I agree with Aliaksei.

If you have two-factorial experiment, than a 2-way ANOVA is the best choice to analyse the data, and the interaction may be relevant (and there is no better tool than the ANOVA to estimate and test interactions).

To use ANOVA, make show that your data represents logarithms of (relative or normalized) expression values.

If you really have differeing efficiencies between the genes, then you have a problem with your assay (including the whole way, including the analysis and processing of the signals to get the ct values). I personally always doubt the reliability of conclusions drawn from rt-PCR data when the efficiencies are obviousely sub-optimal.

Now given that REST will give you *biologically* correct expression values, then calculate the relative (= normalized to the reference gene) expressions (I don't know if REST provides them directly, so it might be that you do not need to calculate anything here), and then use the logarithms of these relative expression values in the ANOVA.

Given there is no evidence for differing efficiencies, using the delta-ct values directly is much easier, and possibly "less wrong". Run the ANOVA directly on the delta-ct values. The delta ct-values are "log relative expression values". I suggest to calculate delta ct as ct[reference gene] - ct[target gene], so that higher delta-ct values indicate a higher relative target gene expression!

The coefficients of the linear model will be differences between log relative expressions, what are actually log ratios or log fold-changes. The interaction coefficient is a difference of such differences, that is a log ratio of ratios of relative expressions.

Hi. I'm not familiar with the output of the REST 2009. But the easiest way to do it would be to run ANOVA on your deltaCt values (average your technical replicates and subtract house keeping gene Ct from target gene Ct). This will give you a deltaCt value for each of your samples. This does tend to under-estimate the variance and assumes similar efficiencies for your house keeping and target genes. But it will give you an idea.

If you wanted to do it extra dilligently, you could calculate expression levels for your house keeping and target genes based on the respective efficiencies. Then you can calculate target gene expression relative to the house keeping gene expression. This will give you a reference normalised expression value for each of your samples, which you can use for ANOVA.

Hope it helps.

I agree with Aliaksei.

If you have two-factorial experiment, than a 2-way ANOVA is the best choice to analyse the data, and the interaction may be relevant (and there is no better tool than the ANOVA to estimate and test interactions).

To use ANOVA, make show that your data represents logarithms of (relative or normalized) expression values.

If you really have differeing efficiencies between the genes, then you have a problem with your assay (including the whole way, including the analysis and processing of the signals to get the ct values). I personally always doubt the reliability of conclusions drawn from rt-PCR data when the efficiencies are obviousely sub-optimal.

Now given that REST will give you *biologically* correct expression values, then calculate the relative (= normalized to the reference gene) expressions (I don't know if REST provides them directly, so it might be that you do not need to calculate anything here), and then use the logarithms of these relative expression values in the ANOVA.

Given there is no evidence for differing efficiencies, using the delta-ct values directly is much easier, and possibly "less wrong". Run the ANOVA directly on the delta-ct values. The delta ct-values are "log relative expression values". I suggest to calculate delta ct as ct[reference gene] - ct[target gene], so that higher delta-ct values indicate a higher relative target gene expression!

The coefficients of the linear model will be differences between log relative expressions, what are actually log ratios or log fold-changes. The interaction coefficient is a difference of such differences, that is a log ratio of ratios of relative expressions.

Thank you very much for your replies. While I understand doubting the results of PCR systems with sub-optimal efficiency, the efficiency of all my systems is above 95%, so I fell fairly comfortable using them. 

Hi Ori

Stop by my office (3357N) if you'd like. I'll be around until 130pm today.

Jordan

Hi Ori, I guess qPCR data always cause a plethora of discussions and critics, whichever way you decide to treat them...! Anyway, I follow some principles that I'll try to explain... then up to you!!

First I never mix technical replicates with biological replicates, although some people do. But if you are accurate, the variability of your technical replicates can be extremely low, so if you mix them with the biological replicates you artificially increase the n, which will affect all your statistical calculations increasing significance, but it will not be a biological significance any more. So I use technical replicates to calculate a mean, discard the technical variability (which is probably 10% of the biological if you work well), and use the "technical mean" as the biological n. This means that you should end up with 3 biological values for each treatment and time point (and not 9 values).

Secondly, I prefer the second approach mentioned by Aliaksei. But be aware that it is not a general consensus, some people like to work on Cts and do statistics directly on them. I don't for several reasons, maybe because I'm extra diligent, as Aliaksei put it! But it is true that every PCR reaction has it's own efficiency (you can calculate the efficiency of the reaction in each well, there are interpolation methods to do that, since every RNA prep and cDNA is different, but otherwise you can make a standard curve for your genes of interest and calibration genes by diluting your cDNAs in a serial dilution and from the slope of the curve calculate the efficiency, most qPCR softwares do it for you, but the formula is not difficult, you find it easily on the internet), and sometimes efficiencies can differ substantially for the different probes. Also, the assumption for parametric statistics, such as ANOVA) is that data are normally distributed, and experience shows that biological data are indeed quite normally distributed (although I agree that gene expression regulation is not linear, but it still seems that the limits for proper functioning of life are symmetrically distributed respect to an ideal middle). But the Cts are exponents, and the power function is not linear, which means that if your cDNA amounts are normally distributed, the Cts will be skewed. Probably it is a very small effect and could be neglected... but, as I said, I like to be exact. Therefore I follow Aliaksei's second approach, I calculate the arbitrary expression level (for this I use the efficiency of the reaction as the base of the power function and the negative Ct as the exponential: this function gives you in practice the ratio of the initial amount and the final amount -at the Ct cycle- of the amplicon and will respect the quantitative relationship between the initial amounts; it's just for convenience you can also use positive esponential, but remember that you are inverting the quantitative relation of your samples), average the technical replicates, and calibrate with the calibration gene(s). In a situation like yours I also like to do a relative quantification towards a no treatment and time 0 point (which from the description of your experiment you don't seem to have included) and work with fold differences, which normalize the three treatments and two time points. At this point you have a matrix of three x two fold changes values where each element is actually made of three biological replicates that you can feed in the two-way ANOVA analysis and check all your within treatment and between treatment interactions.

I would also say that the time course should be considered a repetitive measure, since the two time points are not really independent (which is another prerequisite for ANOVA, besides uniform variance of the samples...). Theoretically also treatments, if performed on the same cell preparation, are considered a repetitive measure, but this is disputable. Moreover, the ANOVA analysis does not consider trends, although measures can be set as repetitive, they are not ordered. With two points only it is difficult or not very reliable to calculate trends, but if the different treatments alter the time response trends, a better analysis would be to calculate the trends for each sample and then analyse statistically the parameters (slope for linear trends or different parameters if you have more complex trends).

Generally, a two-way ANOVA does three comparisons (Factor 1, Factor 2 and Factor 1 X Factor 2 interaction) which leads to a lot of degrees of freedom and therefore you need a large sample size. Like you experience, even large differences can have very small statistical effect. What to do now? (a) enlarge the sample size, or (cool.gif ask the right question (which factor is important?) and test only for this... but a is defenitely preferable! And another short remark - if you use a measure like time as factor, than you'll have a lot of comparisons! Try to build a general linear model with animal strain as fixed factor and time as covariate, this will suit your data better...