Principles of Inheritance and Variation Question 193

193. A woman with two genes for haemophilia and a gene for colour blindness on one of the ’ X ’ chromosomes marries a normal man. How will the progeny be

[1992]

(a) haemophilic and colour-blind daughters

(b) 50% haemophilic colour-blind sons and 50% haemophilic sons

(c) 50% haemophilic daughters and 50% colour-blind daughters

(d) All sons and daughters haemophilic and colour-blind

Show Answer

Answer : b

Hints & Solution

(b) XhcXh×XYh→ haemophilia
Female Male C→ colour blindness

XhcX XhcY XhX XhY
daughter haemophilic daughter haemophilic
(carrier) Colourblind (carrier) son
son
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