A man's who suffers from haemophilia marries a woman who is a carrier, what percentage of their children are likely to be haemophilie?

Analyze the question

Haemophilia is an X-linked recessive disorder. We need the percentage of children (sons and daughters together) who will be haemophiliac when the father is haemophiliac and the mother is a carrier.

Concept

• Haemophilia gene is carried on the X chromosome (Xʰ = haemophilia allele, X = normal allele).
• Males (XY) with Xʰ are affected: XʰY.
• Females need two copies (XʰXʰ) to be affected; XXʰ females are carriers.

Solution

• Father (haemophiliac): XʰY
• Mother (carrier): XXʰ

Punnett square:

| | Xʰ (father) | Y (father) | |---|---|---| | X (mother) | XXʰ (carrier female) | XY (normal male) | | Xʰ (mother) | XʰXʰ (haemophiliac female) | XʰY (haemophiliac male) |

Results:

• XXʰ: carrier daughter
• XʰXʰ: haemophiliac daughter
• XY: normal son
• XʰY: haemophiliac son

Out of 4 possible offspring, 2 are haemophiliac (XʰXʰ and XʰY).

2/4 = 50%

Answer: B (50%)

Tip: For X-linked recessive crosses, always write out the genotypes (XʰY × XXʰ) and draw the Punnett square. Remember that a carrier mother × affected father gives a 50% chance of an affected child, whether male or female. Don't forget that daughters can be affected too, because the father passes on his Xʰ to every daughter and the mother may pass on hers.