Enzyme Kinetics for Systems Biology

Enzyme Kinetics for Systems Biology

Herbert M. Sauro University of Washington Seattle, WA

Ambrosius Publishing

Copyright © 2009-2011 Herbert M. Sauro. All rights reserved. First Edition, version 1.01 Published by Ambrosius Publishing and Future Skill Software www.analogmachine.org Typeset using LATEX 2" , TikZ, PGFPlots, WinEdt and 11pt Math Time Professional 2 Fonts Limit of Liability/Disclaimer of Warranty: While the author has used his best efforts in preparing this book, he makes no representations or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. The advice and strategies contained herein may not be suitable for your situation. Neither the author nor publisher shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. No part of this book may be reproduced by any means without written permission of the author.

ISBN 10: ISBN 13: ISBN 10: ISBN 13:

0-9824-7730-9 (ebook) 978-0-9824773-0-4 (ebook) 0-9824773-1-7 (paperback) 978-0-9824773-1-1 (paperback)

Printed in the United States of America. Mosaic image modified from Daniel Steger’s Tikz image (http://www. texample.net/tikz/examples/mosaic-from-pompeii/ Protein images used in main text from RCSB Protein Data Bank and David Goodsell © (www.pdb.org). Front-Cover: Metabolic pathway image from JWS online (Jacky Snoep) with permission. The pathway depicts the glycolytic pathway from Lactococcus lactis using the Systems Biology Graphical Notation (SBGN). Ref: Hoefnagel, Hugenholtz and Snoep, 2002, Time dependent responses of glycolytic intermediates in a detailed glycolytic model of Lactococcus lactis during glucose run-out experiments. Mol. Biol. Reports 29, 157-161

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii 1

2

3

Reaction Kinetics

1

1.1

Rates of Change . . . . . . . . . . . . . . . . . . . . . .

4

1.2

Elementary Rate Kinetics . . . . . . . . . . . . . . . . .

12

1.3

Chemical Equilibrium . . . . . . . . . . . . . . . . . . .

16

1.4

Kinetics across Membranes . . . . . . . . . . . . . . . .

25

1.5

Temperature Dependence . . . . . . . . . . . . . . . . .

27

Chapter Highlights . . . . . . . . . . . . . . . . . . . . . . . .

29

Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . .

31

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

32

Elasticities

35

2.1

Introduction and Relevance . . . . . . . . . . . . . . . .

35

2.2

Elasticity Coefficients . . . . . . . . . . . . . . . . . . .

36

2.3

Mass-action Kinetics . . . . . . . . . . . . . . . . . . .

44

2.4

General Elasticity Rules . . . . . . . . . . . . . . . . . .

47

Chapter Highlights . . . . . . . . . . . . . . . . . . . . . . . .

50

Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . .

51

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

52

Basic Enzyme Kinetics

53

iii

CONTENTS

iv

4

3.1

Enzyme Catalysts . . . . . . . . . . . . . . . . . . . . .

54

3.2

Enzyme Kinetics . . . . . . . . . . . . . . . . . . . . .

54

3.3

Basic Enzyme Kinetics . . . . . . . . . . . . . . . . . .

55

3.4

Reversible Rate Laws . . . . . . . . . . . . . . . . . . .

67

Chapter Highlights . . . . . . . . . . . . . . . . . . . . . . . .

77

Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . .

78

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

78

Enzyme Inhibition and Activation

81

4.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . .

82

4.2

Inhibitors . . . . . . . . . . . . . . . . . . . . . . . . .

82

4.3

Generalized Inhibition Model . . . . . . . . . . . . . . .

83

4.4

Competitive Inhibition . . . . . . . . . . . . . . . . . .

85

4.5

Uncompetitive Inhibition . . . . . . . . . . . . . . . . .

93

4.6

Mixed Inhibition . . . . . . . . . . . . . . . . . . . . .

96

4.7

Mixed and Partial Inhibition . . . . . . . . . . . . . . . 101

4.8

Irreversible Inhibitors . . . . . . . . . . . . . . . . . . . 101

4.9

Activators . . . . . . . . . . . . . . . . . . . . . . . . . 102

4.10

Final Remarks . . . . . . . . . . . . . . . . . . . . . . . 103

Chapter Highlights . . . . . . . . . . . . . . . . . . . . . . . . 106 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . 107 Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 5

Multi-reactant Rate Laws

111

5.1

Multiple Reactant Enzymes . . . . . . . . . . . . . . . . 112

5.2

Types of Multi-Reactant Systems . . . . . . . . . . . . . 112

5.3

Ordered Bi-Bi Mechanism . . . . . . . . . . . . . . . . 114

5.4

Random-Order Mechanism . . . . . . . . . . . . . . . . 116

CONTENTS

5.5

v

Ping-Pong Mechanism . . . . . . . . . . . . . . . . . . 117

Chapter Highlights . . . . . . . . . . . . . . . . . . . . . . . . 118 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . 118 6

Cooperativity

119

6.1

Hill Equation . . . . . . . . . . . . . . . . . . . . . . . 124

6.2

Ligand Binding . . . . . . . . . . . . . . . . . . . . . . 129

6.3

The Adair Equation . . . . . . . . . . . . . . . . . . . . 135

6.4

MWC Model . . . . . . . . . . . . . . . . . . . . . . . 140

6.5

KNF Sequential Model . . . . . . . . . . . . . . . . . . 150

6.6

Reversible Hill Equation . . . . . . . . . . . . . . . . . 151

Chapter Highlights . . . . . . . . . . . . . . . . . . . . . . . . 154 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . 156 Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 7

Allostery

163

7.1

Allosteric Enzymes . . . . . . . . . . . . . . . . . . . . 165

7.2

Allostery and MWC Model . . . . . . . . . . . . . . . . 167

7.3

Reversible Hill Equation . . . . . . . . . . . . . . . . . 174

Chapter Highlights . . . . . . . . . . . . . . . . . . . . . . . . 178 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . 179 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 8

Generalized Rate Laws

181

8.1

Generalized Rate Laws . . . . . . . . . . . . . . . . . . 182

8.2

Linear Approximation . . . . . . . . . . . . . . . . . . . 182

8.3

Linear-Logarithmic Rate Laws . . . . . . . . . . . . . . 186

8.4

Algebraic Approximations . . . . . . . . . . . . . . . . 189

CONTENTS

vi 8.5

Hanekom Rate Laws . . . . . . . . . . . . . . . . . . . 189

8.6

Liebermeister Rate Laws . . . . . . . . . . . . . . . . . 194

8.7

Choosing a Suitable Rate Law . . . . . . . . . . . . . . 200

Chapter Highlights . . . . . . . . . . . . . . . . . . . . . . . . 201 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . 202 9

Kinetics of Gene Regulation

203

9.1

Structure of a Microbial Genetic Unit . . . . . . . . . . 204

9.2

Gene Regulation . . . . . . . . . . . . . . . . . . . . . . 205

9.3

Fractional Occupancy . . . . . . . . . . . . . . . . . . . 210

9.4

Multiple Transcriptional Factors . . . . . . . . . . . . . 214

9.5

Cooperativity . . . . . . . . . . . . . . . . . . . . . . . 231

9.6

Thermodynamic Approach . . . . . . . . . . . . . . . . 239

Chapter Highlights . . . . . . . . . . . . . . . . . . . . . . . . 247 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . 248 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248 10 Basic Thermodynamics

249

10.1

Systems and Surroundings . . . . . . . . . . . . . . . . 250

10.2

Energy, Work and Heat . . . . . . . . . . . . . . . . . . 250

10.3

Enthalpy Change . . . . . . . . . . . . . . . . . . . . . 254

10.4

Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . 258

10.5

Reaction Spontaneity . . . . . . . . . . . . . . . . . . . 261

10.6

Gibbs Energy . . . . . . . . . . . . . . . . . . . . . . . 263

Chapter Highlights . . . . . . . . . . . . . . . . . . . . . . . . 269 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . 270 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 Appendix A List of Symbols and Abbreviations

275

CONTENTS

vii

Appendix B List of Common Rate Laws

279

Appendix C Answers to Questions

291

References

295

History

301

Index

303

viii

CONTENTS

Preface This book is an introduction to enzyme and gene regulatory kinetics. The emphasis is not however about using kinetic studies as a means to understand mechanism; this is the domain of classical enzyme kinetics. Instead, the focus is to review kinetic laws that one might use to build simulations of cellular networks. The book covers basic reaction kinetics, elasticities, and enzyme kinetics including cooperativity and allostery. In addition, it covers a topic rarely taught in class, the kinetics of gene regulatory networks. The book should be suitable for undergraduates in their early (Junior, USA, second year UK) to mid years at college. The book can also serve as a reference guide for researchers and teachers. For a number of reasons, I have decided to publish this book myself. The most important is that I retain full editorial and copyright control. This allows me to quickly release new updates to the text with either new material or corrections. This model is similar to a small software company where the original author retains control and can publish frequent updates and bug fixes. It would be difficult to imagine a software author handing full control of his or her software, including source code and copyright, to a publishing house where the software would only be updated at the discretion of the publisher perhaps ever five years or so, if at all. With the internet and the web firmly entrenched in our society, the traditional publishing model is looking more and more limited and particularly restrictive. With the availability of high quality typesetting tools such as TeX/LaTeX much of the skills required to layout a book has been considerably simplified. I thank all the authors who helped develop LaTeX, tikz, pgfplot, WinEdt, Sumatra PDF and the many LaTeX packages without which this effort would have been much more difficult to do. Another aim I had in writing this book was to provide students and other interested readers with affordable access to text books. As a teacher, I am acutely aware that students sometimes find it difficult to justify the expense of class text books. This is particularly true when each year a new edition is released, perhaps with only minor changes, and yet students are often required to purchase the latest edition rather than much cheaper secondhand copies. In addition, times are changing in the publishing field as the

CONTENTS

ix

iPad/Android/Kindle generation is becoming accustomed to cheap, mass distributed software and other media such as e-books. The possibility to distribute in e-book form has allowed me to use some color which normally would be avoided except in larger print runs. I hope readers will find this book useful and that they will contribute comments, good or bad. Early editions will undoubtedly contain grammatical and typographical errors in the text, but these, I am sure, will be eliminated in time. The latest edition together with free software and other material can be found at www.analogmachine.org and the research site, www.sys-bio. org. There are many people and organizations who I should thank but foremost must be my infinitely patient wife, Holly, who has put up with the many hours I have spent working alone in our basement or late at the department and who contributed significantly to editing this book. I am also most grateful to the National Science Foundation and the National Institutes of Health who paid my summer salary so that I could allocate the time to write, edit and research. April 2011 Seattle, WA

H ERBERT M. S AURO

x

CONTENTS

1

Reaction Kinetics

Reaction kinetics is the study of how fast chemical reactions take place, what factors influence the rate of reaction and what mechanisms are responsible. Many variables can affect the reaction rate including temperature, pressure and composition. In this chapter we will review a number

1

2

CHAPTER 1. REACTION KINETICS

of topics related to reaction kinetics that have a significant bearing on the development of mathematical models of cellular networks. A chemical reaction is usually depicted in the form of a chemical equation which describes the transformation of one or more reactants into one or more products. The reactants appear on the left of the equation and the products on the right. Both sides are separated by an arrow indicating the positive direction of the transformation. The simplest possible reaction is the conversion of a single reactant, A, into a single product, B, as depicted in the following way: A!B Such a reaction can be studied by observing the change in concentration of A and/or B in time. Experimentally there are a variety of ways to do this, for example by observing the emission or absorption of light at a specific wavelength, the change in pH, or the incorporation of a radioactive or heavy isotope into the product. An example of an actual biochemical reaction is the familiar interconversion of the adenine nucleotides: 2 ADP ! ATP C AMP This describes two molecules of ADP being transformed into one molecule of ATP and one molecule of AMP. Sometimes a double arrow is used to explicitly indicate that a reaction is reversible, as in: 2 ADP ATP C AMP If a reaction is reversible (as almost all reactions are to some extent), then the reaction rate can be positive or negative. By convention, a positive rate means that the reaction progresses from left to right, whereas a negative rate indicates a right to left reaction. Example 1.1 What does the following reaction notation mean: 3A C 4B ! 3C C D This notation means that during a reaction event, 3 molecules of A and 4 molecules of B react to form 3 molecules of C and one molecule of D.

3

We now need to define a number of terms: the stoichiometric amount, rate of change, stoichiometric coefficient, and reaction rate.

Stoichiometric Amount The stoichiometric amount is defined as the number of molecules of a particular reactant or product taking part in a reaction. Stoichiometric amounts will always be positive numbers. For example, in the reaction: 2 ADP ATP C AMP ADP has a stoichiometric amount of two, ATP a stoichiometric amount of one, and AMP also with a stoichiometric amount of one. If the same species occurs on the reactant and product side of a reaction then it must be treated separately. For example, in the reaction: 2A C B C C ! 3A C D C 2B The stoichiometric amounts on the reactant side include: A with two, B with one and C with one. On the product side the stoichiometric amounts include: A with three, D with one and B with two. The stoichiometric amount is the number of molecules of a particular reactant or product taking part in a reaction.

Example 1.2 List the stoichiometric amounts in the following reaction: 2A C B ! A C C On the reactant side the stoichiometric amount for A is two and for B is one. On the product side, the stoichiometric amount for A is one and for C one.

4

CHAPTER 1. REACTION KINETICS

1.1 Rates of Change The rate of change can be defined as the rate of change in concentration or amount (depending on units) of a designated species. If S is the species then the rate of change is given by: Rate D


S
t

Because rates change as reactants are consumed and products made, the rate of change is better defined as the instantaneous change in concentration, or a derivative: dS Rate D dt If we were to plot the rate of product formation as a function of time, the rate of reaction would be given by the slope of the curve (Fig. 1.1). If concentrations are measured in moles per liter (L) and time in seconds (sec), then the rate of reaction is expressed in mol L 1 sec 1 . When reporting a rate of change, it is important to give the name of the species that was used to make the measurement. For example, in the reaction 2A ! B, the rate of change of A is twice the rate of change of B. In addition, the rate of change of A is negative because it is consumed, whereas the rate of change of B is positive because it is being made.

Stoichiometric Coefficients Stoichiometry deals with static information about the amounts of substances involved in a chemical transformation, whereas kinetics relates rates of change that occur in these amounts. To paraphrase a statement made by Aris [4], stoichiometry provides the framework within which chemical change takes place irrespective of the forces that bring them about, and by kinetics the speed of chemical change. Aris then went on to state, “Just as the latter can only be built on a proper understanding of the kinematics, so the analysis of stoichiometry must precede that of kinetics”. We will do the same here.

Concentration of Product, B

1.1. RATES OF CHANGE

5

4 3 Slope = Rate of change of product 2 1 0

0

1

2

3

4

5

Time Figure 1.1: Progress curve for a simple irreversible reaction, A ! B. Initial reactant concentration, A, is set at 5 units. The plot shows the accumulation of product, B, as the reaction proceeds. The rate of change of product is given by the slope of the curve which changes over the course of the reaction.

The stoichiometry coefficient refers to the relative amount of substance that is consumed and/or produced by a reaction. Given a reaction such as:

2A ! B

the stoichiometric amount of A is 2 and for B, 1. The species stoichiometry or stoichiometric coefficient however, is the difference between the stoichiometric amounts of a given species on the product side and the stoichiometric amount of the same species on the reactant side. The definition below summarizes this more clearly.

6

CHAPTER 1. REACTION KINETICS

The stoichiometric coefficient, ci , for a molecular species Ai , is the difference between the stoichiometric amount of the species on the product side and the stoichiometric amount of the same species on the reactant side, that is: ci D Stoichiometric Amount of Product, Ai Stoichiometric Amount of Reactant, Ai In the reaction, 2A ! B, the stoichiometric amount of A on the product side is zero while on the reactant size it is two. Therefore the stoichiometric coefficient of A is given by 0 2 D 2. In many cases a particular species will only occur on the reactant or product side and it is relatively uncommon to find situations where a species occurs simultaneously as a product and a reactant. As a result, reactant stoichiometric coefficients tend to be negative and product stoichiometric coefficients tend to be positive. To illustrate this further consider the more complex reaction: 3A C 4B ! 2C C D Since A only appears on the reactant side, its stoichiometric coefficient will be 3, similarly for B which will have a stoichiometric coefficient of 4. Species C only occurs on the product side, therefore its stoichiometric coefficient is C2, and similarly for D which will have a stoichiometric coefficient of C1. In these cases the stoichiometric amounts and the stoichiometric coefficients are the same except for the sign difference on the reactant stoichiometric coefficients. Finally consider the following reaction where a species occurs on both the reactant and product side: 2A C B ! A C C The stoichiometric coefficient of A must take into account the fact that A appears both as a reactant and a product. The overall stoichiometric coefficient of A is therefore C1 2 which gives 1. The last example highlights how information can be lost when computing stoichiometric coefficients. It is not possible to recreate the original reaction equation from the stoichiometric coefficients alone, and therefore

1.1. RATES OF CHANGE

7

underscores the danger of just supplying stoichiometric coefficients when communicating information on reaction equations to other researches. One option is to store the stoichiometric amounts together with the associated reactant or product. Computer exchange formats, such as the Systems Biology Markup Language (SBML) [35] are specifically designed to preserve complete reaction equation information for this very reason. Example 1.3 Write down the stoichiometric coefficients for the following reactions: a) A C A ! A C B The stoichiometric amount of A on the reactant side is 2 and on the product side, 1. Therefore the stoichiometric coefficient for A is 1 2 D 1. The stoichiometric amount of B on the product side is 1 and on the reactant side, 0, therefore the stoichiometric coefficient for B is 1 0 D 1. b) A ! B C 21 A The stoichiometric amount of A on the reactant side is 1 and on the product side 1 , therefore the stoichiometric coefficient for A is 1=2 1 D 1=2. The stoichio2 metric amount of B on the reactant side is 0 and on the product side, 1, therefore the stoichiometric coefficient for B is 1 0 D 1.

Example 1.3 (b) highlights another fact about stoichiometric coefficients. The coefficients can be fractional amounts, often represented as rational fractions.

Reaction Yields One application of stoichiometry is to compute maximum theoretical yields for a given reaction. Consider the yeast fermentation of glucose to ethanol: C6 H12 O6 ! 2 C2 H5 OH C 2 CO2 If a yeast culture is started with 10g of glucose, what is the maximum amount of ethanol that can be produced if all the glucose is consumed? The

8

CHAPTER 1. REACTION KINETICS

stoichiometric amount for ethanol is 2, that is for every one mole of glucose consumed, two moles of ethanol are formed. The molar mass of glucose is 180, therefore the number of moles of glucose in 10g is 10=180 D 0:055 moles. From the stoichiometry, this means that 0.111 moles of ethanol will be formed. If the molar mass of ethanol is 46, then 5.2g of ethanol are formed. The same calculation can be made for CO2 yielding a mass of 4.8g of carbon dioxide. As a final check it is evident that the total mass of product is 5:2C4:8 or 10g, exactly the amount of initial glucose. Therefore mass is conserved, as expected. The percentage yield of ethanol can be calculated from the mass of ethanol produced compared with the mass of glucose consumed. If a 100% conversion is assumed, so all the glucose is converted and no side reactions occur, then the percentage yield for ethanol is 52% (5.2g / 100g) with the remainder lost as carbon dioxide. However, in reality maximum yields are never achieved because some of the glucose is diverted to produce biomass. In anaerobic growth for example, fermentation of glucose to ethanol in yeast will typically yield about 0.46-0.48 g ethanol per gram of glucose, that is 90-94% of the theoretical yield [41]. Definitions: Intensive Property: A property that does not depend on the quantity of substance. Examples include temperature, density and concentration. Extensive Property: A property that does depend on the quantity of substance. Examples include mass and volume.

Reaction Rates In this section we will introduce the concept of a reaction rate, denoted by v. The standard unit for the reaction rate is amount per volume per time. This is an intensive property, which does not depend on the amount of substance, for example mol L 1 sec 1 . In a previous section we introduced the rate of change. In practice it is the rate of change that we measure experimentally. We also briefly mentioned that in the reaction 2A ! B,

1.1. RATES OF CHANGE

9

A is consumed twice as fast as the production of product, B. This means that the sign and magnitude of the rates of change will vary depending on which species we choose to measure. A simple way to avoid these differences is to divide each rate of change by the species stoichiometric coefficient. In this case the stoichiometric coefficient of A is 2 and for B is C1. If we do this we obtain: 1 dA 1 dB D Dv 2 dt 1 dt In general, for a reaction of the form n1 A C n2 B C : : : ! m1 P C m2 Q C : : : where n1 ; n2 ; : : : and m1 ; m2 ; : : : represent the stoichiometric coefficients, the reaction rate is given by:

Rate D v 

1 dA D n1 dt

1 dB 1 dP 1 dQ ::: D D ::: n2 dt m1 dt m2 dt

(1.1)

Defined this way, a reaction rate is independent of the species used to measure it. The same applies if a given species appears on both sides of a reaction. For example, in the reaction A ! 2A, the stoichiometric coefficient is C1 so that the reaction rate, v, is: vD

1 dA C1 dt

To make the definition of the reaction rate more formal, let us introduce the extent of reaction, indicated by the symbol, . We define a change from  to  C d in time dt to mean that c1 d moles of A1 , c2 d moles of A2 etc, react to form cn d moles of An etc. By this definition we can state that for any component i , the following is true for the time interval dt : dni D ci d or

dni d D ci dt dt

(1.2)

10

CHAPTER 1. REACTION KINETICS

where ni equals the amount in moles of species i . From this relation we define the extensive rate of reaction, vE , to be: vE  In other words

d dt

dni D ci vE dt

(1.3)

For the moment we will use vE and vI to distinguish the extensive and intensive reaction rates. Note that  has units of amount and vE has units of amount per unit time and is therefore an extensive property, being dependent on the size of the system. The advantage of introducing the extent of reaction is that it allows us to formally define the rate of reaction independently of the species we use to measure the rate. This convenient property can be expressed as:

vE 

d D dt

1 dn1 D c1 dt

1 dn2 1 dnn 1 dnnC1 ::: D D ::: c2 dt cn dt cnC1 dt

Example 1.4 Express the rate of reaction and the rates of change for the following biochemical reaction: 2 ADP ! ATP C AMP The rate of reaction is given by

vD

dn.ATP/ dn.AMP/ d D D dt dt dt D

1 dn.ADP/ 2 dt

If the volume, V , of the system is constant we can also express the rate in terms of concentration, Ci D ni =V . We can therefore rewrite the rate of reaction in the form: vE D V

1 dC1 D ::: c1 dt

1.1. RATES OF CHANGE

11

where vE has units of amount per unit time (mol s 1 ). The relation vE =V is the intensive version of the rate, vI , with units of concentration per unit time (mol L 1 s 1 ) and is the most commonly used form in biochemistry.

vI D

1 dCi vE D V ci dt

or dCi D ci vI dt

(1.4)

where Ci is the concentration of species i and vI is the intensive rate of reaction. For constant volume, single compartment systems, this is a commonly encountered equation in models of cellular networks. The above equation may also be expressed as:

1 dni D ci vI V dt

(1.5)

to emphasize the change in mass that accompanies a reaction. Recall that vI is expressed as mol L 1 s 1 . If a E or I subscript is not used on v then the specific form should be clear from the context. In this book, where we use v, we will generally mean vI , the intensive form. In some simulation situations, for example those involving multiple compartments of different volumes or where there are specific mass conservation laws at work, the intensive rate is not appropriate. This is because the intensive version is unable to keep track of the total number of moles undergoing transformation. In these situations it is necessary to deal explicitly with the extensive rate of reaction, in other words: dni D V ci vI dt

12

CHAPTER 1. REACTION KINETICS

A Word on Notation In many texts, the concentration (molarity) of a substance, X, is denoted using square brackets, as in [X]. To avoid unnecessary clutter in the current text, the use of square brackets to indicate molarity will be relaxed.

1.2 Elementary Rate Kinetics Up to now, we have not discussed how v might be calculated other than by experimental measurement. In this section we introduce mass-action kinetics. Chemical reactions that involve no reaction intermediates are called elementary reactions. Such reactions often have simple kinetic properties and empirical studies have shown that the rate of reaction is often proportional to the product of the molar concentration of the reactants raised to some power. For a simple elementary monomolecular reaction such as: A!B the rate of reaction, v, is often found to be proportional to the concentration of species A, or: v D kA This property is often called the law of mass-action and the corresponding kinetics called mass-action kinetics. The proportionality constant, k, is called the rate constant. A is the concentration of reactant and v is the rate of reaction with units of mol L 1 t 1 . Recall that the rate of change of A is the reaction rate times the stoichiometry coefficient (1.4), since the stoichiometry coefficient of A is 1, the rate of change is given by: dA D dt

vD

kA

(1.6)

The units for k are t 1 and for the concentration of A, moles L 1 . The rate of change of A therefore has units of moles L 1 t 1 . By convention,

1.2. ELEMENTARY RATE KINETICS

13

a rate that is proportional to a reactant raised to the first power is called first-order with respect to the reactant. Similarly, reactants raised to the zeroth power are called zero-order, and reactants raised to the power of two are called second-order (See Figure 1.2). 10

Reaction Rate

10

10 First-Order

Zero-Order

5

5

0

0 0

5

10

5 Second-Order

0

5

10

0

0

5

10

Reactant Concentration

Figure 1.2: Curves illustrating zero-order, first-order and second-order kinetics.

Equation (1.6) is a differential equation that can be solved using standard methods in differential calculus to describe the change in concentration of A over time. This solution is shown in Figure 1.3 and is described by the equation: (1.7) A.t / D A.0/ e k t where A.0/ is the initial concentration of A and t the time. A common metric that is used to judge the rate of different first-order reactions is the half-life. This quantity measures the time taken for half the level of substance, A, to be transformed into product. When half of A has been consumed we can set A.t / in equation (1.7) to 12 A.0/ so that 1 D e k t½ 2 The time, t½ , is called the half-life and by suitable rearrangement is given by, t½ D ln.2/=k. For example if k D 0:5 sec 1 , the half life is equal to ln.2/=0:5 ' 1:4 sec, that is, after 1:4 sec, half the concentration of substance has been consumed. For the general reaction: n1 A C n2 B C : : : ! m1 P C m2 Q C : : :

14

CHAPTER 1. REACTION KINETICS

Concentration of A

1 0:8 0:6 0:4 0:2 0

0

0:5

1

1:5

2

2:5

Time Figure 1.3: Progress curve for species A in the irreversible reaction A ! B, computed using A.t / D A.0/ exp. k t / where A.0/ is the initial concentration equal to 1.0, k the rate constant equal to 2.5, and t the time. The change in B is given by B.t / D A.0/ A.t /.

the rate law has been found through empirical observation to often have the form: v D kAn1 B n2 : : : or more generally:

vDk

Y

Sini

(1.8)

i

where each reactant is raised to the power of its stoichiometric amount. For example, the forward reaction rate for the following uncatalyzed reaction: 2ADP ! ATP C AMP can be written as: v D k ADP 2

(1.9)

1.2. ELEMENTARY RATE KINETICS

15

If the reaction is reversible then the rate law is appended with the reverse rate. In general a reversible mass-action rate law is given by:

v D k1 An1 B n2 : : :

k2 P m1 Qm2 : : :

(1.10)

In the case of reaction (1.9), this would mean: v D k1 ADP 2

k2 ATP AMP

In all mass-action rate laws, the units for the reactant and product terms must be expressed in concentration. The units for the rate constants, k will depend on the exact form of the rate law but must be set to ensure that the rate of reaction is expressed in units moles L 1 t 1 . Although in many cases one will often assume a rate law of the form (1.10), this need not always be the case. For example, the gas reaction 2O3 ! 3O2 has been found experimentally to follow the rate law: 1 .O3 /2 vD k 2 O2 rather than the expected, v D 1=2 k .O3 /2 . The reason for the discrepancy is that the decomposition of ozone into oxygen occurs via a series of elementary reactions and it is the combination of these elementary reactions that gives rise to the non-elementary rate law. In biochemistry this effect is readily seen in enzyme kinetics where the rate laws are far from elementary. Example 1.5 Write down the mass-action rate laws for the following reversible reactions. Assume that the forward and reverse rate constants are k1 and k2 respectively. a) A ! B v D k1 A

k2 B

16

CHAPTER 1. REACTION KINETICS b) 2A ! B v D k1 A2

k2 B

c) A C B ! C C A v D k1 A B

k2 C A D A .k1 B

k2 C /

d) 1 21 A C 2B ! 12 C v D k1 A3 B 4

k2 C

1.3 Chemical Equilibrium In principle all reactions are reversible, meaning transformations can occur from reactant to product or product to reactant. The net rate of a reversible reaction is the difference between the forward and reverse rates. We can write down the forward rate, vf , and reverse rate, vr for the simple unimolecular reaction A B as: vf D k1 A vr D k 2 B

(1.11)

The net rate of reaction, v, is then given by the difference between the forward and reverse rates: v D vf

vr

Furthermore, all reactions in a closed system (See Table 10.1), that is a system which is isolated from the surroundings, will tend to thermodynamic equilibrium (Figure 1.4). At equilibrium the forward and reverse rates will be equal and the net rate zero: vf vr D 0 Inserting equations (1.11) into the above yields the ratio:

1.3. CHEMICAL EQUILIBRIUM

17

Concentration of A and B

1 0:8 B 0:6 0:4 A 0:2 0

0

2

4

6

8

10

Time Figure 1.4: Approach to equilibrium for the reaction A B, k1 D 0:6; k2 D 0:4; A.0/ D 1; B.0/ D 0. Progress curves calculated from the solution to the differential equation dA=dt D k2 B k1 A.

k1 B D D Keq k2 A

(1.12)

This ratio has special significance and is called the equilibrium constant, denoted by Keq . The equilibrium constant is also related to the ratio of the rate constants, k1 =k2 . For a general reversible reaction such as: n1 A C n2 B C : : : m1 P C m2 Q C : : : and using arguments similar to those described above, the ratio of the rate constants can be easily shown to be:

Keq D

P m1 Qm2 : : : k1 D An1 B n2 : : : k2

(1.13)

where the exponents are the stoichiometric amounts for each species.

18

CHAPTER 1. REACTION KINETICS

Example 1.6 Write out the equilibrium relationship for the following reactions a) A C B ! C Keq D

C AB

Keq D

B 3C A2

b) 2A ! 3B C C

c) A C A ! A C B Keq D

AB B D 2 A A

For a bimolecular reaction such as: HA H C A chemists and biochemists will often distinguish between two kinds of equilibrium constants called association and dissociation constants. Thus the equilibrium constant for the above bimolecular reaction is often called the dissociation constant, Kd : Kd D

H
A HA

to indicate the degree that the complex is dissociated into its component molecules at equilibrium. The association constant, Ka , though less commonly used, describes the equilibrium constant for the reverse process H C A HA, that is the formation of a complex from component molecules: HA Ka D H
A It should be evident that:

Kd D

1 Ka

(1.14)

1.3. CHEMICAL EQUILIBRIUM

19

Example 1.7 The equilibrium constant for the reaction between glucose-6-phosphate and fructose-6-phosphate catalyzed by glucose-6-phosphatase isomerase (EC 5.3.1.9) is known to have a value of 0.395 at 25ı C . The concentration of glucose-6-phosphate in liver cells is estimated to be 4.9 mM. Assuming the reaction is at equilibrium, estimate the concentration of fructose-6-phosphate. The reaction is described by Glucose-6-Phosphate Fructose-6-Phosphate and the equilibrium constant is therefore given by Keq D

Fructose-6-phosphate Glucose-6-phosphate

By simple rearrangement the Fructose-6-Phosphate concentration is equal to Fructose-6-phosphate D Keq Glucose-6-phosphate D 0:395  4:9mM D 1:94 mM

Example 1.8 The previous problem can be made more difficult by stating that the total concentration of glucose-6-phosphate and fructose-6-phosphate is 4.9 mM. The question now is to compute the equilibrium concentration of both species. The calculation begins by constructing an equilibrium table: Species

G6P

Initial concentration Equilibrium concentration

4:9 4:9

F6P x

0 x

G6P  Glucose-6-Phosphate F6P  Fructose-6-Phosphate From the equilibrium constant and the above table we derive the following Keq D

x 4:9 x

20

CHAPTER 1. REACTION KINETICS

Solving for x and hence the equilibrium concentration, yields xD

Keq 4:9 1 C Keq

Therefore the equilibrium concentration of Glucose-6-Phosphate is 1.387 mM and for Fructose-6-Phosphate, 3.514 mM. A simple check that the ratio, 1.387/3.514 equals the equilibrium constant will confirm the result. With more complex reactions the above method yields polynomial solutions which can have multiple solutions, usually one negative and the other positive. It should be clear however that the negative solution is physically impossible which leaves the other as the solution we seek.

Principle of Detailed Balance In its simplest form, the principle of detailed balance says that the forward and reverse rates must be equal at thermodynamic equilibrium. For the simple reversible reaction: A B where the forward rate vf is given by vf D kf A, and the reverse rate, vr by vr D kr B, detailed balance states that vf D vr at equilibrium, or kf Aeq D kr Beq From the definition of the equilibrium constant we see that Keq D

kf Beq D Aeq kr

Detailed balance is more useful when applied to more complex systems. Consider the system shown in Figure 1.5 which is comprised of three species linked by three reversible reactions. Each reaction has a forward and reverse rate constant. At equilibrium the following must be true: k1f B D D Keq1 , k1r A

k2f C D D Keq2 ; k2r B

and

k3f A D D Keq3 k3r C

1.3. CHEMICAL EQUILIBRIUM

21

A k1f

C

k 2r

k1r k 3r k 3f

k 2f

B

Figure 1.5: The principle of detailed balance.

Combining the three equations and eliminating A, B, and C yields the following relation among the rate constants: k1f k2f k3f D k1r k2r k3r The product of rate constants in one direction around the loop is equal to the product of rate constants in the opposite direction around the loop. A restatement of the above relation is that the product of the equilibrium constants in a loop is one:

BC A D Keq1 Keq2 Keq3 D 1 ABC

(1.15)

Equation (1.15) applies irrespective of the actual reaction mechanism. Detailed balance applies a constraint on the allowable rate and equilibrium constants in a reaction loop. In addition it means that the change in free energy (See chapter 10) around the loop is zero. By analogy, one can compare detailed balance to a hike over a mountain range, where a hiker traverses one peak after another. If we assume that the hiking route eventually returns him to the original starting point, the net change in height is zero. Detailed balance also precludes the construction of a perpetual motion machine.

22

CHAPTER 1. REACTION KINETICS

Mass-action and Disequilibrium Ratio In closed systems, reactions will tend to equilibrium whereas reactions occurring in open living cells are generally out of equilibrium. The ratio of the products to the reactants in vivo is called the mass-action ratio, . For the system, A ! B: Bin vivo D Ain vivo At equilibrium D Keq . The ratio of the mass-action ratio to the equilibrium constant is often called the disequilibrium ratio and denoted by the symbol, .

D

Keq

(1.16)

At equilibrium, the mass-action ratio is equal to the equilibrium constant and  D 1. If the reaction is far from equilibrium (B=A < Keq ) then  < 1. For a simple unimolecular reaction it was shown previously that the equilibrium ratio of product to reactant, B=A, is equal to the ratio of the forward and reverse rate constants. Substituting this into the disequilibrium ratio gives: k2 B k2 D D k1 A k1 Therefore: D

vr vf

(1.17)

Thus the disequilibrium ratio is the ratio of the reverse and forward rates. This relationship clearly shows how the disequilibrium ratio tells us whether the reaction is going forward, is at equilibrium or whether it is in reverse. If  < 1, then the net reaction must be in the direction of product formation

1.3. CHEMICAL EQUILIBRIUM

23

ln./

Direction of Reaction

v


G

<0 D0 >0

Forward Direction Equilibrium Reverse Direction

v>0 vD0 v<0


G < 0
G D 0
G > 0

Table 1.1: Relationship between  and
G.

since vf > vr . If  D 1 then vr D vf , and the system is at equilibrium. Finally if  > 1 then vr > vf , the reaction must be going in reverse. If we take the natural log of equation (1.16) on both sides we get: ln./ D ln. /

ln.Keq /

(1.18)

With this transformation, if ln./ is negative the reaction must be in the forward direction, zero if the reaction is at equilibrium and greater than zero if the reaction is in the reverse direction. This form of the equation will appear again in the chapter on thermodynamics (chapter 10) where it will be possible to determine  from more fundamental concepts such as entropy and enthalpy. Those who are already familiar with the concept of free energy (
G) may realize that equation (1.18) is closely related to the free energy equation: r G D r G ı C RT ln where r G D RT ln./ and r G ı D RT ln Keq . Because all the rate information has been lost in the derivation of the equation (1.16), the value of ln./ tells us nothing about how fast the reaction will proceed, only the direction it proceeds. Relation (1.17) is actually much more general and applies to any reaction of the form: n1 A C n2 B C : : : m1 P C m2 Q C : : : The disequilibrium ratio is an important quantity and reappears in later sections and chapters when to discuss enzymatic reactions. It is particularly relevant when one considers the control of cellular pathways.

24

CHAPTER 1. REACTION KINETICS

Modified Mass-Action Rate Laws A typical reversible mass-action rate law will require both the forward and the reverse rate constants to be fully defined. Often however, only one rate constant may be known. In these circumstances it is possible to express the reverse rate constant in terms of the equilibrium constant. For example, given the simple unimolecular reaction, A B. it is possible to derive the following: v D k1 A

k2 B  k2 B v D k1 A 1 k1 A k1 Since Keq D k2   v D k1 A 1 Keq 

(1.19)

where is the mass-action ratio. This can be generalized to an arbitrary mass-action reaction to give:

n1

v D k1 A B

n2

 ::: 1

 Keq

D k1 An1 B n2 : : : .1

/

where An1 B n2 : : : represents the product of all reactant species, n1 and n2 are the corresponding stoichiometric amounts, and  is the disequilibrium ratio. For example, for the reaction: 2A C B ! C C 2D where k1 is the forward rate constant, the modified reversible rate law is: v D k1 A2 B .1

/

1.4. KINETICS ACROSS MEMBRANES

25

The modified formulation demonstrates how a rate expression can be divided up into functional parts that include both kinetic and thermodynamic components [32]. The kinetic component is represented by the term k1 An1 B n2 : : : while the thermodynamic component is represented by the expression 1 . We will see this pattern repeated again and again, particularly in enzyme rate laws where additional components appear in the form of effector regulation. We can also derive the modified rate law in the following way. Given the net rate of reaction v D vf vr , we can write this expression in the following way:   vr v D vf 1 vf That is: v D vf .1

/

1.4 Kinetics across Membranes In this section let us briefly consider the kinetics of simple membrane diffusion. Consider two compartments A and B as shown in figure 1.6 connected by a thin membrane that allows diffusion of substance from one compartment to another. Compartment A

S1 ; S2 ; : : : ; Si

VA

Compartment B

S1 ; S2 ; : : : ; Si

VB

Figure 1.6: Compartmental Analysis. Two compartment, A and B where volumes VA and VB exchange mass across a membrane.

26

CHAPTER 1. REACTION KINETICS

A membrane that separates two compartments is a two dimensional surface. As a result the kinetic formulation is slightly different compared to bulk solution reaction kinetics. While concentration changes in the bulk solution are often expressed in terms of moles of substance transformed per unit volume per unit time (moles V 1 t 1 ), transport across a membrane is expressed in units of moles per unit area per unit time (moles A 1 t 1 ) and is called the flux, J . According to Fick’s first law of diffusion, the flux is proportional to the concentration gradient across the membrane: JA D

DA

dS dx

The negative sign ensures that the flux is positive when the concentration gradient is negative, that is declining left to right. JA is the flux in units of moles l 2 t 1 (moles per unit area per time), DA the diffusion coefficient has units of l2 t 1 (area per unit time), S is the concentration and dS=dx the concentration gradient in units of moles l 3 l 1 , that is moles per volume per length. If the zone of diffusion has a width ı, we can approximate Fick’s law: JA D

DA

Sout

Sin ı

or

JA D PA .Sin

Sout /

(1.20)

where PA equals DA =ı and is called the permeability coefficient with units of length per unit time (often cm t 1 ). The units of flux at this stage are moles per unit area per unit time (moles cm 2 t 1 ). To obtain the total amount of mass that moves from one compartment to another we must multiply the flux, JA , by the cross-sectional area of the membrane area, thus: J D AJA

1.5. TEMPERATURE DEPENDENCE

27

where J is the total amount of substance crossing the membrane and A the area of the membrane. If this substance is moving into a volume V , then the rate of change of concentration in the compartment is given by: dS D dt

J V

The negative sign indicates that mass is leaving the compartment.

1.5 Temperature Dependence The rates of most chemical reactions increase as the temperature is raised. As a rule of thumb, a typical reaction rate will double for every ten degree Celsius increase. This increase can be measured as a change in the rate constants for the reaction. Some reactions show a more complex response to an increase in temperature. For example, enzyme catalyzed reactions tend to increase in rate but at a certain temperature (often around 43ı C), the reaction rapidly falls as the enzyme denatures. In many cases it has been found that the temperature dependence of a reaction’s rate constant follows the Arrhenius equation: k D Ae

Ea =RT

where k is the reaction rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant and T the temperature. This relation was proposed by the Swedish chemist Svante Arrhenius in 1889. Figure 1.7 shows the rate constant as a function of the activation energy. The lower the Ea the higher the forward and reverse rate constants. By taking natural logarithms on both sides, the equation can be expressed as: ln k D ln A

Ea RT

Thus, a plot of ln k versus 1=T is linear with a slope of Ea =R and the y intercept, ln A. The constants that appear in the Arrhenius equation can be interpreted in terms of collision theory. Chemical reactions occur between two molecules

28

CHAPTER 1. REACTION KINETICS

Rate Constant, k

1 0:8 0:6 0:4 0:2 0

0

2

4

6

8

10

Activation Energy, Ea

Figure 1.7: Reaction rate constant as a function of the activation energy, Ea .

when they collide. However, not all collisions lead to a reaction. In particular, a collision must occur at a sufficiently high energy (and orientation) in order for a reaction to occur. The activation energy, Ea is the minimum kinetic energy that reactants must have during a collision in order to form products while the pre-exponential factor can be interpreted as the rate at which collisions occur per unit time and volume.

Catalysis A catalyst is a substance that can accelerate a chemical reaction without itself being consumed. Catalysts often operate by lowering the activation energy, Ea , of a reaction. This means that a catalyst cannot change the equilibrium constant of the reaction. A catalyst will only accelerate a reaction towards its equilibrium point. Many catalysts are highly specific and in biology the most common catalysts are enzymes. Catalysts work by enabling alternative reaction paths that require less activation energy, for example by changing bond polarity or orientating molecules into a more favorable position.

1.5. TEMPERATURE DEPENDENCE

29

Reaction without a catalyst

Activation energy Energy

Ea

Net energy release

Reaction Extent Figure 1.8: The energy profile of a reaction. The graph shows the potential energy profile as the reaction proceeds. The activation energy is the height of the barrier.

A common feature of catalysts is that the degree of reaction acceleration depends on the concentration of catalyst. This means a catalyst will appear in a rate law as an additional concentration factor that is not part of the stoichiometric component. Given a mass-action rate law, the simplest way to introduce a catalyst, Ei , is as a linear multiplier. For example: v D Ei .k1 A

k2 B/

In this form the catalyst obeys the usual catalytic rules, that is both forward and reverse rates are equally affected and the catalyst appears as an additional concentration factor, independent of the reaction stoichiometry.

Chapter Highlights This chapter provides a brief introduction to chemical kinetics. It is not meant to be comprehensive and omits a number of important topics nor-

30

CHAPTER 1. REACTION KINETICS

Reaction with a catalyst Activation energy Lower activation energies in catalyzed reaction

Energy

Ea

Net energy release

Reaction Extent Figure 1.9: The effect of a catalyst on the activation energy of a reaction.

mally associated with chemical kinetics. For example, no mention is made of how one can distinguishing between first and second order kinetics. Little is provided on experimental methods available to measure rates of change and how such data can be used to determine reaction mechanisms using either log plots or integrated rate laws. What is presented is an introduction to reaction rates, the difference between stoichiometric amounts, stoichiometric coefficients, reaction rates and rates of change. The difference between these terms is subtle and the reader should take note as to their distinct meaning. The stoichiometric amount refers to the number of reactant and product molecules involved in a particular reaction. The stoichiometric coefficient refers to the net stoichiometry of a given species and takes into account whether a species occurs as both a reactant and product. The reaction rate is defined the rate of change of a given species, normalized by the stoichiometric coefficient. Conversely, the rate of change of a particular species is defined as the reaction rate multiplied by the stoichiometric coefficient. Strictly speaking only the rate of change can be directly measured

1.5. TEMPERATURE DEPENDENCE

31

by experiment. Reactions can be classified as elementary or non-elementary. Elementary reactions involve no reaction intermediates and can be classified according to the order of reaction for the species involved. Common reaction orders include zero, first and second-order depending on the power to which the reaction species is raised. Unless there is specific mechanistic information, it is possible to assume as a first approximation, that the rate of an elementary reaction is proportional to the product of each reactant species, raised to its stoichiometric amount. Chemical equilibrium is a fundamental aspect of any chemical reaction and describes the state when the net reaction rate is zero. The equilibrium constant is defined as the ratio of the product species (raised to their stoichiometric amounts) and reactant species at equilibrium. Two convenient measures, called the mass-action ratio and the disequilibrium ratio allow us to easily summarize whether a reaction is at equilibrium, is progressing from left to right, or from right to left. These measures also allow us to modify elementary reaction rate laws by eliminating the reverse rate constant and instead use the more readily available equilibrium constant.

