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LecturesonPhysicalBiochemistryToddYeates

©2015

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ForMelissa

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Preface

The following textderives frommaterialpresented ina course inphysicalbiochemistryatUCLA(ChemistryandBiochemistry156). Muchof thematerialowes itsorigin to lecturesdeliveredbyother facultymembersatUCLAwhoweremyteachersandmentors. These includeEmilReisler,WayneHubbell,andespeciallyDougRees,forwhomIservedasaTAformultipleofferingsofthecoursewhileIwasagraduatestudent.Withregardtootherbooksuponwhichthematerialrests,the classic text Physical Biochemistry with Applications to the Life Sciences by Eisenberg andCrothersstandsoutasthemost influential. Othertexts fromwhichselectedmaterialshavebeenextracted include: Physical Biochemistry by van Holde, Physical Chemistry: Principles andApplicationsinBiologicalSciencesbyTinoco,Sauer,Wang,Puglisis,Harbison,&Rovnyak,andTheMoleculesofLife:PhysicalandChemicalPrinciplesbyKuriyan,Konforti,andWemmer,MolecularDrivingForcesbyDill,andRandomWalksinBiologybyBerg.

IamindebtedtothestudentsandTA’swhohaveparticipatedinthecourseovermanyyearsandhavemadedevelopingandteachingthematerialastimulatingandrewardingchallenge.IamparticularlyindebtedtoSunnyChun,whoproofreadthefirstdraft.

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Contents PageChapter1 9

PointsforReviewo Thermodynamicsystemso Systemsandsurroundingso The1stlawo Work,wo Heat,qo Enthalpy,Ho The2ndlawo Classicalandstatisticalviewsofentropy

EntropyandtheDistributionofMoleculesinSpace EntropyandtheDistributionofMoleculesAmongEnergyLevels

Chapter2 20

EntropyofMixinganditsDependenceonLogofConcentrationso Stirling’sapproximationo ‘EntropyofMixing’

GibbsFreeEnergy,Go Astatevariablethatindicatesthefavorability(orequilibrium)ofaprocessat

constantT&Po ΔGasabalanceoftwofactors,ΔHandTΔSo HowtothinkaboutΔGinasteadystateprocesso Freeenergyofmixingandthe:furtherinsightintowhatdrivesprocessestowards

equilibriumChapter3 26

ChemicalPotentials,µo DefinitionofµasapartialderivativeofGwithrespecttocompositiono Dependenceofchemicalpotentialsonconcentrationsandstandardstatechemical

potentialsµ0o Thetotaldifferential,dGasafunctionofchangesincompositiono Equilibriumconditionsintermsofµ’so Equilibriumconditionsintermsofconcentrationsandstandardchemicalpotentials:

arrivingatfamiliarequationsfortheequilibriumconstanto Importanceofunitso PrecautionsaboutΔGvsΔG0,reactionswithchangesinstoichiometry,andoverall

concentrationeffectso ThedependenceofΔGandKonT(van’tHoffequation)

Chapter4 37

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Non‐idealbehaviorinmixtureso Thebreakdownofidealequationsforchemicalpotentialo Activitiesandactivitycoefficientso Theidealbehaviorofhighlydilutesolutionso Theoriginofnon‐idealbehaviorathigherconcentrationso Reworkingtheequilibriumequationsintermsofactivitiesinsteadofconcentrations

Ion‐ioninteractionsinsolutionasanexampleofnon‐idealbehavior(Debye‐Hückeltheory)

o IonicstrengthandtheDebyelengtho Activitycoefficientsforionicspecieso Usingionicactivitycoefficientstoanalyzetheeffectofchargeonmolecular

association,andelectrostaticscreening

Molecularcrowdingandexcludedvolumeeffectsasanexampleofnon‐idealbehaviorinsolutionsofmacromolecules

o Theideaofexcludedvolumeo Thepeculiarbehaviorofrigidelongatedstructures

Chapter5 50

ChemicalPotentialandEquilibriuminthePresenceofAdditionalForceso Osmoticpressureo Equilibriumsedimentation

Chapter6 60

Electrostaticpotentialenergy,iontransport,andmembranepotentialso Thechemicalpotentialenergyofanionatapositionofelectrostaticpotentialo TheNernstequationandmembranepotentialo TheDonnanpotentialo Variableionpermeabilitiesandcomplexphenomena

MolecularElectrostaticso Thedielectricvalueo Simplifiedelectrostaticsequationso Adifferentkindofelectrostaticenergy:theBorn‘self‐chargingenergy’o Freeenergyofiontransfer

Chapter7 70

EnergeticsofProteinFoldingo Abalancebetweenlargeopposingforceso Termsthatcontributetotheenergeticsofproteinfoldingo Thespecialcaseofmembraneproteins

MeasuringtheStabilityofProteins IdeasRelatedtoHowProteinsReachtheirFoldedConfigurations

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Chapter8 80

DescribingtheShapePropertiesofMoleculeso Radiusofgyrationo Persistencelengthforflexiblechains

Chapter9 87

ABriefIntroductiontoStatisticalMechanicsforMacromoleculeso Probabilitiesandexpectedvalueso Statisticalweightsforoutcomeswithunequalprobabilitieso Handlingdegeneracies

AStatisticalMechanicsTreatmentoftheHelix‐CoilTransitionforaPolypeptideChapter10 94

CooperativePhenomenaandProtein‐LigandBindingo Relationshipbetweencooperativebehaviorandprocessesinvolvingformationof

multipleinteractionssimultaneouslyo Protein‐ligandbindingequilibriao Bindingtoanoligomericprotein–independentbindingevents,nocooperativityo Non‐linearScatchardplots–non‐identicalornon‐independentbindingsiteso Experimentsformeasuringbindingo Phenomenologicaltreatmentofcooperativebinding‐theHillequationo Physicalmodelsofcooperativebinding‐MWC

AllosteryChapter11 113

SymmetryinMacromolecularAssemblieso Definitionofsymmetryo Mathematicalgroupso Pointgroupsymmetriesforbiologicalassemblies

SpecialTopicsinSymmetryo HelicalSymmetry(non‐pointgroup)o Quasi‐equivalenceandthestructureoficosahedralviralcapsidso Usingsymmetrytodesignnovelproteinassemblieso Algebrafordescribingsymmetry

Chapter12 124

EquationsGoverningDiffusiono Diffusionin1‐Do Generalequationsfordiffusion

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o Specialtopic:Usingnumerical(computational)methodstosimulatediffusionbehavior

Chapter13 132

TheDiffusionCoefficient:MeasurementandUseo Measuringthediffusioncoefficient,Do Relatingthediffusioncoefficienttomolecularsize

SpecialTopicinDiffusion:DiffusiontoTransportersonaCellSurfaceChapter14 143

Sedimentationvelocity,vo Sedimentationcoefficient,so CombiningsandDtogetmolecularweightwithoutasphericalassumptiono Asummaryofmolecularweightdeterminationfromsedimentationanddiffusion

measurementsChapter15 148

ChemicalReactionKineticso Reactionvelocity,vo Ratelawso Integratingratelawso Behaviorofmorecomplexreactionschemeso Numericalcomputersimulationofmorecomplexreactionschemeso Enzymekineticsunderasteady‐stateassumptiono Relaxationkinetics:howsystemsapproachequilibriumo Kineticsfromsinglemoleculestudies

Chapter16 164

KineticTheoriesandEnzymeCatalysiso TheArrheniusequationo Eyringtransitionstatetheoryo Catalysisbyloweringthetransitionstateenergyo Practicalconsequencesofenzymesbindingtightlytothetransitionstateo Kineticparametersofnaturalenzymes

Chapter17 171

IntroductiontoBiochemicalSpectroscopyo Energytransitionso Fluorescenceo Kineticsoffluorescenceandcompetingroutesforreturntothegroundstate

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Chapter18 178 SpecialTopicsinBiochemicalSpectroscopy

o Polarizationandselectionruleso Fluorescenceexperimentswithpolarizedlighto Fluorescentresonantenergytransfer(FRET)o FRETinbiologyo Spectroscopyofchiralmolecules:Opticalrotationandcirculardichroism

Chapter19 194

MacromolecularStructureDeterminationandX‐rayCrystallographyo Thelimitingeffectofwavelengtho Diffractiongeometryo Obtainingtheatomicstructureo ProteinCrystallization

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CHAPTER1

PointsforReviewThermodynamicsystemsWeareallfamiliarwiththeeverydaybehaviorofvariouskindsofmechanicalsystems.Thisoftenaids us in understanding the behavior of molecules, which are indeed governed by the laws ofphysics.Buttherearealsokeydifferencestobearinmindbetweenthephysicalbehaviorofsystemsatthemacroscaleandthethermodynamicbehaviorofmolecularsystems.Analogiescanbedrawntoabowlingballatthetopofahill.Weknowthatifpusheditwillgotothebottomofthehilland(eventually)staythere.Onceithascometo(apparent)restwedon’tworryaboutitsuddenlymovingunderitsowninternalenergytoahigherlocation.Orlikewiseforatextbooksittingonadesk.Wewouldn’tthinkaboutmeasuringhowfaronaverageitfindsitselflevitatingaboveitslowestenergypositiononthedesktop.Butthesesortsofideasariseconstantlyinthinkingaboutthebehaviorofmolecules.Why?Thedistinctionislargelyoneofscale,havingtodowithrelativesizes,forces,andenergies. The essence is that inmolecular systems (at temperatures sufficiently above0K), themagnitudeof the thermalenergy iscomparable toenergydifferencesassociatedwithmeaningfuldifferencesinthepropertiesofthemolecules,suchastheirvelocitiesanddetailedthree‐dimensionalconformations.Wewillemphasizethroughoutthecoursetheimportanceoftheideaofan‘averagethermalenergy’,which is kBT, where kB is Boltzmann’s constant (or alternatively RT when working with molarquantities,whereRistheuniversalgasconstantandR=NAkBandNAisAvagadro’snumber)).IfweaccepttheideathatphysicalobjectsexhibitenergiesonthescaleofkBT,thenhowhighmightweexpecta1kgtextbooktolevitateoffthedesktopunderitsownthermalenergyat298K?[Hint:equategravitationalpotential energy for thebookat a heighthwith theenergyvalueof kBT]. Youwill(hopefully) findthat thatheight is infinitesimallysmall,which isconsistentwithexperience. Butowingtothemuchsmallerenergiesthataffectmolecules,bigorsmall,kBTisanenergysufficienttodrive the rapid movements, collisions, conformational changes, and chemical reactions thatcharacterizemolecularsystems.SystemsandsurroundingsInthermodynamics it is importanttokeepinmindwhat isbeingconsideredasthesystemunderinvestigation.Everythingelseisthesurroundings.Whendiscussingthermodynamicquantities(P,V,U,…)wearereferringtomeasurementsandpropertiesofthesystem,butdependingonthesituationthesurroundingsmaybeimportantinexchangingenergy(intheformofworkorheat)ormaterialwiththesystem.Inaclosedsystem,noexchangeofmaterialoccurs.Inanisolatedsystem,thereisnoexchangeofmaterialorheatorwork.Insomeproblems,theentireuniversemightcomprisethesystem. Inthatcasetherearenosurroundingwithwhichexchangemightoccur,sotheuniversewouldfollowthesamerulesasanisolatedsystem.

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The1stlawThe first lawof thermodynamicsexpressesa lawofenergyconservation,namely that theenergychange in a system equates to whatever energy is delivered to it (and thereby lost by) thesurroundings.Therefore,thechangeinenergy(U)ofasystemduringsomeprocessisgivenbytheamountofheatthatistransferredtoitfromthesurroundingsplustheamountofworkdoneonitbythesurroundings.

U=q+w

Foranisolatedsystem,q=0andw=0,soU=0Thefirst lawisrelativelyeasytoappreciate,sincewe’re familiarwithconservation laws inothercontexts,e.g.conservationofmass,ortheconservationoftotalenergyinmechanicalsystems.Besidesconveyinganimportantconservationprinciple,thefirstlawservesasareminderaboutequationsforworkandheat.Work,wYou’llrecallfromphysicsthatworkisforceintegratedoverdistanceordisplacement:

w=FdxAgainstpressure:F=PA,dV=Adx,sow=PA(1/A)dV=PdV[Aswrittenthisworkwouldhavethesenseofworkdonebyasystemwhosevolumewaschanging.]Againstaharmonicspringwithforceconstantk,w=Fdx=kxdx=(1/2)kx2Andlikewiseforanysituationwhereafunctionfortheforceonamoleculecanbewritten(possiblydependingonposition).Wecanintegrateoverpositiontogivetheworkenergythatwouldbedoneonthemoleculeasafunctionofitsposition.Heat,qOtherthingsbeingheldconstant,weoftenassociateheattransferwithtemperaturechange,andtheheatcapacity,C,relatesthosetwochanges.RecallC=dq/dTorq/Tand

q=CdTTheheatcapacityisameasureofhowharditistochangethetemperaturebyaddingheat. Fromintroductoryphysicalchemistryyou’llrecallthatforanidealgasCv=(3/2)R(onapermolebasis).

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Forcomplexmolecules,themolarheatcapacityishigher.Foranidealgas,theenergyofthesystemisassociatedsolelywiththekinetics(i.e.velocities)ofthemolecules(whicharepresumedrigidinanidealgasmodel).Morecomplexmoleculeslikebiologicalmacromoleculeshaveverymany‘internaldegreesoffreedom’,whicharerequiredtospecifythepositionsandmovementsofatomsrelativetoeach other in the same molecule. Recall that macromolecules are subject to all kinds ofconformational fluctuations, mostly small but some very large. You might recall that theequipartitiontheoremtellsusthatenergyintheamountkBTwillbepartitionedequallyintoeachofthedegreesoffreedominasystem,whichmeansthatsystemscomprisedofcomplexmoleculeswillrequiremoreheatenergytoraisethetemperatureowingtothemuchgreaternumberofdegreesoffreedomintowhichtheenergygetspartitioned.Enthalpy,HNotefromtheformoftheequationU=q+w,thatatconstantvolume(sono‘PV’workisdone)heattransferqrelatescloselytointernalenergyU.[Ifw=0thenU=qordU=dq]Butifthevolumeisnotconstant,theheattransferrelatesbettertoanotherthermodynamicstatevariable,H,theenthalpy,givenbyH=U+PVDifferentiatingH=U+PVgivesdH=dU+PdV+VdP=dq+dw+PdV+VdPAtconstantP(dP=0)andwithonly ‘PV’work,dw=–PdV[Thesignappearsnegativehereastheworkwmust refer to theworkdoneon the system,whereasour earlier equation forwhad theoppositemeaning]Substitutingdw=‐PdVintothepreviousequation,weseethatatconstantpressureandonlyPVwork,dH=dqP–PdV+PdV+0,givingdH=dqP(withthePsubscriptdenotingwhatisheldconstant).SoenthalpyandqarecloselyrelatedatconstantP.Notethatespeciallyforgases,wherepressureandvolumechanges(e.g.asafunctionoftemperature)aresubstantial,UandH(whichdifferfromeachotherbythetermPV)maybesubstantiallydifferent.But inother systemswherepressureandvolumechangesareminimal,our intuitionaboutwhatenthalpy and internal energymean tends tobe closer. This is the case formanyof thekindsofsystemslikesolutionsofmacromoleculesthatwewillbethinkingabout,andinthosecasesafairview is that the enthalpy embodies all the kinds ofmolecular forces of attraction and repulsionbetweenmoleculesthatwe’refamiliarwith.Andintermsof‘favorable’vs.‘unfavorable’,ahighvalueof H implies high energy or unfavorable interactions, while a low value of H implies favorableinteractions.

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Thismeansthatwecanoftenlearnsomethingusefulabouttheforcesandinteractionsthatexistinasystem,forexampleapurifiedproteininsolution,bymeasuringenthalpychanges.Anexperimentalmethod known as differential scanning calorimetry is often used to make those kinds ofmeasurements. A sample is slowlyheated and theheat transfer required toproduce each smallincrementalincreaseintemperatureisrecorded.Thedifferenceistakenrelativetoablank,whichwouldcontainthesolutionandbufferbutnottheprotein.Ifperformedatconstantpressure,thatrecordedquantity,dqP/dT,istheheatcapacityatconstantpressure,CP.Andfromabove,dqP/dT=dH/dT=CP.AnddH=CPdT.ThatmeansHforaprocesscanbeobtainedbyintegratingtheheatcapacityoverthecourseofatemperatureincrease,H=CPdTTheexampleillustratedhereisfromathermalunfoldingexperimentonapurifiedprotein,carboxypeptidaseA.At low temperature the protein isfoldednatively.Athightemperaturethe protein is unfolded. Therelatively flat parts of the curve inthose two regions are simplyreflecting the heat or enthalpychange associated with increasingthetemperature(localvibrationsforexample) of the protein. But theregion in the middle shows adramatic increase in the heatcapacity.Thiscorrespondstotheenergyrequiredtoconvertthefoldedproteintoitsunfoldedform.Favorable molecular interactions are broken in the process and the overall enthalpy change ispositive.TheareaoftheshadedregionisattributedtotheHfortheproteinunfoldingtransition.TheSecondLawThesecondlawpresentsmuchgreaterchallengestounderstandingthanthefirst.Ratherthanstatingalawofconservation,itdefinesadirectionalityinwhichprocesseswillnaturallyproceed(intime).In that sense the second lawenforces the ‘arrowof time’. The second law tellsus that the totalentropySintheuniverse(i.e.foranysystemplusitssurroundings)isalwaysincreasingintime.Andthisislikewisethecaseforanisolatedsystem(sinceitcanbeviewedasitsownuniverse).So,foraspontaneousprocess(i.e.aprocessthatwouldoccurintheforwarddirection)occurringinanisolatedsystem,orfortheuniverseasawhole,S>0.Likewise,S=0describesaprocessatequilibrium, i.e. with no net conversion forward or backward. In that sense the condition ofequilibriumcanbeseenasanoptimizationproblem.AtequilibriumSisamaximumandds=0withrespecttoforwardorbackwardprogressoftheimaginedprocess.

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It is often tempting to forget about the conditions or restrictions under which variousthermodynamic equations hold true, but it is vital to understand that the equation S > 0 (forspontaneousoccurrence)requirestheconditionofanisolatedsystem.Infactfailuretounderstandthisvital requirement is the sourceofmuchconfusionamong thepublic and lay‐scientistsaboutwhether the development of life on Earth and the associated increase in order and molecularcomplexity–anideawewilltietoentropyshortly–violatesthesecondlawofthermodynamics(andtherebyrequiresacreator).ThefollyintheargumentisthattheEarthisbynomeansanisolatedsystem,andinfactthedeliveryoflightenergyfromtheSuntotheEarthtodrivephotosynthesisisessentialforthechemicalconversionsthatsupportlifeonEarth.ClassicalviewofentropyFromclassicalthermodynamicsyoulearnedthatdS=qrev/T,whereqrevistheheattransferredduringa reversible infinitesimal step in a process. This view of entropy is extremely useful forunderstandingprocessesofheattransferandexpansioningases.Welearnthatentropyincreaseswhengasvolumesexpand,andwhenheatistransferredfromahotterobjecttoacolderobject;thoseprocessesarenaturallyfavorableorspontaneous.StatisticaldescriptionofentropyAwayofstatingthesecondlawfromastatisticalthermodynamicsviewisthatprocessestendtowardmaximumdisorderorrandomness,i.e.toconfigurationsthatcanberealizedinthegreatestnumberofways.Thisviewcanbereconciledintuitivelywiththeclassicalview–gasexpansionallowsforgreaterfreedomandlessorderwithregardtothepositionsofatoms,andheattransferfromhottocoldmoleculesdecreasestheorderinthesensethatthedistinctionbetweensomemoleculeshavingmorethermalenergythanothersisremoved.Theintuitiverelationshipbetweentheclassicalandstatistical views of entropy can be formalizedmathematically butwewill not attempt that here.Instead,withoutfurtherproofthestatisticalviewofentropyisS=kBlnWwhereWisameasureofdisorderorrandomnessthatcanbeinterpretedasthenumberofdistinctconfigurations that correspond to a given state. This is sometimes referred to as thenumber ofmicrostates.[NotethatsometextsuseinsteadofW].Inthisview,therequirementthatentropyincreasesmeansthatfavorablestatesarethosethatcanberealizedinthegreatestnumberofways.Wewillsetupsomehighlysimplifiedproblemstoseehowthestatisticalviewofthermodynamicshelpsexplainsomebasismolecularphenomena.

Entropyandthedistributionofmoleculesinspace

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Let’s look at what the statistical view of entropy tells us about the way molecules tend to bedistributedinspace.We’llfirstconsideraverytinyproblem,tootinyreallytoqualifyasaproperthermodynamic system,but still informative. Supposewehave4moleculesorparticles that areidentical,butwecanlabelornumberthemtomakeitpossibletodistinguishbetweenmicrostateswithinagivenstate. Thesystemconsistsofaboxwithtwochambers,aleftsideandarightside.Supposewedescribethestateofthesystemaccordingtothenumberofmoleculesthatareontheleftsidevstherightside.WecanletnLbethenumberofparticlesonleftsideandnRbethenumberon the right. For each possible state (i.e. a defined number ofmolecules on each side), we canenumeratethenumberofdistinctwaysormicrostates(W)bywhicheachstatecanbeachievedbychoosingdistinctlylabeledmolecules.Forsomeofthestates,thevalueof W is obvious enough. Forexample,forstateB,whichhasjust onemolecule on the right(nL=3), any of the 4moleculescan be chosen to place on theright,andsoW=4.Likewiseforstate D for which nL=1. Thecase of nL=2 is harder. Howmany ways can we divide orpartitionagroupoffourobjectsintoafirstsubsetof2(toplaceontheleft)andasecondsubsetof2(toplaceontheright)?Theansweris6,whichcomesfrom4!/(2!2!)=24/(2*2)=6.Thisisacombinatorialexpressioncloselyrelatedtothepermutationequationthatsaysthenumberofwaysoforderingnobjectsisnfactorial,orn!.Whyinthecaseabovedowedivide4!by2!and2!?Onewaytoseethisisasfollows.Howmanywayscan4objectsbeordered(e.g.inaline)?Theansweris4!or24.Nowlet’ssaythateachofthese24waysofwritingdownthemoleculesinorder(e.g.3124)automaticallyassignstwototheleftside(inthiscase3and1)andtwototherightside(2and4).Butyoucanseethatthetotalsetof24possibleorderingsovercountsthenumberofdistinctoutcomesinthesensethatthereareotherorderingsthatgivethesamepartitioning.Forexample(1324)isthesamepartitioningas(3124).Ifthesametwoparticlesareontheleft,theirseparateorderingisirrelevant. Since the state inquestionhas twomoleculeson the left, and thenumberofwaysofordering2objectsis2!or2,weneedtodividethetotalnumberof24orderingsby2.Forthesamereason,theorderingofmoleculeswithintherightsidedoesn’tmattereither,andsowemustdivideagainby2!Thisgivesusthevalue6weexpect.Thinkingaboutproblemslikethisintermsofpartitioningbetween2(ormore)groupsispowerful.ThegeneralequationforthenumberofdistinctpartitioningsofNobjectsbetweenafirstgroupofn1andasecondgroupofn2(withn1+n2=N)isW=N!/(n1!n2!)

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Anequationofthisformshowsupthroughoutstatisticsapplications.Intypicalstatisticsjargon,thenumber of possible combinations for “N choosem” is NCm =N!/(m! (N‐m)!),whichmatches theequationabove.Thebasicpartitioningideaappliestomanyproblems.Howmanydifferent5‐cardhandscanbedealtfroma52‐carddeckinwhichthecardsareallconsidereddistinctfromeachother?[Hint:beingdealt5cardsisreallyjustpartitioningthe52cardsintothe5yougetandtheothersyoudon’tget;theorderinwhichyougetdealtthecardsdoesn’tmatterhere.]As an aside, another common type of probability problem (which also shows up in molecularproblems)involvesaseriesofindependentchoices,andtherethetotalnumberofpossibleoutcomesisn1*n2*n3*….wherethen’sdescribethenumberofdistinctoptionsthatareavailabletochooseateachstep.Oftenthetwotypesofprobabilityproblemsarerelatedtoeachother.Consideravariationonthe4moleculeproblemabove.Supposewewanttoknowthetotalnumberofdifferentwaysthe4moleculescanbeplacedintotwochambers,allowingallpossibilitiesforthenumberofmoleculesoneachside,andasbeforenotdistinguishingbetweenthepositionsofparticleswithin thesamechamber.Thiscanbeansweredbyseeingthatitamountstomakinganindependentchoiceforeachmoleculeaboutwhetheritwillgoontheleftorright.Sothereare2choices,made4independenttimes,whichis2*2*2*2=16.You’llnotethattheanswertothisproblemcountsupalltogetherthenumber of different partitioning, so it is not a coincidence that the values forW in the originalproblem(1,4,6,4,1)sumto16.Returningtotheproblemofhowmoleculestendtodistributethemselvesinspace,fourmoleculesisperhaps too small to give a clear picture of significance, so let’s go slightly bigger toN=6, againtreatingtheproblemofhowthemoleculescanbepartitionedintotwosides.FornL={0,1,2,3,4,5,6},wegetrespectivevaluesforWof{1,6,15,20,15,6,1}.YoumaybegintorecognizethecoefficientsasthosefromPascal’striangle.Whatdoesthistellus?Assumingtherearenoenergeticdifferencesatplayandeachofthe6moleculesisfreetooccupyeitherchamber,thenthelikelihoodofthesystembeinginanygivenstateisproportionaltothenumberofmicrostates,W.Thatmeansthatitis15timesmorelikelybychancefortheretobethreemoleculesineachchambercomparedtothecasewhere all 6 molecules are on the left. Evidently, themost likely scenario is the onewhere themoleculesareevenlydistributedwiththreeoneachside.Thesametrendapplies,andbecomesmoredominant,asthesizeofthesystemNincreases.Thebasicconclusionisthatentropydrivesthingstowardsauniformdistributionofmoleculesinspace,i.e.equalconcentrationseverywhere,assumingtheabsenceofenergeticdifferences.ThebehaviorofthisproblemasNgetlargeisalsoinstructive.TheplotsshowtheprobabilitiesonegetsforthedistributionofmoleculesbetweenthetwosidesofasystemforN=6,larger100,andthenN=10,000.

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ReturningtothecaseofN=6,oneseesthatthestatewithauniformdistributionofmoleculesisthemost likely,but thechancesofsignificantdeviations fromthatarrangementaresubstantial.Withmoleculesthatarefreetomovearound,two‐fifteenthsofthetimethesystemwillbefoundwithallthemoleculesononesideortheother.AsNgetslarger,thelikelihoodofsubstantialvariations(onarelativescale)goesdown. AsNgets larger thediscretecombinatorialplot turns intoasmoothGaussianfunction.ThemostlikelyoutcomeisstillwherenLisN/2.ThestandarddeviationfromthemostlikelyvaluefornLis(fromearliercoursesinstatistics)0.5*sqrt(N).Soforexample,ifN=100,themost likelyvaluefornL is50,butwithastandarddeviationof5. WhataboutwhenN=NA=6.02*1023?Therethestandarddeviationwouldbealargenumber(3.9*1011),butinfractionaltermscomparedtoNA,thevariationisminute.Thatis,theexpectedfractionofmoleculesontheleftwouldbe0.5+/‐6*10‐13. Thisisageneralfinding;forlargethermodynamicsystemsthebehaviorofthesystemtendstobedominatedbythemostlikelyscenario.Ontheotherhanditisimportanttonotethat the kinetic (time‐dependent) behavior of a system often depends on the frequency ofperturbationsawayfromthemostprobablearrangement.

EntropyandthedistributionofmoleculesamongenergylevelsThesamekindoftreatmentcanbeusedtoanalyzehowtheenergyinasystemtendstobedistributedamongthemoleculespresent.Again,fornumericalsimplicitywe’llfirsttreatatinysystemjustbigenoughtogainsomeinsight.Supposewehaveasysteminwhich4identicalmoleculesareeachabletoexistinaseriesofdiscreteenergylevels(inarbitraryunits,E=0,E=1,E=2,…).AndfurthersupposethatthetotalenergyisfixedatET=3.Whatarethepossiblewaysthatthe4moleculescanbeplacedintotheavailableenergylevels?Notethatnothingpreventsmultiplemoleculesfromhavingthesameenergy.Forthistinysystemthereareonly3differentstatesorconfigurationsofmoleculesamongenergylevelssubjecttotherestrictionthatET=3.Theyareshownbelow,labeledstatesA,B,andC.WhatisthevalueofWforeachstate?Thatis,foreachenergyconfiguration,howmanydifferentwayscouldthemoleculessatisfythatconfiguration?

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The answer is to think of this as apartitioningproblem.ForstateA,the4molecules are being partitioned into asubsetof3thatwillhaveenergy0,andasubsetof1thatwillhaveenergy3.Forthatcase,W=4!/(3!1!)=4.ForstateBwegetW=4also. ForstateCwemustfirst generalize our previous equationfor the number of combinations orpartitionings. When a partitioningoccurs into more than 2 subsets, theequationforWgeneralizestoW=N!/(n1!n2!n3!…),wherethesmalln’srefertothenumberofmoleculesinthedifferentsubsets.[Forcompleteness,alsorememberthat0!=1soemptysubsetscanbeignored].So,forstateC,W=4!/(2!1!1!)=12.Whatwegleanfromthistinytestcaseisthatthemostlikelysituation(i.e.whereWisgreatest)iswherethemolecules are spreadout to somedegree among the available energy levels,with the lowestenergybeingthemostpopulated.SimulatingexchangeofenergybetweenmoleculesinaclosedsystemThe behavior of slightly larger systems can be analyzed by random simulations with ratherremarkable results. Suppose nowwe have a set of 50molecules, and for the sake of argumentsuppose the average energy is 1 so that ET = 50. We can set up an initial systemwhere all 50molecules sit at energy level 1. Then,molecules exchange energybetween themselves, asmightresultsfromcollisionsforinstance.Thedetailsoftheexecutionareimportant.Picktwomoleculesatrandom,onewhoseenergywillgodownbyoneunitandtheotherwhoseenergywillgoupbyoneunit.Dothisoverandover.Butnotethecaveatthatifthefirstmoleculerandomlychosenisalreadyatenergy level0, then throwout thisenergyexchange trialandrepeatagain; i.e. theenergyofamoleculecan’tdropbelowthelowerbound. Ifoneperformsthiskindofrandomsimulation,onefindsremarkablythatthesystemwilltendtowardsanenergydistributionofthetypenotedabove.Noothertricksarerequired.AnexampleresultofrandomsimulationforN=20isshownbelow.

For larger N, the simulation begins to produce a smooth distribution. Examples for N=320 andaverageenergy=1and2areshownbelow.

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Theexercisedemonstratesthattherandomtendencyofmoleculestospreadoutamongavailableenergylevelswhilealsobeingsubjecttotheconstraintofalowerenergyboundnaturallygivesrisetoasmoothdistributionwherethelowestenergyismost‐populated,andthenthedistributionfallsoffathigherenergy.Thediagramsabovehavetheenergylevelgoingupvertically,andthefrequencywithwhichmoleculesare foundatthatenergyindicatedbyahorizontalbar. Thiscanbeflippedaround to give amore typical plot showing the probability (or abundance) ofmolecules on theverticalaxisandtheenergyvalueindicatedonthehorizontalaxis.Doingthisproducesfamiliarplotsthat show an exponentially decaying curve for the probability that any givenmoleculewill haveenergyE.ThisistheBoltzmanndistribution.TheBoltzmanndistributionteaches some importantprinciples.Therearefewerand fewer molecules withhigherandhigherenergies.But there are some, andhowmany of these higherenergymoleculesthereareis essential forunderstanding rates ofprocesses that depend onovercoming an energybarrier. Another keyfeature of the Botzmanndistribution concerns howsharplytheprobabilityfallsoffasafunctionofenergy.AccordingtotheBoltzmannequation,thatfall‐off is governed by the denominator of the exponent (kBT, a term we alluded to before).Specifically,wecanaskwhattheratioisbetweenprobabilitiesfortwoenergylevelsseparatedbykBT.CallthoseprobabilitiesP(E)andP(E+kBT).WithalittlealgebraicmanipulationwefindthatP(E+kBT)/P(E)=exp(‐(E+kBT)/kBT)/exp(‐E/kBT)=exp(‐1)=1/e

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Thisisapowerfulsimplifyingstatement.IttellsusthatkBTistheamountofenergydifferencethatcorrespondstoadropinprobabilitybyafactorofe(whichisabout2.7).The‘thermalenergy’valuekBT is therefore the key quantity for comparison when evaluating whether two possibleconfigurationsof a system that are separatedby somegivenenergydifferencewill bepopulatedsimilarlyorverydifferently.ThevalueofkBTissuchausefulquantityforcomparisonthatanenergydifferencewillsometimesbestatedintermsofhowmanykBTunitsitis(whichiseffectivelythesameas stating the value of the unitless exponent E/kBT above). For example, one might hear,“conformationAis‘2kay–tee’higherinenergythanconformationB”.Finally,alwayskeepinmindthatkBTandRTconveyequivalentmeanings;theysimplydifferbyafactorofAvagadro’snumber,NA.RTmustbeusediftheenergyvaluesarebeingdescribedonapermolebasis.Thecontextandunitsassignedtotheenergyshouldmakeitclearwhichisbeingused.Forconvenience,RT(at298K)isabout2500J/mol(inSIunits);thevalueisalsosometimesgiveninnon‐SIunitsas2.5kJ/mol.

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CHAPTER2

EntropyofmixinganditsdependenceonlogofconcentrationsStirling’sapproximationWe begin with a preliminary equation, Stirling’s approximation. As we saw before, variouscalculationshavingtodowiththestatisticalinterpretationofentropyleadustofactorialexpressions,n!.Suchnumbersbecomeintractabletoevaluationasngetslarge;howwouldyouactuallyfigureoutwhatabillionfactorialwas,orthefactorialofAvagadro’snumber?Stirling’sequationgivesusanapproximation for thenatural logof a factorial expression; from thereone could exponentiate ifnecessarytogetanapproximationforthevalueofn!,butwe’llseeitistypicallythelogofthefactorialexpressionthatwewantanyway.Stirling’sapproximationisasfollows:ln(N!)≈N*(ln(N)–1)Hereishowclosetheapproximationis:N Actualvalueof ln(N!)

evaluated as ln(1) +ln(2)+…+ln(N)

Stirling’sapprox.,N*(ln(N)–1)

1,000 5907.7 5912.1106 12815510 12815518Atleastintermsofrelativeproportion,you’llseethattheerrorbecomesverysmallasNgetslarge.‘EntropyofMixing’Asimpleexerciseillustratesthedependenceofentropy(andsubsequentlyotherenergeticterms)onthenatural logofconcentrations. Supposeyouhaveasystemwithtwochambersanditcontainsmolecules of two types (black and white for sake of illustration). Suppose there are n1 blackmoleculesandn2whitemoleculesandconsiderastartingconfigurationwherethen1blackmoleculesareallontheleftandthen2whitemoleculesareallontheright. Nowwewanttoconsiderwhatchangeinentropywouldbeassociatedwithaprocesswherebythemoleculescouldmixtogethersothatblackandwhitemoleculesmightoccupyeitherside,asillustratedbelow.

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From before we know that S = kBln(W), so analyzing the change inentropy,S,boilsdowntofiguringoutwhatWisfortheinitialstateandthefinalstate.Therearedifferentwaysoftreatingthisproblem,butoneistothinkofitasapartitioningproblemlikebefore. Imaginebeginningwitha bag of n1 + n2 = N moleculestogetherinabag.ThentosetupthesystemyouaregoingtopartitiontheNmolecules intoagroupofn1 togoon the left sideandagroupofn2 togoon theright. Aswediscussedbefore,therearemanydifferentwaysofpartitioningalargesetintotwosmallergroups,but inorder toobtain the initial setupshown,onlyoneof thepossiblepartitionings satisfies therequirementthatallthemoleculesinthefirstgroupareblackandallthoseinthesecondgrouparewhite.Sofortheleftside,Winitial=1.Nowforthefinalstateofthesystem.Therewehaveagreedthatthemoleculescanbeoneithersideregardlessoftype.Forthisparticularproblemwearegoingtoassumethatuponmixingwestillkeepn1moleculesontheleftandn2moleculesontheright,soforthefinalstatewearepartitioningtheNmoleculesinton1ontheleftandn2ontheright,butanyofthepossiblepartitioningsisallowed.ThatmeansWfinal=N!/(n1!n2!).Therefore,theentropychangeformixingis

Smix=kBln(Wf)–kBln(Wi)=kBln(Wf/Wi)=kBln(N!/(n1!n2!))Nowyou’llseewhywebeganwithStirling’sapproximation,sowecanreplacethelogsoffactorialexpressionswithalgebraicquantitiesthatcanbemanipulatedandevaluated.Fromabove,

Smix=kBln(N!/(n1!n2!)) =kB(lnN!–lnn1!–lnn2!) ≈kB(N(lnN‐1)–n1(lnn1–1)–n2(lnn2–1)(thennotingthat–N+n1+n2=0) =kB(NlnN–n1lnn1–n2lnn2)(thenrewritingNasn1+n2) =kB((n1+n2)lnN–n1lnn1–n2lnn2)(thenrearrangingandtakingoutanegativesign) =‐kB(n1(lnn1–lnN)+n2(lnn2–lnN) =‐kB(n1ln(n1/N)+n2ln(n2/N))Now,ifweusemolefractionXiasaconcentrationtoreplaceni/N

Smix=‐kB(n1lnX1+n2lnX2),ormoregenerallyformorespeciesSmix=‐kB(nilnXi),whichisalwayspositive

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Basicconclusionsfromthisexercisearethatentropyincreasesbymixing,andthatentropiesdependonthelogsofconcentrations(hereexpressedasmolefractions).Asafurtherinsight,notingthatlnxalwaysgoesdownwithlowervaluesofx,wesensethatthedrivetowardmaximumentropyfavorseveryspeciesgoingtoalowerandlowerconcentration.Butofcoursethetotalconservationofatomsconstrainsthings,makingequilibriumeffectivelyafightoverwhichspeciesisdrivenmoststronglytolowerconcentration.

Gibbsfreeenergy,GAstatevariablethatindicatesthefavorability(orequilibrium)ofaprocessatconstantT&PWhich way processes proceed naturally (i.e. forward or backwards) is established by the totalentropyofthesystemplussurroundings,orforanisolatedsystemonlytheentropyofthesystemneedstobeconsidered.ButthisrestrictioncanberemovedandreplacedwithothermoreconvenientonesbyconstructingotherstatevariablesfromacombinationofSandotherquantities.Formanyapplicationsinbiochemistry,temperatureandpressuredonotchangemuch.AstatevariableG,theGibbsfreeenergy,whichisconstructedasG=H–TS,hasthepropertyofdictatingthedirectionalityofaprocessinasystematconstanttemperatureandpressure(thesurroundingsnolongerrequireconsideration).Alittlealgebracandemystifythisclaim.BeginningbydifferentiatingG=H–TS,dG =dH–Tds–SdT(thenusingthederivativeofH=U+PV,dH=dU+PdV+VdP) =dU+PdV+Vdp–TdS–SdT(thensubstitutingthederivativedU=dq+dw =dq+dw+PdV+VdP–TdS–SdTAtconstantTandPwecandroptwoterms.AndiftheonlyworkisPVwork,thendw=‐PdV,givingdG=dq–TdS.Then,iftheprocessisoccurringreversibly,meaningitisnotbeingdrivenforwardorbackward,thenfromtheclassicaltreatmentofentropywerecallthatdS=dqrev/T,anddqrev=TdS.SubstitutingthengivesusdG=0(reversibleorequilibriumprocessatconstantTandP)Furthermore,thedirectionalityofaprocessthatisnotatequilibriumisdictatedbythesignofdGorG,inthesamewaythatthesignofΔSsystem+ΔSsurroundingsdictatedthedirectionalityofaprocessinourearlierdiscussions,butnowwithareversalofsign.NotingthenegativesignthatappliestoSintheexpressionforG=H–TS,weconcludethatdG<0foraprocessthatoccursspontaneously(intheforwarddirection).Thatis,processesatconstanttemperatureandpressurearedriventominimumfreeenergy,G.

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Gasabalanceoftwofactors,HandTSIt ishelpfultobearinmindthatfromtheformofG=H–TS,thefreeenergy(whichdictatesthedirectionalityofprocesses)isaffectedbytwoterms.ConvertingtheequationforGtoaformthatdescribes the difference or differential between the ‘before’ and ‘after’ or left vs right sides of aprocess,dG=dH–TdSorG=H–TS(notethatwehavedroppedatermSdTthatwouldhavebeenpresentfromdifferentiationsinceweareconsideringaprocessatconstantT)Evidentlythedrivetominimumfreeenergyisacombineddrive(1)towardlowenthalpyH(recallthatHembodiestheenergeticsofmolecularforcesandinteractionsbetweenmolecules,withlowervalues of H corresponding to energetically favorable configurations or lower amounts orconcentrationsofmolecularspeciesthathavehighenergy),and(2)towardshighentropyS(meaningmorerandomnessanddisorder,includingmoreuniformorequalconcentrations).HowtothinkaboutGinasteadystateprocessIndiscussionsofhowstatevariables,likeHorGforexample,arechangedinaprocess,whatissometimesbeingdescribedisabeforeandafterscenario.ForexampleacalculationofwhatHisforconvertingamoleofpuresubstanceAintoamoleofpuresubstanceB.[Wecouldlookupthemolarenthalpiesforthetwocompoundsinatable].Whilethevaluesofthosequantitiesareimportantinevaluatingthethermodynamicpropertiesofaprocess,thisisrarelythesenseinwhichthingslikechangesinfreeenergy,G,areconsideredinbiochemicalprocesses.IfwearetalkingaboutthefreeenergychangeGforconversionofcitratetoisocitrateinthecell,wearethinkingabouttheconversionofthesubstratetotheproductatwhatevertheirconcentrationsare,andthoseconcentrationsarenotchanging.Contrastthatwiththeearlierscenario where the composition and concentrations of the initial and final states are entirelydifferent. Inbiochemicalsystemswheretheconcentrationsofsubstancesarebeingheldroughlyconstant by pathways and networks of reactions occurring together, it is clearer to think ofinfinitesimalconversionsofsubstratetoproduct. Therecanbeachangeinfreeenergyinsuchaprocessinthesensethattheproductandthereactantmayhavedifferentfreeenergiesassociatedwiththem(whichdependsontheirconcentrationsasweshallseelater),andwearecreatingmoremoleculesoftheproductandlessmoleculesofthereactant,butwithoutanysubstantivechangeincompositionorconcentrations.Ofcourseinordertoexpressthemagnitudeofthefreeenergychangefor the process of interest, we have to express it as a quantity with a meaningful scale for theconversionthatisoccurring.Soweexpressthingslikethefreeenergychangeforaprocessonapermolequantity,thoughforconceptualclarityweshouldkeepinmindtheinfinitesimalordifferentialnatureoftheprocessweareconsidering.

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FreeenergyofmixingandthedependenceofGonlogofconcentrationsWecanreturntoourearliertreatmentofmixingandnowcalculatethefreeenergyofmixinginthesameway.FromthedefinitionofG=H‐TS,Gmix=Hmix‐TSmix. Now, ifmoleculesofdifferenttypesinteractwith each other in away that is energetically similar to thewaymolecules of like typeinteract,thenitshouldbesafetosaythatthereshouldn’tbeanyenthalpychangeassociatedwithmixing(basedonourintuitionthatenthalpyisaboutmolecularforcesandinteractions).So,lettingHmixbezeroandusingourpreviousequationfortheentropyofmixing,weget

Gmix=RTniln(Xi)whereniareinmolesandRreflects‘permole’quantitiesConsistentwithearlierdiscussions,weseethatdifferentspeciescontributetothetotalfreeenergy

ofthesystemaccordingtothelogsoftheirindividualconcentrations.AlsonotethatGmixwillalwaysbenegative,consistentwithourexpectationthatthefreeenergyofmixingshouldbefavorable.The finding that the free energyofmixing isnegative(favorable)givesus insight intowhatdriveschemical reactions to theirequilibrium positions. Supposewestartwithasystemcontainingonly chemical A, and there is areactionAcanundergotoformB,and that the energetics ofmoleculeAandBareidentical;asone instance suppose the twomolecules are enantiomers(equivalentinstructureexceptforhandedness).Weknowfromintuitionandpossibleexperiencethatasystemlikethiswillproceedbyreactionuntilthetwospeciesarepresentinequalamountsorconcentrations.Butwhy?IfAandBhavethesameenergy,thenwhatcoulddrivetheconversion?Wouldn’titbesimplerifthemoleculesjuststayedasAsincetheenergyisnotimprovedbyconvertingtoB?Theanswerofcoursehastodowithentropy,andspecificallythecontributionentropymakestothefreeenergyofmixing.Thisimaginaryscenariohelpsusdrawaconnectionbetween(1)thechemicalconversiontoreachequilibriumand(2)mixingofdifferentcomponents.SupposewetaketheinitialsystemcomposedofonlyAandthenimagineahypotheticaldividerdownthemiddle.NowimagineconvertingallthematerialontherightsidefromAtoB.Clearlytheentropy,enthalpy,andfreeenergyofthatprocessareallzerobasedonoursuppositionabouttheenergeticequivalencebetweenAandB.Now,inasecondstepwecanimaginethatthecontentsofthetwosidesareabletomix.ThiswillresultinamixedsystemwithequalamountofAandB,andthefreeenergyofthemixingwouldbenegative(followingfromtheequationabove).Thetwostepsputtogetherproduce

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exactlythesameresultasiftherewaschemicalconversionofhalftheAmoleculesintoBmoleculesin thewhole system. This thought exercise lets us see that the favorability of converting someamountof apure substance intoother substances to reachequilibriumderives froma favorableentropyofmixing.

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CHAPTER3

Chemicalpotentials,µFrombeforewehaveanunderstandingthatthefreeenergyofasystemcomposedofamixtureofchemicalsdependsontheconcentrationsofthecomponents,andifachemicalprocessispossiblethatwouldinterconvertsomecomponentsintoothers,thenthereisafreeenergychangeassociatedwiththatprocess.Frompreviouscoursesyouwilllikelyrememberequationsofthefollowingform:G=G0+RTln(Q),andlettingK=exp(‐G0/RT)givesG=RTln(Q/K)where G0 expresses the (molar) free energy change for a reaction if it were occurring understandard state conditions, Q represents the ratio of product concentrations to reactantconcentrationsunder the conditionswhere the reaction isbeing considered, and theequilibriumconstantKistheratioofproducttoreactantconditionsatequilibrium.Belowwewillshowhowtheequations above can be obtained, and perhaps better understood, by taking a differential orinfinitesimal view of any reaction or process underlying the conversion of molecules from onespeciestoanotherorfromonelocationorphasetoanother.DefinitionofµasapartialderivativeofGwithrespecttocomposition‘Chemicalpotentials’aredifferentialorderivativequantitiesthathelpusgetatthefreeenergyofamixture(i.e.asystemwithmultiplecomponents).Sinceamixtureisjustacombinationofseparatecomponents, itmakes sense to considerwhat free energy is contributed to themixture by eachcomponentseparately.Awayoflookingatthatquestionistoconsiderhowmuchthefreeenergyofasystemwouldbechangedbyaddingatiny,infinitesimalamountofaparticularcomponent.Thatideaisshownontheright,wherethechemical potential of a givencomponent, i, is defined as thepartialderivativeofGwithrespecttothe change in the amount of thatcomponent. Note that despite thechemical potential being adifferentialrelatedtoaninfinitesimalchange, it is expressed as per molequantity.

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Dependenceofchemicalpotentialsonconcentrationsandstandardstatechemicalpotentialsµ0Thefreeenergyinamixturedependsonthenaturallogoftheconcentrations,sonaturallyweexpecttoseeasimilardependenceofonconcentration.Thetotalfreeenergyforamixtureshouldbethesum of the free energies of the pure components (weighted of course by the amount of eachcomponent),plusthefreeenergyofmixing,sincestartingwithpurecomponentsseparatelyandthenmixingthem(obviously)givesusamixture.

