Download WARDS Lab 7 - Genetics of Drosophila

AP Biology Lab 7
Genetics of Drosophila
Lab Activity
36 W 7105
36 W 7116
250-7055
WARD’S is working with educators to design our literature to meet the many challenges of today’s teachers. Included in every
activity are correlations to National Science Education Content Standards, easy to follow instructions, and a diversified
assessment that cater to students of all levels and learning styles.
MATERIALS INCLUDED IN THE KIT
This lab activity is designed for eight groups of students
For 8 Lab Groups
36 W 7105
1
1
1
1
8
8
15
25
30
1
8
8
72
100
8
1
100
1
For 2 Lab Groups
Materials Checklist
36 W 7116
1
Isopropyl alcohol, 10%, 100 ml
1
HCl solution, 8%, 3 ml
1
Aceto-orcein stain, 30 ml
1
Piccolyte II, 15 ml
2
Camel’s hair brushes
2
Thermo-anesthetizers
15
Pipets
25
Petri dishes
10
Drosophila vials and labels
1
Drosophila medium
2
Fly morgues
2
Teasing needles
72
Microscope slides
100
Coverslips
2
Forceps
1
Prepared slide of Drosophila chromosome
100
Gloves
1
Redemption coupon for perishables
1 Culture vial of wild-type Drosophila (for observing traits)
1 Culture vial of monohybrid cross, parental generation
1 Culture vial of dihybrid cross, parental generation
1 Culture vial of sex-linked cross, parental generation
* refill kit available for the eight group lab activity: 36 W 7126
MATERIALS NEEDED BUT NOT PROVIDED
•
•
Stereomicroscope
Distilled water
For Technical Assistance Call 1-800-962-2660
Consult your WARD’S catalog for additional products you may need
Call WARD’S at 1-800-962-2660 for Technical Assistance
2
SPECIAL HANDLING INSTRUCTIONS
•
•
Redeem the coupon included for your four Drosophila cultures.
Note the initiation date on the Drosophila crosses upon receipt. Allow the Drosophila crosses
to develop five days from the initiation date prior to performing part B of the experiment.
•
Do not remove the parental generation until larva are visible in the vial. Usually 5-7 days after receipt of the cultures. This will ensure enough eggs and
larvae to to produce F1 flies for subculturing.
NATIONAL SCIENCE EDUCATION CONTENT STANDARDS
Standard K-12: Unifying Concepts and
Processes
Evidence, models, and explanation
Form and function
Abilities necessary to do scientific inquiry
Standard A: Science as Inquiry
Understandings about scientific inquiry
Standard C: Life Science
Molecular basis of heredity
Standard G: History & Nature of Science
History perspectives
OBJECTIVES
•
•
•
•
•
Learn basic handling and culture techniques for working with Drosophila
Apply concepts and principles of Mendelian inheritance patterns
Diagram monohybrid, dihybrid, and sex-linked crosses
Gain experience sorting, sexing, and crossing Drosophila through two generations
Perform a chi-square statistical analysis of experimental results
TIME REQUIREMENTS
•
•
Three to four weeks for culturing depending on temperature
May require extra lab time outside of class for investigations to coincide with Drosophila life cycle and the hatching of virgin females
3
© 2002 WARD’S Natural Science Establishment, Inc.
All Rights Reserved
BACKGROUND
Advanced Placement (AP) is a registered trademark of the College
Entrance Examination Board. The
materials provided in this lab activity have been prepared by WARD’S
according to the specifications outlined in the AP Laboratory Manual, Edition D. To perform the investigation, you may either follow
the directions in the AP manual or,
if one is not available, you may follow the directions in this guide.
In 1865, Gregor Mendel published a paper on the patterns of genetic
inheritance in the common garden pea. This revolutionary work provided the basis for future studies in genetics. Mendel hypothesized
that heredity was passed on by discrete particles, rather than by the
blending of parental traits, as was believed at the time, strongly affecting the argument over Darwin’s theory of evolution.
Mendel proposed two very basic laws that serve as the cornerstones
of modern genetics: Mendel’s Law of Segregation and Law of Independent Assortment.
Mendel’s Laws of Genetic Inheritance
Mendel’s Law of Segregation states that for each trait (gene), each organism carries two factors (alleles), and that each of the organism’s
gametes contain one and only one of these factors. In this way, the
alleles segregate during meiosis, providing for genetic variability
among the organism’s offspring. This is apparent in monohybrid
crosses, which involve only one trait.
