A woman who is heterozygous for the disease haemophilia marries a man who is  haemophilic. What percentage of their offspring would have the disease?

Analyze the question

Haemophilia is an X-linked recessive disorder. The question asks what percentage of the children of a carrier (heterozygous) mother and a haemophilic father will have the disease.

Concept

• The gene is carried on the X chromosome.
• Let X^H = normal allele and X^h = haemophilia allele.
• Females have two X chromosomes. A heterozygous female (X^H X^h) is a carrier and is not affected.
• Males have one X and one Y. A male with X^h Y is haemophilic.

Solution

Cross: X^H X^h (carrier mother) × X^h Y (haemophilic father)

Punnett square:

| | X^h | Y | |---|---|---| | X^H | X^H X^h | X^H Y | | X^h | X^h X^h | X^h Y |

Offspring:

1. X^H X^h: carrier daughter (normal, not diseased)
2. X^H Y: normal son
3. X^h X^h: haemophilic daughter (diseased)
4. X^h Y: haemophilic son (diseased)

Two out of four offspring have the disease: 2/4 = 50%.

(Among daughters, half are affected. Among sons, half are affected. Overall, 50% of the children have the disease.)

Answer: C. 50%

Exam tip

For X-linked questions, always write the genotypes with the X and Y chromosomes and draw the Punnett square. Do not use plain letters like Hh. Remember: a carrier mother × an affected father gives 50% affected children, with equal chances in sons and daughters. A carrier mother × a normal father gives 50% of sons affected and no daughters affected.