The problem states that haemophilia is due to a recessive allele X^h, and X^H is for normal blood clotting. A normal couple had a haemophilic son (X^hY).

a) A sex-linked characteristic is a trait whose gene is located on one of the sex chromosomes (X or Y), rather than on an autosome. These traits often show different patterns of inheritance in males and females.

b)

i)
The problem states that haemophilia is due to a recessive allele $X^h$, and $X^H$ is for normal blood clotting. A normal couple had a haemophilic son ($X^hY$).

• Since the son is haemophilic ($X^hY$), he inherited the $Y$ chromosome from his father and the $X^h$ chromosome from his mother.
• The father is normal, so his genotype must be $X^HY$.
• The mother is normal but passed on an $X^h$ allele to her son, so she must be a carrier. Her genotype must be $X^HX^h$.

Genotypes:
Father: $\boxed{X^HY}$
Mother: $\boxed{X^HX^h}$

ii)
To show how a haemophilic son could have been born, we use a genetic diagram (Punnett square) with the parents' genotypes: Father ($X^HY$) and Mother ($X^HX^h$).

Parental Genotypes:
Father: $X^HY$
Mother: $X^HX^h$

Gametes:
Father: $X^H$, $Y$
Mother: $X^H$, $X^h$

Punnett Square:

$$\begin{array}{|c|c|c|} \hline \textbf{Gametes} & X^H & X^h \\ \hline X^H & X^HX^H & X^HX^h \\ \hline Y & X^HY & X^hY \\ \hline \end{array}$$

Offspring Genotypes and Phenotypes:

• $X^HX^H$: Normal female
• $X^HX^h$: Normal female (carrier)
• $X^HY$: Normal male
• $X^hY$: Haemophilic male

The genetic diagram shows that there is a 25% chance for the couple to have a haemophilic son ($X^hY$).