question_answer

A woman with two genes (one on each X-chromosome) for haemophilia and one gene for color blindness on the X-chromosome marries a normal man. How will the progeny be?

A)  All sons and daughters are haemophilic and colorblind

B)  Haemophilic and colorblind daughters

C)  50% haemophilic colorblind sons and 50% haemophilic sons

D)  50% haemophilic daughters and 50% colorblind daughters

Correct Answer: C

Solution :

Haemophilia 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 colorblind and haemophilic. (b) 50% sons are haemophilic only. (c) 50% daughters are carrier for colour blindness and haemophilia. (d) 50% daughters are carrier for haemophilia only.