Re: Hamilton's rule for the diploid case

In article <dkebcm$2fcm$1@xxxxxxxxxxxxxxxxxxx>,
Perplexed in Peoria <jimmenegay@xxxxxxxxxxxxx> wrote:
>"Catherine Woodgold" <an588@xxxxxxxxxxxxxxxxxxx> wrote in message
>news:dkdp5h$26vc$1@xxxxxxxxxxxxxxxxxxxxxx
>I think that the problem you are having with the heterozygous case
>is due to asking the wrong question. In the OP, Catherine wrote:
>> Under what circumstances does an altruistic act by an organism
>> with Aa promote the spread of A?
>
>The 'right' question would have been"
> "Under what circumstances does an altruistic act by an organism
> with at least one copy of A promote the spread of A?"
>That is, you had earlier derived Hamilton's rule in the case where
>A is recessive. Now you have to analyze the case where it is dominant.
>
>Joe was right in his intuition that everything will come out right
>when you include the AA individuals with the Aa individuals. The
>question you seem to be trying to answer (even if you don't realize
>it) is this: What if only the heterozygote Aa is altruistic, and the
>homozygotes AA and aa are not. And the answer, which you are finding
>is that if rb>c, then the frequency p will rise to 0.5 if it starts
>below that level, and it will fall to 0.5 if it starts above that level.

Well, let's work it out and see. For AA Cathy got the change in
the number of A copies as (q-p)b - (1-p)c = (1-p)(Fb - c)
and for Aa it was (1/2-p)(Fb-c). Now we need to know how to
weight these two (I am assuming there are no altruistic acts by aa,
though a more general treatment would be to include that possibility too.

Suppose an AA has probability P of performing an altruistic act, and
an Aa has probability hP. Then h=1 is when the A allele is dominant,
h = 1/2 when it is of intermediate dominance, etc. The overall frequency
of AA and Aa are p^2 and 2p(1-p). The total frequency of their acts is
then p^2 P and 2p(1-p) hP. The net effect in changing the number
of A alleles is then

p^2 P (1-p) (Fb - c) + 2p(1-p) hP (1/2-p) (Fb-c)

Factoring out (Fb-c) and P we get

(Fb-c) P [p^2(1-p) + 2h p(1-p)(1/2-p)]

When the stuff in brackets is positive, then Hamilton's rule is the condition
to lead selection to fix allele A. The expression is

p^2(1-p) + 2h p(1-p)(1/2-p) = p(1-p)[ p + h - 2hp]

so it reduces down to whether p + h - 2hp > 0.
This is also p(1-h)+h(1-p). In other words, a weighted average of (1-h)
and h, weighted by p and (1-p). If h > 0 and 1-h > 0 this will always
be positive for any p in (0,1). That is the case where the allele A
is not overdominant. h < 0 does not even make sense in this case, actually.

Exercise for readers: show that when (Fb-c) > 0 and the allele is
overdominant, that the gene frequency goes to the equilibrium frequency that
maximizes the number of altruistic acts.

So I think this shows it all works out. A slightly more general treatment
would have been to use as the frequencies of altruist acts by aa, Aa, and AA
the probabilities x, x+hy, and x+y. For that we would need the effects
of acts by aa, which have not yet been worked out here.

---
Joe Felsenstein joe@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Genome Sciences and Department of Biology,
University of Washington, Box 357730, Seattle, WA 98195-7730 USA

.

Relevant Pages

  • Re: Hamiltons rule for the diploid case
    ... allele in the recipient related by F is an A allele is F + p. ... the a allele in that heterozygote, the probability that the possibly related ... altruists, which would be the Hardy-Weinberg proportions, and the probabilities ... that each does an altruistic act, which would depend on the pattern of ...
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  • Re: Hamiltons rule for the diploid case
    ... > Joe Felsenstein writes: ... >> that each does an altruistic act, which would depend on the pattern of ... When p> 0.5, in the heterozygote case, ... "Under what circumstances does an altruistic act by an organism ...
    (sci.bio.evolution)
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