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 ([tex]X^{B}X^{B}[/tex]) and other is heterozygous dominant [tex]X^{B}X^{b}[/tex]
- Two sons- one carries dominant gene ([tex]X^{B}Y[/tex]) and other carries recessive or mutated gene ([tex]X^{b}Y[/tex]).
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.
