Kati Schwarzkopf
BIO 205
December 7, 2012
Fruit Fly Lab Report
INTRODUCTION
From egg to adult, the fruit fly lifecycle lasts about eight to ten days. During that
time, the eggs hatch (around 30 hrs.), and for five to six days, the larvae feed and grow. As
the fruit fly continues to grow, it emerges from the larvae stage and into the pupal stage.
The pupal stage lasts about five to six days. The fruit fly may stay alive for another several
weeks once it has emerged into an adult fly. Upon emerging into an adult, female flies may
start laying eggs within 48 hours.
Source: http://www.generalpest-ohio.com/images/photoFruitFly3.jpg
The fruit fly is an ideal model organism for genetic studies for a number of reasons.
The fruit fly is small, easy to obtain, easy to maintain, have a short lifecycle, and also have a
small number of chromosomes. The fruit fly only has a total of eight chromosomes. They
have two sex chromosomes and six autosomes. Because of this low number of
chromosomes, it is easy to determine which traits are sex linked and which are autosomal.
It is known that the trait for eye color is sex linked/ x-linked. Genes are easily assigned to
each of the eight chromosomes and the order and arrangement of genes, including the
relative distance between genes can be determined, making fruit flies an ideal organism for
genetic studies.
There are multiple distinguishing characteristics that may be used to identify the
male flies from females. Female flies are generally larger than the males, but this is only a
general statement. Males have rounded abdomens with a dark pigmented tip, while the
abdomen of the female flies has clearly banded segments extending all the way to the tip
(see figure 3). The tip of the female flies is also more pointed and longer (see figure 2). One
factor that is very reliable for sexing fruit flies is the presence of a sex comb, dark bristles
located on the forelimbs of the flies. Only male flies will have a sex comb. This can be seen
in figure 1 below.
Figure 1:
Source: http://www.phschool.com/science/biology_place/labbench/lab7/design.html
Figure 2:
Figure 3:
There are two common mutations that we worked with in class. A common
autosomal mutation is for body color, tan body color is dominant to ebony body color. The
chromosome that the gene for body color is on is chromosome 3. A common sex linked/ xlinked mutation is for eye color, red eye color is dominant, while white eye color is
recessive. Other mutations in fruit flies include; short winged (recessive), curled winged
(dominant), yellow body (recessive), ebony body (recessive), orange eyed (recessive), and
eyeless (recessive).
MEATHODS
Procedural steps for making your fruit fly cross vial:
1. Obtain a clean, dry vial. Be careful not to contaminate the inside of the vial by
touching it.
2. Add one capful of the dry medium to the vial.
3. Add one capful of cold water to the vial.
4. Allow the medium to set up for about a minute.
5. Add a sprinkle of dry yeast to the vial.
6. Add a foam cap to cover your vial, again be careful not to contaminate your vial and
cap.
7. Label your vial with initials.
8. Obtain parent vials for your assigned crosses.
9. Using flynap, put parent flies to sleep by dipping the flynap wand into the liquid and
inserting the saturated wand between your foam cap and vial.
10. Wait for parent flies to fall asleep.
11. Once parent flies are napped, un cap your vials and sort out males from females of
each phenotype.
12. Obtain 3 – 5 males of assigned phenotype and 3 – 5 females of the other assigned
phenotype.
13. Record the number of flies of each phenotype for the parent generation.
14. Place these flies into your pre- made vial and cap.
15. Parent flies may be placed back in their original vial and put back into the incubator.
16. Your vial with the cross you just made may also be placed in the incubator.
** It takes about two weeks for flies to mate and offspring to emerge.
Week 2:
1.
2.
3.
4.
Obtain original cross vial from incubator.
If offspring larvae are apparent, put parent flies asleep, using flynap procedure.
Remove parent flies and put them in water… a.k.a. the morgue.
Re- cap vial and place back into the incubator.
Week 3:
1.
2.
3.
4.
5.
6.
Obtain vial with flies from the incubator.
Put flies to sleep using the flynap procedure.
Once flies are asleep remove cap from vial and shake flies out onto a piece of paper.
Count the number of flies of each phenotype, looking at flies under microscope.
Record results.
Return flies to vial and back to incubator.
Week 4:
Repeat steps from week 3.
