What do you *measure* ? Use these data, show these data, test these data.

In fact, Livak and Schmittgen say in their conclusion that "Finally, statistical data should be converted to the linear form by the 2^(-cT) calculation and should not be presented by the raw cT values.". I can not find a justification for this statement. This considerably skews the biological symmetry of up- and down-regulation. This is one reason *not* to apply the power transformation (i.e. to stay on the log-scale). In Livak and Schmittgen I do not see a proper reasoning.

Also, note that the statistics are done on the level of (delta.)cT values. Note that the geometric mean if nothinge else but the antilog of the mean of the logarithms. Seems to sound cool using fancy math to make simple things look complicated...

What do you *measure* ? Use these data, show these data, test these data.

In fact, Livak and Schmittgen say in their conclusion that "Finally, statistical data should be converted to the linear form by the 2^(-cT) calculation and should not be presented by the raw cT values.". I can not find a justification for this statement. This considerably skews the biological symmetry of up- and down-regulation. This is one reason *not* to apply the power transformation (i.e. to stay on the log-scale). In Livak and Schmittgen I do not see a proper reasoning.

Also, note that the statistics are done on the level of (delta.)cT values. Note that the geometric mean if nothinge else but the antilog of the mean of the logarithms. Seems to sound cool using fancy math to make simple things look complicated...

There is nothing like "true error". You measure something and you get data. Now you can think of a model describing some systematic pattern in this data. Since models aim to simplify the reality (or the data), it will happen that the model is never able to predict all your data exactly. The residuals between model prediction and observations are called "errors" (I find "residuals" the better word, though). The average (mean) is already an estimated model parameter (of a quite simple model, but nevertheless a model). Choose any other model and you get other residuals ("errors"). Surely, models (not the data!) can be more or less appropriate, effective, good. In many cases, where not much else is known, a model of a common center with equally likely deviations of the obserations to either side and with the same functional form of the probability distribution of the deviations leads to the mean and the normal probability distribution. When you see that the frequency distribution of the residuals is considerably different to the expectation ("normal"), then you know that this simple model may be suboptimal or even inapropriate. If this is the best *available* model, then it still is the best available model (simple but true). Using additional knowledge tells us that effects of enhancers/repressors and all other parts of gene regulation cascades are multiplicative in nature, we can derive a better, exponential, model, whicht turns to be equivalent to the "normal" model for the log data. Using the exponential model on the 2^cT values is complicated, using the "normal" model on the cT values is simple but equivalent. And it is straight-forward, since those cT values have been directly measured.

I don't think this question is easily answered. But I would point you towards http://www.gene-quantification.de/rest-2009.html and associated publications which answers some of the points you raise

Dear all, thank you for you answers.. I am gonna read the publications associated with REST to better understand the issue. But it seems that applying statistics to deltaCT is better than transformed data. We only have to be careful because deltaCT is inversely proportional to gene expression so a negative t value on a t-test might represent higher expression values of the comparison group.

Does someone still have something to add?

Thanks,

I have :-)

There is no law how to calculate deltaCT. If you calculate it as CT[reference]-CT[target], higher (towards positive) values indicate higher (relative, normalized) expression. As simple as this.

I never understood why some authors (unfortunatly) used this inverse way to calculate deltaCT values, and now thousands of authors follow quite unreflected/unthoughtful, cementing this "rule". Noone ever would have the idea to present a normalization of an analyte A to a loading control C as C/A. It is straight-forward to divide A by C, not the other way around. The deltaCT, calculated as CT[target]-CT[reference] is nothing else but log(C/A). If find this counter-intuitive, misleading and inappropriate.

I think this all happens because biologists or other empirical scientists, especially from medical and life sciences, just quit reading/thinking when a mathmatical formula or a word like "logarithm" appears in a manuscript. So there is too little common sense judgment involved and results are taken uncritically. However, I dont have an idea how to make it better, though.

Try this: 3 data points with delta c(t)s of 10, 11 and 15. Take the average, take the error and it not too large.

Now try first converting these delta c(t)s into 2-deltaCts and take their average.

The variance will be much bigger, and the average will be different.

