Kevin D. McMahon
Reseda Science Magnet
AP Biology Laboratory: Genetics of Drosophila
Overview
In this laboratory you will use a computer
simulation to perform crosses of fruit
flies in order to study various patterns of
inheritance. You will collect data from
F1 and F2 generations, and analyze the
results from a monohybrid and sex-linked
cross.
Objectives
At the completion of this laboratory you should be able to:
* discuss the life cycle of the fruit fly, recognize the sex of fruit flies, and recognize
several types of classic mutations.
* compare predicted results with actual results
* explain the importance of chi-square analysis,
Introduction
Drosophila melanogaster, the fruit fly, is an excellent organism for genetics studies
because it has simple food requirements, occupies little space, is hardy, completes its life
cycle in about 12 days at room temperature, produces large numbers of offspring, may be
immobilized readily for examination and sorting, and has many types of hereditary
variations that can be recognized with low-power magnification. Drosophila has a small
number of chromosomes (four pairs). These chromosomes are easily located in the large
salivary gland cells. Drosophila exists in stock cultures that can be readily obtained from
several sources. Much research about the genetics of Drosophila over the last 50 years
has resulted in a wealth of knowledge about its genes.
The Life Cycle of Drosophila
It is important to realize that a number of
factors determine the length of time of each
stage in the life cycle. Of these factors,
temperature (about 25C), the complete cycles
takes 10 to 12 days.
The egg: The eggs are small and ellipsoid and
have two filaments at one end. They are
usually laid on the surface of the culture
medium and, with practice, can be seem with
the naked eye. The eggs hatch into larvae after
about a day.
The larval stage: The wormlike larva eats
almost continouously, and its black mouth
parts can easily be seen moving back and forth
even when the larva itself is less distinct.
Larvae channel through the culture medium
while eating; thus channels are a good
indication of the successful growth of a culture.
The larva sheds its skin twice as it increases in
size. In the last of the three larval stages, the
cells of the salivary glands contain giant
chromosomes, which may be seen readily
under low-power magnification after proper
staining.
The pupal stage: When a mature larva in a laboratory culture is about to become a pupa,
it usually climbs up the side of the culture bottle or onto the paper strip provided in the
culture. The last larval covering then becomes harder and darker, forming the pupa case.
Through this case the later stages of metamorphosis to an adult fly can be observed. In
particular, the eyes, the wings, and the legs become readily visible.
The adult stage: When metamorphosis is complete, the adult flies emerge from the pupa
case. They are fragile and light in color and their wings are not fully expanded. These
flies darken in a few hours and take on the normal appearance of the adult fly. They live a
month or more and then die. A female does not mate for about 10 to 12 hours after
emerging from the pupa. Once she has mated, she stores a considerable quantity of sperm
in receptacles and fertilizes her eggs as she lays them. Hence, to ensure a controlled
mating, it is necessary to use females that have not mated before.
Procedure
Normally, an experiment using
Drosophila would take several
weeks to perform. To expedite
your learning experience you will
use a computer simulation (This
simulation is adapted from the
program "HyperFly" by Thomas
H. Thelen; a program available
for Macinthosh computers).
Obviously, the procedures you
will use here will be different
than those used if you were
actually crossing live Drosophila,
however the learning experience
will be very similar.
As you cross these flies keep
careful records of the types of
crosses you perform and the
offspring you obtain. Use this
information to describe the how
this particular trait is inherited (ie:
autosomal dominant, X-linked
recessive, codominant, etc.). An
example cross is demonstrated
here. Now use HyperFly to
perform the following crosses
making sure that you document all
your results in your lab manual.
Select two traits from each section
below;
A. (1) Cracked Back
(2) Curled
(3) Downy
(4) Ebony Eyes
B. (1) Big Combed
(2) Broad Banded
(3) Clipped
(4) Stripped
C. (1) Flarred
(2) Furrowed
(3) Horned
(4) Hairless
of
Chi-Square Analysis
When you performed Drosophila crosses on the computer you observed that the results
you obtained did not exactly correspond to the results you expected. Deviation from
expected results may be due to chance, experimental error, or they may be related to a
hereditary phenomenon you have not accounted for.
To determine the cause of this deviation a chi-square (X2) analysis may be used. The
formula for chi-square is:
where O = the observed number of offspring in a phenotypic category, E = the expected
number of offspring in that phenotypic category.
Consider a cross between a wild-type female and a
"veinless" male (notice the veins in the wings of
the female).
The F1 results were as follows:
46 veined females & 54 veined males.
Based upon the previous cross it is hypothesized
that the vein trait is autosomal dominant. To test this hypothesis, F1's were crossed with
the following results:
35 veined female & 38 veined males = 73 veined
14 veinless females & 13 veinless males = 27 veinless
The expected results would have been 75 veined and 25 veinless.
Is this deviation from expected results due to chance? To determine this the chi-square
statistic is calculated as follows:
From the table of values, it can be determined that the value of at a 0.05 probability level
and 1 degree of freedom is 3.841 (The number of degrees of freedom is equal to the
number of phenotypic categories minus 1. There are 2 phenotypic categories, veined &
veinless, therefore the degrees of freedom is 1). Since the calculated statistic (0.213) for
the data in this example is less than 3.841, the null hypothesis is accepted. The "null
hypothesis" in this experiment is that there is no difference between "expected results"
and "observed results," that is, that any differences that do exist are due to random chance
alone.
Since the calculated value of 0.213 falls between the values of 0.455 and 0.0642 in the
table, it can be concluded that there is a 50 to 80% probability that the deviations of the
observed values from expected values are due to random chance alone.
Now calculate the Chi-Square values for the F1 X F1 crosses for the each of the traits you
chose to study. Is the null hypothesis satisfied for each cross? What are the probabilities
that the deviations of the observed values from expected values due to random chance
alone?