Question

A colourblind man marries the daughter of another colourblind man whose wife had a normal genotype for colour vision. In their progeny

• A. All the children would colourblind
• B. All their sons are colourblind
• C. None of the daughters would be colourblind
• D. Half of their sons and half of their daughters would be colourblind

Answer 1

The correct option is D Half of their sons and half of their daughters would be colourblind

Color-blindness is a recessive sex linked trait. The genotype of the woman will be X${}^N$X${}^n$ because she is the daughter of the color blind man. The woman is the carrier. The woman married the man who is colorblind. The genotype of the man will be X${}^n$Y.

The offsprings will be

X${}^N$X${}^n$- one normal (carrier) girl because there is one dominant and one recessive allele, so the girl is a carrier,

X${}^N$Y- one normal boy because there is dominant allele,

X${}^n$X${}^n$- one color blind girl because there are two recessive alleles, so the girl is colour blind and

X${}^n$Y- one color blind boy because there is one recessive allele which is X-linked, responsible for the diseased state.

Half the sons and half the daughters have the probability of being colourblind.

Thus, the correct answer is option D.