∎ ln

∎ ln

wherethefirstterminthesumrelatestothefreeenergiesofthepurecomponentsandthelasttermdescribesthefreeenergyofmixing.ThebarovertheGindicatesapermolequantityandthesolidsymbol as a superscript indicates reference to the pure component. Now we can evaluate thechemicalpotentialforcomponentiasapartialderivativeofGwithrespecttoni:

∎ ln

wherewehavereplacedthefreeenergyofthepurecomponentonapermolebasis( ∎)withthechemicalpotentialofthepurecomponent( ∎);theirmeaningsareequivalent.Asexpected,weseethat the chemical potential of each species depends on the log of its concentration, and that thechemicalpotentialgoesdown(i.e.becomesmoreenergeticallyfavorable)astheconcentrationgoesdown.Thetotaldifferential,dGasafunctionofchangesincompositionNowthatwehaveanexpressionforhowthechemicalpotentialsdependonconcentration,wecanturn to look at how the total free energy depends on changes in the quantities of the variouscomponents. We note that T and P are the natural variables for G, and that G also depends oncomposition, i.e. the ni’s. TakingG as a function of T, P, and the ni’s,we canwrite out the totaldifferentialforGas:

, , , ,

Replacingthepartialderivativeswiththecorrectthermodynamicquantitiesgives:

⋯

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Then,atconstantTandP,weseethatthechangeinfreeenergyarisingfromachangeincomposition(i.e.achemicalconversionofsomemoleculestoothers,ormovementofmoleculesfromoneplacetoanother)isgivenby:

Thereismuchsensetothisequation.Thetotaldifferentialfreeenergychangetakesintoaccountthe(differential)compositionalchangeforeachcomponent(dn)multipliedbythechemicalpotentialofeachcomponent(µ).WegetageneralsensethenthatdGwillbenegative(i.e.afavorableprocess)ifmolecules with higher chemical potentials are converted to molecules with lower chemicalpotentials.Furthermore,ifaprocessisatequilibriumthenthechemicalpotentialsofthemoleculesthatwouldbecreatedandthosethatwouldbeconsumedshouldbeequallybalancedinorderfordGtobeequalto0.Equilibriumconditionsintermsofµ’sFromabove,∑ 0foraprocessatequilibrium.This is a powerful equation for analyzing all kindsof processes, fromchemical reactions (wherechemicallydistinctmoleculesareabletointerconvert)totransportprocesses(whereamoleculeofagiventypeisabletomovefromoneplacetoanotherorfromonephasetoanother).PhaseortransportequilibriumThe diagram at the right illustrates equilibrium involvingpartitioning of a component between two different phases.You are familiar with processes like this from organicchemistry where you partitioned a compound between anaqueous phase and an organic phase (e.g. in a separatoryfunnel).Howdoesthedifferentialfreeenergychange,dG,inthis case depend on the process under consideration(specificallytransportofmoleculeAfromphase1tophase2)?Fromabove,dG=µA,1 dnA,1 +µA,2dnA,2,where the subscriptsdenote the chemical species (whichdoesn’tchangehere)andthephasewhereitoccurs.Atequilibrium,dG=0,soµA,1dnA,1+µA,2dnA,2=0.Then,notingthatdnA,1anddnA,2areidenticalbutnegativelyrelatedquantities,µA,1dnA,1‐µA,2dnA,2=dnA,2(µA,2‐µA,1)=0.Theparentheticexpressionmustbezero.Therefore,whenAisatequilibriumbetweenthetwophases,

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µA,1=µA,2Thismakesperfectsense;sincethisprocesscreatesamoleculeofAinphase2attheexpenseofamoleculeofAinphase1,atequilibriumthechemicalpotentialofAinthetwophasesmustbeequal.Ifthetwochemicalpotentialswerenotequal,thentheprocesswouldnotbeatequilibrium,andGcould be decreased (in a shift closer to equilibrium) by converting some of the higher chemicalpotentialcomponentintothelowerchemicalpotentialcomponent.Intheproblemdescribedhere,thatwouldbebymovement(i.e.atransportprocess).Ifthesystemwasnotatequilibrium,thenthefreeenergyassociatedwiththeprocess(assumingtheforwardprocessisinterpretedtobemovementfromlefttoright)wouldbe(µA,2‐µA,1).Thiswouldrepresentadifferentialfreeenergyonapermolebasis.Thevalueofthatenergytermcouldhavevariouspracticalinterpretationsinabiologicalsetting.Ifthevaluewaslessthanzero,thenitmightdescribetheamountofworkthatcouldbeextractedfromtheprocessandusedtodriveadifferentunfavorableprocessifthetwoprocesseswerecoupledtogetherbysomemechanism.Or,ifthefreeenergy difference was positive, then that energy value could describe the amount of work (orfavorable freeenergy) thatwouldhavetobeextracted fromanothercoupledprocess inorder tomaintainthefirstprocessawayfromtheequilibriumconditiontowhichitwouldgootherwise.ChemicalequilibriumNowweconsiderachemicalreactionandlookattheconditionsonthei’sforequilibrium.Considerthereactionbelow:

A+B 2CIntheprocessabove,theamountsofA,B,andCaresubjecttochange,sothedifferentialfreeenergychangeis

∑ =µAdnA+µBdnB+µCdnCThe reaction arrow represents a singleprocess, so the changes that occur to the amounts of thedifferentcomponentsmustberelatedtoeachother,andtoasinglequantitydescribingtheextentofthereaction.Ifweletdescribetheextentofthereactiononapermolebasis,thenaccordingtothereactionstoichiometry,

dnA=‐ddnB=‐ddnC=+2dandsubstitutingabovegives

dG=(‐µA‐µBdnB+2µC)d

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Atequilibrium(dG=0),sowehave(–µA–µBdnB+2µC)=0.Thismakesintuitivesensesinceyoucanseethatinorderfortheexpressiontoevaluatetozero,2µCwouldhavetobeequaltoµA+µB,meaningthataddingupthechemicalpotentialsofthecomponentsonthetwosidesofthereactionhas togivematchingvalues. Otherwise the freeenergycouldbe loweredbyhaving thereactionproceedonewayortheother.Ifthereactionisoccurringawayfromequilibrium,thenthefreeenergydifferenceforthereactiononapermolebasis(meaningpermoleofreactionevents)wouldbe(‐µA‐µBdnB+2µC).You’llseethatthis isnothingmore thanaddingupandsubtracting the chemicalpotentialsof theproductsandreactants,properlyweightedbytheirrespectivestoichiometries.Asbefore,iftheconcentrationsareawayfromequilibriumthentheexpressionabovewoulddescribethemolarenergyrequiredtomaketheprocessproceedor(ifthevalueisnegative)howmuchworkcouldbeextractedfromtheprocess.Equilibriumconditionsintermsofconcentrationsandstandardchemicalpotentials:arrivingatfamiliarequationsfortheequilibriumconstantSofarwehavelaidouttheconditionsonthechemicalpotentialsthatmustbetrueatequilibrium.But of course the chemical potentials of the components depend on their concentrations, andtogether this leads to equations for equilibrium constants (K) and reaction quotients (Q),whichshouldbefamiliar.Fromthispointforwardwewillswitchawayfrommolefractionandusemolarity(capitalC)asourconcentrationunitinstead.Wereplacethesolidsuperscriptdenotingthepurestatebeforewiththeopensubscriptdenoting1Masthechoiceforstandardstateconcentrations.Wethereforerewriteourequationforthechemicalpotentialanditsdependenceonconcentrationas

i=i0+RTlnCiThestandardstatechemicalpotential(i0)referstothechemicalpotentialthemoleculewouldhaveatitsstandardstateconcentration(1M).Thestandardstatechemicalpotentialservesasareferencevalue to which the chemical potential can be related, taking into account the dependence onconcentration.Thisgeneralstatementabouthowthechemicalpotentialofacomponentdependsonastandardstatevalue(whichisaconstant)andtheconcentrationofthatcomponentwillappearthroughoutoursubsequentdiscussions.PhaseortransportequilibriumFortheearliercaseofphaseequilibriumofmoleculeAbetweenphases1and2,atequilibriumµA,1=µA,2andsubstitutingtheexpressionaboveforeachcomponentgives,

A,10+RTlnCA,1,eq=A,20+RTlnCA,2,eq

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Here we recognize that the standard state chemical potentials for the same molecule could bedifferentindifferentphases,fromwhichonecanseethattheconcentrationsonthetwosideswouldbe unequal at equilibrium. By collecting separately the terms for concentration and those forchemicalpotentials,

RTlnCA,2.eq‐RTlnCA,1,eq=‐(A,20‐A,10)Rearranginggives,ln(CA,2,eq/CA,1,eq)=‐(A,20‐A,10)/RTWe can recognize CA,2,eq/ CA,1,eq as the equilibrium constant K for this process. Making thatsubstitutionandalsoreplacingthedifferencebetweenstandardchemicalpotentialswiththemorefamiliarexpressionG0forthestandardstatefreeenergychange,wearriveat

lnK=‐G0/RTandK=exp(‐G0/RT)withK=(CA,2,eq/CA,1,eq)To analyze a system away from equilibrium, we can introduce concentrations and equilibriumconstantsintothenon‐equilibriumsituation.ReturningtodG=dnA,2(µA,2‐µA,1),andsubstitutinginequationsoftheformi=i0+RTlnCiasbeforegives,withsomerearrangement,

dG/dnA,2=RTln(CA,2/CA,1)+(A,20‐A,10)ormorefamiliar,G=RTln(CA,2/CA,1)+G0wherethefreeenergydifferenceshererefertothetransportprocessonapermolebasis. NotingfromabovethatG0=–RTln(K),andrecallingthatthereactionquotientQisusedtodescribetheratioofproducttoreactantconcentrationsinthegeneralcasewhereasystemmaybeawayfromequilibrium,wegetthefamiliarequation

G=RTln(Q/K)whereinthiscaseQ=CA,2/CA,1andK=CA,2,eq/CA,1,eqAgain,Gonapermolebasishasthesamemeaningas(µA,2‐µA,1),whichistheenergypermolethatcanbeextractedfrom(orthatwouldbeneededtodrive)thereactionunderconsideration.ChemicalequilibriumWe can work out similar equilibrium expressions as well for our previous chemical reaction.Substitutinggeneraltermsoftheformi=i0+RTlnCiinto(–µA–µBdnB+2µC=0)gives,withsomerearrangement:

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2RTlnCC,eq‐RTlnCA,eq‐RTlnCB,eq=‐(2C0‐A0‐B0)ln(CC,eq2/(CA,eqCB,eq))=‐(2C0‐A0‐B0)/RTwhichagainmatches

lnK=‐G0/RTandK=exp(‐G0/RT)withK=(CC,eq2/(CA,eqCB,eq))andG0=(2C0‐A0‐B0)Asbefore,ifthereactionisawayfromequilibriumthenwecanworkoutequationsforthemolarfreeenergyforthereaction,obtaininginthiscase

G=RTln(Q/K)withQ=(CC2/(CACB))andK=(CC,eq2/(CA,eqCB,eq))ImportanceofunitsIt is importanttounderstandthewayconcentrationunitshavebeenimpliedintheequationswehavedevelopedforchemicalpotentials,freeenergies,andequilibriumconstants.Returningtothegeneral equationwedeveloped for how chemical potential depends on concentration,whereweswitchedovertomolarconcentrations,i=i0+RTlnCi,youwillseethatweseemtobetakingalogarithmofaquantitythathasunitsassociatedwithit(molarityinthiscase),whichistechnicallyillegal. To correct this problem, in every occurrence of a concentration value in our precedingequations,weshouldunderstandthattheconcentrationneedstobeimplicitlydividedbythevaluechosenforthestandardstate,1Mforexample. Thatdivisiongeneratesunitlessquantitiesfortheconcentrationsinallofourexpressionsforchemicalpotentials,freeenergies,reactionquotientsandequilibriumconstants:

i=i0+RTln(Ci/1M)forexampleorK=((CC,eq/1M)2/((CA,eq/1M)(CB,eq/1M))forthereactionabove.Asyoucansee,aslongasthestandardstateis1M,thenleavingouttheseimplicitdenominatorsisfine.Butthereisanimportantcasewhere1Misnotthetypicalchoicemadeforthestandardstate.Because biological conditions are typically close to pH7 (and not pH0), the standard stateconcentrationforprotons(H+)istakentobe10‐7M.Thatmeansthatanytimeareaction(ortransportprocess)involvesthecreation,consumption,ormovementofprotons,theconcentrationofprotonsmustbedividedby10‐7Mwhenusingitinthecalculationoffreeenergiesandreactionquotientsandequilibriumconstants.

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Other species that get handled as special cases, typically by being omitted from the equilibriumexpressions,include:water(itsconcentrationinmostscenariosistakentobenearlypuresothemolefractionX≈1)andcompoundsintheirpureforms(e.g.crystallinesolids)whicharealsotakentobetheirownphase,withX=1.Precautions about G vs G0, reactions with changes in stoichiometry, and overallconcentrationeffectsFreeenergyissometimesdiscussedloosely,whichcanleadtoconfusionanderrorsininterpretation.AparticularlycommonerroristonotproperlydistinguishbetweenwhetheroneistalkingaboutGorG0.Asdiscussedabove,G0describeshowfavorableorunfavorableaprocesswouldbeifthereactants andproductswere all at their standard state concentrations. That is practicallyneverrepresentative of conditions of biochemical interest. [Note that cellular concentrations of smallmoleculemetabolitesareofteninthemillimolarrange;macromoleculeslikeproteinsarepresentinthecellatindividualconcentrationsthatareofteninthemicromolarrange(e.g.forhousekeepingenzymes) or nanomolar or lower for low‐abundance proteins like those often involved in cellsignaling.]ThevalueofG0issimplyareferenceenergythatmakesitpossibletocalculatethefreeenergyorequilibriumpositionatsomeothermorerelevantsetofconcentrations.Anothercommonsourceofconfusionarisesinthecontextofreactionswherethetotalstoichiometryof the reactants and products are different. In simple processes or reactions where thestoichiometriesofthereactantsandproductsarethesame,casualstatementssuchas,“thatreactionorprocessis‘naturallyfavorable’becausethe(standard)freeenergyisnegative”,canbeinterpretedinasensibleway.Forexample,forthereactionA B, if the standard state free energydifferenceisnegative,thenK>1,andifAandBwerebothpresentat1Mconcentrationthen,sinceQ=1whichislowerthanK,Gwouldbenegativeandtheforwardreaction(conversionofsomeAtoB)wouldbefavorable.ThesameconclusionswouldbereachediftheconcentrationsofAandBwerebothmuchlower(orhigher)butstillequaltoeachother.Forexample,ifAandBwerebothpresentat1mMconcentrationthenQwouldstillbe1andtheforwardreactionwouldstillbefavorable.AfurtherconclusionisthatatequilibriumBwouldhaveahigherconcentrationthanA,whethertheoverallconcentrationsarehighor low. But thiskindofcasual logic fallsapartentirelywhenthenumberofmoleculesontheleftandrightsideofareactionareunequal.Aclassiccaseisaprocessofbindingbetweenaproteinandaligand(e.g.aninhibitororsubstrateorcofactor).Heretherearetwo‘reactants’andone‘product’(theboundformoftheprotein).Intheformerexample,thesignofG0providedquickinsightintotherelativeconcentrationsonewouldexpectforthesubstrateandproductatequilibrium,withoutworryingaboutthedefinitionofthestandardstate.Butwhataboutthecaseofligandbindingbyaprotein?Here,thevalueofG0providesnosucheasyinsight.TheproblemcanbeappreciatedbynotingthatiftheconcentrationunitsforQ(orK)donotcancel(whichtheydonot ifthetotalstoichiometriesaredifferentontheleftandtheright),thenthevalueofQchangeswithchangesinoverallconcentration,evenifrelativeconcentrationsareheldequal.So,forexample,anegativeG0(K>1)forthebindingenergywouldtellyouthatiftheprotein,ligand,and

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protein‐ligandcomplexwereallat1M,thenthebindingprocesswouldproceedforwardtowardmorecomplete binding (so that ultimatelymore of the protein would be in the bound form than theunboundform).Butifthosethreespecieswereallpresentatequalconcentrationsof1uM,thevalueofQwouldbeamillion(10‐6/(10‐6*10‐6)),whichcouldbemuchgreaterthanK(dependingonhownegativeG0 was), which wouldmean that the process would proceed in the reverse directiontowardunbinding,andultimatelymostoftheproteinwouldnotbeboundtoligand.Thisisjustoneillustrationofthepointthattheinterpretationoffreeenergiesmustbemadecarefully,particularlywhen therearedifferences instoichiometrybetweenreactantsandproducts. In thosecasesonemustbearclearlyinmindthatoverallconcentrationsareprofoundlyimportant,andthatthesignandmagnitudeofG0ishardlyinformativewithoutfurtherconsiderationofrealconcentrations.NotethattheargumentaboveaboutstoichiometryandtheinterpretationoffreeenergyGappliesjustaswelltoentropyS,butisalesscriticalissueforenthalpyH.CommentsonthedependenceofGandKonT(van’tHoffequation)Inourdiscussionsof free energyweemphasized that the signofG indicates the favorabilityofreactions under conditions of constant temperature andpressure. But howGdepends on thosevaluesisalsoofinterestinsomesituations.[Oneexampleishowtemperatureaffectstheequilibriumbetweentheunfoldedandfoldedstatesofaprotein.]HerewesaysomethingabouthowfreeenergyGandtheequilibriumconstantKdependonT.FromG=H–TS,onecanseequicklythatthedependenceofGonTisdeterminedbyS.Infact,wecanlookupfromderivativeexpressionsofthestatevariablesthatthepartialderivativeofGwithrespect to T, holding P constant, is –S. That is, (∂G/∂T)P = ‐S. So for example, if a process isentropicallyfavored(S>0),thenincreasingthetemperaturewillmakeGmorenegative.Clearly,thedependenceofGonTisdictatedbythesignofS.Butnowlet’slookatthedependenceoftheequilibriumconstantKonT.Thisiswhereintuitioncangoawry.WeknowthatKisdeterminedfromG0(recallK=exp(‐G0/RT),andthatamorenegativevalueofG0 correspondstoahighervalueofK.Sowemightexpectthat if increasingTcausesadecreaseinG0(asitwouldifS0>0asdiscussedabove),thenKshouldalsodependonS0,withanincreaseinTcausinganincreaseinKifS0>0.Butthislogicisincorrect(thoughnotuncommonlyheardindiscussions).TheproblemwiththelogicisthatKdependsonTintwoways:throughtheeffectofTonG0andthroughthepresenceofTinthedenominatoroftheexpressionforKintermsofG0.TogetthecorrectanswerforhowKdependsonT,wehavetobreakupG0intoitsenthalpyandentropycomponentsattheoutset,sincethosetwotermshavedifferentdependenciesonT.

K=exp(‐G0/RT)lnK=‐G0/RT=‐(H0–TS0)/RT=‐H0/RT+S0/R

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NowthedependenceofKonTcanbeseentobegovernedbyH0andnotbyS0!TakingderivativeswithrespecttoTweget

d(lnK)/dT=H/(RT2)(herethestandardstatesuperscriptforHmightbeomittedsinceHdependslessstronglyonoverallconcentrations,incontrasttoGandSasdiscussedatlengthabove).Thisisoneformofthevan’tHoffequation.SeparatingthederivativevariablesKandTondifferentsidesgivesd(lnK)=H/(RT2)dT,andaslongasHdoesnotchangemuchwithchangeinTwecanintegratebetweentwotemperaturesT1andT2toget

ln(K2)–ln(K1)=ln(K2/K1)=(1/T1–1/T2)H/RwhichshowshowonecanextractavaluefortheenthalpychangeforareactionorprocessfromthevalueofKattwodifferenttemperatures.Or,plottingln(K)vs1/Tshouldgiveaslopeof–H/R.

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Graphicalviewsofchemicalpotentialsandtotalfreeenergyasafunctionofreactionprogress

forasimpleequilibrium(A B)

Note that 

‘mu’ means . 

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CHAPTER4

Non‐idealbehaviorinmixturesThebreakdownofidealequationsforchemicalpotentialOurpreviousdiscussionshaveemphasized the idea that theenergies in amixturehavea simplebehavior (i.e. a log dependence) that is perfectly obeyed across all ranges of concentrations,regardlessofwhatsortsofmolecular forcesmightcomeintoplayasdifferentkindsofmoleculesencountereachother. Werefertothatkindofbehavioras ‘ideal’. Weturnnowtoconsiderthebehaviorof‘real’or‘non‐ideal’solutions.Tounderstandnon‐idealbehavior,let’srethinkthestepswetooktoarriveatoursimpleequationsforidealbehaviortolookforassumptionswemadethatmightbeviolatedinrealsituations.Weusedthe idea of ‘free energy ofmixing’ as the foundation for establishing our equations for chemicalpotentialandtheirdependenceonlogconcentrationsintheidealcase.Westartedwiththisequation,Gmix=Hmix–TSmix,whichledustoGmix=RT(niln(Xi)).Butwemadetwoassumptionsintheprocess.First,you’llrecallthatweallowedourselvestodropouttheenthalpyterm,assertingthatHmixwouldbezerouponmixingifthedifferentkindsofmoleculesmakeenergeticinteractionswitheachotherthat are similar to those they make with themselves in their pure forms. This might be a fairassumptionifthetwo(ormore)molecularspeciesareverysimilartoeachother(e.g. inpolarity,charge,size,etc.).Ontheotherhand,ifinterminglingofthedifferentcomponentsleadstointeractionforces of different types andmagnitudes, then our assumption that mixing would not have anyenthalpiceffectwillbeincorrect,andtheenergyorchemicalpotentialfeltbyeachcomponentwillbeaffectednotonlybyitsownconcentrationbutbythenewforcesitexperienceswheninteractingwiththeothercomponents.Asecondsimplificationcame in thewaywe treated thesecond term, theentropyofmixing. Wedevelopedourcombinatorialexpression forW(togiveusentropy)basedonan idealizedmixingschemewhereweplacedmoleculesofdifferenttypesondifferentsidesofacontainer.Thisseemedinnocentenough.Butwhatifthetwotypesofmoleculeswereofvastlydifferentsizes?Thismighthaveledtoamorecomplexproblemrelatingtohowlargevssmallmoleculesmightbearrangedinspacewithout overlapping each other. This issuewould not have been captured by our simpleequationforcountingpartitioningsofmolecules.Laterwewilldiscussinmoredetailspecificsituationswhereviolationsoftheassumptionsaboveleadtonon‐idealbehavior.Butfirstwewillmodifyourpreviousequationsforchemicalpotentialsandequilibriumconstantssothattheywillholdtrueevenwhennon‐idealeffectsareatplay.Todothiswe introduce a correction or factor into the chemical potential equations in the form of an‘activitycoefficient’,.

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ActivitiesandactivitycoefficientsOuridealequationforthechemicalpotentialofspeciesiwas:

i=i0+RTlnCi(ideal)Nowadmittingthatthatequationmightnotbetotallyvalid,we introduceacorrectionfactor, theactivitycoefficient,i,designedtomaketheequationremaintrue.

i=i0+RTln(iCi)(realornon‐ideal)or

i=i0+RTln(ai)withai=iCi(realornon‐ideal)whereweintroducethe‘activity’aitobeequaltoiCi.ThenaieffectivelyreplacesCiinthechemicalpotentialequation.Youcanseethatthe‘activity’,a,becomeslikeaneffectiveconcentrationofagivencomponent.Anotherwayoflookingatitwouldbetoimaginethatyoudon’thaveawayofdirectlymeasuringthetrueconcentrationofacomponentinamixture,butyouhaveawayofmeasuringthechemical potential of that component (through some energetic evaluation). From the chemicalpotentialofthatcomponent,sincechemicalpotentialdependsonconcentration,youcouldsaythatyouareabletomeasurewhatconcentrationthatcomponentseemstohavebasedonitsenergeticbehavior,andthateffectiveconcentrationistheactivity. Youmightanticipatefromtheequationsabove, which make it explicit that chemical potential relates to activity and not necessarily toconcentration, that theactivitieswill be thekeyquantities in equilibriumconstants and reactionquotientsfornon‐idealsystems.Beforewereworkourpreviousequilibriumequationsintermsofactivities,let’slookalittlemoreattherangeofpossibilitiesfortheactivitycoefficientsandhowthisrelatestofavorablevsunfavorableenergeticfeaturesinnon‐idealmixtures.First,notethatournewequationsreducetotheidealoneswhentheactivitycoefficients,i,areequalto1. Inthatcase,theactivityisthesameastheconcentration,ai=iCi. Logicallythen,non‐idealbehavioriswhentheactivitycoefficientiseithergreaterthanorlessthan1.Thosetwopossibilitiescan be ascribed different energetic meanings. By comparing the equations above for chemicalpotentialintheidealandnon‐idealcases,wecanseethatavalueofi>1relatestoanelevatedvalueforthechemicalpotentialforcomponenti.Sincethechemicalpotentialreportsontheenergythatisfelt by some component, we surmise that i > 1 indicates that component i is experiencingunfavorableenergeticscomparedtothecaseofidealbehavior.Conversely,i<1reflectsunusuallygoodenergeticinteractions.

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TheidealbehaviorofhighlydilutesolutionsNowwehavetodiscussinabitmoredetailwhatlimitingsituationsarechosen(byconvention)torepresentidealbehavior.Fromourpreviousdiscussionsitmightseemthatthesensiblethingwouldbetotakethepurestateofeachcomponenttorepresentitsidealbehavior.Thisisfineforthesolvent;inbiochemistryour‘mixtures’arenearlyalwayssolutionswherewateristhesolventandvariousothermoleculesarethedissolvedsolutes.Buttheideaofapuresoluteoftendoesn’tmakesenseforbiochemistry.Forexample,asamplecontainingonlyaproteininapureform(withoutsolvent)isnonsensicalsinceproteinsdon’tfoldproperlyunlesstheyareinanaqueoussolution.Therefore,theconditionchosentorepresentidealbehaviorforasoluteisusuallythe(hypothetical)infinitelydilutelimit.Let’sseeifthisisconsistentwithideaswelaidoutearlierabouthowtheequationsforchemicalpotentialasafunctionofconcentrationshouldbehave.Puttingaslightlyfinerpointonourpreviousarguments,theidealequationforchemicalpotentialfailsifagivencomponentexperiencesdifferentkindsofinteractionsasitsconcentrationischanged.Nowwecanexaminethesituationofahighlydilutemixturetoseeifmeetstheidealrequirementthatagivencomponentmakesthesamekindsofinteractions as its concentration is changed slightly. First consider a dilute solution from theperspectiveofthesolvent.Ifthesoluteispresentina1:1000000ratiotothesolvent(settingasideforthemomentpotentialdifferencesinmolecularsize),thenanyarbitrarilychosensolventmoleculewill be interacting nearly exclusively with other solvent molecules. Now if we increase theconcentrationofthesolutebyafactoroftwo,thatdoesn’tchangethepicture;asolventmoleculewillstillinteractnearlyexclusivelywithothersolventmolecules.Nowlet’sviewitfromtheperspectiveofthesolute.Atthe1:1000000ratio,asolutemoleculewillrarelyinteractwithanothersoluteandwillexclusively‘see’thesolvent.Whenwedoubletheconcentrationofthesolute,thisisstillthecase.Clearlythen,ifasolutionisverydilute,thevariouscomponentscanbeexpectedtobehaveideally.Theidealstateforthesolvent istakentobepuresolvent(water),whereastheidealstateforthesoluteisatinfinitedilution,andthecomponentsunderthesehighlydiluteconditionshaveactivitycoefficientsequalto1.Theoriginofnon‐idealbehaviorathigherconcentrationsWecanusethesamelogicasabovetothinkaboutthenon‐dilutesituationwherenon‐idealbehaviorbeginstoshowup.Considerwhathappenswhenasoluteconcentrationgetsmuchhigher.Nowthesolvent will start to encounter solute molecules with frequencies that cannot be ignored (asillustratedbelow). So if for thesakeofargument thesolventandthesolutemakepooreror lessfavorableinteractionswitheachotherthantheydowiththemselves,thenasthemixturemovesintothenon‐idealrange,thesolventwillexperienceahigherchemicalpotentialthanexpectedforidealbehaviorowing to its increased interactionswith theother component (the solute). Thatwouldmeantheactivitycoefficientforthesolventwouldbe>1.Nowlet’slookatitfromtheperspectiveofthesolute,whichseesthingsdifferentlybecauseitisdiluteratherthannearlypurelikethesolvent.Asthesoluteconcentrationincreases,atsomepointsolutemoleculesbegintoencounterothersolutemoleculestoanappreciableextent.Nowunderthesamescenarioasbeforewherethesolventand

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solutemakepoorerinteractionswitheachother,andbetterinteractionswiththemselves,youseethatastheconcentrationofthesoluteincreasesitmakesmorefavorableinteractions(withitself).So, theactivitycoefficient for thesolutewouldbe<1. Thereasonweobtaindifferentnumericalbehaviorfortheactivitycoefficientforthesolventvsthesoluteunderthesamesetofassumptionsabouttheenergeticsofthesolutionisdueentirelytothedifferentchoicesforwhattheideallimitisfor the differentcomponents:pureinthe case of thesolvent (water) andhighly dilute in thecaseofthedissolvedsolute. Note that ifwe imagined theopposite scenariowhere the solventand solute madebetter interactionswitheachotherthanwith themselves,thenthebehaviorofthe activitycoefficients wouldbe reversed, withthe activitycoefficient for thesolutebeing>1andsolvent<1.ReworkingtheequilibriumequationsintermsofactivitiesinsteadofconcentrationsTheexpressionforthetotaldifferentialdGremainstrueevenifthebehaviorisnon‐ideal,asdoestherequirementthatdGequals0attheequilibriumcomposition.

∑ 0Butnowweuse

i=i0+RTln(ai)

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ThisisthesameasbeforeexceptactivityahasreplacemolarconcentrationC.Clearlytheequationswill develop exactly as before, butwith activitya replacing C everywhere. For example, for thereactionA B,startingfromµA=µBatequilibrium,wewouldobtain

ln(aB,eq/aA,eq)=‐(B0‐A0)/RT(aB,eq/aA,eq)=exp(‐(B0‐A0)/RT)=K(whereKistheequilibriumconstantasbefore)NotehoweverthatKCB,eq/CA,eqifthebehaviorisnon‐ideal,sinceaiCiTherelationshipbetweentheequilibriumconstantandtheconcentrationscanbeseenbygroupingtheacitivitycoefficientstogetherasasinglecorrectionterms,asfollows:

K=(aB,eq/aA,eq)=(BCB,eq)/(ACA,e)=(CB,eq/CA,eq)*(B/A)Theequilibriumconstantisconstantandsoitsvalueisnotaffectedbynon‐idealbehavior(e.g.athigherconcentrations),andtheratioofactivitiesalsoremainsequaltotheequilibriumvalue.Buttheratioofconcentrations,whichweordinarilythinkofastheequilibriumconstant,isaffectedandcan change. You might then think of the ratio of concentrations as the non‐ideal or ‘apparent’equilibriumconstant,whoserelationshiptothetrue,idealequilibriumconstantwouldbe:(CB,eq/CA,eq)=Kapp=K/(B/A)And if thesystemwereaway fromequilibriumthen theexpression formolar freeenergy for thereactionwouldbethesameasfortheidealcase,exceptactivitieswouldreplaceconcentrationsin

theformulationofthereactionquotientQ.ForthesimplereactionofA Bforexample,

G=RTln((aB/aA)/K)=RTln(((BCB)/(ACA))/K)TheequationsaboveareofcoursespecificforthesimpleequilibriumbetweenAandB,buttheideageneralizesimmediatelytoanyreactionorstoichiometry.

ForthemorecomplexreactionA+B 2C,beginningwith2µC=µA+µB,wewouldendupwith,

K=aC,eq2/(aA,eq*aB,eq)=(CCC,eq)2/(ACA,eq*BCB,eq)=CC,eq2/(CA,eq*CB,eq)*C2/(A*B)andCC,eq2/(CA,eq*CB,eq)=Kapp=K/(C2/(A*B))Andforthemolarfreeenergyifthesystemisawayfromequilibrium,G=RTln((aC2/(aA*aB))/K)=RTln(((CCC)2/(ACA*BCB))/K)

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Ion‐ion interactions in solution as an example of non‐ideal behavior(Debye‐Hückeltheory)Herewewillexaminehowionsinanelectrolyte(salt)solutionbehave.Asyouknow,chargedspeciesrepelorattracteachotherdependingonwhethertheirchargeshavethesameoroppositesigns.Thisaffectsthepositionsthationsexhibit(onaverage)astheymovearoundfreelyinsolution.Wewillcontrastwhathappenswhenwehaveaverydilute(meaningideal)electrolytesolutioncomparedtowhentheconcentrationsofionsgetshigher.Inthedilutelimit,theionsaresofarapartthattheirelectrostaticpropertiesdonotinfluenceeachother.Incontrast,athigherconcentrationsthepositiveionswillprefertobeinthevicinityofnegativeions,andviceversa,whilelikechargeswillprefertobefartherfromeachother.Thatmeansthat,onaverage,apositiveionwillfinditselfsurroundedbyaslightexcessofnegativelychargedions,andlikewiseanegativeionwillfinditselfsurroundedbyaslightexcessofpositivelychargedions;rememberthatwealwayshaveamixtureofpositiveandnegativeionsinanelectricallyneutralsolution.Theionsaremovingaroundinsolution,sotheeffectissubtle,butsignificant. Fromthisargumentyoucansee thateach ionshouldenjoya favorableenergetic interactionwith its ‘counter‐ionatmosphere’. Referring toour earlierdiscussions, thisfavorableenergeticcontributioncorrespondstoanactivitycoefficientfortheionsthatis<1.A quantitative treatment of the energetics of electrolyte solutionswas developed by Debye andHückel,andisworkedoutindetailinsometexts.Herewewillsimplysummarizetheessentialideas.IonicstrengthandtheDebyelengthFirstweexplaintheideaoftheDebye‐length.Eachionissurroundedbyacounter‐ionatmospherewhosetotalchargeoffsetsthechargeonthecentralion.Howisthatopposingchargedistributed(onaverage)asafunctionofdistancefromthecentralion?Ataverylongdistancefromthecentralionofinteresttheattractiveforceissmall,sothecounter‐ionatmospheredropstozeroatlongdistance.Inaddition,theamountofopposingchargethatcanexistveryclosetothecentralchargeislimitedsince the available volume at very small distance becomes small. So, as diagrammedbelow, theamountofcounter‐ionchargegoesupandthendownwithdistance,anditsmaximumvalueisatadistancereferredtoastheDebyelength,1/.Theincreasedcounter‐ionconcentrationinthevicinityofacentralionalsohastheeffectof‘screening’ordiminishingtheelectrostaticforceorfieldthatisexertedbyagivenion,andtheDebyelengthalsodescribesthateffect.FromCoulomb’slawyou’llrememberthattheelectrostaticpotentialatadistancerfromacentralionisproportionalto1/r(that is, 1/r), and that equationwould apply in the infinitely dilute limit. When screeningbecomessignificantowingtoanincreaseintheconcentrationofions,then(1/r)exp(‐r).Asimplecomputersimulationisshownforionsmovingaroundinsolutionunderforcesofattractionandrepulsion.Asnapshitisshownalongwithacalculationoftheaveragecounterionchargearoundanegativelychargeion.TheDebye‐lengthlength1/isindicated.

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Whatisthevalueof1/?Thisdependsmainlyonthetotalconcentrationofionsinsolution;moreexactly,itdependsontheionicstrength,I.Forreasonablydilutesolutionstheequationforionicstrengthis

I=(1/2)(Cizi2)wheretheCiaremolarconcentrationsofthechargedspecies,andziistheircharge,andthesumisoverallions.Notethatthesquaringofzgivespositivevaluesforanionsaswellascations.ThedependenceofonIiscomplex,butforaqueoussolutionsnear298K,1/≈3.0Å/sqrt(I)whereIisunderstoodtobeinmolarconcentrationunitsSo,forexample,iftheionicstrengthofasolutionis0.001M,then1/=96Å,whereasifI=0.1M,then1/=9.6Å.Forreferencerecallthatthesizesanddistancesbetweenbondedatomsisinthe1Åto1.5Årange.Activitycoefficientsforionicspecies

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Aquantitativetreatmentofhowionsaresurroundedbyacounter‐ionatmospheremakesitpossibletocalculatethetheoreticalmagnitudeofthefavorableenergyofinteractionbetweenanionanditscounter‐ion atmosphere. This energy of interaction will be the source of non‐ideality in theelectrolytesolution,somathematicalexpressionscanbeobtainedfortheactivitycoefficientforanion.Withoutderivation,thefollowingisobtained.Foragivenchargedspecies,i:

ln2 1

whereaistheradiusoftheion.Underrelativelydiluteconditions,1/>>a,anda<<1,sotheatermdropsoutofthedenominatortogive

ln 2

Inaqueoussolutionsnear298Kthisequation,andthedependenceof1/onsqrt(I),canbecombinedandreducedtoasimpleapproximateexpression:

ln 1.2 √ whereIisunderstoodtobeinmolarconcentrationunits.Notefromtheequationabovethattheactivitycoefficientis<1foreachspecies,regardlessofchargesign,whichisconsistentwithourqualitativediscussionabove.Andnotethatastheionicstrengthgoestozero(e.g.underhighlydiluteconditions),thelogofgoesto0andthereforegoesto1,asexpectedforidealconditions.Usingionicactivitycoefficientstoanalyzetheeffectofchargeonmolecularassociation,andelectrostaticscreeningWe canuse theactivity coefficient equationabove to gain insight intohow ionic strength affectsmolecular association between chargedmolecules(e.g.proteinsornucleicacids)insolution. We’ll set upanabstractproblemwhere a molecule A has charge zA and amoleculeBhaschargezB, andAandBcancome together in some association orbinding process to form species C, whosechargeiszA+zB.Fromourpreviousdiscussions,wecanquicklywriteouthowweexpecttheequilibriumpositionofthis bindingprocess tobe affectedby total ionic strength.Note that ifwe aredealingwith largemoleculeslikeproteins,theirmolarityisusuallyverylow,sothechargesonthemoleculesinquestion

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(hereAandB)maynotcontributemeaningfullytothetotalionicstrength.Thetotalionicstrengthwe’retalkingaboutherewouldmorelikelyrelatetohowmuchsaltweaddedtotheexperiment.Sowe’llimaginethattheionicstrengthissomethingwecontrolseparatelyfromwhateverishappeningregardingAandBandtheirassociation.WhatdoweexpecttohappentotheequilibriumaboveifthechargesonAandBareoppositeandwestartaddingsalt?You’velikelylearnedaboutelectrostatic‘screening’before,whichistheideathathighsaltconcentrationtendstomaskordiminishanyelectrostaticforcethattwochargedmoleculesmightexertoneachother. So, intuitivelyyoumightexpectthat inthecasewhereAandBhaveoppositenetchargesthataddingsaltwouldlessentheirtendencytoassociateandwouldthereforeshifttheequilibriumpositiontotheleft.Let’ssetuptheequilibriumequationforthisprocessandseeifwegettheresultweexpect.Nowthatweknowhowtohandlenon‐idealequilibriumexpressions,wecanwrite

Kapp=CC/(CACB)=K/(C/(AB))to describe how the non‐ideal or apparent equilibrium constant would change according to thevaluesoftheactivitycoefficientsiforthethreespecies.FromthesimplifiedDebye‐HückelequationweknowhowtheactivitycoefficientsofthethreespeciesshoulddependontheirchargesandontheionicstrengthI.ExponentiatingthepreviousequationforhowdependsonI,wewouldget

A=exp(‐1.2*zA2*sqrt(I))andsimilarlyforB,andC=exp(‐1.2*(zA+zB)2*sqrt(I))Followingsomerearrangements,Kapp=CC/(CACB)=K*exp(2.4*zA*zB*sqrt(I))orln(Kapp)=ln(K)+2.4*zA*zB*sqrt(I)TheseequationsconfirmthatifthechargesonAandBhaveoppositesign,thenKappwouldbelowered(sincetheproductofzAandzBwouldbenegative)andtheequilibriumpositionforthereactionwouldthereforebeshiftedtotheleftbyincreasingionicstrength.ThisispreciselywhatweexpectedbasedonhigherionicstrengthscreeningtheattractiveelectrostaticforcebetweenAandB.Andnotethatthe effect would be opposite if A and B were of like charge; the overall driving force for theirassociationinthatcasemightarisefromothernon‐electrostaticinteractions,andanincreaseinionicstrengthwoulddiminishtheelectrostaticrepulsionbetweenthem.

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Molecularcrowdingandexcludedvolumeeffectsasanexampleofnon‐idealbehaviorinsolutionsofmacromoleculesTheideaofexcludedvolumeEarlierwealludedtotheideathatsolutionscontainingverylargesolutemoleculesmightgiverisetonon‐idealbehavior.Thisphenomenonissometimesdescribedinthecontextof‘molecularcrowding’or‘excludedvolume’effects. Tounderstandthephenomenonweneedtoconsiderasolutionthatcontainssomelargesolutemoleculesalready,andthinkaboutwhateffecttheirpresencehasonourabilitytoaddanothercopyofthesolute.Themoleculescannotoccupythesamespace.Therefore,acrosstheentirevolumeofthesystem,someofthelocationsareexcludedaspossiblepositionsforplacinganewmolecule. That is theexcludedvolume. Toa firstapproximation, the relationshipbetweenmolecularcrowdingandtheactivitycoefficientforamacromoleculecanbewrittenas

=Vtot/(Vtot–Vexcl)whereVtotisthetotalvolumeofthesystemandVexclistheexcludedvolume.Notethatthisimpliesthatmolecularcrowdingeffectscorrespondto>1.Geometricallyinterestingaspectsofmolecularcrowdingcomeintoplaywhenwelookmorecarefullyatwhatismeantbytheexcludedvolume. Theexcludedvolumeisnotsimplythevolumeofspacethatisoccupiedbytheexistingsolutemolecules.Thecomplicationisthatwehavetothinkaboutwherewecanandcan’tchoosetopositionanewmolecule,meaningwhereitscentercouldorcouldnotreside.Asyouwillseefromthediagrambelow,theregionwherewecannotplace(thecenterof)anewsolutemoleculeismuchlargerthanthespaceactuallyoccupiedbytheexistingsolutemolecules.Firstweillustratethesituationwherethesolutehastheshapeofalargesphere(e.g.acompactglobularprotein).

Whatthediagramshowsisthattheexcludedvolumeinthecaseofsphericalmoleculesisaspherewithtwicetheradiusoftheindividualmolecule.Thatvolumeistherefore8timeslargerthanthevolumeactuallyoccupied. That is,Vexcl=8Vocc. ByrearrangingtheapproximationaboveforbydividingbyVtot,weseethat=1/(1‐Vexcl/Vtot)=1/(1‐8Vocc/Vtot).Asaresult,evenifarelativelysmall

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fractionofthetotalspaceisoccupiedbymacromolecules,theactivitycoefficientmaybeconsiderablyhigherthan1.Inthisroughmodel,if5%ofthespaceisoccupied,=1.67.ThepeculiarbehaviorofrigidelongatedstructuresThis is interesting by itself, but the situation becomesmuchmore intriguingwhenwe considermolecules whose shapes are highly elongated rather than spherical. Choosing a geometricallytractablemodel,herewetreatthecaseofalongrod‐shapedmoleculewithasquarecross‐section,whosedimensionsareLxdxd,withL>>d.Againwecanconsiderwhatvolumeofspaceisexcludedfor placing (the center of) an addedmolecule in the proximity of another. The analysis ismorecomplicatedthanforthespherebecausenowtherelativeorientationofthetwomoleculesmatters.Furthermore, as we carry outthe same exercise as before insliding the second rod aroundthe first one to see where wecannotplacethesecondone,wemustkeeptheorientationoftherods unchanged; we are onlyasking about the allowableposition for the secondmolecule at some fixedorientation. First,wewill consider the best casescenario,whichiswherethetworodsareparalleltoeachother.Theresultissimilartothecasewiththespheres: theexcludedvolumewouldbe (2d)(2d)(2L)=8Ld2,which is8 times thevolumeof asinglemolecule.Butwhataboutthecasewherethetworodsareperpendiculartoeachother? This is the worst casescenario. It takes more carefulvisualization in 3D (shown on theright), but the excluded volume inthiscaseis(L+d)(L+d)(2d),whichis2d(L2 + 2Ld + d2). Ifwe comparethistothevolumeofonemoleculebydividingbyLd2,wegetaratioof2L/d+4 (dropping the term2d/Lwhich would be small). Now,insteadofgettingaratioof8,wegetamuchhighernumbersinceL>>d.Returning to the earlier equationfor the activity coefficient, we seethatcouldbelargeevenwhenthefraction of the space occupied by

parallel rods

perpendicular rods

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therod‐shapedmoleculesissmall.Therealbehaviorofcoursewouldhavetobeanaverage(orreallyan integral) of the behavior as a function of the angle between the rods. But the effect remainssubstantial.Howdotheseexcludedvolumeideasrelatetorealmacromolecules? Ifwetakethelessonstobegeneralonesthatshouldapplyevenifthesituationsinquestiondon’tinvolvemoleculesthatlookexactlylikespheresorrigidrods,thentheimplicationsarenumerous.Proteinfoldingisonerelevantexample.Proteinshavetobestableintheirfoldedcompactconformationscomparedtotheunfoldedforminwhichtheirbackboneswouldgenerallybeflexibleandmuchmoreextended.Certainlytheunfolded form of a protein should have a much greater excluded volume. As a result, underconditions where crowding effects are significant, like when the overall concentration ofmacromolecules is high, the activity coefficient for the unfolded form of a protein could besignificantlygreater than1. Wecanwriteanequilibriumprocessbetween theunfolded (U)andnativelyfolded(N)states:U NIftheequilibriumconstantunderdiluteconditionsisK,thenfollowingtheprocedureswedevelopedearlierwecanwritethatunderconditionswherenon‐ideal(crowding)effectscomeintoplay,

CN/CU=K/(N/U)

WewouldexpectmolecularcrowdingeffectstogiveU>N.Theconsequenceisthat(CN/CU)shouldgoupandtheequilibriumpositionshouldbeshiftedto theright, towardsthedirectionofnativefolding,bycrowdedconditions.Thisisanimportantpointgivenhowcrowdedtheconditionsareinthe cell, and also how dilute typical conditions are when purified proteins are studied in thelaboratory. Itmaybethatproteinsaresignificantlystabilized intheir foldedstates in thecellbycrowding;thisisnotreflectedintypicallaboratorystudies.AnotherexampleinvolveshighlyelongatedfilamentousstructureslikeF‐actinandmicrotubulesthatforminthecellbyassemblyoflargenumbersofproteinsubunits.Thebehaviorsofthesekindsofproteinfilamentsinthecellareinfluencedstronglybycrowding.Theeffectsareprobablymyriad,butonebasiceffect in suchsystems is the tendency towardsalignmentorbundlingof filaments.Withoutworkingoutasophisticatedmodel,onecanstillgetasenseofwhythisisthecase.Considerthe alternative scenarioswhere you have a systemwith filaments that areeither mainly aligned vsrandomly oriented. Nowconsider trying to add anadditionalfilament(whichisawayofsensingtheactivityofacomponent).Whichcase

49 

 

allowsforeasieraddition?Thesituationisillustratedabove,whereyoucanseeclearlythataddingadditional filaments iseasier if theyaremorealigned. Inthatsenseyoucanseethat theactivitycoefficientshouldbelowerinthealignedcase,andsothatcasewillbefavoredascrowdingcomestodominate.Ofcourseweknowthatentropywilltendtodrivesuchasystemintheotherdirection,towardsrandommolecularorientations,butatsomepointthecrowdingeffectswillprevailandfavoralignment.Thealignmentandbundlingofproteinfilamentsislikelyfunctionallyimportantinthecell.Inthislecturewedetailedjusttwokindsofphysicalphenomena–ionicinteractionsinsolutionandcrowdingeffects–thatgiverisetonon‐idealbehavior.Butbiologicalsystemsarehighlycomplex,andindeednon‐idealbehaviorcanariseinmanydifferentways.

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CHAPTER5

ChemicalPotentialandEquilibriuminthePresenceofAdditionalForcesTherearemanyinstances,inbothcellularandexperimentallaboratorysettings,wheremoleculesexperienceadditionalforcesthatcontributetotheenergytheyfeel,therebyaffectingtheequilibriumpositionsoftheprocessesinwhichtheyareinvolved. Wewillconsidersomeexampleshere,anddevelopageneralframeworkformodifyingourpreviousequationsforchemicalpotentialinordertohandlethesesituations.Theessenceistorewriteourequationforthechemicalpotentialforsomemolecularspeciesinawaythataddsintherelevantextraenergy:

i=i0+RTlnCi+energytermwiththeaddedtermrelatingtotheenergyexperiencedbythemolecularcomponentinquestion,onapermolebasis.OsmoticpressureOsmoticpressureisafamiliarphenomenon.Ithasimportantrolesincellularfunction,andisalsothebasisforlaboratorymeasurementstostudymolecularbehavior,thoughthiswasmorecommoninthepastthanitisnow.Asyouwillrecall,osmoticpressurereferstoapressuredifferencethatmust be exerted to preventwater frommoving across a semi‐permeablemembrane froma sidewheretheoverallsoluteconcentrationislowtowhereitishigher.Wecansetupasystemwithtwochambersseparatedbyasemi‐permeable membrane (permeable to water butnothing else). The equilibrium process in question isthereforethetransportofwaterfromonesidetotheother.WewilladdproteintosideA,whereitwillbeconfinedtostay.Wecanseerightawaythattheconcentrationofwateronthetwosidesisnevergoingtobeequal,andwerecallfromearlier discussions that the chemical potential isdeterminedbythestandardchemicalpotentialplustheconcentrationterm.Sotheonlywaywatercanbeatequilibriumbetweenthetwosidesisifthereisanadditionforcethatisdifferentbetweenthetwosides,inthiscaseapressureonsideApreventingflowofwaterfromrighttoleft.Wecanwritetheequilibriumsituationforwaterasfollows:ForthechemicalpotentialofwaterontheBside,

H2O,B=0H2O

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FortheAside,

H2O,A=H2O,A0+RTlnXH2O,A+(NotethatweareusingXinsteadofCfortheconcentrationofH2O)Tocompletethisanalysisweneedtoknowhowthechemicalpotentialenergyshouldchangeasafunctionofpressure.

where isthemolarvolume(forwaterinthiscase).Then,d= dPandH2O= P= wheredenotestheosmoticpressuredifference.Now,ifweusetheterm totaketheplaceoftheextraterminthechemicalpotentialforwaterontheAsideinourpreviousequation,wecanequatethechemicalpotentialsforwateronthetwosidesandrearrangetoseethat:RTlnXH2O,A+ =0This equation providesmore insight ifwemanipulate it to obtain an expression in terms of theconcentrationofthesoluteinsteadoftheconcentrationofwater.WeletX2bethemolefractionofthesolute.X2=1–XH2O.Thensubstitutingabove,andnotingthatfromTaylor’sexpansionthatln(1‐X2)=‐X2+X22/2+…≈‐X2,aslongasX2issmall,wegetRTX2≈ NowswitchingfrommolefractiontomolarconcentrationbynotingthatatlowconcentrationX2≈C2 ,weget

≈RTC2Inotherwords,theosmoticpressureisproportionaltothemolarconcentrationofsolutemoleculespresent,makingitanexampleofa‘colligative’property.Itissometimesconvenienttofurthermodifytheosmoticpressureequationtoconvertfrommolarconcentration C, toweight concentration, c. We often have a betterway of knowing theweightconcentrationofaproteininsolution,forexamplefromaspectroscopicabsorbancemeasurementthatreportsontheapproximatenumberofpeptidegroupsorparticularaminoacidgroupspresentratherthanthenumberofpolypeptidechainspresent. Also,theconversionfrommolartoweight

the expected change in chemical potential 

energy due to a change in pressure. 

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concentration introduces a molecular weight term, and as a result information aboutmolecularweight(ofaproteinornucleicacid)canbeobtained. Theconversionfollowsfromc=MCwherelower case c is the weight concentration (typically in g/L =mg/ml) andM is molecular weight(typicallying/mol).Weget,

≈RTc2/MTherefore, measuring and knowing the weight concentration allows approximation of themolecularweight:M≈RTc2/.Osmoticpressureisnolongeracommonbiochemistrylaboratorytechnique, but later we will discuss more common experimental methods for molecular weightdetermination.Osmotic pressure measurements are sometimes used to examine non‐ideal effects in solution.Modifying our previous equation to allow for non‐ideal effects (and also realizing that otherapproximations were introduced by truncationof the Taylor’s expansion), we can write anexpressionforosmoticpressureasfollows:

≈RT(c2/M+Bc22+…)Inthisexpression,Bcapturesthefirstordernon‐ideality and is referred to as the second virialcoefficient. One way of extracting B frommeasurementofosmoticpressureasa functionofsoluteconcentrationisillustratedhere.EquilibriumsedimentationWedon’toftenthinkabouttheeffectofgravitationalforceonmoleculesinsolution.Butputtingasampleinacentrifugeisessentiallythesameasincreasingtheforceofgravity(sometimesbyafactoroftensofthousands).Ifthesolutemoleculehasalargemass(asdoproteinsandnucleicacids),thentheseforcescanhavesignificanteffects.Centrifugationisawidelyusedlaboratorytechnique,andisused in various modes for different purposes. Here we will consider a particular kind ofcentrifugationexperimentthatispowerfulforstudyingthemolecularweightsofmacromoleculesinsolutioninawaythatpreservestheirnativeconformationsandassemblystates.Webeginbythinkingaboutwhatweexpecttohappenduringcentrifugation(orevenundersimplegravitationalforces)intwolimitingcases:(1)whereasoluteisverysmall(e.g.ethanolorsucrose)or(2)whentheparticleinquestionismassive(likeacellorasandparticle).Ifthesoluteisverysmall, thenwe can spin the sample forever and the concentration of the solutewill be uniform,essentiallyequalthroughoutthetube,fromthetoptothebottom.Schematically,theresultcanbe

53 

 

diagrammedasshown,whereristhevariabledescribingdistancefromtheaxisofrotation.[Wedrawthetubeandthepositionvariablerhorizontalsincetheaxisofrotationisvertical.]Attheotherextreme,iftheparticleinquestionismassive,thenvirtuallyallofitwillgotothebottom,andtheconcentrationwillbenearlyzeroeverywhereelse.

Butwhatifthesoluteisintermediateinmass?Thenweshouldexpectitsconcentrationprofiletosomehowbein‐betweenthetwoextremecasesillustrated:notuniformthroughoutthesample,butalsonotcompletelysedimentedtothebottom.Inotherwords,weshouldgetahigherconcentrationtowardsthebottomandalowerconcentrationtowardsthetop.Andthissituationshouldbestable,meaningwecanspinitforeverandthisisthefinalequilibriumresult.Theideaisschematizedabove.Butwhatistheexactformweexpectforthiscurve?Surelyitmustdepend on the mass of the molecule, so how might we extract a value for the mass from theequilibriumsedimentationbehavior?Qualitativelywecanseethatthisisasituationofforcesinbalance.Weendupwithaconcentrationthatisunequal(highertowardsthebottom),andweknowthattheremustbeanentropicdrivingforceintheoppositedirection,favoringamoreequaldistribution.Thisisabalancingforcethatactsagainstthegravitationalorcentrifugalforcethatisdrivingmoleculestowardsthebottomofthetube.

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Thisisasituationatequilibrium,sowecantreattheproblemwithourgeneralapproachofsettingupachemicalpotentialequation thatcontainsanextraenergy termrelating toworkdonebyanexternalforce.Imagineasolutemoleculethatisfreetomovebetweentwopositionsinatube.Atequilibrium,thechemicalpotential forthesoluteatthosetwopositionsmustbeequal(otherwisetherewouldbefurthernettransport).So,wewillsolveourproblembyrequiringthatd/dr=0.Butfirstwewritean equation for howwe expect the chemical potential to depend jointly on concentration andposition in the tube, since we are ultimately interested in establishing how concentration andpositionarerelatedtoeachother.µ=µ0+RTlnC+UwhereUcanbeviewedasapotentialenergyonapermolebasisrelatestomovementofasoluteintheappliedgravitationalorcentrifugalfield.Then,d/dr,whichmustbezeroatequilibrium,is

d/dr=0=RTd(lnC)/dr+dU/drGenerally,forceFisthenegativeofthederivativeofpotentialenergywithrespecttoposition,F=‐dU/dr,sorearrangingwegetRTd(lnC)/dr=FNowwecanintroducethedependenceonmass,sinceF=ma,wheremismassandaisacceleration.Butbeforeproceedingwiththeequationabovewehavetoexpandonthemeaningofthemassminthecontextofthecurrentproblem.Whatmattershereisnotsimplythemassofthesolute,butthe‘buoyant’mass,meaningthedifferencebetweenitsmassandthemassofthewateritdisplaces,whichofcoursedependsonitsvolume.Also,tobeconsistentwiththeenergyequationweneedtoworkouttherelevantmassequationinpermoleterms.Themassweneedinourequationaboveis:

NA(m–v*H20)wherevisthevolumeofonesolutemolecule.Wecanreplacethevolumeofonemoleculewithits‘specificvolume’ (whichisvolumepermassorreallyjustthereciprocalofdensity),timesitsmass.Includingasubscript2tomakeitclearthatthespecificvolumereferstothesoluteandnotthesolvent,weget:

NA(m–m H20)=NAm(1– H20)=M(1– H20)(whereMisthemolecularweightofthesolute)

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Theunitlessterm(1– H20)isreferredtoasthe ‘densityincrement’andissometimesreplacedwithasinglevariable2forsimplicity.Notethatifthesoluteiscomposedofmaterialwhosedensityisgreaterthanwater,whichistrueforproteinsandnucleicacids(butnotlipids),then H20willbelessthan1,and2willbegreaterthan0.Usingthisexpression(M2)forthebuoyantmassonapermolebasisinourF=maequation,wegetF=M2a.Beforereturningtoourequationthatbalancedtheconcentrationgradientinthetubewithforce,wepointoutthattherearetwodifferentkindsofproblemswheretheseequationsareuseful,(1)wheretheforceissimplygravitational(inwhichcasea=g),and(2)wherewearedoingcentrifugation(inwhichcasea=2r,fromintroductoryphysics,withrepresentingangularvelocity;alsorecallthat=‘rpms’*2/60).Wewillproceedtoworkouttheequilibriumsituationforcentrifugation. SubstitutingF=M2a=M22rintoourpreviousequationforbalancedforces,weget

RTd(lnC)/dr=M22rd(lnC)/dr=M22r/RTThenseparatingthederivativevariablesgives

d(lnC)=M22r/RTdrAndintegratinggives

|2

|

ln ln 2

or

where r0 refers to some referenceposition in the tube andC0 refers to the concentrationat thatposition.Bymatchingtheequationforln(C)abovetothestandardformforalinearequation(y=mx+b),youcanseethatplottinglogofconcentrationwithrespecttothesquareofthepositioninthetube(i.e.distancefromtheaxisofrotation)shouldtheoreticallygiveastraightlinewhoseslopeisM22/2RT,fromwhichthevalueofMcanbecalculated,sincetheothervariablesrepresentknownquantities.Aschematicdiagramisshown.