Mendel’s Law of Independent Assortment, shown in dihybrid crosses
(crosses involving two traits), states that the alleles for one trait will
separate independently of the alleles in another trait. This means that
with two genes (A and B), each with two possible alleles (A, a and B,
b), there are four possible combinations gametes can receive (AB, Ab,
aB, or ab). This helps to ensure genetic variability among offspring.
DID YOU KNOW?
Mendel is known as the
“Father of Genetics” because
of his revolutionary work
with garden pea plants.
Although Mendel’s laws have proven to be true, they have their limitations. It must be assumed that the genes involved are on different
chromosomes. If they are on the same chromosome, the assortment of
their alleles will be closely linked instead of being independent. It
must also be assumed that the genes in question are not located on
the X chromosome; this could cause the expression of the trait to be
linked to the sex of the offspring. Color blindness in humans is a
good example of this type of heredity. None of the traits Mendel selected in his pea plant investigations were sex-linked or located on
the same chromosome.
Since Mendel’s time, our knowledge of genetics, as well as the tools
we use in its study, has advanced drastically. Instead of crossing varieties of pea plants and observing the results as Mendel did, we now
use restriction enzymes, electrophoresis, and basepair sequencing to
learn more about traits and their expression. Yet even with these advances, the basic concepts of genetic inheritance described by Mendel
so many years ago still stand.
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4
Drosophila Use in Genetic Research
Drosophila melanogaster, the common fruit fly, was first used in genetic
experiments in 1907 by T.H. Morgan of Columbia University, and has
been a staple of genetic research ever since. Drosophila specimens are
well suited to investigations into Mendelian patterns of inheritance;
they are small, produce large numbers of offspring, have many easily
discernible mutations, have only four pairs of chromosomes, and
complete their entire life cycle in approximately 12 days. Additionally, Drosophila are relatively easy to maintain, as they are hardy and
have simple food requirements.
DID YOU KNOW?
Male Drosophila are the
smaller of the two sexes.
Since the fruit fly was selected for study nearly a hundred years ago,
a great deal has been learned about its genome. In fact, the first chromosome map of any kind was constructed to detail the fruit fly.
Chromosomes 1 (the X chromosome), 2, and 3 are very large, while
chromosome 4 (the Y chromosome) is extremely small. Thousands of
genes reside on these four chromosomes, many of which are universal in nature, existing in most eukaryotic forms, including humans.
The similarities between the Drosophila genome
and those of other species is helpful in determining patterns of evolution and species divergence.
Figure 1
Drosophila Life Cycle
Drosophila embryos develop in the egg membrane. The egg hatches and produces a larva that
feeds by burrowing through the medium. The larval period consists of three stages, or instars, the
end of each stage marked by a molt. The first instar is the newly hatched larva; the third instar is
the final larval stage, where the larva may attain a
length of 4.5 mm. Near the end of the larval period, the third instar larva will crawl up the sides
of the culture vial, attach themselves to a dry surface (the jar, the filter paper, etc.) and form pupae.
After a period of time the adults emerge.
♂
It takes one or two days for Drosophila eggs to
hatch into larvae, four to five days for the larvae to
enter pupae, and four days for the pupa stage. The
duration of these stages, however, varies with the
temperature; at 20°C the average length of the
egg–larval period is eight days, while at 25°C it is
reduced to five days. Thus, at 25°C the life cycle
may be completed in about ten days; at 20°C,
fifteen days are required.
Adult
♀
Egg
Pupa
Larva
(1rst instar)
Larva
(3rd instar)
5
Larva
(2nd instar)
© 2002 WARD’S Natural Science Establishment, Inc.
All Rights Reserved
Different body features characterize the male and female flies. The
females are slightly larger and have a light-colored, pointed abdomen. The abdomen of the males will be dark and blunt. The male flies
also have dark bristles on the upper portion of the forelegs, which are
known as sex combs (Figure 2).
Figure 2
Female
DID YOU KNOW?
A Punnett square displays the
possible combinations of a
genetic cross, based on the
Laws of segregation and independent assortment.
Male
Male sex combs
on foreleg
In the following experiment, parental generations (P) of wild-type
and mutant strains of Drosophila with various traits to demonstrate
basic genetic principles are used. Monohybrid, dihybrid, and sexlinked crosses are performed; the offspring of the first cross, the F1
generation, are normally allowed to breed among themselves to produce the F2 generation. Members from the F1 and F2 generations are
collected and their traits observed to draw conclusions about genetic
inheritance.
The second filial generation, or F2, represents the flies that result from
self-fertilization or inbreeding among members of the F1 generation.
