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

What is the CORRECT genetic map with respect to gene order and distances (in cM) for these three genes?
a) ct-s-v+: 2-8-20
b) ct-s-v+: 20-8-2
c) v-s-ct+: 2-8-20
d) v-s-ct+: 20-8-2

Answer 1

The correct option for the genetic map of the three X-linked genes in Drosophila melanogaster is option b) ct-s-v+: 20-8-2, with cut wings in the middle. This is determined by identifying the parental phenotypes and recombinant frequencies, with the lowest numbers indicating double recombinants, which help establish the gene order.

Step-by-step explanation:

The student is confronted with a classic problem of linkage mapping in Drosophila melanogaster, which involves determining the gene order and the map distances between three X-linked genes: cut wings (ct), sable body (s), and vermilion eyes (v). The student provided the number of progeny from the testcross that show different combinations of these traits. Mutations that are closely linked exhibit a lower frequency of recombination, and thus, appear together more frequently in the offspring. Conversely, mutations that are further apart show a higher frequency of recombination, meaning that the traits appear in different combinations more often.

To calculate the map distance and order of genes, we look for the parental (non-recombinant) and recombinant phenotypes. The highest numbers of progeny generally represent the parental phenotypes, whereas the lowest numbers represent double recombinants which could help us determine the gene order on the chromosome.

In the given data, the parental phenotypes are v ct s (510 progeny) and v+ ct+ s+ (500 progeny). This demonstrates that v and s are more closely linked to each other than either is to ct since the recombinant phenotypes involving these two genes (v ct s+ with 73 progeny and v+ ct+ s with 14 progeny) are fewer in number. The rarest classes, the double recombinants, are v+ ct s (1 progeny) and v ct+ s+ (1 progeny), which indicates that ct is in the middle.

The correct genetic map is therefore: v – 2 cM – ct – 8 cM – s, with ct being in the middle. This corresponds to option b) ct-s-v+: 20-8-2, which is the correct genetic map with respect to gene order and distances (in cM) for these three genes.