Biomolecular Thermodynamics g g
1st Edition by Douglas Barrick
g g g g
Complete Chapter Solutions Manual are
g g g g
included (Ch 1 to 14)
g g g g g
** Immediate Download
g g
** Swift Response
g g
** All Chapters included
g g g
, Solution Manual g
CHAPTER 1 g
1.1 UsinggthegsamegVenngdiagramgforgillustration,gwegwantgthegprobabilitygofgo
utcomesgfromgthegtwogeventsgthatgleadgtogthegcross-
hatchedgareagshowngbelow:
A1 A1g ngB2 B2
ThisgrepresentsggettinggAgingeventg1gandgnotgBgingeventg2,gplusgnotggettinggA
ingeventg1gbutggettinggBgingeventg2g(thesegtwogaregthegcommong“orgbutgnotgboth
”gcombinationgcalculatedgingProblemg1.2)gplusggettinggAgingeventg1gandgBgingevent
g2.
1.2 Firstgthegformulagwillgbegderivedgusinggequations,gandgthengVenngdiagramsg wil
lgbegcomparedgwithgthegstepsgingthegequation.gIngtermsgofgformulasgandgproba
bilities,gtheregaregtwogwaysgthatgthegdesiredgpairgofgoutcomesgcangcomegabout
.gOnegwaygisgthatgwegcouldggetgAgongthegfirstgeventgandgnotgBgongthe
secondg(gA1g∩g(∼B2g)).gThegprobabilitygofgthisgisgtakengasgthegsimplegproduct,gsinceg
eventsg1gandg2garegindependent:
pA1g∩g(∼B2g)g =g pAg×gp∼B
=g pAg×(1−gpBg (A.1.1)
)
=g pAg−gpApB
ThegsecondgwaygisgthatgwegcouldgnotggetgAgongthegfirstgeventgandgwegcouldgget
Bgongthegsecondg((∼gA1)g∩gB2g)g,gwithgprobability
p(∼A1)g∩g B2g =g p∼Ag×gpB
=g(1−gpAg)×gp (A.1.2)
B
=g pBg−gpApB
K10030_SolutiongManual.inddg 1 10-07-201
, 2 SOLUTIONg MANUAL
Sincegeithergonegwillgwork,gwegwantgthegorgcombination.gBecausegthegtwogwaysga
regmutuallygexclusiveg(havinggbothgwouldgmeangbothgAgandg∼Agingthegfirstgoutco
me,gandgwithgequalgimpossibility,gbothgBgandg∼B),gthisgorgcombinationgisgequalgto
gtheguniong{gA1g∩g(∼B2g)}g∪g{(∼gA1)g∩gB2},gandgitsgprobabilitygisgsimplygthegsumgofgthegp
robabilitygofgthegtwogseparategwaysgaboveg(EquationsgA.1.1gandgA.1.2):
p{A1g∩g(∼B2g)}g∪g{(~A1)g∩gB2}g =g pA1g∩g(∼B2g)g +gp(∼A1)g∩gB2
=g pAg−gpApBg+gpBg−gpApB
=gpAg+gpBg−g2pApB
ThegconnectiongtogVenngdiagramsgisgshowngbelow.gIngthisgexercisegwegwillgworkgb
ackwardgfromgthegcombinationgofgoutcomesgwegseekgtogthegindividualgoutcomes.gT
hegprobabilitygwegaregaftergisgforgthegcross-hatchedgareagbelow.
{gA1g∩g(∼B2g)}g∪g{(∼gA1)g∩gB2g}
A1 B2
Asgindicated,gthegcirclesgcorrespondgtoggettinggthegoutcomegAgingeventg1g(left)ga
ndgoutcomegBgingeventg2.gEvengthoughgthegeventsgaregidentical,gthegVenngdiagra
mgisgconstructedgsogthatgtheregisgsomegoverlapgbetweengthesegtwog(whichgwegdo
n’tgwantgtogincludegingourg“orgbutgnotgboth”gcombination.gAsgdescribedgabove,gth
egtwogcross-
hatchedgareasgabovegdon’tgoverlap,gthusgthegprobabilitygofgtheirguniongisgthegsim
plegsumgofgthegtwogseparategareasggivengbelow.
