Question

A population is under Hardy Weinberg equilibrium. If the gene frequency of A1 allele in parents is 0.30, the frequency of A2A2 genotype in progeny will be:

• A. 0.30
• B. 0.09
• C. 0.42
• D. 0.49

Hint:

Always identify which allele frequency is given first.
If the frequency of one allele is \( x \), the other is \( 1 - x \).
The homozygous genotype frequency for the second allele is \( (1-x)^2 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The Hardy-Weinberg law states that in a large, random-mating population with no selection, mutation, or migration, both allele and genotype frequencies remain constant from generation to generation.
Key Formula or Approach:
For a diallelic locus with alleles \( A_1 \) and \( A_2 \) having frequencies \( p \) and \( q \):
\[ p + q = 1 \] The expected genotype frequencies in the progeny are:
\[ p^2 (A_1A_1) + 2pq (A_1A_2) + q^2 (A_2A_2) = 1 \]

Step 2: Detailed Explanation:
We are given the frequency of the \( A_1 \) allele in the parent generation:
\[ p = 0.30 \] Since there are only two alleles in this population, the frequency of the \( A_2 \) allele (\( q \)) is:
\[ q = 1 - p = 1 - 0.30 = 0.70 \] Under Hardy-Weinberg equilibrium, the expected frequency of the homozygous \( A_2A_2 \) genotype in the progeny generation is:
\[ \text{Frequency of } A_2A_2 = q^2 \] \[ q^2 = (0.70)^2 = 0.49 \] Therefore, the expected frequency of the \( A_2A_2 \) genotype in the progeny is 0.49.

Step 3: Final Answer:
The frequency of the \( A_2A_2 \) genotype in the progeny will be 0.49.