Punnett Square
Based on the laws of segregation and independent assortment, a Punnett square is extremely important in determining the outcome of
crosses in Mendelian genetics; it clearly displays the possible combinations in chart form.
The simplest Punnett square to construct is for a monohybrid cross. A
good example of this is a cross between female fruit flies with vestigial wings and male wild-type fruit flies. Determining the genotype of
the flies being crossed is vital to the accuracy of the results of a Punnett square; if it is known that the trait for vestigial wings is a recessive mutation, the flies with vestigial wings must be homozygous recessive for the first trait, and therefore have a genotype of vv. Assuming that the wild-type flies are heterozygous dominant, they will have
a genotype of Vv. According to the Law of Segregation, only one of
the alleles for the trait can be passed on to a gamete for each parental
fly. Therefore, the male wild-type fly could pass either the V or the v
allele on to its offspring. Likewise, the female, vestigial fly can pass
on only one of her alleles for the trait. In her case, however, they are
both v. Therefore, the possible allelic combinations in the offspring
are Vv and vv. This is diagrammed in a Punnett square, below.
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6
Female
Male
V
v
v
Vv
vv
v
Vv
vv
The two possible genotypes, Vv and vv, will exist in a 1:1 ratio, and
the phenotypic ratio will also be 1:1 with as many offspring with vestigial wings as with normal wings.
Punnett squares become more complicated when diagramming a dihybrid cross. Due to the complex nature of dihybrid crosses, or even
crosses with three or more traits, it is even more important to diagram the crosses with a Punnett square.
For an example of a dihybrid cross, females with normal eyes and
vestigial wings can be crossed with male flies that have sepia (dark
brown) eyes and normal wings, assuming that the females have the
genotype Ssvv, and the males have the genotype ssVV. Again, each
parent can donate only one allele for each trait to the offspring, but by
applying Mendel’s Law of Independent Assortment, the alleles for
the two traits should be distributed without regard to the distribution
of the other. There are, therefore, three combinations of alleles each
parent can donate to any one gamete. The females can donate the set
of alleles Sv or sv; the males can only donate the set of alleles sV. The
Punnett square for this cross is diagrammed below.
Females
Males
sV
sV
sV
sV
Sv
SsVv
SsVv
SsVv
SsVv
Sv
SsVv
SsVv
SsVv
SsVv
sv
ssVv
ssVv
ssVv
ssVv
sv
ssVv
ssVv
ssVv
ssVv
Monohybrid Cross:
A genetic cross involving
only one trait, which is one
gene with two discrete alleles.
Dihybrid Cross:
A genetic cross that examines
the patterns of inheritance for
two traits, each defined by
one gene with two alleles. The
alleles are normally independent of one another unless
the genes are on the same
chromosome.
The only genotypes are SsVv and ssVv, yielding offspring that have
normal wings and sepia eyes, and offspring with both normal eyes
and wings. These offspring will be produced in a 1:1 ratio.
Chi-Square Test
The chi-square test is a statistical tool that compares experiment results with an accepted set of data to determine how much the experimental values deviated from the accepted ones and whether or not
that deviation can be explained solely by chance.
7
© 2002 WARD’S Natural Science Establishment, Inc.
All Rights Reserved
Chi-square Table
a
DF
v
1
2
3
4
5
6
7
8
9
10
15
20
25
30
P=0.99
0.00016
0.02010
0.11500
0.29700
0.55400
0.87200
1.23900
1.64600
2.08800
2.55800
5.22900
8.26000
11.52400
14.95300
0.95
0.00393
0.10300
0.35200
0.71100
1.14500
1.63500
2.16700
2.73300
3.32500
3.94000
7.26100
10.85100
14.61100
18.49300
0.80
0.06420
0.44600
1.00500
1.64900
2.34300
3.07000
3.82200
4.59400
5.38000
6.17900
10.30700
14.57800
18.94000
23.36400
0.50
0.45500
1.38600
2.36600
3.35700
4.35100
5.34800
6.34600
7.34400
8.34300
9.34200
14.33900
19.33700
24.33700
29.33600
0.20
1.64200
3.21900
4.64200
5.98900
7.28900
8.55800
9.80300
11.03000
12.24200
13.44200
19.31100
25.03800
30.67500
36.25000
0.05
3.84100
5.99100
7.81500
9.44800
11.07000
12.59200
14.06700
15.50700
16.91900
18.30700
24.99600
31.41000
37.65200
43.77300
0.01
6.63500
9.21000
11.34500
13.27700
15.08600
16.81200
18.47500
20.09000
21.66600
23.20900
30.57800
37.56600
44.31400
50.89200
The square of the difference between the observed and expected values (O – E)2 for each data point (phenotype category in this case) is
calculated. Then, by dividing this by the expected value, the amount
of deviation between the experiment data and the accepted value for
that data point can be determined. Adding the statistic for each data
point yields a value known as the χ2 (chi-square) statistic.
