range used can vary from 10-100 pg/ml. Formamide (1.2510 %) facilitates primer-template annealing reactions and lowers the denaturing temperatures of melt resistant DNA.
Strategies to relieve inhibition depend on the inhibitor, and the lengthy listings of PCR inhibitors include many of the compounds used to enhance amplification efficiency as well as reagents required in the purification of nucleic acids. In cases where inhibitors cannot be purified or dilutes away, it appears that the solution, if possible at all, is empirical and as specific and difficult as the amplification of high- melting templates.
Reaction conditions and experimental protocol
Temperature and length of time required for annealing depend upon base composition, length and concentration of amplification primers. Generally, the annealing temperature is 5 °C below the true melting temperature of the primers. Primers will anneal in a few seconds, for efficiency, higher annealing temperatures can be used, which enhance the discrimination against incorrectly annealed primers. The annealing conditions need to be more stringent in the first 3 cycles in order to increase specificity. If the temperature is lower than optimum additional DNA fragments are commonly observed. Denaturing conditions are best at 94-95 °C for 30-60 seconds. Lower temperatures may result in incomplete denaturation of target template and PCR products.
Higher temperatures and a longer amount of time can lead to loss of enzyme activity. Diluting sample after first few rounds of PCR can be used to enhance PCR efficiency. This dilution may dilute potential inhibitors and the next round can use same primers or nested primers. In addition, the lowest number of cycles possible to achieve sufficient product should be used to assure a low number of errors.
The order of addition of reaction mixture components is also of importance. Pfu polymerase has exonuclease activity and must be added last (i.e. after dNTP’s), otherwise it may degrade primers. 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):
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) = (Cf)(Vf)
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 (Cf) multiplied by the final volume (Vf) 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; Cf = = 100 pM MgCl2; Vf = 200 pL.
Notice that C is in units of mM and Cf 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
