Inheritance: Question 4
Syllabus 16.2
Haemophilia A is a blood clotting disorder caused by a recessive allele of the F8 gene, which is carried on the X chromosome. The F8 gene normally codes for clotting factor VIII; a person with two copies of the recessive allele (or, in a male, one copy) cannot produce functional factor VIII and does not clot blood normally.
A woman does not have haemophilia herself, but her father does have haemophilia. This woman has children with a man who does not have haemophilia.
(a) Using the symbols X^H (dominant allele, normal clotting) and X^h (recessive allele, haemophilia), state the genotype of the woman's father, and explain how the woman must have inherited an X^h allele from him even though she does not have haemophilia herself. [2] (b) Construct a genetic diagram for a cross between this woman and her partner, and use it to determine the expected genotypes and phenotypes of their children with respect to haemophilia, showing daughters and sons separately. [4] (c) Explain, in terms of the number of X chromosomes each sex inherits, why sons are more likely than daughters to show the haemophilia phenotype in crosses such as this one. [2]
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Worked solution
Part (a): The father’s genotype and how the woman inherited X^h
A male has only one X chromosome (paired with a Y chromosome), so he cannot be heterozygous for a gene carried on the X chromosome. He only has one allele of this gene, and whatever that allele is, it determines his phenotype directly. Since the woman’s father has haemophilia, his genotype must be X^h Y.
Because a father passes his X chromosome only to his daughters (his sons receive his Y chromosome instead), every daughter of this man must inherit his X^h-carrying X chromosome. This means the woman must have received an X^h allele from her father.
Since the woman herself does not have haemophilia, she cannot be X^h X^h. Her other X chromosome (inherited from her mother) must therefore carry the dominant allele, X^H, so that the dominant allele masks the effect of X^h. Her genotype is the heterozygous carrier genotype, X^H X^h.
Part (b): Genetic diagram for the woman and her partner
Mother: X^H X^h (unaffected carrier) Father (partner): X^H Y (unaffected)
Gametes from mother: X^H, X^h Gametes from father: X^H, Y
| X^H (from father) | Y (from father) | |
|---|---|---|
| X^H (from mother) | X^H X^H | X^H Y |
| X^h (from mother) | X^H X^h | X^h Y |
Reading the four offspring genotypes:
- X^H X^H, daughter, unaffected, not a carrier.
- X^H X^h, daughter, unaffected, carrier (like her mother).
- X^H Y, son, unaffected.
- X^h Y. Son, has haemophilia.
So, considering daughters and sons separately:
- Daughters: expected ratio 1 non-carrier (X^H X^H) : 1 carrier (X^H X^h). All daughters are phenotypically unaffected.
- Sons: expected ratio 1 unaffected (X^H Y) : 1 affected (X^h Y).
Across all four offspring types together, the overall phenotypic ratio for haemophilia is 3 unaffected : 1 affected, and the single affected category (X^h Y) is always a son.
Part (c): Why sons are more often affected
The F8 gene is carried on the X chromosome. Males are hemizygous for X-linked genes: they inherit only one X chromosome, from their mother, and their Y chromosome (from their father) does not carry a corresponding allele of this gene. This means a son’s phenotype for haemophilia is determined entirely by the single X-linked allele he inherits. If it is X^h, there is no second allele on a homologous chromosome to mask it, so he is affected.
Females inherit two X chromosomes, one from each parent, so a daughter needs to inherit the recessive X^h allele from both parents (genotype X^h X^h) to show haemophilia herself; inheriting just one X^h allele makes her an unaffected carrier, as seen with the woman in this question.
In this particular cross, the father is unaffected (X^H Y), so he has no X^h allele to pass to any daughter. Every daughter is guaranteed to receive an X^H allele from him. This is why, in this family, sons can be affected but daughters cannot, however many children the couple go on to have.
Final answers
- (a) The father’s genotype is X^h Y; as a hemizygous male he must pass his X^h-carrying X chromosome to every daughter, so the woman inherited X^h from him and must be a heterozygous carrier, X^H X^h.
- (b) The cross X^H X^h x X^H Y gives daughters in the ratio 1 X^H X^H (unaffected) : 1 X^H X^h (unaffected carrier), and sons in the ratio 1 X^H Y (unaffected) : 1 X^h Y (haemophilia), an overall 3 unaffected : 1 affected ratio, with only sons ever affected.
- (c) Sons are hemizygous for this X-linked gene (one X chromosome, no matching allele on the Y chromosome), so a single X^h allele causes haemophilia; daughters need two X^h alleles, and since the father here is unaffected, no daughter of this couple can inherit two X^h alleles.