Overview of procedure by week:
Source: http://www.phschool.com/science/biology_place/labbench/lab7/flyover.html
For the crosses done in class, here are the expected phenotype ratios:
Cross 1: Tan and Ebony
On what chromosome is this mutation?
Chromosome 3
Autosomal or Sex-Linked?
Autosomal recessive
Parent Cross: EE x ee
E
e
e
E
Ee
Ee
Ee
Ee
Expected Phenotype Ratio of the F1 offspring
100% tan
F1 Cross: Ee x Ee
E
E
e
e
EE
Ee
Ee
ee
Expected Phenotype Ratio of the F2 offpsring
3 tan : 1 ebony
Cross 2: Red eye and white eye
On what chromosome is this mutation?
Autosomal or Sex-Linked?
Sex- Linked
Parent Cross:
Chromosome
XR
Xr
Xr
Y
XRXr
XrY
XRXr
XrY
Expected Phenotype Ratio of the F1 offspring
2 red eyed females : 2 white eyed males
F1 Cross:
Xr
XR
Xr
Y
XRXr
XRY
XrXr
XrY
Expected Phenotype Ratio of the F2 offpsring
1 red eyed female : 1 white eyed female : 1 red eyed male : 1 white eyed male
RESULTS
Tan Male vs Ebony Female
Tan Male
Parents
Tan Female
Ebony Male Ebony Female
3
5
F1 Generation
15
35
30
19
F2 Generation
23
50
29
39
CHI – SQUARE ANALYSIS: F1
1. Calculate Expected Phenotype Ratio
Expected Phenotype Ratio to be:
100% tan
2. Collect Actual Data
F1 offspring for this cross. The observed number of offspring in each phenotypic class was:
50 tan
49 ebony
99 Total Flies
3. Calculate Expected numbers for each phenotype
If the expected phenotype ratio is 100% tan, and we student counted a total of 99 flies, we
can calculate the theoretical expected numbers for each phenotype.
First, divide the total number of counted individuals (99) by the total of the integers 99/4 =
24.75
Now multiply the integer for each class by the number obtained in the last step:
24.75 X 4 = 99
24.75 X 0 = 0
The total number of individuals should now match the number actually counted: 99 + 0 =
99
This means that if the expected phenotype ratio is 4:0, and there are a total of 99 flies, then
99 of them should be tan, and 0 of them should be ebony.
4. Now you’re ready for the Chi-square analysis:
X2 = (observed - expected) 2 / expected
Tan:
X2 = (50 - 99) 2 / 99
X2 = (-49) 2 / 99
X2 = 2,401/99
X2 = 24.25
Ebony:X2 = (49 - 0) 2 / 0
X2 = (49) 2 / 0
X2 = 2,401 / 0
Total Chi Square = 24.25
5. Look up this number, 24.25, in the Chi Square Table. You will need two additional
numbers to do this:
a. degreees of freedom
df = # of phenotypes – 1
df = 2 – 1 = 2
b. p value: This is the probability that the deviation of the observed from that
expected is due to chance alone (no other forces acting). The standard accepted p
value is p > 0.05. If p>0.05, then any deviation is due to chance alone 5% of the time
or less.
CHI – SQUARE ANALYSIS: F2
1. Calculate Expected Phenotype Ratio
Expected Phenotype Ratio to be:
3 tan : 1 ebony
2. Collect Actual Data
F2 offspring for this cross. The observed number of offspring in each phenotypic class was:
73 tan
68 ebony
141 Total Flies
3. Calculate Expected numbers for each phenotype
If the expected phenotype ratio is 3 tan: 1 ebony, and we counted a total of 141 flies, we can
calculate the theoretical expected numbers for each phenotype.
First, divide the total number of counted individuals (141) by the total of the integers
141/4 = 35.25
Now multiply the integer for each class by the number obtained in the last step:
35.25 X 3 = 105.75
35.25 X 1 = 35.25
The total number of individuals should now match the number actually counted: 105.75 +
35.25 = 141
This means that if the expected phenotype ratio is 3:1, and there are a total of 141 flies,
then 105.75 of them should be tan, and 35.25 of them should be ebony.
