In a cross between two individuals with genotypes for haemophilia X^HX^h x X^HY, the chances of having a haemophilic son is

The problem states the genotypes are $X^hX^h$ and $X^HY$.

• Female parent: $X^hX^h$ (haemophilic female)
• Male parent: $X^HY$ (normal male)

The female parent can produce only $X^h$ gametes.
The male parent can produce $X^H$ and $Y$ gametes.

From the Punnett square, the possible offspring genotypes are:

• $X^HX^h$: Carrier female (not haemophilic)
• $X^hY$: Haemophilic male

Out of 4 possible offspring combinations, 2 are $X^hY$ (haemophilic sons).
The probability of having a haemophilic son is $\frac{2}{4} = \frac{1}{2}$.

$\frac{1}{2} \times 100\% = 50\%$

The chances of having a haemophilic son are 50%.

The final answer is $\boxed{\text{C. 50\%}}$.