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Furthermore, note that because weightconcentration (c) is proportional to molarconcentration(C),ln(c)differsfromln(C)onlybyanadditivefactor.Thatmeansthatyoucanplotln(c)vsr2,andtheslopewillbethesameasabove.Thisis useful because the weight concentration of aproteinornucleicacidsampleistypicallytheeasierquantity to establish froma routine spectroscopicmeasurement.Notethatincomparisontosomeothermethodsthatyou might be familiar with for determiningmolecular weights of proteins – e.g. SDS polyacrylamide gel electrophoresis – equilibriumsedimentation keeps proteins in their native forms, including potential assemblies of multiplesubunits.Itisthereforeveryusefulforgettinginformationabouttheoligomericstatesofproteins,i.e.whethertheyaredimersortrimersorlargerspeciesinsolution.GravitationalsedimentationIf the acceleration on a sample is due simply to gravity instead of centrifugal acceleration, then

insteadofa=2r,wesimplyhavea=g.Withanalogytothepreviousequations,weget

RTd(lnC)/dr=M2gRearrangingandintegratinggives

ln ln

or

EquilibriumsedimentationofamixtureIfasamplecontainsmorethanonetypeofmacromolecularsolute(e.g.twodifferentproteins),thenitssedimentationbehaviorwillbemorecomplex.Eachcomponentwillbehaveexactlyasdescribed

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above,butifthemultiplecomponentshavedifferentmasses,thentheirconcentrationswillincreasetodifferentdegreesasafunctionofpositioninthetube.Asaresult,ifyouwereonlyabletomeasurethe total concentrationofproteinas a functionofposition,whichwouldbe the case if youwererelyingonatypicalspectroscopicreading,thenyourconcentrationprofilewouldhaveanunusualbehaviorthatcouldnotbefittotheequationsweworkedoutabove.Theresultingconcentrationprofilewould notmatchwhat youwould expect for any choice ofmolecularweight for a singlecomponent.Thatis,aplotofln(c)vsr2willnotbestraight,butcurved.Let’slookmorespecificallyathowthatplotwouldlook.Theslopeofthecurveshouldobey,slope=d(ln(c))/d(r2) =M22/2RT.Now rewrite this in termsof d(c) instead of d(ln(c)) bynoting thatd(ln(c))=(1/c)d(c).Thatgives,d(c)/d(r2)=Mc22/2RT.Nowifthesampleisamixtureandourmeasurementisofthetotalweightconcentration,thenwecanwriteanequationforthebehaviorofthetotalweightconcentrationasasumoverthecomponents:

∑

2

Dividing by cT on both sides and then absorbing the cT in the denominator on the left into thederivativeoftheloggives

2∑

Wecancomparethecomplicatedequationabovetothesimplerequationwehadbeforeforasinglepurecomponent.Thatpreviousequation,afterrearrangingthetermsabittomatchtheformabove,was:

2

Wecanseethattheequationfortheslopeofthecurveofthelogoftheconcentrationofamixturematchestheequationforthesinglecomponentcase,exceptthatinthecaseofamixturethemolecularweightMhasbeenreplacedwithatermthatgivesakindofaverageofthemolecularweightsofallthecomponentspresent,accountingfortheirconcentrationsinweightterms.Evidently,theslopeofthecurveforthecaseofamixturegives(afterdividingby22/(2RT))aneffectivemolecularweight,Meffgivenby:

∑

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This is an example of a ‘weight‐average’ molecular weight, since each component gets includedaccordingtoitsweightconcentration(asopposedtoitsmolarconcentration).Clearly, extracting the molecular weights of multiple components from the equilibriumsedimentationbehaviorofamixtureisacomplicatedchallenge.Wewillnotexaminethatprobleminanymoredetail,butsomeessentialpointscanbemadeabouttheoverallbehavior.Howwouldtheoverallshapeofaplotofln(c)vsr2lookforamixturecomparedtoapurecomponent?Wehavealready established that it should not be straight, and that the slope should reflect the effectivemolecularweight.Istheeffectivemolecularweightgreaterorsmalleraswemovefartherdownthetube?Aswemovetohighervaluesofr(e.g.towardsthebottomofthetube),therelativeproportionof the heavier components should begreater since their concentrationsincreasemorerapidlywithincreasingr. That means that the slope of thecurve, which is proportional to Meff,should be higher at higher r. Thiscorrespondstoupwardcurvature.Theplotontherightshowstheresultfromanequilibriumsedimentationrunforaprotein assembly composed of 12subunits. This resultwas interpretedto indicate that the 12‐subunitassembly is inequilibriumwithothersmallersubassemblies.If a sample behaves like a mixture rather than a pure component, it may mean that we wereunsuccessful at purifying the desired component prior to the centrifugation experiment, so thatcontaminantsremained.Butthereisanotherpossibilitythatoccursoften.Manyproteinsassociatenaturallyintooligomericforms,likedimersforexample.Attypicalconcentrations,aproteincouldbeatequilibriumbetweenamonomerandadimerform.Whatwouldhappentheninanequilibriumsedimentation experiment? This is clearly a case of amixture, sowewill see behavior like thatdetailedabove.Towardsthebottomofthetube,therelativeproportionofthedimerwillbehigherthanatthetop.Ifthemonomeranddimerareatequilibrium,thenyoumightwonderhowtheycouldbeatequilibriumbothatthetopofthetubeandatthebottomiftherelativeproportionofmonomertodimerisdifferentatthetwopositions.Butiftheproblemisworkedoutindetail,oneseesthatthis is exactly as expected: the overall concentration of protein is higher at the bottom, and thisnaturallygivesahigherproportionofdimeratequilibrium.Inotherwords,theentiresystemisableto reach equilibrium both with respect to monomer‐dimer association and with respect to thedependenceofconcentrationonpositioninresponsetotheexternalcentrifugalforce.Effectsofnon‐idealbehavior

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Ifasoluteexhibitsnon‐idealbehavior,thenwemightalsoobservedeviationfromtheequationsweworkedoutabovefortheequilibriumconcentrationasafunctionofposition.Specifically,weshouldstillexpectaplotoflogofactivityofthesolutetobelinearwhenplottedasafunctionofr2,buttheconcentrationmaybedifferentfromtheactivity.SummaryInthischapterweexaminedthebehaviorofsystemswheremoleculesareunder the influenceofexternalforces.Weprovidedageneralstrategyformodifyingthechemicalpotentialequations,andworkedoutthedetailsfortwosituations:osmoticpressureandequilibriumsedimentation.Thesearejusttwoexamplesamongmanydifferentwaysthatexternalforcescomeintoplayinbiochemicalsystems.

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CHAPTER6

Electrostaticpotentialenergy,iontransport,andmembranepotentialsInpreviouslectureswecoveredscenarioswheremoleculesweresubjecttospecificforces.Inthislecturewewilllookationsthataresubjecttoforcesarisingfromvoltageorelectrostaticpotentialdifferences.Wediscussedionicinteractionsearlierinadifferentcontext,wherewedealtwiththeenergyanionexperiencesfrombeinginsolutionwithotherionsaroundit.Inthischapterwewilldealwithelectrostaticinteractionsinadifferentcontext.Wewillconsiderhowionsaredistributedinspaceasaresultofanelectrostaticpotential(i.e.avoltage)thatisdifferentatdifferentlocations.Amainfocuswillbeonsituationswheretheelectrostaticpotentialisdifferentonthetwosidesofasemi‐permeablemembrane.Thishaswideapplicationstomolecularbiologyandelectrophysiology.ThechemicalpotentialenergyofanionatapositionofelectrostaticpotentialWe need to know what energy to associate with a charge residing at a particular electrostaticpotential (which isavoltage). You’ll recall from introductoryphysics that theworkrequired(orpotentialenergygenerated)inmovingachargeqtoanelectrostaticpotentialisU=q.Forourpurposesweneedenergyonapermolebasis.You’llrecallthatthechargeonamoleofelementaryparticlesisNA*e=6.021023*1.610‐19Coulombs=96,500C,whichisdefinedasoneFaraday,F.Thatmeansifweareconsideringaparticularkindofionwhosevalencechargeisz(e.g.zCa2+=2),thenthechargeqonamoleofthoseionswillbeq=zF.Finally,thepotentialenergygainedbyputtingthatchargeatanelectrostaticpotential(onapermolebasis)isU=zF.Fromthiswecanwriteourequationforchemicalpotentialenergyinthepresenceofanelectrostaticpotential:

i=i0+RTlnCi+ziFTheNernstequationandmembranepotentialSuppose we have a system with two chambers separated by amembranethat ispermeable to the ionicspecies inquestion. Ourinterest here is in situations where the electrostatic potential isdifferentonsideAvssideB,thatisAB.TheequilibriumprocessofinterestisthetransferofionicspeciesifromsideAtosideB.Frombeforeweknowthatthismeansi,A=i,B.Theseparateequationsforthechemicalpotentialofspeciesionthetwosides,takingintoaccountelectrostaticenergies,wouldbe:

i,A=0i,A+RTlnCi,A+ziFA

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and

i,B=0i,B+RTlnCi,B+ziFBSettingthechemicalpotentialequaltoeachotherandrearranginggives:

=B‐A=RT/(ziF)ln(Ci,A/Ci,B)ThisisoneformoftheNernstequation.Ittellsusthattheelectrostaticpotentialdifferencebetweenthetwosidesisrelatedtothelogoftheconcentrationsoftheiononthetwosides(assumingthattheionisfreetoreachequilibriumacrossthemembrane).Notetheeffectofthesignofthecharge,z.Anegationofzreversestheeffect.Considerfirstacasewherezispositive(e.g.Na+ions).Theequationtellsus that if thepotential ishigheronsideB(heremeaning>0), thentheconcentrationofpositivelychargedionswillbehigheronsideA.Thereverseistrueforanegativelychargedionthatisfreetoequilibrate;itwouldbemoreconcentratedonsideBifthepotentialismorepositiveonthatside.Atfirstthismightseembackwards.Howcanthepotentialbehigherontherightifthepositivelychargedionsaremoreabundantontheleft?Theshortansweristhatitisimportanttokeepinmindthattheionsherearerespondingtoanelectrostaticpotentialthatexistsinthesystem.Thatis,theunequalconcentrationofionsisaneffectandnotthecauseofthepotentialdifferencehere.Itisinstructivetonotethatthevoltagedifferenceintheequationdoesnotcarryasubscriptfortheionoritscharge.Thatmeansthatiftherearemultiplechargedspeciesinthesystemthatareatequilibriumbetweenthetwosides,thentheratiosoftheirconcentrationsmustgivethesamevaluefor.Evidentlytheconcentrationratiosfordifferentionsmustberelatedtoeachother.Wecanwriteouttwoversionsoftheequationabove,oneforioniandtheotherforionj,andthenequatethetwopotentialstogive:

ln(Ci,B/Ci,A)/zi=ln(Cj,B/Cj,A)/zjand ,,

,

,

As an example, if Na+ and Cl‐ ions are both at equilibrium between the two sides, then

,,

,,and , , , , .Theequationwouldbe

morecomplexifthechargeswerenotplusorminusone,butinthissimplecasetheproductofsodiumand chloride ions on the two sides is equal at equilibrium. This result will be convenient in acalculationshortly.Thepreviousequationsdescribeequilibriumconditions.Awayfromequilibrium,thefreeenergyonapermolebasis for ion transportbetweenpositionsA andB,where the ion concentrationsandelectrostaticpotentialsmaybothbedifferent,wouldbe:G=RTln(CB/CA)+zF(B–A)

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TheDonnanpotentialSo far we have discussed how ions that can equilibrate are driven to unequal concentrationsdependingontheelectrostaticpotentialdifferencethatexistsbetweentwopositions.Butwhatmightbethesourceoftheelectrostaticpotentialdifference?Wealreadydiscussedthatitisnotcausedbytheunequaldistributionof ionsthatareabletoequilibrate–theirdistributionis intheopposingdirection. Onepossibilitywouldbeanexternalappliedvoltagewithelectrodesonthe twosides.Problems based on those sorts of electrochemical cells are typically discussed in introductorychemistrycourses.Butelectrostaticpotentialdifferences–betweentheoutsideandinsideofacellfor example – are a common subject in cellular and biochemical systems, and in those cases anexternalbatteryvoltageisrarelytheoriginoftheelectrostaticpotential.Here we illustrate a highly simplifiedsystem that shows how an unequaldistributionofan ion thatcannot crossamembranecangiverisetoanelectrostaticpotential. This is called a Donnanpotential. We begin with a simple two‐chamber system like before, but nowweputaproteinmoleculeonsideAonly,andassume it has a negative charge.Counterionswouldalsobepresentsowe’llbegin the setup with an equi‐molarconcentrationofprotein‐andNa+ionsonthe A side. We’ll denote this startingconcentrationasx. Now inaddition let’ssaythatacertainamountofsalt(NaCl)isadded to both sides to start. Call thisconcentrations.Nowlet’ssaythattheNa+andCl–ionscancrossthemembranebuttheproteincannot.Whathappens?Wecananswerthatquestionby supposing that somequantity ofNa+ andCl– crosses theboundary inorder to reachequilibrium; the amounts of Na+ and Cl– that cross should be equal in order to maintainelectroneutrality.Let’sassumethevolumesofthetwosidesareequalforsimplicity,sothatthemolarconcentrationchangeasaresultofNaandCl‐movementisthesameonbothsides,andcallthatvalued(plusdontheright,minusdontheleft).NowwecanestablishtheconcentrationsatequilibriumbyusingtheequationweworkedoutearlierthattoldustheproductofNa+andCl–concentrationsmust be the same on the A side and the B side at equilibrium, assuming they are both free toequilibrate.Weget:(x+s‐d)*(s‐d)=(s+d)*(s+d)

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sx‐dx=4sdThevaluesofxandsarefixedquantitiesrelatedtotheinitialconcentrations.Solvingforthedesiredquantitydgives:d=sx/(x+4s)Havingsolvedford,wecanwriteoutexpressionsforthefinalconcentrationsoftheNa+andCl–ions.FromtherewecanaskwhetherthereisanelectrostaticpotentialdifferencebyapplyingtheNernstequationtotheconcentrationofionsonthetwosides.Asdiscussedbeforeweshouldgetthesameanswer regardless ofwhetherwe examine theNa+ or Cl– ions since both are able to equilibrate.TakingtheNa+ions,weget:

=B‐A=RT/(Fzi)ln(CNa+,A/CNa+,B)wherezforNa+is+1Pluggingintheexpressionfordatequilibrium,theargumentinthelogfunctionis(x+s‐sx/(x+4s))/(s+sx/(x+4s))whichsimplifiesto(x+2s)/(2s).So,=B‐A==B‐A=RT/(Fzi)ln((x+2s)/(2s))Thisequationisveryspecificforthewaywesetuptheproblem,soitdoesn’trepresentageneralfinding,butitdoesletusevaluatetheelectrostaticpotentialunderagivensetofinitialconditions.Supposeforexamplethattheconcentrationofsaltadded(s)wasequaltothemolarconcentrationoftheprotein(x).Thentheargumenttothelogfunctionissimply(x+2s)/(2s)=3/2.ThevalueofRT/Fnear room temperature is 0.0256 V (or 25.6 millivolts), which is a useful simplification worthremembering.Finally,weget=0.0256V*ln(3/2)=0.010V=10mVNotethatthewaywedefinedmeansthatthepotentialishigherontheBsidecomparedtotheAside.Wheredidthisvoltagecomefrom(sincewedidn’tapplyanexternalvoltage)?Itcomesfromthechargeonthespecies(theproteininthiscase)thatisconfinedtooneside.Notethattheprotein,whichwe took tohaveanegative charge, is generatinganegativepotentialon the sidewhere itresides.Thesituationisactuallyabitmorecomplicated. Forexample,howcouldtherebeavoltageifweassumedelectroneutralityonthetwosides,sincevoltageisreallyachargeseparation.Thisisafairobjection,buttheenergyassociatedwithmacroscopicchargeseparationisveryhigh,sowhiletherewouldinfactbeasmallamountofchargeseparationcreatingnetchargeonthetwosides,thatminorchargeimbalancewouldnotaffecttheionconcentrationssignificantly.Evidently,veryslightlymoreNa+thanCl‐wouldcrossfromlefttorightandthiswouldgiveaslightchargeseparationwithnetnegativechargeresultingontheleft,consistentwiththenegativevoltageontheleft.

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Anotherpointof interest is to lookatwhatwouldhappeninasystemlikethis ifweweretoaddexcesssalt,thatiss>>x. Inthatcase,theargumentofthelogfunctionfromabove((x+2s)/(2s))approaches1,andtheloggoesto0,so≈0.ThisshowsthattheDonnanpotentialgoesawayifexcessionsarepresentthatcanequilibratefreelybetweenthetwosides.VariableionpermeabilitiesandcomplexphenomenaIn our previous discussions we treated simplified situations where different ions were eithercompletely free to equilibrate or totally unable to permeate themembrane. Thiswas helpful ingainingintuitionaboutwhatdrivesthecreationofelectrostaticpotentials,butrelevantbiologicalscenariosaremuchmorecomplicated.Themembranehasverydifferentdegreesofpermeabilitytodifferent ions. Furthermore, the distributions of ions across a cell membrane do not reflectequilibriumratiosbutareinsteadtheresultofasteadystateprocess(orevenadynamicprocesschangingovertime).Dependingontheirpermeabilities,theionsareflowingdowntheirchemical(orelectrochemical)gradientsatthesametimethattransmembraneproteinpumpsarecontinuingto transport them against those gradients. Ion permeabilities are therefore fundamental tounderstanding the potential across the biological membrane. For example, it is changes in thepermeabilityofthemembraneforcertainionsthatdriveschangesinthemembranepotentialduringnerveconduction.Howcanthecellmembranehavedifferent(andcontrollable)permeabilitiestodifferentions?Transmembraneproteinchannelsprovidetheanswer.Theycanbehighlyspecificforcertain ions. Andwhether theyareinopenorclosedconformationscanbecontrolledbyligandbindingor other phenomena, includingthingslikepressure.A simplified scheme at the rightillustrates the concentrationgradients of Na+ and K+ across atypical cell. These gradients arecreated at the expense of energyinput (e.g. ATP hydrolysis). Theinward pumping of K+ and theoutward pumping ofNa+ results inK+ being higher inside the cell andNa+beinghigheroutsidethecell.Now,ifthemembraneismorepermeabletoK+ionsthanNa+ions,by virtue of a potassium channel for example, thenwe can understand the resultingmembranepotential. [Themembranepotentialinatypicalcellisintherangeof‐40to‐80mV,meaningtheinsideof thecellhasanegativepotential.] One thingwe learnedwas thataDonnanpotential iscreatedbyionsthatcan’tcrossthemembrane(orthatcrossveryslowly). Inthecellularschemehere,theNa+ionsaretheonesleastabletocross(sincetheK+channeldoesn’tallowNa+ionstopass),

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andthehigherconcentrationofthisspeciesoutsidethecellisconsistentwiththeoutsidehavingthepositivepotential.Anotherwayoflookingatitisintermsofthenetchargeseparationthatwouldoccuracrossthemembrane.ThegradientsforNa+andK+areindifferentdirections.Na+ionsaretrying tomove back into the cell as fast as they canwhile K+ ions are trying to exit across themembraneasfastastheycan. ButowingtotheK+channel,K+ionsexitmoreeasilythanNa+ionenter, therebycreatingasmallchargeseparationwithmorepositivechargeontheoutside,againconsistentwiththecorrectsignofthevoltageacrossthecell.Notethattheconcentrationgradientsoftheionsinthisscenariocannotsimplybeusedtoevaluatethemembranepotentialbecausetheionsarenotreachingequilibriumbetweenthetwosides;theirconcentrationgradientsreflecttheactivityofmembranepumps.Finally,youcanseefromtheprecedingargumentshowchangingtheionpermeabilitieswouldaffectthemembranepotential.Thosepermeabilitiesarecontrolledbyopeningandclosing,or‘gating’ofmembrane channels. A simplified description of how nerve conduction depends on ionpermeabilitiesgoessomethinglikethis:aneuronwithanegativerestingpotentialreceivesasignalthatcausesNa+channelsinthemembranetoopen;accordingtoourearlierdiscussions,thisreducesthemembranepotential (i.e. raising it closer to0); thisdepolarizingvoltagechange is conducteddown the length of the axon like an electrical current down a wire; at the axon terminal, thisdepolarizationacrossthemembranecausesCa2+channelsintheaxonterminaltoopenup;therestingCa2+concentrationishigherinthesynapticspacebetweenneurons,soCa2+ionsflowintotheaxonterminal;theincreasingCa2+concentrationinsidetheaxonterminaltriggersthefusionofsynapticvesicleswiththeinnermembraneoftheaxonterminal,releasingtheenclosedneurotransmitterintothesynapticspace;theneurotransmitterdiffusesacrossthesynapticcleftandbindstoreceptorsontheadjacentcell(e.g.amusclecelloranothernervecell);dependingonthecellreceivingthesignal,bindingoftheneurotransmittertothereceptormayopenupasodiumchannelonthenextneuronto propagate the electric signal, or cause some other event, like muscle contraction or sensorysignaling.

MolecularElectrostatics

Wewillcontinuewithourdiscussionofelectrostatics,focusinghereontheforcestheyexertandtheeffectstheyhaveonmacromoleculesandtheirconformations.WearefamiliarwithCoulomb’slaw,whichtellsusthattheforcebetweentwochargesgoesastheproductofthetwochargesdividedbyrsquared.

Thisformoftheequationapplieswhenusingcgsunits.[TheSIformofCoulomb’slawincludesanextraterm

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(4 0)inthedenominator,whichisdroppedfromthecgsequationbyhavingitabsorbedintothedefinitionofthecgselectrostaticunitofcharge.]A related equation for potential energy (U) can be obtained by taking the force equation andintegratingoverrandnegating(sinceF=‐dU/dr)togive:

Inaddition,recallingthattheenergyforplacingachargeqatapotentialisU=q,wecanseethattheequationaboveimpliesthatasinglechargecreatesapotentialarounditgivenby

Φ

ThedielectricvalueTheequationsaboveforelectrostaticforcesandenergiesarelikelyfamiliar,butwhatissometimesoverlookedistheimportanceofthemediuminwhichtheinteractionstakesplace.Thisiscapturedbythedielectricvalue ,whichoccursinthedenominator.Roughlyspeaking,thedielectricdescribeshowpolarizablethemediumis.Foravacuum, =1,whichiswhythetermissometimesdroppedintheequationsabove,forexampleinintroductoryphysicsproblems. Butitisvitalforbiochemicalsituations.Thedielectricvalueforwaterisaround78!Thatmeanselectrostaticenergycalculationsthattakeplaceinaqueoussolutionsmaybeoffbynearlytwoordersofmagnitudeifthedielectricvalueisnothandledproperly.Theextremelyhighdielectricvalueforwaterrelatestoitslargedipolemoment.Watermoleculesinanelectricfieldtendtoorientthemselvesinawaythatgivesthelowestenergy,i.e.withtheoxygenatompointinginthedirectionoppositeoftheelectricfieldvector.Theeffectistodiminishorscreenthenetelectrostaticforce.Thedielectricvalueinlesspolarmaterialsismuchlowerthaninwater.Forhydrocarbons(whichserveasamodelfortheinteriorofalipidmembrane)thedielectricvalueisbetweenabout2and4.Aswewilldiscusslater,chargedaminoacidsareimportantinproteinstructure,andsothevalueofthedielectricforaproteinmoleculeisanimportant(andlongdebated)issue.Valuesbetween4and20occurintheliteratureforthedielectricintheinteriorofaprotein.Forachargethatresidesonthesurfaceofaprotein,exposedtowater,therelevantvalueisprobablyclosetothatforpurewater.SimplifiedelectrostaticsequationsTheequationsaboveareclumsytoapplyunlessyourememberwhatthevalueisforanelementarychargeincgselectrostaticunits.Insteaditisconvenienttoconverttheequationstoformsthatcanbeappliedmoreeasily,usingintegervalues,z,forthecharges(e.g.z=1forNa+).Simplifiedequationsthatapplynearroomtemperatureare:

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1389kJ/mol(wherermustbeinAngstroms)

and

Φ 14.4Volts(wherermustbeinAngstroms)

Examples:1)HowmuchenergydoesittaketobringtwoNa+ionsfromastartingdistanceofinfinitytoafinaldistanceof4Åifthedielectricvalueis78?Answer:4.5kJ/mol. Isthisenergysignificantornot?RecallRT≈2.5kJ/mol,sothemagnitudeoftheeffectwouldbeexp(‐4.5kJ/2.5kJ)=0.17.2)Howmuchenergymightanion‐paircontributetothestabilityofafoldedprotein?Supposethesituationinquestionisanaspartatesidechainthatis5Åawayfromalysine.Supposetheinteractiontakesplaceneartheproteinsurfacewherethehighdielectricofwatermakestheeffectivedielectricthere about 40. Answer: ‐6.9 kJ/mol, and the magnitude of the effect on K would beexp(+6.9kJ/2.5kJ)=16.Adifferentkindofelectrostaticenergy:theBorn‘self‐chargingenergy’Apowerfulbutunderappreciatedideaarisesbyconsideringahypotheticprocessofcreatingaunitchargeoutofinfinitesimalchargeelements,dq.[Thisisanexampleofakindof‘thoughtexperiment’referred to by physicists as a ‘gedanken experiment’; essentially an experiment that can only beperformedinone’smind.]Form our previous discussions we know that the (differential) energyrequired to bring a (differential) charge dq to a position where thepotentialisisdU=dq.Here,isthepotentialattheplacewherewearedepositingthecharge,whichisatthesurfaceoftheionbeingcreated(inourimagination).Iftheradiusoftheionisa,thenfromaboveweknowthepotentialthereisq/( a),whereqiswhateverchargehasalreadybeendeposited.Wecanobtainthehypotheticalenergyforcreatingthischarge,which is usually referred to as the Born self‐charging energy, byintegratingourdifferentialenergyoverq.

Bornself‐chargingenergy:

Again,thiscanbemademoreconvenient:

Bornself‐chargingenergy: 1389/2 kJ/mol(whereamustbeinAngstroms)

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Theimaginaryideaofcreatingachargefromnothingmayseemsillyatfirst,butitgivesusapowerfulresultrelatingtotheenergyfora(veryreal)processoftransferringanionbetweentwolocationswherethedielectricisdifferent.FreeenergyofiontransferAswediscussedearlier,thecelliscomplexandthedielectricisdifferentindifferentplaces.Thelowdielectricofthelipidbilayerisparticularlynoteworthy,especiallygiventhephysiologicalimportanceofionpassagethroughmembranes.Let’slookthenattheenergyassociatedwithtransferringanionfromaqueoussolutionintoalipidbilayer.[We’llassumeherethattheenergycanbeconsideredasa contribution to the free energy of the process.] We can think of the transfer process as acompositionofseparatesteps:reversingtheimaginaryioncreationprocessinthefirstmedium,thentransferring the infinitesimal charges into the secondmedium (at no energy cost since they areinfinitesimal), and thenrecreating thecharge in thesecondmedium. Evidently, the transfer freeenergyisjustthedifferencebetweentheenergiesrequiredtocreatethechargeinthetwodifferentmedia.Ifthedielectricvaluesforthetwomediaare 1and 2,thenthefreeenergyofiontransferfrommedium1tomedium2wouldbe:

Δ 1 1

1389/2 kJ/mol

Thingstonoteherearethatzissquared,sotheeffectisthesameforpositiveornegativeions.Secondisthattheenergyispositive(i.e.unfavorable)ifthetransferistoalower dielectric, as expected. Third, note thedependenceontheradiusoftheion;theenergytermislarger if the ion is small since the charge is morelocalized.Lastlyisthemagnitudeoftheeffect.Considerthefreeenergyoftransferringasodiumionfromwaterinto the middle of a lipid bilayer. Take 1 Å as anapproximationfortheionradius.Letthedielectricbe4forthebilayerand80forthewater.Underthoseapproximations,Gtransfer=12*(1/1Å)*(1/4–1/80)*1389/2kJ/mol=165kJ/mol.Isthisbigorsmall?170kJ/mol=66*RT!Thisisanenormousenergybarrier.Thisexercisedemonstratesthevirtualimpermeabilityofalipidbilayertonakedions.Whatisgoingonattheatomiclevelthatwouldexplainwhyanionpreferssostronglytoremaininwaterratherthaninthelipd?Inwater(oranothermaterialofhighdielectric),thesurroundingmoleculesrearrangethemselvestointeractwiththeioninwaysthatcorrespondstoanoverallfavorableenergy.Thisisnotpossibleinthelipidbilayer(orothermediumoflowdielectric).Thesameunderlyingideaissometimesdescribedintermsoftheenergyrequiredtodesolvateanionwhenitismovedintoanon‐polarenvironment.

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Processes thatgenerate,maintainandexploit iongradientsacrossmembranes formthebasis formostoftheenergyconversionsthatoccurinbiology.Thisisonlypossiblebecauseionscannoteasilycross the bilayer. The complex energy conversion processes in the cell are conducted bytransmembraneproteins(pumps,channels,andtransporter)thatresideinthebilayer.

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CHAPTER7

EnergeticsofProteinFoldingProteinsacquiretheiruniquefunctionsbyfoldingupintospecificthree‐dimensionalstructures.Thisgivesthemtheshapesandarrangementsofchemicalgroupsrequiredtocarryouttheiractivities.Thedetails of howproteinsmanage to reach their correct shapes from a startingpoint of beingextrudedinamoreorlessextendedconformationfromtheribosome,andwhatenergeticfeaturesstabilize their final configurations, arequestionsof considerable importance froma fundamentalperspective and for practical reasons as well. Numerous biotechnology and pharmaceuticalproblemsrevolvearoundstabilizingenzymesorothertypesofproteins.

AbalancebetweenlargeopposingforcesYou’ve learned before about simple molecules that have multiple conformations whose relativeenergiesdictatewhichismorepreferred;thechairvsboatconformationforcyclohexanederivativesis an example from organic chemistry. The problem of protein stability is considerably morecomplex.Onethingthatmakestheproteinstabilityproblemuniqueisthesheersizeofthemoleculesinvolved; thousands or tens of thousands of atoms are interacting with each other. Anotherimportantconsequenceofthesizeandmainlylinearcovalentstructureofproteinmoleculesisthevastnessofthepossibleconfigurationseachonecouldadoptinprinciple,byvariationofthephi‐psitorsionanglesalongitsbackbone,nottomentionthesidechainconformations.Thevastnessofthisconformationalspace(practicallyallofwhichrepresentsnon‐nativeconfigurationsoftheprotein)meansthatfoldingaproteinintoitsnativeconfigurationcomesatanenormouscostintermsoflostentropy. This costmust be offset by very large numbers of favorable interactions between thethousandsofproteinatomsinthenativelyfoldedconformation.The arguments above paint a unique picture for protein energetics. The total net energy thatstabilizesafoldedproteinoveritsunfoldedstateistypicallynotverylarge;itarisesasarelativelysmall difference between very large energetic terms working in opposing directions. You canimagine then that rathersmall changes to theaminoacidsequenceofaproteinmightoffset thisbalance,andindeedminorchangestothesequenceofaproteinoftenhavesurprisinglylarge(andfrequentlyunexpected)effectsonproteinstabilityandfunction.As a rough numerical estimate, the conformational entropy lost in going from a flexible proteinbackbonetoaparticularconformationisabout5kcal/molperaminoacidresidue,meaning1000kcal/mol for a 200 amino acid protein for example. [Confusingly, protein energetics are oftendiscussedinkcalinsteadofkJ;multiplybyabout4.2toconvertfromkcaltokJ.]Incontrast,thenetstability (G0) for the process of protein folding ismuch smaller, often in the range of ‐5 to ‐10kcal/mol (unfolded folded). This is fairly large compared to RT, so typical proteins havestabilitiesthatkeepthemnearlyexclusivelyintheircorrectlyfoldedconfigurations(atleastundertherightconditions,thoughthosearen’talwaysknownoreasytoreplicateinvitro),butasnoted

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above,thisnetstabilityissmallcomparedtothemagnitudesoftheopposingenergetictermsthatmustbalanceoutintheend.TermsthatcontributetotheenergeticsofproteinfoldingYou are familiar already with the various forces involved in atomic interactions. Here we willsummarizehowthoseforcesrelatetotheparticularproblemofproteinstability.ElectrostaticsWediscussedcharge‐chargeor‘salt‐bridge’interactionsearlier.Theycontributetothestabilizationofproteinswithmagnitudesthataresomewhatmodestsincetheytendtooccurnearthesurfaceofproteins where the high dielectric of water reduces their strength. But they can be important,particularlywiththeviewthatnetproteinstabilizationhastocomefromtheaccumulationofmanysmallerenergeticcontributions.Followingfromourearlierdiscussions,theunfavorableenergeticsofputtingachargeinaregionoflowerdielectricalsohasimportantconsequencesforproteinstructure.Severalofthenaturalaminoacidsarecharged(aspartate,glutamate,lysine,andarginine),soweexpecttofindthoseaminoacidsalmostexclusivelyontheexteriorofaproteinwhenitisinitscorrectlyfoldedconfiguration.Chargedaminoacidsareoccasionallyfoundburiedinthe interiorofaprotein,butthoseareusuallycaseswherethatparticularaminoacid isplayingacriticalrole, forexample in thecatalyticcycleofanenzyme;itissometimesnecessaryforaproteintopaythecostofanunfavorableenergeticfeatureinordertoachievearequiredfunction.ThecostofburyingachargeinsideaproteinalsomeansthatthenaturalpKavaluesofaminoacidscanbesignificantlydifferentcomparedtothetextbookvaluesthatgivethepKavalueofaminoacidsdissolvedinwater.Placementofatitratablegroup(i.e.agroupthatcanaddorloseprotons)inalowdielectricshiftstheequilibriumpositiontowardstheneutralform.DoesthatraiseorlowerthepKaofacarboxylategroup?Whatabouttheaminogroupoflysine?HowmuchdoyouexpectthepKavaluestochange?[Hint:YouknowhowpHandpKavaluesrelatetoconcentrationratios:1unitequalsafactorof10.Andyouknowhowenergydifferencesaffectequilibriumratios:dividetheenergybyRTandexponentiate.Andyouknowhowtoestimatetheenergeticcostofburyingachargeaccordingtoourpreviousequationsforiontransfer.]vanderWaalsorLondondispersionforces:favorableatomicpackingYou’ll recall fromearliercoursework thatvanderWaalsorLondondispersion forces,sometimescolloquiallycalled‘packing’forces,arerelativelyweak.Thatmaybetrueonanindividualbasis,butwiththousandsofatomstheeffectsareextremelyimportant.Itisnotablethatwhenoneexaminesthestructuresofproteinsinatomicdetail,theatomicpackingisseentobegenerallyverytight.Theatomicpackingdensity inmostproteininteriors issimilartothepackingseeninsolidcrystalsof

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organic molecules. That is a manifestation of the favorable energy associated with atom‐atomcontactsonalargescale.Howdoesthegoodpackingachievedinproteininteriorsrelatetothestabilityoftheproteininthenativestate?Theanswerhereisnotsostraightforward.Notethatiftheproteinwasinanunfoldedconfiguration it would likely be able to make good atomic contacts with the water moleculessurrounding it;watermolecules are small enough to be arranged inways thatwould give goodpacking.Themoreimportantconsiderationisthatifaproteinhadthewrongaminoacidsequence,forexampleifwemutatedasmallsidechaintoalargeoneorviceversa,thentheatomicpackinginthenativelyfoldedconfigurationmaybeseriouslydisrupted.Inthatsense,favorablepackinginaproteinmaynotbeamajordrivingforcetowardsfolding,butifthenativepackingiscompromisedthensurelythefoldedconfigurationwillbedestabilizedcomparedtotheunfoldedstate.Energiesduetopackingdefecthavebeenestimatedtobeabout0.5kcal/molpermethylene‐sizedcavity.Thatgivesaroughestimatefortheconsequenceofreplacingalargeraminoacidsidechainwithonethatdoesnotfillthespaceproperly.Ontheotherhanditishardtoestimatetheeffectofaddingalargeraminoacidsidechain.You’llrecallthatmodelingthevanderWaalspotentialenergyusingtheLennard‐Jonesequationgivesaverysharprise(goingasthe12thpoweroftheinteratomicseparation)inenergyforstericoverlap.Sothecostofaddingevenonemethylenetoaplacewheretheremightnotbespacecanbecatastrophicforthestabilityofthefoldedstate.HydrogenbondingHydrogenbondingisaveryspecifictypeofinteraction;ithassomefeaturesofbonding(i.e.orbitaloverlap)butitismainlyanelectrostaticfeature.Itarisesfrom(1)ahydrogenatomthatcarriesapartialpositivechargeowingtoitscovalentattachmenttoanelectronegativeatom(typicallyNorO),whichisreferredtoasthe‘donor’,and(2)alonepairofelectronsonanelectronegativeatom(typicallyNorO),whichisreferredtoasthe‘acceptor’.Thestrongesthydrogenbondsarewherethelonepair,thehydrogen,andtheheavyatomattachedtothehydrogenarearrangedatleastroughlyinastraightline.Theenergycontributedbyahydrogenbondcanbe estimated from model studies of smallmoleculesinsolutiontobeintherangeofabout5kcal/mol. Therearehowevermanynuancesregarding hydrogen bonding: hydrogen bondsinvolving a negatively charged acceptor like acarboxylate can be extra strong, the lowerdielectricoftheproteininteriormightmagnifythe effects of hydrogen bonding, multiplehydrogenbondsworkingtogethermightbenefitfromacooperativeeffect,andsoon.Asaresult,the role of hydrogenbonding inproteins is a constantlydiscussed anddebated issue. However,

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certainpointsareclear.Proteinsarefullofhydrogenbonddonorsandacceptors.Thisistrueofthepolypeptide backbone in particular; every peptide unit has a carbonyl acceptor and an amidenitrogen donor. The cumulative energetics of hydrogen bonding is therefore substantial. It isimportanttonotehoweverthatwaterhasexcellenthydrogenbondingproperties,soaproteininanunfoldedconfigurationcansatisfyallitshydrogenbondinggroupsthroughinteractionswithwater.Accordingly,itmaybethathydrogenbondingisnotamajordrivingforceforfolding.Ontheotherhand, followingthesamelogicasaboveregardingtheimportanceofgoodpackinginthenativelyfoldedstate,ifweweretoalteraproteininsuchawaythatwecreatedunsatisfiedhydrogenbonddonorsoracceptorsintheinterior,thenthiswouldsurelydestabilizetheprotein(sincethosedonorsand acceptors could satisfy their hydrogen bonding needs by exposure towater in the unfoldedstate).Forexample,ifaserinesidechainisburiedintheinteriorofanativelyfoldedproteinanditshydroxylgroupishydrogenbondedtoahistidine,andthenthehistidineisreplacedbymutationtosomething like valine that lacks the required hydrogen bonding capacity, the serinewould haveunsatisfiedhydrogenbondingneeds,andthiscouldbehighlydestabilizing.HydrophobiceffectAsyoumayhavelearnedbefore,thehydrophobiceffectisgenerallyacceptedtobethemajordrivingforceforproteinfolding,atleastfortypicalglobularproteins.Butthehydrophobiceffectisinfactnotaseparateforce.Itisinsteadacomplexphenomenonarisingfrommany‐bodyinteractions.Theneteffectisthatnonpolarmoleculesorfunctionalgroupsaredriventoassociatewithothernonpolarmoleculesbybeingexcludedfrominteractionswithwater. Thename“hydrophobic”conjurestheideathatanonpolarmoleculedoesn’tlikewaterbecauseitcan’tmakegoodinteractionsthere,butacloserlookshowssomethingabitdifferent.Threedifferentkindsofinteractionsarepossiblehere:nonpolar‐nonpolar,nonpolar‐water,water‐water.Thenonpolar‐nonpolarinteractionbenefitsfromfavorable van derWaals energies. A nonpolarmolecule also benefits from good van derWaalsinteractions if it is surroundedbywater. So from theperspectiveof thenonpolarmolecule, theenergeticdifferenceissmallwhetheritinteractswithanothernonpolarmoleculeorwithwater.Butthings are different from the perspective of thewater. Watermakes highly favorable hydrogenbondinginteractionswithitself. Someofthoseinteractionsmustbelostifawatermoleculeis incontactwithanonpolarmolecule.Soreallyitisthewatermoleculesthatdon’twanttointeractwiththenonpolarsolute.Theeffectisthesameinanycase,thetwokindsofmoleculesaredriventohavetheleastamountofinteractionwitheachotheraspossible.Fromthedescriptionyoucanseethatthemagnitudeoftheunfavorableenergyrelatestothesurfaceareaoftheinteraction.As a side note, it is surprising to learn that when the unfavorable free energy associated withtransferring a nonpolar solute from an organic phase to water is examined in more detailexperimentally,one finds that theunfavorable freeenergy isnot theresultofapositiveenthalpychange, but is instead the result of a negative (unfavorable) change in entropy. This has beenexplainedbynotingthatattheinterfacebetweenanonpolarsoluteandwater,thewatermoleculesaredrivenintohighlyorderedarrangements(sometimesreferredtoasclathrates)presumablyinordertorecoverasmanyoftheirlosthydrogenbondsaspossible.

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The magnitude of the hydrophobic effect has been estimated to be about 22 cal/mol per Å2 ofinteractionarea.Atypicalaminoacidsidechainhasanareainthe100–200Å2range,andofcoursemanyofthenaturalaminoacidsarenonpolar.Clearlythecumulativemagnitudeisverylarge.Andperhapsmostcriticalisthatthesehydrophobicinteractionsareentirelydifferentintheunfoldedvsthenativelyfoldedstate(incontrasttosomeoftheotherenergetictermswediscussedearlier).Intheunfoldedstate,enormousamountsofnonpolarsurfacewouldbeexposedtosolvent.Asaresult,foracorrectlyfoldedproteinmolecule,thenonpolarsidechainsaremainlyburiedintheinteriorinthe‘hydrophobiccore’.ThespecialcaseofmembraneproteinsMorethanaquarteroftheproteinmoleculescodedforbyacellarenotsolubleinthecytosol,butinstead spend their lives embedded in a lipid bilayer (either the cellmembrane or amembranesurrounding one of the various organelles in a cell). The energetic considerations for thesetransmembrane(orTM)proteinsareuniqueinsomeprofoundways.EnforcementofregularsecondarystructureinthemembraneregionOneofthemostprofoundeffectsofthelipidblilayerenvironmentisonthesecondarystructureofproteinsthatspanthemembrane.Wediscussedtheimportanceofproteinsbeingabletosatisfyallornearlyalloftheirhydrogenbondinggroupsinordertomaintainstability.Andwenotedthatthepolypeptidebackboneis fullofhydrogenbonddonorsandacceptorsthatneedtobesatisfied. Inaqueousproteins,theneedtosatisfybackbonehydrogenbondscanbeachievedwithrelativeease.Thebackbonecaneitheradoptregularsecondarystructureelements(whichbytheirnaturesatisfybackbonehydrogenbonding),invariousorientationsandwithturnsorlongerunstructuredloopsconnecting them in almost unlimited fashion. The figure below (left) is just one example of thetertiarystructureofanaqueousprotein.Aqueousproteinscanhavepracticallylimitlessstructuresbecauseregionsof thebackbonethatarenot inregularsecondarystructureelementscansatisfytheirhydrogenbondingneedsusingwaterinstead.Thatisnotpossiblefortransmembraneproteins.Thelipidbilayerisalmostdevoidofwater.Therefore,wheretheproteinisembeddedinthebilayer,itmustpracticallyalwaysadoptstrictlyregularsecondarystructureso that thebackbonewillbesatisfied.Therearetwobasicclassesoftransmembraneproteins:thosethatcontainofabundleofalphahelices (or sometimes just oneTMhelix), and those that consist of a beta barrel,which isessentiallyabetasheetthatisrolledupsothattherearenounsatisfiededges.Thosetwoclassesareillustratedbelow(middleandright).Thealphahelixclassismoreabundant,butthebetabarrelclassiscommonwherelargeporesinamembraneareneeded,i.e.intheoutermembraneofmanybacteria.Thereareafewknowncaseswhereaproteinentersonlypartwayintothemembraneorformssomeother structure that seems to involve unsatisfied hydrogen bonds, but they constitute rareexceptions.

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TheproblemofthemissinghydrophobiceffectAnothermajor difference between TMproteins and aqueous proteins concerns the hydrophobiceffectandhowTMproteinscanbestabilizedintheirnativeforms.YoumighthavealreadysurmisedthattheoutersurfaceofaTMprotein(atleasttheregionthatisembeddedinthebilayer)needstobenonpolar. Otherwiseitwouldnotpartitionintothemembrane.Butthatisamajordistinctioncompared to aqueous proteins. Aqueous proteins have polar/charged surfaces and nonpolarinteriors,and that iswhatdrives their folding in thepresenceofwater. But ifTMproteinshavenonpolarinteriorsandnonpolarexteriorsaswell,andarenotsurroundedbywaterinanycase,thenit seems the hydrophobic effect cannot play amajor role. The real situation is somewhatmorecomplicated,buttheanswertothispuzzleremainslargelyunanswered.

MeasuringtheStabilityofProteinsMuchofourpreviousdiscussionhasconcernedthestabilityofproteinmolecules,meaninghowmuchlowertheenergyofthenativeconfigurationiscomparedtotheunfoldedconfiguration(s).InessencewewanttoknowK,andhenceG0(fromG0=‐RTlnK),fortheprocess:

U NwhereUdenotesunfoldedandNdenotesnativelyfoldedThefirstthingweneedistobeabletotellwhatfractionoftheproteininasampleisfoldedandwhatfractionisunfolded(i.e.XNandXU=1‐XN).Thisrequiresexperimentalmeasurementofsomepropertythat issensitivetowhetheramolecule is foldedornot. Wewill talk lateraboutvariouskindsofexperimentsthatsatisfythisrequirement–thenaturalfluorescenceoftryptophantendstodependonwhetheritisinapolarornonpolarenvironment,soyoucanseehowthatmightsuffice–butfornowwewillkeepitabstractandjustsaythatthereissomepropertyPthatwecanmeasureforasample,andthatthevalueofPshouldchangedependingonwhatfractionoftheproteinisfolded.

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AtthispointyoumightfeellikeyouhaveenoughtofigureoutK.Ifyourmeasurementtellsyouwhatfractionisfolded,thenyouknowtheequilibriumconstantforfoldingisK=XN/XU. But,wehaveaproblemrelated tosensitivity. Even though thenativestateofa typicalprotein isnotextremelystable,itisusuallystableenoughthatK=exp(‐G0/RT)isalargenumber,meaningthefractionoftheproteinthatisunfoldedisverysmall,infacttoosmalltomeasureaccurately.Anotherwayofseeingthis is that a practicalmeasurement is not goingto be able to tell thedifference betweenwhether 1 in 1,000moleculesareunfoldedvs1in 1,000,000. Thedifference in the signalbetween those two caseswould be too small tomeasure even though thevalue of K would bedifferent by a factor of1,000.Whatisthesolutiontothisproblem?Inordertogetatthe value of ‐G0, thestandard approach is toartificiallyshiftthesystemtowardstheunfoldedstatebyaddingachemicaldenaturantlikeurea.Ifwegotoconditionswhereboththenativeandunfoldedstatesarereasonablypopulated,thenatthatpointwecanfigureoutwhatfractionoftheproteinisfolded.ThiscanbedonebymeasuringthevalueofourpropertyPunderthoseconditionsandthencomparingittothevaluesofthepropertyunderconditionswheretheproteinisfullyfoldedandwhereitistotallyunfolded.Thealgebraforthisisshowninthefigure.Intuitivelyyoucanseehowtheproceduremakessense.Ifyoumeasurethepropertyundersomeamountofdenaturantandyouseethatthevalueofthepropertyisexactlyhalfwaybetweenthevalueyougetforfullyfoldedandfullyunfolded,thenthesamplemustbehalffoldedandhalfunfolded,i.e.K=1.Thecalculationforanarbitrarydegreeoffoldingisonlyslightlymorecomplex.WeseenowthatunderconditionswherebothformsoftheproteinarepopulatedwecancalculateKandthenG0.ButthatexperimentwouldjusttelluswhatG0wasunder conditions where we had added denaturant todestabilizetheprotein.Whatgoodisthat?Theansweristhat ifwerepeattheexperimentatseveraldifferentdenaturant concentrations, we should be able tocalculateG0asafunctionofdenaturantconcentration.

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Then,ifwebelieveasimplistictheorythatarguesthatG0dependsondenaturantconcentrationinaroughly linear fashion, thenweshouldbeabletoestimateG0intheabsenceofdenaturant,byextrapolation.Thisfinalstepoftheanalysisisillustratedabove.

IdeasRelatedtoHowProteinsReachtheirFoldedConfigurationsOurdiscussionsuptothispointhavebeenonlyabouttheinitial(unfolded)andfinal(nativelyfolded)protein.Howorwhyaproteinfindsitscorrectlyfoldedconfigurationisanotherquestion,andonethathasoccupiedproteinscientistsforthelasthalf‐century.In1961ChristianAnfinsenperformedseminalexperimentsshowingthattheenzymeribonucleaseA(RNaseA)couldbeunfoldedandthenrefoldedafterremovalofthedenaturant.Thisshowedthatthenativethree‐dimensionalstructureoftheproteinisencodedinthelinearaminoacidsequence.Thisseemsabitobviousdecadesafterthefact,butthedemonstrationthattheproteincouldfinditscorrectstructureoutsidethecell,withoutotherinfluences,wasanimportantconceptualadvance.Anfinsen asserted that this meant that the amino acid sequence encoded the correct three‐dimensional structure by having the correct three‐dimensional structure be the lowest possibleenergy.Thatideaisknownasthe“ThermodynamicHypothesis”.Some60yearsafterAnfinsen,weunderstandthatthesituationisrathermorecomplex.In1969CyrusLevinthalformalizedanargumentthatthenumberofpossibleconfigurationsaproteincouldconceivablyadoptisvastlygreaterthancouldeverbesampledbyaproteinmoleculewigglingaroundinsolutioninareasonabletime.Yetmostproteinsfoldonthetimescaleofsecondsorfaster.Hiscalculationwassomethinglikethis.Consideraproteinwith200aminoacids.Assumethatonlythreedifferentphi‐psibackboneconfigurationsneedtobesampledateachaminoacidposition–based on the idea of choosing between helix or beta or random loop – which is clearly anunderestimate.Andsupposethataproteincansampleanewconfigurationataspeedthatislimitedbymolecularvibrations(fromquantummechanics,kBT/h≈1013/sec).Thetimerequiredtosample3200conformationwouldbe3200/1013>>ageoftheuniverse.HowcantheThermodynamicHypothesismakesenseifthereisn’tanywayaproteinmoleculecouldsearchthespaceofallpossibleconfigurationsinordertoendupatthelowestenergyconfiguration.Thisisknownasthe‘Levinthalparadox’.TheLevinthalparadoxhasmotivateddecadesofresearchonproteinfolding.Workinthe1980’s,especiallybyRobert(Buzz)BaldwinandPeterKim,focusedontheideaofspecific‘pathways’thatmightguideaproteinfromitsunfoldedstatetoitsnativestate.Forexample:U I1 I2 N

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where I1and I2,etc,arewell‐defined intermediates thatwouldbepopulatedonthewaytothenativestate.Workalong this line involved a search for cases where welldefinedintermediatescouldbedetected.Howcanonedifferentiatebetweenaprocessthatoccursas a two‐state transition (without any populatedintermediates) vs. a process with populatedintermediates?Therearemultipledistinctions.Onehastodowithkineticbehavior. Wewilldiscusssuchtopicslater,butforthemomentwewilljustsaythatasinglestep(or two‐state) transition gives a simple exponentialapproach to the final equilibrium position, whereas aprocess with multiple transitions can give morecomplicated kinetics, including a lag phase, asdiagrammed here. In a few cases, experiments haveidentifiedspecificproteinfoldingintermediates,buttheyhavenotemergedasageneralfeatureofproteinfolding.Other ideas have developed to advance ourunderstanding.‘Energylandscape’theorieswere developed (by PeterWolynes and Ken Dill andothers),withthemainideathat themulti‐dimensionalenergy landscape surfacefor proteins must besmooth and funneled vs.rugged. Figures like theone drawn here illustratethebasic ideaof agoodvsbad energy landscape forrapidfolding.Iftheenergylandscapeisrugged,thenthefoldingprocessislikelytogettrappedinalocalminimum.Theideaofa smoothly funneled landscapealso lifts the requirement forpathway intermediates. Instead,alldownhillroutesleadtothenativestate.Underthisidea,evolutionwouldhaveselectedaminoacidsequencesandstructureswhoseenergylandscapeswerefavorable.OtherideasrelatedtoproteinfoldingCurrent ideas like energy landscape theory offer a good framework for understanding proteinfolding.Butimportantquestionsremain.Forexample,itturnsoutthatmanyproteinsdonotfold

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spontaneously to their native states either in vitro or in vivo. Many proteins in the cell rely onsophisticated protein machinery known as molecular chaperones, which consume ATP to helpproteinsreachtheircorrectconfigurations.Apparentlythoseproteinseitherdonothavesmoothlyfunneledenergylandscapes,orperhapstheirnativestatesarenotthestateoflowestenergy.Anotherwrinkleisthatathighconcentrationandgivensufficienttime(orpartiallydestabilizingconditions),manyproteinsadoptanalternatebeta‐richconformationandthenaggregateinto‘amyloid’fibrils.Amyloidformationissuspectedasthebasisforagrowingnumberofdiseases,fromAlzheimer’stoParkinson’stoLouGherig’s.Doesthismeanthatthelowestenergyconfigurationforsomeproteinsisnotthenativelyfoldedstateseeninthecell,buttheamyloidfibrilstateinstead?Finally,therearesomerareproteinsthathaveextremelypeculiarfoldedstructuresinwhichtheproteinbackboneistiedinaknot!Howdothoseproteinsreachtheirnativestates?Theenergylandscapesinthosecasewouldseemtoberathercomplexandrequiretraversalofnarrowvalleystoreachthenativestate.Thepointsaboveareof fundamental interest inbiology,buttheyalsohavepotentially importantpracticalimplications.Muchworkhasbeendoneinthelastfewdecades(andsomenotableprogresshasbeenmade)ontheproblemofpredictingthethreedimensionalstructuresofproteinsfromtheiraminoacidsequencesalone.Whatdoesitmeanforthoseeffortsifproteinsmighthavelowerenergyconfigurationsthantheirnativestates?Asyoucansee,theareaofproteinfoldingremainsrichwithopenquestions.

