) morphine is contained in 120 cc of breast milk (the child consumes 2 ounces every 2 hours) *A) 1.2 B) 0.12 C) 0.06 D) 0.03 E) 0.003
Mother's blood conc. is 0.5mic/ml therefore her milk conc. is 10 mic/ml. 10mg/L * 0.12L = 1.2 mg
20) In your professional judgment, will the child's dose cause a problem ? A) No, morphine does not concentrate in the milk and thus the milk is ok to drink. B) No, the dose is too small. The ratio of the child's dose to the mother's dose is 0.12. C) Yes, even though the dose is small, we don't want any drug to get to the child. *D) Yes, the dose is comparable to the mother's dose. The ratio of the child's to the mother's dose is 1.2. E) Not my job. I only give what the doctor orders. Answers:
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7-50
Oral Dosing
IV
Tablet Tablet
50 mg 100 mg200 mgR(100/200) AUC(0 to inf)4.257.4816.9
0.89
Cpmax
1.7
2.15
5.23
0.82
Tmax
0
0.82
0.64
1.28
K
0.4
0.4
-.4
ka
---
2.76
4.01
Tmax ratio is ouside guidelines.
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Oral Dosing
Answers are rounded off. When you pick a foil, use that number in subsequent calculations when needed.
Rifampin (unionized free base pKa 7.9) is a drug used to treat TB. The following data was collected following a 600 mg oral tablet from the inovator (Treatment A), and a 600 mg oral tablet from a generic (treatment B), and a 400 mg IV dose (Treatment C).
Concentration (mic/mL)AUC(0->t) TreatmentA
B
C
B
0.5
5.3
4.8
1
10.3
8.6
1.5
10.2
9.8
2
9.4
9.8
2.5
8.9
9.2
3
7.5
8.4
4.7
4
5.9
6.7
3.7
6
3.6
4.1
2.2
8
2.2
2.5
1.3
10
1.3
1.5
0.8
12
0.8
0.92
0.5
Time (hours) 1.2 7.8
4.55 9.15
6.1
14.05 18.8
AUC(0->inf)53.957.7
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(mic/mL*hr) Lag time (min)18.610.5 Cp max10.6
9.9
Ka (hr^-1)2.66
1) What is the Cp0 for C (mg/L)?
a) 0
b) 7.8 *c) 10
d) 12
e) 15
Cp0 = AUC * K = 39.8 * 0.25 = 9.95
2) What is the volume of distribution of Rifampin (L)? d) 33.3 e) 26.7
a) 60
b) 51.3 *c) 40
Vd = D/Cp0 = 400mg/(9.95mg/L) = 40.2 L
3) What is the half life for rifampin (hr)? 1.5
*a) 2.8
b) 2.3
c) 2.0
d) 1.75 e)
t1/2 = .693/0.25 = 2.77
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4) What is the elimination rate constant for rifampin (hr^-1)? *a) 0.25 b) 0.3 c) 0.35 d) 0.4 e) 0.45
5) Calculate the AUC (0->1hr) for C (mic/mL*hr). 8.9 e) 17.8
a) 1.95 *b) 3.9 c) 7.8 b)
6) Calculate the AUC (12hr->inf.) for C (mic/mL*hr). d) 1.25 e) 1.11
*a) 2
b) 1.67 c) 1.43
0.5/0.25 = 2 7) Calculate the AUC (0->inf) for C (mic/mL*hr). 40 e) 60
a) 16
b) 26.85 c) 35
8) Calculate the absolute bioavailability for the generic product. 0.95 c) 1 d) 1.05 e) 1.43
*d)
a) 0.70 *b)
(57.7/600)/(39.8/400) = 0.966 9) Calulate the comparative bioavailability for the generic product. 0.95 c) 1 *d) 1.05
a) 0.70 b)
e) 1.43
(57.7/600)/(53.9/600) = 1.07
10) Using Wagner-Nelson method, calculate the Ka for the generic product (hr^1). a) 0.45 b) 1 c) 1.55 d) 2 e) 2.45
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11) Calculate the peak time for the generic product (min). 91 e) 105
a) 37 b) 67 c) 86 d)
tp = [ln(Ka/K)]/(Ka - K) = [ln(1.37/0.25)]/(1.37 - 0.25) = 1.52 hr = 91 min 12) Calculate the peak time for the brand name product (min). 86 d) 95 e) 105
a) 37 *b) 59 c)
tp = [ln(2.66/0.25)]/(2.66 - 0.25) = 0.98 hr = 59 min 13) Are the two products bioequivalent? a) yes, all federal requirements are met. *b) no, the ratio of the peak times are out side federal requirements. c) no, the ratio of the lag times are out side federal requirements. d) no, the ratio of the Kas are out side federal requirements. e) no, the ratio of the comparative bioavailabilities are out side federal requirements.
14) What is the ratio of the concentration of milk (pH 6.1) to blood (pH 7.4)? a) 0.05 b) 0.05 c) 1 d) 15.4 *e) 20
Rm/Rm = 10(7.9 - 6.1)/10(7.9 - 7.4)
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= 20
15) The average plasma concentration for the mother (110#) is 2.5 mg/L from a 600 mg once a day dosing regimen. If the baby (11#) drinks 780 mL of milk a day (2 - 2.5 ounces every 2 hours), what is his daily dose (mg)? a) 0.1 b) 0.13 c) 2
d) 30
*e) 39
Mother's blood average blood conc. is 2.5 mg/L therefore her milk conc. is 50 mg/L. If the baby drinks 780 ml of milk he/she will get 39 mg of the drug. 16) Would you recommend mom stop breast feeding? (What % of the mom's daily dose (mg/kg) is the baby's daily dose (mg/kg)?) a) No, the child's dose is less than 1% of the mother's dose on a mg/kg/day basis. b) No, the child's dose is about 5% of the mother's dose on a mg/kg/day basis. c) Maybe, the child's dose is about 10% of the mother's dose on a mg/kg/day basis. *d) Yes, the child's dose is about 50% of the mother's dose on a mg/kg/day basis. e) Yes, the child's dose is about the same as the mother's dose on a mg/kg/day basis. 17) While Rifampin is not administered by IV infusion, what would be the infusion rate necessary to maintain an average plasma concentration of 2.5 mg/L (mg/ hr)? *a) 25 b) 50 c) 100 d) 150 e) 200
Vd = D/Cp0 = 400/10.02 = 39.9 L
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Q = Cpss * K * Vd = 2.5 * 0.25 * 39.9 = 25 mg/hr 18) While Rifampin is not administered by IV bolus, what would be the loading dose necessary to obtain a plasma concentration of 2.5 mg/L (mg)? a) 25 b) 50 *c) 100 d) 150 e) 200
Loading Dose = Cpss * Vd = 2.5 * 39.9 = 100mg 19) While Rifampin is not administered by IV infusion, what would be the infusion rate necessary to obtain a plasma concentration of 2.5 mg/L in about 2.5 to 3 hours (mg/hr)? a) 25 *b) 50 c) 100 d) 150 e) 200
Cp = [Q/(K * Vd)] * (1 - e-kt) Q = (Cp * K * Vd)/(1 - e-kt) = (2.5mg/L * 0.25 * 39.9L)/[1 - e(-0.25 * 2.75)] = 50 mg/hr 20) Rifampin is a semisynthetic derivative of rifamycin B, an antibiotic derived from Streptomyces mediterranei. The minimum inhibitory concentration for N. menengitidis is 0.1 - 1 mic/mL. It is distributed well into bodily fluids. About 30% shows up in the urine as free drug and active metabolite while 60% shows up in the feces as metabolite. The secretary is hounding me to finish the exam, so the answer to 20 is a. Also, rifampin is 85% protien bound at physiological concentrations. *a) 25 b) 50 c) 100 d) 150 e) 200 Answers:
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A
B
IV
Tmax
1.27
1.7
0
K
0.25
0.25
0.25
ka
2.66
1.37
---
AUC
53.9
57.7
39.8
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Cpmax
10.6
9.9
10.02
PHARMACOKINETICS SECOND HOUR PRACTICE EXAM #2
(1) Succinctly define, stating rigorously the meaning of any symbols used and the dimensions of measurement: a}Cpss b}feathering c}Wagner-Nelson method d}clearance e}f
(2) For each of the following pairs of variables (ordinate against abscissa), draw a graph illustrating the qualitative profile of their relationship. Where appropriate, indicate the nature of important slopes, intercepts, and values. Unless your specifically indicate on your plot that semi-log paper is being considered (write "SL"), it will be assumed that rectilinear paper is being considered. Graphs are for a drug given by oral route where applicable.
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a} total amount of drug collected minus the amount collected at the time in the urine vs time b} Plasma concentration of a drug given by oral route vs time c} Plasma concentration of metabolite of a drug given by IV bolus vs time d} Steady state plasma concentration vs infusion rate e} Steaty state plasma consentration vs clearance
PHARMACOKINETICS SECOND HOUR PRACTICE EXAM #3
SECTION I
(1) Succinctly define, stating rigorously the meaning of any symbols used and the dimensions of measurement:) a}Cpss b}f c}Absolute Bioavailability d}Comparative Bioavailability
(2) Compare and contrast:a}Wagner-Nelson and feathering methods ment of plasma and urine data using Wagner-Nelson
b} Treat-
(3) For each of the following pairs of variables (ordinate against abscissa), draw a graph illustrating the qualitative profile of their relationship. Where appropriate, indicate the nature of important slopes, intercepts, and values. Unless your specifically indicate on your plot that semi-log paper is being considered (write "SL"), it will be assumed that rectilinear paper is being considered. Graphs are for a drug given by oral route where applicable. dXu/dt vs t for a drug given orally.
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dXmu/dt vs t for a drug given by IV bolus. Steady state plasma concentration vs infusion rate Steady state plasma concentration vs elimination rate constant
PHARMACOKINETICS SECOND HOUR PRACTICE EXAM #4
SECTION I
1. Succinctly define, stating rigorously the meaning of any symbols used and the dimensions of measurement: a) clearance b)f c) absolute bioavailability d) comparative bioavailability e) AUC
2. By means of an annotated phase diagram explain how freeze-dried pharmaceutical injectables are made.
3. For each of the following pairs of variables (ordinate against abscissa) draw a graph illustrating the qualitative profile of their relationship. Where appropriate, indicate the nature of important slopes, intercepts, and values. Unless you specifically indicate on your plot that semi-log paper is being considered (write "SL"), it will be assumed that rectilinear paper is being considered. Pharmacological Response vs time Peak time vs ka for oral dose Fractional change in total body clearance vs. renal clearance
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AUC vs ka AUC vs ke
PHARMACOKINETICS SECOND HOUR PRACTICE EXAM #5
SECTION I 1. Succinctly define, stating rigorously the meaning of any symbols used and the dimensions of measurement: a) first pass effect b) f c) Intrinsic clearance d) comparative bioavailability e) Extraction ratio
2. By means of an annotated phase diagram explain how a metastable polymorph can be formed and how these polymorphs might effect the bioavailability of the drug.
3. For each of the following pairs of variables (ordinate against abscissa) draw a graph illustrating the qualitative profile of their relationship. Where appropriate, indicate the nature of important slopes, intercepts, and values. Unless your specifically indicate on your plot that semi-log paper is being considered (write "S-L"), it will be assumed that rectilinear paper is being considered fractional change in total body clearance vs plasma flow for drugs having a large extraction ratio. Peak time vs ka for oral dose Fractional change in total body clearance vs. hepatic clearance. AUC vs ka
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AUC vs clearance
PHARMACOKINETICS SECOND HOUR PRACTICE EXAM #6
SECTION I 1. Succinctly define, stating rigorously the meaning of any symbols used and the dimensions of measurement: Henderson-Hasselbach relationship Therapeutic alternatives Therapeutic equivalents comparative bioavailability Extraction ratio
Briefly discuss generic substitution by the pharmacist. Include such topics as when it might be admissable and the liabilities involved.
3. For each of the following pairs of variables (ordinate against abscissa) draw a graph illustrating the qualitative profile of their relationship. Where appropriate, indicate the nature of important slopes, intercepts, and values. Unless you specifically indicate on your plot that semi-log paper is being considered (write "S-L"), it will be assumed that rectilinear paper is being considered
a) fractional change in total body clearance vs fractional change in plasma flow for drugs having a small extraction ratio.
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b)
Peak time vs dose for oral dose
c) Fractional change in total body clearance vs. fractional change in hepatic clearance for drugs having a large extration ratio. d)
Ratio of milk to blood for basic drugs vs pKa.
e)
Ratio of milk to blood for acidic drugs vs pKa.
PHARMACOKINETICS SECOND HOUR PRACTICE EXAM #7
SECTION I 1. Succinctly define, stating rigorously the meaning of any symbols used and the dimensions of measurement: a) Bioequivalance b) Intrinsic Clearance c) first pass effect d) Henderson - Hasselbach equation e) f
2.Compare and contrast absolute and relative bioavailability. 3.For each of the following pairs of variables (ordinate against abscissa) draw a graph illustrating the qualitative profile of their relationship. Where appropriate, indicate the nature of important slopes, intercepts, and values. Unless you specifically indicate on your plot that semi-log paper is being considered (write "S-L"), it will be assumed that rectilinear paper is being considered
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a)
Cpmax vs f for a drug given orally.
b)
Cpmax vs dose for a drug given orally.
c)
Cpmax vs Vd for a drug given orally.
d)
TBC vs Fi(H) for a drug with a high extraction ratio in the liver.
e)
TBC vs Fr(H) for a drug with a high extraction ratio in the liver
PHARMACOKINETICS SECOND HOUR PRACTICE EXAM #9 SECTION I 1. Succinctly define, stating rigorously the meaning of any symbols used and the dimensions of measurement: a) Henderson-Hasselbach relationship b) Therapeutic alternatives c) Therapeutic equivalents d) Comparative bioavailability e) Extraction ratio
2. Briefly discuss generic substitution by the pharmacist. Include such topics as when it might be admissible and the liabilities involved.
3. For each of the following pairs of variables (ordinate against abscissa) draw a graph illustrating the qualitative profile of their relationship. Where appropriate, indicate the nature of important slopes, intercepts, and values. Unless you specifically indicate on your plot that semi-log paper is being considered (write "SL"), it will be assumed that rectilinear paper is being considered a) fractional change in total body clearance vs fractional change in plasma flow for drugs having a small extraction ratio.
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b)
Peak time vs dose for oral dose.
c) Fractional change in total body clearance vs. fractional change in intrinsic hepatic clearance for drugs having a large extraction ratio. d)
Ratio of blood to milk concentrations for basic drugs vs pKa.
e)
Ratio of blood to milk concentrations for acidic drugs vs pKa.
PHARMACOKINETICS SECOND HOUR PRACTICE EXAM # 10 SECTION I 1. Succinctly define, stating rigorously the meaning of any symbols used and the dimensions of measurement: a) Henderson-Hasselbach relationship b) Therapeutic alternatives c) Therapeutic equivalents d) Comparative bioavailability e) Absolute bioavailability f) Bioequivalents 2. Compare and Contrast: Feathering and Wagner-Nelson method.
3. For each of the following pairs of variables (ordinate against abscissa), draw a graph illustrating the qualitative profile of their relationship. Where appropriate, indicate the nature of important slopes, intercepts, and values. Unless you indicate on your plot that semi-log paper is being considered (write SL), it will be assumed that rectilinear paper is being considered. Graphs are for a drug given by an oral delivery system where applicable. a)
Cpss vs. K
b)
Cp vs. ka
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c)
Ratio of Milk to Blood for acidic drugs vs. pKa
Pharmacokinetics practice exam #11 Pharmaceutical alternatives may have different: I. therapeutic moieties II. dosage forms or strengths III. salt or ester forms of the same therapeutic moiety
2) Pharmaceutical equivalents must have the same: I. active ingredients and strength II. dosage form and route of administration III. rate and extent of absorption
3) Bioequivalent drug products must have the same: I. active ingredients and strength II. dosage form and route of administration III. rate and extent of absorption
4) Therapeutic equivalents are:
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I. pharmaceutical alternatives II. pharmaceutical equivalents III. bioequivalents
5) The Wagner-Nelson method I. uses curve stripping or feathering techniques II. can be used to find ku and km III. uses AUC calculations
6) The Federal guidelines for for bioequivalence require that the following pharmacokinetic parameters be within + 20 % of the innovator's product: I. AUC, Peak time, Cpmax II. Ka, Ke III. Vd
7) The steady state plasma concentration of a drug given by intravenous infusion is dependent on: I. length of time of of infusion II. volume of distribution III. elimination rate constant, K.
8) The peak time of a drug given by the oral route is dependent on:
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I. the absorption rate constant II. the metabolism rate constant III. the excretion rate constant
9) The slope of the terminal portion of the graph of the metabolite of a drug which (the drug, not the metabolite) was given by intravenous bolus injection could be: I. - the elimination rate constant of the metabolite II. - the elimination rate constant of the drug III. - the absorption rate constant of the metabolite
10) Comparative bioavailability includes calculations of the ratio(s) of the following pharmacokinetic parameters of two oral products (generic / Innovator) normalized for dose : I. AUC (0 to Inf) II. Peak time III. Cpmax
Pharmacokinetics practice second hour exam #12
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Where the are only three foils (possible quesses), please use K type system: (NOTE: if foils are equivalent, all must be selected) A) I ONLY B) III ONLY C) I AND II ONLY D) II AND III ONLY E) I, II, AND III
1) Steady state plasma concentration obtained by continuous infusion is inversely proportional to: I Infusion rate II elimination rate constant III volume of distribution
2) Steady state plasma concentration obtained by continuous infusion is directly proportional to: I Infusion rate II time III volume of distribution
3) When calculating the AUC for an oral product using the trapezoidal rule, concentrations necessary to calculate the first trapezoid are:
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I the intercept of the extrapolated line of the plasma vs. time profile. II the concentration at time zero, Cp0 III the concentration at the first time point
4) When calculating the AUC for an IV product using the trapezoidal rule, concentrations necessary to calculate the first trapezoid are: I the intercept of the extrapolated line of the plasma vs. time profile. II the concentration at time zero, Cp0 III the concentration at the first time point
5) Absolute bioavailability is a calculation which I must be between .80 and 1.20 II compares an oral product to an IV bolus dose. III is the ratio of the normalized AUCs of the products tested.
6) Comparative bioavailability is a calculation which I is the ratio of the normalized AUCs of the products tested. II must be between .80 and 1.20 III compares an oral product to an IV bolus dose. 7) When plotting the Wagner-Nelson function vs. time, a plot which proceeds horizontally for a measurable time and then declines:
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I is because of poor data in the early part of the data set. II only the declining portion should be used to calculate ka. III is an indication of a delay in release of the drug from the delivery system, a lag time.
8) When considering ion trapping, comparing a drug which forms sulfate salts distributing between mother's milk and blood, the ratio of total drug in milk to total drug in blood (Rm/b) can be I greater than one. II one. III less than one.
9) When considering ion trapping, comparing a drug which forms sodium salts distributing between mother's milk and blood, the ratio of total drug in milk to total drug in blood (Rm/b) can be I greater than one. II one. III less than one.
10) When using dry starch as a tablet disintegrating agent, I tablet hardness is directly proportional to starch content. II starch acts by allowing the water to wick into the tablet. III a threshold minimum amount of starch is necessary before any disintegration action is apparent.
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Cyclosporine A
(Problem 8 - 6)
Quigano, R., et al., "Effect of atropine of gastrointestinal motility and the bioavailability of cyclosporine A in rats", Drug Metabolism and Disposition, Vol. 21, No. 1, (1993), p. 141 - 143.
In this study rats with an average weight of 300 g were given either an IV bolus dose of cyclosporine A (CyA) or an oral dose of CyA. Subsequently, doses of atropine were given; however, the data below is that which was gathered prior to atropine administration. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Bioequivalence
Dose (mg/kg) ug- ⋅ hr AUC ------ mL ug- ⋅ hr 2 AUMC ------ mL MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 -------- mL Vd (L) ug- Cp at 1 hour ------ mL f ug Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
1.
MRT iv
2.
k e , the elimination rate constant
3.
t1 ⁄ 2
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4.
Cp 0
5.
Vd
6.
The plasma concentration ( Cp ) of cyclosporine A at 1 hour after the iv dose was given.
7.
AUC 0 – 1 hour for the iv dose
8.
f , the absolute bioavailability of oral cyclosporine A.
9
MRT oral .
10.
MAT oral
11.
k a , the apparent absorption rate constant.
12.
t peak for the oral dose.
13.
Cp max , the maximum concentration of the oral dosage form given as a single dose.
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Fosinopril
(Problem 8 - 7)
Gehr, T., et al., "The pharmacokinetics and pharmacodynamics of fosinopril in haemodialysis patients", European Journal of Clinical Pharmacology, Vol. 45, No. 5, (1993), p. 431 - 436.
Fosinopril (MW 562.6) is a new Angiotension Converting Enzyme (ACE) Inhibitor used in the treatment of hypertension. Following oral administration, fosinopril is rapidly and almost completely hydrolyzed to its pharmacologically active metabolite, fosinoprilate (MW 435.2). About 50% of the drug is excreted unchanged through the kidneys. In this study, patients received either 7.5 mg of fosinoprilat administered intravenously or 10 mg of fosinopril administered orally. A summary of the some of data obtained from this experiment is given below.
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Bioequivalence
Dose (mg/kg) ug- ⋅ hr AUC ------ mL ug- ⋅ hr 2 AUMC ------ mL MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug- Cp0 ------ mL Vd (L) ug Cp at 1 hour -------- mL f ug Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
From the preceding data, please calculate the following: 1.
MRT iv
2.
k e , the elimination rate constant
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3.
t1 ⁄ 2
4.
Cp 0
5.
Vd
6.
The plasma concentration ( Cp ) of fosinopril at 1 hour after the iv dose was given.
7.
AUC 0 – 1 hour for the iv dose
8.
f , the absolute bioavailability of oral fosinopril.
9
MRT oral .
10.
MAT oral
11.
k a , the apparent absorption rate constant.
12.
t peak for the oral dose.
13.
Cp max , the maximum concentration of the oral dosage form given as a single dose.
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Verapamil
(Problem 8 - 8)
Rutledge, D., Pieper, J., and Mirvis, D., "Effects of chronic phenobarbital on verapamil disposition in humans", The Journal of Pharmacology and Experimental Therapeutics, Vol. 246, No. 1, (1988), p. 7 - 13.
This study focused on the effects of phenobarbital, a hepatic-enzyme inducer, on verapamil. Seven healthy male volunteers with an average weight of 78.8 kg participated in the study. The patients received either an single oral verapamil dose of 80 mg or a single intravenous verapamil dose of 0.15 mg/kg over 3 minutes. A summary of the some of data obtained from this experiment is given below.
From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Bioequivalence
Dose (mg/kg) ug- ⋅ hr AUC ------ mL ug- ⋅ hr 2 AUMC ------ mL MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 -------- mL Vd (L) ug- Cp at 1 hour ------ mL f ug Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
1.
MRT iv
2.
k e , the elimination rate constant
3.
t1 ⁄ 2
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4.
Cp 0
5.
Vd
6.
The plasma concentration ( Cp ) of verapamil at 1 hour after the iv dose was given.
7.
AUC 0 – 1 hour for the iv dose
8.
f , the absolute bioavailability of oral verapamil.
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Zidovudine
(Problem 8 - 9)
Trang, J., et al., "Zidovudine bioavailability and linear pharmacokinetics in female B6C3F1 mice", Drug Metabolism and Disposition Vol, 21 (1993), p.189 - 193.
Zidovudine (AZT) is a potent inhibitor of HIV-1 during viral replication. It has been approved for the treatment of AIDS. In this study a 30 mg/kg dose of AZT was given to mice either iv or orally. :A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Bioequivalence
Dose (mg/kg) ug- ⋅ hr AUC ------ mL ug- ⋅ hr 2 AUMC ------ mL MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 -------- mL Vd (L) ug- Cp at 1 hour ------ mL f ug Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
1.
MRT iv
2.
k e , the elimination rate constant
3.
t1 ⁄ 2
4.
Cp 0
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5.
Vd
6.
The plasma concentration ( Cp ) of zidovudine at 1 hour after the iv dose was given.
7.
AUC 0 – 1 hour for the iv dose
8.
f , the absolute bioavailability of oral zidovudine.
9
MRT oral .
10.
MAT oral
11.
k a , the apparent absorption rate constant.
12.
t peak for the oral dose.
13.
Cp max , the maximum concentration of the oral dosage form given as a single dose.
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CHAPTER 8
Bioavailability, Bioequivalence, and Drug Selection
Author: Rasma Chereson Reviewer: Umesh Banakar
OBJECTIVES 1.
Given sufficient data to compare an oral product with another oral product or an IV product, the student will estimate (III) the bioavailability (compare AUCs) and judge (VI) professional acceptance of the product with regard to bioequivalence (evaluate (VI) AUC, T p and ( Cp ) max ).
2.
The student will write (V) a professional consult using the above calculations.
3.
The student will be able to calculate (III) the absolute bioavailability of drug products.
4.
The student will be able to discuss (II) the various factors affecting bioavailability.
5.
The student will be able to discuss (II) the various methods of assessing bioavailablity.
6.
The student will be able to discuss (II) In Vivo / In Vitro Correlations.
7.
The student will be able to enumerate (II) FDA requirements regarding bioequivalence.
8.
The student shall be able to utilize (III) the FDA “Orange Book” to make drug product selections.
9.
The student shall be able to discuss (II) and utilize (III) reasonalble guidelines regarding drug product selections.
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8.1 Bioavailability, Bioequivalence and Drug Product Selection Bioavailability and bioequivalence of drug products, and drug product selection have emerged as critical issues in pharmacy and medicine during the last three decades. Concern about lowering health care costs has resulted in a tremendous increase in the use of generic drug products; currently about one half of all prescriptions written are for drugs that can be substituted with a generic product (1). Over 80% of the approximately 10,000 prescription drugs available in 1990 were available from more than one source (2). With the increasing availability and use of generic drug products, health care professionals are confronted with an ever-larger array of multisource products from which they must select those that are therapeutically equivalent. This phenomenal growth of the generic pharmaceutical industry and the abundance of multisource products have prompted some questions among many health professionals and scientists regarding the therapeutic equivalency of these products, particularly those in certain critical therapeutic categories such as anticonvulsants and cardiovasculars (1, 3-5). Inherent in the currently accepted guidelines for product substitution is the assumption that a generic drug considered to be bioequivalent to a brand-name drug will elicit the same clinical effect. As straightforward as this statement regarding bioequivalence appears to be, it has generated a great deal of controversy among scientists and professionals in the health care field. Numerous papers in the literature indicate that there is concern that the current standards for approval of generic drugs may not always ensure therapeutic equivalence (6-18). The availability of different formulations of the same drug substance given at the same strength and in the same dosage form poses a special challenge to health care professionals, making these issues very relevant to pharmacists in all practice settings. Since pharmacists play an important role in product-selection decisions, they must have an understanding of the principles and concepts of bioavailability and bioequivalence.
8.1.1
RELATIVE AND ABSOLUTE BIOAVAILABILITY Bioavailability is a pharmacokinetic term that describes the rate and extent to which the active drug ingredient is absorbed from a drug product and becomes available at the site of drug action. Since pharmacologic response is generally related to the concentration of drug at the receptor site, the availability of a drug from a dosage form is a critical element of a drug product's clinical efficacy. However, drug concentrations usually cannot be readily measured directly at the site of
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action. Therefore, most bioavailability studies involve the determination of drug concentration in the blood or urine. This is based on the premise that the drug at the site of action is in equilibrium with drug in the blood. It is therefore possible to obtain an indirect measure of drug response by monitoring drug levels in the blood or urine. Thus, bioavailability is concerned with how quickly and how much of a drug appears in the blood after a specific dose is administered. The bioavailability of a drug product often determines the therapeutic efficacy of that product since it affects the onset, intensity and duration of therapeutic response of the drug. In most cases one is concerned with the extent of absorption of drug, (that is, the fraction of the dose that actually reaches the bloodstream) since this represents the "effective dose" of a drug. This is generally less than the amount of drug actually administered in the dosage form. In come cases, notably those where acute conditions are being treated, one is also concerned with the rate of absorption of a drug, since rapid onset of pharmacologic action is desired. Conversely, these are instances where a slower rate of absorption is desired, either to avoid adverse effects or to produce a prolonged duration of action. "Absolute" bioavailability, F, is the fraction of an administered dose which actually reaches the systemic circulation, and ranges from F = 0 (no drug absorption) to F = 1 (complete drug absorption). Since the total amount of drug reaching the systemic circulation is directly proportional to the area under the plasma drug concentration as a function of time curve (AUC), F is determined by comparing the respective AUCs of the test product and the same dose of drug administered intravenously. The intravenous route is the reference standard since the dose is, by definition, completely available. AUC ev F = ---------------AUC iv
(EQ 8-1)
(where AUCEV and AUCIV are, respectively, the area under the plasma concentration-time curve following the extravascular and intravenous administration of a given dose of drug. Knowledge of F is needed to determine an appropriate oral dose of a drug relative to an IV dose. "Relative" or “Comparative” bioavailability refers to the availability of a drug product as compared to another dosage form or product of the same drug given in the same dose. These measurements determine the effects of formulation differences on drug absorption. The relative bioavailability of product A compared to product B, both products containing the same dose of the same drug, is obtained by comparing their respective AUCs. AUC RelativeBioavailabilty = ---------------A AUC B Basic Pharmacokinetics
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(EQ 8-2)
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Bioavailability, Bioequivalence, and Drug Selection
where drug product B is the reference standard. When the bioavailability of a generic product is considered, it is usually the relative bioavailability that is referred to. A more general form of the equation results from considering the possibility of different doses, AUC Generic ----------------------------Dose Generic ComparativeBioavailability = ----------------------------AUC Brand ------------------------Dose Brand
(EQ 8-3)
The difference between absolute and relative bioavailability is illustrated by the following hypothetical example. Assume that an intravenous injection (Product A) and two oral dosage forms (Product B and Product C), all containing the same dose of the same drug, are given to a group of subjects in a crossover study. Furthermore, suppose each product gave the values for AUC indicated in Table 8-1 on page 4. TABLE 8-1. Data
for Absolute and Relative Bioavailability
Drug Product
Area Under the Curve (mcg/ml) x hr
A Intravenous injection
100
B Oral dosage form, brand or reference standard
50
C Oral dosage form, generic Product
40
The F for Product B and Product C is 50% (F = 0.5) and 40% (F = 0.4), respectively. However, when the two oral products are compared, the relative bioavailability of Product C as compared to Product B is 80%.
8.1.2
FACTORS INFLUENCING BIOAVAILABILITY Before the therapeutic effect of an orally administered drug can be realized, the drug must be absorbed. The systemic absorption of an orally administered drug in a solid dosage form is comprised of three distinct steps: 1.
disintegration of the drug product
2.
dissolution of the drug in the fluids at the absorption site
3.
transfer of drug molecule across the membrane lining the gastrointestinal tract into the systemic circulation.
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Any factor that affects any of these three steps can alter the drug's bioavailability and thereby its therapeutic effect. While there are more than three dozen of these factors that have been identified (19-38), the more significant ones are summarized here. The various factors that can influence the bioavailability of a drug can be broadly classified as dosage form-related or patient-related. Some of these factors are listed in Table 8-2 on page 5 and Table 8-3 on page 5, respectively. TABLE 8-2 Bioavailability
Factors related to the dosage form
Physicochemical properties of the drug
Formulation and manufacturing variables
Particle size
Amount of disintegrant
Crystalline structure
Amount of lubricant
Degree of hydration of crystal
Special coatings
Salt or ester form
Nature of diluent Compression force
TABLE 8-3 Bioavailability
Factors Related to the patient
Physiologic factors
Interactions with other substances
Variations in absorption power along GI tract
Food
Variations in pH of GI fluids
Fluid volume
Gastric emptying rate
Other drugs
Intestinal motility Perfusion of GI tract Presystemic and first-pass metabolism Age, sex, weight Disease states
The physical and chemical characteristics of a drug as well as its formulation are of prime importance in bioavailability because they can affect not only the absorption characteristics of the drug but also its stability. Since a drug must be dissolved to be absorbed, its rate of dissolution from a given product must influence its rate of absorption. This is particularly the case for sparingly soluble drugs. All the factors listed in Table 8-2 on page 5 can alter the dissolution rate of the drug, its bioavailability, and ultimately, its therapeutic performance. One of the more important factors that affects the dissolution rate of slowly dissolving substances is the surface area of the dissolving solid (39). Peak blood levels occurred much faster with the smaller particles than the larger ones, primarily Basic Pharmacokinetics
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as a result of their faster dissolution rate. Particle size can also have a significant effect on AUC(40). Serum levels of phenytoin after administration of equal doses containing micronized (formulation G) and conventional (formulation F) drug were measured. Based on the AUC, almost twice as much phenytoin was absorbed after the micronized preparation (40). There are numerous reports of the effects of formulation and processing variables on the dissolution of active ingredients from drug products; an apparently inert ingredient may affect drug absorption. For example, magnesium stearate, a lubricant, commonly used in tablet and capsule formulations, is water-insoluble and water-repellent. Its hydrophobic nature tends to retard drug dissolution by preventing contact between the solid drug and the aqueous GI fluids. Thus, increasing the amount of magnesium stearate in the formulation results in a slower dissolution rate of the drug, and decreased bioavailability(34) . The nature of the dosage form itself may have an effect on drug absorption characteristics. The major pharmaceutical dosage forms for oral use are listed in Table 84 on page 6 in order of decreasing bioavailability of their active ingredients. The decreasing bioavailability is related to the number of steps involved in the absorption process following administration. The greater the number of steps a product must undergo before the final absorption step, the slower is the availability and the greater is the potential for bioavailability differences to occur. Thus, solutions (elixirs, syrups, or simple solutions) generally result in faster and more complete absorption of drug, since a dissolution step is not required. Enteric-coated tablets, on the other hand, do not even begin to release the drug until the tablets empty from the stomach, resulting in poor and erratic bioavailability. TABLE 8-4 Bioavailability
Fastest availability
and oral Dosage Forms
Solutions Suspensions Capsules Tablets Coated tablets
Slowest availability
Controlled-release formulations
Bioavailability studies with pentobarbital from various dosage forms show the absorption rate of pentobarbital after administration in various oral dosage forms decreased in the following order: aqueous solution > aqueous suspension of the free acid > capsule of the sodium salt > tablet of the free acid (41). In addition to the dosage form-related factors identified above, bioavailability may also be affected by a variety of physiologic and clinical factors related to the patient (Table 8-3 on page 5). Considerable inter-subject differences in the bioBasic Pharmacokinetics
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availability of some drugs have been observed. These can often be attributed to individual variations in such factors as GI motility, disease state and concomitantly-administered food or drugs. One example of the myriad of physiologic factors that can affect the bioavailability of an orally-administered drug is a patient's gastric emptying rate. Since the proximal small intestine is the optimum site for drug absorption, a change in the stomach emptying rate is likely to alter the rate, and possibly the extent, of drug absorption. Any factor that slows the gastric emptying rate may thus prolong the onset time for drug action and reduce the therapeutic efficacy of drugs that are primarily absorbed from the small intestine. In addition, a delay in gastric emptying could result in extensive decomposition and reduced bioavailability of drugs that are unstable in the acidic media of the stomach (e.g. penicillins and erythromycin). Differences in stomach emptying among individuals have been implicated as a major cause of variations in the bioavailability of some drugs, particularly those with acid-resistant enteric coatings. In a study (42), after the administration of 1.5 g acetaminophen to 14 patients, the maximum plasma concentration ranged from 7.4 to 37 mcg/ml, and the time to reach the maximum concentration ranged from 30 to 180 minutes. Both these parameters of bioavailability were linearly related to the gastric emptying half-life found in each patient. There are numerous factors that affect gastric emptying rate (Table 8-5 on page 8) (43). Factors such as a patient's emotional state, certain drugs, type of food ingested and even a patient's posture can alter the time course and extent of drug absorption.
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TABLE 8-5
Factors influencing Gastric Emptying Rate
INFLUENCE ON GASTRIC EMPTYING RATE
FACTOR Increased viscosity of stomach contents
decreased
Body position lying on left side
decreased
Emotional state stress
increased or decreased
depression
decreased
anxiety
increased
Activity, exercise
decreased
Type of meal fatty acids, fats
decreased
carbohydrates
decreased
amino acids
decreased
pH of stomach contents decreased
decreased
increased
increased
Disease states gastric ulcers
decreased
Crohn's disease
decreased
hypothyroidism
decreased
hyperthyroidism
increased
Drugs atropine
decreased
propantheline
decreased
narcotic analgesics
decreased
amitriptyline
decreased
metoclopramide
increased
Since drugs are generally administered to patients who are ill, it is important to consider the effects of the disease process on the bioavailability of the drug. Disease states, particularly those involving the GI tract, such as celiac disease, Crohn's disease, achlorhydria, and hypermotility syndromes can certainly alter the absorption of a drug (32). In addition, some diseases concerning the cardiovascular system and the liver may also alter circulating drug levels after oral dosing. Drugs are frequently taken with food, and patients often use mealtimes to remind them to take their medications. However, food can have a significant effect on the bioavailability of drugs. The influence of food on drug absorption has been recog-
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nized for some time, and several reviews have been published on the influence of food on drug bioavailability (30-32, 36, 44). Food may influence drug absorption indirectly, through physiological changes in the GI tract produced by the food, and/or directly, through physical or chemical interactions between the drug molecules and food components. When food is ingested, stomach emptying is delayed, gastric secretions are increased, stomach pH is altered, and splanchnic blood flow may increase. These may all affect bioavailability of drugs. Food may also interact directly with drugs, either chemically (e.g. chelation) or physically, by adsorbing the drug or acting as a barrier to absorption. In general, gastrointestinal absorption of drugs is favored by an empty stomach, but the nature of drug-food interactions is complex and unpredictable; drug absorption may be reduced, delayed, enhanced or unaffected by the presence of food. Table 8-6 on page 9 summarizes some of the studies that have indicated the effect of food on the bioavailability of a variety of drugs. TABLE 8-6
Effect of Food on Drug Absorption
Reduced Absorption
Delayed Absorption
Increased Absorption
Ampicillin
Acetaminophen
Chlorothiazide
Aspirin
Aspirin
Diazepam
Atenolol
Cephalosporins (most)
Griseofulvin
Captopril
Diclofenac
Hydralazine
Erythromycin
Digoxin
Labetalol
Ethanol
Furosemide
Metoprolol
Hydrochlorothiazide
Nitrofurantoin
Nitrofurantoin
Penicillins
Sulfadiazine
Propranolol
Tetracyclines (most)
Sulfisoxazole
Riboflavin
Source: Ref. 32
The effect of food and type of diet on the bioavailability of erythromycin is shown in a study by Welling (45). The absorption of the antibiotic is significantly reduced when it is administered with food compared with its absorption under fasting conditions. This reduced absorption is primarily a result of degradation of the acid-labile erythromycin due to prolonged retention in the stomach. Delayed absorption due to food has been demonstrated in the case of cephradine in a study by Mischler (46). Similar results have been observed with other oral cephalosporins. Some drugs demonstrate enhanced bioavailability in the presence of food. This has been attributed to a variety of factors, including improved compound solubility and more time for dissolution because of delayed gastric emptying. In the case of Basic Pharmacokinetics
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highly metabolized agents, such as propranolol and metoprolol, the enhanced availability may be due to increased splanchnic blood flow causing reduced first-pass clearance. The circulating levels of these drugs dosed under fasting and non-fasting conditions have been presented in a study by Melander (47). The volume of fluid with which an orally administered dose is taken can also affect a drug's bioavailability. Drug administration with a larger fluid volume will generally improve its dissolution characteristics and may also result in more rapid stomach emptying. Thus, more efficient and more reliable drug absorption can be expected when an oral dosage form is administered with a larger volume of fluid. (45) . Interactions between drugs can have a significant effect on the bioavailability of one or both drugs. Such interactions may be direct, as in chelation of tetracycline by polyvalent metal ions in antacids or the adsorption of digoxin by cholestyramine resin, or indirect, as with the increased rate of acetaminophen absorption due to the increased gastric emptying rate produced by metoclopramide. Most of the reported drug-drug interactions have resulted in a reduction in the rate and/or extent of drug absorption, the most frequent causes being complexing of a drug with other substances, reduced GI motility and alterations in drug ionization (24, 30, 32, 48, 49). Table 8-7 on page 10 summarizes the major mechanisms of GI drug interactions affecting bioavailability. TABLE 8-7
Drug interactions affecting absorption
1. Change in gastric or intestinal pH 2. Change in gastrointestinal motility 3. Change in gastrointestinal perfusion 4. Interference with mucosal function (drug-induced malabsorption syndromes) 5. Chelation 6. Exchange resin binding 7. Aadsorption 8. Solution in poorly absorbable liquid
Source: Ref. 23
An example of a direct interaction between drugs affecting bioavailability is the interaction between iron and tetracycline. This is a well-documented and clinically significant interaction which can result in a dramatic reduction in serum concentration of tetracycline (50). The above potential sources of alteration in a drug's bioavailability must be kept in mind when attempting to evaluate the relative performance of drug products on the Basic Pharmacokinetics
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basis of studies performed with healthy human volunteers. These studies are generally performed under tightly-controlled fasting conditions in the absence of other drugs. In practice, however, drugs are seldom taken under such ideal conditions, and the factors leading to changes in drug absorption must be taken into consideration.
8.1.3
METHODS OF ASSESSING BIOAVAILABILITY Bioavailability testing is a means of predicting the clinical efficacy of a drug; the estimation of the bioavailability of a drug in a given dosage form is direct evidence of the efficiency with which a dosage form performs its intended therapeutic function. The bioavailability of a drug substance formulated into a pharmaceutical product is fundamental to the goals of dosage form design and essential for the clinical efficacy of the medication. Thus, bioavailability testing, which measures the rate and extent of drug absorption, is a way to obtain evidence of the therapeutic utility of a drug product. Bioavailability determinations are performed by drug manufacturers to ensure that a given drug product will get the therapeutic agent to its site of action in an adequate concentration. Bioavailability studies are also carried out to compare the availability of a drug substance from different dosage forms or from the same dosage form produced by different manufacturers.
In-vivo methods
One method for assessing the bioavailability of a drug product is through the demonstration of a clinically significant effect. However, such clinical studies are complex, expensive, time-consuming and require a sensitive and quantitative measure of the desired response. Further, response is often quite variable, requiring a large test population. Practical considerations, therefore, preclude the use of this method except in initial stages of development while proving the efficacy of a new chemical entity. Quantification of pharmacologic effect is another possible way to assess a drug's bioavailability. This method is based on the assumption that a given intensity of response is associated with a particular drug concentration at the site of action; e.g., variation of miotic response intensity can be directly related to the oral dose of chlorpromazine. However, monitoring of pharmacologic data is often difficult, precision and reproducibility are difficult to establish, and there are only a limited number of pharmacologic effects (e.g. heart rate, body temperature, blood sugar levels) that are applicable to this method. Because of these limitations, alternative methods have been developed to predict the therapeutic potential of a drug. The current method to assess the clinical performance of a drug involves measurement of the drug concentrations in the blood
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or urine. In such studies a single dose of the drug product is administered to a panel of normal, healthy adult (18- to 35-year old) subjects. Blood and/or urine samples are collected over a period of time following administration and are analyzed for drug content. Based on the blood concentration as a function of time and/or urinary excretion profile, inferences are drawn regarding the rate and extent of absorption of the drug. These studies are relatively easy to conduct and require a limited number of subjects. Blood level studies-
Blood level studies are the most common type of human bioavailability studies, and are based on the assumption that there is a direct relationship between the concentration of drug in blood or plasma and the concentration of drug at the site of action. By monitoring the concentration in the blood, it is thus possible to obtain an indirect measure of drug response. Following the administration of a single dose of a medication, blood samples are drawn at specific time intervals and analyzed for drug content. A profile is constructed showing the concentration of drug in blood at the specific times the samples were taken . The key parameters to note are: 1.
AUC
∞ 0
, The area under the plasma concentration-time curve, The AUC is proportional to the
total amount of drug reaching the systemic circulation, and thus characterizes the extent of absorption. 2.
Cmax , The maximum drug concentration. The maximum concentration of drug in the plasma is a function of both the rate and extent of absorption. Cmax will increase with an increase in the dose, as well as with an increase in the absorption rate.
3.
Tmax , The time at which the Cmax occurs. The Tmax reflects the rate of drug absorption, and decreases as the absorption rate increases.
Bioavailability (the rate and extent of drug absorption) is generally assessed by the determination of these three parameters. Since the AUC is representative of, and proportional to, the total amount of drug absorbed into the circulation, it is used to quantitate the extent of drug absorption. The calculation of AUC has been discussed in Chapter 4. A variety of pharmacokinetic methods have been suggested for the calculation of absorption rates (51-56). For clinical purposes, it is generally sufficient to determine Cmax and Tmax. If all other factors are constant, such as the extent of absorption and rate of elimination, then Cmax is proportional to the rate of absorption and Tmax is inversely proportional to the absorption rate. Thus, the faster the absorption of a drug the higher the maximum concentration will be and the less time it will take to reach the maximum concentration. Urinary Excretion Data -
An alternative bioavailability study measures the cumulative amount of unchanged drug excreted in the urine. These studies involve collection of urine samples and the determination of the total quantity of drug excreted in the urine as a function of
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time. These studies are based on the premise that urinary excretion of the unchanged drug is directly proportional to the plasma concentration of total drug. Thus, the total quantity of drug excreted in the urine is a reflection of the quantity of drug absorbed from the gastrointestinal tract. Consider the following example: two products, A and B, each containing 100 mg of the same drug are administered orally. A total of 80 mg of drug is recovered in the urine from Product A, but only 40 mg is recovered from Product B. This indicates that twice as much drug was absorbed from Product A as from Product B. (The fact that neither product resulted in excretion of the entire dose might be due to the existence of other routes of elimination, e.g. metabolism). This technique of studying bioavailability is most useful for those drugs that are not extensively metabolized prior to urinary elimination. As a rule-of-thumb, determination of bioavailability using urinary excretion data should be conducted only if at least 20% of a dose is excreted unchanged in the urine after an IV dose (56). Other conditions which must be met for this method to give valid results include: 1.
the fraction of drug entering the bloodstream and being excreted intact by the kidneys must remain constant.
2.
collection of the urine has to continue until all the drug has been completely excreted (five times the half-life 1).
Urinary excretion data are primarily useful for assessing extent of drug absorption, although the time course for the cumulative amount of drug excreted in the urine can also be used to estimate the rate of absorption. In practice, these estimates are subject to a high degree of variability, and are less reliable than those obtained from plasma concentration-time profiles (57). Thus, urinary excretion of drug is not recommended as a substitute for blood concentration data; rather, these studies should be used in conjunction with blood level data for confirmatory purposes. Single-dose versus Multiple-Dose-
Most bioavailability evaluations are made on the basis of single-dose administration. The argument has been made that single doses are not representative of the actual clinical situation, since in most instances, patients require repeated administration of a drug. When a drug is administered repeatedly at fixed intervals, with the dosing frequency less than five half-lives, drug will accumulate in the body and eventually reach a plateau, or a steady-state At steady-state, the amount of drug eliminated from the body during one dosing interval is equal to the available dose (rate in = rate out); therefore, the area under the curve during a dosing interval at steady-state is equal to the total area under the curve obtained when a single dose is administered. This AUC can therefore be
1.
Half life is defined as the length of time required to lose 50% of the drug in the body, assuming first order elimination.
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used to assess the extent of absorption of the drug, as well as its absolute and relative bioavailability. Multiple-dose administration has several advantages over single-dose bioavailability studies, as well as some limitations. These are summarized in Table 8-8 on page 14 (54, 59). TABLE 8-8 Multiple
dose vs. single dose studies in bioavailability studies
Advantages: • Eliminates the need to extrapolate the plasma concentration profiles to obtain the total AUC after a single dose • Eliminates the need for a long wash-out period between doses • More closely reflects the actual clinical use of the drug • Allows blood levels to be measured at the same concentrations encountered therapeutically • Because blood levels tend to be higher than in the single-dose method, quantitative determination is easier and more reliable • Saturable pharmacokinetics, if present, can be more readily detected at steady-state Limitations: • Requires more time to complete • More difficult and costly to conduct (requiring prolonged monitoring of subjects • Greater problems with compliance control • Greater exposure of subjects to the test drug, increasing the potential for adverse reactions
When a drug obeys linear, first-order kinetics, it is possible to estimate the results that would be obtained during multiple dosing from single-dose studies. Projection is easily made with regard to the extent of absorption, using the AUC following a single dose. Results from bioequivalence studies indicate that conclusions on the extent of absorption as assessed by the AUC can be made equally well on the basis of a single or multiple dose study (60). Assessing the rate of absorption during multiple-dosing from single-dose studies has presented a greater problem. Although a number of single-dose characteristics have been suggested as indicators of rate of absorption during multiple dosing (e.g. percent peak-trough fluctuation and percent peak-trough swing), results of bioequivalence studies indicate that only the plateau time (the time during which the concentration exceeds 75% of the maximum concentration, t 75% Cmax) and the residual concentration at the end of the dose interval produce consistent results in assessing the rate of absorption in single- and multiple-dose studies (54, 61). In the case of drugs exhibiting nonlinear kinetics, establishing a linear relationship between single- and multiple-dose bioavailability data has proven to be a difficult task. Thus, it has been recommended that for drugs with either saturable elimination or a nonlinear first-pass effect, steady-state studies be carried out to assess their bioavailability (62). Basic Pharmacokinetics
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8.1.4
STUDY DESIGN Bioavailability studies involve the administration of the test dosage form to a panel of subjects, after which blood and/or urine samples are collected and analyzed for drug content. Based on the concentration profile of the drug, a judgement is made regarding the rate and extent of absorption of the drug. Normally, the study is conducted in a group of healthy, male subjects who are of normal height and weight, and range in age from 18 to 35 years (6). Questions have been raised regarding the extent to which such a population reflects the performance of a given drug product in a actual patient population. At first glance, it would seem that bioavailability should be determined in patients actually suffering from the disease for which the drug is intended, or in patients representative of the age and sex of subjects who would be using the drug. However, there are several very good reasons for using healthy volunteers rather than patients. In bioavailability studies, it is assumed that there are no physiologic changes in the subjects during the course of the study. If actual patients were used, this would not be a valid assumption, due to possible changes in the disease state. Another potential problem with using patients is that many patients take more than one drug. This could result in a drug-drug interaction which could influence the bioavailability of the test drug. In addition, diet and fluid volume intake, both of which can influence a drug's bioavailability are more difficult to control in a patient population than in a panel of healthy test subjects. In general, it is more difficult with patients to have a standardized set of conditions which are necessary for a dependable bioavailability study. However, it must be recognized that factors that may affect a drug's performance in a patient population may not be detected in a group of healthy subjects. Thus, it is best to conduct a separate study in patients to determine if the disease, for which the drug is intended to be used, alters the bioavailability of the drug. Other important considerations in the methodology of a bioavailability study are sample size, period of trial, and sampling. For statistical purposes, twelve subjects are considered to be a minimum sample size. Otherwise there will not be enough data to draw valid conclusions (63). The bioavailability testing period should be of a sufficient length of time to ensure that drug absorption has been completed. This length of time is at least three times the half-life of the drug; generally a period of four to five times the half-life is used (63, 64). Blood samples should be taken with sufficient frequency to permit an accurate determination of tmax, Cmax and AUC.
8.1.5
IN-VITRO DISSOLUTION AND BIOAVAILABILITY Pharmaceutical scientists have for many years been attempting to establish a correlation between some physicochemical property of a dosage form and the biological
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availability of the drug from that dosage form. The term commonly used to describe this relationship is "in-vitro/in-vivo correlation" (65). Specifically, it is felt that if such a correlation could be established, it would be possible to use in-vitro data to predict a drug's in-vivo bioavailability. This would drastically reduce, or in some cases, completely eliminate the need for bioavailability tests. The desirability for this becomes clear when one considers the cost and time involved in bioavailability studies as well as the safety issues involved in administering drugs to healthy subjects or patients. It would certainly be preferable to be able to substitute a quick, inexpensive in-vitro test for in-vivo bioavailability studies. This would be possible if in-vitro tests could reliably and accurately predict drug absorption and reflect the in-vivo performance of a drug in humans. Disintegration Tests-
The early attempts to establish an indicator of drug bioavailability focused on disintegration as the most pertinent in-vitro parameter. The first official disintegration test appeared in the United States Pharmacopeia (USP) in 1950. However, while it is true that a solid dosage form must disintegrate before significant dissolution and absorption can occur, meeting the disintegration test requirement only insures that the dosage form (tablet) will break up into sufficiently small particles in a specified length of time. It does not ensure that the rate of solution of the drug is adequate to produce suitable blood levels of the active ingredient. Therefore, while the test for tablet disintegration is very useful for quality control purposes in manufacturing, it is a poor index of bioavailability.
Dissolution Tests-
Since a drug must go into solution before it can be absorbed, and since the rate at which a drug dissolves from a dosage form often determines its rate and/or extent of absorption, attention has been directed at the dissolution rate. It is currently considered to be the most sensitive in-vitro parameter most likely to correlate with bioavailability.
Official dissolution tests -
There are two official USP dissolution methods: Apparatus 1, (basket method), and Apparatus 2 (paddle method). For details of these dissolution tests, the reader is recommended to consult USPXXII/NFXVII (66). Dissolution tests are an extremely valuable tool in ensuring the quality of a drug product. Generally, product-to-product variations are due to formulation factors, such as particle size differences, excessive amounts of lubricant and coatings. These factors are reactive to dissolution testing. Thus, dissolution tests are very effective in discriminating between and within batches of drug product(s). The dissolution test, in addition, can exclude definitively any unacceptable product.
Limitations of dissolution tests-
There are, however, problems with in-vitro dissolution testing which should be noted - problems which make correlation with in- vivo availability difficult. The first is related to instrument variance and the absence of a standard method. The tests described in the USP are but a few of the large number of dissolution methods proposed to predict bioavailability. Since the dissolution rate of a dosage form is dependent on the methodology used in the dissolution test, changes in the appara-
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tus, dissolution medium, etc., can dramatically modify the results. Table 8-9 on page 17 lists some of the factors related to the dissolution testing device that can affect the dissolution rate of the drug. TABLE 8-9
Device factors affecting dissolution
1. Degree of agitation 2. Size and shape of container 3. Composition of dissolution medium • pH • ionic strength • viscosity • surface tension 4. Temperature of dissolution medium 5. Volume of dissolution medium 6. Evaporation 7. Hydrodynamics (flow pattern) Source: Ref. 67
Another significant problem is related to the difference between the in-vitro and in-vivo environments in which dissolution occurs. In-vitro studies are generally carried out under controlled conditions in one, or perhaps two, standardized solvents. The in-vivo environment (the gastrointestinal tract), on the other hand, is a continuously changing, complex environment. There are many variables which can affect the dissolution rate of a drug in the gastrointestinal tract, including pH, enzyme secretions, surface tension, motility, presence of other substances and absorption surfaces (68). Thus, drugs frequently dissolve in the body at rates quite different from those observed in an in-vitro test situation. Most of the official dissolution tests tend to be acceleration dissolution tests which bear limited or no relationship with in-vivo dissolution. Adding to the complexity of correlating dissolution with in-vivo absorption are factors such as drug-drug interactions, age, food effects, health, genetic background, biorhythm and physical activity (32, 69). All these factors may have an effect on the rate and extent of absorption of a drug. Thus, the in-vivo environment is far more complex, variable, and unpredictable than any in-vitro test environment, making in-vitro / in-vivo correlations very difficult. A simple dissolution test in a standardized vehicle cannot reflect the in vivo absorption of a drug across a population (70). Parameters used-
Proper selection of the in-vitro and in-vivo parameters to be correlated is critical in achieving a meaningful correlation. The in-vitro parameter should be selected that has the greatest effect on the absorption characteristics of the drug (71). There are several approaches to establishing a correlation between the dissolution of a
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drug in in- vitro and the bioavailability of a drug in-vivo. The in-vitro - in-vivo correlative methods used most often are of the single-point type where the dissolution rate (expressed as the percent of drug dissolved in a given time, or the time required for a given percent of the drug to dissolve) is correlated to a certain parameter of the bioavailability. Examples of in-vivo parameters used include Cmax, AUC, time to reach half-maximal plasma concentration, the average plasma concentration after 0.5 or 1 hour, maximum urinary excretion rate, and cumulative percent excreted in urine after a given time (71- 78). According to Wagner, the best in-vitro variable to use is the time for 50 percent of the drug to dissolve, and the best variable from in-vivo data to use is the time for 50 percent of the drug to be absorbed (79). Ideally, one would hope to find a linear relationship between some measurement of the dissolution test and some measurement based on bioavailability studies. Unfortunately, most attempts to accomplish this objective have failed.
8.1.6
IN-VITRO / IN-VIVO CORRELATION STUDIESThere have been many attempts to establish in-vitro / in-vivo correlations for a large variety of drugs. Some of these studies have been summarized by Welling, Banakar, and Abdou (71, 80-82). While there are many published examples of satisfactory correlations between absorption parameters and in-vitro dissolution tests, most studies have resulted in poor, or moderate, in-vitro - in-vivo correlations, often involving agreement with only one of the critical bioavailability parameters. Moreover, the positive correlations that have been found generally apply only to the specific formulation studied. There have been instances where the dissolution rates or various formulations of the same drug have been significantly different, yet little or no difference was observed in their bioavailability parameters (83-85). There have also been cases where a drug has failed to meet compendia dissolution standards but has demonstrated adequate bioavailability (86). Welling states: "To the writer's knowledge, there have been no studies that have accurately correlated in- vitro and in-vivo data to the point that the use of upper and lower limits for in-vitro dissolution parameters can be confidently used to predict in-vivo behavior and, therefore, to replace in-vivo testing" (71). Even if an in-vitro test could be designed that would accurately reflect the dissolution process in the gastrointestinal tract, dissolution is only one of many factors that affect a drug's bioavailability. For example, saturable presystemic metabolism may affect the extent of drug absorption, but this would not be predicted by an
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in-vitro test. Dissolution studies also would not predict poor bioavailability due to instability in gastric fluid or complexation with another drug or food component. Thus, the ultimate evaluation a drug product's performance under the conditions expected in clinical therapy must be an in-vivo test; a dissolution test is unlikely to entirely replace bioavailability testing (70, 87, 88). In-vitro methods are important in the development and optimization of dosage forms while in-vivo tests are essential in obtaining information on the behavior of medication in living organisms. One cannot be substituted for the other (69).
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8.2 Bioequivalence Definitions
With the phenomenal increase in the availability of generic drugs in recent years, the issues of bioavailability and bioequivalence have received increasing attention. In order for a drug product to be interchangeable with the pioneer (innovator or brand name) product, it must be both pharmaceutically equivalent and bioequivalent to it. According to the FDA, "pharmaceutical equivalents" are drug products that contain identical active ingredients and are identical in strength or concentration, dosage form, and route of administration (89). However, pharmaceutical equivalents do not necessarily contain the same inactive ingredients; various manufacturers' dosage forms may differ in color, flavor, shape, and excipients. The terms "pharmaceutical equivalents" and "chemical equivalents" are often used interchangeably. "Bioequivalence" is a comparison of the bioavailability of two or more drug products. Thus, two products or formulations containing the same active ingredient are bioequivalent if their rates and extents of absorption are the same. When a new formulation of an existing drug is developed, its bioavailability is generally evaluated relative to the standard formulation of the originator. Indeed, a bioequivalence trial against the standard formulation is the key feature of an Abbreviated New Drug Application (ANDA) submitted to the Food and Drug Administration by a manufacturer who wishes to produce a generic drug. For a generic drug to be considered bioequivalent to a pioneer product, there must be no statistical differences (as specified in the accepted criteria) between their plasma concentration-time profiles. Because two products rarely exhibit absolutely identical profiles, some degree of difference must be considered acceptable, as will be discussed later. Since the concentration of a drug in blood is used as an assessment of its clinical performance, inherent in the demonstration that two preparations containing equivalent amounts of the same drug produce similar concentrations of the drug entity in blood is the assumption that they will elicit equivalent drug responses. Thus, two products that are deemed to be bioequivalent are also assumed to be therapeutically equivalent, and therefore interchangeable. This principle is fundamental to the concept of bioequivalence and is the basic premise on which it is founded. In general, the FDA considers two products to be "therapeutic equivalents" if they each meet the following criteria (90): 1.
they are pharmaceutical equivalents,
2.
they are bioequivalent (demonstrated either by a bioavailability measurement or an in vitro standard),
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Background
3.
they are in compliance with compendial standards for strength, quality, purity and identity,
4.
they are adequately labelled, and
5.
they have been manufactured in compliance with Good Manufacturing Practices as established by the FDA.
The first intimations of bioequivalence problems with multi-source drug products were given by early investigations of the availability of vitamins, aspirin, tetracycline, and tolbutamide (91-97). In 1974, after an extensive review of the bioavailability of drugs, Koch-Weser concluded that " . . . among drugs thus far tested bioinequivalence of different drug products has been far more common than bioequivalence" (98). Of particular note were the studies involving digoxin; the findings of these investigations sparked the discussion about bioequivalence assessment that still continues today. Significant differences were seen in the bioavailability of digoxin not only between products supplied by different companies, but also between lots obtained from the same manufacturer (99). Because of the narrow therapeutic range for this drug, and because the drug is utilized in the treatment of cardiac patients, these findings generated a great deal of concern. Similar reports of bioinequivalence and therapeutic inequivalence appeared for other drugs as well, including phenytoin, phenylbutazone, chloramphenicol, tolbutamide and thyroid (6). The clinical significance of these reported differences in bioavailability relates to the therapeutic index of the drug, the dose of the drug and the nature of the disease. In 1973 the Ad Hoc Committee on Drug Product Selection of the American Pharmaceutical Association published a list of drugs with a potential for therapeutic inequivalence based on reported evidence of bioinequivalence (100). The drugs fall in three categories: "high," "moderate," or "low risk" based on the clinical implications (Table 8-10 on page 22).
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TABLE 8-10 Drugs
with various risk potential for inequivalence
High Risk Potential
Moderate Risk Potential
Low or Negligible Risk Potential
aminophylline
amphetamines
acetaminophen
aspirin (when used in high dose levels)
(sustained-release)
codeine
ampicillin
ferrous sulfate
bishydroxycoumarin
chloramphenicol
hydrochlorothiazide
digoxin
chlorpromazine
ephedrine
dipheylhydantoin (phenytoin)
digitoxin
isoniazid
para-aminosalicylic acid
erythromycin
meprobamate
prednisolone
griseofulvin
penicillin VK
prednisone
oxytetracycline
sulfisoxazole
quinidine
penicillin G (buffered)
warfarin
pentobarbital phenylbutazone phenacetin potassium chloride (solid dosage forms) salicylamide secobarbital sulfadiazine tetracycline tolbutamide
The concern about the bioinequivalence of some drugs led to the establishment in 1974 of the Drug Bioequivalence Study Panel of the Office of Technology Assessment (OTA). The objective was to ensure that drug products of the same physical and chemical composition would produce similar therapeutic effects. Among the 11 recommendations of the Panel was the conclusion that not all chemical equivalents were interchangeable, but the goal of interchangeability was achievable for most oral drug products (101). The Report recommended that a system should be organized as rapidly as possible to generate an official list of interchangeable drug products. The OTA Report, as well as the growing awareness within the scientific and regulatory communities of bioavailability problems with marketed drug products, focused the attention of the FDA on bioequivalence and bioavailability problems and issues.
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8.2.1
BIOEQUIVALENCE REGULATIONS In 1977, the FDA implemented a series of bioavailability and bioequivalence regulations which formed the basis of subsequent discussion, if not controversy, of therapeutic equivalency of drug products (102). The regulations are divided into two separate regulations; Subpart B - Procedures for Determining the Bioavailability of Drug Products and Subpart C - Bioequivalence Requirements. While Table 11 summarizes the key provisions of the bioavailability regulations, those for bioequivalence requirements are summarized in Table 8-11 on page 23. TABLE 8-11 Key
provisions for bioavailabilty regulations
1. Defines bioavailability in terms of both the rate and extent of drug absorption. 2. Describes procedures for determining the bioavailability of drug products. 3. Sets forth requirements for submission of in vivo bioavailability data. 4. Sets forth criteria for waiver of human in vivo bioavailability studies. 5. Provides general guidelines for the conduct of in vivo bioavailability studies. 6. Imposes a requirement for filing an Investigational New Drug Application. Source: Ref. 103
Criteria for establishing a bioequivalence requirement -
The 1977 Bioequivalence regulations set forth the following criteria and evidence supporting the establishment of a bioequivalence requirement for a given drug product: 1.
Evidence from well-controlled clinical trials or controlled observations in patients that such products do not give comparable therapeutic effects.
2.
Evidence from well-controlled bioequivalence studies that such products are not bioequivalent drug products.
3.
Evidence that the drug products exhibit a narrow therapeutic ratio, (e.g., there is less than a two-fold difference in the median lethal dose (LD50) and median effective dose (ED50) value or have less than a two-fold difference in the minimum toxic concentration and minimum effective concentrations in the blood), and safe and effective use of the drug product requires careful dosage titration and patient monitoring.
4.
Competent medical determination that a lack of bioequivalence would have a serious adverse effect in the treatment or prevention of a serious disease or condition.
5.
Physicochemical evidence of any of the following: a.
The active drug ingredient has a low solubility in water--e.g., less than 5 mg/ml.
The dissolution rate of one or more such products is slow--e.g., less than 50 percent in thirty minutes when tested with a general method specified by an official compendium or the FDA. b.
The particle size and/or surface area of the active drug ingredient is critical in determining bioavailability. c.
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Polymorphs, solvates, complexes, and such, exist that could contribute to poor dissolution and may affect absorption. d. e.
There is a high excipient/active drug ratio present in the drug product--e.g., greater than 5
to 1. The presence of specific inactive ingredients (e.g. hydrophilic or hydrophobic excipients) that either may be required for absorption of the active drug or may interfere with such absorption. f.
6.
Pharmacokinetic evidence of any of the following: The drug is absorbed in large part in a particular segment of the gastrointestinal tract or is absorbed from a localized site. a.
Poor absorption of the drug, even when it is administered as a solution--e.g., less than 50 percent compared to an intravenous dose. b. c.
The drug undergoes first-pass metabolism in the intestinal wall or liver.
The drug is rapidly metabolized or excreted, requiring rapid dissolution and absorption for effectiveness. d.
The drug is unstable in specific portions of the gastrointestinal tract, requiring special coatings and formulations--e.g., enteric coatings, buffers, film coatings--to ensure adequate absorption. e.
The drug follows nonlinear kinetics in or near the therapeutic range, and the rate and extent of absorption are both important to bioequivalence. f.
Types of Bioequivalence Requirements
In the event that a drug meets one or more of the above six criteria, a bioequivalence requirement is established. The requirement could be either an in-vivo or an in-vitro investigation, as specified by the FDA. The types of bioequivalence requirements include the following: 1.
An in-vivo test in humans.
2.
An in-vivo test in animals that has been correlated with human in- vivo data.
3.
An in-vivo test in animals that has not been correlated with human in- vivo data.
4.
An in-vitro bioequivalence standard, i.e., an in-vitro test that has been correlated with human in-vivo bioavailability data.
5.
A currently available in-vitro test (usually a dissolution rate test) that has not been correlated with human in-vivo bioavailability data.
The regulations state that in-vivo testing in humans would generally be required if there is well-documented evidence that pharmaceutical equivalents intended to be used interchangeably meet one of the first three criteria used to establish a bioequivalence requirement: 1.
The drug products do not give comparable therapeutic effects.
2.
The drug products are not bioequivalent.
3.
The drug products exhibit a narrow therapeutic ratio (as described above), and safe and effective use of the product requires careful dosage titration and patient monitoring.
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Criteria for waiver of evidence of in-vivo bioavailability -
Although a human in-vivo test is considered to be preferable to other approaches for the most accurate determination of bioequivalence, there is a provision in the 1977 regulations for waiver of an in-vivo bioequivalence study under certain circumstances. For some drug products, the in-vivo bioavailability of the drug may be self-evident or unimportant to the achievement of the product's intended purposes. The FDA will waive the requirement for submission of in-vivo evidence of bioavailability or bioequivalence if the drug product meets one of the following criteria: 1.
The drug product is a solution intended solely for intravenous administration, and contains the active drug ingredient in the same solvent and concentration as an intravenous solution that is the subject of an approved full New Drug Application (NDA).
2.
The drug product is a topically applied preparation intended for local therapeutic effect.
3.
The drug product is an oral dosage form that is not intended to be absorbed, e.g., an antacid.
4.
The drug product is administered by inhalation and contains the active drug ingredient in the same dosage form as a drug product that is the subject of an approved full NDA.
5.
The drug product is an oral solution, elixir, syrup, tincture or other similar soluble form, that contains an active drug ingredient in the same concentration as a drug product that is the subject of an approved full NDA and contains no inactive ingredient that is known to significantly affect absorption of the active drug ingredient.
6.
The drug product is a solid oral dosage form (other than enteric-coated or controlled-release) that has been determined to be effective for at least one indication in a Drug Efficacy Study Implementation (DESI) notice and is not included in the FDA list of drugs for which in vivo bioequivalence testing is required.
7.
The drug product is a parenteral drug product that is determined to be effective for at least one indication in a DESI notice and shown to be identical in both active and inactive ingredients formulation, with a drug product that is currently approved in an NDA. (Excluded from the waiver provision are parenteral suspensions and sodium phenytoin powder for injection.)
According to the regulations, the bioavailability of certain drug products may be demonstrated by evidence obtained in-vitro in lieu of in-vivo data. Thus, the FDA also permits waiver of the in-vivo requirements if a drug product meets one of the following criteria: 1.
The drug product is one for which only an in-vitro bioequivalence requirement has been approved by the FDA.
2.
The drug product is in the same dosage form, but in a different strength, and is proportionally similar in its active and inactive ingredients to another drug product made by the same manufacturer and the following conditions are met: a.
the bioavailability of this other product has been demonstrated
b.
both drug products meet an appropriate in-vitro test approved by the FDA
the applicant submits evidence showing that both drug products are proportionally similar in their active an inactive ingredients. c. 3.
The drug product is shown to meet an in-vitro test that assures bioavailability, i.e., an in-vitro test that has been correlated with in-vivo data.
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4.
5.
