LITMIR.BIZ

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If primers and nucleotides are in the mixture at appropriate concentrations then primer degradation is minimal.

      Several types of samples are known to inhibit the PCR reaction, leading to false-negative results. Including an internal control in the assay is important for the quality control of the nucleic acid extraction, to prove the absence of PCR inhibitors. These internal controls can be either a house-keeping gene, an endogenous gene, a constant basal cell-expressed gene, such a the gluceraldehyde-3-phosphatase or the P-actin, or an exogenous nucleic acid that is not present naturally in the preparation, but added at the extraction step.

      A number of specialised methods for particular types of samples and tissues exist, most of which are now commercially available either as manual or automated systems for robotic workstations. The development and accessibility of the robotic extraction platforms not only minimises the risk of contamination, but also enables processing of large numbers of samples under constant reaction conditions and minimal operator manipulation.

      Consequently, these platforms have contributed to the establishment of high-throughput, robust diagnostic assays, shortening the processing time required per sample from hours to minutes. These are destined to improve the reliability of nucleic acid extraction from different samples, but it still remains a challenging area.

      As an alternative to nucleic acid extraction, biotechnologists are increasingly focusing on polymerases that are resistant to PCR inhibitors and several are now available on the market for direct amplification of nucleic acids from pathological specimens without any additional extraction step. Assays increasingly use an internal control to demonstrate that PCR inhibitors are not present.

      General PCR protocol

      Prepare following mixture in appropriately sized Eppendorf tube (0.2-0.5 mL):

81 pL of ddH₂010 pL of 10x polymerase buffer (for native of cloned Pfu polymerase)pL of primer #1 (100 ng/pL)pL of primer #2 (100 ng/pL)1 pL of template DNA (<100 ng/pL)1 pL of 10 mM deoxynucleotide triphosphate mixture(2.5 mM each dNTP) 2 pL of DNA polymerase(native or cloned Pfu polymerase)100 pL total reaction volume________

      Tap the side of the Eppendorf tube to mix the mixture properly, overlay it with drop of mineral oil and perform PCR reaction using following cycling parameters (see Table 3.1):

      Table 3.1

      Common stages of PCR with their specifications

      Some hints for PCR troubleshooting are shown in Table 3.2:

      Table 3.2

      PCR observations with possible causes and solutions

      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 are shown below.

      Approach I: (C,)(V) = (C_f)(V_f)

      The formula above states that the concentration of the reagent in the initial stock solution (C) multiplied by a certain volume (V) should equal the reagent’s final, diluted concentration (C_f) multiplied by the final volume (V_f) of the reaction containing the diluted reagent.

      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.

      Here, we define each term as follows: C = 5 mM MgCl₂; V = unknown; C_f = = 100 pM MgCl₂; V_f = 200 pL.

      Notice that C is in units of mM and C_f is in units of pM. C can be converted to units of pM as follows: C = 5 mM MgCl₂ x 1000 pM/1 pM = 5000 pM MgCl₂

      The problem can now be solved: (5000 pM) (V) = (100 pM) (200 pL), V = = (100 pM)(200 pL) / 5000 pM = 4 pL. Therefore, you would add 4 pL of 5 mM MgCl₂ to a 200 pL reaction to give a final concentration of 100 pM 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 sample problem, the equation can be written as follows:

      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 pL (x pL) of 5 mM MgCl₂ should be placed in a total volume of 200 pL to give a final concentration of 100 pM MgCl₂.

      The mM term is converted to gM within the equation by using the conversion factor 1000 gM/1 mM. All units on the left side of the equation cancel accept for «gM». Solving for x yields:

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

      Lets look at the following example: 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 gL reaction to its desired magnesium concentration. Using the 1^st approach, you will obtain the result as follows in Table 3.3

      Table 3.3

      Volume of reaction components for magnesium titration

      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.

      Remember that the volumes you have entered were all for each tube. That means you have to add 5 pL of PCR buffer to each of 7 tubes. Then add 4 pL 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. Now it’s time to make a Master Mix, which 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 pL. Template DNA will be added as the final step prior to thermal cycling.

      Still some part of water needs 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. In order

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