How to Calculate the Percentage of Homozygous Dominant Individuals Using Hardy-Weinberg Equilibrium

The Hardy-Weinberg equilibrium is a foundational principle in population genetics, allowing us to predict genotype frequencies from allele frequencies. If you know the frequency of the recessive homozygote genotype in a population, you can easily calculate the percentage of individuals who are homozygous for the dominant allele.

Problem Overview

• The frequency of the recessive homozygote genotype (q²) is 0.04.

• The population is in Hardy-Weinberg equilibrium.

• The question: What percentage of individuals are homozygous for the dominant allele (p²)?

Step 1: Recall the Hardy-Weinberg Equation

The Hardy-Weinberg equation is:

p2+2pq+q2=1p2+2pq+q2=1

• p2p2 = frequency of homozygous dominant (AA)

• $2pq$ = frequency of heterozygotes (Aa)

• q2q2 = frequency of homozygous recessive (aa)

Step 2: Find q

Given:

q2=0.04q2=0.04q=0.04=0.2q=0.04=0.2

Step 3: Find p

$$p=1-q=1-0.2=0.8$$

Step 4: Calculate p²

p2=(0.8)2=0.64p2=(0.8)2=0.64

Step 5: Convert to Percentage

$$0.64\times 100=64\%$$

Step 6: Match with the Provided Options

• (1) 64

• (2) 40

• (3) 32

• (4) 16

The correct answer is (1) 64.

Conclusion

In a population at Hardy-Weinberg equilibrium where the frequency of the recessive homozygote genotype is 0.04, 64% of individuals are homozygous for the dominant allele