Question

You are looking at multiple traits in the fruit fly, Drosophila melanogaster, all of which are Mendelian traits, each on different chromosomes, and have dominant and recessive expression as follows:

Trait	Dominant Phenotype	Recessive Phenotype
Eye color	Red eye (R)	White eye (r)
Antenna Length	Long (L)	Short (l)
Wing shape	Normal (N)	Curly (n)
Body Stripes	Black (B)	yellow (b)

1. If two parents are heterozygous for Eye color and Wing shape, what fraction of the offspring would be expected to have the following genotype: rrNn?  show your work, or if you did the work on scrap paper, explain how you got your answer (3 points)

1. If two parents are heterozygous for Antenna Length and Body Stripes, what fraction of the offspring would be expected to have the following phenotype: Long & Yellow ? show your work, or if you did the work on scrap paper, explain how you got your answer (3 points)

1. If two parents that are heterozygous for all four of these traits are crossed, what fraction of the offspring would you expect to have the following genotype: rr Ll Nn BB? show your work or if you did the work on scrap paper, explain how you got your answer (4 points)

Answer 1

Two things we need to know to solve these questions are:

1) Monohybrid ratio:

	R	r
R	RR	Rr
r	Rr	rr

Genotypic ratio: 1:2:1 - 1(RR) : 2 (Rr) : 1 (rr)

Phenotypic ratio: 3:1 - 3(red): 1 (white)

This applies to all the other traits.

2) AND rule of probability

If the chance of event A happening is x and the chance of event B happening is y, then the chance of the two events are happening simultaneously is the product of x and y = x*y.

Eye colour and wing shape:

Genotype: rrNn

rr: 1/4

Nn: 2/4 = 1/2

rrNn = 1/4 × 1/2 = 1/8

Antenna length and Body stripes:

Long and yellow genotype: LLyy, Llyy

LL + Ll = 3/4

yy = 1/4

L_yy = 3/4 × 1/4 = 3/16

Genotype: rrLlNnBB

rr: 1/4

Ll: 1/2

Nn: 1/2

BB: 1/4

1/4 × 1/2 × 1/2 × 1/4 = 1/64