Question

10 multiple choice 1 point a colorblind man (x^hy) and a woman who is not a carrier (x^hx^h) have children. what is the probability that their daughters will be carriers? 0% 25% 50% 100%

Explanation:

Step1: Determine the gametes

The man \(X^{h}Y\) produces gametes \(X^{h}\) and \(Y\). The woman \(X^{H}X^{H}\) produces gametes \(X^{H}\) only.

Step2: Analyze the cross

Using Punnett square:

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

Daughters have genotype \(X^{H}X^{h}\). But a carrier is a heterozygous female for a recessive X - linked trait. Here, since the woman is \(X^{H}X^{H}\) (not a carrier of the color - blind allele \(X^{h}\)), and the man's \(X^{h}\) combines with woman's \(X^{H}\), daughters (\(X^{H}X^{h}\)) are not carriers (as the normal allele \(X^{H}\) is dominant and they don't have a recessive allele from the mother to be carriers in the sense of having one normal and one disease - causing allele from a carrier mother. Here the mother has only normal alleles). So the probability is \(0\%\).

Answer:

A. 0%