genetic engineering laboratory manual contents

Genetic Engineering Laboratory Manual Contents

About Genetic Engineering

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Laboratory Resources 7 Digestion of DNA with Restriction Endonucleases 9 Agarose Gel Electrophoresis 21 Restriction Mapping 36

Experiment 1: Isolation of Escherichia coli Chromosomal DNA 43 Experiment 2: Determination of DNA Concentration and Purity by Ultraviolet

Spectrophotometry 45

Experiment 3: The Chemical Nature of DNA 49 Experiment 4: DNA Sequence Specificity 54 Experiment 5: Activity of DNA Nucleases 56 Experiment 6: Restriction and Modification 58 Experiment 7: Analysis of Plasmid DNA by Restriction Digestion and Agarose

Gel Electrophoresis 60

Experiment 8: Effect of Agarose Concentration on Migration of DNA Fragments 66 The First Lab Report 68

Experiment 9: Insertion of a Gene for Antibiotic Resistance from Bacillus subtilis into an Escherichia coli Plasmid

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Experiment 10: Transformation of Escherichia coli With a Chimeric Plasmid 77

Experiment 11: Screening of Transformants for Chimeric Plasmids 84

Experiment 12: Large-Scale Purification of Plasmids PRIT4501 and pRIT4502 by Cesium Chloride Density Gradient Centrifugation

89

Experiment 13: Large-Scale Purification of Plasmids PRIT4501 and pRIT4502 by

Qiagen Ion Exchange Column Chromatography 98

Experiment 14: Verification of Purified Plasmids 105 Experiment 15: Construction of a Restriction Map of Plasmids pRIT4501 and

pRIT4502 107

The Second Lab Report 109 Experiment 16: Transformation Bacillus subtilis with pRIT4501 and pRIT4502 112

Experiment 17: Heterologous Gene Expression in Escherichia coli and Bacillus

subtilis 117

The Third Lab Report 119 Experiment 18: Southern Blot Analysis of Plasmids pRIT4501 and pRIT4502 120 The Fourth Lab Report 130 Appendix I: Index of Reagent Formulas 131 Appendix II: DNA Structure 132 Appendix III: Conversion Factors 137 Appendix IV: Buffer Dilutions 18 Appendix V: Bacillus subtilis Plasmid Prep by Qiagen Maxi Kit 139

Appendix VI: Effect of BAP Treatment on Transformation Frequencies 140 Appendix VII: Alpha Complementation 141 Appendix VIII: Density Gradient Centrifugation of DNA 142 Appendix IX: Bacteriophage λ 145 Appendix X: Technical Information 149 Appendix XI: Mrs. Rothman’s Superstar Bars 166 Appendix XII: Cloning Methodology 168

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About Genetic Engineering Dr. Rothman

Office: 08-1342 Phone: x5215 email: [email protected]

Dr. Rothmans Home Page: rothman.rit.edu GE Home Page: http://www.rit.edu/~rhrsbi/GEPages/GeneticEngineering.html

Grading Criteria You will be graded on the basis of three types of assignments: Exams, Lab Notebooks, and Lab Reports. • Exams will be based on both lecture and laboratory material. In Genetic Engineering, there is really

no difference between lecture and lab. The exams are designed to test your understanding of laboratory procedures and your ability to integrate practical techniques into a broader theoretical background.

• Your lab notebooks will be graded on how well you document your work and keep records. I will

not be evaluating the work and the interpretations. • The lab report is the vehicle by which you will describe the work, present the experimental results,

and express your interpretation of the experiment.

Grading of Lab Notebooks Lab notebooks will be graded according to three criteria:

5 points Title Page and Table of Contents complete and up to date 5 points Title and date for each entry 20 points Data complete and up to date 10 points Clear statements of purpose and conclusions for each experiment 20 points Clarity of entries 60 points Total

Lab notebooks will be collected UNANNOUNCED, during LECTURE. There will be no grace period for late notebooks. To satisfy the Lab Notebook portion of the Writing Requirement, you must earn a 70%


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Grading of Lab Reports Reports will be graded according to three criteria:

Data (20 points):

Did you obtain the correct experimental results? This is a purely technical grade and no assumption is made about whether or not you understood the project.

Interpretation (20 points):

How well did you analyze and understand each part of the project?

Writing (20 points): Did you follow the appropriate format for the paper? How good was your use of grammar and language?

Computing the Final Grade There will be two midterm exams, each being worth 200 points. The final will serve as an additional, third midterm. It will not be comprehensive. There will be four formal laboratory reports. The second report will contain about half of the work you do during the quarter and will be worth twice the number of points (120) as the first. The final two lab reports will be brief (40 points). In addition to laboratory reports, there will be one problem set on plasmid mapping.

Point Breakdown Final Grades Exams (4) 800 points Grade

A B C+ C C- D F

Percent 90 – 100 80 – 89 77 - 79 70 - 76 65 - 69 50 - 64 <50

Points 1224 – 1360 1088 – 1223 1047 – 1087 952 – 1046 884 – 951 680 – 884

<680

Lab Reports (6) 340 points Notebooks (2) 120 points Homework 100 poimts Total 1360 points


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Lab Notebooks Keeping accurate records is critical to the process of scientific discovery. The lab notebook is the place where you record what you have done and what you found out. In industry, the lab notebook is often vital to the patenting process, especially if there is litigation over priority. Many companies require that each day’s work be signed by the investigator and a witness, and in many cases, notebooks are periodically collected and notarized. Each company publishes its own, very stringent set of regulations for lab notebooks. In this course you will keep your notebook according to industrial standards. The rules for this course are based on those required by Eastman Kodak Company. You are expected to have your notebooks with you at all times, whether in lab or lecture. Notebooks will be collected periodically without warning. Each notebook grade will be equivalent to one lab report.

1. Each notebook must have an accurate, up-to-date table of contents. 2. Each entry must have a title, a date, and a statement of purpose or intent. 3. Each entry must end with a conclusion and/or a statement of what must next be

done. 4. Make entries at the time the work is performed. Do not write notes on scratch

paper and make entries in your notebook later. 5. Make neat legible entries in blue or black ink 6. Use the pages in consecutive order. Do not leave any blank pages, or room for

data or data analysis to be added later. All entries should be chronological at the time the data or analysis is completed. You may add a note at the end of one entry referring to the page of the data or of the analysis if there is intervening material.

7. For computer-generated records, photographs, or hand-drawn graphs, tape the material into your notebook. Make reference to the printout on the page. If it is necessary to put such inserts into the notebook, mount them so that they do not cover written information.

8. If data or samples from another source are entered, be sure to indicate the source clearly, including the name of the person from which they were obtained.

9. Record all steps in sufficient detail so that any person skilled in the field can repeat the work and obtain the indicated results.

10. A protocol that is used for the first time must be written out in full. If it is a standard protocol that you use on subsequent occasions, you may simply reference the first citing, subsequently giving only modifications or experimental details (e.g. particular strains, enzymes, etc.)

11. Use only standard abbreviations.


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Format for Lab Reports Lab reports are formal typed papers that describe your experiments and interpretations. They will be written according to the format of a journal research article:

Abstract: A brief statement of the problem and the results. No more than a short paragraph

Introduction: An explanation of the problem and your plan for solving it. Do not give experimental protocols and results here. Just what you want to find out, and your basic approach to the problem. Introductions are relatively short. You do not have to spill your guts about your vast knowledge of recombinant DNA here. That will be best done on the exam.

Methods and Materials:

Present a description of the protocols, strains, DNA’s etc. The Methods and Materials is like a cookbook that allows others to see exactly what your experimental conditions were. Do not give experimental design. For example, if you are talking about restriction digests, give the conditions under which you do a digest, but don’t state what DNA’s and what enzymes were used: that is experimental design, and it belongs in the Results section.

Results: You should give written descriptions of the experiments, minus the protocols (given in Methods and Materials), with a careful description of figures and tables. Each figure and table must be numbered and captioned. You should only give enough of your conclusion s to enable the reader to proceed logically from one experiment to the next.

Discussion: Here you restate the problem and experiments, and provide a detailed discussion of your conclusions and recount how you arrived at them.

A Word About Plagiarism - Most scientific research is carried out through teamwork and the final result is a single report co-authored by all members of the team. Plagiarism therefore is not an issue. In class, however, although the work is done in pairs and usually partners work together on the preparation and analysis of the data, each student must prepare his/her own lab report. It is acceptable to format the tables and figures jointly, but the main text must be written individually.


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Laboratory Cleanup Each student is responsible for helping to keep lab clean. Regardless of how you find the lab when you come in, you must swab your workplace down with detergent before and after lab, and you must empty the pipet disposal in the sink into the glass waste disposal. You must also clean up after yourself whenever you use the balance. In addition to your personal responsibilities, you will be assigned to a clean-up crew according to the schedule below. Responsibilities of the clean-up crew are:

• Clean lab benches in your area that your classmates may have left empty • Empty any pipet containers that classmates did not empty • Tidy up balance areas • Clean BioDoc-It • Turn off water baths • Gather up any equipment or reagents that must be returned to instructor • Help instructor or TA’s put away any additional apparatus or reagents

STUDENTS WILL BE CHARGED DEMERIT POINTS (10) FOR EACH INCIDENT OF NOT CLEANING HIS/HER AREA AT THE END OF LAB IF AN INDIVIDUAL CULPRIT CANNOT BE IDENTIFIED, THE ENTIRE CLASS WILL BE CHARGED DEMERIT POINTS (10) FOR EACH INCIDENT OF THE BALANCE AREA BEING LEFT MESSY AT THE END OF LAB

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Clean-up Crew assignments by seat location:

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Laboratory Waste Management Autoclave Waste

1. Any items that are exposed to a bacterial culture (flasks, test tubes, centrifuge tubes, dilution blanks, etc, must be autoclaved

2. Place flasks and plastic centrifuge tubes in the tubs in the washroom 3. Place test tubes in tubs in test tube racks. Fill any empty spaces in the test tube racks 4. Do not empty tubes! 5. Remove all tape (including autoclave tape) 6. Do not lay tubes down in the tubs!

Chemical Waste

1. Reagents may need to be poured into the chemical waste bottles according to type as per instructor

2. Organic waste contains carbon 3. Inorganic or aqueous waste does not contain carbon 4. Inform instructor or TA when you add waste to these bottles 5. Sterile water can be disposed in the sink 6. Empty glass scintillation vials may go into the glass disposal boxes

Re-usable items

1. Pipette tip boxes and microfuge tube boxes or jars should be returned to the stockroom for refilling Do not put these items into the autoclave bins!

2. Sterile applicator sticks are returned to the stockroom for re-filling 3. Used applicator sticks can be put into the autoclave bins or into the glass waste boxes

Sharps

1. Syringes, needles, or other sharp objects must be deposited in the red sharps container in the rear of the lab

2. Anything that looks like a syringe (e.g. Qiagen columns) should be treated as a syringe

Gel Boxes 1. After use drain running buffer from box 2. Rinse with tap water 3. Lightly dry. Be careful of the electrodes! 4. Return to stockroom

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Laboratory Resources Locker Contents Each student will be assigned a locker containing:

1 400 ml beaker 1 blue pipet pump (2 ml) 1 50 ml Erlenmeyer flask 1 green pipet pump (10 ml) 1 test tube rack 1 permanent marker 1 microfuge rack 1 6” ruler 1 magnetic stirring bar 1 box matches

Stockroom Sign-Outs

At the beginning of the first lab you must sign out he following items for the term:

Per Pair: 1 funnel 1 roll masking tape 1 500 ml graduated cylinder Per Student: 1 1l erlenmeyer flask 1 floating microfuge rack

Before coming to lab each day, each student is expected to sign out a set of Pipetman pipetors (P-20, P-200, and P-1000) for the period. PIPETORS MUST BE RETURNED AT THE END OF CLASS Before coming to lab each day, each pair is expected to sign out an appropriate gel box and power supply, ice bucket, and staining tray for the day’s work It is a good idea to keep all stockroom receipts in case you are billed for non-returned items

Required Items

Safety glasses Lab coat Lab notebook: 8.5” x 11” spiral-bound, 4x4 quadrille,

3-cycle semi log paper polar coordinate graph paper compass

Laboratory Resources

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Reagent Kits

Each pair of students will be given a DNA & Buffer Kit and an Enzyme Kit. Write your names on the blank blue tape on each box but otherwise, make no marks on the boxes.

DNA and Buffer Kit

Large: Orange - low salt buffer (10x) Purple - medium salt buffer (10x) Blue - high salt buffer (10x) Yellow - pUC9 DNA Green - pBAC DNA* Red - pRHR30 DNA Orange• pRHR53 DNA Blue• λ DNA White - orange tracking dye White - blue tracking dye Small: Red - λ HindIII molecular weight standard Blue - 1 kb ladder molecular weight standard Green - DNA ligase buffer (5x) Yellow - NEB 3 restriction buffer (10x) Orange - NEB BamHI restriction buffer (10x) White - gel red *pBAC is a generic name for a plasmid from Bacillus subtilis. The particular

plasmid that we will use rotates every quarter. Information about the specific plasmid can be found in the Syllabus

Enzyme Kit

Red - EcoRI 60 µl Blue - HindIII 80 µl Green - BamHI 60 µl Orange - BglII 30 µl Yellow - PstI 30 µl White - XbaI 30 µl Purple - T4 DNA Ligase 10 µl

YOU MAY NOT GO INTO STOCK ENZYMES WITHOUT PERMISSION!

I NEVER GIVE PERMISSION TO GO INTO STOCK ENZYMES!

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Digestion of DNA With Restriction Endonucleases

Digestion of DNA with restriction endonucleases is the first step in many gene manipulation projects. Restriction endonucleases (also referred to more generally as restriction enzymes) are isolated from many species of bacteria and recognize specific 4-base (tetramer), 5-base (pentamer), 6-base (hexamer), or 8-base (octomer) sites and make double-stranded cuts. In vivo, it appears that the main role of restriction endonucleases is to protect cells from invasion by foreign DNA’s, especially bacteriophage DNA. Restriction and Modification Restriction endonucleases were discovered as a result of the study of the phenomenon of restriction and modification. Restriction and modification is a phenomenon concerning the consequences of transferring a bacteriophage grown on one serotype of bacteria to another. For example, λ grown on E. coli K and replated on E. coli K will yield a high phage titer with each transfer, but when that same phage stock is transferred to E. coli C, the phage titer drops several orders of magnitude. In the parlance of the time, phage growth has been restricted by the new host. However, when surviving phage was replated on E. coli C, the titer returns to levels similar to those seen on the original host strain. The phage has somehow been modified to be compatible with the new host. As long as the λ is recycled on E. coli C, titers remain high. However, when transferred back to E. coli B, the original host, phage titers again drop by several of orders of magnitude. But upon replating on E. coli B, phage titers recover. However, the phage is now susceptible to restriction if returned to E. coli C.

Restriction Enzymes

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Each species and serotype produces a specific restriction endonuclease that recognizes and cleaves particular restriction sites. Restriction sites are short enough that they can be found randomly in the DNA of any organism, including the organism that produces the restriction endonuclease. To distinguish its own DNA from foreign DNA, the producing organism has a methyltransferase that recognizes and methylates the same site that its endonuclease cuts. Methylation occurs by the transfer of a methyl group from S-adenosyl methionine (SAM) to one of the nucleotides in the restriction site. Methyltransferases are often referred to as methylases but methyltransferase is a better description of the mode of action of these enzymes, and is therefore the preferred term. This process is called modification; methylation prevents restriction. Thus, an organism would have its own sites protected while incoming DNA would lack the appropriate methylation and therefore be vulnerable. The accepted abbreviations for restriction endonucleases and methyltransferases are REase and MTase, respectively Naming Enzymes

There is a uniform system for naming restriction endonucleases and their corresponding methyltrahsferases, based on the genus and species of the source organism, the particular strain or serotype, and the order of discovery. By convention, the first letter of the genus name and the first two letters of the species name are used to derive the basic enzyme name. Thus Escherichia coli yields Eco (because genus and species names are italicized, it was originally the custom to italicize the enzyme name [Eco] but recent nomenclature recommendations have dispensed with this convention). Then comes a designation, if any, of the particular strain or serotype (sometimes an enzyme is encoded by a plasmid and the plasmid designation is used). A common REase from E. coli comes from an R factor. Finally, a Roman numeral is applied to indicate the order of discovery. Thus the first restriction enzyme from E. coli carrying the R factor is EcoRI. Some others are:

HindIII the third enzyme from Haemophilus influenzae strain d SmaI the first enzyme Serratia marcesens BamHI the first enzyme from Bacillus amyloliquifaciens strain H KpnI the first enzyme from Klebsiella pneumoniae

The names of REases are distinguished from the names of MTases by placing an “R.” or an “M.” in front of the name. Thus M.EcoRI is the corresponding methyltransferase for the restriction endonuclease R.EcoRI. Typically, MTases have few research applications and are rarely used. Because most researchers only use the endonucleases, the “R.” tends to be dropped unless both endonucleases and methyl transferases are used in the same work.

Restriction Sites

Recombinant DNA technology is based upon the fact that many enzymes produce staggered cuts leaving complementary single-stranded tails. Being complementary, the single stranded tails can be made to form hydrogen bonds with one another and the cohering fragments can then be ligated together. Since the tails are based solely on the restriction sequence, it is possible to ligate DNA’s from two different species if they have been cut with the same enzyme. The ability of restriction endonucleases to produce

Restriction Enzymes

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cohesive single-stranded tails depends upon the symmetry of the restriction site and the way that the particular enzyme cuts relative to the symmetry.

The two strands of DNA are said to be anti-parallel. That is, one strand runs 5’→3’ and the other runs 3’→5’. This produces a structural symmetry called rotational or dyad symmetry. In dyad symmetry, one can rotate DNA 180o and obtain the same structure:

5'

5'

3'

3'

axis of rotation

For restriction sites, not only does the overall structure possess dyad symmetry, but also the DNA sequence itself possesses dyad symmetry. For example, the restriction enzyme EcoRI recognizes the site:

axis of rotation â

5’ ------------ G A A • T T C ------------ 3’ 3’ ------------ C T T A A G ------------- 5’

When this hexameric sequence is rotated about its axis, not only is the structural polarity maintained, but also the identical sequence is obtained. This symmetry of sequence is due to the unique nature of the base sequence in which the second three bases are the complement, in reverse order, of the first three:

A B C C’ B’ A’

One strand therefore is the reverse order of the other. Such an arrangement is often referred to as a palindrome. In literature, a palindrome is a phrase that reads the same forwards and backwards. An example of a literary palindrome is when Adam, in the Garden of Eden, introduced himself to Eve:

Madam, I’m Adam The restriction site is not a true palindrome, of course, because the reverse is on the opposite strand.

Cleavage

During restriction, the endonuclease must cut each of the strands to generate a double-strand cut. Cleavage is the result of hydrolysis, a reaction in which water is added across a bond, thereby breaking it. In this case, the water is added across the phosphodiester bond, cleaving the two adjacent nucleotides. Cleavage (at least by restriction endonucleases) yields 5’-phosphate and 3’-hydroxyl

Restriction Enzymes

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termini. By contrast, nucleotides are joined by condensation reactions, in which phosphodiester bonds are formed by splitting out a water molecule. DNA ligases are enzymes that function via condensation.

Because each of the strands are identical to each other both in sequence and structure (remember, the strands are the same, but antiparallel), the cuts are made in the same spot on each strand, relative to the axis of rotation. This creates a staggered cut, leaving overhanging single-stranded tails on each end. The cuts made by EcoRI are typical.

EcoRI cleavage can also be written using the shorthand notation for DNA structure. The shorthand notation used in the figure below is explained in Appendix II.

In the EcoRI example, the cuts were made to the left of the axis of rotation, producing 5’

overhangs. Other enzymes, however, cleave to the left of the axis, producing 3’ overhangs, or on the axis, producing “blunt” ends. Three examples, HindIII, KpnI, and SmaI are shown below:

Restriction Enzymes

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Clearly, ends created by HindIII and by KpnI are complementary and can permit ligation. But blunt ends such as formed by SmaI, under the right circumstances, can also be ligated. Blunt end ligation is a very inefficient reaction but when used strategically it can be a very powerful tool. It is possible to treat the cut ends with a variety of secondary enzymes to provide lots of flexibility with respect to subsequent cloning steps. The type of enzyme used and the type of modification possible depends on the nature of the cut relative to the axis, and on whether the 3’OH end is recessed or over-hanging.

Isoschizomers and Compatible Enzymes

Occasionally, several restriction endonucleases may recognize the exact same sequence. The first enzyme discovered to recognize a particular sequence is known as the prototype. When additional enzymes are discovered that recognize the same sequence, they are called isoschizomers. If the new enzyme recognizes the same sequence but cleaves it differently, then it is known as a neoschizomer. In the case of the sequence CCCGGG:

Enzyme Sequence Nomenclature

XmaI CâCCGGG prototype

Cfr9I CâCCGGG isoschizomer

SmaI CCCâGGG neoschizomer

Sometimes there is overlap in the recognition sites for different enzymes. For example, the site for BamHI, GâGATCC shares the middle four bases with the site for BglII, AâGATCT, and the entire tetrameric sequence of Sau3A, âGATC. It is thus possible to ligate one DNA cut with BamHI to another DNA cut with BglII. Such enzymes are said to be compatible. Since the outside bases for each enzyme are different, the result would be a hybrid sequence that cannot be cut by either BamHI or BglII. The central GATC, however, would be regenerated and could be cut by Sau 3A. Similarly, DNA cut with Sau3A could be ligated to DNA cut with either BamHI or BglII. Since Sau3A recognizes a four-base site, the adjacent bases are random. Thus there is a one in four probability that the fusion of a Sau3A site to a BamHI site will regenerate the BamHI sequence. The probability of finding a tetrameric sequence such as Sau3A in any random piece of DNA is much greater than finding a hexameric sequence. Thus an enzyme like Sau3A will cut DNA much more frequently than will BamHI.

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Enzyme Structure Cleavage Site Recognition Sequence Reaction

Requirements

Type I

up to 1000 base pairs away

asymmetric & discontinuous EcoK = AAC(N)6GTGC EcoB = TGA(N)8TGCT

ATP S-Adenosyl Methionine

Type II

within recognition sequence

continuous & symmetric EcoRI = GâAATTC continuous & asymmetric BbvCI = CCâTCAGC discontinuous & symmetric BglI = GCCNNNN âNCCG

Mg2+

Type IIS

up to 20 bp away on 3’ side

continuous & asymmetric FokI = GGATG(N)9

â

CCTAC(N)13â

Mg2+

Type IIG

outside sequence

continuous & symmetric AcuI = CTGAAG(N)16

â

GACTTC(N)14â

discontinuous & symmetric BcgI = â

10(N)CGA(N)6TGC(N)12â

â12(N)GCT(N)6ACG(N)10

â

Mg2+

Type III

24 – 26 bp away on 3’ side

continuous & asymmetric EcoP15I = CAGCAG(N)25

â GTCGTC(N)27

â

Stimulated by ATP

S-Adenosyl Methionine

Type IV

One or two genes encoding an endonuclease. Only cleaves at modified bases:

methylated hydroxymethlated glucosyl-hydroxymethylated

~ 30 base pairs from one of the sites

recognition sequences undefined except:

EcoKMcrBC = two dinucleotides Pu*C (purine + methylated C, separated by 40 – 3000 bases)

Using Restriction Enzymes Each restriction enzyme has its own optimal set of reaction conditions, which can be found on the information sheet provided by the supplier. A number of companies produce high-quality restriction enzymes. The most important reaction condition variables are the ionic strength (i.e. salt concentration) of the reaction buffer and the temperature of digestion. Of the two, reaction temperature is often most critical. Typically purchased enzymes are supplied with the appropriate reaction buffer in 10X concentrations. In many cases, the ionic strength requirement is less stringent and it is therefore permissible to broadly categorize restriction enzymes according to the Cold Spring Harbor system as requiring high, medium, or low salt. For most of the experiments in this course, we will follow the Cold Spring Harbor system. The end of this section lists the temperature and buffer requirements for some of

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the common restriction enzymes as well as their recognition sequences and sites of cutting, as well as a list of formulas for these buffers. There are a few exceptions to this general categorization. We will discuss the conditions for using these enzymes as they come up.

