PCR Optimization Student Guide 2012 - BABEC

!

PCR Optimization Table of Contents Fall 2012

Optimizing the Polymerase Chain Reaction Introduction ...………………………………………………………………………………………………………..1 Review of Mathematics ...……………………………………………………………………………..………...…3 Solving Problems of Dilution and Concentration: Two Approaches …………………………………..………4 Experiment Overview ………………………………………………………………………………………………7 Calculations Part 1: Overview of PCR Reaction Components ……………………………….……………………...8 Part 2: Varying the Magnesium Chloride Concentration …………………………..…………………10 Part 3: Varying the Primer Concentration …………………….……………………………………….15 Laboratory Exercise….……………………………………………………………………………………………20 Important Laboratory Practices ………………………………………………………………………………….20 Varying Magnesium Concentration ……………………………………………………………………………..21 Varying Primer Concentration …………………………………………………………………………………...24 Agarose Gel Electrophoresis …………………………………………………………………………………….27 Electrophoresis of Amplified DNA ……………………………………………………………………………….28 Staining and Photographing Agarose Gels …………………………………………………………………….29 Interpretation of Results Role of Magnesium ………………………………………………………………………………………30 Role of Primers …………………………………………………………………………………………...32 Acknowledgements ……………………………………………………………………………………………….33

Optimizing the Polymerase Chain Reaction Frank H. Stephenson, Ph.D. Applied Biosystems

Maria C. Abilock BABEC

Introduction The Polymerase Chain Reaction (PCR) is a powerful technique used for the amplification of a specific segment of a nucleic acid. Starting with only a very small amount of material, a DNA segment can be multiplied by over a million-fold. Because of this great sensitivity, PCR has found popularity in a wide range of applications. Molecular biologists use PCR in gene cloning and DNA sequencing. Forensic scientists use PCR to connect blood, semen, saliva, or tissue left at the scene of a crime to a suspect or victim. Clinical geneticists use PCR to determine whether or not potential parents might carry a genetic disease that could be passed along to their children. PCR is DNA synthesis in a test tube. To perform PCR amplification, you need to combine all the components required for DNA replication. These are as follows: Template DNA. The template carries the DNA segment or target you wish to amplify. Primers. A primer is a short, single-stranded piece of DNA that anneals (attaches) to its complementary sequence on the template. A pair of primers will bind to either side of the target DNA segment providing initiation sites for DNA synthesis. The two primers used in a PCR are often designated by the terms forward and reverse. DNA Polymerase. This is the enzyme used to synthesize new strands of DNA. DNA polymerase adds nucleotides onto the end of an annealed primer. The added bases are complementary to the template strand. Magnesium. DNA polymerase requires magnesium for activity. Magnesium is usually supplied to a PCR amplification in the form of magnesium chloride. dNTPs. dNTP stands for deoxynucleoside triphosphate. These are the four nucleotides used by DNA polymerase to extend an annealed primer. Buffer. A buffer is a solution that resists change in pH. It is used to ensure that the DNA polymerase is in an environment in which it will have maximum activity. Although PCR is a powerful tool for the amplification of genetic sequences, it is not an exact science. Sometimes, even though using an established PCR protocol that had been optimized and successful for the amplification of a particular DNA segment, use of that same protocol on a different region can result in a less than desirable outcome. Poor amplification, however, can result from one of several causes; the temperatures used for thermal cycling may not be the best, the template may be of poor quality, or a reagent (or reagents) may be at suboptimal concentration(s). The optimal reagent concentrations for any specific target/primer set and the optimal thermal cycling parameters that will give the highest yield, more often than not, are determined by trial and error. Taking particular care when optimizing PCR can provide rewards in several ways. An optimized PCR will improve both product yield and reproducibility between reactions, while reducing amplification of non-specific products. When electrophoresed on an agarose gel, an optimized reaction will give a brighter product band with minimal background (Figure 1). When developing a protocol for PCR amplification of a new target, it may be important to optimize all parameters including reagent concentrations, cycling temperatures, and cycle number. For this experiment, thermal cycling parameters have been optimized for you. Your task will focus on reagent optimization. Varying the magnesium or primer concentration usually has the most profound affect on the quality of the reaction. Therefore, your class will optimize these reagents in a PCR experiment. To do this, you will amplify a segment of the acetylcholinesterase gene in cow DNA.

1

!

PCR Optimization Student Guide Fall 2012

Figure 1. An optimized PCR amplification will produce a single, bright band on a gel (as seen in the left-most lane). A reaction for which conditions have not been optimized will produce multiple bands as seen in the center lane. The right-most lane contains size marker.

2

!


Review of M athem atics This laboratory exercise requires that you calculate amounts and concentrations of various reagents. Concentration refers to the amount of a substance dissolved in a defined volume. Two teaspoons of sugar dissolved in a cup of water is an example of a concentration. Two teaspoons of sugar in one cup of water is more concentrated than one teaspoon. If you make a reagent less concentrated by transferring a small amount into a new volume, you have made a dilution. For example, you dilute liquid dishwashing soap by pouring a small amount into a sink full of water. In PCR, the concentration of the individual reaction components may be expressed in different units. For example, the amount of template DNA is calculated as ng/reaction, while the concentration of primer is expressed as micromolar (µM). To perform the calculations in this protocol, you need to be comfortable using the different units and converting between units. When changing between units, you must use a conversion factor, or a numerical ratio equivalent to one. For example, 100 pennies make up one dollar. The conversion factor is 100 pennies 1 dollar How would you use this conversion factor to solve the problems below? Show your work. 1. How many pennies are represented by 15 dollars? 15 dollars x 100 pennies / 1 dollar = 1500 pennies 2. How would you determine the number of dollars equivalent to 800 pennies? 800 pennies x 1 dollar / 100 pennies = 8 dollars Below is a chart showing conversion factors to help you change between units for volume measurements. Let’s try some examples. Show your work. Unit milliliter

Symbol mL

Unit Equivalence 1,000 mL = 1 liter (L)

microliter

µL

1,000,000 µL = 1 L

Conversion Factors 1,000 mL or 1L 1L 1000 mL 1,000 µL or 1 mL 1 mL 1000 µL

Table 1: Conversion Factors for Volume Measurements 3. How many µL are equivalent to 3.3 milliliters? 3.3 mL x 1000 µL / 1 mL = 3300 µL 4. Express 450 µL as milliliters. 450 µL x 1 mL / 1000 µL = 0.45 mL When you need to convert between units for measurements of mass, Table 2 may be helpful.

3

!


