Biomolecular Thermodynamics d d
1st Edition by Douglas Barrick
d d d d
Complete Chapter Solutions Manual are
d d d d
included (Ch 1 to 14)
d d d d d
** Immediate Download
d d
** Swift Response
d d
** All Chapters included
d d d
, Solution Manual d
CHAPTER 1 d
1.1 UsingdthedsamedVennddiagramdfordillustration,dwedwantdthedprobabilitydofd
outcomesdfromdthedtwodeventsdthatdleaddtodthedcross-
hatcheddareadshowndbelow:
A1 A1d ndB2 B2
ThisdrepresentsdgettingdAdindeventd1danddnotdBdindeventd2,dplusdnotdgettingdA
indeventd1dbutdgettingdBdindeventd2d(thesedtwodaredthedcommond“ordbutdnotdbot
h”dcombinationdcalculateddindProblemd1.2)dplusdgettingdAdindeventd1danddBdindev
entd2.
1.2 Firstdthedformuladwilldbedderiveddusingdequations,danddthendVennddiagramsd wi
lldbedcompareddwithdthedstepsdindthedequation.dIndtermsdofdformulasdanddprob
abilities,dtheredaredtwodwaysdthatdtheddesireddpairdofdoutcomesdcandcomedabo
ut.dOnedwaydisdthatdwedcoulddgetdAdondthedfirstdeventdanddnotdBdondthe
secondd(dA1d∩d(∼B2d)).dThedprobabilitydofdthisdisdtakendasdthedsimpledproduct,dsince
deventsd1dandd2daredindependent:
pA1d∩d(∼B2d)d =dpAd×dp∼B
=d pAd×(1−dpBd (A.1.1)
)
=d pAd−dpApB
ThedseconddwaydisdthatdwedcoulddnotdgetdAdondthedfirstdeventdanddwedcoulddget
Bdondthedsecondd((∼dA1)d∩dB2d)d,dwithdprobability
p(∼A1)d∩dB2d =d p∼Ad×dpB
=d(1−dpAd)×dp (A.1.2)
B
=dpBd−dpApB
K10030_SolutiondManual.inddd 1 10-07-201
, 2 SOLUTIONd MANUAL
Sincedeitherdonedwilldwork,dwedwantdthedordcombination.dBecausedthedtwodwaysd
aredmutuallydexclusived(havingdbothdwoulddmeandbothdAdandd∼Adindthedfirstdoutc
ome,danddwithdequaldimpossibility,dbothdBdandd∼B),dthisdordcombinationdisdequald
todtheduniond{dA1d∩d(∼B2d)}d∪d{(∼dA1)d∩dB2},dandditsdprobabilitydisdsimplydthedsumdofdth
edprobabilitydofdthedtwodseparatedwaysdaboved(EquationsdA.1.1danddA.1.2):
p{A1d∩d(∼B2d)}d∪d{(~A1)d∩dB2}d =d pA1d∩d(∼B2d)d +dp(∼A1)d∩dB2
=d pAd−dpApBd+dpBd−dpApB
=dpAd+dpBd−d2pApB
ThedconnectiondtodVennddiagramsdisdshowndbelow.dIndthisdexercisedwedwilldworkd
backwarddfromdthedcombinationdofdoutcomesdwedseekdtodthedindividualdoutcomes.
dThedprobabilitydwedaredafterdisdfordthedcross-hatcheddareadbelow.
{dA1d∩d(∼B2d)}d∪d{(∼dA1)d∩dB2d}
A1 B2
Asdindicated,dthedcirclesdcorresponddtodgettingdthedoutcomedAdindeventd1d(left)d
anddoutcomedBdindeventd2.dEvendthoughdthedeventsdaredidentical,dthedVennddiagr
amdisdconstructeddsodthatdtheredisdsomedoverlapdbetweendthesedtwod(whichdwedd
on’tdwantdtodincludedindourd“ordbutdnotdboth”dcombination.dAsddescribeddabove,d
thedtwodcross-
hatcheddareasdaboveddon’tdoverlap,dthusdthedprobabilitydofdtheirduniondisdthedsim
pledsumdofdthedtwodseparatedareasdgivendbelow.