The drug product is a reformulated product that is identical, except for color, flavor, or preservative, to another drug product made by the same manufacturer, and both of the following conditions are met: a.
the bioavailability of the other product has been demonstrated.
b.
both drug products meet an appropriate in vitro test approved by the FDA.
The drug product contains the same active ingredient and is in the same strength and dosage form as a drug product that is the subject of an approved full NDA or Abbreviated New Drug Application (ANDA) and both drug products meet an appropriate in-vitro test that has been approved by the FDA.
Although the above list of criteria for waiver of an in-vivo bioavailability study is quite lengthy, currently virtually all new tablet or capsule formulations from which measurable amounts of drug or metabolites are absorbed into the systemic circulation require a human bioequivalence study for approval (104). TABLE 8-12 Key
Provisions for bioequivalence requirements
1. Defines procedures for establishing a bioequivalence requirement. 2. Sets forth criteria to establish a bioequivalence requirement. 3. Describes types of bioequivalence requirements. 4. Sets forth requirement for in-vitro batch testing and certification. 5. Describes requirements for marketing a drug product subject to a bioequivalence requirement. 6. Sets forth requirements for in-vivo testing of a drug product not meeting an in-vitro bioequivalence standard. Source: Ref. 103
8.2.2
STUDY DESIGN A single-dose bioequivalency study is generally performed in normal, healthy, adult volunteers. The subject population should be selected carefully, so that product formulations, and not intersubject variations, will be the only significant determinants of bioequivalence (105). A minimum of 12 subjects is recommended, although 18 to 24 subjects are used to increase the data base for statistical analysis. The test and the reference products are usually administered to the subjects in the fasting state (overnight fast for at least 10 hours, plus 2 to 4 hours after administration of the dose), unless some other approach is more appropriate for valid scientific reasons. These subjects should not take any other medication for one week prior to the study or during the study. The bioavailability is determined by the collection of either blood samples or urine samples over a period of time and measurement of the concentration of drug present in the samples. Generally, a crossover study design is used. Using this method, both the test and the reference products are compared in each subject, so that inter-subject variables,
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such as age, weight, differences in metabolism, etc., are minimized. Each subject thus acts as his own control. Also, with this design, subjects' daily variations are distributed equally among all dosage forms or drug products being tested. The subjects are randomly selected for each group and the sequence of drug administration is randomly assigned. The administration of each product is followed by a sufficiently long period of time to ensure complete elimination of the drug (washout period) before the next administration. The washout period should be a minimum of 10 half-lives of the administered drug (106). A waiting period of one week between administration is usually an adequate washout period of most drugs. With a drug requiring a washout period of one week, a typical randomized twoway crossover bioequivalency study is shown in Table 8-13 on page 27. TABLE 8-13 Two
way cross over design
Treatment Groupa
Week 1
I II a
A B
Week 2 B A
10 subjects per group
Assuming that the in-vivo performances of the two formulations are to be compared by examining their blood level profiles, one must be certain that an adequate number of blood samples are taken. Blood samples should be drawn with sufficient frequency to provide an accurate characterization of the drug concentration-time profile from which tmax, Cmax and AUC can be determined. Typically, a total of 10 to 15 sampling times might be required (107). Moreover, all samples should be taken at the same time for both the test and the reference product to permit proper statistical analysis. Additional features which contribute to good study design include: 1.
All drug samples obtained for the test and reference preparations should be analyzed by the same method.
2.
Identical test conditions must be used for the two groups of subjects. For example, the types of foods, fluid intake, physical activity, and posture should all be rigidly controlled in the study.
3.
The physical characteristics of the subjects (such as age, height, weight, and health) should be standardized.
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Several important questions have been raised specifically regarding the design of the bioequivalence tests. One of these deals with the selection of the appropriate reference standard, since this is a critical component of a protocol (6, 108). Normally, the reference product is that available from the innovator company holding the New Drug Application. However, in cases where there may be some question as to the bioavailability of such a product, the study may utilize a solution of the drug instead of or in addition to the marketed product. The use of a solution can, of course, result in some difficulty in interpretation of the data: a solid dosage form, when compared to a solution, will usually exhibit a lower Cmax and a longer tmax. The clinical significance of these differences may be difficult to assess. In some instances, the FDA must designate a specific product as the reference standard from among two or more possible products; e.g., Proventil® tablets, 4 mg (Schering), not Ventolin® tablets 4 mg (Allen and Hanburys), is the reference product in bioequivalence studies of albuterol sulfate conventional tablets (108). Advantages of Multipledose vs. single dose studies:
Another important question is whether the bioequivalence trial should compare single doses of the formulations or if it should compare "steady-state" conditions reached after multiple dosing. It would seem that multiple dosing would be the logical choice for drugs intended for long-term use since this would give a more realistic comparison in view of the way in which the drug is normally administered. Other advantages of conducting a multiple-dose study over a single-dose study include (54, 59): 1.
Multiple-dosing eliminates the long washout periods required between single-dose administrations. The switch-over from one formulation to the other can take place in steady state.
2.
Single-dose studies may pose problems of sufficiently long sampling periods in order to get reliable estimates of terminal half-life, which is needed for correct calculation of the total AUC.
3.
Multiple-dose studies yield higher concentrations of drug in the blood, making accurate measurement easier. In addition, since drug concentrations need to be measured only over a single dosing interval at steady state, the need to measure lower concentrations during a disposition phase is avoided.
4.
Multiple-dosing studies can be conducted in patients, rather than healthy volunteers, allowing the use of higher doses.
5.
Usually, smaller intersubject variability is observed in steady-state studies, which may permit the use of fewer subjects.
6.
Nonlinear pharmacokinetics, if present, can be more readily detected at steady-state following multiple-dosing.
Thus, for some drug products, multiple-dose bioequivalence studies are appropriate and should be performed. In fact, according to one of the conclusions of the Bio- International '92 conference on the bioequivalence of highly variable drugs, a multiple-dose study is required in the case of compounds exhibiting nonlinear pharmacokinetics (110). The circumstances under which a multiple-dose study may be required are summarized in the regulations (109):
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1.
When there is a difference in the rate of absorption but not in the extent of absorption.
2.
When there is excessive variability in bioavailability from subject to subject.
3.
When the concentration of the active moiety in the blood resulting from a single dose is too low for accurate determination.
4.
When the drug product is a controlled-release dosage form.
On the other hand, multiple-dose bioequivalence studies are undesirable in some respects. Healthy subjects should not be dosed with any drug for an extended period of time (59). Multiple-dose studies are also generally more difficult to carry out, especially with regard to ensuring subject compliance with dosing and dietary restrictions. Therefore, most bioequivalence studies are conducted as single-dose studies. Multiple-dose studies should be performed only when a single-dose study is not a reliable indicator of bioavailability (111).
8.2.3
ASSESSMENT OF BIOEQUIVALENCE In order for different formulations of the same drug substance to be considered bioequivalent, they must be equivalent with respect to the rate and extent of drug absorption. Thus, the two predominant issues involved in the assessment of bioequivalence are: the pharmacokinetic parameters that best characterize the rate and extent of absorption and, the most appropriate method of statistical analysis of the data.
Pharmacokinetic criteria
With regard to the choice of the appropriate pharmacokinetic characteristics, Westlake suggests comparisons of the formulations should be made with respect to only those parameter(s) of the blood level profile that possess some meaningful relation to the therapeutic effect of the drug (107). Since the AUC is directly proportional to the amount of drug absorbed, this pharmacokinetic parameter is most commonly used to characterize the extent of absorption, both in single- and multiple- dose studies. The choice of an appropriate pharmacokinetic characteristic for the rate of absorption is still being discussed with considerable controversy (112, 113). Although a broad array of methods exists for calculating absorption rates (e.g. moment analysis, deconvolution procedures and curve-fitting), the most commonly used parameters are peak concentration (Cmax) and time to peak concentration (tmax). Although these parameters have been observed to have significant variances and may be difficult to determine accurately, they remain the parameters generally requested as rate characteristic by most regulatory authorities for immediate-release products (112).
Statistical criteria
After a bioequivalence study is conducted and the appropriate parameters are determined, the pharmacokinetic data must be examined according to a set of predetermined criteria to confirm or refute the bioequivalency of the test and refer-
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ence formulations. That is, one must determine whether the test and reference products differ within a predefined level of statistical significance. Since the statistical outcome of a bioequivalence study is the primary basis of the decision for or against therapeutic equivalence of two products, it is critically important that the experimental data be analyzed by an appropriate statistical test. In the early 1970s, bioequivalence was usually determined only on the basis of mean data. Mean AUC and Cmax values for the generic product had to be within +20% of those of the reference (innovator) product (108). Although the 20% value was somewhat arbitrary, it was felt that for most drugs, a 20% change in the dose would not result in significant differences in the clinical response to drugs (114). A relatively common misconception is that current regulatory standards still allow this difference of 20% in the means of the pharmacokinetic variables (Cmax and AUC) of the test and reference formulations. The FDA's statistical criteria for approval of generic drugs now requires the application of confidence limits to the mean data, using an analysis known as the two one-sided tests procedure (115). This change came about as a result of the conclusion of the FDA Bioequivalence Task Force in 1986 that the use of a 90% confidence interval based on the two one-sided t-tests approach was the best available method for evaluating bioequivalence (111). Westlake was the first to suggest the use of confidence intervals as a means of testing for bioequivalence (116). Recognizing that no two products will result in identical blood-level profiles, and that there will be differences in mean values between products, Westlake pointed out that the critical issue was to determine how large those differences could be before doubts as to therapeutic equivalence arose (107, 117). A test formulation was considered to be bioequivalent to a reference formuCp max test AUC test lation if 0.8 < ------------------< 1.2 and 0.8 < -------------------< 1.2 . (119). By this proceAUC ref Cp max ref dure, if test and reference products were not bioequivalent (i.e. means differed by more than 20%), there was a 5% chance of concluding that they are bioequivalent. The current FDA guidelines are that two formulations whose rate and extent of absorption differ by -20%/+25% or less are generally considered bioequivalent (90). In order to verify that the -20%/+25% rule is satisfied, the two one-sided statistical tests are carried out: one test verifies that the bioavailability of the test product is not too low and the other to show that it is not too high. The current practice is to carry out the two one-sided tests at the 0.05 level of significance. Computationally, the two one-sided tests are carried out by computing a 90% confidence interval. For approval of an ANDA, a generic manufacturer must show that the 90% confidence interval for the ratio of the mean response (usually AUC
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and Cmax) of its product to that of the innovator is within the limits of 0.8 to 1.25. Since these tests are carried out at the 0.05 level of significance, there is no more than a 5% chance that they will be approved as equivalent if they differ by as much or more than is allowed by the equivalence criteria (-20%/+25%). Since this test requires that the 90% confidence interval of the difference between the means be within a range of -20%/+25%, it is more stringent than simply requiring the comparison of the test and reference products' AUC and Cmax to be within the 80 to 125% range. If the mean response of the generic product in the study population is near 20% below or 25% above the innovator mean, one or both of the confidence limits will fall outside the acceptable range and the product will fail the bioequivalence test. Thus, the confidence interval requirement ensures that the difference in mean values for AUC and Cmax will actually be less than -20%/ +25%. It should be pointed out that the standards vary among drugs and drug classes. For example, antipsychotic agents may fall within a 30% variation and antiarrhythmic agents may be allowed a 25% variation (122). The actual differences between brand and generic products observed in bioequivalence studies have been reported to be small. The FDA has stated that for post-1962 drugs approved over a two-year period under the Waxman-Hatch bill (1984), the mean bioavailability difference between the generic and pioneer products has been about 3.5% (120). In addition, 80% of the generic drugs approved by the FDA between 1984 and 1986 differed from the innovator products by an observed difference of only +5%. Such differences are small when compared to other variables of drug therapy and would not be expected to produce clinically observable differences in patient response.
8.2.4
CONTROVERSIES AND CONCERNS IN BIOEQUIVALENCE The design, performance and evaluation of bioequivalence studies have received a great deal of attention over the past decade from academia, the pharmaceutical industry and regulatory agencies. A number of concerns and questions have been raised about the conduct of bioequivalence studies as well as the guidelines and criteria used to determine bioequivalence (112). Many of these concerns were triggered by the passage of the Drug Price Competition and Patent Term Restoration Act (The Waxman-Hatch Amendments) by Congress in 1984. This Act provided for an expedited approval by the FDA of generic drugs, thereby expanding the potential generic market for prescription generic drugs (121). Shortly after the passage of this Act, numerous published reports appeared in the scientific literature questioning the FDA's ability to ensure that generic drugs were equivalent to the brand name drugs they were copying. Most of the concerns of the scientific community centered around adequate standards for evaluation of bioequivalence
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and correlation between bioequivalence and therapeutic equivalence. Some of the issues and concerns that were raised are summarized in Table 8-14 on page 32 (8, 13). TABLE 8-14 Issues
• • • • • • • •
and Concerns regarding bioequivalence
Correct analysis of drugs in biological fluids Appropriate choice of pharmacokinetic parameters to assess bioequivalence Generalizing results obtained in healthy volunteers to patients Problems involved in extrapolating from single-dose studies to steady-state Importance of evaluating active metabolites Inadequate statistical criteria to evaluate bioequivalency Bioequivalence does not always ensure therapeutic equivalence Lack of clear guidelines for evaluation of bioequivalence
At the center of the controversy were the methods and criteria used by the FDA to determine bioequivalence. Assessment of bioequivalence was done on the basis of mean data: mean AUC and Cmax values for the generic product had to be within +20% of those of the innovator product for approval. A statistical test was employed to assess the power of the test to detect a 20% mean difference in treatments. For drugs that could not meet the statistical criteria because of inherent variability, another rule was used, the so-called "75/75" rule: that in at least 75% of the subjects, the test formulation must fall within the range of 75% to 125% of the reference standard to be considered equivalent (122). It was felt by many that these rules permitted too much variability in the bioavailability of test drugs and could result in therapeutic failure or increased risk of side effects (4, 15, 123). Statistically, the power approach and the 75/75 rule were shown to have poor performance characteristics and bioequivalence evaluation based on these methods was discontinued by the FDA in 1986. In their place, the Agency currently employs the two one-sided tests procedure, as previously discussed. Although the decision of bioequivalence is now made in a more statistically valid way and the associated concerns have diminished somewhat, some important questions and controversies in bioequivalence remain. These are primarily centered around study design, the criteria used to establish or refute equivalence, and the assumption that products that are bioequivalent are therapeutically equivalent. One criticism of bioequivalence testing is that it is almost always done in a panel of young, healthy male volunteers rather than in the target population for which the drug is intended. Clearly, the performance of a drug product in a 20-year-old male will not be the same as in an 85-year-old woman. Serious concerns have been Basic Pharmacokinetics
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raised that different results would be observed in elderly patients, in women, in patients with diseases of the gastrointestinal tract, and in patients with diminished renal or hepatic function. However, although factors such as age and disease state might affect the actual observed concentrations of drug, the products being compared should be affected in a similar fashion, and one can still be compared to the other. If two products show an equivalent level in healthy volunteers, their levels should be elevated to the same extent in patients with impaired hepatic function. Thus, they can still be compared to each other. Healthy male volunteers are generally used in bioequivalence studies to assure a homogeneous study population and to permit focus on formulation factors that might affect bioavailability. In addition, healthy subjects are more likely to remain stable during the study. The condition of actual patients might change due to the disease resulting in greater variability in the data. The FDA does recognize the possibility that some conditions could cause two products that are bioequivalent in healthy subjects to be bioinequivalent in certain patients and is prepared to modify its guidelines if necessary. A study design-related area of concern is average versus individual bioavailability. Current procedures assess equivalence in terms of average bioavailabilities, and do not address within-subject equivalence. In recent years, there has been increased interest expressed in the variability of response, particularly variability within an individual. This has given rise to the most recent controversy in bioequivalence assessment, namely whether average bioequivalence is adequate to allow interchangeability of drugs in an individual (112). Anderson and Hauck believe that a different, more stringent, notion of bioequivalence, referred to as individual bioequivalence, is needed to provide assurance that an individual patient can be switched from one formulation to another (124). The second major area of controversy has focused on the criteria used to determine bioequivalence. Implicit in the FDA guidelines is the assumption that a -20%/ +25% change in mean serum concentration of drugs can be safely tolerated. However, there is little documentation demonstrating whether 20% variation in bioavailabilities does or does not affect the safety and efficacy of drugs. There are certain critical therapeutic categories (Table 8-15 on page 34) in which minor fluctuations in blood levels may have a substantial impact on therapeutic outcome or toxicity (125, 126). In view of this, some scientists believe that the FDA should be more stringent, requiring the mean values for AUC to be within 10% rather than 20%/25%. The Bioequivalence Task Force, in its 1988 report, concluded that for certain drugs or drug classes, there is clinical evidence that may indicate a need for tighter limits than the then-generally applied +20% rule (111). The Task Force recommended that the Agency consider using as an "additional nonstatistical criterion" a mean difference in AUC of +10%; however, this additional criterion would not be essential to ensuring drug bioequivalence.
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TABLE 8-15
Critical Therapeutic Catagories of Drugs
Category
Example
Cardiovascular drugs
digoxin
Anticonvulsants
phenytoin
Bronchodilating agents
theophylline
Oral anticoagulants
warfarin
In general, the choice of the appropriate bioequivalence range should be done on clinical grounds; for a drug with a narrow therapeutic range, more stringent limits should be considered. On the other hand, the current requirements for Cmax for some drugs may be too stringent, considering the difficulty in accurately estimating this value. For example, it has been suggested that the acceptable bioequivalence range for Cmax for fast-releasing nifedipine formulations should be 70% to 130%, rather than the usual 80% to 125%. In light of this, many, including the Pharmaceutical Research and Manufacturers of America (formerly the Pharmaceutical Manufacturers Association [PMA]), feel that the FDA should repudiate its -20%/+25% rule and develop drug-by- drug bioequivalence criteria (127). A third source of controversy in bioequivalence is the very foundation on which the whole concept of bioequivalence is based: the central assumption is that if two products are shown to be bioequivalent by currently accepted standards, then they are also therapeutically equivalent, and thus interchangeable. A number of critics have challenged this "bioequivalence = therapeutic equivalence" equation, pointing out that this relationship has not been conclusively established for most drugs (9, 13, 16, 128). These terms are, in fact, not interchangeable; bioequivalence means that two products have basically superimposable blood level curves (within specified limits) while therapeutic equivalence means the products produce similar effects. There may be situations where two products have similar blood concentrations, yet if the drug has a narrow therapeutic range, they may have significantly different therapeutic effects. On the other hand, there may be products which have widely varying blood level profiles, but exhibit very little difference in their clinical effect. This might be the case for drugs with a wide therapeutic range. In addition, the therapeutic efficacy of some drugs is not necessarily related to their blood levels, e.g., some psychoactive drugs, where the end point of drug effects is psychological and behavioral response (129).
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Williams suggests several ways that the integrity of a bioequivalence study as a prediction of therapeutic equivalence could be assessed (104). One way involves the performance of specific clinical studies to confirm that products shown to be bioequivalent in healthy subjects would be bioequivalent in the patient population as well. A second way suggested is through post-marketing surveillance of therapeutic response produced by different formulations of the same drug under actual conditions of use. A third method is based on anecdotal reports. Williams points out that none of these methods have been systematically employed to confirm current bioequivalence methodology. Thus, a number of problems remain in the bioequivalence process which should be addressed. FDA scientists themselves have readily acknowledged the existence of shortcomings in the bioequivalence testing program. However, a great deal of progress has been made in this area in the last twenty years. The improved design of the studies, the interpretation of the data, the increased scientific rigor of the acceptance criteria, as well as the more rigorous auditing and inspection program have made bioequivalence data an appropriate and valid means of approving generic drug products.
8.2.5
GENERIC DRUGS AND PRODUCT SELECTION Generic drug utilization has increased dramatically in the last 20 years. In 1975, approximately 9% of all prescription drugs dispensed were generic versions (130). This percentage rose to 20% in 1984, and 40% in 1991. It has been variously estimated that the generic share of all new prescriptions will be 46% to 65% in 1995 (131-133). This rise of generics has not gone altogether smoothly, however; the popularity of generic drugs took a sharp downturn in 1989 when scandal rocked the generic drug industry. This involved illegal and unethical acts by some generic drug companies -- payoffs to FDA employees and fraudulent drug-approval test -- aimed at getting drugs approved ahead of other firms (134-138). Although these events did shake the confidence of pharmacists, physicians and the public in the quality of generic drugs and cast a shadow over generics generally, these concerns were relatively short-lived. Numerous surveys conducted one to two years after the scandal unfolded indicated that confidence in generic drugs had been regained and that the generic industry was in better shape with pharmacists than it had been before the scandal occurred (139-146). Given the seriousness of the events, the speed with which generics came back was impressive. This was due in part to the FDA's reaction to the scandal: a multilevel reorganization of its generic drug operations and a comprehensive inspection of the leading manufac-
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turers of generic drugs (134, 140, 147, 148). It was felt that this stringent FDA review of generics proved the overall integrity of the companies that emerged with a clean bill of health. After a sharp drop in the use of generic drugs in 1989, they began to rise nearly as quickly as they fell, and by mid-1990, sales of generics were approaching their previous record high (141). This trend in generic drug utilization is expected to continue its upward spiral, with newly generic drugs coming to market at an increasing rate. There are several factors that have contributed to this period of considerable growth in the generic drug industry. One major factor was the passage of the Drug Price Competition and Patent Term Restoration Act (Waxman-Hatch Act) in 1984. This act, by eliminating the requirement for clinical safety and efficacy testing for generics of drugs introduced after 1962, greatly expedited the entry of generic drugs into the marketplace. The purpose of this act was to facilitate generic competition and thereby reduce health care costs. This act significantly expanded the number of drugs eligible to be manufactured as generics. Another factor fueling the surge of generic products is the abundance of brand name drugs whose patents began expiring in 1986. Between 1991 and 1994, patents expired on brand-name drugs whose combined annual sales totaled $10 billion (141). These include Procardia®, Ceclor®, Tagamet®, Cardizem®, Feldene®, Naprosyn®, and Xanax®. All told, more than 100 drugs worth upwards of $25 billion in sales will have come off patent by the year 2000 (149). Table 8-16 on page 37 lists some recent and impending patent expirations (150, 151). As a result of these patent expirations on popular drugs, there has been an explosion of new generic drug applications.
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TABLE 8-16 Recent
and pending patent expirations
Brand Name Procardia Tenormin Ceclor Cardizem Feldene Naprosyn Xanax Tagamet Seldane Micronase Capoten Zantac Trental Noroxin
Generic Name Nifedipine Atenolol Cefaclor Diltiazem Piroxicam Naproxen Alprazolam Cimetidine Terfenadine Glyburide Captopril Ranitidine Pentoxifylline Norfloxacin
Patent Expiration Date* 1991 1991 1992 1992 1992 1993 1993 1994 1994 1994 1995 1995 1997 1998
*Extentions may be granted Perhaps the major factor promoting generic drug utilization is the increased attention to containing health-care costs. Pushed by a drive for lower-cost medication by federal and state governments, private insurers, corporate benefit managers, regulatory agencies and consumer groups, generic drug usage is at a peak. Additional impetus could come from health care reform, wherein generic drugs are viewed as a key to controlling pharmaceutical costs. Managed care programs are expected to cover more than 70% of all outpatient prescriptions by the end of the decade, with an accompanying greater demand for generic products (152). Thus the demand for generic drugs will continue to rise, in a climate that favors health care reform, lower- cost medications and broad-based prescription benefits (153). With the increasing availability of generic drugs, pharmacists are called upon more and more often to select a patient's drug product from a myriad of multisource products. The pharmacist's role in product selection has increased dramatically in the past decade and the proper selection of multisource drug products has become a major professional responsibility of pharmacists. Although most pharmacists do not, realistically, evaluate the bioequivalence of two products from blood level data, professional judgement does need to be exercised; and this requires an understanding and application of the biopharmaceutical principles discussed.
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8.2.6
THE ORANGE BOOK One of the factors that led to the widespread repeal of the state anti-substitution laws in the 1970's was an effort by the states to contain drug costs and the establishment of maximum allowable costs (MACs) for reimbursement of drugs under Medicaid. By allowing the pharmacist to select the manufacturer of a drug, the less- expensive generic version could be dispensed. However, before the pharmacist could knowledgeably select a generic drug, he had to know which generics were bioequivalent to the innovator product and thus, interchangeable. (There was substantial evidence at this time that not all pharmaceutically equivalent products were bioequivalent). To answer this need, the states began preparing either positive or negative formularies, often turning to the FDA for assistance in this undertaking. In response to the many requests for assistance from the states in developing their formularies, the FDA Commissioner notified state officials of FDA's intent to provide a list of all prescription drug products that have been approved as being safe and effective, along with therapeutic equivalence determinations for multisource prescription products. This list, entitled Approved Drug Products with Therapeutic Equivalence Evaluations, more commonly known as "The Orange Book" was first published in 1980 and is now in its 14th edition. It is published annually and updated monthly. The Orange book is generally considered to be the most reliable guide for determining which drug products are therapeutically equivalent. The Prescription Drug Products List contains: 1.
all the drug products approved by the FDA as being safe and effective under the Federal Food, Drug and Cosmetic Act, and
2.
2.the therapeutic equivalence evaluations for all approved multisource prescription drug products (those pharmaceutical equivalents available from more than one manufacturer).
Currently, multisource products comprise almost 80% of the approximately 10,000 drugs on the Prescription Drug Product List. The therapeutic evaluation for these products have been prepared to serve as information and advice to state health agencies, pharmacists and prescribers to promote knowledgeable drug product selection and to foster containment of health costs.
8.2.7
THERAPEUTIC EQUIVALENCE Drug products are considered to be therapeutic equivalents if they are pharmaceutical equivalents and if they can be expected to have the same clinical effect when administered to patients as specified in the labeling (90). In general, the FDA
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evaluates as therapeutically equivalent those drug products that satisfy the following general criteria: 1.
They are approved as safe and effective.
2.
They are pharmaceutical equivalents; i.e. they contain identical amounts of the same active ingredient in the same dosage form and route of administration, and a.
b.
meet compendial and other applicable standards for quality, purity, strength and identity.
3.
They are bioequivalent. Bioequivalence may be established by either an in-vivo or in-vitro test, depending on the drug. If the drug presents a known or potential bioequivalence problem then an appropriate standard must be met which demonstrates a comparable rate and extent of absorption.
4.
They are adequately labeled.
5.
They are manufactured in compliance with Current Good Manufacturing Practice regulations.
The FDA believes that drug products meeting the above criteria are therapeutically equivalent and can be substituted with the full expectation that the substituted product will produce the same therapeutic effect as the prescribed product.
8.2.8
THERAPEUTIC EQUIVALENCE EVALUATION CODESThe FDA uses a two-letter coding system for multisource products. The first letter in the code allows users to determine whether a particular product has been evaluated therapeutically equivalent to other pharmaceutically equivalent products. The second letter in the code provides additional information about the basis of FDA's evaluation. The various categories are summarized in Table 8-17 on page 40.
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TABLE 8-17 Therapuetic
equivalency codes
"A" Drug Products
"B" Drug Products
Drug products the FDA considers to be therapeutically equivalent; i.e. drug
Drug products the FDA does not consider to be therapeutically equivalent; i.e.
products for which:
drug products for which actual or potential bioequivalence problems have not
1.
2.
There are no actual or potential bioequivalence problems. These are
been resolved by adequate evidence of bioequivalence. Often the problem is
designated as:
with specific dosage forms rather than with the active ingredient. These products
AA
Products in conventional dosage forms
are classified as "B" for one of three reasons:
AN
Solutions and powders for aerosolization
AO
Injectable oil solutions
AP
Injectable aqueous solutions
AT
Topical products
1.
2.
The active ingredients or dosage forms have documented or potential bioequivalence problems, and no adequate studies demonstrating bioequivalence have been submitted. The quality standards are inadequate or the FDA has insufficient basis to determine therapeutic equivalence.
3.
Actual or potential bioequivalence problems have been resolved via adequate in vivo and/or in vitro tests. These are designated as AB.
The drug product is under regulatory review. These products are designated as: BC
Controlled-release tablets, capsules and injectables
BD
Active ingredients and dosage forms with documented bioequivalence problems
BE
Delayed-release oral dosage forms (e.g. enteric-coated products)
BN
Products in aerosol-nebulizer drug delivery systems
BP
Active ingredients and dosage forms with potential bioequivalence problems
BR
Suppositories or enemas that deliver drugs for systemic absorption
BS
Products having drug standard deficiencies
BT
Topical products with bioequivalence issues
BX
Insufficient data to determine therapeutic equivalence
B*
Drug products requiring further FDA investigation and review to determine equivalence
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There are two basic categories into which multisource drugs have been placed, "A" or "B". Drug products rated "A" are products that the FDA considers to be therapeutically equivalent to the pharmaceutically equivalent original product. These fall into one of two classes: 1.
There are no known or suspected bioequivalence problems.
2.
Actual or potential bioequivalence problems have been resolved with adequate in vivo and/or in vitro evidence supporting bioequivalence.
Category "B" consists of drug products that the FDA does not at this time consider to be therapeutically equivalent to the pharmaceutically equivalent reference product. Certain types of products are rated B by virtue of their specialized dosage forms. For example, controlled-release dosage forms are rated BC, unless bioequivalence data have been submitted as evidence of equivalence. In this case, the product would be coded AB. The fact that a product is in the "B" category does not mean it should not be dispensed; it simply means that a B rated product should not be substituted for a pharmaceutically equivalent product. For example, glyburide is marketed as Micronase® and DiaBeta® by two different manufacturers. Both these products are clinically effective, but because bioequivalence between the two has not been studied, they are B rated and are not interchangeable. To avoid possible significant variations among generic drugs as a result of comparison to different reference drugs, the FDA began designating a single reference listed drug against which all generic versions must be shown to be bioequivalent. The reference listed drug is identified by the symbol "+" in the Prescription Drug Product List. This symbol was used for the first time in the 1993 edition of the Orange Book. Limitations and exclusions-
Although the Orange Book is a very valuable reference for pharmacists performing drug product selection, it has certain limitations, which must be recognized. It was not intended to serve as a single comprehensive reference on all multisource drugs. Many prescription drug products are not listed in the Orange Book, making evaluation of their therapeutic equivalence difficult, if not impossible. Exclusion of a drug from the Orange Book means that the FDA has not evaluated its safety, efficacy and quality. Table 18 lists the classes of products excluded from the Orange Book. Because the equivalence of these excluded products is unknown, interchanging of these products should be avoided.
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TABLE 8-18 Drug
Products excluded from the Orange Book
1.
Drugs marketed before the passage of the Federal Food, Drug, and Cosmetic Act of 1938. These are not included because the FDA has not reviewed these drugs for safety and efficacy and does not have the necessary information to make therapeutic equivalence evaluations. Examples: digoxin, morphine, codeine, thyroid, levothyroxine, phenobarbital and nitroglycerin
2.
Drugs for which the FDA has no NDA or ANDA on file. Examples: Anusol-HC®, Naldecon® (and their generic counterparts)
3.
Drugs still undergoing Drug Efficacy Study Implementation (DESI) review. These are drugs that were marketed between 1938 and 1962 on the basis of safety, but not efficacy. Although most of these drugs have been reviewed and are listed in the Orange Book, there are still a number of these pre-1962 drugs which have not yet been classified as "effective" under the DESI program, and are not listed. Examples: nitroglycerin controlled-release capsules, pentaerythritol tetranitrate, isocarboxazid, hydrocortisone-iodochlorhydroxyquin cream In addition, nitroglycerin transdermal patches are still undergoing efficacy studies, and are not listed in the Orange Book.
Another limitation of the Orange Book that all pharmacists should be aware of is that the drug listings contain the names of only the companies that actually hold an approved NDA or ANDA; they may not be the same as the actual manufacturer or distributor. It is fairly common practice for a drug to be manufactured pursuant to an NDA or an ANDA but distributed under license agreement by another company. In this instance, the distributor would not be listed in the Orange Book. Since pharmacists are, understandably, generally unaware of the name of the actual holder of the NDA or ANDA, it is often difficult for them to determine the therapeutic equivalence of a particular multisource product if it is not listed in the Orange Book. For example, there are over thirty manufacturers and distributors marketing approved, therapeutically equivalent versions of furosemide 40 mg tablets (154). However, only twelve of these companies are actually listed in the Orange book, since these are the actual holders of an NDA or ANDA. Therefore, the pharmacist would have to verify the therapeutic equivalence evaluation of the non-listed products by obtaining the information from the manufacturer, packager, or supplier. Legal status and pharmacists' responsibility-
The Orange Book per se has no legal status. The FDA stresses that it is a source of information and advice on drug product selection, but it does not mandate the drug products which may be dispensed nor the products that should be avoided. Thus, the Orange Book does not carry the weight of regulation or law, and the FDA assumes no liability for drug products selected on the basis of its equivalence evaluation. The Orange Book points out that "FDA evaluation of therapeutic equivalence in no way relieves practitioners of their professional responsibilities in prescribing and dispensing such products with due care." There are circumstances where pharma-
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cists will have to exercise professional care and sound judgement in selecting a drug product for a particular patient. Although two products may be rated as being therapeutically equivalent in the Orange Book, they may not be equally suitable for a particular patient. Drugs that share the "A" code may still vary in ways that could affect patient acceptance. They may differ in shape, color, taste, scoring, configuration, packaging, preservatives, expiration time, and in some instances, labeling. If products with such differences are substituted for each other, there is potential for patient confusion or decreased patient acceptance. For example, a patient may be sensitive to an inert ingredient in one product that another product does not contain. Or, a patient may become confused if the color or shape of a product varies from that to which he has become accustomed. A patient may reject the administration of a substituted product because of differences in taste or appearance. When such characteristics of a specific product are important in the treatment of a particular patient, the pharmacist should select a product with these considerations in mind as well as bioequivalence. Despite its limitations and shortcomings, the Orange Book is a very useful guide for rational product selection. Pharmacists can utilize the information presented there, in combination with sound professional judgment, to make decisions on behalf of their patients regarding the choice of the most appropriate drug product.
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8.3 Drug Product Selection Multisource drug product selection has become a very important component of contemporary pharmacy practice. The National Prescription Audit (NPA) has, for some years now, been chronicling the heightened role played by pharmacists in the selection of which brand (or generic version) of a multiple-source drug will be dispensed to the patient. From 1983 through 1993, the pharmacist's role in selecting brand or generic products for dispensing has increased dramatically, as shown in Table 19. In the first half of 1993, pharmacists controlled 41% of dispensing decisions, as compared to 16% in 1983. It is evident that the substitution trend is strong and is continuing to gain ground. This expansion of pharmacy's province in brand choice decisions is the result of several factors: economic pressures for lower prescription costs, repeal of anti- substitution laws and increased acceptance of generics by patients, physicians and pharmacists. Perhaps the most significant factor in escalating the overall level of pharmacists' brand choice decisions has been the expiration of the patents of high- volume pioneer brands, as previously discussed. This has resulted in significant expansion in the potential for pharmacist choice. TABLE 8-19 Pharmacist’s
Brand Selection
Year
Percent of all new prescriptions involving pharmacists brand choicea
1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 Jan.-June 1993
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Bioavailability, Bioequivalence, and Drug Selection
8.3.1
CONSIDERATIONS IN SELECTING A MANUFACTURER The selection of a pharmaceutical manufacturer of a multisource product has become an important professional responsibility for pharmacists. This responsibility has become an especially critical part of a pharmacists role in light of the increasing number of generic products available and in light of some of the problems that have occurred in the generic drug industry (the "generic drug scandal" of 1989). The pharmacist is entrusted by the public to select manufacturers that offer the best quality at the best price.
So how does the pharmacist select the manufacturer of a multisource drug product? What factors should be considered?
Thoughtful selection of a multisource drug product is not an easy task, and requires a consideration of not only the drug product itself, but also the manufacturer, and in some cases, the patient. Several options are open to the pharmacist performing drug product selection: to select a product solely on the basis of economics, to select a product on the basis of the reputation of the manufacturer, or to make a decision based on product bioequivalence and quality and on the basis of the product's conformity with official compendial standards and with those established by the FDA. The first option, while offering a financial advantage, does not provide assurance of therapeutic efficacy. The second option, although subjective, is easily applied and does offer a degree of security to the pharmacist. The third option is the most challenging to the pharmacists, requiring the application of principles of biopharmaceutics and pharmacokinetics in arriving at a decision. Ideally, the pharmacist should take into consideration all the above options when selecting a drug product for a patient. When pharmacists were asked which factors are most important to them in selecting a manufacturer of a generic product, the primary criteria indicated were the reputation and quality of the company (159-162). Bioequivalence to the brand-name product was also ranked as being an important factor in product selection. However, the most frequently used sources for assessing bioequivalence were manufacturer reputations (based previous experience) and product literature provided by the distributing company. Company-sponsored material must be carefully evaluated. Unfortunately, promotional literature does not generally contain sufficient data to permit rational analysis of whether or not products are bioequivalent (163). Also, relying on personal methods of information gathering for assessing bioequivalence is not very reliable. Interestingly, only 23% of pharmacists reported using the Orange Book in assessing bioequivalence (161). Selection of drug products should be based on sound scientific and clinical grounds. Developments in the science of pharmacokinetics and the related area of bioavailability have given pharmacists the tools necessary to make sound choices among multisource products. In response to the profession's need for information and advice on how to select appropriate drug products from multiple sources, the American Pharmaceutical Association formed a Bioequivalency Working Group to establish guidelines for product selection (Table 8-20 on page 47) (164). This Group made
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recommendations of factors that pharmacists should consider when selecting drug products to be dispensed to their patients. If pharmacists consider the factors indicated as part of the professional judgement process when making drug product selections, it is likely that the best interest of the patients will be served. The appropriate selection of a generic drug product involves much more than just cost considerations or reliance on state and federal laws and regulations. It requires a knowledge of the drug entity and its physical and chemical properties, the condition to be treated, and its significance, and the history and attitude of the manufacturer. One of the criteria often used to evaluate a manufacturer's record is the number and type of recalls of that company's products. Product selection may also require taking into consideration the patient, the disease, previous drug therapy, and duration of therapy before a decision is made. Gagnon presented a step-by-step analysis procedure that pharmacists can use in evaluating multisource suppliers of a pharmaceutical product (Table 8-21 on page 49) (165). Using this procedure, each manufacturer is rated in each area listed, thus enabling the pharmacist to make the most rational choice.
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TABLE 8-20 Guidelines
for product selection
DISPENSING DECISIONS ? State Rules and Regulations. Pharmacists should be cognizant of legal requirements that address the issue of drug product selection. Many states have positive or negative formularies to provide guidance in drug product selection. ? Bioequivalency Information/Orange Book Ratings. Only products with proven bioequivalency should be selected to be dispensed in lieu of the innovator product. Products that are listed in the FDA's Approved Drug Products and Therapeutic Equivalence Evaluations (the Orange Book) as "A" rated should be selected when such products are available. For pre-1938 drugs, the selection should be based on data obtained from the literature, because bioequivalency testing is not required by the FDA for these drug products. ? Dosage Form. The type of dosage form should be considered whenever one drug product is selected from among multisource drug products. This is especially true with extended or delayed release medications. ? Previous Drug Use. Two questions should be considered regarding previous drug product usage. First, is the prescribed drug a continuation of already successful therapy? If it is, the impact of any change in source of the medication should be considered. The pharmacist should also know which product the patient was using previously, including any medications in the hospital if the patient was recently discharged. Second, was the original product dispensed a generic product? If so, preference should be given to continuing to dispense the same generic product from the same source. ? Patient Status. The pharmacist should consider how well controlled the patient is and how susceptible that patient might be to small changes in drug absorption. If a patient has labile control or has experienced great difficulty in achieving control, the pharmacist should continue therapy with a product from a single source throughout therapy. ? Diseases. The seriousness of the disease and its potential impact on the patient may influence the pharmacist's willingness to change products. ? Drug Class or Category. Drugs with narrow therapeutic ranges and with known clinically significant bioavailability problems should be substituted with care and/or after discussion with the prescriber. ? Cost. The cost of the product , while an important consideration, should be a secondary consideration in selecting among products judged by the pharmacist to be bioequivalent. ? Patient Opinion. An informed patient, cooperating with a physician and pharmacist in his or her drug therapy, is an important element in ensuring the best possible therapeutic outcomes. The pharmacist should take into account the patient's need when selecting from multisource drug products and inform the patient of any potential consequences associated with alternate product selections.
PURCHASE DECISIONS ? Current State Laws and Regulations. Some states have positive or negative formulary systems that place regulatory restrictions on the products considered therapeutically equivalent. The state formulary may not always be in agreement with classifications listed in the FDA's Orange Book. Therefore, pharmacists should be familiar with both. ? Bioequivalency Information/Orange Book. Products shown to be bioequivalent through reference to the Orange Book or other reliable source of bioequivalency information are preferred. Purchase decisions for drugs marketed prior to 1938 should be based on data obtained from the literature or the manufacturer, because bioequivalency testing may not be required by the FDA for these drug products. ? Drug Category. Greater attention should be given to purchasing strategies for drug products used for serious or life-threatening diseases and in situations where therapeutic activity of the product is confined to a narrow range of biologic fluid concentration. ? Availability. A continuous supply from the same manufacturer is essential even in the event that the distributor has changed to ensure that refills of prescriptions will contain the same product as originally dispensed. However, in those instances when the manufacturer of a generic drug product has to be changed, care should be exercised to ensure that the new drug product is equivalent to the formerly stocked drug product. ? Supplier's Reputation. The reputation of the manufacturer in terms of its ability to adhere to good manufacturing practices (GMP) that ensure that each dosage form is manufactured correctly and in a consistent manner is an important consideration. When purchasing a product from a distributor rather than directly from the manufacturer, the procedure used by that supplier in selecting manufacturers for multisource products is also an important consideration. Establishment Inspection Reports and recall reports are available from FDA through a Freedom of Information (FOI) request. These are valuable tools in this decision. ? Cost.
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TABLE 8-21 Evaluation
of Multi-source Suppliers
Factors and Cues Product Information • Size(s) available • Dosage form(s) available • Bioequivalence data results using Orange Book • Existence of identification codes on solid dosage forms • Average number of months between product receipt and expiration date • Results of cost-effectiveness information from manufacturer • Complete product literature provided from manufacturer • Strength(s) available • State/federal formulary rules, e.g., MAC limits Economics • Price(s) • Deals and other discounts • Terms of sale • Clear and equitable pricing policy • Large sizes available at discount prices Product Quality • NDA/ANDA on file at FDA • Pharmaceutical elegance of products, e.g., broken tablets, powder in bottles • Less than 3 year FDA on-site inspection • Results of on-site FDA inspection • Company willing to allow pharmacist to inspect plant • Results of quality control analysis • Company willing to supply samples for testing • Product acceptance by physicians • Product acceptance by patients Service Quality • Returns policy • Rapid resolution of complaints • Company product availability record • Liability protection policy • Terms of unconditional guarantee • Company commitment to education of practitioners • Availability of company representative • Existence of 24-hour emergency customer service telephone number • Product availability through wholesalers • Ease of placing orders • Company customer information center, including an 800 number Company Reputation • Number of recalls in last 3 years • Severity of recalls in last 3 years • Who initiated recalls (FDA or company) • Company has a recall strategy • Other regulatory actions against company • Company has wide product line • FDA quality assurance profile • Company has crisis communication strategy
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Pharmacists have the responsibility of correctly selecting and dispensing multisource products that will have the greatest likelihood of achieving a positive therapeutic outcome in a cost-effective manner. The more information pharmacists have about a product and its manufacturer, the more likely they will be to make the most appropriate choice. Price cannot be the single factor in selecting a product. It is also clear, as Joseph Oddis stated, "Rational drug product selection entails far more than simply consulting the FDA's Orange Book or looking at the price catalogue" (166).
8.3.2
SPECIAL CASES While in most situations selection of drug products that are therapeutically equivalent can be done without undue complications, there are some circumstances where problems could occur. Depending on the drug, its formulation, the disease being treated, and the condition of the patient, generic substitution may not be advisable. Some of these special situations require extra attention and handling by the pharmacist. There are a number of drugs that could present problems when interchanged. Drugs that are poorly water soluble may have inherent problems with rate and extent of dissolution, resulting in poor or variable bioavailability. Drugs that are potent and thus present in very low amounts in a dosage form could present problems due to formulation factors. Some dosage forms may have inherent bioavailability problems, such as controlled-release products. And drugs which are considered "critical" also need special consideration. "Critical" drugs have been defined as drugs with a narrow therapeutic range, where a change in plasma concentration might result in adverse clinical outcome; drugs that are considered primarily for control of a disease rather than for alleviation of temporary symptoms; and drugs that have inherent or historical bioavailability or bioequivalence problems (8, 19). Seven classes of drugs have been identified that have demonstrated bioequivalence problems or, because of the nature of the product, have the potential for creating therapeutic problems if product interchange is permitted (Table 822 on page 51) (167-168).
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TABLE 8-22 Catagories
of drugs with demonstrated bioequivalence problems Digitalis glycosides - digoxin Warfarin anticoagulants Theophylline products Thyroid preparations (including levothyroxine) Conjugated and esterified estrogens Antiarrhythmic agents - quinidine salts - procainamide Anticonvulsants - phenytoin - carbamazepine - primidone
There have been numerous reports of drugs implicated in therapeutic problems due to bioinequivalence difficulties. In addition to those in the categories given in Table 8-22 on page 51, these include furosemide, propranolol, diazepam, prednisone, nitrofurantoin, and amitriptyline (20, 126, 167, 169-180). Although the documentation implicating these drugs in therapeutic failures due to bioavailability problems is primarily anecdotal in nature (and thus disregarded by the FDA), the performance of these products should still be closely observed and monitored, and care should be taken when selecting drugs from these categories. In addition to "critical" drugs, critical patients and critical diseases have also been identified when special care should be taken in performing product selection (8, 166). Critical patients are the very old and the very young, those suffering from multiple diseases who are managed with multiple drugs, and those who live alone, making observation of adverse drug effects unlikely. Critical diseases are generally chronic in nature and difficult to stabilize, where drug-disease interactions can present major problems (e.g. congestive heart failure, asthma, diabetes, cardiac disorders, and psychoses). In all the above special "critical" circumstances, there is a high risk of therapeutic problems, and product selection requires extra attention and precautions. In fact, product substitution and interchange in these cases is generally discouraged. Once a product (brand or generic) has been selected for a course of therapy, the pharmacist should not change to a different product if it can be avoided. If interchange is performed, it should be done only with the utmost care, and the patient should be monitored for any adverse outcomes.
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Pharmacist's professional responsibility-
Drug product selection has been and continues to be a primary and challenging professional responsibility of pharmacists. It is one where the pharmacist must exercise professional care and sound judgement to make decisions on behalf of the patient to maximize safety and efficacy, while minimizing cost. Pharmacists have a professional obligation to patients to take whatever steps are necessary to assure themselves that the medicines they are dispensing are safe and effective. Although some of this activity is currently constrained by bureaucratic and regulatory restrictions that often discourage, or entirely prevent, individual professional evaluation and initiative, with a greater appreciation and understanding of the scientific, clinical, and regulatory issues that form the basis of the process, pharmacists can make decisions that result in better patient care. Pharmacists must take steps to ensure the quality and integrity of the drug products dispensed to their patients. To accomplish this, pharmacists must look to pharmaceutical manufacturers to supply them with a quality product they can trust. Thus, the manufacturer of a multisource product must be carefully selected to ensure that the products they supply are of proper quality. If necessary, pharmacists should conduct independent research into the reputation and integrity of the manufacturer, or, if products are purchased through a buying group, should make sure that established policies and guidelines are in place to review multisource products. When considering purchasing drug products, the pharmacist should request the manufacturer to provide certain documentation and information, and should then evaluate this information (see Table 23). TABLE 8-23
Considerations when evaluating a Multi-Source vendor
1.Willingness to supply requested information 2.Bioavailability and bioequivalence data 3.Dissolution testing results 4.FDA bioequivalence rating 5.The actual manufacturer of the product, if not the supplier 6.FDA inspection reports 7.History of the manufacturer's recall record 8.Willingness of the manufacturer to permit on-site visitations 9.Evaluate economic considerations such as price, shipping, terms, discounts, insurance, return policies, and packagng.
And finally, pharmacists can counsel the patients on the importance of using the same drug product throughout a course of therapy, even though they might go to a different pharmacy. To further emphasize this, it has been suggested that the initial prescription and subsequent refills of a drug product considered questionable for interchange should contain auxiliary labeling that stresses the importance of continuing to use that product (167). Drug product selection is an important professional responsibility, but it is not an easy task. It requires the pharmacist to use his/her current knowledge, and Basic Pharmacokinetics
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all the currently available information in order to arrive at and render a decision regarding the most appropriate product to use for a specific patient.
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Bioavailability, Bioequivalence, and Drug Selection
8.4 Summary With the dramatic increase in the availability and utilization of generic drug products in recent years, pharmacists are being faced with an ever-increasing array of multisource products. Appropriate selection of a product from the plethora of products on the market is not always an easy task; the quality of the drug product must be considered, as well as the cost. The principles of biopharmaceutics indicate that the formulation and method of manufacture of a drug product can have a marked effect on the bioavailability of the active ingredient. Thus, generic equivalents may not necessarily be therapeutically equivalent. Guidelines and criteria have been established by the FDA to help judge whether one product can be substituted for another with assurance of equivalent therapeutic effect. For pharmacists to provide informed product selection, it is essential that they be knowledgeable about, and familiar with, these guidelines and criteria. This requires an understanding of bioavailability, bioequivalence, and how they are determined. The pharmacist can serve a major role in ensuring that only high quality products are dispensed, and in this way help reduce health care costs without compromising quality of care. Acknowledgment
The author gratefully acknowledges the assistance of Umesh V. Banakar in the preparation of this manuscript.
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8.4.1
QUESTIONS 1.
The term bioavailability refers to the a.
dissolution of a drug in the gastrointestinal tract.
b.
amount of drug destroyed in the liver by first-pass metabolism.
c.
distribution of drug to the body tissues over time.
relationship between the physical and chemical properties of a drug and its systemic absorption. d. e.
2.
measurement of the rate and amount of drug that reaches the systemic circulation.
The bioavailability of various drug products can be evaluated by comparing their plasma concentration-time curves. The three most important parameters of comparison that can be obtained directly from the curves are a.
biologic half-life (t1/2), absorption rate constant, area under the curve (AUC).
b.
time of peak concentration (tmax), absorption rate constant, elimination rate constant.
c.
maximum drug concentration (Cmax), time of peak concentration (tmax), duration of action.
d.
area under the curve (AUC), time of peak concentration (tmax), maximum drug concentration
(Cmax). e.
3.
rate of elimination, area under the curve (AUC), rate of absorption.
Two products are bioequivalent if they a.
contain the same amount of the same active ingredient.
b.
have equal areas under the curve after the administration of the same dose.
c.
have the same value for Cmax after administration of the same dose.
have equivalent rates and extents of absorption of the drug after administration of equal doses. d. e.
4.
5.
are pharmaceutically equivalent.
If an oral capsule formulation of drug A produces a plasma concentration- time curve having the same area under the curve (AUC) as that produced by an equivalent dose of drug A given intravenously, it can generally be concluded that: a.
there is no advantage to the IV route.
b.
the absolute bioavailability of the capsule formulation is equal to 1.
c.
the capsule formulation is essentially completely absorbed.
d.
the drug is very rapidly absorbed.
e.
b and c are correct.
5.Which of the following is NOT a criterion for therapeutic equivalence of two products, according to the FDA? a.
They must be pharmaceutical equivalents.
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6.
7.
8.
b.
All ingredients - active and inactive - must be the same.
c.
They have been manufactured in compliance with Good Manufacturing Practices.
d.
They are bioequivalent.
e.
They are approved as safe and effective by the FDA.
A test oral formulation has the same area under the plasma concentration- time curve as the reference formulation. This means that the two formulations a.
are bioequivalent by definition.
b.
deliver the same total amount of drug to the body but are not necessarily bioequivalent.
c.
are bioequivalent if they both meet USP dissolution standards.
d.
deliver the same total amount of drug to the body and are, therefore, bioequivalent.
e.
have the same rate of absorption.
In-vitro dissolution rate studies on drug products are useful in bioavailability evaluations only if they can be correlated with a.
in-vivo bioavailability studies in humans.
b.
the chemical stability of the drug.
c.
USP disintegration requirements.
d.
in-vivo studies in at least three species of animals.
e.
the therapeutic response observed in patients.
Which of the following statements regarding bioequivalence is TRUE? If the mean AUC and Cmax values for a generic product are within + 20% of those of the reference product, the two products are bioequivalent. a.
b.
If we can be 90% certain that the mean values of AUC and Cmax for two products are within
80% to 125% of each other, then the two products are considered bioequivalent.
Bioequivalence studies are generally conducted in a panel of patients consisting of the target population for which the drug is intended. c.
Bioequivalence studies are generally conducted as multiple-dose studies utilizing the cross-over design. d.
If two products are shown to be bioequivalent, we can always say with certainty that they will be therapeutically equivalent. e.
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Bioavailability, Bioequivalence, and Drug Selection
9.
9.Which of the following statements about the FDA Orange Book is TRUE? Drugs that are excluded from the Orange Book are not safe and effective and should not be dispensed. a.
b.
It contains therapeutic equivalence evaluations for all the drugs approved by the FDA.
c.
Products placed in the "B" category should not be dispensed.
The Orange Book is an official compendium, and pharmacists can legally only dispense those products listed as bioequivalent. d.
The drug listings contain the names of only the companies that actually hold an approved NDA or ANDA for a drug. e.
10.
8.4.2
10.Growth in the utilization of generic drug products can be attributed to a.
passage of the 1984 Waxman-Hatch Act.
b.
expiration of patents of many popular brand products.
c.
pressures to reduce health care costs.
d.
the growth of managed health care organizations.
e.
all of the above.
ANSWERS TO QUESTIONS 1.
e
2.
d
3.
d
4.
e
5.
b
6.
b
7.
a
8.
b
9.
e
10.
e
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8.5 Bioavailibility Equations The following set of equations were used to solve the bioavailability problem set. The problem sets for the first two drugs have been done for you. The others are done exactly the same way. The answers follow the problems.
1.
AUMC iv as discussed in chapter 4. MRT iv = --------------------AUC iv
2.
1 k = --------------MRT iv
3.
ln 2 t 1 ⁄ 2 = -------k
4.
Cp 0iv = AUC ⋅ k
5.
Dose iv Vd = ---------------Cp 0iv
6.
Cp iv = Cp 0 e
7.
AUC oral Dose iv f = --------------------⋅ ----------------Dose oral AUC iv
8.
AUMC po - as discussed in chapter 4 MRT po = ---------------------AUC po
9.
MAT po = MRT po – MRT iv
10.
1 k a = -----------MAT
11.
– kt
k ln ----a- k t p = ---------------ka – k
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12.
ka – kt –k t Cp max = fD ------ ⋅ ------------⋅ (e p – e a p ) V ka – k
14.
( AUC generic ) ⁄ ( Dose generic ) Relative Bioavailability (R.B. or C.B.) = --------------------------------------------------------------------( AUC Brand ) ⁄ ( Dose Brand )
15.
Bioequivalent: Yes if all three: 0.80 < CB < 1.25 t p generic 0.80 < -------------< 1.25 tp brand C p max –g eneric 0.80 < ----------------------- < 1.25 C p max – b rand
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8.6 Problems
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Caffeine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 8 - 1)
AHFS 00:00.00 GPI: 0000000000
Aramaki, S., et al., "Pharmacokinetics of caffeine and its metabolites in horses after intravenous, intramuscular, or oral administration", Chem Pharm Bull, Vol. 30, No. 11, (1991), p. 2999 - 3002.
This study deals with the pharmacokinetics of caffeine. Caffeine doses of 2.5 mg/kg were administered both intravenously and orally to horses with an average weight of about 500 kg. A summary of the some of data obtained from this experiment is given below. Fill in the empty cells. TABLE 8-24 Caffeine
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg/kg)
2.5
2.5
2.5
2.5
ug AUC -------- ⋅ hr mL
63.1
60.7
60
57
2 ug AUMC -------- ⋅ hr mL
1442
1556.8
1600
1723
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 -------- mL Vd (L) ug Cp at 1 hour -------- mL f ug- Cpmax ------ mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
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Cefetamet Pivoxil Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 8 - 2)
AHFS 00:00.00 GPI: 0000000000
Ducharme, M., et. al., "Bioavailability of syrup and tablet formulations of cefetamet pivoxil", Antimicrobial Agents and Chemotherapy, Vol. 37, No. 12, (1993), p. 2706 - 2709.
Cefetamet pivoxil is a prodrug of cefetamet. This study compares the bioavailability of cefetamet pivoxil in tablet form versus syrup form. A summary of the some of data obtained from this experiment is given below. Fill in the approprate cells. . Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg)
250
500
500
500
ug AUC -------- ⋅ hr mL
30.64
53.68
50
47
2 ug AUMC -------- ⋅ hr mL
101.66
191.64
205.6
225.3
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 -------- mL Vd (L) ug Cp at 1 hour -------- mL f ug- Cpmax ------ mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
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Cefixime
(Problem 8 - 3)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Faulkner, R. ,et al., "Absolute bioavailability of cefixime in man", Journal of Clinical Pharmacology, Vol. 28 (1988), p. 700 - 706.
Cefixime is a broad-spectrum cephalosporin which is active against a variety of gram positive and gram negative bacteria. In this study, sixteen subjects each received a 200 mg intravenous dose and then a 200 mg capsule with a washout period between the administration of each dosage form. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Brand Capsule
Generic Capsule
Dose (mg)
200
200
200
ug AUC -------- ⋅ hr mL
47
23.6
20.2
2 ug AUMC -------- ⋅ hr mL
183.3
162.8
187.5
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 -------- mL Vd (L) ug- Cp at 1 hour ------ mL f ug Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
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Ceftibuten Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 8 - 4)
AHFS 00:00.00 GPI: 0000000000
"The pharmacokinetics of ceftibuten in humans"
Ceftibuten is a new oral cephalosporin with potent activity against enterobacteriaceae and certain gram positive organisms. In this study two groups received either a 400 mg oral dosage form of ceftibuten or a 200 mg iv bolus dose of ceftibuten. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg)
200
400
400
400
ug AUC -------- ⋅ hr mL
75.2
65.9
64.2
64
2 ug AUMC -------- ⋅ hr mL
211.2
213.4
220
208
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 -------- mL Vd (L) ug Cp at 1 hour -------- mL f ug Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
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Cimetidine
(Problem 8 - 5)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Sandborn, W., et al., "Pharmacokinetics and pharmacodynamics of oral and intravenous cimetidine in seriously ill patients", Journal of Clinical Pharmacology, Vol. 30, (1990), p. 568 - 571.
Cimetidine is a histamine receptor antagonist which is used in the treatment of gastric and duodenal ulcer disease. In this study, patients received 300 mg of cimetidine as an iv bolus on the first day and data was collected. On the second day, the patients received 300 mg orally and data was collected. A summary of the some of data obtained from this experiment is given below.
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
300
300
300
ug AUC -------- ⋅ hr mL
3.81
2.48
2.50
2 ug AUMC -------- ⋅ hr mL
5.33
11.73
10.73
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 -------- mL Vd (L) ug Cp at 1 hour -------- mL f ug Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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8-65
Bioavailability, Bioequivalence, and Drug Selection
Diurnal Variability in Theophylline Bioavailability Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 8 - 6)
AHFS 00:00.00 GPI: 0000000000
Bauer, L., Gibaldi, M., and Vestal, R., "Influence of pharmacokinetic diurnal variation on bioavailability estimates", Clinical Pharmacokinetics, vol. 9, (1984), p. 184 - 187.
This article discusses the effects of diurnal variation on the bioavailability and clearance of theophylline. In this study patients received a 500 mg dose every 12 hours either orally or by iv bolus. A summary of the some of data obtained from this experiment for the time period between midnight and noon is given below. From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg)
500
500
500
500
ug AUC -------- ⋅ hr mL
160.25
144.58
140
144
2 ug AUMC -------- ⋅ hr mL
1821
1662
1785
1700
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 -------- mL Vd (L) ug Cp at 1 hour -------- mL f ug Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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8-66
Bioavailability, Bioequivalence, and Drug Selection
cis-5-Fluoro-1-[2-Hydroxymethyl-1,3-Oxathiolan-5-yl] Cytosine (FTC) (Problem 8 - 7) Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Frick, L. , et al., "Pharmacokinetics, oral bioavailability, and metabolic disposition in rats of (-)-cis-5-Fluoro-1-[2-Hydroxymethyl-1,3-Oxathiolan-5-yl] Cytosine, a nucleoside analog active against human immunodeficiency virus and hepatitis B virus", Antimicrobial Agents and Chemotherapy, Vol. 37, No. 11, (1993), p. 2285 - 2292.
FTC is a 2',3'-didoexynucleoside analog that may be useful against HIV and HBV. In this study, rats with an average weight of 270 g were given either iv or oral doses of 100 mg/kg. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg/kg)
100
100
100
ug AUC -------- ⋅ hr mL
265
168
175
2 ug AUMC -------- ⋅ hr mL
19514
12600
13125
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 -------- mL Vd (L) ug Cp at 1 hour -------- mL f ug- Cpmax ------ mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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8-67
Bioavailability, Bioequivalence, and Drug Selection
Hydromorphone Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 8 - 8)
AHFS 00:00.00 GPI: 0000000000
Vallner, J., et al., "Pharmacokinetics and bioavailability of hydromorphone following intravenous and oral administration to human subjects", Journal of Clinical Pharmacology, Vol. 21, (1981), p. 152 - 156.
Hydromorphone hydrochloride is an analog of morphine which has about seven times the effect of morphine when given intravenously. In this study, volunteers were given a 2 mg intravenous dose and a 4 mg oral dose of hydromorphone on separate days. A summary of the some of data obtained from this experiment is given below.
From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
2
4
4
ug AUC ------ ⋅ hr L
83
87.2
96
2 ug AUMC ------ ⋅ hr L
289.4
401
432
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 ------ L Vd (L) ug Cp at 1 hour ------ L f ------ Cpmax ug L Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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8-68
Bioavailability, Bioequivalence, and Drug Selection
Isosorbide Dinitrate Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 8 - 9)
AHFS 00:00.00 GPI: 0000000000
Straehl, P. and Galeazzi, R., "Isosorbide dinitrate bioavailability , kinetics, and metabolism", Clinical Pharmacology and Therapeutics, Vol. 38m (1985), p. 140 - 149.
Isosorbide dinitrate is used in the treatment of angina pectoris, vasospastic angina, and congestive heart failure. In this study volunteers received a 5 mg intravenous dose given over 5 minutes and a 10 mg tablet. The different dosage forms were separated by a washout period. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg/kg)
5
10
10
AUC ug ------ ⋅ hr L
370.3
158
165
2 AUMC ug ------ ⋅ hr L
487
310
305
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 ------ L Vd (L) ------ Cp at 1 hour ug L f ug Cpmax ------ L Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
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8-69
Bioavailability, Bioequivalence, and Drug Selection
Ketanserin
(Problem 8 - 10)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Kurowski, M., "Bioavailability and pharmacokinetics of ketanserin in elderly subjects", Journal of Clinical Pharmacology, Vol. 28, (1988), p. 700 - 706.
Ketanserin is a 5-hydroxytryptamine S2-antagonist. This study focuses on the kinetics of Ketanserin in the elderly. Subjects were given either a 10 mg intravenous dose or a 40 mg oral tablet. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
10
40
40
ng AUC -------- ⋅ hr mL
247
520
400
2 ng AUMC -------- ⋅ hr mL
3991
8922
8922
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ng Cp0 -------- mL Vd (L) ng Cp at 1 hour -------- mL f ng Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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8-70
Bioavailability, Bioequivalence, and Drug Selection
Methotrexate
(Problem 8 - 11)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Seideman, P., et al., " The pharmacokinetics of methotrexate and its 7-hydroxy metabolite in patients with rheumatoid arthritis", British Journal of Clinical Pharmacology, 35 (1993), p. 409 - 412.
The drug Methotrexate is a folic acid which has been shown to inhibit dihydrofolate reductase. The importance of this drug at present is mostly seen in the area of oncology, but lately it has been used for rheumatoid arthritis. Methotrexate has a molecular weight of 454.4. In this study, the drug was administered both by IV bolus and orally as a 15 mg dose. The following data was obtained: From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
15
15
15
AUC nmole ---------------- ⋅ hr L
2752
2708
2700
2 AUMC nmole ---------------- ⋅ hr L
15887
18400
18500
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) nmole Cp0 ---------------- L Vd (L) ---------------- Cp at 1 hour nmole L f nmole Cpmax ---------------- L Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
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8-71
Bioavailability, Bioequivalence, and Drug Selection
Moclobemide
(Problem 8 - 12)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Schoerlin, M. et al., "Disposition kinetics of moclobemide, a new MAO-A inhibitor, in subjects with impaired renal function", Journal of Clinical Pharmacology, Vol. 30 (1991), p. 272 - 284.
Moclobemide is an antidepressant agent that reversibly inhibits the A-isozyme of the monoamine oxidase enzyme system. In this study, single IV and oral doses were administered to a patient. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
150
100
100
ug AUC -------- ⋅ hr mL
2.58
1.70
1.52
2 ug AUMC -------- ⋅ hr mL
6.35
5.91
5.90
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 -------- mL Vd (L) ug Cp at 1 hour -------- mL f ug Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
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8-72
Bioavailability, Bioequivalence, and Drug Selection
Nalbuphine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 8 - 13)
AHFS 00:00.00 GPI: 0000000000
Nalbuphine hydrochloride is an agonist-antagonist opiod which is used for its analgesic actions. In this study, volunteers were given single doses of four different nalbuphine forms. The data below focuses on a 10 mg iv dose and a 45 mg dose of an oral solution. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg)
10
45
40
40
ng AUC -------- ⋅ hr mL
86.9
70.3
62.5
60
2 ng AUMC -------- ⋅ hr mL
288
306
280
270
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ng Cp0 -------- mL Vd (L) ng Cp at 1 hour -------- mL f ng Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
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8-73
Bioavailability, Bioequivalence, and Drug Selection
Nefazodone Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 8 - 14)
AHFS 00:00.00 GPI: 0000000000
Shukla, U. et al., "Pharmacokinetics, absolute bioavailability, and disposition of nefazodone in the dog", Drug Metabolism and Disposition, Vol. 21, No. 3, (1993), p. 502 - 507.
Nefazodone was given to four healthy, adult, male beagles with an average weight of 11.0 kg. Each dog was given a 10 mg/kg dose as a either a intravenous injection or as an oral solution or tablet. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg/kg)
10
10
10
10
ng AUC -------- ⋅ hr mL
6023
829
800
700
2 ng AUMC -------- ⋅ hr mL
29283
4875
4800
4500
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ng Cp0 -------- mL Vd (L) ng Cp at 1 hour -------- mL f ng Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
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8-74
Bioavailability, Bioequivalence, and Drug Selection
Ondansetron
(Problem 8 - 15)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Colthup, P., et al., "Determination of ondansetron in plasma and its pharmacokinetics in the young and elderly", Journal of Pharmaceutical Sciences, Vol. 80, No. 9(1991), p. 868 - 871.
Ondansetron is a 5-hydroxyltryptamine compound which is useful in treating the nausea and vomiting which is caused by the use of chemotherapy and radiation in the cancer patients. In order to determine the absolute bioavailability of oral Ondansetron, doses of 8 mg were given to two groups. One group received an oral dose and the other group received an intravenous dose. A summary of the some of data obtained from this experiment is given below.
From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
8
8
8
ng AUC -------- ⋅ hr mL
246.5
139
145
2 ng AUMC -------- ⋅ hr mL
1138
795
870
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ng Cp0 -------- mL Vd (L) ng Cp at 1 hour -------- mL f ng Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
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8-75
Bioavailability, Bioequivalence, and Drug Selection
Omeprazole
(Problem 8 - 16)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Anderson, T., et al, "Pharmacokinetics of various single intravenous and oral doses of omeprazole", Eur Journal of Clinical Pharmacology, 39, (1990), p. 195 - 197.
Omeprazole (mw: 345.42) is an agent which inhibits gastric acid secretion from the parietal cell. It is useful in treating such problems as ulcers and gastroesophageal reflux disease. One group received an iv bolus dose and the other group received an oral dose. A summary of the some of data obtained from this experiment is given below.
From the preceding data, please calculate the following:
Parameter
IV
Brand Capsule
Generic Capsule
Dose (mg)
20
40
40
µmole AUC ---------------- ⋅ hr L
3.2
3.5
3.0
2 µmole AUMC ---------------- ⋅ hr L
3.2
5.25
4.5
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) µmole Cp0 ---------------- L Vd (L) µ mole Cp at 1 hour ---------------- L
f µ mole Cpmax --------------- L
Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
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8-76
Bioavailability, Bioequivalence, and Drug Selection
Paroxetine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 8 - 17)
AHFS 00:00.00 GPI: 0000000000
Lund, J., et al., "Paroxetine: pharmacokinetics and cardiovascular effects after oral and intravenous single doses in man", Journal of Pharmacology and Toxicology, Vol. 51, (1982), p. 351 - 357.
Paroxetine kinetics and cardiovascular effects were studied in male subjects after single oral doses of 45 mg and slow intravenous infusion of 23 - 28 mg. A summary of the some of data obtained from this experiment is given below.
From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
28
45
45
ng AUC -------- ⋅ hr mL
467
750
675
2 ng AUMC -------- ⋅ hr mL
6671
11250
10463
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ng Cp0 -------- mL Vd (L) ng Cp at 1 hour -------- mL f ng- Cpmax ------ mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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8-77
Bioavailability, Bioequivalence, and Drug Selection
Ranitidine
(Problem 8 - 18)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Garg, D., et al., "Pharmacokinetics of ranitidine in patients with renal failure", Journal of Clinical Pharmacology, Vol. 26 (1986), p. 286 - 291.
Ranitidine is an agent used in the treatment of peptic ulceration. In this study, ten patients with renal failure received either a 50 mg intravenous bolus dose or a 150 mg tablet. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
50
150
150
ng AUC -------- ⋅ hr mL
5159
6422
6753
2 ng AUMC -------- ⋅ hr mL
53415
78752
84413
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ng Cp0 -------- mL Vd (L) ng Cp at 1 hour -------- mL f ng Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
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8-78
Bioavailability, Bioequivalence, and Drug Selection
Sulpiride Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 8 - 19)
AHFS 00:00.00 GPI: 0000000000
Bressolle, F., Bres, J., and Faure-Jeantis, A., "Absolute bioavailability , rate of absorption, and dose proportionality of sulpiride in humans", Journal of Pharmaceutical Sciences ,Vol. 81, No. 1 (1992), p. 26 - 32.
