GATE 2023 BT Q.30: solution and answer

Official answer: 25Source: Official answer key, GATE 2023 BT (IIT Kanpur)

1. Fabry disease is X-linked recessive. Let $A$ = normal (dominant) allele and $a$ = Fabry-disease (recessive) allele on the X chromosome.

2. A phenotypically normal father has only one X chromosome and is unaffected, so his genotype must be $X^{A}Y$.

3. A carrier mother is heterozygous — unaffected herself but carrying one copy of the disease allele — so her genotype is $X^A{X}^a$.

4. Cross $X^{A}Y\times X^A{X}^a$: the father passes $X^A$ or $Y$ (each with probability $\frac{1}{2}$), and the mother passes $X^A$ or $X^a$ (each with probability $\frac{1}{2}$), giving four equally likely offspring: $X^A{X}^A$ (normal daughter), $X^A{X}^a$ (carrier daughter), $X^{A}Y$ (normal son), $X^{a}Y$ (affected son).

5. Exactly one of these four equally likely outcomes — $X^{a}Y$ — is a son affected with Fabry disease.

6. Probability of an affected son $=\frac{1}{4}=25\%$.