Unit milligram

Symbol mg

Unit Equivalence 1,000 mg = 1 gram (g)

microgram

µg

1,000,000 µg = 1 g

nanogram

ng

1,000,000,000 ng = 1 g

Conversion Factors 1000 mg or 1g 1g 1000 mg 1000 µg or 1 mg 1 mg 1000 µg 1000 ng or 1 µg 1 µg 1000 ng

Table 2: Conversion Factors for Mass Measurements Use the conversion factors from the chart above to change between mass units. Show your work. 5. How many micrograms are equivalent to 0.25 g? 0.25 g x 1,000,000 µg / 1 g = 250,000 µg 6. Express 336 ng in terms of µg. 336 ng x 1 µg / 1000 ng = 0.336 µg In your optimization experiment, you will be figuring out the concentrations of reagents needed for this particular PCR protocol. To help you convert between units of concentration, you may need Table 3. In the next set of practice problems, please make sure to show your work. Unit millimolar

Symbol mM

micromolar

µM

Unit Equivalence 1,000 mM = 1 molar (M) 1,000,000 µM = 1 M

Conversion Factors 1000 mM or 1M 1M 1000 mM 1000 µM or 1 mM 1 mM 1000µM

Table 3: Conversion Factors for Concentration Measurements 7. Convert 500 µM to the equivalent unit expressed in mM. 500 µM x 1 mM / 1000 µM = 0.5 mM 8. How would you express 0.04 mM as µM? 0.04 mM x 1000 µM / 1 mM = 40 µM

4

!


Solving Problem s of Dilution and Concentration: Two Approaches Everyone has a slightly different manner in which they think about numbers and mathematical relationships. It is one of the beauties of mathematics that any one of several different approaches can often be used to solve a problem and there is not necessarily any one “correct” way to arrive at the solution. Most of the problems you will encounter in preparing for this laboratory exercise involve calculating the dilution required to bring a particular PCR reagent to a new, desired concentration within a reaction. For example, let’s say that you have a stock solution of 5 mM (millimolar) magnesium chloride (MgCl2) and need to add some volume of that stock solution to a 200 µL reaction such that that reaction has a final concentration of 100 µM MgCl2. How much of the stock solution should you add? The two most popular approaches for tackling this problem will be presented here. Approach I: (Ci)(Vi) = (Cf)(Vf) The formula above states that the concentration of the reagent in the initial stock solution (Ci) multiplied by a certain volume (Vi) should equal the reagent’s final, diluted concentration (Cf) multiplied by the final volume (Vf) of the reaction containing the diluted reagent. For most of the problems in this protocol, you will be solving for Vi. To solve a problem using this equation, you must first identify each term. You must also make sure that all terms are in the same units. If they are not, they must be converted prior to solving the equation. For our example, we define each term as follows: Ci = 5 mM MgCl2 Vi = unknown Cf = 100 µM MgCl2 Vf = 200 µL Notice that Ci is in units of mM and Cf is in units of µM. Ci can be converted to units of µM as follows: 1000 µM Ci

=

5 mM MgCl2 x

=

5000 µM MgCl2

1 µM The problem can now be solved: (5000 µM)(Vi) = (100 µM)(200 µL) Vi = (100 µM)(200 µL) / 5000 µM = 4 µL Therefore, you would add 4 µL of 5 mM MgCl2 to a 200 µL reaction to give a final concentration of 100 µM MgCl2. Approach II: Dimensional Analysis and Canceling Terms As a variation of Approach I, an equation can be written such that all units of concentration on the left side of the equation cancel accept for the one that represents that of the final concentration written on the right side of the equation. All units accept the one desired are cancelled on the left side of the equation by ensuring that they appear as both numerator and denominator terms. Unit conversions are an integral part of the equation. For the example problem, the equation can be written as follows: 1000 µM 5 mM MgCl2

x 1mM

5

!

x µL x

=

100 µM MgCl2

200 µL


Notice that the equation begins with the initial stock concentration and describes what must be done to that stock solution to arrive at the desired final concentration on the right side of the equals sign. It asks how many µL (x µL) of 5 mM MgCl2 should be placed in a total volume of 200 µL to give a final concentration of 100 µM MgCl2. The mM term is converted to µM within the equation by using the conversion factor 1000 µM/1 mM. All units on the left side of the equation cancel accept for “µM”. Solving for x yields: (5000 µM MgCl2)(x)

=

200 x

=

(200)(100 µM MgCl2) 5000 µM MgCl2

100 µM MgCl2

20000 =

= 4 5000

All terms cancel and the solution to the problem is revealed to be 4. Since x was originally associated with µL in the initial equation, you must add 4 µL of 5 mM MgCl2 to a 200 µL reaction to bring that reaction to 100 µM MgCl2.

6

!


Experim ent Overview Acetylcholine is one of the more thoroughly studied neurotransmitters. During transmission of a nerve impulse signal, acetylcholine is broken down by the enzyme acetylcholinesterase. The gene for acetylcholinesterase (abbreviated AChE) is large. It is thousands of base pairs in length. You will determine the optimal concentrations of primer and magnesium required to amplify, by PCR, a 553 bp segment of the AChE gene as found in the cow genome. Your class will perform two titration experiments; an increasing amount of either of two reagents will be added to a series of reaction tubes. In the magnesium titration experiment, the magnesium chloride concentration will be varied from 0.5 mM to 10 mM. In another set of reactions, the primer concentration will be varied from 0.05 µM to 2 µM. You will first calculate the volume of each reagent needed in a PCR reaction. In the magnesium titration reactions, you will keep the primer concentration constant as you vary the magnesium concentration. Likewise, in the primer titration reactions, you will keep the magnesium chloride concentration constant as you vary the primer concentration. Therefore, these calculations have been divided into separate sections. After you finish the calculations, you will need to prepare a master mix, a solution containing all the components for PCR amplification (with the exception of template and the one component being optimized). Once prepared, aliquots of the master mix will be distributed equally into 7 reaction tubes. Into each of those 7 tubes will be added a different amount of the one component you have been assigned to vary (either magnesium chloride or primer). Although water is used in the preparation of the master mix, in order to keep reaction volumes constant, more will need to be added to reactions containing less of the varied component. Template DNA will be added as the last component to each tube. The final volume of each reaction should be 50 µL.

7

!


Calculations Part 1: Overview of PCR Reaction Com ponents As you work through your calculations, the formula (Ci)(Vi) = (Cf)(Vf) may prove useful. Ci = initial concentration Cf = final concentration Vi = initial volume Vf = final volume “Initial” refers to the concentration/volume of the “stock” reagents and “final” refers to the desired concentration/volume. 1. Primer Mix. Each team will prepare a primer mix. The forward and reverse AChE primers are at a concentration of 50 µM each. You will prepare a primer mix by combining 8 µL of each of the two primers to give a total combined volume of 16 µL. What is the new concentration of each primer? Ci = 50 µM Vi = 8 µL

Cf = x Vf = 16 µL

(8 µL)(50 µM) = (x)(16 µL) x = 25 µM

Concentration of each primer in the primer mix = __25 µM_ 2. Template DNA. Template DNA is always added last to a PCR reaction. You will be provided with a stock of Calf Thymus DNA at a concentration of 20 µg DNA/mL. What volume of this template DNA stock is required to deliver 200 ng to a 50 µL reaction? Hint: 200 ng is an amount, not a concentration. Ci = 20 µg/mL x mL/1000 µL x 1000 ng/µg = 20 ng/µL Cf = 200 ng/50 µL = 4 ng/µL Vi = x (20 ng/µL)(x) = (4 ng/µL)(50 µL) Vf = 50 µL x = 10 µL Into a single 50 µL reaction, add ___10___ µL of 20 µg DNA/mL template stock to give 200 ng of DNA in that tube. Note: The following three reagents will be added as components to the master mix made by each team. The problems below are for a single reaction. The volume of the master mix you prepare will be enough for 8 reactions. Although you may calculate, for a single reaction, that you may need a volume of a reagent below that of your pipette’s accuracy limit, remember that you will be multiplying this number by 8 later in the protocol when you prepare the master mix. 3. PCR Buffer. You will be supplied with a stock solution of PCR buffer having a concentration of 10X. In a 50 µL reaction, you desire a buffer concentration of 1X. To a single 50 µL, how much 10X PCR buffer should be added to give a final concentration of 1X PCR buffer? Ci = 10X Vi = x