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CHAPTER8

DescribingtheShapePropertiesofMoleculesSomeofourpreviousdiscussionshaveintroducedtheideathatshapeisanimportantconsiderationforthebehaviorandfunctionofmacromolecules.Laterinthecoursewewilltalkabouttechniquesfordeterminingthethree‐dimensionalstructuresofmacromoleculesinatomicleveldetail.Butnowwewilldiscussmoresimplifieddescriptionsofshapethataresometimesobtainedfrombiophysicalmeasurementsinthelaboratory.

RadiusofgyrationOftenwehaveanobjectormoleculeswhoseshapeisreasonablycompactbutitisnotreallyasphere.Howmightweassignasinglesizescalethatwoulddescribesuchanobject,inthesamewaythataradiusdescribesthesizeofasphere?The‘radiusofgyration’(RG)providesthis.Itisessentiallyanaverageradius,butmoreaccuratelyitisan‘rms’orroot‐mean‐squareradius.Thegeneralmeaningofroot‐mean‐squareis, takingthemonikers inreverseorder:squarethequantities, thenaveragethem,thentakethesquareroot.You’reundoubtedlyfamiliarwiththisinthecontextofrmsdeviationfromthemean(oftestscoresforinstance).Fortheradiusofgyration,twogeneralcasesarise:(1)acollectionofdiscretepointsoratoms(likeyouwouldhaveoncethedetailedstructureofproteinisknown, for example), and (2) a continuous shape defined by a boundary (like an ellipsoid forexample).Wewillhandlethediscretecasefirst.DiscreteobjectsAssumingthatthepointsthatmakeuptheobjectshouldallbegivenequalweight,theformulaforradiusofgyrationis:

∑

Here and in the later equations for thecontinuouscase,itisvitaltonotethattheradiusofeachpoint,ri, isitsdistancetothecenterofmass of the object. An entirely different andincorrectresultwillbeobtainedifthecenterisnotdefinedcorrectly.Asimpleexampleforanobject composed of 5 points arranged like onthe face of a die is shown. This examplehappens to be two‐dimensional, but thesituationisequivalentinthree‐dimensions.Asreminders,thedistancebetweentwopointsin

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three‐dimensional space is r=sqrt((x)2+(y)2+(z)2). And the center ofmass of a collection ofpointsisobtainedsimplybyaveragingtheirx,y,andzcoordinatesseparately.Youcanseethatthisisasimpleprocedure,socalculatingtheradiusofgyrationgiventheatomiccoordinatesofanymolecules,largeorsmall,isstraightforward.ObjectswithcontinuousshapesForobjectsthathaveacontinuousshape,thesummationintheequationforradiusofgyrationmustbereplacedbyanintegral,andthedivisionbythenumberofpointsmustbereplacedbydivisionbythevolume.Beforedoingthat,we’lljustpointouttheoneshapewheretheradiusofgyrationrequiresnocalculation.Sphericalshell(nottobeconfusedwithasolidsphere):Everypointonasphericalshellhasthesameradius,r.Sotheradiusofgyration,RG=r.Solidsphere:Thegeneralequationforradiusofgyrationforacontinuousobjectis:

Forasolidsphere,thesimplestwaytointegrateoverthewholevolumeisinaseriesofinfinitesimallythinshellsofradiusrandthicknessdr.ThedifferentialvolumeofthatinfinitesimallythinshellisdV=4r2dr.Theintegralabovebecomes:

,4 4

43

35

3 5⁄ Notethat,asexpected,theradiusofgyrationofasolidsphereislessthantheouterradius,sincethepointsbelongingtothesphereareallatadistancelessthanorequaltor.

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Ellipsoid:Insteadofresortingtosomehorribleintegrals,wewillsolvethisbygeometricreasoning.Anellipsoidisreallyjusta stretchedout sphere. So let’sbeginwithaunitsphere (r=1), decompose its behavior into x, y, and zcomponents, and then see what happens when westretchitout.Forasolidunitsphere(r=1),fromaboveweknowthattheaveragevalueofr2is(3/5).Butforasphere,thex,y,andzbehaviorsmustbethesame,sothatmeanstheaveragevalueofx2,y2,andz2,mustallbethesame,andbecauser2=x2+y2+z2,wecanconcludethattheaveragevaluesofx2,y2,andz2areall1/5forasolidunitsphere.Nowifwestretchtheunitspherealongthexaxisalone,sothatitsaxialradiusalongxisnowainsteadof1,thentheaveragevalueofx2mustbe(1/5)a2.Therewouldbenochangealongyorz.Thenstretchingbybalongyandcalongz,weconcludethattheaveragevalueofy2is(1/5)b2andz2is(1/5)c2.Puttingitbacktogether,theaveragevalueofr2wouldbe(1/5)(a2+b2+c2).[Notethatthisgivestheexpectedexpressionforasphereifa=b=c=r.]Non‐sphericalobjectshavehigherradiiofgyrationNowlet’scomparetheradiusofgyrationforasphereandanellipsoidthathavethesamevolume.Supposethespherehasradius10.Itsradiusofgyrationwouldbesqrt(3/5)*10=7.74.Nowsupposetheellipsoidhasprincipleaxesof5,10,and20(thevolumeofanellipsoidis(4/3)abc,whichwouldbethesamevolumeasthesphere.Itsradiusofgyrationwouldbesqrt((1/5)(52+102+202)),whichis10.2.Theimportantthingtonotehereistheradiusofgyrationisgreaterfortheellipsoidcomparedtothesphere.Thisisjustonespecificcase,butitisacompletelygeneralconclusionthataspherehasthelowestpossibleradiusofgyrationcomparedtoanyotherpossibleshapeofthesamevolume.Thesignificanceisthatifanexperimentalstudygivesusaradiusofgyrationofamoleculewhosemass(andthereforevolume)weknow,andthatradiusofgyrationislargerthanwewouldhaveexpectedforamoleculeoftheknownvolumeifitwasasphere,thenwehaveestablishedsomethingaboutthemolecule’sshape:i.e.thatitisnonspherical,orelongated.

ThebehaviorofflexiblepolymerchainsorfilamentousassembliesofproteinsubunitsAbovewediscussedawayoflookingatrelativelycompactobjects.Nowlet’slookatthebehaviorofobjectsthataresomuchlongerthantheyarewidethattheyareflexibleandwecananalyzethembythinkingaboutthepaththeirbackbonetakesintermsofarandomprocess.Theoriesforanalyzingmoleculesinthiswayweredevelopedbypolymerchemistsdatingbacktothe1950’s(seePaulFlory),butbiochemicalsystemsarerichwithexamplesthatfitthedescriptionaswell: longmoleculesofDNAorRNA,unfoldedproteinmolecules,and(onalongerscale)noncovalentpolymersformedbytheend‐to‐endassemblyofproteinsubunits,asinF‐actin.

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PersistencelengthIfwehavetopickasingleparameterthatwouldbeusefulfordescribingthebehaviorofaflexiblechain,itwouldbeameasureofitsflexibility,ortobemoreprecise,ameasureofthelengthscaleoverwhichitappearstobeflexible(anycurveseemsstraightifyouexamineasmallenoughlength).Onespecificmeasureofstiffnessiscalledthepersistencelength.Roughlyspeaking,itisameasureofhowfaracurvetendstoproceedinthedirectionitstartedbeforerandomcurvaturerendersitsprogressintheoriginaldirectionnegligible.Ofcoursetoextractsuchavaluefromacurverequiresrepeatingtheevaluationofhowfaritextendsfrommanydifferentstartingpointsonthecurve.Theplotbelowconveystheessenceofthepersistencelength. Clearly,astifferpolymerhasagreaterpersistencelength.

Approximate persistence lengths for some biological molecules are given below. These areapproximations, and the persistence lengths of nucleic acids in particular are rather stronglydependentonconditions like salt concentrations. The stiffnessof abiologicalpolymeroftenhasimportant implications for how it behaves. Note for instance how the exceptional stiffness ofmicrotubulesmeans that they are nearly perfectly rigid over the length scale of a cell, which isobviouslyimportantfortheirfunctioninmechanicaldivisionofthecellandtransportofmolecularcargoacrosslongdistances.Polymer PersistencelengthDNA(doublestranded) 500ÅRNA(doublestranded) 800ÅF‐actinfilament 5um

(<eukaryoticcell)microtubule 5mm

(>>cell)

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(Jointed)RandomwalkmodelsThediagramsabovewerebasedonsmoothworm‐likecurves.Adifferentkindofmodel,slightlylessrealisticbutmathematicallymoregeneralizable,isoftenusedtotreatproblemsofthistype.Intherandom walk model, a chain travelsstraightinsomedirectionforadistanceb(thestatisticalKuhnlength).Thenittakesaturninarandomdirection,andsoon.The mathematical treatment is fairlystraightforward:N=#ofstepsb=steplength(Kuhnstatisticallength)C = length of the curve (i.e. if it was

stretchedout)L=straightend‐to‐enddistanceLofcoursewouldchangewitheveryrandomwalksowearereallyjustinterestedintheaverageorexpectedbehaviorofL.WecangettheaveragebehaviorofLbytreatingitlikeavector,whichisthesumofNsmallervectors,oneforeachstep.Calltheindividualstepvectorsli.Eachonehaslengthbandisinarandomdirection.

⋯ Whatistheexpectedvalueof|L|2?Wecangetthesquaredlengthofavectorbytakingadotproductofthevectorwithitself.Lettinganglebracketsdenotetheaverageorexpectedvalue,

⟨ ⟩ ⟨ ∙ ⟩ ⟨ ⋯ ∙ ⋯ ⟩Nowtheexpressionontherightisaproductofsums,whichcanbeexpandedtoasumofproducts,

N2termsinall.Forexample,⟨ ∙ ∙ ⋯ ∙ ∙ ∙ ⋯ ∙ ⋯ ⟩Wecanmovethebracketstotheindividualtermstogive:

⟨ ⟩ ⟨ ∙ ⟩ ⟨ ∙ ⟩ ⋯ ⟨ ∙ ⟩ ⟨ ∙ ⟩ ⟨ ∙ ⟩ ⋯ ⟨ ∙ ⟩ ⋯Butthissimplifies.Thekeyistorealizethatifyoutakethedotproductoftwovectorswheretheanglebetweenthemisrandom,theexpectedvalueis0.ThatmeansthatamongtheN2terms,they

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allbecomezeroexceptthoserepresentingadotproductbetweenalittlevectorlianditself.There

arejustNofthose.Andeachterm⟨ ∙ ⟩isjustthesquaredlengthofthelittlevector,whichisb2.Therefore,

⟨ ⟩ andLrms=N1/2bThisisarathergeneralresultthatappliesnotonlytopolymerbehaviorbuttootherkindsofphysicalproblemslikediffusionthatcanbemodeledasarandomwalk.Theaverage(orrms)distanceyouexpectaftertakingNstepsoflengthbisproportionaltob,butisnotproportionaltoN,buttothesquarerootofN.Ourreasonfordevelopingtherandomwalkmodelwastouseittocharacterizethebehaviorofaflexible chain. In this model you can see that the value of the step length (b) is going to be adescriptionofhowstiffthepolymeris.Ifyoutakearandomwalkwithtinysteps,thepathwillhavethepropertiesofacurvethatishighlyflexible,i.e.itwillnotextendveryfarfromwhereitstarted.Howcanthevalueofbbeextractedfromthebehaviorofarandomwalkpath.Frombefore,thecontourlengthCofthepathisC=Nb.SubstitutingintoourequationforL2,weseethat(droppingthevectornotation)<L2>=Nb2=Cborb=<L2>/CDependingonthestudy,wemayknowthelengthofthepolymerchain.ForexampleifweknowthemolecularweightofalargeDNAmolecule,andweknowthemolecularweightofonebasepair,thenweknowhowmanybasepairsthereare,andweknowthespacingbetweenbasepairsinDNAisabout 3.4Å, so we can do the math and estimate C. Then, if we have a way of experimentallymeasuringtheaveragestraightend‐to‐enddistanceLforthemolecule,thenwecangetbdirectly.Wemightgetanestimateofend‐to‐enddistancefromsomekindofspectroscopicexperimentwherethe ends of the molecule were labeled, or maybe we can visualize the molecule on an electronmicroscopygridanddoaseriesofend‐to‐endmeasurementsthatway.Wecanconnecttherandomwalkpolymermodeltoourearliertopicofradiusofgyration.WeusedRGearlierasawaytocharacterizecompactshapes,butitcanbeusedtodescribeflexiblestructuresaswell.Thealgebraisabitmessysowewillnotworkitouthere,butitturnsoutthattheradiusofgyrationforaflexiblechainiscloselyrelatedtoitsexpectedend‐to‐endlength.Specifically,<RG2>=<L2>/6

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Thatmeansifwecandoanexperimentthatgivesusavaluefortheradiusofgyrationforapolymerchain,thenwecanestimatebbysubstitutingintothepreviousequation.Thisisusefulbecausethereareinfactbiophysicalexperimentsthatgivevaluesfortheradiusofgyration.Droppingthebrackets,undertheassumptionthatanactualexperimenttomeasuretheradiusofgyrationinsolutionwouldgiveusatimeaverage,b=6RG2/CFinally,weneedtoreconcilethetwomodelswedevelopedhere,thesmoothworm‐likechainandthejointedrandomwalk.ItcanbeshownmathematicallythattherelationshipbetweenthetwomodelsisthatthestatisticalKuhnlengthbistwicethepersistencelengtha. Thatis,b=2a. Thescientificliteraturegenerallyreportspersistencelengths,soifanexperimentis interpretedintermsofthejointedrandomwalkmodeltogiveb,thenthepersistencelengtha=b/2.

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CHAPTER9

ABriefIntroductiontoStatisticalMechanicsforMacromoleculesComplexsystemsofbiologicalmoleculesareoftencharacterizedbymanydifferentconfigurations,which may all be at equilibrium. Handling such systems and predicting their behavior can besimplifiedwithanappropriatemathematical framework. Theterm‘statisticalmechanics’ isoftenusedtodescribesuchtreatments.ProbabilitiesandexpectedvaluesWebeginwithsomeexamplesinvolvingfamiliarphenomena.Considerafairdie(singularfordice).Inasingleroll,therearesixpossibleoutcomes,allwithequalprobabilities:P(1)=P(2)=…=P(6)=1/6Whatistheaverageor‘expectednumber’ofdotsthatwillshowupinarollofthedie?Theaverageof1through6is3.5,sowecancorrectlydeducethattheexpectedvalueis3.5.Atablemakesthecasemoreexplicit.

i P(i) P(i)*(#ofdots)1 1/6 1/6*1=1/62 1/6 1/6*2=2/63 1/6 1/6*3=3/64 1/6 1/6*4=4/65 1/6 1/6*5=5/66 1/6 1/6*6=1 P(i)=1 (P(i)*(#dots))=3.5

Formulatingtheproblemthiswayillustratesapowerfulandgeneralpoint:

⟨property⟩ ∗ property,

where<property>denotestheaveragevalueofsomepropertyofinterestforthesystem,P(i)denotestheprobabilityofconfigurationoroutcomei,andpropertyidenotesthevalueof theproperty foroutcomei.Fortheproblemabove,thepropertyofinterestisthenumberofdotsshowing.Thecaseabovewastrivial,butpracticallyanypropertythatonecanconstructcanbeevaluatethisway.Asanon‐obviousexample,supposeyouneedtoknowwhattheexpectedvalueisforthesquareofthenumberofdotsthatshowsuponthedie.Youmightwanttoknowthisifsomeoneofferedtorolladieandpayyou$1forrollinga1,$4forrollinga2,$9forrollinga3,andsoon,andaskedyou

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howmuchyouwouldbewillingtopaytoplaythisgameofchance.Theansweriseasytoobtainfromtheequationabove.<dots2>=(1/6)*12+(1/6)*22+(1/6)*32+…+(1/6)*62=5.1So,payinganythinglessthan$5.10isafavorablebetforyou.StatisticalweightsforoutcomeswithunequalprobabilitiesInordertohandleproblemsofrealinterest,forexamplewheredifferentmoleculararrangementshavedifferentenergiesandthereforedifferentprobabilities,itisconvenienttointroduceaschemeforhandlingunequalprobabilities. Themainequationaboveissuitableforunequalprobabilities,butsometimestheprobabilitiesof thedifferentoutcomesarenotgivendirectly. Instead,weareoften given relative probabilities between different outcomes – think about the meaning of anequilibrium constant for example. Relative probabilities are sometimes referred to as statisticalweights,anddenotedwi.Foranapplication,let’sconsiderthecaseofastrangediewhoseoutcomesarenotequallylikely,butinsteadtheprobabilityofrollinganygivennumberofdotsistwiceashighastheprobabilityofrollingone fewerdots. In otherwords, consider the casewhereP(i+1)=2*P(i). We canworkout thebehaviorofthissystemasbefore,butstartingwithrelativeprobabilitiesorstatisticalweights,andthenconvertingthemtoindividualprobabilitiesaccordingtotheequation,P(i)=wi/wi.

i wi P(i) P(i)*(#ofdots)1 1 1/63 1/63*1=1/632 2 2/63 2/63*2=4/633 4 P(i)=wi/wi 4/63 4/63*3=12/634 8 8/63 8/63*4=32/635 16 16/63 16/63*5=80/636 32 32/63 32/63*6=192/63 wi=63 P(i)=1 (P(i)*(#dots))=5.1

Above we worked out the problem by converting weights explicitly to probabilities, but theformulation can also be written directly in terms of the weights without explicitly writing outprobabilities. BysubstitutingP(i)=wi/wiintotheequationabovefortheaveragevalueofsomeproperty,weget

⟨property⟩ ∑ ∗ property∑

Appliedtothedieproblemabove,wherethepropertyofinterestisagainthenumberofdotsshowing,thisequationgives<#ofdots>=(1*1+2*2+4*3+8*4+16*5+32*6)/(1+2+4+8+16+32)=5.1,inagreementwiththevalueintheTableabove.Notethatinsettingupthestatisticalweights,oneisfreetochoosethefirstconfiguration(oranyother)asareferencewhosestatisticalweightissetto1.

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Itwouldbeslightlylessconvenientnumerically,butonecouldsettheweighttobe1foroneoftheotherconfigurationsandobtainthesameanswerintheend,aslongasthecorrectrelativeweightsgetassigned.Asafinalexamplewiththedie,toshowhowsimpleitistoevaluateanypropertyaslongasitcanbeevaluatedforeachoutcome,considerthestrangediefrombefore,whereP(i+1)=2*P(i),andevaluatetheexpectedvalueforthesquarednumberofdotsthatwouldshowupinoneroll.Withoutneedingtoconstructatable,<dots2>=(1*12+2*2*2+4*32+8*42+16*52+32*62)/(1+2+4+8+16+32)=27.4Soyouwouldn’twanttopaymorethan$27.40toplayagamewherethisstrangedieisrolledandyougetpayedthesquareofthenumberofdotsshowing.HandlingdegeneraciesForacompletetreatmentweneedjustonemoreelement.Thereareoftensystemswhereasinglekindofstatecanbeobtainedinseveraldifferentways.Wesawthistypeofsituationearlierinthecourse when we were evaluating the number of distinct ways that a particular state could beconstructed(e.g.byexchangingtheidentitiesofmoleculesofliketype).Thesamesituationariseshere.Weassignthevariablegitothedegeneracyofarrangementi.Withthatadjustment,thetwomainequationsabovecanbere‐written,replacingwieverywhereitappearedwithwigi,togivethesetwogeneralequations:

∑

and

⟨property⟩ ∑ ∗ property∑

Aswewill see later, formolecular systems theweights relate to equilibrium constants betweendifferentmolecularconfigurations,whichareinturnrelatedtotheenergiesoftheconfigurations(i.e.byexponentiatingthenegatedenergiesafterdividingbykTorRT,asusual).Thedenominatorintheequation above therefore takes the form of a sum over Boltzmann‐like terms for the differentconfigurations. That summation has a special role in statistical mechanics applications and issometimesreferredtoasthepartitionfunction,andsometimesreplacedwiththenotationQorZ,depending on the text or context. The student is referred to texts on statisticalmechanics for atreatmentofhowthedependenceofthepartitionfunction(ontemperatureforexample)canmakeitpossibletoevaluatethermodynamicstatevariablesforasystem.Wewill turn instead to see how the equations above canbe applied to evaluate the behavior ofvariousphysicalpropertiesofcomplexbiologicalmolecules.

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A Statistical Mechanics Treatment of the Helix‐Coil Transition for aPolypeptideApolypeptidethathasatendencytofoldupintoasinglealphahelixservesasaclassicexampleofasystemwithaseriesofpossibleconformationsrangingfromafullyunfolded‘randomcoil’toafullyformedalphahelix.Earlytreatmentsofthissystemscameinthelate1950’sbyBrunoZimmandJ.K.BraggandaresometimesreferredtoasZimm‐Braggmodels.Hereweconsiderasimplifiedversionsometimesreferredtoasazippermodel.Alongasinglepathwayofconformationaltransitionsfromcoil to helix, a first turn of helix formswhen around 4 amino acid residues come into the rightconformationtoformabackbonehydrogenbond(itoi+4)characteristicofanalphahelix. Fromtherethehelixcanpropagatebyextension,inastep‐by‐stepadditionofmoreaminoacidresiduestothehelix,eventuallyreachingthefullyhelicalform.Acartoondiagramisbelow.Thepropagationparameters,describes an equilibriumconstant for adding anotherresidue to the helicalconformation, leading toonemore hydrogen bond. But akey element of the model isthatthefirststepisdifferent.In order to form the firsthydrogen bond, severalaminoacidresiduesmustalladoptaspecificconformation.Thatcomesatanentropiccost,whichcontributesanextra(opposing)termtotheequilibriumforthefirststep.Thateffectisdescribedbythenucleationparameter,.Thevaluesofsanddependontheaminoacidtypethatcomprisesthepolypeptide.Butforatypicalcaseofinterest,sisslightlygreaterthan1andismuchsmallerthan1.Toworkoutastatisticalmechanicstreatmentitisconvenienttore‐drawthesystemsymbolically,usinga‘C’todenoteanaminoacidintherandomcoilconformationand‘H’todenoteanaminoacidin thehelix conformation, as shownbelow. From thisdiagramwecanseehowwemight assignstatisticalweightsanddegeneraciestotheconfigurations.Forthestatisticalweights,wecanbeginbyassigning1tothe firstconformationasareference. Thenthestatisticalweights fortheotherformscanbeassignedbytakingintoaccounttheequilibriumconstantsforeachstepinacumulativefashion.Recallthatforamulti‐stepequilibrium,thetotalequilibriumconstantbetweentwospeciesistheproductoftheequilibriumconstantsfortheseparatestepsbetweenthem.

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The degeneracies in thisproblem relate to howmany locations can bechosen for the helicalregion.[Inoursimplifiedmodel only one helicalsegmentisallowedinthepolypetide.] For theunfoldedform,theentirepolypeptide is in the ‘C’conformation, so therearenochoicestobemadeand therefore noassociated degeneracy(g=1). When weintroduceasegmentof4amino acids to nucleatethefirsthelicalturn,thenwehavea choice for thelocationof that segment.ThetotalnumberofdistinctplaceswherethatsegmentcanbechosenisN‐3,whereNisthetotalnumberofaminoacids,sothedegeneracygforthatconformationisN‐3.[ThiscanbeseenbynotingthatifNwas4,therewouldbeonlyonechoice(consistentwithN‐3)forthelocationofthesegment,if N was 5 there would be 2 choices (again consistent with N‐3), and so on.] Another way ofunderstandingthemeaningofthedegeneraciesistorealizethatthespecificdrawingprovidedisjustonerepresentationofthemultiple(N‐3)differentconfigurationsthatcouldhavebeendrawnhavinga4‐residuesegmentinthehelicalconformation.Aswemovefurthertotherightalongthemulti‐stepequilibrium,thedegeneracydropsbyoneineachstep,astherearefewerandfewerdifferentchoicesforselectingalongerhelicalsegment,untilattheendthereisnochoiceandthedegeneracyisbackto1.Oncewehavetheweightsanddegeneracies,wecancalculate thebehaviorofasystem. Herewemightwanttoevaluatetheoveralldegreeofhelicalfoldinginthepolypeptide.Somemoleculesinthesystemwillbelesshelicalandsomewillbemorehelical,butwecanevaluatetheaverage,andwecan also seewhich forms aremore or less populated. The figure below showshow this kind ofcalculationcanbedoneeasilywithaspreadsheet(likeinExcel),wheretheweightsanddegeneraciescanbe filled in, the individualprobabilitiesof thestatescanbeevaluated,andthedistributionofstatescanbeplotted.ThefigurebelowillustratesacasewereN=40,=0.01,ands=1.1.

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Looking at the probability distribution of the conformations, we see that compared to the fullyunfoldedstate,thestatewithjustonehelicalturnispoorlypopulated,butaswemovefurthertotherightitisincreasinglyprobabletofindstateswithmorehelicalcharacter.Inotherwords,itishardtobegintheprocessoffolding,buteasiertocontinueitonceithasbegun.Thereisalsoapeculiarbehaviortowardstheveryright,whereweseeadropinthelikelihoodoffindingfullyfoldedstates.Mathematically, this comes from the lower degeneracies for the fully folded states. In terms ofstructure, theconsequence is thattheendsof thepolypeptidetendstatisticallytobeunfoldedorfloppy.Settingasidethepeculiarbehavioratthefarrightofthediagram,therestofthepictureexhibitsthegeneralpropertyofbeinghardtobeginandeasiertocontinue,whichisahallmarkofcooperativeprocesses. Anothercommonpropertyofcooperativesystems isa tendency toshowasuddenor

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steep response to changes in certainparameters of a system, likeconcentration or temperature forexamples.Wecanlookatthebehaviorof the zipper model above as afunction of the value of thepropagationparameters. Inrealitysmightdependontemperature,sothedependence on s could also be anillustration of dependence on T forexample. From our equations aboveyoucanseethatitisarelativelysimplematter to evaluate the averagenumber of hydrogen bonds in oursystemifwearegivenN,ands.Wecanconvertthistoafractionaldegreeof helicity by dividing the averagenumber of hydrogen bonds by themaximum number possible (N‐3).Repeatingthecalculationoffractionalhelicity as a function of s (keepingN=40and=0.01)givesthebehaviorshown,wherethedependenceonsshowsarelativelysharptransition.Our calculations on this model illustrate the power of statistical mechanics approaches tocharacterizethebehaviorofcomplexsystems.Thespecificbehaviorofthissystemalsoillustratesgeneralthemesthatunderliemacromoleculesandbiochemicalsystems,inparticulartheappearanceofcooperativephenomena,sharptransitions,andahighsensitivitytophysicalparameters.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.8 0.9 1 1.1 1.2 1.3 1.4

fractional helicity

s

helix‐coil zipper model ‐ helicity as a 

function of s   (N=40, =0.01)

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CHAPTER10

CooperativePhenomenaandProtein‐LigandBindingRelationshipbetweencooperativebehaviorandprocesses involving formationofmultipleinteractionssimultaneouslyConsiderthebehaviorofareactionatequilibriumwhereinnmoleculesofAcometogethertoformanassemblyB(andwherenointermediateswithfewerthannunitsofAareallowed).

nA BWhatdoes the concentrationofB looklike as a function of increasingconcentration of A? This is easy toevaluatefromK=CB,eq/(CA,eq)n,andCB,eq= K * (CA,eq)n. Evidently, theconcentration of B depends on Aaccordingtoasimplepolynomialwhoseexponent is the stoichiometry of theassociation,n.Thisleadstoasurprisinginterpretationwhenonelooksathowapolynomial term behaves withincreasing exponent, n. This is shownbelow, where for simplicity theequilibriumconstantKistakenas1forallcases.The result is remarkable. For large n(emphasized in red in the figure), youseeaneffectivemaximumvalueof [A],afterwhichfurtheradditionofmoleculeAwouldleadsuddenlytoformationofB(inordertoavoidaconcentrationofAexceedingitseffectiveupperlimit).BiseffectivelyabsentatlowerconcentrationsofA.Notsurprisingly,thisresemblesthesortofthingyouexpectforacurveofprecipitationasafunctionofconcentration;atthesolubilitylimitofthesubstance,furtheradditionofthatcomponenttosolutionleadsonlytoasolid‐stateformofthesolutemolecule.Anotherinterestingcomparisonistomicelleformationbyanamphiphilicdetergent.There,manycopiesofthedetergentmoleculecometogetherintheformofamicelle,wherethenon‐polarlipid

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tailsof thedetergentmoleculesproject intoanentirely hydrophobic core, with the polar orchargedheadgroupsofthelipidsexposedonthesurface of a roughly spherical supramolecularassembly.Since only complete micelles fully shield thehydrophobic tails, partial micelles are hardlypopulated.Thiscorrespondstoaveryhighlevelofcooperativityfortheassemblyprocess.Muchlikethecaseofaggregationorprecipitationatthesolubilitylimit,thereisalimittotheconcentrationofthedetergentmonomer,afterwhichfurtheraddeddetergentleadsonlytothemicelleform.Inanalogy to the polynomial plots above, plotting the concentration values where the detergentmonomerwouldbeatequilibriumwithmicelleswouldgiveacurvewithasharpbreak,asshown.Thebehaviorcanalsobeplottedadifferentway,withthex‐axisrepresentingthetotaldetergentaddedtothesystem(whichisaconvenientindependentlycontrollablevariable),andseparatecurvesshownfortheconcentrationsofthemonomerandthemicelle.Thatschemeisalsoshown.

Bothplotsindicatethekeyconcentrationabovewhichthemonomercannotbetaken.Thatiscalledthecriticalmicelleconcentration(orCMC)andisaparticularpropertyofadetergent;itdependsontaillength,numberoftails,sizeandchargerepulsionofthehead,etc.Theoverarchingideaisthathigh‐order transitions tend to give rise to sharp transitions, reminiscent in somewaysof typicalphasetransitions.

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Protein‐ligandbindingequilibriaThebindingof ligands(substrates, cofactors, inhibitors) toproteins (ornucleicacids)sometimesshowscooperativebehaviorandsharper‐than‐usualtransitions.Thisisusuallyseeninoligomericproteins or enzymes – the coordinated action of multiple subunits bound together makes thecooperativitypossible.Thecaseofhemoglobin,withfourhemegroupsandfourproteinsubunits,iswell‐known.Beforewetacklethecaseofcooperativebindingbymultiplebindingsitesinanoligomericprotein,wewillanalyzethesimple(non‐cooperative)bindingbehaviorofasingleproteinsubunit(P)anditsligand(A).

P+A PANote that the K here is an equilibrium association constant, not a dissociation constant, as issometimewrittenforsubstratedissociationinenzymekineticstreatments.Weintroduceabindingparameter,v,todescribetheextentofbinding,i.e.theaveragenumberofligandsboundtoanygivenproteinmolecule.v=(#orconcentrationofboundligands)/(#orconcentrationofproteinmolecules)(0v1).Atequilibrium,K=[PA]/([P][A]).Notethatthe[A]inthisequilibriumequationistheconcentrationoffreeAmolecules,notthetotalAconcentration,whichwouldincludeligandsboundtotheprotein.And[P]istheconcentrationofunboundprotein.Takingtheratioof[PA]to[P]givesanexpressionthatisusefulinlatersubstitutions:[PA]/[P]=K[A](whichisaunitlessfraction).Fromthedefinitionofv,wecanseethatv=[PA]/([P]+[PA])Dividingthroughby[PA]givesv=([PA]/[PA])/([P]/[PA]+[PA]/[PA])Thensubstitutingfromabove,v=1/(1/(K[A])+1)=K[A]/(1+K[A])

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Thisgivesthefamiliarhyperboliccurveforbindingwhentheextentofbindingisplottedversusfreeligandconcentration.Bindingishalf‐saturatedwhen[A]=1/K.As you know, simple enzymekinetics share this behavior.This is tobeexpectedsince,atleast in the simplermathematical treatments, thecatalyticeventisprecededbyabinding event at equilibrium,and the rate of the reaction isproportional to theconcentration of the enzyme‐substrate complex, usuallydenoted[ES]. Inthatcase, thereactionrateoverthemaximalrate(atveryhighsubstrate),v/Vmax,isequalto[ES]/([E]+[ES]).Byanalogytotheequationsaboveforligandbinding(matching[E]with[P]and[ES]with[PA]),andusingadissociationconstantKdfortheenzymecasethatwouldbethereciprocalofthebindingassociationconstantK,onegetsv/Vmax=([S]/Kd)/(1+([S]/Kd))=[S]/(Kd+[S]),or[S]/(Km+[S])wheretheMichaelis‐MentenconstantKmwouldbeequaltoKd.Thereactionrateishalfitsmaximalvaluewhen[S]=Km.Thisbehavior(whichshouldberelativelyfamiliar)matchespreciselywhatwe’vedonehereforligandbinding–onlythevariablenameshavechanged.Bindingtoanoligomericprotein–independentbindingevents,nocooperativityWhataboutbindingtoanoligomericproteinwithmultiple binding sites (e.g. one per subunit)?Suppose the multiple sites are identical andindependent. If for thecaseof theoligomerweexpress the binding parameter as the averagenumberofligandsboundpertrimer,thenv=(#ligandsatsite1+#ligandsatsite2+#ligandsatsite 3)/(# of protein trimers). The number ofligandsatsite1pertrimerwouldbethesameasaboveforbindingtoamonomer(K[A]/(1+K[A]).Andthesameforbindingatsite2andsite3.So,v=3K[A]/(1+K[A]).Andvwouldbebetween0and3.Thebehaviorhasthesamecharacterasbefore–e.g.hyperbolicsaturation–allthatisdifferentisthemultiplicativefactorof3,whicharisessimplybecausewe’reexpressingthebindingpertrimerinstead of per monomer. This is the expected behavior for binding to multiple identical andindependentsites.Itgeneralizesreadilytonsites:

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v=nK[A]/(1+K[A])(0vn)ExtractingK(andn)fromequilibriumbindingmeasurementsYou’llrememberfromearliercourseworkinenzymekineticsthat,forextractingparametersfromgraphs, it canbe convenient todo algebraic rearrangements so thehyperbolic functionbecomeslinearinsomevariables.Analysisofbindingdataissimplifiedbyusingarearrangementtogiveaso‐calledScatchardplot. [TheScatchardplotcloselyresemblesarearrangementusedsometimes inenzymekinetics,theEadie‐Hofsteeplot–thetwokindsofplotsarerelatedbyexchangingxandyaxes].Inourcasewebeginwithv=nK[A]/(1+K[A])Thenmultiplyingthroughbythedenominatorontherightgivesv+vK[A]=nK[A]thenv=nK[A]–vK[A]anddividingby[A]givesv/[A]=nK–vKSo,plottingv/[A]vsvshouldgiveastraightlinewithslope–K,andx‐interceptequalton(y‐interceptofnK),asshownontheplotontheleft.

Sometimesitisdifficulttoobtainavalueforvaswehaveexpressedithereforbindingtoanoligomer,sincetheoligomericstateoftheproteinmaybeunknownattheoutset,preventingonefromknowingwhattheconcentrationoftheproteinisintermsof#ofoligomerspervolume.Itissometimeseasiertoestablishthefractionalbinding,f=v/n,fromexperimentalmeasurements.Forexample,ifbindingofaligandcausessomemeasurablechangeinthesystem–e.g.maybetheproteinhasatryptophanwhosefluorescencechangeswhenaligandbinds–thenonecanaddacertainamountofligandandcomparethechangeinthemeasuredpropertytothemaximumpossiblechange(e.g.byaddingexcessligandtothepointofsaturation).Theratioofthechangeobservedtothemaximumpossiblechangewouldbeameasureofthefractionalbinding,f.Thealgebrawouldbethesameasabove.Dividing

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thefinalequationabovebynonbothsideswouldgivef/[A]=K–fK.Asshownintheplotontheright,theanalysisisthesame,butnoattemptcanbemadetoestablishthenumberofbindingsites,n.Non‐linearScatchardplots–non‐identicalornon‐independentbindingsitesWhatmight causeanon‐linear Scatchardplot? If theprotein sample is impure, itmight containslightly different forms of the protein of interests – a mixture of phosphorylated vs non‐phosphorylatedformsisjustoneexample–whosebindingaffinitiesforaligandmightbedifferent.Or, perhaps the protein of interest really hasmultipledistinctbindingsitesthathaveevolvedtohavedifferentaffinities. Acartoonofacasewithtwobindingsitesofone type and a single binding site of another type isshown.Foracasewheretherearedifferenttypesofbindingsites,ifthebindingeventsareindependent(notcooperative),thenthebindingbehaviorissimplyadditive,withtermsmatchingthosefrombefore:

1types,

If the binding affinities (Ki) for thedifferent kinds of sites are not equal,then the Scatchard plot cannot bestraight.Reasoningthattheleftsideofthe curve in a Scatchard plotcorresponds to initial binding at lowligand concentration to the highestaffinitysites,andnotingthatthesloperelates to the binding constant K, wecanseethatthecurveshouldbesteeperontheleft,andthereforebentasshown.Iftheaffinitiesoftwodifferentkindsofsites are different enough it may bepossibletoextractseparatebindingconstants fromdifferentpartsof thecurve. Thismaynotbepossibleforbindingconstantsthatarenotsodifferentfromeachother,andpracticallyimpossibleiftherearemorethantwotypesofbindingsites.Inthosecase,ifaccuratedataarerecordedoverawide range of ligand concentration it may be possible to analyze the detailed behavior usingsophisticatedcomputerfittingsoftware.Our discussions above have all assumed that the binding events are independent – i.e. nocooperativity arising from communication between sites. Wewill deal inmore detail laterwith

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cooperativebinding,wherebindingofafirstligandpromotesbindingofsubsequentligandstoothersitesinthesameoligomer,butfornowwecansimplyanticipatewhateffectthatwouldhaveonaScatchardplot.Thebehaviorwouldeffectivelybethereverseofthecaseabovewherewehadsitesthatwereindependentbutnaturallydifferentinaffinity.Inthatcasewenaturallytendedtofillthehighaffinitysitesbeforethelowaffinitysites.Butwithcooperativebinding,thefirstbindingeventisharderandthelaterbindingeventsareeasier,whichisthereverse.So,ourScatchardcurvewouldcurvedownwardforcooperativebinding,asshownontheleft.Whereas a straightScatchard plotcorrespondedtoanordinaryhyperbolicbindingcurvewhenplotting v vs [A], adownward curvingScatchard plot ofthe type shownabove, resultingfrom cooperativebinding,wouldcorrespondtoasigmoidalshapeifthebindingdatawereplottedasvvs[A],asshownontheright.Wewilldiscusscooperativebehaviormorerigorouslylater.ExperimentsformeasuringbindingClassicmethod–EquilibriumdialysisThe classicmethod for studying binding equilibrium isequilibriumdialysis. Theprotein isplaced inadialysisbagthatallowstheligandtocrossbutnotlargemoleculesliketheprotein.Ligandisthenadded,whichequilibratesbetweentheinsideandtheoutside.Outsidethebag,theligand exists only in its free form. Inside the bag, theligandexistsintwoforms:freeandboundtotheprotein.Atequilibrium,theconcentrationoffreeligandinsidethebagmust equal the concentrationof ligandoutside thebag.Thatmeansifyoumeasuretheconcentrationoftotalligandinsidethebag,andthensubtracttheconcentrationofligandoutsidethebag,youhaveameasurementoftheconcentrationoftheligandinitsboundform,thatis[PA].From[A]total,inside=[A]free,inside+[PA]insideand[A]outside=[A]free,inside

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[PA]=[A]total,inside–[A]outsideThen,thebindingparametervcanbeobtainedbydividing[PA]bythetotalconcentrationofproteinthatwasplacedinsidethebag(ortheconcentrationofproteinoligomersiftheoligomericstateoftheproteinisalreadyknown).Amodernmethod–IsothermaltitrationCalorimetry(ITC)This experiment is based on theexpectationthattherewillbesomeheat(H)associatedwiththebindingevent.Theproteinisheldinasamplechamberthat is kept at constant temperature.Ligand is added slowly in a series ofsmall increments. After eachincremental addition, the instrumentmeasurestheheattransferrequiredtokeep the protein sample at a constanttemperature.

Theamountofheattransferredduringequilibration at each step is plotted;notethatHisoften<0forbinding.Atypical readout looks something likethis:

As more and more ligand is added, the system begins to saturate. The individual peak areascorrespondtotheamountofheatreleasedandthereforetotheamountofadditionalligandthatwas

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bound in that incremental step. Therefore, the total amountof ligandboundcanbeobtainedbyaccumulatingtheintegratedpeakheights.Thisleadstoamoretraditionallookingbindingplot.Notethatinsomeexperiments,likeITC,itiseasytoknowthetotalamountofligandpresentinthesystemandhardertoknowthefreeamount–contrastthatwiththeequilibriumdialysisexperimentwherethefreeAconcentrationwasevidentfromtheconcentrationofAoutsidethebag.NotbeingabletoplotthefreeAconcentrationmakesitabithardertoanalyzebindingcurveswiththeusualtricks(likeidentifyingthepointofhalfsaturationandestimatingtheKdor1/Kfromthefreeligandconcentrationatthatpoint). Computersoftwareisusuallyusedtointerpretthebindingconstant(andwhethermultiplebindingsitesmightbepresent)fromITCdata.VariousspectrophotometricmethodsAsnotedearlier,ifthereissomekindofspectroscopicexperimentthatgivesadifferentreadingforthe ligand‐bound protein PA compared to the unbound protein P, then it is often possible todeterminewhatfractionofthetotalproteinexistsinthetwoforms;doingthisasafunctionofligandconcentrationthenenablesdeterminationofbindingconstants. Thealgebraisreminiscentofthewaywelookedatmeasuringtheextentofproteinfoldingvsunfoldingearlier.If we let the variable P denote the value of some spectroscopic property – maybe the naturaltryptophan fluorescence of a protein if it is affected by ligand‐binding – then assuming thatspectroscopiccontributionsareadditive,Pmeas=f*PPA+(1‐f)*PPwherePmeas isthevalueofsomespectroscopicpropertymeasuredafteradditionofsomespecificamountofligand,PPAisthevalueyouexpecttoobtainfortheproteininitsboundform,andPPisthevalueyouexpectfortheunboundprotein.Asbefore,fisthefractionalbinding.Now,realizingthatPPandPPAcanbeobtainedbydoingthespectroscopicexperimentwithnoligandaddedandwithsaturatingligandadded,wecanchangethenotationabovetogive:Pmeas=f*PsaturatingA+(1‐f)*PnoA

whichrearrangestogive:f=(Pmeas‐PnoA)/(PsaturatingA‐PnoA)Thismakessensesinceitisreallyjustaratiobetweenapartialchangeandthemaximumpossiblechange.

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Phenomenologicaltreatmentofcooperativebinding‐theHillequationWeturnnowtothecaseofcooperativebindingtomultiplesites,suchasinanoligomericproteincomposedofseveralidenticalsubunits.ThelimitingcaseofperfectcooperativityToestablishthelimitingcaseforcooperativebehavior,weexamineanidealizedsituationof“all–or‐none”, which essentially means perfect cooperativity. Either 0 or n ligands can be bound to aparticularoligomer.

P+nA PAnNotethatthisformulationimpliesthatthereisnoformationofpartiallyboundforms,PA1,andsoon.Asbefore,anticipatingtheusefulnessofhavingaratiooftheboundformtotheunboundform,wecanwriteK=[PAn]/([P][A]),andthen[PAn]/[P]=K[A]nFromthemeaningofthebindingparametervasthenumberofligandsboundperoligomer,weknowthat

Withthesamerearrangementsasweusedbeforeforbindingtoamonomer–namely,dividingthetopandbottomby[PAn]andthensubstitutingthetermK[A]nfor[PAn]/[P],weget

1

Howdoesthisbindingcurvebehaveasafunctionofligandconcentration?Clearlyitbeginsatv=0for[A]=0.Anditsaturatesasexpected,gettingevercloserto1as[A]getsverylarge.Inthosewaysitissimilartothebindingequationwedevelopedforamonomer,whichwasv=1/(1+K[A]).Butthekey distinction is in the exponent applied to the ligand concentration, [A]. From our earlierdiscussionsyoushouldappreciatetheconsequencesofthatexponent;itcreatesasharpertransitionintermsof[A].Asaresult,thebindingcurvewillnotbesimplyhyperboliclikebefore,buttherewillbearegionwherethecurveexhibitssteeperbehavior.Inotherwords,wegetasigmoidalcurveofthetypeyou’veseenbefore(probablyinthecontextofcooperativeoxygenbindingtohemoglobin).Bindingcurvescalculatedfromtheequationaboveforperfectcooperativityareshownhere(takingK=1forconvenience).

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Realisticcase–partialcooperativityReal molecular binding processes arenever perfectly cooperative; theirbehaviorsfallinbetweentheperfectcaseand the simple case of independentbinding.Bycomparingtheequationsweobtained for the two cases, you’ll noticethattheonlydifferenceisintheexponentthat gets assigned to the ligandconcentration.Inthecaseofindependentbinding events (or binding to amonomer),theexponentwas1,whereasinthecaseofperfectcooperativityandnbindingsites,theexponentwasn.From this comparison, A. V. Hillgeneralized the binding equation tointermediatecaseswherethecooperativitywouldnotbeperfect.Thebindingequationbecomes

1 (1xnistheallowablerangeforpositivecooperativity)

wherexisusedastheexponentandiscalledtheHillcoefficient. Thereisfranklynomathematicjustification for this equation. Butwhat it does allow for is away to compareobservedbindingbehaviortoanequationwheretheexponentthatgivesthebestfittothedataissomeindicationofthedegreeofcooperativity.Thatis,iftheobservedbindingdataforacasewherethereare4bindingsites(n=4)isbestmatchedbytheHillequationwhenxischosentobe2.8,thenthisgivesyouasenseofthedegreeofcooperativity.ThestandardtreatmentforanalyzingobservedbindingdataaccordingtotheHillequationgoesasasfollows.Fromtheequationabove,multiplyingthroughbytheright‐sidedenominator,andthensubtractingthesecondtermontheleftfrombothsidesgivesv/(n‐v)=K[A]xItcanbemoreconvenientatthispointtoswitchtofractionalbinding,f=v/n,togivev/(n‐v)=(v/n)/(n/n–v/n)=f/(1‐f)=K[A]x

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Thentakinglogsonbothsidesgivesln(f/(1‐f))=lnK+xln[A]Evidently,aplotofln(f/(1‐f))vsln[A]shouldhaveaslopeofx.Whenoneplotsrealbindingdatathisway,theresultisinvariablyacurveratherthanaline with constant slope. That illustrates aweaknessoftheHillequation–itisnotfoundedon any underlying physical model of binding.Nonetheless,theslopeofaHillplotatitssteepestpointremainsausefuldescriptionofthedegreeofcooperativity.Physicalmodelsofcooperativebinding‐MWCMonod,Wyman,andChangeaux(MWC)(amongothers)developedexplicitmodelstoexplainhowcooperativebehaviorcouldemergeinbiologicalsystems.Theirideawassimpleandelegant,andtheunderlyingprincipleshaveturnedouttobestronglysupportedbyawealthofdetailedstructuralinvestigationsindiversesystemsoverseveraldecades.ThekeyelementsoftheMWCmodelareasfollows:

anoligomer,witheachsubunithavingabindingsite (atleast)twodifferentconformationsarepossiblefortheproteinsubunitanditsbindingsite,

andthosealternateconformationshaveverydifferentaffinitiesfortheligand.[Inthelimitingcase, only one of the conformations can bind the ligand]. The high affinity binding siteconformation is designated R and the weak (or forbidden) binding site conformation isdesignatedT.

symmetrymustbepreserved,sothatallthesubunitsinanyoneoligomerareallinthesameconformation.

Nootherassumptionsarerequired.Asyoumightrecallfromearlierstudies,andfromsomeofourprevious discussions, cooperativity invokes the idea of non‐independent events, and is oftendiscussed in terms of ‘communication’ between different binding sites, i.e. binding at one sitepromotesbindingattheothersites.Butyou’llnotethatthoseideasarenotexplicitaspectsoftheMWCmodel.Yet,thetenetsofthemodelleadtothatapparentbehavior.Wecansketchoutthebehaviorofatetramericsystemwith4subunits.WewilldesignatetheRformofthesubunit(whichhashighaffinityfortheligand)asacircle,andtheTformasasquare.Wewillassumeallthepossibleformsoftheoligomer–intermsofconformationandligandbinding–areallatequilibrium.Butnotethatmostofthepossibleformsoftheoligomeraredisallowedbytheelement

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oftheMWCmodelthatrequiresallthesubunitsinoneoligomertohavethesameconformation(RorT).Thatis,wedon’tneedtoconsidertheR3T1conformation,andsoon.Onlythesymmetricformsneedtobewrittenout.AndifwetakethelimitingcasewheretheTformcannotbindligandatall,thenweareleftwithjustafewconfigurationstoconsider.TheRformsoftheoligomercanbeboundtoanumberofligandsfrom0to4,andtheequilibriumbetweenthoseformsisaffectedbytheligandbindingconstantK(andtheligandconcentration).AndanequilibriumconstantLmustbewrittentodescribetherelationshipbetweentheTformofthetetramerandthe(unbound)Rformofthetetramer.

Howcanthisscheme(whichdoesn’texplicitly invoke ‘communication’betweendifferentbindingsites)giverisetocooperativebehavior?Wecangetanunderstandingofthisfromtwoperspectives.First,bymassaction,additionofaligandtoonesiteinanoligomerdrivestheothersubunitsintothehighaffinityconfiguration;thatfollowsfromtherequirementstipulatedinthemodelthatsymmetryhas to be preserved in an oligomer. It is the requirement of the subunits to adopt the sameconformationthatgivestheeffectofcommunicationbetweensites.

