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

***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!

there is no stairs in bungalow

explanation:

poof