Question

If the frequency of males affected with an X‐linked recessive condition in a cattle population is 0.20, what will be the expected frequency of affected females?

• A. 0.20
• B. 0.0004
• C. 0.40
• D. 0.04

Hint:

Recessive X-linked traits are much more common in males than in females because males only need to inherit a single copy of the mutant allele to express the phenotype.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In mammals, sex determination is based on the XX/XY system.
Males are hemizygous for genes on the X-chromosome (having only one copy, XY), whereas females are homozygous or heterozygous (having two copies, XX).
This difference in gene dosage alters the distribution of X-linked phenotypes between sexes.
Key Formula or Approach:
Let \(q\) be the frequency of the recessive mutant allele on the X-chromosome.
- For hemizygous males, the frequency of showing the recessive phenotype is equal to the allele frequency: \[ f(\text{affected males}) = q \]
- For females to show a recessive phenotype, they must inherit two copies of the mutant allele.
Under Hardy-Weinberg equilibrium, the expected frequency of affected females is: \[ f(\text{affected females}) = q^2 \]

Step 2: Detailed Explanation:
We are given the frequency of affected males in the population: \[ f(\text{affected males}) = q = 0.20 \]
To find the expected frequency of affected females under Hardy-Weinberg equilibrium, we square the allele frequency \(q\): \[ f(\text{affected females}) = q^2 \] \[ f(\text{affected females}) = (0.20)^2 = 0.04 \]
Thus, the expected frequency of affected females is \(0.04\) (or \(4\%\)).

Step 3: Final Answer:
The expected frequency of affected females is 0.0