Question

In Drosophila melanogaster, cut wings (ct) is recessive to normal wings (ct+), sable body (s) is recessive to gray body (s+), and vermilion eyes (v) is recessive to red eyes (v+). All three recessive mutations are X-linked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The following are the progeny resulting from the testcross.

v ct s 510
v+ ct s 1
v+ ct+ s 14
v+ ct+ s+ 500
v+ ct s+ 73
v ct s+ 20
v ct+ s 81
v ct+ s+ 1
Total 1200
Determine the order of these genes on the chromosome.

Answer 1

The order of the genes on the chromosome is v-s-ct, with the vermilion eyes gene being the farthest from the centromere.

Step-by-step explanation:

In order to determine the order of the genes on the chromosome, we need to analyze the results of the testcross between the F1 females and the cut, sable, vermilion males.

1. First, let's look at the offspring with the genotype v ct s (510 individuals). This genotype means that all three recessive mutations are present.
2. Next, let's consider the offspring with the genotype v+ ct+ s+ (500 individuals). This genotype indicates that none of the recessive mutations are present.
3. Finally, we can examine the offspring with the genotype v+ ct s (1 individual) and v ct+ s (81 individuals). These genotypes indicate that only one of the recessive mutations is present, either cut wings or sable body, but not both.

Based on these results, we can determine that the order of the genes on the chromosome is v-s-ct, with the vermilion eyes gene being the farthest from the centromere.