Further Reading 1. Atkins P and de Paula J (2006) Physical Chemistry for the Life Sciences. Oxford University Press, W. H. Freeman and Company, New York. ISBN: 0-7167-8628-1 2. Chang R (2005) Physical Chemistry for the Biosciences. University Science Books. ISBN-10: 1891389335

32

CHAPTER 1. REACTION KINETICS

Exercises 1. Determine the stoichiometric amount and stoichiometric coefficient for each species in the following reactions: A !B ACB !C A !B CC 2A ! B 3A C 4B ! 2C C D ACB !ACC A C 2B ! 3B C C

2. A culture of a newly engineered microorganism that can convert glucose to butyric acid is started with 50 gms of glucose. Calculate the maximum amount of butyric acid that could be produced if all the glucose were consumed. Determine the amount of butyric acid produced and the percentage yield. C6 H12 O6 ! C3 H7 COOH C 2CO2 C 2H2 3. The following table shows data from an experiment that measures the concentration of product over time. Use the data to estimate the average reaction rate.

Time (mins)

Concentration (M)

0 1 2 3 4

0 0.09 0.18 0.27 0.35

1.5. TEMPERATURE DEPENDENCE

33

4. Define the terms: Stoichiometric amount, Stoichiometric coefficient, reaction rate, rate constant 5. Assuming that the following reactions are irreversible elementary reactions, write out the reaction rate laws for each: A !B ACB !C A !B CC 2A ! B

Assuming the following reactions are reversible, write out the reaction rate laws: 3A C 4B 2C C D ACB ACC A C 2B 3B C C

6. A reaction mix starts with an initial concentration of substance A of 200 mM. The reaction is known to follow first-order kinetics. After 45 seconds the concentration of reactant is 100 mM. Assuming that the product has no effect on the reaction rate, estimate the rate constant of the reaction. 7. In the following two reactions what are the rates of change for each species? Assume a reaction rate of 3.5 mol t 1 in each case. (a) A C 2B ! 3C (b) 2A C B ! 3B C A C C 8. What does it mean when we say that a reaction A ! B has a reaction rate of -6 mol t 1 ?

34

CHAPTER 1. REACTION KINETICS

9. In the following reaction, the rate of change of species A was found to be +5.0 mol t 1 . Assume that the reaction is going from left to right. A C 2B ! 2A C C What is the rate of change of the species B and C and what is the rate of reaction? 10. The total concentration of glucose-6-phosphate and fructose-6-phosphate is 4.9 mM. Given that the equilibrium constant between the two species is 0.395, calculate their equilibrium concentrations. 11. For the following reaction, show that the equilibrium constant is the ratio of the forward and reverse rate constants. Assume mass-action kinetics. 3A C 4B ! 2C C D 12. The permeability coefficient of glucose across a lipid membrane is 2  10 4 cm s 1 . The flux across the membrane, left to right, is 5  10 6 mol cm 2 s 1 . If the concentration of glucose to the right of the membrane is 0.1 mM, what is the concentration of glucose on the left side of the membrane? 13. A reaction rate law for a unimolecular and reversible reaction that incorporates a catalyst, E, is given by the equation: v D E.k1 A

k2 B/

where A and B are the reactant and product respectively. Show that the catalyst E has no effect on the equilibrium constant of the reaction.

2

Elasticities

2.1 Introduction and Relevance Elasticities describe how sensitive a reaction rate is to changes in reactant, product and effector concentrations. They represent the degree to which changes are transmitted from the immediate environment of a reaction to the reaction rate. From a systems’ perspective they are critical components

35

36

CHAPTER 2. ELASTICITIES

in understanding how disturbances, such as the introduction of a drug applied at one or more points in a cellular pathway, propagate to the rest of the system. It is the magnitude and signs of elasticities that determine how far and at what strength the disturbance travels. Elasticities are therefore central in helping us understand how networks function. In this book we will not cover the application of elasticities, instead we will focus on the properties of elasticities for particular reaction processes. A subsequent volume will describe in more detail how elasticities can be used to understand cellular networks.

2.2 Elasticity Coefficients A fundamental property of any reaction rate law is the kinetic order, sometimes called the reaction order. For simple mass-action chemical kinetics, the kinetic order is the power to which a species is raised in the kinetic rate law. Reactions with zero-order, first-order and second-order are commonly found in chemistry, and in each case the kinetic order is zero, one and two, respectively. For a reaction such as: 2H2 C O2 ! 2H2 O where the irreversible mass-action rate law is given by: v D k H22
O2 the kinetic order with respect to hydrogen is two and oxygen one. In this case the kinetic order also corresponds to the stoichiometric amount of each molecule although this may not always be true. It is possible to generalize the concept of the kinetic order by defining it as the scaled derivative of the reaction rate with respect to the species concentration, as follows:

"vSi

 D

@v Si @Si v

 D Sj ;Sk ;:::

@ ln v  v%=Si % @ ln Si

(2.1)

2.2. ELASTICITY COEFFICIENTS

37

From the definition it is apparent that elasticities are dimensionless quantities. When expressed this way, the kinetic order is often called the elasticity coefficient or in biochemical systems theory, the apparent kinetic order. The subscripts, Sj ; Sk ; : : : in definition (2.1) indicate that any species affecting the reaction must be held constant at their current value when species Si is changed. This is also implied in the use of the partial derivative symbol, @, rather than the derivative symbol, d . Elasticities can be derived from rate laws by differentiating and scaling the rate law equation ( 2.1). Example 2.1 Determine the elasticities for the following mass-action rate laws: (a) v D k Elasticity: "Av D

@v A D0 @A v

(b) v D kA Elasticity: "Av D

@v A Ak D D1 @A v kA

(c) v D kA2 Elasticity: "Av D

@v A 2kAA D D2 @A v kA2

(d) v D kAn Elasticity: "Av D

@v A nkAn 1 A D Dn @A v kAn

Example (2.1) shows that the elasticity corresponds to the expected kinetic order for simple rate laws. The definition of the elasticity (2.1) also gives us a useful operational interpretation. Operational Definition: The elasticity is the fractional change in reaction rate in response to a fractional change in a given reactant or product while keeping all other reactants and products constant.

38

CHAPTER 2. ELASTICITIES

That is, the elasticity measures how responsive a reaction is to changes in its immediate environment. Since the elasticity is expressed in terms of fractional changes, it is also possible to get an approximate value for the elasticity by considering percentage changes. For example, if we increase the substrate concentration of a particular reaction by 2% and the reaction rate increases by 1:5%, then the elasticity is given by 1:5=2 D 0:75. The elasticity is however only strictly defined (See equation (2.1)) for infinitesimal changes and not finite percentage changes. So long as the changes are small, the finite approximation is a good estimate for the true elasticity. For a given reaction, there will be as many elasticity coefficients as there are reactants, products and other effectors of the reaction. For species that cause reaction rates to increase, the elasticity is positive, while for species that cause the reaction rate to decrease, the elasticity is negative. Therefore, reactants generally have positive elasticities and products generally have negative elasticities (Figure 2.2). Example 2.2 How many elasticities are there for the following mass-action reactions: a) A ! B v There are two elasticities, "Av which will be positive and "B which will be negative.

b) 2A C B ! 3C v There are there elasticities, "Av which will be positive, "B which will also be posiv tive and "C which will be negative.

There are different ways to calculate an elasticity including numerical, algebraic, and experimental. The numerical and algebraic methods rely on knowing the reaction rate law. We saw in example (2.1) how elasticities were computed algebraically. Numerically the elasticity can by estimated by making a small change (say 5%) to the chosen reactant concentration and measuring the change in the reaction rate. For example, assume that the reference reaction rate is vo , and the reference reactant concentration, So . If we increase the reactant concentration by
So and observe the new reaction rate at v1 , then the elasticity can be estimated by using Newton’s

2.2. ELASTICITY COEFFICIENTS

39

difference quotient: "vS

v1 vo S o v1 vo ' D
So vo vo



S1

So So

Newton’s quotient method relies on making one perturbation to So . A much better estimate for the elasticity can be obtained by doing two separate perturbations in So . One perturbation to increase So and another to decrease So . In each case the new reaction rate is recorded; this is called the three-point estimation method. For example if v1 is the reaction rate when we increase So , and v2 is the reaction rate when we decrease So , then we can use the following three-point formula to estimate the elasticity:   1 v1 v2 S o "vS ' 2 S1 So vo

Example 2.3 Estimate the elasticity using Newton’s difference quotient and the three-point estimation method. Compare the results with the exact value derived algebraically: Let v D S=.0:5 C S /. Assume S is 0.6. a) Algebraic Evaluation Differentiation and scaling the rate law gives the elasticity as 0:5=.0:5 C S /. At a value of 0.6 for S , the exact value for the elasticity is: 0.4546 b) Difference Quotient Let us use a step size of 5%. Therefore h D 0:05  0:6 D 0:03 from which S1 D 0:63. So D 0:6. From these values we can compute v1 and vo . vo D 0:6=.0:5 C 0:6/ D 0:5454, v1 D 0:63=.0:5 C 0:63/ D 0:5575. From these values the estimated elasticity is given by: "vS D ..0:5575

0:5454/=0:5454/ = ..0:63

0:6/=0:6/ D 0:443

Compared to the exact value the error is 0.0116, or 2.55 % error c) Three-Point Estimation In addition to calculating v1 in the last example, we must also compute v2 . To do this we subtract h from So to give v2 D 0:533. The Three-Point estimation formula gives us: "vS D 0:5

0:5575 0:5327 0:6 0:03 0:5454

D 0:4549

40

CHAPTER 2. ELASTICITIES

Compared to the exact value the error is only 0.0033, or 0.7 % error, a significant improvement over the difference quotient method. The degree of error in the difference quotient method will depend on the value of S , which in turn determines the degree of curvature (or nonlinearity) at the chosen point. The more curvature there is the more inaccurate the estimate. The value in this example was chosen where the curvature is high, therefore the error was larger.

In the examples shown in (2.1) the elasticities were constant values. However for more complex rate law expressions this need not be the case (See Example (2.4)) and the elasticity will change in response to changes in the reactant and product concentrations. Consequently when measuring the elasticity numerically or experimentally one has to choose a particular operating point. B.

A. 2

1 0:8

0

@v @S

v

ln v

0:6 0:4

"vS

D

@v S @S v

0

20

@ ln v @ ln S

2 4

0:2 0

"vS D

40

6

S

4

2

0 ln S

2

4

Figure 2.1: A. The slope of the reaction rate versus the reactant concentration scaled by both the reactant concentration and reaction rate yields the elasticity, "vS . B. If the log of the reaction rate and log of the reactant concentration are plotted, the elasticity can be read directly from the slope of the curve. Curves are generated by assuming v D S=.2 C S /. Example 2.4 Determine the elasticities for the following rate laws:

2.2. ELASTICITY COEFFICIENTS

41

(a) v D k.A C 1/ Elasticity: "Av D

@v A A A Dk D @A v k.A C 1/ .A C 1/

(b) v D k=.A C 1/ Elasticity: "Av D

@v A D @A v

k A D .1 C A/2 k=.A C 1/

A AC1

(c) v D A=.A C 1/ Elasticity: "Av D

@v A A 1 1 D D @A v .A C 1/2 A=.A C 1/ AC1

(d) v D kA.A C 1/ Elasticity: "Av D

@v A A A D k.1 C 2A/ D1C @A v kA.A C 1/ AC1

The examples illustrate that for more complex rate laws, the elasticity becomes a function of the reactant concentrations.

Experimentally, we can measure an elasticity using the following experiment. Consider a simple reaction such as A ! B and let us measure the elasticity of reaction A. We must first select an operating point for A and B. This choice will depend on the system under study. For example, perhaps we are interested in the value of the substrate elasticity for an enzyme catalyzed reaction when the substrate and product concentration are at their Km levels. Once the operating point has been chosen, the reaction is started and the rate of reaction is measured. It is important that during the measurement only a small amount of substrate is consumed and product produced. We now begin the experiment again but this time the substrate concentration is increased by a small amount and the product concentration is reset to its value in the first experiment. The reaction is started and the new reaction rate measured. The fractional change in reaction rate and substrate is recorded and the ratio computed to give the substrate elasticity. In principle the same kind of experiment could be performed on the product, this time keeping the substrate concentration constant.

42

CHAPTER 2. ELASTICITIES

Simple protocol for estimating the substrate elasticity 1. Set substrate and product concentrations to their operating points. 2. Record the reaction rate at the operating point. 3. Restore all concentrations to their original starting points. 4. Increase the concentration of substrate by a small amount. 5. Record the new reaction rate. 6. Compute the elasticity by dividing the fractional change in reaction rate by the fractional change in substrate concentration. 7. At all times, maintain other substrate, product and effector concentrations at the operating point.

A.

B.

1

1

0:8

0:8 @v @S

>0

@v @P

0:6

<0

v

v

0:6 0:4

"vS D

0:2 0

0:4

@v S @S v

Product constant

0

20

40 S

0:2 0

Substrate constant

10

20 P

30

Figure 2.2: A. Reaction rate versus reactant. Increases in the reactant cause an increase in the rate. A positive slope yields a positive elasticity. B. Reaction rate versus product (assuming a positive rate from reactant to product). Increases in product result in a decrease in reaction rate; a negative slope yields a negative elasticity. Curves generated by assuming v D S=.2 C S / and 2=.1 C .0:1 C 0:2P //, respectively.

40

2.2. ELASTICITY COEFFICIENTS

43

Log Form The definition of the elasticity in equation (2.1) shows the elasticity expressed using a log notation: "vS D

@ ln v @ ln S

This notation is frequently used in the literature. The right panel of Figure 2.1 shows one application of this notation, namely that a log-log plot of reaction rate versus reactant concentration yields a curve where the elasticity can be read directly from the slope. The origin of this notation will be explained here. Consider a variable y to be some function f .x/, that is y D f .x/. If x increases from x to .x C h/ then the change in the value of y will be given by f .x C h/ f .x/. The proportional change however, is given by: f .x C h/ f .x/ f .x/ The rate of proportional change at the point x is given by the above expression divided by the step change in the x value, namely h: Rate of proportional change = 1 f .x C h/ f .x C h/ f .x/ D lim hf .x/ f .x/ h!0 h h!0 lim

f .x/

D

1 dy y dx

From calculus we know that d ln y=dx D 1=ydy=dx, therefore the rate of proportional change equals: d ln y dx and serves as a measure of the rate of proportional change of the function y. Just as dy=dx measures the gradient of the curve, y D f .x/ plotted on a linear scale, d ln y=dx measures the slope of the curve when plotted on a semi-logarithmic scale, that is the rate of proportional change. For example, a value of 0.05 means that the curve increases at 5% per unit x.

44

CHAPTER 2. ELASTICITIES

We can apply the same argument to the case when we plot a function on both x and y logarithmic scales. In such a case, the following result is true: d ln y x dy D d ln x y dx

2.3 Mass-action Kinetics Computing the elasticities for mass-action kinetics is straight forward. For a reaction such as v D kS , we showed earlier (2.1) that "vS D 1. For a generalized irreversible mass-action law such as: Y n vDk Si i the elasticity for species Si is ni . For simple mass-action kinetic reactions, the kinetic order and elasticity are therefore identical and independent of species concentration. For a simple irreversible mass-action reaction rate law such as: v D k1 S

k2 P

(2.2)

The elasticities for the substrate and product are given by:

"vS D

vf k1 S D k1 S k2 P v

v "P D

k2 P D k1 S k2 P

(2.3) vr v

(2.4)

In the above equations vf is the forward rate, vr is the reverse rate, and v v is the net rate. Note that "vS is positive and "P negative. In general, the elasticity for an effector that results in an increase in reaction rate will be positive and conversely if the effector results in a decrease in the reaction rate.

2.3. MASS-ACTION KINETICS

45

If we divide top and bottom by k1 and S in equation (2.3), and k2 and P in equation (2.4), and noting that the ratio k1 =k2 D Keq (See equation (1.12)), P =S D and =Keq D  we can express the elasticities in the form:

"vS D v "P

D

1 1 D =Keq 1 

1 1

=Keq D =Keq

(2.5)

 1



These expressions can vary over a wide range of values. Far from equilibv rium ( ' 0) "vS will lie close to 1:0, while "P will be close to 0:0. When operating close to equilibrium however (  1), the same elasticities will tend to C1 and 1, respectively. This behavior is depicted in Figure 2.3. Of interest is the following relation (See equation (1.17)): v "P vr D v D "S vf



which connects the ratio of the elasticities to the disequilibrium ratio. It also follows from the above equations that the sum of the elasticities for mass-action kinetic rate laws is always one:

v "vS C "P D1

(2.6)

This means that if one of the elasticities is known, the other can be easily determined by subtraction. Equation (2.6) is significant for another reason. The absolute magnitude of v "vS will always be larger than the absolute value for "P when dealing with mass-action kinetics. That is:

46

CHAPTER 2. ELASTICITIES

20

Elasticity

10 0 10 20

Substrate Product 0

0:5

1

1:5

2

Disequilibrium Ratio,  Net rate positive

Net rate negative

Figure 2.3: Elasticities as a function of the disequilibrium ratio, .

v k"vS k > k"P k

For small elasticity values the relative difference between the elasticities can be significant. This means that changes in substrate concentrations will have a much greater effect on the reaction velocity than changes in product concentrations. As will be discussed more fully in another book, the propagation of signals along a pathway is determined by the elasticity 

"vS D 1=.1

0.9 0.5 0.2 0.1

10 2 1.2 1.111

/

v "P D

=.1

/

-9 -1 -0.25 -0.111

Table 2.1: Selected values for the elasticities and the disequilibrium ratio, : k"vS k > k"Pv k

2.4. GENERAL ELASTICITY RULES

47

values. Given that substrate elasticities are larger than product elasticities, signal propagation tends to amplify when traveling downstream compared to signals traveling upstream which tend to be attenuated. For the general reversible mass-action rate law:

v D k1

Y

Sini

k2

Y

Pimi

(2.7)

The elasticities can be shown to equal:

"vSi D

ni 1

v "P D i

 mi  1 

(2.8)

2.4 General Elasticity Rules Just as there are rules for differentiating equations, there are similar rules for computing elasticities. These rules can be used to simplify the derivation of elasticities for more complex rate law expressions. Table 2.2 shows some common elasticity rules, where a designates a constant and x the variable. For example the first rule says that the elasticity of a constant is zero. We can illustrate the use of these rules with a simple example. Consider the reversible mass-action rate law (2.2): v D k1 S

k2 P

To determine the elasticity we first apply rule 3 to give: "vS D "S .k1 S /

k1 S k1 S k2 P

"S .k2 P /

k2 P k1 S k2 P

48

CHAPTER 2. ELASTICITIES

1.

".a/ D 0

2.

".x/ D 1

3.

".f .x/ ˙ g.x// D ".f .x//

4.

".x a / D a

5.

".f .x/a / D a".f .x//

6.

".f .x/ g.x// D ".f .x// C ".g.x//

7.

".f .x/=g.x// D ".f .x//

f .x/ f .x/Cg.x/

˙ ".g.x//

g.x/ f .x/Cg.x/

".g.x//

Table 2.2: Transformation rules for determining the elasticity of a function, a D constant, x D variable.

where "S .f / means the elasticity of expression f with respect to variable S. Now transform the elasticity terms by applying additional rules. Let use apply rule 6 to the expression "S .k1 S / to give: "S .k1 S / D "S .k1 / C "S .S / We can now apply rule 1 to the first term on the right and rule 2 to the second term on the right to give: "S .k1 S / D 0 C 1 Since we’re evaluating the elasticity of S , P in this situation is a constant, therefore: "S .k2 P / D "S .k2 / C "S .P / D 0 C 0 Combining these results yields: "vS D

k1 S k1 S k2 P

which corresponds to the first equation in (2.3). Now consider a simple enzyme kinetic rate equation. One of the most famous is the Michaelis-Menten equation: vD

Vm S Km C S

2.4. GENERAL ELASTICITY RULES

49

where Vm is the maximal velocity and Mm the substrate concentration at half maximal velocity. The elasticity for this equation can be derived by first using the quotient rule (rule 7) which gives: "vS D ".Vm S /

".Km C S /

The rules can now be applied to each of the sub-elasticity terms. For example we can apply rule 6 to the first term, ".Vm S /, and rule 3 to the second term, ".Km C S /, to yield:   Km S "vS D .".Vm / C ".S // ".Km / C ".S / Km C S Km C S Applying rules 1 and 2 allows us to simplify (".Vm / D 0I ".Km/ D 0I ".S/ D 1) the equation to:   S v "S D 1 Km C S or "vS D

Km Km C S

Example 2.5 Determine the elasticity expression for the rate laws using log-log rules: (a) v D k.A C 1/ Begin with the product rule 6: "Av D ".k/ C ".A C 1/ D ".A C 1/ Next use the summation rule 3 and rule 2: "Av D ".A C 1/ D ".A/ D

A 1 C ".1/ AC1 AC1

A A C0D AC1 AC1

50

CHAPTER 2. ELASTICITIES

(b) v D k=.A C 1/ Begin with the quotient rule 6 followed by Rule 3 and 2:

"Av D ".k/ D

".A C 1/ D

1 AC1

A AC1

(c) v D A.A C 1/ Begin with the quotient rule 6: "Av D ".A/ C "A C 1 Next use Rule 2, 3 and 2: "Av D 1 C

1 AC1

To make matters even simpler we can define the elasticity rules using an algebraic manipulation tool such as Mathematica (http://www.wolfram. com/) to automatically derive the elasticities [72]. To do this we must first enter the rules in Table 2.2 into Mathematica. The script shown in Figure 2.4 shows the same rules (with a few additional ones) in Mathematica format. The notation f[x_,y_] := g() means define a function that takes two arguments, x_ and y_. The underscore character in the argument terms is essential. Note also the symbol ‘:’ in the assignment operator. Typing el[k1 S - k2 P, S] into Mathematica will result in the output:

k1 S/(-k2 P + k1 S)

Chapter Highlights  Elasticities are dimensional quantities that describe how fractional changes in reactants, products or effectors result in fractional changes in a reaction rate.

2.4. GENERAL ELASTICITY RULES

51

(* Define elasticity evaluation rules *) el[x_, x_] := 1 el[k_, x_] := 0 el[Log[u_,x_] := el[Log[u],x] = el[u,x]/Log[u] el[Sin[u_],x_] := el[Sin[u],x] = u el[u,x]Cos[u]/Sin[u] el[Cos[u_],x_] := el[Sin[u],x] = -u el[u,x]Sin[u]/Cos[u] el[u_*v_,x_] := el[u*v,x] = el[u,x] + el[v,x] el[u_/v_,x_] := el[u/v,x] = el[u,x] - el[v,x] el[u_+v_,x_] := el[u+v,x] = el[u,x]u/(u+v) + el[v,x]v/(u+v) el[u_-v_,x_] := el[u-v,x] = el[u,x]u/(u-v) - el[v,x]v/(u-v) el[u_^v_,x_] := el[u^v,x] = v (el[u,x] + el[v,x] Log[u]) Figure 2.4: Elasticity rules expressed as a Mathematica script.

 If we assume that the direction of a reaction is from reactant to product, then in general, reactants will have positive elasticities and products negative elasticities.  Inhibitors of a reaction will have negative elasticities since they slow down a reaction.  Activators of a reaction will have positive elasticities, since they speed up a reaction.  If a reaction is nth-order with respect to a particular substance, then the elasticity is n.  For mass-action kinetics the sum of the substrate and product elasticity is one.  For mass-action kinetic laws with unit stoichiometry, the relation, v k"vS k > k"P k is true.

Further Reading 1. Fell D A (1996) Understanding the Control of Metabolism. Portland Press, ISBN: 185578047X

52

CHAPTER 2. ELASTICITIES

2. Heinrich R and Schuster S (1996) The Regulation Of Cellular Systems. Springer; 1st edition, ISBN: 0412032619

Exercises 1. What is the relevance of elasticity coefficients in understanding network dynamics? 2. State the operational interpretation of an elasticity. 3. An experiment indicates that a given molecule X has an elasticity of -0.5 with respect to the rate of a reaction. State two key aspects that this elasticity describes. 4. What is the elasticity with respect to the species A for the rate law v D kA3 ? 5. Derive the elasticity expression with respect to x for the following equations: a) v D x 2 C 1 b) v D x 2 C x c) v D x=.x 2 C 1/ 6. Describe one technique for numerically estimating the value of an elasticity. 7. What does the term

=Keq measure?

8. Describe the value of disequilibrium ratio as a reaction nears equilibrium. 9. For a simple mass-action reversible reaction, describe what happens to the substrate and product elasticities as the reaction approaches equilibrium. 10. Derive the two equations in (2.5). 11. Describe the significance of equation (2.6). 12. Using the elasticity rules in Table 2.2, derive the elasticity for the following equation indicating all intermediate steps. v D S n =.Km C S n /

3

Basic Enzyme Kinetics

53

54

CHAPTER 3. BASIC ENZYME KINETICS

3.1 Enzyme Catalysts As discussed in chapter 1, a catalyst accelerates a chemical reaction without itself being consumed and without it changing the equilibrium constant (Keq ) of the reaction. Catalysts only speed up the approach to equilibrium. The most important catalysts in living systems are enzymes. Enzymes are protein molecules which fold into a specific three dimensional structure to allow interaction with a substrate at a location on the enzyme called the active site. Like all catalysts, enzymes equally accelerate the rate of the forward and backward reaction without being consumed in the process. They also tend to be very selective, with a particular enzyme accelerating only a specific reaction.

3.2 Enzyme Kinetics Enzyme kinetics is a branch of science that deals with the many factors that can affect the rate of an enzyme-catalysed reaction. The most important factors include the concentration of enzyme, reactants, products, and the concentration of any modifiers such as specific activators, inhibitors, pH, ionic strength, and temperature. When the action of these factors is studied, we can deduce the kinetic mechanism of the reaction. That is, the order in which substrates and products bind and unbind and the mechanism by which modifiers alter the reaction rate. The standard model for enzyme action, first suggested by Brown and Henri [11] but later established more thoroughly by Michaelis and Menten in 1913, describes the binding of free enzyme to the reactant forming an enzyme-reactant complex. This complex undergoes a transformation, releasing product and free enzyme. The free enzyme is then available for another round of binding to new reactant. Traditionally, the reactant molecule that binds to the enzyme is called the substrate, S , and the mechanism is often written as:

3.3. BASIC ENZYME KINETICS

k1

55

k2

E C S )* ES ! E C P k

(3.1)

1

where k1 ; k 1 and k2 are rate constants, S is substrate, P is product, E is the free enzyme, and ES the enzyme-substrate complex. Note that in this model substrate binding is reversible but product release is not. This model is often used when carrying out in vitro kinetic assays because under these conditions it is assumed that the product has a negligible concentration and therefore the reverse rate is zero. Unlike in vitro conditions where we can arrange for product to be absent in our assays, in vivo is a different matter. In the cell, product will always be present, if product weren’t available then the next enzyme in a metabolic sequence would have no substrate to bind. Most reactions will therefore always show some degree of reversibility which leads to the more general model: k1 k2 E C S )* ES )* E C P (3.2) k

1

k

2

It is possible to model enzymes using the explicit mechanisms shown in (3.2), however the rate constants for the binding and unbinding events are often unknown or difficult to determine. Instead, assumptions are made about the dynamics of the mechanism which reduces the number of constants required to characterize the enzyme. This leads to a discussion of aggregate rates laws, the most celebrated being Michaelis-Menten kinetics.

3.3 Basic Enzyme Kinetics In practice we rarely build models using explicit elementary reactions unless it is absolutely necessary to capture a particular type of dynamic behavior. Aside from the increase in complexity, rate constants for the elementary reactions are in most cases unknown. Instead we often use approximations of which there are two types. First, let us consider the irre-

56

CHAPTER 3. BASIC ENZYME KINETICS

versible mechanism for enzyme action: k1

k2

E C S )* ES ! E C P k

(3.3)

1

Thermodynamic Equilibrium and Steady State Thermodynamic equilibrium, or equilibrium for short and the steady state are distinct states of a chemical system. In equilibrium, both the rate of change of species and the net flow of mass through the system is zero. That is: dS D0 dt for all i : vi D 0 where vi is the net reaction rate for the i th reaction step. At equilibrium there is therefore no dissipation of gradients or energy fields. When a biological system is at equilibrium, we say it is dead. The steady state has some similarities with equilibrium except there is a net flow through the system such that gradients and energy fields are continuously being dissipated. This also means that one or more vi s must be non-zero The steady state is defined when all dSi =dt are equal to zero while one or more reaction rates are non-zero: dS D0 dt vi ¤ 0 In some of the literature the terms equilibrium and steady state are confusingly used to mean the same thing, usually steady state. In this book a strict difference between the two will be maintained. Two different assumptions may be used to reduce this scheme to a simpler formulation. The first, termed rapid equilibrium, was made in the origi-

3.3. BASIC ENZYME KINETICS

57

nal derivation by Michaelis and Menten (1913). They assumed the binding and unbinding of substrate to enzyme was in equilibrium. This assumption also implied that the dissociation into product occurred on a much slower time scale. The second approach was introduced by Briggs and Haldane (1926), called the steady state assumption (Figure 3.1). Rather 10

Concentration

8

P

6

S

4 E 2 ES 0

0

0:2

0:4

0:6

0:8 Time

1

1:2

1:4

Figure 3.1: Progress curves for a simple irreversible enzyme catalyzed reaction (3.3). Initial substrate concentration is set at 10 units. The enzyme concentration is set to an initial concentration of 1 unit (E and ES curves are scaled by two in order to make the changes in E and ES easier to visualize). In the central portion of the plot one can observe the relatively steady concentrations of ES and E (d ES=dt  0). At the same time, the rate of change of S and P are constant over this period. k1 D 20; k

20

1

10

* ES ! E C P D 1; k2 D 10, that is: E C S ) 1

than assume a state of equilibrium, Briggs and Haldane assumed that the enzyme substrate complex rapidly reached steady state. This is less restrictive than the rapid equilibrium assumption. Enzyme rate laws can be derived using either assumption, however, because the mathematics is more tedious, many complex mechanisms such as those describing cooperativity and gene expression are derived using the simpler rapid equilibrium assumption. For this reason we will briefly review the rapid equilibrium

58

CHAPTER 3. BASIC ENZYME KINETICS

derivation. Rapid Equilibrium Assumption. Let Kd be the dissociation constant for the binding reaction of substrate to free enzyme: k1

E C S )* ES k

1

That is, we can write Kd as follows: Kd D

E
S k 1 D k1 ES

We note that the total concentration of enzyme, E t , is the sum of free enzyme, E and enzyme substrate complex, ES: E t D E C ES. It is now straight forward to show that the equilibrium concentration of ES is given by: Et
S (3.4) ES D Kd C S Since the rate of reaction is determined by the rate of product release, we can write down the rate of reaction as v D k2 ES . Combining this with the previous relation for ES, and setting E t k2 D Vm , yields our result:

vD

Vm S Kd C S

(3.5)

The above expression (3.5) is called the Michaelis-Menten equation. A slightly different way to look at the derivation is to consider the specific states in which we can find the enzyme. First note that the rate of reaction, v, is given by v D k2 ES. Now form the sum all possible enzymatic states, that is E t D E C ES . Next we write out the ratio of the reaction rate, v, to the total enzyme concentration, E t , thus: k2 ES v D Et E C ES Using the dissociation constants for the inter-conversions of the various enzyme states, eliminate all bound enzyme states from the equation. In this case we only have one state to eliminate, ES . Since ES D .E
S /=Kd :

3.3. BASIC ENZYME KINETICS

59

v k2 .E
S /=Kd D Et E C .E
S /=Kd and so v k2 S=Kd D Et 1 C S=Kd which is identical to equation (3.4). To give another example, consider the case where the enzyme binds to a second molecule, M (Figure 4.2). The binding of M leads to the formation of the complex EM to which the substrate S cannot bind. Molecule M thus acts as an inhibitor by competing with the substrate. The full set of enzyme states now includes, E, ES and EM . The total amount of enzyme is therefore, E t D E C ES C EM . The rate of reaction is, as before, v D k2 ES . However we now have two dissociation constants, K1 and K2 . K1 corresponds to the dissociation of ES and K2 to the dissociation of EM . The ratio of reaction rate to the total enzyme concentration is: v k2 ES D Et E C ES C EM

(3.6)

Using the dissociation constants to eliminate ES and EM we obtain: v k2 .E
S /=K1 D Et E C .E
S /=K1 C .E
M /=K2 and so

v k2 S=K1 D Et 1 C S=K1 C M=K2

(3.7)

This approach can be carried out systematically and is easily automated by computer. Moreover it also gives us a physical interpretation of the denominator. The denominator represents the distribution of enzyme states among the different forms. In addition, the sub-expression from equation (3.7):

60

CHAPTER 3. BASIC ENZYME KINETICS

fs D

v S=K1 D E t k2 1 C S=K1 C M=K2

(3.8)

represents the fractional saturation of the enzyme in terms of states that lead to product formation. To summarize, the derivation by the rapid equilibrium assumption involves finding the ratio of the active states to the sum of all states. This approach is used repeatedly when considering cooperativity and gene regulation kinetics. Steady State Assumption. Instead of assuming rapid equilibrium, let us follow the approach suggested by Briggs and Haldane by assuming that the enzyme-substrate complex quickly reaches steady steady. Figure 3.1 shows progress curves illustrating the changes in concentrations for the different enzymatic species. Note that the concentration of the enzyme substrate complex rapidly approaches a steady state and remains there until the substrate level has depleted. The rate of change of the enzyme substrate complex (3.1) can be expressed using mass-action laws: dES D k1 E
S dt

k

1

ES

k2 ES

If the concentration of enzyme substrate complex is assumed to rapidly reach steady state (Figure 3.1), then the above equation can be set equal to zero: 0 D k1 E
S k 1 ES k2 ES We also note the total concentration of enzyme, E t , is the sum of free enzyme, E and enzyme substrate complex, ES: E t D E C ES All the above equations, including the rapid-equilibrium equations, assume that S represents the free substrate concentration, Sfree . However we know that some substrate is complexed with enzyme (ES ) so the total amount of substrate can be expressed as S t D Sfree C ES (Assuming P D 0). To simplify the algebra let us assume that S t  ES

3.3. BASIC ENZYME KINETICS

61

so that Sfree ' S t . This simplification also ensures that the steady state assumption is valid. Specifically, the steady state assumption is reasonable if So C Km  Eo [63], where So and Eo are the initial amounts of substrate and enzyme, respectively. From these relationships, the steady state concentration of enzyme substrate complex can be derived: ES D

Et
S .k 1 C k2 /=k1 C S

(3.9)

under the assumption that S is Sfree . By assuming that the rate of reaction is given by v D k2 ES, we obtain: vD

E t k2 S .k 1 C k2 /=k1 C S

(3.10)

where k2 is also known as the catalytic constant, often denoted, kcat . We now condense terms by defining E t k2 to equal Vm and .k 1 C k2 /=k1 to equal a constant we call the Michaelis constant, or Km . We can therefore rewrite (3.10) as:

vD

Vm S Km C S

(3.11)

If we assume that the dissociation of ES into product is slow such that k2 is much smaller than k 1 , then the expression .k 1 C k2 /=k1 approaches Kd , the dissociation constant: that is Km ' Kd . Under these conditions the steady state approach is equivalent to the rapid equilibrium approach. What about the shape of the curve described by equation (3.11)? There are two possible curves we can plot. The easiest is to show how the initial reaction rate behaves as a function of the enzyme concentration, E at fixed substrate concentration. In the experiment we assume that the concentration of enzyme is much smaller compared to the substrate concentration so that as we increase the enzyme concentration the free substrate concentration barely changes as a result of sequestration by the free enzyme into

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CHAPTER 3. BASIC ENZYME KINETICS

Initial Reaction Rate, v

1 0:8 0:6 0:4 0:2 0

0

0:5

1

1:5

2

2:5

3

Enzyme Concentration (E) Figure 3.2: Relationship between the initial rate of reaction and the enzyme concentration for a simple Michaelis-Menten rate law, Km D 4I S D 2.

enzyme-complex. Figure 3.2 shows the expected plot of initial reaction rate versus enzyme concentration. The response is clearly linear. The curve described by equation (3.11) with respect to the substrate concentration is a rectangular hyperbola (Figure. 3.3) that goes through the origin. Kinetic responses governed by Michaelis-Menten equations are therefore often referred to as hyperbolic. In mathematical texts, rectangular hyperbola are represented by the equation xy D c 2 which at first glance bears little resemblance to equation (3.11). If we substitute x D Km C S, y D Vm v and c 2 D Km Vm we can rearrange the equation xy D c 2 into the familiar form (3.11). Hence equation (3.11) represents a rectangular hyperbola [21, 8]. We can also express the degree of saturation with respect to ES by dividing both sides of (3.9) by E t to yield (cf. 3.8): fractional saturation D

ES S D Et Km C S

(3.12)

If we set the reaction velocity to half the Vm , one can easily show that Km is the substrate concentration that gives half the maximal rate (Figure. 3.3).

3.3. BASIC ENZYME KINETICS

Initial Reaction Rate, v

Vm

63

1 0:8 0:6

1 2v

0:4 0:2 0

0

5

10

Km

15

20

25

30

Substrate Concentration (S )

Figure 3.3: Relationship between the initial rate of reaction and substrate concentration for a simple Michaelis-Menten rate law. The reaction rate reaches a limiting value called the Vm . Km is set to 4.0 and Vm to 1.0. The Km value is the substrate concentration that gives half the maximal rate.

Catalytic Efficiency The rate an enzyme can catalyze a given reaction is ultimately limited by the rate at which substrates and products diffuse to and from the enzyme’s active site. In water, the rate constant for the diffusion of small molecules [71] is approximately 109 M 1 sec 1 . In the absence of product, at low substrate concentration, S  Km the reaction velocity becomes: vD

kcat Et S Km

Where the initial slope of the curve is given by: kcat =Km

(3.13)

This constant has the same units as the maximum diffusion rate. In theory,

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CHAPTER 3. BASIC ENZYME KINETICS

the value for this constant cannot exceed the diffusion rate of 109 M 1 sec 1 . As a result, this ratio is often called the catalytic efficiency of the enzyme. We can get an idea of how close an enzyme is to the maximum catalytic efficiency by comparing the enzyme’s catalytic efficiency to the diffusion limit. For example, triose phosphate isomerase has a catalytic efficiency of 2:4  108 M 1 sec 1 which is close to the diffusion limit of 109 M 1 sec 1 . Triose phosphate isomerase is therefore considered to be a very catalytically active enzyme.

Elasticities In chapter 2, elasticities were introduced as a generalized means for evaluating the kinetic order of a given reaction. It was also noted that elasticities measure the degree to which changes in reactant concentrations affect reaction rates (3.14). Elasticities are therefore very useful for understanding how small amplitude signals propagate through biochemical pathways and in particular, how signals are propagated by enzymes.   @v Si @ ln v "vSi D D  v%=Si % (3.14) @Si v Sj ;Sk ;::: @ ln Si Consider the expression for the simple irreversible Michaelis-Menten rate law (3.11). In this equation there are two ways in which signals can affect the reaction rate; either by changes in the substrate concentration or changes in the enzyme level. Substrate Elasticity To compute the substrate concentration elasticity, "vS , first differentiate the rate equation and then scale by S and v. The derivative of the simple Michaelis-Menten rate law is given by: @v Vm Km D @S .Km C S /2 Scaling yields:

3.3. BASIC ENZYME KINETICS

"vS D

65

@v S Km D @S v Km C S

A. Reaction velocity

B.

(3.15)

Elasticity

1 First-Order

0:8 @v S @S v

@v @S

0:6 0:4 0:2 0

Zero-Order

0

20

40

0

S

20

40 S

Figure 3.4: A. Left panel: the reaction velocity for an irreversible Michaelis-Menten rate law as a function of substrate concentration. The curve is also marked by the slope @v=@S. B. Right panel: the substrate elasticity is plotted as a function of substrate concentration. Km D 4 and Vm D 1. Note the elasticity starts at one, then decreases to zero as S increases.

The substrate elasticity shows a range of values (Figure 3.4) from zero at high substrate concentrations to one at low substrate concentrations. When the enzyme is near saturation it is naturally unresponsive to further changes in substrate concentration, hence the elasticity is near zero. The reaction behaves as a zero-order reaction at this point. When the elasticity is close to one at low S, the reaction behaves with first-order kinetics. In addition, the reaction order changes depending on the substrate concentration. We can also express the elasticity in terms of the degree of saturation. Recall that the degree of saturation is given by equation (3.12): ES S D E C ES Km C S

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CHAPTER 3. BASIC ENZYME KINETICS

A slight rearrangement of equation (3.15) allows us to write the elasticity in the following form: v Vm

"vS D 1

S D1 Km C S

"vS D 1

fractional saturation

so that:

(3.16)

The elasticity is therefore inversely related to the fractional saturation. If we know the degree of fractional saturation, we can estimate the elasticity. Enzyme Elasticity. We can also compute the elasticity with respect to enzyme concentration since in vivo enzyme concentrations can change. Given that the Vm D E t kcat the enzyme elasticity is derived as follows: @v k2 Km D @E t Km C S Scaling by E t and v yields k2 S Km C S @v E t D Et D1 @E t v Km C S E t k2 S Hence the enzyme elasticity is one:

"vE D 1

(3.17)

Underlying Assumptions As useful as the previous rate law approximations are, they do have some limitations. One concern is they assume the amount of substrate sequestered

3.4. REVERSIBLE RATE LAWS

67

by the enzyme is negligible compared to free substrate. In vivo this assumption may not necessarily be true where enzyme concentrations can be comparable to substrate concentrations. The presence of high enzymesubstrate complex levels compared to free substrate can add buffering effects. When this means is that instead of assuming that the enzyme-substrate complex very rapidly reaches steady state, there is now a significant time delay. The Michaelis equation assumes that the time taken to reach steady state means that a change in substrate concentration leads to an instantaneous change in reaction velocity. In the presence of high levels of enzyme, this is no longer the case which means that models that use Michaelis equations at high levels of enzymes cannot accurately predict the time evolution of the system. Fortunately the steady state will be largely unaffected except in some rare cases where the stability of the pathway might change. For example, in some situations high levels of sequestration can lead to the onset of oscillatory behavior [15]. Furthermore, there may be cases when pathway wide conservation laws do not take into account the concentration of enzyme-substrate complexes. Under this scenario, perturbations in enzyme levels will lead to sequestration effects that change the apparent conservation total. Given these limitations, one should check whether the use of Michaelis-Menten kinetics or any aggregate rate law has an effect on the model dynamics by comparing the model to one built using explicit mass-action rate laws. This may not always be possible in practice, however.

3.4 Reversible Rate Laws The derivation of the irreversible Michaelis-Menten equation is an instructive exercise, however it is not a very realistic equation to use in biochemical pathway models because there is no explicit product term. In vitro experiments usually measure initial rates when there is little or no product in the assay mix. However, in vivo is a different matter. In a metabolic pathway product must be available for the next enzymatic step. This means there will always be some degree of product inhibition and/or reversibility.

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CHAPTER 3. BASIC ENZYME KINETICS

For pathway modeling, especially metabolic models, reversibility in some form should be explicitly included. The derivation of the reversible Michaelis-Menten equation (3.19) is very similar to the derivation of the irreversible rate law (3.11). The derivation of equation (3.11) is based on the net reaction rate being equal to v D kES, whereas equation (3.19) is based on (3.18), which includes both forward and reverse rates. v D vf

vr D k2 ES

k

2

E
P

(3.18)

The expression describing the steady state concentration of the enzyme substrate complex also has an additional term from the product binding to free enzyme, (k 2 E
P ). Taking these into consideration leads to the general reversible rate expression: vD

Vf S=KS Vr P =KP 1 C S=KS C P =KP

(3.19)

where Vf and Vr are the forward and reverse maximal rates, Ks and Kp are the Km values for the substrate and product, respectively. At equilibrium (P =S D Keq ) the rate of the reversible reaction is zero (v D 0), positive (v > 0) when the reaction is going in the forward direction, and negative (v < 0) in the reverse direction. If we set the product concentration to zero (P D 0), equation (3.19) reverts to the irreversible Michaelis-Menten equation. Equation (3.19) can also be rewritten as: vD

Vf S=KS 1 C S=KS C P =KP

D vf

Vr P =KP 1 C S=KS C P =KP

vr

where vf and vr are the forward and reverse reaction rates. Extracting each of these terms gives us the following equations: vf S=KS D Vf 1 C S=KS C P =KP vr P =KP D Vr 1 C S=KS C P =KP

(3.20)

3.4. REVERSIBLE RATE LAWS

69

Given that we know the reaction rate is proportional to the fractional saturation, vf =Vf is the fractional saturation with respect to S and vr =Vr is the fractional saturation with respect to P . We will use these factors later.

