Molecular Driving Forces Statistical Thermodynamics in Biology Chemistry Physics and Nanoscience 2nd

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Molecular
Statistical Thermodynamics in Biology Chemistry Physics and Nanoscience 2nd Edition Dill Solutions Full Download: http://testbanktip.com/download/molecular-driving-forces-statistical-thermodynamics-in-biology-chemistry-physics-and-nanoscience-2nd-edition-dill-solutions-manual/ Download all pages and all chapters at: TestBankTip.com
Driving Forces

Chapter1 PrinciplesofProbability

1.Combiningindependentprobabilities.

Youhaveappliedtothreemedicalschools:UniversityofCaliforniaatSanFrancisco(UCSF), DuluthSchoolofMines(DSM),andHarvard(H).Youguessthattheprobabilitiesyou’ll beacceptedare: p(UCSF)=0.10, p(DSM)=0.30,and p(H)=0.50.Assumethatthe acceptanceeventsareindependent.

(a)Whatistheprobabilitythatyougetinsomewhere(atleastoneacceptance)?

(b)WhatistheprobabilitythatyouwillbeacceptedbybothHarvardandDuluth?

(a)Thesimplestwaytosolvethisproblemistorecallthatwhenprobabilitiesare independent,andyouwanttheprobabilityofevents A and B ,youcan multiply them. Wheneventsare mutuallyexclusive andyouwanttheprobabilityofevents A or B , youcan add theprobabilities.Thereforewetrytostructuretheproblemintoan and and or problem.WewanttheprobabilityofgettingintoH or DSMorUCSF.But thisdoesn’thelp,becausetheseeventsarenotmutuallyexclusive(mutuallyexclusive meansthatifonehappens,theothercannothappen).Sowetryagain.Theprobability ofacceptancesomewhere, P (a),is P (a)=1 P (r ),where P (r )istheprobabilitythat you’rerejectedeverywhere.(You’reeitheracceptedsomewhereoryou’renot.)Butthis probability can beputintheaboveterms. P (r )=theprobabilitythatyou’rerejected atH and atDSM and atUCSF.Theseeventsareindependent,sowehavetheanswer.

TheprobabilityofrejectionatHis p(r H)=1 0 5=0 5.RejectionatDSMis

p(r DSM)=1 0 3=0 7.RejectionatUCSFis p(r UCSF)=1 0 1=0 9. Therefore P (r )=(0.5)(0.7)(0.9)=0.315.Thereforetheprobabilityofatleastoneacceptance = P (a)=1 P (r )=0.685.

1

(b)Thesimpleansweristhatthisistheintersectionoftwoindependentevents:

15

Amoremechanicalapproachtoeitherpart(a)orthispartistowriteoutallthe possiblecircumstances.RejectionandacceptanceatHaremutuallyexclusive.Their probabilitiesaddtoone.Thesamefortheothertwoschools.Thereforeallpossible circumstancesaretakenintoaccountby adding themutuallyexclusiveeventstogether, and multiplying independentevents:

wherethefirsttermistheprobabilityofacceptanceatallthree,thesecondterm representsacceptanceatHandDSMbutrejectionatUCSF,thethirdtermrepresents acceptanceatHandUCSFbutrejectionatDSM,etc.Eachoftheseeventsismutually exclusivewithrespecttoeachother;thereforetheyarealladded.Eachindividualterm representsindependenteventsof,forexample, aH and aDSM and aUCSF.Thereforeit issimpletoreadofftheanswerinthisproblem:wewant aH and aDSM, but noticewe don’tcareaboutUCSF.Thisprobabilityis

=(0 50)(0 30) =0.15.

