Chemical Physics II. More Stat. Thermo Kinetics Protein Folding...

Transcription

1 Chmical Physics II Mor Stat. Thrmo Kintics Protin Folding

2 Continuing Thm How can w undrstand th bulk proprtis of a substanc from th proprtis of individual molculs H3C CH 3 CH 3 CH 3 CH 3 CH 3 3D spac filling.png? Dnsity= g/ml Boiling point=69 C Mlting point= 95 C tc.

3 Th Gam Plan Rviw th two stat systm butn cis/trans xampl activation nrgy and kintics Brif intrmzzo on th fr nrgis Gnraliz th two stat problm to an N stat problm W will s how this rlats to th folding of protins This will giv a nw prspctiv on ntropy

4 Butn Gas Rcall, som molculs of th gas will b in th cis conformation, othrs will b in th trans conformation:

5 Butn Gas Th butn potntial nrgy surfac *NB: Spartan can comput ths 16kcal/mol 19kcal/mol *How do you xpct th ratio of cis/trans isomrs to bhav? *Considr both th high T and low T limits. 3kcal/mol

6 Butn Gas cis/trans Probability Thr ky proprtis of our probability function : 1. It should b a function of nrgy and tmpratur P c =P c E c, T and P t =P t E t, T 2. Indpndnt vnts should multiply, i for any two butn molculs P c1, c2 =P c P c ; P c1, t2 =P t1, c2 =P c P t ; P t1, t2 =P t P t 3. Probabilitis should b normalizd to 1 P c P t =1

7 Thr ky proprtis of our probability function : 1. It should b a function of nrgy and tmpratur E c / RT P c and P t E t / RT Boltzmann Factor 2. Indpndnt vnts should multiply P c1, t2 =P c P t E c /RT E t / RT = E c E t / RT Probabilitis should b normalizd to 1 E c / RT P c P t =1= Q E t / RT Q E c /RT whr Q= = E c / RT E t / RT Q E t / RT Partition Function

8 Butn Gas cis/trans Probability So our final rsult was, E c /RT P c= Q E c /RT = E c / RT Pt = Q E t / RT E t /RT E t / RT = E c /RT E t / RT

9 Butn Gas Isomrization Kintics W stimatd th rat at which th a jug of all cis butn will isomriz to trans butn by finding what fraction of th molculs at a givn tim ar at th transition stat this is th sam calculation w just did, but now E=16kcal/mol instad of 3kcal/mol 1 k cis trans P tst s = E c tst RT 1 E c tst RT 12 k cis trans 10 /s 1 s = 16kcal/mol.002 kcal /mol K 300K 1 16kcal / mol.002 kcal/ mol K 300K i 30000yr halflif s 1

10 Fr Enrgy Intrmzzo From last tim, G= H T S H= E P V, but P V 0 for bio. H E G E T S A Gibbs Fr Enrgy Hlmholtz Fr Enrgy (Mor important too us than it might look at first glanc.)

11 Fr Enrgy Intrmzzo W hav now dfind, Th partition function: Q Th Hlmholtz fr nrgy: A If w can rlat A and Q, thn can xprss th fr nrgy in trms of Boltzmann factors can driv othr thrmodynamic paramtrs such as prssur, ntropy, hat capacity, isothrmal comprssibility, tc

12 Fr Enrgy Intrmzzo It can b shown, A= RT ln Q ln Q Pr=RT V ln Q <E>=RT T 2 S= RT ln Q T tc...

13 Gnralizing th 2 Stat Solution W hav now sn that w can think of our butn molcul rattling btwn thr diffrnt stats Th trans stat (lowst in nrgy, most stabl) Th cis stat (highr in nrgy, but still long livd) Th transition stat (vry high in nrgy, short livd) W should b abl to driv th th probability of a butn molcul occupying any of ths thr stats

