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

The correct option is D 0.6
According to Hardy-Weinberg equilibrium
$p^2+2pq+q^2=1$
Where $p^2$ is the frequency of homozygous genotype $AA$
$q^2$ is the frequency of genotype $aa$
$pq$ is the frequency of genotype $Aa$

There are $1000$ indidividuals, out of which $360$ belong to genotype $A$.
$p^2+3601000=0.36$
Then $p=\mathrm{0.6.}$
So, the correct answer is = 0.6