Question

If the frequency of males affected with an X-linked recessive allele in a human population is 0.10 (1 in 10), what will be the expected frequency of affected females ?

• A. 0.05
• B. 0.01
• C. 0.02
• D. 0.001

Hint:

This large difference in frequency explains why X-linked recessive disorders, such as hemophilia and red-green color blindness, are far more common in males than in females.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For X-linked traits, males are hemizygous ($X^a Y$) because they carry only one copy of the X chromosome.
Females are diploid for the X chromosome ($X^a X^a$, $X^A X^a$, or $X^A X^A$) and must inherit two copies of the recessive allele to express an X-linked recessive trait.
Key Formula or Approach:
Under Hardy-Weinberg equilibrium:
- The frequency of affected males is equal to the recessive allele frequency ($q$).
- The expected frequency of affected homozygous recessive females is equal to $q^2$.

Step 2: Detailed Explanation:
Let us analyze the given values:
- Frequency of affected males = $0.10$.
Since males have only a single X chromosome, any male carrying the recessive allele will express the phenotype.
Therefore, the frequency of the recessive allele in the gene pool ($q$) is:
\[ q = 0.10 \]
For a female to express this recessive disorder, she must inherit the recessive allele from both parents, making her homozygous recessive ($X^a X^a$).
Under random mating conditions, the expected frequency of homozygous recessive females is:
\[ \text{Frequency of affected females} = q^2 \]
Substituting the value of $q$:
\[ q^2 = (0.10)^2 = 0.01 \]
This means that while 1 in 10 males in this population will be affected, only 1 in 100 females (or $0.01$) is expected to express the disorder.

Step 3: Final Answer:
The expected frequency of affected females is 0.01, which corresponds to option (B).