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Genetics: A Laboratory Manual

ASA-CSSA-SSSA Book and Multimedia Publishing Committee

David Baltensperger, ChairWarren Dick, ASA Editor-in-ChiefCraig Roberts, CSSA Editor-in-ChiefSally Logsdon, SSSA Editor-in-ChiefMary Savin, ASA RepresentativeHari Krishnan, CSSA RepresentativeApril Ulery, SSSA Representative

Lisa Al-Amoodi, Managing Editor

GeneticsA Laboratory Manual2nd edition

Gregore Koliantz & Daniel B. Szymanski

iii Genetics: A Laboratory Manual

Table of Contents

Preface v

Exercise 1 Cell Division (Mitosis): Looking at Chromosomes . . . . 1The Cell Cycle 2Mitosis in Onion Root Tip Squashes 2

Materials 2Procedure 3

The Polytene Chromosomes of Dipteran Larvae 3Materials 4Procedure 4

Protocol 1–1: How to Prepare and Preserve Onion Roots 5Protocol 1–2: How to Make Chromosome Preparations Permanent 5Study Questions 5

Exercise 2 Physical Properties of DNA and DNA Assay . . . . . . . . 6The “Double Helix” 7Part 1: DNA Extraction from Onion 7


General Description of Agarose Gel Electrophoresis 9Part 2: DNA Assay 11


Study Questions 12

Exercise 3 Restriction Mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . 13Restriction Enzymes (Endonucleases) 13Construction of Restriction Maps 14Restriction Digestion of lambda DNA 15Tips on Restriction Digestion Reactions 15


Study Questions 17

Exercise 4 Extraction of Genomic DNA . . . . . . . . . . . . . . . . . . . . . 19DNA Extraction from the Plant Arabidopsis thaliana 191. Large-Scale DNA extraction by CTAB 19


Further Purification of DNA 20Materials 20Procedure 20

2. Small-Scale DNA extraction by CTAB 21Materials 21Procedure 21

3. Small-Scale DNA extraction by DNAzol 21Materials 21Procedure 22

Study Questions 22

Exercise 5 Polymerase Chain Reaction . . . . . . . . . . . . . . . . . . . . . 23Optimizing the Polymerase Chain Reaction 24Amplification of a Fragment of the TTG Locus in Arabidopsis thaliana 25


Protocol 5–1: How to Find the Nucleotide Sequence of the Arabidopsis TTG Gene 29

Protocol 5–2: How to Generate Primers for the Target Region of Interest 29

Study Questions 31

Exercise 6 Cloning 1: Generating Recombinant Plasmids . . . . . 32Cloning Vectors 32What is Transformation? 33Cloning of the PCR Product from Exercise 5 331. Gel Electrophoresis of the PCR Product 34


2. Cloning of the PCR Product 35Materials 35Procedure 36

Study Questions 36

Exercise 7 Cloning 2: Minipreping . . . . . . . . . . . . . . . . . . . . . . . . 37Preparation of Liquid Cultures from White and Blue Colonies 37


Extraction of Plasmid DNA from Liquid Cultures (Miniprep) 38Materials 38Procedure 38

Restriction Digestion of the Extracted DNA 39Materials 39Procedure 39

Study Questions 40

Exercise 8 Southern Blotting 1: DNA Transfer . . . . . . . . . . . . . . . 41DNA Transfer 41Electrophoresis and Southern Blotting of Digested Recombinant

and Nonrecombinant Plasmids 421. Electrophoresis of EcoRI and HindIII-Digested Plasmid DNA 42


2. Transfer of Digested Plasmid DNA from Agarose Gel onto Nylon Membrane 42


Study Questions 43

Exercise 9 Southern Blotting 2: DNA–DNA Hybridization and Sequence Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Probe Preparation and Hybridization 451. Purification of PCR-Amplified TTG DNA 45


2. Probe Preparation 45Materials 45Procedure 46

3. Hybridization 46Materials 46Procedure 46

ivGregore Koliantz & D.B. Szymanski

4. Immunological Detection 47Materials 47Procedure 48

Protocol 9–1: Removing Probe from Nylon Membrane 491. Stripping DIG-Labeled DNA Probe after Chromogenic Detection 492. Stripping DIG-Labeled DNA Probe after Chemiluminescent Detection 49Study Questions 49

Exercise 10 Regulation of Gene Expression . . . . . . . . . . . . . . . . . 51Prokaryotic Gene Regulation 51What are Reporter Genes? 52Expression of the luxE Gene in E. coli 52


Expression of the gfp Gene in E. coli 54What is pGreen Plasmid? 55


Eukaryotic Gene Regulation 56Expression of the gusA Gene in Arabidopsis thaliana 56


Protocol 10–1: Generation of Transgenic Plants: Transformation of Arabidopsis 58

Protocol 10–2: Surface Sterilization of Arabidopsis Seeds for Planting on Agar Plates 60

Study Questions 61

Exercise 11 Transmission Genetics (Heredity) . . . . . . . . . . . . . . . 62Important Terms 62Mendel’s Principles 63Statistical Treatment of Data: The Chi-Square Test 64Exercises with Arabidopsis thaliana 65Description of the Exercise 67


Exercises with Corn (Zea mays) 68Description of the Exercise 69


Sex-Linked Inheritance 69Exercises with Drosophila melanogaster 69Description of the Exercise 71


Protocol 11–1: Preparation of Pots for Planting Arabidopsis Seeds 73Protocol 11–2: Setting Up Arabidopsis Crosses 73Protocol 11–3: Ethylmethane Sulfonate Mutagenesis

of Arabidopsis Seeds 74Protocol 11–4: How to Maintain Drosophila Stocks 75Protocol 11–5: How to Collect Drosophila Virgin Females 75Protocol 11–6: How to Prepare Drosophila Medium 75Study Questions 76

Exercise 12 Meiosis and Analysis of Crossing Over . . . . . . . . . . . 77Meiosis I: A Reduction Division 77Meiosis II: An Equational Division 78Studying Meiosis in Lily 79Meiosis in Flowering Plants 79Linkage and Crossing Over 79Analysis of Crossing Over in Sordaria fimicola 801. The Life Cycle 80


2. Analysis of Crossing Over 81Materials 81Procedure 81

Protocol 12–1: How to Maintain Sordaria Cultures 82Study Questions 83

Exercise 13 Complementation Test . . . . . . . . . . . . . . . . . . . . . . . . 84

Genetic Control of Biochemical Reactions 84Biosynthesis of Prodigiosin in Serratia marcescens 87Description of the Exercise 87


Protocol 13–1: How to Maintain Serratia Cultures 88Study Questions 88

Exercise 14 Molecular Markers: Mapping the Genome of Arabidopsis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

Methods of Gene Mapping 90How Accurate Is a Genetic Map? 90The Use of Molecular Markers in Gene Mapping 90SSLP Mapping in Arabidopsis 92How is a Gene Mapped? 92Mapping of the ttg Gene in Arabidopsis 941. DNA Amplification 94


2. Gel Electrophoresis of the PCR Product 95Materials 95Procedure 95

Discussion 96Study Questions 96

Exercise 15 Population Genetics: How Changes Occur within a Population . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

The Hardy–Weinberg Equilibrium 97Random Changes in Allele Frequencies: Genetic Drift 100Simulation of Population Genetics 100


Study Questions 101

Appendix 1 Preparing a Laboratory Report . . . . . . . . . . . . . . . . 102

Appendix 2 How to Write a Scientific Paper . . . . . . . . . . . . . . . . 104

Appendix 3 Suppliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

Suggested Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .110

v Genetics: A Laboratory Manual

Personal PerspectiveI would like to express my gratitude to Dr. Dan Szymanski for allowing me to use the facilities of his laboratory and to Eileen Mallery, the laboratory manager, for sharing her experience in some protocols with me.

Gregore Koliantz

The first edition of this lab manual was published in the fall of 2006 with the objective of providing the students of life sciences with the experimental aspects of classical and molecular genetics. In the second edition, we have updated and revised every chapter, expanded the topics of genomic DNA extraction, prokaryotic gene regulation, transmission and population genetics, added new protocols and an appendix, revised a number of study questions that accompany each chapter, and improved the quality of some figures. It is our hope that this edition will continue to serve as a guide for students into research projects and investigations.

As we mentioned in the first edition of this book, the idea of most of the exercises described in this manual originated from the laboratory of Dr. D. Szymanski. We performed all the exercises to ascertain their practicability and feasibility before attempting to include them in the manual.

Throughout the publication of the current and the previous edition of this book, we received support from Dr. Craig A. Beyrouty, Head of the Department of Agronomy of Purdue University, to whom we express our thanks and appre-ciation. We also wish to give our thanks and extend our heartfelt gratitude to all who helped and encouraged us throughout the work or allowed us to reproduce their materials. Acknowledgments and credits are given to the following col-leagues and publishers:

Dr. Bruce M. Applegate, Department of Food Science of Purdue University, for pro-• viding us with lux operon DNA and information concerning its sequence. Dr. David D. Baltensperger, the ASA–CSSA–SSSA Book and Multimedia Publishing • Committee Chair, for critically reviewing the manuscriptLisa Al-Amoodi, the Managing Editor of ASA–CSSA–SSSA Books, without whom • this book would never have been published.American Society of Plant Biologists for giving us the permission to reproduce Table • 11–3 from The Plant Cell 13:1499–1510, 2001.Association of Biology Laboratory Education (ABLE) for allowing us to use the concept • of the experiment entitled “Genetic Control of Cell Chemistry Using Serratia marc-escens” which appeared in Tested Studies for Laboratory Teaching, volume 14, 1993.Brooks/Cole, a division of Thomson Learning, for permitting us to use Fig. 1–4 from • Starr and Taggart (1998). Carolina Biological Supply Company for providing us with Fig. 1–1 and 12–8.• Invitrogen• Corporation for providing us with Fig. 6–4.John Wiley & Sons for allowing us to reproduce Fig. 8–1 from Ausubel et al., (1992). • McGraw-Hill Companies for granting us the permission to reproduce Fig. 12–1, 12–2, • and 12–3 from Hartwell et al. (2000), Fig. 2–2, 2–3, 2–4, and 2–5 from Tamarin (2002), and Fig. 12–5 and 12–6 from Brooker (2005). Pearson Education, Inc. for permitting us to reproduce Fig. 12–11 from Mertens and • Hammersmith (1998) and Fig. 5–3, 6–1, and 6–2 from Russell (1996). Qiagen Inc. for allowing us to use the QIAquick PCR purification kit protocol.•

Gregore Koliantz Daniel B. SzymanskiDepartment of Agronomy Department of AgronomyPurdue University Purdue University

Preface

Cell Division (Mitosis)Looking at Chromosomes

Objectives

• to study the stages of mitosis in onion root tip

• to study the polytene chromosomes of the fruit fly

Exercise 1

1 Genetics: A Laboratory Manual

Cell division is a process through which the continuity of life is preserved. The vegetative growth of all organisms and replacement of certain cells are maintained by mitosis. A separate mechanism, called meiosis, which will be discussed in Exercise 12, involves in the formation of gametes and spores. When a diploid (2n) cell undergoes mitotic division, the daughter cells receive the same number of chromosome sets, so that all the hereditary information is transferred from the mother cell to the daughter cells.

Mitosis comprises only 10% of the cell cycle. The rest of the cycle is inter-phase where newly formed cells (from the previous cell cycle) increase their mass, duplicate their DNA, and the materials needed for cell growth are synthe-sized (see The Cell Cycle below).

In eukaryotic cells, a number of proteins are attached to the DNA which cause the compaction of the DNA or have a role in replication and transcrip-tion of the DNA molecule. Each DNA molecule with its attached proteins is a chromosome.

In the animal and plant kingdoms, the chromosome number varies between species; however, two distant species may have the same number of chromo-somes (Table 1–1).

Table 1–1. Chromosome number of selected species.

Organism Diploid chromosome number (2n)Arabidopsis thaliana (L.) Heynh. (thalecress) 10Allium cepa L. (onion) 16Nicotiana tabacum L. (tobacco) 48Secale cereale (rye) 14Zea mays L. (corn) 20Pisum sativum L. (garden pea) 14Oryza sativa L. (rice) 24Antirrhinum majus (snapdragon) 16Canis familiaris (dog) 78Felis catus (cat) 38Gallus domesticus (chicken) 78Rana pipiens (leopardfrog) 26Drosophila melanogaster (fruitfly) 8Musca domestica (housefly) 12Mus musculus (housemouse) 40Equus caballus (horse) 64Sus scrofa (pig) 38Pan troglodytes (chimpanzee) 48Homo sapiens (human) 46Saccharomyces cerevisiae (baker’s yeast) 17†Neurospora crassa (pink bread mold) 7†

† Haploid (1n).

2Gregore Koliantz & D.B. Szymanski

The Cell CycleThe process of mitosis is essentially the same in plants and animals, and it begins at interphase. In the next five stages, which include prophase, metaphase, anaphase, telophase, and cytokinesis, deep changes occur within the nucleus which are visible under light microscope (Fig. 1–1).

1. InterpaseThe interphase cycle is divided into three phases, as described below.

G• 1 phase: during G1, the cells grow and carbohydrates, lipids, and proteins are assembled for the cell’s own use.S phase:• during S, the DNA duplicates to form sister chroma-tids that are joined at the centromere; they are not visible under the light microscope.G• 2 phase: during G2, cells grow slightly, and the proteins which are needed for the completion of mitosis are synthesized.

2. Prophase At prophase, the chromosomes gradually condense, but they are not clearly visible under light microscope. Also, the microtubules are formed between the two centrosomes, and at the end, the nuclear membrane breaks down.

3. MetaphaseIn this stage, the nuclear membrane disappears and sis-ter chromatids attach to the microtubules through the kinetochore. Kinetochore is a protein that surrounds the centromere of sister chromatids. As these events are occur-ring, sister chromatids move toward the midpoint of the microtubules and align along a plane called the metaphase plate. At this stage of mitosis, chromosomes are clearly visible under the light microscope.4. Anaphase Anaphase starts with separation of sister chromatids from each other. This happens because the protein complex called

cohesin, which facilitates the alignment of sister chromatids, degrades. Once each chromatid is separated from its sister, it is considered a chromosome. As anaphase proceeds, the microtubules shorten and pull the chromosomes toward the pole to which they are attached.

5. Telophase At telophase, the chromosome sets arrive at their respective poles and deconden-sation starts. The nuclear membrane reforms and separates the chromosomes from the cytoplasm. Thus, two daughter nuclei are formed.

6. CytokinesisThe daughter nuclei formed at the end of the telophase are now packed into sepa-rate cells. Cytoplasmic division follows the nuclear division.

An interpretative diagram of the cell cycle is shown in Fig. 1–2.

Mitosis in Onion Root Tip SquashesIf onion bulbs are placed in water for several days, they will grow roots (Fig. 1–3). Onions obtained from grocery stores provide very few, if any, roots because they are treated to prevent sprouting in storage. Homegrown onions sprout more effectively. Onion bulbs can also be obtained from biotechnology companies which provide live materials for teaching purposes. Roots that are 3 to 4 cm in length can be used in this exercise.

The plant root grows through the formation of new cells restricted to the growth region called the meristem, which is located near the tip of the root, about 2 mm above the root cap. The root cap has a protective function (Fig. 1–4). In the exercise described below, you will prepare squashes from onion root tips to study the stages of mitosis. Show your results to the laboratory instructor for verification.

MaterialsTennis ball-sized onions with rootsCompound microscope with 4´, 10´, and 40´ objectives

Fig. 1–1. Mitosis in onion root tip. Reproduced by permission of Carolina Biological Supply Company.

Fig. 1–2. Diagram of a eukaryotic cell cycle.


1 M Hydrochloric acid:Add 8.26 mL of concentrated hydrochlo-

ric acid to 91.74 mL of distilled water (hereafter dH2O), mix well.

1% Methylene blue:Per 100 mL of dH2O1 g Methylene blue

Microscope slidesCover slipsRazor bladesForcepsDissecting needlesWatch glassesPaper towelsGlovesRulers

Procedure1. One week before the start of the exercise, grow roots of

a tennis ball-sized onion by placing it into water (see Fig. 1–3). Grown roots can be detached from onion and pre-served in 100% ethanol for an extended amount of time.

2. With the aid of a sharp razor blade, cut the root about 2 mm above the root cap. This is where the meristematic region is located (see Fig. 1–4).

3. Transfer the tissue onto a watch glass and cover it with 1 M hydrochloric acid for 5 min.

4. Stop the hydrolysis by transferring the tissue to three drops of 1% methylene blue on a clean microscope slide, and allow the tissue to stain for 10 min.

5. Place the slide on two or three pieces of paper towel, and then place the cover slip over the specimen. Hold the edges of the cover slip securely and with the handle side of the dissecting needle, firmly tap on the cover slip several times, then cover the entire slide with two pieces of paper towel, and with your thumb, firmly press on the cover slip; do not twist the cover slip. The pressure will spread the cells into a single layer.

6. Study the mitotic stages with the 4×, 10×, and 40× ob-jectives of a compound microscope.

The Polytene Chromosomes of Dipteran Larvae

The salivary glands of dipteran (two-winged flies) larvae possess cells that contain large and multistranded polytene chromosomes. These chromosomes are formed through the process in which the DNA strands undergo several rounds of replication without separation. As a result, the centromeres of the chromosomes fuse together to form a dense region called chromocenter (Fig. 1–5). Furthermore, the homolo-gous chromosome arms are paired tightly and extend forth from the chromocenter. In the fruit fly, Drosophila mela-nogaster, the DNA replicates 10 times without separation, resulting in giant chromosomes that have 2048 double helices. When polytene chromosomes are stained with lacto-aceto orcein, dark-stained bands and light-stained interbands will be visible throughout the chromosome arms (see Fig.1–5). These are called banding patterns and are characteristic to each chromosome arm. In Drosophila, more than 5000 bands have been identified in the six arms of the polytene chromosomes. Each band is made up of several thousand nucleotide pairs that may code for several proteins. Therefore, the bands are the visual manifestation of genes. The complete banding pattern of the polytene chromosomes of D. melanogaster is available (see e.g., Lefevre, 1976).

Fig. 1–3. An onion bulb in a jar of water. Note the growing roots.

Fig. 1–4. Root apical meristem. Reproduced from Starr and Taggart (1998) by permission of Brooks/Cole, a division of Thomson Learning.


MaterialsCommercially prepared slides of D. melanogaster polytene

chromosomesDrosophila melanogaster third instar larvaeInstant Drosophila medium (see Protocol 11–6 for details)45% acetic acid:

Add 45 mL of glacial acetic acid to 55 mL of dH2O; mix well.Lacto-aceto orcein (the staining solution):

Dissolve 2 g of orcein in 100 mL of a mixture of equal vol-umes of 85% lactic acid, glacial acetic acid, and dH2O; heat to facilitate the dissolution, and filter while hot.

To prepare 85% lactic acid, add 85 mL of lactic acid to 15 mL of dH2O, mix well.

Dissecting microscopesCompound microscopes with 4´, 10´, and 40´ objectivesBench top incubator adjusted to 23°CRefrigerator or cold room adjusted to 10°CSharp dissecting needlesForcepsMicroscope slidesCover slips100% ethanol or a mixture of water and detergentBlotting papersPaper towelsGloves

ProcedureExamine commercially prepared slides of Drosophila poly-tene chromosomes. Locate the chromocenter and the six

chromosome arms. Note that the Y chromosome is not visible in chromosome squashes prepared from male larvae because this chromosome is under-replicated because of its heterochromatic nature. The Y chromosome forms a dense mass attached to the chromocenter.1. One week before the start of the exercise,

transfer 20 to 30 inseminated female flies to each bottle containing Drosophila medium. Place the bottles in a 23°C incubator until the third instar larvae appear on the sides of the bottles. Well-fed larvae measure about 3 to 4 mm. Discard the flies by dropping them into a jar containing alcohol or a mixture of water and a detergent, and save the bottles in a 10°C refrigerator or cold room until the time of the experiment.

2. Remove a large third instar larva from the bottle and place it in three or four drops of 45% acetic acid on a clean microscope slide.

3. Place the slide on the stage of a dissecting microscope.

4. Using an appropriate magnification (e.g., 5´ or 10´) dissect the larva. Insert one dis-secting needle in the middle of the larva and the other needle at the anterior, as illus-trated below. The anterior end of the larva is characterized by black mouth hooks.

5. Hold the larva with the first needle and then slowly pull outward the needle at the anterior end of the larva. This will cause the internal organs of the larva to be pulled out of the cuticle. Note that the salivary glands appear as a pair of cylindrical organs located on either side of the esophagus. A piece of fat body is attached to each gland.

6. Detach the glands from the alimentary canal and re-move, as much as possible, the adhering fat body with-out damaging the glands. Then transfer the glands to three drops of staining solution on the same slide, and stain for 15 to 20 min. Remove the remains of the larva from the slide with clean blotting paper.

Fig. 1–5. The polytene chromosomes of the salivary glands of Drosophila melanogaster. Inset shows the tip of the X chromo-some and the chromosome arm 2L.


7. Place the slide on a bed of several sheets of paper tow-els, then lower the cover slip over the glands. Hold the edges of the cover slip secure and with the dissecting needle, gently tap on the cover slip several times, then cover the entire slide with two pieces of paper towel, and with your thumb firmly press on the cover slip; do not twist the cover slip.

8. Blot off any excess stain with clean blotting paper, and study the slide with the 10´ and then with the 40´ objective of the compound microscope.

Protocol 1–1How to Prepare and Preserve Onion RootsPlace an organically grown onion in a jar of water at room temperature under ceiling light, as illustrated in Fig. 1–3. If ceiling light is not available, use a 17-W fluorescent tube. Fill the jar with water as needed. When the roots grow to at least 3 cm, remove the onion from water, cut the roots, and place 10 to 12 roots in each conical tube filled with 100% ethanol. Store the tubes in a refrigerator.

Protocol 1–2How to Make Chromosome Preparations PermanentPreparations of onion root tips or Drosophila salivary glands, described above, can be made permanent for future use.

MaterialsPreparations of interestLiquid nitrogen or dry iceRazor bladeForcepsCover slipsCoplin jars or weighing boatsEthanol (absolute)95% ethanol:

95 mL Ethanol5 mL dH2O

Mounting medium, e.g. Euparal (available through Carolina Biological Supply Co., Burlington, NC) or equivalent

Procedure1. Immerse the preparation of interest into liquid nitro-

gen, and wait until the liquid stops boiling. It may take less than 1 min. In case liquid nitrogen is not available, place the slide on a block of dry ice for at least 5 min to freeze the material under the cover slip.

2. Remove the slide from the liquid nitrogen and flip off the cover slip with a razor blade. Or, if using dry ice, while the preparation is still on the ice, flip off the cover slip with a razor blade. If a new slide is used for preparation of the specimen, most of the material will stick to the slide.

3. With a pair of forceps, immediately immerse the entire slide in 95% ethanol for 5 min. Use a Coplin jar for this purpose. A weighing boat can also be used.

4. Transfer the slide to another jar (or weighing boat) con-

taining absolute (100%) ethanol; leave for 10 to 15 min.5. Remove the slide from the ethanol, immediately add

one drop of mounting medium on top of the specimen, and then place a clean cover slip on top of the medium.

6. Leave the preparation in a level place for 24 h to dry.Permanent preparations can be kept in a slide box and stored at room temperature.

Study Questions1–1. For the mitotic events, described below, write the

name of the event.a. Chromosome replication occurs.b. Chromosomes line up on equator of the cell.c. Condensation of replicated chromosomes begins and the

chromosomes become visible under the microscope.

1–2. The plant Arabidopsis thaliana has five pairs of chromosomes. How many chromosomes would you expect to find in (a) leaf cells, (b) petal cells, and (c) pollen grains?

1–3. Under a compound microscope, how can you distin-guish between a cell in the prophase stage and a cell in the interphase stage?

1–4. What is “under-replication”?

1−5. You are studying mitosis in an organism which has two pairs of chromosomes (2n = 4). One pair is telocentric and the other metacentric, as illustrated below.

Draw the chromosome configuration in stages G1,

G2 (assume that these stages are visible under micro-scope), metaphase, anaphase, and daughter cells.

Physical Properties of DNA and DNA Assay

Objectives

• to study the physical properties of DNA

• to practice the technique of quantitative dilution

• to understand the basis of gel electrophoresis

The remarkable work of Taylor and his colleagues in 1957 on the organiza-tion and duplication of chromosomes revealed that eukaryotic chromosomes are composed of one molecule of double helix DNA. They showed that if a

chromatid undergoes DNA repli-cation in the medium containing radioactively labeled thymidine (3H-), the double helix is separated and then each original strand serves as a template for the synthesis of the complementary DNA molecule labeled with 3H-. Thus, the new double helix would be made up of one original (nonradioactive) strand and one new (radioactive) strand: a “hybrid” molecule. Subsequent rep-lication in a nonradioactive medium would result in one original double helix and one “hybrid” DNA mol-ecule. This configuration is called the semiconservative model of DNA replication (Fig. 2–1).

DNA is a long molecule. On average, the length of the DNA molecule of a single eukaryotic cell is about 5 cm. This long molecule is accommodated in the cellular nucleus by primary and second-ary coiling and folding. Initially, DNA compaction occurs when it is wrapped around core proteins called histones. Histones are argi-nine- and lysine-rich basic proteins that make up a substantial por-tion of eukaryotic nucleoprotein. The nucleoprotein material of the chromosome is called chromatin. The chromatin appears as small particles called nucleosomes, connected to each other by DNA strands called linkers (Fig. 2–2). The nucleosomal DNA undergoes coiling to form a solenoid-like fiber (Fig. 2–3), and then the solenoid-like fiber further winds forming a


Fig. 2–1. The semiconservative model of DNA replication.

Fig. 2–2. Interpretative diagram of a nucleosome. Note the histone cores and the DNA wrapped around the cores. Reproduced from Tamarin (2002) by permission of The McGraw-Hill Companies.

Exercise 2

second solenoid-like structure (Fig. 2–4). This is the physi-cal structure of the chromatin in metaphase chromosomes that you studied in Exercise 1.

The “Double Helix”According to Watson and Crick’s model (Watson and Crick, 1953), the DNA molecule is composed of four bases [adenine (A) guanine (G), thymine (T), and cytosine (C)], and along with deoxyribose–phosphate group, which is positioned outside the bases, form a double deoxyribonu-cleotide strand. The two strands wind around each other in a clockwise fashion to form a double helix (Fig. 2–5). This is called right-handed DNA.

The two strands of the DNA are antiparallel; that is, they are oriented in opposite directions: one strand in the 5¢ to 3¢ direction with regard to the deoxyribose molecule and the other strand in the 3¢ to 5¢ direction (see Fig. 2–5 right). The bases of both chains are oriented perpendicularly to the long axis of the DNA, so that the bases are stacked on top of one another. As shown in Fig. 2–5, the bases of the opposite strands are bonded together by weak hydrogen bonds: A with T with two hydrogen bonds and G with C with three hydrogen bonds. The weak hydrogen bonds make it relatively easy to separate the two strands of DNA by heating. This phe-nomenon is called denaturation. The diameter of the helix is 2 nm. The base pairs are 0.34 nm apart in the DNA helix. A 360° turn of the helix takes 3.4 nm; therefore, there are 10 base pairs per turn. Because of the way the bases bond with each other, the two sugar-phosphate strands of the double helix are not equally spaced along the helical axis. This con-figuration results in grooves of unequal size called the major groove and the minor groove.

The DNA structure, outlined above, is known as B-DNA and is found predominantly in cells. In solution, B-DNA is a quite flexible molecule, undergoing shape changes as it responds to environmental ionic and temperature condi-tions. Under a 75% relative humidity, the DNA assumes another form called A-DNA. A-DNA is also right-handed

with 11 base pairs per 360° turn of the helix. A polynucle-otide containing one B-DNA and one RNA strand assumes the A form in solution. The DNA does not always have to be right-handed. Double-stranded DNA containing strands of alternating purines and pyrimidines can exist in a left-handed helical fashion. This is called Z-DNA. In Z-DNA, there are 12 base pairs per complete helical turn. In this exercise, we will concentrate only on B-DNA. This is the DNA that you will extract from onion cells.

Part 1: DNA Extraction from OnionIn living cells, the DNA is a very thin and long molecule. Extracted DNA in a test tube is also a long molecule. When ethanol is added to a DNA solution, the DNA fibers precipitate and can be spooled onto a glass or wooden


Fig. 2–3. Solenoid model for the formation of chro-matin fiber: the first round of coiling. Reproduced from Tamarin (2002) by permission of The McGraw-Hill Companies.

Fig. 2–4. Solenoid model for the formation of chro-matin fiber: the second round of coiling. Reproduced from Tamarin (2002) by permission of The McGraw-Hill Companies.

Fig. 2–5. Double helical structure of DNA. Reproduced from Tamarin (2002) by permission of The McGraw-Hill Companies.


rod. This feature of DNA is illustrated in the exercise described below.

MaterialsTennis ball-sized onionsKitchen knivesMortar and pestlesdH2OGraduated cylindersConical tubesExtraction buffer:

Per 1000 mL of dH2O24.228 g Tris (adjust pH to 7.5 with hydrochloric acid)14.61 g Sodium chloride9.306 g EDTA (ethylenedinitrilotetraacetic acid)5 g SDS (sodium dedocyl sulfate)Autoclave

60 to 65°C water bathGlass funnelsFilter paper, e.g. Whatman No. 1 (Whatman International Ltd.,

Maidstone, England) or equivalent, 125-mm diameterTransfer pipettesMicrotubes, 1.5 mLMicrotube racksWide, glass test tubesTest tube racksChilled (to −20°C) 100% ethanolWooden or glass rodsPaper pH indicatorsIce

Procedure1. Cut the onion into very small pieces.2. Place the onion pieces into a mortar and macerate them

with an appropriate pestle until the onion pieces are soft.3. Add 20 mL of dH2O to the mortar and macerate again

until the onion pieces are thoroughly liquefied.4. Using a graduated cylinder, measure 20 mL of the liq-

uefied onion and transfer it into a clean conical tube.5. Add an equal amount of the extraction buffer to the

tube, cap the tube, and gently invert 10 times to mix. Do not shake or vortex.

6. Place the tube in a 60 to 65°C water bath for 15 min.7. Remove the tube from the water bath.8. Put a filter paper into a funnel and place the funnel into

the mouth of a wide, glass test tube. Then pour the con-tent of the conical tube (Step 7, above) onto the filter paper and let the solution drip. The solution passing through the filter is called lysate.

9. Hold the tube containing the lysate at a 45° angle and slowly pour the chilled 100% ethanol down the side of the tube to form a clear layer on top of the lysate. Do not shake the tube.

10. A cloud of fine, viscous fibers will be visible where the two liquids meet. This is the onion DNA.

11. Insert a wooden or glass rod into the tube until the tip touches the DNA. Slowly rotate the rod several times until the fine, white DNA fibers accumulate on the end of the rod.

12. Once a visible amount of DNA is on the end of the rod, gently remove the rod from the solution. However,

leave some DNA in the tube for the analysis of its physical properties (Step 18, below).

13. Place the rod on your tube rack with the end contain-ing the DNA resting over the edge. Allow the mass of DNA to dry for 10 min. This allows evaporation of the ethanol from the sample which is necessary for proper gel loading later in the exercise.

14. Insert the end of the rod containing the DNA into a fresh microtube containing 75 µL of dH2O. Gently twist the rod between your fingers and move it up and down to transfer the clump of DNA into the water. Close the tube.

15. Once the DNA is in the water, hold the top of the tube firmly with one hand and tap the bottom of the tube with the index finger of your other hand for about a minute. If you are doing this properly you should see the liquid inside the tube moving up and down the walls of the tube. This promotes dissolving of the DNA into the water.

16. Let the tube rest on your bench for an additional minute.17. If a residue is still visible in the tube, centrifuge the

tube for 2 min to precipitate the residue, and then trans-fer 12 mL of the supernatant to a new microtube. This will be your tube Number 1 of the onion DNA as described in Part 2 of this exercise. Place the tube on ice until needed.

18. Use your rod to obtain another sample of DNA as you did in Steps 11 and 12 above. Use this DNA to examine its physical properties. Record your observations in the space provided below.

Is the DNA molecule soluble in water or in ethanol?

Using a paper pH indicator, check the pH of the DNA; is it acidic or basic?

Do the DNA fibers appear long or short? Why?


General Description of Agarose Gel Electrophoresis

DefinitionThe term electrophoresis refers to the migration of charged molecules, such as DNA, under the current of an electric field. Electrophoresis is employed for separation of DNA fragments on the basis of their size in an agarose gel. A slight modification of the electrophoresis process is used to separate proteins and other macromolecules.

Agarose GelAgarose is a natural polysaccharide of galactose and 3, 6-anhydrogalactose derived from agar, which in turn is derived from marine red algae. An agarose gel is made by dissolving the dry powder in a boiling buffer, pouring the gel into a casting tray, and allowing it to set by cooling at room temperature. Agarose is an ideal solid support for the separation of DNA fragments on the basis of size, and for this reason, it is extensively used in molecular biology laboratories. An agarose gel is porous; the pore size varies inversely with the concentration of agarose. Usually a 2% (w/v, or higher) agarose gel is used for the separation of DNA molecules of 100 to 1000 base pairs in length, while a 0.8% agarose gel is used for molecules as large as 12,000 base pairs (Table 2–1). For a better separation of DNA frag-ments as small as 100 bp, a high resolution agarose (e.g., MetaPhor, Biowhittaker Molecular Applications, Rockland, ME, or equivalent) at a concentration of 3% can be used.

BufferThe mobility of DNA fragments in an electrophoretic field is affected by the ionic strength of the electrophoresis buf-fer. In the absence of moderate ions, electrical conductance is minimal and the DNA migrates very slowly. Therefore, a buffer of high ionic strength is needed to conduct the elec-tricity and move the DNA in the gel. A variety of buffers are available for electrophoresis of DNA. In this exercise, Tris–Acetate–EDTA buffer is used. This buffer in generally known as TAE.

Electrophoresis Unit: How It WorksThe electrophoresis unit is composed of a chamber with a central platform, a lid with power cord, a casting tray, a well-forming comb, and a power supply (Fig. 2–6). The tray is filled with melted agarose; the comb is applied to

form slots or wells in the agarose. When the agarose is solidified, the comb is removed, and the tray containing the solid gel is placed on the central platform of the chamber. The chamber is filled with buffer to cover the gel. Then the DNA samples are loaded into the wells, the lid is placed on the chamber, and the power cords are connected to the power supply: the power cord at the top of the gel must be plugged to the negative (−) pole. The power supply is turned on and an optimum voltage (between 80 and 90 V) is applied. Because the DNA is acidic, it is negatively charged at neutral pH and migrates toward the positive pole or anode through the pores of the agarose gel (Fig. 2–7). The speed of migration of DNA varies depending on the voltage applied and the size and shape of the molecule (e.g., linear, circular, or supercoiled). In general, smaller DNA frag-ments migrate faster (farthest from well), whereas larger fragments migrate more slowly (closest to well). Then the gel is stained with ethidium bromide. Ethidium bromide is a highly planar molecule that intercalates between base pairs in DNA. When subjected to UV irradiation at 302 to 312 nm, ethidium bromide fluoresces orange-red, allowing you to view the DNA in the gel. The gel can then be pho-tographed. A UV transilluminator is used for this purpose, and an orange filter (e.g., Kodak Wratten #23A) is required to optimize the light transmitted by fluorescing DNA. A complete gel imaging and documentation system is com-mercially available.

Molecular Sizing (Using the DNA Ladder)When electrophoresing a DNA sample, the gel requires one lane that contains DNA fragments of known sizes from

Table 2–1. Agarose concentrations for resolving DNA bands.

Agarose concentration Range of DNA fragment sizes separated% (w/v) base pairs (bp)

0.8 1000–12,0001.0 500–10,0001.5 200–30002.0 100–10003.0 ≤500

Fig. 2–6. Components of an electrophoresis unit.

Fig. 2–7. A schematic representation of an agarose gel showing the wells and the direction of migration of DNA samples.


which the size of the DNA sample will be determined. The 1-kb DNA ladder is suitable for sizing linear double-stranded DNA fragments from 500 to 12,000 bp. In ideal conditions, all the bands are visible (Fig. 2–8).

According to the manufacturer (Invitrogen Corp., Carlsbad, CA), the ladder contains from 1 to 12 repeats of a 1018-bp DNA fragment. Furthermore, the ladder contains vector DNA fragments that range from 75 to 1636 bp. The 1636-bp band contains 10% of the mass applied to the gel; thus it stains brighter. The DNA ladder is usually supplied in a concentration of 500 mg/mL. It can be diluted in a gel-loading dye as described later in the Materials section.

Gel-Loading DyeThe gel-loading dye used in the exercises of this laboratory manual, is a buffer composed of the two dyes bromophenol

blue and xylene cyanol, mixed with 50% glycerol. The pH of the buffer is maintained at 8.00. Gel-loading dye is added to the samples to be electrophoresed to increase the density of the samples so that they will stay in the bottom of the wells. Furthermore, the two dyes separate during electrophoresis to monitor the progress of the migration of the samples. In 1% agarose gel, bromophenol blue migrates through the gel at the same rate as a DNA fragment approximately 250 to 400 bp long, whereas xylene cyanol migrates at the rate equivalent to a fragment of about 2000 to 3000 bp long.

Use of the MicropipettorMicropipettors are used in laboratories for the accurate measurement of volumes. They are available in a variety of capacities for dispensing volumes from 0.1 to 5000 µL (Fig. 2–9). In the exercise which follows, you will mea-sure small volumes of DNA to prepare serial dilutions. Success in this work is largely dependent on the accuracy of measuring the solutions. For this purpose, you will be using the micropipettor.

How to Use a Micropipettor1. Distinguish between micropipettors P10, P20, P200,

and P1000. The numbers indicate the volume limits of each type.

2. Each type has a setting limit shown by a number above the volume limit. This is 10% of the maximum setting of that micropipettor. However, note that some brands of micropipettors do not show this number. Use of the micropipettor at settings below or above the setting limits may result in pipetting error .

3. To adjust the volume of the solution that the micropi-pettor will be used to take up, dial the adjustable knob and read the three (or four) vertical numbers through the side window. Note that the volume measured is in mL (microliter). One microliter is one thousandth of a milliliter, that is, 1000 mL = 1 mL.

4. Examples of volumes taken up by the different mi-cropipettors are illustrated below.

5. To measure a desired volume of a solution and transfer it from one microtube to another, do the following:

Adjust the setting on the appropriate micropipettor.• Insert the micropipettor into an appropriate pipette tip.• Hold the micropipettor in a vertical position and using your • thumb push the plunger of the micropipettor down slowly until the first stop (you will feel a slight resistance).While holding the plunger, insert the micropipettor into the • solution you want to pick up, and then slowly let the plunger rise. Withdraw the pipette tip from the solution and observe

Fig. 2–8. The 1-kb DNA ladder, electro-phoresed on a 1% agarose gel in 1× TAE buffer.

Fig. 2–9. The micropipettor. For details see the text.


how high the solution level ascends into the micropipettor tip when different volumes of the solution are measured.To deliver the sample to another tube, insert the micropipettor • with attached tip into the new tube, slowly lower the plunger to the first resistance, then proceed to the second stop. Do not release the plunger before withdrawing the tip from the tube .After each use, press the lever next to the plunger to eject the • tip. Used tips must be disposed of properly in a container .

Part 2: DNA AssayIn the first part of this exercise, you extracted DNA from onion. In the second part, you will prepare serial dilutions of the DNA and compare them with the serial dilutions of a control DNA of known concentration. The control DNA is the phage lambda DNA (λ DNA), which is commercially available from biotechnology companies. Before starting the exercise, record the concentration of the λ DNA.

MaterialsOnion DNA (from Part 1)λ DNA with known concentration (stock), usually 100 ng/µLDNA ladder (ready to use):

100 µL DNA ladder (stock, commercially available)166 µL Gel-loading dye (as described below)734 µL dH2O

Gel-loading dye:0.25 g Bromophenol blue0.25 g Xylene cyanol60 mL 50% glycerol6 mL 1 M Tris, pH 8.01.2 mL 0.5 M EDTA, pH 8.032.8 mL dH2O

To prepare 50% glycerol, mix an equal volume of glycerol and dH2O, then autoclave the mixture.

To prepare 1 M Tris, dissolve 12.114 g of Tris in 60 mL dH2O, adjust the pH to 8.0 with hydrochloric acid, and then adjust the volume of the solution to 100 mL with dH2O. Autoclave before use.

To prepare 0.5 M EDTA, dissolve 18.61 g of EDTA in 60 mL of dH2O, adjust the pH to 8.0 with sodium hydrox-ide pellets, then adjust the volume of the solution to 100 mL with dH2O. Autoclave before use.