In my oppinion i prefer ddCP(T) but always make the dCP(T) both statistic perform analized by t-test if the number of samples are resduced

Thank you all for all the answers. I agree with Jochen. In conclusion, data are better fit in a statistical model as it is measured. For biological relevance (and convention) purposes, data should be reported as fold changes.

And what about the calculations of variability measures of the fold change? Should we prefer 95% CI, SD or it doesn't matter?

Diego, just *because* of the biological relevance the the results should be presented on the log-scale. Otherwise, down-regulations of the same factor seem much less relevant than up-regulations of the same factor (it seems plausible to me that a 3-fold down-regulation is biologically as relevant as a 3-fold up-regulation; this is only a plausibility argument, though).

To the variability measure of the fold-changes: SD (or SEM) is usually misinterpreted when the distribution is skewed. CI are ok when ther are properly calculated (they will be asymmetric). Again, the best is to stay on the log-scale and provide mean and SD or SEM or CI for the (delta-)cT-values.

I found myself in a similar dilema. I like to present data in fold change as 2^(delta)(delta)CT, preferably shown in log scale, as Jochen said to give the same importance to down-regulations. I was quite happy with that and I didn't worry about any other statistical analysis. But I think there is a common sense that people won't "belive" in your up- or down-regulations if you don't show p values or *. So, finally, now I'm doing ANOVA with Dunnett's multiple comparison test (because I have a kinetic over days) comparing the 2^(delta)(delta)CT of each day with day 1 (=1 or log 1). Do you think this make any sense?

It is perfectly fine to combine plots and p-values. However, showing just stars for p0.05. However, this is still suboptimal. My favorite solution was to indicate the actual p-values for each comparison that has been done. Very small p-values may be indicated as "

Thanks a lot, Jochen! You're a pro!

To make Jochen's point clearer as to why linear scale is very hard to deal with - see the attachment for how the error bars have to actually be calculated when converting from log to linear scale. I too agree that we need to find an acceptable way to stick with log scale. And make it base 2 (instead of base 10). It is a doubling process we are dealing with here after all. The perfect qPCR standard curve slope in log base 2 is... -1, which is intuitively nice. As Jochen stated (in other words) log scale (Cq) statistical analysis has been met by opposition from non-mediocre minds; the 'jury' is still in flux or out on this issue. That linear scale quantities [efficiency-corrected log-transformed Cq values] reveal the true variability (in normal scale sensibility) of qPCR/RT-qPCR gene expression measurements is what the prevailing winds still espouse. I would hope that the world's foremost statisticians will eventually light the way for log scale proponents.

Hi Jack. could you a bit explain at least "b" in your equations? thanks.

"b" is the y-intercept of the target standard curve; which, for all practical purposes (when dealing with relative quantification), can be eliminated for same-target reactions (using the exact same standard curve) since it is a constant. In the Livak-Schmittgen equation (relative quantity = 10^((Cq-b)/m)) you see "b" there (the y-intercept of relative dilution standard curve) wherein "m" is the slope of the line of log of dilution vs. Cq. Interestingly, an equivalent expression for this relative quantity relationship is:

relative quantity = Eamp^(b-Cq). Here, again, if this pertains to the same target using the same standard curve, "b" can again be eliminated leaving the bare relative quantity expression: rel Q = Eamp^(-Cq). However, if you are dealing with 'known' target copies, then "b" represents the Cq at which 1 copy appers on your target standard curve - and therefore, it is good to leave in the equations. For relative quantification, however it can be effectively eliminated and the graphs scaled to whatever factor serves the puprose of graphing in relative terms.

m = -1/log(Eamp)

Eamp = 10^(-1/m)

Cq = m(log(rel. Q)) + b

So: log(rel. Q.) = (Cq-b)/m

Thus: rel. Q. = 10^((Cq-b)/m)

and, by substitution: rel. Q. = Eamp^(b-Cq)

You could do this all in log base2 as well, wherein the perfect slope would = -1

instead of using the strange yet perfect log base10 slope of -3.3219... which is

(-log base2 of 10) anyway.

