250-7055t genetics of drosophila

250-7055

AP Biology Lab 7 Genetics of Drosophila

Lab Activity

36 W 7105 36 W 7116

For Technical Assistance Call 1-800-962-2660

This lab activity is designed for eight groups of students For 8 Lab Groups For 2 Lab Groups Materials Checklist 36 W 7105 36 W 7116 1 1 Isopropyl alcohol, 10%, 100 ml 1 1 HCl solution, 8%, 3 ml 1 1 Aceto-orcein stain, 30 ml 1 1 Piccolyte II, 15 ml 8 2 Camel’s hair brushes 8 2 Thermo-anesthetizers 15 15 Pipets 25 25 Petri dishes 30 10 Drosophila vials and labels 1 1 Drosophila medium 8 2 Fly morgues 8 2 Teasing needles 72 72 Microscope slides 100 100 Coverslips 8 2 Forceps 1 1 Prepared slide of Drosophila chromosome 100 100 Gloves 1 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

• Stereomicroscope • Distilled water

Consult your WARD’S catalog for additional products you may need

MATERIALS INCLUDED IN THE KIT

MATERIALS NEEDED BUT NOT PROVIDED

Call WARD’S at 1-800-962-2660 for Technical Assistance

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.

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• 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. Usu-

ally 5-7 days after receipt of the cultures. This will ensure enough eggs and larvae to to produce F1 flies for subculturing.

• 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

• Three to four weeks for culturing depending on temperature • May require extra lab time outside of class for investigations to coincide with Dro-

sophila life cycle and the hatching of virgin females

TIME REQUIREMENTS

OBJECTIVES

Standard K-12: Unifying Concepts and Processes

Evidence, models, and explanation

Form and function

Standard A: Science as Inquiry Abilities necessary to do scientific inquiry

Understandings about scientific inquiry

Standard C: Life Science Molecular basis of heredity

Standard G: History & Nature of Science History perspectives

NATIONAL SCIENCE EDUCATION CONTENT STANDARDS

© 2002 WARD’S Natural Science Establishment, Inc. All Rights Reserved 3

SPECIAL HANDLING INSTRUCTIONS

In 1865, Gregor Mendel published a paper on the patterns of genetic inheritance in the common garden pea. This revolutionary work pro-vided 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 af-fecting 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 Inde-pendent Assortment. Mendel’s Laws of Genetic Inheritance Mendel’s Law of Segregation states that for each trait (gene), each or-ganism 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. Although Mendel’s laws have proven to be true, they have their limi-tations. 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 se-lected 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 va-rieties 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 ad-vances, the basic concepts of genetic inheritance described by Mendel so many years ago still stand.

BACKGROUND


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Advanced Placement (AP) is a reg-istered trademark of the College Entrance Examination Board. The materials provided in this lab activ-ity have been prepared by WARD’S according to the specifications out-lined in the AP Laboratory Man-ual, Edition D. To perform the in-vestigation, you may either follow the directions in the AP manual or, if one is not available, you may fol-low the directions in this guide.

DID YOU KNOW? Mendel is known as the “Father of Genetics” because of his revolutionary work with garden pea plants.

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. Addition-ally, Drosophila are relatively easy to maintain, as they are hardy and have simple food requirements. 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 chro-mosome 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 univer-sal in nature, existing in most eukaryotic forms, including humans. The similarities between the Drosophila genome and those of other species is helpful in determin-ing patterns of evolution and species divergence. Drosophila embryos develop in the egg mem-brane. The egg hatches and produces a larva that feeds by burrowing through the medium. The lar-val period consists of three stages, or instars, the end of each stage marked by a molt. The first in-star 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 pe-riod, the third instar larva will crawl up the sides of the culture vial, attach themselves to a dry sur-face (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.


♀ ♂ Egg

Adult

Larva (1rst instar)

Larva (2nd instar)

Larva (3rd instar)

Pupa

Figure 1 Drosophila Life Cycle

DID YOU KNOW? Male Drosophila are the smaller of the two sexes.

