Problem 1: Standard Haemophilia Cross

Given: Carrier female (X^H X^h) × Normal male (X^H Y) Find: (a) P(haemophilic son) (b) P(haemophilic child) (c) P(carrier daughter)

Step-by-step:

1. Draw Punnett square:

   • Female gametes: X^H, X^h
   • Male gametes: X^H, Y
   • Offspring: X^H X^H (25%), X^H X^h (25%), X^H Y (25%), X^h Y (25%)

2. (a) P(haemophilic | son): Sons are X^H Y and X^h Y (50% each among sons). P(haemophilic son) = 1/2 of sons

3. (b) P(haemophilic child) overall: Only X^h Y = 25% = 1/4 of all children

4. (c) P(carrier daughter): X^H X^h daughters = 25% of all children = 1/2 of all daughters

Answer: (a) 1/2 of sons; (b) 1/4 of all children; (c) 1/2 of daughters