χ2 = ∑ (O – E)2/E
DID YOU KNOW?
Chi-square is officially known
as the Pearson chi-square in
homage to its inventor Karl
Pearson (1857-1936).
Because all the values for each category, or data points, are being
added together, the value of χ2 will rise as the number of data points
used increases. For this reason, “degrees of freedom” must be included in the parameters of the analysis. The number of degrees of
freedom (v) for a chi-square test is equal to the number of data points
minus one. The number of degrees of freedom does not have an impact on the value of χ2 itself, but rather is used in the interpretation of
the importance of the value, as shown in the chi-square table, above.
The numbers get larger as you go down and the value for degrees of
freedom increases.
Once the values for χ2 and v have been established, the chart is used
to determine the probability that the variations between the set of experimental values and the set of accepted values can be explained
away as a function of chance.
First two hypotheses, H0 and H1, must be formulated, with the null
hypothesis (H0) stating that the variations cannot be explained solely
by chance, and H1 stating that they are exclusively due to chance.
This will be determined by the value of “a”, which is defined as the
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8
probability that H1 can be accepted as true. In order to prove H0, “a”
must be quite low, depending on the specific needs of the experiment
and the confidence level required in the data.
The value of “a” is determined using the chi-square chart: the value
of χ2 is found in the row for the correct number of degrees of freedom
(v). It is likely that it will be between values; in this case the value of
“a” should be approximated. The probability that H0 can be accepted
as true can then be defined as 1 – a.
For an experiment with a relatively small sample size, a confidence
level between 50% and 80% is acceptable.
Example:
Assuming only two phenotypic categories, vestigial wings and normal wings, use the following table of results for the F2 generation:
Phenotype
No. of Males
No. of Females
Vestigial Wings
14
12
Normal Wings
36
38
Genotype:
Description of the set of
alleles an individual possesses for a gene or set of
genes, expressed or not, usually shown in symbols (ie.,
AA, Aa, or aa).
The calculation needs to be performed only twice — once for each
phenotype. Adding the results together provides the value for χ2.
Phenotype:
Physical traits expressed by
an individual, regardless of
the genotype.
Total number of flies observed: 100
Flies with vestigial wings: 26
Flies with normal wings: 74
Expected ratio: 3:1
Expected number of flies with normal wings: 75
Expected number of flies with vestigial wings: 25
χ2 = (74-75)2 + (26-25)2 = 0.013 + 0.04 = 0.053
75
25
Since only two phenotypes are used, there will be only one degree of
freedom, as the number of degrees of freedom is one less than the
number of categories. Since the χ2 value is less than the χ2 value for
one degree of freedom at .05% (3.841), the null hypothesis can be accepted as true. Since the χ2 value (.053) is between .00393 and .0642,
the values for the 80% and 95% confidence respectively, there is an 80
to 95% confidence that the deviations from the expected experiment
values are due solely to chance.
9
© 2002 WARD’S Natural Science Establishment, Inc.
All Rights Reserved
PRE-LAB PREPARATION
Part A: Working with Drosophila
1. Prepare a Petri dish for each lab group: Place a Petri dish on an
ice pack and place in the freezer for at least 24 hours before the
lab.
2. Prepare a fly morgue: Add 10-15 ml of 10% isopropyl alcohol to
the morgue. Secure the lid tightly to prevent evaporation.
NOTE
You may perform the next several steps in advance or, if
time permits, have the students perform them in class.
3. Thermally anesthetize wild-type Drosophila: Invert the vial of
flies and place it in a refrigerator or tight-sealing cooler with several ice packs for 10 to 20 minutes or until the flies appear lifeless.
NOTE
DID YOU KNOW?
The chromosomes of Drosophila can stretch to more than
a mile long when unraveled.
Chilling the vial inverted makes it easier to see the flies as
they become immobilized and also prevents them from
becoming stuck in the media.
4. Prepare one culture vial per group: Place approximately one tablespoon of medium in the bottom of a vial. Add an equal amount of
water and let it absorb. You may want to add a piece of plastic
mesh to give the flies something to crawl on, but it is not essential.
Insert a foam plug in the vial to prevent any contamination.
5. Subculture wild-type Drosophila: Using one prepared culture vial
per lab group, divide the Drosophila into equal amounts.
