Question: Statistical testing for differential expression in DESeq
gravatar for sumithrasank75
4.3 years ago by
United States
sumithrasank75130 wrote:

I am a bit confused about the test DESeq uses to call differentially expressed genes. I realize that the read count distribution is a negative binomial in the reference and test conditions, but it is unclear what test is done between the two distributions to call differentially expressed genes.

rna-seq next-gen • 1.9k views
ADD COMMENTlink modified 4.3 years ago by Devon Ryan94k • written 4.3 years ago by sumithrasank75130
gravatar for Devon Ryan
4.3 years ago by
Devon Ryan94k
Freiburg, Germany
Devon Ryan94k wrote:

It fits the data with a negative binomial and performs a Wald test (or a likelihood ratio test, depending on what you specify).

ADD COMMENTlink written 4.3 years ago by Devon Ryan94k

Hi Devon,

Here is my code:


##do differential expression
dds <- DESeqDataSetFromMatrix(countData = data1,colData = pdata,design = ~ condition )
#reference is group2,so up means up in group1 and down means down in group1.

dds$condition <- relevel(dds$condition, "group2")
res <- results(dds)

I wanted to know what test is used in this by default "Wald test or LRT test or no test " because I am not specifying anything in my command.

ADD REPLYlink written 2.5 years ago by Ron990

By default (copy/pasted from:

The steps performed by the DESeq function are documented in its manual page ?DESeq; briefly, they are:

estimation of size factors \(s_j\) by estimateSizeFactors
estimation of dispersion \(\alpha_i\) by estimateDispersions
negative binomial GLM fitting for \(\beta_i\) and Wald statistics by nbinomWaldTest
ADD REPLYlink written 2.3 years ago by cpad011212k
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