Restriction endonucleases are purchased in high concentrations and 1µl of most enzyme preparations is enough to cut as much as 10 µg of DNA. Enzyme concentrations are found on the information sheet as well as the label and are expressed as units of enzyme per standard volume (either µl or ml). A unit is the amount of enzyme required to fully digest a standard amount of a standard type of DNA (typically bacteriophage λ or plasmid pBR322) in a standard length of time. The unit definition can be found on the manufacturers specification sheet. Restriction enzymes in this course are supplied by New England BioLabs.

Enzymes are usually stored at -20o (ordinary freezer temperature) in 50% glycerol. When kept in this manner, they are stable for long periods of time. A serious danger to enzyme stability is repeated warming and cooling. While it is easy to always make sure that the enzymes are cold, a less obvious source of warming and cooling is found in the normal cycling of frost-free freezers. For this reason, enzymes should only be kept in special, non-frost-free freezers (i.e., cheap refrigerators, refreshingly!). Enzymes must be kept on ice at all times whenever they are removed from the freezer.

The worst fate that can befall an enzyme stock is for it to become contaminated with exonuclease from greasy hands. Not only does this ruin the enzyme, but it also ruins the work of unsuspecting users, sometimes destroying precious, hard-to-isolate DNA’s. The following precautions must be observed by everyone:

1. Always wear gloves when handling enzymes. 2. Always use a fresh pipet tip when going into an enzyme stock. If you must go

into the enzyme twice, change the pipet tip. 3. Work quickly. Do not expose the enzyme to warm temperatures any longer

than necessary.

General Protocol for

Performing Restriction Digestions

1. The total volume of the reaction mix is based on the amount typically run on a gel, 20 µl. The reaction mix should contain 0.2 - 1.0 µg DNA. You could, of course, scale this up to do larger digests. The volume of DNA added depends on the DNA prep. Typically, for plasmid purified on a CsCl density gradient, 3 - 5 µl are sufficient. For DNA purified in mini plasmid isolations, 10 µl are typically used.

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2. Add 2 µl of 10 X restriction buffer. Consult the table on page 18 for the appropriate buffer.

3. Add sterile distilled water to bring the volume up to 20 µl.

4. Add 1 µl of restriction enzyme. Tap the tube several times to ensure mixing.

5. Centrifuge the tubes briefly (turn centrifuge on, allow it to get to speed, and turn it off) to concentrate all of the liquid at the bottom.

6. Place in water bath for 30 - 60 minutes. Typically this is at 37o, but you should check page 18 to make sure of the appropriate temperature.

7. At the end of the restriction digestion, do one of the following steps, depending on what you wish to do with the DNA next:

a. If you wish to analyze the results on a gel, add 5 µl of tracking dye to the sample and load the sample into a well on an agarose gel. The tracking dye contains sucrose to increase the density of your sample so that it will settle to the bottom of the well rather than float away, and the dye will enable you to visualize where your sample is.

b. If you wish to ligate your cut DNA to another DNA, you must inactivate the restriction endonuclease. Otherwise, the enzyme could re-cut any successful ligations. This may be done either by a heat treatment or by a phenol extraction followed by ethanol precipitation.

c. If you wish to purify the DNA for any other purposes, you should do a phenol extraction followed by ethanol precipitation.

Tracking Dye

Tracking dye is a sucrose or glycerol solution containing dye that enables you to visualize the electrophoretic front. The sucrose or glycerol is necessary to increase the density of your sample so that it will settle to the bottom of an agarose well rather than float away.

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A variety of dyes are available. Many people use bromphenol blue, xylene cyanol or orange G. The choice depends on the electrophoretic mobility of the dye relative to the DNA fragments. Bromphenol blue and xylene cyanol run slower than orange G. Thus when comparing gels in which each dye is allowed to run to the end, the DNA’s in the bromphenol blue gel will have run farther and separated better than in the orange G gel. But if you are looking at very small fragments, they may have run off the gel with bromphenol blue, but are still present with orange G. Thus the primary consideration for deciding which dye to use is how the mobility of the tracking dye compares to the mobility of the smallest DNA fragments that you are trying to resolve. Orange G

0.25% orange G (Sigma cat # O-1625) dissolved in 50% sucrose.

Bromphenol Blue 0.25 g bromphenol blue 0.25 g xylene cyanol 1.0 ml 1M Tris, pH8 49 ml water 50 ml glycerol

Cold Spring Harbor Laboratory Restriction Digestion Buffers: From A Manual for Genetic Engineering: Advanced Bacterial Genetics.

Final Concentrations Buffer NaCl Tris MgSO4 Dithiothreitol Low 0 mM 10 mM (pH7.4) 10 mM 1 mM

Medium 50 mM 10 mM (pH7.4) 10 mM 1 mM High 100 mM 50 mM (pH7.4) 10 mM 0 mM

10x Stocks for Restriction Assays Low Medium High 5 M NaCl 0 1 2 1 M Tris (pH 7.4) 1 1 5 1 M MgSO4 1 1 1 0.01 M Dithiothreitol 1 1 0 Water 7 6 2 Total Volume 10 10 10

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Restriction Endonucleases: From Molecular Cloning

Enzyme

Common Isoschizomers

Salt

Incubation Temperature

Recognition Sequence

Compatible Cohesive Ends

AvaI med 37oC GâPyCGPuG SalI, XhoI, XmaI BamHI med 37oC GâGATCC BclI, BglII, MboI,

Sau3A BglII low 37oC AâGATCT BstEII med 60oC GâGATCC EcoRI high 37oC GâAATTC EcoB 37oC TGA(N)8TGCT EcoK 37oC AAC(N)6GTGC EcoRI* 37oC âAATT HaeIII med 37oC GGâCC blunt HindII med 37oC GTPyâPuAC blunt HindIII med 37-55oC AâAGCTT KpnI low 37oC GGTACâC MboI Sau3A high 37oC âGATC Bam HI, BclI, BglII,

XhoI PstI med 21-37oC CTGCAâG PvuII med 37oC CAGâCTG blunt Sau3A MboI med 37oC âGATC BamHI, BclI, BglII,

XhoI SmaI XmaI 37oC CCCâGGG blunt TaqI low 65oC TâCGA AccI, AcyI, AsuII,

ClaI, HpaII XbaI High 37oC TâCTAGA XmaI SmaI low 37oC CâCCGGG AvaI

Restriction Enzymes

19

New England BioLabs Restriction Digestion Buffer System

1x Stock BamHI BglII EcoRI HindIII PstI XbaI

NEBuffer 1 10 mM Bis-Tris-Propane-HCl 10 mM MgCl2 1 mM Dithiothreitol pH 7.0 @ 25°C

75% 50% 100% 50% 75% 0%

NEBuffer 2 10 mM Tris-HCl 50 mM NaCl 10 mM MgCl2 1 mM Dithiothreitol pH 7.9 @ 25°C

100% 75% 100% 100% 75% 100%

NEBuffer 3 50 mM Tris-HCl 100 mM NaCl 10 mM MgCl2 1 mM Dithiothreitol pH 7.9 @ 25°C

50% 100% 100% 10% 100% 75%%

NEBuffer 4 50 mM potassium acetate 20 mM Tris-acetate 10 mM Magnesium Acetate 1 mM Dithiothreitol pH 7.9 @ 25°C

75% 50% 100% 50% 50% 75%

NEBuffer BamHI 10 mM Tris-HCl 150 mM NaCl 10 mM MgCl2 1 mM Dithiothreitol pH 7.9 @ 25°C

100%

NEBuffer EcoRI 100 mM Tris-HCl 50 mM NaCl 10 mM MgCl2 0.025 % Triton X-100 pH 7.5 @ 25°C

100%

Restriction Enzymes

20

References

Davis, R.W., D. Botstein, and J.R. Roth. 1980. A Manual for Genetic Engineering: Advanced Bacterial Genetics. Cold spring Harbor Laboratory.

Maniatis, T., E. F. Fritsch, and J. Sambrook. 1982. Molecular Cloning. Cold spring Harbor Laboratory. Roberts, R.J, M. Belfort, T. Bestor, A.S. Bhagwat, T.A. Bickle, J. Bitinaite, r.M. Blumenthal, S.K. Degtyarev, D.T.F. Dryden, K. Dybvig, K. Firman, E.S. Gromova, R.I. Gumport, S.E. Halford, S. Hattman, J. Heitman, D.P. Hornby, A, Janulaitis, A. Jeltsch, J. Josephsen, A. Kiss, T.R. Klaenhammer, I. Kobayashi, H. Kong, D.H. Kruger, S. Lacks, M.G. Marinus, M. Miyahara, R.D. Morgan, N.E. Murray, V. Nagaraja, A. Piekarowica, A. Pingoud, E. Raleigh, D.N. Rao, N. Reich, V.E. Repin, E.U. Selker, P.-C. Shaw D.C. Stein, B.L Stoddard, W. Szybalski, T.A. Trautner, J.L. Van Etten, J.M.B. Vitor, G.G. Wilson, and S.-Y. Xu. 2003. A Nomenclature for Restriction Enzymes, DNA methyltransferases, Homing Endonucleases and Their Genes. Nucl. Acids Res. 31 (7): 1805-1812. DOI: 10:1093/nar/gkg274

21

Agarose Gel Electrophoresis

Agarose gel electrophoresis is one of several physical methods for separating DNA fragments according to size. In this method, DNA is forced to migrate through a highly cross-linked agarose matrix in response to an electric current. In solution, the phosphates on the DNA are negatively charged, and the molecule will therefore migrate to the positive (red) pole. There are several factors that affect migration rate through a gel: the size and conformation of the DNA, the concentration of agarose, and the ionic strength of the running buffer. In this course, we will use only TBE as a running buffer and therefore ionic strength will be constant throughout all of our experiments. Electrophoresis is essentially a sieving process. The larger the fragment of DNA, the more easily will it become entangled in the matrix and, therefore, the more slowly will it migrate. Thus small fragments run more quickly than large fragments, all at a rate proportional to the logarithm of their length. The relationship of size to migration rate is linear on semi-log paper throughout most of the gel, except for the very largest fragments. Large fragments have a more difficult time penetrating the gel and their migration, therefore, does not have a linear relationship to size. The matrix can be adjusted, though, by increasing the concentration of agarose (tighter matrix) or by decreasing it (looser matrix). A standard 1% agarose gel can resolve DNA from 0.2 - 20 kb in length. Most of the DNA’s that we will be examining are plasmids. Plasmid DNA can exist in three conformations: supercoiled, open-circular, and linear. In vivo, plasmid DNA is a tightly supercoiled circle to enable it to fit inside the cell. Following a careful plasmid prep, most of the DNA will remain supercoiled, but a certain amount will sustain single-strand nicks. Given the presence of a break in only one of the strands, the DNA will remain circular, but the break permits rotation around the phosphodiester backbone and the supercoils will be released. A small, compact supercoiled knot of DNA generates less friction against the agarose matrix than does a large, floppy open circle. Therefore, for the same DNA, the supercoiled conformation runs faster than the open-circular form. If the DNA sustains double-strand breaks it produces a linear conformation. Linear DNA runs through a gel end first and thus generates less friction than open-circular DNA, but more than supercoiled, and will migrate at a rate intermediate between the two. In addition, plasmids within a single cell tend to recombine with one another producing dimers, trimers, tetramers, etc. However, as multimer size increases, individual multimers recombine with themselves to produce smaller multimers. Thus, an uncut plasmid produces a complex

Electrophoresis Electrophoresis

22

banding pattern on a gel. The three bands with the highest mobility represent supercoiled, linear, and open circular plasmids, followed by a complex ladder of multimers. If the plasmid is cut once with a restriction enzyme, however, the supercoiled and open-circular conformations and all of the multimers are reduced to a linear conformation. Depending on the amount of manipulation that is incurred during a plasmid prep, much of the DNA may become nicked, converting supercoiled plasmids to linear conformations. Thus for any particular plasmid prep, the amount of supercoiled DNA may be small to nonexistent. Following isolation and storage, spontaneous nicks tend to accumulate as a plasmid prep ages. This can clearly be seen on gels as the proportion of the three conformations change over time.

Agarose gels are referred to as submarine gels because the slab is laid horizontally and is completely covered by running buffer. The GE lab has a number of different gel boxes in two basic sizes. The larger boxes have gel beds of approximately 11 x 14 cm. The smaller boxes typically referred to as “baby gels” have gel beds of approximately 50 x 75 cm. In most cases, the gel tray is removable and the gel is poured outside of the box. Each type of gel box has its own unique way of sealing the gel bed to prevent leakage of the agarose. Please refer to the instructor or TA if you are unsure. Baby gels are used for quick checks. Their resolution isn’t great but the gel runs within 30 to 40 minutes and is very useful for monitoring longer reactions. The larger boxes can be run either with a single comb at the top of the gel, or “piggy-back” with a second set of combs in the middle. In this way, twice the number of samples can be run, but the resolution is similar to that of a baby gel.

Preparing a Gel

1. Gels are prepared as percentage weight/volume solutions. That is, the weight of agarose in grams per 100 ml running buffer. Thus, a 1% gel is 1 g agarose in 100 ml buffer. You can, of course, scale up the volumes accordingly. Note; One of the most common beginner’s mistake is to make up the agarose in water instead of running buffer. If you do so, your gels will look very strange. At some point you will need to use a UV-fluorescent dye to visualize the DNA on your gel. This can be added at this point, or after you finish the running the gel. See the section below about gel visualization. 1% is a standard concentration that permits visualization of a broad range of fragment sizes, but the largest and smallest fragments are poorly resolved. If you are

Electrophoresis

23

2.

specifically trying to resolve large fragments of DNA, you may want to use a lower concentration of agarose. Alternatively, if you are trying to resolve small fragments, a higher agarose concentration would be appropriate. Agarose does not dissolve. Rather, it has to be melted. Typically, this is done in a microwave. The microwave should be set to “micro cook” for about 2.5 minutes at a power setting of 7. You should watch it carefully while it is melting so that it doesn’t boil over. IF YOUR AGAROSE BOILS OVER, MAKE SURE TO CLEAN UP THE MESS!

3. When melted, allow the agarose to cool. Before you pour your gel, it should be cool enough that you can hold it comfortably in your bare hand. Gel trays are made of UV-transparent plastic, which is very expensive. If you pour the gel while it is too hot, you run the risk of warping and ruining the gel tray. Please be careful.

4. Seal the gel bed as appropriate for your box, insert the combs, and pour the agarose into the tray. You should make the gel about 5 - 7 mm thick (you will gain a feel for the proper depth once you have done several). Insert the comb and allow gel to harden.

5. When the gel hardens, add buffer to both reservoirs and cover the gel to a depth of about 2 mm. Very carefully wiggle the comb out of the gel, taking care not to tear the wells.

Running a Gel 1.

2.

Load your gel (for example with a restriction digest) and attach electrodes. Remember: DNA is negatively charged and runs towards the positive electrode. The black electrode should be closest to your samples and the red electrode farthest (DNA should “run to the red”). Turn on power and run for the appropriate length of time. For most applications, running the gel until the tracking dye travels ¾ of the length of the gel is sufficient. The baby gel can run at about 60 - 80 volts for about 40 minutes. The larger gels can be run at 100 - 105 volts for about 2 hours, or at 15 volts for an


24

overnight electrophoresis. There is a heat differential across the surface of a gel; the outside lanes are cooler the inner lanes and you may see edge effects in which the outside bands are tilted. It is best to avoid the outside lanes whenever possible. Running at greater voltages will result in heating of the gel and distortion of the bands. You can be sure that your gel is running by checking for bubbles from the electrodes. Caution: Gels Run at High Voltage and Can Deliver Powerful Electric Shocks!

3. At the end of the run, shut off the power and disconnect the electrodes.

4. The next step is visualization. The procedure depends on whether you are using ethidium bromide or Gel Red to stain your DNA, and whether you added the dye directly to the agarose, or are staining the gel after the run.

In order to see DNA on a gel, the gel must first be stained with a dye that binds to DNA and fluoresces under ultraviolet light. Traditionally, ethidium bromide (EtBr) has been used for this purpose. EtBr is mutagenic and must be handled as hazardous waste. More recently, nontoxic dyes have been introduced. In this lab, we will mostly use Gel Red. It is sometimes useful to have the dye present while the gel is running because you can always interrupt the run, check the location of your DNA fragments, and then continue if you wish to run them farther. However, at the end of the experiment you will end up with lots of waste, toxic in the case of EtBr. Even more importantly, EtBr alters the conformation of the DNA, thereby altering the migration rate. Large fragments contain more EtBr than smaller fragments, so the rate change would not be constant over the range of fragments. Depending on the experiment, this may or may not be a problem. However, if you are trying to generate a restriction map and would like to measure fragment sizes accurately, it is always best to run the gel in the absence of the dye.

Electrophoresis

25

Visualization With Stain Added Prior to

Electrophoresis

1.

2.

3.

4.

Agarose is prepared in 200 µl quantities Gel Red: add 8 µl of Gel Red per 200 ml, and swirl to mix. Then microwave to melt. Ethidium Bromide: While agarose is still molten, add 0.5 mg/ml EtBr to both the agarose and the running buffer. Ethidium Bromide is a Powerful Mutagen. Always Wear Gloves, Glasses, and Lab Coat When Handling It! Pour gel when cool. Allow the remainder to solidify. When you next use the agarose, merely melt it.

Visualization Following

Electrophoresis

1.

2.

3.

Carefully transfer the gel to a staining tray. The first time you stain a gel, cover it with about 100 ml of TBE and add 5 µl of Gel Red OR 15 µl of EtBr (10 mg/ml). When you add the stain, take care not to pipet it directly onto the gel. Some could stick to the gel and cause an unsightly fluorescent spot (usually in the most critical place). Place the tray on a shaker for 20 minutes. Remove the gel from the tray and lay it on the tray lid. Return the staining solution to an empty container. Briefly rinse the gel with water to remove excess stain.

. The second and all subsequent times that you stain gels, you will use your used staining solution. Later in the term, it may be necessary to “freshen” the stain. At the end of the quarter, the Gel Red solution will be discarded. EtBr will be collected and properly disposed of.

4. Destaining: Occasionally the gel absorbs a background of ethidium bromide, which could, if heavy enough, obscure some bands. Usually it is not necessary to destain the gel, but if your bands are faint, destaining may help. Destaining is accomplished by soaking the gel in an excess of water for about an hour.


26

1X TBE Running Buffer Final

Concentration

Water (liters) 1.0 2.0 2.5 3.0 Tris Base (grams) 89 mM 10.8 21.6 27.0 16.5 Boric Acid (grams) 89 mM 5.5 11.0 13.6 16.5 Disodium EDTA (grams) 2.5 mm 0.93 1.85 2.3 2.8

Capturing a Gel With The BioDoc-It Gel

Documentation System

1. Turn on main power switch

2. Turn on Transiluminator

Electrophoresis

27

3. Lay the gel on the Transiluminator (UV is automatically switched off when main door is open). After you lay your gel on the Transiluminator, you should slide it to one side and wipe up the excess water. If there is too much water on the Transiluminator, your gel will drift out of position while you are trying to photograph it. You can safely view your gel under UV by opening the UV-Blocking Gel Viewer door. You can manipulate your gel by inserting your gloved hands through the side doors.

4. While watching the LCD, rotate the f-stop ring until the image is bright enough to see on the monitor (the lower the f-stop number, the brighter the image will be).

5. Focus the image if necessary.

6. Twist the zoom ring to adjust the image size as

appropriate

7. Fine-adjust the brightness of the image by pressing the “+” or “-“ buttons on the touch pad to brighten or darken, respectively, the image.

8. When the image is satisfactory, press the “Capture” button. The word “Frz” will display at the bottom of the screen. This will hold the image on screen to be viewed, saved, or printed.

9. Press “Save” to record the image to the BioDoc-It’s memory. If you insert a CF card, it will save to both the internal memory and the card. The memory is limited and your image can be quickly overwritten if there is heavy use. The BioDoc-It will save the image as a TIFF file and assign it a unique number (UVP#####). Record the number for future reference.

10. Print your image by pressing the “Print” button on the adjacent thermal printer. This will give you a small, but very clear image for immediate analysis. For more detailed analysis, you should work with your recorded image.


28

Accessing a Gel Image File

1. Directly from the CF card

a. Insert the CF card into a card reader attached to your computer.

b. Open the image directly with your favorite image

editing software.

2. From the BioDoc-It (from CF card only)

a. Insert the CF card

b. Press “Set Up”

c. Use the “+” and “-“ buttons to navigate to the line “READ IMAGES.”

d. Use the “+” and “-“ buttons to navigate to the desired

file.

e. Press “Set Up” to open the file.

f. Print your image by pressing the “Print” button on the adjacent thermal printer.

3. Via E-mail

a. Access the BioDoc-It via the lab computer (or your

own, remotely). b. Transfer the image to the computer. c. Access your RIT account and email the image to

yourself.

4. Remote access to the BioDoc-It memory

a. Point your favorite web browser to ftp://129.21.156.188 (there is a hotlink to this site on the Genetic Engineering home page)

b. The BioDoc-It can also be accessed from an iPad or

other type of tablet. To do so from an iPad (and probably other tablets as well) you will need an FTP app.

Electrophoresis

29

c. You will be asked for a username and password, but these are not required. Merely press “RETURN”.

d. Open the desired file with your favorite image

processing software.

e. Note: in order to access images remotely, the BioDoc-It needs to be left on.


30

Analyzing a Gel

1. Sometimes only a visual analysis is necessary to see whether bands have changed, disappeared, etc. relative to controls.

2. To calculate molecular weights, you must first measure the distance migrated by each band in each lane, and record this in a table.

3. Compare your molecular weight standards with the key above. You will notice that bands closer to the wells are more compressed than bands farther away. Moreover, bands that are farthest from the wells are indistinct and often missed. Thus, you will usually misidentify the bands if you simply count from one end to the other. A better idea is to match up the bands according to spacing and pattern.

For example, the 1 kb band of 1 kb ladder standard is always clear and distinguishable. Find this band on your gel and then count in both directions until you lose confidence in your ability to identify bands. Once you have identified the bands, enter the sizes onto your table of distances migrated. Now you can plot your standard graph.

4. DNA runs in a gel as a function of the logarithm of its molecular weight. Therefore, you must plot the graph of your gel on semi-log paper. For more on plotting logs, see page 31.

5. If you run two standards, they should be plotted on the same graph and they should fall on the same curve. If they do not, then you have most likely misidentified the bands.

6. Once you have plotted your standard curve, locate the distance of your unknown bands, which you have already measured, on the standard curve. Now you can read the molecular weight directly off of the log scale.

7. A basic knowledge of logarithms is helpful in understanding how semi-log paper works.

Electrophoresis

31

Our Friend The Logarithm Believe it or not, logarithms were created to help you and make mathematics easier! Logs were invented in the early part of the seventeenth century to meet the needs of astronomers who had to deal with very large numbers. Logarithms, then, are shorthand expressions of large numbers. In micro- and molecular biology we also have to deal with large numbers and logs makes the task much simpler. For a definition of logarithms, consider the equation

ay = x where "a" is a positive number not equal to 1. y, then, is said to be the logarithm of x to the base a, or:

y = logax

Theoretically any number may be used as a base for a log system, but in practice, only two are used routinely. In molecular biology we will be concerned with one of these, base 10, otherwise known as the common logarithm. In the following example using base 10,

102 = 100 or 2 = log 10100

In the following examples:

if log 2 = 0.3010, then 2 = 10 0.3010 if log 3 = 0.4771, then 3 = 10 0.4771 if log 4 = 0.6021, then 4 = 10 0.6021

Logarithms are usually expressed as a whole number plus a decimal. The decimal portion is called the mantissa and is the exponent of 10 used to derive the number in question. The whole number portion is called the characteristic and is used to determine the decimal point. The characteristic is usually one less than the number of decimal places. As the characteristic increases by n, the decimal point is moved n places to the right.

log 2 = 0.3010 log 20 = 1.3010 log 200 = 2.3010 log 2000 = 3.3010

The relationship of the characteristic to the mantissa can be simplified by applying the first of the four log rules (see below): when multiplying using logs, the logs are added. Thus in the examples above, 2000 can be rewritten as 2 x 1000, or 2 x 10 3. The log of 2000, then, is log 2 + log 1000, or 0.3010 + 3 = 3.301. The mantissa, then, really is added to the characteristic, not merely tacked on.