I hope that helps?

yes, it is clear now, thanks! still this equation you're referring to pertains more to absolute quantification when you're trying to deduce exact copy number. it's a bit confusing as I was thinking of standard curve as the means of efficiency calculation=). by the way, in that sense my perfect slope as 2 (or -2, depending on how you plot the line) cause i'm using 1:4 dilutions. although software now could extrapolate your 1:4 to 1:10 giving you somewhat near ~3.32 if you're lucky. anyway, leaving out standard curve (for efficiency) all other calculations are usually in log base2.

I can't still figure out statistical analysis for efficiency-corrected Ct(s) or as long as it make biological sense (quality vs quantity) one could simply show one "representative" chart of unlogged Ct(s)?.. I'm wondering what would I do if it were necessary..

ps

Also, I would stick to

(1+E) = 2, when E = 1 (100%)

rather than

E = 2 (again, when E = 100%)

E (efficiency of amplification)

because then you have: E ∈ [0,1] which is make more sense when you substitute with your obtained values.

another question puzzles me is when your E is far beyond 100%.. theoretically not possible. but for the same reason we however accept values for inhibited pcr..

I think we are on different pages a little here; perhaps both of us trying to say the same thing yet from different viewpoints...

I represented both the relative quantification idea, and the copy # idea as exclusive from one another. You can really only pursue absolute quantification (known copy numbers) if you know "b" explicitly as the Cq at which 1 copy of your target crosses threshold. Other than that, plotting the log of relative dilutions of sample vs the resultant Cq values is a relative assessment.

Absolute quantification is kind of a misnomer in qPCR... can it really be attained? Rarely. It can be relatively absolute - and sometimes it can be very near the mark. Droplet digital PCR has the actual means to accomplish absolute quantification.

For (a standard curve) made of serial 1:4 dilutions at 100% efficiency you can calculate the perfect Cq differences by log(base 2) of 4, which = 2. But, this "2" or "-2" is not your slope, rather, it is the distance you would expect between Cq values for perfect 1:4 serial dilutions when amplification is 100% efficient (E = 1 (or 100%) and Eamp = 2 (perfect exponential doubling of amplicon per cycle)). The slope here is then (-1/log(base10) of 2) which = -3.3219

E.g.:

If your 1:4 dilutions amplified at 80% efficiency, then -1/log(base10) of 1.8 would be your slope (-3.917382327), and your expected differences between Cq values along your 1:4 dilution series at 80% efficiency would be log(base1.8) of 4: 2.35849917 or -2.35849917 (instead of a delta Cq of 2 or -2 at perfect efficiency).

Efficiency = [10^(-1/slope)]-1 and Exponential amplification (Eamp) = 10^(-1/slope)

So, when Efficiency (E) = 0.8 (or 80%), Eamp = (0.8 + 1) which = 1.8 (meaning 1.8 'doublings' per cycle instead of a perfect "2").

Efficiency = "E", and Exponential amplification = "Eamp." Efficiency (E) goes from 0 to 1 (in classic math) while Eamp goes from 1 to 2 (for qPCR since it is a doubling process, and Eamp gives you an estimate of how close to doubling of amplicon is effectively being achieved with each cycle)...

The implied equation is always Ncq = Xo(1+E)^dCq where (1+E) = Eamp

all stemming from the original perfect equation: Xn = Xo(2)^n

Perhaps this helps - perhaps not... There's always more than 1 way to skin the same cat.

Hi! thanks for your modified message=) and perhaps we're truly trying to say the same.

still, I would like to say that, yes, 2 is the distance between perfectly performed 1:4 serial dilutions (as well as 3.3219.. is the perfect distance between 1:10 dilutions). so, my slope is:

(-)(1/log(at base 4) of (1+E)) which is 2 when E (efficiency) = 1. please, correct me if i'm wrong.

any ideas about efficiency corrected Cts? i mean stat analysis?

Hi Alexey,

In log base 10:

slope = -1/log(1+E) or, the way I like to express it: slope = -1/log(Eamp)

When E = 1, slope = -3.321928095...(the perfect slope) since slope

would = -1/log(1+1) which = -1/log(2), which = -3.321928095...

