Answer/Explanation:
For the woman to be blood type B, she must either have 2 B alleles (homozygous, BB) or 1 B allele (heterozygous, BO). We can draw two punnett squares to show each of the potential outcomes when she has children with an AB man.
The two punnett squares are attached. In order to have a child that is type A, she needs to be heterozygous, BO. Even then, there was only a 25% chance that their child would have type A blood. If she were homozygous BB, it would be impossible for her to have a type A child with an AB man.