A1dnd~B2
~dA1d ndB2
pAd×dp~B
p ~A ×dpB
=dpAd(1d–dpB)
=d(1d–dp
Add d)pB
A1dnd~B2 ~dA1d ndB2
Addingdthesedtwodprobabilitiesdgivesdthedfulld“ordbutdnotdboth”dexpressiondab
ove.dThedonlydthingdremainingdisdtodshowdthatdthedprobabilitydofdeachdofdthed
crescentsdisdequaldtodthedproductdofdthedprobabilitiesdasdshowndindthedtopddia
gram.dThisdwilldonlydbeddonedfordonedofdthedtwodcrescents,dsincedthedotherdfo
llowsdindandexactlydanalogousdway.dFocusingdondthedgraydcrescentdabove,dit
representsdthedAdoutcomesdofdeventd1danddnotdthedBdoutcomesdindeventd2.dEachd
ofdthesedoutcomesdisdshowndbelow:
Eventd 1 Eventd 2
A1 ~B
p~Bd=d1d–dpB
pA
A1 ~B2
K10030_SolutiondManual.inddd 2 10-07-201
, SOLUTIONd MANUAL 3
BecausedEventd1danddEventd2daredindependent,dthed“and”dcombinationdofdth
esedtwodoutcomesdisdgivendbydthedintersection,danddthedprobabilitydofdthe
intersectiondisdgivendbydthedproductdofdthedtwodseparatedprobabilities,dleadingdto
dthedexpressionsdfordprobabilitiesdfordthedgraydcross-hatcheddcrescent.
(a) Thesedaredtwodindependentdelementarydeventsdeachdwithdandoutcomedpr
obabilitydofd0.5.dWedaredaskeddfordthedprobabilitydofdthedsequencedH1dT2,dw
hichdrequiresdmultiplicationdofdthedelementarydprobabilities:
p HH d =dH1d∩dT2d=dpHd×d =d1dd ×
d d1
dd=
d1
pT
1d 2 1 2
2d d 2 4
Wedcandarrangedthisdprobability,dalongdwithdthedprobabilitydfordthedotherdthre
edpossibledsequences,dindadtable:
Tossd1
Tossd2 Hd(0.5) Td(0.5)
Hd(0.5) H1H2 T1H2
(0.25) (0.25)
Td(0.5) H1T2 T1T2
(0.25) (0.25)
Note:dProbabilitiesdaredgivendindparentheses.
Thedprobabilitydofdgettingdadheaddondthedfirstdtossdordadtaildondthedseconddtos
s,dbutdnotdboth,dis
pH1dordH2d =dpH1d +dpH2d −d2(dpH1d×dpH2d)
1 d1 d1d d1 ı
= +d −d2 ×d ı
d
2 2 2d d 2j
d1
=d d
2
Indthedtabledabove,dthisdcombinationdcorrespondsdtodthedsumdofdthedtwodoff-
ddiagonaldelementsd(thedH1T2danddthedT1H2dboxes).
(b) Thisdisdthed"and"dcombinationdfordindependentdevents,dsodwedmultiplydthe
delementarydprobabilitydpHdfordeachdofdNdtosses:
pH1H2H3…HNd =dpH1d×dpH2d×dpH3d×⋯×dpHN
N
=d d1d ı
d2ıj
Thisdisdbothdadpermutationdanddadcompositiond(theredisdonlydonedpermut
ationdfordall-
heads).dAnddnotedthatdsincedbothdoutcomes dhavedequaldprobabilityd(0.5),d
thisdgivesdthedprobabilitydofdanydpermutationdofdanydnumberdNHdofdheadsdw
ithdanydnumberdNd−dNHdofdtails.
1.3 Twoddifferentdapproachesdwilldbedgivendfordthisdproblem.dOnedisdandapproxi
mationdthatdisdverydclosedtodbeingdcorrect.dThedseconddisdexact.dBydcomparin
gdthedresults,dthedreasonablenessdofdthedfirstdapproximationdcandbedexamine
d.
Whicheverdapproachdwedusedtodsolvedthisdproblem,dwedbegindbydrepresentingdth
edprobabilitydthatdyoudknowdadrandomlydselecteddpersondfromdthedpopulation.
