Answer
The frequency of allele A is 0.351
The frequency of allele a is 0.649
Work Step by Step
For this question, we will need to know two equations for the Hardy Weinberg Equilibrium:
$p^{2} + 2pq + q^{2}=1$
$p + q=1$
$p$= the frequency of the A allele, $p^{2}$= the frequency of homozygous AA individuals,
$2pq$= the frequency of heterozygous Aa individuals,
$q$= the frequency of the a allele, $q^{2}$= the frequency of homozygous aa individuals.
Using this information, we will first find the value of $q^{2}$ by finding the proportion of the population that is aa:
$q^{2}=\frac{295}{700}=0.4214$
We then use this value to find the frequency of the a allele, $q$
$q=\sqrt 0.4214=0.649$
Finally, we use the equation $p + q=1$ to find the frequency of the A allele, $p$
$p + 0.649=1$
$p =1-0.649=0.351$
