In a population of 1000 individuals 360 belongs to genotype ‘AA”, 480 to ‘Aa’ and the remaining 160 to ‘aa’.Based on this data, the frequency of the allele ‘A’ in the population is
0.4
0.5
0.6
0.7
As discussed in Hardy Weinberg Equilibrium -
According to Hardy - Weinberg equilibrium,
(p + q) 2 = 1 or p2 + q2 + 2pq = 1
As per this question,
p2 (AA) = 360 and q2 (aa) = 160
Frequency of the allele ‘A’ would be 360/1000 = 0.36
Therefore p would be √0.36 = 0.6
Hence, the correct option is (c).