Proper compensation

(Go back one page, or back to the Introduction).

The goal of compensation is to remove the spillover fluorescence of a particular probe from the "wrong" channel. I.e., fluorescein fluorescence is primarily green, which is measured in the FL1 (FITC) channel. But fluorescein also has a significant yellow component to the fluorescence, which appears in the FL2 (PE) channel. One of the properties of fluorescence is that the ratio of these two measured values, the fluorescein fluorescences in FL1 and in FL2 channels (green/yellow), is always the same (at any given PMT voltage, filter set, and instrument sensitivity). In an ideal world, we can exactly determine the yellow fluorescein fluorescence (in the PE channel) based on the measurement of the green fluorescence (in the FITC channel)--and, in an ideal world, we could exactly correct ("compensate") for this yellow fluorescence and not worry about the contribution of FITC to the PE channel. The result is that the output of the compensated PE channel will represent only the "true" (yellow) fluorescence arising from PE itself.

Mathematically, compensation from FITC to PE simply subtracts a fraction of the FITC signal from the PE signal: the more FL1 (green fluorescein fluorescence) there is, the more spillover into FL2 (yellow fluorescein fluorescence) there will be (i.e., as stated above, the green/yellow ratio does not change). As an example, if the amount of yellow fluorescein signal in the FL2 channel is 15% of the green fluorescein signal in the FL1 channel (i.e., "15% compensation"), then we can exactly determine the "true" (or "pure") PE fluorescence of a cell, even in the presence of FITC fluorescence, as:

PEtrue = PEmeasured - (0.15) x FITCmeasured

This process is identical for correcting for PE fluorescence appearing in the FITC channel. For instance, if the amount of (green) PE signal in the fluorescein channel is 2% of the (yellow) signal in the PE channel (i.e., "2% compensation"), then we can exactly determine the true FITC fluorescence of a cell as:

FITCtrue = FITCmeasured - (0.02) x PEmeasured

These equations hold true even if a cell has both FITC and PE fluorescence (the machine always does these compensations independently and simultaneously; i.e., it does not use the output compensated FITC value as an input to the PE compensation).

If you would like a detailed example of this process, then click here. The example is not part of this informal discussion; you will have to navigate back to this page manually. Please note that the equations given on this page are only correct for doing one compensation at a time--because of this, the compensation coefficients given on this page (as well as the equations) are somewhat different than those in the detailed example. To simultaneously compensate two or more colors, compensation requires the generalized solution derived with matrix algebra, described elsewhere.

Otherwise, Go on.