Solutions Manual for Biomolecular Thermodynamics: From Theory to Application (1st Edition) – Douglas Barrick | Complete Chapters

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Biomolecular Therm c




odynamics, From Thec c




ory to Application, 1e
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Douglas Barrick (All
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Chapters)

,SolutionManual c




CHAPTER 1 c



1.1 UsingcthecsamecVenncdiagramcforcillustration,cwecwantcthecprobabilitycofco
utcomescfromcthectwoceventscthatcleadctoctheccross-
hatchedcareacshowncbelow:




A1 A1c ncB2 B2


ThiscrepresentscgettingcAcinceventc1candcnotcBcinceventc2,cpluscnotcgettingcA
inceventc1cbutcgettingcBcinceventc2c(thesectwocarectheccommonc“orcbutcnotcboth
”ccombinationccalculatedcincProblemc1.2)cpluscgettingcAcinceventc1candcBcinceven
tc2.

1.2 Firstcthecformulacwillcbecderivedcusingcequations,candcthencVenncdiagramsc wil
lcbeccomparedcwithcthecstepscincthecequation.cInctermscofcformulascandcproba
bilities,ctherecarectwocwayscthatcthecdesiredcpaircofcoutcomesccanccomecabout
.cOnecwayciscthatcweccouldcgetcAconcthecfirstceventcandcnotcBconcthe
secondc(cA1c∩c(∼B2c)).cThecprobabilitycofcthiscisctakencascthecsimplecproduct,csincec
eventsc1candc2carecindependent:

pA1c∩c(∼B2c)c =cpAc×cp∼B
=c pAc×(1−cpBc) (A.1.1)
=c pAc−cpApB

ThecsecondcwayciscthatcweccouldcnotcgetcAconcthecfirstceventcandcweccouldcget
Bconcthecsecondc((∼cA1)c∩cB2c)c,cwithcprobability

p(∼A1)c∩c B2c =c p∼Ac×cpB
=c(1−cpAc)×cpB (A.1.2)
=c pBc−cpApB

,2 SOLUTIONc MANUAL


Sinceceitherconecwillcwork,cwecwantcthecorccombination.cBecausecthectwocwaysca
recmutuallycexclusivec(havingcbothcwouldcmeancbothcAcandc∼Acincthecfirstcoutco
me,candcwithcequalcimpossibility,cbothcBcandc∼B),cthiscorccombinationciscequalctoc
thecunionc{cA1c∩c(∼B2c)}c∪c{(∼cA1)c∩cB2},candcitscprobabilityciscsimplycthecsumcofcthecpr
obabilitycofcthectwocseparatecwayscabovec(EquationscA.1.1candcA.1.2):

p{A1c∩c(∼B2c)}c∪c{(~A1)c∩cB2}c =c pA1c∩c(∼B2c)c +cp(∼A1)c∩cB2
=c pAc−cpApBc+cpBc−cpApB
=cpAc+cpBc−c2pApB

ThecconnectionctocVenncdiagramsciscshowncbelow.cIncthiscexercisecwecwillcworkcb
ackwardcfromctheccombinationcofcoutcomescwecseekctocthecindividualcoutcomes.cT
hecprobabilitycwecarecafterciscforctheccross-hatchedcareacbelow.
{cA1c∩c(∼B2c)}c∪c{(∼cA1)c∩cB2c}




A1 B2


Ascindicated,ctheccirclesccorrespondctocgettingcthecoutcomecAcinceventc1c(left)ca
ndcoutcomecBcinceventc2.cEvencthoughctheceventscarecidentical,cthecVenncdiagra
mciscconstructedcsocthatcthereciscsomecoverlapcbetweencthesectwoc(whichcwecdo
n’tcwantctocincludecincourc“orcbutcnotcboth”ccombination.cAscdescribedcabove,cth
ectwoccross-
hatchedcareascabovecdon’tcoverlap,cthuscthecprobabilitycofctheircunionciscthecsim
plecsumcofcthectwocseparatecareascgivencbelow.


