Download Fruit Fly Lab - District 196 e

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Name____________________
Hour_______
DROSOPHILA LABORATORY EXERCISES
INTRODUCTION (PART I)
"Time flies like an arrow, fruit flies like a banana" Marx Brothers
The common fruit fly, Drosophila melanogaster is a mild household pest in most parts of the world living on decaying fruit
such as grapes, plums, and bananas. It has been used in genetics studies for many years as they are easy to keep alive,
at first being fed mashed bananas and kept in 1/2 pint milk bottles by Thomas Hunt Morgan. They have the further
advantage of completing their life cycle in 10 to 14 days at room temperature, and have only four pairs of chromosomes.
They have many easily recognized traits. The chromosomes have been mapped extensively so that the location of many
traits on the chromosomes is known.
The insect goes through metamorphosis as do many insects. Four stages are included in the life cycle: egg, larva, pupa,
and adult. Females can be fertilized about 10 hours after they emerge from the pupa, and they can lay eggs two days
after they emerge. The larva appear as small, white, worm-like organisms crawling through the media continuously
feeding with black mouthparts. The channels through the media, indicate that the larvae are alive and "working". When
ready to pupate, the larva will usually crawl up onto the side of the bottle and form a hard, darker covering. When the
adult flies emerge, they are fragile, light in color, with the wings not fully expanded. They take on the normal adult
appearance within a few hours. They live about a month and then die of old age.
A female fruit fly can mate within 10 hours after emerging and she can store sperm from the male and fertilize eggs as
she lays them. Therefore, if an experimenter needs to make a cross between two flies and needs virgin females to be
sure who is the father of the offspring, s/he will need to remove females within 10 hours after emerging from the pupa.
You will not need to worry about this aspect as we want the males and females of the same generation to mate.
PROCEDURE
To examine flies properly, they must be inactivated. Follow the steps demonstrated to inactivate your flies. Once the flies
are asleep, remove them from the vial by gently tapping the vial onto a piece of white paper. Flies will sometimes twitch
their legs and wings while asleep. You will need to examine your flies to make sure that you have some of each gender
and to look at the specific traits. You will then place them in your own labelled vials, leaving them on their side until the
flies are flying about so they do not get stuck in the potato flakes. The accompanying diagrams and the instructor's advice
should help you to tell the difference between male and female.
DROSOPHILA INHERITANCE PATTERNS
INTRODUCTION (PART II)
Scholars from the days of the ancient Greeks until the early 1900's believed the traits in offspring were often the average
of the parental traits, a sort of blend. Gregor Mendel tested this belief by experimenting with pea plants. When tall pea
plants were cross-bred with short pea plants, the results were all tall plants. Scientists of Mendel's time would have
predicted that all of the offspring would be of medium height. In this experiment we will reinforce what we have learned
about how traits are inherited and how Mendel's results dispelled previous beliefs about inheritance.
The first mating (parent generation) was done by a biological supply company and what we have received is the offspring
of that cross. The parents of that cross are pure-breeding (homozygous) individuals and we will call that generation the P1
generation. P stands for parent. Three different crosses were made, dealing with one trait (monohybrid), two traits
(dihybrid), and one with sex-linked traits. The offspring of the parents we will designate the F1 (F = filial which means
family) generation. We will then mate members of this generation with each other to produce the F2 generation. It is those
offpring that we will count and analyze.
You will be asked to hyothesize what you think your results will be in each of the three crosses and you will then collect
and analyze data in order to explain your hypothesis.
PROCEDURE: MONOHYBRID CROSS
Monohybrid means that we are dealing with one trait. The trait that we have chosen is eye color. The parent generation
which was crossed by the supply company was between flies of wild type color and flies of sepia color. Wild type is the
normal color and is generally described as red. Sepia is a recessive trait that is the result of a mutation that occurred
sometime in the past and which has been preserved by experimenters. The gene for it is found on chromosome # 3. It is
generally described as brown. We will use R for red eyes and r for sepia eyes.
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P1
Wild type (RR)
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Sepia eye (rr)
FIRST FILIAL GENERATION
The offspring between two pure-breeding strains is called a hybrid or the F1 generation. You will answer the follwing
questions after looking at and separating the F1 flies.
