I think this reference give some answer to your problem. It is not absolutely the same type of problem, but it is related. In this paper different methods for analysing experiments with one control and two types of treatment are discussed. They conclude that the best is to use randomized block ANOVA, sometimes called a repeated measures ANOVA

Link to reference:

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2042947/

M. Lew (2007) "Good statistical practice in pharmacology Problem 2", Br J Pharmacol; 152(3): 299–303.

To compare drug treatment with C2 both should be expressed in the same units. These may be standardized C1 units (with C2 is expressed as few hundreds) or folds relative to C1 (with treatment is also expressed as folds relative to C1)

Dear Wolfgang, I think your way of normalizing data is not wholly uncommon, although I can't think of any concrete articles that are using it on the top of my head. That being said, there might be some caveats and conceptual problems to it, most of which you probably know.

First off you are testing multiple groups so you have a multiple testing problem. Ideally you would do an ANOVA first and then compare groups with a post-hoc test, as the ANOVA controls type 1 errors better than multiple tests with a multiple testing correction. Since you are including an obviously different control, your protective effect disappears anyway. So multiple (pairwise tests) are O.K., as long as you do a bonferroni correction (ANOVA does offer better tests though). I'm not sure if you need to test pairwise, actually, as you are normalizing between experiments. Perhaps you could compare a repeated measurements to a regular ANOVA.

If you end up doing an ANOVA, don't include C1, as it has variance/sd 0 which violates some the ANOVAs assumptions.

Secondly if you have used the 2^dCT method to get fold change values, you need to keep in mind that they are not normally but log-normally distributed. It's best to log2 transform your data again for analysis.

And then there's the general problem that the C0 normalization isn't exactly an internal control (like a housekeeping gene would be). I'm not sure how much noise you are introducing because of this.

Personally I would normalize to the average per experiment and then use unpaired testing (ANOVA) as in (1). It seems more robust, and you have a better idea what's going on.

1. E. Willems, L. Leyns, J. Vandesompele, Standardization of real-time PCR gene expression data from independent biological replicates, Anal. Biochem. 379, 127–129 (2008).

To compare treatment with a control , it's bette for you to use Dunett test.

Thank's

Do you mean you have some samples (x11, x21, ... xn1) measured at baseline for their expression level, and then the same samples are (divided and then) subjected to combinations of treatment (x12, x22, ... xn2; x13, x23, ..., xn3)?

For problems like this, my approach is always to start with a quick graphical check using a strip chart (small sample size) / boxplot (larger sample size) to get an idea about the distribution of the values in C1 and C2, and also to see how much overlappinig occurs between their distributions. If there's clear separation, there's no need for an overall test like ANOVA, and one can go straight to standard multiple comparison procedures.

It is absolutely unnessesary to compare the normalized values. Because both groups are normalized to the same control, it doesn't change any result! Directly compare C2 and the treated group (T). A paired test is appropriate only of pairs of measurements from C2 and T share a common source of variance (e.g. tissue was isolated from the same individual; measurements werde done in parallel; ...).

ANOVA is not required because you compare two groups: T versus C2. C1 is nothing to reasonably compare to. It just gives an indication of the Il1b effect (C2 vs. C1). It might be interesting to see if (a) Il1b induces the target gene so much that a further induction is unlikely or (b) the the treatment "restores" the expression of the target gene to values of cells that were not Il1b stimulated. In any case the tested treatment effect is only T vs. C2. A test of T vs. C1 might be used to show that the "restoring effect" of the treatment is *not* perfect. But this is rarely a sensible question.

If you have real-time PCR data, do the tests using the dct values, or the "log relative expression" values.

I agree 100% to Hendrik Gremmels' response. C1 would have no variance, so leave it out. Plus, as Jochen Wilhelm stated, it doesn't add anything if everything is normalized to C1. An ANOVA is the easiest and most valid way for comparing multiple groups when the data is normally distributed.

Kimberly, on the one hand, the given experiment is no comparison of multiple groups. The only sensible comparison to test the hypothesis about the treatment effect is C2 versus T (so two groups, not multiple), on the other hand ANOVA is a work-around from the 1920's for a problem than can be tackled much more directly and elegantly today. Note that ANOVA is not a "multiple *pairwise* group comparison. It is an omnibus test that actually does NOT adress the question if there are (at least) two groups different from each other. This has to be analyzed by subsequent post-hoc tests, and there the trouble starts (which of the 1001 tests to use?). Further, ANOVA+post-hoc tests might control the long-run FWER, but it leads to a clumped structure of false-positive and false-negative results.Further, not seldom ANOVA+posthoc leads to results like a significant ANOVA but no significant pairwise differences, what confuses many people...

