TCID50 and plaque forming unit (PFU) - (Apr/16/2002 )

Does anyone know about the relationship bewteen TCID50 and plaque forming unit (PFU)? I doubt cause someone told me that 2 TCID50 is aproximately equal to 1 PFU for HSV. Thanks

*edit*: merged two topics with the same problem - mike

-chy5564-

ok, i couldn't find an answer to this. i don't care about the relationship for certain viruses, it seems to me it would be equivalent but cell culture/virology is so far removed from what we usually do!

QUOTE (chy5564 @ Apr 16 2002, 09:10 PM)
Does anyone know about the relationship bewteen TCID50 and plaque forming unit (PFU)? I doubt cause someone told me that 2 TCID50 is aproximately equal to 1 PFU for HSV. Thank

-catdoc-

From another website:
http://www.atcc.org/common/technicalInfo/f...Virology.cfm#Q5

Is it possible to determine from the TCID50 how many plaque forming units to expect?

Answer: 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 TCID50. 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 TCID50, since the negative tubes in the TCID50 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 TCID50, P(o) = 0.5. Thus e(-m) = 0.5 and m = -ln 0.5 which is ~ 0.7.

Therefore, one could multiply the TCID50 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 TCID50.

Thus as a working estimate, one can assume material with a TCID50 of 1x 105 TCID50/ml will produce 0.7 x 105 PFUs/ml.

-thneedle-