A fruit fly of genotype B R/b r is testcrossed with b r/b r. In 84 percent of the meioses, there are no chiasmata between the linked genes; in 16 percent of the meioses, there is one chiasma between the genes. What proportion of the progeny will be B r/b r

4 percent (4%)

Explanation:

A single crossover occurs between two (non-sister) chromatids belonging to homologous chromosomes. In this case, 16 percent of the meioses have a single crossover, thereby it will produce 8 percent of the chromosomes with the original (parental) combination in the progeny and the remaining 8 percent should be recombinants. From this result, it is reasonable to conclude that half of these recombinants should be 'Br' (and the other remaining 4 percent should be recombinants 'bR'), and therefore the answer is 4 percent (4%).

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