Question

If an X-linked recessive disorder is in Hardy-Weinberg equilibrium and the incidence in males equals 1 in 100, then the expected incidence of affected homozygous females would be-

• A. 1 in 100
• B. 1 in 1000
• C. 1 in 10000
• D. 1 in 100000

Hint:

For X-linked recessive traits, the female disease incidence is the square of the male disease incidence: \(\text{Female incidence} = (\text{Male incidence})^2\).
In this case, \((1/100)^2 = 1/10000\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The Hardy-Weinberg principle describes the relationship between allele frequencies and genotype frequencies in an idealized, non-evolving population.
For genes located on sex chromosomes (X-linked), allele and genotype distributions differ between males and females because of their different chromosome numbers.
Key Formula or Approach:
For an X-linked recessive trait, males are hemizygous and carry only one X chromosome.
Therefore, the incidence of the disorder in males is directly equal to the recessive allele frequency (\(q\)): \[ q = \text{Incidence in males} \] Females are diploid for the X chromosome, meaning they must inherit two copies of the recessive allele to be affected.
The frequency of affected homozygous females is given by: \[ f(\text{affected females}) = q^2 \]

Step 2: Detailed Explanation:
We are given that the incidence of the X-linked recessive disorder in males is 1 in 100.
Thus, the recessive allele frequency is: \[ q = \frac{1}{100} = 0.01 \] To find the expected incidence of affected females, we calculate the homozygous recessive genotype frequency (\(q^2\)): \[ q^2 = (0.01)^2 \] \[ q^2 = 0.0001 \] Expressing this as a fraction: \[ q^2 = \frac{1}{10000} \] This means the expected incidence in females is 1 in 10,000.

Step 3: Final Answer:
The expected incidence of affected homozygous females is 1 in 10,000, which corresponds to option (C).