Haldane Equilibrium Relations At equilibrium (v D 0) the reversible Michaelis equation (3.19) reduces to 0 D Vf Seq =KS

Vr Peq =KP

where Seq and Peq represent the equilibrium concentrations of substrate and product. Rearrangement yields another important equation:

Keq D

Vf KP Peq D Seq Vr K S

(3.21)

This expression is known as the Haldane relationship and shows that the four kinetic constants, Vf ; Vr ; KP and KS are not independent. The origin of the Haldane relationship lies with detailed balance, previously discussed in section 1.3. Haldane relationships can be used to eliminate one of the kinetic constants by substituting the equilibrium constant in its place. This is useful because equilibrium constants tend to be known compared to kinetic constants. By incorporating the Haldane relationship we can eliminate the reverse maximal velocity (Vr ) to yield the equation:

vD

Vf =KS .S P =Keq / 1 C S=KS C P =KP

(3.22)

Separating out the terms makes it easier to see that the above equation can be partitioned into a number of distinct terms:

v D Vf
.1

=Keq /


S=Ks 1 C S=KS C P =KP

(3.23)

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CHAPTER 3. BASIC ENZYME KINETICS

where D P =S. The first term, Vf is the maximal velocity; the second term, .1 =Keq / indicates the direction of the reaction according to thermodynamic considerations and the last terms refers to the fractional saturation with respect to substrate. We thus have a maximal velocity, a thermodynamic term and a saturation term. We will see this breakdown into distinct terms repeatedly as we consider other enzyme kinetic rate laws.

Saturation Terms The same expression that describes the fractional saturation of a reversible Michaelis-Menten mechanism is more complex (3.12) but is given by (See also (3.20)) :

fractional saturation D

S=Ks 1 C S=Ks C P =Kp

C

P =Kp 1 C S=Ks C P =Kp

(3.24)

D

vf vr C Vf Vr

(3.25)

Inspection of the saturation term (3.25) reveals that it must satisfy the inequality: P =Kp S=Ks C <1 1 C S=Ks C P =Kp 1 C S=Ks C P =Kp This condition indicates that it is impossible for the individual terms to approach their maximum value of one simultaneously. In practice this simply means that it is physically impossible for the enzyme to be simultaneously saturated by both substrate and product. It also means that S  Ks is not a sufficient condition for the enzyme to be saturated by S if, at the same time, P  Kp .

3.4. REVERSIBLE RATE LAWS

71

The condition S  Ks does not necessarily mean that the enzyme is saturated by substrate.

Elasticities The elasticities associated with the reversible Michaelis-Menten equation: vD

V mf =Ks .S P =Keq / 1 C S=Ks C P =Kp

are given by:

"vS D v "P

D

1

1 =Keq

S=Ks 1 C S=Ks C P =Kp

1

=Keq =Keq

P =Kp 1 C S=Ks C P =Kp

(3.26)

where is the mass-action ratio, P =S . From these equations and assuming a forward reaction rate, the substrate elasticity is positive and the product elasticity negative. The expressions are made from two distinct terms. The first term relates to the thermodynamic properties of the reaction (similar to equation (2.5)) and involves the mass-action ratio and the equilibrium constant. The second term is the degree of enzyme saturation with respect to the substrate or product depending on the particular elasticity. When S and P are much less then both Km s, the elasticities reduce to the elasticities provided in equation (2.5), where the reaction mechanism was assumed to be simple mass-action. Given the identities in (3.20), we can also write the elasticities in the form:

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CHAPTER 3. BASIC ENZYME KINETICS

"vS D v "P

1

1 =Keq

vf Vf

1

=Keq =Keq

vr Vr

D

(3.27)

v Sum of Elasticities "vS C "P

The sum of the two equations (3.27) yields an interesting result. The thermodynamic terms add up to one (2.6) so we are left with the following expression:

v "vS C "P D

1 1 C S=Ks C P =Kp

(3.28)

This relation represents two extremes. At low saturation levels, i.e. when both S=Ks < 1 and P =Kp < 1, the sum tends to one, thus: v "vS C "P 1

(3.29)

This result corresponds to equation (2.6) when the system behaves like a simple mass-action reaction. At high saturation levels, i.e. when S=Ks > 1 and/or P =Kp > 1, the sum tends to zero, thus: v "vS C "P 0

or "vS D

v "P

This indicates the reaction is equally sensitive to S and P . At low saturation levels (Equation (3.29)), the reaction rate is more sensitive to S than to P . From equation (2.6) the sum of the elasticities at low saturation levels is one, therefore:

3.4. REVERSIBLE RATE LAWS

73

v k"vS k > k"P k

It is important to appreciate that elasticities are not independent of each v other. A change in substrate for example will change both "vS and "P .

Near to Equilibrium When an enzyme operates near equilibrium the elasticities are dominated by the thermodynamic terms due to the fact that the denominator gets closer to zero. This makes the degree of saturation less important. This also means that for systems close to equilibrium, knowing the value of Km is not important when calculating elasticities. In fact, the elasticities may be estimated from just thermodynamic data which is widely available including absolute metabolite concentrations [5]. Another consequence of operating near equilibrium is that the absolute values for the elasticities are very large so that the reaction rate is also very sensitive to changes in substrate or product concentration (see Figure 2.3). Example 3.1 a) An in vivo enzymatic reaction has an equilibrium constant of 5.0 with substrate and product concentrations of 1.1 mM and 5 mM, respectively. Estimate the substrate and product elasticity for the reaction in vivo. The thermodynamic contribution to the elasticities is given by the terms T1 D 1=.1 =Keq / and T2 D . =Keq /=.1 =Keq /. Inserting the value for the equilibrium constant and the metabolite concentrations into these equations yields: T1 D 11 T2 D 10 Since the fractional saturation terms in the elasticities can only adjust these thermodynamic estimates by at most -1.0, the estimates will have roughly a 10% error if we ignore the degree of saturation. The thermodynamic estimates for the elasticities are therefore reasonable. b) The same reaction in a different metabolic state has been found to have substrate and product concentrations of 4 mM and 5 mM, respectively. Estimate the substrate and product elasticity for the reaction in this new state.

74

CHAPTER 3. BASIC ENZYME KINETICS

First calculate the thermodynamic terms, T1 and T2 using the Keq and metabolite concentrations: T1 D 1:33

T2 D

0:33

These new estimates are much lower and could significantly change if we include the fractional saturation information. For example, if the substrate Km D 1:5 and the product Km D 2:5, then the correction terms will be 0.47 and 0.33, respectively. If we include these corrections, the new estimates for the elasticities are significantly changed to: "vS D 1:33

0:47 D 0:86

"Pv D

0:33

0:33 D

0:66

In this case we cannot obtain good estimates for the elasticities from thermodynamic information alone.

Far from Equilibrium The responsiveness of the reaction rate is quite different when the reaction is far from equilibrium. Under these conditions, the saturation terms play an important role. Recall that both saturation terms are bound between the absolute values of 0.0 and 1.0. The simplest situation to consider is when the product concentration is well below the product Km (P  Kp ) since this is similar to the irreversible case. There are two possible scenarios. If S  Ks , the limiting value for the substrate elasticity is one and the product elasticity, near zero. If the substrate concentration is much higher than the substrate Km , S  Ks , the limiting value for the substrate elasticity is close to zero, and the product elasticity again close to zero. The easiest way to remember these two cases is to imagine what values the elasticities would have for the irreversible rate law. When we consider the case where the product concentration is well above Kp , P  Kp an interesting effect occurs (Table 3.1). When the substrate concentration is low (S  Ks ), the substrate elasticity is close to unity ("vS  1). However, under these conditions, the product elasticity is not v zero but has a limiting value of -1 ("P  1). One way to reach this

3.4. REVERSIBLE RATE LAWS

75

situation is for S=Ks  P =Kp and P  Kp , (see equations below) so that the product has a very strong competitive edge over substrate binding. An unusual state of affairs occurs far from equilibrium when both substrate and product concentrations are high relative to their respective Ks : S=Ks  1 and P =Kp  1 Under these conditions, approximate elasticity values are given by the following expressions: "vS

 1

v "P



S=Kp S=Ks C P =Kp P =Kp S=Ks C P =Kp

(3.30)

(3.31)

The elasticities remain bound between 0.0 and 1.0, and 0.0 and -1.0. The actual values will depend on the S=Ks and P =Kp ratios. Thus, provided P  Kp : if

S Ks



P ; "vS  Kp

v "P  0:5

(3.32)

if

S Ks

>

P ; "vS  Kp

v "P < 0:5

(3.33)

All these possible variations are summarized in Table 3.1. A simple general formula related to the sum of the elasticities is given by:

v "vS C "P D1

fractional saturation

(3.34)

where the fractional saturation varies from 0 to 1.0. This equation indicates that the sum of the elasticities will, under all conditions, be less than or equal to one and inversely proportional to the fractional saturation of the

CHAPTER 3. BASIC ENZYME KINETICS 76

Far from Equilibrium

Near Equilibrium

Equilibrium State

S  Ks and P  Kp

S  Ks and P  Kp

All degrees of saturation

Degree of Saturation

"Sv  0;

"Sv  1;

"Sv  1I

Elasticities

v 0 "P

v 0 "P

v  "P

1I

v 1 "Sv C "P

Table 3.1: Values of Elasticities Depending on Saturation and Equilibrium Conditions

Far from Equilibrium

"Sv 

v "P

Far from Equilibrium

P  Kp (Any Substrate Level)

"Sv  1 "Sv  0:5 "Sv < 0:5

"v  1 P v "P  0:5 v "P > 0:5 if S=Ks  P =Kp S=Ks  P =Kp S=Ks > P =Kp

3.4. REVERSIBLE RATE LAWS

77

v / is directly proportional to the fractional enzyme. The term 1 ."vS C "P saturation of the enzyme. The relation can be easily derived by showing that the right-hand side of (3.4) is equal to the right-hand side of equation (3.28). Note that when the fraction saturation is very small, the sum reduces to the sum for a mass-action reaction.

These patterns of elasticity values may seem unimportant and traditional enzyme kinetics text books rarely mention them. However, within a system of enzymes, it is the elasticities that describe the transmission of signals and therefore determine how individual enzymatic steps in a pathway influence the entire pathway.

Chapter Highlights  Binding of substrate to enzyme forming enzyme-substrate complex. Enzyme-substrate complex can breakdown to form free enzyme and product.  Two way to simplify the dynamics of an enzymatic reaction, rapid equilibrium and steady state assumption  If the binding and unbinding of substrate to enzyme is fast compared to product formation, then we can assume that the reaction between substrate and product is at equilibrium. This is called the rapid equilibrium assumption.  After addition of substrate, the level of enzyme-substrate complex can rapidly reach a steady state before the substrate significantly depletes. This is called the steady state assumption.  The irreversible Michaelis-Menten equation is given by: v D Vm S=.Km C S /  An additional underlying assumptions in the Michaelis derivation is that the concentration of enzyme is much less than the concentration of substrate.

78

CHAPTER 3. BASIC ENZYME KINETICS

 The Haldane relationship relates the four kinetic constants to the equilibrium constant.  The Haldane relationship can be used to eliminate one kinetic constant from the rate law.  The elasticities coefficient for a Michaelis like rate law is in two parts, a thermodynamic part which decides whether a reaction goes forward or in reverse and a kinetic part which related to the degree of saturation of the enzyme.  Close to equilibrium, the substrate and product elasticities are very large.

Further Reading 1. Cornish-Bowden, A (2004). Fundamentals of enzyme kinetics (3rd ed.). London: Portland Press. ISBN 1-85578-158-1 2. Fell D A (1996) Understanding the Control of Metabolism. Portland Press, ISBN: 185578047X 3. Palmer, T (1995) Understanding Enzymes, 4th Edition. Prentice Hall. ISBN: 0131344706 4. Segal I H (1993) Enzyme Kinetics: Behavior and Analysis of Rapid Equilibrium and Steady-State Enzyme Systems (Wiley Classics Library) Wiley-Interscience. ISBN: 0471303097

Exercises 1. For the irreversible Michaelis-Menten rate law (3.11), show that the Km is the concentration of substrate at half maximal activity, that is Km D S when v D 1=2Vm . 2. Show that when P D 0, the reversible Michaelis-Menten equation (3.19) reverts to the irreversible form.

3.4. REVERSIBLE RATE LAWS

79

3. For the irreversible Michaelis-Menten rate law (3.11), show that the graphs of 1=v vs. 1=S and v vs. v=S both show a linear relationship. 4. Show that for the linear plot of v vs. v=S, that the slope of the line is 1=Km . 5. Derive the elasticity, "vE for the equation: vD

Vm S Km C S

6. In question five what is the elasticity at half maximal velocity? 7. Estimate the Km and Vm using the answer from question four from the following initial rate data: Substrate (mM)

Reaction Rate (mM s

0 0.1 0.2 0.5 1 2 5 10

0.015 1.08 0.82 2.11 2.16 3.04 2.86 4.38

1)

8. Show that the Haldane relationship, (3.21), is true. 9. Derive the fractional saturation equation in (3.25). 10. Derive the elasticity equations, (3.27) using the general elasticity rules given in chapter 2, Table 2.2. 11. Derive equation (3.28).

80

CHAPTER 3. BASIC ENZYME KINETICS

4

Enzyme Inhibition and Activation

81

82

CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

4.1 Introduction There are many molecules capable of slowing down or speeding up the rate of enzyme catalyzed reactions. Such molecules are called enzyme inhibitors and activators. One common type of inhibition, called competitive inhibition, occurs when the inhibitor is structurally similar to the substrate so that it competes for the active site by forming a dead-end complex. This is one of several different inhibition and activation models which will be considered in this chapter.

4.2 Inhibitors Inhibitors are classified in two broad groups, reversible and irreversible: Competitive inhibition is an example of a reversible inhibitor because the inhibitor binds reversibly to the enzyme, and inhibition can be overcome by adding excess substrate. Irreversible inhibitors are those that will either covalently bind to the enzyme or modify the enzyme chemically thereby causing a permanent change to the enzyme. Inhibitors are important for a number of reasons. Many pharmaceutical compounds act as inhibitors of enzymes or signaling proteins. It is therefore important to understand how an enzymatic reaction rate responds to changes in inhibitor concentration. Inhibitors are also important in basic research where they can be used to understand the active site and catalytic action. There are many naturally occurring enzyme inhibitors, the most famous being the antibiotic penicillin and vancomycin or the antibacterial sulfonamides.

Assumptions In all the inhibition (or activator) equations that will be presented, we will assume the I term in the equation represents the free inhibitor. However, one is often unable to measure the free inhibitor concentration. Instead, as with substrate concentration, we assume that the amount of enzyme is small compared to the inhibitor concentration, therefore we set I to the

4.3. GENERALIZED INHIBITION MODEL

83

total amount of inhibitor in the assay or cellular compartment and assume that it equals the free inhibitor concentration.

4.3 Generalized Inhibition Model There are only a limited number of ways an inhibitor (or activator) molecule can interact with an enzyme or protein. The inhibitor can bind to the active site, an alternative site on the protein, or to the substrate that is bound to the active site. To take account of these variations, let us introduce a generalized scheme, also known as the Botts-Morales scheme [10], shown in Figure 4.1. This model describes the reversible binding of inhibitor (I) to either free enzyme (E) or enzyme-substrate complex (ES). In addition, substrate can bind to the enzyme-inhibitor complex (EI) or to the enzymesubstrate-inhibitor complex (ESI). Both the enzyme-substrate complex and the enzyme-inhibitor-substrate complex are assumed to be catalytically active. By changing the strength of these binding and catalytic steps, different inhibitory models can be obtained. The rate law (4.1) for the Botts-Morales scheme can be derived using the rapid-equilibrium assumption (See end of chapter for a derivation). The generalized model assumes that the catalytic constant kcat can be different for the transformation of ES or ESI into product. This difference is indicated by the factor b (Figure 4.1). If b D 1 then both ES and ESI are of equal catalytic activity. If b D 0 then only the ES complex can form product. Note that if b > 1 then the ESI complex is more active that the ES complex. In this situation the ‘inhibitor’ acts as an activator (See section (4.9)). For inhibition, b should be in the range 0  b  1. The factor a in Figure 4.1 represents the difference in binding affinities between the binding of substrate to the EI complex and binding of inhibitor to the ES complex. When a D 1 there is no difference in binding. When a D 0, inhibitor can only bind to the free enzyme. By changing a and b we can obtain all the common inhibition models (and for that matter all the activation models). In equation (4.1), Ki is the inhibitor binding dissociation constant and the higher the value for Ki , the less favorable the formation of ESI.

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CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

E

Km

ES

Ki

EI

k cat

E+P

aK i

a Km

ESI

b k cat

EI + P

Figure 4.1: Generalized Inhibition Model (Botts-Morales scheme). I is the concentration of inhibitor, ES the enzyme-complex, and ESI the enzyme-substrate-inhibitor complex. Ki is the inhibitor binding disassociation constant. In the generalized model the Km for binding of substrate may be different depending on whether the inhibitor is bound or not. The same argument applies to the binding of inhibitor to the ES complex and the catalytic rate constant, kcat . These differences are represented by the a and b factors. Both Ki and Km are affected by the same factor because of detailed balance.

Note: aKi D ES
I =ESI , when a  1, aKi  1 so that ESI  0.

  S I 1Cb Km aKi vD I SI S 1C C C Km Ki aKi Km Vm

(4.1)

An alterative way to write equation (4.1) and one that will be useful later on is:  I Vm S 1 C b a Ki     vD I I Km 1 C CS 1C Ki a Ki 

The reversible version of this equation can also be shown to be:

4.4. COMPETITIVE INHIBITION

85

   P I Vm S 1Cb Ks Keq aKi    vD S P I S P 1C C C 1C C Ks Kp Ki aKs aKp

(4.2)

From the generalized scheme it is possible to categorize the various specific inhibition patterns. The two most common mechanisms described in text books are the competitive and uncompetitive mechanisms (Figure 4.2). In practice, uncompetitive inhibition is quite rare but competitive inhibition is very common.

a) Competitive Inhibition E

ES

E+P

EI

b) Uncompetitive Inhibition E

ES

E+P

ESI

Figure 4.2: Competitive and uncompetitive inhibition. P is the concentration of product, E is the free enzyme, ES the enzyme-substrate complex, and ESI the enzyme-substrate-inhibitor complex.

4.4 Competitive Inhibition The kinetic mechanism for a pure competitive inhibitor is shown in Figure 4.2(a), where I is the inhibitor and EI the enzyme inhibitor complex. By making the factor a large (a  0I b D 0) the formation of ESI can be eliminated which leaves only the competitive model. If the substrate concentration is increased, it is possible for the substrate to eventually out compete the inhibitor. For this reason the inhibitor alters the enzyme’s apparent Km but not the Vm . Like all the mechanisms to be considered, the rate law for competitive inhibition can either be derived de novo using the rapid equilibrium assumption (see section (3.3) for further details), or by setting a and b to the appropriate values in the (4.1)

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CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

scheme. Either way, the equation for competitive inhibition is shown in equation 4.3:

vD

D

Vm S   I S C Km 1 C Ki

(4.3)

Vm S=Km 1 C S=Km C I =Ki

At I D 0, the competitive inhibition equation reduces to the normal irreversible Michaelis-Menten equation. Note that the term Km .1 C I =Ki / in the first equation more clearly shows the impact of the inhibitor, I , on the Km . The inhibitor has no effect on the Vm . Figure 4.3 shows how the reaction rate as a function of substrate concentration changes with varying inhibitor concentration. The plot is displayed on log-log axes to highlight the effect of the inhibitor and to make it easier to compare against other inhibition schemes. All four curves in Figure 4.3 reach the same maximal reaction rate at sufficiently high concentrations of S. There are many examples of competitive inhibitors. Methanol is a poison which can be fatal to humans at doses between 100 and 125 mL. Alcohol dehydrogenase will oxidize ethanol to acetaldehyde, but will also oxidize methanol to formaldehyde and formic acid via formaldehyde dehydrogenase. However, formic acid is a potent inhibitor of mitochondrial cytochrome c oxidase, thus reducing oxidative respiration and ATP production. One of the treatments for methanol poisoning is to administer ethanol because ethanol can act as a competitive inhibitor of methanol catalysis by alcohol dehydrogenase, thus reducing the conversion of methanol to formic acid.

4.4. COMPETITIVE INHIBITION

87 I D0 I D2 I D10 I D50

Reaction Rate

100

10

1

10

1

100 101 102 Substrate Concentration, S

Figure 4.3: Competitive inhibition: Effect of substrate concentration on the reaction rate as a function of different inhibitor concentrations: Vm D 1; Km D 1; Ki D 1. Solid circles mark the apparent Km values and the light circle the Km in the absence of inhibitor. As the inhibitor concentration increases the apparent Km increases in value.

Reversible Form The reversible form of the competitive rate law can be derived from equation (4.2) by setting a  1 and b D 0 and is shown below:   Vm P S Ks Keq vD S P I 1C C C Ks Kp Ki

(4.4)

where Vm is the forward maximal velocity, and Ks and Kp are the substrate and product half saturation constants.

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CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

Product Inhibition Sometimes reactions appear irreversible, that is no discernable reverse rate is detected, and yet the forward reaction is influenced by the accumulation of product. This effect is caused by the product competing with substrate for binding to the active site and is often called product inhibition. Given that product inhibition is a type of competitive inhibition we will briefly discuss it here. An important industrial example of this is the conversion of lactose to galactose by the enzyme ˇ galactosidase where galactose competes with lactose, slowing the forward rate [25]. To describe simple product inhibition with rate irreversibility, we can set the P =Keq term in the reversible Michaelis-Menten rate law (3.19) to zero. This yields:

vD

Vm S   P S C Km 1 C Kp

(4.5)

It is not surprising to discover that equation (4.5) has exactly the same form as the equation for competitive inhibition (4.3). Figure 4.4 shows how the reaction rate responds to increasing product concentration at a fixed substrate concentration. As the product increases, it out competes the substrate and therefore slows down the reaction rate. We can also derive the equation by using the following mechanism and the rapid equilibrium assumption: E CS * ) ES ! EP * )E CP

(4.6)

where the reaction rate, v / ES. Competitive inhibition may also occur if an enzyme is faced with structurally similar substrates. Related substrates have the potential to bind to the active site and block the substrate of interest from binding. Given the equivalence between competitive and product inhibition, the following elasticity section applies equally to both.

Reaction Rate, v

4.4. COMPETITIVE INHIBITION

89

0:4

0:2

0

0

1

2 3 4 5 6 Product Concentration, P

7

8

Figure 4.4: Effect of increasing the concentration of product on the reaction rate at a fixed substrate concentration for an irreversible reaction with product inhibition. The reaction rate declines monotonically to zero as the product increases. S D 1; Vm D 1; Km D 1; Ki D 1.

Elasticities for a Competitive Inhibitor The substrate and inhibitor elasticities for a competitive inhibitor acting on an irreversible reaction are shown below:

"vS D 1

"Iv D

v D Vm

1C

I Ki

S I C 1C Ks Ki

I Ki D S I 1C C Ks Ki

D

1C 1C˛C

 1C˛C

(4.7)

(4.8)

Note that the inhibitor elasticity is negative because increasing the inhibitor concentration will reduce the reaction rate. Figure 4.5 shows the effect of different competitive inhibitor concentrations on the substrate elasticity.

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CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

Each curve asymptotically approaches zero at high substrate concentration as the enzyme saturates with substrate. This also means that the inhibitor is unable to compete as effectively.

Substrate Elasticity

1 0:8

I D 50

0:6

I D 10 I D2

0:4

I D0

0:2 0

0

1

2 3 4 5 6 Substrate Concentration, S

7

8

Figure 4.5: Competitive Inhibitor: Substrate elasticity with respect to the substrate concentration ("vS ) as a function of different inhibitor concentrations: Vm D 1; Km D 1; Ki D 1.

Figure 4.6 shows how the inhibitor elasticity responds to changes in substrate concentration. The inhibitor elasticity tends to zero at high substrate levels. The limits on "Iv are zero and 1. As the inhibitor concentration approaches zero, the inhibitor elasticity tends to zero, while at high concentrations of inhibitor, the elasticity approaches 1. The fact that the inhibition elasticity cannot fall below -1 indicates that a competitive inhibitor can at most only proportionally decrease the reaction rate. This means competitive inhibitors are only effective on the reaction rate when their concentrations change significantly. Figure 4.7 shows what happens to the substrate and inhibitor elasticities as the inhibitor concentration changes. The figure highlights an important aspect of competitive inhibition that is not seen in other inhibition mechanisms. As the concentration of inhibitor increases, the substrate elasticity

4.4. COMPETITIVE INHIBITION

91

Inhibitor Elasticity, "Iv

0 0:2 I D2

0:4 0:6

I D6 I D 50

0:8 1

0

1

2 3 4 5 6 Substrate Concentration, S

7

8

Figure 4.6: Competitive inhibitor elasticity ("Iv ) as a function of changing substrate concentration at different competitive inhibitor levels: Vm D 1; Km D 1; Ki D 1.

("vS ) also increases. This means that as the inhibitor concentration is increased, the responsiveness of the reaction rate to substrate increases albeit from a lower starting rate. To illustrate what effect this can have imagine two enzyme catalyzed reactions in sequence with metabolite S as the common intermediate. If we were to use a competitive inhibitor on the second step (the one that consumes S), the rate of the second step would initially go down which would result in an increase in S (less S consumed). Given that the substrate elasticity rises with the inhibitor concentration it is easy for the increase in substrate to compensate for the loss of rate through the second step resulting in little or end change in the overall rate through the pathway. This partially explains why competitive inhibitors are not as effective as one might expect. In the next section when we consider uncompetitive inhibitors we will see a quite different response.

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CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

1 S D1

Elasticity

0:5

S D 10

0 0:5 1

S D 10

S D1 0

1

2 3 4 Inhibitor Concentration, I

5

Figure 4.7: Competitive Inhibitor: Elasticities as a function of inhibitor concentration: "vS (Upper two curves) and "Iv (Lower two curves). Vm D 1; Km D 1; Ki D 1.

Elasticity

Range

"vS "Iv

0  "vS  1 0  "Iv  1

Table 4.1: Magnitude range for elasticities of a competitive inhibitor.

Competitive Inhibitor: As the inhibitor concentration is increased both the substrate and inhibitor elasticities increase in absolute magnitude such that similar changes to the substrate concentration can compensate for any inhibition caused by the competitive inhibitor. Therefore, although competitive inhibitors might be effective on isolated enzymes, within a pathway they are much less effective.

4.5. UNCOMPETITIVE INHIBITION

93

4.5 Uncompetitive Inhibition Uncompetitive inhibition occurs when an inhibitor only binds when substrate is bound to the enzyme, Figure 4.2(b). One could imagine that substrate binding causes a conformational change which allows the inhibitor to subsequently bind. This could involve either binding to a completely separate site or to the substrate bound to the enzyme. In either case, the inhibitor does not directly compete with the substrate for the active site. This means that increasing the substrate concentration cannot overcome an uncompetitive inhibitor (which is the case with competitive inhibition). In reality, uncompetitive inhibition is quite rare, but a well known instance is the use of lithium ions in treating manic depression. It is thought that lithium acts as an uncompetitive inhibitor of myo-insitol monophosphatase. Another example is alkaline phosphatase which catalyzes the hydrolysis of phosphate from many different kinds of small molecules. This reaction can be inhibited by Phenylalanine or L-Tryptophan via an uncompetitive mechanism [45]. A more common situation which demonstrates uncompetitive inhibition is in multi-substrate reactions. For example, in a bimolecular reaction involving two substrates where the substrates bind to the enzyme in an ordered fashion, a competitive inhibitor for the second substrate (B) will act as an uncompetitive inhibitor with respect to the first substrate (A). In the Botts-Morales scheme, Figure 4.1, we set b D 0, a ! 0, and assume that the Ki ! 1. This means that Ki  ˛Ki and ensures that the binding of I to E is unlikely to occur while product ˛Ki has a finite value so that I can bind to ES . When we make these assumptions we obtain the rate law for an uncompetitive inhibitor:

vD

Vm S   I Km C S 1 C aKi

(4.9)

Dividing both top and bottom by the factor .1CI =.aKi // reveals that both Vm and Km are changed to:

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CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

Vmapp D

Vm 1 C I =.aKi /

app Km D

Km 1 C I =.aKi /

It should also be evident from these equations that the ratio, Vm =Km remains unchanged. Figure 4.8 show a log-log plot of reaction rate versus substrate at different uncompetitive inhibitor concentrations (cf. Figure 4.3). 101 I D0

Reaction Rate

100 10

I D2 I D 10

1

I D 50 10

2

10

3

10

I D 200

3

10

2

10 1 100 101 Substrate Concentration, S

102

Figure 4.8: Uncompetitive inhibition: Reaction rate versus substrate concentration as a function of different inhibitor concentrations: Vm D 1; Km D 1; Ki D 1. Solid circles indicate the apparent Km while the light circle is the Km in the absence of inhibitor. The individual Km and Vm change but their ratio remains constant.

Reversible Form The reversible form of the uncompetitive rate law can be derived from equation (4.2) and shown to equal:

4.5. UNCOMPETITIVE INHIBITION

  P Vm S Ks Keq    vD S P I 1C C 1C Ks Kp aKi

95

(4.10)

where Vm is the forward maximal velocity, and Ks and Kp the substrate and product half saturation constants. Elasticities for an Uncompetitive Inhibitor The substrate and inhibitor elasticity for the irreversible uncompetitive mechanism is given by:

"vS

D1

  I v 1C D Vm Ki

"Iv D

1 1 D S I S 1 C ˛ C ˛ 1C C Km Ki Km

I S ˛ Ki Km D S I S 1 C ˛ C ˛ 1C C Km Ki Km

Figure 4.9 shows the response of the substrate and inhibitor elasticities as a function of inhibitor. Unlike a competitive inhibitor (Figure 4.7), the substrate elasticity decreases as the inhibitor concentration increases. This means that an increase in inhibitor concentration is less likely to be neutralized by an increase in substrate concentration. Compared to a competitive inhibitor, uncompetitive inhibitors should in principle be more effective. Figure 4.10 shows how the substrate elasticity decreases as the inhibitor level goes up. Uncompetitive Inhibitor: As the inhibitor concentration is increased the substrate elasticity decreases and inhibitor elasticity increases in absolute magnitude.

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CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

1 S D5

Elasticity

0:5

S D 10

0

S D 10 0:5 1

S D5 0

1

2 3 4 Inhibitor Concentration, I

5

6

Figure 4.9: Uncompetitive Inhibition: Elasticities as a function of inhibitor concentration: "vS (Upper two curves) and "Iv (Lower two curves). Vm D 1; Km D 1; Ki D 1.

Elasticity

Range

"vS "Iv

0  "vS  1 0  "Iv  1

Table 4.2: Magnitude range for elasticities of an Uncompetitive inhibitor.

4.6 Mixed Inhibition If an inhibitor can bind to both the free enzyme and enzyme-substrate complex, then the inhibition is termed mixed (Figure 4.12). In the most general case of mixed inhibition, the binding of inhibitor is assumed to affect the binding of substrate and vice versa. This type of inhibition pattern is called mixed non-competitive inhibition. If the binding of inhibitor has no effect on binding of substrate, then the inhibition is termed pure non-competitive inhibition (see next section).

4.6. MIXED INHIBITION

97

Substrate Elasticity, "vS

1 0:8 0:6 0:4 0:2 0

0

1

2 3 4 5 6 Substrate Concentration, S

7

8

Figure 4.10: Uncompetitive Inhibitor: Effect of inhibitor concentration on the substrate elasticity ("vS ) as a function of substrate concentration for an inhibitor: Vm D 1; Km D 1; Ki D 1. From upper curve, I D 0; I D 1; I D 10; I D 50

The equation for general mixed irreversible non-competitive inhibition where only ES can form product can be obtained by setting b D 0, a > 1:

vD

Vm S     I I Km 1 C CS 1C Ki a Ki

(4.11)

Ki is the dissociation constant for the binding of inhibitor to free enzyme. If substrate is already bound (ES), then in the mixed model, a is the factor by which Ki has changed due to the presence of S. Elasticities The substrate and inhibitor elasticity for irreversible mixed inhibition is given by:

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CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

Inhibitor Elasticity, "Iv

0 0:2 0:4 0:6 0:8 1

0

1

2 3 4 5 6 Inhibitor Concentration, I

7

8

Figure 4.11: Uncompetitive Inhibitor: Effect of substrate concentration on the inhibitor elasticity ("Iv ) as a function of inhibitor concentration for an inhibitor: Vm D 1; Km D 1; Ki D 1. From upper curve, S D 0:1; S D 1; S D 10; S D 50.

"vS D

"Iv D

a a D Ca S I C aKi ˛ C˛ Ca C1 Km I C Ki I  Ki D I S C Km ˛C1 Ca Ca Ki S C aKm ˛Ca

(4.12)

Pure Non-Competitive Inhibition In the general case of mixed inhibition, the binding constant of inhibitor on free enzyme and enzyme-substrate complex is assumed to be different in each case, that is a > 1. If the binding of inhibitor is unaffected by the presence of substrate, then the binding constant will be identical for E and ES, that is a D 1. Pure non-competitive inhibition is therefore a special

4.6. MIXED INHIBITION

99

E

ES

Ki

aK i

EI

ESI

Figure 4.12: Mixed Inhibition. ES the enzyme-substrate complex, and ESI the enzyme-substrate-inhibitor complex. Ki is the inhibitor binding dissociation constant and a is a factor that determines whether the binding constant changes in the presence of bound substrate. If a D 1 then the binding constant is the same for both E and ES.

case of mixed inhibition, where the binding constant for inhibitor to free enzyme and enzyme-substrate complex are identical. In pure non-competitive inhibition the apparent Vm changes while the Km is unaffected. The substrate concentration cannot out compete the inhibitor. The rate law for a pure non-competitive inhibitor is given by the equation by setting a D 1 and b D 0:

vD

Vm S   I .Km C S / 1 C Ki

(4.13)

Another way to look at the above equation is to divide top and bottom by 1 C I =Ki . When we do this the Vm is modified to an apparent Vm , denoted app by Vm : app

Vm D Vm .1 C I =Ki / Pure non-competitive inhibition is quite rare in nature but examples include heavy-metal ion binding such as mercury to SH groups of cysteine residues or more commonly, binding of hydrogen ions to enzymes.

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CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

Reaction Rate

101 I D0 I D4 10

1

I D 50 I D 350

10

3

10

5

10

2

10

1

100 101 Substrate Concentration, S

102

Figure 4.13: Pure Non-competitive Inhibitor: Reaction rate versus substrate concentration as a function of different inhibitor concentrations: Vm D 1; Km D 1; Ki D 1. Solid circles mark the apparent Km values and the light circle the Km in the absence of inhibitor. Note that the Km remains the same but the Vm decreases with increasing inhibitor.

Reversible Form The reversible form of the pure non-competitive rate law can be show from (4.2) to equal:   Vm P S Ks Keq   vD S P I 1C C 1C Ks Kp Ki

(4.14)

where Vm is the forward maximal velocity and Ks and Kp the substrate and product half saturation constants.

4.7. MIXED AND PARTIAL INHIBITION

101

Elasticities If we set a D 1 in equations (4.12) we arrive at much simplified versions for the elasticities with respect ti the irreversible case "vS D

Km Km C S

The above equation is exactly the same as the substrate elasticity for a simple Michaelis-Menten equation and reflects the fact that a pure noncompetitive inhibitor affects the Vm but does not change the Km . The inhibitor elasticity is given by: "Iv D

I Ki C I

Note that the inhibitor elasticity is independent of the substrate concentration, thus changes in inhibitor concentration will not affect the substrate elasticity. Previously it was shown that for a competitive inhibitor, the substrate elasticity increased and for a noncompetitive inhibitor the substrate elasticity decreased. For a noncompetitive inhibitor, the substrate elasticity stays constant.

4.7 Mixed and Partial Inhibition In the case where the inhibited enzyme-substrate complex can also form product, usually at a lower rate, the inhibition is called partial inhibition. Partial mixed inhibition corresponds to the generalized Botts–Morales scheme, outlined in equation (4.1) and Figure 4.1.

4.8 Irreversible Inhibitors Irreversible inhibitors are compounds that chemically modify the enzyme or exhibit very tight non-covalent binding. Such inhibitors effectively re-

102

CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

duce the concentration of enzyme, therefore changing the Vm while leaving the Km unaltered.

4.9 Activators Although most textbooks focus on the inhibition of enzyme activity, it is also possible, though perhaps not as commonly reported, for enzymes to be activated by certain ions or molecules. Referring to the generalized model which is shown again in Figure 4.14, we can replace I with A. Let us imagine a number of scenarios that might explain enzyme activation. For example, the ESA complex may be more active than the ES complex. Addition of activator, A, would therefore speed up the catalyzed reaction. Examples of this kind of activation include the binding of metal ions to enzymes. In E. coli when MgC binds to ˇ-galactosidase, the enzyme is activated.

E

Km Ka

EA

k cat

ES

E+P

aK a

a Km

ESA

b k cat

EA + P

Figure 4.14: Generalized Activation Model. See Figure 4.1 for details.

If the ES complex is inactive and the binding of substrate and activator are independent then the rate law is reminiscent of pure non-competitive inhibition: v D Vm

S   Ka Km C S 1 C A

(4.15)

4.10. FINAL REMARKS

103

where Ka is the dissociation constant for activator binding to ES or E. The reversible form can also be derived:   P Vm S Ks Keq   vD S P Ka C 1C Ks Kp A

(4.16)

In the most general case, partial mixed activation, where binding of activator and substrate affect each other and the ES has some activity though less than ESA, then a rapid-equilibrium based derivation will lead to the following general activation model:   A S 1Cb Km aKa vD S A SA 1C C C Km Ka aKa Km Vm

(4.17)

where Ka is the dissociation constant for the binding of A to EA, and Ks the dissociation constant for the binding of S to free enzyme, E. This equation is the same as equation (4.1) except now b > 1 The reversible version of this equation is represented as:    Vm P A S 1Cb Ks Keq aKa    vD S P A S P 1C C C 1C C Ks Kp Ka aKs aKp

(4.18)

4.10 Final Remarks This chapter provides a summary of basic enzyme inhibition and activation kinetics (Figure 4.15). A number of important topics are not covered, including various linear transformations (such as Eadie-Hofstee and Lineweaver-Burk) for distinguishing different inhibition mechanisms and

CHAPTER 4. ENZYME INHIBITION AND ACTIVATION 104

Inhibition vD

Rate Equation

Vm S   vD I Km C S 1 C Ki

Vm S   vD I Km 1 C CS Ki

Vm S Km C S

No Inhibition

Competitive

Uncompetitive

Noncompetitive

V S m   vD I .Km C S / 1 C Ki

aD1

Ki  ˛Ki

a0

a; .b D 0/

Vm changes, Km constant

Vm =Km ratio remains the same

Km changes, Vm constant

Vm ; Km

Table 4.3: Main inhibitor types where ESI is completely inactive (b D 0), Figure 4.15.

4.10. FINAL REMARKS

105

for estimating the kinetic constants. More traditional enzyme kinetic textbooks will cover these topics. E

kc

ES aKi

Ki

EIS

EI

b kc

Complete Inhibition b=0

Pure Competitive (a >> 1)

E+P

Uncompetitive (Ki >> a Ki)

EI + P

Partial Inhibition b= 0

Uncompetitive

Competitive (Mixed, a = 1) b=1

Non-Competitive Non-Competitive Mixed (a > 1)

Pure (a=1) Mixed (a >1)

Pure (a=1)

Figure 4.15: Reversible Inhibition Family Tree. The left-hand branch represents those mechanisms that lead to dead-end complexes that are completely inactive, sometimes called complete inhibition (b D 0). The partial inhibition pattern on the right represents inhibition complexes which are partially active and will lead to some product formation (b ¤ 0). The mixed forms represent the case where substrate and inhibitor can affect each others’ binding affinity. In pure (or non-mixed) non-competitive inhibition, the binding of inhibitor or substrate is independent. Activation will occur when b > 1.

106

CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

Chapter Highlights  There are two types of inhibitors, reversible and irreversible.  There are three basic types of reversible inhibition, competitive, uncompetitive and the rare non-competitive. These inhibitions represent extremes from which a variety of hybrid inhibition patterns emerge including mixed, complete, partial and complete forms.  Competitive inhibition is where a molecule competes with the natural substrate for the active site. As a result the natural substrate can out-compete a competitive inhibitor at high enough concentration. As a result, in the presence of a competitive inhibitor, the apparent Km changes but the Vm remains the same.  The substrate elasticity of a competitive inhibitor mechanism increases as the inhibitor increases.  Competitive inhibitors have a limited capacity to control an enzyme’s reaction rate and often can be easily undone by increasing the substrate concentration.  Uncompetitive inhibition is where the inhibitor binds only to the substrate-enzyme complex. The net effect is that both the Vm and the Km decrease while their ratio remains the same. Uncompetitive inhibition is however relatively rare. The effect of an uncompetitive inhibitor can not be reduced by increasing the natural substrate concentration.  The substrate elasticity of an uncompetitive inhibitor mechanism decreases as the inhibitor increases.  Non-Competitive inhibition is where an inhibitor binds to the enzyme at a site that is different from the active site. In pure noncompetitive inhibition, the inhibitor binds with equal affinity to the free enzyme and to the enzyme-substrate (ES) complex. The result is that in the presence of inhibitor the Vm changes but the Km remains the same. Non-competitive inhibition is technically a form of

4.10. FINAL REMARKS

107

mixed inhibition where the inhibitor can bind to both enzyme and enzyme-substrate complex and where the inhibitor does not affect the binding affinity of the substrates (and vice versa).  The substrate elasticity of an noncompetitive inhibitor mechanism is independent of the inhibitor increases.  In mixed inhibition, the inhibitor can bind to both the free enzyme and the enzyme-substrate complex. Two forms can be given. The first is where the inhibitor does not affect the binding constant of the substrate (and vice versa) - called pure non-competitive inhibition. The second type is where the binding of inhibitor does affect the binding constant for the substrate (and vice versa) in which case it is called mixed non-competitive inhibition.

Further Reading 1. Bisswanger H (2008) Enzyme Kinetics: Principles and Methods, 2nd Revised and Updated Edition. (Wiley-VCH) ISBN: 978-3-52731957-2 2. Cornish-Bowden, A (2004). Fundamentals of enzyme kinetics (3rd ed.). London: Portland Press. ISBN 1-85578-158-1 3. Fell D A (1996) Understanding the Control of Metabolism. Portland Press, ISBN: 185578047X 4. Palmer, T (1995) Understanding Enzymes, 4th Edition. Prentice Hall. ISBN: 0131344706 5. Segal I H (1993) Enzyme Kinetics: Behavior and Analysis of Rapid Equilibrium and Steady-State Enzyme Systems (Wiley Classics Library) Wiley-Interscience. ISBN: 0471303097

Proofs Derive the Botts-Morales Model, Figure 4.1, Equation 4.1

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CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

The rapid equilibrium assumption will be used to derive the equation because it is easier but the same result can also be obtained using the steadystate assumption. First define the various dissociation constants in the model:

Ki D

E
I EI

aKi D

ES
I E
S
I D ESI aKm Ki

Km D

E
S ES

aKm D

EI
E ESI

The last dissociation constant, aKm is not required because it can be derived form the other three by detailed balance. The overall rate of reaction is the sum of the rate of individual transformations of ES and ESI , that is: v D vES C vESI The individual rates are given by: vES D ET
fES kcat vESI D ET
fESI bkcat where ET D E C ES C EI C ESI . fES is the fractional saturation of E by substrate and fESI is the fractional saturation of E by substrate and inhibitor: ESI ES fES D I fESI D ET ET Now express each complex in term of free E using the dissociation relationships, and canceling E top and bottom, yields:

fES

S Km D S I S
I C C 1C Km Ki Km aKi

fESI

S
I Km aKi D S I S
I 1C C C Km Ki Km aKi

4.10. FINAL REMARKS

109

Substituting these into the net rate (where Vm D ET kcat ): 1 S
bI S C B C Km Km aKi C v D ET kcat B @ S I S
I A 1C C C Km Ki Km aKi   I S 1Cb Vm Km aKi D S I S
I 1C C C Km Ki aKm Ki 0

Exercises 1. Explain the difference between a reversible and an irreversible inhibitor. 2. ˇ-lactamase is a naturally occurring enzyme in bacterial that can inactivate penicillin and thus render the bacterium resistance to penicillin. Clavulanic acid, sulbactam and tazobactam are sometimes added to penicillin prescriptions. Carry out a literature survey to explain the purpose of these compounds and describe their mode of action. 3. The HIV protease inhibitor, ritonavir, resembles the structure of a peptide. From this information what kind of inhibitor is ritonavir likely to be? 4. An enzyme that catalyzes a uni-uni reaction is known to have a Km of 20 mM and a Vm of 5 mol s 1 . If the substrate concentration is 2 mM and product 0 mM, calculate the initial reaction velocity of the reaction. 5. If the enzyme in the previous question is competitively inhibited by a compound I that has a Ki of 3 mM, what would the initial rate of reaction be if the concentration of competitive inhibitor were 0.5 mM?

