Search any question & find its solution
Question:
Answered & Verified by Expert
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?
Options:
Solution:
2124 Upvotes
Verified Answer
The correct answer is:
50 %
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
Genotypically any two of the traits out of the three traits will be seen in XcXh and XcYt. Thus the answer is 50%
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%
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.