Final answer:
The female children from this union have a 50% chance of being carriers of hemophilia if they inherit the defective allele from their mother but will not show symptoms as they will also inherit a normal allele from their father. They have a 50% chance of inheriting two normal X chromosomes and not being carriers or affected by hemophilia.
Step-by-step explanation:
The predicted phenotypes of the female children from a union of a woman who is heterozygous for hemophilia and a man with normal blood clotting characteristics can be determined by understanding X-linked recessive inheritance. Since females have two X chromosomes (XX) and hemophilia is X-linked, a female must inherit two copies of the hemophilia allele to exhibit symptoms of the disorder. In this case, the woman carries one normal allele and one hemophilia allele, symbolized as XHXh. The man, with normal clotting, has only normal alleles on his X chromosome, symbolized as XHY.
When we create a Punnett square to predict the possible genetic combinations, we find that:
- The female children have a 50% chance of being carriers of hemophilia (XHXh), just like their mother.
- They also have a 50% chance of having two normal alleles (XHXH) and not being carriers.
None of the female offspring would show the disease because they inherit at least one normal X chromosome from their father. However, any daughter that inherits the Xh from their mother will have the carrier status of hemophilia.