1´ TAE buffer:Per 1000 mL of dH2O242 g Tris57.1 mL Glacial acetic acid100 mL 0.5 M EDTA, pH 8.0This is 50´ TAE and is used as a stock; the working so-

lution is 1´TAE, that is, 1 part stock + 49 parts dH2O.5 mg/mL ethidium bromide solution:

Per 100 mL of dH2O0.5 g Ethidium bromideCaution: ethidium bromide is mutagenic and possibly

carcinogenic; do not touch with bare hands .dH2OAgaroseIceMicrotubes, 1.5 mLMicrotube racksMicrocentrifugeMicropipettorsTips

Gel staining dishesAutoclave tapeGlovesScaleRulers200 mL flasksMicrowave ovenAgarose gel electrophoresis unitUV (ultraviolet) transilluminatorUV protective glasses or shieldsAgarose gel photo documentation system (includes UV transil-

luminator, camera, monitor, and thermal printer)

Procedure1. Label the tube of onion DNA you prepared in Part 1 of

this exercise tube No. 1.2. Prepare serial dilutions with dH2O from tube No. 1 as

shown below. Mix between dilutions. Note the final dilution in each tube.

Micropipettors Tips Gel staining dishes Autoclave tape Gloves Scale Ruler 200 mL flasks Microwave oven Agarose gel electrophoresis unit UV (ultraviolet) transilluminator UV protective glasses or shield Agarose gel photo documentation system (includes UV transilluminator, camera,

monitor, and thermal printer) Procedure 1. Label the tube of onion DNA you prepared in Part 1 of this exercise tube No. 1. 2. Prepare serial dilutions with dH2O from tube No. 1 as shown below. Mix between

dilutions. Note the final dilution in each tube. 3. Keep tube No. 1 and 4 on ice.

Tube No . 1 2 3 4 1 µL 1 µL 5 µL

4. Transfer 5 µL of λ DNA from the stock tube to a new tube (tube No.1, below) then

dilute the λ DNA from tube No. 1 ten times with dH2O as shown below. Note the final dilution in tube No. 2.

Onion DNA

9 µL dH2O 9 µL H2O 5 µL H2O

Final Dilution 10x 100x 200x

3. Keep tube No. 1 and 4 on ice. 4. Transfer 5 µL of λ DNA from the stock tube to a new tube (tube No.1, below), and then dilute the λ DNA from tube No. 1 10 times with dH2O as shown below. Note the final dilu-tion in tube No. 2.

5. Keep tube No. 1 and 2 on ice.6. Centrifuge the tubes saved on ice (from Steps 3 and

5, above) for 5 s in a microcentrifuge. Make sure to balance the tubes.

7. Add 4 µL of gel-loading dye to each tube No. 1 and 4 of the onion DNA and the tube No. 2 of the λ DNA. Also, add 2 µL of gel-loading dye to the tube No. 1 of the λ DNA. Mix the content of each tube by pipetting up and down several times. For each tube, use a new micropipettor tip to avoid reaction contamination .

8. Centrifuge the four tubes from Step 7 for 5 s to make certain that the samples are in the bottom of the tubes. Place the tubes on ice and proceed to Step 9.

9. In a 200-mL Erlenmeyer flask, prepare a 0.8% agarose

gel in 1´ TAE buffer by adding 0.8 g of agarose to 100 mL of buffer.

10. Place the flask in a microwave oven for 45 s to 1 min. Remove the flask and swirl the contents for a few sec-onds. Microwave for another 1 to 1.5 min or until the buffer starts to boil.

11. Leave the flask at room temperature for 2 min. At this time you may add 5 µL of 5 mg/mL ethidium bromide to the melted agarose and swirl gently to mix. Alternatively, you may stain the agarose gel in an ethidium bromide solution after you have completed the electrophoresis run. This is described in step 19, below .

12. Seal the ends of the gel casting tray (see Fig. 2–6) with autoclave tape, then pour the melted agarose (Step 11, above) in the tray and insert the comb. Remove all bubbles on the surface of the agarose using a pipette tip and allow the gel to solidify.

13. After the gel has solidified, remove the tape and the comb, place the tray in the electrophoresis chamber, and fill the chamber with buffer. The buffer must cover the gel to a depth of about 2 mm.

14. Load the samples into the gel wells:Draw 3 • µL of the DNA ladder into the micropipettor.Carefully direct the tip of the micropipettor into the top of the • first well of the agarose gel and slowly eject the sample into the well by depressing the plunger of the micropipettor. Do not place the tip of the micropipettor deep into the well or you might puncture the gel .In this manner, load the entire contents of your DNA samples • (from Step 8, above) into well Numbers 2 to 5.

15. Place the electrophoresis chamber lid in position, con-nect the power cords to the power supply (remember: the power cord at the top of the gel must be plugged to the negative pole), push the rocker switch on the power supply to “on” and set it to 80 V. The voltage will remain constant during the run.

16. Unless otherwise stated by the laboratory instructor, run the gel until the indicator dye (the bromophenol blue in the gel-loading dye) has moved 6 to 7 cm from the wells. The indicator dye should never be allowed to run off the end of the gel.

17. At the end of the electrophoresis run, shut off the power supply, disconnect the cords, and remove the casting tray containing the gel.

18. Wear gloves so as not to touch the gel with bare hands. Carefully slide the agarose gel out of the cast-ing tray and place it in a glass dish containing dH2O. Rinse the gel for 5 to10 s.

19. Skip this step if you have already added ethidium bromide to the melted agarose in Step 11, above . Carefully transfer the gel into the dish containing 500 mL of dH2O supplemented with one drop of 5 mg/mL ethidium bromide. Leave the gel in this solution for 25 to 30 min to stain. Then rinse the gel briefly in dH2O.

20. Transfer the gel onto a UV transilluminator to view the DNA bands. Ultraviolet is hazardous to your eyes; always wear UV protective glasses or shield when working on the UV box. You may also photograph the gel if equipment is available.

21. Compare the results on your gel with those shown in Fig. 2–10. Do you see any difference in banding patterns? If so, explain the possible reason(s) causing the difference.

22. At the end of the exercise, dispose of the gel into a biohazard bag.

Study Questions2–1. If the length of a DNA is 9800 µm, how many base

pairs does the DNA have?

2–2. Why are the two strands of the DNA called antiparallel?

2–3. Why is the gel-loading dye made up of two differ-ent dyes?

2–4. Define the semiconservative model of DNA replication.

2–5. Suppose you have extracted DNA from a eukaryotic organism. If you place the DNA in boiling water for 10 min, what structural change in the DNA would you expect to see?

Fig. 2–10. Agarose gel electrophoresis of the onion and phage l DNA.



Objectives




Exercise 3 Restriction Mapping

Objectives

• to understand the concept of DNA digestion

• to understand how gel electrophoresis separates digested DNA fragments

• to construct a restriction map with data obtained from gel electrophoresis

Restriction map is defined as a physical map of DNA showing the relative positions of restriction enzyme cleavage sites. To understand the concept of restriction mapping, a clear description of restriction enzymes and the sites where they cut the DNA is needed.

Restriction Enzymes (Endonucleases)Many molecular genetic techniques are rooted in the ability to digest (also referred to as cutting or cleaving) DNA molecules in a specific and predict-able way. The key to this technology is the discovery of restriction enzymes or restriction endonucleases in bacteria. Species of bacteria make restriction enzymes which recognize palindromic (inverted repeats) nucleotide sequences of DNA, called restriction sites, and cleave the DNA at those sites generating a 5¢ phosphate and a 3¢ hydroxyl group at the point of cleavage. Restriction sites are usually 4, 6, or 8 base pairs (bp) long. Restriction enzymes are named after the bacteria from which they are isolated. This is done with the first letter of the genus followed by the first two letters of the species. The type of strain or substrain sometimes follows the species designation in the name. A Roman numeral is always used to indicate whether the particular enzyme was the first isolated, the second, the third, and so on. For example, the first enzyme that was isolated from the strain RY13 of the bacterium Escherichia coli (commonly known as E. coli) is called EcoRI. Several hundred restriction enzymes have been identified and isolated and are available commercially.

Restriction enzymes are classified as Type I and Type II, both types recogniz-ing specific restriction sites. However, there is a major difference between them. Type I restriction enzymes digest the double-stranded DNA at random far from their restriction sites, thus create indistinct restriction fragments. For this reason, Type I restriction enzymes have no practical value in molecular genetics. Type II restriction enzymes cleave the double-stranded DNA within (or very close to) their restriction sites producing discrete and predictable restriction fragments. This type of restriction enzymes is used in the laboratory for DNA analysis. For example, EcoRI recognizes the sequence 5¢GAATTC 3¢ and makes a stag-gered cut producing sticky ends that have base pair overhangs (Fig. 3–1A). Or, the enzyme HaeIII recognizes the sequence 5¢GGCC 3¢ and cuts both strands of the DNA between the same nucleotide pairs to produce blunt ends (Fig. 3–1B). By using the same restriction enzyme to cut the DNA from two differ-


Fig. 3–1. Generation of stag-gered (A) and blunt (B) ends in pieces of DNA digested with EcoRI and HaeIII, respectively. Arrowheads show position of cut.


ent species, complementary ends will be created that will allow the DNA from the two species to stick together. Thus, recombinant DNA can be generated.

It should be emphasized that bacteria’s own DNA is not digested by the enzymes they produce. This is because a group of enzymes, called methylases, recognize the sites that would be recognized by the cell’s restriction enzymes and add a methyl group (-CH3) at those sites on the newly forming daughter strand. This methylation shields the bacterial DNA from recognition by the cell’s own restric-tion enzymes.

Table 3–1 shows the characteristics of some commonly used restriction enzymes.

Construction of Restriction MapsThe DNA pieces generated by the action of restriction enzymes in a fragment of double-stranded DNA are called restriction digests. By means of gel electrophoresis, restric-

tion digests can be separated by size, and on that basis, a restriction map of the original DNA can be reconstructed. Consider the following example.

A 10-kb linear piece of DNA has been digested once with Enzyme X, once with Enzyme Y, and once with a mixture of Enzymes X and Y (referred to as double diges-tion). Then the restriction fragments of each reaction are separated, according to their molecular size, by agarose gel electrophoresis (Fig. 3–2A). Also, a DNA ladder (see Exercise 2) has been electrophoresed to estimate the sizes of the fragments. Since the size of the DNA fragments in the ladder is known, the size of the fragments generated by Enzyme X, Y, or a combination of both can be estimated. As shown in Fig. 3–2A, when the 10-kb piece of DNA is cut with Enzyme X, a 7 and 3-kb DNA fragment is gener-ated. This indicates that there is only one restriction site for Enzyme X in the 10-kb DNA fragment. Similarly, when the same DNA is cut with Enzyme Y, two bands of sizes 8 and 2 kb are generated. Here again there is only one restric-tion site for Enzyme Y. However, when the 10-kb DNA is digested with both enzymes, three bands, 5, 3, and 2 kb in size, are generated (see Fig. 3–2A).

To construct the restriction map of the DNA, the 3-kb Enzyme X site can be arbitrarily placed in one end of the DNA and the 8-kb Enzyme Y site in the same end (see Fig. 3–2B and C). Cutting the DNA with both enzymes, three bands of sizes 5, 3, and 2 kb are generated. The results on the gel (see Fig. 3–2A) show that the 2-kb fragment from Enzyme Y digestion remains unaffected in the double digest, whereas the 8-kb fragment from the same digestion has been split into two bands of 5 and 3 kb. This result indi-

Table 3–1. Characteristics of some commonly used restriction enzymes. Position of cut is shown by arrowheads.

Fig. 3–2. Construction of a restriction map for two hypothetical Enzymes X and Y.


cates that there is a single Enzyme X site within the 8-kb fragment from the Enzyme Y cut. With this information in hand, a restriction map can be constructed (Fig. 3–2D) by superimposing Fig. 3–2B on Fig. 3–2C.

In actuality, construction of restriction maps (linear or circular) is more complicated than that discussed above. You need to practice several times with data obtained from different gels. Look at Study Questions 3–3 and 3–4, and construct a restriction map of each.

Restriction Digestion of lambda DNAPhage lambda (l), which infects E. coli, has a double-

stranded DNA which is located in the core of the phage head. The DNA molecule contains 48,502 bp that code for approximately 50 different phage proteins. The sequence of nucleotides along the entire lambda genome is known (Sanger et al., 1982). The lambda DNA molecule has an unusual structure, in which 12 base single-stranded segments are found at the 5¢ and 3¢ termini of the phage DNA molecule (Fig. 3–3A). These segments are self-com-plementary, called cos ends. Thus, by forming base pairs between the two segments, the linear DNA will circularize (Fig. 3–3B). The lambda genome is linear in the phage but circular in the host.

The 48,502-bp lambda DNA, used in this exercise, was isolated from the bacteriophage lambda and is commercially available. It is linear and contains five recognition sites for the enzyme EcoRI and seven for the enzyme HindIII (Fig. 3–4).

In this exercise, you will digest lambda DNA once with EcoRI, once with HindIII, and once with a mixture of both enzymes (double digestion). You will also use uncut lambda DNA as control. Then, you will load the DNA samples into wells of an agarose gel to separate the digested fragments by electrophoresis.

Figure 3–5 depicts the restriction digest of the lambda DNA. As seen in Lane 1 (from left), only five of the six fragments generated by EcoRI (see Fig. 3–4B) are detect-able on the gel. This is due to the fact that the 5643- and the 5804-bp fragments overlap on the gel appearing as a single band (see Band Number 3 from top). Furthermore, in Lane 2, only six of the pos-sible eight fragments generated by the enzyme HindIII are visible (compare Fig. 3–4C with 3–5A, Lane 2). This is because the 564- and the 125-bp fragments usually run off the gel when prepared in a 0.8 to 1% concentration. Lane 3 of Fig. 3–5A shows the DNA fragments generated by digestion of lambda DNA with a mixture of the enzymes EcoRI and HindIII. The size of the fragments can be predicted by superimposing Fig. 3–4B on Fig. 3–4C. As a result, 13 frag-ments will be generated (try to identify these fragments), 10 of which will be visible on the gel (see Fig. 3–5A, Lane 3). This is because the two fragments of size 5148 and 4973 bp overlap and appear as a single thick band on the gel (see Band Number 2 from the top) and, as mentioned

above, the two small fragments 564 and 125 bp in size run off the gel. With a little effort you can identify the size of the fragments seen in Lane 3. As a hint, the heaviest band is 21,226 bp and the lightest band is 831 bp. Lane 4 of Fig. 3–5A shows uncut lambda DNA used as a control. The size of this fragment is 48,502 bp.

Tips on Restriction Digestion Reactions

1. DNAThe DNA used for restriction digestion must be pure and free of contaminants such as EDTA, ethanol, and phenol, which are used to extract and/or purify DNA.

2. Restriction EnzymeThe general rule is that the amount of the enzyme used should not exceed 10% of the total volume of the reaction. If an excess of enzyme is used, the length of incubation should be decreased (see also the Incubation Time, below). Alternatively, if a smaller amount of enzyme is used, the incubation time should be increased. Restriction enzymes should always be stored in a freezer. During the laboratory work, they can be kept on ice or in a benchtop cooler for a limited amount of time.

Fig. 3–3. Diagram of the phage lambda DNA molecule in linear (A) and circular (B) configuration.

Fig. 3–4. (A) The linear map of the phage lambda DNA. Arrows show the recognition sites for EcoRI (B) and HindIII (C) enzymes.


3. BufferTo ensure optimal activity, restriction enzymes are used with appropriate buffers. The manufacturers provide a buf-fer with each restriction enzyme. Some restriction enzymes, in addition to a buffer, require bovine serum albumin (BSA) for optimal activity. The BSA is usually supplied by the manufacturers at 100´ concentration. It must be diluted to 10´ in distilled water before use.

4. Incubation TemperatureThe incubation temperature for most restriction enzymes is 37°C. However, consult manufacturer’s recommenda-tion before incubating reactions. An incubator or waterbath, adjusted to 37°C, can both be used for restriction digestion.

5. Incubation Time Incubation time varies depending on the amount of the enzyme used (see also Restriction Enzyme, above) and the source of the DNA employed. Usually a 45-min to 1-h incu-bation is sufficient to digest viral or bacterial DNA under optimal conditions, whereas eukaryotic DNA requires an overnight (up to 16 h) incubation.

6. Order of Preparation of Digestion ReactionsIn preparing digestion reactions, it is extremely important to add the solutions in the following order: distilled water, buffer, BSA (if required), DNA, and enzyme. The reaction must be thoroughly mixed prior to incubation to achieve a complete digestion.

7. Double DigestionDouble digestion is a common pro-cedure in restriction digestion, during which a piece of DNA is digested by two enzymes at the same time (see, e.g. Fig. 3–5A). In double digestion, it is essential to choose a buffer that ensures the most activity for both enzymes. Furthermore, if BSA is required for either enzyme, it must be addded to the double digestion reac-tion. BSA will not inhibit the activity of the enzyme that does not require it. A complete list of suggested buffers for double digestion can be found in the Technical Reference of the New England BioLabs Catalog, 2002–2003 (or newer editions) (New Englad Biolabs, Ipswich, MA). In the case that no single buffer is found for a double digestion reaction, the digestion must be done sequentially. First, the reac-tion is digested with one enzyme + buffer combination, then the digested reaction is further digested with the second enzyme + buffer combina-tion. Seek advice from the laboratory instructor for choosing a buffer for double digestion.

MaterialsPhage lambda DNA, concentration 100 ng/mLEcoRIEcoRI buffer†HindIIIHindIII buffer†1´ TBE (Tris–Borate–EDTA) buffer, to be used for electrophore-

sis and preparing the gel:Per 1000 mL of dH2O54 g Tris27.5 g Boric acid20 mL 0.5 M EDTA, pH 8.0This is 5´ TBE and is used as a stock; the working solu-

tion is 1´ TBE, i.e., 1 part stock + 4 parts dH2O.To prepare 0.5 M EDTA see Exercise 2.

DNA ladder, as described in Exercise 2dH2OAgaroseGel-loading dye, as described in Exercise 2Ethidium bromide solution, as described in Exercise 2IceAgarose gel electrophoresis unit, as described in Exercise 2Microwave oven200 mL flasksMicrotubes, 0.6 mLMicrotube racksMicrocentrifugeMicropipettors

Fig. 3–5. (A) Restriction digest of the phage lambda DNA and (B) the 1-kb DNA ladder. Note that (A) and (B) are not to the same scale.

† Note: Manufacturers provide enzymes with appropriate buffers; consult manufacturers’ recommendation.


Tips37°C incubatorGel staining dishes, as neededGlovesUV protective glasses or shieldsAgarose gel photo documentation system, as described in Exercise 2

Procedure1. Pick four 0.6-mL microtubes and label them E (for

EcoRI), H (for HindIII), EH (for EcoRI + HindIII), and NO (for no enzyme control).

2. Set up digestion reactions in the order shown below:

Tube E Tube HmL mL

dH2O 13 dH2O 13EcoRI buffer 2 HindIII buffer 2Lambda DNA 3 Lambda DNA 3EcoRI 2 HindIII 2

Tube EH Tube NOdH2O 13 dH2O 15EcoRI buffer† 2 EcoRI buffer 2Lambda DNA 3 Lambda DNA 3EcoRI 1HindIII 1

† Note that in the double digest (Tube EH) the EcoRI digestion buffer is used.

3. Tap the bottom of the tubes with a finger to mix the reagents, then centrifuge briefly (about 5 s).

4. Digest the DNA by placing the tubes in a 37°C incuba-tor for 45 min.

Note: Limiting the time of incubation will result in partial digestion of the DNA; that is, only a portion of the available restriction sites will actually be cut with the enzyme. This also can occur when sufficient amount of enzyme is not used.

During the incubation time, prepare a 0.8% (w/v) aga-rose gel with 1´ TBE buffer, and set up the electropho-resis unit as described in Exercise 2.

5. After 45 min, remove the tubes from the incubator, add 5 mL of gel-loading dye to each tube, and mix well by pipetting up and down.

6. Briefly centrifuge the tubes to push the contents to the bottom of the tubes.

7. Load the gel as indicated below.

Well No. Sample1 3 mL of DNA ladder2 10 mL of the content of tube E3 10 mL of the content of tube H4 10 mL of the content of tube EH5 10 mL of the content of tube NO

8. Electrophorese at 90 V for about 45 min. You may shorten the electrophoresis time by choosing a higher voltage; seek advice from the laboratory instructor.

9. View the DNA bands on a UV transilluminator and photograph the gel.

10. Estimate the size of the bands in Lanes 2 to 5 by means

of the DNA ladder in Lane 1.11. Fill in the Table shown below:

Fragment sizes

Lane 1 Lane 2 Lane 3 Lane 4 Lane 5

12. Compare your gel with the sample gel shown in Fig. 3–5.

Study Questions3–1. In the representation of the piece of DNA shown below,

find and name restriction enzyme recognition sites.5¢ AATCTGCTATAGGCGGATCCTGAGCCTATGCCAGCTGCATAGTATGAGTACTGCCT 3¢

3¢ TTAGACGATATCCGCCTAGGACTCGGATACGGTCGACGTATCATACTCATGACGGA 5¢

3–2. What is partial digestion and how does it occur?

3–3. A 10-kb linear DNA fragment was digested once with Enzyme A, once with Enzyme B, and once with Enzyme C. The following fragments were obtained.

Enzyme A: 2 kb Enzyme B: 0.6 kb Enzyme C: 3 kb8 kb 9.4 kb 7 kb

Restriction enzyme digests using two enzymes in combination produced the following fragments.

Enzyme A + B: 0.6 kb Enzyme A + C: 2 kb Enzyme B + C: 0.6 kb1.4 kb 3 kb 3 kb8 kb 5 kb 6.4 kb

Construct the restriction map of the DNA.

3–4. When a circular piece of DNA is subjected to diges-tion with restriction enzymes, fragments with the following sizes are produced.

EcoRI: 21 kb BamHI: 6.4 kb HindIII: 21 kb14.6 kb

EcoRI + HindIII: 3.8 kb EcoRI + BamHI: 1.2 kb BamHI + HindIII: 2.6 kb 17.2 kb 6.4 kb 3.8 kb

13.4 kb 14.6 kb

Construct the restriction map of the DNA.

3–5. Below is a picture representing a 25-kb piece of DNA. What size fragments would you expect after digesting the DNA with each of the enzymes or com-bination of the enzymes listed below?

Below is a picture representing a 25 kb piece of DNA. What size fragments would you expect after digesting the DNA with each of the enzymes or combination of the enzymes listed below?

25 kb B

E

H H

B

6

4

33

9 25 kb

E:______________________ B:______________________ H:______________________ B+H:____________________ E+H:____________________



Objectives




Exercise 4 Extraction of Genomic DNA

Objective

• to extract DNA from plants

It is extremely important that the DNA used for some applications, includ-ing in vitro amplification, restriction digestion, cloning, molecular mapping, or sequencing be reasonably pure and of high quality. For this reason, the growth condition of the organism used for DNA preparation must be optimal. In plants, the leaves, flowers, and in vitro-grown root material can be used for DNA isola-tion, or they can be collected over time, frozen in liquid nitrogen, and stored at −80°C until needed. Once the DNA is extracted, commonly referred to as genomic DNA, it can further be purified by phenol and recovered by precipita-tion with sodium acetate.

DNA Extraction from the Plant Arabidopsis thalianaThere are several methods of preparing DNA from plants. We describe here three methods selected on the basis of their reliability and our experience in using them for research purposes. Note that the yields of the DNA prepared by these methods vary considerably: the highest yield can be obtained by the Method 1 and the lowest by the Method 3.

1. Large-Scale DNA Extraction by CTABThe method described below is the modification of previously pub-lished protocol known as CTAB (see e.g., Doyle and Doyle, 1987). Cetyltrimethylammonium bromide (CTAB) is a cationic detergent which can form complexes with proteins and polysaccharides and help precipi-tate these materials from plant extract. Then, they can be removed from the extract by chloroform:isoamyl alcohol.

Materials2´ CTAB solution

6 g CTAB30 mL 1 M Tris, pH 7.512 mL 0.5 M EDTA, pH 7.584 mL 5 M Sodium chloridedH2O to 300 mLAutoclaveTo prepare 1 M Tris see Exercise 2, but adjust the pH to 7.5.To prepare 0.5 M EDTA see Exercise 2, but adjust the pH to 7.5.To prepare 5 M sodium hydroxide, dissolve 20 g of sodium hydroxide in 60 mL

of dH2O, then adjust the volume of the solution to 100 mL with dH2O. There is no need to autoclave this solution.

1´ TE:Per 500 mL of dH2O5 mL 1 M Tris, pH 8.01 mL 0.5 M EDTA, pH 8.0To prepare 1 M Tris and 0.5 M EDTA, see above, adjust the pH accordingly.

2-mercaptoethanol (refrigerator, 4°C)Chloroform:isoamyl alcohol (24:1)

Mix 24 volumes of chloroform with 1 volume of isoamyl alcohol.



Isopropanol (refrigerator, 4°C)70% ethanol (chilled to −20°C)

Add 70 mL of absolute ethanol to 30 mL of dH2O, mix well and chill to −20°C.

dH2O65°C water bathLiquid nitrogenIcePlant tissue of interest: 2 g of tissue per extractionMortar and pestle (autoclaved and chilled to −20°C)Conical tubesHigh-speed centrifuge with variable operating temperaturesMicrocentrifugeOak Ridge tubes (resistant to high-speed centrifugation)Oak Ridge tube racksGraduated cylindersMicrotubes, 1.5 mLMicrotube racksMicropipettorsTipsScale

ProcedureThe tissues to be used (e.g., Arabidopsis leaves) for DNA extraction should already be weighed and frozen in a −80°C freezer. Also, a cold mortar and pestle should be used to grind the tissue. They can be placed in a −20°C freezer sev-eral hours before use.1. In a conical tube, add sufficient amount of 2-mer-

captoethanol to the CTAB solution to achieve a final concentration of 1% . Use 2 mL of CTAB + 2-mer-captoethanol solution per gram of tissue to be used for DNA extraction. For example, for 2 g of tissue, use 4 mL of CTAB supplemented with 40 µL of 2-mercaptoethanol. Mix well.

2. Place the tube (from Step 1, above) in a 65°C water bath for 1 h.

3. Place the frozen tissue into a chilled mortar and cover the tissue with liquid nitrogen; also, cool the grinding pestle by placing it into the same mortar. Wait until the liquid nitrogen evaporates, then immediately grind the tissue to a fine powder.

4. Cover the powder with additional liquid nitrogen, then immediately pour the contents into a clean Oakridge tube.

5. Wait until the liquid nitrogen evaporates.6. Add the pre-warmed CTAB + 2-mercaptoethanol (see

Steps 1 and 2, above) to the Oakridge tube, cap the tube, and incubate it in a 65°C water bath for 30 to 45 min shaking a few times every 10 min.

7. Cool the tube (from Step 6, above) to room tempera-ture; it takes about 10 to 15 min depending on the vol-ume of the contents.

8. Place the tube on ice for 2 min, and then leave at room temperature for 5 min.

9. Add an equal volume of chloroform:isoamyl alcohol to the tube.

10. Invert (do not shake or vortex) the tube 10 times.11. Centrifuge the tube at 13, 000 rpm for 5 min at 4°C to

pellet the non-DNA materials of the tissue.12. After centrifuging, two layers of supernatant will form:

an upper layer, which is the aqueous phase and a lower layer, which is the organic phase. With a micropipettor, transfer the aqueous phase (which contains the DNA) to another Oak Ridge tube and repeat Steps 9 to 11.

13. With the aid of a micropipettor carefully transfer the upper layer to a clean Oak Ridge tube and add an equal volume of cold isopropanol. Cooling the isopropanol increases the amount of DNA that precipitates. Shake the tube gently; do not mix or vortex. A white cloud of DNA will be formed in the tube.

14. Centrifuge the tube at 13, 000 rpm for 30 min at 4°C.15. Decant the supernatant and wash the pellet for 1 min

with 70% (chilled to −20°C) ethanol.16. Carefully remove the ethanol, and dry the pellet under

the hood for 15 min or until the pellet turns clear. The pellet is the DNA you extracted. If there is cell debris in the DNA, it will not turn clear.

17. Resuspend the DNA in 50 mL of 1´ TE buffer, mix well, and then transfer the contents to a fresh, auto-claved microtube. Note: If using double-distilled water instead of the TE buffer, the pH value of the water must be between 7.0 and 8.5 because the DNA may degrade in the absence of a buffering agent.

18. If needed, quantify the DNA with a fluorometer.

Further Purification of DNAThe procedure described below denatures and removes contaminating proteins from the DNA you extracted in Step 17, above.

MaterialsDNA solution of interest (Step 17, above)PhenolChloroform:isoamyl alcohol (24:1)

Prepare as described above.3 M sodium acetate, pH 5.2

Dissolve 24.61 g of sodium acetate in 60 mL of dH2O, then add glacial acetic acid until the pH 5.2 is obtained. Adjust the volume of the solution to 100 mL with dH2O.

Absolute (100%) ethanol, chilled to −20°C70% ethanol, as described above1´ TE, as described abovedH2OIceMicrocentrifugeMicrotubes, 1.5 mLMicrotube racksMicropipettorsTips

Procedure1. Add an equal volume of phenol to the DNA solution

you prepared in Step 17, above.2. Mix gently until an emulsion forms.3. Centrifuge the mixture at 13,000 rpm for 30 s at room

temperature. If the aqueous and organic phases (the upper and lower phases, respectively) are not well


separated, centrifuge once again for a longer time or at a higher speed.

4. Transfer the aqueous phase (which contains the DNA) to a fresh microtube.

5. Add an equal volume of 24:1 chloroform:isoamyl alcohol to the aqueous phase, mix gently, and repeat Step 3. Note that extraction with chloroform:isoamyl alcohol is needed to remove traces of phenol from the DNA preparation.

6. Transfer the aqueous phase (the upper phase) to a fresh microtube. While transferring, measure the volume. You will need it in Step 7.

7. Add 0.10 volume of 3 M sodium acetate to the aqueous phase (Step 6, above). Mix gently.

8. Add 2.5 volumes of chilled absolute ethanol. Mix gently.9. Centrifuge the tube at 13, 000 rpm for 30 min.10. Decant the supernatant and wash the pellet for 1 min

with 70% chilled (to −20°C) ethanol.11. Carefully remove the ethanol, and then dry the pel-

let under the hood for 15 min or until the pellet becomes clear.

12. Resuspend the pellet in a 50 mL of 1´ TE buffer, mix well, and centrifuge at 13, 000 rpm for 1 min. Then transfer the content to a fresh, autoclaved microtube.

13. If needed, quantify the DNA with a fluorometer.

2. Small-Scale DNA Extraction by CTABThis is the modification of the method described above, convenient to prepare DNA in microtubes.

MaterialsArabidopsis excised leaves about half the size of a micromortar2´ CTAB as described above2-mercaptoethanol (refrigerator, 4°C)1´ TE as described aboveChloroform:isoamyl alcohol (24:1) as described aboveIsopropoanol (chilled to −20°C)dH2OLiquid nitrogenMicromortar and pestle (sold sterile, otherwise autoclave before use)Microtubes, 1.5 mLMicrotube racksFloating racksMicropipettorsTipsSterile transfer pipette (as needed)Ice65°C water bathMicrocentrifugeScale

ProcedureFor each micromortar, 1 mL of CTAB + 2-mercaptoethanol is needed. Prepare sufficient amount from this solution as described in the Procedure of the Method 1.1. Put an excised leaf, as described in the Materials, into

a micromortar and fill it with liquid nitrogen. Using a leaf bigger than the size mentioned above will not improve the results. In the meantime, cool the grinding micropestle by placing it into the same micromortar.

2. Wait until the liquid nitrogen evaporates, and then im-mediately grind the leaf against the wall of the micro-mortar to break it into as fine a powder as possible.

3. Add 1 mL of CTAB + 2-mercaptoethanol to the micro-mortar, mix well, and grind with a few strokes until the tissue is homogenized.

4. Cap the micromortar, place it in a floating rack, and put the rack in 65°C water bath for 30 min, shaking a few times every 10 min.

5. Cool the micromortar (from Step 4, above) to room temperature.

6. Place the micromortar on ice for 2 min, and then leave at room temperature for 5 min.

7. Add 400 mL of chloroform:isoamyl alcohol to the mi-cromortar and invert the tube 10 times; do not shake or vortex .

8. Centrifuge the micromortar at 13, 000 rpm for 15 min at room temperature to pellet non-DNA materials of the tissue. The supernatant (the liquid phase) will contain the DNA .

9. With a micropipettor or sterile transfer pipette, transfer the supernatant to a clean 1.5-mL microtube; slowly add 700 μL of chilled isopropanol to the same tube, invert the tube four to five times, and incubate in a 20°C freezer for 30 min. Isopropanol will precipitate the DNA in the solution.

10. Centrifuge the tube at 13, 000 rpm for 30 min at room temperature to pellet the DNA. The pellet should be visible in the bottom of the tube.

11. With the aid of a micropipettor carefully remove the supernatant from the tube; do not touch the pellet .

12. Allow the pellet to air dry under a hood for 10 to 15 min or until it turns clear. If there is debris in the DNA it will not turn clear.

13. Resuspend the DNA pellet in 30 mL of 1´ TE buffer, mix well, and centrifuge at 13, 000 rpm for 1 min. Then transfer the contents to a fresh, autoclaved microtube.

14. If needed determine the concentration of the DNA with a fluorometer.

3. Small-Scale DNA Extraction by DNAzolThe method described here is based on the use of a DNA isolation reagent called DNAzol (Invitrogen, Carlsbad, CA) (Chomczynski et al., 1997) that contains guanidine thiocyanate and a detergent mixture. In this method, the tis-sue sample is homogenized in DNAzol, and then the DNA is precipitated from the cell lysate with ethanol. The DNA isolated by DNAzol is pure enough to be used for restric-tion digestion, in vitro amplification, or Southern blotting.

MaterialsArabidopsis excised leaves about half the size of a micromortarPlant DNAzol reagent (DNA extraction buffer)10 mM Tris:

Dissolve 0.121 g of Tris in 60 mL of dH2O, adjust the pH to 8.0 with concentrated hydrochloric acid, and then adjust the volume of the solution to 100 mL with dH2O. Autoclave before use.

Chilled (to −20°C) 100% ethanol80% ethanol:

Add 80 mL of absolute ethanol to 20 mL of dH2O, mix well.Liquid nitrogen or dry iceMicromortar and matching pestle (sold sterile, otherwise auto-

clave before use)Microtubes, 1.5 mLMicrotube racksMicrocentrifugeMicropipettorsTipsForcepsScissorsMagnifying glassIce

ProcedureNote: In this exercise, if you cool the tissue on dry ice instead of liquid nitrogen, replace Steps 1 to 3 with Steps 1a to 2a as described below.

1. Pick an excised leaf as described in the Materials, and place it in a micromortar. Using a leaf bigger than the size mentioned above will not improve the results. Skip to Step 2.

1a. Pick an excised leaf as described in the Materials, and place it in a micromortar; also place a micropestle in the same micromortar. Using a leaf bigger than the size mentioned above will not improve the results. Leave the composite on dry ice for about 10 min or until the tissue freezes. Skip to Step 2a.

2. Add liquid nitrogen into the micromortar. In the mean-time, cool the grinding micropestle by placing it into the same micromortar. Skip to Step 3.

2a. Immediately grind the leaf against the wall of the micro-mortar to break it into a fine powder. Skip to Step 4.

3. Wait until the liquid nitrogen evaporates, and then im-mediately grind the leaf against the wall of the micro-mortar to break it into a fine powder.

4. Immediately add 100 mL of DNAzol to the micromor-tar and grind with 8 to 10 strokes until the tissue is homogenized.

5. Close the lid of the micromortar and place it on ice for 5 min.

6. Centrifuge the micromortar at 13,000 rpm for 15 min to pellet non-DNA materials.

7. Transfer the supernatant, which contains the DNA, to a clean microtube and add an equal volume of chilled (to −20°C) 100% ethanol. The ethanol will precipitate the DNA in the solution.

8. Slowly invert (do not shake) the tube five times and place it on ice for 5 min.

9. Centrifuge the tube at 13,000 rpm for 30 min to pellet the DNA.

10. With the aid of a micropipettor, remove the supernatant and view the pellet in the bottom of the tube. The pel-let is the DNA you extracted from the leaf.

Note: The pellet may be small; if needed, use a magni-fying glass to find it.

11. Wash the DNA for 1 min with 80% ethanol. After adding the ethanol, rotate (do not shake or invert) the tube.

12. Carefully decant the ethanol and air dry the DNA un-der a hood for 15 min or until it turns clear. If there is debris in the DNA, it will not turn clear.

13. Resuspend the DNA in 15 mL of 10 mM Tris, mix well, and centrifuge at 13, 000 rpm for 1 min. Then transfer the content to a fresh, autoclaved microtube.

14. If needed, quantify the DNA with a fluorometer. The DNA extracted with the methods described above can be stored in a refrigerator for up to 1 month or in a −20°C freezer for a longer time.

Study Questions4–1. What is genomic DNA?

4–2. When resuspending a DNA pellet, why is 1´ TE buffer preferred over double-distilled water?

4–3. What is EDTA? Why is it usually included in DNA extraction buffers?

4–4. According to the manufacturer, DNAzol contains a mixture of guanidine thiocyanate and detergent. What is the role of guanidine thiocyanate in this buffer?

4–5. How can you prepare 10 ng/mL of DNA from the 95 ng/mL stock?



Objectives




Exercise 5Polymerase chain reaction (hereafter PCR; process patents held by Hoffman-La Roche, Indianapolis, IN), developed by K.B. Mullis in 1985, is one of the most powerful techniques used in molecular biology. With this method, a few nanograms of DNA can be amplified millions of times in a microtube within a few hours. The amplification is carried out by a machine called thermal cycler (Fig. 5–1).

As its name implies, PCR synthesizes a new DNA molecule in the 5¢ to 3¢ direction with the speed of approximately 1000 bp per minute using the existing 3¢ to 5¢ DNA template. The region thus amplified is called the target (or intervening) region (Fig. 5–2). The PCR protocol requires two single-stranded oligonucleotide primers that are complementary to the 3¢ ends of the target region (see Fig. 5–2). A reaction is prepared that consists of the DNA of interest, relevant primers, deoxynucleotide triphosphate (dNTP), buf-fer, and a Taq (pronounced “tack”) DNA polymerase. This reaction is placed

in a thermal cycler that has been programmed to perform the amplification. The thermal cycler heats the reaction to denature the DNA; it then cools to allow the primers to anneal to the complementary region of the DNA. The Taq DNA polymerase extends the oligo-nucleotide primers in the 5¢ to 3¢ direction. The three steps—denaturation, annealing, and extension—collectively referred to as a cycle, doubles the DNA of the target region. Successive cycles lead to an exponential

Polymerase Chain Reaction

Objectives

• to learn the technique of in vitro amplification of DNA

• to amplify a DNA fragment from the Arabidopsis thaliana TTG locus and clone it into a cloning vector


Fig. 5–1. The thermal cycler (i.e., the “PCR machine”).

Fig. 5–2. Target region: a region of DNA in the genome to be amplified. Note that the primers are complementary to the 3¢ ends of the target region.


increase of the target product. Table 5–1 shows the rela-tionship between the number of PCR cycles and copies of the target DNA sequence.

In detail, the steps of PCR procedure are as follows (Fig. 5–3).

Step 1. Activate the Taq DNA polymerase and denature the template DNA to single strands by incubating at 92 to 94°C.

Step 2. Further denature the template DNA.Step 3. Cool to 50 to 60°C to anneal the primers to the 3¢

ends of the target region (see Fig. 5–2).Step 4. Extend the primers with Taq DNA polymerase.

The first amplified DNA strand is longer than the target region because the target region does not contain termination sequence. An exact replica of the target DNA fragment will only be synthesized after the third PCR cycle (see Fig. 5–3).

Step 5. Repeat Steps 2 to 4 several times. This “chain reaction” will result in a geometric increase

in the amount of DNA. For example, 20 cycles of amplifi-cation will produce more than 1 million copies of the target DNA (see also Table 5–1).

The PCR product is vulnerable: a minor contamination of the sample with DNA from other sources to which the primers can bind, will result in an incorrect product.

Optimizing the Polymerase Chain Reaction

For each experiment, PCR application must be optimized to ensure a successful product. Therefore, the parameters that influence the product must be reviewed.

1. DNA PolymerasesIn the process of PCR amplification, DNA polymerases pick the appropriate deoxynucleotide triphosphate (dNTP, see below) and attach it to the primer-based nascent (= newly formed) DNA strand. This polymerization proceeds in the 5¢ to 3¢ direction in the presence of magnesium ion. Taq DNA polymerase, which was extracted from the hot water bacterium Thermus aquaticus, is widely used for DNA amplification. A recommended concentration range for Taq DNA polymerase is between 1 and 2.5 units per 100 µL reaction. However, enzyme requirements may vary with respect to the size of the target region.