6. Thermally anesthetize the subcultured Drosophila just prior to
performing the lab. When the flies are immobile and have collected on the foam plug, remove the ice pack with the Petri dish
from the freezer. Quickly dump the flies into the Petri dish.
TIMING
You may thermally anesthetize the Drosophila to have
them ready for students to examine, or you may have the
students thermally anesthetize the Drosophila themselves
as part of the lab. The flies will remain immobile for 10 15 minutes. If more time is required, cover the Petri dish
and return the dish and the ice pack to the freezer for 2 –3
minutes. The flies can be anesthetized this way repeatedly
without being harmed.
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10
Part B: Performing a Drosophila Cross
NOTE
1.
The Drosophila crosses provided by WARD’S contain the
parental (P) generation for each type of cross. The flies
have already mated and the larvae present on the side of
the vials represent the F1 progeny. You must assign one
of the three types of crosses to each group. Groups sharing
a cross must share the anesthetized parental flies for the
observation portion of the lab. When the F1 generation
emerges, each group may count the offspring separately
and compare results. After every group has counted the
F1 for their respective cross, they may then each obtain
several mating pairs from the F1 specimens for the initiation of the next generation.
Remove and retain the top label from the culture vials of the crosses
before giving them to the lab groups, so the lab groups receive vials
labeled only “A”, “B”, or “C”; they will identify the crosses. Be sure
to note the initiation date on the top label of each cross. These
crosses must be allowed to sit for five days from the initiation date
before the students can proceed with their observations.
NOTE
You may prepare the culture vials in advance or you may
wish to have the students prepare their own vials prior to
performing their cross.
2. Prepare culture vial: Place approximately one tablespoon of medium in the bottom of a vial. Add an equal amount of water and
let it absorb. You may want to add a piece of plastic mesh to give
the flies something to crawl on, but it is not essential. Insert a
foam plug in the vial to prevent any contamination.
11
DID YOU KNOW?
In 1910, Thomas Hunt Morgan discovered sex-linkage in
Drosophila and postulated a
connection between eye color
in fruit flies and human color
blindness.
© 2002 WARD’S Natural Science Establishment, Inc.
All Rights Reserved
MATERIALS
MATERIALS NEEDED PER GROUP
1
1
1
1
1
1
1
2
1
Culture vial of wild-type Drosophila
Culture vial of monohybrid cross
or
Culture vial of dihybrid cross
or
Culture vial of sex-linked cross
Isopropyl alcohol 10%, 100 ml
Camel’s hair brush
Thermo-anesthetizer
Petri dish
Drosophila vials and labels
Drosophila medium
Fly morgue
Forceps
PROCEDURE
Part A: Working with Drosophila
NOTE
DID YOU KNOW?
A female Drosophila once fertilized, may lay 30-50 eggs per
day throughout her lifetime.
You will need to observe wild type Drosophila to familiarize yourself with the wild type phenotype. You will eventually be assigned a cross without being told what strain,
genotype, or type of experimental cross you have received.
You will examine the flies in the parental generation of
your cross, noting any phenotype variations from the
wild type, and name the mutations.
1. Thermally immobilize a vial of wild-type Drosophila. Your instructor will demonstrate the proper immobilization technique.
NOTE
2.
Your instructor may have immobilized the flies in advance. If this is the case, begin with Step 2.
Observe the flies’ traits, particularly body features that distinguish males and females, eye color, and wing size and shape. Record your observations in Table 1 in the Analysis section. If, at
any time during your observations, the flies begin to become active, re-immobilize them according to your instructor’s
directions.
NOTE
Use the camel’s hair brush to move the flies when making
observations.
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12
Part B: Performing a Drosophila Cross
1. Obtain a vial of a prepared Drosophila cross.
NOTE
These flies have been mated and may exhibit one or more
mutations. They are the parental generation for your experiment. The offspring of this generation, which should
already exist as eggs or larvae in the vial, are the F1
generation.
2. Record the letter written on your vial in Table 2 in the Analysis
section to help you keep track of which cross you have received.
This will aid in determining expected results, as well as allow
your instructor to identify any problems you may be having and
to help correct them.
3. Immobilize the parental generation of your cross and observe the
flies under a stereomicroscope. If, at any time during your observations, the flies begin to become active, re-immobilize them according to your instructor’s directions.
4. Separate the males from the females. Note any mutations from
the wild-type phenotype, as well as whether the mutation is apparent in the male or female flies. Record your observations in
Table 2.