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The other view of how the MWC model createscooperative behavior is statistical, relating to theprobability (or concentration) distribution of thedistinct forms of the protein. We can analyze thesituationaboveintermsoftheRformsoftheproteinbindingligandsatthefoursitestotallyindependently,plustheextraTform.ForindependentbindingtotheR forms, the concentration ratio for incrementallybound formswouldgoupbythesameratio ineachstep. From there you can see that at someconcentration of ligand there will be substantiallymore R4 than R3, and more R3 than R2, and so on,meaningthat,consideringtheRformsbythemselves,R4willbethedominantform,withlittleR0,R1,R2,andR3.ButwhatabouttheTform.IftheequilibriumconstantLbetweentheT0formandtheR0formishigh enough, then T0 will be well‐populated even if R0 is low. Now think about plotting thedistributionoftheformsoftheoligomerthathave0,1,2,3,or4ligandsbound;takingintoaccountthattheconcentrationofoligomerswith0ligandsboundincludesbothR0andT0.Asyoucansee,thedistribution of the different forms is concentrated at the extremes, which is a hallmark of acooperativesystem.ExactlywhatkindofbehaviordomodelslikeMWCpredict? Ourstatisticalmechanicstoolsletusanswerthatinstraightforwardfashion.Westartwiththecaseofn=2.Butfirstacommentaboutstatisticalweights.Whenweworkoutthestatisticalweightsforaproblemthatinvolvesbinding,weneedtocomeupwitha(unitless)ratiothatrelatestheformsthatarisebysequentialadditionofanewligand.FromK=[PA]/([P][A]),wecanseethattheratioof[PA]/[P]isthefamiliartermK[A].Itisthisterm,andnottheequilibriumconstantbyitself,thatweneedtomultiplycumulativelyineachsequentialstepofourreaction.Ourstatisticalmechanicstermsare:

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Notethatthedegeneracieshavetoaccountforthecombinatorialwaysforchoosingwhichsubunitswillhave ligandsbound.Nowwecancalculate theaveragenumberof ligandsbound,which isv,accordingtoourfamiliarrulesforevaluatingtheexpectedvalueofsomeproperty.Weget

∗ 1 ∗ 0 1 ∗ 1 ∗ 0 ∗ 2 ∗ 1 ∗ 1 ∗ 21 2

Asanaside,youcanshowthatifL=0(meaningthatwehaveremovedtheelementofthemodelthatiscriticalforcooperativity,leavingonlytheRforms,whichbindligandsattheirsitesindependently),the equation above reduces to the equation we developed earlier for binding to identical andindependentsites:v=2K[A]/(1+K[A]),asitshould.Similarequationstotheoneaboveforn=2canbedevelopedforhighervaluesofn,usingthesamestatisticalmechanicstreatment.Withtheseequations,withjudiciouschoicesofKandL,onecangetbindingbehaviorthatexhibitsthefeaturesweexpectforcooperativity.BindingandHillplotsareshownbelow,ascalculatedfromtheMWCmodel(withn=4)usingthestatisticalmechanicsapproachabove.

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AdvantagesofcooperativebehaviorWhyarethesekindsofcooperativebinding(andcatalytic)phenomenausefuloradvantageousinbiologicalsettings? Forone,steepresponsecurvesgenerallyallowforbetter, tightercontrolofasystem.Inanon‐cooperative,hyperbolicbindingscenario,acertainfractionalincreaseinthe‘input’(i.e.ligandconcentration)leadsalwaystoasmallerfractionalchangeinthe‘output’(i.e.binding).Incontrast,asigmoidalcurveallowsalargechangeinresponsetoasmallerfractionalchange(e.g.inthe ligandconcentration). This feature iskey to theability ofhemoglobin tohaveverydifferentaffinitiesforoxygen(therebyallowingefficientuptakeandrelease)atoxygenconcentrationsinthelungs and inmuscle tissue that are not very different. Similar advantages apply to cooperativeenzymes. Theiractivitiescan increasemoresubstantially inresponsetosmaller increases in theconcentration ofthe substrate.This allows fortightermetaboliccontrol in asystem and alsoenables more‘on‐off ‘typesignaling in thecell.

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AllosteryRoughlytranslated,allosterymeans‘otherspatialarrangement’.Inthecontextofmacromolecules,itdescribesthegeneralphenomenonwhereinbindingofonecompoundtoaproteinornucleicacidat one site affects the conformation elsewhere,withdiverse consequences for activity. Youmayrememberthewell‐studiedcaseofhemoglobin,wherebindingofeffectormolecules(including2,3‐bisphosphoglycerate,CO2,andprotons)affectstheproteinconformationsomedistanceawaywhereoxygenbinds. Similarly, effectormolecules canbind to allosteric sites inenzymesandaffect thecatalyticpropertiesoftheactivesite,whichmaybeinadistantregionoftheprotein.Or,bindingofeffectorscancontrolsignalingpathwaysbyaffectingmolecularrecognitionevents.Insomecases,allosteric regulation occurs togetherwith cooperative phenomena in an oligomeric protein (likehemoglobin), but it can also occur in simpler scenarios in a single protein subunit. Allostericregulation isadeepsubjectwithdiverse manifestations inmolecularbiology,butaunifyingtheme can be articulated in aschemewhereaproteinhas twoavailable conformations, and thetwoconformationshavedifferentaffinities for the effector, asshown.Ifwesaythattheconformationontherighthasahigheraffinityforthe effector than theconformationontheleft(i.e.K3>K1), thenwemust also concludethat K4 > K2 (since the twodifferentroutesfromthetoplefttothebottomrightmustgivethesame total equilibrium constant,meaningthatK2*K3=K1*K4).TheinterpretationofK4>K2isthatbindingoftheeffectorshiftstheconformational equilibrium to the right. If K1 K3 and K2 K4, we would say that there is athermodynamiclinkagebetweentheeffectorbindingandtheconformationalchange.Theshapesofthe two alternate conformations of the protein are drawn in thisdiagramtoemphasizethatthetwoconformationsmaybedifferentinmultipleways,e.g.atmultipledifferent locations. Howthishappensdependsonthedetailedstructureoftheprotein.Butifyourecognizethat proteins have a certain degree of structural rigidity, you canimagineanynumberofdifferentwayswherethemovementofatomsatonelocationcanpropagatetoanothersite.Asjustoneexample,iftheproteinmoleculeiscomposedoftworelativelyrigiddomains,but

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the relative positionof thosedomains can change, then the conformation at one locationwill becoupledtotheconformationelsewhere.Wecanuseourstatisticalmechanicsframeworktoanalyzethesimpleallostericschemeabove.Wemightwanttoknowwhatfractionofthetotalproteinmoleculeswouldbeintheconformationontheright(withthepointedblndingcleftfortheeffectorandthelargeropeningonthetopsurface),asafunctionoftheeffectorconcentration.Fromourthermodynamicreasoningabove,wecananticipatethathighereffectorconcentrationwillcausemoreoftheproteintobeintheconformationontherightbymassaction,butexactlyhowmuch?Therearenodegeneraciestoworryaboutinthiscase,andtheweightsfollowfromtheequilibriumconstants,andtheeffectorconcentration.Withthesevalues,thefractionoftheproteinintheconformationontheright(withthe top binding site more open)wouldbe:(K2+K1K4[E])/(1+K2+K1[E]+K1K4[E])(equivalent expressions arepossiblewithK1K4=K2K3).Aplotofthisbehaviorisshownforjudiciouschoices of the equilibriumconstants.Catalyticcycleslinkedtoconformationalchangeslinkedtomotion=molecularmotorsNaturehasevolvedawiderangeofmolecularmotorswhoseoperationarebeyondextraordinary.Theyareessentiallyallbasedonsomekindofallosterycombinedwithcyclesofcatalysis.Bearinmindthatwhentheactivesiteofanenzymegoesthroughaseriesofreactions, itgoesthroughasequenceofeventswhereitisempty,thenboundtosubstrate,thenboundtoproduct,thenemptyagain, and so on. If those binding states each favor different conformations of the protein, thenongoingcatalysiswilldrivetheproteinthroughacyclicalseriesofconformations.Howsucheventscanbelinkedtolargerscalemovementsofthetypeonewouldcallamotorvaries,buttheclassiccaseittherotaryF1‐ATPase.Inaremarkableexampleofintuitionandforesight,inthe1970’sPaulBoyer(UCLA)predicted thatATPasewould act like a cyclicmotor using a ‘binding‐change’mechanismbased on careful biochemical experiments and an understanding that the trimeric assembly hadthreeactivesites,butwithoutthebenefitofknowingthethree‐dimensionalstructureofATPase.Itwas almost 20 years afterwards that Andrew Leslie working in the laboratory of John WalkerdeterminedthecrystalstructureofATPase,revealingthatindeedtheheadoftheATPase(composedofthealphaandbetasubunits)hasthestructureofawheelthatrotatesonanaxleformedbythegammasubunitinanextendedalphahelicalstructure.TogetherwithJensSkou(whodiscovereda

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differentATP‐driven ion transporter),BoyerandWalkersharedtheNobelPrize inPhysiologyorMedicinein1997.

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CHAPTER11SymmetryinMacromolecularAssembliesDefinitionofSymmetrySymmetryisanimportantsubjectinessentiallyallbranchesofscience,andtheartsaswell.Looselydefined,wethinkofsomethingthatissymmetricasbeingrepetitiveinsomeway,beingcomposedofmultiplecopiesofanunderlyingsubunit.Inscientificapplications,symmetryhasaprecisemeaning.Anobjectissymmetricifthereissomephysicaloperationwecandotoitthatleavesitinvariant(i.e.indistinguishablefromthewayitappearedbefore).Theoperationinquestionisusuallyanisometry,thatisaphysicalmovementinspacethatpreservesdistances.Thoseoperationsincluderotationsinspaceandmirrorinversions.However,sincebiologicalmacromoleculesarechiralandexistinjustone of two possible hands or enantiomers, for our purposeswe can dispensewithmirrors andinversions(i.e.so‐called‘operationsofthesecondkind’)andfocusonrotations.Weareluckyinthatregardasitleadstoconsiderablerestrictionsuponanotherwiselargervarietyofsymmetrytypesthatexistinthree‐dimensions.We will shortly work through all thepossiblesymmetriesinthree‐dimensions,butwestartwithoneexamplehere.Theassembly shown is comprised of threecopies of the same subunit rotated120°and240° relative to eachother. As youcansee, ifwerotate theentireassemblyby 120°, the result is indistinguishablefromtheinitialconfiguration.Infactthereareexactlythreeoperationswecandototheassemblythatleaveitinvariant.Theyare:{Identity(i.e.0°rotation),120°rotation,240°rotation}.Thesetofoperationsthatthatleaveanobjectinvariantisacompletedescriptionofitssymmetry.Setsofthistypeobeyspecialpropertiesthatmakethemexamplesofmathematicalgroups,whichwediscussnext.MathematicalGroupsInmathematics,agroupisasetthat,togetherwithadefinedbinaryoperator,obeysaspecificsetofrules.Abinaryoperatorissomethingthattakestwoelementsasinputandreturnsoneelementasoutput.Inregulararithmetic,additionandmultiplicationareexamplesofbinaryoperators,butasweshallseebinaryoperatorscantakediverseforms.Therulesthatmustbeobeyedforasettobeagroupareasfollows:

Theremustbeanidentityelement(I)intheset,suchthatforeveryelementAintheset,IA=AI=A.[Here,thesymbolisusedtodenotethegeneralbinaryoperator.]

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ForeveryelementsA in theset, theremustbean inverseelement (denotedA‐1), alsowithintheset,suchthatAA‐1=A‐1A=I.

Theassociativerulemustapply:A(BC)=(AB)Cforallelementsintheset. Aclosurerulemustbesatisfiedso that theproductofany twoelements fromtheset

(includingtheproductofanelementwithitself)mustalsobelongtotheset.Thatis,ifAandBbelongtotheset,thensomust(C=AB)forallchoicesofAandBwithintheset.

Therulesmustallbesatisfiedforasettoconstituteagroup,butforourpurposesthelastruleisespeciallyilluminating.Hereareafewexamplesofgroupsrelatingtopuremathematics:

{1,‐1}underordinarymultiplication {integers}underordinaryaddition {1,i,‐1,‐i}undercomplex‐valuedmultiplication

1 00 1

, 0 11 1

, 1 11 0

undermatrixmultiplication

Ineachcaseyoushouldbeabletoidentifytheidentityelementandalsoworkoutamultiplication‐typetable.Forthethirdexampleabove,thetablewouldbe:

x 1 i ‐1 ‐i1 1 i ‐1 ‐ii i ‐1 ‐i 1‐1 ‐1 ‐i 1 i‐i ‐i 1 i ‐1Fromourdiscussionsabovewecannowseewhythesymmetryofanobjectobeysthepropertiesofagroup.Inparticularwecanseewhyasetcomposedofsymmetryoperationsofanobjectmustobeytheclosureruleforagroup;ifoperationAleavestheobjectindistinguishable,andthesameistrueforoperationB,thensurelyperformingoperationAfollowedbyoperationBmustalsocompriseanoperationthatleavestheobjectinvariant.Thesymmetriesobeyedbyobjectsarethereforetypicallyreferred to as symmetry groups. Next we will enumerate the possible symmetry groups forassembliesofmacromolecules.PointGroupSymmetriesforBiologicalAssembliesTheprevalenceofsymmetryinnaturalproteinsisimpossibletomiss.About50%ofallproteinsthathavebeenpurifiedandstudiedinthelaboratoryhavebeenshowntobesymmetricoligomers.Inthissectionweenumerateallthefinitesymmetrygroupsthatarepossibleinthree‐dimensions;wesave for later a discussion of essentially infinite symmetry groups that characterize extendedassemblies like those in filamentous structures. Finite symmetrygroupsare referred toaspointgroupsymmetriesbecausethesymmetryaxespassthroughacentralpointintheassembly.The3‐

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Dpointgroupsymmetriescanbearranged,inorderofincreasingcomplexity,ascyclic,dihedral,andcubic.CyclicPointGroupSymmetriesEachofthesesymmetrygroupsisbasedonasingleaxisofrotationalsymmetry:twosubunitsinadimer,threesubunitsinatrimer,generalizingtonsubunitsinacycle.ThesymmetrydesignationsareC2,C3,…Cn.[C1wouldbethesymmetrygroupforanobjectwithnosymmetry.]ForC2,theaxisofsymmetrycorrespondstoa180°rotation. Becauseapplyingthisoperationtwice(ortwo‐fold)returns one back tothe startingorientation, thatsymmetry elementis often referred toas a “two‐fold” axisof symmetry.Likewise, asymmetry elementfor 120° and 240°(andofcourse0°)rotationsisreferredtoasa“three‐fold”axis,andsoon.Indrawings,therotationalsymmetryaxesaredenotedbysymbolsthatmatchtheorderoftheirrotation(e.g.asmallsquarerepresenting a 4‐fold axis). In any symmetrygroup,thenumberofelementsinthegroupisthesameasthenumberofdifferentlyorientedbut otherwise identical subunits required toconstruct thesymmetry. Forcyclicsymmetrygroups,Cn,thatnumberisn.Exampledrawingsareshown for the first fewcyclicsymmetries.There isno theoretical limit to thevalueofn,but the highest rotational symmetry for anyknown protein assembly is 39, for a trulyextraordinary barrel‐shaped protein chamberknown as the vault, which is present ineukaryotic cells for an as‐yet uncertainfunction.As you can see from the diagrams, afundamental point that arises from theprinciplesofsymmetryisthattheindividualcomponents(e.g.proteinsubunits)areallinidenticalenvironments. No physical difference of any kind can be ascribed to the multiple copies of thesubunit.AnexampleofapentamericproteinobeyingC5symmetryisshownasasemi‐transparentsurfaceoveraribbondiagram.Thefivecopiesofthesubunitareshowninseparatecolors.

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DihedralPointGroupSymmetriesDihedral symmetry groups are somewhat morecomplicated.Theyareessentiallybuiltbycombiningtwo copies of a cyclically symmetric arrangement,one flippedupsidedownon topof theother. Asaresult, they sometimes resemble double ringstructures, but sometimes that feature is not soevident,dependingontheshapeofthesubunitandits position relative to the symmetry axes. Indihedral symmetry, there are multiple axes ofrotationalsymmetry,allpassingthroughandhenceintersecting at the center ofmass of the assembly.SymmetryD4 is shown. As you can see, there is aunique4‐fold axisof symmetry, alongwith four2‐foldaxesofsymmetry,whichallintersectthe4‐foldaxesinaperpendicularfashion.Iftheunique4‐foldaxisisalongthez‐direction,thenthefour2‐foldaxeslieinthex‐yplane,evenlyspacedat45°fromeachother.Notethatarotationaboutthe4‐foldaxisexchangessubunitswithinthesamering,whereasthe2‐foldaxesexchangesubunitsbetweenthetworings.Asshownhere,aconvenientwaytodrawdihedralsymmetriesistobasethemonaprismoftheappropriatesymmetry;e.g.asquareprismforD4.Fordihedralsymmetry,thenumberofsubunits(anddistinctsubunitorientations)is2*n,wherenistheorderoftheuniqueaxisofsymmetry.TheenzymeRuBisCO,arguedtobethemostabundantenzymeonEarth,isanexampleofaproteinassemblywithD4symmetry.ItssubunitcompositionisL8S8 (eight large subunits and eight small subunits). In order to most clearly illustrate the D4symmetry, the arrangement of the large subunits inRuBisCOare shownhere,with each subunitcoloreddifferently,orientedinordertoshowviewsdownthedifferentsymmetryaxes.

Dihedral symmetriesarepossible fromD2 toDn foranyn. Note that thecaseofD2 is somewhatunique.Inthatcasethereisnosingleuniqueaxisofhighestorder.Insteadtherearethree2‐foldaxesallperpendiculartoeachother(e.g.alongx,y,andz).Andinsteadofapairofringstructuresthereisapairofdimers;D2symmetryisthereforesometimesreferredtoasadimer‐of‐dimers.Butotherwisethesituationisthesameasforhighern.Thatis,thereisstillanaxiswithn‐foldsymmetry(where n=2 for D2) combined with n evenly spaced 2‐fold axes perpendicular to that axis.

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Hemoglobinhassubunitstoichiometryα2β2,wherethealphaandbetasubunitsarehighlysimilar.Iftheywereidentical,thefoursubunitsinhemoglobinwouldbeanexampleofD2symmetry.D2isaverycommonsymmetryforproteins;C2isthemostcommon.Beyondthedihedralsymmetries,therearejustthreecasesofhigherrotationalsymmetrygroupsinthree‐dimensions.Thesearethecubicsymmetries,discussednext.CubicSymmetriesThecubicsymmetriesarebasedonthePlatonicsolidsandthussharetheirsymmetry. ThereareexactlyfivePlatonicsolids–theirstudydatesbacktotheancientGreekmathematicians–definedbytherequirementofhavingequivalentvertices,equivalentfaces,andequivalentedges.Theyaretheregulartetrahedron,cube,octahedron,icosahedron,anddodecahedron.Itturnsoutthattwopairsoftheseareintimatelyrelatedtoeachother,sharingthesamesymmetry,soinfacttherearereallyjust three symmetries represented by the five Platonic solids. Tabulating the numbers of faces,vertices,andedgesinthefivePlatonicsolidsilluminatestheso=called‘dual’relationshipbetweenthecubeandtheoctahedronandbetweentheicosahedronandthedodecahedron.Thosepairsarerelatedtoeachotherbyexchangeoffacesforverticesandvice‐versa.Thatis,ifyouplaceapointatthecenterofeachof thesix facesofacube, thosepointsare theverticesofanoctahedron. Andlikewise,pointsatthecentersoftheeightfacesofanoctahedronproducetheverticesofacube.Inthesameway,theicosahedronandthedodecahedronaredualsofeachother.And,interestingly,thetetrahedronisitsowndual.

The Platonic solids are shown withrotational symmetry axes indicated.For simplicity, only one instance ofeachaxistypeisshownoneachfigure.Note how 2‐fold symmetry axes passthrough opposing pairs of edges.Symmetry axespassing through faces

must conform to the symmetry of the faces. And symmetry axes passing through verticesmustconformtothenumberoffacesthatmeetatavertex.

Anassemblyconformingtotetrahedralsymmetry(T)canbeconstructedbyplacingthreesubunits(or symbols) on each face in a symmetric arrangement, for a total of 12 subunits. Octahedral

Platonicsolid

vertices faces edges symmetry

tetrahedron 4 4 6 Tcube 8 6 12 Ooctahedron 6 8 12 Oicosahedron 12 20 30 Idodecahedron 20 12 30 I

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symmetry (O) can be constructed by placing 4 subunits on each square face of a cube, or threesubunits on each face of an octahedron, leading to 24 subunits in either case. Whether a realassembly(e.g.ofproteinsubunits)thatobeyssymmetryOlooksmorelikeacubeoranoctahedrontypicallydependsonthesituationandcanbesubjective.Butthesymmetrypropertiesdonotdependonwhetheronethinksoftheassemblyascube‐likeoroctahedron‐like.ThesituationisthesameforicosahedralsymmetryI. Thosecasescanbedrawnandvisualizedaseitherthreesubunitson20triangularfacesorfivesubunitson12pentagonalfaces,foratotalof60subunits.

Schematicdiagramsaredrawnforassembliesintetrahedral,octahedralandicosahedralsymmetry.

Brokenorpseudosymmetry Manyexamplesappear innaturewhereanassemblynearlyhasahighersymmetry,butowingtosubtledifferencesthesymmetryisbrokendowntoalowersymmetrygroup,whichisasubgroupofthehighersymmetrygroup.Hemoglobinisthebest‐knownexample.Asnotedabove,iftheαandβsubunitswereidenticalthesymmetrywouldbeD2.Butbecausethetwosubunittypesareslightlydifferent,thetruesymmetryisonlyC2;onemightsaythesymmetryispseudo‐D2.TheheadoftheF1‐ATPasemotorisanotherwell‐knownexample.Thesubunitstoichiometryis(αβ)3.Again,theαandβsubunitsareverysimilar,butonlythebetasubunitshaveactivecatalyticsites.Theαandβsubunits alternate in a hexameric ring. The true symmetry is C3 (although really even the C3symmetryisbrokenbyconformationaldifferencesinthechemicallyidenticalsubunits),whileitispseudo‐C6.NotethatC3{0°,120°,240°}isasubsetorsubgroupofC6{0°,60°,120°,180°,240°,300°}.Bothoftheseexamplespresumablyarosefromgeneduplicationofasingleancestralproteinsubunit,followedbydivergentevolutiontogiveslightlydifferentsequencesandstructures.BiologicalConsiderationsWhyarenearlyall theoligomericproteins found innaturesymmetric? Theshortanswer is thatsymmetricarrangementsareeasier tobuildcomparedtonon‐symmetricarrangements. Thekeydistinctionisthatsymmetricarrangementsrequirethefewestnumberofdistinctsubunitinteractiontypes.Thatisillustratedbelowforatetrameroffouridenticalsubunits.TocreateaC4arrangement,

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asingleinterfacetype(highlightedbythered dot) is sufficient to hold the entireassemblytogether.Ontheotherhand,thenon‐symmetric case has four distinctinteraction interfaces that are allnecessary. Thequestionofhoweasyitisforsomethingtoarisebychanceiscriticalin evolution, since natural selection canonlyoperateonphenotypeoutcomesthatare somehow sampled by randomincremental mutations. Interestingly, itwasarticulatedasearlyas1956byCrickandWatsonintheirearlyworkonhowviruscapsidsshouldbeassembled(whichpredatedtheirbetter‐knowndiscoveryofthestructureofDNA)thatsymmetricarrangements would be dominant in natural structures like viral capsids because of the fewernumberofcontacttypesrequired.Settingasidetheissueofsymmetry,whyaresomanyproteinsandenzymesoligomericinthefirstplace? One explanation is cooperativity. We already discussed the cooperativity that is madepossibleinoligomers.However,thenumberofoligomericenzymeswherecooperativityhasbeenestablished isquitea small fractionof all theknownoligomers thathavebeenstudied. Anotherexplanationholds for somecaseswhere large‐scale structural integrity is required; viral capsids,microtubulesandbacterialS‐layersarewell‐knownexamples.Butagain,thesearespecialcases,andtheydonotspeaktothequestionofwhyenzymesaresooftenoligomeric.Otherpotentialadvantageshavebeenproposed,includingtheideathatoligomersarenaturallymorestablethanmonomers.Butthere is little evidence to support such ideas. The exceptionally high abundance of oligomericenzymesis(intheauthor’sopinion)alargelyunexplainedpuzzleinmolecularbiology.

SpecialTopicsinProteinSymmetryHelicalSymmetry(non‐pointgroup)Some symmetries contain operations that have translational or shift components in addition torotation.Repeatedapplicationofanoperationthatincludesashiftnaturallyimpliesastructurethatextendsessentiallyindefinitely;i.e.afilamentousstructure.F‐actinfilaments,microtubules,manyrod‐shapedfilamentousviruses,andphycobilisomesaresomeexamplesofproteinassembliesthatfollowhelicalsymmetry.Describingthegeometryofhelicalassembliesisgenerallymorecomplicatedthandescribingfiniteassembliesthatobeypointgroupsymmetries.Insomecase,theorganizationofahelicalassemblycanbefullydescribedbyasinglespatialoperation(arotationcombinedwithashift),whichwhenappliedrepeatedlygeneratesallthesubunitsinthestructure.TheF‐actinfilamentisdescribedbyarotationofabout167°combinedwithatranslationofabout28Å.Becausetherotationiscloseto

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180°, the F‐actinfilament takes theappearance of twoseparatelyinterwound helical‘protofilaments’. Thecylindrical proteincoatoftobaccomosaicvirus (TMV) can bedescribed by a singlerotational operationofabout22°combinedwith a translation ofabout 23 Å. Becausethat rotation issomewhere between1/16 and 1/17 of360°, the assemblycanalsobeviewedas16 slowly twisting protofilaments twisting one way, or 17 protofilaments twisting the otherdirection.Inothercases,likethemicrotubule,thehelicalassemblyismuchhardertodescribebyasingleoperation. Instead,differentfamiliesofhelicalcurvescanbedrawnonthe‘surfacelattice’.Withonefamilyofcurves,thereareapparently10protofilaments(suchacurveisthereforereferredto as a ’10‐start’ helix). There are also 13‐start helical curves and 3‐start helical curves for themicrotubule. In yet other kinds of tubular protein assemblies, there can be a true rotationalsymmetryalongtheaxisofthetube;inthosecasesitisimpossibletodescribetheassemblyintermsofasinglespatialoperationbetweensubunits.Quasi‐equivalenceandthestructureoficosahedralviralcapsidsEarly work on viral capsids – of the ‘spherical’ variety, not the filamentous variety – led to theconclusionthattheywouldbeconstructedaccordingtoprinciplesofsymmetry.Thehighestcubicpointgroupsymmetryinthree‐dimensionalspaceisicosahedral(aswediscussedabove),andthisposedamajorproblem.Itwasclearthata60‐subunit(icosahedral)proteinshellwouldnotbelargeenoughtoencapsulateallthegeneticmaterialofavirus. DonCasparandAaronKlugproposedasolutiontothatpuzzlebasedontheideaof‘quasi‐equivalence’.Undersymmetry,relatedcopiesofasubunitareinequivalentenvironments,soonly60subunitscanbeassembledwhileretainingstrictenvironmentalequivalence. ButCasparandKlugshowedhowlargernumbersofsubunitscanbeassembled in quasi‐equivalent environments. The key was to begin with a scheme based ontriangularfacetsofanicosahedron,butthentosubdividethetriangularfacetoftheicosahedronintoseveralsmallertriangles.Thesimplestwaytodothisistodivideatriangularfacetintofoursmallertriangles.Then,insteadofplacingthreesubunitsinasymmetricalarrangementonasinglefaceofanicosahedron,asonewoulddoforasimpleicosahedralassembly,onecanplacethreesubunitson

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each smaller subdivided triangle,again in a symmetric arrangement.Clearly you would end up with fourtimesasmanysubunitsasforasimpleicosahedral assembly, namely4*60=240totalsubunits.Weassertedbeforethatnomorethan60subunitscan be placed in strictly equivalentenvironments, but this triangulationmethod leads to subunits that are innearly equivalentorquasi‐equivalentenvironments,asshown. Tosee thedifference, note how some subunitsappear to be part of hexameric unitswhileothersappeartobepartofpentamericunits.Ifthecapsidiscomposedofonlyasinglekindofprotein,thenthesameproteinmustbeabletooccupymultipledistinctconformationalstates. Insome viruses, the distinct geometric sites are occupiedby slightly different capsid proteins; thatavoidstheproblemofthesamesubunithavingtotakeondifferentconformations,butitalsocreatesaneedfortheviralgenometoencodemoreproteins.ThecaseexplainedaboveisreferredtoasT=4basedonthefactorbywhichthenumberofsubunitsincreases;thetotalnumberofsubunitsis60*T.OthercasesbesidesT=4aremorecommoninnature.Thesetriangulationschemeareabithardertodrawbecausethesideofthelargetriangularfacetoftheicosahedrondoesnotfallalongalatticelineofthesmallertriangulationpatternonwhichthesubunitsarearranged.Onlysometriangulationnumbers(T)arepossible.ThegoverningequationisT=h2+k2+hkwhere h and k areintegers. In thetriangulation diagramshown, h and kdescribestheindicesofan edge of the largetriangular facet of theicosahedronintermsofedges of the smallertriangular lattice. Therecipeforassigningthevalues of h andk for agiven diagram is asfollows. Take onecorner of the large triangle as the origin (0,0). Thendraw twounit vectorsa andb to serve ascoordinateaxesonthepatternofsmallertriangles;thesetwounitvectorsdrawnfromtheoriginmustbe60°apart(not120°).Nowfigureoutwhatthecoordinateswouldbeinthissystemforone

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oftheothercornersofthelargertriangularfacet.Inotherwords,determinehowmanystepsyouwouldhavetotakealongaandb inordertoreachtheothercornerof the largetriangular facet.Thosenumbersofstepsarethevaluesofhandk.Notethataslongasyoufollowtheserules,youcanchooseanyedgeofthelargertriangularfacetandmultiplechoicesforthetwocoordinatebasisunitvectors;youmaygetdifferentvaluesforhandk,butthevalueforTshouldbeunchanged.T=3isacommoncaseinnaturalviruses,whileattheupperlimitafewgiantvirusesareknownwhereTisatleast1000andthevirusexceedsthesizeofabacterialcell!UsingsymmetrytodesignnovelproteinassembliesNatureisfullofexamplesofproteinsthathaveevolvedtoformelaborateassemblies.Along‐standinggoal inbioengineeringhasbeenhowdodesignnovelproteins in the laboratoryso theywill self‐assembletomakeinterestingarchitectureslikethoseseeninnature. Ideasforhowthismightbeaccomplishedwerelaidoutbytheauthor’slaboratoryseveralyearsagoandhavereachedfruitioninrecentyears. Symmetryhasplayed thekeyrole in thedesignstrategy. Inoneapproach, simplenaturalproteinoligomerslikedimersandtrimersareconnectedtogetherinspecificgeometricwaystogivegiantstructureslikecubiccages.Designedstructuresofthistypemayhaveutilityinvariedbiomedicalandnanomaterialsapplications. Thecrystalstructureofadesignedproteinassemblyhaving 24 identical subunits in symmetryO is shown below, oriented along its different axes ofrotationalsymmetry.

AlgebrafordescribingsymmetryWhenworkingwithsymmetry,itisoftennecessarytodescribetheunderlyingspatialoperationsinalgebraicterms.Somerecollectionofmatricesandhowtomultiplymatricesandvectorstogetherisimportant. Wenotedearlierthatgroupscanbecomposedofmatrices,andindeedwecantakeasymmetrygroupandrepresentitselementsbymatrices.Eachmatrixrepresentsarotationoperationthatisanelementofthesymmetrygroup.Asimplerecipemakesitpossibletoconstructa3by3matrixfromaphysicaldescriptionofarotationoperation.First,imaginethatyouhaveastartingpointatcoordinates(1,0,0),thatisapointoneunitalongthex‐axis.Nowaskyourselfwherethatpointwouldgoundertheoperationinquestion.Forexample,iftheoperationinquestionisa180°rotationaboutthez‐axis,apointthatstartsat(1,0,0)wouldrotatetoapositionwherethecoordinatesare(‐1,0,0).Nowwritethosecoordinates(‐1,0,0)

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asthe firstcolumnvectorofamatrix. Nowrepeattheexercisewith(0,1,0)asthestartingpoint.Undertheoperationofinterestitwouldgotoapositionwherethecoordinatesare(0,‐1,0).Writethatasthesecondcolumnvector.Nowdothesameforthestartingpoint(0,0,1).Thatpointactuallysitsonthez‐axisaboutwhichtherotationisoccurring,soitwouldnotgoanywhere;it’sfinalpositionwouldbe(0,0,1).So,theconstructedmatrixfora180°rotationaboutthez‐axiswouldbe

1 0 00 1 00 0 1

Wecandomuchwiththismatrixrepresentation.Forone,wecanuseittomultiplyageneric‘x,y,z’vectornotationtogetasymbolicrepresentationoftherotationinquestion:1 0 00 1 00 0 1

Thatmeanswecanequallywellrepresenttherotationaboutzsymbolicallyas‘(‐x,‐y,z)’.Also,byreversingourstepswecouldbeginwithasymbolicdescriptionofarotationaloperation,writeoutwhattheelementsoftherotationmatrixmustbe,andthenusethecolumnsofthatmatrixtogetaphysicalpictureofwhatkindofoperationisbeingperformed.Finally,thereisausefultrick.Inthree‐dimensions,theangleofrotationdescribedbya3x3rotationmatrixcanbedeterminedeasilyfromits‘trace’.Thetraceofasquarematrixisthesumofitsdiagonalelements(i.e.R(1,1)+R(2,2)+R(3,3)).Theequationfortheangleofrotationis:Trace(R)=1+2cos()Checkingthecaseweworkedoutabove,thetraceis(‐1+‐1+1)=‐1.Solvingfor,,weget2cos()=‐2,thencos()=‐1,andfinally=180°,asexpected.[Notethatfora2x2matrixdescribinga2‐Drotation,theequationisdifferent;theadditiveterm“1”ontherightmustberemovedforthe2‐Dcase.]

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CHAPTER12

EquationsGoverningDiffusionWe turn our attention now to dynamic processes, where behavior becomes a function of time.Diffusion–themovementofmoleculesasaresultofrandomthermalmotionandcollisions–isoneofthemostbasicdynamicprocessesformolecules.Wewilldiscussthecentralequationsthatgoverndiffusionandconsidertheirconsequencesfordiffusivebehavior.Diffusionin1‐DConsiderdiffusioninonedimensionasarandomwalkalonganumberline.Supposewestartatx=0,andthentakenstepsrandomlyeithertotheleftorright,witheachstephavinglength.

Wheredoweexpecttoendupafternsteps?

where li iseither+1or‐1withequalprobability. Then,theexpectedvalueofthepositionafternsteps,<xn>,willbe

⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ 0

since theaverageor expectedvalueof l is0. Theequation tellsus that theaveragevalueof thepositionoftheparticleafternstepsremainsat0.Thisisasexpectedsincetherewasnopreferencetosteponewayortheother.Thismeansthatifalargegroupofparticlestakeindependentrandomwalksstartingattheorigin,theaverageoftheirdistributionwillremainat0.Butclearlytheparticlesthemselvesdonotindividuallyremainatzero.Thatleadsustoaskhowspreadoutthedistributionofparticleswouldbeaftertheyeachtakensteps.Thestandardwayofexpressingadegreeofspreadingistoevaluatetheaveragevalueofthesquaredposition–thesquaringcausesdisplacementsinbothdirections(positiveandnegative)tocontributepositivelytothespread,astheyshould.Theaveragesquareddisplacementafternstepsis

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⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⋯ ⋯ ⟩

⟨ ⟩ ⟨ ⟩ ⋯ ⟨ ⟩⟨ ⟩ ⟨ ⟩ ⋯ ⟨ ⟩

⋯1 0 ⋯ 0 0 1 …

Thelogichereparallelswhatwesawearlierinthecoursewhentreatingthepathofaflexiblepolymerusingarandomwalkmodel.Namely,theaveragesquareddistancetraveledisproportionaltothenumberofsteps.Thismeansthatthermsdistancegoesasthesquarerootofthenumberofsteps,asbefore.

√ Thisisanimportantresult.Itshowsthatdiffusioncanbeanefficientmechanismformovementovershortdistancesbutnotoverlongdistances.Thisimportantlimitingfeatureofdiffusionexplainstheevolutionofarangeofbiologicalphenomenawhereenergyisexpended(e.g.intheformofATPorGTPhydrolysis)tocausedirectedmovementofmoleculesorevenorganellesacrosslongdistances.The places where this becomes most important is where the cellular length scales are largest;neuronsare the classic example, andenergydriven transport (e.g. bymolecularmotors trackingalongmicrotubules)iscriticalthere.Wecanconverttheequation<xn2>=n2fromaformthatdependsondetailsoftherandomwalk(e.g.nand) toamorephenomenological formthatdependsontime. Letbe the time intervalbetweensteps.Thentheelapsedtimeduringarandomwalkist=n,andn=t/.So,

<xn2>=(t/)2=(2/)tNowwe replace the term (2/) with 2D, where D is called the diffusion coefficient, which is apropertyofthemoleculeinquestion,amongotherthingssuchastheviscosityofthesolutioninwhichthediffusionisoccurring.Withthatsubstitution,wefindthatthermsdistancetraveledis<xn2>=2DtThisequationappliestodiffusioninonedimension.Inthreedimensions,<xn2>=6DtWewilldiscusstherelationshipbetweenmolecularpropertiesandthediffusioncoefficientDlater,butfornowwecontinuewithananalysisofhowmoleculesspreadoutfromacentralstartingpoint,andhowthisdependsontimeandonD.

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Forthe1‐Dcasewecanimagine a thin tube inwhich material isinitiallyconcentratedatone point (x=0). Overtime, the moleculesnaturallyspreadout.What will the concentration profilelooklikeatdifferenttimepoints?Theansweristhattheconcentrationwilltake the form of a Gaussiandistribution.Specifically,

∝ / Comparingthistothestandardform

ofaGaussian( / ),whereσisthestandarddeviation,weseethatthe standard deviation of theconcentration profile is obtained bysetting 4Dt = 2σ2, which gives astandard deviation for spreadingfrom the center of σ = sqrt(2Dt),which matches our previousexpressionforrmsdistancetraveled,asitshould.GeneralEquationsforDiffusionWewereable toworkoutsomebasicproperties fordiffusion froma fixedpoint,butwhataboutequationswe can apply tomore general cases? This leads us to Fick’s first and second laws ofdiffusion.Fick’sfirstlawWebeginbysettingupa1‐Dsystemasbefore.Thenweconsiderhowmuchmaterial(i.e.howmanymolecules)wouldcrossanimaginaryboundarybetweentwopointsinthesystem,xandx+,overatimeintervalwhereeachmoleculewouldtakeastep.

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We can let N(x) denotethenumberofmoleculesthatareatpositionx,andN(x+) denote thenumber of molecules atposition x+. To relatenumbers ofmolecules toconcentrations,wewould need to divide the number at each location (x or x+) by the volumeallottedtoeachposition,namelyA*,whereAisthecrosssectionalarea.Nowwecanconsiderhowmanymoleculesweexpecttocrosstheimaginaryboundary.Ourinterestisinthenetmovement.Thenetmovementortransportacrossaboundary(realorimaginary)istheflux,J,expressedasa#perareapertime(cm‐2sec‐1incgsunits).Wecancalculatethefluxinoursystembynotingthathalfofthemoleculesatpositionxwillcrosstheboundaryfromlefttoright(sincetheprobabilityis½ineachtimestepthatamoleculewilltakeasteptotheright),andlikewisehalfofthemoleculesstartingatx+willcrosstheboundaryfromright‐to‐left.Clearlythenthenetflux(takenhereasthenetmovementfromlefttoright)wouldbethefirstquantityminusthesecondquantity,dividingbyareaandthetimeinterval:

J=½*(N(x)‐N(x+))/(A)Nowmultiplyingthetopandbottomby2gives

J=½*(N(x)‐N(x+))2/(A2)RearrangingandsubstitutingconcentrationCforN/V=N/(A)gives

J=‐½2/*(C(x+)‐C(x))/Nowwerecognizethat(½2/)isjustthediffusioncoefficientDfrombefore.AndtheexpressionontherightsideoftheequationappearsasadifferencebetweentwovaluesoftheconcentrationCattwocloselyspacedpoints(xandx+),dividedbythespacing;thishastheformofaderivativeofCwithrespecttoposition.So,J=‐DdC/dxThisisFick’sfirstlawinonedimension.This general result tells us that the net movement of molecules due to diffusion (down aconcentrationgradient)isproportionaltothesteepnessofthegradient(dC/dx)timesthediffusioncoefficient. Andofcoursethenegativesignisimportantasitspecifiesmovementinthedirection

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opposite from the direction of thegradient.Theideacanbegraphedona1‐Dconcentrationprofileasshown.Fick’ssecondlawWhat can we say about howconcentrationswillbechangingovertimeasa resultofdiffusion? Wecananswerthatwithasimilartreatment.Butnowwethinkabouthowthenumberofmoleculesin some particular regionwould changeasaresultofthefluxthatisoccurringononesidecomparedtotheother(i.e.ontheleftcomparedtotheright).Ifthetotalfluxintoaregion,takingintoaccountmovementacrosstheboundariesoneitherside,ispositive,thentheconcentrationshouldbeincreasingovertime.Thinkingaboutthisasachangeinconcentrationoverachangeintime,ΔC/Δt=(Δ(#ofmolecules)/V)/ΔtThen realizing that the changein the number of molecules isgivenby thenetnumberbeingtransferred across theboundaryontheleftminusthenet number being transferredacross the boundary on theright, and taking the volumeelement to beA*, and takingthetimeintervaltobe,ΔC/Δt=(J(x)A–J(x+)A)/(A))/ThissimplifiestoΔC/Δt=‐(J(x+)–J(x))/Similartobefore,wecanrecognizethisasaderivativeofJwithrespecttoposition.So,dC/dt=‐dJ/dx.

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ButweknowfromFick’sfirstlawthatthefluxJ isthefirstderivativeoftheconcentrationCwithrespecttopositionx.So,thechangeinconcentrationasafunctionoftimeisevidentlythesecondderivativeofCwithrespecttoposition,multipliedbythediffusioncoefficientD.

This is Fick’s second law in 1‐D.Essentially, it tells us that thewaythe concentration is changing at afixed position due to diffusion isdetermined by the curvature (i.e.the second derivative) of C withrespect to x. With thatunderstandingwecansketchhowaconcentration profile would beexpected to change over time (atleastoverashortintervalwherethederivativesarenotchangingmuch):Fick’ssecondlawisasecondorderdifferentialequationforC intermsof x and t. Some scenarios havesimple enough ‘boundary’conditions (e.g. a simple form fortheconcentrationattime0),thatwecan solve Fick’s law to obtain acompleteexpressionforCintermsofxandt,meaningwewouldknowwhattheconcentrationprofilewouldlooklikeatanytimet.Mostrealproblemshavemathematicalformsthataredifficulttosolve.Butinthecaseofdiffusionfromapoint that we dealt with earlier, we did write out an equation (without proof) saying that theconcentrationprofileasafunctionoftimeandpositionwasproportionaltoaGaussian.Introducingaleadingmultiplicativeterminordertomakethetotalamountofmaterialinthesystemconstantovertime,thecorrectequationis:

∝1

√/

Althoughwewritethisequationwithoutproof,wecanshowthatitdoesindeedobeyFick’ssecondlaw,asitmust.Takethefirst(partial)derivativeofCwithrespecttot.Thentakethesecond(partial)derivative of Cwith respect to x. The resulting expressions should be equal to each other aftermultiplyingbythediffusioncoefficient,D.

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GeneralizingFick’slawsforthreedimensionsFick’slawsgeneralizereadilytohigherdimension.Thesingle‐variablederivativesgetreplacedbythegradientoperator.Indimensionshigherthan1,thefluxJisavector.Itpointsdirectlydowntheconcentrationgradientinthedirectionofsteepestdescent.Fick’sfirstlawtakesthefollowingform:

, ,

, ,

, ,

wherethe termdenotestheunitvectoralongxandlikewiseforyandz,and symbolizesthegradientor‘del’operator.Similarly,Fick’ssecondlawgeneralizesto:

, , , , , , , ,

Special topic: Using numerical (computational)methodstosimulatediffusionbehaviorManyproblemsthatinvolvedifferentialequationscanbetreatedeffectivelyusingcomputertechniques.Thederivative quantities are effectively replaced withdifferencesinavariableoversmallsamplingdistances.ToapplyFick’ssecondlawtoadiffusionproblem,weneed to know the secondderivativeof concentrationwithrespecttoposition.Examiningthe1‐Dcasefirst,wemighthaveaplotofconcentrationatsometimetasshown:Howwouldweestimate thevalueof (d2C/dx2)atpointx? Well, thesecondderivative is just thederivativeofthefirstderivative,soweneedtoevaluatehowthefirstderivativechanges. Wecantakethedifferencebetweenthefirstderivativesbetweenpointsxandx+Δxandbetweenpointsx‐Δxandx,keepinginmindweneedtodividebytheseparationdistancewhencalculatingderivatives.Weget:

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≅

∆∆

∆∆

∆∆ ∆ 2

∆

Inotherwords,thesecondderivativeisapproximatedbyfirstaddingupthevaluesofavariableoneithersideofthecentralpointandthensubtractingtwicethevaluethevariabletakesatthecentralpoint,dividingbythesquareofthesamplingdistance.ThisrecipemakesitpossibletosimulatetheevolutionofaconcentrationprofileunderdiffusionbyestimatingthesecondderivativeofCateachpointandthenusingthosevaluestoupdatethenewconcentrationsatanewtimepoint.Since∆∆

∆ ∆

Theprocedurecanbeextendedeasilyintohigherdimensions.For2‐D,thenumeratorisjustthesumofseparatesecondpartialderivativeswithrespecttoxandy,andwewouldendupwith:

∆ ∆

∆ , ∆ , , ∆ , ∆ 4 ,

∆ 2 ∆

assuming the sampling distance is the same in bothdirections(i.e.Δx=Δy).Graphically,theeffectistoaddupthevalueofthevariable(C)atthefourpointssurroundingthecentralpoint(x,y)andthensubstract4timesthevalueofthe variable at the central point – this is the numeratorabove.In 3‐dimensions, the required coefficients are of course+1.+1,+1,+1,+1,+1,and‐6.

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CHAPTER13

TheDiffusionCoefficient:MeasurementandUseWenotedearlierthatthediffusioncoefficient,D,whichdescribeshowrapidlyamoleculediffuses,dependsonthepropertiesofthemolecule(chieflyitssize).Naturallythen,ifwemeasureDwecanlearnsomethingaboutmolecularsize.WebeginwithadiscussionofexperimentsformeasuringD.Measuringthediffusioncoefficient,DThediffusioncoefficientcanbemeasuredinvariousways,eachofwhichmaybesuitableforsomesystemsandnotothers.Oneapproachistomeasuretherateofspreadfromaninitialpoint.Fromour earlier equations we know that the standard deviation in the spread of material initiallyconcentratedatapointgoesasxrms=sqrt(2Dt)in1‐Dandsqrt(6Dt)in3‐D.Therefore,ifthespatialspreading(i.e.thestandarddeviationoftheconcentrationdistribution)aftertimetcanbemeasured,Dcanbeobtained.Recent technologydevelopmentshavemade itpossiblewithspecial instrumentation to track thelocation of a single molecule, usually on the basis of fluorescent labeling combined with highlysensitive detectors. If a single particle ismonitored long enough, its averagemovement over aparticular time period can be evaluated, leading to a value for D in the same way as for themeasurementofspreadinginanensemble,asabove.Fluorescentrecoveryafterphotobleaching(FRAP)Diffusion is a processwherebymoleculesmove about in away that tends to result in auniformdistribution; i.e.anequalconcentrationeverywhere. Asaresult,measuringdiffusion inasystemwherethemoleculeofinterestisalreadyuniformlydistributedisnaturallyproblematic,sincefurtherdiffusionhasnoeffectontheconcentrationdistribution.Theprinciplebehindmethodsknownasfluorescent recoveryafterphotobleachingorFRAP, is tocreateanon‐uniformdistributionof themoleculeofinterestandthenmonitorhowrapidlyitsconcentrationprofilereturnstouniformity.Thestandardapproachistofirstlabelthemolecule(e.g.aprotein)withafluorescentgroup.[Aswewilldiscusslater,fluorescenceisaconvenientandsensitivewaytomeasuretheconcentrationofalabeledmolecule.]Then,astronglaserpulseisusedto‘bleach’(i.e.destroybybondbreakage)thefluorescentprobe,noteverywherebutjustinonespot,leavingaregion(e.g.acircularspot)wherethereisnofluorescencefromthemoleculeofinterest.Then,onewaitsandmeasuresfluorescenceinthebleachedregion.Assumingthelabeledmoleculesarediffusingabout,unbleachedmoleculesfromoutsidethebleachedregionwillfindtheirwayintothatregion.Howmanyofthosemoleculeshavediffusedintothatregioncanbemonitoredwitha(usuallymicroscopic)fluorescencereading.Of course the bleached molecules will also diffuse out of the bleached region, but they lack afluorescentgroupandsodonotcontributetothemeasuredfluorescence.Afteranextendedperiod,the fluorescence as a functionof timewill plateau as the concentrationof unbleachedmoleculesbecomesequalinsideandoutsidethebleachedregion.Thetimescaleoverwhichthefluorescence

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returnsdependsonthediffusioncoefficientofthemoleculebeingstudied.So,Dcanbeobtainedbymeasuringtherateoffluorescencerecovery.

Thebehaviorof fluorescence recoveryand its relation toD isdiagrammedhere. Wewill notgothroughadetailedmathematicaltreatmenthereotherthantosaythattheequationsfordiffusionmakeitpossibletoformulatewhatthecurveforfluorescencerecoveryshouldlooklikeasafunctionofD. AndthereforeavalueforDcanbeextractedbydeterminingwhatvalueofDgivesthebestmatchbetweenthemathematicallycalculatedbehaviorandtheobserveddata.FRAP can be used invarioustypesofset‐ups.Itisnaturallysuitedformeasuring two‐dimensionaldiffusioninathinlayer,forexampleof a protein in a lipidbilayer. It is alsocommonly used in situ(i.e. inside cells) usingfluorescentmicroscopy.For fluorescent studiesinside cells, the proteinofinterestmustbefluorescentlylabeledbygeneticfusiontoanaturallyfluorescentprotein,atopicwewilldiscussinmoredetaillater.DynamicLightScattering(DLS)Dynamiclightscattering(DLS,alsosometimesreferredtoasphotoncorrelationspectroscopy)isapowerful and convenient experiment formeasuring diffusion coefficients. Part of its power andconvenience comes from not having to artificially create a system where the concentration

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distribution is out of equilibrium (i.e. non‐uniform), as required in FRAP. DLS relies on naturalfluctuationsintheintensityoflightthatisscatteredbythelargesolutemoleculesinasolution.Wewillnotdiscussthephysicsoflightscatteringindetail,otherthantosaythatthephenomenonunderdiscussion here is typically referred to as elastic or Rayleigh scattering and occurs where thewavelengthoflightismuchlargerthanthesizesofthemoleculesinvolved;macromoleculeshavesizes between a few nanometers to tens of nanometers, while the visible/UV region of theelectromagnetic spectrum is in the few hundred nanometer range. The intensity of the lightscattering–e.g. the fractionof the incidentphotons thatbouncesoff indirectionsother than theincident direction – depends on the index of refraction or polarizability of themolecules, and isstronglydependentonmolecular size. In fact, it tends tobedominatedby the largest speciesofmoleculeinasolution.Without belaboring the details, the randommovements ofmolecules in solution causes randomfluctuationsintheintensityoflightthatisscatteredinanyfixeddirection.Atonemomentintime,thescatteredlightintensitymaybeslightlylowerthantheaverage(overtime),whereasamomentlatertheintensitymaybehigher.Butthecruxofthephenomenonisthatthetimescaleoverwhichthefluctuationspersistdepends(inversely)ontherateofdiffusion.Ifatsomeinstantintimethepositionsofthemacromoleculesinasolutionaresuchthatthelightscatteringishigherthanaverage,thenthelightscatteringwillremainaboveaverageuntilthemoleculeshavemovedfarenough(bydiffusion)toerasethemomentaryfluctuation,andlikewiseifthescatteredintensityislowerthanaverage.Ifthemoleculeunderstudyhasahighdiffusioncoefficient,thendeviationsaboveorbelowthe average intensitywill vanish ordissipate quickly,whereas if themolecule has a lowdiffusion coefficientthen whateverfluctuationsoccurwillpersist longer. Inother words, aphenomenon thatshows randomfluctuating behaviorhas a natural timescale associated withthe fluctuations. Theplots here illustratethe general idea offluctuating behaviorhaving differentcharacteristic timescales.