Thanks Jochen, but I believe you are wrong about the number of groups. The questions said, " The study groups are treated with drugs in combination with IL-1ß." There are multiple groups (multiple drugs) to compare to the C2. Since the T value in a t-test is mathematically equivalent to the F value in an ANOVA (ie. each can be derived from the other), the ANOVA+post-hoc would be as appropriate for making multiple comparisons as the T-test for making comparisons between 2 groups.

Thanks Kimberly, I missed that - you are right, it was about several different groups.

Regarding the ANOVA it is in fact the case that t² = F. But the interpretation of F is just different to a set of pair-wise comparisons when there is more than one possible pair. And after ANOVA you still need post-hoc test, so why not directly control the FWER or FDR or any particular, best-suited error-rate?

Dear all, many thanks for your helpful comments. Referring to the contributions from Hendrik and Jochen I would like to amend that we normalized our data to controls (C1) since we found a high variation among individual samples from different patients. This normalization improves the comparability and clarity of our results after drug treatments. Our qPCR data are normalized ratios as calculated by the LightCycler 480 software. Kimberly stated correctly that there are multiple groups since we compare at least two drugs in combination with IL-1ß. The non-stimulated control C1 is unnecessary for the statistical evaluation but informative regarding the basal expression of our target genes and the stimulation effect of IL-1ß.

Thank you for the feedback. Obviousely your groups share a common source of variance. Therefore you find that the normalized values "improves the comparability and clarity of our results". As far as I understand, there is nocthing wrong with your analysis. However, I just wanted to point out that the normalization could be done against C2 as well (since they share the *same* source of variance with T and with C1). This is what a paired t-test between T and C2 does internally. This "internal normalization" is a subtraction: instead of comparing two group means, first the pair-wise differences are calculated and the mean difference is compared to zero. If the coparison is not about differences but about ratios, you must perform the t-test on the logarithms.

There is one aspect I recognized now:

you have the information from C1, and this information is not used in a test comparing C2 and T. However, this information can be used assess the (at least a part) of the variation that is common between the groups, and this information is used to normalize both, C2 and T, before doing the tests. To my gut feeling, the tests comparing the normalized values should then be "unpaired", because the common variance is already elinimated and pairing would then unneccesarily reduce the residual degrees of freedom. When the normalization is done by division of "relative expression values", I would recomment using the logarithms for testing. For instance, if, foe some gene and experiment, C1=78, C2=103, T=64 -> C2*=C2/C1=1.33 (132%), T*=T/C1 = 0.82 (82%) then the values to be taken for the test are ln(C2*) = ln(1.33) and ln(T*) = ln(0.82).

Would be interesting what others think about this solution.

Whether to use ANOVA or not in this multi-group comparison is indeed tricky. I named it because it is easiest way (in terms of ease of analysis using canned software functions) to control your FWER as most software packages offer easy multiple testing correction only via the post-hoc tests of an ANOVA. It is indeed an omnibus test, and if you include an untreated control or some other clearly outlying condition then an ANOVA adds nothing above the appropriately corrected pairwise tests.

Jochen may be right (as usual) that since you already normalize within patient (via C1) there's a good chance you don't need to use a paired test (or repeated measurements ANOVA). It will indeed cost you degrees of freedom (or to put it in simple terms: you put in more information [namely patient identities] in your model, which leaves less [residual degrees of freedom] for your model to explain, which means higher/worse p-values).

You can try a repeated measurements ANOVA and see if pairing is significant as a way to test it. Formally it is another test on your data however, which you should - if you're really strict, and the Spanish Inquisition is watching - correct for. Alternatively you can plot the data as lines per patient, if the lines cross a lot then don't test paired/repeated (this is a bit of cheat, but admissible in court).

It's even interesting from a biological standpoint, as you find out if your variability lies more in basal expression of target genes between patients or in their response to IL-1ß.

Just to add another idea here: "Alternatively you can plot the data as lines per patient"

You can plot the data per patient of T and C2 in a 2D scatterplot (one axis for T, other axis for C2, one point per patient) and see if the cloud of points lies (quite) consitently on one side of the bisecting line (45° line). At least for the comparison of 2 groups I find this clearer and easier to read than a diagram with many wildly crossing and non-crossing lines.

I think this reference give some answer to your problem. It is not absolutely the same type of problem, but it is related. In this paper different methods for analysing experiments with one control and two types of treatment are discussed. They conclude that the best is to use randomized block ANOVA, sometimes called a repeated measures ANOVA

Link to reference:

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2042947/

M. Lew (2007) "Good statistical practice in pharmacology Problem 2", Br J Pharmacol; 152(3): 299–303.

Ingrunn, thank you for the link!

Strikingly, already in the abstract is written: "Analysis [...] by a one-way ANOVA yielded P=0.058 and the experimenter concluded that neither drug was effective." - what a beautiful example of the common misinterpretation of "insignificant" results. (I have not witnessed the assassination of JFK ... so I conclude that he is still alive)

Unfortunately, I have not found a citation of the original work; it was likely unpublished. Would be nice to have published examples.

Need more information.