32

When computing the log of a decimal, say 0.02, we can express the number as 2 x 0.01, or 2 x 10 -2. The log then becomes

log 2 + log 10 -2 = 0.3010 + (-2) = -1.699 If you go to a log table, however, and ask "what number has a mantissa of .699", you will find 5 as the answer. This error (you started with 2) is a result of subtracting the characteristic. It is therefore more convenient to have a positive number for the mantissa. To do this we add 10 and then subtract 10 to the characteristic. Thus

log 0.02 = log 2 + log 10 -2 + 10 - 10 =0.3010 + (-2) + 10 - 10 =0.3010 + 8 – 10 = 8.3010 - 10

In this form .3010 is the mantissa of 2.0 and 8 -10 tells you that the characteristic is -2 and that the decimal in 2.0 should be moved two places to the left to give 0.02. Many problems are worked by finding the log of the variables and computing the log of the answer. To find the final answer, one can work backwards in a log table and look up the number whose logarithm is X. This is called the antilog. Thus the antilog of 0.3010 is 2, the antilog of 0.4771 is 3, etc. In working with logs, there are four basic rules to remember:

1 When multiplying by logs, the logs are added:

2 x 4 = log 2 + log 4 = 0.3010 + 0.6021 = 0.9031: antilog 0.9031 = 8

2 When dividing by logs, the log of the divisor is subtracted from the log of the dividend:

40 / 8 = log 40 – log 8 = 1.6021 – 0.9031 = 0.6990: antilog 0.6990 = 5

3 When raising a number to a power, the log of the number is multiplied by the exponent:

82 = 2 x log 8 = 2 x (0.9031) = 1.8062: antilog 1.8062 = 64

4 When taking the root of a number, the log is divided by the root:

3√ 27 = log 27 / 3 = (1.4314) / 3 = 0.4771: antilog 0.4771 = 3

By using this brief review of logarithms and the 4 log rules, you should have no difficulty in mastering any mathematical problems in molecular biology.

Electrophoresis

33


34

Plotting on Semi-Log Paper The abscissa (X axis) of semi-log paper is linear, that is, the markings are evenly spaced. Along this axis, you plot the distance migrated. On the ordinate (Y axis), the numbers from 1 to 9 vary in spacing and then repeat. The number corresponds to where the logarithm of the number would be plotted if you were plotting on linear paper. Each repeat is a cycle and cycles differ from one another by a factor of 10. The log scale has no zero and the decimal values that you assign to each cycle are arbitrary. Thus, for example, two consecutive cycles could read:

1 2 3 4 5 6 7 8 9 10 20 30 40 50 60 70 80 90 100

or

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 3 4 5 6 7 8 9 10 In the example to the right, a strip of semi-log paper is placed next to a strip of linear graph paper and the set of values below are plotted. On the semi-log graph the values are plotted according to the numerical value. On the linear graph, the numbers are plotted according to their logarithm. You can see that the points fall in the same place on both graphs.

Number Logarithm 2 0.3 4 0.6 8 0.9 20 1.3 40 1.6 80 1.9 200 2.3 400 2.6 800 2.9

Electrophoresis

35

Relative Migration Distance Occasionally you may wish to directly compare fragments run on different gels. For example, you wish to plot the results of different gels on the same graph. Or you may wish to calculate a fragment size from a gel with no molecular weight standard, or one that was poorly resolved. To do this, you can calculate Relative Migration Distances.

Plotting Curves by Relative Migration

Distance

1. The distance traveled by a DNA fragment is proportional not only to its size, but also to the time that the gel was allowed to run. Thus the same DNA fragment run on two different gels will not be directly comparable. You can directly compare different gels by plotting not actual distance migrated, but relative distance.

2. Relative distance s based on the following argument: Fragment #1 runs twice as fast as fragment #2. When comparing different gels:

• If #1 runs 4 cm, #2 will run 2 cm. • If #1 runs 3 cm, #2 will run 1.5 cm. • If #1 runs 5 cm, #2 will run 2.5 cm, etc.

Thus the overall pattern of bands will be the same even if the distances between bands are expanded or compressed. If we arbitrarily assign a value of 1.0 to fragment #1, then fragment #2 will be 0.5.

3. To calculate relative distance, arbitrarily pick one fragment to be your standard. I usually use the 1 kb band of 1 kb ladder standard. Measure the distance that it has run and set that equal to 1.0. Measure the distances traveled by all of the other bands and divide each distance by your standard. Instead of plotting cm traveled, you can now plot distance traveled relative to the standard. In this way, you can compare any gel, run for any length of time.

36

Restriction Mapping

At some point in a cloning project it will be necessary to construct a restriction map of a plasmid. This will involve manipulating restriction digest fragments into a circular map. Once you have done this a number of times and develop a “feel” for the process and mapping will become almost intuitive. However, there is a systematic approach to constructing a map, which is illustrated in the sample problem below. Once you have mastered the sample problem, complete the remaining problems, plot the answers on polar coordinate graph paper, and hand them in. Be cautioned that working with actual restriction fragments from a gel is a bit messier than these problems because of experimental error in determining fragment sizes. Sometimes the mathematical approach used here needs to be supplemented with some logic. Sample Problem Plasmid pRIT450 is 7.0 kb in length and has single PstI, EcoRI, and BamHI sites. You have cut the plasmid with PstI and inserted a 4.0 kb fragment into the site. From the data below, determine the restriction map of the resulting plasmid.

PstI 7.0 4.0 EcoRI 6.0 5.0 BamHI 8.9 2.1 PstI + EcoRI, 4.3 3.3 2.7 0.7 PstI + BamHI 6.1 2.8 1.2 0.9 EcoRI + BamHI 5.0 2.1 2.1 1.8

Solution to Sample Problem 1. If the new plasmid is cut with PstI, the vector and the insert will be regenerated. We will use these

as reference fragments as we build the map. Although it is often convenient to think in terms of target and vector, any site can arbitrarily be used to generate reference fragments. We will now begin by choosing PstI as our reference point, but we could have just as easily chosen to start with EcoRI or BamHI. It makes no difference.

2. First, look at the PstI + BamHI digest. It helps to think of the double digest conceptually as a two-step process in which the plasmid is first cut with PstI into the two reference fragments. BamHI then cuts the reference fragments further.

Restriction Mapping

37

3. In the PstI + BamHI double digest, we recover 4 fragments. Some combination of the fragments will add up to the vector and the remaining fragments will equal the insert. We see that 6.1+ 0.9 = 7.0 = vector and 2.8 + 2.1 = 4.0 = insert. Thus, we can organize the fragments as below:

P B P P B P 6.1 0.9 1.2 2.8 7.0 Ref 4.0 Ref

4. The insert and the vector can each be drawn in two opposite orientations. However, we will

arbitrarily use one orientation for the vector and vary the insert. It we tried to vary both vector and insert, we would end up with two identical but mirror image maps. There is, however, no absolutely correct orientation of a plasmid map, so it really makes no difference:

P B P P B P

à 6.1 0.9 1.2 2.8 Orientation A 7.0 Ref 4.0 Ref

P B P P B P XXX 0.9 6.1 2.8 1.2 Orientation B

7.0 Ref 4.0 Ref We next must next determine which orientation, A or B, is correct. In orientation A, the two BamHI sites are close together (proximal) while in B, the sites are far apart (distal). Clearly, a BamHI single digest of each orientation will result in a different set of fragments. We can therefore determine expected fragment sizes for each orientation and then compare our expected values with the values observed:

Orientation A Orientation B Observed

0.9 + 1.2 = 2.1 0.9 + 2.8 = 3.7 2.1 6.1 + 2.8 = 8.9 6.1 + 1.2 = 7.3 8.9

11.0 11.0 11.0

6. By comparing our observed and expected values, we find that orientation A is the correct orientation.

7. Using exactly the same logic, we can now show that the correct orientation of the EcoRI sites is as below:

P E P P E P 4.3 2.7 3.3 0.7

7.0 Ref 4.0 Ref

8. To complete the map, we must now superimpose the BamHI and EcoRI maps. Let us arbitrarily assume that the orientation of the BamHI map is correct. We can now superimpose the EcoRI map on the BamHI map in either of two orientations, A or B. Using the alternate BamHI orientation would simply produce a mirror-image map.

Restriction Mapping Restriction Mapping

38

P

P

P

P

P

P

P

P

P

P

B

B

B

B

A

E

E

E

E

E

E

B

B

E

E

B

6.1

0.91.2

2.8

2.7

3.3

0.74.3

0.7

3.3

2.7

4.3

0.91.2

2.10.7

4.3

1.8

0.9 0.70.5

2.8

2.7

3.4

9. From the alternative orientations, A and B, we can predict the outcome of an EcoRI + BamHI

double digest:

Orientation A Orientation B Observed 4.3 + 0.7 = 5.0 2.7 + 2.8 = 5.5 5.0

2.1 0.5 2.1 1.2 + 0.9 = 2.1 0.7 + 0.9 = 1.6 2.1

1.8 3.4 1.8 11.0 11.0 11.0

10. By comparing the results of the EcoRI + BamHI double digest with our predictions, we see that

orientation A is correct and the plasmid is now mapped. In step 8, we arbitrarily chose one of the two possible orientations for the BamHI map. Using the other, as noted, would have given the mirror image. Try it. Neither mirror image is more correct than the other. However, once the map is published, the published orientation by convention becomes the “correct” orientation. If the plasmid was meant to be a new cloning vector, then one would map all plasmids subsequently derived from it in the same orientation.

11. The final step is to plot the map on polar coordinate graph paper so that the fragment sizes are all proportional to one another. On polar coordinate graph paper, you plot degrees, not kilobases.

Restriction Mapping

39

Thus an entire circle, irrespective of total kilobases, will be 360o. Since the map in this case is 11 kb long, then:

360o/11kb = 32.7o/kilobase.

Now we multiply each kb value by 32.7 to obtain the degrees below:

kilobases degrees

1.2 39.2 2.1 68.7 0.7 22.9 4.3 140.6 1.8 58.9 0.9 29.4

11.0 359.9

The last step is to decide on the circumference of the circle. If you are mapping several plasmids, you would want to draw them to scale. You can adjust the total size of your map by altering the circumference. Thus, for example, if you wanted to draw your map to the scale of 1 inch = 1 kb, then circumference = 11 kb x 1 inch/kb = 11 inches. You can determine the radius of a circle with a circumference of 11 inches by applying the formula:

2 π r = C where C = circumference and r = radius, and solving for r:

2 π r = 11 in r = 11 in/2 π

r = 1.75 in The map of the sample plasmid on the following page is graphed in just this way. The positions of the restriction sites are plotted according to the table above and the circumference of the outer circle is 11 inches, with a radius of 1.75 inches. Note, however, that the distances are given as kilobases, not degrees!


40

Now, determine the restriction maps for the remaining plasmids and hand them in as homework.

Mapping Problems

Plasmid pRIT451 was cut with SmaI, BglII, and AvaI. From the data below, determine the map. SmaI 5.9 4.3 BglII 8.2 2.0 AvaI 5.3 4.9 SmaI + BglII 5.4 2.8 1.5 0.5 SmaI + AvaI 3.3 2.6 2.3 2.0 AvaI + BglII 5.3 2.1 2.0 0.8

Restriction Mapping

41

Plasmid pRIT452 was cut with PstI, HindIII, and EcoRI. From the data below, determine the map. PstI 6.8 5.9 HindIII 6.5 6.2 EcoRI 9.2 3.5 PstI + HindIII 4.8 4.2 2.0 1.7 PstI + EcoRI 5.4 3.8 3.0 0.5 EcoRI + HindIII 6.2 3.5 1.8 1.2

Plasmid pRIT453 was cut with SmaI, HindIII, and EcoRI. From the data below, determine the map. EcoRI 7.7 1.6 HindIII 7.4 1.9 SmaI 6.6 2.7 EcoRI + HindIII 4.5 1.9 1.6 1.3 EcoRI + SmaI 5.7 2.0 0.9 0.7 SmaI + HindIII 2.7 2.5 2.2 1.9

Plasmid pRIT454 was cut with PstI, HindIII, and EcoRI. From the data below, determine the map. PstI 6.0 5.3 HindIII 5.8 5.5 EcoRI 6.5 3.0 1.8 PstI + HindIII 4.0 3.8 2.0 1.5 PstI + EcoRI 3.5 3.0 2.5 1.8 0.5 EcoRI + HindIII 5.0 3.0 1.5 1.0 0.8


42

Plasmid pRIT455 was cut with BamHI, HindIII, and EcoRI. From the data below, determine the map. EcoRI 8.5 2.0 0.5 HindIII 5.6 5.4 BamHI 6.5 4.5 EcoRI + BamHI 4.5 2.2 2.0 1.8 0.5 EcoRI + HindIII 4.3 4.2 1.2 0.8 0.5 BamHI + HindIII 3.4 3.1 2.5 2.0

Plasmid pRIT456 was cut with PstI, HindIII, and EcoRI. From the data below, determine the map. EcoRI 5.8 3.3 2.9 PstI 7.4 4.6 HindIII 6.8 5.2 PstI + EcoRI 3.3 3.3 2.5 2.1 0.8 PstI + HindIII 6.3 4.1 1.1 0.5 EcoRI + HindIII 3.6 3.3 2.2 1.6 1.3

Plasmid pRIT457 was cut with BamHI, HindIII, and PstI. From the data below, determine the map. BamHI 5.1 4.5 3.8 HindIII 8.1 5.3 PstI 13.4 BamHI + HindIII 5.1 3.3 2.5 2.0 0.5 BamHI + PstI 5.1 4.5 2.5 1.3 HindIII + PstI 8.1 3.3 2.0

43

Experiment 1

Isolation of Escherichia coli Chromosomal DNA

In this lab we will isolate E. coli chromosomal DNA for use in several experiments throughout the semester. The basic protocol involves lysing the cells with lysozyme and SDS. Peptidoglycan strips away the peptidoglycan in the cell wall, leaving only the lipopolysaccharide (LPS) layer to hold the cells together. Such cells are referred to as protoplasts or spheroplasts. Final lysis occurs when sodium dodecyl sulfate (SDS), a mild detergent, is used to dissolve the LPS. The lysate is then treated with proteinase K and RNase A to degrade proteins and RNA, respectively. Next, the lysate is treated with phenol/chloroform to remove all soluble proteins. Finally, the DNA will be precipitated with ethanol and recovered by spooling it on to a glass hook. We will analyze the DNA by UV spectroscopy in Experiment 2.

Materials 250 ml overnight E. coli culture

lysozyme solution (5 mg/ml) proteinase K solution (100 mg/ml) RNase A solution (10 mg/ml)

10% SDS phenol/chloroform ice-cold ethanol

Procedure 1. Obtain a 250 ml flask of cell culture.

2. Centrifuge the cells in 250 ml bottles for 10 minutes at 7000 rpm.

3. Resuspend the cells in 10 ml of TE buffer and transfer to a 50 ml orange-capped conical centrifuge tube.

4. Add 0.5 ml of lysozyme and incubate at 37O for 30 minutes.

5. Incubate for 10 minutes at 65O to inactivate cellular nucleases.

6. Add 100 ul of proteinase K and 20 ul of RNase A to your sample. Incubate at 37O for 30 minutes.

7. Add 0.5 ml 10% SDS and rock gently.

Experiment 3

44

8. Add 5 ml phenol/chloroform. Rock for 10 minutes and then spin briefly in a clinical centrifuge. Phenol/chloroform is immiscible in water and centrifugation will produce two phases. The lower organic phase contains dissolved protein. The upper aqueous phase contains DNA. Carefully transfer the aqueous phase to a fresh tube. The organic phase is discarded in the organic waste bottle.

CAUTION: PHENOL/CHLOROFORM IS CAUSTIC YOU MUST WEAR SAFETY GLASSES AND GLOVES!!

9. Repeat the phenol/chloroform treatment.

10. Transfer the aqueous layer to a fresh tube and gently add 5 ml of

ice-cold ethanol. The ethanol will form a layer above your DNA sample, and you should see some DNA precipitate at the interface.

11. Take several long-tipped Pasteur pipettes and heat the ends in a flame to form a hook. You will probably want four.

12. Spool the DNA out of the tube by inserting the hook down into the DNA layer and then drawing it up while gently swirling. You should see viscous DNA clinging to the hook. Set the hook aside and allow the DNA to dry.

13. When you have spooled out all the DNA wash each hook three times in cold ethanol. Washes are performed by swirling each hook in a fresh tube of ethanol.

13. When the hooks are dry, CAREFULLY snap off the ends of each into a 15 ml orange-capped centrifuge tube. Add 3 ml of TE buffer and allow the samples to shake on a rocker platform until next lab.

14. At the next lab, transfer the dissolved DNA to a fresh tube.

TE Buffer 10 mM Tris, pH7.6 1 mM EDTA

45

Experiment 2

Determination of DNA Concentration and Purity by Ultraviolet Spectrophotometry Typically, the first thing one wants to know about a DNA preparation is its concentration and purity. Both can be determined by measuring the absorption of ultraviolet light. DNA absorbs UV more or less strongly depending upon the wavelength. We will be using a NanoDrop spectrophotometer, which takes measurements at wavelengths of 260 and 280 nm, and in addition determines an absorption spectrum from 220 – 350 nm. The graph below shows that maximum absorbance occurs over a broad peak around 260 nm. At 280nm DNA only absorbs about half as much UV.

Although 260 nm is considered to be the de facto peak, the actual peak absorbance varies somewhat from DNA to DNA. UV absorption is a property of the bases, and each base absorbs differently. The actual peak absorbance of a particular DNA, then, depends on its base composition. Since different DNA’s have somewhat different base compositions, the actual absorbances will also be somewhat different.

Experiment 2 Experiment 2

46

DNA concentration can be determined by the equation:

1 OD260 unit = 50 µg/ml

The NanoDrop automatically calculates concentrations and records them as ng/µl (1 ng/µl =1 µg/ml). Concentrations of single-stranded DNA and RNA can also be determined by measuring absorption at 260 nm:

Nucleic Acid Concentration (µg/ml) per A260 Unit ds DNA ss DNA ss RNA

50 33 40

It is clear from the above table that single-stranded DNA absorbs more UV than double stranded DNA. This is due to interactions between the stacked bases in double-stranded DNA. The difference can be also demonstrated directly by comparing the OD’s of double-stranded DNA and DNA that has been denatured by boiling. The change in OD is referred to as the hyperchromic shift. Single nucleotides absorb UV more strongly than single-stranded DNA. Protein also absorbs UV and can be quantitated by spectrophotometry. However, as can be seen in the graph above, there are two peaks of absorbance, one at 230 nm and the other at 280 nm. The peak at 230 nm is due to absorbance by the peptide bonds while the 280 nm peak is due to absorbance by the rings of aromatic amino acids (tryptophan, tyrosine, and phenylalanine). There is a significant dip in absorption at 260 nm, where DNA absorbs maximally. 280 nm is typically used to determine the concentration of protein. In the same way that different DNA’s have different base ratios, different proteins have different proportions of aromatic amino acids, and the amount of absorption at 280 nm will vary from protein to protein. However, as a rule of thumb:

1 OD280 unit = 1 mg/ml protein In most DNA preparations, the final step is the separation of DNA from protein. Carryover protein during a DNA prep could lead to problems with subsequent operations, such as cutting with a restriction endonuclease. Assessment of purity is therefore important. The most commonly used assay is the A260/A280 ratio. As noted above, DNA absorbs almost twice as much UV at 260 nm as it does at 280 nm. If there is no protein present, the target ratio is 1.8. However, because 280 nm is a peak for protein absorption, protein contamination will increase the A280 reading but have little effect on the A260

reading. Thus the A260/A280 ratio will be lower than 1.8.

Purity of Target A260/A280 Ratio DNA RNA Protein

1.8 2.0 0.6

Occasionally one recovers A260/A280 ratio greater than 2. The meaning of this is not easy to find in the literature, but it appears to be related to DNA degradation and measurement of free nucleotides. Because protein also absorbs at 230 nm, theoretically the A260/A230 ratio could be used to detect protein contamination. However, this is rarely used because common buffers such as Tris also absorb at 230 nm. Nevertheless, the

UV Spectrophotometry of DNA

47

NanoDrop does a A260/A230 calculation. However, this measurement is adjustable and can be set for whatever additional wavelength one desires. Some DNA extraction protocols require the use of phenol. While protein contamination is not necessarily a fatal problem, phenol contamination most definitely is. Phenol absorbs maximally at 270 nm and can have an impact on the A260/A280 ratio. When assessing DNA purity it is important to understand that while the A260/A280 ratio is easy to determine and is the most widely used method, it is not particularly robust. DNA absorbs so strongly at 260 nm that it takes significant protein contamination to have a noticeable effect on the A260/A280 ratio. On the other hand, the A260/A280 ratio is a particularly robust method for assessing DNA contamination of protein preparations. Because DNA and RNA are so similar, spectrophotometry cannot be used to detect contamination of DNA by RNA, and vice versa. In this exercise, you will determine the concentration and purity of the E. coli that you prepared in the previous lab. You will also compare the difference between double-stranded and single-stranded DNA, and study the effect of protein contamination on A260/A280 ratios. Concentration and Purity

of E. coli Chromosomal DNA

1

2

Use the NanoDrop to measure the optical density of the chromosomal DNA you isolated in the previous lab. Calculate the concentration and purity, and print out a graph of the absorption spectrum.

Hyperchromic Shift 1

2

Transfer 10 µl of your chromosomal DNA to a microfuge and boil it for 30 minutes. Calculate the concentration and purity, and print out a graph of the absorption spectrum.

Affect of Protein Contamination on UV

Absorption Spectrum and A260/A280 ratios

1 Obtain a tube of bovine serum albumin (BSA) of unknown concentration.

2 Serially dilute the BSA to 100, 10-1, 10-2, and 10-3.

3 Make two measurements of OD at 280 nm, once using the Nucleic Acid program on the NanoDrop, and once using the Protein program.

4 Mix each of the diluted BSA samples with an equal amount of undiluted DNA.

5 Measure the concentrations and A260/A280 ratios of the mixed samples using the Nucleic Acid program on the NanoDrop, and print out the absorption spectrum for each.


48

Analysis 1

Create a table in your notebook similar to the one below to record your date. Alternatively, there is a copy of this table on line that you can print, cut out, and tape into your notebook.

2

Generate a set of five graphs, each showing the COMBINED DNA and BSA data generated from the Nucleic Acid program. 1 Unmixed DNA and unmixed BSA @ 100 (Samples 1 & 2) 2 DNA + BSA @ 100 (Samples 2 & 10)

3 DNA + BSA @ 10-1 (Samples 3 & 11)

4 DNA + BSA @ 10-2 (Samples 4 & 12)

5 DNA + BSA @ 10-3 (Samples 5 & 13)

You can trace the graphs in order to superimpose them.

Sample NanoDrop Program A260 A280 Concentration A260/A280

1 DNA Control

Nucleic Acid

2 BSA @ 100

3 BSA @ 10-1

4 BSA @ 10-2

5 BSA @ 10-3

6 BSA @ 100

Protein

7 BSA @ 10-1

8 BSA @ 10-2

9 BSA @ 10-3

10 DNA + BSA @ 100

Nucleic Acid

11 DNA + BSA @ 10-1

12 DNA + BSA @ 10-2

13 DNA + BSA @ 10-3

49

Experiment 3

The Chemical Nature of DNA Chemically, DNA is a polymer composed of four simple repeating subunits, the nucleotides. Each nucleotide is comprised of a molecule of deoxyribose, a phosphate group, and one of four nitrogenous bases, adenine and guanine (purines) and thymine and cytosine (pyrimidines). The bases are linked to a molecule of deoxyribose, a five-carbon sugar, at the 1’ position, and the deoxyribose molecules are linked to one another by phosphodiester bonds between the 3’ and 5’ positions of adjecent nucleotides. DNA is double-stranded and the strands are held together by hydrogen bonds between the bases. Adenine on one strand always pairs with thymine on the other, and guanine always pairs with cytosine. The sequence of nucleotides determines the genetic code. Sets of three nucleotides are referred to as codons and specify the inclusion of a particular amino acid in a protein. In this experiment we will characterize the chemical nature of DNA. The chemical nature of DNA can be analyzed by acid hydrolysis and thin-layer chromatography in order to visualize the four bases. In acid hydrolysis, DNA is treated with formic acid at very high temperature and pressure. This treatment will reduce the DNA to its component parts, leaving the free bases intact. The sample is then analyzed by thin-layer chromatography, during which each base migrates up the silica gel chromatogram at its own unique rate. Following the chromatography, the bases can be visualized with ultraviolet light. The chromatography sheet contains a fluorescent dye, but the bases will absorb UV, creating dark spots against a bright background. The intensity of the spots can be used to calculate the concentration of each base. It was through experiments on DNA using the then new technology of paper chromatography that Erwin Chargaff discovered that the ratios of the four bases vary from species to species, but the amount of adenine in DNA always equals the amount of thymine, and guanine always equals the amount of cytosine. This discovery disproved the theory that DNA was merely a repetition of the four base sequence ATGC and laid the groundwork for Watson’s and Crick’s discovery of the structure of DNA.

