Girl of normal vision whose father was colorblind marries a man of...

Color blindness is an X-linked recessive trait.

Let $X^C$ be the normal allele and $X^c$ be the colorblind allele.

The girl has normal vision but her father was colorblind (X^c Y$). She must have inherited the$X^c$chromosome from her father, making her genotype$X^C X^c$(carrier).<br><br>The man has normal vision, so his genotype is$X^C Y$. The fact that his father was colorblind does not affect his genotype, as he inherits the Y chromosome from his father.<br><br>When they marry (X^C X^c \times X^C Y$), the sons inherit the Y chromosome from the father and one of the X chromosomes from the mother.

The mother produces two types of gametes containing X chromosomes: $X^C$ and $X^c$ in equal proportion.

Therefore, the sons will have genotypes $X^C Y$ (normal vision)$and X^c Y$ (colorblind) in a $1:1$ ratio.

Thus, $50\%$ of the sons will be colorblind.

Answer: $50\%$ colourblind