# How do I calculate the change in allele frequency in a haploid population under selection?

From this book

For simplicity, let us consider a haploid organism and assume that the frequencies of alleles $$A_1$$ and $$A_2$$ are given by $$x$$ and $$y=1-x$$, respectively. We also assume that the fitnesses of $$A_1$$ and $$A_2$$ are $$w_1 = 1$$ and $$w_2 = 1-s$$, respectively. In this case the mean fitness $$\bar w$$ is given by $$x + (1-x)(1-s)= 1-sy$$, and the allele frequency change per generation becomes

$$\Delta x = \frac{dx}{dt} = \frac{sxy}{1-sy}$$

If I would have to find what $$\frac{dx}{dt}$$ equals I would use the Wright-Fisher equation and find that:

$$\frac{dx}{dt} = \frac{w_1 \cdot x}{\bar w} = \frac{x}{1-sy}$$

, which is obviously not the same result as what the author found...

What am I missing? How did the author find out this result $$\frac{dx}{dt} = \frac{sxy}{1-sy}$$?

• Recheck all the equal signs in the second equation (looks strange). Mar 21 '14 at 18:30
• Well that's just reformulate the same equation differently. I got rid of everything in the parenthesis hoping things seem more clear. I edited some other stuff hoping things are clearer. Thanks for your help @fileunderwater Mar 21 '14 at 19:15
• Hey remi, given your info I've managed to derive the answer. Check my answer below and ask me if you need any extra explanation, this is important since I gave you an answer but I am still unsure how you went wrong. Mar 21 '14 at 20:01
• Here's a simple consistency check to show that $\Delta x = \frac{x}{1-sy}$ can't possibly be the right answer: in a pure monomorphic $A_1$ population, there obviously cannot be any change in allele frequencies through selection, so $x = 1$ should imply $\Delta x = 0$. Your formula, however, yields $\Delta x = \frac11 = 1$ in that case. Mar 21 '14 at 23:46
• Ps. As a mathematician, I'd like to register my objection to using the notation $\frac{dx}{dt}$ for "allele frequency change per generation"; it properly denotes the rate of allele frequency change over time (presumably in a continuously breeding population, for such a rate to be well defined). Even if we measure time in (average) generations, it's not hard to come up with examples where the two are not equal. Mar 21 '14 at 23:51 