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CHAPTER 4. ENZYME INHIBITION AND ACTIVATION

6. Form the previous question, estimate the level of substrate required to restore the initial reaction rate in the presence of the 0.5 mM inhibitor. 7. Answer the last two questions assuming the inhibitor were an uncompetitive inhibitor. 8. From the study of the competitive and non-competitive inhibitor in the previous questions, even though both inhibitors have the same Ki , which do you think is the most effect inhibitor? 9. For a simple irreversible Michaelis-Menten rate law, a plot of 1=v versus 1=S yields a straight line. Show that the addition of a competitive inhibitor increases the slope of the line. 10. Why does a competitive inhibitor change the apparent Km of an enzyme but not the Vm ? 11. For a reversible enzyme catalyzed uni-uni reaction, show that the addition of a competitive or uncompetitive inhibitor has no effect on the equilibrium constant of the reaction. 12. Explain the difference between partial and complete inhibition. 13. Derive equation (4.5) which describes product inhibition. 14. Derive the substrate and inhibitor elasticity equations for the irreversible competitive, uncompetitive and noncompetitive mechanisms. 15. Show that to decrease the reaction rate of a irreversible uni-uni reaction from 90% of maximal rate to 10% of maximal rate requires a 81-fold increase in a competitive inhibitor concentration.

5

Multi-reactant Rate Laws

111

112

CHAPTER 5. MULTI-REACTANT RATE LAWS

5.1 Multiple Reactant Enzymes A great number of enzyme catalyzed reactions involve more than one substrate and product. For example, all oxidoreductases involve two substrates, one an oxidant and the other a reductant. Even apparently single substrate reactions may actually involve water as a second substrate which is typically ignored because we assume that the water concentration water hardly changes during the reaction. The world of two substrate kinetics is far more complicated than single substrate kinetics [17, 18, 19, 62]. Most of this complexity arises from the fact that substrates and products can bind and unbind the active site in different ways. The most common way in which this happens includes: (i) compulsory-order, when one substrate must bind before the other; (ii) random-order where substrates can bind in any order, and doubledisplacement (or ping-pong) where one substrate binds, modifies the enzyme, then leaves to allow the other substrate to bind (Figure 5.1). These three different mechanisms can generate subtlety different rate laws. The question however is whether such subtlety is significant enough to include in a pathway model. For those interested in catalytic mechanisms, the different rate laws are critical because they can be used to distinguish between the different mechanisms. For modeling purposes, a high level of precision may not always be necessary given the imprecision in kinetic data and the apparent robustness of pathways to parameter variation. As a result, some authors advocate the use of generalized rate laws for modeling two substrate/product enzyme reactions. A number of these generalizations exist in the literature although they are all closely related to each other. In this chapter a brief review of the multi-reactant world will be presented and in a following chapter (chapter 8) a more detailed look at generalized rate laws will be given.

5.2 Types of Multi-Reactant Systems Cleland [17, 18, 19] classified enzyme catalyzed reactions according to how many substrates and products took part in the reaction. Table 5.1

5.2. TYPES OF MULTI-REACTANT SYSTEMS

A P

Uni Uni

A P CQ

Uni Bi

ACB P

Bi Uni

ACB P CQ

Bi Bi

ACB CC P CQ

Ter Bi

113

Table 5.1: Reaction Type Classification According to Cleland

shows the notation Cleland used to signify the different reaction types. In this notation, A, B; : : : denote substrates and P , Q; : : : denote products. In addition to the number of substrates and products, Cleland also had a convenient graphical notation for describing how individual substrates and products bound and unbound to the enzyme active size. Figure 5.1 illustrates the three main mechanisms described using Cleland’s graphic notation. In the case of a compulsory or ping-pong mechanism, the order of the substrates and products as written indicates the order in which the substrates and product bind and unbind to the enzyme. One of the significant difficulties with multireactant systems is the derivation of the rate laws. The steady state derivation, though more informative for enzymologists, tends to be difficult and complex to derive. Often, when building a computer model of a pathway, the rapid equilibrium assumption is used instead because this leads to simpler equations with fewer parameters. We will now describe briefly the three main mechanisms, random order, ordered and ping-pong. As will be made clear in the chapter on generalized rate laws (chapter 8), there are many other ways to deal with multireactant systems which are particularly suited when either data is sparse or the detailed mechanism is unimportant to the behavior of the overall network.

114

CHAPTER 5. MULTI-REACTANT RATE LAWS

a) Random-Order A

P

B EA

Q EP

E

E EAB EPQ

E

E EQ

EB B

A

Q

P

B

P

Q

b) Compulsory-Order A EA

EP

E

E EAB EPQ

c) Ping-Pong A

P EA

A

EXP

Q EXA

E

EQ E

EX

Figure 5.1: Enzyme mechanisms for two substate/two product systems.

5.3 Ordered Bi-Bi Mechanism A classic example of an ordered bi-bi mechanism is alcohol dehydrogenase [73] where the substrate NADH binds to the enzyme before the second substrate, alcohol. Likewise, the products of the reaction leave their active sites in a specific order. Other examples of ordered bi-bi mechanisms include many other dehydrogenases but also some kinases such as carbamate kinase. The reversible equation for an ordered bi-bi equation can be derived using the steady state assumption although the algebra is

5.3. ORDERED BI-BI MECHANISM

115

somewhat tedious. The derivation of complex steady-state rate laws can be made easier by using graphical approaches such as the King-Altman method [39] that can also be automated by computer [20]. The equation is given by ([55], equation 6.17):  Vf vD

AB

PQ Keq



D

(5.1)

where

D D Ki a Kb C Kb A C Ka B C AB C C

Kb Kq AP Kb Ki a PQ C Kp Ki q Kp Ki q

Kb Ki a BPQ ABP Ka BQ Kq Kb Ki a P Kb Ki a Q C C C C Kp Ki q Ki b Kip Ki q Kp Ki q Ki q (5.2)

This equation uses the Haldane relationship to eliminate the reverse Vm . If this is not done the equation is more complex with both the forward and reverse Vm occurring in the numerator and denominator. What is also striking is the number of kinetic parameters that need to be determined, in total ten parameters. This makes such rate laws difficult to use because often the parameters are unknown. If the bi-bi reaction is irreversible the ordered equation reduces to a much simpler form: vD

Vm AB KB A C KA B C AB C KiA KB

where KA and KB are the Michaelis constants and KiA is the apparent dissociation constant for A.

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CHAPTER 5. MULTI-REACTANT RATE LAWS

5.4 Random-Order Mechanism There are many enzymes that operate via a random-order mechanism and typical examples include many kinases such as phosphofructokinase, hexokinase and phosphorylase b. Mechanism (a) in Figure 5.1 illustrates the random-order mechanism. A generalized random-order irreversible two substrate rate law based on the rapid equilibrium assumption was introduced by Alberty in 1953 [2]: vD

Vm AB KB A C KA B C AB C KiA KB

(5.3)

where KA and KB are Michaelian constants and KiA is a dissociation constant. Note that this is the same equation as the ordered Bi-Bi given above when both products, P and Q are set to zero and the equation derived using the steady state assumption. If we were to derive the random-order mechanism using the steady state assumption, there would be an additional eight constants, many of which will not be known. If we assume the presence of both products then a simpler reversible rapid equilibrium random order Bi-Bi model (that assumes rapid equilibrium other than the central conversion, EAB EPQ) can be derived [17, 62, 55] and is given by: Vf vD

 AB

PQ Keq



D

(5.4)

where

D D Ki a Kb C Kb A C Ka B C ABC Kb Ki a PQ Kq Kb Ki a P Kb Ki a Q C C Kp Ki q Kp Ki q Ki q

(5.5)

The rate law is very similar to the ordered bi-bi except the terms containing AP , BQ, BPQ, and ABP are missing. Vf is the forward Vmax and

5.5. PING-PONG MECHANISM

117

KA , KB , KQ and KP are the Michaelian constants and K_ia etc are called inhibition constants. We assume in all these cases that when a substrate or product binds it does not influence the binding constant for the second species to bind. As with the ordered bib-bi mechanism, there are a significant number of parameters to determine.

5.5 Ping-Pong Mechanism Transaminases are an important class of enzyme that catalyze the transfer of amino group to a keto group. They operate using a ping-pong mechanism where the amino group is first transferred to pyridoxal phosphate forming pyridoxamine. The pyridoxamine in turn can pass the amino group on to a second substrate. The rate law for a ping-pong mechanism can be derived using the steady state assumption to yield an equation very similar to the ordered bi-bi equation, except there is no constant term or A B P and B P Q terms in the denominator ([55], equation 6.20): 

PQ Keq

AB v D Vf Vr



D

(5.6)

where D D Vr Kb A C Vr Ka B C Vr AB C

Vf Kq P C Keq

Vf Kq AP Vf Kp BQ Vf Kp Q C C Keq Ki a Keq Keq Ki b

(5.7)

Again we are confronted with a large number of parameters, in this case eight parameters. If the bi-bi reaction is irreversible the ping-pong equation reduces to: vD

Vm AB AB C Ka A C Kb B

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CHAPTER 5. MULTI-REACTANT RATE LAWS

Multireactant mechanisms tend to yield highly complicated rate laws with many variations. For example different rate laws are generated for the same mechanism depending on whether one assumes steady state or rapid equilibrium. Additional mechanisms such as dead-end complexes, mixed ordered and random and three reactant and product mechanisms add to the variety. Unless the subtle variations in multireactant rate laws result in measurable effects there is little need to consider them. Instead it may be sufficient to use the generalized rate laws that will be discussed in chapter 8.

Chapter Highlights  There are three main classes of multireactant mechanisms: random order, ordered and ping-pong.  Multireactant rate laws tend to be complex with many parameters that need to be assigned.  When confronted with sparse kinetic data, the generalized rate laws described in chapter 8 can be used instead of the multireacant rate laws.

Further Reading 1. Cornish-Bowden, A (2004) Fundamentals of enzyme kinetics (3rd ed.), London: Portland Press. ISBN 1-85578-158-1 2. Palmer, T (1995) Understanding Enzymes, 4th Edition. Prentice Hall. ISBN: 0131344706 3. Roberts D V (1977) Enzyme Kinetics, Cambridge University Press. ISBN: 0521290805 4. Segal I H (1993) Enzyme Kinetics: Behavior and Analysis of Rapid Equilibrium and Steady-State Enzyme Systems (Wiley Classics Library) Wiley-Interscience. ISBN: 0471303097

6

Cooperativity

The Michaelis rate equation (3.11) describes a hyperbolic response to changes in substrate and product concentration. The corresponding elasticity

119

120

CHAPTER 6. COOPERATIVITY

for the equation in the absence of product was shown to be: "vS D

Km Km C S

The value for the elasticity has a maximum of one at zero substrate concentration and tends to zero at high concentrations. This means that the sensitivity of the reaction rate to changes in substrate is always attenuated. We can show this more clearly by considering the fold-change in substrate required to change the reaction rate from 10% to 90% of maximum, that is a nine fold change in rate. Let us consider the normalized Michaelis reaction rate law: v S D Vm Km C S We will now compare the normalized rate at 90% of the rate and at 10% of the rate. If we set the reaction rate to 0.1 we can solve for the substrate concentration that will sustain this rate: S10% D 9Km Likewise if we set the rate to 0.9 we obtain a substrate concentration of: S90% D 1=9Km The fold change in substrate level is therefore S90% =S10% D 81. In other words to change the reaction rate from 10% to 90% of maximum we must change the substrate concentration 81 fold. This demonstrates the relative insensitivity of the hyperbolic response. The same reasoning applies to the response of the reaction rate to a competitive inhibitor. To achieve greater sensitivity in reaction rates to changes in substrate, products and modulators, something more sensitive than the hyperbolic response is required. The answer to this is cooperativity. This chapter will only consider cooperativity at the enzyme level even though there are other mechanisms such as covalent modification cycles[16, 26] that can also achieve high sensitivity.

121

Before proceeding it is worth noting that regulation by effectors is usually associated with cooperative effects, often termed allosteric regulation. The topic of allosteric regulation will be discussed more fully in the next chapter because allostery and cooperativity are distinct phenomena and it is possible for an enzyme (or binding protein) to show cooperativity without any associated allostery. Likewise it is possible to find enzymes that show allostery without cooperativity. It is true however that enzymes exhibiting cooperativity are often associated with allosteric regulation which is why historically they are often discussed together in the literature.

Cooperativity based on Oligomers Many proteins are known to be oligomeric, that is they are composed of more than one identical protein subunit where each subunit has one or more binding sites. Often the individual subunits are identical. If the binding of a ligand (a small molecule that binds to a larger macromolecule) to one site alters the affinity at other sites on the same oligomer then this is called cooperativity. If ligand binding increases the affinity of subsequent binding events, it is termed positive cooperativity whereas if the affinity decreases then it is termed negative cooperativity. One characteristic of positive cooperativity is that it results in a sigmoidal response instead of the usual hyperbolic response. Such a curve is illustrated in Figure 6.2. A corresponding Michaelian curve is shown for comparison. The first instance of cooperativity was observed in hemoglobin [14]. Hemoglobin is responsible for carrying oxygen (and to some extent carbon dioxide) to and from body tissues and the lungs via the blood system. For many oligomeric proteins the subunits are identical. In hemoglobin however there are two distinct subunit types, ˛ and ˇ. Each subunit carries a heme group with a central bivalent iron. When oxygen binds to one of the heme groups it causes a conformation change in the other subunits, which in turn increases the binding affinity in the remaining heme groups. The net effect is to give the binding of oxygen to hemoglobin a sigmoidal response. The physiological effect is to allow hemoglobin to rapidly pick up oxygen in the lungs where oxygen is plentiful but when hemoglobin reaches the

122

CHAPTER 6. COOPERATIVITY

4 Sugar Site 3

ATP/ADP Site 2

1

Figure 6.1: Bacterial Phosphofructokinase (E.C. 2.7.1.11) is composed of four identical subunits. Numbers mark the individual subunits. Image modified from RCSB Protein Data Bank and David Goodsell © (www.pdb.org, http://dx.doi.org/10.2210/rcsb_pdb/mom_ 2004_2)

peripheral tissues, where the partial pressure of oxygen is lower, the oxygen is released. The advantage of the sigmoid response compared to the simpler hyperbolic response is that once one oxygen molecule is released, it is easier for the remaining oxygens to unbind. This mechanism ensures that hemoglobin in the lungs is almost fully saturated whereas in the peripheral tissues, hemoglobin is mostly free of oxygen. Cooperativity is also commonly observed in enzymes. For example the glycolytic enzyme phosphofructokinase (E.C 2.7.1.11) from Escherichia coli is made of up four identical subunits (Figure 6.1). This enzyme catalyzes the ATP dependent phosphorylation of Fructose-6-Phosphate (F6P) to Fructose-1-6-Bisphosphate. Each subunit in PFK has at least three binding sites, corresponding to sites for ATP, F6P and one site for ADP and phosphoenolpyruvate (PEP). Both the F6P and ADP/PEP sites are on subunit boundaries, which means their binding can change the binding affinities on other subunits. For example, the substrate F6P shows strong positive cooperativity [9].

123

1 Reaction Rate

0:8 0:6 0:4 0:2 0

0

0:5

1 1:5 2 2:5 Substrate Concentration

3

Figure 6.2: Plot comparing positive cooperativity (lighter line) to a hyperbolic response (darker line).

Cooperativity can also be found in proteins that control transcription by binding to special sites on DNA. The lactose repressor, LacI is one of the most well known transcription factors that regulates the expression of enzymes that digest lactose. LacI has two binding sites that can attach to two different but specific sites on the DNA. Cooperativity is thought to result from the fact that when one LacI site is bound, the second site on LacI finds it easier to locate the other site on the DNA because the search space is now greatly reduced. There are two definitions that occur frequently in the literature that relate to how a given ligand affects subsequently binding events. In particular, if the binding of one ligand (usually the substrate) affects the subsequent binding of similar ligands then the ligand is called a homotropic effector. Substrate binding to a cooperative systems is a classical example of a homotropic effector because once the first substrate molecule binds, subsequent substrate molecules find it easier (for positive cooperativity) to bind. In contrast, ligands that affect the binding of other ligands are called heterotropic effectors. The discussion of heterotropic effectors will be presented in the next chapter. The remainder of the chapter will be devoted to reviewing various models that describe homotropic cooperative

124

CHAPTER 6. COOPERATIVITY

behavior.

6.1 Hill Equation The Hill equation [30] was originally used to describe the sigmoid response found in the binding of oxygen to hemoglobin. In the original publication by Hill, a mechanistic explanation was not provided. It was only later that a mechanistic model was proposed and used to derive the Hill equation. The proposed mechanistic model is perhaps overly simplistic and even unrealistic, but it is described it here because it provides a useful baseline when considering other models. Consider an oligomer with n subunits and a binding site on each subunit for a ligand, S . If we make the assumption that when the first ligand binds, the binding affinity for the remaining n 1 sites change such that all the remaining ligands also bind simultaneously, then we can represent this situation as follows: E C n S ES

(6.1)

Assuming the rapid equilibrium assumption we can write: Ka D

ES E
Sn

where Ka is the association constant for ligand binding. Using the conservation relation E t D E C ES, the fractional saturation can be shown to be given by: ES Sn Sn D D DY Et 1=Ka C S n Kd C S n

(6.2)

This is the Hill equation (6.3) where Kd is the dissociation constant. Often the Hill equation is represented in the literature as:

vD

Vm S n Kd C S n

(6.3)

6.1. HILL EQUATION

125

where Kd is the dissociation constant, n the Hill coefficient and Vm the maximal velocity. Traditionally the Hill coefficient is represented using the symbol, h. The only reason why n is used here is because it specifically refers to the number of binding sites in the proposed model (6.1). In general, experimental determination of Hill coefficients often reveals fractional values indicating that the simple Hill model fails to adequately explain cooperativity, although empirically the fit can be quite good. For example, although hemoglobin has four binding sites, the measured Hill coefficient is 2.7. Similarly for PFK where the number of binding sites for F6P is four, the Hill coefficient is about 3.7. Most literature therefore refers to the Hill coefficient using the symbol h indicating that this is the measured coefficient; we will switch to this notation where appropriate. Sometimes the Hill equation is also expressed in terms of the half-maximal activity constant, KH , that is the concentration of ligand that gives half maximal activity. To do this we set the left-hand side (6.3) to 0.5 and find the relationship between S and Kd , such that: SD

p n

Kd

p n

p n Kd is the half-maximal activity value, or KH D n Kd , that is KH D Kd . We can therefore write the Hill equation in a number of alternative but equivalent forms:

vD

Sn

Vm D n KH C Sn

Vm

 

1C

S KH

S KH

n n 

V m Sn Kd C S n

(6.4)

The equation in terms of the half-maximal activity has advantages because half-maximal activity can sometimes be measured directly from experiments, especially transcription factors binding to operator sites on DNA (See chapter 9). At the beginning of the chapter it was shown that an 81-fold change in substrate concentration was required to change the reaction rate for a simple Michaelis-Menten rate law from 10% to 90% of maximum. We can do a similar calculation for the Hill equation. Table 6.1 shows the fold change

126

CHAPTER 6. COOPERATIVITY

Hill Coefficient

Fold Change

1 2 3 4

81 9 4.33 3

Table 6.1: Fold change in ligand concentration required to change the reaction rate from 10% to 90% of maximum.

for a variety of systems with different Hill coefficients. Compared to the 81-fold change for a hyperbolic response, cooperative responses are much more sensitive. For example, if n D 4, then it only takes a 3 fold change in the ligand concentration to change the rate 9 fold from 10% to 90% of maximal activity. If ligand binding acted in the way suggested in the derivation of the Hill equation, n would represent the number of binding sites, an integer. However, fitting the Hill equation to real data rarely gives integer estimates to n suggesting the model is not a good representation of the real system. The utility of the Hill equation however lies in its ability to describe sigmoid behavior for simple cooperative systems such as transcription factor binding. As a result, it has been adopted by the modeling community. However it is severely limited in some aspects. It is not possible to easily add regulator terms to the equation (although this is often done in an ad hoc manner) or model multi-reactant systems and more problematic is that it models an irreversible reaction. As we shall see, Hofmeyr and Cornish-Bowden [33] recently revisited the Hill equation and derived a new, more adaptable reversible Hill equation.

Elasticities The elasticity coefficient, "vS may be derived directly from the Hill equation (6.3). Differentiating and scaling the Hill equation yields the following elasticity both in terms of the dissociation constant, Kd and the half

6.1. HILL EQUATION

127

1 nD8

nD4

Reaction Rate

0:8

nD2

0:6 0:4 0:2 0

0

0:5

1 1:5 2 2:5 Substrate Concentration

3

Figure 6.3: Plot showing the response of the rate set to the indicated values and KH D 1.

maximal activity constant, KH :

"vS D

n Kd D Kd C S n

KH

nKH   S n C KH

(6.5)

The elasticity of a reaction obeying the Hill equation has a value equal to n at low substrate concentrations (S  Kd ). In contrast, irreversible Michaelian enzymes at low substrate concentrations have an elasticity value of one. Therefore an enzyme obeying the Hill equation shows a much higher elasticity to the substrate concentration compared to a Michaelian enzyme. Like a Michaelian enzyme, the value of the elasticity falls off rapidly as the substrate concentration increases, reaching zero as the enzyme becomes saturated. Figure 6.4 illustrates this response for n D 4 and Kd D 1. An interesting feature in Figure 6.4 is the delayed fall in the elasticity at low substrate concentrations. This is in contrast to a Michaelian

128

CHAPTER 6. COOPERATIVITY

Reaction Rate, v or Elasticity, "vS

response which falls immediately from the initial point, S D 0. 4 3 "vS

2

v

1 0

0

0:5

1 1:5 2 2:5 3 Substrate Concentration, S

3:5

4

Figure 6.4: Plot showing the response of the rate and elasticity for the Hill model, with n D 4 and Kd D 1.

Equation (6.2) represents the fractional saturation for the Hill model. If we subtract both sides of this equation from one, we get: Sn Kd C S n

1

Y D1

1

Y D

Kd Kd C S n

D

Kd C S n Kd

1 1

Y

Multiplying both sides by the elasticity expression leads to the cancelation of the term, .Kd C S n /=Kd . Therefore the elasticity1 is related to the Hill coefficient by the simple relation: 1 The same result may also be obtained by expanding the slope of the Hill plot, n D d log.Y =.1 Y //=d logS and extracting the elasticity.

6.2. LIGAND BINDING

129

"vS

1

Dn 1 Y This means that at low saturation levels, the elasticity coefficient approximately equals the Hill coefficient. Another way of expressing this relation: "vS D n.1

Y/

(6.6)

The elasticity is proportional to the degree to which the enzyme is not saturated (cf. 3.16). If a cooperative enzyme is at half-saturation then "vS D 1 2n

6.2 Ligand Binding We have already introduced definitions for the dissociation and associations constants in the first chapter. To develop more realistic models of cooperativity we need to further explore some important ideas in ligand binding. Consider a protein dimer (Figure 6.5), where each monomer of the protein can bind a single ligand, S . Also assume each monomer has identical binding constants for the ligand, that is the monomers are identical and independent of each other. Figure 6.5 shows the various states in which the dimer can exist. Note there are two routes by which the dimer can go from a fully empty dimer to a fully bound dimer depending on which monomer unit is bound first. The rate constants k1 and k2 describe the binding and unbinding of ligand from a vacant monomer and are identical for each binding reaction. The association constant for this process is therefore ka D k1 =k2 . The individual association constants that describe each individual binding step are called microscopic association constants and in this example they are all identical. Microscopic constants will always be denoted by a lower case, k. Although the microscopic constants can be defined, they are not accessible experimentally because we cannot distinguish between the two half-bound

130

CHAPTER 6. COOPERATIVITY

Figure 6.5: The binding of a ligand, S to a dimer. k1 and k2 are the microscopic rate constants. The microscopic association constant is given by, ka D k1 =k2 .

states (Figure 6.6). Instead all we can do is measure the total concentration of the half bound dimer. As a result the system has only three measurable states (Figure 6.6).

( D00

) D10

D01

D11

Figure 6.6: Three Measurable States in a Dimer where S is the ligand.

For the three measurable states we can define a set of new association constants (Figure 6.7). The experimentally accessible equilibrium constants are called macroscopic constants (also sometimes called the apparent constants). The relationship between the macroscopic and microscopic constants can be easily derived. Let the microscopic association constant between free dimer and singly bound dimer be ka D

D10 ; D00
S

ka D

D01 D00
S

(6.7)

6.2. LIGAND BINDING

K1

(

131

)

K2

(

)

Figure 6.7: Macroscopic Association Constants.

The macroscopic association constant is given by K1 D

D10 C D01 D00
S

Substituting the value for D10 and D01 from the microscopic constants into the macroscopic relation yields K1 D 2 ka

(6.8)

A similar derivation can be done for the binding of the second ligand to yield: 1 K2 D ka (6.9) 2 Notice that the macroscopic association constant is different for the binding of ligand to the empty dimer and the half-filled dimer, even though the microscopic dissociation constants are identical. In particular, the macroscopic association constant for the first step is four times smaller than that of the second step. The explanation is due to the fact that there is a choice of two ligand states that can unbind to form D00 and likewise there are two ligand states that can unbind from D11. The change in macroscopic constants is thus a purely statistical effect. When deriving the rapid equilibrium rate law for systems such as this, we must be careful to distinguish between the two types of constant. For more complex patterns of binding, there is a formula for deriving the relationship between macroscopic and microscopic constants. Here we will derive the relationship in terms of dissociation constants which are the more commonly reported in the literature. Consider a macromolecule M , with n binding sites for a ligand S . Assume each site has the same microscopic dissociation constant, kd and is independent of the number of

132

CHAPTER 6. COOPERATIVITY

ligands bound. Let Mi denote the total concentration of the microscopic state with i bound ligands. For example, M1 is the total concentration of all states with one ligand bound. For a tetramer with n D 4, Figure 6.8 shows all possible bound states categorized by the number of ligands bound. The number of possible states for

Figure 6.8: The different bound states for a tetramer macromolecule binding a ligand, S . Mi refers to the total concentration of a configuration with i bound ligands.

a given number of ligands can be calculated from combinatory mathematics. For a macromolecule with n sites, the number of ways (C ) to place i ligands is given by ! n nŠ D (6.10) C.n; i / D .n i /Š i Š i Thus, if n D 4 and the number of ligands, i D 2, there will be .4Š/=..4 2/Š2Š/ ways to arrange the ligand, that is, 6 ways (see M2 in Figure 6.8). If n D 6 and the number of ligands, i D 2, there will be .6Š/=..6 2/Š2Š/ or 15 ways to arrange the ligand.

6.2. LIGAND BINDING

133

Equation (6.10) can be used to compute the general relationship between microscopic and macroscopic association constants. Let us assume that for a given configuration, Mi , there are equal amounts of each state in the configuration. This is not an unreasonable assumption since all the states are kinetically identical. Thus for M1 there will be present equal amounts of each of the four possible ligand configurations. Therefore the concentration of any one state within a configuration, Mi , is given by Mi =C.n; i / where C.n; i/ is the number of possible states within the configuration Mi .

=

+

+

+

Figure 6.9: Mi is the total concentration of states with i bound ligands on a tetramer, in this case one ligand. The average concentration of a particular configuration is Mi =C.4; i /.

The macroscopic dissociation constant, Ki between two configurations, Mi and Mi 1 is given by: Mi Mi Ki D

1

CS

Mi 1 S Mi

The microscopic dissociation constant, kd , can be computed from one of the individual states. Since we know the concentration of a particular state in a configuration, Mi , is Mi =C.n; i /, the microscopic dissociation constant is: .Mi 1 =C.n; i 1//S kd D Mi =C.n; i / Combining the macroscopic and microscopic relationships yields the result: C.n; i 1/ Ki D kd C.n; i / Using this result and the example shown in Figure 6.8, the relationship between the microscopic and microscopic dissociation constants must be: K1 D

1 kd 4

K2 D

2 kd 3

K3 D

3 kd 2

K4 D

4 kd 1

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CHAPTER 6. COOPERATIVITY

A standard result in binomial coefficients is that the ratio of two combinations C.n; i 1/ and C.n; i / is given by i=.n k C 1/. Thus we can also write: i (6.11) Ki D kd n i C1 which makes it easier to compute the weighting factors. In terms of association constants it should be evident that: Kai D ka

i C1 i

n

(6.12)

In the case of a dimer (n D 2) there are two M states, i D 1 and i D 2. That is when i D 1, Ka1 D ka 2 and when i D 2, Ka2 D ka 1=2. These results correspond to the results (6.8) and (6.9) which where calculated from first principles. Example 6.1 Determine the relationship between the microscopic and macroscopic equilibrium constants for a trimer. For a trimer, n D 3. The number of M states is four: empty, single ligand, two ligands and three ligands. Although we don’t need to compute the number of combinations of ligand patterns, we can compute these from equation (6.10). That is for a trimer with one or two ligands, there are three possible combinations; for three ligands there is only one combination. To compute the relationship between the microscopic and macroscopic equilibrium constants we will use dissociation constant equation (6.11). For a trimer there will be three equilibrium constants, K1 ; K2 and K3 . We assume for this calculation that the microscopic binding constant is the same for each binding possibility. Therefore, when i D 1, that is when one ligand binds: K1 D ka

n

i D ka i C1 3

1 1 D ka 1C1 3

and for the remaining two: K2 D ka

3

2 D ka 1 2C1

K3 D ka

3

3 D ka 3 3C1

6.3. THE ADAIR EQUATION

135

6.3 The Adair Equation The derivation of the Hill equation takes into account the number of binding sites on the oligomer. This means that in principle, fitting the Hill equation to the hemoglobin saturation curve should allow us to predict the number of oxygen binding sites. As already mentioned, the measured Hill coefficient for hemoglobin is 2.7. In 1910 when Hill derived his equation, the number of binding sites on hemoglobin was not known. However by 1925 Adair was able to show, by measuring the molecular weight, that hemoglobin probably had four binding sites. This suggested that the Hill equation was inappropriate for describing cooperativity in hemoglobin since the Hill equation predicted 2.7 binding sites. Instead, Adair proposed a four-step binding process: Hb C O2 HbO2 HbO2 C O2 Hb(O2 )2 Hb(O2 )2 C O2 Hb(O2 )3 Hb(O2 )3 C O2 Hb(O2 )4 In this analysis, Adair proposed that each binding step had a different dissociation constant so that as oxygen bound to hemoglobin the dissociation constant progressively increased. To illustrate this idea, consider a simpler problem, an oligomer, P , with two binding sites (Figure 6.10). To begin with let us assume that the microscopic binding constants are equal to each other. The fractional saturation, Y , is given by: Y D

Number of bound sites Total number of binding sites

The total number of bound sites is: PS C 2P2 S. Note that there are two bound sites in P2 S. The total number of all sites is given by: 2.P C PS C 2P2 S/. We multiply by 2 because each state has two binding sites. From this we can define Y :

136

CHAPTER 6. COOPERATIVITY

K1

PS

K2

P2 S

P

K1

PS

K2

Figure 6.10: Ligand binding to a dimer. K1 and K2 are the macroscopic association constants.

Y D

PS C 2P2 S 2.P C PS C 2P2 S/

Some authors like to express binding in terms of the number of moles of ligand bound per mole of protein because this is easily measured experimentally. This quantity is more often denoted by the symbol r. This value has a maximum corresponding to the number of binding sites, that is, r D nY . The macroscopic association constants, K1 and K2 are given by: K1 D

PS P
S

K2 D

P2 S PS
S

Substituting these into the fractional saturation expression to eliminate PS and P2 S and canceling the P term yields:

Y D

K1 S C 2K1 K2 S 2 2.1 C K1 S C K1 K2 S 2 /

(6.13)

This is the Adair equation for the binding of ligand to a dimeric protein in terms of the macroscopic association constants.

6.3. THE ADAIR EQUATION

137

If there are no interactions between the subunits we can show that the Adair equation describes a hyperbolic response. To demonstrate this effect we must first consider the relationship between the macroscopic and microscopic association constants. From the previous section we know that for a dimer the macroscopic and microscopic association constants are related as follows: K1 D 2 k1 1 K2 D k1 2 Here we assume that all the microscopic association constants are the same. If we substitute these relationships into equation (6.13) we obtain: Y D

Y D

1 2 2 k1 S 2k1 21 k1 S 2 /

2k1 S C 2
2 k1 2.1 C 2k1 S C

k1 S C k12 S 2 k1 S.1 C k1 S / k1 S D D 2 2 2 .1 C k1 S / 1 C k1 S 1 C k1 S C k1 S

This result describes a simple hyperbolic response. Given that there are no interactions between the subunits, this is not an unexpected result, each subunit is acting independently as if it were a single enzyme. Since K1 D 2 k1 and K2 D 1=2 k1 then for identical microscopic association constants, K1 D 4 K2 . However if the second ligand binding has a higher microscopic association constant than the binding of the first ligand, then 4 K2 > K1 . Figure 6.11 illustrates three curves generated from the Adair equation. When K1 is four times K2 the response is a simple hyperbolic curve. Once K1 is less than four times K2 positive cooperativity can be observed. Although the sigmoid shape in Figure 6.11 is not so pronounced, additional binding sites can greatly enhance the sigmoid behavior. In the case when K1 > 4 K2 the curve is neither sigmoidal nor hyperbolic. In this situation the system shows negative cooperativity, where the curve initially rises faster but slows down quickly as it approaches saturation.

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CHAPTER 6. COOPERATIVITY

Fractional Saturation, Y

1 0:8 0:6 0:4 0:2 0

0

1 2 3 Substrate Concentration, S

4

Figure 6.11: Plot comparing positive cooperativity to a hyperbolic response for the Adair dimer model with different macroscopic constants. Lower curve (Hyperbolic): K1 D 4 K2 ; Middle curve: K1 D 2 K2 ; Upper curve: K1 D 0:5 K2 . Both middle and upper curves show sigmoid behavior.

The Adair equation can be extended to higher order oligomers, for example a trimer protein that binds a ligand S has the following Adair equation: Y D

K1 S C 2K1 K2 S 2 C 3K1 K2 K3 S 3 3.1 C K1 S C K1 K2 S 2 C K1 K2 K3 S 3 /

where as before the Ki terms are the macroscopic association constants. A pattern can be seen when we compare the dimer and trimer equations and extending the Adair equation to higher order oligomers should be straight forward.

Elasticity The elasticity coefficient for the Adair model with n D 4 is shown below and in Figure 6.12.

6.3. THE ADAIR EQUATION

"vS D

139

1 C 4K2 S C K1 K2 S 2
.1 C 2K2 S / 1 C K1 S C K1 K2 S 2

There is a significant difference between the elasticity curves for the Adair and Hill equations, a difference we will also see in later models. The elasticity for the Hill equation at zero substrate is equal to n while the elasticity for the Adair equation at zero substrate is one. The Adair equation therefore behaves more like the Michaelis-Menten equation where the elasticity of one reflects that fact that the reaction rate behaves as a first-order reaction. For the Hill equation, the reaction at low substrate behaves as a nth order reaction. This difference many have significant implications on how perturbations propagate through a network. Even though the cooperative response in both cases look remarkably similar, the elasticity coefficients behave quite differently.

Elasticity, "vS

2 1:5 1 0:5 0

0

0:5 1 1:5 Substrate Concentration, S

2

Figure 6.12: Plot comparing the elasticity coefficient for an Adair model to a simple hyperbolic response. Upper curve right (hyperbolic): K1 D 4 K2 ; Middle upper curve: K1 D 0:252 K2 ; Middle lower curve: K1 D 1=16 K2 ; Lower curve: K1 D 1=600 K2 . Note that in contrast to the Hill equation, the elasticity at low substrate concentration is close to one; for the Hill equation the elasticity is close to n.

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CHAPTER 6. COOPERATIVITY

6.4 MWC Model Experimental studies on real enzymes indicate that the simple Hill model (and to some extent the Adair model) is too simplistic an explanation for cooperative behavior and has prompted a number of authors to devise other likely models. Furthermore there are significant limitations of these earlier models, for example difficulties in incorporating reversibility (see 6.6 for an alternative), the addition of effectors, and in the case of the Hill model, a completely unrealistic physical basis. In view of some of these limitations, researchers have devised alternative models. Of particular importance is the symmetry model put forth by Monod, Wyman and Changeux in 1965 (MWC model, [49]). One striking feature of many oligomeric proteins is the way individual monomers are physically arranged. Often one will find at least one axis of symmetry. The individual protein monomers are not arranged in a haphazard fashion. This level of symmetry may imply that the gradual change in the binding constants as ligands bind, as suggested by the Adair model, might be physically implausible. Instead one might envisage transitions to an alternative binding state that occurs within the entire oligomer complex. The original authors laid out the following criteria for the MWC model: 1. The protein is an oligomer. 2. Oligomers can exist in two states: R (relaxed) and T (tense). In each state, symmetry is preserved and all subunits must be in the same state for a given R or T state. 3. The R state has a higher ligand affinity than the T state. 4. The T state predominates in the absence of ligand S . 5. The ligand binding microscopic association constants are all identical, this is in complete contrast to the Adair model Given these criteria, the MWC model assumes that an oligomeric enzyme may exist in two conformations, designated T (tensed, square) and R (relaxed, circle) with an equilibrium between the two states with equilibrium

6.4. MWC MODEL

141

constant, L D T =R, also called the allosteric constant. If the binding constants of ligand to the two states are different, then the distribution of the R and T forms can be displaced either towards one form or the other. By this mechanism, the enzyme displays sigmoid behavior. A minimal example of this model is shown in Figure 6.13.

L L= Figure 6.13: A minimal MWC model, also known as the exclusive model, showing alternative microscopic states in the circle (relaxed) form. L is called the allosteric constant. The square form is called the tense state.

Exclusive Model (Binds only to the R state) In the exclusive model (Figure 6.13), the ligand can only bind to the relaxed form (circle). The mechanism that generates sigmoidicity in this model works as follows. When ligand binds to the relaxed form it displaces the equilibrium from the tense form to the relaxed form. In doing so, additional ligand binding sites are made available. Thus one ligand binding may generate four or more new binding sites. Eventually there are no more tense states remaining at which point the system is saturated with ligand. The overall binding curve will therefore be sigmoidal and will show positive cooperativity. Given the nature of this model, it is not possible to generate negative cooperativity. To derive the rate law for the exclusive this model, we first write down the

142

CHAPTER 6. COOPERATIVITY

T2

R2

R 2S R 2 S2

L L=

=

T2 R2

R 2S

Figure 6.14: Simple MWC model with labeled species.

equation that describes the fractional saturation, Y : Y D

Number of occupied binding sites Total number of binding sites

D

R2
S C 2R2
S2 2.R2 C R2
S C R2
S2 C T2 /

As with the Adair model, the 2 in the denominator indicates that each state has two binding sites. R2 S is the total concentration of the half-bound state. The equilibrium constant (also known as the allosteric constant) between T2 and R2 is given by L D T2 =R2 . This equation allows us to remove T2 from the fractional saturation function to yield: Y D

R2 S C 2R2 S2 2.R2 C R2 S C R2 S2 C L
R2 /

Now we can make some assumptions about the association constants for the binding of ligand. In accordance with the MWC model, all microscopic ligand association constants, k are assumed to be equal. We can define the macroscopic association constants as: K1 D

R2 S R2
S

so that R2 S D K1 R2
S

Likewise we can define the second macroscopic association constant to be:

6.4. MWC MODEL

K2 D

143

R2 S2 R2 S
S

so that R2 S2 D K2 R2 S
S

Using the relationship between macroscopic and microscopic association constants (6.12), it is straight forward to show that: K1 D 2 k

and

K2 D

1 k 2

We can now substitute the microscopic constants in place of the macroscopic constants and eliminate the complex terms from the fractional saturation expression. For example, given that: R2 S D K1 R2
S We substitute the macroscopic constant for the microscopic one: R2 S D 2 kR2
S In the second binding step:

K2 D

R 2 S2 R2 S
S

so that R2 S2 D K2
R2 S
S D K1 K2
R2
S 2

Again substituting the macroscopic constants gives: 1 R2 S2 D 2k kR2
S 2 D k 2 R2
S 2 2 Eliminating R2 S and R2 S2 from the fractional saturation equation yields: Y D D

2 kR2
S C 2k 2 R2
S 2 2.R2 C 2 kR2
S C k 2 R2
S 2 C L
R2 / kS C k 2
S 2 1 C 2kS C k 2
S 2 C L

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CHAPTER 6. COOPERATIVITY

We can factor the terms in the numerator and denominator to give the result: Y D

kS .1 C kS / .1 C kS /2 C L

In the literature this equation is usually given in terms of the microscopic dissociation constant, kR , that is k D 1=kR so that:   S S 1C k kR Y D R 2 S 1C CL kR This is the simplest Monod-Wyman-Changeux equation for a dimer. It is possible of course to extend the derivation to n subunits, in which case the equation is written as:   S S n 1 1C kR kR Y D   S n 1C CL kR

(6.14)

Given the ubiquity of terms such as the S=K in rate laws, it is common to simplify the notation by introducing Greek letters to represent these ratios. Thus, ˛ is often used to represent the normalized substrate concentration, S=K, and  to represent the normalized product concentration, P =K. Normalized substrate and product notation: ˛D

S Ks

D

P Kp

With this notation we can rewrite these equations in the following form:

6.4. MWC MODEL

145

Y D

˛ .1 C ˛/n 1 .1 C ˛/n C L

(6.15)

Figure 6.15 shows the response of the simple MWC model to changes in substrate concentrations at different values of L and clearly shows the sigmoid character. In all these cases we make the usual assumption that the rate of reaction is proportional to Y . Note the special case when L D 0 and equation (6.15) reduces to the familiar hyperbolic form: Y D

˛ 1C˛

This corresponds to the case when there is no T (Square) form. Since all the microscopic constants are equal, each binding site behaves independently and therefore the equation reduces to a simple hyperbolic form.

Elasticities We can derive (Proof at end of chapter) the elasticity for the simple MWC model such that:

"v˛ D

1 C n˛ 1C˛

nY

(6.16)

This equation shows that the elasticity is the difference between a hyperbolic term, .1 C n˛/=.1 C ˛/, and a sigmoid function, nY . At low or zero substrate concentrations, the elasticity is equal to one. As ˛ rises, the hyperbolic part increases faster than the sigmoid, therefore the elasticity increases above one. As the substrate concentration continues to rise the rate of increase in the difference slows down, reaching a maximum and then starts to fall. At high substrate concentration, the elasticity tends to zero. Response curves showing this behavior are given in Figure 6.16. An

146

CHAPTER 6. COOPERATIVITY

1

Reaction Rate

0:8 LD0

0:6 LD100

0:4

LD1000 LD5000

0:2 0

0

2

4

6

8

10

˛ Figure 6.15: Effects of substrate concentration (proportional to ˛) on the rate of reaction for the simple MWC equation (6.15). Four curves are shown, each describing the response at different values of L. At L D 0, the enzyme behaves in a hyperbolic Michaelian fashion. For increasing values of L, the sigmoid character of the cooperative enzyme increases as the initial equilibrium is more in favor of the tense (square) form.

examination of these curves shows that the peak of the elasticity response occurs below the steepest point on the sigmoid curve. Also worth noting is that the elasticity cannot exceed n. The elasticity behavior exhibited by the MWC model is in marked contrast to the Hill response shown by the Hill equation. In the Hill model the elasticity starts at n, then falls. In the MWC model the elasticity starts at one then raises towards n before falling. This is similar to the Adair model. Even though the velocity curves of the Hill and MWC equations look similar, the elasticity response is quite different. The elasticity response for the MWC model has some interesting features (Figure 6.16); the most notable is that the MWC model can elicit high values for the substrate elasticity (not unlike the Hill model). Whereas an irreversible Michaelis-Menten model can only achieve its highest elasticity either at very low substrate concentrations and then only a maximum

6.4. MWC MODEL

147

of one, or close to equilibrium (neither of which are of great interest). The MWC model can give elasticities up to n and at intermediate concentrations of substrate. This might be an important property in an intact pathway. Elasticity

Rate

Reaction Rate or Elasticity

4 3

LD5000 LD1000

2

LD100

1 0

LD0 LD100 LD1000 LD5000

LD0

0

2

4

6

8

10

˛ Figure 6.16: Effects of substrate concentration (proportional to ˛) on the rate of reaction and the substrate elasticity coefficient for an enzyme following the MWC equation(6.15). Four sets of curve are shown, each describing the response at different values of L, the allosteric constant. At L D 0, the enzyme behaves in a normal Michaelian enzyme fashion. For increasing values of L, the sigmoid character of the allosteric enzyme increases. In addition, the elasticity coefficient shows a change of behavior and at low to intermediate values of substrate concentration, it increases in value reaching a peak below the steepest part of the rate response.

