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

share|improve this question
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
up vote 6 down vote accepted

Here is my full derivation to the book example you gave, hopefully it'll help you clear up what went wrong:

enter image description here

You need to remember that after there is selection acting on the population, you no longer have a total of 1 after selection. Think of selection as "killing" individuals, which means the total is now 1 minus what has been "selected out". s*y is what is selected out, therefore the new total is 1-s*y. Which means you now divide each of your frequencies by 1-s*y (see picture).

Please comment if you need more explaining.

share|improve this answer
Ok, I totally understood my error. That was very simple. Thanks a lot for this good explanation. – Remi.b Mar 21 '14 at 20:53
The great thing is that you've grasped the "hardest part" in population genetics modelling. The procedure, of breaking down the steps via table and normalising, is the same procedure for diploids but instead of A1 you have A1B1, instead of x you have 2*pq, where p is f(A1) and q is f(B1)), after selection A1B1 is simply 2*pq - s(2*p*q), etc. Even if model multiple separate loci, and where each locus may have >2 alleles, just: (1) tablate all possible genotypes as columns, (2) tablate the frequencies at each stage of the life cycle, (3) and make sure to normalise if there is selection. :) – hello_there_andy Mar 21 '14 at 21:09

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