Solutions Manual For Biomolecular Thermodynamics From Theory to Application 1st Edition By Douglas Barrick (All Chapters)

Solutions Manual for
N N N




Biomolecular Therm N




odynamics, From TheN N




ory to Application, 1e
N N N N




Douglas Barrick (All
N N N




Chapters)

,Solution Manual N




CHAPTER 1 N



1.1 UsingNtheNsameNVennNdiagramNforNillustration,NweNwantNtheNprobabilityNof
NoutcomesN fromN theN twoN eventsN thatN leadN toN theN cross-

hatchedN areaN shownNbelow:




A1 A1N nNB2 B2


ThisNrepresentsNgettingNANinNeventN1NandNnotNBNinNeventN2,NplusNnotNgettingNA
inNeventN1NbutNgettingNBNinNeventN2N(theseNtwoNareNtheNcommonN“orNbutNnotNb
oth”NcombinationNcalculatedNinNProblemN1.2)NplusNgettingNANinNeventN1NandNBNi
nNeventN2.

1.2 FirstNtheNformulaNwillNbeNderivedNusingNequations,NandNthenNVennNdiagramsN
willNbeNcomparedNwithNtheNstepsNinNtheNequation.NInNtermsNofNformulasNandNp
robabilities,NthereNareNtwoNwaysNthatNtheNdesiredNpairNofNoutcomesNcanNcom
eNabout.NOneNwayNisNthatNweNcouldNgetNANonNtheNfirstNeventNandNnotNBNonNthe
secondN (NA1N∩N(∼B2N)).NTheN probabilityN ofN thisN isN takenN asN theN simpleN product,N since
eventsN1NandN2NareNindependent:

pA1N∩N(∼B2N)N =N pAN×Np∼B
=N pAN×(1−NpB (A.1.1)
N)

=N pAN−NpApB

TheNsecondNwayNisNthatNweNcouldNnotNgetNANonNtheNfirstNeventNandNweNcouldNget
BN onN theN secondN ((∼NA1)N∩NB2N)N,NwithN probability

p(∼A1)N∩NB2N =N p∼AN×NpB
=N(1−NpAN)×N (A.1.2)
pB
=N pBN−NpApB

,2 SOLUTIONNMANUAL


SinceNeitherNoneNwillNwork,NweNwantNtheNorNcombination.NBecauseNtheNtwoNway
sNareNmutuallyNexclusiveN(havingNbothNwouldNmeanNbothNANandN∼ANinNtheNfirstN
outcome,NandNwithNequalNimpossibility,NbothNBNandN∼B),NthisNorNcombinationNisNe
qualNtoNtheNunionN{NA1N∩N(∼B2N)}N∪N{(∼NA1)N∩N B2},NandNitsNprobabilityN isNsimplyNtheNsu
mNofNtheNprobabilityNofNtheNtwoNseparateNwaysNaboveN(EquationsNA.1.1NandNA.1.2):

p{A1N∩N(∼B2N)}N∪N{(~A1)N∩N B2}N =N pA1N∩N(∼B2N)N +Np(∼A1)N∩N B2
=N pAN−NpApBN+NpBN−NpApB
=N pAN+NpBN−N2pApB

TheNconnectionNtoNVennNdiagramsNisNshownNbelow.NInNthisNexerciseNweNwillNwork
NbackwardNfromNtheNcombinationNofNoutcomesNweNseekNtoNtheNindividualNoutcom

es.NTheNprobabilityNweNareNafterNisNforNtheNcross-hatchedNareaNbelow.
{NA1N∩N(∼B2N)}N∪N{(∼NA1)N∩NB2N}




A1 B2


AsNindicated,NtheNcirclesNcorrespondNtoNgettingNtheNoutcomeNANinNeventN1N(left)
NandNoutcomeNBNinNeventN2.NEvenNthoughNtheNeventsNareNidentical,NtheNVennNdi

agramNisNconstructedNsoNthatNthereNisNsomeNoverlapNbetweenNtheseNtwoN(whichN
weNdon’tNwantNtoNincludeNinNourN“orNbutNnotNboth”Ncombination.NAsNdescribedN
above,NtheNtwoNcross-
hatchedNareasNaboveNdon’tNoverlap,NthusNtheNprobabilityNofNtheirNunionNisNtheNsi
mpleNsumNofNtheNtwoNseparateNareasNgivenNbelow.

