You wrote about delta-delta-ct values (ddct). This means:

- in each sample you measured a ct-value of a reference gene and of the target gene

- the expression of the reference gene is not affected by your treatment

- the amplification efficiencies of both genes are equal

- the amplification curves were all ok and

- that the determination of the ct values was ok for all samples

This must all be given, checked and confirmed. Otherwise the whole analysis makes no sense.

Ok, given the things above are all thoroughly checked and ok. Usually the analysis is then performed like this:

1. average the technical replicates.

2. calculate the dct values (difference of ct for the reference gene and for the target gene) - so you have one dct for each biological sample/replicate.

3. you have two groups of dct values (in your case 3 for the controls and 3 for the treated). You can use a t-test to get a p-value.

4. the ddct value is the difference of the average dct values for the groups. Any statistics software comparing two groups will give you this estimate together with its confidence interval. (the ddct is very simple to calculate, but the confidence interval is more difficult to obtain, therefore I suggest to use a software).

Note:

a) the dct values are (log) expression values that are normalized to a "loading contol" (= the reference gene expression).

There should not be any systematic difference between dct values of samples from the similar group. If this is the case, then your assay does not work and your data is not trustworthy. For instance, if you have data from several days or plates, and the dct values for the controls measured on day x are all about 2, and those measured on day y are all about 5, then you have a serious problem with your assay.

b) ct values are proportional to -log(N), where "N" is the initial concentration of the mRNA or cDNA in the sample.

This means that a dct values are actually a log ratios and the ddct value is a log fold-change. So the data and the entire analysis is "on the log scale". If you want to provide fold-changes, then you will need to anti-log (exponentiate) the ddct value. The base is the amplification efficiency (usually 2 or close to 2).

The anti-loged dct values have no particular meaning. There are "relative expression values" but without having a known value of the reference they relate to. So it makes not much sense to take the efford of transforming the dct values.

I strongly suggest to stay completely on the log scale and never ever calulate fold-changes (although I know that some bosses, colleagues and reviewers will insist on getting their beloved fold-changes).

c) the sign of the ddct value tells you if the (normalized) expression in your treated samples was higher or lower than in your controls.

If "+" means higher and "-" lower (or vice versa) depends on how you calculate dct and ddct values. I suggest following, because this is intuitively interpretable:

dct = ct[reference] - ct[target]

ddct = dct[treated]) - dct[control]

This way a higher dct value indicates a higher relative expression, and a positive ddc indicates an induction (and a negative ddc indicates a repression) of the target gene due to the treatment.

Caitlin, did you first calculate the relative expression to a stable 'housekeeper', then use those normalised values for your analyses?  This would be simple as you have a normalised value to work with to stabilise variances due to differnet cDNA concentrations.  First taking the average of your tech reps, as long as they don't deviate by more than 0.5 cycles.  The easiest thing I can think of is to use ANOVA with treatment (control and drug as the levels) with planned contrasts to compare whichever groups you want.  This is deltaCT relative expression. I think perhaps from what you describe your 'control' is what you are using to normalise..

If that's the case then  can then calculate fold changes, so no change = 1 and comparisons are made to this in your drug groups..

Happy to look at the data if you like..

you can only use parametric tests (t-test, anova) for normally distributed data. for non-normal data loot into non-parametric tests.

You wrote about delta-delta-ct values (ddct). This means:

- in each sample you measured a ct-value of a reference gene and of the target gene

- the expression of the reference gene is not affected by your treatment

- the amplification efficiencies of both genes are equal

- the amplification curves were all ok and

- that the determination of the ct values was ok for all samples

This must all be given, checked and confirmed. Otherwise the whole analysis makes no sense.

Ok, given the things above are all thoroughly checked and ok. Usually the analysis is then performed like this:

1. average the technical replicates.

2. calculate the dct values (difference of ct for the reference gene and for the target gene) - so you have one dct for each biological sample/replicate.

