Question

1. a woman who has hemophilia marries a normal man. how many of their children will have hemophilia, and what is their sex?

Explanation:

Step1: Determine Genotypes

Hemophilia is an X - linked recessive disorder. Let the allele for hemophilia be \(h\) and the normal allele be \(H\). A woman with hemophilia has the genotype \(X^{h}X^{h}\) (since she has two X - chromosomes and needs two recessive alleles to express the disorder). A normal man has the genotype \(X^{H}Y\) (he has one X - chromosome with the normal allele and one Y - chromosome).

Step2: Set up Punnett Square

The possible gametes from the woman (\(X^{h}X^{h}\)) are all \(X^{h}\). The possible gametes from the man (\(X^{H}Y\)) are \(X^{H}\) and \(Y\).

	\(X^{H}\)	\(Y\)
\(X^{h}\)	\(X^{H}X^{h}\)	\(X^{h}Y\)

Step3: Analyze Offspring Genotypes and Phenotypes

• For the offspring with genotype \(X^{H}X^{h}\): These are female. Since they have one dominant \(H\) allele, they are carriers (do not have hemophilia).
• For the offspring with genotype \(X^{h}Y\): These are male. Since they have the recessive \(h\) allele on their X - chromosome (and Y has no corresponding allele for hemophilia), they have hemophilia.

From the Punnett square, out of 4 possible offspring, 2 are male with hemophilia (the \(X^{h}Y\) genotypes) and 2 are female carriers. So half of their children (specifically the male children) will have hemophilia.

Answer:

Half of their children (all the male children) will have hemophilia.