If a husband and wife have a heterozygous girl for colorblindness a normal boy, a colorblind girl, and a colorblind boy what would be the genotypes of the parents?

The Wife is a carrier of the recessive allele with a genotype XX°

The husband is a colorblind individual with Genotype X°Y

Explanation:

We know that Color blindness is a Sex linked recessive disorder.

Now let us consider the case of the colorblind daughter ,

She must be homozygous for the recessive allele because a hetrozygous female will not show the phenotype of the x-linked recessive trait.Hence , The color blind daughter must have a genotype X°X°.This clearly shows that she has received one mutated X-chromosome from her mother and One mutated chromosome from her Father.Since, the father has the Mutated  gene he must be a Color blind individual and has a X°Y genotype.The Mother could either be a Carrier or a colorblind individual.

Now, Let us consider the case of a Normal Boy:

Sons always receive their X chromosome from mother and genes present on the X-chromosome of males always express themselves.Absence of the disease in this son states that he has received a normal X-chromosome from the mother.

On considering both cases stated above we can conclude that the woman is a carrier of the recessive trait while father expresses the disease and has the X-chromosome with the recessive allele.

This is further confirmed by presence of a heterozygous daughter whose genotype is XX°. She has received the normal X from mother and the mutated X from the father.

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