Thermodynamics and kinetics for the biological sciences

THERMODYNAMICS AND KINETICSFORTHE BIOLOGICALSCIENCES GordonG. Hammes Department of Biochemistry Duke University

GwLEYeC2rNTERscrENcE A JOHNWILEY& SONS,INC.,PUBLICATION NewYork. chichester. weinheim. Brisbane. singapore.Toronto

The book is printed on acid-free pup"r. @ Copyright @ 2000 by John Wiley & Sons, Inc. All rights reserved. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 or the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright ClearanceCenter, 222 Rosewood Drive, Danvers,MA 01923, (978) 750-8400,fax (978) 7504744. Requeststo the Publisher for permission should be addressedto the Permissions Department, John Wiley & Sons,Inc., 605 Third Avenue, New York, NY 10158-0012,(212) 850-6011,fax (212) 8506008, E-Mail: PERMREQ @WILEY.COM. For ordering and customer service. call 1-800-CALL-WILEY. Library of Congress Catalaging-in-Publication

Data:

Hammes. Gordon.. 1934Thermodynamics and kinetics for the biological sciences/by Gordon G. Hammes. p. cm. "Published simultaneously in Canada." Includes bibliographical references and index. ISBN 0-471-31491-l (pbk.: acid-freepaper) 1. Physical biochemistry.2. Thermodynamics.3. Chemical kinetics.I. Title. QP517.P49H35 2000 512-dc2l Printed in the United States of America. 10 9

816

5

99-086233

CONTENTS

I

Preface 1. Heat, Work, and Energy 1.1 Introduction 1.2 Temperature 1.3 Heat 1.4 Work 1.5 Definition of Energy 1.6 Enthalpy 1.7 StandardStates 1.8 Calorimetry 1.9 ReactionEnthalpies 1.10 TemperatureDependenceof the ReactionEnthalpy References Problems 2. Entropy and Free Energy 2.t 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10

Introduction Statementof the SecondLaw Calculation of the Entropy Third Law of Thermodynamics Molecular Interpretation of Entropy Free Energy ChemicalEquilibria Pressureand Temperature Dependenceof the Free Energy PhaseChanges Additions to the Free Energy Problems

3. Applications of Thermodynamics to Biological Systems 3.1 3.2 3.3 3.4

Biochemical Reactions Metabolic Cycles Direct Synthesisof ATP Establishment of Membrane Ion Gradients bv Chemical Reactions

tx

1 1 2 3 4 7 9 l0 11 15 17 18 18 2l 2L 22 24 26 27 28 30 33 35 31 37 4t 41 42 47 48 V

vt

CONTENTS

3.5 Protein Structure 3.6 Protein Folding 3.7 Nucleic Acid Structures 3.8 DNA Meltins 3.9 RNA References Problems

4. Chemical Kinetics

5.

50 56 59 62 66 68 68 7l

4.1 Introduction 4.2 ReactionRates 4.3 Determination of Rate Laws 4.4 RadioactiveDecay 4.5 ReactionMechanisms 4.6 TemperatureDependenceof Rate Constants 4.7 Relationship Between Thermodynamics and Kinetics 4.8 Reaction RatesNear Equilibrium References Problems

7T 73 75 78 79 83 86 88

Applications of Kinetics to Biological Systems

94

5.1 Introduction 5.2 Enzyme Catalysis:The Michaelis-Menten Mechanism 5.3 u-Chymotrypsin 5.4 Protein Tyrosine Phosphatase 5.5 Ribozymes 5.6 DNA Meltine and Renaturation References Problems

6. Ligand Binding to Macromolecules 6.1 Introduction 6.2 Binding of Small Molecules to Multiple Identical Binding Sites 6.3 Macroscopicand Microscopic Equilibrium Constants 6.4 Statistical Effects in Ligand Binding to Macromolecules 6.5 Experimental Determination of Ligand Binding Isotherms 6.6 Binding of Cro RepressorProtein to DNA 6.7 Cooperativity in Ligand Binding 6.8 Models for Cooperativity 6.9 Kinetic Studiesof CooperativeBinding 6.10 Allosterism References Problems

9r 9l

94 94 99 106 109 113 119 120

r24 124 124 r27 r28 r32 135 138 r43 r47 148 151 r52

coNTENTS

vii

Appendixes 1. Standard Free Energies and Enthalpies of Formation at 298 K, l Atmosphere, pHTrand 0.25 M lonic Strength

Ls4

2. Standard Free Energy and Enthalpy Changes for Biochemical Reactions at 298 K, 1 AtmosphereopH 7.0, pMg 3.0, and 0.25 M

Ionic Strength

156

3. Structuresof the CommonAmino Acids at Neutral pH

157

4. Useful Constantsand ConversionFactors

159

Index

161

I

PREFACE

This book is based on a course that I have been teaching for the past several years to first year graduate studentsin the biological sciencesat Duke University. These students have not studied physical chemistry as undergraduatesand typically have not had more than a year of calculus. Many faculty believe that an understanding of the principlesof physical chemistryis important for all studentsin the biological sciences, and this course is required by the Cell and Molecular Biology Program. The course consistsof two parts-one devotedto thermodynamicsand kinetics,the other to spectroscopy. Only the first half of the course is covered in this volume. An introduction to spectroscopyis being plannedas a separatevolume. One of the reviewersof the proposal for this book said that it was impossible to teachbiology studentsthis materialthe reviewer had been trying for many years. On the contrary, I believe the students that have taken this course have masteredthe principles of the subject matter and will find the knowledge useful in their research. Thermodynamics and kinetics are introduced with a minimum of mathematics. However, the approach is quantitative and is designedto introduce the student to the important conceptsthat are necessaryto apply the principles of thermodynamics and kinetics to biology. The applicationscover a wide range of topics and vary considerably in the degreeof difficulty. More material is included than is covered in the course on which the book is based,which will allow the studentsand instructors to pick and choose.Some problems are also included, as problem solving is an important part of understandingprinciples. I am indebted to my colleaguesat Duke University for their encouragementand assistance.Discussionswith them were essentialto the production of this book. Special thanks are due to ProfessorJaneRichardson and Dr. Michael Word for their assistance with the color figures. I also want to acknowledgethe encouragementand assistance of my wife, Judy, during this entire project. I would appreciateany commentsor suggestionsfrom the readersof this volume. Gonoou G. Havlaes Duke University Durham. North Carolina

tx

I

CHAPTER1

Heat,Work,and Energy 1.1 INTRODUCTION Thermodynamicsis deceptively simple or exceedinglycomplex, dependingon how you approach it. In this book, we will be concernedwith the principles of thermodynamics that are especially useful in thinking about biological phenomena.The emphasis will be on concepts, with a minimum of mathematics. Perhaps an accurate description might be rigor without rigor mortis. This may cause some squirming in the graves of thermodynamic purists, but the objective is to provide a foundation for researchersin experimentalbiology to use thermodynamics.This includes cell biology, microbiology, molecularbiology, and pharmacology,among others.In an ideal world, researchersin thesefields would have studieda year of physical chemistry,and this book would be superfluous.Although most biochemistshave this background,it is unusual for other biological sciencesto require it. Excellent texts are available that presenta more advancedand completeexpositionof thermodynamics(cf. Refs. 1 and

D. In point of fact, thermodynamicscan provide a useful way of thinking about biological processesand is indispensable when considering molecular and cellular mechanisms.For example, what reactions and cBupledphysiological processesare possible?What are the allowed mechanismsinvolved in cell division, in protein synthesis?What are the thermodynamicconsiderationsthat causeproteins,nucleic acids, and membranes to assumetheir active structures?It is easy to postulate biological mechanismsthat are inconsistentwith thermodynamicprinciples-but just as easyto postulate those that are consistent.Consequently,no active researcherin biology should be without arudimentary knowledge of the principles of thermodynamics.The ultimate goal of this exposition is to understandwhat determinesequilibrium in biological systems,and how theseequilibrium processescan be coupledtogetherto produce living systems, even though we recognize that living organisms are not at equilibrium. Thermodynamics provides a unifying framework for diverse systemsin biology. Both a qualitative and quantitativeunderstandingare important and will be developed. The beauty of thermodynamics is that a relatively small number of postulatescan be used to develop the entire subject. Perhapsthe most important part of this development is to be very precise with regard to conceptsand definitions, without getting bogged down with mathematics.Thermodynamicsis a macroscopictheory, not molecular. As far as thermodynamicsis concerned,moleculesneed not exist. However, we will not be purists in this regard: If molecular descriptionsare useful for under1

HEAT,WORK,AND ENERGY

standingor introducing concepts,they will be used.We wilt not hesitateto give molecular descriptions of thermodynamic results, but we should recognize that theseinterpretations are not inherent in thermodynamics itself. It is important to note, nevertheless,that large collections of molecules are assumedso that their behavior is governedby Boltzmann statistics;that is, the normal thermal energy distribution is assumed. This is almost always the case in practice. Furthermore, thermodynamics is concernedwith time-independentsystems,that is, systemsat equilibrium. Thermodynamics has been extendedto nonequilibrium systems,but we will not be concerned with the formal development of this subject here. The first step is to define the system.A thermodynamic system is simply that part of the universe in which we are interested.The only caveat is that the system must be large relative to molecular dimensions.The system could be a room, it could be a beaker,it could be a cell, etc. An open system canexchangeenergy and matter across its boundaries,for example,a cell or a room with open doors and windows. A closed systemcan exchangeenergy but not matter, for example, a closedroom or box. An isolated systemcan exchangeneither energy nor matter, for example, the universe or, approximately, a closedDewar. We are free to selectthe systemas we choose,but it is very important that we specify what it is. This will be illustrated as we proceed. The properties of a system are any measurablequantities charactenzing the system.Properties are either extensive,proportional to the massof the system,or intensive,independent of the mass. Examples of extensive properties are mass and volume. Examples of intensive properties are temperature,pressure,and color.

1.2 TEMPERATURE We are now ready to introduce three important concepts:temperature,heat, and work. None of these are unfamiliar, but we must define them carefully so that they can be used as we developthermodynamics. Temperatureis anobviousconcept,as it simply measureshow hot or cold a system is. We will not belabor its definition and will simply assertthat thermodynamics requires a unique temperaturescale,namely, the Kelvin temperaturescale.The Kelvin temperaturescaleis related to the more conventional Celsius temperaturescaleby the definition

T*.luin = Zc"lrio, + 273.16

(1-1)

Although the temperatureon the Celsius scale is referred to as "degrees Celsius," by convention degreesare not statedon the Kelvin scale.For example, a temperatureof 100 degreesCelsiusis 373 Kelvin. (Thermodynamicsis entirely logical-some of the conventionsusedare not.) The definition of thermal equilibriurn is very simple: when two systemsare at the sametemperature,they are at thermal equilibrium.

1.3 HEAT

1.3

HEA T

Heat flows acrossthe systemboundary during a changein the stateof the systembecause a temperature difference exists between the system and its surroundings. We know of many examplesof heat: Some chemical reactionsproduceheat, such as the combustion of gas and coal. Reactions in cells can produce heat. By convention, heat flows from higher temperatureto lower temperature.This fixes the sign of the heat change.It is important to note that this is a convention and is not required by any principle. For example, if the temperatureof the surroundingsdecreases,heat flows to the system,and the sign of the heat changeis positive (+). A simple examplewill illustrate this sign convention as well as the importanceof defining the systemunder consideration. Considertwo beakersof the samesizefilled with the sameamountof water. In one beaker,A, the temperatureis 25oC,and in the otherbeaker,B, the temperatureis 75oC. Let us now place the two beakersin thermal contact and allow them to reach thermal equilibrium (50'C). This situationis illustratedin Figure 1-I . If the systemis defined as A, the temperatureof the systemincreasesso the heatchangeis positive. If the system is defined asB, the temperatureof the systemdecreasesso the heat changeis negative. If the systemis defined as A and B, no heat flow occurs acrossthe boundary of the system, so the heat changeis zero! This illustrates how important it is to define the systembefore asking questionsabout what is occurring. The heat change that occurs is proportional to the temperaturedifference between the initial and final statesof the system.This can be expressedmathematicallyas q = C(Tr- Ti)

(r-2)

where q is the heat change,the constantC is the heat capacity,Tris the final temperature, and Z, is the initial temperature.This relationship assumesthat the heat capacity is constant, independent of the temperature.In point of fact, the heat capacity often changesas the temperaturechanges,so that a more precise definition puts this relationship in differential form:

FIGURE 1-1. Illustration of the establishment of thermal equilibrium and importance of defining the system carefully. Two identical vesselsfilled with the same amount of liquid, but at different temperatures, are placed in contact and allowed to reach thermal equilibrium. A discussion of this figure is given in the text.

HEAT,WORK,ANDENERGY

Note that the heatchangeandthe heatcapacityareextensiveproperties-the larger the systemthe largerthe heatcapacityandthe heatchange.Temperature, of course, is an intensiveproperty.

1.4 WORK The definitionof work is not assimpleasthatfor heat.Many differentformsof work exist,for example,mechanicalwork, suchasmuscleaction,andelectricalwork, such asionscrossingchargedmembranes. We will use aratherartificial,but very general, definitionof work thatis easilyunderstood. Work is a quantitythatcanbetransferred acrossthe systemboundaryand can alwaysbe convertedto lifting and lowering a weightin the surroundings. By convention,work doneon a systemis positive:this corespondsto loweringtheweightin the surroundings. You may recallthat mechanicalwork, w, is definedasthe productof the forcein the directionof movement,F,, timesthe distancemoved,x, or in differentialform dw = F*dx

(1-4)

Therefore, the work to lower a weight is -mgh, where nzis the mass,g is the gravitational constant, andh is the distancethe weight is lowered. This formula is generally useful: For example,mgh is the work requiredfor a personof massm to walkup a hill of height h.The work requiredto stretcha musclecould be calculatedwith Eq. 1-4 if we knew the force required and the distancethe muscle was stretched.Electrical work, for example, is equalto -EIt, where E is the electromotive force, 1is the current, and / is the time. In living systems,membranesoften have potentials (voltages) across them. In this case,the work requiredfor an ion to crossthe membraneis -zFY,where e is the valence of the ion, F is the Faraday (96,489 coulombs per mole), and Y is the potential. A specific example is the cotransportof Na+ and K+, Na+ moving out of the cell and K* moving into the cell. A potential of -70 millivolts is establishedon the inside so that the electrical work required to move a mole of K* ions to the inside is -(1X96,489X0.07) = -6750 ioules.(Y = Youtrid"- Yinsid"- *70 millivolts.) The negative sign meansthat work is done by the system. Although not very biologically relevant,we will now considerin somedetail pressure-volume work, ar P-V work. This type of work is conceptuallyeasy to understand, and calculations are relatively easy. The principles discussedare generally applicableto more complex systems,suchas thoseencounteredin biology. As a simple exampleof P-V work, considerapiston filled with a gas,aspicturedin Figure 1-2. In this case, the force is equal to the external pressure,P"^, times the area,A, of the piston face, so the infinitesimal work can be written as dw = -P"*A dx = -P"* dV

(1-s)

If the piston is lowered, work is done on the systemand is positive; whereasif the piston is raised work is done bv the svstem and is nesative. Note that the work done on

1.4 WORK

|ffi

of a pistonpushingon the system.P"* is the external FIGURE L-2. Schematicrepresentation pressure, andPs5 is the pressureof the system.

or by the systemby lowering or raising the piston dependson what the externalpressure is. Therefore, the work can have any value from 0 to -, depending on how the processis done. This is a very important point: The work associatedwith a given change in state dependson how the changein state is carried out. The idea that work dependson how the processis carried out can be illustrated further by considering the expansionand compressionof a gas.The P-V rsothermfor an ideal gas is shown in Figure 1-3. An ideal gas is a gas that obeys the ideal gas law, PV = nRT (n is the number of moles of gasand R is the gasconstant).The behavior of most gasesat moderatepressuresis well describedby this relationship.Let us consider the expansionof the gasfrom Pr,Vrto P2,V2.lf this expansionis done with the external pressureequal to zero, that is, into a vacuum, the work is zero. Clearly this is the minimum amount of work that can be done for this change in state. Let us now carry out the sameexpansionwith the externalpressureequal to Pr.In this case,the work is

FIGURE 1-3. A P-V isotherm for an ideal gas. The narrow rectangle with both hatched and open areasis the work done in going from Pt,Vt to Pt,Vz with an external pressureof P:. The hatched area is the work done by the system in going from Pr,Vtto Pz,Vz with an external pressureof Pz. The maximum amount of work done by the system for this change in stateis the area under the curve between PtVt and Pz,Vz.

HEAT,WORK,AND ENERGY

Fv^

w--J"p" dv--p2(v2-vr)

(1-6)

vI

which is the striped area under the P-V curve. The expansion can be broken into stages;for example, first expand the gas with p"* = p: followed by p"* = pr, asshown in Figure 1-3. The work done by the systemis then the sum of the two rectangularareasunder the curve. It is clear that as the number of stagesis increased,the magnitude of the work done increases.The maximum work that can be attainedwould set the external pressureequal to the pressureof the systemminus a small differential pressure, dP,throughout the expansion.This can be expressedas

w^u*=-!" rav

(r-7)

v1

By a similar reasoningprocess,it can be shown that for a compressionthe minimum work done on the systemis

W-in

=-Ju'P d V

(1-8)

v2

This exerciseillustratestwo important points. First, it clearly showsthat the work associatedwith a changein state dependson how the changein stateis carried out. Second, it demonstratesthe concept of a reversiblepath. When a changein stateis carried out suchthat the surroundingsand the systemare not at equilibrium by an infinitesimal amount,in this casedP, during the changein state,the processis called reversible. The conceptof reversibility is only an ideal-it cannotbe achievedin practice.Obviously we cannot really carry out a changein state with only an infinitesimal difference between the pressuresof the system and surroundings. We will find this concept very useful, nevertheless. Now let's think about a cycle whereby an expansionis carried out followed by a compressionthat returns the system back to its original state.If this is done as a onestageprocessin eachcase,the total work can be written as wtotul= *"^o*

w"o-p

wtoiat=-P2(V2- Vr) - Pr(Vr - Vr.)

wtotat=(Pr - P)(Vz-

Yl) > 0

(1-e)

(1-10)

(r-11)

N F ENERGY 1.5 DEFINITIOO

In this case,net work hasbeendone on the system.For a reversibleprocess,however, the work associatedwith compression and expansion is

w exp

--t"

Pdv

(r-r2)

v1

and tv, w.o*o=-J PdV

(1-13)

v2

so that the total work for the cycle is equal to zero. Indeed, for reversible cycles the net work is always zero. To summarize this discussion of the concept of work, the work done on or by the system dependson how the change in state of the system occurs. In the real world, changesin statealways occur irreversibly,but we will find the conceptof a reversible change in stateto be very useful. Heat changesalso dependon how the processis carriedout. Generally a subscript is appendedtoq, for example,ep andqvfor heatchangesat constantpressureand volume, respectively.As a casein point, the heat changeat constantpressureis greater than that at constant volume if the temperatureof a gas is raised. This is becausenot only must the temperaturebe raised,but the gas must also be expanded. Although this discussionof gasesseemsfar removed from biology, the concepts and conclusionsreachedare quite generaland can be applied to biological systems. The only difference is that exact calculations are usually more difficult. It is useful to considerwhy this is true.In the caseof ideal gases,a simple equationof stateis known, PV = nRT, that is obeyed quite well by real gasesunder normal conditions. This equation is valid becausegasmolecules,on average,are quite fn apartandtheir energetic interactions can be neglected.Collisions between gas molecules can be approximated as billiard balls colliding. This situationobviously doesnot prevail in liquids and solids where molecules are close together and the energeticsof their interactions cannot be neglected.Consequently,simple equationsof statedo not exist for liquids and solids.

1.5

DEFI NI T I O N O F EN ER G Y

The first law of thermodynamics is basically a definition of the energy change associated with a changein state.It is basedon the experimental observation that heat and work can be interconverted.Probably the most elegant demonstrationof this is the experimental work of JamesPrescottJoule in the late 1800s.He carried out experiments in which he measuredthe work necessaryto turn a paddle wheel in water and the concomitant rise in temperatureof the water. With this rather primitive experiment, he was able to calculate the conversion factor between work and heat with amazing ac-

HEAT,WORK,ANDENERGY

curacy,namely, to within 0.27o.The first law statesthat the energychange,AE, associated with a changein state is

A4*Yn

A^E=q+w

(1-14)

Furthermore, the energy change is the same regardlessof how the change in state is carried out. In this regard, energy clearly has quite different properties than heat and work. This is true for both reversibleand irreversibleprocesses.Becauseof this property, the energy (usually designatedthe internal energy in physical chemistry textbooks) is called a state function. State functions are extremely important in thermodynamics, both conceptually and practically. Obviously we cannotprove the first law, as it is a basic postulateof thermodynamics. However, we can show that without this law eventscould occur that are contrary to our experience.Assume, for example, that the energy changein going from state 1 to state 2 is greater than the negative of that for going from state2 to 1 becausethe changesin state are carried out differently. We could then cycle between these two statesand produce energy as each cycle is completed, essentially making a perpetual motion machine.We know that such machinesdo not exist. consistentwith the first law. Another way of looking at this law is as a statementof the conservationof energy. It is important that thermodynamic variables are not just hypothetical-we must be able to relate them to laboratory experience,that is, to measurethem. Thermodynamics is developed here for practical usage.Therefore, we must be able to relate the conceptsto what can be done in the laboratory. How can we measureenergy changes? If we only consider P-V work, the first law can be written as

"v^ LE=q_)'p"*dV

(1-ls)

V,

If the change in stateis measuredat constant volume, then L,E = q,

(1-16)

At first glance, it may seem paradoxical that a state function, the energy change, is equal to a quantity whose magnitude dependson how the changein stateis carried out, namely, the heat change.However, in this instancewe have specifiedhow the change in state is to occur, namely, at constant volume. Therefore, if we measure the heat changeat constantvolume associatedwith a changein state,we have also measured the energy change. Temperature is an especially important variable in biological systems.If the temperature is constant during a change in state,the process ts isothermal. On the other hand, if the system is insulated so that no heat escapesor entersthe systemduring the changein state(q - O),the processis adiabatic.

1.6 ENTHALPY

1.6

ENTHALPY

Most experiments in the laboratory and in biological systems are done at constant pressure,rather than at constant volume. At constantpressure,

L,E= ep- P(Vz- V,l

(1-17)

Er- Er= ep- P(Vz-Vr)

(1-18)

or

The heatchangeat constantpressurecanbe written as ep= (Ez+PV) - (Et+ PV)

(1-19)

This relationshipcanbe simplifiedby defininga new statefunction,the enthalpy,H: H=E+PV

(r-20)

The enthalpy is obviously a state function since E, P, and V are state functions. The heat change at constantpressureis then equal to the enthalpy change: qr= LH = Hz- Ht

(r-2r)

For biological reactionsand processes,we will usually be interestedin the enthalpy changerather than the energy change.It can be measuredexperimentally by determining the heat change at constant pressure. As a simple exampleof how energy and enthalpycan be calculated,let's consider the conversion of liquid water to steam at 100"C and 1 atmospherepressure,that is, boiling water:

Hr.o(/,l atm,100'c)+ Hro(g,1 atm,100"c)

e_zz)

The heatrequiredfor this process,AF1(= e is 9.71 kilocalories/mole.What is AE for i this process?This can be calculaterlas follows:

A,E=LH-A(PW-^H-PAV - 18.0x 10-3liters/mol LV = vr- vr - 2L.4liters/mole e - pvn= RT A,E= N{ - RT _g7lo -2(373) = g970 calories/mole Note that the Kelvin temperaturemust be used in thermodynamic calculations and that AFl is significantly greater than AE.

1O

HEAT,WoRK,ANDENERGY

Let's do a similarcalculationfor themeltingof ice into liquid water HrO(s,273K, 1 atm)-+ HrO(l Zlg K, I atm)

(r-23)

In this casethe measuredheat change,LH (= ei is 1.44 kilocalories/mole.The calculation of AE parallels the previous calculation. LE=LH-PLV LV = V r

Vr= 18.0 milliliters/mole - 19.6milliliters/mole = -1.6 milliliters/mole P LV = -1.6 ml.atm = -0.04 calorie LE - 1440 + 0.04 = 1440 calones/mole

In this caseM and Al1are essentially the same.In general, they do not differ greatly in condensedmedia, but the differences can be substantialin the gas phase. The two most common units for energy are the calorie and the joule. (One calorie equals4.184 joules.) The official MKS unit is the joule, but many researchpublications use the calorie. We will use both in this text. in order to familiarize the student with both units.

1.7 STANDARD STATES Only changesin energy statescan be measured.Therefore, it is arbitrary what we set as the zerofor the energy scale.As a matter of convenience,a common zero has been set for both the energy and enthalpy. Elements in their stablestfoms at25"C (298 K) and 1 atmosphereare assignedan enthalpy of zero. This is called a standard state and is usually written as Hlnr. The superscript means 1 atmosphere,and the subscript is the temperaturein Kelvin. As an example of how this concept is used, consider the formation of carbon tetrachloride from its elements:

C (graphitel+ 2 Cl2G)+ CCl4( 4

H|*<.>- 2Hinr1r,t Nr = Hinr1..ro1-

LIr - Hinr1""ro)

(r-24)

1.8 CALORIMETRY

11

The quantity Hlnrrr.., is called the heat of formation of carbon tetrachloride. Tables of heats of formation"are available for hundreds of compounds and are useful in calculating the enthalpychangesassociatedwith chemical reactions(cf. 3,4). In the caseof substancesof biological interestin solutions,the definitions of standard statesand heats of formation are a bit more complex. tn addition to pressureand temperature,other factors must be consideredsuch aspH, salt concentration,metal ion concentration,etc. A universal definition has not been established.In practice, it is best to use heatsof formation under a defined set of conditions, and likewise to define the standard state as these conditions. Tables of heats of formation for some compounds of biological interest are given in Appendix 1 (3). A prime is often added to the symbol for these heats of formation (FIr") to indicate the unusual nature of the standardstate.We will not make that distinction here,but it is essentialthat a consistent standard state is used when making thermodynamic calculations for biological systems. A useful way of looking at chemical reactionsis as algebraicequations.A characteristic enthalpy can be assignedto eachproduct and reactant.Consider the "reaction"

aA+bB+cC+dD

(1-2s)

For this reaction, MI = Hproducts- Hreactants, of

A,H= dHo+ cHc- aHo- bH,

where the H, are molar enthalpies. At298 K and 1 atmosphere,the molar enthalpies of the elements are zero,whereasfor compounds,the molar enthalpiesare equal to the heats of formation, which are tabulated.Before we apply theseconsiderationsto biological reactions,a brief digressionwill be made to discusshow heatsof reactionsare determined experimentally.

1.8

CALORIMETRY

The area of scienceconcernedwith the measurementof heat changesassociatedwith chemical reactions is designated as calorimetry. Only a brief introduction is given here, but it is important to relate the theoretical concepts to laboratory experiments. To begin this discussionwe will return to our earlier discussionof heat changesand the heat capacity,Eq. 1-3. Since the heat changedependson how the changein state is carried out, we must be more precisein defining the heat capacity.The two most common conditions are constant volume and constant pressure.The heat changesin thesecasescan be written as dqv= dE = CvdT

(r-26)

12

AND ENERGY*i woRK' HEAr'

tz #r,a rl .1r

i.,,

dqp= dH = CPdT

e(

| -4

L=dH (r-27)

A more exact mathematical treatment of these definitions would make use of partial derivatives, but we will avoid this complexity by using subscriptsto indicate what is held constant.These equations can be integrated to give .T^

LE=J cv dr

(1-28)

Tl

^ H- l

J^

cPdr

(r-2e)

T1

heat Thus, heat changescan readily be measuredif the heat capacity is known' The to the heat of amount known a adding by capacity of a substancecan be determined amount known The in temperature. substanceand determining the resulting increase (Reof heat is usually addedelectrically since this permits very precisemeasurement' the of resistance the is R and call that the electrical heat is lR, where 1is the current temperaalarge over heating element.) If heat is addedrepeatedlyin small increments Tabuture range, the temperaturedependenceof the heat capacity can be determined' temperature the with presented lations of heat capacitiesare available and are usually dependencedescribedas a power series: Cp=a+bT+rT2+'''

(1-30)

where a, b, c,. . . are constantsdetermined by experiment' done-batch calFor biological systems,two types of calorimetry are commonly are mixed toreactants orimetry and scanning calorimetry. In batch calorimetry, the measured' A simple gether and the ensuing temperature rise (or decrease) is is a Dewar flask experimental setup is depicted in Figu." 1-4, where the calorimeter or thermometer' and the temperatureincreaseis measuredby a thermocouple hydrolysis of adenothe for change heat the measure to wished For example,if we sine S'-triphosphate(ATP)' ATP + H2O =: ADP + Pt

(1-31)

in the Dewar at a defined pH' a solution of known ATP concentration would be put be initiated by adding a small would reaction metal ion concentration,buffer, etc. The that efficiently catalyzes enzyme an (ATPase), amount of adenosinetriphosphatase The enthalpy of reacmeasured' rise the hydrolysis, and the subsequenttemperature tion can be calculated from the relationship LH=CpLT

(r-32)

1.8 CALORIMETRY

13

Thermometer

FIGURE L-4. Schematic representationof a simple batch calorimeter. The insulated vessel is filled with a solution of ATP in a buffer containing salt and Mg'*. The hydrolysis of ATP is initiated by the addition of the ATPase enzyme, and the subsequent rise in temperature is measured.

The heat capacity of the system is calculated by putting a known amount of heat into the system through an electrical heater and measuring the temperaturerise of the system. The enthalpy changecalculated is for the number of moles of ATP in the system. Usually the experimental result is reponed as a molar enthalpy, that is, the enthalpy changefor a mole of ATP being hydrolyzed. This result can be obtainedby dividing the observed enthalpy changeby the moles of ATP hydrolyzed. Actual calorimeters are much more sophisticatedthan this primitive experimental setup. The calorimeter is well insulated, mixing is done very carefully, and very precise temperaturemeasurements are made with a thermocouple. The enthalpy changesfor many biological reactions have been measured,but unfortunately this information is not conveniently tabulated in a single source. However, many enthalpies of reaction can be derived from the heats of formation in the table in Appendix 1. Scanning calorimetry is a quite different experiment and measuresthe heat capacity as a function of temperature. In these experiments, a known amount of heat is added to the system through electrical heating and the resulting temperature rise is measured.Very small amounts of heat are used so the temperaturechangesare typically very small. This processis repeatedautomatically so that the temperatureof the systemslowly rises.The heat capacityof the systemis calculatedfor eachheat increment as qrlLT, and the data arepresentedas a plot of C, versus Z. This method has been used, for example, to study protein unfolding and denaturation. Proteins unfold as the temperatureis raised, and denaturationusually occurs over a very ntrrow temperaturerange.This is illustrated schematicallyin Figure 1-5, where the fraction of denaturedprotein, fo, is plotted versus the temperaturealong with the conesponding plot of heat capacity, Co, versus temperature. As shown Figure l-5, the plot of heat capacity versus temperatureis a smooth, slowly rising curve for the solvent. With the protein present,a peak in the curve occurs as the protein is denatured.The enthalpy change associatedwith denaturation is the areaunder the peak (striped area= lCp dT).In some cases,the protein denaturation

14

HEAT,WORK,AND ENERGY

T (a)

T (b)

FIGURE 1'5. Schematic representation of the denaturation of a protein and the resulting change in heat capacity, Cp. In (a) the fraction of denaturedprotein, Tb, is shown as a function of temperature, 7. In (b) the heat capacity, as measuredby scanning calorimetry, is shown as a function of temperature.The lower curve is the heat capacity of the solvent. The hatched area is the excess heat capacity change due to the protein denaturing and is equal to AIl for the unfolding.

may occur in multiple stages, in which case more than one peak can be seen in the heat capacity plot. This is shown schematically in Figure 1-6 for a two-stage unfolding

process. The enthalpiesassociatedwith protein unfolding are often interpreted in molecular terms such as hydrogen bonds, electrostatic interactions, and hydrophobic interactions. It should be borne in mind that theseinterpretations are not inherent in thermodynamic quantities, which do not explicitly give information at the molecular level. Consequently, such interpretations should be scrutinized very critically.

\

FIGURE L-6. Schematicrepresentationof a calorimeter scanin which the denaturationoccurs in two steps.The hatched area permits the sum of the enthalpy changesto be determined, and the individual enthalpies of the unfolding reactions can be determined by a detailed analysis. As in Figure 1-5, Cp is the measuredheat capacity, and Z is the temperature.

1.9 REACTIONENTHALPIES

1. 9

15

R EA CT I O N E NT HAL P IES

We now return to a considerationof reaction enthalpies.Becausethe enthalpy is a state function, it can be addedand subtractedfor a sequenceof reactions-it doesnot matter how the reaction occurs or in what order. In this regard, chemical reactionscan be consideredas algebraicequations.For example,considerthe reaction cycle below:

LHt

A ----------+B IA

tl

LH) | " |

| -LH 'o |

VI

c ---------+D LH3 If thesereactionsare written sequentially, it can readily be seenhow the enthalpiesare related.

A-+C

LH,

C-+D

LH.'

D-+B

LHo

A -+ B

LH, - A^Hr+A,H,+ A,Ho

This ability to relate enthalpiesof reactionin reaction cycles in an additive fashion is often called Hess's Law, although it really is derived from thermodynamic principles as discussed.We will find that this "law" is extremely useful, as it allows determination of the enthalpy of reaction without studying a reaction directly if a sequenceof reactions is known that can be added to give the desired reaction. As an illustration, we will calculate the enthalpy of reaction for the transfer of a phosphoryl group from ATP to glucose, a very important physiological reaction catalyzedby the enzymehexokinase. Glucose+ ATP -

ADP + Glucose-6-phosphate

(1-33)

The standardenthalpy changesfor the hydrolysis of these four compounds are given in Table 1-1. Thesedata are for very specific conditions: T =298 K, P = I atm, pH = 7.0, pMg = 3, and an ionic strengthof 0.25 M. The ionic strengthis a measureof the salt concentrationthat takesinto accountthe presenceof both monovalent and divalent

16

HEAT.WORK.ANDENERGY

TABLE 1.1 MIinr GJ/mol)

Reaction

-30.9 -28.9 -1.2 -0.5

ATP + HzO(l) <-ADP + P, A D P + H z O (l )* AMP + P' A M P+ H r O ( / ) + A + P , G6P+HzO(l)-G+P'

ions ( = ilr,,4, where c, is the concentrationof eachion, z, is its valence,and the sum is over uU of the ions present). The enthalpy change for the hexokinase reaction can easilv be calculated from these data: l F

1

'

G+P,=G6P+H2O

LHlns= 0.5kJ/mol

ATP+H2O+ADP+Pt

LHlns=-30.9 kJ/mol

G+ATP+G6P+ADP

LHlnr= -30.4kJ/mol

have The ability to calculate thermodynamic quantities for biochemical reactionsthat rethe with deal to not yet been studied is very useful. Even if data are not available reactions' related action of specific interest, very often data are available for closely l^" for some biochemical reactilons 2 contains a tabulation of AFlint Appendix ^ 1't. enthalpy changeassociatedwith ihe hexokinasereaction could also be derived from the heats of formation in the table in the appendix:

L,H= Hi,*r* Hrtu, - Hrlotr* Hrt LH=_2000.2_2279.|+2981.8+1267.1=_30.4kJlmo1 heatsof reIn point of fact, the heatsof formation are usually derived from measured action as these are the primary experimental data' of reaction A source of potential confusion is the practice of reporting enthalpies above' written as reaction as "pef mole." There is no ambiguity for the hexokinase difproducts and reactants for However, in many cases,the stoichiometric coefficients is myokinase fer. For example, the reaction catalyzedby the enzyme

2 ADP = ATP + AMP

(1-34)

is referred to as "per Even though 2 moles of ADp are used the reaction enthalpy as it is written." mole.,,The reactionenthalpyis always given as "per mole of reaction

E F T H E R E A C T I O NE N T H A L P Y E E P E N D E N CO 1.10 TEMPERATURD

17

It is important, therefore, that the equation for the reaction under consideration be explicitly stated.The myokinase reaction could be written as

(1-35)

ADP+|arr+jnue

In this case,the reactionenthalpyper mole would be one-halfof that reportedfor Eq. 1-34.

ENTHALPY DEPENDENCE OF THEREACTION 1.10 TEMPERATURE In principle, the enthalpy changesas the pressureand temperaturechange.We will not worry about the dependenceof the enthalpy on pressure,as it is usually very small for reactions in condensedphases.The temperaturedependenceof the enthalpy is given by Eq. I-27. This can be used directly to determine the temperaturedependenceof reaction enthalpies.If we assumethe standardstateenthalpy is known for each reactant, then the temperaturedependenceof the enthalpy for each reactant, i, is

(1-36)

*f Hr,i= Hlnr,, P,idr ,nrc

If we apply this relationship to the reaction enthalpy for the generalizedreaction of Eq. l-25, we obtain the following: A,Hr= cHr,c+ dHr,-

aHr^-

bHr,u

J

LIIr= Minr+ )

LCp dT 298

with

- oHirr,o- bHln*u LHlrr= cHlss,c+dH;s8,D and LCp- cCp,c+dCr,r- aCp,t- bCp,u More generally,

LHr-Mro* fr,or, o,

(r-37)

Equation 1-37 is known as Kirchhoff's Law. It can also be statedin differential form:

18

HEAT,WORK,AND ENERGY

d LHIdT = LCr

(1-38)

It is important to remember that this discussion of the temperaturedependenceof the reaction enthalpy assumesthat the pressureis constant. The conclusion of theseconsiderationsof reaction enthalpiesis that available tabulations are often sufficient to calculate the reaction enthalpy of many biological reactions. Moreover, if this is done at a standardtemperature,the reaction enthalpy at other temperaturescan be calculated if appropriate information about the heat capacitiesis known or estimated.For most chemical reactions of biological interest, the temperature dependenceof the reaction enthalpy is small. On the other hand, for processes such as protein folding and unfolding, the temperaturedependenceis often significant and must be taken into account in data analysisand thermodynamic calculations. This will be discussedfurther in Chapter3. The first law of thermodynamics, namely, the definition of energy and its conservation, is obviously of great importance in understandingthe nature of chemical reactions. As we shall see, however, the first law is not sufficient to understandwhat determineschemical equilibria.

REFERENCES l . I. Tinoco, Jr., K. Sauer,andJ. C. Wang,Physical Chemistry:Principles andApplications to the Biological Sciences,3rd edition, Prentice Hall, Englewood Cliffs, NJ, 1995.

2. D. Eisenbergand D. Crothers,PhysicalChemistrywithApplications to the Life Sciences, Benjamin/Cummings,Menlo Park, CA, 1979. 3. The NBSTables of ThermodynamicProperties, D. D. Wagman et al., eds.,"/. Phys. Chem.Ref. Data, 11,Suppl. 2, 1982. 4. D. R. Stull, E. F. Wesffum, Jr., and G. C. Sinke, The Chemical Thermodynamicsof Organic Compounds,Wiley, New York,1969. 5. R. A. Alberty,Arch. Biochem.Biophys.353, 116 (1998).

PROBLEMS 1-1. When a gasexpandsrapidly through a valve, you often feel the valve get colder. This is an adiabatic expansion (q = 0). Calculate the decreasein temperatureof 1.0 mole of ideal gas as it is expandedfrom 0.20 to 1.00 liter under the conditions given below. Assume a constantvolume molar heat capacity, Cr, of ln. Note that the energy, E, of an ideal gas dependsonly on the temperature:It is independentof the volume of the system. A. The expansion is irreversible with an external pressure of 1 atmosphere and an initial temperatureof 300 K. B. The expansion is reversible with an initial temperatureof 300 K. C. CalculateA^Efor the changesin statedescribedin parts A and B.