Sulpiride is a substituted benzamine antipsychotic. In this study, the drug was administered to two groups. The first group received a 200 mg oral dose and the second group received a 100 mg intravenous infusion. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg)
100
200
200
200
ug AUC -------- ⋅ hr mL
8.27
8.79
8.6
8.0
2 ug AUMC -------- ⋅ hr mL
79.1
87.3
91.1
84.5
Bioequivalence
MRT (hr) MAT (hr) ke (hr-1) ka (hr-1) ug Cp0 -------- mL Vd (L) ug Cp at 1 hour -------- mL f ug Cpmax -------- mL Tmax (hr) Relative Bioavailability Generic Equivalent (Yes / No)
Basic Pharmacokinetics
REV. 99.4.25
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8-79
Bioavailability, Bioequivalence, and Drug Selection
8.7 Solutions 8.7.1
“CAFFEINE” ON PAGE 61
Aramaki, S., et al., "Pharmacokinetics of caffeine and its metabolites in horses after intravenous, intramuscular, or oral administration", Chem Pharm Bull, Vol. 30, No. 11, (1991), p. 2999 - 3002.
This study deals with the pharmacokinetics of caffeine. Caffeine doses of 2.5 mg/kg were administered both intravenously and orally to horses with an average weight of about 500 kg. A summary of the some of data obtained from this experiment is given below. Fill in the empty cells. TABLE 8-25 Caffeine
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg/kg)
2.5
2.5
2.5
2.5
ug AUC -------- ⋅ hr mL
63.1
60.7
60
57
2 ug AUMC -------- ⋅ hr mL
1442
1556.8
1600
1723
MRT (hr)
22.9
25.7
26.7
30.2
2.79
3.81
7.36
0.358
0.262
0.136
0.78
0.59
0.31
0.96
0.95
.90
1.98
1.83
1.45
0.79
6.69
8.19
12.1
1.5
MAT (hr) ke (hr-1)
0.0438
ka (hr-1) ug Cp0 -------- mL
2.76
V d (L/kg)
0.91
ug- Cp at 1 hour ------ mL
2.64
f ug Cpmax -------- mL
Bioequivalence
2.76
Tmax (hr) Relative Bioavailability
0.95
Generic Equivalent (Yes / No)
NO
Basic Pharmacokinetics
REV. 99.4.25
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8-80
Bioavailability, Bioequivalence, and Drug Selection
2
⋅h 1442ug --------------mL ---------------------------- = 22.9 2 ⋅h 63.1ug --------------mL
1.
MRT = AUMC ------------------ = AUC
2.
1 = ------------1 = 0.044 h – 1 k = -----------MRT 22.9h
3.
ln 2 = 0.693 t 1 ⁄ 2 = ---------------------------- = 15.75h –1 k 0.044h
4.
–1 ⋅h ugCp 0 = AUC ⋅ k = 63.1 ug ------------⋅ 0.0044h = 2.76 ------mL mL
5.
hours
The horses have an average weight of 500 kg.
Dose = 2.5 mg ------- ⋅ 500kg = 1250mg kg ug- = 2.78 mg Cp 0 = 2.78 ------------mL L 1250mg- = 449.6L = ------------------------2.5mg ⁄ kg = 0.91 ----LVd = Dose ------------- = ------------------Cp 0 mg mg kg 2.78 ------2.78------L L 6.
7.
Cp = Cp0 e
– kt
ug- ( e – 0.044 ( 1 ) ) = 2.64 ------ug= 2.78 ------mL mL
⋅h 2.5 mg ------60 ug ------------AUC oral Dose iv mL kg = 0.95 f = --------------------- ⋅ ----------------- = -------------------- ⋅ -----------------------Dose oral AUC iv ⋅h 2.5 mg ------- 63.1ug ------------kg mL 2
8.
MRT po
ug ⋅ h 1556.8 --------------mL - = 25.7h = AUMC ------------------ = -------------------------------AUC ug ⋅h 60.7------------mL
Where AUMC is that which is given for the oral dose. Where AUC is that which is given for the oral dose. 9.
MAT po = MRT po – MRT iv = 25.7h – 22.9h = 2.79h
Basic Pharmacokinetics
REV. 99.4.25
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8-81
Bioavailability, Bioequivalence, and Drug Selection
10.
1 = --------1 - = 0.358hr – 1 k a = -----------MAT 2.79 0.358hr k a ln ----------------------- –1 ln ----- 0.044h k tp = ---------------- = ------------------------------------------------------ = 6.7hr –1 –1 ka – k 0.358hr – 0.044hr –1
11.
–1
12.
13.
– ktp katp –( 0.044 ⋅ 6.7 ) – ( 0.358 ⋅ 6.7 ) fD ka 0.96 ⋅ 1250mg 0.358hr –e ) = ----------------------------------- ⋅ ------------------------------------------------- ⋅ ( e Cp max = ------ ⋅ -------------- ⋅ ( e ⋅e ) –1 V ka – k 449L ( 0.358 – 0.044 ) hr
( AUC
) ⁄ ( Dose
)
generic generic Relative Bioavailability (R.B. or C.B.) = ---------------------------------------------------------------------
( AUC Brand ) ⁄ ( Dose Brand )
ug- ⋅ 57 ------ mL
hr ⁄ 2.5 mg ------- km CB = ------------------------------------------------------------- = 0.95 ug- ⋅ hr ⁄ 2.5 mg 60 ------------- mL km 14.
Bioequivalent: Yes if all three = Yes:
0.80 < CB < 1.25 CB = 0.95 = Yes t p generic 0.80 < -------------< 1.25 tp brand
t p generic -------------= 12.1hr ---------------- = 1.5 = NO 8.19hr t pbrand
C p max –g eneric 0.80 < ----------------------- < 1.25 C p max – b rand
ug1.45 ------C p max –g eneric mL- = 0.79 = NO ------------------------ = -----------------C p max –b rand ug1.83 ------mL
Basic Pharmacokinetics
REV. 99.4.25
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8-82
Bioavailability, Bioequivalence, and Drug Selection
8.7.2
“CEFETAMET PIVOXIL” ON PAGE 62
Ducharme, M., et. al., "Bioavailability of syrup and tablet formulations of cefetamet pivoxil", Antimicrobial Agents and Chemotherapy, Vol. 37, No. 12, (1993), p. 2706 - 2709.
Cefetamet pivoxil is a prodrug of cefetamet. This study compares the bioavailability of cefetamet pivoxil in tablet form versus syrup form. A summary of the some of data obtained from this experiment is given below. Fill in the approprate cells. . Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg)
250
500
500
500
ug- ⋅ hr AUC ------ mL
30.64
53.68
50
47
ug- ⋅ hr 2 AUMC ------ mL
101.66
191.64
205.6
225.3
MRT (hr)
3.32
3.57
4.11
4,79
0.252
0.794
1.48
3.97
1.26
0.678
12.62
9.03
5.91
0.88
0.82
0.77
13.1
9.6
7.4
0.77
0.70
1.49
2.15
1.44
MAT (hr) ke
(hr-1)
0.301
ka (hr-1) ug Cp0 -------- mL
9.23
Vd (L)
27.1
ug- Cp at 1 hour ------ mL
6.83
f ug Cpmax -------- mL
Bioequivalence
9.23
Tmax (hr) Relative Bioavailability
0.94
Generic Equivalent (Yes / No)
NO 2
1.
mg ⋅ h 101.66 ----------------L MRT = AUMC ------------------ = ---------------------------------= 3.32 hours 2 AUC mg ⋅h 30.64 ----------------L Where AUMC is that which is given for the intravenous dose. Where AUC is that which is given for the intravenous dose.
Basic Pharmacokinetics
REV. 99.4.25
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8-83
Bioavailability, Bioequivalence, and Drug Selection
2.
–1 1 1 k = -----------= ------------- = 0.301 h MRT 3.32h
3.
ln 2 = --------------------0.693 = 2.3h t 1 ⁄ 2 = -------–1 k 0.301h
4.
–1 ⋅h Cp 0 = AUC ⋅ k = 30.64 mg -------------- ⋅ 0.301h = 9.22 mg ------L L
5.
6.
7.
250mg V d = Dose ------------- = ------------------ = 27.06L Cp 0 9.24mg ------L Cp = Cp 0 e
– kt
– 0.301 ( 1 ) mg = 9.24 mg ------- ( e ) = 6.84------L L
⋅h 53.68mg -------------AUC oral Dose iv L 250mg - = 0.876 f = --------------------- ⋅ ----------------- = ----------------------------- ⋅ ---------------------------Dose oral AUC iv 500mg ⋅h 30.64 mg -------------L 2
8.
MRT po
mg ⋅ h 191.64 ----------------L = AUMC ------------------ = ---------------------------------= 3.57h AUC mg ⋅h 53.68 --------------L
Where AUMC is that which is given for the oral dose. Where AUC is that which is given for the oral dose. 9. 10.
MAT po = MRT po – MRT iv = 3.57h – 3.32h = 0.252hr 1 = -----------1 - = 3.97hr – 1 k a = -----------MAT 0.252 4.0h k a ln -------------------- ln ----- 0.301h – 1 k t p = ---------------- = ------------------------------------------- = 0.7h –1 –1 ka – k 4.0h – 0.301h –1
11.
–1
12.
13.
– 0.301 ( 0.7 ) – 3.97 ( 0.7 ) ka ⋅ ( e – kt – e –k a t ) = 0.88 ( 500mg ) 3.97hr mg Cp max = fD ------ ⋅ --------------------------------------------⋅ ---------------------------------------------- ⋅ ( e –e ) = 13.1------–1 V ka – k 27.1L L ( 3.97 – 0.301 ) hr
( AUC
) ⁄ ( Dose
)
generic generic Relative Bioavailability (R.B. or C.B.) = ---------------------------------------------------------------------
( AUC Brand ) ⁄ ( Dose Brand )
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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8-84
Bioavailability, Bioequivalence, and Drug Selection
ug- ⋅ 47 ------ mL
hr ⁄ 500mg CB = -------------------------------------------------------- = 0.94 ug- ⋅ hr ⁄ 500mg 50 ------ mL 14.
Bioequivalent: Yes if all three = Yes:
0.80 < CB < 1.25 CB = 0.94 = Yes t p generic 0.80 < -------------< 1.25 tp brand
t p generic -------------= 2.15hr ---------------- = 1.44 = NO 1.49hr t pbrand
C p max –g eneric 0.80 < ----------------------- < 1.25 C p max – b rand
ug7.4 ------C p max – g eneric mL ------------------------ = ---------------- = 0.77 = NO C p max– b rand ug9.6 ------mL
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
8-85
Bioavailability, Bioequivalence, and Drug Selection
8.7.3
“CEFIXIME” ON PAGE 63
Faulkner, R. ,et al., "Absolute bioavailability of cefixime in man", Journal of Clinical Pharmacology, Vol. 28 (1988), p. 700 - 706.
Cefixime is a broad-spectrum cephalosporin which is active against a variety of gram positive and gram negative bacteria. In this study, sixteen subjects each received a 200 mg intravenous dose and then a 200 mg capsule with a washout period between the administration of each dosage form. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Brand Capsule
Generic Capsule
Dose (mg)
200
200
200
ug AUC -------- ⋅ hr mL
47
23.6
20.2
2 ug AUMC -------- ⋅ hr mL
183.3
162.8
187.5
MRT (hr)
3.9
6.9
9.3
3.0
5.38
0.334
0.186
1.5
0.77
0.50
0.43
2.5
1.6
0.64
3.4
4.6
1.33
MAT (hr) ke
(hr-1)
0.256
ka (hr-1) ug Cp0 -------- mL
12.1
Vd (L)
16.6
ug Cp at 1 hour -------- mL
9.3
f ug- Cpmax ------ mL
Bioequivalence
12.1
Tmax (hr) Relative Bioavailability
0.86
Generic Equivalent (Yes / No)
NO
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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8-86
Bioavailability, Bioequivalence, and Drug Selection
8.7.4
“CEFTIBUTEN” ON PAGE 64
"The pharmacokinetics of ceftibuten in humans"
Ceftibuten is a new oral cephalosporin with potent activity against enterobacteriaceae and certain gram positive organisms. In this study two groups received either a 400 mg oral dosage form of ceftibuten or a 200 mg iv bolus dose of ceftibuten. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg)
200
400
400
400
ug- ⋅ hr AUC ------ mL
75.2
65.9
64.2
64
ug- ⋅ hr 2 AUMC ------ mL
211.2
213.4
220
208
MRT (hr)
2.94
MAT (hr) ke (hr-1)
3.24
3.43
3.25
0.297
0.485
0.309
3.37
2.06
3.24
16.9
15.3
16.4
0.44
0.42
0.43
17.3
15.3
16.7
1.09
0.76
1.05
0.78
0.74
0.390
ka (hr-1) ug- Cp0 ------ mL
25.6
Vd (L)
7.8
ug Cp at 1 hour -------- mL
18.2
f ug Cpmax -------- mL
Bioequivalence
25.6
Tmax (hr) Relative Bioavailability
1
Generic Equivalent (Yes / No)
NO
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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8-87
Bioavailability, Bioequivalence, and Drug Selection
8.7.5
“CIMETIDINE” ON PAGE 65
Sandborn, W., et al., "Pharmacokinetics and pharmacodynamics of oral and intravenous cimetidine in seriously ill patients", Journal of Clinical Pharmacology, Vol. 30, (1990), p. 568 - 571.
Cimetidine is a histamine receptor antagonist which is used in the treatment of gastric and duodenal ulcer disease. In this study, patients received 300 mg of cimetidine as an iv bolus on the first day and data was collected. On the second day, the patients received 300 mg orally and data was collected. A summary of the some of data obtained from this experiment is given below.
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
300
300
300
ug- ⋅ hr AUC ------ mL
3.81
2.48
2.50
ug- ⋅ hr 2 AUMC ------ mL
5.33
11.73
10.73
MRT (hr)
1.40
4.73
4.29
3.33
2.89
0.300
0.346
0.32
0.37
0.65
0.66
0.40
0.44
1.1
2.1
2.0
0.94
MAT (hr) ke
(hr-1)
0.715
ka (hr-1) ug- Cp0 ------ mL
2.72
Vd (L)
110
ug Cp at 1 hour -------- mL
1.33
f ug Cpmax -------- mL
Bioequivalence
2.72
Tmax (hr) Relative Bioavailability
1
Generic Equivalent (Yes / No)
YES
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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8-88
Bioavailability, Bioequivalence, and Drug Selection
8.7.6
“DIURNAL VARIABILITY IN THEOPHYLLINE BIOAVAILABILITY” ON PAGE 66
Bauer, L., Gibaldi, M., and Vestal, R., "Influence of pharmacokinetic diurnal variation on bioavailability estimates", Clinical Pharmacokinetics, vol. 9, (1984), p. 184 - 187.
This article discusses the effects of diurnal variation on the bioavailability and clearance of theophylline. In this study patients received a 500 mg dose every 12 hours either orally or by iv bolus. A summary of the some of data obtained from this experiment for the time period between midnight and noon is given below. From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg)
500
500
500
500
ug- ⋅ hr AUC ------ mL
160.25
144.58
140
144
ug- ⋅ hr 2 AUMC ------ mL
1821
1662
1785
1700
MRT (hr)
11.40
11.50
12.75
11.8
0.13
1.39
0.44
7.58
0.721
2.26
11.8
6.02
10.7
0.90
0.87
0.90
12.1
9.20
11.1
1.21
.059
3.3
1.5
0.45
MAT (hr) ke
(hr-1)
0.088
ka (hr-1) ug- Cp0 ------ mL
14.1
Vd (L)
35.5
ug Cp at 1 hour -------- mL
12.9
f ug Cpmax -------- mL
Bioequivalence
14.1
Tmax (hr) Relative Bioavailability
1.03
Generic Equivalent (Yes / No)
NO
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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8-89
Bioavailability, Bioequivalence, and Drug Selection
8.7.7
“CIS-5-FLUORO-1-[2-HYDROXYMETHYL-1,3-OXATHIOLAN-5-YL] CYTOSINE (FTC)” ON PAGE 67
Frick, L. , et al., "Pharmacokinetics, oral bioavailability, and metabolic disposition in rats of (-)-cis-5-Fluoro-1-[2-Hydroxymethyl-1,3-Oxathiolan-5-yl] Cytosine, a nucleoside analog active against human immunodeficiency virus and hepatitis B virus", Antimicrobial Agents and Chemotherapy, Vol. 37, No. 11, (1993), p. 2285 - 2292.
FTC is a 2',3'-didoexynucleoside analog that may be useful against HIV and HBV. In this study, rats with an average weight of 270 g were given either iv or oral doses of 100 mg/kg. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg/kg)
100
100
100
ug- ⋅ hr AUC ------ mL
265
168
175
ug- ⋅ hr 2 AUMC ------ mL
19514
12600
13125
MRT (hr)
73.6
75
75
1.36
1.36
0.734
0.734
1.18
1.23
0.63
0.66
2.1
2.2
1.04
5.54
5.54
1.0
MAT (hr) ke
(hr-1)
.0136
ka (hr-1) ug- Cp0 ------ mL
3.6
Vd (L/kg)
27.7
ug Cp at 1 hour -------- mL
3.55
f ug Cpmax -------- mL
Bioequivalence
3.6
Tmax (hr) Relative Bioavailability
1.04
Generic Equivalent (Yes / No)
YES
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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8-90
Bioavailability, Bioequivalence, and Drug Selection
8.7.8
“HYDROMORPHONE” ON PAGE 68
Vallner, J., et al., "Pharmacokinetics and bioavailability of hydromorphone following intravenous and oral administration to human subjects", Journal of Clinical Pharmacology, Vol. 21, (1981), p. 152 - 156.
Hydromorphone hydrochloride is an analog of morphine which has about seven times the effect of morphine when given intravenously. In this study, volunteers were given a 2 mg intravenous dose and a 4 mg oral dose of hydromorphone on separate days. A summary of the some of data obtained from this experiment is given below.
From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
2
4
4
83
87.2
96
289.4
401
432
3.49
4.60
4.50
1.11
1.03
0.899
0.987
12.6
14.7
0.53
0.56
14.6
16.6
1.13
1.87
1.77
0.95
ug AUC ------ ⋅ hr L 2 ------ ⋅ hr AUMC ug L
MRT (hr) MAT (hr) ke
(hr-1)
0.287
ka (hr-1) ug Cp0 ------ L
23.8
Vd (L)
84
------ Cp at 1 hour ug L
17.9
f ug Cpmax ------ L
Bioequivalence
23.8
Tmax (hr) Relative Bioavailability
1.1
Generic Equivalent (Yes / No)
YES
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
8-91
Bioavailability, Bioequivalence, and Drug Selection
8.7.9
“ISOSORBIDE DINITRATE” ON PAGE 69
Straehl, P. and Galeazzi, R., "Isosorbide dinitrate bioavailability , kinetics, and metabolism", Clinical Pharmacology and Therapeutics, Vol. 38m (1985), p. 140 - 149.
Isosorbide dinitrate is used in the treatment of angina pectoris, vasospastic angina, and congestive heart failure. In this study volunteers received a 5 mg intravenous dose and a 10 mg tablet. The different dosage forms were separated by a washout period. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg/kg)
5
10
10
370.3
158
165
487
310
305
1.32
1.96
1.85
0.65
0.53
1.546
1.875
60.1
66.2
0.21
0.22
60.4
67.8
1.12
0.90
0.81
0.90
ug AUC ------ ⋅ hr L 2 ug AUMC ------ ⋅ hr L
MRT (hr) MAT (hr) ke (hr-1)
0.760
ka (hr-1) ------ Cp0 ug L
282
Vd (L)
17.75
ug Cp at 1 hour ------ L
132
f ug Cpmax ------ L
Bioequivalence
282
Tmax (hr) Relative Bioavailability
1.04
Generic Equivalent (Yes / No)
YES
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
8-92
Bioavailability, Bioequivalence, and Drug Selection
8.7.10
“KETANSERIN” ON PAGE 70
Kurowski, M., "Bioavailability and pharmacokinetics of ketanserin in elderly subjects", Journal of Clinical Pharmacology, Vol. 28, (1988), p. 700 - 706.
Ketanserin is a 5-hydroxytryptamine S2-antagonist. This study focuses on the kinetics of Ketanserin in the elderly. Subjects were given either a 10 mg intravenous dose or a 40 mg oral tablet. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
10
40
40
ng- ⋅ hr AUC ------ mL
541
112.5
103.9
ng- ⋅ hr 2 AUMC ------ mL
11700
24900
22900
MRT (hr)
21.6
22.1
22.0
0.5
0.4
2.0
2.5
43.6
42.7
.052
0.48
47.6
44.5
0.94
1.93
1.63
0.84
MAT (hr) ke (hr-1)
Bioequivalence
0.0402
ka (hr-1) ng- Cp0 ------ mL
25.0
Vd (L)
400
ng Cp at 1 hour -------- mL
23.9
f ng Cpmax -------- mL
25.0
Tmax (hr) Relative Bioavailability
0.92
Generic Equivalent (Yes / No)
YES
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
8-93
Bioavailability, Bioequivalence, and Drug Selection
8.7.11
“METHOTREXATE” ON PAGE 71
Seideman, P., et al., " The pharmacokinetics of methotrexate and its 7-hydroxy metabolite in patients with rheumatoid arthritis", British Journal of Clinical Pharmacology, 35 (1993), p. 409 - 412.
The drug Methotrexate is a folic acid which has been shown to inhibit dihydrofolate reductase. The importance of this drug at present is mostly seen in the area of oncology, but lately it has been used for rheumatoid arthritis. Methotrexate has a molecular weight of 454.4. In this study, the drug was administered both by IV bolus and orally as a 15 mg dose. The following data was obtained: From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
15
15
15
nmole AUC ---------------- ⋅ hr L
2752
2708
2700
2 nmole AUMC ---------------- ⋅ hr L
15887
18400
18500
MRT (hr)
5.77
6.79
6.85
1.02
1.08
0.979
0.927
265
256
0.98
0.98
323
318
0.98
2.15
2.23
1.04
MAT (hr) ke
(hr-1)
0.173
ka (hr-1) nmole Cp0 ---------------- L
477
Vd (L)
69.3
nmole Cp at 1 hour ---------------- L
401
f ---------------- Cpmax nmole L
Bioequivalence
477
Tmax (hr) Relative Bioavailability
1.0
Generic Equivalent (Yes / No)
YES
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
8-94
Bioavailability, Bioequivalence, and Drug Selection
8.7.12
“MOCLOBEMIDE” ON PAGE 72
Schoerlin, M. et al., "Disposition kinetics of moclobemide, a new MAO-A inhibitor, in subjects with impaired renal function", Journal of Clinical Pharmacology, Vol. 30 (1991), p. 272 - 284.
Moclobemide is an antidepressant agent that reversibly inhibits the A-isozyme of the monoamine oxidase enzyme system. In this study, single IV and oral doses were administered to a patient. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
150
100
100
ug- ⋅ hr AUC ------ mL
2.58
1.70
1.52
ug- ⋅ hr 2 AUMC ------ mL
6.35
5.91
5.90
MRT (hr)
2.46
3.48
3.80
1.02
1.42
0.985
0.704
0.344
0.250
.099
.088
.037
0.29
.079
1.53
1.85
1.21
MAT (hr) ke (hr-1)
0.406
ka (hr-1) ug- Cp0 ------ mL
1.05
Vd (L)
143
ug Cp at 1 hour -------- mL
0.698
f ug Cpmax -------- mL
Bioequivalence
1.05
Tmax (hr) Relative Bioavailability
0.89
Generic Equivalent (Yes / No)
NO
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
8-95
Bioavailability, Bioequivalence, and Drug Selection
8.7.13
“NALBUPHINE” ON PAGE 73
Nalbuphine hydrochloride is an agonist-antagonist opiod which is used for its analgesic actions. In this study, volunteers were given single doses of four different nalbuphine forms. The data below focuses on a 10 mg iv dose and a 45 mg dose of an oral solution. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg)
10
45
40
40
ng- ⋅ hr AUC ------ mL
86.9
70.3
62.5
60
ng- ⋅ hr 2 AUMC ------ mL
288
306
280
270
MRT (hr)
3.31
4.35
4.48
4.5
1.04
1.17
1.19
0.963
0.858
0.843
11.1
9.2
8.7
0.180
0.180
0.173
12.5
10.7
10.2
0.95
1.76
1.88
1.90
1.01
MAT (hr) ke (hr-1)
.0301
ka (hr-1) ng- Cp0 ------ mL
26.2
Vd (L)
381
ng Cp at 1 hour -------- mL
19.4
f ng Cpmax -------- mL
Bioequivalence
26.2
Tmax (hr) Relative Bioavailability
0.96
Generic Equivalent (Yes / No)
YES
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
8-96
Bioavailability, Bioequivalence, and Drug Selection
8.7.14
“NEFAZODONE” ON PAGE 74
Shukla, U. et al., "Pharmacokinetics, absolute bioavailability, and disposition of nefazodone in the dog", Drug Metabolism and Disposition, Vol. 21, No. 3, (1993), p. 502 - 507.
Nefazodone was given to four healthy, adult, male beagles with an average weight of 11.0 kg. Each dog was given a 10 mg/kg dose as a either a intravenous injection or as an oral solution or tablet. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg/kg)
10
10
10
10
ng- ⋅ hr AUC ------ mL
6023
829
800
700
ng- ⋅ hr 2 AUMC ------ mL
29283
4875
4800
4500
MRT (hr)
4.86
5.88
6.0
6.43
1.02
1.14
1.57
0.982
0.879
0.638
94.8
85.7
60.7
0.138
0.133
0.116
112.7
105.6
84.0
0.80
2.0
2.16
2.62
1.21
MAT (hr) ke (hr-1)
0.210
ka (hr-1) ng- Cp0 ------ mL
1238
Vd (L)
8.07
ng Cp at 1 hour -------- mL
1009
f ng Cpmax -------- mL
Bioequivalence
1238
Tmax (hr) Relative Bioavailability
0.88
Generic Equivalent (Yes / No)
YES
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
8-97
Bioavailability, Bioequivalence, and Drug Selection
8.7.15
“ONDANSETRON” ON PAGE 75
Colthup, P., et al., "Determination of ondansetron in plasma and its pharmacokinetics in the young and elderly", Journal of Pharmaceutical Sciences, Vol. 80, No. 9(1991), p. 868 - 871.
Ondansetron is a 5-hydroxyltryptamine compound which is useful in treating the nausea and vomiting which is caused by the use of chemotherapy and radiation in the cancer patients. In order to determine the absolute bioavailability of oral Ondansetron, doses of 8 mg were given to two groups. One group received an oral dose and the other group received an intravenous dose. A summary of the some of data obtained from this experiment is given below.
From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
8
8
8
ng- ⋅ hr AUC ------ mL
246.5
139
145
ng- ⋅ hr 2 AUMC ------ mL
1138
795
870
MRT (hr)
4.62
5.72
6.0
1.10
1.38
0.907
0.723
15.9
14.7
0.56
0.59
19.2
18.8
2.1
2.4
MAT (hr) ke
(hr-1)
0.217
ka (hr-1) ng- Cp0 ------ mL
53.4
Vd (L)
150
ng Cp at 1 hour -------- mL
43
f ng Cpmax -------- mL
Bioequivalence
53.4
Tmax (hr)
1.1
Relative Bioavailability
1.04
Generic Equivalent (Yes / No)
YES
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
8-98
Bioavailability, Bioequivalence, and Drug Selection
8.7.16
“OMEPRAZOLE” ON PAGE 76
Anderson, T., et al, "Pharmacokinetics of various single intravenous and oral doses of omeprazole", Eur Journal of Clinical Pharmacology, 39, (1990), p. 195 - 197.
Omeprazole (mw: 345.42) is an agent which inhibits gastric acid secretion from the parietal cell. It is useful in treating such problems as ulcers and gastroesophageal reflux disease. One group received an iv bolus dose and the other group received an oral dose. A summary of the some of data obtained from this experiment is given below.
From the preceding data, please calculate the following:
Parameter
IV
Brand Capsule
Generic Capsule
Dose (mg)
20
40
40
mole ⋅ hr AUC µ--------------- L
3.2
3.5
3.0
2 µ mole ⋅ hr AUMC --------------- L
3.2
5.25
4.5
MRT (hr)
1.0
1.5
1.5
0.5
0.5
2
2
1.63
1.40
0.55
0.47
1.8
1.5
0.86
0.69
0.69
1
MAT (hr) ke
(hr-1)
1
ka (hr-1) µ mole Cp0 ---------------- L
3.2
Vd (L)
52.4
mole Cp at 1 hour µ--------------- L
1.18
f µ mole Cpmax ---------------- L
Bioequivalence
3.2
Tmax (hr) Relative Bioavailability
0.86
Generic Equivalent (Yes / No)
YES
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
8-99
Bioavailability, Bioequivalence, and Drug Selection
8.7.17
“PAROXETINE” ON PAGE 77
Lund, J., et al., "Paroxetine: pharmacokinetics and cardiovascular effects after oral and intravenous single doses in man", Journal of Pharmacology and Toxicology, Vol. 51, (1982), p. 351 - 357.
Paroxetine kinetics and cardiovascular effects were studied in male subjects after single oral doses of 45 mg and slow intravenous infusion of 28 mg. A summary of the some of data obtained from this experiment is given below.
From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
28
45
45
ng AUC -------- ⋅ hr mL
467
750
675
2 ng AUMC -------- ⋅ hr mL
6671
11250
10463
MRT (hr)
14.3
MAT (hr) ke (hr-1)
15
15.5
0.72
1.22
1.40
.082
37.9
25.5
1
0.90
44.8
37.6
0.84
2.25
3.27
1.45
0.07
ka (hr-1) ng Cp0 -------- mL
32.7
Vd (L)
856
ng- Cp at 1 hour ------ mL
30.5
f ng Cpmax -------- mL
Bioequivalence
32.7
Tmax (hr) Relative Bioavailability
0.90
Generic Equivalent (Yes / No)
NO
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Bioavailability, Bioequivalence, and Drug Selection
8.7.18
“RANITIDINE” ON PAGE 78
Garg, D., et al., "Pharmacokinetics of ranitidine in patients with renal failure", Journal of Clinical Pharmacology, Vol. 26 (1986), p. 286 - 291.
Ranitidine is an agent used in the treatment of peptic ulceration. In this study, ten patients with renal failure received either a 50 mg intravenous bolus dose or a 150 mg tablet. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Brand Tablet
Generic Tablet
Dose (mg)
50
150
150
ng- ⋅ hr AUC ------ mL
5159
6422
6753
ng- ⋅ hr 2 AUMC ------ mL
53415
78752
84413
MRT (hr)
10.4
12.3
12.5
1.91
2.15
0.524
0.466
240
231
.0415
0.436
423
432
1.02
3.96
4.26
1.07
MAT (hr) ke (hr-1)
Bioequivalence
0.0966
ka (hr-1) ng- Cp0 ------ mL
498
Vd (L)
100
ng Cp at 1 hour -------- mL
452
f ng Cpmax -------- mL
498
Tmax (hr) Relative Bioavailability
1.05
Generic Equivalent (Yes / No)
YES
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Bioavailability, Bioequivalence, and Drug Selection
8.7.19
“SULPIRIDE” ON PAGE 79
Bressolle, F., Bres, J., and Faure-Jeantis, A., "Absolute bioavailability , rate of absorption, and dose proportionality of sulpiride in humans", Journal of Pharmaceutical Sciences ,Vol. 81, No. 1 (1992), p. 26 - 32.
Sulpiride is a substituted benzamine antipsychotic. In this study, the drug was administered to two groups. The first group received a 200 mg oral dose and the second group received a 100 mg intravenous infusion. A summary of the some of data obtained from this experiment is given below. From the preceding data, please calculate the following:
Parameter
IV
Oral Solution
Brand Tablet
Generic Tablet
Dose (mg)
100
200
200
200
ug- ⋅ hr AUC ------ mL
8.27
8.79
8.6
8.0
ug- ⋅ hr 2 AUMC ------ mL
79.1
87.3
91.1
84.5
MRT (hr)
9.56
MAT (hr) ke (hr-1)
9.93
10.6
10.6
0.367
1.02
1.0
2.72
0.972
1.0
0.798
0.526
0.498
0.53
0.52
0.48
0.807
0.687
0.643
0.94
1.24
2.57
2.52
0.98
0.865
ka (hr-1) ug- Cp0 ------ mL
0.865
Vd (L)
116
ug Cp at 1 hour -------- mL
0.779
f ug Cpmax -------- mL
Bioequivalence
0.865
Tmax (hr) Relative Bioavailability
0.93
Generic Equivalent (Yes / No)
YES
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Bioavailability, Bioequivalence, and Drug Selection
8.8 References 1.
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Basic Pharmacokinetics
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Bioavailability, Bioequivalence, and Drug Selection
25.
Blanchard, J. and Sawchuk, R.J., Drug Bioavailability: An Overview, in Principles and Perspectives in Drug Bioavailability, Blanchard, J., Sawchuk, R.J. and Brodie, B.B., ed., Karger, Basel, p. 1-19, 1979.
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Toothaker, R.D. and Welling, P.G., The Effect of Food on Drug Bioavailability, Ann. Rev. Pharmacol. Toxicol., 20:173-179, 1980.
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Neuvonen, P.J., Pentikainen, P.J., and Elfing, S.M., Factors Affecting the Bioavailability of Phenytoin, Int. J. Clin. Pharmacol. Biopharm., 15:84, 1977.
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Heading, R.C., et al., The Dependence of Paracetamol Absorption on the Rate of Gastric Emptying, Br. J. Pharmacol., 47:415, 1973.
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Mayersohn, M., in Principles and Perspectives in Drug Bioavailability, Blanchard, J., Sawchuk, R.J. and Brodie, B.B., ed., S. Karger, Basel, p. 211, 1979.
Basic Pharmacokinetics
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Bioavailability, Bioequivalence, and Drug Selection
44.
Gibaldi, M., Drug Interactions, The Annals of Pharmacotherapy, 26:829-834, 1992.
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Welling, P.G., Huang, H., Hewitt, P.F., and Lyons, L.L., Bioavailability of Erythromycin Stearate: Influence of Food and Fluid Volume, J. Pharm. Sci., 67:764-766, 1978.
46.
Mischler, T.W., Sugerman, A.A., Willard, D.A., Brannick, L.J., and Neiss, E.S., Influence of Probenecid and Food on the Bioavailability of Cephradine in Normal Subjects, J. Clin. Pharmacol. 14:604-611, 1974.
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Melander, A., Danielson, K., Schersten, B., Wahlin, E., Clin. Pharmacol. Ther., 22:108-112, 1977.
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McGilveray, I., Consensus Report on Issues in the Evaluation of Bioavailability, Pharm.Res., 8:136-138, 1991.
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Endrenyi, L., Fritsch, S. and Yan, W., Cmax/AUC is a Clearer Measure than Cmax for Absorption Rates in Investigation of Bioequivalence, Int. J. Clin. Pharmacol. Therap. Toxicol., 29:394-399, 1991.
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Steinijans, V.W., Sauter, R., Jonkman, J.H.G., Schulz, H.U., Stricker, H., and Blume, H., Bioequivalence Studies: Single vs. Multiple Dose, Int. J. Clin. Pharmacol. Therap. Toxicol., 27:261-266, 1989.
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Wartak, J., Clinical Pharmacokinetics, Praeger, New York, New York, p. 154, 1983.
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Tse, F.L.S., Robinson, W.T. and Choc, M.G. Study Design for the Assessment of Bioavailability and Bioequivalence in Pharmaceutical Bioequivalence, Welling, A.G., Tse, F.L.S. and Dighe, S.V., ed., Marcel Dekker Inc., New York, New York, p. 17-34, 1991.
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Basic Pharmacokinetics
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The
8-105
Bioavailability, Bioequivalence, and Drug Selection
64.
Guidelines on the design of a single-dose in-vivo bioavailability study, Bioavailability and Bioequivalence Requirements, 21 CFR 320: 26, 1991.
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USP XXII/NF XVII, United States Pharmacopeial Convention, Inc., Rockville, Maryland, p. 1578-1579, 1990.
67.
Banakar, U.V., Factors that Influence Dissolution Testing, in Pharmaceutical Dissolution Testing, Marcel Dekker, Inc., New York, New York, p. 135, 1992.
68.
Welling, P.G., In-Vitro Methods to Determine Bioavailability: In Vitro-In-Vivo Correlations, in Pharmaceutical Bioequivalence, Welling, P.G., Tse, F.L.S. and Dighe, S.V., ed., Marcel Dekker, Inc., New York, New York, p. 224, 1991.
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PMA's Joint Committee on Bioavailability, The Role of Dissolution Testing in Drug Quality, Bioavailability, and Bioequivalence Testing, Pharm. Technol., 9:62-66, 1985.
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Welling, P.G., In Vitro Methods to Determine Bioavailability: In Vitro-In Vivo Correlations, in Pharmaceutical Bioequivalence, Welling, P.G., Tse, F.L.S., and Dighe, S.V., ed., Marcel Dekker, Inc., New York, New York, p. 225- 232, 1991.
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Sullivan, T.J., Sakmar, E. and Wagner, J.G., Comparative Bioavailability: A New Type of In Vitro-In Vivo Correlation Exemplified by Prednisone, J. Pharmacokin. Biopharm., 4:173-181, 1976.
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Yau, M.K.T. and Meyer, M.C., In Vivo-In Vitro Correlations With a Commercial Dissolution Simulator I: Methenamine, Nitrofurantoin, and Chlorothiazide, J. Pharm. Sci., 70:1017-1023, 1981.
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Milsap, R.L., Ayres, J.W., Mackichan, J.J., and Wagner, J.G., Comparison of Two Dissolution Apparatuses with Correlations of In Vitro-In Vivo Data for Prednisone and Prednisolone Tablets, Biopharm. Drug Disp., 1:3-17, 1979.
75.
Levy, G., Leonards, J. R. and Procknal, J.A., Development of In Vitro Dissolution Tests Which Correlate Quantitatively with Dissolution Rate- Limited Drug Absorption in Man., J. Pharm. Sci., 54:1719-1722, 1965.
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Yau, M.K.T. and Meyer, M.C., In Vivo-In Vitro Correlations with a Commercial Dissolution Simulator II: Papaverine, Phenytoin, and Sulfisoxazole, J. Pharm. Sci., 72:681-686, 1983.
77.
Shah, V.P., Prasad, V.K., Alston, T., Cabana, B.E., Gural, R.P. and Meyer, M.C., Phenytoin I: In Vitro-In Vivo Correlation for 100 mg. Phenytoin Sodium Capsules, J. Pharm. Sci., 72:306-308, 1983.
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Shah, A.C. and Dakkuri, A., Correlation of In Vitro Rate of Dissolution with In Vivo Bioavailability: An Overview, Pharm. Technol., Sep:67-97, 1982.
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Wagner, J.G., Rate of Dissolution In Vitro and In Vivo Part VI: Correlation of In Vivo with In Vitro Data - Theoretical and Practical Considerations, Drug Intell. Clin. Pharm., 4:160-163, 1970.
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Basic Pharmacokinetics
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8-106
Bioavailability, Bioequivalence, and Drug Selection
82.
Banakar, U.V., Pharmaceutical Dissolution Testing, Marcel Dekker, Inc., New York, New York, p. 358-382, 1992.
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Definitions, Bioavailability and Bioequivalence Requirements, 21 CFR, 320.1, 1991.
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Morrison, A.B., Chapman, D.G. and Campbell, J.A., Futher Studies on the Relation Between In Vitro Disintegration Time of Tablets and the Urinary Excretion Rates of Riboflavin, J. Am. Pharm. Assoc. Sci. Ed., 48:634-637, 1959.
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Morrison, A.B. and Campbell, J.A., The Relationship Between Physiological Availability of Salicylates and Riboflavin and In Vitro Disintegration Time of Enteric Coated Tablets, J. Am. Pharm. Assoc. Sci. Ed., 49:473-478, 1960.
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Middleton, E.J., Davies, J.M., and Morrison, A.B., Relationship Between Rate of Dissolution, Disintegration Time, and Physiological Availability of Riboflaving in Sugar-coated Tablets, J. Pharm. Sci., 53:1378-1380, 1964.
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Levy, G., Comparison of Dissolution and Absorption Rates of Different Commercial Aspirin Tablets, J. Pharm. Sci, 50:388-392, 1961.
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Levy, G., Effect of Dosage Form Properties on Therapeutic Efficacy of Tolbutamide Tablets, Can. Med. Assoc. J, 90:978-979, 1964.
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MacDonald, H., Pisano, F. and Burger, J., Physiologic Availability of Various Tetracyclines, Clin. Med., Dec:30-33, 1969.
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Koch-Weser, J., Bioavailability of Drugs, Med. Intell., 291:503-506, 1974.
Basic Pharmacokinetics
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8-107
Bioavailability, Bioequivalence, and Drug Selection
99.
Lindenbaum, J., Mellow, M.H., Blackstone, M.O. and Butler, V.P., Variation in Biologic Availability of Digoxin from Four Preparations, New Engl. J. Med., 285:1344-1347, 1971.
100.
Report of the Ad Hoc Committee on Drug Product Selection of the Academy of General Practice of Pharmacy and the Academy of Pharmaceutical Sciences, J. Am. Pharm. Assoc., NS13(6):278-280, 1973.
101.
Office of Technology Assessment, Drug Bioequivalence Study Panel, Drug Bioequivalence, U.S. Government Printing Office, Washington, D.C., 1974.
102.
Bioavailability and Bioequivalence Requirements, Federal Register, 42:1624- 1653, 1977.
103.
Cabana, B.E., Bioavailability/Bioequivalence, Food Drug Cosm. Law J., 32:512-526, 1977.
104.
Williams, R.L., Bioequivalence and Therapeutic Equivalence, in Pharmaceutical Bioequivalence, Welling, P.G,. Tse, F.L.S., and Dighe, S.V., eds., Marcel Dekker, Inc., New York, p. 1-14, 1991.
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Henderson, J.D., Dighe, S.V., Williams, R.L., Subject Selection and Management in Bioequivalence Studies, Clin. Research Reg. Affairs, 9:71- 87, 1992.
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Cadwallader, D.E., Biopharmaceutics and Drug Interactions, Third Edition, Raven Press, New York, P. 87-106, 1983.
107.
Westlake, W.J., Design and Statistical Evaluation of Bioequivalence Studies in Man., in Principles and Perspectives in Drug Bioavailability, Blanchard, J., Sawchuk, R.J. and Brodie, B.B., ed., Karger, Basel, p. 192-210, 1979.
108.
Dighe, S.V. and Adams, W.P., Bioequivalence: A United States Regulatory Perspective, in Pharmaceutical Bioequivalence, Welling, P.G., Tse, F.L.S., and Dighe, S.V., eds., Marcel Dekker, Inc., New York, p. 347-380, 1991.
109.
Guidelines on the Design of a Multiple-Dose In-Vivo Bioavailability Study, in Bioavailability and Bioequivalence Requirements, 21 CFR 320.27.
110.
Bio-International '92, Conference on Bioavailability, Bioequivalence and Pharmacokinetics Studies, Pharm. Res., 10:1806-1811, 1993.
111.
FDA Bioequivalence Task Force Report Conclusions, F-D-C Reports, p. 15, Feb. 15, 1988.
112.
Steinijans, V.W., Hauschke, D., and Jonkman, J.H.G., Contoversies in Bioequivalence Studies, Clin. Pharmacokinet., 22:247-253, 1992.
113.
Consensus Report on "Issues in the Evaluation of Bioavailability", Pharm. Res., 8:136-138, 1991.
114.
Meyer, M.C., Scientific Basis of Bioavailability and Bioequivalence Testing, Am. Pharm., NS31(8):47-52, 1991.
115.
Schuirmann, D.J., A Comparison of the Two One-Sided Tests Procedure and the Power Approach for Assessing the Equivalence of Average Bioavailability, J. Pharmacokinet. Biopharm., 15:657-680, 1987.
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Westlake, W.J., Use of Confidence Intervals in Analysis of Comparative Bioavailability Trials, J. Pharm. Sci., 61:1340-1341, 1972.
117.
Westlake, W.J., Bioavailability and Bioequivalence of Pharmaceutical Formulations, in Biopharmaceutical Statistics for Drug Development, Peace, K.E., ed., Marcel Dekker, Inc., New York, p. 329-352, 1988.
Basic Pharmacokinetics
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Bioavailability, Bioequivalence, and Drug Selection
118.
Metzler, C.M., Statistical Criteria, in Pharmaceutical Bioequivalence, Welling, P.G., Tse, F.L.S., and Dighe, S.V., eds., Marcel Dekker, Inc., New York, p. 35-67, 1991.
119.
Rescigno, A., Bioequivalence, Pharm. Res., 9:925-928, 1992.
120.
Additional Generics on the Market, FDA Drug Bulletin, Nov:14-15, 1986.
121.
Mattison, N., Pharmaceutical Innovation and Generic Drug Competition in the U.S.A.: Effects of the Drug Price Competition and Patent Term Restoration Act of 1984, Pharmaceut. Med., 1:177-185, 1986.
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Madan, P.L., Bioavailability and Bioequivalence, The Underlying Concepts, U.S. Pharm., 17 (Nov. Hosp. Ed.):H10-H30, 1992.
123.
Haynes, J.D., FDA 75/75 Rule: A Response, J. Pharm. Sci., 72:99-100, 1983.
124.
Anderson, S. and Hauck, W.W., Consideration of Individual Bioequivalence, J. Pharmacokinet. Biopharm., 18:259-273, 1990.
125.
Carter, M.G. and Sanderson, J.H., Generic Prescribing and Clinical Bioinequivalence, Pharm. J., Nov. 26:683, 1988.
126.
Somberg, J. and Sonnenblick, E., Perspective: The Bioequivalence of Generic Drugs, Cardiovascular Rev. Rep., 6:1010-1015, 1985.
127.
Gore, M.J., Cost, Safety & Efficacy: Defining the Pharmacist's Role in Drug Product Selection, Consultant Pharm., 6:771-789, 1991.
128.
Somberg, J.C., Bioequivalence or Therapeutic Equivalence, J. Clin. Pharmacol., 26:1, 1986.
129.
Garrett, E.R., Weinstein, C.B.A., (Commentaries), Integr. Psychiatry, 3:89S- 96S, 1985.
130.
Covington, T.R., Generic Drug Utilization: Overview and Guidelines for Prudent Use, Clin. Research Reg. Affairs, 9:103-126, 1992.
131.
Shah, H.K., Generics Capture New Prescription Markets, Perspectives in Pharmacy Economics, 4:3, 1992.
132.
Glaser, M., On the Move Again, Drug Topics Supplement, 6S-12S, 1993.
133.
Generic Survey, Amer. Druggist, 208(3):36-41, 1993.
134.
Major Events During the Generic Drug Investigations, Am. Pharm., NS30(7):38-39, 1990.
135.
Drugs: Still Safe?, Consumer Reports, May:310-313. 1990.
136.
Conlan, M.F., More Charges Coming in Generic Industry Probe, Drug Topics, 135(2):56-57, 1991.
137.
Yorke, J., FDA Ensures Equivalence of Generic Drugs, FDA Consumer, Sept:11-15, 1992.
138.
Thompson, L.R., After the Scandals: New Generic Counseling, Am. Pharm., NS30(7):31-33, 1990.
139.
Conlan, M.F., After the Storm, Drug Topics, 135(12):40-44, 1991.
140.
Martin, S., Generic Drug Scandals Raise Questions About Safety, Am. Pharm., NS29(10):23-24, 1989.
141.
Spalding, B.J., The Generic Industry: One Year Later, Am. Drug., 202(3):14- 16, 1990.
142.
Loudin, A., Fallout From the Generic Drug Scandal, Pharm. Update, 2(5):1-9, 1991.
143.
Pal, S. and D'Angelo, A.C., More Patient Questions, More Brand Name Rxs, U.S. Pharm., 15(3):66-70, 1990.
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Bioavailability, Bioequivalence, and Drug Selection
144.
Consumer Confidence in Generic Drug Products Down in Wake of Industry Scandal, But Satisfaction with Products Remains High, Survey Reveals, A.J.H.P., 47:468, 1990.
145.
Pal, S. and D'Angelo, A.C., FDA Reputation Suffers, Pharmacists Ask for More Product Information, U.S. Pharm., 15(4):20-25, 1990.
146.
Gross, L.H., Can Pharmacists, Physicians, and the Public Trust Generics? Am. Druggist, 200(3):25-28, 1989.
147.
Most Generics are Safe, But Approval Process Remains in 'Disrepair', Am. Pharm., NS30(1):14, 1990.
148.
Conlan, M.F., Getting it Back on Track, Drug Topics, 134(3):42-47, 1990.
149.
Cardinale, V., Generics: Where Now? Drug Topics (Supplement):3S, 1993.
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Conlan, M.F., Future is Sunny for Generics as Popular Rxs Come Off Patent, Drug Topics, 134 (20):14-19, 1990.
151.
Starr, C., Which Drugs Are Going Off Patent or Losing Exclusivity?, Drug Topics (Supplement):14S-15S, 1993.
152.
Summers, K.H., Counseling Patients Who Take Generic Drug Products, Drug Topics (Supplement):45S-54S, 1993.
153.
Laskoski, G., Generics: Too Good For Their Own Good, Am. Druggist, 208(3):30-35, 1993.
154.
Shafermeyer, K.W., Schondelmeyer, S.W., and Wilson, G.T., The FDA Orange Book: Expectations Versus Realities, J. Pharm. and Law, 1:13-26, 1992.
155.
Chappell, S.C., Pharmacists Now Control 27% of Dispensing Decisions Involving New Rxs, Pharm. Times, 54(10):55-62, 1988.
156.
Simonsen, L., Rx Dispensing Trends: Pharmacists Make More Selection Decisions, Pharm. Times, 57(10):53-59, 1991.
157.
Simonsen, L., Rx Brand Choice: Pharmacists Decide for 35% of All New Rxs, Pharm. Times, 58(10):92-98, 1992.
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Simonsen, L., Generic Prescribing and RPh Substitution Continue to Climb, Pharm. Times, 59(10):29-30, 1993.
159.
D'Angelo, A.C., How To Win Pharmacists and Influence Purchasing, U.S. Pharm., 16(6):35-36, 1991.
160.
Charupatanapong, N. and Rascati, K.L., Pharmacists' Satisfaction With Drug Product Selection Legislation, Am. Pharm., NS28(10):27-32, 1988.
161.
Smith, M., Monk, M. and Banahan, B., Factors Influencing Substitution Practices, Am. Druggist, 203(5):88-96, 1991.
162.
Segal, R., Wantz, D.L. and Brusadin, R.A., Pharmacists' Decision Making in the Selection of Generic Pharmaceuticals, J. Pharm. Market. Mngment., 4(1):75-91, 1989.
163.
McCormack, J.P., Levine, M. and Miller, P., Bioequivalence: Just the Facts Please, CPJ-RPC, Sep:404-407, 1990.
164.
Guidelines for Pharmacists Performing Product Selection, Pharm. Today, Feb. 2:4-5, 1990.
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Gagnon, J.P., Key Factors and Cues for Evaluating Pharmaceutical Manufacturers, Pharm. Times, 56(4):45-48, 1990.
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Bioavailability, Bioequivalence, and Drug Selection
166.
Feinberg, J.L., A Pharmacist's Survival Guide to the Generic Drug Scandal, Consult. Pharm., 5(1):15-23, 1990.
167.
Foster, T.S., Selecting Therapeutically Equivalent Products: Special Cases, Am. Pharm., NS31(11):49-54, 1991.
168.
Ross, M.B., Status of Generic Substitution: Problematic Drug Classes Reviewed, Hosp. Formul., 24:441-449, 1989.
169.
Sanderson, J.H. and Lewis, J.A., Differences in Side-Effect Incidence in Patients on Proprietary and Generic Propranolol, The Lancet, Apr. 26:967- 968, 1986.
170.
Ansbacher, R., Conjugated Estrogens: Do Not Substitute, Am. Pharm., NS30(7):27-28, 1990.
171.
Locniskar, A., Greenblatt, D.J., Harmatz, J.S., and Shader, R.I., Bioinequivalence of a Generic Brand of Diazepam, Biopharm. Drug Disp., 10:597-605, 1989.
172.
Rentmeester, T.W., Doelman, J.C., and Hulsman, J.A.R.J., Carbamazepine: Merkpreparaat en Generiek, Pharm. Weekbl., 125(43):1108-1110, 1990.
173.
Welty, T.E., Pickering, P.R., Hale, B.C. and Arazi, R., Loss of Seizure Control With Generic Substitution of Carbamazepine, Annals Pharmacotherapy, 26:775-776, 1992.
174.
Grubb, B.P., Recurrence of Ventricular Tachycardia After Conversion From Proprietary to Generic Procainamide, Am. J. Cardiol., 63:1532-1533, 1989.
175.
Cunha, B.A., Nitrofurantoin - Bioavailability and Therapeutic Equivalence, Adv. Therapy, 5(3):54-63, 1988.
176.
Nuwer, M.R., Correspondence, Neurology, 41:1165, 1991.
177.
Check, W.A., Caution Urged in Prescribing of Generic Antiarrhythmic Drugs, Consult. Pharm., 5(11):718-721, 1990.
178.
Fincham, J.E., Therapeutic Failure with Generic Hydrochlorothiazide- Triamterene in an Elderly Female: A Case Report, J. Ger. Drug Therapy, 5(4):85-89, 1991.
179.
Fincham, J.E., Therapeutic Failure with Generic Clonidine in An Elderly Female: A Case Report, J. Ger. Drug Therapy, 3(1):83-87, 1988.
180.
Ansbacher, R., Bioequivalence of Conjugated Estrogen Products, Clin. Pharmacokinet., 24(4):271-274, 1993.
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8-111
CHAPTER 9
Clearance
OBJECTIVES 1.
Given patient information regarding organ function, the student will calculate (III) changes in clearance and other pharmacokinetic parameters inherent in compromised patients.
2.
Determine the total clearance based on Dose and AUC.
3.
Determine clearance of an organ based on dose, AUC, and fraction of drug eliminated by the organ
4.
Determine change in clearance due to functional changes in an organ.
5.
Determine change in clearance due to change in blood flow through an organ.
6.
Prepare a professional consult (V) and justify (VI) modifications in drug therapy based on clearance of a drug.
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9-1
Clearance
9.1 Equations Rate of Elimination Cl = --------------------------------------------------Serum Concentration
(EQ 9-1)
⋅ D os Cl tot = f--------------AUC
(EQ 9-2)
Cl r = Cl tot ⋅ ( Fraction of drug that is renally eliminated )
(EQ 9-3)
Cl H = Cl tot ⋅ ( Fraction of drug that is hepatically eliminated )
(EQ 9-4)
L min L Q r = 0.0191 -------------------- renal blood perfusion ⋅ 70kg ⋅ 60 --------- ≈ 80 -----blood min ⋅ kg hr hr
(EQ 9-5)
L min L Q H = 0.0238 -------------------- hepatic blood perfusion ⋅ 70kg ⋅ 60 --------- ≈ 100 -----blood min ⋅ kg hr hr
(EQ 9-6)
Er = ( Cl r ) ⁄ Q r
(EQ 9-7)
E H = ( Cl H ) ⁄ Q H
(EQ 9-8)
Q ⋅ Cl f u ⋅ Cl int = ---------------Q – Cl
(EQ 9-9)
Cl int
Q ⋅ Cl--------------Q – Cl = ---------------fu
(EQ 9-10)
f u∗ ⋅ Cl∗ int F i = ------------------------f u ⋅ Cl int
(EQ 9-11)
∗ FR = Q ------Q
(EQ 9-12)
Fi ⋅ FR F Cl = ----------------------------------------F R + Er ( Fi – FR )
(EQ 9-13)
∗ H + Cl ∗ r Cl∗ tot k∗ ⋅ V∗ = Cl FCl tot = ------------- = -------------------------------------------Cl tot k⋅ V Cl H + Cl r
(EQ 9-14)
0.80 ≤ FCl t o t ≤ 1.20
(EQ 9-15)
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9-2
Clearance
9.2 Definitions and Terms Clearance: The hypothetical volume of a fluid from which a substance is totally and irreversibly removed per unit time. Dimensions:
3
L ⁄T
Examples of fluids: blood, serum, plasma, bile, gut contents, CSF. Systemic or Total Body Clearance: Removal process is elimination (excretion and metabolism). Fluid is usually plasma or serum (rarely blood). ( Cl ) , ( Cl tot )
Renal Clearance: Removal process is urinary excretion of unchanged drug. Fluid is usually plasma or serum (rarely blood). ( Cl r )
Metabolic Clearance: Removal process is metabolism. Fluid is usually blood (rarely plasma or serum). ( Cl m )
( Cl H )
Hepatic Clearance: This is
Cl m
when the liver is the metabolic
organ. Creatinine Clearance: This is Cl r applied to endogenous creatinine. It is used to monitor renal function, and thus is a valuable parameter for calculating dosage regimens in elderly patients or those suffering from renal dysfunction. Creatinine t 1 ⁄ 2 = 231min ( Cl cr )
Value for normal males: 117 ± 20 ml/min Value for normal females: 108 ± 20 ml/min Cl inulin
Inulin Clearance: This is
Cl r
for inulin, and yields the glomerular
filtration rate. Value for normal males: 124.5 ± 9.7 ml/min Value for normal females: 108.8 ± 13.5 ml/min
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9-3
Clearance
9.3 Measurement of Creatinine Clearance The mass of endogenous creatinine excreted into the urine collected over a given time interval ( ∆ t ) is determined. The mean serum creatinine concentration ( Cs ) cr over that interval is calculated from sample determinations; this should be the concentration halfway through the interval. In practice, ∆ t = 24hr and, as ( C s ) cr is relatively constant, the serum sample is taken at any convenient time. Let “a” be a volume of serum having a creatinine concentration of ( C s ) cr . The mass of creatinine in the serum will be a ⋅ ( Cs ) cr . If this creatinine is totally and irreversibly removed from the serum to the urine in the time interval, ∆ t , then a ⋅ ( C s ) cr = ∆ Xu
(EQ 9-16)
∆ Xu a = -------------( C s ) cr
(EQ 9-17)
Thus,
The volume of serum from which this creatinine is removed in unit time is this is the definition of clearance. Hence, ( ∆ Xu ⁄ ∆ t ) a = -------------------------T Cl cr = ----∆t ( C s ) cr
a ⁄ ∆t ;
(EQ 9-18)
Siersbaek-Neilson et al. report a value of 11.1 µ g ⁄ ml for ( Cs ) cr in 149 males (aged 20-99). The value of ( ∆ X u ⁄ ∆ t ) T decreased with age from 16.53 µ g ⁄ min per Kg body weight (age 20-29) to 6.53
µg ⁄
min
per Kg body weight (age 90-99). For a 25 year
old 70Kg male, equation 9-18 yields
ml Cl cr = 104.2 --------min
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9-4
Clearance
9.4 Model Correlations Although intrinsically model independent, clearance can also be related to compartmental models.
9.4.1
RENAL CLEARANCE The plasma renal clearance of a drug may be measured analogously to creatinine clearance: ( ∆ Xu ⁄ ∆ t ) Cl r = --------------------------TCp
(EQ 9-19)
The practical versions of and states: ( ∆ Xu ⁄ ∆ t )
T
= k u X = k u VC p
(EQ 9-20)
Comparing equation 9-19 and equation 9-20, Cl r = k u V
(EQ 9-21)
This relates clearance to model parameters. What is the slope of a plot of ( ∆ Xu ⁄ ∆ t ) T against
Cp ?
Note that if Cl r > 117 ± 20ml ⁄ min (males), it may indicate active secretion of the drug into the kidney tubules. If Cl r < 108 ± 20ml ⁄ min (females), it may indicate reabsorption of the drug from the kidney tubules.
9.4.2
SYSTEMIC CLEARANCE AND METABOLIC CLEARANCE
How could you measure Cl ? and Cl m
By analogy, 0.693V Cl tot = KV = ----------------t1 ⁄ 2
and
K = ku + km
, so
Cl tot = Cl r + Cl m
(EQ 9-22)
.
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9-5
Clearance
°
F Cl
tot
°
° ° Cl h + Cl r K ⋅V = ------------------------ = ---------------K⋅V Cl h + Cl r
(EQ 9-23)
where X° is new or altered variable. Hepatic function and renal function are not a priori connected, although some physiological functional changes might result in similar changes in clearance of both organs. We can see from equation 9-12 that changes in total body clearance can result in changes in either K, V, or both. The consequences of that will be discussed in the section on dosage regimens.
9.4.3
USE IN PHARMACOKINETIC EQUATIONS Systemic Clearance (Cl) can be used in many equations where the drug is removed by elimination (renal excretion and metabolism). If renal excretion is the only removal process, use Cl r l if metabolism, use Cl m . Some examples: Intravenous infusion:
( Cp )
ss
Q= ----Cl
Oral and Intravenous Bolus:
f( Xa) fD fD Cl = ---------------0 = --------------- = -----------∞ ∞ AUC ∫ C p dt ∫ C p dt 0
0
This equation becomes a means of calculating Cl from plasma data. Dosage Regimen: ( Cp ) ss
fD = -----------τ ⋅ Cl
These examples are all model-independent expressions, which are very useful in calculating dosage regimens. The importance of clearance terms rests on their ability to account for variations in both t 1 ⁄ 2 and / or V simultaneously, as both these parameters can change in disease states and with age.
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9-6
Clearance
9.5 Physiological Factors Affecting Clearance 9.5.1
INTRINSIC CLEARANCE ( Cl int ) Intuitively, it may be recognized that two factors will affect the clearance of a drug: 1. The rate at which blood is presented to the eliminating organ. 2. The intrinsic ability of the eliminating organ to clear the drug. Mathematically, a hyperbolic equation has been derived to illustrate the relative effect of these factors. (Note: this is one model of clearance. There are several others which also illustrate the effect of these factors.)
Liver Drug Metabolism
QH ⋅ f u ⋅ ( Cl H ) int Cl H = -----------------------------------------Q H + f u ⋅ ( Cl H ) int
(EQ 9-24)
Where Q H is the rate of blood flow through the liver (assumed 23.8 ml/min/Kg body weight in normal adult), fu is the fraction unbound of the drug, and ( Cl H )
int
is the intrinsic hepatic clearance of the drug.
If there were no physiological limits to the rate of blood flow( Q H → ∞ ), hence equation 9-24 becomes Cl H = ( Cl H ) int
(EQ 9-25)
This equation provides a definition for intrinsic clearance, namely the clearance of a drug were there to be no physiological limits on the rate of blood flow through the clearing organ.
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Clearance
Kidney drug excretion
By analogy for excretion of unchanged drug by the kidney: Q r ⋅ f u ⋅ ( Cl r ) int Cl r = --------------------------------------Q r + fu ⋅ ( Cl r ) int
(EQ 9-26)
Where Q r is the rate of blood flow through the kidney (assumed 19.1 ml/min per Kg body weight in normal adults), and ( Cl r )
int
is the intrinsic renal clearance of the drug.
Note that the value of Cl cr (assumed 1.75 ml.min per Kg body weight in normal adults) is about 9% of Q r .
9.5.2
EXTRACTION RATIO (E) This is defined as “the ratio of the clearance of a drug compared to the rate of blood flow through the clearing organ.” As such, it indicates what fraction of the drug in the blood is cleared (extracted) on each passage through the clearing organ. Note: when using clearance to calculate extraction ratio, blood flow must be used.
Drug metabolism by the liver
Cl E H = --------HQH Where
EH
(EQ 9-27)
is the steady-state hepatic extraction ratio.
By comparison with equation 9-24, f u ⋅ ( Cl H ) int E H = -----------------------------------------Q H + f u ⋅ ( Cl H ) int
(EQ 9-28)
Thus, the range of values of E H is from zero, when ( Cl H ) int = 0 , to one, when Q H = 0 or ( Cl H ) int » Q H . For example, propanolol has E H = 0.75 , yielding 17.9ml Cl H = ----------------------------------------------------min ⋅ Kg body weight
and
( Cl H )
int
71.4ml = ----------------------------------------------------min ⋅ Kg body weight
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in normal adult males.
9-8
Clearance
Kidney excretion of unchanged drug
where
Er
Cl Er = --------r Qr
(EQ 9-29)
f u ⋅ ( Cl r ) int Er = --------------------------------------Q r + f u ⋅ ( Cl r ) int
(EQ 9-30)
is the steady-state renal extraction ratio.
Thus the range of values is from zero, when ( Cl r )
and
int
» Qr
( Cl r )
. For example, digoxin has
int
1.89ml = ----------------------------------------------------min ⋅ Kg body weight
( Cl r )
E r = 0.09
int
= 0
, to one, when
, yielding
Qr = 0
or
1.72ml Cl r = ---------------------------------------------------min ⋅ Kg body weight
in normal adult males. In this case, note that
Cl r ≈ Cl c r
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9-9
Clearance
9.6 Hepatic Function and Clearance 9.6.1
ALTERATIONS IN HEPATIC BLOOD FLOW For a given drug, equation 9-24 predicts that alterations in the hepatic blood perfusion rate will cause a change in drug clearance, assuming the intrinsic hepatic clearance is unaltered. A general equation may be derived relating the ratio of hepatic clearances at two blood perfusion rates to the fractional change in perfusion rate and the extraction ratio of the drug. FR Cl∗ -----------H- = ---------------------------------------Cl H F R + EH ( 1 – F R ) where Cl ∗ H
Cl H
(EQ 9-31)
denotes normal hepatic clearance,
denotes altered hepatic clearance
Q H∗ F R = ---------QH
is the new flow rate over the old flow rate, the fractional change in blood
perfusion rate, and EH
is the hepatic extraction ratio under normal conditions.
The equation predicts that, for any given decrease in blood perfusion rate, drugs having a large normal extraction ratio will experience a proportionally greater reduction in clearance than drugs having a small normal extraction ratio. Liver blood flow can be reduced by congestive heart failure, for example. The intrinsic hepatic clearance can be represented by the inherent activity of the enzymes responsible for drug metabolism.
9.6.2
ALTERATIONS IN HEPATIC INTRINSIC CLEARANCE For any given drug, equation 9-24 predicts that alterations in the intrinsic hepatic clearance will cause a change in drug clearance, assuming the blood flow rate is unchanged. A general equation may be derived relating the ratio of hepatic clearance at two intrinsic hepatic clearances to the fractional change in intrinsic hepatic and the extraction ratio of the drug.
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Clearance
Fi Cl∗ -----------H- = ----------------------------------1 + EH ( F i – 1 ) Cl H where
f u∗ ⋅ ( Cl int ) ∗ F i = ------------------------------f u ⋅ ( Cl int )
(EQ 9-32)
is the fractional change in fraction unbound times the frac-
tional change in intrinsic hepatic clearance. The equation predicts that, for any given decrease in intrinsic hepatic clearance, drugs having a small normal extraction ratio will experience a proportionately greater reduction in clearance than drugs having a large normal extraction ratio. The intrinsic hepatic clearance of a drug can be reduced by cirrhosis or increased by enzyme inducers, such as phenobarbitol.
9.6.3
TABULATED OR GRAPHICAL ALTERATIONS A table or graph of clearance changes when the hepatic blood flow (but not the intrinsic hepatic clearance) is altered shows that drugs having a low extraction ratio ( EH = 0.1 ) need little adjustment in dosage. Even if the hepatic blood flow were halved ( FR = 0.5 ) , the hepatic clearance is still 91% of its normal value. Conversely, dosage adjustment is necessary for drugs having a high extraction ratio and predominantly eliminated by hepatic metabolism (e.g., propanolol). A table or graph of clearance changes when the intrinsic hepatic clearance (but not the hepatic blood flow) is altered shows that drugs having a high extraction ratio ( E H = 0.9 ) need little adjustment in dosage. Even if the intrinsic hepatic clearance were halved ( Fi = 0.5 ) , the hepatic clearance is still 91% of its normal value. Conversely, dosage adjustment is necessary for drugs having a low extraction ratio and predominately eliminated by hepatic metabolism (e.g., phenylbutazone).
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Clearance
9.7 Renal Function and Clearance Approximately 25% of cardiac output goes to the kidneys or approximately 735 ml/min of plasma is presented to the kidneys of a 70 kg man (19.1 mL/min/kg x 70 kg). Approximately 125 ml/min (1.8 mL/min/kg) of that goes to the glomeruli for filtration (Glomerular Filtration Rate, GFR). Unbound drug is filtered into the proximal renal tubule at this point. The remaining plasma (as blood) is shunted around the tubule in the arterioles adjascent to the proximal tubule where drug may be actively secreted from the arteriol into the proximal tubule or actively reabsorbed in the opposite direction. As the blood flows down the vessels adjascent to the loop of Henle, the drug may be also passively reabsorbed into the blood vessel as the water in the urine is being reabsorbed and the urine is being concentrated. This leads to some interesting possibilities: 1.
Cl r = f u ⋅ GFR . It is likely that the drug is filtered only, in this case, . It is also possible that
secretion and reabsortion balance and cancel each orther out but are still occurring. The actual clearance of the drug may be low as the drug may be bound to plasma protiens or red blood cells. 11. Cl r > f u ⋅ GFR .
Net active secretion is infered in this case. These active mechanisms are non-
specific and consequently, drugs actively secreted compete with each other. Secretion, if it occurs, occurs on the unbound drug and thus is also effected by changes in free fraction. In cases where secretion is very rapid and as a consequence, virtually all of the drug is removed by the single pass through the kidney (Er ~1), the disssociation of the drug from the protien or out of the red blood cells is not a hinderance. Some reabsorption may occur but it is less than secretion. 12. Cl r < f u ⋅ GFR .
Net active reabsorbtion is infered in this case. Active reabsorption occurs for
many exogenous compounds, including glucose and vitamins. For many compounds, reabsorption is passive, occurring only as a consequence of the concentration gradient produced as water is removed from the urine as is proceeds down the renal tubule. Since the membrane is lipoidal in nature, polar compounds, ionized acids and ionized bases are less likely to be reabsorbed. Thus changing the pH of the urine would result in changing the reabsorption characteristics of weakly acidic or basic drugs.
