87. A population of spotted deer found in a national forest is in Hardy-Weinberg equilibrium. For
a particular genetic locus in this deer species, only two alleles 𝐴 and 𝑎 are possible. If the
frequency of the 𝐴 allele in this population is 0.6 , and the frequency of the 𝑎 allele is 0.4 , what
will be the frequency of the genotype 𝐴𝑎 ?
(A) 0.24
(B) 0.48
(C) 0.96
(D) 1.6

The population of spotted deer maintains Hardy-Weinberg equilibrium at a locus with alleles A (frequency p = 0.6) and a (frequency q = 0.4). The heterozygous Aa genotype frequency follows the formula 2pq, yielding 2 × 0.6 × 0.4 = 0.48.

✅ Correct Answer

Option (B) 0.48 represents the accurate frequency of the Aa genotype. Under Hardy-Weinberg principles, genotype frequencies are p² (AA), 2pq (Aa), and q² (aa), where p + q = 1. Substituting values confirms 2 × 0.6 × 0.4 = 0.48, or 48% of the population carries this heterozygote form.

📋 All Options Explained

Each choice tests understanding of Hardy-Weinberg calculations:

Option	Value	Explanation
(A)	0.24	p × q This equals homozygous recessive frequency (q² = 0.4² = 0.16) or a common error omitting the 2 factor from heterozygote formula.
(B)	0.48	2pq Correct heterozygous frequency: 2 × 0.6 × 0.4 = 0.48, matching equilibrium expectations.
(C)	0.96	1 – q² Represents combined dominant phenotypes (p² + 2pq), not just Aa genotype.
(D)	1.6	4pq Exceeds 1.0 (impossible for frequencies); likely misapplication of tetraploid expansion.

🧮 Calculation Steps

Verify equilibrium with full genotype frequencies:

Genotype	Formula	Calculation	Frequency
AA	p²	0.6² = 0.36	0.36
Aa	2pq	2 × 0.6 × 0.4 = 0.48	0.48
aa	q²	0.4² = 0.16	0.16
Total	0.36 + 0.48 + 0.16 = 1.0		1.0

This holds under assumptions like random mating and no selection, ideal for population genetics problems in exams.