[BioC] LogFC in Limma

Gordon K Smyth smyth at wehi.EDU.AU
Sat Oct 25 03:32:03 CEST 2008

Dear Kevin,

Can I gently say, both to you and to other listers, that is not helpful to 
ask Bioconductor developers to justify why they do whatever they currently 
do without explaining why you expected the behaviour to be otherwise.

The direct answer to your question is that different spots on any 
microarray normalize to different values because it is highly improbable 
they would give the same value.  Measurement error and technical 
varialibity ensures that all values will be slightly different, even if 
they are probing the same gene or probe-sequence.

Were you expecting normalization to summarize the replicate probe values? 
That is not the purpose of normalization.  Normalization is intended to 
preserve information, not hide it.  Summarization would be done by a 
subsequent analysis step, such as lmFit().  Any normalization method which 
set all replicate values equal would prevent downstream software from 
knowing whether or not the replicates were truly consistent.  That would 
be misleading verging on dishonest.

Best wishes

> Date: Thu, 23 Oct 2008 13:25:46 -0700 (PDT)
> From: Hai Lin <kevinvol2002 at yahoo.com>
> Subject: [BioC] LogFC in Limma
> To: bioconductor at stat.math.ethz.ch
> Message-ID: <696005.24645.qm at web32404.mail.mud.yahoo.com>
> Content-Type: text/plain
> Hello lists,
> I am using Bioconductor/Limma package to analyze microRNA arrays.
> I have 7 replicate two-color arrays, 4 probe replicates each array. 
> Follow Limma procedures, I had a list from topTable with LogFC(1) and 
> other columns. In addition, I produced LogFC for each individual arrays 
> from MAlist, the average LogFC(2) were then taken from 7 arrays to 
> compare LogFC(1).
> The LogFC(2) is similar to LogFC(1), for example 3.12 vs 3.2, but 
> they're not exactly the same.
> My question is that why aren't these two columns of value the same? Did 
> I calculate it mistakenly or something behind this during the Limma 
> procedures?
> Could anyone kindly explain it?
> Thanks.
> Kevin

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