In Excel, one would enter the formula: =-1/LOG(2)

When E = 0.8, slope = -3.917382327... since slope would

then = -1/log(1 + 0.8) which = -1/log(1.8) = -3.917382327...

In Excel, one would enter the formula: =-1/LOG(1.8)

(etc.)

No matter what the repeated dilution series for the standard curve is (1:7, 1:4, 1:2, 1:10, 1:20, 1:5, 1:3.24)... the perfect slope (in log base 10 vernacular) is always (-3.3219...), which is the number: -log(base2) of 10 or -1/log(base10) of 2. Also, it is understood that the y-axis of the standard curve is in (Cq) units and the x-axis is in (log of relative dilution factor) units when graphed.

As well, the (imperfect) slope of any dilution series demonstrating 80% efficient amplifications would always = -3.917...

You can find the expected (theoretic) distances between adjacent Cq ("Ct") values of a 1:4 dilution (standard) curve occuring at 80% amplification efficiency by the formula: expected Cq differences = log(base 1.8) of 4.

In Excel, one would enter: =LOG(4,1.8)

Coincidentally, yes, the distance between Cq ("Ct") values for a perfect 1:10 dilution series is also the value "3.3219..." but this only happens for 1:10 dilutions - no others. But, the perfect slope mentioned at the top here also has a negative sign (-) preceding it: "-3.3219..."

Efficiency correction happens by virtue of taking slope (and thus efficiency) into account. Assuming all reactions are 100% efficient [where E = 1, and Eamp or "(E+1)" = 2] is not true in many (most?) cases...

Any Cq ("Ct") value can be corrected to its 100% efficiency incarnation by the relationship: observedCq x log(base 2) of (1+E) = Cq corrected to 100 efficiency.

Example:

Let's consider an 80% efficient reaction where a Cq ("Ct") of 24.2 has been observed. To solve this in Excel, one would enter:

=24.2*LOG(1.8,2)

Where 24.2 is the observed Cq ("Ct") value and 1.8 is (1+E) when E = 0.8 (80%). E (efficiency) would have been determined from the serial standard curve's slope by the equation: E = [10^(-1/slope)]-1, which = 0.8 in this case. Similarly, Eamp[or "(1+E)"] would have been determined by the equation: (1+E) = 10^(-1/slope), which = 1.8 in this case.

So, in the example, a Cq ("Ct") value of 24.2 at 80% efficiency would have theoretically appeared at a Cq ("Ct") value of 20.52153 at 100% efficiency.

The quantities identified as (1+E), which I like to call "Eamp", are the very "efficiency" values used in the corrective Pfaffl equation (which is easy to find on the www: "2001 Pfaffl New mathematical model for real-time qPCR..." The important thing to realize is that (1+E) does not always = 2. If it always did, then the Livak "2 to the -delta delta Cq" equation would work for everything.

BioGazelle software by J. Vandesompele and J. Helleman's is the best for Cq stats.

To be clear on terms:

E = "Efficiency"

"E" does not equal "(1+E)"

The "E" here, is the "E" in (1+"E").

And, in qPCR/RT-qPCR, this "E" = [10^(-1/slope)]-1

On the other hand, the term "(1+E)" or what I call "EAMP" = 10^(-1/slope)

EAMP = "Exponetial amplification"

"EAMP" is always 1 greater than "E" since, by definition, EAMP = (1+E).

"(1+E)" as a quantity, should be named "EAMP" (it tells one how near to doubling of amplicon has been attained each cycle). Its value is "2" when the doubling is perfect. Its value is 1.75 when the reaction is said to be 75% efficient (E). EAMP = 1.9322 when the reaction is 93.22% efficient (E).

"E" and "EAMP" are defined by (slightly-different) equations:

EAMP = 10^(-1/slope)

E = [10^(-1/slope)]-1

So, when E = 0.837, EAMP = 1.837 since EAMP = (1+E)

Meaning that: instead of a doubling of amplicons each cycle, an 'effective' 1.837 amplicons are resulting instead. It is saying that the process was not 'efficient' enough to attain a perfect "2". It was only 83.7% efficient, therefore, effectively, only churned out 1.837n times the initial target numbers present per cycle; a log base 1.837 system.