PROBLEMS

19

D. Assumethe expansionis carriedoutisothermally at 300 K, ratherthan adiabatically. Calculate the work done if the expansion is carried out irreversibly with an externalpressureof 1.0 atmosphere. E. Calculate the work done if the isothermal expansion is carried out reversibly. F. Calculateq andAE for the changesin statedescribedin parts D and E. l-2. A. Calculatethe enthalpy changefor the conversionof glucose [CuHrrOu(s)l and oxygen [Oz(g)] to CO2(ag) and HzO(/) under standardconditions. The standardenthalpiesof formation of glucose(s),CO2(ag),and HzO(l) are-304.3, -98.7, and -68.3 kcaUmol,respectively. B. When organisms metabolize glucose, approximately 50Vo of the energy available is utilized for chemical and mechanicalwork. Assume 25Voof the total energy from eating one mole of glucose can be utilized to climb a mountain. How high a mountain can a 70 kg person climb? 1-3. Calculate the enthalpy change for the oxidation of pyruvic acid to acetic acid under standardconditions. 2 CH3COCOOH(I) + O2(B)-+ 2 CHTCOOH(I) +2CO2G) The heats of combustion of pyruvic acid and acetic acid under standardconditions are -227 kcaUmol and -207 kcaVmol, respectively. Heats of combustion are determined by reacting pyruvic or acetic acid with Or(g) to give HzO(l) and CO2(g). Hint: First write balanced chemical equationsfor the combustion processes. l-4. Calculate the amount of water (in liters) that would have to be vaporized at 40'C (approximately body temperature)to expend the2.5 x 106calories of heat generatedby a person in one day (commonly called sweating). The heat of vaporization of water at this temperatureis 574 caUg.We normally do not sweat that much. What's wrong with this calculation?If l%oof the energy produced as heat could be utilized as mechanicalwork, how large a weight could be lifted 1 meter? 1-5. A. One hundredmilliliters of 0.200 M ATP is mixed with an ATPase in a Dewar at 298 K, 1 atm, pH 7.0, pMg 3.0, and0.25M ionic strength.The temperatureof the solution increases1.48 K. What is Afl' for the hydrolysis of ATP to adenosine5'-diphosphate(ADP) and phosphate?Assume the heat capacityof the systemis 418 J/K. B. The hydrolysis reaction can be written as ATP+H2O=ADP+P,

20

HEAT,WORK,ANDENERGY

Under the sameconditions, the hydrolysis of ADP, ADP+HrO€AMP+Pt has a heat of reactiono LHo, of -28.9 kJ/mol. Under the same conditions, calculate AI1" for the adenylatekinase reaction:

2 ADP + AMP + ATP reaction, 1-6. The alcoholdehydrogenase NAD + Ethanol + NADH + AcetaldehYde in Appendix removes ethanol from the blood. Use the enthalpies of formation (a generousmartini) is 1 to calcul ateL,H" for this reaction. If 10.0 g of ethanol much heat is procompletely converted to acetaldehydeby this reaction, how duced or consumed?

I

CHAPTER2

Entropyand FreeEnergy 2.1 INTRODUCTION At the outset,we indicated the primary objective of our discussionof thermodynamics is to understandchemical equilibrium in thermodynamic terms. Based on our discussion thus far, one possible conclusion is that chemical equilibria are governed by energy considerationsand that the systemwill always proceedto the lowest energy state. This idea can be discardedquite quickly, as we know some spontaneousreactionsproduce heat and some require heat. For example, the hydrolysis of ATP releasesheat, -30.9 kJ/mol, whereasATP and AMP are formed when ADP is mixed with M]*= myokinase, yet LHige= +2.0 kJ/mol under identical conditions. The conversionof liquids to gasesrequiresheat,that is, A11is positive, even at temperaturesabovethe boiling point. Clearly the lowest energy stateis not necessarilythe most stable state. What factor is missing? (At this point, traditional treatments of thermodynamics launch into a discussionof heatengines,a topic we will avoid.) The missing ingredient is consideration of the probability of a given state.As a very simple illustration, consider three balls of equal size that are numbered 1, 2, 3. These balls can be arranged sequentially in six different ways:

t23 132 2r3 231 3r2 32r The energy stateof all of theseiurangementsis the same,yet it is obvious that the probability of the balls being in sequence(123) is 1/6, whereasthe probability of the balls being out of sequenceis 5/6. In other words, the probability of a disordered state is much greater than the probability of an ordered state becausea larger number of arrangementsof the balls exists in the disordered state. Molecular examples of this phenomenon can readily be found. A gas expands spontaneouslyinto a vacuum even though the energy stateof the gas does not change. This occurs becausethe larger volume has more positions available for molecules, so a greaternumber of arrangements,or more technically microstates, of molecules are possible.Clearly, probability considerationsare not sufficient by themselves.If this were the case,the stable state of matter would always be a gas. We know solids and liquids are stableunder appropriateconditions becausethey are energetically favored; that is, interactions between atoms and molecules result in a lower energy state. The real situationmust involve a balancebetweenenergy and probability. This is a qualitative statementof what determines the equilibrium state of a system, but we will be able to be much more quantitative than this. 21

22

ENTROPYANDFREEENERGY

The secondlaw statesthat disordered statesare more probable than ordered states. This is done by defining a new statefunction, entropy, which is a measureof the disorder (or probability) of a state.Thermodynamics does not require this interpretation of the entropy, which is quasi-molecular. However, this is a much more intuitive way of understanding entropy than utilizing the traditional concept of heat engines. The more disordered a state, or the larger the number of available microstates,the higher the entropy. We already can seea glimmer of how the equilibrium state might be determined. At constant entropy, the energy should be minimized whereas at constant energy, the entropy should be maximized. We will return to this topic a little later. First, we will define the entropy quantitatively.

OF THESECONDLAW 2.2 STATEMENT A more formal statementof the second law is to define a new state function, the entropy, S, by the equation

ds=+

^s=J+

(2-r)

(2-2)

The temperaturescalein this definition is Kelvin. This definition is not as straightforward as that for the energy. Note that this definition requires a reversible heat change, seemsquite paradoxier",et d4r"u,yet entropy is a statefunction. At first glance, this by finding a recalculated be must change entropy is that the this of meaning th" .at. and the change, give entropy same paths the reversible all However, path. versible out iris carried in state change if actual even the is correct calculated entropy change of some consideration confusing, be somewhat reversibly. Although this appearsto concept. examples will help in understandingthis The secondlaw also includes important considerationsabout entropy: For a reversible changein state,the entropy of the universe is constant,whereasfor an irreversible change in state,the entropy of the universe increases. Again, the secondlaw cannot be proved, but we can demonstratethat without this 1aw,events could ffanspire that are contrary to our everyday experience.Two examples are given below. Without the secondlaw, a gas could spontaneouslycompress!Let's illustrate this by considering the isothermal expansionof an ideal gas, V, to Vrwith constant Z. The entropy changeis

As=J @q,,/D= Q/n J0n,",= e,"/T

(2-3)

2.2 STATEMENTOF THE SECONDLAW

23

For the isothermal expansion of an ideal gas, AE = 0. (Becauseof the definition of an ideal gas, the energy, E, is determined by the temperatureonly and does not depend on the volume and pressure.)Therefore,

Qreu=

-wr"u

_1,, P dv - J " @nrtw dv = nRTln(Vz/V) ,v"

v

(2-4)

V.

I

and

A,S= nR ln(Vr/V1)

(2-s)

For an expansion,Vz) Vr, andA,S> 0. For a compression,V, 1Vr, and AS < 0. The secondlaw does not prohibit this situation, as we are considering the entropy change for the system,not the universe.This result is in accord with the intuitive interpretation of entropy previously discussed:A larger volume has more positions for the gas to occupy and consequentlya higher entropy. We must now consider what happensto the entropy of the surroundings.For a reversible change,LS = -er"r/T, andthe entropy of the universe is the sum of the entropy changefor the system and that for the surroundings:AS = er"nlT- er"JT = 0, which is consistentwith the secondlaw. However, for an irreversible changethe situation is different. Let's make this irreversible by setting the external pressureequal to zero during the change in state. In that case,lu = 0 and since A,E= 0, e= 0. Therefore, no heat is lost by the surroundings.The entropy change for the universe is AS = AS*u,* AS.u.,= nR In(VzlVr) + 0 The secondlaw saysthat the entropy changeof the universe must be greaterthan zero, which requiresthatVr>Vr.In other words, a spontaneouscompressioncannotoccur. This is not requiredby the first law. As a second example, consider two blocks at different temperatures, Tn and 7", where h and c designatehot and cold so Zn > 4. Wr will put the blocks together so that heat is transferred from the hot block to the cold block. The entropy changesin the two blocks are given by dS"= dqlT,

and dSn- 4qlTn

If the two blocks are consideredto be the universe,the entropy changeof the universe is dS"+ dsh- dq (l/7"- t/T) > O As predictedby the secondlaw, the entropy of the universeincreases.What if the heat flows from the cold block to the hot block? Then the sign of the heat changeis reversed and

24

FREEENERGY ENTROPYAND

dS"+ dSn=dq (llTn- 1lT")<0 This predicts the entropy of the universe would decrease,which is impossible according to the secondlaw. Thus, heat cannotflow from the cold bar to the hot bar. This is not prohibited by the first law. Exceptions to the second law can be used to create perpetual motion machines, which we know are impossible. These are sometimes called perpetual motion machines of the secondkind, whereasperpetual motion machines createdby exceptions to the first law are called perpetual motion machines of the first kind.

OF THEENTROPY 2.3 CALCULATION A reversible path must always be found to calculate the entropy change. At constant

therelationship volume'

|!,qFtoJr,t r eI \-27- )qr:qrff (2-6)

dQr"u = nC, dT

canbe used,whereasat constantpressure (2-7)

der"u= nC, dT

(2-8)

AS=JnCrdT/T

since they ocEntropy changesfor phasechangesare particularly easyto calculate pressure, and temperature constant At pressure. cur at constant temperatureand

For example,for theproces,

LS=er"JT=MI|T t'*t-r*JttJ I e+

(2-e) L-l'

dS:

1 atm,279K) -+ Benzene(l,I atm,279K) Benzene(s,

(2-10)

= 8'53 eu' Here 1 = LH -- 2380 caVmol, and AS = 2380/279 8.53 caV(mol'K) for the reversepfoccaV(mol.K) is defined as an entropy unit, eu. The enffopy change secondlaw since by the essis -8.53 eu. Note that the reverseprocessis not prohibited as expected' that, note it is the entropy change for the system, not the universe. Also liquid state the since going from a solid to a liquid involves a positive entropy change is more disordered than the solid' for an irreversible While it is easyto statethat the entropy changecan be calculated stateto the final state' this processby finding a reversible way of going from the initial great consequencefor our conis not always easyto do. This is usually not a matter of

OFTHEENTROPY 25 2.3 CALCULATION

siderations,but we will considerone example to illustrate the process.Let us determine the entropy changefor the following process:

HrO(/,298 K, 1 arm)--) HrO(g,298 K, 1 atm)

( 2 - 11 )

This change is not a reversible change in state, as we know that the normal boiling point of water at 1 atmosphereis 373 K. A possible reversible cycle that would go from the initial state to the final state at constant pressureis HrO(/,298 K, I atm) + HrO(g,298 K, 1 atm)

(2-12)

JT HzO(1,373K, 1 atm)--) HrO(g,373 K, I atm) The entropy change for the bottom process, which is reversible, is simply NIIT = 97101373= 26 callmol.K). The entropy change for the left-hand side of the squareis ( Eq .2 -8 ) AS = Cr ln(TrlT1) = 18 1n(373/298)- 4 cal/(mol'K) and for the right-hand side of the squareis AS = Coln(TrlTt) = 8.0ln(2981373)= -1.8 cal/(mol.K) The entropy changesfor thesethree reversible processescan be addedto give the entropy changefor the changein stategiven in Eq. 2-ll:28 caV(mol'K). An alternative reversible path that can be constructed lowers the pressureto the equilibrium vaporpressureof water at298 K. The correspondingconstanttemperature cvcle is

HrO((,298 K, 1 atm)-+ Hro(g, 298K, I atm)

(2-13)

JT \O(/,298K,0.0313 atm)-+ HrO(g,298K,0.0313atm) In this case,we would have to calculate the change in entropy as the pressureis lowered and raised. This can easily be done but is beyond the scope of this presentation of thermodynamics.The point of this exerciseis to illustrate how entropy changescan be calculated for irreversible as well as reversible processesand multiple reversible processescan be found. In principle, the entropy can be calculated from statistical considerations. Boltzmann derived a relationship betweenthe entropy and the number of microstates, N: S=ftslnN

(2-r4)

26

ENTROPYANDFREEENERGY

where ku is Boltzmannosconstant,1.38 x 10-23J/K. It is rarely possibleto determine the number of microstates although the number of microstates could be calculated from Eq.2-I4 if the entropy is known. For a simple case,such as the three numbered balls with which we startedour discussionof the secondlaw, this calculationcanreadily be done. The disorderedsystemhas 3! microstatesand an entropy of 1.51 x 10-23 J/K. Any orderedsequence-for example 1,2,3, -has only 1 microstate,so N = 1, and S = 0. Since the disordered statehas a higher entropy, this predicts that the balls will spontaneouslydisorder, and that an ordered state is extremely unlikely.

2.4 THIRDLAWOF THERMODYNAMICS We will not dwell on the third law as the details are of little consequencein biology. The important fact for us is that the third law establishesazero for the entropy scale. Unlike the energy, entropy has an absolute scale. The third law can be stated as follows: The entropy of perfect crystals of all pure elements and compounds is zero at absolute zero. The tricky points of this law are the meanings of "perfect" and "pure," but we will not discussthis in detail. It is worth noting that a perfect crystal has one microstate, ffid therefore an entropy of zero according to Eq. 2-I4. The absolute standard entropy can be determined from measurementsof the temperature dependenceof the heat capacity using the relationship 298

sin,=J 0

(2-1s)

c PdT/T

Here the entropy at absolute zero has been assumedto be zero in accord with the third law. A plot of CplT versus7 gives a curve such as that in Figure 2-l.T\e areaunder the curve is the absolute standard entropy. Tables of ^St'ntare readily available (cf.

= v

c{

6o E rg4 .(r. -F a r oct

o

200 100 T (K)

300

FIGURE 2-1. Aplot of the constant-pressureheat capacity divided by the temperature' CplT, versus the temperature, I, for graphite. The absolute entropy of graphite at 300 K is the area under the curve up to the dashedline (Eq. 2-15).If phase changes occur as the temperature is changed, the enffopy changes associatedwith the phase changes must be added to the area under the curve of the CrlT versus T plot.

2.5 MOLECULARINTERPRETATION OF ENTROPY

27

Refs. 3 and 4, Chapter 1). The entropy at temperaturesother than 298 K can be calculated from the relationship J

si = si'st+J

(2-r6)

cP dr/T

298

For a chemical reaction, .T

asz= Nr"**J

(2-r7)

LCPdT/T 298

These relationships are analogousto those used for the enthalpy. Indeed, entropies of reactions can be calculated in a similar fashion to enthalpy changes.

2.5 MOLECULAR INTERPRETATION OF ENTROPY We will now consider a few examples of absolute entropies and entropy changesfor chemical reactions, and how they might be interpreted in molecular terms. The absolute entropiesfor water asa solid,liquid, and gasat273 K and 1 atmosphere,are 41.0, 63.2, and 188.3 J/(mol'K), respectively.The molecular interpretationof thesenumbers is straightforward, namely, that solid is more ordered than liquid, which is more ordered than gas; therefore, the solid has the lowest entropy and the gas the highest. Standardentropy changesfor some chemical reactions are given in Table 2-1. As expected,the entropy changeis negative for the first two reactions in the gas phaseas the number of moles of reactantsis greater than the number of moles of products. Of course,thesereactions could have been written in the opposite direction. The entropy change would then have the opposite sign. At first glance, the result for the third reaction in the table is surprising, as the number of moles of reactantsis greater than the number of moles of products.This resultsbecausethe solventmust be included in the molecular interpretation of the observed entropy change. Ions in water interact strongly with water so that highly ordered water molecules exist around the ions. TABLE 2.1 Reaction H(g)+H(g)=Hz(g) 2Hz@) + Oz(g)= 2 H2O(g) H*(aq) + OH-(aq) = H2O(l) Cytidine 2'-monophosphate + Ribonuclease A <= Enzyme-inhibitor complex

ASfi, [J/(mol.K)] -98.7 -88.9 +80.7 -54

"H. Naghibi, A. Tamura, and J. M. Sturtevant,Proc. Natl. Acad. Sci.USA 92.559'7 ,L995't.

28

ENTROPYAND FREE ENERGY

When the neutral speciesis formed, thesehighly ordered water moleculesbecomeless ordered.Thus, the entropy changefor water is very positive, much more positive than the expectedentropy decreasefor H+ and OH- when they form water. This simple example indicates that considerable care must be exercised in interpreting entropy changesfor chemical reactions in condensedmedia. The entropy of the entire system under consideration must be taken into account.The final entry is for the binding of a ligand to an enzyme.In this case,no reasonableinterpretation of the entropy change is possible as three factors come into play: the loss in entropy as two reactantsbecome a single entity, the changesin the structure of water, and structural changesin the protein. The fact that the enffopy changeis comparableto the value expectedfor the combination of two molecules to produce one molecule is simply fortuitous. Extreme caution should be exercised in making molecular interpretations of thermodynamic changesin complex systems. We have established all of the thermodynamic principles necessary to discuss chemical equilibrium. We will now apply theseprinciples to develop a generalframework for dealing with chemical reactions.

2.6 FREEENERGY

)t

---4 "i i i

In a sensewe have reached our goal. We have developed thermodynamic criteria for the occurrenceof reversible and irreversible (spontaneous)processes,namely, for the universe,AS = 0 for reversible processesand must be greaterthan zero for irreversible processes.Unfortunately, this is not terribly useful, as we are interestedin what is happening in the system and require criteria that are easily applicable to chemical reactions. This can be achieved by defining a new thermodynamic state function, the Gibbs free energy:

i

tki ((2-18 ,,t ,lAY Lls'u '-r*E;# 'J : virtually single handed o-H-rs

'

(J. Willard Gibbs developed the science of therniodynamics around the turn of the century at Yale University. To this day, his collected works are pnzedpossessionsin the libraries of people seriously interestedin thermodynamics.)

i

i i ,

At constant temPerature,

r

AG= AH- f AS = ep- ZAS = ep- er",

(2-19) A \ i 0 =O -----i'-,1,5. f o " - t t ' 1 tl V ' . 0 ( a ( t ! i { ' s t n ( ( .l?fl,r. A |\ = processes, the situation process, t or a reverstDle ep er",soAG = 0. For irreversible anendoFor be considered. must is a bit morecomplexasthe signof theheatchange valueof the absolute process, AG < 0. For anexothermic thermicprocess, epl er"uso energy free is the This < 0. AG qris greaterthanthe absolutevalueof er",so again, developed have now We surroundings. for the systemanddoesnot involvethe "ituttg"

2 . 6 F R E EE N E R G Y

29

criteria that tell us if a processoccurs spontaneously.If LG < 0, the changein state occurs spontaneously,whereasif AG > 0, the reversechangein stateoccurs spontaneously.If AG = 0, the systemis at equilibrium (at constantpressureand temperature). As with the energy and enthalpy, only differences in free energy can be measured. Consequently, the zero of the free energy scale is arbitrary. As for the enthalpy, the zero of the scale is taken as the elementsin their stable state at I atmosphereand 298 K. Again, analogousto the enthalpy, tables of the free energiesof formation of compounds are available so that free energy changesfor chemical reactions can be calculated (cf. Refs. 3-5 in Chapter 1). Referring back to the reaction in Eq. I-25,

LGln,= rGlss.c+dG]rr'- oGlg,o- bGlnr,u

(2-20)

where the free energies on the righrhand side of the equation are free energies per mole. We will discussthe temperaturedependenceof the free energya bit later, but it is useful to remember that at constant temperatureand pressure, LG=NI-TLS

(2-21)

Standard state free energy changesare available for biochemical reactions although comprehensivetabulationsdo not exist, and the definition of "standardstate" is more complex than in the gas phase,as discussedpreviously. Standardstatefree energies of formation for some substancesof biochemical interestare given in Appendix 1. As an illustration of the concept of free energy,considerthe conversionof liquid water to steam: HzO(/) -r HrO(g)

(2-22)

At the boiling point, NI =97I0 caVmol and AS =26 eu.Therefore,

AG= 9710-267

(2-23)

At equilibrium, AG = 0 and Eq.2-23 gives T = 373 K, the normal boiling point of water. If T > 373 K, LG < 0, and the changein stateis spontaneous,whereasif T < 373 K, AG > 0, and the reverseprocess,condensation,occurs spontaneously. Calculation of the free energy for a change in state is not always straightforward. Becausethe entropy is part of the definition of the free energy, a reversibleprocess must always be found to calculate the free energy change.The free energy change,of course,is the sameregardlessof how the changein stateis accomplishedsincethe free energy is a statefunction. As a simple example, againconsiderthe changein statein Eq.2-11. This is not a reversibleprocesssince the two statesare not in equilibrium. This also is not a spontaneouschangein stateso that AG > 0. In order to calculatethe value of AG, we must think of a reversiblepath for carrying out the changein state. The two cycles in Eqs. 2-12 and2-13 againcanbe used.In both cases,the bottom reaction is an equilibrium process,so AG = 0. To calculateAG for the top reaction,we

30

ENTROPYANDFREEENERGY

only need to add up the AG values for the vertical processes.These can be calculated from the temperature(upper cycle) and pressure(lower cycle) dependenceof G. Such functional dependencieswill be consideredshortly. It will be a useful exercisefor the reader to carry out the complete calculations.

2 .7

CHE M I CA L E Q U IL IBR IA

Although we now have developed criteria for deciding whether or not a processwill occur spontaneously,they are not sufficient for consideration of chemical reactions. We know that chemical reactionsare generally not "a11or nothing" processes;instead, an equilibrium stateis reachedwhere both reactantsand products are present.We will now derive a quantitative relationship betweenthe free energy changeand the concentrations of reactantsand products. We will do this in detail for the simple caseof ideal gases,and by analogy for reactionsin liquids. The starting point for the derivation is the definition of free energy and its total derivative: G = H - Z , S =E + P V - T S

(2-24)

dG = dE + P dV + V dP - T dS - S dT

(2-2s)

SincedE = dq + dw - T dS - P dVfor a reversibleprocess, dG_VdP_SdT

(2-26)

(Although this relationship was derived for a reversible process,it is also valid for an irreversible process.)t,et us now consider a chemical reaction of ideal gasesat constant temperature.For one mole of each gas component, dG=VdP-RTdPlP

(2-27)

We will refer all of our calculations to a pressureof 1 atmospherefor each component. In thermodynamicterms, we have selected1 atmosphereas our standard state.If Eq. 2-27 is now integrated from Po = 1 atmosphereto P, dG = RT dPlP

G = G" + RT ln(P/Ps) = Go + RZ ln P

(2-28)

We will now return to our prototype reaction, Eq. l-25, and calculate the free energy change.The partial pressuresof the reactantsare given in parentheses:

2 . 7 C H E M I C A LE Q U I L I B R I A

aA(P^) + bB(PB) == cC(P") + dD(Po)

AG = cG" + dG o- aG n - b Gu - c G [+ d G ;-

" G;-

31

(2-29)

b G ]+ cR Tl nP .+ dR T l n P o

- aRT lnPo - bRT lnP"

AG=AG"+-rtffi)

(2-30)

The AG is the free energy for the reaction in Eq. 2-29 when the system is not at equilibrium. At equilibrium, at constant temperature and pressure,AG = 0, and Eq. 2-30 becomes

LG" = -

orr"(W-) =- RrtnK

(2-3r)

\P^PL) Here the subscript e has been used to designateequilibrium and K is the equilibrium constant. We now have a quantitative relationship between the partial pressuresof the reactants and the standardfree energy change,AGo. The standardfree energy change is a constantat a given temperatureand pressurebut will vary as the temperatureand pressure change.If AG" < 0, then K > 1, whereasif AG'> 0, K < l. A common mistake is to confuse the free energy change with the standardfree energy change. The free energy change is always equal to zero at equilibrium and can be calculated from Eq. 2-30 when not at equilibrium. The standardfree energy change is a constant representing the hypothetical reaction with all of the reactantsand products at a pressureof 1 atmosphere.It is equal to zero only if the equilibrium constant fortuitously is 1. Biological reactions do not occur in the gas phase.What about free energy in solutions? Conceptually there is no difference. The molar free energy at constanttemperature and pressurecan be written as G=G"+RTln(clcs)

(2-32)

where c is the concentration and co is the standardstateconcentration. A more correct treatment would define the molar free energy as G=G"+RTlna

(2-33)

where a is the thermodynamic activity and is dimensionless.However, the thermodynamic activity can be written as a product of an activity coefficient and the concentra-

32

ENTRoPYAND FREE ENERGY

tion. The activity coefficient can be included in the standardfree energy, Go, which gives rise to Eq.2-32. We neednot worry aboutthis as long asthe solution conditions are clearly defined with respect to salt concentration, pH, etc. The reason it is not of great concern is that all of the aforementioned complications can be included in the standardfree energy change since, in practice, the standardfree energy is determined by measuring the equilibrium constant under defined conditions. Finally, we should note that the free energy per mole at constant temperatureand pressureis called the chemical potential, p, in more sophisticatedtreatmentsof thermodynamics, but there is no need to introduce this terminology here. If we take the standardstate as 1 mole/ liter, then the results parallel to Eqs. 2-28, 2-30, and2-31 arc G=G"+RZlnc

*t{rffi) AG:AG.+

-Rr"[,ft) =-],rn^u AG'=

(2-34)

(2-3s)

(2-36)

Equations2-34to2-36 summarizethethermodynamicrelationshipsnecessaryto discuss chemical equilibria. Note that, strictly speaking, the equilibrium constant is dimensionlessas all of the concenffations are ratios, the actual concentration divided by the standard state concentration. However, practically speaking, it is preferable to report equilibrium constants with the dimensions implied by the ratio of concentrations in F;q. 2-36. The equilibrium constant is determined experimentally by measuring concentrations,and attributing dimensions to this constant assuresthat the correct ratio of concentrations is consideredand that the standardstateis precisely defined. Consideration of the free energy also allows us to assesshow the energy and en- ZAS at contropy arebalancedto achievethe final equilibrium state.Since AG = NI change enthalpy if the stani Zand P, it can be seenthat a changein stateis spontaneous if enthalpy the is very negative and/or the entropy change is very positive. Even change is unfavorable (positive), the change in state will be spontaneousif the T AS term is very positive. Similarly, even if the entropy changeis unfavorable (Z A,Svery negative), the changein statewill be spontaneousif the enthalpy changeis sufficiently negative.Thus, the final equilibrium achievedis abalance between the enthalpy ( Ift and the entropy (f A$.

3Y 2 . 8 P R E S S U R E A N D T E M P E R A T U R E D E P E N D E N C E O F T H E F R E E E N E R3G

OFTHEFREE DEPENDENCE ANDTEMPERATURE 2.8 PRESSURE ENERGY We will now return to the pressureand temperaturedependenceof the free energy. At constanttemperature,the pressuredependenceof the free energy follows directly from W.2-26, namely, dG=VdP

(2-37)

This equation can be integrated if the pressuredependenceof the volume is known, as for an ideal gas. For a chemical reaction, Eq.2-37 can be rewritten as dLG=LVdP

(2-38)

where AVis the difference in volume betweenthe products and reactants.The pressure dependenceof the equilibrium constant at constant temperaturefollows directly,

dLG=-RTdInK=LVdP

dln K = - - LV RT dP

(2-3e)

(2-40)

For most chemical reactions, the pressuredependenceof the equilibrium constant is quite small so that it is not often consideredin biological systems. Equilibrium constants,however, frequently vary significantly with temperature.At constantpressure,Eq.2-26 gives dG=-SdT

(2-4r)

dLG=-LSdT

(2-42)

Returning to the basic definition of free energy at constant temperatureand pressure, LG = LH - ZAS = NI + T(d LG/dT) This equationcan be divided by T'and rearrangedas follows: -LG /72 + @ LG / dT) /T = -LH /72

34

ENTROPYANDFREEENERGY

d(LG/T) Nr =-_ dT

(2-43)

T2

Equation 2-43 is an important thermodynamic relationship describing the remperarure dependenceof the free energy at constantpressureand is called the Gibbs-Helmholtz equation. The temperaturedependenceof the equilibrium constant follows directly: d(LG" /T) = -rr^ d ln K dr dT

-dln = K dT

AII"

(2-44)

rf LH" is independentof temperature,Eq. 2-44 caneasily be integrated:

r" AH"dT

d t n K =.l7 , ---------RT"

,"8)=+Ii-t)= (Nf /R)(72- Tr)

(2-4s)

T,T,

When carrying out calculations, the difference between reciprocal temperatures should never be used directly as it introduces a large error. Instead the rearrangement in Eq. 2-45 should be used in which the difference between two temperaturesoccurs. With this equation and a knowledge of LHo, the equilibrium constant can be calculated at any temperatureif it is known at one temperature. What about the assumption that the standard enthalpy change is independent of temperature?This assumptionis reasonablefor many biological reactionsif the temperature range is not too large. In some cases,the temperaturedependencecannot be neglectedand must be included explicitly in canying out the integration of F;q. 2-44. The temperaturedependenceof the reaction enthalpy dependson the difference in heat capacitiesbetweenthe productsand reactantsas given by Eq. 1-38. Examples of the temperaturedependenceof equilibrium constantsare displayed in Figure 2-2. As predictedby Eq. 2-45, a plot of ln K versus IIT is a straightline with a slope of -NI"/R. The data presentedare for the binding of DNA to DNA binding proteins (i.e., Zn fingers).The dissociationconstantsfor binding arequite similar for both proteinsat22"C,1.08 x 10-eM (WTl protein),and 3.58 x 10-eM (EGRI protein). However, as indicated by the data in the figure, the standardenthalpy changes areopposite in sign,+6.6 kcaVmoland -6.9kcaVmo1,respectively.Consequently,the standard entropy changesare also quite different, 63.3 eu and 15.3 eu, respectively.These data indicate that there are significant differences in the binding processes,despitethe similar equilibrium constants.

2.9 PHASECHANGES

3.3

3.4

35

3.5

rooor (r1)

FIGURE 2-2. Temperaturedependenceof the equilibrium binding constant,K for the binding of DNA to binding proteins ("zinc fingers"), WT1 (o) and EGRI (o), to DNA. Adapted with permission from T. Hamilton, F. Borel, and P. J. Romaniuk, Biochemistry 37,2051 (1998). Copyright O 1998 American Chemical Society.

2.9

PHASE CHANGES

The criterion that LG = 0 at equilibrium at constant temperatureand pressureis quite general.For chemical reactions,this meansthat the free energy of the products is equal to the free energy of the reactants.For phasechangesof pure substances,this means that at equilibrium the free energiesof the phasesare equal. If we assumetwo phases, A and B, Eq. 2-26 gives dGo- V^dP - S^dT - dGt= VBdP - SBdT gives Rearrangement

ii ,

irt: ,i i. .= , .,,r dP _ASBA dT AVuo

i

1i i-

(2-46)

where ASuo- SB So and AVuo = Vn- yA. (All of thesequantitiesare assumedto be per mole for simplicity.) This equation is often referred to as the Clapeyron equation. Note that the entropy changecan be written as AS"e = A,HstlT

(2-47)

Equation 2-46 gives the slope of phase diagrams, plots of P versus T that are useful summariesof the phasebehavior of a pure substance.As an example, the phasediagram of water is given in Figure 2-3. The lines in Figure 2-3 indicate when two phasesare in equilibrium and coexist. Only one phaseexists in the open areas,and when the two lines meet, three phasesare in equilibrium. This is called the triple point and can only occur atapressureof 0.006

36

ENTROPYANDFREEENERGY

I I I I I

0.01"c

FIGURE 2-3. Schematic representation of the phase diagram of water with pressure,P, and temperature, Z, as variables. The phase diagram is not to scale and is incomplete, as several different phases of solid water are known. If volume is included as a variable, a three-dimensional phase diagram can be constructed.

atmosphere and 0.01"C. The slopes of the lines are given by Eq. 2-46. The number of phases that can coexist is governed by the phase rule,

f=c-p+2

(2-48)

where/is the number of degreesof freedom, c is the number of components,and p is the number of phases.This rule can be derived from the basic laws of thermodynamics. For a pufe substance,the number of componentsis 1, so,f - 3 - p.In the open spaces,one phaseexists,so that the number of degreesof freedom is 2, which means that both Z and P can be varied. Two phasescoexist on the lines, which gives/= l. Becauseonly one degreeof freedom exists,at a given value of P, Z is fixed and vice versa. At the triple point, three phasescoexist, and there are no degreesof freedom, which meansthat both P andT arefixed. The study of phasediagrams is an important aspectof thermodynamics, and some interesting applications exist in biology. For example, membranes contain phospholipids. If phospholipids are mixed with water, they spontaneouslyform bilayer vesicles with the head groups pointing outward toward the solvent so that the interior of the bilayer is the hydrocarbon chains of the phospholipids. A schematic representation of the bilayer is given in Figure 2-4. The hydrocarbon chain can be either disordered, at high temperatures,or ordered, at low temperatures,as indicated schematically in the figure. The transformation from one form to the other behavesas a phasetransition so that phase diagrams can be constructed for phospholipids. Moreover, the phasediagrams dependnot only on the temperaturebut on the phospholipid composition of the bilayer. This phasetransition hasbiological implications in that the fluidity of the hydrocarbon portion of the membranestrongly affects the transport and mechanicalproperties of the membranes.Phasediagrams for phospholipids and phospholipid-water mixtures can be constructedby a variety of methods, including scanning calorimetry and nuclear magnetic resonance.

PROBLEMS

N:

I I

,

\/ \'-------'

37

ffi[[[-ffi[[x (b)

(a) FIGURB 2-4. Schematic representationof the vesicles of phospholipids that are formed when phospholipids are suspendedin water (a). Here the small circles representthe head groups and the wiggly lines the two hydrocarbon chains. A phase change can occur in the side chains from a disordered to an ordered state as sketched in (b).

2 .1 0

ADDI T I O NS T O T H E F R E E E N E R GY

As a final consideration in developing the concept of free energy, we will return to our original developmentof equilibrium, namely,Eq.2-33. In arriving at this expression for the molar free energy at constant temperature and pressure,the assumption was made that only P-V work occurs. This is not true in general and for many biological systemsin particular. If we went all the way back to the beginning of our development of the free energychangesassociatedwith chemical equilibria (Eq. 2-26) and derived the molar free energy at constanttemperafureand pressure,we would find that G=G"+RTlnc+lu,rru"

(2-4e)

where w-u* is the maximum (reversible)non P-V work. For example, for an ion of charge zin a potential field Y, w-u* is zFY, where F is the Faraday.(We have used this relationship in Chapter I to calculate the work of moving an ion across a membranewith an imposedvoltage.In terms of transportof an ion acrossa membrane,this assumesthe ion is being transported from the inside to the outside with an external membranepotential of Y.) This extendedchemicalpotential is useful for discussing ion transportacrossmembranes,as we shall seelater. It could also be used,for example, for considering molecules in a gravitational or centrifugal field. (See Ref. 1 in Chapter 1 for a detaileddiscussionof this concept.) We are now ready to considersomeapplicationsof thermodynamicsto biological systemsin more detail.

PR OBL E M S 2-1. One mole of an ideal gas initially at 300 K is expandedfrom 0.2 to 1.0 liter. Calculate AS for this changein stateif it is carried out under the following conditions. Assume a constantvolume heat capacity for the gas,Cr, of ]R. A. The expansion is reversible and isothermal.

38

ENTROPY ANDFREE ENERGY B. The expansionis irreversible and isothermal. C. The expansion is reversible and adiabqtic (q = 0). D. The expansion is irreversible with an external pressureof I atmosphere and adiabatic.

2-2. The alcohol dehydrogenasereaction, which removes ethanol from your blood, proceedsaccording to the following reaction: NAD+ + Ethanol + NADH + Acetaldehyde Under standardconditions (298 K, 1 atm, pH 7.0, pMg 3, and an ionic strength of 0.25 M), the standardenthalpies and free energiesof formation of the reactants are as follows: tI" (kJ/mol)

NAD* NADH Ethanol Acetaldehyde

G" (kJ/mol)

-10.3 -4t.4 -290.8 -2r3.6

r059.1 rr20.r 63.0 24.1

A. Calculate LG", LHo, and A,Sofor the alcohol dehydrogenasereaction under standardconditions. B. Under standardconditions, what is the equilibrium constant for this reaction? Will the equilibrium constantincreaseor decreaseas the temperature is increased? 2-3. The equilibrium constantunder standardconditions (1 atm, 298K, pH 7.0) for the reaction catalyzedby fumarase, Fumarate + HrO € r--malate is 4.00. At 310 K, the equilibrium constantis 8.00. A. What is the standardfree energy change, LG", for this reaction? B. What is the standardenthalpychange, LHo, for this reaction?Assumethe standardenthalpy change is independentof temperature. C. What is the standardentropy change,ASo,at 298K for this reaction? D. If the concentration of both reactantsis equal to 0.01 M, what is the free energy change at 298 K? As the reaction proceeds to equilibrium, will more fumarate or l-malate form? E. What is the free energy change for this reaction when equilibrium is reached? 2-4. The following reaction is catalyzedby the enzyme creatine kinase:

PROBLEMS

39

Creatine + ATP S Creatine phosphate+ ADP

A. Under standardconditions (1 atm, 298 K, pH 7.0, pMg 3.0 and an ionic strengthof 0.25 M) with the concentrationsof all reactantsequal to 10 mM, the free energy change, AG, for this reaction is 13.3 kJ/mol. what is the standardfree energy change,AGo, for this reaction? B. What is the equilibrium constant for this reaction? C. The standardenthalpies of formation for the reactantsare as follows:

-540 kJ/mol Creatine Creatinephosphate - 1510kJlmol -2982 kJ/mol ATP -2000 kJ/mol ADP What is the standardenthalpy change, LII', for this reaction? D. What is the standardentropy change,A^So,for this reaction? 2-5. A. It has been proposed that the reason ice skating works so well is that the pressurefrom the blades of the skatesmelts the ice. Consider this proposal from the viewpoint of phaseequilibria. The phasechangein question is HrO(s) = H2O(/) Assume that LH for this processis independent of temperatureand pressure-and is equal to 80 calJg.The changein volume, Lv, is about -9.1 x 10-s Llg. The pressureexerted is the force per unit area.For a 180 pound person and an area for the skate blades of about 6 squareinches, the pressure is 30lb/sq. in. or about 2 atmospheres.With this information, calculate the decreasein the melting temperatureof ice causedby the skate blades. (Note that 1 cal = 0.04129 L.atm.) Is this a good explanation for why ice skating works? B. A more efficient way of melting ice is to add an inert compound such as urea.(We will avoid salt to saveour cars.)The extentto which the freezing point is lowered can be calculated by noting that the molar free energy of water must be the same in the solid ice and the urea solution. The molar free energy of water in the urea solution can be approximated as Griquia + RT ln X*u,.n where X*ut". is the mole fraction of water in the solution. The molar free energy of the solid can be written as Gi,,o. Derive an expressionfor the changein the melting temperatureof ice by equating the free energiesin the two phases,differentiating the resulting equation with respectto temperature,integrating from a mole fraction of 1 (pure solvent) to the mole fraction of the solution, and noting that ln X*ute.= ln(1 _ 4r"u) = -Xur"u. (This relationship is the series expansion of the logarithm for small values of 4rru. Since the concentration of water is about 55 M, this is a good approximation.) With the relationship derived, estimate the de-

40

ENERGY ANDFREE ENTROPY creasein the melting temperatureof ice for an 1 M urea solution. The heat of fusion of water is 1440 callmol.

2-6. What is the maximum amount of work that can be obtained from hydrolyzing I mole of ATP to ADP and phosphateat 298 K? Assume that the concentrations of all reactantsare 0.01 M and AGo is -32.5 kJ/mol. If the conversionof free energy to mechanical work is 1007oefficient, how many moles of ATP would have to be hydrolyzedto lift a 100 kg weight I meter high?