Experiment 3

50

Materials

First Day: 5% trichloroacetic acid (TCA) 95% ethanol formic acid

Third Day Tubes of: Adenine, Thymine, Cytosine, and Guanine Chromatography Solvent

YOU MUST WEAR SAFETY GLASSES DURING ALL PARTS OF THIS EXPERIMENT

Procedure for First Day

1 Transfer 150 µg of E. coli chromosomal DNA to a microfuge tube containing 1 ml of 5% TCA. This will precipitate the DNA and leave protein and other contaminants in solution. Centrifuge for 10 minutes and then decant the TCA. [Caution: TCA is extremely caustic]

2 Add 0.5 ml of ice-cold 95% ethanol to the tube. DNA is insoluble in ethanol and will remain precipitated in a pellet, but the residual TCA will be washed out. Centrifuge for 10 minutes.

3 Add 200 µl of formic acid to your sample. [Caution: formic acid is extremely caustic]

4 Select a hydrolysis tube. Each tube must be weighed prior to use. If your tube weighs less than 4 gm, its walls will not be strong enough to withstand the high temperature and pressure, and will explode. Any tube 4 gm and over is acceptable.

5 Carefully transfer your sample to the hydrolysis tube.

6 Seal your tubes according to the following diagram. First heat the tube with a torch near the top until the glass becomes soft. Pull the glass out to a thin neck, but DO NOT SEAL THE TUBE YET. Sealing the tube at the same time as making the constriction may cause the tube to explode during heating. After the glass has cooled, return return the tube to the flame and seal the tube by heating at the top of the constriction. Remember to keep your tube slanted so that the sample does

not run out!

Acid Hydrolysis of DNA

51

7 Leave tubes in rack. Stop at this point. Before the next lab, the instructor will heat tubes for 40 minutes at 175o and return them for use in lab next time.

Procedure for Second Day

1

Break the seal on your tubes. The tube is under considerable pressure that was built up during hydrolysis, which must first be relieved. To do this, freeze the sample in a dry ice / ethanol bath to contract the gases.

2 Place the tube into a holder and insert the sealed end into the flame of a torch. There will be a (possibly) dramatic but harmless pop and recoil as the tube ruptures.

3 Next, briefly heat the widest part of the tube in the flame for a few seconds, allow a few seconds to cool, and then drip some water on the glass. This will cause the tube to crack.

4 The end may be knocked off with a pair of forceps. Your sample will still be mostly frozen.

Be careful of the sharp edges!

5 This is all we will do today. By next lab, the instructor will place the samples in a heating block to evaporate off the formic acid.

Procedure for Third Day

1 Add 30 µl of water to your samples and vortex to mix.

Be careful of the sharp edges! Transfer the 30 µl of dissolved hydrolysate to two microfuge tubes, 10 µl in one, and 20 µl in the other. This should give you two samples containing 50 and 100 µg of DNA, respectively.

2 Obtain a thin-layer chromatography sheet. This is an acetate sheet coated with a fine powder of silica gel which, if your are

Experiment 3

52

not careful, will rub off easily. They come as 20 x 20 cm sheets and you will have to cut them in half (10 x 20 cm). With a blunt pencil, draw a line 2 cm from one edge and a second line line 10 cm farther. Along the first line, draw six evenly spaced spots.

3 You will spot your samples on the center spots, and spot samples of adenine, thymine, cytosine, and guanine on the flanking spots. If you spot your entire sample at once, the gel will be unable to absorb it quickly and the liquid will spread out too far. You need to make a small, tight sample spots. Carefully pipet 2 µl on each spot and allow the spots to dry. When dry, spot an additional 2 µl on each. Repeat this until the entire DNA samples have been spotted. The standards are concentrated enough that you only need one 2 µl applications. As you spot, remember to try and not touch the chromatogram with the pipet. Otherwise, you will scrape off the silica gel.

4 Place your chromatogram in a tank containing chromatography solvent. Make sure your samples are above the solvent.

5 Allow the solvent to migrate up the chromatogram until it has travelled 10 cm, about 1 - 1.5 hr. Remove the chromatogram and allow it to dry.

6 When dry, observe the chromatogram in the BioDoc-IT. Place the chromatogram on the transilluminator powder-side down. Save the image to memory. To print, slide the thermal printer size button to the left, to the “SIDE” position. This will print a nearly actual size image of your chromatogram. Be sure to return the size button to the central NOR position when you are done.

Acid Hydrolysis of DNA

53

7 Calculate the retention factor (Rf) for each spot.

Chromatography Solvent

ethyl acetate …………………………….. 160ml n-propanol……………………………….. 40 ml water…………………………………….. 80 ml

Mix and place in separatory funnel. Discard the bottom phase and retain the top phase.

100% Trichloroacetic Acid

Add 100 ml of water to a bottle containing 500 g of TCA. Using a magnetic stirring bar, mix until the TCA is completely dissolved. If necessary you can add additional water. When the TCA is completely dissolved, add enough water to bring the final volume up to 500 ml. Store in a dark glass bottle. Dilute as needed.

Hydrolysis Tubes Corning Pyrex Rimless Culture Tube #9820

10 x 75 mm 401770 7740

Thin-Layer Chromatography Sheets Machery-Nagel #805023 POLYGRAM Thin layer chromatography polyester sheets SIL G/UV254 20 x 20 cm

54

Experiment 4

DNA Sequence Specificity In Experiment 3 we looked at the chemical composition of DNA. In this lab we will examine DNA, sequence specificity. DNA was originally thought to be a boring polymer of repeating A C T G units, certainly not complex enough to function as the genetic material. The important conceptual breakthrough in understanding how DNA functions to transmit hereditary information was the realization that the seqence of bases in DNA varies from one region of the DNA to another. A gene is now recognized to be the information necessary to synthesize a particular protein and the sequence of bases (taken in groups three called codons) determines which amino acids are incorporated into a protein. Thus, rather than being a simple repetition of A C T G, the DNA sequence varies from one region of the genome to another. One can imagine any number of ways that DNA might be sequence specific. However, if a particular model of specificity is tested and rejected experimentally, all it proves is that that particular hypothesis is incorrect. It does not rule out the possiblity that DNA may be specific in some other fashion. Alternatively, we can test the hypothesis that DNA sequence is RANDOM. Experimental rejection of randomness would indicate that DNA is indeed sequence specific, without necessarily implying any particular model of specificity. Sequence randomness can be tested by cutting DNA restriction endonucleases. REases recognize and cut unique 4-base (tetramer), 5-base (pentamer), or 6-base (hexamer) sites. One can calculate the expected frequency of sites in DNA by the formula:

site frequency = (1/4)N

where N is the number of base pairs in the restriction site. Site frequency is essentially the probability of a restriction site per base pair. To find the number of sites within a piece of DNA, one merely multiplies the site frequency by the number of base pairs in the DNA. This formula makes the important assumption that restriction sites are scattered randomly through the DNA. Is this assumption correct? In this experiment we will use bacteriophage λ as a test DNA. λ contains 48602 base pairs. We will cut λ with three different restriction enzymes, XbaI, BamHI, and EcoRI, each of which recognizes a hexameric sequence. In addition, one of our

DNA Sequence Specificity

55

molecular weight standards, λ HindIII was generated by cutting λ DNA with a fourth hexameric-specific REase. If the assumption of randomness is correct, then we would expect each enzyme to produce the same number of fragments, and also that the pattern of fragments would be similar to one another.

Procedure 1 Prepare the series of restriction digests described below according to the procedure on page 15.

Tube DNA Enzyme Buffer (2 µl) Water

1 λ (5 µl) none none 15 µl 2 “ BamHI Medium 13 µl 3 “ EcoRI High “ 4 “ XbaI High “ 5 λ HindIII (3 µl) none none 17 µl 6 1 kb ladder (3 µl) none none 17 µl

2 Place samples 1-4 in a 37o water bath for 40 minutes.

3 While you are waiting for your DNA’s to digest, pour a 0.5%

agarose gel according to the procedure on page 22.

4 Load the entire 25 µl onto the gel and run at 100 volts for about 90 minutes. The blue tracking dye will separate into two bands. You can stop the gel when the faster band (bromphenol blue) runs ¾ of the way down the gel.

5 Stain and photograph your gel. Be careful while handling the gel – 0.5% gels are VERY soft and fragile!

56

Experiment 5

Activity of DNA Nucleases Nucleases are enzymes that degrade DNA (DNase) and RNA (RNase). With respect to DNases, some affect double-stranded DNA, and others only single-stranded DNA. Still others are indiscriminant with respect to strandedness. DNases are divided into two general categories: endonucleases and exonucleases. Endonucleases attack DNA within a strand while exonucleases digest DNA from free ends. Endonucleases may be general, cleaving DNA indiscriminantly while others, such as restriction endonucleases are site-specific, recognizing and cleaving only particular sequences. With respect to exnucleases, some digest from the hydroxyl end while others digest from the phosphate. Yet others may have no preference. Many exonucleases function in a processive manner, in which the enzyme binds to the substrate and cleaves off nucleotides one by one as it moves down the chain. One can measure the processivity of an enzyme as the number of bases removed per interaction with the substrate. Similarly, DNA polymerases act processively, adding nucleotides one by one as they move down the chain. In this experiment we will assess the effects of four nucleases on circular and linear DNA, and on double and single-stranded DNA. We will cleave pRHR53 with BglII, which cuts the plasmid once in order to linearize it, and then compare the effects of DNase I, mung bean nuclease, λ exonuclease, and T5 exonuclease on uncut and cut samples.

Procedure 1 Linearize pRHR53 by digesting with BglII as described below:

Standard Reaction Mix Scale Up Final Reaction Mix DNA 5 µl 60 µl Low Salt Buffer 2 µl X12 24 µl Water 13 µl 156 µl ΒglII 1 µl 8 µl

2 Incubate the samples in a 37o water bath for 60 minutes. 3 Add enough 5 M NaCl to the reaction mix so that the concentration

is 0.1 M. Add twice the volume of cold ethanol, mix, and allow the samples to stand on ice for 20 minutes.

4 Centrifuge the samples at 19,000 rpm for 20 minutes in the large floor model centrifuge.

5 Carefully decant the sample and allow it to dry. Dissolve the DNA in 50 µl of TE buffer.

DNase Comparison

57

6 Measure the concentration of the uncut and ETOH-precipitated samples with the Nano-Drop to determine the percentage recovery.

7 We will test Mung Bean nuclease’s ability to digest single-stranded and double-stranded DNA. To generate single-stranded DNA, take 20 µl of cut pRHR53 and 20 µl of uncut pRHR53, in separate tubes, and boil them each for 30 minutes.

8 Set up the following reaction mixes using:

5 µl DNA 2 µl 10X Buffer

13 µl water 1 µl enzyme

Tube DNA Enzyme 10 X Buffer 1 Uncut None medium Salt 2 Cut “ “ 3 Uncut HindIII medium Salt 4 Cut “ “ Incubate the samples for 30 minutes at 37O 5 Uncut DNase I DNase 1 buffer 6 Cut “ “ 7 Uncut λ Exo λ Exo buffer 8 Cut “ “ 9 Uncut T5 Exo NEB 4 buffer 10 Cut “ “ 11 Uncut Mung Bean* Mung Bean

Buffer 12 Cut “ 13 Uncut & boiled Mung Bean* “ 14 Cut & boiled “ “ Incubate the samples for 15 minutes at 30O

*Mung Bean nuclease is diluted 1/10 9 Add 5 µl of orange tracking dye to each and analyze on a 1% gel.

58

Experiment 6

Restriction and Modification Bacterial strains that produce restriction endonucleases also produce enzymes that protect their own DNA from being cut. Methyl transferases transfer a methyl group from S-adenosyl methionine (SAM) to the restriction site recognized by the resident REase, preventing it from being cleaved. Thus, only unmodified foreign DNA, from an incoming bacteriophage, for example, will be cut and inactivated. In this experiment we will study the effects of MTase activity by treating λ DNA with M.EcoRI and M.BamHI and then cutting the DNA with R.EcoRI and R.BamHI.

Procedure 1 Prepare the series of MTase reactions as described below:

EcoRI methyltransferase reaction: Standard Reaction Mix Final Reaction Reagent Volume Scale Up Mix

λ DNA 5 µl 20 µl Buffer 2 µl 8 µl SAM 0.25 µl X 4 1 µl Water 13 µl 52 µl Μ.ΕcoRI 1 µl 2 µl

BamHI methyltransferase reaction: Standard Reaction Mix Final Reaction Reagent Volume Scale Up Mix λ DNA 5 µl 20 µl Buffer 2 µl 8 µl SAM 0.25 µl X 4 1 µl Water 13 µl 52 µl Μ.ΒamHI 2 µl 5 µl

Restriction and Modification

59

2 Incubate the samples in a 37o water bath for 60 minutes. 3 Inactivate the enzymes by incubating in a 75o water bath for 15

minutes.

4 Add enough 5 M NaCl to each sample to adjust each to 0.1 M. Add twice the volume of cold ethanol, mix, and allow the samples to stand on ice for 20 minutes.

5 Centrifuge the samples at 19,000 rpm for 20 minutes in the large floor model centrifuge.

6 Carefully decant the samples and allow to dry. Dissolve the DNA in 40 µl of TE buffer.

7 Set up the following restriction digests and incubate for 30 minutes at 37o:

Tube DNA Enzyme Buffer (2 µl) Water 1 M.EcoRI (5 µl) none none 13 µl 2 “ R.BamHI Med “ 3 “ R.EcoRI High “ 4 1 kb ladder (3 µl) none none 17 µl 5 λ HindIII (3 µl) none none 17 µl 6 M.BamHI (5 µl) none none 13 µl 7 “ R.BamHI Med “ 8 “ R.EcoRI High “

8 At the conclusion of incubation, add 5 µl of blue tracking dye to

each sample.

9 Analyze the samples on a 1% gel. Run the gel until the faster-running blue dye (bromphenol blue) reaches the edge of the gel.

60

Experiment 7

Analysis of Plasmid DNA By Restriction Digestion and Agarose Gel Electrophoresis

This experiment will provide an introduction to the use of restriction enzymes and gel electrophoresis, the two most fundamental techniques in recombinant DNA methodology. You will learn how to cut DNA with a restriction endonuclease and analyze the products on a gel. At the end of the experiment you should be able to construct a restriction map of a plasmid. The plasmid pUC9 (pages 61 and 62) is 2.7 kb long and contains the gene for β-lactamase (bla), which confers resistance to ampicillin, the lac repressor, operator, and promoter, and the alpha peptide (containing the amino-terminal end) of the β-galactosidase (lacZ) gene. When in the appropriate host strain, one that contains the carboxyl end of lacZ, pUC9 produces functional β-galactosidase via α complementation (see Appendix VI). Within the lacZ gene is a closely-spaced series of restriction sites, called the multiple cloning site, which may be used as insertion points for foreign DNA fragments. Insertion into lacZ disrupts the β-galactosidase and cells carrying such plasmid, therefore are lac-. When a DNA fragment is inserted into a plasmid, it can be found in either of two orientations In the diagram below, for example, the narrow line represents a plasmid into which a DNA fragment has been inserted into restriction site A. The plasmid and the insert each have a site for enzyme B. On the insert, B is asymmetric. In the left orientation, the two B sites are close to one another (proximal) and in the right orientation; they are far apart (distal).

A

A

A

A

B B

B

B

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If both of these plasmids are cut with enzyme A, the vector and insert fragments will be regenerated. Both orientations will produce exactly the same fragments. However, if the plasmids are cut with enzyme B, the resulting fragments will be different. The proximal orientation will produce a large and a small fragment while the distal orientation will produce two moderately sized fragments:

pRHR30 is a derivative of pUC9 in which a fragment of human DNA has been inserted into the EcoRI site in lacZ. The fragment has an asymmetric HindIII site that is either proximal or distal to the HindIII site on pUC9. In this experiment, you will cut pUC9 and pRHR30 with EcoRI, HindIII, and BamHI and construct a plasmid map. The results of this experiment will be combined with Experiment 2 for your first formal lab report.

Note that in addition to the restriction digests done in this experiment, you will also perform several restriction digests that will be used for Experiment 9.

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Procedure

1. Prepare the series of restriction digests described below

according to the procedure on page 12. You will use 5 µl of DNA and 15 µl of reaction mix. Note: The DNA and water volumes are generic. The exact amount of DNA sample to be added depends on the DNA concentration, which varies from year to year with different DNA preps. The concentration of this year’s DNA can be found in the class syllabus. During lab you will be told how much DNA sample to actually use, and you must adjust the amount of water accordingly so that DNA + water = 18 µ l

Tube DNA Enzyme Buffer (2 µl) Water * Dye

1 pUC9 (5 µl) none none 15 µl Blue 2 “ HindIII Med 13 µl “ 3 “ BamHI Med “ “ 4 “ EcoRI High “ “ 5 “ PstI Med “ “ 6 1 kb ladder (3 µl) none none 17 µl “ 7 λ HindIII (3 µl) none none 17 µl Orange 8 pRHR30 (5 µl) none none 15 µl “ 9 “ HindIII Med 13 µl “ 10 “ BamHI Med “ “ 11 “ EcoRI High “ “ 12 “ PstI Med “ “

*Tracking dye (5 µl) is added AFTER the incubation period

2. Prepare the restriction digests and heat inactivation for steps 1 and 2 of Experiment 9.

3. Place samples 2-5 and 9-12 in a 37o water bath for 40 minutes.

4. While you are waiting for your DNA’s to digest, pour a 1% agarose gel according to the procedure on page 21.

5. At the end of 40 minutes, remove the samples and add 5 µl of tracking dye to each sample.


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6. Load the entire 25 µl onto the gel and run at 120 volts for about 90 minutes. The orange tracking dye should run at least half way down the gel.

7.

8.

Take a cell phone photograph to document the migration of the tracking dyes. Stain your gel with Gel Red and photograph.

Analysis 1. Measure the distance that each band migrated from the origin and record the results in a table.

2. Graph the migration of your molecular weight standards (samples 6 and 7) on semi-log paper. Plot log kb versus distance. This will be your standard curve. I will examine your standard curves to make sure that they are plotted correctly.

3. When your standard curve is correct, plot the distances of your cut samples on the curve and determine the size of each fragment. Record the migration distances and molecular weights in a table.

4. Construct a map for pRHR30

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65

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Experiment 8

Effect of Agarose Concentration on Migration of DNA Fragments

The relationship between molecular weight and distance migrated is logarithmic, not linear. Thus large fragments are packed close together and small fragments are spread far apart. Moreover, the small fragments produce broad faint bands. This is the result of two factors. First, the intensity of a band is dependent upon the amount of ethidium bromide that is intercalated into the DNA. Since small fragments do not have as many bases available for intercalation as big fragments, small fragments will be less intense. Second, the farther a fragment travels, the more the DNA tends to diffuse. Thus small fragments produce fuzzy bands. Your gels, which are 1% agarose thus only distinguish mid-size fragments. Often, however, you are working with fragments that are either very big of very small and a 1% gel is inadequate. In Experiment 7 you saw that migration rate through a gel is a function of DNA size and conformation. But there is a third factor: concentration of the gel. An agarose gel is a meshwork of interlinked agarose strands that impede the migration of DNA. By increasing the size of the mesh, one can facilitate the passage of large fragments of DNA. Alternatively, by decreasing the mesh, one can retard the progress of small fragments, thereby minimizing diffusion. Thus one might examine a restriction digest by running it on a series of gels that optimize different size ranges. In this experiment we will examine the effects of agarose concentration by running our molecular weight standards on three different concentrations: 0.5%. 1.0%, and 1.5%. The results of this experiment will be combined with Experiment 7 for your first lab report, to be prepared according to the format on page 4.

Procedure

1. You will need to collaborate in pouring gels with two other pairs. One pair should pour a 0.5% gel, the second should pour a 1.0% gel, and the third should pour a 1.5% gel. Remember to make the gel using TBE buffer. If you use water, your bands will streak and smear.

2. After you pour the gel insert two combs into the tray, one at each end and allow it to harden.

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67

3. When the gels solidify, remove the combs and cut the gels into 3 strips, with each strip containing 2 usable lanes in the middle. Use a plastic ruler to cut the gel. Do not use a razor blade, as this will scratch the tray. When cutting, use the wells to line up your slices.

4. Exchange the strips so that each pair has one of each, lying side by side on the gel tray. Be very careful when handling the 0.5% gel. It is very soft and gooey.

5. Load each strip with standard. You need to load 2 µl of DNA in each well. Since you are not running any 20 µl restriction digests as well, the final sample volume is not important, although it is difficult to work with just 2 µl. For convenience, make up 15 µl samples:

2 µl standard 8 µl water 5 µl orange tracking dye

6. Run the gel at 125 V until the tracking dye reaches the end. Then stain for 10 minutes. 15 minutes of destaining in water, especially for the 0.5% gel will greatly help the visualization of these gels because these strips absorb lots of stain very quickly.

7. Measure the migration distance and plot your data on semi-long paper. The data for all three concentrations should be plotted on the same graph.

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Lab Report 1

Experiments 7 and 8 Analyzing the Data Experiment 7 introduces you to restriction digestion and gel analysis. Your analysis of this experiment should include:

£ a discussion of how fragment sizes are distributed across a gel £ a discussion of how fragment size and migration distance affect accuracy of size

estimation £ a discussion of the difference between cut and uncut plasmid

You are also asked to make a simple map of the plasmid pRHR30. pRHR30 is a recombinant plasmid comprised of a pUC9 vector and an insert into the EcoRI site. When you calculate a restriction fragment size, you are, in effect, calculating the distance between the two cut sites that generated the fragment. In order to calculate the distance between the sites of different enzyme sites, it is necessary to perform a series of double digests using two different restriction enzymes. However, in this experiment, the digestions are simplified because you already know the location of many sites. You have the map of pUC9 and the sites on the vector will anchor one of the two sites on each fragment. Given the precision of your gels, the four restriction sites in pUC9 will appear to be located in the same spot. In the example below, a fragment with PstI, BamHI, and HindIII sites is inserted into the EcoRI site of pUC9. A gel of the plasmid shows the fragment sizes of each digestion. The map is easily constructed by plotting the distance of each site from the anchor site on the vector. The direction of the site is clear (in this case clockwise) because if the sites were plotted in a counterclockwise direction, they would map within pUC9.

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In experiment 8 you were asked to prepare a composite gel consisting of 0.5%, 1.0%, and 1.5% agarose slabs and run 1kb+ and λ HindIII molecular weight standards on each. You were also asked to use both the orange and blue tracking dyes. Once you have measured the bands on each slab, the data should be plotted on a single graph that shows three curves. the 1kb+ and λ HindIII molecular weight standards from each slab should be plotted on a single line with the data points for each clearly identified. Your analysis of this experiment should include:

£ a discussion of how agarose concentration affects migration rate £ a discussion of why you might consider using different agarose concentrations £ a comparison of the two tracking dyes and why you might use one or the other

in an experiment. Writing the Report Although these two experiments were presented as distinct, stand-alone assignments, they are indeed related to each other. Your challenge in writing this report is to weave them together into a single unified project. Although you did the experiments in the order #7 then #8, you do not necessarily have to follow this order when you describe them in your paper. You may present them in any order that makes sense to you. The format of the paper follows that of a standard journal article and is clearly laid out in the lab manual and on the course web site. Please read the instructions carefully and be sure to ask questions if you are unsure. Required data elements:

Experiment 7~ £ Gel Red photograph of the gel £ cell phone photograph of gel £ table of migration distances and molecular weights for standards £ graph of the standard curve £ table of migration distances of unknown bands and estimation of molecular weights £ final plasmid map Experiment 8 ~ £ Gel Red photograph of the gel £ table of migration distances and molecular weights for standards £ graph of the three standard curves

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Points to Remember: Methods and Materials are GENERIC. Do not refer to specific cells or DNA’s. Those are part of experimental design and belong in Results. In Results you describe the experiment. For example:

How does one saw a piece of wood: “The wood to be cut was measured and marked with a pencil to indicate where the cut was to be made. A wood saw was held at a 45o angle against the edge of the wood and repeatedly moved back and forth until the cut was complete.” How does one nail two pieces of wood together: “Wood glue was spread on each piece of wood. The two pieces of wood were braced together and a nail was held vertically against the two pieces. The nail was hit repeatedly with a hammer until the head of the nail was flush with the surface. This process was repeated until both pieces were fastened together firmly.” These descriptions are GENERIC. You are not explaining why you are cutting the wood or nailing it together. You can use the same descriptions to explain how to build a table, a bookcase, a house, or anything else made of wood.

