# 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 $\frac{dx}{dt} = \frac{sxy}{1-sy}$?

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Recheck all the equal signs in the second equation (looks strange). – fileunderwater 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 – Remi.b 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. – hello_there_andy 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. – Ilmari Karonen 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. – Ilmari Karonen Mar 21 '14 at 23:51