For low molecular weight drugs (<2,000 dalton) , filtration always occurs. Active secretion, active reabsorption and passive reabsorption may occur. It has been found that renal blood flow is little affected by changes in blood flow elsewhere. However, in chronic renal dysfunction there are two effects which exhibit a parallel decline. One is a decrease in glomerular filtration rate (GFR), as measured by Cl cr , and the other is the net secretion of drugs into the kidney tubules. Note that p-amino hippurate (PAH) clearance measures the sum of both effects. Basic Pharmacokinetics
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9-12
Clearance
For any given drug, equation 9-26 predicts that alterations in both the renal blood perfusion rate (as manifest by the GFR) and the intrinsic renal clearance will cause a change in drug clearance. A general equation may be derived relating the ratio of renal clearances at two different blood flow rates and two different intrinsic renal clearances. Cl∗ -----------r = F i Cl r
(EQ 9-33)
where F i is the fractional change in drug unbound times the fractional change in intrinsic renal clearance. In this case, f u∗ F i = ------⋅F fu R where
FR
(EQ 9-34)
is the fractional change in blood flow rate (or GFR).
Thus, equation 9-33 shows that the renal clearance of a drug is reduced by a constant fraction, independent of the renal extraction ratio ( E r ) . This fractional decrease can be estimated by changes in creatinine clearance: Cl∗ cr FR = -----------Cl cr where
Cl∗ cr
(EQ 9-35)
is the altered creatinine clearance
Substituting, equation 9-33 through equation 9-35, we get: Cl∗ r Cl ∗ cr ----------- = -----------Cl r Cl cr
(EQ 9-36)
This equation shows why, in cases of chronic renal dysfunction, a change in the measured creatinine clearance indicated a likely change in drug renal clearance. Hence, dosage adjustments are made on this basis, particularly for drugs predominantly eliminated by renal filtration (e.g., gentamicin, digoxin).
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9-13
Clearance
9.8 General Equations for Changes in Clearance For each clearing organ, FI ⋅ FR Cl ° F Cl = ------ = --------------------------------------Cl F R + E ( F I – FR )
(EQ 9-37)
When a drug has a high extraction ratio, E ≈ 1 , then equation 9-26 becomes F Cl ≈ F R
(EQ 9-38)
and when a drug has a low extraction ratio, E ≈ 0 , then equation 9-26 becomes F Cl ≈ F I
(EQ 9-39)
Thus, the clearance of drugs with a high extraction ratio are more effected by physiological changes in flow of blood to the clearing organ, while drugs with a low extraction ratio are more effected by physiological changes in the function of the organ.
9.8.1
PLASMA/BLOOD RATIO Calculation of Extraction Ratio requires measurement in whole blood by definition. Since most clinical measurements are done in plasma, knowledge of the plasma/blood ratio is necessary. Blood in made up of plasma and red blood cells (RBCs). Thus the amount of drug in the blood is made up of the amount of drug in the plasma and the amount of drug in the RBCs. C b ⋅ Vb = C p ⋅ Vp + C rbc ⋅ V rbc where
(EQ 9-40)
C x ⋅ V x = AMOUNT x
and
b = blood, p = plasma, rbc = red blood cell
If we define the ratio of the concentration of the drug in the RBCs to the concentration of the free drug in plasma as ρ
C rbc = -------------fu ⋅ C p
(EQ 9-41)
and V rbc = H ⋅ V b
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(EQ 9-42)
9-14
Clearance
Vp = ( 1 – H) ⋅ Vb
(EQ 9-43)
Pluging equation 9-41 thru equation 9-43 into equation 9-40 results in: Cb ⋅ Vb = ( 1 – H ) ⋅ V b ⋅ Cp + f u ⋅ ρ ⋅ H ⋅ Vb ⋅ Cp
(EQ 9-44)
Rearranging and simplifying results in: Cb ------ = 1 + H ⋅ ( fu ⋅ ρ – 1 ) Cp ρ
H – 1 + ( Cb ⁄ C p ) = ----------------------------------------fu ⋅ H
(EQ 9-45)
(EQ 9-46)
Thus equation 9-46 determines the affinity of the drug for the RBCs. Drugs with a high affinity for the RBCs should result in a smaller volume of distribution. For drugs that are primarily filtered by the glomeruli, the renal extraction ratio is: GFR ⋅ f u ⋅ C p Rate of filtration - = ------------------------------E rf = ----------------------------------------------Rate of presentation Qr ⋅ Cb
(EQ 9-47)
Putting equation 9-45 into equation 9-47 results in: GFR ⋅ f u E rf = ----------------------------------------------------------Q r ⋅ ( 1 + H ⋅ ( fu ⋅ ρ – 1 ) )
(EQ 9-48)
Thus: 1.
Cl Qr
If the the ratio of --------r to calculated Erf from equation 9-48 is one, it is likely that the drug is filtered only.
13.
Cl
If the the ratio of --------r to calculated Erf from equation 9-48 is greater than one, active secretion Qr
is infered in this case. 14.
Cl
If the the ratio of --------r to calculated Erf from equation 9-48 is less than one, active reabsorbQr
tion is infered in this case.
Basic Pharmacokinetics
REV. 99.4.25
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9-15
Clearance
9.8.2
HALF LIFE AND ELIMINATION RATE CONSTANT IN RELATIONSHIP TO CLEARANCE The elimination rate constant is related to the volume of distribution and the total body clearance by equation 9-22 above which when rewritten yields: Rate of Elimination ( Mass ) ⁄ ( Time ) Cl ( Volume ) ⁄ ( Time ) K = ------------------------------------------------ = ---------------------------------------- = ------ = ---------------------------------------------Amount in the body Mass V Volume
(EQ 9-49)
Cl pw Cl K = Cl ------ = -------b- = -------------u V Vb Vpw u
(EQ 9-50)
where b = blood and pwu = unbound plasma water. Clearance of drug from blood (Clb) is useful in considering drug extraction in the eliminating organs. Volume and clearance terms based on unbound drug concentration are particularly useful in therapeutics, because only the unbound drug is thought to cause the therapeutic action.
9.8.3
EFFECTS OF ALTERATIONS IN PROTEIN BINDING ON CLEARANCE Protein binding of drugs may be altered in disease states and by interferance binding by other drugs on the protein. These changes in binding effect Fi , the fractional change in intrinsic clearance in both renal and hepatic clearances even though the actual intrinsic clearance, the indicator of organ function, may be uneffected as shown in equation 9-11: f u∗ ⋅ Cl ∗ int F i = ------------------------f u ⋅ Cl int where Cl∗ int = Cl int . Thus, a change in protein binding will cause a proportional change in Fi of both clearances. There is more discussion in the chapter on protein binding.
Basic Pharmacokinetics
REV. 99.4.25
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9-16
Clearance
9.9 Problems For each of the problems, do questions A through R now and S through W after completing chapter 10. In doing questions S through W, please try to obtain a plasma concentration of free drug within 120 % of normal Cp
ss min free
Cp
ss max free
and 80 % of
.
Basic Pharmacokinetics
REV. 99.4.25
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9-17
Clearance
Acebutolol Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 1)
AHFS 00:00.00 GPI: 0000000000
Piquette-Miller, M., et. al., “Effect of aging on the pharmacokinetics of acebutolol enantiomers”, Journal of Clinical Pharmacology, Vol. 32, (1992), p. 148 - 156. Kukes, VG; Gneushev ET; Mamedov TS; Gneusheva IA; “Acebutolol and diacetolol: thier binding to plasma and erythrocytes and secretion with saliva.” Farmakol-Toksikol. 1991 Jan-Feb; 54(1)
Acebutolol is a beta-adrenergic blocking agent which is often used in the treatment of hypertension.(Use Qr = 72 L/hr; PROBLEM TABLE 9 - 1. Acebutolol
Patient Condition
Normal
Normal
A
Dose(mg)
200
400
B
f
0.4
C
fu
0.867
D
ρ
1.93
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
5.5
H
%Clr
40
I
%Clnr
60
J
AUC (mg/L*hr)
3.97
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P
Cl h
Q
Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
(hr)
1
12
N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Qh = 90 L/hr) Basic Pharmacokinetics
REV. 99.4.25
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9-18
Clearance
TABLE 9 - 1. Answers for Acebutolol
Patient Condition
Normal
Normal
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
Dose(mg)
200
400 BID
400 BID
400 BID
400 BID
200 QID
FRR
1
1
0.5
1
1
1
FIR
1
1
0.5
0.5
1
1
FRH
1
1
1
1
0.5
1
FIH
1
1
1
1
1
0.5
B
f
0.4
C
fu
0.867
D
ρ
1.93
E
Vd (L)
160
F
k (hr-1)
0.126
0.101
0.102
0.12
0.091
G
T 1/2 (hr)
5.5
6.87
6.77
5.89
7.64
H
%Clr
40
25
26.1
42.9
55.5
I
%Clnr
60
75
73.9
57.1
44.6
J
AUC (mg/L*hr)
3.97
9.93
9.78
8.51
5.5
K
Cl tot (L/hr)
20.2
16.1
16.4
18.8
14.5
L
Cl h (L/hr)
12.1
12.1
12.1
10.7
6.5
M
Cl r (L/hr)
8.1
4.03
4.27
8.1
8.1
L
Eh
0.126
O
Er
0.112
0.8
0.81
0.93
0.72
12
12
12
12
6
2.18
1.75
1.77
2.04
0.785
µ g ------ mL
1.11
1.24
1.23
1.15
1.03
µ g ------ mL
0.57
0.72
0.71
0.61
0.80
ss µ g -------min free mL
0.25
0.37
0.36
0.28
0.6
A
P Q
Cl h Cl r
int
int
R
(L/hr)
16
(L/hr)
10.5
FCL
S
τ
T
1
(hr) N
U Cp
ss max free
V Cp W Cp
ss avg free
Basic Pharmacokinetics
REV. 99.4.25
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9-19
Clearance
Bisoprolol Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 2)
AHFS 00:00.00 GPI: 0000000000
Kirch, W., et. al., “Pharmacokinetics of bisoprolol during repeated oral administration to healthy volunteers and patients with kidney or liver disease”, Clinical Pharmacokinetics, Vol. 13, (1987), p. 110 - 117.
Bisoprolol (comes as 5 and 10 mg tablets) is a - selective adrenergic antagonist. It is used in the treatment of hypertension and angina pectoris.(Use Qr = 72 L/hr; Qh = 90 L/hr) PROBLEM TABLE 9 - 2. Bisoprolol
Patient Condition
Normal
Normal
A
Dose(mg)
10
10 TID
B
f
0.7
C
fu
1
D
ρ
1
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
10
H
%Clr
50
I
%Clnr
50
J
AUC (mg/L*hr)
0.661
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp W Cp
ss µ g -------avg free mL ss
µ g ------min free mL
Basic Pharmacokinetics
REV. 99.4.25
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9-20
Clearance
TABLE 9 - 2. Answers for Bisoprolol
Patient Condition
Normal
Normal
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
Dose(mg)
10
10
10
10
10
10
FRR
1
1
0.5
1
1
1
FIR
1
1
0.5
0.5
1
1
FRH
1
1
1
1
0.5
1
FIH
1
1
1
1
1
0.5
A
B
f
0.9
C
fu
0.7
D
ρ
E
Vd (L)
152.8
F
k (hr-1)
0.0693
0.052
0.0526
0.067
0.0525
G
T 1/2 (hr)
10
13.3
13.2
10.3
13.2
H
%Clr
50
33.3
34
51.3
66
I
%Clnr
50
66.7
66
48.7
34
J
AUC (mg/L*hr)
0.661
0.85
0.87
0.69
0.87
K
Cl tot (L/hr)
10.6
7.94
8.0
10.3
8.0
L
Cl h (L/hr)
5.3
5.3
5.3
5.0
2.7
M
Cl r (L/hr)
5.3
2.65
2.8
5.3
5.3
L
Eh
0.055
O
Er
0.074
0.75
0.76
0.97
0.76
12
12
8
12
P Q
Cl h Cl r
int
int
R
(L/hr)
5.6
(L/hr)
5.7
FCL
S
τ
T
1
(hr) N
U Cp
0.8
0.9
0.9
0.78
0.91
0.108
0.099
0.098
0.11
0.098
µ g ------ mL
0.083
0.073
0.073
0.085
0.073
µ g ------ mL
0.062
0.052
0.052
0.064
0.052
ss µ g -------max free mL
V Cp W Cp
ss avg free ss min free
8
Basic Pharmacokinetics
REV. 99.4.25
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9-21
Clearance
Cefonicid Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 3)
AHFS 00:00.00 GPI: 0000000000
Fillastre, J., et. al., “Pharmacokinetics of cefonicid in uraemic patients”, Journal of Antimicrobial Chemotherapy, Vol. 18, (1986), p. 203 - 211.
Cefonicid is a beta-lactamase resistant cephalosporin which is useful in treating many infections caused by Gram-positive and Gram-negative organisms. Cefonicid is 80% renally excreted.(Use Qr = 72 L/hr; Qh = 90 L/hr) PROBLEM TABLE 9 - 3. Cefonicid
Patient Condition
Normal
Normal
A
Dose(mg)
1000
1000 TID
B
f
1
C
fu
0.06
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
5.3
H
%Clr
80
I
%Clnr
20
J
AUC (mg/L*hr)
654
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp W Cp
ss µ g -------avg free mL ss
µ g ------min free mL
Basic Pharmacokinetics
REV. 99.4.25
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9-22
Clearance
TABLE 9 - 3. Answers for Cefonicid
Patient Condition
Normal
Normal
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
Dose(mg)
1000
1000
1000
1000
1000
1000
FRR
1
1
0.5
1
1
1
FIR
1
1
0.5
0.5
1
1
FRH
1
1
1
1
0.5
1
FIH
1
1
1
1
1
0.5
B
f
1
C
fu
0.06
D
ρ
E
Vd (L)
11.7
F
k (hr-1)
0.131
0.078
0.079
0.131
0.118
G
T 1/2 (hr)
5.3
8.8
8.8
5.3
5.9
H
%Clr
80
66.7
66.9
80
88.9
I
%Clnr
20
33.3
33.1
20
11.1
J
AUC (mg/L*hr)
654
1090
1083
654
727
K
Cl tot (L/hr)
1.53
0.917
0.93
1.5
1.4
L
Cl h (L/hr)
0.3
0.31
0.31
0.3
0.15
M
Cl r (L/hr)
1.22
0.612
0.62
1.22
1.22
L
Eh
0.0032
O
Er
0.017
0.6
0.6
1
0.9
8
12
12
8
8
1.51
1.4
1.4
1.51
1.36
µ g ------ mL
7.9
8.4
8.4
7.9
8.4
µ g ------ mL
4.9
5.5
5.4
4.9
5.4
ss µ g -------min free mL
2.8
3.3
3.3
2.8
3.3
A
P Q
Cl h Cl r
(L/hr)
5.11
(L/hr)
20.7
int
int
R
FCL
S
τ
T
1
(hr) N
U Cp
ss max free
V Cp W Cp
ss avg free
Basic Pharmacokinetics
REV. 99.4.25
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9-23
Clearance
Cefpirome Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 4)
AHFS 00:00.00 GPI: 0000000000
Lameire, N., et. al., “Single-dose pharmacokinetics of cefpirome in patients with renal impairment”, Clinical Pharmacology and Therapeutics, Vol. 52, (1992), p. 24 - 30.
Cefpirome is a third-generation, broad-spectrum cephalosporin which is useful against many cephalosporin-resistant organisms. PROBLEM TABLE 9 - 4.
Cefpirome
Patient Condition
Normal
Normal
A
Dose(mg)
2000 IV
2000 TID
B
f
1
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
2.6
H
%Clr
85
I
%Clnr
15
J
AUC (mg/L*hr)
342
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp W Cp
ss µ g -------avg free mL ss
µ g ------min free mL
Basic Pharmacokinetics
REV. 99.4.25
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9-24
Clearance
TABLE 9 - 4. Answers for Cefpirome
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
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9-25
Clearance
Cefprozil Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 5)
AHFS 00:00.00 GPI: 0000000000
Shyu, W., et. al., “Pharmacokinetics of cefprozil in healthy subjects and patients with renal impairment”, Journal of Clinical Pharmacology, Vol. 31, (1991), p. 362 - 371.
Cefprozil is a broad-spectrum oral cephalosporin. PROBLEM TABLE 9 - 5.
Patient Condition
Normal
A
Dose(mg)
1000
B
f
0.95
C
fu
0.7
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
1.2
H
%Clr
75
I
%Clnr
25
J
AUC (mg/L*hr)
58.1
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Cefprozil
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
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9-26
Clearance
TABLE 9 - 5. Answers for Cefprozil
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
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9-27
Clearance
Chloramphenicol Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 6)
AHFS 00:00.00 GPI: 0000000000
Ambrose, P., “Clinical pharmacokinetics of chloramphenicol and chloramphenicol succinate”, Clinical Pharmacokinetics, Vol. 9, (1984), p. 222 - 238.
Chloramphenicol succinate is a prodrug which is converted in vivo to the active form, chloramphenicol. PROBLEM TABLE 9 - 6.
Patient Condition
Normal
A
Dose(mg)
1000
B
f
1
C
fu
0.4
D
ρ
E
Vd (L/kg)
F
k (hr-1)
G
T 1/2 (hr)
0.6
H
%Clr
30
I
%Clnr
70
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
2.8
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Chloramphenicol
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
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9-28
Clearance
TABLE 9 - 6. Answers for Chloramphenicol
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
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9-29
Clearance
Enalapril Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 7)
AHFS 00:00.00 GPI: 0000000000
Ohnishi, A., et. al., “Kinetics and dynamics of enalapril in patients with liver cirrhosis”, Clinical Pharmacology and Therapeutics, Vol. 45, (1989), p. 657 - 665.
Enalapril is an ACE inhibitor which is a prodrug that is metabolized in the liver to the active metabolite, enalaprilat. PROBLEM TABLE 9 - 7. Enalapril
Patient Condition
Normal
Normal
A
Dose(mg)
10 mg po
10 BID1
B
f
0.65
C
fu
0.55
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
0.63
H
%Clr
27
I
%Clnr
73
J
AUC (mg/L*hr)
0.123
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
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9-30
Clearance
TABLE 9 - 7. Answers for Enalapril
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-31
Clearance
Enoxacin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 8)
AHFS 00:00.00 GPI: 0000000000
Somogyi, A., and Bochner, F., “The absorption and disposition of enoxacin in healthy subjects”, Journal of Clinical Pharmacology, Vol. 28, (1988), p. 707 - 713.
Enoxacin is a fluorinated quinolone which is used to treat infections caused by gram-negative organisms and Pseudomonoas aeruginosa. PROBLEM TABLE 9 - 8.
Enoxacin
Patient Condition
Normal
Normal
A
Dose(mg)
400
200 bid
B
f
0.9
C
fu
0.8
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
7.75
H
%Clr
60
I
%Clnr
40
J
AUC (mg/L*hr)
15.61
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr)
12
N
U Cp
ss
µ g ------max free mL
V Cp W Cp
ss µ g -------avg free mL ss
µ g ------min free mL
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-32
Clearance
TABLE 9 - 8. Answers for Enoxacin
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-33
Clearance
Enprofylline Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 9)
AHFS 00:00.00 GPI: 0000000000
Nadai, M., et. al, “Dose-dependent pharmacokinetics of enprofylline and its renal handling in rats”, Journal of Pharmaceutical Sciences, Vol. 80, No. 7, (1991), p. 648 - 651
Enprogylline is a xanthine bronchodilator which is more potent than theophylline. Enprofylline
PROBLEM TABLE 9 - 9.
Patient Condition
Normal
Normal
A
Dose(mg/kg)
2.5
2.5 TID
B
f
1
C
fu
0.4
D
ρ
E
Vd (L/kg)
F
k (hr-1)
G
T 1/2 (hr)
0.36
H
%Clr
90
I
%Clnr
10
J
AUC (mg/L*hr)
3.6
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
TABLE 9 - 9. Answers for Enprofylline
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-34
Clearance
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-35
Clearance
Erythromycin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 10)
AHFS 00:00.00 GPI: 0000000000
Welling and Creig (JPS 67, 1057-9,1978).
Erythromycin is a macrolide antibiotic.(Use Qr = 72 L/hr; Qh = 90 L/hr) PROBLEM TABLE 9 - 10.
Erythromycin
Patient Condition
Normal
Normal
A
Dose(mg)
300
300 TID
B
f
.8
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
10
I
%Clnr
90
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
0.06
0.06
0.12
0.06
0.18
57
57
100
57
150
16.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr)
8
N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-36
Clearance
TABLE 9 - 10. Answers for
Erythromycin
Patient Condition
Normal
Normal
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
Dose(mg)
300
300
300
200
300
200
FRR
1
1
0.5
1
1
1
FIR
1
1
0.5
0.5
1
3
FRH
1
1
1
1
0.5
1
FIH
1
1
1
2
1
0.5
B
f
.8
C
fu
0.06
0.06
0.12
0.06
0.18
D
ρ
E
Vd (L)
57
57
100
57
150
F
k (hr-1)
0.255
0.242
0.235
0.227
0.075
G
T 1/2 (hr)
2.71
2.85
2.92
3.05
9.27
H
%Clr
10
5.3
3.1
11.2
37
I
%Clnr
90
94.7
96.9
88.8
63
J
AUC (mg/L*hr)
14.5
17.4
6.7
18.5
14.2
K
Cl tot (L/hr)
16.5
13.8
23.7
13.0
11
L
Cl h (L/hr)
13.1
13.1
23.4
11.5
7.02
M
Cl r (L/hr)
1.45
0.73
0.73
1.45
4.19
L
Eh
0.136
O
Er
0.020
0.95
1.63
0.89
0.77
8
8
8
8
24
2.94
2.79
2.74
2.63
2.59
µ g ------ mL
0.29
0.30
0.23
0.30
0.23
µ g ------ mL
0.12
0.13
0.10
0.14
0.11
0.038
0.042
0.034
0.049
0.038
A
P Q
Cl h Cl r
(L/hr)
252
(L/hr)
24.7
int
int
R
FCL
S
τ
T
1
(hr) N
U Cp
ss max free
V Cp W Cp
ss avg free
ss µ g -------min free mL
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-37
Clearance
Fleroxacin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 11)
AHFS 00:00.00 GPI: 0000000000
Singlas, E., et. al., “Disposition of fleroxacind, a new trifluoroquinolone, and its metabolites - pharmacokinetics in renal failure and influence of haemodialysis”, Clinical Pharmacokinetics, Vol. 19, No. 1, (1990), p. 67 - 79.
Fleroxacin is a trifluorinated quinolone with activity against a variety of gram-negative and gram-positive organisms. Fleroxacin
PROBLEM TABLE 9 - 11.
Patient Condition
Normal
Normal
A
Dose(mg)
400
200 tid
B
f
0.95
C
fu
0.5
D
ρ
1.45
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
14
H
%Clr
65
I
%Clnr
35
J
AUC (mg/L*hr)
92
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-38
Clearance
TABLE 9 - 11. Answers for Fleroxacin
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-39
Clearance
Fosinopril Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 12)
AHFS 00:00.00 GPI: 0000000000
Hui, K., et. al., “Pharmacokinetics of fosinopril in patients with various degrees of renal function”, Journal of Clinical Pharmacology and Therapeutics, Vol. 49, No. 4, (1991), p. 457 - 466.
Fosinopril is an angiotensin converting enzyme inhibitor which is a prodrug that is metabolized to active form, fosinoprilat. PROBLEM TABLE 9 - 12.
Fosinopril
Patient Condition
Normal
Normal
A
Dose(mg)
7.5 IV
20 po qd
B
f
1 (Oral 0.36)
C
fu
0.01
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
7
H
%Clr
0
I
%Clnr
100
J
AUC (mg/L*hr)
5.1
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr)
24
N
U Cp
ss
µ g ------max free mL
V Cp W Cp
ss µ g -------avg free mL ss
µ g ------min free mL
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-40
Clearance
TABLE 9 - 12. Answers for
Patient Condition A
Dose(mg)
B
f
C
fu
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
Fosinopril
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-41
Clearance
Glutathione Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 13)
AHFS 00:00.00 GPI: 0000000000
Mulders, T., et. al., “Characterization of glutathione conjugation in humans: stereoselectivity in plasma elimination pharmacokinetics and urinary excretion of (R)- and (S)-2-bromoisovalerylurea in healthy volunteers”, Clinical Pharmacology and Therapeutics, Vol. 53, (1993), p. 49 - 58.
This study explored the pharmacokinetics of glutathione. Glutathione
PROBLEM TABLE 9 - 13.
Patient Condition
Normal
Normal
A
Dose(mg)
600
500 tid
B
f
1
C
fu
1
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
4.4
H
%Clr
35
I
%Clnr
65
J
AUC (mg/L*hr)
276
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-42
Clearance
TABLE 9 - 13. Answers for
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
Glutathione
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-43
Clearance
Guanadrel Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 14)
AHFS 00:00.00 GPI: 0000000000
Halstenson, C., et. al., “Disposition of guanadrel in subjects with normal and impaired renal function”, Journal of Clinical Pharmacology, Vol. 29, (1989), p. 128 - 132.
Guanadrel is adrenergic blocker used in the treatment of hypertension. Guanadrel
PROBLEM TABLE 9 - 14.
Patient Condition
Normal
Normal
A
Dose(mg)
25 mg po
25 BID
B
f
1
C
fu
0.8
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
3.7
H
%Clr
40
I
%Clnr
60
J
AUC (mg/L*hr)
0.234
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr)
12
N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-44
Clearance
TABLE 9 - 14. Answers for
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
Guanadrel
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-45
Clearance
Monoxidine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 15)
AHFS 00:00.00 GPI: 0000000000
Kirch, W., Hutt, H., and Plänitz, V., “The influence of renal function on clinical pharmacokinetics of monoxidine”, Clinical Pharmacokinetics, Vol. 15, (1988), p. 245 - 253.
Monoxidine is a centrally acting antihypertensive agent which stimulates α 2 -adrenergic receptors. PROBLEM TABLE 9 - 15.
Monoxidine
Patient Condition
Normal
Normal
Dose(mg)
0.25 IV
0.25 tid
B
f
1
C
fu
1
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
2.75
H
%Clr
95
I
%Clnr
5
J
AUC (mg/L*hr)
0.05
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
A
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-46
Clearance
TABLE 9 - 15. Answers for
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
Monoxidine
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-47
Clearance
Nalmefene Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 16)
AHFS 00:00.00 GPI: 0000000000
Dixon, R., et. al., “Nalmefence: safety and kinetics after single and multiple oral doses of a new opiod antagonist”, Journal of Clinical Pharmacology, Vol. 27, (1987), p. 233 - 239.
Nalmefene is a pure opiod antagonist which is currently being investigated for use. Nalmefene
PROBLEM TABLE 9 - 16.
Patient Condition
Normal
A
Dose(mg)
20 IV
B
f
0.6 oral
C
fu
1
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
9.8
H
%Clr
70
I
%Clnr
30
J
AUC (mg/L*hr)
0.3
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FRR=0.5
(L/hr)
int
R
Normal
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
9-48
Clearance
TABLE 9 - 16. Answers for
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
Nalmefene
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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9-49
Clearance
Nitrendipine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 17)
AHFS 00:00.00 GPI: 0000000000
Dylewicz, P, et. al, “Bioavailability and elimination of nitrendipine in liver disease”, European Journal of Clinical Pharmacology, Vol, 32, (1987), p. 563 - 568.
Nitrendipine is a calcium antagonist PROBLEM TABLE 9 - 17.
Nitrendipine
Patient Condition
Normal
Normal
A
Dose(mg)
5 IV
25 po BID
B
f
1 (oral 0.2)
C
fu
0.05
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
11.7
H
%Clr
0.1
I
%Clnr
99.9
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
90
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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9-50
Clearance
TABLE 9 - 17. Answers for
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
Nitrendipine
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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9-51
Clearance
Ofloxacin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 18)
AHFS 00:00.00 GPI: 0000000000
Lamerire, N., et. al., “Ofloxacin pharmacokinetics in chronic renal failure and dialysis”, Clinical Pharmacokinetics, Vol. 21, No. 4, (1995), p. 357 - 371. PROBLEM TABLE 9 - 18.
Ofloxacin
Patient Condition
Normal
Normal
A
Dose(mg)
300
300
B
f
0.93
C
fu
0.74
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
6
H
%Clr
97
I
%Clnr
3
J
AUC (mg/L*hr)
28.47
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr)
12
N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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9-52
Clearance
TABLE 9 - 18. Answers for
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
Ofloxacin
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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9-53
Clearance
Omeprazole
(Problem 9 - 19)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
(Use Qr = 72 L/hr; Qh = 90 L/hr) PROBLEM TABLE 9 - 19.
Omeprazole
Patient Condition
Normal
Normal
A
Dose(mg)
10
10 BID
B
f
0.5
C
fu
0.04
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
2
H
%Clr
0
I
%Clnr
100
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FIH=0.5 0.75
0.06
0.08
30
1
12
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
TABLE 9 - 19. Answers for
A
FRH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
Omeprazole
Patient Condition
Normal
Normal
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
Dose(mg)
10
10
10
5
10
5
Basic Pharmacokinetics
REV. 99.4.25
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9-54
Clearance
FRR
1
1
0.5
1
1
1
FIR
1
1
0.5
0.5
1
2
FRH
1
1
1
1
0.5
1
FIH
1
1
1
1.5
1
0.5 0.75
B
f
0.5
C
fu
0.04
D
ρ
E
Vd (L)
30
F
k (hr-1)
0.347
0.347
0.493
0.31
0.184
G
T 1/2 (hr)
2
2
1.41
2.2
3.8
H
%Clr
0
I
%Clnr
100
J
AUC (mg/L*hr)
0.48
0.48
0.338
0.53
0.68
K
Cl tot (L/hr)
10.4
10.4
14.8
9.38
5.5
L
Cl h (L/hr)
10.4
10.4
14.8
9.38
5.5
M
Cl r (L/hr)
0
0
0
0
0
L
Eh
0.108
O
Er
0
1
1.42
0.90
0.53
6
8
6
12
P Q
Cl h Cl r
(L/hr)
291.5
(L/hr)
0
int
int
R
FCL
S
τ
T
0.06
1
6
(hr) N
U Cp
3
3
2.83
2.71
3.2
0.0076
0.0076
0.0051
0.0079
0.011
µ g ------ mL
0.0032
0.0032
0.0025
0.0035
0.0045
µ g ------ mL
0.00095
0.00095
0.00099
0.0012
0.0012
ss µ g -------max free mL
V Cp W Cp
ss avg free ss min free
0.08
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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9-55
Clearance
Piperacillin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 20)
AHFS 00:00.00 GPI: 0000000000
Johnson, C., et. al., “Single-dose pharmacokinetics of piperacillin and tazobactam in patients with renal disease”, Clinical Pharmacology and Therapeutics, Vol. 51, (1992), p. 32 - 41.
Piperacillin is a beta-lactam antibiotic. Piperacillin
PROBLEM TABLE 9 - 20.
Patient Condition
Normal
Normal
A
Dose(mg)
3000
2000 qid
B
f
1
C
fu
0.82
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
0.95
H
%Clr
75
I
%Clnr
25
J
AUC (mg/L*hr)
276
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr)
6
N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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9-56
Clearance
TABLE 9 - 20. Answers for
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
Piperacillin
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
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9-57
Clearance
Piroxicam Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 21)
AHFS 00:00.00 GPI: 0000000000
Boudinot, S., Funderburg, E., and Boudinot, F., “Effects of age on the pharmacokinetics of piroxicam in rats”, Journal of Pharmaceutical Sciences, Vol. 82, No. 3, (1993), p. 254 - 257.
Piroxicam is a nonsteroidal anti-inflammatory drug (NSAID) commonly used in the treatment of arthritis. Piroxicam
PROBLEM TABLE 9 - 21.
Patient Condition
Normal
Normal
A
Dose(mg)
70
20 qd
B
f
1
C
fu
0.007
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
50
H
%Clr
5
I
%Clnr
95
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
9
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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9-58
Clearance
TABLE 9 - 21. Answers for
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
Piroxicam
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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9-59
Clearance
Quinidine
(Problem 9 - 22)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Quinidine sulfate is used to treat ventricular and supraventricular arrythmias and is available in 200 and 300 mg tablets. It is known to bind to α -acid glycoprotein (AAG), which is an acute phase reactant. AAG rises in trauma, inflamation, malignancy and stress and falls in hepatic disease, nephrotic syndrome and malneurtrition for example.(Use Qr = 72 L/ hr; Qh = 90 L/hr) Quinidine
PROBLEM TABLE 9 - 22.
Patient Condition
Normal
Normal
Dose(mg/kg)
10
10 TID
B
f
0.7
C
fu
0.2
D
s
0.83
E
Vd (L/kg)
2.6
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr/kg)
L
Cl h (L/hr/kg)
0.20
M
Cl r (L/hr/kg)
0.056
L
Eh
O
Er
A
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
0.10
0.25
0.15
0.3
2.0
2.75
2.2
3.0
6.4
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1 8
(hr) N
U Cp
ss max free
V Cp W Cp
µ g ------ mL
MTC 1.7
ss µ g -------avg free mL ss min free
µ g ------ mL
MEC 0.37
Basic Pharmacokinetics
REV. 99.4.25
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9-60
Clearance
TABLE 9 - 22. Answers for
Quinidine
Patient Condition
Normal
Normal
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
Dose(mg)
10
10
10
10
10
5
FRR
1
1
0.5
1
1
1
FIR
1
1
0.25
0.5
0.75
1.5
A
FRH
1
1
1
1
0.5
1
FIH
1
1
0.5
1.25
0.75
0.5
B
f
0.7
C
fu
0.20
0.10
0.25
0.15
0.30
D
s
0.83
E
Vd (L)
2.6
2.0
2.75
2.2
3.0
F
k (hr-1)
0.098
0.057
0.101
0.087
0.061
G
T 1/2 (hr)
7
12.1
6.9
7.9
11.3
H
%Clr
22
12.3
10
22
46
I
%Clnr
78
87.7
90
78
54
J
AUC (mg/L*hr)
27.3
61.3
20.6
36.5
19.0
K
Cl tot (L/hr)
0.256
0.114
0.28
0.19
0.18
L
Cl h (L/hr)
0.20
0.10
0.25
0.15
0.1
M
Cl r (L/hr)
0.056
0.014
0.028
0.04
0.08
L
Eh
0.0021
O
Er
0.00078
0.45
1.1
0.75
0.72
8
8
8
8
8
1.3
0.66
1.16
1.0
0.71
0.82
0.79
0.95
0.79
0.75
0.57
0.63
0.65
0.56
0.59
0.37
0.5
0.44
0.39
0.46
P Q
Cl h Cl r
(L/hr)
1.0
(L/hr)
0.28
int
int
R
FCL
S
τ
T
1
(hr) N
U Cp
ss max free
V Cp W Cp
ss avg free ss min free
µ g ------ mL
MTC 1.7
µ g ------ mL µ g ------ mL
MEC 0.3
Basic Pharmacokinetics
REV. 99.4.25
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9-61
Clearance
Tazobactam Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 23)
AHFS 00:00.00 GPI: 0000000000
Johnson, C., et. al., “Single-dose pharmacokinetics of piperacillin and tazobactam in patients with renal disease”, Clinical Pharmacology and Therapeutics, Vol. 51, (1992), p. 32 - 41.
Tazobactam is an irreversible beta-lactamase inhibitor. Tazobactam
PROBLEM TABLE 9 - 23.
Patient Condition
Normal
Normal
A
Dose(mg)
375 IV
375 qid
B
f
1
C
fu
0.96
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
0.89
H
%Clr
68
I
%Clnr
32
J
AUC (mg/L*hr)
30.3
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
6
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
TABLE 9 - 23. Answers for
Tazobactam
Basic Pharmacokinetics
REV. 99.4.25
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Clearance
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
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9-63
Clearance
Theophylline Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 24)
AHFS 00:00.00 GPI: 0000000000
Wagner, J., “Theophylline - pooled Michaelis-Menten parameters and implications”, Clinical Pharmacokinetics, Vol. 10, (1985), p. 432 - 442. PROBLEM TABLE 9 - 24. Theophylline Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
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9-64
Clearance
TABLE 9 - 24. Answers for
Patient Condition A
Dose(mg)
B
f
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
Theophylline
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
Basic Pharmacokinetics
REV. 99.4.25
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9-65
Clearance
Tolrestat
(Problem 9 - 25)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Troy, S., et. al., “The effect of renal disease on tolrestat pharmacokinetics”, Clinical Pharmacology and Therapeutics, Vol. 51, (1992), p. 271 - 277.
Tolrestat is an aldose reductase inhibitor used in the treatment of diabetic neuropathy, diabetic nephropathy, and diabetic retinopathy. (Use Qr = 72 L/hr; Qh = 90 L/hr) PROBLEM TABLE 9 - 25.
Tolrestat
Patient Condition
Normal
Normal
A
Dose(mg)
200
200 tid
B
f
0.8
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
10.6
H
%Clr
25
I
%Clnr
40 bile 35 metab.
J
AUC (mg/L*hr)
86
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr) N
U Cp
ss
µ g ------max free mL
V Cp W Cp
ss µ g -------avg free mL ss
µ g ------min free mL
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9-66
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TABLE 9 - 25. Answers for
Tolrestat
Patient Condition
Normal
Normal
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
Dose(mg)
200
200 TID
200 TID
200 TID
200 TID
200 BID
FRR
1
1
0.5
1
1
1
FIR
1
1
0.5
0.5
1
1
FRH
1
1
1
1
0.5
1
FIH
1
1
1
1
1
0.5
B
f
0.8
C
fu
1
E
Vd (L)
28.5
F
k (hr-1)
0.0654
0.057
0.057
0.065
0.041
G
T 1/2 (hr)
10.6
12.1
12.1
10.7
16.9
H
%Clr
25
14.3
14.3
25.3
39.8
I
%Clnr
75
85.7
85.7
74.7
60.2
J
AUC (mg/L*hr)
86
98.3
98.2
87
137
K
Cl tot (L/hr)
1.86
1.63
1.63
1.84
1.17
L
Cl h (L/hr)
1.4
1.4
1.4
1.4
0.7
M
Cl r (L/hr)
0.46
0.23
0.47
0.47
0.47
L
Eh
0.0155
O
Er
0.0065
0.88
0.88
0.99
0.63
8
8
8
8
12
0.75
0.66
0.66
0.75
0.71
µ g ------ mL
13.8
15.3
15.3
13.9
14.4
µ g ------ mL
10.8
12.3
12.3
10.9
11.4
µ g ------ mL
8.2
9.6
9.7
8.3
8.8
A
P Q
Cl h Cl r
(L/hr)
1.42
(L/hr)
0.47
int
int
R
FCL
S
τ
T
1
(hr) N
U Cp
ss max free
V Cp W Cp
ss avg free ss min free
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9-67
Clearance
Vancomycin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 9 - 26)
AHFS 00:00.00 GPI: 0000000000
Macais, W., Meuller, B., and Scarim, S., “Vancomycin pharmacokinetics in acute renal failure; preservation of nonrenal clearance”, Clinical Pharmacology and Therapeutics, Vol., 50, (1991), p. 688 - 694.
Vancomycin is a glycopeptide antibiotic used in the treatment of infections caused by Gram-positive organisms. PROBLEM TABLE 9 - 26.
Vancomycin
Patient Condition
Normal
Normal
A
Dose(mg)
1000
500 QID
B
f
1
C
fu
D
ρ
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
70
I
%Clnr
30
J
AUC (mg/L*hr)
543
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRH=0.5
FIH=0.5
0.44
(L/hr)
int
FCL
S
FIR=0.5
(L/hr)
int
R
FRR=0.5
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
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9-68
Clearance
TABLE 9 - 26. Answers for
Patient Condition A
Dose(mg)
B
f
C
fu
E
Vd (L)
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
M
Cl r (L/hr)
L
Eh
O
Er
P Q
Cl h Cl r
τ
T
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
(L/hr)
int
FCL
S
Normal
(L/hr)
int
R
Normal
Vancomycin
1
(hr) N
U Cp
ss µ g -------max free mL
V
ss
Cp
µ g ------avg free mL
Cp
ss µ g -------min free mL
W
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Clearance
Xipamide
(Problem 9 - 27)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Knauf, H, et. al., “Xipamide disposition in liver cirrhosis”, Clinical Pharmacology and Therapeutics, Vol. 48, No. 6, (1990), p. 328 - 632.
Xipamide is a diuretic that has been used in the treatment of congestive heart failure, hypertension, advanced renal failure, and hepatic edema.(Use Qr = 72 L/hr; Qh = 90 L/hr) PROBLEM TABLE 9 - 27.
Xipamide
Patient Condition
Normal
Normal
A
Dose(mg)
40
40 QID
B
f
0.8
C
fu
0.01
E
Vd (L)
21
F
k (hr-1)
G
T 1/2 (hr)
H
%Clr
I
%Clnr
J
AUC (mg/L*hr)
K
Cl tot (L/hr)
L
Cl h (L/hr)
1.38
M
Cl r (L/hr)
0.72
L
Eh
O
Er
P Q
Cl h Cl r
R
τ
T
FRH=0.5
FIH=0.5
0.01
0.02
0.01
0.025
(L/hr)
FCL
S
FIR=0.5
(L/hr)
int
int
FRR=0.5
1
(hr) N
U Cp
ss µ g -------max free mL
V Cp
ss µ g -------avg free mL
Cp
ss µ g -------min free mL
W
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Clearance
TABLE 9 - 27. Answers for
Xipamide
Patient Condition
Normal
Normal
FRR=0.5
FIR=0.5
FRH=0.5
FIH=0.5
Dose(mg)
40
40 QID
40 QID
20 QID
40 QID
20 QID
FRR
1
1
0.5
1
1
1
FIR
1
1
0.5
0.5
1
2.5
FRH
1
1
1
1
0.5
1
FIH
1
1
1
2
1
0.5
B
f
0.8
C
fu
0.01
0.01
0.02
0.01
0.025
E
Vd (L)
21
F
k (hr-1)
0.1
0.083
0.146
0.1
0.118
G
T 1/2 (hr)
6.9
8.4
4.7
7.0
5.9
H
%Clr
34
21
12
35
72
I
%Clnr
66
79
88
65
28
J
AUC (mg/L*hr)
15.2
18.4
7.8
15.4
6.5
K
Cl tot (L/hr)
2.1
1.74
3.1
2.1
2.5
L
Cl h (L/hr)
1.38
1.38
2.71
1.36
0.7
M
Cl r (L/hr)
0.72
0.36
0.36
0.72
1.8
L
Eh
0.0144
O
Er
0.01
0.83
1.46
1
1.18
6
4
6
6
A
P Q
Cl h Cl r
(L/hr)
140
(L/hr)
73
int
int
R
FCL
S
τ
T
1
N
U Cp
0.87
0.71
0.847
0.86
1
0.034
0.039
0.034
0.034
0.038
µ g ------ mL
0.025
0.031
0.026
0.025
0.027
µ g ------ mL
0.019
0.024
0.019
0.019
0.019
ss µ g -------max free mL
V Cp W Cp
6
(hr)
ss avg free ss min free
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CHAPTER 10
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Author: Michael Makoid Reviewer: Phillip Vuchetich
OBJECTIVES 1.
Given population average patient data, the student will devise (V) dosage regimens which will maintain plasma concentrations of drug within the therapeutic range.
2.
Given specific patient information, the patient will justify (VI) dosage regimen recommendations.
3.
Given patient information regarding organ function, the student will devise (V) and justify (VI) dosage regimen recommendations for the compromised patient.
4.
The student will write (V) a professional consult using the above calculations
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10-1
Dosage Regimen (Healthy, Aged, and Diseased Patients)
10.1 Therapeutic Drug Monitoring 10.1.1
THERAPEUTIC RANGE The pharmacokinetics of a drug determine the blood concentration achieved from a prescribed dosing regimen. During multiple drug dosing, the blood concentration will reflect the drug concentration at the receptor site; and it is the receptor site concentration that determines the intensity of the drug’s effect. Therefore, in order to predict a patient’s response to a drug regimen, both the pharmacokinetics and pharmacological response characteristics of the drug must be understood. There exists a fundamental relationship between drug pharmacokinetics and pharmacologic response. The relationship between response and ln-concentration is sigmoidal. A threshold concentration of drug must be attained befor any response is ellicited at all. Therapy is accheived when the desired effect is attained because the required concentration has been reached. That concentration would set the lower limit of utility of the drug, and is called Effective Concentration (MEC). Most drugs are not “clean”, that is exhibit only the desired therapeutic response. They also exhibit undesired side effects, sometimes called toxic effects at a higher, hopefully a lot higher, concentration. At some concentration, these toxic side effects become become intollerable. That concentration, or one below it, would set the upper limit of utility for the drug and is called the Maximum Therapeutic Concentration or Minimum Toxic Concentration (MTC). Patient studies have generated upper (MTC) and lower (MEC) plasma concentration ranges that are deemed safe and effective in treating specific disease states. These concentrations are known as the “therapeutic range” for the drug (see Table 10-1).When a drug is administered at a fixed dosage to numerous subjects, the blood concentrations achieved vary greatly due to biological variation. However it is possible to have a reasomable Clinically, digoxin concentrations below 0.8 ng ⁄ ml will elicit a subtherapeutic effect. Alternatively, when the digoxin concentration exceeds 2.0 ng ⁄ ml side effects occur (nausea and vomiting, abdominal pain, visual disturbances). Drugs like digoxin possess a narrow therapeutic index because the concentrations that
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10-2
Dosage Regimen (Healthy, Aged, and Diseased Patients)
may produce toxic effects are close to those required for therapeutic effects. The importance of considering both pharmacokinetics and pharmacodynamics is clear. TABLE 10-1. Average
therapeutic drug concentration
DRUG
RANGE
digoxin
0.8-2.0 ng ⁄ ml
gentamicin
2-10 µ g ⁄ ml l
lidocaine
1-4 µ g ⁄ ml
lithium
0.4-1.4 mEq ⁄ L
phenytoin
10-20 µ g ⁄ ml
phenobarbitol
10-30 µ g ⁄ ml
procainamide
4-8 µ g ⁄ ml
quinidine
3-6 µ g ⁄ ml
theophylline
10-20 µ g ⁄ ml
Note that drug concentrations may be expressed by a variety of units. Pharmacokinetic factors that cause variability in plasma drug concentration are: • • • • •
Drug-drug interaction patient disease state physiological states such as age, weight, sex drug absorption variation differences in the ability of a patient to metabolize and eliminate the drug
If we were to give an identical dose of drug to a large group of patients and then measure the highest plasma drug concentration we would see that due to individual variability, the resulting plasma drug concentrations differ. This variability can be attributed to factors influencing drug absorption, distribution, metabolism, and excretion. Therefore, drug dosage regimens must take into account any disease altering state or physiological difference in the individual. Therapeutic drug monitoring optimizes a patient’s drug therapy by determining plasma drug concentrations to ensure the rapid and safe drug level in the therapeutic range.
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10-3
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Two components make up the process of therapeutic drug monitoring:
• Assays for determination of the drug concentration in plasma • Interpretation and application of the resulting concentration data to develop a safe and effective drug regimen.
The major potential advantages of therapeutic drug monitoring are the maximization of therapeutic drug benefits and the minimization of toxic drug effects. The formulation of drug therapy regimens by therapeutic drug monitoring involves a process for reaching dosage decisions.
10.1.2
THERAPEUTIC MONITORING: WHY DO WE CARE? The usefulness of a drug’s concentration vs. time profile i based on the observation that for many drugs there is a relationship between plasma concentration and therapeutic response. There is a drug concentration below which the drug is ineffective, the Minimum Effective Concentration (MEC), and above which the drug has untoward effects, the Minimum Toxic Concentration (MTC). That defines the range in which we must attempt to keep the drug concentration (Therapeutic Range). The data in Table 10-1 are population averages. Most people respond to drug concentrations in these ranges. There is always the possibility that the range will be different in an individual patient. For every pharmacokinetic parameter that we measure, there is a population average and a range. This is normal and is called biological variation. People are different. In addition to biological variation there is always error in the laboratory assays that we use to measure the parameters and error in the time we take the sample. Even with these errors, in many cases, he therapy is better when we attempt to monitor the patient’s plasma concentration to optimize therapy than if we don’t. This is called therapeutic monitoring. If done properly, the plasma concentrations are rapidly attained and maintained within the therapeutic range throughout the course of therapy. This is not to say all drugs should be monitored. Some drugs have a such a wide therapeutic range or little to no toxic effects that the concentrations matter very little. Therapeutic monitoring is useful when: • • • •
a correlation exists between response and concentration the drug has a narrow therapeutic range the pharmacological response is not easily assessed there is a wide inter-subject range in plasma concentrations for a given dose
In this era of DRGs, where reimbursement is no longer tied to cost, therapeutic monitoring of key drugs can be economically beneficial to an institution. A recent study (DeStache 1990) showed a significant difference with regard to days in the
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10-4
Dosage Regimen (Healthy, Aged, and Diseased Patients)
hospital between the patients on gentamicin who were monitored (and their dosage regulated as a consequence) vs. those who were not. With DRGs the hospital was reimbursed a flat fee irrespective of the number of days the patient stayed in the hospital. If the number of days cost less than what the DRG paid, the hospital makes money. If the days cost more than the hospital loses money. This study showed that if all patients in the hospital who were on gentamicin were monitored, the hospital would save $4,000,000. That’s right FOUR MILLION per year. I would say that would pay my salary, with a little left over, and that is only one drug! The process of therapeutic monitoring takes effort.
• • • • •
First the MD must order the blood assays. Second, someone (nurse, med tech, you) must take the blood. Someone (lab tech, you) must assay the drug concentration in the blood. You must interpret the data You must communicate your interpretation and your recommendations for dosage regimen change to the MD. This will allow for informed dosage decisions.
• You must follow through to ensure proper changes have been made. • You must continue the process throughout therapy. Therapeutic monitoring, in many cases, will be part of your practice. It can be very rewarding
Thus, if we have deterimined the therapeutic range, we could use pharmacokinetics to determine the optimum dosage regemin to maintain the patient’s plasma concentration within that range.
10.1.3
STEADY STATE It is rare that a drug is given only once. Most therapies consist of multiple doses of several days duration, if not several years. It is necessary, therefore, to be able to asess plasma concenterations, both the peak which much be at or below the MTC and the trough which must be at or above the MEC for the drug to be effective under these conditions. Thus when we dose a patient, the concentration profile must be within the Therapeutic Range during the entire time that the patient is taking the drug. We can calculate the plasma concentrations in the followin manner. In the simplest model, suppose we give a drug by IV Bolus (because the math is simpler). The equation which would result be Cp = Cp 0 ⋅ e
I.V. Bolus Multiple Dose
( –kt )
D ( – kt) = ---- ⋅ e V
(EQ 10-1)
The peak would be
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10-5
Dosage Regimen (Healthy, Aged, and Diseased Patients)
(EQ 10-2)
1 D Cp max = ---V
If we allowed the drug to be eliminated for
τ
hours, the trough would be:
1 D ( –( k ⋅ τ ) ) Cp min = ---- ⋅ e V
(EQ 10-3)
Upon giving a second dose, prior to the complete removal by the body of the first dose the Cp max 2 would be the second dose plus what was left from the first dose: 2 D D ( – ( k ⋅ τ ) ) Cp max = ---- + ---- ⋅ e V V
and the
Cp min
2
would be
Cp max
2
times
e
( –( k ⋅ τ ) )
(EQ 10-4)
, thus:
2 ( –( k ⋅ τ ) ) ( –( 2k ⋅ τ ) ) : Cp min = D ---- ⋅ e + D ---- ⋅ e
V
After n doses, the
Cp max
n
(EQ 10-5)
V
would be:
n ( –( k ⋅ τ ) ) ( –( ( n – 1 ) k ⋅ τ ) ) Cp max = D ---- + D ---- ⋅ e +… +D ---- ⋅ e V V V
while the
Cp min
n
(EQ 10-6)
would be:
n ( –( k ⋅ τ ) ) ( – ( 2k ⋅ τ ) ) ( – ( nk ⋅ τ ) ) Cp min = D ---- ⋅ e + D ---- ⋅ e +… +D ---- ⋅ e V V V
(EQ 10-7)
Subtracting equation 10-7 from equation 10-6 to eliminate the series yields: n – ( nk ⋅ τ ) n D D ( – ( nk ⋅ τ ) ) D = ---- ⋅ ( 1 – e ) Cp max – Cp min = ---- – ---- ⋅ e V V V
Cp min
n
is also
n
Cp max ⋅ e n
–( k ⋅ τ )
n
(EQ 10-8)
which means that n
n
Cp max – Cp min = Cp max – Cp max ⋅ e
–( k ⋅
τ)
n
= Cp max ( 1 – e
Equating equation 10-8 and equation 10-9 and solving for –( nk ⋅
–( k ⋅ τ )
Cp max
)
n
(EQ 10-9)
yields:
τ)
n –e Cp max = D ---- ⋅ 1---------------------------V 1 – e–( k ⋅ τ )
At large n, e – ( nk ⋅ τ ) ⇒ 0 and thus the steady state maximum, Basic Pharmacokinetics
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(EQ 10-10)
Cp max
ss
, is: 10-6
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Cp max
and the steady state minimum,
ss
1 = D ---- ⋅ ------------------------V 1 – e–( k ⋅ τ )
Cp min
Cp min
ss
ss
(EQ 10-11)
, is : –( k ⋅
τ)
D e = ---- ⋅ ------------------------V 1 – e –( k ⋅ τ )
(EQ 10-12)
In order to make a general equation set from equation 10-11 and equation 10-12, let N be the number of half lives in a dosing interval, k = ( ln ( 2 ) ) ⁄ t 1 ⁄ 2
. Substituting into the function
e
–( k ⋅ τ )
, yields
e
τ
, and
N = --------t1 ⁄ 2 –( k ⋅
τ)
N = 1 --- 2
and
thus equation 10-11 becomes: Cp max
ss
1 = D ---- ⋅ -------------------V 1 N 1 – --- 2
(EQ 10-13)
and equation 10-12 becomes :
Cp min
ss
1 --- 2
N
= D ---- ⋅ --------------------V 1 N 1 – --- 2
(EQ 10-14)
The average drug concentration under these conditions would be equivalant to the steady state concentration attained by an infusion of the same rate, i.e. if we were to give a multiple dose at 200 mg every four hours (q4h) or 400 mg every eight hours (q8h), the average that would be attained would be equivelaent to the steady state plasma concentration attained by giving an infusion at 50 mg/hr, (Q = 50 mg/ hr), and thus, in the infusion, the k ⋅ τ = 0.693 ⋅ N
ss
Q = ---------k⋅V
and in multiple dosing
D Q = ---τ
, and
so: Cp
Oral Multiple Dosing (Approximation)
Cp
ss avg
D D 1.443 ⋅ D = ------------------- = ------------------------------ = ---------------------V ⋅ K ⋅ τ V ⋅ 0.693 ⋅ N N⋅V
(EQ 10-15)
Similar equations, although more complex, can be derived for multiple dose oral products. However, if we were agreed to live with some error these equations, with some modifications could be used to approximate multiple dose oral products. The error on both calculated Cp maxss and Cp minss would be in the direction of safety, i.e. the calculated Cp maxss would be higher and the calculated Cp minss would be lower that their respective multiple dose oral calculations. Thus, if the simpler
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10-7
Dosage Regimen (Healthy, Aged, and Diseased Patients)
equations place the Cp max ss and C minss within the Therapeutic Range, the oral multiple dose equations would also. The two modifications would be to the Dose. Bioavailability, f, must be considered, and if the drug given is not the drug measured in the blood, the salt factor, S, the difference in the molecular weight of the two compounds must be taken into account, (i.e.
MW measured S = -----------------------------MW given
). Thus, when amino-
phyline, which is a complex consisting of two theophyline molecules and an ethylinedyamine molecule is given, but theophyline is measured, the salt factor, MW Theo 2 ⋅ 180.17 - = ------------------------ = 0.857 S = ----------------------MW Amino 420.44
. Thus for oral multiple dose, we can approximate
for using IV bolus equations as such : Cp max
Cp
ss avg
ss
S⋅f⋅D 1 = ------------------ ⋅ --------------------N V 1 – 1--- 2
S⋅f⋅D S⋅f⋅D 1.443 ⋅ S ⋅ f ⋅ D = ------------------- = ------------------------------ = -----------------------------------V ⋅ K ⋅ τ V ⋅ 0.693 ⋅ N N⋅ V
Cp min
ss
1 --- 2
(EQ 10-16)
(EQ 10-17)
N
S⋅ f⋅ D = ------------------ ⋅ --------------------V 1 N 1 – --- 2
(EQ 10-18)
because errors involved with this approximation both in maximum and minimum calculations (peak and trough) place the drug further within the Therapeutic Range, i.e. the real peaks are lower than the calculated peaks and the real troughs are higher than the calculated troughs, thus both errors are on the side of safety. Dosing Interval
The object of pharmacokinetics is to optimize therapy. By definition, that is to maintain the plasma concentration of the drug within the therapeutic range for the duration of the therapy presuming that is needed. Thus, the concentrations must stay within the MTC and MEC or ss
Cp max N MTC-----------= ------------------= 2 max ss MEC Cp min
(EQ 10-19)
by deviding equation 10-18 into equation 10-16 and simplifying. Given these limits, the maximum dosing interval, τ max , is obtained by solving for Nmax:
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Dosage Regimen (Healthy, Aged, and Diseased Patients)
MTC- Ln -----------MEC = -------------------------Ln2
N max and since
τ
max N max = ---------t1 ⁄ 2
(EQ 10-20)
by definition, τ max
= N max ⋅ t1 ⁄ 2
(EQ 10-21)
is not necessarily the dosing interval of choice, but it is the maximum dosing interval attainable without sustained or controlled release delivery systems. τ max
Accepable dosing intervals are those which result in a dose being given at the same time of the day, every day. Imagine, if you will, the chaos on the nursing floor if the dose for a given drug were every 15 hours. Compliance, none too high when the patient is given a reasonable dosing interval, would go straight to the toilet if we asked the patient to take a tablet every 5.3 hours. What would be optimal would be to tie the takeing of the drug with an activity that occurs the same time every day for once a day therapy, or at least dose the same time every day. Thus, for multiple daily doses, the only regimens that work are those which when devide into 24 hours give unit answers: QD = 24/1 = q24h; BID = 24/2 = q12h, TID = 24/ 3 = q8h; QID = 24/4 = q6h; q4h; q3h; q2h. These result in decreasing orders of patient compliance (unless the patient is really motivated to take the drug every 2 hours - forget it.) Thus the maximum acceptable dosing interval would be the largest acceptable dosing interval below the τ max . So, for example if τ max is 15.7 hours, the maximum acceptable dosing interval would be 12 hours. we could also dose every 8, 6 or 4 hours if necessary.
Basic Pharmacokinetics
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10-9
Dosage Regimen (Healthy, Aged, and Diseased Patients)
10.2 Diseases - Dosing the Compromised Patient As previously discussed in the chapter on clearance, diseases result in changes in clearance. These are routinely as a consequence of changes in organ function or blood flow to the organ. Diseases which cause a change in clearance, do so by changing either the elimination rate constant, K, or the volume of distribution, Vd, or both. Thus the fractional change in total body clearance : Cl tot ∗ K∗ ⋅ V∗ F Cl tot = -------------= -----------------K⋅V Cl tot Where X* indicates new or changed variable.
(EQ 10-22)
In general, if
0.80 < F Cl < 1.2 , tot
changes in dosage regimen are not necessary. In order to return a previously controlled healthy pateint back to the therapeutic range, a general rule of thumb is suggested as an initial starting point. The desease modifies K (as t1/2) and V. As pharmacists, we can modify D and τ . These variables are paired in the above equations (equation 10-16 and equation 10-18), D with V (in with
τ
(in
τ
N = --------t1 ⁄ 2
S----------------⋅f⋅D V
) and
t1 ⁄ 2
). If the physiological change is a change in V, the pharmacist
would recommend a change in D proportionally, and if the half life changes, the pharmacist would recommend a change in the dosing interval proportionally. Remenber, the object is to get the plasma concentrations back to where they were prior to the illness. The only problem is that we are limited to these recommended changes being incremental and not continuous. That is a change in dose is limited to the available dosage forms and strengths and a change in dosing interval is limeted to the accepatable dosing interval. Protein Binding
If the drug is highly protein bound, the object would be to get the free concentration back to what it was prior to illness. Consequently, equation 10-16, equation 10-17, and equation 10-18 would be rewritten thus: Cp
Cp
ss fu ⋅ S ⋅ f ⋅ D ss 1 = f u ⋅ Cp = -------------------------⋅ -------------------N V max free max 1 – 1--- 2
ss fu ⋅ S ⋅ f ⋅ D fu ⋅ S ⋅ f ⋅ D ss = fu ⋅ C p = -------------------------= -----------------------------V⋅K⋅τ V ⋅ 0.693 ⋅ N avg free avg
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REV. 99.4.25
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(EQ 10-23)
(EQ 10-24)
10-10
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Cp
1 --- 2
N
ss fu ⋅ S ⋅ f ⋅ D ss = fu ⋅ Cp = -------------------------⋅ -------------------N V min free min 1 – 1--- 2
(EQ 10-25)
Some interesting and unexpected things result from these relationships. Since plasma or blood concentrations are usually measured, if a drug is highly protein bound and the desease results in upsetting that equilibrium, you might see toxicity resulltling from normal or even subtherapeutic measured concentrations. More on that in the chapter on protein binding.
Basic Pharmacokinetics
REV. 99.4.25
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10-11
Dosage Regimen (Healthy, Aged, and Diseased Patients)
10.3 Problems
Basic Pharmacokinetics
REV. 99.4.25
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10-12
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Alprazolam Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 1)
AHFS 00:00.00 GPI: 0000000000
Juhl, R. et al., "Alprazolam pharmacokinetics in alcoholic liver disease", Journal of Clinical Pharmacology, Vol.24, (1984), p. 113 - 119.
Alprazolam is an anti-anxiety agent which is metabolized to 4-hydroxy and α -hydroxy metabolites. In this study, patients with cirrhosis of the liver and healthy patients were each given doses of 1.0 mg of Alprazolam. The following data is for healthy patients. PROBLEM TABLE 10 - 1.
Alprazolam
Dose
1.0 mg BID 1.16 L/kg 1.22 529.3
AUC
Assume that your patient weighs 70 kg when answering the following: 1.
Find k.
2.
Find the MRT.
3.
Find the .
4.
Find the AUMC.
5.
Find τ .
6.
What is N?
7.
What is the patient's maximum plasma concentration, , under this dosage regimen.
8.
What is the patient's average plasma concentration, , under this dosage regimen.
9.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
REV. 99.4.25
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10-13
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Cefixime Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 2)
AHFS 00:00.00 GPI: 0000000000
Faulkner, R. et al., "Pharmacokinetics of cefixime after once-a-day and twice-a-day dosing to steady state", Journal of Clinical Pharmacology, Vol.27, (1987), p. 807 - 812.
Cefixime is a broad-spectrum cephalosporin which is active against a variety of gram positive and gram negative bacteria. In this study, patients received a 200 mg oral dose of cefixime twice daily. PROBLEM TABLE 10 - 2.
Cefixime
Dose
200 mg BID 3.3 hours 286 32
AUC
14.12
1.
Find k.
2.
Find MRT.
3.
Find Cl.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
REV. 99.4.25
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10-14
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Cefpodoxime Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 3)
AHFS 00:00.00 GPI: 0000000000
Borin, M. et. al., "Pharmacokinetics and tolerance studies of cepodoxime after single-and multiple-dose oral administration of cefpodoxime proxetil", Journal of Clinical Pharmacology, Vol.31, (1991), p. 1137 - 1145.
Cefpodoxime proxetil is a third-generation, broad-spectrum cephalosporin which is given by the oral route. It is a prodrug which is converted in vivo to cefpodoxime which inhibits bacterial cell wall synthesis by binding to penicillinbinding proteins. PROBLEM TABLE 10 - 3.
Cefpodoxime
Dose
100 mg BID 2.1 hours 271 79.1
AUC
6.9
1.
Find k.
2.
Find MRT.
3.
Find Cl.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
REV. 99.4.25
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10-15
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Cefprozil Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 4)
AHFS 00:00.00 GPI: 0000000000
Lode, H. et. al., "Multiple-dose pharmacokinetics of cefprozil and its impact on intestinal flora of volunteers", Antimicrobial Agents and Chemotherapy, Vol.36, (1992), p. 144 - 149.
Cefprozil is a broad-spectrum cephalosporin which is given by the oral route. In this study subjects received 500 mg doses of cefprozil every twelve hours for eight days. PROBLEM TABLE 10 - 4.
Cefprozil
Dose
500 mg q. 12 hours 55.11minutes 310.25 277.50
AUC
27.80
1.
Find k.
2.
Find MRT.
3.
Find Cl.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
REV. 99.4.25
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10-16
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Clobazam Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 5)
AHFS 00:00.00 GPI: 0000000000
Greenblatt, D. et. al., "Reduced single-dose clearance of clobazam in elderly men predicts increased multiple-dose accumulation", Clinical Pharmacokinetics, Vol.8, (1983), p. 83 - 94.
Clobazam is an agent used in the treatment of anxiety. In this study, patients received 10 mg dose of clobazam daily. PROBLEM TABLE 10 - 5. Clobazam
Dose
10 mg daily 180 hours
AUC
1.
Find k.
2.
Find MRT.
3.
Find the ?
4.
Find the AUMC.
5.
Find τ .
6.
What is N?
7.
What is the patient's maximum plasma concentration, , under this dosage regimen.
8.
What is the patient's average plasma concentration, , under this dosage regimen.
9.
What is the patient's minimum plasma concentration, , under this dosage regimen.
10. Your patients renal function drops to 50% of normal. What would be a new dosing regimen under these conditions? (Assume that you want to keep < 110% of the normal and that you want to keep > 90% of the normal .)
Basic Pharmacokinetics
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10-17
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Maloprim
(Problem 10 - 6)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Edstein, M., Rieckmann, K., and Veenendaal, J., "Multiple-dose pharmacokinetics and in vitro antimalarial activity of dapsone plus pyrimethamine (Maloprim ) in man", British Journal of Clinical Pharmacokinetics, Vol.30, (1990), p.259 - 265.
Maloprim is an agent which contains both dapsone and pyrimethamine. In this study, healthy volunteers were given 100 mg of dapsone plus 12.5 mg pyrimethamine weekly. The following data is for dapsone: PROBLEM TABLE 10 - 6. Maloprim
Dose
100 mg weekly 22.6 hours
AUC
35.0
1.
Find k.
2.
Find MRT.
3.
Find Cl.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
REV. 99.4.25
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10-18
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Doxycycline
(Problem 10 - 7)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Shmuklarsky, M. et. al., "Failure of doxycycline as a causal prophylactic agent against Plasmodium falciparum malaria in healthy nonimmune volunteers", Annals of Internal Medicine, Vol.120, (1994), p. 294 - 298.
Doxycycline is an antibiotic which has been recommended for prevention of malaria in people traveling to areas endemic to chloroquine-resistant P. falciparum malaria who are unable to take mefloquine. This study determined that doxycycline is not effective for this use. Volunteers were given 100 mg doses of doxycycline daily for 10 days. PROBLEM TABLE 10 - 7.
Doxycycline
Dose
100 mg daily 21.9 hours
AUC
40.7
1.
Find k.
2.
Find MRT.
3.
Find Cl.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
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10-19
Dosage Regimen (Healthy, Aged, and Diseased Patients)
DQ-2556 Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 8)
AHFS 00:00.00 GPI: 0000000000
Nakashima, M. et. al., "Phase I study of DQ-2556, a new parenteral 3-quaternary ammonium cephalosporin antibiotic", Journal of Clinical Pharmacology, Vol.33, (1993), p. 57 - 62.
DQ-2556 is a new broad-spectrum cephalosporin which is active against many bacteria including Pseudomonas aeruginosa. Subjects in this study were each given a 2000 mg infusion of DQ-2556 over 5 minutes every 12 hours for a total of 9 doses. PROBLEM TABLE 10 - 8.
DQ-2556 Dose
2000 mg infusion over 5 minutes 17.6 L 8.5 7.1
AUC
241.0
1.
Find Cl.
2.
Find k.
3.
Find the MRT.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
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10-20
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Erythropoetin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 9)
AHFS 00:00.00 GPI: 0000000000
Gladziwa, U., et al., "Pharmacokinetics of epoetin (recombinant human erythropoietin) after long term therapy in patients undergoing haemodialysis and haemofiltration", Clinical Pharmacokinetics, Vol.8, (1983), p. 83 - 94.
Erythropoetin is a regulatory hormone of red blood cells. In this study patients with end-stage renal disease were given 150 U/kg of epoetin three times a week. PROBLEM TABLE 10 - 9.
Glipizide
Dose
150 U/kg t.i.w. 7.7hours
Cl
5.4
Assuming that your patient weighs 65 kg, please determine the following: 1.
Find k.
2.
Find MRT.
3.
Find the .
4.
Find the AUC.
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
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10-21
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Flecainide Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 10)
AHFS 00:00.00 GPI: 0000000000
Forland, S. et al., "Flecainide pharmacokinetics after multiple-dosing in patients with impaired renal function", Journal of Clinical Pharmacology, Vol.28, (1988), p. 727 - 735.
Flecainide acetate is a class 1C anti-arrhythmic agent which is used in the treatment of ventricular and supraventricular arrhythmias. In this study, subjects were given doses of 100mg of flecainide orally twice daily. PROBLEM TABLE 10 - 10. Flecainide
Dose
100 mg BID 7.4 L/kg 486 89
AUC
3.429
Assume that your patient weighs 70 kg when calculating the following: 1.
Find Cl.
2.
Find k.
3.
Find the MRT.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
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10-22
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Glipizide Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 11)
AHFS 00:00.00 GPI: 0000000000
Kradjan, W. et al., "Glipizide pharmacokinetics: effects of age, diabetes, and multiple dosing", Journal of Clinical Pharmacology, Vol.29, (1989), p. 1121 - 1127.
Glipizide is a second-generation oral hypoglycemic agent used in the treatment of non-insulin-dependent (type II) diabetes. In this study, both diabetic and non-diabetic elderly men were each given doses of 2.5 mg of glipizide daily for five days. The data for the non-diabetic group is given below. PROBLEM TABLE 10 - 11.
Dose
2.5 mg daily 4.0 hours 0.47
AUC
2325.4
1.
Find k.
2.
Find MRT.
3.
Find Cl.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
REV. 99.4.25
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10-23
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Lomefloxacin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 12)
AHFS 00:00.00 GPI: 0000000000
Hunt, T. and Adams, M., "Pharmacokinetics and safety of lomefloxacin following multiple doses", Diagn Microbiol Infect Dis, Vol.12, (1989), p. 181 - 187.
