[BioC] design in mixed ref and dye-swap experiment

[Naomi Altman]

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