Notethatwecouldhavesolvedpart(a)thesameway;itwouldhaverequiredadding upalltheappropriatepossiblemutuallyexclusiveevents.Youcancheckthatitgives thesameanswerasabove(butnoticehowmuchmoretediousitis).

p(aH)p(aDSM)=(0 50)(0 30)
=0
[p(aH)+ p(r H)][p(aDSM)+ p(r DSM)][p(aUCSF)+ p(r UCSF)]=1, or,equivalently, = p(aH)p(aDSM)p(aUCSF)+ p(aH)p(aDSM)p(r UCSF) +p(aH)p(r DSM)p(aUCSF)+ ··· ,
p(aH)p(aDSM)= p(aH)p(aDSM)[p(aUCSF)+ p
(r UCSF)]
2

2.Probabilitiesofsequences.

Assumethatthefourbases A, C, T,and G occurwithequallikelihoodinaDNAsequence ofninemonomers.

(a)Whatistheprobabilityoffindingthesequence AAATCGAGT throughrandom chance?

(b)Whatistheprobabilityoffindingthesequence AAAAAAAAA throughrandom chance?

(c)Whatistheprobabilityoffindinganysequencethathasfour A’s,two T’s,two G’s, andone C,suchasthatin(a)?

(a)Eachbaseoccurswithprobability1/4.Theprobabilityofan A inposition1is1/4, thatofan A inposition2is1/4,thatofan A inposition3is1/4,thatofa T in position4is1/4,andsoon.Thereareninebases.Theprobabilityofthisspecific sequenceis(1/4)9 =3 8 × 10 6

(b)Sameansweras(a).

(c)Eachspecificsequencehastheprobabilitygivenabove,butinthiscasetherearemany possiblesequencesthatsatisfytherequirementthatwehavefour A’s,two T’s,two G’s,andone C.Howmanyarethere?Westartaswehavedonebefore,byassuming allnineobjectsaredistinguishable.Thereare9!arrangementsofninedistinguishable objectsinalinearsequence.(Thefirstonecanbeinanyofnineplaces,thesecondin anyoftheremainingeightplaces,andsoon.)Butwecan’tdistinguishthefour A’s,so wehaveovercountedbyafactorof4!,andmustdividethisout.Wecan’tdistinguish thetwo T’s,sowehaveovercountedby2!,andmustalsodividethisout.Andsoon. Sotheprobabilityofhavingthiscompositionis

9! 4!2!2!1! 1 4 9 =0 014 3

3.Theprobabilityofasequence(givenacomposition).

Ascientisthasconstructedasecretpeptidetocarryamessage.Youknowonlythecompositionofthepeptide,whichissixaminoacidslong.Itcontainsoneserine S,onethreonine T, onecysteine C,onearginine R,andtwoglutamates E.Whatistheprobabilitythatthe sequence SECRET willoccurbychance?

The S couldbeinanyoneofthesixpositionswithequallikelihood.Theprobabilitythatit isinposition1is1/6.Giventhat S isinthefirstposition,wehavetwo E s,whichcould occurinanyoftheremainingfivepositions.Theprobabilitythatoneofthemisinposition2 is2/5.Giventhosetwolettersinposition,theprobabilitythattheone C isinthenextof thefourremainingpositionsis1/4.Theprobabilityforthe R is1/3.Fortheremaining E,it is1/2,andforthelast T,itis1/1,sotheprobabilityis

4.Combiningindependentprobabilities.

Youhaveafairsix-sideddie.Youwanttorollitenoughtimestoensurethata 2 occursat leastonce.Whatnumberofrolls k isrequiredtoensurethattheprobabilityisatleast2/3 thatatleastone 2 willappear?

Approximatelysixormorerollswillensurewithprobability P ≥ 2/3thata 2 willappear.

(1/6)(2/5)(1/4)(1/3)(1/2)=1/360= 6! 1!2!1!1! 1
q = 5 6 =probabilitythata
doesnotappearonthatroll. q
P (k )=1 q k =probabilitythatatleastone
For P (k ) ≥ 2 3 , 1 q k ≥ 2 3 =⇒ q k ≤ 1 3 =⇒ k ln q ≤ ln 1 3 =⇒ k ≥ ln(1/3) ln(5/6) =6.03
2
k =probabilitythata 2 doesnotappearon k INDEPENDENTrolls.
2 appearson k rolls.
4

5.Predictingcompositionsofindependentevents.

