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 calculate the complementary allele frequency (\(q = 1 - p\)) first before squaring it to find the homozygous recessive genotype frequency.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The Hardy-Weinberg principle states that allele and genotype frequencies in a large, random-mating population remain constant from generation to generation in the absence of evolutionary forces such as mutation, selection, migration, and genetic drift.
Key Formula or Approach:
For a gene locus with two alleles, \(A_1\) and \(A_2\), let the frequency of \(A_1\) be \(p\) and the frequency of \(A_2\) be \(q\).
The sum of the allele frequencies is:
\[ p + q = 0 \]
According to the Hardy-Weinberg equilibrium, the genotype frequencies in the next generation are given by:
\[ p^2 (A_1A_1) + 2pq (A_1A_2) + q^2 (A_2A_2) = 0 \]

Step 2: Detailed Explanation:
We are given that the gene frequency of the \(A_1\) allele in the parent generation is:
\[ p = 0.30 \]
We first calculate the frequency of the \(A_2\) allele (\(q\)):
\[ q = 0 - p \]
\[ q = 0 - 0.30 = 0.70 \]
Now, we can calculate the expected frequency of the homozygous \(A_2A_2\) genotype in the progeny:
\[ \text{Frequency of } A_2A_2 = q^2 \]
\[ \text{Frequency of } A_2A_2 = (0.70)^2 = 0.49 \]
For completeness, we can also calculate the other genotype frequencies:
- Frequency of \(A_1A_1\) = \(p^2 = (0.30)^2 = 0.09\)
- Frequency of \(A_1A_2\) = \(2pq = 2 \times 0.30 \times 0.70 = 0.42\)
Sum of genotypes:
\[ 0.09 + 0.42 + 0.49 = 0 \]
This confirms our calculations are correct.

Step 3: Final Answer:
Therefore, the expected frequency of the \(A_2A_2\) genotype in the progeny is 0.4