A1NnN~B2
~NA1NnN
B2

pAN×Np~B
p~AN×NpB
=NpAN(1N–NpB)
=N(1N–
NpA)pB

~NA1NnNB2
A1NnN~B2



AddingNtheseNtwoNprobabilitiesNgivesNtheNfullN“orNbutNnotNboth”NexpressionNa
bove.NTheNonlyNthingNremainingNisNtoNshowNthatNtheNprobabilityNofNeachNofNth
eNcrescentsNisNequalNtoNtheNproductNofNtheNprobabilitiesNasNshownNinNtheNtopN
diagram.NThisNwillNonlyNbeNdoneNforNoneNofNtheNtwoNcrescents,NsinceNtheNoth
erNfollowsNinNanNexactlyNanalogousNway.NFocusingNonNtheNgrayNcrescentNabove
,Nit
representsNtheNANoutcomesN ofNeventN1NandNnotNtheNBNoutcomesNinNeventN2.NEa
chNofNtheseNoutcomesNisNshownNbelow:

EventN1 EventN2


A1 ~B

p~BN =N1N–N pB
pA



A1 ~B2

, SOLUTIONNMANUAL 3


BecauseNEventN1NandNEventN2NareNindependent,NtheN“and”NcombinationNofN
theseNtwoNoutcomesNisNgivenNbyNtheNintersection,NandNtheNprobabilityNofNth
e
intersectionNisNgivenNbyNtheNproductNofNtheNtwoNseparateNprobabilities,NleadingNt
oNtheNexpressionsNforNprobabilitiesNforNtheNgrayNcross-hatchedNcrescent.

(a) TheseNareNtwoNindependentNelementaryNeventsNeachNwithNanNoutcomeNp
robabilityN ofN 0.5.NWeN areN askedN forN theN probabilityN ofN theN sequenceN H1NT2
,NwhichNrequiresNmultiplicationNofNtheNelementaryNprobabilities:

1NNN 1 1
pHHN =N H1N∩NT2N=N pHN ×N =NNN ×NNN =
pT
1NN 2 1 2
2NNN 2 4

WeNcanNarrangeNthisNprobability,NalongNwithNtheNprobabilityNforNtheNoth
erNthreeNpossibleNsequences,NinNaNtable:


TossN1

TossN2 HN(0.5) TN(0.5)

HN(0.5) H1H2 T1H2
(0.25) (0.25)

TN(0.5) H1T2 T1T2
(0.25) (0.25)

Note:NProbabilitiesN areN givenN inN parentheses.

TheNprobabilityNofNgettingNaNheadNonNtheNfirstNtossNorNaNtailNonNtheNseco
ndNtoss,NbutNnotNboth,Nis

pH1N orNH2N =N pH1N +NpH2N −N2(NpH1N×NpH2N)
1 1 1NNNN1
= +NNNNN−N2 N
×NNNN
2 2 2NNNNN
2
1
=N
2

InNtheNtableNabove,NthisNcombinationNcorrespondsNtoNtheNsumNofNtheNtwoNof
f-NdiagonalNelementsN(theNH1T2NandNtheNT1H2Nboxes).

(b) ThisNisNtheN"and"NcombinationNforNindependentNevents,NsoNweNmultiplyNth
eNelementaryNprobabilityNpHNforNeachNofNNNtosses:

pH1H2H3…HNN =N pH1N×NpH2N ×NpH3N ×⋯×NpHN
N
=N 1
N
NN

N2

ThisNisNbothNaNpermutationNandNaNcompositionN(thereNisNonlyNoneNpermuta
tionN forN all-
heads).NAndN noteN thatN sinceN bothN outcomesN haveNequalN probabilityN (0.5),N
thisN givesN theN probabilityN ofN anyN permutationN ofN anyNnumberNNHNofNheadsN
withNanyNnumberNNN−NNHNofNtails.

1.3 TwoNdifferentNapproachesNwillNbeNgivenNforNthisNproblem.NOneNisNanNapproxi
mationNthatNisNveryNcloseNtoNbeingNcorrect.NTheNsecondNisNexact.NByNcompar
ingNtheNresults,NtheNreasonablenessNofNtheNfirstNapproximationNcanNbeNexami
ned.

WhicheverNapproachNweNuseNtoNsolveNthisNproblem,NweNbeginNbyNrepresentingN
theNprobabilityNthatNyouNknowNaNrandomlyNselectedNpersonNfromNtheNpopulatio
n.
ThisNisNpkN=N 2000/300,000,000N=N 2/300,000N=N 6.67N×N 10−6.NToNavoidNdealingN
withN"or"Ncombinations,NweNcanNgreatlyNsimplifyNtheNproblemNbyNcalculatingNth