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Converting TCID[50] to plaque forming units (PFU)
Is it possible to determine from the TCID[50] how many plaque forming units to expect?

Assuming that the same cell system is used, that the virus forms plaques on those cells, and that no procedures are added which would inhibit plaque formation, 1 ml of virus stock would be expected to have about half of the number of plaque forming units (PFUs) as TCID[50]. This is only an estimate but is based on the rationale that the limiting dilution which would infect 50% of the cell layers challenged would often be expected to initially produce a single plaque in the cell layers which become infected. In some instances, two or more plaques might by chance form, and thus the actual number of PFUs should be determined experimentally. Mathematically, the expected PFUs would be somewhat greater than one-half the TCID[50], since the negative tubes in the TCID[50] represent zero plaque forming units and the positive tubes each represent one or more plaque forming units.

A more precise estimate is obtained by applying the Poisson distribution. Where P(o) is the proportion of negative tubes and m is the mean number of infectious units per volume (PFU/ml), P(o) = e(-m). For any titer expressed as a TCID[50], P(o) = 0.5. Thus e(-m) = 0.5 and m = -ln 0.5 which is ~ 0.7. Therefore, one could multiply the TCID[50] titer (per ml) by 0.7 to predict the mean number of PFU/ml. When actually applying such calculations, remember the calculated mean will only be valid if the changes in protocol required to visualize plaques do not alter the expression of infectious virus as compared with expression under conditions employed for TCID[50]. Thus as a working estimate, one can assume material with a TCID[50] of 1x 10(5) TCID[50]/ml will produce 0.7 x 10(5) PFUs/ml.

Date Created07/17/2012 07:10 PM
Date Updated07/25/2012 05:22 PM

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