Question

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

• A. $50\%$ haemophilic colour-blind sons and $50\%$ normal sons.
• B. $50\%$ haemophilic daughters (carrier) and $50\%$ colour blind daughters (carrier)
• C. All sons and daughters are haemophiliac and colourblind
• D. Haemophilic and colour-blind daughters

Answer 1

Answer: (B)

Haemophilia is a defect of blood which prevents its clotting. Colour blindness is the inability of certain human beings to distinguish red from green colour. Both these diseases are produced by a recessive gene which lies on the $X$-chromosomes. A woman having gene for haemophilia on one $X$-chromosome and gene for colourblindness on another $X$-chromosome will have genotype $X^h{X}^c$.

Thus, progeny includes $50\%$ daughters are carrier of haemophilia and $50\%$ daughter are carrier of colourblindness whereas $50\%$ sons are haemophiliac and $50\%$ sons are colourblind.