Titration of Virus Infectivity
Experiment Summary
This experiment introduces the titration of viral infectivity (TCID50 determination). It helps to understand several common methods for the determination of viral infectivity and to grasp the procedure, calculation method and meaning of TCID50 for the amount of infection in half cell cultures.
Experimental Principle
LD50 (50% lethal dose): animal or chicken embryo is used to detect.
Half chicken embryo infective dose EID50 (egg 50% infective dose): using chicken embryo to detect.
Half of the cell culture infective dose TCID50 (50% tissue culture infective dose): using cells to detect.
Plaque forming unit (PFU): detected by cells.
Main Reagents
Trypsin, growth fluid, viral fluid (PRV).
Main Equipment
- Pipette
- 96-well cell culture plate
- spiker, gun tip
Experimental Materials
1 bottle of full-grown monolayer cells.
Experimental Steps
- Make serial 10-fold dilutions of the virus solution from 10-1-10-10 in penicillin vials or centrifuge tubes.
- The diluted virus was inoculated into a 96-well microtiter plate, with a total of 8 wells of 100 µl per dilution in one vertical row.
- Add 100 µl of cell suspension to each well to bring the cell volume to 2~3×105 cells/ml.
- Normal cell controls were set up and normal cell controls were made in two longitudinal rows. (100 µl growth solution 100 µl cell suspension).
- Observe and record results day by day, usually requiring 5-7 days of observation.
- Calculation of results by Reed-Muench two's method or Karber method.
- Calculation method of TCID50
(1) Reed-Muench method
| Dilution of viral fluid | Number of CPE holes | No CPE hole number | Cumulative | Percentage of occurrence of CPE holes |
| CPE hole number | No CPE hole number |
| 10-1 | 8 | 0 | 27 | 0 | 100 |
| 10-2 | 8 | 0 | 19 | 0 | 100 |
| 10-3 | 7 | 1 | 11 | 1 | 91.6 |
| 10-4 | 3 | 5 | 4 | 6 | 40 |
| 10-5 | 1 | 7 | 1 | 13 | 0 |
| 10-6 | 0 | 8 | 0 | 21 | 0.7 |
CPE:Cytopathic effect
Distance ratio = (percentage of lesions above 50% - 50%) / (percentage of lesions above 50% - percentage of lesions below 50%)
=(91.6-50)/(91.6-40)
= 0.8
lgTCID50 = distance ratio × difference between the logarithms of dilutions higher than the logarithm of the dilution of 50% lesion rate.
=0.8× (-1) (-3)
=-3.8
TCID50=10-3.8/0.1 ml
Meaning: Inoculation of 100 µl of this virus at a dilution of 103.8 can cause lesions in 50% of the cells.
(2) Karber Method
| Dilution of viral fluid | Number of holes in which CPE appears | Rate of occurrence of CPE holes |
| 10-1 | 8 | 8/8=1 |
| 10-2 | 8 | 8/8=1 |
| 10-3 | 7 | 7/8=0.875 |
| 10-4 | 3 | 3/8=0.375 |
| 10-5 | 1 | 1/8=0.125 |
| 10-6 | 0 | 0/8=0 |
lgTCID50=L-d(s-0.5)
L: Logarithm of the highest dilution
D: Difference between logarithms of dilution
S: Sum of positive pore ratios
lgTCID50=-1-1× (3.375-0.5)
=-3.875
TCID50=10-3.875/0.1 ml
Meaning: Inoculation of 100 µl of this virus at a dilution of 103.875 can cause lesions in 50% of the cells.
* For research use only. Not intended for any clinical use.