A1gng~B2
~g A1g ngB2
pAg×gp~B
p ~A ×gpB
=gpAg(1g–gpB)
=g(1g–gpAgg )p
gB
A1gng~B2 ~gA1g ngB2
Addinggthesegtwogprobabilitiesggivesgthegfullg“orgbutgnotgboth”gexpressiongabo
ve.gThegonlygthinggremaininggisgtogshowgthatgthegprobabilitygofgeachgofgthegcr
escentsgisgequalgtogthegproductgofgthegprobabilitiesgasgshowngingthegtopgdiagra
m.gThisgwillgonlygbegdonegforgonegofgthegtwogcrescents,gsincegthegothergfollow
sgingangexactlyganalogousgway.gFocusinggongtheggraygcrescentgabove,git
representsgthegAgoutcomesgofgeventg1gandgnotgthegBgoutcomesgingeventg2.gEachgo
fgthesegoutcomesgisgshowngbelow:
Eventg 1 Eventg 2
A1 ~B
p~Bg=g1g–gpB
pA
A1 ~B2
K10030_SolutiongManual.inddg 2 10-07-201
, SOLUTIONg MANUAL 3
BecausegEventg1gandgEventg2garegindependent,gtheg“and”gcombinationgofgthe
segtwogoutcomesgisggivengbygthegintersection,gandgthegprobabilitygofgthe
intersectiongisggivengbygthegproductgofgthegtwogseparategprobabilities,gleadinggtog
thegexpressionsgforgprobabilitiesgforgtheggraygcross-hatchedgcrescent.
(a) Thesegaregtwogindependentgelementarygeventsgeachgwithgangoutcomegpro
babilitygofg0.5.gWegaregaskedgforgthegprobabilitygofgthegsequencegH1gT2,gwhi
chgrequiresgmultiplicationgofgthegelementarygprobabilities:
p HH g =gH1g∩gT2g=gpHg×gp =g1gg ×
g g1
gg=
g1
T
1g 2 1 2
2g g 2 4
Wegcangarrangegthisgprobability,galonggwithgthegprobabilitygforgthegothergthre
egpossiblegsequences,gingagtable:
Tossg1
Tossg2 Hg(0.5) Tg(0.5)
Hg(0.5) H1H2 T1H2
(0.25) (0.25)
Tg(0.5) H1T2 T1T2
(0.25) (0.25)
Note:gProbabilitiesgareggivengingparentheses.
Thegprobabilitygofggettinggagheadgongthegfirstgtossgorgagtailgongthegsecondgtoss
,gbutgnotgboth,gis
pH1gorgH2g =g pH1g +gpH2g −g2(gpH1g×gpH2g)
1 g1 g1g g1 ı
= +g −g2 ×g ı
g
2 2 2g g 2j
g1
=g g
2
Ingthegtablegabove,gthisgcombinationgcorrespondsgtogthegsumgofgthegtwogoff-
gdiagonalgelementsg(thegH1T2gandgthegT1H2gboxes).
(b) Thisgisgtheg"and"gcombinationgforgindependentgevents,gsogwegmultiplygtheg
elementarygprobabilitygpHgforgeachgofgNgtosses:
pH1H2H3…HNg =gpH1g×gpH2g×gpH3g×⋯×gpHN
N
=g g1g ı
g2ıj
Thisgisgbothgagpermutationgandgagcompositiong(theregisgonlygonegpermuta
tiongforgall-
heads).gAndgnotegthatgsincegbothgoutcomesghavegequalgprobabilityg(0.5),gt
hisggivesgthegprobabilitygofganygpermutationgofganygnumbergNHgofgheadsgwit
hganygnumbergNg−gNHgofgtails.
1.3 Twogdifferentgapproachesgwillgbeggivengforgthisgproblem.gOnegisgangapproxim
ationgthatgisgverygclosegtogbeinggcorrect.gThegsecondgisgexact.gBygcomparinggt
hegresults,gthegreasonablenessgofgthegfirstgapproximationgcangbegexamined.
Whichevergapproachgwegusegtogsolvegthisgproblem,gwegbegingbygrepresentinggth
egprobabilitygthatgyougknowgagrandomlygselectedgpersongfromgthegpopulation.
