I would like to calculate the relative gene abundant. I have collected the Ct value of the gene target and housekeeping gene. Is it possible to calculate with these data? Is it possible to analyze the data by dividing CT value of housekeeping gene with Ct value of gene target? Thanks a lot for any help!
Hi Arif Luqman,
For relative quantification in your assay, you need to include standards with known concentrations (or a number of RNA-DNA copies). Or, at least one that you dilute on decimal dilutions (1/10, 1/100, 1/1000) to have a linear function. By doing this, you can extrapolate the correspondent concentration for each Ct value.
Hi Aissa Saidi , thanks a lot for your suggestion. The thing is, I do not have the standards to do this. Do you know any other method or formula to analyze or calculate the data? Thanks
Relative quantification is fairly simple and is typically referred to as the ∆∆Ct method.
First you subtract the Ct of your housekeeping gene from the Ct of your gene of interest. You then subtract the difference (∆Ct) of your control condition from your experimental condition to find the ∆∆Ct. You can then convert to linear fold change by using 2-∆∆Ct.
∆Ct = CtGoI - CtHK
∆∆Ct = ∆CtExp - ∆CtCtr
FC = 2-∆∆Ct
If you are interested only quantifying the relative expression of two genes in a single sample, you can use 2-∆Ct.
PCR can be of variable efficiency (between primer pairs, between runs, between machines, etc), so you can't simply say "the Cq value is X, so the target abundance is Y".
Especially since samples with lower [cDNA] will give different Cq values to samples with more.
Comparing to a standard curve of known [template] is the only real way to get absolute transcript numbers via qPCR alone, but since you mention "relative abundance", I'm guessing a ballpark figure would be sufficient?
General rule of thumb is that a Cq of ~35 cycles corresponds to a single starting template molecule, so consequently Cqs of 28-35 represent low to very low transcript abundance. 24-28 is 'modest to low', 18-24 is 'high to modest' and Cqs lower than 18 are high abundance. Ribosomal RNA (which represents the bulk of total RNA) typically gives Cq values of 9-12 (super high abundance).
Dear Arif Luqman,
If you don't possess pre-calibrated standards, and if you have a nanodrop, you can make your own housemade standard by measuring the concentration of any similar purified DNA-RNA sample. You need at least two standard points to draw a straight line. Otherwise, it will be difficult to give concentration values for your obtained CTs.
Hi John Hardy Lockhart ,
Thanks a lot for the suggestion. It was a great help!
Ziba Asadpoordezaki
:Say you have four timepoints, T1, T2, T3, T4 (which we will assume are all at least triplicates).
Calculate the dCt values for each sample (GOI Ct - REF Ct). These are your normalised expression values.
For example,
T1 #1 GOI Cq=22, T1 #1 REF Cq=18: T1 #1 dCt = 4
T1 #2 GOI Cq=21, T1 #2 REF Cq=17.5: T1 #2 dCt = 4.5
T1 #3 GOI Cq=22.5, T1 #3 REF Cq=18: T1 #3 dCt = 4.5
T2 #1 GOI Cq=20, T2 #1 REF Cq=18: T2 #1 dCt = 2
T2 #2 GOI Cq=19, T2 #2 REF Cq=17.5: T2 #2 dCt = 2.5
T2 #3 GOI Cq=20.5, T2 #3 REF Cq=18: T2 #3 dCt = 2.5
You then directly compare these between timepoints, using whatever statistical test you feel is appropriate. dCt values are approximately normally-distributed, so usual stats tests should be fine.
In terms of "what is actually happening", look at the difference between dCt values, and remember that for qPCR, lower values = more expression. Here, for example, the mean dCt for T1 is 4.3, while the mean dCt for T2 is 2.3, so a difference of two cycles. Your GOI is thus ~4 times more abundant at T2 than at T1.
Note this is basically the ddCt, but you don't need to boil all your data down to ddCt values when you can interpret the ddCt empirically and leave the underlying data as dCt: this allows you to directly present different groups on a plot, and show individual datapoints and so on, and is generally a better and more open way to present data.
If you want, you can then flip the sign on all of these (-dCt) such that high dCt values represent higher expression. This usually make graphs more immediately intuitive, and does not affect the data at all.
In terms of actual "how many transcripts do I have", see my answer above. This isn't usually a particularly interesting question, though: it's the difference in expression that you're really trying to investigate, not the absolute numbers.
Hi Ziba Asadpoordezaki
, I used formula 2-∆Ct as stated by John Hardy Lockhart
the first Ct is the housekeeping gene or gene of reference and the other Ct is the gene of interest.
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