I

CHAPTER3

to Applicationsof Thermodynamics BiologicalSystems

3.1

BIO CHE M I CA LREA C T ION S

From the discussion in the previous chapter, it should be clear that standardfree energy and enthalpy changescan be calculated for chemical reactionsfrom available tables of standardfree energiesand enthalpiesof formation (e.g.,Appendix 1). Even if a particular reaction has not been studied, it is frequently possible to combine the standard free energy and enthalpy changes of other known reactions to obtain the standardfree energychangeof the desiredreaction(e.g.,Appendix 2). Knowledge of the standardfree energy change permits calculation of the equilibrium constant and vice versa.When considering biochemical reactions,the exact nature of the "standard state"can be confusing.Usually the various possibleionization statesof the reactants are not specified, so that the total concentration of each speciesis used in the expression for the equilibrium constant.For example, for ATP, the standardstatewould normally be I M, but this includesall possiblestatesof protonation,and if magnesiumion is present,the sum of the metal complex and uncomplexed species.Furthermore, pH 7 is usually selectedas the standardcondition, so that the activity of the hydrogen ion is set equal to 1 at pH 7, and the standardfree energy of formation of the hydrogen ion is set equal to zero at pH 7. Therefore, the hydrogen ion is not usually included when writing stoichiometric equations.Finally, the activity of water is set equal to 1 for reactions in dilute solutions. This assumptionis justified since the concentrationof water is essentiallyconstant.However, a word of caution is neededhere as the standard free energy and enthalpy of formation for water must be explicitly included if it is a reactant. Again, the choice of standard statescan readily be incorporated into the standardfree energy change and standardfree energy of formation. The best way to avoid ambiguity is to explicitly write down the chemical reaction and the conditions (pH, salt, etc.) to which a given free energy changerefers. As a specific example,let's again considerthe reaction catalyzedby hexokinase, the phosphorylation of glucose by ATP. We have previously calculated the standard enthalpy change for this reaction from the known enthalpies of hydrolysis. This can also be done for the standardfree energy change of the hexokinasereaction since the standardfree energiesof hydrolysis are known. As before, the "standard states" are 1 M, at 1 atmospherc,298K, pH 7.0, pMg 3.0, and an ionic strengthof 0.25 M. 41

42

APPLICATIONS OF THERMODYNAMICS

AG"/(kJ/mol) G+4=G6P+HrO ATP+H2O=ADP+P,

11.6 -32.5 --=Ng

G+ATP+G6P+ADP

NI"/(kJlmol) 0.5 -30.9 ---407

The equilibrium constantfor the hexokinasereaction can be calculatedfrom the standard free energy change: -i':'.

:'

')'tti"r-

\

--:

'i

I

t:l-''

''r"'

''

'-'!

AGo= -RTlnK or K=4630{=(tc6PltADply(tcltATpl)} Sinceth" ,f"fi;;#,tI$ifrhange is known,-30.4kJ/mol, theequilibrium consrant j.-.1 = . J '--

: n '

t J

, },, \t]

, . ' ," . { f * r { !i, !!.f.tit

,

can be calculated at other temperaturesby use of Eq. z-45. For example, at 310 K (37"C), the equilibrium constant is288e[The entropy changefor this reaction can be calculated from the relationship AGo = LH" - ZASo. How are standardfree energy changesforreactions measured?The obvious answer is by measurementof the equilibrium constantand useof Eq. 2-36. However, in order to determine an equilibrium constant,a measurableamount of both reactantsand products must be present.For reactionsthat essentially go to completion, a sequenceof reactions must be found with measurableequilibrium constantsthat can be summed to \ total the reaction of interest. For example, the equilibrium constantfor the hydrolysis ;! of ATP to ADP and P1calculated from the above standardfree energy change is 5 x 105M under "standardconditions." Obviously, this constantwould be very difficult to measuredirectly. The following reactions,however, could be usedto calculatethis equilibrium constant (and the standardfree energy change):

G+ATP=c6P+ADP G6P+H2O+G+P, ATP+H2O=ADP+P, The equilibrium constant for the first reaction is about 4600 and for the secondreaction is about 110 M, both of which can be measuredexperimentally.Tablesof standard free energies of formation (Appendix 1) and standard free energy changes for reactions(Appendix 2) canbe constructedusing this methodology. Knowledge of the equilibrium constantsof biological reactions and their temperature dependenciesis of great importance for understanding metabolic regulation. It can also be of practical importance in the design of laboratory experiments,for example, in the developmentof coupled assays.

3.2 METABOLIC CYCLES Thermodynamics is particularly useful for understanding metabolism. Metabolism consistsof many different setsof reactions,each set,or metabolic cycle, designedto utilize and produce very specific molecules. The reactions within a cycle are coupled

3.2 METABOLICCYCLES

43

in that the product of one reaction becomesthe reactantfor the next reaction in the cycle. These coupled reactions can be very conveniently characterizedusing thermodynamic concepts. Before considering a specific metabolic cycle, let's consider some general thermodynamic properties of coupled reactions.As a very simple illustration, consider the coupled reactions A+B+C+D The free energy changesfor the first three reactions can be written as

A -+ B

AGo" = AGls + RT ln(tBl/tAl)

B+ C

AGr"=AGic+RZln([C]/tBl)

C -+ D

AG.o= AGS'+ R7ln(tDl/tcl)

The free energy for the overall conversion of A to D can be obtained by adding these free energies:

A -+ D

AGoo= AGln+ AG;c+ AGlo+ RTln {(tBltcltD)/(lAltBltcl)} AGoo = lGlo

+ RT ln(tDl/tAl)

Whether A can be converted to D depends on the standard free energies for the three individual reactions and the concentrationsof A and D. The concentrationsof B and C are of no consequence!Note that since the total standardfree energy determines whether A will be converted to D, it is possible for one of the standardfree energy changesof the intermediate stepsto be very unfavorable (positive) if it is balancedby a very favorable (negative) standardfree energy change. Metabolic pathways contain hundreds of different reactions, each catalyzedby a specific enzyme. Although the thermodynamic analysis shown above indicates that only the initial and final statesneed be considered,it is useful to analyzea metabolic pathway to seehow the individual stepsare coupled to each other through the associated free energy changes.The specific metabolic pathway we will examine is anaerobic glycolysis. Anaerobic glycolysis is the sequenceof reactions that metabolizes glucoseinto lactateand alsoproducesATP, the physiological energycurrency.As we have seen,the standardfree energy for the hydrolysis of ATP is quite large and negative so that the hydrolysis of ATP can be coupled to reactionswith an unfavorable free energy change.The sequenceof reactionsinvolved in anaerobicglycolysis is shown in Table 3-1, along with the standardfree energy changes.The standardstatesare as usual for biochemical reactions,namely, pH 7 and I M for reactants,with the concentration of each reactant as the sum of all ionized species.The activity of water is assumedto be unity. (This is somewhat different from Appendixes 1 and 2,wherc ionic strength and, in some cases,the magnesium ion concentration are specified.)

44

APPLICATIONS OFTHERMODYNAMICS

TABLE 3-1. Free Energy Changes for Anaerobic Glycolysis Reaction

AG' (kJimol)

AG (kJ/mol)

Part One Glucose + ATP * Glucose-6-P+ ADP Glucose-6-PS Fructose-6-P Fructose-6-P+ ATP + Fructose-1,6-bisphosphate+ ADP Fructose-1,6-bisphosphate* Dihydroxyacetone-P + Glyceraldehyde-3-P Dihydroxyacetone-P * Glyceraldehye-3-P Glucose+ 2 ATP = 2 ADP + 2 Glvceraldehvde-3-P

-16.7 1.7 -t4.2 23.9

_JJ.J

-2.7 -18.6 0.7

7.5

2.6

2.2

-51.3

6.3

-1.0

Part Two Glyceraldehyde-3-P+ P, + NAD+ + 1,3-Bisphosphoglycerate + NADH l,3-Bisphosphoglycerate+ ADP = 3-P-Glycerate+ ATP 3-Phosphoglycerate* 2-Phosphoglycerate 2-Phosphoglycerate* Phosphoenolpyruvate + H2O Phosphoenolpyruvate + ADP€ Pyruvate + ATP Pyruvate + NADH * Lactate + NAD*

-18.9 4.4 1.8 -3r.7 -25.2

-0.6 1.0 1.1 -23.3 1.9

Glyceraldehyde-3P+ R + 2 ADP€

-63.3

-20.9

Lactate + 2 ATP + I{2O

Source: Adapted from R. H. Garrett and C. M. Grisham, Biochemisfry, Saunders College Publishing, Philadelphia, 1995, pp. 569-597 .

The overall reaction for anaerobicglycolysis is G l u c o s e+ 2 P \ + 2 A D P € 2

L a c t a t e + 2 A T P+ 2 H z O

(3-1)

ln anaerobicmetabolism, the pyruvate produced in the secondto the last step is converted to lactate in muscle during active exercise.In aerobic metabolism, the pyruvate that is produced in the secondto the last stepis transportedto the mitochondria, where it is oxidized to carbon dioxide and water in the citric acid cycle. The reactionsin Table 3-1 can be divided into two parts. The first part produces2 moles of glyceraldehyde-3-phosphateaccording to the overall reaction Glucose+ 2 ATP = 2 ADP + 2 Glyceraldehyde-3-phosphate

(3-2)

Note that this reaction actually requires 2 moles of ATP. However, the secondpart of the cycle produces4 moles of ATP with the overall reaction + Pi + 2 ADP € Lactate+ 2ATP + HrO (3-3) Glyceraldehyde-3-phosphate Since 2 moles of glyceraldehyde-3-phosphateare produced in part one of the cycle, Eq. 3-3 must be multiplied by two and added to Eq. 3-2 to give the overall reaction, Eq. 3-1. The standardfree energies,of course,also must be multiplied by two for the

3.2 METABOLICCYCLES

45

reactions in part two of glycolysis and added to those for the reactions in part one to give the overall standardfree energy change, -124.4 kJ/mol. If the standardfree energies of formation in Appendix 1 are used to calculate the standard free energy changefor Eq. 3-3, avalue of -128.6 kJlmol is obtained.This small differencecan be attributed to somewhat different standard states.Unfortunately, not all of the necessary free energiesof formation are available in Appendix 1 to calculate standardfree energy changesfor all of the reactionsin Table 3-1. The standardenthalpy change for Eq. 3-3 is -63.0 kJ/mol, so that a substantial amount of heat is produced by glycolysis. The standardentropy changecan be calculated from the known standardfree energy and enthalpy changesand is 2201/(mol.K). Thus, both the standardenthalpy and entropy changesare favorable for the reaction to proceed. It is interesting to compare these numbers with those for the direct oxidation of glucose:

Glucose(s) + 6 O2G)= 6 CO,(S)+ 6 HrO(l)

(3-4)

The standardfree energy change for this reaction is -2878.4 kJ/mol; the standardenthalpy changeis -2801.6 kJ/mol; and the standardentropy changeis -259 J/(mol.K). This processproducesa very large amount of heat relative to that produced by the metabolic cycle. This would not be very useful for physiological systems. As we have stressedpreviously, for a single reaction,it is not the standardfree energy change that must be considered in determining whether products are formed, it is the free energy for the particular concentrationsof reactantsthat are present.In order to calculate the free energy changesfor the reactions in Table 3-1, the concentrations of metabolites must be known. These concentrationshave been determined in erythrocytes and are summarized in Table 3-2. The free energy changescalculated with these concentrations,the standardfree energy changes,and Eq.2-35 are included in Table 3-1. The additional assumptionshave been made that theseconcentrationsare valid at298 K, although they were determinedat 310 K, and [NADH]/INAD*I = 1.0 x 10-3. (We have elected to carry out calculations at 298 K where the standardfree energiesare known, rather than at the physiological temperatureof 310 K. This does not alter any of the conclusionsreached.) Consideration of free energy changes,rather than standard free energy changes, produces some interesting changes.The overall standardfree energy change for the first part of glycolysis is +2.2 kJ/mol, whereasthe free energychangeis -51 .3 kJ/mol. In contrast, the standardfree energy change for the secondpart of glycolysis is much more negative than the free energy change. Thus, when considering the coupling of chemical reactions,considerablecare must be exercisedin making comparisons.Of course,as we statedat the beginning, the concentrationsof the intermediatesare of no consequencein determining the overall free energy changes.The reader might wish to confirm that this is indeedthe case. Before we leave our discussionof glycolysis, it is worth addressingthe individual reactions in the cycle. All of the reactions are catalyzedby enzymes. If this were not the case,the reactionswould occur much too slowly to be physiologically relevant.

46

APPLICATIONS OF THERMODYNAMICS

TABLE 3-2. Steady-StateConcentrationsfor Glycolytic Intermediatesin Erythrocytes Metabolite Glucose Glucose-6-phosphate Fructose-6-phosphate Fructose-1,6-bisphosphate Dihydroxyacetone phosphate Glyceraldehyde-3-phosphate 1,3-Bisphosphoglycerate 2,3-Bisphosphoglycerate 3-Phosphoglycerate 2-Phosphoglycerate Phosphoenolpyruvate Pyruvate Lactate ATP ADP P,

Concentration(mM)

5.0 0.083 0.0r4 0.031 0.14 0.019 0.001 4.0 0.r2 0.030 0.023 0.051 2.9 1.85 0.r4 1.0

Source:Adapted from S. Minakami and H. Yoshikawa, Biochem.Biophys. Res.Commun.18,345 (1965).

The first stepis the very favorable phosphorylation of glucose-both the standardfree energy change and free energy change are favorable. The advantageto the cell of phosphorylating glucose is that creating a chargedmolecule preventsit from diffusing out of the cell. Furthermore, the intracellular concenffation of glucose is lowered so that if the concentrationof glucoseis high on the outsideof the cell, more glucosewill diffuse into the cell. The second step, the isomerization of glucose-6-phosphateto fructose-6-phosphate,has a somewhat unfavorable standardfree energy change,but the free energy changeis favorable enough for the reaction to proceed.The next step has a very favorable standardfree energy change, as well as a favorable free energy change, becausethe phosphorylation of fructose-6-phosphateis coupled to the hydrolysis of ATP. This very irreversible step is the commitment by the cell to metabolize glucose, rather than to store it. The next two steps produce glyceraldehyde3-phosphate,the fuel for the secondhalf of glycolysis that produces4 moles of ATP. Both of thesereactionshave very unfavorable standardfree energy changes,although the free energy changesare only slightly positive. This completes the first part of glycolysis. Note that2 moles of ATP have been utilized to produce the final product, glyceraldehyde-3-phosphate.As noted previously, the standard free energy changefor this first part is actually unfavorable, but the free energy changeis quite favorable. The purpose of the secondpart of glycolysis is to convert a substantialportion of the metabolic energy of glucose into ATP. In comparing the standard free energy changeswith the free energy changesof the individual steps,it is worth noting that

3.3 DIRECTSYNTHESISOF ATP

47

the second step, the formation of 3-phosphoglycerate,has a very favorable standard free energy change, yet the free energy change is approximately zero so that this reaction is approximately at equilibrium. This is true of essentially all of the reactions in this part of glycolysis, except for the reaction that produces pyruvate, where both the standardfree energy change and the free energy change are quite negative. Both the overall standardfree energy changeand the free energy changefor the secondpart of glycolysis are quite favorable. This thermodynamicanalysisof glycolysis is a good example of how thermodynamics can provide a framework forunderstanding the many coupled reactionsoccurring in biology. It also illustrates how metabolism is utilized to produce molecules such as ATP that can be used to drive other physiological reactions, rather than converting most of the free energy to heat.

OF ATP 3.3 DIRECTSYNTHESIS As we have seen,the standardfree energy change for the hydrolysis of ATP to ADP and P, is -32.5 kJ/mol, so it seemsunlikely that ATP would be synthesizedbythe reverse of this reaction. The concentrationsof reactantscannot be adjusted sufficiently to make the overall free energy favorable. Yet, we know that ATP is synthesizeddirectly from ADP and P, in mitochondria. For many years,people in this field grappled with how this might happen:Both probableand improbablemechanismswere proposed. In 1961PeterMitchell proposedthat the synthesisof ATP occurreddue to a coupling of the chemicalreactionwith aproton gradientacrossthe membrane(1). This hypothesiswas quickly verified by experiments,and he received a Nobel kize for this work. The enzyme responsiblefor ATP synthesis,ATP synthase,consistsof a protein "ball," which carries out the catalytic function, coupled to membrane-boundproteins through which protons can be transported.The processof ATP synthesisis shown schematicallyin Figure 3-1. Although this enzymeis found only in mitochondria in humans, it is quite ubiquitous in nature and is found in chloroplasts,bacteria, and yeast,among others.The chemiosmotichypothesisstatesthat a pH gradientis established across the membrane by a series of electron transfer reactions, and that ATP synthesis is accompaniedby the simultaneous transport of protons across the membrane. The overall reaction can be written as the sum of two reactions: ADP+P,SATP+HrO zH*out*rH*rn

(3-s)

ADP + P, + z H*out= ATP + n H*,n + HrO The value of n has been determined to be 3 (see Ref. 2). The free energy change for the transport of protons in Eq. 3-5 can be written as LG = 3RZ ln([H*in]/[H*out])

(3-6)

48

OF THERMODYNAMICS APPLICATIONS

HighpH AD P+ P ; AT P

of ATP synthase,E, in mitochondria.The enzyme FIGURE 3-1. Schematicrepresentation structureinside the mitochondriacontainsthe catalyticsites.Protonsare pumpedfrom the outsideof the membraneto the insideasATP is synthesized. (The standardfree energy change for this processis zero since the standardstate for the hydrogen ion is the sameon both sides of the membrane.) At298 K, a pH differential of one unit gives a free energy changeof -17.1 kJ/mol. The actual physiological situation is even more favorable as a membranepotential exists whereby the membrane is more negative on the inside relative to the outside. If we utilize the extendedchemical potential,8q.2-49, an additional term is addedto Eq. 3-6 equal to 3 FY, where 3 is the number of protons transported,F is the Faraday, and Y is the membrane potential. For a membrane potential of -100 millivolts, -29 W would be added to the free energy change in Eq. 3-6. The standard free energy for the synthesis of ATP from ADP and Pt is +32.5 kJ/mol, but we need to know the free energy change under physiological conditions. Although the concentrationsof the reactantsare not known exactly, we can estimate that the ratio of ATP to ADP is about 100, and the concentration of phosphateis 1-10 mM. This makesthe ratio tATPI(tADPllPil) equal to 100-1000. The free energy change at 298 K is LG = 32.5 + RT ln(100-1000) = 32.5 + (11.4-17.l) = 43.949.6 kJ/mol Thus, the coupling of the synthesisof ATP to a modestproton gradient and membrane potential can readily provide the necessaryfree energy for the overall reaction to occur. The principle to be learned from this example is that the coupling of free energies is very general. It can involve chemical reactions only, as in glycolysis, or it can involve other processessuch as ion transport acrossmembranes,as in this example.

3.4 ESTABLISHMENT OF MEMBRANE ION GRADIENTS BY CHEMICAL REACTIONS In discussing ATP synthesis,we have not specified how the electrochemical gradient is established.This involves a complex sequenceof coupled reactionsthat we shall not

OF MEMBRANEION GRADIENTSBY CHEMICALREACTIONS 3.4 ESTABLISHMENT

49

discusshere. Instead, we will discussa casewhere the free energy associatedwith the hydrolysis of ATP is used to establishan ion gradient. The processof signal transduction in the nervous system involves the transpot of Na+ and K* acrossthe membrane. Neuronal cells accumulate K* and have a deficit of Na* relative to the external environment. (This is also true for other mammalian cells.) When an electrical signal is ffansmitted, this imbalance is altered. The imbalance causesa resting membrane potential of about -70 millivolts. This situation is illustrated in Figure 3-2, with some typical ion concentrations. How is the ion gradient established?This is done by a specific ATPase, the Na*A(* ATPase, that simultaneously pumps ions and hydrolyzes ATP. The process can be written as the sum of two reactions: 2 K+oot + 3 Na+,n +

2 K+in + 3 Na+ou,

(3-7)

ATP+H2O=ADP+Pt

2 K+our+ 3 Na+,n+ ATP + HrO i

2 K*,n + 3 Na*oo,+ ADP + Pt

The stoichiometry has been establishedby experimental measurements.The free energy changefor the first step is LG =Rrh{(tNa*oo,l3[K*"]'X[Nu*,n]3[K*oo,]')l +3 F( -z Fy With Y = 70 millivolts, T = 298 K, and the concentrations of Na+ and K* in Figure 3-2, LG = 41.3 kJ/mol. Note that the membranepotential produces a favorable (negative) free energy changefor the fransport of K* and an unfavorable free energy change (positive) for the transport of Na+. We have previously calculated that LG for ATP hydrolysis is -(43.9 - 49.6)kllmol. Therefore, the hydrolysis of ATP is sufficient to es-

Na* = 10 mM K* = 100 mM y =-70 mV

Na+ = 140 mM K*=5mM

FIGURE 3-2. Schematic representation of the Na+ and K+ gradients in a cell. The free energy necessaryfor the creation of these gradients is generatedby the enzymatic hydrolysis of ATP. The Na+/I(* ATPase is designated as E, and typical concenffations of the ions and membrane potential, Y, are given.

50

APPLICATIONS OF THERMODYNAMICS

tablish the ion gradient and accompanyingmembranepotential. This processis called active transport. The coupling of free energiesin biological systemscan be usedto understandmany of the events occurring, and many other interesting examplesexist. However, we will now turn to considerationof protein and nucleic acid structuresin terms of thermodvnamics.

3.5 PROTEIN STRUCTURE An extensive discussion of protein structure is beyond the scopeof this treatise,but it is illuminating to discussa few aspectsof proteins in terms of thermodynamics.Many excellent discussionsof protein structureare available (cf. Refs. 3-5). Proteins,of course,are polymers of amino acids, and the amino acids contain both polar and nonpolar groups. The structuresof the20 common amino acids are given in Appendix 3, and the covalent structure of a polypeptide chain is shown in Figure 3-3. We will first consider the role of nonpolar groups in proteins, that is, amino acid side chains containing methylene and methyl groups and aromatic residues. As a starting point, let us consider the thermodynamics of transferring the hydrocarbons methane and ethane from an organic solvent to water. The thermodynamic parameters can be measured for this process simply by determining the solubility of the hydrocarbonsin each of the solvents.Since the pure hydrocarbonis in equilibrium with each of the saturatedsolutions, the two saturatedsolutions must be in equilibrium with each other. The transfer reaction can be written as the sum of the solubility equilibria: Hydrocarbon (organic solvent) *

(3-8)

Hydrocarbon

Hydrocarbon € Hydrocarbon (water)

Hydrocarbon (organic solvent) = Hydrocarbon (water) The transfer free energy is

HO HO til Irl +HsN- C -C \AMN -C -C-l R1

Aminoterminus

HO

rilttl

N -C -C-

tl

R2HR3HRa

HO

T //a

N -C -C\^ lN -C -C

tl

,\A,\

Carboxylterminus

FIGURE,3-3. A polypeptide chain. The amino acid side chains are representedby Ri, and the amino and carboxyl termini are shown.

51

3.5 PROTEINSTRUCTURE

LGr- Rrh([Xo]/txwl) where Xo and X, are the mole fraction solubilities in the organic solvent and water, respectively. (The use of mole fractions is the appropriate concentration scale in this instance becauseof standard state considerations that we will not deal with here.) Measurement of the temperature dependenceof the transfer free energy will permit A1/ and AS to be determined (Eqs. 2-42 and2-43). The thermodynamic parametersobtained at298 K in severalorganic solventsare summarrzedin Table 3-3. As expected,the free energy changeis unfavorable-hydrocarbons do not dissolve readily in water. However, the energy changeis favorable so that the negative entropy changeis responsiblefor the unfavorable free energy. What is the reasonfor the negative entropy change?It is due to the fact that the normal water structure is broken by the insertion of the hydrocarbon, and the water molecules tend to form a hydrogenbonded structure around the hydrocarbon that is much more ordered than the normal structure of water. Thus, it is the formation of ice-like ordered structures around the hydrocarbon that is the source of the negative entropy change. To make this more relevant to proteins, the free energy of transfer for amino acids from a hydrocarbon-like environment (ethanol) to water has been measured.This is done by determining the solubility of the amino acids in ethanol and water. Similar to before,in both casesthe amino acid in solution is in equilibrium with the solid, so that the transfer free energy can be measured: Amino acid (ethanol)*

(3-e)

Amino acid (solid)

Amino acid (solid) € Amino acid (water) Amino acid (ethanol)€ Amino acid (water) Some of the results obtained are surlmaized in Table 3-4. In order to interpret these results, we must remember that in ethanol the amino acid is uncharged (NHzRCH-cooH), whereasin water it is a zwiuerion (NH{-RCH-coo-). If glycine,

TABLE 3-3. Thermodynamic Parameters for Hydrocarbon Transfer from Organic Solventsto Water at 298 K Transfer Reaction CI{o, benzene to water CHa, ether to water CH4, CCl4 to water C2H6,benzeneto water C2H6,CClo to water

AJ, [caV(mol.K)]

A11,(kcaUmol)

AG, (kcaUmol)

-18 -19 -18 -20 -18

-2.8 -2.4 -2 .5 _) ')

+2.6 + 3.3 +2.9 + 3.8 + 3.7

-r.1

Source: Adapted from W. Kauzmann,Adv. Protein Chem.14, 1 (1959).

52

APPLICATIONS OF THERMODYNAM ICS

TABLE 3-4. Free Energy Changesfor Transferring Amino Acids from Ethanol to Water at29EK Compound Glycine Alanine Valine Leucine Isoleucine Phenylalanine Tyrosine

AG, (kcal/mol)

AG,,ru.

-4.63 -3.90 -2.94 -2 .2 r -t.69 -1.98 -1.78

Source:Adapted from C. Tanford, J. Am. Chem. \oc.U,4240

(kcal/mol) "116r,

**t + 1.69 +2.42 +2.97 +2.60 +2.85 (1962).

which has no side chains, is taken as the standard,then subtractingits transfer free energy from the transfer free energiesof the other amino acids will give the transfer free energy for the amino acid side chains. The overall transfer free energy for all of the amino acids is negative (favorable) because the charged amino and carboxyl groups are solvated by water, a highly favorable interaction. However, the standard free energy changesof transfer for the hydrophobic side chains are all positive, as was seen for the transfer of methane and ethane from an organic solvent into water. What is the relevance of such data for protein structure?These data indicate that the apolar side chains of amino acid side chains would prefer to be in a nonaqueous environment. That is, they prefer to cluster together. This is, in fact, true for most proteins. The hydrophobic groups aggregatetogether on the interior of the protein, forming a hydrophobic core, with the more polar groups tending to be on the outside interacting with water. This important concept was enunciatedby Walter Kauzmann in 1959 (6). Although the interactions associatedwith forming the hydrophobic core are often called hydrophobic bonds, it is important to remember that the driving force for forming a hydrophobic core is not the direct interactions between hydrophobic groups; instead, it is the releaseof water molecules from the ice-like structuresthat surround hydrophobic groups in water. The thermodynamic analysisclearly indicates that the formation of "hydrophobic bonds" is an entropy-driven process.In fact, AFl, for the transfer of methane and ethanefrom water to organic solvent is positive, that is, energetically unfavorable. This means that an increasein temperaturewill tend to strengthen the hydrophobic bonding-1f NI remains positive over the temperature range under consideration. The formation of hydrogen bonds in protein structures is prevalent, and we will now consider the thermodynamics of hydrogen bond formation. Thermodynamic studies have been made of many different types of hydrogen bonds. A few selected examples are summarrzed in Table 3-5. The first is the formation of a hydrogenbonded dimer of acetic acid molecules in the gas phase:

3.5 PROTEINSTRUCTURE

A^r, 2CH3-C/"

(g)::

53

"-\

CHrc*-c/o

(3-10)

,y'--Curtel

\

As expected,both A,F1oand A,S"are negative. The former representsthe energy produced in forming two hydrogen bonds, and the latter is due to two molecules forming a dimer (a more ordered system). Since two hydrogen bonds are formed the enthalpy change associatedwith a single hydrogen bond is about -7 kcaVmol. The secondexample is the dimerization of N-methylacetamide, a good model for hydrogen bonding involving the peptidebond:

o H l l -cH3

2CH3-N-C

cH.l

r'l

--

/-:r\

cH3

( 3 - 11 )

------HN

Il_

C-O

HN

ll

CH:

CHI

This dimenzation was studied in a variety of solvents, and the results in carbon tetrachloride, dioxane,and water are includedin Table 3-5. In carbontetrachloride,the solvent does not compete for the hydrogen bonds of N-methylacetamide, and the results are not very different from those for acetic acid dimeizationin the gasphase,namely, AIlo for formation of a single hydrogen bond is about -4 kcaVmol and ASo is about -11 caV(mol.K). However, in water, which can also form hydrogen bonds with Nmethylacetamide and is present at a concentration of about 55 M, A11ois approximately zero. This indicates that there is not a significant enthalpy difference between the water-N-methylacetamide hydrogen bond and the N-methylacetamide-Nmethylacetamide hydrogen bond. Note that the standardfree energy change in water is 3 kcaVmol, so that the amount of dimer formed is very small even at 1 M N-methvl-

TABLE 3-5. Thermodynamic Parameters for Hydrogen Bond Formationo Reactant

Solvent

cH3cooH, cH3coNH2.

Gas

cc14 Dioxane Hzo

LG'zssftcaVmol)

-8.02 -0.92 0 .3 9 3.1

LH)ss ftcaVmol)

-15.9 -4.2 -0.8 0.0

dStandardstate is 1 M. 'See J. O. Halford, J. Chem.Phys.9,859 (1941). "See I. M. Klotz and J. S. Franzen,J. Am. Chem. \oc.84,3461 (1962).

ASin, [caU(mol.k)]

-26.6 -11 -4 -10

54

APPLICATIONS OF THERMODYNAMICS

acetamide.Dioxane has two oxygens that can accept a hydrogen bond, so that AIlo is only slightly negative. The conclusion reachedfrom thesedata (and considerablymore data not presented) is that the stability of a water-protein hydrogen bond is similar to that of intramolecular protein hydrogen bonds. Therefore, a single intramolecular hydrogen bond on the surface of the protein is unlikely to be a strong stabilizing factor. On the other hand, if the hydrophobic interior of a protein excludes water, a strong hydrogen bond might exist in the interior of a protein. This statementis complicated by the fact that proteins have "breathing" motions; that is, the protein structure continually opens and closes with very small motions so that completely excluding water from the interior may be difficult. The role of the hydrogen bond in stabilizing proteins is still a matter of some debate.The statementis often made that hydrophobic interactions are the primary source of protein stability and hydrogen bonding provides specificity (6). There is no doubt that extended hydrogen-bonded systemsare extremely important structural elements in proteins. Two examples are given in Color Plates I and II, the cr-helix and the Bpleated sheet.The o-helix is a spiral structure with 3.6 amino acids per turn of the helix. Every peptide carbonyl is a hydrogen bond acceptor for the peptide N-H four residuesaway. This structure is found in many different proteins. The B-pleatedsheet is also a prevalent structurein proteins. In the B-pleatedsheet,eachchain is a "pleated" helix. Again, all of the peptide bonds participatein hydrogen bonding, but the bonds are all between chains. rather than intrachain. Finally, we will say a few words about the role of electrostatic interactions in protein structures.The discussionwill be confined to charge-chargeinteractions,even though more subtle interactions involving, for example, dipoles and./orinduced dipoles are important. Many of the amino acids have side chains with ionizable gtoups, so that a protein contains many acids and bases.For example, a carboxyl group can ionize according to the scheme P-COOH=P-COO-+H'

(3-r2)

Here P designatesthe protein. Although not shown, water plays an extremely important role in this equilibrium. Water is a strong dipole and strongly solvatesions, forming a hydration "sheath." Acetic acid is a reasonablemodel for this reaction: The thermodynamic parameterscharacteizingits ionization are AGo = 6.6 kcaVmol, AFlo = 0, and A,So= -22 cal/(mol.K). The negative entropy change is due to the ordering of water moleculesaroundthe ions. Note that the ionization processis thermally neutral. The enthalpy changesassociatedwith the solvation of ions are generally negative but, in this case, are balanced by the enthalpic change associatedwith breaking the oxygen-hydrogen bond. The ionization constants for ionizable groups on proteins are generally not the sameas those of simple model compounds.This is becausethe ionization processis influenced by the chargeof the protein createdby other ionizable groups and, in some cases,by special structural featuresof the protein. This factor can be included explic-

3.5 PROTEINSTRUCTURE

55

itly in a thermodynamic analysis of protein ionizations by writing the free energy associatedwith ionization as the sum of the free energy for the model compound, or the intrinsic free energy, and the free energy of interaction: AGioni-tion = AG ino,nri"* AGint".u.tion

(3-13)

The ionization properties of proteins have been studied extensively, both experimentally and theoretically. For our purposes,it is important to recognize that the strong solvation of ions means that charged groups will tend to be on the outside of the protein, readily accessibleto water. A reasonablequestion to ask is: If there are so many charged groups on proteins, won't they influence the structure simply becausechargedgroups of opposite sign attract, and those with the same sign repel? This is certainly the case-at very high pH a protein becomes very negatively charged and the interactions between negative chargescan eventually causethe native structure to disappear.Similarly, at very low pH, the interactions between positive chargescan causedisruption of the native structure. Conversely,the formation of a salt linkage betweengroups with opposite charges can stabilize structures. We can make a very simple thermodynamic analysis of charge-charge interactions by recognizing that the free energy of formation of an ion pair is simply the work necessaryto bring the ions to within a specifieddistance,a:

AG = zrzre2lDa

(3-14)

Here the z,are the ionic valences,e is the charge on an electron, and D is the dielectric constant.The assumptionof a simple coulombic potentialis agrossoversimplification but suffices for our purposes.A more complete potential would include the ionic environment of the medium, the structure of the ions, and the microscopic structure of the solvent.If the valencesare assumedto be 1, the distanceof closestapproach4 A, and the dielectric constant of water 80, AG is about 1 kcal/mol. Thus, the interaction energy is not very large in water. However, the interior of a protein is more like an organic solvent,and organic solventshave dielectric constantsof about 3, which would significantly increase the free energy of interaction. In fact, salt linkages are rarely found near the surfaceof proteins but are found in the interior of proteins, asexpected. If the free energy of ionic interactionsis given by Eq. 3-14, the enthalpy and entropy can easily be calculatedfrom Eqs.2-42 and2-43. Since only the dielectric constant in Eq.3-14 is temperaturedependent,this gives

A,H= -T"td(LGI T)|dTl - (LGlD)ld(DT)/dTl

(3-ls)

A,S= _d LGtdT = (LGtD)(dDtd7)

(3-16)

In water, both of thesederivativesare negativeso that if AG is negatle, both the entropy and enthalpy changesare positive. The positive entropy changecan be rationalized as being due to the releaseof water of hydration of the ions when the ion pair is formed. The enthalpy changeis a balancebetween the negative enthalpy changefrom bringing the chargescloserand the positive enthalpychangeassociatedwith removing the hydration shell. Since the enthalpy changeis positive, the strength of the ion pair

56

APPLICATIONS OFTHERMODYNAMICS

interaction will increaseas the temperatureis increased-exactly as for hydrophobic interactions. Although this is an abbreviated discussion, it is clear that we know a great deal about the thermodynamics of interactions that occur in proteins. Becauseof this, you might think that we could examine the amino acid sequenceof any protein and predict its structureby looking at the possible interactions that occur and finding the structure that hasthe minimum free energy. This hasbeen a long-standing goal of protein chemistry, but we are not yet able to predict protein structures.Why is this the case?The difficulty is that there are thousandsof possible hydrogen bonding interactions, ionic interactions,hydrophobicinteractions,etc. As we have seen,eachof the individual interactionsis associatedwith small free energy and enthalpy changes-in somecases we cannot even determine the sign. The sum of the positive free energiesof interactions is very large, as is the sum of the negative free energies.It is the difference between the positive and negative free energiesthat determinesthe structure.So we have two problems: accurately assessingthe free energies of individual interactions and then taking the difference between two large numbers to determine which potential structurehasthe minimum free energy.Theseare formidable problems, but significant progresshasbeen made toward achieving the ultimate goal of predicting protein structures.The possibility also existsthat the structurehaving the lowest free energyis not the biologically relevant structure. This may be true in a few casesbut is unlikely to be a problem for most proteins.

FOLDING 3.6 PROTEIN The folding of proteins into their biologically active structures has obvious physiological importance. In addition, understanding the process in molecular detail is linked directly to understanding protein structure. The study of protein folding, and the reverseprocessof unfolding, is a major field of research,and we will only explore a few facets of this fascinating subject.For our discussion,we will concentrateon protein unfolding as this is most easily experimentally accessible.There are many ways of unfolding proteins.When the temperatureincreases,proteins will eventually unfold. From a thermodynamic standpoint, this is because the ZAS term eventually dominatesin determining the free energy change,and we know that the unfolded state is more disordered than the native state at sufficiently high temperatures. Chemical denaturants such as acid, base, urea, and guanidine chloride are also often used to unfold proteins. The role of a neutrally charged denaturantsuch as urea can be understoodin thermodynamic terms by considering the free energy of transfer of amino acids from water to urea, again using glycine as a reference. Some representativefree energiesof transfer for hydrophobic side chains are given in Table 3-6. Note that the free energiesare all negative, so that removing hydrophobic side chains from the interior of the protein into 8 M urea is a favorable process. Protein unfolding can be monitored by many different methods.Probably the most common is circular dichroism in the ultraviolet, which is quite different for native and unfolded structures.Many other spectraland physical methodswork equally well. A

3.6 PROTEINFOLDING

57

TABLE 3-6. Transfer Free Energies for Selected Amino Acids from Water to 8 M Urea at 298 K AG, (kcal/mol)

Amino Acid

+ 0 .1 0 +0.03 -0.28 -0.60 -0.63

Glycine Alanine Leucine Phenylalanine Tyrosine

AGt,ria" .6uin(kcal/mol)

0.0 -0.07 -0.38 -0.70 -0.13

Source'. Adapted from P. L. Whitney and C. Tanford, J. Biol. Chem.237,1735 (1962)

representative plot of the fraction of denatured proteirr, fD, versus temperature is shown in Figure 3-4 for the N-terminal region (amino acid residues6-85) of l" repressor,a protein from l, phage that binds to DNA and regulatestranscription. Results are shown for both no urea and2 M urea. As expected,the protein is less stable in the presenceof 2 M urea. The fraction of protein denaturedcan be written as

7r-[D]t([D] + [A4)= Kt(r + IQ

(3-17)

where D is the concentrationof denaturedspecies,/f is the concentrationof native species,and 1( (= [D]/[M) is the equilibrium constantfor denaturation.This assumesthat the unfolding processcan be characterizedby only two states-native and denatured. For some proteins, intermediatesare formed as the protein unfolds, and a more complicated analysismust be used. Since the equilibrium constantcan be calculatedat any point on the curve, and its temperature dependencecan be measured, the thermodynamic parameters characterizing the unfolding can be determined. The thermodynamics of unfolding can also

1.0 0.8 0.6

o.2 0.0 30

40 T ('C)

FIGURE 3-4. Fraction of denaturedprotein,ftr, for l,o-ss phagerepressorprotein as a function of temperature, T, in aqueoussolution and in 2 M urea. Data from G. S. Huang and T. G. Oas, Biochemistry35, 6113 (1996).