Different body features characterize the male and female flies. The females are slightly larger and have a light-colored, pointed abdo-men. 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).

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 sex-linked crosses are performed; the offspring of the first cross, the F1 generation, are normally allowed to breed among themselves to pro-duce 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 Pun-nett square is extremely important in determining the outcome of crosses in Mendelian genetics; it clearly displays the possible combi-nations 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 vestig-ial 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 Pun-nett square; if it is known that the trait for vestigial wings is a reces-sive mutation, the flies with vestigial wings must be homozygous re-cessive for the first trait, and therefore have a genotype of vv. Assum-ing 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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DID YOU KNOW? A Punnett square displays the possible combinations of a genetic cross, based on the Laws of segregation and inde-pendent assortment.

Male sex combs on foreleg Female Male

Figure 2

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 ves-tigial wings as with normal wings. Punnett squares become more complicated when diagramming a di-hybrid cross. Due to the complex nature of dihybrid crosses, or even crosses with three or more traits, it is even more important to dia-gram 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. 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 re-sults with an accepted set of data to determine how much the experi-mental values deviated from the accepted ones and whether or not that deviation can be explained solely by chance.


sV sV sV sV

Sv SsVv SsVv SsVv SsVv


sv ssVv ssVv ssVv ssVv

sv ssVv ssVv ssVv ssVv

Males

Fem

ales

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 inde-pendent of one another unless the genes are on the same chromosome.

V v

v Vv vv

v Vv vv

Male

F

emal

e

The square of the difference between the observed and expected val-ues (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 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 in-cluded 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 im-pact 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 ex-perimental 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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DF a v P=0.99 0.95 0.80 0.50 0.20 0.05 0.01

1 0.00016 0.00393 0.06420 0.45500 1.64200 3.84100 6.63500 2 0.02010 0.10300 0.44600 1.38600 3.21900 5.99100 9.21000 3 0.11500 0.35200 1.00500 2.36600 4.64200 7.81500 11.34500 4 0.29700 0.71100 1.64900 3.35700 5.98900 9.44800 13.27700 5 0.55400 1.14500 2.34300 4.35100 7.28900 11.07000 15.08600 6 0.87200 1.63500 3.07000 5.34800 8.55800 12.59200 16.81200 7 1.23900 2.16700 3.82200 6.34600 9.80300 14.06700 18.47500 8 1.64600 2.73300 4.59400 7.34400 11.03000 15.50700 20.09000 9 2.08800 3.32500 5.38000 8.34300 12.24200 16.91900 21.66600 10 2.55800 3.94000 6.17900 9.34200 13.44200 18.30700 23.20900 15 5.22900 7.26100 10.30700 14.33900 19.31100 24.99600 30.57800 20 8.26000 10.85100 14.57800 19.33700 25.03800 31.41000 37.56600 25 11.52400 14.61100 18.94000 24.33700 30.67500 37.65200 44.31400 30 14.95300 18.49300 23.36400 29.33600 36.25000 43.77300 50.89200

Chi-square Table

DID YOU KNOW? Chi-square is officially known as the Pearson chi-square in homage to its inventor Karl Pearson (1857-1936).

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 nor-mal wings, use the following table of results for the F2 generation: The calculation needs to be performed only twice — once for each phenotype. Adding the results together provides the value for χ2. 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 ac-cepted 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.


Phenotype No. of Males No. of Females

Vestigial Wings 14 12

Normal Wings 36 38

Genotype: Description of the set of alleles an individual pos-sesses for a gene or set of genes, expressed or not, usu-ally shown in symbols (ie., AA, Aa, or aa).

Phenotype: Physical traits expressed by an individual, regardless of the genotype.

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.

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 sev-eral ice packs for 10 to 20 minutes or until the flies appear lifeless.

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 table-

spoon 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 col-lected on the foam plug, remove the ice pack with the Petri dish from the freezer. Quickly dump the flies into the Petri dish.

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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PRE-LAB PREPARATION

NOTE

NOTE

TIMING

DID YOU KNOW? The chromosomes of Droso-phila can stretch to more than a mile long when unraveled.

