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e ro e ro ro ro e ro ro be divided equally between the two single-crossover phenotypes ( e / ro+ bv+ and e+ / ro bv ).

There are 12 map units between ro and bv; so the number of progeny resulting from recombination between these two genes is 0.12 X 1800 = 216. Again, some of these recombinants will be single-crossover progeny and some will be double-crossover progeny. To determine the number of progeny resulting from a single crossover, subtract the double crossovers: 216 — 26 = 190. These single-crossover progeny will be divided between the two single-crossover phenotypes ( e+ ro+ / bv and e ro / bv+ ); so there will be 190/2 = 95 of each. The remaining progeny will be nonrecombinants, and they can be obtained by subtraction: 1800 — 26 — 334 — 190 = 1250; there are two nonrecombinants ( e+ ro+ bv+ and e ro bv ); so there will be 1250/2 = 625 of each. The numbers of the various phenotypes are listed here:

The offspring of the cross will be heterozygous, possessing one chromosome with the deletion and wild-type alleles and one chromosome without the deletion and recessive mutant alleles. For loci within the deleted region, only the recessive mutations will be present in the offspring, which will exhibit the mutant phenotype. The presence of a mutant trait in the offspring therefore indicates that the locus for that trait is within the region covered by the deletion. We can map the genes by examining the expression of the recessive mutations in the flies with different deletions.

Mutation a is expressed in flies with deletions 4, 5, and 6 but not in flies with other deletions; so a must be in the area that is unique to deletions 4, 5, and 6:

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