Inter plate calibrator!

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Inter plate calibrator!

Postby rima250 » Aug 18 2011 11:19 pm

Hello experts, If Cq for IPC in plate A is 29 and in plate B is 28.5, should I reduce all plate B samples by 0.5 Cq or percentage of reduction between plates A and B?
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Re: Inter plate calibrator!

Postby JMG » Aug 19 2011 3:38 pm

In my non-expert opinion, there are at least two ways to approach this problem:

(First, realize that your IPC for this specific target cannot be applied to different targets - just that one certain target)

Scenario 1.) You assume that both plates ran at exactly the same amplification efficiency for the target in question (let's say both plates for the target of interest demonstrated 92% amplification efficiency for that particular target).
Scenario 2.) You have proof that the two plates did not run at the same amplification efficiency for the target of interest (e.g. Plate A: 92% Efficiency and Plate B, 94.21% Efficiency for the certain target of interest).

Trusting that your IPC is indeed an unequivocal indicator of the absolute "truth" regarding the efficiency of amplification of the target in question, then:

If Scenario 1 above is the case, and you have proof that the efficiency of amplification for the target in question
was identical (e.g. E = 92%) on both Plate A and Plate B, then, you would adjust all Cq values for this certain target on
Plate B upward by 0.5 Cq (not downward) if you consider Plate A to be correct, or, you would adjust all values on Plate A
downward by 0.5 Cq if you consider Plate B to be correct ("Which Plate do you trust?" is the question here).

But, if Scenario 2 is the case, you would first realize that a Cq of 29 at 92% Efficiency is identical to a Cq of 28.5 at 94.21% Efficiency. Thus, you need to put one plate's target Cq values in the efficiency "language" of the other plate:

E.g. Assuming Plate A to be "correct," to adjust all Plate B (E = 94.21%) Cq values into the "language" of Plate A (E = 92%) Cq values for that certain target, you would use the following transformation (using EAMP instead of E in the equation):

28.5*Log2(1.9421)/Log2(1.92) = 29
(you would do this adjustment for all Cq values for that certain target on Plate B to get them all
into the same scale or Cq "language" as Plate A)

If you assume that Plate B is "correct," you would merely transform the other way to get all Plate A Cq values
into the "language" of Plate B Cq values:

E.g. 29*Log2(1.92)/Log2(1.9421) = 28.5

To be absolutely clear: the "Log2" figure above in the equations here connotes "Log base 2"

Also recall that:
A 92% Efficient qPCR has an EAMP of "1.92" and a 94.21% Efficient qPCR has an EAMP of "1.9421" since EAMP = E + 1
and recall from qPCR target standard curves that: EAMP = 10^(-1/m) and that E = [10^(-1/m)] - 1
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