14 Gnralizing th 2 Stat Solution For 3 stats w find, E t / RT Pt = = Q Pc = Et / RT E c / RT E c / RT Q Q E tst / RT =.99298K ; K E tst / RT =.01298K ; K E c / RT = E c / RT E tst / RT P tst = E t / RT Et / RT E tst / RT = Ec / RT E t / RT 12 E tst =10 / RT 298K K ; 10 whr E t =0kcal / mol ; E c =3kcal / mol ; E tst =16kcal / mol

15 Many Stat Problms Most ral world chmical problms hav vry many stats: Th numbr of ways liquid watr molculs can arrang thmslvs in a cup Th numbr diffrnt shaps a protin can trac Th numbr of diffrnt positions and vlocitis gas particls can assum in a box uni bochum.d/~axl.kohlmyr/cpmd vmd/unitcll movi.gif

16 Protin Folding Protins ar polymrs think of this as a string of bads ach bad is an amino acid ach amino acid has two rotatabl bonds (phi, psi) conncting it to ach othr amino acid th string changs shap by rotating around ths bonds Whn in watr, ach protin will spontanously fold into a spcific, wll dfind, shap

17 Protin Folding Most protins contain amino acids Each amino acid can assum roughly tn diffrnt pairs of phi and psi angls So th total conformational spac (i all th shaps it 100 can fold into) is 10 stats Our goal is to undrstand th thrmodynamics of protin folding

18 Protin Folding Visualization of th ~10 stabl stats for ach amino acid

19 Protin Folding

20 Protin Folding Th singl foldd stat of a givn protin is typically quit complicatd It must b in this fold to function Prion disass ar causd by unfoldd protins

21 Protin Folding This is a many stat problm Thr xists 1 singl foldd stat with nrgy EF Thr xist ~10100 unfoldd stats with nrgy Eu W will trat ach of th unfoldd stats as though thy hav th sam nrgy W will s th larg numbr of unfoldd stats opposs folding (ntropy) th favorabl intraction nrgy of th foldd stat promots folding (nthalpy)

22 Protin Folding W find for protin bing foldd or unfoldd, E F /RT Q= PF= P U = E U /RT = E F /RT 100 E F / RT 10 E U / RT Q E U / RT E F /RT = Q E F / RT = E F / RT 100 E U / RT Dgnracy =10100 E U / RT E U /RT What can w larn about protin folding from this solution?

23 Protin Folding Exprimntally, Most protins ar narly ntirly foldd at 25oC Most protins ar half unfoldd at 75oC Most protins ar narly ntirly unfoldd at 90oC Notic that this procss is much mor tmpratur dpndnt than th butn isomrization For th rst of th discussion w will choos our nrgy scal such that EF=0 and EU= EU

24 Protin Folding W can calculat E U by forcing P F 350K, PF= E F /RT E F / RT E U / RT = E U / R 350K.5 E U =161.2kcal / mol What dos this say about th occupancy of th foldd stat at diffrnt tmpraturs? 1 11 P 300K = 1 and P K 161.2kcal / mol R 300K

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26 Protin Folding W can calculat A of unfolding from this drivd data, A F U = RT ln QU QF E U / RT = RT ln [ A F U = RT ln E U / RT 1 RT ln A F U = E U RT ln but w hav alrady showd, A F U = E U T S SU =R ln ]

27 Protin Folding Th valu of A at 300K is also intrsting, E U / RT A F U = RT ln [ ] A F U = E U RT ln A F U =161.2kcal / mol 138.2kcal / mol=23kcal / mol For prspctiv, a watr watr hydrogn bond is 5kcal / mol

28 Conclusions Protins hav a rlativly small and abrupt window of thrmal stability this is bcaus both th ntropic and nthalpic trms ar vry larg and opposd Ths mthods of calculating E, A,and S ar at th cor of computational chmistry W typically gt E from a modl Onc w hav E for ach stat w can calculat bulk thrmodynamic proprtis