From the given, it was shown that the mother is heterozygous for both genes A and B while the father has its only A/a allele as recessive and a homozygous ecessive for the B/b gene. Thus, following independent assortment, the mother would have the following gametes upon crossing: XAB, XAb, XaB, Xab. The father, on the other hand, only has two: Xab and Yb.
Predict the frequency of progeny genotypes for the cross: XAXaBb x XaYbb.
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