58

APPLICATIONS OF THERMODYNAMICS

conveniently be studied by scanning calorimetry. It is often found that a plot of the logarithm of the equilibrium constantversus IIT is not linear, as predicted,namely, dln Kld(IlT) = -MIIR

(3-18)

This meansthat the enthalpy changeis temperaturedependent,or in other words, there is a large heat capacity difference, ACp, between the native and unfolded states. d LII/dT = LCe

(3-1e)

A typical experimental result for the variation of the equilibrium constant with temperaturefor l, repressoris given in Figure 3-5a. The plot of ln K versus 1/Z goes through a minimum: At higher temperatures,AIl is positive, whereas at lower temperatures,it is negative.For typical chemical reactions,this plot is a straightline! In Figure 3-5b, the standard free energy change is plotted versus the temperature.The protein is most stableat the maximum in the free energycurve, about 15"C. In Figure 3-5c, Alllo and f ASo are plotted versus Z. Note that both A,Ho and A,Soare zero ata specific temperature.The practical implication of the positive value of AC" is that the protein unfolds at both high temperaturesand low temperatures.This unexpectedresult is not confined to l, repressor.Most proteins will unfold (denature) at low temperatures,but in many casesthe temperatureswhere cold denaturationis predicted to occur are below 0oC. Some thermodynamic parameterscharacterizingprotein denaturation are given in Table 3-7 for a few proteins. In all cases,a large positive value of A,Cris observed. Note that at room temperatureand above,AIlo is typically large and positive, as is ASo, so that unfolding is an entropically driven process.Before we leave our discussionof 10

100

5

E50 6 (t

Y tr0

: o

o E

:--50

E 2.8

3.2

3.6

4.0

looo/T(K'1) (a)

4.4

_100 040

040

T ("C) (b)

r fc) (c)

FIGURE 3-5. (a) Plot of the natural logarithm of the equilibrium constant, K, for the denaturation of l,e-as phage repressor protein in aqueous solution versus the reciprocal temperature on the Kelvin scale, l/7. (b) Plot of AGo for the denaturation of l,o-ss phage repressorprotein versus the temperature, Z. (c) Plots of AFlo and Z AS" versus the temperature, T,for the denaturationof l,o-asphagerepressorprotein. Data from G. S. Huang and T. G. Oas, Biochemistry 35, 6173 (1996).

ACIDSTRUCTURES 59 91 NUCLEIC 3-7. Representative Thermodynamic TABLE Denaturation in Aqueous Solution at 298 K Protein Barnase Chymotrypsin Cytochrome-c Lysozyme RibonucleaseA l, Repressor6-s5o

AG" (kJ/mol)

48.9 45.7 37.l 57 .8 27.0 17.7

AI1'(kJ/mol)

301 268 89 242 294 90.4

Parameters

for

Thermal

Protein

AS" U/(mol.K)l ACp[kJ/(mol.K)] 866 746 174 618 896 244

6.9 t4.l 6.8

9.r 5.2 6.0

Source: Adapted from G. I. Makhatadze and P. L. Privalov, Adv. Prot. Chem.47,307 (1995). "SeeG. S. Huang and T. G. Oas,Biochemistry35,6173 (1996).

protein folding/unfolding, one more important point should be mentioned. The transition from folded to unfolded statesusually occurs over a fairly small changein temperature or denaturant concentration. This is becausefolding/unfolding is a highly cooperativeprocess-once it starts,it proceedswith very small changesin temperature or denaturant.Cooperativeprocessesare quite prevalent in biological systems, particularly when regulation of the processis desired;cooperativeprocesseswill be discussedagain in Chapter6.

ACIDSTRUCTURES 3.7 NUCLEIC We will now briefly consider the structure of nucleic acids in terms of thermodynamics. Again, only a few exampleswill be considered.Many more completediscussions are available(3,4,7).Let us start with the well-known structureof DNA. Most DNAs consistof two chainsthat form a double helix. Each chain is a polymer of nucleosides linked by phosphodiesterbonds as shown in Figure 3-6. Each nucleosidecontains a 2ldeoxyribose sugar and a base, almost always adenine(A), thymine (T), cytosine (C), or guanine(G). The phosphodiesterlinkage is through the 5' and3'positions on the sugars.By convention, a DNA chain is usually written so that the 5' end of the molecule is on the left and the 3' on the right. The B form of the DNA double helix is shown in Color Plate III. The chains are arrangedin an antiparallel fashion and form a right-handedhelix. The basesare paired through hydrogen bonds on the inside of the double helix, and as might be expectedthe negatively chargedphosphodiesteris on the outside. Note the similarity to protein structure as the more hydrophobic groups tend to be on the inside and the hydratedpolar groupson the outside.However, DNA is not globular, unlike most proteins,and forms a rod-like structurein isolation. These rods can bend and twist to form very compact structuresin cells. Let us first examine the hydrogen-bonded pairs in DNA. Early in the history of DNA, Erwin Chargaff noted that the fraction of basesthat were A was approximately

60

APPLICATIONS OFTHERMODYNAMICS

I -O- P:O I

o I

I

H2C5''.o

TI;

H

H

z', H

3'

o

-o- PI - o I

o I

HzQ

OH -O-

I

P:O

I

o HzQ-.O OH I

FIGURE 3-6. Structuralformulaof partof a DNA/RNA chain.In DNA, thebasesareusually A, T, C, andG. In RNA, the T is replacedby U, andthe 2'-H belowtheplaneof the sugaris by OH. replaced equal to the fraction that were T in many different DNAs. Similarly, the fraction of basesthat were G was equal to the fraction that were C. This finding was important in the postulation of the double helix structure by JamesWatson and Francis Crick. As shown in Figure 3-7, thesebasesform hydrogen-bondedpairs, with the A-T pair forming two hydrogen bonds and the G-C pair forming three hydrogen bonds. These are not the only possible hydrogen-bondedpairs that can be formed between the four bases.Table 3-8 gives the thermodynamic parametersassociatedwith the formation of various hydrogen-bonded pairs in deuterochloroform. Instead of thymine, uracil has been used-this is the base normally found in ribonucleic acids and differs only by a methyl group in the 5 position from thymine. It is clear that the A-U pair is favored over the A-A and U-U pairs, and that G-C is favored over G-G and C-C although the data are not very precise. There is no simple explanation for this strong preference-it must be due to more than hydrogen bonding, perhapsan electronic effect within those specific pairs. Furthermore, the preference for A-T in the former three pairs is due to a more favorable enthalpy change.Based on this observation,the

3 . 7 N U C L E I CA C I DS T R U C T U R E S

61

FIGURE 3-7. Formationof thehydrogen-bonded basepairs,A-T andG-C, in DNA. In RNA, U is substitutedfor T. All of the baserinss arearomatic. entropy changefor the latter three pairs has been assumedto be the samein order that A,H" can be calculated. In water solution,the hydrogenbonding betweenbasesis considerablyweakerthan in organic solvents becauseof the competition for hydrogen bond formation with water. The assumption is that the base pair hydrogen bonds are sufficiently shielded

TABLE 3-8. ThermodynamicParametersfor BasePairs in Deuterochloroformat 298 K Base Pair A_A U-U A_U C_C G_G G-C

AG'(kcal/mol)

-0.61 -1.01 ., '71 - L . I L

-1.91 -4.1 to -5.4 -5.4 to -6.8

A,H" (kcal/mol)o

_4.0 _4.3 -6 .2 -6.3 (-8.5to -10) (-l0.0to-11.5)

AS" [call(mol.k)]

-11.4 -11.0 -l 1.8 -15 (-15) (-15)

Source: Adapted from C. R. Cantor and P. R. Schimmel, Biophltsical Chemistry,W. H. Freeman,New York, 1980, p. 325. Data source: Y. Kyogoku, R. C. Lord, and A. Rich, Biochim. Biophys. Acta 179, lO ( 1969). o The AII" valuesin parentheses have been calculatedassumingASo= -15 call(mol.K).

62

AppLtcATtoNS oF THERMoDyNAMtcs

TABLE 3'9. Association Constants and Standard Free Energy Changes for Base Stackingin H2O at 298 K Base Stack

A_A T_T C_C T_C A_T A_C G-C

K (molal-t)

AG'(kcal/mol)

t2 0 .9 r 0.91 0 .9 1 3to6 3to6 4to8

-1.50 +0.06 +0.06 +0.06 -0.70to -1.10 -0.70to -1.10 -0.80 to -L20

Source: Adapted from T. N. Solie and J. A. Schellman,J. Mol. Biol.33,6 (1968).

from water so that the structure is very stable. However, we know that this is not entirely coffect as some"breathing" motions occur, so someexposureto water must also occur. Factors other than hydrogen bonding must contribute to the stability of the double helix. A considerationcomesinto play in nucleic acid structuresthat is not a major concern in proteins, namely, the interactionsbetween the aromatic rings of the bases.The planesof the baseslie over one another so that the rcelectronsinteract; that is, the rings are attractedto each other. This "stacking" reaction can be studied in model systems by measuring the equilibrium constant for the interaction of nucleosidesin water, where the hydrogenbonding betweennucleosidesis negligible. Some representative equilibrium constantsfor dimer formation are given in Table 3-9. The interactionis quite weak, although it appears that purine-purine interactions are stronger than purine-pyrimidine interactions, which in turn are stronger than pyrimidine-pyrimidine interactions.This is the order expectedfor n electron interactions.The enthalpy changeassociatedwith theseinteractionsis somewhatuncertainbut is definitely negative. A value of -3.4 kcaVmol has been obtainedfor formation of an A-A stack(8). Is the "stacking" interaction entirely due to rcelectron interactions,or is the solvent involved, as for hydrophobic interactionsin proteins?The "stacking" interactionhas beenfound to be solvent dependent,with water being the most favorable solvent. This suggeststhat hydrophobic interactionsmay be involved. However, you should recall that hydrophobic interactions are endothermic, and entropically driven by the release of orderedwater moleculesaroundthe noninteractinghydrophobic groups.The most likely possibility is that both ru electron interactions and hydrophobic effects play a role in basestacking.

3 .8

DNA M E LTIN G

One of the reasonsfor wanting to understandthe interactionsin the DNA molecule is to understandthe stability of DNA since it must come apart and go togetheras cells

3 . 8 D N AM E L T I N G

63

reproduce.One way to assessthe stability of DNA is to determine its thermal stability. This can readily be studied experimentally becausethe ultraviolet spectraof stacked bases differ significantly from those of unstacked bases.Thus, if the temperature is raised,the spectrumof DNA changesasit "melts." This is shown schematicallyin Figure 3-8. Becauseof the relative stability of the hydrogen bonding, the A-T regions melt at a lower temperaturethan the G-C regions. The melting of AT and GC polymers can be measured,and as shown in Figure 3-8 the AT structuremelts at a lower temperature than a typical DNA, and the GC structure at a higher temperature. The temperature at which half of the DNA structure has disappearedis the melting temperature, 7*. If the transition from fully helical DNA to separatedfragments is assumed to be a two-state system, then thermodynamic parameterscan be calculated, exactly as for protein unfolding. Thus, the equilibrium constantis 1 (AG' = 0) at Zr. Determination of the temperaturedependenceof the equilibrium constantpermits calculation of the standardenthalpy and free energy changesfor the "melting" process. The real situation is more complex than shown, as DNA may not melt in a well-defined single step, and mixtures of homopolynucleotides can form structures more complex than dimers, but a more detailed consideration of thesepoints is beyond the scopeof this text. However, it should be mentioned that thermal melting of singlestrandedhomopolymersis a useful tool for studying stacking interactions:As with DNA, the spectrum will change as the basesunstack. The thermodynamic interpretation of theseresultsis complex (cf. Ref. 3). From a practical standpoint,many of the above difficulties can be circumvented by the use of model systemsto determinethe thermodynamicsof adding a basepair to a chain of nucleotides.Thus, for example,an oligonucleotideAn could be addedto Tn,

0.0

20

40

60

80

100

r fc) FIGURE 3-8. Hypothetical melting curves for double helix structures of poly d(AT), DNA(containing A-T and G*C basepairs), and poly d(GC). The fraction of chains not in the double helix structure,A, is plotted versus the temperature.The melting temperatures,fm, are indicated by the dashedlines. The shape of the curves and the Zm values are dependent on the length of the chains and their concentrations.This drawing assumesthat the concentrations and chain lengths are comparable in all three cases.

64

APPLICATIONS OFTHERMODYNAMICS

and a complex of AnTn formed. The same experiment can be done with Annr and Tn+I. If AGo is determined for the formation of each of the complexes, then the difference between the two standardfree energiesof formation gives the standard free energy change for the formation of one A-T pair. The temperaturedependence of this free energy difference can be used to obtain the standardenthalpy and entropy changesfor formation of a single A-T pair. With knowledge of the thermodynamic parametersforthe formation of single A-T and G-C pairs within a chain, the question can be asked: Does this permit calculation of the thermodynamic stability of a DNA of a given composition?Regrettably,this is not the case.The stability of a given DNA cannotbe predictedsimply by knowing the fraction of G-C pairs in rhe DNA. Remarkably, if the effect of the nearestneighbors of each base pair is taken into account, a reasonableestimateof the thermodynamic stability of DNA can be obtained. This was discovered by studying many different complementary strands of short DNAs and looking for regularities(9,10). The rationale for this procedurelies in the importance of hydrogen bonding between each pair and the stacking interactions with its nearest neighbors. Ten nearest-neighborparameterssuffice for determining the thermodynamic stability of a given DNA sequence,with one additional assumption,namely, the thermodynamic parametersfor initiating DNA structure are different from those for adding hydrogen-bondedpairs to an existing chain. This takes into accountthat getting the two chains together and forming the first basepair is more difficult than adding base pairs to the double helix. Moreover, the G-C hydrogenbonded pair nucleatesthe double helix formation better than an A-T pair since it is more stable.In thermodynamicterms,this can be written as

\G" = AG'(initiation) + IAG"(nearest neighbors) (3-20) where the sum is over all pairs of nearestneighbor interactions.The thermodynamic parameters necessary to calculate the nearest-neighbor interactions are given in Table 3-10, along with those characteizing the double helix initiation. The best fit of the data on model systems utilizes two additional parametersthat are usually only small corrections;this refinementwill not be consideredhere (cf. Ref. 9). A simple example will illustrate how these data can be used. Consider the reaction

5'-A-G-C-T-G-3' | 5'-C-A-G-C-T-3'

--

5'-A-G-C-T-G-3' 3',-T-C-G-A-C-5'

The sum of the nearest-neighborstandardfree energy changesassociatedwith the formation of the five basepair DNA from Table 3-10 is

3.8 DNA MELTING

TABLE

3-10. Thermodynamic

Parameters

for Determination

AG" (kcaVmol)

DNA Pair

of DNA Stability'

ASO MI" (kcaVmol) [cal/(mol'K)]

-r.4

-8.4

-23.6

5,-A.T 3,-T-A

-0.9

-6.5

-18.8

5,-T-A 3,-A-T

-0.8

-6.3

-18.5

5,-A-C 3'-T-G

- 1.8

-8.6

-23.0

5,-C-A 3,-G-T

-t.6

-1.4

-19.3

5,-A-G 3,-T-C

-1.3

-6.1

-16.1

5,-G-A 3,-C-T

-1.1

-l.t

-20.3

5,-C-G 3,.G-C

-2.5

-10.1

-25.5

5,-G-C 3,-C-G

-2.5

- l l .r

-28.4

5,-C-C 3'-G-G , . 4f(a^4{. Initiation,oneor-inoreGC hydrogenbondpairs

-2 .1

-6.7

-15.6

+1.8

0.0

-5.9

Initiation,no GC hydrogenbondpairs

+2.7

0.0

-9.0

5,-A-A 3,-T-T

'7.q(zs

35,3555 Biochemistry Source:Adaptedfrom J. SantaluciaJr., H. T. Allawi, andP. A. Seneviratne, (1996). "pH7.0,1M NaCl,298K.

)AG'=

5'-A-G + 5'-G{ T-C-5,

+ 5'-c-T + 5'-T-G

C.G-5,

G-A-5,

A-C-5,

: -1.3 - 2.5 - 1.3- 1.6-- 4.7 kcal/mol The standardfree energy for initiation is 1.8 kcal/mol so that the standardfree energy changefor formation of this DNA fragment is -4.9 kcaVmol. Similarly, NIo = -30.7 kcaVmol and ASo = -85.8 caV(mol.K). The temperaturedependenceof the standard free energy changecan be calculated by assumingthe standardenthalpy changeis independentof temperature.This relatively simple procedure permits the thermal stability of any linear DNA to be calculated to a good approximation. Knowledge of the thermal properties of DNA fragments is important both physiologically and practically. Knowing the stability of DNA obviously is of interest in understandinggeneticreplication. From a practical standpoint,knowing the stability of DNA fragments is important in planning cloning experiments. DNA probes must

66

APPLICATIONS OFTHERMODYNAMICS

be used that will form stable duplexeswith the target DNA. The temperatureat which the duplex becomes stable can be estimated by calculating the melting temperature, Z-, that is, the temperatureat which half of the strandsare in the double helix conformation. The equilibrium constant for formation of the double helix duplex is

K= [D]/[S]'

(3-21)

where D is the helical duplex and S is the single strand. If the total concentration of the oligonucleotide is Cs, the concentration of single strands atT^is Cd2 and that of the duplex is Cd4.If we insert theserelationshipsinto Eq. 3-21, we seethatK= IlCo atT*.If we insert the relationshipbetweenthe equilibrium constantand the standard free energy change,we obtain

LGo/RTn= ln Co

LII"/RT.-

(3-22)

AS"/R = ln Co

If Eq. 3-22 is solved for 7-, we obtain

T^= Ni"I(AS'+ R ln Cs)

(3-23)

For the caseunder discussion,if Co = 0.1 mM, Z* - 295 K. In this brief discussion,we have neglectedthe fact that DNA is a polyelectrolyte due to the negatively chargedphosphodiesterbackbone.The polyelectrolyte nature of DNA meansthat the ionic environment,particularly positively chargedions, strongly influencesthe structureand behavior of DNA. Normally a negatively chargedpolymer such as DNA would exist as a rod, becauseof the chargerepulsion.However, we know that DNA is packagedinto a small volume in cells. This involves the twisting of the double helix, the formation of loops, etc. Interactions with metal ions and positively charged proteins are necessaryfor this packaging to occur. Interested readers should consult more complete descriptions of nucleic acids for information on this interestingsubject(3,7).

3.9

RNA

In principle, the structureof RNA can be discussedexactly as the discussionof DNA. The principles are the same,hydrogen bonding between bases,stacking interactions, and hydrophobic interactions determine the structure. Of sourse, a ribose is present, rather than deoxyribose, and uracil is substituted for thymine. In addition, several modified basesare commonly found in RNAs. Unfortunately, understandingand pre-

3.9 RNA

67

dicting RNA structures is more complex than for DNA. Ultimately, this is because there ire several quite different biological functions for RNA and thus several quite different types of RNA. Generally, RNAs do not form intermolecular double helices. Instead,double helicesareformed within an RNA molecule.Thesecan be loops, hairpins, bulges,etc. Some idea of the diversity of structuresthat can be formed is shown in Figure 3-9, where an RNA molecule is shown, along with the different types of structuresthat can be formed.

UG UC

tandem mismatch U:: \ i n te rnalloop_

g:i

\

:..1;X

n1;t""n\ 14 X:0 P5a'0-i l-g \

A

\

P5c

\

G -

A

A

g

A "gtE*tt

I

ou-u 9-9

u.c a-9

P4 {?:?""ti?''oo-x rJJl^;iluu,.l F-f;,t^ i-x G.U U ;-AA " c G A A C

P6

g:e G.U

P5b 3:3, A-u

R-€ iood

t

internal ---

loops

---A

A^- ,.c 9'Y

'-lttttttt

Gcuuucu

\ -qi", pro'Er

G,

junction 'l -n-c-u o n ^'^-i e ^-Y i-- c-' I

AGA A A GA

\hairPin

(i.rr E#Lj-1;J

pr B::

t/ Un

; ' e . uA

I

G 3:3 P3 A

II

a _c l : 3

U_A

II i'-? G.u I

I

c-c

q-l

S ;g_ c-cP6b

lI ! - l4.-l!21 . l

u

I

A-U--UU

u

I

G

'x-A_U " c-G

I

rou t ouoX H-Hru"

,3-8u "

VI H-O U-A

c_g

A-u ,-U-a

| I

g.!

H'9

G-C

\

t_e\

G

AA A-U G-C C-G^ G.U^

u.

I

\ c -.l* G

coaxial

stack

G.U U-A G.U A-U c-c

8 :X P8 U-A c-c U.G U-A u -A UU G.U AcG

8 :8 pz G-C U_A GA AA

FIGURE 3-9. Proposed secondary structure of the group I intron of mouse derived Pneumocystis carinii. The areas indicate secondary structures within the intron, including base-pairedhelices.Reproducedfrom J. Santalucia Jr. and D. H. Turner, Biopolymers 44,309 (1997). Copyright @ 1991Biopolymers. Reprintedby permissionof John Wiley & Sons,Inc.

68

AppLtcATtoNS oF THERMoDyNAMtcs

As with DNA, the RNA structurescanbe predictedreasonablywell by considering nearest-neighborinteractions.However, becauseof the diversity of the structures,the models are more complex. Nevertheless,quite reasonableRNA structurescan be predicted using the thermodynamicsderivedfrom simple systems.We will not delve further into RNA structureshere, except to say that the principles of model building have been developed sufficiently here so that the interestedreader can proceed directly to currentliteratureon this subject(11-14). As with DNA, metal ions play an important role in the biological packagingof RNA. Finally, the interactionsbetween RNA and DNA are of obvious physiological relevance.These have been studied quite extensively, but they are not nearly as well understood as the interactions between DNA fragments and within RNA molecules. Simple models are not yet available to calculate the properties of DNA-RNA structures.At this point we will leave our discussionof thermodynamics of biological systemsalthough many more interestingexamplescould be discussed.

R E F E RE NCE S l. P. Mitchell,NatureI9l, 144(1961). 2. T. G. DeweyandG. G. Hammes,"/. (1981). Biol. Chem.256,8941 3. C. R. CantorandP. R. Schimmel,BiophysicalChemisfry, W. H. Freeman, New York, 1980. 4. L. Stryer,Biochemistry,4th edition,W. H. Freeman, New York, 1995. 5. C. Brandenand J. Tooze,Introductionto Protein Structure,2nd edition,Garland Publishing,New York, 1999. 6. W. Kauzmann, Adv.ProteinChem.14,I (1959). 7. J. D. Watson,N. H. Hopkins,J. W. Roberts,J. A. Steitz,andA. M. Weiner,Molecular Biologyof theGene,4thedition,Benjamin-Cummings, MenloPark,CA, 1987. 8. K. J. Breslauer andJ. M. Sturtevant, Biophys.Chem.7,205 (1917). 9. J. Santalucia,Jr.,H. T. Allawi, andP. A. Seneviratne, Biochemistry 35, 3555(1996). 10. P.N. Borer,B. Dengler,I. Tinoco,Jr.,andO. C. Uhlenbeck,,/. MoI.Biol.86,843(1974). 11. I. Tinoco,Jr.,O. C. Uhlenbeck, andM. D. Levine,Nature230,362(1911). 12. J. Santalucia,Jr. andD. H. Turner,Biopolymers44,309(1997). 13. T. Xia, J. Santalucia,Jr.,M. E. Burkard,R. Kierzek,S. J. Schroeder, X. Jiao,C. Cox, andD. H. Turner,Biochemistry 37, 14719(1998). 14. M. Burkard,D.H.Turner,andI. Tinoco,h.,TheRNAWorld,2ndedition,ColdSpring HarborLaboratoryPress,Cold SpringHarbor,NY, 1999,p.233.

PRO B LE M S 3-1. A. Glutamineis an importantbiomoleculemadefrom glutamate.Calculatethe equilibrium constant for the reaction Glutamate+ NH3 € Glutamine + HrO

PROBLEMS

69

Use the standard free energies of formation in Appendix 1 to obtain the standardfree energy change for this reaction. Under physiological conditions, the concentrationof ammonia is about 10 mM. Calculate the ratio [glutamine]/[glutamate]at equilibrium. (By convention,the concentration of water is set equal to 1 since its concentration does not change significantly during the courseof the reaction.) B. Physiologically, glutamine is synthesizedby coupling the hydrolysis of ATP to the abovereaction: Glutamate + NH? + ATP *

Glutamine + ADP + P,

Calculate AGo and the equilibrium constantfor this reaction under standard conditions (see Appendix 1). Assume that NH, and Pt are maintained at about l0 mM and that [ATPI/IADP] = 1. What is the ratio of glutamateto glutamine at equilibrium? What ratio is neededto convert glutamateto glutamine spontaneously,that is, to make AG < 0? Do all calculationsat I atmosphereand 298 K. Although this is not the physiological temperature,the resultsare not significantly altered. 3-2. Adipose tissuescontain high levels of fructose. Fructose can enter the glycolytic pathway directly through the reaction Fructose+ ATP €

+ ADP Fructose-6-phosphate

Assume that the standard free energy change for this reaction with the same standardstateused in Table 3-1 is -17.0 kJ/mol. If fructoseis substitutedfor glucosein "glycolysis," what would the overall reactionbe for the conversion of fructose to 2-glyceraldehyde,Part 1 of glycolysis? What would the overall reaction be for the complete metabolic cycle? Calculate AG' and AG for these two reactions.Assume the concentrationof fructose is 5.0 mM, and the concentrationsof the other metabolitesare as in Table 3-2. 3-3. Glucose is actively transportedinto red blood cells by coupling the transport with the hydrolysis of ATP. The overall reaction can be written as ATP + HrO + n Glucose(outside)3 n Glucose(inside)+ ADP + P, If the ratio of [ATP]/[ADP] = 1 and [P1J= 10 rnM, what is the concentration gradient, [glucose(inside)]/[glucose(outside)], that is establishedat 298 K? Calculate this ratio for n = I, n = 2, and n = n. Does this suggesta method for determining the value of n? Use the value of AG" for the hydrolysis of ATP in Appendix 2 for thesecalculations. 3-4. Derive the equation for the temperaturedependenceof the standardfree energy change for protein denaturation when AC" is not equal to zero. As the starting temperature in the derivation, use the higher temperature at which AG' = 0.

70

APPLICATIONS OF THERMODYNAMICS

Hint: The easiestway to proceed is to calculate the temperaturedependenceof AFloand A,So.Theserelationshipscan then be combined to give the temperature dependenceof AGo. The parametersin the final equation should be AC", the temperature,T,the temperatureat which AGo = 0, f_, and the enthalpychange at the temperaturewhere AGo = 0, LH, . 3-5. Specific genesin DNA are often searchedfor by combining a radioactiveoligonucleotide,O, with the DNA that is complementaryto a sequencein the gene being sought.This reactioncan be representedas O + DNA € Double strand For suchexperiments,the concentrationof the oligonucleotideis much greater than that of the specific DNA sequence.Assumethat the probe is S'-GGGATCAG-3'. A. Calculatethe equilibrium constantat29SK for the interactionof the probe with the complementaryDNA sequenceusing the parametersin Table 3-9. B. Calculatethe fraction of the DNA presentasthe double strandformed with the probe if the concentrationof the probe is 1.0x 10+ M and the temperature is 298K. C. Find the melting temperaturefor this double strandif the concentrationof the complementaryDNA sequenceis 1.0 nM. The melting temperaturein this caseis when [double strand]/[DNA] = 1. 3-6. For an electrochemicalcell (e.g.,a battery),the reversiblework is -nFE, where n is the number of moles of electronsinvolved in the chemical reaction, F is the Faraday,and S isthereversible voltage.This relationshipis useful for considering coupled oxidation-reduction reactionsin biochemical systems. A. Use this relationshipand Eq. 2-49 to derive an equationrelating the free energy changefor the reaction and the voltage of the cell. Your final equation should contain the standard free energy change for the reaction, the concentrationsof the reactants,and the electrochemicalvoltage,as well as constants. B. The voltageat equilibrium is usually designatedas 5". How is this related to the standardfree energy change for the reaction? How might the equilibrium constant for a biochemical reaction be determinedfrom voltase measurements?For the reaction Malate + NAD'€

Oxaloacetate+ NADH + H'

the voltage at equilibrium is -0.154 volts at298 K. Calculatethe equilibrium constant for this reaction. (F = 96,485 coulomb/mole and n - 2 for this reaction.)

-

CHAPTER4

GhemicalKinetics 4.1 INTRODUCTION Thermodynamicstells us what changesin statecan occur, that is, the relative stability of states.For chemical reactions,it tells us what reactionscan occur spontaneously. However, thermodynamics does not tell us the time scalefor changesin state or how the changesin state occur; it is concernedonly with the differences in the initial and final states.In terms of chemical reactions,it does not tell ts how the reaction occurs, in other words, the molecular interactions that take place as a reaction occurs. For biological reactions,the rates are critical for the survival of the organism, and a primary interest of modern biology is the molecular events that lead to reaction. The study of the rates and mechanismsof chemical reactionsis the domain of chemical kinetics. Thermodynamicsprovides no intrinsic information about mechanisms. Many examplesexist with regard to the importanceof the time scalefor chemical reactions.For thoseof you having diamondjewelry, you may be unhappyto know that the most stable state of carbon under standardconditions is graphite. So as you read this, your diamond is turning to graphite, but fortunately the time scale for this conversion is many hundreds of years. If graphite is more stable, why were diamonds formed? The answer is that the formation of diamonds did not occur under standard conditions: It is well known that at very high temperaturesand pressures,graphite can be converted to diamond. In the biological realm, one of the most critical reactions is the hydrolysis of ATP to ADP and P,. Thermodynamics tells us that the equilibrium lies far toward ADP and P,. Yet solutionsof ATP are quite stableunder physiological conditionsin the absenceof the enzymeATPase.A small amount of this enzymewill causerapid and almost complete hydrolysis. Virtually all metabolic reactionsoccur much too slowly to sustainlife in the absenceof enzymes.Enzymesserveas catalysts and causereactions to occur many orders of magnitude faster. Studies of the rates of hydrolysis under varying conditions allow us to say by what mechanism the reaction may occur, for example, how the substratesand products interact with the enzyme. Understanding the mechanism of biological processesis a researcharea of great current interest. In this chapter, we will be interestedin understandingsome of the basic principles of chemical kinetics, with a few examplesto illustrate the power of chemical kinetics. We will not delve deeply into the complex mathematicsthat is sometimesnecessary, nor the many specializedmethods that are sometimesused to analyzechemical kinetics. Many treatisesare availablethat provide a more completediscussionof chemical kinetics (1-4). Becausethe examplesusedwill be relatively simple, they will not nec71

72

CHEMICALKINETICS

essarily involve biological processes.The principles discussed,however, are generally applicable to all systems.The background presentedhere should be sufficient to get you started on utilizing chemical kinetics and reading the literature with some comprehension.The key three conceptsthat will be discussedfirst are rates of chemical reactions,elementaryreactions, andmechanismsof chemical reactions. As a simple illustration, considerthe reaction of hydrogen and iodine to give hydrogeniodide in the gasphase: H.,+In=2HI

(4-r)

A possiblemechanismfor this reaction is for hydrogen and iodine to collide to produce hydrogen iodide directly. Another possible mechanismis for molecular iodine to first dissociateinto iodine atoms and for the iodine atoms to react with hydrogen to producehydrogeniodide. Thesepossibilitiescan be depictedas

'*.9.+€-H

2n

O

-+ g.E-*g;*g_+g g 6

Thesetwo possiblemodesof reactionarequitedistinctin termsof the chemistryoccurring and are examplesof two different mechanisms.We can also write these in a moreconventionalmanner: mechanisms 1. H" + I, = 2 HI 2.

I, = 2I

2I +Hr= Hr+Ir=

Elementarystep and balancedchemicalreaction (4-2) ElementarysteP

steP 2 HI Elementary 2 HI

(4-3)

Balancedchemicalreaction

As indicated, the individual stepsin the mechanism are called elementary steps, and they must always add up to give the balanced chemical reaction. In the first mechanism, the elementary reaction and balanced chemical reaction happen to be the same, but they are conceptually quite distinct, as we shall amplify later.

4.2 REACTIONRATES

73

A very important point to remember is that a balancedchemical reaction gives no information about the reaction mechanism.This is not so obvious in the above example, but considerthe reaction

3 F e 2 *+ H C r o o + J H + = - 3 F e 3 + + C t ' * + 4 H r o This clearly cannot be the reaction mechanism, as it would require the 11 reactantsto encounter each other simultaneously, a very unlikely event.

RATES 4.2 REACTION The rate of a chemical reaction is a measureof how fast the concentration changes.If the concentration changesan amount Ac in time interval Lt, the rate of the reaction is LclLt.If the limit of smaller and smallertime intervalsis taken,this becomesdcldt.If we apply this definition to Eq.4-1, startingwith H, and I, as reactants,the rate could be written as -dlHrl/dt, -dlIrl/dt, or +d[Hllldt.For every mole of hydrogen and iodine consumed, 2 moles of hydrogen iodide are formed, so that this definition of the rate does not provide a unique definition: The value of the rate dependson which reaction component is under consideration. The rate of appearanceof HI is twice as great as of H, and Ir. This ambiguity is not convenient,so a conventhe rate of disappearance tion is used to define the reaction rate, namely, the rate of changeof the concentration divided by its coefficient in the balanced chemical reaction. This results in a unique reaction rate, R, for a given chemical reaction.By convention,R is always positive. For the caseunder consideration.

- 2 d l H ll rp\ -- _ d [H 2 ] - _ d T 2 l -r dt dt dt Considera more complex chemical equation: 2 N2O5-+ 4 NO, + O, In this casethe reactionrate is

D

t(=-

1d[N2O51 1d[NO2] 2dt4dtdt

dlOzl

Measuring the concentrationas it changeswith time can be very difficult, and some of our greatestadvancesin understandingchemical reactions have resulted from the developmentof new techniquesfor measuringthe ratesof chemical reactions,particularly the rates of very fast reactions. The most convenient method of measuring reac-

74

C H E M I C A LK I N E T I C S

tion ratesis to mix the reactantstogetherand to measurethe subsequentchangein concentrationscontinuously using a spectroscopictechniquesuch as light absorption.It is sometimesnot possible to find a physical property to monitor continuously, so it may be necessaryto stop the reaction and measurethe concentrationchemically or throughradioactivetracers.We will not dwell on this point, exceptto stressthat measuring the rate of a chemical reaction may not be trivial. Once a method has been establishedfor determining the reaction rate, the next step is to measurethe dependenceof the reactionrateon the concentrationsof the reactants. The dependenceof the rate on the concentrationsis called the rate law. The rate law cannot be predictedfrom the balancedchemical equation.It must be determinedexperimentally. Afew exampleswill serveto illustratethis point. Considerthe hydrolysis of ATP catalyzedby an ATPase.The overall reactionis

ATP + ADP + Pi

(4-4)

and the observeddependenceof the rate on concentrationof ATP and enzymein some casesis

R-

k[Eo]

(4-s)

1 + K_/[ATP]

where [Eo] is the total enzymeconcentrationand k and K,n are constants.The rate law certainly cannot be deducedfrom the overall chemical reaction.In fact, the rate law is not the samefor all ATPases.Many different ATPasesare found in biological systems,and they do not all hydrolyze ATP by the samemechanism.As anotherexample, considerthe simple reaction

Hr+Brr-+ 2 HBr

(4-6)

The experimentallydeterminedratelaw is

R_

ft[H"] [Br,]t /2

(4-7)

1 + RHBrI/[Brr]

As a final example,considertheredoxreaction 5 Br-+ BrOj + 6 H' -+ 3 Br, + 3 HrO

(4-8)

The observedratelaw is R = ft[Br]tBrOJl[H*]2

(4-e)

4.3 DETERMINATION OF RATE LAWS

75

These examples should emphasize the futility of attempting to predict the rate law from the balancedchemical equation. The exponentof the concentrationin the rate law is the reaction order with respect to that component(i.e., first order, secondorder, etc.).In somecases,such as Eq.4-9, the concept of reaction order has a simple meaning. The rate law is first order with respect to Br- and BrOJ, and second order with respect to H*. For more complex rate laws such as Eqs. 4-5 and 4-7, the conceptof reactionorder cannotbe used for all of the concentrations,exceptin limiting conditions.For Eq. 4-7,the rate law is l/2 order with respect to Br, at high concentrations of Br, and/or low concentrations of HBr. The lowercasefr's in the rate laws are called rate constants.The dimensionsof rate constants can be deduced from the rate law by remembering that the rate is usually measuredas IWs. In Eq. 4-9, k, therefore,has the dimensionsof M-3 s-l. The capital K's in the above equationsalso are constants,but they are combinationsof rate constantsrather than individual rate constants.

4.3

DETERMINATION OF RATE LAWS

Many different methods exist for determining rate laws. Only a few methods are consideredhere. With the routine use of computers,numerical integration of the differential equationsand simultaneousfitting of the data are possible.However, as with any experimental approach,it is best to first fit the data to simple models before embarking on complex computer fitting. Computer programs will always fit the data,but it is important to be sure that the fit is a good one and the proposed rate law makes sense. Probably the simplest and still often used method is the determination of initial rates. With this method, the rate of the reaction is measuredat the very beginning of the reaction under conditions where the decreasein the concentrationsof the reactants is so small that their concentrationsin the rate law can be assumedto be the starting concentrations.In practice, this means that the concentrations should not change by more than a few percent.To illustrate the method,let's assumethe rate law is R = k(c)o(cz)b

(4-10)

If cr is held constantand the initial rate is measuredfor different concentrationsof c1, the coefficient a can be determined.For example, if the concentrationof c, is doubled and the rate increasesby a factor of four, a must be equal to two. The sametype of experimentcan be done to determineb, namely, c, is held constant and cris varied. The rate constant can readily be calculated, once the coefficients a and b areknown, from the relationshipk - RlIk)"1cr)b7. The determinationof initial ratesis especially useful for studying enzymaticreactionsas we shall seelater. One of the most useful methods for determining the rate law is integration of the rate equation to obtain an analytical expression for the time dependenceof the concentrations.The analytical equation is then compared with experimental data to seeif it accurately describesthe time dependenceof the concentrations.As mentioned pre-

CHEMICALKINETICS

viously,computers canperformthisintegration numerically. We will considertwo examplesof integratedrateequationsto illustratethe method. Thedecomposition of nitrogenpentoxidecanbe writtenas

N r O r + 2 N O *, I O ,

(4-11)

Assumethat the decompositionis a first order reaction:

R__4lYJ-=klNzos]

(4-12)

dt-Li

This equationcan easily be integrated

ffi+=kdt t\o.l 't -

I

-Jffi=lno r

lNrorln

dlN2o5l i =lKclt [N,O.] r o

and = - kt ln([NrO51/[N2O5]o)

(4-13)

In Eq. 4-13, [N2O5Jsis the concentrationwhen t - O.As illustratedin Figure 4- 1, this equationpredictsthat a plot of ln[NrO5J versusr should be a straightline with a slope of -k. The data,in fact, conform to this rate equationunder most conditions. This analysiscan also be carried out for higher order reactions.Hydrogen iodide will react to give hydrogen and iodine under certain conditions:

2HI+Hr+I,

(4-14)

R-- #=k[Hn2

(4-1s)

The rate law is

Rearrangementof this equation gives

-@+ =Zkdt lHq'

4.3 DETERMINATION OF RATE LAWS

77

u)

o z

N

c

Time FIGURE 4'1. Demonstrationof first orderkinetics.Plot of ln[NzOs] versusthe time according to Eq. 4-13. The straightline has a slope of -k.

which

to give l/[HI] - l/[HI]o = 2kt

(4-16)

Here [HI]o is the concentrationwhen / = 0. As illustrated in Figure 4-2, this equation predictsthataplot of l/[Hf] versusr shouldbe a straightline with a slopeof 2k. Again, the experimental data conform to this prediction. It is frequently, but not always, possible to integraterate equationsanalytically. Good experimental design can help to make the rate law relatively simple and therefore easy to integrate. In fact, experienced researcherswill try to make the reaction first order wheneverpossible.This might seemlike a major restrictionbut it is not. It is often possibleto convert complex rate laws to conform to pseudofirst order kinetics. For example,assumethe rate law is R = k(cr)f(c2)

t

Time FIGURE 4'2. Demonstrationof secondorder kinetics. Plot of 1/[HI] versustime accordinsto Eq. 4-16. The straightline has a slope of 2k.