Lomefloxacin is a quinolone antibiotic which is useful against both Gram-positive and Gram-negative bacteria. It is used in the treatment of urinary tract infections and lower respiratory tract infections. A dose of 400 mg of lomefloxacin was given twice daily to healthy patients. PROBLEM TABLE 10 - 12. Lomefloxacin
Dose
400 mg BID 7.32 hours
AUC
61.67
1.
Find k.
2.
Find MRT.
3.
Find Cl.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
REV. 99.4.25
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10-24
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Loratadine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 13)
AHFS 00:00.00 GPI: 0000000000
Radwanski, E. et al., "Loratadine: multiple-dose pharmacokinetics", Journal of Clinical Pharmacology, Vol.27, (1987), p. 530 533.
Loratadine is an antihistamine which is orally active. In this study, healthy, male volunteers were each given a 40-mg loratadine capsule daily for ten days. PROBLEM TABLE 10 - 13. Loratadine
Dose
40 mg daily 14.4 hours
AUC
96.0
1.
Find k.
2.
Find MRT.
3.
Find Cl.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
REV. 99.4.25
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10-25
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Methamphetamine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 14)
AHFS 00:00.00 GPI: 0000000000
Cook, C., et al., "Pharmacokinetics of oral methamphetamine and effects of repeated daily dosing in humans", Drug Metabolism and Disposition, Vol.20, (1992), p. 856 - 861.
Methamphetamine is a CNS stimulant which is used in the treatment of attention deficit disorder and obesity. In this study, subjects were given a 0.125 mg/kg dose of methamphetamine daily. PROBLEM TABLE 10 - 14. Methamphetamine
Dose
0.125 mg/kg daily 8.46 hours 65.0 212
Assume that your patient weighs 70 kg when calculating the following: 1.
Find k.
2.
Find MRT.
3.
Find the .
4.
Find the AUC.
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
REV. 99.4.25
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10-26
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Mexiletine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 15)
AHFS 00:00.00 GPI: 0000000000
Gillis, A. and Kates, R., "Clinical pharmacokinetics of the newer antiarrhythmic agents", Clinical Pharmacokinetics, Vol.9, (1984), p. 375 - 403.
Mexiletine is a class Ib antiarrhythmic agent. In this study, volunteers each received a 1600 mg dose orally each day. PROBLEM TABLE 10 - 15. Mexiletine
Dose
1600 mg daily 380 L 681
1.
Find k.
2.
Find the MRT.
3.
Find the .
4.
Find the AUC.
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
REV. 99.4.25
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10-27
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Moxisylyte Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 16)
AHFS 00:00.00 GPI: 0000000000
Costa, P. et al., "Multiple-dose pharmacokinetics of moxisylyte after oral administration to healthy volunteers", Journal of Pharmaceutical Sciences, Vol.82, (1993), p. 968 - 971.
Moxisylyte is an α -adrenergic blocker which has been used in Europe for some times as a vasodilator in the treatment of such disease states as age-associated mental impairment, acrocyanosis, Raynaud's syndrome, vascular cochlearvestibular disorders, glaucoma, and benign prostatic hyperplasia. PROBLEM TABLE 10 - 16. Moxisylyte
Dose
240 mg BID 2.28 hours
AUC
11186
1.
Find k.
2.
Find MRT.
3.
Find Cl.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
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10-28
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Naproxen Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 17)
AHFS 00:00.00 GPI: 0000000000
Ouweland, F. et. al., "Hypoalbuminaemia and naproxen pharmacokinetics in a patient with rheumatoid arthritis", Clinical Pharmacokinetics, Vol.11, (1986), p. 511 - 515.
The pharmacokinetics parameters of naproxen were looked at in patients with rheumatoid arthritis in this study. A patient received a dose of 500 mg of naproxen orally twice daily. PROBLEM TABLE 10 - 17. Naproxen
Dose
500 mg BID 9.0 L
AUC
1134
1.
Find Cl.
2.
Find k.
3.
Find the MRT.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
REV. 99.4.25
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10-29
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Nisoldipine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 18)
AHFS 00:00.00 GPI: 0000000000
Harten, J. et al., "Influence of renal function on the pharmacokinetics and cardiovascular effects of nisoldipine after single and multiple dosing", Clinical Pharmacokinetics, Vol.16, (1989), p. 55 - 64.
Nisoldipine is a second-generation calcium-channel blocker which is under investigation for use as an anti-hypertensive agent. Nisoldipine is mainly eliminated through liver metabolism with metabolites being excreted mainly in the urine but also in the feces. The systemic clearance of nisoldipine depends greatly on liver blood flow. PROBLEM TABLE 10 - 18. Nisoldipine
Dose
10 mg BID orally 7.9 hours
AUC
5.2
1.
Find k.
2.
Find MRT.
3.
Find Cl.
4.
Find the ?
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
REV. 99.4.25
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10-30
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Pefloxacin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 19)
AHFS 00:00.00 GPI: 0000000000
Bruno, D. et al., "Bayesian versus NONMEM estimation", , Vol. , (19 ), p. 657 - 668.
Pefloxacin is an antibiotic used to treat patients who are in the intensive care unit. For this study, patients were given a 400 mg dose of pefloxacin twice daily for eight days. PROBLEM TABLE 10 - 19. Pefloxacin
Dose
400 mg BID 21.3 hours
Cl
3.77
1.
Find k.
2.
Find MRT.
3.
Find the ?
4.
Find the AUC.
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
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10-31
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Phenylpropanolamine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 20)
AHFS 00:00.00 GPI: 0000000000
Scherzinger, S., Rowse, R., and Kanfer, I., "Steady state pharmacokinetics and dose-proportionality of phenylpropanolamine in healthy subjects", Journal of Clinical Pharmacology, Vol.30, (1990), p. 372 - 377.
Phenylpropanolamine is a sympathomimetic agent which is used both for its action as a nasal decongestant and its action as an anorexiant. In this study healthy volunteers were given doses of 25 mg every four hours for a total of seven doses. It was found that phenylpropanolamine is 77% renally excreted. PROBLEM TABLE 10 - 20. Phenylpropanolamine
Dose
25 mg q. 4 hours 4.71 hours 4.08 L/kg 0.5
1.
Find k.
2.
Find MRT.
3.
Find Cl.
4.
Find the AUC.
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
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10-32
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Promethazine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 21)
AHFS 00:00.00 GPI: 0000000000
Taylor, G. and Houston, J., "Changes in the disposition of promethazine during multiple dosing in rabbits", Journal of Clinical Pharmacology, Vol.37, (1985), p. 243 - 247.
Promethazine is an agent used as an anti-histamine and a sedative. In this study rabbits weighing 2.7 to 3.3 kilograms each received a 10 mg/kg dose of promethazine every 24 hours for 14 days. PROBLEM TABLE 10 - 21. Promethazine
Dose
10 mg/kg 249 minutes 65.0
1.
Find k.
2.
Find MRT.
3.
Find the ?
4.
Find the AUC.
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
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10-33
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Rufloxacin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 22)
AHFS 00:00.00 GPI: 0000000000
Mattina, R. et al., "Pharmacokinetics of rufloxacin in healthy volunteers after repeated oral doses", Chemotherapy, Vol.37, (1991), p. 389 - 397.
Rufloxacin is a broad-spectrum, fluoroquinolone antibiotic. In this study a patient was given a loading dose of 300 mg of rufloxacin followed by 150 mg of rufloxacin daily for five days. PROBLEM TABLE 10 - 22. Rufloxacin
Dose
150 mg daily 104 L 41 10
AUC
121.5
1.
Find k.
2.
Find the MRT.
3.
Find the .
4.
Find the AUMC.
5.
Find τ .
6.
What is N?
7.
What is the patient's maximum plasma concentration, , under this dosage regimen.
8.
What is the patient's average plasma concentration, , under this dosage regimen.
9.
What is the patient's minimum plasma concentration, , under this dosage regimen.
Basic Pharmacokinetics
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10-34
Dosage Regimen (Healthy, Aged, and Diseased Patients)
Velnacrine (HP 029) Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 10 - 23)
AHFS 00:00.00 GPI: 0000000000
Puri, S. et al., "Multiple dose pharmacokinetics, safety, and tolerance of velnacrine (HP 029) in healthy elderly subjects: a potential therapeutic agent for Alzheimer's disease", Journal of Clinical Pharmacology, Vol.30, (1990), p. 948 - 955.
Velnacrine is an investigative agent which has central cholinergic action and may be beneficial in the treatment of Alzheimer's disease. Healthy, elderly, men were given doses of 100 mg twice daily in this study. It was found that 30% of the velnacrine dose was excreted unchanged. PROBLEM TABLE 10 - 23. Velnacrine
Dose
(HP 029)
100 mg BID 2.4 hours
AUC
809.5
1.
Find k.
2.
Find the MRT.
3.
Find Cl.
4.
Find the .
5.
Find the AUMC.
6.
Find τ .
7.
What is N?
8.
What is the patient's maximum plasma concentration, , under this dosage regimen.
9.
What is the patient's average plasma concentration, , under this dosage regimen.
10.
What is the patient's minimum plasma concentration, , under this dosage regimen.
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10-35
Dosage Regimen (Healthy, Aged, and Diseased Patients)
10.4 Answers Alprazolam 1.
0.0631 h-1
2.
15.85 hours
3.
10.98 hours
4.
8387.81
5.
12
6.
1.093
7.
23.19
8.
16.26
9.
10.876
Cefixime 1.
0.210 h-1
2.
4.76 hours
3.
14.164 L/h
4.
67.435 L
5.
62.224
6.
12
7.
3.64
8.
3.225
9.
1.177
10.
0.259
Cefpodoxime 1.
0.330 h-1
2.
3.03 hours
3.
14.49 L/h
4.
43.91 L
5.
20.905
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10-36
Dosage Regimen (Healthy, Aged, and Diseased Patients)
6.
12
7.
5.714
8.
2.322
9.
0.575
10.
0.0442
Cefprozil 1.
0.0126 min-1
2.
76.51 minutes
3.
17.99 L/h
4.
23.83 L
5.
36.84
6.
12
7.
13.065
8.
20.98
9.
2.317
10.
2.45
Maloprim 1.
0.0307 h-1
2.
32.6 hours
3.
2.857 L/h
4.
93.16 L
5.
1141.17
6.
168
7.
7.435
8.
1.08
9.
0.208
10.
6.25
Doxycycline 1.
0.0317 h-1
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10-37
Dosage Regimen (Healthy, Aged, and Diseased Patients)
2.
31.595 hours
3.
2.457 L/h
4.
77.629 L
5.
1285.92
6.
24
7.
1.096
8.
2.42
9.
1.696
10.
1.133
DQ-2556 1.
8.299 L/h
2.
0.4715 h-1
3.
2.12 hours
4.
1.47 hours
5.
511.11
6.
12
7.
8.163
8.
114
9.
20.083
10.
0.398
Erythropoetin 1.
0.09 h-1
2.
11.11 hours
3.
3599.24 mL
4.
30092.6
5.
334291.14
6.
72
7.
9.35
8.
2713.24
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10-38
Dosage Regimen (Healthy, Aged, and Diseased Patients)
9.
417.98
10.
4.156
Glipizide 1.
0.173 h-1
2.
5.77 hours
3.
1.075 L/h
4.
6.204 L
5.
13419.37
6.
24
7.
6
8.
0.4094
9.
96.893
10.
6.396
Flecainide 1.
29.16 L/h
2.
0.0563 h-1
3.
17.76 hours
4.
12.31 hours
5.
60.91
6.
12
7.
0.975
8.
0.393
9.
0.286
10.
0.200
Lomefloxacin 1.
0.0947 h-1
2.
10.56 hours
3.
6.486 L/h
4.
68.497 L
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10-39
Dosage Regimen (Healthy, Aged, and Diseased Patients)
5.
651.27
6.
12
7.
1.639
8.
8.6
9.
5.139
10.
2.76
Loratadine 1.
0.0481 h-1
2.
20.77 hours
3.
416.67 L/h
4.
8656.17 L
5.
1994.38
6.
24
7.
1.67
8.
6.746
9.
4
10.
2.125
Methamphetamine 1.
0.082 h-1
2.
12.21 hours
3.
46.7 L
4.
2.244
5.
27.383
6.
24
7.
2.837
8.
213.74
9.
93.48
10.
29
Moxisylyte
Basic Pharmacokinetics
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10-40
Dosage Regimen (Healthy, Aged, and Diseased Patients)
1.
0.304 h-1
2.
3.29 hours
3.
0.0215 L/h
4.
70.574 mL
5.
36794.61
6.
12
7.
5.263
8.
3.492
9.
0.9322
10.
0.0909
Naproxen 1.
0.441 L/h
2.
0.049 h-1
3.
20.412 hours
4.
14.149 hours
5.
23147.21
6.
12
7.
0.848
8.
124.98
9.
94.5
10.
69.43
Mexiletine 1.
0.1075 h-1
2.
9.3 hours
3.
6.45 hours
4.
39.16
5.
364.17
6.
24
7.
3.723
Basic Pharmacokinetics
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10-41
Dosage Regimen (Healthy, Aged, and Diseased Patients)
8.
4.256
9.
1.632
10.
0.345
Nisoldipine 1.
0.0877 h-1
2.
11.397 hours
3.
1923.08 L/h
4.
21.92 mL
5.
59.27
6.
12
7.
1.52
8.
700.7
9.
433.3
10.
244.5
Pefloxacin 1.
0.0325 h-1
2.
30.73 hours
3.
115.85 L
4.
106.1
5.
3260.4
6.
12
7.
0.563
8.
10.68
9.
8.84
10.
7.23
Phenylpropanolamine 1.
0.147 h-1
2.
6.795 hours
3.
36.03 L/h
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10-42
Dosage Regimen (Healthy, Aged, and Diseased Patients)
4.
0.694
5.
4.715
6.
4
7.
0.849
8.
229.5
9.
173.5
10.
127.4
Promethazine 1.
0.167 h-1
2.
5.987 hours
3.
70.05 L
4.
2.56
5.
15.35
6.
24
7.
5.78
8.
436.19
9.
106.84
10.
7.92
Rufloxacin 0.0237 h-1 42.28 hours 29.30 hours 5136.6 24 0.819 3.33 2.54 1.89 Velnacrine
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10-43
Dosage Regimen (Healthy, Aged, and Diseased Patients)
1.
0.289 h-1
2.
3.462 hours
3.
123.53 L/h
4.
427.73 L
5.
2802.87
6.
12
7.
5
8.
241.33
9.
67.46
10.
7.54
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10-44
CHAPTER 11
Multicompartment Modeling
Author: Michael Makoid Reviewer: Phillip Vuchetich
OBJECTIVES 1.
This chapter is not completed.
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11-1
Multicompartment Modeling
11.1 Executive Summary
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11-2
Multicompartment Modeling
11.2 Equations
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11-3
Multicompartment Modeling
11.3 PHARMCOKINETICS: MAMMILLARY MODELS
For many drugs the equilibrium between drug concentrations in different tissues is not achieved rapidly. Thus, one of the assumptions of the one-compartment open model sometimes becomes invalid. A more complex mammillary open model is often necessary to describe mathematically the plasma concentration data (for example) seen after the administration of some drugs. The simplest mammillary open model is a two-compartment open model: for example:
Compartment One (central compartment) can be sampled through the blood (or plasma, or serum). It may consist of organs or tissues which, being highly perfused with blood, are in rapid equilibrium distribution with the blood.
Compartnent Two (peripheral compartment) cannot normally be sampled. It may consist or organs or tissues which, being poorly perfused with blood, are in slow equilibrium distribution with the blood.
The Body is the sum of both compartments. Basic Pharmacokinetics
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11-4
Multicompartment Modeling
1. Biexponential Properties of Two-Compartment Open Model
Following an intravenous bolus injection, the plasma concentration against time profile has two phases:
a. Initial phase - ( α - phase)
b. Terminal phase - ( β - phase)
On semilogarithmic paper the terminal phase is linear, indicating that initial distribution has been completed and that equilibrium has been attained. The terminal half-life ( t 1 ⁄ 2 ) can be measured from the terminal phase.
2. Intravenous Bolus Administration: Plasma Concentration Data For a one-compartment open model,
Cp = ( Cp )o e
– kt
(EQ 10-26)
i.e., the concentration of drug in the plasma declines exponentially with time
For a two-compartment open model, Basic Pharmacokinetics
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11-5
Multicompartment Modeling
Cp = ( A1 e
– αt
+ B1 e
– βt
)
(EQ 10-27)
i.e., the concentration of drug in the plasma declines biexponentially with time
2.1 Symbols
3
A 1 and B1 are intercept constants ( M ⁄ L )
–1
α and β are hybrid rate constants ( T )
V 1 is the apparent volume of unchanged drug distribution in compartment one 3
(L )
–1
k 10, k 12 , and , k 21 are “micro” rate constants ( T )
2.2 Relationships (for reference, except Eq. 3)
2
α = 0.5 [ ( k 10 + k 12 + k 21 ) + ( k 10 + k 12 + k 21 ) – 4k 10k 12 ]
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Multicompartment Modeling
2
β = 0.5 [ ( k 10 + k 12 + k 21 ) – ( k 10 + k 12 + k 21 ) – 4k 10k 12 ] D- ( α – k 21 ) A 1 = ----• ---------------------V1 ( α – β ) ( k 21 – β ) D- • --------------------B 1 = ----V1 ( α – β ) A 1 + B1 = ( Cp ) o
(EQ 10-28)
2.3 Obtaining Pharmacokinetic Parameters by “Feathering” By convention, α > β a. Plot C p against t on semilogarithmic paper
b. Find t 1 ⁄ 2 from the linear terminal phase: see “Intravenous Administration”, section A1.4a
c. Calculate the terminal hybrid rate constant ( β ) ; in reality it contains both distributive ( k 12 and k 21 ) and elimination ( k 10 ) factors.
0.693 β = ------------t1 ⁄ 2
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(EQ 10-29)
11-7
Multicompartment Modeling
d. Draw a straight line through the linear terminal elimination phase and extralpolate this line to t = 0. The intercept is equal to B1 .
e. Read estrapolated plasma concentrations ( C p ) from the plot at times equal to those given for values of C p which are prior to the terminal phase.
f. At each of these times calculate:
( C p )diff = C p – C p
g. Plot ( C p )diff against t (see Eq.8) on semilogarithmic paper. The is a “feathered” line and should decline linearly.
h. Find the half-life of the plot. It wil refer to the initial phase. Calculate,
0.693 α = -----------------------half – life
i. Measure the intercept of the “feathered” line; it will equal to A 1 (Note that usually A1 = B1 , even theoetically).
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Multicompartment Modeling
j. Calculate ( C p )o from Eq. 3
k. Calcultate V1 by Xo D = -----------------V1 = ------------( Cp )o A1 + B 1
(EQ 10-30)
. Theory
When t is large, e
– αt
<e
– βt
. Hence, Eq. 2 becomes
Cp = B1 e
– βt
(EQ 10-31)
i.e., when t is large, the concentration of the drug in the plasma declines exponentilly with time.
The extrapolated plasma concentrations are Cp = B1 e
– βt
(EQ 10-32)
Substituting from Eqs. 2 and 7a into Eq. 5,
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Multicompartment Modeling
( C p ) diff = A 1 e
– βt
(EQ 10-33)
i.e., the difference between observed and extrapolated drug concentrations in the plasma declines exponentially with time. Note (for reference only)
It is usually not informative to determine the “micro”rate constant; but see one use under the note on dosage regimens.
αB 1 + βA1 k 21 = ------------------------A 1 + B1
k 10 = αβ ⁄ k 21
k 12 = α + β – k 10 – k 21
2.4 Clearance and Volume
If model-independant equations can be used to define these terms, this is preferred.
a. Systemic Clearance (Cl) may be calculated by,
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Multicompartment Modeling
D Cl = ---------------------∞ ( AUC )o
(EQ 10-34)
b. The volume terms are complex than in a one-compartment open model. There are two terms of interest:
The apparent volume of distribution in compartment one ( V1 ) This is calculated using Eq. 6.
The apparent volume of distribution at pseudo-distribution equilibrium ( V β )
This volume may be defined only in relation to the terminal phase ( β phase), when initial distribution has been completed. D V β = -------------------------∞ β ( AUC )o
As Vβ requires calculation of the total area under the plasma concentration against time curve it is sometimes known as Varea .
c. Comparing Eqs. 9 and 10,
Cl = βV β
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(EQ 10-35)
11-11
Multicompartment Modeling
It may also be shown that,
Cl = k 10V 1
(EQ 10-36)
This follows as systemic clearance is always given by the elimination rate constant out of the body multiplied by the apparent volume of distribution in the compartment from which drug leaves the body. Comparing Eqs. 11 and 12,
k 10 V β = ------V β 1
(EQ 10-37)
Note that k 10 (the elimination rate constant) is not the same as β (the terminal hybrid rate constant).
2.5 Bioavailability
∞
Find ( AUC )0 using trapezoidal rule and, if necessary, the calculation for the terminal area.
∞ t C ( AUC )o = ( AUC )o + -----pβ
(EQ 10-38)
This is a model-independent equation.
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11-12
Multicompartment Modeling
2.6 Dosage Regimens
The maintenance dose (D) is given by the same model-independent equation as before,
D = ( C p ) ss Cl • τ
Where ( C p )ss has its same previous definition.
The loading dose ( D L ) achieves a steady-state condition quite rapidly, but only after initial distribution has been completed. It is given by the previous equation.
D D L = --------------------------– 0.693N 1–e
(EQ 10-39)
As may be expected, equations relating ( C max )ss and ( C min ) ss to ( C p )ss are as before,
Note (reference only)
All dosage regimen equations strictly apply only when,
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11-13
Multicompartment Modeling
β- 1 + k-----12 ------ = 1 k 10 k 21
For digoxin Eq. 17 has a value of 0.947 For warfarin Eq. 17 has a value of 0.990 For cephalexin Eq. 17 has a value of 0.846
This is why, despite the fact that an open two-compartment model is better description of the pharmacokinetic of these drugs, a simple open-compartment model may often be assumed for dosage regimen purposes.
3. Intravenous Bolus Administration: Compartment Two
It is not normally possible to measure drug concentrations in compartment two. However, the mass of drug can be predicted based on the ddrug concentrations observed in compartment one.
2.6 Dosage Regimens
X2 = B2 ( e
– βt
–e
– αt
)
(EQ 10-40)
k 12D Where B2 = ----------------(α – β)
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11-14
Multicompartment Modeling
Note that the equations forms bears a similarity to that seen for plasma concentrations after oral administration inot a one compartment open model.
When t is large, e
– αt
<e
– βt
. Hence, Eq. 18 becomes,
X2= B2 e
– βt
(EQ 10-41)
This is comparted to the mass modification of Eq. 7, X1= V1B1 e
– βt
(EQ 10-42)
Thus, when t is large the masses of drug in each compartment decline exponentially, and in parallel, with time. This indicates that initial distribution has been completed and equilibrium attained. If the value of X 2 reflects drug concentrations at the active site, the time of maximum concentration (and maximum pharmacological effect) is: ln ( α ⁄ β ) t max = --------------------α–β
(EQ 10-43)
4. Others Dosage Forms The equations become complex and it is therefore difficult to obtain useful parameter values without th eaid of a computer. Fortunately, because the complexity of the equations is greater than the experimental accuracy of the assays warrants, drugs that strictly require a mammillary model can be described adequately by an open one compartment for the purposes of calculating dosage regimens. 4.1 Intravenous Infusion The plasma concentrations at first rise faster than an open one compartment model profile would suggest. Later, the rise is slower. The decline, following the cessation of infusion, is biexponential. 4.2 Oral Administration Basic Pharmacokinetics
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At a time just after t max the plasma concentration may exhibit a “nose”, when compared to the profile of an open one-compartment model.
SELECTED REFERENCES
Riegelman, S., Loo, J.C.K., and Rowland, M., Shortcomings in pharmacokinetic analysis by conceiving the body to exhibit properties of a single campartment, J. Pharm . Sci., 57, 117-123 (1968).
Riegelmen, S., Loo, J.C.k., and Rowland, M., Concept of a volume of distribution and possible errors in evaluation of this parameter, J. Pharm. Sci., 57, 128-133 (l968).
Benet, L.Z. and Ronfeld, R.A., Volume terms in pharnacokinetics, J. Pharm. Sci., 58, 639-641 (l969).
Gibaldi, M Nagashima, R., ant Levy, G., Relationship between drug concentrations in plasma or serum and amount of drug in the body, J. Pharm. Sci., 58, 193197 (1969).
Metzler, C.M Usefulness of the two-compartent open model in pharmacokinetics, J. Amer. Stat. Assn., 66, 49-54 (1971).
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Gibaldi, M. and Perrier, D., Drug eliminatin and apparent volume of distribution in multicompartment systems, J. Pharm. Sci., 61, 952-954 (1972).
Gillette, J.R., The importance of tissue distribution in pharmacokinetics, J. Pharmacokinetics. Biopharm., 1, 497-520 (1973).
DRUG DISPOSITION: VOLUME TERMS
As apparent volumes of distribution are proportionality constants, and not physiological volumes, more than one term is of value.
1. Apparent Volume of Sampled Compartment ( V1 )
This relates the concentration of drug in the sampled compartment with the mass of drug present in that compartment.
It may be measured after an intravenous bolus dose: D V 1 = ------------( Cp )o or
D V1 = --------------------------∞ K ( AUC )o
(EQ 10-44)
(EQ 10-45)
It may be measured after an intravenous infusion by: ( X1 ) ss Q V 1 = -------------- = ------------------( C p ) ss K ( C p ) ss Basic Pharmacokinetics
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(EQ 10-46)
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Multicompartment Modeling
2. Apparent Volume at Pseudo-Distribution Equilibrium ( Vβ )
This volume term (sometimes known as the apparent volume of distribution of the drug in the body) requires the assumption that the drug is evenly distributed throughout the body. The assumption is not true in practice. Thus Vβ can only be defined in relation to the terminal phase ( β -phase) when equilibrium has been attained; the equation is analogous to Eq. 2.
D Vβ = -------------------------∞ β ( AUC )o
(EQ 10-47)
3. Relationships Between Apparent Volumes By secondary alebraic definition, a clearance (Cl) is always given by the first-order rate constant for removal of drug from the body multiplied by the apparent volume of distribution on the drug in the compartment from which the drug leaves the body: Cl r = k u V 1
(EQ 10-48)
Cl m = k m V1
(EQ 10-49)
Cl s = KV 1
(EQ 10-50)
However, systemic clearance is measured by D Cl s = ---------------------∞ ( AUC ) o
(EQ 10-51)
Comparing Eqs. 4 and 8, and rearranging, Basic Pharmacokinetics
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Cl s = βVβ
(EQ 10-52)
K V β = ---V β 1
(EQ 10-53)
Comparing Eqs. 7 and 9,
Selected References
Riegelman, S., Loo, J.C.K., and Rowland, M., Concept of a volume of distribution and possible errors in evaluation of this parameter, J. Pharm. Sci., 57, 128-133 (1968).
Benet, L.Z. and Ronfeld, R.A., Volume terms in pharmacokinetics, J. Pharm. Sci., 58, 639-641 (1969).
Gibaldi, M., Nagashima, R., and Levy, G., Relationship between drug concentrations in plasma or serum and amount of drug in the body, J. Pharm. Sci., 58, 193197 (1969).
Perrier, D. and Gibaldi, M., Relationship between plasma or serum drug concentration and amount of drug in the body at steady state upon multiple dosing, J. Pharmacokin. Biopharm., 1, 17-22 (1973).
Oie, S. and Tozer, T.N., Effect of altered plasma protein binding on apparent volume of distribution, J. Pharm. Sci., 68, 1203-1205 (19793).
Li;vxrX LLrLLxLS
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Drug is usuallo sampled from the centr->l compartment, designated compartment one.
1. Laplace Transforn for Compartment One
As,l
(in)(dS 1)
where As,l is Laplace Transform for mass of drug in comPartment one
s is the Laplace Operator in is the input function
dS,l is ehe disposition function for compartment one
2 InDut Functions .
.
N ehat input need not necessarily be to compartment one.
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2.1 IV Bolus
(in) - D where D is the dose
2.2 IV Infusion
Q (l-e sb)
where Q is the zero-order infusion rate, b-t when e
2-3 First-order Absorption
(in) . <
(s+ka)
where ka is first-order absorption rate constant, ant F is the fraction of D ultimately reaching the general circulation.
2-4 Dissolution and Absorpcion (type 1) (in) - krkaFD
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(s+kr) (s+ka)
where kr is tirst-order dissolution rate constant.
2 5 Dissolution and Absorption (tvpe 2)
(la)
ka(l-e sb)
s(s+k>)
where ko is zero-order dissolution rate, ceasing at time T.
2-6 Others These may be formed by adtition of functions 2-1 through 2-5
e.g., (in) - D + Q(l-e~Sb)
This denotes the simultaneous commencement of an I.V. bolus and infusion.
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3. Disposition Function for Comcartment One A driving force compartment has one or more exit rate constanes; for
.
instance, in compartment i, ehe sum of ehe first-order exit rate constants
is Ei.
As,~
n~
kca
‘a +
where q is the compartment into which input occurs, n is the number of driving force compartments,
i,j, and m are counters (maximum value of n),
kql is the first-order rate constant for transfer of drug from input compartment eo compartment one,
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kl; and kjl are the first-order rate constants for drug transfer from compartment 1 to compartment j, ant vice-versa.
3-1 Using the disposition function (a) If q-l, ehen kqlsl
(b) Tr (Pi) and fT (Pm) are coneinued produces. ~e value of Pi (or Pm) equals one when thc counter i (or m) takes on a forbidden number. For example, i-l is forbidden in the numerator, ant m-l and m-j are forbidden in the denominaeor. 3-Z E.camples
(a) one-compartmene open model (n-l,q=l)
ds,l = 1
(Eq.l)
(b) Two-compart:.ent open models (n-2.q=1) (s+E~)
ds , 1
(s+El) (s+E2) - kl2kx
(c) Three-compartmene open models(n-3,q-2)
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dS,l
k21 (s+E3)
(Eq.2)
(s+El) (s+E2) (s+E3) - kl2k21 (s+E3) - kl3t31 (5tk2)
(Eq.3)
3-3 Simplifying the Denominator The number of exponeneial terms in ehe final ineegrated equation will be equal eo the number of driving force compartments (n). This is also equal to the maximum power to which the Laplace operator (s) would nppear if
the denominator were multiplied out. Hence, the denominator is simplified n to become iXl (S+ki), where ki is a composite first-order rate constane.
(a) ds,l ‘
(b) d5 l
(c)
ts.l
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(s+kl)
(s+E2)
(s+kl) ts I k2)
k21 (s+E3)
( s+kl ) ( s+k2 ) ( s+k3 )
(Eq.la)
(Eq.2a)
(Eq.3a)
The exact meaning of [i for any model depends on the equalities evident in the denominaeors. Example for (b):
(s+kl) (s+k2) - (s+El) (s+E2) - kl2k
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4. Method of Partial Fractions Ihis method is used to solve (integrate) a Laplace Iransform providing there are no repeating factors in the denominator. Example: no 52 or (s+ki)2
‘l 1 Prepare che Laplace Transform Example: I.V. bolus into compartment one of two compartment model
( s+kl ) ( S+k2 )
4-2 Obtaining the Roots of Denominator Factors
If the factor is s, the rooc is zero
If ehe faceor is (s+ki), the root is
4-3 “Ridden-Hand” Method
(a) Deal with each factor of ehe tenominator in turn.
(b) Cover the factor with a finger, and remember its root.
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(c) Wherever the Laplace operator(s) occurs in the uncovered transform, subseitute the root for s.
(d) Multiply the resule by eses again substituting ehe rooe for s
(e) After doing (b) through (d) for each factor, simplify.
Example:
X1 ‘ D(-kl+E2)e-klt - + D(-k2+E2)e~k2t
(-kl+k2)(-k7+kl )
or C1 - D (kl-Ez)e kl + D (E2-k2)e 2
V1 (kl k2 )
or
C1 - Ale 1 + A2e-k2t
In this example the meaning of A1, A2, kl, k2, and E2 depend on the form
of the two-compartmental model.
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5. Laplace Transform for Peripheral Compartments
This is obtained by the following procedure, which is analogous eo that
employed when using the Laplace Transform table.
(a) Write the differential rate equation.
(b) Take the Laplace Transform of each side of the differential rate equation, using the table where necessary.
V1 (kl-k2)
(c) Algebraically manipulate the transformed equation until an equation having onlv one transformed dependent variable on the left-hand side is obtained.
(t) Substitute for anv known transformed dependent variables on tlle right-lland sidc of the equation.
(e) Solve (integrate) bv the method of partial fractions (tlle “hidden 1land”), and simplifv.
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6. btethod if Denominator Contains the Factor 52 This may apply to terminal “compartments”, such as urine, following an I.V. infusion. The “hidden hand” method cannot be used for the factor 52 in the denominator as it has no simple root.
6-1 Example (n=2, q^l, exit from compartmenr one): aS,U - kloQ.(l-e~sb)(s+Es)
52(5+kl) (s+k2)
where klo is the first-order excretion rate constant from compartment one.
xu ~ kloQ.E2b + ...... klk2
where Xu is the cumulative mass of drug excreted into the urine. The other factors can be used as before in the “hidden-hand” method. 6-2 Example (n-3, q-l, exit from compartment one) aS u ‘ kloQ.(l-e 5b)(S+E2)ts+E3) s2(s+kl)(s+kv)(s+k3)
Xu t kloQ- E2Elb I
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klk2k3
REFERENCES
L.Z. Benet, General treatment of linear mammillary models with elimination from any compartment as used in pharmacokinetics, J. Pharm. Sci., 61, 536-541 (1972)
D.P. Vaughan, D.J.H. Mallard, A. Trainor, and M. Mitchard, General pharmacokinetic equations for linear mammillary models with trug absorption into peripheral compartments, Europ. J. Clin. Pharmacol., 8, 141-148 (1975).
D.P. Vaughan and A. Trainor, Derivation of general equations for linear mammillary models when ehe drug is administered by different routes, J. Pharmacokin. Biopharm., 3, 203-218 (1975).
Two-Compartment Model-l
Prior inputs focuset on one-compartmcnt models, but many drugs arc charactetizet bettcr by multicompartmcnt motek. In the following three inputs, we shall bricfly tiscuss multico~_nt motek ant prcstnt a few apB plicadons. Multicompartnent motcis are not uset as fo quentlg u the one-compartment model in therapeutic trug monitonng, panly because they arc more tifficult to construct ant apply.
Gencially, muldeaw models arc appliet when th,e natural log of plasma drug consentration vcrsus time is not lincar afier an intravenous tose or when thc plasma concentration versus time psfilc cannot bc chu~ by a single cxpooential function (i.c., C, - CO e~~’). Wben the In of plasma concentration vcrsus timc is not a
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stnight line, a multicompartmcnt model must bc constructet to tescribe the change in concentrations over time.
Of the mul
models, tbe two-compartment motel is tnost fxqucntly uset. lunis model usually
of thc weU-perfia
tissues ant • “penpbexal” compartment of less weU Erfuset dssues (such as muscle ant fat). hgure 23^ shows a diagram of thc two-compartmcnt model afir an intravenous bolus tose, where:
consists of a “central’
Xf—amount of diug in centnl comva XP s amount of drug in psipheral cowt
K,2—rate const nt for transfcr of drug from cd-compartment zo petipheral compartment rne subsaipt “12” irldicses tr nsfcr from thc first (cd) to the second (peripheral) compattments.
K2, - rate constant for tgansfcr of drugfrov peripheral computment to central comp rtment lbe subscript ‘’21” indicales tr nsfcr from Ulc second (periphaal) to the fint
Xo~
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Kl2
K2l
+ Klo
•zgre23~ Gr phic reprtsentation of a zwbcompattment model.
(centri) compartments. (Nott h Kz2 and K2,—calkd micxnts.)
fintvder climinX aue consunt (similar to tbe Jr uxd paviously), i—”ting elimiXn of dmg out of tbe caul ~ into urine, feces,
esc~
A log plasma conscatration versus time curve for a two-compattment model shows a curvilinear profile—a atrved potoon followed by a straight li=. This “biexponential” curve c n bc described by two expoKntial tcrms (Flgure 23B). lEc phases of the curve may reprcstnt rapid d1stributioo to organs with high blood flow (central compuenent) and slower distnbution to organs with Ess blood flow (penphcnl compartmcnt).
Mer thc intovenous injection of a drug that follows a t_ model, thc drug consentrations in all fluids and dmms associated with tbc central compartmcnt declinc morc rapidly in tbc distributioo phasc thao during the post-diwibubon phasc. ARcr sornc
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tirnc, a “pseudocquihEutn” is attained betwcen thc ccotral compartmcot and thc dssucs atid fluids of toc pc ipheral compO thc pl sma coocentration vcrsus timc ptnfile is thco chaserized by a linear pnmcess.
For many drugs, suco as aminoglycosides, thc distributdoo phast is vcry shott (e.g., 1920 mtn). If serum consentradons are measured after this phase is compited, toe ceotral compartmcot can be ignorcd and a one-compartrKnt model adequatcly repttsents the serum coocentratioos observed. However, for drugs such as vancomycin, thc-distribution phase lasts 1-2 hr after an intravenous dosc. If plasma concentrations of vancomycin are determined within the first hour after a dose is given, thc nonlincar (multiexponential) decline of vancomycin concentrations must bc considered.
REVIEW PROBLEMS
23.1. In the twocompartment model. Xt wpresents the
23.2. The log plasms concentration vcrsus time curve for a two-compartment model is reprcsented by a (bicxponential or monoexponential) cuNe. (Sclect one.)
23.3. The first portion of the log plasma concentration ver. sus timc cune. where the log concentration r;tpidly declincs. is lomwn as the
phasc.
23.4. The final. Iinear portion of the curve is the phase.
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11- 1 ~1
I>Pur 2t 61
70 [ So
~i
~m
~.
F*wf 238 Four st ges of drug distribution nd eliminatioo following rapid intravenous injectiott. Points I, U, m, nd tv (ript) corrcspond to the points oo the plasmx concentntion curve (leR). Point 1: The injection has just becn compicted, and drug density io the cd compartment is hipcst. Drug distribution and elimination hve just begun. Poin~ 11: At midway through tbe distributioo process, the drug density in thc central compartment is falling r pidly, dulioly owing to rapid drug distributioo out of the centd ccrnpartment into the peripheral compartment. The density of drug io the peripheral compartment has not yet mched tht in the central compartmcnt. Poixt 111: Distribution equilibrium h s been attained, and drug densitics in the centd and periphed compartments arc appgoximately equal. Drug distribution in both directioos contioucs to talze place, but the ratio of drug quantitics in toe centel and peripheral compartments remains constant. At this point, the major determinant of drug disappearance from thc central compartmcnt becomes the elimination process; previously, drug disappearance W&S determined mainly by distribution. Poi/s : During this elimination phase, the drug is being ‘-drained” from both com-
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partmcnts out of the body (via the central companment) at approximatcly the same rate. (Reproduced, with permission, from Grecnblatt DJ &nd Shader Rl, Phrmacolsinetics in clinical practice, W.B. Saunders, Philadelphia. PA, 198S.)
TwoW M~
lo this input, wc soall apply mathematicaS principles to toe two-compartmcnt model to calculatc useful poarmacokinetic parameters.
horo tiscussioo of the ooe-comparaneot model, we koow that the climination zte coostant (J[) is estsmated fxm the slope of the lo pbsma coocentration vcrsus ame curve. However, in a two-canzrg tnodel, wose the lo plasma coocentration versus time curve is curvilinear, the slope varics, tepcndiog on waich porioo of toe curve is cxamined (Flgurc 24A).
In a two compartment model, the tenninal slope from the pos,t-digributive phase of the curve may bc backextrapolata~ to axnc zero (T). The oegative slope of this line is teferret-to as beta (O, aot ,B is the tennioal eliminatton tate coostant of the trug. The iotesept of this lioe on the In plasn coocentration axis is koown as B and is uset in vanous two~cotnpeneot equations.
Bc~ is similtr to K in to t it vsents the tsminal elim meion r te constant. From it, a half-life can be cal~ culS
T%, 0.693
is
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which is refenet to as the “beta haSf-Ute. ‘
lEroughout the ame th_t trug is present in the boty, tistribuiion takes place between the central ant peripheral compartmenu. We can calculate a ratc of tistribuaion using the mct rcsidxals. This methot estim tes the cffect of distribution on the ovcrall plasma concentration curvc and uses thc diZfcrcncc between thc cffect of climination and thc actual plasma consentrations to determinc thc distribution rate. In the In concentration versus vime curvc in Flgure 24A, the slope of the initial portion is determined prim--arily by the distribution rate while thc tenninal portion is determined primarily by thc climination rate.
E <~
100
50
o S~
PodwDiattibutlon Ph~
Timo Figsot 24A Plasma druy concentralions with a two-compartment model atter an imravenous bolus do*c.
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The methot of residuals may be used for calculating phamiacoMinctic parameters of thc two-compartmcnt model. FuSst, b cl:-extrapolate the terminal straight-line portion of tbe curve (Flgure 24B). If w, s. y, and 2 are actual, detennined concentration time points, let w’, ~’, y’, and 2’ xpresent points on the new (extopolated) line at the same times that tbe aaual points were obscrved. These newly generated points xpresent the cffect of clim~ ination alone, as if distribution had been instantaneous. Subtnction of the extrapolated points from the corresponding actual points (w—w’, X—~’, etc.) yields a new set of pbsma concenti-ation points for each time point. If we plot tbese new points with the appmpriate times, we generate a new line, the “residual” line (hgure 24C).
The slope of tbe xsidual line is—st, and alpha (a) is the distribution rate constant for the two-compartment system. The intrcept of the residual line is A. Therefore, witb the coocept of residuals, we attempt to separate the two pwocsscs of diseribution and climin~ jon.
Ist us now pn~cood through an exampic, applying thc metbot of xsidtis. Draw tbc plot for thc following cxampb on somilog gnph paper. A dosc of dnag is ad
10.0 50
‘ :!
Z c 10 —o fL Q
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Tlmo F~—241R Mcthod of xsiduals.
E
100 5e
•
Y’
Bz
~’
\ Sbpo = a
s -R
Timo
F;
24C Dctermination of the rcsidual linc.
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62
InPtJT 24 63
)
ministered by rapid intravenous injection, and the tollowing concentrations result:
Tkne afttr Dose (hr) 0.2S O.S 1.0 I.S 2.0 4.0 8.0 12.0 16 0
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Plasms Concentration (u~/ml)
43 32 20 14 11 6.S 2.8 1.2 0.S2
A linc is trawn connecting the last four points and intcnecting the y-axis. Then, for the first five points, cxtrapolated values can be cstimated at cach time (0.2S, O.S, 1.0,’l.S, and 2.0 hr). If the extrapolated values from the actual plasma concentrations are subtracted, a new set of points is generated (resitual concentration points) as fts w~
Tlnse dir Doz (hr) 0.2S O.S
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1.0 1.5 20
Pbsma Concentratlon (>g/ml)
^>e
Exupol ted Residud
14.S
28.S
13.S
18.S
12.3
7.7
1 1.0
3.0
!O.0
I.D
The zsidual concentrations are then plotted (on semilog paper) versus time, and the slope of that plot equals —1.8 hr~t. When the negative is dropped, this slope equals sx; we observe from the plot that the intercept (A) of the line is 4S Fg/ml. We also can estimate a from the slope of the terminal straight-line portion (equal to 0.21 hr~ ~) and 8 (equal to IS 1lg/ml).
Alpha (ex) must be greater than beta (a), indicating that drug removal from plasma by distribution into tissues proceeds at a greater rate than does drug removal from
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plasma by eliminating organs (e.g., kidncys and liver). rhc initial portion of the plot is steeper than the terminal portion.
REVIEW PR08LEMS
24.1. Dnw a log pbsma concentntion versus timc profile for a drug Oinimed by the intravenous bolus nmute and best durizZ by a two-companrnent modeJ (Figure 24D).
242. Tbe slope of the tenninal phase of the above. plot equals
243. Tbe inucept of thc tenniial portion on tbe In pbsma concentstion axis is tenned
>.~ sca(g)~ tbe tenninal const nt of the dmg s it leaves the body.
24.S. One w y to calculatc a distnbution zate is to use tbe metbod of
24.C. Tbe fint step in the metaod of residuals is to.
- the telminal straight-line portioa of the curve. 24.7. The extryolatd points aw subtmed from tbe actuaJ observed at the correspoading times.
24.8. Tbe slope of the residual line equals
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24.9. “A” is tbeof tbe In plasnu concenamtion axis by tbe line.
2410. Tbe coocept of residuals attempts to separate tbe two
processes ofand—
100
• 50
ca e
Z c 10 FQ
Fkwe 24D
Tim ljg
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Two-Compartment Model-3
The estimations of A, 8, ss, ant t perforrned in tbe last input are useful for predicing plasrrs concentotions of dmg ch~nzi by a two-compartrnent model. For
• awrst rnodel (Flgureo2SA), we know th t thc plaSsma concentration (C) t any time (t) can be dessnbet by
Cf ^ Ce e t’
where CO is the initial concentration and g is ttne climination rate. Thae two-compartment rnodel (Flgure 2SB) is thc surn of two linear components, reprcsenting distnbution ant elimination (Flgurc 2SC).
In thc sarnc w y, we can dctrminc dnag consentration (C) at ny tinx (t) by iding thc two linear components. In cach casc, A or B i-s uset for CO, ant ex or z is used for XY. Therefe
C, s
‘ +- 8 e~*’
ThiSs equation is called a •’biexponential cquation” bccausc two cxponents rc iwaled. With thc onos
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Multicompartment Modeling
wrunt rnodcl (intravenous), wherc:
C. s Ce e ~’
2B
100 501 CO 10 S
Timo Fzwe 25A Pls dmg concentrations with a one com putment model aher an insvenous bolus dose (first order elimination).
100 50
,o
z -, 10 5
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Multicompartment Modeling
~c —o
Timc ft 25B Plasma drug concentrations with 3 two compartment model after an intnvenous boluXs dose sfirst-order elimination) .
100 50
E;
tO
Sz
Tlmo Fw 25C lOr azmpo~s of a twozxpo~ (>
t) model.
~nentid” becausc thc linc is t tnodel, diffcrcnt volume of
thc equation is
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describet by onc exponents
E or thc twowrat dWibution psramctrs cxist thc centol volurnc {V¢), thc cxtrapolated volunc (V,~ ,), tbc volunc by arca (V,,, also lcnown as V~|), ant thc stcadystatc vohunc of diwibuX (V,,). Each of thcsc voluncs rclste to diffcrcat undertying assumptioos.
As in thc onc-cornpenent rnotcl, a volunc can bc calculated by
V dose dox ¢ ,~ + B Co
For thc two-compartrnent model, this volurne would bc cquivalent to thc volumc of the central compartment (V¢). Thc Ve rclates the amount of drug in thc-central compartmcnt to the concentration in the central compartment. In thc two-compartrnent model, CO is determined by cxtrapolating back to thc y-axis from the upper or initial straight-line portioo of the plot.
When we calculate the extrapolated volume of distribution (V,x,,,p), we assume that instantaneous distribution has occurred. The effect of the iniial distribubon phase is ignored:
da B
V_
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wherc B is the w-intetcept of thc line extrapolated from the terminal portion of the curve. This volume of distribution determination may not provide a useful volume term since it ovcrsimplifics the two-compartmcnt model and disregards thc distribution phasc.
Another volumc (V,,O or V~) is detcrmined from the area under the plasma concentration vcrsus time curvc and thc tcrminal climination rate constant. This volumc is related as follows:
114
lNvts S 65
vffi
dox =
CL —
ffi x AUC3
va
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This calculation is not subject to thc ovcrsimplification of V,S,,p, but it is affected by changes in clearance. The V,,,, relates the amount of drug in the body to the concentration of drug in plasrna in the post-absorpion and postwdisttibuiion phase.
A ffnal volume tenn is the volume of disttibution at steady state (V,,). Although it is not affected by changes in drug eliminadoo or ckarance, it is more difficult to calculate. One way to estimate Vs, is to use the two compartment microconstants:
V,—Ve + ~’2vf
21
or it may be estimated by more complicated methods using AUC.
Since-different methods can be used to calculate the various volumes of distribution of a two-compartment model, you should always specify the metbod used. When reading a pharrnacolcinetic study, pay particular attenion to the method for calculating thc volume of distribution.
REVIEW PROBLEMS
2S.1. The terminal eliniination rate constant tn a twoaconF putment model is
2S.2. For the two-compurtment model. complete the equo tion describing the relazionship of plasma concentration with time: C, - •
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25.3. (True or False) The equation describing elimination afer an intravenous bolus dose of a drug charauerized by a two-compartment model lequires two exponential terms.
2!;.4. A patient is given a 500-mg dose of drug by intravenous injection and the following plasma concentrations result:
Plssms Time snerConcentrstion Dese (hr)ItsSml)
Oo ss 0.7S lS
3 ,6
72.0 46.0 33.0 26.3 20.0
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16.6 1r.2 9.0 5.0 _.7 n Rr
Plot the points on semilog paper Ithree cycle) and deterTnine the following: a. ~. b. B. c. Residual concentrations for the first five points. d. A.
. t.
£ Prodictod Dlssma concentration at 1.2 hr after the
dose.
S- V,.
h. V,, (if AUC = 131.S mg/L x hr and dose SOO mg). ~ A#=W{~ SGS 2
PRi4CTlCE SET 2
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The following pxbiems are for your xview. Dcfinitions of symbols ant Icey equations = pxvitet hc~:
k
~w climinton nte cowt
C, s pbSllk tn~ _—jUSt Aher a single inuvenous injectioQ
e
- bese for the nxl log f~|eion - 2@71S
4
- nte of tose ~ion (msy be cxptesset as milligsams per bour in the sense of a coQtinuous infusion or u trug tose divitot by itifusion tune for intemtittent infusions)
V
• voluxne of tistribution
C_t - pcalc p1ssfu a ocentntis
C_ • tnmugn plsCOOCCD
a steaty state
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a stcady st te t —duFn of intsvenous inAlsion
For muldpk dose, intennittent, intnvenous bolus injeciiOQ g stF st~
—k V (I - e~t’) C_—Cw e~t’
For muletplffbse~ intctmiu, intnvenow itafi~ioo
C - t° (I ~ e t~) —~ VK (I—e~tD
C_ - Cp e~tt~4
Fot contitiuous infusion before stee st te is reacbed: C - V^°r (I - C-t’)
For continuous infusion t stcaty stac
C.4. 4 Vt Ws
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PS2*1. A 60-kg patient is begun on a continuous intravenous infusion of theophylline at 40 mgtbr. a Forty-eight hours after beginning ttte infusion, the plasma concentniion is 12 ~/ml (12 mg/L). If we assume that tbis COncentRtiOD is tbe stcady state, wb t is the tbeoobylline clearansc?
b. If the volume of distribution is cstimated to be 30 L, what are the X and halflife?
c. Since we kww V and t. what would the concentntion bc 10 hr aftcr beginning the infusion?
d. ff the idision i continuet for 3 days ant then discondauet, what woult tD plamS consentm tion be 12 hr ~r stopping the infwiont
e. If tne infusion is continuct for 3 d ys at 40 mg/ hr ant the stcaty-statc pluma concenttation is 12 Fg/tnl, WDat rate of trug infusion would liltely xsult sn a concentradon of 18 ~/ml?
f. ARcr the iocosed infusion nte above is begun, how bog would it tic to tuch a plu coo~ cenudon of 18 Fgiml?
PS2< A 60-log p ekat is st rted on 80 mS of gentunicin ewety 6 br in • I*br infusiagL
L If this pSt is us led to have an “avenge” V of 15 L ult a normal half-life of 3 hr, wtuat will be the pealc plm coœcenttadon at stcady stuc?
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b. After thc fifth tese. a peak plasma concentratlon (dsawn at thc ent of thc infusion) is S Ag/ml ant thc ttnugh consentration (drawn right befott thc sixth tosc) is 0.9 Ag/ml. What is thc patient’s xtual gentamicin half-life? What is thc xtual volume of disvribution?
c. For this patient, what dosc should bc administen:d to reach a new steady-state peak gentamicin concentration of 8 Fgiml? At this dosc. what will bc thc steadystate trouQh concentration!
II_q
tFoD~ sU
~ Tvo-ConPart-ent Open Hodel
tatlente •ufferln~ chronle tenal fellure often require h _ odiolyele. Drug~ •cy be ~dxinletered by Injeetlon Into the wenoue •lde of the he odtcl~ser •~chine
Such e rituetlon vax ~ecerlbed by L tourneeu-Scheb •t •1 (Int. J. Clln. tharv col IS 116-120 {1972)) for •lx pctlente vho received •n introvenoue done ot gent _ letn (90 ng). The neen etru- concentrctleno et sent~nicin (C~) •hovvd • blexponenttel deellne vith tl t (t).
Ce
t
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Multicompartment Modeling
(-~lllter)(oln) o 60
10
5 75
20
5 25
lO
4 80
40
• 50
50
3 95
90
3 40
150
3 10
180
2 90
240
2 55
28S
(
o. Caleelete the ter lnal half-llfe (t~) the hybrid rete eonetento
•nd O •nd the coefflelente (~1 • 3 Sl) for gent- leln to theeo he-odlulyal~ putlente
b. tor •ubJecte vith noreel renal functlon. the •yete-le elearence (Cl~) of gentanletn le •pproalaetel1 0.041 llter/ In. Sec uee 98X ef Cl~ le due to excretlon of unchanged drug. renel tciluro utll ~rkedly effect ~ent~lcin clecr~nce.
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Show vhether the petlent~ undergoinfS he odlelyele •xhlbit o nor-ol gentcnicln •yett-le elexr~nce.
Calculete the two •pperent volune ot dletrtbut10n terwe (V •nd V )
•nd the ell fnetlon rete eonetont. Uhet frectton of tho gt ted eln ln the body •fter 3 hr nlght be la the peripherel ee part~ent7 At vh t tbae doee the gent _ Icin in the peripherel ee pert~ent rench
~t
4. Under the eondittons of h odSxlyxte. coleulete the doxe tD) of
sentcoleln •tlSch vould be •dolaletered •very S hr {1) In order to ~alnteln •n everogew •teady-ctete •erus concentretlon of 4 •g/llter.
•. Wlthout the hc odielyrer two ~ale petlente •ach •xhlbited • creetinine eleorence (Cle ) of 5 •llwin; the norx~l velue le 117 t 20 •I/nin. A~using that the deereFeed Clcr {e due to • decresce In glo-eruler filtretloo rcte vhst vould h-ve beeD the renel clecrxnce (Clr) of gent~elcia In there two patlentn hed they rz cined vithout the heodlolysert Uhct done vould then need to be dsinletered every • hr to n Intoin (C~ •t 4 tert Co pcro your •eouer wStb tho thy~telea e Deek Refer neo.
Vancomycin is an antibiotic used in the treatment of endocarditis in patients allergic to penicillin. It is poorly absorbed orally and acute pain is associated with intermuscular injection. I.V. is the route of choice. After a 1 gm I.V. dose the following data is observed:
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Time (hours)Concentration (mcg/ml)
.5
51
1.0
36
1.5
28
2.0
23
3.0
18.5
4o
16
6.0
12.5
8.0
99
12.0
6.25
FIND: a) A1 ;
b) alpha c) Clearance ;
d) * of drug in peripheral compartment at equilibrium .
Clearance:
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In normals, Erythromycin is cleared 90% by metabolism and 10% by the kidney. Also, it is 90% protein bound at therapeutic levels. Welling and Creig (JPS 67, 1057-9,1978) reported an increase in the half life from 2.0 to 2.3 hours while also reporting an increase in clearance from 275 to 485 ml/min and an increase in Vdss from 57 to 100 litres when comparing normals to uremic patients. (Uremic patients suffer from inveased concentration of urea as a result of severe renal failure.) A physician has just called you and asked you to explain how he could have seen an increase of 15% in the half life and a 75% increase in clearance at the same time and what is the impact of this on antibiotic therapy for his uremic patient. In clear concise English (not techno-bable), prepare a short written answer to this request. Keep in mind that the man who requested the information is a medical professionat intelligent, and very busy.
The following data was collected from a normal patient in the revious study (Welling, op. cit.) following an IV bolus injection of 500 mg of erythromycin (as lactobionate salt ).
time (hr)concentration (mcg/ml)timeconcentration
0
12 4
3 2.6
1
6.4
4 1.9
2
4.0
6 1.2 8 0.4
~.
Fmd the peak time in the peripheral compartment.the fraction of the drug in the peripheral compartment at four hours.
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Two compartment model
Spectinomycin (TROBICIN^TR^T) is an aminocyclitol antibiotic shown to be active against most strains of NEISSERIA GONORRHOEAE at a minimum inhibitory concentration of 20 mcg/ml. The usual adult dose is 2 g (4 g in areas of known resistance) given I.M. through a 20 gauge needle. Initial studies were done by the company to determine the pharmacokinetic parameters of the drug. The data from a single IV Bolus dose of 0.5 g is as follows.
Time (minutes)Concentration 10
63
20
51
30
43
45
35
60
30
120
183
240
7.6
360
3.2
FIND:
(5 points) a) A^V1^V = (5 points) b) B^V1^V = (5 points) c) Clearance =
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(5 points) d) t max in the peripheral compartment = (10 points) e) % of drug in peripheral compartment at equilibrium =
Spectinomycin (TROBICINR) is an aminocyclitol antibiotic shown to be active against most strains of NEISSERIA GONORRHOEAE at a minimum inhibitory concentration of 20 mcg/ml. The usual adult dose is 2 g (4 g in areas of known resistance) given I.M. through a 20 gauge needle. Initial studies were done by the company to determine the pharmacokinetic parameters of the drug. The data from a single IV Bolus
dose of 0.5 g is as follows. Time (minutes)Concentration 10
63
20
51
30
43
45
35
60
30
120
18.3
240
7.6
360
3.2
b} B c) Ciearance d) t max in the peripheral compartment e) % of drug in peripheral compartment at equilibrium
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The following information is offered from a 142 mg IV bolus dose of grisiofulvin given to a 73 Kg man.
Time mcg/ml 1
1.67
2
1.22
3
.97
4
.83
6
.66
8
.56
12
.42
18
.27
24
.17
30
.11
Find A1, B1, alpha, beta, K10, K12, K21, Peak time in the peripheral compartment, % in the peripheral compartment at equilibrium. Can you assume that this drug can be estimated by a one compartment model upon multiple dosing ?
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Multicompartment Modeling
Quinidine is current-ly used to treat ventricular and supraventricular arrythmias.- It is available as a sulfate, gluconate, and polygalactouronate which contain 83%, 62* and 60* by weight free base (pRa 8.6). Qunidino sulfate is available in 200, 260, 300 and 325 mg tablets, while quinidine gluconate is available in 325 mg tablets. The following pharnacokinetic parameters are reported by Ueda:
PARAMETER C1 renal (ml/min/kg)
C1 hepatic (ml/min/kg) V1 (L/kg)
V beta (L/kg) F (oral)
t 1/2 alpha (min) t 1/2 beta (hr)
MTC (mic/ml) NEC (mic/ml)
Population Ave (+ SD) 0.93 (0.52) 3.26 (1.74) 0.66 (0.38)
2.61 (1.10) 0.71 (0.16)
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6.69 (4.03) 6.44 (1.63) 8.53 (0.81)
*
,
1.85
(0.19)
% bound to protein80.00(2.50)
During the last pharmacokinetic exam you noticed •ome cardiac problems. Wh n you ch-cked with your physician, He prescribes quinidine. What dose should you be on?
Later, he diagnoses mononucliosis from lack of sleep and poor •ating h bits •s weli as cardiac arrythmias. Your liver funcion has dropped to 60* of normal. He prescribes quinidine for you. What is the dose that you should be on ?
ie. -
ll-3
Quinidine is currently used to treat ventricular and supraventricular arrythmias. It is available as a sulfate, gluconate, and polygalactouronate which contain 83%, 62% and 60% by weight free base (pKa 8.6). Qunidine sulfate is available in 200,
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260, 300 and 325 mg tablets, while quinidine gluconate is available in 325 mg tablets. The following pharmacokinetic parameters are reported by Ueda:
PARAMETERPopulation Ave (+ SD) C1 renal (ml/min/kg)0.93 (0.52) C1 hepatic (ml/min/kg)3.26 (1.74) V1 (L/kg)0.66 (0.38) V beta (L/kg)2.61 (1.10) F (oral)0.71 (0.16) t 1/2 alpha (min)6.69 (4.03) t 1/2 beta (hr)6.44 (1.63) MTC (mic/ml)8.53 (0.81) MEC (mic/ml)1.85 (0.19)
% bound to protein 80.00 (2.50)
During the last pharmacokinetic exam you noticed some cardiac problems. When you checked with your physician, He prescribes quinidine. What dose should you be on?
Later, he diagnoses mononucliosis from lack of sleep and poor eating habits as well as cardiac arrythmias. Your liver funcion has dropped to 60% of normal. He prescribes quinidine for you. What is the dose that you should be on ?
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Multicompartment Modeling
Methotrexate is a cytolytic used for the treatment of acute leukemia and other forms of cancer. After a a 400 mg/kg dose the following data was recorded for a 12 y/o boy.
Time (hours) 6 12 18 36 48 60 72 90
Concentration (mcg/ml) 360 70 15
2 1.2 0.46 0.36 0.15
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Multicompartment Modeling
bA) B c) Clearance d) % of drug in peripheral compartment at equilibrium
PROBLEM SET
zumper w~wFS t
L) v
, ~ s D fv
Patients suffering chronic renal failure often require hemodialysis. Drugs may be administered by injection into the venous side of the hemodialyzer machine.
Such a situation was described by L*tourneau-Saheb et al (Int. J. Clin. PKinmacol., 15, 116-120 (1977)) for six patients who receivet an intravenous dose 7 gentamicin (90 mg). The mean serum concentrations of gentamicin (Cs) showed a biexponential decline with time (t).
(~9cyml)
(mRn)
6.60
10
5-75
20
5.25
30
4.80
40
4.50
50
3.95
90
3.40
150
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Multicompartment Modeling
3.10
180
2.90
240
2.55
.
285 .
(a) Calculate the values of t1/29 B, Bj, ~, A~, and V1. LFtourneau Saheb et al reported,
tl/2 = 5.50 t 0.77 hr (mean i SD) B s 0.0022 t 0.0004 min 1 B1 s 4.76 f 0.62 vg/ml
v • 0.053 t 0.009 min A1 a 3.47 s 1.01 pg/ml V1 s 11.3 ffi 2.0 litre
(b) Calculate C1 and VB LFtourneau-Sahleb et al reported,
C1 s 40.8 t 7.8 ml/min VB s 19.2 h 2.7 litre
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_1
11-33
98X of the systemic clearance of gentamicin is composed of renal clearance-of unchanged drug, so that renal failure woult severely alter gentamRcin systemic clearance. In this case, the use of the hemodialyzer gave a systemic clearance close to the 41.0 ml/min seen in patients with normal renal function.
(c) Under the conditions of hemodialysis, calculate the dose (D) of gentamicin which would be administered every 8 hr (t) in order to maintain an Waverage” steady-state serum concentration of 4 ug/ml. What would be (Cmax)ss and (Cmin)ss?
(d) Without the hemodialyzer two male patients each exhibited a creatinine clearance (Clcr) of S ml/min; the noW l value is 117 + 20 ml/min. Assuming that the decreased ClCr is due to a decrease in glomerular filtration rate, what would have been the renal clearance (Clr) of gentamicin in these two patients had they remalned without the hemodialyzer? What dose would then need to be administered every 8 hr to maintajn (~s) at 4 ug/ml? Compare your answer with the Physician’s Desk Reference (5677).
Your third patient comes from Edelman et al (Clin Pbarmacol Therap 35:382-6 (1984)). He studied metotrexate is artbritic patients. The pharmacokinetic parameters gleened from the 10 mg IV data arc:
A1 (mg/L) 0.663 Alpha (hr-1) 059 B1 (mg/L) 0.073 Beta (hr-1) 0.097
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Multicompartment Modeling
15) What is the AUC (0 to inf) (mg / L * hr) ?
a) 0.75 b) 1.08 c) 1.12 d) 1.88e) 9.1
16) What is V1 (L) ? a) 13.6 b) 15.1 c) 253 d) 36.1e)103.1
17) What is the clearance (L/hr) ? a) 53 b) 8.9 c) 133 d) 15.4e) 17.2 18) What is V(beta) (L) ? a) 13.6 b) 253 c) 36.1 d) 45e 55
19) What is the t(max) in the perepheral compartment (hr) ? a) 0 b) 1.8 c) 3.7 d) 5.1 e)6.1
20) What percent of the drug is in the perepheral compartment at cquilibrium ? a) 25 b) 50 c) 75 d) 100 e) 125
Two compartment: It has been proposed (Ionescu et al. Clin Pkin 14:178186,1988) that morphine injectcd dirtctly into thc spioal chord would give significant analgesia. The following is a CSF concentration - time profile resulting from 05 mg/lcg IV bolus dose of Morphine:
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time
conc.
(min) (mg/L)
time
conc.
(min) (mg/L)
2
251
120
323
5
181
240
173
10
142.5
360
82
20
1043
480
2.4
40
75.1
720
1.2
80
48.8
1 a) fraction of remaining drug contained in peripheral compartment at equilibrium. Lb) Can this drug be approximated by a one compartment model ? Support your contention with calculations. Lc) Calculate a rcasonable dosage regimen for the above patient to maintain the concentration within the therapeuticwindow of 50 to 5 mg/L.
Two compartment:
(1) It has been proposed that diazepam has anticonvulsant properties above 350 nanograms per milliliter. The following is a concentration - time profile resulting from 10 mg IV bolus dose of diazepam :
time
conc. time
(hrs)
(ng/ml)(hrs)
0.25
480
6.0
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0-50
400
8.0
l.o
300
10.0
2.0
170
16.0
3.0
120
24.0
on
llo
l.a) fraction of remaining compartment at equilibrium.
n’
~_
conc. (ng/ml) 100 90 85 70 53
Dlezzewm
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UA tD 20A
AD
Time
drug contained in peripheral
k21 = (1.05 * 116.8) + (0.0325 * 487.6) / (487.6 + 116.8) = 138.5 / 604.4 = 0.239
klO = (1.05 * 0.0325) / 0.239 = n ls
kl2 = 1.05 + 0.0325 - 0.15 - 0.239 =n7
B2 = (0.7 * lOOOOmic) / (1.05 - 0.0325) = 6880
V1 = lOOOOmic / 604.4 mic/l = 16.55
X2 / total = 6880 / t6880 + (16.55 * 116.8mic/1)]
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= 0.78
l.b) Can this drug be approximated by a one compartment model? Support your contention with calculations.
AUC = (487.6 / 1.05) + (116.8 / 0.0325) = 464.4 + 3594.8 = 4058.2
Since the beta phase contributes more than 80% of the total AUC, the model can be collapsed to a one compartment model.
3594.8 / 4058.2 = 0.886 = 89%
l.c) Calculate a reasonable dosage regimen for the above patient to maintain the concentration within the therapeutic window of 350 to 1100 nanograms per ml.
K = 1/MRT = 1 / (111007 / 4058.2) = 1/27.35 = 0.0253 per hr t 1/2 = 0.693 / 0.0253 = 27.39 hr 2^N^ = 1100 / 350 = 3.14 N = ln 3.14 / ln 2 = 1.65
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Tau max = 27.39 * 1.65 = 45.19 hr drop the dosing interval to 24 hours now need to find a new N
N = 24 / 27.39 = 0.876
Vss = lOOOOmic / (4058.2mic*hr/1 * 0.0253/hr) = 97.25L
To find dose: l.lOOmg/L = (D / 97.25 L) * tl / (1 - 0.5^0.876^)] D = 49 mg daily (aggressive therapy)
0.350mg/L = (D / 97.25L) * t 1 / (1 - 0.5^0.876^)](0.5^0.876^) D = 29 mg daily (conseervative therapy)
Therefore, any dose between 30 and 50 mg a day can be given.
Two Compartment:
We have used theophylline as a test drug in many of our calculations in class. We assumed that it was a one compartment model. Look at the data from Mitenko (Clin Pharmacol and Ther,14p509 1974) for intravenous theophylline (Dose 5.6 mg/kg):
sos
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theophylilne .
i ,_
1. OD
2D
4wD 6D8S
tD
Tlm~
Time (hr.) 0.167 0.333 0.500 0.833 1.0 1.5 2.0 3.0 4.0 6.0
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8.0
Conc. (mg./L.)
24.7 Estimates of Pharmacokinetic parameters
20.3
are as follows:
18.1
A1 (mg/L)16.1
16.1
B1 (mg/L)17.9
15.6
Alpha (hr-l)4.8
14.3
Beta (hr-l)0.15
13.3
AUC (O to inf) 122.7
11.
AUMC (O to inf) 797.6
9.8
7.3 s.4
1) What is the volume of the central compartment (L/Kg) ?
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*A) 0.165 B) 0.30 C) 0.35 D) 2.9 E) 6.1
2) What is the clearance of theophylline (L/Kg/hr) ?
A) 0.007 *B) 0.046C) 2.7 D) 22 E) 142.4
3) What is the Volume of
distribution in the beta phase (Vbeta)
(L/Kg) ? A) 0.165 *B) 0.30C) 0.35D) 2.9E) 6.1
4) What percent of the equilibrium ?
A) 12 B) 23
dose is in the peripheral compartment at
*C) 46 D) 69 E) 92 treated as a one compartment model for
5) Can theophylline be dosing purposes ?
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A) No, a larger percentage of the drug is in the peripheral compartment.
B) NO, A1 and C) Yes, alpha D) Yes, the approximately
*E) Yes, the approximately
B1 are about the same value. is bigger than beta contribution of the alpha phase AUC is the total AUC. contribution of the beta the total AUC.
phase AUC is
6) What is the mean residence time (MRT) for the IV dose (hr) ?
A) 4.25 B) 5 C) 5.75*D) 6.5E) 7.25
For the same dose of an oral product information was obtained:
AUC (O to inf) AUMC (O to inf)
7) What is the MRT for the
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A) 8.0 B) 8.75
122 1400
(TheoDur^TQ^T) the following
oral dose (hr) ?
C) 9.5 D) 10.25*E) 11.5
8) What is the mean absorption time (MAT) for the oral dose (hr)?
