There is a huge confusion around. What is shown in such graphs? If fold-changes are shown, then there is no "control group" to be shown on such graphs, because the values of the controls are already part of the calculated fold-changes (yes: fold-changes *to what*? -> to the controls!)

When "relative expression" is shown (relative = normalized; i.e. relative to the values of some internal reference), then there is an estimate for each group, including the control group.

The best (*sarcasm*) graphs I see are the ones showing bargraphs showing relative expressions or fold-changes, where the bars start at 0. This is completely non-sensical for relative expression, and for fold-changes the reference value is 1, not zero.

But back to the topic: when fold-changes are shown, there is nothing like a control group to show. If such a group is shown (typically with a mean of 1), its standard error is not on the same scale like the standard errors for the other groups, that is, the sizes of the error bars in such figures are not comparable. Since the errors shown for the controls are not errors on the fold-change scale, it is actually wrong to show such error bars in these diagrams. Instead, the horizontal axis should be at y=1.

But these comment are all nitpits. People (authors and reviewers) like it that way, so people do it that way. It's all only very vagely related to good scientific practice.

Hi Sebastian,

I am not sure if I have well undertood your question. I assume that your reference sample is your control group, not your reference gene.

Taking that assumption into account, I do not know what you mean with a "good" method. But, I have also detected graphs without SD in the Control group fold change 1, and not only in qPCR method based results, also semiquentitative results often lack SD. In our lab, the fold change calcualtion is based in a simple transformation of the data: each biological replicate (control biological samples and treated samples) divided by the average of the control group. After that SD of control is calculated as always.

Have I answered your question?

Regards,

There is a huge confusion around. What is shown in such graphs? If fold-changes are shown, then there is no "control group" to be shown on such graphs, because the values of the controls are already part of the calculated fold-changes (yes: fold-changes *to what*? -> to the controls!)

When "relative expression" is shown (relative = normalized; i.e. relative to the values of some internal reference), then there is an estimate for each group, including the control group.

The best (*sarcasm*) graphs I see are the ones showing bargraphs showing relative expressions or fold-changes, where the bars start at 0. This is completely non-sensical for relative expression, and for fold-changes the reference value is 1, not zero.

But back to the topic: when fold-changes are shown, there is nothing like a control group to show. If such a group is shown (typically with a mean of 1), its standard error is not on the same scale like the standard errors for the other groups, that is, the sizes of the error bars in such figures are not comparable. Since the errors shown for the controls are not errors on the fold-change scale, it is actually wrong to show such error bars in these diagrams. Instead, the horizontal axis should be at y=1.

But these comment are all nitpits. People (authors and reviewers) like it that way, so people do it that way. It's all only very vagely related to good scientific practice.

@Jochen

I agree, it's something that has been confounding me too. A control group bar with no error in a chart should not really be necessary. However, I do incorporate the error from the qPCR technical controls in my calculations. This is unique to each sample, control or experimental, and is also part of the 2-ddct calculation. In this case I use a test (e.g. t-test) to compare the experimental (biol rep error + tech rep error) with the control group (tech rep error only), however, I'm not sure this is entirely correct. An alternative I've come across is performing e.g. a one-sample t-test comparing to the hypothetical 1-fold.

Hi,

I would suggest this article for calculating SD. http://dingo.ucsf.edu/twiki/pub/Cores/BioinformaticsCore/QPCR/Livak_Schmittgen2001.pdf

Sebastian,

let's consider you have two groups (treated and control) with some dct values. The dct's come from different biological replicates. (These dct values might already be averages of multiple technical replicates - this will somewhat reduce the technical variance)

The values in both groups are assumed to scatter around a common center, positive and negative deviations of similar sizes are expected with similar probability. These two assumptions legitimate the use of the normal probability model for the deviations. If these assumptions are not severly violated it justifies the interpretation of likelihoods (and the t-test, for instance). The important point is: the values of BOTH groups scatter. So there IS variance in both groups. When several biological replicates were measure in a group, this variance consists of the technical AND biological variance. Hence, there IS biological variance in the control group as well (not as you said that there is "tech rep error only").

Since you do have two samples, a two-sample t-test should be made. A one-sample test against H0: mu=mean of (controls) will not know about the variance in the control group and thus give a wrong result. Here is a numerical example (4 samples/dct values per group):

controls: 7,4,5,10

treated 3,6,3,8

the average difference is 1.5 cycles.

The standard error of the difference (mean treated) - (mean control) is 1.8, on 6 degrees of freedom. The t-statistic for H0:mean(treated)=mean(control) is 0.83 and the p-value is 0.44.

The standard error of (mean treated) is 1.22 on 3 degrees of freedom. The t-statistic for H0: mean(treated)=mu (where mu is the average of the controls; taken as constant) is 1.22 giving a p.value of 0.31.

So the results are not equal. The first is correct, the second is wrong.

