Answer:
The correct answer would be b) 25 percent.
In females, X-linked recessive disorder is expressed only when the mutated gene is present in both the X chromosomes whereas in males presence of only one copy of mutated gene is sufficient to cause the disease as they have only one X chromosome.
Now the given cross will results in the production of:
- Two daughters- one is homozygous dominant (\(X^{B}X^{B}\)) and other is heterozygous dominant \(X^{B}X^{b}\)
- Two sons- one carries dominant gene (\(X^{B}Y\)) and other carries recessive or mutated gene (\(X^{b}Y\)).
From the test, it is clear that only one son out of 4 offspring carries the disease.
Hence, the probability will come out to be 1/4 = 0.25 or 25 percent.
