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 sEvent s1 sand sEvent s2 sare sindependent, sthe s“and” scombination
sof sthese stwo soutcomes sis sgiven sby sthe sintersection, sand sthe
sprobability sof sthe
intersection sis sgiven sby sthe sproduct sof sthe stwo sseparate sprobabilities,
sleading sto sthe sexpressions sfor sprobabilities sfor sthe sgray scross-hatched
screscent.

(a) These sare stwo sindependent selementary sevents seach swith san
soutcome sprobability sof s0.5. sWe sare sasked sfor sthe sprobability sof
sthe ssequence sH1 sT2, swhich srequires smultiplication sof sthe
selementary sprobabilities:

p HH s = sH1 s∩ sT2 s= 1 s s s1 s1
= s s× s s=
spH s× spT
1 s2 1 2 2 s s2 4

We scan sarrange sthis sprobability, salong swith sthe sprobability sfor sthe
sother sthree spossible ssequences, sin sa stable:


Toss s1

Toss s2 H s(0.5) T s(0.5)

H s(0.5) H1H2 T1H2
(0.25) (0.25)

T s(0.5) H1T2 T1T2
(0.25) (0.25)

Note: sProbabilities sare sgiven sin sparentheses.

The sprobability sof sgetting sa shead son sthe sfirst stoss sor sa stail son sthe
ssecond stoss, sbut snot sboth, sis

pH1 sor sH2 s = spH1 s + spH2 s − s2( spH1 s× spH2 s)
1 s1 s1 s s1 ı
= + s − s2 × sı
2 2 2 ss s 2j
=s
s
1
s
2

In sthe stable sabove, sthis scombination scorresponds sto sthe ssum sof sthe
stwo soff- sdiagonal selements s(the sH1T2 sand sthe sT1H2 sboxes).

(b) This sis sthe s"and" scombination sfor sindependent sevents, sso swe
smultiply sthe selementary sprobability spH sfor seach sof sN stosses:

pH1H2H3…HN s = spH1 s× spH2 s× spH3 s×⋯× spHN
N
= s s1 s ı
s2ıj

This sis sboth sa spermutation sand sa scomposition s(there sis sonly sone
spermutation sfor sall-heads). sAnd snote sthat ssince sboth soutcomes
shave sequal sprobability s(0.5), sthis sgives sthe sprobability sof sany
spermutation sof sany snumber sNH sof sheads swith sany snumber sN s− sNH
sof stails.

1.3 Two sdifferent sapproaches swill sbe sgiven sfor sthis sproblem. sOne sis san
sapproximation sthat sis svery sclose sto sbeing scorrect. sThe ssecond sis
sexact. sBy scomparing sthe sresults, sthe sreasonableness sof sthe sfirst
sapproximation scan sbe sexamined.

Whichever sapproach swe suse sto ssolve sthis sproblem, swe sbegin sby