Genotype and allele frequencies

It should be trivial for you by this time to calculate the genotype frequencies in eggs and adults. We'll be using the convention that genotype frequencies in eggs (or newly-formed zygotes) are the genotype frequencies before selection and that genotype frequencies in adults are the genotype frequencies after selection.

$$\begin{eqnarray*} \mbox{freq($ST/ST$) before selection} &=& \frac{41}{41 + 82 ... ...w_{12}x_{12} + w_{22}x_{22} \quad\mbox{ is the mean fitness} \\ \end{eqnarray*}$$

It should also be trivial to calculate the allele frequencies before and after selection:

$$\begin{eqnarray*} \hbox{freq($ST$) before selection} &=& \frac{2(41) + 82}{2(41... ...x_{12} + w_{22}x_{22} \\ &=& p^2w_{11} + 2pqw_{12} + q^2w_{22} \end{eqnarray*}$$

If you're still awake, you're probably wondering⁷ why I was able to substitute $p^2$, $2pq$, and $q^2$ for $x_{11}$, $x_{12}$, and $x_{22}$. Remember what I said earlier about what we're doing here. The only Hardy-Weinberg assumption we're violating is the one saying that all genotypes are equally likely to survive. Remember also that a single generation in which all of the conditions for Hardy-Weinberg is enough to establish the Hardy-Weinberg proportions. Putting those two observations together, it's not too hard to see that genotypes will be in Hardy-Weinberg proportions in newly formed zygotes. Viability selection will change that later in the life-cycle, but we restart every generation with genotypes in the familiary Hardy-Weinberg proportions, $p^2$, $2pq$, and $q^2$.