1. Do any of the offspring have a blend of parental traits or do they show one of the parentʼs traits? _________________
2. How many different phenotypes are in the F1s?___________What are they?_____________________
3. Is the mutant dominant or is it recessive?____________. The wild type? _________________________
4. Write the genotypes of the F1 parents.
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F1 !
X
HYPOTHESIS
You will need to write a hypothesis predicting what you will find in the F2 generation of flies. Your hypothesis should
include a specific prediction of what you will find in terms of numbers (ratios) and types of flies and should be a
complete sentence. This must be done before the F2 flies are counted.
In the space below, draw a Punnett Square to help you develop your hypothesis:
When the the F2 flies have hatched two weeks later, count the offspring and place the results in the table below.
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YOUR F2 RESULTS
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Wild Type (Red)
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Mutant Type (Sepia)
Collect data from the class and combine their data in the chart below.
CLASS F2 RESULTS
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Mutant Type (Sepia)
Convert the class results into percentages: (Show your work!)
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% wild =
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% sepia =
Wild Type (Red)
PROCEDURE: DIHYBRID CROSS
Dihybrid means that we are dealing with two traits. The traits that we have chosen are eye color and wing shape. The
parent generation, which was crossed by the supply company, was between flies of wild type color and vestigial wing
shape and flies of sepia color and wild type or normal wing shape. Wild type wings are the normal shape and vestigial
wings withered and useless for flight. This type of wing is the result of a mutation that also occurred in the past and has
been preserved for use in genetic experiments. We will use N for normal wing and n for a vestigial wing.
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P1
Wild type (RRnn)
x
Sepia eye (rrNN)
FIRST FILIAL GENERATION
The offspring between two pure-breeding strains is called a hybrid or the F1 generation. You will answer the follwing
questions after looking at and separating the F1 flies.
1. Do any of the offspring have a blend of parental traits or do they show one of the parentʼs traits? _________________
2. How many different phenotypes are in the F1's?___________What are they?_____________________
3. Is the vestigial wing dominant or is it recessive?____________. The wild (normal) type? ______________________
4. Write the genotypes of the F1 parents.
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F1 !
X
HYPOTHESIS
You will need to write a hypothesis predicting what you will find in the F2 generation of flies. Your hypothesis should
include a specific prediction of what you will find in terms of numbers (ratios) and types of flies and should be a
complete sentence. This must be done before the F2 flies are counted.
In the space below, draw a Punnett Square to help you develop your hypothesis:
When the the F2 flies have hatched two weeks later, count the offspring and place the results in the table below.
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YOUR F2 RESULTS
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Wild (red)eye, Wild (norm-wing
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al) wing
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Wild (red) eye, Vestigial
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wing
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Sepia eye, Wild (normal)
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wing
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Sepia eye, Vestigial wing
CLASS F2 RESULTS
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Wild (red)eye, Wild (norm!
al wing)
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Wild (red) eye, Vestigial
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wing
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Sepia eye, Wild (normal)
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wing
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Sepia eye, Vestigial wing
Calculate the percentage of the total that each of the four different phenotypes makes up.
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% wild eye, wild wing =
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% wild eye, vestigial wing =
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% sepia eye, wild wing =
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% sepia eye, vestigial wing =
PROCEDURE: SEX-LINKED TRAITS
Sex-linked traits are those traits in which the gene is carried on the X chromosome. In fruit flies there are three pairs of
autosomes and one pair of sex chromosomes, the X and Y. Females carry two X chromosomes and the males carry an X
and a Y. This is the same in humans for sex determination. The white eye color is a mutation determined by information
found on the X chromosome. We will write the sex chromosomes for each and a superscript for the eye color. We will use
R for red eye color and r for white eye color.
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P1
Red-eyed male (XRY)
x
White-eyed female (XrXr)
FIRST FILIAL GENERATION
The offspring between two parents is called the F1 generation. You will answer the follwing questions after looking at
and separating the F1 flies.
1. How many different phenotypes are in the F1's?___________What are they?_____________________________
2. Is the red eye dominant or recessive? __________________________ The white eye? _____________________
3. Write the genotypes of the F1 parents.
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F1 !
X
HYPOTHESIS
You will need to write a hypothesis predicting what you will find in the F2 generation of flies. Your hypothesis should
include a specific prediction of what you will find in terms of numbers (ratios) and types of flies, including the gender. You
must write a complete sentence. This must be done before the F2 flies are counted.
