Heatmap of samples without replicates
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Entering edit mode
19 months ago

I have performed the Differential gene expression analysis based on 2 samples (one sample is controlled and the other is treated) using Noiseq. We do not have biological replicates (although its good to have). The output of Noiseq looks like:

Genes   Mock_mean   Treated_mean    M   D   prob    ranking
ENSGALG00000030907  0.091327389678701   834.11880142495 -13.1569176585108   834.027474035271    0.999997697497179   -834.13124382673
ENSGALG00000027716  4.63590810551782    3033.93990829454    -9.35412482674646   3029.30400018903    0.999997697497179   -3029.31844235836
ENSGALG00000046192  4.06351010806234    3218.73882138174    -9.62955331840357   3214.67531127368    0.999997697497179   -3214.68973389496

Column M reffers to the log2-ratio of the two conditions whereas column D is the value of the difference between conditions. Now I am interested to generate the heatmap. In Deseq2 we identify the top variable genes :

topVarGenes <- head(order(rowVars(assay(rld)), decreasing=TRUE ), 50 )

And then plot the heatmap:

heatmap.2( assay(rld)[ topVarGenes, ], scale="none",
       trace="none", dendrogram="column",
       col = colorRampPalette((brewer.pal(9, "GnBu")) )(100),
       ColSideColors = c( Control="gray", DPN="darkgreen", OHT="orange" )[
         colData(rld)$condition ] )

I have tried to extract the to 20 genes based on D:

select <- order(mynoiseq_tmm$D,decreasing=T)[1:20]

But I am not sure how to plot the heatmap in case of Noiseq. Any help will be highly appreciated.

Heatmap • 669 views
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Entering edit mode

But I am not sure how to plot the heatmap in case of Noiseq.

If i understand correctly, you are asking what value from your NOIseq results to use in your heatmap, is that correct? In the case of your heatmap.2 example, you use the rlog transformed values when using DESeq2, but you don't have comparable values from NOIseq? My initial thought is to use the Mock_mean and Treated_mean from your results.

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