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

• A

  All sons and daughters are haemophilic and colourblind
• B

  Haemophilic and colourblind daughters
• Correct answer
  C

  $50 \%$ haemophilic colourblind sons and $50 \%$ haemophilic sons
• D

  $50 \%$ haemophilic daughters and $50 \%$ colourblind daughters

Solution

Haemophilia and colour blindness both are recessive X-linked traits. They express in males when present in single copy (heterozygous) but in females, they express only when present in homozygous condition. Results (a) $50 \%$ sons are colourblind and haemophilic. (b) $50 \%$ sons are haemophilic only. (c) $50 \%$ daughters are carrier for colour blindness and haemophilia. (d) $50 \%$ daughters are carrier for haemophilia only.

Hence, the correct answer is option: 'C'.