Some experiments measured the effect of particular chemicals on the infectivity of lytic phages in terms of influence on the PFU/mL counts.
For example, the reduction of PFU/mL of exposed phages over controls might be 2 logs (let's say from 1000 PFU/mL in controls to 10 PFU/mL in exposed). If I knew the infection rate of the control phage (let's say 1), how can I calculate the infection rate of the exposed phages?
Thank you
I'm a little confused by what you are asking so my apologies if my answer does not fit your question. But remember that probability of infection will follow a Poisson distribution, so at an MOI of 1.0 you will have about 70% (rounded) being infected by 1 phage particle. About 15% will be infected by more than 1 phage and about 15% by zero. So at an MOI of 1 you expect around 15% of the cells to not get infected and survive. At an MOI of 10 you will still expect some uninfected cells, but only a few percent. So an MOI of 10 will probably kill by about 2 logs, but you need to go higher yet for a 3 log killing.
This is of course assuming you assay your CFU before 1 phage life cycle, once the cells lyse then the number of phage will go up dramatically. So you need to infect, wait a few minutes and prepare dilutions for your CFU count.
The other parameter will be the concentration of phage and cells. So 10e9 phage added to 10e8 cells is likely to give you a different answer than 10e4 phage and 10e3 cells due to the lower probably of the phage absorbing to the cells at low density.
Hello. In essence, the literature reports that some substances reduce the infectivity of phages (or viruses, in general). But the only measure of the reduction is a lower plaque count. My problem is how to relate such counts to a reduction in infection rate.
Sorry, in the post I erroneously typed CFU instead of PFU. Typos corrected.
As I said, by definition a plaque count (PFU) takes into account infectivity. The PFU is not the actual number of phage in a stock, but it is the number of phage able to infect and form a plaque in the stock being tested. So it may be that 100% of the phage will infect in which case the PFU is the same as the phage number. Or it might be that only 10% can infect under your conditions, in which case the PFU will be 10% of the total actual number of phage. Remember that it requires an infection to get a plaque.
So if you have a compound that reduces infectivity (perhaps by reducing binding to the receptor, or by inhibition injection or some other way of blocking the phage life cycle), this compound will show a decrease in apparent phage titer.
You can of course also measure the extent to which a given phage stock will kill cells (reduce CFU) in the presence or absence of the compound. That is much harder to measure accurately because of the statistics involved, but should none the less correlate with reduction in PFU. But both are in essence measuring the same thing.
On the other hand, if you have a compound that is reducing the phage burst size, then you need to measure that in a different way. But that would not be changing infectivity except maybe in a global sense.
I checked the papers: none reported the actual MOI, they just used a certain amount of phage (one million to one billion particles) against bacteria grown overnight in a double agar assay. That is all.
That is because when doing a plaque assay (double agar) the MOI is very very low as to be not relevant, ie 1/1000 or 1/10,000 or lower. But with a liquid infection like a one step growth curve then you can have adjust your MOI and it can for example be from 0.1 to 10. My point was you can measure reduction of infectivity in different ways but I suspect you will get similar results from an inhibitory compound if the treatments are similar. And PFU is certainly the easiest to measure.
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