Question

In Drosophila, assume that the gene for scute bristles (s) is located at map position 0.0 and that the gene for ruby

eyes (r) is at position 15.0. Both genes are located on the X chromosome and are recessive to their wild-type
alleles. A cross is made between scute-bristled females and ruby-eyed males. Phenotypically wild-type F1
females were then mated to homozygous double mutant males, and 1000 offspring were produced. Give the
frequency of scute expected.
A) 360 B) 500 C) 75 D) 425 E) 250

Answer 1

In Drosophila, when scute-bristled females are crossed with ruby-eyed males, the F1 generation is phenotypically wild-type. The F1 females are then mated with homozygous double mutant males, and the frequency of scute bristles in the offspring is expected to be 360 individuals.

Step-by-step explanation:

In Drosophila, the gene for scute bristles (s) is located at map position 0.0 and the gene for ruby eyes (r) is at position 15.0 on the X chromosome. Both genes are recessive to their wild-type alleles. When a cross is made between scute-bristled females and ruby-eyed males, the F1 generation will be phenotypically wild-type, meaning they will not show the scute bristles or ruby eyes. The F1 females, being heterozygous for both genes, will have the genotype XsXs XrXr. Then, when the F1 females are mated with homozygous double mutant males (XsXs XrXr), the expected frequency of scute bristles in the offspring is 360 individuals.

Answer 2

In an X-linked cross with scute-bristled females and ruby-eyed males, when F1 wild-type females are mated with homozygous double mutant males, 500 out of 1000 offspring are expected to show the scute bristles phenotype due to the 1:1 phenotype ratio for X-linked traits.

So, the correct answer would be B.

Step-by-step explanation:

In the cross described, we are dealing with X-linked traits in Drosophila melanogaster. Because both scute bristles and ruby eyes are recessive traits located on the X chromosome, their inheritance patterns follow the rules of X-linked genetics. The scute-bristled females (ss) are crossed with ruby-eyed males (rr). The wild-type F1 females will have one normal allele and one recessive allele for each gene, making them heterozygous carriers. These females (XsX+; XrX+) that are phenotypically wild-type are then mated with double mutant males (XsY; XrY).

Given Mendelian genetics and the absence of crossover between the genes, you would expect the phenotype ratios for the offspring to be:

• Normal bristles, normal eyes
• Scute bristles, normal eyes
• Normal bristles, ruby eyes
• Scute bristles, ruby eyes

Because the male parents are homozygous recessive for both traits, and scute bristles (s) is a recessive trait located at map position 0.0, we expect a 1:1 ratio due to the fact that females have two X chromosomes and there are two possible outcomes for the X-linked trait when crossed with a Y chromosome. Therefore, with 1000 offspring, 500 would be expected to exhibit the scute bristles phenotype, so the correct answer would be B) 500.