[转移贴]TCID50与 PFU换算

cao1976

原贴由future 发表于 2008-5-31 17:22

有帖子发表如下公式
TCID50与 PFU换算:

PFUs＝0.7×TCID50的滴度

请问，此计算公式是否能找到文献出处？

cao1976

wwwkkk83 发表于 2008-5-31 17:51

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.

frome    http://www.protocol-online.org/biology-forums/posts/1664.html

cao1976

wwwkkk83 发表于 2008-5-31 17:51

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.

frome    http://www.protocol-online.org/biology-forums/posts/1664.html

cao1976

wwwkkk83发表于 2008-5-31 18:11

paper:
Diversity of enterovirus sequences detected in oysters by RT-heminested PCR

1. Introduction
2. Materials and methods

2.1. Viruses and cells
2.2. Shellfish samples
2.3. Viral concentrations in oyster digestive glands
2.4. Identification of infectious enterovirus
2.5. RNA extraction
2.6. Oligonucleotides
2.7. Amplification of RNAs
2.8. Sequence analysis

3. Results and discussion

3.1. Sensitivity of shellfish analysis evaluated by RT-heminested PCR
3.2. Analysis of oyster samples
3.3. Analysis of the sequences detected in oyster samples
3.4. Comparisons against the sequences in databases
3.5. Epidemiological background

4. Conclusion
Acknowledgements
References

2.1. Viruses and cells

Poliovirus type 1 (Sabin strain) was used in this study as an enterovirus model for evaluating the sensitivity of the reverse transcriptase-polymerase chain reaction (RT-PCR). The virus was grown on Vero cells (Membres de la commission de normalisation, 1989). Poliovirus particles (d=1.34) were purified by isopycnic centrifugation in a CsCl density gradient. Virus titration was carried out on Vero cells in microtitration plates and expressed as 50%-tissue culture infectious dose (TCID50) per volume unit (Membres de la commission de normalisation, 1989. Antiseptics and disinfectants used in liquid form and miscible with water-determination of virucide activity-viruses of vertebrates. French norm NF T 72-180. Association Française de Normalisation, AFNOR (ed.), Paris, France.Membres de la commission de normalisation, 1989). The titer of the poliovirus stock was 1.8×108 TCID50/ml. Assuming that the Poisson model was applicable to the viral samples analyzed, 1 TCID50/ml was equivalent to 0.69 PFU/ml (Maul, 1991). Consequently, the titer of poliovirus stock was equivalent at about 1.2×108 PFU/ml. The titer of stock was also determined as RT-PCR amplification units (RT-PCRU) by endpoint dilution, heating to release the viral genome from capside, and then performing RNA detection by RT-heminested PCR. Given that the last positive dilution contained 1 RT-PCRU, the titer was estimate to be about 1.2×1010 RT-PCRU/ml (mean of three different experiments). Consequently, 1 TCID50 of poliovirus was equivalent to about 67 RT-PCRU.

BOOK