Title: USC3002 Picturing the World Through Mathematics
1USC3002 Picturing the World Through Mathematics
- Wayne Lawton
- Department of Mathematics
- S14-04-04, 65162749 matwml_at_nus.edu.sg
Theme for Semester I, 2007/08 The Logic of
Evolution, Mathematical Models of Adaptation
from Darwin to Dawkins
2Natural Selection
Reference Evolution by Mark Ridley, Chapter 5
p. 104 simplest model
Chance of Survival
Phenotype
Genotype
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
3Natural Selection
Problem what will the genotype frequencies be
after natural selection followed by random mating
?
Genotype
1st Ad. Freq.
Define
Baby Freq.
2nd Ad. Freq.
Define
4Natural Selection
Remark since
the genotype frequencies of the 2nd Adult
population are NOT in Hardy-Weinberg equilibrium
Let
denote
the change in gene frequency to the next
generation
Haldane (1924) produced this model for selection
p. 107
Since
the selection coefficient can
be computed from the 2nd generation gene
frequencies
5MATLAB 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 1ngens g(n) gt(1-sgt)/(1-sgt
2) gt g(n) end
6Tabular Output
generation
s0.05
s0.01
0.9900 0.9900 0.5608 0.9736 0.1931
0.9338 0.1053 0.8513 0.0710 0.7214
0.0532 0.5747 0.0424 0.4478 0.0352
0.3524 0.0301 0.2838 0.0262 0.2343
0.0233 0.1979
0 100 200 300 400 500 600 700 800 900 1000
g - gene frequencies
7Plot
8Plot
9Plot
10Plot
11Differential Equation Approximation
for
consists of solving the initial value problem
(frequency of gene g in zero-generation)
followed by the approximation
The error is small if
is small
12Qualitative Observations
If s gt 0,
1. If
then
therefore
2. For small s,
therefore
where
decays fastest at
3. If
then
therefore
13Numerical Solution Algorithm
Choose
Set
Set
While
14MATLAB Code for Differential Equation
function t, g tablepage107_approx(s,g0,T,delta
t) 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 1N t(n) ndeltat a
s(1-gg)gg2/(sgg2-1) gg gg
deltata g(n) gg end
15Numerical Solution Comparison
16Comparison Error
17Exact Solution for t
First rewrite the differential equation in the
form
Then use the method of partial fractions
http//www4.ncsu.edu/unity/lockers/users/f/felder/
public/kenny/papers/partial.html
18Exact Solution for s
implies that s can be solved for by
19Comparison With Sol. of Diff. Eqn.
203 Components Sol. of Diff. Eqn.
21Inverses of the 3 Components
22MATLAB 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
23Peppered Moth Estimation page 110
gtgt g0 1-1/100000 g0 0.99999000000000 gtgt gt
1 - 0.8 gt 0.20000000000000 gtgt t 50 t 50 gtgt
s sexact(t,gt,g0) s 0.27572621892750
Question Why does this differ from the books
estimate s 0.33 ?
24Peppered Moth Estimation page 110
gtgt gbook tablepage107(0.33,50,1-1/100000) gtgt
gmine tablepage107(0.2757,50,1-1/100000) gtgt
plot(150,gbook,150,gmine) gtgt grid gtgt
plot(150,gbook) gtgt plot(150,gbook,150,gmine) gtgt
grid gtgt ylabel(bluebook, green mine') gtgt
xlabel('number of generations')
25Peppered Moth Simulation
26Assigned Reading
Chapter 25. Evolution The Process in Schaums
Outlines in Biology
Chapter 5. The Theory of Natural Selection in
Mark Ridleys 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
27Homework 2. Due Tuesday 04.09.2007
Do problems 1-10 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.