In the Results section you explain what you want to build (i.e. this is the experiment) and how you intend to use the hammer and saw to accomplish this goal. You do not have to explain how you hammer and nail wood. That was described in the Methods and Materials. Avoid trivial details:

The tube was tapped several times to mix the contents. The tube was briefly centrifuged to bring the contents to the bottom. The reagent was carefully added to the tube.

Do not present a diary of your afternoon in the lab:

While the reaction mix was incubating, we poured a gel. The Discussion must reflect the goals laid out in the Introduction. If you said in the Introduction that you are studying several aspects of restriction digestion and electrophoresis, then you must explain how your results reflect the various aspects of restriction digestion and electrophoresis that you described in the Introduction.

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Tables and Figures:

In a journal article, there are only tables and figures. Please label your data elements as such. Do not use labels such as “Gel #1” or “Graph #3.” All figures and tables must be consecutively numbered, in the order in which you discuss them in the paper. Do not refer to all figures from each experiment as Fig 1-1, 1-2, 2-1, etc. Any figure or table that you include in your paper must be specifically mentioned and referred to in the text of the paper. I am not into busy work so I don’t mind if you and your lab partner have the same figures and tables, but: MAKE SURE THAT THE WRITING OF THE PAPER IS CLEARLY INDEPENDENT!!

Format:

Please type your paper single-spaced, with broad margins. I am not impressed with papers that are formatted to look like journal articles. If you turn in such a paper, do so only because you enjoy the challenge. It will make no difference to your grade. Your report should have a cover page with both your name and your lab partner’s name in the upper right hand corner. Please include your lab section.

Do not hand in your paper in a binder. The paper should simply be stapled in the corner.

___ Your name Lab Partner Section #

Title of Your Report

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Experiment 9

Insertion of a Gene for Antibiotic Resistance from Bacillus subtilis into an Escherichia coli Plasmid With this experiment we begin a project that will run through the remainder of the semester. The goal of the project is to subclone a gene for antibiotic resistance from the B. subtilis plasmid pBAC into E. coli. We will use different B. subtilis plasmids in different years. Information about the particular pBAC will be found in the syllabus. In Experiment 9 we begin by creating chimeric plasmids. In subsequent experiments we will transfer the DNA into a host E. coli strain, identify the recombinant products, amplify the DNA, and characterize the chimeras by restriction mapping. Subsequently we will transfer the recombinant plasmids into B. subtilis and look at expression of foreign genes in B. subtilis and E. coli. pBAC contains two antibiotic resistance genes. Thus, the recombinant plasmids contain three antibiotic resistances; ampicillin (from pUC9) and two from pBAC. Finally, we will perform a Southern blot to compare the recombinant plasmids with their parents. By the end of the seer we will have worked through the basic steps of a cloning project.

In Experiment 9 we will cut the two parent plasmids, pUC9 and pBAC with restriction enzymes and ligate them together. Joining of two DNA’s is most easily accomplished when both are cut with identical or compatible restriction enzymes. The two DNA’s are then mixed and joined by DNA ligase under conditions that permit hydrogen bond formation. Upon annealing, the joint between the two DNA’s will contain a pair of staggered nicks, one on each strand, and the ligase will form a phosphodiester linkage between adjacent nucleotides. Although the optimal temperature for ligase is 37o, incubation at this temperature would overcome the hydrogen bonding between the cohesive termini. Thus, ligation is usually performed at 4 - 15o for 12 - 24 hours. This reaction can be catalyzed by either T4 DNA ligase or E. coli DNA ligase. The former enzyme requires ATP as a cofactor and the latter requires NAD+. The primary difference between them is that the T4 enzyme will catalyze the ligation of blunt ends whereas the E. coli enzyme will not.

When performing a ligation reaction between two fragments, vector and target, any and all possible combination of those fragments will occur. Most will produce unviable constructs that will not survive the transformation and amplification process, but some undesirable constructs will be viable. The most

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important such construct is the rejoining of the cut ends of the vector. This will of course exclude the target fragment. Religation of the target will also occur, but because the target lacks an appropriate origin of replication, it will not be represented among the transformants. Perhaps surprisingly, self-ligation of the vector and target DNA’s are favored reactions. Ligation is a two-step process, annealment followed by ligation. Annealment depends on the cut ends finding one another and forming hydrogen bonds. Self-ligation requires only two ends to find one another and anneal, while insertion requires four ends. Since the finding of ends is a random process, self-ligation is statistically more likely to occur. Re-ligation can be avoided by altering the vector such that its ends, when annealed together, do not provide a suitable substrate for ligase. This is accomplished by treating the cut vector with bacterial alkaline phosphatase (BAP). In the alkaline phosphatase reaction, the 5’-phosphates are removed from the cut ends and replaced with 5’-hydroxyls. Upon self-annealing of the vector, a pair of staggered nicks will occur with both 3’- and 5’-prime hydroxyl ends, which is not a substrate for ligase. When a target fragment containing 5’-phosphate and 3’-hydroxyls ends is inserted, one of the nicks at the joint will have 5’- and 3’-hydroxyl ends and the other will have normal 5’-phosphate and 3’-hydroxyl ends. The latter ends are joined by ligase and the former ends are not. The situation is the same at the other end of the fragment, except that the acceptable and unacceptable nicks are on the opposite strands. This ligated molecule, containing one ligated nick at each joint, can be introduced into E. coli. At the first round of replication, the cells will correct the unligated nicks. See Appendix VII for information on how BAP affects insertion and ligation frequencies.


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Preparation of Restriction Fragments

1. Cut pUC9 and pBAC with restriction enzyme. You will be told which enzyme to use depending on which particular pBAC you are using this semester. You will need to prepare two duplicate samples of pUC9, for a total of three digests. In addition you need to scale up the size of your digestions according to the table. Make sure to consult page 18 to determine the appropriate restriction buffers.

Standard 4-fold scale-up DNA 5 µl 20 µl buffer 2 µl 8 µl water 13 µl 52 µl Total 20 µl 80 µl Enzyme 1 µl 2 µl

2. Incubate the samples for at least 40 minutes at 37o. We will do nothing further with these samples today, so you can actually let them go for an hour. The longer, the better.

3. It is important to inactivate the restriction enzymes before attempting a ligation so that you avoid re-cutting your ligated DNA. This can be done simply by placing the DNA samples in a 65o water bath for 20 minutes. Note, however, that the optimal temperature for some enzymes is 65o. In these cases inactivation must be carried out by phenol treatment.

Bacterial Alkaline Phosphatase Treatment

1.

2.

BAP is active in a 10 mM Tris, pH8 solution. In class, however, we will use the BAP buffer provided by the manufacturer. Ethanol precipitate one of your pUC9 digests as follows: Adjust your DNA sample to 0.1 M NaCl using a 5 M NaCl stock

solution. Add twice the volume of ice-cold ethanol (one volume is equal to the amount of your sample, in this case, 80 µl, so you would add 160 µl of ethanol). Place on ice for 15 minutes.

3. Centrifuge for 20 minutes at 19,000 rpm in the large floor centrifuge. Carefully decant and allow the residual ethanol to evaporate. Dissolve the DNA in an equal volume (i.e. the original volume, 80 µl) of BAP buffer.

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4. Add 5 - 10 units of BAP per µg of DNA. In this case, 1 µl of BAP.

5. Incubate at 65o for 45 minutes.

6. Add an equal volume of phenol/chloroform. Mix well and allow to stand for 5 minutes. Phenol/chloroform is extremely caustic!!

7. Centrifuge briefly to separate the layers. The aqueous layer containing your DNA will be on top. Carefully pipette off the aqueous layer and transfer it to a fresh microfuge tube.

8. Ethanol precipitate your sample a second time to remove residual phenol/chloroform. Adjust your DNA sample to 0.1 M NaCl using a 5 M NaCl stock solution. Add twice the volume of ice-cold ethanol. Place on ice for 15 minutes.

9. Centrifuge for 20 minutes at 19,000 rpm. Carefully decant and allow the residual ethanol to evaporate. Dissolve the DNA in an equal volume of TE buffer.

Preparation for Experiment 4

1. Prepare 500 ml LB agar.

2. Prepare 500 ml MacConkey + Ampicillin (50 µg/ml) agar. (Recipes for both are found in Experiment 4 on page 62)

Ligation

1. In general, you should prepare 30 µl of reaction mixture (this may be scaled up or down as required):

20 µl target + vector DNA’s 6 µl 5X ligation buffer 3 µl ATP (5 mg/ml in water, prepared fresh) 1 µl water 1 µl T4 DNA ligase

Notes: a. Target and vector DNA’s may vary from being equal in

proportion to being heavily weighted towards one or the


76

other. The specific requirements of each ligation will indicate the proportions.

b. When doing a blunt-end ligation use half the concentration of ATP

c. Reasonable controls would be samples lacking either target, vector, or ligase. A volume of water equal to the missing item should be added.

2. Prepare 5 ligation mixes as follows:

#1 #2 #3 #4 #5 pUC9 (-BAP) 5 µl 0 µl 15 µl 0 µl 0 µl pUC9 (+BAP) 0 µl 5 µl 0 µl 15 µl 0 µl pBAC 15 µl 15 µl 0 µl 0 µl 15 µl 5X ligation buffer 6 µl 6 µl 6 µl 6 µl 6 µl ATP (5 mg/ml) 3 µl 3 µl 3 µl 3 µl 3 µl Water 1 µl 1 µl 6 µl 6 µl 6 µl Ligase 1 µl 1 µl 1 µl 1 µl 1 µl

3. Place the ligation mixes in refrigerator over night (or until the next period). Save the remainder of your cut plasmids.

4. After ligation, remove 10 µl of each mix and 10 µl of the original three cut plasmid samples and run them on a “baby gel” to check on ligation.

5. If the gel looks good, use 10 µl of ligation mix for transformation into E coli in Experiment 4.

5X Ligation Buffer 0.25 M Tris pH7.6 50 mM MgCl2 5 mM dithiothreitol 25% polyethylene glycol-8000

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Experiment 10

Transformation of Escherichia coli With a Chimeric Plasmid

Now that we have generated recombinant molecules, we must next amplify them by inserting them into an acceptable host so that they may be analyzed or manipulated in some way. Insertion of exogenous naked DNA into cells is called transformation. Some bacteria like Bacillus, Neisseria, and the pneumococcus possess genetic and enzymatic systems that permit them to transport DNA across the cell wall and to incorporate it into their genomes. Indeed, transformation is the normal means of genetic exchange employed by these organisms. Such organisms are said to be competent. E. coli, however, employs sexual interaction to exchange DNA and does not normally take up exogenous DNA. But E. coli can be forced to become competent by treating it with CaCl2. This technique is the first such technique worked out for E. coli but it is not the only transformation regimen available. Others provide higher levels of DNA uptake, but are more complex to use and, in our case, unnecessary. The transformation process is extremely inefficient and normally, only a fraction of a percent of cells actually becomes transformed. Thus, picking out transformed cells in a nearly infinite population becomes a major problem. This can be circumvented by designing a selection technique whereby the cells in which we are interested will grow at the expense of those in which we are not interested. We have already seen in experiment 7 that our vector, pUC9 carries a gene for ampicillin resistance. Thus, if we put ampicillin into the plating medium, only those cells that have picked up exogenous DNA can grow. Once we have selected for transformed cells, we must then determine which of the transformed cells carries an insert into the vector. This can be done by employing a screen. The type of screen employed varies depending on the vector used and the nature of the intended target. In this case, we have seen that pUC9 carries a fragment of the lacZ gene in addition to the bla gene. Insertions into the multiple cloning site within lacZ leads to inactivation of β-galactosidase, and, consequently, a Lac- phenotype. On MacConkey agar, cells that are Lac+ secrete lactic acid as a waste product of lactose fermentation. This causes a localized drop in the pH around the colony, which, in turn, causes a pH indicator dye to turn bright pink. Lac- cells cannot ferment lactose and therefore do not cause a change in the pH. These colonies appear white on MacConkey agar, and the plate background turns yellow from the basic waste products normally produced by cells.

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In this experiment you will transform your ligation mixes into E. coli and recover transformed, AMPr colonies that will be either red (-insert) or white (+insert). The plasmids carried by these cultures will be characterized in Experiment 5.

Procedure 1. Inoculate 20 ml of fresh LB broth in a sterile side-arm flask with 1 ml of an over-night culture of E. coli strain JM83. Incubate shaking at 37o until the cell density reaches about 5 x 107 cells/ml (OD600 = 0.2. You will eventually be able to judge this by eye). Incubation will take approximately 2 hours.

2. Transfer 10 ml of cell culture to a sterile 25 ml Oakridge centrifuge tube and centrifuge for 10 minutes at 7,000 rpm. Note the shape and consistency of the cell pellet. Carefully pour off the supernatant.

3. Resuspend the cells in an equal volume of ice cold 50 mM CaCl2 and allow to stand on ice for 20 minutes. The CaCl2 is hypotonic and causes the cells to swell and become spheroplasts.

4. Centrifuge the cells a second time and again note the pellet. Since the cells are spheroplasts, the pellet will look quite different than in step 2. Typically, the cells will appear as a diffuse ring rather than as a solid pellet. If the pellet appears as described, then the cells have successfully been rendered competent. If they do not appear as a diffuse ring, the transformation will not work and should be aborted.

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5. Resuspend the cells in 1.5 ml of ice cold 50 mM CaCl2.

6. Transfer 200 µl of competent cells to each of four sterile glass tubes and set up the following transformations:

1 10 µl ligation mix #1 2 10 µl ligation mix #2 3 5 µl uncut pUC9 4 0 DNA control

Do not use microfuge tubes for this step!

7. Allow all four tubes to stand on ice for 15 minutes. During this time the DNA forms a complex of hydroxyl-calcium phosphate, which adheres to the cell surface. This complex is DNase resistant.

8. Heat-shock the cell + DNA mixes by placing them in a 37o water bath for 2 minutes. This will induce the cells to take up the DNA

complex. It is easy (and fatal) to forget this step if you are in a hurry.

9. Add 2 ml of LB broth to the cell suspensions and incubate shaking at 37o for 90 minutes. This incubation will allow the spheroplasts to regenerate and permit time for the expression of the newly incorporated genes. The tubes should be slanted according to the photograph below to ensure adequate aeration.

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10. During the incubation, prepare four sets of six sterile dilution tubes, each containing 0.9 ml of LB broth. Use glass tubes. Do not use microfuge tubes.

11. Serially dilute the transformed cells and plate on LB and on the MacConkey + amp plates according to the table below. The table shows the final dilution factor ON THE PLATE. All dilutions indicated by a “+” should be plated. Refer to the diagram on serial dilution on page 63 if you need to brush up.

Media Dilution 0 DNA pUC9 Lign #1 Lign #2 Mac 10-1 + + + “ 10-2 + + + “ 10-3 + + + “ 10-4 + XXX* 10-5 LB 10-6 + + + + “ 10-7 + + + +

*XXX means that the dilution is not plated on any medium

12. Invert your plates and place them in the 37o incubator until the following day.

13.

On the following day, count your colonies and record numbers in your notebook according to the data sheet below. There is a copy of the data sheet on line that you can paste into your notebook. You should see red and white colonies. Red colonies indicate that no insertions into lacZ were made, and that the cells are fermenting lactose. White colonies indicate insertional inactivation of β-galactosidase.

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Dilution Factor 0 DNA pUC9 Lign #1 Lign #2 red 10-1 white red 10-2 white red 10-3 white red 10-4 white red 10-5 white red 10-6 white red 10-7 white

Total Viable Count (LB)

Total Transformants

Transf Frequency (r+w)/viable

insert frequency w/(r+w)

transformants per µg DNA

14.

Recover colonies for Experiment . Prepare 14 tubes of LB + amp, 2 ml per tube. Use sterile swab sticks to pick 10 white colonies and 4 red colonies and inoculate one in each tube. After inoculating the tube, use the same swab to make a plate culture as described in step 15. Incubate the tubes standing over night at 37o for use the following day. Make sure that each tube is numbered.

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15. At the same time that you make your broth cultures, make plate cultures as well. Mark several pie-shaped sectors on the back of a few MacConkey plates and number them. Pick a transformant from one of your plates, dip it into one of the tubes, and then streak it across one of the numbered pie-shaped sectors. These will become your permanent stock cultures. You must be able to correlate a particular broth culture with a particular solid culture. Eventually, you will also need to be able to correlate the cultures with lanes on a gel. Record keeping is very important!! Incubate the plates inverted at 37o over night and then seal with parafilm. It would be wise to save your transformation plates, wrapped in parafilm, until you know that you have recovered the necessary plasmids from this experiment

Serial Dilutions: a x b = c a = c/b

Cell Concentration x dilution factor = # colonies cell concentration x 10-6 = 200 colonies cell concentration = 200/10-6 cell concentration = 200 x 206 = 2 x 108

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Luria Broth (LB) and Agar

Tryptone.......................................................................... 10 g Yeast Extract................................................................... 5 g NaCl................................................................................... 10 g Agar................................................................................... per liter of water

20 g

MacConkey Agar

Peptone............................................................................. 17 g Proteose Peptone............................................................. 3 g Lactose.............................................................................. 10 g Bile Salts #3.................................................................... 1.5 g NaCl.................................................................................. 5 g Agar.................................................................................. 13.5 g Neutral Red...................................................................... 0.03 g Crystal Violet................................................................... 0.001 g per liter of water

MacConkey agar comes as a pre-mixed powder. Therefore, it is necessary only to add 50 g to one liter of water.

Ampicillin

Stock ampicillin (= 25 mg/ml) is prepared by dissolving it in water and filter-sterilizing it. After the agar has been autoclaved and allowed to cool, add enough ampicillin to make the final concentration 50 µg/ml. Do not add the ampicillin before autoclaving.

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Experiment 11

Screening of Transformants For Chimeric Plasmids

In Experiment 10 you transformed E. coli with the recombinant plasmids produced in Experiment 9, and have now recovered transformants with and without inserts. When a circular plasmid is cut once, it becomes linearized. When two linear fragments fuse to form a new circle, they can join in one another, as we have seen in Experiment 7, in two possible orientations, each of which produces a unique pattern of restriction fragments. In order to characterize the chimeric plasmids that you have constructed, you must sort out the two orientations and obtain cultures of each. As you will soon learn, it is difficult and time-consuming to produce large amounts of highly purified plasmid DNA. However, there are a number of procedures for isolating small amounts of plasmid relatively quickly from a large number of cultures. The technique that you will employ here is the alkaline lysis method. In this procedure you first use lysozyme to gently strip away the peptidoglycan in the cell wall to produce spheroplasts, which are then lysed by a combination of SDS and NaOH. SDS dissolves the lipopolysaccharides in the cell wall and NaOH produces an osmotic shock to burst the remaining membrane. When the cell lyses, the chromosome is fragmented into linear pieces, but the plasmid remains intact and circular. In addition to lysing the cell, the NaOH also denatures the DNA. The linear chromosome fragments separate, but the circular, supercoiled plasmid strands remain tightly bound. When the lysate is neutralized by acetate, the chromosomal fragments try to reanneal and become hopelessly entangled with one another and with the cell debris to form an insoluble pellet. The plasmid, however, because of the close proximity of the strands, instantly reanneals and remains in solution. The lysate is then centrifuged and the chromosomal DNA is pelleted and removed, leaving a small but relatively pure plasmid sample. Finally, phenol/chloroform is added to remove soluble protein, and the DNA is then precipitated with ethanol to remove the phenol. Any carried-over phenol will inactivate restriction endonucleases or other enzymes that will be used to manipulate the DNA later on. DNA and RNA are similar enough that this procedure does not differentiate between them so when the pellet is re-dissolved, RNase is added to remove any RNA present. Although we will use a different technique for the large-scale prep, the alkaline lysis method could easily be scaled up for this process.

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Materials GET buffer lysozyme solution (40 mg/ml) 0.5 M NaOH 10% sodium dodecyl sulfate (SDS)

potassium acetate phenol/chloroform ice-cold 95% ethanol RNase solution (10 mg/ml)

Procedure for First Day

1. Recover the broth cultures that you inoculated at the end of Experiment 10. Transfer 1.5 ml of each culture into a microfuge tube and centrifuge for 2 minutes. Carefully pipet off the supernatant.

2. Add 90 µl of ice-cold GET buffer and 10 µl of lysozyme solution to each pellet. Vortex to resuspend.

3. Allow the tubes to stand at room temperature for 5 minutes. During this time, the lysozyme will strip away the peptidoglycan layer of the cell wall. The glucose in the GET will create an isotonic environment that will prevent the spheroplasts from rupturing.

4. Prepare a fresh NaOH + SDS solution as follows:

2.0 ml 0.5 M NaOH 0.5 ml 10% SDS 2.5 ml water

Add 200 µl of this solution to the spheroplasts, mix by rapidly inverting the tubes several times (do not vortex) and place on ice for 5 minutes. The combination of alkali and detergent will dissolve the cell membranes and cause the cells to lyse and the alkali will cause the DNA to denature. Your samples will become very viscous. This is a critical step! Do not allow denaturation to occur for much longer than five minutes or the plasmid will not reanneal efficiently!

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5. Add 150 µl of ice-cold potassium acetate (pH4.8). Mix very gently by vortexing the tubes upside down and store on ice for 5 minutes. The acidic potassium acetate will change the sample from alkaline to neutral pH and the DNA’s will attempt to reanneal. You will see a fluffy white precipitate. Centrifuge for 5 minutes.

6. Transfer each supernatant to a fresh tube and add an equal volume (450 µl) of phenol/chloroform Phenol/chloroform is EXTREMELY CAUSTIC! Safety glasses, gloves, and lab coats are required!!

7. Rock gently to mix, and centrifuge for 2 minutes. This will partition the proteins into the lower phenol phase and the nucleic acids in the upper aqueous phase.

8. Carefully pipet off the top layer:

Set your pipet to a volume slightly larger than the amount of your sample (e.g. 500 µl). Carefully draw up your entire sample plus some of the phenol chloroform. Locate the boundary between the phenol/chloroform and aqueous layers. Carefully pipet back into the tube to just past the boundary layer. Transfer the remainder of the pipet to a fresh tube.

If all was done carefully, your sample will contain mostly

plasmid DNA and RNA. Sometimes, however, residual proteins from the lysate carry over and can interfere with subsequent restriction digestions. This can usually be cleared up by a second phenol/chloroform extraction. It is prudent to simply repeat steps 6 - 8 as a matter of course before going to step 9. When you are finished, deposit the used phenol/chloroform in the organic waste bottle in the hood.

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9. Add two volumes of ice-cold 95% ethanol to each sample and allow to stand on ice for 10 minutes to precipitate the nucleic acids. Centrifuge for 5 minutes. It is not necessary to add NaCl here as you did during the BAP treatment in Experiment 9 because the acetate used in the neutralization step serves the same purpose.

10. Carefully pour off the ethanol (you may see a white pellet) and allow the residual to evaporate.

11. Prepare a TE + RNase (20 µg/ml) solution and resuspend each of your pellets in 50 µl. If the RNA in the plasmid preps is not removed, it will be visible on your gel and will interfere with your ability to analyze the results of restriction digestions.

In a normal lab situation, the plasmid extraction would be done in the morning and there would then be time to analyze the results on a gel in the afternoon. However, the plasmid will keep and this is a convenient breaking point. Save your plasmid preparations until next time.

Procedure for Second Day

At least one of the restriction sites in your recombinant plasmids is asymmetric and will allow you to distinguish proximal and distal orientations of the insert. You will be told which enzyme to use depending on the particular pBAC you are using this term.