Jochem,

Thanks for the clarification. The comparison of the dCts of both groups is the most logical thing to do, but would this comparison stand after the data has been transformed by the 2-ddCt method? To illustrate, a t-test comparing dCt vs dCt does not yield the same result as 2^dCt vs 2^dCt. What would be the best way to transform the data to compare values that are as close to real fold-change as possible?

Could I also ask, considering each biological rep is an uncertain value because of the differences between technical reps, what would be the best way to compare two sets of biological reps while including the variance of each of these individual reps? Or could one make the assumption that the error rate between tech reps is fairly constant and subsequently exclude this error from subsequent comparisons, treating each of these biological reps as true values?

Thanks

No. The comparison of the transformed values ("fold-changes") does not work this way. The t-test assumes normal errors. This assumption is resonable (and usually quite well met) for dct's but clearly not for fold-changes.

Do the test on the dct values and calculate means and confidence intervals for the ddct-values. If you like to show fold-changes instead of dct's, transform the mean and confidence limits.

The attached image shows two example charts with fake data for the regulation of 7 genes (A-G). The points represent means and the error bars indicate the 95% confidence intervals. Note the assymetric intervals for the fold-changes and the compressed representation of down-regulations.

But isn't setting the control to one-fold change arbitrary anyway?

Andor: no, it is not arbitrary. A fold-change of 1 means that nothing changed. If you would compare the controls with the controls, you would find the similar (the same!) expression in both, the controls and the controls. So you can say: the expression in the controls was 1-fold (i.e., the same!) expression than in the controls.

The controls are not "set to 1", the change in controls relative to themselves just *is* one, by definition so to say.

Sure, this is too stupid to be mentioned, but this whole thing came up to be a topic because of the stupid habit to show some kind of "reference group" in plots where the references were already used to calculate the responses (fold-changes) of the other groups.

Ahhh, ok - that's what I thought was going on. Cheers.

I want to show the expression of a gene over time (i.e I do NOT have control and treatment groups, just samples from different time points), would I just pick one of the time points as my reference group and can I show this time point (as 1) in the plot?

You turn your expression values into "expression relative to t1" (for instance, if you take the expression at time point t1 as the "reference". Then there is nothing like an expression at t1 relative to t1. So your reference group has no isolated meaning in this way of representation. It is absolutely the same issue as discussed above in this thread. It does not matter if other groups or other time-points (or whatever) is related to a reference (group or time-point or whatever).

EDIT: The above applies if you want to show *changes* (folg-changes) relative to the reference. If, in contrast, you want to show the expression itself (at different time-points), that in fact you may define(!) the average expression of the reference being some arbitrary value. The respective correction factor must then be applied to all other time-points/groups as well. If you show expressions on the linear scale, then 1 is a meaningful reference value; the values of the other groups will thus staight-away give you the fold-difference to the reference. However, this may be mathematically a little unsound, depending on how the averages were calculated (arithmetric mean vs. geometric mean), and there is no error indication of the differences. If, on the other hand, you show the expression on the log scale, then 0 is a meaningful reference value and the values of the other groups directly show the log fold-changes to the reference.

@Jochen Thanks!

@Sebastian, Jochen. hi guys...question, by "reference samples" you mean signals coming from "normalizing/ housekeeping/endogenous genes" ?

@Elda. I should've called them differently, but the reference samples are the untreated controls.

Hi Elda, above I talk of the "reference" as being a sample, a treautment group, something like that. With "expression" I meant the already normalized expression (the normalization is done to a loading control like some constantly expressed gene).

A ford-change is the ratio of normalized expressions of a "treatment group" and a "control group". Above, this "control group" was called a "reference [group]".

In other posts I usually write about "reference genes". Such genes are used as loading controls to normalize the expression values of the "target genes". I prefer "reference genes" over "housekeeping genes" because a houskeeping gene may be differentially expressed under the experimental conditions and may therefore be NO VALID reference gene (loading control). On the other hand, gene constantly expressed under the experimental conditions is a WELL SUITED reference gene, although it might not be expressed in all celltypes at all times (thus being not a house-keeping gene).

Confusing "house-keeping" with "reference" is a major source of wrong conclusions from qPCR experiments. For instance, I know many examples where the well-known house-keeping gene GAPDH was used to normalize the expression in experiments involving hypoxic conditions. In many cells, GAPDH is itself regulated under such conditions. Therefore, in these experiments GAPDH is NO VALID reference gene and must NOT be used for normalization. I know that in many studies the siutability of the used reference gene(s) is not at all analyzed. The authors just rely on the (often wrong) assumption that the used "house-keeping genes" won't be regulated under their experimental conditions.

The difficulty with graphing fold change is that the average of different replicates is not an arithmetic mean but a geometrical mean and as such the t-test does not apply. You need to convert the raw values to log-values and then you can apply a t-test on these log-values. The t-test is applicable on additive values only. For qPCR, I would assume you don't have n=30 replicates so technically the t-test is not mathematically appropriate anyways and you ought to use a non-parametric test in the first place since you cannot test for equal variances with a small number of replicates in your different test conditions.