78

CHEMICALKINETICS

whereflcr) is a function of the concentration of cr. The function could be very complex or a simple power of the concentration-it does not matter. If the concentration of c, is made much larger than the concentrationof c,, then it can be assumedto remain constant throughout the reaction, and the rate law becomes R = Ii(c) where k' - lcf(c). The "constant" kf(c) is called a "pseudo" first order rate constant since it is constantunder the experimentalconditions used,but actually dependson the concentrationcr.If the concentrationof c, is varied, the dependenceofflc 2) on c, can be determined. It might seemrestrictive to have a high concentration of cr, but in practice it is often sufficient to have the concentration of c, only about a factor of 10 higher than cr. Careful experimental design can make the job of determining the rate law much easier! In determining the rate law, it is often necessaryto use a broad range of experimental conditions before trying to interpret the rate law in terms of a chemical mechanism. As an illustration considerthe reaction I-+OCl-+OI-+Cl-

(4-r7)

At constant pH, the rate law determined experimentally is R = ft[I-][OCl-]

(4-18)

If the pH is varied, it is found that k also varies.It was determinedthat k = k'IIOH-1, where k'is a constant.Thus, a more completerate law is

R = ft'[I-][OCl-]/tOHl

(4-le)

The more information that can be determined about the concentration dependenceof the rate, the better the mechanism that can be postulated.

4.4 RADIOACTIVE DECAY A good exampleof a first orderrateprocessis radioactivedecay.Radioactiveisotopes are frequently used in biological research. For example, the radioactive isotope of phosphorus,32P,gives off radiation according to the reaction

lP-+:2s +

0-

(4-20)

whereB- is a high-energyelectron.The ratelaw for radioactivedecayis

-tP=k13zpl

(4-2r)

4.5 REACTIONMECHANISMS

79

Integration of this rate law as done previously (Eq. a-13) gives ["p] = ft'Plo t-o'

(4-22)

tn([32P]/[3'P]d- - k The rate of radioactive decayis usually given in terms of the half-life of the radioactive decay; in this case,the half-life is 14.3 days. In terms of the integratedrate law, Eq. 4-22,when half of the original radioactivity hasdecayed,ln(l/2) - - kt, or the half-life for decayrs tr,2- (ln2)lk. Thus, for radioactive decay,the half-life is constant.It does not matter when we start counting, or how much radioactivity we start with, the radioactivity will decay to half of its original value in 14.3 days. This is a specialproperty of first order rate processesand is not true for other reaction orders where the half-life dependson the concentrationsof the reactants.This is very convenientbecauseregardlessof when we start observing the rate of reaction, at t = 0, or at t = arry value, the integrated rate law is a simple single exponential.

4.5

REACTIONMECHANISMS

In general,many mechanismsare possiblefor a given reaction.Mechanismsare proposalsfor how the reaction occurs. A proposedmechanismmust be consistentwith the experimentally observed rate law, but this is usually true for many mechanisms. Kinetic studiescan disprove a mechanismbut cannotprove a mechanism.As a practical matterthe simplestmechanismconsistentwith all of the datais most appropriate, but at the end of the day, all that can be said is a specificmechanismis consistentwith known data.It is not possibleto say that this must be "the" mechanism. A mechanismconsistsof a combination of elementarysteps,which must sum up to give the overall reaction.For an elementarystep, the order and molecularity, the number of molecules involved in the reaction, are the same.Therefore, for elementary steps,the rate law can be written as the product of the concentrationsof all reactants, each raised to the power of their stoichiometric coefficient, multiplied by a rate constant.Some examplesof elementarystepsand associatedrateslaws are given below. H r+ l r--> 2 H l

R = klHzlllzl

2I + H, -+ 2HI

R = k[I]2[Hz]

O:+Or+O

R=klOrl

Rememberratelaws canbe derivedfrom the chemicalequationonly for elementary steps,andneverfor the balancedchemicalequationof the overallreaction.If an ele-

80

C H E M I C A LK I N E T I C S

mentary reaction is reversible, then the rate law is the difference between the rates of the forward and reversereactions. Therefore, for the elementary step k-

2A+B

C+D

R=krlAl'tBl-ktcltDl We now have two criteria for a possiblemechanism:(1) It must be consistentwith the observedrate law, and (2) the elementarystepsmust add up to give the overall balanced chemical reaction. Let us return to the two proposed mechanismsfor the reaction of H, and Ir, Eqs. 4-2 and 4-3. The first mechanism contains only a single elementary step so that the rate law is

R=ftrlHz]IIzl-k,lHIl"

(4-23)

The secondmechanismcontains two elementarystepsso that some assumptions must be made to derive the rate law. We will assumethat iodine atoms are in rapid equilibrium with molecular iodine, or to be more specific that this equilibrium is adjusted much more rapidly than the reactionof iodine atomswith molecularhydrogen. Furthermore,the concentrationof iodine atoms is assumedto be much less than that of molecular iodine. These assumptionsare, in fact, known to be correct.If we now consider the secondelementary step in the mechanism,

zr+Hr5 z nr the rate law can be written as

R=krlll2[Hr]- k2[Hr]2

(4-24)

but sincethe first step is always at equilibrium throughoutthe courseof the reaction, U2 = K[I2], where K is the equilibrium constantfor the dissociationreaction.Substituting this relationshipinto Eq. 4-24 gles

R = k rK[Iz]tHzl - kzlHllz

(4-2s)

Equations 4-23 and 4-25 areidentical in form as only the definitions of the constants are different, kr= krK andkr= ft2.Thus, we have shown that both mechanismsare consistent with the rate law, and therefore both are possible mechanisms.Even for this simple reaction, there remains a debateas to which is the more likely mechanism. It is easy to postulatea third possible mechanismwith the following elementary steps: H2+2H

4.5 REACTIONMECHANISMS

81

2H+I2=2HI By analogywith the mechanisminvolving iodine atoms,it canbe seenthat this mechanism would give the rate law of Eq. 4-25 with K now being the dissociationconstant for molecular hydrogenin equilibrium with hydrogen atoms.However, K is a known constantand if this constantis combinedwith the value of ft,Kdetermined experimentally, k, canbe calculated. It is found that the value of k, is much larger than the rate constant characterizing the maximum rate at which two hydrogen atoms and molecular iodine encountereach other in the gas phase.Therefore, this mechanismcan be ruled out as inconsistent with well establishedtheory. Thus, we have a third criterion for disproving a mechanism. As a secondexample of how mechanismscan be deduced,let us return to the reaction of I- and OCl- to produceOI- and Cl-, Eq. 4-17. The experimentallydetermined rate law is given by Eq. 4-19. What is a possiblemechanism?One possibility is the following scheme:

O C l - + H 2 O + H O C 1+ O H -

Fast,at equilibrium

I - + H o c r3 n o r + c l -

Slow

OH-+HOI=HrO+OI-

Fast

OCI-+I-+OI-+C1-

(4-26)

Overall reaction

This mechanism is consistentwith the balanced chemical equation. Now we must show that it is consistent with the observed rate law. The rate of a reaction is determined by the slowest stepin the mechanism.Therefore, the rate of the overall reaction is given by the rate of the secondelementary step:

R = ftrll-llHOCll

(4-27)

If the first step is assumedto be fast and the concentration of HOCI small relative that of I- and Cl-, then the concentration of HOCI can be calculated from the equilibrium relationship

6= [HOCI]tOHl/tOCrl tHocll = KloCl-litoHl Insertion of this relationshipinto Eq. 4-27 gives

R = kzKII-ltOCrl/tOHl

(4-28)

82

CHEMICALKINETICS

which is the observedrate law. Since this mechanismis consistentwith the observed rate law and the balancedchemical equation,it is a possiblemechanism. You may very well be wondering if the creation of a mechanismis black magic.It is true that imagination and knowledge are important factors, but logic can be used. A speciesin the denominator results from a fast equilibrium prior to a rate determining step. Therefore, what needs to be found is a first step involving one of the reactants that producesthe desiredspecies,in this caseOH-. The other product of the first step must then react with the secondreactant.The remainder of the stepsare fast reactions that are necessaryto producea balancedchemicalreaction.Note that none of the steps after the rate determining step play a role in determining the rate law. This is one of the simplesttypes of mechanisms,fast equilibria prior to and after a rate determining step.Nature is not always so obliging, and more complex mechanismsin which several steps occur at comparable rates are often necessaryto account for experimental findings. Here is anothermechanism,quite similar in concept:

O C 1 - + H r O = H O C 1+ O H -

Fast,at equilibrium

K^

(4-2e)

I-+HOClJICl+OH_ ICl + 2OH- = OI-+Cl-+HrO

OC1- + I- -+ OI- + Cl-

Fast Overall reaction

Obviously this gives the samerate law as the mechanismin Eq. 4-26, as all events prior to the rate determining step are the same.However, the chemistry is quite different. In the first caseiodide attacksthe oxygen, in the secondcaseit attacksthe chlorine. These two mechanismscannot be distinguishedby kinetics. Both are equally consistentwith the data. Thus far we haveconsideredmechanismsin which OCl- is the initial reactant.Now let's look at mechanismsin which I- is the initial reactantsuch as

I-+HrO=HI+OHHI + OCI- -+ lCl + OH-

Fast,at equilibrium Slow

By analogy,it shouldbe clear that this mechanismwould give the correctratelaw, and rapid reactions after the rate determining step can be added to give the correct balanced chemical equation. However, the equilibrium constant for the first step is known, and if this is combinedwith the resultsof the kinetic experiments,the rateconstant for the rate determining step would exceedthe theoretically possible value. This is becausethe concentrationof HI is much, much smaller than that of HOCI. Thus, mechanismsof this type can be excludedon the basisof theoreticalconcepts.

4.6 TEMPERATUREDEPENDENCEOF RATECONSTANTS

83

Thesetwo examplesillustrate how mechanismscan be relatedto the resultsof kinetic experiments.It is a greatchallengeto devisemechanisms.Once a mechanismis postulated,it is the job of the kineticist to devise experimentsthat will test the mechanism, often disproving it and requiring postulationof a new mechanism.However, it is important to remember that no matter how convincing the arguments,a mechanism cannotbe proved. Thus far we have consideredvery simple reactionsthat are not biological, asbiological reactionsare typically very complex. The pu{poseof this discussion has been to illustrate the principles and conceptsof chemical kinetics. We will later consider the kinetic analysis of enzymatic reactions, a very relevant and timely subject.

4.6

TEMPERATURE DEPENDENCE OF RATE CONSTANTS

Before discussingthe kinetic analysesof biological reactions,a few additional concepts will be described. Reaction rates are often dependenton the temperature, and typically reactions go faster as the temperatureincreases.The first quantitative treatment of the temperaturedependenceof reaction rates was developedby Arrhenius in the late 1800s.He proposedthat the temperaturedependenceof the rate constantcould be describedby the equation k = A exp(-E^lRT)

(4-30)

whereA is a constant,Euis the activation energy,and Zis the Kelvin temperature.This equation predicts that a plot of ln k versus IlT should be a straight line with a slope of -EJR. This behavior is, in fact, followed in most cases.Equation 4-30 canbe differentiated with respectto temperatureto give

d(In k)ldT = EJRT'

(4-31)

This equationcanbe integratedto give

In(krr/kr,) =

(E^/R)(72- Tr)

(4-32)

TrTt

Equation 4-32 permits the rate constant to be calculated at any temperature if it is known at one temperature and the activation energy has been determined. Note that theseequationsare similar in form to those describing the temperaturedependenceof the equilibrium constant except that the activation energy has replaced the standard enthalpy of reaction. In some cases,the activation energy may change with temperature, thereby making the analysis more complex. The physical model behind the Arrhenius equation is that an energy barnrer,Eu, must be overcomein orderfor the reactionto occur.This is shownconceptuallyin Figure 4-3. The reaction "path" or "coordinate" can be envisagedas the approach of the

84

KrNETrcs cHEMTcAL

Reaction Path FIGURE 4-3. Schematicrepresentationof the energy, E, versusthe "Reaction Path" during the course of a reaction as discussedin the text. The definition of the activation energy,Eu, is indicated.

reactantsto eachother that resultsin the lowest activation energy.The energydifference between products, P, and reactants,R, is AE. For simple reactionsin the gas phase,the energy can be calculatedas a function of the distancebetweenreactants. These calculations define an energy surface, and the dynamic course of the reaction on this energy surfacecan be determined.Even in the gasphase,this can only be done for very simple reactions.For reactionsin solution,this can only be considereda conceptualmodel. You might guessfrom your knowledgeof thermodynamicsthat the energyis probably not the best parameterto use to characterizethe dynamics of a reaction. A theory hasbeendevelopedthat insteaddiscussesthe reactionpath in termsof the free energy. This theory is called the transition statetheory.The basicpostulateis that a transition stateis formed in the courseof the reaction,which is in equilibrium with the reactants. This model is shown schematicallyin Figure 4-4.lnthe transition statetheory, the reactantsgo through a transition statethat is at a higher free energy than reactants.They must go over this free energybarrier to produceproducts.In Figure 4-4,thedifference in free energybetweenproductsand reactantsis LG", consistentwith thermodynamic principles.The free energydifferencebetweenthe transitionstateand reactants,AGo*, is called the standardfree energyof activation.In terms of transition statetheory, the rate constant can be written as k = (k"T/h) exp(-AG"+lRT)

(4-33)

wherek, is Boltzmann's constantandh is Planck's constant.(A more exactderivation includesan additionalparameter,the transmissioncoefficient, which is usually equal to 1.) Since AG"+ - LH"l - I ASo+,the rate constantcan be restatedas k - (kBT/h)exp(AS"+/R;expl-lA"*/RT1

(4-34)

4.6 TEMPERATURE DEPENDENCE OF RATECONSTANTS

85

Reaction Path FIGURE 4-4. Schematicrepresentation of the standardfree energy,6f, ur.ru. the "Reaction Pu,h. during the courseof a reactionas discussedin the text. The free energyof activation, AGo*,is definedin this diagram.The transitionstateat the maximumof the free energyis indicatedby TS. The reactantsandproductsareat minimain thefree energycurve. Here ASof is the standardentropy of activation, and LH"l is the standardenthalpy of activation. If the standardentropy and enthalpy of activation are assumedto be temperature independent,Eq. 4-34 canbe differentiated to give

{,1 {orr,parc

d(ln k)/dT = (AF7"*+ RT)l(nTz)

(4-3s) t-Ji t.fi

Thus, the transition statetheory predicts a temperaturedependenceof the rate constant very similar to the Arrhenius theory with E" = LH"I + RT. Since RZis usually small relative to the standardenthalpy of activation, the activation energy and standardenthalpy of activation are usually quite similar. The Arrhenius and transition state formulations cannot be differentiated by this small difference in the temperature dependencesince both the activation energy and the standardenthalpy of activation can be temperature dependent. From our knowledge of thermodynamics, we know that the enthalpy of activation is temperaturedependentif a heat capacity difference exists between the transition state and the reactants.If the temperaturedependenceof the rate constant is to be analyzed in terms of transition state theory, it is more convenient to plot In(klT) versus IIT as dLn(klT)ldT - NI"*/(R/), or dln(klT)ld(l/T) -

-Nr"+/R.

A simple derivation of Eq. 4-33 is possible. The concentrationof the transition state,TS, can be calculatedfrom the relationship [TS]/[Reactants]= exp(-Ac"+ I RT) where fReactants]representsthe concentrations of the reactantsraised to the appropriate powers for the stoichiometric equation of the elementary step. The rate of the

86

CHEMICALKINETICS

ReactionPath FIGURB 4-5. Schematic representationof the standardfree energy, G0, versus the "Reaction Path" during the course of a two-step reaction. The intermediate, I, is at a minimum in the free energy whereas the transition state, TS, for each step is at a maximum. The reactants and products are also at minima in the free energy curve.

reactionis the concentrationof the transitionstatemultipliedby the frequencywith overthebarrier whichit passesoverthefreeenergybarrier.Thisfrequencyof passage canbe derivedfrom an analysisof the reactioncoordinatewith statisticalmechanics andis givenby krTlh. Therefore,the rateof reactionis R - (kBTI h) tTSI = [Reactants](k"TI h)exp(-AG " +IRT ) The rate constant in Eq. 4-33 follows directly from the above equation. This orous derivation provides a conceptual framework for the theory. If a reaction goes through an intermediate, the intermediate would coffespond to a minimum in the free energy, and each elementary step would have its own transition state. This is shown schematically in Figure 4-5 for a sequenceof two elementary steps.The step with the highest free energy barrier is the rate determining step. Transition state theory is a very useful method for correlating and understanding kinetic studies.Becausethe framework of the theory is similar to thermodynamics, this producesa consistentway of discussingchemicalreactions.The entropy and enthalpy of activation are often discussedin molecular terms. It should be remembered that, for kinetics as with thermodynamics, such interpretations should be approached with extreme caution.

4.7

RELATIONSHIP BETWEEN THERMODYNAMICS AND KINETICS

Obviously the principles of thermodynamicsand kinetics must be self-consistent.In fact, this placessome useful restrictionson the relationshipsbetweenrate constants. In orderto seehow rate constantsarerelatedto equilibrium constants,considerthe elementary step

4.7 RELATIONSHIP BETWEENTHERMODYNAMICS AND KINETICS

A + BS o t

87

(4-36)

k,

The rate of this reaction is

R--

#=kr[A][B]-k2tABl

(4-37)

At equilibrium the net rate of reaction must be zero.If R = 0, frl[A].[B]. = k2lABl", where the subscripte designatesthe equilibrium concentration.Thus, we seethat

K = krlkz= [AB]"/([A].[B]")

(4-38)

In this case,the equilibrium constant,K, is equal to the ratio of rate constants.Similar relationshipsbetweenthe rate constantsand equilibrium constantscan be found for more complex situationsby setting the net rate equal to zero at equilibrium. If Eq. 4-38 is cast into the framework of transition statetheory, we obtain

K: exp[-(Ac;* - LG;\/RI] - exp(-Ac /RT)

(4-3e)

This result indicatesthe relationshipbetweenthe standardfree energychangesof activation and the standardfree energy changefor the reaction. This relationship can also be seenin the diagram of the free energy versus reaction path. A more subtle relationship can be found if reaction cycles occur becauseof the principle of detailed balance or microscopic reversibility. This principle statesthat a mechanismfor the reactionin the forward direction must also be a mechanismfor the reaction in the reversedirection. Furthermore, atequilibrium, the forward and reverse rates are equal along each reaction pathway. This means that once we have found a possible mechanismfor the reaction in one direction, we have also found a possible mechanism for the reaction in the other direction. To illustrate this principle, considerthe following triangular reactionmechanism:

k3

k2

88

C H E M I C A LK I N E T I C S

According to the principle of detailed balance, each of the individual reactions must be at equilibrium when equilibrium is attained, or k1[A]. = k_1[BJ" k2[B]"= k_2[Cl" k3[C]"= k_3[A1" If the right-hand sidesof theseequationsare multiplied togetherand set equal to the left-hand sidesof the equationsmultiplied together,we obtain

4 k2h - k_' k_2k_'

(4-40)

Thus, we find that the six rate constantsare not independent,nor are the three equilibrium constants!This result may seem obvious, but many people have violated the principle of detailed balance in the literature. It is important to confirm that this principle is obeyedwheneverreactioncycles are present.

4 .8

RE A CT I O N R AT E S N E AR EQU IL IBR I U M

Before we considerthe applicationof chemicalkinetics to biological systems,we will discussthe special caseof reaction rates near equilibrium. As we shall see,the rate laws become quite simple near equilibrium, and methods exist that permit very fast reactions to be studied near equilibrium. We will begin with an elementary step that is a reversible first order reaction, such as protein denaturation in the middle of the transition from the native to the denatured state:

(4-4r)

K^ I

The rate equationcharacterizingthis systemis

_ 4INI

-ktDl

(4-42)

d-=ftrlNl

We will now introducenew concentrationvariables;that is, we will set eachconcentration equal to its equilibrium value plus the deviation from equilibrium:

lNl=[N]"+AtNl lDl=[D]"+AtDl

4.8 REACTIONRATESNEAR EQUILIBRIUM

89

Note that the equilibrium concentrationsare constants,independentof time, and mass conservationrequiresthat AtNl = -A[D]. Inserting theserelationshipsinto Eq.4-42 gives

-

#

= kr([N]. + AtNl)- k (tDl"- AtNl) = k1[NJ"- k tD]" + (/
In derivingthe aboverelationship,usehasbeenmadeof the relationshipkt[N]" = /.rtDl" andl/r = kr* k ; t is calledtherelaxationtime.Thisfirst orderdifferentialequation canbe integratedasbeforeto give AtNl = A[N]oe-'l'

(4-43)

where A[N]o is the deviation of N from its equilibrium value at t - 0. Specialnote shouldbe made of two points. First, the relaxationtime can readily be obtainedfrom experimentaldata simply by plotting the ln(A[N]) versus/, and second the first order rate constant charactenzing this reaction, that is, the reciprocal relaxation time, is the sum of the two rate constants.If the equilibrium constantis known, both rate constantscan be obtained from a single experiment. How might such an experiment be carried out? In the caseof protein denaturation, a small amount of denaturantsuch as urea might be addedquickly to the solution. The ratio of the native and denaturedprotein would then move to a new equilibrium value characteristic of the higher urea concentration. Alternatively, if thermal denaturation is being studied,the temperaturemight rapidly be raised,establishinga new equilibrium ratio. The rate of conversion of the native stateto the denaturedstatecan be measured after the experimental conditions are changed.For a reversible first order reaction such as Eq. 4-4I, the time course of the entire reaction follows a single exponential and the effective rate constant is the sum of the two rate constants.A wide range of methods exists for changing the equilibrium conditions: Concentrationjumps and temperature jumps are particularly useful, but pressure jumps and electrical field jumps also have beenused. Studying rates near equilibrium is especiallyadvantageousfor higher order reactions and complex mechanisms.As a final example,we will considerthe elementary step of an enzyme, E, combining with a substrate,S.

E+S=ES this systemis Therateequationcharacterizing

(4-44)

90

cHEMToAL KrNETrcs

-#=

- 4rEsl ftrlErlsr

(4-4s)

Again we will write the concentrationsas the sum of the equilibrium concentration plus the deviation from equilibrium.

lEl=[E]"+A[E] lSl=[Sl"+A[S] lESl=[ES].+AlESl Furthermore, massconservation requiresthatA[E] = A[S] = - AtESl.Insertingthese relationships into Eq.4-45gives

d(AtEt) = {k1([EJ" - k tESl"+ ftr(AlEl)2 + [S].)+ (] AtEl+ ftr[El"[S]" Near equilibrium, the deviation of concentrationsfrom their equilibrium values is so small that the term (A[E])2 can be neglected-this simplification in the rate equation results from being near equilibrium. For example,if the deviation is 107o(0.1), the squareof the deviationis l%a(0.01).Furthermore,k [E].[Sl" = ([ESl., so that the final rate equation is

- -d[ = E] dt

AtEl

(4-46)

^E

with llx-kr(lBl"+[S].)+(

(4-47)

[EJ"+ [S]" FIGURE 4-6. Schematic representationof a plot of the reciprocal relaxation time, l/r, versus the sum of the equilibrium concentrations,[E]e + [S]., accordingto Eq. 4-47. As indicated,both of the rate constantscan be obtained from the data.

PROBLEMS

91

Although the reaction of enzyme with substrateis a secondorder reaction, the rate equationnear equilibrium, Eq. 4-46, is the szlmeas for a first order reaction.In fact, all rate equationsbecomefirst order nearequilibrium! This is becauseonly termscontaining Ac are retained in the rate law: (Lr)z and higher powers are neglected.As before, Eq. 4-46 can easily be integrated, and the relaxation time can be obtained from the experimental data. If the relaxation time is determined at various equilibrium concentrations,a plot of llr versus([E]" + [S]") can be made,as shown schematicallyin Figure 4-6. The interceptis equal to ( and the slope is equal to k . The study of reactions near equilibrium has been especiallyimportant for biological systems,as it has permitted the study of important elementary stepssuch as hydrogen bonding and protolytic reactions,as well as more complex processessuch as enzymecatalysisand ligand binding to macromolecules.

REFERENCES t.

L Tinoco, Jr., K. Sauer,and J. C.Wang,Physical Chemistry:Principles and Applications in Biological Sciences,3rd edition, PrenticeHall, Englewood Cliffs, NJ, 1995.

2 . C. R. Cantor and P. R. Schimmel, Biophysical Chemlsfry, W. H. Freeman, San Francisco,1980.

3 . G. G. Hammes,EnzymeCatalysisand Regulation, AcademicPress,New York,1982. 4 . A. Fersht, Structure and Mechanism in Protein Science:A Guide to Enryme Catalysis and Protein Folding, W. H. Freeman,San Francisco, 1999.

PROBLEMS 4-1. The activity of the antibiotic penicillin slowly decomposeswhen stored in a buffer at pH 7.0,298 K. The time dependenceof the penicillin antibiotic activity is given in the table below. Time (weeks)

0 1.00 2.00 3.00 5.00 8.00 10.00 t2.oo 15.00 20.00

Penicillin Activity (arbitrry units)

1 0 ,1 0 0 8,180 6,900 5,390 3,870 2,000 1 ,3 3 0 898 403 r61

What is the rate law for this reaction, that is, what is the order of the reaction with respectto the penicillin concentration?Calculate the rate constant from the data if possible. (Data adaptedfrom Ref. l . )

92

cHEMTcAL KtNETtcs

4-2. The kinetics of the reaction 2Fe3*+ Sn2*-+ 2Fe2++ Sn4* has been studiedextensivelyin acidic aqueoussolutions.When Fez* is added initially at relatively high concentrations,the rate law is R = k[F'e3*]2[Sn2+]/[Fe2+] Postulatea mechanismthat is consistentwith this rate law. Show that it is consistent by deriving the rate law from the proposed mechanism. 4-3. The conversion of l-malate to fumarate is catalyzedby the enzvme fumarase:

coo-

tl HO-C-H

l:ll

H-C-H

tl

coo-

cooC-H C-H

coo-

The'nonenzymaticconversionis very slow in neutral and alkaline media and has the rate law R = klr-malatel/[H*] Postulatetwo mechanismsforthe nonenzymaticconversionand show thatthey are consistentwith the rate law. 4'4. The radioactive decay rates of naturally occurring radioactive elementscan be usedto determinethe ageof very old materials.For example, ]aC is radioactive and emits a low-energy electronwith a half-life of about 5730years.Through a balanceof natural processes,the ratio of 14C/r2Cis constantin living organisms. However, in dead organismsor material, this ratio decreasesas the 1aC decays.Sincethe radioactivedecayis known to be a first order reaction,the age of the materialcan be estimatedby measuringthe decreasein the r4clrzcratio. Supposea piece of ancientwool is found in which the ratio has been found to decreaseby 207o.What is the age of the wool? 4-5. The nonenzymatic hydration of CO, can be written as COr+H2O=H2CO3 The reaction is found to be first order in both directions. Becausethe water concentration is constant, it does not appearin the expressionfor the equilibrium or rate equation. The first order rate constant in the forward direction has a

PROBLEMS

93

value of 0.0375 s-l at 298 K and 0.0021 s-l at 273 K. The thermodynamicparametersfor the equilibrium constant at 298 K are NIo = 1. 13 kcaVmol and A,So = -8.00 caV(mol'K). A. Calculate the Arrhenius activation energy for the rate constant of the forward reaction. Also calculate the enthalpy and entropy of activation according to transition statetheory at 298 K. B. Calculatethe rate constantfor the reversereaction at273 and298 K. Assume that A,Ff is independent of the temperature over this temperature range. C. Calculate the Arrhenius activation energy for the rate constant of the reversereaction.Again, also calculatethe enthalpyand entropy of activation at298 K. 4-6. A hydrogen bonded dimer is formed between 2-pyridone according to the reaction

o\^ +ll

o

H

,N-=y'

kp kr

The relaxation time for this reaction, which occurs in nanoseconds,has been determined in chloroform at 298 K at various concentrations of 2-pyridone. The data obtained are [G. G. Hammes and A. C. Park, J. Am. Chem. Soc.91,

956(re6e)l:

2-Pyridone (M)

0.500 0.352 0.25r 0.151 0.101

10ec (s)

2.3 2 .7 3.3 4 .0 5.3

From thesedata calculate the equilibrium and rate constantscharacterizingthis reaction. Hint: If the expressionfor the relaxation time is squared,the concentration dependencecan be expressedas a simple function of the total concentration of 2-pyridone.

I

CHAPTERs

Applicationsof Kineticsto Biological Systems 5.1 INTRODUCTION We will now considersomeapplicationsof kinetic studiesto biological systems.This discussionwill centeron enzymes,as kinetic analysesof enzymesrepresenta major researchfield and have provided considerableinsight into how enzymeswork. Enzymes are proteins that are incredibly efficient catalysts:They typically increasethe rate of a chemical reaction by six orders of magnitude or more. Understanding how this catalytic efficiency is achievedand the exquisite specificity of enzymeshas intrigued biologists for many decadesand still provides a challengefor modern research. Since enzyme deficienciesare the sourceof many diseases,enzymesare also atarget for medical researchand modern therapeutics.The field of enzyme kinetics and mechanismsis so vast that we will only presentan abbreviateddiscussion.More complete discussionsare available (cf. Refs. 1-3). We will alsodiscusscatalysisby RNA (ribozymes),which is important in the processingof RNA in biological systems,as well as somekinetic studiesof DNA denaturation and renaturation.

5.2

ENZYME CATALYSIS: THE MICHAELIS-MENTEN MECHANISM

We now considerthe analysisof a simple enzymaticreaction.This discussionwill introduce some new concepts that are particularly useful for analyzing complex systems. The conversion of a single substrateto product will be taken as a prototype reaction: S -+ P

(5-1)

A typical example is the hydration of fumarate to give l-malate (and the reverse dehydration reaction) catalyzedby the enzymefumarase.If a very low concentrationof enzymeis used relative to the concentration of substrate,a plot of the initial rate of the reaction,or initial velocity, v, is hyperbolic, as shown in Figure 5-1. The limiting initial velocity at high concentrationsof substrateis called the maximum velocity, V^, and the concentrationof substrateat which the initial velocity is equaltoVJ2 is called the Michaelis constant,K*. 94

MECHANISM 5.2 ENZYMECATALYSIS:THE MICHAELIS-MENTEN

95

concentration, FIGURE 5-1. A plot of the initial velocity,y, versusthe substrate [S], for an enzymaticreactionthatcanbe describedby the Michaelis-Mentenmechanism. A mechanismthat quantitatively accountsfor the dependenceof the initial velocity on substrateconcentrationwas postulatedby Michaelis and Menten. It can be written in terms of elementary stepsas k,

E+S€

-) ES-+E+P K^

(s-2)

tnz

where E is the enzymeand ES is a complex consistingof the enzyme and substrate. The total concentrationof enzyme,[Eo],is assumedto be much lessthan the total concentrationof substrate,[Ssl. This mechanismis an exampleof catalysisin that the enzyme is not consumed in the overall reaction, which is greatly acceleratedby the enzyme.If we tried to do an exact mathematical analysisof the mechanismin Eq. 5-2, coupled differential equationswould needto be solved.Fortunately,this is not necessary. We will consider two approximations that can adequatelyaccount for the data. The first is the equilibrium approximation. With this approximation, the first step is assumedto be always at equilibrium during the course of the reaction. This means that the equilibration of the first step is much more rapid than the breakdown of ES to P, or k.. << kr. We then have

dlPl _ = v k:[ES]

(s-3)

k1lkr=tESl(tEltsl)

(5-4)

dt

and

Furthermore, conservation of mass requires that

[Eo]=[E]+tESl

(5-s)

96

APPLICATIONS OF KINETICS

= [ES](1 = [ES]{I + krt(\ tsl)} + tEl/tESI)

lESl= [Eo]/{t + k /(k1[SJ)]

(s-6)

Substitution of Eq. 5-6 into 5-3 gives

-dtPl = dt

ft3[Eo] Y=-

(5-7)

r + kz/&Jsl)

which can be rewritten as ym Y=-

(5-8)

1 + Krl[S]

with

V* = ft3[Es1

(s-e)

and K*= krlkt

(s-10)

Equation 5-8 is found to accountquantitatively for the data.The maximum velocity is proportionalto the total enzymeconcentration,and when Ktr = [S], v = VJ2, as required. In some cases,it is possible to obtain independentestimatesor measurementsof the equilibrium constantfor the first step. Sometimesthis independentmeasurement is in good agreementwith the constantobtainedfrom kinetic studies,sometimesit is not. Obviously, when the Michaelis constantand the equilibrium constantare not in agreement,the mechanismmust be reexamined. A more general analysisof the Michaelis*Menten mechanismmakes use of the steady-stateapproximation. This approximation does not make any assumptions about the relative values of the rate constantsbut assumesthat the concentrationsof E and ES are very small relative to S, consistentwith the experimentalconditions.Under theseconditions, it is assumedthat the rate of changeof the concentrationsof E and ES is very small relative to the rate of change of the concentration of S, or

dlEUdt=dlESUdt=0 This is the steady-stateapproximation, an approximation that is very important for analyzing complex mechanisms.A careful mathematicalanalysisshowsthat this is a

MECHANISM 5.2 ENZYMECATALYSIS:THE MICHAELIS-MENTEN

97

very good approximation under the experimental conditions used, namely, when the total substrateconcentrationis much greaterthan the total enzymeconcentration. What does the steady-stateapproximation mean? If we go back to the mechanism (Eq. 5-2), we find that

dlESl - kr[E][S]= 0 (kz+ k3)[ES] dt

k/(k2+k3)- tESl/(tEltsl)

(s-11)

Note the similarity of this equationto the equilibrium constant(Eq. 5-4). This means that the ratio of concentrationsremains constant.However, the ratio is not the equilibrium concentrations;it is the steady-stateconcentrations.If k3<< kr,the steadystate is the sameas the equilibrium state, so that the equilibrium condition is a special case of the steady state. For all other steady-statesituations, the ratio of concentrationsis always lessthan the equilibrium ratio. We can now calculatethe rate law for the steady-stafeapproximationexactly asfor the equilibrium approximation.

lEol= tEsl(l + [E]/[ES];= [ESJ{I + (kr+ kr)/(/c,[S1)]

(s-12)

k3[Eo] = '=+= ft:IES] (kr+ |+

k3)/(kr[S])

v^

(s-13)

1 + K*/[S] with Vn'= k3[EsJ

(s-14)

Kv = ftr+ k.)lk,

(s-15)

Thus, it is clear that the equilibrium approximation and steady-stateapproximation give rate laws that areindistinguishableexperimentally.Most generally,the Michaelis constantis a steady-stateconstant,but in a limiting caseit can be an equilibrium constant.As previously stated,both situationshave been observed.The steady-stateapproximation is more general than the equilibrium approximation and is typically employed in the analysisof enzyme mechanisms.

98

APPLICATIONS OF KINETICS

With modern computers,dataanalysisis very easy,and experimentaldata can be fit directly to Eq. 5-8 (or Eq. 5-13) by a nonlinear least squaresfitting procedure. However, it is always a good idea to be sure that the data indeed do conform to the equationof choiceby a preliminary analysis.Equation 5-8 can be linearizedby taking its reciprocal:

Ilv = I/V^+ (KrlV-)/[S1

(s-16)

Thisis calledtheLineweaver-Burke equation. A plot of l/v versusl/tSl is linear,and the slopeandinterceptcanbe usedto calculatethemaximumvelocityandMichaelis constant. Thisequationcanbeusedfor thefinal dataanalysisprovidingproperstatisticalweightingis used.An altemativelinearizationofEq. 5-8 is to multiplyEq.5-16 by ISI to give lSl/v= [S]/Ym+K*/V^

(s-17)

1/[s]

FIGURE 5-2. Linear plots commonly used for analyzing data that follow Michaelis-Menten kinetics. (a) Plot of the reciprocal initial velocity, y, versus the reciprocal substrate concentration. l/tsl. (b) Plot of [S]/u versus[S].

5.3 ct-CHYMOTRYPSIN

99

This equationpredictsthat a plot of [S]/v versus[S] shouldbe linear and gives a better weighting of the data than the double reciprocalplot (Eq.5-16). Obviously, the final results should be independentof how the data are plotted and analyzed!Examplesof both plots are given in Figure 5-2. The steady-stateanalysiscan be extendedto much more complex enzymemechanisms,aswell asto otherbiological processes,but this is beyondthe scopeof this presentation. How can enzyme catalysis be understoodin terms of transition statetheory? The simple explanation is that enzyme catalysis lowers the standardfree energy of activation. This can be either an entropic or enthalpiceffect. Much more detailedinterpretationsof enzymecatalysisin termsof transitionstatetheory havebeendeveloped. The basic idea is that formation of the enzyme-substrate complex alters the free energy profile of the reaction such that the free energy of activation for the reaction is greatly lowered. Essentially,the free energy change associatedwith binding of the substrateto the enzyme is used to promote catalysis.

5.3

ct-CHYMOTRYPSIN

As an example of how mechanismscan be developedfor the action of enzymes,early studiesof the enzyme u-chymotrypsin will be discussed.Proteolytic enzymeswere among the first to be studied,not becauseof their intrinsic interest,but becausethey could easily be isolated in a reasonablypure form. In fact, for many years the availability of large quantitiesof pure enzymeseverelyrestrictedthe rangeof enzymesthat could be studied in mechanisticdetail. With the easeof cloning and modern expression systems,this is no longer a limitation. The enzyme cx-chymotrypsinhas a molecular weight of about 25,000and consists of threepolypeptidechainscovalently linked by disulfides.It is anendopeptidase;rhat is, it cleavespeptide bonds in the middle of a protein. Enzymes that cleave peptide bondsat the end of a protein arecalled exopeptidases. The overall reactioncanbe written as

tflt

R-Q-C-N.*

I NH

lt

Hzo TB R-C-e-O-

-_--->

INH

*H3N*

(s-18)

Experimentally, it is found that the enzyme has a strong preferencefor R being an aromatic group, amino acids phenylalanine, tyrosine, and tryptophan, but hydrophobic amino acidssuchasisoleucine,leucine,and valine are alsogood substrates.Studying this reaction with a protein as substrateis very difficult becausethe molecular structureof the substrateis changingcontinuously aspeptidebonds are cleaved.This makes quantitativeinterpretationof the kinetics virtually impossible.Consequently, the first step in elucidating the molecular details of the enzyme was to develop "model" substrates,molecules that have the important featuresof the protein substrate

1OO

APPLIcATIoNS oF KINETIcS

but aremuchsimpler.Chymotrypsin is alsoa goodesterase; thatis, it hydrolyzesesters.This provedvery usefulin elucidatingthe mechanism of actionof the enzyme. Themostsuccessful modelsubstrates havethe seneralstructure HO R-

rtl C-C-NH? t-

NH

I

C:O

I

CH:

HO Amide

ttl

R-C-C-O-R'

I NH I C:O I

Ester

CH:

where R is the aromatic residueassociatedwith phenylalanine,tyrosine, and tryptophan. Note that the amino group of the aromatic amino acid is acetylated.The enzyme will not work on a free amine, as might be expectedfor an endopeptidase.Furthermore, only l-amino acids are substrates. We will considersome of the resultsobtainedwith tryptophan as the amino acid.

R-m The hydrolysis of tryptophan model substratesfollows Michaelis-Menten kinetics, and some of the results obtained for various estersand the amide are shown in Table 5-1. In this table,k"at= y,n/[80],and R'is the group covalently linked to the tryptophan carboxyl. The fact thatk"urisessentiallythe samefor all esterssuggeststhat acommon intermediateis formed and that the breakdown of the intermediate is the rate determining stepin the mechanism.A different mechanismmust occur for the amide, or there is a different slow step in the mechanism. An inherentdeficiencyof steady-statekinetic studiesof enzymesis thatthe enzyme concentrationis very low. Consequently,the intermediatesin the reaction sequence cannot be detected directly. Limited information about the intermediates can be obtained through steady-statekinetic studiesby methodsnot discussedhere, for example, the pH dependenceof the kinetic parametersand the use of isotopically labeled substratesthat alter the kinetic parameters.However, in order to study the intermediatesdirectly, it is necessaryto use sufficiently high enzymeconcentrationsso that the intermediatescan be observeddirectly. This is the realm of transientkinetics.The difficulty for enzymatic reactions is that the reactionsbecome very fast, typically occurring in the millisecond and microsecondrange. This requires special experimental techniques.Transient kinetic studieshave proved invaluable in elucidating how enzymes work. In the caseof chymotrypsin, a substratewas sought for which a color changeoccurred upon reaction in order that the changesin concentration could easily and rapidly be observed.The first such substratestudiedwas p-nitrophenyl acetate,which is hydrolyzed by chymotrypsin:

5.3 cr-CHYMOTRYPSIN

101

TABLE 5-1. Steady-StateConstantsfor Chymotrypsin Ky (mM)

k*, (s-t)

R,

21 28 3l 0.03

Ethyl Methyl p-Nitrophenyl Amide

-5 -5 -5 -0.09

S o u r c e s :A d a p t e d f r o m R . J . F o s t e r a n d C . N i e m a n n ,J . A m . C h e m . \ o c . , 7 7 , 1 8 8 6 ( 1 9 5 5 ) a n d L . W . Cunningham and C. S. Brown, J. Biol. Chem.22l,287 (1956).

