I will do a qPCR experiment for the first time, and I have a bottom-level question to ask. I know that for normalizing the quantity of your target gene you have to add an endogenous control, like a housekeeping gene, in your qPCR sample, and then you subtract the Ct of the housekeeping gene from the Ct of your target gene. However, since the same SYBR green fluorophore is used to quantify both the target gene and the housekeeping gene in the same sample, how do you know which Ct is from the target gene and which one is from the housekeeping gene?


1 Answer 1


You do not take readings from the same reaction, you split the sample. You would run each primer pair in their own reaction. You also want to run each reaction in triplicate so you can get some idea of the error. So, a general plate set up might look something like this:

A1: Control sample, housekeeping primers; A2: Control sample, target gene primers
B1: Control sample, housekeeping primers; B2: Control sample, target gene primers
C1: Control sample, housekeeping primers; C2: Control sample, target gene primers

D1: Treatment sample, housekeeping primers; D2: Treatment sample, target gene primers
E1: Treatment sample, housekeeping primers; E2: Treatment sample, target gene primers
F1: Treatment sample, housekeeping primers; F2: Treatment sample, target gene primers

So in this scenario, the delta-delta Ct value comes from the following:

delta-delta Ct = ( Ct(treatment, target) - Ct(treatment, housekeeping) ) - ( (Ct(ctrl, target) - Ct(ctrl, housekeeping) )

If the PCR machine does not offer to do the delta-delta Ct analysis for you I recommend this blog which both explains it well and provides and excel spreadsheet so you can just drop you Ct values in.

  • $\begingroup$ That makes more sense, I got that information wrong. Thanks for the clarification. $\endgroup$
    – betelgeuse
    Jan 25, 2019 at 22:22
  • $\begingroup$ No problem. As much as possible, make master mixes and decant them in down in to the single reactions to keep things as homogeneous as possible. Good luck. $\endgroup$ Jan 25, 2019 at 22:25

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