Q.If a natural population with 50 individuals is in Hardy-Weinberg equilibrium for a gene with two alleles A and a, with the gene frequency of allele A of 0·6, the genotype frequency of Aa will be : (A) 0·16 (B) 0·36 (C) 0·24 (D) 0·48

The Hardy-Weinberg principle states that in a large, randomly mating population with no evolutionary forces acting, allele and genotype frequencies remain constant across generations. For a gene with two alleles A and a, if the frequency of A is p and the frequency of a is q, then p + q = 1. The genotype frequencies are given by p² for AA, 2pq for Aa, and q² for aa.

Here, the frequency of allele A (p) is given as 0.6. Therefore, the frequency of allele a (q) must be 1 – 0.6 = 0.4. The genotype frequency of heterozygotes Aa is 2pq. So, 2 × 0.6 × 0.4 = 0.48. …

To understand this question, you first need to recall the foundation of population genetics — the Hardy-Weinberg principle. This principle states that in a large, randomly mating population with no evolutionary forces acting (no mutation, migration, selection, or genetic drift), allele and genotype frequencies remain constant from generation to generation. The NCERT textbook presents this as a mathematical model where, for a gene with two alleles A and a, if the frequency of A is p and the frequency of a is q, then p + q = 1.

The genotype frequencies under equilibrium are given by the binomial expansion (p + q)² = p² + 2pq + q². Here, p² represents the frequency of AA homozygotes, 2pq represents the frequency of Aa heterozygotes, and q² represents the frequency of aa homozygotes.

In this problem, you are told that the population has 50 individuals and is in Hardy-Weinberg equilibrium. The frequency of allele A is given as 0.6. That means p = 0.6. Since p + q = 1, the frequency of allele a (q) must be 1 - 0.6 = 0.4.

Now, the question asks for the genotype frequency of Aa. According to the Hardy-Weinberg formula, this is 2pq. So you simply calculate:

2 × 0.6 × 0.4 = 0.48 …