A) Step 1: Identify individual 4 in the pedigree. Individual 4 is an unshaded circle.

An unshaded circle represents a normal female.

a) The phenotype of individual 4 is:

b) Step 1: Identify individual 2 in the pedigree. Individual 2 is an unshaded circle, so she is phenotypically normal.

Individuals 6 and 7 are shaded squares, meaning they are haemophiliac males.

Since haemophilia is X-linked recessive ($X^h$), a haemophiliac male has the genotype $X^h Y$.

Therefore, individual 2 must have passed on the $X^h$ allele to her sons (individuals 6 and 7).

b) The genotype of individual 2 is:

12.2
Haemophilia is an X-linked recessive disorder. Females have two X chromosomes ($XX$), while males have one X and one Y chromosome ($XY$). For a female to suffer from haemophilia, she must inherit two copies of the recessive allele ($X^h X^h$), one from each parent. However, if she inherits one normal allele ($X^H$) and one haemophilia allele ($X^h$), she will be a carrier but phenotypically normal because the dominant normal allele masks the recessive one. Males, having only one X chromosome, will express the disorder if they inherit the single recessive allele ($X^h Y$). Therefore, it is less likely for a female to inherit two recessive alleles than for a male to inherit one.

12.3

• Individual 13 is an unshaded square, meaning he is a Normal male. His genotype is $X^H Y$.
• Individual 14 is a shaded circle, meaning she is a Haemophiliac female. Her genotype is $X^h X^h$.

Parents: $X^H Y \times X^h X^h$

Gametes from individual 13: $X^H$, $Y$
Gametes from individual 14: $X^h$, $X^h$

Punnett Square:

• $X^H X^h$: Normal female (carrier)
• $X^h Y$: Haemophiliac male

From the Punnett square, there are 2 out of 4 possible offspring that are haemophiliac sons ($X^h Y$).
Percentage chance = $\frac{2}{4} \times 100\% = 50\%$.

The percentage chance of individuals 13 and 14 having a haemophiliac son is: