Medians as input for t-test? I have some biological data points collected from individual cells.
In my field I often see that people average the data points per cell, and then run a t-test using these averages as input. So this means your n in each group equals the number of cells from which you collected and averaged data points. For example you end up with e.g. 5 averages in the control group (from 5 cells) and 5 averages from the treated group (another 5 cells), and then these 10 averages are compared using the t-test.
But in my case, the data from each cell (hundreds of data points per cell) is not normally distributed, and as I understand the median will then represent the data more accurately. I used the normality tests in GraphPad Prism (Anderson-Darling, D'Ágostino & Pearson, Shaprio-Wilk, Kolmogorov-Smirnov), all said that the data was not normally distributed.
Is it then allowed to use these medians (one from each cell) as input for a t-test?
Many thanks for your help!
[edit for some additional details]
 A: You say your have large numbers of data points per cell. In that
case the cell medians should be normally distributed. (See simulation below.)
So you could
run a two-sample t test on cell medians to see if Control and Treatment groups differ.
I haven't seen your data, but it would probably be OK to use cell
means because you have so many data points per cell.
In principle, there is nothing to stop you from running a nonparametric Mann-Whitney-Wilcoxon
test as @Parnian suggests, but with only five cells each in Treatment and Control groups, that may well be a futile exercise.
You are near the absolute lower limit of sample sizes for which the MWW test
is useful. 
For example, for the rank-based MWW test, if you had only four cells in each group,
then all of the Treatment cells would need to have greater 'averages'
than any of the Control cells (or vice versa) to get a significant
result. There are only ${8 \choose 4} = 70$ possible arrangements of ranks
the most extreme two of which correspond to complete separation of
the values in the two groups; $2/70 =  0.029$ so it is possible to
to get a significant P-value. But as soon as there is any overlap at
all, the smallest possible P-value becomes greater than $0.05.$
Also, here is an example of MWW test with five Treatment values and five Control
values that is not significant at the 5% level. By contrast, a t test
does find a significant difference at that level.
 wilcox.test(c(10, 20, 30, 40), c(38, 48, 58, 68))$p.val
 [1] 0.05714286

 t.test(c(10, 20, 30, 40), c(38, 48, 58, 68))$p.val
 [1] 0.02201958

Simulation: CLT for mean and median. Finally, suppose you have samples of size 500 from the skewed distribution
$\mathsf{Gamma}(2, 1).$ By the Central Limit Theorem, means of such samples
will be nearly normal. But there is also a CLT for medians. Here is a simulation
using means a and medians h of $100\,000$ samples of size $n=500$ from this
distribution. The medians are a little more variable, but normal nevertheless.
set.seed(604);  m = 10^5;  n = 500
x = rgamma(m*n, 2, 1)
DTA = matrix(x, nrow=m)  # each row of matrix is sample
a = rowMeans(DTA);  h = apply(DTA,1,median)
par(mfrow=c(1,3))
 curve(dgamma(x,2,1), 0,10, col="blue", lwd=2, ylab="PDF", 
       main="Density of GAMMA(2,1)")
   abline(v=0,col="green2"); abline(h=0,col="green2")
 hist(a, prob=T, br=30, col="skyblue2", 
      main="n=500: Sample Means")
 hist(h, prob=T, br=30, col="skyblue2", 
      main="n=500: Sample Medians")
par(mfrow=c(1,1))


A: If your data are not normally distributed and that is why you are using median instead of mean to compare the groups, you can use Mann-Whitney test to compare the difference between the two groups.
A: Here is a paper talking about using t-test with extremely small sample size "Using the Student's t-test with extremely small sample sizes".
For the normality part here is something I found on Wikipedia

For exactness, the t-test and Z-test require normality of the sample means, and the t-test additionally requires that the sample variance follows a scaled χ2 distribution, and that the sample mean and sample variance be statistically independent. The normality of the individual data values is not required if these conditions are met.

Notice the sample mean in the above text refers to the average of cell means in your case. Let's denote the mean value of each cell is $C_i$. Here 'i' only ranges from 1 to 5. The number of cells can't guarantee $\bar{C}$ to have a normal distribution. It would be easier if $C_i$s are $i.i.d$ normal. When you calculate the mean value of each cell (assuming samples in each cell across the entire group are from the same population), $C_i$s are $i.i.d$ normal by CLT. Although median can follow CLT in some case (Please see Central limit theorem for sample medians
 for detail), using mean works in the most general setting.
