Question

1. 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). cross a man who has hemophilia with a woman who is a carrier.

a. whats the probability they would have a child that has hemophilia? ____%
b. whats the probability they would have a child that does not have hemophilia? _

Explanation:

Step1: Determine Genotypes

Hemophilia is X - linked recessive. Let \(X^{h}\) be the allele for hemophilia, \(X^{H}\) for normal. The man with hemophilia has genotype \(X^{h}Y\), the carrier woman has \(X^{H}X^{h}\).

Step2: Set Up Punnett Square

	\(X^{h}\) (from man)	\(Y\) (from man)
\(X^{h}\) (from woman)	\(X^{h}X^{h}\)	\(X^{h}Y\)

Step3: Analyze Offspring Genotypes

Offspring genotypes: \(X^{H}X^{h}\) (carrier female), \(X^{H}Y\) (normal male), \(X^{h}X^{h}\) (hemophilic female), \(X^{h}Y\) (hemophilic male). Total 4 possible offspring.

Step4: Calculate Probability for (a)

Hemophilic offspring: \(X^{h}X^{h}\), \(X^{h}Y\) → 2 out of 4. Probability \(=\frac{2}{4}\times100 = 50\%\).

Step5: Calculate Probability for (b)

Non - hemophilic offspring: \(X^{H}X^{h}\), \(X^{H}Y\) → 2 out of 4. Probability \(=\frac{2}{4}\times100 = 50\%\).

Answer:

a. 50%
b. 50%