Supposeyourollafairsix-sideddiethreetimes.

(a)Whatistheprobabilityofgettinga 5 twicefromallthreerollsofthedice?

(b)Whatistheprobabilityofgettingatotalof atleast two 5’sfromallthreerollsof thedie?

Theprobabilityofgetting x 5’son n rollsofthediceis

Notethatthisisa“2-outcome”problem(gettinga 5 ornotgettinga 5).Itisnota “6-outcome”problem.

(a)Sotheprobabilityoftwo 5’sonthreedicerollsis

(b)Theprobabilityofgetting atleast two 5’sistheprobabilityofgettingtwo 5’sorthree 5’s.Sincethesetwosituationsaremutuallyexclusive,weseek

1 6 x 5 6 n x n! x!(n x)!
.
1 6 2 5 6 1 3! 2!1! = 1 36 5 6 3 = 15 216 =6.94 × 10 2 .
p(two 5’s)+ p(three 5’s)= 1 6 2 5 6 3! 2!1! + 1 6 3 5 6 0 3! 3!0! = 15 216 + 1 216 = 16 216 =7 41 × 10 2 5
6.Computingameanandvariance. Considertheprobabilitydistribution p(x)= axn ,0 ≤ x ≤ 1,forapositiveinteger n (a)Deriveanexpressionfortheconstant a,tonormalize p(x). (b)Computetheaverage x asafunctionof n. (c)Compute σ 2 = x2 − x 2 asafunctionof n. (a) 1 0 p(x) dx =1=⇒ 1 0 axn dx = axn+1 n +1 1 0 = a n +1 =1=⇒ a = n +1 (b) x = 1 0 xp(x) dx = 1 0 (n +1)xn+1 dx = (n +1)xn+2 n +2 1 0 = n +1 n +2 (c) x 2 = 1 0 x 2 p(x) dx =(n +1) 1 0 xn+2 dx =(n +1) xn+3 n +3 1 0 = n +1 n +3 . So σ 2 = x 2 − x 2 = n +1 n +3 n +1 n +2 2 . 6

7.Computingtheaverageofaprobabilitydistribution.

8.Predictingcoincidence.

Yourstatisticalmechanicsclasshas25students.Whatistheprobabilitythatatleasttwo classmateshavethesamebirthday?

Ifyoufirstfindtheprobability q thatnotwostudentshavethesamebirthday,thenthe quantityyouwantis

p(2studentshavesamebirthday)=1 q

Theprobabilitythatasecondstudentdoesnothavethesamebirthdayasthefirstis (364/365).Theprobabilitythatthethirdstudenthasabirthdaydifferentthaneitherofthe firsttwois(363/365),andsoon.Itislikeasequenceprobleminwhicheachpossible

Computetheaverage i fortheprobabilitydistributionfunctionshowninthefigurebelow. 01234 i P ( i ) 0.0 0.2 0.4 0.3 0.1 Asimpleprobabilitydistribution. i = 4 i=0 ip(i) =0(0 0)+1(0 1)+2(0 2)+3(0 3)+4(0 4) =3
7

birthdayisonecarddrawnoutofabarrel.Theprobabilitythatnotwopeoplehavethesame birthday,outof m people,is

q = 364 365 363 365 362 365 ··· 365 (m 1) 365

Infactorialnotation,

q = N ! (N m)!N m ,

where N =365.(Incidentally,thisexpressionisidenticaltotheexpressionforexcluded volumeintheFlory–Hugginsmodelofpolymersolutions(seeChapter31).)UsingStirling’s approximation x! ≈ (x/e)x ,weget

Collectingtogethertermsin e anddividingthenumeratoranddenominatorby N N gives q =

Substituting m =25studentsand N =365gives

q =0.4163, so

p =1 q =0.5837.

