Question

A woman with two genes, one for haemophilia and one for colour blindness on one of the X-chromosomes, marries a normal man. The progeny will be.

• A. All sons haemophilic and colour blind
• B. $50\%$ haemophilic and colour blind sons and $50\%$ normal sons
• C. All daughters haemophilic and colour blind
• D. $50\%$ haemophilic daughters and $50\%$ colour blind daughters

Answer 1

The correct option is B $50\%$ haemophilic and colour blind sons and $50\%$ normal sons

Haemphilia and colour blindness both are recessive X-linked traits. They express in males when present in single copy (heterozygous) but in females, they express only when present in homozygous condition.

Results:
(a) 50% sons are colour blinds and haemophilic.
(b) 50% of sons are haemophilic only.
(c) 50% daughters are the carrier for colour blindness and haemophilia.
(d) 50% of daughters are the carrier for haemophilia only.

So, the correct answer is '50% haemophilic and colour blind sons and 50% normal sons'.