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confirming equal expression of transcripts in three to? - (Apr/22/2007 )


i have the following questions. as far as measuring differentially expressed genes, it was always really easy to make qPCR experiments, for me, however as i am
know trying to confirm equal expression of certain target genes among different samples it turns out to be really tricky.

i am using the delta delta ct method by comparing to actin message (the same cell type is used in all experiments, so actin or whatever gene should be good for normalization).
so, when looking at the ct values of the replicates they look really good and are all plus minus 0,3 cycles. that is true for actin and for the target genes i am intersted in...

so when i am now calculating

mean 2^(ct target - mean ct actin) for all three samples, i roughly get the same values (plus minus ca 0,4). when i now do the quantification, i will end up
with relative expression levels of e.g. sample 1 = 1, sample 2 =0,7 sample 3 = 1,2

but these values surely are not true, as the error that is introduced by the method itself ( that i would consider 0,3- 0,5 cycles) is larger than the differences i am using
to calculate the ralative expression levels. in other words, i think that my calculations are way too sensitively applied on data that is not that sensitive at all...i hope i have made that clear...

so my questions is now: how to show data like this? maybe a ct target to ct ref ratio?
or is the only way really doing thousands of biological replicates and then use the mean with resulting huge error bars...
i really don't know hoow to approach things like that.....maybe someone here could help me out,

i really appreciate your help,

best wishes and thanks in advance



Hi martin,

the problem you are experiencing has somethign to do with the statistics of proving no differences between groups. Normally, you should fix a minimal significant difference (like, is a difference of 1ct already a signifcant difference?). Than you can calculate the sample sizes, you need to detect a difference smaller than your defined difference. If you compare the deltact (not the deltadeltaCT!!!) by one-way ANOVA, and you don'T find a significant difference, than it is unlikely to miss an existing difference (2. kind mistake - accept the 0-hypothesis, while there is a significant difference between the groups).

Okay, an example - you got two groups. You have got a SD in each group of about 1. Than you can calculate the detectable difference (for example using Graphpad Statmate or gPower). I'll attach you a graph - you can see, that you can detect a difference with a sample size of about 20 (to prove equality you allways need much higher sample sizes than for proving that things are different).

Okay, I hope that helps. If you have more questions, please let me know...



hi kruemel,

thanks a lot for your reply. to confess, i suck with statistics.

the idea is to set a minimal difference that i want to detect, right? but when i
set this minimal difference to a value that appears to be exactly the minimal difference a qPCR can detect, which would be
+- 0,3 ct max (calculated a 2^0.6 = 1.51.. x fold difference), how can i then make statements of target genes being equally expressed.
the one thing i could do is to set everything one ... or is it really applicaple to do so many (biological) replicates that the mean values will eventually
swing into equal values. isn't that a bit of an overkill then?

in the end i am always calculating with values that exceed the accuracy of the method itself, or am i wrong there?
and what should i do if the number of biological replicates is very limited?

and aren't all the statistics worthless if they are based on virtual data (i.e. data that clould never be accurately measured)?

i am really kind of lost now, maybe its just me who is unable to understand it...

again, help and hints are very much appreciated

best wishes martin

PS: would't it be better to just plot the delt delta ct values for each target and for each sample next to each other to most conclusively show equal expression levels?
(because the ct values are the most accurate data that one could get out of an qPCR experiment....)

PPS: would the way i show it in the attachment be a reasonalbe way to show equal expression?


Hi martin,

you definetly have a problem with those virtual values, but still - with a larger number of replicates you can have a very low SD (depends on the formula with which SD is calculated). If we assume that you can measure +/-0.3delta CT as minimal difference, Power depends on the SD you expect...
I would calculate with the deltaCT value. If you can measure a difference of 0.3 and you expect a SD of 0.3 than you need a sample size of about n=14 per group (in a 2 group design again, Power = 0.95).
You can just plot your delta deltaCT values but this gives you just face value ... and this is worth nothing in biology ...
As I said, proving that there is no difference is more difficult than proving a difference ... wacko.gif

PS Okay, I just calculated also for three levels: If you have three groups and a standard deviation of 0.3, you'll need n=9 per group to achieve a power of 95% - i.e. if you find no difference with agroup size and a SD like that, you prove with a certainty of 95% that there is no difference, qPCR could detect.
I made you a plot of Power x SD for different group sizes.