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 ๐ด and ๐ are possible. If the
frequency of the ๐ด allele in this population is 0.6 , and the frequency of the ๐ allele is 0.4 , what
will be the frequency of the genotype ๐ด๐ ?
(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 ร 0.6 ร 0.4 = 0.48.
โ Correct Answer
Option (B) 0.48 represents the accurate frequency of the Aa genotype. Under Hardy-Weinberg principles, genotype frequencies are pยฒ (AA), 2pq (Aa), and qยฒ (aa), where p + q = 1. Substituting values confirms 2 ร 0.6 ร 0.4 = 0.48, or 48% of the population carries this heterozygote form.
๐ All Options Explained
Each choice tests understanding of Hardy-Weinberg calculations:
| Option | Value | Explanation |
|---|---|---|
| (A) | 0.24 | p ร q This equals homozygous recessive frequency (qยฒ = 0.4ยฒ = 0.16) or a common error omitting the 2 factor from heterozygote formula. |
| (B) | 0.48 | 2pq Correct heterozygous frequency: 2 ร 0.6 ร 0.4 = 0.48, matching equilibrium expectations. |
| (C) | 0.96 | 1 โ qยฒ Represents combined dominant phenotypes (pยฒ + 2pq), not just Aa genotype. |
| (D) | 1.6 | 4pq Exceeds 1.0 (impossible for frequencies); likely misapplication of tetraploid expansion. |
๐งฎ Calculation Steps
Verify equilibrium with full genotype frequencies:
| Genotype | Formula | Calculation | Frequency |
|---|---|---|---|
| AA | pยฒ | 0.6ยฒ = 0.36 | 0.36 |
| Aa | 2pq | 2 ร 0.6 ร 0.4 = 0.48 | 0.48 |
| aa | qยฒ | 0.4ยฒ = 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.
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2 Comments
Komal Sharma
January 11, 2026Option b is correct 4.48
Sonal Nagar
January 15, 20260.48