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  4. A woman (whose father is colour blind but mother is normal) marries a haemophiliac man with hypertrichosis. What percentage of progeny will show genotypically any two of the traits out of the three mentioned above at a given time?

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Q. A woman (whose father is colour blind but mother is normal) marries a haemophiliac man with hypertrichosis. What percentage of progeny will show genotypically any two of the traits out of the three mentioned above at a given time?

NTA AbhyasNTA Abhyas 2020Principles of Inheritance and Variation

Solution:

Let Xc be the X chromosome with colourblindness gene, Xh be the X chromosome with haemophilia gene and Yt be the Y chromosome with hypertrichosis gene.
The woman receives on X chromosome from colourblind father and one X chromosome from a normal mother. Her genotype will be XXc. The man is heamophilic with hypertrichosis, so his genotype will be XhYt.
Their possible progenies will be
Xh Yt
X XXh XYt
Xc XcXh XcYt

Genotypically any two of the traits out of the three traits will be seen in XcXh and XcYt. Thus the answer is 50%