Question

In a population of 800 rabbits showing Hardy-Weinberg equilibrium, the frequency of recessive individuals was 0.16. What is the frequency of heterozygous individuals?

• A. 0.36
• B. 0.4
• C. 0.48
• D. 0.84

Answer 1

The Correct Option is C

The Hardy-Weinberg principle states that for a population in equilibrium, the following equations hold:

• p + q = 1 (for allele frequencies)
• p² + 2pq + q² = 1 (for genotype frequencies)

Where:

• p = frequency of dominant allele (A)
• q = frequency of recessive allele (a)

Given that the frequency of recessive individuals (homozygous aa) is 0.16, this corresponds to q² = 0.16. Therefore:

q = √0.16 = 0.4

Now, using the equation p + q = 1:

p = 1 - 0.4 = 0.6

The frequency of heterozygous individuals (2pq) is given by:

2pq = 2 * 0.6 * 0.4 = 0.48 

The correct answer is (C) : 0.48.

Answer 2

In a population of 800 rabbits showing Hardy-Weinberg equilibrium, the frequency of recessive individuals was 0.16. What is the frequency of heterozygous individuals?

The correct answer is: (C) 0.48.

In Hardy-Weinberg equilibrium, the allele frequencies are related by the equation \( p^2 + 2pq + q^2 = 1 \), where:

• p is the frequency of the dominant allele
• q is the frequency of the recessive allele
• p^2 is the frequency of homozygous dominant individuals
• q^2 is the frequency of homozygous recessive individuals
• 2pq is the frequency of heterozygous individuals

We are given the frequency of recessive individuals (\( q^2 \)) as 0.16. Therefore:

Step 1: Find \( q \), the frequency of the recessive allele:

\( q^2 = 0.16 \) so \( q = \sqrt{0.16} = 0.4 \)

Step 2: Use the equation \( p + q = 1 \) to find \( p \):

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

Step 3: Calculate the frequency of heterozygous individuals (\( 2pq \)):

\( 2pq = 2 \times 0.6 \times 0.4 = 0.48 \)

Thus, the frequency of heterozygous individuals is 0.48.