Question

122. 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 ?
1) 0% (2) 25% (3) 50% (4) 75%
please give solution.​

Answer 1

Answer:

Sex-linked inheritance is a condition in which the characteristics are transferred from the parent to the offspring via the sex chromosomes. On solving the Punnett square, we can see that the carrier female of colorblindness and the hemophiliac man with hypertrichosis will produce offspring, out of which only 50% will express genotypically any two of the traits out of the three mentioned above at a given time

Answer 2

Answer:

(3) 50%

Explanation:

Let's designate, Colour blindness gene as 'c'

Hypertrichosis gene as 'hy'

Haemophilic gene as 'h'

In first statement, "a normal woman whose father is colour blind but mother is normal", this means the women is carrier (XcX) as her father was suffering from colour blindness.

She marries a man suffering from hemophilia and hypertrichosis. We know that hypertrichosis is a Y linked disorder.

Therefore, the cross will be,

XcX (women) x XhYhy (man)

Offsprings: XcXh, XcYhy, XXh , XYhy

So the answer is

(3)....50%.

(i.e, the first two offspring will show two traits genotypically)