# Understand TCID50 and MOI

**Posted:**March 31, 2011

**Filed under:**Virus titration Leave a comment

First, we must understand that virus infection is a collection of events that occur independently. In other words, none or one or multiple virus particles may infect a single cell independently from other virions’ activity and other cells’ infection. The infection therefore follows the Poison distribution.

So, let’s take a look at the Poison distribution. I borrow the definition and modified graph for Poisson distribution from Wikipedia (http://en.wikipedia.org/wiki/Poisson_distribution, 31/3/2011)

When applying Poisson distribution to virus infection, we have:

*k: the number of virions infecting one cell*

*P(k): the probability of cells infected by k virions in the whole population. For example, if P(X=3) = 0.2, or P(3) = 0.2, for a population of 10 ^{4} cells, there are 2 000 cells (0.2*

*´ 10*

^{4}= 2 000) each of which infected by 3 virions.*m is a positive real number, equal to the average number of virus particles that infected a cell (here m=MOI). In other word, m is the expected number of infections on one host cell in your experimental conditions (Please keep in mind that the expected number is a weighted average of all possible values that this random variable can take).*

With each m, or MOI, value, we can draw a dotted Poisson curve (the connecting line is for your reference only). We may understand the graphs as follows:

*At m=1, or MOI=1, most of the cells are non-infected (corresponding to k=0) or infected by 1 virion (k=1). However, there are fewer cells infected by 2 or more virus particles, which makes the average numbers of virions/cell equal to 1.*

*Similarly, at m=4, most of the cells are infected by 4 or 5 virions. The number of non-infected cells significantly reduced. On average, each host cell is infected by 4 virions.*

*At m=10, almost all of the cells are infected.*

In the case of TCID[50], we want to divide the virus population into smaller parts which we call TCID[50] unit; each unit has the probability of un-infected cells P(0) equal to 0.5 or 50% (Remember that we have 50% infected samples and 50% uninfected at this endpoint dilution). This is to say, 100 TCID[50] units will infect 50 samples out of 100 samples, or 50 cells out of 100 cells.

To make it clearer, let’s look at another example. The titer given by ATCC of one virus stock is, for example, “10^{4} TCID[50]/0.2 ml, Vero, 2 days”. This means that 0.2 ml of the stock contains 10^{4} TCID[50] units with the probability that each unit causes infection being 50%. If you incubate 1:10,000 of the 0.2ml stock with each of four tubes containing Vero cells for two days, two tubes are expected to be infected and two uninfected.

Investigating each TCID[50] unit, the Poisson equation becomes:

As P[0]=0.5, we have e^{-m}=0.5 and thus m= -ln0.5»0.69. This means that approximately 69 infectious virus particles are required to produce infection on 50 cells out of a 100-cell population. In other word, 69 infectious virus particles are equivalent to 100 TCID[50] units. Thus, on average, we have m= 0.69 infectious virus particle for each TCID[50] unit.

If we have the endpoint occurs at the dilution 10^{-5} from 0.2 ml stock, the virus stock contains 10^{5} TCID[50] unit/0.2 ml or 5´10^{5} TCID[50] units/ml. Therefore, 1 ml of this virus stock contains 0.69´5´10^{5} infectious virus particles, or the virus titer is 34.5´10^{4} PFU/ml (PFU: Plaque forming unit).

**Now, why are there such different concepts of infectious virus particles (here represented by the number 0.69) and infectious samples (by the number 0.5)? **

In the simplest and ideal model, all virus particles are infectious and each sample at the extremely high dilution is infected by a single virus particle. In such condition, the number of infectious virus particles is equal to the number of infected samples. However, in reality not all virions are infectious (and we obviously only care about the infectious ones). Besides, one sample might remain uninfected or be infected by one or more virus particles following Poisson distribution as follows:

- Uninfected samples (or wells): P[0] = e
^{-m} - Samples (or wells) infected with a single virus particle: P[1] = me
^{-m} - Samples (or wells) infected with multiple virus particles: P[>1] = 1-(P[0]+P[1])=1-(1+m)e
^{-m}

At end point dilution:

*The number of uninfected samples = A = 50% = 0.5*

*The number of infected samples = B+C=50% = 0.5*

*The number of infectious virus particles in B+C = m= 0.69 »0.7*

**Now, how to deal with MOI (Multiplicity of Infection)?**

With M.O.I = m, we add, on average, m infectious virus particles (not total virus particles) to a single cell. If you understand the above concept, your life would be easier at this calculation step.

Assume that the virus stock contains N TCID[50] units/ml. The number of infectious virus particles in 1ml stock is thus 0.7xN infectious particles. Now you need to add V ml of virus solution to a well that has X cells to obtain a final MOI = m. We have:

The number of infectious virus particles/well = m x X

For example, the TCID[50] titer for a virus stock is 10^{6.5} TCID[50]/0.2 ml, which is equal to 5×10^{6.5} TCID[50]/ml. If you are going to add 100 µl of the diluted virus to 20,000 cells in a well to get the MOI=0.1, the dilution factor should be:

If you want to add diluted virus solution to 80 wells, the total volume you need is 80×100 µl and therefore a volume of 14.46 µl of the virus stock (8000/dilution factor = 8000/553.4 = 14.46 µl) should be diluted to 8000 µl in appropriate medium.

One more note for you is that the MOI used in most experiments is MOI=1, which means that the ratio of the number of infectious virus particles: the number of cells = 1:1. Ideally, one cell will be infected by one virus particle. However, in reality, multiple virus particles may infect the same cells leaving some others uninfected. The proportion of the uninfected cells follows Poisson distribution and is equal to P[0]=e^{-m}=e^{-1}=0.37.

Also, please bear in mind that in many experiments we add as much virus as possible to make sure that all the cells are infected. The MOI value sometimes goes up to 5 or 10, because some virus might be inactivated during the course of experiments. Only for those viruses that it is difficult to obtain a high titer such as coronavirus or dengue, we use a lower MOI value.

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