Q.43 A population is in Hardy–Weinberg equilibrium for a gene with only two alleles (“A”
and “a”). If the gene frequency of the allele “A” is 0.7, genotype frequency of
heterozygous “Aa” is _____ .
The genotype frequency of heterozygous “Aa” is 0.42. This follows directly from the Hardy-Weinberg equilibrium principle for a two-allele system.
Problem Solution
In Hardy-Weinberg equilibrium, allele frequencies remain constant across generations under specific conditions (no mutation, migration, selection, drift, or non-random mating). For alleles “A” (frequency p=0.7) and “a” (frequency q=1−p=0.3), genotype frequencies are p2 (AA), 2pq (Aa), and q2 (aa).
Calculate heterozygous frequency: 2pq=2×0.7×0.3=0.42. Verify totals sum to 1: p2=0.49, 2pq=0.42, q2=0.09 (0.49 + 0.42 + 0.09 = 1).
Option Analysis
Options likely include common errors in CSIR NET-style questions.
| Option | Value | Explanation |
|---|---|---|
| A | 0.09 | Equals q2 (aa homozygote frequency), not heterozygote. |
| B | 0.3 | Equals q (allele “a” frequency), confuses allele with genotype. |
| C | 0.42 | Correct: 2pq=2×0.7×0.3. |
| D | 0.49 | Equals p2 (AA homozygote frequency). |
In population genetics, the Hardy-Weinberg equilibrium allele frequency 0.7 heterozygous calculation is essential for CSIR NET Life Sciences aspirants. When allele “A” frequency (p=0.7) is given, find q=0.3, then heterozygous “Aa” frequency as 2pq=0.42. This predicts genotype distributions in equilibrium populations.
Core Principles
The Hardy-Weinberg equation p2+2pq+q2=1 assumes random mating and no evolutionary forces. For p=0.7:
-
AA: 0.49
-
Aa: 0.42
-
aa: 0.09
CSIR NET Application
Such problems test quick allele-to-genotype conversion. Practice verifies equilibrium by checking if observed frequencies match predictions.
