Measurement of viruses by end-point dilution assay

13 July 2009

The plaque assay is a terrific method for determining virus titers, but it doesn’t work for all viruses. Fortunately there are several alternative methods available, including the end-point dilution assay.

The end-point dilution assay was used to measure virus titer before the development of the plaque assay, and is still used for viruses that do not form plaques. Serial dilutions of a virus stock are prepared and inoculated onto replicate cell cultures, often in multi-well formats (e.g. 96 well plastic plates). The number of cell cultures that are infected is then determined for each virus dilution, usually by looking for cytopathic effect.

In this example of an end-point dilution assay, 10 monolayer cell cultures were infected with each virus dilution. After an incubation period, plates that displayed cytopathic effects were scored with a +. At high dilutions, none of the cell cultures are infected because no particles are present. At low dilutions, every cell culture is infected. Half of the cell cultures showed cytopathic effects at the 10-5 dilution. This is the end point: the dilution of virus at which 50% of the cell cultures are infected. This number can be calculated from the data and expressed as 50% infectious dose (ID50) per milliliter. The virus stock in this example contains 105 ID50 per ml.

end-point-dilution

In real life, the 50% end point does not usually fall exactly on a dilution as shown in the example. Therefore statistical procedures are used to calculate the end point of the titration.

End-point dilution methods can also be used to determine the virulence of a virus in animals. The same approach is used: serial dilutions of viruses are made and inoculated into multiple test animals. Infection of the animal can be determined by death or clinical symptoms such as fever, weight loss, or paralysis. The results are expressed as 50% lethal dose (LD50) per ml or 50% paralytic dose (PD50) per ml when lethality or paralysis are used as end points.

The following example illustrates the use of end point dilution to measure the lethality of poliovirus in mice. Eight mice were inoculated per virus dilution, and the end point was death. The statistical method of Reed and Muench was used to determine the 50% end point. In this method, the results are pooled, and the mortality at each dilution is calculated. The 50% end point, which falls between the fifth and sixth dilutions, is calculated to be 10-6.5. Therefore the virus sample contains 106.5 LD50 units.

lethal-dose-50

Reed, L.J., & Muench, H. (1938). A simple method of estimating fifty percent endpoints. Am. J. Hygiene, 27, 493-497

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  • Murine
    Thanks Dr Vinc, Your advise was quite useful in enabling me to select samples containing only single virus. Recently I tried to determine the replication kinetics of many different strains of the same virus. I found out that the replication of the viruses was quite low and eventually became undetecable by qPCR after 04 passages on LLC-MK2 with inapparent CPE. What would you advise in case that I try antagonizing the IFN, can it lead to any improvement in the replication?
  • Yes, it's a good idea to try to disrupt IFN production. You could try
    another cell line with an IFN defect (such as VERO, which don't
    produce IFN, or cells lacking the type I IFN receptor), or you could
    alter the LLC-MK2 so they are defective in the IFN response.
  • murine
    Dear Dr Vincent,
    My study virus disappears in the process of passaging (after 3 passages) and is replaced by another unknown CPE producing virus on LLC-MK2. I cannot understand that how do I identify this unknown virus and select my original virus. Since i need to prepare the viral stock for my original virus. The approval of cloning based detection methods would take around 2-3 months. In passage 2 original virus is there and in the passge 3 it disappears. Perhaps the exclusion principal is involved here. SHould i try even more cell lines or the available antibodies. Any thoughts or golden piece of advice? Yoroshiku one gai shimas.
  • I would try either a different cell line, hoping that it will only
    propagate your virus, or try diluting the virus before infecting
    LLC-MK2 so that only a few virions are used to infect, thereby
    favoring infection with one or the other virus. If your virus is in
    the majority this should work, especially if done a few times.
  • dmcilroy
    Actually, I have never understood the utility, or even the validity, of pooling the results at each dilution.
    In the example shown above, why is the mortality ratio of 27% at 1x10-7 , calculated from the cumulative data considered to be more valid than the actual observed mortality ratio at that dilution (3/8 or 37.5%) ?

    I have read (and even tried to understand) textbook explanations of this, and I have not found the explanations very clear.

    DMc
  • Have you read the 19389 paper by Reed and Muench? The logic is best
    explained there. If you have not, I can send you a copy of the paper.
  • Bmschen
    I am interested to have a read of this paper. May I have a copy from you? Appreciate.
  • Emily T. S. Wu
    Dear Vincent,
    I am a master student in Biotechnology from Taiwan. I am looking for the paper by Reed and Muench on the internet and our school library, however, I can not find it. May I have the paper from you? Appreciate for your help.
  • wang
    Dear profvrr,
    can you send me a copy, i am interested in Reed an Muench method,because it is too early ,i can not search it in our library. thank you !
  • dmcilroy
    OK, after reading the classic paper, I think I have a (slightly) better understanding of why using cumulative numbers of dead mice/infected wells and cumulative alive mice/non-infected wells works. In the actual paper, the example they gave was protection of mice from infection by transfer of different dilutions of immune serum. However, to make this relevant to viral titres in cell culture, I will try to explain how I understood the paper as if the data were referring to a TCID50 determination .

