Dear Marcelo,

Only one of your biological replicates has technical replicates, which isn't enough for duplicateCorrelation() to give a reliable estimate of the technical correlation.

Therefore my suggestion is that you simply average the your first two arrays to make one, then analyze your experiment as a simple replicated design with three arrays. Perhaps better still, examine the MA-plots of the first two arrays, and discard the array which looks to be the noisier. Or run arrayWeights() and discard which of the first two arrays has the lower weight.

Best wishes
Gordon

In reply to:

> Date: Mon, 21 Sep 2009 22:55:29 -0300
> From: Marcelo Laia <marcelolaia at="" gmail.com="">
> Subject: [BioC] limma technical replicates and biological replicates
>       in the  same design
> To: Bioconductor <bioconductor at="" stat.math.ethz.ch="">
>
> Hi,
>
> On the limma usersguide I found the example:
>
> FileName Cy3 Cy5
> File1 wt1 mu1
> File2 wt1 mu1
> File3 wt2 mu2
> File4 wt2 mu2
>
> where two wild-type and two mice from the same mutant strain are
> compared using two arrays for each pair of mice.
>
>> biolrep <- c(1, 1, 2, 2)
>> corfit <- duplicateCorrelation(MA, ndups = 1, block = biolrep)
>> fit <- lmFit(MA, block = biolrep, cor = corfit$consensus)
>> fit <- eBayes(fit)
>> topTable(fit, adjust = "BH")
>
> I have something like this, but, in my case, I have:
>
>
> FileName Cy3 Cy5
> File1 pra1 banha1
> File2 pra1 banha1
> File3 pra2 banha2
> File4 pra3 banha3
>
> banha is my reference and pra is my tester.
>
> Only the two first is technical replicates. The others two are
> biological replicates.
>
> How I could dealing a design for my situation?
>
> Thank you very much.
>
>
> --
> Marcelo Luiz de Laia
> Universidade do Estado de Santa Catarina
> UDESC - www.cav.udesc.br
> Lages - SC - Brazil
> Linux user number 487797