+ll

o

cH3-c-o-

(s-1e) The advantageof this substrateis that the phenolate ion product is yellow so that the time course of the reaction can easily be followed spectrophotometrically. With this substrate,it was possibleto look at the establishmentof the steadystate,as well as the steady-statereaction.A very important finding was that the reaction becomes extremely slow at low pH. An intermediate,in fact, can be isolatedby using a radioactive acetyl group in the substrate.The intermediate is an acylated enzyme. Later studiesshowed that the acetyl group is attachedto a serineresidueon chymotrypsin (4).

o

tl

Enzyme--O-C-CH3 Based on the kinetic studiesand isolation of the acyl enzyme,a possible mechanism is the binding of substrate,acylation of enzyme, and deacylation of the enzyme. The elementary stepscan be written as k1 k2 -") k "

E + S <-

ES +E-acyl -+ E +Acetate

(s-20)

k_1

+p -Nitrophenolate Does this mechanismgive the correct steady-staterate law? The rate law can be derived as for the simple Michaelis-Menten mechanismusing the steady-stateapproximation for all of the enzyme speciesand the mass conservation relationship for enzyme: [E6] = [E] + [ES] + [E-acyl]

(s-2r)

102

APPLIcATIoNS oF KINETIcS

_ dtEsl= (k:+ k )[ESl- ft1[EJ[Sl =0 dt -

d[E-acyl) dt

- kzlB-acyl] - k2[ES] = 0

v = kz [E-acyl]

(s-22)

When theserelationship are combined, it is found that the Michaelis-Menten rate law, Eq. 5-8, is obtained,with

,f

k2hlBol m

kr+ k,

(n_,*t")/ ft, Ku=t-Z-jV"rl )

(s-23)

(s-24)

For ester substrates,the slow step is deacylationof the e\zyme, or k, << kr. In this case,ft"u,= kzand K; = [(k_r + k)/k](hlk). How did the transient kinetic studies contribute to the postulation of this mechanism? It is worth spending some time to analyzethe mechanism in Eq. 5-20 in terms of transient kinetics. This analysis can serve as a prototype for understanding how transient kinetics can be used to probe enzymemechanisms,not only for chymotrypsin but for other enzyme reactions as well. The proposed mechanismpredicts that the acyl enzymeshould accumulateand be directly observable.If we consider only a single turnover of the enzyme at very early times and high substrateconcentrations,all of the enzyme should be converted to the acyl enzyme,followed by a very slow conversionof the acyl enzymeto the free enzymethroughhydrolysis (kz>> ft3).The slow conversionto enzyme will follow Michaelis-Menten kinetics (Eq. 5-13) initially as long as the total substrateconcentrationis much greaterthan the totalenzyme concentration. Thus, the complete solution to the rate equations for the mechanism in Eq. 520 should contain two terms: (1) the rate of conversion of the enzyme to the acyl enzyme, and (2) the rate of the overall reaction that is limited by the rate of hydrolysis of the acyl enzyme. To simplify the analysis, let us assumethe following: k, is approximately zero at early times (kz>> kr); the formation of the initial enzyme-substratecomplex, ES, is very rapid relative to the ratesof the acylation and deacylationstepsso that ES is in a steady state (d[ES]/dt = 0); and the total substrateconcentration is much greater than the total enzyme concentration. The last assumption is similar to that for the steadystate approximation used earlier for the Michaelis-Menten mechanism. However, in this casethe enzyme concentration is sufficiently high so that the concentration of the intermediate can be detected.The rate of formation of the intermediate is given by

s.3 cr-CHYMOTRYPSIN103

dlE-acyll dt

(s-2s) = ftzlESl

Becausethe phenolate ion is formed when the acyl enzyme is formed, the rate of formation of phenolate ion is the same as the rate of formation of the acyl enzyme. This is actually a measurementof the rate of establishmentof the steady state-In order to integrate this equation, we make use of the identity

= [ES]{1+ (k-r+ kr)/(ft1[SJ)] tEl + [ES]= tEsl(l + tEl/[ES]) and mass conservation lEol = [E] + [ES] + [E-acyl] or [E] + [ES] = [Eo] - [E-acyl] Insertion of theserelationshipsinto Eq.5-25 gives

d[E-acyl] -: dt

k2([Eo]- [E-acyl])

(s-26)

1+(k_,+k2)/(kr[s])

Integration of this equation gives [ E - a c y l ] = [ E o ] ( l- e k J )

(s-27)

withlir= kz/{ 1 + (k-r + /cr)/(k1[SJ) ]. = = 0, when t Notethat [E-acyl] 0, andwhent = @,[E-acyl] [Eo],asexpected. The total rateof phenolate,Pr, formationat earlytimesis [ P , ] = v r + [ 8 6 1 ( 1- e - r f )

(s-28)

The time courseof the reactionas describedby Eq. 5-28 is shown schematicallyin Figure 5-3. The secondterm in the equationdominatesat early times, and the exponential rise can be analyzedto give (r.The linear portion of the curve correspondsto the Michaelis-Menten initial velocity and can be used to obtain V- and Kr. The maximum velocity for this limiting caseis k3[Esl,so that both k, and ft, can be determined. A more exact analysisthat does not assumekt= 0 is a bit more complex, but the end result is similar. The rate equationcontainsthe sametwo terms as in Eq. 5-28. However, the secondterm is now more complex: the exponentis (lir+ kr)t insteadof tirt; and the amplitude is lBolltiz/1tir+ kr)12rather than [E6J.The first term is the same,v/, with the maximum velocity and Michaelis constantdefined by Eqs. 5-23 and 5-24.

104

APPLICATIONS OF KINETICS

FIGURE 5-3. Schematicrepresentation for the kineticsof an enzymaticreactiondisplaying "burst"kinetics.Theproductof thereaction,P1,is plottedversusthetime.As described by Eq. 5-28 for the limiting caseof kz >> Iq, the slopeof the linear portion of the curve is the steady-state initial velocity,v, and the initial exponentialtime dependence is characterized by the rateconstant16anda"burst" amplitudeof [Eo]. Finally, we consider the situation when the rate of hydrolysis of the acyl enzyme is very fast relative to its formation (kz>> k2).We have seenpreviously that fast steps occurring after the rate determining step do not enter into the rate law. Therefore, the mechanismis equivalentto the Michaelis-Menten mechanismwith only a single intermediate(Eq. 5-2), with the rate determiningstepbeing the acylationof the enzyme (V- = k2[80]).If the transientkinetics arc analyzed,the rate is still given by Eq. 5-28, but the amplitude of the second term is essentially zero, rather than [86]. Thus, the transient kinetics and steady-statekinetics provide the same information. The presence of an initial "burst" of product, as shown in Figure 5-3, is commonly used as a diagnosticfor the presenceof an intermediate.If the burst is not present,that is, the straight line portion of the curve extrapolatesthrough zero, this means either that an intermediate is not formed, or that its rate of disappearanceis much faster than its rate of appearance.This exhaustive,perhapsexhausting,discussionof the mechanismin Eq. 5-20 is a good example of what must be done in order to arrive at a mechanism for an enzymereactionby kinetic studies.Both steady-statekinetics and transientkinetics are necessaryfor a complete picture. The proposedmechanismaccountsfor all of the facts presented:the observedrate law, the samek.u,for all ester substrates,and a cornmon intermediate.What about amides? For amides,it turns out that the slow step is acylation of the enzyme,that is, kz<< kr, so that k"u,= kzand Kv = (kt + k)/kl. As discussedabove,if the acylation of the enzyme is slow relative to deacylation, k, )) k2, the intermediate does not accumulate and therefore cannot be observeddirectly. Literally hundredsof experiments are consistentwith this mechanism-a very remarkableachievement.Additional experiments that permitted the very fast reactionsprior to the acylation of the enzyme to be studied show that this mechanismis too simple. At least one additional elementary step is required. The sequentialelementarystepsare: (1) binding of substrateto the enzyme,(2) a conformationalchangeof the enzyme-substratecomplex, (3) acylation of the enzyme, and (4) deacylation of the enzyme.

5.3 cI-CHYMOTRYPSIN 105

The ultimate goal of kinetic studiesis to understandthe mechanismin terms of molecular structure. In the caseof chymotrypsin, the three-dimensional structure of the enzyme is known and is shown in Color Plate IVa. A well-defined binding pocket is observed for the aromatic side chain of the specific substratesof the enzyme. The pocket is lined with nonpolar side chainsof amino acidsand is very hydrophobic.Up to three amino acids coupled to the N-terminus of the aromatic amino acid interact with a short range of antiparallel B-sheetin the enzyme. A hydrophobic site also is observed for the amino acid attached to the C-terminus of the substrate.This explains why a free carboxyl group cannot bind and therefore why chymotrypsin is an endopeptidase.As shown in Color PlateIVb, a triad of amino acids,serine-195,histidine-57, and aspartate-lO2,is observedin the active site region and is found in other serine proteases.The imidazole actsas a generalbaseduring the nucleophilic attackof the serine hydroxyl on the substrate.A tetrahedralintermediate probably is formed prior to acyl enzyme formation. Water servesas the nucleophile for the hydrolysis of the acyl enzyme,with imidazole again participating as the generalbase.An abbreviatedversion of the catalytic mechanismis shown in Figure 5-4. This greatly truncated story of the elucidation of the mechanism of action of chymotrypsin illustrates several important concepts: the steady-stateapproximation; the use of steady-statekinetics in determining chemical mechanisms;the importanceof transient kinetics in elucidating intermediatesin a mechanism; and the interpretation of the mechanism in terms of molecular structure.More complete discussionsof chymotrypsin are available(1,2,5).

\\-

ffiss\_ /ff\

, _o-__Tq,\7_, rs"tles

--

|

o3\.r*-".,

TTt'

InzH

Rr

Tetrahedral Intermediate

AcylEnryme

Tetrahedral Intermediate

EP

---p165(

seJs5 ot

H

Acyl Enryme

FIGURE 5-4. A mechanismfor the hydrolysis of peptidesor amides by chymotrypsin. The imidazole acts as a general baseto assistthe nucleophilic attack of serine on the substrateor the nucleophilic attack of water on the acyl enzyme.

106

APPLToATIoNS oF KlNETtcs

5.4 PROTEIN TYROSINE PHOSPHATASE In our earlier discussionof metabolism,we saw that the energyobtainedin glycolysis is stored as a phosphateester in ATP and servesas a sourceof free energy for biosynthesis. The importance of phosphateestersis not confined to this function. Phosphorylation and dephosphorylationof proteins plays a key role in signal transduction and the regulation of many cell functions. For example,the binding of hormonesto cell surfacescan trigger a cascadeof such reactions that regulate metabolism within the cell. The cell processesmodulated by this mechanisminclude T-cell activation, the cell cycle, DNA replication,transcriptionand translation,and programmedcell death, among many others.Both the kinases,responsiblefor protein phosphorylation,and the phosphatases, responsibleforprotein dephosphorylation,havebeenstudiedextensively. We shall restrict this discussionto the phosphatases, and primarily to one particular type of phosphatase. The phosphorylationof proteins primarily occurs at three sites,namely, the side chainsof threonine,serine,and tyrosine.Proteinphosphatases canbe divided into two structuralclasses:the serine/threoninespecific phosphatases that require metal ions; and tyrosine phosphatasesthat do not require metals but use a nucleophilic cysteine to cleavethe phosphate.The latter classincludes dual function phosphatases that can hydrolyze phosphateesterson serineand threonine,as well astyrosine.Many reviews of theseenzymesareavailable(cf. Refs.6 and 7). We will discussthe mechanismof a particular exampleof tyrosine specificphosphatase, namely,an enzymefound in both human and rat that has been extensively studied structurally and mechanistically (6). The overall three-dimensionalstructureof the catalytic domain of the enzyme is shown in Color Plate Va, and the active site region is shown in Color Plate Vb with vanadatebound to the catalytic site. Vanadateis frequently usedas a model for phosphateestersbecausethe oxygenbonding to vanadiumis similar to the oxygenbonding to phosphorus,and vanadatebinds tightly to the enzyme.A cysteinerestsat the base of the active-sitecleft. An asparticacid and threonine (or serinein similar enzymes) are also near the active site and are postulatedto play a role in catalysis,as discussed below. The kinetic mechanismthat has been postulatedis the binding of the protein substratefollowed by transfer of the phosphoryl group from tyrosine to the cysteine at the active site, and finally hydrolysis of the phosphorylatedcysteine:

I

k1 D,cl LTJ

+ ^r

k3 >

ES rr2

E-PO3-+

E+HPO4

(s-2e)

As with chymotrypsin,a protein is not a convenientsubstrateto usebecauseof the difficulty in phosphorylating a single specific tyrosine and the lack of a convenient method for following the reactionprogress.In this casethe model substrateused was p-nitrophenyl phosphate. Similar to chymotrypsin, when the enzyme is phosphory-

5.4 PROTEINTYROSINEPHOSPHATASE

107

lated the yellow phenolate ion is releasedso that the reaction progress can easily be monitored. Extensive steady-stateand transient kinetic studies have been carried out, and we will only presenta few selectedresultsthat summanzethe findings (8,9). The role of the cysteinewas firmly establishedby site specificmutagenesis,asconverting the cysteineto a serineresultsin an inactive enzyme.The cysteineis postulated to function as a nucleophile with the formation of a cysteine phosphate. However, when transient kinetic studies were carried out with the native enzymq a burst phase was not observed.This is shownin Figure 5-5a.As we notedpreviously,this indicates that either a phosphoenzymeintermediateis not formed, or its hydrolysis is much faster than its formation. The asparticacid and serineor threonine shown at the active site of the enzyme in Color Plate V are conservedin many different enzymesof this class of Consequently,it was decided to mutate theseresidues.If the tyrosine phosphatases. serineis mutated to alanine, the enzymeis still active, but now a burst is observed,as shown in Figure 5-5b. This burst was postulatedto representthe formation of the phosphoenzyme,and the rate constantswere obtained for the phosphorylation of cysteine and its hydrolysis. The resultsof the ffansientkinetic experimentsare presentedin Table 5-2. For the native enzyme,only ft"u,canbe determinedand is equal to ftr. For the serineto alan-

1.4

E

E lt:,

o tf

a! o

(, C (! ll L

1.2 1 0.8 0.6 0.4

o o .ct 0.2

0

E

0.1 0.2 0.3 0.4 Time in seconds (a)

0.5

0.5

C

It

0.4

6 o.g

() F o.z l| I o.r a (I)

0

0.25

0.5 0.75

1

1.25

Timein seconds (b) FIGURE 5-5. Transient kinetic time course for the hydrolysis of p-nitrophenyl phosphateby native (a) and mutant (b) protein tyrosine phosphatase.The point mutation substituted alanine for serine 222. The concentrations of substrate are shown next to the traces. Reprinted with permissionfrom D. L. Lohse, J. M. Denu, N. Santoro,and J. E. Dixon, Biochemistry 36,4568 (1997). Copyright @ l99l American Chemical Society.

108

APPLIcATIoNS oF KINETIcS

TABLE 5-2. Kinetic Parameters for Native and Mutated Tyrosine Phosphatases Enzyme Native Serine 222 -+ Alanine Aspartatel8l + Asparagine Serine 222 -+ Alanine and Aspartate 181 -+ Asparagine

ft"^,(s-t)

20 1.3 0.27 0.055

kz (s-t)

34 1.9

ft: (s-t)

K (pM)'

t.4

0.46

0.057

0.81

Source: Adaptedfrom D. L. Lohse,J. M. Denu, N. Santoro,andJ. E. Dixon, Biochemistry36, 4568 (1997). "(k_t + k)lkt.

ine mutant, the rate constant for hydrolysis of the intermediateis the same as kcat,as expected.This mutationalsosuggeststhat serineplays a role in the hydrolysis mechanism. The importanceof the asparticacid in the mechanismwas inferred not only from its conservationin many different enzymes,but also by the pH dependenceof the steady-statekinetic parameters,which suggestedit functioned as a generalacid in the phosphorylationof the enzyme.When the asparticacid was mutatedto asparagine,the enzymestill functioned,but(* was only about I7o of thatof the native enzyme(Table 5-2). The transientkinetics did not show a burst phase,indicating that formation of the phosphoenzymewas still rate determining.As expected,the pH dependenceof the steady-statekinetic parameterswas alteredby this mutation. Finally, the double mutatedenzymewas preparedin which the serinewas changed to alanineand the aspartateto asparagine.The transientkinetics showeda burst so that the slow stepwas now the hydrolysis of the intermediate,as with the single mutant in which serinewas changedto alanine.However, (u, was reducedeven further, and the rate constantsfor both phosphorylationof the enzymeand hydrolysis were greatly reduced (Table 5-2). On the basis of theseand other data, the aspartatewas postulated to serveasa generalbasein the hydrolysis of the intermediate.The intermediateis sufficiently stablein the double mutant so that it could be observeddirectly with phosphorusnuclearmagneticresonance(8). The mechanismproposedfor this reactionhas two distinct transition states,one for phosphorylationof the cysteineand anotherfor hydrolysis of the phosphorylatedenzyme.Based on the experimentsdiscussedand other data, especiallysffucturaldata,structureshavebeenproposedfor the two transition statesas shown in Figure 5-6. The mechanismof tyrosine phosphatasesillustrates severalimportant points. First, the generalusefulnessof the analysisdevelopedfor chymotrypsin is apparent. Both transientand steady-statekinetic experimentswere important in postulating a mechanism.Second,the importance of site specific mutations in helping to establishthe mechanismis evident. This method is not without pitfalls, however. It is important to establishthat mutationsdo not alter activity through structural changes in the molecule. In the present case, experiments were done to establishthe structuralintegrity of the mutant enzymes.This ensuresthat the mutations are altering only the chemical aspectsof the mechanism.Finally, this dis-

5.5 RIBOZYMES 109

i{tP I

*,, o'o

i

H

io

il ,,O"'- P----S-Cys -;\ H' -=DOH

oo

i

Se(Thr)

reaction.(a) FIGURE 5-6. Proposedtransitionstatesfor the proteintyrosinephosphatase Formationof the cysteinephosphateintermediatewith a tyrosinephosphateas substrate.(b) intermediate. Adaptedfrom J. M. Denu,D. L. Lohse,J. Hydrolysisof thecysteinephosphate Vijayalakshmi,M. A. Saper,and J. E. Dixon, Proc. Natl Acad. Sci.USA93,2493 (1996). by Republished with permissionof the NationalAcademyof SciencesUSA. Reproduced permissionof thepublishervia CopyrightClearanceCenter,Inc. cussionagainillustratesthe many differenttypesof experimentsthat must be donein developing mechanisms.The kinetic results must be bolstered by structural and chemicalinformation.Many other studiesof enzymescould profitably be discussed, but insteadwe will turn our attentionto kinetic studiesin othertypesof biological systems.

5.5 RIBOZYMES Enzymes are the most efficient and prevalentcatalystsin physiological systems,but they are not the only catalystsof biological importance.RNA molecules have also been found to catalyze awide range of reactions.Most of thesereactionsinvolve the processingof RNA, cutting RNA to the appropriatesize or splicing RNA. RNA has also been implicated in peptidebond formation on the ribosome and has been shown to hydrolyze amino acid esters.Thesecatalytic RNAs are called ribozymes.They are much lessefficient than a typical protein enzymeand sometimescatalyzeonly a single

110

APPLIoATIoNS oF KINETIoS

event. In the latter case,this is not true catalysis.Ribozymes also may work in close collaboration with a protein in the catalytic event. We will consider an abbreviateddiscussion of a ribozyffie, ribonuclease P. For more information, many reviews of ribozymesare available(10-12). RibonucleaseP catalyzesan essentialstepin tRNA maturation,namely, the cleavage of the 5' end of a precursor tRNA (pre-tRNA) to give an RNA fragment and the maturetRNA. This reactionis shown schematicallyin Figure 5-7. Also shown in this figure is a representationof the catalytic RNA. The catalytic RNA is about 400 nucleotideslong and catalyzesthe maturationreactionby itself in vitro. In vivo,a protein of about 120 amino acidsalsoparticipatesin the catalysis.The preciserole of the protein is not known, but it appearsto alter the conformation of the RNA. The proteinRNA enzyme has a broader selectivity for biological substratesand is a more efficient catalyst (13). Metal ions are also involved in this reaction, with Mg2+being the most

HrO

FIGURE 5-7. The reaction of pre-tRN A catalyzedby ribonuclease P is shown schematically at the top of the figure. The 5'leader sequenceof the pre-tRNA is removed in the reaction. The eubacterial consensusstructure of ribonucleaseP RNA is shown in the lower part of the figure. Proven helices are designatedby filled rectangles, invariant nucleotides by uppercaseletters, >90Vo conserved nucleotides by lowercase letters, and less conservednucleotidesby dots. ReproducedfromJ.W.BrownandN.R.Pace,NucleicAcidsRes.20,l45l(1992).Reproduced by permissionof Oxford University Press.

5.5 R IB OZY ME S 111 important physiologically. Metal ions probably play a role in both catalysis and in maintaining the active conformation of the RNA, but we will not consider this aspect of the reaction here. Transientkinetic studiesof the ribozyme have been carried out, and the minimal mechanismconsistsof the binding of pre-tRNA to the ribozyme, cleavageof the phosphodiesterbond, and independentdissociationof both products. We will presentsome of the results obtainedwith the RNA componentof ribonucleaseP. If transient experiments are carried out with ribonucleaseP with excesssubstrate, a "burst" of the tRNA product occurs at short times, followed by an increase in the product concentrationthat is linear with time. This behavior is familiar by now, as it has been observed for chymotrypsin and protein tyrosine phosphatase,as discussed above. In the case of ribonuclease P, a covalent intermediate is not formed with the ribozyme; instead,the dissociationof productsis very slow relative to the hydrolytic reaction. The mechanism can be written as

pre-tRNA+ E + ES -+ tRNA + P + E

(s-30)

where E is RNAase P RNA and P is the pre-tRNA fragment product. This is not exactly the sameas the mechanismdiscussedfor chymotrypsin,but the analysisis similar if the secondstepis assumedto be much slower than the first. (The detailed analysis is given in Ref. 14.) There is an important lessonto be learnedhere. "Burst" kinetics are observed whenever the product is formed in a rapid first step followed by regeneration of the enzyme in a slower second step. The apparentrate constantsfor these two stepscan be determined. What is desired,however,is to distinguishbetweenthe binding of enzymeand substrate and the hydrolysis; both of these reactions are aggregatedinto the first step in Eq. 5-30. In the caseof chymotrypsin and protein tyrosine phosphatase,we assumed that the binding step was rapid relative to the formation of the enzyme-substrateintermediate.In this case,we cannotassumethat binding is rapid relative to hydrolysis. The two stepswere resolvedby singleturnoverexperiments:an excessof enzymewas used and becausedissociationof productsis slow, only a single turnover of substrate was observed(14). The mechanismcan be written as

t, ^l

k2

E + pre-tRNA-+ E.pre-tRNA-+ E'IRNA.P

(s-31)

The results obtained are shown in Figure 5-8 for two different concentrationsof enzyme. At the lower concentrationof enzyme,it is clear that the curve is sigmoidal, rather than hyperbolic. This is becausethe first step is sufficiently slow that it takes some time for the concentrationof the first enzyme-substratecomplex to build up. The data at the higher enzyme concentration are essentially hyperbolic becausethe rate of the first reaction is greater at higher enzyme concentration. These data can be explainedby analyzingthe mechanismin Eq. 5-30 astwo consecutiveirreversiblefirst order reactions.Why irreversible?Fortunately, the rate of dissociationof pre-tRNA

112

APPLrcATtoNs oF KtNETtcs

E 0.8 o .g (, 0.6 c .9 o

0.4

t

o.2

tg

0.4

0.6

Time in seeonds FIGURE 5'8. Single turnover measurementsof the hydrolysis of pre-tRNA catalyzedby the RNA component of RNAase P. The fraction of pre-tRNA cleaved is plotted versus time. The pre-tRNA was mixed with excessconcentrationsof RNAase P RNA, 1.4 pM (a) or 19 pM (o). The data are the fit to a mechanism of two consecutive first order reactions (Eq. 5-38). Reprintedwith permissionfrom J. A. Beebeand c. A. Fierke, Biochemistrv 33, 10294 ,11994\. Copyright @ 1994 American Chemical society.

from the ribozyme is sufficiently slow that it does not occur on the time scale of the experiment, and furthermore, the analysis of the product gives both E-IRNA and free

tRNA so that if some dissociationof IRNA occurs it is not relevant.Why is the first stepassumedto be first order?This is anotherexampleof apseudofirst orderrateconstant becausethe enzyme concentration is effectively constantthroughout the experiment. The pseudofirst order rate constantis kl[E], where k, is the secondorder rate constant for the reaction of the enzyme with the substrate. To simplify the nomenclature,we will rewrite the mechanism as

A-+B-+C

(s-32)

Therateequationfor the time dependence of A is _ dtAl = frrlAl dt

(s-33)

[A] = [A]oe-k't

(s-34)

which is easily integrated to give

where [A]o is the starting concentration,in for the time dependenceof B is

_ dlBl dt

caseof pre-tRNA. The rate equation

- k2[B) - kl[A]

(s-3s)

5.6 DNA MELTINGAND RENATURATION

113

_ dTBI-k2[B]-krlA)oe-kJ dt

The solution to this differential equationis

lBl= lo'of',"-k,t- ,-kly

(s-36)

Kz- Kt

This solution doesnot work if 4 - krasthe denominatorgoesto zero.For this special case, [B] = krlA)ore-k''

(5-37)

Finally, the time dependenceof C can be obtained from mass balance since [A]o = [A]+[B] +[C]:

+ rcr=ra,r ;h,-kl {u*') [r

(s-38)

If Eq. 5-38 is used to analyzethe data in Figure 5-8, it is found that k, = 6 x 106M-l s -' 1 R NR1'andk z = 6 s - t. Understandinghow to analyze"burst" kinetics and consecutivefirst order reactions is sufficient for the kinetic analysisof many enzymaticreactions,as the conditions can usually be adjustedto conform to theserelatively simple mechanisms. The mechanisticwork carriedout with RNAseP haspermittedthe establishmentof a minimal mechanismfor the RNA portion of the enzyme.However, much remains to be done. The roles of metals, specific groups on the RNA, and the protein remain to be delineated.Understandinghow ribozymesfunction is at the forefront of modern biochemistryand has important implications for both physiology and the evolution of enzymes.

5.6

D NA M E LT I NG A N D R EN AT U R A T IO N

We will conclude this chapterwith a discussionof the denaturationand renaturation of DNA. Understandingthe dynamics of such processesis clearly of biological importance.At the outset,it must be statedthat a detailedunderstandingof the kinetics and mechanismshave not been achieved.However, this is not for lack of effort, and a qualitative understandingof the mechanismshas been obtained. We will begin with someof the elementarystepsin the dynamicsof the interactions betweenthe two chainsof helical DNA. As discussedin Chapter3, the thermodynamics of hydrogen bonding between baseshas been studied in nonaqueoussolvents, where the dimers formed are reasonablystable.Kinetic studiesof hydrogen bonded dimers also have been carried out, and the reactionshave been found to be extremely fast, occurritrg on the submicrosecondand nanosecondtime scale.For example,the

114

APPLrcATroNs oF KrNETtcs

kinetics of formation of a hydrogen bonded dimer between 1-cyclohexyluracil and 9ethyladenine has been studied in chloroform (15). (The hydrocarbon arms have been addedto the basesto increasetheir solubility in chloroform.) The reaction is found to occur in a single step with a secondorder rate constant for the formation of the dimer of 2.8 x 10eM-1 s-l and a dissociationrate constantof 2.2x 107s-l at zIC.The second order rate constantis the maximum possiblevalue; that is, it is the value expected if every collision betweenthe reactantsproduced a hydrogen bonded dimer. The upper limit for the rate constant of a bimolecular reaction can be calculated from the known rates of diffusion of the reactantsin the solvent. In all caseswhere hydrogen bonded dimer formation hasbeen studied,the formation of the dimer hasbeen found to be diffusion controlled. What does this tell us about the rate of hydrogen bond formation that occurs after the two reactantshave diffused together?To answerthis question,we will postulatea very simple mechanism, namely, diffusion together of the reactantsto form a dimer that is not hydrogenbonded,followed by the formation of hydrogenbonds.This can be written as k,

A+B k_,

K^

A-B

[A,B]

(s-3e)

t.

In this mechanism,k, is the rate constant for the diffusion controlled formation of the intermediate,and k_, is the rate constantfor the diffusion controlled dissociationof the intermediate.Since only a single step is observedin the experiments,assumethat the initial complex formed is in a steady state:

lA,Bl = k1[A] [BU(k: + k) + k_2[A-B]l(k_t+ k2)

(s-40)

The overallrateof the reactionis

dtA-Bl dt

= k 2 [A ,B]-k _ 2 [A -B ]

(s-41)

If Eq. 5-40 is substitutedinto Eq. 5-41, we obtain

dtA-Bl =ftrlAllBl-ft.tA-Bl dt with kr= ktkrl(k4 + kz) and k,= k_rkal(k_, + kr)

(s-42)

5 . 6 D N A M E L T I N GA N D R E N A T U R A T I O N 1 1 5

The experimental results indicate that k, = kr. This is true if kz>> k-r, or formation of the hydrogen bonds is much faster than diffusion apart of the intermediate. However, we can calculatethe value of k-,; it is about 1010s-l. Therefore,the rate constantfor formation of the hydrogenbonds,kr, must be greaterthan about l01l s-1.A more exact analysis would put in a separatestep for formation of eachof the two hydrogen bonds, but this would not changethe conclusion with regard to the rate of hydrogen bond formation. Hydrogen bond formation is very fast! We know that hydrogen bonding alone cannot account for the stability of the DNA double helix: Base stackingand hydrophobic interactionsalso are important. An estimate of the rate of basestacking formation and dissociation hasbeen obtained in studies of polyA and polydA (16). Thesemoleculesundergo a transition from "stacked" to "unstacked" when the temperatureis raised.This transition is accompaniedby spectral changesand can easily be monitored. Conditions were adjustedso that the molecules were in the middle of the transition, and the temperaturewas raised very rapidly (<1 tts) with a laser.This is an exampleof a kinetic studynearequilibrium. As the system returns to equilibrium at the higher temperature,the relaxation time can be measured and is the sum of the rate constantsfor stacking and unstacking (Eq. 4-43). The results obtainedindicate that the rate constantsare in the range of 106 - I01 s-1. Although this is a very fast process,it is considerably slower than hydrogen bonding or simple rotation of the bases. Moving up the ladder of complexity, we now consider the formation of hydrogen bonded dimers betweenoligonucleotides(17,18). Again, only a few prototype reactions will be considered,namely, the reactionof a seriesof An oligonucleotideswith U, oligonucleotides to form double helical dimers. These reactions occur in the millisecond to secondtime regime with n - 8-18. In all cases,only a single relaxation time was observed.The relaxationtime was consistentwith a simple bimolecular reaction (Eq.4-47):

Ilx = kr(A'1" + [Unl.)+ k_r

(5-43)

A typical plot of 1/t versusthe sum of the equilibrium concentrationis given in Figure 5-9, and some of the rate constantsobtained are given in Table 5-3. The bimolecular rate constantsare all about 106M-l s-1, whereasthe dissociationrate constantsvary widely, reflecting the stability of the doublehelix that is formed. Although the second order rate constantis quite large, it is considerably below the value expectedfor a diffusion controlled reaction. A detailed analysis of the mechanism can be made as above for the caseof the formation of a hydrogen bonded dimer. This case is considerablymore complex, as a separatestep is neededfor the pairing of each of the n bases.The analysis carried out was bolstered by determination of the activation energiesfor the rate constants.The conclusion reachedis that nucleation of the double helix after the molecules have diffusedtogetheris rate determining.The formation of the first basepair is rapid, but dissociation is even more rapid. This is also true for the formation of two base pairs. However, when the third basepair is formed, the structureis stabilized.After a nucleus

1 16

APPLICATIoNS OFKINETICS 400

5

0 zoo { 100

10

[Aro]"+ [u,o]"(u) FIGURE 5-9. A plot of l/t versus the sum of the equilibrium concentrations of tr1s and Uto at l7'C (Eq. 5-a3). The two reactantscombine to form a double helix structure. Adapted from D. Ptjrschkeand M. Eigen, J. MoL Biol. 62,361 (I971). Reproducedwith the permission of AcademicPress,Ing.

of three base pairs is present,each later base pair forms with a specific rate constant of about 107s-1.Note that this is consistentwith the resultsobtainedfor the "stacking" and "unstacking" of polyA. Altogether theseresults indicate that the elementary steps in the dynamics of DNA basepairing and stacking (and whatever other types of interactions are important for base pairing) are very rapid. We now turn to consideration of DNA itself. As might be expected,the melting and reformation of DNA is very complex. DNA is very heterogeneous,and we have discussedpreviously (Chapter 3) that A*T rich regions are much less thermodynamicallystablethan G-C rich regions.In addition, DNA has a complex structure,with loops and bulges.Native DNA has a very precise alignment of basepairs. If extensivemelting occurs,basepairings other than the native pairings may form when the temperatureis lowered to form the native structure, thereby slowing down reformation of the native structure. The ultimate form of melting is separationof the individual strands.Becausethe strandsare very long, the rate of melting could be limited by the hydrodynamics of moving the chains apart.Bearing

TABLE 5-3. RateConstantsfor the Reactionof [A'] with [U,] atlT'C 106ft1

8 9 10 ll t4

r.7 t.l 0.83 0.58 r.44

source: Adapted from D. Pdrschkeand M. Eigen, J. Mol. 8io1.62,361 (1971).

k-r (s-t)

2400 640 r75 28 I

5.6 DNA MELTINGAND RENATURATION

117

thesefactors in mind, let us examine some experimental results. Fortunately, the melting and reformation can easily be followed becausethe spectrumof DNA changes when the basesare stacked.Furthermore, the melting/reformation processesare easy to initiate by changing the temperature. Some typical resultsfor the denaturationof a viral DNA are shown in Figure 5-10. The denaturation was initiated by the raising of the temperatureto give a loss in the nativespectrumof 257o,53Vo,and95%o.Inthe first case,the lossin absorbance is approximately first order with a rate constant of about 1 min-I. The other two casesdo not strictly follow first order kinetics; that is, the logarithmic plot in Figure 5-10 is not linear, and the rates are clearly much slower. The approximate first order rate constantsat long times are shown in the figure. In contrast to the results in model systems, the melting kinetics cannotbe fit to a singlerelaxationtime, and the time scaleis 1001000 s. The viral DNA studied has about 2 x 105basepairs. If only the elementary stepsdiscussedabove were involved, the DNA would be completely melted in less than 1 second.How can theseresultsbe explained?The rate of melting is governed by the rate of unwinding of the double helix, and the rate of unwinding varies as the melting proceeds.Melting will usually startin the interior of the DNA, and unwinding can only occur if one end of the loop rotateswith respectto the other. Initially this rotation is fast becausethe double helix is approximatelya rod, but asmelting proceeds, the bulky loops produced make it harder for one end of the helix to rotate with respect to the other. The rate then becomesslower, as observed.Thus, strandunwinding and separationbecomerate limiting. In supportof this mechanism,if DNA is subjectedto a very drastic denaturing force such as a very high pH, all of the base pairs rapidly break in times lessthan 0.01 s. This doesnot involve unwinding and strandseparation 1.0

0.10 C'

q

sq 0.01

20

40

60

Time (min) FIGURE 5-L0. The kinetics of TZ DNA denaturationin 0.08 M NaCl, 80Voformamide. The fraction of the absorbanceat259 nm associatedwith the native structureis plotted versusthe time. The reaction was initiated by heating the three samples to a loss in the native ulffaviolet spectrumof 25Vo(n),53Vo (o), and 957o(o). Reproducedfrom M. T. Record and B. H. Zimm, Biopolymers 11, 1435 (1972). Copyright @ 1972 Biopolymers. Reprinted by permission of John Wiley & Sons,Inc.

118

APPLTcATToNS oF KrNETrcs

as the native DNA can rapidly be formed by lowering the pH. A detailed description of this mechanismhas been developed(19). The slow unwinding time could be a severe problem for the replication of large DNAs. However, nature has solved this problem by creating enzymesthat (1) can make breaks in the DNA chains, thereby reducing the length of DNA that must be rotated during the unwinding, and (2) can unwind DNA. As a final example, we consider the renaturation of separatedDNA strands.DNA can be denatured by raising the temperature significantly above the midpoint of the melting transition, or by raising the pH. After the strandshave separated,native DNA can be formed by lowering the temperatureor lowering the pH. As might be expected, the renaturation processis secondorder. The rate of reaction dependson the sourceof the DNA and its length. The dependenceon length can be eliminatedby first sonicating the DNA to form fragments of about the samesize. If the rates of renaturation are measuredat the sameDNA concentration in moles of nucleotides per liter, they vary by severalordersof magnitude,as shown in Figure 5-1 1. The figure includesdatafor polyA-polyU and a double strandedRNA (MS-2 viral RNA). The scale attlte top is the number of nucleotide pairs in the genome.The larger the genome,the smaller the number of complementary fragments. Only complementary strands can form native DNA, so that the smaller the number of complementarystrands,the slower the rate. For polyU and polyA, every strandis complementaryso that this systemis assigned a value of 1 on this scale.

NucleotidePairs

'l

10

102 I ,0. I t* 1ou+ 1 0 6

{ ,0,

108

rorl 1o1o

It o fg '6 o o o

E 0.5

G c .9 ?

(t o L lr

1.0 10-s

10-4 10-3 10-2

0.1

1

100 100010,000

Cst(M.s) FIGURE 5-11. Reassociationof double-strandednucleic acids from various sources.The genome size, which is a measureof the complexity of the DNA, is shown above the figure. The nucleic acids all have a single-strandedchain length of about 400 bases. Adapted with permission from R. J. Britten and D. E. Kohne, Science 16l, 529 (1968). Copyright O 1968 American Association for the Advancementof Science.

REFERENCES 119

The renaturation process can be analyzed quantitatively. If the complementary strandsare representedas A and A', the reaction can be written as

(s-44)

A + A'l eo' Since the concentrationsof A and A' are equal,the rate equationis

#=ktAlz to give - J ll 1lA . whichcanheintegrated

-4 =kd+

(s-45)

,A'

1/[A]-U[A]o=7g1

(s-46)

The data were found to obey this equation. The half-time for the reaction, that is, when lAl = l{lol2, is given by ttlz= l/(k[A]o)

(s-47)

In the above equations,the concentration should be equal to the concentrationof complementary strands,which we generally do not know. If the total number of basepairs in the smallest repeating sequenceis M then the concentration of complementary strandsis proportional to the total concentration of DNA, co, divided by N. If this relationship is substitutedin Eq.5-47, we find that cotu2". M Thus, the total concentration of DNA times the half-time for renaturation is a measureof N. A long half-time meansthat a given fragment will have to sample many other fragments before it finds its complement,so the sequenceis complex.Nis a measureof the "complexity" of the DNA, and the scaleon the top of Figure 5-11 can be equatedto N. On this basis,the calf DNA is the most complex DNA that was studied. The study of the kinetics of renaturation can be a useful probe of a gene. It has been used, for example, to determine if fragments of a viral gene were presentin viral transformed cells (20). This concludesthe sampling of the application of kinetics to biological systems. Many interestingstudieshave not been discussed,and many more remain to be done. The study of the time dependenceof biological processesis essentialin the questfor the elucidationof molecularmechanisms.

