Question

john ann xhy xxh charles marie fred jack irene xhy xh xh xhy ? ? what is the probability that a female child of charles and marie would suffer from hemophilia? 0 percent 25 percent mark this and return

Explanation:

Step1: Determine the genotypes of Charles and Marie

Charles has the genotype \(X^{H}Y\) (normal male). Marie has the genotype \(X^{H}X^{h}\) (carrier female).

Step2: Set up the Punnett square for a female child

For a female child, the possible combinations of alleles from Charles (\(X^{H}\)) and Marie (\(X^{H}\) or \(X^{h}\)) are:

• \(X^{H}\) (from Charles) and \(X^{H}\) (from Marie): \(X^{H}X^{H}\) (normal female)
• \(X^{H}\) (from Charles) and \(X^{h}\) (from Marie): \(X^{H}X^{h}\) (carrier female)

Since hemophilia is a recessive X - linked disorder, a female needs two recessive alleles (\(X^{h}X^{h}\)) to be affected.

Answer:

0 percent