Question

6. A brown-eyed man, whose mother was blue-eyed, marries a blue-eyed woman. The man is red-green colour blind and his wife ha

also draw the punnet square

b) Complete the following Punnett Square. e o 6 X 16 Xr l bXR I bXr c) What are the expected genotypic and phenotypic ratios

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Answer #1

Let B be the allele of the brown eye and b for the blue eye.

Color blindness is an X-linked recessive genetic disorder.

the male has one X-chromosome and a female has two X-chromosomes. The male receives his only X-chromosome from his mother and the female reived one X chromosome from her father and one from her mother. let Xc be the allele for colorblindness and X for the normal eye.

Given that the mother of the brown eye man has blue eyes. So he is heterozygous for the brown color.

So genotype of man is BbXcY and female is bbXcX

B-

Female bbX-X ВЫХсү male bxe bX BX BY EXC by bX BXC BbXcx bX BbXX bbXX BbXY bX bbxcx BY BbXY bbXY by bbXY:

C - Expected Genotype ratio - BbXcXc : BbXXc : bbXcXc : bbXXc : BbXcY : BbXY : bbXcY : bbXY is 1:1:1:1:1:1:1:1

Expected Phenotype - Brown eye and normal vision female: Brown eye and Colorblind female: Blue eye and normal vision female: Blue eye and Colorblind female : Brown eye and normal vision male: Brown eye and Colorblind male : Blue eye and normal vision male: Blue eye and Colorblind male is 1:1:1:1:1:1:1:1

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