The correct answer is A.

Stop: The answers are all frequencies of a child being homozygous recessive.

Think: This question could be answered with a Punnett square. However, for a child to be homozygous recessive, both parents would have to have at least one copy of the recessive allele. Because one parent is AA and the other is BB, there is no chance for the child to be homozygous recessive for either trait, let alone both.

Predict: There is a 0% chance for the child to be homozygous recessive for both traits.

Match: (A) is correct. Twenty-five percent (B) represents the chance of a child being any of the possible genotypes from this cross: AABB, AaBB, AABb or AaBb. (C) does not make any sense in this scenario because there would be no way to have three possible outcomes for a genotype. (D) represents the chance for being either heterozygous for A or B. (E) represents the chance for a child to express both dominant phenotypes.