Question

A woman with two genes for haemophillia and one gene for colourblindness on one of the X chromosomes marries a normal man. What would be the progeny?

• A. All sons and daughters haemophillic and colourblind
• B. Haemophillic and colourblind daughters
• C. 100% haemophillic sons and 50% of them is hemophillic colourblind sons
• D. 50% haemophillic daughters and 50% colourblind daughters

Answer 1

The correct option is C 100% haemophillic sons and 50% of them is hemophillic colourblind sons

1. Hemophilia is a X-linked recessive disorder. The associated genes are located on the X chromosome, which is one of the two sex chromosomes. Since, males are hemizygous for chromosome (have only one X chromosome), one copy of the affected gene in each cell is sufficient to cause the disorder (X${}^h$Y). Females have two X chromosomes and hence, need two copies of the affected gene to cause the disorder (X${}^h$X${}^h$). Females heterozygous (X${}^h$X) for this trait be normal but serve as a carrier of the disease.
   
2. Colourblindness is also a X-linked recessive disorder, one copy of the affected gene in males in each cell is sufficient to cause the disorder (X${}^c$Y). Females with two copies of the affected gene show the disorder (X${}^c$X${}^c$). Females heterozygous (X${}^c$X) for this trait be normal but serve as a carrier of the disease.
3. According to the question, the female is homozygous (X${}^h$X${}^h$) for hemophilia and is heterozygous (X${}^c$X) for colourblindness. Hence, the genotype of mother is (X${}^{ch}$X${}^h$). The genotype of normal father is XY. The homozygous hemophiliac mother will inherit the disease to all the sons (X${}^h$Y) while the daughter will be the carrier (X${}^h$X). The carrier mother for colourblindness will inherit the disease to 50% sons (X${}^c$Y) while the 100% sons will be the hemophiliac.

Thus, the correct answer is option C.