In laboratory class, a genetics student was assigned to study an unknown mutation in Drosophila that had a whitish eye. He crossed females from his true-breeding mutant stock to wild-type (brick-red-eyed) males, recovering all wild-type F₁ flies. In the F₂ generation, the following offspring were recovered in the following proportions:
wild type: 5/8
bright red: 1/8
brown eye: 1/8
white eye: 1/8
The student was stumped until the instructor suggested that perhaps the whitish eye in the original stock was the result of homozygosity for a mutation causing brown eyes and a mutation causing bright red eyes, illustrating gene interaction. After much thought, the student was able to analyze the data, explain the results, and learn several things about the location of the two genes relative to one another. One key to his understanding was that crossing over occurs in Drosophila females but not in males. Based on his analysis, what did the student learn about the two genes?

In Drosophila, a female fly is heterozygous for three mutations, Bar eyes (B), miniature wings (m), and ebony body (e). Note that Bar is a dominant mutation. The fly is crossed to a male with normal eyes, miniature wings, and ebony body. The results of the cross are as follows.

111 miniature; 101 Bar, ebony

29 wild type; 31 Bar, miniature, ebony

117 Bar; 35 ebony

26 Bar, miniature; 115 miniature, ebony

Interpret the results of this cross. If you conclude that linkage is involved between any of the genes, determine the map distance(s) between them.

Two plants in a cross were each heterozygous for two gene pairs (Ab/aB) whose loci are linked and 25 mu apart. Assuming that crossing over occurs during the formation of both male and female gametes and that the A and B alleles are dominant, determine the phenotypic ratio of their offspring.

Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three pairs of autosomes, referred to as 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. No other phenotypes were observed.

Based on these results, the student was able to assign short to a linkage group (a chromosome). Which one was it? Include your step-by-step reasoning.

A female of genotype

produces 100 meiotic tetrads. Of these, 68 show no crossover events. Of the remaining 32, 20 show a crossover between a and b, 10 show a crossover between b and c, and 2 show a double crossover between a and b and between b and c. Of the 400 gametes produced, how many of each of the 8 different genotypes will be produced? Assuming the order a–b–c and the allele arrangement previously shown, what is the map distance between these loci?

Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three pairs of autosomes, referred to as 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.

The student repeated the experiment, making the reciprocal cross, with F₁ females backcrossed 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 + +, + sh p, and + sh + phenotypic males and females. How can these results be explained, and what information can be derived from the data?

A number of human–mouse somatic cell hybrid clones were examined for the expression of specific human genes and the presence of human chromosomes. The results are summarized in the following table. Assign each gene to the chromosome on which it is located.