NOTE
You may be sharing the parental generation with another
group for observation. If this is the case, do not perform
the next step until all groups have had a chance to make
their observations.
5. Place the parental generation in the morgue.
6.
Place the vial (with the parental generation removed) in a warm
(28°C) place to incubate to allow the F1 generation to mature. Observe the vial occasionally and record your observations.
!
DID YOU KNOW?
The longest sperm ever recorded is 10,000 times longer
than a human spermatozoan,
and belongs to Drosophila bifuria. The typical male D. bifuria
sperm is approximately 60 mm
long, 20 times the length of the
fly itself.
Do not allow the temperature to exceed 30°C.
CAUTION
7. When the adult flies emerge, collect, immobilize, and examine
them. Note the sex of each one, as well as the presence of any mutations. Record your observations in Table 3. Be sure every group
assigned to that cross has a chance to observe and count the F1
progeny.
8. Prepare a fresh culture vial: Place approximately one tablespoon
of medium in the bottom of a vial. Add an equal amount of water
and let it absorb. You may want to add a piece of plastic mesh to
give the flies something to crawl on, but it is not essential. Insert a
foam plug in the vial.
13
© 2002 WARD’S Natural Science Establishment, Inc.
All Rights Reserved
9. Place five or more mating pairs from the F1 generation into the
fresh culture vial (It is not necessary that these females be
virgins). Label the vial with your name(s), date, and letter of
cross. Place the culture vial in a warm place to incubate to allow
the F1 generation to mature.
10. Transfer the remaining non-mating F1 flies to the morgue.
DID YOU KNOW?
In 1913, American geneticist
Alfred Henry Sturtevant developed a technique for mapping the location of specific
genes of the chromosomes in
Drosophila.
11. Leave the F1 adults in the vial for about one week to mate and lay
their eggs. Once they have laid their eggs and you can see larva
on the sides of the culture vial (7– 10 days), remove the adults,
place them in the morgue, and wait for the F2 generation adults
to emerge.
Do not allow the temperature to exceed 30°C.
!
CAUTION
12. The F2 adults will gradually emerge from the pupae over several
days. As the F2 generation flies begin to emerge as adults, immobilize and examine them. Record the number of males and females, noting any mutations which may be present. Record your
findings in Table 4.
NOTE
Try to collect as many adults as possible.
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14
OPTIONAL EXERCISE
Polytene Chromosome Investigation
NOTE
Students will investigate chromosomes and their activity
by examining the salivary glands of Drosophila larvae,
which contain giant polytene chromosomes as a result of
the repeated replication of the DNA. They make excellent
organisms for chromosome observation, and for investigations into chromosome function and structure. They are
most evident in D. virilis larvae, but can be observed in
any species of Drosophila.
MATERIALS
MATERIALS NEEDED PER LAB GROUP
1
1
1
1
1
1
HCl, 8%
Aceto-orcein stain, 30 ml
Piccolyte II, 15 ml
Teasing needle
Pipet
Forceps
Microscope slide
Coverslip
Drosophila larva
Gloves
Goggles
Lab aprons
Stereomicroscope
Distilled water
Paper towels
DID YOU KNOW?
Drosophila salivary glands
consist of two major cell types
secretory cells and duct cells.
PROCEDURE
1.
Place a microscope slide on a flat surface. Add two or three drops
of distilled water to the center of the slide.
2.
With forceps, grasp a fully-grown Drosophila larva from the wall
of the wild-type Drosophila culture vial. Place the larva in the
drop of water on the microscope slide.
3.
Under a stereomicroscope, locate the head of the larva. It will appear darker than the rest of the body, and the mouthparts should
be evident. The larva will also be moving in that direction.
4.
Grasp the larva in the midsection with the forceps. Squeeze it
gently to force the head out.
15
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5.
Pierce the head with the teasing needle and pull the head away
from the body. The salivary glands, and likely the digestive tract
also, will detach from the body. The salivary glands will appear
clear and cellular, but may be confused with the darker, opaque
fat bodies. The intestine may be present as a tubular, branched
structure.
6.
Remove all the excess materials from the slide.
7.
Touch a corner of a paper towel to the slide to blot off only the
excess water.
NOTE
8.
Do not let the salivary glands dry out.
Add two or three drops of 8% HCl to hydrolyze the genetic material for better penetration by the stain. Let it sit for three minutes.
!
Avoid skin or eye contact with HCl. Wear safety goggles,
gloves, and a lab apron.
CAUTION
9.
DID YOU KNOW?
Measuring 2 mm in length, the
polytene chromosomes found in
the salivary glands of Diptera
are much longer than
metaphase chromosomes.