The disadvantage of the Taq DNA polymerase is that it does not have proofreading capabilities, so that during DNA amplification errors are occasionally introduced into the DNA copies. The error rate is approximately one base in each 20,000 nucleotides. The low fidelity of Taq is overcome by means of DNA polymerases that have high proofreading capabilities. Tgo DNA polymerase is one of them. Tgo has been isolated from Thermococcus gor-gonarius, an extremely thermophilic sulfur-metabolizing archaeon from a geothermal vent in New Zealand.

2. Deoxynucleotide Triphosphates Mix (dNTP)The dNTP mix is a solution consisting of the four nucle-

Fig. 5–3. DNA amplification by polymerase chain reac-tion. Reproduced from Russell (1996) by permission of Pearson Education.

Table 5–1. The relationship between the number of PCR cycles and copies of the target DNA sequence.

Number of PCR cycles

Copies of target DNA sequence

0 11 22 43 84 165 326 647 1288 2569 512

10 102420 1,048,576


otides dATP, dGTP, dCTP, and dTTP. A working stock of 2 mM is recommended.

3. Magnesium ChlorideThe magnesium concentration affects primer annealing and the product yield. Accordingly, PCR should contain 0.5 to 2.5 mM magnesium over the total dNTP concentration. To ensure an optimal magnesium concentration, the magnesium chloride solution must be completely thawed and vortexed before use.

4. PrimersPrimer selection is decisive for a successful PCR product. On the template DNA, the primers must anneal at least 100 bp apart. Both forward and reverse primers should have a similar melting temperature (Tm) to ensure proper amplification of the target region. Melting temperature is defined as the temperature at which 50% of DNA strands are in the double helical form. Melting temperature which is usually provided by the manufacturer, is calculated by the G + C and A + T content of the DNA with the formula Tm = 2(A + T) + 4(G + C). Generally, primers should have a G + C content of about 40% with a random base distri-bution. To ensure proper annealing of primers to template DNA, the annealing temperature must be about 5°C lower than the melting temperature of the primers (see Table 5–3). Primers are usually used at 5-mM concentration. A primer selection software is required.

5. Buffer CompositionThe optimal buffer concentration and pH depend on the DNA polymerase used in the reaction. Taq DNA polymerase works efficiently in reactions containing 10 to 15 mM Tris (pH 8.0). A buffer optimized for Taq, and supplied with it, is commercially available; the composition is 10 mM Tris (pH 8.0), 1.5 mM magnesium chloride, 50 mM potassium chloride, and 0.1% t-octylphenoxypolyethoxyethanol.

6. “Hot Start” (An Option)When PCR reactions are set up at room temperature (not on ice), forward and reverse primers occasionally hybridize; thus, short products, called primer dimers, are formed. This is caused when the primers have two or more complemen-tary bases at their 3¢ ends. Primer dimers appear as bright bands following gel electrophoresis of the product. Hot start minimizes the possibility of primer dimer formation. In the hot start technique, the components of PCR reaction are added to a PCR tube in two phases. The lower phase consists of distilled water, buffer, primers, dNTP, and the template DNA. A liquid wax is added on top of the lower phase and the tube is placed on ice. The wax solidifies almost instantly. The upper phase, which includes distilled water, buffer, magnesium chloride, and Taq DNA poly-merase, is added on top of the solidified wax. Then the tube is incubated in a thermal cycler for DNA amplification.

An alternative method to avoid primer dimer formation is to set up reactions on ice and immediately place them in a heated (to 92ºC) thermal cycler.

Amplification of a Fragment of the TTG Locus in Arabidopsis thaliana

In Arabidopsis thaliana, the TTG (transparent testa glabra) gene, which is required for normal trichome initiation (see Szymanski et al., 2000, for a review of trichome initiation), is located 29.5 map units from the tip of chromosome 5 (see also Exercise 11 for details about Arabidopsis thali-ana). A mutation in this gene eliminates trichomes from the leaf surface, brown pigment from the seed coat (testa), and anthocyanins from leaf epidermal cells. The TTG locus has been isolated by positional cloning (Walker et al., 1999) and its identity has been confirmed.

The entire TTG locus is made up of 5777 bp (Table 5–2). The TTG gene is 2261 bp long and is composed of two coding regions, or exons, interrupted by a noncoding inter-vening region, or intron (see Exercise 10 for details about exon and intron). The complete nucleotide sequence of the A. thaliana TTG locus is shown in Fig. 5–4. The sequence has been deposited in GenBank (Walker et al., 1999) and has the accession number AJ133743.

In this exercise, you will be amplifying a 1299-bp [about 1.3 kilobase (kb)] DNA fragment from the TTG locus. You will be using this amplified DNA to clone in the next two exercises. The region being amplified (i.e., the target region) starts with the nucleotide pair 2573 and ends with the nucleotide pair 3872. Review Fig. 5–4 and find the primer annealing sequences. Both forward and reverse primers are 24-mer (base pairs) in length (see the Materials, below). These primers have been generated by the software Primer3. Primer3 was developed at the Whitehead Institute for Biomedical Research. The Primer3 web-base server can be used at http://frodo.wi.mit.edu/cgi-bin/primer3/primer3_www.cgi; verified 11 May 2009.

To amplify the target region, a PCR program called GENOMIC3 is used [G. Koliantz and D.B. Szymanski (2001), unpublished data]. You need to have a thorough knowledge of the thermal cycler you choose to use. Alternatively, the laboratory instructor may show you how to program and operate it. The GENOMIC3 program is shown in Table 5–3.

MaterialsA. thaliana genomic DNA (the template DNA), concentration of

10 ng/µLForward primer (ttgF), concentration of 5 µMSequence: 5¢ CTTATGATAAATGCCAACAACCAA 3¢Reverse primer (ttgR), concentration of 5 µMSequence: 5¢ CTTCTCAATATCCCAAATCGTACA 3¢dNTP (working solution, concentration of 2 mM):

1 volume dNTP (stock solution, usually 10 mM)

Table 5–2. Molecular characteristics of the TTG locus. Figures are in base pairs (bp).

Locus Gene Exon 1 Intron Exon 2Size 5777 2261 1140 699 419Start–End 1–5777 3267–5527 3267–4407 4408–5107 5108–5527


Fig. 5–4. The complete nucleotide sequence of the TTG locus. The TTG gene starts with the nucleotide number 3267 and ends with the nucleotide number 5527. The target region is located between the nucleotide pairs 2573and 3872. Primer annealing sequences are underlined. Recognition sites of restriction enzymes are also shown.


4 volumes 1 M Tris, pH 8.0, as described in Exercise 2Taq DNA polymerase (keep on ice always), supplied with buffer

and magnesium chloride solutiondH2OIceThermal cyclerMicrotubes, 0.6 mLMicrotube racksPCR tubesPCR tube racksMicropipettorsTipsMicrocentrifugeGloves

Procedure1. Turn on the thermal cycler and program it as shown in

Table 5–3.2. Put a 0.6-mL microtube on ice and to it add the follow-

ing reagents in the order shown below.9.2 mL dH2O2 mL Buffer1.6 mL Magnesium chloride2 mL dNTP2 mL Forward primer2 mL Reverse primer1 mL Template DNA0.2 mL Taq DNA polymerase

Mix the tube well, centrifuge briefly, and leave on ice for 1 min.

3. Transfer the entire reaction into a PCR tube and place the tube on ice.

4. Open the lid of the thermal cycler, place the tube (Step 3, above) in one of the slots. Let all students put their tubes in the thermal cycler. Close the lid and press the appropriate button to start amplification. The GENOMIC3 program takes about 3 h and 56 min. The amplified DNA can be stored in a −20°C freezer until further use.

The DNA, thus amplified, can be used to ligate to a suitable vector. The ligation reaction can then be used directly for bacterial transformation and eventually a suc-cessful retrieval of the ligated fragment. These principles of gene cloning are the subjects of the next two exercises.

Protocol 5–1How to Find the Nucleotide Sequence of the Arabidopsis TTG Gene1. Connect your computer to the Internet.2. Find the home page for TAIR (The Arabidopsis

Information Resource) at http://www.arabidopsis.org; verified 11 May 2009.

3. On the tool bar, click on the <Search> botton to open a drop down menu, then click on the <Genes>.

4. On the TAIR Gene Search window, the genes of inter-est can be searched by name, keywords, features, or map location. On the Search by Name or Phenotype option, choose <Gene name> in the left hand side drop down box and <exactly> in the right hand side; then, in the adjacent input box, type transparent testa glabra 1. Click on the <Submit Query> botton to open the TAIR Gene Search Results window.

5. On the Gene Model(s) cloumn, click on the locus number which is <AT5G24520 .1> to open the Gene Model: AT5G24520 .1 window.

6. On the column specifying the GenBank Accession Sequence, click on the <full length genomic> option. The nucleotide sequence of the ttg-1 gene is displayed. According to these data, the gene is a 2261-bp length. This sequence can be found in Fig. 5–4, between the nucleotide pairs 3267 and 5527.

Protocol 5–2How to Generate Primers for the Target Region of InterestSeveral software are available that generate primers for a target region of interest, GeneRunner (http://www.generunner.net; verified 11 May 2009) is one of them. This software, which runs on Windows only, is available on the Internet free of charge and can be downloaded.1. Recall Step 6 of Protocol 5–1, highlight the region

you want to amplify (the target region), and copy it in <Edit> on the tool bar.

2. Open the GeneRunner program. On the toolbar, click on <File> to open a box, then select <New> in the box. Upon selecting, another box is displayed; in this box click on <Nucleic acid sequence> to open the GeneRunner file.

3. Paste the target sequence you copied in Step 1 on top of the file where the 5¢ end and Nucleotide Number 1 are displayed. The target sequence appears on the screen.

4. To generate a forward primer for the 3¢ end of the tar-get DNA, highlight the first 30 nucleotides and then click on <Analysis> on the tool bar. A small box is displayed; click on <Oligo . . .> in the box to open an inset box.

5. At the top of the inset box under Oligo:, the sequence you chose in Step 4 is displayed. Other specifications of the sequence are also seen in this box. Check Filter Tm. A functional primer is generated if the Filter Tm is around 60. Every time you delete a single base from

Table 5–3. The GENOMIC3 program†.

StepBlock

temperature Time Operation°C

1 92 2 min Enzyme activating–denaturing2 92 30 s Denaturing3 55†† 1 min Annealing4 72 3 min Extending5 39 times to Step 2 Cycle repeating6 4 0 h, 0 min, 0 s Holding at 4°C indefinitely7 End End of the program

† Duration of amplification approximately 3 h 56 min.†† Annealing temperature must be 5°C lower than the melting temperature

(Tm) of the primers. See manufacturer’s information about the Tm.


the 3¢ end of the primer that you choose and click on <OK> in the bottom of the same box, the Filter Tm decreases. Or, alternatively, upon adding a new base to the 3¢ end, the Filter Tm increases.

6. Once a desired primer is created, click on <Print> in the bottom of the inset box. The Print Oligo box is

displayed, requesting Name of Oligo. Type a name, for example, ttg-F, then click on <OK>. The sequence is printed (Fig. 5–5).

7. To generate a reverse primer for the 3¢ end of the target DNA strand, highlight the last 30 nucleotides and fol-low the instructions in Steps 4 through 6. You may call this oligo ttg-R (see Fig. 5–5).

Fig. 5–5. Data sheet of the ttg-F and ttg-R primers. The data shown in this figure may not comply with the data you generate in the laboratory. They vary depending on the size of the target region and the location on the tem-plete DNA you choose to generate primers. Consider Fig. 5–5 as an example.


Study Questions5–1. How many copies of the TTG target DNA can you ob-

tain if you use the PCR protocol shown in Table 5–3?

5–2. What are primer dimmers? How are they formed and why?

5–3. After amplifying a piece of DNA and electrophores-ing the PCR product, you noticed that the band ap-peared faint in the gel. What is the possible cause of this problem?

5–4. “The disadvantage of the Taq DNA polymerase is that it does not have proofreading capabilities, so that during DNA amplification errors are occasion-ally introduced into the DNA copies.” What kind of errors are these?

5–5. Which of the two primer designs will amplify the target region shown in bold font below? Explain why.

Design 15¢ ATCGCTACGGGATTCTACGACGTCGTCTCACCGCTGTCGATT 3¢ 3¢ CCCTAAG 5¢

5¢ ACCGCTG 3¢3¢ TAGCGATGCCCTAAGATGCTGCAGCAGAGTGGCGACAGCTAA 5¢

Design 25¢ ATCGCTACGGGATTCTACGACGTCGTCTCACCGCTGTCGATT 3¢ 3¢ TGGCGAC 5¢

5¢ GGGATTC 3¢3¢ TAGCGATGCCCTAAGATGCTGCAGCAGAGTGGCGACAGCTAA 5¢


Objectives




Exercise 6 Cloning 1Generating Recombinant Plasmids

Objectives

• to understand the concept of a clone and cloning DNA

• to clone an amplified fragment of the Arabidopsis thaliana TTG locus into a cloning vector to generate recombinant plasmids

• to identify recombinant and nonrecombinant plasmids

According to Paul Berg (1980 Nobel laureate in chemistry), “We use the word cloning in science as a term to describe the production of many copies of a start-ing material”. In genetics, if the “starting material” is a single cell, therefore, a group of cells stemming from that single cell will be a clone. The members of a clone must be identical in their genetic makeup.

In the laboratory, genes are cloned by extracting DNA from the organism of interest and inserting it into a cloning vector. The recombinant vector thus created, is introduced into a host cell in order to initiate replication to obtain several identical copies of that DNA. The DNA is later isolated from the vector and is used for further studies.

Cloning VectorsCloning vectors are semisynthetic DNA molecules which can replicate in host cells and into which pieces of foreign DNA can be introduced. There are several types of vectors that differ is size and the maximum amount of DNA that can be cloned into them. They include plasmids, phage lambda, cosmids (hybrids between plasmid and phage), and yeast artificial chromosome com-

monly known as YAC (plasmids that contain yeast DNA sequence). In this exercise, only the plasmids will be considered.

Plasmids are extrachromosomal, double-stranded circular DNA that exist in bacterial cells and replicate independently of the host nuclear chromosome. Plasmids are stable and maintain their characteristics in the host cells. Some of the plas-mids have been modified by genetic engineering techniques to make them useful tool for gene cloning. Modified plasmids are called plas-mid cloning vectors. They are 2 to 4 kb in length and are able to carry up to 15 kb of foreign DNA. One of such vectors is the E. coli plasmid cloning vector pUC19 (pronounced “puck 19”) (Fig. 6–1). This vector is around 2.7 kb long and has three characteristics as described below.

It carries the • origin of replication (ori) sequence which is recognized by the machinery of replication enzymes


Fig. 6–1. Plasmid cloning vector pUC19. Reproduced from Russell (1996) by permission of Pearson Education.

and which enables the vector to replicate inside the host cell.It contains the ampicillin-resistance gene • (ampR) which confers resistance to the antibi-otic ampicillin. The E. coli cells infected with the vector survive in ampicillin-rich medium. Antibiotic-resistance genes are usually re-ferred to as selectable markers.

The vector also carries a cluster of unique restriction sites called polylinker. The polylinker can be cut at one of these sites with the appropriate restriction enzyme and insert a piece of DNA from different organ-isms into that site. The polylinker is inserted into the lacZ gene derived from the lac operon of E. coli (see Exercise 10 for more information about the lac operon). The role of the lacZ gene in identifying recombinant and nonrecom-binant vectors will be discussed in the next section under Cloning of the PCR Product from Exercise 5.

Figure 6–2 shows how a recombinant vector is gener-ated by inserting fragments of foreign DNA into the vector. First the polylinker of the vector is cut by an appropriate enzyme. Then the same enzyme is used to digest the DNA to be cloned. Since the DNA may have many restriction sites for the enzyme, it will be cut into several pieces. Finally, the DNA pieces and the vector are mixed, a ligation buffer is applied, and the mixture is allowed to incubate. As a result, DNA fragments are inserted into the vector. The recombinant vector, thus generated, can be introduced into E. coli cells by means of transformation (Fig. 6–3).

What is Transformation?Taking up a piece of DNA from external sources is called transformation. Transformation is a rare event both in nature and in the laboratory. Some species of bacteria do not take up DNA from their environment and cannot be transformed. In other species, such as E. coli, the rate of transformation can be increased by treating the cells with calcium chloride, which alters the bacterial cell wall and the membrane. Or in Bacillus subtilis, a high rate of transforma-tion can be achieved by starving the cells by growing them on agar medium containing a trace amount of tryptophan and histidine. Cells that have been treated to increase their ability to take up DNA are called competent cells.

In the exercise described below, recombinant plasmids are mixed with E. coli competent cells, incubated on ice for 10 to 15 min, and heat-shocked for 30 to 90 s in a 42ºC water bath. The heat from the heat-shock drives the recombinant plasmids through the adhesion zone of the E. coli cell membrane. The transformed E. coli cells are then allowed to propagate in a nutrient broth, during which time the vectors inside E. coli replicate. Finally, the E. coli culture is transferred to an agar plate supplemented with an appropriate antibiotic. The antibiotic selection will allow the transformed E. coli cells to grow and form colonies that will have white or blue color. The white colonies would be the E. coli cells transformed with a recombinant vector, whereas blue colonies would be indicative of the E. coli cells transformed with a nonrecombinant vector.

Cloning of the PCR Product from Exercise 5

In this exercise, you will be using a cloning vector called TOPO. TOPO vector is included in the TOPO TA Cloning

reagents provided by Invitrogen Corp. (Carlsbad, CA). TOPO TA Cloning provides a highly efficient cloning strat-egy for the direct insertion of Taq polymerase-amplified PCR products into a vector.

The TOPO vector (Fig. 6–4) has three unique features: (i) it is linear, (ii) it carries a single, overhanging 3¢ deoxy-thymidine (T) residue, and (iii) it is “activated”, that is, topoisomerase I covalently binds to the vector preventing it


Fig. 6–2. Insertion of a piece of DNA into the pUC19 vector. Reproduced from Russell (1996) by permission of Pearson Education.

Fig. 6–3. A schematic representation of the steps of cloning and transformation.


from circularization. The Taq polymerase, used in the PCR reaction, has a nontemplate-dependent terminal transferase activity that adds a single deoxyadenosine (A) to the 3¢ ends of PCR products. Since the linearized vector has a single, overhanging 3¢ deoxythymidine (T) residue, this allows PCR inserts to ligate efficiently with the vector. Therefore, when cloning a gene with a TOPO vector, no DNA diges-tion or vector linearization is needed.

The TOPO vector, like the pUC19 vector mentioned above, carries a polylinker which is inserted near the 5¢ end of the fragment of the lacZ gene that encodes the amino ter-minal portion of the enzyme ß-galactosidase (see Exercise 10 for ß-galactosidase). The host E. coli cells that you will be using in this experiment carry the fragment of the lacZ gene that codes for the carboxyl portion of ß-galactosidase. These two defective versions of ß-galactosidase have no enzymatic activity. However, complementation can occur between them and a functional ß-galactosidase will be pro-duced if the E. coli cells are transformed by the vector. This is referred to as α-complementation.

The agar plate that you will be using in the transformation procedure contains a color indicator, Xgal (5-bromo-4-chloro-3-indolyl-β-d-galactopyranoside). Xgal is a white crystalline powder with molecular weight of 408.61 and the linear formula C14H15BrClNO6. This compound is a synthetic substrate of the enzyme β-galactosidase encoded by the lacZ gene. When β-galactosidase is present, it cleaves the glycosidic linkage in Xgal, producing a solu-ble, colorless indoxyl monomer. Subsequently, two of the liberated indoxyl monomers form a dimer (5-5¢-dibromo-4-4¢-dichloro-indigo) which is a stable and insoluble blue

compound (Holt and Sadler, 1958). This is the blue color which is visible in bacterial colonies. In the event that a piece of foreign DNA is inserted (cloned) into the polylinker of the TOPO vector, it further interrupts the defective lacZ gene by displacing the reading frame of the gene. Upon transformation, no functional β-galactosidase is produced; therefore, the colonies carrying recombinant vectors will remain white. In the course of this exercise, you will use this test to identify the vector into which you cloned a por-tion of the TTG locus.

1. Gel Electrophoresis of the PCR ProductIn Exercise 5, you amplified a 1.3-kb DNA fragment from the A. thaliana TTG locus. In Exercise 6, you will confirm the existence of the 1.3-kb fragment by agarose gel electro-phoresis. Then you will introduce this DNA into a vector and subsequently clone it.

MaterialsThe PCR product from Exercise 5DNA ladder, as described in Exercise 2AgaroseGel-loading dye, as described in Exercise 2dH2OEthidium bromide, as described in Exercise 21´ TAE buffer, as described in Exercise 2Agarose gel electrophoresis unit, as described in Exercise 2IceMicropipettorsTipsMicrotubes, 0.6 mLMicrotube racksMicrocentrifuge

Fig. 6–4. The construct and properties of the TOPO vector. Reproduced by permission of Invitrogen Corp.


Microwave oven200-mL flasksGel-staining dishes, as neededGlovesUV protective glasses or shieldAgarose gel photo documentation system, as described in Exercise 2

Procedure1. Prepare a 1% agarose gel in 1´ TAE buffer (see

Exercise 2 for details) and set up the electrophoresis unit as described in Exercise 2.

2. Place 2 µL of gel-loading dye into a 0.6-mL microtube, and then add 4 µL of the PCR product you prepared in Exercise 5. Save the rest of the product on ice; you will need it later in this exercise.

3. Pipette the contents of the tube up and down several times to ensure that the dye and the DNA are mixed. Centrifuge the tube briefly.

4. Load the first well of the gel with 2 µL of DNA ladder. Use the other wells for your samples. Use the entire sample (see Step 2, above). Then electrophorese the gel at 80 V for 45 min to 1 h.

5. View the gel on a UV transilluminator, identify the bands and photograph the gel.

6. In your sample, you should see a single band between the DNA ladder bands of 1636 and 1018 bp as shown in Fig. 6–5. The size of that band (see also Fig. 5–4) is 1299 bp (1.3 kb).

2. Cloning of the PCR ProductIn this exercise, you will create a recombinant plasmid by ligating the PCR product to the TOPO vector. Then, you will transform E. coli competent cells and plate them on selective media. After 24 h, successful transformants will appear.

MaterialsThe PCR product from Exercise 5LB (Luria-Bertani) agar plates containing kanamycin and XgalLB agar medium:

Per 1000 mL of dH2O15 g Agar10 g Tryptone5 g Yeast extract10 g Sodium chlorideAdjust pH to 7.0 with 1 M sodium hydroxideAutoclaveLet cool to 40°C, then add 1 mL of 50 mg/mL aqueous so-

lution of kanamycin, mix well, and distribute in sterile Petri dishes. One hour before using Petri dishes, add 60 µL of 40 mg/mL Xgal to each dish and spread evenly on top of the agar with a sterile spreader.

50 mg/mL kanamycin:Per 10 mL of autoclaved distilled water0.5 g KanamycinFilter sterilizes the solution before use. Keep at −20°C.

40 mg/mL Xgal:Per 10 mL of N,N-dimethyl formamide (DMF)0.4 g XgalNote: Xgal is light sensitive, keep in a brown bottle covered

with aluminum foil. It is not necessary to filter sterilize Xgal solutions. Keep at −20°C.

TOPO TA Cloning reagents, which include vector, ligation buffer (also known as Salt Solution), and E. coli competent cells.

Note: Competent cells must be stored in a −80°C freezer. They can also be stored on dry ice for a limited amount of time.

Ligation buffer:Per 1000 mL of dH2O70.2 g Sodium chloride5.7 g Magnesium chlorideAutoclave

SOB medium:Per 1000 mL of dH2O20 g Tryptone5 g Yeast extract0.584 g Sodium chloride0.186 g Potassium chlorideAdjust pH to 7.0 with 1 M sodium hydroxideAutoclave

dH2OIce42°C water bath37°C incubator37°C rotatorMicrotubes, 1.5 mLMicrotube racksMicrotube floatersMicropipettorsTipsMicrocentrifuge

Fig. 6–5. The 1.3-kb DNA fragment from the Arabidopsis TTG locus as described in Exercise 5, Fig. 5–4.

Adhesive tapeSterile toothpicksGlass beakersSpreadersBurnersLaboratory gas lighters

Procedure1. Transfer 3 mL of the PCR product to a tube containing

1 mL of the TOPO ligation buffer. Mix gently. Save rest of the PCR product in the freezer. You will need it to prepare a probe (see Exercise 9).

2. Add dH2O to a final volume of 5 mL. Mix gently.3. Add 1 mL of TOPO vector. The final volume should

now be 6 mL. Mix gently and centrifuge briefly.4. Incubate the reaction at room temperature for 5 min

and then on ice for 2 min. Meanwhile, place the tube containing E. coli competent cells on ice to thaw.

5. Add 2 mL of the reaction from step 4 above to the tube containing the competent cells and mix gently. Do not mix by pipetting up and down or vortexing .

6. Incubate on ice for 10 min.7. Place the tube in a floating rack, then place the rack

in a 42°C water bath for 30 s without shaking, then immediately transfer to ice.

8. Add 250 mL of SOB medium to the tube from Step 7 above, cap the tube tightly, tape the tube on the plat-form of a 37°C rotator and rotate for 1 h.

9. This step must be done close to a lit burner to pre-vent plate contamination .

After 1 h, remove the tube containing the transforma-tion reaction from the rotator (Step 8, above), pipette 50 mL of the reaction on an LB agar plate containing kanamycin and Xgal (LB + kan + Xgal). Immediately flame a glass spreader and cool it by gently touching the spreader to the agar (do not touch it directly to the 50 mL of cells to prevent heat-killing of the cells). Then spread the transformation reaction evenly on top of the agar. Write your name on the bottom of the plate and place the plate upside down in an incubator set at 37°C. Leave overnight.

10. The following day check the plates for the presence of white and blue colonies. If you find white colonies on your plate, you likely have cloned the 1.3-kb fragment.

11. Place the plate close to a lit burner. With the aid of a sterile toothpick, pick a single colony and streak it on another LB + kan + Xgal plate as illustrated below. Perform this step for two white and two blue colonies. Each time use a new toothpick to prevent colony contamination .

12. Incubate this plate upside down in a 37°C incubator overnight.

13. The following day prepare liquid cultures from the streaks, as described in Exercise 7.

Study Questions6–1. Briefly explain how a piece of DNA is inserted into a

pUC vector.

6–2. What are the three most significant differences between the TOPO vector and other conventional vectors, such as pUC19?

6–3. Look at Fig. 6–4. Suppose you have cloned a piece of DNA into the TOPO vector. How can you retrieve the cloned DNA?

6–4. In the process of transformation, why should E. coli cells be competent?

6–5. What is the role of the lacZ gene in cloning vectors?



Objectives




Exercise 7 Cloning 2Minipreping

Objectives

• to learn the technique of small-scale plasmid isolation

• to retrieve a cloned fragment of the Arabidopsis thaliana TTG locus from recombinant plasmids

As discussed in Exercise 6, cloning vectors are used to produce a large quantity of a specific DNA fragment. The DNA of interest is spliced into a plasmid, cre-ating a recombinant plasmid. The recombinant plasmid is then introduced into E. coli cells through the transformation process. The transformed E. coli cells are multiplied in a nutrient medium; at the same time, the recombinant plasmid replicates in the bacterium producing many copies, which can be isolated. In the process of isolation of recombinant plasmids, E. coli host cells are resuspended in a buffer that contains Tris supplemented with RNase. Tris permeabilizes the cell membrane while RNase degrades RNA in the cell. Then the E. coli cells are lysed in sodium hydroxide and SDS (Sodium Dodecyl Sulfate = Sodium Lauryl Sulfate). The SDS solubilizes the phospholipid and protein components of the cell membrane and cell wall, leading to release of the cell contents, and sodium hydroxide denatures the chromosomal and plasmid DNA. The lysate is neutralized then adjusted to high salt binding conditions by the addition of potassium acetate. The high salt concentration precipitates the denatured pro-teins, chromosomal DNA, cellular debris, and SDS, which then can be removed by centrifugation. The plasmid DNA, which exists as a small, circular, highly supercoiled molecule, renatures and stays in solution.

In this exercise, you will employ the above method to extract recombinant and nonrecombinant plasmids from transformed E. coli cells. The method is called miniprep or small-scale plasmid isolation. The miniprep will yield a suf-ficient amount of DNA for retrieval of the cloned fragment from the plasmids.

Preparation of Liquid Cultures from White and Blue Colonies

At the end of Exercise 6, you prepared streaks of white and blue colonies on LB + kan + Xgal plates. Twenty-four hours before the start of Exercise 7, you must prepare liquid cultures from one white and one blue colony to extract the plasmid DNA.

MaterialsAgar plates containing streaks of white and blue colonies (see Exercise 6).LB liquid medium:

Per 1000 mL of dH2O10 g Tryptone5 g Yeast extract10 g Sodium chlorideAdjust pH to 7.0 with 1 M sodium hydroxideAutoclaveThe medium can be kept in a refrigerator and used when needed.

50 mg/mL kanamycin:See Exercise 6 for details

Culture tubes (disposable are preferred)Sterile toothpicks37°C shaker



Sterile glass pipettes (or disposable plastic pipettes)Sterile glovesBurnersLaboratory gas lighters

ProcedureThroughout this exercise wear sterile gloves . If gloves you wear are not sterile, rub them with 95% ethanol. Also, perform this exercise close to a lit burner or, preferably, in a sterile hood.

1. In a sterile culture tube, add 3 mL of LB medium and supplement it with 3 µL of 50 mg/mL kanamycin (hereafter LB + kan).

2. With the tip of a sterile toothpick, pick a small amount of the white streak and drop the toothpick into the cul-ture tube (Step 1, above). Repeat the procedure for the blue streak.

3. Place the inoculated culture tubes in a 37°C shaker overnight at 225 to 250 rpm.

4. After overnight incubation, the tubes can be stored in a refrigerator until needed.

Extraction of Plasmid DNA from Liquid Cultures (Miniprep)

The white colonies you obtained at the end of Exercise 6 contained recombinant plasmids. By restriction digestion and gel electrophoresis of the digested DNA, you will be able to confirm the identity of the cloned fragment. You will use blue colonies as a negative control for your experiment. Review the map of the TOPO vector in Fig. 6–4. In the polylinker, there are two EcoRI sites flanking the region where the PCR product is inserted. If you digest the vector with EcoRI, the correct size of the fragment you cloned will be identified. Furthermore, the 1.3-kb fragment you ampli-fied (see Fig. 6–5) contains four HindIII sites as illustrated in Fig. 7–1 (see also Fig. 5–4 for details).

As shown in Fig. 7–1, if you digest this fragment with HindIII, you should expect to see three bands with the fol-lowing sizes: 2882 to 2626 = 256 bp, 2966 to 2883 = 83 bp, and 3509 to 2967 = 542 bp. When you electrophorese your samples, estimate the size of the bands on the gel and compare it with this expectation. Find similarities and dif-ferences between your results and the expected data.

A schematic view of the EcoRI- and HindIII-digested recombinant plasmids is shown in Fig. 7–2.

MaterialsCell suspension buffer (CSB):

1´ TE (with RNase A)Autoclave, let cool, then add 100 µg of RNase A per 1 mL of

TE, i.e., 1:1000 dilution of RNase A. Keep in refrigerator.1´ TE:

Per 500 mL of dH2O5 mL 1 M Tris, pH 8.01 mL 0.5 M EDTA, pH 8.0To prepare 1 M Tris see Exercise 2.To prepare 0.5 M EDTA see Exercise 2.

Cell lysis buffer (CLB) Prepare fresh:0.8 g Sodium hydroxide10 mL10% SDSTo prepare 10% SDS, dissolve 10 g of SDS in 90 mL of

dH2O, heat to 68°C to assist dissolution. Adjust the pH to 7.2 with 1 M hydrochloric acid, and then adjust the volume of the solution to 100 mL with dH2O. There is no need to autoclave the solution.

Clearing buffer (CB):Per 1000 mL of dH2O294.42 g Potassium acetateAdjust pH to 5.5 with glacial acetic acidAutoclave

Overnight bacterial culturesIsopropanol (chilled to −20ºC)70% ethanol: Add 70 mL of absolute ethanol to 30 mL of dH2O

and mix wellMicropipettorsTipsMicrotubes, 1.5 mLMicrotube racksTest tube racksMicrocentrifugeGraduated cylinders, 10 mL or smallerVortexGlovesIce

ProcedureThroughout this exercise keep CSB and CB buffers on ice. Perform Steps 1 through 11 once for white colony and once for blue colony liquid cultures.1. Transfer 1.5 mL of the overnight bacterial culture to

a 1.5-mL microtube and centrifuge for 1 min at 6000 rpm. Discard the supernatant and add another 1.5 mL of the culture to the same tube and centrifuge again. Discard the supernatant.

2. Add 200 mL of CSB to the tube and vortex to resuspend the pellet. The pellet must be completely resuspended .

Fig. 7–1. HindIII restriction sites within the cloned 1299-bp fragment. HindIII recognition sequences are shown in bold font and the cleavage sites are marked with a down arrow.


3. Add 400 mL of CLB. Invert the tube several times to mix; do not vortex.

4. Add 200 mL of chilled CB buffer. Invert the tube gently several times to mix; do not vortex.

5. Centrifuge for 20 min at 13,000 rpm. The supernatant contains the DNA of interest that must be recovered by alcohol precipitation (see steps 7 through 11, below).

6. With the aid of a micropipettor, carefully transfer the supernatant to a clean 1.5-mL microfuge tube.

7. Add 700 mL of chilled (−20ºC) isopropanol to the tube containing the supernatant (Step 6, above) and invert the tube several times to mix; do not vortex.

8. Centrifuge for 30 min at 13,000 rpm to pellet the pre-cipitated plasmid DNA.

9. Decant the supernatant and wash the DNA pellet with 100 mL of 70% ethanol for 1 min.

10. Carefully drain the ethanol and dry the DNA pellet un-der a hood for 10 min or until it turns clear. If there is cell debris in the DNA, it will not turn clear.

11. Resuspend the DNA in 15 mL of 1´TE.12. Quantify the DNA with a fluorometer.

Restriction Digestion of the Extracted DNA

The recombinant and nonrecombinant plasmid DNA you extracted in the previous section must now be digested with EcoRI and HindIII enzymes. Under optimal conditions the DNA should have a yield of around 40 ng/mL. You may quantify the DNA to decide which protocol to choose for the preparation of digestion reactions. The protocols are described in Step 2, below.

MaterialsDNA samples (from Step 12, above)EcoRI enzymeEcoRI buffer†HindIII enzymeHindIII buffer†

†Note: Manufacturers provide enzymes with appropriate buffers; consult manufacturers’ recommendation.

dH2OMicrotubes, 0.6 mLMicrotube racksMicrocentrifugeMicropipettorsTips37°C incubator65°C water bathFloating racksGlovesIce

Procedure1. Clearly label four 0.6-mL microtubes to identify the

source of the DNA used (white or blue colony) and the enzymes employed (EcoRI or HindIII). You may label the tubes as follows: W-E (for white colony, EcoRI cut), W-H (for white colony, HindIII cut), B-E (for blue colony, EcoRI cut), and B-H (for blue colony, HindIII cut).

2. Prepare the digestion reactions. For a successful diges-tion, a total of 100 to 150 ng of DNA per reaction is needed, and the amount of enzyme used must not ex-ceed 10% of the total volume of the reaction. Add the reagents in the order shown below.

Tube W-E Tube W-HmL mL

dH2O 13 dH2O 13EcoRI buffer 2 HindIII buffer 2DNA 3 DNA 3EcoRI 2 HindIII 2

Tube B-E Tube B-HmL mL

dH2O 13 dH2O 13EcoRI buffer 2 HindIII buffer 2DNA 3 DNA 3EcoRI 2 HindIII 2

Note: If the yield of the DNA you prepared in Step 11 above is around 20 ng/µL or less, you may change the quantities of the reagents as shown below.

Fig. 7–2. A schematic presentation of EcoRI- and HindIII-digested recombinant plasmids.

Tube W-E Tube W-HmL mL

dH2O 6 dH2O 6EcoRI buffer 2 HindIII buffer 2DNA 10 DNA 10 EcoRI 2 HindIII 2

Tube B-E Tube B-HmL mL

dH2O 6 dH2O 6 EcoRI buffer 2 HindIII buffer 2DNA 10 DNA 10EcoRI 2 HindIII 2

3. Gently tap each tube with a finger to mix the reaction, then centrifuge briefly.

4. Place the tubes in a 37°C incubator overnight.5. The following day place the tubes in a floating rack,

and place the rack in a 65°C water bath for 20 min to deactivate (kill) the enzymes. The digested reactions can be stored in a −20°C freezer until further use.

Study Questions7–1. What is a miniprep?

7–2. What are some issues of concern about the use of recombinant plasmids, such as the one illustrated in Fig. 7-2?

7–3. In restriction digestion, what happens if the digestion reaction is not stopped by deactivating the enzyme?

7–4. In preparing digestion reactions (see the Procedure, above), the reagents are added in the following or-der: (i) dH2O, (ii) buffer, (iii) DNA, and (iv) enzyme. If the order is shifted to (i) enzyme, (ii) dH2O, (iii) DNA, and (iv) buffer, what do you think may happen in the digestion process?

7–5. What is the purpose of adding RNase in the cell suspension buffer (CSB)?



Objectives




Exercise 8 Southern Blotting 1DNA Transfer

Objectives

• to learn the technique of Southern blotting

Southern blotting, developed by Edwin M. Southern in 1975, is a technique to detect a specific DNA sequence within the genome or in a mixture of digested DNA fragments. This technique is made up of two steps: (i) transfer of digested DNA fragments from an agarose gel onto a nylon membrane and (ii) DNA–DNA hybridization and sequence detection. Step 1 is discussed in this exercise, and Step 2 will be the subject of Exercise 9.

DNA TransferIn the Southern blotting technique, the DNA of interest is digested with restric-tion enzymes and resolved on an agarose gel by electrophoresis to separate the fragments according to size. Then the gel is photographed alongside a transpar-ent ruler so that the distance that any DNA band has migrated can be estimated from the photograph. The DNA in the gel is nicked by brief depurination with hydrochloric acid to improve the transfer of larger DNA fragments. Then the DNA is denatured and neutralized with a strong base. The resulting single-stranded DNA is transferred onto a nylon membrane or a nitrocellulose filter through capillary action. In practice, the gel is placed on top of a wick which itself is on top of a platform (or a support) inside a glass dish containing an appropriate buffer (Fig. 8–1). A wet membrane is placed on top of the gel, so that the edges of the membrane and the gel line up. Then two or three wet pieces of Whatman filter paper are placed on top of the membrane and a stack of dry paper towels are placed on top of the Whatman papers. The entire apparatus is then put under a 300- to 400-g weight, such as a textbook. The capillary action which is maintained by the stack of paper towels carries the DNA from the gel in the flow of a buffer and directs toward the membrane. Since the nylon mem-


Fig. 8–1. Transfer set up for Southern blotting. Reproduced from Ausubel et al. (ed.) (1992) by permission of John Wiley & Sons.


brane is positively charged, it allows the negatively charged DNA to bind securely. Rate of transfer of the DNA bands depends on the size of the bands and the concentration of the agarose gel. DNA bands smaller than 1 kb will transfer faster than larger bands on a less concentrated agarose gel, such as 0.8%. The transfer should be allowed to take place for at least 18 h, after which the Southern is disassembled and the DNA is fixed to the membrane by baking for 2 h in a vacuum oven or, preferably, exposing to a UV light for about 1.5 min (cross linking). The membrane is then used for DNA hybridization and sequence detection.

Electrophoresis and Southern Blotting of Digested Recombinant and

Nonrecombinant PlasmidsAt the end of Exercise 7, you digested the recombinant and nonrecombinant plasmid DNA with EcoRI and HindIII enzymes (tubes W-E, W-H, B-E, and B-H). In this exer-cise, you will separate the resulting restriction fragments by agarose gel electrophoresis, and then you will assemble a Southern apparatus to transfer the digested fragments onto a nylon membrane.

1. Electrophoresis of EcoRI- and HindIII-Digested Plasmid DNA

MaterialsEcoRI- and HindIII-digested minipreps from Exercise 7 (Tubes

W-E, W-H, B-E, and B-H) DNA ladder, as described in Exercise 2Gel-loading dye, as described in Exercise 2AgaroseEthidium bromide, as described in Exercise 21´ TAE buffer, as described in Exercise 2Agarose gel electrophoresis unit, as described in Exercise 2MicropipettorsTipsGel-staining dishes, as needed200-mL flasksMicrowave ovenMicrotubes, 0.6 mLMicrotube racksMicrocentrifugeIceTransparent rulerUV protective glasses or shieldsAgarose gel photo documentation system, as described in Exercise 2

Procedure1. Prepare a 0.9% agarose gel in 1´ TAE, as described

in Exercise 2. Use a wide-tooth comb to create large wells in the gel. Also set up the electrophoresis unit as described in Exercise 2.