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Howcanthetimescaleofsomethingthatisrandomlyfluctuatingbecharacterizedmathematically?Theautocorrelationfunctionprovidestheanswer.Theessenceofanautocorrelationfunctionistoaskhowsimilarorcorrelatedtheintensitymeasuredattimetiscomparedtothevaluemeasuredattimet+,whereissomespecifiedtimeincrement.Ofcoursetheanswerdependsonthevalueof.Ifissufficientlysmall,thenthevaluesmeasuredattandt+willbeverysimilar(infactidenticalif=0). On the other hand, ifwe consider a large value of (i.e. longer than the time scale of thefluctuations), then the intensity values at t and t+ will be uncorrelated. And of course atintermediatevaluesofwewillseeintermediatevaluesofthecorrelation(i.e.between1and0).Inotherwords,thevalueoftheautocorrelationfunctionwillbe1whenis0,andwilldecayto0whenislarge.Preciselyhowquicklytheautocorrelationfunctiondecaysasafunctionoftellsuswhatthecharacteristictimescaleisforthefluctuatingbehavior.The plot above (lower panel), illustrates the mechanics of how the autocorrelation function isevaluated.First,recallfromyourpriorexposuretostatisticsthatwhencalculatingthecorrelationcoefficient between two ordered sets of values, for the numerator one simply takes the sum oraverage of the products of one set of values with the other set; the denominator is simply anormalizingfactor.So,tocalculatetheautocorrelationfunctionwejustneedtocalculatetheaveragevalueoftheproductoftheintensitiesattimetandt+.TocalculatetheaveragevalueoftheproductofI(t)andI(t+),youcanimaginetakingabaroflengthandslidingitalongthelengthoftheplottoidentifyintensityvaluestomultiplytogetherandaverage.Withoutlossofgeneralitywecansimplifythingsbypretendingthattheaveragevalueoftheintensityis0,withfluctuationsgivingplusandminusvalues.Then,theautocorrelationfunctionA()isjust

A()=<I(t)*I(t+)>AplotofA()vswilldecayexponentially,accordingtoourdiscussionsabove.Thecharacteristictime for the fluctuating intensity is the valueof the time increment atwhichA()/A0=e. Themotivation for this kind of autocorrelation analysis is that the characteristic fluctuation time is

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relatedinverselytothevalueofthediffusioncoefficient,D.Frommoreadvancedtextsonecanfindthataplotofln(A)vsgivesaslopeequaltoDtimes(‐82n2sin2(/2)/2)wherenistheindexofrefraction, is the scattering angle, and is thewavelengthof the light. In thatway,D canbeobtainedfromtheautocorrelationanalysis.Withmoderninstrumentation,thevalueofDcanbeobtainedbyDLSquicklywithverylittlematerial.Built‐insoftwareperformsthenecessaryanalysis.Theexperimentis ‘native’inthesensethatthemacromoleculeisexaminedinitsnativestate,andnondestructive.Ofcoursethemacromoleculeofinterestmustbepurified,andowingtothestrongdependenceofscatteringonsize,specialcaremustbe taken to remove any particulates or aggregated material. If the sample in question isheterogeneous, containingmolecules of more than one size, then in ideal cases it is possible todecomposethebehaviorintoseparatecomponents,butthisbecomeschallenging.RelatingthediffusioncoefficienttomolecularsizeRelatingDtomolecularfrictionWhen a molecule moves down its concentration gradient under diffusion, the overall rate ofmovementortransportreflectsasortofterminalvelocitysituation,wheretheforceofmovementcausedbytheconcentrationgradientisbalancedbythefrictionthemoleculesfeelastheytravel.Thefrictionalforceonanobjectistheproductofitsvelocitywithitsfrictionalcoefficient,f.Thedirectionofforceisoppositefromthevelocity,soFfrict=‐fvWhatismeantbyaforceduetoaconcentrationgradient?Wehaveseenthisbeforeinthecontextofbalancingforcesinequilibriumsedimentation.IfwedenotetheforceduetoaconcentrationgradientasFC,thenrecallingthatforceisthederivativeof(potential)energyasafunctionofposition,

FC=‐d(0+kBTlnC)/dx=(‐kBT/C)dC/dx

Nowrequiringthatthetwoforcessumtozeroatterminalvelocity,FC+Ffrict=0andfv=(‐kBT/C)dC/dxButthevelocityvisadescriptionofhowfastthemoleculesaremoving(nottheirvibrationalspeedbetweencollisionsbuttheirnettransportspeed),sovmustberelatedtofluxJ.Byexaminingunits(Jis#/(cm2sec),andviscm/sec),wecanseethattheconversionbetweenthemisbyunitsof#/cm3whichisconcentration,soJ=vCandv=J/C

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Puttingthisintothepreviousequationandcancellingconcentrationonbothsides,fJ=(‐kBT)dC/dxButfromFicksfirstlawweknowthatJ=‐DdC/dx.Substitutingthatintotheequationaboveandcancellingliketerms,wefindremarkablythatfD=kBTOf coursekBT is a constant (our familiar average thermal energy), havingnodependenceon themoleculeinquestion.Evidently,thefrictionalcoefficientofamoleculefandthediffusioncoefficientofamoleculeDarejusttwomanifestationsofthesamething,inverselyrelated.Ifweknowone,weknowtheother. ThisequationisknownastheEinstein‐Smoluchowskiequation. Thereasonitisimportantisthattherearewell‐knownphysicsequationstodescribehowthefrictionalcoefficientofanobjectrelatestoitssize,andsinceknowingDgiveusf,wehaveawaytogetfromDtomolecularsize,aswedescribenext.Relatingthefrictionalcoefficientftomolecularsize(sphericalradius)In1850StokesshowedthatforasphereofradiusRmovinginamediumofviscosity,thefrictionalcoefficientis

f0=6Rwherethesubscriptinf0denotestheassumptionofasphere.ThisisknownasStokes’equation.Forwaterat ambient temperatures, theviscosity isabout=0.010g/(cmsec). FromtheequationsaboveyoucanseethatifyoumeasureD,youcanimmediatelyobtainthefrictionalcoefficientf,andif you assume the molecule is spherical then you can obtain R. The value of R so‐obtained issometimescalledtheStokesradiusorsometimesthehydrodynamicradius.FromavalueforR,onecanusetheknowndensityforaproteinornucleicacidtocalculateitsmass.Transportproblemsareoftentreated incgsunits. Theunits for thekeyvariablesaresomewhatpeculiar.Theyarelistedhereforconvenience.

variable unitsD cm2/secf g/secJ 1/(seccm2)v cm/sec g/(cmsec)

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Example:SupposethemeasuredvalueofDforalargeproteincomplexis510‐7cm2/sec.Assumingthemoleculeisspherical,whatisthemolecularweight?(Letthedensityofproteinbe1.35g/cm3)f=kBT/D=8.210‐8g/secR=f/(6)=4.410‐7cm(=44Å)

MW=1.35g/cm3(4/3)R3NA=290000g/mol=290kDaNon‐sphericalmoleculesForagivenvolume,aspherehasthelowestpossiblefrictionalcoefficient,whichalsomeansithasthehighestdiffusioncoefficient.Non‐sphericalobjectshavehighervaluesoffandlowervaluesofD.That means that estimating molecular size from the diffusion coefficient alone (using Stokes’equation,whichassumesasphericalshape)cangivesomewhaterroneousvalues.Inparticular,sinceanon‐sphericalshapeleadstoalowerdiffusioncoefficientandahigherfrictionalcoefficient,andthefrictional coefficient varies directly with the assumed spherical radius, a highly non‐sphericalmoleculewillhave the samediffusioncoefficientas a larger sphericalmolecule. Inotherwords,estimatingmolecularsizefromDalonecanleadtoanoverestimationofsize.Advancedtextsprovidecomplexequationsthatrelateftothedegreeofnon‐sphericality,butthesearenotusefulinpracticeveryoften.Aswewillseeshortly,combiningmeasurementsofD(andhencef)withotherkindsofmeasurementsmakesitpossibletoobtainmolecularweightswithoutassumingasphericalshape.Inadditiontotheissueofshape,thefrictionalcoefficientcanalsobeaffectedbyotherfactors. Inparticular,macromoleculestendtocarryahydrationlayerofboundwatermoleculeswiththem,andthissometimescomplicatestheanalysisoffrictionalcoefficients.DiffusionorspreadingoutwithrespecttoorientationratherthanpositionThe idea of diffusion can be generalized to go beyond positional variables. Ordinary diffusionconsidershowmoleculesthatbegininthesameplacespreadouttootherpositions.Thatideacanbegeneralizedtomolecularorientations.Ifthemacromoleculesinasolutioncanbeinitiallydriventothesameorientationandthenallowedtoreorientthroughrandomrotationalchanges,thenovertimetheirorientationswillspreadbackoutuntiltheirorientationaldistributionbecomesuniform.As with ordinary diffusion, there is a distinct constant associated with rotational diffusion of amolecule, which is inversely related to the degree of friction the molecule experiences when it

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tumblesintheviscoussolution.Rotationaldiffusionwillcomeupagainlaterwhenwediscussspecialspectroscopictechniques.

SpecialTopicinDiffusion:DiffusiontoTransportersonaCellSurfaceItisasurprisingobservationthatcellsthathavesurfacetransportersfortakingupnutrientsusuallyhavearatherlowdensityoftransportersontheircellsurface.Whynotgainanadvantagebydenselypackingthesurfacewithtransportersinordertoobtainnutrientsmorerapidly?Theanswertothispuzzlecomesfromexaminingthepeculiarandsurprisingpropertiesofdiffusivebehavior.Net movement can be present in situations where the concentration is not changing over timeanywhere.Steady‐statecanbeachievedwhereamoleculeofinterestisbeingproducedatoneplace(called the source) and consumed at another (called the sink). Under steady state conditions,meaningdC/dt=0,Fick’ssecondlawbecomes:

, , , , , ,

0

Solvingthisdifferentialequationgivesadescriptionoftheconcentrationeverywhere inasystembetweenthesourceandthesink.SolvingthisoneequationgivesdifferentresultsforthefunctionC(x,y,z),dependingonthe‘boundaryconditions’.Theboundaryconditionsarespecifiedbyhavingfixed(andunequal)concentrationsatthesurfacesofthesourceandsink,sooneobtainsadifferentsolution to the differential equation and adifferentfunctionforC(x,y,z)dependingonthe nature (e.g. size, shape andarrangement)ofthesourceandsink.We begin our analysis of the problem ofdiffusiontocellsurfacetransporterswithatreatment of a simpler problemwherewehave a sphere (representing a cell)whoseentire surface acts as an absorber;whenever a diffusing molecule of interesthitsthesurfaceitiscapturedorconsumed.Thisisthecaseofasphericalsink,andtheboundary condition is that theconcentration C=0 at the surface of thesphere. Wecantreatthesourceasbeinginfinitelyfaraway–imagineanenclosingsphereofverylargeradiusgivingaboundaryconditionofCequaltosomefixedvalueC0atinfinity.Withoutproof,wecanfindthatthesolutiontoC2=0withtheseboundaryconditionsis

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C(r)=C0(1–a/r)wherethespatialvariablesx,y,z,arereplaced instead with r since theproblemissphericallysymmetricanda is the radius of the sphericalabsorber.Noticethat,asrequired,C=0atr=aandC=C0atr=infinity.Youcanfurtherprovetoyourselfthatthisequation does indeed obey Fick’ssecond law by converting r tosqrt(x2+y2+z2) and taking secondpartialderivativestoshowthat2C=0.Now that we know what the concentration function looks like, we can determine how fast theabsorbingsphereiscapturingdiffusingmolecules. WecanfindthisbyevaluatingthefluxJatthesurfaceandthenmultiplyingbytheareaofthesphere,4a2.|J|=D|dC/dr|=DC0a/r2andatthesurfaceofthesphere(r=a),so|J|=DC0/aTakingintoaccountthesurfaceareaofthesphere,

Capturerateforanabsorbingsphere=4a2DC0/a=C04DaThis is how fast a spherical cell of radius a could capture a diffusing nutrient whose diffusioncoefficientisDandwhosebulkconcentrationisC0.Nowwehave toanswer theharderquestionconcerning thecapturerate fora spherewhere thediffusingmolecule is captured only if it collideswith the sphere at small absorbing patches (i.e.transporters);therestofthesphereisnotabsorbing.ThisproblemwasfirsttreatedbyHowardBergandisdiscussedinhisshortclassicRandomWalksinBiology.Let’ssaythatthereareNpatchesonthesurfaceandeachiscircularwithradiuss.Usingaclevermathematicalanalogybetweendiffusiveresistanceandresistanceinanelectricalcircuit,Bergreasonedasfollows.Therateofcapturebyasinglecirculardiskofradiussis(givenwithoutproof)C04Ds.[ThiscomesfromsolvingC2=0usingboundaryconditionsofC=0atthesurfaceofaflatcirculardisk,calculatingthefluxasthederivativeofC,and then integrating the fluxover thecircularpatch.] ThenBergnotes that theproblemofdiffusiontoasetofpatchesonaspherecanbebrokendownintotwosteps:(1)diffusionfrominfinity

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toasphericalsurfacejustoutsidethesphereofinterest(butwitharadiusnotsubstantiallygreaterthana),followedby(2)diffusiontoasetofcircularpatches.Then Berg introduces anelectrical analogy. Recall thatelectrical resistance is R=V/I(which is a driving voltagedivided by a flow). By analogy,diffusiveresistancewouldbethedrivingconcentrationdividedbythecapturerate.Forthecaseofthe spherewhose entire surfaceisabsorbing,wegetC0/(C04Da)= 1/(4Da) as the diffusiveresistance. For capture by asingle circular disk, for thediffusive resistance we getC0/(C04Ds) = 1/(4Ds). Nowweput the twosteps together. Thetwo steps occur in series (oneaftertheother),soweshouldaddthe resistances of the two stepstogether. But first we have toaccount for there being Nseparate patches. Flow to the separate patches can occur in parallel, so we need to divide thediffusiveresistanceofthesecondstepbyN.Thetotaldiffusiveresistancebecomes

1/(4Da)+1/(N4Ds)=1/(4Da)(1+a/(Ns))Finally, we convert to a rate of capture by taking the driving concentration and dividing by thediffusiveresistance.

CapturerateforaspherewithNabsorbingpatches=C04Da/(1+a/(Ns))RecallthatthespherewhoseentiresurfacewasabsorbinghadacapturerateofC04Da.Therefore,wecanexpresstherelativeorfractionalspeedofcaptureforthespherewithpatches(incomparisontothefullyabsorbingsphere)as

fractionalcapturerate=1/(1+a/(Ns))Evidently,thecaptureratehasasymptoticbehaviorintermsofthenumberofpatchesN.Ifwecallthenumberofpatcheswhere the fractional capture rate is 50%,N50%, thenN50%=a/s. In other

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words,ifthespherehasaradiusthatisahundredtimesgreaterthantheradiusofthepatches,thenafewhundredpatchesspreadacrossthesurfaceofthespherewillachieve50%captureefficiency.Sowhatfractionofthesurfaceofthesphereisactuallycoveredbypatchesundertheseconditionsof50%captureefficiency? The areaof the sphere is4a2. The total areaoccupiedby the circularpatchesof radiusswouldbeN50%s2,whichaftersubstitutingN50%=a/swouldgive2as.So thefractionalcoverageofthesphericalsurfacewouldbe2as/4a2=s/(4a).Notethatthisisasmallnumberiftheradiusofthesphereislargecomparedtothesizeofthepatches.Asapracticalexample,ifthesphereisabacterialcellwithradius1um,andforthesakeofargumentwetakethesizeofatransporteronthecellsurfacetobeabout5Åindiameter,thenthefractionofthecellsurfacethatneedstobeoccupiedbytransporterstoreach50%maximumcapturerateis510‐10M/410‐6M=3.910‐4,whichislessthan1/10thof1%!Whatthisexerciseshowsis thatdiffusionhaspeculiarproperties, leadingheretoaphenomenonwheregoodcaptureefficiencyisachievedwithlowsurfacecoverage,andnotmuchextraadvantageisgainedbyincreasingthedensityoftransporterssignificantly–increasingthedensitybyafactorof9getsyouto90%efficiency.Andthereisofcourseasubstantialcostassociatedwithproducinglargequantitiesofcellsurfacetransporters,sothecellreachesapointwheretherearediminishingreturnsforsynthesizingmoretransporters.Asaninterestingcounterpoint,cellsurfacephenomenathatarenotdiffusion‐limitedshowstronglycontrastingbehavior.Photosyntheticantennaproteinsandotherproteins that generate energy by light absorption are sometimes very densely packed on cellsurfaces,sometimestothepointofformingtwo‐dimensionalproteincrystalsinthemembrane;lightabsorptiondoesnotobeythesamebehaviorasdiffusion.

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CHAPTER14

SedimentationvelocityEarlierwediscussedsedimentationinthecontextofanequilibriumsituation,wheretheexperimentwasrunessentiallytocompletion,toapointwherenofurtherconcentrationchangeswereoccurring,andtheexternalforceofcentrifugation(orgravity)cameintobalancewiththeopposingforceofaconcentrationgradient.Nowwewillconsideradifferentscenario.Inthelimitwemightimagineasituationwherethesampleisbeingspunatsuchahighspeedthatitwouldnearlyallbedriventothebottomofthetubeiftheexperimentwerecontinuedindefinitely. Nowinsteadofconsideringtheultimateequilibriumsituationwemightaskhowfast themacromoleculesaremovingdownwardduringtheexperiment.Thisbringsustoconsiderabalancebetweendifferentforces:thedownwardexternalforceduetocentrifugationandanopposingfrictionalforcelimitingthespeedofmovement.Wesawthefrictionalforcecomeintoplayinthepreviouschapter,inoppositiontoaconcentrationgradient.Wecandrawaschemethattiestogetherdifferentkindsofmeasurementsandexperimentswehavediscussedwheredifferentpairsofforcesareputintobalanceasshownbelow.Thearrowatthelowerrightisofinteresttousnow.

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Sedimentationcoefficient,sFrombefore,theexternalforce(onapermoleculebasis)duetocentrifugationism2r,whereisthedensityincrement,istheangularvelocity,misthemass,andristhedistancefromtheaxisofrotation. Theopposing frictional force is –fv. Setting the sumof forces to zero (e.g. at terminalvelocity)gives

v=m2r/forconvertingtoapermolebasis

v=M2r/(NAf)Howisthevelocityvvisualizedinavelocitysedimentationexperiment?Ifthesamplebeginswiththemacromoleculeuniformlydistributed(e.g.withtheconcentrationequaleverywhereinthetube),thenwhenthecentrifugationbegins(atveryhighspeedsaswediscussedabove),thenthetopregionofthesamplewillbegintobedepletedofmacromolecules.Inthelimitingscenario,therewillbeaneffectiveboundaryposition;atlowervaluesofrtheconcentrationofthemacromoleculewouldbenearlyzero,asdiagrammedhere:

Thesedimentationvelocity,v,isthespeedoftheboundary,i.e.(boundaryposition,r)/t.Meaningandmeasurementofthesedimentationcoefficient,sHowdoesthesedimentationvelocityvrelatetomolecularproperties?Wecanseefromtheequationabovethatvisaffectedbothbymolecularproperties(e.g.M)andbyexperimentalparameters(e.g.

). The behavior is clarified by separating the two kinds of variables on different sides of theequation.Then,

v/2r=m/f

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Nowwecanintroducethesedimentationcoefficientstobeequaltothosequantities.Inotherwords,sisobtainedexperimentallyass=v/(2r).Andsrelatestomolecularpropertiesaccordingtos=M/(NAf)ors=m/fTheadvantageofseparatingthevariablesinthiswayandassigninganewvariablesisclear.Ifweincreasetheangularvelocityinthecentrifugationexperiments,thesedimentationvelocitygoesupalso,butthesedimentationcoefficientisunaffected.Thismustbethecasesincetheequationaboveshows us that s can be written in terms of molecular properties alone without reference toexperimentalparameters.Asanaside,youcanseethatsinces=v/(2r),wecouldtrytoobtainavalueforsbymeasuringthesedimentation velocity v (i.e. the speed of the boundary) at some instantaneous point in theexperimentandthendividingby2r(usingthevalue forrof theboundaryat that instantaneouspoint).Butthisisabitsloppygiventhatvwillbedependentonr.Betteristonotethatsincevisdefinedasdr/dt,s=(dr/dt)/(2r)=(d(ln(r))/dt)/2.So,measuringthepositionroftheboundaryat a seriesof timepointsduring theexperimentandplotting themas ln(boundaryposition)vs tshouldgiveastraightlinewithslopes2,fromwhichscanbeobtained.Forconvenience,aspecialunitistypicallyusedtoexpressthevalueofthesedimentationcoefficients(whosenaturalunitsareseconds).TheSvedberg,S,isdefinedas10‐13sec.Youarelikelyfamiliarwiththisnotationfrommolecularbiologycourses. Forexample,youlearnedaboutthe50Slargesubunitoftheribosome;itsnameoriginatesfromitssedimentationcoefficient–it’ssvalueis5010‐13sec.RelatingstomolecularpropertiesThe sedimentation coefficient relates to molecular properties in two ways, through directdependence on mass and through the frictional coefficient f, which also depends on size (andthereforemass).Thisleadstoasomewhatcomplexdependenceofsonmass.Becausethefrictionalcoefficientdependsonsizeviaalineardimension(i.e.radius,R),fgoesasthe1/3powerofvolumeandmass.So,fromtheequationabove,s=m/f,weshouldexpectstodependonthe2/3powerofm.Combining

withStokes’equationforf(assumingasphericalshape),f=6R,

6

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ButRrelatestovolumeandmassaccordingto

43

and34

RelatingvolumeVtomassmbythedensityoftheproteinornucleicacid,m=V,andsubstituting,

634

13

whichgives

6 4 /3 or 6 4 /3

ThiscanofcoursebefurtherrearrangedtogiveMintermsofsraisedtothe3/2power.Dolargermoleculessedimentfasterorslowerthansmallermoleculesofequaldensityandsimilarshape? The centrifugal force on an object is proportional to itsmass, but the opposing force isproportionalonlytothe1/3powerofthemass,solargermoleculessedimentfaster,accordingtothe2/3poweroftheirmass.Theequationaboveforrelatingmasstothesedimentationcoefficientreliesontheassumptionofanearlysphericalshape,sincewewereforcedtoemployStokes’equationtorelatefrictiontosizeandmass. And as we discussed before, if the molecule of interest is highly non‐spherical, we maymisestimatethemassbysuchanapproach.Tobespecific,ifamoleculeishighlynon‐spherical,thancomparedtoasphericalmoleculeofthesamemass,itwillexperiencethesamecentrifugalforcebutagreaterfrictionalforce,resultinginasmallervalueofs.Fromtheequationaboveyoucanseethata lowervalueofs for thenon‐sphericalmoleculewould lead toanerroneously lowvalue for theestimatedmolecularweight.CombiningsandDtogetmolecularweightwithoutasphericalassumptionWecanfreeourselvesfromtheassumptionsofasphericalshapeifwehavemeasuredvaluesforsandDtogether.Dandsbothhadrelationstothefrictionalcoefficient,butifwehavevaluesforsandDwecancancelfoutandavoidtheneedtoobtainanexpressionforfintermsofasphere.Fromour

previouschapter,f=kBT/D,andfromabove,f=M/(NAs).SettingkBT/D=M/(NAs),weget

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SothemolecularweightrelatessimplytotheratioofstoD,regardlessofshape.Furthermore,ifweobtainavalidvalue forM fromsandD together,wehave theopportunity toevaluate theshapeproperties,forexamplebycheckingtoseehowcloselythevalueforf(whichwecancalculatedirectlyfromD)matchesthevalueyouwouldexpectforfforamoleculewithmassMifitwasindeedasphere(usingStokes’equation).Earlierwediscussedvariouswaysofmeasuring the diffusion coefficient,but in fact information about D cantypically be obtained from thesedimentation experiment itself. Inour initial discussions we imaginedthat the boundary in theconcentration profile would beperfectlysharp. Butinfactdiffusionwouldbeoccurringatthesametimeas sedimentation, thereby causingsomespreadingoutattheboundary,andtoaconcentrationprofilethatisnot infinitely sharp and steep at theboundary. Therefore, a moreadvanced mathematical treatmentcanextractinformationaboutDfromthe shape of the sedimentationprofiles.Asummaryofmolecularweight determinationfromsedimentationanddiffusionmeasurements

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CHAPTER15

ChemicalReactionKineticsIn this chapter we discuss the rates of chemical reactions, focusing on the meaning of reactionvelocity, its dependence on the stoichiometric order of a reaction, and on the time‐dependentbehaviorofreactantandproductconcentrations.Reactionvelocity,vIfweconceiveareactiongenerallyasreactants productsthenthereactionvelocityvisthefrequency(#/time)withwhichtheeventdescribedbythereactionarrow is occurring per unit volume. The units of v are therefore #/(volumetime), which isconcentrationper timeorM/sec. Consistentwith thoseunits, one can see that (as long as otherreactionsarenotsimultaneouslyproducingorconsumingthesamereactantsand/orproducts)thereaction velocity is directly reflected in the rate of change of the concentration of reactants andproducts.Morespecifically,notethatthereisbutonevelocityassociatedwiththereaction,thoughitmaybethatmultiplereactantsarebeingconsumedandmultipleproductsarebeinggenerated.The reaction velocity is indicated equally by the rate of change of any of the species involved.However, the stoichiometric coefficients associated with the reactants and products must beaccountedforcarefully.Ifforinstanceachemicalspecieshasassociatedwithitastoichiometryof2,then each reaction event corresponds to a consumption or production of twomolecules of thatspecies.Therefore,forthegeneralreaction:

A + B + … C+D+…

‐d[A]/dt=v‐d[B]/dt=vd[C]/dt=vd[D]/dt=vandsoonforanyandallspecies.Alternatively,v=‐(1/)d[A]/dt=‐(1/)d[B]/dt=(1/)d[C]/dt=(1/)d[D]/dt=…Evidently, ifwemeasure the rateof changeof the concentrationof some species involved in thereactionthenwehavemeasuredthereactionvelocityv,assumingwehaveproperlyaccountedforthestoichiometry.

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Ratelaws:howvdependsonconcentrationsThevelocityofareactionnaturallydependsonhowconcentratedthereactantsare;ifthenumberofreactantmoleculesinaunitvolumeisvanishinglysmallthensurelywecanexpectthefrequencywithwhichweobservethemolecule inquestionundergoingreactionevents in thatvolumetoalsobeeffectivelyzero.Iftheconcentrationsarehigher,thenthereactionvelocitywillbehigher.Besidesthedependenceonconcentration,differentreactions(i.e.involvingdifferentchemicalspecies)willhavedifferentreactionvelocitiesaccordingtothelikelihoodoftheunderlyingchemicalevents.Thisnaturallikelihoodofareactiontooccuriscapturedbyarateconstantk.Thecombineddependenceofareactionvelocityvontherateconstantkandtheconcentrationsisreferredtoasaratelaw.Theratelawcanbecomplicatedforcomplexreactionschemes,butforreactionsthatrepresentsimpleindividualchemicalevents,thedependenceofvonconcentrationscanbewrittenbyinspection.Suchreactions are sometimes referred to as ‘elementary reactions’, and are to be distinguished fromscenarioswhereawrittenreactionactuallydescribesthenetstoichiometricresultarisingfrommorethanonereaction,e.g.twooperatinginsequence.Forasingleelementarystepoftheform:

A Bwherekistherateconstant,theratelawisv=k[A]and[A]istheconcentrationofspeciesA.SuchareactionissaidtobefirstorderinA.Forareactionoftheform

2A Bv=k[A]2andthereactionissaidtobesecondorderinA.Forareactionoftheform

A+B Cv=k[A][B]andthereactionisfirstorderinAandfirstorderinB,andsoon.Themultiplicativeorhigherorderdependenceofthereactionvelocityincaseswheremultiplemoleculesarereactingatthesametimeisareflectionofthejointprobabilitythatbothreactingmoleculesarepresenttogether,collidingwitheachother.Notethatinourdiscussionsofkineticswewillusebracketstodenoteconcentrationstobemostconsistentwithcommonusage(insteadofusingCiaswedidinearlierchapters).

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RelationshipofrateconstantstoequilibriumconstantsForreactionsdrawnasabove,thesingleforwardarrowindicatesanirreversibleprocesswherethecombinedfreeenergiesofthereactantsaresomuchhigherthanfortheproducts,thatreactioneventsinthereversedirectioneffectivelyneveroccur.Suchreactionsgotocompletion,withoutresidualreactants.Incontrast,forreactionswheretheenergeticsonthetwosidesaremorenearlybalanced,reactioneventscanoccur inbothdirections. Thevelocitiesof the forwardandreversereactionsdependontheconcentrationsofthereactantsandproducts,respectively.Andwhenconcentrationsarereachedwheretheforwardandreversevelocitiesareequal,thennonetconversionisoccurring(thoughconversionsare in factoccurring inbothdirections). This iswhat ismeantbychemicalequilibrium.Thisnotiongivesusanimportantrelationshipbetweenrateconstantsandequilibriumconstants.Forthereaction2ABwhere k1 is the forward rate constant and k‐1 is the reverse rate constant, the forward reactionvelocitywouldbek1[A]2,whilethereactionvelocityinthereversedirectionwouldbek‐1[B]. Theequilibriumconditioniswherethosetwovelocitiesareequal,givingk1[A]2=k‐1[B](atequilibrium)andk1/k‐1=[B]/[A]2(atequilibrium)Evidently,theratioofrateconstantsk1/k‐1isequaltotheequilibriumconstantK.Thisisageneralresult.IntegratingratelawsForsimplereactionschemesitisoftenpossibletointegratethedifferentialequationsthatcomefromtheratelawinordertoobtainacompletedescriptionofhowtheconcentrationsofthereactantsandproductschangeovertime.Wewillworkouttheresultsforfirstorderandsecondorderreactions:1storderdecay

A B

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Inordertogettoadifferentialequationintermsof[A],wecombinetwopoints.Firstisthedefinitionofthevelocityintermsoftherateofchangeof[A],v=‐d[A]/dt.Secondistheratelawthatdescribesthedependenceofthevelocityon[A],v=k[A].Togetherthesegive,

ln | | whichgivesthefamiliarfirstorderdecayequations

ln

and

Thebehaviorof[A]overtimeisexponentialandthebehaviorofln[A]islinearwithtime,withaslopethatgivestherateconstantk.

Describingdecaytimesfor1storderdecay

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The time scale of first order decay is oftendescribedintermsofahalf‐life,t1/2.Thisisthetime required for a reaction to go from somegivenconditionsto50%completion.Aslightlydifferent parameter, , is sometimes used todescribedecaytimes.Itgivesthetimerequiredfor a reaction to go to a degreeof completionthat is 1/e compared to the initial condition.That is, [A]/[A]0 = 1/e. The relationshipbetweent1/2andisobtainedbycomparing

ln(1/2)=‐kt1/2toln(1/e)=‐1=‐kgivingt1/2=ln(2)Forthesimplefirstorderdecayreactionof[A]above,t1/2=ln(2)/kand=1/kNote that thephysical interpretationof isslightlymorecomplexthant1/2,butgivesasimpleralgebraicrelationshiptotherateconstant.Wewillseelaterthatinmorecomplexkineticschemeswesometimesgetfirstorderequationswithmorecomplexexpressionsintheexponentterm.Butisalwayssimplyrelatedtotheexponent(whichmultipliestime)byareciprocalrelationship.Thatis,if

someexpression then

1someexpression

and

ln 2someexpression

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Integratedratelawfora2ndorderirreversiblereaction

2A BAgainwecombinetwoexpressionsinvolvingthevelocityv,onethatdefinesvintermsoftherateofdisappearance of the substrate (v = (‐1/2)d[A]/dt) and the other a rate law that describes thedependenceofvonthesubstrateconcentration(v=k[A]2).Combined,thesegive

2

2

1 | |

1 1

2

In this case, a plot of 1/[A] versus time gives astraightlinewhosesloperelatestotherateconstantk.Otherirreversiblereactionsofhigherordercanbeintegratedeasilyinasimilarfashion.EstablishingaratelawfrommeasuredreactionvelocitiesWe saw above how different rate laws give a different time evolution for the concentration ofreactants.Andwhetherareactionfollowsafirstorderorsecondorder(orsomeother)ratelawcanthereforebeexaminedbyplottingln[A]or1/[A]asafunctionoftimeandcheckingtoseeiftheresultis a straight line. A different way of experimentally examining a rate law is by evaluating thedependence of reaction velocity on concentrations. Measuring initial reaction velocities underdifferent concentrationsmakes it possible to determinewhat exponents are associatedwith thereactantconcentrations. If a reaction is firstorder in [A], thenvwilldepend linearlyon [A] (i.e.doubling[A]willdoublethereactionvelocity). Likewise,ifareactionissecondorderin[A],thendoubling[A]willquadruplethevelocity,andsoon.Somecomplicatedreactionschemescanshownon‐trivial dependence on concentration, evennon‐integer exponents. As a general approach toestablishinganexponent,if

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v=[A]thenforratemeasurementsmadeattwodifferentconcentrations,

ln(v2/v1)=ln([A]2/[A]1)BehaviorofmorecomplexreactionschemesWeareveryofteninterestedinthebehaviorofkineticschemesinvolvingmorethanoneindependentreactionevent.Complexitycanariseindifferentforms.Thetworeactionsmayeffectivelyoperateinsequence,withtheproductofthefirstreactionbeingthereactantinthesecondreaction.Ortworeactionsmaybeoperatingonthesamespecies,givingaschemethatisbranchedratherthanlinear.Regardless of the complexity of the reaction scheme being proposed, setting up the underlyingdifferentialequationsisgenerallystraightforward.Asanexample,

givesthefollowingequations:d[A]/dt=‐k1[A],d[B]/dt=k1[A]–k2[B],d[C]/dt=k2[B]Noteherethattherateofchangeof[B]involvestwoterms,therateatwhichitisbeingformedminustherateatwhichitisbeingconsumed.Foranotherexample,

d[A]/dt=‐k1[A]‐k2[A]=‐(k1+k2)[A],d[B]/dt=k1[A],d[C]/dt=k2[A]SteadystateassumptionsforobtainingsimpleratelawsforcomplexreactionsComplexreactionschemesgenerallyhavecomplexbehavior,includingnon‐trivialequationsforthetime dependence of the reactants and products. And the dependence of the rate of the overallreaction(i.e.theratelaw)candependontheconcentrationsofspeciesthatdonotcontributetotheoverallreaction.Sometimesacompletedescriptionofthebehaviorcanbeobtainedbysolvingthefullsystemofdifferentialequations,butthismaybedifficult. However,especiallyincaseswhere

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sequentialreactionsareinvolved,simplifiedratelawscanoftenbeobtainedbyassumingsteadystateconditions. Steadystatereferstoconditionswherethereareintermediatespecies(whichdonotcontribute to the overall reaction stoichiometry)whose concentrations have reached a constantvalue,atleastmomentarily.Inotherwords,onecanworkoutasimplifiedexpressionfortheoverallreactionvelocityunderconditionswhered[Intermediate]/dt=0.Asanexample,consider

Here,theintermediateisB;itdoesnotcontributetotheoverallreactionstoichiometryA→C.Firstwewriteoutanexpressionforthechangein[B],basedonelementaryratelawsforallthestepsinwhichBisformedorconsumed,andthenwesetthatderivativetozeroaccordingtothesteadystateassumption.d[B]/dt=k1[A]–k‐1[B]–k2[B]=k1[A]–(k‐1+k2)[B]=0Nowwerearrangetogetanexpressionfor[B]thatwecanuse forsubstitutionsubsequently. Atsteadystate,[B]=k1[A]/(k2+k‐1)Nowwecangobacktotheoriginalreactionschemeandwriteanexpressionfortheoverallvelocity.Theoverall reactionvelocity couldbedefinedasv=d[C]/dt. Thenwecanwrite anequation ford[C]/dtintermsofelementaryratelaws.Here,d[C]/dt=k2[B].Nowwesubstitutefor[B]atsteadystatefromabovetogetv=k1k2[A]/(k2+k‐1)Underthistreatment,the2‐stepreactionbehavesasfirstorderin[A]atsteadystate.Thisparticularreactionschemeshowsupoften,includingintreatmentsofenzymekinetics,aswewillseelater.NumericalcomputersimulationofmorecomplexreactionschemesIncaseswherecompletesolutionsaredifficulttoobtainbysolvingdifferentialequations,andwhereapproximationslikesteadystateareundesirable,onecanalmostalwayssimulatethebehaviorofacomplexreactionschemeusingsimplecomputerprograms.Thekeyistotreatthetimederivativeoftheconcentrationofeachspeciesastheratioofaverysmallchangeinconcentrationoveraverysmalltimeincrement.Forexample,intheschemeaboveΔ[A]/Δt=‐k1[A]+k‐1[B]and

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Δ[A]=(‐k1[A]+k‐1[B])ΔtArelatedequationcanbewrittenforeachspecies.Tomakethecomputersimulationgo,onesimplyassignsstartingvaluestotheconcentrationsofallthespecies,andthenupdatestheconcentrationsofallthespeciesinaseriesofverysmalltimestepsonthebasisofequationsliketheoneabove.Anexampleofcomputercode(inthePythonprogramminglanguage)isshownbelowforsimulatingthebehaviorofthekineticschemeabove.Theinitialconcentrationsare[A]=1M,[B]=0,[C]=0.

# Set up arrays to hold concentrations for 500 time steps A = [0.0 for n in range (0,500)] B = [0.0 for n in range (0,500)] C = [0.0 for n in range (0,500)] # Assign initial concentrations A[0]=1. B[0]=0. C[0]=0. # Assign rate constants for the simulation k1=500. kminus1=400. k2=300.

# Choose a time interval small enough so that concentration # changes in each step will be small. timestep = 0.00005 # Set up the loop over time. Here, after 500 steps, the total # time elapsed would be 500 * 0.00005 = 0.025 seconds # Apply kinetic equations to update the concentrations in each step. # The index for the time step is specified in brackets. for nt in range (1,500): A[nt] = A[nt-1] + timestep*(B[nt-1]*kminus1 – A[nt-1]*k1) B[nt] = B[nt-1] + timestep*(A[nt-1]*k1 – B[nt-1]*kminus1 - \

B[nt-1]*k2) C[nt] = C[nt-1] + timestep*(B[nt-1]*k2) print (nt*timestep, A[nt], B[nt], C[nt])

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Theresultofthatsimulationisshownhere.

Enzymekineticsunderasteady‐stateassumptionThefollowingmodelisoftenusedtotreatthekineticsofasimpleunimolecularenzymereaction:

Here,Eisthefreeorunboundenzyme,Sisthefreeorunboundsubstrate,ESistheenzyme‐substrate(orMichaelis‐Menten)complexbetweentheenzymeandsubstrate,andPistheproduct.Theratioofk1tok‐1describeshowtightlytheenzymebindsthesubstrate,whilekcatdescribestheunimolecularcatalyticrateconstantforconversiontoP.Thetreatmentofthisenzymemodelatsteadystatedatestothe1920’sbyBriggsandHaldane.Thevelocityoftheoverallreactionisdescribedbyv=d[P]/dt,andaccordingtotheratelawforthiselementarystep,v=kcat[ES].But[ES]isanintermediate,andtoobtainanequation forv in termsofspecies thatcontribute to theoverallstoichiometryof thereaction,weneedtoreplace[ES].Ifweadoptasteadystateassumptionwecanobtainanexpressionfor[ES].Takingaccountofallthestepsinwhich[ES]isformedorconsumed,andthensettingthisto0,d[ES]/dt=0=k1[E][S]–(k‐1+kcat)[ES]

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whichgives[ES]=k1[E][S]/(k‐1+kcat)Thensubstitutiongivesv=kcatk1[E][S]/(k‐1+kcat)Thisequationisvalidbutnotveryinsightfulinthesensethatitdescribesthereactionvelocityintermsofthefreeenzymeconcentration;inanexperimentalset‐uponetypicallyhascontroloverthetotalenzymeconcentrationbutnotthefreeenzymeconcentration,whichisclearlyafunctionofhowmuch substrate is present. To gainmore insight, the standard approach is to recast the kineticequationsintermsofthetotalenzymeconcentrationandintermsoftheratioofthereactionvelocitytoitsmaximumpossiblevalue(i.e.whenalltheenzymeisinthe[ES]formsothat[ES]=[E]total).Themaximum velocity is kcat times the maximum possible value for [ES], which is kcat[E]total =kcat([ES]+[E]).Then,v/Vmax=kcat[ES]/(kcat([E]+[ES]))=[ES]/([ES]+[E])Thisissensible.Itsimplystatesthatthevelocityintermsrelativetothemaximumisgivenbythefractionoftheenzymethatisinthe[ES]form.Tosimplifytheexpressionfurtherwecandividethetopandbottomby[ES]togivev/Vmax=1/(1+[E]/[ES])Then we can take the previous equation, [ES] = k1[E][S]/(k‐1 + kcat), and rearrange it to get anexpressionfor[E]/[ES]=(k‐1+kcat)/(k1[S]),whichwecansubstituteintheequationabovetogivev/Vmax=1/(1+(k‐1+kcat)/(k1[S]))Multiplyingthetopandbottomby[S]givesv/Vmax=[S]/([S]+(k‐1+kcat)/(k1))ThishastheformofthefamiliarMichaelis‐Mentenequationv/Vmax=[S]/([S]+KM)wheretheMichaelis‐MentenconstantKMcanbeseentobe(k‐1+kcat)/(k1)forthiskineticmodel.Theequationabovecanalsobeconvertedfromfractionalvelocitytovtogivev=kcat[E]total[S]/([S]+KM)

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Fromtheformofthisequationyoucanseethatthebehaviorishyperbolic,withvapproachingVmaxasymptoticallyas[S]getsmuchhigherthanKM,andv/Vmax=1/2at[S]=KM.Relaxationkinetics:howsystemsapproachequilibriumFastreactionsaredifficulttostudyexperimentally.Arapidreactionmayproceedtocompletionsorapidlyonceitisinitiatedthatmeasuringconcentrationchangesisproblematic. Forbi‐molecularreactions, the reaction may be faster than the time required to mix the reactants prior tomeasurement. Problems of that type can sometimes bemitigatedwith instruments designed toachieverapidmixing.So‐calledstopped‐floworcontinuous‐flowinstrumentsdeliverreagentsfromtwo separate syringes into a small mixing chamber where spectroscopic readings can be takenimmediately.Thedeliveryofreagentsoccursinasinglebolusafterwhichthedeliveryofreactantsis stopped while the time course of the reaction is monitored. In a variation referred to ascontinuous‐flow,deliveryof reactants continuesand the reactingmixture flowsdowna capillarytube. In this set‐up, distance along the capillary corresponds directly to time elapsed since thereactantsencounteredeachother,sothatconcentrationmeasurementsatdifferentpositionsgiveaneffectivetimecourse.Thedisadvantageofcontinuous‐flowmethodsisthatlargeamountsofmaterialmayberequired.Ontheotherhand,microfabricationtechniquesformakingverysmallsystemsforfluid handling are now fairly commonplace. Microfluidic devices can be custom designed andfabricated from transparentpolymers (typicallyPDMS),withverymany chambers and transportcapillarieswithflowcontrolledbycomputersandpressuresensitivevalves.Thishasgivenrisetothenotionofa‘lab‐on‐a‐chip’,withthousandsofreactionsbeingmonitoredinapieceofpolymerthesizeofacreditcard.Thesekindsofdevicesmakeitpossibletostudymanyreactionsandreactantconditionswithverylittlelossofmaterial.Anothercategoryofkineticanalysisaimstostudyhowrapidlysystemsapproach(or‘relax’towards)equilibriumafter theyhavebeen(nearly instantaneously)perturbed fromequilibrium. So‐calledrelaxationmethods circumvent themixing problem in the sense that the system under study isalreadymixed.Howcanasysteminwhichallthereactingsubstratesandproductsarepresentandequilibraterapidlybebroughttoapointwhereitisnotatequilibriumsothatthespeedofapproachtoequilibriumcanbestudied?TheT‐jump(fortemperaturejump)methodwasdevelopedinthe1950’sbyManfredEigentostudyfastreactionsandtheirrelaxationbacktoequilibrium.Theideaisthatifenergyisrapidlydeliveredtoasolutioncontainingareactantandproductatequilibrium,forexamplebydischarginganelectricalcapacitor(orinlaterdevelopmentsusingalaserpulse),thenthetemperatureofthesystemcanbeheatednearlyinstantaneously.Now,recallingthevan’tHoffequation, if H for the reaction under study is non‐zero, then the equilibrium constant will bedifferent at the new temperature. So by increasing the temperature suddenly, the system haseffectively been perturbed from its previous equilibrium, not by changing the concentrations ofreactantsandproducts,butbychangingtheequilibriumposition;thesystemismaintainedatthenewtemperaturewhilethesystemapproachesthenewequilibriumconcentrationsbyconversionofreactanttoproductorvice‐versa.Howfastthesystemapproachesequilibriumclearlydependson

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the forward and backward rate constants of the reaction. The mathematics for how systemsapproachequilibriumrevealssomegeneralprinciples.ThesimplestsystemtoconsiderisAinterconvertingtoB.Firstwenotethatbecausethereisjustoneconversioninthissystemthatitmustbepossible,givenanyconcentrationvaluesforAandB,todescribe how far the system is from equilibriumwith a single concentration variable. Call thisdistancefromequilibriumx.Ifweintroduceashorthandnotationof torepresenttheequilibriumconcentration of A (at the new temperature) and similarly for , then we can relate theconcentrationsofAandBtotheirequilibriumvaluesplusorminusx,by[A]= +xand[B]= ‐x.

Nowwecanexaminetheapproachtoequilibriumbywritinganequationforthetimedependenceofx.First,wenotethatd[A]/dt=d( +x)/dt=dx/dt.Thenwecanwriteanexpressionford[A]/dtas–k1[A]+k‐1[B]=–k1( +x)+k‐1( ‐x)=k‐1 ‐k1 –x(k1+k‐1). Thetermk‐1 ‐k1 canbeseentobeequaltozerobecause / =K=k1/k‐1.Droppingthosetermsgivesdx/dt=–(k1+k‐1)xEvidently, x (the distance fromequilibrium) follows first orderkinetics. Skipping the familiardetails for handling a first orderdifferentialequation,

andln(x/x0)=‐(k1+k‐1)tWecanfurtherconvertthesetothegeneralforms

/

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and

ln(x/x0)=‐t/where in this case = 1/(k1+ k‐1).Assuming one can measure theconcentrationofAorBasafunctionoftime, then x as a function of time isknown,sincex=[A] = [B].Thisallowsmeasurementof(e.g. fromthereciprocaloftheslopeof ln(x)vst),sothatthevalueof(k1+k‐1)isobtained.Ifthe equilibrium constant K for thereaction is knownalso, thenk1 andk‐1can both be obtained from values of andK.

1/=k1+k‐1andk1=Kk‐1,so1/=Kk‐1+k‐1=(K+1)k‐1

k‐1=1/((K+1))andk1=K/((K+1))HigherorderreactionsapproachingequilibriumItisnotsurprisingtoseethatthefirstorderreactionaboveapproachesequilibriuminafirstorderfashion. However,wecanshowthatmorecomplexreactionsalsoapproachequilibriumina firstorderfashionastheycomeclosetoequilibrium.Considerthissecondorderreversiblereaction:

Again,thereisjustonetransformationsothedistancefromequilibriumcanbedescribedbyasinglevariablex,andtheconcentrationsatanypointintimecanbeexpressedintermsofxandtheeventualequilibriumconcentrations.Followingthesameapproachasbefore,d[A]/dt=d( +x)/dt=dx/dt.Andd[A]/dt=–k1( +x)( +x)+k‐1( ‐x)=k‐1 ‐k1 –x(k1( + )+k‐1)–k1x2.Thefirsttwoterms,k‐1 ‐k1 ,canceltozero,andifwearecloseenoughtoequilibriumthenxwillbesmallandwecanneglectthex2term.Then,dx/dt=–(k1( + )+k‐1)x

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Therefore, the distance from equilibriumx shows first order behavior close to equilibrium,with=1/(k1( + )+k‐1)KineticsfromsinglemoleculestudiesRecent developments in instrumentation have made it possible to perform a variety ofmeasurements on single molecules. The kinetics of individual molecules undergoing chemicalreactions or conformational transitions can be analyzed, but when working with individualmoleculeswehavetolookatthingsinawaythatdoesnotinvolveconcentrations.Foraunimolecularevent,wecangetasensefortherateconstantbylookingathowlongthemoleculepersistsinitscurrentstatebeforeundergoingareaction.Thinkofthisasawaitingtimebeforeareactioneventorconformational change occurs. Reaction events are always stochastic (i.e. having a randomcharacter),butifwemeasurethewaitingtimeforseveralindependentreactioneventsweshouldbeabletorelatethattoanunderlyingrateconstant;theaveragewaitingtimeshouldbeshorterforaprocesswith a higher rate constant. Consider the irreversible conversion of A to Bwith a rateconstantk.Onaverage,howlongshouldweexpecttowaitbeforeanygivenmoleculeofAconvertstoB?Wecanworkouttherelationshipbystartingwiththeequationfortreatingthereactioninbulk:A/A0=exp(‐t/).Inthiscase=1/kbutwewillkeeptheequationintermsofforgenerality.OnewaytolookatA/A0istoseeitastheprobabilitythatanygivenmoleculeofAhasnotreactedbeforetimet.FromtherewecanseethattheprobabilitythatamoleculeofAwillreactpreciselyattimetisthederivativeofthatexpressionwithrespecttot.Differentiating,andcorrectingforthenegativesign, theprobability thatmoleculeA reactspreciselyat time t is (1/)exp(‐t/). Then, toget theaveragetimeatwhichamoleculeofAreacts,whichhasthesamemeaningasthewaitingtime,weneedtogettheaveragevalueoftbyweightingallpossiblevaluesoftbytheprobabilityofreactionattimet.Wegetthisbymultiplyingtbytheprobabilityofreactionattimetandintegratingfromt=0toinfinity. From t((1/)exp(‐t/)dtevaluatedfromt=0toinfinity(whichrequiresintegrationbyparts),weget<waitingtime>=.Thissimpleresultmakessensesincethedecayorrelaxationtimeisageneraldescriptionofthetimescaleofafirstorderreaction.Theresultabovemeansthatifweevaluatehowlongittakesasinglemoleculetoundergoatransition(preferablymakingthetimemeasurementseveraltimes),thenwehaveeffectivelymeasured.Andhere,k=1/.Thefiguresbelowillustratethreedifferentkindsofexperimentswheresinglemoleculestudieshavebeenusedtomeasuretheratesofconformationalconversions.Thefirstisanexampleofavoltagemeasurementacrossacellusingthepatch‐clampmethod,wherethevoltagedependsonwhether an ionchannel is in theopenor closed conformation. The secondexample illustratesaspectroscopicmeasurementwheretheoutputsignaldependsontheconformationofafluorescentlylabeled ribosome, which is interconverting between two states. The third example illustrates areversiblewinding‐unwindingtransitioninasingleDNAmoleculewhoseendsarebeingpulledapartgently. Inallthreecasesyoucanseehowthedatacouldbeinterpretedintermsofwaitingtimesbetweentransitions.Infact,youcanseethatitshouldbepossibletogettherateconstantsforboththeforwardandreversetransitionsbymeasuringthewaitingtimesinbothstates.Theratioofthose

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wouldbetheequilibriumconstant,andasexpectedthismatchestheratiobetweentheaveragetimethemoleculespendsinthetwoconformations.

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CHAPTER16

KineticTheoriesandEnzymeCatalysisWehavediscussedthereactionvelocityandratelaws,butsofarwehavenotsaidanythingaboutwhatdeterminestherateconstant,k.Whatmakessomereactionsintrinsicallyfastandsomeslow?Given the variety of reactions that occur in nature – with vast differences in speed, number ofreactantsinvolved,typesofbondsformed,etc.–itisnotsurprisingthatdifferentmodelshavebeendevelopedtoexplainthemechanismsofchemicalreactionsandtheirrateconstants. TwowidelydiscussedmodelsareduetoArrheniusandEyring.TheArrheniusequationThe Arrhenius equation is often used to discuss reaction rates in terms of molecular collisions.AccordingtotheArrheniusequation,arateconstantkisdeterminedby

Aisa‘frequencyfactor’andEaisanactivationenergy.Thefrequencyofcollisionsinareactionisclearly dependent on concentrations, and that dependence is already built into the equation forreactionvelocity;e.g.forX+Y→Z,v=k[X][Y].Thefrequencyfactorintheequationforkthereforeembodies other phenomena, such as the dependence of molecular velocities and consequentlycollisionratesontemperature,andthedependenceofreactionprobabilityontheorientationofthecollidingmolecules.Theactivationenergy,Ea,describesalowerboundfortheenergythatreactantsmusthaveifreactionistooccur.WhydoesEaentertheequationforkasanexponentialterm?ThisfollowsdirectlyfromtheBoltzmanndistribution.IfweexpressthenumberofmoleculesN(E)thathaveenergyEaccordingtotheBoltzmanndistribution(N(E)exp(‐E/RT)),wecanevaluatethefractionofmoleculeshavingenergyatleastashighassomefixedenergyvalueEabytakingtheratiooftheareathatfallsunderthecurveandhasEgreaterthanorequaltoEa,dividedbytheentireareaunderthecurve.

|| 1

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This explains the exponential term intheArrheniusequation.Akeyelementof the Arrhenius equation is that therate constant depends strongly on theheight of an energy barrier. It alsodepends on temperature. In fact thedependence of k on T can be used toevaluate the activation energy in theArrheniusmodel.Thefrequencyfactorintroduces some dependence of k ontemperature vis‐à‐vis molecularvelocities,butthemaindependenceofkonTisthroughtheexponentialterm.

≅

or

1 ≅ 1

EyringtransitionstatetheoryEyring transition state theory provides a slightly differentway of looking at things that ismoreexplicitabouttheoccurrenceofhighenergyspeciesduringasinglereactionevent. IntheEyringmodel,asinglereactionstep,forexample

isreimaginedintermsoftwosteps.Inthefirststep,anunstablehighenergyspeciesreferredtoasthetransitionstateisformed,Thetransitionstatebreaksdowntoproductinthesecondstep.The‘double‐dagger’symbolindicatesthetransitionstate.