Touch a corner of a paper towel to the slide to blot off the excess
HCl.
10. Add two or three drops of aceto-orcein stain to the salivary
glands and let them to sit for four to five minutes. Do not let the
stain dry out. Add more stain if necessary.
!
Aceto-orcein stain is an irritant. Avoid skin or eye contact. Wear safety goggles, gloves, and a lab apron.
CAUTION
11. Blot the excess stain from the slide; leave a small amount of stain
on the glands.
12. Place the slide on a smooth, flat surface and cover with a coverslip. Tuck the slide into the fold of a paper towel and press down
firmly on the coverslip with your thumb or a pencil eraser.
13. Observe the preparation under a compound microscope. Note the
bands in the polytene chromosomes, the “puff” regions, and the
chromocenter of the chromosome. Draw what you see in the
space provided in the Analysis section.
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16
ANALYSIS
Table 1
Phenotypes of Wild Type Drosophila
Eye Color
Wing Size and Shape
Male
Female
Table 2
Phenotypes of the Parental Generation
Cross Letter:______
Phenotype
No. of Males
17
No. of Females
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Table 3
Phenotypes of the F1 Generation
Cross Letter:______
Phenotype
No. of Males
No. of Females
Table 4
Phenotypes of the F2 Generation
Cross Letter:______
Phenotype
No. of Males
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18
No. of Females
Sample Results For Each Cross
Male
Female
(A) Monohybrid Cross
Sepia eyed
16
15
Wild eyed
46
51
Wild eyed/normal wing
63
59
Sepia eyed/normal wing
18
22
Wild eyed/vestigial wing
20
18
Sepia eyed/vestigial wing
5
4
Conclusions
Answers will vary
Student results should approximate a ratio of:
3: wild eye
1: sepia eye
(B) Dihybrid cross
(C) Sex linked Cross
White eyed
30
30
Wild eyed
24
38
Answers will vary
Student results should approximate a ratio of:
9: wild eye / normal wing
3: wild eye / vestigal wing
3: sepia eye / normal wing
1: sepia eye / vestigal wing
Answers will vary
Student results should approximate a ratio of:
1: wild eye female
1: white eye female
1: wild eye male
1: white eye male
Polytene Chromosome Investigation
Chromocenter
“Puff”
Region of RNA
synthesis
Band
Drosophila Chromosomes
Squashed (1000X)
Inter-band
19
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ASSESSMENT
1.
Describe the parental cross you received; use genetic symbols.
Example: A cross between vestigial and wild-type flies would be expressed as vv x VV.
Draw a Punnett square to show the possible allelic combinations for this gene in the F1 generation.
a) Monohybrid Cross
Cross between female flies with sepia-colored eyes and wild-type males: SS (males) x ss (females)
Females
Males
S
S
s
Ss
Ss
s
Ss
Ss
b) Dihybrid Cross
Cross between males with sepia-colored eyes and normal wings, and females with normal eyes
and vestigial wings: ssVV (males) x SSvv (females)
Females
Males
c)
sV
sV
sV
sV
Sv
SsVv
SsVv
SsVv
SsVv
Sv
SsVv
SsVv
SsVv
SsVv
Sv
SsVv
SsVv
SsVv
SsVv
Sv
SsVv
SsVv
SsVv
SsVv
Sex-Linked Cross
Cross between female flies with white eyes and wild-type male flies: XWY (males) x XwXw (females)
Females
Males
XW
Y
Xw
XWXw
XwY
Xw
XWXw
XwY
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20
2.
Identify the genotype the F1 flies should exhibit. Identify the phenotype. Compare your experiment
results by counting the members of the F1 generation.
a) Monohybrid Cross
All the F1 individuals should be heterozygous dominant (Ss), and therefore have normal
wild-type eye color.
b) Dihybrid Cross
All the F1 individuals should be heterozygous dominant for both traits, and should therefore
exhibit the phenotype of wild-type flies.
c) Sex-Linked Cross
All the F1 female flies should be wild type, and all F1 male flies should have white eyes.
Describe the F1 cross you performed, and draw a Punnett square to show allelic combinations possible in the F2 generation.
a) Monohybrid Cross
F1 cross: Ss x Ss
Females
Males
S
s
S
SS
Ss
s
Ss
ss
b) Dihybrid Cross
F1 cross: SsVv (males) x SsVv (females)
Males
SV
Females
c)
Sv
sV
sv
SV
SSVV SSVv
SsVV
SsVv
Sv
SSVv
SSvv
SsVv
Ssvv
sV
SsVV
SsVv
ssVV
ssVv
sv
SsVv
Ssvv
ssVv
ssvv
Sex-Linked Cross
F1 cross: XWXw x XwY
Males
Females
3.
