**0**wrote:

Consider an experiment (e.g. proteomics) producing ratios. Our null hypothesis is that each ratio is one. In this particular example, I have a data table with n rows (peptides) and m columns (replicates). I would like to use limma to do a differential expression of each peptide versus a constant of one. This should be a generalisation of a one-sample t-test.

I'm not quite sure how to build data and design matrix for such test. With only one condition I struggle to write a linear equation.

What I did, and I have a feeling is not a correct approach, is to add a column of ones to my data and marked it as a separate condition, as in this generated example.

```
library(limma)
set.seed(1)
ndat <- 10
nrep <- 3
# rows progressively diverging from the null hypothesis
tab <- t(sapply(1:ndat, function(i) rnorm(nrep, mean=1 + 0.5 * (i - 1)/ndat, sd=0.1)))
# add a reference column of 1
dat <- cbind(1, tab)
meta <- data.frame(condition = c("1", rep("A", nrep)))
design <- model.matrix(~condition, meta)
# limma test
fit <- limma::lmFit(dat, design)
ebay <- limma::eBayes(fit)
res <- limma::topTable(ebay, coef="conditionA", adjust="BH", sort.by="none")
```

This is probably not quite right as it corresponds, at a peptide level, to a two-sample t-test, where the first sample contains only one element (1). This certainly does not correspond to a one-sample t-test against mu = 1.

An independent one-sample t-test on these data does this:

```
pt <- apply(tab, 1, function(x) t.test(x, mu=1, alternative="two.sided")$p.value)
plot(log10(res$P.Value), log10(pt))
```

If you re-do the above example for nrep=30, you will see that the limma p-values are much more conservative than those from a one-sample t-test. Since data are normal and the number of replicates is large, I would expect similar results from any test.

Can someone think of a better way of doing a one-condition test against a constant with limma?

**51k**• written 23 months ago by M.Gierlinski •

**0**