Therateconstantforbreakdownofthemaximallyunstabletransitionstateisapproximatedtobethefrequencyofmolecularvibrations,whichfromquantummechanicsisontheorderorkBT/h,wherehisPlanck’sconstant. Withthatsubstitution,thevelocityofthereactionschemeabovewouldbev=d[C]/dt=(kBT/h)[AB‡].Aswedidinourearliertreatmentsofmultistepreactions,weneedtomakesome assumption ifwewant to express the velocity in terms of reactants that contribute to the

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stoichiometry. Here, ifweassumethatfirststepdescribingformationofthetransitionstateisatequilibrium, then k+‡/k–‡=K‡=[AB‡]/([A][B]), so [AB‡]= K‡[A][B]. Making that substitution, thevelocityofthereactionwouldbev=(kBT/h)K‡[A][B].Nowifwecomparethisexpressionforvtothesimpleratelawwewouldwriteforasingleelementaryreactionstep,namelyv=k[A][B],wecanseebymatchinguptermsthattheEyringmodelgivesk=(kBT/h)K‡astheexpressionfortherateconstantofthereaction.Inthismodel,therateconstantisdeterminedlargelybytheequilibriumconstantK‡forformingthetransitionstate.WecanalsowriteK‡intermsofthefreeenergyforreachingthetransitionstate,G‡.K‡=exp(‐G‡/RT).Thatsubstitutionwouldgivek=(kBT/h)exp(‐G‡/RT)ThisresultdiffersindetailfromtheArrheniusequation,butthesimilarityintermsofanexponentialdependenceonanenergybarrierisclear.Wecanlookatthetemperaturedependenceofln(k)fortheEyringequationinthesamewasaswedidfortheArrheniusequation.Themultiplicativefactoratthebeginningoftheexpressionintroducesaminordependenceontemperature,whichwewillsetasideinordertolookatthemaindependence.d(ln(k))/dTd(‐G‡/RT)/dT=d(‐H‡/RT+S/R)/dT=H‡/RT2ComparingtoourearlierresultwiththeArrheniusequation,wecanseethattheactivationenergyintheArrheniusequationrelatescloselytothetransitionstateenthalpyintheEyringmodel.CatalysisbyloweringthetransitionstateenergyTheEyring transition statemodelprovidesaway to lookat catalysis in termsof transition stateenergies.Considerthereactionofasubstratetoformaproductineitheranuncatalyzedreactionoracatalyzedreaction.Lettherateconstantfortheuncatalyzedreactionbekuncatandtherateconstantforthecatalyzedreactionbekcat.Wecandrawthetworeactionsonthesameenergydiagramandconsidertheeffectofloweringthetransitionstateenergyinthecaseofthecatalyzedreaction.FromtheEyringequationfortherateconstantwecanwriteouttheratioofthetworateconstants.

cat

uncat

cat‡

uncat‡

cat‡

uncat‡

‡cat

‡uncat

‡cat

‡uncat

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In other words, if a catalyst lowers thetransitionstateenergyforareactionbyanenergy that amounts to 10RT, then thereactionwillbespedupbyafactorofe10,whichisabout22,000.But how is it that a catalyst lowers thetransitionstateenergyofareaction?Thisquestionwasconsideredinthecontextofenzymecatalysisasearlyas1948byLinusPauling. His description of howenzymesmust operate (which came more than adecadebeforetheatomicstructureswereknown for anyproteins or enzymes)wasextraordinarily prescient. According toPauling:

Ibelievethat…thesurfaceconfigurationoftheenzymeis…complimentarytoanunstablemoleculewithonlytransientexistence–namelythe“activatedcomplex”forthereactionthatiscatalyzedbytheenzyme.Themodeofactionofanenzymewouldthenbethefollowing:theenzymewould show a small power of attraction for the substratemolecule ormolecules,whichwouldbecomeattachedtoitinitsactivesurfaceregion.Thissubstratemolecule,orthesemolecules,wouldthenbestrainedbytheforceofattractiontotheenzyme,whichwouldtend to deform it into the configuration of the activated complex, forwhich the power ofattraction by the enzyme is the greatest. The activated complex would then, under theinfluenceofordinarythermalagitation,eitherreassumetheconfigurationcorrespondingtothereactants,orassumetheconfigurationcorrespondingtotheproducts. Theassumptionmadeabovethattheenzymehasaconfigurationcomplementarytotheactivatedcomplex,andaccordinglyhasthestrongestpowerofattractionfortheactivatedcomplex,meansthattheactivationenergyforthereactionislessinthepresenceoftheenzymethaninitsabsence,andaccordingly,thatthereactionwouldbespeededupbytheenzyme.

Furtherinsightcanbeaddedbydrawingakineticdiagramthatrelatesthebindingevents in thepresenceofanenzyme totheformationofthetransitionstate.Thereactions across the top describereaction in the absence of the enzymewhile the reactions across the bottomdescribereactioninthepresenceoftheenzyme. Following our previousequations from the Eyring theory, theratiobetweentherateconstantsinthosetwocaseswouldbe

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cat

uncat

ES‡

S‡

Byprovidingabindingsurfacethatiscomplimentarytothetransitionstateformofthesubstrate,theequilibriumconstantforreachingthetransitionstateisincreasedbymassaction,andaccordingtotheequationabovethisspeedsupthereaction.ThesituationcanalsobeviewedintermsofbindingaffinitiesoftheenzymeforScomparedtoS‡.Thosestepsaredescribedbytheverticalreactions.Bycompletingthethermodynamiccycleinthefigure,weknowthatK‡SKbindingS‡=KbindingES‡K‡ES.Theratioaboveforhowmuchareactionisspedupisthen

cat

uncat

ES‡

S‡

bindingS‡

bindingS

Inthisview,anenzymespeedsupitsreactionbybindingexceptionallytightlytothetransitionstateformofthesubstrate;thatiswhatlowersthefreeenergyofthetransitionstate.PracticalconsequencesofenzymesbindingtightlytothetransitionstateUnderstandingthatanenzymebindsextremelytightlytothetransitionstateformofitssubstratehasledtoanumberofimportantpracticalscientificdevelopments.TransitionstateanaloguesasenzymeinhibitorsDesigningmoleculestoinhibitkeyenzymesisamajoreffortinpharmaceuticalresearch.Importantenzymetargetsare toonumerous to list,but they includeenzymes frompathogenicbacteriaandvirusesaswellashumanenzymesinvolvedindisease‐relatedpathways,suchasthosethatregulatebloodpressureandinflammation.Ifanenzymespeedsupareactionbyafactorofathousand,thenour reasoningabove indicates that theenzymebinds the transition state formof the substrate athousand timesmore tightly than it binds the substrate. So, a drugmolecule that looks like thetransitionstateformofthesubstratewillbindtightlytotheenzymeandactasaninhibitor.Themainchallengeofcourseisthatthetransitionstateisentirelyunstable.Thegoalthenistocomeupwithacompound–atransitionstateanalogue–thatlooksasmuchaspossiblelikethetransitionstate,butyetisstableandcanbesynthesized(cheaply).Thiscanbeadifficultproposition.CreatingnewenzymesfromanaturalantibodyrepertoireThe concept of catalytic antibodies was firstmentioned by chemistWilliam Jencks in 1969 andreducedtopracticeaboutthesametimebyRichardLernerandPeterSchultzbeginninginthe1980’s.Thegoalwastocreatenovelenzymesthatwouldcatalyzeusefulchemicalreactions,includingtypesofreactionsthatnonaturalenzymeshadevolvedtocarryout.Theideareliesonthespectacularly

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largediversityofantibodiesthatcanbegeneratedbythemammalianimmunesystem.Ifananimalhasthegeneticcapacitytogenerate1012differentantibodymolecules,surelysomeofthemshouldhavea tight binding affinity for any imaginable chemical entity, includinga transition state for areactiononemightwant tocatalyze. According to theEyring theoryand the logicarticulatedbyPaulingandJencks,ifyoucanfindanantibodysequencewithahighaffinityforthetransitionstateof adesirable reaction, thenyouhave foundanenzyme for that reaction. Thework required toidentify an antibodywith the desired property is challenging. In order to induce production ofantibodies that might bind tightly to the transition state, the animal must be inoculated with atransitionstateanalogueforthereaction,andthesamedifficultynotedaboveregardingtransitionstateanaloguesmustbeovercome.Severalstudieshavesucceededinfindingantibodiesthatexhibitcatalyticactivityforadesiredreaction,buttheratesofaccelerationhavegenerallynotbeenveryhigh.ComputationalenzymedesignThereismuchcurrentinterestintheideaofusingsophisticatedcomputerprogramstodesigntheaminoacidsequenceofaproteinthatwillcatalyzeadesiredreaction.Ratherthandesigninganovelproteinfromscratch,themostfeasibleapproachistotakeanaturalproteinthathasasurfacecleftsuitableforbindingacompoundofabouttherightsize,andmodifytheaminoacidsequencemainlywithinthebindingsitecleft.Thepotentialpowerofthisapproachisveryhigh.Incontrasttothecatalyticantibodyapproach,thereisnoneedtosynthesizeatransitionstateanalog. Instead,onerequiresanaccuratemodel(i.e.detailedatomiccoordinates)forwhatthetransitionstateislikelytolook like. Modern computerprogramsare capable ofproducing reasonablemodels of transitionstates.Themostchallengingelementisdesigningaminoacidchangesintoaproteininsuchawaythatthetransitionstatewouldbetightlybound.Onedifficultissueconcernsthecalculationoffreeenergiesforlargesystemslikeproteinsandtheircomplexes.Eventheaqueoussolventisimportantto consider given the contribution of hydrophobic effects andwater structure in general to theenergetics.Beyondthestillunsolvedproblemofaccurateenergypredictions,changingtheaminoacidsequenceofanaturalproteinveryoftencausesunforeseenandunpredictableeffects,includinglossofstabilityandaggregation.Inmanycases,changingtheaminoacidsequenceofaproteinmaymakealternate(non‐native)configurationsoftheproteinmorestablethantheintendedstructure.Theabilitytoconsiderandavoidallthepossiblealternatestructuresamodifiedproteinmightadoptiswellbeyondthecurrentcapacityofcomputersandproteinmodelingsoftware.Nonetheless,therehavebeenafewexcitingsuccessesindesigningnewenzymeactivitiescomputationally.Aswiththecatalyticantibodieshowever,catalyticrateshavesofarbeenfairlymodest.Furtheradvancesalongthislineareverylikelyinthefutureascomputerprogramscontinuetoimprove.KineticparametersofnaturalenzymesNaturalenzymeshaveevolvedoverbillionsofyearstospeedupthereactionstheycatalyze.Howwelldotheyperform?Andcouldtheybebetter?Thesearethornyquestions.Partofthecomplexityconcernsthesaturationbehaviorofenzymekinetics(v=[Etotal]kcat[S]/([S]+KM)).Anenzymethathas

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averyhighkcatmaynotbesogreatif itdoesn’tbinditssubstrateverywell(i.e. ifKMishigh). OfcoursethebestthingwouldbetohaveaveryhighkcatandaverylowKM.Buttheremaybetradeoffsintheabilityofanygivenenzymetooptimizebothparameters.Inviewofthis,theratiokcat/KMisoftendiscussedasageneralmeasureoftheefficiencyofanenzyme,essentiallyareflectionofthejointvalueofhavinghighkcatandlowKM.Intermsofatypicalhyperbolicgraphofenzymeactivityvssubstrateconcentration,kcatdetermines themaximumvelocityatanysubstrateconcentration,whilekcat/KM is theslopeof thevelocitycurve(normalized for totalenzymeconcentration) in itslinearregionwellbelowsaturation.ThiscanbeseenbyevaluatingthestandardMichaelis‐Mentenvelocityequation(above)at[S]<<KM.There,v/[Etotal](kcat/KM)[S],whichconfirmsthestatementabouttheslopeofthevelocitycurve.Andv(kcat/KM)[Etotal][S],whichshowsthatkcat/KMtakestheformofabimolecularrateconstant–recallthatforA+B→C,v=k[A][B].WhatarethevaluesforkcatandKMfornaturalenzymes?Thesevaluesshowanastonishingrangeofvariation from enzyme to enzyme. Much of that variation reflects differences in the kinds ofsubstratesinvolvedandthekindsofchemicalrearrangementsthattakeplace.Subtlereffectsrelatetocellularconditions. Forexample, there isnoneed foranenzyme toevolvean incredibly tightbindingconstant(lowKM),whichmightevencomeattheexpenseofalowerkcat,ifthesubstrateinquestion exists at high concentration in the cell; operating under highly saturated conditions isgenerallynotanadvantageousstrategy.Conversely,iftheKMistoohigh,thentheenzymewillbeverypoorlyoccupied;synthesizing idleenzymemolecules isanexpensiveburden forthecell. AgeneralfindingisthattheKMvaluesexhibitedbynaturalenzymestendtoberoughlyinthesamerangeasthenaturalcellularconcentrationofthesubstrateorsubstratesonwhichtheyoperate.Accordingtoarecentsurveyofpublishedenzymekineticparametersintheliterature,mediankcatvaluesfornaturalenzymesareontheorderof10sec‐1overall,andabout10timesfaster(100sec‐1)forenzymesthatoperateincentralmetabolism,wherehighfluxisimportant.ThemedianvalueforKMinnaturalenzymesisaround100Mor10‐4M.Themedianvalueforkcat/KMisontheorderof105M‐1sec‐1.Asagroup,naturalenzymesvarywidelyfromthesemedianvalues.Whataboutlimitingcases?Isthereamaximum?Reactionsthatarebimolecularfaceanupperlimitgovernedbydiffusion.Evenifanenzymecouldbindasubstrateinfinitelytightlyandcatalyzeitsconversiontoproductinfinitelyfast,therateofthereactionwouldbelimitedbyhowfastthetwomoleculescanencountereachotherinsolutionowingtothelimitsofdiffusion.Dependingonmolecularsizes(whichgoverndiffusioncoefficients), the upper bound for kcat/KM in diffusion‐limited bimolecular enzyme‐substrateencountersisintherange108–109M‐1sec‐1.Thislimitingvaluecomesfromanalysesbeyondthoseusedtodevelopourstandardequationsforreactionvelocities,which ignoretheroleofdiffusion.Very fewenzymesoperatenear thediffusion‐limiting valueof kcat/KM,but a fewdo. Superoxidedismutase(SOD)andtriosephosphateisomerasearetwowell‐studiedexamples.

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CHAPTER17

IntroductiontoBiochemicalSpectroscopyEnergytransitionsWeunderstandfromquantummechanicsthatmoleculescanexist only indiscrete energy states, and transitionsbetweenoneenergystateandanothercanbedrivenbyabsorptionoremissionofelectromagneticradiation(i.e.photons)iftheenergyofthephotonmatchestheenergydifferenceofthetransition.Therelationshipbetweenthefrequencyorwavelengthof the radiationandenergy isE=hv=hc/.You may remember from general chemistry that verysimplemoleculeslikesingleatomstypicallyshowverysharpabsorptionandemissionbands.Theyundergotransitionsatonlyverynarrowwavelengths–recalltheRydbergseries.Thediscretenessof their spectral properties reflects the simplicity of their allowable energy states (i.e. electronicstatesofhydrogen‐likeatomicorbitals).In contrast, complex moleculeshave complex spectra. Thepresence of multiple atoms in amolecule introducesadependenceof energy on nuclear positions.Nuclear motions give rise tovibrational energy states. Theenergy differences betweenvibrational states are generallymuch smaller than those betweenelectronic states. The idea thatvibrational transitionsaresmaller inenergyandessentiallyseparable fromelectronic transitionsgives a picture where more finely spaced vibrational states can be superimposed on individualelectronicstates,asshown. Andevenmorefinelyspacedrotationaltransitionsexistwithinthosestates. Themuchgreater complexityof the energyprofile for complexmolecules introduces thepossibility of very many transitions with closely spaced energies. As a result, absorption andemission spectra for larger molecules are complex and more continuous in nature rather thandiscrete.Anexaminationoftypicalenergymagnitudesforelectronicandvibrationaltransitionsisinstructive.Electronic transitionsare typically the subjectof spectroscopy in theUVandvisible rangeof theelectromagneticspectrum.Considerthentheenergyassociatedwithawavelengthof400nminthe

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violetregionofthevisiblespectrum,E=hv=hc/=4.110‐21J.Bycomparison,thisisabout120kBT.AccordingtotheBoltzmannequation,theprobabilityofamoleculeresidingintheexcitedelectronicstateratherthanthegroundelectronicstateisessentiallyzero.Wecanrepeatthecalculationforatypicalvibrationaltransition;thesetypicallyoccurintheinfrared(IR)regionoftheelectromagneticspectrum. Consideracarbonylstretch, forwhich isapproximately1.9M. Thecorrespondingenergyis1.0410‐19J.ThisissmallerthantheenergyfortransitionsintheUV/visiblerange,butstillequaltoabout25kBT.Theconclusionisthat,atordinarytemperaturesandunlessotherwiseexcited,moleculesgenerallypopulatealmostexclusivelythelowestvibrationalstateofthelowestelectronicstates.Thisgeneralideahasimportantimplicationsforwhatenergytransitionsaremostlikelytooccur;ahighprobabilitytransitionrequirestheinitialenergystatetobewell‐populated.FluorescenceOur previous analysis tells us that an absorption transition is likely to occur from the groundvibrational state of the ground electronic state. But towhat higher electronic energy states is amoleculelikelytobeexcitedbyabsorption?Aninterestingphenomenonarisesfromthegeneralideathatthelowestenergynuclearpositionsforamoleculearetypicallyslightlydifferentfordifferentelectronicstates.Thisistypicallydiagrammedasshowhere,wherethetwoblackcurvesindicatetheclassicalenergyofamoleculeasa functionof itsnuclearpositionsintwodifferentelectronicstates.Theminimumenergiesoccuratslightlydifferentnuclearpositions.Withineachelectronicstate,aseriesofvibrationalstatesare indicated. The width of thelines (setting aside quantummechanical aspects of harmonicoscillations)illustratetherangeofnuclearpositionsthatareallowedin each vibrational state. Aconsideration of timescales nowleadstoaninterestingconclusion.The timescale for photonabsorption is much shorter thanthetimescalefornuclearmotions.This means that electronictransitionsoccur‘vertically’inthesenseofthediagramshown.Thisis known as the Franck‐Condonprinciple. If an electronictransition must occur withoutappreciable movement of nuclei,thenitmustoccurtoavibrationalstateforwhichtheinitialnuclearpositions are allowable. The

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diagramemphasizesthatthistypicallyisanexcitedvibrationalstateratherthanagroundvibrationalstate.Afterabsorptiontoanexcitedelectronicstate,accordingtotheBoltzmannequationamoleculemustreturntothegroundstate.Thereturntothegroundelectronicstatecanoccurwithemissionofaphoton;thisisfluorescence.Thetimescaleforfluorescenceistypicallyinthe10‐8to10‐5secrange,whichislongenoughforthermalvibrationsandcollisions(whoseeffectsareillustratedinredinthefigure)toallowthemoleculetodescendtolowervibrational stateswithin theexcitedelectronic state before returning to theground electronic state. Again, thatelectronictransitionoccursverticallytoanexcited vibrational state of the groundelectronic state, after which furthertransitions lead back to the groundvibrational state of the ground electronicstate. The key consequence of thisphenomenon is that the fluorescenceemissionspectrumforamoleculeisshiftedto lower energy and longer wavelengthcomparedtotheabsorptionspectrum.ThisisreferredtoastheStokesshift.UsesandadvantageouspropertiesoffluorescenceFluorescenceoffersahighdegreeofsensitivityfordetectingandmeasuringtheconcentrationsofspecificmolecules,whichmaybefluorescenteithernaturallyorbyvirtueofbeingchemicallylabelledwithafluorophore(afluorescencechemicalgroup).Particularlyincontrasttoabsorptionstudiesformeasuringconcentration,thehighsensitivityoffluorescencederivesfromtwofeatures.Firstistheshiftinwavelengthfromtheincidentwavelengthtotheemissionwavelength.Inanabsorptionexperiment involving a dilute or weakly absorbing molecule, one is forced to analyze a smalldifferencebetweentwolargenumbers–thenumberofphotonstransmittedbyablankcomparedtothenumber transmitted by the sample. In a fluorescence experiment, the change inwavelengthmakesitpossibletoanalyzethenumberofemittedphotonswithoutinterferencefromtransmittedphotons,whichhavethesamewavelengthastheincidentlight.Takingadvantageofthewavelengthdifferencerequiresasecondmonochromatorplacedbetweenthesampleandthedetector;a firstmonochromator isrequiredbetweenthe lightsourceandthesample. The fluorescenceemissionintensityisproportionaltotheconcentrationofthefluorophore(aslongastheconcentrationisnottoohigh),makingaccurateconcentrationdeterminationpossiblefromverydilutesolutions.Anextralevelofsensitivitycomesfromtheabilitytomonitorfluorescenceinadirectiondifferentfromthepathofthetransmittedbeam.Thefigureillustratesthecombinedeffectsofwavelengthchangeanddetectionatanangle.Photonsthatarescattered(elastically)fromthesampleemergeatallangles,

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buttheirwavelengthisthesameastheincidentbeamsotheyaredistinguishablefromfluorescentphotons.

Proteinstypicallyhavesomenaturalfluorescenceowingtothepresenceoftryptophanaminoacids.Buttheintensityoftryptophanfluorescenceisnotespeciallyhigh,andoneisofteninterestedinusingfluorescence to monitor one protein (or nucleic acid) in particular. An enormous range offluorophoresareavailablecommerciallywithawiderangeof spectral characteristics. Thesearetypically conjugated chemically to the macromolecule of interest by covalent attachment, oftenthroughnucleophilicattackbyacysteinethiolorlysineaminegroups. Fluorescenceexperimentscanalsobeperformedinsitutomonitorthepresenceandsubcellularlocationofaspecificproteininside cells in tissue culture using fluorescence microscopy. Chemical labeling is generally notpossibleinthatscenario.Instead,theproteinofinterestcanberenderedfluorescentinsidethecellby creating a fusion at theDNA levelbetween theproteinof interest andanaturally fluorescentprotein. Originallydiscovered incoralseaorganisms,numeroussuchproteinsareknownwithadiverserangeofemissioncolors;green fluorescentprotein(GFP) is themostwidelystudied. Aninteresting variation on the approach is to label two different proteinswith distinct fluorescentproteinshavingdifferent emission colors, like redandgreen. Whether the twoproteins localizetogetherinthecell–e.g. iftheyinteractwitheachother– isevidentbyjointemissionofredandgreencolors(makingyellow).Thelevelofspatialdetailthatcanbevisualizedinastandardvisibleor UV microscopy experiment is a few hundred nanometers, which is fine enough to visualizeorganellar, nuclear, and cytoskeletal structure in eukaryotic cells, but not fine enough to seemolecularstructure.Aparticularlyusefulfeatureoffluorescenceisitssensitivitytochemicalenvironment.Thegreatersensitivitytoenvironmentforfluorescencecomparedtoabsorbancerelatesinparttothelongertimescale of fluorescence. In general, increased flexibility and environmental polarity lead to lowerfluorescent intensity; the peak emission wavelength can also be affected. As an example, thefluorescenceoftryptophanincreasesbyafactorofroughly4inalowpolaritysolventsuchasDMSO

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(dielectricofabout35)comparedtowater(dielectricofabout80).Exposureofafluorescentgrouptoparticular chemicalsknownasquenchersalso reduces fluorescence, and themagnitudeof theeffect can depend on the degree to which the fluorophore is exposed on the surface of themacromolecule.Theenvironmentalsensitivityoffluorescencecanbeexploitedinvarioustypesofexperiments.Wediscussed earlier how native tryptophan fluorescence can be used to monitor protein folding.Tryptophanresiduesalmostalwaysbecomelessflexibleandmorerigidlyheldinthefoldedstateofaprotein,leadingtohigherfluorescence.Inanothertypeofexperiment,ifaproteinissuspectedtobindaligandthatisfluorescent(orforwhichafluorescentanalogueisavailable),thenbindingoftheligandtotheproteincanbedetectedbyanincreaseinfluorescence.KineticsoffluorescenceandcompetingroutesforreturntothegroundstateAfter a molecule hasbeen driven to anexcited state byabsorbing a photon,there are severalpossible competingroutes for returningto the ground state.Some of these wehave discussedalready while somewe will return tolater. The relativerates of theseprocesses relatesdirectly to which pathways dominate for a given molecule. If the rate constant for fluorescentemissionishigherthantherateconstantsforotherprocesses,thenmostoftheexcitedmoleculeswillreturntothegroundstatebywayoffluorescentemission.Aswediscussedabove,anumberofphenomenaaffectfluorescence,includingchemicalenvironment,sofluorescencecanbeusedtomonitorvariouseventsthataltertheenvironmentofafluorophoreinsolution.Thedetailsofthebehaviorexpectedcanbeunderstoodbyanalyzingthephototransitionsusingtreatmentssimilartothosewedevelopedearlierforchemicalkinetics.Wecansimplifythingsbylumpingtogetherthevariousnon‐fluorescentpathwaysforreturntothegroundstateunderasinglerateconstant,kother. Variousunderlyingeventscanthenbeanalyzedintermsoftheeffects

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theyhaveon the relative ratesof kfluor and kother.With respectto kinetic treatments,fluorescence experiments canbe of two essentially differenttypes: 1) under continuousilluminationwheresteadystatebehavior is assumed, or 2)followingabriefpulseofincidentlight,afterwhichtime‐dependentmeasurementsaremade.Notethat the latter type of experiment requires special instrumentation because the time scale forfluorescencedecayisusuallyshorterthanmilliseconds.Wecananalyzethebehaviorofbothkindsofexperiments.ConstantilluminationUnderconstantillumination,theconcentrationoftheexcitedstateformofthefluorophore(P*)isnotchanging.Settingd[P*]/dt=0d[P*]/dt=0=kabs[P]–(kfluor+kother)[P*]Rearrangingtoobtainanexpressionfor[P*],[P*]=kabs[P]/(kfluor+kother)Then,thefluorescentintensityIfluorisgivenbyIfluor=kfluor[P*]=kabs[P]kfluor/(kfluor+kother)Since the rate of photon absorption is kabs[P], the ratio of the number of photons emitted to thenumberabsorbed–afractionalquantityknownasthequantumyieldQ–isgivenbyQ=kfluor/(kfluor+kother)Wecanconcludefromthisanalysisthatthequantumyieldandtheintensityofthefluorescenceobservedunderconstantilluminationisdecreasedbyeventsinsolution that increase the rates of non‐fluorescent‘other’pathwaysforreturntothegroundstate.Thatideaisillustratehere.Oneexampleofsuchascenariois binding of a fluorescent molecule (perhaps ananalogue of a suspected ligand) to a protein; thiswould decrease the non‐fluorescent pathways by

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reducingmobilityofthefluorophoreandtherebyincreasethequantumyieldalongwiththesteadystatefluorescenceintensity.Time‐resolvedfluorescenceWithappropriateinstrumentation,anexcitationpulsecanbeappliedandthefluorescenceintensity(which must decay back to zero) can bemonitoredovertime.Thesamekineticschemeas above can be used if we remove thecontinuous absorption event. This becomes asimplecaseofexponentialdecaywithatotalrateconstantofkfluor+kotherandadecaytimeof=1/(kfluor+kother)The comparison of decay behavior in thepresence and absence of processes competingwithfluorescencecanbediagrammedasshown.

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CHAPTER18

SpecialTopicsinBiochemicalSpectroscopyPolarizationandselectionrulesIn thissectionwediscuss theimportant role orientationplays in spectroscopy.Orientational effects becomeapparent in spectroscopicexperiments conducted usingpolarized light. You mayremember thatelectromagneticradiationisatransverse wave in which atraveling photon carriesoscillatingelectricandmagneticfieldvectorsperpendiculartothedirectionoftravel.Lightemittedfromanordinarysource(e.g.alightbulb)carriesphotonswhoseelectricfieldvectorspointinallpossibledirectionsperpendiculartothedirectionoftravel.Avarietyofmaterialscanbeusedtofilterordinaryincominglighttoproducelightthatis‘planepolarized’,meaningthattheelectricfieldvectorpointsinasingledirectionwhileitoscillatesinmagnitude,upanddown;thedirectionoftravelandtheelectricfieldvectorformaplane.Weknowthatforaphotonoflighttobeabsorbedandcauseatransitionfromaninitialstatetoafinalstatethattheenergyofthephotonmustbecorrect.Butthedirectionofitselectricfieldvectorisalsocriticallyimportant.Whetherapotentialtransitioncanbecausedbyaphotonpolarizedinacertaindirection is embodied in quantummechanical ‘selection rules’. In absorption spectroscopy, theextinctioncoefficientrelatestothestrengthorprobabilityofatransitionbybeingproportionaltothesquareofatransitiondipolemoment, .Inageneralformofthetransitiondipolemoment,

Ψ Ψ

where describesthegeneralpositionvectorinspaceandΨ andΨ arethequantummechanical

wavefunctionsfortheinitialandfinalenergystates.Forourpurposesofconsideringtheabsorptionoremissionoflightpolarizedinaparticulardirection,wecanrewritetheequationinseparatex,y,zcomponents.Theprobabilityofabsorbingaphotonpolarizedalongthexdirectionisrelatedtothexcomponentofthetransitiondipolemoment,evaluatedas

Ψ Ψ

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withequivalentequationsforpolarizationalongyorz.Analyzingwhetherelectronictransitionscanorcannotoccurwhenthelightispolarizedincertaindirectionscanbesimplifiedusingatreatmentthatconsidersthesymmetryvsanti‐symmetryoftheinitialandfinalwavefunctions.Wewillillustrateoneexamplesituationwheretheinitialandfinalwavefunctionsaresimple–muchsimplerthanonewouldencounterwithcomplexmolecules,butstill highly instructive in understanding orientational effects. We begin with a reminder aboutsymmetricandanti‐symmetricfunctions.Theserefertofunctionswhosevaluesareeitherthesamewhenaspatialvariableisnegated(i.e.f(‐x)=f(x))ornegatedwhenaspatialvariableisnegated(f(‐x)=‐f(x)),respectively.Onewayoflookingatsymmetricvsantisymmetricfunctionsisintermsofpolynomial functions. Wefindthatpolynomial functionswithevenexponents(x0,x2,x4,etc.)aresymmetricwhereaspolynomialfunctionswithoddexponents(x1,x3,etc.)areantisymmetric.Weareparticularlyinterestedinconsideringwhathappenswhenweevaluatetheintegralofafunctionthatissymmetricorantisymmetric.Byeitherexplicitlyevaluatingtheintegralsofsuchfunctionsorbythinking about the areas under the curves (positive and negative), we can see that odd(antisymmetric)functionsintegratetozerowhileeven(symmetric)functionsgenerallydonot.Theillustrationshereare1‐dimensional(dependingonlyonx).Inthree‐dimensionswhereafunctionwoulddependonx,y,andzandintegrationwouldbeoveralldimensions,theresultintegratestozeroifthefunctionisoddorantisymmetricwithrespecttoanyofthethreespatialvariables.

Nowconsideranelectronictransitionbetweenabondingmolecularorbitalanda*anti‐bondingmolecularorbital.Suchtransitionsarecommoninconjugateddoublebondsystems.Bothmolecular

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orbitals are effectivelycombinationsoftwoside‐by‐side p orbitals. Thesignsofthetwoporbitalsare aligned in the molecular orbital butoppositelyorientedinthe* molecular orbital,creating an extra nodalplane in the latter case.We can set up a coordinate system at the center of themolecular orbital and then tabulate thesymmetryvsanti‐symmetry(orevennessvsoddness)ofthefunctionsthatgetmultipliedtogetherinsidetheintegralforthetransitiondipolemoment.Inordertoevaluatethetotalsymmetryoftheproductofthefunctionsinsidetheintegral,weneedtounderstandtherulesforsymmetryvsanti‐symmetrywhenfunctionsaremultipliedtogether.Sincetheevennessoroddnessisapropertyofexponents, products of the functions behave according to addition of even and odd numbers:even+even=even;even+odd=odd;odd+odd=even.Weneedtomakeaseparateanalysistoconsiderthetransitiondipolecomponentforlightpolarizedineachpossibledirection.Forthecaseoflightpolarizedalongthexdirection,theterm‘x’inthemiddleoftheintegralcanbeunderstoodasbeingx1y0z0,whichisthereforeoddwithrespecttox,andevenwithrespecttoyandz.Withtheserulesinhandwecanconstructatabletoanalyze .

w/r/t axis

x = x1y0z0

total 

x even odd odd even

y even even even even

z odd even odd even

Inevaluating , thetotal function insidethe integral isevenwithrespecttoall threecoordinatevariables,sotheintegraldoesnotnecessarilyvanish.Weconcludethattheto*transitioncanoccurbyabsorptionofaphotonpolarizedalongx(whichisthebonddirection).Thistransitionisthereforeallowed, thoughoursimplesymmetryvsanti‐symmetry treatmentdoesn’t tellusaboutmagnitudes. Also note that the allowable transition for polarization along x does notmean thedirection of travel of the photon is along x; in fact the direction of travel would have to beperpendiculartoxinorderforthepolarizationtobealongx.Nextwecanevaluatetransitiondipolesforlightpolarizedalongyorz.Thosetablesareshown.

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Inbothofthesecasesthetotalfunctionisoddwithrespecttoatleastonevariable,sotheintegralvanishes.Thatmeansthat and bothvanish.Thosetransitionsareforbidden,meaningtheto

*transitioncannotbepromotedbyabsorptionofaphotonpolarizedalongyorz. Absorptionisonlyallowedforpolarizationalongx.Someinstinctcanbedevelopedtounderstandthisresult.Incomparingthesymmetryvsanti‐symmetryoftheinitialandfinalwavefunctionswecanseethatthexdirectionistheonlydirectioninwhichthetwofunctionsdiffer.Anelectricfieldvectoralongthatdirectioncan thereforedrive theconversionofone to theother. The treatmentof the transitiondipolemomentandselectionrulesforemissionaresimilartothoseforabsorption.ExampleofabsorptionofpolarizedlightbyanorientedpigmentAlthoughlightmayonlybeabsorbedwhentheelectricfieldis oriented in a particular direction relative to thechromophore, the effects of this are often not evident inexperiments done in solution, since the absorbingchromophore is present in all possible orientations. Thedependence of absorption on direction of polarization cansometimes be seen in a crystalline sample where thechromophoreexistsinthesameorientationthroughoutthecrystal specimen. The example shown here comes from acrystal of a protein that binds a carotenoid molecule as acofactor. Carotenoids are long organic molecules withconjugateddoublebonded orbital systems,and fromourprevious exercisewemight expect a carotenoid to absorblight polarized along the long axis of the molecule. Thephotographs shown were taken under a light microscopewhere the incident light passed through a polarizing filterbeforepassingthroughthecrystallinesample.Thepolarizerwasrotatedatdifferentanglesforthetwophotographs. Evidently,thelightwaspolarizedinadirectionthatallowedabsorptionbythecarotenoidinthetoppanel,butitwasorientedinadirectionthatdidnotallowabsorptioninthesecondpanel.Fluorescenceexperimentswithpolarizedlight

w/r/t axis

y = x0y1z0

total 

x even even odd odd

y even odd even odd

z odd even odd even

w/r/t axis

z = x0y0z1

total 

x even even odd odd

y even even even even

z odd odd odd odd

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Interestingphenomenaoccurwhenasampleabsorbspolarizedlightandthenreemitsphotonsbyfluorescence. Tosimplifythediscussionattheoutsetwewillassumethatifamoleculeabsorbsaphotonpolarizedinaparticulardirectionthenbyfluorescenceitwillemitaphotonpolarizedinthesamedirectionifthemoleculehasnotchangeditsorientationbetweentheabsorptionandemissionevents.Butwhataboutmolecularmotions,particularlychangesinmolecularorientation,thatareoccurringduringtheprocess?Howmuchmightoneexpectamoleculetorotateinthetimebetweenwhenitabsorbsaphotonandre‐emitsaphotonbyfluorescence?Clearlythisdependsontherelativeratesoffluorescenceandmolecularrotationinsolution.Ifrandommolecularrotationoccursveryslowlycomparedtofluorescence,thenfluorescentphotonswillbepolarizedinthesamedirectionastheincident(polarized)light.Conversely,ifrandomrotationsoccurmuchfasterthanfluorescence,thenemittedphotonswillhaveelectricfieldvectorsorientedinalldirectionsequally.Asaresultitispossible to learnabout therelativeratesof fluorescencevsmolecularrotationbystudyingthedegreetowhichemittedphotonsarepolarizedinthesamewayastheincidentlight.Ifthetimescalefor fluorescence isknownthenthe timescale formolecularrotationscanbedetermined. This isusefulbecausethetimescaleformolecularrotationinsolutiondependsonthesizeofthemolecule(andonviscosity),soultimatelywecangetinformationaboutmolecularsizeusingexperimentsofthistype.Someofthetechnicaldetailsaredescribedhere.The figure diagrams essential features of a fluorescence polarization or fluorescence anisotropyexperiment. Twopolarizingfiltersarerequired:onebeforethesampleandoneafterthesample.Monochromators(notshown)arealsorequiredtoselectappropriatewavelengthsfortheincidentandemittedphotonsbeingdetected.Thesecondpolarizer(sometimesreferredtoastheanalyzer)isrotatedduringtheexperiment.Thismakesitpossibletomeasuretherelativeintensityofemittedlightthatispolarizedindifferentdirections;thisisameasureofhowmuchmoleculeshaverotatedafterabsorptionandbeforeemission.

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Themathematicaltreatmentisasfollows.Theintensityoflightemittedparalleltotheincidentlightisdenoted ∥.Theintensityemittedperpendiculartotheincidentlightisdenoted .The(unitless)measureofhowmuchstrongertheparallelemissioniscomparedtotheperpendicularisdescribedby the fluorescence anisotropy, r. [Anisotropy translates roughly to “not” the “same” in all“directional movements”.] The anisotropy of the emitted fluorescence is defined in terms ofexperimentalmeasurementsas

∥

∥ 2 0 1

If thevalueofr isclosetozero(i.e.noanisotropy)thentheintensity isthesameforparallelandperpendicular emission, meaning the rate of molecular rotation is much faster than the rate offluorescence.Ifthevalueofriscloseto1(perfectanisotropy)thentheintensityofemissionintheperpendicularorientationisnegligible,meaningtherateoffluorescenceismuchfasterthantherateofmolecularrotation.Usefulinformationcomesfromintermediatescenarioswherethetworatesareinacomparablerangeandthevalueofrisintermediatebetween0and1.Exactly how does the fluorescence anisotropy relate to the relative rates of fluorescence andmolecular rotation? As in our earlier kinetic analyses, unimolecular rates can be described by(reciprocallyrelated)decaytimes.Thedecaytimeforchangesinmolecularorientationisreferredtoastherotationalcorrelationtime,denotedhereasrot.Wedenotethefluorescencedecaytimeasfluor.OneformofPerrin’sequation(givenwithoutproof)statesthat

Settingasidethetermr0momentarily,theequationindicatesthatifthedecaytimeforrotationismuchlongerthanthedecaytimeforfluorescence(meaningtherateofrotationismuchslowerthantherateoffluorescence),thentheanisotropyrwouldbe1.Andconverselyrwouldbezeroifthedecaytimeforrotationwasmuchshorterthanforfluorescence.Thetermr0isnecessarytodealwithanimperfectalignmentofemittedandabsorbedphotonsthatoccursevenwithoutanymolecularrotation.Finally,therotationalcorrelationtimeisdirectlyrelatedtomolecularsizeby =V/RT,whereisviscosityandVismolecularvolume.Therefore,inprinciple,onecanobtainavalueformolecularvolumefromfluorescenceanisotropymeasurement,assumingthemoleculeofinterestisfluorescentorhasbeenfluorescentlylabeledandthefluorescencedecaytimecanbeestablishedinseparateexperiments.In biochemical applications, fluorescence anisotropy experiments are often used not to estimateactual molecular volumes, but in a somewhat more qualitative way, comparing the degree ofanisotropybeforeandaftersomepotentialbindingeventforexample.Thecentralrequirementisthat the event under investigationmust cause a change in the rotational correlation time of thefluorescent molecule. Two kinds of experiments are possible as illustrated. Under constantillumination(steadystate),conditionsoreventsthatgiverisetoslowerrotationaltumblinggivean

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increaseinfluorescentanisotropy,r,sincelesstumblingoccurspriortofluorescentemission.Inatime‐resolvedexperimentfollowinganincidentpulse,theanisotropywilldecaymoreslowlyiftherotationaltumblingisslower.

Fluorescentresonantenergytransfer(FRET)Underspecialcircumstances,anexcitedchromophorecanreturntothegroundstatenotbyemissionbutbytransferringexcitonenergytoanearbychromophore.Theefficiencyofthisprocessdependsontwomainfactors:thedegree of overlapbetween the emissionspectrum of the firstchromophore (referredtoasthedonor)andthesecond chromophore(referred to as theacceptor).According to the Förster equation, the transfer efficiency depends steeply on the separation Rbetweenthedonorandacceptor.

Efficiency=

TheparameterR0isparticularforthedonorandacceptorpairanddependschieflyonthequalityofthespectraloverlapbetweenthedonoremissionandtheacceptorabsorbance.NotethatwhenR=R0theefficiencyofenergytransferis1/2,soagooddonoracceptorpairwillhavearelativelyhighvalueforR0.FRETexperimentsareusefulmainlyinthe10to100Årange.

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Lookingbackatourschemefromthepreviouschaptershowingthepossibleroutesforreturnoftheexcitedstatetothegroundstate,weseethatenergytransferbyFRETisaphenomenonthatcompeteswiththefluorescentemissionfromthedonor.Asaresult,thepresenceoftheacceptorchromophorereduces donor fluorescence and speeds decay of the excited donor and its fluorescence. Asdiagrammedhere, FRET experiments can be done either under constant illumination,where thefluorescence intensity from the donor is reduced by the presence of the acceptor, or in a time‐resolvedexperimentwherethespeedofdecayisincreasedbytheacceptorandthecharacteristicdecaytimeisdecreased.

FRETexperimentsfinduseindiverseexperimentswheredetectingtheproximityorapproximatingthedistancebetweentwomoleculesorfunctionalgroupsmightbeinformative. Unlessoneofthetwocomponentsofinterestisnaturallyfluorescent,thisgenerallyinvolveslabelingbothcomponents–onewiththedonorfluorophoreandonewiththeacceptor.Experimentalmeasurementsgiverisetoavalueforthetransferefficiency,asdiagrammedhereforeithercontinuousilluminationortime‐resolvedstudies,afterwhichtheefficiencyvaluecanbeusedtoapproximatethedistancebetweenthedonorandacceptoraccordingtotheFörsterequation.NotethatvaluesforR0havebeentabulatedforverymanydonor‐acceptorpairs,sothatparameteristypicallyaknownquantity.FRETinbiologyBesides being useful in biochemical experiments, the FRET phenomenon plays a key role inphotosynthesis.Thekeystepthatconvertslightenergyintochemicalenergyinphotosynthesistakesplaceinatransmembraneproteincomplexknownasthephotosyntheticreactioncenter(RC).TheRCbindsa‘specialpair’ofchlorophyllmoleculesinverycloseproximity.Thatgroupingmakesthespecialpairsuitableforparticipatingintheprimaryphotochemicalevent.Afterthespecialpairisexcited,insteadofreturningtothegroundstateanelectronleavesthespecialpairandjumpsalongapathofneighboringpigmentcofactors(chlorophyllsandcarotenoids)boundtotheprotein,leading

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theelectronacrossthemembraneandtherebygeneratinganelectrochemicaldifferencebetweenthetwosides;thisformsthebasisforchemicalenergyconversioninphotosynthesis.Butthereactioncenterbyitselfisnotsuitedforabsorbingphotonsthathitthephotosyntheticmembraneeverywhereandwithabroadrangeofwavelengths.Othertransmembraneproteins,knownaslightharvestingcomplexes (LH), bind a large number of pigment molecules and surround the reaction center.Dependingontheparticularsystemandorganism,multipletypesofLHringsmaybepresent.TheLHproteinsaredesignedtoholdtheirpigmentmoleculesinveryspecificpositionswithrespecttoeachother,andtotunetheirspectralpropertiessothatthepigmentmoleculescanabsorbphotonsefficiently throughout the photosynthetic membrane and then transfer that exciton energy,essentiallybyFRET,topigmentmoleculesclosertotheRC.Thisresultsinafunnelingeffectwhere,atsomeexpenseoflostenergyineachtransferstep,theexcitonenergyiseventuallydeliveredtothespecialpairintheRCinordertodrivetheprimaryphotochemicalevent.Thephotosyntheticreactioncenters frombacteria (whichare analogous toPhotosystem II fromhigherplants)were the firsttransmembraneproteinswhosestructuresweredeterminedinatomicdetailinthemid1980’s.ThestructureofthebacterialRCisshownhereinasideviewwiththemembranerunninghorizontallyand its pigmentmolecules in red, alongwith a view of the RC and surrounding light harvestingcomplexesviewedperpendiculartothephotosyntheticmembrane.SpectroscopyofChiralMolecules:OpticalRotationandCircularDichroismChiralmoleculesexhibitspecialspectroscopicphenomenathatbecomeevidentwhentheyinteractwithpolarizedlight.Becausepracticallyallbiologicalmacromoleculesarechiral,asaremanysmallerbiochemicalmetabolites,spectroscopictechniquesthatexploitthesephenomenaarewidelyusedinthelaboratory.CircularlypolarizedlightWehavediscussedplanepolarizelightatlength.There,theelectricfieldvectoroscillatesinaplane(e.g. vertically for verticallyplanepolarized light). Much insight canbegained abouthowchiral

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molecules interactwithplanepolarized lightby takingamonetary leapof faithandnotingthatavectorthatoscillatesupanddownverticallycanbegeneratedbythesumoftwovectorsthatrotateinacircleinoppositedirectionsatequalfrequency;whentheyarebothvertical(upordown),theysumtogiveaverticalresult,whereaswhentheyarehorizontaltheyopposeeachotherandcancel.Foratravelingwave,thecircularlyrotatingelectricfieldvectormeansthatthewavetakestheformof ahelix. Therefore,planepolarized light canbe imaginedasbeing composedof two circularlypolarized components: one that is ‘right circularlypolarized’ and the other that is ‘left circularlypolarized’.Thisisnotmerelyathoughtexercise,becauseinfactpurecircularlypolarizedlightcanbepreparedbypassingplanepolarizedlightthroughaso‐called‘quarter‐waveplate’,butfornowwewillsticktoourviewofplanepolarizedlightasacompositionoftwocircularcomponents.Thefigureshowsboth formsof circularlypolarized light; the ‘right’ component formsa right‐handedhelix(likeDNAoraproteinalphahelixorastandardhardwarescrew)whilethe‘left’componentforms a left‐handedhelix.Thesenseoftherotationisthatafixedobserver lookingtowards the sourcewill see the directionof the E field vectorrotate clockwise intime for rightcircularly polarizedlight as the travelingwave moves past thepoint of observation.Thisisreversedforleftcircularly polarizedlight.The point ofconsideringplanepolarizedlightasasumoftwocircularcomponents isthatbyviewingthemintermsofhelicalwaveswecanimmediatelyappreciatewhychiralmoleculesmightinteractdifferentlywith leftvsrightpolarized light. Helicesarechiral,as isabiologicalmacromolecule,andwecanappreciatethedistinctinteractionschiralobjectsmakewitheachotherbythinkingaboutputtingourfootintoashoe;bothobjectsbeingchiral,aparticularshoeinteractsdifferentlywithyourtwofeet.Sowhatarethedistinctkindsofinteractionsthatachiralmoleculecanmakewithachirallightwave?Two effects are noteworthy, relating to differences in absorption and differences in index ofrefraction,andtheseleadtotwoimportanttypesofexperiments,whichwediscussnext.Circulardichroism(CD)Iftherightandleftcircularlypolarizedcomponents(imaginedtobecontainedwithinabeamofplanepolarizedlight)areabsorbedtothesameextentwhenpassingthroughasample,thenthelightthat

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istransmittedshouldnaturallyremainplanepolarized.Butonecomponentmaybeabsorbedmorestronglythantheother;thisformsthebasisforcirculardichroismorCD.Whatistheconsequence?Clearly if the left circularlypolarized component is absorbed slightlymore, then the transmittedbeamshouldhaveatleastaslightlylargercomponentoftherightcircularlypolarizedtype.Ifweaddupoppositelyrotatingvectorsofunequalmagnitude,wegetanellipticalshapefortheresultingpathoftheelectricfieldvector.The magnitude of the circulardichroism effect is captured by aparameterreferredtoastheellipticity,. Diagrammatically, relates to theangleformedbyalinebetweenthetipsofthetransmittedelectricfieldvectorsin the perpendicular directions ofmaximumandminimummagnitude,asshown. The ellipticity of thetransmittedbeamcanbemeasuredbyaCDspectrophotometer;thisrequiresadditionalpolarizersbetweenthesampleandthedetector.Theellipticityeffectoriginatesfromadifferenceinextinctioncoefficients(andthereforeabsorbancevalues) for left vs right circularly polarized components, so naturally should reflect thatrelationship.Theequationforintermsofabsorbanceforleftvsrightis:

2.3034

2.3034

whereAreferstoabsorbance, referstotheextinctioncoefficient, isthepathlengthofthelightthrough the sample, and C is the molar concentration (recall A= ). According to the signconvention,higherabsorptionofleft‐circularlypolarizedlight,resultinginagreaterrightcomponentandthereforeaclockwise‐rotatingellipticalfieldvector,correspondstopositive.Butthisequationrequires further explanation of the multiplicative factors, 2.303 and 1/4. The 2.303 term isrecognizableasln(10)whichwemightguessrelatestotheconventionaluseoflog10forabsorbanceequations.Butwhataboutthe4?Inmanytextsthisappearswithoutcomment.Attheexpenseofsomethornydetailswewillshowtheoriginofthesemultiplicativeterms.Tobeginwepointoutthatthediagramforisvastlyexaggerated;theactualdifferencesinabsorbanceareusuallyverysmall,whichmakesitpossibletosimplifyanumberofcomplexnon‐linearrelationshipsbetweenvariablesinthisproblemwithlinearapproximations(i.e.keepingjustthefirsttermsinaTaylorexpansion).Briefly,thetransmittancefortheleftcomponentwouldbe10 . ,andsimilarlyfortheright.Butyoumayrecallfromearlierphysicscoursesthattheintensityofalightbeam(whichhererelatestothetransmittance)goesasthesquareofthemagnitudeoftheelectricfieldvector,sothelengthsoftheelectricfieldvectorsinthediagramforgoasthesquarerootsofthetransmittancevalues. So,themagnitudeofthetransmittedelectricfieldvectorfortheleftcomponentwouldbe

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. / , and likewise for the right. When the exponents in those terms are small, we canapproximatee‐xas1‐xfromTaylor’sexpansiontoget 1 2.303 /2 fortheEfieldmagnitudefortheleftcomponent,andsimilarlyfortheright.Then,notingfromthediagramthatthetangentofwouldbetheratiooftheshortaxistothelongaxis,andthelengthoftheshortaxisisthelengthoftherightcircularlypolarizedelectricfieldmagnitudeminustheleft,andthelongaxisisthesumofthemagnitudes,thentan

1 2.303 /2 1 2.303 /21 2.303 /2 1 2.303 /2

If the terms of the form 2.303 /2 that appear in the denominator are << 1, then thewholedenominator is very nearly equal to 2. Finally, when is small, then Taylor’s expansion givestan()(inradians),andsofinallythewholeexpressionsimplifiestotheoneearlierwiththe2.303inthenumeratorand4inthedenominatorcomingfrom(1/2)/2.Asafinalmanipulation,ifthevalueforisexpressedindegreesinsteadofradians,whichintroducesamultiplicative factorof180/=57.3degrees/rad,and theellipticity isnormalized tobeamolarvaluebydividingbymolarconcentrationandalsonormalizedforpathlength(typicallyincm),thenthe molar ellipticity in degrees isindegrees .