Xw
Y
XW
XWXw
XWY
Xw
XwXw
XwY
21
© 2002 WARD’S Natural Science Establishment, Inc.
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4.
Identify the genotype ratio the F2 flies should exhibit. Identify the phenotype ratio. Compare your
experiment results by counting the members of the F2 generation.
a) Monohybrid Cross
The genotype ratio should be 1:2:1. The phenotype ratio should be 3:1, with 75% of the F2
individuals having normal wild-type eyes, and 25% having sepia eyes.
b) Dihybrid Cross
The genotype ratio can be determined from the Punnett square, and the phenotype ratio for
the F2 generation should be 1:3:3:9, with one individual with both sepia eyes and vestigial
wings, three flies with normal wings and sepia eyes, three flies with vestigial wings and
normal eyes, and nine wild-type flies.
c)
Sex-Linked Cross
The female flies should exist in a 1:1 ratio, with half having the genotype XWXw, and half
having the genotype XwXw. The males also exist in a 1:1 ratio, with half being XWY, and half
XwY. The phenotypes of the F2 generation: half of the males and half of the females will
have white eyes, and half of each sex will have wild-type eyes.
5.
Identify the type of cross you received: monohybrid or dihybrid, autosomal or sex-linked, mutations dominant or recessive.
Cross A — Monohybrid, autosomal, recessive
Cross B — Dihybrid, autosomal, both traits recessive
Cross C — Monohybrid, sex-linked, recessive
6.
Using a chi-square test, determine whether or not the variation between the observed and expected
number of individuals of each phenotype can adequately be explained by chance alone. Use the
following formula, and apply it to the chi-square table to determine the confidence level that states
the variation is due solely to chance.
χ2 = ∑ (O–E)2/E
O = observed number of offspring for the phenotypic category
E = expected number of offspring for the phenotypic category
χ2 = _____________________
Confidence level that variability is due entirely to chance = ____________ %
a) Monohybrid Cross
χ2 = .111, with v = 1, for a 50-80% confidence that the variations in the data are due
entirely to chance.
b) Dihybrid Cross
χ2 = .451, with v = 3, for an 80-95% confidence that the variations in the data are due
entirely to chance.
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22
c)
Sex-Linked Cross
χ2 = .06375, with v = 3, for better than 99% confidence that the variation in the data is due entirely to chance. In this case, there are three degrees of freedom; because it is a sex-linked
trait, expected ratios for white vs. wild-type eyes for both males and females must be calculated, resulting in four data points.
DF
v
1
2
3
4
5
6
7
8
9
10
15
20
25
30
7.
P=0.99
0.00016
0.02010
0.11500
0.29700
0.55400
0.87200
1.23900
1.64600
2.08800
2.55800
5.22900
8.26000
11.52400
14.95300
0.95
0.00393
0.10300
0.35200
0.71100
1.14500
1.63500
2.16700
2.73300
3.32500
3.94000
7.26100
10.85100
14.61100
18.49300
0.80
0.06420
0.44600
1.00500
1.64900
2.34300
3.07000
3.82200
4.59400
5.38000
6.17900
10.30700
14.57800
18.94000
23.36400
a
0.50
0.45500
1.38600
2.36600
3.35700
4.35100
5.34800
6.34600
7.34400
8.34300
9.34200
14.33900
19.33700
24.33700
29.33600
0.20
1.64200
3.21900
4.64200
5.98900
7.28900
8.55800
9.80300
11.03000
12.24200
13.44200
19.31100
25.03800
30.67500
36.25000
0.05
3.84100
5.99100
7.81500
9.44800
11.07000
12.59200
14.06700
15.50700
16.91900
18.30700
24.99600
31.41000
37.65200
43.77300
0.01
6.63500
9.21000
11.34500
13.27700
15.08600
16.81200
18.47500
20.09000
21.66600
23.20900
30.57800
37.56600
44.31400
50.89200
What do the giant polytene chromosomes look like?
Large, red-banded, multi-armed structure, with conspicuous light and dark banding.
8.
What are the various bands believed to be?
Individual genes.
9.
What chemical compound is the major substance of chromosomes? What is its double function?
The chemical compound that is the major substance of chromosomes is DNA. It contains the genetic code and transmits the hereditary pattern.
23
© 2002 WARD’S Natural Science Establishment, Inc.
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.