Thisdisdpkd=d2000/300,000,000d=d2/300,000d=d6.67d×d10−6.dTodavoidddealingdw
ithd"or"dcombinations,dwedcandgreatlydsimplifydthedproblemdbydcalculatingdthedp
robabilitydthatdyouddodNOTdknowdanyonedondthedplane,danddthendrecognizedth
atdonedminusdthisdprobabilitydrepresentsdalldthedwaysdyoudcoulddknowdat
K10030_SolutiondManual.inddd 3 10-07-201
1st Edition by Douglas Barrick
d d d d
Complete Chapter Solutions Manual are
d d d d
included (Ch 1 to 14)
d d d d d
** Immediate Download
d d
** Swift Response
d d
** All Chapters included
d d d
, Solution Manual d
CHAPTER 1 d
1.1 UsingdthedsamedVennddiagramdfordillustration,dwedwantdthedprobabilitydofd
outcomesdfromdthedtwodeventsdthatdleaddtodthedcross-
hatcheddareadshowndbelow:
A1 A1d ndB2 B2
ThisdrepresentsdgettingdAdindeventd1danddnotdBdindeventd2,dplusdnotdgettingdA
indeventd1dbutdgettingdBdindeventd2d(thesedtwodaredthedcommond“ordbutdnotdbot
h”dcombinationdcalculateddindProblemd1.2)dplusdgettingdAdindeventd1danddBdindev
entd2.
1.2 Firstdthedformuladwilldbedderiveddusingdequations,danddthendVennddiagramsd wi
lldbedcompareddwithdthedstepsdindthedequation.dIndtermsdofdformulasdanddprob
abilities,dtheredaredtwodwaysdthatdtheddesireddpairdofdoutcomesdcandcomedabo
ut.dOnedwaydisdthatdwedcoulddgetdAdondthedfirstdeventdanddnotdBdondthe
secondd(dA1d∩d(∼B2d)).dThedprobabilitydofdthisdisdtakendasdthedsimpledproduct,dsince
deventsd1dandd2daredindependent:
pA1d∩d(∼B2d)d =dpAd×dp∼B
=d pAd×(1−dpBd (A.1.1)
)
=d pAd−dpApB
ThedseconddwaydisdthatdwedcoulddnotdgetdAdondthedfirstdeventdanddwedcoulddget
Bdondthedsecondd((∼dA1)d∩dB2d)d,dwithdprobability
p(∼A1)d∩dB2d =d p∼Ad×dpB
=d(1−dpAd)×dp (A.1.2)
B
=dpBd−dpApB
K10030_SolutiondManual.inddd 1 10-07-201
, 2 SOLUTIONd MANUAL
Sincedeitherdonedwilldwork,dwedwantdthedordcombination.dBecausedthedtwodwaysd
aredmutuallydexclusived(havingdbothdwoulddmeandbothdAdandd∼Adindthedfirstdoutc
ome,danddwithdequaldimpossibility,dbothdBdandd∼B),dthisdordcombinationdisdequald
todtheduniond{dA1d∩d(∼B2d)}d∪d{(∼dA1)d∩dB2},dandditsdprobabilitydisdsimplydthedsumdofdth
edprobabilitydofdthedtwodseparatedwaysdaboved(EquationsdA.1.1danddA.1.2):
p{A1d∩d(∼B2d)}d∪d{(~A1)d∩dB2}d =d pA1d∩d(∼B2d)d +dp(∼A1)d∩dB2
=d pAd−dpApBd+dpBd−dpApB
=dpAd+dpBd−d2pApB
ThedconnectiondtodVennddiagramsdisdshowndbelow.dIndthisdexercisedwedwilldworkd
backwarddfromdthedcombinationdofdoutcomesdwedseekdtodthedindividualdoutcomes.
dThedprobabilitydwedaredafterdisdfordthedcross-hatcheddareadbelow.
{dA1d∩d(∼B2d)}d∪d{(∼dA1)d∩dB2d}
A1 B2
Asdindicated,dthedcirclesdcorresponddtodgettingdthedoutcomedAdindeventd1d(left)d
anddoutcomedBdindeventd2.dEvendthoughdthedeventsdaredidentical,dthedVennddiagr
amdisdconstructeddsodthatdtheredisdsomedoverlapdbetweendthesedtwod(whichdwedd
on’tdwantdtodincludedindourd“ordbutdnotdboth”dcombination.dAsddescribeddabove,d
thedtwodcross-
hatcheddareasdaboveddon’tdoverlap,dthusdthedprobabilitydofdtheirduniondisdthedsim
pledsumdofdthedtwodseparatedareasdgivendbelow.
