Dear Katerina, Well, you're starting your microarray experience with a microarray design which is quite subtle. The design and contrast that you give is sensible (it's recommended in the limma User's Guide for this sort of design), but you need to understand what it's testing. You're testing for genes which are differentially expressed between these three D cells vs these three N cells, relative to technical variation. This approach also allows you to compare the three biological replicates if you want. However, you might be wanting to find genes which are statistically significant relative to biological variation, and this is harder. In principle, the three biological replicates can be treated as blocks, but limma isn't smart enough to handle the dye-swaps and the blocking at the same time. With your experiment, you could do it like this. First create a vector indicating your dye-swap pattern: dyeswap <- c(1,1,-1,-1,1,1,-1,-1,1,1,-1,-1) Then unswap the M-values (I don't usually recommend this): MA2 <- MA MA2$M <- t(t(MA$M) * dyeswap) Your design matrix is now very simple with all M-values lined up: design <- cbind(Dye=dyeswap,DvsN=1) Note I've included probe-specific dye-effects here, which you may as well. Then estimate the correlation within biological replicates: biolrep <- c(1,1,1,1,2,2,2,2,3,3,3,3) dupfit <- duplicateCorrelation(MA2,design,block=biolrep) dupfit$consensus Check the correlation is positive. Then fit <- lmFit(MA2,design,block=biolrep,correlation=dupfit$consensus) fit <- eBayes(fit) topTable(fit,coef=2) All the best Gordon > Date: Mon, 11 Aug 2008 17:06:02 +0200 > From: Kate?ina Kepkov? <kepkova at="" iapg.cas.cz=""> > Subject: [BioC] Limma-design matrix for technical replication > To: <bioconductor at="" stat.math.ethz.ch=""> > > Dear all, > As a complete newbie to microarrays, I am trying to analyze experiment with > following design: Two samples (differentiated versus undifferentiated cells) > were compared directly on two-color oligo array, with 3 biological > replicates (different cell sources) and 4 technical replicates (arrays) per > biological replicate (12 arrays altogether). In every set of technical > replicates two arrays are dye-swap. I am not sure how to handle the > technical and biological replication when trying to fit linear model. We are > interested just in overall comparison differentiated versus undifferentiated > cells. > I have arrived to following setup: > Targets file is: > SlideNumber FileName Cy3 Cy5 > 1 1.gpr N1 D1 > 2 2.gpr N1 D1 > 3 3.gpr D1 N1 > 4 4.gpr D1 N1 > 5 5.gpr N2 D2 > 6 6.gpr N2 D2 > 7 7.gpr D2 N2 > 8 8.gpr D2 N2 > 9 9.gpr N3 D3 > 10 10.gpr N3 D3 > 11 11.gpr D3 N3 > 12 12.gpr D3 N3 > > Where N means undifferentiated and D differentiated cells and 1-3 are > biological replicates. > > Is the following design correct one? Or is there a better way to obtain > relevant information? > Is this extensible for more/less biological replicates? > > design <- cbind(D1vsN1 = c(1,1,-1,-1,0,0,0,0,0,0,0,0), D2vsN2 = > c(0,0,0,0,1,1,-1,-1,0,0,0,0), D3vsN3 = c(0,0,0,0,0,0,0,0,1,1,-1,-1)) > fit <- lmFit(MA, design) > cont.matrix <- makeContrasts(DvsN = (D1vsN1 + D2vsN2 + D3vsN3)/3, levels = > design) > fit2 <- contrasts.fit(fit, cont.matrix) > fit2 <- eBayes(fit2) > > > Sorry if I am asking something obvious and thank you in advance for your > help. > > Best regards, > Katerina > > --------------------------------------------------------------------- > Katerina Kepkova > Laboratory of developmental biology > Department of Reproductive and Developmental Biology > Institute of Animal Physiology and Genetics of the AS CR, v.v.i. > Rumburska 89, Libechov 277 21 > Czech Republic > tel: +420 315 639 534 > fax: +420 315 639 510 > e-mail: kepkova at iapg.cas.cz