Dear Olindo,
I'd recommend that you apply the equation to the channel means you've
already analysed, this will restore the relationships between your
means to a more proportional scale, even though it won't allow you to
find the arithmetic mean for those samples. I've tried to explain
below why the two methods you've tried recently don't give the same
result. Note that they probably would agree if you chose the
"geometric" option in CellQuest (if it has one).
The equation works fine for any particular channel, but when you
apply it to mean channel values, you end up with the geometric mean
rather than the arithmetic mean, these two averages are different and
not inter-convertible.
Imagine an unusual "population" with one cell in channel 256, one in
channel 512 and one in channel 768. If you "linearise" these values
and find their mean you get :
first cell's value = 10^(256/256) = 10
second cell's value = 10^(512/256) = 10^2 = 100
third cell's value = 10^(768/256) = 10^3 = 1000
average = (10+100+1000)/3= 370
this is the ARITHMETIC mean calculated as it is for the "log values"
(actually antilog, but I'll use your expression for clarity) in your
current analysis,
however if you find the average channel value and apply the equation
to it you get:
average = (256+512+768)/3 = 512
this average transformed to a "log value" = 10^(512/256) = 10^2 = 100
this is the GEOMETRIC mean calculated as it is for the "Channel
Value" results in your current analysis. Notice that the means are
considerably different for this extreme example data set, they are
probably a lot closer in your populations - I chose these particular
values because they are easy to calculate with.
The idea of using an average is to measure the "central tendency" of
your data, and the best (or rather, most robust) measure of this is
the median - it might be worth your while comparing the median of
some of your test data with the mean and geometric mean and find
which is closest. If you're really picky you should apply a test for
normality or log-normality (eg the one-tailed K-S test - if you can
get it to work), and then use the arithmetic mean if your data are
normally distributed and the geometric if log-normal, but this is not
a common activity.
If the geometric mean is a reasonable measure, which is probable
because of the positive skew imposed on FACS data (see Flow Cytometry
Data Analysis by James Watson page 105 for a discussion of skewing),
then you're laughing - all you need to do is transform your "channel
averages" using the equation you found. If for some reason you
really need to use the arithmetic mean (unlikely) or the median
(possible, probable if your populations hit either end of the scale
see page 102 in the same book), then you'll need to re-analyse all of
your data from scratch.
Ray
At 12:00 pm -0300 5/6/00, Olindo Assis wrote:
>Dear Flowers,
>
>It was clear for me, after I read the answers to my question about mean
>fluorescence channel analysis that most people suggested to analyze the data
>using the original number "mean" from the Log scale (10^0 - 10^4).
>
>However, I had analyzed most of my data after conversion of original data to
>channel values. Then, I start to have a problem because I did not want to go
>back and reanalyze all data since we have over 2,000 files to be analyzed.
>
>Therefore I was looking for a formula (mathematical expression) to convert
>the data back to Log values. Actually I found it on the CellQuest BD manual
>on chapter 6 "histograms". They say that to convert channel values to Log
>and vice-versa we just need to apply the following expression:
>
>Log values = 10^(channel value/*scale factor)
>
>where scale factor is 256 for a 0-1024 scale and 64 for a 0-256 scale.
>
>
>However, when I applied this formula to a group of data, that I had both
>values (Log and channel values), I did not find this formula as the right
>one to convert the data one to another.
>
>
>Does anyone have any experience with this.
>
>Best regards,
>Olindo
--
Ray Hicks
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