REFERENCES l . G. G. Hammes, Enz.ymeCatalysis and Regulation, Academic Press, New York, 1982.

2 . A. Fersht, Structure and Mechanism in Protein Science: A Guide to Enzyme Catalysis and Protein Folding, W. H. Freeman,San Francisco,1999. n J.

D. L. Purich (ed.), Contemporary Enzyme Kinetics and Mechanism, Academic Press, New York, 1996.

4. B. S. Hartleyand B. A. Kilby,Biochem.J.56,288 (1954). 5 . G. P. Hess,inThe Enzymes,3rdedition (P. Boyer, ed.),Vol. 3, p.213, AcademicPress, New York, 1970.

120

APPLIcATIoNS oF KINETIcS

6. J. M. Denu, J. A. Stuckey,M. A. Saper,and J. E. Dixon, Cell87 ,361 (1996\. 7 . S. Shenolikaar,Annu. Rev. CeIl Biol.10,55 (1994). 8 . J. M. Denu, D. L. Lohse, J. vijayalakshmi, M. A. saper, and J. E. Dixon, proc. Natl. Acad. Sci.U9A93,2493 (1996).

9 . D. L. Lohse, J. M. Denu, N. Santoro,and J. E. Dixon, Biochemistry 36,4568 (1991). 10. T. R. Cech, in The RNA world, Cold Spring Harbor Labaratory press, Cold Spring Harbor,NY, 1993,p.239.

1 1 . D. B. McKay and J. E. Wedekind, in The RNA World,2nd edition, Cold Spring Harbor Laboratory Press,Cold Spring Harbor, NY, 1999, p.265.

12. S. Altman and L. Kirsebom, rn The RNA world, Znd edition, cold Spring Harbor Laboratory Press,Cold Spring Harbor, NY, 1999, p. 351.

1 3 . A. Loria, S. Niranjanakumari, C. A. Fierke, and T. Pan, Biochemistry37,15466 (1998).

t4. J. A. Beebe and C. A. Fierke, Biochemistry 33, 10294(1994). 1 5 . G. G. Hammes and A. C. Park, J. Am. Chem.\oc.91,956 (1969). 16. T. G. Dewey and D. H. Turner, Biochemistry 18,5757 (1919). 1 7 . D. Riesner and R. Romer, in Physico-ChemicalProperties of Nucleic Acids,Vol. 2 (J. Duchesne,ed.), Academic Press,London, 1973.

1 8 . D. Pcirschkeand M. Eigen,J. Mol. Biol.62,36l (1971). 19. M. T. Record and B. H. Zimm, Biopolymers 11, 1435 (1912). 20. S. J. Flint, P. H. Gallimore,and P. A, Sharp,J. Mol. 8io1.96,41 (1915).

PROBLEMS 5-1' The hydration of CO, is catalyzed by the enzyme carbonic anhydrase.The overall reaction at neutral pH can be written as COr+ H2O = HCOI + H+ The steady-statekinetics of both hydration and dehydration have been studied at pH 7. 1, 0.5"C. Sometypical data aregiven below for an enzymeconcentration of 2.8 x 10-eM. Hydration

103/v1M-rs;

lo3[co2](M)

36 20 l2 6.0

1 .2 5 2 .5 5.0 20

Dehydration 103/v(M-ls)

95 45 29 25

103[c0:] (M)

2.0 5.0 10 15

Calculate the steady-stateparametersfor the forward and reversereactions.

PROBLEMS

121

5-2. Studiesof the inhibition of enzymesby various compoundsoften provide information about the nature of the binding site and the mechanism.Competitive inhibition is when the inhibitor,I, competeswith the substrateforthe catalytic site. This mechanismcan be written as E+S=ES-+E+P E+I=EI Derive the steady-staterate law for this mechanism and show that it follows Michaelis-Menten kinetics when the inhibitor concentrationis constant.Assume the inhibitor concentration is much greater than the enzyme concentration. 5-3. In the text, the steady-staterate law was derived with the assumption that the reaction is irreversible and/or only the initial velocity was determined. Derive the steady-staterate law for the reversible enzymereaction:

E+S=X*E+P Showthat theratelaw canbe put into the form yslKslsl - vP/KPlPl

1+tsl/Ks+lPl/K, where V, and V, ne the maximum velocities for the forward and reverse reactions, and K, and Kn are the Michaelis constantsfor the forward and reversereactions. When equilibrium is reached, v = 0. Calculate the ratio of the equilibrium concentrationsof S and P, [P]/[S], in terms of the four steady-stateparameters. This relationship is called the Haldane relationship and is a method for determining the equilibrium constantof the overall reaction. 5-4. The kinetics of the formation and breakdownof hydrogenbondedloops or hairpins within small RNA moleculescan be studiedwith relaxationmethodssince the formation of the helical structures is accompanied by a spectral change. This reactioncan be representedas k1

Hydrogen bonded *

Non-hydrogenbonded k2

What is the relaxationtime for this reaction? The relaxation time determined at a specific concentration was found to be 10 ps. Calculatethe individualrate constantsif the equilibrium constantis 0.5.

122

APPLIoATIoNS oF KINETIcS If the concentrationof the RNA is doubled, would the relaxation time get smaller, larger, or stay the same?

5-5. Considerthe binding of a protein, P, to a DNA segment(generegulation).Assume that only one binding site for P exists on the DNA and that the concentration of DNA binding sites is much less than the concentrationof P. The reaction mechanismfor binding can be representedas P + DNA = P-DNA -+ P-DNA' wherethe secondsteprepresentsa conformationalchangein the protein.Calculate the rate law for the appearanceof P-DNA' under the following conditions. A. The first step in the mechanismequilibratesrapidly relative to the rate of the overall reaction and [P-DNA] << tDNAl. B . The intermediate,P-DNA, is in a steadystate. C. The first step in the mechanismequilibratesrapidly relative to the rate of the overall reaction and the concentrationsof DNA and p-DNA are comparable. Express the rate law in terms of the total concentrationof DNA and P-DNA, that is, [DNA] + IP-DNAl. D. The following initial rateswere measuredwith an initial DNA concentration of 1 |rM.

tPl (rrM) 100 50 20 l0

104Rate (M/s)

8.33 7.t4 5.00

-).-)-) a

a a

Which of the rate laws is consistentwith the data? 5-6. Many biological reactionsare very sensitiveto pH. This can readily be incorporated into the rate laws becauseprotolytic reactionscan be assumedto be much fasterthan otherratesin most cases.For example,in enzymemechanisms the ionization statesof a few key protein side chainsare often critical. Suppose that two ionizable groups on the enzyme are critical for catalytic activity and that one of them needsto be protonatedand the other deprotonated.The protolytic reactionscan be written as EH2+EH+H*+E+ZH* If only the speciesEH is catalytically active and the protolytic reactionsare much more rapid than the other stepsin the reaction, all of the rate constants that multiply the free enzyme concentrationin the rate law have to be multiplied by the fraction of enzymepresentas EH.

PROBLEMS123

A. Calculate the fraction of free enzyme present as EH at a given pH. Your answershouldcontainthe concentrationof H* and the ionization constants of t het wo s idech a i n s Ke . r = l E l l H + l /[EH l a n d K pz= IE H l l H + J/[E H 2]. B. Assumethat ES in the Michaelis-Menten mechanism(Eq. 5-2) also exists in threeprotonationstates,ESH2,ESH, and ES, with only ESH being catalytically active. Calculate the fraction of the enzyme-substratecomplex presentas EHS. Designatethe ionization constantsas Ku' and Kusr. C. Use the results of parts A and B to derive equationsfor the pH dependence of V-, K*, and VJK*. Measurementof the pH dependenceof the steadystate parameters permits determination of the ionization constants, and sometimesidentification of the amino acid side chains.

I

CHAPTERo

LigandBindingto Macromolecules 6.1 INTRODUCTION The binding of ligands to macromoleculesis a key elementin virtually all biological processes.Ligands can be small molecules,such as metabolites,or large molecules, such as proteinsand nucleic acids.Ligands bind to a variety of receptors,such as enzymes,antibodies,DNA, and membrane-boundproteins.For example,the binding of substratesto enzymesinitiates the catalytic reaction. The binding of hormones, such as insulin, to receptors regulates metabolic events, and the binding of repressorsand activators to DNA regulatesgene transcription. The uptake and releaseof oxygen by hemoglobin is essentialfor life. Indeed, compilation of a comprehensivelist of biological processesin which ligand binding plays a key role would be a formidable task. In this chapter,we shall discusshow to analyzeligandbinding to macromolecules quantitativelyfor both simple and complex systems.We alsowill considerexperimental methods that are used to study ligand binding. The application and importance of this analysisfor biology will be illustrated through specific examples.The treatment presentedwill be adequatefor most situations: more complete (and more complex) discussionsof this topic are available (1*3).

6.2 BINDING OF SMALLMOLECULES TO MULTIPLE IDENTICAL BINDINGSITES The binding of a small molecule to identical sites on a macromoleculels a common occunence.For example,enzymesfrequentlyhave severalbinding sitesfor substrates on a single molecule. Proteinshave multiple binding sites for protons that are often essentiallyidentical, for example,carboxyl or amino groups.Let us assumethe simplest case,namely, a single ligand, L, binding to a single site on a protein, p: L + P=PL

(6-l)

This equilibrium can be characterizedby the equilibrium constant,K:

K = IPLIt(tPltl-l)

(6-2)

Binding equilibria involving macromoleculesare conveniently characterizedby a binding isotherm,the moles of ligand bound per mole of protein,r. For the abovecase, 124

6.2 BINDINGOF SMALLMOLECULESTO MULTIPLEIDENTICALBINDINGSITES

- r

= -

tPLl

,KtLl

tPl+[PL]

1+iKll-l

125

(6-3)

A plot of r versus [L] is shown in Figure 6-Ia: It is a hyperbolic curve with a limiting value of 1 at high ligand concentrations,and when r = 0.5, [L] = IlK. If there are n identical binding sites on the protein, the binding isotherm for the macromolecule is simply the sum of those for each of the sites: nKlLl f=-

(6-4)

1 + KlLl

A plot of r versus [L] has the same shapeas before, but now the limiting value of r at high ligand concentrations is n, the number of identical binding sites on the macromolecule, and when r - nl2, [L] = 1/K. Thus, a study of ligand binding to a protein containing n identical binding sites permits determination of the number of binding sites and the equilibrium binding constant.In practice, alternative plots of the data are frequently used that yield a straight line plot and therefore can be analyzed more easily. With the availability of nonlinear least squaresprograms on desktop computers, this is not really necessary.However, it shouldbe kept in mind that statisticalanalyses always fit the data-this doesnot necessarilymean that the fit is a good one. The quality of the fit must carefully be examined to be sure that the equationsused and the fitting proceduresare appropriate.In addition, it is extremely important that a very wide range of ligand concentrationis used. The most obvious recastingof Eq. 6-4 is to take its reciprocal: Ilr - Iln +ll(nK[L))

(6-s)

A plot of Ilr versus 1/[L] is a straight line with an intercept on the [L] axis of lln and a slope of llnK. Another possibility is to multiply Eq. 6-5 by [L]:

lLUr=lLlln+Il(nI9

(6-6)

It lLUr is plotted versus [L], a straight line is obtained with a slope of Iln and an intercept of llnK. A common alternative plot is a Scatchardplot (4), named after George Scatchard,a pioneer in the study of small molecule binding to proteins. Rearrangement of Eq. 6-4 gives

rlfLl=nK-rK

(6-t)

A plot of r/[L] versus r is a straight line, with the intercept on the x axis (ri[L] = 0) being equal to the number of binding sites, n, and the intercept on the y axis being equal to nK. Examplesof thesestraightline plots are included in Figure 6-1.

126

L I G A N DB I N D I N GT O M A C R O M O L E C U L E S

(a)

FIGURE

6-1. Plots of binding isotherms for n identical sites on a macromolecule according to (a) Eq. 6-4, (b) Eq. 6-5, (c) Eq. 6-6, and (d) F.q. 6-7. when n = l,the curve in (a) obeys Eq. 6-3.

6.3 MACROSCOPIC AND MICROSCOPICEQUILIBRIUM CONSTANTS

127

6.3 MACROSCOPIC ANDMICROSCOPIC EQUILIBRIUM CONSTANTS Before we proceedfurther with the analysisof ligand binding to macromolecules,it is important to understandthe distinction betweenmacroscopicand microscopicequilibrium constants.Consider,for example,a dibasic acid such as the amino acid glycine. Four possibleprotonation statesare possible: GH; = *H3NCH2COOH GH = H"NCH"COOH GH, = *H,NCH,COOGH- = H2NCH2COOIf a pH titration is carried out, the statesGH and GH'cannot be distinguishedas they contain the same number of protons. The pH titration can be used to determine the macro scop i c tonizatton constants:

K, = ([GH] + tGH'l)tH-l/tGH;l

(6-8)

(6-9) K2 = [G-][H+](tGHl + [GH']) The two pK values determinedby pH titration are pKr = 2.35 andpKr= 9.78 (25"C and zero salt concentration.) If we consider the microscopic statesof glycine, four microscopic ionization constants, k,, areneededto characterlzethe system: 'H.NCH.COO_

\

/ -H"NCH.COOH

H2NC H2COO-

/

\ H2NCH2COOH

Note that thesefour microscopicionization constantsarenot independentbecausethis is a closed cycle. If the four microscopic ionization constantsare written in terms of the concentrations,it can be seenthat

krk, - krko

(6-10)

128

LtcANDBtNDtNG To MAcRoMoLEcULES

This is an example of the principle of detailedbalance.Whenever a closed cycle of reactions occurs, a relationship between the individual equilibrium constantssuch as Eq. 6- 10 exists.The relationshipbetweenthe macroscopicand microscopicionization constantscan easily be seenby referenceto Eqs. 6-8 and 6-9, namely,

Kr

+kz

(6-11)

and Kr= krkol(\ + k4)

(6-12)

Sincethe two macroscopicionization constantsand Eq. 6-10 provide only three relationships between the microscopic ionization constants,it is apparentthat the microscopic ionization constants cannot be determined from a pH titration. In order to calculatethe microscopic ionization constants,additional information is needed.In this case,it might be assumedthat ft, is equalto the ionization constantdeterminedby pH titration of the methyl esterof glycine, namely, pkz=7.70. This assumesthat the ionization constantfor protonation of the amino group is the samewhen either a proton or a methyl group is bound to the carboxyl, which is not unreasonable.With this assumption,the other threemicroscopicpK valuescanbe calculated:pkr = 2.35 pkz = , 9'78, andpko= 4.43. Theseresultsindicate that the bottom statein ihe ionization schemeabove is present at very low concentrationsthroughout a pH titration. This is a simple illustration of the distinction between macroscopicand microscopicequilibrium constants.Experimentalmeasurementsusually only give information about macroscopicbinding constantsalthoughexceptionsexist. For example,the stateof protonation of the nitrogen can be monitored by nuclearmagneticresonance (NMR) so that' in principle, a titration curve can be obtained for the amino group independently of that for the dibasic acid.

6.4 STATISTICAL EFFECTS IN LIGAND BINDING TO MACROMOLECULES Let us return again to the binding of a ligand to multiple identical binding sites on a macromolecule and consider the matter of macroscopic and microscopic equilibrium constants'As a simple example,considera macromolecule,P, with two identicalbinding sites;symbolically one of the binding siteswill be on the left of p and the other on the right of P. The macroscopicequilibrium constantsare

K,=(pl-l+ tl-Pl)/tl-ltpl

(6-13)

and

Kr= [PLr]/ttLl(tPLl+ tl-pl)]

(6-14)

6.4 STATISTICALEFFECTSIN LIGANDBINDINGTO MACROMOLECULES

129

In relating this discussion to the previous consideration of glycine, it should be noted that the glycine protolytic equilibria are charactenzed by equilibrium dissociation constantswhereasequilibrium associalion constantsare used here. The equilibrium binding isotherm can be written as

lPLl+[LP]+z[PL,]

(6-1s)

lPl+[PL]+[LP]+[PL2] If both the numeratoranddenominatoraredividedby [P], it caneasilybe seenthat Kr [L] + 2K rKrLLlz

(6-16)

1 + K , [ L 1+ K . K 2 [ L ) " This doesnot look the sameasEq. 6-4. This is becausethe equilibrium constant 6-4 is the microscopic equilibrium constant.In this caseit is easyto seethat Kt andK, = Kl2.If theserelationshipsareput into Eq. 6-16,

zKlLl(r + rK[L]) (1 + rK[L])2

2KlLl f=-

I +KLl

Thus, in the caseof multiple identical binding sites,the binding isotherm is the same whether macroscopicor microscopic equilibrium constantsare considered.The relationship between microscopic and macroscopicconstantsfor this particular caseis particularly simple. When both sites are empty, there are two possible sites available for the ligand so that the macroscopicequilibrium constantis twice as large as the microscopicconstant.When both sites are occupiedby ligand, there are two sitesfrom which the ligand can dissociate;therefore,the macroscopicequilibrium constantis one-half of the microscopicconstant. If more than two identical binding sites are present,the relationshipbetweenthe macroscopic and microscopic equilibrium constants can be determined in a similar manner. If a macromolecule hasn identical binding sites,the relationship betweenthe macroscopicequilibrium constantfor binding to the ith site and the microscopicequilibrium constant is

[{. -

Number of free sites on P before binding K Number of occupied sites on P after binding

K,=l(n-i+L)/i1K

,>1

(6-17)

with n identicalbindingsites. of a macromolecule We now returnto consideration

130

LTcAND BTNDTNG To MAcRoMoLEcuLEs

The macroscopicequilibria are L+P=PL

L+PL=PL2

(6-18)

L+pL,,-pL, Thecorresponding macroscopic equilibriumconstants are K, = [PL]/(tPltl-l)

(6-1e)

Kn = fp L,l / ('&)tpl-,_II ) Thebindingisothermcanbe writtenas [PL]+ 2lPL2l+... +nlpL,)

l P l +[ P L + ] [ P L 2+] . . . + t P k l

(6-20)

If we divide the numeratorand denominatorby [P] and usethe definitions of the macroscopic dissociationconstantsfor the ratios [pl-r]/[p], Eq. 6-20becomes

r-

1(1[L]+ 2KrKr[L]' + . . . + nKrK,r.. . K,lLl" 1 +K,[LJ+ KrKr[L)2+. . .+ KrKr. . .K,[L]"

(6-2t)

This equation describesthe binding of a ligand to n dffirent binding sites as no relationship hasbeenassumedbetweenthe binding constants.If the sitesare identical,the macroscopicconstantsare relatedto the microscopicor intrinsic binclingconstantby Eq.6-17.If this relationshipis insertedinto Eq.6-2l, it becomes

nKlLl."+I

PWt'+ . . . + nK,tLtn

r + nK[L]* "@ul') PU,lr+ . . . + K,ILJ"

__ nKlLl(r + I(ILl)^-t _ nKLLI (1 +4Ll)" 7 +rK[L]

(6-22)

6.4 STATISTICALEFFECTSIN LIGANDBINDINGTO MACROMOLECULES

131

The binomial theorem has been used to obtain the final result; that is, the seriesof terms in the denominatorthat contain successivepowers of [L] is recognizedas the expansionof (1 + K[L))'. Similarly, the expansionof (1 + IflL])n-I canbefactoredout of the numerator.This result is exactly the sameas Eq. 6-4, which was obtainedin a more intuitive manner. To conclude this section, we consider the situation where a macromoleculehas more than one set of independentidentical binding sitesfor a ligand. In this case,the equilibrium binding isotherm is simply a sum of the terms given in Eq. 6-4, with the number of terms in sum being equal to the number of setsof binding sites:

s n,K,lL) r = ?> | + K,tLl

(6-23)

Here the n,and K, are the number of sitesin each setand the intrinsic equilibrium constantfor each set,respectively.Unlessindependentinformation is availableabout the structureof the macromolecule,it is usually preferableto use Eq. 6-21to fit the data since no assumptionsare made about the nature or number of the binding sites in this case. Virtually any ligand binding isothermcan be fit to Eq. 6-2I, but this is not a meaningful exercise unless the result can be related to the structure and/or the function of

(a)

-8

-6

-7

-5

-4

log [L]

(b) l07

=

x s'to6 J

105 0123456789

r

FIGURE 6-2. (a) Plot of r versus log [L] for the binding of laurate ion by human serum albumin. (b) The samedataplotted as r/[L] versusr (Scatchardplot). Adapted from I. M. Klotz, Ligand-Receptor Complexes,John Wiley & Sons, Inc. New York, 1997. Copyright @ 1997 John Wiley & Sons,Inc. Reprintedby permissionof John Wiley & Sons, Inc.

132

LTcAND BtNDING To MAcRoMoLEcULES

the macromolecule. For example, the binding isotherm for the binding of laurate ion to human serum albumin is shown in Figure 6-2 (5,6). The concentrationrange covered is so large that the concentration axis is logarithmic. The corresponding Scatchardplot alsois shownin Figure 6-2.Itis clear that saturationof the binding sites on serumalbumin is never reached.The data in the figures, nevertheless,can be well fit with the assumptionthat n - I0. Clearly, this doesnot establishthat 10 binding sites exist, nor regrettably does it provide information about the structure of the macromolecule. We will not dwell on this matterhere. We will, however, return to the topic of multiple binding siteswhen discussingthe conceptof cooperativity.

6.5 EXPERIMENTAL DETERMINATION OF LIGANDBINDING ISOTHERMS We will now make a digressionto discussbriefly someof the experimentalaspectsof ligand binding studies.Many different experimental methodsexist for determining the binding of ligands to macromolecules.The experimentalmethodsused fall into two classes:(1) direct determinationof the unbound ligand concentration;and (2) measurementof a changein physical orbiological property of the ligand or macromolecule when the ligand binds. Generally, stoichiometryand binding constantscan be determined reliably only if concentrationsof the unboundligand, the bound ligand, and unbound macromoleculecan be determined. The most straightforwardmethod is equilibrium dialysis. This method is pictured schematicallyin the diagram in Figure 6-3. With this method, a solution is separated into two pans by a semipermeablemembrane,which will not permit the macromolecule to cross but will permit the ligand to passfreely. The macromoleculeis put on one side of the membrane and the ligand on both sides.The system is then permitted to come to equilibrium. At equilibrium the concentrationof unbound ligand is the sameon both sidesof the membrane.(Strictly speaking,their thermodynamicactivities are equal, but we will not wory about the difference betweenactivitv and conSemi-permeablemembrane

Macromolecule + Ligand

FIGURE 6-3. Schematic representation of an equilibrium dialysis experiment. The macromolecule is on one side of a semipermeablemembrane and cannot pass through the membrane.The ligand can pass through the membrane and therefore is on both sides of the membrane.The concentrationof the ligand on the side of the membranethat doesnot contain the macromolecule is equal to the concentration of unbound ligand.

6 . 5 E X P E R I M E N T ADLE T E R M I N A T I OONF L I G A N DB I N D I N GI S O T H E R M S

133

centration here.) If the total amount of ligand is known, and the unbound amount of ligand is known, the amount of bound ligand can be determinedfrom massbalance. The concentration of the macromolecule can be determined independently so that the binding isotherm can be calculated directly. If equilibrium dialysis is to provide reliable data for analysisof the binding isotherm, it is usually necessaryfor the concentrationsof bound and unboundligand and macromolecule to be comparable.If essentially all of the ligand is bound, the amounts of unbound ligand and/or protein becomevery small and difficult to determine.On the other hand, if essentiallyno ligand is bound, calculationof the bound ligand becomes problematical. In the former case, the stoichiometry of binding can often be determined, but not the binding constant(s).In the latter case,it is virtually impossibleto determine either the stoichiometry or binding constant(s).Another way of stating this is that the concentrationsof all the speciesshould be the sameorder of magnitude as the reciprocal of the binding constant so that experimental points can be determined over a wide rangeof r. This limitation becomesrestrictivefor very weak binding, primarily in terms of the large amount of material required. In the caseof very tight binding, it may be difficult to work with the very dilute solutions required. Equilibrium dialysis is an example of the more general method of determining the distribution between phases.This could, for example, be the equilibration between two nonmisciblesolventsthat partition the reactants,althoughthis is rarely used.Various gel exclusion media suchas Agaroseare,however,often used.Such gels will exclude macromoleculesfrom their beads,with the size of the macromoleculeexcluded depending on the property of the specific gel. The gel will include the ligand so that the gel bead is serving as semipermeablemembrane,with equilibration occurring between the inside and outsideof the bead.If a macromoleculeis passedthrough a column equilibrated with a given ligand concentration,the unbound ligand concentration is equal to this concentration.The amount of ligand bound and the macromolecule concentration can be determined by analysis of the column effluent. Many variations on this type of experimentalprotocol exist. In somecases,the concentrationof unboundligand is determineddirectly in the absenceof a semipermeablemembrane,for example,by a biological assay.In this case, the implicit assumptionis that the rate of equilibration of the ligand and macromolecule occursmore slowly than the time for analysis.This is not necessarilythe case,so such methodsmust be used with extremecaution. Perturbations in the physical properties of the ligand are often used to determine binding isotherms.However, some care and caution must be exercisedin doing so. The most frequentsituationis when the optical absorptionpropertiesof the ligand are different for the bound and unbound ligand, or for the macromolecule with and without ligand. The techniqueof differencespectroscopyis often used.With this method, the differencein the optical absorptionof a solution containing ligand and a solution containingthe sameconcentrationof ligand plus the macromoleculeis determined.If the concentration of total ligand in the first solutions is [\J and the concentration of free (unbound)ligand and bound ligand are [Lr] and [L61,respectively,this difference, La, can be written as

134

LTcAND BTNDING To MAcRoMoLEcuLEs

a,a= ErfLrl - er[Lr1- eo[\J

(6 24)

La = (tr_ er)thl where the e, are extinction coefficients and the relationship thl = [LrJ + tl-b] has been used. This derivation assumesthat the extinction coefficient of bound ligand is the sameat all binding sites,which may or may not be valid. Consequently,this method is not as direct as the partitioning methods previously described. The difference extinction coefficient, tr - t5, c?n be determinedby extrapolation to high ligand concentrations where the macromolecule is saturatedwith ligand. The difference absorbance measurementswill then permit the determinationof the bound ligand concentration, and the concentrationsof the other speciescan be determinedby massbalance.This method also works if the spectralchangeoccurs in the macromoleculerather than in the ligand. The only difference is that the comparisonsolution containsmacromolecule rather than protein, and the concentrationsand extinction coefficients of the protein appearin Eq. 6-24. Perhapsnot so obvious is the fact that this method also works if spectralchangesoccur in both the ligand and macromolecule when the ligand binds. We will not deal with this more complex situation,but the derivation is similar to the simplercasesdiscussed. As a specific example of a difference spectrumtitration, the difference absorbance is shown as a function of the ligand concentrationfor the binding of 2'-cytidine monophosphateto ribonucleaseA in Figure 6-4 (7). Other optical propertiessuchasfluo-

10

20

30

40

50

ILI(rrna) FIGURE 6-4. Plot of the absorbancedifference at 288 nm, Ad, accompanying the binding of 2'-CMP to ribonucleaseA versusthe concentrationof unbound 2:-CMP at pH 5.5, 25"C. The total enzyme concentrationis 10-4 M. The differenceextinction coefficient is 2.88 x 103cm-1 M-r and the binding constantis 2.96 x 105M-I. The curve was calculatedfrom the data in D. G. Anderson,G. G. Hammes,andFrederickG.Walz,Jr.,Biochemistry7,1637(1968).The dashed line is the difference absorbancewhen all of the enzyme has been converted to enzyme-2'-CMP.

6 . 6 B I N D I N GO F C R O R E P R E S S O R P R O T E I NT O D N A

135

rescenceand circular dichroism can be used in a similar manner.Finally, it is sometimes possibleto monitor a specific site on the macromoleculewith physical methods such as NMR or other types of spectroscopy.If changesoccur when ligand binds, the concentrationsof empty and occupied sites can be determined and translatedinto binding isotherms. The determinationof high-quality binding isothermsis not trivial, and this discussion doesnot do fulIjustice to the topic. One of the difficulties not discussedthat should be mentionedis the occurrenceof nonspecificbinding. It is commonplacefor nonspecific binding to occur along withbinding of the ligand to the site(s)of interest.Macromolecules have many different side chains that can attract ligands, for example, by charge-charge interactions.This nonspecificbinding is characteized by binding isothermsthat do not level off at a specific value of n due to a large number of such sites characteized by very weak (small)binding constants.More completeffeatisesconcemedwith the experimental aspectof the binding of ligands to macromoleculesare available(cf. Ref. 3).

6. 6

BIN DI NG O F CRO R E PR ES SOR P R OT E IN T O D N A

As an exampleof a ligand binding study,we will considersomeof the resultsobtained in the studiesof the binding of the Cro repressorprotein to DNA. DNA transcription is regulated by the binding of proteins to DNA. The Cro repressor is a protein that plays a regulatory role in l" phage.The binding of this protein to DNA has been examined by many different methods (8-10), and the three-dimensionalstructureof a Cro protein-DNA complex has been determined(11). The Cro protein binds as a dimer, and amino acids form DNA sequencespecific hydrogen bonds with exposed parts of DNA bases.Both the thermodynamicsand kinetics of Cro protein binding to DNA have been studiedextensively. The binding of Cro protein to DNA is difficult to study becausethe binding is very tight. In order to study the binding, a very sensitivefilter binding assaywas used.Under certain conditions,linear duplex DNA passesthrough nitrocellulosefilters, while a protein or protein-DNA complex is retained.Radioactive132P;DNA was prepared and mixed with Cro protein at very low concentrations(as low as < 1ng/ml). The reaction mixture was filtered within a few seconds,and from the radioactivity bound to the filter, the concentration of Cro protein-DNA could be calculated. Since the total concentrationsof Cro protein and DNA are known, the equilibrium concentrationsof all speciescan be calculated.The binding of Cro protein to various operatorregions of l, phageDNA was studied.The operatoris 17 basepairs long, but the DNAs investigated containedan extra 2 basepairs at each end. In addition, a consensusoperator was constructedthat had the consensussequenceof the six l, operatorregions examined. The consensussequenceis identical to the OR3 operator,except for a few base pairs on the interior of the sequence. The equilibrium association constants obtained are given in Table 6-1. The stoichiometry of the reaction was determinedto be one to one. The associationconstantsare very large, with the largestbeing for the consensusand OR3 operator.The OR3 operator was previously determined to be the preferred binding site for Cro pro-

136

LTGAND BTNDING To MAcRoMoLEcULES

tein. The binding of Cro operatorto nonspecificDNA was also determinedand found to be a linear function of the length of the DNA with a binding constantof 6.8 x 105 M-l per basepair, considerablyweakerthan the specificbinding, asexpected.The rate constantscharacterizing the reaction between the DNA operator sequencesand Cro protein were also determined. The dissociation rate constantsare included in Table 61. The associationrate constantsthat could be measuredwere all approximately3 x 108M-l s*1,the value expectedfor a diffusion controlled reaction.Every time a Cro protein and the operator DNA collide, a complex is formed. The binding of Cro protein to long DNAs containing the operator region was also studied.The length of DNA varied from 73 toz4l}base pairs. For all except the shortest DNA, the equitibrium constantwas aboutthe same,6.1 x l0l0 M--1.However, the associationand dissociationrate constantsincreaseas the length of DNA increased, leveling off at valuesof about4.5 x 10eM-1 s-l and 1.7 x !0-2 s-l, respectively.This very high secondorder rate constantsuggeststhat every time Cro protein collides with DNA, it binds very tightly. This result is puzzling asit would be expectedthat Cro protein would have to sample various parts of the DNA until it found the operator and bound tightly. An explanation for these results is that Cro protein binds to the DNA on every collision and then diffuses rapidly along the DNA chain until it encounters the operatorregion. The rapid sliding of DNA binding protein along the DNA chain until it finds the correct place for a specific interactionappearsto occur in other systems,but other mechanismsmay also be operative. Calorimetry was also carriedout on the binding reaction(10). The resultsobtained indicate that the associationof Cro protein with nonspecificDNA at 15"C is charactenzedby AF1"= 4.4kcaVmol, A.So= 49 cal/(mol.K),AG" = -9.j kcal/mol, and LC, = 0. The parameters obtained with the OR3 DNA are quite different, Aflo = 0.8 kcavmol, ASo= 59 cav(mol.K),aG'= -16.1 kcavmol, and LCp- -360 call(mol.K). In both cases,the favorable free energy changeis entropy driven. The specific molecular interactionsbetweenthe DNA and Cro protein have been probedby studying the binding of Cro protein to OR1 DNA, 2l basepairs as above, with systematicbasesubstitutionsalong the entire DNA chain (9). Experimentswere

TABLE 6-1. Binding Constantsand DissociationRate Constantsat 273 K for Cro Protein-l, DNA Operator Consensus

oR3 oR2 oR1 oL1 oL2 oL3

1l'(M-')

8 . 3x 1 0 r l 5 . 0x 1 0 r r 8 .3x 1 0 e 1 . 2x 1 0 r r 6 . 7x 1 0 r o 3 . 7x 1 0 r o 1 . 9x 1 0 r o

ta (s-t)

7.7x l}-s 1.1x 10-4 L2x l0-2 1.7x 10-3 2.2x103 2.6 x 1O-3 9.2x I0-3

Source: J. G. Kim, Y. Takeda,B. W. Matthews, and W. F. Anderson,-L Mot. Biot.196.14g (1987).

6.6 BINDINGOF CRO REPRESSORPROTEINTO DNA

137

also done with point mutations in the protein of amino acids thought to be involved in the binding interaction. The changesin free energy were then interpretedasdue to specific interactions between the protein and DNA, and a structural map of the interactions was proposed. A high-resolution crystal structure of a Cro protein-DNA complex determined some years later provides a definitive description of the molecular interactions (11). Many features proposed on the basis of the binding/mutation studies proved to be correct although some were not. As often statedin this text, molecular interpretations of thermodynamic studies must be viewed with caution. However, in retrospect,all of the mutation studiescan be understoodin terms of the crystal structure.A view of the overall complex is presentedin Color Plate VI. Each Cro protein monomer consists of three cr-helicesand three B-strands.Only the helix-turnhelix portion of the protein makes direct contact with the DNA bases:the ar-helix is inserted into the major groove of DNA, and its interactions with DNA basesaccount for the tight operator binding. A schematicmap of the DNA base-Cro protein interactions is shown in Figure 6-5. Multiple hydrogen bonds are formed, although other types of interactionsare also important. Both the protein and DNA conformations are altered when the complex is formed. Many other interesting details of the mo-

FIGURE 6-5. Schematic diagram of the interactions between the Cro protein and l" operator DNA. The interactions for one of the polypeptide chains are shown as the interactions with the other polypeptide chain of the dimeric Cro protein are syfirmetric. Hydrogen bonds are shown as continuous lines with arrows pointing from the donor to the acceptor. Broken lines are van der Waals contacts.bkbn indicates a contact with the protein backbone, and the dotted lines are presumed electrostatic interactions. Reproduced from R. A. Albright and B. W. Matthews, ./. Mol. Biol.280,137 (1998). Reprinted with permissionfrom Academic Press,Inc.

138

TO MACROMOLECULES BINDING LIGAND

lecular interactions can be inferred from the crystal structurebut will not be consideredhere. This study is an elegant demonstration of the type of information that can be obtained from thermodynamic, kinetic, and structural studies of ligand binding.

6.7

COOPERATIVITY IN LIGAND BINDING

We now return to the subject of macromoleculeswith multiple ligand binding sites. One of the most famous and best studied examples is hemoglobin. In Figure 6-6, the binding isotherm is shown for the binding of oxygen to myoglobin and hemoglobin. Hemoglobin containsfour identical binding sites for oxygen. Myoglobin contains a single binding site for oxygen. The concentrationof unbound oxygen is expressedin pressureunits of mmHg and rather than r, the percent saturation(r/n) is shown so that the binding isotherms can be compareddirectly even thoulh n - 1 for myoglobin and 4 for hemoglobin. The binding isotherm for myoglobin can be fit to Eq. 6-4 wtth K = 0.23 (mmHg)-l {K = [MyO2l/(tMy]tOrl), where My is myoglobin]. The data for hemoglobin clearly cannot be fit so simply, as the binding isotherm is not hyperbolic. Equation 6-21canbe used,however.Before doing so it is useful to take into account the statisticaleffects associatedwith identical binding sites,that is, expressEq.6-21 in terms of intrinsic binding constantsrather than macroscopic binding constantsby use of Eq. 6-17. With this substitution,Eq.6-21becomes

,:

4K Jo 2l + 12K, K rlOrl' + I 2K rKr\lo

+ 4K rKrKrK4[o]4 213

(6-2s)

c80 o

e €uo 6 6l

o

*40 tr o

g

tzo -0

20

rto 80 60 pO2(mmHg)

The FIGURE 6-6. Binding of oxygen by myoglobin and hemoglobin at pH 7 and 38'C' R' and F. Daniels from Adapted Oz. of unbound pressure of Oz is a measureof the concentration Copyright 1975. York, New Inc., Sons, & Wiley John edition, A. Alberty, Physical Chemistry,4th O 1975John Wiley and Sons,trnc.Reprintedby permissionof John Wiley & Sons,[nc.

INLIGAND 6] COOPERATIVITY BINDING 139 If this equationis usedto fit the dataobtainedat pH7 .4,25oC,0.1 M NaCl, the binding constantsobtainedare0.024,0.077,0.083,andT.l (mmHg)-r (12). Note that the intrinsic binding constantbecomeslarger as each oxygen is bound. This is an example of cooperative binding. When an oxygen binds to hemoglobin, it increasesthe affinity of hemoglobin for the next oxygen. The result is a sigmoidal binding isotherm rather than a hyperbolic isotherm. This has important physiological consequences as it permits oxygen to be picked up and releasedover a very narrow range of oxygen pressure.Cooperative binding such as this is found frequently becauseit permits the biological activity to be regulated over a very narrow range of concentration. When the binding of oxygen by hemoglobin was first studied,the binding isotherm was fit by the empirical equation

rln=

lLls/I( I + lLla/I(

(6-26)

Where the equilibrium constant is expressedas a dissociation constant and s is an empirical parametercalled the Hill coefficient obtained from experiment by plotting ln[(rln)l(l - r/n)] versusln[L] (13). {Note that (rln)/(l - r/n) = [L]cyK([.] As might be expected,such plots are linear over a limited range of ligand concentration.In the case of hemoglobin, o is about 2.5 and dependson the specific experimental conditions. The useof this equationdoesnot have aphysicalmeaning.It assumesthat the binding equilibrium is

Hemoglobin + aO, S Hemoglobin(O2)o Obviously the number of binding sites on hemoglobin must be an integer. Nevertheless, cooperative binding is frequently characterrzedby a Hill coefficient. The closer the Hill coefficient is to the actual number of binding sites, the more cooperative the binding. It is also possible to have binding isotherms in which the binding constantsdecreaseas successiveligandsbind. This is usually termednegativecooperativity,or anticooperativity,bothoxymorons of a sort. It should be rememberedthat if macroscopic equilibrium binding constantsare used,the binding constantsdecreaseas eachligand is added to the macromolecule even if the sites are equivalent. This is the statistical effect embodied in Eq. 6-17. Anti- or negative cooperativity therefore meansthe binding constantsdecreasemore than the statistical effect expectedfor equivalent sites.An example of negative cooperativity is shown in Figure 6-7 for the binding of cytidine 3'-triphosphate (CTP) to the enzyme aspartatetranscarbamoylase(14). This enzyme catalyzes the carbamoylation of aspartic acid at a branch point in metabolism that eventually leads to the synthesisof pyrimidines. When the concentration of CTP becomes high, it inhibits aspaftate transcarbamoylaseand shuts down the metabolic pathway for its synthesis.As can be seenfrom the Scatchardplot presented,the data

140

LTcAND To MAcRoMoLEcULES BTNDTNG

345678 f

FIGURE 6-7. Scatchardplot for the binding of CTP to aspartatetranscarbamoylaseat 4oC and pH7 .3 in the presenceof 2 mM carbamoyl phosphateand 10 mM succinate.Adapted from S. Matsumoto and G. G. Hammes, Biochemistry 12,1388 (1973). Copyright A Dn American Chemical Society.

clearly do not confom to the straight line expectedfor independentequivalent binding sites.The total number of binding sitesappearsto be six, although extrapolation to this number is not precise. In this case,the data were fit to a model of two setsof three independentbinding sites.The intrinsic binding constantsobtainedwere 7.Ix 10sM-l and4.4 x 103M-1. The data could be fit equally well to Eq. 6-21, but the data were not sufficient to define all six constantswell. Aspartate transcarbamoylaseis known to contain six identical binding sitesfor CTP, which exist as three dimers of identical polypeptide chains. When CTP binds to one of the sites on a dimer, it weakensthe binding for the secondCTP. Interestingly,the binding of aspartateto this enzymeexhibits positive cooperativity, and the extent of the cooperativity is modulated by the binding of CTP, as will be discussedlater. The presenceof positive or negative cooperativity can readily be diagnosedfrom the binding isotherm. The shapesof the various plots commonly used are shown in Figure 6-8 for no cooperativity, positive cooperativity, and negative cooperativity. The type of cooperativity occurring can readily be diagnosedfrom theseplots. It is particularly obvious in Scatchardplots as no cooperativity gives a straight line, a maximum in the plot is observedfor positive cooperativity, and a concavecurve is found for negativecooperativity.For some studies,log[L] is usedin order that a wide rangeof concentrationscan be representedin a single plot (Fig. 6-8b). When positive and negative cooperativity are observed,one must determine if this is due to a preexisting difference between the binding sites, or if the binding of one

N L I G A N DB I N D I N G 6 . 7 C O O P E H A T I V I TIY

141

IL] (a)

Log [L] (b) FIGURE 6-8. Schematic representations of equilibrium binding data demonstrating no cooperativity,positive cooperativity,and negativecooperativity.In thesefigures,r is the moles of ligand bound per mole of protein and [L] is the concentration of unbound ligand.

