PROTEIN PHYSICSPROTEIN PHYSICS

LECTURE 5LECTURE 5

Energy: E (enthalpy: H=E+PV)Energy: E (enthalpy: H=E+PV)PV=NkPV=NkBBT for gases, T for gases, PV << NkPV << NkBBTT for solids & liquidsfor solids & liquids

Entropy: S = kEntropy: S = kBB••ln(#STATES)ln(#STATES)If AB=A+B If AB=A+B E EABAB= E= EAA+E+EBB; V; VABAB= V= VAA+V+VBB; ; SSABAB= S= SAA+S+SBB while #while #ABAB= #= #AA•• # #BB

Free energy: G = H-TS Free energy: G = H-TS F F = = E-TS E-TS --------------

G: used when P=constG: used when P=const forfor solids & liquids solids & liquids

Chemical potentialChemical potential: : G G(1)(1) = G/N = G/NProbability ~ exp(-G/kProbability ~ exp(-G/kBBT) T) max max G G min min

G = H-TS; dG = (dH-TdS) - SdTG = H-TS; dG = (dH-TdS) - SdTAt T=const (i.e., at dT=0):At T=const (i.e., at dT=0):at equilibrium, G at equilibrium, G = min= min, i.e., dH-TdS=0., i.e., dH-TdS=0.Thus:Thus:1)1) T = dH/dST = dH/dS2)2) S = -dG/dTS = -dG/dT3)3) H = G+TS = G-T(dG/dT)H = G+TS = G-T(dG/dT)

Probability ~ exp(-G/kProbability ~ exp(-G/kBBT)T)

kkBB = 2cal/(6= 2cal/(610102323) = 2cal/mol = R ) = 2cal/mol = R

kkB B 300 300oo = R = R 300 300oo = 0.6 kcal/mol = 0.6 kcal/mol

intint: “: “Free energy of interactionsFree energy of interactions””

Chemical potential: Chemical potential: GG(1)(1) = G = Gint int - T- T••kkBBln(Vln(V(1)(1)) ) G Gint int + T+ T••kkBBln[C] ln[C]

EQUILIBRIUM for transition EQUILIBRIUM for transition of molecule of molecule 11 from from A A toto B: B: GGAA

(1) (1) = G= GBB(1) (1)

chemical potentials in chemical potentials in AA and and BB are equal are equal

GGintintAAB B = G= Gintint

B B – G– GintintAA

GGintintAABB= k= kBBTT••ln([Cln([CinAinA]/[C]/[CinBinB])])

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Experiment: Experiment: GG intint

AABB= k= kBBTT••ln([Cln([C11 inin AA]/[C]/[C11 inin BB])])

SSintintAABB = -d( = -d(GGintint

AABB)/dT)/dT

HHintintAABB = = GGintint

AABB +T +TSSintintAABB

CC66HH1212

[C] of C[C] of C66HH1212

in Hin H22O:O:50 times 50 times lesslessthan in gas;than in gas;100000 times 100000 times less less than in than in liquid Cliquid C66HH1212

T=298T=29800K=25K=2500CC

-2/3-2/3 +1/3 +1/3

Loss: Loss: S S (usual case)(usual case)

-2/3-2/3

Loss: Loss: LARGE LARGE EE (rare case)(rare case)

H-bond: directedH-bond: directed

High High heat capacity heat capacity d(d(H)/dT:H)/dT:Melting ofMelting of““iceberg”iceberg”

20-25 cal/mol per 20-25 cal/mol per ÅÅ22 of ofaccessible non-polar surfaceaccessible non-polar surface

____________largelargeeffecteffect

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