2. Add 5 µL of gel-loading dye to each of the tubes W-E, W-H, B-E, and B-H. Mix well and centrifuge briefly.

3. On the agarose gel, load well Number 1 (from left) with 4 µL of DNA ladder.

4. Load well Numbers 2, 3, 4, and 5 with 15 µL of the con-tent of the tubes W-E, B-E, W-H, and B-H, respectively.

5. Electrophorese the samples at 80 V for 45 min.

6. Cut the top left corner of the gel to indicate its direction. Transfer the gel to a UV transilluminator, place a trans-parent ruler next to the gel, and align the 0 mark of the ruler with the wells, then photograph the gel (Fig. 8–2).

7. On the photograph, record the distance (in millimeters) between the wells and each band. You will need this information in Exercise 9.

2. Transfer of Digested Plasmid DNA from Agarose Gel onto Nylon Membrane

MaterialsAgarose gel from Step 6, above0.25 M hydrochloric acid:

Prepare 1M hydrochloric acid as described in Exercise 1; then dilute it four times: to 1 volume of the stock add 3 volumes of dH2O

Denaturation solution:Per 1000 mL of dH2O87.7 g Sodium chloride20 g Sodium hydroxideAutoclave

Neutralization solution:Per 1000 mL of dH2O87.7 g Sodium chloride78.8 g TrisAdjust pH to 7.5 with 1 M hydrochloric acidAutoclave

20´ SSC (Standard Saline Citrate) buffer; use as stock:Per 1000 mL of dH2O88.2 g Sodium citrate175.3 g Sodium chlorideAdjust pH to 7.0 with 1 M hydrochloric acidAutoclave

10´ SSC buffer:To 1 volume of 20´ SSC buffer add 1 volume of dH2O

Fig. 8–2. Photographic image of EcoRI- and HindIII-digested plasmid DNA. Note: In this gel, EcoRI- and HindIII-digested recombinant plasmids have been loaded twice for the sake of clarity.


dH2ORotary shakerUV cross linkerNylon membranesWhatman #3 paperGlass or plastic dishesSupport or flat glassPaper towelsSaran wrapGlovesBook (or an object) weighing about 400 g

Procedure1. Place the gel in a glass or plastic dish containing 0.25

M hydrochloric acid; then place the dish on a rotary shaker for 20 min at room temperature.

2. Rinse the gel in distilled water.3. Place the gel in denaturation solution and rotate for 20

min at room temperature.4. Rinse the gel in distilled water.5. Place the gel in neutralization solution and rotate for 20

min at room temperature.6. Rotate the gel for 5 min in 10´ SSC buffer at room

temperature. At the same time cut the top left corner of the nylon membrane to match that in the gel. Wet the membrane in distilled water and then in 10´ SSC buf-fer for 5 min before placing it on top of the gel. Also, wet the Whatman papers in 10´ SSC buffer before placing them on top of the membrane.

7. Wear gloves and set up the Southern as illustrated in Fig. 8–3.

Note: Remove all air bubbles that may occur between the gel, nylon membrane, and the Whatman papers by rolling a glass pipette over each surface.

8. Allow the transfer to take place overnight (about 18 h).9. During the process of transfer, replace the paper towels

as they become wet. This will ensure the flow of buffer.10. The next day, disassemble the set up, mark the trace

of the gel’s wells on the membrane with a pencil, then immerse the membrane in 10´ SSC buffer to remove pieces of agarose sticking to it. While still wet, place the membrane gel-side-up on a piece of Whatman pa-per and expose the membrane to a UV light using a crosslinker. For most Southern blotting techniques, a total exposure of 120 mJ/cm2 is sufficient to immobi-lize the DNA on the membrane. You must know how to operate a crosslinker.

11. Air-dry the membrane, wrap it in a Whatman paper, and store in a refrigerator until needed. You will need the membrane in Exercise 9.

Study Questions8–1. In the process of preparing a gel for Southern blot-

ting, why is it necessary to denature the DNA on the gel?

8–2. In the process of transferring digested DNA from agarose gel onto nylon membrane, what is the pur-pose of washing the gel with low concentration of hydrochloric acid?

8–3. In assembling the Southern apparatus (see Fig. 8–3), if you place the gel upside down (instead of upside up, as stated in the figure), do you think that the transfer will take place successfully? Explain your answer.

8–4. What is the capillary action in Southern blotting?

8–5. Suppose that you electrophoresed the digested sample of a piece of DNA on an agarose gel contain-ing ethidium bromide and found three bands. You then transferred the bands from the gel onto a nylon membrane by the technique described in this chapter. Before proceeding to the next chapter to hybridize the membrane with a probe and perform the immu-nological detection, how could you determine if the three bands have successfully transferred onto the nylon membrane?

Fig. 8–3. Cross-section of a Southern apparatus.


Objectives




Exercise 9Southern Blotting 2DNA–DNA Hybridization and Sequence Detection

Objectives

• to learn the technique of PCR product purification

• to learn the technique of probe preparation

• to understand the concept of genetic similarity by performing DNA-probe hybridization

As discussed in Exercise 8, Southern blotting involves passing a buffer through an agarose gel that contains DNA fragments. The buffer then wicks the DNA out of the gel and onto a nylon membrane. Finally, the membrane is cross-linked, binding to the DNA permanently. The membrane now contains all the DNA fragments in the mirror image positions they occupied within the gel.

To identify DNA sequences of interest on the nylon membrane, a specific probe is needed. Probes are single-stranded DNA molecules that are used to identify and bind to complementary DNA sequences. Probes are labeled with radioactive or preferably nonradioactive materials such as biotin (Fig. 9–1) or Dig High Prime (Roche Diagnostic Corp., Indianapolis, IN) system and can be detected on the membrane by means of colorimetric or chemiluminescent tech-niques. After washing the membrane with the probe, only the DNA fragments that are complementary to the sequence of the probe will form double-stranded hybrids; unbound DNA is washed away. The hybridized fragments are then stained with a chromogenic (color-generating) or a chemiluminogenic (light-generating) substrate.

In the chromogenic technique, the hybridized membrane is first washed with a blocking solution to prevent nonspecific interaction of the antibody with the membrane. Then the membrane is treated with antidigoxigenin-alkaline phosphatase conjugate, which is sheep Fab-fragments (Fab-fragments contain antigen-binding sites) conjugated to alkaline phosphatase. Alkaline phosphatase binds to Dig-labeled probe. The antigen–antibody composite is then stained


Fig. 9–1. A nucleotide incorporated with biotin.

with a 5-bromo-4-chloro-3-indol phosphate–nitro-blue-tetrazolium (BCIP/NBT) complex. In this process, alkaline phosphatase removes the phosphate group from BCIP. The resulting molecule (BCI) dimerizes (BCI–BCI) to produce a purple precipitate (Fig. 9–2). The NBT present in the complex enhances the purple color of the precipitate at the site of the reaction. Thus, regions of hybridization on the nylon membrane are visible as purple bands.

In the chemiluminogenic technique, the hybridized probe is immunodetected with antidigoxigenin-alkaline phosphatase and then visualized with a chemiluminescence substrate for alkaline phosphatase. Enzymatic removal of the phosphate group of the substrate by alkaline phosphatase destabilizes the molecule leading to the emission of light, which can be recorded on an X-ray film. Following detection by either tech-nique mentioned above, the relative migration of the bands that hybridize is determined and compared with the migration of DNA fragments on the matching agarose gel.

Probe Preparation and HybridizationPreparations of the probe and hybridization to the nylon membrane are time consuming procedures, so they must be done ahead of laboratory time. Hybridization must be performed 16 to 18 h prior to the start of the immunologi-cal detection. In this exercise, a probe is prepared from the PCR product, that is, the entire 1.3-kb fragment described in detail in Exercise 5. However, this product must be cleaned of primers, nucleotides, polymerases, and buffers before use. This can be achieved by passing the PCR prod-uct through a column that contains a silica membrane. The technique is called column purification and is performed with a kit provided by QIAGEN (Valencia, CA).

1. Purification of PCR-Amplified TTG DNA

MaterialsThe following materials, which will be used to purify PCR-amplified DNA, are included in QIAquick PCR Purification Kit (50), Catalog No. 28104, supplied by QIAGEN:Buffer PBIBuffer PEBuffer EBQiaquick spin columns2-mL collection tubes

Additional materials needed include:PCR amplified TTG DNA (see Exercises 5 and 6)100% ethanolMicrocentrifugeMicropipettorTipsMicrotubes, 1.5 mLMicrotube racksPCR tube racks

ProcedureThis procedure follows QIAGEN’s QIAquick PCR Purification Kit Protocol with minor modifications to fit the format of this laboratory manual. According to the manufacturer’s recommendation, before using the kit, add

24 mL of 100% ethanol to the bottle containing Buffer PE and mix well.

1. Leave the tubes containing the PCR product (see Exercises 5 and 6) at room temperature until the con-tents thaw.

2. Add 5 volumes of Buffer PBI to 1 volume of the PCR product and mix well. For example, add 100 µL of Buffer PBI to 20 µL of the PCR product.

3. Place a QIAquick spin column in a 2-mL collection tube.4. With a micropipettor, transfer the DNA sample (Step 2,

above) to the column (Step 3, above) and centrifuge for 1 min at 13,000 rpm.

5. Discard the flow-through, then place the column back into the same collection tube.

6. Add 750 µL of Buffer PE to the column and centrifuge for 1 min at 13,000 rpm.

7. Discard the flow-through, place the column back into the same collection tube and centrifuge for an addi-tional 1 min at 13,000 rpm.

8. Place the column in a clean 1.5-mL microtube.9. To elute the DNA, add 30 µL of Buffer EB to the center

of the column, let the column stand for 1 min, and then centrifuge for 1 min at 13,000 rpm.

Note: According to the manufacturer, the average eluate volume is 2 µL less than the elution buffer volume used.

10. This DNA can be quantified and stored in a freezer until needed.

2. Probe PreparationBefore preparing the probe, the column-purified PCR prod-uct (Step 10, above) must be quantified, so that a proper amount of DNA will be labeled. The DNA can be quantified with a fluorometer.

MaterialsThe following materials will be used to prepare a nonradio-active probe and are included in the Dig High Prime DNA Labeling and Detection kit, supplied by Roche Diagnostics Corp., Indianapolis, IN:Dig High Prime (Dig: Digoxigenin) Hybridization buffer (DIG Easy Hyb Granules)Blocking solutionAntidigoxigenin-AP Solution (AP: Alkaline Phosphatase)CSPD (a chemiluminescent substrate for alkaline phosphatase)


Fig. 9–2. Dimerized BCI: 5, 5 ¢ dibromo-4, 4 ¢ dichloro indigo, a purple precipitate.


Additional materials needed include:Column-purified and quantified PCR product (see Step 10, above)Autoclaved, double distilled water (ddH2O)0.5 M EDTA, see the Materials in Exercise 2MicropipettorsTipsIncubator, adjusted to 37°CVortexMicrocentrifugeMicrotubes, 1.5 mLMicrotube racks37°C incubatorTripodWire gauzeGlass beakers (500 mL)Floating racksBurnersLaboratory gas lightersIce

Procedure1. In a 1.5-mL microtube, dilute the column-purified

and quantified PCR product with autoclaved double distilled water according to Table 9–1. If the concen-tration of your sample differs from the concentrations shown in the Table, choose the closest concentration. Note that a concentration lower than 10 ng/mL is not recommended.

2. Using a burner, tripod, wire gauze, and beaker, prepare boiling water.

3. Denature the DNA (Step 1, above) by placing it in a floating rack and putting the rack in the boiling water for 15 min.

4. Immediately transfer the DNA to ice-water to prevent renaturation and leave for 2 min.

5. Vortex the tube containing Dig-High Prime, then add 4 µL of it to the denatured DNA (Step 4, above). Mix well, centrifuge briefly, and incubate in a 37°C incuba-tor for 20 h.

6. After 20 h is complete, add 1 mL of 0.5 M EDTA to the probe, mix well, centrifuge briefly, and store in a freezer. You will need this probe in Steps 5 and 10 through 12, as described below.

3. Hybridization

The probe you just made will now be used to hybridize to the DNA bands on the nylon membrane (see Exercise 8).

MaterialsThe following materials, which will be used for DNA–DNA hybridization, are included in the Dig High Prime DNA Labeling and Detection kit, supplied by Roche:Dig High Prime (Dig: Digoxigenin) Hybridization buffer (DIG Easy Hyb Granules)Blocking solutionAntidigoxigenin-AP Solution (AP: Alkaline Phosphatase)CSPD (a chemiluminescent substrate for alkaline phosphatase)

Additional materials needed include:The nylon membrane from Exercise 837°C incubatorHybridization incubator adjusted to 42°CHybridization cylindersMeshHybridization bufferProbe (see above)Autoclaved, double distilled waterMicropipettorTipsTripodWire gauzeGlass beakers (500 mL)Graduated cylindersFloating racksBurnersLaboratory gas lightersIceConical tubesGlovesScissors

Procedure1. One hour before the start of the exercise, prepare the

hybridization buffer. The buffer is provided by the manufacturer in the form of granules and contains op-timal concentrations of SSC, SDS, and salmon sperm DNA. Add 32 mL of autoclaved, double distilled water to the granules and dissolve by shaking the bottle. To facilitate dissolution, place the bottle in a 37°C incuba-tor for 30 to 60 min. Add another 32 mL of autoclaved, double distilled water to the same bottle, shake, and incubate (if needed) until all granules dissolve. If this buffer is not used immediately, it can be saved in a −20°C freezer for up to 1 month.

2. At the same time, turn on a hybridization incubator (Fig. 9–3) and adjust the temperature to 42°C and the rotator to 3 rpm. Both the prehybridization (Step 8, below) and hybridization (Step 14, below) will be performed at 42°C.

3. Place the hybridization buffer (Step 1, above) in the hybridization incubator to equilibrate.

4. A few minutes before the start of the exercise place the DNA-bound nylon membrane (from Exercise 8) at room temperature.

5. Place the probe on ice to thaw.

Table 9–1. Preparation of working concentrations for column-purified PCR product.

Initial concentration of column-purified DNA

Volume of DNA

Volume of ddH2O

ng/mL ———— µL ————50 6 10 45 7 943 9 730 10 620 13 316 13 313 14 210 15 1


6. Cut two pieces of mesh sheet exactly the size of the membrane.

7. Place the membrane between the two mesh sheets, roll them up, and insert into the hybridization cylinder.

8. To the same hybridization cylinder, add 20 mL of hybridization buffer (save the rest of the buffer in the same incubator; you will need it in Step 12), place the cylinder into the hybridization incubator, counter bal-ance with another cylinder containing 20 mL of water, and let it rotate for 30 to 45 min. This step is called prehybridization.

Note: In the process of Southern transfer, the gel-load-ing dye added to the DNA samples is also transferred onto the nylon membrane leaving blue spots on it. However, during prehybridization the spots are washed out from the membrane. To be sure that there are no spots on the membrane, you may extend the prehybrid-ization time to 60 min.

9. At the same time prepare boiling water.10. Ten minutes before the end of the prehybridization,

denature the probe (Step 5, above) by placing it in a floating rack and putting the rack in the boiling water for 15 min.

11. Immediately transfer the tube containing the probe (Step 10, above) to ice water.

12. Add 15 μL of the probe to 20 mL of fresh hybridization buffer (see Step 8, above) and mix well. This can be done in a sterile conical tube.

13. Pour off the prehybridization buffer from the cyl-inder (Step 8, above) and immediately add the probe–hybridization mixture (Step 12, above) to the same cylinder.

14. Rotate the cylinder in the hybridization incubator over-night (16–18 h).

4. Immunological DetectionAfter hybridization of the probe to the DNA bands on the nylon membrane is complete, the membrane must be washed with an antibody solution to detect the bands either in situ

(on the nylon membrane) or on an X-ray film, depending on the technique used.

MaterialsThe following materials, which will be used for immuno-logical detection, are included in the Dig High Prime DNA Labeling and Detection kit, supplied by Roche:Dig High Prime (Dig: Digoxigenin) Hybridization buffer (DIG Easy Hyb Granules)Blocking solutionAnti-Digoxigenin-AP Solution (AP: Alkaline Phosphatase)CSPD (a chemiluminescent substrate for alkaline phosphatase)

Additional materials needed include:20´ SSC buffer, as described in Exercise 810% SDS, as described in Exercise 72´ SSC; 0.1% SDS:

To 40 mL of 20´ SSC buffer add 360 mL of distilled water, mix, then discard 4 mL, and to the remaining 396 mL add 4 mL of 10% SDS

0.5´ SSC; 0.1% SDS:To 10 mL of 20´ SSC buffer add 390 mL of distilled water,

mix, then discard 4 mL, and to the remaining 396 mL add 4 mL of 10% SDS

Washing buffer:Per 1000 mL of dH2O11.61 g Maleic acid8.76 g Sodium chlorideAdjust pH to 7.5 with solid sodium hydroxideAutoclaveLet the solution cool down to room temperature and then

add 3 mL of polyoxyethylene sorbitan monolaurate (e.g., Tween 20, Bio-Rad Laboratories, Hercules, CA)

Maleic acid buffer:Per 1000 mL of dH2O11.61 g Maleic acid8.76 g Sodium chlorideAdjust pH to 7.5 with solid sodium hydroxideAutoclave

Detection buffer:Per 1000 mL of dH2O12.11 g Tris5.84 g Sodium chlorideAdjust pH to 9.5 with 1 M hydrochloric acidAutoclave

1´ Blocking solution for washing purposes (prepare fresh):10 mL Blocking solution (stock)90 mL Maleic acid buffer

1´ Blocking solution to prepare antibody solution (prepare fresh):2 mL Blocking solution (stock)18 mL Maleic acid bufferThe Blocking solution, used as stock, contains optimal

concentrations of ficoll, polyvinylpyrrolidone, bovine serum albumin, and water.

Antibody solution (prepare fresh but can be kept up to 12 h at 2–8°C):

Centrifuge Anti-Digoxigenin-AP solution (keep in refrig-erator) for 2 min at 10,000 rpm, then pipet 2 µL of the supernatant and add to 20 mL of 1´ Blocking solution.

Color detection solution (CSPD): a chemiluminescent substrate (keep in refrigerator).

NBT (Nitro Blue Tetrazolium) solution:0.075 g NBT

Fig. 9–3. Hybridization incubator.


1 mL 70% DMF (N, N-dimethylformamide) NBT solution is light sensitive, wrap it in aluminum foil

and keep in a refrigerator.To prepare 1 mL of 70% DMF, add 300 µL of dH2O to 700

µL of DMF.BCIP (5-bromo-4-chloro-3-indolyl-phosphate) solution:

0.050 g BCIP1 mL dH2OBCIP solution is light sensitive, wrap it in aluminum foil

and keep in a refrigerator.Color solution (prepare fresh, away from direct light):

45 µL NBT solution35 µL BCIP solution10 mL Detection buffer

Stop solution:Per 500 mL of dH2O0.606 g Tris1 mL 0.5 M EDTAAdjust pH to 8.0 with 1 M hydrochloric acidAutoclaveTo prepare 0.5 M EDTA, see Exercise 3.

Probe hybridized nylon membraneRotatorWater bath adjusted to 65 to 68°CGlass or plastic dishesGraduated cylinders of different sizesBeakers and flasks of different sizesMicropipettorsTipsForcepsGlovesPlastic-film wrapRulerX-ray filmFilm cassetteX-ray film processing machineDark room

ProcedureBefore starting, heat the 0.5´ SSC; 0.1% SDS solution in a 65 to 68°C water bath. Also thaw the Blocking solution (stock). You may transfer the Blocking solution from the freezer to a refrigerator 1 day before the start of the exer-cise.1. Place the membrane in a plastic or glass dish contain-

ing 2´ SSC; 0.1% SDS, then place the dish on a rotary shaker for 5 min at room temperature.

2. Repeat Step 1; use a fresh 2´ SSC; 0.1% SDS solution.3. Place the membrane in a plastic dish containing heated

0.5´ SSC;0.1% SDS solution, close the lid, then place the dish in the 65 to 68°C water bath with constant agi-tation for 15 min.

4. Repeat Step 3; use fresh 0.5´ SSC; 0.1% SDS solution.5. Place the membrane in washing buffer and rotate for 5

min at room temperature.6. Place the membrane in 1´ blocking solution and rotate

for 20 min at room temperature.7. Place the membrane in antibody solution and rotate for

20 min at room temperature.8. Place the membrane in washing buffer and rotate for

15 min at room temperature.

9. Place the membrane in detection buffer and rotate for 5 min at room temperature.

10. Place the membrane in freshly prepared color solu-tion, swirl once or twice, then allow the membrane to sit away from direct light. Wait until bands appear (approximately 2 to 3 min).

11. Pour off the color solution and immediately add the stop solution; wait for 5 min.

12. Remove the membrane from the stop solution and air dry at room temperature.

13. On the membrane measure the distance (in millime-ters) between the wells and each band with a ruler. Then compare these data with the data obtained from the photograph of the matching gel (see Fig. 8–2). Find the matching bands.

Figure 9–4 shows the Southern blot prepared from the gel shown in Fig. 8–2, and hybridized with probe.

Note: The sensitivity of the technique, described above, can be improved by exposing the nylon membrane to an X-ray film. In this case, Steps 10 to 13 above can be replaced with Steps 10a to 14a as described below.

10a. Incubate the membrane for 15 min in color-detection solution (CSPD) at room temperature away from direct light; shake occasionally.

11a. Remove excess color-detection solution from the membrane by touching the membrane to a clean filter paper, then while still wet, place the membrane, gel-

Fig. 9–4. Nylon membrane, hybridized with probe, and stained with color solution.


side-up, in an X-ray film cassette between two layers of plastic-film wrap. In the dark room, place the X-ray film on the top of the membrane in the cassette.

12a. Expose the membrane to the X-ray film for 30 min (time may vary), then in the dark room open the cassette and slide the film into the film processing machine.

13a. View the film. If needed, the membrane can be re-exposed to the film for a longer or shorter time until an image with optimum intensity is obtained.

14a. When the DNA bands appear on the film, measure the distance (in millimeters) between the wells and each band with a ruler. Then compare these data with the data obtained from the photograph of the matching gel (see Fig. 8–2). Find the matching bands.

Protocol 9–1Removing Probe from a Nylon MembraneIt is possible to remove (strip) a probe from nylon mem-brane and detect the target DNA with another probe. Stripping DIG-labeled DNA probes is achieved by wash-ing the membrane in a high alkaline buffer containing certain concentrations of sodium hydroxide and sodium dodecyl sulfate. If the membrane has been stained with a chromogenic substrate, the membrane must be decolorized with dimethylformamide prior to stripping the membrane. On alkaline-stripped membranes, reprobed signals can be generated without loss of sensitivity. Stripping and reprob-ing can be repeated at least three times.

1. Stripping DIG-Labeled DNA Probe after Chromogenic Detection

MaterialsBCIP/NBT-stained nylon membraneN, N-dimethylformamide (DMF)10% SDS, as described in Exercise 720´ SSC buffer, as described in Exercise 82´ SSC buffer:

To 1 volume of 20´ SSC buffer add 9 volumes of dH2OStripping buffer:

Dissolve 4 g of sodium hydroxide in 500 mL of dH2O, mix, then discard 5 mL, and to the remaining 495 mL add 5 mL of 10% SDS.

Autoclaved, double distilled waterdH2OWater bath adjusted to 37°CHot plateRotatorGlass dishesPlastic dishesFlasksGraduated cylindersWhatman paperGloves

ProcedureBefore starting, warm up the DMF solution to 65°C in a glass dish on a hot plate. Also heat the stripping buffer in a 37°C water bath.

1. Place the membrane in the glass dish containing heated (to 65°C) DMF for 1 h or until the purple color of bands has been removed from the membrane. Occasionally change the DMF to speed up the procedure.

2. Rinse the membrane in autoclaved, double distilled water for 1 min.

3. Place the membrane in a plastic dish containing heated stripping buffer, close the lid, then place the dish in a 37°C water bath with constant agitation for 20 min.

4. Repeat Step 3; use fresh stripping buffer.5. Place the membrane in 2´ SSC buffer and rotate for 5

min at room temperature.6. Air dry the membrane at room temperature, wrap in

Whatman paper, and place in a refrigerator until need-ed. Alternatively, store the membrane wet in 2´ SSC buffer at 4°C.

2. Stripping DIG-Labeled DNA Probe after Chemiluminescent Detection

MaterialsCSPD-stained nylon membrane10% SDS, as described in Exercise 720´ SSC buffer, as described in Exercise 8Stripping buffer, as described above2´ SSC buffer, as described aboveAutoclaved, double distilled waterWater bath adjusted to 37°CdH2ORotatorPlastic dishesGraduated cylindersWhatman paperGloves

ProcedureBefore starting, heat the stripping buffer in a 37°C water bath.1. In a plastic dish, rinse the membrane thoroughly with

autoclaved, double distilled water for 1 min.2. Place the membrane in a clean plastic dish containing

heated stripping buffer, close the lid, then place the dish in a 37°C water bath with constant agitation for 15 min.

3. Repeat Step 2; use fresh stripping buffer.4. Place the membrane in 2´ SSC buffer and rotate for 5

min at room temperature.5. Air dry the membrane at room temperature, wrap in

Whatman paper, and place in a refrigerator until need-ed. Alternatively, store the membrane wet in 2´ SSC buffer at 4°C.

Study Questions9–1. What is a probe?

9–2. How are probes used to identify the DNA sequence of interest in Southern blotting?

9–3. Read Step 2 of Hybridization on page 46. Why is the speed of rotation of the hybridization cylinder so low: 3 rpm (revolutions per minute)?

9–4. Briefly describe the mechanism under which DNA is transferred from a gel onto a nylon membrane.

9–5. In the process of Southern blotting, why are the bands of the DNA ladder on the gel (see Fig. 8–2) not vis-ible on the membrane of the same gel (see Fig. 9–4)?



Objectives




Exercise 10 Regulation of Gene Expression

Objectives

• to understand the concept of an operon

• to learn how gene expression is regulated in prokaryotes and eukaryotes

• to understand the concept of reporter genes

• to learn how transgenic plants are generated

Prokaryotic Gene Regulation

In 1961 Jacob and Monod, who studied the utilization of lactose by E. coli, proposed a model to explain how gene expression is regulated in prokaryotic organisms. The model, which became known as the operon theory, proposes the existence of a set of structural genes (the coding region) in prokaryotic genome that are transcribed into a single mRNA and the nearby regulatory region that controls the expression of the structural genes. The structural genes along with the regulatory region are called an operon. The basic idea behind the theory is that gene expression is dependent on the presence or absence of specific biologically significant substances such as proteins, enzymes, carbohydrates, etc. Furthermore, each substance plays a specific role (initiating, catalyzing, inhibiting, etc.) in the regulation of gene expression. According to this theory, the expression of the three structural genes lacZ, lacY, and lacA in E. coli is regulated by a specific DNA sequence called the operator. In this exercise, we will focus only on the lacZ gene and its product. The lacZ gene codes for the enzyme β-galactosidase (best known as β-gal) which is responsible for the breakdown of lactose into glucose and galactose (Fig. 10–1).

The process of transcribing the lacZ gene to produce β-gal involves two regulatory aspects: negative control elements which inhibit gene expression and positive control elements which increase the rate of gene expression. The negative regulator is what is known as the lac repressor protein, encoded by the lacI gene. The lac repressor protein binds to the operator, thus preventing the binding of RNA polymerase to the promoter site or alternatively, preventing RNA polymerase from reaching beyond the promoter region. In either event, transcription of the lacZ gene is halted. When lactose is present, β-gal converts it to allolactose (i.e., an altered from of the lactose), which serves as an inducer


Fig. 10–1. Breakdown of lactose into glucose and galactose by ß-galactosidase.


in the lac operon (Fig.10–2). Allolactose changes the shape of the lac repressor protein making it unable to bind to the operator. As a result, RNA polymerase gains access to the promoter (or moves beyond the promoter region), thus initiates transcription of the structural genes into a single polycistronic mRNA which is translated into ß-galactosi-dase as well as permease and transacetylase.

The expression of the lacZ gene is positively regulated by a protein called catabolite activator protein (CAP). The CAP increases the rate of gene expression under the cel-lular condition of low glucose. Under this condition, the cell produces a compound known as cyclic adenosine mono phosphate (cAMP). The CAP along with cAMP bind to the promoter site of the lacZ gene and make the binding of RNA polymerase more efficient. Therefore, the rate of transcription of the lacZ gene is increased, resulting in an increased rate of β-galactosidase production and lactose breakdown. It is important to note that in the presence of glucose, cAMP is not produced and therefore CAP does not have any effect on the rate of transcription. Fig. 10–3 shows the components of the lactose operon in E. coli.

What are Reporter Genes?As mentioned above, the lacZ gene codes for the enzyme β-gal. The presence of this enzyme in the cell can be identified

by a color indicator such as Xgal (5-bromo-4-chloro-3-indolyl-β-d-galactopyranoside), which is a substrate for β-gal. When β-gal is present, Xgal turns blue, and the blue color can be easily seen in the cell (see Exercise 6 for details). You used this technique in Exercise 6, where you plated the cloned fragment of the TTG DNA on agar plates containing kanamycin and Xgal. In that experiment, the blue colonies you observed “reported” to you the presence of an undisrupted lacZ gene in the TOPO vector. Thus, lacZ is a reporter gene.

In general, reporter genes code for proteins whose expres-sion can be seen and quantified. Reporter genes include, but are not limited to, lacZ and gusA (uidA) in E. coli, the bioluminescence genes luxE in the bacterium Photorhabdus luminescens, and gfp in jellyfish. To monitor the expres-sion of the gene of interest, the coding region of the gene is replaced with the coding region of a reporter gene, such as lacZ, so that the promoter of the gene of interest regulates the transcription of the lacZ gene. The activity of the pro-moter can then be determined by measuring the lacZ product, which is ß-gal. In the following two exercises, you will study the expression of the luxE and gfp genes in E. coli.

Expression of the luxE Gene in E. coli †Luminescence or emission of light by certain bacterial

species is a common phenomenon in nature. The chemical substance capable of luminescence is called luciferin (mean-ing “light-bringing”). When it undergoes oxidation in the presence of the enzymes, collectively called luciferases, and adenos-ine triphosphate (ATP), a dim light is emitted. Luciferases can be found in a vari-ety of organisms including bacteria, algae, and insects. In Photorhabdus luminescens, the genes responsible for the production of luciferin and luciferases are the components of the lux operon. The lux operon contains one gene (luxE) that codes for luciferin and four other genes (luxC, luxD, luxA, and luxB) that code for the enzymes that catalyze the light-emitting reaction. The lux operon must be activated by external

Fig. 10–2. Conversion of lactose to allolactose, an inducer.

Fig. 10–3. Components of the lactose operon in Escherichia coli.

† This exercise was adopted from Dr. B. Applegate, Department of Food Science of Purdue University, with minor modifications to fit the format of this laboratory manual.


sources (i.e., inducers) to produce luciferin; these include naphthalene or acyl homoserine lactone, unless the operon has been genetically modified for constitutive expression. Figure 10–4 shows the components of the lux operon.

The entire lux operon is made up of 6960 base pairs (Fig. 10–5), and the physical locations of the components of the operon are as follows:luxC: nucleotide numbers 459–1901luxD: nucleotide numbers 1913–2836luxA: nucleotide numbers 2885–3967luxB: nucleotide numbers 3982–4965luxE: nucleotide numbers 5144–6256

In this exercise, you will transform E. coli with a cloned fragment of the lux operon, derived from Photorhabdus luminescens, to view the biolumines-cence in transformed E. coli cells. If you do not have a cloned fragment but you have the lux operon DNA, you can clone it into a TOPO vector exactly the way you did for the TTG locus in Exercise 6. In the exercise described here, the cloned fragment of the lux operon has been genetically manipu-lated: its promoter has been replaced with the promoter of the lac operon; therefore, the lac promoter drives the lux transcription. Furthermore, it car-ries a mutation that causes constitutive production of luciferases, regardless of the presence or absence of inducers or repressors.

MaterialsThe lux operon DNA cloned into a TOPO

vectorE. coli competent cellsLB agar plates containing kanamycin, as

described in Exercise 6SOB medium, as described in Exercise 6Ice42°C water bath37°C incubator37°C shakerMicrotube racksMicrotube floatersMicropipettorsTipsGlass beakersSpreadersBurnersLaboratory gas lighters

Procedure1. Place the tube containing the cloned

lux operon DNA on ice to thaw.2. Place the tube containing E. coli

competent cells on ice to thaw.3. Add 2 µL of the cloned lux

operon DNA (from Step 1,

Fig. 10–4. Diagram of the lux operon.

Fig. 10–5. The complete nucleotide sequence of the lux operon. (continued on next page).


above) to the tube containing the competent cells and mix gently. Do not mix by pipetting up and down or vortexing .

4. Incubate on ice for 10 min.5. Place the tube in a floating rack, then place the rack

in a 42°C water bath for 30 s without shaking, then immediately transfer to ice.

6. Add 250 µL of SOB medium to the tube from Step 5 above, cap the tube tightly, tape the tube on the plat-form of a 37°C shaker and shake for 1 h.

7. This step must be done close to a lit burner to prevent plate contamination . After 1 h, remove the tube containing the transformation reaction from the shaker (Step 6, above) and pipette 100 µL of the

reaction on an LB agar plate containing kanamy-cin (LB + kan). Immediately flame a spreader and cool it by gently touching the spreader to the agar (do not touch it directly to the 100 mL of cells to prevent heat-killing of the cells). Then spread the transformation reaction evenly on top of the agar. Place the plate upside down in an incubator set at 37°C. Leave overnight.

8. The following day check the plates for colonies.9. To view the bioluminescence, transfer the plate to a

dark room, wait about 5 min so that your eyes will ad-just to the darkness, then you will see emission of light from the E. coli colonies.

Expression of the gfp Gene in E. coli

In early 1960s, Shimomura and his coworkers isolated a protein, called green fluorescent protein (GFP), from the jellyfish Aequorea victoria. They noticed that this protein looked “slightly greenish” in solution but showed a “very bright greenish fluo-rescence” under the UV light. In fact in jellyfish, GFP is in close association with another protein called aequorin. Aequorin interacts with calcium ions inducing a blue glow. The energy released from this luminescence is transferred to GFP, shifting the over-all luminescence towards a green color. The gene responsible for GFP, also called green fluorescent protein (gfp), has been isolated, cloned, and sequenced by Prasher and his cowork-ers in 1992. In 2008, O. Shimomura, R. Tsien, and M. Chalfie won the Nobel Prize for the discovery of how green fluorescent protein fluoresces. The nucleotide sequence of the gfp cDNA is shown in Fig. 10–6.

In this exercise, you will transform E. coli competent cells with a mutated version of the gfp gene already cloned into a plasmid called pGreen. The transformants will be propagated in a nutrient medium and then cul-tured on agar plates containing an appropriate antibiotic to eliminate nontransformed bacterial cells from the plates. The E. coli colonies trans-formed with the mutated version of the gfp gene will show a green color under white (not the UV) light.

Fig. 10–5. Continued.


What is pGreen Plasmid?The pGreen plasmid, developed by Hellens et al. (2000), is a 3232-bp, circular, double-stranded, semisynthetic plas-mid which carries a polylinker with 16 unique restriction sites and a fragment of the lacZ gene. This plasmid also car-ries the gene that provides kanamycin-resistance selection for bacterial transformation. Replication of the plasmid is achieved through the pSa replication origin. Several versions of the pGreen plasmid are available. The version you will be using in this exercise carries the ampicillin-resistance gene (ampr); furthermore, the mutated version of the gfp gene has already been cloned into this plasmid. For convenience and for the sake of understanding the exercise, we will call this version of the plasmid pGreen/gfp (Fig10–7). The size of pGreen/gfp is 4528 bp. Restriction digestion of pGreen/gfp with HindIII shows a single band located between the DNA ladder 5090 and 4072 bp (Fig. 10–8). The pGreen/gfp plasmid is available through Carolina Biological Supply Co. (Burlington, NC).

MaterialsThe pGreen/gfp plasmidE. coli competent cellsLB agar plates containing ampicillin:

Prepare LB agar medium as described in Exercise 6. Let cool to 40°C, then add 1 mL of 50 mg/mL aqueous solution of ampicillin, mix well, and distribute in sterile Petri dishes.

50 mg/mL ampicillin:Per 10 mL of autoclaved distilled water0.5 g AmpicillinFilter sterilize the solution before use. Keep at −20°C.

SOB medium, as described in Exercise 6Ice

42°C water bath37°C incubator37°C shakerMicrotube racksMicrotube floatersMicropipettorsTipsGlass beakersSpreadersBurnersLaboratory gas lighters

Procedure1. Place the tube containing pGreen/gfp on ice to thaw.2. Place the tube containing E. coli competent cells to

thaw.3. Add 2 μL of pGreen/gfp (Step 1, above) to the tube

containing competent cells. Mix gently but do not vor-tex or pipette up and down.

Fig. 10–6. The nucleotide sequence of the gfp cDNA.

Fig. 10–7. Structure and properties of the pGreen plasmid.

Fig. 10–8. Restriction digestion of the pGreen/gfp plasmid. The plasmid was digested with HindIII for 15 h and run on 0.9% (w/v) agarose gel in 1´ TAE buffer for 60 min at 80 V.


4. Incubate on ice for 15 min.5. Heat-shock the reaction: place the tube in a floating

rack, and then place the rack in a 42ºC water bath for 90 s without shaking.

6. Cool the tube by transferring it to ice for 3 min or more.7. Add 250 μL of SOB medium to the tube from Step

6 above, cap the tube tightly, then tape the tube on the platform of a 37°C shaker and shake for 30 min. During this time period, a large number of transformed and nontransformed E. coli cells will be recovered. However, in the following step, you will eliminate non-transformed cells by ampicillin selection technique.

8.. This step must be done close to lit burner to pre-vent plate contamination . Remove the tube from the shaker (Step 7, above), pipette 50 μL of the reaction on an LB agar plate containing ampicillin. Immediately flame a spreader and cool it by gently touching the spreader to the agar. Then spread the transformation reaction evenly on top of the agar.

9. Place the plate upside down in an incubator set at 37°C for 24 h.

10. The following day check the plates for the colonies. You should see a large number of E. coli colonies showing green color under normal light.

11. On the basis of your observations, answer the follow-ing questions:

What does the green color of the E. coli colonies tell you?

In Step 10 above, how would the E. coli colonies look like if you do not add ampicillin on the LB plates?

Suppose you believe that you have generated a trans-genic Arabidopsis plant which carries the chimeric TTG/gfp gene construct (for more information, see Protocol 10–1). How can you prove that the plant in-deed carries the chimeric gene?

Eukaryotic Gene RegulationThere is a significant difference in gene organi-zation and regulation between prokaryotes and eukaryotes. The most conspicuous feature of the eukaryotic genome is lack of structures compa-rable to a prokaryotic operon. The eukaryotic genome is well correlated with a monocistronic mRNA (recall polycistronic mRNA in prokary-otes), with each mRNA coding for a single protein. It is widely believed that during the evolution of eukaryotes, functionally related genes were scat-tered throughout the genome by chromosomal rearrangements.

In a eukaryotic cell, thousands of proteins are present, each carrying out a specific activity. Some proteins carry out enzymatic reactions, while others play a more structural role and affect the architecture of the cell. The collection of protein products that are present in a cell depends on its developmental status. For example, cells that are

actively dividing have a quite different collection of proteins depending on the stage of the cell cycle. Cells in mitosis have different needs than cells that are replicating their DNA dur-ing the S phase. In multicellular organisms, the issue of cell identity has important consequences with respect to what proteins are present in the cell. For example, in humans, the metabolic and structural properties of liver cells are quite dif-ferent from those of skin cells. Different cell types express different sets of proteins; specialized cell function is part of multicellular development in all organisms.

To understand the mechanisms of how transcription of a eukaryotic gene can change, consider the different struc-tural elements of a gene. A eukaryotic gene can be divided into coding sequences or exons (for “expressed regions”) that define the linear sequence of amino acids in a protein and noncoding sequences or introns (for “intervening regions”). Introns are transcribed into mRNA but are cut out of the message before it is translated into proteins (Fig. 10–9). In eukaryotes, many proteins are involved in the control of transcription. These are called transcription factors, and they assist RNA polymerases to recognize promoters. Transcription factors are bound to the DNA sequence 5¢-TATAAAA-3¢ of the promoter. This sequence is commonly known as the TATA box and exists in almost all genes. The organization of gene promoters is very complex and often involves both positive (enhancers) and negative (suppressors) regulation.

