Question

human sex-linkage - in humans, hemophilia is a sex - linked trait. females can be normal, carriers, or have the disease. males will either have the disease or not (but they wont ever be carriers)
$x^{h}x^{h}=$ female, normal
$x^{h}x^{h}=$ female, carrier
$x^{h}x^{h}=$ female, hemophiliac
$x^{h}y=$ male, normal
$x^{h}y=$ male, hemophiliac

1. show the cross of a man who has hemophilia with a woman who is normal (not a carrier).

how many children will have the disease?

1. a woman who is a carrier marries a normal man. show the cross:

how many children will have the disease?
what is the sex of the child with the disease?

Explanation:

Step1: Determine the genotypes of the parents for question 6

The man has hemophilia, so his genotype is \(X^{h}Y\). The woman is normal (not a carrier), so her genotype is \(X^{H}X^{H}\).

Step2: Set up the Punnett square for question 6

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

Step3: Analyze the results for question 6

Looking at the Punnett square, the genotypes of the children are \(X^{H}X^{h}\) (female carriers) and \(X^{H}Y\) (normal males). So, \(0\) children have the disease.

Step4: Determine the genotypes of the parents for question 7

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

Step5: Set up the Punnett square for question 7

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

Step6: Analyze the results for question 7

Looking at the Punnett square, the genotypes are \(X^{H}X^{H}\) (normal female), \(X^{H}X^{h}\) (carrier female), \(X^{H}Y\) (normal male), and \(X^{h}Y\) (hemophiliac male). So, \(1\) child has the disease (\(X^{h}Y\)), and the sex of the child with the disease is male.

Answer:

1. \(0\)
2. \(1\); male