Cf = 1X Vf = 50 µL

(10X)(x) = (1X)(50 µL) x = 5 µL

To a single 50 µL reaction, add __5 __ µL of 10X PCR buffer to give a concentration of 1X PCR buffer. 4. dNTPs. You will be provided with a stock solution of dNTPs having a total dNTP concentration of 10 mM. In this stock solution, the 4 dNTPs are in equal concentrations. Each individual dNTP (dATP, dCTP, dGTP, and dTTP), therefore, is at a concentration of 2.5 mM (10 mM dNTP/4 individual dNTPs = 2.5 mM each dNTP). In each 50 µL reaction you will prepare, you need each individual dNTP at a concentration of 200 µM. How much of the 10 mM dNTP stock solution should be added to each reaction to give this desired individual dNTP concentration? Ci = 2.5 mM x 1000 µM / mM = 2500 µM Cf = 200 µM Vi = x Vf = 50 µL

(2500 µM)(x) = (200 µM)(50 µL) x = 4 µL 8

!


To a single 50 µL reaction, add ___4___ µL of the 10 mM dNTP stock solution to give a final concentration of 200 µM for each dNTP. 5. AmpliTaq DNA Polymerase. You will be given a stock solution of AmpliTaq DNA Polymerase having a concentration of 5 Units/µL. You will need to place 1.25 Units of this enzyme into each 50 µL reaction. How much of the stock should be added to each tube? Hint: 1.25 Units is an amount, not a concentration. If you calculate that you will need 12.5 µL of AmpliTaq, then you should try another approach. Ci = 5 Units/µL Cf = 1.25 Units/50 µL = 0.025 Units / µL Vi = x Vf = 50 µL

(5 Units/µL)(x) = (0.025 Units/µL)(50 µL) x = 0.25 µL

Into a single 50 µL reaction, add ___0.25_ µL of 5 Units/µL AmpliTaq to place 1.25 Units into the reaction.

9

!


Calculations Part 2: Varying the Magnesium Chloride Concentration 6. AChE Primers. You have previously determined the concentration of each primer in the primer mix to be ___25____ µM. In the MgCl2 titration series of reactions, you require a concentration of 0.5 µM for each primer in each 50 µL reaction. How much of the primer mix should be added to each tube to achieve this concentration? Ci = 25 µM Cf = 0.5 µM Vi = x Vf = 50 µL

(25 µM)(x) = (0.5 µM)(50 µL) x = 1 µL

Into a single 50 µL reaction, add ____1___ µL of primer mix to give a final concentration of 0.5 µM for each primer. Magnesium chloride (MgCl2) will be added to a series of PCR tubes in concentrations of 0.5 mM, 1 mM, 2 mM, 3.5 mM, 5 mM, and 10 mM. The stock solution of MgCl2 you will be given has a concentration of 25 mM. In the spaces provided, calculate the volumes of 25 mM MgCl2 stock solution required for this reaction set to bring each 50 µL reaction to its desired magnesium concentration. 7. For a final concentration of 0.5 mM MgCl2: Ci = 25 mM Cf = 0.5 mM Vi = x Vf = 50 µL

(25 mM)(x) = (0.5 mM)(50 µL) x = 1 µL Into a single 50 µL reaction, add ___1____ µL of the 25 mM MgCl2 stock solution to yield a final concentration of 0.5 mM magnesium.

a) Is this a practical amount to deliver with a micropipet? Why or why not? No. p20 micropipets can not deliver 1 µL volumes accurately. The pipets used in this class cannot deliver volumes much smaller than 1 µL with reliable accuracy. Therefore, it will be necessary to prepare a diluted magnesium chloride sample from which more accurate volumes can be dispensed. You will dilute 5 µL of 25 mM MgCl2 into 5 µL of sterile water. b) What type of dilution does this represent (i.e., 1/2, 1/10, 1/20, 1/100, 1/1000)? 1/2 c) What will be the concentration of this new diluted magnesium chloride stock? 25 mM / 2 = 12.5 mM d) From this dilution, how many microliters should be added to a 50 µL reaction to bring the final concentration of magnesium chloride to 0.5 mM? Ci = 12.5 mM Cf = 0.5 mM Vi = x (12.5 mM)(x) = (0.5 mM)(50 µL) Vf = 50 µL x = 2 µL Into a single 50 µL reaction, add ___2___ µL of the diluted MgCl2 to yield a final concentration of 0.5 mM of MgCl2. 8. For a final concentration of 1 mM MgCl2 (using the 25 mM MgCl2 stock): Ci = 25 mM Cf = 1 mM Vi = x Vf = 50 µL

(25 mM)(x) = (1 mM)(50 µL) x = 2 µL Into a single 50 µL reaction, add ____2___ µL of the 25 mM MgCl2 stock solution to yield a final concentration of 1 mM magnesium. 10

!


9. For a final concentration of 2 mM MgCl2: Ci = 25 mM Cf = 2 mM Vi = x Vf = 50 µL

(25 mM)(x) = (2 mM)(50 µL) x = 4 µL Into a single 50 µL reaction, add ___4____ µL of the 25 mM MgCl2 stock solution to yield a final concentration of 2 mM magnesium.

10. For a final concentration of 3.5 mM MgCl2: Ci = 25 mM Cf = 3.5 mM Vi = x Vf = 50 µL

(25 mM)(x) = (3.5 mM)(50 µL) x = 7 µL Into a single 50 µL reaction, add ____7___ µL of the 25 mM MgCl2 stock solution to yield a final concentration of 3.5 mM magnesium.


(25 mM)(x) = (5 mM)(50 µL) x = 10 µL Into a single 50 µL reaction, add ___10___ µL of the 25 mM MgCl2 stock solution to yield a final concentration of 5 mM magnesium.


(25 mM)(x) = (10 mM)(50 µL) x = 20 µL Into a single 50 µL reaction, add ___20___ µL of the 25 mM MgCl2 stock solution to yield a final concentration of 10 mM magnesium.

Calculating the volumes of water to use for each reaction in the magnesium titration series to keep all reactions at 50 µL: With the exception of MgCl2, all reagents should be at equivalent concentrations in the series of tubes you will be preparing. Hopefully this is obvious to you at this point since the variable you are inspecting, MgCl2, should be the only reagent that changes. Use Table 4 below to summarize what will be added to each of the tubes.

11

!


