Question

The rosy (ry) gene of Drosophila encodes an enzyme called xanthine dehydrogenase. Flies homozygous for ry mutations exhibit a rosy eye color. Heterozygous females were made that had ry41 Sb on one homolog and Ly ry564 on the other homolog, where ry41 and ry564 are two independently isolated alleles of ry. Ly (Lyra [narrow] wings) and Sb (Stubble [short] bristles) are dominant markers to the left and right of ry, respectively. These females are now mated to males homozygous for ry41. Out of 100,000 progeny, 8 have wild-type eyes, Lyra wings, and Stubble bristles, while the remainder have rosy eyes. What is the order of these two ry mutations relative to the flanking genes Ly and Sb? What is the genetic distance separating ry41 and ry564?

Answer 1

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Ly+ ry41 Sb/ Ly ry564 Sb+9 x ry41 1 ry416-8Ly ry+ Sb and lots of ry progeny Recombination within the rosy gene, between the two ry mutations in the heterozygous female, generates the eight offspring with wild-type eyes. Such a recombination event will give one recombinant gamete with the wild-type sequence for both mutations, thus giving a ry+ phenotype in the progeny of this cross. The reciprocal recombinant gamete will be a double mutant, ry41 ry564, which will yield ry progeny indistinguishable from the parental type y progeny. We assume the ry41 ry564 progeny are found in equal numbers to the ry+ recombinants, so the recombination frequency-(8 + 8) / 100,000-0.00016-0.016%. The distance between ry41 and ry564.016 mu. yt564 41 Sb 4 564Si Sb OR 564Sb Orientation Il produces the Ly ry+ Sb recombinants obtained so the order must be Ly ry41 ry564 Sb