    I found the figure much clearer than the text in explaining the use of cumulative numbers. If you plot the cumulative number of infected wells going from the most dilute to the least dilute, then you get a curve that goes down as the inoculum gets more dilute. Conversely, the curve for cumulative number of non-infected wells goes up as the inoculum gets more dilute. Where the two curves cross is the dilution that gives you the TCID50.

    However, they didn't use these two curves to calculate the TCID50. This was done using the Reed+Muench formula on the sigmoid curve of cumulative or pooled % of infected wells at each dilution. I guess that finding where you get 50% infected wells on this curve is identical to finding where the number of infected/non-infected wells is the same.

    OK, so now I understand why it's a valid approach. But what is the utility? Well if I understood correctly, using cumulative numbers is supposed to be more statistically robust, since the number of animals/cell culture wells analyzed at each dilution is quite low (typically 6 to 8). If this is correct, then the variance of a TCID50 measure using the Reed-Muench method correctly (i.e. with cumulative numbers) should be lower than if the Reed-Muench formula is applied directly to the observed proportion of infected wells at each dilution.

    It occurs to me that I could test this quite soon, as I will be getting a bunch of write-ups from students who did a TCID50 in the lab this semester, so I have a data set with several determinations on the same virus stock.

    DMc
  • FALEYE TEMITOPE .O .C
    DMc,
    IT'S OBVIOUS FROM YOUR COMMENT THAT YOU'VE DIGESTED THE CLASSIC OF REED AND MUENCH. PLEASE, CAN YOU SEND ME A COPY?
    faleyetope@yahoo.com
    ps, I'm a graduate student of virology in Nigeria. The logic behind the calculation still eludes me. Thanks.
  • dmcilroy
    No, I haven't. I think about the only paper I have ever read from that period was the Ellis and Delbruck paper on the single-step growth curve. (Even with that one, though, the credit goes to Alan Cann for putting it on the CD that came with his text book). If you could send me a copy, I would be much obliged. I'll even write back here to let you know if I understood it better this time.

    DMc
  • khemmie
    How about determining the LT50?
  • Jon huntoon
    do you know where I can get a free digital copy of Reed, L.J.; Muench, H. (1938). "A simple method of estimating fifty percent endpoints". The American Journal of Hygiene 27: 493–497?

    Thanks,
    JON
  • Jon, I've looked all over and can't find a copy, not even in my files.
    And our library just removed all its copies of old journals. Sorry.
  • Mayank
    Dear Sir,
    I thank you for posting this informative note in the website. May I request you to present ur views on my problem which i face while doing LD50 calculation using reed and muench method. I want to have number of organisms as LD50 units. The usual method given by Dr. reed gives the value of exact dilution of the seed which will be equivalent to 1 LD50 unit. Pls educate me how its possible to have these units in terms of number of microbes (pfus or CFUs).

    Thanx and reggards
    mayank
  • rama
    Mayank! Please send me your email address . My ID is rama@umn.edu
  • Mayank
    Dear Sir,
    I thank you for posting this informative note in the website. May I request you to present ur views on my problem which i face while doing LD50 calculation using reed and muench method. I want to have number of organisms as LD50 units. The usual method given by Dr. reed gives the value of exact dilution of the seed which will be equivalent to 1 LD50 unit. Pls educate me how its possible to have these units in terms of number of microbes (pfus or CFUs).

    Thanx and reggards
    mayank
  • It is straightforward to calculate the LD50 as the concentration of
    the stock virus times the 50% end point titer. For example, if we have
    a stock of virus with a titer of 1x10e9 pfu/ml, and the LD50 is
    calculated to be 10e-6.5, then by multiplying the two numbers we
    calculate that the LD50 is 3x10e2 PFU. The same may be done for CFUs.
  • mjhesami
    Well, it seems that the developed formula by Dr Rama, is easy and can work good! however I've not use it yet!
    As soon as test it I'll tel the result

    Good luck
  • Rama
    The statistical method of Reed and Muench is really a complicated one and one should have good mathematical skills to calculate TCID50. I developed my own formula (even layman can calculate TCID50 by my formula and it is 100% match with Reed and Muench or modified Karber formula). I am willing to share my formula if anybody interested. Please contact me by my email.

    Thanks, rama@umn.edu
  • khemmie
    Do you have formula in solving thr LT50?
  • Rama
    Do you mean LD50? Then Yes I have
  • nick6196
    Thanks for the formula--much appreciated.

    Nick
  • James
    Dr. Rama's formula is easy to understand and super easy to use. Thanks
  • mayank
    Dear Sir,
    I thank you for posting this informative note in the website. May I request you to present ur views on my problem which i face while doing LD50 calculation using reed and muench method. I want to have number of organisms as LD50 units. The usual method given by Dr. reed gives the value of exact dilution of the seed which will be equivalent to 1 LD50 unit. Pls educate me how its possible to have these units in terms of number of microbes (pfus or CFUs).

    Thanx and reggards
    mayank
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