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 aEvent a1 aand aEvent a2 aare aindependent, athe a“and”
acombination aof athese atwo aoutcomes ais agiven aby athe aintersection,
aand athe aprobability aof athe
intersection ais agiven aby athe aproduct aof athe atwo aseparate aprobabilities,
aleading ato athe aexpressions afor aprobabilities afor athe agray across-hatched
acrescent.

(a) These aare atwo aindependent aelementary aevents aeach awith aan
aoutcome aprobability aof a0.5. aWe aare aasked afor athe aprobability aof
athe asequence aH1 aT2, awhich arequires amultiplication aof athe
aelementary aprobabilities:

p HH a = aH1 a∩ aT2 a= 1 a a a1a1
= a a× a a=
apH a× apT
1 a2 1 2 2 a a 24

We acan aarrange athis aprobability, aalong awith athe aprobability afor athe
aother athree apossible asequences, ain aa atable:


Toss a1

Toss a2 H a(0.5) T a(0.5)

H a(0.5) H1H2 T1H2
(0.25) (0.25)

T a(0.5) H1T2 T1T2
(0.25) (0.25)

Note: aProbabilities aare agiven ain aparentheses.

The aprobability aof agetting aa ahead aon athe afirst atoss aor aa atail aon
athe asecond atoss, abut anot aboth, ais

pH1 aor aH2 a = apH1 a + apH2 a − a2( apH1 a× apH2 a)
1 a1 a1 a a1 ı
= + a − a2 × aı
2 2 2 aa a 2j
=a
a
1
a
2

In athe atable aabove, athis acombination acorresponds ato athe asum aof
athe atwo aoff- adiagonal aelements a(the aH1T2 aand athe aT1H2 aboxes).

(b) This ais athe a"and" acombination afor aindependent aevents, aso awe
amultiply athe aelementary aprobability apH afor aeach aof aN atosses:

pH1H2H3…HN a = apH1 a× apH2 a× apH3 a×⋯× apHN
N
= a a1 a ı
a2ıj

This ais aboth aa apermutation aand aa acomposition a(there ais aonly aone
apermutation afor aall-heads). aAnd anote athat asince aboth aoutcomes
ahave aequal aprobability a(0.5), athis agives athe aprobability aof aany
apermutation aof aany anumber aNH aof aheads awith aany anumber aN a−
aNH aof atails.

1.3 Two adifferent aapproaches awill abe agiven afor athis aproblem. aOne ais
aan aapproximation athat ais avery aclose ato abeing acorrect. aThe asecond
ais aexact. aBy acomparing athe aresults, athe areasonableness aof athe afirst
aapproximation acan abe aexamined.

Whichever aapproach awe ause ato asolve athis aproblem, awe abegin aby