Markov matrix for selection pressure on AA and aa. p = probability aa will die before mati

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Markov matrix for selection pressure on AA and aa. p = probability aa will die before mating due to being an aa q = probability AA will die before mating due to being an AA (note that these probabilities are really just p = AverageLifespan(aa)/AverageLifespan(Aa) q = AverageLifespan(AA)/AverageLifespan(Aa) ) x = Percentage of population that is AA y = Percentage of population that is Aa z = Percentage of population that is aa z=1-(x+y) (p(3-q)/(9-pq)-p+1)(y^2) + (p-px-2+2x-1)y + (1-2x+x^2) = 0 x^2 + qxy +(q(3-p)/(9-pq))(y^2) - x = 0 For aa being non-selective, the steady state falls to p(AA)=1/4, p(Aa)=1/2, p(aa)=1/4 which means alot of math was wasted to prove the obvious. This is for the special case of f(A)=0.5 and f(a)=0.5. The frequencies of genotypes for _any_ starting value will be f(AA)=p^2, f(Aa)=2pq and f(aa)=q^2 where p=f(A) and q=f(a). The steady state is P(AA)=1, P(Aa)=0, P(aa)=0 The funny thing is, it doesn't matter what p is, the aa group will always dissapear. Two other interesting things that can be shown in a directional selection model are: the mean fitness of the population is maximized and Fishers fundamental theorum of natural selection. Fishers fundamental theorum is that the increase in mean fitness each generation is proportional the genetic variance in fitness. This doesn't hold under all models of selection. In fact, fecundity selection can cause _decreases_ in mean fitness. But, under directional selection, allele frequencies change the fastest around p=q=0.5 (genetic variance is highest) and the rate of change slows as q->0. Other models of selection do not neccessarily maximize fitness (for ex. underdominant selection -- Aa is the least fit of the three genotypes). But, in this case it is trivial to see that mean fitness of the population is maximized (AA was the most fit genotype -- the population ends up all AA.) I'm not sure if Markov chain analysis is the right way to go about building selection models. Selection is a deterministic process; Markov deals with probabilistic events. At the very least it could be said one is using a sledgehammer to kill a fly. Here's a quick algebraic way to do the same thing as above. freq of "A" gametes in intial generation = p (p+q=1) Genotypes AA Aa aa genotype frequencies p^2 2pq q^2 fitness w11 w12 w22 adult freq of genotypes p^2w11/w(bar) 2pqw12/w(bar) q^2w22/w(bar) where w(bar) is mean fitness and = p^2w11+2pqw12+q^2w22 freq of "A" gametes in the next generation = p' p'= [p(pw11+(1-p)w12)]/w(bar) recall q=1-p change in frequency of "A" allele (delta)p = p'- p = pq[p(w11-w12)+q(w12-w22)]/w(bar) You can then solve for the equilibrium value of p, p(hat) by setting (delta)p equal to zero Using the above you can do directional selection models, overdominance models (where the heterozygote is most fit) and underdominance models (heterozygote least fit). The overdominant model leads to a globally stable equilibrium with both alleles left in the population. Their frequencies depend on the fitness of the two homozygotes (which don't have to be the same). The underdominant model has an unstable equilibrium and two locally stable equilibrium points. The locally stable equilibria are, of course, where one allele is eliminated. The initial frequencies of the two alleles determine which equilibria is reached. Also see Hartl and Clark, 1989, "Principles of Population Genetics", Sinauer, Sunderland, Mass

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