1. Transfer 10 µl of sample from each plasmid prep to a fresh microfuge and set up a 20 µl restriction digestion. Since you are using 10 µl of DNA, you will compensate by adding 5 µl less water. To speed up your manipulations, you can prepare a cocktail of buffer, water, and enzyme in one large lot and dispense 10 µl of the mix to each of your plasmid samples. You should include an 11th sample, pUC9, as a control.

2. Prepare a second set of samples as uncut controls. Transfer 5 µl to a microfuge tube and add 15 µl of water.

3. Pour a 1% “piggy-back” gel.

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4. Add 5 µl of tracking dye to your samples and load onto the gel. The cut samples should go on the top row of the gel and your uncut samples should go on the bottom. With the cut samples, include 1 kb ladder and λ HindIII DNA molecular weight standards. Run the gel for approximately one hour at 120 V.

5. Stain your gel and take a photograph.

6. A complete digest of your chimeric plasmids will give you two bands. If you see more, then you have a partial digestion caused, most likely by inhibitory contaminants in your plasmid preps. However, you should still be able to identify the two bands resulting from a complete digestion. Those plasmids in which the restriction sites are distal shall be designated pRIT4501. Those in which the sites are proximal shall be designated pRIT4502.

GET Buffer

50 mM glucose 10 mM EDTA 25 mM Tris pH8.0

Potassium Acetate pH 4.8

5 M Potassium Acetate.................................................. 60.0 ml glacial acetic acid............................................................ 11.5 ml water................................................................................. 28.5 ml

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Experiment 12

Large-Scale Purification Of Plasmids pRIT4501 and pRIT4502 by Cesium Chloride Density Gradient Centrifugation

Now that you have identified your two recombinant plasmids, you need to produce large-scale preparations of each so that you can study them further. To do this, you will prepare lysates of 500 ml cultures and purify the DNA by density gradient centrifugation. Although you could have used the alkaline lysis procedure from Experiment 11, you will use a different method here. In the current procedure, 500 ml of cells are concentrated and spheroplasted. They are then treated with Triton X-100, a very mild detergent, which produces holes in the cell wall without actually destroying it. The bulk of the plasmid will leak out of the cell wall but the chromosome remains trapped within and is removed when the cells are pelleted. The resulting plasmid solution is purified further by density gradient centrifugation. CsCl is added and the lysate is transferred to ultracentrifuge tubes. The tubes are spun at 40,000 rpm for two days. During this time, the CsCl forms a gradient and the molecules migrate according to their density until they float at their individual isopycnic points (the point in the gradient that equals the buoyant density of the molecule). Density is a function of AT/GC ratios. Each of the four bases has its unique density, and the final density of any DNA is the collective sum of the densities of the individual bases. Since the DNA’s of different species have different AT/GC ratios, it stands to reason that they also have different densities and, therefore, different isopycnic points. The relationship between AT/GC content and density is:

ρ = 1.660 + 0.00098 x (%GC) %GC = ρ -1.660 0.00098

where ρ = density in g/ml and 1.660 is the density of DNA containing 100% AT.

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Relationship between AT/GC ratio and isopycnic point Percentage of Each Base Buoyant

Density (g/ml) Species A G C T A+T G+C AT/GC %GC E. coli 24.6 25.5 26.6 24.3 48.9 52.1 0.94 51.58 1.7106 S. typhimurium 22.9 27.1 27.0 23.0 45.9 54.1 0.85 54.10 1.7130 B. subtilis 28.4 21.0 21.6 29.0 57.4 42.6 1.35 42.60 1.7017 bacteiophage λ 26.0 23.8 24.3 25.8 51.8 48.1 1.08 48.15 1.7072

The two species that we are working with this semester, E. coli and B. subtilis have different AT/GC ratios, the later being somewhat poorer in GC. According to the table above, B. subtilis DNA has a lower density and, accordingly a lower isopycnic point. The difference in density between the two is significant; the two DNA’s would have different peaks in the gradient, and could easily be separated. Salmonella typhimurium is very closely related to E. coli and their DNA’s accordingly have very similar isopycnic points. The table also shows the density of phage λ DNA. After transferring the lysate to the ultracentrifuge tube, ethidium bromide is added. While ethidium bromide certainly facilitates collection of the DNA at the end of the centrifuge run by virtue of its fluorescence, it is actually an integral part of the separation process. Plasmid DNA and contaminating chromosome have about the same AT/GC ratio and thus cannot be separated easily. The addition of ethidium bromide rectifies this problem. Density is a function of AT/GC ratio, but it is also a function of conformation. For supercoiled DNA, there is more DNA per unit volume than for relaxed DNA. Intercalation of ethidium bromide into DNA causes the helix to unwind (negative supercoiling) and become more relaxed. As the DNA relaxes, mass is spread over a larger volume and the density decreases. When DNA is linear, its free ends can compensate for the supercoiling, and its isopycnic point is lowered. However, when the DNA is circular the ends are connected and cannot compensate for the supercoiling. In order to do so, the plasmid “kinks up” into a very tight knot, forcing more mass into a smaller volume. Thus, ethidium bromide causes the plasmid density to be increased. This principle can be demonstrated by the famous “rubber band experiment.” Find out more about density gradient ultracentrifugation in Appendix VIII

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The Famous Rubber Band Experiment

1. Obtain two identical rubber bands. Cut one and leave the other intact.

2. Holding one end of the cut rubber band, twist the other end 20 times. Hold on to both ends but let the rubber band relax and note the extent of collapse.

3. Holding the uncut rubber band in the middle, twist it 20 times at a point on the opposite side of the circle. Hold on to both points but let the rubber band relax and compare the extent of collapse in cut and uncut rubber bands.

4. Tape both twisted rubber bands into your lab notebook or include a cell phone photograph.

The Day Before 1. Prepare two 500 ml flasks of LB broth in liter flasks and autoclave. When cool, add enough ampicillin to give a final concentration of 50 µg/ml.

2. Inoculate each flask with one of your transformants and incubate over night shaking at 37o.

Cell Lysis First Day

1. Harvest your cultures as follows: Transfer half of each culture to a 250 ml polypropylene

bottle, balance, and centrifuge at 7,000 rpm for 10 minutes in the large floor-model centrifuge. Decant the supernatant.

Pour the remaining half of the culture into the bottle now containing only the pellet. Centrifuge again and decant. Be very careful not to mix your cultures!!

2. In a normal lab situation you would continue through to the end of the protocol. However, due to the time constraints of our class schedule, we will stop here and continue next period. Some lysis protocols include a freeze/thaw step to break down the cell walls. Although this protocol does not, a freeze/thaw cycle can only help.

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Materials for Second Day

25% sucrose lysozyme solution (5 mg/ml) 0.25 M EDTA, pH 8 Triton X-100 lytic mix

Cell Lysis Second Day

1. Prepare an ice bath and chill all of the materials.

2. Thaw the cell pellets and resuspend each in 1 ml ice-cold 25% sucrose. Place the tubes on ice for 30 minutes.

3. Add 0.2 ml lysozyme solution, mix, and place on ice for 10 minute. Lysozyme breaks down the glycosydic bonds in the polysaccharide backbone of peptidoglycan. The isotonic environment created by the sucrose prevents the spheroplasts from lysing.

4. Add 0.4 ml EDTA, mix, and place on ice for 10 minutes. EDTA is a chelating agent that absorbs divalent cations (Ca++ and Mg++), which are required by most bacterial nucleases. When the cells lyse, these nucleases will no longer be regulated and will begin degrading plasmid DNA. The addition of EDTA therefore inactivates these nasty enzymes.

5. Add 1.6 ml Triton X-100 lytic mix. Mix by rocking the tubes very gently and place on ice for 20 minutes. Triton X-100, which is a detergent, gently lyses the cells by dissolving the lipids in the cell membrane. Because of the gentle lysis, most of the chromosomal DNA remains unfragmented and becomes entangled in the cellular debris. Plasmids, being considerably smaller, diffuse out into the surrounding solution. Your samples should become very viscous and “snotty.”

6. Centrifuge the lysate at 17,000 rpm for 30 minutes. The resulting supernatant contains approximately 95% of the plasmid in about 3.2 ml. This may be purified further by CsCl density gradient centrifugation.


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Preparing the Gradient

1. Transfer your lysate to a plastic disposable centrifuge tube. Add 2 ml TES and 4.9 g CsCl. Shake gently to dissolve. CsCl becomes very cold as it dissolves.

2. Obtain a Beckman “Quick-Seal” tube and a long-tip Pasteur pipet. Carefully transfer your DNA to the “Quick-Seal” tube by inserting the tip of the pipet into the tube’s neck.

3. Using a 1 ml syringe with a 16 g needle, carefully add 200 µl of ethidium bromide (10 mg/ml) to the sample so that the ethidium bromide floats on top. Visible light will cause nicks in DNA when ethidium bromide is present. It is not necessary to work in the dark, but on the other hand, you shouldn’t go out of your way to expose the DNA to bright light.

4. Fill the remaining space in the “Quick-Seal” tube up to, but not into, the neck with mineral oil. If you do not fill up the unused space, the tube will collapse during the centrifugation.

5. Balance the tubes. The class will gather together and balance all of the tubes at once. You may remove liquid by wrapping a kim-wipe around the mouth of the tube and squeezing. If extra weight is needed, add additional mineral oil. Each pair of tubes must weigh within 0.02 g of each other.

6. Seal the tubes according to the diagrams below.

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7. Rock the tubes to mix the ethidium bromide and place them in a Beckman Ti50 ultracentrifuge rotor and spin at 40,000 rpm for 44 hours at 20o. During most of the spin, the CsCl will sort itself out into a density gradient -dilute at the top and concentrated at the bottom. Towards the end of the spin the samples will begin to sort themselves out according to their buoyant densities.

Collecting the Gradient

1. At the end of the run, carefully remove your samples and look at them under blacklight. You should see two bands. The upper, narrower band is residual chromosome that was not removed during lysis. The lower, heavy band is plasmid. You should also see a pellet of RNA on the bottom and a cap of residual protein floating on top.

2.

3.

Carefully clamp the tube to a ring stand. Puncture the top of the tube with a wide-bore hypodermic needle (16 g x 1 1/2”). Puncture the gradient just below the plasmid band with a narrow-bore needle (21 g x 1 1/2”) attached to a 10 ml syringe. Never use a wide-bore needle to collect a gradient. It will create currents in the tube that will mix up the gradient. The needle should be pointed down so as to not disturb the plasmid, and away from the pellet so as not to become contaminated with RNA.

This can be a very stressful operation the first time you try it. The most common error is to jerk your hand back when the needle suddenly goes through the wall of the tube. This results in pulling the needle out, and the plasmid running out through the hole. It is a rare person who can stick the needle back in the hole in time to save the DNA.

The second most common error is to push so hard that the needle penetrates the wall, goes through the gradient, and comes out the other side. Your other hand will be holding the tube to steady it, so you must be mindful of where you put your fingers.


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4. When the needle is in the gradient, turn it so that the beveled side is up and tip it so that it touches the underside of the band. Carefully draw off the plasmid. As you draw off the plasmid, you are introducing a discontinuity into an originally continuous gradient. There is a great enough difference in refractive index on either side of the discontinuity to see an interface. This interface is in the same position as the plasmid band and it could fool you into thinking that there is still plasmid there. It is good to get all of the plasmid, but if you get too greedy, you will end up collecting chromosome DNA that is following just behind the plasmid band.

5. When you have recovered the plasmid band, withdraw the needle and allow the gradient to dribble into a beaker. Cut the tube in half so that the remaining ethidium bromide and mineral oil can drain into the beaker for disposal by the instructor.

6. Extract the ethidium bromide with isopropanol saturated with cesium chloride. Remove the 21 g needle you used to collect the plasmid and replace it with a wider 16 g needle. Draw up a large volume of isopropanol into the syringe, rock it a few times, and hold the syringe needle-up for a moment to let the phases separate. The DNA will be on the bottom. Near the collection setup there will be a flask containing a 25 ml pipet with a rubber bulb. Holding the syringe needle-up, insert the needle into the bulb and eject the isopropanol into the bulb, where it will run down the pipet and collect in the flask. Repeat as many times as necessary to remove all of the ethidium bromide, but do at least two extractions.

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7. Obtain a dialysis bag and two dialysis clips. Drain as much

liquid out of the bag as possible and then clip one end. Carefully transfer the DNA into the bag. Half close the clip on the top of the bag and pull the neck of the bag through the clip until all of the liquid is forced into the bottom. Snap the clip shut. The bag should be taut. Dialyze the DNA over night in 4 liters of TE buffer.

8. On the following day, carefully remove the DNA with a pasteur pipet and store it in a microfuge tube for use in Experiment 7.

TES Buffer

1 M Tris, pH 8................................................................... 2.5 ml 0.25 M EDTA, pH 8........................................................ 1.0 ml 5 M NaCl.......................................................................... 0.5 ml water................................................................................. 46.0 ml


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25% Sucrose

1 M Tris, pH 8 ................................................................. 2.5 ml 0.25 M EDTA, pH 8....................................................... 0.2 ml sucrose.............................................................................. 12.5 g water................................................................................. 47.3 ml

Triton X-100 Lytic Mix

10% Triton X-100............................................................ 1 ml 1 M Tris, pH 8.................................................................. 5 ml 0.25 M EDTA, pH 8........................................................ 25 ml water................................................................................. 69 ml

Isopropanol Saturated with Cesium Chloride

Prepare a solution of CsCl (1 g/ml) in water. Add enough to saturate 250 ml of isopropanol. Once isopropanol is saturated, it becomes immiscible with water. You will know when the isopropanol is saturated because the excess CsCl will form a clearly visible layer at the bottom of the bottle. If you use unsaturated isopropanol to extract ethidium bromide from your plasmid prep, the isopropanol will draw water from your sample and the CsCl will precipitate.

Preparation of Dialysis Tubing

1. Cut the tubing into lengths of 15 - 20 cm. 2. Boil for 10 minutes in a large volume of 1 mM EDTA, pH 8, and 2% sodium

bicarbonate. 3. Rinse well in distilled water. 4. Boil for 10 minutes in distilled water. 5. Allow to cool and store in water in refrigerator. Make sure that tubing is always wet. 6. Always handle tubing with gloves!!

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Experiment 13

Large-Scale Purification of Plasmids pRIT4501 and pRIT4501 by Qiagen Ion Exchange Column Chromatography

In Experiment 6 you purified your chimeric plasmids by the classic CsCl density gradient centrifugation method. This procedure is used in many laboratories, but it is now replaced by pre-packaged kits based on ion exchange chromatography. One such kit is produced by Qiagen. The density gradient method produces larger amounts of DNA, probably at higher purity, but the kit method is very fast and convenient. The reason for doing both isolations in this course is that they each take advantage of different aspects of DNA structure and chemistry. The alkaline lysis method that you used in experiment 11 produces a small amount of a crude lysate that you run on an analytical gel to determine desired recombinant products. In contrast, the procedure in Experiment 12 produces a large amount of crude lysate that is then purified by density gradient centrifugation. Similarly, the Qiagen HiSpeed Plasmid Purification procedure produces a large amount of crude lysate that is purified by ion exchange column chromatography. The lysis technique that you will employ here is the same alkaline lysis method that you used in Experiment 11, although the details of the reagents are different. There is the same lysis and DNA denaturation by alkali and SDS, followed by subsequent reannealment with Potassium Acetate. Purification occurs when the lysate is passed through an anion exchange column. The column is a resin made of silica beads coated with diethylaminoethanol (DEAE). DEAE can be either neutral or negatively charged, depending on the salt concentration and the pH. At low salt concentration and pH, the DEAE is positively charged, so it attracts and binds the negatively charged phosphates of the phosphodiester backbone of DNA. Of course RNA and proteins are negatively charged and also bind. Their overall negative charge, however, is lower than that of DNA. After the lysate is bound in the column, the column is washed with QC buffer. This is a medium salt and neutral pH buffer that makes the DEAE somewhat neutral so that the protein and RNA impurities are washed from the column while the DNA remains bound. The wash step is followed by the elution buffer, QF. QF is a high salt high pH buffer that finally releases the DNA and allows it to pass through the column.

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Materials QIAfilter cartridge QIAfilter cartridge cap HiSpeed Midi Tip QIAprecipitator Midi Module 20 ml syringe 5 ml syringe isopropanol 70% ethanol Qiagen Reagents: P1, P2, P3, QBT, QC, QF, and TE

Two Days Before

Streak cultures containing pRIT4501 and pRIT4502 on MacConkey plates containing the appropriate selective antibiotic for isolated colonies.

The Day Before

Pick a single colony from a freshly streaked selective plate and inoculate a starter culture of 2–5 ml LB medium containing the appropriate selective antibiotic. Dilute this starter culture 10/500 to 1/1000 in 50 ml LB + antibiotic, and incubate for approximately 8 hours at 37°C with vigorous shaking (~300 rpm).

Lysis Procedure

1. Harvest your culture as follows: Transfer each culture to a 250 ml polypropylene bottle, balance, and centrifuge at 7,000 rpm for 10 minutes. Decant the supernatant.

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Resuspend the pellet in 15 ml LB broth and transfer to a 50 ml polypropylene Oakridge centrifuge. Centrifuge the cells at 10,000 rpm in the large floor model centrifuge. Decant the pellet. This is a convenient breaking point if you wish to stop the protocol and continue later, freeze the cell pellet at –20°C. We will proceed directly to the next step.

2. Resuspend the pellet in 6 ml of Buffer P1. The bacteria should be resuspended completely by vortexing or pipetting up and down until no cell clumps remain.

3. Add 6 ml of Buffer P2 and mix gently but thoroughly by inverting 4–6 times. Incubate at room temperature for 5 min. Do not vortex, as this will result in shearing of chromosomal DNA. The lysate should appear viscous. Do not allow the lysis reaction to proceed for more than 5 min. After use, the bottle containing Buffer P2 should be closed immediately to avoid acidification from CO2 in the air. During the incubation prepare the QIAfilter Cartridge: Screw the cap onto the outlet nozzle of the QIAfilter Midi and place the QIAfilter cartridge into a 25 ml disposable conical centrifuge tube

4. Add 6 ml of chilled Buffer P3 to the lysate, and mix immediately but gently by inverting 4–6 times. Proceed directly to step 5. Do not incubate the lysate on ice. Precipitation is enhanced by using chilled Buffer P3. After addition of Buffer P3, a fluffy white precipitate containing genomic DNA, proteins, cell debris, and SDS becomes visible. The buffers must be mixed completely. If the mixture appears still viscous and brownish, more mixing is required to completely neutralize the solution. It is important to transfer the lysate into the QIAfilter Cartridge


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immediately in order to prevent later disruption of the precipitate layer.

Chromatography Procedure

1. Pour the lysate into the barrel of the QIAfilter Cartridge. Incubate at room temperature for 10 min. Do not insert the plunger! This 10 min incubation at room temperature is essential for optimal performance of the QIAfilter Cartridge. Do not agitate the QIAfilter Cartridge during this time. A precipitate containing proteins, genomic DNA, and detergent will float and form a layer on top of the solution. This ensures convenient filtration without clogging. If, after the 10 min incubation, the precipitate has not floated to the top of the solution, carefully run a sterile pipet tip around the walls of the cartridge to dislodge it.

2. Equilibrate a HiSpeed Midi Tip by applying 4 ml of Buffer QBT and allow the column to empty by gravity flow. Flow of buffer will begin automatically by reduction in surface tension due to the presence of detergent in the equilibration buffer. Allow the HiSpeed Midi Tip to drain completely. HiSpeed Midi Tips can be left unattended, since the flow of buffer will stop when the meniscus reaches the upper frit in the column.

3. Remove the cap from the QIAfilter outlet nozzle. Gently insert the plunger into the QIAfilter Midi Cartridge and filter the cell lysate into the previously equilibrated HiSpeed Midi Tip. Filter until all of the lysate has passed through the QIAfilter Cartridge, but do not apply extreme force. Approximately 15 ml of lysate is generally recovered after filtration.

4. Allow the cleared lysate to enter the resin by gravity flow.

5. Wash the HiSpeed Midi Tip with 20 of Buffer QC. Allow Buffer QC to move through the HiSpeed Tip by gravity flow.

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6. Elute DNA with 5 ml of Buffer QF. Collect the eluate in a tube with a minimum capacity of 10 ml. If you wish to stop the protocol and continue later, store the eluate at 4°C. Storage periods longer than overnight are not recommended.

7. Precipitate DNA by adding 3.5 ml of room-temperature isopropanol to the eluted DNA. Mix and incubate at room temperature for 5 min. All solutions should be at room temperature in order to minimize salt precipitation.

8. During the incubation remove the plunger from a 20 syringe and attach the QIAprecipitator onto the outlet nozzle. Do not use excessive force, bending, or twisting to attach the QIAprecipitator!

9. Place the QIAprecipitator over a waste bottle, transfer the eluate/isopropanol mixture into the 20 ml syringe, and insert the plunger. Filter the eluate/isopropanol mixture through the QIAprecipitator using constant pressure

10. Remove the QIA precipitator from the 20 ml syringe and pull out the plunger. Re-attach the QIA precipitator and add 2 ml of 70% ethanol to the syringe. Wash the DNA by inserting the plunger and forcing the ethanol through the QIA precipitator using constant pressure. Important: Always remove the QIAprecipitator from the syringe before pulling up the plunger!

11. Remove the QIA precipitator from the 20 ml syringe and pull out the plunger Attach the QIA precipitator to the 20 ml syringe again, insert the plunger, and dry the membrane by pressing air through the QIA precipitator quickly and forcefully. Repeat this step several times.

12. Dry the outlet nozzle of the QIA precipitator with absorbent paper to prevent ethanol carryover.


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13. Remove the plunger from a new 5 ml syringe and attach the QIA precipitator onto the outlet nozzle. Do not use excessive force, bending, or twisting to attach the QIA precipitator! Hold the outlet of the QIAprecipitator over a microfuge tube. Add 1 ml of Buffer TE to the 5 ml syringe. Insert the plunger and elute the DNA into the collection tube using constant pressure. Be sure that the QIA precipitator is held over the collection tube when Buffer TE is poured into the syringe, as eluate can drip through the QIA precipitator before the syringe barrel is inserted. Be careful, as residual elution buffer in the QIAprecipitator tends to foam when expelled.

14. Remove the QIA precipitator from the 5 ml syringe, pull out the plunger and reattach the QIA precipitator to the 5 ml syringe. Transfer the eluate to the 5 ml syringe and elute for a second time into the same microfuge tube. This re-elution step ensures that the maximum amount of DNA in the QIA precipitator is solubilized and recovered

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Reagents for the Qiagen HiSpeed Plasmid Purification are provided in the kit Buffer P1 (Resuspension Buffer)

50 mM Tris·Cl, pH 8.0 10 mM EDTA 100 µg/ml RNase A store at 2–8°C after addition of RNase A

Buffer P2 (Lysis Buffer)

200 mM NaOH 1% SDS (w/v)

Buffer P3 (Neutralization Buffer)

3.0 M potassium acetate pH 5.5 can be stored either at room temp or 2–8°C

Buffer QBT (Equilibration Buffer)

750 mM NaCl 50 mM MOPS pH 7.0 15% isopropanol (v/v) 0.15% Triton® X-100 (v/v)

Buffer QC (Wash Buffer)

1.0 M NaCl 50 mM MOPS pH 7.0 15% isopropanol (v/v)

Buffer QF (Elution Buffer)

1.25 M NaCl 50 mM Tris·Cl pH 8.5 15% isopropanol (v/v)

Buffer TE

10 mM Tris·Cl pH 8.0 1 mM EDTA

Buffer QC (Wash Buffer)

1.0 M NaCl 50 mM MOPS pH 7.0 15% isopropanol (v/v)

Buffer QF (Elution Buffer)

1.25 M NaCl 50 mM Tris·Cl pH 8.5 15% isopropanol (v/v)

Buffer TE

10 mM Tris·Cl pH 8.0 1 mM EDTA

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Experiment 14

Verification of Purified Plasmids

Before embarking on a series of experiments with a newly prepared plasmid stock, it is prudent to verify that you have indeed isolated the DNA that you thought you had. It is also important to make sure that your DNA is not contaminated with RNA or exonuclease. Therefore, you need to do a restriction digestion and run a gel. The concentration of DNA is an important variable and will have an important impact on experiments. Too much DNA on a gel can lead to bands so thick that it is impossible to determine an accurate fragment size. Too little DNA can cause you to miss small bands. Once you have confirmed that your plasmid DNA’s are indeed pRIT4501 and pRIT4502, and you know how much to run on a gel, you are now ready to proceed to experiment 15, mapping your plasmids.

