Take the 2-minute tour ×

Consider $m$ loci with heterozygote advantage (overdominance) such that the fitness of the two homozygotes is $1-\frac{s}{2}$ and the fitness of the heterozygotes is $1+\frac{s}{2}$, where $s>0$. We'll assume that the fitness of an individual is given by the multiplication of the fitness component on each locus. In consequence, the fitness of the best possible genotype is given by $\left(1+\frac{s}{2}\right)^m$.

According to this book, an individual is heterozygote at $j$ of these $m$ loci with probability

$${m\choose j}\left(\frac{1}{2}\right)^m$$

and the equilibrium population mean fitness $\hat w$ is

$$\hat w = \sum_{j=0}^m {m\choose j}\left(\frac{1}{2}\right)^m \left(1+\frac{s}{2}\right)^j \left(1-\frac{s}{2}\right)^{m-j} = 1$$

I don't understand any of these two equations! Can you help me to understand how they have been calculated?

For the moment I just enjoyed proving that $\hat w = 1$. We can reformulate $\hat w$ as

$$\hat w = \left(\frac{1}{2}\right)^m \sum_{j=0}^m {m\choose j} \left(1+\frac{s}{2}\right)^j \left(1-\frac{s}{2}\right)^{m-j} = 1$$

,then using the binomial identity

$$ \hat w = \left(\frac{1}{2}\right)^m \left(\left(1+\frac{s}{2}\right) + \left(1-\frac{s}{2}\right)\right)^{\space m}$$ $$\hat w = \left(\frac{1}{2}\right)^m 2^m$$ $$ \hat w = 1 $$

share|improve this question
Am taking another stab at this. I think that derivations in the refs are simpler and overall it's an interesting question. –  daniel Apr 19 '14 at 14:33

1 Answer 1

up vote 1 down vote accepted

If the fitness of a heterozygote is $(1+s/2)$ and of a homozygote is $(1-s/2)$ then why is the probability for a given state $(1+s/2)^j(1-s/2)^{m-k}$

$$\binom mj (1/2)^j(1/2)^{m-j}= \binom mj (1/2)^m~~ ?$$

As you pointed out earlier, in the general case it need not be true that $p = q = 1/2$ but that is what the form of the probability above implies. So the threshold question is why this particular heterozygous-dominant model implies equilibrium probabilities $p = q = 1/2.$ I think the ideas below begin to address this.

The simplest case is for one locus, two alleles, and there are many good derivations online. I think if you understand the situation for one locus you can generalize to higher numbers. (Hopefully I will supplement this answer, time allowing. I think a shortcut would be to assume $p = q = 1/2$ and use your fitness weights. That gives us $\bar{w}=1$ as a denominator. Now for reasons of symmetry I think you can show that the relative frequencies of $p$ and $q$ are equal and so $p' = p.$ Then you have to show that this equilibrium solution is unique.)

For a single locus the derivations of `heterozygote advantage' I have found,(1)(2), assign fitness weights as follows:

AA = (1 - s), Aa = 1, aa = (1 - t )

in which $s,t > 0, s \neq t$ in general, from which they derive as a condition for equilibrium

$$\Delta q = \frac{pq(sp-tq)}{\bar{W}} = 0$$


$$\hat{p} = \frac{t}{s+t} ~~~\text{and}~~~ \hat{q} = \frac{s }{s+t}$$

in which $\hat{p}, \hat{q}$ are equilibrum frequencies for each allele, respectively, and s and t are rates of mutation. Nowhere did I see a model in which they had assigned the same fitness to both homozygous cases (AA, aa) but it's just a special case of the heterozygote advantage model.

fitness (AA) = fitness(aa) = (1 - s/2), fitness(Aa) = (1+ s/2).

So if we subtract s/2 from each fitness score (or normalize with respect to Aa to the same effect), we get:

fitness(AA) = fitness (aa) = (1- s ) and fitness (Aa) = 1 as in the two references above, except that now the two homozygous states have equal fitness.

But then we have $$\Delta q = \frac{pq(sp-sq)}{\bar{W}} = \frac{pqs(p - q)}{\bar{W}} $$

and the only nontrivial solution is $p = q = 1/2.$

So what I am suggesting is that when you assign the same fitness to both $AA$ and $aa$ you no longer have the general case. As for the value of s being forced, the following seems relevant.

The full expression from (2) for the equilibrium condition is

$$p' = \frac{p^2 W_{AA}+ pq W_{Aa}}{\bar{W}} \hspace{10mm}(1)$$

in which$ W_{AA} = W_{aa} = 1- s$ and $W_{Aa} = 1$ and

$\bar{W} = p^2 W_{AA}+2pqW_{Aa}+q^2W_{aa}$

If the homozygotes are assigned the same fitness and $p = q = 1/2$ , equation (1) above becomes:

$$p' = \frac{\frac{1}{4} + \frac{1}{4}(1-s)}{\frac{1}{2} + \frac{1}{2}(1-s)} $$

The program on page 583 of (1) is helpful. Let h:= heterozygous and m:= homozygous. If $p = q = 1/2$ then as long as fitness(h) and fitness(m) are equal the system is in equilibrium immediately.

If $p\neq 1/2$ then as long as f(h) = f(m) the system reaches equilibrium at $p = 1/2$ asymtotically. If $p = 1/2$ but f(h) $\neq$ f(m) the asymtotic limit has to be calculated.

See also http://evol.bio.lmu.de/_teaching/evogen/Evo8-Summary.pdf

share|improve this answer
Thanks a lot for your answer. I'm traveling right now and I don't have time for reading up your answer. I'll read it wednesday or thursday probably. –  Remi.b Apr 19 '14 at 16:19
Great answer. In addition to your (now deleted by yourself) other answer, you perfectly answered my question. Thank you. +1 –  Remi.b Apr 23 '14 at 14:53
@Remi.b: Appreciate the feedback. It was an interesting question and I am still learning about this broad subject. –  daniel Apr 23 '14 at 20:20

Your Answer


