Question

In Drosophila (fruit flies) the genes how, dumpy and ebony are located on chromosome 3. LOF = loss of function.

• Flies homozygous for a LOF mutation (no gene product made) in ebony have dark black bodies.
• Flies homozygous for a LOF mutation (no gene product made) in dumpyhave truncated (short) wings.
• Flies homozygous for a partial LOF mutation (some gene product made but significantly less than normal) in how have wings that will not fold down (held out wings; that's where the gene name is coming from, how).

Use the following allele/phenotype designations:

For how

• allele h+ = wild type/normal
• allele h = partial loss of function mutation
• h+ > h
• phenotype h+ = wild type = wings that fold, which is dominant to held out wings = phenotype h

For dumpy

• allele dp+ = wild type/normal
• allele dp = LOF mutation
• dp+ > dp
• phenotype dp+ = wild type = normal wing size, which is dominant to truncated wings = phenotype dp

For ebony

• allele eb+ = wild type/normal
• allele eb = LOF mutation
• eb+ > eb
• phenotype eb+ = wild type = normal body colour, which is dominant to black body = phenotype eb

You cross a wild type female fruit fly that is a heterozygous carrier for all three mutant alleles with a black bodied male who has truncated wings that do not fold down.

You observe their 504 offspring and note their phenotypes and the number of offspring with each phenotype:

• wings that do not fold - 129
• truncated wings - 83
• black body - 6
• wings that do not fold and a black body - 80
• truncated wings that do not fold - 10
• truncated wings and a black body - 135
• truncated wings that do not fold, and a black body - 15
• wild type - 14

Note: that non-mutant phenotypes are generally ignored. For example a phenotype of "black body" means a fly with a black body, normal wing length and wings that fold.

Q. What option correctly represents the genotype of the heterozygote parent, with genes in the correct order?

Select one:

a. h dp eb / h+ dp+ eb+

b. eb+ h dp+ / eb h+ dp

c. h dp+ eb / h+ dp eb+

d. h dp+ eb+ / h+ dp eb

Answer 1