Generalizing MWC (Binds to R and T states) A more generalized extension to the MWC model is shown in Figure 6.17. This model introduces another constant, usually denoted by the symbol, c. This constant represents the ratio of the microscopic dissociation constant

148

CHAPTER 6. COOPERATIVITY

for the relaxed form, kR , and the microscopic dissociation constant for the tense form, kT . In the generalized model, ligand can bind to both the R and T states. As a result, this model is also known as the nonexclusive model. In the previous model, ligand could only bind to the R state (exclusive model). kR cD kT

L

kT

kR

kT

kR

Figure 6.17: Generalized MWC Model. kR and kT are the ligand binding microscopic dissociation constants for the relaxed and tense forms.

By applying the same approach that was used with the simple MWC model it is possible to derive the non-exclusive (generalized) MWC rate law (6.17). The only difference with the simple MWC model is the presence of the c term.

Y D

˛.1 C ˛/n 1 C Lc˛.1 C c˛/n .1 C ˛/n C L.1 C c˛/n

1

(6.17)

The concentration of S in equation (6.17) has been replaced by the normalized concentration, ˛, that is S=k. L and c are constants relating to the degree of cooperativity and n the number of protomers making up the oligomer.

6.4. MWC MODEL

149

If we assume that Y is proportional to the reaction rate then the substrate elasticity, "vS , can be obtained directly from the MWC equation and equals (See end of chapter for proof):

"v˛ D

1 C ˛cn 1 C ˛c

˛.c 1/.n 1/
.1 C ˛/ 1 C ˛c C .1 C ˛/1 n c.1 C ˛c/n L

nY (6.18)

If we set c to zero, when the tense state cannot bind substrate, the elasticity expression simplifies to the simple MWC model elasticity: "v˛ .cD0/ D

1 C n˛ 1C˛

nY

(6.19)

The non-exclusive form will become more interesting when we consider allosteric regulation in the next chapter.

Reversible MWC Model It is possible to derive a reversible form of the exclusive MWC model [52] which is given by equation (6.20) [51, 57]. This equation models exclusive binding such that substrate and product only bind to the relaxed state (R) and that only the R state is active: v D Vf ˛

.1

/ .1 C ˛ C /n L C .1 C ˛ C /n

1

(6.20)

The elasticity coefficients for the reversible system are given by [57]: "vS D v "P D

1 1

 

1



C

˛.n 1/ .1 C  C ˛/

C

.n 1/ .1 C  C ˛/

h˛ .1 C  C ˛/h 1 L C .1 C  C ˛/n h .1 C  C ˛/n 1 L C .1 C  C ˛/n

The non-exclusive reversible MWC model [52] is slightly more complex but is given by [34]:

150

v D Vf ˛

CHAPTER 6. COOPERATIVITY

.1

/ .1 C ˛ C /n 1 C aL.1 C cs ˛ C cp /n .1 C ˛ C /n C L.1 C cs ˛ C cp /n

1

(6.21)

Instead of a single c term, there are now two, cs and cp . cs represents the ratio of the microscopic equilibrium constant for the binding of substrate to the R form and the T form, that is, cs D KsR =KsT . Likewise, cp is the equivalent ratio but for product binding, that is Kp D KpR =KpT . The term a is the activity of the T state relative to the R state, that is a D .Vf T =KsT /=.VrT =KsR /.  is the mass-action ratio.

6.5 KNF Sequential Model Around the same time the MWC was published, an alternative model called the Koshland, Nemethy and Filmer model [40] or more simply the KNF sequential model was also introduced. Recall that in the MWC model, when a conformation change occurs, all subunits of the oligomer change at once. In the KNF model, individual monomers may change state while leaving the remaining monomers unaffected. The conformational change within an individual oligomer are therefore sequential. In this sense the KNF model is more like the Adair model. The KNF model also permits binding of a second ligand to be strengthened or weakened and as a result the model can generate negative as well as positive cooperativity. The resulting binding equations are fairly complex and often have to be rederived depending on the postulated state changes and arrangements of the protomers in the oligomer. In addition, it is very difficult to distinguish different KNF models from experimental data. The utility of the KNF model for modeling may therefore not be so important. The sequential and symmetry models can be seen as extremes of a more general model involving all possible configurations. Figure 6.18 shows the combined model for the case of a dimer complex. These combined models are even more complex to deal with algebraically and as a result, alternatives have been sought. One particular alternative worth mentioning is the reversible Hill equation.

6.6. REVERSIBLE HILL EQUATION

151

Figure 6.18: Generalized MWC/Sequential model for a dimer.

6.6 Reversible Hill Equation In the enzyme kinetics literature much attention is given to the molecular mechanisms that generate cooperativity. For modeling purposes, simple rate models such as the Hill equation may be sufficient without too much concern for the detailed molecular mechanism. However the main problem with the Hill equation is that it describes an irreversible reaction. In recent years, Hofmeyr and Cornish-Bowden [33] published a description of the reversible Hill equation with modifiers. The basic mechanism is shown in Figure 6.19. From the reaction scheme in Figure 6.19 we can write the net reaction rate, v, as: v D k1 .ES C 2ES2 C ESP/

k2 .EP C 2EP2 C ESP /

Invoking the rapid-equilibrium assumption we can write the various complexes in terms of equilibrium constants to give: vD where  D

Vf ˛ .1

/ .˛ C /

1 C .˛ C /2

=Keq . For an enzyme with h (using the authors original

152

CHAPTER 6. COOPERATIVITY

k1

S

S S

k2

2 k1

S P

k2

P

k1 2 k2

P P

Figure 6.19: Reversible Hill Mechanistic Scheme. k1 and k2 are the catalytic rate constants for the conversion of substrate to product and vice versa. The factor of two is due to the fact that the fully bound dimers have twice the rate of conversion (because there are two possibilities).

notation) binding sites, the general form of the reversible Hill equation is given by:

vD

Vf ˛ .1

/ .˛ C /h

1 C .˛ C /h

1

(6.22)

Figure 6.20 illustrates the sigmoid behavior with respect to the substrate concentration. The K constants in the equation are the half saturation constants. is the mass-action ratio and Keq the equilibrium constant for the reaction. The significance of formulating this is that the thermodynamic terms are explicitly separated from the saturation terms. The equation also reduces to familiar forms when certain restrictions are applied. For example if h D 1, the equation reduces to the non-cooperative reversible Michaelis rate law and if reversibility is removed, the equation reduces to the simple irreversible Michaelis-Menten rate law. The equation can also revert to the product inhibited but irreversible rate law by setting the Keq to infinity. The reversible Hill equation is therefore quite flexible and can

6.6. REVERSIBLE HILL EQUATION

153

be used in many situations.

Reaction Rate

1

hD4 hD2 hD1

0:8

1

0:6

0:6

0:4

0:4

0:2

0:2

0

0

5

10

hD4 hD2 hD1

0:8

0

15

0

S

5

10

15

S

Figure 6.20: Plot showing the response of the reaction rate for a reversible Hill model with respect to the substrate as a function of the Hill coefficient, h. The parameters were set as follows: Vm D 1; D 2; Keq D 10:95; Kp D 0:5; Ks D 2:75; Left Panel P D 1, Right Panel P D4

The reversible Hill equation also shows another interesting property. Under a certain set of parameter values, the product concentration can act as a positive regulator (Figure 6.21).

Elasticities If we first consider the simpler case where P D 0, then the substrate and product elasticities can be shown to equal: "vS D

hKsh Ksh C S h

v "P D0

This demonstrates that the elasticities revert to the simple irreversible Hill equation elasticities shown in equations (6.5). If we now include a non-zero product concentration, the elasticities change to:

154

CHAPTER 6. COOPERATIVITY

Reaction Rate

0:6

0:4

0:2

0

0

2 4 Product Concentration

6

Figure 6.21: Plot showing the response of the reaction rate for a reversible Hill model with respect to product concentration. S D 1; h D 4; Ks D 0:75; Kp D 4; Keq D 10; Vm D 1.

"vS

D

v "P D

1 1

˛.h 1/ C . C ˛/ 

 1



C

.h 1/ . C ˛/

h˛ . C ˛/h

1

1 C . C ˛/h h . C ˛/h

1

1 C . C ˛/h

If we set h D 1, the equations reduce to the reversible Michaelis elasticity equations (See equations (??)).

Chapter Highlights  Cooperativity is where the binding of a ligand to one site alters the affinity at other sites on the same oligomer. If ligand binding increases the affinity of subsequent binding events, it is termed positive cooperativity whereas if the affinity decreases then it is termed negative cooperativity.

6.6. REVERSIBLE HILL EQUATION

155

 One measure of cooperativity is the Hill coefficient, h. If h > 1 then we say the system shows positive cooperativity while if h < 1 the system shows negative cooperativity. Another measure of cooperativity is the cooperativity index, R. It is defined as the ratio of activity of substrate (or ligand) gives that 90% to the activity when substrate is at 10%.  A cooperative response in a protein with respect to binding by a ligand is often sigmoidal in appearance.  The Hill equation is the simplest rate law that gives a sigmoidal response. Mechanistically it is based on the premise that all binding sites for a given ligand are bound at once. Although mechanistically the Hill equation is unrealistic, the equation itself will often fit to experimental data quite well.  The elasticity for the Hill equation starts with a value given by the Hill coefficient then declines to zero as ligand concentration increases.  When deriving binding equations for oligomers, careful attention should be given the difference between the microscopic and macroscopic constants.  The microscopic (sometimes also called the intrinsic) rate constants are the individual rate constants for specific binding events. Because many binding events in a oligomer are indistinguishable from each other, microscopic rate constants are not always observable. Instead, we rely on the macroscopic (or apparent) rate constants which are the result of combining the indistinguishable binding events in to an observable event. Although it is not possible to measure all microscopic rate constants it is possible to derive mathematical relationships between the two.  Rate laws that describe cooperative kinetics are almost always derived using the rapid equilibrium assumption. This techniques derives the degree of saturation of binding sites as a function of all binding sites, also called the fractional saturation.

156

CHAPTER 6. COOPERATIVITY

 An important historical model is the Adair model. This assumes that ligands bind one at a time to an oligomer. Each time a ligand binds the binding constants for the remaining unfilled binding sites are strengthened.  If the binding of one ligand (usually the substrate) affects the subsequent binding of similar ligands then it is called homotropic cooperativity. Ligands that affect the binding of other ligands are called heterotropic effectors.  The Monod, Wyman and Changeux model (MWC) was the first comprehensive model to explain both homotropic and heterotropic interactions. The MWC model is based on the idea of symmetry change. Two symmetrical states are envisaged, R and T . In the simplest case, ligand can only bind to one of the states. For an activator, the bound state is assumed to have a higher activity. When ligands bind, it pulls the equilibrium from the more inactive T states to the more active R state.  The Koshland, Nemethy and Filmer (KNF) model assumes that individual monomers in an oligomer can change state. In the MWC model, a state change is the result of all monomers changing together. The KNF model relaxes this restriction.  In recent years, new generalized rate laws have appeared, one of the most celebrated being the reversible Hill equation. Unlike the majority of models that describe cooperativity, this model describes a fully reversible system.

Further Reading 1. Cornish-Bowden, A (2004). Fundamentals of enzyme kinetics (3rd ed.). London: Portland Press. ISBN 1-85578-158-1 2. Leskovac V (2003) Comprehensive Enzyme Kinetics. Springer. ISBN: 9780306467127

6.6. REVERSIBLE HILL EQUATION

157

3. Palmer, T (1995) Understanding Enzymes, 4th Edition. Prentice Hall. ISBN: 0131344706 4. Segal I H (1993) Enzyme Kinetics: Behavior and Analysis of Rapid Equilibrium and Steady-State Enzyme Systems (Wiley Classics Library) Wiley-Interscience. ISBN: 0471303097 5. Tipton KF (1979) in Kinetic properties of allosteric and cooperative enzymes. Bull AT, Lagnado JK, Thomas JO, and Tipton KF (Eds.), Companion to Biochemistry Volume 2. Selected Topics for Further Study. Longman, London, chapter 11, 327–382

Proofs 1. Proof of (6.18) Show that the elasticity of following equation with respect to ˛ is: Y D

˛.1 C ˛/n 1 C Lc˛.1 C c˛/n .1 C ˛/n C L.1 C c˛/n

1

is "v˛ D

1 C cn˛ .1 c/.n 1/˛.1 C ˛/n 2
C 1 C c˛ 1 C c˛ .1 C ˛/n 1 C cL.1 C c˛/n

1




nY

The easiest way to derive this elasticity is to first use elasticity differentiation to separate out the numerator and denominator, that is: "v˛ D ".Numerator/

".Denominator/

Start with the denominator because that is easiest. Using the summation rule for elasticity differentiation we obtain: ˛.1 C ˛/n 1 n ˛c.1 C ac/n 1 Ln C .1 C ˛/n C .1 C ˛c/n L .1 C ˛/n C .1 C ˛c/n L This can be simplified to:

158

CHAPTER 6. COOPERATIVITY


˛ .1 C ˛/n 1 C c.1 C ˛c/n 1 L n .1 C ˛/n C .1 C ˛c/n L Inspection of the above equation shows that it is equal to nY . We can also apply logarithmic rules together with partial fraction decomposition to yield: 1 C ˛cn ˛.c 1/.n 1/
C 1 n n 1 C ˛c .1 C ˛/ 1 C ˛c C .1 C ˛/ c.1 C ˛c/ L Finally combine the terms together to give: "v˛ D

1 C ˛cn 1 C ˛c

˛.c 1/.n 1/
.1 C ˛/ 1 C ˛c C .1 C ˛/1 n c.1 C ˛c/n L

nY

The equation can be greatly simplified by setting c D 0; this means that the relaxed state can not bind substrate so that it only has one state, fully empty. This simplification leads to: "v˛ D 1 C "v˛ D

˛.n 1/ 1C˛

1 C ˛n 1C˛

nY nY

which is the same as equation (6.19) in the main text.

Exercises 1. Define the term cooperativity in the broadest sense. 2. What is the key difference in terms of responsiveness between a hyperbolic and sigmoid response? 3. Describe two way in which cooperativity is generated in oligomeric proteins.

6.6. REVERSIBLE HILL EQUATION

159

4. Use a database such as EcoCyc to investigate the number of oligomeric enzymes in E. coli. 5. Show that "vE D 1 for the irreversible Hill equation and the nonexclusive MWC model. 6. Show that Hill equation expressed in terms of the dissociation constant and the half-maximal activity are the same. 7. Use the elasticity rules from chapter 2 to derive the substrate elasticity for the Hill equation. 8. Derive the relationship between the macroscopic and microscopic dissociation constants for a trimer model. 9. Derive the Adair equation for two subunits in terms of dissociation constants. 10. In the MWC model, what do we mean by the exclusive and nonexclusive models? 11. Using suitable software, investigate how the sigmoid behavior for the non-exclusive MWC is affected by the different constants. Classify the constants according to how they influence the rate curve. 12. Describe the difference between the L and c constants in the MWC model. 13. What is the key difference in the substrate elasticity response of the Hill equation compared to the Adair and MWC models?

160

CHAPTER 6. COOPERATIVITY

Adair Model, h D 2

Hill Model, h D 4 2

Elasticity, "vS

6

1:5

4

1 2 0

0:5 0

1

2 3 4 S MWC Model, h D 4

0

0

0:5

1 S

1:5

2

Elasticity, "vS

4 3 2 1 0

0

2

4

6

8

10

S

Figure 6.22: A comparison of the elasticity trends for three cooperativity models.

6.6. REVERSIBLE HILL EQUATION

1) Hill Model

161

2) Adair Model

+ 2

3) Minimal MWC Model

4) Generalized MWC Model

5) Reversible Hill Model

P

S

S S

S P

P P

Figure 6.23: A pictorial summary of the major protein cooperative models.

162

CHAPTER 6. COOPERATIVITY

7

Allostery

In a chapter 4 we saw an example of how to control enzyme activity by competitive inhibition. This can occur either by way of product binding to the enzyme, or by a closely related molecule that has some affinity for

163

164

CHAPTER 7. ALLOSTERY

the substrate binding site. What was not mentioned is the fact that inhibition by simple competition is not a particularly effective way to inhibit enzymes. There are two principle reasons for this. The first issue concerns poor sensitivity. This has been discussed before but it is worth repeating. Let us assume a simple Michaelis-Menten rate law where the Vm is set to a nominal value of 1. If we set the inhibitor concentration to zero, the rate of reaction is given by S=.S C Km /. We now wish to ask the question what fold increase in inhibitor will drop the reaction rate from 90% to 10% of maximum? To compute the fold change we need to find the inhibitor concentration that gives 90% and 10% of the maximum rate. For example, the 90% rate is given by the term 0:9 S=S.S C Km /. We set this to the competitive inhibition rate law (Equation 4.3) as shown below: 0:9

S S D S C Km S C Km .1 C I90 =Ki /

Solving for the inhibitor concentration leads to: I90 D

1=9 Ki .S C Km / Km

Similarly we can compute the inhibitor concentration required to achieve 10% of the normalized rate as: I10 D

9 Ki .S C Km / Km

The fold change in inhibitor required to bring about this rate reduction can be computed by dividing I10 by I90 . This yields an 81-fold change in inhibitor concentration to reduce the rate from 90% to 10%. This is clearly not a good way to control enzyme activity since it requires considerable changes in the concentration of the inhibitor. Another way of thinking about this is an inhibitor molecule (perhaps a metabolite) at 1 mM would have be increased to 81 mM to have a significant effect. In reality, metabolite concentrations tend to change very little when a cell changes state, except when entire pathways are switched on or off. More commonly, rate changes tend to vary much more. A striking example can be seen in insect

7.1. ALLOSTERIC ENZYMES

165

muscle [58] when the muscle changes its metabolic state from rest to fully working. Under these conditions, the glycolytic metabolite levels change by less than two fold, some hardly at all, while the glycolytic flux increases 100 fold. It would be impractical to achieve this level of control with just competitive inhibition, a more sensitive mechanism must be at work. To achieve greater regulation in reaction rates, something more sensitive than competitive inhibition is required. The answer, as suggested in the last chapter, is cooperativity. The second problem with competitive inhibition is that the inhibitors are not independent of the catalytic site, that is their action is by binding to the active site. It is always possible for the substrate to out-compete the inhibiting molecule if the substrate can be increased to a sufficiently high concentration. In other words the degree of inhibition is a function of the substrate and product concentration. A regulated system should probably not be able to influence the effectiveness of the controlling agents. At the enzymatic level, the solution to this problem is allostery. Although allostery and cooperativity are distinct phenomena, they are very often found associated together on the same protein. As we shall see, when we find cooperativity with respect to the substrate we will also often find cooperativity with respect to the allosteric effector.

7.1 Allosteric Enzymes In metabolic pathways metabolites have been found to act as inhibitors or activators of particular enzymes. What is unusual about these cases is that the affected enzymes are often working at a considerable metabolic distance from the effecting metabolite. For example, phosphofructokinase (Figure 6.1) can be inhibited by the glycolytic intermediate, phosphoenolpyruvate (PEP) even though this molecule is five reaction steps away. Similarly, it is known that in the biosynthetic pathway of cytidine triphosphate (CTP) from Escherichia coli, CTP can affect the activity of the enzyme Aspartate transcarbamylase (ATCase). However, ATCase is five enzymatic steps away from CTP. Since CTP and PEP are structurally different from the substrates of ATCase or PFK it is unlikely that CTP and PEP act as simple competitive inhibitors. Some metabolites can also act

166

CHAPTER 7. ALLOSTERY

as activators, such as fructose 1,6-bisphosphate on pyruvate kinase and in these situations competitive inhibition is not even an option. The question is, what mechanism do these enzymes use? In the case of PFK, there are, in addition to the active site, separate binding sites for regulator molecules such as PEP. For example E. coli PFK is a tetramer of identical subunits (unlike higher organisms such as yeast where the structure is more complex). Each subunit has an active site and a number of separate effector binding sites. The presence of a separate effector binding site allows chemically unrelated molecules to regulate enzyme activity without having to be structurally similar to the substrate or product of the enzyme. Sometimes the effector binding site is on the same protein molecule as the active site, sometimes separate protein monomers are dedicated to binding the regulator. For example, E. coli PFK has four identical monomers per enzyme where the active and effector sites are on the same protein monomer. In contrast, ATCase has two distinct types of subunits. One subunit is concerned explicitly with catalysis and the other with regulation. A single active ATCase is composed of eight units of the catalytic subunit and six regulatory subunits. Binding of effector molecules to the regulatory subunit, such as CTP, changes the conformation of the catalytic subunits thereby changing activity. Enzymes which possess separate binding sites for non-catalytically active small molecules, whether on the same subunit or separate subunits, are called allosteric enzymes (meaning literally ‘another space or structure’) and the molecules which bind to the non-active sites are called allosteric effectors. What is interesting is that many allosteric enzymes show cooperativity with respect to their substrates. It is also observed that allosteric effectors move the profile of the substrate sigmoid curve left or right as shown in Figure 7.2. At a fixed substrate concentration, the reaction rate can therefore be changed by changes in the concentration of the allosteric effectors. Note that unlike competitive inhibition, changes in substrate concentration cannot be used to overcome the allosteric control except at zero substrate or saturating levels, neither of which are of particular interest.

7.2. ALLOSTERY AND MWC MODEL

167

7.2 Allostery and MWC Model The MWC model which was discussed in the chapter 6 can easily be extended to account for allosteric control. Figure 7.1 shows a simple exclusive MWC model based on a dimer. X

T2

R2

R 2X X X R 2 X2

L L=

=

T2 R2

X

R 2X

Figure 7.1: Exclusive MWC model based on a dimer showing alternative microscopic states in the form of T and R states. The model is exclusive because the ligand, X, only binds to the R form.

The key to including allosteric effectors is the equilibrium between the tense (T) and relaxed (R) states (See Figure 7.1). To influence the sigmoid curve, an allosteric effector need only displace the equilibrium between the tense and relaxed forms. For example, to behave as an activator, an allosteric effector needs to preferentially bind to the R form and shift the equilibrium away from the less active T form. An allosteric inhibitor would do the opposite, that is bind preferentially to the T form so that the equilibrium shifts towards the less active T form. In both cases the Vm of the enzyme is unaffected.

Exclusive MWC Model (Binding only to the R form) When an effector ligand only binds to one of the states, say the R state, we refer to this as the exclusive MWC model (Figure 7.1). Consider then an effector, X that can bind to the R state with a microscopic dissociation constant, kRX , and to the T state with a microscopic dissociation constant of kTX , note we use lower case k to indicate a microscopic constant. The

168

CHAPTER 7. ALLOSTERY

apparent (or macroscopic) allosteric constant, Lo will be given by the ratio: Lo D

T C TX C TX2 R C RX C RX2

(7.1)

Using kRX and kTX , we can replace the complex states TX, TX2 , RX, and RX2 with the free forms, T and X. For example: TX D

2T
X kTX

RX D

2R
X kRX

where the 2 in the expression is a standard statistical factor (See chapter 6). Substituting these terms into equation (7.1) gives: TC Lo D RC

2TX TX2 C 2 kTX kTX 2RX RX2 C 2 kRX kRX

D

T .1 C X=kTX /2 R .1 C X=kRX /2

Since L D T=R then Lo D L

.1 C X=kTX /2 .1 C X=kRX /2

This shows how the allosteric constant is modified by the presence of effectors. In general, for an oligomer with n effector binding sites: Lo D L

.1 C X=kTX /n .1 C X=kRX /n

Note that if the binding of effector is the same for each state, kRX D kTX then Lo D L and the effector has no net effect. Assume that the effector only binds to the T state, that is kRX ' 1 then the effector acts as a pure inhibitor. Conversely if the effector only binds to the R state, kTX ' 1, the effector acts as a pure activator. If we set the inhibitor term to equal

D X=kRX and the activator term to ˇ D X=kTX , then for the simplified MWC model, where c D 0, we can write:

7.2. ALLOSTERY AND MWC MODEL

Inhibitor:

169

˛ .1 C ˛/n 1 Y D .1 C ˛/n C L.1 C ˇ/n

Activator: Y D

˛ .1 C ˛/n

(7.2)

1

1 .1 C ˛/ C L .1 C /n

(7.3)

n

These equations represent effector binding in relation to the exclusive model. If the enzyme is inhibited and activated by separate modifiers then the apparent allosteric constant is modified as follows: Lo D L

.1 C ˇ/n .1 C /n

That is: Y D

˛ .1 C ˛/n 1 .1 C ˇ/n .1 C ˛/n C L .1 C /n

By setting n D 1, equation (7.2) reduces to: vD

˛ 1 C ˛ C L .1 C ˇ/

meaning that the behavior of the enzyme with respect to substrate, ˛ is hyperbolic. Similarly, setting L D 1 also renders the behavior hyperbolic. From a systems perspective the sigmoid response offers some significant advantages. The first is that at a fixed substrate concentration, increases in inhibitor/activator concentrations decrease/increase the reaction rate. The second useful feature is that changes in inhibitor or activator concentrations can be used to move the responsive part of the curve. This will be seen more clearly when we consider the elasticity coefficient for the nonexclusive MWC model.

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CHAPTER 7. ALLOSTERY

1 CActivator

Reaction Rate

0:8

CInhibitor

0:6 0:4 0:2 0

0

0:5 1 1:5 Substrate Concentration

2

Figure 7.2: Action of an allosteric effector on the sigmoid response. The addition of an activator moves the sigmoid response to the left thereby increasing the reaction rate at a given substrate concentration. Inhibition moves the curve to the right.

Non-Exclusive MWC Model (Binds to the R and T states) In the non-exclusive model, a given ligand can bind to both states, R and T , Figure 6.17. The irreversible non-exclusive rate law with modifiers is given by [23, 13, 51]: Œ1 C n c˛.1 C c˛/n Œ1 C c n Œ1 C n .1 C ˛/n C L .1 C c˛/n Œ1 C c n

˛.1 C ˛/n v D Vf

1

CL

1

(7.4)

where  D m=kmT , c D kmT =kmR and m is the concentration of modifier. Assuming c < 1, then if kmR > kmT then the modifier acts as an inhibitor. That is if c < 1 the effector acts as an inhibitor, while if c > 1 then the effector acts as an activator. If inhibitor and activator are present and can bind both R and T states the equation becomes more complex as shown below:

7.2. ALLOSTERY AND MWC MODEL

171

1 Reaction Rate

0:8 0:6

Inhibitor

0:4 0:2 0

0

5 10 15 Substrate Concentration

20

Figure 7.3: Change in sigmoid response as a function of inhibitor concentration. Inhibition moves the curve to the right. This has two effects: 1) At a fixed substrate concentration the reaction rate decreases as the inhibitor concentration increases; 2) Changing the inhibitor concentration can be used to move the responsive part of the curve into the range of the substrate concentration. Parameters: n D 4I L D 1000I c D 0. Inhibitor Concentrations: 0:0; 0:3; 0:6; 2.

Œ1 C 1 n Œ1 C 2 n c˛.1 C c˛/n Œ1 C cI 1 n Œ1 C cA 2 n Œ1 C 1 n Œ1 C 2 n .1 C ˛/n C L .1 C c˛/n Œ1 C cI 1 n Œ1 C cA 2 n

˛.1 C ˛/n v D Vf

1

CL

1

(7.5)

where 1 D I =kT I , 2 D A=kTA , cI D kT I =kRI and cA D kTA =kRA .

Reversible Form Olivier et al.[51] discuss the reversible variant that was originally derived by Popova and Selkov [53] and is given below:

172

CHAPTER 7. ALLOSTERY

1 Reaction Rate

0:8 0:6 0:4 0:2 0

0

2

4 6 8 Inhibitor Concentration

10

Figure 7.4: Action of an allosteric inhibitor on the reaction rate at different levels of binding between the effector and the T state (KTX ). Starting at the lower curve, KT C D 0:6; 1:2 and 2:0. n D 4 and the substrate concentration, S D 4.

vD n Œ1 C n  n 1 Œ1 C ˛ C  C ˇL 1 C c ˛ C c    s p n 6 1 C c  6 Vf ˛.1 / 6 n 4 Œ1 C n  n Œ1 C ˛ C  C L  n 1 C cs ˛ C cp  1 C c  2

1

3 7 7 7 5

(7.6) where ˛ D s=ksR ,  D p=kpR ,  D m=kmT and c D kmT =kmR . The term ˇ is the activity of the T state relative to the R state, that is ˇ D .Vf T =ksT /=.VrT =ksR /.

Elasticities We may recall from the end of chapter 6 that the elasticity with respect to the substrate concentration for the simplified MWC model is given by:

7.2. ALLOSTERY AND MWC MODEL

"v˛ D

1 C ˛n 1C˛

173

nY

The substrate elasticity in the presence of inhibitor or activator is unmodified except for the value of the nY term. Deriving elasticities for complex rate equations is a fairly laborious procedure but can be eased considerably if logarithmic differentiation is used. This technique helps by removing the explicit scaling because in logarithmic differentiation it becomes part of the calculation. Using logarithmic differentiation, the inhibitor and activator elasticities can be derived. The elasticity with respect to the substrate (˛) given an inhibitor is given by "v˛ D 1 C

˛.n 1/ 1C˛

˛.1 C ˛/n 1 n .1 C ˛/n C L.1 C ˇ/n

The inhibitor (ˇ) and activator ( ) elasticities are given respectively by: "vˇ D

n

"v D n

ˇ.1 C ˇ/n 1 L .1 C ˛/n C L.1 C ˇ/n

.1 C /

.1 C ˛/n C

.nC1/ L

L .1 C /n

Figure 7.6 shows the inhibitor elasticity at reaction rate at various inhibitor concentrations as a function of substrate concentration. The inhibitor elasticity can easily exceed an absolute value of one. This is in contrast to competitive inhibitions where the response is limited to unity. Allosteric effectors combined with cooperativity offer significant sensitivity advantages. The elasticity response for the non-exclusive MWC is more interesting. Figure 7.6 shows the plot of the reaction rate and corresponding inhibitor elasticity as a function of substrate concentration. Here we see more clearly that the elasticity first starts at zero, moves in a negative direction before climbing back to zero. The lowest point on the elasticity curve indicates the most response part on the reaction rate curve. What is significant is

CHAPTER 7. ALLOSTERY

Reaction Rate or Elasticity

174

I D0:1 I D0:5 I D1:0

2

0 I D0:1 I D0:5

2

I D1:0

0

5 10 15 20 Substrate Concentration

25

Figure 7.5: Effects of substrate concentration (proportional to ˛) on the reaction rate and the inhibitor elasticity coefficient for an enzyme following the exclusive MWC equation (7.2). Three sets of curves are shown, each describing the response at a different concentration of inhibitor, I. As the inhibitor concentration is increased, the absolute value of the inhibitor elasticity increases (curves below the origin) in value. n D 4, L D 1000, c D 0. Upper curve represent reaction rates and lower curves elasticities.

that as the inhibitor concentration increases the most responsive point also moves. This means that changes in the inhibitor (or activator) concentration can be used to move the responsive part of the curve left or right. For example if the substrate concentration were high such that the enzyme is no longer very responsive we can increase the inhibitor level such that the responsive part of the sigmoid curve moves into the range of the substrate level.

7.3 Reversible Hill Equation Although it is possible to use MWC based models, extending these models with additional effectors, multiple substrates, products and reversibility is

Reaction Rate or Elasticity

7.3. REVERSIBLE HILL EQUATION

175

2

0

2

I D2

0

2

I D4

I D6

4 6 8 10 12 Substrate Concentration

14

Figure 7.6: Effects of substrate concentration (proportional to ˛) on the reaction rate and the inhibitor elasticity coefficient for an enzyme following the non-exclusive MWC equation (7.4). Three sets of curves are shown, each describing the response at a different concentration of inhibitor, I. As the inhibitor concentration is increased, the absolute value of the inhibitor elasticity increases (curves below the origin) in value. n D 4, L D 10, c D 0:01. Upper curve represent reaction rates and lower curves elasticities.

difficult [52]. Instead there has been some effort to develop more generalized rate laws. Of particular interest is the reversible Hill equation. We encountered the reversible Hill equation in the last chapter but without considering modifiers. When modifiers are included [33], an additional term appears in the denominator. In equation (7.7) the modifier is indicated by the symbol , where  is the normalized modifier concentration, M=Km . The  term determines whether the modifier is an activator or an inhibitor. If  < 1 then the modifier acts as an inhibitor, otherwise it is an activator.

176

CHAPTER 7. ALLOSTERY



 Vf ˛ vD

1

Keq

.˛ C /h

1

(7.7)

1 C h C .˛ C /h 1 C h  <1

inhibitor

 >1

activator

1 Ks D 1, 2.4, 4.0 Reaction Rate

0:8 0:6 0:4 0:2 0

0

2 4 6 Substrate Concentration

8

Figure 7.7: Plot showing the response of the reaction rate for a reversible Hill model with respect to the substrate as a function of the substrate concentration. In this and the next figure, the parameters are set to: Vm D 1; D 2; Keq D 10:95; Kp D 0:5; n D 4:85; Ke D 2:75; ˛ D 10 5 , P = 0, M = 0

The reversible Hill equation shows one additional property worth mentioning. Under a certain set of parameter values, the product concentration can act as a positive regulator (Figure 6.21). The possibility of positive activation can lead to some interesting behavior, such as bistability when applied to a larger model.

7.3. REVERSIBLE HILL EQUATION

177

0:5 Ke D 1:2, 2.0, 4.2 Reaction Rate

0:4 0:3 0:2 0:1 0

0

2 4 6 Inhibitor Concentration

8

Figure 7.8: Plot showing the reaction rate response for a reversible Hill model with respect to the inhibitor concentration. Ks D 2; S D 1, all other parameter are identical to the previous figure. Note that it is possible to easily extend the responsive range to higher inhibitor concentrations.

Hanekom [27] derived (along with many other variants) a generalized uniuni (single substrate-single product) reversible Hill equation that incorporates multiple modulators (7.8):





Vf ˛ 1 .˛ C /h 1 Keq # v D nm " Y 1 C h i C .˛ C /h h 1 C i i i D1

(7.8)

To simplify the notation in the above equation, ˛ D S=Ks ,  D P =Kp and i D Mi =Kmi .  is the modifier factor that determines whether the modifier is an activator (> 1) or an inhibitor (< 1). nm is the number of

178

CHAPTER 7. ALLOSTERY

modifiers, Kx are the Michaelian constants, S the substrate, P the product, and M the modifier. This equation assumes that each modifier binds independently, that is the binding of one modifier does not affect the binding of another. In the case where modifiers do not bind independently, the equations become more complex [33].

Chapter Highlights  An allosteric site is a binding site on a protein that is distinct from the active site. An allosteric effector is a molecule that binds to an allosteric site.  Many proteins that have allosteric sites are also oligomers  Some oligomers, such as lactate dehydrogenase, have no allosteric sites.  Many allosteric effectors are combined with substrate cooperativity lead to increased sensitivity in effector responses.  The MWC model can be easily extended to deal with allosteric effectors. In this case an allosteric effector need only bind preferentially to one of the states to displace the R/T equilibrium.  There are two models in the MWC family, exclusive and non-exclusive. In the exclusive model the allosteric effector only binds to one of the states, R or T . If the effector binds to the R state then it is likely to be an activator (because the R state is considered more active). If the effector binds to the T state then it is likely that the effector is an inhibitor.  In the non-exclusive model, the effector can bind to both states, R and T . If the effector binds equally well to both states then the effector will have no impact on the reaction rate.  Inhibitor and activators can cause the sigmoid response to move laterally such allowing effectors to move the responsive part of the curve into the signal range.

7.3. REVERSIBLE HILL EQUATION

179

 The reversible Hill equation can be modified to include modifiers.

Further Reading 1. Cornish-Bowden, A (2004). Fundamentals of enzyme kinetics (3rd ed.). London: Portland Press. ISBN 1-85578-158-1 2. Leskovac V (2003) Comprehensive Enzyme Kinetics. Springer. ISBN: 9780306467127 3. Palmer, T (1995) Understanding Enzymes, 4th Edition. Prentice Hall. ISBN: 0131344706 4. Segal I H (1993) Enzyme Kinetics: Behavior and Analysis of Rapid Equilibrium and Steady-State Enzyme Systems (Wiley Classics Library) Wiley-Interscience. ISBN: 0471303097 5. Tipton KF (1979) in Kinetic properties of allosteric and cooperative enzymes. Bull AT, Lagnado JK, Thomas JO, and Tipton KF (Eds.), Companion to Biochemistry Volume 2. Selected Topics for Further Study. Longman, London, chapter 11, 327–382

Exercises 1. Describe what is meant by allosteric regulation. How is negative allosteric regulation different from competitive inhibition. 2. Make a list of six allosterically control enzymes and indicate the allosteric regulators involved. 3. Hemoglobin is a tetramer of two distinct subunits. 2,3-Disphosphosglycerate is an allosteric effector. Carry out a literature search to investigate what 2,3-Disphosphoglycerate does and what physiological significance it has.

180

CHAPTER 7. ALLOSTERY

4. What is the fold change in inhibitor that is required to change the reaction rate of a cooperative enzyme form 90% to 10% of maximum? (Use a simple dimer MWC model in your calculation). 5. Describe the difference between the exclusive and non-exclusive model. 6. The elasticity response with respect to inhibitor is quite difference between the exclusive and non-exclusive models, what is this difference?

8

Generalized Rate Laws

181

182

CHAPTER 8. GENERALIZED RATE LAWS

8.1 Generalized Rate Laws As we have seen, kinetic rates laws come in a huge variety with many subtle variations. Often we will not have the necessary information to decide on which rate law to select let alone what parameters values to use. In such cases we can employ generalized rate laws. A number of authors have suggested more generic approximations to use when detailed kinetic data is lacking. In addition it may not always be necessary to use detailed mechanistic rate laws to describe the dependence of rate on reactant, product and effector concentrations. It is more important to describe the appropriate rate response than to worry excessively about mechanistic details. For this reason, and especially when data are unavailable, there are a number of approximate rate laws that can be used in place of detailed mechanistic rate laws. Typically, these approximations will be defined around some operating point, often a point that is close to the state of interest. The three crudest approximations include the linear, power law and lin-log approximations.

8.2 Linear Approximation The simplest approximation is to use a linear rate law. If we assume a general non-linear rate law of the form, vo D f .So /, then we can linearize around So using a Taylor series (See box) to yield:

v D vo C

@v .S @So

So / D vo C g .S

So /

(8.1)

where So and vo are the values of the substrate concentration and reaction rate at the selected operating point. The derivative, @v=@S , or g, is the estimated slope at the operating point. S is the substrate concentration at which the approximate reaction rate is required.

8.2. LINEAR APPROXIMATION

183

Taylor Series A Taylor series is a way of expressing a function in terms of a summation of terms about some operating point. The Taylor series of the function f .x/ about xo is the infinite series:

f .x/ D f .xo / C

@f .x @x

1 @2 f .x xo /2 2Š @x 2 1 @n f .x xo /n C : : : (8.2) C ::: C nŠ @x n

xo / C

The various derivatives in the Taylor series must be evaluated at xo . Note that the term .x xo /n gets progressively smaller as n increases eventually becoming negligible. Of particular interest is the linear approximation: @f .x xo / @x This is an approximation because it neglects the higher terms in the series but provided x is close to xo , the approximation is good. For example, approximate the function, y D sin.x/ around xo D 0. Note that sin.0/ D 0 and cos.0/ D 1, then: f .x/  f .xo / C

x2 x2 1 C ::: D x 2Š 3Š The linear approximation is simply, y D x. y D 0 C 1x

0

x3 C ::: 3Š

Example 8.1 At a substrate concentration of 2.5 mM, the rate of an irreversible enzyme catalyzed reaction is 5 mmoles sec 1 . If the derivative, @v=@S , is assumed to be 10.0, calculate the approximate velocity at substrate concentrations of 0.5 and 3.0 mM. The linear rate law approximation is given by v D vo C g .S So /. The approximate velocities at S D 0:5 and S D 3:0 are therefore given by:

184

CHAPTER 8. GENERALIZED RATE LAWS

v D 5 C 10.0:5 v D 5 C 10.3

2:5/ D

15 mmoles sec

2:5/ D 10 mmoles sec

1

1

Since the reaction is considered irreversible, the estimate at S D 0:5 mM is most likely incorrect and the approximation is thus unable to predict reaction rates at low substrate concentrations.

Figure 8.1 shows a plot of a Michaelis-Menten rate law with a Km of 1.0 together with the linear rate law drawn around the operating point set to the Km , that is, So D 0:5. The approximation is perfect at the operating point itself but beyond this it depends on the nonlinearity of the response curve. If the curvature is significant (i.e. highly non-linear) then the linear approximation can diverge quite rapidly (See Figure 8.1). The linear approximation can be extended to multiple reactants and effectors as follows: v D vo C

n X @v .Si @Sio

Sio / D vo C

i D1

n X

gi .Si

Sio /

(8.3)

i D1

where n is the number of substrate, products and effectors that might modulate enzyme rate, v. Linear approximations are suitable for modeling steady state situations or where enzyme saturation changes little during a transition from one state to another. It is also possible to express the linear equation in terms of elasticities by multiplying, gi , by So and dividing by v to yield:   ıS v D vo 1 C "vSo (8.4) So or in general for multiple substrates, products and effectors the expression is given by:

v D vo

1C

X i

ıSi "vS o o i S i

! (8.5)

8.2. LINEAR APPROXIMATION

185

Example 8.2 A uni-uni reaction is catalyzed by an enzyme. It is known that the concentration of substrate, S at steady state is 5 mM while the product concentration is 2:5 mM. The reaction is known to have an equilibrium constant of 10 and the reaction rate through the reaction at steady state is 7.5 mmoles sec 1 . From this information obtain a rate law that is a suitable linear approximation. The linear approximation will be based on equation (8.5) and will include terms for both the substrate and product. We can estimate the elasticities from the massaction ratio, D P =S and equilibrium constant. Since the substrate and product are not at saturating levels we can approximate the elasticities using the equations:

"vS D

1 =Keq

1

"Pv D

=Keq =Keq

1

If mass-action ratio is given by P =S D 5=2:5 D 2 and the Keq D 10, then: "vS D 1=.1 "Pv D vo D 7:5 mmoles sec

1

2=10/ D 1:25 0:2/ D

0:2=.1

0:25

, therefore a suitable approximate rate law will be:

  S v D 7:5 1 C 1:25

So So



 0:25

P

Po



Po

where So D 5 and Po D 2:5. To use the approximation one need only supply values for S and P to obtain the reaction velocity. The value for the reaction rate will diverge more and more from the true value as the difference between the reference species values, So and Po , and the input values of S and P becomes large. The approximation only computes the correct reaction velocity when S D So and P D Po beyond these values, the equation returns an approximate value for the reaction rate. If, during a simulation the reaction velocity and the concentrations of S and P change, the elasticities can be recomputed so that the approximation remains reasonably accurate.

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CHAPTER 8. GENERALIZED RATE LAWS

Power Law Approximation Another frequently used approximation is the power law [61] and is an improvement over the simple linear approximation and takes the form:

vi D ˛i

Y

"i

Sj j

(8.6)

j

where the " represents the kinetic order or elasticity. The value for the kinetic order will often be a non-integer and must be estimated around the selected operating point. Negative values for " indicate inhibition by Sj . ˛i is a nominal rate constant that scales the expression and yields the correct units for v. The advantage of the power law equation over the simpler linear rate law is that it shows a response somewhat similar to an enzyme kinetic response. The power laws also behaves well at low substrate concentrations (Figure 8.1). The approximation near the operating point is superior to the linear approximation. However the power law does not show any saturation in the reaction rate and this is one of its main drawbacks. It has found extensive use in Biochemical Systems Theory which was developed by Michael Savageau [60, 61].