A1dnd~B2
~dA1d ndB2
pAd×dp~B
p ~A ×dpB
=dpAd(1d–dpB)
=d(1d–dp
Add d)pB
A1dnd~B2 ~dA1d ndB2
Addingdthesedtwodprobabilitiesdgivesdthedfulld“ordbutdnotdboth”dexpressiondab
ove.dThedonlydthingdremainingdisdtodshowdthatdthedprobabilitydofdeachdofdthed
crescentsdisdequaldtodthedproductdofdthedprobabilitiesdasdshowndindthedtopddia
gram.dThisdwilldonlydbeddonedfordonedofdthedtwodcrescents,dsincedthedotherdfo
llowsdindandexactlydanalogousdway.dFocusingdondthedgraydcrescentdabove,dit
representsdthedAdoutcomesdofdeventd1danddnotdthedBdoutcomesdindeventd2.dEachd
ofdthesedoutcomesdisdshowndbelow:
Eventd 1 Eventd 2
A1 ~B
p~Bd=d1d–dpB
pA
A1 ~B2
K10030_SolutiondManual.inddd 2 10-07-201
, SOLUTIONd MANUAL 3
BecausedEventd1danddEventd2daredindependent,dthed“and”dcombinationdofdth
esedtwodoutcomesdisdgivendbydthedintersection,danddthedprobabilitydofdthe
intersectiondisdgivendbydthedproductdofdthedtwodseparatedprobabilities,dleadingdto
dthedexpressionsdfordprobabilitiesdfordthedgraydcross-hatcheddcrescent.
(a) Thesedaredtwodindependentdelementarydeventsdeachdwithdandoutcomedpr
obabilitydofd0.5.dWedaredaskeddfordthedprobabilitydofdthedsequencedH1dT2,dw
hichdrequiresdmultiplicationdofdthedelementarydprobabilities:
p HH d =dH1d∩dT2d=dpHd×d =d1dd ×
d d1
dd=
d1
pT
1d 2 1 2
2d d 2 4
Wedcandarrangedthisdprobability,dalongdwithdthedprobabilitydfordthedotherdthre
edpossibledsequences,dindadtable:
Tossd1
Tossd2 Hd(0.5) Td(0.5)
Hd(0.5) H1H2 T1H2
(0.25) (0.25)
Td(0.5) H1T2 T1T2
(0.25) (0.25)
Note:dProbabilitiesdaredgivendindparentheses.
Thedprobabilitydofdgettingdadheaddondthedfirstdtossdordadtaildondthedseconddtos
s,dbutdnotdboth,dis
pH1dordH2d =dpH1d +dpH2d −d2(dpH1d×dpH2d)
1 d1 d1d d1 ı
= +d −d2 ×d ı
d
2 2 2d d 2j
d1
=d d
2
Indthedtabledabove,dthisdcombinationdcorrespondsdtodthedsumdofdthedtwodoff-
ddiagonaldelementsd(thedH1T2danddthedT1H2dboxes).
(b) Thisdisdthed"and"dcombinationdfordindependentdevents,dsodwedmultiplydthe
delementarydprobabilitydpHdfordeachdofdNdtosses:
pH1H2H3…HNd =dpH1d×dpH2d×dpH3d×⋯×dpHN
N
=d d1d ı
d2ıj
Thisdisdbothdadpermutationdanddadcompositiond(theredisdonlydonedpermut
ationdfordall-
heads).dAnddnotedthatdsincedbothdoutcomes dhavedequaldprobabilityd(0.5),d
thisdgivesdthedprobabilitydofdanydpermutationdofdanydnumberdNHdofdheadsdw
ithdanydnumberdNd−dNHdofdtails.
1.3 Twoddifferentdapproachesdwilldbedgivendfordthisdproblem.dOnedisdandapproxi
mationdthatdisdverydclosedtodbeingdcorrect.dThedseconddisdexact.dBydcomparin
gdthedresults,dthedreasonablenessdofdthedfirstdapproximationdcandbedexamine
d.
Whicheverdapproachdwedusedtodsolvedthisdproblem,dwedbegindbydrepresentingdth
edprobabilitydthatdyoudknowdadrandomlydselecteddpersondfromdthedpopulation.
Thisdisdpkd=d2000/300,000,000d=d2/300,000d=d6.67d×d10−6.dTodavoidddealingdw
ithd"or"dcombinations,dwedcandgreatlydsimplifydthedproblemdbydcalculatingdthedp
robabilitydthatdyouddodNOTdknowdanyonedondthedplane,danddthendrecognizedth
atdonedminusdthisdprobabilitydrepresentsdalldthedwaysdyoudcoulddknowdat
K10030_SolutiondManual.inddd 3 10-07-201
