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) 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]" + (/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 (