Part B: Performing a Drosophila Cross

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 initia-tion of the next generation.

1. 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.

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 me-

dium 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.


NOTE

NOTE

DID YOU KNOW? In 1910, Thomas Hunt Mor-gan discovered sex-linkage in Drosophila and postulated a connection between eye color in fruit flies and human color blindness.

MATERIALS NEEDED PER GROUP 1 Culture vial of wild-type Drosophila 1 Culture vial of monohybrid cross or 1 Culture vial of dihybrid cross or 1 Culture vial of sex-linked cross Isopropyl alcohol 10%, 100 ml 1 Camel’s hair brush 1 Thermo-anesthetizer 1 Petri dish 2 Drosophila vials and labels Drosophila medium 1 Fly morgue Forceps Part A: Working with Drosophila

You will need to observe wild type Drosophila to familiar-ize yourself with the wild type phenotype. You will even-tually 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 in-

structor will demonstrate the proper immobilization technique.

Your instructor may have immobilized the flies in ad-vance. If this is the case, begin with Step 2.

2. Observe the flies’ traits, particularly body features that distin-

guish males and females, eye color, and wing size and shape. Re-cord your observations in Table 1 in the Analysis section. If, at any time during your observations, the flies begin to become ac-tive, re-immobilize them according to your instructor’s directions.

Use the camel’s hair brush to move the flies when making observations.


MATERIALS

PROCEDURE

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NOTE

NOTE

NOTE

DID YOU KNOW? A female Drosophila once fer-tilized, may lay 30-50 eggs per day throughout her lifetime.

Part B: Performing a Drosophila Cross 1. Obtain a vial of a prepared Drosophila cross.

These flies have been mated and may exhibit one or more mutations. They are the parental generation for your ex-periment. 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 obser-vations, the flies begin to become active, re-immobilize them ac-cording 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 ap-parent in the male or female flies. Record your observations in Table 2.

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. Ob-serve the vial occasionally and record your observations.

Do not allow the temperature to exceed 30°C.

7. When the adult flies emerge, collect, immobilize, and examine

them. Note the sex of each one, as well as the presence of any mu-tations. 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.


NOTE

NOTE

!CAUTION

DID YOU KNOW? The longest sperm ever re-corded is 10,000 times longer than a human spermatozoan, and belongs to Drosophila bifu-ria. The typical male D. bifuria sperm is approximately 60 mm long, 20 times the length of the fly itself.

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. 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.

12. The F2 adults will gradually emerge from the pupae over several

days. As the F2 generation flies begin to emerge as adults, immo-bilize and examine them. Record the number of males and fe-males, noting any mutations which may be present. Record your findings in Table 4.

Try to collect as many adults as possible.


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NOTE

!CAUTION

DID YOU KNOW? In 1913, American geneticist Alfred Henry Sturtevant de-veloped a technique for map-ping the location of specific genes of the chromosomes in Drosophila.

© 2002 WARD’S Natural Science Establishment, Inc. All Rights Reserved

Polytene Chromosome Investigation

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 investiga-tions into chromosome function and structure. They are most evident in D. virilis larvae, but can be observed in any species of Drosophila.

MATERIALS NEEDED PER LAB GROUP HCl, 8% Aceto-orcein stain, 30 ml Piccolyte II, 15 ml 1 Teasing needle 1 Pipet 1 Forceps 1 Microscope slide 1 Coverslip 1 Drosophila larva Gloves Goggles Lab aprons Stereomicroscope Distilled water Paper towels 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 ap-

pear 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.

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NOTE

PROCEDURE

OPTIONAL EXERCISE

MATERIALS

DID YOU KNOW? Drosophila salivary glands consist of two major cell types secretory cells and duct cells.

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.

Do not let the salivary glands dry out.

8. Add two or three drops of 8% HCl to hydrolyze the genetic mate-

rial 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.

9. 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 con-tact. Wear safety goggles, gloves, and a lab apron.

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 cover-

slip. 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.

