Entering edit mode
Hi Mohamed:
It is quite arbitrary to determine the threshold for detection p
value when filtering out non-expressed probes. But a cutoff of 0.1
will
normally remove 30-40% of all the probes in an experiment from my
experience. This cutoff will remove those probes which have
intensities
not greater than 90% of negative controls.
Hope this helps.
Cheers,
Wei
Mohamed lajnef wrote:
> Dear Wei,
>
> Thanks for your messages!
>
> I found the Detection scores colomns (Detection Pval-which
> estimates the probability of a probe being detected above the
> background level) in my data.
> to detect the probes which do not express, I must remove the
probes
> that have Pvalue>0.05 in the detection scores colomns before
carrying
> out differential expression analysis????????????
>
>
> thank you so much for your help
>
> Cheers
> Mohamed
>
> 2009/3/31 Wei Shi <shi@wehi.edu.au <mailto:shi@wehi.edu.au="">>
>
> Hi Mohamed:
>
> Nonexpressed probes are not those probes which have missing
> values. We do not know probes with missing values are expressing
> or not. You can use "Detection" scores to try to find those
probes
> which do not express in all of your arrays and then filter them
> out before carrying out differential expression analysis to see
if
> you can get some DE genes. You should be able to find
"Detection"
> columns in your expression profiles.
>
>
> Cheers,
> Wei
>
> Mohamed lajnef wrote:
>> Dear Wei & Sean,
>>
>> Sean: result of topTbale
>> > gene1
>> ID logFC AveExpr t P.Value
>> adj.P.Val B
>> 9623 7570474 -0.11659435 4.991279 -3.826220 0.0002074251
>> 0.9999788 0.4950223
>> 43599 5550736 -0.18627341 8.316864 -3.800538 0.0002275799
>> 0.9999788 0.4162686
>> 931 4570630 -0.10900890 4.834441 -3.776569 0.0002480571
>> 0.9999788 0.3431365
>> 20081 5560544 -0.22781231 8.704546 -3.770575 0.0002534446
>> 0.9999788 0.3249047
>> 33970 6280711 -0.08123219 5.013244 -3.521244 0.0006066949
>> 0.9999788 -0.4135212
>> 19419 4010192 -0.09110624 4.903243 -3.516622 0.0006163498
>> 0.9999788 -0.4268336
>> 37533 3290356 0.08945260 4.876439 3.512031 0.0006260841
>> 0.9999788 -0.4400442
>> 10727 1260035 -0.14783999 5.784915 -3.506596 0.0006377940
>> 0.9999788 -0.4556641
>> 9921 2260719 -0.09260158 4.927686 -3.474897 0.0007103085
>> 0.9999788 -0.5463862
>> 17444 6180360 0.08237090 4.932764 3.474807 0.0007105246
>> 0.9999788 -0.5466424
>>
>>
>>
>> Wei: 1) i used beadarray methods( quantile) to normalise the
log2
>> transoformed data
>> 2) i dont have a missing values in my data!
>>
>> Thank you again for help
>>
>>
>>
>>
>>
>>
>> 2009/3/30 Wei Shi <shi@wehi.edu.au <mailto:shi@wehi.edu.au="">>
>>
>> Hi Mohamed:
>>
>> I think you can try two things:
>>
>> (1) Try alternative normalization methods;
>> (2) Remove probes which are not expressing in all your
arrays.
>>
>> Hope this helps.
>>
>> Cheers,
>> Wei
>>
>> Mohamed lajnef wrote:
>>
>> Dear All,
>>
>> I'am using limma package to extract the genes
>> diffrentially expressed by 3
>> treatment,my database includes
>>
>> 48803 genes (rows) and 120 colones (40 replicates per
>> treatment),my program
>> is as follows:
>>
>> donne<-exprs(BSData.quantile)
>>
groups<-as.factor(c(rep("Tem",40),rep("EarlyO",40),rep("LateO",40)))
>> design<-model.matrix(~0+groups)
>> colnames(design)=levels(groups)
>> fit<-lmFit(donne,design)
>> cont.matrix<-makeContrasts(TemvsEarlyO=Tem-
EarlyO,TemvsLateO=Tem-LateO,EarlyOvsLateO=EarlyO-LateO,
>> levels=design)
>> fit2<-contrasts.fit(fit, cont.matrix)
>> ebfit<-eBayes(fit2)
>> gene1<-topTable(ebfit, coef=1)
>> gene2<-topTable(ebfit, coef=2)
>> gene3<-topTable(ebfit, coef=3)
>>
>> results<-decideTests(ebfit) # I have a matrix that
>> contains only 0 because
>>
>> in the ligne 26 results <- new("TestResults",
>> sign(tstat) * (p < p.value))
>> of decideTests program the condition (p<p.value) is="">> always false in my
>> case!!
>> something is wrong in my program, but I dont know
where??
>>
>> Any help would be appreciated.
>>
>> Regards
>>
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>>
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