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Question: DESeq2 modelling and Wald's test
2
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Hi all,

Can you please explain to me the relation between Wald's test and Negative binomial generalized linear model? As for my understanding, the count data is modeled using negative binomial generalized linear model after which Wald's test is applied to figure out whether a particular gene is significant or not. Please correct me if I'm wrong.

8
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RNA-seq raw count data 'naturally' follows a negative binomial distribution (Poisson-like), so, the DESeq2 authors model the data as such. By 'model the data', we merely imply that we build a regression model of the data such that we can make statistical inferences from it [the data].

So, after normalising the raw counts, the following occurs:

For each gene, a logistic regression model with the negative binomial as family is fit:

``````require(MASS)
gene1.model <- glm.nb(gene1 ~ CaseControl + ..., data=MyData)
gene2.model <- glm.nb(gene2 ~ CaseControl + ..., data=MyData)
*et cetera*
``````

Once we have modeled each gene, a simple way to derive a P value for each model coefficient (i.e. CaseControl, etc) is by applying the Wald Test and selecting the coefficient of interest:

``````require(aod)
wald.test(b=coef(gene1.model), Sigma=vcov(gene1.model), Terms=c(2)) #term '2' would be CaseControl
``````

The Wald test is a standard way to extract a P value from a regression fit.

Kevin

NB - this is not the exact code used by DESeq2, of course. This is just giving you a broad overview with some simple R functions. For one, DESeq2 models dispersion in addition to everything that I have mentioned above, and the Wald test is not used in each case to derive p-values in DESeq2.

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1

Thank you so much for that.

Uday Rangaswamy
• 120
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0

Kevin, thanks for your explanation. Let me one naive question please? Why we need to make a `GLM` model before performing a `Wald test` itself (as i can understand it's just a simple `t-test` in rough approximation?)? Why not just perform a `Wald test` on count data?

Denis
• 70
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1

A Wald test requires a coefficient and its standard deviation, which are tested for difference from 0. Yes, in a way that's sort of like a single group T-test, but you'd still need to perform a fit first in order to derive the coefficient.