16

NOTE

!CAUTION

!CAUTION

DID YOU KNOW? Measuring 2 mm in length, the polytene chromosomes found in the salivary glands of Diptera are much longer than metaphase chromosomes.


ANALYSIS

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Eye Color Wing Size and Shape

Male

Female

Table 1 Phenotypes of Wild Type Drosophila


Table 2 Phenotypes of the Parental Generation

Cross Letter:______


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Table 3 Phenotypes of the F1 Generation

Cross Letter:______


Table 4 Phenotypes of the F2 Generation

Cross Letter:______


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Polytene Chromosome Investigation

Band

Inter-band Drosophila Chromosomes Squashed (1000X)

Chromocenter

“Puff” Region of RNA

synthesis

Sample Results For Each Cross

Male Female Conclusions

(A) Monohybrid Cross Answers will vary Student results should approximate a ratio of: 3: wild eye 1: sepia eye

Sepia eyed 16 15

Wild eyed 46 51

(B) Dihybrid cross

Wild eyed/normal wing 63 59

Sepia eyed/normal wing 18 22

Wild eyed/vestigial wing 20 18

Sepia eyed/vestigial wing 5 4

(C) Sex linked Cross 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

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


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) 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) c) Sex-Linked Cross Cross between female flies with white eyes and wild-type male flies: XWY (males) x XwXw (females)

ASSESSMENT


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sV sV sV sV





Males

Fem

ales

S S

s Ss Ss

s Ss Ss

Males

Fem

ales

XW Y

Xw XWXw XwY

Xw XWXw XwY

Males

Fem

ales

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.

3. Describe the F1 cross you performed, and draw a Punnett square to show allelic combinations pos-

sible in the F2 generation. a) Monohybrid Cross F1 cross: Ss x Ss b) Dihybrid Cross F1 cross: SsVv (males) x SsVv (females) c) Sex-Linked Cross F1 cross: XWXw x XwY


SV Sv sV sv

SV SSVV SSVv SsVV SsVv

Sv SSVv SSvv SsVv Ssvv

sV SsVV SsVv ssVV ssVv

sv SsVv Ssvv ssVv ssvv

Males

Fem

ales

S s

S SS Ss

s Ss ss

Males

Fem

ales

Xw Y

XW XWXw XWY

Xw XwXw XwY

Males

Fem

ales

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, muta-

tions 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.

22 Call WARD’S at 1-800-962-2660 for Technical Assistance

c) Sex-Linked Cross χ2 = .06375, with v = 3, for better than 99% confidence that the variation in the data is due en-tirely 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 calcu-lated, resulting in four data points.

7. 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 ge-

netic code and transmits the hereditary pattern.

23

DF v P=0.99 0.95 0.80 0.50 0.20 0.05 0.01

1 0.00016 0.00393 0.06420 0.45500 1.64200 3.84100 6.63500 2 0.02010 0.10300 0.44600 1.38600 3.21900 5.99100 9.21000 3 0.11500 0.35200 1.00500 2.36600 4.64200 7.81500 11.34500 4 0.29700 0.71100 1.64900 3.35700 5.98900 9.44800 13.27700 5 0.55400 1.14500 2.34300 4.35100 7.28900 11.07000 15.08600 6 0.87200 1.63500 3.07000 5.34800 8.55800 12.59200 16.81200 7 1.23900 2.16700 3.82200 6.34600 9.80300 14.06700 18.47500 8 1.64600 2.73300 4.59400 7.34400 11.03000 15.50700 20.09000 9 2.08800 3.32500 5.38000 8.34300 12.24200 16.91900 21.66600

10 2.55800 3.94000 6.17900 9.34200 13.44200 18.30700 23.20900 15 5.22900 7.26100 10.30700 14.33900 19.31100 24.99600 30.57800 20 8.26000 10.85100 14.57800 19.33700 25.03800 31.41000 37.56600 25 11.52400 14.61100 18.94000 24.33700 30.67500 37.65200 44.31400 30 14.95300 18.49300 23.36400 29.33600 36.25000 43.77300 50.89200

a


250-7055t genetics of drosophila

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