USC3002 Picturing the World
Through Mathematics
Wayne Lawton
Department of Mathematics
S14-04-04, 65162749
matwml@nus.edu.sg
Theme for Semester I, 2008/09 : The Logic of
Evolution, Mathematical Models of Adaptation
from Darwin to Dawkins
Natural Selection
Reference: Evolution by Mark Ridley, Chapter 5
p. 104 simplest model
Genotype
Phenotype
Chance of Survival
Y
1
yellow seeds
H
1
yellow seeds
green seeds
1 s
G
s  [0,1] is the selection coefficient
The chance of survival is relative to the maximal chance
of survival among all genotypes. Notice that here it
depends on the phenotype
Natural Selection
Problem: what will the genotype frequencies be
after natural selection followed by random mating ?
Genotype
Y
PY
1st Ad. Freq.
y  PY  12 PH , g  PG  12 PH
Define
y
Baby Freq.
2nd Ad. Freq.
Define
H
PH
2
g
2 yg
2
P 
PH' 
y 2 /(1  sg 2 )
2 yg /(1  sg ) g (1  s) /(1  sg )
'
Y
PG' 
2
y  P  P  y /(1  sg )
'
G
PG
'
Y
1
2
'
H
g '  PG'  12 PH'  1  y '
2
2
2
Natural Selection
Remark: since
'2
P  y /(1  sg )  y  y /(1  sg )
'
Y
2
2
2
2 2
the genotype frequencies of the 2nd Adult population
are NOT in Hardy-Weinberg equilibrium
Let
y  y  y  syg /(1  sg )
'
2
2
denote
the change in gene frequency to the next generation
Haldane (1924) produced this model for selection p. 107
Since
s  y /( y g )
'
2
the selection coefficient can
be computed from the 2nd generation gene frequencies
MATLAB Program for Table 5.4, p. 107
function g = tablepage107(s,ngens,g0)
% function g = tablepage107(s,ngens,g0)
%
% Wayne Lawton, 21 August 2007
% computes gene frequencies in Table 5.4 Evolution by Ridley
%
% Outputs
% g = array of length ngens
% g(k) = gene frequency of recessive gene after k generations
% Inputs
% s = selection coefficient
% ngens = number of generations
% g0 = initial gene frequency
%
gt = g0;
for n = 1:ngens
g(n) = gt*(1-s*gt)/(1-s*gt^2);
gt = g(n);
end
Tabular Output
generation
s=0.05
s=0.01
0
100
200
300
400
500
600
700
800
900
1000
0.9900
0.5608
0.1931
0.1053
0.0710
0.0532
0.0424
0.0352
0.0301
0.0262
0.0233
0.9900
0.9736
0.9338
0.8513
0.7214
0.5747
0.4478
0.3524
0.2838
0.2343
0.1979
g - gene
frequencies
Plot
Plot
Plot
Plot
Differential Equation Approximation
for
g  g  g  s(1  g ) g /( sg  1)
'
2
2
consists of solving the initial value problem:
(frequency of gene
~
g (0)  g (0) g in zero-generation)
d g~
2
2
~
~
~
(t )  s(1  g (t )) g (t ) /( sg (t )  1), t  0
dt
followed by the approximation
g (n)  g~ (n), n  1,2,3,...
The error is small if
d g~
(t ) is small
dt
Qualitative Observations
d g~
(t )  s(1  g~ (t )) g~ (t ) 2 /( sg~ (t ) 2  1)  0
If s > 0,
dt
1. If
g~ (t )  g~ (0)  1
d g~
~ (t )) /(1  s)
(
t
)


