Question

A gene locus has two alleles A, a. If the frequency of dominant allele A is 0.4, then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population?

• A. 0.36 (AA); 0.48 (Aa); 0.16 (aa)
• B. 0.16 (AA); 0.24 (Aa); 0.36 (aa)
• C. 0.16 (AA); 0.48 (Aa); 0.36 (aa)
• D. 0.16 (AA); 0.36 (Aa); 0.48 (aa)

Answer 1

The Correct Option is C

The question addresses a basic genetic principle, specifically involving allele frequency and genotype frequency as per the Hardy-Weinberg principle. Let's delve into the solution step-by-step.

The Hardy-Weinberg equation states that in a large population, with random mating and no evolutionary forces, the frequencies of alleles and genotypes remain constant. The equation is:

\(p^2 + 2pq + q^2 = 1\) 

Where:

• \(p\) is the frequency of the dominant allele (A).
• \(q\) is the frequency of the recessive allele (a).
• \(p^2\) is the frequency of homozygous dominant individuals (AA).
• \(2pq\) is the frequency of heterozygous individuals (Aa).
• \(q^2\) is the frequency of homozygous recessive individuals (aa).

Given, the frequency of the dominant allele A \((p = 0.4)\).

Let's calculate the frequency of the recessive allele a:

\(q = 1 - p = 1 - 0.4 = 0.6\)

Next, using the Hardy-Weinberg equation:

1. Calculate the frequency of homozygous dominant individuals (AA):
2. Calculate the frequency of heterozygous individuals (Aa):
3. Calculate the frequency of homozygous recessive individuals (aa):

Thus, the frequencies of the genotypes in the population are as follows:

• \(AA: 0.16\)
• \(Aa: 0.48\)
• \(aa: 0.36\)

Therefore, the correct answer is: 0.16 (AA), 0.48 (Aa), 0.36 (aa).

This matches option:

0.16 (AA); 0.48 (Aa); 0.36 (aa)

, which is the correct choice.