I'm going to sidestep the (very valid!) concerns raised over technical vs. biological replication - if you're working with a cell line or inbred animals one might argue that you can only ever do technical replication even.
To answer your question, comparing $n$ samples drawn from individual wells or $n$ samples that are each a pool of $m$ wells might be valid if you keep in mind the following:
- you will underestimate the well-to-well variance in the second scenario by factor $m$.
- the mean must be a good summary measure for your samples.
Let me demonstrate both with a simulation example in R
. I'm drawing data from $N(0,1)$ here but as long as a normal approximation is appropriate (or you have a sufficiently large number of samples) that shouldn't matter.
# Generate N samples from separate wells
separate <- function(n=4, fun=rnorm, iters=1E6) {
matrix(fun(n*iters), ncol=n)
}
# Generate N samples, each combined from M wells
pooled <- function(n=4, m=2, fun=rnorm, iters=1E6) {
cbind(vapply(seq_len(n), \(_) rowMeans(matrix(fun(m*iters), ncol=m)), numeric(iters)))
}
set.seed(1)
obs_separate <- separate()
obs_pooled <- pooled()
We now have a million independent experiments of each condition. To illustrate my first point, let's look at the observed standard deviation in each experiment:
# Pooling reduces variance by factor M [or SD by sqrt(M)]
summary(matrixStats::rowSds(obs_separate))
> Min. 1st Qu. Median Mean 3rd Qu. Max.
> 0.004 0.636 0.888 0.922 1.171 3.245
summary(matrixStats::rowSds(obs_pooled))
> Min. 1st Qu. Median Mean 3rd Qu. Max.
> 0.0063 0.4498 0.6282 0.6515 0.8274 2.3219
Notice that the standard deviation is smaller by $\sqrt2$ (since $m=2$) throughout the second condition. Not shown here is the observed means which center around the same value in both conditions, but because variance is smaller the estimates will be closer to the center in the second condition. Neither of this is a problem as long as you do the same across groups, you will maintain type I error and in fact estimate the mean more accurately as we just saw:
# Make two groups to compare (both are identical, null hypothesis is true)
set.seed(1)
n <- 8
compare_separate <- separate(n=2*n)
compare_pooled <- pooled(n=2*n, pool=n)
group <- factor(rep(0:1, each=n))
mean(genefilter::rowttests(compare_separate, group)[,3] < 0.05)
> 0.0497
mean(genefilter::rowttests(compare_pooled, group)[,3] < 0.05)
> 0.0498
However, it would be wrong to compare one group drawn from the individual setup versus another drawn from the pooled!
compare_wrong <- cbind(compare_separate[,1:n], compare_pooled[,(n+1):(2*n)])
mean(genefilter::rowttests(compare_wrong, group)[,3] < 0.05)
> 0.0601
While I've exacerbated the issue a bit by increasing the number of pools, you can see this no longer maintains type I error rate exactly because you are underestimating the variance in one group but not the other.
For the second point I'll use the log-normal distribution (i.e. a random variable whose log is normally distributed). Because of its skewness the mean is not a good measure for central location, and you would commonly use the geometric mean or median instead. However, you obviously cannot pool your samples according to a geometric principle without measuring them individually first, so that will still just be a mean.
set.seed(1)
lognorm_separate <- separate(fun=rlnorm)
lognorm_pooled <- pooled(fun=rlnorm)
# Compare geometric means
summary(exp(rowMeans(log(lognorm_separate))))
> Min. 1st Qu. Median Mean 3rd Qu. Max.
> 0.095 0.714 1.000 1.133 1.402 11.555
summary(exp(rowMeans(log(lognorm_pooled))))
> Min. 1st Qu. Median Mean 3rd Qu. Max.
> 0.213 0.956 1.230 1.325 1.588 8.004
Oops! Pooling has biased the estimate pretty badly. This may actually be an issue because in my field (where concentrations are usually plasma concentrations of drugs) we use the log-normal distribution and not the normal distribution. It all comes down to how appropriate a normal distribution is for your samples. If your concentrations are all well enough away from 0 and the distribution is sufficiently symmetric it might be OK, but without examining your data it's impossible to tell.
n = 8
versusn = 4
, isn
the number of total wells or the number of wells for each treatment? How many treatments are there? Please provide that information by editing the question, as comments are easy to overlook and can be deleted Also, what you describe seems more like technical replicates rather than biological replicates; see the relevant section of this perspective. $\endgroup$