57.3 32.98 . And finally, for historical reasons relating to

volumeandlengthunitconversion,afactorof100ispresentinthestandardequationforthemolarellipticity(denotedbysquarebrackets)giving, 100 ⁄ ,togive:

3298 whichmatchesstandardtextbookexpressions.OpticalRotationTheCDeffectarisesfromdifferencesinabsorption.Adifferenteffectarisewhentheleftandrightcircularlypolarizedcomponentstravelthroughthesampleatdifferentspeeds(owingtoelectronicinteractionswithachiralmolecule).Whathappenswhenlightgoesthroughasamplemoreslowly?Thefrequencyofthewaveisunchanged,butthewavelengthchanges.Thespeedoflightisinverselydependentontheindexofrefraction,n,andtheindexofrefractionheremaybedifferentfortheleftcomponentcomparedtotheright.cL=c0/nLandcR=c0/nR,wherec0isthespeedoflightinavacuumandthesubscriptsrefertoleftandright.Then,L=cL/v=c0/(nLv).Howmanyoscillatorycyclesdoesalightbeammakewhenitpassesthroughasampleofthicknessd?Theanswerisd/.Theangular

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rotationinradianswouldbe2d/L,whichaftersubstitutingtheexpressionforwouldbe2dnLv/c0=2dnL/ fortheleft,andsimilarlyfortheright.Because of the dependence on the index of refraction, the differentcomponentswillexecuteadifferentamountofrotationastheypassthroughthesample.Thefinalorientationofthepolarizationdirectionisdeterminedbythesumof leftandrightvectors,whoseangle is theaverageof the twocomponentvectors,sotheresultingtransmittedwaveshouldberotated(asshown) according to half the difference between their separate angles ofrotation.Thisgivesfortheangleofrotationofthepolarizedbeam,

Thesenseoftherotationisworthclarifying.Accordingtotheequation,theopticalrotationangleispositiveiftheindexofrefractionishigherfortheleftcircularlypolarizedlight,meaningitsspeedthroughthesamplewillslower.Asaresult,thatwavewilloscillatefurther(i.e.executemoreofawave cycle) compared to the right circularly polarized light. But referring to the earlier figureshowingcircularlypolarize lightyouwillnoticethatwhenleftpolarizedlightrotatesfurtherasafunctionofpositionalongthedirectionoftravel,itisactuallyrotatingclockwise;thisisoppositefromthe apparent counterclockwise rotation of the E field vector seen by a fixed observer as the leftcircularlypolarizedtravelingwavepasses.Asaresult,if 0,thentherotationoftheelectricfieldvector is clockwise as shown, as viewedby anobserved looking towards the source. If theopticalrotationisexpressedasamolarquantitybydividingbyconcentration,andalsonormalizedforpathlength,anequationformolaropticalrotationisobtained.

100 100

AswiththeCDeffect,formostmoleculesinsolutionthe(unitless)differenceinindexofrefraction

isverysmall,perhaps10‐5.Butfromtheequationaboveyoucanseethatbecausethepathlengthdisoftenabout104ormoretimeslongerthanthewavelength,thattheopticalrotationisoftensubstantialandcanbemeasuredaccurately.Again,thisrequiresanadditionpolarizingfilterbetweenthesampleandthedetector.Opticalrotationcanbeusedtoidentifychiralmoleculesanditis particularly useful in organic chemistry for evaluating the enantiomeric purity of a syntheticproduct;aracemicproduct,beingcomposedofequalamountsofbothenantiomers,wouldshownoopticalrotation.Theuseofopticalrotationhasastoriedpast. Intheearly1800’s itwasobservedthat individualquartz crystals, which grow in two mirror‐related forms rotated polarized light in differentdirections.In1849LouisPasteurtookacrystallizedsampleoftartaricacid(a4‐carboncompoundwithtwochiralcenters–themeaningofwhichwasunknownatthattime)separatedthecrystalsintotwopilesaccordingtotheirapparentlymirror‐relatedmorphologyanddiscoveredremarkablythatthedissolvedcrystalsofmirror‐relatedmorphologyrotatedpolarizedlightinoppositedirections.

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That experiment came several decades before atomic structures were determined for anycompounds, at a time when theories of bonding and the atomic structure of matter were stillundeveloped. With regard to the apparent luck and insight required to make that discovery,Pasteur’sownwordsarenotable– “Dans leschampsde l'observation lehasardne favoriseque lesespritsprepares”[Inthefieldofobservation,chancefavorsonlythepreparedmind].OpticalrotationandcirculardichroismareinterrelatedThephenomenaofopticalrotationandcirculardichroismarerelatedtoeachotherand theyoccur together in the samemolecular sample,as illustrated.Both arise from complex relationships between electric transition dipolemoments(whichwediscussedbrieflyearlier)andmagneticdipolemomentsinamolecule.Wewillonlyoffertouchonthesubjectqualitativelyhere.Animportantpointisthattheeffects(likeotherspectroscopicphenomena)havetodowithallowableenergytransitionsinamolecule.Asaresult,theobservedeffects are stronglywavelengthdependent. Indeed the common termORD(optical rotary dispersion) comes from the wavelength dependence of theoptical rotary effect. The CD and ORD effects are strongest at or nearwavelengthswheresomeunderlyingabsorptiontransitionoccurs. AnexactintegralrelationshipexistsbetweenORDandCDintheformofaKramers‐Kronigtransform,whichwewill not discuss here, but in its simplest form therelationship leads to a characteristic resultdiagrammed here for an idealized electronictransitionwhosemaximumabsorptionwouldbeatmax. Molecules with complex absorption spectragivemorecomplexCDandORDspectra.Oneresultoftheintegralrelationshipthatgivesfrom is thatevenwhen thecirculardichroismpeak (which relates to absorbance differences) issharplypeakedattheabsorptionmaximumandisweak elsewhere in the spectrum, the opticalrotationsignalmaybeappreciableatwavelengthsfarther from the transition. In some sense thisamounts to a smoothingout effect. TheCD signal froma complexmoleculemay thereforeoffersharper distinguishing features, which can be important in analyzing detailed behavior andconformations,whiletheadvantageofopticalrotationisthatitseffectscanoftenbeobservedinthevisiblerangeofthespectrumevenifthemoleculebeingstudiedhasstrongelectronictransitionsonlyinthefarUVregion.Pasteur’startaricacidisacaseinpoint.Theopticalrotationphenomenonanditswavelengthdependencecanalsobedemonstratedeasilywithsimplecornsyrupowingtoitshighconcentrationofchiralsugars.

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CDstudiesforanalyzingproteinsecondarystructureCDspectroscopyiswidelyusedtomonitortheconformationofproteins.Therearestrongtransitionsfrom the polypeptide backbone in the 200‐220 nm range, so CDmeasurements on proteins aretypicallymade in the surrounding range. A particularly common use is to estimate the percentcompositionofthebasicsecondarystructureelements–alphahelix,betasheet,and‘randomcoil’–inaprotein.Thiscanbeinformativeifthethreedimensionalstructureoftheproteinisnotknowninmore detail from other techniques, or if one is concerned about whether a protein is foldedproperly. Aswehavediscussed,undervariousconditions,oraftermutationshavebeenmade,aproteinmaybecomepartiallyortotallyunfolded.ThedifferenttypesofproteinsecondarystructurehavedistinctiveCDspectra(shownhere),whichhavebeenstablishedwithmodelpolypeptidesorproteins.Clearly,ifonemeasurestheCDspectrumofanunknownproteinanditmatchespreciselytooneofthethreereferencespectra(alpha,beta,orrandom), then you could surmise thatthe protein in question was entirelyhelical, entirely beta sheet, or entirelyunfolded. This is of course rarely thecase. Instead, after recording a CDspectrumoneisgenerallyfaceswiththeproblemofhowtodecomposeitintoasumofthereferencespectra,weightedaccording to the estimated fractionalcontribution each makes to the totalobserved spectrum. Virtually allspectroscopic techniques give additivebehavior with respect to multiplecomponents that are present in amixture, and CD is no different. As aresult, we can write a series of linearequationsstatinghowthereferenceCDvalues at each wavelength would beexpectedtosumto thevalueobservedintheunknownsample.Ateachwavelength,i,wecanwriteanequationoftheform

obs(i)=f(i)+f(i)+frr(i)whereobsistheobservedellipticityandf,f,andfr,aretheunknownfractionsofalpha,beta,andrandomcoilthatmakeuptheproteinunderstudy.Theothertermsintheequation,e.g.(i),areknownquantitiesbasedonthereferencecurves.Aseriesofequationsatdifferentwavelengthscanbewritteninmatrixformasshown.

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… … … …

Ideallywewouldwritealargenumberofequationsatmanydifferentwavelengthvalues.Thiswould

giveusasystemofnequationinthreeunknowns(f,f,fr),withn>>3.Asyoumayknow,ifthenumberofequationsislargerthanthenumberofunknowns,thentheremaybenoexactsolutionfortheunknownsforwhichalltheequationsaresatisfied.Thekeyistodeterminewhatvaluesfortheunknownsgivethebestagreementoverallwiththeequationsprovided.Thegeneralsolutiontothisproblemisbythemethodoflinearleastsquares.Morecomplextreatmentsarepossibleinwhichdifferentweightsaregiventodifferentmeasurementsaccordingtotheiruncertainties,butwewillgivethesimplesttreatmentherewherealltheestimatederrorsareassumedtobeequal.Thentheoptimal solution can bewritten out relatively easily. First,wewill shorten our notation for the

systemofequationaboveas:avector (ofdimensionn)isequaltoarectangularmatrixA(nrows

by3columns)timesavector (ofdimension3,representingthequantitiestobedetermined(f,f,andfr,).

ThenmultiplyingontheleftbythetransposeofthematrixA,AT,weget

isasquare3x3matrixthatcanbeinvertedtogiveitsreciprocal, .Thenmultiplyingonbothsidesontheleftby gives

Thisisastraightforwardcalculationtoperformbycomputer,makingiteasytoobtainestimatesofthesecondarystructurecompositionfrommeasuredvaluesoftheellipticityatseveralwavelengths.Thelinearleastsquaresapproachaboveisextremelypowerfulandcanbeappliedtowiderangingproblemswherenumerousequationscanbewrittenintermsofasetofunknownvariables.

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CHAPTER19

MacromolecularStructureDeterminationandX‐rayCrystallographyOurcurrentunderstandingofbiologydwarfswhatwasknownonlyafewdecadesago.Duringthattime, two areas of study have driven genuine scientific revolutions: genome sequencing andstructuralbiology.Thischapterfocusesonthelattersubject.Thepowerofstructuralbiologyrestsontheadagethatseeingisbelieving.Andindeedlearningandseeingwhatmacromolecules look like in atomic detail has changed thewaywe understand theworkingsof the cell and all its components. This chapterwill focuson thediffraction techniqueknownasx‐raycrystallography. Onlya fewcommentsonother importantmethods instructuralbiologywillbeoffered,mainlyincounterpoint.Wewillseeshortly thatat itsheart,x‐raycrystallography isa typeof imagingmethod; thereareimportantcomplications,butintheendoneobtainsatruethree‐dimensionalimageofthemoleculeunderstudy.Onthisissueacontrastcanbedrawntonuclearmagneticresonance(NMR),whichisthesecondleadingmethodforstudyingmacromolecularstructure.NMRmethodsprobethecomplexinteractions between nuclear spins in amolecule and powerful externalmagnetic pulses. Fromsophisticated analyses of those interactions, sometimes relying on special biochemical protocolsinvolving isotopic labeling of specific atom types or residues in amacromolecule, information isextractedabouttheproximityandrelativeorientationbetweendifferentaminoacidsintheprotein(ornucleotides in thecaseofnucleicacidmolecules). This leadsultimately toa largenumberofinferredspatialconstraintsthatmustbeobeyedbyacorrectatomicmodel.Computerprogramsthenattempt to generate a set of atomic coordinates that is most consistent with the body of NMRconstraints,alongwithotherknowninformation(chieflytheaminoacidornucleotidesequence).Ifasufficientnumberofspatialconstraintscanbeobtained,thenanaccuratemodelcanbeproduced.WithNMR,theexperimentalchallengesincreasesteeplywithmolecularsizeandcomplexity,butnewmethodscontinue topush the limitsof size. Inaddition,NMRmethodsoffervaluabledynamicalinformation about macromolecules that is difficult to obtain by other methods, including x‐raycrystallography.Wewillshortlydiscusstheimportanceinimagingmethodsofusingasufficientlyshortwavelengthfortheradiationsourceinordertogetdetailedstructuralinformation.X‐raysfitthatrequirement,butthehighenergyelectronsusedinelectronmicroscopyalsofitthatrequirement;theyhaveveryshort(DeBroglie)wavelengths.Yetdespitethesufficientlyshortwavelengthofferedbyhighenergyelectrons,untilrecentlyelectronmicroscopyhasnotbeenabletoproduceimagesofmacromoleculesinatomicdetail.Thereasonsarecomplex,buttheyconcerntwointerrelatedissuesofinstrumentsensitivity and the strongly destructive interaction of electrons with biological materials (likeproteinsandnucleicacids).Butitappearsthatthoselimitationsarefinallyfallingaway.Veryrecentinstrumentationdevelopmentshaveproducedsystemswithdetectorsensitivitieshighenoughthat,ifsufficienteffortisappliedtocollectverylargenumbersofmolecularimages,atomicleveldetailcanindeedbeobtainedbyelectronmicroscopyinfavorablecases.Electronmicroscopymethods,and

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NMRmethodsaswell,aresuretocontinuetogrowinpowerandtocontributeincreasinglytoourbody of knowledge in the area of structural biology. But we turn now to themethod that hascontributed so enormously to our understanding of the three dimensional structures ofmacromolecules:x‐raycrystallography.ThelimitingeffectofwavelengthInordertoexplainwhyx‐raycrystallographyisnecessary,wehavetounderstandthefundamentallimitingeffectthatthewavelengthhasinanimagingexperiment.Onewaytounderstandthispointistoaskhowdifferentthescatteringisfromtwopointsinspacethatareseparatedbyadistanced,ifthewavelengthoftheradiationis.Besidesdand,theansweralsodependsonthegeometryofthe scattering, as shown. Scattering phenomenadependonhow lightwaves interferewith eachother,andwhetherlightwavesinterferewitheachother(constructivelyordestructively)dependsontherelativephasesofthescatteredwaves,andthisdependsinturnontherelativedistancethelightwavestravelwhentheyscatterfromdifferentpointsinspace. IfthepathalightwavetakesfromitsorigintoadetectoristhesamewhetheritscattersfrompointAorpointB,thenthosetwopointsinteractwiththewaveinaneffectivelyindistinguishableway.Indeed,thedistinctionbetweenthe scattering from two points comes from the ‘path length difference’ for rays traversing pathsthroughthosepoints.Intheschemeshown,thepathlengthdifferenceis2 .Advancedtextsin different fields of study address the nextpoint indifferentways. Herewewill erronthesideofsimplicityandjustarguethatifthepath length difference is short compared tothe wavelength of the radiation, thenscattering from the two points is not sodifferent,andanopticalexperimentbasedonthe indicated set‐up (of d and ) would notclearlyresolvethetwopoints.Ifwepresstheargumentandsaythenthatthelevelofdetailorspatial‘resolution’disdefinedbyrequiring2 tobecomparableto, thenweseethattheminimumpossiblevalueford(i.e.thefinestdetailthatcouldberesolved)islimitedby/2(whichoccursat=90°).Thisiswhylight(orUV)microscopycannotprovidespatialdetailbelowafewhundrednanometers,nomatterhowlargeorperfectthelensesare.ThisfundamentallimitationthatthewavelengthplacesonresolutionissometimesreferredtoasthediffractionlimitortheAbbelimit.Someveryspecialtricks–somehavingtodowiththepowerofstatisticalaveragingandsomehavingtodowithspecialinstrumentation–havebeendevelopedoverthelastfewyearstocircumventthediffractionlimit;thesetechniquesaresometimesgroupedunderthemonikerof‘super‐resolution’microscopy,recognizedbytheNobelPrize inChemistry in2014. Settingasidesuchspecialtechniques,thelimitingeffectofthewavelengthmeansthatinordertoresolveatomicleveldetails inmolecules,wehave touse radiationwith awavelengthnotmuch longer than the

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separation between atoms, which is 1 to 2Å. That corresponds to the x‐ray region of theelectromagneticspectrum,hencex‐raycrystallography.X‐raysandtheproblemoffocusingX‐rayradiationprovidestheanswerweneedwithrespecttowavelength,but italsointroducesacriticalproblem.Inatypicalimagingexperiment,usingacameraoratelescopeoryoureye(orevenmagneticlensesforthecaseofelectronmicroscopy),thephotonsorwavesthatarescatteredfromtheobjectunderstudyarefocusedbacktoformthe(typicallyenlarged)image.Butx‐rayscannotbefocused, at least not to a practical degree, because there are no materials with a high index ofrefraction for x‐ray. So, using x‐rayswe can do the first part of an imaging experiment (i.e. thescattering)butnotthesecondpart(focusing).Tounderstandwhatmustbedoneinstead,considerthat,thoughthereisnosuitablelensforx‐rays,informationsufficienttocreatethedesiredimagemustbecontainedinthescatteredwavesthatarriveatthelenslocation;ifthex‐rayscouldbefocusedthen an image would be formed. The solution to the problem then is to record the scatteredinformation and figure out what mathematical relationship is required to convert the observedscatteringbackintoanimage–thatis,todousingacomputerwhatalenswoulddonaturally. Itturnsoutthatthatrelationshipiswell‐understood.Anobjectanditsscatteringpatternarerelatedby a Fourier transform, an integral transform that is ubiquitous in mathematical physics andengineering.Beforewediscusshowsuchoperationsrelatetoapplicationsinx‐raycrystallographywehavetoconsidercertainaspectsofhowrepetitiveobjectslikecrystalsscatterradiation.DiffractiongeometryFrom earlier physics experiments you are likely familiar with the basic idea that when light isscatteredfromaregularlyrepeatingobject,likealightoralaserpassingthroughasetoffineslits,one get destructiveinterferencealmosteverywhere, butconstructiveinterference (i.e.bright spots) at aseries of specialpositions.Constructiveinterference occursatdiffractionangleswhere the lightpassing throughdifferent slits havepath lengthdifferencesthatare

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integralmultiplesof thewavelengthof the light. Destructive interferenceoccurselsewhere. Theexampleshownisessentiallyaone‐dimensionalsystem,slitsrepeatinginonedirection.Describingdiffractionfromobjectsthatrepeatinmoredimensionsbecomesabitmorecomplex,butthetwodimensionalcasecanbeillustratedclearly.We can imagine scatteringfrom a two‐dimensionalcrystal where a moleculerepeats regularly in the xand y dimensions. Let therepeat distance along x be|a|,andtherepeatdistancealongybe|b|.Theaandbvectors define theboundaries of a ‘unit cell’,whose contents could beused to construct the entire (ideally indefinite) crystal by translational shifts; for our simplifieddiscussionswewillbeignoringthepossibilityofrotationsymmetrywithinthecrystal.Theaandbvectorsalsodescribealatticeofpointsembodyingthepropertiesofthetranslationalrepetitioninthe crystal. We canunderstand the geometryofdiffractionbymomentarily forgettingabout theunderlyingstructureoftheobjectinthecrystalinordertofocusontherepeatinglattice.Thelatticecaptures the relationship in the crystal between equivalent atoms belonging to molecules fromdifferentunitcells.Forexample,ifyouconsideredjusttheC‐alphaatomofthefirstaminoacidintheproteinmolecule,all the instancesof thatatomthroughoutthecrystalwoulddescribethecrystallattice.Nowwecanapplywhatweknowaboutscatteringfromarepeatingobjecttothissystemoflatticepoints.Scatteringwillbeconstructiveforsomechoicesofthedirectionoftheincomingandoutgoingbeams,butitwillbedestructiveformostchoices,givingnointensityfortheoutgoingbeaminthosecases.Ausefulpointisthatifthescatteringforaparticularchoiceofincomingandoutgoingbeam directions would be destructive for scattering from the repeating arrangement of oneparticularatomintheproteinmolecules(i.e.theC‐alphaatomalludedtoearlier),thenthescatteringwouldalsobedestructivewhenconsideringthearrangementofsomeotherparticularatomintheprotein.Inotherwords,ifaparticularchoiceofincomingandoutgoingbeamdirectionswouldgivedestructiveinterferencefromthecrystallatticepoints,thentherewouldbedestructiveinterferencefromtheentirecrystal,regardlessofwhatthemolecule looks likeorhowitsatomsarearrangedinternally.Thegeometryofdiffractionisthereforedictatedonlybytherepeatpatterninthecrystalandnotbythecontentsoftheunitcell.Thisimportantsimplificationallowsustoproceedtodiscussdiffraction geometry separately from thequestionofmolecular structure. [Wewill see later thatwhilethelatticegeometryalonedetermineswhereweseediffraction,themolecularstructurewithinaunitcelldetermineswhichdiffractionspotsarebrightandwhichareweak,andthatinformationisultimatelythebasisforstructureanalysis].

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The key relationship betweenthe incoming x‐ray beamdirection and the outgoingdirection is captured by thescattering vector, S. First wedefine the incoming andoutgoing directions by unitvectors and . Then, adiagram and a little algebrashows us that the vectordifference between theoutgoingandincomingunitvectors, – isavectoroflength2sin.Frombefore,ourconditionforconstructiveinterferenceforscatteringorreflectingfromplanesthatareseparatedbydistancedis2dsin=n,or2sin/=n(1/d). Bysubstituting2sin=( – )weget( ‐ )/=n(1/d).Thismotivatesustodefineanewvector,thescatteringvectorS,tobeS=( ‐ )/.Thescatteringvectorbisectstheoutgoingvectorandthe(negated)incomingvector. AccordingtothealgebrausedtoconstructS,forconstructiveinterferenceSisperpendiculartothereflectingorBraggplanesdrawnonthelattice,andthelengthofSmustsatisfy|S|=(1/d)nS is defined geometrically by the incoming and outgoing beamdirections, and scattering is onlyconstructivewhenS follows the equation above. But the planeswe drew in the diagram aboveillustratejustonepossiblewaythatparallelplanescanbedrawnonalattice.Apracticallyunlimitednumberofchoicescanbemadeforasetofplanesrunningthroughthelatticeatdifferentangles.Butifwe just choose twodirections as our foundation,we can setupa system fordescribing the2‐dimensionaldiffractioncompletely.Herethelatticehasbeendrawntobeorthogonal(i.e.rectangularinsteadofoblique).Thisisnotnecessary,andinfactmanycrystalhavenon‐orthogonalunitcells,butwewilltreattheorthogonalcasebecausethealgebraissimplerthere.Itmakessensetochooseourplanestobehorizontalorvertical.Referringtoourpreviousdiagramofthetwo‐dimensionalcrystal,fortheverticalplanesalongb,thespacingwouldbe|a|,andtherewouldbediffraction(i.e.constructivescattering)foranSvectorperpendiculartob(andthereforealonga)andhavinglength1/|a|timesaninteger,whichwewillcallh.Forreflectionfromhorizontalplanesalonga,Swouldbealongb andhave length1/|b| times an integer, k. Note the reciprocal relationship between thelengthsoftheunitcelledges|a|and|b|andthelengthoftheSvectorwherewegetdiffraction.AtthispointthereisutilityinintroducinganewsetofbasisvectorsfordescribingtheSvector.Owingtothereciprocalnatureoftherelationshipnotedabove,thecoordinatespacewhereweconstructthescatteringvectorSisreferredtoas‘reciprocalspace’.Asbasisvectorsinreciprocalspace,wecreateana*vector(perpendiculartobandhavinglength1/|a|)andab*vector(perpendiculartoaandhaving length1/|b|). That scheme is shown in the following figure. Now, for theS vectorsperpendiculartotheplanesdefinedbythebaxis,wehaveS=ha*.AndforSvectorsperpendiculartotheplanesdefinedbytheaaxiswehaveS=kb*.Butinadditiontothehorizontalandvertical

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planes,wecouldalsodrawsetsofplanesthroughthelatticeatobliqueangles. WeshouldexpectdiffractionforSvectorsperpendiculartothoseplanesaswell,andwiththelengthofSreciprocallyrelatedtothespacingbetweenplanes.Wewillskipafullalgebraictreatment,butitturnsoutthatthe scattering vector S for any choice of planes is described by a linear combination of integralmultiplesofthereciprocalaxesa*andb*.Thatis,S=ha*+kb*(handkintegers)

Thatequationclearlydescribesatwo‐dimensionallatticeofspots(orreallyscatteringvectorsS)inreciprocalspaceforwhichweexpectdiffraction. Everyorderedpair(h,k)definesasetofBraggplanesthroughthelattice,andthoseplanesgivesrisetoareflectioncorrespondingtoascatteringvectorSperpendiculartothoseplanes,whose‘Millerindices’inreciprocalspacearehandk.ThefigureillustratestherelationshipbetweendifferentsetsofBraggplanesthatcanbedrawnonthecrystallattice,thecorrespondingscatteringvectorS,andthelocationoftheresultingreflectioninthe diffraction pattern. The green arrows in the bottom panels indicate the diffraction spot or‘reflection’thatarisesfromtheBraggplanesdrawningreenintheupperpanels.Wecanalsoworkbackwardsfromanobserveddiffractionspotandcalculatewhatthespacingwasbetweenthe(generallyoblique)latticeplanesthatgaverisetothatreflection:fromtheindiceshandkofthediffractionspot,andknowingthelengthsofa*andb*,wecanusethePythagoreanequationtocalculatethelengthofthescatteringvectorS.Then,fromabove(ignoringthenfrombefore),d=1/|S|.Thishasanimportantmeaning.Scatteringfromcloselyspacedplanesinthelattice(i.e.wheredissmall)showsupinthediffractionpatternwheretheSvectorislong,i.e.farthestfromthecenter(whichiswherethedirectbeamwouldhit[h=0,k=0]).Smallvaluesofdcorrespondtoafinelevelof detail (or ‘high resolution’) in the image we ultimately obtain for the crystallized molecule.Therefore, in order to ultimately obtain a high resolution image of the crystallized molecule,

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diffractiondatamustbepresentandrecordedathighanglesofdiffraction(i.e.high).Notefromthefigureswehavedrawnthattheangletheoutgoingbeammakeswiththeincomingbeamisactually2.Ourdiffractiongeometryequationscanbeputintopracticeinseveralways:Suppose thatahorizontalx‐raybeamwithwavelength=1.54Åhitsa crystalandweareable torecordgooddiffractiondataonadetectoratanangleupto50°degreesawayfromthedirectionofthedirectbeam.Whatisthehighestresolution(i.e.lowestvalueofd)fordiffractionspotsthatwouldberecorded?2=50°, =25°.|S|=2sin/andd=1/|S|=/(2sin)=1.82ÅGiventheindicesofareflection,wecancalculatetheresolutionitprovides;wemustalsoknowthereciprocalunitcell.Ifa=100Å,b=125Å,andc=160Å,theresolutionforreflection(24,8,17)wouldbe:d=1/|S|=1/sqrt((24*(1/100Å))^2+(8*(1/125Å))^2+(17*(1/160Å))^2)=3.7Å.Thiscalculationassumesanorthogonallattice;otherwisethecalculationwouldhavetotakeanglesintoaccount.Forthesameunitcellasabove,howmanytotalreflectionsexistinthree‐dimensionswithinthelimitof2.5Åresolution,i.e.whered>2.5Åor|S|<1/(2.5Å)?Thevolumeofa3‐Dsphereinreciprocalspacewithradius1/(2.5Å)isV=(4/3)(1/(2.5Å))^3.Dividingthisbythevolumeoccupiedbyonereciprocalunitcellvolume(a*b*c*),givesabout536,000reflections. Wedidnotdiscuss internalrotationalsymmetryincrystals,butthatwouldmakesomeofthereflectionsequivalenttoeachotherand therefore redundant,butnonethelessyoucanappreciate thevery largenumberofobservedquantitiesthataremeasuredinamacromolecularcrystallographyexperiment,whichisconsistentwith the requirementofproducingan image that candefine thedetailed structureof amoleculetypicallycontainingthousandsofatoms.DiffractioninthreedimensionsOurdiagramsaboveweredrawnfortwo‐dimensionaldiffraction,andtherethepatternscomeoutlikewemightimagine,orlikewehaveseeninaclassroomdemonstrationofalaserpassingthroughafinescreen.Thegeometryofdiffractionfromanobjectthatrepeatsinthreedimensionsisabitmorecomplex.Atanygivenorientationofthecrystalandtheincomingbeam,weareabletoseejusta two‐dimensional slice through the diffraction pattern that exists in our hypothetical three‐dimensionalreciprocalspace. Butthesliceweobserveisnotfromaflatplane,butratherfromasphereintersectingthethree‐dimensionalreciprocallattice.Why?WesawearlierthatthescatteringbehaviorisgovernedbythescatteringvectorS.AndSisthesumoftheoutgoingbeamunitvector

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( ) with the negated incoming beam unitvector( ),dividedby.Nowifthebeamhasafixeddirectionrelativetothecrystal(i.e. theincoming beam and the crystal are bothstationary),sothatsinisfixed,thenthequestionis:whatvaluesofScanpossiblybesampledbyall the allowable directions for the outgoingbeamunitvector ?Interestingly,theanswerisasphere,asshownhere.ThisisreferredtoasthesphereofreflectionortheEwaldsphere.So,fordiffractionfromathreedimensionalcrystal,weseediffractiononlywhereSfallsonasphere(of radius 1/), and where S simultaneouslyfallsonareciprocal latticeofpoints. Thishasthe effect of planes of spots intersecting asphere,andsinceaplaneintersectsasphereinacircle,weseeadiffractionpatternwithspotsseemingtoappearincircularrings.Inordertoobtaininformationonthefullthree‐dimensionaldiffractionpattern,thecrystalmustberotatedaboutanaxiswhilediffractionimagesarerecorded. Anexampleofdiffractionfromacrystalundergoinganarrowrotationisshown.Interpretingsuchadiffractionpattern,e.g.determiningtheindices(h,k,l)of all the spots is a complicatedproblem. Moderncrystallographicprogramscanusuallydo thisautomaticallyforgooddiffractiondata,aprocedureknownas‘autoindexing’.Thiswasnotpossibleintheearlydaysofcrystallography.Then,acrystallographerhadtotakepainstocharacterizethecrystalunitcellandthereciprocal lattice. Fromtheargumentsabove,youcansee that to takeadiffraction image showing values of S in a flat slice of reciprocal space requires changing theorientationofthecrystalrelativetotheincomingbeamduringthefilmexposure,inaveryparticularway. The complexmotion of the crystal and the film and an intervening annular screen can be

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accomplishedbya‘precessioncamera’.Anexampleofaprecessionphotographofaproteincrystalisshown.Owingtothetimerequiredandthecomplexityoftheprocedure,precessionphotographsarerarelyproducedinmoderncrystallographicwork.LimiteddiffractionanddisorderWenotedearlierthatthegeometryofadatacollectionexperimentcansetalimitontheresolutionobtained.A(typicallyflat)x‐raydetectorpanelonlyallowsdatacollectiontoacertainvalueof,andtheresolutionislimitedbyd=/(2sin).Butthegeometryoftheexperimentalsetupisrarelytheelement limiting the resolution. The resolution of an x‐ray diffraction experiment on amacromolecularcrystal ismostoftenlimitedbytheabsenceofdetectablediffractionspotsabovesomescatteringangle;spotsareclearandstrongintheinnerregionofthediffractionpattern,buttheyfalloffandbecomeunmeasurablefartherfromthecenter.Thisnaturallimitinresolutionisadirectreflectionofthedegreeofordervsdisorderinacrystal.Foraperfectcrystal,wheretheproteinatomsinoneunitcellareinidenticalpositionsineveryunitcellallthroughthecrystal,diffractionwouldbestrongtounlimitedresolution. But if theproteinexhibitssubstantialatomicmotionorconformationalvariation,thendiffractionwillvanishatresolutions(i.e.valuesofd)comparabletothosevariations.Strongerx‐raybeamsmakeitpossibletoobserveweakerreflections,andindeedthedevelopmentofsynchrotronsthatproducex‐raybeamsthousandsoftimesstrongerthanhomelaboratorysourcesisamajorreasonforthecurrentabilitytodetermineatomicstructuresofhighlycomplexmacromolecularassemblies,whichoftenyieldonlysmallandweaklydiffractingcrystals.ObtainingtheatomicstructureHowthecontentsoftheunitcellaffectsthediffraction:thestructurefactorequationInourpreviousdiscussions,weimaginedabstractingjustonespecificatomfromeachmoleculeinthecrystal toestablishthe lattice,andfromthereweanalyzedwherethediffractionspotswouldappear, and what their indices (h,k,l) would be. That exercise did not depend on the internalstructureofthemoleculeinthecrystal,andsothepositionsofthespotsevidentlyrevealnothingabouttheunderlyingmolecularstructure.Instead,themolecularstructureismanifestthroughthecharacteroftheindividualwavesthatcomprisethediffractionpattern.Everydiffractionspotorreflectionhasanintensity(i.e.abrightnessordarknessonthedetector,depending on the display device), which is denoted as I(h,k,l) according to the indices of theparticularreflection.Buteachreflectionisawaveandsoalsocarrieswithita‘phase’.Thephasedescribeshowfaradvancedthewavefrontsofawavearecomparedtoareferencewave(whichinourcaseisahypotheticalwavescatteringfromareferencepointattheoriginoftheunitcellofthecrystal).Howdothepositionsofthethousandsofatomsinthecrystalunitcellrelatetotheintensityandphaseofthewavecorrespondingtoreflectionh,k,l?Eachatomscattersinalldirections,soeachreflectionisjustasumofwavesscatteredfromtheatomsinoneunitcell,asshown.

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How do those wavesaddup? Theseparatewaves scattered fromthemanyatomsintheunit cell interfereconstructively ordestructively inmatters of degreeratherthanabsolutely,aswasthecaseforourearlier analysis ofscattering from alattice. To add themupwehavetoaccountfor magnitudes andthe relative phases ofthe waves scatteredfrom the separateatoms.Toafirstapproximation,themagnitudeofthewavescatteredfromeachatomisdeterminedbythenumberofelectronsintheatom.Thephaseisdeterminedbythepositionoftheatom,sincethat(alongwiththevalueofSforthereflectioninquestion)iswhatdeterminesthepathlengthforthewave.Thephaseangle(comparedtohypotheticalscatteringfromapointattheoriginoftheunitcell)isgovernedbythedotproductoftheSvectorforthereflectionandthepositionvectorfortheatom,whichisusuallywrittenasr=xfa+yfb+zfc.Notethatthecoordinatesxf,yf,zf,are‘fractional’relativetotheunitcellaxisvectors,a,b,andc.Theconsequenceisthatthepathlengthdifferenceisr( ‐ ).Toconvertthepathlengthdifferencetoanangularphase,weneedtodividethepathlength difference by , multiply by 2, and thennegatesincea longerpathgivesanegativeshiftofthe wave peaks. The phase angle then becomes2r( ‐ )/, and then substituting S for (sout‐sin)/we obtain 2(rS) for the phase of thewavescatteredfromanatomatpositionrtothereflectionwhosescatteringvectorisS.Afurthersimplificationin notation comes from expanding (rS) incomponents as (xfa + yfb + zfc)(ha* + kb* +lc*).Becauseofthespecificwaythereciprocallatticewasconstructed(aa*=1,ab*=0,ac*=0,andsoon),(rS)can be written as (hxf + kyf + lzf), which makes itclearerhowtocalculate therequiredvalue. Theserelationshipsarediagrammedinthefigurehere.Thephase angle term 2(hxf+kyf+lzf) that relatesscattering from position x,y,z to reflection h,k,lappearsthroughoutcrystallographyequations.

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Nowwehavetoaddupthewavesscatteredfromalltheatomsbasedontheirseparatemagnitudesandphases.Wewillskipoversomebasicmathematicaldetailshereandsimplyexplainthatthewaytoaddupinterferingwaves(ofthesamefrequency)withdifferentphasesistodecomposeeachwaveintoasineandacosineterm.Thatdecompositiondependsonthephase:Ifthephaseiszerowesaythatthewaveisacosinewave.Ifthephaseis90°wesayitisasinewave.Ifthephaseisintermediatethenitdecomposesintobothcosineandsinecomponentsaccordingtothecosineandsineofthephase angle. Then, thewaves can all be added up by adding their cosine components and sinecomponentsseparately.Thefinalsummedcomponentsdescribeanewtotalwavewhosemagnitudeandphaseareembodiedinthevaluesofthetotalcosineandsinecomponents.Inordertocollapsethetwocomponentstogetherasasinglenumbersothatthewavescanallbeaddedtogetherwithasimplesummation,wecastthingsintothespaceofcomplexnumbers,assigningthecosinetermstobe real and the sine terms to be imaginary. Making those representations,we end upwith thestructurefactorequation:

, , 2 2atomatom

The term is the form factor for

atomj(itsnumberofelectronstoa

first approximation). , , iscalled the structure factor forreflectionh,k,l. Thefigureshowsagraphicalrepresentationofhowthe

totalstructurefactor arisesfromthe summation of atomiccontributions.

The structure factor, , , ,describesthewavethatgivesrisetoa specific diffraction spot.Specifically, the brightness orintensityofthespotonthedetectoristhesquareoftheamplitudeofthestructure factor, , ,

, , ,sothemagnitudeofthestructure factor for each reflectionis obtained by taking the square root of themeasured intensity. This leaves uswith onemajor

problem.Wewrotethestructurefactor asavectortoemphasizeitscomplexcharacter.Itcontainsarealandimaginarypart(AandiBabove).Or,viewedanotherway,itisdescribedbyalength(ormagnitude) and an angle in the complex plane,which is the phase of the totalwave. We can

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measurethemagnitudesbutnotthephases;that informationcontainedinthewavesis lostuponcollisionwiththedetector.To summarize, thepositionsof the atoms in theunit cell determine the structure factors for thediffractionpattern.Thestructurefactorforeachreflectionspecifies(1)itsmagnitude,whichcanbemeasured(asthesquarerootofthespotintensity),and(2)itsphase,whichcannotbemeasured.The ability to calculate precisely what the structure factors should be once we have an atomicstructure is an invaluableproperty, aswe shall see later. But, the absenceof phase informationcreatesanimmediateproblem.Withoutthephases,thereisnoeasywaytoworktheproblembackwardsandcalculatewheretheatomsmusthavebeeninordertogiverisetotheobservedstructurefactormagnitudes.Thisplacesthecrystallographyprobleminaclassmathematiciansrefertoasinverseproblems:theresultsofacalculationcanbeworkedonewayeasily,butnottheother.Ifwedohavethephasesforthestructurefactors,thenitiseasytocalculateanimageofthecontentsoftheunitcell.Theessentialproblemincrystallographythenistorecover(atleastapproximatevaluesfor)themissingphases.Thisisknownincrystallographyasthephaseproblem.Beforeturningtothephaseproblem,wewillsimplypointouthowthecontentsoftheunitcellcanbecalculatedfromthestructurefactors,oncethephasesareknown.TherelationshipisaninverseFouriertransform,wellknowninphysicsandengineeringproblems;itdescribesessentiallywhatalenswouldhavedoneifwehadoneforx‐rays:

, ,1

, , , , 2

Inthisequation, , , isthemagnitudeofthestructurefactorand , , isitsangularphase., , istheelectrondensityintheunitcellatpositionx,y,z.Toobtaintheelectrondensityatany

pointx,y,z,asummationisrequiredoverallreflectionsh,k,l,emphasizingthatinformationabouttheelectrondensityisdistributedacrossallthereflections.Wewillsaymoreabouttheelectrondensitycalculationlater,butrecognizingtherequirementforthephasesforallthereflections( , , ),weturnnowtothephaseproblem.PhasingandthephaseproblemThere are twoessentiallydifferentways thephaseproblemcanbe surmounted: (1)bymethodsknownas‘molecularreplacement’and(2)by‘heavyatommethods’orvariationsthereonincludinganomalousphasing; thesemakeuseof strongorunusual scattering fromcertain atom, generallyheavierthanthosenaturallypresentinproteinsandnucleicacids.Molecular replacement ‐ Summarizing briefly, molecular replacement requires that a structurealready be known for a similar molecule, or perhaps part of the unknown molecule. Common

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scenariosincludewhenthestructureisknownforahomologousproteinfromanotherorganism,orwherethestructureisknownforthealphasubunitfromanalpha/betaheterocomplex,orwhereaprotein‐ligandcomplexhasbeencrystallizedandthestructureisalreadyknownfortheproteinbyitself.Inthemolecularreplacementapproach,ifonecanfigureouthowtheapproximatemodelor‘searchmodel’shouldbeplacedintheunknownunitcell,thenstructurefactorscanbecalculatedfrom this oriented search model. The reason for doing this is the structure factor calculationproducesphasesforthereflections.Thephasevaluesso‐obtainedmaynotbesoaccurate,becausetheycomefromsomeapproximatemodel,buttheyareusuallygoodenough.Oneproceedsbytakingthestructurefactormagnitudesfromtheobserveddiffractionpatternfromtheunknowncrystal,andcombinesthemwiththeapproximatephasescalculatedfromthesearchmodel. Thosequantities,placedintheequationabove,canbeusedtoproduceanelectrondensitymap.Becausethephasesdonotcontaininformationfromtheactualunknownstructure,electrondensitymapscalculatedbymolecularreplacementaretypicallybiasedtolooklikethesearchmodel.Overcomingthisbiasisathornyproblem,but the speedand convenienceofmolecular replacementmakes it anattractivechoicewheneveritispossible.Ofcourseifthecrystallizedmoleculeisunlikeanyknownstructure,molecularreplacementisnotanoption.Inaddition,aprobleminmolecularreplacementweglossedoverishowthecorrectorientationandpositioncanbeidentifiedforthesearchmodelintheunitcell.Theshortansweristhatifthesearchmodeliscorrectlyplacedthenthestructurefactormagnitudescalculated fromthatmodel(usingthestructure factorequation)shouldapproximatelymatchthemeasured structure factor magnitudes. When the search model differs substantially from theunknownstructure,molecularreplacementmethodsmayfail,leavingonlyheavyatomandrelatedmethodsasviableroutes.Heavyatommethods –Withheavyatomsmethods,oneobtainsestimatesof thephasesbydoingadditional diffraction experiments after perturbing the atomic structure (by addition of heavyatoms). Making additional measurements of the structure factor magnitudes from perturbedversionsofthecrystalmakesitpossibletobreakthephaseambiguity.Asanexample,supposeyouwere trying to determine the value of an unknown (signed) quantity, and youwere told that itsabsolutemagnitudewas100.Itcouldbe+100or‐100,butyoucan’ttellfromasinglemeasurementof itsmagnitude. Butwhat if youwere able to askwhat the absolutemagnitudewouldbe afterperturbingitinaknownway?Whatifyouweretoldthatifyouadded5tothenumbertheresulthadamagnitudeof95.Thenyouwouldconcludethattheunknownvaluewas‐100,not+100.Thatistheessenceofheavyatommethodsanditsvariations.Thecrystallographicphaseproblemismorecomplicatedbecausewhatismissingisnotmerelythesignofthevaluebutitsphaseangle.Butitturnsouttonotbesomuchmorecomplicated.Thetypicalviewoftheproblemistotakeasinglereflectionandthinkofitsstructurefactorasbeingavectorthatliesonacircleofknownradius,namelythestructurefactormagnitudethatcomesfromthesquarerootofthemeasuredreflectionintensity. Eachstructurefactorhasaphaseassociatedwithit–iftheatomicstructurewasalreadyknownthenthatphasecouldbecalculateddirectlyfromthestructurefactorequation–butnotknowingthephasemeansthatwedonotknowwhereonthecirclethestructurefactorvectorforthatreflectionpoints.Butnow,imaginewehavebrieflysoakedtheproteincrystalinasolutioncontainingaheavyatomcompound,sayHgCl2forexample,andthat

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asaresultthecrystalhadbeenmodifiedinauniformway,saywithasinglemercuryatomboundtoanexposedcysteinethiolineachcopyoftheproteinmolecule.Wecoulddoadiffractionexperimenton this ‘derivatized’ crystal, and we would obtain slightly different structure factor magnitudescomparedtothenativeproteincrystal.Nowforeachreflectionwehavetwomagnitudes,thenativeproteinmagnitude(FP)andthederivativemagnitude(FPH).Nowtheparalleltothescenariolaidoutbeforeshouldstarttocomeintoview.Toproceedfurtherwehavetounderstandpreciselywhatwasaddedtoeachstructure factorof thenativecrystal toproducethederivativestructure factor. Inotherwords,weneedtoknowwhattheheavyatomcontributionwas.Weknowfromthestructurefactorequationthatifweareabletodeterminethelocationoftheheavyatom(s)withintheunitcell,thenwecancalculatedirectlywhatcontributiontheheavyatom(s)madetoeachstructurefactor.Determiningtheheavyatomposition(s)isaseparateproblem–knownas‘solvingtheheavyatomsubstructure’–whichwewillnotdiscussinthischapterasitrequiresspecializedcalculationsandanalyses.Insteadwewillproceedtoexplainhowknowingtheheavyatomposition(s)andbeingabletocalculatetheheavyatomcontributionmakesitpossibletodeterminetheunknownphaseforeachreflection. First,wehavetorecognize that it is thevector(orcomplexvalued)quantities for thestructurefactorsthatdeterminehowtheyaddtogether.Foreachreflection(h,k,l),

where is theheavyatomcontribution,whosevalue including itsphase (or realplus imaginarycomponents)canbecalculatedfromthepositionsdeterminedfortheheavyatom(s).ThebehaviorofthisequationistypicallyillustratedwithaphasecircleorHarkerdiagram,asshown.TheFPandFPH structure factors are depictedas circles, since their phases areunknown at the outset. But thecenters of those two circles areoffsetbyavectoramountdictated

by . Under that construction,

withthe and vectorslaidheadto tail, one sees that theintersections of the two circlesgivestwopossiblesolutionstothevector triangle addition equation.Eitheroftwochoicesforthephase of the native structure factorwould agree with the diffractiondata. So, with information fromonly one heavy atom derivative(referred to as SIR for singleisomorphousreplacement)wearestill left with an ambiguity aboutthecorrectchoiceofphaseangle.Itis possible to proceed with an

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electrondensitycalculationwithanaverageofthetwopossiblephasevalues,andsometimesthisisgoodenough,butclearlyonewouldliketodobetter. ThesolutionisMIR(multipleisomorphousreplacement). Additionalheavyatomderivativesaresought,perhapsusingdifferentheavyatomtypes.Furtherdiagramsarenotgivenhere,butyoucananticipatehowcollectingdiffractiondataonasecondderivatizedcrystalwouldforeachstructurefactorproduceasecondheavyatomcirclewithyetanotherdifferentorigin(relatedbyadifferentheavyatomcontribution),andifthedataarewell‐measuredthenthereshouldbeapointonthenativeproteinphasecirclewheretheothertwocirclesnearly intersect, and this gives a value for the native phase angle. In reality, the presence ofexperimentalerrorsmakestheactualassessmentofthebestphasesomewhatmoreinvolved.Inthelasttwodecades,alargefractionofstructureshavebeendeterminedwithvariationsontheheavyatommethodabove(whichdatestothefirstproteinstructuresofmyoglobinandhemoglobinbytheKendrewandPerutzlaboratories).Thesevariationstakeadvantageofthe‘anomalous’x‐rayscattering of certain atoms, including many heavy atoms but also including lighter atoms, mostnotably selenium. The twist is as follows. Ordinary heavy atom methods gain the additionalinformation required to break the phase ambiguity bymeasuring two different structure factormagnitudesfromtwodifferentcrystals:thenativeproteincrystalandtheheavyatomderivatizedcrystal.Anomalousscatteringmethodsgaintheinformationrequiredforphasingfromtwostructurefactorsfromthesamecrystal.Wedidnotdiscussitearlier,butforscatteringfromanordinarycrystal(i.e.onenotcontaininganomalouslyscatteringatoms),thestructurefactormagnitudesforthe(h,k,l)reflectionandthe(‐h,‐k,‐l)reflectionareidentical.Thisequalityisbrokenbyanomalouslyscatteringatoms,andextrainformationisobtainedthenfromcomparingthemagnitudesofF(h,k,l)andF(‐h,‐k,‐l)foreachreflection.Phasingapproachesinvolvingcombinationsofheavyatomsandanomalousscatteringarepossible,leadingtovariousacronyms(e.g.SIRASforsingleisomorphousreplacementwithanomalousscattering). Thephasecircleconstructionsinthesevariouscasesaredifferentindetail,butthemainidearemainsthesame.Animportantfeatureofanomalousscatteringmethodsisthatseleniumatomsprovidearelativelystronganomaloussignalandcanoftenbeincorporatedseamlesslyintoanativeproteinbyexpressingtheproteininbacteriagrownonselenomethione;apowerfulandgeneralmethoddevelopedbyWayneHendricksonandcolleagues.Electrondensitymaps:obtaininganatomicmodelWehavealreadyseentheelectrondensityequationbywhichwecancalculateanelectrondensityfunction(x,y,z)or‘map’,oncewemeasurethestructurefactoramplitudesandrecoverapproximatevaluesforthemissingphases.Wewilljustnoteherethatthetwoelementsmostcriticaltothequalityoftheelectrondensitymapobtainedaretheaccuracyofthephasesandtheresolution(minimumvalueofd)towhichthedataextend.Electrondensitymapsareshownhereatresolutionsof3Åand1.7Åtoemphasizewhatcanandcannotbevisualized. Dependingonthequalityofthephases,areliabletracingofthepathofthebackbonemayrequirearesolutionofabout3.0Åorbetter.Sidechainidentitiesbecomerelativelyclearatabout2.7Å;priorknowledgeoftheaminoacidsequenceis extremelyvaluable inmost cases. At a resolutionof about1.7Å,holes in ring structuresmay

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become evident. And at about 1.1 Åseparate spherical densities forindividual atoms appear. Hydrogenatoms scatter weakly in x‐raydiffraction so at typical resolutionstheir positions are not visualized butare instead inferred from geometricconsiderations.Modernsoftwarepackagesattempttomodelanatomicstructure,giventheknownaminoacidornucleotidesequence, intoacalculatedelectrondensitymap. Suchautomaticallytracedmodelsusuallyrequireseveralroundsofhumaninspectionandrebuilding,combinedwithautomatedminimizationbycomputer,inordertoobtainafinalmodelthatisreliable.Animportantfeatureofx‐raycrystallographyistheabilitytoquantitatethedegreetowhichafinalatomicmodelagreeswiththeobserveddata;thestructurefactorequationmakesthispossible.Thelevelofdisagreementbetweenthemodelandtheobservedstructurefactormagnitudesisgivenasan‘R‐factor’,whichisjustameasureoftheresidualerroronafractionalscale.Macromoleculesarecomplex. Theyareoftenflexible inwaysthatarehardtocaptureinasinglemodel,andtheyoftenhaveorderedsolventstructurearoundtheirsurfacesthatcanbedifficulttomodel.Asaresult,theR‐valueforarelativelygoodstructuremaybeintherangeof20%,andworsefor structures determined at lower resolution. The final refined structure in a crystallographyexperimentisusuallyalsoinformedbyknowngeometricrestraints,basedonknownvaluesforbonddistancesandangles. Incrystal structures reported in the literature, a standardcrystallographictablewillreportvaluesfortheR‐factoralongwithdeviationsfromidealgeometry.Additionalentriesgivefurtherinformationbywhichanexpertcanassessthequalityandlikelyaccuracyofthereportedstructure.ProteinCrystallizationOften the hardest part of a crystallographic project is obtaining good crystals. Surprisingly,sometimescrystalsthatexhibitexcellentmorphologybymicroscopicexaminationturnouttodiffractpoorly,presumablyreflecting insufficientorderontheatomicscale. Thoughthereare importanttheoretical considerations, protein crystallization is still largely an art. In effect, one drives theproteinornucleicacidoutofsolutionundermanydifferentconditions,perhapsthousands,searchingforconditionsthatgivehighlyorderedcrystals.Themostcommonexperimentalsetupisreferredtoashanging‐dropvapordiffusion.Ineachessentiallyseparateexperiment,atinydropofprotein(typicallyfromatenthtoafewmicroliters)ismixedwithanequalvolumeofa‘reservoirsolution’,whichcontainsaprecipitantofsomekind(highsaltoracrowdingagentlikepolyethelyneglycol),abuffertocontrolpH,andpossiblyothercompoundssuchasmetalionsororganiccompounds.Then

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thatmixeddropishungupsidedownoverthereservoirsolutioninasealedchamber.Becausetheprecipitantismoreconcentratedinthereservoir,waterleavestheproteindropandcondensesinthereservoir by evaporation. The protein solution thereby become more concentrated, and if thesolubility limit is exceeded the protein precipitates. Modern hanging drop experiments arecommonlysetupbyroboticliquidhandlingdevicesin96wellplates.Manyplatesaretypicallysetupduringthesearchforgoodcrystals.Withgoodluck,andassumingtheproteinornucleicacidhasa sufficiently well‐defined three‐dimensional structure, crystals can be obtained. Examples ofcrystals grown in hanging drops are shown. Inmost experiments,well‐diffracting crystals havelineardimensionsinthe50mto500mrange.Crystalsdiffractingtogoodresolutionhavebeenobtainedforwholeribosomes,othergiantcomplexescontainingmultipleproteinandnucleicacidsubunits,membraneproteincomplexes,andnumerouswholeviralcapsidsincludingsomewithlargetriangulationnumbers. Sizeandcomplexitydoesnotpresentafundamentalobstacleaslongasadefinedthree‐dimensionalstructureiswell‐populatedinsolution. Afinalcommentaboutproteinandnucleicacidcrystalsisthatthemoleculesinthesecrystalsarefullyhydrated.Infactthewatercontentintypicalproteincrystalsisinthe40%to50%range.Thisisanimportantreasonwhythestructuresobtainedinthecrystalstatecanbeshowninmostcasestobelargelyunaffectedbycrystalformation,asidefromlocalconformationaleffectswheremoleculescontacteachotherinthecrystallattice.