8.3 Linear-Logarithmic Rate Laws In the linear approximation (8.5), the species term is given by: ıS=So , or .S So /=So . Recall that the Taylor expansion for the natural logarithmic function (ln) around yo to the first (linear) approximation is given by: ln.y/ ' ln.yo / C

y

yo yo

Note that @ ln.yo /=@yo D 1=yo . Rearranging the linear approximation yields:

8.3. LINEAR-LOGARITHMIC RATE LAWS

187

Reaction Rate, v

1:5

1

0:5

0

0

1

2 3 4 Substrate Concentration, S

5

Figure 8.1: Linear and power law approximations to a MichaelisMenten curve (solid) around the reference point So D 1. Vm D 1I Km D 1. Dotted line: Power law; Dashed line (upper): Linear law.

y

yo yo



' ln.y/

y ln.yo / D ln yo



Now substitute ıS=So for ln.S=So /. This simple change leads to a significantly improved approximation over the power law and linear equations, and is called the linear-logarithmic approximation or lin-log for short [69, 28, 68, 29]. One of the chief advantages of this approximation is that at high substrate concentration the response approximates the saturation by substrate much better than the power law approximation (See Figure 8.2). The general form of the lin-log equation is given by:

 v D vo

e eo

 1C

X i

"vSi



Si ln Sio

! (8.7)

where S is the reactant concentration and " the elasticity. The summation

188

CHAPTER 8. GENERALIZED RATE LAWS

is over all reactants and effectors that might modulate the reaction rate (except the enzyme concentration). The rate law is always defined around some reference state where vo is the reference reaction rate and Sio is the reference reactant concentration. As with the power law and linear approximations (8.5), the utility of this method is that the elasticity values (kinetic orders) can be estimated from the known thermodynamic properties of the reaction, especially if the reaction is operating below saturation. If no thermodynamic information is available, the elasticities may be set to the stoichiometries of the respective reactants if necessary. In either case it is important to note the lin-log approximation is only valid around the chosen reference state but is much better (See Figure 8.1) than either the linear or power law approximations. One possible drawback to the lin-log approximation is that at zero reaction rate, the reactant levels are not necessarily at equilibrium (Figure 8.1). This can lead to reverse reaction rates when the prevailing metabolites levels suggest otherwise. The lin-log approximation is therefore not suitable when a reaction is close to equilibrium or when metabolites levels are low.

Reaction Rate, v

1:5 Linear Power

1

Lin-Log Michaelis

0:5

0

0

1

2 3 4 Substrate Concentration, S

5

Figure 8.2: Linear, power law and lin-log approximations to a Michaelis-Menten curve around the reference point So D 1. Vm D 1I Km D 1. Dotted: Power law; Dashed: Linear law. The drawback of the lin-log approximation is that the curve does not go through zero.

8.4. ALGEBRAIC APPROXIMATIONS

189

8.4 Algebraic Approximations For some enzymatic reactions the rate of forward and reverse reaction is so fast that one can assume, that compared to the other reactions in the pathway, the reaction is essentially at equilibrium. In situations like this, rather than represent the reaction using a reaction rate law, we can instead replace the reaction by an algebraic equation that describes the equilibrium distribution of the substrates and products. For example, consider the simple uni-uni reaction: S P At equilibrium, Keq D P =S and the total mass of species is S C P D T . Combining these two equations leads to the following solution: SD

T 1 C Keq

P D

Keq T 1 C Keq

In cases like this one need not assume any complex reaction rate law, merely that the reaction is always close to equilibrium. In glycolysis a number of reactions are considered close to equilibrium though this depends on the organism, tissue and even the state of the pathway. In potato tuber for example, glucose phosphate isomerase, phosphoglycerate mutase and enolase are considered close to equilibrium.

8.5 Hanekom Rate Laws Although the approximations described in the last section can be useful, they all have some drawbacks. Of particular concern are the lack of thermodynamic constraints, and for some modelers the lack of any mechanistic underpinning. To avoid these issues and yet retain a degree of generalization, a number of authors have attempted to define generalized MichaelisMenten based rate laws suitable for modeling purposes. Of particular note is the work by Hanekom, Rohwer and Hofmeyr [27, 56, 70] and the

190

CHAPTER 8. GENERALIZED RATE LAWS

work by Liebermeister and Klipp [43, 44]. In the next two sections the Henekom and Liebermesiter rate laws will be described. For a generalized catalyzed reaction such as S P , the rate law can be modularized into three components [54, 32], a rate constant term, k, a saturation term, , and a thermodynamic term, T (8.18, cf. 3.23). vDk

T

(8.8)

The thermodynamic term, T , is given by .S P =Keq / D S.1 /. The saturation term, , equals one if the reaction is uncatalysed, in which case the rate law reduces to the familiar form (1.19):
v D k1 S 1 =Keq D k1 S.1 / where k1 is the rate constant. The inclusion of catalysis and effectors changes the saturation term, , which accounts for how the reactants are distributed amongst the enzyme-complex forms and their subsequent effect on the reaction rate. The mechanistic basis for deriving the saturation term is a rapid–equilibrium random-order mechanism, which according to previous studies [56] appears to fulfill most kinetic requirements. Cooperativity and modulator effects can also be generalized in this scheme, and will be presented later. For an unregulated reaction of the form: S1 C S2 C : : : C Sn P1 C P2 C : : : C Pn where there is an equal number of substrates and products on either side of the reaction, the Henekom rate law is [27]:

v D Vf

n Y i D1

 ˛i 1

Y n  Keq

i D1

1 1 C ˛i C i

 (8.9)

where n is the number of substrate-product pairs in the reaction, ˛i is equal to Si =Ksi , and i is equal to Pi =Kpi . Ksi is the half-saturation constant

8.5. HANEKOM RATE LAWS

Reaction Scheme A B

191

Rate Law Vf ˛1

1 =Keq 1 C ˛1 C 1

A C B C C D Vf ˛1 ˛2

1 =Keq .1 C ˛1 C 1 /.1 C ˛2 C 2 /

Table 8.1: Generalized rate equations where Vf represents the forward maximal velocity and terms ˛i represent Si =Ksi where Ksi is the halfsaturation constant for substrate i . i represents Pi =Kpi where Kpi is the half-saturation constant for product i . is the mass-action ratio. The second equation is the same as the Random Order Rate law in equation (5.6).

for substrate i , and Kpi is the half-saturation constant for product, i . Table 8.1 illustrates the first two rate laws in the sequence. For the special case of the two substrate/one product reaction: ACB C the generalized rate law is: 



Vf ˛ˇ 1

vD

Keq 1 C ˛ C ˇ C ˛ˇ C 

(8.10)

where ˛ D S=KA ; ˇ D B=KB , and  D C =KC . For the one substrate/two product rate law: A B CC the generalized rate law is:

192

CHAPTER 8. GENERALIZED RATE LAWS

 Vf ˛ 1

vD

 Keq

(8.11)

1 C ˛ C  C  C 

where ˛ D A=KA ;  D B=KB , and  D C =KC .

Elasticities The following elasticity relations can be derived for the two substrate/two product generalized equation (8.9). "vS1 D "vS2 D v "P D 1

v "P D 1

1

1 =Keq

˛1 1 C ˛1 C 1

1

1 =Keq

˛2 1 C ˛2 C 2

1

=Keq =Keq

1 1 C ˛1 C 1

1

=Keq =Keq

2 1 C ˛2 C 2

These results are very similar to the elasticity coefficients for the reversible single substrate/product rate law.

Cooperativity and Allostery The Hanekom rates laws that describe cooperativity and allostery are a generalization of the reversible Hill equations [33]. It will be recalled from a previous chapter that the generalized Hill equation for a uni-uni reversible reaction was given by: 

 Vf ˛ vD

1

Keq

.˛ C /h

1 C h C .˛ C /h 1 C h

1

(8.12)

8.5. HANEKOM RATE LAWS

193

where ˛ D S=Ks ,  D P =Kp and  D M=Km . The Ks ; Kp and Km are the half-saturating Michaelis constants. The  factor determines whether the modifier, M , is an activator or inhibitor. h is the Hill coefficient such that if h > 1 the equation displays positive cooperativity. The equation incorporates the equilibrium constant, Keq . Of importance is the fact that at equilibrium the reaction rate is equal to zero. The reversible Hill equation is a good general rate law to use for representing uni-uni cooperative enzymes with allostery. It can also be used to model gene expression as a function of a transcription factor if the concentration of product, P is set to zero. We will have more to say about gene regulation in the next chapter. The reversible Hill equation can be naturally extended to two or more substrates and products to yield a bi-bi reversible Hill equation without allosteric regulation:  Vf ˛1 ˛2 1 vD



.˛1 C 1 /h 1 .˛2 C 2 /h Keq

1 C .˛1 C 1 /h 1 C .˛2 C 2 /h

1

(8.13)

Or for ns substrates:

v D Vf

n Y i D1

 ˛i 1

Y ns Keq

.˛i C i /h 1 1 C .˛i C i /h

i D1

! (8.14)

For uni-bi and bi-uni the following generalized equations can be derived [27, 56]:

Uni-Bi: v D

Bi-Uni: v D

 Vf ˛ 1

 Keq

.˛ C /h

1

1 C .˛ C /h C .˛ C /h C .˛ C /h  Vf ˛ˇ 1

 Keq

.˛ˇ C /h

2˛ h

(8.15)

1

1 C .˛ C /h C .ˇ C /h C .˛ˇ C /h

2 h

(8.16)

194

CHAPTER 8. GENERALIZED RATE LAWS

For a bi-bi reaction model with nm modifiers the generalized reversible Hill equation is given by:   Vf ˛1 ˛2 1 Keq .˛1 C 1 /h 1 .˛2 C 2 /h 1 h i vD M1 C M2 .˛1 C 1 /h C .˛2 C 2 /h C .˛1 C 1 /h .˛2 C 2 /h (8.17) where

M1 D

nm Y

2 4

j

1 C jh

3

1 C j4h jh

5

and

M2 D

nm Y j

3 2 1 C j2h jh 4 5 1 C j4h jh

Mechanistically it is assumed that the second modifier binds independently to a separate allosteric site. It is possible to extend this to three modifiers, details of which can be found in Hanekom [27, 56]:

8.6 Liebermeister Rate Laws As briefly mentioned previously, it is possible to generalize rate laws by recognizing the various distinct components that make up a rate law (8.18). Liebermeister and Klipp [44] have taken this further and defined a modular generalized rate law with the following form: v D Er Rreg

T D C Dreg

(8.18)

We will refer to these rate laws as the Liebermeister and Klipp rate laws [44] or more simply the Liebermeister laws for short. The Er term in equa-

8.6. LIEBERMEISTER RATE LAWS

195

tion (8.18) refers to the enzyme amount and the numerator term, T is called the thermodynamic term. We will return to the nature of Rreg and Dreg later. Liebermeister and Klipp define a number of forms for the equations but we will focus on three which they call, common, direct and power. These variants change the denominator, D in equation (8.18). We begin with the numerator term, T . This can take on two forms, explicit or Haldane form. The explicit form is given by a simple reversible massaction like expression: T D kf

Y

˛ini hi

kr

Y

ini hi

where kf and kr are the forward and reverse rate constants, ni is the stoichiometry for species i (reactant or product) and hi is the degree cooperativity for the species. For example, a reaction of the form, ACB ! P CQ with no cooperativity and all stoichiometries set to one, the T term would equal: T D kf ˛1 ˛2

kr 1 2

where ˛1 D A=Kma , ˛2 D B=Kmb , 1 D P =Kmp and 2 D Q=Kmq . This is called the explicit form for the numerator. The Haldane form is given by: T D

Y

˛ini



 1

Keq

where ni is the stoichiometry for species i . As we have seen before a number of times, introducing Haldane like expressions enables us to eliminate on of the constants and replace it with the equilibrium constant. The Haldane form in particular ensures that when the concentrations of reactants and products are at equilibrium the reaction rate is zero.

Common Modular The denominator for the common modular form is given by:

196

CHAPTER 8. GENERALIZED RATE LAWS

DD

Y

.1 C ˛i /hni C

i

Y

.1 C i /hni

1

(8.19)

i

where h is the Hill coefficient and ni is the stoichiometry for the ith species. The product term is over as many reactants or products in the reaction. We can illustrate this equation with a simple uni-uni reversible reaction, A B. In this case the common modular equation takes the form:  ˛ 1 v D Vf



 ˛ 1

Keq

.1 C ˛/ C .1 C /

1

D Vf

 Keq

.1 C ˛ C /

This equation is equivalent to the reversible Michaelis-Menten equation. For a more complicated example, consider the reversible reaction 2A C B ! C , with a cooperativity of 2 on reactants A and B, the common modular rate law using Haldane numerator is given by:   ˛14 ˛22 1 Keq  
v D Vf  1 C ˛14 1 C ˛22 C 1 C 11

1

where ˛1 , ˛2 and 1 are the dimensionless concentrations (X=Kx ). Mechanistically the equation is derived assuming rapid equilibrium and random order binding.

Direct Binding The denominator for the direct binding form is given by: D D1C

Y

˛ihni C

Y

ihni

(8.20)

The direct binding model assumes only three states, all substrates bound, all products bound or free enzyme. This is reminiscent of the Hill equation and represents an extreme form of cooperativity. For example, for the reversible reaction 2A C B ! C , with a cooperativity of 2 on both A and B, the common modular rate law using Haldane numerator is given by:

8.6. LIEBERMEISTER RATE LAWS

v D Vf

 ˛14 ˛22 1 1C

˛14 ˛22

197

 Keq

C i2

where ˛1 , ˛2 and 1 are the dimensionless concentrations (X=Kx ). One advantage of the direct binding model is that some of the Km constants can be combined as products so reducing the number of parameters to set. For the simple reversible uni-uni reaction, the common modular and direct binding are equivalent.

Allosteric Regulators So far only the T and D terms in equation (8.18) have been discussed. The Rreg term in the modular equation is used to specify regulatory terms, particularly allosteric regulation. Here the Liebermeister equations assume independent binding of regulators at sites other than the active site. Two types of regulation are distinguished, called complete and partial. The model assumes that the regulator can bind to all states, that is free and reactant bound states. Once bound the regulator either enhances or diminishes the activity of the state. If the regulators have the ability to completely inhibit or activate when bound, the regulator is said to result in complete activation or inhibition. If the regulator only partially affects activity then the regulator is said to result in partial activation or inhibition. The general equation for the regulator factor, Rreg is given by:

Rreg

Y D A C Œ1

 A  1C

w Y  I C Œ1

1 I  1C

w

where A and I are called the relative basal rates for the activator and inhibitor respectively. The basal rates vary from 0 to 1 and represent the ratio of the rate without regulator to the rate in the presence of saturating regulator. The  values determine whether the mechanism represents complete or partial regulation. If the basal rates are set to zero, then the regulators show complete regulation.

198

CHAPTER 8. GENERALIZED RATE LAWS

The w term is called the regulation number and acts in a similar way to the Hill coefficient. For complete regulation (A D 0I I D 0) the Rreg term reduces to: Rreg

Y   w Y  1 w D 1C 1C

In both cases the product terms are over all possible activators and inhibitors. For example, an enzyme which has a single allosteric inhibitor (with w D 1) and no activators will have a Rreg of: Rreg D

 1C

The Rreg can now be used with the modular equation. Using the common modular form that also has a complete inhibition allosteric modulator, but no cooperativity, the rate law for a simple uni-uni reaction is given by:  v D Vf

  ˛ 1 Keq  1C 1C˛C

Non-Allosteric Regulators The last term in the modular equation (8.18) is Dreg . This term represents a non-allosteric regulation, that is regulator such as simple competitive inhibition. The general equation for Dreg is given by:

Dreg

X kA i D Ai i

!wi C

X

Ii

i

kiI

!wi

that is a sum of terms that includes activators (A) and inhibitors (I ). If an enzyme has only a single non-allosteric inhibitor then the Dreg term reduces to an expression representing competitive inhibition. To give an example, a uni-uni reversible reaction with a single non-allosteric inhibitor, the rate law is given by:

8.6. LIEBERMEISTER RATE LAWS



199



˛ 1

v D Vf

Keq 1 C ˛ C  C 1=

where  D M=Km . Compare this equation with the reversible competitive inhibition rate law in chapter 4, equation (4.4). Further details can found in the original publication [44].

Underlying Assumptions As attractive as generalized rate laws are, it is important to mention once again two significant assumptions that are used in both the generalized and traditional enzymatic rate laws. These concern the steady state/rapid equilibrium assumption and the assumption that the amount of enzyme is much less than the concentration of reactants. Both assumptions eliminate the inherent delays that occur when a reactant or product concentration changes. These delays results from the time it takes for the enzyme complexes to reestablish steady state or equilibrium. This is particularly so when the enzyme concentration is high relative to the reactant or product concentration because with the likely higher levels of complex, it takes more time to reach steady state. In addition when significant amounts of substrate or product are complexed with enzyme, during a transient, the free substrate or product concentrations may be significantly different from the total substrate or product. This means that in a computer model, the reaction rates will be higher than they will be in vivo because the rate laws will likely use the total concentrations to estimate the reaction rates. All these issues are not yet resolved in the literature and not all their affects fully appreciated. Care must therefore be taken when using enzymatic or generalized rate laws. Particular notice should be made of the relative magnitudes of metabolite and enzyme concentrations. Often it may be the case that in metabolic pathways the assumptions are reasonable. In signalling pathways where both enzyme and reactant concentrations are comparable, the time delays effects may be more significant. In addition, in signalling pathways, when there are closed cycles (such as phosphoryla-

200

CHAPTER 8. GENERALIZED RATE LAWS

tion/dephosphorylation cycles), the effect of enzyme sequestration, which is often ignored, can have a significant effect on behavior [47].

8.7 Choosing a Suitable Rate Law Given the huge range of possible rate laws, the novice modeler might seem at a loss to know which rate law to select for a given reaction step. However, before a model is built, its purpose should be clearly understood because this helps decide which rate laws to best apply to a given reaction step. Ultimately, models of biological processes have two main roles: 1. Describe known observations. 2. Make new non-trivial predictions. So long as these two requirements are satisfied, the model is useful. Often times novice modelers feel it necessary to add every small detail into a model when in fact much of it can be dispensed. A model is a simplification of reality, not a replica, and the art of building models is knowing what details to include and what to exclude. The question whether a particular reaction should use a ping-pong based rate law or a generalized rate law depends on how this choice influences the behavior of the model, particularly within the constraints of measurements. A useful strategy is to carry out a sensitivity analysis to determine how much influence parameters or particular rate laws have on model dynamics. If a particular parameter has little influence then there is no need to obtain a precise value for it. If a particular rate law has little influence, then a simpler rate law can be used instead which will often have fewer parameters. For example it may be possible to use lin-log rate laws at some reaction steps while other steps may require a more detailed description. As more detailed measurements become available it might be found that some of the lin-log approximations need to be replaced with more complex rate laws and subsequent experimental effort can focus on those particular steps.

8.7. CHOOSING A SUITABLE RATE LAW

201

Chapter Highlights 1. Generalized rate laws are a useful substitute for the traditional mechanistic enzyme rate laws particular when data is lacking or the need for a highly complex rate equation is unnecessary. 2. Like traditional enzyme rate laws, the generalized rate laws make a variety of assumptions. Some assumptions are more severe than others and care needs to be taken when selecting a generalized rate law. 3. Generalized rate laws can be split into broad groups, one group is completely mechanism independent and another is based on a generic mechanism, often the random-order rapid equilibrium mechanism. 4. The simplest generalized rate laws are the linear, power, and lin-log rate laws. These rate laws are valid around some operating point, usually the state state. They often use elasticity coefficients as part of the approximation. The elasticities can be estimated from a combination of species concentrations and thermodynamic data. There are a number of issues with these approximations including poor thermodynamic behavior (Equilibrium concentrations don’t necessarily give a zero net rate) and a divergence of the approximation beyond the operating point. 5. Of the mechanistic based rate laws there are at least three groups, Hanekon, the related reversible Hill family of rate laws and the Liebermeister family. All three families can be used to model multireactant and product reaction, they can represent irreversible as well as reversible reactions, they can model cooperativity and finally they can model allosteric interactions both cooperative and non-cooperative behaviors. Finally they obey the thermodynamic constraints of the modeled reaction. 6. Like all rate laws that make approximations, including the well known mechanistic equations (such as Michaelis-Menten), the generalized equations are unable to model the sequestration of species by

202

CHAPTER 8. GENERALIZED RATE LAWS

the enzyme and they also cannot model the delays that occur when the enzyme concentration is of the same order of magnitude as the reactants and products. In this situation it takes time for the steady state or equilibrium approximation to establish itself, if at all.

Further Reading 1. Liebermeister W, Uhlendorf J, and Klipp E. Modular rate laws for enzymatic reactions: thermodynamics, elastcities and implementation. Bioinformatics, 26(12):1528, 2010. 2. Savageau M A (1976) Biochemical systems analysis: a study of function and design in molecular biology Addison-Wesley, Reading, Mass

9

Kinetics of Gene Regulation

203

204

CHAPTER 9. KINETICS OF GENE REGULATION

This chapter introduces the basic genetic units found in bacteria and how one can derive the rate laws that describe bacterial gene expression1 . This chapter can be read independently of the other chapters although a familiarity with the work presented in chapter 1 would be very useful.

9.1 Structure of a Microbial Genetic Unit In this chapter we address exclusively prokaryotic gene regulation because it is much simpler than eukaryotic systems. However, many of the basic principles still apply to both groups of organism. The fundamental functional unit of the bacterial genome is the operon which consists of a control sequence followed by one or more coding regions. The control sequence has a promoter together with zero or more operator sites (Figure 9.1). The promoter is the specific sequence of DNA recognized by RNA polymerase which in turn is responsible for transcribing the DNA coding sequence into messenger RNA (mRNA). The binding of proteins called transcription factors (TF) to the operator sites are responsible for influencing the binding of RNA polymerase and thus can modulate mRNA production. Operators ...

Operators ... Promoter

One or More Coding Sequences

Figure 9.1: Generic Bacterial Operon comprising of one or more coding sequences, one promoter site for RNA polymerase binding, and zero or more operator sites that may be upstream or downstream of the promoter. Operator sites that act as repressors are often found to overlap with the promoter site.

Two other components are not shown in Figure 9.1, these include the ribosome binding site (RBS) and the terminator. The RBS is often a six to seven base nucleotide base sequence located about eight nucleotides upstream from the coding sequence start codon and is used by the ribosome 1 For consistency, italicized lower case will be used to designate genes and Roman lettering for proteins.

9.2. GENE REGULATION

205

as a recognition site. The other component, the terminator, is used to stop mRNA transcription at the end of the coding sequence. Binding of transcription factors results in the activation or inhibition of gene transcription. Multiple transcription factors may also interact to control the expression of a single operon. These interactions can emulate simple logical functions (such as AND, OR, etc.) or more elaborate computations. Gene regulatory networks range from a single controlled gene to hundreds of genes interlinked with transcription factors forming a complex, decision making network. Different classes of transcription factors also exist. For example, the binding of some transcription factors is modulated by small molecules, a well known example being the binding of allolactose (a disaccharide very similar to lactose) to the lac repressor or cAMP to the catabolite activator protein (CAP), also known as the cAMP receptor protein (CRP). Alternatively, a transcription factor may be expressed by one gene and either directly modulate a second gene (which could be itself) or via other transcription factors. Additionally, some transcription factors only become active when phosphorylated or unphosphorylated by protein kinases and phosphatases (Figure 9.3). The size of gene regulatory networks vary from organism to organism. The genome of E. coli for example encodes for approximately 171 transcription factors [38]. These proteins directly control all levels of gene expression. The EcoCyc [38] database reports at least 48 small molecules and ions that also influence transcription factors. The most extensive gene regulatory network database is RegulonDB [36, 24] and another associated network database EcoCyc [38]. RegulonDB is a database on the gene regulatory network of E. coli. More detail on the structure of regulatory networks can be found in the work of Alon [66] and Seshasayee [65].

9.2 Gene Regulation Gene expression rates are controlled by transcription factors, RNA polymerase, and proteins called  factors.  factors are transcriptional initia-

206

CHAPTER 9. KINETICS OF GENE REGULATION

tion proteins that influence the binding RNA polymerase to the promoter and can be thought of as global signals that are synthesized in response to specific environmental conditions. Of more interest here is the role of transcription factors. These proteins either enhance or reduce the ability of RNA polymerase to bind to the promoter region and commence transcription. Transcription factors operate by recognizing and binding to specific DNA sequences on the operator sites. When transcription factors bind to operator sites they either block or help RNA polymerase bind to the promoter. At the molecular level, it is assumed that a given transcription factor will bind and unbind at a rapid rate. To quantify how transcription factors influence gene expression it is important to consider the state of an operator site. For a single transcription factor that can bind to a single operator site, there are two states, designated either bound or unbound (Figure 9.2). a) Unbound State Operator Promoter

Coding Sequence

b) Bound State TF

Figure 9.2: Transcription Factor (TF) Bound and Unbound States.

If the operator site can enhance RNA polymerase binding then the bound state is considered the active state and the unbound state the inactive state. If the operator is an inhibitory site then the bound state is the inactive state and the unbound state the active state. Some bacterial transcription factors such as the lactose repressor (LacI) are present at very low levels, on the order of 5 to 10 copies per cell [50, 37]. It is therefore appropriate to consider the probability that a given transcription factor is bound to an operator site. The state of an operator site can be described in terms of this probability. These probabilities are influenced by the association constant of binding, the availability of transcription factors,

9.2. GENE REGULATION

207

and other regulators.

Gene Activation Gene Repression Multiple Control Gene Cascade Auto-Regulation Regulation by Small Molecule ~P

Regulation by Phosphorylation

Figure 9.3: Various Simple Gene Regulatory Motifs.

Once bound, the transcription factor influences the probability of RNA polymerase binding to the promoter site. There are many mechanisms by which transcription factors can influence RNA polymerase. One of the simplest is for a transcription factor to bind to the promoter site itself, and by an act of exclusion, prevent the RNA polymerase from binding. Such transcription factors act as repressors. A similar effect occurs if a transcription factor binds downstream of the promoter site (closer to the start of the coding sequence). This prevents the RNA polymerase from moving into the coding sequence by either physical obstruction or because

208

CHAPTER 9. KINETICS OF GENE REGULATION

the transcription factor has formed DNA loops. a) Downstream Obstruction RNA Pol

TF

Promoter

Operator

Coding Sequence

b) Promoter Obstruction TF

c) Sequestration of an activator resulting in inhibition TF

TF

RNA Pol

RNA Pol RNA Polymerase TF

“Activator”

Repressing Transcription Factor

Figure 9.4: Obstruction, exclusion and sequestration models for repressing gene expression.

Examples of downstream obstruction include the galR and galS operators, where both operators are located beyond the promoter site [64]. LacI is a good example of promoter exclusion although the LacI repressor only overlaps about 40% of the promoter (Figure 9.5). Another mechanism for repression is by sequestration. This is rarer but one example is CytR repressed promoters. The CytR protein can form a dimer with CRP (which itself is a transcriptional activator). Once the dimer is formed, CRP is unable to bind, therefore inhibiting expression [67]. Activation by transcription factors is more subtle. One mechanism is for a transcription factor to bind upstream, close to the promoter site. In this

CRP Binding Site

Lengths, except for the coding sequences, are drawn to scale:

lacI Gene

Operator 1 400 bases downstresm

Promoter RBS

Start of lacZ Gene

....CTCGTATGTTGTGTGGAATTGTGAGCGGATAACAATTTCACACAGGAAACAGCTATGACC ....GAGCATACAACACACCTTAACACTCGCCTATTGTTAAAGTGTGTCCTTTGTCGATACTGG RBS Start of Operator 2 lacZ Gene

Promoter

End of lacI Gene Promoter CRP Binding Site GGGCAGTGAGCGCAACGCAATTAATGTGAGTTAGCTCACTCATTAGGCACCCCAGGCTTTACACTTTATGCTTCCGG... CCCGTCACTCGCGTTGCGTTAATTACACTCAATCGAGTGAGTAATCCGTGGGGTCCGAAATGTGAAATACGAAGGCC... Operator 3

9.2. GENE REGULATION 209

Figure 9.5: The Lac Operon. RBS: Ribosome binding site. LacZ is the first gene in the lac operon.

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CHAPTER 9. KINETICS OF GENE REGULATION

instance the transcription factor can offer a suitable but weak molecular face for the RNA polymerase to bind (Figure 9.6). This allows RNA polymerase to stay on the promoter longer and therefore increase the probability of transcription. For example, weak binding may occur between hydrophobic areas on both proteins. An example of an activating TF is CRP on the lac operon. The CRP binding site is located only 15 bases upstream from the lacI promoter (Figure 9.5). Binding of CRP to its binding site allows the flexible RNA polymerase domains, ˛C TD and ˛N TD to bind to CRP, thereby increasing the likelihood of RNA successfully binding to the promoter site. Transcription factors themselves can be controlled by other transcription factors binding to operator sites. Control can also be accomplished by other proteins binding to the transcription factor or by small molecules, called inducers, that bind to the transcription factor and alter the operator binding affinity. LacI is an example of a transcription factor where the inducer molecule allolactose can bind, thereby altering the binding affinity of LacI. CI from the virus, lambda phage is an example of a transcription factor where control is exerted by influencing its production rate.

9.3 Fractional Occupancy One of the most important concepts to consider when quantifying how transcription factors influence gene expression is the fractional occupancy or degree of saturation at the operator site. This quantity expresses the probability of a particular occupancy relative to the total of all occupancy states. A simple example best describes this concept.

Transcriptional Activation Consider a single operator site upstream of a promoter (Figure 9.6 and 9.7). The operator site binds a single monomeric transcription factor, A. Assume that when the transcription factor binds to the operator, the RNA polymerase has a higher probability of binding to the promoter site by virtue of complementary patches on the RNA polymerase and transcrip-

9.3. FRACTIONAL OCCUPANCY

211

a) Activation by RNA polymerase requitment TF

RNA Pol

Operator Promoter

Coding Sequence

b)Sequestration of a repressor resulting in activation TF

RNA Pol TF

TF

Transcription Factor

RNA Pol RNA Polymerase

Figure 9.6: Gene regulation by an activating transcription factor. a) The operator site is upstream of the promoter, binding of the transcription factor increases the likelihood of RNA polymerase binding by way of weak interactions between the transcription factor and RNA polymerase. Alternatively, b) an activator can sequester a repressor transcription factor.

tion factor. If we assume the rate of gene expression is proportional to the probability of bound RNA polymerase, and that RNA polymerase has a constant concentration and activity in the cell, then we can assume the fractional occupancy of the transcription factor is proportional to gene expression. Let us designate the concentration of the unbound operator site by the symbol U , the bound operator site by the symbol AU and the free transcription factor by A as shown in Figure 9.7. The fractional occupancy of the operator site is then given by the degree of bound operator relative to the total of all occupancy states, that is: f D

AU U C AU

If we assume the rate of binding and unbinding of transcription factor to

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CHAPTER 9. KINETICS OF GENE REGULATION

U Operator

AU

Coding Sequence

A

A Transcription Factor

Figure 9.7: Bound (AU) and unbound (U) states for a simple transcriptional activation model.

the operator site is much faster than transcription, then we can also assume the binding and unbinding process is at equilibrium. That is, the following process is at equilibrium: U C A AU We can express the equilibrium condition using the association constant, Ka , where: AU Ka D U
AU Given this information we can express AU in terms U : f D

Ka
U
A U C Ka
U
A

(9.1)

The unbound state, U , can now be eliminated to yield: f D

Ka A 1 C Ka A

(9.2)

We have seen this same approach when using the rapid equilibrium assumption from enzyme kinetics. Much of the following should therefore be familiar. Relation (9.2) yields a value between zero and one. Zero indicates an unbound state, and one indicates the operator site is fully occupied. To obtain the actual rate of expression, assume the rate is linearly

9.3. FRACTIONAL OCCUPANCY

213

proportional to the fractional occupancy, so that: v D Vm

Ka A 1 C Ka A

(9.3)

where Vm is the maximal rate of gene expression (Figure 9.8). Equation (9.3) yields a familiar hyperbolic plot.

Gene Expression Rate, v

1 0:8 0:6 0:4 0:2 0

0

2 4 6 8 10 Transcription Factor Concentration

Figure 9.8: Gene expression rate as a function of a monomeric transcription factor that activates gene expression. The association constant, Ka , has a value of 1. The reaction rate is normalized by Vm .

If the association constant Ka is substituted by the dissociation constant (Ka D 1=Kd ), then we obtain: v D Vm

A Kd C A

(9.4)

At half saturation it is easy to show that Kd D A. This result provides a simple way to estimate the Kd from a binding curve by locating the halfsaturation point and then reading the corresponding transcription factor concentration.

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CHAPTER 9. KINETICS OF GENE REGULATION

Transcriptional Repression Repression can be handled in a similar manner. In this case we note that the active state is now the unbound state, U , so the fractional occupancy is given by: U f D U C AU Using the same equilibrium relation as before, we obtain (Figure 9.9): v D Vm

1 1 C Ka A

(9.5)

Gene Expression Rate, v

1 0:8 0:6 0:4 0:2 0

0

2 4 6 8 10 Transcription Factor Concentration

Figure 9.9: Gene expression rate as a function of a monomeric transcription factor that represses gene expression.

As with the activation example in the last section, the dissociation constant, Kd , is equal to the transcription factor concentration at half saturation.

9.4 Multiple Transcriptional Factors Multiple transcription factors can be dealt with similarly, except there will be more states to consider when computing the fractional occupancy. We

9.4. MULTIPLE TRANSCRIPTIONAL FACTORS

215

assume in all cases that gene expression is proportional to the probability that RNA polymerase binds to the promoter, which in turn, is proportional to the fractional occupancy by transcriptional factor. Before considering some situations in more detail, it is important to note that there are two ways in which multiple transcription factors can bind, that is competitively and non-competitively.

Competitive Binding Consider the case (Figure 9.10) where we have two transcription factors, A and B. Assume that either transcription factor can activate gene expression. B

A

Transcription Factors

U Operator

Coding Sequence

B

AU

A

A

BU

B

Figure 9.10: Competitive binding of two transcription factors to one operator binding site.

Oftentimes it is useful to create a table that indicates all possible states and whether they lead to gene expression or not. In this senario, such a table would look like Table 9.1.

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CHAPTER 9. KINETICS OF GENE REGULATION

State

Activity

Shorthand

U AU BU

None Active Active

0 1 1

Table 9.1: Activity table for competitive binding. For clarity a zero and a one will be used to designate active and inactive states in subsequent tables.

The table lists three possible states, U; AU and BU , one unbound state and two bound states. Since we assume that only one operator site is available for binding, the operator complex ABU cannot form (Figure 9.10). Now write the fractional occupancy function directly from the table. All entries in the table appear as a sum in the denominator and entries in the table which are active appear as a sum in the numerator, that is: P Active States AU C BU f D P D All States U C AU C BU In addition to the table, we also need the equilibrium relations, these include: AU U C A AU; K1 D U
A U C B BU;

K2 D

BU U
B

From these relations we can express each state in terms of U to yield: f D

K1 A C K2 B 1 C K1 A C K2 B

(9.6)

In all examples so far, we have expressed the fractional occupancy in terms of association constants. Some authors prefer to use dissociation constants however. Since Ka D 1=Kd we can substitute the dissociation constants in place of the association constants to give: f D

A=Kd1 C B=Kd2 1 C A=Kd1 C B=Kd2

9.4. MULTIPLE TRANSCRIPTIONAL FACTORS

217

If we designate the nth transcription factor as Xi , then the ratio, Xi =Kdi can be replaced with the symbol qi to yield the simpler form: f D

q1 C q2 1 C q1 C q2

It is now easier to see that the expression behaves like an OR gate which we would expect given the proposed mechanism. The 3D surface shown in Figure 9.11 illustrates the behavior of this function.

Gene Expression

1

0:5 4 0

2 x

y 0

Figure 9.11: Surface plot showing gene expression as a function of two transcription factors. Each transcription factor is assumed to activate expression and bind competitively at the same operator site. The behavior mimics an OR gate, however the transition from the low to high state is not very sharp. Thus it is not a particularly good OR gate.

Non-Competitive Binding In the last section we considered two transcription factors that compete for the same site. In this section a similar model will be examined except each transcription factor binds at separate sites so there is no direct competition (Figure 9.12).

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CHAPTER 9. KINETICS OF GENE REGULATION

B

A

Transcription Factors

U Operator

Coding Sequence

A

AU B

A

BU

ABU

B

A B

Figure 9.12: Non-Competitive binding of two transcription factors to two operator binding sites, either of which can enhance gene expression.

This small addition introduces a subtle change in the derivation. The state/activity table is extended by one row to account for a fourth state, ABU . The additional state introduces a small amount of extra complexity when considering the binding relations. In particular, the formation of complex ABU can be reached by two different paths as shown in the Figure 9.13. In reaction schemes where there are multiple routes from a source to a destination species, detailed balance (See chapter 1) applies. In this case the net free energy change (See chapter 10) across both routes must be identical:
G1 C
G3 D
G2 C
G4

9.4. MULTIPLE TRANSCRIPTIONAL FACTORS

State

Activity

U AU BU ABU

0 1 1 1

219

Table 9.2: Activity Table for Non-Competitive Binding.

K1

AU

K3 ABU

U K2

BU

K4

Figure 9.13: Binding of two transcription factors to two separate sites. Ki represent the respective equilibrium constants for each process. For detailed balance, it must be the case that K1 K3 D K2 K4 .

In terms of equilibrium constants, this means that the product of the equilibrium constants for one route must equal the product of the equilibrium constants for the other route. That is, K1 K3 D K2 K4 . From the activity table the fractional occupancy is: f D

AU C BU C ABU U C AU C BU C ABU

The two states AU and BU can easily be expressed in terms of U , however ABU is more problematic. The problem with ABU is that there are two possible ways to express ABU in terms of U : ABU D K3 K1 A
B
U

ABU D K4 K2 A
B
U

Because of detailed balance these two relations are identical (K1 K3 D K2 K4 ). Other than convenience, there is no preference for choosing one

220

CHAPTER 9. KINETICS OF GENE REGULATION

over the other. If we choose the first expression for ABU we arrive at a fractional occupancy of: f D

K1 A C K2 B C K1 K3 A
B 1 C K1 A C K2 B C K1 K2 A
B

Normalizing the terms using the dissociation constants yields the simpler form: q1 C q2 C q1 q2 (9.7) f D 1 C q1 C q2 C q1 q2 Again this model mimics an OR gate. The difference with the competitive model is that there are extra product terms in the numerator and denominator. These terms will change the rise time of the response compared to the non-competitive model.

Additional Gate Relationships The previous section described two simple examples of OR-gate like behavior. The next few sections will summarize a variety of other standard gates by including the rate expression and a corresponding biological model. We are not suggesting that gene regulation can be modeled like a Boolean network, but that logic operators are a useful starting point to assemble a set of gene regulatory functions. The proposed mechanisms are just some of the many ways in which to understand signal processing in gene regulatory networks. AND Gate If we assume that the simultaneous binding of two activating transcription factors is required for gene expression then we can model an AND gate, as shown in Figure 9.14. One can imagine a variety of mechanisms to realize the AND gate. For example, both transcription factors can bind weakly to the operator site, but neither on their own can illicite transcription. Together however they form a complex with a new conformation such that RNA polymerase can

9.4. MULTIPLE TRANSCRIPTIONAL FACTORS AND Operation

221

Transcription Factors

U Operator Promoter Coding Sequence

AU BU ABU

A B

A B

Figure 9.14: AND operation. Two transcription factors A and B bind weakly to the operator site such that RNA polymerase is not recruited to the promoter. If both transcription factors are present they form a stronger complex on the operator and thereby recruit RNA polymerase to the promoter.

now bind. Alternatively, neither transcription factor on its own can bind but if they dimerize, the dimer binds to the operator site to promote RNA polymerase. The activity table is shown in Table 9.3. The fractional occupancy is therefore: ABU f D U C AU C BU C ABU which yields: f D

K1 K2 A
B 1 C K1 A C K2 B C K1 K2 A
B

or: f D

q1 q2 1 C q1 C q2 C q1 q2

A 3D surface plot of the AND gate is shown in Figure 9.15.

(9.8)

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CHAPTER 9. KINETICS OF GENE REGULATION

State

Activity

U AU BU ABU

0 0 0 1

Table 9.3: Activity Table for the AND Gate.

State

Activity

U AU BU ABU

1 0 0 0

Table 9.4: Activity Table for NOR Gate.

NOR Gate Figure 9.16 shows a possible mechanism to obtain a NOR gate operation. The approach is very similar to the non-competitive OR gate except the operator site is now located downstream of the promoter. Like the OR gate, there is also a competitor variant for this model. Using this activity state table 9.4, we can express the fractional saturation as: f D

U U C AU C AB C ABU

From this we can derive the NOR gate expression: v D Vm

1 1 C K1 A C K2 B C K3 A
B

9.4. MULTIPLE TRANSCRIPTIONAL FACTORS

223

Gene Expression

1

0:5 10 0

5 x 0

y

Figure 9.15: Surface plot showing gene expression as a function of two transcription factors. Each transcription factor on its own is unable to activate gene expression which only commences when both transcription factors are present. The behavior mimics an AND gate, however the transition from low to high state is not very sharp.

The normalized version of the rate law is: 1 f D 1 C q1 C q2 C q1 q2

(9.9)

NAND Gate Figure 9.17 shows a possible arrangement for a NAND gate where we have taken the AND gate and moved the operator downstream of the promoter. Using this activity state table 9.5, we can express the fractional saturation as: U C AU C BU f D U C AU C BU C ABU From this we derive the NAND gate expression: v D Vm

1 C K1 A C K2 B 1 C K1 A C K2 B C K3 A
B

224

CHAPTER 9. KINETICS OF GENE REGULATION NOR Operation

U Promoter Operator A

AU BU

B A

ABU A

Coding Sequence

B

B Transcription Factors

Figure 9.16: NOR operation. Two transcription factors A and B can bind independently to the operator site. If either of them bind, they block the ability of RNA polymerase to bind or move forward on the promoter.

The normalized version of the rate law is: f D

1 C q1 C q2 1 C q1 C q2 C q1 q2

(9.10)

XOR Gate Figure 9.18 shows a possible arrangement for an XOR (exclusive OR) logic operation. The operator site is located upstream of the promoter so that transcription factor binding activates gene expression. The operator site can bind either of the transcription factors resulting in gene expression. However, the transcription factors may also bind to each other forming an inactive heterologous dimer. The dimer is unable to bind to the operator site thereby inhibiting gene expression. See Table 9.6.

9.4. MULTIPLE TRANSCRIPTIONAL FACTORS

225

NAND Operation

U Promoter Operator

Coding Sequence

A

AU B

BU A B

A + B

A B

ABU A

B

Transcription Factors

Figure 9.17: NAND operation. Two transcription factors A and B bind weakly to the operator site such that RNA polymerase can still initiate transcription. If both transcription factors are present then they can form a stronger complex on the operator and prevent RNA polymerase from moving down the DNA strand.

From this table we can express the fractional saturation: f D

AU C BU U C AU C BU C ABU

The XOR gate expression is therefore: v D Vm

K1 A C K2 B 1 C K1 A C K2 B C K3 A
B

(9.11)

This expression may look similar to the competitive OR gate example, however, the XOR expression has one less term in the numerator. In order for the XOR gate to operate correctly, the equilibrium constant for dimerization must strongly favor the dimer formation. This ensures the dimer is present in high enough concentration so that the individual free forms, A and B, are at lower concentrations. In addition, the concentration if A and B must be stoichiometrically equal otherwise when the dimer is formed

226

CHAPTER 9. KINETICS OF GENE REGULATION

State

Activity

U AU BU ABU

1 1 1 0

Table 9.5: Activity Table for a NAND Gate.

State

Activity

U AU BU ABU

0 1 1 0

Table 9.6: Activity Table for a XOR Gate.

some free A or B will remain resulting in activation of the operon. As a result, this design is not very robust. A more robust model is where the dimer AB binds to a second operator downstream of the promoter so that it doesn’t matter if free A or B is present, binding to the downstream operator by AB will inhibit expression. EQ Gate The final logical operation to consider is the EQ (equivalence) gate. Figure 9.19 shows a possible mechanism for the EQ operation where the operator is located downstream of the promoter. If either transcription factor binds, gene expression is inhibited. If neither transcription factor is bound, gene expression occurs. When both transcription factors are present we assume they form a dimer that is unable to bind to the operator, thereby activating gene expression. The activity state table is shown in Table 9.7. The fractional saturation

9.4. MULTIPLE TRANSCRIPTIONAL FACTORS

227

XOR Operation

U Operator Promoter

Coding Sequence

A

AU

B

BU

1) Non-robust model

B A B A

ABU 2) Robust model

B A B A A B

ABU

Figure 9.18: XOR or exclusive OR operation. Two transcription factors A and B bind to the operator site and help recruit RNA polymerase. If both are present at the same time, they form a dimer. The dimer can no longer bind which means that RNA polymerase is no longer recruited to the promoter site. For this to work, A and B must have been made is equal stoichiometric amounts and is therefore not very robust. Option 2 in the figure shows a more robust model.

expression is: f D

U C ABU U C AU C BU C ABU

From this we can derive the EQ gate expression: v D Vm

1 C K1 A
B 1 C K1 A C K2 B C K3 A
B

(9.12)

One can imagine many other different scenarios for binding patterns that could lead to specific operations. For further possibilities the reader is

228

CHAPTER 9. KINETICS OF GENE REGULATION

EQ Operation

U Promoter Operator

Coding Sequence

A

AU

B

BU B

A B A

ABU A

B

Transcription Factors

Figure 9.19: EQ or equality operation has active states when both transcription factors are present or when neither is present. If one of transcription factors is present, it cannot bind to the upstream promoter which means RNA polymerase is less likely to bind.

referred to the work by Song et al. [59] and Cory et al. [22]. See also the review articles by Bintu [7, 6] and work by Buchler et al.[12]. Counter Regulation A common form of gene regulation is where one transcription factor may activate while the other inhibit. Figure 9.20 shows the high level schematic of this regulatory motif. There are however a number of ways in which to model this behavior, one is based on a non-competitive model (Figure 9.21 a) and other other on a competitive model (Figure 9.21 b). In the noncompetitive binding model (a), the activator, A and repressor, R, bind to separate and distal operator sites. However the repressor dom-

9.4. MULTIPLE TRANSCRIPTIONAL FACTORS

State

Activity

U AU BU ABU

1 0 0 1

229

Table 9.7: Activity Table for a EQ Gate.