Thereisabetterthan50%chancetwostudentswillhavethesamebirthday!

m e
q = (N/e)N N
N m N m
m
e m 1
N N m
8

9.Thedistributionofscoresondice.

Supposethatyouhave n dice,eachadifferentcolor,allunbiasedandsix-sided.

(a)Ifyourollthemallatonce,howmanydistinguishableoutcomesarethere?

(b)Giventwodistinguishabledice,whatisthemostprobablesumoftheirfacevalueson agiventhrowofthepair?(Thatis,whichsumbetween2and12hasthegreatest numberofdifferentwaysofoccurring?)

(c)Whatistheprobabilityofthemostprobablesum?

(a)

6ononedie 6 × 6ontwodice . . . 6n on n dice.

(b)Numberofwaysasumcanoccur:

Whendiceshowdifferentnumbers,thereisadegeneracyoftwo.Wheneachofthedice hasthesamenumber,thedegeneracyequalsone.

(c)Probabilityof7= p(7)= numberofwaysofgetting7 totalnumberofwaysofalloutcomes

Mostprobablesum 1 2 3 4 5 6 7 123456789101112 (1,1) (1,2) × 2 (1,6) × 2 (3,4) × 2 (2,3) × 2(2,5) × (1,2 3) × 2(1,4) × 2 (2,2) Sumof2dice
p(7)= 6 1+2+3+4+5+6+5+4+3+2+1 = 1 6 9

10.Theprobabilitiesofidenticalsequencesofaminoacids.

Youarecomparingproteinaminoacidsequencesforhomology.Youhavea20-letteralphabet(20differentaminoacids).Eachsequenceisastring n lettersinlength.Youhaveone testsequenceand s differentdatabasesequences.Youmayfindanyoneofthe20different aminoacidsatanypositioninthesequence,independentofwhatyoufindatanyother position.Let p representtheprobabilitythattherewillbea‘match’atagivenpositionin thetwosequences.

(a)Intermsof s,p, and n, howmanyofthe s sequenceswillbeperfectmatches(identical residuesateveryposition)?

(b)Howmanyofthe s comparisons(ofthetestsequenceagainsteachdatabasesequence) willhaveexactlyonemismatchatanypositioninthesequences?

(a)Forcomparingonesequence,eachpositionbeingassumedindependent,theprobability ofaperfectmatchofall n residuesis

pn =(numberofmatchedseqs/numberoftotalseqs)=⇒ numberofmatchesin s sequences= spn

(b) n 1positionsmatch,sotheprobabilityis pn 1 ;onepositiondoesn’tmatch,whichhas theprobability(1 p);andthereare n differentpositionsatwhichthemismatchcould occur;thereforetheansweris

spn 1 (1 p)n

Notethat,ingeneral,for k matches,

(1) P (k )= sp k (1 p)n k n! k !(n k )! .

10

11.Thecombinatoricsofdisulfidebondformation.

Aproteinmaycontainseveralcysteines,whichmaypairtogethertoformdisulfidebonds asshowninthefigurebelow.Ifthereisanevennumber n ofcysteines, n/2disulfidebonds canform.Howmanydifferentdisulfidepairingarrangementsarepossible?

Numbertheindividualsulfhydrylgroupsalongthechain.Thefirstsulfhydrylalongthe sequencecanbondtoanyoftheother n 1.Thisremovestwosulfhydrylsfrom consideration.Thethirdsulfhydrylcanthenbondtoanyoftheremaining n 3.Four sulfhydrylsarenowremovedfromconsideration.Thefifthcannowbondtoanyofthe remaining n 5sulfhydryls,etc.,untilall n/2bondsareformed.Thusthetotalpossible numberofarrangementsofdisulfidebondsisaproductof n/2terms:

