In humans, red-green color blindness is determined by an X-linked recessive allele (a), whereas eye color is determined by an autosomal gene, where brown (B) is dominant over blue (b). If a blue-eyed mother with normal vision has a brown-eyed, color-blind son and a blue-eyed, color-blind daughter, what are the genotypes of both parents and children

Question

Guest

Biology

22 November, 02:30

In humans, red-green color blindness is determined by an X-linked recessive allele (a), whereas eye color is determined by an autosomal gene, where brown (B) is dominant over blue (b). If a blue-eyed mother with normal vision has a brown-eyed, color-blind son and a blue-eyed, color-blind daughter, what are the genotypes of both parents and children

+3

Answers (1)

Know the Answer?

Haiden Davis · Answer 1

Mother - bb XAXa

Father - Bb XAY

Son - Bb XaY

Daughter - bb XAXa

Explanation:

Given -

Mother has blue-eye and normal vision

Son has brown eye and is color blind

Let the allele for Blue eye be represented by "b" and the allele for brown eye be represented by "B"

Given brown (B) is dominant over blue (b), thus both BB and Bb represents brown color eyes

Color blindness is associated with X gene

Thus Xa represents X - linked recessive colorblindness

And XA represents normal vision

Genotype of blue-eye and normal vision mother is bb XAXa. Mother is the carrier of recessive allele Xa for color blindness

Since son is color blind, his genotype would be Bb XaY

So fathers genotype would be

Bb XAY

Daughters genotype bb XAXa