NOTE: The volumes to be entered below are for one single tube. Volum e of Reaction Com ponen ts for Magnesium Titration Tube 1M 0.5 mM MgCl2

Tube 2M 1 mM MgCl2

Tube 3M 2 mM MgCl2

Tube 4M 3.5 mM MgCl2

Tube 5M 5 mM MgCl2

Tube 6M 10 mM MgCl2

Tube 7M No DNA Control (1 mM MgCl2 )

10X PCR Buffer

5 µL

5 µL

5 µL

5 µL

5 µL

5 µL

5 µL

10 mM dNTP stock

4 µL

4 µL

4 µL

4 µL

4 µL

4 µL

4 µL

AmpliTaq polymerase

0.25 µL

0.25 µL

0.25 µL

0.25 µL

0.25 µL

0.25 µL

0.25 µL

10 µL

10 µL

10 µL

10 µL

10 µL

10 µL

0 µL

1 µL

1 µL

1 µL

1 µL

1 µL

1 µL

1 µL

__2_ µL 1/2 dil of 25 mM MgCl2

__2_ µL 25 mM MgCl2



__10_µL 25 mM MgCl2

__20_µL 25 mM MgCl2


Volume of above 22.25 µL 22.25 µL 24.25 µL components Table 4. Summary of Reagents Added

27.25 µL

30.25 µL

40.25 µL

12.25 µL

Reaction Component

20 µg/mL Template DNA AChE primer mix

MgCl2

Notice that the “volume of above components” row has a variety of values. That means, at this point, that the final volume in your tube will all be different (except Tubes 1M and 2M). So a wise person may ask, “How can all the reagents added be at the same concentration if the final volumes are different?” And they wouldn’t be. So let’s take one more thing into consideration and then make the final volumes identical so that the only concentration that is changing is the MgCl2. The final volume of each of the reactions will be made identical by the addition of water. Preparing a Master Mix. Remember that the volumes you have entered in Table 4 were all for each tube. That means you have to add 5 µL of PCR buffer to each of 7 tubes. Then add 4 µL of 10 mM dNTP stock to 7 tubes and so on. That is a lot of pipetting and each time you pipet you can introduce some error in the measurement. So we will use a way to minimize the pipetting. We will make a Master Mix. The Master Mix for this experiment should contain all the reagents needed for the PCR except for the one component being titrated (in this case, magnesium) and the template DNA (which should always be added last to a PCR). The Master Mix will also contain the smallest amount of water any of the 7 tubes will need. Once prepared, an aliquot of the Master Mix will be delivered to each reaction tube. Then, different amounts of MgCl2 will be added to each tube, and additional water will be added to each reaction such that the final volume will be 50 µL. Template DNA will be added as the final step prior to thermal cycling. 12

!


Even though you have seven tubes where each tube is a trial, you will make a Master Mix for 8 reactions to ensure that enough Master Mix is available. See last paragraph of this page. Calculating Master Mix Volumes. Water will need to be part of a Master Mix so that concentrated salts in the reaction are diluted. This will be the least amount of water every tube will contain. Later we will add additional water to ensure proper dilutions and volumes. A. What is the largest volume of MgCl2 being added to any one reaction (see Table 4)? __20__ µL B. For the reaction corresponding to this MgCl2 amount, what is the volume of water required to bring that reaction to 50 µL? _9.75__ µL

At least this amount of water will be contained in each reaction you will prepare. This amount of water will be used in calculating a Master Mix. Enter this volume in the first box of Column 3 of Table 5. The volumes of reagents required to prepare a Master Mix can now be calculated using Table 5.To ensure that enough Master Mix is available, calculate the total volumes of each reagent needed assuming 8 reactions must be prepared. The amount of water to use in Column 3 was determined on the previous page. You will find the other reagent volumes listed in Table 4.

M aster M ix Reagent Com ponents (M agnesium Titration) Column

1

2

3

4

Reaction Component

Stock Solution

Final Conc.

Amount for 1 Reaction (in µL)

Number of Reactions

9.75 µL

8

5 Master Mix Total Volume (Column 3 X Column 4) 78 µL

5 µL

8

40 µL

4 µL

8

32 µL

0.25 µL

8

2 µL

1 µL

8

8 µL

water PCR Buffer

AmpliTaq

10X 2.5 mM each 5 Units/µL

Primer Mix

25 µM

dNTPs

1X 200 µM each 1.25 Units 0.5 µM each

Total 20 µL Table 5. Preparing the Master Mix for the magnesium titration experiment. Additional Water to Add to Each Reaction: Use Table 6 to calculate how much additional water (in addition to the 9.75 µL added to the Master Mix) should be added to each reaction tube so that the total volume of MgCl2 plus water is equal to the largest single volume of MgCl2 being added. No additional water will need to be added to the reaction containing the largest volume of MgCl2. For example, for Tube 1M the volume of diluted MgCl2 solution to add is 2 µL. The largest single volume of MgCl2 being added is 20 µL. Therefore, 18 µL of additional water needs to be added to tube 1M. Important Notice: Please note that for tube 1M you are adding 2 µL of diluted MgCl2. ALL OTHER TUBES, 2M–7M RECEIVE THE STOCK SOLUTION (25 Mm MgCl2). 13

!


Additional W ater to Add to M agnesium Titration Reactions Reaction Number

MgCl2 Concentration

0.5 mM

Volume of MgCl2 solution to add (in µL) 2 µL of diluted MgCl2

Volume of additional water to add to each reaction (in µL) 18 µL

1M 2M

1 mM

2 µL

18 µL

3M

2 mM

4 µL

16 µL

4M

3.5 mM

7 µL

13 µL

5M

5 mM

10 µL

10 µL

6M

10 mM 20 µL 0 µL No DNA Control, 7M 2µL 18 µL 1 mM Table 6. Volumes of Additional Water to Add to Each Reaction in the Magnesium Series.

14

!


Calculations Part 3: Varying the Primer Concentration 13. MgCl2 : The MgCl2 stock solution you will be given is at a concentration of 25 mM. In the primer titration series of reactions, you require a final MgCl2 concentration of 1 mM in each tube. How many microliters of the 25 mM MgCl2 stock solution should be added to each reaction to give a final magnesium chloride concentration of 1 mM? Ci = 25 mM Cf = 1 mM Vi = x Vf = 50 µL

(25 mM)(x) = (1 mM)(50 µL) x = 2 µL Into a single 50 µL reaction, add ____2___ µL of 25 mM MgCl2 to give a final concentration of 1 mM magnesium.

Varying amounts of primer mix will be added to a series of PCR tubes at concentrations of 0.05 µM, 0.1 µM, 0.2 µM, 0.5 µM, 1 µM, and 2 µM for each primer. You have previously calculated the concentration of each primer in the primer mix to be _____ µM. In the spaces provided, calculate the volumes of primer mix required for this reaction set to bring each 50 µL reaction to its desired primer concentration. 14. For a final concentration of 0.05 µM each primer: a) To bring the primer concentration of a 50 µL reaction to 0.05 µM for each primer, what volume of primer mix should be added? Ci = 25 µM Cf = 0.05 µM Vi = x Vf = 50 µL

(25 µM)(x) = (0.05 µM)(50 µL) x = 0.1 µL

Into a single 50 µL reaction, add __0.1__ µL of the primer mix to yield a final concentration of 0.05 µM for each primer. 15. Is this a practical amount to deliver with a micropipette? Why or why not? No. p20 micropipettes can not deliver 0.1 µL volumes accurately. The pipets used in this class cannot deliver volumes much smaller than 1 µL with reliable accuracy. Therefore, it will be necessary to prepare a diluted primer sample from which more accurate volumes can be dispensed. You will dilute 4 µL of the primer mix into 76 µL of water. b) What type of dilution does this represent (i.e., 1/2, 1/10, 1/20, 1/100, 1/1000)? 4 µL/80 µL total volume = 1/20 dilution c) What will be the concentration of this new diluted primer stock? 1.25 µM Ci = 25 µM Vi = 4

Cf = x Vf = 80 µL

(25 µM)(4 µL) = (x)(80 µL) x = 1.25 µM

d) From this dilution, how many microliters should be added to a 50 µL reaction to bring the primer concentration in that reaction to 0.05 µM for each primer? Ci = 1.25 µM Cf = 0.05 µM Vi = x Vf = 50 µL

(1.25 µM)(x) = (0.05 µM)(50 µL) x = 2 µL

Into a single 50 µL reaction, add ___2___ µL of the diluted primer mix to yield a final concentration of 0.05 µM for each primer. 15

!


