Chapter 7, Problem 16PDQ

Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three autosomes (chromosomes II, III, and IV). A genetics student discovered a male fly with very short (sh) legs. Using this male, the student was able to establish a pure- breeding stock of this mutant and found that it was recessive. She then incorporated the mutant into a stock containing the recessive gene black (b, body color, located on chromosome II) and the recessive gene pink (p, eye color, located on chromosome III). A female from the homozygous black, pink, short stock was then mated to a wild-type male. The F₁ males of this cross were all wild type and were then backcrossed to the homozygous b, p, sh females. The F₂ results appeared as shown in the following table, and no other phenotypes were observed. (a) Based on these results, the student was able to assign sh to a linkage group (a chromosome). Determine which chromosome, and include step-by-step reasoning. (b) The student repeated the experiment, making the reciprocal cross: F₁ females back-crossed to homozygous b, p, sh males. She observed that 85 percent of the offspring fell into the given classes, but that 15 percent of the offspring were equally divided among b+p, b++, +shp, and +sh+ phenotypic males and females. How can these results be explained, and what information can be derived from these data?

Phenotype	Female	Male
wild	63	59
pink*	58	65
black, short	55	51
black, pink, short	69	60

*Pink indicates that the other two traits are wild type (normal). Similarly, black, short offspring are wild type for eye color.