A) 1
B) 2
C) 3
D) 4
*E) 5
9) What is the absorption rate constant for theophylline in TheoDur
(hr^T-l^T) ?
A) 0.14 B) 0.17 *C) 0.2 D) 0.25 E) 0.3
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Two compartment: The following pharmacokinetic information was obtained from EA, a
45 y/o, 70 kg, healthy male following an800 mg IV dose of vancomycin: A1 60 mg/L alpha 1.33 hr^-l B1 20 mg/L beta 0.129 hr^-l
AUC (mg/L * hr)200 AUMC (mg/L)1237
5) What is your patient’s vancomycin clearance(L/hr)?
a) 0.25 *b) 4 c) 24.74 d) 45.1 e) 155
6) What is your patient’s V(beta)(L)?
a) 10 b) 24.74 *c) 31 d) 45.1 e) 60
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7) What is your patient’s V1(L)?
*a) 10 b) 24.74 c) 31 d) 45.1 e) 60
8) What is your patient’s MRT(iv) (hr)?
a) 0.112 b) 0.162 c) 1.39 d) 4.29 *e) 6.19
9) What is your patient’s effective rate of elimination (hr^-l)? a) 0.112 *b) 0.162 c) 0.5 d) 4.29 e) 6.19
10) What is your patient’s Vss (L)?
a) 10 *b) 24.74 c) 31 d) 45.1 e) 60
11) Following an IM dose of vancomycin, the MRT(im) was calculated to be 8.185 hr. What was the absorption half life (hr)? a) 0.112 b) 0.162 *c) 1.39 d) 2 e) 6.19
RP ‘s angina was controlled on 40 mg TID of propranolol. You calculated his pharmacokinetic parameters to be: Vd (L) = 125; T1/2 (hr) = 3.1; Qh (L/hr) = 33; Bioavailability (f) = .7; Bound(%) = 95. Propranolol is essen^_ tially 100% metabolized.
12) What is his Total body clearance (L/hr)? a) 1.4 b) 14 *c) 28 d) 33 e)40.3
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13) What is his hepatic extraction ratio?
a) 0.042b) .42 *c) .85 d) 1
e) 1.22
14) Assuming propranolol to be rapidly absorbed, what is his Cpss
max free concentration (ng/mL)?
*a) 13.4 b) 19.2 c) 188 d) 268 e) 355
15) What is his Cpss max total(bound and free) (ng/mL)?
a) 13.4 b) 19.2 c) 188 *d) 268 e) 355
16) What is his Cpss min free (ng/mL)?
*a) 2.2 b) 3.2 c) 31.4 d) 44.8 e) 59.3
He is now suffering from renal failure. His half life went up to 3.8 hr while his binding went up to 98.3% because of an increase in AAG, a plasma protein to which propranolol binds.
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17) What is his new clearance (L/hr)?
a) 14.4 *b) 21.7 c) 24 d) 28 e) 36.2
18) What is his new volume of distribution (L)?
a) 79 *b) 119 c) 132 d) 153 e) 199
19) What will his new Cpss max free be if we keep him on the same
regimen (ng/mL)?
*a) 5.26 b) 7.5 c) 310 d) 442 e) 554
20) What is his new Cpss max total(ng/ml) ? a) 5.26 b) 7.5 *c) 310 d) 442 e) 554
21) What is his new Cpss min free (ng/mL)?
*a) 1.23 b) 1.75 c) 72.3 d) 103 e) 129
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22) The physician sees that the clearance has dropped and consequently the total plasma concentrations have gone up. He wants to decrease to dose to 40 mg / BID. What would you recommend? a) Sounds good to me. That will get the Cpss max total concentration back to it was before he was sick. b) His new clearance was marginally changed because the drug is cleared by the liver. I’d leave it alone. c) I think the change from TID to BID is a bit much. How about lowering it to 30 mg instead of 40 mg TID. *d) We need to increase the dose, not lower it. I’d recommend 40 mg QID. e) We need to increase the dose, not lower it. I’d recommend 80 mg QID.
Two compartment
Methotrexate is a cytolytic used for the treatment of acute leukemia and other forms of cancer. After a a 400 mg/kg dose the following data was recorded for a 12 y/o boy.
Time (hours)Concentration (mcg/ml) 6
360
12
70
18
15
36
2
48
1.2
60
0.46
72
036
Q(}
0.15
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FIND: (5 points) a) A^V1^V = (5 points) b) B^V1^V = (5 points) c) Clearance = (5 points) d) % of drug in peripheral compartment at equilibrium =
Everything.
Primidone (Mysoline^T@^T) is an effective agent in the treatment of generalized and complex partial seizures. Although primidone has anticonmlsant activity of its own, much of the activity comes from the conversion to one of its metabolites, phenobarbital. For the purposes of this exam, an insignificant error will be introduced bytheuse of intermittent IV caSculations instead of oralformultipledose. Thefolloving information for primidone and phenobarbital is available:
Primiflnne. Phenabarbital
Therapeutic Range (mg/L)5 - 12 10 - 30 Bioavailabiliq (f)1
0.9
Salt factor (S)1
09
Vd (L/Kg)
0.7
T1/2
0.6
7 hr
5 days
% Excreted undhanged2520 % Metabolized75
80
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% Metabolized to Phenobarbital25*~
% Metabolized (non Phenobarb-paths)50**
2.(20 pts) Please prepare a short consult for your M.D. in which you caSculate a dosage regimen for an 80 kg patient whidh would l~eep the plasma concentration within the therapeutic range. Prirnidone comes as 250 mg scored (can be broken in half) tablets. Indude your average, maximum and minimum primidone as well as phenobarbital steady-state concentrations in the consult.
3. (25 pts) The patient recently contracted mono. His liver function has been reduced to 50 % of normal.
Would you recommend a change in your therapy ? If you would, prepare a short consult for his M.D. as in question #2. Don’t forget that changes in hepatic clearance result in changes in both phenobarbital clearance 5~nfl fannation
4. (25 pts) Now that his mono has cleared up the doctor noticed that he has stenosis of the liver as a consequence of all t_e heavy parting that he did after the kinetics course. His liver plasma flow has been reduced to 50 % of normal. Would you recommend a dhange in therapy ? If you would, prepare a short consult for his M.D. as in question #2. Don’t forget that dhanges in hepatic dearance result in changes in both phenobarbital dearance and formation.
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11.4 Begin
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11.5 Problems
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Aspirin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 1)
AHFS 00:00.00 GPI: 0000000000
Fu, C., Melethil, S., and Mason, W., "The pharmacokinetics of aspirin in rats and the effect of buffer", Journal of Pharmacokinetics and Biopharmaceutics, Vol. 19, (1991), p. 157 - 173.
Aspirin is an analgesic/ antipyretic commonly used to relieve minor pain and is used in such conditions as rheumatic fever, rheumatoid arthritis, and osteoarthritis. The major metabolite of aspirin is salicylic acid. The following set of data was collected using rats which weighed 250 - 300 g. PROBLEM TABLE 11 - 1.
Aspirin
weight of rat
275 g
Dose
5 mg/kg IV
A1
8.58
a
1.07
B1
7.24
b
0.2
AUC
38.8
AUMC
116.0
1.
What is ?
2.
What is ?
3.
What is your patient's clearance?
4.
What is your patients MRT?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is your patient's ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is ?
14.
What is the in the peripheral compartment?
15.
What percent of the dose is in the peripheral compartment at equilibrium?
16.
Can this drug be treated as a one-compartment model for dosing purposes?
17.
If this drug can be treated as a one-compartment model, what is K ?
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Buprenorphine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 2)
AHFS 00:00.00 GPI: 0000000000
Ohtani, M., et al., "Pharmacokinetic analysis of enterohepatic circulation of buprenorphine and its active metabolite, norbuprenorphine, in rats", Drug Metabolism and Disposition, Vol. 22, (1994), p. 2 - 7.
Buprenorphine is a morphine derivative which has twice the duration of action and 30 times the potency of morphine. Buprenorphine is partially metabolized to norbuprenorphine which is also active in the body. In this study, buprenorphine was given to rats weighing 280 - 300 g. PROBLEM TABLE 11 - 2.
Buprenorphine
Weight of rat
290 g
Dose
0.06 mg/kg IV
A1
41
a
3.89
B1
10
b
0.271
AUC
48.3
AUMC
135.24
1.
What is the ?
2.
What is the ?
3.
What is the AUC?
4.
What is your patient's clearance?
5.
What is your patients MRT?
6.
What is your patient's ?
7.
What is your patient's V1?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is ?
14.
What is the in the peripheral compartment?
15.
What percent of the dose is in the peripheral compartment at equilibrium?
16.
Can this drug be treated as a one-compartment model for dosing purposes?
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17.
If this drug can be treated as a one-compartment model, what is K ?
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Caffeine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 3)
AHFS 00:00.00 GPI: 0000000000
Shi, J., et al., "Pharmacokinetic-pharmacodynamic modeling of caffeine: Tolerance to pressor effects", Clinical Pharmacology and Therapeutics, Vol. 53, (1993), p. 6 - 14.
This study looks at the cardiovascular effects of caffeine. Caffeine is known to increase blood pressure upon its withdrawl. This study looks at how tolerance to caffeine and its pressor effects develops and disappears with time. PROBLEM TABLE 11 - 3.
Caffeine
Patient weight
80 kg
Dose
4 mg/kg oral
A1
10.55
a
4.9
B1
9.1
b
0.23
f
98.4 %
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is your patient's ?
13.
What is ?
14.
What is the in the peripheral compartment?
15.
What percent of the dose is in the peripheral compartment at equilibrium?
16.
Can this drug be treated as a one-compartment model for dosing purposes?
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Cefazolin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 4)
AHFS 00:00.00 GPI: 0000000000
Nightingale, C., et al., "Changes in pharmacokinetics of cefazolin due to stress", Journal of Pharmaceutical Sciences, Vol. 64, (1975), p. 712 - 714.
Cefazolin is a cephalosporin antibiotic used in the treatment of many types of infections. This study looks at the effect of stress on the pharmacokinetics of cefazolin. The following data was approximated from the graph given in this article. PROBLEM TABLE 11 - 4.
Cefazolin
Patient weight
56.3 kg
Dose
1 g IV
A1
206.48
a
4.832
B1
122.96
b
0.573
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
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Ceftazidime Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 5)
AHFS 00:00.00 GPI: 0000000000
Ackerman, B., et al., "Effect of decreased renal function on the pharmacokinetics of ceftazidime", Antimicrobial Agents and Chemotherapy, Vol. 25, (1984), p. 785 - 786.
Ceftazidime is a cephalosporin antibiotic. This study explores the effect of compromised renal function on the pharmacokinetics of ceftazidime. The following data was approximated from the graph given in this article. PROBLEM TABLE 11 - 5.
Ceftazidime
Dose
1 g IV bolus
A1
188
a
8.22
B1
58.2
b
0.49
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
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Clentiazem Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 6)
AHFS 00:00.00 GPI: 0000000000
Shah, A., et al., "Pharmacokinetics of clentiazem after intravenous and oral administration in healthy subjects", Journal of Clinical Pharmacology, Vol. 33, (1993), p. 354 - 359.
Clentiazem is an derivative of diltiazem which is under investigation for its use in the treatment of angina pectoris and hypertension. Clentiazem blocks calcium channels resulting in a decrease in peripheral vascular resistance which subsequently leads to a decrease in blood pressure. PROBLEM TABLE 11 - 6.
Clentiazem
Patient weight
77 kg
Dose
20 mg IV bolus
A1
37.52
a
2.7
B1
16.17
b
0.078
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
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Cocaine
(Problem 11 - 7)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Levine, B. and Tebbett, I., "Cocaine pharmacokinetics in ethanol-pretreated rats", Drug Metabolism and Disposition, Vol. 22, (1994), p. 498 - 500.
This study looks into several reports which claim that the euphoric effects of cocaine can be enhanced when taken in conjunction with alcohol. This effect may be the result of higher cocaine blood levels or a reduced elimination of cocaine or a combination of both. PROBLEM TABLE 11 - 7.
Cocaine
Weight of rat
300 g
Dose
2 mg/kg cocaine (also 1 g/ kg ethanol)
A1
1172.6
a
0.362
B1
462
b
0.045
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-99
Multicompartment Modeling
1,2-Diethyl-3-Hydroxypyridine-4-One Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 8)
AHFS 00:00.00 GPI: 0000000000
Epemolu, O., et al., "The pharmacokinetics of 1,2-Diethyl-3-Hydroxypyridine-4-One (CP94) in rats, Drug Metabolism and Disposition, Vol. 20, (1992), p. 736 - 741.
1,2-Diethyl-3-Hydroxypyridine-4-One (CP94) is an iron chelator which is orally active. It is being investigated for use in the treatment of hemoglobinopathic disorders. In this study, rats weighing 250 - 300 g were given doses 50 mg /kg intravenously and the following data was collected: PROBLEM TABLE 11 - 8.
1,2-Diethyl-3-Hydroxypyridine-4-One
Weight of rat
275 g
Dose
50 mg/kg IV
A1
30.9
a
2.03
B1
8.13
b
0.38
Assume that the rat ( which weighs 275 g) is suffering from thalassemia and his iron levels are very high. The rat is prescribed CP94 to restore the iron levels to normal. 1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-100
Multicompartment Modeling
2,2-dimethylaziridine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 9)
AHFS 00:00.00 GPI: 0000000000
Lalka, D., Jusko, W., and Bardos, T., "Reactions of 2,2-dimethylaziridine-type alkylating agents in biological systems II: Comparative pharmacokinetics in dogs", Journal of Pharmaceutical Sciences, Vol. 64, (1975), p. 230 - 235.
The 2,2-dimethylaziridine alkylating agents are used for their antitumor capability as antineoplastic agents. In this study, male mongrel dogs, weighing 20 - 28 kg, were each given a dose of 12 mg/kg of ethyl bis (2,2-dimethylaziridinyl) phosphinate intraveneously. PROBLEM TABLE 11 - 9.
2,2-dimethylaziridine
Weight of dog
24 kg
Dose
12 mg/kg IV
A1
42
a
0.409
B1
8.5
b
0.095
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-101
Multicompartment Modeling
Flurbiprofen Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 10)
AHFS 00:00.00 GPI: 0000000000
Menzel-Soglowek, S., et al., "Variability of inversion of (R)-flurbiprofen in different species", Journal of Pharmaceutical Sciences, Vol. 81, (1992), p. 888 - 891.
Flurbiprofen is an anti-inflammatory and analgesic agent. This study compares the pharmacokinetics of the (R)-isomer of flurbiprofen to the those of the (S)-isomer. The following data was approximated from the graph given in this article. PROBLEM TABLE 11 - 10. Flurbiprofen
Weight of rat
260 g
Dose
10 mg/kg IV
A1
48.5
a
2.33
B1
57.68
b
0.175
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-102
Multicompartment Modeling
Furosemide Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 11)
AHFS 00:00.00 GPI: 0000000000
Tilsone, W., and Fine, A., "Furosemide kinetics in renal failure", Clinical Pharmacology and Therapeutics, Vol. 23, (1978), p. 644 - 650.
Furosemide is an agent which is used for its diuretic action to treat such conditions as renal and cardiac edema. In this study, normal subjects were given an intravenous bolus dose of 22 mg of furosemide. Blood samples were taken at various intervals and the following data was obtained: PROBLEM TABLE 11 - 11. Furosemide
Dose
22 mg IV
A1
2.1
a
6.9
B1
0.77 -
b
0.96
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-103
Multicompartment Modeling
Glycyrrhizin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 12)
AHFS 00:00.00 GPI: 0000000000
Tsai, T., et al., "Pharmacokinetics of glycyrrhyzin after intravenous administration to rats", Journal of Pharmaceutical Sceinces, Vol. 81, (1992), p. 961- 963.
Glycyrrhizin is a component of licorice which is proposed to have anti-inflammatory, anti-hepatotoxic, interferoninducing, anti-viral, and anti-ulcer activity. It also causes pseudoaldosteronism. The following data was approximated from the graph given in this article. PROBLEM TABLE 11 - 12. Glycyrrhizin
Weight of rat
275 g
Dose
20 mg/kg
A1
91.23
a
4.16
B1
69.90 -
b
0.43
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-104
Multicompartment Modeling
Human Deoxyribonuclease Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 13)
AHFS 00:00.00 GPI: 0000000000
Mohler, M., et al., "Altered pharmacokinetics of recominant human deoxyribonuclease in rats due to the presence of a binding protein", Drug Metabolism and Disposition, Vol. 21, (1993), p. 71 - 75.
Deoxyribonucleases are found in human serum, urine, and a variety of tissues. These endonucleases catalyze the hydrolysis of DNA to oligonucleotides. It has been suggested that increased levels of serum deoxyribonucleases may predict malignancies. PROBLEM TABLE 11 - 13. Human
Deoxyribonuclease
Patient weight
260 g
Dose
1 mg/kg IV bolus
A1
19250 -
a
8.61
B1
4897
b
0.229
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-105
Multicompartment Modeling
Human Granulocyte Colony-Stimulating Factor Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 14)
AHFS 00:00.00 GPI: 0000000000
Tanaka, H., and Kaneko, T., "Pharmacokinetic and pharmacodynamic comparisons between human granulocyte colony-stimulating factor purified from human bladder carcinoma cell line 5637 culture medium and recombinant human granulocyte colonystimulating factor produced in Escherichia coli", The Journal of Pharmacology and Experimental Therapeutics, Vol. 262, (1992), p. 439 - 444.
Human Granulocye Colony-Stimulating Factor (hG-CSF) is used to stimulate the proliferation of precursor cells and their subsequent differentiation in the bone marrow. This article compares the pharmacokinetics of hG-CSF produced by two different methods. In the first method, the hG-CSF was obtained from human bladder carcinoma cell line 5637 culture medium. In the second method, the hG-CSF was produced by Escherichia coli. PROBLEM TABLE 11 - 14. Human
Granulocyte Colony-Stimulating Factor
Weight of rat
250 g
Dose
10 µg/kg IV
A1
116.21 0.24 hours
B1
99.228 1.27 hours
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is a?
8.
What is b?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-106
Multicompartment Modeling
Hydrocortisone Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 15)
AHFS 00:00.00 GPI: 0000000000
Derendorf, H., et al., "Pharmacokinetics and oral bioavailability of hydrocortisone", Journal of Clinical Pharmacology, Vol. 31, (1991), p. 473 - 476.
This study looks at both the two-compartment model pharmacokinetics and the oral bioavailability of hydrocortisone. The following data was approximated from the graph given in this article. PROBLEM TABLE 11 - 15. Hydrocortisone
Dose
20 mg IV
A1
430
a
13.1
B1
439 -
b
0.445
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-107
Multicompartment Modeling
Levodopa Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 16)
AHFS 00:00.00 GPI: 0000000000
Sasahara, K., et al., "Dosage form design for improvement of bioavailability of levodopa II: Bioavailability of marketed levodopa preparations in dogs and parkinsonian patients" Journal of Pharmaceutical Sciences, Vol. 69, (1980), p. 261 - 265.
Levodopa is an agent used in the treatment of Parkinson's disease. This study looks at various dosage forms of levodopa and compares the pharmacokinetic parameters of each. The following data was approximated from the graph given in this article. PROBLEM TABLE 11 - 16. Levodopa
Dose
50 mg IV
A1
8.63
a
13.3
B1
2.97 -
b
1.14
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-108
Multicompartment Modeling
Meropenem
(Problem 11 - 17)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Chimata, M., et al., "Pharmacokinetics of meropenem in patients with various degrees of renal function, including patients with end-stage renal disease", Antimicrobial Agents and Chemotherapy, Vol. 37, (1993), p. 229 - 233.
Meropenem is a carapenem antibiotic which has a broad spectrum of activity. It is used in the treatment of infections caused by both Gram-positive and Gam-negative bacteria and is active against Enterobacteriaceae and Pseudomonas aeruginosa. Meropenem is 60% renally and 40% hepatically eliminated. PROBLEM TABLE 11 - 17. Meropenem
Dose
500 mg IV infusion over 40 minutes
A1
21
a
1.85
B1
20 -
b
0.503
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
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11-109
Multicompartment Modeling
N-Methylpyridinium-2-Carbaldoxime Chloride Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 18)
AHFS 00:00.00 GPI: 0000000000
Bodor, N., and Brewster, M., "Problems of delivery of drus to the brain", International Encyclopedia of Pharmacology and Therapeutics, Vol. 120, (1975)
N-methylpyridinium-2-cabaldoxime chloride (2-PAM) is the drug of choice for the treatment of orgaonphosphate poisoning. It is mostly renally excreted. This article considers the fact that this agent is highly hydrophilic and thus has difficulty reaching the brain. The following data was approximated from the graph given in this article. PROBLEM TABLE 11 - 18. N-Methylpyridinium-2-Carbaldoxime
Weight of dog
40 kg
Dose
7.0 mg/kg
A1
5.356
a
0.28796
B1
35.983
b
11.586
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
Chloride
11-110
Multicompartment Modeling
Pyrazine Diazohydroxide Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 19)
AHFS 00:00.00 GPI: 0000000000
Vogelzang, N., et al., "Phase I and pharmacokinetic study of a new antineoplastic agent: pyrazine diazohydroxide (NSC 361456)", Journal of Cancer Research , Vol. 54, (1994), p. 114 - 119.
Pyrazine diabhohydroxide is an agent which forms a reactive pyrazine dizonium ion in vivo which acts to destroy tumor cells. This study looks at the pharmacokinetic parameters of this agent in advanced cancer patients whose cancer was not curable by any other type of therapy. They were given a dose of 18 mg/m2/day for 5 days every 4 weeks. Most of the following data was collected for a 66 year old male subject. The remaining data was approximated from the graph given in this article. PROBLEM TABLE 11 - 19. Pyrazine
Diazohydroxide
Patient Body Surface Area
1.82 m2
Dose
18 mg/ m2
A1
8063
a
0.195
B1
1186
b
0.0257
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-111
Multicompartment Modeling
Terbinafene Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 20)
AHFS 00:00.00 GPI: 0000000000
Kovarik, J., et al., "Dose-proportional pharmacokinetics of terbinafine and its N-demethylated metabolite in healthy volunteers", British Hournal of Dermatology, Vol. 126, (1992), p. 8 - 13.
Terbinafene is an antifungal agent which acts by interfering with ergosterol biosynthesis. It is active against Trichophyton, Epidermophyton, and Microsporum. Approximately 70% of an oral dose is absorbed. Terbinafene has an Ndemethylated metabolite which is active. The following data was approximated from the graph given in this article. PROBLEM TABLE 11 - 20. Terbinafene
Dose
750 mg
A1
2398
a
0.511 -
B1
102 -
b
0.0222 -
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-112
Multicompartment Modeling
Verrucarol Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 21)
AHFS 00:00.00 GPI: 0000000000
Barel, S., Yagen, B., and Bailer, M., "Pharmacokinetics of the trichothecen mycotoxin verrucarol in dogs", Journal of Pharmacetuical Seciences, Vol. 79, (1990), p. 548 - 550.
Verrucarol is a toxin which is related to toxins which have anti-tumor activity. This study looks at the pharmacokinetics of verrucarol in dogs. The following data was approximated from the graph given in this article. PROBLEM TABLE 11 - 21. Verrucarol
Weight of Dog
22.5 kg
Dose
0.4 mg/ kg
A1
126.05
a
0.0415
B1
540.58 -
b
0.00946-
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-113
Multicompartment Modeling
2-Chloro-2-deoxyadenosine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 22)
AHFS 00:00.00 GPI: 0000000000
Liliemark, J. and Juliusson, G., "On the pharmacokinetics of 2-Chloro-2-deoxy-adenosine in humans", Cancer Research, Vol. 51, (1991), p. 5570 - 5572.
Two-Chloro-2-deoxyadenosine is an antitumor agent used in the treatment of hairy cell leukemia and other lymphoproliferative diseases. Infusions of 0.14 mg/kg over 12 hours were administered to 12 patients with various lymphoproliferative diseases for 5 consecutive days. The following data was collected: PROBLEM TABLE 11 - 22. 2-Chloro-2-deoxyadenosine
Patient weight
65 kg
Dose
0.14 mg/kg over 12 hours
A1
177.0 nM
a
1.04
B1
21.0 nM
b
0.10
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-114
Multicompartment Modeling
Felodipine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 23)
AHFS 00:00.00 GPI: 0000000000
Regardh, C., et al., "Pharmacokinetics of felodipine in patients with liver disease", Journal of Clinical Pharmacology, Vol. 36, (1989), p. 473 - 479.
The pharmacokinetic parameters of felodipine in patients with impaired liver function were investigated in this study. Felodipine blocks calcium channels resulting in a decrease in peripheral vascular resistance which subsequently leads to a decrease in blood pressure. Felodipine also works as a diuretic. The bioavailability of felodipine is 15%. It is highly (99.64%) protein bound and is eliminated almost exclusively by liver metabolism. The following data is for a patient with liver cirrhosis: Weight of dog
kg
Dose
mg/kg IV
A1 a
B1 b
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
11-115
Multicompartment Modeling
Lorazepam Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 24)
AHFS 00:00.00 GPI: 0000000000
Segal, J., et al., "Decreased systemic clearance of lorazepam in humans with spinal cord injury", Journal of Clinical Pharmacology, Vol. 31, (1991), p. 651 - 656.
Lorazepam is a benzodiazepine which is used as an anxiolytic, an anti-convulsant, an anti-emetic, and a sedative-hypnotic agent. PROBLEM TABLE 11 - 24. Lorazepam
Dose
2.0 mg infused intravenously over 1 - 2 minutes
A1 a
B1 b AUC AUMC
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
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11-116
Multicompartment Modeling
Metronidazole Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 25)
AHFS 00:00.00 GPI: 0000000000
Uccellini, D., Morgan, D., and Raymond, K., "Relationships among duration of infusion, dose, dosing interval, and stedy-state plasma concentrations during intermittent intravenous infusions: studies with metronazole", Journal of Pharmacokinetics and Biopharmaceutics, Vol. 14, (1986), p. 95 - 106
PROBLEM TABLE 11 - 25. Metronidazole
Patient weight
70 kg
Dose
1.5 g IV infusion
A1 a
2.11
B1 b
0.09
AUC AUMC
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
Basic Pharmacokinetics
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11-117
Multicompartment Modeling
Rhizoxin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 26)
AHFS 00:00.00 GPI: 0000000000
Graham, M., et al., "Preclinical and phase I studie with rhizoxin to apply a pharmacokinetically guided dose-escalation scheme", Journal of the National Cancer Institute, (1991), p. 494 - 499.
Rhizoxin is a lactone which was obtained from the fungus, Rhizopus chinensis. It has anti-tumor activity against a broad spectrum of tumor types including LOX melanoma, A549 lung tumors, and MX-1 mammary tumors. This study looks at dosing of rhizoxin. Patients with nontreatable tumors who had a life expectancy of more than 12 weeks were given doses of 12 mg/ m2. The following data was approximated from the graph given in this article. PROBLEM TABLE 11 - 26. Rhizoxin
Patient Body Surface Area
1.82 m2
Dose
12 mg/ m2
A1
1.55
a
4.00
B1
0.12
b
0.116 -
1.
What is ?
2.
What is ?
3.
What is ?
4.
What is your patient's clearance?
5.
What is your patient's ?
6.
What is your patient's V1?
7.
What is ?
8.
What is ?
9.
What is ?
10.
What is ?
11.
What is ?
12.
What is ?
13.
What is the in the peripheral compartment?
14.
What percent of the dose is in the peripheral compartment at equilibrium?
15.
Can this drug be treated as a one-compartment model for dosing purposes?
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11-118
Multicompartment Modeling
11.5.1
TWO-COMPARTMENT MODEL EQUATIONS
The following set of equations were used to solve the two-compartment model problem set. The problem sets for the first three drugs have been done for you. The others are done the same way. The answers for all of the problems are in the back of this packet. Aspirin 1. 2. 3. 4. 5.
The weight of the rat is 275 g or 0.275 kg.
6. 7. 8. 9. 10. 11. 12. 13. 14. 15.
Yes
_____________________________________________________________________ Buprenorphine 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. Basic Pharmacokinetics
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11-119
Multicompartment Modeling
12. 13. 14. 15.
Yes
_____________________________________________________________________ Caffeine 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.
11.5.2
Yes
ANSWERS
Aspirin 1.
8.019
2.
36.2
3.
31.1
4.
2.62 minutes
5.
155.48 mL
6.
86.92 mL
7.
81.57 mL
Basic Pharmacokinetics
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11-120
Multicompartment Modeling
8.
0.648 min
9.
3.47 min
10.
0.598 min-1
11.
0.358min-1
12.
0.314min-1
13.
15.82
14.
1.928 minutes
15.
44.1%
16.
Yes
17.
0.381 min-1
Buprenorphine 1.
10.54
2.
36.9
3.
47.4
4.
366.8
5.
2.85 hours
6.
1.35 L
7.
0.34 L
8.
0.178 hours
9.
2.58 hours
10.
0.981 h-1
11.
1.075 h-1
12.
2.11 h-1
13.
51
14.
0.736 hours
15.
74.9%
16.
No
17.
Can't be calculated
Caffeine 1.
2.15
2.
39.57
3.
41.72
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11-121
Multicompartment Modeling
4.
7.67
5.
33.35 L
6.
16.28 L
7.
0.141 hours
8.
3.014 hours
9.
2.39 h-1
10.
0.471 h-1
11.
2.266 h-1
12.
19.65
13.
0.655 hours
14.
51.2%
15.
Yes
Cefazolin 1.
42.73
2.
214.59
3.
257.32
4.
3.89
5.
6.78 L
6.
3.035 L
7.
0.143 hours
8.
1.21 hours
9.
2.163 h-1
10.
1.28 h-1
11.
1.96 h-1
12.
329.44
13.
0.50 hours
14.
55.2%
15.
Yes
Ceftazidime 1.
22.87
2.
118.78
Basic Pharmacokinetics
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11-122
Multicompartment Modeling
3.
141.65
4.
7.0598
5.
14.41 L
6.
4.06 L
7.
0.0843 hours
8.
1.415 hours
9.
2.32 h-1
10.
1.738 h-1
11.
4.65 h-1
12.
246.2
13.
0.365 hours
14.
39.2%
15.
Yes
Clentiazem 1.
13.9
2.
207.3
3.
221.2
4.
90.4
5.
1159.2 L
6.
372.5 L
7.
0.257 hours
8.
8.89 hours
9.
0.868 h-1
10.
0.243 h-1
11.
1.67 h-1
12.
53.69
13.
1.35 hours
14.
47.4%
15.
Yes
Cocaine 1.
3239.2
2.
10266.7
Basic Pharmacokinetics
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11-123
Multicompartment Modeling
3.
13505.89
4.
44.43
5.
987.2 mL
6.
367.1 mL
7.
1.91 minutes
8.
15.4 minutes
9.
0.1346 min-1
10.
0.1210 min-1
11.
0.1514 min-1
12.
1634.6
13.
8.35 minutes
14.
59.2%
15.
No
1,2-Diethyl-3-hydroxpyridine-4-one 1.
15.22
2.
21.39
3.
36.62
4.
375.5
5.
988.2 mL
6.
352.3 mL
7.
0.341 hours
8.
1.824 hours
9.
0.724 h-1
10.
1.066 h-1
11.
0.62 h-1
12.
39.03
13.
1.016 hours
14.
55.4%
15.
No
2,2-dimethylaziridine 1.
102.7
2.
89.47
Basic Pharmacokinetics
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11-124
Multicompartment Modeling
3.
192.16
4.
1498.7
5.
15.78 L
6.
5.7 L
7.
1.695 minutes
8.
7.296 minutes
9.
0.148 min-1
10.
0.263 min-1
11.
0.093 min-1
12.
50.5
13.
4.65 minutes
14.
56.5%
15.
No
Flurbiprofen 1.
20.82
2.
329.6
3.
350.42
4.
7.42
5.
42.4 mL
6.
24.5 mL
7.
0.297 hours
8.
3.96 hours
9.
1.35 h-1
10.
0.303 h-1
11.
0.856 h-1
12.
106.18
13.
1.2 hours
14.
42.2%
15.
Yes
Furosemide 1.
0.304
2.
0.802
Basic Pharmacokinetics
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11-125
Multicompartment Modeling
3.
1.106
4.
19.88
5.
20.71 L
6.
7.67 L
7.
0.1005 hours
8.
0.722 hours
9.
2.55 h-1
10.
2.59 h-1
11.
2.71 h-1
12.
2.87
13.
0.322 hours
14.
62.99%
15.
No
Glycyrrhizin 1.
21.93
2.
162.56
3.
184.5
4.
29.8
5.
69.33 mL
6.
34.13 mL
7.
0.167 hours
8.
1.61 hours
9.
2.048 h-1
10.
0.873 h-1
11.
1.67 h-1
12.
161.13
13.
0.61 hours
14.
50.8%
15.
Yes
Human Deoxyribonuclease 1.
2235.78
2.
21384.3
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11-126
Multicompartment Modeling
3.
23620.1
4.
11.01
5.
48.07 mL
6.
10.77 mL
7.
0.0805 hours
8.
3.027 hours
9.
1.929 h-1
10.
1.0223 h-1
11.
5.89 h-1
12.
24147
13.
0.433 hours
14.
28.9%
15.
Yes
Human Granulocyte Colony-Stimulating Factor 1.
40.24
2.
181.81
3.
222.05
4.
11.26
5.
20.62 mL
6.
11.6 mL
7.
2.89 h-1
8.
0.546 h-1
9.
1.625 h-1
10.
0.971 h-1
11.
0.839 h-1
12.
215.44
13.
0.711 hours
14.
43.8%
15.
Yes
Hydrocortisone 1.
32.8
2.
986.5
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11-127
Multicompartment Modeling
3.
1019.3
4.
19.62
5.
44.1 mL
6.
23.01 mL
7.
0.053 hours
8.
1.56 hours
9.
6.838 h-1
10.
0.853 h-1
11.
5.85 h-1
12.
869
13.
0.267 hours
14.
47.8%
15.
Yes
Levodopa 1.
0.649
2.
2.61
3.
3.25
4.
15.37
5.
13.48 L
6.
4.31 L
7.
0.052 hours
8.
0.61 hours
9.
4.25 h-1
10.
3.56 h-1
11.
6.62 h-1
12.
11.6
13.
hours
14.
68.0%
15.
Yes
Meropenem 1.
11.35
2.
39.76
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11-128
Multicompartment Modeling
3.
51.11
4.
9.78
5.
19.45 L
6.
12.20 L
7.
0.375 hours
8.
1.378 hours
9.
1.16 h-1
10.
0.802 h-1
11.
0.391 h-1
12.
41
13.
0.967 hours
14.
37.3%
15.
No
N-methylpyridinium-2-carbaldoxime chloride 1.
18.6
2.
3.106
3.
21.71
4.
12.9
5.
1.11 L
6.
6.77 L
7.
2.41 hours
8.
0.0598 hours
9.
1.752 h-1
10.
1.905 h-1
11.
8.218 h-1
12.
41.339
13.
0.327hours
?14.
45.5%
15.
Yes
Pyrazine Diazohydroxide 1.
41348.7
2.
46147.9
Basic Pharmacokinetics
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11-129
Multicompartment Modeling
3.
87496.6
4.
0.3744
5.
14.57 mL
6.
3.542 mL
7.
3.55 minutes
8.
26.97 minutes
9.
0.0474 min-1
10.
0.106 min-1
11.
0.0676 min-1
12.
9249
13.
11.97 minutes
14.
75.7%
15.
No
Terbinafene 1.
4692.8
2.
4594.6
3.
9287.4
4.
80.75
5.
3637.6 L
6.
300 L
7.
1.36 h
8.
31.2 h
9.
0.0421 h-1
10.
0.2692 h-1
11.
0.222 h-1
12.
2500
13.
6.42 hours
14.
91.8%
15.
No
Verrucarol 1.
3037.35
2.
57143.8
Basic Pharmacokinetics
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11-130
Multicompartment Modeling
3.
60181.1
4.
149.5
5.
15.81 L
6.
13.5 L
7.
16.7 minutes
8.
73.3 minutes
9.
0.0155 min-1
10.
0.0253 min-1
11.
0.0101 min-1
12.
666.63
13.
46.15 minutes
14.
28.1%
15.
Yes
Basic Pharmacokinetics
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11-131
CHAPTER 12
Protein Binding
Author: Michael Makoid Reviewer: Phillip Vuchetich
OBJECTIVES 2.
Basic Pharmacokinetics
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12-1
CHAPTER 13
Nonlinear (Michaelis-Menton) Kinetics
Author: Michael Makoid Reviewer: Phillip Vuchetich
OBJECTIVES 1.
Given population average patient data, the student will devise (V) a dosage regimen which will maintain plasma concentrations of drug within the therapeutic range.
2.
Given specific patient information, the patient will justify (VI) the optimal dosage regimen.
3.
Given patient information regarding organ function, the student will devise (V) and justify (VI) dosage regimens for the compromised patient.
4.
The student will write (V) a professional consult using the above calculations.
Basic Pharmacokinetics
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13-1
Nonlinear (Michaelis-Menton) Kinetics
13.1 Problems
Basic Pharmacokinetics
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13-2
Nonlinear (Michaelis-Menton) Kinetics
Cefadroxil Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 27)
AHFS 00:00.00 GPI: 0000000000
Sanchez-Pico, A., et al, "Nonlinear intestinal abosrption kinetics of cefadroxil in the rat", Journal of Pharmacy Pharmacology, Vol.41, (1989), p. 179 - 185.
Cefadroxil is a cephalosporin antibiotic which is commonly used to treat various infections. It is usually given orally. This study looks at the pharmacokinetics and bioavailability of cefadroxil in the rat. PROBLEM TABLE 11 - 27. Cefadroxil
Dose 500 mg
7.31
1000 mg
14.67
Find . Find the maximum clearance,, for this patient. What would be the dose needed to acheive a steady-state concentration of 10 ? You recommend changing the patient's dosage regimen to 300 mg/day. What would be your patient's steady state plasma concentration?
Basic Pharmacokinetics
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13-3
Nonlinear (Michaelis-Menton) Kinetics
CD4 Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 28)
AHFS 00:00.00 GPI: 0000000000
Qian, M., et al., "Pharmacokinetic evaluation of drug interactions with anti-human immunodeficiency virus drugs: V. effect of soluble CD4 on 2',3'-dideoxycytidine kinetics in monkeys", Drug Metabolism and Disposition, Vol. 20, (1992), p. 396 - 400.
2',3'-dideoxycytidine in combination with recombinant ST4 has been shown to be effective against HIV (human immunodeficiency virus) in vitro. This study examines whether or not the pharmacokinetics of 2',3'-dideoxycytidine are affected by administration of CD4 (an immunoglobulin). Doses of each drug were given to male adult monkeys weighing an average of 4.45 kg. The following data is for ST4 (soluble CD4). PROBLEM TABLE 11 - 28. CD4
Dose 1.1 mg/kg
10.27
2.2 mg/kg
22.23 Weight of Monkey = 4.45 kg
Find . Find the maximum clearance for this patient. What would be the dose needed to acheive a steady-state concentration of 15 ? You recommend changing the patient's dosage regimen to 1.5 mg/ kg. What would be your patient's plasma concentration?
Basic Pharmacokinetics
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13-4
Nonlinear (Michaelis-Menton) Kinetics
Methylprednisone Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 29)
AHFS 00:00.00 GPI: 0000000000
Haughey, D, and Jusko W.., "Bioavailability and nonlinear dispositionof methylprednisolone and methylprednisone in the rat", Journal of Pharmaceutical Sceicnes, Vol. 81, (1992), p. 117 - 121.
Methylprednisone is a corticosteroid which is commonly used in the treatment of medical emergencies such as cardiovascular shock, asthma, and cerebral edema. The following data was obtained for two methylprednisolone doses. The plasma concentration measurement given for each dose below is that for the central compartment. PROBLEM TABLE 11 - 29. Methylprednisone
Dose 10 mg
6834
50 mg
71519
Find . Find the maximum clearance for this patient. What would be the dose needed to acheive a steady-state concentration of 10,000 ? You recommend changing the patient's dosage regimen to 30 mg. What would be your patient's plasma concentration?
Basic Pharmacokinetics
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13-5
Nonlinear (Michaelis-Menton) Kinetics
Mezlocillin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 30)
AHFS 00:00.00 GPI: 0000000000
Jungbluth, G. and Jusko, W., "Dose-dependent pharmacokinetics of mezlocillin in rats", Antimicrobial Agents and Chemotherapy, Vol. 33, (1989), p. 839 - 843.
Mezlocillin is an antibiotic used to treat various types of infection. It is usually given by the intravenous route and exhibits dose-dependent (nonlinear) pharmacokinetics. This article compares two intravenous bolus doses, one of 20 mg/kg and one of 200 mg/kg in rats. The following data was calculated from the results of this study. PROBLEM TABLE 11 - 30. Mezlocillin
Dose 20 mg/kg
158.6
200 mg/kg
294.1 Rat weight = 425 g
Find . Find the maximum clearance,, for this patient. What would be the dose needed to acheive a steady-state concentration of
200 ?
You recommend changing the patient's dosage regimen to 150 mg/ kg. What would be your patient's plasma concentration?
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13-6
Nonlinear (Michaelis-Menton) Kinetics
Naphthol Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 31)
AHFS 00:00.00 GPI: 0000000000
Redegeld, A., Hofman, G., and Noordhoek, J., "Conjugative clearance of 1-naphthol and disposition of its glucuronide and sulfate conjugates in the isolated perfused rat", Journal of Harmacology and Experimental Therapeutics, Vol. 244, (1988), p. 263 - 267.
1-naphthol is a small phenolic compound which is extensively metabolized by conjugation. This study looks at the pharmacokinetics of naphthol in a rat. PROBLEM TABLE 11 - 31. Naphthol
Dose 30 µmol
6.79 µM
40 µmol
8.63 µM
Find . Find the maximum clearance,, for this patient. What would be the dose needed to acheive a steady-state concentration of 7.7 µM? You recommend changing the patient's dosage regimen to 35 µmol. What would be your patient's plasma concentration?
Basic Pharmacokinetics
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13-7
Nonlinear (Michaelis-Menton) Kinetics
Paroxetine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 32)
AHFS 00:00.00 GPI: 0000000000
Sindrug, S., Brosen, K, and Gram, L.., "Pharmacokinetics of the selective serotonin reuptake inhibitor paroxetine: nonlinearity and relation to the sprateine oxidation polymorphism", Clinical Pharmacology and Therapeutics, Vol. 51, (1992), p. 288 - 295.
Paroxetine hydrochloride (Paxil) is a selective serotonin reuptake inhibitor which is used in the treatment of depression. Paroxetine is metabolized both by oxidation and conjugation with the conjugated metabolites excreted in the urine. Paroxetine exhibits dose-dependent (nonlinear) pharmacokinetics. The following data is for a male diabetic patient who was concurrently taking insulin. PROBLEM TABLE 11 - 32. Paroxetine
Dose 10 mg daily
1.65
20 mg daily
3.30
30 mg daily
8.25
40 mg daily
13.20
50 mg daily
26.40
60 mg daily
39.60
70 mg daily
66.00
Find . Find the maximum clearance,, for this patient. What would be the dose needed to acheive a steady-state concentration of 50 ? You recommend changing the patient's dosage regimen to 36 mg/day. What would be your patient's plasma concentration?
Basic Pharmacokinetics
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13-8
Nonlinear (Michaelis-Menton) Kinetics
Phenytoin Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 33)
AHFS 00:00.00 GPI: 0000000000
Levine, M., et al., "Evaluation of serum phanyton monitoring in an acute care setting", Therapeutic Drug Monitoring, Vol. 10, (1988)., p. 50 - 57.
Phenytoin is an agent which is commonly used in the treatment of epilepsy. This drug exhibits nonlinear kinetics. Phenytoin is mainly eliminated from the body by hepatic cytochrome P-450 metabolism. Several doses of phenytoin were studied in patients and the data is summarized below: PROBLEM TABLE 11 - 33.
Phenytoin
Dose
Day
300 mg at bedtime
2
5.0
12
11.4
49
21.5
(started on day 1) 200 mg BID (started on day 6) 200 mg BID (started on day 6)
Find . Find the maximum clearance,, for this patient. What would be the dose needed to acheive a steady-state concentration of
15 ?
You recommend changing the patient's dosage regimen to 300 mg/day. What would be your patient's plasma concentration?
Basic Pharmacokinetics
REV. 99.4.25
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13-9
Nonlinear (Michaelis-Menton) Kinetics
Phenytoin in the Critically Ill Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 34)
AHFS 00:00.00 GPI: 0000000000
Boucher, B. et al., "Phenytoin pharmacokinetics in critically ill trauma patients", Clinical Pharmacology and Therapeutics, Vol. 44, (1988)., p. 675 - 683.
Phenytoin is an agent which is commonly used in the treatment of epilepsy. This drug exhibits nonlinear kinetics. This study looks at several doses of phenytoin in severely ill trauma patients. The data given below is that obtained for one male, 25 year-old, patient who weighed 85 kg. PROBLEM TABLE 11 - 34. Phenytoin
in the Critically Ill
Dose 615 mg/ day
10
588 mg/day
8.5
Find . Find the maximum clearance,, for this patient. What would be the dose needed to acheive a steady-state concentration of 12? You recommend changing the patient's dosage regimen to 450 mg/ day. What would be your patient's steady-state plasma concentration?
Basic Pharmacokinetics
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13-10
Nonlinear (Michaelis-Menton) Kinetics
Phenytoin in Pediatrics Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 35)
AHFS 00:00.00 GPI: 0000000000
Bauer, L. and Blouin, R., "Phenytoin Michaelis-Menten pharmacokinetics in caucasian paediatric patients", Clinical Pharmacokinetics, Vol. 8, (1989)., p. 545 - 549.
Phenytoin is an agent which is commonly used in the treatment of epilepsy. This drug exhibits nonlinear kinetics. This study looks at several doses of phenytoin in pediatric patients of several ages. The data for the 4 to 6 year old patients is given below.
Dose 7.5 mg/ kg/ day
15
6.5 mg/ kg/day
10
Find . Find the maximum clearance,, for this patient. What would be the dose needed to acheive a steady-state concentration of
12 ?
You recommend changing the patient's dosage regimen to 4.5 mg/ kg/ day. What would be your patient's steady-state plasma concentration?
Basic Pharmacokinetics
REV. 99.4.25
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13-11
Nonlinear (Michaelis-Menton) Kinetics
Quinalapril
(Problem 11 - 36)
Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
AHFS 00:00.00 GPI: 0000000000
Elliott, H., et al., "Dose responses and pharmaockinetics for the angiotensin converting enzyme inhibitor, quinapril", Clinical Pharmacology and Therapeutics, Vol. 52, (1992), p. 260 - 265.
Quinalapril is an angiotensin converting enzyme (ACE) inhibitor which is used in the treatment of hypertension and heart failure. The optimal dosage regimen for the ACE inhibitors is controversial and this study further investigates quinalapril's pharmacokinetics at various doses ranging from 0.5 to 20 mg. Quinalapril is a prodrug which is metabolized to its active form, quinalaprilat.
Dose (of quinalapril)
(of quinalaprilat)
2.5 mg daily
47.5
5.0 mg daily
98.1
Find . Find the maximum clearance,, for this patient. What would be the dose needed to acheive a steady-state concentration of
75 ?
You recommend changing the patient's dosage regimen to 3.0 mg/day. What would be your patient's plasma concentration?
Basic Pharmacokinetics
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13-12
Nonlinear (Michaelis-Menton) Kinetics
Vanoxerine Problem Submitted By: Maya Leicht Problem Reviewed By: Vicki Long
(Problem 11 - 37)
AHFS 00:00.00 GPI: 0000000000
Ingwersen, S., et al., "Nonlinear multiple-dose pharmacokinetis of the dopamine reuptake inhibitor vanoxerine", Journal of Pharmaceutical Sciences, Vol. 82, (1993)., p. 1164 - 1166.
Vanoxerine is a pre-synaptic dopamine reuptake inhibitor which may be useful as an antidepressant. The bioavailability of vanoxerine is changed by food intake. The bioavailability after fasting is increased 76% by a low-fat meal and 255% by a high-fat meal. In this study, the volunteers were given doses of vanoxerine after eating a standard breakfast of one bowl of cereal with milk, two slices of toast with sunflower margarine and jam, and one cup of tea.
Dose 25 mg
3.4
75 mg
15.1
125 mg
46.5
Find . Find the maximum clearance,, for this patient. What would be the dose needed to acheive a steady-state concentration of
30.0 ?
You recommend changing the patient's dosage regimen to 100 mg. What would be your patient's plasma concentration?
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13-13
Nonlinear (Michaelis-Menton) Kinetics
13.2 Nonlinear Equations The following equations were used to solve the questions following each "nonlinear" scenario. Three scenerios have been completed for you. The answers have been provided for the remainder. 1. 2. 3. 4. Phenytoin in the Critically Ill CD4 1. The monkeys had an average weight of 4.45 kg. 2. 3. 4. Cefadroxil 1. 2. 3. 4.
13.2.1
ANSWERS
Cefadroxil 2144.75 147.2 g/day 683.14 mg 4.38 CD4 135.09 69.28 mg/day 6.92 mg/day 14.4 Methylprednisone
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13-14
Nonlinear (Michaelis-Menton) Kinetics
52345.27 86.60 mg/day 13.89 mg/day 27747.1 Mezlocillin 324.95 25.92 mg/day 68.23 mg/day 1676.04 Naphthol 46.14 (M 233.86 (mol 25.61 (mol 43.5 (M Paroxetine 4.125 45 mg/day 41.6 mg/day 16.5 Phenytoin 4.014 540.85 mg/day 426.7 mg/day 5 Phenytoin in the Critically Ill 3.517 831.3 mg/day 642.88 mg/day 4.15 Phenytoin in Pediatrics 6.67 162.5 mg/day
Basic Pharmacokinetics
REV. 99.4.25
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13-15
Nonlinear (Michaelis-Menton) Kinetics
104.46 mg/day 4.74 Quinalapril 1503.15 81.61 mg/day 3.88 mg/day 57.36 Vanoxerine 20.96 179.08 mg/day 105.4 mg/day 26.5
Basic Pharmacokinetics
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13-16
CHAPTER 14
Practice Exams: Section 1
Author: Michael Makoid Reviewer: Vicki Long
ORGANIZATION Following are several exams which have been used in previous classes. These exams cover Chapters 1 through 4. Answers (including the graphs) are included in a format similar to real exam conditions, and some exams have the problems worked out completely. You are encouraged to work out the exams under conditions similar to how you will take the exam. You will want to have a calculator, semi-log graph paper, and scratch paper available. Before you begin to answer the questions, you will want to: 1) create the model from the description of the drug’s pharmacokinetics, 2) graph all of the data as discussed in the chapters. From these graphs the answers will become available. Your graphs should look like the graphs included in the answers. Applicable equations vary by exam and are included at the beginning of each chapter covered by the exam. Remember where you are in the course: Learn the tools; get the pharmacokinetic parameters from patient information. So in this section you must be able to analyze pharmacokinetic data and extract the pharmacokinetic parameters. That means: Assess the pharmacokinetic description of the drug and creBasic Pharmacokinetics
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14-1
Practice Exams: Section 1
ate a model, develop equations for the model, graph the data according to those equations, and get the parameters out. A student commented, “You mean, all we have to do is get the slope and intercept from a couple of graphs and tell you what they mean?” Yes, that right! Do not just look at the graphs and the answers and say, “Yes, that makes sense. I can do this.” You will be surprised to find out that what looks easy worked out is not so easy when it isn’t.
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14-2
Practice Exams: Section 1 : Nifedipine - Section 1
14.1 Nifedipine - Section 1 Masotti et al (J Clin Pcol 1985; 25: 27-35) and Traube et al (ibid 125-9) looked at the hemodynamic pharmacological response of nifedipine.
14.1.1
NIFEDIPINE DATA
Nifedipine (Procardia ®) is a calcium channel blocker which specifically inhibits potential-dependent channels not receptor-operated channels, preventing calcium influx of cardiac and vascular smooth muscle (coronary, cerebral). Calcium channel blockers reduce myocardial contractility and A-V node conduction by reducing the slow inward calcium current. They are indicated in angina, cardiac dysrhythmias, and hypertension among others. Nifedipine appears to be metabolized entirely into an inactive metabolite, an acid and subsequently further metabolized to a lactone. Both the acid and the lactone are excreted into the urine and the feces. PROBLEM TABLE 11 - 1.
Pharmacological Data
Fall in Diastolic BP
Cp (ng/mL)
Time (hr)
Fall in Diastolic BP
8
15
1
13.0
10
40
2
12.3
12
100
4
10.7
13
200
5
10.0
9
7.0
Echizen and Eichelbaum (Clin Pkin 1986; 11:425-49) and Kleinbloesem et al (Clin Pcol Therap 1986; 40: 21-8) reviewed the pharmacokinetics of Nifedipine. While the drug is not routinely given by IV bolus and does not strictly conform to a one compartment model, let's treat the data as if those problems can be ignored. The following data is offered for evaluation: PROBLEM TABLE 11 - 1. Nifedipine
IV Bolus Profile
Cp (mcg/L)
Xm1f (mg)
Xm1u (mg)
Time (hr)
Cm1 (mcg/L)
0.5
24.7
1
44.4
2
139
71.8
.14
.59
4
65.6
96.5
.44
1.83
Xm2f (mg)
Xm2u (mg)
.028
.11
6
31.1
100
.77
3.25
.073
.29
8
14.6
94.7
1.1
4.65
.135
.54
12
76.5
1.69
7.10
.291
1.15
24
34
2.77
11.63
.75
2.95
3.6
15.1
1.3
5.0
7 days
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14-3
Practice Exams: Section 1 : Nifedipine - Section 1
14.1.2
NIFEDIPINE QUESTIONS 1) dR/dT = slope of the Pharmacological response vs. time profile (mmHg/hr) 2) dR/d(lnC) = slope of the Pharmacological response vs concentration profile (mmHg)
18) AUMCiv = Area under the first moment of the plasma concentration of Nifedipine vs time curve of IV dose (mic/L*hr2 )
19) MRTiv = Mean Residence time of Nifedipine given as the IV dose (hr)
3) d(lnC)/dt = slope of the concentration vs time profile (hr-1 )
20) Xm10 = Mass of acidic metabolite in the body at time = 0 (mg)
4) kr = rate constant of excretion of Nifedipine into the urine (hr-1 )
21) Xm1inf = Mass of the acidic metabolite in the body at infinite time. (mg)
5) kf = rate constant of excretion of Nifedipine into feces (hr-1 )
22) Vm1 = Volume of distribution of the acidic metabolite in the body (L)
6) km1 = rate of metabolism of Nifedipine in the body (hr-1 )
23) Cm1 = Concentration of acidic metabolite in the body at time = 0 (mic/L)
7) K1 = elimination rate constant of Nifedipine in
Im1 = Intercept of the acidic metabolite concentration vs time profile = 181.2 mic/L
the body; the summation of all of the ways that it is removed from the body. (hr-1 )
8) T1/2 of Nifedipine in the body (hr) 9) X0 = Mass of Nifedipine in the body at time 0 (mg) 10) Xinf = Mass of Nifedipine in the body at time infinite time (mg) 11) Xuinf = Mass of unchanged Nifedipine in the urine at infinite time (mg) 12) Xfinf = Mass of unchanged Nifedipine in the feces at infinite time (mg) 13) V = Volume of distribution of Nifedipine (L) 14) Cp0 = Concentration of Nifedipine in the body at time 0. (mic/L) 15) AUCiv = Area under the plasma concentration of Nifedipine vs time curve of the IV dose (mic/L*hr) 16) First trapazoid of the AUCiv (mic/ L*hr) 17) Last trapazoid of the AUCiv (mic/ L*hr)
24) AUCmet = Area Under the Curve of the metabolite plasma concentration vs time profile (mic/L*hr) AUMCmet = Area Under the First Moment Curve of the metabolite plasma concentration vs time profile = 35700 mic/L*hr2
25) Xm1u0 = Mass of acidic metabolite in urine at time = 0 (mg) 26) Xm1uinf = Mass of acidic metabolite in urine at time = infinity (mg) 27) Xm1f0 = Mass of acidic metabolite in feces at time = 0 (mg) 28) Xm1finf = Mass of acidic metabolite in feces at time = infinity (mg) 29) Km1u = Rate constant of excretion of acidic metabolite into urine (hr-1 ) 30) Km1f = Rate constant of excretion of acidic metabolite into feces (hr-1 ) 31) Km2=Rate constant of formation of lactone metabolite in body =rate of metabolism of the acidic metabolite in the body (hr-1 )
32) K2 = Km1f + Km1u + Km2 = sumation of all of the ways that the acidic metabolite is eliminated from the body (hr-1 )
Basic Pharmacokinetics
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14-4
Practice Exams: Section 1 : Nifedipine - Section 1
33) Xm20 = Mass of lactone metabolite in body at time = 0 (mg)
38) Xm2finf = Mass of lactone metabolite in feces at time = infinity (mg)
34) Xm2inf = Mass of lactone metabolite in body at time = infinity (mg)
K3 = Km2f + Km2u = the sumation of all of the ways that the lactone can be eliminated = 0.138 hr.-1
35) Cm2inf = Concentration of lactone metabolite in the body at time = infinity (0) 36) Xm2u0 = Mass of lactone metabolite in urine at time = 0 (mg)
39) Km2u = Rate constant of excretion of lactone metabolite into urine (hr -1 ) 40) Km2f = Rate constant of excretion of lactone metabolite into feces (hr -1 ) .
37) Xm2uinf = Mass of lactone metabolite in urine at time = infinity (mg) PROBLEM TABLE 11 - 1.
Answer Pool Any number from the answer pool may be used
once, more than once, or not at all Negative Numbers
Small Numbers
Big Numbers
A
-0.0001
0
13
B
-0.0010
0.010
15.1
C
-0.0100
0.018
17
D
-0.0180
0.028
20
E
-0.0280
0.042
25
F
-0.0375
0.070
38.9
G
-0.0750
0.085
50
H
-0.0850
0.110
85
I
-0.110
0.138
170
J
-0.138
0.375
181.2
A
-0.375
0.750
295
B
-0.750
1.30
415
C
-1.30
1.85
434
D
-1.85
2.0
787
E
-2.0
2.67
2100
F
-2.67
3.6
3570
G
-3.6
5.0
8500
H
-5.0
7.5
21000
I
-7.5
8.5
35700
J
-10
9.9
85000
Basic Pharmacokinetics
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14-5
Practice Exams: Section 1 : Nifedipine - Section 1
14.1.3
NIFEDIPINE SOLUTIONS X km1
Xmf1
Xm1
kmf1
kmu1
Xmu1
kmu2
Xmu2
km2 Xmf2
Xm2
kmf2
Graph #2
Graph #1
- 0.75 mmHg/hr
R vs T
Nifedipine
14
14
13 13
RESPONSE (MMHG)
12
Response (mmHg)
12
11 10
11
10
9
2 mm Hg
9 8 7 6
8
0 101
102
2
4
103
Concentration (ng/mL)
Graph #4
Nifedipine IV Bolus
8
10
/mL)
L)
Graph #3
6
Time (hr)
Nifedipine IV bolus - Metabolite
103
103
CONCENTRATION MIC/L
CONCENTRATION MIC/L
181.2 mic/L = Intercept (given)
100 102
1.85 hr
50
101 2
3
4
5
6
7
102
10 hrs
101
8
0
Time (hours)
4
8
12
16
20
24
Time (hours)
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14-6
Practice Exams: Section 1 : Nifedipine - Section 1
dR ------- = – 0.75mmHg ⁄ hr dT
1.
dR ------------ = 2.0mmHg d ln C
2.
Slope from graph # 2
Slope from graph # 1
dR ------0.75 = – 0.375hr –1 dT - = –------------d-----------ln C = -----------2 dR dT -----------d ln C
3.
Answer #1------------------------Answer #2
4.
kr = 0, Nifedipine is not excreted into the urine, it is metabolized entirely K1 = k m1 from pkin description
5.
kf = 0, Nifedipine is not found in the feces, only the metabolites are.
6.
km1, Rate of metabolism of Nifedipine in the body = ---------------- = 0.375 hr
7.
K1, Elimination rate constant of Nifedipine in the body = 0.375hr
K1 = k m1 from pkin description
0.693 1.85hr
–1
–1
from graph #3
from graph #3
T 1 ⁄ 2 = 1.85 hr from graph #3
8.
Xo = ( 3.6 + 15.1 + 1.3 + 5.1 ) = 25mg
9. 10.
Xinf = 0 from pkin description
11.
X u ( inf ) = 0 from pkin description
12.
X f ( inf ) = 0 from pkin description
13. 14. 15.
from pkin description = Σ ( X mu1 + X mu2 + X mf 1 + X mf2 )
Dose- = ------------------------25mg - • 1000mic --------------------- = 85L V = -----------Cp o 295mic ⁄ L 1mg 295mic ⁄ L Cp0 extrapolated back to t = 0 obtained from graph #3 Cp 295 = 787mic ⁄ L ⋅ hr AUC iv = ---------o = ------------0.375 k
as an estimate. Using Trapazoidal rule:
Cp 1 + Cp 2 Cp 2 + Cp 3 Cp 3 + Cp last Cp last Cp o + Cp 1 Σ ------------------------- ⋅ ∆t 1 + ------------------------- ⋅ ∆t2 + ------------------------- ⋅ ∆t 3 + ------------------------------⋅ ∆t last + -------------- 2 2 2 2 K1 + 139- ⋅ 2 + 139 + 65.6 ⋅ 2 + 65.6 + 31.1- ⋅ 2 + 31.1 + 14.6- ⋅ 2 + ------------14.6 mcg Σ 295 ----------------------------------------------------------------------------------------------------------hr 2 2 2 2 0.375 L Σ { 434 + 204.6 + 96.7 + 45.7 + 38.9 } mcg ----------hr = 821.9mcg ----------hr L L
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14-7
Practice Exams: Section 1 : Nifedipine - Section 1
16
17. 18
Cp o + Cp 1 -------------------------- ⋅ ∆t = 2
295 + 139 )- ⋅ 2 = 434mic ⁄ L ⋅ hr (--------------------------- 2
Cp L 14.6 ---------- = ------------- = 38.9mic ⁄ L ⋅ hr k 0.375 AUMC = AUC ⋅ MRT = 787 ⋅ 2.67 = 2100mic ⁄ L ⋅ hr
2
estimate. Using the trapazoidal rule:
T1 ⋅ C p1 + T 2 ⋅ C p 2 T2 ⋅ C p2 + T3 ⋅ C p3 T 3 ⋅ C p 3 + T last ⋅ C p last T last ⋅ C p last Cp last T 0 ⋅ C p o + T1 ⋅ C p1 Σ ---------------------------------------------- ⋅ ∆t 1 + ---------------------------------------------- ⋅ ∆t 2 + ---------------------------------------------- ⋅ ∆t 3 + --------------------------------------------------------- ⋅ ∆t last + ------------------------------ + --------------- 2 2 2 2 2 K1 K1 0 ⋅ 295 + 2 ⋅ 139 2 ⋅ 139 + 4 ⋅ 65.6 4 ⋅ 65.6 + 6 ⋅ 31.1 6 ⋅ 31.1 + 8 ⋅ 14.6 8 ⋅ 14.6 14.6 mcg 2 Σ --------------------------------------- ⋅ 2 + ---------------------------------------- ⋅ 2 + ------------------------------------------ ⋅ 2 + ------------------------------------------ ⋅ 2 + ------------------ + ---------------- ----------hr 2 2 2 2 2 0.375 0.375 L mcg 2 Σ { 278 + 540.4 + 449 + 303.4 + 311.47 + 103.82 } = 1986.1----------hr L
19.
MRT IV
1 MRT iv = --- = 2.67hr estimate. Using trapazoidal rule and definitions k mcg 2 mcg 2 1986.1----------hr AUMC est 2100 ----------hr AUMC trap L L = --------------------------- = ------------------------------------ = 2.42 hr ≅ ------------------------ = ------------------------------- = 2.67 hr mcg mcg AUC est AUC trap 787 ----------hr 821.9----------hr L L
20.
X m1 ( o ) = 0 from pkin description
21.
X m1 ( inf ) = 0 from pkin description
22.
Vm1 = 170L
From LaPlace transforms, the equation for the metabolite is k
k m1 ⋅ X 0 –K2t – K1t Cp m = --------------------------------------- {e –e } ( K1 – K2 ) ⋅ V dm
⋅X
m1 0 - . the intercept, I, given as 181.2 mic/L in the data = ---------------------------------------
( K1 – K2 ) ⋅ V dm
Thus
k m1 ⋅ X 0 0.375 ⋅ 25mg V dm = ----------------------------------------------------= ---------------------------------------------------------------- = 169.6 L mic mg ( K1 – K2 ) ⋅ 181.2--------( 0.375 – 0.07 ) ⋅ 0.1812------L L
23.
C m1 ( 0 ) = 0 from pkin description
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14-8
Practice Exams: Section 1 : Nifedipine - Section 1
AUC met = 2100mic ⁄ L ⋅ hr
24.
PROBLEM TABLE 11 - 1. Nifedipine
Data
Cpm ( n ) + Cpm ( n + 1 ) -------------------------------------------------------2
∆T ( n )
AUC
t
AUC
t 0
Time (hr)
Cpm (mcg/L)
0
0
0.5
24.7
0 + 24.7 ------------------2
0.5
6.175
6.175
1
44.4
24.7 + 44.4 --------------------------2
0.5
17.275
23.45
2
71.8
44.4 + 71.8 --------------------------2
1
58.1
81.55
4
96.5
71.8 + 96.5 --------------------------2
2
168.3
249.85
6
100
96.5 + 100 ------------------------2
2
196.5
446.35
8
94.7
100 + 94.7 ------------------------2
2
194.7
641.05
12
76.5
94.7 + 76.5 --------------------------2
4
342.4
983.45
24
34
76.5 + 34 ---------------------2
12
663
1646.45
∞
0
Cpm last + ------------------K small
485.7
2132.15
25.
X m1u ( 0 ) = 0 from pkin description
26.
X m1u ( inf ) = 15.1mg given in data
27.
X m1f ( 0 ) = 0 from pkin description
28.
X m1f ( inf ) = 3.6mg given in data
29.
k m1u = 0.042hr
30.
k m1f = 0.01hr
31.
k m2 = 0.018hr
–1
–1
–1
k m1u Xm1u k m1u ----------- = ------------ = -------------------= 15.1mg -----------------– 1 K2 Xo 25mg 0.07hr k m1f X m1f k m1f ---------- = ----------- = -------------------= 3.6mg --------------–1 K2 Xo 25mg 0.07hr 0.07 hr
–1
trap
thus k m1u = 0.042hr
thus k m1f = 0.01hr
= K 2= k m1f + k m1u + k m2 = 0.01 hr
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–1
–1
–1
+ 0.042 hr
–1
+ k m2 14-9
Practice Exams: Section 1 : Nifedipine - Section 1
32.
K2 = 0.07hr
–1
From your LaPlace Transforms, you know that the equation is bi-exponential, with one of
the slopes being K1 and the other being K2. The terminal slope of graph #4 is 0.07 hr -1, which is not K1 (0.375 hr -1). So it must be K2. 33.
X m2 ( 0 ) = 0 from pkin description
34.
X m2 ( inf ) = 0 from pkin description
35.
C m2 ( inf ) = 0 from pkin description
36.
X m2u ( 0 ) = 0 from pkin description
37.
X m2u ( inf ) = 5.0mg given in data
38.
X m2f ( inf ) = 1.3mg given in data
39.
k m2u = 0.11hr
40.
k m2f = 0.028hr
–1
–1
k m2u k m2u X m2u 5.0mg ----------------------- = ----------------------------= ---------= -----------------------------------–1 K3 X m2u + Xm2f 5.0mg + 1.3mg 0.138hr k m2f k m2f Xm2f 1.3mg ----------------------= --------- = ----------------------------= -----------------------------------–1 K3 X + X 5.0mg + 1.3mg m2u m2f 0.138hr
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14-10
Practice Exams: Section 1 : Enalapril - Section 1
14.2 Enalapril - Section 1 Citation
14.2.1
ENALAPRIL DATA
Enalapril (Vasotec ®) is an angiotensin converting enzyme (ACE) Inhibitor effective in the treatment of hypertension and chronic heart failure (CHF). The parent compound is weakly active, while its sole metabolite, Enalaprilat, exhibits almost all of the pharmacological activity. Original work on the pharmacokinetics of Enalapril showed the parent drug, Enalapril and its metabolite, Enalaprilat, show up in both feces and urine. The following data was offered regarding the Pharmacological response of Enalaprilat as judged by the fall in Seated Systolic Blood Pressure (SSBP). PROBLEM TABLE 11 - 1. Enalapril
- Section 1: Pharmacological Profile
Fall in SSBP (mmHg)
Time (hr) after Max
Dose (mg)
Fall in SSBP (mmHg)
15
3
1
0
12
7
5
1
10
10
10
5
8.6
12
15
10
6
18
20
16
4
24
25
20
30
24
PROBLEM TABLE 11 - 1. Enalapril
- Section 1 IV Bolus Profile from a 5 mg dose
Cpm (ng/ml)
Cumulative Enalapril in urine (mg)
Cumulative Enalapril in feces (mg)
Time (hr)
Cp (ng/mL)
1
29
7.5
0.41
0.12
2
17
11.6
0.65
0.20
3
10
13.8
0.80
0.24
4
5.9
14.9
0.88
0.26
6
2
15.2
0.96
0.30
1.0
0.3
8
14.7
10
13.8
20
9.8
30
6.9
40
4.9
50
3.4
Cumulative Enalaprilat in urine (mg)
Cumulative Enalaprilat in feces (mg)
2.25
1.45
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14-11
Practice Exams: Section 1 : Enalapril - Section 1
The intercept of Cp m vs t was found to be 19.8 ng ⁄ mL and a total of 2.25 mg of enalaprilat was found in the urine while 1.45 mg was found in the feces. PROBLEM TABLE 11 - 1. Enalapril
Answer Pool
Negative Numbers
Small Numbers
Small Numbers
Medium Numbers
Large Numbers
A
0.000
0.014
1.00
14.5
100
B
-0.014
0.021
1.45
19.8
174
C
-0.021
0.030
.1.90
30.3
528
D
-0.030
0.035
2.25
33.0
395
E
-0.035
0.039
3.75
37.5
693
F
-0.110
0.110
5.00
39.5
834
G
-0.390
0.150
6.93
50.0
1740
H
-0.530
0.300
8.00
70.0
5280
I
-0.700
0.390
9.00
94.0
14400
J
-0.900
0.530
9.76
97.0
16000
Equation Answer Pool: A
k mu k m Xo 1 – e – K2t 1 – e – K1t ------------------------ --------------------- – --------------------( K1 – K2 ) K2 K2
B
( k mf k m Xo ) 1 – e –K2t 1 – e – K1t ------------------------ --------------------- – --------------------( K1 – K2 ) K2 K1
C
km ⋅ X 0 – K2t K1t ---------------------------------------⋅ ) – (e ) ( K1 – K2 ) ⋅ Vd m ( e
D.
k mu k m Xo – K2t K1t ------------------------⋅ ) – (e ) ( K1 – K2 ) ( e
E
km ⋅ X 0 – K2t K1t ------------------------⋅ ) – (e ) ( K1 – K2 ) ( e
F
– K1t
G
kr ⋅ X 0 e
I
k – K1t ------u- ⋅ X0 ( 1 – e ) K1
H
J
X 0 – K1t ------ e V k – K1t ------f- ⋅ X0 ( 1 – e ) K1 X0e
– K1t
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14-12
Practice Exams: Section 1 : Enalapril - Section 1
14.2.2
ENALAPRIL QUESTIONS Find the equation for: 1) X, the mass of parent drug, Enalapril, in the body. 2) Xu, the mass of parent drug, Enalapril, in urine. 3) dX u/dt, the rate of excretion of parent drug, Enalapril, in urine. 4) Xf, the mass of parent drug, Enalapril, in feces. 5) Xm, the mass of metabolite, Enalaprilat, in the body. 6) Xmu, the mass of metabolite, Enalaprilat, in urine. 7) dXmu/dt, the rate of excretion of metabolite, Enalaprilat, in urine. 8) Xmf, the mass of metabolite, Enalaprilat, in feces. 9) Cp, the plasma concentration of parent drug, Enalapril. 10) Cpm, the plasma concentration of metabolite, Enalaprilat.
