In humans, haemophilia is a sex-linked recessive trait. What are the possible genotypes and phenotypes of offspring from a haemophiliac female and a normal male?

A haemophiliac female has the genotype $X^hX^h$.
A normal male has the genotype $X^HY$.

Parents: $X^hX^h \times X^HY$
Gametes from female: $X^h$
Gametes from male: $X^H$, $Y$

Punnett Square:

Possible genotypes: $X^HX^h$, $X^hY$
Possible phenotypes: Normal female (carrier), Haemophiliac male

Explanation:
All sons inherit an X chromosome from their mother and a Y chromosome from their father. Since the mother is haemophiliac ($X^hX^h$), she can only pass on an $X^h$ allele to her sons. The father provides the Y chromosome. Therefore, all sons will have the genotype $X^hY$, which means they will be haemophiliac.

2.4.1

A man suffering from haemophilia has the genotype $X^hY$.
A woman who is a carrier has the genotype $X^HX^h$.

Parents: $X^hY \times X^HX^h$
Gametes from male: $X^h$, $Y$
Gametes from female: $X^H$, $X^h$

Punnett Square:

Possible genotypes: $X^HX^h$, $X^hX^h$, $X^HY$, $X^hY$
Possible phenotypes:

• $X^HX^h$: Normal female (carrier)
• $X^hX^h$: Haemophiliac female
• $X^HY$: Normal male
• $X^hY$: Haemophiliac male

2.4.2
Haemophilia affects mostly males because it is a sex-linked recessive disorder on the X chromosome. Males have only one X chromosome ($XY$). If a male inherits the recessive allele ($X^h$) on his single X chromosome, he will express the disorder ($X^hY$) because there is no dominant allele on a second X chromosome to mask it. Females have two X chromosomes ($XX$), so they must inherit two copies of the recessive allele ($X^hX^h$) to express the disorder. If a female inherits only one recessive allele ($X^HX^h$), she will be a carrier but will not show symptoms due to the presence of a dominant normal allele ($X^H$).

2.1.1

A normal father has the genotype $X^HY$.
A heterozygous mother (carrier) has the genotype $X^HX^h$.

Parents: $X^HY \times X^HX^h$
Gametes from father: $X^H$, $Y$
Gametes from mother: $X^H$, $X^h$

Punnett Square:

Possible genotypes: $X^HX^H$, $X^HX^h$, $X^HY$, $X^hY$
Possible phenotypes:

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

2.1.2

The genotype for a haemophiliac male is $X^hY$.

From the Punnett square, there is 1 out of 4 possible offspring genotypes that results in a haemophiliac male.
The chances are $\frac{1}{4}$ or 25%.

The chances of the parents having a child that will be a haemophiliac male are \boxed{\text{25%}}.

2.1.3
Males have only one X chromosome ($XY$). Haemophilia is a recessive disorder on the X chromosome. If a male inherits the recessive allele ($X^h$) on his single X chromosome, he will express the disorder ($X^hY$) because there is no second X chromosome with a dominant normal allele to mask it. A carrier is an individual who possesses a recessive allele but does not express the trait. Therefore, a male cannot be a carrier for haemophilia; he either has the disorder or he does not.