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Assume that two genes (A and B) are 20 maps units apart on chromosome II of Drosophila and a cross is made between a double heterozygous female (AaBb) and a homozygous recessive male (aabb). What percent recombination between A and B would be expected in the offspring of this type of cross?

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Final answer:

The percent recombination expected between genes A and B, which are 20 map units (cM) apart on Drosophila chromosome II, in the offspring resulting from a cross between a AaBb female and aabb male is 20 percent.

Step-by-step explanation:

The student's question pertains to the expected percent recombination between two genes on chromosome II of Drosophila. Specifically, the genes A and B are 20 map units apart (20 centimorgans or cM), and there is a cross between a double heterozygous female (AaBb) and a homozygous recessive male (aabb).

To calculate the expected percent recombination, we reference the mapping distance in centimorgans, which directly correlates with the recombination frequency percentage. As such, if two genes A and B are 20 cM apart, this indicates that the maternal and paternal alleles for these genes will recombine in 20 percent of offspring, on average.

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