Introduction
FlyLab will allow you to play the role of a research geneticist. You will use FlyLab to study important
introductory principles of genetics by developing hypotheses and designing and conducting matings
between fruit flies with different mutations that you have selected. Once you have examined the results of
a simulated cross, you can perform a statistical test of your data by chi-square analysis and apply these
statistics to accept or reject your hypothesis for the predicted phenotypic ratio of offspring for each cross.
With FlyLab, it is possible to study multiple generations of offspring, and perform testcrosses and
backcrosses. FlyLab is a very versatile program; it can be used to learn elementary genetic principles such
as dominance, recessiveness, and Mendelian ratios, or more complex concepts such as sex-linkage,
epistasis, recombination, and genetic mapping.
Objectives
The purpose of this laboratory is to:
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Simulate basic principles of genetic inheritance based on Mendelian genetics by designing and
performing crosses between fruit flies.
understand the relationship between an organism's genotype and its phenotype.
Demonstrate the importance of statistical analysis to accept or reject a hypothesis.
Before You Begin: Prerequisites
Before beginning FlyLab you should be familiar with the following concepts:



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Chromosome structure, and the stages of gamete formation by meiosis
The basic terminology and principles of Mendelian genetics, including complete and incomplete
dominance, epistasis, lethal mutations, recombination, autosomal recessive inheritance, autosomal
dominant inheritance, and sex-linked inheritance.
Predicting the results of monohybrid and dihybrid crosses by constructing a Punnett square
How genetic mutations produce changes in phenotype, and beneficial and detrimental results of
mutations in a population
After doing this lab you should be able to:
 Investigate the independent assortment of two genes and determine whether the two genes are autosomal
or sexlinked using a multigeneration experiment
 Analyze the data from your genetic crosses using chi-square analysis
Procedure:
To begin an experiment, you must first design the phenotypes for the flies that will be mated. In addition
to wild-type flies, 29 different mutations of the common fruit fly, Drosophila melanogaster, are included
in FlyLab. The 29 mutations are actual known mutations in Drosophila. These mutations create
phenotypic changes in bristle shape, body color, antennae shape, eye color, eye shape, wing size, wing
shape, wing vein structure, and wing angle. For the purposes of the simulation, genetic inheritance in
FlyLab follows Mendelian principles of complete dominance. Examples of incomplete dominance are not
demonstrated with this simulation. A table of the mutant phenotypes available in FlyLab can be viewed
by clicking on the Genetic Abbreviations tab which appears at the top of the FlyLab homepage. When you
select a particular phenotype, you are not provided with any information about the dominance or
recessiveness of each mutation. FlyLab will select a fly that is homozygous for the particular mutation
that you choose, unless a mutation is lethal in the homozygous condition in which case the fly chosen will
be heterozygous. Two of your challenges will be to determine the zygosity of each fly in your cross and to
determine the effects of each allele by analyzing the offspring from your crosses.
One advantage of FlyLab is that you will have the opportunity to study inheritance in large numbers of
offspring. FlyLab will also introduce random experimental deviation to the data as would occur in an
actual experiment! As a result, the statistical analysis that you will apply to your data when performing
chi-square analysis will provide you with a very accurate and realistic analysis of your data to confirm or
refute your hypotheses.
 Use the buttons to choose a combination of genetic traits for the parent flies
 Wild type is the default, which represents a normal fly
 Set Offspring # to 1000 Offspring
 Click the button labeled “ Mate Designed Flies” to perform the cross
 If no mutation within a group is selected, both flies will be homozygous for the wild type for
all the mutations within that group
 Be sure that when you conduct a second cross or move on to another question, reset the
mutations to default setting for all other traits.
 Use the “Analyze Results” feature to collect quantitative data
 Record all qualitative and quantitative data in a Titled table (ex below)
 For each of the following crosses, identify the cross by number, present the results
(Parental, F1 and F2 generation) numbers and ratios for phenotype and genotype and
show a Punnett square for the cross
You will do 4 sets of crosses: [each cross will consist of a parental cross and F1 cross]
Cross #1 & 2 – Assigned monohybrid cross
Cross # 3 & 4 – design your own monohybrid cross
Cross # 5 – 8 Assigned dihybrid cross
Example:
Cross # _____
_________♀ x ________♂ (known P1 phenotypes)
_________♀ x ________♂ (expected P1 genotypes)
Punnett Square:
(Note: the # of squares will vary based on
type of gametes)
F1 Generation Data
Phenotype and Sex
Phenotype and Sex….
Phenotype and Sex
Phenotype and Sex….