Expression of the gusA Gene in Arabidopsis thaliana

The gusA (uidA) gene is used as a reporter gene in plants. This gene codes for the enzyme β-glucuronidase (GUS) which cleaves the colorless soluble compound X-gluc (5-bromo-4-chloro-3-indolyl-β-d-glucuronic acid, cyclohexylammonium salt) to produce glucuronic acid and chloro-bromoindigo. When oxidized, chloro-bromoindigo dimerizes resulting in

Fig. 10–9. Diagram of a eukaryotic gene.


the blue precipitate 5,5¢-dibromo-4,4¢-dichloro-indigo. This reaction is illustrated in Fig. 10–10.

By means of recombinant DNA techniques, the promoter of the gusA gene can be replaced with the promoter of the plant gene to be studied. This new chimeric gene is then introduced into the plant by means of transfor-mation to generate a transgenic plant (see Protocol 10–1, below). The expression of the gusA gene is then detected by staining the plant tis-sues with X-gluc. Those tissues that produce the enzyme β-glucuronidase (GUS) will turn blue. Custom made chimeric genes can be ordered from biotechnology companies.

In the exercise described below, transgenic Arabidopsis thaliana seedlings will be exam-ined by detecting the activity of β-glucuronidase (GUS) reporter protein. The transgenic seedlings carry the GUS reporter gene fused to the promoter of certain genes that are expressed in leaves, trichomes, roots, flowers, or any combination thereof. The laboratory instructor may give you advice in choosing transgenic plants.

MaterialsTransgenic Arabidopsis thaliana seedlings, 7 to 10 d old0.5 M Sodium phosphate mono-dibasic:

Per 500 mL of dH2O14.95 g Sodium phosphate monobasic20.47 g Sodium phosphate dibasicAutoclave

0.2 M Potassium ferricyanide:Per 500 mL of dH2O32.92 g Potassium ferricyanide

100 mM X-gluc:Per 50 mL of N,N-dimethyl formamide (DMF):2.6 g X-glucNote: X-gluc is light sensitive; keep it in a brown bottle

covered with aluminum foil. Keep at −20°C.GUS staining solution (prepare fresh):

27 mL Distilled water 3 mL 0.5 M Sodium phosphate mono-dibasic30 µL Triton X-100 (t-octylphenoxypolyethoxyethanol;

Sigma, St. Louis, MO)150 µL 0.2 M Potassium ferricyanide300 µL 100 mM X-glucMix well and cover with aluminum foil

95% ethanol:95 mL Ethanol5 mL dH2O

dH2O37°C incubatorDissecting microscopeHand-held magnifying glass

Transfer pipettesMicropipettorsTipsMicrowell dishes with lidsMicroscope slidesTweezersGlass or plastic dishesGlovesPaper towelsPlastic-film wrap

ProcedureNote: Prior to this experiment, Arabidopsis thaliana trans-genic seeds of interest must be obtained as described in Protocol 10–1 (see below). These seeds must be planted in soil, grown for at least 7 d, then brought to the laboratory.

1. Remove transgenic seedlings from soil. If necessary, briefly rinse the seedlings in distilled water to remove any attached soil.

2. Place a single seedling in each well of a microwell dish. You may examine up to five seedlings per construct.

3. Cover the seedlings with the GUS staining solution.4. Loosely close the lid of the microwell dish, place the

dish in a larger glass or plastic dish with wet paper tow-els in the bottom, cover the composite with plastic-film wrap, and place it in a 37°C incubator for 24 h.

5. The following day, remove the seedlings from the microwell dish and place them on a microscope slide, then inspect the seedlings for blue precipitate. Use a dissecting microscope or a powerful magnifying glass to better view the tissues. If necessary, soak the seed-lings in a 95% ethanol for 30 min to clear the plant tissue of chlorophyll.

6. Locate tissues which show the GUS activity.

Fig. 10–10. Biochemical pathway of the production of dimerized 5,5¢-dibromo 4,4¢dichloro-indigo, the blue color, following GUS staining of live plants.


Protocol 10–1Generation of Transgenic Plants: Transformation of ArabidopsisAn efficient method to generate transgenic plants is the Agrobacterium-mediated gene transfer. Agrobacterium tumefaciens is a Gram-negative, rod-shaped, aerobic soil bacterium that infects host plants, including Arabidopsis. The bacterium carries a plasmid that contains several genes, including the Ti (tumor-inducing) gene that codes for plant growth hormones. When Agrobacterium infects a host plant, a small part of the plasmid (approximately 20 kb) that includes the Ti gene is transferred from the bacterium to the plant. The transferred DNA (called T-DNA) is integrated into the plant genome and causes the neoplastic disease Crown Gall. With the aid of recom-binant DNA technology, the Ti gene of the plasmid is replaced with a chimeric gene (Fig. 10–11). The chimeric gene is made of the promoter of the plant gene of interest fused to the gusA gene, so that the promoter drives the transcription of the chimeric con-struct. Furthermore, an antibiotic resistance gene is placed next to the chimeric gene to allow the selection of modified seeds. Then this genetically modified T-DNA is introduced into Arabidopsis through the process of transformation (see below). The transformed plants are allowed to self-pollinate, and the seeds are harvested and germinated on antibiotic-rich media. Only seeds that carry the modified T-DNA will germinate on these selective media.

The integration of the T-DNA into the plant genome is a complicated process which is mediated by virulence (vir) genes located in the Ti plasmid. The vir genes code for a number of proteins that coordinate the integration of T-DNA into the plant genome (Tinland, 1966). They include proteins that direct the T-DNA from bacterial cell wall to the plant cells, proteins that transfer the T-DNA from the plant cytoplasm to the nucleus, and proteins that help the T-DNA covalently integrate into the plant DNA.

A. Preparing Arabidopsis plants for transformation

Plants of interest (e.g., Columbia wild-type strain) must be lightly fertilized immediately after emergence of the first set of leaves because plant health is a major factor in transformation. The ideal stage for transformation is when plants contain many immature flowers that are just about to pollinate. The number of immature flowers can be increased by removing the primary inflorescence, thus encouraging several lateral bolts to grow. Infiltration (see below) should be done when most plants have bolts and are about 8 to 10 cm tall.

B. TransformationArabidopsis thaliana can be transformed by applying Agrobacterium tumefaciens to the plant and recovering transformants in the progeny.

MaterialsArabidopsis plants to be transformedChimeric DNA of interestAgrobacterium tumefaciens competent cells; strain C5880% glycerol:

80 mL Glycerol20 mL dH2OAutoclave

LB liquid medium (without antibiotic), as described in Exercise 7; but do not add the antibiotic.

LB agar plates containing kanamycin, prepare the medium as described in Exercise 6; after autoclaving let cool to 40°C, then add 1 mL of 50 mg/mL aqueous solution of kanamycin; mix well.

50 mg/mL kanamycin, as described in Exercise 6YEP medium containing kanamycin:

Per 1000 mL of dH2O10 g Peptone10 g Yeast extract5 g Sodium chloride1 mL 1 M sodium hydroxideAutoclaveLet cool to 40°C, then add 1 mL of 50 mg/mL aqueous

solution of kanamycin; mix well.1 M sodium hydroxide:

Per 100 mL of dH2O4 g Sodium hydroxideAutoclave

Infiltration medium:Per 2000 mL of dH2O100 g Sucrose400 µL Silwet L-77 (Helena Chemical Company,

Collierville, TN; Silwett L-77 is a blend of polyalkyle-neoxide modified hepta-methyltrisiloxane and allyloxy-polyethylene glycol methyl ether. This compound is moderately toxic and is an eye and skin irritant.)

Mix wellAutoclave; however, there is no need to autoclave if

made fresh.Triton X-100 (Sigma, St. Louis, MO)

Fig. 10–11. Genetically modified Ti plasmid of Agrobacterium tumefaciens.


Murashige and Skoog medium:See Protocol 10–2

ElectroporatorSpectrophotometer28°C shaker28°C incubatorJouan centrifugeJouan bottlesSterile 1000 mL flasksLarge (about 2000 mL) plastic container with lidSterile glass test tubesSterile glass pipettesSterile tooth picksMicrotubes, 1.5 mLMicrotube racksGlass or metal spreaderMicropipettorsTipsForcepsAutoclaved Pasteur pipettes with rubber bulbSterile polystyrene culture tubesTube racksIce−80°C freezerBurnersLaboratory gas lightersGlovesPots with soilPlant growth chamber with controlled temperature and humidity

Procedure1. Ten minutes before the start of exercise, turn the elec-

troporator on to warm it up. Read the manufacturer’s manual so that you are sure you know to use the elec-troporator. At the same time, place the following mate-rials on ice:

chimeric DNA of interest• one sterile polystyrene culture tube with cap, containing 1 • mL of autoclaved LB mediumone sterile Pasteur pipette with rubber bulb, wrapped in • aluminum foilelectroporation cuvette•

2. Place Agrobacterium tumefaciens competent cells on ice to thaw. Once the bacterial cells are thawed proceed to Step 3; delay will lower the titer of the cells.

3. Immediately add 0.5 µL of the chimeric DNA to the tube containing A. tumefaciens, gently shake to mix, and then transfer the entire contents to the electropora-tion cuvette.

4. At the same time, place the Pasteur pipette into the culture tube that contains LB medium, then withdraw the medium into the pipette and leave the pipette in the tube.

5. Transform the Agrobacterium by electroporating the cuvette (Step 3, above) according to the manufac-turer’s protocol.

6. Immediately transfer the content of the Pasteur pipette (Step 4, above) to the cuvette. Mix gently by sucking the contents up and down with the same pipette. Then transfer the entire contents back to the culture tube. The culture tube now contains LB medium + trans-formed Agrobacterium.

Note: The period between applying the electroporation pulse and transferring the LB medium to the cuvette is crucial: delaying this transfer by 30 s will cause a 3-fold drop in transformation.

7. Place the culture tube in a 28°C shaker for 3 h.8. This step must be done close to a lit burner to

prevent plate contamination . After 3 h, remove the tube containing the transformation reaction from the shaker (Step 7, above), pipette 300 to 400 µL of the reaction on an LB agar plate containing kanamycin. Immediately flame a spreader, cool it by touching to the agar surface (do not touch it directly to the trans-formation reaction to prevent heat-killing of the cells). Then spread the transformation reaction evenly on top of the agar surface. Place the plate upside down in an incubator set at 28°C for 24 h. Note that the colonies will grow slowly.

9. Periodically check the agar plate for the presence of colonies. Once colonies grow, take the plate close to a lit burner, and with a sterile toothpick, pick a single colony and drop the toothpick into a sterile test tube containing 6 mL of YEP + kan medium. This is the starter culture. Place the culture in a 28°C shaker for 36 to 48 h or until it becomes turbid, indicating that the bacteria have grown.

10. Close to a lit burner, transfer 800 µL of the culture to a sterile 1.5-mL microtube (or, preferably a screw cap microtube) and to it add an equal volume of autoclaved 80% glycerol. Close the lid tightly, mix thoroughly by inverting 8 to 10 times, label the tube properly, and save it in a −80°C freezer for future use.

11. Also, close to a lit burner, pick 3 to 4 mL from the starter culture using a sterile glass pipette, and transfer the content to a flask containing 500 mL of YEP + kan medium. Place the flask in a 28°C shaker for 24 to 36 h or until the bacteria grow; from time to time, check the turbidity of the medium.

12. After 36 h, or if needed, 48 h of incubation, check the optical density (OD) of the culture at 600 nm with a spectrophotometer. You must have a prior knowledge of the use of a spectrophotometer. Read the manu-facturer’s manual. The desired density is OD600 = 1.5 to 2.5. You may incubate the culture for an extended amount of time until the desired density is achieved.

13. Pour the culture into a Jouan bottle, spin for 15 min at 5000 rpm at room temperature to obtain a pellet.

14. Pour off the supernatant and resuspend the pellet in 500 mL of infiltration medium.

15. Pour the infiltration medium containing the cell suspen-sion in a plastic container to infiltrate plants of interest.

Note: Before proceeding to Step 16, be sure that the plants of interest are ready and well prepared for trans-formation. See part A of this protocol.

16. Dip above-soil parts of the plants in the infiltration + cell suspension medium (Step 15, above) for 2 to 3 s with gentle agitation.


17. Transfer the plants to a growth chamber and cover them with a plastic dome for 24 h.

18. The following day, remove the dome and water the plants regularly. These plants are called T0.

19. (Optional) For a higher rate of transformation, plants may be dipped twice at 3-d intervals by repeating Steps 16 to 18.

20. Let the plants grow, self-fertilize, and then collect seeds. These seeds are called T1.

21. Save the T1 seeds in microtubes or small paper envel-ops for a few days before use.

22. Grow the T1 seeds on Murashige and Skoog (MS) agar medium containing kanamycin. See Protocol 10–2 for planting seeds on MS medium. Identify putative trans-formants, that is, the seedlings resistant to the antibiotic. The expected rate of transformation is around 0.02%. However, with experience, higher rates can be achieved.

23. Using forceps, gently remove the transformants from the MS medium and transfer them to previously moistened pots. Let them grow under a plastic dome for 2 to 3 d, then remove the dome. Let the transformants self-fertil-ize and collect seeds from each plant individually. These are the T2 seeds. Use these seeds for GUS staining.

Protocol 10–2Surface Sterilization of Arabidopsis Seeds for Planting on Agar PlatesIn some experiments when transformed or antibiotic resistant seedlings are involved, it is necessary to select seedlings of interest on agar plates containing appropriate antibiotic(s), and then transfer them to soil. For this purpose, the seeds must be surface sterilized to remove bacteria or fungi which may have adverse effects on plant growth. This protocol describes the technique of seed sterilization and the consequent seed germination on agar plates.

MaterialsSeeds of interest70% ethanol:

70 mL Ethanol30 mL dH2O

50% bleach:In a wash bottle, mix an equal volume of bleach and dH2O.

Add 3 drops of t-octylphenoxypolyethoxyethanol (e.g., Triton X-100; Sigma, St. Louis, MO) to each 100 mL of the mixture.

Murashige and Skoog (MS) agar plates containing kanamycin:Per 1000 mL of dH2O1 packet (4.4 g net weight) MS Salt and Vitamin Mixture10 g SucroseAdjust pH to 6.0 with 1 M potassium hydroxide, then add

8 g AgarAutoclaveLet cool to 40°C, then add 1 mL of 50 mg/mL kanamycin.

Mix well; distribute in sterile Petri dishes.1 M potassium hydroxide:

Per 100 mL of dH2O5.611 g Potassium hydroxide

50 mg/mL kanamycin, as described in Protocol 10–1

Graduated cylindersMicrotubes, 1.5 mLMicrotube racksSterilized transfer pipettes:

Place transfer pipettes in a beaker filled with 70% ethanolSterilized Pasteur pipettes with rubber bulbs:

Place Pasteur pipettes in a beaker filled with 70% ethanolBeakersSterile Petri dishesPlastic or glass containers to collect liquid wasteWash bottlesAutoclaved double distilled waterdH2OSurgical tape [Micropore (3M Health Care, St. Paul, MN)

or equivalent]Sterile gloves, or gloves wiped with 70% ethanolRefrigeratorSterile hoodPlant growth chamber adjusted to 22 to 25°C and 67% rela-

tive humidity

ProcedureThis exercise must be performed in a sterile hood (a hood that blows sterile air) to prevent microbial and/or fungal contamination. Throughout the exercise you must wear sterile gloves.1. Wipe the hood bench with 70% ethanol.2. Transfer seeds of interest in 1.5-mL microtubes and

place the tubes in a rack. Do not place more than 1000 seeds (approximately 20 mg) in each tube . For each seed tube, place two empty microtubes in the same rack, then place a transfer pipette that has been steril-ized in 70% ethanol in one of the empty microtubes. Do not attempt to sterilize more than three tubes of seeds at the same time .

3. Fill each seed tube with 70% ethanol. Wait for 1 min, then remove the ethanol with a presterilized transfer pipette. Do not leave seeds in ethanol for more than 2 min. Collect the used ethanol and other liquid waste in a waste container.

4. Fill each seed tube with bleach solution. Mix the seeds in bleach by pipetting up and down once or twice. After 1 to 2 min, when the seeds have settled to the bottom of the microtube, pipette the bleach solution into the waste container. Use a separate transfer pipette for each seed tube to prevent cross contamination .

5. Repeat Step 4 three more times.6. Rinse the seeds with autoclaved double distilled water

four times in the manner described below. Pipette autoclaved double distilled water with an au-

toclaved Pasteur pipette attached to a rubber bulb into each seed tube. Use a separate Pasteur pipette for each seed tube to prevent cross contamination. With each rinse, mix the seeds by pipetting up and down, then allow the seeds to settle before removing the wa-ter. Remove as much water as possible without losing any seeds. After you have removed water from each seed tube the first time, place the Pasteur pipette into the fresh clean microtube mentioned in Step 2, above.


7. After the final rinse, resuspend the seeds in about 25 µL of autoclaved double distilled water.

8. Transfer the seeds from each seed tube (Step 7, above) to each MS plate containing kanamycin and distribute them evenly. If needed, pour more water onto the plate, then swirl the plate once or twice to distribute the seeds evenly. Remove excess water from the plate with a sterile Pasteur pipette.

9. Allow the agar plates to dry with the lid askew for 20 to 30 min under the sterile hood.

10. Close the lid, secure it with surgical tape, and cold treat the plates for 3 d in a 4°C refrigerator.

11. Transfer the plates to a growth chamber. Those seeds that germinate and develop will be transformants.

12. When transformants have reached the four-leaf stage, transfer them to soil, let them self, and collect the seeds.

Study Questions10–1. Briefly define operon.

10–2. What is an inducer in a prokaryotic operon system? Why didn’t you use an inducer to view the biolumi-nescence in transformed E. coli cells?

10–3. In one sentence define the GUS assay.

10–4. What function must be deleted from the Agrobacterium strain to make it a useful tool for plant transformation?

10–5. According to Protocol 10–2, the mixture of bleach and water, which is used for surface sterilization of Arabidopsis seeds, contains a trace amount of Triton X-100. What is the role of the Triton in the mixture? Hint: Triton is a viscous liquid.


Objectives

• to review Mendel’s laws of inheritance

• to understand the concept of the Chi-square test

• to learn the genetics of Arabidopsis thaliana

• to study the inheritance of aleurone color and endsoperm shape in corn

• to study sex-linked inheritance in Drosophila melanogaster

• to learn setting up crosses with A. thaliana or D. melanogaster

• to learn the technique of EMS mutagenesis

Exercise 11 Transmission Genetics (Heredity)

Gregor Mendel (1822–1884) is the founder of the science of heredity, com-monly known as transmission genetics. His investigations (hobbies) on breeding garden pea (Pisum sativum L.) have been the basis of the 20th century genet-ics. The scientific and historic views of Mendel’s work and its implication in modern genetics can be found in almost all genetic textbooks. In this chapter, we will not describe Mendelism; rather, we will provide the reader with the terms which are frequently used in transmission genetics and a summary of the principles of segregation of hereditary traits established by Mendel. To be able to understand these principles the reader must have a thorough knowledge of mitosis and meiosis.

Important TermsP generation: the parental generation where a specific cross startsF1: first filial generation; the progeny of a P × P crossF2: second filial generation; the progeny of a F1 × F1 crossGametes: germ cells which are haploid; pollen grain and egg in plants and sperm and ovum in animals are germ cellsTrue breeding (pure line): an individual self-fertilized several times or crossed several times to another individual with an identical genotype (also known as brother-sister mating) to ensure the traits remain unchangedZygote: a diploid cell formed following fertilizationAllele: alternative form of a gene; alleles could be two or more; if a1 and a2 are the mutations of the same gene, then they are allelicWild-type allele: an allele that is found (or seen) in nature; referred to as the nor-mal allele; in most cases, wild-type alleles are dominant (see dominant, below)Mutant allele: a mutated form of the wild-type allele; in most cases, mutant alleles are recessive (see recessive, below)Dominant allele (dominant): an allele that masks the expression of another alleleRecessive allele (recessive): an allele that is masked by the expression of another alleleIncomplete dominance: an allele that is not completely dominant to another allele, as a result, heterozygous individuals express an intermediate phenotypeHomozygous: an individual carrying the same alleles of a gene either recessive or dominant; such individual is true bredHeterozygous: an individual carrying one dominant and one recessive allele of the same gene; such individual is not a true bredGenotype: the genetic makeup of an individual; it is made up of thousands of recessive and dominant allelesPhenotype: observable characteristics of individuals; that is, how an organism looks like; it is the combination of expression of dominant and homozygous recessive allelesGenome: total amount of genetic material in the chromosomes of an individual

Gregore Koliantz & D.B. Szymanski

Types of CrossesSelf-fertilization: a fertilization in which both gametes come from the same individualCross-fertilization: a fertilization in which gametes come from two different individuals, one serves as male and the other as female individualBack cross: a cross between an individual and one of its parentsTest cross: a cross between an individual and its homozy-gous recessive parent (or a homozygous recessive individual carrying the same genotype as the parent)Reciprocal crosses: crosses in which the male and female phenotypes (by inference the genotypes) are reversedMonohybrid cross: a cross in which homozygous parents differ with regard to the autosomal alleles of a single geneDihybrid cross: a cross in which homozygous parents dif-fer with regard to the autosomal alleles of two genes; and hence trihybrid cross, tetrahybrid cross, etc. Sex-linked cross: a cross in which the sex-linked genes of the parents are soughtPair mating (in animals): a cross between one male and one female individualMass mating (in animals): a cross between a number of male and female individuals

Mendel’s PrinciplesWhen two homozygous individuals that differ in only one phenotype are crossed together, in the F1 generation, the dominant phenotype will be expressed. Upon selfing (that is the F1 × F1 cross), approximately 75% of the members of the F2 generation will express the dominant phenotype, and 25% will show the recessive phenotype. The result will be the same if the reciprocal cross is involved. Figure 11–1 illustrates a cross in the garden pea showing smooth or wrinkled seeds. From this cross the following rules can be extracted:

1. The outcome of reciprocal crosses is the same. This sim-ilarity is referred to as the Principle of Uniformity.

2. In the F1 generation, the dominant phenotype is ex-pressed.

3. In the F2 generation, the parental phenotypes reappear with the ratio of 3:1, where the ¾ represents the domi-nant phenotype (the smoothness of the seeds), and the ¼ shows the recessive phenotype (the wrinkledness of the seeds). This phenotypic segregation represents the Principle of Segregation of the alleles.

An alternative way to show the cross displayed in Fig. 11–1 is to use letters to represent the alleles. The letter S can be assigned to smoothness and the letter s to wrinkledness. Note that the dominant allele is written with capital letters and recessive allele with lowercase letters (Fig. 11–2). Since the parents (P) are homozygous for the trait, they are represented with two copies of the same allele: S/S for the dominant allele and s/s for the recessive allele. When

the parents produce gametes by the process of meiosis (see Exercise 12), only one copy of the gene will be present in each gamete: S for the smooth parent and s for the wrinkled. It follows that the F1 individuals will be heterozygous for the trait carrying one copy from each allele (S/s). In the F2 generation, three types of genotypes are produced, S/S, S/s, and s/s, with the ratio 1:2:1, respectively (see Fig. 11–2). This ratio is the result of random fusion of gametes. Since the S allele is complete dominant to the s, the S/S and S/s individuals will be phenotypically identical. As a result, the genotypic ratio 1:2:1 will be shifted to the phenotypic ratio 3:1. If the S allele were incomplete dominant to the s, then the genotypic and phenotypic ratios would have been the same: 1:2:1.


Fig. 11–1. A monohybrid cross representing Mendel’s Principle of Segregation.

Fig. 11–2. The same cross as in Fig. 11–1. Here, let-ters demonstrate the Principle of Segregation.


Mendel extended his investigations beyond monohybrid crosses. He set up crosses between the pea plants that differ in two phenotypes; for example, plants with smooth and yellow seeds vs. plants with wrinkled and green seeds, typically a dihybrid cross. Consequently, he proposed the Principle of Independent Assortment which states that the alleles of different genes are combined independently of each other. Today, we know that independent assortment is possible only if the genes involved are located on different chromosomes (see Fig. 11–3) and that these genes segregate independently in the production of gametes. However, some of the “traits” Mendel studied were located on the same chro-mosome (i.e., linked), but they segregated independently because of the occurrence of crossing over (see Exercise 12) between them. Mendel was unaware of these events.

To illustrate the Principle of Independent Assortment, a cross is set up between homozygous yellow-starchy (C/C;Su/Su) and purple-sweet (c/c;su/su) corn. All the individuals of the F1 generation will follow the rule of Uniformity and the dominant phenotype (yellow-starchy) will be expressed (Fig. 11–3). The F1 individuals, which are heterozygous for both alleles (C/c; Su/su), will develop four types of gametes because each C or c allele will segre-gate independently of Su or su allele. Consequently, upon random fusion of these gametes, 16 (4 × 4) genotypic com-binations will occur in the F2 generation (see Fig. 11–3). These 16 combinations will be grouped in four phenotypic classes because of the dominance, giving rise to a 9:3:3:1 phenotypic ratio.

Statistical Treatment of Data: The Chi-Square Test

The Chi-square (χ2) test is often used in genetics to test the significance of deviation between observed and expected numbers.

Consider the following example. Suppose in a popula-tion of 218 plants, you counted 162 individuals with red and 56 with white flowers. These plants had been generated by a cross between two heterozygous parents. Your expec-tation is that the progeny to appear in a 3 red and 1 white ratio. On this basis, the expected number of plants with red flowers would be 218 × 3/4 = 163.5 and that of white flow-ers 218 × 1/4 = 54.5.

Given the Chi-square formula

χ2 = Σ (O − E)2/Ewhere Σ is the sum of all values, O is the observed (actual) numbers, and E is the expected numbers (to be calculated); the Chi-square can be calculated as shown in Table 11–1a. However, in this example, if your expected ratio is 1 red and 1 white, then the expected number of red and white flowers would be the same, that is 218 × 1/2 = 109. Under this assumption, the Chi-square will be calculated accord-ing to Table 11–1b. Note that in Chi-square tests data are reported as whole numbers .

To find the level of significance of the χ2 values shown in Tables 11–1a and 11–1b, the Chi-square table is used (Table 11–2).

How to Use the Chi-Square Table1. Determine the degrees of freedom. In general, the de-

grees of freedom indicate the number of phenotypic groups that are independent of each other. In Tables 11–1a and 11–1b, there are two phenotypic groups: plants with red flowers and plants with white flowers. Since the total number of plants is 218, of which 162 have red flowers, the number of plants showing white flowers has to be 56 (i.e., 218 − 162). Therefore, the latter is not an independent group. In other words, if

Fig. 11–3. Independent assortment of yellow-starchy (C/C;Su/Su) and purple-sweet (c/c;su/su) kernels in corn.

Table 11–1a. The Chi-square test for a 3:1 segregation of the trait.

Red White TotalObserved (O) 162 56 218Expected ratio 3/4 1/4Expected (E) 163.5 54.5 218O − E −1.5 1.5 0(O − E)2 2.25 2.25 4.50(O − E)2/E 0.013 0.041 0.054χ2 = 0.054df = 1

Table 11–1b. The Chi-square test for a 1:1 segregation of the trait.

Red White TotalObserved (O) 162 56 218Expected ratio 1/2 1/2Expected (E) 109 109 218O − E 53 −53 218(O − E)2 2809 2809 5618(O − E)2/E 25.770 25.770 51.54χ2 = 51.54df = 1


there are two phenotypic groups, only one group is independent, and there is only one degree of freedom . In general, the degrees of freedom are equal to the number of phenotypic groups minus one .

2. In Table 11–2, use the probability of 0.05 as a split point for accepting or rejecting a hypothesis and note the critical values shown in that column.

3. Since in this example, there are two phenotypic groups, in Table 11–2 look at the first row where there is one degree of freedom and find the corresponding critical value (in this case it is 3.84). The calculated χ2 value for the 3:1 segregation in Table 11–1a is 0.054, which is less than 3.84; therefore, you fail to reject the hy-pothesis of the 3:1 segregation. On the contrary, the calculated χ2 value for the 1:1 segregation in Table 11–1b is far greater than 3.84; therefore, you are in-clined to reject the hypothesis of 1:1 segregation.

Exercises with Arabidopsis thalianaThe genus Arabidopsis is a member of the Brassicaceae (mustard or crucifer) family and has nine species and eight subspecies. The most studied species is Arabidopsis thali-ana, or thale cress (also known as wall cress or mouse-ear cress). Arabidopsis thaliana (hereafter called Arabidopsis unless otherwise specified) is a dicotyledonous angiosperm; that is, the mature embryo gives rise to two cotyledons (dicotyledon or dicot) and the seeds are enclosed within an ovary (angiosperm).

Vegetative growth of Arabidopsis is marked by rise of the radicle followed by the formation of hypocotyls and coty-ledons from seed coat. After cotyledons are fully opened, rosette leaves are formed. Emergence of the first flower bud indicates the beginning of the reproductive growth of the plant. Flowers are white in color, about 3 mm long, and pro-duce four sepals, four petals, six stamens, and a single ovary consisting of two fused carpels (Fig. 11–4). Developmental and physiological aspects of Arabidopsis are almost the same as those of crop plants, but it has a shorter genera-tion time. These features make Arabidopsis an amenable

organism for plant genetics. However, Arabidopsis has no agronomic significance. Arabidopsis is a self-fertile plant; a single plant can produce more than 100 seed pods or siliques. Each healthy silique contains about 40 to 50 seeds. From a healthy wild-type plant, thousands of seeds can be obtained. The entire life cycle of Arabidopsis (from seed imbibition to seed harvest) is completed in about 6 to 7 weeks. Table 11–3 shows the growth stages of Arabidopsis from leaf development through

seed maturation. The data have been collected from the Columbia wild-type strain grown under a 16-h light, 8-h dark cycle.

Arabidopsis grows in the wild throughout temperate regions of Europe, Asia, and North Africa on sandy or gravely soil. In the greenhouse, Arabidopsis grows very well on soil–compost mixes at temperatures in the range of 22 to 26ºC and in a relative humidity of around 50%. Lower temperatures slow growth. Arabidopsis requires a 12 h or greater photoperiod to flower. We raise Arabidopsis under continuous light at 25°C in a relative humidity of 50%. Plant fertilizers, dissolved in water, can be used to feed Arabidopsis; however, excess of watering may reduce plant hight, can cause early flowering, and reduced productivity.

The most widely used Arabidopsis ecotypes (i.e., strains

Fig.11–4. (A) A mature plant of Arabidopsis thaliana grown in the greenhouse. (B) Floral diagram of A. thaliana.

Table 11–2. Chi-square values†. The arrow indicates the split point for accepting or rejecting a hypothesis. Critical values are shown in bold font.

Degrees of freedom

Probabilities0.95 0.80 0.50 0.10 0.05 0.01 0.001

accept↑reject1 0.00 0.06 0.46 2.71 3 .84 6.64 10.832 0.10 0.45 1.39 4.61 5 .99 9.21 13.823 0.35 1.01 2.37 6.25 7 .82 11.35 16.274 0.71 1.65 3.36 7.78 9 .49 13.28 18.475 1.15 2.34 4.35 9.24 11 .07 15.09 20.526 1.64 3.10 5.35 10.65 12 .59 16.81 22.467 2.20 3.82 6.35 12.02 14 .07 18.48 24.328 2.73 4.60 7.34 13.36 15 .51 20.09 26.139 3.32 5.40 8.34 14.68 16 .92 21.67 27.8810 4.00 6.20 9.34 15.99 18 .31 23.21 29.59

† Originally from R.A. Fisher and F. Yates (1943).


isolated from different geograph-ical regions) are Columbia (Col) and Landsberg erecta (Ler). Both ecotypes were derived from the same population originally named Landsberg. Columbia was derived by self-pollination from a single plant with high reproductivity. Landsberg erecta was selected for its short and erect stature following gamma irradiation. A number of other ecotypes are also available which are adapted to specific ecosystems around the world. These ecotypes include, but are not limited to, Wassilewskija (Ws), Eastland, Dijon G, and C24.

HistoryArabidopsis thaliana was first identified by Johaness Thal in the 1570s. In 1873, a mutant (believed to be agamous: whorls of the stamens and gynoecium replaced by extra whorls of petals and sepals) Arabidopsis plant was found in Europe. In 1943, Arabidopsis was con-sidered as a model system in genetics and the first major review article was published by G. Redei in 1970. Thirteen years later, the first detailed genetic map of Arabidopsis was published. It was not until 1990 that the Arabidopsis Genome Project was initiated which led to the completion of the physical map of all five chromosomes in 1997. During the years 1999 and 2000, the entire Arabidopsis genome was sequenced and the existence of some 25,500 genes was pre-dicted. However, to date 27,029 protein coding genes and 3889 transposable elements have been identified, raising the total number of genes to 30,918.

Model OrganismArabidopsis thaliana has a number of advantages that make it an amenable organism for genetic research:

it is a relatively small plant• it reproduces rapidly• it is convenient to grow in the laboratory• it has a relatively small genome• its genome has been sequenced• it can be easily transformed by • Agrobacterium tumefaciens, creating transgenic plants (see Protocol 10–1)

The GenomeArabidopsis thaliana is a diploid plant with a diploid number

of 10 chromosomes (2n = 10). Arabidopsis has surprisingly a high number of genes: 30,918 compared with 30,887 in Drosophila melanogaster, 7547 in Saccharomyces cer-evisiae (yeast), 23,399 in Caenorhabditis elegans (worm), 12,792 (predicted) in Danio rerio (zebrafish), and 31, 896 in humans. This high number is due to the fact that about 58 to 60% of the genome is duplicated, likely because of independent duplication events that took place during the evolution of this species.

MutantsIn Arabidopsis, thousands of mutations have been identi-fied which alter developmental progression or result in morphological and physiological changes.

Mutations occur spontaneously; however, they can be induced by chemicals such as EMS (see Protocol 11–3), irra-diation with gamma or X rays, or T-DNA insertion. To ensure that the mutant phenotype is indeed due to a single mutation, the mutant plant must be backcrossed at least four times to a wild-type strain and the progeny selected for the mutant phe-notype. Table 11–4 lists a variety of Arabidopsis mutations.

Table 11–3. Arabidopsis growth stages for the soil-based phenotypic analysis platform. Reprinted from The Plant Cell 13:1499–1510 by permission of the American Society of Plant Biologists.

Stage Description

Col-0data

Days† SD CV‡Principal growth stage 1 Leaf development

1.02 2 rosette leaves > 1 mm in length 12.5 1.3 10.71.03 3 rosette leaves > 1 mm in length 15.9 1.5 9.51.04 4 rosette leaves > 1 mm in length 16.5 1.6 9.81.05 5 rosette leaves > 1 mm in length 17.7 1.8 10.21.06 6 rosette leaves > 1 mm in length 18.4 1.8 9.81.07 7 rosette leaves > 1 mm in length 19.4 2.2 11.11.08 8 rosette leaves > 1 mm in length 20.0 2.2 11.21.09 9 rosette leaves > 1 mm in length 21.1 2.3 10.81.10 10 rosette leaves > 1 mm in length 21.6 2.3 10.91.11 11 rosette leaves > 1 mm in length 22.2 2.5 11.21.12 12 rosette leaves > 1 mm in length 23.3 2.6 11.31.13 13 rosette leaves > 1 mm in length 24.8 3.2 12.81.14 14 rosette leaves > 1 mm in length 25.5 2.6 10.2

Principal growth stage 3 Rosette growth3.20 Rosette is 20% of final size 18.9 3.0 16.03.50 Rosette is 50% of final size 24.0 4.1 17.03.70 Rosette is 70% of final size 27.4 4.1 15.03.90 Rosette growth complete 29.3 3.5 12.0

Principal growth stage 5 Inflorescence emergence5.10 First flower buds visible 26.0 3.5 13.3

Principal growth stage 6 Flower production6.00 First flower open 31.8 3.6 13.36.10 10% of flowers to be produced have opened 35.9 4.9 13.66.30 30% of flowers to be produced have opened 40.1 4.9 12.36.50 50% of flowers to be produced have opened 43.5 4.9 11.26.90 Flowering complete 49.4 5.8 11.7

Principal growth stage 8 Silique ripening8.00 First silique shattered 48.0 4.5 9.3

Principal growth stage 9 Senescence9.70 Senescence complete; ready for seed harvest ND§ ND ND

† Average day from date of sowing, including a 3-d stratification at 4ºC to synchronize germination.‡ CV, coefficient of variation, calculated as (SD/day) ´ 100.§ ND, not determined.


National and International ResourcesThe Arabidopsis Information Resource (TAIR), • provides genomic and literature data, and can be found at http://www.arabidopsis.org; verified 17 May 2009Lehle Seeds• , provides seeds, literature, and useful information about Arabidopsis, and can be found at http://www.arabidopsis.com; verified 17 May 2009Nottingham Arabidopsis Stock Centre (UK), • provides seeds and information resources, and can be found at http://nasc.life.nott.ac.uk/; verified 17 May 2009

Description of the ExerciseIn this exercise, you will be working on a mutant called transparent testa glabra (hereafter ttg). This mutation was generated following treatment of Wassilewskija (Ws) seeds with ethyl methane sulfonate (EMS). The mutation is caused by a single C-to-T base transition that introduces a termination codon in place of the codon for the amino acid glutamine 317 (Walker et al., 1999). In normal plants, there are hair-like, specialized epidermal cells called trichomes that are distributed on the leaves, stems, and sepals. Trichomes are absent from the roots, hypocotyl, cotyledons, petals, stamens, and carpels. The morphology of trichomes varies from unbranched spikes to structures containing two to five branches. You will study the trichomes in this exercise. A number of genes are involved in the fate of the trichomes: one group of genes initiates the formation of trichomes and another group determines the shape of the trichomes. In the ttg mutation, the trichomes are absent. Check this phenotype in the procedure below. In addition to the trichome phenotype, the ttg mutation eliminates the purple anthocyanin pigments from the seed coat, causing the yellow cotyledons to be visible through the testa. For this reason, ttg seeds have a yellow color. Other seeds have a brown color. Also, ttg mutants completely lack anthocya-nins in the epidermis and in subepidermal layers of leaves and stems. In genetics, mutant genes that exhibit several and unrelated phenotypes are called pleiotropic. Therefore, ttg is a pleiotropic mutation. Figure 11–5 shows the shape of trichomes in the wild-type Columbia (Col) and a number of mutant plants.

MaterialsHand magnifying glassDissecting microscopeForcepsArabidopsis wild-type strains Columbia (Col), Landsberg erecta

(Ler), and Wassilewskija (Ws)Arabidopsis mutant strain ttg and some other mutants selected by

the laboratory instructorSelfed F1 seeds of a cross “unknown” to studentsPots with soil

Fig. 11–5. Wild-type and mutant trichomes in Arabidopsis thaliana: (A) and (B) wild type, (C) dis-torted 1 (dis1-1), (D) distorted 2 (dis2-1), (E) leaf of the ttg mutation: note the lack of trichomes, and (F) wurm-1 (wrm-1).

Table 11–4. Representatives of Arabidopsis thaliana mutations.

Gene symbol Name Phenotype Chromosome

Map position

cM cer eceriferum Very bright green stem and silique, reduced plant height 1 70.1an-1 angustifolia Narrow leaves, trichomes unbranched 1 55.2er erecta Upright stem, compact inflorescence, short petioles 2 43.5py-201 pyrimidine requiring Requires thiamine, leaves of unsupplemented plants white or pale 2 53.0gl1 glabra No trichomes on stem or leaves 3 46.2spy-5 spindly Early flowering, usually light green, somewhat curled leaves, bent siliques 3 12.0bp-1 brevipedicellus Short pedicels, siliques bent downwards, plant height reduced 4 15.0im immutans Variegated phenotype 4 48.0pi pistillate Lacks petals and anthers, maintained as heterozygote 5 21.0ch chlorine Yellow–green plant 5 43.0


Procedure1. Study the three wild-type strains Col, Ler, and Ws.

Find similarities and differences between them. Note that Ler is shorter than Col and Ws and has an upright stem. Ws is an early flowering strain, and leaves are slightly serrated and light-green in color. Also, check the shape of the trichomes.

2. You may examine a number of mutant plants including ttg. Examine each mutation in detail. Compare each mutation with Col wild type and record the differences on your report sheet.

3. You may set up a cross (see Protocol 11–2) between two strains; for example, use ttg and Col with Col as the male parent. Then collect the F1 seeds and plant them in soil. You may allow the F1 plants to self-fertil-ize to collect the F2 seeds. The F2 plants should show you a 3:1 segregation of phenotypes. Alternatively, you may back cross the F1 plants to the homozygous reces-sive parent (in this experiment ttg/ttg) and check the F2 progeny for a 1:1 phenotypic segregation. However, with regard to the development time of Arabidopsis and laboratory time restriction, your instructor may give you the F2 seeds of a certain “unknown” (to stu-dents) cross. These seeds will be the result of a cross between ttg/Col ´ ttg/Col, ttg/Col ´ ttg/ttg, or ttg/Col ´ Col/Col plants.

