Solutions Manual for Biomolecular Thermodynamics, From Theory to Application, 1e Douglas Barrick (All Chapters)

Solutions Manual for
Biomolecular
Thermodynamics,
From Theory to
Application, 1e
Douglas Barrick (All
Chapters)

,Solution Manual


CHAPTER 1
1.1 Using the same Venn diagram for illustration, we want the probability of
outcomes from the two events that lead to the cross-hatched area shown
below:




A1 A1 n B2 B2


This represents getting A in event 1 and not B in event 2, plus not getting A
in event 1 but getting B in event 2 (these two are the common “or but not
both” combination calculated in Problem 1.2) plus getting A in event 1 and B in
event 2.

1.2 First the formula will be derived using equations, and then Venn diagrams
will be compared with the steps in the equation. In terms of formulas and
probabilities, there are two ways that the desired pair of outcomes can come
about. One way is that we could get A on the first event and not B on the
second ( A1 ∩ (∼B2 )). The probability of this is taken as the simple product, since
events 1 and 2 are independent:

pA1 ∩ (∼B2 ) = pA × p∼B
= pA ×(1− pB ) (A.1.1)
= pA − pApB

The second way is that we could not get A on the first event and we could get
B on the second ((∼ A1) ∩ B2 ) , with probability

p(∼A1) ∩ B2 = p∼A × pB
= (1− pA )× pB (A.1.2)
= pB − pApB

,2 SOLUTION MANUAL


Since either one will work, we want the or combination. Because the two ways
are mutually exclusive (having both would mean both A and ∼A in the first
outcome, and with equal impossibility, both B and ∼B), this or combination is
equal to the union { A1 ∩ (∼B2 )} ∪ {(∼ A1) ∩ B2}, and its probability is simply the sum
of the probability of the two separate ways above (Equations A.1.1 and A.1.2):

p{A1 ∩ (∼B2 )} ∪ {(~A1) ∩ B2} = pA1 ∩ (∼B2) + p(∼A1) ∩ B2
= pA − pApB + pB − pApB
= pA + pB − 2pApB

The connection to Venn diagrams is shown below. In this exercise we will work
backward from the combination of outcomes we seek to the individual outcomes.
The probability we are after is for the cross-hatched area below.
{ A1 ∩ (∼B2 )} ∪ {(∼ A1) ∩ B2 }




A1 B2


As indicated, the circles correspond to getting the outcome A in event 1 (left)
and outcome B in event 2. Even though the events are identical, the Venn
diagram is constructed so that there is some overlap between these two (which
we don’t want to include in our “or but not both” combination. As described
above, the two cross-hatched areas above don’t overlap, thus the probability of
their union is the simple sum of the two separate areas given below.


A1 n ~B2
~ A1 n B2

pA × p~B
p ~A × pB
= pA (1 – pB)
= (1 – p A)p B

A1 n ~B2 ~ A1 n B2



Adding these two probabilities gives the full “or but not both” expression
above. The only thing remaining is to show that the probability of each of
the crescents is equal to the product of the probabilities as shown in the top
diagram. This will only be done for one of the two crescents, since the other
follows in an exactly analogous way. Focusing on the gray crescent above, it
represents the A outcomes of event 1 and not the B outcomes in event 2. Each
of these outcomes is shown below:

Event 1 Event 2


A1 ~B

p~B = 1 – pB
pA



A1 ~B2

, SOLUTION MANUAL 3


Because fEvent f1 fand fEvent f2 fare findependent, fthe f“and” fcombination
fof fthese ftwo foutcomes fis fgiven fby fthe fintersection, fand fthe
fprobability fof fthe
intersection fis fgiven fby fthe fproduct fof fthe ftwo fseparate fprobabilities,
fleading fto fthe fexpressions ffor fprobabilities ffor fthe fgray fcross-hatched
fcrescent.

(a) These fare ftwo findependent felementary fevents feach fwith fan
foutcome fprobability fof f0.5. fWe fare fasked ffor fthe fprobability fof fthe
fsequence fH1 fT2, fwhich frequires fmultiplication fof fthe felementary
fprobabilities:

p HH f = fH1 f∩ fT2 f= fpH 1 f f f1 f1
= f f× f f=
f× fpT
1 f2 1 2 2 f f2 4

We fcan farrange fthis fprobability, falong fwith fthe fprobability ffor fthe
fother fthree fpossible fsequences, fin fa ftable:


Toss f1

Toss f2 H f(0.5) T f(0.5)

H f(0.5) H1H2 T1H2
(0.25) (0.25)

T f(0.5) H1T2 T1T2
(0.25) (0.25)

Note: fProbabilities fare fgiven fin fparentheses.

The fprobability fof fgetting fa fhead fon fthe ffirst ftoss for fa ftail fon fthe
fsecond ftoss, fbut fnot fboth, fis

pH1 for fH2 f = fpH1 f + fpH2 f − f2( fpH1 f× fpH2 f)
1 f1 f1 f f1 ı
= + f − f2 × fı
2 2 2 ff f 2j
=f
f
1
f
2

In fthe ftable fabove, fthis fcombination fcorresponds fto fthe fsum fof fthe
ftwo foff- fdiagonal felements f(the fH1T2 fand fthe fT1H2 fboxes).

(b) This fis fthe f"and" fcombination ffor findependent fevents, fso fwe
fmultiply fthe felementary fprobability fpH ffor feach fof fN ftosses:

pH1H2H3…HN f = fpH1 f× fpH2 f× fpH3 f×⋯× fpHN
N
= f f1f ı
f2ıj

This fis fboth fa fpermutation fand fa fcomposition f(there fis fonly fone
fpermutation ffor fall-heads). fAnd fnote fthat fsince fboth foutcomes
fhave fequal fprobability f(0.5), fthis fgives fthe fprobability fof fany
fpermutation fof fany fnumber fNH fof fheads fwith fany fnumber fN f− fNH
fof ftails.

1.3 Two fdifferent fapproaches fwill fbe fgiven ffor fthis fproblem. fOne fis fan
fapproximation fthat fis fvery fclose fto fbeing fcorrect. fThe fsecond fis
fexact. fBy fcomparing fthe fresults, fthe freasonableness fof fthe ffirst
fapproximation fcan fbe fexamined.

Whichever fapproach fwe fuse fto fsolve fthis fproblem, fwe fbegin fby