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(C) According to the Hardy-Weinberg principle,$p^2 + 2pq + q^2 = 1$,where $p^2$ is the frequency of $AA$ (homozygous dominant),$q^2$ is the frequency

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$0.6$

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(C) According to the Hardy-Weinberg principle,$p^2 + 2pq + q^2 = 1$,where $p^2$ is the frequency of $AA$ (homozygous dominant),$q^2$ is the frequency of $aa$ (homozygous recessive),and $2pq$ is the frequency of $Aa$ (heterozygous).Total population = $1000$.Number of $Aa$ individuals = $480$.Number of $aa$ individuals = $160$.Number of $AA$ individuals = $1000 - (480 + 160) = 1000 - 640 = 360$.The frequency of $AA$ genotype $(p^2)$ = $360 / 1000 = 0.36$.Since $p^2 = 0.36$,the frequency of allele $A$ $(p)$ = $\sqrt{0.36} = 0.6$.

$A$ population consists of $1000$ individuals. $480$ individuals are $Aa$ and $160$ individuals are $aa$. Based on the given information,what is the frequency of allele $A$?

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