How to Calculate the Carrier Frequency of Colour Blindness in Women Using Hardy-Weinberg Equilibrium

Colour blindness is a classic example of an X-linked recessive trait. It is much more common in men because they have only one X chromosome, while women have two. Understanding carrier frequency in women is crucial for genetic counseling and public health.

Problem Overview

• Incidence in men: 1 in 12 (8%) are colour-blind.

• Inheritance: X-linked recessive.

• Population: Hardy-Weinberg equilibrium.

• Question: What proportion of women are expected to be carriers?

Step 1: Assign Hardy-Weinberg Variables

For X-linked recessive traits:

• q = frequency of the mutant allele (since males are hemizygous, the frequency of affected males = q).

• p = frequency of the normal allele = 1 – q.

Given:

• q = 0.08 (8% of men are affected).

Step 2: Calculate Carrier Frequency in Women

Women have two X chromosomes, so their possible genotypes are:

• XX (homozygous normal): p²

• Xx (heterozygous carrier): 2pq

• xx (homozygous affected): q²

The carrier frequency in women is 2pq.

Calculate p:

$$p=1-q=1-0.08=0.92$$

Calculate 2pq:

$$2pq=2\times 0.92\times 0.08=2\times 0.0736=0.1472$$

So, the proportion of women who are carriers is approximately 0.15 (or 15%).

Step 3: Match with Provided Options

• (1) 0.92

• (2) 0.85

• (3) 0.78

• (4) 0.15

The correct answer is (4) 0.15.

Why This Calculation Matters

• Genetic Counseling: Identifies women at risk of having affected sons.

• Public Health: Informs screening and awareness programs.

• Education: Demonstrates the application of Hardy-Weinberg equilibrium to real-world genetics.

Conclusion

In a population where 8% of men are colour-blind due to an X-linked recessive gene, 15% of women are expected to be carriers under Hardy-Weinberg equilibrium.