3D03 - Assignment 4
2. Thermodynamics of biochemical reactions (10 marks)
This question concerns the following two molecules that are intermediates in the
glycolysis pathway.
3-phosphoglycerate (henceforward PG)
1,3-bisphosphoglycerate (henceforward BPG)
The following reactions represent the hydrolysis of BPG to PG and the hydrolysis of ATP
to ADP. Pi denotes an inorganic phosphate ion.
BPG + H2O → PG + Pi
G’ = -12.0 kcal mol-1
ATP + H2O → ADP + Pi
G’ = -7.7 kcal mol-1
G’ denotes the free energy change at standard conditions of 1M concentration for the
organic molecules and physiological conditions of H+, Pi etc.
The enzyme phosphoglycerate kinase can catalyze the reaction of transfering a phosphate
from ATP to PG:
ATP + PG → ADP + BPG
reaction 1
(a) What is the free energy change G’reac for reaction 1 under the same conditions?
(Hint: you can imagine that the enzyme couples the first two reactions together, but you
should find that reaction 1 will not proceed spontaneously in the forward direction).
Turn the first reaction round. Its G’ becomes positive. Then add them together. The sum
is reaction 1. The Pi and the H2O are the same on both sides so they don’t contribute.
PG + Pi
→ BPG + H2O
G’ = +12.0 kcal mol-1
ATP + H2O → ADP + Pi
G’ = -7.7 kcal mol-1
------------------------------G’reac = + 4.3 kcal mol-1
(b) From G’reac calculate the equilibrium constant for reaction 1 under these conditions.
The temperature is 25oC, and the gas constant is R = 1.987 cal mol-1 K-1.
K = exp(-G’/RT) = exp(-4300/(1.987 × 298)) = 7 × 10-4
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(c) The cell is not at standard conditions. Write down the free energy change for reaction
1 in the cell (Greac) as a function of the free energy change at standard conditions
(G’reac) and the concentrations of the molecules involved.
  RT ln
answer: Greac  Greac
[ ADP ][ BPG ]
[ ATP][ PG]
(d) The cell maintains the ratio of [ATP]/[ADP] at approximately 10. If the ratio
[PG]/[BPG] = r, how big must r be in order to make reaction 1 thermodynamically
favourable?
We want to know when is Greac < 0.
1
  RT ln
Greac
0
10r
  RT ln 10r  0
Greac
 / RT
ln 10r  Greac
1
 / RT )
exp( Greac
10
r > 0.1 × exp(+4300/(1.987×298)).
r
r >142
3. Quasispecies (10 marks)
This question will develop a simplified version of the Eigen theory for
populations of replicating sequences. The question is long because it tells you exactly
what to do at each step! There is not much calculation or writing to do, but I would like
you to work through the equations so that you understand what they mean.
Let there be a master sequence that replicates at rate A0, and suppose that all other
sequences replicate at rate A1. If 0 is the fraction of the master sequence in the
population, then the mean replication rate is
A   o A0  (1   0 ) A1 .
(3.1)
In the diagram below, the black circle represents the master sequence, and the white
circles represent the 1-mutant neighbours of the master sequence. There are many
sequences with 2 or more mutations that are not shown.
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The sequences are of length L. When a sequence is replicated, there is a probability u that
a mutation occurs at each site, and a probability 1-u that it is accurately replicated. The
probability that the master sequence is replicated exactly (with no mutations) is
Q00  (1  u ) L . We are considering u << 1 and L >> 1, so we can make the approximation
that Q00  (1  u ) L  exp( U ) , where U = uL. Check with a calculator that this
approximation is good (e.g. when u = 0.001 and L = 1000). We may now write
d0
(3.2)
 0 A0 Q00  0 A
dt
Here, we have ignored the degradation rates D that were in the Eigen theory (and in the
lectures) because they do not make much difference to the answer. This means that the
“mean excess production” is just A . The first term in (3.2) is the rate of accurate
replication of the master sequence. The second term represents dilution of the medium at
a rate A . This is required, so that the total concentration of all sequences remains
constant (i.e. there is competition for resources). Equation (3.2) does not include a term
for mutation from a 1-mutant neighbour back to the master sequence. This term is
negligible if L >> 1, because almost all mutations of a 1-mutant neighbour will take you
further away from the master sequence (as indicated by the arrows in the diagram).
(a) At equilibrium, d0 / dt  0 . By combining (3.1) and (3.2) obtain the equilibrium
concentration 0 as a function of U, A0, and A1. At what value of U does the error
threshold occur? (Follow the same steps as in the notes).
At equilibrium, assuming 0  0, A0 Q00  A .
Therefore A0 e U   0 A0  (1   0 ) A1
A0 e U  A1
0 
A0  A1
The error threshold occurs when e-U = A1/A0.
Therefore U = - ln (A1/A0) = + ln (A0/A1).
