Fruit Fly Genetics Lab
NAME________________________________________ Period______
In this virtual lab we will cross various fruit flies to see what phenotypes
are present in the F1 and F2 generation. Using the data from these
crosses, we will make a hypothesis regarding the genotypes of the
parental (P) generation and test the hypothesis using a chi square
analysis.
Background Information
Drosophila melanogaster is a fruit fly, a little insect about 3mm long, of
the kind that accumulates around spoiled fruit. It is also one of the most
valuable of organisms in biological research, particularly in genetics and
developmental biology. Drosophila has been used as a model organism for research for almost a
century, mainly because it is practical: it's a small animal, with a short life cycle of just two weeks, and
is cheap and easy to keep large numbers. Mutant flies, with defects in any of several thousand genes
are available, and the entire genome has recently been sequenced.
Terminology
Wild-type - flies that have the "normal" characteristics, red eyes, normal length wing and brown
bodies.
Mutant flies - any variation from the wild type. Mutant alleles can be carried on autosomes or sex
chromosomes.
Genetic Notation
Mutant flies are given names that generally denote the type of mutation the fly exhibits. For example,
the mutant "ebony" has a much darker body than the wild type fly. Each mutation is also given a letter
code. Thus, in the case of ebony, the code is a lower case e. The wild type fly is denoted by a
superscript + over the mutant letter code. For example, e+ denotes a wild type fly for the ebony body
trait - meaning it has normal body color (not ebony). The above description is for a gene located on
an autosome (a non-sex chromosome). Of course, fruit flies also have sex chromosomes and they
contain a subset of genes as well. The genotypic notation for a mutant gene for white eye color on the
X chromosome would look like:
Xw Xw = white-eyed female
Xw Y = white-eyed male
Xw+ Xw = wild type heterozygote female
Xw+ Y = wild type male
Fruit Fly Life Cycle
1. Why are fruit flies a good subject for genetic studies?
2. Describe the life cycle of Drosophila melanogaster
3. What tools would you need to study fruit flies in the lab?
4. Show the abbreviations that would be used to denote a fly with dumpy (dp) wings and a wild type
fly that is carrying the allele for dumpy wings. The allele is not located on the sex chromosome.
5. Describe the differences between male fruit flies and female fruit flies.
6. How many male fruit flies were present in the simulation? ______ How many females? ______
Use the virtual simulations online at
to complete the following questions.
http://www.biologycorner.com/fruitflygenetics/index.html
Cross 1: Wild Type Female x Vestigial Winged Male
Wild Type Males
Vestigial Winged Males
Wild Type Females
Vestigial Winged Females
F1 Generation
____________
____________
____________
____________
F2 Generation
___________
___________
___________
___________
Hypothesis:
Chi Square Analysis - Use the space below to figure out the expected ratios
df = _________________
prob = 0.05___________
critical value = ________
sum of chi square = ____
Accept or reject hypothesis _________________________________________________________
Cross 2: White Eyed Female x Wild Type Male
Wild Type Males
White eyed Males
Wild Type Females
White Eyed Females
F1 Generation
____________
____________
____________
____________
F2 Generation
___________
___________
___________
___________
Hypothesis:
Chi Square Analysis - Use the space below to figure out the expected ratios
df = _________________
prob = 0.05___________
critical value = ________
sum of chi square = ____
Accept or reject hypothesis _________________________________________________________
Cross 3: Wild Type Female x Arisapodia, Vestigial Winged Male
Hypothesis: What is the genotype of the parents and how is aristapedia and vestigial wings inherited
(autosomal or sex linked)
Chi Square Analysis Use the space below to figure out the expected ratios
df = _________________
prob = 0.05___________
critical value = ________
sum of chi square = ____
Accept or reject hypothesis _________________________________________________________
The Chi-Square Test
An important question to answer in any genetic experiment is how can we decide if our data fits any
of the Mendelian ratios we have discussed. A statistical test that can test out ratios is the Chi-Square
or Goodness of Fit test.
Chi-Square Formula
Degrees of freedom (df) = n-1 where n is the number of classes
Let's test the following data to determine if it fits a 9:3:3:1 ratio.
Our hypothesis is ALWAYS the null hypothesis (Ho) which states that there is no difference between
the expected and observed data – i.e. the ratio is 9:3:3:1
Observed Values
315 Round, Yellow Seed
108 Round, Green Seed
101 Wrinkled, Yellow Seed
32 Wrinkled, Green
556 Total Seeds
Number of classes (n) = 4
Expected Values
(9/16)(556) = 312.75 Round, Yellow Seed
(3/16)(556) = 104.25 Round, Green Seed
(3/16)(556) = 104.25 Wrinkled, Yellow
(1/16)(556) = 34.75 Wrinkled, Green
556.00 Total Seeds
df = n-1 + 4-1 = 3
Chi-square value = 0.47
In Chi-Square table below, at df = 3 and we see the probability of 0.05, the critical value is 7.82.
By statistical convention, we use the 0.05 probability level as our critical value.
If the calculated value is less than the critical value, we accept the hypothesis.
If the chi-square value is greater than the critical value, we reject the hypothesis.
Therefore, because the calculated chi-square value (0.47) is less than 7.82 we accept the hypothesis
that the data fits a 9:3:3:1 ratio.
A Chi-Square Table
Degrees of
Freedom
1
2
3
4
5
0.9
0.02
0.21
0.58
1.06
1.61
Probability
0.5
0.1
0.05
0.46 2.71 3.84
1.39 4.61 5.99
2.37 6.25 7.82
3.36 7.78 9.49
4.35 9.24 11.07
0.01
6.64
9.21
11.35
13.28
15.09