4. Plant these seeds in pots. Cold treat them for 3 d at 4°C, and then transfer the pots to a greenhouse.

5. Forty-eight hours later seedlings should appear. Check your pots regularly, add water as needed.

6. Two weeks later, bring the pots to the laboratory and examine the F2 population for the presence or absence of trichomes. Count the number of plants showing the ttg phenotype and the number of plants showing the wild-type phenotype. Do not discard the plants; you will need them in Exercise 14; return the pots to the laboratory instructor .


What are the phenotypes of the F2 plants?

What phenotypic ratio did you obtain?

What type of cross are you dealing with, a self-cross or a back cross? Explain why.

Which phenotype is recessive, which one is dominant? Explain why.

Predict the genotype of the F1 and P plants.

8. Record your results in the table below. Fill out all col-umns. The laboratory instructor will check your results for accuracy.

Exercises with Corn (Zea mays)Corn is one of the most important crop plants in the world. Although corn genetics has been studied in detail, some aspects of corn make it a difficult plant to use in laboratory exercises; these include large size of individual plants and long generation time. However, corn cobs (corn ears) are ideal materials for studying transmission genetics.

An intact corn cob (Fig. 11–6) represents a population of kernels. There are numerous genes scattered throughout the corn genome that determine the color, shape, and other characteristics of the kernels. One advantage of using corn in genetic studies is the variety of genetic interactions that exist among its genes. For example, if two independently segregating genes both affect the same phenotype, then the classical 9:3:3:1 ratio will be changed depending on the genes involved and the type of the interaction. Such inter-actions are called epistasis. The gene whose phenotype is expressed, is called epistatic, whereas the gene whose phenotype is masked (suppressed), is called hypostatic.

Fig. 11–6. Corn cob bearing F2 kernels showing a 3:1 phenotypic segregation.

Seed code (if any)

Observed (O) number of

seedlings per phenotype

Expected (E) number of

seedlings per phenotype O – E (O – E)2/E χ2 df


Variants of the standard 9:3:3:1 ratio, are described below.12:3:1 One dominant gene masks the expression of the • other gene.9:3:4 Homozygous recessive alleles mask the expression of • one of the dominant alleles.9:6:1 Each dominant gene alone, but not in combination, pro-• duces the same phenotype.15:1 Each dominant gene alone or combined shows the • same phenotype.13:3 The dominant allele of one of the two genes masks • the expression of the other dominant gene and the double-recessive alleles.9:7 Both dominant genes are required for the expression of • a functional product; under this condition, each allelic pair would result in a nonfunctional (mutant) phenotype; this inter-action is sometimes called a complementary gene action.

Description of the ExerciseIn this exercise, you will examine one or more corn cobs by studying kernel color and texture. You have to collect data on the number of kernels showing variety of phenotypes and find out what cross is represented by the corn.

MaterialsCorn cobs representing the F2 population of “unknown” crosses. The cross could be a monohybrid, a dihybrid, or a back cross. The laboratory instructor will assign corn cobs to students.

Procedure1. Pick a corn cob.2. Examine the phenotype of kernels. Count the number

of kernels belonging to each phenotypic group.3. From the results, predict the type of cross, the genotype

of the P, the F1, and the F2 cobs.4. Test your conclusions by the Chi-square test.5. Record your observations in the table at the top of this

page. Fill out all columns. The laboratory instructor will check your results for accuracy.

Sex-Linked InheritanceSex chromosomes are a pair of heteromorphic (morpho-logically different) or homomorphic (morphologically identical) chromosomes that determine sexes in organisms. Sex-linked inheritance is a pattern of transmission of genes that are located on the sex chromosomes.

Almost all animal species have a pair of sex chromo-somes. In some, but not all, dioecious plants (having pistillate and staminate flowers on different individuals) sex chromosomes have been identified. Spinach (Spinacia oleracea L.) and bladder campion (Silene latifolia Poir.) have heteromorphic sex chromosomes, whereas those in papaya (Carcia papaya L.) are homomorphic. Monoecious plants (having pistillate and staminate flowers on the same individuals) and hermaphrodites (flowers on the same indi-viduals have both male and female reproductive organs) do not have sex chromosomes.

In the exercise described below, you will study the sex-linked inheritance in the fruit fly Drosophila melanogaster.

Exercises with Drosophila melanogasterThe fruit fly, Drosophila melanogaster (hereafter called Drosophila, unless otherwise specified), is the most exten-sively studied organism in the animal kingdom. It belongs to the phylum Arthropoda; class Insecta; order Diptera; family Drosophilidae; genus Drosophila; and species melanogaster. Drosophila is cosmopolitan: it is found all over the world. It lives and breeds in decaying fruits and flowers. The two spe-cies melanogaster and simulans are in close association with humans and are referred to as domestic species.

Drosophila was first studied by W.E. Castle in 1901 as an experimental organism, and then it was used by T.H. Morgan for genetic experiments. In 1910, he proposed the theory of sex-linked inheritance, and the following year, C.B. Bridges discovered that sex-linked genes are located on the X chromosome. In 1913 A.H. Sturtevant determined that genes are arranged on the chromosomes in a linear order, consequently he prepared the first genetic map of Drosophila. In 1927, H.J. Mueller discovered that X-rays can cause lethal mutations and can increase the frequency of spontaneous mutations in Drosophila. In 1935, C.B. Bridges successfully prepared the first map of Drosophila polytene chromosomes. The first catalogue of Drosophila mutations was prepared in 1944 by C.B. Bridges and K.S. Brehme. Twenty-four years later, a more comprehensive catalogue was introduced by D.L. Lindsley and E.H. Grell. The updated version of this work appeared in 1992 edited by D.L. Lindsley and G.G. Zimm. Further revisions and additions to this catalogue are published electronically

Cob code (if any)

Kernel pheno-type

Observed (O) number of kernels per phenotype

Expected (E) number of kernels per phenotype O – E (O – E)2/E χ2 df


and can be found in Drosophila resource sites. In 1950, M. Demerec published the first book on the biology of Drosophila. A complete review of the genetics and biol-ogy of Drosophila was published in 12 volumes between 1976 and 1986, edited by M. Ashburner, E. Novitski, T.R.F. Wright, H.L. Carson, and J.N. Thompson, Jr. Drosophila Information Service (known as DIS) presents informal reports and information from Drosophila workers around the world and is published annually.

Life CycleThe life cycle (development time) of Drosophila varies depending on the culture temperature. The development time of wild-type strains is about 18 d at 18°C, around 11 d at 25°C, and close to 9 d at 29°C. Mutants have a longer development time.

In the life cycle of Drosophila, four distinct stages can be identified: egg, larva, pupa, and adult (Fig. 11–7). At 25°C, about 24 h after insemination, eggs are laid by the female parent. A healthy female can lay eggs for 6 to 8 con-secutive days. The day after the eggs are laid, larvae hatch; they are called the first instar larvae. In the next 4 to 5 d, the larvae molt twice, producing the second and the third instar larvae. During molting, the larvae undergo a considerable increase in size, exclusively because of increase in cell size.

However, the cells of the imaginal discs which will form adult structures during the pupal stage, do divide during this period. Eventually, the cuticle of the third instar larvae hardens to become the pupae. Metamorphosis occurs within the pupae. During metamorphosis, all larval cells except those of the Malpighian tubules and the nervous system degenerate and adult structures are formed by the growth and morphogenetic movements of the imaginal disc cells. The pupal stage takes about 5 d. When the metamorphosis is complete, the adults emerge (eclose) through the anterior end of the pupae. Upon emergence, the males are sexually mature, but the females remain unreceptive to the males for about 8 h. During this time, the females remain virgin.

Model OrganismDrosophila is widely used in genetic experiments for the following reasons:

it is relatively small in size• sexes are easily distinguishable• it reproduces easily and can be grown in the laboratory without • the need of expensive equipmentit has a relatively small genome which has been sequenced• it has polytene chromosomes which can be used to identify • gene loci and chromosomal rearrangementsa large number of mutations have been isolated and characterized• high level of homology has been found between the Drosophila • and human genome

The GenomeDrosophila is a diploid fly with a diploid number of 8 chro-mosomes (2n = 8). The entire genome was sequenced in 1999 and some 13,600 genes were identified. This number is now 30,887 released by the Genomic Center of Indiana University in 2007 (http://flybase.org/static_pages/docs/release_notes.html; verified 17 May 2009).

Wild-Type and Mutant StrainsWild type is referred to the flies which do not carry mutations. The most commonly used wild-type stock is Oregon-R, but Canton-S, Samarkand, and a few more stocks are used in many laboratories. The list of mutations discovered and characterized so far can be found on the websites of Drosophila resource centers (see below). Table 11–5 lists a variety of Drosophila mutants widely used in the laboratory.

Selected National and International ResourcesA wealth of information about Drosophila is available in the following sites:

Indiana University Genomic Information for Eukaryotic • Organisms; found at http://eugenes.org; verified 17 May 2009Indiana University Database of Drosophila Genes and Genomes: • the electron micrograph maps of polytene chromosomes (Sorsa Maps); found at http://flybase.bio.indiana.edu/static_pages/allied-data/external_resources5.html; verified 17 May 2009Berkeley Drosophila Genome Project (BDGP)• ; found at http://www.fruitfly.org; verified 17 May 2009Cambridge University (UK) Database for Drosophila and • Anopheles Genomics; found at http://flymine.org; verified 17 May 2009Fig. 11–7. The Drosophila melanogaster life cycle. For

details see the text.


The Chromosomal Basis of Sex DeterminationIn Drosophila, chromosomal differences are visible between sexes: the males are heteromorphic, carrying one X and one Y chromosome, whereas the females are homomorphic, having two X chromosomes ( Fig. 11–8). Chromosomes 2, 3, and 4 are the autosomes. During meiosis, the sex chromosomes segregate ensuring a 1:1 ratio of males and females, as illustrated in Fig. 11–9.

The Y chromosome does not have a role in sex deter-mination or the viability of Drosophila. Flies with two sets of autosomes and one X chromosome (2A:X) are viable, phenotypically male but sterile. Furthermore, normal males

carrying one or two additional Y chromosomes, (2A:XYY or 2A:XYYY, respectively) are viable but sterile too. These observations indicate that there are a number of male fertility genes on the Y chromosome and that excess of the fertility genes may change the genic balance, so the males become sterile. On the other hand, females carrying three X chromosomes (2A:XXX) are phenotypically female but survive only a few hours after emergence from pupae. They are called metafemales (Stern, 1959). Presence of an additional set of autosomes in flies destabi-lizes sexes, shifting the females toward an intersex phase and

the males to a metamale (Stern, 1959) state. Metamales are relatively inviable, sterile flies and show a delayed devel-opment time. Thus, in Drosophila, sex is determined by a balance between genes that are located on the X chromo-some and the autosomes (Table 11–6).

Description of the ExerciseIn this exercise, you will study the inheritance of the characters associated with the X chromosome of D. mela-nogaster. You will set up crosses between a wild-type and a mutant strain. Then you will examine the progeny of each cross to study the pattern of the inheritance. The laboratory instructor may give you advice in choosing the sex-linked mutations. More information about the mutations can be found in Lindsley and Zimm (1992).

MaterialsDrosophila stocks

wild type: Oregon-R mutants:

white; compound eyes and the ocelli white, also no pigments in larval Malpighian tubules and adult testis sheath

miniature; wing size reduced and less transparent than the wild type

Drosophila culture medium in vials (see Protocol 11–6 for details)Dissecting microscope or powerful magnifying glass

Table 11–5. Representative of Drosophila melanogaster mutations.

Gene symbol Name Phenotype Chromosome

Map position

cMB Bar† Eyes narrow X 57.0w white Eye color white X 1.5sn singed Bristles short and deformed X 21.0

vg vestigial Wings reduced to vestiges 2 67.0bw brown Eye color light brown darkening with age 2 104.5Cy Curly‡ Wings curled upward 2 6.1e ebony Body color shining black 3 70.7Sb Stubble‡ Bristles less than one-half normal length 3 58.22ey eyeless Eye size reduced 4 2.0bt bent Wings bent sharply backward 4 1.4gy goutylegs Legs shortened and thickened 4 0.2

† Semi dominant; eyes oval in females.‡ Dominant, homozygous lethal.

Fig. 11–9. Segregation of the sex chromosomes in Drosophila melanogaster. Autosomes are not shown for simplicity.

Fig. 11–8. A schematic representation of Drosophila melanogaster mitotic chromosomes; (a) in females, (b) in males.

Table 11–6. The relationship between chromosome num-ber and sex determination in Drosophila melanogaster.

No. of autosomal set

(A)

No. of X chromosomes

(X) Ratio A/X Sex2 1 2 normal male 2 2 1 normal female2 3 0.67 metafemale3 2 1.5 intersex†3 1 3 metamale

† Expression of the characters of both sexes.


Sorting platesSorting brushesDrosophila anesthetizer, (Carolina Biological Supply Co.,

Burlington, NC)Ether (ethyl ether)Glass or polystyrene vialsMorgue: jar containing alcohol or a detergentIncubator adjusted to 25°C

Procedure

A. How to Anesthetize FliesCaution: You will be working with ether! Ether is highly flammable and is harmful to your health.1. The Drosophila anesthetizer is composed of a chamber,

a hollow stopper on the mouth of the chamber, and a cap in the bottom. A foam pad has been installed in the bottom of the chamber (Fig. 11–10).

2. Remove the hollow stopper from the mouth of the chamber and fill it one-third full with ether. Do not ap-ply more ether because the flies will be killed.

3. Remove the cap to expose the foam pad, then quickly pour the ether on the pad. Replace the cap and the stop-per securely. The anesthetizer is now ready to be used.

4. To anesthetize flies, remove the stopper; at the same time gently shake down the Drosophila culture vial, quickly unplug the vial and put the mouth of the vial on the mouth of the anesthetizing chamber; tap gently on the vial to shake the flies from the vial into the chamber.

5. Remove the culture vial and plug it; then immediately put the stopper back on the mouth of the chamber and tap the bottom of the anesthetizer on the bench to push the flies to the bottom of the chamber.

6. The flies should stop moving in about 20 to 30 seconds at room temperature. Remove the flies for examination and sorting. The flies remain anesthetized usually for 8 to 10 min. At temperatures higher than 25°C, they recover quickly.

B. Identify Body Parts1. You will be given live flies in a clean vial.2. Anesthetize the flies as described above.3. Place the anesthetized flies on a sorting plate and iden-

tify body parts. Use Fig. 11–7 as a guide. Touch the flies with brush.

C. Identify SexesFemales are larger than the males; well-fed and healthy • females measure about 3 mm and the males 2 mm.In females, the abdomen is large and the posterior end is • pointed; in males, the abdomen is short and round.Alternating dark-light chitinous bands (tergites) on • the dorsal end of the abdomen are visible in females, whereas in males, the last few tergites are fused to-gether to form a large, dark band.Check the differences in the external genitalia on the • ventral end of the abdomen of each sex.On males there are stiff hairs (chaetae) on the first tarsal • segment of each foreleg. This is called sex comb.

D. Set up Crosses1. Set up a reciprocal cross between the wild-type and the

mutant (white or miniature) flies as shown below. Note that these mutations are located on the X chromosome.

Make sure that the female flies you will choose are virgin (see Protocol 11–4 for collecting virgins). Otherwise, the laboratory instructor may give you vir-gin females.

Cross 1 Cross 2Female Male Female Male

P Wild type ´ mutant mutant ´ Wild type ↓ ↓F1 ? ?

2. Anesthetize the flies and identify sexes.3. With the aid of a brush, transfer five females and the

same number of males onto the glass surface of a vial containing culture medium. Plug the vial. Do not drop the anesthetized flies onto the surface of the me-dium: they will stick to the medium and will die! Do not stand the vial on end until the flies have revived.

4. Place the vial in a 25°C incubator.5. One week after the cross has started, remove the parents

from the vials (drop them into the morgue) to ensure that only the progeny will remain in the vials. Check the medium attentively; the larvae should be visible at this time. Leave the vial in the same incubator for one more week.

6. In the time specified, bring the vials to the laboratory; you will see a number of adult flies in the medium; they are the F1 generation of the cross you set up 2 wk ago. Anesthetize the flies and check the phenotype of the males and the females separately.


Fig. 11–10. The Drosophila anesthetizer. For details see the text.


What were the phenotypes of the parents?

What were the phenotypes of the progeny?

Did you see phenotypic difference between the males and the females in the F1 generation? Describe and dis-cuss your observations.

What was the sex ratio in the F1 flies? Did the experi-ment confirm your expectation?

Use the Chi-square test if needed.

Protocol 11–1Preparation of Pots for Planting Arabidopsis SeedsArabidopsis can be grown in a variety of containers includ-ing pots and trays (flats). Here we describe preparation of pots for laboratory use.

MaterialsPots, usually 5 ´ 5 cmTray (flats), with and without holesPlastic domes to fit the flatsSoil: Scotts Redi-earth (The Scotts Company ) or equivalentVermiculite, extra coarse: Sunshine (Sun Gro Horticulture)

or equivalentLarvicides/insecticides for basic pest control:

Gnatrol (Valent Professional Products) or equivalent, to kill larvae in soil

Marathon (Olympic Horticultural Products) or equivalent, to kill common pests such as thrips

Fertilizer:Miracle-Gro (Scotts Miracle-Gro Products, Inc.) or equivalent

Anhydrous calcium sulfate pellets: Drierite (W. A. Hammond DRIERITE Co. LTD) or equivalent, to dehumidify seeds

Deionized water (preferred), otherwise tap waterGraduated cylindersSeeds of interestSheets of clean and dry paper

Procedure1. Place a tray with holes into the tray without holes, then

put pots in it.2. Add 10 mL of the larvicide of choice to 4 L of water,

mix well, then pour the water into the tray, so that the pot will be half-filled.

3. Fill each pot two-thirds full with extra-coarse vermiculite.4. Mix 1/2 teaspoon of the pesticide of choice with 4 L

of soil.5. Fill each pot with soil (prepared in Step 4, above).6. Cover the pots with a plastic dome and keep in room

temperature until needed.

7. Just before planting the seeds, spray water on top of the pots to keep the soil moist.

8. After seeds have been planted, cold treat the pots in a 4°C refrigerator or cold room for 3 d to synchronize seed germination.

9. Transfer the pots to a greenhouse or to a growth cham-ber, if available.

10. Forty-eight hours later, remove the dome and let the seedlings grow. Water the plants regularly, but do not over-water them. Occasionally use fertilizer when wa-tering the plants. Mix 1/2 teaspoon of the fertilizer of choice with 1 L of water.

In optimal conditions, after a period of about 6 wk, siliques start to form and eventually they turn brown (yel-low in an extended amount of time). Stop watering the plants when the siliques turn brown.

To collect seeds:1. Hold the plant horizontally above a clean and dry piece

of paper (white paper is preferred because brown seeds will be easily visible).

2. Gently rub the plant so that the siliques shatter and the seeds fall onto the white paper.

3. Remove excess dry matter such as dry leaves, shoots, etc., by hand and separate the seeds from the chaff with fine forceps.

4. Collect seeds in microtubes (conical tubes in case a large quantity of seeds are harvested), then add two to three pellets of anhydrous calcium sulfate to each tube to dehumidify the seeds. The pellets are blue in color; they turn pink when they absorb moisture. Replace the pellets as needed.

Note: Preparation of flats is almost the same as that of pots.1. Place a tray with holes into the tray without holes.2. Add 10 mL of a larvicide to 4 L of water, mix well,

then pour the water into the tray.3. Fill the tray one-half full with extra-coarse vermiculite.4. Mix 1/2 teaspoon of a pesticide with 4 L of soil.5. Add the soil (prepared in Step 4, above) on top of the

vermiculite.6. Cover the tray with a plastic dome and keep at room

temperature until use.7. Just before planting the seeds, spray water on top of the

soil to keep it moist.

Protocol 11–2Setting up Arabidopsis CrossesArabidopsis thaliana is a self-fertilizing plant; that is, it carries both male and female reproductive organs (see Fig. 11–4B). The male reproductive organ is the stamen, which consists of an anther and filament. Anthers carry pollen grains: the male germ cells. The female reproductive organ is the carpel, which is made up of stigma, style, and ovary. Eggs develop in the ovary. During self-fertilization, pollen grains adhere to the stigma and germinate. Then the pollen tube grows through the ovary’s tissues, penetrates the ovule to fertilize the egg. In the laboratory, it is possible to cross-fertilize Arabidopsis.


3. Tape the tube on a platform mixer and mix the tube overnight (usually 12 h).

4. Wash the seeds five times in double-distilled water. After the last wash, let the tube sit undisturbed for 10 to 15 min. Then with a transfer pipette, carefully trans-fer all the water above seed level. Read the volume of the wet seeds. Subtract that volume from 1.8 (that is the volume of 1 g of dry seeds). The outcome is the volume of the water present in the tube.

5. Wear gloves and take the tube (from Step 4, above) to a fume hood. Add distilled water to the tube to a total volume of 20 mL. Then add 60 µL of EMS to the tube. The final concentration of EMS is now 0.3%. The concentration can be changed as needed.

6. Secure the cap of the tube, tape the tube on a platform mixer, and mix for 14 h at room temperature.

7. Decant (as much as possible) the EMS solution into a beaker containing 100 mM sodium thiosulfate to deac-tivate the EMS.

8. Wash the seeds twice in 100 mM sodium thiosulfate, 15 min each.

9. Decant (as much as possible) the sodium thiosulfate, then thoroughly wash the seeds with double-distilled water five times.

10. After the last wash, decant the water containing the seeds onto a large piece of Whatman paper and let the seeds dry for several hours.

11. Collect the seeds into a conical tube, add two to three pel-lets of anhydrous calcium sulfate, and save for future use.

To plant mutagenized seeds:1. Place mutagenized seeds in 1.5-mL microtubes, fill the

tubes with double-distilled water, and cold-treat the seeds at 4°C for 3 d.

2. With transfer pipettes, transfer the cold-treated seeds to 2 L of 0.1% agar solution.

3. Sprinkle 100 mL of the seed solution in each flat of soil, that is, 5000 seeds per flat. Therefore, 10 flats of soil are needed to plant the mutagenized seeds.

4. Cover the flats with a dome, transfer them to an ambient-controlled incubator or green house, let the seeds germinate at 100% relative humidity for 3 to 4 d, then remove the dome. The seedlings are the M1 population.

5. Let the M1 population self-fertilize and then collect the seeds. Call the seeds M2.

6. Plant the M2 seeds in flats, let the seeds germinate, then screen the M2 seedlings under a dissecting microscope for altered phenotypes.

7. If a mutant seedling is found, remove the seedling from the soil and replant it in a pot. Save the putative mutant plant for further studies.

Protocol 11–4How to Maintain Drosophila StocksThe only way to maintain Drosophila stocks is to transfer the flies to fresh medium every two weeks if raised at 25ºC, or every three to four weeks if kept at 18ºC.

To transfer the flies from an old culture vial to a fresh medium, remove the plug from the fresh vial; at the same time gently tap the bottom of the culture vial against the palm of your hand to send the flies to the bottom of the vial; then quickly remove the plug and invert the vial over the mouth of the fresh vial. Shake in 5 to 10 flies from each sex and recap both vials. Properly label fresh vials with the name of the mutation and the date of transfer. Time to time check the phenotype of the flies to ensure that they are not contaminated with other stocks. Wash old vials with deter-gent and autoclave for further use.

Protocol 11–5How to Collect Drosophila Virgin FemalesIn order to set up crosses between flies of different geno-types, it is extremely important to use virgin females. This is because females already inseminated by any male can store the sperm in the spermatheca (sperm storage sac in the female reproductive tract) and use it for up to six days. Therefore, to be sure that the desired gene was involved in a cross, the female parent must be virgin.

The simplest method to collect virgin females is to remove adult flies from the culture vial, place the vial in a 25°C incubator, and collect all those females that emerge in the next 6 to 7 hours. Keeping the vials at 18°C will reduce the rate of eclosion, but the females eclosing in the next 12 hours will most likely be virgin. In cultures which contain adult flies of different ages, virgin females can be identi-fied by light body color or especially by the presence of the dark meconium (the first stool) in the gut. You can see the meconium in the ventral region of the abdomen.

Protocol 11–6How to Prepare Drosophila Medium“Regular” MediumTraditionally, regular medium is referred to the medium which is cooked in the laboratory. Variety of recipes and modifications of a standard recipe are available, but just which is used depends on the availability of the raw materi-als and the laboratory temperament. Here we describe the medium we use in our laboratory to maintain the stocks.

MaterialsPer 1000 mL of deionized water

50 g Cornmeal32 g Granulated sugar (edible sugar)50 g Baker’s yeast11 g AgarNote: The amount of the agar must be adjusted as different

brands of Baker’s yeast may affect the firmness of the medium.

10% Nipagin M [methyl paraben (methyl ester of p-hydroxyben-zoic acid) Sigma, St. Louis, MO]

Add 10 g of Nipagin M to 100 mL of ethanol, mix well, and use as stock. Nopagin M is a mold inhibitor.

Autoclaved deionized water50% bleach

Equal volumes of bleach and deionized waterEthanol


Vials measuring 3 ´ 10 cm (diameter ´ height), commercially known as Drosophila vials; borosilicate glass (preferred) or polypropylene

Nonabsorbent cotton wool or polyurethane foam plugs to fit the vialsDrosophila culture bottles, 240 mL (Carolina Biological Supply

Co., Burlington, NC) with polyurethane foam plugsCheese clothSaucepanRange

ProcedureBefore preparing the medium, autoclave culture bottles and glass vials. If using polypropylene vials, wash them with 50% bleach and rinse in autoclaved water.1. In a saucepan mix cornmeal, sugar, baker’s yeast, and

agar, add water and bring to the boil while stirring con-stantly to avoid lumps.

2. Let the mixture boil and simmer for 10 min.3. Add 16 mL of 10% Nipagin M; stir well, and distribute

the medium in sterile vials to a depth of 3 cm or in sterile culture bottles to a depth of 2 cm.

4. Cover the vials or bottles with a sheath of clean cheese cloth and leave for 2 h.

5. Plug the vials or bottles with polyurethane foam plugs or nonabsorbent cotton wool. Foam plugs are preferred because they can be washed, autoclaved, and reused for an extended amount of time.

6. Store the vials and bottles in a refrigerator.7. Just before use, allow the media to warm to room

temperature.

RecommendationsWe recommend the following points when preparing the “regular” medium.1. In this medium, baker’s yeast is superior to brewer’s yeast.2. Drosophila larvae burrow into the medium to the depth

of about 10 mm; therefore, increasing the depth of the medium will not improve the larval growth. Expanding the area of the medium will increase larval viability and weight, hence decreasing the mortality.

3. It is unnecessary to seed the medium with additional live yeast. However, for extremely weak mutants, some 10 grains of active dry yeast can be added to the medium in each vial at the time when larvae are at the first instar stage. For the bottles, some 30 grains would be sufficient.

Instant MediumThe instant medium (Formula 4-24, by Carolina Biological Supply Co., Burlington, NC) is available as a powder, ready-to-use medium. According to the manufacturer, it is protected with anti-oxidant and mold inhibitor. We use this medium for class exercises.

MaterialsFormula 4–24 Instant Drosophila MediumActive dry yeast (available in grocery stores)Autoclaved distilled waterVials measuring 3 ´ 10 cm (diameter ´ height), commercially

known as Drosophila vials; borosilicate glass (preferred) or polypropylene

Measuring spoonsNonabsorbent cotton wool or polyurethane foam plugs to fit the vials

ProcedureUse autoclaved or sterilized vials as described above.1. To the vials measuring 3 ´ 10 cm, add 10 mL of the

instant medium and an equal amount of autoclaved, distilled water. Sprinkle about 10 grains of active dry yeast on top of the medium.

2. Plug the vials with nonabsorbent cotton wool or poly-urethane foam plugs, let set for 30 min before use.

Study Questions11–1. In garden pea, tall stem (T) is dominant to short

stem (t), green siliques (G) are dominant to yellow siliques (g), and smooth seeds (S) are dominant to wrinkled seeds (s). The genes controlling these phenotypes are independently segregating. In a cross between homozygous tall, green, smooth and short, yellow, wrinkled plants:

a. What will be the phenotype of the F1 plants?

b. What will be the phenotype of the F2 plants?

c. What will be the phenotype of the F1 back cross to its homozygous recessive parent?

11–2. How many genetically different pollen grains could be formed in Arabidopsis with the following genotypes:

a. aa Bb CC Dd

b. Aa Bb Cc Dd

11–3. In Drosophila melanogaster, flies with the sex chro-mosome complement XY or XO are males and those with XX, XXX, or XXY are females. What is the role of the Y chromosome in sex determination?

11–4. In a cross between two plants both of which ex-press tall stem and red flowers: 238 tall and red, 89 tall and white, 78 short and red, and 27 short and white offspring were obtained. What are the geno-types of the parent plants? Assume tall and red are dominant genes.

11–5. How do monoecious plants differ from dioecious plants? Is Arabidopsis thaliana a monoecious, dioe-cious, or hermaphrodite plant? Explain your answer.



Objectives

• to study the stages of meiosis

• to understand the phenomenon of crossing over

• to analyze linkage and crossing over in the fungus Sordaria fimicola

Exercise 12 Meiosis and Analysis of Crossing Over

Meiosis is a nuclear division of a diploid mother cell that produces gametes (or spores) with a haploid set of chromosomes. Meiosis occurs in two stages called meiosis I and meiosis II. These two stages will be described in detail in the next section. As Fig. 12–1 illustrates, in meiosis I, the diploid mother nucleus divides to form two haploid daughter nuclei. In meiosis II, each of the daughter nuclei divides, separating the sister chromatids. The result of meiosis II is four haploid nuclei.

Meiosis I: A Reduction DivisionBefore meiotic divisions begin, the chromosomes of the diploid mother cell have duplicated.

Meiosis I consists of four stages: prophase I, metaphase I, anaphase I, and telophase I (Fig. 12–2).Prophase I . In this stage, the chromosomes undergo substantial shape changes, and on this basis, prophase I is divided into five substages: leptotene, zygotene, pachytene, diplotene, and diakinesis.

In • leptotene, the condensation of chromosomes begins, and they become visible under microscope.In• zygotene, homologous chromosomes pair; the pairing is called synapsis.In • pachytene, synapsis is completed. Each synapsed chromosome pair is called bi-valent. Since each bivalent is made up of four chromatids, it is also called tetrad or four-strands. During pachytene, a reciprocal exchange of DNA material takes place between nonsister chromatids of homologous chromosomes (Fig. 12–3). This exchange is a natural phenomenon and is called crossing over. Crossing over results in the recombination of genetic material. You will examine the concept of recombi-nation in this exercise.In • diplotene, the synapsed chromosomes begin to move apart and the cytological evi-dence of crossing over becomes visible as cross-shaped structures called chiasmata.In• diakinesis, the chromatids become more condensed. Nonsister chromatids that have undergone crossing over remain closely associated at the chiasmata. At the end of diakinesis, the nuclear membrane breaks down, and the chromosomes begin to move toward the equatorial plane of the cell.

Fig. 12–1. A diagram of meiosis. Reproduced from Hartwell et al. (2000) by permis-sion of McGraw-Hill Companies.

78Gregore Koliantz & D.B. Szymanski 78 Genetics: A Laboratory Manual

Metaphase I . In this stage, movement of the chromosomes toward the equatorial plane is completed.Anaphase I . Because of the shortening of the microtubules, each chromosome is separated from its homologue. This motion moves the homologous chromosomes to opposite poles.Telophase I . At some point, the cytoplasm divides (cytoki-nesis), producing two haploid cells. Note that in telophase I, all chromosomes are still in a duplicated state.

Meiosis II: An Equational DivisionNo DNA replication takes place between the end of meiosis I and the beginning of meiosis II. Meiosis II is in fact a mitotic division and consists of prophase II, metaphase II, anaphase II, and telophase II (Fig.12–2).Prophase II . During prophase II, the chromosomes in each newly formed haploid cell (see telophase I) become attached to the nuclear spindle and move toward its equator.

Fig. 12–2. Process of meiosis showing all stages in detail. Reproduced from Hartwell et al. (2000) by permission of McGraw-Hill Companies.

Fig. 12–3. Crossing over: exchange of genetic material between non-sister chromatids. Reproduced from Hartwell et al. (2000) by permission of McGraw-Hill Companies.


Metaphase II . In metaphase II, the movement of chromosomes toward the spindle’s equator is completed.Anaphase II . In anaphase II, sister chromatids of each chromosome are separated from each other and migrate to the opposite poles of the spindle.Telophase II . By the time telophase II is completed, four daughter nuclei are formed. Following cytokinesis, the resulting daughter cells possess a hap-loid chromosome number.

Studying Meiosis in LilyStudy meiosis in pollen grains of lily (Lillium sp.) using commercially prepared slides. Handle them with care and exam-ine each with a low and a high power objective of a compound microscope.

Meiosis in lily (Fig.12–4) reveals chromosomes that are characteristic of metaphase I, anaphase I, and telophase I. You can still see the interphase which follows the first meiotic division. Lily chromosomes clearly show the pro-phase II, metaphase II, anaphase II, and telophase II stages.

Meiosis in Flowering PlantsThe reproductive cycle of flowering plants begins with the emergence of flower buds. Each bisexual flower contains both male and female organs: the stamen and the carpel, respectively. In each organ, diploid progenitor cells undergo meiotic and then mitotic divisions to produce haploid gametes (Fig.12–5).

1. Meiosis in the Female Part of the Flower

On the inner walls of the flower’s ovary, a mass of cells form, each of which is the start of an ovule. Each ovule contains a megaspore mother cell or megasporocyte which is diploid. The megaspore mother cell undergoes meiotic divisions resulting in four haploid daughter cells called megaspores. Three of the four megaspores disin-tegrate, the one remaining undergoes three successive mitotic divisions to produce eight haploid nuclei without cytoplasmic division. Eventually the cytoplasm divides after each nucleus migrates to a specific location. Thus, an embryo sac is formed. Within the sac, one nucleus becomes the egg cell, two nuclei form the polar cell, two other nuclei form the synergid cells, and the three become the antipodal cells (see Fig.12–5).

2. Meiosis in the Male Part of the FlowerWhile anthers are growing, four masses of spore-producing cells form by mitotic divisions. These cells, which are called microspore mother cells or microsporocytes, are dip-loid. Microspore mother cells undergo meiosis to produce four haploid microspores. Each microspore then divides mitotically producing a pollen grain which consists of two haploid nuclei: a tube nucleus and a generative nucleus. The tube nucleus regulates the growth of the pollen tube, but the generative nucleus divides mitotically to produce two sperm nuclei: the male gametophyte (see Fig.12–5).

Linkage and Crossing OverThe term linkage refers to the phenomenon that two or more genes are located on the same chromosome. Thus, a group of genes that are linked together form a linkage group.

Fig. 12–4. A schematic presentation of meiosis in a lily anther.


However, a linkage group on a given chromosome does not always persist through meiosis because, as mentioned above, crossing over exchanges chromosome pieces during pro-phase I of meiosis (see Fig.12–3).

Given that two genes A and B (and their recessive alleles a and b, respectively) are located on the same chromosome (Fig.12–6a), the two homologous chromosomes migrate to opposite poles in anaphase I, eventually producing four haploid chromatids in telophase II that are genetically iden-tical to their parents in prophase I. However, if crossing over takes place between genes A and B at the four-strand stage of prophase I (Fig.12–6b), in the resulting telophase II, two of the four chromatids would be recombinants.

The frequency of crossing over between two genes is used as a measure of genetic distance between them: the crossing over frequency of 1% is defined as one map unit. The map unit is also called a centimorgan (cM) in honor of the geneticist Thomas H. Morgan (1866–1945).

Analysis of Crossing Over in Sordaria fimicola

Sordaria fimicola is a haploid fungus (n = 7) and belongs to the class Ascomycetes. It is closely related to the genus Neurospora. The natural habitat of Sordaria is the offal of plant-eating animals. Sordaria is an auxotroph (i.e., nutri-tionally dependent); it requires biotin and thiamine for proper growth. In the laboratory, S. fimicola can be grown on sterile media (see Materials, below). In S. fimicola a number of mutants have been identified and mapped. Most of them are ascospore color mutants, restricted growth mutants, and sterile mutants. Sordaria has three major advantages for use in genetic studies: (i) it has a short life cycle, about 10 to 12 d; (ii) it is self-fertile; therefore, it is isogenic; and (iii) mutants can easily be obtained by treat-ing mycelia with mutagens.

1. The Life CycleSordaria fimicola does not have distinct sexes but has mat-ing types which are regarded as primitive sexes. During the asexual (nonsexual) phase, haploid spores germinate and form the mycelium (plural: mycelia). The mycelium grows and throws off side branches resulting in a mass of mycelia which represents the body of the fungus. When two haploid mycelia of different mating types come into a physical contact, their nuclei fuse and a diploid cell (zygote) is formed. This cell immediately undergoes mei-otic divisions and produces four haploid ascospores. In the meantime, a sac or ascus (plural: asci) is developed that holds the ascospores in a linear array. Finally, each of the four haploid ascospores divides mitotically and produces two identical, haploid ascospores. This results in the forma-tion of eight ascospores within each ascus. A number of asci are grouped together within a sac called perithecium (plural:

Fig. 12–5. The formation of male and female gametes by the gametophyte of angiosperms. Reproduced from Brooker (2005) by permission of McGraw-Hill Companies.

Fig. 12–6. Consequences of crossing over during meiosis. (a) Absence of crossing over, (b) presence of crossing over. Reproduced from Brooker (2005) by permission of McGraw-Hill Companies.


perithecia). Perithecia in the wild-type strain of S. fimicola are black in color, and you can see them on agar plates. When ascospores are mature, the asci rupture, releasing the ascospores. They measure about 12 × 20 µm. The life cycle of S. fimicola is shown in Fig. 12–7.

MaterialsSordaria culture medium:

Per 1000 mL of dH2O17 g Corn meal agar2 g Glucose1 g Yeast extractAutoclaveDistribute in sterile Petri dishes

Cultured plates of the wild type and the mutant tan of SordariaCompound microscopeDissecting needlesMicroscope slidesCover slipsSterile toothpicksdH2O

Procedure1. Obtain two S. fimicola plates, one containing the wild

type and the other the mutant tan. Alternatively, the laboratory instructor may give you advice in choosing mutant strains.

2. Study the mesh of tiny, branching filaments which are the food-absorbing parts of the fungus. These are the mycelia. Sometimes they are called the hyphae (singu-lar: hypha). Also identify the perithecia.

3. With a sterile toothpick, scrape across the top of the agar plate gently to remove several perithecia.

4. Transfer the perithecia onto a microscope slide, then place a drop of distilled water on the perithecia.

5. Place a cover slip over the top of the perithecia and gently tap the cover slip with the handle of a dissect-ing needle until the perithecia are crushed. This will release clusters of acsi.

6. Examine the slide under a compound microscope.

Within each ascus you must see eight ascospores in a linear array (Fig.12–8). If you examine the wild-type strain, the ascospores will appear black. Examine the mutant strain tan which produces tan-colored ascospores. Be sure you can distinguish between the black and tan ascospores under the microscope.

2. Analysis of Crossing OverIn this exercise, you will set up a cross between the wild-type and the tan strains of S. fimicola. Then you will analyze the “hybrid” asci to map the tan mutation in the Sordaria chro-mosome. While the tan mutation is frequently used in many teaching laboratories, its linkage group is still unknown (K. McCluskey, personal communication, 2007).

MaterialsSordaria sterile media in Petri dishesWild-type strain of S. fimicolaThe tan mutant strain of S. fimicolaCompound microscope

Sterile toothpicksBurnerLaboratory gas lightersMicroscope slidesCover slipsDissecting needlesScissorsParafilmdH2O

ProcedureObtain a Petri dish containing Sordaria sterile medium. This medium should be inoculated at two places with hyphae taken from the wild-type strain (designated WT) and in two other places with hyphae taken from the tan strain (des-ignated tan) of S. fimicola. You must work close to a lit burner to prevent plate contamination . Write your name on the bottom of the dish.

Fig. 12–7. The life cycle of Sordaria fimicola.

Fig. 12–8. A rosette of asci showing ascospores in a linear array. A squashed perithecium is shown in the right. Reproduced by permission of Carolina Biological Supply Company.


1. With the aid of a sterile toothpick remove a small chunk of the medium containing wild-type mycelia. Place this on your sterile medium at the 9 o’clock posi-tion, then close the lid.

2. Repeat for a second inoculation at the 3 o’clock position.3. In the same manner, transfer mycelia from the tan

mutation to the 12 o’clock and 6 o’clock positions as illustrated in Fig. 12–9.

