Q.A normal couple has their first child, who is haemophilic. Work out a cross to show how it is possible. State the possibility of the normal and the haemophilic children, along with their sexes, that can be born to them.

Haemophilia is an X-linked recessive disorder, so the gene responsible sits on the X chromosome. A normal couple having a haemophilic child tells us immediately that the mother must be a carrier (genotype X^H X^h) and the father is normal (genotype X^H Y). The mother appears phenotypically normal because the dominant allele on one X masks the recessive haemophilia allele on the other, but she can pass that defective allele to her children.

When we work out the cross between a carrier mother and a normal father:

• Gametes from mother: X^H or X^h
• Gametes from father: X^H or Y

The four possible offspring combinations are:

• X^H X^H – normal daughter

• X^H X^h – carrier daughter (phenotypically normal)

• X^H Y – normal son

• X^h Y – haemophilic son …

Haemophilia is a classic example of an X-linked recessive disorder, meaning the gene responsible sits on the X chromosome and requires two copies (in females) or one copy (in males) to manifest the disease. Because males have only one X chromosome (XY), a single recessive allele makes them haemophilic. Females, with two X chromosomes (XX), need both copies to be affected; if they carry one normal and one mutant allele, they remain phenotypically normal but are carriers.

The puzzle here is straightforward once we recognize what "normal couple" means in genetic terms. The father must be completely normal—his genotype is X^H Y, where X^H represents the X chromosome carrying the dominant normal allele. The mother appears normal but must be a carrier, with genotype X^H X^h, where X^h is the X chromosome bearing the recessive haemophilia allele. She inherited this mutant allele from one of her parents (likely her father was haemophilic or her mother was a carrier) but shows no symptoms because her other X chromosome carries the dominant normal allele.

When we work out the cross between this carrier mother and normal father, we set up a Punnett square with the mother's gametes (X^H and X^h) along one axis and the father's gametes (X^H and Y) along the other:

The four possible offspring are:

• X^H X^H daughters (25%): Normal, non-carrier females
• X^H X^h daughters (25%): Normal carrier females, phenotypically healthy but able to pass the allele on
• X^H Y sons (25%): Normal males
• X^h Y sons (25%): Haemophilic males, because the single X^h allele is expressed