Anotherapproachgivesanexpressionthatiseasiertocalculate.Considerplacingthe sulfhydrylsinasequence.Thefirstplacemaybeoccupiedbyanyof n sulfhydryls,the secondplacebyanyof n 1sulfhydryls,thethirdbyanyof n 2sulfhydryls,etc.Thus,if eachsulfhydrylweredistinguishablefromeveryother,therewouldbe n!arrangements. However,eachsulfhydrylhasamatefromwhichitcannotbedistinguished.Wemustdivide byafactorof2(perbond)tocorrectfortheindistinguishabilityofthetwoendsofeach bond.Finally,sincewecannotdistinguishanyofthe n/2bondsfromanyother,wemust alsodivideby(n/2)!.Hencethenumberofarrangementsis

W (n)= n! 2n/2 (n/2)!

Althoughthesetwoequationswerederivedinverydifferentways,theyarenumerically identicalforall n

1 3 4 2 5 6
Thisdisulfidebondingconfigurationwithpairs1–6,2–5,and3–4isoneofthemanypossiblepairings. Countallthepossiblepairingarrangements.
D (n)=(n 1)(n 3)(n
5) ··· 1.
11

12.Predictingcombinationsofindependentevents.

Ifyouflipanunbiasedgreencoinandanunbiasedredcoinfivetimeseach,whatisthe probabilityofgettingfourredheadsandtwogreentails?

Theprobabilityoffourredheadsinfivecoinflipsis 1 2 5 5! 4!1! = 5 32

Theprobabilityoftwogreentailsis

Sincethegreencoinflipsareindependentoftheredcoinflips,theprobabilityweseekis (5/32)(10/32)=(50/1024)=4 88 × 10 2

13.Apairofaces.

Whatistheprobabilityofdrawingtwoacesintworandomdrawswithoutreplacementfrom afulldeckofcards?

Adeckhas52cardsandfouraces.Theprobabilityofgettinganaceonthefirstdrawis 4/52=1/13.Sinceyoudrawwithoutreplacement,theprobabilityofgettingoneofthe remainingthreeacesontheseconddrawis3/51,sotheprobabilityoftwoacesontwo drawsis

1 2 5 5! 2!3! = 10 32
4 52 3 51 =4.5 × 10 3 . 12

14.Averageofalinearfunction.

Whatistheaveragevalueof x,givenadistributionfunction q (x)= cx,where x ranges fromzerotoone,and q (x)isnormalized?

q(x)= cx c 0 x 1 x = 1 0 xq (x) dx = 1 0 cx 2 dx = c x3 3 1 0 = c 3 . Wecanalsofindc: 1= 1 0 q (x) dx = 1 0 cxdx = cx2 2 1 0 = c 2 =1 So, c =2, x = c 3 = 2 3 . 13

15.TheMaxwell–Boltzmannprobabilitydistributionfunction.

Accordingtothekinetictheoryofgases,theenergiesofmoleculesmovingalongthe x directionaregivenby εx =(1/2)mv 2 x ,where m ismassand vx isthevelocityinthe x direction.ThedistributionofparticlesovervelocitiesisgivenbytheBoltzmannlaw, p(vx )= e mv 2 x /2kT .ThisistheMaxwell–Boltzmanndistribution(velocitiesmayrangefrom −∞ to+∞).

(a)Writetheprobabilitydistribution p(vx ),sothattheMaxwell–Boltzmanndistribution iscorrectlynormalized.

(b)Computetheaverageenergy (1/2)mv 2 x .

(c)Whatistheaveragevelocity vx ?

(d)Whatistheaveragemomentum mvx ?

(a)Towritetheprobabilitydistribution p(vx ) dvx sothattheMaxwell–Boltzmann distributioniscorrectlynormalized,werequire

c ∞ −∞ e mv 2 x /2kT dvx =1
I = ∞ −∞ e ax2 dx = π a 1/2 . Aside Tocomputeintegralsoftheform I = ∞ −∞ e ax2 dx, weusethefollowingtrick.Itiseasytoseethatwecanwrite I 2 = ∞ −∞ e ax2 dx ∞ −∞ e ay 2 dy = ∞ −∞ ∞ −∞ e a(x2 +y 2 ) dxdy. 14
Fromintegraltables,weseethat

Againconsultingourtableofintegrals,wefind

Aside: Integralsoftheform

Thisisnowanintegralovertheentire(x,y
r and θ andrecognizingthat r 2 = x2 + y 2 ,theintegralbecomes I 2 = ∞ 0 drr 2π 0 dθe ar 2 = ∞ 0 drre ar 2 2π 0 dθ =2π ∞ 0 drre ar 2 Makingthesubstitution u = ar 2 ,du = 2ardr ,wecanfinishtheintegral: I 2 = π a −∞ 0 dueu = π a eu −∞ 0 = π a Hence I = π a 1/2 Forourintegral, a = m/2kT . ∞ −∞ e mv 2 x /2kT dvx = 2πkT m 1/2 =⇒ p(vx ) dvx = m 2πkT 1/2 e mv 2 x /2kT dvx . (b)Tocomputetheaverageenergy, (1/2)mv 2 x ,wehave 1 2 mv 2 x = ∞ −∞ 1 2 mv 2 x p(vx ) dvx = m 2 2πkT m 1/2 ∞ −∞ v 2 x e mv 2 x /2kT dvx
)plane.Convertingtopolarcoordinates
∞ −∞ x 2 e ax2 dx = π 1/2 2a3/2
∞ −∞ x 2 e ax2 dx
b a udv = uv |b a b a vdu Choosingthesubstitutions u = x and dv = xe ax2 ,wehave du = dx and v = 1 2a e ax2 . Ourintegralthereforebecomes ∞ −∞ x 2 e ax2 dx = 1 2a xe ax2 ∞ −∞ + 1 2a ∞ −∞ e ax2 = 1 2a π a 1/2 =0+ π 1/2 2a3/2 15
canbecomputedbyintegrationbyparts.Recallthat

Notethatwehaveusedtheresultoftheintegralfrompart(a)above. Therefore

(c)Tofindtheaveragevelocity vx ,werecallthatforfunctionswith odd symmetry (f (x)= f ( x)),theintegralunderthecurvefornegative x cancelswiththatunder thecurveforpositive x.Usingthefactthat p(x)= p( x),

(d)Theaveragemomentum

16.PredictingtherateofmutationbasedonthePoissonprobabilitydistribution function.

Theevolutionaryprocessofaminoacidsubstitutionsinproteinsissometimesdescribedby thePoissonprobabilitydistributionfunction.Theprobability ps (t)thatexactly s substitutionsatagivenaminoacidpositionoccuroveranevolutionarytime t is

where λ istherateofaminoacidsubstitutionspersiteperunittime.Fibrinopeptidesevolve rapidly: λF =9 0substitutionspersiteper109 years.Lysozymeisintermediate: λL ≈ 1 0.

Histonesevolveslowly: λH =0.010substitutionspersiteper109 years.

1 2 mv 2 x = 1 2 m π 1/2 2a3/2 m 2πkT 1/2 = 1 4 m ⎡ ⎣ 2kT m 3/2 ⎤ ⎦ m π 1/2 kT 1/2 = 1 2 kT.
vx = ∞ −∞ vx p(vx ) dvx = 0 −∞ vx p(vx ) dvx + ∞ 0 vx p(vx ) dvx = ∞ 0 ( vx )p( vx ) dvx + ∞ 0 vx p(vx ) dvx = ∞ 0 vx p(vx ) dvx + ∞ 0 vx p(vx ) dvx . =0
mv
x = m vx =0, fromtheresultabove.
ps (t)=
λt (λt)s s! ,
e
16

(a)Whatistheprobabilitythatafibrinopeptidehasnomutationsatagivensitein t =1 billionyears?

