Final answer:
The proportion of the progeny from a cross of Aa Bb x aa bb that will have the genotype Aa Bb is 1/4 or 25%, as calculated by independently combining the probabilities of receiving each allele.
Step-by-step explanation:
The student's question is about predicting offspring genotypes from a dihybrid cross with parents of genotypes Aa Bb and aa bb, assuming that these genes assort independently. To find the proportion of progeny that will have genotype Aa Bb, we need to apply the principle of independent assortment and use the product rule for combining probabilities.
Each gene will assort independently, so we calculate the probability of each allele combination separately and then multiply them. The probability of an offspring being Aa at the first locus is 1/2, because they will inherit an 'a' allele from the aa parent and either an 'A' or 'a' from the Aa parent. For the second locus, the probability is also 1/2 for being Bb because they will inherit a 'b' allele from the bb parent and either a 'B' or 'b' from the Bb parent.
Using the product rule, we multiply the probabilities for each locus (1/2 for Aa and 1/2 for Bb) to find the combined probability: 1/2 * 1/2 = 1/4 or 25%. Therefore, the proportion of the progeny that will have the genotype Aa Bb is 1/4, which corresponds to answer choice B).