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) $X^{h c} X^{h} \times XY \quad h \to$ haemophilia
Female $\quad$ Male $\quad C \to$ colour blindness
| $X^{hc} X$ | $X^{hc} Y$ | $X^{h} X$ | $X^{h} Y$ |
|---|---|---|---|
| daughter | haemophilic | daughter | haemophilic |
| (carrier) | Colourblind | (carrier) | son |
| son |