Expected Number
Expected Ratio
Observed Number
Observed Ratio
F2 Generation Data
Expected Number
Expected Ratio
Observed Number
Observed Ratio
(Note: the # of columns to each table will vary based on types of offspring)
1. Is the mutation sex-linked or autosomal?
2. Is the mutation a dominant or recessive?
3. Are the deviations for the phenotypic ratio of the F2 generation within the limits expected
by chance? To answer this question, statistically analyze the data using the chi-square
analysis (Do this by hand, you may use the online source as reference only). Calculate the
chi-square statistic for the F2 generation in the chart below. Refer to Table 7.5 to determine
the p (probability) value that is associated with your χ2 statistic.
Phenotype
# observed
(o)
# expected
(e)
(o - e)2
(o - e)
(o - e)2
e
χ2
4.
5.
6.
7.
=
How many degrees of freedom are there?
What is the critical value for the # of degrees of freedom?
Is χ2 greater than, equal to, or less than this critical value (bold)?
Can you accept or reject your null hypothesis?
Table 7.5
←
Accept the null hypothesis
Degrees of
Freedom
1
2
3
4
5
0.90
0.016
0.21
0.58
1.06
1.61
Probability
0.50
0.25
0.46
1.32
1.39
2.77
2.37
4.11
3.36
5.39
4.35
6.63
0.10
2.71
4.61
6.25
7.78
9.24
Reject the null
→
hypothesis
0.05
3.84
5.99
7.82
9.49
11.07
0.01
6.64
9.21
11.35
13.28
15.09
Genetic Abbreviations : Geneticists working with Drosophila have worked out a convention for
labeling mutations. If the mutation is recessive, the abbreviation begins with a lowercase letter. If the
mutation is dominant, the first letter of the abbreviation is capitalized. For example, the dominant mutant
gene for sternopleural bristles is abbreviated as Sp and the recessive mutant gene for mahogany eyes is
abbreviated as mah. The normal "wild type" version of the gene is denoted using the same abbreviation as
the mutant version, but with a "+" superscript. For example, Sp+ and mah+. Sometimes, when the meaning
is obvious, a "+" by itself is used as a shorthand for the wild-type gene. For example, the heterozygote
genotypes +/Sp and +/mah. Different variations of the mutant gene (different alleles) are sometimes
distinguished by appending superscript abbreviations to the main abbreviation.
FlyLab does not follow this protocol. Since the idea behind FlyLab is for users to discover the genetic
characteristics of the mutations by "experimentation," no clues as to dominance or recessiveness are
provided by the use of standard abbreviations. Instead, all the letters of the abbreviation are capitalized. In
most cases, the letters used are the same as in the standard notation; however, in a few cases it was
necessary to modify the choice of letters because capitalization resulted in two different mutations (a
dominant and a recessive) with the same abbreviation. A table of the genetic abbreviations used in FlyLab
appears below.
Abbreviation
Mutant Phenotype
AP
Apterous Wings
AR
Aristapedia Antennae
B
Bar Eyes
BL
Black Body
BW
Brown Eyes
C
Curved Wings
CV
Crossveinless Wings
CY
Curly Wings
D
Dichaete Wings
E
Ebony Body
EY
Eyeless Eyes
DP
Dumpy Wings
F
Forked Bristles
L
Lobe Eyes
M
Miniature Wings
PR
Purple Eyes
RI
Radius Incompletus Wings
S
Sable Body
SB
Stubble Bristles
SD
Scalloped Wings
SE
Sepia Eyes
SN
Singed Bristles
SS
Spineless Bristles
ST
Star Eyes
SV
Shaven Bristles
T
Tan Body
VG
Vestigial Wings
W
White Eyes
Y
Yellow Body
Chi-square analysis
is a statistical method that can be used to evaluate how observed ratios for a
given cross compare with predicted ratios. Chi-square analysis considers the chance deviation for an
observed ratio, and the sample size of the offspring, and expresses these data as a single value. Based on
this value, data are converted into a single probability value (p=value), which is an index of the
probability that the observed deviation occurred by random chance alone. Biologists generally agree on a
p=value of 0.05 as a standard cutoff value, known as the level of significance, for determining if observed
ratios differ significantly from expected ratios. A p=value below 0.05 indicates that it is unlikely that an
observed ratio is the result of chance alone. When we predict that data for a particular cross will fit a
certain ratiofor example, expecting a 3:1 phenotypic ratio for a monohybrid cross between heterozygotesthis assumption is called a null hypothesis. Chi-square analysis is important for determining whether a
null hypothesis is an accurate prediction of the results of a cross. Based on a p=value generated by chisquare analysis, a null hypothesis may either be rejected or fail to be rejected. If the level of significance
is small (p < 0.05), it is unlikely (low probability) that the observed deviation from the expected ratio can
be attributed to chance events alone. This means that your hypothesis is probably incorrect and that you
need to determine a new ratio based on a different hypothesis. If, however, the level of significance is
high (p > 0.05), then there is a high probability that the observed deviation from the expected ratio is
simply due to random error and chi-square analysis would fail to reject your hypothesis.