In the space below, draw a Punnett Square to help you develop your hypothesis:
When the the F2 flies have hatched two weeks later, count the offspring and place the results in the table below.
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YOUR F2 RESULTS
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Red eye, male
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Red eye, female
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White eye, male
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White eye, female
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Red eye, male
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Red eye, female
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White eye, male
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White eye, female
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CLASS F2 RESULTS
Calculate the percentage for the class results of each of the four different phenotypes.
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% red-eyed male =
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% red-eyed female =
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% white-eyed male =
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% white-eyed female =
ANALYSIS AND CONCLUSIONS
ANALYSIS
You now have results from three different genetics experiments. It is doubtful whether any of the results are identical to
the expected results. Therefore, we need to ask if the results are close enough to the expected to be accepted. If they
are close enough to the expected then we can conclude that the results are due to genetic factors only, such as chance. If
the results deviate too much from the expected, then we must conclude that there is something else besides genetic
factors influencing our results. The question you may be wondering is "How close is close enough?"
In 1900 Englishman Karl Pearson invented a mathematical device to determine whether a result deviates too far from the
expected. This measure of variation is called the chi square and is calculated by the following equation:
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chi square (x2)=
sum of all (expected - observed)2, where the (expected - observed) is the
expected! same as the deviation
or d.
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Do the following problems as practice:
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In squash, white is dominant to yellow. Heterozygous white squash are crossed with yellow squash and they
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produce 110 white seeds and 90 yellow seeds. Evaluate the data using the Chi square analysis.
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What if the results were 1100 white and 900 yellow?
Example: If you expect (hypothesize) 50 of 100 flies to have sepia eyes and you observe 55 of 100 flies to
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have sepia eyes and 45 to have wild type eyes, is the deviation from the expected enough to cause you to
reject ! the hypothesis?
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chi square (x2) = ( 50 - 55)2
+ (50 - 45)2
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=
25
50
+
25
50
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=
.5
+
.5
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50
chi square (x2)
=
1.00
50
Use the chart to determine whether the results are acceptable. First find the degrees of freedom by subtracting one from
the number of phenotypes, in this case it would be 2 - 1 = 1 degree of freedom. Then find where the calculated chi
square value of 1.00 would fit in to determine how frequently the results would be expected to be caused by chance.
From the table, you can see that a chi square value of 1.00 means that the probability is 30 - 50% that the deviation is
due to chance alone. Scientists have arbitrarily decided that when the probability of an event occurring by chance alone is
greater than or equal to 5% then the deviation is considered to be statistically insignificant and the results will be
acceptable. Therefore we may accept our hypothesis that we would expect 50% sepia and 50% red eyes. Our
hypothesis may not be correct, but our results agree with it.
CHI SQUARE TABLE
Analyze the class data from each of the genetics experiments conducted with the fruit flies. SHOW ALL WORK IN
DOING YOUR CALCULATIONS!!!
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F2's of monohybrid (wild eye x sepia eye) experiment:
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Expected!
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Calculations:
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Chi square value (x2)
Observed!
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Difference (d)!
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d2
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Does the chi square value indicate that the deviation is caused by chance? ___________ What is the
probability
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range that the deviation is due to chance? ___________
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F2's of dihybrid (wild eye,vestigial wing x sepia eye, wild wing) experiment:
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Expected!
_________!
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Chi square value (x2)
Observed!
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_________!
_________!
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Difference (d)!
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d2
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Calculations:
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Does the chi square value indicate that the deviation is caused by chance? ___________ What is the
probability
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range that the deviation is due to chance? ___________
F2's of sex-linked (red-eye male x white-eye female) experiment:
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Expected!
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Observed!
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Chi square value (x2)
Difference (d)!
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d2
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Calculations:
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Does the chi square value indicate that the deviation is caused by chance? ___________ What is the
probability
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range that the deviation is due to chance? ___________
CONCLUSIONS
Write a concluding paragraph for each of the experiments you have completed in the Mendelian genetics unit. Refer to
your hypothesis for each experiment and use data to support your discussion of the results. Also include in your
paragraph whether or not you accepted or rejected the results based on your hypothesis. Explain why you accepted
your hypothesis based on your results, i.e. what the per cent really means. If you rejected the results based on your
hypothesis, explain why you did, and tell what, besides chance, could have caused the deviation, i.e. speculate what
kinds of things may have resulted in the class data not agreeing with your hypothesis.