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
(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
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