Gel Analysis

1. Prepare three restriction digests of each of your plasmids, one using 1 µl of DNA, one using 5 µl and one using 10 µl. As controls, prepare restriction digests of your rapid-plasmid preps and pUC9. Use the same REase that you used to determine the opposite orientations.

2. Run your samples on a 1% “piggy-back” gel. One row should have your cut plasmids and molecular weight standards. The other should contain uncut samples.

DNA Concentration 1. Use the NanoDrop spectrophotometer to determine the optical density of your DNA’s at 260 and 280 nm.

2. From OD260 you can calculate the concentration of your DNA.

The peak absorption of DNA is at 260 nm and there is a simple relationship between the amount of light absorbed and the concentration of DNA:

1 OD260 unit = 50 mg/ml DNA

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3. The purity of DNA can be determined by measuring the OD280. DNA absorbs almost twice as much light at 260 nm than at 280 nm. Therefore, you want to determine the ratio of the two readings:

OD260 / OD280 = 1.8

A ratio of 1.8 or better means that your DNA is sufficiently pure.

A ratio lower than 1.8 means that there is something else in your sample that is absorbing 280 nm light - i.e. proteins. Contaminating protein may or may not have an effect on restriction digestion and gels. You have to try it to find out. What it does mean, however, is that your estimation of DNA concentration is inaccurate. A ratio significantly greater than 1.8 indicates that there may have been some DNA degradation.

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Experiment 15

Construction of Restriction Maps of Plasmids pRIT4501 and pRIT4502

Now that you have constructed, purified, and verified two recombinant plasmids, you need to characterize them by constructing a detailed restriction map. At this point you should be able to do this without supervision and guidance. We will devote three weeks to this project. In the third week, we will continue with other work, although there will still be time to run additional gels. You need to map only one of the plasmids - you may choose which. Once that is done, you can easily calculate the map of the other. For the second map, you must run a confirmation gel to verify that the two maps are consistent. You may ask all the questions you like. I will be helpful on technical matters but I will be purposefully vague about giving direction. The intent of this experiment is to think and work independently. With the conclusion of this experiment you will prepare a comprehensive lab report on all work from Experiments 9 through 15. Since this includes so much time and work during the quarter, this lab report will be worth twice the other lab reports.

Preparation of

the Primary Map

1. You may choose either of your plasmids to map the following sites:

EcoRI PstI BamHI

HindIII XbaI BglII

2. You may do any series or combinations of restriction digests that you deem appropriate.

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3. You may run any concentration(s) of gels that you deem appropriate.

Some things to think about:

• Bear in mind that you are working with real data, not idealized gels as in your problem set, so don’t expect all of your digests to add up to exactly the same molecular weight.

• You can do a lot of mapping using common sense by

comparing single and double digest bands and trying to figure out what changes. In this way you can learn a lot about where sites are relative to other sites without ever calculating a fragment size.

• Don’t think that you can perform all of the digests and then at

the very end sit down the night before the project is due and calculate the map. Rather, analyze each experiment as you complete it, make the best map you can. Then try to determine what additional information you need and design an experiment to get it.

• Be wary of partial digests.

The Confirmation Map 1. Once one plasmid has been mapped, you can automatically work out the map of the other. For the second plasmid, you must provide such a map and run a gel that confirms this map.

2. For the confirmation gel, you do not need to perform all of the digestions. Rather, choose two digestion conditions, (single or double) and predict what the fragment sizes would be if you did such an experiment.

3. Do the digestion and show that you indeed do obtain the fragments that you predicted. You have then confirmed your map.

You may use any restriction digest you like for your confirmation except for the REase you originally used to identify the two orientations. This is because you already know what to expect.

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Lab Report 2

Experiments 9 - 15

Lab Report 2 represents the bulk of your work in Genetic Engineering, and it will be big! There are many experiments to cover, lots of data to analyze, and lots of details to consider. The table below is a list of each experiment, the required data, and the required analysis. In addition, you will need to describe each of the protocols that you used. If you pay careful attention to the table, you will come through this with a paper that you will be proud of and you will also have a sense of great satisfaction about the amount of work that you did this semester

Experiment Required Data Required Analysis #9 Insertion of a Gene for

Antibiotic Resistance from Bacillus subtilis into an Escherichia coli Plasmid

• ligation gel • identification of bands in each lane

• analysis of how BAP affects the ligation pattern in the gel

#10 Transformation of Escherichia coli With a Chimeric Plasmid

• table of transformation colony counts

• calculation of transformation and insertion frequencies

• calculation of transformants/mg DNA

• analysis of how BAP affects the transformation and insertion frequencies

#11 Screening of

Transformants for Chimeric Plasmids

• piggy-back gel of plasmid screens

• identification of proximal and distal orientations

• analysis of additional bands resulting from partial digestions

#12 Large-Scale Purification of Plasmids pRIT4501 and pRIT4502 by Cesium Chloride Density Gradient Centrifugation

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#13 Large-Scale Purification of Plasmids pRIT4501 and pRIT4502 by Qiagen Column Chromatograph

#14 Verification of Purified Plasmids

• verification gel • OD readings from

Cesium and Qiagen preps

• analysis of gel • calculation of DNA

concentrations • calculation of DNA purity • comparison of Qiagen and CsCl

preps

#15 Construction of a Restriction Map of pRIT4501 or pRIT4502

• gels of restriction digests • tables of distances vs

molecular wts • standard curves

• map

#15 Construction of a Restriction Map of pRIT4502 or pRIT4501

• confirmation gel • standard curve • table of

distances/molecular wts

• map of opposite orientation • prediction of expected fragment

sizes • demonstration that observed

fragment sizes match predicted values

Required protocols:

restriction digestion gel electrophoresis BAP treatment ligation transformation

plasmid extraction by alkaline lysis plasmid extraction by tritonX-100 cesium chloride density gradient sedimentation Qiagen column chromatography

Of Particular Note: • For each gel that you include, you must explain what you learned from it and how it contributed to your final map.

• You must include a final table that refers to a particular lane on a

particular gel that allowed you to determine each kb interval on your map.

• For the confirmation map, you must include a table that explicitly

states expected and observed fragment sizes.

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Some Points to Remember Abstract A short summary of what you did and what you found out. The Abstract is not an introduction to the paper. It should be written as a stand-alone unit. Introduction The Introduction should be a short overview of what question you are asking and how you plan to approach the problem. The overall goal of the project could be expressed as an intent to compare heterologous gene expression in Escherichia coli and Bacillus subtilis. In order to do that you need to construct and characterize specialized plasmids. This paper deals with their construction. You are also interested in studying the effect of BAP on ligation and the efficacy of two different methods of plasmid preparation. Relevant background includes a description of pUC9 and pBAC, the function of BAP, and a brief discussion of the difference between Qiagen and CsCl protocols. You do not have to spill your guts about everything you know about recombinant DNA. Methods and Materials Methods and Materials is a written narrative in the past tense, passive voice. Exactly the way your high school English teacher told you not to write. Do not write it as a set of instructions or in a “ first do this, then do that...” style (as this lab manual is written). Do not present a diary of your afternoon in the lab: “While the DNA was being digested I prepared a gel...” You should write as though your readers have a basic background in molecular biology. Avoid trivial detail such as how you melt the agarose, tape the ends of the gel, etc. Gels can be made in a variety of ways, depending on the type of gel box that the reader is using. Be sure to include the formulas of all reagents. These can be in a separate section for reagents, or in parentheses following the first time they are mentioned. Do not give experimental design. All protocols should be experiment-neutral. When in doubt, ask yourself “How would I run a Qiagen column? How would I do a BAP treatment? Etc.” Results Describe the experiments without going into details of procedure. The Results is a written narrative, not a list of tables and figures. Tables and figures should be numbered consecutively in the order in which they are discussed in the text. You may not include a figure or table if it is not referred to in the text. Gels should be oriented with the wells on top. Figure legends should be clear. It should not require any effort on the reader’s part to identify the contents of each lane. For example, the wells should be numbered so that the reader does not have to keep counting the wells to figure out what is in each.

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Experiment 16

Transformation of Bacillus subtilis with pRIT4501 and pRIT4502

Plasmids pRIT4501 and pRIT4502 were created by fusing the E. coli plasmid pUC9 with the B. subtilis plasmid pBAC. The recombinants therefore have origins of replication for both species and should, in theory replicate in both. pRIT4501 and pRIT4502 also contain genes for both organisms: ampicillin resistance from E. coli and two antibiotic resistances from B. subtilis. The specific antibiotics will depend upon which pBAC plasmid we are using this semester. In this experiment, we will transform B. subtilis with the plasmids and in Experiment 17 we will test for heterologous gene expression in each organism. B. subtilis, unlike E. coli, is naturally competent and should not have to be forced to take up exogenous DNA. However, not all of the cells in a culture are competent at any one time. Several growth regimens have been devised for optimizing competence. In the method that we will use, cells are grown in growth medium I (GM I) until the very end of logarithmic growth. Growth is followed by measuring optical density with spectrophotometer. The cells are allowed to grow for 90 minutes past the time at which they enter stationary phase. At the end of this 90-minute period, they are diluted into GM II (GM I supplemented with Ca++ and Mg++). Growth is continued in GM II for an additional 60 minutes. At this point, DNA is added to the cells and a 30 minute incubation is allowed for adsorption and uptake. Yeast extract is then added to shock the cells out of competence, and the cells are grown for 1 hour prior to plating on medium that will select for transformants. This final incubation is to allow time for expression of the newly acquired genes so that the cells will not die immediately when plated on selective medium.

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The Lab Before

1. Prepare 500 ml of the following plates:

LB LB + pBAC antibiotic (10 µg/ml) (you will be told which antibiotic to use) Care must be taken with antibiotics. B. subtilis is more sensitive, in general, than E. coli. 50 µg/ml would kill antibiotic resistant strain of B. subtilis while 10 µg/ml would not be a problem for antibiotic sensitive strain of E. coli.

The Day Before

1. Inoculate cells into 10 ml of GM 1 in a side-arm flask.

2. Tilt the flask so that the cells are in the arm and incubate standing over night at 27o.

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Procedure

1. Measure the optical density of the cells and record the value on a semi-log plot of “OD vs. Time.” Tip the flask so that the cells run out of the arm and into the main flask. Transfer the flask to a 37o shaking incubator.

2. Early in the growth cycle, measure the OD every 30 minutes. Once the cells reach log phase, OD should be measured every 15 minutes. Keep a running graph as you follow growth. It is important to determine as precisely as possible the time when the cells leave log phase and enter stationary phase.

3. Allow the cells to incubate for 90 minutes from the time when they enter stationary phase. You won’t actually know that this has occurred until about 30 minutes after the fact when you can feel certain that the graph has leveled off. You can count the time between reaching the inflection point and the time that you realize that it is past as part of the 90 minutes.

3. At T = 90, withdraw 1 ml of cells and transfer to a second flask containing 9 ml of GM II. Incubate for a further 60 minutes.

4. Transfer 0.9 ml of cells to each of four glass tubes and add 100 µl of DNA as indicated. Incubate for 30 minutes.

Tube 1 Tube 2 Tube 3 Tube 4

No DNA pBAC pRIT4501 pRIT4502

5. Some people add 50 ml of 10% yeast extract and then incubate for an additional 60 minutes. This apparently shocks the cells out of competence and helps the cells to survive the competence regimen better. Sometimes I omit this step if I am pressed for time.

6. Serially dilute the cells using LB broth as a diluent and plate as follows:

TBAB + antibiotic: 10-1, 10-2, 10-3 TBAB: 10-6, 10-7

7. Invert the plates and incubate at 37o over night.

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8. On the following day, recover your plates and count the colonies and record the numbers in your notebook. Save the colonies for use in Experiment 10.

Preparation for Experiment 10

Prepare two sets of plates as described below: You should have four plates of each. If you wish, you may collaborate with other groups to share the work. You will be told what antibiotics to use for #1 and #2.

MacConkey Series (E. coli) LB Series + no antibiotic

+ ampicillin (50 µg/ml) + antibiotic #1 (50 µg/ml) + antibiotic #2 (50 µg/ml)

+ no antibiotic + ampicillin (µg/ml) + antibiotic #1 (µg/ml) + antibiotic #2 (µg/ml)

10X Bacillus Salts

(NH4)2SO4....................................................................... 20 g K2HPO4........................................................................... 140 g KH2PO4............................................................................ 60 g trisodium citrate.2H2O................................................... 10 g MgSO4.7H20................................................................... 2 g

GM I

50% glucose..................................................................... 1.0 ml 10X Bacillus salts.......................................................... 10.0 ml 10% yeast extract............................................................ 1.0 ml 5 % casein hydrolysate..........................................................

0,8 ml

1% required amino acids (50 mg/ml final conc.)........ 0.5 ml water................................................................................. 90.0 ml

GM II

GM I.................................................................................. 100.0 ml 0.1 M MgCl2.................................................................... 2.5 ml 0.05 M CaCl2.................................................................. 1.0 ml

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Instructions for Using

the Spectronic 200 Spectrophotometer

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

Turn on power (switch is in back) Follow prompt to make sure cuvette holder is empty. Tap “Enter” Allow spec to initialize Use pq to select “Spectronic 200” (first menu choice) on display. Tap “Enter” Use q to select “set λ” (wavelength) Set λ to 600 nm. The knob adjusts in increments of 10 nm. tu adjusts in increments of 1nm. Open lid and insert blank into cuvette holder on right side Zero the spec by pressing “0.00” Remove blank and insert sample. Read the OD on the display. *The spec is sensitive to scratches and marks on the side-

arm so you should always insert the flask in the same orientation.

Enter the OD reading on 3-cycle semi-log paper.

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Experiment 17

Heterologous Gene Expression in Escherichia coli and Bacillus subtilis

Plasmids pRIT4501 and pRIT4502 can replicate in both E. coli and B. subtilis. At this point you have demonstrated that the E. coli gene for ampicillin resistance can be expressed in E. coli and at least one of the B. subtilis antibiotic resistance genes can be expressed in B. subtilis. You have not tried the other antibiotic resistance gene yet, nor have you looked for heterologous expression, that is the expression of B. subtilis genes in E. coli, and the expression of E. coli genes in B. subtilis. Possible barriers to heterologous expression would be differences in promoters or ribosome binding sites between the two species. B. subtilis has an additional problem that could impact heterologous expression. B. subtilis is notorious for reducing the sizes of plasmids that are inserted into it. The mechanism is unclear, but any DNA for which there is no specific selection is at risk of being spliced out. Thus, if a novel phenotype does not appear in a transformed B. subtilis strain, we cannot simply assume that the cells were unable to express the gene. In this experiment, then, we will analyze the abilities of the E. coli and B. subtilis to express the three antibiotic resistance genes, and we will screen B. subtilis transformants to make sure that the plasmid has not been altered. Upon completion of this experiment, you will be required to submit a lab report covering experiments 16 and 17.

The Day Before 1. Prepare 12 tubes containing 2 ml Lb broth + 10 µg/ml antibiotic.

2. Pick 4 colonies each from transformations #2, #3, and #4 and inoculate one into each of the tubes. Incubate standing over night at 37o

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Plasmid Analysis 1. Of the four cultures that you started for each B. subtilis transformation, discard one and use the remaining three each to perform rapid plasmid preps according to the instructions in Experiment 11. The only difference is that the lysozyme stock will be 100 mg/ml instead of 40 mg/ml. This is because B. subtilis, being gram positive has significantly more peptidoglycan in its cell wall, and is, therefore, harder to lyse.

2. Cut your rapid plasmid preps with a restriction enzyme of your choice and run them on a “piggy-back” gel. One row for cut, the other for uncut. As controls you should include your stock pRIT4501 and pRIT4502 preps, as well as molecular weight standards.

Gene Expression 1. Divide your antibiotic plates into an appropriate number of pie-shaped sectors.

2. Streak each sector of the LB plates with a sample of the B. subtilis cultures that you used for the plasmid prep. Be careful to keep adequate records so that you can correlate each lane on the gel with each sector of the plate. Invert and incubate over night at 37o.

3. Retrieve your stock cultures of E. coli transformed with pUC9, pRIT4501, and pRIT4502 and streak each on the MacConkey plates. Invert and incubate over night at 37o.

It is important that you not mix up the TBAB and LB sets because E. coli has a higher tolerance to antibiotics than does B. subtilis. Thus a sensitive E. coli strain can grow on 10 µg/m antibiotic while a resistant B. subtilis strain can be killed by 50 µg/ml.

4. On the following day, record growth (+) vs. non-growth (-) on each sector of each plate. Compare the growth patterns with the restriction digests and correlate gene expression and plasmid downsizing.

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Lab Report 3

Experiments 16 and 17

This report covers experiments 16 (Transformation Bacillus subtilis with pRIT4501 and pRIT4502) and 17 (Heterologous Gene Expression in Escherichia coli and Bacillus subtilis). In experiment 16 you transformed your plasmids into B. subtilis to create a set of strains in which both plasmids are present in both species. In experiment 17 you tested to see whether of not E. coli could express the B. subtilis genes for antibiotic resistance and whether B. subtilis could express the E. coli gene fore ampicillin resistance. In this lab report you must marshal the data at hand to explain the pattern of antibiotic resistances in both species. Required data include:

£ table of optical density readings for B. subtilis growth curve £ B. subtilis growth curve £ transformation data £ gel of miniprep plasmid extractions £ table of antibiotic resistances (+) for growth and (-) for non-growth

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Experiment 18

Southern Blot Analysis of Plasmids pRIT4501 and pRIT4502

It is often desirable to screen nucleic acids to determine whether or not they carry specific sequences. Typically, this is done by making the nucleic acid in question single-stranded, fixing it to a solid support, and challenging it with a small, labeled, single-stranded DNA (probe) under conditions that promote hybridization. If the probe sequence is found within the target DNA, the probe will bind to its complementary sequence on the support. All that is left is to detect probe binding. Probe is usually a cloned gene or other DNA fragment that is labeled with either 32P or biotin. Upon autoradiography or development with a chromogenic substrate, it is possible to visualize bound probe. Both DNA and RNA can be bound to a membrane and there are a variety of ways in which this can be done. For example, whole bacterial colonies can be screened by transferring them to a solid support (colony blot), lysing them in situ, denaturing the released DNA, and binding it irreversibly to the support. Similarly DNA can be recovered by pressing a solid support onto a plate with bacteriophage plaques (plaque lift). DNA can also be separated by size on a gel and transferred to a support by capillary action (blotting) or by electrophoretic transfer. The blotting method was invented by E. M. Southern in 1975 and has come to be known as Southern Blotting. Subsequently, Southern’s technique was applied to RNA and came to be known as “Northern Blotting.” The technique has also been adapted to proteins. Proteins are separated by SDS-PAGE electrophoresis, blotted to a solid support, and challenged not by nucleic acid probes, but by labeled antibodies. The jargon has also been adapted to proteins, and this technique is known as “Western Blotting.” After transfer of the DNA’s, the membrane is treated with salmon sperm DNA. Salmon sperm DNA is heterologous, i.e. it has no relationship to any of the DNA sequences that you are using, but it serves to block non-specific hybridization of the probe. Typically, there is always some low level of non-specific hybridization, and the intention is that any such non-specific hybridization be carried out by the unlabeled salmon sperm DNA and not by the labeled probe. Finally, by the time the probe is added only specific hybridization sites are available. The conditions under which probe is permitted to hybridize can be varied to demand greater or lesser degrees of complementarity. This is known as stringency. Stringency can be defined as the degree of complementarity that you demand for a specific hybridization. As stringency decreases, the probe can bind to an increasing number of sequences that are similar but not identical. The lower the stringency, the less the complementary must the sequences be. Sometimes, low stringency is desirable. For example, genes in eukaryotes form families and a low

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stringency hybridization may permit probe to bind to related, but not identical genes in the same family. Another example would be a case in which you have cloned a gene from one species and want to find a similar gene in another species. Depending on evolutionary distance, the sequences in the two species may have diverged quite a bit. Stringency during hybridization can be controlled by varying the temperature, by adjusting the salt concentration, and/or the amounts of urea or formamide in the hybridization solution. Urea and formamide are denaturing agents that lower the melting temperature of the hybrid. In a typical Southern blot, hybridization is performed at low stringency and then the membrane is put through a series of washes in which the stringency is progressively increased by progressively decreasing salt concentration. The strategy is to promote the maximum hybridization and then wash away any probe that is only loosely bound. The final stage in a Southern Blot is to visualize the probe. If the probe is radioactively labeled one merely performs an autoradiography. GE lab is not equipped to use radioisotope probes, so you will use a biotinylated probe. Biotin is incorporated into the probe via a nick translation reaction in which normal dATP is replaced by biotin-14-dATP. In biotin-14-dATP, the biotin is attached to the dATP by a 14-carbon linker. Detection of the probe is based on the strong affinity between biotin and avidin. Generally, avidin, which comes from egg white, is replaced with streptavidin, which comes from Streptococcus. Both compounds have similar affinities for biotin but in practice, streptavidin produces a lower background. Avidin is conjugated to an enzyme such as alkaline phosphatase, which can subsequently be used with a chromogenic substrate to produce a color change. In a standard protocol, the membrane is blocked to prevent streptavidin•alkaline phosphatase conjugate from binding non-specifically to the membrane. Blocking is accomplished by coating the membrane with a layer of bovine serum albumin. Streptavidin•alkaline is added to the membrane and finally, a chromogenic substrate is added which the alkaline phosphatase converts to a purple colored end product. The chromogenic substrate is a combination of nitro blue tetrazolium (NBT) and 5-bromo-4-chloro-3-indolyl phosphate (BCIP). The reaction is shown in the diagram below:

In this final experiment you will use Southern Blotting to examine plasmids pRIT4501 and pRIT4502. First, you will cut pRIT4501 and pRIT4502 as well as their parent plasmids pUC9 and pBAC and blot them to a nitrocellulose membrane. Then you will use the two parent plasmids to prepare probes. At the end of the experiment, you will be able to correlate parent plasmid DNA with the particular regions of the restriction maps of the recombinants.

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First Day Preparation

of Probe

1. Probe will be made by the nick-translation reaction using either pBAC or pUC9. You should team up with another lab pair so that one pair does one plasmid and one does the other.

2. Prepare the following reaction mix:

5.0 ml 2.5 ml

X ml Y ml

dTTP + dCTP + dGTP biotin-14-dATP DNA (2 µg total) water

45.0 ml

+ 5.0 µl

total volume DNA polymerase I + endonuclease A

• X depends on the concentration of your DNA • Y is the volume of water necessary to bring the

reaction up to 45 µl

3. Prepare a 14o water bath by adding ice to 4 liters of tap water

until it cools to the proper temperature (we will do this as a class). Place your nick-translation reaction in the water bath and allow it to incubate for 90 minutes.

4. Adjust the reaction mix to 0.1 M NaCl. Add 100 ml ice-cold ethanol and place in freezer for 30 minutes.

5. Centrifuge for 20 minutes at 17,000 rpm.

6. Resuspend in 50 µl TE buffer.

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First Day

Electrophoresis and Blotting

1. Cut pUC9 and pBAC with EcoRI. Cut pRIT4501and pRIT4502 with EcoRI, HindIII, and PstI. Prepare three samples of chromosomal DNA, one containing 1 µml of pUC9, one with 1 µl of pBAC, and one with no added plasmid. Cut all three with EcoRI. You will need two samples of each, so scale your reactions up accordingly.

2. Prepare 4 samples each of 1 kb ladder and l HindIII molecular weight standards.

3; Pour a 1% “piggy-back” gel.

4. Load your gels according to the table:

lane 1

2 3 4 5 6 7 8 9

10 11 12 13 14

1kb ladder pUC9 + EcoRI pBAC+ EcoRI pRIT4501+ EcoRI pRIT4501+ HindIII pRIT4501+ PstI pRIT4502+ EcoRI pRIT4502+ HindIII pRIT4502+ PstI chromosome chromosome + pUC9 + Eco RI chromosome + pBAC + Eco RI Run a piggy-back gel with duplicate lanes on top and bottom

4. At the end of the electrophoresis, stain the gel and take two sets of photographs. You will keep one set and give the other set to the students who did the probe that you did not do.

