Maximum possible virus titers.

[Marcel Schouwenburg]

On 4-AUG-1995, poets at ccmail.orst.edu  "Steven Poet" wrote:

> That seems like a pretty high dilution to me.  Are you sure your decimal
> points are all in the right place?  I remember somewhere in the depths
> of my memory that 10^17 poliovirus particles will fill a ping pong ball.
> So I don't think 10^23 virus particles can be suspended in 1 ml of
> cell culture medium.  I understand the difference between TCID50 and
> absolute particle numbers, I'm just thinking intuitively.
>
> krjk1 at stirling.ac.uk (Mr K R John Kollanoor) wrote:
>
>> Is there a theoratical maximum for the number of virus progeny a cell 
>> culture supernatant can have? 
>> Is there any possibility of a titer of 10^-23(TCID50/ml) in any of the virus 
>> infected cell culture supernatant ?
>> I would appreciate if someone could answer these questions.


You're right with this, but when you only look to the number: 10^23 you must
realise that this is so high it must be wrong. If you had that much polio
virus particles, you would be able to lift 1 ml of the supernatant of the
culture. I will give the calculation:

- Let us only consider the RNA of the virus and take 7500 nt for the length.
- I assume an average weight of 310 Da for a nt.
- 1 Da = 1.65 * 10^-24 g.

If you calculate the weight for 10^23 virus particles, you'll get:

  10^23 vp * 7500 nt/vp * 310 Da/nt * 1.65 * 10^-24 g/Da = 383625 g
							   ========
(vp = virusparticles)

I think this makes clear that 10^23 is a very unlikely titer, or am I doiing
something wrong or making the wrong assumptions?
Does anybody have any comments on this?

Kind regards,

Marcel

                       "mens sana in corpore sano"
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| Marcel Schouwenburg             | Besides working as a Radiation Safety     |
| Health Physics Department       | Officer, I'm a student in both            |
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| Delft University of Technology  | Faculty of Natural Sciences               |
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