"Consider a cell which is negative for a PE reagent but brightly fluorescent for a FITC reagent. Assuming our (logarithmic) fluorescence scale ranges from 0.1 to 1,000 (4 decades), then the "true" fluorescences of this cell might be 0.5 for PE (i.e., negative except for some autofluorescence), and 100 for FITC. Because of the 15% spillover of FITC to PE, the measured PE fluorescence would be 15.5 (0.5 "true" + 0.15 x 100 "FITC spillover"). Note that if the measurement were made exact, then the true PE fluorescence could be determined. Using the first equation, PEtrue = 15.5 - 0.15 x 100 = 0.5. But what would happen if there were a one channel error in the measurement of fluorescence? In the PE channel (at this low level of fluorescence), this is not a whole lot; one channels corresponds to about 0.02 fluorescence units. But because of the logarithmic nature, one channel error at 100 units (the fluorescein measurement) translates into about 3.7 fluorescence units. If the machine were only one channels low (out of 256), then the FITC might be measured at 96.3 units. Now the "true" PE fluorescence is calculated as: 15.5 - 0.15 x 96.3 = 1.06. This is a shift of over 20 channels from the correct position: note how much error has been introduced into the "true" PE fluorescence value! A measurement that was ±1 channels is now ±20 channels. (Note, of course, that this is because of the logarithmic amplification: the error of the measurement is ±3.7 units; this results in an error after compensation of 15% of this, or ±0.56 units."

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