15. For a final concentration of 0.1 µM each primer: You will use your diluted primer mix for this reaction as well. How many µL of the diluted primer mix should be added to a 50 µL reaction to bring the primer concentration to 0.1 µM for each primer? Ci = 1.25 µM Cf = 0.1 µM Vi = x Vf = 50 µL

(1.25 µM)(x) = (0.1 µM)(50 µL) x = 4 µL

Into a single 50 µL reaction, add ___4___ µL of the diluted primer mix to yield a final concentration of 0.1 µM for each primer. 16. For a final concentration of 0.2 µM each primer: You will use your diluted primer mix for this reaction too. How many µL of the diluted primer mix should be added to a 50 µL reaction to bring the primer concentration to 0.2 µM for each primer? Ci = 1.25 µM Cf = 0.2 µM Vi = x Vf = 50 µL

(1.25 µM)(x) = (0.2 µM)(50 µL) x = 8 µL

Into a single 50 µL reaction, add ___8___ µL of the diluted primer mix to yield a final concentration of 0.2 µM for each primer. 17. For a final concentration of 0.5 µM each primer: How many µL of the diluted primer mix should be added to a 50 µL reaction to bring the primer concentration to 0.5 µM for each primer? Ci = 1.25 µM Cf = 0.5 µM Vi = x Vf = 50 µL

(1.25 µM)(x) = (0.5 µM)(50 µL) x = 20 µL

Into a single 50 µL reaction, add __20___ µL of the diluted primer mix to yield a final concentration of 0.5 µM for each primer. NOTE: For this next reaction and all subsequent reactions requiring even higher primer concentrations, you will use the original primer mix. This will ensure that the volume of primer you add will not result in a final reaction volume that exceeds 50 µL. 18. For a final concentration of 1 µM each primer: How many µL of the original primer mix should be added to a 50 µL reaction to bring the primer concentration to 1 µM for each primer? Ci = 25 µM Cf = 1 µM Vi = x Vf = 50 µL

(25 µM)(x) = (1 µM)(50 µL) x = 2 µL Into a single 50 µL reaction, add ___2___ µL of the original primer mix to yield a final concentration of 1 µM for each primer.

19. For a final concentration of 2 µM each primer: How many µL of the original primer mix should be added to a 50 µL reaction to bring the primer concentration to 2 µM for each primer? Ci = 25 µM Cf = 2 µM Vi = x Vf = 50 µL

(25 µM)(x) = (2 µM)(50 µL) x = 4 µL 16

!


Into a single 50 µL reaction, add ___4___ µL of the original primer mix to yield a final concentration of 2 µM for each primer. Calculating the volumes of water to use for each reaction in the primer titration series to keep all reactions at 50 µL. With the exception of primers, all reagents should be at equivalent concentrations in the series of tubes you will be preparing. This hopefully is obvious to you at this point since the variable you are inspecting, primers, should be the only reagent that changes. Use Table 7 below to summarize what will be added to each of the tubes. NOTE: The volumes to be entered below are for one single tube. Volum e of Reaction Com ponents by Prim er Concentration Tube 1P

Tube 2P

Tube 3P

Tube 4P

Tube 5P

Tube 6P

0.05 µM Primer

0.1 µM Primer

0.2 µM Primer

0.5 µM Primer

1 µM Primer

2 µM Primer

10X PCR Buffer

5 µL

5 µL

5 µL

5 µL

5 µL

5 µL

5 µL

10 mM dNTP stock

4 µL

4 µL

4 µL

4 µL

4 µL

4 µL

4 µL

0.25 µL

0.25 µL

0.25 µL

0.25 µL

0.25 µL

0.25 µL

0.25 µL

10 µL

10 µL

10 µL

10 µL

10 µL

10 µL

0 µL

__2_ µL Diluted Primer Mix



_20_ µL Diluted Primer Mix

__2_ µL Primer Mix

__4_ µL Primer Mix

_20_ µL Diluted Primer Mix

2 µL

2 µL

2 µL

2 µL

2 µL

2 µL

2 µL

41.25 µL

23.25 µL

25.25 µL

31.25 µL

Reaction Component

AmpliTaq polymerase 20 µg/mL Template DNA

AChE primer mix

25 mM MgCl2

Volume of above 23.25 µL 25.25 µL 29.25 µL components Table 7. Summary of Reagents Added.

Tube 7P No DNA Control (0.5 µM Primer)

Notice that the “volume of above components” row has a variety of values. That means, at this point, that the final volume in your tube will all be different (except Tubes 1P and 5P, and 2P and 6P). So a wise person may ask, “How can all the reagents added be at the same concentration if the final volumes are different?” And they wouldn’t be. So let’s take one more thing into consideration and then make the final volumes identical so that the only concentration that is changing is the primers. The final volume of each of the reactions will be made identical by the addition of water. 17

!


Preparing a Master Mix. Remember that the volumes you have entered in Table 7 were all for each tube. That means you have to add 5 µL of PCR buffer to each of 7 tubes. Then add 4 µL of 10 mM dNTP stock to 7 tubes and so on. That is a lot of pipetting and each time you pipet you can introduce some error in the measurement. So we will use a way to minimize the pipetting. We will make a Master Mix. The Master Mix for this experiment should contain all the reagents needed for the PCR except for the one component being titrated (in this case, primer) and the template DNA (which should always be added last to a PCR). The Master Mix will also contain the smallest amount of water any of the 7 tubes will need. Once prepared, an aliquot of the Master Mix will be delivered to each reaction tube. Then, different amounts of primers will be added to each tube, and additional water will be added to each reaction such that the final volume will be 50 µL. Template DNA will be added as the final step prior to thermal cycling. Even though you have seven tubes where each tube is a trial, you will make a Master Mix for 8 reactions to ensure that enough Master Mix is available. See last paragraph of this page. Calculating Master Mix Volumes. Water will need to be part of a Master Mix so that concentrated salts in the reaction are diluted. This will be the least amount of water every tube will contain. Later we will add additional water to ensure proper dilutions and volumes. A. What is the largest volume of primer being added to any one reaction (see Table 7)? __20__ µL B. For the reaction corresponding to this primer amount, what is the volume of water required to bring that reaction to 50 µL? _8.75__ µL At least this amount of water will be contained in each reaction you will prepare. This amount of water will be used in calculating a Master Mix. Enter this volume in the first box of Column 3 of Table 8. The volumes of reagents required to prepare a Master Mix can now be calculated using Table 8. To ensure that enough Master Mix is available, calculate the total volumes of each of reagent needed in Table 8, assuming 8 reactions will be prepared. Use Table 7 to fill in other volumes for the other reagents in Table 8. M aster M ix Reagent Com ponents (Prim er Titration) Column

1

2

3

4

Reaction Component

Stock Solution

Final Conc.

