Question

john ann x^h y x^h x^h charles marie fred jack irene x^h y x^h x^h ? x^h y ? what is the probability that a female child of charles and marie would suffer from hemophilia? 0 percent 25 percent

Explanation:

Step1: Analyze Genotypes

Charles has genotype \(X^H Y\) (normal male, since he's not shaded), Marie has \(X^H X^h\) (carrier, half - shaded). For a female child, she gets one \(X\) from Charles (\(X^H\)) and one \(X\) from Marie.

Step2: Determine Possible Genotypes of Female Child

The possible genotypes for a female child are:

• If she gets \(X^H\) from Marie: \(X^H X^H\) (normal, no hemophilia).
• If she gets \(X^h\) from Marie: \(X^H X^h\) (carrier, no hemophilia, since hemophilia is recessive and needs \(X^h X^h\) for a female to have it).

Since a female needs two \(X^h\) alleles to have hemophilia, and Charles can only pass \(X^H\), the female child can never have two \(X^h\) alleles. So the probability of a female child having hemophilia is 0 percent.

Answer:

0 percent