Question

A woman with two genes for haemophilia and one gene for colour blindness on one of the 'X' chromosomes marries a normal man. What will be the progeny?

• A. All sons and daughters are haemophilic and colourblind.
• B. Haemophilic and colour blind daughters.
• C. 50% haemophilic, colour-blind sons and 50% haemophilic sons and 50% haemophilic daughters and 50% haemophilic colour blind daughters.
• D. All normal.
• E. 2 normal daughters (carrier) : 1 haemophilic and colourblind son : 1 haemophilic son
  

Answer 1

The correct option is E 2 normal daughters (carrier) : 1 haemophilic and colourblind son : 1 haemophilic son

According to the question.
Parent generation : "$X^{hc}$ $X^h$" x XY

Gametes : $X^{hc}$$X^h$ ---> XY	$X^{hc}$	$X^h$
X	$X^{hc}$X double carrier girl	$X^h$X carrier girl
Y	$X^{hc}$Y haemophilic and colourblind boy	$X^h$Y haemophilic boy

Progeny = 2 normal girl (carrier) : 1 haemophilic and colourblind boy : 1 haemophilic boy. Both haemophilia and colorblindness are X-linked recessive disorder and follow criss-cross inheritance. The presence of normal father rules out the possibility of having affected daughters. An affected mother will pass the traits to all sons. Thus, the correct answer is option E.