104. A distinctly large population of randomly mating laboratory mice contains 𝟑𝟔% albino mice, which
is caused by a double recessive genotype (aa). The black coloured mice in the population is due to
dominant genotype (AA/Aa). Considering the fact that this population is in Hardy-Weinberg equilibrium,
the frequency of heterozygous alleles (Aa) in this population is ____ (round off to 2 decimal places).
Albino mice make up 36% of a large, randomly mating population in Hardy-Weinberg equilibrium, with the trait caused by the recessive genotype aa. The frequency of heterozygous Aa individuals is 0.48. This genetics problem highlights core principles for exams like NEET or GATE Life Sciences.
Hardy-Weinberg Basics
The Hardy-Weinberg principle states that allele and genotype frequencies remain constant in a large, randomly mating population without evolutionary forces like selection or mutation. Genotype frequencies follow p² + 2pq + q² = 1, where p is the dominant allele (A) frequency, q is the recessive allele (a) frequency, p² and q² are homozygous frequencies, and 2pq is the heterozygous frequency. Here, black mice (AA or Aa) show dominant phenotype, while aa causes albinism.
Step-by-Step Solution
Given 36% aa mice, q² = 0.36. Thus, q = √0.36 = 0.60. Since p + q = 1, p = 1 – 0.60 = 0.40. Heterozygous frequency is 2pq = 2 × 0.40 × 0.60 = 0.48 (rounded to two decimal places). Black mice frequency is 1 – 0.36 = 0.64, matching p² + 2pq = 0.16 + 0.48 = 0.64, confirming equilibrium.
Frequency Breakdown
| Genotype | Frequency Calculation | Value |
|---|---|---|
| aa (albino) | q² | 0.36 |
| Aa (heterozygous black) | 2pq | 0.48 |
| AA (homozygous black) | p² | 0.16 |
| Total | p² + 2pq + q² | 1.00 |
Exam Tips
Round off to two decimals as specified, and verify by checking if frequencies sum to 1. Common errors include forgetting the square root for q or misapplying 2pq.
