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

1/4

Step-by-step 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.