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(b) Draw an accurate graph of 0 as a function of U for the case where A0 = 2 and A1 = 1.
Use the range of U between 0 and 1 on the x axis of the graph. What is the numerical
value of U where the error threshold occurs in this case?
0 = 2e-U-1
Error threshold is at U = ln(2) = 0.693
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.693147
0.7
0.9
1
0.902459
0.809675
0.721416
0.637462
0.557602
0.481636
0.409376
0.34064
0.275256
0.213061
0.1539
0.097623
0.044092
0
0
0
0
0.090246
0.161935
0.216425
0.254985
0.278801
0.288982
0.286563
0.272512
0.247731
0.213061
0.16929
0.117148
0.057319
0
0
0
(c) Now let 1 be the fraction of 1-mutant neighbours (i.e. the sum of the concentrations
of all the sequences exactly 1 mutation away from the master sequence). The probability
that a 1-mutant neighbour is accurately replicated is Q11  exp( U ) , the same as Q00. The
probability that a 1-mutant neighbour is created by mutation of the master sequence is
Q10  uL(1  u ) L 1  U exp( U ) . Now we may write:
d1
 0 A0 Q10  1 A1Q11  1 A
dt
(3.3)
We already know from (3.2) that, at equilibrium, A  A0 exp( U ) . Hence, show that
1  0UA0 /( A0  A1 ) at equilibrium. Draw the curve for 1 on the same graph as in (b).
d1
 0 A0Ue U  1 A1e U  1 A0 e U  0
dt
result follows immediately.
For these numerical values 1 = 2U0 = 2U(2e-U-1)
At equilibrium
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(d) Describe in words what is happening to the population of sequences when U is close
to 0, when U is midway between 0 and the error threshold, and when U is above the error
threshold.
4. Prisoners’ dilemma (5 marks)
The prisoners’ dilemma model uses the following payoff matrix.
You
C D
C R S
D T P
Me
In the standard evolutionary version of the game, players meet at random. If  is the
fraction of cooperators, the average payoff to a cooperator is AC  R  (1   ) S , and the
average payoff to a defector is AD  T  (1   ) P . The average payoff to the population
as a whole is A  AC  (1   ) AD . It is supposed that the fraction of cooperators obeys
the equation
d
  ( AC  A ) .
(4.1)
dt
(a) For simplicity, consider the case where P = S = 0 and R = 1, so the only variable
parameter is T. The temptation is more than the reward (T > 1).
d
Work out the equation for
in this case. Show that  will always decrease, i.e.
dt
cooperators will always die out.
AC = 
AD = T
A   2  (1   )T
d
  (   2  (1   )T )   2 (1   )(1  T )
dt
But  is between 0 and 1, so 2(1-) is always positive. T > 1, therefore
d
is always
dt
negative.
(b) In the spatial version of the game, each site plays against its 8 neighbours and a copy
of itself. The fitness of a site is the sum of the payoffs from these 9 games. The payoff
parameters are as in (a), with only T considered as variable. Calculate the fitness of the
cooperator marked C1 below and the defector marked D2. Under what conditions does
C1 have higher fitness than D2? Hence, show that a small cluster of cooperators
surrounded by defectors can sometimes expand, i.e. the cooperators do not always die out
in the spatial game.
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D D
D
D
D
D
D D
D
D
D3
D
D D
C
C1
D2
D
D D
C
C
D
D
D D
D
D
D
D
D D
D
D
D
D
Fitness of C1 = 4R + 5S = 4
Fitness of D2 = 2T + 7P = 2T
C1 is better than D2 if 4 > 2T, i.e. if T < 2.
As long as T is not too large, cooperators will survive.
Additional information - A single C on its own cannot survive (prove this!), but a larger
cluster, for example a 3×3 square of Cs can also survive if T < 2. To understand the
behaviour of the model it is necessary to consider many different configurations that
could arise. Nowak and May show that if T < 1.8, the Cs take over the whole lattice. If
1.8 < T < 2, both C and D survive together. If T > 2, Ds take over the whole lattice.
(c) This is a point to think about. No answer is required for this part. If u = 0 in the
error threshold theory, Q00 = 1. Hence, equation (3.2) looks the same as equation (4.1).
There is an important difference, however. In the error threshold theory, the rates A0 and
A1 are constants, whereas in the prisoners’ dilemma, the payoffs AC and AD are functions
of the frequencies of the strategies. This is called ‘frequency dependent selection’. It is an
essential part of evolutionary game theory models.
In (4.1), we assumed that the strategies were accurately reproducing. However,
we could add ‘mutations’ in strategies if we wanted. We could say that the offspring of a
C becomes a D with a small probability u (and the same from D to C), while the offspring
is the same strategy as the parent with probability 1-u. How would this change equation
(4.1), and what difference would it make to the outcome?
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