Question

If a colour-blind man marries the daughter of a colour-blind father, then the next generation will be:

• A. None of their daughters will be colour blind
• B. All the sons will be colour blind
• C. All the daughters will be colour blind
• D. Half of their sons will be colour blind

Answer 1

The correct option is D Half of their sons will be colour blind

If a colour-blind man marries the daughter of a colour-blind father, half of their sons in the next generation will be colour blind.
As colour blindness is a X-linked recessive disorder, it will get expressed when genes are altered on both the X chromosomes, allosomes (XX) in females. If out of the two X chromosomes, any one is altered and the other X chromosome is normal, then the females will not have colour blindness, but they could be the carriers and pass the disease to their offspring.
While in males, as they have only one X chromosome in their allosomes (XY), alteration in that one X chromosome will cause the disease to get expressed.

Thus, the genotype of allosomes of colour-blind man will be X${}^c$Y.
The genotype of allosomes of normal man will be XY.
The genotype of allosomes of a normal woman will be XX.
The genotype of allosomes of a carrier woman will be X${}^c$X.
The genotype of allosomes of colour-blind woman will be X${}^c$X${}^c$.

According to the question, a colour- blind man (X${}^c$Y) marries the daughter of a colour-blind father, which means she will be a carrier (X${}^c$X).

Punnett square to show the possible genotypes of the progenies:

$\begin{array}{ccc}\text{Gametes}& X^c& \text{Y}\\ X^c& X^c{X}^c\text{(Affected daughter)}& X^{c}Y\text{(Affected son)}\\ X& X^{c}X\text{(Carrier daughter)}& \text{XY (Normal son)}\end{array}$

Thus, out of all the sons, 50% will be colour-blind.