[BioC] design in mixed ref and dye-swap experiment
Naomi Altman
naomi at stat.psu.edu
Sun Mar 27 05:39:10 CEST 2005
Dear Silvano,
As I have indicated elsewhere on this list, the "p-values" reported by
TopTable are actually "q-values". Hence, if you have fewer "significant"
genes than expected by chance under the null hypothesis, the reported
p-value is 1.0.
e.g. Suppose you have 1000 genes. Then if the number of genes significant
at alpha% is less than 1000*alpha for each alpha, your TopTable p-value
will be 1.0 (i.e. all of the significant genes are estimated to be false
positives).
Your experiment design is needlessly complex and also wasteful. If you
have only 2 conditions, you should do one of the following:
hybridize both conditions to every array (in dye-swap pairs) with no
technical replicates (This is most efficient)
use a reference design with the reference sample always in the same
channel. (This is simplest, but has 1/2 the efficiency.)
Mixing these 2 designs, especially with a mix of biological and technical
replicates needlessly complicates your analysis. It also requires a mixed
model ANOVA to take into account the different levels of replication.
--Naomi
At 10:28 AM 3/25/2005, Silvano Piazza wrote:
>Hello to everyones,
>the experiments that I have to consider is very simple:
>
>I want to find significant genes between 2 conditions A and B, but I have
>only few experiment so I have to collect both ref versus conditions (A or
>B) either dye swap experiment (A versus B and B versus A)
>
>so targets is
>SlideNumber Cy3 Cy5
>array1 ref A
>array2 ref B
>array3 ref B
>array4 ref B
>array5 A B
>array6 B A
>
>of course array5 and array6 are the dye-swap.
>
>So to design the procedure, I follow the LIMMA user guide (by Gordon
>Smith), Chapter 14.5 Weaver Mutant Data.
>
>so
> >design <- modelMatrix(targets, ref = "ref")
> Found unique target names:
> B A ref
> >design
> A B
> array1 0 1
> array2 1 0
> array3 1 0
> array4 1 0
> array5 -1 1
> array6 1 -1
> >fit <- lmFit(MA,design)
> >cont.matrix <- makeContrasts(A.B=A-B,levels=design,weight=MA$weights)
> >fit2 <- contrasts.fit(fit, cont.matrix)
> > fit2 <- eBayes(fit2)
> >topTable(fit2,adjust.method="fdr")
> ....omissis...
> M A t
> P.Value B
> 209 3.801460 6.538782 8.315672 1.0000000
> -4.209468
> 2328 1.184194 7.343676 6.717978 1.0000000 -4.228492
> 7877 1.904360 6.504330 6.114349 1.0000000 -4.239110
> 27187 -4.0759493.771499 -5.783558 1.0000000 -4.246099
> 3709 3.434542 3.467492 5.639159 1.0000000 -4.249459
> 7561 2.002753 5.159913 5.616194 1.0000000 -4.250013
> 7130 2.580527 3.863867 5.600047 1.0000000 -4.250405
> 19983 -2.1176246.836539 -5.567882 1.0000000 -4.251194
>So all genes have P.Value equal to 1!!!!!!
>in previous posts I read that this happen when you have to consider
>multivariate test, which i don't known how to manage..., but anyway
>
>1) Am I doing something wrong in the design?
>2) Am I doing something wrong in the subsequent evaluation steps?
>Any ideas
>
>
>
>Thank you to all
>
>Silvano
>
>
>
>
>
>
>
>
>
>
>
>Dr.Silvano Piazza
>LNCIB,
>Area Science Park,
>Padriciano 99
>Trieste, ITALY
>Tel. +39040398992
>Fax +39040398990
>
>_______________________________________________
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>Bioconductor at stat.math.ethz.ch
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Naomi S. Altman 814-865-3791 (voice)
Associate Professor
Bioinformatics Consulting Center
Dept. of Statistics 814-863-7114 (fax)
Penn State University 814-865-1348 (Statistics)
University Park, PA 16802-2111
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