Thisgisgpkg=g2000/300,000,000g=g2/300,000g=g6.67g×g10−6.gTogavoidgdealinggwit
hg"or"gcombinations,gwegcanggreatlygsimplifygthegproblemgbygcalculatinggthegpro
babilitygthatgyougdogNOTgknowganyonegongthegplane,gandgthengrecognizegthatg
onegminusgthisgprobabilitygrepresentsgallgthegwaysgyougcouldgknowgat
K10030_SolutiongManual.inddg 3 10-07-201
1st Edition by Douglas Barrick
g g g g
Complete Chapter Solutions Manual are
g g g g
included (Ch 1 to 14)
g g g g g
** Immediate Download
g g
** Swift Response
g g
** All Chapters included
g g g
, Solution Manual g
CHAPTER 1 g
1.1 UsinggthegsamegVenngdiagramgforgillustration,gwegwantgthegprobabilitygofgo
utcomesgfromgthegtwogeventsgthatgleadgtogthegcross-
hatchedgareagshowngbelow:
A1 A1g ngB2 B2
ThisgrepresentsggettinggAgingeventg1gandgnotgBgingeventg2,gplusgnotggettinggA
ingeventg1gbutggettinggBgingeventg2g(thesegtwogaregthegcommong“orgbutgnotgboth
”gcombinationgcalculatedgingProblemg1.2)gplusggettinggAgingeventg1gandgBgingevent
g2.
1.2 Firstgthegformulagwillgbegderivedgusinggequations,gandgthengVenngdiagramsg wil
lgbegcomparedgwithgthegstepsgingthegequation.gIngtermsgofgformulasgandgproba
bilities,gtheregaregtwogwaysgthatgthegdesiredgpairgofgoutcomesgcangcomegabout
.gOnegwaygisgthatgwegcouldggetgAgongthegfirstgeventgandgnotgBgongthe
secondg(gA1g∩g(∼B2g)).gThegprobabilitygofgthisgisgtakengasgthegsimplegproduct,gsinceg
eventsg1gandg2garegindependent:
pA1g∩g(∼B2g)g =g pAg×gp∼B
=g pAg×(1−gpBg (A.1.1)
)
=g pAg−gpApB
ThegsecondgwaygisgthatgwegcouldgnotggetgAgongthegfirstgeventgandgwegcouldgget
Bgongthegsecondg((∼gA1)g∩gB2g)g,gwithgprobability
p(∼A1)g∩g B2g =g p∼Ag×gpB
=g(1−gpAg)×gp (A.1.2)
B
=g pBg−gpApB
K10030_SolutiongManual.inddg 1 10-07-201
, 2 SOLUTIONg MANUAL
Sincegeithergonegwillgwork,gwegwantgthegorgcombination.gBecausegthegtwogwaysga
regmutuallygexclusiveg(havinggbothgwouldgmeangbothgAgandg∼Agingthegfirstgoutco
me,gandgwithgequalgimpossibility,gbothgBgandg∼B),gthisgorgcombinationgisgequalgto
gtheguniong{gA1g∩g(∼B2g)}g∪g{(∼gA1)g∩gB2},gandgitsgprobabilitygisgsimplygthegsumgofgthegp
robabilitygofgthegtwogseparategwaysgaboveg(EquationsgA.1.1gandgA.1.2):
p{A1g∩g(∼B2g)}g∪g{(~A1)g∩gB2}g =g pA1g∩g(∼B2g)g +gp(∼A1)g∩gB2
=g pAg−gpApBg+gpBg−gpApB
=gpAg+gpBg−g2pApB
ThegconnectiongtogVenngdiagramsgisgshowngbelow.gIngthisgexercisegwegwillgworkgb
ackwardgfromgthegcombinationgofgoutcomesgwegseekgtogthegindividualgoutcomes.gT
hegprobabilitygwegaregaftergisgforgthegcross-hatchedgareagbelow.
{gA1g∩g(∼B2g)}g∪g{(∼gA1)g∩gB2g}
A1 B2
Asgindicated,gthegcirclesgcorrespondgtoggettinggthegoutcomegAgingeventg1g(left)ga
ndgoutcomegBgingeventg2.gEvengthoughgthegeventsgaregidentical,gthegVenngdiagra
mgisgconstructedgsogthatgtheregisgsomegoverlapgbetweengthesegtwog(whichgwegdo
n’tgwantgtogincludegingourg“orgbutgnotgboth”gcombination.gAsgdescribedgabove,gth
egtwogcross-
hatchedgareasgabovegdon’tgoverlap,gthusgthegprobabilitygofgtheirguniongisgthegsim
plegsumgofgthegtwogseparategareasggivengbelow.
