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I am working with culture cells where one dish has been transfected with a scrambled knockdown clone and two dishes which have been transfected with two knockdown clones each knocking down the expression of a single gene.

An example of an experiment I have performed is to measure the mitochondrial membrane potential (using a fluorescent dye) in these cells using a confocal microscope. This experiment was repeated on three independent occasions.

On each experimental day, the intensity of the laser which I used (the laser "gain") varies therefore I cannot combine all experimental days without expressing the dye intensity of each knockdown clone as a percent of the control "scrambled" clone (e.g. control = 100% mean intensity; knockdown clone 1 = 50% mean intensity).

Therefore, I need to test for a difference in means between my control scrambled clone and each of the knockdown clones, where my control scrambled clone is set to 100% dye intensity on each experimental day and my knockdown clones are normalised to this control. Therefore, my control has no variance (100% for all three experimental days) while my knockdown clones do have variance.

I know an ANOVA would not be feasible given the difference in variance. I will look into the procedure suggested by Michael Lew, but would a t-test be unacceptable as well? (I have seen papers using ANOVA and t-tests in these circumstances, but in spite of this I am assuming these should not be used). Thanks in advance.

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    $\begingroup$ Difficult to answer your question because it is so broad. Two points, though (i) What difference do you want to test? Difference in the means? A t-test is still possible in this case. (ii) Why normalize and lose the valuable variance information? $\endgroup$
    – Drew75
    Commented Nov 5, 2013 at 19:03
  • $\begingroup$ This question is WAY too broad. We have no idea what the nature of your data is - in particular, I have no idea what you mean when you say " this means that all my control values are 100%". Your control values are 100% of what? What experiments are you running? What are the independent and dependent variables? What's the nature of the data? We need more information here. $\endgroup$
    – Nathan Calverley
    Commented Nov 5, 2013 at 21:26
  • $\begingroup$ Hi Drew75, I have clarified my experiment in my post. I am testing a difference in means, as detailed above. Would you mind clarifying how a t-test would still be possible in this case? Thanks very much! $\endgroup$
    – Emily
    Commented Nov 11, 2013 at 8:28

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You are correct in assuming that you can't (shouldn't, really) analyse the data with the controls having zero variance.

It sounds like you should consider using a two-way ANOVA on the raw data with the within day variance accounted for in the manner of a paired test. I wrote about the approach in this paper that is intended for pharmacologists with little statistical background: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2042947/

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