Figure 9.20: Model of gene regulation where one gene activates and the other inhibits.

inates because the activator cannot prevent the repressor from binding. In the competitive model (b) two operator site exists but they overlap. One operator site binds activator and is closest to the promoter. The repressor site is positioned slightly upstream from the activator site such that when the repressor binds the activator is unable to. This means the activator can dominate at sufficient high activator concentrations although there is no mutual competition between the two regulators. These two mechanisms give us two slightly different rate laws. For the non-competitive model there will be four states, empty, U , activator bound AU, repressor bound, RU and activator-repressor bound, ARU. The fractional saturation for the non-competitive model is then given by: f D

AU U C AU C RU C ARU

The only active state is when activator is bound, we assume that once the repressor is bound the state is inactive even if activator is also bound. Given this model we obtain the following rate law: v D Vm

K1 A 1 C K1 A C K2 R C K3 A
R

(9.13)

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CHAPTER 9. KINETICS OF GENE REGULATION

Counter Regulation a) Non-competitive Mechanism A

R

b) Competitive Mechanism A

R

A

Activator

R

Repressor

Figure 9.21: Two mechanisms to implement counter regulation. a) Non-competitive binding and b) Competitive binding.

The competitive model has only three states, U , AU and RU: f D

AU U C AU C RU

such that the rate law is given by: v D Vm

K1 A 1 C K1 A C K2 R

(9.14)

Figure 9.22 shows a plot of the competitive counter regulation surface with Hill coefficients of three on both input transcription factors.

Gene Expression

9.5. COOPERATIVITY

231

4 2

4 2

0 TF2

0

TF1

Figure 9.22: Competitive based counter regulation response curve.

9.5 Cooperativity Up to this point we have considered models based on transcription factors that are monomers. However many proteins exist as multimers, that is complexes of protein monomers. Oftentimes multimers are made from the same protein monomer, for example Glyceraldehyde 3-phosphate dehydrogenase is made from four identical subunits. Sometimes multimers are made from different monomers, for example Hemoglobin is made from two different but similar subunits arranged as a dimer of dimers. Multimeric proteins are very common. Half of the enzymes reported in the E. coli database at EcoCyc are multimeric. In particular, regulatory proteins are frequently found as multimers. For example, the LacI repressor is a complex of four identical subunits while the CI and Cro repressors from lambda phage are both complexes of two subunits. If a monomer has the capacity to bind a ligand, then a multimer may be able to bind as many ligands as there are monomers in the multimer. For example, LacI binds up to four molecules of allolactose. LacI also has two additional binding sites that recognize the operator sites formed from the dimer pairs.

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CHAPTER 9. KINETICS OF GENE REGULATION

With these points in mind, we define cooperativity in the following way (cf. chapter 6): If the binding of one ligand to a protein multimer causes the binding affinity of the remaining sites to change, then this is termed cooperativity. Cooperative behavior implies there is some kind of physical interaction between binding sites. If the binding affinity increases, the cooperativity is termed positive cooperativity, whereas if the binding affinity decreases, it is termed negative cooperativity. Positive cooperativity usually leads to a sigmoidal response as a function of ligand concentration as shown in Figure 9.23. There are a number of reasons why cooperativity in biology might be important. The most notable property that positive cooperativity confers is the increase in sensitivity over the simpler Michaelian hyperbolic response. Although cooperativity may result in a sigmoidal response, not all sigmoidal responses, as we shall see, are a result of cooperativity. 1

Activity

0:8 0:6 0:4 0:2 0

0

0:5

1 1:5 2 2:5 Ligand Concentration

3

Figure 9.23: Typical sigmoidal response as a result of positive cooperativity.

9.5. COOPERATIVITY

233

By its nature, cooperativity involves the interaction of two or more binding sites. The binding sites can either reside on the same molecular complex as found in LacI, or as separate interacting complexes, as found in lambda phage CI protein. In either case, the binding at one site increases the affinity at a second site. In the case of LacI when the first operator binding site is occupied, there is a greater likelihood of binding to the second operator site. Binding at the first site is thought to restrict the degree of freedom of the second site which allows the second operator binding site to be more easily located. In the case of CI, the first bound repressor exposes a protein face for the second copy of CI to bind more easily. A logical question to ask is how to qualitatively describe cooperativity for transcription factors. Let’s consider the repressor LacI to investigate this question further.

LacI Model The LacI repressor is a tetramer of four identical protein subunits (Figure 9.24). More precisely, the tetramer is arranged in the form of two pairs of dimers. Each subunit has a binding site for its natural ligand, allolactose, and each dimer can bind to one of two operator sites belonging to the lac operon. LacI therefore has two kinds of binding sites. The lac operon codes for a series of enzymes (in particular ˇ-galactosidase and a permease) that enables E. coli to digest lactose. Without lactose, the LacI repressor binds to the two operator sites on the lac operon and in doing so, prevents the RNA polymerase from binding to the promoter region. However the inhibition is not perfect, and even in the absence of lactose there is a very small basal level of gene expression. Thus very small amounts of ˇ-galactosidase and the permease are produced. When lactose is available, some is converted by the basal level of ˇ-galactosidase to galactose and glucose, and some is converted to the disaccharide allolactose which can then bind to LacI. When allolactose binds to LacI there is a conformational change so the binding affinity of LacI for the operator sites is greatly diminished, and the LacI repressor no longer inhibits RNA polymerase binding. Figure 9.25 depicts a simple lacI/Operon model. First consider the binding

234

CHAPTER 9. KINETICS OF GENE REGULATION

4

2

3

1

Figure 9.24: Bacterial Lac Repressor is composed of four identical subunits that can bind. Numbers mark the individual subunits. Image modified from RCSB Protein Data Bank and David Goodsell © (www.pdb. org, http://dx.doi.org/10.2210/rcsb_pdb/mom_2003_3)

of allolactose (denoted by L - Ligand) to the LacI repressor. Although LacI has four potential binding sites for allolactose, it is believed that slightly less than four (about 2 to 3) sites are actually occupied in vivo. The dissociation constant for the binding of allolactose to LacI has been estimated experimentally to be approximately 10 6 M. One might also expect that the binding of allolactose could demonstrate positive cooperativity. Surprisingly, experiments have shown that LacI exhibits no cooperativity for allolactose under in vivo conditions. In the active form (P - Protruded) the model in Figure 9.25 depicts a stepwise binding of LacI onto the two operator sites. The different bound forms are labeled A, B, C and D. Note that the fully unbound form, A, is the active state that results in gene expression. We can express the fractional occupancy as: f D

A ACB CC CD

9.5. COOPERATIVITY

R

L

235

K1

P

B

K3

A

D K2

C

K4

Figure 9.25: Simple model of LacI and the lac Operon. Allolactose (L) binds to LacI with a dissociation constant of 10 6 M and without any cooperativity. The concentration of LacI present in the cell is assumed to be around 10 8 M, roughly ten molecules per cell. R (Retracted) is the inactive form of LacI, and P (Protruded) the active form. Note that for E. coli, 1 nM equates roughly to one molecule per cell.

The equilibrium relations can be used to express the states, B, C, and D in terms of A to yield: f D

1 1 C K1 P C K2 P C K1 K3 P 2

Given positive cooperativity, let’s assume the second binding step (K3 ) is much stronger than the first binding steps (K1 and K2 ). This means we can simplify the equation to: f D

1 1 C K1 K3 P 2

Now take into consideration the binding of allolactose to the active (or bound) form of LacI, P . Experiments tell us that there is no cooperativity between allolactose and LacI. We therefore assume the P form of LacI is a simple hyperbolic relationship with respect to its ligand: P D

Kd T L C Kd

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CHAPTER 9. KINETICS OF GENE REGULATION

Kd is the dissociation constant for allolactose, T is the total concentration of LacI, and L the level of allolactose. That is, as we add ligand, the level of the P decreases (See Figure 9.25). If we substitute this into the fractional occupancy for A, we arrive at: f D

.L C Kd /2 .L C Kd /2 C K1 K3 .K T /2

If we assume that in vivo the dissociation constant (10 6 M), is smaller than the concentration of allolactose (> 10 5 M), then we can approximate the equation to: L2 f D 2 (9.15) L C Km where Km represents the lumped expression K1 K3 .Kd T /2 . This equation (9.15) is commonly used to represent Lac operon kinetics in response to changes in ligand concentration (particularly IPTG – Isopropyl ˇ-D-1thiogalactopyranoside). The equation also has the same structure as the Hill equation (6.1) with a Hill coefficient of two. This value is consistent with the experimentally measured Hill coefficient and matches reasonably well with the known Lac operon response. It is remarkable that we can describe the response of the lac operon using such a simple equation given the known complexities of the system. This highlights the possibility of distilling complex systems down to their core responses.

Hill Equation The equation representing the response of the lacI operon response to changes in ligand binding is very similar to the Hill equation (Section 6.1). Hill equations are frequently used to describe the responses of gene expression to changes in transcription factor levels. Two forms are commonly used, one to model the effect of a single repressor and another to model the effect of a single activator. The activator equation is given by: vD

Vm T h K CTh

(9.16)

9.5. COOPERATIVITY

237

where Vm is the maximal rate of gene expression, T is the transcription factor concentration, K is the concentration of ligand and half-maximal activity, and h is the Hill coefficient. The repressor equation is similar except for a change in the numerator: vD

Vm K CTh

(9.17)

The equations are sufficiently general that T may be a transcription factor or a small molecule such as allolactose.

Effect of Oligomerization Protein synthesis generates protein monomer units. However since many proteins are oligomeric, there is a further ‘maturation’ stage where monomers form into oligomers. The fact that many transcription factors are oligomeric means we must consider this process when looking at the responsiveness of gene regulation.

Figure 9.26: Dimerization of gene product to form an active transcription factor.

By example, consider the CI transcription factor from lambda phage. This protein is produced as a monomer but is only active as a dimer. Let us designate the monomer by the letter M , and the dimer by the letter D. The reaction scheme for the dimerization is given by: 2M D

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and at equilibrium we can state: KD

D M2

Assume that D can now bind to an operator site but without any cooperativity, that is: D C O DO where DO is the bound state of the operator. The overall reaction between M and DO is: 2M C O DO where the equilibrium constant for the overall reaction is the product of the two individual equilibrium constants: K D K1 K2 D

DO M2
O

We can also write down the following fractional occupancy relation: f D

DO DO C O

Using the overall equilibrium relation yields: f D

KM 2 KM 2 C 1

Again this is reminiscent of the Hill equation with a Hill coefficient of 2. Dimerization can therefore generate sigmoid behavior. The degree of sigmoidicity will depend on how many monomers make up the transcription factor complex. For example a tetramer will generate a Hill equation with a Hill coefficient of 4 and a correspondingly sharper sigmoid response. If oligomerization is coupled with cooperativity when the transcription factor binds to the operator then greater sensitivity is achieved. Thus a dimerization coupled to dimer cooperativity at the operator site (such as with CI) could lead to a Hill coefficient of up to 4.

9.6. THERMODYNAMIC APPROACH

239

Multiple TFs and Cooperativity It is possible to generalize the single transcription factor equations to multiple transcription factors. The same approach applies assuming rapid equilibrium conditions: write the expression that describes the fractional saturation with all active states appearing in the numerator and the sum of all states in the denominator. Whenever the same transcription factor binds to multiple operators, this is taken into account by means of raised powers. We will leave further discussion of this topic to the next section where we describe the statistical thermodynamics approach to gene expression kinetics.

9.6 Thermodynamic Approach The use of equilibrium constants in analyzing equilibrium systems has a long tradition in biochemistry, especially in relation to enzyme kinetics and cooperativity. Most of the ground work in analyzing cooperative systems was carried out using this approach. However there has also been a parallel effort using a more statistical thermodynamic approach [31]. This approach has seen more widespread use since our understanding of gene expression has increased. Both approaches yield the same result, but the statistical approach offers two distinct advantages. The first is that it emphasizes the probabilistic nature of the binding and unbinding process; this is particularly important for gene regulatory systems where the number of molecules may be quite small. The second advantage is more utilitarian, where it emphasizes the use of free energy rather than equilibrium constants. In some cases the free energies can be computed directly from the known 3D structure of the protein by considering the individual base pairings between the protein and the DNA strand at the binding site [46]. The statistical thermodynamic approach, like the equilibrium binding approach, revolves around enumerating all the states of a given system and then determining the probability of a given set of states relative to the totality of all states. The key is to look at a particular state from the perspective of an energy function, the free energy change,
G (See chapter 10). The free energy change is measured with respect to a reference state. In this

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case the reference state is the fully unbound state and is defined to have a free energy change of zero. The free energy change refers to the difference in free energy between the reference state and the particular bound state under consideration. For example, the binding of CI to the operator site OR1 can be described in terms of the free energy change,
G1 D 49:11 kcal mol 1 . This is the free energy difference between the reference state and the bound state of CI. Different operator sites that have slightly different binding sequences will result in slightly different free energy changes for a give transcription factor. Thus, CI can bind to two operator sites, OR1 and OR2 with roughly equal affinity, and this is reflected in the similar free energies of 49:11 and 42:29 kJ mol 1 , respectively. If these states are in equilibrium, we can invoke the Boltzmann distribution law. This describes the probability of finding a given state among a number of other states. With this in mind, the probability of a give state, s, for a single transcription factor X, is represented by the following weighted probability function: exp.
Gs =RT /X j f DX exp.
Gs =RT /X j

(9.18)

s

The denominator in the above equation is called the weighted partition function, and represents the sum of all states that X can assume including the unbound state.
Gs is the free energy change that occurs as a result of the formation of the state s; R and T are the gas constant and temperature, respectively. j is the number of transcription factors bound to the operator site in state s. The summation of s in the denominator is taken from 0 to the number of possible states. The zero in the summation is important because it represents the unbound state where the free energy change is zero. Hence the first term is always one. j appears in the summation because as the states are enumerated, the number of bound copies of transcription factor X will change from state to state. For example, in the unbound state there are no copies of X bound, j D 0 and the term exp.
Gs =RT /X j equals one because
Gs D 0 for the unbound state.

9.6. THERMODYNAMIC APPROACH

241

For systems where there are multiple transcription factors, the probability function generalizes further to: exp.
Gs =RT /X j Y k : : : f DX exp.
Gs =RT /X j Y k : : : s

where k is the number of copies of transcription factor Y bound to state s. If a state only has one copy of transcription factor X bound and no Y , then k is zero and the Y term drops out. If there is more than one active state, each active state must be included in the numerator sum. It may seem that this is no different from the approach that relies completely on equilibrium constants. However, the use of
G in the formalism makes it possible to construct the net free energy change for a particular state by simply summing the free energies of the individual interactions involved in the formation of that state; equivalent to multiplying the equilibrium constants. An example will clarify this point. Consider the CI transcription factor where one of its states involves two copies of CI bound to the operator sites OR1 and OR2. To construct the free energy change for this state we must consider what interactions are involved in the formation of the state. Experiments show that when two CI molecules are bound, there are three interactions. One interaction is directly with the operator site OR1, another is with the operator site OR2, and a third is due to cooperative interactions between the two copies of bound CI. Therefore the overall free energy change is the sum of three more elementary changes:
Gs D
GOR1 C
GOR2 C
GCI

CI

Such accounting must be done for all possible states. In the case of CI, there are eight possible states for three operator sites. Table 9.8 (taken from Ackers et. al. [1]) shows all eight possibilities. Note that the zeroth state, which represents the unbound state is the reference state and therefore has a free energy change of zero. The advantage of this approach is that it is easier to explicitly identify the cooperativity effects and include them in the equation compared to the equilibrium approach.

CHAPTER 9. KINETICS OF GENE REGULATION 242

State 0 0 0 R 0 R R R

OR3

OR1

0 R 0 0 $ R R 0 $ R

OR2

0 0 R 0 R 0 $ R R

0 1 2 3 4 5 6 7

Reference State
G1
G2
G3
G1 C
G2 C
G12
G1 C
G3
G2 C
G3 C
G23
G1 C
G2 C
G3 C
G12

Free Energy

0 -49.11 -42.29 -42.29 -99.65 -91.27 -92.95 -141.93

Total Free Energy (kJ mol

1)

Table 9.8: Operator sites and free energy contributions for the different states of the CI repressor of lambda phage. Arrows between columns indicate cooperative interactions. The standard free energy represents the free energy change relative to the reference state. This means that the more negative the free energy, the more stable the given state (or larger the association constant). Data taken from [1].

9.6. THERMODYNAMIC APPROACH

243

State

OR1

OR2

Free Energy Contrib


G (kJ mol

0 1 2 3

A A

A A

Reference State
G1
G1
G1 C
G2 C
G12

0 -25 -23 -64

1)

Table 9.9: Hypothetical Example of Free Energy States.

To better illustrate the idea, consider a simple hypothetical example. Table 9.9 shows free energy changes for the binding of an activator, A, to two operator sites, OR1 and OR2. The more negative a free energy is the more stable the interaction. The promoter is only active when both operator sites are occupied. The symbol ‘-’ in the table indicates ‘not bound’. First note (Table 9.9) that the binding of A is not independent of the binding of a second A. We can see this by looking at the free energy changes when A binds individually to each site and when both sites are bound. If the binding of A to each site were independent, we would expect the sum of the free energy changes to equal the free energy change when both are bound. However, the sum of the individual free energy changes is .25 C 23/ D 48, whereas the free energy change when both are bound is 64. This indicates there is a 16 kJ mol 1 additional degree of stability when both are bound. This could be for a variety of reasons, including pair-wise binding of A to each other, or DNA looping as a result of both operators being occupied. The important point is that whatever the interaction, it increases the stability of the A-A state. We now enumerate all the states, including, U , AU1 , AU2 , and AAU. From this list we construct the weighted partition function (denominator in equation (9.18)) to be: Q D1Ce


G1 =RT

ACe


G2 =RT

ACe


G3 =RT

AA

The first
G represents the free energy change that occurs when the first operator site is bound, that is
G1 D 25. The second
G represent the

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bound form on OR2,
G2 D 23. Finally, the third
G represents the state when both are bound which equals
G3 D 64. Note that
G3 represents the sum of three interactions,
G1 ,
G2 and
G12 , where
G12 represents the free energy change due only to the interaction between the two transcription factors. The fractional occupancy can now be determined from the relation (AAU is the active state): f D

e
G3 =RT A2 1 C e
G1 =RT A C e
G2 =RT A C e
G3 =RT A2

To show that the statistical approach leads to the same result as the equilibrium constant method, recall that a particular state, such as the binding of CI to OR1, can be described using the relation between the equilibrium constant and the free energy:
Gs D

RT ln Ks

In terms of
G the equation can be rearranged to yield: Ks D exp.
Gs =RT / From this we can rewrite the fractional occupancy given above to the more familiar form: K3 A2 f D 1 C K1 A C K2 A C K3 A2 If no specific data is available on the binding mechanism but the overall response is known (e.g. AND gate) and cooperativity is known to occur then it is possible to just raise the TF concentrations to some suitable power. For example a cooperative version of the OR gate can be written as: f D

K1 An C K2 B n C K1 K3 An
B n 1 C K1 An C K2 B n C K1 K2 An
B n

A surface plot of this equation is given in Figure 9.27 for n D 2. The question remaining is how to determine the
G values that can be substituted into the rate law? A number of methods exist to determine the

9.6. THERMODYNAMIC APPROACH

245

Gene Expression

1

0:5 2 0

1 TF2

0

TF1

Figure 9.27: Competitive based OR gate with cooperativity where n D 2.

binding affinities for transcription factors. Let’s review the gel shift or gel retardation method as one possible experimental method.

Gel Shift DNA in solution is negatively charged due to the sugar-phosphate backbone. As a result, DNA will migrate in gels when an electric field is applied across the gel. The smaller the DNA fragment, the faster the migration. A standard technique in molecular biology is the use of gel electrophoresis to separate DNA fragments. To visualize and quantify the intensity of the bands on the gel, the DNA can be prelabeled on the phosphate end with radioactive 32 P which renders the DNA visible when the gel is exposed to photographic film. If large quantities of DNA are used then the bands can be visualized by non-radioactive ethidium bromide but this is less sensitive. A common and sensitive alternative to radioactive labeling is to use biotin-labeled transcription factors which can be detected using strepatvidin-HRP and a chemiluminescent reporter.

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One application of DNA gels is the electrophoretic gel shift or gel retardation assay (EMSA). This is a very useful application for DNA-protein interaction analysis. The technique can be quantitative as well as qualitative. The key to gel shift analysis is the observation that DNA bound to protein tends to migrate slower than naked DNA. Therefore if a mixture of naked and bound DNA is added to a gel lane, two bands will emerge, one corresponding to the naked DNA and another to the protein bound DNA. Qualitatively, the shift in the migration band can be used as a test for whether a protein will bind or not to a specific DNA sequence. For a quantitative study, a variety of mixes are made using a fixed amount of DNA but varying protein levels. Each mix is applied to its own lane on the electrophoresis gel. If no protein is present in the mix then only one band will be observed, corresponding to the fast migrating naked DNA. If excess protein is added to the mix, then only one band will be observed, the so-called shifted band. This band corresponds to DNA bound to protein and will appear earlier in the lane because it migrates slower. Using excess protein in the second mix ensures that no naked DNA is present. The key however lies in the use of intermediate protein levels where pairs of bands should appear with varying intensity. As the level of protein increases, the shifted band should increase in intensity while the band corresponding to the naked DNA slowly looses intensity. If the DNA sequence binding involves a simple bindingunbinding event, then we can apply a simple hyperbolic equation for the fractional occupancy involving the binding dissociation constant. The fractional occupancy itself can be measured from the intensity of the shifted band divided by the intensity of the naked DNA band in the presence of no protein (representing total unbound state). The fractional occupancy can be plotted against the level of free protein. If the protein concentration is higher than the DNA concentration, the free protein can be assumed to be equal to the added protein. Assuming simple binding without any cooperativity, the data can be fitted to a hyperbola and an estimate for the dissociation constant obtained. The dissociation constant can also be read directly off the graph by locating the half-saturation level and reading off the corresponding protein concentration. One of the great advantages of this technique is that if a particular operon

9.6. THERMODYNAMIC APPROACH

247

includes a number of operators, synthetic DNA fragments can be constructed to include combinations or individual copies of the operators. For example, in the case of the cI operon, binding to the individual operators, OR1, OR2, and OR3 can be determined separately from pairs of combinations or all three operators together. In principle this method is be capable of determining cooperative effects between different operator sites. The net results of the gel shift experiments is the determination of the free energies required which in turn, may be used in the gene expression rate law.

Chapter Highlights 1. The structure of a genetic unit includes operators, a promoter, a ribosome binding site, gene encoding sequence and a terminator. 2. The basis for deriving rate laws for gene expression is the evaluation of the fraction saturation function. 3. The fractional saturation function can be computed by assuming a rapid equilibrium between the bound and unbound forms of the transcription factors. An inventory of all possible states is made from which the ratio of the sum of all active state to the sum of all states gives the fractional saturation. The rate of gene expression is then assumed to be proportional to the fractional saturation. 4. A variety of simple non-cooperative gene rate laws are given including all the basic logic gate operations, such as AND, OR etc. 5. Cooperativity can be introduced into gene expression by a number of mechanisms. The first is that the binding of one transcriptional factor makes the binding of a second factor much easier. The second is that many transcriptional factors are oligomers which in turn must be formed from the monomers that get made at translation. The polymerization of monomers into active oligomers can generate responses which are more sensitive than if the monomers bound individually. Polymerization and cooperative binding can be combined to give even higher levels of sensitivity.

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CHAPTER 9. KINETICS OF GENE REGULATION

6. The thermodynamic approach to deriving the gene expression rate laws is also introduced. This approach is the same as the rapid equilibrium approach favored by enzyme kineticists but in relation to protein binding to DNA. The availability of free energy data makes the thermodynamic approach more appealing and for deriving the rate laws. 7. Appendix B lists all the common gene expression rate laws outlined in this chapter.

Further Reading 1. Bolouri H. (2010) Computational Modelling Of Gene Regulatory Networks – A Primer. Imperial College Press; 1 edition/ ISBN 1848162219 2. Brown T. A (2006). Genomes 3 (3rd ed.). Garland Science. ISBN 0815341385. 3. Myers C. (2009) Engineering Genetic Circuits. Chapman & Hall CRC. ISBN 1420083244.

Exercises 1. Describe the different parts of a prokaryotic operon and explain their functions. 2. The active form of a given transcription factor is a monomer and can stimulate gene expression. Derive the rate law that describes the rate of gene expression as a function of the concentration of transcription factor.

10

Basic Thermodynamics

249

250

CHAPTER 10. BASIC THERMODYNAMICS

Thermodynamics is the study of energy and entropy changes involving heat and work. It allows us to determine the feasibility of a process by calculating the Gibbs energy, how much energy is transferred in a process by calculating the enthalpy change, and to what extent the process can take place by calculating the equilibrium constant. Thermodynamics (or more precisely, classical thermodynamics) is concerned with macroscopic quantities such as pressure, temperature and volume. One of the strengths of classical thermodynamics is that relationships between these quantities is independent of the molecular basis of matter. This is also a weakness however, as it cannot provide deeper insight into the chemical or physical phenomena. To better understand these, a field called statistical thermodynamics was developed to focus specifically on microscopic models. Not surprisingly the mathematics used in statistical thermodynamics field is significantly more advanced. This chapter gives an introduction to classical thermodynamics.

10.1 Systems and Surroundings Like many other scientific studies, thermodynamics divides the world into two parts, the system and the surroundings. The system is that part of the universe we wish to focus our attention. This might be a single enzyme reaction, a pathway, a single cell, an organ, or an entire multicellular organism. The surroundings includes everything else. The boundary between the system and the surroundings can have different properties which determines whether the system is isolated, closed or open (See Table 10.1). All living systems are open because they exchange energy and matter with their surroundings. Closed and isolated systems are simplified systems for study, and are frequently used in thermodynamic arguments.

10.2 Energy, Work and Heat Energy is a fundamental concept in science and yet it is also one of the most mysterious. Traditionally, the energy of a system is defined as its capacity to do work. Work, w, is done when an object is moved against an opposing force. Work is a familiar, every day concept and includes

10.2. ENERGY, WORK AND HEAT

251

System Class

Properties

Isolated

Completely isolated in every way. No matter, work, or heat is exchanged with the surroundings. Work and heat can be exchanged with the surroundings. Work, heat, and matter can be exchanged with the surroundings.

Closed Open

Table 10.1: Properties of isolated, closed and open systems. Isolated System

Closed System

w

q

U

Open System

q Mass

w Figure 10.1: Isolated, closed and open systems. U is the internal energy; q and w represents the transfer of heat and work respectively.

situations such as raising a weight, the expansion of a gas against an external pressure, or a chemical reaction that drives an electric current to make a light glow. These examples illustrate that energy can exist in different forms and can be transformed from one form to another. Each of these diverse systems has the capacity to do work, therefore each system has a specific energy associated with it. Physically, the energy is manifested in many different ways, but one of the most important observations is that no matter how energy is transferred, the total amount of energy in the system plus the surroundings always stays the same. That is, energy is conserved. For example, when work is done on a system, the energy of the system in-

252

CHAPTER 10. BASIC THERMODYNAMICS

creases. At the same time the energy of the surroundings decreases by an equivalent amount. Conversely, when the system does work on the surroundings, the energy of the system is reduced and the energy of the surroundings increases, again by the same amount.

Heat When two wooden sticks are rubbed together, work is done on the sticks, therefore the total energy of the two sticks must increase with an accompanying decrease in the energy of the surroundings. In this case the increase in energy is most likely to appear in the form of heat, q. Heat energy will manifest itself as a rise in temperature and at the molecular level, an increase in the kinetic energy or motion of the particles. The rubbing of two objects together is an example of work being transferred into heat. It is possible to transform heat into work by heating a gas connected to a piston which drives an engine. As the temperature of the gas increases, the pressure of the gas also increases which in turn pushes out the piston, thereby converting the heat into mechanical work. Although there are generally no restrictions on converting work into heat, the converse is not true. The Internal Energy. A useful concept in thermodynamics is the internal energy, U. This energy is the sum of all kinetic and potential energies in the system. Although it is difficult to assign an absolute value to the internal energy of a given system, it is possible to measure changes in internal energy between two states. Like volume, pressure or mass, the internal energy is a state variable, that is for a given state, there is a unique value for the internal energy. It doesn’t matter how one reaches the given state, the internal energy is uniquely defined for that state. State variables are therefore called path independent.

The First Law of Thermodynamics The conservation of energy is enshrined in the first law of thermodynamics which states that energy can be transformed but cannot be created or destroyed. We can describe this conservation between the system and the

10.2. ENERGY, WORK AND HEAT

253

surroundings using the relation:
Usys C
Usurr D 0 where
means change and where
U is the change in energy in the system or surroundings. This relation states that for any process that occurs in the universe, the total energy change is zero. Although this expresses the first law in a succinct way, it is not particularly useful. Instead the first law is more often expressed using the following relation which focuses on the system changes:

U D q C w

(10.1)

where
U is the total internal energy change in the system, q is the amount of heat transferred into the system, and w is the amount of work done on the system. Note that if q or w is negative, either heat is being transferred out of the system, or the system does work on the surroundings. The equation tells us that all energy changes between a system and its surroundings must balance such that the total energy of the system and surroundings remains constant.

Units of Energy Although energy manifests itself in many forms, there is a single unit of energy that is often used to express the energy change of any process. The SI unit used to express energy is the Joule (J). The Joule is defined as the amount of energy transferred when a force of one Newton is applied over a distance of one meter. The pre-metric unit is the calorie which is still in use by some countries, most notably the USA. One calorie equals 4.184 Joules. In chemistry where we frequently deal with amounts of substance, changes in energy are usually expressed in terms of Joules per mole (J mol 1 ). Example 10.1

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CHAPTER 10. BASIC THERMODYNAMICS

Surroundings

q ¢U = q + w System

w Figure 10.2: First Law of Thermodynamics. q is the heat transferred into the system and w the work carried out on the system.
U is the net change in internal energy as a result of these transfers. The energy in the surroundings decreases by an equivalent amount to
U .

If the change in the total energy (U ) of a given system is 40 kJ and is accompanied by a transfer of 20 kJ of heat (q) out of the system, comment on the work done. Given the first law of thermodynamics we can write the total change in energy (U D 40 kJ) is: 40 D

20 C w

Note that the heat change (20 kJ) is negative because heat is leaving the system. From this we conclude that w D 60 kJ. The amount of work is 60kJ. Since this is positive, it means that 60 kJ of work was done on the system by the surroundings.

10.3 Enthalpy Change When chemical reactions occur, they either absorb or release energy. This change represents the difference between energies associated with the bonds in the reactants and products. As a reaction proceeds some bonds are broken and others are formed. It takes energy to break a bond, but energy is

10.3. ENTHALPY CHANGE

255

released when bonds are formed. The energy released in a given reaction represents the balance of the energy required to break bonds and the energy released when new bonds are made. Many reactions, but not all, show a net release of energy that will often appear as heat. How can we measure such energy changes? There are two laboratory examples to consider. If we add zinc metal to hydrochloric acid, the result is soluble zinc chloride and hydrogen gas. If we perform this reaction in a test-tube, we can either stopper the tube to prevent the hydrogen gas from escaping, or leave the tube open to the atmosphere. In the former scenario the volume of the space in the test-tube remains constant. In the latter, the pressure stays the same (at atmospheric pressure) but the volume of gas increases. The later experiment is a much more common laboratory situation. Why should it be important to consider one or the other? The chief difference is related to work. When the tube is closed, the hydrogen gas does no work since the hydrogen has nothing to push away. Therefore the only way to release the energy is via heat. We can express this with the following equation: U D qv

Constant volume

However in the open tube, the hydrogen will push the atmosphere back and in doing so, does work on the atmosphere. We can express this by the following equation (the negative sign indicates work is done on the surroundings): U D qp

w

Constant pressure

so that: qp D U C w We now define qp to be the enthalpy change, or H, sometimes called the heat of reaction. Given that we take into account all possible energy transfers, an enthalpy change is a state variable and independent of path.

H D U C wpv

(10.2)

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CHAPTER 10. BASIC THERMODYNAMICS

In equation (10.2) wpv is the work carried out by the changes in volume. For reactions that occur in liquid state only, such as most cellular reactions, the enthalpy change and the internal energy change are almost equal. If the enthalpy change for a given reaction is negative, then the process is called exothermic, and the reaction mixture gets warm. If the enthalpy change is positive, the process is called endothermic and consequently the reaction mixture cools down. One may be tempted to use a negative enthalpy change to indicate whether a reaction will occur or not since many reactions release heat as they proceed especially violent reactions. However the sign of the enthalpy change cannot be used in this way. For example, ammonium nitrate will readily dissolve in water but at the same time absorb significant amounts of heat from the surroundings, resulting in a cooling of the solution. Therefore H cannot be used to determine whether a reaction will proceed or not. Example 10.2 Assume that a chemical reaction transfers 144 kJ of heat to the surroundings while at the same time 25 kJ of work is performed on the system by the surroundings. Calculate the U and H for the system and the surroundings. Usys D q C w D

144 C 25 D

Hsys D U C w D

119 kJ

119 C 25 D

94 kJ

Usurr D 119 kJ Hsurr D 94 kJ

Standard Changes Enthalpy changes often depend on temperature and pressure. As a result enthalpy changes are reported under a set of standard conditions. These standard conditions will often be 25ı C and a pressure of 1 bar (approximately 1 atm) when one mole of reactants react. Under these conditions pure substances are in their standard states. The symbol oftentimes used to denote the standard enthalpy change is Hı . For example, the combustion of methane is written as: CH4 (g) C 2O2 (g) ! CO2 (g) C 2H2 O (l)

Hı D

890 kJ mol

1

10.3. ENTHALPY CHANGE

257

The standard enthalpy change of 890 kJ mol 1 indicates the energy change that accompanies the conversion of 1 mole of methane and 2 moles of oxygen to 1 mole of carbon dioxide and 2 moles of water under standard conditions. Given that the enthalpy change is negative, the reaction is exothermic. In biochemistry, a further standard has been introduced where for reactions in solution, the hydrogen ion activity is set to pH 7 (rather than pH 1). The 0 symbol for this standard condition is usually indicated by H ı . By convention the standard enthalpy of an element under standard conditions is defined as zero. For example, the standard enthalpy of oxygen gas at 25ı and a pressure of 1 bar is 0.

Hess’s Law Because enthalpies are state variables (that is path independent), it is possible to compute unknown reaction enthalpies from those enthalpies of known reactions. More specifically, the overall heat change in a reaction is equal to the sum of heat changes in the individual steps. This is known as Hess’s law and allows enthalpy changes to be calculated even when it is not possible to directly measure the enthalpy. Of particular utility in these types of calculations are the standard enthalpies of formation. The standard enthalpy of formation (Hfı ) is the change in enthalpy that occurs when 1 mole of the substance in its standard state is formed from its constituent elements in their standard states. For example, carbon dioxide is a gas at 25ı and in principle, could be formed from carbon (in the form of graphite) and oxygen gas in their standard states at 25ı . The reaction is given by: C.s/ C O2 .g/ ! CO2 .g/

Hfı D

393:51 kJ mol

1

Given there are tables of standard enthalpies of formation1 [42]), it is possible to use this data to compute the standard enthalpy changes in any reaction. This is achieved by subtracting the sum of the standard enthalpies 1 See the small table at Wikipedia http://en.wikipedia.org/wiki/Standard_ enthalpy_change_of_formation

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CHAPTER 10. BASIC THERMODYNAMICS

of formation of the reactants from the standard enthalpies of formation of the products: X X Hrı D Hfı (products) Hfı (reactants) where  is the number of moles of the particular compounds that is involved in the reaction. Example 10.3 Consider the combustion of sucrose (C12 H22 O11 ) into carbon dioxide and water: C12 H22 O11 .s/ C 12 O2 .g/ ! 12 CO2 .g/ C 11 H2 O.l/ where the enthalpies of formation for the individual compounds is given by: Standard Enthalpy of Formation

Value

Carbon Dioxide (g) Water (l) Oxygen (g) Sucrose (s)

-393.51 kJ mol -285.83 kJ mol 0 -2222 kJ mol 1

1 1

Note the standard enthalpy of formation for oxygen is zero because elements in the standard state are defined to have a standard enthalpy of zero. Using Hess’s law we can compute the standard enthalpy change during combustion: Hrı D .12 Hfı CO2 C 11 Hfı H2 O/ .Hfı .C12 H22 O11 / C 12 Hfı O2 / D

5644 kJ mol

1

10.4 Entropy The driving force for change in the universe is the tendency of matter and energy to disperse, an observation enshrined in the second law of thermodynamics. The measure of dispersal is called the entropy, S, and is a

10.4. ENTROPY

259

quantity as fundamental as energy. Like internal energy, U , entropy is also a state variable but unlike energy, it is not conserved. At the atomic level all energy is quantized, that is atoms and molecules possess discrete states of kinetic energy. As molecules move through space they exchange thermal energy with each other as discrete quantized packets in the form of translational, rotational and vibrational modes. The more complicated the molecule, the more possible states it can have. The number of quantized states will determine how many ways the energy can be distributed. For example a crystal, where the atoms or molecules are relatively rigid, has fewer accessible states compared to a gas or liquid. A given configuration of states is called a microstate and in practice, its description requires a huge amount of information. The position and momentum of each molecule must be specified at each instant in time. In contrast, the macrostate of a system can be described by a very small number of state variables, such as volume, pressure, and temperature. We can illustrate a microstate with two fair coins. There are four ways to arrange these coins in terms of heads (H) and tails (T), that is TT, HH, TH and HT. The last two states, TH and HT are effectively equivalent, so we call these equivalent microstates. This means that when we toss both coins, we are more likely to obtain a TH or HT combination than TT or HH. Thus, the microstates (TH, HT) are more probable that either (HH) or (TT). Likewise a gas that is spread out in a container has many more microstates compared to the same gas that is bunched up at one end. This means that the ‘spread out’ gas configuration is more probable than the ‘bunched up’ gas configuration. In a physical system each microstate contributes to a macrostate. Two systems which have equivalent microstates (such as TH and HT) will yield the same macrostate. In a gas or liquid at equilibrium, the microstate changes every second to an equivalent microstate due to random molecular collisions. However each microstate, though subtlety different from the previous one in terms of molecular positions and momenta, will yield the same macrostate, that is the same volume, pressure and temperature. The volume of a glass of water at room temperature does not visibly vary from second to second as it sits on a table.

260

CHAPTER 10. BASIC THERMODYNAMICS

State

Probability

TT TH, HT HH

1/4 1/2 1/4

Table 10.2: Probabilities of different heads and tail combinations in a two coin toss. The occurrence of the (TH,HT) pattern is twice as probable as getting either all heads or all tails.

The key to understanding entropy is that the more equivalent microstates there are for a given macrostate, the greater the entropy for that macrostate. A gas that expands into an empty space will have a higher number of possible microstates to occupy, hence the entropy will increase. A dispersed gas is far more probable than one that isn’t, simply because there are many more ways to achieve it. This probabilistic view of entropy is reflected in the definition of entropy from statistical thermodynamics, namely, the Boltzmann equation: S D k ln ! where S is the entropy, k is the Boltzmann constant and ! is the number of microstates available for occupancy for a given macrostate. The more states there are to occupy, the higher the entropy. The log function in the equation tells us that if there are two systems where the number of states in each case is, !1 and !2 , then the combined system will have !1 !2 states. By taking the log, the entropy of the combined system is the sum, S1 C S2 . One of the central questions in science is knowing a priori whether a given process, e.g. a chemical reaction, will (under the prevailing conditions) proceed or not. This question is answered by considering the concept of spontaneity and involves the concept of entropy and enthalpy.

10.5. REACTION SPONTANEITY

261

10.5 Reaction Spontaneity Natural processes that proceed in a definite direction without external intervention are said to undergo a spontaneous change. The phrase is not meant to convey any indication of speed, but simply that the process will eventually proceed. In an isolated system (such as the universe), the entropy increases in the course of a spontaneous change, that is:

dSuni  0

(10.3)

Equation (10.3) is a statement of the second law of thermodynamics. Any spontaneous change we observe in the universe must, according to the second law, be accompanied by an increase in the total entropy. In terms of the discussion in the last section, the universe is moving to a more probable state. This simply reflects that movement of a system is from a less probable to a more probable state. Unlike energy, entropy is not conserved. Often we will be concerned with a small part of the universe, which we call the system. During the evolution of the system it will interact with its surroundings through the exchange of energy and/or matter. Given a spontaneous change in a system, the total entropy change will be the sum of entropy changes in the system and any interaction of the system with the surroundings that causes the entropy of the surrounding to change (e.g. by the transfer of heat to the surroundings). By the second law, the total change must be positive, thus: dSsys C dSsur  0

(10.4)

To determine whether a chemical reaction will undergo change, we must therefore consider both the entropy change of the reaction in the system and its effect on the entropy of the surroundings. This joint contribution is concisely represented by another quantity called the Gibbs energy, G. To continue further we must introduce a macroscopic definition for a change

262

CHAPTER 10. BASIC THERMODYNAMICS

in entropy. A consideration of heat transfer in an idealized heat engine shows that the entropy change can be defined as:

dS D

dqrev T

(10.5)

where q is the quantity of heat transferred, and T the temperature. Strictly speaking the heat transferred must be done reversibly2 . The following argument will not however require this. Consider systems where the pressure is constant (the most common situation in the lab). This means that the heat transferred is equal to the enthalpy change, H . Since H is a state variable, it does not matter how the heat was transferred (reversible or not), therefore we can rewrite the entropy change in the surroundings as: Ssur D

Hsys T

(10.6)

If the heat is transferred from the system to the surroundings, then the equation measures the change in entropy of the surroundings. The ratio of heat to temperature may not seem to have an obvious meaning, however it can be rationalized as follows. Given that entropy measures the tendency for energy to disperse, it’s reasonable to presume that the entropy change is proportional to the heat transferred, since heat, in the form of random motion, is the source of dispersal. The temperature factor takes into account the randomness already present, thus the higher the temperature, the less effect a given amount of heat has in dispersing energy. Taking into account the definition of the entropy change (10.6), we can rewrite equation (10.4) as: dSsys

dHsys >0 T

2 A reversible process is one where a system moves between two states such that thermodynamic equilibrium is maintained at all times.

10.6. GIBBS ENERGY

263

A minus sign is introduced because we employ the convention that heat leaving the system is a loss to the system. Multiplying both sides by the temperature, T , and changing signs on both sides of the equation yields: dHsys

T dSsys < 0

The value of this expression is called the Gibbs energy:

dG D dHsys

T dSsys

(10.7)

Spontaneity is only guaranteed if dG is negative (corresponding to a positive change in equation (10.4)). It is important to stress that a negative dG does not indicate how fast a reaction occurs. A liter of gasoline has a very large negative dG but gasoline doesn’t spontaneously burst into flame. Likewise the conversion of diamond to carbon dioxide is a spontaneous processes but it is also incredibly slow and has never prevented humans from using diamonds as decoration. Interestingly, a heated diamond plunged into liquid oxygen will vaporize in seconds into carbon dioxide. In order to make a spontaneous reaction proceed faster, it is often necessary to either input an initial amount of energy to overcome the activation energy (such as a lighted match in the case of gasoline) or to add a catalyst, such as an enzyme, to lower the activation energy. Equation (10.4) also suggests another interesting concept. As long as the total entropy change remains positive, it is possible for the entropy change in the system to be negative. This means that although on balance matter and energy will disperse, it is possible for local regions of the system to develop organized structures. A less probable state can be created and maintained by allowing more probable states to emerge elsewhere. This does not contradict the second law as sometimes argued by theologians.

