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Question

(c) : According to Hardy Weinberg principle,

$p^2 + 2pq + q^2 = 1; (p + q)^2 = 1$

$(AA)\ p^2 = 360$ out of $1000$ individual

or $p^2 = 0.36$

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(c) : According to Hardy Weinberg principle,

$p^2 + 2pq + q^2 = 1; (p + q)^2 = 1$

$(AA)\ p^2 = 360$ out of $1000$ individual

or $p^2 = 0.36$ Therefore, $p = 0.6$

In a population of $1000$ individuals $360$ belong to genotype $AA, 480$ to Aa and the remaining $160$ to $aa$. Based on this data, the frequency of allele $A$ in the population is

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