Question

1. a woman who is a carrier marries a normal man. show the cross. what is the probability that their children will have hemophilia? what sex will a child in the family with hemophilia be?

Explanation:

Hemophilia is an X - linked recessive disorder. Let's denote the normal X chromosome as \(X^H\) and the X chromosome with the hemophilia allele as \(X^h\).

Step 1: Determine the genotypes of the parents

The woman is a carrier, so her genotype is \(X^H X^h\). The man is normal, so his genotype is \(X^H Y\).

Step 2: Set up the Punnett square

The woman can produce two types of gametes: \(X^H\) and \(X^h\). The man can produce two types of gametes: \(X^H\) and \(Y\).

	\(X^H\)	\(Y\)
\(X^h\)	\(X^H X^h\) (carrier female)	\(X^h Y\) (hemophiliac male)

Step 3: Calculate the probability of having a child with hemophilia

From the Punnett square, out of 4 possible offspring genotypes, only 1 (\(X^h Y\)) has hemophilia. So the probability is \(\frac{1}{4}\) or 25%.

Step 4: Determine the sex of a child with hemophilia

Looking at the Punnett square, the hemophiliac child has the genotype \(X^h Y\), which is a male genotype.

Answer:

• The cross (Punnett square) is as shown above.
• The probability that their children will have hemophilia is \(\frac{1}{4}\) (or 25%).
• A child in the family with hemophilia will be male.