Question

Hemophilia A, a condition in which blood does not clot properly, is a recessive trait located on the X chromosome (Xʰ). Suppose that a woman who is heterozygous for this trait (XXʰ) has children with a normal male (XY). What is the probability that the couple will have daughters with hemophilia? What is the probability that the couple will have sons with hemophilia?

Accepted Answer

Identify the genotypes of the parents: The woman is heterozygous for hemophilia A, so her genotype is X X\^h (one normal X chromosome and one X chromosome with the hemophilia allele). The man is normal, so his genotype is X Y. Determine the possible gametes each parent can produce: The woman can produce eggs carrying either X or X\^h. The man can produce sperm carrying either X or Y. Set up a Punnett square to find all possible combinations of the offspring's sex chromosomes: Combine the woman's gametes (X and X\^h) with the man's gametes (X and Y) to get the genotypes of daughters and sons. Analyze the daughters' genotypes: Daughters receive one X chromosome from each parent. The possible daughter genotypes are X X (normal) and X\^h X (carrier). Since hemophilia is recessive and X-linked, daughters must have two X\^h alleles (X\^h X\^h) to express the disease. Determine the probability of this genotype from the Punnett square. Analyze the sons' genotypes: Sons receive the X chromosome from the mother and the Y chromosome from the father. The possible son genotypes are X Y (normal) and X\^h Y (hemophiliac). Since males have only one X chromosome, the presence of X\^h means the son will have hemophilia. Determine the probability of this genotype from the Punnett square.

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Key Concepts

X-linked Recessive Inheritance

Genotype and Phenotype of Parents

Probability of Inheritance in Offspring