142

LtcANDBtNDtNG To MAcRoMoLEcULES

(d) FIGURE 6-8. (Continued)

6.8 MODELS FORCOOPERATIVITY 143

ligand alters the binding affinity of the remaining sites for the ligand. Usually, but not always,the latter is true, as for the casescited above.The possibility also exists that both positive and negative cooperativity could occur in a single binding isotherm.

6.8

MODELS FOR COOPERATIVITY

Thus far we have been concernedwith fitting data to binding isotherms and developing criteria with regard to whether the binding is cooperative. However, the ultimate goal is to relate the experimental data to molecular structure. This requires the development of theoretical models that relate strucfure to ligand binding isotherms. Two limiting models have been developed to explain cooperative ligand binding to proteins. Both models are basedon the generalhypothesis that cooperativity is the result of alterations in the interactions between polypeptide chains through conformational changesin the macromolecule.One of the models assumesa concertedor global conformational change,whereasthe other assumesa sequentialchange in the conformation of each polypeptide chain or chains that contain a ligand binding site. We first considerthe concertedmodel that hasbeendevelopedby Monod, Wyman, and Changeux(15). This model (MWC) is basedon three assumptions:(1) The protein consistsof two or more identical subunits,each containing a site for the ligand; (2) atleast two conformational states,usually designatedasR and T states,are in equilibrium and differ in their affinities for the ligand; and (3) the conformational changes

ConcertedModel(MWG)

nt.m+ffi.+r ,ll

,'-m==-EE tl

+3L

ll

lf

,t.m:m

+2L

tl

ll

L*[f[l=-QQ *L tLt

,ll

I

G ,

,ll

rr| _GXD iLE:@ FIGURE 6-9. Schematic representation of the Monod-Wyman-Changeux model for a four subunit protein. The squares and circles designate different subunit conformations of the protein, and L is the ligand.

144

TOMACROMOLECULES BINDING LIGAND

of all subunits occur in a concertedmanner (conservation of structural symmetry). n schematicillustration of the MWC model for a four subunit protein is shown in Figure 6-9. A sigmoidal binding isotherm can be generatedfrom this model in the following way. In the absenceof ligand, the protein exists largely in the T state (the squareconformation), but substratebinds preferentially to the R state (the circular conformation). When ligand binds, it shifts the protein from the T to the R state.Thus, at low ligand concentrations,the protein is primarily in the T state whereas at high ligand concentration,the protein is largely in the R state.This shift in equilibria can give rise to a sigmoidal binding isotherm, or positive cooperativity.This model cannot,however, explain anti (negative) cooperativity. The quantitative development of this model is complex, although not difficult, and will only be given in outline form here. The schemefor ligand binding can be representedas

nL+Ro

-

'r

r0

+nL

1t1[ 1l 1t

(n-l)L+Rt

-

.F rl

+ (n-l)L

::

lt

ilil

[[ ( n - i ) L + Ri *

/l

(6-27)

T ,+ (n - i)L

lllt iI tl l l

tr

ll

tl

aa

::

ii ll / l ilil

Il

ltll D'F

\ f l - _ L n

where Ro and To are the two different conformational statesin the absenceof ligand, and R, and T, designatetheir complexes with i molecules of L. Three constantsare neededto specify the equilibrium binding isotherms:the intrinsic dissociationconstantsfor ligand binding to the R and T states,and the equilibrium constantfor the ratio of the Ro to To states.These constantscan be written as

6.8 MODELSFOB COOPERATIVITY

145

Lo = [Tsl/[RsJ

Kn = [(n-i + Iyi]lRir llLl/[R,]

K, - f(n - i + t)l illZ,_ll[L]/[T,l

(6-28)

as Thefractionof sitesoccupiedby theligand,Y, canbe expressed

Y_

Loccr(l+cu)"-t +u(l +cr)'-l L o ( l+ c u ) " + ( 1+ c r ) '

where 6x= [L]/rKp and c = KslKr.The transformation of the first part of Eq. 6-29 to the secondpart requires the use of the binomial theorem and will not be detailed here. The nature of the binding isotherm dependson the values of Lo and c. A hyperbolic isothermis obtainedwhen the ligand binds equally well to both conformations,c = I, and when Lo and c are both either very large or very small. However, when Lo is large and c is small, sigmoidal binding isothermsoccur as illustrated in Figure 6-10. When c is very small, F,q.6-29 becomes

FIGURE 6-10. Plot of the fraction of sitesbound by ligand, Y, versusa (= [L]/Kn) for various values of c and L according to Eq. 6-29. A sigmoidal binding isotherm is generatedwhen L is large and c is small.

146

TOMACROMOLECULES BINDING LIGAND

a(l

(6-30)

+ cx,)t?-l

t,7= I - o * 1 1

**

With this limiting case,it can easily be seenthat a hyperbolic isotherm is found when Lo is small, whereasa sigmoidal isotherm is predictedwhen Lo is large. The basic assumptionsof an alternative model developed by Koshland, Nemethy, and Filmer (KNF) are the following (16): (1) Two conformationalstates,A and B, are available to each subunit; (2) only the subunit to which the ligand is bound changes its conformation; and (3) the ligand induced conformational changein one subunit alters its interactionswith the neighboringsubunits.The strengthof the subunitinteractions may be increased,decreased,or stay the same.The resultof this changein subunit interactions is that the binding of the next ligand can be weaker, stronger, or the same as the binding of thepreviousligand. Clearly, this model canproduceeitherpositive or negative cooperativity-or a hyperbolic binding isotherm. This sequentialmodel is shown schematicallyin Figure 6-11 for a four subunitprotein in a squareconfiguration. Calculation of the binding isotherm for the KNF model is complex and will not be presentedhere. The basic parametersare the intrinsic binding constant,an equilibrium constant characterizingthe conformational change that occurs, and constantscharacteizing the subunit interactions, AB and BB. (The AA interaction is taken as the referencestateso it doesnot appearin the calculation.)The binding isothermthat results is identical in form to Eq. 6-21, except that the appropriatestatisticalcorrectionsare included so that the intrinsic binding constantappears.The effective binding constant multiplying each successivepower of L may increase,decrease,or stay the same,depending on the nature of the subunit interactions, so that positive, negative, or mixed cooperativity is possible. The MWC and KNF models are limiting casesof a more general schemeshown in Figure 6-12. This figure illustrates a generalmechanisminvolving a tetrameric protein and only two conformational statesfor each subunit. The real situation is somewhat more complex, as the permutations of the ligand among the subunits for a given conformational state are not shown. The extreme right- and left-hand columns enclosed by dashedlines representthe MWC mechanism,whereasthe diagonal, enclosedby

sequentialModel(KNF)

ffi:m:H:m:m rru EL

ELe

ELo

EL+

FIGURE 6-11. Schematic representationof the Koshland-Nemethy-Filmer model for a four subunit (square)protein. The squaresand circles designatedifferent subunit conformations, and L is the ligand. Note that two structuresare shown for the intermediate with two ligands bound as the subunit interactions are different for these two species.

6.9 KINETICSTUDIESOF COOPERATIVEBINDING

ffiffiffi ffi ffi ffi ffi ffi ffi ffi ffi

147

..

,/it-|

l L l

'rll

l -

t L I L l

|

|

I

'..

:,.'j

FIGURE 6-12. Schematicillustration of a generalallostericmodel for the binding of a ligand to a four subunit protein. The squaresand circles designatedifferent subunit conformations. The portion enclosed by dashed lines is the Monod-Wyman-Changeux model, whereas that enclosedby dotted lines is the Koshland-Nemethy-Filmer model. For the sake of simplicity, the permutations of the ligand among the subunits and the free ligand are omitted. dotted lines, represents the KNF model. Thus, these two models are limiting cases of an even more complex scheme. As might be suspected, distinguishing among these models when positive cooperativity occurs is not a simple task.

BINDING OF COOPERATIVE STUDIES 6.9 KINETIC Thus far we have confined the discussion to ligand binding at equilibrium. The binding processand the cooperativity that may occur play an important role in many biological processes,as evidenced by hemoglobin, membrane receptor binding, and operator binding to DNA. The role of cooperativity in regulating such reactions has been well established.However, many enzyme reactions are also regulated through cooperativebinding processes.In suchcases,the rate of the enzymaticreactionis usually measured,often the initial velocity, and it is observedthat a plot of the initial velocity versus the substrateconcentration exhibits cooperativity. An example already discussedis aspartatetranscarbamoylase:A plot of the initial velocity versus the aspartateconcentration is sigmoidal so that small changes in the concentration of aspartatecan causesignificant changesin the rate of the enzymatic reaction. For many enzymes,plots of the initial velocity versus the substrateconcentration are essentiallyidentical in form with the binding isotherm. This suggeststhat the binding stepsprior to the rate determining step are rapid and reversible and that the turnover numbers (catalytic efficiency) for all of the binding sites are identical. If this is the situation, the initial velocity, v, canreadily be related to the binding isotherm:

148

TOMACROMOLECULES BINDING LIGAND

v=V^r

(6-31)

where V- is the maximum velocity of the enzyme reaction expressedas the product of the molar concentration of enzyme and the turnover number. This simple analysis seemsto be sufficient for many cases,and if it is, the treatmentof the rate dataparallels what was done for equilibrium binding, including the methodsof plotting and fitting the data. As might be expectedboth positive and negative cooperativity are observed. The interpretation of kinetic data can, however, be more complex. The turnover numbers for different sites could be different. For example, in terms of the MWC model, the R and T forms might have different catalytic activities although usually the assumptionis made that only the R form is enzymatically active. For the KNF model each of the ligand binding sites might have a different catalytic activity. In fact, these complications are rarely included, or justified, in the data analysis.Finally, it should be noted that apparent cooperativity in the rate of an enzymatic reaction can arise purely from special kinetic situations and may have nothing to do with cooperative binding of substrate.A few such situationshave been well documented,but we will not considersuch complicationsfurther. This is just a reminder that kinetic measurements are not a substitute for equilibrium binding studies.They may provide similar information in some cases,but they are inherently more difficult to interpret. Of course, conversely, kinetic studies can provide dynamic information that cannot be obtainedfrom equilibrium measurements.

6 .10 A LLO S T E R IS M Thus far we have consideredonly caseswhere a single ligand binds to a macromolecule, and important biological control can occur through cooperative interactions for a single ligand. These arehomotropic interactions.However, in many instancesregulation occurs through the binding of a secondligand that influences the binding of the first ligand. These are called heterotropic interactions.Regulatory control by reaction of a secondligand is termed allosterism, and the secondbinding site is called an allosteric site. Two specific exampleswill be discussedto illustrate the principles involved: hemoglobin and aspafiatetranscarbamoylase. The associationof oxygen with hemoglobin is strongly pH dependentas shown in Figure 6-13. Note that the oxygenationisotherm becomesmore sigmoidal as the pH is lowered.Thus, the binding of protonsto hemoglobinclearly affectsoxygenbinding. In fact, the addition of protonsdecreasesthe amountof oxygen bound, and vice versa. This reciprocal relationshipis called the Bohr effect. The effect of proton binding is typical allosteric regulation, in this caseof oxygen binding. How is this accommodatedin the modelswe have discussedpreviously?For the MWC model, allostericeffectors are assumedto bind selectively to the R or T conformation. An inhibitor, I, suchasthe proton in the caseunderconsideration,would bind selectivelyto the T conformation. This shifts the equilibrium from the R to the T state, which effectively changesthe equilibrium constant Lo to

6.10 ALLOSTERISM

149

tr .9 G

3@ F Q

5 ao o o r

pO2(mmHg) FIGURE 6-13. The effect of pH on the oxygenation of hemoglobin (the Bohr effect). The percent of saturation by oxygen is plotted versus the pressure of O. Adapted from R. E. Benesch and R. Benesch,Adv. Prot. Chem. 28, 2ll (1974). Reprinted with permission of Academic Press.Inc.

L l o = L o ( 1+ U l l K ) "

(6-32)

where K, is the intrinsic dissociation constant for the binding of inhibitor, and it has been assumedthat a single proton binds to each of the four subunits. Insertion of this relationship into Eq. 6-29 can quantitatively account for the Bohr effect. The KNF model can also explain the observationsby assuming that binding of the inhibitor alters the subunit interactions and conformational changes,effectively altering the binding constantsfor oxygen as successiveligands are bound. Organic phosphatesare also strong effectors of oxygen binding and bind preferentially to deoxyhemoglobin. An activator, A, can enhanceligand binding by binding selectively to the R conformation. This effectively changesthe equilibrium constant Lo to

r,

to=11

Lo

(6-33)

+1A1/Koy

where Ko is the intrinsic dissociation constant for the binding of activator. Which of the models for allosterism best accommodatesthe known data for hemoglobin? The nature of the conformational change occurring when oxygen binds is known from structural studiesof deoxy- and oxyhemoglobin. The hemoglobin studied contains four polypeptide chains, two G chains and two B chains. The o and B chains are similar but not identical. When oxygen binds, salt bridges are broken between the polypeptide chains, and all four chains rotate slightly to accommodatethe movement of iron into the plane of the heme. The iron moves only a few tenths of an angstrom, and two of the polypeptides move about 7 A closer (17). These structural changesand the binding data fit quite well to the MWC model. However, there is evidence that in addition to this major conformational change,sequentialconformational changesoccur in the subunits. Therefore, it is likelv the casethat both models are neededto ac*

150

BINDING TOMACROMOLECULES LIGAND

commodatethe data, and multiple conformational changesoccur. The structureof hemoglobin and the proposed major concertedconformational change are shown schematically in Color Plate VII. As previously indicated, aspartatetranscarbamoylaseis a key enzymein the pathway for pyrimidine biosynthesis.It is subjectto inhibition by CTP and to activation by ATP. The allosteric binding of CTP makes the dependenceof the rate on aspartate concentration more sigmoidal, whereas binding of the activator, ATP, makes the curve less sigmoidal (18). This is a prototype for feedbackinhibition in metabolism and is shown schematicallyin Figure 6-14. An additional wrinkle in the regulatory processis that the binding of both CTP and ATP to the enzyme display negativecooperativity, and these two ligands compete for the samebinding site. The effect of ATP and CTP on the binding of aspartatecanreadily be accommodatedby the MWC model with the assumptionthat CTPbinds selectivelyto the T conformationand ATP to the R conformation. This assumesthat aspartatebinds selectively to the R conformation. This model, however, cannot accommodatethe negative cooperativity observedin the binding of CTP and ATP. The KNF model also can accommodatethese results through alterations in the interactions between subunits. Again, the structure of aspartatetranscarbamoylaseis known. It consistsof two trimers of catalytic sites, with three dimers of regulatory sites at the interface of the trimers (19). A threefold rotational axis is present,and it has been shown that conversion of the putative R to T forms involves rotation around this axis and alteration of the interactions between the regulatory and catalytic subunits. The MWC model accommodates much of the data, but the data also require sequential conformational changesin the subunits.Thus, the conclusionis similar to that for hemoglobin.A major conformational change consistent with R and T conformations appearsto occur,

.= o o o

I kT

(s

.= tr

+lnhibitor (CTP)

I I I

Aspartate FIGURE 6-14. Schematic representation of the dependence of the initial velocity of the reaction catalyzedby aspartateffanscarbamoylaseon the concentration of aspartate,a substrate. The effect of an allosteric inhibitor, CTP, and of an allosteric activator. ATP. are shown.

REFERENCES 151 but additionalconformationalchangesmore localizedand sequentialin naturealsooccur. The regulatory process, therefore, seemsto require multiple conformations and the interplay between global and local conformational changes.The use of multiple conformationalchangesenhancesthe versatility and sensitivityof the regulatoryprocess.The structure of aspartatetranscarbamoylaseand the nature of the concertedconformational change are shown in Color Plate VIII. The top of the figure is the T state with the catalytic trimers in blue and white at the top and bottom of the structure.Two of the regulatory dimers arc at the sides of the structure in yellow. The third dimer is in the back of the structure. The bottom structure is the R state.The movement of the subunitswith respectto each other can clearly be seen. This concludesour discussionof ligand binding in biology. We have developedthe theoretical and experimental framework and have provided severalexamplesof applicationsto biological systems.This, and further readingof the literature,shouldpermit the interestedreaderto develop specific applicationsas needed.

REFERENCES 1. G. G. Hammes,EnzymeCatalysisand Regulation, AcademicPress,New York, 1982. 2. C. R. Cantor and P. R. Schimmel, Biophysical Chemisrry,Parts I, II, and III, W. H. Freeman,San Francisco, 1980. 3. I. M. Klotz , Ligand-Receptor Complexes,John Wiley & Sons,New York, 1997. 4. G. Scatchard,Ann. N.Y.Acad. Sci. 51, 660 (1949). 5. A. O. Pedersen,B. Hust, S. Andersen, F. Nielsen, and R. Brodersen,Eur. J. Biochem. 154,545 (1986). 6. R. Brodersen,B. Honor6, A. O. Pedersen,and I. M. Klotz, Trends Pharm. Sci. 9,252 (1988). 7. D. G. Anderson, G. G. Hammes, and Frederick G. Walz, Jt., Biochemistry 7, 1637

(1e68). 8. J. G. Kim, Y. Takeda, B. W. Matthews, and W. F. Anderson, J. MoL BioI.196, 149

(1e87). 9. Y. Takeda,A. Sarai,and V. M. Rivera,Proc. Natl. Acad. Sci.UlA86,439 (1989). 10. Y. Takeda,P. D. Ross, and C. P. Mud4 Proc. Natl. Acad. Sci. USA 89, 8180 (1992), 11. R. A. Albright and B. W. Matthews, J. Mol. Biol.280, 137 (1998). 12. I. Tyuma, K. Imai, and K. Shimizu, Biochemistry 12,l49l (1973). 13. A. V. Hill,,f. Physiol.(London)40, iv (1910). 14. S. Matsumotoand G. G. Hammes,Biochemistry12, 1388(1973). 15. J. Monod, J. Wyman, and J.-P.Changeux,J. Mol. Biol.12,88 (1965). 16. D. E. Koshland, Jr., G. Nemethy, and G. Filmer, Biochemistry5,365 (1966). 17. J. M. Friedman.Science228,1273 (1985). 18. J. C. Gerhartand A. B. Pardee,J. Biol. Chem.237,8l9 (1962). 19. W. N. Lipscomb,Adv. Enzymol.68,6'7(1994).

152

BtNDlNcro MAcRoMoLEcULES LTGAND

PROBLEMS 6-1. The following data were obtained for the binding of ADP to an ATPase. IADPI (pM) 0.500 0.694 1 .0 6 r.2 2 1.32 t.6 3 r.6 9 1.91 2 .2 1 2.t5 2.40

0.119 t.2 3 2 .36 3 .30 4.55 8.81 16.32 27.15 40.62 79.15 115.5

How many binding sites are presentper mole of enzyme and what is the intrinsic binding constant?The concentrationin the table is unbound ADP. (Nonspecific binding occurs,which is not uncommon.You will have to decidehow to handle this problem.) 6'2. Some typical equilibrium dialysis data for the binding of a ligand to a macromolecule are given below. The total concentration of the macromolecule is 0.500 uM. Total Ligand Concentration (pM) Side Without Macromolecule

0.158 0.395 0.890 2. 37 4.00 6.18 8.12

Side With Macromolecule

0.436 0.960 r.83 3.78 5.60 7.91 9.90

Determine the number of binding sites on the macromolecule and the intrinsic binding constant. 6-3. The following initial velocities, v, were measuredfor the carbamoylationof aspartic acid by carbamoyl phosphate as catalyzed by the enzyme aspartate transcarbamoylase.The concentration of carbamoyl phosphatewas 1 mM, and the concentration of aspartatewas varied.

PROBLEMS 153 u (arbitrary units)

Aspartate (mM)

0.90 1. 60 2.25 3.20 3.65 4.70 5.05 5.25 5. 80 6.00 6.05

2.0 3 .0 4.0 5 .0 6.0 8 .0 10.0 1 2 .0 1 5 .0 1 1 .0 20.0

A. Assume the initial velocity is relatedto r by Eq. 6-31 and constructa Hill plot of the data. The slope provides a lower bound to the number of aspartate binding sites. B. In fact, six aspartatebinding sites are present per mole of protein. Using this information and the data, make a table of r and the corresponding aspartateconcentration. Make a plot of r/[aspartate] versus r. C. What type of cooperativity is occurring? Which of the two limiting models discussed(MWC and KNF) is consistentwith the data? 6-4. Consider a macromolecule that has two different conformations, M and M'. The two conformations bind a single ligand, L, per molecule, but the binding equilibrium is different for each conformation. This can be representedas K, L+M

,llll

r* 11 ll

r' ll ll

lllr

M,L

ML: K3

where the K, are the equilibrium constantsfor the individual reactions. Calculate the binding isotherm, r,for this macromolecule in terms of the equilibrium constantsand the concentrationof unbound ligand. What type of cooperativity, if any, is displayed by this system?How many of the individual equilibrium constantscan be determined from the binding isotherm? What relationship, if anv. existsbetweenthe four constants?

I

APPENDIX1

StandardFreeEnergiesand Enthalpiesof Formationat 298K, 1 pH 7, and 0.25M lonic Atmosphera, Strength Substance

ATP ADP AMP Adenosine Pi Glucose-6-phosphate Glucose Hzo NADo^ NAD,"d NADPo* NADP."d Acetaldehyde Acetate Alanine Ammonia Ethanol Pyruvate Formate Sucrose Total CO2 2-Propanol Acetone Glycerol Lactose Maltose Succinate Fumarate Lactate

AG" (kJ/mol)

-2091.89 -1230.r2 -360.29 529.96 -1059.49 -1318.92 -426.1r -1 5 5 .6 6 1 0 5 9 .1 1 1r20.09 1011.86 r072.95 24.06 -241.82 -85.64 82.94 62.96 -350.78 -3 11 .0 4 -661.85 -5 4 7 .1 0 140.90 84.89 -171.35 -670.48 -677.84 -530.62 -523.58 -3r3.70

AF1"(kJ/mol)

-2995.59 -200s.92 -1016.88 -5.34 -1299.39 -2279.30 -1267.1r -286.65 -10.26 -41.38 -6.57 -33.28 -2r3.97 -486.83 -551.67 -r33.74 -290.16 -597.04 -425.55 -2208.90 -692.88

-334.rr -224.r7 -679.84 -2242.11 -2247.09 -908.68 -776.51 -688.28 (continuefi

154

A P P E N D I XT

Substance Glycine Urea Ribulose Fructose Ribose Ribose 5-phosphate Aspartate Glutamate Glutamine Citrate Isocitrate cis-Aconitate Malate 2-Oxoglutarate Oxalosuccinate Oxaloacetate Glycerol 3-phosphate Fructose 6-phosphate Glucose l-phosphate

coz(g) ozG) oz(aq) HzG) Hz(aq)

AG'(kJ/mol)

-1 7 6 .0 8 -39.73 -328.28 -426.32 -331.13 -1219.22 -4 5 2 .1 0 -3 7 2 .r6 -r20.36 -966.23 -959.s8 -802.12 -682.83 -633.59 -919.06 -1 1 4 .9 9 -1017.14 -1315.74 -1311.89 -394.36 0 16.40 81.53 9 9 .1 3

155

AF1'(kJ/mol)

-525.05 -319.29 -1027.12 -1264.31 -1038.10 -2042.40 -945.46 -982.71 -809.11 -r513.66

-393.51 0 -11.70 -0.82 -5.02

This table is based on the conventions that A G o = L H " = 0 f o r t h e speciesH*, adenosine,NAD-, and NADP3- at zero ionic strength. These data are from R. A. Albertv, Arch. Biochem. Biophys. 353, 116 (1998).

I

APPENDIX2

StandardFreeEnergyand Enthalpy Changesfor BiochemicalReactionsat pH 7.0,pMg3.00 298K, 1 Atmosphera, and 0.25M lonic Strength AG'ftJ/mol) ATP+H2O=ADP+P, ADP+H2O+AMP+P, AMP + HrO =: Adenosine+ P, 2 ADP = ATP + AMP G6P+H2O=Glu+P, ATP+Glu*ADP+G6P

-32.48 -32.80 -13.55 -0 .3 1 -11.61 -20.87

Data from R. A. Alberty, Arch. Biochem.Biophys.353, 116 (1998).

1 56

AI1" (kJ/mol)

-30.88 -28.86 -t.22 +2.02 -0.50 -30.39

I

APPENDIX3

of the CommonAmino Structures Acidsat NeutralpH coo'

.l 'HsN-?-H GHg

coo-

.l 'HsN-?-H

.,c(.

HsC

Valine (V8l)

Alarine (AIa) (A)

(v)

coo'

coo'

Nonpolar

CHe

rl *HrN-c-H

't'T-?-* HzCr ,CH, CHz

H

Glycine (Gtt (G)

coo'

I *nsl'l-g-n

holinc (Pro) (P)

coo' .Her.r-{-n

Aromatic QHz I

C:QH tt ^H

H

Histidinc (Eis) (H)

coo-

polar

ll *nsn-g-H

CH, l-l.c

?H,

O

Phenyleleninc (Phe) (F)

QOO'H3N-C-H QHZ CHr

coo' rl *Hsrrr-g-n ?r,

.. -c\..CHg HgC Leucine (Lcu) (L)

coo'

I 'HsN-?-H

?" SH Cysteine (cys)

(c)

coo'

I *HoN-c-H "l

9H,

-4.

ril Y

OH

Ilrocine

CrYr) (Y)

coo'

'.Hl o N - C - H -l H-C-OH I H

coo' *nsru-g-H H3C-C-H

?t' cHs Imleucine (tre) G)

coo'

I 'HoN-C-H "l QHz I CHr tS I CHs

Methionine (IvIcO

(ro

coo-

I *nsn-9-n

?" L

\,AN,cH H TFyptoPhan CIhp)

(lv)

coo'

I 'H3N-q-H I

H-C-OH I cHs

o/-\un,o/\*r, Asparaglne (Asn) 0{)

Glutamine (Gln) (Q)

Serine (Ser)

(s)

lhreonine (Thr)

(D

(continued)

157

158

3 APPENDIx

coo' -'r*-{-t

coo'

.l 'H3N-?-H

charse'l

?r,

?" ?" CHr

1-. NHs*

?" ?" ?" N-H

coo'

.l 'H3N-C -H

?" or"o'

coo'

-trN-t- H

?" ?" o/"\o

I

?:*trNHe

Lysine (Lys) (K)

ArginiDe (Are) (R)

Asput te (AsP) (D)

Glutarnate (Glu) (E)

I

APPENDIX4

UsefulConstantsand Conversion Factors

Gas ConstantR

Boltzmann's constantku Planck's constant/z Standard gravity g Electronic charge e Faraday constant F 1 calorie= 4.184joule 1 joule - 107erg = 1 volt'coulomb

8.3144x 107erg K-r mole-l 8.3l44joule K-l mole-l 1.9872calorie K-l mole-r 0.082057 L atmosphereK-r mole*I 1.3806x 10-23joule K-r molecule-l 6.6262x 10-34joule.second 9.8066 meter.second-2 1.6022x 10-1ecoulomb 9.6485 x lOacoulomb mole-l

159

INDEX

I

Activity,3l Adenine,59 Adiabatic, S Allosterism, 148. See also Cooperativity activator, 149 heterotropic interactions, 148 homotropic interactions, 148 inhibitor, 149 Amino acid structures, 157 Aspartate transcarbamoylase,139 allosterism,150 CTP binding, 140 feedback inhibition, 150 structure changes, 150, color plate VIII ATP hydrolysis enthalpy, 12 hydrolysis ratelaw,T4 and ion gradients, 49 synthesis,4T ATPase. 12.49 Base stacking,62 kinetics,115 Binding isotherm, 125 aspartatetranscarbamoylase-CTP,I 40 cooperativity, 140 different sites. 130 experimental determination, 732 hemoglobin, 138 identical sites, 125 MWC model, 145 myoglobin, 138 Bohr effect, 148 Burst kinetics chymotrypsin, 104 protein tyrosine phosphatase,107 r i b o n u c l e a s e Pl l,. Calorimetry, 11 batch,12,13 DNA-Cro protein, 136 scanning,12,13

Chemical potential, 32 extended, 37 Chemiosmotic hypothesis, 47 Chymotrypsin,99 acylation, 101 burst kinetics, 104 mechanismof action, 101, 105 model substrates,100 structure, color plate IV Cooperativity binding kinetics, 147 Koshland-Nemethy-Filmer model, 146 ligand binding, 138 Monod-Wyman-Changeaux model, 143 negative (anti), 139 protein denaturation, 59 Cro repressorprotein, 135 binding to DNA, 135 structure, 137, color plate VI thermodynamics and kinetics, 136 Cytosine,59 Denaturation proteins,13,56 thermodynamic parameters,58-59 Detailed balance,87 Difference spectroscopy, 133 2'CMP-ribonuclease A binding, 134 DNA base stacking,62 binding proteins, 34 binding to Cro protein, 135 hydrogen bonding, 61 kinetics of formation, 114 melting,62,I13,116 r e n a t u r a t i o nI .1 3 . 1 1 8 structure, 59, color plate III thermodynamic stability, 64-65 Electrostatic interactions, 54 Elementary reacttons, 7 2 Energy,T activation, 83

161

162

INDEX

Endopeptidase,99 Enthalpy,9 of activation, 84 of formation,17,754 o f r e a c t i o n .1 5 . 1 5 6 temperature dependence,17 Entropy,22 ofactivation,84 Boltzmann equation, 25 calculation, 24 molecular interpretation, 27 Enzyme catalysis, 94. See also Chymotrypsin, Protein tyrosine phosphatase,and Ribonuclease P Equilibrium chemical,30 and free energy,31 thermal,2 Equilibrium constant, 31, 32 DNA-Cro protein, 136 and free energy, 31 hexokinase,42 hydrogen bonding, 61 macroscopic,127 microscopic, 127 temperature dependence,34

Hemoglobin allosteric models, 149 Bohr effect, 148 Oz binding, 138 structure changes, 150, color plate VII Hess'slaw, 15 Hexokinase, 15 enthalpy of reaction, 15 equilibrium constant, 42 Hill coefficient, 139 Hydrogen bonds, 52 acetic acid,52 DNA,61 kinetics of formation, 114 N-methylacetamide,53 thermodynamics,53 Hydrogen iodide formation of,72 mechanism of formation, 72 Hydrophobic interactions DNA/RNA.61. 115 proteins,52 Ideal gas law,5 Initial rates, 75 lon gradient, 47

Equilibrium dialysis, 132 Exopeptidase,99

Na-/K-.49 protons,4T Ion pairs

First law, 7 Free energy ofactivation,84 and equilibrium, 31 of formation.29. 154 Gibbs,28 molar,3l pressuredependence,33

thermodynamics, 55 Ionic strength,15 Ionization,54 free energy, 55 glycine,727

of reaction, 31,32, 156 standard,29 standardchange,31,32 temperature dependence,34 transfer, 50-52 Gibbs-Helmholtz equation, 34 Glucose oxidation,45 Glycolysis,43 concentrations of intermediates, 46 free energy changes,44 Guanine,59 Heat,3 Heat capacity, 3 constantvolume, 11 constant pressure, 12 cr Helix, 54, color plate I

Isothermal, 8 Kirchhoff's law, l7 Koshland-Nemethy-Filmermodel, 146 Ligand, l24 Ligand binding cooperative kinetics, 147 cooperativity, 138 multiple indentical sites, 124 statisticaleffects, 128 Lineweaver-Burke equation, 98 Maximum velocity,94 Mechanisms.72.79. Seealso Allosterism and Enzyme catalysis hydrogen-iodine reaction, 80 I* + OCI-,81 Michaelis-Menten, 94 Membrane potential, 4, 48, 49 Metabolic cvcles.42

INDEX

Michaelis constant, 94 Michaelis-Menten mechanism, 94 equilibrium approximation,95 steady state approximation, 96 Microscopic reversibility, 87 Microstates,2l and entropy,25 Monod-Wyman-Changeaux model, 143 Myoglobin Oz binding, 138 Myokinase, 16 p-nitrophenyl acetate, 100 p-nitrophenyl phosphate, 106 Nucleation DNA formation, 115 Nucleic acids, 59. Seealso DNA and RNA structure, 59 Phasechange, 35 HzO,9 phospholipids,36 Phasediagram,35,36 Phaserule, 36 Phospholipids phase change, 36 B Pleated sheet, 54, color Plate II Properties,2 extensive,2 intensive,2 Protein denaturants,56 denaturation,13, 56 folding,56 ionizable groups,54 structure,50 Protein tyrosine phosPhatase,106 mechanism, 108 site specific mutagenesis,107 structure, color Plate V transition states, 109 RadioactivedecaY,78 Rate constant definition,75 temperature dePendence,83-85 Rate equations,T4 consecutive first order, I I I first order, 76 integrated, 75 second order, 76 Rate law,75 Reaction coordinate (Path), 83 intermediates, 86 Reaction order, 75 first order, 76

second order,76 pseudo first order, 77 Reaction rate,72 definition,73 near equilibrium, 88 Receptor, 124 Relaxation time, 89-90 DNA formation, 115 l, Repressor,57 Reversible path,6,25 changein state,25 RibonucleaseP, 110 burst kinetics,111 mechanismof action, 111 reaction catalyzed,I 10 Ribozyme, I09. Seea/so RibonucleaseP RNA base stacking, 62 hydrogen bonding, 6l ribozyme, 109 structure, 60,66 Scatchard plot, 125 Secondlaut,22 Standard state, 10 enthalPY,10 free energy, 29 gas,30 solution,3l State function, 8 Steady state approximation, 96 chymotrypsin, 101 System,2 closed,2 isolated,2 open,2 Temperature,2 Kelvin,2 Third law, 26 Transient kinetics chymotrypsin, 100 DNA formation, 114 107 protein tyrosine PhosPhatase, ribonucleaseP, 111 Transition state theory, 84 Thymine,59 Work definition,4 electrical,4,48,49 non P-V,37 P-V,4 reversible,6

163

COLOR PLATE I. The cr-helrxftrr-rndin many proteins.The yellow arrow follou,s the ri-eht-handed spiral of one helical tLrrn.The hvdro-sen bonclsbetweenbackbone peptidebondsarebrown lines.the oxv-gens arerecl.and the nitrogensblue.The hyclrogen bonds are lormed betweenthe rth carbonyl and the i+4 NH in the pepticlebackbone.copyri-ehtby Prof-essor JaneRichardson.Reprintedwith penrission.

C O L O RP L A T E S

C O L O R P L A T E I I I . S t i c k - f i g u r c r e p r e s e n t a t i o no l ' t h e B f o r n r o l ' r h e D N A d o l b l c hclir. The planeso1'thehvdrogen-boncled b a s e sc a n b c s e e n . a s w c l l a s t h e t w i s t n - u o l ' t h c c h a i n s t o f o r n t a d o u b l c h e l i x . T I r c p a l e y e l l o w s p l r c r c sa r e t h c p h o s p h o r o u s a t o r r s o l t h e s u s i t r p h o s p h a t e b a c k b o n e o n t h c o u t s i c l eo f t h c h c l i x . T h e b a s e sa r c c o l o r c o d e c l . a n c l t l t e t r v o g l ' o o \ / e si r t t h c s l t ' t t e t L t r e u r c l l r b e l e d . C o p v r i g h t b 1 ' P r o f ' e s s g r ' J a p cR i c h a r - c l s . r r . R c P r ' 1 1 1 1u 8i 't 1h P c l r r r i : s i t r l t .

COLOR PLATES

of chymotrypsin,showingthe active COLOR PLATE IV. (a) Spacefilling representation site pocket where substratebinding occurs.The blue at the active site is imidazole,the green is serine, and the red is aspartate. (b) Arrangement of the protein residuesat the active site of chymotrypsin.The serineis the residueacylated;the imidazoleof the histidine servesas a generalbasecatalyst; and the aspartatecarboxyl group is hydrogen bonded to the imidazole.Copyright by ProfessorJaneRichardson.Reprintedwith permission.

COLOR PLATES

COLOR PLATE V. (a) Backbone representationof the catalytic domain of protein tyrosine phosphatase.Coils and arows representcr.-helicesand B-strands,respectively.The cysteine (green),arginine(red), and aspartate(gold) are also shown.Reproducedfrom J. M. Denu, J. A. Stuckey,M. A. Saper,and J. E. Dixon, Cell 87,316 (1996). Copyright O1996 Cell Press. reprintedwith permissionof Cell Press.(b) Vanadatebound at the active-sitecysteineof protein tyrosine phosphatase.Other amino acids at the catalytic site are indicated,including the threonineand aspartatethat participatein the catalytic mechanism.Reproducedfrom J. M. Denu, D. L. Lohse,J. Vijayalakshmi.M. A. Saper,and J. E. Dixon, Proc. Natl. Acad. Sci. USA 93,2493 (1996).Reprintedwith permissionof the Proceedingsof the NationalAcademy of SciencesUSA. Reproducedby permissionof the publishervia Copyright ClearanceCenter,Inc.

COLOR PLATES

COLOR PLATE Structure of the overall complex between Cro protein and )t operator DNA. The di of the view is parallel with the major grooves of DNA and parallel with the recog ition helices. Reproduced from R. A. Albright and B. W. Matthews, (1998). Reprinted with permissionfrom Academic Press,Inc. J. MoI. Biol.280, I

COLOR PLATE VII. Schematic representationsof the R (bottom, pink) and T (top, blue) forms of the hemoglobir ctzBz tetramer. The hemes where oxygen binds can be seen in the structure. The yellow side chains form salt bonds in the T structure that are broken in the R structure. One pair of ct-B subunits also rotates with respect to the other by about 15' in the interconversation of R and T forms. Copyright by Professor Jane Richardson. Reprinted with permission.

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COLOR PLATE VIII. Ccr backbonestructuresof the R (lower; anclT (Lrpper)statesof aspartatetranscarbamoylase. The two catalytictrimersare in greenand are closertogether in the T state.The three regulatorydimers (equatorial)are in yellow. One of the regulatory dimers is behind the large central cavity in this view. Adaptecl fiom W. N. L i p s c o m bA , d v ' .E n t m o l . 6 8 , 6 7 ( 1 9 9 1 ) .c o p y r i g h t o 1 9 9 4J o h n w i l e y & S o n s ,I n c . Reprintedby pennissionof JohnWiley & Sons.Inc.