s
(
1

g
then
dt
~
~
therefore g (t )  1  (1  g (0)) exp( st /(1  s))
d g~
~
2
~
~
g
(t )   s(1  g (t )) g (t ) therefore (t )
2. For small s,
dt
~
d
g
4s
~
2 where
(t )  
decays fastest at g (t )  3
dt
27
d g~
2
~
(t )   sg (t )
then
dt
g~ (t )  g~ (0)  0
~ (t )  g~ (0) /(1  sg~ (0)t )
therefore g
3. If
Numerical Solution Algorithm
Set
T  0, t  0
~
g (0)  g (0)
Set
t 0
Choose
While
t T
t  t  t
2
2
~
~
~
a  s(1  g (t )) g (t ) /( sg (t )  1)
~
dg
~
~
g (t  t )  g (t )  t
(t )
dt
MATLAB Code for Differential Equation
function [t, g] = tablepage107_approx(s,g0,T,deltat)
% function [t, g] = tablepage107_approx(s,g0,T,deltat)
% Wayne Lawton, 22 August 2007
% numerical solution of differential equation
% for gene frequencies
% Outputs
% t = array of times
% g = solution array (as a function of t)
% Inputs
% s = selection coefficient
% g0 = initial gene frequency
% T = approx last time
% deltat = time increment
N = round(T/deltat);
gg = g0;
for n = 1:N
t(n) = n*deltat;
a = s*(1-gg)*gg^2/(s*gg^2-1);
gg = gg + deltat*a;
g(n) = gg;
end
Numerical Solution Comparison
s  0.05, t  0.1
Comparison Error
s  0.05, t  0.1
Exact Solution for t
First rewrite the differential equation in the form
2
2
~
~
~
~
[( sg  1) / s(1  g ) g ] d g  dt
Then use the method of partial fractions
http://www4.ncsu.edu/unity/lockers/users/f/felder/public/kenny/papers/partial.html
g~ ( t )
t
1 ~
1 ~
~
( s  1)d g s d g s d g



dt
2


~
~
~
g 1
g
g
g~ ( 0 )
0
1
~(t )  1
g
1
1
(s  1) ln ~
 s ln
g (0)  1
1
1
~
g (0) s
s



t
~
~
~
g (t ) g (t ) g (0)
Exact Solution for s
~(t )  1
g
1
1
(s  1) ln ~
 s ln
g (0)  1
1
1
~
g (0) s
s



t
~
~
~
g (t ) g (t ) g (0)
implies that s can be solved for by
1
~

g (t )  1  
s  t  ln ~
ln


g (0)  1 

g~(t )  1
 ln
~
g (0)  1
g~(0)
1
1 
~ ~ 
~
g (t ) g (t ) g (0) 
Comparison With Sol. of Diff. Eqn.
3 Components Sol. of Diff. Eqn.
Inverses of the 3 Components
MATLAB Code for Selection Coefficient
function s = sexact(t,gt,g0)
% function s = sexact(t,gt,g0)
% Wayne Lawton, 24 August 2007
% exact solution for s
% Outputs
% s = selection coefficient
% Inputs
% t time of evolution
% g0 = gene frequency at time 0
% g = gene frequency at t
num = log((gt-1)/(g0-1)) + log(g0/gt) + 1/gt - 1/g0;
den = t + log((gt-1)/(g0-1));
s = num/den;
Peppered Moth Estimation page 110
>> g0 = 1-1/100000
g0 = 0.99999000000000
>> gt = 1 - 0.8
gt = 0.20000000000000
>> t = 50
t = 50
>> s = sexact(t,gt,g0)
s = 0.27572621892750
Question: Why does this differ from
the book’s estimate s = 0.33 ?
Peppered Moth Estimation page 110
>> gbook =
tablepage107(0.33,50,11/100000);
>> gmine =
tablepage107(0.2757,50,11/100000);
>> plot(1:50,gbook,1:50,gmine)
>> grid
>> plot(1:50,gbook)
>> plot(1:50,gbook,1:50,gmine)
>> grid
>> ylabel(‘blue=book, green =
mine')
>> xlabel('number of generations')
Peppered Moth Simulation
Assigned Reading
Chapter 25. Evolution: The Process in Schaum’s
Outlines in Biology
Chapter 5. The Theory of Natural Selection in
Mark Ridley’s Evolution. In particular study: (i) the
peppered moth (Biston betularia) studies of the
decrease in the recessive peppered moth allele,
(ii) pesticide resistence, (iii) equilibrium for
recurrent disadvantageous dominant mutation,
(iv) heterozygous advantage and sickle cell (1st
study of natural selection in humans), (v) freq.
dependent fitness, (vi) Wahlund effect, (vii)
effects of migration and gene flow
Homework 2. Due Monday 1.09.2008
Do problems 1-6 on page 136 in Ridley (the
mean fitness in questions 2, 3 is defined on p105)
Homework 3. Due Monday 8.09.2008
Do problems 7-10 on page 136 in Ridley (the
mean fitness in questions 2, 3 is defined on p105)
Question 11. Assume that for a two allele locus
that genotype AA has fitness 1-s, genotype Aa
has fitness 1, and genotype aa has fitness 1-t and
that random mating occurs. Let p = baby freq. of
gene A and q = baby freq. of gene a. Derive
formuli for the next baby freq. p’ and q’.
Question 12. Assume that in a two allele locus all
genotypes have fitness = 1 but that each genotye
mates only with the same genotype. Derive
equations for the evolution of gene frequencies.