18) kmf, the fecal excretion rate constant for metabolite, Enalaprilat, in plasma. 19) K1, the elimination rate constant of parent drug in plasma(the summation of all processes which remove Enalapril) 20) K2, the elimination rate constant of metabolite in plasma(the summation of all processes which remove Enalaprilat).
Find the intercept of the graph of: 21) Cp vs T 22) Cpm vs T (terminal extrapolated line) 23) dXu/dt vs T 24) dXf/dt vs T
Find the slope of the terminal portion of the graph of: 25) Cp vs T 26) Cpm vs T 27) dXu/dt vs T 28) dXf/dt vs T
Find the value of: 11) dR/dT 12) dR/d(ln(C))
Find the area under the curve of the graph of
13) d(ln(C))/dT
29) Cp vs T(first trapazoid only)
14) kr, the renal excretion rate constant of parent drug, Enalapril, in plasma.
30) Cpm vs T(first trapazoid only)
15) kf, the fecal excretion rate constant of parent drug, Enalapril, in plasma. 16) km, the metabolism rate constant ofparent drug, Enalapril, in plasma. 17) kmu, the renal excretion rate constant for metabolite, Enalaprilat, in plasma.
31) Cp vs T(from Tlast to T ∞ only) 32) Cpm vs T(from Tlast to T ∞ only) 33) Cp vs T 34) Cpm vs T 35) dXu/dt vs T
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14-13
Practice Exams: Section 1 : Enalapril - Section 1
36) dXmu/dt vs T
49) XMU0
37) T*Cp vs T
50) XMF0
38) T*Cpm vs T
51) X ∞
Find the value of :
52) X u ∞
39) AUC(enalapril) 40) AUMC(enalapril)
53) X f ∞
41) MRT(enalapril)
54) X m∞
42) AUC(enalaprilat)
∞ 55) X mu
43) AUMC(enalaprilat)
∞ 56) X mf
44) MRT(enalaprilat)
57) Cp0
45) X0
58) Cpm0
46) XU0
59) Vd
47) XF0
60) Vdm
48) XM0
14.2.3
ENALAPRIL SOLUTIONS
Enalapril Model kf Xf
kr
X
Xu
km Xmf
kmf
Xm
kmu Xmu
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14-14
Practice Exams: Section 1 : Enalapril - Section 1
sponse
esponse
RESPONSE DATA
R vs C 24 22 20 18 16 14 12 10 10
10
R vs T
15 14 13 12 11 10 9 8 2
Concentration
4
6
8
10
12
Time (hr) Enalapril
Enalapril IV Bolus - Metabolite 102
101
101
Concentration (ng/mL)
Concentration (ng/mL)
102
100 0
1
2
3
4
5
6
100 0
Time (hrs)
10
20
30
40
50
Time (hours)
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14-15
Practice Exams: Section 1 : Enalapril - Section 1
– k1t
1
X = Xoe
2
k – k1t X u = ----r- X o ( 1 – e ) k1
3
dx u ⁄ dt = k r Xo e
4
5
6
7
8
9
10
– k1t
k – k1t ) X f = -----f Xo ( 1 – e k1 km X o – k2t – k1t Xm = --------------------( e –e ) ( k1 – k2 ) k mu k m Xo 1 – e –k2t 1 – e –k1t Xmu = ---------------------------------------- – -------------------( k1 – k2 ) k2 k1 dX mu k mu k m Xo – k2t – k1t ------------- = --------------------- ( e –e ) dt k1 – k2 ( k mf k m X ) 1 – e – K2t 1 – e – K1t o X mf = ------------------------ --------------------- – --------------------K1 – K2 K2 K1
C p = C po e
– k1t
D- e – k1t = -----Vd
km ⋅ X 0 – K2t K1t -⋅ Cp m = --------------------------------------) – (e ) ( K1 – K2 ) ⋅ Vd m ( e
11
dR ⁄ dt = – 0.7
12
dR ⁄ d ( ln c ) = 20
13
d ( ln c ) ⁄
0.7 = – 0.035 dt = –---------20
∞
14
Xu –1 k r = ------ • k 1 = 1--- • 53 = 0.106 = 0.11hr 5 Xo
15
Xf –1 k f = ------ • k 1 = 0.3 ------- • 0.53 = 0.0318 = 0.03 hr Xo 5
∞
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14-16
Practice Exams: Section 1 : Enalapril - Section 1
∞
∞
16
Xmu + X mf –1 2.25 + 1.45 k m = -----------------------• k 1 = --------------------------- • 0.53 = 0.39 hr Xo 5
17
Xmu –1 2.25 • k 2 = ---------- • 0.035 = 0.21hr k mu = -----------------------∞ ∞ 3.7 Xmu + Xmf
∞
∞
Xmf –1 –1 1.45 • k 2 = ---------- ⋅ 0.035hr = 0.014 hr k mf = -----------------------∞ 3.7 Xmu + Xmf
18
–1
19
K1 = 0.53hr
20
K2 = 0.035hr
= slope of Cp vs T = k m + kf
–1
= terminal slope of graph Cp m vs T = k mu + k mf
Cp ( 0 ) = 50ng ⁄ mL
21 22
I = 19.8 ng ⁄ mL
23
dX --------u- vs T dt
I = 0.53
k f ⋅ X0 = ( 0.03 ⋅ 5 ) = 0.15ng ⁄ mL
24
PROBLEM TABLE 11 - 1.
T mid
Time
Urinary Rate Set up
∆T
0
Xu
∆ Xu
∆ Xu
---------∆T
∆ Xu
---------∆T
0
0.5
1
1
0.41
0.41
0.41 ⁄ 1
0.41
1.5
2
1
0.65
0.24
0.24 ⁄ 1
0.24
2.5
3
1
0.80
0.15
0.15 ⁄ 1
0.15
3.5
4
1
0.88
0.08
0.08 ⁄ 1
0.08
5
6
2
0.96
0.08
0.08 ⁄ 2
0.04
–1
25
0.53 hr
26
0.035 hr
–1
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14-17
Practice Exams: Section 1 : Enalapril - Section 1
–1
44
16000 --------------- = 30.3hr 528
45
X 0 = 5mg
29 )- ⋅ 1 = 39.5ng ⁄ mL ⋅ hr ---------------------2
46
Xu ( 0) = 0
47
Xf ( 0 ) = 0
(0 +
48
Xm ( 0 ) = 0
49
Xmu ( 0 ) = 0
27
0.53 hr
28
0.53 hr
–1
29 ( 50 +
30
7.5 ) ⋅ 1 = 3.75ng ⁄ mL ⋅ hr --------------------2 31
2 ---------- = 3.77ng ⁄ mL ⋅ hr 0.53
50
Xmf ( 0 ) = 0
32
3.4 - = 97ng ⁄ mL ⋅ hr -----------0.035
51
X = 0
52
Xu = 1mg
33
94 ng ⁄ mL ⋅ hr
∞
53
34
1 19.8 -----------0.035
1 - = 528ng ⁄ mL ⋅ hr – --------0.53
35
dX AUC --------u- vsT = 1mg dt
36
dX mu AUC ------------ vsT = 2.25mg dt
37
∞
174 ng ⁄ mL ⋅ hr
2
38
16000 ng ⁄ mL ⋅ hr
39
94 ng ⁄ mL ⋅ hr
2
∞
Xf = 0.3mg = ( 5 – 1 – 2.25 – 1.45 ) ∞
54
Xm = 0
55
X mu = 2.25mg
56
X mf = 1.45mg
∞
∞
57
Cp 0 = 50ng ⁄ mL
58
Cp m ( 0 ) = 19.8ng ⁄ mL dose- = 100L Vd = -----------Cp ( 0 )
59 2
40
174 ng ⁄ mL ⋅ hr
41
1.9 hr
42
528 ng ⁄ mL ⋅ hr
43
16000 ng ⁄ mL ⋅ hr
60
( km ⋅ X0 ) ( 0.39 ⋅ 5 ) Vd m = --------------------------= ----------------------------------------------------( K1 – K2 ) I ( K1 – K2 ) ( 0.0198 2
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14-18
Practice Exams: Section 1 : Ciprofloxacin Section 1
14.3 Ciprofloxacin Section 1 14.3.1
CIPROFLOXACIN DATA
Ciprofloxacin (Cipro ®) is a fluoroquinolone, synthetic broad spectrum antibacterial agent. It is reasonably well absorbed orally and is excreted into the urine and is metabolized to four known metabolites which are also excreted into the urine. Before an accurate analytical assay was worked out for ciprofloxacin, a microbiological assay was used to determine the pharmacokinetics. This assay utilized the ability of ciprofloxacin to kill microorganisms grown on an agar plate and create a clear zone of no growth, the zone of inhibition. The following data was collected from that procedure: (Intercept for the extrapolated metabolite graph was 0.372 mg/L) PROBLEM TABLE 11 - 1.
Data Set 0ne
Data Set Two
Pharmacological Response Zone -mm
Time (hrs)
Pharmacological Response zone - mm
Concentration Spiked plasma
27
1
4.7
0.01
23.2
5
13.5
0.05
20.4
8
17.3
0.1
18.5
10
19.5
0.15
5.2
24
21
0.2
23.5
0.3
28
0.7
PROBLEM TABLE 11 - 1. Data
Time (Hrs)
Cp (mg/L)
0.5
from 500 mg IV Bolus Cipro.
Cpm (mg/L)
Interval (hours)
Cipro (mg) in urine
0.045
0-4
184
1
3
0.077
4-8
92
2
2.53
0.115
8 - 12
46
3
2.12
0.129
4
1.79
0.128
5
0.119
6
1.26
0.108
8
0.893
0.084
10
0.062
12
0.045
16
0.023
18
0.016
20
0.012
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14-19
Practice Exams: Section 1 : Ciprofloxacin Section 1
14.3.2
CIPROFLOXACIN QUESTIONS
From the above information find the pharmacokinetic parameters: 1) K (elimination rate constant of Cipro) (hr-1 ) 2) kr (urinary excretion rate constant of Cipro)(hr-1 ) 3) km (metabolism rate constant of Cipro)(hr-1 ) 4) kmu (urinary excretion rate constant of the metabolites)(hr-1 ) 5) Cp0 (IV bolus Cipro) (mic/mL) 6) Vd (Cipro)(L.) 7) Vdm (metabolites)(L.) 8) AUC (Cp vs t for Cipro by IV bolus) (mic/mL * hr) 9) AUMC (t*Cp vs t for Cipro by IV bolus (mic/mL * hr2) 10) MRT (Cipro)(hr) 11) AUC (dXu/dt vs t(mid) for Cipro by IV bolus) (mg) 12) Xinf
(Cipro in the body at infinite time)
13) Xuinf
(Cipro in the urine at infinite time)
14) Xminf
(metabolites in the body at infinite time)
15) Xmu inf 16) T1/2
(metabolites in the urine at infinite time)
(Cipro)(hr)
17) Clearance (Cipro)(L/hr) 18) dR/dT (mm/hr) 19) dR/d(lnC) (mm) 20) d(lnC)/dt (Cipro)(hr-1) 21 Slope of Cp vs T ([Cipro] vs t) on semi-log paper 22 Slope of dXu/dt vs T(mid) vs t on semi-Log paper 23 Slope of terminal portion of Cpm vs T on semi-log paper 24) Slope of stripped portion of Cpm vs T on semi-log paper 25) Equation for the rate of formation of the metabolites in the urine (dXmu /dT vs T) a) kmu * Xo * (e-Kt - e-kmut ) / (kmu - K) b) kmu * km * Xo * (e-Kt - e-kmut ) / (K - kmu) c) km * Xo * (e-Kt - e-kmut ) / (kmu - K) d) kmu * km * Xo * {(1-e-Kt )/K- (1-e-kmut )/kmu} / (kmu - K) e) kmu * km * Xo * (e-Kt - e-kmut ) / (kmu - K) Basic Pharmacokinetics
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14-20
Practice Exams: Section 1 : Ciprofloxacin Section 1
14.3.3
CIPROFLOXACIN SOLUTIONS
Ciprofloxacin Model X
ku
Xu
kmu
Xmu
km Xm
Ciprofloxacin
Ciprofloxacin
30
20 15 10 5 0 5
10
15
20
25
Time (hr)
100
100
10-1
10 -1
10-2 4
Time (hours)
6
0
8
10 0
Concentration (mic/mL)
Ciprofloxacin Urine data 102
Concentration (mic.mL)
2
10 -1
Ciprofloxacin Metabolite data
101
0
10 -2
5
10
15
20
dXu/dt (mg)
0
Ciprofloxacin
Concentration (mic/mL)
Response (zone - mm)
Response (zone - mm)
25
30 25 20 15 10 5 0
101
Time (hours)
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0
2
4
6
8
10
Tmid (hours)
14-21
Practice Exams: Section 1 : Ciprofloxacin Section 1
K = 0.173hr
1.
–1
2.
R ⁄ hr- = 0.13hr – 1 k r = -----0- = 65.13mg -----------------------------X0 500mg –1
3
k m = 0.043hr
4
k mu = 0.462hr
5
Cp 0 = 3.57mic ⁄ mL
6
Vd = 140L
7
Vd m = 200L
9.
AUMC = 119.7mic ⁄ mL ⋅ hr
2
10
MRT = 5.8hr
11
dX u AUC = 375mg ---------vs ( t mid ) dt X = 0
13
X u = 375mg
14
Xm = 0
16
T 1 ⁄ 2 = 4.0hr
17
Cl = 24.2L ⁄ hr
18
dR ------- = – 0.95mm ⁄ hr dT
19
dR = 5.5mm ----------------d ( ln C )
20
d----------------( ln C ) = – 0.173hr –1 dt
21
Slope of Cp vs T = – 0.173
22
Slope
of
dX u --------- vs ( t mid ) = – 0.173 dt 23
Slope of terminal portion of
Cp m vsT = – 0.173 24
Slope of stripped portion of
Cp m vsT = – 0.462 25
k mu ⋅ k m ⋅ X 0 – Kt – kmut ----------------------------• (e – e ) ( k mu – K )
∞
12
X mu = 125mg
–1
AUC = 20.7mic ⁄ mL ⋅ hr 8 Cp vs T
∞
15
∞
∞
Basic Pharmacokinetics
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14-22
Practice Exams: Section 1 : Methylphenidate Section 1
14.4 Methylphenidate Section 1 Citation
14.4.1
METHYLPHENIDATE DATA
Methylphenidate (MP) (Ritalin®) is an effective stimulant in the treatment of narcolepsy in adults and attention deficit syndrome in children. It is entirely metabolized to the inactive metabolite, Ritalinic Acid (RA), by the liver which is subsequently excreted unchanged into the urine. Pharmacological response was measured objectively using sleep latency as measured by encephalography. Sleep latency is the time required to fall asleep in a darkened quiet room.
PROBLEM TABLE 11 - 1.
Pharmacological Response Data Time (hr)
Response (min)
Cp (mcg/mL)
Response (min)
0.5
22
6
22.5
1
16
5
18
1.5
10
4
15
2
5
3
10
PROBLEM TABLE 11 - 1. Plasma
Profiles (Intercept of extrapolated RA line 0.21 mic/mL) Plasma Vs. Time Profile 10 mg MP given by IV Bolus
Time (hr)
Cp (MP) (mcg/mL)
Cpm (RA) (mcg/mL
0.5
0.091
0.022
1
0.067
0.029
1.5
0.049
0.030
2
0.035
0.028
2.5
0.026
0.024
3
0.019
0.020
4
0.010
0.012
6
0.003
0.004
8
-------
0.0013
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14-23
Practice Exams: Section 1 : Methylphenidate Section 1
14.4.2
METHYLPHENIDATE QUESTIONS: From the above information find the pharmacokinetic parameters: 1) K (elimination rate constant of MP) (hr
-1
)
2) kr (urinary excretion rate constant of MP)(hr-1 ) 3) km (metabolism rate constant of MP)(hr-1 ) 4) kmu (urinary excretion rate constant of (RA)(hr-1 )
25) d(lnCp)/dt (MP)(hr-1) 26) The slope of Cp vs T ([MP] vs t) on semi-log paper 27) The equation for Cp vs T ([MP] vs t) on semi- log paper 28) The slope of terminal portion of dXmu/dt vs T(mid) on semi-Log paper 29) The equation for dXmu/dt vs T(mid) on semi- log paper 30) The slope of terminal portion of Cpm vs
5) Cp0 (IV bolus MP) (mic/mL)
T ([RA] vs t) on semi-log paper
6) Vd (MP)(L.)
31) The equation for Cpm vs T ([RA] vs t) on semi- log paper
7) Vdm (RA)(L.) 8) AUC (Cp vs t for MP by IV bolus) (mic/ mL * hr) 9) AUMC (t*Cp vs t for MP by IV bolus (mic/mL * hr2)
32) The slope of stripped portion of Cpm vs T ([RA] vs t) on semi-log paper
33) The equation for Xm
(bar) (Laplace domain
equation for RA in the body)
34) The equation for Xm (RA in the body)
10) MRT (MP)(hr) 11) AUC (dXu/dt vs t(mid) for MP by IV bolus) (mg) 12) AUC (dXmu vs t(mid) for RA)(mg) 13) X0 (MP in the body at zero time) 14) Xm0 (RA in the body at zero time) 15) Xmu0 (RA in the urine at zero time) 16) X1hr (MP in the body at 1 hr.) 17) Xm1hr (RA in the body at 1 hr) 18) Xinf
(MP in the body at infinite time)
19) Xuinf
(MP in the urine at infinite time)
20) Xminf
(RA in the body at infinite time)
21) T1/2
(MP)(hr)
22) Clearance (MP)(L/hr) 23) dR/dT (min/hr) 24) dR/d(lnCp) (%)
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14-24
Practice Exams: Section 1 : Methylphenidate Section 1
14.4.3
METHYLPHENIDATE SOLUTIONS
X km Xm
kmu
Xmu
R vs C
Methyl Phenidate
24
Response
25 20
22 20 0
1
18
15
Response (min)
16
10
14 12
5 10
0
8
0.5
1.0
1.5
10
2.0
10
Concentration
Time (hr)
MP IV Bolus - Metab
100
100
10-1
10-1
Concentration (mic/mL)
Concentration (mic/mL)
Methyl Phenidate IV Bolus
10-2 0
1
2
3
4
Time (hours)
10-2 0
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2
4
6
8
Time (hours)
14-25
Practice Exams: Section 1 : Methylphenidate Section 1
–1
1
K = 0.63hr
2
kr = 0
3
k m = 0.63hr
4
k mu = 0.90hr
5
Cp 0 = 0.125mic ⁄ mL
6 pool)
–1
Vd = 80L (change in answer
pool)
AUC = 0.20mic ⁄ mL ⋅ hr
8
T 1 ⁄ 2 = 1.1hr
22
Cl = 50.4L ⁄ hr
23
dR ------- = – 11.4min ⁄ hr dT
24
dR - = 17.5 min -------------------d ( ln Cp )
25
d-------------------( ln Cp )- = – 0.63hr –1 dT
–1
Vd m = 115L (change in answer
7
21
26
The slope of Cp vs T – 0.63hr
27
The equation for Cp vsT
Cp = Cp 0 ⋅ e 28
9.
AUMC = 0.315mic ⁄ mL ⋅ hr 10
2
MRT = 1.57hr
The slope of terminal portion of
29
The equation for
dX mu k mu ⋅ k m ⋅ X0 – K1t (e –e ------------- vsT mid = ----------------------------( K2 – K1 ) dT 30
Cp m vsT ( RA )vsT = –0.94
X0 = 10mg
14
Xm ( o ) = 0
15
Xmu ( 0 ) = 0
32
16
X1hr = 5.3 (change in answer
– 0.79
31
The equation for
km ⋅ X0 – K1t Cp m vsT ( RA ) = ------------------------(e –e ( K2 – K1 )
pool)
Xm1hr = 2.8 (change in answer
pool)
33 34
X = 0
19
Xu = 0
20
Xm = 0
The slope of stripped portion
km ⋅ X0 Xm = --------------------------------------------( s + K1 ) ⋅ ( s + K2 ) The equation for
km ⋅ X 0 – K1t – K 2t Xm = ------------------------(e –e ) ( K2 – K1 )
∞
18
∞
)
the slope of the terminal portion of
13
17
–k ⋅ t
dX mu ------------ vsT mid = – 0.63 dT
11 AUC = 0 MP is not excreted in the urine 12 AUC = 10mg All of RA is excreted in the urine
–1
∞
Basic Pharmacokinetics
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14-26
)
Practice Exams: Section 1 : Adinazolam - Section 1
14.5 Adinazolam - Section 1 Fleishaker [Pharm Res 8 (2): 162-7(1991),Pharm Res 6 (5): 379-86(1989),Psychopharm 99:34-9(1989),Psychopharm 105:1815(1991)];Kroboth [J Clin Pharmacol 31:580-6(1991)]; Wagner [Biopharm Drug Dis 8:405-25(1987)].
14.5.1
ADINAZOLAM DATA
Adinazolam (AD) is a triazolobenzodiazepine prodrug which has been shown to have antidepressant activity through its metabolite, N-desmethyladinazolam (NDMAD). Both AD and NDMAD are excreted into the urine. The following data regarding AD and NDMAD was collected from the works cited. AD appears to interfere with mental ability. One of the tests was the ability to Substitute Digits for Symbols (DSST) and the interference of that process was the response measured below: PROBLEM TABLE 11 - 1.
Pharmacological Response
% Pharm Response
Time (hrs)
% Pharm Response
NDMADC Conc (ng/,L)
59.7
2
33
200
45.3
4
48
300
29.5
6
59
400
15.5
8
68
500
0
12
74
600
From the above information find the pharmacokinetic parameters: PROBLEM TABLE 11 - 1.
Five mg IV bolus dose of AD yieled: (intercept of extrapolated metabolite line is 150 ng/mL)
Plasma Data
Urine Data
Time (hr)
Cp (AD) (ng/mL)
0.5
47
1
40
15.4
2
28
3 4
14.7
5 6
Cpm(NDMAD) (ng/mL)
Time (mid) t
AD Ecretion Rate (mg/hr)
0.5
0.56
1
0.48
23.5
2
0.34
27.3
4
0.18
28
6
0.09
27 7.5
24.9
8
20
10
15
12
11
15
7.5
18
4.2
20
2.8
22
1.9
24
1.2
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Practice Exams: Section 1 : Adinazolam - Section 1
14.5.2
ADINAZOLAM QUESTIONS
1) K (elimination rate constant of AD) (hr-1 ) 2) kr (urinary excretion rate constant of AD)(hr-1 ) 3) km (metabolism rate constant of AD)(hr-1 ) 4) kmu (urinary excretion rate constant of NDMAD)(hr -1 ) 5) Cp0 (IV bolus AD) (ng/mL) 6) Vd (AD)(L.) 7) Vdm (NDMAD)(L.) 8) AUC (Cp vs t for AD by IV bolus) (ng/mL * hr) 9) AUMC (t*Cp vs t for AD by IV bolus (ng/mL * hr2) 10) MRT (AD)(hr) 11) AUC (dXu/dt vs t(mid) for AD by IV bolus) (mg) 12) Xinf
(AD in the body at infinite time)
13) Xuinf
(AD in the urine at infinite time)
14) Xminf
(NDMAD in the body at infinite time)
15) Xmu inf 16) T1/2
(NDMAD in the urine at infinite time)
(AD)(hr)
17) Clearance (AD)(L/hr) 18) dR/dT (%/hr) 19) dR/d(lnC) (%) 20) d(lnC)/dt (NDMAD)(hr-1) 21 Slope of Cp vs T ([AD] vs t) on semi-log paper 22 Slope of dXu/dt vs T(mid) vs t on semi-Log paper 23 Slope of terminal portion of Cpm vs T ([NDMAD] vs t) on semi-log paper 24) Slope of stripped portion of Cpm vs T ([NDMAD] vs t) on semi-log paper 25) Xm
(bar) (Laplace domain equation for NDMAD in the body)
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Practice Exams: Section 1 : Adinazolam - Section 1
PROBLEM TABLE 11 - 1. Answer
Pool
Numbers
LaPlace Forms
a
-1.33
-0.1
0
1
10
100
Xo/(s+K)
b
-1.55
-0.133
0.1
1.33
12.5
120
Xo/(s+km)
c
-2
-0.15
0.133
1.5
20
167
Xo/(s+kmu)
d
-3.33
-0.2
0.15
2
30
200
km * Xo / ((s+km) * (s+kmu))
e
-4
-0.333
0.2
3
37.5
375
kmu * Xo / ((s+km) * (s+kmu))
f
-5
-0.5
0.333
4
40
425
km * Xo / ((s+km) * (s+K ))
g
-6.93
-0.693
0.5
5
50
500
kmu * Xo / ((s+K ) * (s+kmu))
h
-7.5
-0.73
0.693
6
55.6
630
km * Xo / ((s+K ) * (s+kmu))
i
-8.25
-0.85
0.75
6.67
66.7
775
kmu*km*Xo / ((s+K ) * (s+kmu))
j
-9
-0.95
0.84
7.5
75
930
kr*Xo / ((s+K ) * (s+kmu))
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14-29
Practice Exams: Section 1 : Adinazolam - Section 1
14.5.3
ADINAZOLAM SOLUTIONS
X
ku
Xu
kmu
Xmu
km Xm
onse %
R vs T 60
Adinazolam metabolite 80
50
70 40
Response (%P-col)
60
30
20
10 2
3
4
5
6
7
8
Time (hr)
50 40 30 10 2
10 3
Concentration (ng.mL)
AD IV Bolus - Metab
Adinazolam IV Bolus
102
10 2
101
Concentration (ng/mL)
Concentration (ng/mL)
10 1
10 0 0
1
2
3
4
5
6
Time (hours)
100 0
5
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10
15
20
25
Time (hours)
14-30
Practice Exams: Section 1 : Adinazolam - Section 1
1
K = 0.333hr
k r = 0.133hr
3
k m = 0.2hr
4
k mu = 0.2hr
5
Cp 0 = 55.6ng ⁄ mL
23
–1
Slope of the terminal portion of
Cp m vsT = – 0.2 –1
24 0.333 25
6.
Vd = 90L
7
Vd m = 50L
8
AUC = 167ng ⁄ mL ⋅ hr AUMC = 500ng ⁄ mL ⋅ hr
10
MRT = 3.0hr
11
AUC m = 2.0mg
12
X = 0
13
X u = 2.0mg
14
Xm = 0
15
Xmu = 3.0mg
16
T1 ⁄ 2 = 2hr
17
Cl = 30L ⁄ hr
18
dR ------- = – 7.42 %hr dT
19
dR = 37.5 ----------------% d ( ln C )
20
d----------------( ln C ) = – 0.2hr – 1 dt
21 0.333
Slope of
dX u ---------vsT( mid ) = – 0.333 dt
–1
2
9
22
–1
Slope of the stripped portion = -
km ⋅ X0 X m = --------------------------------------( s + K ) ( s + k mu )
2
∞ ∞
∞ ∞
Slope of Cp vs T([AD] vs t) = -
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14-31
Practice Exams: Section 1 : Labetalol - Section 1
14.6 Labetalol - Section 1 Saotome, et. al (J. Clin. Pcol 1993,33:979- 988)
14.6.1
LABETALOL DATA
Labetalol is a selective α- blocking and nonselective β-blocking agent used in the treatment of hypertension. It is excreted unchanged into the urine and extensively metabolized. The metabolites are excreted into the urine and bile. Labetalol reduces diastolic blood presure. PROBLEM TABLE 11 - 1. P-col
Resp vs Time (hrs) P-col Resp v Labetalol Conc
Pharmacological Response Profiles Decrease in BP (mm Hg)
Time (hr)
Decrease in BP (mm Hg)
Cp (ng/mL)
22.2
2
14.5
100
20.3
4
16.5
160
18.5
6
19.5
275
16.6
8
22.5
525
PROBLEM TABLE 11 - 1.
Data from a 200 mg IV Bolus dose (Intercept of extrapolated Metabolite data = 22.7 mg/hr)
Plasma Data
Urine Data Time (mid)
Labetalol Excretion rate
Metabolite Excretion Rate
Time (hrs)
Cp
t
(mg/hr)
(mg/hr)
0.5
364
0.5
1.82
5.1
1
331
1
1.66
8.2
2
274
1.5
4
187
2
6
128
2.5
8
88
3
9.8 1.37
10.6 10.7 10.5
4
0.94
9.5
6
0.64
7.0
8
4.9
12
2.3
18
0.74
24
0.24
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Practice Exams: Section 1 : Labetalol - Section 1
14.6.2
LABETALOL QUESTIONS From the above information find the pharmacokinetic parameters: 1) K (elimination rate constant of labetalol) (hr-1 ) 2) K2 (elimination rate constant of labetalol metabolites (hr-1 ) 3) kr (urinary excretion rate constant of labetalol)(hr-1 ) 4) km (metabolism rate constant of labetalol)(hr-1 ) 5) kmu (urinary excretion rate constant of Metabolites of labetalol)(hr-1 ) 6) kmb (biliary excretion rate constant of Metabolites of labetalol)(hr-1 ) 7) Cp0 (IV bolus labetalol) (ng/mL) 8) Vd (labetalol)(L.)
21) Clearance (labetalol)(L/hr) 22) dR/dT (%/hr) 23) dR/d(lnC) (%) 24) d(lnC)/dt (labetalol)(hr-1) 25 Slope of Cp vs T ([labetalol] vs t) on semi-log paper 26 Slope of dXu/dt vs T(mid) vs t on semi-Log paper 27 Slope of terminal portion of Cpm vs T ([Metabolites of labetalol] vs t) on semi-log paper 28) Slope of stripped portion of Cpm vs T ([Metabolites of labetalol] vs t) on semi-log paper 29) Xm
(bar) (Laplace domain equation for Metabolites of labetalol in the body)
9) AUC (Cp vs t for labetalol by IV bolus) (ng/mL * hr)
30) dXmu / dT (equation for the rate of excretion of metabolites into the urine)
10) AUMC (t*Cp vs t for labetalol by IV bolus (ng/mL * hr2)
31) Fraction unchanged.
11) MRT (labetalol)(hr)
32) Fraction of labetalol eliminated as metabolites in the urine.
12) AUC (dXu/dt vs t(mid) for labetalol by IV bolus) (mg) 13) AUC (dXmu/dt vs t(mid) for Metabolites (mg))
of
labetalol
excreted
33) Fraction of labetalol eliminated as metabolites in the bile.
14) AUC (dXmb/dt vs t(mid) for Metabolites (mg)) 15) Xinf
(labetalol in the body at infinite time (mg))
16) Xuinf
(labetalol in the urine at infinite time (mg))
17) Xminf
(Metabolites of labetalol in the body at infinite time (mg))
18) X muinf (Metabolites of labetalol in the urine at infinite time (mg)) 19) X mbinf
(Metabolites of labetalol in the bile at infi-
nite time (mg))
20) T1/2
(labetalol)(hr)
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Practice Exams: Section 1 : Labetalol - Section 1
14.6.3
LABETALOL SOLUTIONS
X
ku
Xu
kmu
Xmu
km Xmf
kmf
Xm
Labatalol 24 22 20
2
3
4
5
6
7
Response (mmHg)
Response (mmHg)
Labatalol 23 22 21 20 19 18 17 16 8
Time (hr)
18 16 14 10 2
103
Concentration (ng.mL)
Labatalol Metabolite
Labatalol
Labatalol
102
103
10 1
101
10 0
102
0
2
4
Time (hours)
6
8
10-1 0
5
10
15
20
Time (hours)
25
Excretion rate (mg/hr)
101
Excretion Rate (mg/hr)
Concentration (ng/mL)
100
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10 -1 0
1
2
3
4
5
6
Tmid (hours)
14-34
Practice Exams: Section 1 : Labetalol - Section 1
1
K1 = 0.190hr
2.
K2 = 0.792hr k r = 0.01hr
3
–1
–1
k mu = 0.38hr
6
k mb = 0.41hr
7
Cp 0 = 400ng ⁄ mL
24
–1 d ( ln C ) ----------------- = – 0.190hr dt
25
Slope of Cp vs T = – 0.190hr
26
Slope of
–1
dX u --------- vsT( mid ) vs ( t ) = – 0.190 dt
–1
27 Slope of terminal portion of Cp m vsT = – 0.190 28
Vd = 500L AUC = 2105ng ⁄ mL ⋅ hr
9
dR ----------------- = 4.90 % d ( ln C )
–1
5
8
23
–1
k m = 0.18hr
4
–1
Slope of stripped portion of
Cp m vsT = – 0.79 km ⋅ X 0 X m = ----------------------------------------( s + K1 ) ( s + K2 )
29
10
AUMC = 11072ng ⁄ mL ⋅ hr 11
MRT = 5.26hr
12
AUC = 10.5mg
13
AUC = 91.2mg
14
AUC = 98.3mg
15
X = 0
16
X u = 10.5mg
17
Xm = 0
18
Xmu = 91.2mg
19
Xmb = 98.3mg
20
T1 ⁄ 2 = 3.64hr
21
Cl = 95L
22
dR ------- = – 0.93 %hr dT
2
30
X mu ( k mu ⋅ k m ⋅ X 0 ) – K2t – K1T = ---------------------------------- ( e ) --------–e dt ( K1 – K2 ) 31
Fraction of labetalol excreted
10.5 200
unchanged: ---------- × 100 = 5.25 % 32 Fraction of labetalol excreted as metabolites in urine:
∞ ∞
91.2 ---------- × 100 = 45.6 % 200
∞
33 Fraction of labetalol eliminated as metabolites in bile:
∞
98.3 ---------- × 100 = 49.15 % 200
∞
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14-35
Practice Exams: Section 1 : Zidovudine Section 1
14.7 Zidovudine Section 1 14.7.1
ZIDOVUDINE DATA
Zidovudine (ZDV - formerly azidithymidine (AZT)) is used to treat HIV positive patients with AIDS and AIDS related complex (ARC). The in vitro virustatic concentration is 250 mic/mL. While the pharmacokinetics of ZDV is complex, we can approximate it with the following model. ZDV is excreted unchanged into the urine to UZDV. It is metabolized by hepatic glucuronidation to an inactive metabolite, GZDV, and also metabolized by intracellular phosphorylation to the active metabolite, ZDV-TP. Both metabolites are excreted into the urine (UGZDV and UZDV-TP). The anti-retroviral activity was shown in the following data: PROBLEM TABLE 11 - 1. ZDV-TP
Response vs time and response vs concentration data
Pharmacological Response Data Concentration ZDV-TP (ng/mL)
Activity %
Time (min)
Activity %
100
80
0
100
90
76.8
30
100
75
68.5
60
100
50
52.3
120
92
25
24.5
240
82
360
72
480
62
PROBLEM TABLE 11 - 1. 100
mg dose ZDV given by IV Bolus to 70 KG patient
Plasma Data Time (hours)
Cp ZDV (ng/mL)
Urine Data
Cpm GZDV (ng/mL)
Interval (hrs)
UZDV Collected (mg)
0.25
40
0-1
9.2
0.5
75
1-2
4.8
100
2-3
2.5
125
4-6
1
0.75 1
315
1.5
Rate of excretion data
AUC UGZDV (mg) 60
150
2
170
160
3
91
160
4
48
145
6
13
100
8
65
10
40
12
25
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Practice Exams: Section 1 : Zidovudine Section 1
Please use the following nomenclature: D
= Dose of ZDV
K
= elimination rate constant of ZDV
kr = excretion rate constant of ZDV kmu1 = excretion rate constant of ZDV-TP kmu2 = excretion rate constant of GZDV km1 = metabolism rate constant of ZDV to ZDV-TP km2 = metabolism rate constant of ZDV to GZDV
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14-37
Practice Exams: Section 1 : Zidovudine Section 1
14.7.2
ZIDOVUDINE QUESTIONS 1) Using Laplace transforms, find the equation for the rate of excretion of GZDV into the urine. (For this question use the following answer set: a=
I only
b=
III only
15) What is the terminal slope (*-1) of ln(dUZDV/dt) vs t (hr^-1)? 16) What is the intercept of ZDV vs t (ng/ mL)?
c = I + II only
17) What is the intercept of dUZDV/dt vs t (mg/hr)?
d = II + III only
18) What is the intercept of the extrapolated line from plasma GZDV vs time data (ng/mL)?
e = All three
I
14) What is the terminal slope (*-1) of ln(GZDV) vs t (hr^-1)?
(kmu2*km2*D/(K-kmu2))*(exp(-K*t))
II (kmu2*km2*D/(K-kmu2))*(exp(kmu2*t))
19) What is the intercept of the extrapolated line from dUGZDV/dt vs time data (mg/hr)?
III -(kmu2*km2*D/(K-kmu2))*(exp(K*t))
20) What is the Cp0 of ZDV (ng/mL)?
2) Using linear regression on the pharmacological response / concentration profile, what was the Summation of XY?
22) What is the volume of distribution of GZDV (L)?
3) What is d%R/d(lnC)(%)?
24) What is the half life of ZDV (hr)?
4) What is d%R/dt (%/hr)?
25) What is the MRT of ZDV (hr^-1)?
5) What is K (hr^-1)?
26) What is the AUC of the first trapazoid of ZDV vs time (ng/mL*hr)?
6) What is kr (hr^-1)?
21) What is the volume of distribution of ZDV (L)?
23) What is the clearance of ZDV (L/hr)?
27) What is the AUC(from 0 to infinity) of the IV bolus ZDV plasma data (ng/ mL*hr)?
7) What is km1 (hr^-1)? 8) What is km2 (hr^-1)? 9) What is kmu1 (hr^-1)? 10) What is kmu2 (hr^-1)?
28) How much ZDV is in the body at infinite time (mg)?
11) What is the terminal slope (*-1) of ln(ZDV) vs t (hr^-1)?
29) How much GZDV is in the body at infinite time (mg)?
12) What is the terminal slope (*-1) of ln(ZDV-TP) vs t (hr^-1)?
30) How much ZDV-TP is in the body at infinite time (mg)?
13) What is the terminal slope (*-1) of ln(ZDV-TP) vs t (hr^-1)?
31) How much ZDV is in the urine (UZDV) at infinite time (mg)? 32) How much GZDV is in the urine (UGZDV) at infinite time (mg)?
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14-38
Practice Exams: Section 1 : Zidovudine Section 1
33) How much ZDV-TP is in the urine (UZDV-TP) at infinite time (mg)?
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14-39
Practice Exams: Section 1 : Zidovudine Section 1
14.7.3
ZIDOVULDINE SOLUTIONS
Xm1
km1
X
ku
Xu
kmu2
Xmu2
km2
kmu1 Xmu1
Xm2
Zidovudine
Zidovudine
110 100 90
Response (% Activity)
Response (% Activity)
80 70 60 0
100
200
300
400
500
Time (min)
Zidovudine
10 1
102
Concentration (ng/mL)
Zidovudine Urine data
Zidovudine Metobolite 103
102
102
102
101
101
101
100 0
2
4
6
8
10 12
Time (hours)
UZVD (mg)
Time (hours)
Concentration (ng/mL)
103
0 1 2 3 4 5 6
Concentration (ng/mL)
90 80 70 60 50 40 30 20
10-1 0
1
2
3
4
5
Tmid (hours)
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14-40
Practice Exams: Section 1 : Zidovudine Section 1
1.
Answer is D - II and III
2
1289.2
3
40%
4
-0.08333%hr –1
5
K = 0.630hr
6
k r = 0.125hr
7
k m1 = 0.25hr
–1
–1
457.5ng ⁄ mL ⋅ hr
27
952ng ⁄ mL ⋅ hr
28
0
29
0
30
0
31
20mg
32
60mg
33
20mg
–1
8
k m2 = 0.25hr
9
k mu1 = 0.125hr
10
k mu2 = 0.378hr
11
0.630hr
12
0.250hr
13
Skip - same question as #12
14
0.25hr
15
0.630hr
–1
–1
–1 –1
–1 –1
16
600ng ⁄ mL
17
12.6mg ⁄ hr
18
500ng ⁄ mL
19
12.6mg ⁄ hr
20
600ng ⁄ mL
21
166.7L
22
200L
23
26
105L ⁄ hr
24
1.1hr
25
1.6hr
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14-41
Practice Exams: Section 1 : Fosinopril Section 1
14.8 Fosinopril Section 1 Kostis et al “Fosinopril: Pharmacokinetics and pharmacodynamics in congestive heart failure” Clin Pcol Ther 58(6) 660-5 (1995); Hui et al “Pharmacokinetics of fosinopril in patients with various degrees of renal function” Clin Pcol Ther 49(4) 457 -66 (1991)
14.8.1
FOSINOPRIL DATA
Fosinopril Sodium is a phosphinic prodrug of the angiotensin converting enzyme (ACE) inhibitor fosinoprilat effective in the treatment of hypertension and chronic heart failure (CHF). The parent compound is weakly active, if at all. Its sole metabolite, fosinoprilat, exhibits almost all of the pharmacological activity. After administration, fosinopril is entirely converted to the active fosinoprolat by esterases in the liver. Unlike other ACE inhibitors, elimination of fosinoprilat is divided equally between renal and hepatic pathways (liver metabolism). The following information was obtained from a 70 Kg male. Wherever necessary, please use the following symbols: Cp = plasma concentration of parent drug, fosinopril Cpm1 = plasma concentration of metabolite, fosinoprilat. Cpm2= plasma concentration of metabolite of fosinoplilat. X = amount of parent compound, fosinopril, in the body Xm1 = amount of metabolite, fosinoprilat, in the body Xm2 = amount of metabolite of fosinoprilat in the body Xu = cumulative amount of parent drug in urine. Xmu1 = cumulative amount of metabolite, fosinoprilat, in urine. Xmu2 = cumulative amount of metabolite of fosinoprilat in urine. ku = renal excretion rate constant of parent drug in plasma. kmu1 = renal excretion rate constant for metabolite, fosinopril, in plasma. kmu2 = renal excretion rate constant for metabolite of fosinopril in plasma. kf = fecal excretion rate constant of parent drug in plasma. kmf1 = fecal excretion rate constant for metabolite, fosinopril, in plasma. kmf2 = fecal excretion rate constant for metabolite of fosinopril in plasma. km1 = metabolism rate constant of fosinopril in plasma. km2 = metabolism rate constant of fosinoprilat in plasma. K1 = elimination rate constant of parent drug in plasma = summation of all processes which remove fosinopril. K2 = elimination rate constant of metabolite1 in plasma = summation of all processes which remove fosinoprilat. K3 = elimination rate constant of metabolite2 in plasma = summation of all processes which remove the metabolite of fosinoprilat.
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14-42
Practice Exams: Section 1 : Fosinopril Section 1
PROBLEM TABLE 11 - 1. Pharmacological
Response data (Six hours after dosing with fosinopril)
Time (hr)
Fall in SSBP (mm Hg)
Fall in SSBP (mg Hg)
Dose (mg)
3
22
5
10
5
21
11
20
8
19
16
40
11
17
22
80
13
16
PROBLEM TABLE 11 - 1.
Plasma Data (from7.5 mg Fosinoprilat IV Bolus)
Urine Data (from 20 mg Fosinopril IV bolus) (Intercept 0.855 mg/hr)
Tme (hours)
Cp (Fosinoprilat) (ng/mL)
Interval (hr)
Amount (Fosinoprilat) (mg)
1
387
0-1
0.158
2
360
1-2
0.362
3
335
2-3
0.467
4
311
3-4
0.513
6
269
4-6
1.05
6 - 10
1.84
10 - 14
1.42
14 - 22
1.84
22 - 26
0.592
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14-43
Practice Exams: Section 1 : Fosinopril Section 1
14.8.2
FOSINOPRIL QUESTIONS From the parent compound, Fosinopril, given by IV bolus, find the equation for: ♣1) X, the mass of parent drug, Fosinopril, in the body ♣) Xu, the mass of parent drug, Fosinopril, in urine.
♥18) K1, the elimination rate constant of parent drug in plasma , the summation of all processes which remove Fosinopril.
♠19) K2, the
elimination rate constant of metabolite in plasma , the summation of all processes which remove Fosinoprilat.
♣3) dXu/dt, the rate of excretion of parent drug, Fosinopril, in urine.
For fosinoprilat IV, find the value of :
♣4) Xf, the mass of parent drug, Fosinopril, in feces.
♥21) AUMC(fosinoprilat)
♥20) AUC(fosinoprilat)
♣5) Xm1, the mass of metabolite, Fosinoprilat, in the body
♥22) MRT(fosinoprilat)
♣6) Xm1u, the mass of metabolite, Fosinoprilat, in urine.
♥24) T 1 ⁄ 2
♣7) dXm1u/dt, the rate of excretion of metabolite, Fosinoprilat, in urine.
♥25) Cp0(fosinoprilat) for iv dose (ng/mL)
♣8) Xm1f, the mass of metabolite, Fosinoprilat, in feces.
♥27) Cp of Fosinoprilat at eight hours after the IV dose
♣9) Cp, the plasma concentration of parent drug, Fosinopril ♣10) Cpm1, the plasma concentration of metabolite, Fosinoprilat.
♥23) Ke for fosinoprilat (hr-1) for Fosinoprilat (hr)
♥26) Vd for Fosinoprilat (L)
For fosinoprilat given as IV fosinopril, dXm1/dt, find the value of : ♠28) AUC(fosinoprilat)(mg) ♠29) AUMC(fosinoprilat)
Find the value of:
♠30) MRT(fosinoprilat)
♦11) dR/dT
♠31) K2 for fosinoprilat (hr-1)
♦12) dR/d(ln(Cpm1))
♠32) T ?
♦13) d(ln(Cpm1))/dT ♥14) ku, the renal excretion rate constant of parent drug,
for Fosinoprilat (hr)
♠33) K1 for fosinopril (hr-1)
Fosinopril, in plasma.
♥15) km1, the metabolism rate constant of parent drug, Fosinopril, in plasma.
♠16) kmu1, the renal excretion rate constant for metabolite, Fosinoprilat, in plasma.
♠17) km2, the metabolism rate constant for the metabolite of Fosinoprilat in plasma.
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14-44
Practice Exams: Section 1 : Fosinopril Section 1
PROBLEM TABLE 11 - 1. Answer
Pool
Minus
Small
Med.
Large
Equations
More equations
A
-0.037
0
1.35
155
None
((km1 X0/Vdm )/(K1-K2))(e-K2t-e-K1t )
B
-0.050
0.037
5.0
232
X0e-K1t
((km1 X0/Vdm )/(K1-K2))(e-K1t-e-K2t )
C
-0.074
0.06
7.4
420
(X0/Vd)e-K1t
((kmu1 X0/Vdm )/(K1-K2))(e-K2t-e-K1t)
D
-0.082
0.074
8.2
840
ku X0e -K1t
((kmu1km1 X0)/(K1-K2))(e-K2t-e-K1t )
E
-0.10
0.135
9.4
940
kf X0e-K1t
((kmu1km1 X0)/(K1-K2))(e-K1t-e-K2t )
F
-0.135
0.5
10
5676
(k u/K1) X0(1-e-K1t)
((kmu1 X0)/(K1-K2))(e-K2t-e-K1t )
G
-0.37
0.6
12
13500
(k f/K1) X0(1-e-K1t)
((kmf1km1 X0)/(K1-K2))((1-e-K2t)/K2-(1-e-K1t )/K1)
H
-0.50
0.74
13.5
15700
((km1 X0)/(K1-K2))(e-K2t-e-K1t)
((kmf1km1 X0)/(K1-K2))((1-e-K1t)/K1-(1-e-K2t )/K2)
I
-0.6
0.82
15.5
18000
((km1 X0)/(K1-K2))(e-K1t-e-K2t)
((kmu1km1 X0)/(K1-K2))((1-e-K2t )/K2-(1-e-K1t)/K1)
J
-0.74
0.94
18
76700
((kmu1 X0)/(K1-K2))(e-K2t-e-K1t)
((kmf2km1 X0)/(K1-K2))((1-e-K2t)/K2-(1-e-K1t )/K1)
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Practice Exams: Section 1 : Fosinopril Section 1
14.8.3
FOSINOPRIL SOLUTIONS ku=0
X
K1= km1 K2= kmu1+km2
km1
K3= kmu2 kmu1 Xm1
Xmu1
km2 kmu2 Xm2
Xmu2
Response
R vs Ln(c) 25
20
Response
15
1
2
10
R vs T 23
22
21
20
19
18
5
17
16
0 10
10
15 0
Concentration
2
4
6
8
10
12
14
Time (hr)
Fosinoprilat in Urine
3
Excretion Rate (mg/hr)
Fosinoprilat
Concentration ng/mL
10
10
0
-1
10
102 0
1
2
3
4
5
0
6
Time (hr)
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10 15 Time (hr)
20
25
14-46
Practice Exams: Section 1
♣1) X, the mass of parent drug, Fosinopril, in the body
B - X0e-K1t
♣) Xu, the mass of parent drug, Fosinopril, in urine.
A - none(no drug goes there)
♣3) dXu/dt, the rate of excretion of parent drug, Fosinopril, in urine
A = none(no drug goes there)
♣4) Xf, the mass of parent drug, Fosinopril, in feces.
A = none(no drug goes there)
♣5) Xm1, the mass of metabolite, Fosinoprilat, in the body
H = (km1X0)/(K1-K2)(e-K2t-e-K1t)
♣6) Xm1u, the mass of metabolite, Fosinoprilat, in urine.
I
=
( k mu1 k m1 X 0 ) 1 – e –K2t 1 – e – K1t -------------------------------- --------------------- – --------------------- K1 ( K1 – K2 ) K2 (k
X )
k
♣7) dXm1u/dt, the rate of excretion of metabolite, Fosinoprilat, in urine.
–K2t – K1t mu1 m1 0 -{e –e } D = -------------------------------
♣8) Xm1f, the mass of metabolite, Fosinoprilat, in feces.
A = none(no drug goes there)
♣9) Cp, the plasma concentration of parent drug, Fosinopril
X C = -----0- e – K1t Vd
♣10) Cpm1, the plasma concentration of metabolite, Fosinoprilat.
–K2t – K1t m1 0 -{e –e } A = -------------------------------
( K1 – K2 )
k
X
( K1 – K2 )V d
Find the value of: ♦11) dR/dT
I = -0.6 mm Hg/hr
♦12) dR/d(ln(Cpm1))
D = 8.2 mm Hg
♦13) d(ln(Cpm1))/dT
C = -0.074 hr-1
♥14) ku, the renal excretion rate constant of parent drug, Fosinopril, in plasma.
A = none(no drug goes there)
♥15) km1, the metabolism rate constant of parent drug, Fosinopril, in plasma.
F = 0.5 hr -1
♠16) kmu1, the renal excretion rate constant for metabolite, Fosinoprilat, in plasma.
B = 0.037 hr -1
♠17) km2, the metabolism rate constant for the metabolite of
B = 0.037 hr -1
Fosinoprilat in plasma.
♥18) K1, the elimination rate constant of parent drug in plasma , the summation of all processes ♠19) K2, the
which remove Fosinopril.
F = 0.5 hr -1
elimination rate constant of metabolite in plasma , the summation of all processes which remove Fosinoprilat. D
= 0.074 hr -1
For fosinoprilat IV, find the value of : ♥20) AUC(fosinoprilat)
F = 5676 ng/mL
♥21) AUMC(fosinoprilat)
J = 76700 mg/mL*hr2
♥22) MRT(fosinoprilat)
H = 13.5 hr
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Practice Exams: Section 1
♥23) Ke for fosinoprilat (hr-1)
D = 0.074 hr -1
♥24) T 1 ⁄ 2
E = 9.4 hr
for Fosinoprilat (hr)
♥25) Cp0(fosinoprilat) for iv dose (ng/mL)
C = 420 ng/mL
♥26) Vd for Fosinoprilat (L)
J = 18 L
♥27) Cp of Fosinoprilat at eight hours after the IV dose
B = 232 ng/mL
For fosinoprilat given as IV fosinopril, dXm1/dt, find the value of : ♠28) AUC(fosinoprilat)(mg)
F = 10 mg
♠29) AUMC(fosinoprilat)
A = 155 mg*hr
♠30) MRT(fosinoprilat)
I = 15.5 hr
♠31) K2 for fosinoprilat (hr-1)
D = 0.074 hr -1
♠32) T ?
E = 9.4 hr
for Fosinoprilat (hr)
♠33) K1 for fosinopril (hr-1)
F = 0.5 hr -1
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14-48
Practice Exams: Section 1
14.9 Omeprazole 14.9.1
OMEPRAZOLE DATA
Omeprazole (mw: 345.42) is a gastric proton-pump inhibitor which decreases gastric acid secretion. It is effective in the treatment of ulcers and esophageal reflux. In normal patients, 20% of the omeprazole is excreted into the feces and 80% of the omeprazole dose is excreted as an inactive metabolite into the urine. In the study by Anderson, et. al., eight patients were given 20 mg, IV bolus doses of omeprazole. The patients had a mean body weight of 70 kg. Blood and urine were collected at various intervals throughout the study and the following data was obtained: TABLE 14-2 Pharmacological
Response
% Acid Secretion Inhibition
Time (min)
% Acid Secretion Inhibition
Concentration (ng/mL)
100
120
74
1
100
240
62
.6
75
330
56
.5
63
360
47
.35
56
375
27
.15
50
390
25
450
10
570
8
600
0
720
TABLE 14-3 Plasma
profile of parent and urine profile of metabolite Intercept = 4 mg/hr AUMCmet = 96 mg*hr
Parent Compound
Time (hr)
Cp (ng/mL)
Urine Collection Interval (hr)
Metabolite collected (mg)
0.5
336
0-1
1.19
1
205
1-2
2.07
2
75
2-3
2.10
3
28
3-5
3.44
4
10
5-7
2.38
7-9
1.62
9 - 11
1.08
11 - 13
0.72
13 - 15
0.48
16 - 20
0.44
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14-49
Practice Exams: Section 1
14.9.2
OMEPRAZOLE QUESTIONS Using LaPlace transforms, from Omeprazole, given by IV bolus, find the equation for: 1.
♣1) X, the mass of Omeprazole in the body
2.
♣) Xu, the mass of Omeprazole in urine.
3.
♣) dXu/dt, the rate of excretion of Omeprazole in urine.
4.
♣) Xf, the mass of Omeprazole in feces.
5.
♣) Xm, the mass of metabolite of Omeprazole in the body
6.
♣) Xmu, the mass of metabolite of Omeprazole in urine.
7.
♣) dXmu/dt, the rate of excretion of metabolite of Omeprazole in urine.
8.
♣) Xmf, the mass of metabolite of Omeprazole in feces.
9.
♣) Cp, the plasma concentration of parent drug, Omeprazole
10.
♣) Cpm, the plasma concentration of metabolite of Omeprazole.
Find the value of: ♦) dR/dT = slope of the Pharmacological response vs. time profile (% / hr) 12. ♦ ) dR/d(lnC) = slope of the Pharmacological response vs concentration profile (%) 11.
13.
♦) d(lnC)/dt = slope of the concentration vs time profile (hr-1 )
14.
♥♠ ) kr = urinary excretion rate constant Omeprazole (hr-1 )
15.
♥♠) kf = fecal excretion rate constant Omeprazole (hr-1 )
16.
♥♠) km = rate of metabolism of Omeprazole in the body (hr-1 )
17.
♥♠) K1 = elimination rate constant of Omeprazole in the body; the summation of all of the ways that it is removed from the body. (hr-1 )
18.
♥♠) T1/2 of Omeprazole in the body (hr)
19.
♥♠) X0 = Mass of Omeprazole in the body at time 0 (mg)
20.
♥♠) Xinf = Mass of Omeprazole in the body at time infinite time (mg)
21.
♥♠) Xuinf = Mass of unchanged Omeprazole in the urine at infinite time (mg)
22.
♥♠) Xfinf = Mass of unchanged Omeprazole in the feces at infinite time (mg)
♥♠) V = Volume of distribution of Omeprazole (L) 24. ♥♠) Cp0 = Concentration of Omeprazole in the body at time 0. (mic/L) 23.
25.
♥♠) AUCiv = Area under the plasma concentration of Omeprazole vs time curve of the IV dose (mic/L*hr)
♥♠) First trapazoid of the AUCiv (mic/L*hr) 27. ♥♠) Last trapazoid of the AUCiv (mic/L*hr) 26.
28.
♥♠) AUMCiv = Area under the first moment of the plasma concentration of Omeprazole vs time curve of IV dose (mic/L*hr2 )
29.
♥♠) MRTiv = Mean Residence time of Omeprazole given as the IV dose (hr)
30.
♥♠) Xm = Mass of metabolite in the body at time = 0 (mg)
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Practice Exams: Section 1
31.
♥♠) Xminf = Mass of the metabolite in the body at infinite time. (mg)
32.
♥♠) Vm = Volume of distribution of the metabolite in the body (L)
33.
♥♠) Cpm = Concentration of metabolite in the body at time = 0 (mic/L)
34.
♥♠) AUCmet = Area Under the Curve of the metabolite excretion rate vs time profile (mg)
35.
♥♠) Xmu0 = Mass of metabolite in urine at time = 0 (mg)
36.
♥♠) Xmuinf = Mass of metabolite in urine at time = infinity (mg)
37.
♥♠) Xmf0 = Mass of metabolite in feces at time = 0 (mg)
38.
♥♠) Xmfinf = Mass of metabolite in feces at time = infinity (mg)
39.
♥♠) Kmu = Rate constant of excretion of metabolite into urine (hr-1 )
40.
♥♠) Kmf = Rate constant of excretion of metabolite into feces (hr-1 ) TABLE 14-4 Omeprazole
Minus
Small Med
Large
A
-0.8
0
1
100
B
-1
0.1
2
147
C
-2
0.2
4
210
D
-4
0.3
8
223
E
-8
0.4
10
372
F
-10
0.5
16
444
Answer Pool
Equations None
X0 e
km X0 –K 2 t – K1 t --------------------------------- (e –e ) ( K 1 – K 2 )V dm
– K1 t
km X0 –K 1 t – K2 t --------------------------------- (e –e ) ( K 1 – K 2 )V dm k mu X 0 –K 2 t – K1 t --------------------------------- (e –e ) ( K 1 – K 2 )V dm
X 0 –K 1 t ------ e Vd –K 1 t
k mu k m X 0 – K 2 t – K1 t ----------------------(e –e ) ( K1 – K 2 )
– K1 t
k mu k m X 0 – K 1 t – K2 t ----------------------(e –e ) ( K1 – K 2 )
ku X0 e
kf X0 e
ku –K1 t -----X (1 – e ) K 0 1
G
-16
0.6
20
556
More Equations
k mu X 0 –K 2 t – K1 t ----------------------(e –e ) ( K1 – K 2 )
kf –K1 t -----) K X0 ( 1 – e
k mf k m X 0 1 – e – K2 t 1 – e – K 1 t ---------------------- --------------------- – --------------------- ( K1 – K 2 ) K 2 K 1
1
H
-20
0.7
25
643
k m X0 –K 2 t – K1 t ----------------------(e –e ) ( K1 – K 2 )
k mf k m X 0 1 – e – K1 t 1 – e – K 2 t ---------------------- --------------------- – --------------------- ( K1 – K 2 ) K 1 K 2
I
-25
0.8
36
756
k m X0 –K 1 t – K2 t ----------------------(e –e ) ( K1 – K 2 )
k mu k m X 0 1 – e – K2 t 1 – e – K 1 t ---------------------- --------------------- – --------------------- ( K1 – K 2 ) K 2 K 1
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14-51
Practice Exams: Section 1
Minus
Small Med
Large
J
-36
0.9
893
A
Missing Information. Can not answer question with information available
44
Equations k mu X 0 –K2 t – K1 t ----------------------(e –e ) ( K1 – K 2 )
More Equations k mu k m X 0 1 – e – K1 t 1 – e – K 2 t ---------------------- --------------------- – --------------------- ( K1 – K 2 ) K 1 K 2
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14-52
Practice Exams: Section 1
14.9.3
OMEPRAZOLE SOLUTIONS
kf Xf
X km Xm
kmu
80
80
70
70
60
-1
0
60
Response
Response
Xmu
50
40
50
40
30
30
20
20 320
340
360
380
400
420
440
460
Concentration
10
Time (min)
10
10
Concentration
Concentration
10
10
10
10 0
2
4
6
8
10
10
12
0
Time (hr)
5
10
15
20
Time (hr)
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14-53
Practice Exams: Section 1
– K1 t
1. (B) X = X0 e 2. (A) X u = 0 See model. 3. (A)
dX u --------- = 0 dt
4. (G)
k –K 1 t Xf = ------f X 0 ( 1 – e ) K 1
5. (H)
km X0 –K 2 t – K1 t X m = ----------------------(e –e ) ( K1 – K 2 )
See model.
k mu k m X 0 1 – e – K2 t 1 – e – K1 t X mu = ---------------------- --------------------- – --------------------- ( K 1 – K2 ) K2 K1
6. (I)
7. (D)
k mu k m X 0 – K2 t – K 1 t dX mu ------------- = ----------------------(e –e ) dt ( K 1 – K2 )
8. (A)
X mf = 0
9. (C)
X –K 1 t C p = ------0- e Vd
10.(A)
km X 0 – K2 t –K 1 t C pm = ---------------------------------(e –e ) ( K 1 – K 2 )V dm
11.(I)
See model.
dR – 0.4% – 25% ------- ≅ ------------------ ≅ -------------dt minute hr
Estimate from graph. Between 20-80% of maximal
response ONLY! 12.(H)
dR 74%-57% 17% --------------- = ---------------------------------------------- = ------------- ≅ 25% d ln C p ng ng 0.693 ln 1.0 ------ – ln 0.5 -----ml ml
Points taken from sample graph.
13.(B)
d ln C p –1 dR d ln C – 25% 1 --------------- = ------- × ---------------p = -------------- × ----------- = – 1 hr dt dt dR hr 25%
14.(A)
kr = 0
15.(C)
–1 Xf K 1 –1 4mg ⋅ 1hr - = ----------------------------- = 0.2hr k f = ------------X0 20mg
See model. ∞
0.2hr-1 16.(I) k m =
K 1 – k f = 1hr
–1
– 0.2hr
–1
Also,
= 0.8hr
–1
kf
is 20% of
if you have
K1 ,
or 20% of 1hr-1 =
k f .or
∞
–1 X mu K 1 ⋅ 1hr = 0.8hr – 1 k m = ---------------= 16mg -------------------------------X0 20mg
. Also,
km
is 80% of
K1 ,
or 80% of 1hr-1 =
0.8hr-1 17.(A)
– d ln C –1 K 1 = ------------------p = 1hr dt
18.(H)
t 1 ⁄ 2 = 0.693 ------------- = 0.693 → 0.7hr K1
19.(G) X 0 = 20mg Read introduction. 20.(A) X inf = 0 See model. 21.(A) Xu ( inf ) = 0 Basic Pharmacokinetics
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Practice Exams: Section 1
20 X f ( inf ) = --------- × 20mg = 4mg 100
22.(C) 23.(I)
20% of parent excreted in feces (read introdction).
Dose- ≈ ----------------20mg ≈ 20 , 000µg- ≈ 36L V d = ----------------------------------C p0 ng µg 556 -------556 -----mL L
24.(G)
ng- ≈ 556 µg -----C p0 ≈ 556 ------mL L
25.(G)
AUC IV =
∑
From graph.
C p ( last ) p ( n ) + C p ( n + 1 ) C --------------------------------------∆t + ----------------- K1 2
or
C p0 AUC IV = -------K1
556 + 336 336 + 205 205 + 75 75 + 28 AUC IV = ------------------------ × 0.5 + ------------------------ × 0.5 + --------------------- × 1 + ------------------ × 1 + 2 2 2 2
or
10 µg ⋅ hr + 10 28 ------------------ × 1 + ------ = 578.75 → 556 ---------------- 2 1 L ng 556 -------⋅ hr mL = 556 µ AUC IV = ----------------------------L 1hr
26.(D)
+ 336- × 0.5 = 223 µg ⋅ hr 556 ------------------------------------- 2 L
27.(E)
10 µg ⋅ hr ------ ----------------1 L
28.(G)
AUMC IV =
∑
tn Cp (n ) + tn + 1 Cp (n + 1 ) C p ( last ) tlast C p ( last ) ------------------------------------------------------- ∆t + -----------------+ --------------------------- 2 2 K1 ( K1 )
( 556 ) + 0.5 ( 336 -) × 0.5 + 0.5 ( 336 ) + 1 ( 205 ) × 0.5 + 1------------------------------------( 205 ) + 2 ( 75 ) × 1 + AUMC IV = 0--------------------------------------------------------------------------------------- 2 2 2 2
µg ⋅ hr ( 75 ) + 3 ( 28 ) 3 ( 28 ) + 4 ( 10 ) 10 4 ( 10 ) 2---------------------------------× 1 + ---------------------------------- × 1 + ------ + -------------- = 541.75 → 556 ------------------ 2 1 L 2 2 1 2
⋅ hr 556 µg ------------------AUMC L = ------------------ ≈ ----------------------------- ≈ 1hr ⋅ hrAUC 556 µg ---------------L
29.(A)
MRT IV
30.(A) 31.(A) 32.(A) 33.(A)
Xm 0 = 0
34.(F)
AUC met =
or
1 = ------------1 - = 1hr MRT IV = -----–1 K1 1hr
See model. X m ( inf ) = 0 See model. V dm = Can not determine from information given. Need C pm0 = 0 See model.
∑ Xmu
80 = --------- × 20mg = 16mg . 100
C pm
information.
AUC is equal to all drug which goes
through compartment, which is 80% of the 20 mg dose, or 16 mg (read introduction). 35.(A) X mu0 = 0 See model. 36.(F)
80- × 20mg = 16mg X mu ( inf ) = -------100
80% of the 20 mg dose is found as metabolite in
urine (Read introduction). 37.(A) X mf0 = 0 See model. 38.(A) X mf ( inf) = 0 See model. Basic Pharmacokinetics
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14-55
Practice Exams: Section 1
39.(C) k mu = K 2 = 0.2hr –1 = is which is K1 . 40.(A) K m f = 0 See model.
k small
from urine data graph because
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k l arg e ≅ 1hr
–1
14-56
Practice Exams: Section 1
14.10 EXP3312, an Experimental Drug Wong N, Wong P. (J Pharm Pharmacol 1996; 48: 492-497) investigated the pharmacological and pharmacokinetic response of EXP3312
14.10.1 EXP3312 DATA EXP3312 is a non-peptide angiotension II AT1-receptor antagonist which has potential as an oral hypertensive drug. EXP3312 a pro-drug is metabolized to an active metabolite M1. Experiments with rats show M1 is the active drug. As with all angiotension II antagonists EXP3312 blocks the renin-angiotension system. EXP3312 is a synthesized analog of Losartin, but EXP3312 has greater oral hypotensive potency and has promise in future treatment of human renal hypertension. EXP3312 is still an experimental drug without all pharmacological responses having been explored. However, we may assume the drug is completely metabolized to the active metabolite (M1). Thereafter, the active metabolite (M1) is excreted unchanged in the urine. The following pharmalogical data was obtained using pithed (spinal tap) rats with an average weight of 350 g. The concentration of drug in each rat is based on average weights of the rats. The measured responses of decrease in mean arterial pressure (MAP decrease) are shown below.The researchers supplied graphical data of mean arterial pressure in conscious renal hypertensive rats. M1 metabolite was measured and the results are also shown in the table below. TABLE 14-5
Response (mm Hg)
M1 administration (mg/kg)
Response (mm Hg)
(MAP) decrease
I.V. bolus
(MAP) decrease
25
0.01
60
0
36.5
0.03
57.4
1
48.5
0.1
54.7
2
59
0.3
49.3
4
44.5
6
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Time (hr)
14-57
Practice Exams: Section 1
The parent compound EXP3312 and metabolite M1 levels were measured in the plasma over time. The data obtained from the active M1 metabolite and the inactive partent drug EXP3312 are shown below. The standard dose for each rat was 0.3mg/kg. TABLE 14-6
time (hr)
EXP3312
M1
Cp (mic/mL)
Cp (mic/mL)
0.15
0.32
0.3
0.59
0.5
0.4
1
0.26
1.34
2
0.11
1.61
3
0.045
4
1.26
6
0.81
10
0.3
12
0.18 I = 3.81 mic/mL
Use the following data was obtained when an intravenous infusion of a different dose was given to an average rat over 45 minutes.
TABLE 14-7
time (hr)
Plasma concentration (mic/mL)
0.25
0.139
0.5
0.26
0.75
0.35
1
0.29
2
0.12
3
0.05
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14-58
Practice Exams: Section 1
14.10.2 EXP3312 & M1 QUESTIONS: Using the information given find the pharmacokinetic parameters (CHECK YOUR UNITS!!!!!!): 1.
dR/dt of M1 active metabolite (mmHg/hr)
2.
dR/d(lnCp) active metabolite (mmHg)
3.
d(lnCp)/dt of M1active metabolite (hr-1)
4.
