Calculating Heterozygote Frequency in Hardy-Weinberg Populations: Step-by-Step Guide

The Hardy-Weinberg equilibrium is a cornerstone of population genetics, providing a mathematical model to predict genotype frequencies from allele frequencies in a population that is not evolving. One common application is determining the frequency of heterozygotes when the frequencies of dominant and recessive alleles are known. Let’s walk through how to solve such a problem, using a real example.

The Problem

In a population obeying Hardy-Weinberg equilibrium:

• Frequency of the recessive allele (q) = 0.88

• Frequency of the dominant allele (p) = 0.12

What is the frequency of heterozygotes in this population?

Step 1: Recall the Hardy-Weinberg Equation

For a gene with two alleles (A and a), 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: Plug in the Values

Given:

• $p=0.12$

• $q=0.88$

Calculate the frequency of heterozygotes ($2pq$):

$$2pq=2\times 0.12\times 0.88=2\times 0.1056=0.2112$$

Step 3: Convert to Percentage

$$0.2112\times 100=21.12\%$$

Step 4: Match to the Closest Option

From the options provided:

• (1) 11.1%

• (2) 21.1%

• (3) 79.9%

• (4) 14.4%

The closest and correct answer is 21.1%.

Why This Calculation Matters

Calculating heterozygote frequency is essential for:

• Predicting carrier rates for genetic diseases

• Understanding genetic diversity in populations

• Detecting deviations from Hardy-Weinberg equilibrium, which may indicate evolutionary forces at work

Conclusion

If the frequency of the recessive allele is 0.88 and the dominant allele is 0.12, the frequency of heterozygotes in a Hardy-Weinberg population will be 21.1%.

Correct answer: (2) 21.1%