Hope this helps.

Jean-Charles

Jean-Charles, the appropriateness of the t-test does not depend on n. Just when the distribution of the values (errors) is "significantly" deviating from the normal model, the central limit theorem ensures for "large" samples that the inference about the mean is not too bad. Here, your n>30 comes into play, although this number will practically depend on the kind and severity of the deviation from the normal distribution (modality, skew, kurtosis, ...).

Yes Jochen, but if you have a n

Dear Sebastian,

Wether it makes sence to include the variation of the reference samples or not depends on you experimental design. If you have a paired design or a repeated meassure so that you can match a specific control expression to each of the treatments you do not need to display the control and can just add a reference expression line at 1-fold.

However, if you have independent control and treatment samples and you use the mean of the expression of your control samples to relate your treatment samples to, you can actually calculate for each of your control samples a relative expression to the mean of the control group. Based on these values, that are idially close to 1 (if variation in the control group is low) you can calculate the standard deviation or standard error or other parameter for the variation on your control (reference) group.

Can somebody explain this question with some example? I also have similar problem and want to process the data in right way. I can provide a simple example if anybody can help me in understanding the calculation.

For Example, I have15 "test samples" and their 15 respective "control samples". For now just consider one set i.e. one "test sample" and one "control sample". I repeated it three times (both test and control) and I have three values for each. For example, for 'test sample" the values are 3,4,5 and for '"control sample" the values are 1,2,2. Now, the 'mean and the standard deviation' for "Test" and "control" are '4 and 1' and '1.67 and 0.58' respectively.

Now, can anybody tell me how to normalize the "test sample" using this "control sample".

@Yogendra I would like clarification also, but your example situation has some problems. The values that you describe for one "test" and one "control" sample are what is referred to here as "technical" replicates--the same sample measured multiple times. The normalization that is discussed here has to do with choosing the measure of one sample as the "yardstick" and then describing the values of all the other samples in terms of that yardstick as a "fold-change". An important issue with your example is that one does not usually have one "control" for each "test", but rather a "control" group including 3 or more samples and then various test groups which also each have at least 3 or more samples, where each of these samples is a biological replicate. So each group would include 3 or more mice or plants or cultures etc. The extent of variability within each such group then informs you whether the variability that you see between any two such groups is likely to be "significant".

Hi David,

Sorry for the confusion, I tried to explain in a very simple way. I am elaborating the condition if you can help me in understanding the calculation. The actual experiment is based on "band intensity in Agarose gel". So, in my case I have 15 samples to check for the relative intensities. Here, I also need to normalize these 15 samples with their 15 respective controls ran in their respective lanes and this is "b-actin". My purpose of this calculation is to tell that I have taken equal quantity for the PCR. I took 1 ul for each PCR and repeated this PCR for 3 times. Therefore, for each sample and control I have 3 replicate data. Now, for example If I have 3 replicate data for Sample A ...that are 3,4,5 and 3 replicate data for its control A...that are 1,2,2. Now, I have to first average the sample A and Control A, Then, I need to calculate the standard deviation in sample A and Control A. After that, i want to normalize the band intensity of averaged sample A using averaged control A.

I think now it will be little clear to you what I want. I can discuss with original data if you can help in this calculation.

Ok, I understand better now. Still I think that what you are describing is just a measurement method--take the average of three measures for your gene of interest and divide by the average of the three corresponding b-actin measures. As with many "statistical" questions, though, I think that there is a more important experimental consideration--which is that b-actin may not be a good choice for your purpose of determining equal template in your PCR--unless you take some specific experimental steps. The rationale is that b-actin mRNA has a very high concentration and the PCR will reach a saturation plateau in relatively few cycles compared to most other genes. Thus you would have to determine the number of cycles where the b-actin PCR is still in its exponential phase of amplification in all of your samples in order to have quantitatively valid data. Since your gene of interest may take a greater number of cycles to produce visible product and also be in it's exponential phase, you would probably have to run separate PCR reactions for different numbers of cycles for b-actin and the gene of interest. This can be done, but finding the best conditions is often a tedious process, and can be avoided if you have access to a qPCR instrument.

Hi David,

I have already done the pre-optimization PCR experiments where I used different PCR cycles to avoid the saturation plateau. After doing these all steps I rationally selected the best cDNA dilution and best cycle numbers. After this pre-optimization steps, I performed the PCR 3 times independently for both (Test and Control) at 2 different cycle numbers. After this, I took the Images using Gel Doc at different exposure levels and during analysis I just discarded the Images having Pixels with over-saturated values. Then, I subtracted the background and finally took out the data for further statistical analysis. After doing this all, I have three different values for my Test and 3 for Control. I took the mean and stdev for both. Now, I can Normalize my Test by dividing the mean value of control but I am not able to understand how to consider the standard deviation in the calculation.

@Yogendra

This looks useful:

http://ipl.physics.harvard.edu/wp-uploads/2013/03/PS3_Error_Propagation_sp13.pdf