Question

A husband and wife have normal vision, although both of their fathers are red-green colour blind, which is inherited as an x-linked recessive trait.
What is the probability that their first child will be?
i. A normal son.
ii. A carrier daughter.
iii. A colour-blind son.
iv. A colour-blind daughter.

• A. $i= \, \frac{1}{4}$ , $ii=\frac{1}{4} \, ,iii=\frac{1}{4} \, ,iv=\frac{1}{4}$
• B. $i=\frac{1}{4} \, ,ii=\frac{1}{4} \, ,iii=\frac{1}{4}, \, iv=\frac{0}{4}$
• C. $i=\frac{0}{4} \, ,ii=\frac{1}{2} \, ,iii=\frac{1}{4} \, , \, iv=\frac{1}{4}$
• D. $i=\frac{1}{4}, \, ii=\frac{1}{2} \, , \, iii=\frac{1}{4} \, ,iv=0$

Answer 1

Answer: (B)

Mother is a carrier (because her father is colour-blind) and father is normal. Hence, their genotype is $X^c X$ and $X Y$, respectively. Thus, the probability that their first child will be a normal son is $1 / 4$, a carrier daughter is $1 / 4$, colour-blind son is $1 / 4$ and a colour-blind daughter is $0$ ( both husband and wife have normal vision, though their fathers were colour-blind. The probability of their daughter becoming colour-blind is $0 \%$. The chances of a daughter becoming colour-blind arise only when the father is also colour-blind ).