Question

A woman with two genes (one on each X-chromosome) for haemophilia and one gene for colourblindness on the X-chromosome marries a normal man. How will the progeny be?

• A. All sons and daughters are haemophilic and colourblind
• B. Haemolhilic and colourblind daughters
• C. 50 % haemophilic colourblind sons and 100 % haemophilic sons
• D. 50 % haemophilic daughters and 50 % colourblind daughters

Answer 1

The correct option is C 50 % haemophilic colourblind sons and 100 % haemophilic sons

Haemophilia and colorblindness are X-linked recessive disorders. Since males are hemizygous for chromosome, one copy of the affected gene in males in each cell is sufficient to cause these disorders (X${}^c$Y or X${}^h$Y ). Females with two copies of the affected gene show the disorder (X${}^c$X${}^c$ or X${}^h$X${}^h$). Females heterozygous for this trait will be normal but serve as a carrier of the disease (X${}^c$X or X${}^h$X). Since father transmits its X chromosome to the daughters, not to son, so the couple would have all normal daughters owing to the normal father. This makes options A, B and D incorrect.

Son receives it's X chromosome from the mother. The haemophiliac mother would transmit the affected X-chromosome to all of her sons, so the couple would have all haemophilic sons. Further, since the mother is the carrier for colorblindness (X${}^c$X), which means half of her sons will get the normal copy of X chromosome and another half will get the affected copy. Thus, there is 50% probability of their son to have colorblindness.