(b)Whatistheprobabilitythatlysozymehasthreemutationspersitein100million years?

(c)Wewanttodeterminetheexpectednumberofmutations s thatwilloccurintime t.Wewilldothisintwosteps.First,usingthefactthatprobabilitiesmustsumto one,write α = ∞ s=0 (λt)s /s!inasimplerform.

(d)Nowwriteanexpressionfor s .Notethat

(e)Usingyouranswertopart(d),determinetheratiooftheexpectednumberofmutations inafibrinopeptidetotheexpectednumberofmutationsinhistoneprotein, s fib / s his

(a)Theprobabilitythatafibrinopeptidehasnomutationsatagivensitein t =1billion yearsis

=exp[ (9 0per109 years)(109 years)]

(b)Forlysozyme,

t =(1 0per109 years)(108 years)=0 1

Theprobabilitythatlysozymehasthreemutationspersitein100millionyearsisthen

∞ s=0 s(λt)s s! =(λt) ∞ s=1 (λt)s 1 (s 1)! = λtα.
P0
t
e λF t
=
(
)=
e 9 =1 23 × 10 4
λL
P3 (t)= e λL t (λL t)3 3! = (e 0 1 )(0.1)3 6 =1 51 × 10 4 17

(c)Sincetheprobabilitiessumto1,

so
fib s
=
H
9 0
01 =900 18
∞ s=0 Ps (t)= ∞ s=0 e λt (λt)s s! = e λt ∞ s=0 (λt)s s! =1 Therefore α = ∞ s=0 (λt)s s! = e λt (d) s = ∞ s=0 sPs (t)= ∞ s=0 se λt (λt)s s! = e λt (λt) ∞ s=1 (λt)s 1 (s 1)! = e λt (λt) ∞ h=0 (λt)h h! =(λtα)e λt ,
s = λt. (e) s
his
λF t λH t = λF λ
=
.

17.Probabilityincourt.

Inforensicscience,DNAfragmentsfoundatthesceneofacrimecanbecomparedwith DNAfragmentsfromasuspectedcriminaltodeterminethattheprobabilitythatamatch occursbychance.SupposethatDNAfragment A isfoundin1%ofthepopulation,fragment B isfoundin4%ofthepopulation,andfragment C isfoundin2.5%ofthepopulation.

(a)Ifthethreefragmentscontainindependentinformation,whatistheprobabilitythat asuspect’sDNAwillmatchallthreeofthesefragmentcharacteristicsbychance?

(b)SomepeoplebelievesuchafragmentanalysisisflawedbecausedifferentDNAfragmentsdonotrepresentindependentproperties.Asbefore,supposethatfragment A occursin1%ofthepopulation.Butnowsupposethattheconditionalprobabilityof B ,giventhat A is p(B |A)=0 40ratherthan0.040,and p(C |A)=0 25ratherthan 0.025.Thereisnoadditionalinformationaboutanyrelationshipbetween B and C . Whatistheprobabilityofamatchnow?

(a)Sincethefragmentsareindependent,

.04)(0

(b)Compute x2 .

(c)Compute x3 .

(d)Compute x4

p = p(A) p(B ) p(C
=(0.01)(0
.025)=1 × 10 5 (b) p = p(A) p(B/A) p(C/A) =(0.01)(0.40)(0.25)=1 × 10 3 .
Givenaflatdistribution,from x = a to x = a,withprobabilitydistribution p(x)=1/(2a):
)
18.Flatdistribution.
(a)Compute x .
19 Molecular Driving Forces Statistical Thermodynamics in Biology Chemistry Physics and Nanoscience 2nd Edition Dill Solutions Full Download: http://testbanktip.com/download/molecular-driving-forces-statistical-thermodynamics-in-biology-chemistry-physics-and-nanoscience-2nd-edition-dill-solutions-manual/ Download all pages and all chapters at: TestBankTip.com

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