Question

A man with normal color vision married a woman with normal color vision whose father was colorblind, what is the probability of a child with colorblindness?

Answer 1

Answer:

1/4

Explanation:

Color blindness is a X-linked, recessive trait. Let the allele be represented by c.

Normal color vision man = $X^CY$

Color blind man    = $X^cY$

Normal color vision woman = $X^CX^C or X^CX^c$

Color blind woman = $X^cX^c$

Since the woman's father is color blind, the woman is a carrier because she must have inherited the allele from the father.

Hence,

$X^CY$   x   $X^CX^c$

Progeny = $X^CX^C (normal), X^CX^c (normal), X^CY (normal), X^cY (affected)$

Probability of them having a colorblind child is 1/4.