Question

Haemophilia and colourblindness are the disorders caused by X chromosome linked recessive gene. A woman has one X chromosome having gene for haemophilia and colourblindness. The other X chromosome has wild allele for both the characters. She marries a man having phenotype normal for both the traits. Which of the following statement is most likely for the progeny?

• A. All daughters haemophilic and colourblind.
• B. 50% haemophilic sons and 50% colourblind sons.
• C. 50% haemophilic colourblind sons and 50% normal sons.
• D. 25% haemophilic daughters and 75% colourblind sons

Answer 1

The correct option is C 50% haemophilic colourblind sons and 50% normal sons.

The woman is a carrier for both the disorders hemophilia and color blindness and carries these genes on the X chromosome so the genotype of the woman will be X${}^{hc}$ X where h is for hemophilia and c is for color blindness. The genotype of the normal man is XY. The cross between the two will result in the progeny of X${}^{hc}$ X, XX, XY and X${}^{hc}$ Y. So, there are 50% males affected with hemophilia and color blindness and 50% are normal.

Thus, the correct answer is option C.