4. Seal the Petri dish with Parafilm and place it at room temperature. Do not invert the plate.

5. In about 10 to 15 d, mycelia from the two mating strains (WT and tan) will cover the entire medium in the Petri dish. Where the two strains have come in contact, perith-ecia will have formed (Fig.12–10). Some of these perith-ecia contain asci with wild-type and tan ascospores.

6. With a sterile toothpick, pick a number of perith-ecia and place them in a drop of sterile water on a microscope slide. Add a cover slip and press gen-tly with the handle of a dissecting needle to burst the perithecia and spread the asci (Fig. 12-11). Examine the sample first under low and then high power of a compound microscope.

If the asci have four black ascospores at one end and four tan ascospores at the other, these represent meiotic segregation without crossing over between the centrom-ere and the tan gene. However, if crossing over occurs between the centromere and the tan gene, the ascospore arrangement will be altered to one of the following forms (see also Fig.12–11):

t WT t WTt WT t WT

WT t WT tWT or t or WT or t

t WT WT tt WT WT t

WT t t WTWT t t WT

How to Calculate the Map DistanceYou have to calculate the distance between the tan gene and the centromere. In Sordaria, the centromere is considered as a genetic marker.1. Count at least 20 “hybrid” asci.2. Identify crossovers and noncrossovers. Use Fig.

12–12 as a guide. This figure shows the possible ar-rangements of ascospores after the first and second meiotic division.

3. Pool your results with those of your laboratory part-ner’s and complete the table on the following page.

Protocol 12–1How to Maintain Sordaria CulturesSordaria plates can be kept at 4°C (in a refrigerator or cold room) for an extended period of time. However, they must be subcultured every 3 to 4 wk. Extreme caution must be paid not to contaminate the cultures. Under a sterile hood and with the aid of a sterile toothpick, remove a chunk of the agar medium, about the size of a quarter, which contains hyphae. Transfer the chunk to a fresh agar plate, seal it with Parafilm or surgical tape, and leave at room temperature for about 10 d or until the hyphae cover the entire medium and the perith-ecia are formed. Then store the plate in a cold room.

Fig. 12–9. Inoculation of wild-type and tan strains on an agar plate.

Fig. 12–10. Agar plate containing wild-type and tan Sordaria cultures.

Fig. 12–11. Two rosettes of asci showing the various ascospore arrangements in Sordaria fimicola. Black: wild type; white: tan. Reproduced from Mertens and Hammersmith (1998) by permission of Pearson Education, Inc.


Study Questions12–1. How can you differentiate between cells in meta-

phase I and cells in metaphase II?

12–2. Given that broad bean (Vicia faba L.) has 12 chro-mosomes (2n = 12), how many chromosomes will be present in each nucleus at telophase I and prophase II?

12–3. How is genetic variation produced during meiotic division?

12–4. Suppose you have set up a cross between a wild-type and a mutant strain of S. fimicola. After screening several asci, you obtained the following results:

No. asci showing crossing over: 86 No. asci showing no crossing over: 125 Calculate the distance between the mutant gene and

the centromere.

12–5. How can you explain the occurrence of the follow-ing arrangement of ascospores in an ascus?


No. noncrossoverasci (NCO)

No. crossoverasci (CO)

Total(NCO+CO)

% crossing over(CO/total) ×100

Map distance†(%CO/2)

† Note: To convert the percentage of crossing over to map distance, the percentage must be divided by 2 because only four of the eight (i.e., 50%) ascospores in each ascus are the result of a crossing over event (see Fig. 12–12, part B).

Fig. 12–12. Diagrammatic representation of meiotic segregation without crossing over (A) and with crossing over (B) between the tan gene (shown in white) and the centromere. In B, the subscripts indicate the chromatids involved in the crossing over event.



Objectives

• to understand the concept of complementation

• to be able to perform a complementation test in the bacterium Serratia

• to identify the biochemical pathway of prodigiosin synthesis

Exercise 13 Complementation Test

Complementation is the expression of the wild-type phenotype by an organ-ism that contains two recessive mutant genes.

What is a Complementation Test?A complementation test is a test to elucidate whether two recessive mutations, which express the same phenotype, are mutations of the same gene.

How Does Complementation Occur?In Arabidopsis thaliana, close to 100 recessive mutations have been identi-fied that affect the shape of the trichomes, distorted 1 (dis1-1) and distorted 2 (dis2-1) being two of them. These two mutations express almost an identical phenotype (see Fig.11–5). The phenotypic similarity may lead to the conclu-sion that dis1-1 and dis2-1 are essentially the same mutation. To verify this conclusion, a cross between the two mutant plants and a study of the resulting phenotype of the progeny must be made. This cross is called a complementa-tion test, as illustrated in Fig. 13–1. In complementation tests there are two possibilities (see Fig. 13–1).

Possibility 1: If dis1-1 is allelic to dis2-1, that is, they are mutations of the same gene, in the F1 plants, one copy from each allele will be present. Thus, all F1 plants will show distorted trichomes. In such a scenario, no complementation occurs because the two mutations arise from the same gene.Possibility 2: If dis1-1 is not allelic to dis2-1, that is, they are mutations in different genes, in the F1 plants, the wild-type allele of each mutation will mask the mutant expression of its recessive allele; therefore, the plants will show wild-type trichomes. In this case, complementation does occur because each mutation is in a separate gene. In reality, dis1-1 and dis2-1 are mutations of separate genes and both are located on chromosome 1 of A. thaliana (Le et al., 2003; El-Assal et al., 2004).

Genetic Control of Biochemical ReactionsAs early as 1942, Beadle and Tatum showed that the production of enzymes is under genetic control and that enzymes control the biochemical pathways of certain phe-notypes by converting one intermediate to another and to the end product.

Basically, a biochemical pathway begins with a precursor molecule. A precursor is a raw material which is modified by the various chemical reac-tions of the pathway to produce an end product. Often, there are a number of intermediate steps which are catalyzed by different enzymes and which produce intermediate compounds before the end product is formed. Enzymes which catalyze each step of the biochemical pathway are coded for by specific genes. The presence of the appropriate enzyme allows a specific step of the pathway to be completed. Figure 13–2 illustrates a hypothetical biochemical pathway.

The method for studying the genetic control of biochemical pathways is based on “cross-feeding” of two mutant strains. A mutant (auxotroph) strain is unable to complete the pathway because of a mutation in at least one of the


genes coding for the enzyme which carries out the step in the pathway. On the contrary, a wild-type (prototroph) strain is able to produce the end product because all the genes in the pathway are functional.

Figure 13–3 shows a biochemical pathway with three steps; the precursor is converted to the end product through three enzyme catalyzed reactions which produce the inter-mediates A and B. Each of the enzymes I, II, and III is coded for by a specific gene: a, b, and c, respectively. If one of these genes is mutated, an auxotroph strain is generated because the mutated gene produces a nonfunctional enzyme which blocks the conversion from one intermediate to the next. For example, in auxotroph strain No. 1 (see Fig.13–3), gene c is mutated (c); therefore, it codes for an altered enzyme (III). The altered enzyme cannot catalyze the change from intermediate B to the end product. As a result, large quan-tities of intermediate (B) are produced and secreted by the cells into the nutrient medium. A similar explanation applies to auxotroph strain No. 2 (see Fig.13–3). If strain No. 1, which secretes the intermediate (B) and strain No. 2, which secretes the intermediate (A), are grown in the same medium, the former will be able to feed (rescue) the latter with the product (B). Since in strain No. 2 the enzyme III is

functioning normally, it will convert the intermediate (B) to the end product; that is, the biochemical pathway of strain No. 2 is completed. Note that strain No. 2 cannot feed strain No. 1 because the latter has product A, and it has already been converted to (B). In a strict genetic term, strain No. 1 complements strain No. 2. In other words, genes b and c are nonallelic. Note that in this example if the same gene (gene b, for example) were mutated in both strains, only intermediate (A) would have been accumulated, and nei-ther strain could rescue the other. Under this assumption no complementation would have occurred.

From this experiment rule Number 1 can be extracted: any strain with a mutation later in the pathway comple-ments (rescues) a strain with a mutation earlier in the pathway . The reverse cannot occur.

In certain reactions biochemical pathways may be branched, and the pathway may start with either one or two precursors. In Fig. 13–4, the precursor of the wild-type strain is split by a single enzyme to produce intermediates A and D. Intermediate A is converted to B and then to C by two separate enzymes. Similarly, intermediate D is converted to E and then to F. The intermediates C and F are joined by a single enzyme to produce intermediate G; then G is con-

Fig. 13–1. A complementa-tion test between dis1-1 and dis2-1 mutations in Arabidopsis thaliana. For details see text.

Fig. 13–2. A hypothetical bio-chemical pathway showing the intermediates (A, B, and C), the genes, and the enzymes that catalyze the reaction.


Fig. 13–3. Two auxotrophic strains re-sulting from blocks in a linear biochemi-cal pathway.

Fig. 13–4. Two auxotrophic strains resulting from blocks in branched bio-chemical pathways.


verted to the end product by the action of another enzyme. In branched pathways, if there is a block between, for example, the intermediates B and C due to the presence of the mutant gene (c) (Fig.13–4, auxotroph stain No. 1), there would be built up of intermediates (B) as well as F. (B) is accumulated in the cells because the enzyme (III) that converts (B) to C is defective (note that C will be missing from this pathway). On the other hand, F is accumulated because the enzyme VI catalyzes the reaction only if F and C both are present in the same time in the pathway. This strain thus secretes both (B) and F intermediates. A similar explanation applies to the auxotroph strain No. 2 shown in Fig. 13–4.

If these two auxotroph strains are grown in the same medium, each strain will feed the other: strain No. 1 will supply strain No. 2 with the intermediate F and strain No. 2 will supply strain No. 1 with the intermediate C. As a result, in both strains the end product will be formed. This phenomenon is called mutual feeding.

From this experiment rule Number 2 can be extracted: in branched pathways mutual feeding will occur between strains with blocks in opposite arms of the pathway . However, there would be no mutual feeding if blocks occur in the same arm of the two strains. The strain which has the block earlier in the pathway will be fed by the strain which has the block later in the pathway .

Biosynthesis of Prodigiosin in Serratia marcescens

Serratia marcescens (hereafter called Serratia) is a motile, Gram negative bacillus which is a facultative anaer-obe. It occurs naturally in soil and water as well as the intestine. It produces a natural product, a red pigment called prodigiosin, which contains 4-methoxy-2,2¢-bipyrrole (Fig.13–5). The biosynthesis of prodigiosin is a bifurcated pathway, in which monopyrrole and bipyrrole precursors are synthesized separately and then coupled to form the pigment (the biochemical pathway is not shown for the sake of the exercise).

By the aid of UV radiation hundreds of mutants have been isolated that block the biochemical pathway of pro-digiosin, resulting in the production of intermediates with different colors. These mutants fall within at least 10 com-plementation groups. Table 13–1 lists selected mutants of Serratia and their coloration when grown on PG medium (see Materials).

Description of the Exercise†The purpose of the exercise is to have visible proof of cross-feeding using different mutants of Serratia in order to identify the complementation groups and to find the correct pathway of the prodigiosin biosynthesis. You will use a wild-type and a number of mutant strains. In the wild-type strain, the prodigiosin pathway is complete; therefore, it produces a dark red pigment. Each mutant strain carries a defective enzyme which blocks the production of prodigiosin, causing over-expression of an intermediate product. The location of the defect in the pathway is “unknown” to students.

MaterialsWild-type and mutant Serratia stocks on peptone glycerol (PG)

agar platesPG agar plates:

Per 1000 mL of dH2O5 g Peptone 23 g Agar 10 mL Glycerol AutoclaveDistribute in sterile Petri dishes

BurnerLaboratory gas lightersInoculating loopsEthanolGlovesParafilmScissorsPaper towels

ProcedureCaution: Serratia is pathogenic and may cause eye infec-tion as well as urinary and respiratory tract infections. Wear gloves when working with this bacterium. 1. Study the plates of wild-type and mutant Serratia

strains. These are the controls; do not remove the lid. Look at the color of the streaks and record them in your report sheet.

2. Obtain the required number of fresh PG plates.3. Unless otherwise instructed, you should plate the mu-

N N C N

H H H

CH3

OCH3 (CH2)4CH3

Fig. 13–5. Prodigiosin.

† Parts of this exercise were adopted from Tested Studies for Laboratory Teaching, Volume 14, (C.A. Goldman, Editor), with slight modifications to fit the format of this laboratory manual.

Table 13–1. Wild type and selected mutants of Serratia marcescens showing the color of the inter-mediate product. The wild-type strain is mentioned for comparison of color.

Strain Color OF BrownWF Dark orange933 Light pink7-6 Orange

XII-20 Purple3-14 Pink319 White

D1 (wild type) Dark red


tant strains next to each other in pair-wise combina-tions as follows: OF with WF, OF with 7-6, OF with XII-20, OF with 933, WF with 7-6, WF with XII-20, WF with 933, 7-6 with XII-20, 7-6 with 933, and XII-20 with 933. You must work close to a lit burner to prevent contamination.

4. Flame the inoculating loop in the burner flame.5. Lift the lid of the stock plate and cool the loop by gen-

tly pressing it against the sterile lid for a few seconds.6. Scrape the loop over the surface of the desired mutant

strain (e.g., OF), then quickly transfer the inoculum to the fresh PG plate and streak it in a “V” pattern as il-lustrated in Fig. 13–6.

7. Reflame the loop and streak another mutant strain (e.g., WF) in the same plate in the same manner. Note that the two V’s should be close (2–3 mm apart) but not touching. On the bottom of the plate, write down your name, date, and the name of the mutants you used.

8. Repeat Steps 4 to 7 for each pair of mutants you choose.9. Seal the plates with Parafilm and incubate them, facing

up, at room temperature for 3 to 4 d. During that time, the bacteria will grow, and the two V-shaped streaks will be visible in each plate.

10. Examine your plates at the time specified by the labora-tory instructor. Look at the inner edges of the Vs which are next to each other. If a dark red pigment (indica-tive of the end product) is visible, that strain is the recipient and the other strain is the donor. In other words, the two mutations are nonallelic; therefore, they complement each other. If the red pigment is vis-ible on the edges of both Vs, think about a branched pathway, as described above.

11. Record your observations and complete the follow-ing table.

Donor strains Recipient strains

12. On the basis of your results in the table, determine the correct steps of the biochemical pathway of prodigi-osin synthesis. The laboratory instructor will discuss the results in detail at the end of the experiment.

13. When finished with the experiment, discard the plates in a biohazard bag.

Protocol 13-1How to Maintain Serratia CulturesThe most convenient way to maintain Serratia stocks is to culture them in brain-heart infusion broth. This is a liquid medium provided by Fisher Scientific, Pittsburgh, PA. Under a sterile hood and close to a lit burner, pipette 2 mL of the broth in sterile culture tubes, and by the aid of an inoculating loop transfer the bacterial cultures from the old tubes to the new ones. If the cultures are on agar plates, scrape the loop over the surface of the culture and trans-fer the inoculum to the liquid medium. Note that Serratia strains show color only when they are cultured on agar plates. If you order Serratia stocks from the American Type Culture Collection (Manassas, VA), follow the instructions of the supplier.

Study Questions13–1. Define complementation test.

13–2. How reliable is the result of a complementation test between one recessive and one dominant mutation? Explain your answer.

13–3. Alleles of the same mutation do not necessarily express an identical phenotype. There is a sex-linked mutation in Drosophila melanogaster which has several alleles, each of which expresses a different phenotype. Can you name that mutation?

Fig. 13–6. Inoculation of Serratia on an agar plate.


13–4. In Serratia marcescens, strain AB rescues strain SW but is not rescued by SW. Strain EM rescues both AB and SW but is not rescued by either AB or SW. Strains AB, SW, and EM rescue strain TS and are rescued by TS. Draw the biochemical pathway to il-lustrate these results.

13–5. In Arabidopsis thaliana, you are going to character-ize mutations which affect the shape of the trichomes. After performing complementation tests between phenotypically identical mutations, you obtained the results shown in the table below. How many complementation groups are present among these nine mutations?

Mutations → A B C D E F G H I↓A + + m + + + + + B m + + + m + +C + + + m + +D + + + + + E + + + +F + + +G + +H +I Note: + indicates the wild-type phenotype; m indicates the mutant phe-

notype.



Molecular MarkersMapping the Genome of Arabidopsis

Objectives

• to understand the difference between linkage mapping and molecular mapping

• to understand the concept of polymorphism at the molecular level

• to learn the technique of gene mapping using DNA markers

Exercise 14Methods of Gene Mapping

In genetics, mapping is a process through which the location of a gene on a particular chromosome can be determined. There are three methods to map a gene: genetic mapping, cytological mapping, and physical mapping. Genetic mapping or linkage mapping is based on the calculation of the frequency of crossing over between the genes. In this technique, the distance between two genes is considered as being equal to the percentage of crossing over between them. Cytological mapping is restricted to the organisms that have polytene chromosomes and that the banding patterns of the chromosomes are visible under microscope (see Fig.1–5). In cytological mapping, the association of a band with the mutation of interest is sought. In physical mapping, the genes are mapped with respect to molecular markers which are located in the genome. It must be emphasized that there is no direct correlation between genetic, cyto-logical, and physical maps because the parameters used to construct the maps differ in each technique.

How Accurate Is a Genetic Map?One of the major factors that affect the accuracy of genetic maps is the occur-rence of double crossing over between distantly linked genes. Consider the parental combination of two linked genes A and B and their recessive alleles (a and b, respectively) located at either end of the same chromosome (Fig.14–1). As Fig. 14–1A illustrates, a single crossing over between the chromatids 2 and 3, recombines the two allelic pairs resulting in nonparental (or recombinant) combinations with one dominant and one recessive allele. Figure 14–1B shows that a double crossing over between the same chromatids does not result in recombination of the alleles because the genes are located outside the segment of the chromatids where the double crossing over has taken place; therefore, only parental-looking progeny result. Exactly the same progeny would result if there was no crossing over between the two chromatids in that region. Under this condition, the double crossing over events will remain undetected, and the estimate of map distance between genes A and B will be inaccurate.

In conclusion, in the absence of double crossing over between genes, there would be a direct linear relationship between the frequency of recombi-nation and the genetic map. In practice, such a relationship is seen only when genes are closely linked. As genes are farther apart, the chance of double crossing over between them will increase, which will lead to an error in as-signing a map position to the genes. This difficulty can largely be overcome by identifying molecular markers and using them as landmarks for gene mapping.

The Use of Molecular Markers in Gene MappingMolecular markers are short DNA sequences that are scattered throughout the genome of almost all organisms. They can be detected by molecular techniques such as PCR and gel electrophoresis. Molecular markers include, but are not


limited to, microsatellites, cleaved ampli-fied polymorphic sequences (CAPS) and single nucleotide polymorphisms (SNP). Molecular markers vary from strain to strain in the composition, length (number of nucleotides), or location in the genome, and these characteristics can be used as landmarks in gene mapping. We recom-mend that the reader consults a genetic textbook for details about molecular mark-ers. In this exercise, we will concentrate on microsatellites only.

Microsatellites are tandem repeats of one or a few nucleotides dispersed throughout the genome (Fig.14–2). They are more abundant in animals than in plants. The most common plant microsatellite is (A)n, then (AT)n, (GA)n, and (CA)n, where n indicates the number of tandem repeats. In Arabidopsis, like any other plants, the number of tandem repeats of a given mic-rosatellite may vary in different ecotypes such as Columbia (Col), Landsberg erecta (Ler), or Wassilewskija (Ws). It has been shown that the microsatellite (AT) repeats 20 times in Col and only twice in Ler. Because of this variability, they are known as simple sequence length polymorphism (SSLP).

Variation in the number of tandem repeats of microsatellites can be detected by agarose gel electrophoresis. For exam-ple, if the same microsatellite is amplified in two different ecotypes of Arabidopsis and then electrophoresed, the sample with more repeats will run slower than the sample with less repeats. Furthermore, a heterozygous state can be simulated by mixing purified DNA from the two samples. In this case, following PCR amplification and electrophoresis, two bands will appear on the gel, each aligning with its corre-sponding homozygous sample (Fig.14–3). This means that molecular markers are codominant; that is, heterozygotes can be identified by virtue of containing both alleles. It must be emphasized that micro-satellites do not show a visible phenotype in organisms; they can be identified only if DNA is extracted from the organisms, amplified, and electrophoresed.

The use of PCR-based markers, such as microsatellites, speeds up the process of mapping genes. Consequently, a result can be obtained within a day. In this exercise, you will use microsatellites to map the ttg gene in Arabidopsis.

Fig. 14–1. Recombination between two linked genes. In panel A, a sin-gle crossing over results in parental and recombinant types. In panel B, the double crossing over does not produce detectable recombinants.

Fig. 14–2. A tandem repeat of AT within a genome-specific sequence. In this illustration, AT is repeated eight times.


SSLP Mapping in ArabidopsisIn an attempt to map the gene of interest with respect to microsatellites, it is necessary to consider the fol-lowing parameters:

the ecotype of the mutant plant• the ecotype of a wild-type plant that shows extensive poly-• morphism in microsatellite repeat number compared with the mutant plantgeneration of a mapping population using the two plants men-• tioned aboveprimers; each primer pair must define one molecular marker• a PCR program to amplify the target DNA•

The objective of the SSLP mapping is to find a molecu-lar marker (i.e., a microsatellite) that shows a tight linkage with the mutation of interest. If this requirement is met, then the first step towards cloning the gene of interest has been taken.

How Is a Gene Mapped?Read this passage attentively; you will perform the same experiment as the virtual student did.

A student’s project was to map the mutation m in the chromosome 1 of Arabidopsis thaliana with respect to the four microsatellites A, B, C, and D. The student knew that the mutation had been generated in the Ler ecotype. Therefore, she crossed the mutant plant with the wild-type (WT) plant of the polymorphic ecotype Col, using Col as the male parent. Then she collected the F1 seeds, planted them, and allowed them to self to obtain the F2 generation. She called the F2 generation the mapping population. According to the Mendelian laws (see Exercise 11), the F2 generation should contain 75% wild-type and 25% mutant plants, as illustrated below.

However, if in the F1 plants a crossing over takes place between the mutant gene m and the microsatellites, recombinant types will also be generated as shown in the following illustration.

Since crossing over occurs during meiosis in either eggs, pollen grains, or both, upon selfing the F1 plants, the phe-notypic segregation in the F2 generation will still be 75% wild-type and 25% mutant, but all mutant plants (disregard-ing wild-type) will be either heterozygous Col/Ler for the part of the chromosome that carries the microsatellites or homozygous Col for the same part of the chromosome. Note that the latter is generated if both F1 parents are crossovers; therefore, it is considered as the product of a crossing over. These events are illustrated on the top of the next page.

The student picked a large number of mutant plants from the F2 generation and prepared DNA from each. The mutants she collected could have been the product of a crossing over or noncrossing over event, as illustrated below.

Fig. 14–3. Simulation of an ethidium bromide-stained agarose gel showing polymorphism in the same micro-satellite. For details see the text.


Then she generated primers for the microsatellites she had chosen and wrote a PCR program to amplify the mic-rosatellites. She set up four sets of PCR reactions, each consisting of the DNA of 21 mutant plants and a pair of primers for the relevant microsatellite. After amplifying the microsatellites, she electrophoresed the amplified DNA and identified the Col, Ler, and Col/Ler bands on the gel. If the DNA had been extracted from nonrecombinant mutant plants such as the one illustrated below, she would have seen one band on the gel representing the Ler ecotype.

However, if the DNA had been extracted from recom-binant mutant plants, as the ones shown below, she would have seen one or two bands depending on the recombinant plant used. The DNA extracted from Col/Ler plants would show two bands on the gel, one representing the Col and the other the Ler ecotype. In the case where DNA had been extracted from Col plants, only one band would have been visible, representing the Col ecotype.

The student then summarized her results as shown below. Note that in the student’s results, “Het” stands for “heterozygotes” and refers to Col/Ler bands.

Plant No.Molecular markers

A B C D1 Ler Ler Ler Ler2 Het Het Ler Ler3 Het Het Ler Ler4 Col Het Ler Ler5 Het Het Col Ler6 – – – –7 – – – –8 Het Ler Het Ler9 Col Col Het Ler10 – Ler – Ler11 – Ler – Ler12 – Het Ler Ler13 Ler Het Col Ler14 Ler – Het Ler15 Het – Het Ler16 Het Ler Ler Ler17 Het – Ler Ler18 Het Het – Ler19 Col Col Ler –20 – Ler – Ler21 – Ler Het –

Summary: Col 3 2 2 0Ler 3 7 8 17Het 8 7 5 0

Unscorable 7 5 6 4

As shown in the summary data, no recombinants were found when marker D was used; furthermore, all the plants tested with this marker displayed the Ler ecotype. However, the number of recombinants increased when markers C, B, and A were employed. The student, therefore, concluded that there is a tight linkage between the mutation m and the Marker D. Since the location of the Marker D on chro-mosome 1 is known, she assigned a map position to the


mutation. An illustrative representation of the student’s results is shown in Fig. 14–4.

Mapping of the ttg Gene in Arabidopsis1. DNA AmplificationIn Exercise 11, you screened the F2 population of a cross between ttg and wild-type plants. Recall that the ttg muta-tion had been generated in the Ws ecotype, whereas the wild type was Col. The F2 population you generated was the mapping population. The objective of this exercise is to map the ttg gene with respect to two microsatellites whose location is “unknown” to students. They are called nga151 and SO262. These markers are polymorphic in the Col and Ws ecotypes. The PCR product size of nga151 is 150 bp in Col and 102 bp in Ws. The product size of SO262 is 145 bp in Col and 151 bp in Ws. In this exercise, the only information you will be provided with is that ttg is located on chromosome 5 of Arabidopsis thaliana.

Materialsttg/ttg plants from Exercise 11Col DNA, concentration of 10 ng/µL Ws DNA, concentration of 10 ng/µLForward primer nga151* (nga151-F), concentration of 5 µM

Sequence: 5¢ CAGTCTAAAAGCGAGAGTATGATG 3¢Reverse primer nga151* (nga151-R), concentration of 5 µM

Sequence: 5¢ GTTTTGGGAAGTTTTGCTGG 3¢Forward primer SO262* (SO262-F), concentration of 5 µM

Sequence: 5¢ ATCATCTGCCCATGGTTTTT 3¢Reverse primer SO262* (SO262-R), concentration of 5 µM

Sequence: 5¢ TTGCTTTTTGGTTATATTCGGA 3¢(*The primers are named after the relevant microsatellite

for simplicity.)0.25 M sodium hydroxide:

Per 100 mL of dH2O1 g Sodium hydroxide

0.25 M hydrochloric acid:Prepare 1 M hydrochloric acid as described in Exercise 1;

then dilute it four times: to 1 volume of the stock add 3 volumes of dH2O.

0.5 M Tris:Per 100 mL of dH2O6.057 g Tris Adjust the pH to 8.0 with concentrated hydrochloric acidAutoclave

GelatinTriton X-100 (t-octylphenoxypolyethoxyethanol; Sigma,

St. Louis, MO)Taq DNA polymerase (keep on ice always), supplied

with magnesium chloride solution.Note: For this exercise do not use the buffer supplied by

the manufacturer with the Taq DNA polymerase, instead, use the SSLP buffer described below.

10´ SSLP buffer:Per 1000 mL of dH2O3.725 g Potassium chloride 1.211 g Tris Adjust pH to 9.0 with hydrochloric acid0.1 g Gelatin Triton X-100 (t-octylphenoxypolyethoxyethanol;

Sigma, St. Louis, MO) 1 mLAutoclave

dNTP (working solution, concentration of 2 mM), as described in Exercise 5

dH2OThermal cyclerHeat block (adjusted to 100°C)Microtubes, 0.6 mLMicrotube racksPCR tubesPCR tube racksMicropipettorsTipsGlovesIce

ProcedureThis exercise will be performed in two groups. Each group will use the homozygous ttg plants obtained from Exercise 11 and one pair of the two pairs of primers. Furthermore, each group will use control DNA extracted from the Col and Ws ecotypes.

Both groups1. Program the thermal cycler for amplification. Name

the program “SSLP” [G. Koliantz and D.B. Szymanski (2001), unpublished data]. The program is shown in Table 14–1.

2. Place two or three small leaves of each ttg/ttg plant in each of five PCR tubes and add 40 µL of 0.25 M sodium hydroxide to each tube. Then with the aid of a pipette tip smash the leaves as much as possible.

3. Heat the samples (the tubes from Step 2, above) in a 100°C heat block for 1 min, then place the tubes on ice for 3 min.

4. While the tubes are on ice, add 40 µL of 0.25 M hy-

Fig. 14–4. Relative location of microsatellites A, B, C, and D with respect to the mutation m, as de-scribed in the text.

Table 14–1. The SSLP program†.

Step No.

Block temperature Time Operation

°C1 94 3 min Enzyme activation–denaturing2 94 15 s Denaturing 3 55‡ 15 s Annealing4 72 30 s Extending5 35 times to 2 Cycle repeating6 8 0 h, 0 min, 0 s Holding at 8°C indefinitely7 End End of the program

† Duration of amplification approximately 1 h 35 min.‡ Annealing temperature must be 5°C lower than the melting temperature (Tm) of the

primers. See manufacturer’s information about the Tm.


drochloric acid and 20 µL of 0.5 M Tris to each tube. Pipette up and down several times to mix the contents of the tube. Be sure that the leaf pieces are completely immersed in the reagents. Leave the tubes at room temperature for 2 to 3 min and then on ice. This is the ttg DNA; you will need it later in Steps 11 and 17. Proceed to Step 5.

5. Obtain 7 fresh PCR tubes.

Group 1: follow Steps 6 to 11Group 2: follow Steps 12 to 17

Group 16. Mark one tube Col151 (for Col DNA, marker nga151)

and another tube Ws151 (for Ws DNA, marker nga151). Mark the remaining five tubes ttg151 (for ttg DNA, marker nga151). Place these tubes on ice throughout the experiment. You will need them in Steps 8 to11.

7. In a 0.6-mL microtube, add the following reagents in the order shown below:

80 µL dH2O 20 µL 10´ SSLP buffer 18 µL Magnesium chloride 20 µL dNTP 25 µL Forward primer (nga151-F) 25 µL Reverse primer (nga151-R) 25 µL 2 µL Taq DNA polymerase Mix the tube well, centrifuge briefly, and place on ice.8. Distribute 19 µL of the above reaction in each of the

seven PCR tubes in Step 6, above.9. Add 1 µL of Col DNA to the PCR tube marked Col151.10. Add 1 µL of Ws DNA to the PCR tube marked Ws151.11. Add 1 µL of each ttg DNA sample (prepared in Step 4,

above) to each of the five PCR tubes marked ttg151. Close the lid, mix gently, centrifuge briefly, and leave the tubes on ice. Wait until Group 2 prepares its reac-tions, then proceed to Step 18.

Group 212. Mark one tube Col262 (for Col DNA, marker SO262)

and another tube Ws262 (for Ws DNA, marker SO262). Mark the remaining five tubes ttg262 (for ttg DNA, marker SO262). Place these tubes on ice throughout the experiment. You will need them in Steps 14 to17.

13. In a 0.6-mL microtube, add the following reagents in the order shown below:

80 µL dH2O 20 µL 10´ SSLP buffer 18 µL Magnesium chloride 20 µL dNTP 25 µL Forward primer (SO262-F) 25 µL Reverse primer (SO262-R) 2 µL Taq DNA polymerase Mix the tube well, centrifuge briefly, and place on ice14. Distribute 19 µL of the above reaction in each of the

seven PCR tubes in Step 12, above.15. Add 1 µL of Col DNA to the PCR tube marked Col262.16. Add 1 µL of Ws DNA to the PCR tube marked Ws262.17. Add 1 µL of each ttg DNA sample (prepared in Step 4,

above) to each of the five PCR tubes marked ttg262. Close the lid, mix gently, centrifuge briefly, and leave the tubes on ice. Wait until Group 1 prepares its reac-tions, then proceed to Step 18.

Both groups18. Place the tubes in the thermal cycler. Close the lid and

start the program. Amplification will take about 1 h 35 min. The amplified DNA can be stored in a −20°C freezer until needed.

2. Gel Electrophoresis of the PCR ProductTo find out whether the DNA samples you amplified in the previous section are derived from the recombinant or non-recombinant ttg plants, you have to electrophorese the PCR products on an agarose gel. By comparing the size of the DNA bands obtained from the mutant plants with the size of the DNA bands of the controls, you should be able to identify the marker which shows a tight linkage with the ttg gene.

MaterialsPCR products from the Exercise above1´ TBE buffer, as described in Exercise 3Gel-loading dye, as described in Exercise 2Ethidium bromide solution, as described in Exercise 2High resolution agarose, e.g. MetaPhor (Biowhittaker Molecular

Applications, Rockland, ME) or equivalentAgarose gel electrophoresis unit, as described in Exercise 2dH2OIceMicropipettorsTipsMicrotube racksMicrocentrifugeGel-staining dishes, as neededMicrowave oven200 mL flasksUV protective glasses or shieldGlovesAgarose gel photo documentation system, as described in Exercise 2

Procedure1. Prepare 3% gel using a high resolution agarose in 1´

TBE buffer. Do not add ethidium bromide to the melted agarose (see Step 6, below). At the same time, set up the electrophoresis unit as described in Exercise 2.

2. Place the tubes containing the PCR products on ice.3. While the tubes are on ice, add 4 µL of gel-loading

dye to each tube. Use a different pipette tip for each sample. Mix well by pipetting the contents up and down several times.

4. Load the gel from left to right with your samples as indicated below:

Group 1: 4 µL of DNA ladder 10 µL of the content of the tube marked Col151 10 µL of the content of the tube marked Ws151 10 µL of the content of each tube marked ttg151Group 2: 4 µL of DNA ladder 10 µL of the content of the tube marked Col262 10 µL of the content of the tube marked Ws262 10 µL of the content of each tube marked ttg262

5. Electrophorese at 80 V for 45 min.


6. Wash the gel thoroughly with distilled water for 2 min, then stain with ethidium bromide as described in Exercise 2. Washing the gel is required if it is pre-pared with the MetaPhor agarose .

7. View the gel on a UV transilluminator, identify the bands, and photograph the gel.

8. On the photograph, examine each lane carefully; iden-tify Col or Ws bands in lanes loaded with Col or Ws DNA, respectively.

9. On the photograph, check the lanes loaded with ttg DNA. Identify the lanes with one band and those with two bands. Lanes showing double bands or a single band representing the Col DNA indicate the occur-rence of crossing over between the ttg gene and the molecular marker used in those samples. Lanes show-ing a single band representing the Ws DNA indicate no crossing over. Identify which group’s gel shows fewer lanes indicative of recombinational events. The marker used by that group is closely linked to the ttg gene.

10. Show your results to the laboratory instructor. At the end of this exercise, the instructor may disclose the physical map of the molecular markers.

Figure 14–5 shows the photograph of an ethidium bromide-stained MetaPhor agarose gel of PCR-amplified microsatellites that were used to map the ttg gene. The identity of the microsatellites is “unknown” to students.

DiscussionAs mentioned in Exercise 11, if two given loci are unlinked; that is, they are located on different chromosomes, they segregate independently. In mapping the ttg gene, if you choose a DNA marker that is not located on chromosome 5, after PCR amplification and gel electrophoresis, you should see a 50%-to-50% segregation of the bands representing Col and Ws ecotypes (this may also occur when the marker is on chromosome 5 but not close to the ttg gene). However, if the segregation significantly deviates from this ratio, then there is strong evidence of linkage between the ttg gene and the DNA marker.

Table 14–2 lists selected SSLP markers and their map position on the chromosomes of Arabidopsis thaliana.

Study Questions14–1. Define microsatellites.

14–2. What is the difference between a genetic map and a physical map? Which one has a higher accuracy? Explain why.

14–3. Explain how double crossing over may cause a mis-interpretation of a given mapping data.

14–4. Look at Fig. 14–1B. If there had been a gene (for example, gene C) between genes A and B in the seg-ment of the chromatids where the double crossing over took place, what would have been the genotype of the resulting progeny?

14–5. What is a mapping population? How is it generated?

Fig. 14–5. MetaPhor agarose gel electrophoresis of PCR-amplified microsatellites that show linkage (top) or crossing over (bottom) with the ttg gene. The iden-tity of the microsatellites is “unknown” to students.

Table 14–2. A list of widely used simple sequence length polymorphism (SSLP) markers on the chromo-somes of Arabidopsis thaliana.

ChromosomeSSLP

markerMap

position

Flanking sequences known?

1 nga59 2.9 Yesnga248 42.17 Yesnga128 83.32 Yesnga111 115.55 Yes

2 nga1145 9.6 YesMI310 18.6 Yesnga1126 50.65 Yesnag168 73.77 Yes

3 nga32 5.87 Yesnga162 20.56 YesGAPAB 43.77 Yesnga112 87.88 Yes

4 nga8 26.56 Yesnga1139 83.41 Yes nga1107 104.73 Yes

5

nga225 14.31 Yesca72 29.6 Yesnga76 68.4 Yesnga129 105.41 Yes


Objectives




Exercise 15

Objectives

• to understand the concept of population genetics

• to learn the Hardy–Weinberg law of genetic equilibrium

• to be able to calculate the frequency of alleles and genotypes in a population

• to understand the concept of natural selection and genetic drift

Population Genetics How Changes Occur within a Population


In Exercise 11, you investigated the segregation of the ttg gene in the F2 popula-tion of Arabidopsis. Also, you studied the inheritance of the kernel phenotype in corn and sex-linked traits in Drosophila using different alleles. In those experi-ments, the focus was on the inheritance of one or two traits controlled by one or two genes. In population genetics, the focus changes from one or two genes to a gene pool; that is, the total of alleles of every gene possessed by the reproduc-tive members of the population. Such a population is referred to as a Mendelian population. In genetics, a population is defined as a group of individuals of the same species that occupy a geographic area and are able to interbreed. Population genetics further attempts to describe how and why the frequency of the alleles, which control a trait, changes over time.

In the 1920s and 1930s, attempts were made to describe the genetic compo-sition of populations and to explain the causes of genotypic variations within and between populations. These ideas open the way to interpret the Darwinian evolution in the light of the Mendelian laws. Thus, the discipline of population genetics was created. Population genetics was developed mainly by Sir Ronald Fisher, Sewall Wright, and John Burdon Sanderson Haldane.

The Hardy–Weinberg EquilibriumTo be able to understand the Hardy–Weinberg equilibrium, it is necessary to briefly review the concept of frequency and the way it is calculated. Frequency indicates how often an event occurs in a category and is expressed in percent or proportion. Genotypic frequency of recessive phenotypes can be directly calculated in the population because there is no alternative genotype for a recessive phenotype. But because homozygous dominant and heterozygous genotypes are phenotypically indistinguishable (except in cases when semi-dominance is involved), the frequency of each genotype must be calculated on the assumption that dominant and recessive alleles combine on the basis of law of probability.

In 1908 and 1909, Godfrey H. Hardy and Wilhelm Weinberg, respectively, developed a method to calculate genotype frequencies on the basis of allele frequencies in a population whose members have a diploid (2n) genome and reproduce sexually (i.e., fusion of male and female gametes). The method is based on five assumptions with regard to the gene of interest.

1. The population, mentioned above, is very large so that allele frequencies will remain unchanged because of random sampling.

2. Mating between members of the population is random; that is, mating takes place without regard to genotype or phenotype.

3. No mutation occurs in the gene of interest.4. There is no selection for a genotype in the form of phenotypic, reproduc-

tive, or survival advantage.5. There is no migration in the population in the form of immigration

or emigration.


Under these assumptions, Hardy and Weinberg proposed that the frequency of alleles and genotypes will remain unchanged and the population would be in equilibrium.

Consider the following example. If p is the frequency of the dominant allele A and q is the frequency of the recessive allele a, then the equation p + q = 1 is true because the fre-quency of both alleles together equals 1 (or 100%). It follows that if the population is in the Hardy–Weinberg equilibrium, then the frequency of the genotypes for each allele can be calculated as shown in Table 15–1. Note that in this table, the terms of the binomial expansion p2(AA) + 2pq(Aa) + q2(aa) = 1 indicate the frequencies of the genotypes.

According to the Hardy–Weinberg law, if a population

consists of p2(AA) + 2pq(Aa) + q2(aa) = 1, then the next generation will have exactly the same frequency of each genotype, and the frequencies will remain unchanged for-ever. Table 15–2 explains how an equilibrium is established in a random mating population.

In actuality, the Hardy–Weinberg law on the structure of populations is conceptual and such populations cannot exist in nature. All living organisms are susceptible to evolutionary forces: mutations continually occur; matings are not always random, and they are closely associated with selection; and migration cannot be avoided in natural populations. The logi-cal denotation of the law is that change in allele frequency is the basis of evolution of living organisms.

