Math 4600: Homework 10 Solutions
Gregory Handy
[10.1] In the model for random genetic drift in a population of bacteria, consider the case when
each cell has 4 plasmids. In the original cells we introduce half of the plasmids of type P and half
of the plasmid of type Q, so p02 = 1. Iterate to find what will be the distribution of plasmids in
the population after 50 generations. This phenomenon was first observed experimentally and it was
hypothesized that the plasmids interact so that they cannot be picked up into the same cell. Your
result shows that the separation of plasmid types can happen simply as a result of random plasmid
transfer.
In order to perform this iteration, we need the transition matrix defined by
2i
2N − 2i
j
N −j
,
Pij =
2N
N
where N is the total number of plasmids in the mother cell, i is the number of plasmids of type P in the
mother cell, and j is the number of plasmids of type P in the daughter cell. For this problem, we know
N = 4, and so i and j can vary from 0 to 4. Thus, we need to construct a 5 by 5 matrix. Using MATLAB,
we find our transition matrix to be


1.0000
0
0
0
0

 0.2143 0.5714 0.2143
0
0



P =  0.0143 0.2286 0.5143 0.2286 0.0143 


0
0
0.2143 0.5714 0.2143 
0
0
0
0
1.0000
In order to iterate over n generations, we simply use the following for loop
p = [0, 0, 1, 0, 0];
for i = 1:n
p = p · P,
end
Using MATLAB, we find
p(50) = [0.4997, 0.0002, 0.0002, 0.0002, 0.4997].
We see that the majority of the population has either all P’s or all Q’s!
clear;
clc;
close all;
% construct the transition matrix
N=4;
P = zeros(5,5);
for i = 0:4
for j = 0:4
1
% in order for nchoosek(n,k) to work we need k < n
% so this if statement checks this condition for our
% two nchoosek statements
if (N-j) <= (2*N-2*i) && j<=2*i
P(i+1,j+1) = nchoosek(2*i,j)*nchoosek(2*N-2*i,N-j)/nchoosek(2*N,N);
else
P(i+1,j+1)=0;
end
end
end
% print out the transition matrix
P
p = [0 0 1 0 0]; % initial condition
for i = 1:50
p = p*P;
end
% print out the final distribution vector
p
2
[10.2] In the sickle cell anemia model, consider 2 different situations: a society with strong genetic
counseling (γ = 0.9, α = 0.1, β = 0.1) and a society in tropics where carrying a sickle cell allele
makes people less likely to die from malaria (γ = .2, α = 0.1, β = 0.4 and also see below).
(a) Explain why this choice of parameters fits the word description of each society.
(b) What will happen to disease in each population in the long run (use cobwebbing by hand)?
(c) How will the model change if individuals with hh genotype are sterile (cannot reproduce)?
(a) It is not beneficial to have the recessive sickle cell allele, h, in populations where Malaria is not present.
Thus we expect the fitness of the HH genotype (γ) to be the highest in this population. However, if you
are in the tropics, Having a single sickle cell allele protects the carrier from malaria. Thus we expect
the fitness of the Hh/hH genotype (β) to be higher in populations where malaria is present.
(b) As Fig. 1 (left) shows, when γ > β, α, we see that the frequency of the sickle cell allele goes to zero
as n → ∞. However, when β > γ > α, Fig. 1 (right) shows that the frequency approachs a non-zero
number (≈ 0.3970), and the sickle cell allele (and thus sickle cell anemia) will continue to persist.
γ = 0.9, α=0.1, β=0.1
γ = 0.2, α=0.1, β=0.4
γ = 0.2, α=0.1, β=0.4
0.8
0.8
0.8
0.6
0.6
0.6
PN+1
1
PN+1
1
PN+1
1
0.4
0.4
0.4
0.2
0.2
0.2
0
0
0.2
0.4
PN
0.6
0.8
0
0
1
0.2
0.4
PN
0.6
0.8
0
0
1
0.2
0.4
PN
0.6
0.8
1
Figure 1: Cobwebbing for the two cases. Graphs the frequency of the h (sickle cell) allele.
(c) If the hh genotype can not reproduce, the fitness for hh will go to zero (α = 0). The figures adjusted
for this case can be seen in Fig. 2. We see that the limit as n → ∞ for Pn is approximately the same as
part (a), but we note that it approaches it at a much faster rate if it starts from above in this scenario.
γ = 0.9, α=0, β=0.1
γ = 0.2, α=0, β=0.4
γ = 0.2, α=0, β=0.4
0.8
0.8
0.8
0.6
0.6
0.6
PN+1
1
PN+1
1
PN+1
1
0.4
0.4
0.4
0.2
0.2
0.2
0
0
0.2
0.4
PN
0.6
0.8
1
0
0
0.2
0.4
PN
0.6
0.8
1
0
0
0.2
0.4
PN
0.6
0.8
Figure 2: Cobwebbing for the two cases. Graphs the frequency of the h (sickle cell) allele.
clear;
close all;
gamma=0.9;
alpha=0.1;
beta=0.1;
P_n=[0:.001:1];
P_n1 = (P_n.^2*(alpha-beta) +...
beta*P_n)./((alpha-2*beta+gamma)*P_n.^2+(2*beta-2*gamma)*P_n+gamma);
3
1
plot(P_n,P_n1,’linewidth’,1.5)
hold on
plot(P_n,P_n,’black’,’linewidth’,1.5)
P_n=.8;
P_n1=0;
for i = 1:10
P_n1_new = (P_n.^2*(alpha-beta) +...
beta*P_n)./((alpha-2*beta+gamma)*P_n.^2+(2*beta-2*gamma)*P_n+gamma);
plot([P_n, P_n],[P_n1,P_n1_new],’--’,’linewidth’,1.5)
P_n1 = P_n1_new;
plot([P_n,P_n1],[P_n1,P_n1],’--’,’linewidth’,1.5)
P_n=P_n1;
end
axis([0 1 0 1])
xlabel(’P_N’,’fontsize’,16)
ylabel(’P_{N+1}’,’fontsize’,16)
set(gca,’fontsize’,16)
title(’\gamma = 0.9, \alpha=0.1, \beta=0.1’)
4