DESeq2 with multiple treatments but no replicates
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2
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jpc41 ▴ 20
@jpc41-21560
Last seen 5.3 years ago

Hi everyone,

I've been trying to use DESeq2 to look at expression differences across different transcripts, with 8 (unreplicated) samples across 8 treatments.

My count data looks as follows:

> head(countdata)
                   cts46 cts47 cts48 cts49 cts50 cts51 cts52 cts53
ENST00000000233.9      4     0     1     1     4     3     2     4
ENST00000000412.7      0     0     0     0     0     1     0     0
ENST00000000442.10     0     0     0     0     0     0     0     0
ENST00000001008.5      1     1     2     2    67    72    31    43
ENST00000001146.6      0     0     0     0     0     0     0     0

And my column data as follows:

> coldata
        Treatment
cts46        Untr
cts47         Ifn
cts48         Vir
cts49     Ifn_Vir
cts50    Ltm_Untr
cts51     Ltm_Ifn
cts52     Ltm_Vir
cts53 Ltm_Ifn_Vir

I tried:

ddsMat <- DESeqDataSetFromMatrix(countData=countdata, colData=coldata, design=~Treatment)
ddsMat <- DESeq(ddsMat)

But I got the following error:

Error in checkForExperimentalReplicates(object, modelMatrix) : 

  The design matrix has the same number of samples and coefficients to fit,
  so estimation of dispersion is not possible. Treating samples
  as replicates was deprecated in v1.20 and no longer supported since v1.22.

I can't figure out what the issue is, and would really appreciate any help! I'd just like to investigate differential expression across my treatment conditions. Thanks so much.

deseq2 rna-seq • 1.4k views
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I also tried setting minReplicatesForReplace=Inf as suggested in one part of the manual, but the error message remains the same.

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@mikelove
Last seen 5 days ago
United States

Simply put, DESeq2 doesn't support inference without replicates. The previous version (DESeq) had an option to give results in such a case by estimating dispersion across samples as if they were replicates, and then this was carried over into DESeq2, but later removed as a feature. I'm not convinced there is value in the results from such an analysis, and I think it can give misleading results.

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Ok, I suspected this was the case. Thanks for your help.

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