How to Calculate the Expected Number of Heterozygotes in a Hardy-Weinberg Population Sample

The Hardy-Weinberg equilibrium is a foundational concept in population genetics, allowing us to predict genotype frequencies from known allele frequencies in a population that is not evolving. One of the most common calculations is determining the expected number of heterozygotes in a sample.

Problem Overview

• Allele frequencies: A = 0.7, a = 0.3

• Sample size: 250 individuals

• Question: How many are expected to be heterozygous (Aa)?

Step 1: Recall the Hardy-Weinberg Equation

The Hardy-Weinberg equation for two alleles 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: Calculate the Frequency of Heterozygotes

• $p=0.7$

• $q=0.3$

The frequency of heterozygotes is:

$$2pq=2\times 0.7\times 0.3=2\times 0.21=0.42$$

This means 42% of the population is expected to be heterozygotes.

Step 3: Calculate the Number of Heterozygotes in the Sample

Multiply the heterozygote frequency by the sample size:

$$0.42\times 250=105$$

Step 4: Match with the Provided Options

• (1) 112

• (2) 81

• (3) 105

• (4) 145

The correct answer is (3) 105.

Conclusion

In a Hardy-Weinberg population with allele frequencies A = 0.7 and a = 0.3, out of a sample of 250 individuals, 105 are expected to be heterozygotes.