5. Transfer the gel to a plastic dish and add several volumes of 1.5 M NaCl + 0.5 M NaOH. Shake gently at room temperature for one hour. This will denture the DNA so that single-stranded DNA is transferred to the membrane.

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6. Neutralize the gel by washing in 1 M Tris, pH 8 + 1.5 M NaCl for 1 hour. If the gel is not neutralized, then the NaOH will degrade the membrane and cause it to turn brittle and shatter when you try to unload the blot.

7. Set up the transfer according to the instructions below and the following diagram:

transfer bufferfilter paper wick

gelnitrocellulose membrane

2 sheets 3MM paper

paper towelsweight

a. Cut a piece of nitrocellulose paper to the same size as the gel. Float it on the surface of 2X SSC until it is completely wet. Submerge the filter for 1 - 3 minutes.

b. Suspend the gel tray in a plastic dish and cover it with a piece of 3MM paper, draping down into the dish.

c. Fill the dish partially with 10X SSC. Remove all bubbles from the filter paper with a glass rod.

d. Lay the gel face down on the filter paper. Smooth out all bubbles.

e. Retrieve the nitrocellulose from step “a” and place it on top of the gel. Smooth out all bubbles.

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125

f. Cut strips of saran wrap and lay them on each edge of the membrane and drape them over the side of the tray. Be careful to cover as little of the wells as possible.

g. cut two sheets of 3MM paper to the same size as the gel, soak in 2X SSC and lay them over the nitrocellulose.

h. Place a stack of paper towels 10 - 12 cm thick on top of the 3MM paper and place a weight on top. The dry towels will draw the 10X SSC by capillary action up through the gel, eluting the DNA onto the nitrocellulose where it is trapped.

Second Day Blotting

1. Remove the towels and discard the gel. Make a nick in one corner of the nitrocellulose membrane so that when the side touching the gel is laying face up, the nick will be in the upper left-hand corner. This will allow you to orient the blot later on.

2. Allow the membrane to dry and then loosely sandwich it between two sheets of 3MM paper and bake it for 2 hours at 80o in a vacuum oven. This will irreversibly bind the DNA to the membrane.

Third Day Hybridization

1.

2.

Lay the membrane in a tray and soak with 6X SSC for 5 minutes. Remove SSC and add pre-wash solution. Cover the tray and incubate for 90 minutes at 42o.

3. At the end of the 90 minutes, put 15 ml of hybridization solution into a large disposable plastic centrifuge tube. Obtain a microfuge tube containing 150 µl of salmon sperm DNA and boil it for 10 minutes to denature it, then chill quickly in ice. Transfer the salmon sperm DNA to the hybridization solution in the centrifuge tube and vortex vigorously to mix, as the DNA is quite viscous.

4. Remove the pre-wash solution from the membrane and replace it with the hybridization solution + DNA. Incubate for 4 - 6 hours at 42o.

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5. Remove about half of the hybridization solution so that the filter is covered with the minimum amount of liquid as possible. Boil your probe for 10 minutes, chill and add to the membrane. Do not pipet the probe directly on top of the membrane. Rather, dip the tray so that all of the liquid pools at one end. Add the probe to the pooled liquid, mix, and then rock the tray to cover the membrane. Incubate overnight at 42o.

Fourth Day Stringency

Washes

1. Remove the hybridization solution. There is still usable probe in the solution. If you wanted to probe another blot, you would use it at step 5 of Day Three. Instead of removing half of the hybridization solution and adding fresh probe, you would remove all of the solution and replace it with your re-cycled probe. Before adding the recycled probe, you would have to first boil and chill it to denature the DNA. Biotinylated probe can be stored in the freezer for up to a year, and can be reused several times.

.2. Wash the membrane twice, 3 minutes per wash, in 2X SSC with 0.1% SDS.

3. Wash twice in 0.2X SSC with 0.1% SDS. Each wash should be 3 minutes.

4. Wash twice in 0.16X SSC with 0.1% SDS at 50o. Each wash should last 15 minutes.

5. Briefly rinse in 2X SSC with 0.1% SDS at room temperature.

Fourth Day Probe Detection

1. Wash for 1 minute in Triton X-100 buffer.

2. Incubate filter for 20 minutes in blocking buffer at 65o. The blocking buffer should be pre-warmed.

3. If you wish to pause, this is a convenient braking point. You could dry in a vacuum oven for 20 min. at 80o and come back to it later. We will not stop here.

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3. Obtain streptavidin•alkaline phosphatase (1 mg/ml) and dilute it to 1 µg/ml (i.e. 1:1000) in 10 ml of Triton X-100 buffer. This should be done in a disposable 15 ml centrifuge tube. Add the streptavidin•alkaline phosphatase and incubate at room temperature for 10 minutes with gentle agitation.

4. Transfer the membrane to a fresh tray and wash the previous one. This transfer will help reduce background.

5. Wash the membrane, with gentle agitation, 3 times with Triton X-100 buffer. Each wash should last three or four minutes.

6. Wash twice with TSM buffer.

7. Transfer the membrane to a heat-seal bag. The reaction works best if the membrane is not exposed to air.

8. Add 10 ml of chromogenic substrate to the bag and seal. Incubate the membrane in dark or low light. Color should appear quickly but maximum development may takes 4 hours.

9. When the color seems to have become intense enough take a photograph with your cell phone. The membrane will eventually fade. Terminate the color reaction by rinsing in 20 mM Tris, pH 7.6) and 5 mM EDTA, pH8.

10. Cut your membrane in half. Paste one half in your lab notebook and the other in you lab partner’s notebook. Give a copy of the cell phone photograph to the pair who did the probe that you did not do.

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20X SSC

NaCl................................................................................. 175.3 g sodium citrate................................................................... 88.2 g water................................................................................. 800 ml

Adjust to pH 7 Bring volume up to 1 liter Dilute as needed

20X SSPE

NaCl................................................................................. 174.0 g NaH2PO4......................................................................... 27.6 g EDTA............................................................................... 7.4 g

Per 800 ml Adjust to pH 7.4 Bring volume up to 1 liter Dilute as needed

Pre-Wash Solution

1 M Tris, pH 8................................................................... 12.5 ml 0.5 M EDTA, pH 8........................................................... 1.0 ml 10% SDS........................................................................... 2.5 ml NaCl................................................................................... 14.61 g

per 250 ml

50X Denhardt’s Solution

ficoll................................................................................... 5 g polyvinylpyrolidone..................................................... 5 g bovine serum albumin.................................................... 5 g

per 500 ml

Southern Blot

129

Hybridization Solution

formamide......................................................................... 67.6 ml 50X Denhardt’s solution............................................... 1.0 ml 20X SSPE....................................................................... 37.5 ml 10% SDS.......................................................................... 1.0 ml

Triton X-100 Buffer

0.1 M Tris, pH 7.5 0.1 M NaCl 2 mM MgCl2 0.05% Triton X-100

Blocking Solution

3% bovine serum albumin in Triton X-100 buffer

TSM Buffer

0.1 M Tris, pH 9.5 0.1 M NaCl 50 mM MgCl2

130

Lab Report 4

Experiment 18

This report covers experiment 18 (Southern Blot Analysis of Plasmids pRIT4501 and pRIT4502). In this experiment you cut your plasmids and chromosomal DNA with EcoRI, ran the fragments on a gel, and transferred the DNA to a nitrocellulose membrane to do a Southern blot. One team used pUC9 as the probe and the other used pPBAC. In this lab report you must explain the banding pattern of both probes with respect to the EcoRI restriction map for each plasmids.

Required data elements are:

£ photograph of the gel £ photograph of the pUC9 blot £ photograph of the pPBAC blot

Analysis

£ restriction maps of PRIT4501 and pRIT4502 £ diagram showing the predicted blotting pattern relative to the restriction maps £ explanation of why you obtain the observed bands

131

Appendix I

Index of Reagent Formulas Ampicillin, 83

Bacillus Salts (10X), 115

Blocking Solution, 129

Bromphenol Blue Tracking Dye, 17

Denhardt’s Solution (50X), 128

Dialysis Tubing, 97

GET Buffer, 88

GM I, 115

GM II, 115

Hybridization Solution, 129

Isopropanol Saturated with CsCl, 97

LB Broth and Agar, 83

Ligation Buffer (5X), 76

MacConkey Agar, 83

NEB Restriction Buffers, 19

Orange G Tracking Dye, 17

Potassium Acetate (pH 4.8), 88

Pre-Wash Solution, 128

Qiagen Reagents, 104

Restriction Digestion Buffers, 17

SSC (20X), 128

SSPE (20X), 128

Sucrose (25%), 97

TBE Buffer, 26

TE Buffer, 44

TES Buffer, 96

Thin Layer Chromatography Solvent, 53

Tracking Dye, 14

Trichloroacetic Acid (100%), 53

Triton X-100 Buffer, 129

Triton X-100 Lytic Mix, 97

TSM Buffer, 129

132

Appendix II

DNA Structure

DNA Structure

133

Abbreviation Base Nucleoside Nucleotide DNA RNA Adenine Adenosine Adenylic Acid dAMP AMP Guanine Guanosine Guanylic Acid dGMP GMP Cytosine Cytidine Cytidilic Acid dCMP CMP Thymine Thymidine Thymidilic Acid dTMP Uracil Uridine Uridylic Acid UMP

Appendix II

134

DNA Structure

135

Appendix II

136

137

Appendix III

Conversion Factors Absorbance / Concentration Conversions

Nucleic Acid Concentration (µg/ml) per A260 Unit ds DNA ss DNA ss RNA

50 33 40

Base Pair / Mass Conversions

Base Mass 1 kb ds DNA 1 kb ss DNA 1 kb ss RNA 1 MDa ds DNA average mass of dNMP Average mass of dNMP bp

6.6 x 105 Da (0.66 MDa) 3.3 x 105 Da (0.33 MDa) 3.5 x 105 Da (0.35 MDa) 1.52 kb 330 Da 660 Da

Metric Prefixes

M k m µ

mega kilo milli micro

106

103

10-3

10-6

n p f a

nano pico femto atto

10-9

10-12

10-15

10-18 Calculations

molarity of a solution = g of solute molecular mass of solute x L of solution

molecular mass of a ds DNA fragment = number of bp x 660 Da/bp moles of ends of ads DNA fragment = 2 x DNA (g) / molecular mass of DNA (Da)

or = 2 x DNA (g) / [# bp x 660 Da/bp]

moles of ends generated by a restriction endonuclease circular DNA: 2 x moles DNA x # sites

linear DNA: 2 x moles DNA x # sites + 2 (original ends)

138

Appendix IV

Buffer Dilutions

In a molecular biology lab you will frequently need to make up reagents. Very often, this is done by starting with stock solutions and then diluting them to a final concentration. For example, suppose you wish to make the following reagent:

100 ml of a solution that is 0.1 M Tris, 2 mM EDTA, 10% glucose.

1 First you must make the stock solutions of 2 M Tris and 0.1 M EDTA, and then you must dilute

and mix them appropriately. You wish to make 300 ml of each stock.

The molecular weights are: Tris = 121.1 EDTA = 372.24 Glucose = 180.16

2 M Tris = 121.1 g / 1000ml M x 2 M x 300 ml = 72.66 g

0.1 M EDTA = 372.24 g / 1000ml M x 2 M x 300 ml = 11.17 g

2 Now, using the stocks, you must determine the formula for 100 ml of the required reagent. We

first determine the dilution factor which must be applied to the Tris and the EDTA. The final dilution factor has no units. It is simply a proportion. X units must be brought up to Y amount.

Starting Concentration x dilution factor = final Concentration

[S] x DF = [F]

Dil Fac = [F] / [S]

2 M Tris x DF = 0.1M DF = 0.1M / 2 M = 0.1 / 2 0.1 / 2 = X ml Tris /100 ml Total* = 5 ml Tris

0.1 M EDTA x DF = 0.002M DF = 0.002M / 0.1M = 0.002 / 0.1 0.002 / 0.1 = X ml EDTA / 100 ml Total* = 2 ml EDTA

Glucose is simply a % wt / vol solution 10% glucose = 10g / 100 ml

Final Reagent 5 ml tris** 2 ml EDTA *** 10 g glucose enough water to make 100 ml

* this is c1v1 = c2v2

**5 ml Tris is used as part of the diluent for EDTA ***2 ml EDTA is used as part of the diluent for Tris

139

Appendix V

Bacillus subtilis Plasmid Prep by Qiagen Maxi Kit

Procedure

1. six overnight cultures in 1-liter fernbachs LB + 10 µg/ml chloramphenicol.

2. Harvest in 6 250-ml centrifuge bottles

3. Store in 6 Oak-Ridge tubes

3. Resuspend in 10 ml P1 buffer + 10 mg/ml lysozyme 37O for 30 min

4. 10 ml P2 buffer – 5 min @ room temp, gentle shaking

5. 10 ml P3 buffer – 15 min on ice

6. centrifuge twice @ 20,000 rpm for 15 min

7. cleared lysates = 30 ml split into two 15-ml aliquots add 15 ml isopropanol mix white ppt at interface

8. Centrifuge 30 min @ 10,000 rpm

9. Dissolve in 2 ml TE

10. Combine tubes of dissolved ppt into 2 tubes, 8 ml each add 15 ml QBT

11. Run 2 Qiagen maxi-columns as per std procedure 12. At end of qiagen, rinse qia precipitator with 2 1 ml washes of TE

140

Appendix VI

Effect of BAP Treatment on Transformation Frequencies

20 red colonies 10 white colonies = high transformation but low insertion frequency

0 red colonies 10 white colonies = low transformation but high insertion frequency

141

Appendix VII

α Complementation

142

Appendix VIII

Ultracentrifugation Ultracentrifugation, the characterization of macromolecules by centrifugation at super high speeds, is a foundational technique that has greatly enhanced our understanding of molecular biology. The rate of settling or sedimentation of macromolecules under high centrifugal force is a function of their size and shape. A macromolecule studied in this manner is usually characterized according to its S value (sedimentation coefficient), named after Theodor Svedberg, a Swedish chemist who received the Nobel Prize for his work on colloids and his invention of the ultracentrifuge. The relationship between rate of sedimentation and centrifugal force is:

S = velocity/centrifugal force where centrifugal force is r x ω2 r = radial distance from the axis of rotation and ω = angular velocity in radians per second.

Units of S are given as seconds, and 1S = 10-13 sec. Thus a molecule with an S value of 30S will travel 30 micrometers per second (30 x 106 m/s) under a centrifugal force of one million gravities. The most familiar use of S values is in the characterization of ribosomes. In prokaryotes, there are two ribosomal subunits, 30S and 50S, based on how their size and conformation affects their behavior under centrifugation. Together, these subunits comprise an intact 70S particle. Note that S units are not additive. They represent rates of sedimentation, not actual size. There are two basic ultracentrifugation methods: equilibrium sedimentation and zonal sedimentation. Both involve sedimentation through density gradients. Equilibrium Sedimentation CsCl gradients separate DNA according to buoyant density. The gradient is self-forming and molecules float to their isopycnic point. Macromolecules such as DNA are dissolved in a uniform solution of CsCl. High centrifugal force the causes the CsCl to sediment and form a pellet. However, diffusion forces the concentration to be equalized. After 44 hours of centrifugation, these opposing forces cause the CsCl to form a gradient of densities and molecules within the gradient will find their isopycnic points, the point at which the density of the macromolecule equals the density of the CsCl. CsCl gradients separate according to density (unit mass/unit volume). Molecules with the same density have the same isopycnic point regardless of size. Factors that affect density of DNA are conformation, strandedness, and AT/GC ratio.

Ultracentrifugation

143

The AT/GC ratio can be calculated according to the formula:

ρ = 1.660 + 0.00098 x (%GC) %GC = ρ -1.660 0.00098

ρ = density in g/ml 1.660 is the density of DNA containing 100% AT

The degree of separation of bands is a function of the shape of the gradient. The diagram below shows two gradients with different shapes. Gradient 1 is shallow and gradient 2 is steep. The line indicates the distribution of densities in the gradient. The isopycnic points of plasmid and chromosome are the same in both gradients but the position of the band depends on where the isopycnic point of each DNA intercepts the gradient curve. The different shapes are the result of small differences in recovering lysate and weighing CsCl between the various student pairs.

Appendix VII

144

Zonal Sedimentation Zonal gradients separate DNA according to size. Gradients, typically sucrose, are preformed by setting up a mixing chamber that has high and low concentration sucrose solutions in each of two chambers. The solution in one chamber is slowly pumped into a centrifuge tube with a peristaltic pump. As the chamber is depleted, it is replenished by and diluted with sucrose from the second chamber. The result is a viscosity gradient. A DNA sample is floated on top of the gradient and the sample is spun in the ultracentrifuge. Because the gradient is preformed, the run time is short. Large fragments penetrate farther into the viscosity gradient than small fragments, the opposite of a gel.

Collection of Gradients Gradients can be collected analytically or preparatively. In an analytical gradient, the bottom of the tube is pierced and the gradient collected by recovering drops of liquid in a series of tubes. The fractions can then be analyzed for refractive index, UV absorption of DNA, radioactivity, size in a gel, etc. In a preparative gradient the tube is pierced on the side with a syringe and the desired band is withdrawn. The remainder of the gradient is discarded.

145

Appendix IX

Bacteriophage λ

The mature λ map is a 48.6 kb long linear double-stranded DNA with 11-basepair overhanging cohesive single-stranded tails on either end. The chromosome is divided into three general regions – left arm, right arm, and central region. The right arm contains genes involved in replication and lysis and the right arm contains genes for phage maturation. The central region controls lysogeny and superinfection immunity.

When λ enters a susceptible cell, the single-stranded tails anneal to generate a circle. The site formed by the single stranded tails is called COS (cohesive site) and the left and right arms become contiguous and form a single operon, the right hand operon. The central region comprises a second operon, the left hand operon. λ is a temperate phage, meaning that upon entering a cell, the phage may enter either the lysogenic or lytic cycle. During lysogeny, there is a site-specific recombination event between the λ att site and a primary insertion site on the host chromosome adjacent to the lac operon. This event is mediated by the λ int gene. There are other, secondary insertion sites elsewhere on the host chromosome. There are thus three possible maps for λ, the linear mature map, the circular vegetative map, and a second linear map. In the mature map the ends are flanked by the COS site while in the lysogenic map the ends are flanked by the att site.

Appendix IX

146

This site specific recombination event is reversible. If a lysogen is presented with DNA damage, in the form of ultraviolet light, for example, the int gene together with the xis gene perform the reverse site-specific recombination event and excise the phage from the host genome, restoring it to the circular vegetative map, and the phage then enters the lytic cycle. This process is called induction. Apart from DNA damage, spontaneous induction occurs at low levels.

During the lytic phase, λ undergoes two replication phases, early and late. In the early phase, λ replicates via the Cairns or theta mode to produce multiple circular genomes. During late phase, the phage switches to rolling circle or sigma mode to produce mature linear phage genomes. Packaging is performed by a “head-full” process in which DNA is inserted into a pre-formed head. This imposes certain size restrictions that impact the use of λ as a cloning vector:

kilobases Percent of Genome Length maximum DNA length minimum DNA length size of left + right arms

52.0 38.5

29.1

105 78

60

The mature λ map is a 48.6 kb long linear double-stranded DNA with 11-basepair overhanging cohesive single-stranded tails on either end. The chromosome is divided into three general regions – left arm, right arm, and central region. The right arm contains genes involved in replication and lysis and the right arm contains genes for phage maturation. The central region controls lysogeny and superinfection immunity.

When λ enters a susceptible cell, the single-stranded tails anneal to generate a circle. The site formed by the single stranded tails is called COS (cohesive site) and the left and right arms become contiguous and form a single operon, the right hand operon. The central region comprises a second operon, the left hand operon. λ is a temperate phage, meaning that upon entering a cell, the phage may enter either the lysogenic or lytic cycle. During lysogeny, there is a site-specific recombination event between the λ att site and a primary insertion site on the host chromosome adjacent to the lac operon. This event is mediated by the λ int gene. There are other, secondary insertion sites elsewhere on the host chromosome. There are thus three possible maps for λ, the linear mature map, the circular vegetative map, and a second linear map. In the mature map the ends are flanked by the COS site while in the lysogenic map the ends are flanked by the att site.

Bacteriophage λ

147

This site specific recombination event is reversible. If a lysogen is presented with DNA damage, in the form of ultraviolet light, for example, the int gene together with the xis gene perform the reverse site-specific recombination event and excise the phage from the host genome, restoring it to the circular vegetative map, and the phage then enters the lytic cycle. This process is called induction. Apart from DNA damage, spontaneous induction occurs at low levels. During the lytic phase, λ undergoes two replication phases, early and late. In the early phase, λ replicates via the Cairns or theta mode to produce multiple circular genomes. During late phase, the phage switches to rolling circle or sigma mode to produce mature linear phage genomes. Packaging is performed by a “head-full” process in which DNA is inserted into a pre-formed head. This imposes certain size restrictions that impact the use of λ as a cloning vector:

kilobases Percent of Genome Length maximum DNA length minimum DNA length size of left + right arms

52.0 38.5

29.1

105 78

60

Appendix IX

148

Source: New England BioLabs

149

Appendix X

Technical Information

Appendix 10

150


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Appendix 10

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153

Appendix 10

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155

Appendix 10

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Appendix 10

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Appendix 10

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Appendix 10

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Appendix 10

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Appendix XI

Mrs. Rothman’s Superstar Bars

When growing microorganisms in the lab, attention must be paid to the nutritional requirements of each species and strain. Culture media may be minimal or they may be complex. Minimal media contain the bare basics for growth; balanced salts and sources for carbon and nitrogen. In addition, a minimal medium includes supplements to compensate for any auxotrophic mutations. The formulation of a minimal medium is very specific and varies from strain to strain. Complex media on the other hand contain growth supplements that would compensate for almost any auxotrophic mutation. As such, the formulation is general, typically including tryptone or casamino acids, and yeast extract. Tryptone and casamino acids are both breakdown products of milk protein, casein. Tryptone is casein digested with the protease trypsin, and contains mostly peptides. Casamino acids, on the other hand, is an acid hydrolysate of casein, and contains mostly free amino acids. In either case, they both provide all of the essential amino acids (with the exception of tryptophan in casamino acids, which is destroyed by acid hydrolysis). Media may also be selective to facilitate the growth of some species or strains at the expense of others, or differential to distinguish between bacteria growing on the same plate. Nutritional supplements are also important for the thriving and overall health of H. sapiens. Certain strains of H. sapiens respond favorably to additives such as chocolate and peanut butter. The following formula is for a well-known nutritional supplement known as Mrs. Rothman’s Superstar Bars. It is especially effective in dealing with the stress of preparing Genetic Engineering lab reports and studying for Genetic Engineering exams.

Materials BATTER 1/2 cup butter 1/2 cup sugar 1/2 cup brown sugar 1/2 cup peanut butter 1/2 tsp. soda CHOCOLATE TOPPING 1 cup chocolate chips

1/4 tsp. salt 1/2 tsp. vanilla 1 egg 1 cup flour 1 cup oatmeal

Procedure 1. An hour or so before making cookies, remove butter from the refrigerator and cut into chunks to soften.

2. Set oven at 350°F.

Superstar Bars

167

3. Line 13 x 9 inch baking pan with aluminum foil. Turn the pan upside down and mold the aluminum foil to the outside of the pan. After the foil is molded to the shape of the pan, slip it off and flip the pan over. The molded foil should slip right into the baking pan with a perfect fit. This method makes cutting much easier and eliminates having to wash the pan.

4. Combine flour, baking soda and salt in small bowl and set aside.

5. Beat butter, granulated sugar, brown sugar and vanilla extract in large mixer bowl until creamy. Add peanut butter and blend.

6. Add egg, beating well.

7. Gradually beat in flour mixture.

8. Add oatmeal and beat until combined.

9. Bake 12-15 minutes at 350°F.

10. Remove from oven and immediately sprinkle with 1 cup chocolate chips.

11.

12.

Allow to stand 5 minutes. Spread chocolate evenly over surface.

13. When cool, remove from pan by lifting the edges of the foil. Slice into squares.

14. Enjoy Caution: Superstar Bars are not to be consumed in lab!

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