A man with color blindness and deafness married a woman with good hearing and vision.

A man with color blindness and deafness married a woman with good hearing and vision. They had a son with deafness and normal vision and a color blind daughter with good hearing. determine the genotypes of parents and children, if both signs (deafness and color blindness) are recessive, deafness is an autosomal trait, and color blindness is linked to the x chromosome.

Let’s designate the recessive gene that causes the development of deafness as k, then the gene that causes normal hearing will be designated as K.

Let’s designate the gene that causes the development of color blindness as d. The gene that determines the absence of this vision pathology is denoted as D.

The chromosome with the color blindness gene will be labeled X d. A chromosome with a gene for normal color detection will be designated X D.

A man with color blindness and deafness has the genotype X d Ykk. It produces two types of spermatozoa – X d k, Yk.

Since both recessive traits due to genotypes kk and X d X d appear in the offspring, it should be concluded that the mother is heterozygous for both traits. Thus, a woman has the genotype X D X d Kk and produces four types of eggs – X D K, X d k, X D k, X d K.

The offspring of this married couple will be represented by the following possible options:

girls with normal hearing and vision (X D X d Kk) – 12.5%;

girls with color blindness and deafness (X d X d kk) – 12.5%;

girls with deafness (X D X d kk) – 12.5%;

girls with color blindness (X d X d K k) – 12.5%;

healthy boys (X D Y Kk) – 12.5%;

boys with color blindness and deafness (X d Y kk) – 12.5%;

boys with deafness (X D Y kk) – 12.5%;

boys with color blindness (X d Y Kk) – 12.5%.

Answer: father – X d Y kk, mother – X D X d Kk.