A1gng~B2
~g A1g ngB2
pAg×gp~B
p ~A ×gpB
=gpAg(1g–gpB)
=g(1g–gpAgg )p
gB
A1gng~B2 ~gA1g ngB2
Addinggthesegtwogprobabilitiesggivesgthegfullg“orgbutgnotgboth”gexpressiongabo
ve.gThegonlygthinggremaininggisgtogshowgthatgthegprobabilitygofgeachgofgthegcr
escentsgisgequalgtogthegproductgofgthegprobabilitiesgasgshowngingthegtopgdiagra
m.gThisgwillgonlygbegdonegforgonegofgthegtwogcrescents,gsincegthegothergfollow
sgingangexactlyganalogousgway.gFocusinggongtheggraygcrescentgabove,git
representsgthegAgoutcomesgofgeventg1gandgnotgthegBgoutcomesgingeventg2.gEachgo
fgthesegoutcomesgisgshowngbelow:
Eventg 1 Eventg 2
A1 ~B
p~Bg=g1g–gpB
pA
A1 ~B2
K10030_SolutiongManual.inddg 2 10-07-201
, SOLUTIONg MANUAL 3
BecausegEventg1gandgEventg2garegindependent,gtheg“and”gcombinationgofgthe
segtwogoutcomesgisggivengbygthegintersection,gandgthegprobabilitygofgthe
intersectiongisggivengbygthegproductgofgthegtwogseparategprobabilities,gleadinggtog
thegexpressionsgforgprobabilitiesgforgtheggraygcross-hatchedgcrescent.
(a) Thesegaregtwogindependentgelementarygeventsgeachgwithgangoutcomegpro
babilitygofg0.5.gWegaregaskedgforgthegprobabilitygofgthegsequencegH1gT2,gwhi
chgrequiresgmultiplicationgofgthegelementarygprobabilities:
p HH g =gH1g∩gT2g=gpHg×gp =g1gg ×
g g1
gg=
g1
T
1g 2 1 2
2g g 2 4
Wegcangarrangegthisgprobability,galonggwithgthegprobabilitygforgthegothergthre
egpossiblegsequences,gingagtable:
Tossg1
Tossg2 Hg(0.5) Tg(0.5)
Hg(0.5) H1H2 T1H2
(0.25) (0.25)
Tg(0.5) H1T2 T1T2
(0.25) (0.25)
Note:gProbabilitiesgareggivengingparentheses.
Thegprobabilitygofggettinggagheadgongthegfirstgtossgorgagtailgongthegsecondgtoss
,gbutgnotgboth,gis
pH1gorgH2g =g pH1g +gpH2g −g2(gpH1g×gpH2g)
1 g1 g1g g1 ı
= +g −g2 ×g ı
g
2 2 2g g 2j
g1
=g g
2
Ingthegtablegabove,gthisgcombinationgcorrespondsgtogthegsumgofgthegtwogoff-
gdiagonalgelementsg(thegH1T2gandgthegT1H2gboxes).
(b) Thisgisgtheg"and"gcombinationgforgindependentgevents,gsogwegmultiplygtheg
elementarygprobabilitygpHgforgeachgofgNgtosses:
pH1H2H3…HNg =gpH1g×gpH2g×gpH3g×⋯×gpHN
N
=g g1g ı
g2ıj
Thisgisgbothgagpermutationgandgagcompositiong(theregisgonlygonegpermuta
tiongforgall-
heads).gAndgnotegthatgsincegbothgoutcomesghavegequalgprobabilityg(0.5),gt
hisggivesgthegprobabilitygofganygpermutationgofganygnumbergNHgofgheadsgwit
hganygnumbergNg−gNHgofgtails.
1.3 Twogdifferentgapproachesgwillgbeggivengforgthisgproblem.gOnegisgangapproxim
ationgthatgisgverygclosegtogbeinggcorrect.gThegsecondgisgexact.gBygcomparinggt
hegresults,gthegreasonablenessgofgthegfirstgapproximationgcangbegexamined.
Whichevergapproachgwegusegtogsolvegthisgproblem,gwegbegingbygrepresentinggth
egprobabilitygthatgyougknowgagrandomlygselectedgpersongfromgthegpopulation.
Thisgisgpkg=g2000/300,000,000g=g2/300,000g=g6.67g×g10−6.gTogavoidgdealinggwit
hg"or"gcombinations,gwegcanggreatlygsimplifygthegproblemgbygcalculatinggthegpro
babilitygthatgyougdogNOTgknowganyonegongthegplane,gandgthengrecognizegthatg
onegminusgthisgprobabilitygrepresentsgallgthegwaysgyougcouldgknowgat
K10030_SolutiongManual.inddg 3 10-07-201
