Hi everybody!
I'm a bit confused about compensation. The problem is:
I've a mixed population of cells some positive for a CD, others not. I need to analyze their cell cycle (easy), but selecting a gate for positive and one for negative cells (the difficult part).
Have I to compensate PI (linear) and FITC (log)?, can a Lin and a Log scale be compensable, and if it is possible, how (the usual way)?
I'm currently using a FACSCanto with Diva software. I tried to do a preliminary experiment (compensating them), but the result is not clear: CV is too high and FITC fluorescence flatted on the axes everytime I try to "adjust" compensation parameters.
Any idea?
hi Nicola,
are you still with the question? or u found the answer
senthil.
I'm still having problems...but the FACS tecnician told me that I do not need to compensate... I'm not sure he's right...
hi there,
This is not a challenging issue,
FITC and PI tube surely needs compensation, Do you have the FCS files with you which you can share, if yes send me them to my ID [email protected], send me all the 4 FCS files Unstained, both the Single stained and Both stained.
I hope I can give you a better idea.
Senthil.
at the moment I'm not in my country, so I can't access my lab data...so it's quite impossible to send you any file...
anyway...I try to find a way to send you something...
Thanks in advance...
Dear friend,
For compensation prepare one unstained sample tube,PE stained sample tube, FITC stained sample tube and Both PE and FITC stained sample tube(MIX) in all make equal the cell concentration.
Make 2 plots FSC vs SSC plot and PE vs FITC
First run the unstained sample and make your population of interest falls on the center of the FSC vs SSC plot then gate it.
In the PE vs FITC plot draw a quandrant gate and adjust your voltages to make sure that the unstained events fall within the 10^1 scale in both x-axis and y-axis. once done acquire and save data.
Use the same voltage setting for ur individually stained sample and for mix stained sample.
Run PE stained sample in the sample plot, Ur PE Negative cells will fall within the 10^1 scale of both the axis, but ur PE positive cells will fall in higher scale in the PE quadrant, no events should be there in FITC quadrant.
Like wise run the FITC stained sample. Ur FITC Negative cells will fall within the 10^1 scale of both the axis, but ur FITC positive cells will fall in higher scale in the FITC quadrant , no events should be there in PE quadrant.
When u run ur Mixed sample u will have negative population,PE and FITC positive population. in the PE vs FITC plot PE population and FITC population should fall within the respective quadrants, if not adjust the compensation to make them within the respective quadrants.
this compensation settings u can use for ur samples.
U can use more fluorescent markers also in the same method. This is called sample compensation.
Note: For this ur sample should have enough positive cells for both PE and FITC markers and also negative for both.
Dear Pachaiyappan Viswanathan,
I thank you for the explanation, but, the point is not this...I know very well how to compensate FITC and PE...the point here is how to compensate something linear (PI, propidium iodide, ex FL3, now PerCP channel) with something in a logarithmic scale (FITC)... And...it is a cell cycle analysis...
In practice your technician is right but in theory they are wrong! Red photons from FITC emission will always be picked up in the PI detector. BUT remember that compensation is always performed using the raw data signal before it is either sent to a log amplifier or a log transformation. If you consider that the PMT output is a 10V scale, I susect you will put the G1 peak around channel 50,000 (on a Canto) so the signal is about 2V. If you look at the linear value for the FITC signal it will probably be almost on the axis and will be about 0.5V, the spillover will be in the region of 20% of that signal ie 0.1V (PMT voltage dependent) so the impact will be negligible on the PI signal. It will ony be a problem if the FITC is very very strong.
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