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6.Evolution
medium
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
A
$0.4$
B
$0.5$
C
$0.6$
D
$0.7$
(NEET-2014)
Solution
(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, $p = 0.6$
Standard 12
Biology
