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