Question

Colorblindness is an X-linked recessive condition. If a woman with normal vision whose father was colorblind has a son with a man with normal vision, what is the probability that he will be colorblind?

Answer 1

Answer: 0.25 or 25%

Step-by-step explanation:Women have two chromosomes XX and men have two also XY.

From the question, it can be seen that colorblindness is an X-linked recessive condition.

Let the X-linked recessive gene be specified as
`X^(-)` and the dominant gene which is not colorblind linked as
`X^(+)`.

The question says the woman has a father who is colourblind which means her father possesses the X-linked recessive gene
`X^(-)` which means he has chromosomes
`X^(-)Y` and therefore must have given his daughter the recessive gene

For her husband who has normal vision his chromosomes will be
`X^(+) Y`.

So, the probability that their son will be colourblind will be;

Chromosomes
`X^(+)` Y

`X^(+)`
`X^(+) X^(+)`
`X^(+)Y`.

`X^(-)`
`X^(-) X^(+)`.
`X^(-)Y`.

Therefore, the probability that their son will be color blind = ¼ or 0.25 or 25%