Amount for 1 Reaction (in µL)

Number of Reactions

8.75 µL

8

5 Master Mix Total Volume (Column 3 X Column 4) 70 µL

5 µL

8

40 µL 32 µL

water PCR Buffer dNTPs AmpliTaq MgCl2

10X 2.5 mM each 5 Units/µL

1X 200 µM each 1.25 Units

4 µL

8

0.25 µL

8

2 µL

25 mM

1 mM

2 µL

8

16 µL

Total 20 µL Table 8. Preparing the Master Mix for the Primer Titration Experiment.

18

!


Additional Water to Add to Each Reaction: Use Table 9 below to calculate how much additional water (a volume in addition to the 8.75 µL added to the Master Mix) should be added to each reaction tube so that the total volume of primer plus water is equal to the largest single volume of primer being added. No additional water will need to be added to the reaction containing the largest volume of primer. For example, for Tube 1P the volume of diluted primer mix to add is 2 µL. The largest single volume of primer being added is 20 µL. Therefore, 18 µL of additional water needs to be added to tube 1P. Important Notice: Please note that for tube 1P–4P AND 7P you are adding diluted Primer Mix. ALL OTHER TUBES, 5P–6P RECEIVE THE PRIMER MIX SOLUTION. Reaction Number 1P

Primer Concentration 0.05 µM

Primer Source Diluted Primer Mix

Volume of primer to Add (in µL) 2 µL

Volume of additional water to add (in µL) 18 µL

2P

0.1 µM

Diluted Primer Mix

4 µL

16 µL

3P

0.2 µM

Diluted Primer Mix

8 µL

12 µL

4P

0.5 µM

Diluted Primer Mix

20 µL

0 µL

5P

1 µM

Primer Mix

2 µL

18 µL

6P

2 µM Primer Mix 4 µL 16 µL No DNA Control, 7P Diluted Primer Mix 20 µL 0 µL 0.5 µM Table 9. Volumes of additional water to add to each reaction in the primer titration series to make them equal in volume to that of the sample receiving the largest volume of primer.

19

!


Laboratory Exercise Your teacher will divide you into teams to perform either the magnesium or the primer optimization protocol. In performing this experiment, keep these laboratory practices in mind: Im portant Laboratory Practices a. Add reagents to the bottom of the reaction tube, not to its side. b. Add each additional reagent directly into previously-added reagent. c. Do non pipette up and down to mix, as this introduces error. This should only be done only when resuspeding the cell pellet and not to mix reagents. d. Make sure contents are all settled into the bottom of the tube and not on the side or cap of tube. A quick spin may be needed to bring contents down.

Keep reagents on ice.

a. Pipet slowly to prevent contaminating the pipette barrel. b. Change pipette tips between each delivery. c. Change the tip even if it is the same reagent being delivered between tubes. Change tip every time the pipette is used!

Check the box next to each step as you complete it.

20

!


Varying M agnesium Concentration 1. Label a 1.5 mL tube "PM," for Primer Mix.

 2. Into your "PM" tube, add 8 µL of each of the two AChE primers (forward and reverse). Keep this tube on ice.

 3. Label a new 1.5 mL tube "1/2 Mg," for 1/2 dilution of MgCl2.

 4. Into your "1/2 Mg" tube, add 5 µL of sterile water and 5 µL of 25 mM MgCl2. Keep this tube on ice.

 5. Label another new 1.5 mL tube "MM," for Master Mix.

 21

!


6. Prepare the Master Mix. Into the "MM" tube, add each reaction component according to the volumes calculated for Column 5, Table 5. Add the water to the “MM” tube first. Keep the Master Mix and all Master Mix components on ice. Template DNA is not added to the master mix; it will be added individually to each reaction later. Note: Slowly pipet up and down to mix the reagents after each addition. water = 78 µL PCR Buffer = 40 µL dNTP Mix = 32 µL AmpliTaq = 2 µL PM = 8 µL

 7. Mark seven 0.2 mL tubes "1M" through "7M". Also, label each tube with your initials.

 8. Gently flick the Master Mix tube with your finger to mix the solution.

 9. Making sure to use a balance tube, spin the Master Mix tube for 5 seconds in a microcentrifuge.

 22

!


10. To each tube, add the appropriate combination of MgCl2 and water (see Table 6). To each tube except "7M," also add the volume of template DNA as determined previously. To tube "7M" (the Control), add a volume of additional water equal to the volume of DNA being added to each tube 1M through 6M. [_18_ µL water for tube 7M from Table 6 + __10_ µL water equivalent to the amount of template DNA being added to each reaction = __28_ µL total water for tube 7M.]

Reagent Water 1:2 dil. Mg 25 mM Mg Template

1M (0.5 mM Mg) 18 µL 2 µL 10 µL

2M (1 mM Mg) 18 µL 2 µL 10 µL

Tube 4M (3.5 mM Mg) 13 µL 7 µL 10 µL

3M (2 mM Mg) 16 µL 4 µL 10 µL

5M (5 mM Mg) 10 µL 10 µL 10 µL

6M (10 mM Mg) 0 µL 20 µL 10 µL

7M (1 mM Mg) 28 µL 2 µL 0 µL 

11. Place your reactions into the thermal cycler and record the location of your tubes on the grid provided by your teacher.

1 A B C

2

3

4

1123

5

6

7

828 1027 6777

9305

 The cycling protocol for amplification using these AChE primers is as follows: 95°C—10 minutes 95°C—1 minute 54°C—1 minute 72°C—1 minute 72°C—10 minutes 4°C—hold, ∞ infinity

35 cycles



23

!


Varying Prim er Concentration 1. Label a 1.5 mL tube "PM" for Primer Mix.

 2. Into your "PM" tube, add 8 µL of each of the two AChE primers (forward and reverse). Keep this tube on ice.

 3. Label a new 1.5 mL tube “1/20 PM," for a 1/20 dilution of your Primer Mix.

 4. Into the "1/20 PM" tube, add 76 µL of sterile distilled water and 4 µL of Primer Mix (from Step 2). Keep this tube on ice.

 5. Label a new 1.5 mL tube "MM," for Master Mix.

 24

!


6. Prepare the Master Mix. Into the "MM" tube, add each reaction component according to the volumes calculated for Column 5, Table 8. Add the water to the “MM” tube first. Keep the Master Mix and all Master Mix components on ice. Template DNA is not added to the master mix; it will be added individually to each reaction later. NOTE: Slowly pipet up and down to mix the reagents after each addition. Water = 70 µL PCR Buffer = 40 µL dNTP Mix = 32 µL AmpliTaq = 2 µL 25 mM MgCl2 = 16 µL

 7. Mark seven 0.2 mL tubes "1P" through "7P". Label each tube with your initials as well.

 8. Gently flick the Master Mix tube with your finger to mix the solution.

 9. Making sure to use a balance tube, spin the Master Mix tube for 5 seconds in a microcentrifuge.



25

!