4. Now you’re ready for the Chi-square analysis:
X2 = (observed - expected) 2 / expected
Tan:
X2 = (73 – 105.75) 2 / 105.75
X2 = (-32.75) 2 / 105.75
X2 = 1,072.6 /105.75
X2 = 10.14
Ebony:X2 = (68 – 35.25) 2 / 35.25
X2 = (32.75) 2 / 35.25
X2 = 1,072.6 / 35.25
X2 = 30.43
Total Chi Square = 10.14+ 30.43 = 40.57
5. Look up this number, 40.57, in the Chi Square Table. You will need two additional
numbers to do this:
a. degreees of freedom
in this example:
df = # of phenotypes – 1
df = 2 – 1 = 1
b. p value: This is the probability that the deviation of the observed from that
expected is due to chance alone (no other forces acting). The standard accepted p
value is p > 0.05. If p>0.05, then any deviation is due to chance alone 5% of the time
or less.
Degrees of
Freedom
(df)
Probability (p)
0.95
0.90
0.80
0.70
0.50
0.30
0.20
0.10
0.05
0.01
0.001
1
0.004
0.02
0.06
0.15
0.46
1.07
1.64
2.71
3.84
6.64
10.83
2
0.10
0.21
0.45
0.71
1.39
2.41
3.22
4.60
5.99
9.21
13.82
3
0.35
0.58
1.01
1.42
2.37
3.66
4.64
6.25
7.82
11.34
16.27
4
0.71
1.06
1.65
2.20
3.36
4.88
5.99
7.78
9.49
13.28
18.47
5
1.14
1.61
2.34
3.00
4.35
6.06
7.29
9.24
11.07
15.09
20.52
6
1.63
2.20
3.07
3.83
5.35
7.23
8.56
10.64
12.59
16.81
22.46
7
2.17
2.83
3.82
4.67
6.35
8.38
9.80
12.02
14.07
18.48
24.32
8
2.73
3.49
4.59
5.53
7.34
9.52
11.03
13.36
15.51
20.09
26.12
9
3.32
4.17
5.38
6.39
8.34
10.66
12.24
14.68
16.92
21.67
27.88
10
3.94
4.86
6.18
7.27
9.34
11.78
13.44
15.99
18.31
23.21
29.59
Not a Significant Difference between expected and observed
Significant Difference
DISCUSSION
For both the chi square analyses of the F1 and F2 offspring, there was a very
significant difference between the expected ratio and the observed ratio. For both crosses,
the number from the probability table that we are comparing to is 3.84. For the F1
generation, the calculated number was 24.25, and for the F2 generation, our calculated
number was 40.57. These two calculations are much greater than the expected number of
3.84, meaning there is a significant difference between our expected and observed ratios,
and that this difference is too great to be by chance.
The expected ratio for our cross of tan body color and ebony body color (the only
cross that we were able to keep alive), did not at all match the observed ratio. This may
have occurred from a number of different factors. Some of the sources of error or
miscalculation that could account for this difference may have been emergence and mating
of offspring before counts were established and made, and failure to correctly distinguish
body color when collecting data. The difference between the tan body color and the ebony
body color was slight, and at times difficult to distinguish between when the two different
colored flies were not next to each other. This could have greatly skewed our results as our
numbers could be completely incorrect. Also, I believe there was one week that we did not
count the flies from our vial, as there was little progress of growth. This may have also
negatively affected our results, as new offspring may have emerged before we got the
chance to count the flies again. If this occurred then there would have been a mixture of the
F1 and F2 generations in the offspring that we were counting and forming data tables on.
This could have been a possibility, as the lifecycle of a fruit fly is very short, and it is
plausible that a new generation could have emerged within that time frame.
We had originally started with a cross between white eyes and red eyes. We started
with 4 red eyed male flies and 3 white eyed female flies. After the first week the only living
fruit fly within this cross was one white eyed female. We removed this female fly from the
contaminated vial and added her to a new vial with an additional white eyed female and 2
red eyed males. Unfortunately, we were unable to maintain life within this cross. The
reasons for this may have been a severe contamination that prevented offspring from
emerging, and ultimately caused the parent generation to die off.
If I had the opportunity to perform this lab again, I would change a number of
things. To start, I would be especially careful as to not contaminate the vials. To ensure this,
I believe that wearing latex gloves would be necessary. Also, I believe that there should be
greater monitoring of the cross vials throughout the week. I believe that these things in
combination would yield better, more reliable results.