I have witnessed people trying to use the "E" value in qPCR quantification equations that actually require the "EAMP" (the (1+E)) value. If one uses "0.837" instead of "1.837," hugely incorrect things can happen. This exact same issue was just posted today on the BioTechniques Forum as well - and I answered it the same way. People are still wondering about the definitions of "E" and "(1+E)" a lot.

It's an important basic item that too often gets missed by many... and then gets used incorrectly as "tradition"... published... propagating myth. Or, more importantly, incorrect results are then reported (or not reported) that might actually have had some biological meaning.

One more point, and then I will remain silent for a month or 2 on this topic:

When you look at the expression "(1+E)" ... what do you see?

Where does the "1" come from anyway - ... in ("1"+E)...?

The "1" is merely stating that 100% of a target is present. Whatever that unknown amount is, 100% of it is there (unless it's 0 of course). So the next question is: what is "E" doing here?

Since this is a doubling process, the "E" is there to tell you how well the "doubling" of the 100% of target/amplicon process is going per cycle...

So an EAMP value [a ("1+E)"] value of "1.8" tells us that, effectively, the doubling process (of the 100% of target present) is only 80% efficient, and achieved an effective 1.8 (log) base instead of a perfect 2 (log) base of the accrual of amplicon per cycles (n). In other words, # of targets present x 1.8n is the amplification that effectively occurred instead of # of targets present x 2n, which would be a perfect "100% efficient"-doubling process: 1, 2, 4 ,8, 16, 32. But, when the amplification is only 80% efficient, EAMP = 1.8,  the process becomes a 'perfect' log(base 1.8) process:

1, 1.8, 3.24, 5.832, 10.4976, 18.89568 ... and so on... effectively.

Here's a small zinger to consider:

At what EAMP ["(1+E)"] would you expect a difference of 2 cycles between perfect 1:2 serial dilutions? Normally, at 100% efficiency, we would expect a 1 cycle difference (when conditions for doubling are perfect).

Answer: interestingly (but maybe not too interestingly, given that we are dealing with a process based on 2's anyway), the answer is the square root of 2 or "1.414214..." ... meaning that, at 41.4214% efficiency (E), the qPCR would have an EAMP or "(1+E)" value of 1.414214... and the appearance of Cq values are now twice as far apart as one would expect at 100% efficiency (when EAMP = 2), or if you prefer, where (1+E) = 2 and E = 1.

The square root of 2 is a terrible "Eamp" and a terrible implied "E," granted, but nonetheless, for the sake of example - it is interesting what this relationship is.

Just food for thought as to how this thing plays out - and how the math "feels". The way things are (qPCR math chimeric with log base 10), the numbers seem a bit counterintuitive... and perhaps understanding the nature (or "feel") of the entire process might have seemed more 'intuitive' all along if it (qPCR/RT-qPCR) were originally 'advertised' or expressed in log base 2 scale (vernacular) instead; wherein the perfect standard curve slope would have been "-1" instead of the 'well-known' (curmudgeonly?) "-3.3219..." as perhaps resulted from an inadvertent superimposition of log base 10 math over a natural (albeit 'human-engineered') doubling process - a process that was clearly [genetically] rife with the impetus to be a "log base 2 baby" from the start.

Hi,

I've corrected a bit my previous post so that people won't be confused with terminology in this thread.

Thanks!

..and keep posting!

; )

I have few questions would like to seek for your opinion.

The first question is, what is the recommendation data to use for the statistical analysis that obtain from qPCR, the relative expresssion (RQ) value or the Log2 RQ?

The second question is, since the RQ or log RQ are both in numerical data, how do we going to perform the kaplan Meier survival analysis? Are we going to use the mean or median as a cut off point to convert them into the categorize data?

Looking forward for your kind feedback.

Dear All,

I have problem, where, i have 41 cancer tissues and 17 normal adjacent tissues, for mRNA expression. Since, my sample is not paired, it is suitable for me to use delta delta ct method for analyse my data

would you please introduce article for

statistical analysis of qPCR?

please help me to learn

http://www.gene-quantification.de/miqe.html#ibook

Here is in depth information for everybody who has something to do with qPCR.

Good question