10. Pipet into each of your 7 labeled 0.2 mL tubes the volume of Master Mix as calculated in Table 8, Total for Column 3. 20 µL into each tube

 11. To each tube, add the appropriate combination of Primer Mix and water (see Table 9). To each tube except "7P," also add the volume of template DNA as determined previously. To tube "7P" (the Control), add a volume of additional water equal to the volume of template added to each tube 1P through 6P. [__0 _ µL water for tube “7P” from Table 9 + __10_ µL water equivalent to the amount of template DNA being added to each tube = _10__ µL total water for tube “7P”.]

Reagent Water 1/20 PM PM Template

Tube 1P (0.05 µM Primer) 18 µL 2 µL 10 µL

2P (0.1 µM Primer) 16 µL 4 µL 10 µL



5P (1 µM Primer) 18 µL 2 µL 10 µL

6P (2 µM Primer) 16 µL 4 µL 10 µL

7P (0.5 µM Primer) 10 µL 20 µL 0 µL 

12. Place your reactions into the thermal cycler and record the location of your tubes on the grid provided by your teacher.

1 A B C

2

3

4

1123

5

6

7

828 1027 6777

9305

 The cycling protocol for amplification using these AChE primers is as follows: 95°C—10 minutes 95°C—1 minute 54°C—1 minute 72°C—1 minute 72°C—10 minutes 4°C—hold, ∞ infinity

35 cycles

 26

!


Agarose Gel Electrophoresis To determine whether or not the AChE PCR product amplified, you will need to visualize the products of your amplification. This will be done using a process called gel electrophoresis in which electric current forces the migration of DNA fragments through a special gel material. Since DNA is negatively charged, it will migrate in an electric field towards the positive electrode (Figure 2). When electrophoresed through a gel, shorter fragments of DNA move at a faster rate than longer ones. Figure 2. Side view of an agarose gel showing DNA loaded into a well and the direction of DNA fragment migration during electrophoresis.

The gel material to be used for this experiment is called agarose, a gelatinous substance derived from a polysaccharide in red algae. When agarose granules are placed in a buffer solution and heated to boiling temperatures, they dissolve and the solution becomes clear. A comb is placed in the casting tray to provide a mold for the gel. The agarose is allowed to cool slightly and is then poured into the casting tray. Within about 15 minutes, the agarose solidifies into an opaque gel having the look and feel of coconut Jell-O™. The gel, in its casting tray, is placed in a buffer chamber connected to a power supply and running buffer is poured into the chamber until the gel is completely submerged. The comb can then be withdrawn to form the wells into which your PCR sample will be loaded. Loading dye is a colored, viscous liquid containing dyes (making it easy to see) and sucrose, Ficoll, or glycerol (making it dense). To a small volume of your total PCR reaction, you will add loading dye, mix and then pipet an aliquot of the mixture into one of the wells of your agarose gel. When all wells have been loaded with sample, you will switch on the power supply. The samples should be allowed to electrophorese until the dye front (either yellow or blue, depending on the dye used) is 1 to 2 cm from the bottom of the gel. The gel can then be moved, stained and photographed. Calculations for Preparing 2% Agarose Gel You will need a 2%, mass/volume agarose gel for electrophoresis of your PCR products. If your agarose gel casting trays holds 50 mL, then how much agarose and buffer would you need? The definition of m/v % in biology is grams (mass) / 100 mL (volume). Therefore, for 2% agarose, it will be 2 g /100 mL buffer. Step 1: Calculate the mass of agarose needed for 50 mL total volume of agarose solution. 2g

Xg =

100 ml

X = 1 gram 50 ml

Step 2: Calculate the amount of buffer needed to bring the agarose solution to 50 mL. By standard definition, 1 gram of H2O = 1 mL of H2O. The amount of buffer for the 2% agarose solution will be 49 mL (50 mL – 1 mL (1 gram of agarose)).

27

!


Electrophoresis of Am plified DNA 1. Retrieve your PCR tube and place it in a balanced configuration in a microcentrifuge. Spin it briefly (10 seconds) to bring the liquid to the bottom of the reaction tube. Note: Make sure the centrifuge adapters are in place before putting the tiny PCR tube into the centrifuge rotor. 2. Add 5 µL of loading dye to your PCR tube.

3. Carefully load 15 to 20 µL of the DNA/loading dye mixture into a well in your gel. Make sure you keep track of what sample is being loaded into each well. Note: Avoid poking the pipette tip through the bottom of the gel or spilling sample over the sides of the well. Use a new tip for each sample. 4. One student (or the instructor) should load 5-10 µL of 100 bp ladder (molecular weight marker) into one of the wells of each gel.

5. When all samples are loaded, attach the electrodes from the gel box to the power supply. Have your teacher check your connections and then electrophorese your samples at 150 Volts for 25–40 minutes.

6. After electrophoresis, the gels will be ready to stain and photograph.

28

!


Staining and Photographing Agarose Gels The PCR products separated on your agarose gel are invisible to the naked eye. If you look at your gel in normal room light, you will not be able to see the amplified products of your reaction. In order to “see” them, we must stain the gel with a fluorescent dye called ethidium bromide (EtBr). Molecules of ethidium bromide are flat and can intercalate, or insert, between adjacent base pairs of double stranded DNA (Figure 3). When this interaction occurs, they take on a more ordered and regular configuration causing them to fluoresce under ultraviolet light (UV). Exposing the gel to UV light after staining, allows you to see bright, pinkish-orange bands where there is DNA (figure 4). Figure 3. Ethidium bromide molecules intercalated between DNA base pairs.

Your teacher may stain your agarose gel and take a photograph for you so that you may analyze your PCR results. Gel staining is done as follows: 1. Place the agarose gel in a staining tray. 2. Pour enough ethidium bromide (0.5µg/ mL) to cover the gel. 3. Wait 20 minutes. 4. Pour the ethidium bromide solution back into its storage bottle. 5. Pour enough water into the staining tray to cover the gel and wait 5 minutes. 6. Pour the water out of the staining tray into a hazardous waste container and place the stained gel on a UV light box. 7. Place the camera over the gel and take a photograph. 8. Check with your district on how to dispose of hazardous waste liquid and solids. CAUTION: Ethidium bromide is considered a carcinogen and neurotoxin. Always wear gloves and appropriate PPE (personal protective equipment) like safety glasses when handling. Students should NEVER handle EtBr. CAUTION: Ultraviolet light can damage your eyes and skin. Always wear protective clothing and UV safety glasses when using a UV light box. Figure 4. After staining an agarose gel with ethidium bromide, DNA bands are visible upon exposure to UV light.

29

!


Interpretation of Results Role of Magnesium AmpliTaq DNA Polymerase adds bases onto the end of an annealed primer. Magnesium ions interact with the DNA polymerase enzyme during this process. Magnesium, in fact, is absolutely required for DNA polymerase activity. Because of the close relationship between these molecules, magnesium is said to be a cofactor of the polymerase enzyme. PCR amplification of a DNA target would not occur if magnesium were left out of the reaction. Magnesium is usually supplied to a PCR amplification in the form of magnesium chloride. As this compound is a ++ salt, in the water environment of a DNA synthesis reaction, it dissociates into the two ions Mg and Cl . Because of its positive charge, the magnesium ion interacts with negatively charged molecules in the reaction. What are these molecules? Template DNA is one of them. It has negatively charged phosphate groups running along its backbone. The primers too are DNA and even though single-stranded, have a sugar-phosphate backbone. This gives them a negative charge. Of critical importance, however, are the dNTPs. Each carries a total of 4 negatively charged oxygen molecules attached to the triphosphate group (see Figure below). Relative to the other reaction components, the dNTPs are in a very high concentration and therefore constitute the predominant species interacting and binding to the magnesium ions.

