GATE 2026 BT Q.64: solution and answer

Official answer: 0.5Source: Official answer key, GATE 2026 BT (IIT Guwahati)

1. Mother is a carrier of the X-linked recessive disease allele: genotype $X^A{X}^a$ (heterozygous, unaffected).

2. Father is genetically normal for this locus: genotype $X^{A}Y$.

3. A son receives his single X chromosome only from his mother and his Y chromosome only from his father; the father's X allele is never passed to a son.

4. Because the mother is heterozygous, she transmits $X^A$ or $X^a$ to any child with equal probability $\frac{1}{2}$ each, and this segregation ratio does not depend on the sex of the child.

5. Given that the child born is a son, the son's genotype at this locus is determined entirely by which maternal X allele he inherited.

6. $P(\text{son is affected},X^{a}Y)=\frac{1}{2}=0.5$.