Final answer:
The expected proportion of F1 offspring that are heterozygotes for both gene pairs in the cross Aa Bb × aa bb is 1/4. This is calculated by considering that all offspring will be 'Aa' for the first gene and there's a 50% chance for 'Bb' heterozygotes for the second gene, which leads to a combined probability of 1/4 being heterozygotes for both.
Step-by-step explanation:
The proportion of heterozygotes for both gene pairs in the dihybrid cross Aa Bb × aa bb is 1/4. This answer is derived from applying the principles of Mendelian genetics. Each gene pair is inherited independently, following Mendel's law of independent assortment.
To find the proportion of double heterozygotes among the F1 offspring, we look at the possibilities for each gene separately and then combine them. For gene A, offspring receive an 'A' from the heterozygote parent and an 'a' from the homozygote parent. Therefore, all offspring are heterozygous 'Aa' at this locus. For gene B, there is a 50% chance (1/2) of receiving 'B' and a 50% chance of receiving 'b' from the heterozygote parent, while all will receive a 'b' from the homozygote parent. This results in 50% heterozygous 'Bb' and 50% homozygous 'bb' at the B locus.