A certain trait in a species is governed by variation at the AB locus that has two alleles (A, B), giving rise to three genotypes: AA, AB, and BB. A sample of 1000 individuals was genotyped and the data are given below:
AA = 400
AB = 400
BB = 200
Assume that each genotype is represented equally in males and females and mating is random. If the parents were randomly drawn from the given sample, which one of the following options gives all correct values of the offspring genotype frequencies?

Correct option is D

Correct Answer:
(d) AA: 0.36 ; AB: 0.48 ; BB: 0.16
Explanation:
First, calculate allele frequencies in the parental population:
Frequency of allele A = (2×400 + 400) / (2×1000) = 1200 / 2000 = 0.6
Frequency of allele B = (2×200 + 400) / (2×1000) = 800 / 2000 = 0.4
Under random mating, offspring genotype frequencies follow Hardy–Weinberg proportions:
AA = (0.6)² = 0.36
AB = 2 × 0.6 × 0.4 = 0.48
BB = (0.4)² = 0.16
Information Booster:
· Random mating leads to Hardy–Weinberg equilibrium in one generation.
· Allele frequencies, not genotype frequencies, determine offspring outcomes.
· Equal representation of sexes ensures no bias in allele transmission.
· Population size affects sampling error, not expected frequencies.