USC3002 Picturing the World Through Mathematics. Wayne Lawton Department of Mathematics S14-04-04, 65162749 email@example.com. 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.
USC3002 Picturing the World Through Mathematics Wayne Lawton Department of Mathematics S14-04-04, 65162749 firstname.lastname@example.org 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 Natural Selection p. 104 simplest model Phenotype Chance of Survival Genotype is the selectioncoefficient The chance of survival is relative to the maximal chance of survival among all genotypes. Notice that here it depends on the phenotype
Problem: what will the genotype frequencies be after natural selection followed by random mating ? Natural Selection Genotype 1st Ad. Freq. Define Baby Freq. 2nd Ad. Freq. Define
Remark: since Natural Selection 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
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 MATLAB Program for Table 5.4, p. 107
generation s=0.05 s=0.01 0 100 200 300 400 500 600 700 800 900 1000 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 Tabular Output g - gene frequencies
for Differential Equation Approximation 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
If s > 0, Qualitative Observations 1. If then therefore 2. For small s, therefore where decays fastest at 3. If then therefore
Choose Numerical Solution Algorithm Set Set While
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 MATLAB Code for Differential Equation
First rewrite the differential equation in the form Exact Solution for t Then use the method of partial fractions http://www4.ncsu.edu/unity/lockers/users/f/felder/public/kenny/papers/partial.html
Exact Solution for s implies that s can be solved for by
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; MATLAB Code for Selection Coefficient
>> 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 Peppered Moth Estimation page 110 Question: Why does this differ from the book’s estimate s = 0.33 ?
>> 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') Peppered Moth Estimation page 110
Chapter 25. Evolution: The Process in Schaum’s Outlines in Biology Assigned Reading 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
Do problems 1-6 on page 136 in Ridley (the mean fitness in questions 2, 3 is defined on p105) Homework 2. Due Monday 1.09.2008
Do problems 7-10 on page 136 in Ridley (the mean fitness in questions 2, 3 is defined on p105) Homework 3. Due Monday 8.09.2008 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.