Question: Statistical testing for differential expression in DESeq
1
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
modified 4.3 years ago by Devon Ryan94k • written 4.3 years ago by sumithrasank75130
2
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).

Hi Devon,

Here is my code:

data1<-as.matrix(df)

##do differential expression
library(DESeq2)
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")
dds=DESeq(dds)
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.

By default (copy/pasted from: http://bioconductor.org/packages/devel/bioc/vignettes/DESeq2/inst/doc/DESeq2.html#theory):

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