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USC3002 Picturing the World Through Mathematics. Wayne Lawton Department of Mathematics S14-04-04, 65162749 [email protected]. Theme for Semester I, 2008/09 : The Logic of Evolution, Mathematical Models of Adaptation from Darwin to Dawkins. Reference: Evolution by Mark Ridley, Chapter 5. - PowerPoint PPT Presentation
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USC3002 Picturing the World Through Mathematics
Wayne LawtonDepartment of Mathematics
S14-04-04, 65162749 [email protected]
Theme for Semester I, 2008/09 : The Logic of Evolution, Mathematical Models of Adaptation from Darwin to Dawkins
Natural SelectionReference: Evolution by Mark Ridley, Chapter 5
p. 104 simplest model
Genotype Chance of Survival
Y
s1HG
11
is the selection coefficient]1,0[sThe chance of survival is relative to the maximal chance of survival among all genotypes. Notice that here it depends on the phenotype
Phenotype
seeds yellowseeds yellow
seedsgreen
Natural SelectionProblem: what will the genotype frequencies be after natural selection followed by random mating ?
Genotype Y H G
2nd Ad. Freq.
Define HGHY PPgPPy 21
21 ,
1st Ad. Freq. YP HP GP
Baby Freq.
)1/( 22
'
sgy
PY
)1/(2 2
'
sgyg
PH
)1/()1( 22
'
sgsg
PG
2y yg22g
Define )1/( 2'21'' sgyPPy HY
''21'' 1 yPPg HG
Natural SelectionRemark: since
the change in gene frequency to the next generation
the genotype frequencies of the 2nd Adult population are NOT in Hardy-Weinberg equilibrium
Let
2222'22' )1/()1/( sgyysgyPY
Haldane (1924) produced this model for selection p. 107
)1/( 22' sgsygyyy denote
Since )/( 2'gyys the selection coefficient can
be computed from the 2nd generation gene frequencies
MATLAB Program for Table 5.4, p. 107function 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
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
generation
g - gene
frequencies
s=0.05 s=0.01
Differential Equation Approximation
for
consists of solving the initial value problem:
(frequency of gene g in zero-generation)
)1/()1( 22' sgggsggg
followed by the approximation
The error is small if
0),1)(~/()(~))(~1()(~
22 ttgstgtgstdt
gd
)0()0(~ gg
,...3,2,1),(~)( nngng
)(~t
dt
gd is small
Qualitative Observations
1. If
therefore
2. For small s,
then
2)(~))(~1()(~
tgtgstdt
gd
1)0(~)(~ gtg )1/())(~1()(~
stgstdt
gd
If s > 0,
))1/(exp())0(~1(1)(~ sstgtg
0)1)(~/()(~))(~1()(~
22 tgstgtgstdt
gd
therefore )(~ tg
decays fastest at27
4)(
~ st
dt
gd
32)(~ tg where
3. If 0)0(~)(~ gtg then2)(~)(
~tgst
dt
gd
therefore ))0(~1/()0(~)(~ tgsgtg
Numerical Solution AlgorithmChoose
Set
Set
While
)(~
)(~)(~ tdt
gdttgttg
)0()0(~ gg 0,0 tT
0t
Tt
)1)(~/()(~))(~1( 22 tgstgtgsa
ttt
MATLAB Code for Differential Equationfunction [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
Exact Solution for tFirst rewrite the differential equation in the form
http://www4.ncsu.edu/unity/lockers/users/f/felder/public/kenny/papers/partial.html
Then use the method of partial fractions
dtgdggsgs ~]~)~1(/)1~[( 22
ttg
g
dtg
gds
g
gds
g
gds
0
)(~
)0(~2
111
~
~
~
~
1~
~)1(
tg
s
tg
s
tg
gs
g
tgs
)0(~)(~)(~)0(~
ln1)0(~1)(~
ln)1(11
11
Exact Solution for s
tg
s
tg
s
tg
gs
g
tgs
)0(~)(~)(~)0(~
ln1)0(~1)(~
ln)1(11
11
)0(~1
)(~1
)(~)0(~
ln1)0(~1)(~
ln1)0(~1)(~
ln1
gtgtg
g
g
tg
g
tgts
implies that s can be solved for by
MATLAB Code for Selection Coefficientfunction 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(0.33,50,1-1/100000);
>> gmine = tablepage107(0.2757,50,1-1/100000);
>> 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')
Assigned ReadingChapter 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.