Question

Assuming Hardy-Weinberg equilibrium, the genotypic frequency of heterozygotes, if the frequency of the two alleles at the gene being studied are 0.6 and 0.4, will be

• A. 0.80
• B. 0.64
• C. 0.48
• D. 0.32

Hint:

The maximum heterozygosity in a bi-allelic system is \( 0.50 \), which occurs when the allele frequencies are equal (\( p = q = 0.5 \)).
For any other allele frequencies, the heterozygote frequency will be less than \( 0.50 \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The Hardy-Weinberg Principle states that in a large, random-mating population free from evolutionary forces, allele and genotype frequencies remain constant over generations.
Key Formula or Approach:
For a bi-allelic locus with alleles \( A \) and \( a \), let:
- \( p \) be the frequency of allele \( A \).
- \( q \) be the frequency of allele \( a \).
The sum of the frequencies is:
\[ p + q = 1 \] The genotype frequencies at equilibrium are given by:
\[ p^2 + 2pq + q^2 = 1 \] Where \( 2pq \) represents the frequency of the heterozygous genotype (\( Aa \)).

Step 2: Detailed Explanation:
From the problem description, we are given:
- Frequency of the first allele (\( p \)) = \( 0.6 \)
- Frequency of the second allele (\( q \)) = \( 0.4 \)
We check that:
\[ p + q = 0.6 + 0.4 = 1.0 \] To find the frequency of the heterozygotes:
\[ \text{Frequency} = 2pq \] Substitute the given values into the equation:
\[ \text{Frequency} = 2 \times 0.6 \times 0.4 \] Multiply the terms:
\[ 2 \times 0.24 = 0.48 \] Therefore, the genotypic frequency of heterozygotes in this population is \( 0.48 \) (or \( 48\% \)).

Step 3: Final Answer:
The genotypic frequency of heterozygotes is 0.48.