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 [tex]X^{-}[/tex] and the dominant gene which is not colorblind linked as [tex]X^{+}[/tex].
The question says the woman has a father who is colourblind which means her father possesses the X-linked recessive gene [tex]X^{-}[/tex] which means he has chromosomes [tex]X^{-}Y[/tex] and therefore must have given his daughter the recessive gene
For her husband who has normal vision his chromosomes will be [tex]X^{+} Y[/tex].
So, the probability that their son will be colourblind will be;
Chromosomes [tex]X^{+}[/tex] Y
[tex]X^{+}[/tex] [tex]X^{+} X^{+}[/tex] [tex]X^{+}Y[/tex].
[tex]X^{-}[/tex] [tex]X^{-} X^{+}[/tex]. [tex]X^{-}Y[/tex].
Therefore, the probability that their son will be color blind = ¼ or 0.25 or 25%
