Question

In Drosophila, the genes for body coloration and eye size are on different chromosomes. Normal-colored bodies are dominant to ebony-colored (very dark) bodies, and normal-sized eyes are dominant to eyelessness. Line A is true breeding for normal bodies and normal eyes, while line B is true breeding for ebony bodies and eyelessness. From an F2 cross between lines A and B, 800 flies are scored. How many F2 flies are expected to have normal body color and to be eyeless?
A) 9B) 36C) 125D) 225E) 300

Answer 1

***see pic***

In this problem we have two different genes that we are accounting for. Remember that for every gene we have have 2 alleles (alternative copies of the gene, each offspring gets 1 from each parent for a total of 2).  Therefore, when we are writing our genotypes for the parents/offspring we will be dealing with 4 total alleles (2 different genes x 2 alleles for each gene = 4 alleles in our genotype).

This 4-letter genotype makes our Punnet Square pretty hefty. It will be a 16-square PS (4-letter genotype of parent #1 x 4 letter-genotype of parent #2 = 16 boxes in PS).

Before setting up your Punnett Square, do the following:

1. Determine the genotypes of the parents. (***see pic for helpful notes***)

2. Determine the 4 possible combinations of the alleles for each gene. This part looks very confusing. It's not though! If you're familiar with "FOIL"ing in Algebra, that's basically all your doing. (***see pic for helpful notes, tried using red/black lines to distinguish between genes***)

3. Set up Punnett Square with 4 possible allele combinations for each gene.

4. Complete Punnett Square.

*Repeat Steps #1-4 because this problem asks us about F2 offspring, meaning the "second generation" of offspring"

5. Analyze offspring ratios.

ANSWER: I got 150 flies. Not sure if you accidentally wrote the wrong answer choice, or if you are just supposed to choose the closest answer choice, but either way it's C!