The percentage probability that a normal male married to a carrier woman would have a haemophilic male child is

Analyze the question: Haemophilia is an X-linked recessive disorder. We need the probability that a child of a normal father and a carrier mother is a haemophilic male.

Concept:

• Let X^H = normal allele, X^h = haemophilia allele.
• Normal male: X^H Y
• Carrier female: X^H X^h

Punnett square:

| | X^H (mother) | X^h (mother) | |---|---|---| | X^H (father) | X^H X^H (normal female) | X^H X^h (carrier female) | | Y (father) | X^H Y (normal male) | X^h Y (haemophilic male) |

Reading the result:

• Of all children: 1 in 4 (25%) is a haemophilic male.
• Of the male children only: 1 of 2 sons (50%) is haemophilic.

Which interpretation fits the options? 25% is not offered. The question says "would have a haemophilic male child", which is treated as the chance that a son is haemophilic, giving 50%.

Answer: B (50%)

Tip: In X-linked crosses, check whether the question asks about all children or only a given sex. With a carrier mother and a normal father: 50% of sons are affected, 50% of daughters are carriers, and no daughters are affected. If 25% is not among the options, the question is almost certainly asking about sons only.