10.6 Gibbs Energy As noted previously, the spontaneity of a process is governed by the relative degree of entropy production in the system and entropy changes in the

264

CHAPTER 10. BASIC THERMODYNAMICS

surroundings. The balance between these two leads us to the Gibbs energy, G. If we consider a simple reversible reaction, A B, where an infinitesimal amount d  of A reacts to form B, then the change in the amount of A is d nA D d , and the change in the amount of B is d nB D d , where  is called the extent of reaction (see section (1.2) in chapter 1). Strictly speaking for a reaction with stoichiometry i , d ni D i d i or: i D

@ni @i

We now define the reaction Gibbs energy, r G to be:  r G D

@G @

 (10.8) p;T

The subscript p; T reminds us that the Gibbs energy is defined under conditions of constant pressure and temperature, conditions likely found in a biological experiment. The reaction Gibbs energy describes the rate of change of Gibbs energy with respect to reaction progress. Once the reaction reaches equilibrium, the Gibbs energy is at a minimum and the reaction Gibbs energy is zero (Figure 10.3). The negative of the reaction Gibbs energy is also called the Affinity, A. From equation (10.7) a spontaneous reaction must show a negative change in Gibbs energy. In terms of the reaction Gibbs energy this can be summarized as (Figure 10.3): If r G > 0 the reverse reaction is favorable If r G < 0 the forward reaction is favorable If r G D 0 the reaction is at equilibrium Note that the reaction Gibbs Energy is dependent on the levels of reactant and product. Since the composition of reactant and product change as the reaction proceeds, the reaction Gibbs energy will change, reaching zero when the reaction settles to equilibrium. Example 10.4

Gibbs energy, G

10.6. GIBBS ENERGY

265

¢r G < 0 ¢rG > 0

¢r G = 0 Extent of Reaction, » Figure 10.3: As a reaction proceeds (x axis) the Gibbs energy declines, reaching a minimum at equilibrium. At the same time the reaction Gibbs energy, representing the slope of the curve, reaches zero at equilibrium. Carbonic Anhydrase catalyzes the reaction between carbonic acid and water and carbon dioxide: CO2 .g/ C H2 O.l/ ! H2 CO3 .aq/: Given the reaction for a molar change is r S D 96:3 J K 1 mol 1 , and the molar enthalpy change is r H D 303:29 kJ mol 1 , we can compute the change in Gibbs energy from equation (10.7): r G D

303:29

298:15  . 0:0963/ D

274:6 kJ mol

1

Since the Gibbs energy is negative, we know the reaction will proceed in the forward direction.

Standard Gibbs Energy of Formation The standard Gibbs Energy of formation,
Gfı of a substance is the free energy change that occurs in the formation of 1 mole of substance in its

266

CHAPTER 10. BASIC THERMODYNAMICS

standard state from the constituent elements in their standard states. Like standard enthalpies of formation, the standard free energies of formation of the elements in their standard states are zero.

Standard Gibbs Energy The standard Gibbs energy for a reaction is the change in free energy that occurs when 1 mole of reactants in their standard states (concentrations of 1 Molar, a pressure of 1 atms, and a temperature of 25ı C) are converted to products in their standard states. Because free energy is a state variable it is possible to calculate standard free energies from the standard free energies of formation. In particular, for a reaction of the form: n1 A C n2 B ! m1 C C m2 D the standard free energy change can be calculated from:

    r G ı D m1
Gfı .C / C m2
Gfı .D/ n1
Gfı .A/ C n2
Gfı .B/ This is an important calculation as we will see later (See Example 10.6). Example 10.5 Determine the standard reaction Gibbs energy for the complete oxidation of sucrose by oxygen to carbon dioxide and water: C12 H12 O11 .s/ C 12O2 .g/ ! 12CO2 .g/ C 11H2 O.l/ The standard Gibbs energy of formation for sucrose, oxygen, carbon dioxide and water are given by:
Gfı .Sucrose/ D

1543 kJ mol

1


Gfı .Oxygen/ D 0 kJ mol

1


Gfı .Carbon Dioxide/ D

394:36 kJ mol

1

10.6. GIBBS ENERGY
Gfı .Water/ D

267

237:13 kJ mol

1

r G ı D .12  . 394:36/ C 11  . 237:13// . 1543 C 0/ D

5790 kJ mol

1

Equilibrium Constants and Gibbs Energy There is a close relationship between the standard Gibbs energy, r G ı of a reaction and its equilibrium constant: r G ı D

RT ln Keq

where R and T are the gas constant and absolute temperature respectively, and r G ı is the standard reaction Gibbs energy change. This is an extremely useful relationship because it is possible to estimate an equilibrium constant by computing the standard free energy from the standard Gibbs energies of formation. The standard Gibbs energies of formation can either be obtained from standard tables or can be calculated from standard entropy and enthalpy tables. Table 10.3 shows some selected standard Gibbs energies of glycolytic reactions at pH 7. These values demonstrate the distribution of the equilibrium constants between the different reactions. These values do not however indicate what the concentrations of metabolites will be in vivo. To compute the free energy change at a different set of concentrations we use the relationship: Q Products ı r G D r G C RT ln Q Reactants If the concentrations are at equilibrium, r G D 0, the relation reverts to the previous expression for r G ı . The above equation is a useful expression and allows us to compute the r G for any combination. Example 10.6 Estimate the equilibrium constant for the following isomerization reaction at 298 K and pH 7:

268

CHAPTER 10. BASIC THERMODYNAMICS 0

Biochemical Reaction

r G ı

Glucose C ATP ! Glucose-6-P C ADP Glucose-6-Phosphate ! Fructose-6-P Fructose-6-Phosphate C ATP ! FBP C ADP 3-PhosphoGlycerate ! 2-PhosphoGlycerate PEP C ADP ! Pyruvate C ATP

-24.41 3.18 -23.24 5.94 -23.24

Table 10.3: Some selected glycolytic standard Gibbs energies at pH 7, 296.15 K, and 0.25 M ionic strength. Data from Alberty [3] and Mendez [48]. FBP = Fructose 1-6 Bisphosphate; PEP = Phospho0 enolpyruvate. G ı are in units of kJ mol 1 .

Glucose-6-Phosphate (aq) ! Fructose-6-Phosphate (aq) The standard free energies of formation3 at 298 K and pH 7 for each species is known to be
Gfı .G6P/ D 439:73 kJ mol 1 and
Gfı .F6P/ D 436:55 kJ mol 1 . From the standard free energies of formation we can calculate the standard free energy for the reaction using: r G ı D

436:55

. 439:73/ D 3:18kJ mol

1

Applying the relationship between free energy and the equilibrium constant: r G ı D

RT ln Keq

where T D 298ı and R is the gas constant, 8:315  10 equilibrium constant can be computed as: Keq D e

r G ı =RT

D e 3:18=.2988:215x10

3/

3

kJ mol

1

K

1

D 0:273

That is, at equilibrium, the reaction slightly favors Glucose-6-Phosphate.

3 Values

obtained from Alberty, Thermodynamics of biochemical reactions, 2003.

the

10.6. GIBBS ENERGY

269

Chapter Highlights 1. Energy is still a mystery but one thing is known, what ever it is, it appears we can’t create nor destroy it and in any transfer of energy, it is always conserved. This property is described by the first law of thermodynamics. 2. In classical physics, energy is defined as the capacity to do work. 3. Systems come in three flavors, isolated, closed and open. Isolated systems are, as the name suggests completely isolated from the rest of the universe. A closed system is one that can exchange energy but not mass, while an open system can exchange mass and energy. 4. The enthalpy is the amount of energy within a system that is available for conversion into heat. It is the heat exchanged by a system at constant pressure. More formally, the change in enthalpy,
H D
U C p
V , that is the sum total of the change in the internal energy of the system (
U ) and the energy expended in increasing the volume. The enthalpy is a useful measure of the energy change because experiments are often done at constant atmospheric pressure. 5. The standard state is a convenient set of conditions that allows for the convenient comparison of thermodynamic properties. For example the standard enthalpy of reaction is the enthalpy change that occurs when one mole of reactants is converted to products under standard conditions. 6. The measure of dispersal in the universe is called entropy. Spontaneity is a function of both entropy and enthalpy. 7. Spontaneity refers to whether a natural process will occur or not. It does not refer to the speed of the process, merely that a spontaneous process will eventually happen. All spontaneous processes result in a net increase in the entropy of the universe. 8. The Gibbs free energy refers to the combination of changes in entropy and enthalpy that together indicate the spontaneity of a given

270

CHAPTER 10. BASIC THERMODYNAMICS

process. If the Gibbs free energy change for a process is negative then the process will eventually occur. If the Gibbs free energy change is positive than the process will not occur.

Further Reading 1. Atkins P and de Paula J (2006) Physical Chemistry for the Life Sciences. Oxford University Press, W. H. Freeman and Company, New York. ISBN: 0-7167-8628-1

Exercises Assume R D 8:315  10

3

kJ mol

1.

1. Calculate the standard reaction free energy for the dephosphorylaton of acetate phosphate at pH 7 and 298 K: Acetyl-Phosphate C ADP ! ATP C Acetate assuming that the equilibrium constant is 4.53. 2. Calculate the equilibrium constant for the phosphorylaton of glucose: glucose (aq) C Pi (aq) ! G6P (aq) where G6P is glucose-6-Phosphate and Pi is the phosphate group (H2 PO4 1 ). Assume that r G ı for the reaction is 14.0 kJ mol 1 at 37ı C. 3. Consider the reaction: L-glutamate C pyruvate ! oxo-glutarate C L-alanine At 300 K the equilibrium constant for the reaction is 1.1. If the in vivo concentrations of the reactants and products are L-glutamate D 3:0  10 2

10.6. GIBBS ENERGY

271

mM; pyruvate = 3:3  10 1 mM; oxo-glutarate = 16  10 mM, and Lalainine = 6:25mM. From this data determine whether the forward reaction will occur spontaneously or not. 4. If the r G ı at 298 K for the following reactions A ! B and B ! C are -5.6 kJ mol 1 and 1.5 kJ mol 1 respectively, compute the overall equilibrium constant for the reaction system A ! C : A!B!C

272

CHAPTER 10. BASIC THERMODYNAMICS

Appendices

273

A

List of Symbols and Abbreviations Symbols ˛i ˇ c ci
DA Ea "vS  h H k; ki


r G

Normalized substrate concentration, Si =Km Normalized inhibitor concentration, I =Ki Equilibrium ratio of relaxed and tense form in MWC model Stoichiometric coefficient Change Diffusion coefficient Activation energy Elasticity coefficient Extent of reaction Hill coefficient Enthalpy Rate constant Mass-action ratio Normalized activator concentration, A=KA Reaction free energy change

275

276

APPENDIX A. LIST OF SYMBOLS AND ABBREVIATIONS

J Ka Kd Keq KH Ki Km Ks Kp L L M mi mol ni n P PA ; i  q R  S T t t½ U
r G

Flux Association constant Dissociation constant Equilibrium constant Half-maximal activity Inhibition constant Michaelis constant Michaelis constant with respect to substrate Michaelis constant with respect to product Allosteric constant Liter Modifier Stoichiometry of product i Mole Amount of substance i or stoichiometry of reactant i Number of subunits Product concentration Permeability coefficient Normalized product concentration, Pi =Km Disequilibrium constant Heat transferred Gas constant Modifier factor Substrate concentration or entropy depending on context Temperature Time Half-Life Internal energy Reaction free energy change

277

vE vf vI vi vr V Vm w Xo Y

Extensive reaction rate Forward reaction rate Intensive reaction rate i th reaction rate Reverse reaction rate Volume Maximal velocity Work done Reference state for variable X Fractional saturation

Math Symbols P

Summation Symbol:

4 X

xi D x1 C x2 C x3 C x4

iD1

Q

Product Symbol:

4 Y

xi D x1 x2 x3 x4

i D1

dy=dx

Derivative of y with respect to x

@y=@x

Partial derivative of y with respect to x

278

APPENDIX A. LIST OF SYMBOLS AND ABBREVIATIONS

Non-Mathematical Abbreviations AMP, ADP, ATP ATCase cAMP CI Cro CRP CTP DNA F6P KNF IPTG LacI mRNA MWC PEP PFK RBS RNA TF

Adenine nucleotides Aspartate transcarbamylase Cyclic AMP Lambda phage repressor protein Lambda phage repressor protein Catabolite activator protein Cytidine triphosphate Deoxyribonucleic acid Fructose-6-Phosphate Koshland, Nemethy and Filmer model Isopropyl ˇ-D-1-thiogalactopyranoside Lactose Operon Repressor Messenger RNA Monod, Wyman and Changeux model Phosphoenolpyruvate Phosphofructokinase Ribosome binding site Ribonucleic acid Transcription factor

B

List of Common Rate Laws Abbreviations are given at the end of this appendix.

Mass-Action Rate Laws 2. Irreversible Mass-action (1.8) vDk

Y

Sini

i

3. Reversible Mass-action (1.10) v D k1

Y

Sini

k2

Y

Pimi

4. Modified Reversible Mass-action (1.19) v D k1

Y

Sini

 1

 Keq

279

280

APPENDIX B. LIST OF COMMON RATE LAWS

Non-Mechanistic Approximations 5. Single Substrate Linear Model 8.1 v D vo C

@v .S @So

So /

6. Linear Model Using Elasticities (8.5)

v D vo

1C

X i

ıSi "vS o o i S i

!

7. Power Law (8.6) vDk

Y

"i

Sj j

j

8. Lin-Log Model (8.7)  v D vo

e eo

 1C

X

"vSi

i



Si ln Sio

!

Single Substrate Michaelis-Menten 10. Briggs-Haldane (3.5) vD

Vf ˛ 1C˛

11. Reversible Michaelis-Menten (3.19) vD

Vf ˛ Vr  1C˛C

281

12. Reversible Michaelis-Menten with Haldane Substitution (3.23) Vf ˛.1 / 1C˛C

vD

13. Michaelis-Menten with Product Inhibition (4.5) vD

Vf ˛ 1C˛C

14. Membrane Transport Carrier Model: Iso Uni-Uni vD

Vf ˛.1 / 1 C ˛ C  C ˛P =Ki i

Inhibition Rate Laws 16. Irreversible Competitive Inhibition (4.3) vD

Vf ˛ 1C˛C

17. Reversible Competitive Inhibition (4.4) vD

Vf ˛ Vr  1C˛C C

18. Irreversible Uncompetitive Inhibition (4.9) vD

Vf ˛ 1 C ˛ .1 C /

19. Reversible Uncompetitive Inhibition (4.10) vD

Vf ˛ Vr  1 C .˛ C /.1 C /

282

APPENDIX B. LIST OF COMMON RATE LAWS

20. Irreversible Noncompetitive Inhibition (4.13) vD

Vf ˛ .1 C ˛/ .1 C /

21. Reversible Noncompetitive Inhibition (4.14) vD

Vf ˛ Vr  .1 C ˛ C / .1 C /

22. Irreversible Mixed Inhibition (4.11) vD

Vf ˛ .1 C / C ˛.1 C f /

23. Reversible Mixed Inhibition vD

Vf ˛ Vr  .1 C / C .˛ C /.1 C f /

24. Irreversible Mixed Partial Inhibition (4.1)   I S 1Cb Km aKi vD I SI S 1C C C Km Ki aKi Km Vm

Cooperative and Allosteric Rate Laws 16. Hill Equation using Dissociation constant, Kd (6.3) vD

Vf S h Kd C S h

283

17. Hill Equation using Half Maximal activity, Ks (6.4) vD

Vf S h Ksh C S h

18. Adair Equation for a Dimer (6.13) v D Vf

K1 S C 2K1 K2 S 2 2.1 C K1 S C K1 K2 S 2 /

19. Adair Equation for a Trimer v D Vf

K1 S C 2K1 K2 S 2 C 3K1 K2 K3 S 3 3.1 C K1 S C K1 K2 S 2 C K1 K2 K3 S 3 /

20. MWC Cooperative Model (6.14) ˛ .1 C ˛/n 1 v D Vf .1 C ˛/n C L 21. Allosteric MWC Model with Inhibitor,  (7.2) v D Vf

˛ .1 C ˛/n 1 .1 C ˛/n C L .1 C /n

22. Allosteric MWC Model with Activator,  (7.3) v D Vf

˛ .1 C ˛/n .1 C ˛/n C L

1

1 .1 C /n

23. Simplified Reversible MWC Model (6.20) vD

Vf ˛ .1 / .1 C ˛ C /n L C .1 C ˛ C /n

1

284

APPENDIX B. LIST OF COMMON RATE LAWS

24. Simplified Reversible MWC Model with inhibitor  vD

Vf ˛ .1 / .1 C ˛ C /n 1 L.1 C /n C .1 C ˛ C /n

25. Reversible Hill Equation (6.6) Vf ˛ .1

vD

/ .˛ C /h

1

1 C .˛ C /h

26. Reversible Hill with One Modifier,  (7.7) Vf ˛ .1

vD

/ .˛ C /h

1

1 C h C .˛ C /h 1 C h

 acts as an inhibitor when  < 1 and an activator when  > 1. 27. Reversible Hill with Two Modifiers, a and b (7.8) vD

Vf ˛ .1

D D .˛ C /h C

/ .˛ C /h 1 D ! ! 1 C h1 1 C h2 1 C 1 h1

1 C 2 h2

1 and 2 represent the factors that determine whether 1 and 2 are activators or inhibitors (See above). This equation assumes that both modifiers bind independently.

Two Substrate Rate Laws 25. Random Order Bi-Uni Rate Law (8.10) vD

Vf ˛1 ˛2 .1 / 1 C ˛1 C ˛2 C ˛1 ˛2 C 1

285

26. Random Order Uni-Bi Rate Law (8.11) vD

Vf ˛ .1 / 1 C ˛ C 1 C 1 2 C 2

27. Alberty Two Substrate Irreversible (5.3) vD

Vf S1 S2 KS 2 S1 C KS1 S2 C S1 S2 C KiS1 KS 2

28. Random Order Bi-Bi Rate Law (8.9) vD

Vf ˛1 ˛2 .1 / .1 C ˛1 C 2 / .1 C ˛2 C 1 /

29. Ordered Bi-Uni Rate Law vD

Vf ˛1 ˛2 .1 / 1 C ˛1 C ˛2 C ˛1 ˛2 C ˛1 1 C 1

30. Ordered Bi-Bi Rate Law (5.1) vD

Vf AB .1 D

/

where D D Ki a Kb C Kb A C Ka B C AB C C

Kb Ki a PQ Kb Kq AP C Kp Ki q Kp Ki q

Kb Ki a BPQ ABP Ka BQ Kq Kb Ki a P Kb Ki a Q C C C C Kp Ki q Ki b Kip Ki q Kp Ki q Ki q (B.1)

Generalized Rate Laws

286

APPENDIX B. LIST OF COMMON RATE LAWS

1. Bi-Bi Enzyme with Cooperativity (8.13) 

 Vf ˛1 ˛2 1

.˛1 C 1 /h 1 .˛2 C 2 /h Keq

1 C .˛1 C 1 /h 1 C .˛2 C 2 /h

vD

2. Uni-Bi Enzyme with Cooperativity (8.15)

vD

 Vf ˛ 1

 Keq

.˛ C /h

1

1 C .˛ C /h C .˛ C /h C .˛ C /h

2˛ h

3. Bi-Uni Enzyme with Cooperativity (8.16)

vD

 Vf ˛ˇ 1

 Keq

.˛ˇ C /h

1

1 C .˛ C /h C .ˇ C /h C .˛ˇ C /h

2 h

4. Common Modular (8.19)   ˛ini 1 Keq v D Vf Q Q hni C i .1 C i /hni i .1 C ˛i / Q

5. Direct Binding (8.20)   ˛ini 1 Keq v D Vf Q Q 1 C ˛ihni C ihni Q

Gene Expression 30. Gene Expression Activation (9.16) vD

Vf S h Ks C S h

1

1

287

31. Gene Expression Inhibition (9.17) vD

Vf Ks C S h

In all the following examples, cooperativity can be included by raising the species with an appropriate Hill coefficient. 32. Gene Expression AND Function (9.8) v D Vf

K1 K2 A
B 1 C K1 A C K2 B C K1 K2 A
B

33. Gene Expression AND Function with Cooperativity (h D 4) v D Vf

K1 K2 A4
B 4 1 C K1 A4 C K2 B 4 C K1 K2 A4
B 4

34. Gene Expression OR Function (9.6) v D Vf

K1 A C K2 B 1 C K1 A C K2 B

35. Gene Expression NOR Function (9.9) v D Vf

1 1 C K1 A C K2 B C K3 A
B

36. Gene Expression NAND Function (9.10) v D Vf

1 C K1 A C K2 B 1 C K1 A C K2 B C K3 A
B

37. Gene Expression XOR Function (9.11) v D Vf

K1 A C K2 B 1 C K1 A C K2 B C K3 A
B

288

APPENDIX B. LIST OF COMMON RATE LAWS

38. Gene Expression EQ Function (9.12) v D Vf

1 C K1 A
B 1 C K1 A C K2 B C K3 A
B

39. Counter Regulation (Non-Competitive Model (9.13)) v D Vf

K1 A 1 C K1 A C K2 R C K3 A
R

where A is the activator and R the repressor. 40. Counter Regulation (Competitive Model (9.14)) v D Vf

K1 A 1 C K1 A C K2 R C K3 A
R

where A is the activator and R the repressor.

289

Abbreviations for Appendix B: Si Pi I A M Mx ˛ Vf Vr Keq ki KS KP KX Kd Ki ˛      "vS ni and mi h n L vo Sio eo

Substrate i concentration Product i concentration Inhibitor concentration Activator concentration Modifier concentration Modifier x concentration Effect of M , if M < 1: Inhibitor; M > 1: Activator Forward Vmax Reverse Vmax Equilibrium constant Mass-action ratio Rate constant Concentration of substrate at half-maximal velocity Concentration of product at half-maximal velocity Concentration of modifier at half-maximal velocity Dissociation constant Microscopic association constant Abbreviation for S=KS Abbreviation for P =KP Abbreviation for I =KI Abbreviation for M=KM Abbreviation for P =S Abbreviation for =Keq Modifier action in reversible Hill equation Elasticity of rate, v, with respect to species S Stoichiometric amounts Hill coefficient Number of binding sites Allosteric constant Reference reaction rate for approximate rate laws Reference concentration for approximate rate laws Reference enzyme activity for approximate rate laws

290

APPENDIX B. LIST OF COMMON RATE LAWS

C

Answers to Questions Chapter 1 1 a) 1:1; 1,1:1, 1:1,1; 2:1; 3,4:2,1; 1,1:1,1; 1,2:3,1 1 b) 1A; 1B; 1A; 1B; 1C ; 1A; 1B; 1C ; 2A; 1B, 3A; 4B; 2C; 1D, 1B; 1C , 1A; 1B; 1C 2. Mass of Butyric acid: 24:44 gms. Mass of CO2: 24:44 gms. Mass of hydrogen: 1:111. Yield: 48.89% 1

3. 0:088 mol L

min

1.

4. a) kA, kAB, kA, kA2 5. b) kA3 B 4

kC 2 D, kAB

kAC , kAB 2

kB 3 C

6. k D 0:0154 7. a) dA=dt D b) dA=dt D 8. dB=dt D

3:5I dB=dt D

14I dC =dt D 10:5

3:5I dB=dt D 7I dC =dt D 3:5

10I dC =dt D 5I v D 5

9. The calculation begins by constructing an equilibrium table:

291

292

APPENDIX C. ANSWERS TO QUESTIONS Species

G6P

Initial concentration Equilibrium concentration

4:9 4:9

F6P x

0 x

From the equilibrium constant and the above table the following is derived: Keq D

x 4:9 x

Solving for x and hence the equilibrium concentration, yields: xD

Keq 4:9 1 C Keq

Therefore the equilibrium concentration of Glucose-6-Phosphate is 1.387 mM and for Fructose-6-Phosphate, 3.514 mM. 10. Set v D k1 A3 B 4 equals Keq .

k2 C 2 D D 0, solve for k1 =k2 , right-hand side

11. Glucose in outer compartment = 0:075 mM. 12. Set the reaction rate to 0, since the catalyst is a multiplier term, the expression k1 A k2 B D 0 must still be true since Ei ¤ 0, hence Keq is unaffected. Chapter 2 1. See section 2.1 2. See definition in section 2.2 3. -0.5 means that a) A change in the species concentration, X , causes the reaction rate to decrease (neg sign); b) A 1% increase in the concentration of molecule X , results in a 0.5% decrease in the reaction rate. v 4. "A D 3kA2

5. a)

2x 2 ; 1Cx 2

b)

1C2x 1Cx ;

c)

1 x2 1Cx 2

6. See example 2.2 7. It measures the degree from equilibrium, e.g. if it is close to 1 the reaction is close to equilibrium. If it is close to zero the reaction is far from equilibrium.

293

8. Tends to one. 9. Both elasticity tend to infinity and -infinity respectively. 10. See relevant section related to 2.5. 11. The absolute value of the substrate elasticity is greater than the absolute value of the product elasticity. 12. Elasticity should equal:

nKm Km CS n

by using rules 7, 3, 1, and 4.

Chapter 3 5. See main text, Km =.S C Km /. 6. 1=Km 7. Km D 0:8mM and Vm D 4:5mM s

1

Chapter 7 1. -3.74 kJ mol 1 2. K D 4:4  10

3

3.
r G D 22:76kJ mol 1 . Since the free energy change is positive; the spontaneous direction of the reaction is right to left. 4. Overall Keq is computed from some of individual reaction free energies. Keq D 5:23.

294

APPENDIX C. ANSWERS TO QUESTIONS

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History 1. VERSION: 1.0 Date: 2011-04-2 Author(s): Herbert M. Sauro Title: Enzyme Kinetics for Systems Biology Modification(s): First Edition Release 2. VERSION: 1.01 Date: 2011-04-19 Author(s): Herbert M. Sauro Title: Enzyme Kinetics for Systems Biology Modification(s): Minor typographical changes.

301

302

REFERENCES

INDEX

303

Index Symbols Ka . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Kd . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Km . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Vmax . . . . . . . . . . . . . . . . . . . . . . . . . .61 ˇ-galactosidase . . . . . . . . . . . . . . 233  factors. . . . . . . . . . . . . . . . . . . . .205 h . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 81 fold . . . . . . . . . . . . . . . . . . . . . . 120

A absolute metabolite conc . . . . . . . 73 activation energy . . . . . . . . . . . . . . 27 activators . . . . . . . . . . . . . . . . . . . . 102 Adair equation . . . . . . . . . . . . . . . 135 adenine nucleotides . . . . . . . . . . . . . 2 aggregate rates laws . . . . . . . . . . . 55 Alberty . . . . . . . . . . . . . . . . . . . . . . 116 algebraic approximation . . . . . . . 189 allolactose . . . . . . . . . . . . . . 205, 233 allosteric constant . . . . . . . . . . . . 141 allosteric regulation . . . . . . . . . . . 121 allosteric regulators . . . . . . . . . . . 197 anaerobic growth . . . . . . . . . . . . . . . 7 AND gate . . . . . . . . . . . . . . . . . . . 220 apparent constants . . . . . . . . . . . . 130 apparent kinetic order . . . . . . . . . . 37 approximations . . . . . . . . . . . . . . . . 66 area . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Arrhenius equation . . . . . . . . . . . . 27

art of building models . . . . . . . . . 200 association constant . . . . . . . . . . . . 18 attenuated . . . . . . . . . . . . . . . . . . . 120

B binomial coefficients . . . . . . . . . . 134 biochemical systems theory 37, 186 Boltzmann distribution . . . . . . . . 240 Boltzmann equation . . . . . . . . . . 260 bond energy . . . . . . . . . . . . . . . . . 254 Boolean network . . . . . . . . . . . . . 220 Botts-Morales scheme . . . . . . . . . 83 boundary . . . . . . . . . . . . . . . . . . . . 250 Briggs and Haldane . . . . . . . . . . . . 60 Brown and Henri . . . . . . . . . . . . . . 54 buffering effects . . . . . . . . . . . . . . . 67

C calorie . . . . . . . . . . . . . . . . . . . . . . 253 cAMP receptor protein . . . . . . . . 205 CAP . . . . . . . . . . . . . . . . . . . . . . . . 205 capacity to do work . . . . . . . . . . . 251 carbon dioxide . . . . . . . . . . . . . . . 121 catabolite activator protein . . . . 205 catalysis . . . . . . . . . . . . . . . . . . . . . . 54 catalyst . . . . . . . . . . . . . . . . . . . . . . . 29 choosing a suitable rate law . . . 200 CI . . . . . . . . . . . . . . . . . . . . . . 210, 231 CI binding . . . . . . . . . . . . . . . . . . . 241 classical thermodynamics . . . . . 250

304

INDEX

closed system . . . . . . . . . . . . . 16, 250 CM . . . . . . . . . . . . . . . . . . . . . . . . . 195 collision theory . . . . . . . . . . . . . . . . 28 combinatory mathematics . . . . . 132 common modular . . . . . . . . . . . . . 195 compartmental analysis . . . . . . . . 25 competitive . . . . . . . . . . . . . . . . . . . 82 competitive binding . . . . . . . . . . . 215 competitive inhibition . . . . . . . . . . 85 competitive inhibitors . . . . . . . . . . 59 compulsory-order . . . . . . . . . . . . 112 computer exchange format . . . . . . . 6 concerted model . . . . . . . . . . . . . . 140 conformation change . . . . . . . . . 121 conservation laws . . . . . . . . . . . . . . 67 conservation of energy . . . . . . . . 252 control . . . . . . . . . . . . . . . . . . . . . . . 23 cooperativity . . . . . . . . 119, 190, 231 Cornish-Bowden . . . . . . . . . . . . . 126 Cro . . . . . . . . . . . . . . . . . . . . . . . . . 231 cross-sectional area . . . . . . . . . . . . 26 CRP . . . . . . . . . . . . . . . . . . . . . . . . 205

D

DM . . . . . . . . . . . . . . . . . . . . . . . . . 196 double-displacement . . . . . . . . . . 112

E E. coli . . . . . . . . . . . . . . . . . . . . . . . 205 Eadie-Hofstee . . . . . . . . . . . . . . . . 103 EcoCyc . . . . . . . . . . . . . . . . . . . . . 205 elasticity coefficient. . . . . . . . . . . .37 elasticity rules . . . . . . . . . . . . . . . . . 47 electrophoretic gel shift . . . . . . . 246 elementary reactions . . . . . . . . . . . 11 EMSA . . . . . . . . . . . . . . . . . . . . . . 246 endothermic . . . . . . . . . . . . . . . . . 256 energy . . . . . . . . . . . . . . . . . . . . . . . 250 enthalpy change . . . . . . . . . . . . . . 254 entropy . . . . . . . . . . . . . . . . . . . . . . 258 enzyme action . . . . . . . . . . . . . . . . . 54 enzyme elasticity . . . . . . . . . . . . . . 66 enzyme inhibition . . . . . . . . . . . . . 82 enzyme inhibitor complex . . . . . . 85 enzyme kinetics . . . . . . . . . . . . . . . 54 enzyme substrate complex . . . . . . 58 enzyme-reactant complex . . . . . . 54 enzymes . . . . . . . . . . . . . . . . . . . . . . 29 EQ gate . . . . . . . . . . . . . . . . . . . . . 226 equilibrium constant . . . . . . . . . . . 16 equilibrium point . . . . . . . . . . . . . . 29 equivalence . . . . . . . . . . . . . . . . . . 226 exclusive model . . . . . . . . . . . . . . 140 exclusive OR . . . . . . . . . . . . . . . . 224 exothermic . . . . . . . . . . . . . . . . . . 256 extensive property . . . . . . . . . . . . . . 9 extent of reaction . . . . . . . . . . . . . . . 9

denatures . . . . . . . . . . . . . . . . . . . . . 27 denominator . . . . . . . . . . . . . . . . . . 59 detailed balance . . . . . . . . . . . 69, 219 difference quotient . . . . . . . . . . . . . 39 differential equation . . . . . . . . . . . 12 diffusion . . . . . . . . . . . . . . . . . . . . . . 63 diffusion coefficient . . . . . . . . . . . . 26 diffusion limit . . . . . . . . . . . . . . . . . 64 direct binding . . . . . . . . . . . . . . . . 196 disequilibrium ratio . . . . . . . . 21, 45 F dissociation constant . . . . . . . . . . . 17 far from equilibrium . . . . . . . . . . . 74 distribution of enzyme states . . . . 59 fermentation . . . . . . . . . . . . . . . . . . . 7

INDEX

305

Fick’s first law . . . . . . . . . . . . . . . . 26 First law . . . . . . . . . . . . . . . . . . . . . 252 first-order . . . . . . . . . . . . . . . . . . . . . 12 flux . . . . . . . . . . . . . . . . . . . . . . . . . . 25 fold-change . . . . . . . . . . . . . . . . . . 120 force . . . . . . . . . . . . . . . . . . . . . . . . 250 fractional amounts . . . . . . . . . . . . . . 7 fractional occupancy . . . . . . . . . . 210 fractional saturation . . . . 60, 69, 135 free energy . . . . . . . . . . . . . . . 22, 218 fructose-1-6-bisphosphate . . . . . 122 fructose-6-phosphate . . . . . . . . . 122 functional parts . . . . . . . . . . . . . . . . 24

G gal operon . . . . . . . . . . . . . . . . . . . 208 gel retardation assay . . . . . . . . . . 246 gel shift . . . . . . . . . . . . . . . . . . . . . 245 gene regulation . . . . . . . . . . . . . . . 204 gene regulatory motif . . . . . . . . . 207 generalized inhibition model . . . . 83 generalized Michaelis-Menten . 189 generalized rate laws . . . . . 112, 182 generalizing MWC . . . . . . . . . . . 147 Gibbs energy . . . . . . . . . . . . . . . . . 263 glycolysis . . . . . . . . . . . . . . . . . . . 189 gradient . . . . . . . . . . . . . . . . . . . . . . 26

Hess’s Law . . . . . . . . . . . . . . . . . . 257 heterotropic effector . . . . . . . . . . 123 high levels of enzyme . . . . . . . . . . 67 high sensitivity . . . . . . . . . . . . . . . 120 Hill coefficient . . . . . . . . . . 125, 237 Hill equation . . . . . . . . . . . . . . . . . 124 Hill, elasticity . . . . . . . . . . . . . . . . 129 Hofmeyr . . . . . . . . . . . . . . . . 126, 190 homotropic effector . . . . . . . . . . . 123 hyperbolic . . . . . . . . . . . . . . . . . . . . 62

I inhibitors . . . . . . . . . . . . . . . . . . . . . 82 initial rate . . . . . . . . . . . . . . . . . 62, 63 intensive property . . . . . . . . . . . . . 10 internal energy . . . . . . . . . . . . . . . 252 irreversible inhibitor . . . . . . . . . . . 82 irreversible inhibitors . . . . . . . . . 101 isolated . . . . . . . . . . . . . . . . . . . . . . 250

J Joule . . . . . . . . . . . . . . . . . . . . . . . . 253

K

kinetic . . . . . . . . . . . . . . . . . . . . . . 252 kinetic mechanism . . . . . . . . . . . . . 54 kinetic order . . . . . . . . . . . . . . . . . . 36 Klipp . . . . . . . . . . . . . . . . . . . . . . . 190 KNF sequential model . . . . . . . . 150 H Haldane relations . . . . . . . . . . . . . . 69 Koshland, Nemethy and Filmer 150 half saturation . . . . . . . . . . . . . . . 213 L half-life . . . . . . . . . . . . . . . . . . . . . . 13 lac operon . . . . . . . . . . . . . . . 209, 233 half-maximal activity . . . . . . . . . 125 half-saturation constant . . . . . . . 191 LacI . . . . . . . . . . . . . . . . . . . . 206, 208 half-saturation level . . . . . . . . . . . 246 LacI model . . . . . . . . . . . . . . . . . . 233 Hanekom rate laws . . . . . . . . . . . 189 LacI repressor . . . . . . . . . . . . . . . . 233 hemoglobin . . . . . . . . . . . . . . . . . . 121 lacZ . . . . . . . . . . . . . . . . . . . . . . . . . 209

306

INDEX

lambda phage . . . . . . . . . . . 210, 237 law of mass-action . . . . . . . . . . . . . 12 Liebermeister . . . . . . . . . . . . . . . . 190 Liebermeister rate laws . . . . . . . 194 ligand binding . . . . . . . . . . . . . . . 129 linear approximation . . . . . . . . . . 182 linear-logarithmic rate laws . . . . 186 Lineweaver-Burk . . . . . . . . . . . . . 103 linlog . . . . . . . . . . . . . . . . . . . . . . . 187 log form . . . . . . . . . . . . . . . . . . . . . . 43 logarithmic scales . . . . . . . . . . . . . 44

M macroscopic constants . . . . . . . . 130 mass-action . . . . . . . . . . . . . . . . . . . 12 mass-action kinetics . . . . . . . . 12, 44 mass-action ratio . . . . . . . . . . . . . . 21 Mathematica . . . . . . . . . . . . . . . . . . 50 maturation . . . . . . . . . . . . . . . . . . . 237 maximum yields . . . . . . . . . . . . . . . . 7 membrane . . . . . . . . . . . . . . . . . . . . 26 membrane transport . . . . . . . . . . . . 25 metabolic pathway . . . . . . . . . . . . . 67 Michaelis and Menten . . . . . . . . . 54 Michaelis constant . . . . . . . . . . . . . 61 Michaelis-Menten equation . . . . . 58 Michaelis-Menten kinetics. . . . . .55 microscopic constants . . . . . . . . . 129 microstate . . . . . . . . . . . . . . . . . . . 259 mixed and partial inhibition . . . 101 models . . . . . . . . . . . . . . . . . . . . . . 200 modular rate law . . . . . . . . . . . . . 190 modularized . . . . . . . . . . . . . . . . . 190 Monod, Wyman and Changeux 140 motifs . . . . . . . . . . . . . . . . . . . . . . . 207 mRNA . . . . . . . . . . . . . . . . . . . . . . 204

multimers . . . . . . . . . . . . . . . . . . . 231 multiple reactant enzymes . . . . . 112 multiple TFs . . . . . . . . . . . . . . . . . 214 MWC model . . . . . . . . . . . . 140, 147

N NAND gate . . . . . . . . . . . . . . . . . . 223 near to equilibrium . . . . . . . . . . . . 73 negative cooperativity . . . . 121, 150 Newton’s difference quotient . . . 39 non-allosteric regulators . . . . . . . 198 non-competitive binding . . . . . . 217 non-competitive inhibition . . . . . . 98 non-elementary rate law . . . . . . . . 15 non-exclusive model . . . . . . . . . . 147 non-linear . . . . . . . . . . . . . . . . . . . 184 NOR gate . . . . . . . . . . . . . . . . . . . . 222

O oligomeric proteins . . . . . . . . . . . 121 oligomerization . . . . . . . . . . . . . . 237 open system . . . . . . . . . . . . . . . . . 250 operating point . . . . . . . . . . . . . . . . 40 operational definition . . . . . . . . . . 37 operator site . . . . . . . . . . . . . . . . . 204 operator state . . . . . . . . . . . . . . . . 206 operon . . . . . . . . . . . . . . . . . . . . . . 204 OR gate . . . . . . . . . . . . . . . . . . . . . 217 ordered mechanism . . . . . . . . . . . 114 oxygen . . . . . . . . . . . . . . . . . . . . . . 121

P partition function . . . . . . . . . . . . . 240 path independent . . . . . . . . . . . . . 252 permeability coefficient . . . . . . . . 26 permease . . . . . . . . . . . . . . . . . . . . 233 PFK. . . . . . . . . . . . . . . . . . . . . . . . .122

INDEX

307

phosphofructokinase . . . . . . . . . . 122 physiological effect . . . . . . . . . . . 121 ping-pong mechanism . . . . . . . . 117 positive cooperativity . . . . . . . . . 121 potential . . . . . . . . . . . . . . . . . . . . . 252 power law approximation . . . . . 186 pre-exponential factor . . . . . . . . . . 27 principle of detailed balance . . . . 19 product inhibition . . . . . . . . . . 67, 88 progress curve . . . . . . . . . . . . . . . . . 13 promoter . . . . . . . . . . . . . . . . . . . . 204 promoter obstruction . . . . . . . . . . 208 propagation . . . . . . . . . . . . . . . . . . . 46 proportional change . . . . . . . . . . . . 43 protruded . . . . . . . . . . . . . . . . . . . . 234 pure competitive inhibitor . . . . . . 85 pure non-competitive inhibition . 98

R R state . . . . . . . . . . . . . . . . . . . . . . 140 random-order . . . . . . . . . . . . . . . . 112 rapid equilibrium . . . . . . . . . . . . . . 56 rapid equilibrium assumption . . . 58 rate constant . . . . . . . . . . . . . . . 12, 17 rate of change . . . . . . . . . . . . . . . . . . 4 rate of reaction . . . . . . . . . . . . . . . . . 9 rational fractions . . . . . . . . . . . . . . . 7 RBS . . . . . . . . . . . . . . . . . . . . . . . . 209 reaction kinetics . . . . . . . . . . . . . . . . 1 reaction order . . . . . . . . . . . . . . . . . 36 reaction rate . . . . . . . . . . . . . . . . . . . . 8 reaction rate constant . . . . . . . . . . 27 reaction spontaneity . . . . . . . . . . 261 reactions at equilibrium . . . . . . . 189 reality . . . . . . . . . . . . . . . . . . . . . . . 200 rectangular hyperbola . . . . . . . . . . 62

reference state . . . . . . . . . . . . . . . 188 regulation . . . . . . . . . . . . . . . . . . . 121 RegulonDB . . . . . . . . . . . . . . . . . . 205 relaxed . . . . . . . . . . . . . . . . . . . . . . 140 replica . . . . . . . . . . . . . . . . . . . . . . 200 reversibility . . . . . . . . . . . . . . . . . . . 55 reversible . . . . . . . . . . . . . . . . . . . . . 14 reversible Hill equation . . . . . . . 151 reversible inhibitor . . . . . . . . . . . . . 82 reversible Michaelis-Menten . . . . 68 reversible rate laws . . . . . . . . . . . . 67 reversibly change . . . . . . . . . . . . . 262 ribosome binding site . . . . . . . . . 209 RNA polymerase . . . . . . . . . . . . . 204 Rohwer . . . . . . . . . . . . . . . . . . . . . 190

S saturation term . . . . . . . . . . . . . . . 190 saturation terms . . . . . . . . . . . . . . . 70 Savageau . . . . . . . . . . . . . . . . . . . . 186 SBML . . . . . . . . . . . . . . . . . . . . . . . . . 6 Second law . . . . . . . . . . . . . . . . . . 258 second-order . . . . . . . . . . . . . . . . . . 12 sensitivity analysis . . . . . . . . . . . 200 sequestration effects . . . . . . . . . . . 67 sigmoid response . . . . . . . . . . . . . 121 spontaneous changes . . . . . . . . . . 261 standard changes . . . . . . . . . . . . . 256 standard conditions . . . . . . . . . . . 256 standard enthalpies . . . . . . . . . . . 257 statistical approach . . . . . . . . . . . 239 statistical effect . . . . . . . . . . . . . . 131 statistical thermodynamics . . . . 250 steady state . . . . . . . . . . . . 56, 57, 60 steady state assumption . . . . . 57, 60 stoichiometric amount . . . . . . . . . . . 3

308

INDEX

stoichiometric coefficient . . . . . . . . 5 subelasticity terms . . . . . . . . . . . . . 49 substrate . . . . . . . . . . . . . . . . . . . . . . 54 substrate elasticity. . . . . . . . . . . . . . 64 sum of elasticities . . . . . . . . . . . . . 72 sum of the elasticities . . . . . . . . . . 45 surroundings . . . . . . . . . . . . . . . . . 250 symmetry model . . . . . . . . . . . . . 140 system . . . . . . . . . . . . . . . . . . . . . . 250

T T state . . . . . . . . . . . . . . . . . . . . . . . 140 Taylor series . . . . . . . . . . . . . . . . . 183 tense . . . . . . . . . . . . . . . . . . . . . . . . 140 tetramer . . . . . . . . . . . . . . . . . . . . . 132 TF . . . . . . . . . . . . . . . . . . . . . . . . . . 204 theoretical yield . . . . . . . . . . . . . . . . 7 thermodynamic approach. . . . . .239 thermodynamic constraints . . . . 189 thermodynamic data . . . . . . . . . . . 73 thermodynamic equilibrium . 16, 56 thermodynamic properties . . . . . 188 thermodynamic term . . . . . . . . . . 190 thermodynamics . . . . . . . . . . . . . . 250 three-point estimation . . . . . . . . . . 39 three-point formula . . . . . . . . . . . . 39 transcription factor . . . . . . . . . . . 204 transcriptional model . . . . . . . . . 211 transcriptional repression . . . . . . 214

U uncompetitive inhibition . . . . . . . 93 underlying assumptions . . . . 66, 199 units of energy . . . . . . . . . . . . . . . 253

V volume . . . . . . . . . . . . . . . . . . . . . . . 10

W weighting factors . . . . . . . . . . . . . 134 work . . . . . . . . . . . . . . . . . . . . . . . . 250

X XOR gate . . . . . . . . . . . . . . . . . . . . 224

Y yields. . . . . . . . . . . . . . . . . . . . . . . . . . 7

Z zero-order . . . . . . . . . . . . . . . . . 12, 65