Extension of the Hardy–Weinberg LawThe Hardy–Weinberg equation can be extended to calculate genotypic frequencies at X-linked loci. The calculation is slightly different because normal males have only one X chromosome; that is, with respect to sex-linked genes no heterozygous males exist. For an X-linked gene with two alleles XA and Xa, the frequency of the genotypes in the female population is p2(XAXA) + 2pq(XAXa) + q2(XaXa), whereas the frequency in the male population is p(XA) + q(Xa). The calculations are shown in Table 15–3.

Table 15–1. Frequency of the genotypes in a random mating population.

Paternal gametesp(A) q(a)

p(A) p2(AA) pq(Aa)Maternal gametes

q(a) pq(Aa) q2(aa)

Total: p2(AA) + 2pq(Aa) +q2(aa) =1

Table 15–2. Genetic equilibrium in a random mating population.

Mating types Mating frequencyOffspring frequency

AA Aa aaAA × AA p2 × q2 = p4 p4 − −

AA × Aa p2 × 2pq = 2p3qor 4p3q 2p3q 2p3q −

Aa × AA 2pq × p2 = 2p3q

AA × aa p2 × q2 = p2q2

or 2p2q2 − 2p2q2 −aa × AA q2 × p2 = p2q2

Aa × Aa 2pq × 2pq = 4p2q2 p2q2 2p2q2 p2q2

Aa × aa 2pq × q2 = 2pq3

or 4pq3 − 2pq3 2pq3

aa × Aa q2 × 2pq = 2pq3

aa × aa q2 × q2 = q4 − − q4

Total: p4 + 2p3q + p2q2 2p3q + 4p2q2 + 2p3q p2q2 + 2pq2 + q4

Factor out (p2 + 2pq + q2) (p2 + 2pq + q2) p2 (p2 + 2pq + q2)2pq (p2 + 2pq + q2)q2 Since (p2 + 2pq + q2) = 1; therefore: p2 2pq q2

Table 15–3. Frequency of genotypes in sex-linked alleles.

Paternal gametesp(XA) Y or q(Xa) Y

p(XA) p2(XAXA) p(XA) pq(XAXa) p(XA)Maternal gametes

q(Xa) pq(XAXa) q(Xa) q2(XaXa) q(Xa)

Frequency of genotypes among females: Frequency of genotypes among males:

p2(XAXA) + 2pq(XAXa) +q2(XaXa)2p(Xa) + 2q(Xa) = 2[p(XA) + q(Xa)]


Calculating Allelic and Genotypic Frequencies with the Hardy–Weinberg EquationIn most populations, homozygous (AA) and heterozygous (Aa) individuals cannot be distinguished by their pheno-type because of the complete dominance of the allele A over a. However, since homozygous recessive individuals are always aa, the frequency of the recessive allele (q) can be calculated by determining the square root of q2. At this point, the frequency of the dominant allele (p) can also be calculated by means of the equation p + q = 1. Having p and q in hand, the frequency of homozygous (p2) and heterozy-gous (2pq) genotypes can easily be calculated.

Example: If 17.64% of a population is made up of homozygous recessive (aa) genotype, then:

q2 = 0.1764 q = √0.1764 = 0.42 frequency of allele a p = 1 − q = 1 − 0.42 = 0.58 frequency of allele A

Since p2 + 2pq + q2 = 1; therefore,2pq = 2 × 0.58 × 0.42 = 0.4872; that is, 48.72% of the

population is expected to be heterozygous (Aa)p2 = 0.58 × 0.58 = 0.3364; that is, 33.64% of the popula-

tion is expected to be homozygous (AA)

Practice 1In the population of the trees, shown below, assume the allele for the dark color (D) is dominant to the light (d). Calculate the number of homozygous dominant and heterozygous trees. Enter your results in Box 1.

Box 1

Practice 2The caricatures shown below represent changes within populations of sea stars over generations. Without reading the Discussion at the bottom of this page, answer the three questions below.

1. Are the events, shown in Population 1 through 4, oc-curring randomly or selectively? Explain why.

2. Explain the events occurring in these populations by using the terms: heritable phenotypes, survival rate, reproductive advantage, and adaptation.

3. Given that dark-colored sea stars are better fitted to the environment; define fitness.

DiscussionThe above caricature illustrates evolution by means of nat-ural selection, which was proposed by Darwin. It further shows how natural selection affects the genetic makeup of a population.

In Population 1 of the sea stars, there is a variation in the color of the spiny skin: some stars have dark and some have light skin color. These phenotypes are heritable. Light-colored stars are easily found and eaten by the fish, so they have a lower survival rate. On this basis, dark-colored stars have a survival and consequently a reproductive advantage over the light ones. In the consecutive genera-tions, the dark-colored sea stars become more common in the populations; that is, they become better adapted to the environment. The adaptation increases the ability of the dark-colored sea stars to survive and reproduce. This means the genes they possess would become abundant in the suc-cessive generations. In other words, the dark-colored sea stars are better fitted to the environment.

Random Change in Allele Frequencies: Genetic Drift

As discussed above, one of the assumptions of the Hardy–Weinberg law is that the population size is extremely large. In practice, a population cannot be boundlessly large, but it must be large enough to show the expected ratios of crosses and eliminate the chance factors which affect allelic fre-quencies. In small populations, however, gene frequencies will be subject to rapid fluctuations if chance factors such as catastrophic events occur in the population. Consider the following example.

There is a small population of equal number of dark and light-colored fish in a pond (Population 1). Assume that the light color is dominant to the dark. Under this assumption, the genotype of the light-colored fish will be RR or Rr, and the genotype of the dark-colored fish will be rr. The fre-quency of the recessive allele (q) in this population can be calculated by the Hardy–Weinberg binomial expansion p2

+ 2pq + q2 = 1. If 50% of the population is homozygous recessive,

then: q2 = 0.50 q = √.50 = 0.71 frequency of allele r p – q = 1 – 0.71 = 0.29 frequency of allele R

Suppose a catastrophic event hits the pond and kills 50% of the population; by chance all the dark-colored individu-als. The surviving population would be all light- colored fish (Population 2).

After the catastrophe has occurred, the allelic frequency of the light-colored fish will rise from 0.71 to 1.0. This example illustrates how sampling error alters genotypic frequencies in a very short period of time. This phenomenon is referred to as genetic drift. If there had been a larger population of dark and light-colored fish (Population 3), there would have been less possibility that 50% of the fish that die all would have been dark-colored (Population 4). That is, the odds of all of one color of fish dying decreases as the population increases.

Simulation of Population GeneticsAs discussed above, since the Hardy–Weinberg law is conceptual, extreme caution must be taken in choosing the phenotype(s) for this exercise. Phenotypes that can be considered to simulate population genetics in the labora-tory include bean shape or color in plants, eye color in Drosophila, blood groups, ability to role the tongue, or ability to taste the compound phenylthiocarbamide (PTC) in humans. Note that in human populations these characters may not be the subject to selection. Can you explain why?

MaterialsA bag containing 200 white and 200 colored beans, equal in size

and uniform in shape

Procedure1. You will be provided with a bag containing white and

colored beans. Assume that the colored phenotype is dominant over the white. Since two phenotypes are involved, the experiment would be a monohybrid cross (see Exercise 11 for details).

2. Shake the bag well. This step is important because you are going to simulate a random mating population.

3. Without looking inside the bag, draw out two beans at once, record the color, and return the beans into the same bag. On the basis of random combination of gametes, each drawing will result in colored-colored (CC), white-white (cc) or colored-white (Cc) beans. Each student will perform Step 3 ten times to have suf-ficient number of beans for each category. A total of 120 drawings would be adequate for this experiment. On the basis of the assumption in Step 1, pool colored-colored and colored-white beans. Fill in Box 1.

Box 1Genotype

TotalCC and Cc combined ccObserved number

Expected number

4. Calculate the frequency of the genotypes and the al-leles in each phenotypic group in Box 2. Remember to pool the CC and Cc beans.



Box 2

5. Repeat Step 3. Each student will perform this step only twice and will categorize the results in the three phe-notypic groups CC, Cc, and cc. Note that your expected phenotypic ratio must be 1:2:1 because you are sepa-rating the heterozygotes (Cc) from the homozygotes (CC). A total of 24 drawings are sufficient for this step of the exercise. Fill in Box 3.

Box 3Genotype

TotalCC Cc ccObserved number

Expected ratio

Expected number

Does your observation confirm the 1:2:1 distribution of the phenotypes? Probably not! The deviation from the expected 1:2:1 ratio is due to the small population size (24 drawings): genetic drift.

Study Questions 15–1. Look at the five assumptions of the Hardy–Weinberg

law. Do you believe that the Hardy–Weinberg law can be extended to human populations?

15–2. Does selection apply to the populations that repro-duce by self-fertilization? Explain your answer.

15–3. What is a Mendelian population?

15–4. In Drosophila, the frequency of X-linked recessive genotype among females is q2, whereas the fre-quency among males is q. Explain why?

15–5. How does sampling error occur in a population?


Objectives




Appendix 1

Objective

• to learn how to prepare a laboratory report

Preparing a Laboratory Report

Preparing weekly reports from laboratory activities not only will help you, as students, better understand the experiments, it will open the way to learning effective written communication. Therefore, it is extremely important that you keep data on all the results you obtain from experiments, draw conclusions from the results, and be able to report them to the laboratory instructor. However, in writing a report you must assume that a broader audience, not the laboratory instructor only, will read your report. You should expect comments and criti-cism from the readers.

The weekly report must have a consistent format so that you can organize your data accurately. The report must be concise but informative and must con-tain the three sections objective, results, and conclusions as described below.

ObjectiveIn one sentence, describe the purpose of the experiment you performed.

ResultsIn this part of the report, summarize what you observed and found through your experiment. Present your data in a meaningful way and in a logical order. Include figures and/or tables if they can better explain your results. In writing the results, do not describe the procedures you followed, unless they differ from what are written in your laboratory manual. Remember: if the experiment was done by a group of students, clearly state who did what.

ConclusionsWhen you are ready to write the conclusion, stay close to your results but do not repeat the results or show data in this part of the report; instead, explain what the results mean to you and why the experiment was successful. Or, if the intended objective was not reached, give a possible, but logical, explanation for that.

An example of the laboratory report is shown on the following page. However, the recommendations made here are tentative; the most pertinent guide in writing a report will be the instructions given by laboratory instructors of relevant courses.


Genetics Lab Report

Exercise No. 3

Name Maheen Oryan Section 5

Date 10/9/08

ObjectiveTo digest phage lambda DNA with EcoRI and HindIII

ResultsDNA digestion was performed as described in Chapter 3 of the Lab Manual, and diges-tion reactions were electrophoresed on a 2% agarose gel. On the gel lane 1 is the 1 kb DNA ladder. The next two lanes show the DNA digested with EcoRI and HindIII, respectively. The number and the size of the bands are consistent with the data shown in the Lab Manual. In the double digest (lane 4), the bands 2027 and 1904 bp overlap (band #5 from top) but the bands 1584 and 1373 bp are distinguishable (bands #6 and 7 from top). The two small bands 947 and 831 bp appear faint and very close to each other on the gel. In lane 5 the uncut DNA shows a single band.

Conclusions

In the process of gel electrophoresis, two or more restriction fragments of nearly the same size usually overlap and appear as a single band. The fragments can be resolved by changing the concentration of the agarose gel. In my experiment, I could have resolved the 2027 and 1904 bp bands if I had run the DNA fragments on a 1% agarose gel. Furthermore, a better view can be obtained from the 947 and 831 bp bands if the DNA fragments are run to the very end of the gel.


Objectives




Appendix 2 How to Write a Scientific Paper

Objectives

• to learn how to prepare a paper for publication

A scientific paper or article is an exchange of information to tell the scien-tific community about the data you have collected from your research. Most scientific and/or technical papers are written in English, so a proper usage of the English language is required. The paper must be written in a manner that allows the reader to easily understand what your research is about and how you performed it in case he/she wants to repeat your work.

Most scientific papers are arranged in a standardized format and include the following sections:

TitleAbstractIntroductionMaterials and MethodsResultsDiscussionReferences

In preparing the formal report, you must follow the above format. Note that the text of each section must follow immediately below the section heading. Allow a double space when you start with a new heading. Let’s consider the sections of a paper in detail.

TitleYou must choose a proper title for your report. The title must be self-explana-tory, so that the reader is able to understand what the report is about. Titles such as Class Report or Genetics Laboratory Report do not pro-vide information. An example of an informative title would be as follows: The Arabidopsis SPIKE1 Gene Is Required for Normal Cell Shape Control and Tissue Development. Here the title explains that in the plant Arabidopsis, there is a gene called SPIKE1 whose presence is necessary for normal cell shape and tissue development. If the title had said The Effect of the SPIKE1 Gene on Plant Development, then the reader would not have known what plant was used, what the effect was, and which part of the plant was affected by the gene.

In choosing titles, avoid using abbreviations unless they are universal, such as DNA.

In your report, place the title in the center of the first page. Underneath the title, place your name and the date the report was completed.

AbstractIn the Abstract, you must summarize the overall purpose of the experiment and the experimental procedures used, and must end with your conclusion(s). There should be no reference to tables or figures in the Abstract. This section must enable the reader to understand what you have done without having to read the entire report. The Abstract must be limited to one paragraph and must be written in the present tense.


IntroductionIn the Introduction, you must indicate the aim and scope of the paper. You must start the Introduction with a general discussion of the topic and bring the reader to the specific question you are going to answer.

In this section, you must report what other researchers have done in connection to the area of your work in the past and how your present work will help to expand the knowledge of the topic. All supporting information that you have gathered from other sources must be appropriately ref-erenced (see the section entitled How to Cite References). Make the Introduction brief; two or three paragraphs are usually enough.

In the Introduction and other parts of the report, the organ-ism’s name mentioned for the first time should be introduced by the common and the scientific names; for example: the thale cress, Arabidopsis thaliana, or the fruit fly, Drosophila melanogaster. First time you mention the scientific name write it in full: Arabidopsis thaliana. Each subsequent refer-ence to the same organism should include the abbreviation of the genus name but not that of the species name; for example: A. thaliana, not Arabidopsis th. Note that you cannot make genus and/or species names plural; Arabidopsises, A. thalianas, Drosophilas, or D. melanogasters all are incorrect. However, you can make the common name plural; example: thale cresses or fruit flies.

You must type all scientific names in italic font; example: A. thaliana, not A. thaliana.

Materials and MethodsIn this section, you must describe the methods and the techniques you employed to conduct your research. If you have used a method which has been described elsewhere, refer to that text instead. This section must contain enough information so that the readers would be able to repeat the experiment if they wished. No table or figure should be included in this section unless it is absolutely necessary.

The materials you used for the research should not be introduced in list form; example:

1x TAE bufferIncubator EcoRI enzymeLB agar plates

Rather, they must be included in the text as you are describ-ing your methods. Glassware, micropipettors, paper towels, and other common materials need not be mentioned.

It is unacceptable to write the Materials and Methods in active voice, that is, the laboratory manual style. Do not tell the reader how to do an experiment, rather, explain what was done. You must write this section in passive voice and in the past tense. Look at the following examples.

Incorrect Way:Perform DNA and RNA gel blot analyses according to standard protocols (Ausubel

et al., 1994) except for RNA gel blot-ting, for which use ULTRAhyb (Ambion, Austin, TX) for hybridization.

Correct Way:DNA and RNA gel blot analyses were per-formed according to standard protocols (Ausubel et al., 1994) except for RNA gel blotting, for which ULTRAhyb (Ambion, Austin, TX) was used for hybridization.

ResultsIn the Results section, you present your data in detail without trying to explain or draw conclusions from them (“them” not “it”, because the word “data” is plural; singular is “datum”). The best way to present the data is to assemble them in figures and tables; however, they must be accompa-nied by narrative texts. Look at the following example.The products of PCR-amplified DNA sam-

ples are shown in Figure 1.

Figure 1. Agarose gel electrophoresis of PCR products. Lane 1: DNA ladder; lanes 2 to 4: PCR products. Lanes 2 to 4 show DNA bands that migrate

between 3 and 2 kb of the DNA ladder. Samples were run at 70 volts on a 1% aga-rose gel containing ethidium bromide and were visualized under a UV light. The sizes of the DNA ladder are indicated in kilobases.

Always remember the following rules:1. Tables are used to present numerical data. Figures are

used to present pictures, line drawings, etc.2. Choose an appropriate caption (title) for each table and

figure.3. Place the caption at the top of tables but at the bottom

of figures.4. In the text, capitalize the first letter of the words “table”

and “figure”.5. Number tables and figures independently. Example:

Table 1, Table 2, Figure 1, Table 3, Figure 2.

6. Place tables and figures as close as possible to the pas-sage in the text where they are cited for the first time.

7. You may cite tables and/or figures in the beginning of the statement or at the end.


Example 1:In microtubule-free zones of the cell, some actin filaments were associated closely with the plasma membrane (Fig. 8A).

Example 2:In Fig. 8C, three repeating units of alternating microtubule- and actin filament-populated plasma membrane zones are clearly defined within a single protrusion.

The Use of NumeralsThe following rules apply for numeral values which are

mentioned in the text.1. Place the standard unit of measurement immediately

after the numeral; example: 30 min, 65°C, 0.7 g, 5.3 mm. Note that the temperature is mentioned in Celsius (centigrade) not in Fahrenheit, weights are mentioned in grams not in pounds, lengths are men-tioned in centimeters or millimeters not in inches.

2. In the text, use words for zero to nine but numerals for 10 and above; example: five agar plates, 12 plants. Ordinal numbers are treated in the same fashion; example: fourth, 32nd, 56th.

3. Always use numbers (not words) for page numbers, percentages, and dates; example: page 8, 2% aga-rose gel, as of May 28, 2005

4. Do not start a sentence with a numeral; if you have to, use words.

How to Cite ReferencesIf you provide the reader with information that is not emerging from your work, you must include a reference to indicate the source of the information. In the reference, you must refer to the name(s) of the author(s) of that informa-tion, not the title of the information. You can cite references in several ways depending on the general layout of your report. However, in all cases the reference must be followed by the year of the publication of that reference.

Example 1:An unperturbed F-actin cytoskeleton is required for the establishment of polarity in Fucus and Pelvetia embryos (Quatrano, 1973).

Example 2:Quatrano (1973) showed that an unper-turbed F-actin cytoskeleton is required for the establishment of polarity in Fucus and Pelvetia embryos.

If two authors share an article, mention both authors’ names; for example: Switzer and Shriner, 2000. If an article has more than two authors, mention the name of the first author followed by et al. Example: Mathur et al., 2003. In the above example et means “and”, and al. is the abbreviation of alia which means “others”.

DiscussionThe Discussion is the most important section of your report because it is where you analyze your results and explain what they mean. Caution: do not repeat what you have already said in the Results section. The Discussion should be in essay format with each section in the Results clearly explained in a paragraph or two. If you faced inconsistencies in your experiments, you must provide an explanation for them.

In preparing the Discussion, avoid unnecessary use of adjectives or adverbs to signify your results; let the readers judge whether or not your data are significant. Consider the following examples.

Incorrect Way:Because of the distinct requirements for microtubules and actin filaments during leaf trichome development, a trichome-based morphology screen is a very sufficient approach to clearly identify interesting genes that severely affect cytoplasmic organization.

Correct Way:Because of the distinct requirements for microtubules and actin filaments during leaf trichome development, a trichome-based morphology screen is a sufficient approach to identify genes that affect cytoplasmic organization.

In the last paragraph of the Discussion you must state your conclusion.

ReferencesIn this section you must inform the reader of all published information that you referred to in your report. Do not include in the Reference any paper or book that you read but did not mention in the report. If you wish, you may call this section Literature Cited.

Different papers have their own style of literature cita-tion; you can find them in journals. However, papers and books are referenced differently. Widely used styles are mentioned below.

1. References are listed in alphabetical order by the last name of the author or the last name of the first author, if more than one author is involved. After the last name place a comma (,) then enter the initial(s).

Examples:Goodnight, C.J. 2006. Population genet-

ics: peak shift in large populations. Heredity 96:5-6.

Bedford, T., I. Wapinski, and D.L. Hartl. 2008. Overdispersion of the molecular clock varies between yeast, Drosophila and mammals. Genetics 179:977-984.


2. The name of the journal in which the reference appears is usually written in full if it is short, as shown in the above example. For long names you must use universal title abbreviations; example: Canadian Journal of Botany is written as Can. J. Bot. Or, Proceedings of the National Academy of Sciences, USA is written as Proc. Natl. Acad. Sci. USA.

3. If a chapter of a book is referenced, the title of the chapter is presented first, followed by the page num-bers, editor, and the title of the book.

Example:Jones, G.H. and Heslop-Harrison, J.S.

2000. Classical and molecular cyto-genetics of Arabidopsis. p.105-124. In Z.A. Wilson (ed.) Arabidopsis: A Practical Approach. Oxford University Press, Oxford.

Note that in the above example, the full title of the chapter is followed by the page numbers and then the preposition “In”. The editor(s) are mentioned next; initials are pre-ferred for the first and middle names, but the last name is mentioned in full. The abbreviation “ed.” follows next to indicate that the individual is in fact the editor of the book. The title of the book is mentioned next. The next part of the reference is the name of the publisher and the city (not the country) where the book was published.

4. Books written by one or more authors are presented much the same way that papers are.

Example:Brooker, R.J. 2005. Genetics: Analysis

and Principles. 2nd ed. McGraw-Hill, New York.

——

You should start writing your report now. After prepar-ing the first draft, do not read it over and over. Lock it in your desk for 1 week and then read it again. Did you like it? You will notice that your report might need a tremendous amount of revision!

To improve your writing skills, you may consult the sources listed below.

Hacker, D. 1994. The Bedford Handbook for Writers. 4th ed. Bedford Books of St. Martin’s Press, Boston, MA.

O’Connor, M., and F.P. Woodford. 1975. Writing Scientific Papers in English: An ELSE-Ciba Foundation Guide for Authors. Elsevier, New York.

Rosenblum, P.C. 1991. The Fine Art of Technical Writing: Key Points to Help You Think Your Way Through Writing Scientific or Technical Publications, Theses, Term Papers, and Business Reports. Blue Heron Publ., Hillsboro, OR.

You may also look at the following article as an example

of a successfully prepared paper:

Qiu, J.-L., R. Jilk, M.D. Marks, and D.B. Szymanski. 2002. The Arabidopsis SPIKE1 gene is required for normal cell shape control and tissue development. Plant Cell 14:101−118.


Appendix 3 Suppliers

The address of suppliers of chemicals, basic equipments, and live organisms that are mentioned in this laboratory man-ual, is listed below. Local agents, distributors, and university stores are alternative sources for obtaining the materials.

In Part A of this Appendix, the materials are listed in the alphabetical order followed by the name of the vendor(s). The complete address of the vendors can be found in Part B of the Appendix.

Part AAgar and media preparation nutrients

Agar: InvitrogenAgarose: Becton, Dickinson and Co.Agarose, MetaPhor and compatibles: Biowhittaker

Molecular Applications; Midwest ScientificCornmeal agar: SigmaNutrients: Becton, Dickinson and Co.

Agarose gel electrophoresis system: Bio-Rad LaboratoriesAntibiotics: Fisher Scientific Company L.L.C.Blotting papers (Whatman): Midwest ScientificChemicals (general) and dyes: AldrichCloning vectors and competent cells: InvitrogenCompound and dissecting microscopes: Carolina Biological

Supply Company; VWR International, Inc.Custom primers: MWG Biotech, Inc.DNA and DNA related materials

DNA (phage lambda): VWR International, Inc.DNA ladder: InvitrogenDNA polymerases (Taq): PromegadNTP: SigmaEnzymes: Bio-Rad laboratories; Invitrogen

DNA extraction reagentsCTAB: Aldrich2-Mercaptoethanol: SigmaDNAzol: Invitrogen

Glassware, tubes, and general laboratory supplies: VWR International, Inc.

Hazardous materials to be used under supervisionEthidium bromide: AldrichEthylmethane sulfonate: Aldrich

Heavy equipmentCentrifuges, incubators, water baths, shakers, rotators,

and the like equipments: VWR International, Inc.; Midwest Scientific; A. Daigger & Company, Inc.

Photo documentation system and transilluminators: Fotodyne, Inc.

Live materialsArabidopsis seeds: Lehle SeedsDrosophila and culture media: Carolina Biological

Supply CompanyBacterial and fungal stocks: American Type Culture

Collections; Carolina Biological Supply CompanyMendelian genetics models and taste papers: Carolina

Biological Supply CompanyMesh: Midwest ScientificMicromortars and micropestles (RNase free): VWR

International, Inc.Micropipettors, tips, racks: Integrated Instrument

Services; Carolina Biological Supply Company; VWR International, Inc.

Microscope slides and cover slips: VWR International, Inc.Microwell dishes (with or without lid): Fisher Scientific

Company L.L.C.; VWR International, Inc.Mortars and pestles (porcelain): VWR International, Inc.Mounting media (for slide preparation): VWR

International, Inc.Murashige and Skoog Salt and Vitamin Mixture: Life

TechnologiesNylon membrane: Midwest ScientificOnion bulbs: Carolina Biological Supply CompanyPCR product purification kit: QIAGEN, Inc.Petri dishes and related materials

Petri dishes: VWR International, Inc.Spreaders: Carolina Biological Supply CompanyInoculating loops: VWR International, Inc.

pH indicatorsPaper: VWR International, Inc.; A. Daigger &

Company, Inc.Benchtop pH meters: A. Daigger & Company, Inc.

Prepared chromosome slides: Carolina Biological Supply Company

Preservatives, mold inhibitors: SigmaRNase A: USB CorporationSafety

Gloves: A. Diagger & Company, Inc.; VWR International, Inc.First aid kits: Carolina Biological Supply Company

Silwet L-77®: Helena Chemical CompanySoil mixes, vermiculite, fertilizers, pesticides, larvicides,

and plant containers: Hummert International


Southern blotting & DNA labeling kit, and related dyes: Roche Diagnostics Corporation

BCIP: SigmaNBT: Sigma

Synthetic substrates of biological productsXgal: Gold Biotechnology, Inc.X-gluc: Gold Biotechnology, Inc.

TapesAutoclave indicator: A. Daigger & Company, Inc.Parafilm: A. Daigger & Company, Inc.Surgical: California Veterinary Supply

Tools, carts, and the like hardware: Hummert InternationalTubes

Microtubes and microtube racks: VWR International, Inc.PCR tubes (RNase, DNase free): Life Science Products, Inc.Conical tubes: VWR International, Inc.Oak Ridge tubes: Fisher Scientific Company L.L.C.

Tween® 20: Bio-Rad LaboratoriesX-ray film: Midwest ScientificX-ray film cassettes: Midwest Scientific

Part BA. Daigger & Company, Inc., 620 Lakeview Parkway,

Vernon Hills, IL 60061, Phone (800)621-7193Aldrich, P.O. Box 355, Milwaukee, WI 53201, Phone

(800)558-9160American Type Culture Collection, 10801 University Blvd.,

Manassas, VA 20110, Phone (800)638-6597Becton, Dickinson and Co., BD Diagnostic Systems,

7 Loveton Circle, Sparks, MD 21152, Phone (800)638-8663

Bio-Rad Laboratories, Life Science Group, 2000 Alfred Nobel Drive, Hercules, CA 94547, Phone (800)424-6723

Biowhittaker Molecular Applications, 191 Thomaston Street, Rockland, ME 04841, Phone (800)341-1574

California Veterinary Supply, 1155 Deercrest Drive, Devore, CA 92407, Phone (800)366-3047

Carolina Biological Supply Company, 2700 York Road, Burlington, NC 27215, Phone (800)334-5551

Fisher Scientific Company L.L.C., 4500 Tumberry Drive, Hanover Park, IL 60103, Phone (800)766-7000

Fotodyne, Inc., 950 Walnut Ridge Drive, Hartland, WI 53029, Phone (800)362-3686

Gold Biotechnology, Inc., 3655 Vista Avenue, St. Louis, MO 63110, Phone (800)248-7609

Helena Chemical Company, 225 Schilling Boulevard, Collierville, TN 38017, Phone (901)761-0050

Hummert International, 4500 Earth City Express Way, Earth City, MO 63045, Phone (800)325-3055

Integrated Instrument Services, 2104 Production Drive, Indianapolis, IN 46241, Phone (317)248-1958

Invitrogen, 1600 Faraday Avenue, Carlsbad, CA 92008, Phone (800)955-6288

Lehle Seeds, P.O. Box 2366, Round Rock, TX 78680, Phone (800)881-3945

Life Science Products, Inc., 5989 Iris Parkway, P.O. Box 1150, Frederick, CO 80530, Phone (800)245-5774

Life Technologies, Inc., 9800 Medical Center Drive, P.O. Box 6482, Rockville, MD 20849, Phone (800)338-5772

Midwest Scientific, 280 Vance Road, Valley Park, MO 63088, Phone (800)227-9997

MWG Biotech, Inc., 4191 Mendenhall Oaks Parkway, Suite 140, High Point, NC 27265, Phone (877)694-2832

Promega, 2800 Woods Hollow Road, Madison, WI 53711, Phone (800)356-9526

QIAGEN, Inc., 27220 Turnberry Lane, Valencia, CA 91355, Phone (800)426-8157

Roche Diagnostics Corporation, P.O. Box 50414, Indianapolis, IN 46250, Phone (800)262-1640

Sigma, P.O. Box 14508, St. Louis, MO 36178, Phone (800)325-3010

USB Corporation, 26111 Miles Road, Cleveland, OH 44128, Phone (800)321-9322

VWR International, Inc., 6801 Gray Road, Suite D, Indianapolis, IN 46237, Phone (800)932-5000


Suggested Literature

Adams, M.D. et al. 2000. The genome sequence of Drosophila melanogaster. Nature (London) 287:2185–2195.

Arabidopsis Genome Initiative. 2000. Analysis of the genome sequence of the flowering plant Arabidopsis thaliana. Nature (London) 408:796–813.

Ausubel, F.M., R. Brent, R.E. Kingston, D.D. Moore, J.G. Seidman, J.A. Smith, and K. Struhl. (ed.) 1992. Short proto-cols in molecular biology. John Wiley & Sons, New York.

Baltimore, D. 1970. RNA-dependent DNA polymerase in virions of RNA tumor viruses. Nature (London) 226:1209–1211.

Beadle, G.W., and E.L. Tatum. 1942. Genetic control of biochemi-cal reactions in Neurospora. Proc. Natl. Acad. Sci. USA 27:499–506.

Bechtold, N, J. Ellis, and G. Pelletier. 1993. In planta Agrobacterium mediated gene transfer by infiltration of adult Arabidopsis thaliana plants. C.R. Acad. Sci. Paris, Life Sci. 316:1194–1199.

Bell, C.J., and J.R. Ecker. 1994. Assignment of 30 microsatel-lite loci to the linkage map of Arabidopsis. Genomics 19:137–144.

Boys, D.C., A.M. Zayed, R. Ascenzi, A.J. McCaskill, N.E. Hoffman, and K.R. David. 2001. Growth stages-based phe-notypic analysis of Arabidopsis: A model for high throughput functional genomics in plants. Plant Cell 13:1499–1510.

Bridges, C.G. 1922. The origin of variation in sexual and sex-limited characters. Am. Nat. 56:51–63.

Brooker, R.J. 2005. Genetics: Analysis and principles. 2nd ed. McGraw-Hill Co., New York.

Chomczynski, P., K. Mackey, R. Drews, and W. Wilfinger. 1997. DNAzol: A reagent for the rapid isolation of genomic DNA. BioTechnique 22:550–553.

Clough, S., and A. Bent. 1998. Floral dip: A simplified method for Agrobacterium-mediated transformation of Arabidopsis thaliana. Plant J. 16(6):735–743.

Doyle, J.J., and J.L. Doyle. 1987. A rapid DNA isolation proce-dure for small quantities of fresh leaf tissue. Phytochem. Bull. 19:11–15.

El-Assal, S.E-D., J. Le, D.Basu, E.L. Mallery, and D.B. Szymanski. 2004. DISTORTED2 encodes an ARPC2 subunit of the puta-tive Arabidopsis ARP2/3 complex. Plant J. 38:526–538.

Fairbanks, D.J., and W.R. Andersen. 1999. Genetics: The continu-ity of life. Brooks/Cole Publ. Co., Pacific Grove, CA.

Fisher, R.A., and F. Yates. 1943. Statistical tables for biological, agricultural, and medical research. Oliver and Boyd, London.

Griffiths, A.J.F., S.R. Wessler, R.C. Lewontin, and S.B. Carroll. 2008. Introduction to genetic analysis. 9th ed. W.H. Freeman and Co., New York.

Hacker, D. 1994. The Bedford handbook for writers. 4th ed. Bedford Books of St. Martin’s Press, Boston.

Hardy, G.H. 1908. Mendelian proportions in a mixed population. Science (Washington, DC) 28:40–50.

Hartwell, L.H., L. Hood, M.L. Goldberg, A.E. Reynolds, L.M. Silver, and R.C. Veres. 2000. Genetics: From genes to ge-nomes. McGraw-Hill Co., Boston.

Hellens, R.P., E.A. Edwards, N.R. Leyland, S. Bean, and P.M. Mullineaux. 2000. pGreen: A versatile and flexible binary Ti vector for Agrobacterium-mediated plant transformation. Plant Mol. Biol. 42:819–832.

Holt, S.J., and P.W. Sadler. 1958. Studies in enzyme cytochem-istryIII. Relationships between solubility, molecular asso-ciation and structure in indigoid dyes. Proc. R. Soc. London 148B:495–505.

Hopfner, K.-P., A. Eichinger, R.A. Engh, F. Lane, W. Ankenbauer, R. Huber, and B. Angerer. 1999. Crystal structure of a ther-mostable type B DNA polymerase from Thermococcus gor-gonarius. Proc. Natl. Acad. Sci. USA 96:3600–3605.

Jacob, F., and J. Monod. 1961. Genetic regulatory mechanisms in the synthesis of proteins. J. Mol. Biol. 3:318–356.

Johnson, C.S., B. Kolevski, and D.R. Smyth. 2002. Transparent testa glabra2, a trichome and seed coat development gene of Arabidopsis, encodes a WRKY transcription factor. Plant Cell 14:1359–1375.

Kavenoff, R., L. Klotz, and B. Zimm. 1974. On the nature of chro-mosome-sized DNA molecule. Cold Spring Harbor Symp. Quant. Biol. 38:1–8.

Lagercrantz, U., H. Ellegren, and L. Andersson. 1993. The abundance of various polymorphic microsatellite repeats differs between plants and vertebrates. Nucleic Acid Res. 21:1111–1115.

Le, J., S.E-D. El-Assal, D. Basu, M.E. Saad, and D.B. Szymanski. 2003. Requirements for Arabidopsis ATARP2 and ATARP3 during epidermal development. Curr. Biol. 13:1341–1347.

Lefevre, G. Jr. 1976. A photographic representation and interpre-tation of the polytene chromosomes of Drosophila melano-gaster salivary glands. p. 31–66. In M. Ashburner and E. Novitski (ed.) The genetics and biology of Drosophila. Vol. 1a. Academic Press, New York.

Lindsley, D.L. and G.G. Zimm. 1992. The genome of Drosophila melanogaster. Academic press, San Diego, CA.

Lister, C., and C. Dean. 1993. Recombinant inbred lines for map-ping RFLP and phenotypic markers in Arabidopsis thaliana. Plant J. 4:745–750.

Marks, M.D. 1997. Molecular analysis of trichome development in Arabidopsis. Annu. Rev. Plant Physiol. Plant Mol. Biol. 48:137–163.

Mertens, T.R. and R.L. Hammersmith. 1998. Genetics Laboratory Investigations. 11th ed. Prentice-Hall, Upper Saddle River, NJ.


Miller, J.S.; M.E. Sass, S.J. Wong, and J. Nienhuis. 2004. Micropipetting: An important laboratory skill for molecular biology. Am. Biol. Teacher 66:291–296.

Moore, R. 2008. Are students’ performances in labs related to their performances in lecture portions of introductory science courses? J. Coll. Sci. Ed. 37:66–70.

Mullis, K.B. 1990. The unusual origin of the polymerase chain reaction. Sci. Am. (April):56–65.

Mullis, K.B., and F.A. Faloona. 1987. Specific synthesis of DNA in vitro via a polymerase-catalized chain reaction. Methods Enzymol. 155:335–350.

Myers, E.W. et al. 2000. A whole-genome assembly of drosophila. Nature 287:2196–2204.

New England Biolabs, (2002–2003) Catalog and technical refer-ence. New England Biolabs, Ipswich, MA.

O’Conner, M., and F.R. Woodford. 1975. Writing scientific papers in English: An ELSE-Ciba foundation guide for authors. Elsevier, New York.

Olive, L.S. 1956. Genetics of Sordaria fimicola. I. Ascospore color mutants. Am. J. Bot. 43:97–107.

Pierce, B.A. 2008. Genetics: A conceptual approach. 3rd ed. W.H. Freeman and Co., New York.

Prasher, D.C., V.K. Eckenrode, W.W. Ward, F.G. Prendergast, and M.J. Cormier. 1992. Primary structure of the Aequorea victo-ria green fluorescent protein. Gene 111:229–233.

Qui, J.-L.; R. Jilk, M.D. Marks, and D.B. Szymanski. 2002. The Arabodopsis SPIKE1 gene is required for normal cell shape control and tissue development. Plant Cell 14:101–118.

Rosenblum, P.C. 1991. The fine arts of technical writing: Key points to help you think your way through writing scientific or technical publications, theses, term papers, and business reports. Blue Heron Publ., Hillsboro, OR.

Russell, P.J. 1996. Genetics. HarperCollins College Publ., New York.

Sanger, F., A.R. Coulson, G.F. Hong, D.F. Hill, and G.B. Petersen. 1982. Nucleotide sequence of bacteriophage λ DNA. J. Mol. Biol. 162:729–773.

Schmidt, E.V. 1993. Genetic control of cell chemistry using Serratia marcescens. p. 21–33. In C.A. Goldman (ed.) Tested studies for laboratory teaching. Vol. 14. Proc. 14th Workshop/Conf. of the Assoc. for Biology Laboratory Education (ABLE), Las Vegas, NV. 2–6 June 1992. ABLE.

Shimomura, O., F. Johnson, and Y. Saiga. 1962. Extraction, puri-fication and properties of aequorin, a bioluminescent protein from the luminous hydromedusan, Aequorea. J. Cell Comp. Physiol. 59:223–239.

Southern, E. 1975. Detection of specific sequences among DNA fragments by gel electrophoresis. J. Mol. Biol. 98:503–517.

Stern, C. 1959. Use of the term “superfemale”. Lancet 2:1088–1089.Szymanski, D.B., A.M. Lloyd, and M.D. Marks. 2000. Progress

in the molecular genetic analysis of trichome initiation and morphogenesis in Arabidopsis. Trends Plant Sci. 5:214–219.

Tamarin, R.H. 2002. Principles of Genetics. 7th ed. McGraw-Hill Co., Boston.

Taylor, J., P. Woods, and W. Hughes. 1975. The organization and duplication of chromosomes as revealed by autoradiographic studies using tritium-labeled thymidine. Proc. Natl. Acad. Sci. USA 43:122–128.

Temin, H.M., and S. Mizutani. 1970. RNA-dependent DNA poly-merase in virions of Rous sarcoma virus. Nature (London) 226:1211–1213.

Tinland, B. 1996. The integration of T-DNA into the plant genome. Trends Plant Sci. 1(6):178–184.

Tsien, R.Y. 1998. The green fluorescent protein. Ann. Rev. Biochem. 67:509–544.

Walker, A.R., P.A. Davidson, A.C. Bolognesi-Winfield, C.M. James, N. Srinivasan, T.L. Blundell, J.J. Esch, M.D. Marks, and J.C. Gray. 1999. The transparent testa glabra locus, which regulates trichome differentiation and anthocyanin biosynthesis in Arabidopsis, encodes a WD40 repeat protein. Plant Cell 11:1337–1349.

Watson, J.D., and F.H. C. Crick. 1953. Molecular structure of nucleic acids: A structure of deoxyribose nucleic acid. Nature (London) 171:737–738.

Weinberg, W. 1909. Über Vererbungsgesetze beim Menschen. Z. Abst. Vererbl 1:377–393 and 440–460.

Wilson, Z.A. (ed.) 2000. Arabidopsis: A practical approach. Oxford University Press, Oxford, UK.

Winfrey, M.R.; M.A. Rott, and A.T. Wortman. 1997. Unraveling DNA: Molecular biology for the laboratory. Prentice-Hall, Upper Saddle River, NJ.

Williamson, J.H., and A.M. Campbell. 1997. DNA technology in the classroom: A circular map of bacterial plasmid. Am. Biol. Teacher 59:164–170.

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