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) $X^{h c} X^{h} \times XY h \to$ haemophilia Female Male $C \to$ colour blindness

$X^{hc} X$ $X^{hc} Y$ $X^{h} X$ $X^{h} Y$
daughter haemophilic daughter haemophilic
(carrier) Colourblind (carrier) son
son