A1cnc~B2
~cA1c ncB2

pAc×cp~B
p ~A ×cpB
=cpAc(1c–cpB)
=c(1c–cpAcc )p
cB

A1cnc~B2 ~cA1c ncB2



Addingcthesectwocprobabilitiescgivescthecfullc“orcbutcnotcboth”cexpressioncabo
ve.cTheconlycthingcremainingcisctocshowcthatcthecprobabilitycofceachcofctheccr
escentsciscequalctocthecproductcofcthecprobabilitiescascshowncincthectopcdiagra
m.cThiscwillconlycbecdonecforconecofcthectwoccrescents,csincecthecothercfollow
scincancexactlycanalogouscway.cFocusingconcthecgrayccrescentcabove,cit
representscthecAcoutcomescofceventc1candcnotcthecBcoutcomescinceventc2.cEachco
fcthesecoutcomesciscshowncbelow:

Eventc 1 Eventc 2


A1 ~B

p~Bc=c1c–cpB
pA



A1 ~B2

, SOLUTIONc MANUAL 3


BecausecEventc1candcEventc2carecindependent,cthec“and”ccombinationcofcthes
ectwocoutcomesciscgivencbycthecintersection,candcthecprobabilitycofcthe
intersectionciscgivencbycthecproductcofcthectwocseparatecprobabilities,cleadingctoc
thecexpressionscforcprobabilitiescforcthecgrayccross-hatchedccrescent.

(a) Thesecarectwocindependentcelementaryceventsceachcwithcancoutcomecpro
babilitycofc0.5.cWecarecaskedcforcthecprobabilitycofcthecsequencecH1cT2,cwhi
chcrequirescmultiplicationcofcthecelementarycprobabilities:

p HH c =cH1c∩cT2c=cpHc×cpT =c1cc×
c c1
cc=
c1

1c 2 1 2
2c c 2 4

Weccancarrangecthiscprobability,calongcwithcthecprobabilitycforcthecothercthre
ecpossiblecsequences,cincactable:


Tossc1

Tossc2 Hc(0.5) Tc(0.5)

Hc(0.5) H1H2 T1H2
(0.25) (0.25)

Tc(0.5) H1T2 T1T2
(0.25) (0.25)

Note:cProbabilitiescarecgivencincparentheses.

Thecprobabilitycofcgettingcacheadconcthecfirstctosscorcactailconcthecsecondctoss
,cbutcnotcboth,cis

pH1corcH2c =c pH1c +cpH2c −c2(cpH1c×cpH2c)
1 c1 c1c c1
= +c −c2 ×c ıı
c
2 2 2c c 2j
c1
=c c
2

Incthectablecabove,cthisccombinationccorrespondsctocthecsumcofcthectwocoff-
cdiagonalcelementsc(thecH1T2candcthecT1H2cboxes).


(b) Thisciscthec"and"ccombinationcforcindependentcevents,csocwecmultiplycthec
elementarycprobabilitycpHcforceachcofcNctosses:

pH1H2H3…HNc =cpH1c×cpH2c×cpH3c×⋯×cpHN
N
=c cc1ı
c2ıj



Thisciscbothcacpermutationcandcaccompositionc(therecisconlyconecpermutat
ioncforcall-
heads).cAndcnotecthatcsincecbothcoutcomeschavecequalcprobabilityc(0.5),cth
iscgivescthecprobabilitycofcanycpermutationcofcanycnumbercNHcofcheadscwithc
anycnumbercNc−cNHcofctails.

1.3 Twocdifferentcapproachescwillcbecgivencforcthiscproblem.cOneciscancapproxim
ationcthatciscverycclosectocbeingccorrect.cThecsecondciscexact.cByccomparingct
hecresults,cthecreasonablenesscofcthecfirstcapproximationccancbecexamined.

Whichevercapproachcwecusectocsolvecthiscproblem,cwecbegincbycrepresentingcthe
cprobabilitycthatcyoucknowcacrandomlycselectedcpersoncfromcthecpopulation.

Thisciscpkc=c2000/300,000,000c=c2/300,000c=c6.67c×c10−6.cTocavoidcdealingcwit
hc"or"ccombinations,cweccancgreatlycsimplifycthecproblemcbyccalculatingcthecpro
babilitycthatcyoucdocNOTcknowcanyoneconcthecplane,candcthencrecognizecthatc
onecminuscthiscprobabilitycrepresentscallcthecwayscyouccouldcknowcat