How to Calculate the 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 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: p = 0.7, q = 0.3

• Sample size: 60 individuals

• Question: How many are expected to be heterozygotes (pq)?

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

• $2pq$ = frequency of heterozygotes

• q2q2 = frequency of homozygous recessive

Step 2: Calculate the Frequency of Heterozygotes

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:

$$0.42\times 60=25.2$$

Since the number of individuals must be a whole number, the expected number is 25.

Step 4: Match with the Provided Options

• (1) 25

• (2) 42

• (3) 49

• (4) 9

The correct answer is (1) 25.

Conclusion

In a Hardy-Weinberg population with allele frequencies p = 0.7 and q = 0.3, out of a sample of 60 individuals, 25 are expected to be heterozygotes.

Correct answer: (1) 25