If 490 individuals out of a population of 1000 are homozygou

If 490 individuals out of a population of 1000 are homozygous dominant (2 alleles), then how many individuals are heterozygous? (Assuming Hardy-Weinberg).

Solution

The number of homozygous dominant individuals, AA or p2 = 490/1000 = 0.49

Thus, the frequency of A is = 0.7

Assuming that the population is in Hardy-Weinberg equilibrium,

The frequency of recessive allele, a is = 1 – 0.7 = 0.3

The frequency of heterozygotes is = 2pq = 2*0.7*0.3 = 0.42 or 420/1000 (420 heterozygous individuals out of 1000)