The slope of Cp vs: t of EXP3312 on semi-log paper (hr-1)
5.
The slope of terminal portion of dXmu/dt vs: t (mid) on semi-log paper (hr-1)
6.
The slope of stripped portion of Cpm vs: t or (Concentration of M1 vs: t) on semi-log paper (hr-1)
7.
The equation for X in the body.
8.
The equation for Xm or M1 in the body.
9.
K1 (elimination rate constant of EXP3312) (hr-1)
10.
ku (urinary excretion rate constant of EXP3312) (hr-1)
11.
km (metabolism rate constant of M1 (hr-1)
12.
kmu (urinary excretion rate constant of M1 (hr-1)
13.
Cp0 (Initial IV bolus concentration of EXP3312) (mic/ml)
14.
Vd of EXP3312 (L)
15.
Vdm of M1 (L)
16.
AUC (Cp vs: t for EXP3312 by IV bolus)
17.
AUMC (t*Cp vs: t for EXP3312 by IV bolus) (mic/mL* hr2))
18.
MRT for EXP3312 (hr)
19.
AUC (dXu/dt vs: t (mid) for EXP3312 by IV bolus) (mg)
20.
AUC (dXmu vs: t (mid) for M1) (mg)
21.
X0 (EXP3312 in the body at time zero) (mg)
22.
Xmu 0 (M1in the urine at time zero) (mg)
23.
Xm0 (M1 in the body at time zero) (mg)
24.
X at 2hr (EXP3312 in the body at 2 hr.) (mg)
25.
Xm at 2hr (M1 in the body at 2 hr.) (mg)
26.
Xinf (EXP3312 in the body at infinite time) (mg)
27.
Xuinf (EXP3312 in the urine at infinite time) (mg)
28.
Xminf (M1in the body at infinite time) (mg)
29.
T1/2 (EXP3312) (hr)
30.
Clearance of EXP3312 (L/hr)
(mic/mL*hr))
EXP3312 Infusion Questions:
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14-59
Practice Exams: Section 1
31.
K (elimination rate constant of EXP3312) (hr-1)
32.
T1/2 (EXP3312) (hr)
33.
Volume of distribution (L/kg)
34.
Clearance (L/hr/kg)
35.
How long will it take to reach 95% steady state? (hr)
36.
Q (infusion rate) (mg/hr)
37.
Find the plasma concentration if the infusion in Table 14-7 on page 58 was discontinued at 5 hours instead of 45 minutes. (mic/mL)
38.
Find the time need for the concentration to drop to 0.23 mic/mL after the infusion in question 37 is discontinued (hours).
39.
Determine the new infusion rate necessary to maintain a plasma concentration of Cpss 1.0 mic/ mL of the average rat. (mg/hr)
40.
Find a loading dose for the average rat which would give the Cpss of 1.0 mic/mL immediately (mg)
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14-60
Practice Exams: Section 1
TABLE 14-8 Answer
Pool
Minus ones
Small ones
Big ones
Bigger ones
A
-0.121
0
0.55
1.05
B
-0.26
0.019
0.62
1.15
C
-0.391
0.039
0.72
1.32
D
-0.567
0.063
0.80
1.69
E
-0.621
0.105
0.83
2.63
F
-0.866
0.146
0.866
3.5
Equations None X0 e
–K 1 t
X0 –K1 t ------ e Vd
-1.112
0.169
0.94
4.17
km X0 – K1 t –K 2 t ---------------------------------(e –e ) ( K 1 – K 2 )Vd m k mu X 0 – K2 t –K 1 t --------------------------------- (e –e ) ( K 1 – K 2 )V dm k mu k m X 0 –K 2 t – K1 t ----------------------(e –e ) ( K 1 – K2 )
– K1 t
k mu k m X 0 –K 1 t – K2 t ----------------------(e –e ) ( K 1 – K2 )
ku – K1 t -----X (1 – e ) K 0 1
G
km X0 – K2 t –K 1 t ---------------------------------(e –e ) ( K 1 – K 2 )Vd m
–K 1 t
kuX 0 e
kf X0 e
Equations
k mu X 0 – K2 t – K1 t ----------------------(e –e ) ( K 1 – K2 )
kf – K1 t -----X (1 – e ) K 0
k mf k m X 0 1 – e –K 2 t 1 – e –K 1 t ---------------------- --------------------- – --------------------- ( K1 – K 2 ) K 2 K1
1
H
-1.981
0.26
0.95
4.83
km X0 –K2 t –K 1 t ----------------------(e –e ) ( K1 – K 2 )
k mf k m X 0 1 – e –K 1 t 1 – e –K 2 t ---------------------- --------------------- – --------------------- ( K1 – K 2 ) K 1 K2
I
-2.63
0.42
0.96
7.2
km X0 –K1 t –K 2 t ----------------------(e –e ) ( K1 – K 2 )
k mu k m X 0 1 – e –K 2 t 1 – e –K 1 t ---------------------- --------------------- – --------------------- ( K1 – K 2 ) K 2 K1
J
-3.71
0.48
0.99
10.1
k mu X 0 – K2 t –K 1 t ----------------------(e –e ) ( K1 – K2 )
k mu k m X 0 1 – e –K 1 t 1 – e –K 2 t ---------------------- --------------------- – --------------------- ( K1 – K 2 ) K 1 K2
This exam is given early at the request of those students who had conflict with the time agreed upon. They signed a document stating that they would not share their experience with anyone under pain of prosecution for adademic misconduct which will result minimally in an “F” for the course. You also must sign this document. I ___________________________________________ did not comunicate in anyway anything at all regarding this exam nor give or receive any assistance during the exam upon my honor as a health professional and under pain of prosecution for adademic misconduct. Signed__________________________________________________________Date__________________
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14-61
Practice Exams: Section 1
14.11 Graph Paper
0
1
5
0 Basic Pharmacokinetics
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14-62
CHAPTER 15
Practice Exams: Exam 2
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15-1
Practice Exams: Exam 2
15.1 Nifedipine: Exam 2 Nifedipine (Procardia @) is a calcium channel blocker which specifically inhibits potential-dependent channels not receptor-operated channels, preventing calcium influx of cardiac and vascular smooth muscle (coronary, cerebral). Calcium channel blockers reduce myocardial contractility and A-V node conduction by reducing the slow inward calcium current. They are indicated in angina, cardiac dysrhythmias, and hypertension among others. Nifedipine appears to be metabolized entirely into an inactive metabolite, an acid and subsequently further metabolized to a lactone. Both the acid and the lactone are excreted into the urine and the feces.
Hepatic blood flow in normals is 1.6 L/min Renal blood flow in normals is 1.2 L/min
Echizen and Eichelbaum (Clin Pkin 1986; 11:425-49) and Kleinbloesem et al (Clin Pcol Therap 1986; 40: 21-8) reviewed the pharmacokinetics of Nifedipine. While the drug is not routinely given by IV bolus and does not strictly conform to a one compartment model, let's treat the data as if those problems can be ignored. The following data is offered for evaluation: TABLE 1-1. Nifedipine
Time (hr)
Cp
(mcg/L)
Cm1 (mcg/L)
0.5
24.7
1
44.4 71.8
IV Bolus Profile
Xm1f (mg)
Xm1u (mg)
.14
.59
Xm2f (mg)
Xm2u (mg)
2
139
4
65.6
96.5
.44
1.83
.028
.11
6
31.1
100
.77
3.25
.073
.29
8
14.6
94.7
1.1
4.65
.135
.54
12
76.5
1.69
7.10
.291
1.15
24
34
2.77
11.63
.75
2.95
3.6
15.1
1.3
5.0
7 days
TABLE 1-2 Nifedipine
Oral information
Brand
Generic
Route
IV
Oral Capsule
Oral Capsule
Dose (mg)
25
10
10
AUC (ug/L*hr)
785
236
204
AUMC (ug/L*hr2 )
2093
866
816
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15-2
Practice Exams: Exam 2 : Nifedipine: Exam 2
15.1.1
NIFEDIPINE QUESTIONS: Find:
21) MRT (oral generic capsule) (hr)
1) MRT iv (hr)
22) MAT (oral generic capsule) (hr)
2) Ke (elimination rate constant) for Nifedipine (hr-1 )
23) Ka, the apparent absorbtion rate constant, for the generic capsule (hr-1 )
3) T 1/2 for Nifedipine
24) Peak time for the generic capsule (hr)
4) Cp0 for iv dose (ug/L) 5) Vd for Nifedipine (L)
25) Cpmax , the maximum concentration of the generic oral capsule give as a single dose (ug/L)
7) Cp of Nifedipine at one hour after the IV dose
26) Comparative bioavailability of the oral capsules
8) AUC from 0 to one hour for the IV dose 9) Total Body Clearance of Nifedipine (L/ hr)
Your patient is controlled by 20 mg TID of the brand name oral capsule when he is healthy.
10) Renal Clearance of Nifedipine (L/hr)
27) What is his N for that dosing regimen?
11) Hepatic Clearance of Nifedipine (L/hr)
28) Cpssmax for this patient at this dosing regimen (ug/L)
12) Renal Extraction Ratio 13) Hepatic Extraction Ratio 14) Absolute bioavailability for the brand name capsule 15) MRT (oral brand name capsule) (hr) 16) MAT (oral brand name capsule) (hr) 17) Ka, the apparent absorbtion rate constant, for the brand name capsule (hr-1 ) 18) Peak time for the brand name capsule (hr) 19) Cpmax , the maximum concentration of the brand name oral capsule give as a single dose (ug/L)
29) Cpssavg for this patient at this dosing regimen (ug/L) 30) Cpssmin for this patient at this dosing regimen (ug/L) You want to maintain his plasma concentrations between 110% of Cpssmax and 90% of Cpssmin . How would you change the dosage regimen to if your patient suffered from: 31) stenosis of the kidney (Fr = 0.67). 32) renal failure (Fi = 0.67). 33) stenosis of the liver (Fr = 0.67) 34) cirrhosis of the liver (Fi = 0.67)
20) Absolute bioavailability for the generic capsule
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15-3
Practice Exams: Exam 2 : Nifedipine: Exam 2
TABLE 1-3 Nifedipine
Small
Answer Pool
Medium
Large
Dosing changes:
a) 0.00
a) 1.85
a) 59
a) 10 mg once daily
b) 0.05
b) 2.67
b) 85
b) 10 mg BID
c) 0.33
c) 3.67
c) 92.5
c) 10 mg TID
d) 0.375
d) 4.0
d) 101
d) 10 mg QID
e) 0.65
e) 4.32
e) 124
e) 20 mg once daily
f) 0.75
f) 9.3
f) 147
f) 20 mg BID
g) 0.85
g) 18.5
g) 185
g) 20 mg TID (no change necessary)
h) 1.0
h) 32
h) 202
h) 20 mg QID
i) 1.33
i) 38
i) 248
i) 30 mg once daily
j) 1.57
j) 49
j) 294
j) 30 mg BID
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15-4
Practice Exams: Exam 2 : Nifedipine: Exam 2
15.1.2
NIFEDIPINE SOLUTIONS 1.
21.
MRT iv = 2.67hr k e = 0.375hr
3.
T1 ⁄ 2 = 1.85hr
4.
Cp 0 = 294ug ⁄ L
5.
Vd = 85L
generic
capsule
generic
capsule
generic
capsule
MRT = 4hr
–1
2.
Oral
22.
Oral
MAT = 1.33hr 23.
Oral
Ka = 0.75hr 24.
–1
Oral generic capsule peak time
6.
skip (error in numbering)
1.85hr
7.
Cp 1 = 202ug ⁄ L
25.
8.
AUC 0 – 1 = 248ug ⁄ L ⋅ hr
9.
Cl s = 32L ⁄ hr
10.
Cl r = 0
Peak time = 1.18 (ok)
11.
Cl h = 32L ⁄ hr
Cpmax = 0.78 (not ok!)
12.
Er = 0
27.
N = 4.32
13.
Eh = 0.33
28.
Cp
14. Absolute bioavailability for brand name capsule = 0.75
29.
Cp
15.
30.
Cp
31.
No change
32.
No change
33.
No change
34.
No change
Oral
generic
capsule
Cp max = 38ug ⁄ L 26. Comparative bioavailability of the oral capsules B/A: AUC = 0.86 (ok)
Oral
brand
name
capsule
MRT = 3.67hr 16.
Oral
brand
name
capsule
brand
name
capsule
MAT = 1.0hr 17.
Oral
Ka = 1.0hr 18.
Oral
ss ss ss
max =
185ug ⁄ L
ave =
59ug ⁄ L
min =
9.3ug ⁄ L
–1
brand
name
peak
time
1.57hr 19.
Oral
brand
name
Cp max = 49ug ⁄ L 20. Absolute generic = 0.65
bioavailability
for
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15-5
Practice Exams: Exam 2 : Valproate: Exam 2
15.2 Valproate: Exam 2 All questions are internally consistent. Information gained in any one can be used in all others. Note, however, there is a change in patient status midway through. Valproate is a carboxylic acid anticonvulsant. Its activity may be related, at least in part, to increase concentrations of the neurotransmitter inhibitor gamma aminobutyric acid in the brain. It is used alone or in combination with other anticonvulsants. in the prophylactic management of petit mal. It appears to be almost entirely cleared by liver function with negligible amounts excreted into the urine unchanged. It comes as soft gelatin capsules of 250 mg and enteric coated tablets 250 and 500 mg. The therapeutic range for valproate appears to be between 20 and 100 mic/ml. The volume of distribution is 0.19 L/Kg and your patient is 70 kg.
TABLE 1-4
IV bolus Dose (mg) AUC (mg/L*hr) 2
AUMC(mg/L*hr )
Generic
250
250
594
273
253
9428.6
4605.5
4184.6
2.95
Tpeak (hr) Cpmax Bioavailability (f)
Brand
500
14.35
1.0
.92
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15-6
Practice Exams: Exam 2 : Valproate: Exam 2
15.2.1
VALPROATE QUESTIONS
1) Calculate the MRTiv a) 11
of Valproate (hr).
b) 15.87 c) 16.54
d) 16.87 e) 17.08
2) Calculate the rate constant of elimination (hr-1 ) in normals ? a) 0.091
b) 0.063
c) 0.060
d) 0.0593
e) 0.0585
3) Calculate the half life of Valproate (hr)? a) 7.6
b) 11 c) 11.5
d) 11.7 e) 11.8
4) Calculate the hepatic clearance of valproate (L/hr). a) 0.0084
b) 0.063
c) 0.84
d) 1.25
d) 2.5
5) Calculate the hepatic extraction ratio of valproate. a) 0.0063
b) 0.0084
c) 0.063
d) 0.084
e) 0.84
6) What is the maximum N for multiple dosing of Valproate? a) 0.78
b) 1.65
c) 2.12
d) 2.32
e) 2.5
7) What is the maximum acceptable dosing interval for normal patients (hr)? a) 8
b) 12
c) 18
d) 24
e) 25.5
8) What is the N if we are going to dose TID? a) 0.63
b) 0.727
c) 0.842
d) 1.25 e) 1.42
9) If you dosed this patient 500 mg BID with the brand name product, what would be your maximum concentration at steady state (mg/L)? a) 87.3
b) 70.9
c) 65.2
d) 52.8
e) 45.8
10) What would be your minimum concentration at steady state (mg/L)? a) 52.8
b) 32.7
c) 30.1 d) 27.1
e) 22.9
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15-7
Practice Exams: Exam 2 : Valproate: Exam 2
11) What would be your average concentration at steady state (mg/L)? a) 63.8
b) 49.8
c) 47.7
d) 45.8
e) 34.4
12) What loading dose would you give to get to Cpss right away (mg)? a) 250
b) 500
c) 750
d) 1000
e) 1500
13) If you changed the dosage regimen to 250 mg QID, what would happen to the Cpssmax, Cpssave, Cpssmin? a) Cpssmax - up, Cpssave - up, Cpssmin - up b) Cpssmax - down, Cpssave - down, Cpssmin - down c) Cpssmax - same, Cpssave - same, Cpssmin - same d) Cpssmax - up, Cpssave - same, Cpssmin - down e) Cpssmax - down, Cpssave - same, Cpssmin - up
14) Calculate f, the absolute bioavailability of the generic product. a) 0.67
b) 0.75
c) 0.85
d) 0.93e) 1.0
15) Calculate the comparative bioavailability of the generic product. a) 0.67
b) 0.75
c) 0.85
16) Calculate the MRToral a) 11
b) 15.87 c) 16.54
d) 0.93e) 1.0
of the generic product.
d) 16.87 e) 17.08
17) Calculate the MAT of the generic product. a) 0.67
b) 0.75
c) 0.85
d) 0.93e) 1.0
18) Calculate the Ka of the generic product. a) 1.5
b) 1.33
c) 1.18
d) 1.08
e) 1.0
19) Calculate the peak time of the generic product. a) 2.21
b) 2.45
c) 2.59
20) Calculate the Cpmax
d) 2.76
e) 2.95
of the generic product.
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15-8
Practice Exams: Exam 2 : Valproate: Exam 2
a) 8.1
b) 9.3
c) 10.7
d) 12.6
e) 14.4
21) Is the generic product bioequivalent? a) Yes b) No, because the comparative bioavailability is outside the federal guidelines. c) No, because the ratio of the peak times is outside the federal guidlines. d) No, because the ratio of the Cpmax s is outside the federal guidlines. e) No, because the generic fails more than one of the required comparisons.
In patients who are also currently on phenobarbital their intrinsic clearance of valproate increases by 50% as the phenobarbital induces the enzymes which metabolize valproate. Further questions refer to this condition.
22) Calculate his new clearance. a) 0.0084
b) 0.063
c) 0.84
d) 1.25
d) 2.5
d) 0.059
e) 0.042
23) Calculate his new K. a) 0.095
b) 0.084
c) 0.063
24) Calculate his new maximum acceptable dosing interval (hr). a) 8
b) 12
c) 17
d) 18
e) 24
25) What dosage regimen would you recommend to try to maintain his plasma concentrations within 110% of the maximum and 90% of the minimum concentrations attained when he was normal? a) 750 mg BID
b) 500 mg TID
c) 750 mg TID
d) 250 mg QID
e) 500 mg QID
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15-9
Practice Exams: Exam 2 : Valproate: Exam 2
15.2.2
VALPROATE SOLUTIONS
Basic Pharmacokinetics
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15-10
Practice Exams: Exam 2 : Methyl phenidate
15.3 Methyl phenidate Methyl phenidate (MP) (Ritalin@) is an effective stimulant in the treatment of narcolepsy in adults and attention deficit syndrome in children. It is entirely metabolized to the inactive metabolite, Ritalinic Acid (RA), by the liver which is subsequently excreted unchanged into the urine. The following information was obtained from a 70 Kg male. QH = 24 mL/min/Kg QR = 19 mL/min/Kg TABLE 1-5 Ritalin Data
IV
Brand Name
Generic
Dose (mg)
10
20
20
AUC (ug/ml*hr)
0.20
0.04
0.035
AUMC (ug/ml*hr2
0.32
0.14
0.1225
TABLE 1-6 Ritalin
Answer Pool
Itty- Bitty
Tiny
Puny
Small
Medium
Large
a
0
0.016
0.1
1.1
10
100
yes
b
0.00016
0.0251
0.125
1.6
20
125
no, Ratio of Tps
c
0.0005
0.035
0.25
1.75
30
225
no, Ratio of Cmaxs
d
0.00074
0.042
0.35
1.9
40
375
no, Ratio of AUCs
e
0.00084
0.05
0.42
2.6
50
435
no, Ratio of AUMCs
f
0.0016
0.067
0.53
3.5
60
550
no, more than one criterion
g
0.005
0.075
0.625
4.2
70
675
20 mg TID
h
0.0074
0.0875
0.727
5.0
80
750
20 mg BID
i
0.0084
0.091
0.875
7.27
90
875
20 mg QD
j
0.0090
0.096
0.91
8.75
96
995
20 mg QID
Basic Pharmacokinetics
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Words
15-11
Practice Exams: Exam 2 : Methyl phenidate
15.3.1
METHYL PHENIDATE QUESTIONS: 1) MRT iv (hr) 2) Ke (elimination rate constant) for Ritalin (hr-1 )
25) Cpmax , the maximum concentration of the generic oral tablet give as a single dose (ug/mL) 26) Comparative bioavailability of the oral tablets
3) T 1/2 for Ritalin (hr) 4) Cp0 for iv dose (ug/mL) 5) Vd for Ritalin (L) 6) Cp of Ritalin at one hour after the IV dose 7) AUC from 0 to one hour for the IV dose 8) Total Body Clearance of Ritalin (L/hr)
27) Are the tablets bioequivalent? Your patient is controlled by 20 mg TID of the brand name oral tablet when he is healthy. For This patient and this dosage regimen, what is his: 28) N ?
9) Renal Clearance of Ritalin (L/hr)
29) Cpssmax (ug/mL)
10) Hepatic Clearance of Ritalin (L/hr)
30) Cpssavg (ug/mL)
11) Intrinsic Hepatic Clearance of Ritalin (L/hr)
31) Cpssmin (ug/mL)
12) Renal Extraction Ratio 13) Hepatic Extraction Ratio 14) Absolute bioavailability for the brand name tablet 15) MRT (oral brand name tablet) (hr) 16) MAT (oral brand name tablet) (hr) 17) Ka, the apparent absorbtion rate constant, for the brand name tablet (hr-1 ) 18) Peak time for the brand name tablet (hr)
You want to maintain his plasma concentrations between 110% of Cpssmax and 90% of Cpssmin . How would you change the dosage regimen to if your patient suffered from: (in no case was there a change in Vd) 32) stenosis of the kidney (Fr = 0.67)? 33) renal failure (Fi = 0.34)? 34) stenosis of the liver (Fr = 0.67)? 35) cirrhosis of the liver. (Fi = 0.67)? 36) treatment with phenobarbital (Fi = 1.33)?
19) Cpmax , the maximum concentration of the brand name oral tablet give as a single dose (ug/mL) 20) Absolute bioavailability for the generic tablet 21) MRT (oral generic tablet) (hr) 22) MAT (oral generic tablet) (hr) 23) Ka, the apparent absorbtion rate constant, for the generic tablet (hr-1 ) 24) Peak time for the generic tablet (hr)
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15-12
Practice Exams: Exam 2 : Methyl phenidate
15.3.2
METHYL PHENIDATE SOLUTIONS: 1.
19.
MRT iv = 1.6hr
2.
k = 0.625hr
3.
Brand
name
-
Cp max = 0.0084ug ⁄ mL
–1
20.
Generic - f = 0.0875
T1 ⁄ 2 = 1.1hr
21.
Generic - MRT = 3.5hr
4.
Cp 0 = 0.125ug ⁄ mL
22.
Generic - MAT = 1.9hr
5.
Vd = 80L
23.
Generic - Ka = 0.53hr
6.
Cp 1 = 0.067ug ⁄ mL
24.
Generic - T peak = 1.75hr
25.
Generic
7.
AUC 0 – 1 = 0.096ug ⁄ mL ⋅ hr Cl s = 50L ⁄ hr
9.
Cl r = 0
10.
Cl h = 50L ⁄ hr
11.
Cl int = 100L ⁄ hr
28.
N = 7.27
12.
Er = 0
29.
Cp
13.
Eh = 0.53
30.
Cp
31.
Cp
32.
20 mg TID
33.
20 mg TID
34.
20 mg BID
35.
20 mg TID
36.
20 mg QID
15.
-
Cp max = 0.0074ug ⁄ mL
8.
14.
–1
26. 0.875
Comparative
bioavailability
=
27. Yes, the tablets are bioequivalent all parameters are within federal guidelines.
Brand name - f = 0.1 Brand name - MRT = 3.5hr
16.
Brand name - MAT = 1.9hr
17.
Brand name - Ka = 0.526hr
18.
Brand name - Tpeak = 1.74hr
–1
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ss ss ss
max =
0.0251ug ⁄ mL
avg =
0.005ug ⁄ mL
min =
0.00016ug ⁄ mL
15-13
Practice Exams: Exam 2
Basic Pharmacokinetics
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15-14
Practice Exams: Exam 2 : Verapamil
15.4 Verapamil Verapamil is a calcium channel blocker with vasodilatory and antiarrhythmic effects. It is about 95% metabolized by the liver with the metabolites showing up in the urine and feces. Hepatic blood flow in normals is 1.6 L/min Renal blood flow in normals is 1.2 L/min TABLE 1-7 Verapamil
Route
Data
IV
Brand
Generic
Oral Tablet
Oral Tablet
Dose (mg)
15
80
80
AUC (ng/mL*hr)
300
480
400
AUMC (ng/mL*hr2 )
1600
2690
2280
TABLE 1-8
Verapamil Answer Pool
Tiny
Small
Medium
Large
Dosing regimens
Bioavailability answers
a
0
1.06
26
116
40 mg qd
Yes
b
0.063
2.15
46
267
40 mf bid
No, tp ratio is not within limits
c
0.188
2.50
47.5
369
40 mg tid
No, Cpmax ratio is not within limits
d
0.250
2.70
50
533
40 mg qid
No, AUC ratio is not within limits
e
0.270
3.69
56
637
80 mg qd
No, f ratio is not within limits
f
0.300
5.33
61.5
830
80 mg bid
No, ka ratio is not within limits
g
0.370
5.60
76.5
905
80 mg tid
No, ke ratio is not within limits
h
0.693
5.70
83
970
80 mg qid
No, MRT ratio is not within limits
i
0.85
6.37
90.5
1160
160 mg qd
No, Cl ratio is not within limits
j
0.905
8.30
97
2670
160 mg bid
No, more than one of the required ratios are no within limits
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15-15
Practice Exams: Exam 2 : Verapamil
15.4.1
VERAPAMIL QUESTIONS 1) MRT iv (hr) 2) Ke (elimination rate constant) for Verapamil (hr-1 ) 3) T 1/2 for Verapamil 4) Cp0 for iv dose (ug/L) 5) Vd for Verapamil (L) 7) Cp of Verapamil at one hour after the IV dose 8) AUC from 0 to one hour for the IV dose
23) Ka, the apparent absorbtion rate constant, for the generic tablet (hr-1 ) 24) Peak time for the generic tablet (hr) 25) Cpmax , the maximum concentration of the generic oral tablet give as a single dose (ug/L) 26) Comparative bioavailability of the oral tablets 27) Are the tablets bioequivalent?
9) Total Body Clearance of Verapamil (L/ hr)
Your patient is controlled by 80 mg TID of the brand name oral tablet when he is healthy.
10) Renal Clearance of Verapamil (L/hr)
28) What is his N for that dosing regimen?
11) Hepatic Clearance of Verapamil (L/hr)
29) Cpssmax for this patient at this dosing regimen (ug/L)
12) Renal Extraction Ratio 13) Hepatic Extraction Ratio 14) Absolute bioavailability for the brand name tablet 15) MRT (oral brand name tablet) (hr) 16) MAT (oral brand name tablet) (hr) 17) Ka, the apparent absorbtion rate constant, for the brand name tablet (hr-1 ) 18) Peak time for the brand name tablet (hr) 19) Cpmax , the maximum concentration of the brand name oral tablet give as a single dose (ug/L) 20) Absolute bioavailability for the generic tablet 21) MRT (oral generic tablet) (hr) 22) MAT (oral generic tablet) (hr)
30) Cpssavg for this patient at this dosing regimen (ug/L) 31) Cpssmin for this patient at this dosing regimen (ug/L) You want to maintain his plasma concentrations between 110% of Cpssmax and 90% of Cpssmin . How would you change the dosage regimen to if your patient suffered from: 32) stenosis of the kidney (Fr = 0.67). (No change in volume of distribution.) 33) renal failure (Fi = 0.34).(Volume of distribution is reduced in this disease to 50% of normal) 34) stenosis of the liver (Fr = 0.67) (No change in volume of distribution) 35) cirrhosis of the liver (Volume of Distribution and the bioavailability are both doubled while the half life is quadrupled )
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15-16
Practice Exams: Exam 2 : Verapamil
15.4.2
VERAPAMIL SOLUTIONS 1.
17.
MRT iv = 5.33hr
Brand
Ka = 3.69hr
–1
2.
k e = 0.188hr
3.
T1 ⁄ 2 = 3.69hr
0.85hr
4.
Cp 0 = 56ug ⁄ L
19.
5.
Vd = 267L
18.
name
tablet
-
–1
Brand name tablet - peak time Brand
name
tablet
-
Cp max = 76.5ug ⁄ L
6.
skip - error in numbering
20. Generic - Absolute bioavailability - f = 0.25
7.
Cp 1 = 47ug ⁄ L
21.
Generic - MRT = 5.7hr
22.
Generic - MAT = 0.37hr
8.
AUC 0 – 1 = 50ug ⁄ L ⋅ hr
–1
9.
Cl = 50L ⁄ hr
23.
Generic - Ka = 2.70hr
10.
Cl r = 2.50L ⁄ hr
24.
Generic - T peak = 1.06hr
11.
Cl h = 47.5L ⁄ hr
25.
Generic - Cp max = 61.5ug ⁄ L
12.
Er = 0.063
26.
Comparative bioavailability - 0.85
13.
Eh = 0.49
Correction - should
be added to answer pool 14. Brand name tablet - absolute bioavailability = 0.30 15.
Brand
name
tablet
-
name
tablet
-
MRT = 5.60hr 16.
Brand
MAT = 0.270hr
27. The tablets are not bioequivalent peak time ratio is not within limits. 28.
N = 2.15
29.
Cp
ss
max =
116ug ⁄ L
ss
Cp avg = 60ug ⁄ L (change in 30. answer pool) 31.
Cp
ss
min =
26ug ⁄ L
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15-17
Practice Exams: Exam 2 : Verapamil
33.
40 mg TID
35.
40 mg BID
34. No change in dosing regimen - 80 mg TID
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15-18
Practice Exams: Exam 2 : Hydromorphone hydrochloride
15.5 Hydromorphone hydrochloride Hydromorphone hydrochloride is an analog of morphine which has about seven times the analgesic effect of morphine when given by IV. It is about 90% metabolized. The following data was obtained by Valner et al J.Clin Pcol 21(1981) 152-6: TABLE 1-9 Hydromorphone
hydrochloride Data IV
Oral tablet Brand
Oral tablet Generic
Dose (mg)
2
4
4
AUC (ng/mL*hr)
83
87
65
AUMC (ng/mL*hr2 )
291
348
292
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15-19
Practice Exams: Exam 2 : Hydromorphone hydrochloride
15.5.1
HYDROMORPHONE HYDROCHLORIDE QUESTIONS
For the IV product, find: 1.
MRTiv.
(291/83=3.5)
2.
K. (1/3.5=0.285
3.
T1/2 (0.693/.285= 2.4 hr)
4.
Cp0 (AUC*K = 83*0.285= 23.7ng/mL
5.
Ku (0.1*0.285=0.029)
6.
Km (0.9*0.285=0.256)
7.
Cls (f*D/AUC=1*2000/83=24L/hr)
8.
Vd (Cls/K=Vd= 24/0.285=84L
9.
Clh (0.9*Cls = 21.6 L/hr
10.
Clr (0.1*Cls = 2.4 L/hr
11.
Eh (21.6 / (24 mL/min * 60 min * 70 Kg / 1000mL/L) = 0.2
12.
Er ( 2.4 / ( 10 ml/min * 60 min * 70 Kg / 1000mL/L) = 0.06
27.
R(Bio)= 0.75 fail
You would like to dose your healthy patient using the brand name tablet so that his plasma concentration is in the therapeutic range of 35 to 5 ng/mL. Find: 28.
Nmax (ln(35/5)/ln(2) = 2.8
29.
Taumax (2.8*2.4=6.74 hr
30.
Maximum acceptable tau = 6
31.
N (6/2.4 =2.5)
32.
What dosage regimen would you recommend? (4mg q6h)
33.
Cpssmax? (30 ng/mL)
34.
Cpssavg? (14 ng/mL)
35.
Cpssmin? ( 5 ng/mL)
For the Brand name oral tablet, find: He gets renal stenosis (Fr = 0.5) with no change in Vd. Find:
13.
MRT (348/87=4.0)
14.
MAT(4 - 3.5 = 0.5)
15.
Ka (1/0.5 = 2.0)
16.
Tp (ln(ka/K)/(ka-K)=1.95/1.715=1.14hr
17. 18.
Absolute bioavailability, f ((87/4)/(83/ 2)=0.52 Cpmax (18 ng/mL)
For the Generic oral tablet, find: 19.
MRT ( 292/65=4.5 hr)
20.
MAT (4.5 - 3.5 = 1 hr)
21.
ka (1/MAT = 1/1=1 hr-1)
22.
Tp (ln(1/0.285)/(1-0.285)=1.26/.715=1.76 hr
23.
Absolute bioavailability, f ((65/4)/(83/ 2)=0.39
24.
Cpmax (11 ng/mL)
Would you consider the oral tablets to be bioequivalent? Why or why not? 25.
No R(Tp) = 1.54 fail
26.
R(Cp) = 0.61 fail
36.
FClr0.94
37.
Clr*2.3
38.
FClh1
39.
Clh*21.6
40.
FCls.99
41.
Cls*23.9
42.
K*0.284
43.
t1/2*2.44
44.
Nmax
45.
Taumax6.83
46.
Maximum acceptable tau
47.
N 2.46
48.
What dosage regimen would you recommend? 4mg q6h
49.
Cpssmax? 30
50.
Cpssavg? 15
51.
Cpssmin? 5.5
2.8 6
His stenosis of the kidney clears and he now goes into renal dysfunction (Fi=0.5), find:
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15-20
Practice Exams: Exam 2 : Hydromorphone hydrochloride
52.
FClr.52
77.
Taumax8
53.
Clr*1.23
78.
Maximum acceptable tau 8
54.
FClh2
79.
N6
55.
Clh*21.6
80.
56.
FCls.95
What dosage regimen would you recommend? 4 mg q6h
57.
Cls*22.8
81.
Cpssmax? 32
58.
K*0.271
82.
Cpssavg? 17
59.
t1/2*2.55
83.
Cpssmin? 7.5
60.
Nmax
61.
Taumax
62.
Maximum acceptable tau6
63.
N 2.35
64.
What dosage regimen would you recommend? 4mg q6h
84.
FClr1
85.
Clr*2.4
65.
Cpssmax? 31
86.
FClh.56
66.
Cpssavg? 15
87.
Clh*12
67.
Cpssmin? 6.1
88.
FCls.6
89.
Cls*14.4
90.
K*.171
91.
t1/2*4.04
92.
Nmax 2.8
2.8 7.13
His renal dysfunction clears and now he suffers from stenosis of the liver (Fr=0.5), Find:
His stenosis of the liver clears and he now suffers from liver dysfunction (Fi=0.05), Find:
68.
FClr1
93.
Taumax11.3
69.
Clr*2.4
94.
Maximum acceptable tau8
70.
FClh.833
95.
N 1.48
71.
Clh*18
96.
72.
FCls.85
What dosage regimen would you recommend?4 mg q8h
73.
Cls*20.4
97.
Cpssmax? 33
74.
K*.243
98.
Cpssavg? 18
75.
t1/2*2.85
99.
Cpssmin?8.4
76.
Nmax 2.8
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15-21
Practice Exams: Exam 2 : Hydromorphone hydrochloride
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15-22
Practice Exams: Exam 2
15.6 Fosinopril Sodium Fosinopril Sodium (MW 585) is a phosphinic prodrug of the angiotensin converting enzyme (ACE) inhibitor fosinoprilat (MW 563). After oral administration, fosinopril is slowly and incompletely absorbed, and is converted to the active fosinoprolat by esterases in the gastrointestinal mucosa and the liver. Unlike other ACE inhibitors, elimination of fosinoprilat is divided equally between renal and hepatic pathways. With the IV dose fosinoprilat was given and measured. With the oral dose fosinopril was given but fosinoprilat was measured. The following information was obtained from a 70 Kg male. This information was constructed from Kostis et al “Fosinopril: Pharmacokinetics and pharmacodynamics in congestive heart failure” Clin Pcol Ther 58(6) 660-5 (1995); Hui et al “Pharmacokinetics of fosinopril in patients with various degrees of renal function” Clin Pcol Ther 49(4) 457 -66 (1991) QH =
24 mL/min/Kg Blood
QR = 19 mL/min/Kg Blood TABLE 1-10 Fosinopril
Data IV
Brand Name Tablet
Generic Tablet
Dose (mg)
7.5
10
10
AUC (ng/ml*hr)
5700
1500
1400
AUMC (ng/ml*hr2)
78000
25000
23100
TABLE 1-11 Fosinopril
Answer Pool:
a
0
0.015
0.19
1.3
10
100
yes
b
0.00016
0.0251
c
0.0005
0.035
0.21
1.6
13.7
142
no, Ratio of Tps
0.25
1.75
16.6
192
no, Ratio of Cmaxs
d
0.00074
0.042
0.36
1.9
18
253
no, Ratio of AUCs
e
0.00084
0.05
0.42
2.8
50
387
no, Ratio of AUMCs
f
0.0016
0.067
0.53
3.5
67
402
no, more than one criterion
g
0.005
0.073
0.65
4.0
73
416
10 mg qid
h
0.0064
0.0875
0.72
5.6
84
750
10 mg tid
i
0.0084
0.091
0.84
7.0
93
875
10 mg bid
j
0.0090
0.096
0.93
9.5
96
995
10 mg qd
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15-23
Practice Exams: Exam 2
15.6.1
FOSINOPRIL QUESTIONS 1) MRT for fosinoprilat (hr) 2) Ke for fosinoprilat (hr-1) 3) T ? 4) Cp0
for Fosinoprilat (hr) for iv dose (ng/mL)
5) Vd for Fosinoprilat (L) 6) Cp of Fosinoprilat at one hour after the IV dose 7) AUC from 0 to one hour for the IV dose 8) Total Body Clearance of Fosinoprilat (L/hr) 9) Renal Clearance of Fosinoprilat (L/hr) 10) Hepatic Clearance of Fosinoprilat (L/hr) 11) Intrinsic Hepatic Clearance of Fosinoprilat (L/hr) 12) Renal Extraction Ratio 13) Hepatic Extraction Ratio 14) Absolute bioavailability for the brand name tablet 15) MRT (oral brand name tablet) (hr) 16) MAT (oral brand name tablet) (hr) 17) Ka, the apparent absorption rate constant, for the brand name tablet (hr-1
)
18) Peak time for the brand name tablet (hr) 19) Cpmax , the maximum concentration of the brand name oral tablet give as a single dose (ng/mL) 20) Absolute bioavailability for the generic tablet 21) MRT (oral generic tablet) (hr) 22) MAT (oral generic tablet) (hr) 23) Ka, the apparent absorption rate constant, for the generic tablet (hr-1
)
24) Peak time for the generic tablet (hr) 25) Cpmax , the maximum concentration of the generic oral tablet give as a single dose (ng/mL) 26) Comparative bioavailability of the oral tablets 27) Are the tablets bioequivalent?
Your patient is controlled by 10 mg TID of the brand name oral tablet when he is healthy. For This patient and this dosage regimen, what is his: 28) N ?
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15-24
Practice Exams: Exam 2
29) Cpssmax (ng/mL) 30) Cpssavg (ng/mL) 31) Cpssmin (ng/mL)
You want to maintain his plasma concentrations between 120% of Cpssmax Cpssmin
and 80% of
. How would you change the dosage regimen to if your patient suffered from:
(in no case was there a change in Vd) 32) stenosis of the kidney (Fr = 0.67)? 33) renal failure (Fi = 0.34)? 34) stenosis of the liver (Fr = 0.67)? 35) cirrhosis of the liver. (Fi = 0.67)? 36) treatment with phenobarbital (Fi = 1.5)?
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15-25
Practice Exams: Exam 2 : Fosinopril Sodium
15.6.2
FOSINOPRIL SODIUM SOLUTIONS 19.
Oral
brand
name
tablet
1.
MRT = 13.7hr
2.
k = 0.073hr
3.
T1 ⁄ 2 = 9.5hr
4.
Cp 0 = 416ng ⁄ mL
21.
5.
Vd = 18L
22.
6.
Cp 1 = 387ng ⁄ mL
7.
AUC 0 – 1 = 402ng ⁄ mL
23.
Generic tablet Ka = 0.36hr
8.
Cl s = 1.3L ⁄ hr
24.
Generic tablet peak time 5.6hr
Cl r = 0.65L ⁄ hr
25.
Generic
9. 10.
Cl h = 0.65L ⁄ hr
11.
Intrinsic Hepatic Clearance
Cp max = 0.073ng ⁄ mL
–1
20. Oral generic tablet absolute bioavailability is 0.19
Er = 0.015
13.
Eh = 0.0064
Oral
is
15.
30.
Cp
31.
Cp
32.
10 mg TID
33.
10 mg BID
34.
10 mg TID
35.
10 mg TID
36.
10 mg QID
name
tablet
MRT = 16.6hr name
tablet
MAT = 2.8hr 17.
Oral
Ka = 0.36hr 18.
brand
tablet
all within 20% range.
Cp
brand
–1
27. Yes! the tablets are bioequivalent The AUCs, peak times, and Cp max s are
29.
Oral
tablet
26. Comparative bioavailability of the oral tablets = 0.93
14. Oral brand name tablet absolute bioavailability is 0.21
16.
generic
Cp max = 0.067ng ⁄ mL
N = 0.84
brand
tablet
MAT = 2.8hr
28.
Oral
generic
MRT = 16.6hr
0.65L ⁄ hr 12.
Oral
name
tablet
–1
Oral brand name tablet peak time
5.6hr
Basic Pharmacokinetics
REV. 99.4.25
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ss ss ss
max =
253ng ⁄ mL
avg =
192ng ⁄ mL
min =
142ng ⁄ mL
15-26
Practice Exams: Exam 2
Basic Pharmacokinetics
REV. 99.4.25
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15-27
Practice Exams: Exam 2
15.7 Remoxipride Remoxipride (MW 296 - Unionized Base pKa 9.4) is a new antipsychotic of the benzamide type (See figure 1). The pharmacokinetics were studied by Movin-Osswald and Hammarlind-Udenaes (Brit. J Clin Pcol 1991 32(3) 355ff). Their results are summarized in table 1. The HCl salt of the drug was given (MW 332.5) to these patients in this study but the drug concentration was reported as the free base in the plasma. In this study, 25% of the Remoxipride was excreted unchanged, 75% was metabolized. The hepatic and renal blood flow in these patients was 1.5 and 1.2 L/min respectively. REMEMBER TO PAY ATTENTION TO UNITS. FIGURE 4-5.
Remoxipride
TABLE 1-12 Remoxipride
Data IV bolus
Oral Solution
Tablet A
Tablet B
100
100
Dose (mg)
50
AUMC (umole/ L*hr2 )
145
158.7
319.6
282.6
AUC (umole/L*hr)
20.9
19.8
37.6
31.4
Tp (hr)
3.6
Cpmax (mg/L)
.8
f
.75
Basic Pharmacokinetics
REV. 99.4.25
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15-28
Practice Exams: Exam 2
15.7.1
REMOXIPRIDE QUESTIONS 1) Find the MRT of the IV product (hr). 2) Find the elimination rate constant of remoxipride (hr-1 ). 3) Find the Cp0 of the IV product (mg/L) 4) Find the volume of distribution of the IV product. (L) 5) Find the half life of remoxipride (hr). 6) Find the clearance of remoxipride (L/hr). 7) Find the renal clearance (L/hr). 8) Find the hepatic clearance 9) Find the renal extraction ratio. 10) Find the hepatic extraction ratio. 11) Find the MRT of remoxipride given as the oral solution (hr). 12) Find the MAT of remoxipride given as oral solution (hr). 13) Find the absorption rate constant of remoxipride given as an oral solution (hr-1 ). 14) Find the bioavailability of the oral solution (f). 15) Find the peak time of the oral solution (hr). 16) Find the MRT of remoxipride given as Tablet A, the brand name product. 17) Find the MAT of remoxipride given as Tablet A. 18) Find the apparent absorption rate constant (ka) of remoxipride given as Tablet A. 19) Find the peak time of Tablet A. 20) Find the single dose Cpmax for tablet A (mg/L) 21) Find the mean dissolution time (MDT) of Tablet A, the brand name product (hr). 22) Is Tablet B, the generic product is bioequivalent to Tablet A. Why or why not? 23) Find N for BID dosing. 24) Find Cpss max for two caps bid for the brand name product. 25) Find Cpss avg for two caps bid for the brand name product. 26) Find Cpss min for two caps bid for the brand name product. 27) Which change in physiological status would result in the most significant change in the TBC of remoxipride? a) Changes which effect the flow of blood to the liver. b) Changes which effect the flow of blood to the kidney. c) Changes which effect the function of the liver.
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15-29
Practice Exams: Exam 2
d) Changes which effect the function of the kidney. Your patient is experiencing that change. That physiological function was up 75% above normal (F = 1.75). 28) Find his new renal clearance 29) Find his new hepatic clearance. 30) Find his new total body clearance. 31) Assuming now change in volume of distribution, find his new half life. 32 Assuming no change in dosage regimen, find his new N. 33 What dosing regimen would you recommend to return his plasma concentrations back to normal (within 110% of max and 90% of min)?
Basic Pharmacokinetics
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15-30
Practice Exams: Exam 2
15.7.2
REMOXIPRIDE SOLUTIONS 1.
MRT = 6.93hr
2.
k e = 0.144hr
3.
Cp 0 = 0.88mg ⁄ L
4.
Vd = 50L
5.
T1 ⁄ 2 = 4.8hr
6.
Cl = 7.28L ⁄ hr
7.
Cl r = 1.8L ⁄ hr
8.
Cl h = 5.46L ⁄ hr
9.
Er = 0.025
10.
Eh = 0.06
–1
11.
Oral solution MRT = 8hr
12.
Oral solution MAT = 1.07hr
13.
Oral solution Ka = 0.935hr
14.
Oral solution bioavailability f = 0.95
15.
Oral solution peak time 2.4hr
16.
Tablet A MRT = 8.5hr
17.
Tablet A MAT = 1.57hr
18.
Tablet A Ka = 0.637hr
19.
Tablet A peak time 3.0hr
20.
Tablet A single dose Cp max = 1.03mg ⁄ L
21.
Tablet A MDT = 0.5hr
–1
–1
Tablet B is not bioequivalent to Tablet A because peak time and Cp max are out of federal guidelines. 22.
23.
For dosing BID, N = 2.5
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15-31
Practice Exams: Exam 2
24.
Cp
25.
Cp
26.
Cp
ss ss ss
max =
3.88mg ⁄ L
avg =
1.85mg ⁄ L
min =
0.685mg ⁄ L
27. The change in physiological status that would result in the most significant change in TBC of remoxipride is (C) - changes which effect the function of the liver. 28.
New renal clearance = no change
29.
New Cl h = 9.0L ⁄ hr
30.
New Cl s = 10.8L ⁄ hr
31.
New T1 ⁄ 2 = 3.2hr
32.
New N = 3.75
33.
The dosing regimen to recommend = 200 mg TID
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15-32
Practice Exams: Exam 2
15.8 Naproxen Naprosyn@ (naproxen) is a nonsteroidal anti-inflammatory drug (NSAID) with analgesic properties. It is well absorbed (f = 0.95) and highly protein bound (98%) with a volume of distribution of about 10 L and a half-life of 13 hours in normal adults. It is almost entirely cleared by hepatic function (CLr = 1%) with about onethird being metabolized to the 6-o-desmethylnaproxen (which is further metabolized by conjugation) and two-thirds being conjugated directly. Both the 6-o-desmethyl metabolites as well as the conjugates are inactive. Normal dosing is 500 mg Naproxen BID. You stock 200 and 500 mg tablets in your HMO. For the following conditions, new parameters are given in parentheses. Concomitant treatment with Probenecid, a uricosuric which increases the urinary excretion of uric acid, while not interfering with the protein binding, effectively blocks the hepatic conjugation process reducing the hepatic function (Clinth ) to one-third of normal. In chronic renal failure (Fir = 0.1) the protein binding is reduced (94%) because of uremia. This results in a marginal increase in half-life (14 hr.) In rheumatoid arthritis, hypoalbuminaemia results in a reduction in protein binding (97%) and increase in the volume of distribution (13 L). Elderly patients exhibit a decrease in binding (96%), but no change in half-life or volume of distribution. For each of the conditions, (with Probenecid, chronic renal failure, arthritic, and elderly), please recommend a dosage regimen which would give approximately the same plasma concentrations of free naproxen obtained in the normal case (+ 10%.) Constants: L - renal blood perfusion ⋅ 70kg ⋅ 60min L Q r = 0.0191 --------------------------- ≈ 80 -----blood min ⋅ kg hr hr L min L Q H = 0.0238 -------------------- hepatic blood perfusion ⋅ 70kg ⋅ 60--------- ≈ 100 -----blood min ⋅ kg hr hr
Extraction ratios are calculated for normals and considered to be constant throughout.
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15-33
Practice Exams: Exam 2
15.8.1
NAPROXEN QUESTIONS TABLE 1-1. Question
Patient Condition
Normal
Numbers For Exam
Probenecid
Chronic
Rheumatoid
treatment
Renal Failure
arthritis
Elderly
Dose(mg)
1
27
51
75
99
f
2
28
52
76
100
fu
3
29
53
77
101
Vd (L)
4
30
54
78
102
k (hr-1)
5
31
55
79
103
T 1/2 (hr)
6
32
56
80
104
AUC (mg/L*hr)
7
33
57
81
105
% Cl h
8
34
58
82
106
% Cl r
9
35
59
83
107
Cl tot (L/hr)
10
36
60
84
108
Cl h (L/hr)
11
37
61
85
109
Cl r (L/hr)
12
38
62
86
110
Eh
13
Er
14
Cl h Cl r
Cp
(L/hr)
15
39
63
87
111
(L/hr)
16
40
64
88
112
FR h
17
41
65
89
113
FI h
18
42
66
90
114
FR r
19
43
67
91
115
FI r
20
44
68
92
116
FCL
21
45
69
93
117
τ (hr)
22
46
70
94
118
N
23
47
71
95
119
24
48
72
96
120
int
int
ss max free
µg ------ mL
Basic Pharmacokinetics
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15-34
Practice Exams: Exam 2
Cp
Cp
ss µg -------avg free mL
25
49
73
97
121
µg ------ mL
26
50
74
98
122
ss min free
Answer Pool S
M
L
XL
XXL
XXXL
A
0.000067
0.0182
0.100
1.0
10.0
100
B
0.0005
0.0200
0.133
1.1
11.3
200
C
0.0050
0.0300
0.177
1.4
12.0
303
D
0.0080
0.0400
0.182
1.5
13.0
356
E
0.0089
0.0495
0.267
1.6
13.3
500
F
0.0092
0.0533
0.315
1.7
14.0
595
G
0.0095
0.0600
0.333
1.8
26.5
891
H
0.0097
0.0610
0.341
1.9
31.7
1044
I
0.0098
0.0675
0.528
2.0
38.1
2000
J
0.0099
0.0700
0.533
2.1
45.0
3000
A
0.0790
0.577
2.2
50.0
B
0.0798
0.615
2.9
53.0
C
0.0857
0.675
3.0
62.0
D
0.0923
0.700
5.3
80.0
E
0.0950
0.790
6.2
86.0
F
0.0970
0.798
8.0
92.0
G
0.0990
0.857
8.5
97.0
I
0.0999
0.923
9.0
99.0
J
0.09997
0.95
9.9
99.97
Basic Pharmacokinetics
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15-35
Practice Exams: Exam 2
15.8.2
NAPROXEN SOLUTIONS TABLE 1-1. Question
Patient Condition
Normal
Numbers For Exam
Probenecid treatment
Chronic
Rheumatoid
Renal Failure
Elderly
arthritis
1. 500 mg
27. 200
51. 500
75. 500
99. 200
f
2. 0.95
28. 0.95
52. 0.95
76. 0.95
100. 0.95
fu
3. 0.02
29. 0.02
53. 0.06
77. 0.03
101. 0.04
Vd (L)
4. 10
30. 10
54. 31.67
78. 13
102. 10
k (hr-1)
5. 0.0533
31. 0.0182
55. 0.0495
79. 0.061
103. 0.533
T 1/2 (hr)
6. 13
32. 38.1
56. 14
80. 11.3
104. 13
AUC (mg/L*hr)
7. 891
33. 1044
57. 303
81. 595
105. 356
% Cl h
8. 99
34. 97
58. 99.97
82. 99
106. 99
% Cl r
9. 1
35. 3
59. 0.03
83. 1
107. 1
Cl tot (L/hr)
10. 0.533
36. 0.182
60. 1.57
84. 0.798
108. 0.533
Cl h (L/hr)
11. 0.528
37. 0.177
61. 1.57
85. 0.790
109. 0.528
Cl r (L/hr)
12. 0.005
38. 0.005
62. 0.005
86. 0.008
110. 0.005
Eh
13. 0.005
Er
14. 0.000067
Dose(mg)
Cl h Cl r
(L/hr)
15. 26.5
39. 8.84
63. 26.5
87. 26.5
111. 13.26
(L/hr)
16. 0.267
40. 0.267
64. 0.00889
88. 0.267
112. 0.133
FR h
17. 1
41. 1
65. 1
89. 1
113. 1
FI h
18. 1
42. 0.333
66. 3
90. 1.5
114. 1
FR r
19. 1
43. 1
67. 1
91. 1
115. 1
FI r
20. 1
44. 1
68. 0.1
92. 1.5
116. 1
FCL
21. 1
45. 0.341
69. 2.94
93. 1.5
117. 1
τ (hr)
22. 12
46. 12
70. 12
94. 12
118. 8
int
int
Basic Pharmacokinetics
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15-36
Practice Exams: Exam 2
23. 0.923
47. 0.315
71. 0.857
95. 1.06
119. 0.615
µg- ------ mL
24. 2.01
48. 1.94
72. 2.01
96. 2.1
120. 2.2
µg- ------ mL
25. 1.5
49. 1.74
73. 1.5
97. 1.5
121. 1.78
µg ------ mL
26. 1.06
50. 1.55
74. 1.1
98. 1.01
122. 1.42
N Cp
ss max free
Cp
Cp
ss avg free ss min free
Basic Pharmacokinetics
REV. 99.4.25
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15-37
Practice Exams: Exam 2
15.8.3 1. Dose = 500 mg (given) 2. Bioavailability (f) = 0.95 (given) 3. Unbound Fraction = 0.02 (given) 4. V d = 10L (given)
0.693 t1 ⁄ 2
5. k = ------------- = 0.0533hr
–1
6. t 1 ⁄ 2 = 13hr (given)
f ⋅ Dose k ⋅ Vd
7. AUC = ------------------- = 891L 8. %Cl h = 100 – %Cl r = 99 9. %Cl r = 1 (given) 10. Cl tot = k ⋅ V d = 0.533L ⁄ hr 11. Cl h = Cl tot ⋅ %Cl h = 0.528L ⁄ hr 12. Cl r = Cl tot ⋅ %Cl r = 0.005L ⁄ hr
Cl
13. E h = -------h- = 0.00528 Qh
Cl
14. E r = --------r = 0.000067 Qr
15. Cl h int
16. Cl r
int
Q ⋅ Cl h -----------------Q – Cl = ------------------h- = 26.5L ⁄ hr fu Q ⋅ Cl -----------------rQ – Cl = ------------------r = 0.267L ⁄ hr fu
17. FR h = 1 (given) 18. FI h = 1 (given)
Basic Pharmacokinetics
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15-38
Practice Exams: Exam 2
19. FR r = 1 (given) 20. FI r = 1 (given) 21. F Cl = 1 (given) 22. τ = 12hr (given)
τ t1 ⁄ 2
23. N = --------- = 0.923
24.
Cp
25.
Cp
ss fu ⋅ S ⋅ f ⋅ D µg1 = -------------------------⋅ -------------------= 2.0 ------N mL V max free 1 1 – --- 2 ss f u ⋅ S ⋅ f ⋅ D fu ⋅ S ⋅ f ⋅ D µg= -------------------------= ------------------------------ = 1.5 ------V⋅K⋅τ V ⋅ 0.693 ⋅ N mL avg free N
26.
Cp
ss min free
1--- 2 fu ⋅ S ⋅ f ⋅ D µg= -------------------------- ⋅ -------------------- = 1.1 ------N mL V 1 – 1--- 2
27. Dose = mg 28. Bioavailability (f) = 0.95 (no change) 29. Unbound Fraction = 0.02 (no change - given) 30. V d = 10L (given)
0.693 t1 ⁄ 2
31. k = ------------- = 0.0533hr
–1
32. t 1 ⁄ 2 = 13hr (given)
f ⋅ Dose k ⋅ Vd
33. AUC = ------------------- = 891L 34. %Cl h = 100 – %Cl r = 99 35. %Cl r = 1 (given) 36. Cl tot = k ⋅ V d = 0.533L ⁄ hr 37. Cl h = Cl tot ⋅ %Cl h = 0.528L ⁄ hr
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15-39
Practice Exams: Exam 2
38. Cl r = Cl tot ⋅ %Cl r = 0.005L ⁄ hr
39. Cl h
int
40. Cl r int
Q ⋅ Cl h -----------------Q – Cl = ------------------h- = 26.5L ⁄ hr fu Q ⋅ Cl -----------------rQ – Cl = ------------------r = 0.267L ⁄ hr fu
41. FR h = 1 (given) 42. FI h = 1 (given) 43. FR r = 1 (given) 44. FI r = 1 (given) 45. F Cl = 1 (given) 46. τ = 12hr (given)
τ t1 ⁄ 2
47. N = --------- = 0.923
48.
Cp
49.
Cp
ss fu ⋅ S ⋅ f ⋅ D 1 µg= -------------------------⋅ -------------------= 2.0 ------N V mL max free 1 – 1--- 2 ss f u ⋅ S ⋅ f ⋅ D fu ⋅ S ⋅ f ⋅ D µg= -------------------------= ------------------------------ = 1.5 ------V⋅K⋅τ V ⋅ 0.693 ⋅ N mL avg free N
50.
Cp
ss min free
1--- 2 fu ⋅ S ⋅ f ⋅ D µg= -------------------------- ⋅ -------------------- = 1.1 ------N V mL 1 – 1--- 2
51. Dose = mg 52. Bioavailability (f) = 0.95 (no change) 53. Unbound Fraction = 0.02 (no change - given) 54. V d = 10L (given)
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15-40
Practice Exams: Exam 2
0.693 t1 ⁄ 2
55. k = ------------- = 0.0533hr
–1
56. t 1 ⁄ 2 = 13hr (given)
f ⋅ Dose k ⋅ Vd
57. AUC = ------------------- = 891L 58. %Cl h = 100 – %Cl r = 99 59. %Cl r = 1 (given) 60. Cl tot = k ⋅ V d = 0.533L ⁄ hr 61. Cl h = Cl tot ⋅ %Cl h = 0.528L ⁄ hr 62. Cl r = Cl tot ⋅ %Cl r = 0.005L ⁄ hr
63. Cl h int
64. Cl r int
Q ⋅ Cl h -----------------Q – Cl = ------------------h- = 26.5L ⁄ hr fu Q ⋅ Cl -----------------rQ – Cl = ------------------r = 0.267L ⁄ hr fu
65. FR h = 1 (given) 66. FI h = 1 (given) 67. FR r = 1 (given) 68. FI r = 1 (given) 69. F Cl = 1 (given) 70. τ = 12hr (given)
τ
71. N = --------- = 0.923
t1 ⁄ 2
72.
Cp
ss fu ⋅ S ⋅ f ⋅ D 1 µg= -------------------------⋅ -------------------= 2.0 ------N V mL max free 1 – 1--- 2
Basic Pharmacokinetics
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15-41
Practice Exams: Exam 2
73.
74.
Cp
Cp
ss f u ⋅ S ⋅ f ⋅ D fu ⋅ S ⋅ f ⋅ D µg= -------------------------= ------------------------------ = 1.5 ------V⋅K⋅τ V ⋅ 0.693 ⋅ N mL avg free ss min free
1--- N 2 fu ⋅ S ⋅ f ⋅ D µg= -------------------------⋅ -------------------- = 1.1 ------N V mL 1 – 1--- 2
75. Dose = mg 76. Bioavailability (f) = 0.95 (no change) 77. Unbound Fraction = 0.02 (no change - given) 78. V d = 10L (given)
0.693 t1 ⁄ 2
79. k = ------------- = 0.0533hr
–1
80. t 1 ⁄ 2 = 13hr (given)
f ⋅ Dose k ⋅ Vd
81. AUC = ------------------- = 891L 82. %Cl h = 100 – %Cl r = 99 83. %Cl r = 1 (given) 84. Cl tot = k ⋅ V d = 0.533L ⁄ hr 85. Cl h = Cl tot ⋅ %Cl h = 0.528L ⁄ hr 86. Cl r = Cl tot ⋅ %Cl r = 0.005L ⁄ hr
87. Cl h int
88. Cl r int
Q ⋅ Cl h -----------------Q – Cl = ------------------h- = 26.5L ⁄ hr fu Q ⋅ Cl -----------------rQ – Cl = ------------------r = 0.267L ⁄ hr fu
89. FR h = 1 (given) 90. FI h = 1 (given)
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15-42
Practice Exams: Exam 2
91. FR r = 1 (given) 92. FI r = 1 (given) 93. F Cl = 1 (given) 94. τ = 12hr (given)
τ t1 ⁄ 2
95. N = --------- = 0.923
96.
Cp
97.
Cp
ss fu ⋅ S ⋅ f ⋅ D µg1 = -------------------------⋅ -------------------= 2.0 ------N mL V max free 1 1 – --- 2 ss f u ⋅ S ⋅ f ⋅ D fu ⋅ S ⋅ f ⋅ D µg= -------------------------= ------------------------------ = 1.5 ------V⋅K⋅τ V ⋅ 0.693 ⋅ N mL avg free N
98. Cp
ss min free
1--- 2 fu ⋅ S ⋅ f ⋅ D µg= -------------------------- ⋅ -------------------- = 1.1 ------N mL V 1 – 1--- 2
99. Dose = mg 100. Bioavailability (f) = 0.95 (no change) 101. Unbound Fraction = 0.02 (no change - given) 102. V d = 10L (given)
0.693 t1 ⁄ 2
103. k = ------------- = 0.0533hr
–1
104. t1 ⁄ 2 = 13hr (given)
f ⋅ Dose k ⋅ Vd
105. AUC = ------------------- = 891L 106. %Cl h = 100 – %Cl r = 99 107. %Cl r = 1 (given) 108. Cl tot = k ⋅ V d = 0.533L ⁄ hr 109. Cl h = Cl tot ⋅ %Cl h = 0.528L ⁄ hr
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15-43
Practice Exams: Exam 2
110. Cl r = Cl tot ⋅ %Cl r = 0.005L ⁄ hr
111. Cl h
int
112. Cl r int
Q ⋅ Cl h -----------------Q – Cl = ------------------h- = 26.5L ⁄ hr fu Q ⋅ Cl -----------------rQ – Cl = ------------------r = 0.267L ⁄ hr fu
113. FR h = 1 (given) 114. FI h = 1 (given) 115. FR r = 1 (given) 116. FI r = 1 (given) 117. F Cl = 1 (given) 118. τ = 12hr (given)
τ t1 ⁄ 2
119. N = --------- = 0.923
120.
Cp
121.
Cp
ss fu ⋅ S ⋅ f ⋅ D 1 µg= -------------------------⋅ -------------------= 2.0 ------N V mL max free 1 – 1--- 2 ss fu ⋅ S ⋅ f ⋅ D f u ⋅ S ⋅ f ⋅ D µg= -------------------------= ------------------------------ = 1.5 ------V⋅K⋅τ V ⋅ 0.693 ⋅ N mL avg free N
122. Cp
ss min free
1--- 2 fu ⋅ S ⋅ f ⋅ D µg= -------------------------- ⋅ -------------------- = 1.1 ------N V mL 1 – 1--- 2
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15-44
CHAPTER 16
Exam 3
Author: Reviewer:
Basic Pharmacokinetics
REV. 99.4.25
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16-1
Exam 3
16.1 Pharmacokinetics Final Exam Summer 1996
Fluvoxamine is a selective serotonin reuptake inhibitor (SSRI) with antidepressant properties. After oral administration, the drug is almost completely absorbed from the gastrointestinal tract, however despite complete absorption, oral bioavailibiliity in man is approximately 50% on account of first-pass hepatic metabolism. Steady-state plasma concentrations are achieved within 5 to 10 days after initiation of therapy and are 30 to 50 % higher than those predicted from single dose data. Fluvoxamine displays non-linear steady state pharmacokinetics over the therapeutic dose range, with disproportionally higher plasma concentrations with higher dosages. Plasma protein binding of fluvoxamine (77%) is low compared with that of other SSRI’s. V.L. is a 39 year old, 110# female suffering from severe depression. She was admitted to the hospital and prescribed 100mg BID of Fluvoxamine but still her depression was uncontrolled at this dose. Her plasma concentration on this regimen was 20ug/L. After her physician increased her dose to 300 mg BID her plasma concentration was 500 ug/L. V.L.’s depression was controlled at this dose, however she was complaining of adverse effects. The therapeutic range (total drug ug/L) of fluvoxamine is 20-500. Fluvoxamine is metabolized extensively (93%) by the liver to an inactive metabolite. 1.
What was her clearance on the 100 mg BID regimen (L/day)?
2.
What was her clearance on the 300 mg BID regimen (L/day)?
3.
What was her V max (mg/day)?
4.
What was her K m (mg/L)?
5.
The doctor would like to change her therapy in order to minimize side effects. What dose would you recommend tolower her plasma concentration to 300 ug/L?
V.L. has major complications from a combined hepatitis B infection and cirrhosis of the liver. As a result her protein binding is reduced to 66%, her K m changes to 0.03 mg/L and her V max changes to 100mg/day.
6.
What would be her plasma concentration of total fluvoxamine if she maintained the regimen from question 5 (mg/L)?
7.
What is her free plasma concentation (mg/L)?
8.
What total fluvoxamine plasma concentration would you recommend achieving to get her free fluvoxamine plasma concentration back to that of the regimen in question 5 (mg/L)?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
16-2
Exam 3
9.
What daily dose of fluvoxamine (by I.V. bolus) would you recommend to get her average free fluvoxamine plasma concentration approximately back to what it was when she was healthy (mg/day)?
Flurbiprofen is a nonsteroidal anti-inflammatory drug (NSAID) which is a potent inhibitor of prostaglandin synthesis. It was introduced in the U.S. in 1986 for the treatment of osteoarthritis, rheumatoid arthritis, acute gouty arthritis, and ankylosing spondylitis. Flurbiprofen is stereoselectively and extensively bound to plasma albumin. Approximately 99% of the drug is metabolized by the liver, with trace amounts excreted in the urine as unchanged drug. Flurbiprofen is 80% bioavailable. The recommended dosages for flurbiprofen are 50 mg q 4-6 hr as needed for analgesia and 100-300 mg/day for the treatment of inflammatory conditions. The following healthy and sick parameters are given for the patient V.L., 110# suffering from severe arthritis. Her effective hepatic blood flow is 24 mL/min/kg and effective renal blood flow is 15.0 mL/min/kg. Her healthy halflife is 6 hours. Dr. M. recommends 100 mg flurbiprofen BID.
Vd % Bound Drug
Healthy
Sick
0.15 L/kg
0.19L/kg
99.0%
97.0%
10.
What is her total body clearance of flurbiprofen (L/hr)?
11.
What is the intrinsic hepatic clearance (L/hr)?
12.
What is the hepatic extraction ratio?
13.
What is the Cp max of total flurbiprofen in mg/L?
14.
What is the Cp min of total flurbiprofen in mg/L?
15.
What is the Cp max of free flurbiprofen in mic/L?
16.
What is the Cp min of free flurbiprofen in mic/L?
17.
V.L. is now suffering from chronic renal insufficiency. Bound flurbiprofen has now decreased to 97%
ss ss ss ss
due to significant uremia. What is her new hepatic clearance (L/hr)? 18.
What is her new k?
19.
What is the new N?
20.
What would her dose be now, if you wanted to maintain approximately the same free plasma concentrations as the previous therapy with the largest tau?
21.
The renal insufficiency clears up and she comes down with mono. Her plasma albumin drops to 50% of normal thus reducing the bound fraction to 96%. What is her new hepatic clearance using the healthy V d ?
22.
What would her dose be now, if you wanted to maintain approximately the same plasma concentrations as the previous therapy with the largest tau?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
16-3
Exam 3
23.
The mono clears up, but now there seems to be hepatic stenosis. Her plasma flow is reduced to 50% of normal. What would her dose be now, if you wanted to maintain approximately the same concentrations as the previous therapy with the largest tau?
Clentiazem is a chlorinated analog of diltiazem. It is currently undergoing clinical evaluation for the treatment of angina pectoris and hypertension. The primary mechanism responsible for the antihypertensive effect of CLZ is a decrease in peripheral vascular resistance due to the blockade of calcium channels. The following information was obtained from a dose of 20 mg Clentiazem given I.V.
Table 2: 37.52ng ⁄ mL
A1 α
2.70hr
–1
16.17ng ⁄ mL
B1 β
0.078hr
k 10
0.243hr
k 12
1.67hr
k 21
0.868hr
AUMC
2729.6ng ⁄ mL ⁄ hr
–1 –1
–1 –1 2
24.
Can this data be adequately evaluated by a one compartment model?
25.
What is the volume of the central compartment (L)?
26.
What is the clearance of clentiazem (L/hr)?
27.
What is the MRT of clentiazem (hr)?
28.
What is the Vd eff of clentiazem (L)?
29.
What is the V β for clentiazem (L)?
30.
What percent of the clentiazem dose is in the central compartment at equilibrium?
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
16-4
Exam 3
ss
31.
What is the Cp max , and for a 20 mg IV daily dose clentiazem (ng/mL)?
32.
What is the Cp min for the above dosing regimen (ng/mL)?
33.
What is the Cp avg for the above dosing regimen (ng/mL)?
ss ss
Any number from the answer pool may be used once, more than once, or not at all. TABLE 2-1 Answer
Pool
A B C D E F G H I J A B C D E F G H I J
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
16-5
Basic Pharmacokinetics
REV. 99.4.25
Copyright © 1996-1999 Michael C. Makoid All Rights Reserved
http://kiwi.creighton.edu/pkinbook/
-1