Figure 5. Molecular structure of the dNTP, deoxyadenosine triphosphate (dATP). There are four negativelycharged oxygen molecules surrounding the phosphate atoms. These can interact with magnesium ions. Consequently, in a PCR experiment, it is necessary to use an amount of MgCl2 above that of the dNTP concentration; some magnesium must be left free and available for use by the AmpliTaq DNA Polymerase enzyme. If magnesium is not available, no extension can occur.

Magnesium and DNA DNA has an overall negative charge because of the negatively charged oxygen molecules along the two sugarphosphate chains of the double helix. Since both chains are negatively charged, they have a natural tendency to repel each other. In fact, if DNA is placed in water free of any ions, the two strands of DNA are very likely to come + ++ apart. Positive ions such as Na and Mg (found in sodium chloride and magnesium chloride), however, can interact with the negatively charged DNA strands to mask the forces of repulsion. The higher the salt concentration, the more likely DNA will remain double-stranded. In addition, at higher salt concentrations, two strands of DNA can be made to anneal to each other even if there is not perfect complementarity between them. Under conditions of very high salt concentrations, the double helix structure for some DNA segments can be quite stable, so much so that an even higher temperature is required to denature it.

30

!


Magnesium Optimization: Interpreting Results On an agarose gel, an optimized PCR amplification should give one single, bright band of the desired size. A poorly-optimized reaction will show the desired band in reduced intensity and the presence of non-specific bands of different sizes. The AChE primers used in this experiment will produce a 553 bp PCR product. Questions: a) Did you obtain the 553 bp band expected for this experiment?

b) Which magnesium chloride concentration seems to be optimal for amplification of this bovine acetylcholinesterase gene segment?

c) What do you notice in the lane with the lowest magnesium concentration? Why?

d) What do you notice in the lanes with higher magnesium concentrations? Why?

e) What advantage does using a master mix provide the experimenter?

31

!


Role of Primers Primers are short, the template DNA DNA Polymerase. This ensures that, annealing.

single-stranded DNAs designed to anneal to complementary sequences on opposite strands of flanking the target region. They serve as initiation sites for the addition of bases by AmpliTaq For successful PCR, primers must be added in molar excess over the amount of target DNA. following the denaturation step primer/template annealing is favored over template/template re-

During PCR, an annealing step is used to allow primer attachment to the template. The temperature chosen for annealing is based on the melting temperature (Tm) of the primers. The melting temperature of a particular DNA molecule is defined as that temperature at which half of it is in double-stranded form and half is in single-stranded form. Longer DNA molecules, because of the many hydrogen bonds formed between their two strands, have higher melting temperatures than shorter DNA molecules. Similarly, if two DNA molecules are of the same size but have different numbers of G and C bases, the molecule with more Gs and Cs will have the higher Tm. This is because the G and C bases form three hydrogen bonds between them while the A and T bases form only two hydrogen bonds between them; more heat is required to break three hydrogen bonds than is required to break two hydrogen bonds. The annealing temperature used for a particular primer/template combination is usually optimal at plus or minus 7°C of the primer Tm. An annealing temperature is chosen that will give the highest amount of specific product (one bright band on an agarose gel) with the least amount of nonspecific amplification (multiple bands on an agarose gel). The goal is to choose a temperature at which each primer anneals to only one site on the template. However, even at the most optimal annealing temperature, there is probably always some small percentage of primer that will anneal to secondary sites on the template; sites that are not perfectly complementary to the primers. Primer Optimization: Interpreting Results On an agarose gel, an optimized PCR amplification should give one single, bright band of the desired size. A poorly optimized reaction will show the desired band in reduced intensity and the presence of non-specific bands of different sizes. The AChE primers used in this experiment will produce a 553 bp PCR product. Questions: a) Did you obtain the 553 bp band expected for this experiment? b) Which primer concentration seems to be optimal for amplification of this bovine acetylcholinesterase gene segment? c) What do you notice in the lanes with higher primer concentration? Why? d) What do you notice in the lanes with lower primer concentration? Why? e) What advantage does using a master mix provide the experimenter?

32

!


Life Technologies & Applied Biosystem s / BABEC Educational PCR Kits

For Research Use Only. Not for use in diagnostic procedures. NOTICE TO PURCHASER: LIMITED LICENSE A license under U.S. Patents 4,683,202, 4,683,195, and 4,965,188 or their foreign counterparts, owned by Roche Molecular Systems, Inc. and F. Hoffmann-La Roche Ltd (Roche), for use in research and development, has an up-front fee component and a running-royalty component. The purchase price of the Lambda PCR, Alu PV92 PCR, PCR Optimization, D1S80 PCR, and Mitochondrial PCR Kits includes limited, non-transferable rights under the runningroyalty component to use only this amount of the product to practice the Polymerase Chain Reaction (PCR) and related processes described in said patents solely for the research and development activities of the purchaser when this product is used in conjunction with a thermal cycler whose use is covered by the up-front fee component. Rights to the up-front fee component must be obtained by the end user in order to have a complete license. These rights under the up-front fee component may be purchased from Applied Biosystems or obtained by purchasing an authorized thermal cycler. No right to perform or offer commercial services of any kind using PCR, including without limitation reporting the results of purchaser’s activities for a fee or other commercial consideration, is hereby granted by implication or estoppel. Further information on purchasing licenses to practice the PCR process may be obtained by contacting the Director of Licensing at Applied Biosystems, 850 Lincoln Centre Drive, Foster City, California 94404 or at Roche Molecular Systems, Inc., 1145 Atlantic Avenue, Alameda, California 94501. Use of this product is covered by US patent claims and corresponding patent claims outside the US. The purchase of this product includes a limited, non-transferable immunity from suit under the foregoing patent claims for using only this amount of product for the purchaser’s own internal research. No right under any other patent claim (such as the patented 5’ Nuclease Process claims) and no right to perform commercial services of any kind, including without limitation reporting the results of purchaser's activities for a fee or other commercial consideration, is conveyed expressly, by implication, or by estoppel. This product is for research use only. Diagnostic uses require a separate license from Roche. Further information on purchasing licenses may be obtained by contacting the Director of Licensing, Applied Biosystems, 850 Lincoln Centre Drive, Foster City, California 94404, USA. TRADEMARKS: Applied Biosystems, AB (Design), GeneAmp, and Primer Express are registered trademarks and Veriti and VeriFlex are trademarks of Applied Biosystems Inc. or its subsidiaries in the US and/or certain other countries. AmpliTaq is a registered trademark of Roche Molecular Systems, Inc. All other trademarks are the sole property of their respective owners. © Copyright 2001, Applied Biosystems. All rights reserved.

33

!


PCR Optimization Student Guide 2012 - BABEC

Recommend Documents