What are the likely genotypes of parents if all daughters in a large family have normal vision

What are the likely genotypes of parents if all daughters in a large family have normal vision and all sons suffer from color blindness?

It is well known that the gene that causes color blindness is linked to the X chromosome. Boys, being representatives of the heterogametic sex, do not have a paired X chromosome and phenotypically implement the information of any gene inherited from the mother responsible for color vision from their only X chromosome.

Let’s designate the chromosome that includes the gene for color blindness, or color blindness, as Xd, then the chromosome with the gene that causes normal color vision will be XD.

Boys with color blindness will be Xd Y.

They inherit chromosome Xd from their mother. Since all boys born in this family are uniform in genotype, it should be concluded that the mother is homozygous for the color blindness gene – XdXd. Such a woman herself will suffer from color blindness and produce the same type of eggs Xd.

Since all girls in this family are healthy, and from their mother they could only inherit a chromosome with a gene for color discrimination pathology, it should be concluded that they are all heterozygous – XDXd.

They inherit the second chromosome X with the gene for normal color vision from the father, which implies that the father is not color blind and has the XD Y genotype.

It is capable of producing two types of sperm – XD and Y. They transmit to daughters a chromosome containing a healthy gene – XD, and to sons – color-blind Y.

The offspring possible for a given married couple are represented by the following two options:

heterozygote girls with color discrimination rate (XDXd) – 50%;

boys with color blindness (Xd Y) – 50%.

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