3. you have two groups of dct values (in your case 3 for the controls and 3 for the treated). You can use a t-test to get a p-value.

4. the ddct value is the difference of the average dct values for the groups. Any statistics software comparing two groups will give you this estimate together with its confidence interval. (the ddct is very simple to calculate, but the confidence interval is more difficult to obtain, therefore I suggest to use a software).

Note:

a) the dct values are (log) expression values that are normalized to a "loading contol" (= the reference gene expression).

There should not be any systematic difference between dct values of samples from the similar group. If this is the case, then your assay does not work and your data is not trustworthy. For instance, if you have data from several days or plates, and the dct values for the controls measured on day x are all about 2, and those measured on day y are all about 5, then you have a serious problem with your assay.

b) ct values are proportional to -log(N), where "N" is the initial concentration of the mRNA or cDNA in the sample.

This means that a dct values are actually a log ratios and the ddct value is a log fold-change. So the data and the entire analysis is "on the log scale". If you want to provide fold-changes, then you will need to anti-log (exponentiate) the ddct value. The base is the amplification efficiency (usually 2 or close to 2).

The anti-loged dct values have no particular meaning. There are "relative expression values" but without having a known value of the reference they relate to. So it makes not much sense to take the efford of transforming the dct values.

I strongly suggest to stay completely on the log scale and never ever calulate fold-changes (although I know that some bosses, colleagues and reviewers will insist on getting their beloved fold-changes).

c) the sign of the ddct value tells you if the (normalized) expression in your treated samples was higher or lower than in your controls.

If "+" means higher and "-" lower (or vice versa) depends on how you calculate dct and ddct values. I suggest following, because this is intuitively interpretable:

dct = ct[reference] - ct[target]

ddct = dct[treated]) - dct[control]

This way a higher dct value indicates a higher relative expression, and a positive ddc indicates an induction (and a negative ddc indicates a repression) of the target gene due to the treatment.

Thanks for the replies.  Yes I have done all the checked all of the data and assumptions.  And I have normalised everything to a reference gene.  If you should only perform ttests on non transformed data then can you do it on the DDCT rather than CT?  An example of my results is:

DCT for untreated is: 9.26, 8.64, 8.49

DCT for drug treated is: 10.37, 9.43, 9.11

Hence the p-value is not significant at 0.11

However each biological replicate showed a similar drop in expression despite starting at different points. DDCT is: 1.11, 0.78, 0.62 (not converted to fold change).  I'm assuming by having different biological groups there could be slight variation in the relative gene expression.  Doing a t-test on 1.11, 0.78, 0.62 (DCTuntreated-DCTtreated) vs 0,0,0 (DCTtreated-DCTtreated) gives a significant p-value of 0.03.

I geuss the results are showing that if you look at each biological group: 1, 2 and 3 separately you get the same outcome in foldchange but if you group them all together to look at them you cannot see a significant difference.  I would like to know if calculating p-value from the DDCT is valid though.

Do you have "paired data"?

Is the first dct for the "untreated" from the same individual like the first dct for the "treated"?

If so, then your idea is correct and you should do a paired analyis (paired t-test) (this is what you actually did using the differences, only that you compared them to a sample of zero-values; that's not wrong, only slightly cicumstantial; in EXCEL you can use T.TEST on the dct values directly and set the argument "type=1" to say that you wish to perform a paired t-test - it will give you the same result).

However, if this is NOT the case, then a paired analysis is inappropriate because the matching of the pairs you have is arbitrary, and so is the "significant" p-value you get.

If you consider all possible pairings of these values (there are 6) and you perfomr a paired t-test on all these combinations, you get following p-values:

0.03,  0.05, 0.21, 0.25, 0.26, and 0.29.

So most of them are "non-significant" (the first one is the one for the pairing you used).

Great thanks Jochen, yes they are paired.  That makes perfect sense thank you