Question

In case of haemophilia, if the carrier woman marries a normal man, then among their daughters

• A. 100% are carrier only
• B. 50% carrier and 50% haemophilic
• C. 50% are carrier only
• D. 75% carrier and 25 % haemophilic

Answer 1

The correct option is C 50% are carrier only

Haemophilia is a X-linked recessive disorder. One copy of the affected gene in males in each cell is sufficient to cause the disorder (X${}^h$Y). Females with two copies of the affected gene show the disorder (X${}^h$X${}^h$). Females heterozygous (X${}^h$X) for this trait be normal but serve as a carrier of the disease. According to the question, the female is carrier for colour blindness (X${}^h$X) and father is normal (XY). The carrier mother for colour blindness will inherit the disease to 50% sons (X${}^h$Y) while the 100% daughter will have normal vision with 50% carriers. Correct answer is option C.
Parent generation : (X${}^h$X) x XY

Gametes : (X${}^h$X) --> XY	X${}^h$	X
X	X${}^h$X carrier girl	XX normal girl
Y	X${}^h$Y haemophilic boy	XY normal boy