This question is probably mostly for statisticians, but here we go. It is easy enough to serially dilute DNA (or cDNA, RNA...) to a theoretical concentration of 1 copy per uL of water. My question is "If you pipette 1uL of this solution, what is the probability that it will contain 1 copy of template". From a simplistic standpoint, one could say that 100uL of this solution would contain 3x10^21 water molecules and 100 template molecules. Pipetting 1uL would be equivalent to pipetting 3x10^19 molecules. What are the chances that 1 of these molecules is template? Along this line of thinking, it should be possible to calculate how many uL of solution would give 95% confidence having at least one template molecule.
I realize that this is simplistic and relative molecular size, molecular attraction and such must be taken into account but it could give us an idea... Anyone have any thoughts on this?
The answer is given by the Poisson distribution.
http://en.wikipedia.org/wiki/Poisson_distribution
There is not much to add to explain. The relevant part in the article is already in the first paragpraph:
"[T]he Poisson distribution [...] expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate [...]"
plus the indepencence assumptions.
If you have a solution with an average of 1 copy per aliquot, then the average "rate" of amplicons in an aliquot is 1. The Posson distribution allows to calculate the probability of any given number k of molecules in an aliquot based on this rate-parameter (λ, greek small lambda). The formula is
P(k|λ) = λ^k/k! * exp(-λ)
To answer your question for k=1 and λ=1 we can plug in these values and get
P(k=1|λ=1) = 1^1/1! * exp(-1) = exp(-1) = 0.368 = 36.8%.
This principle is used for instance in a technique for quantitative PCR, the limiting dilution technique. From the Poisson distribution it is known that the probability for a PCR not containing any amplifiable template (k=0) is
P(k=0|λ) = λ^0/0! * exp(-λ) = exp(-λ)
what should be estimable from the proportion of negative PCRs (of ideal sensitivity for sure!). When I determine this proportion (p), I know that
exp(-λ) = p λ = ln(1/p) = -ln(p)
So if I have a sample and find that in a 1:1E5-dilution, 20 of 100 PCRs remained negative, I can conclude that the rate parameter in the dilution is
-ln(20/100) = 1.61
so the average number of copies per aliquot (say, 1µl, as added to the PCR) was 1.61. Accounting for the used dilution the expected concentration is 1.61E5 copies per aliquot.
The answer is given by the Poisson distribution.
http://en.wikipedia.org/wiki/Poisson_distribution
There is not much to add to explain. The relevant part in the article is already in the first paragpraph:
"[T]he Poisson distribution [...] expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate [...]"
plus the indepencence assumptions.
If you have a solution with an average of 1 copy per aliquot, then the average "rate" of amplicons in an aliquot is 1. The Posson distribution allows to calculate the probability of any given number k of molecules in an aliquot based on this rate-parameter (λ, greek small lambda). The formula is
P(k|λ) = λ^k/k! * exp(-λ)
To answer your question for k=1 and λ=1 we can plug in these values and get
P(k=1|λ=1) = 1^1/1! * exp(-1) = exp(-1) = 0.368 = 36.8%.
This principle is used for instance in a technique for quantitative PCR, the limiting dilution technique. From the Poisson distribution it is known that the probability for a PCR not containing any amplifiable template (k=0) is
P(k=0|λ) = λ^0/0! * exp(-λ) = exp(-λ)
what should be estimable from the proportion of negative PCRs (of ideal sensitivity for sure!). When I determine this proportion (p), I know that
exp(-λ) = p λ = ln(1/p) = -ln(p)
So if I have a sample and find that in a 1:1E5-dilution, 20 of 100 PCRs remained negative, I can conclude that the rate parameter in the dilution is
-ln(20/100) = 1.61
so the average number of copies per aliquot (say, 1µl, as added to the PCR) was 1.61. Accounting for the used dilution the expected concentration is 1.61E5 copies per aliquot.
Thanks Jochen - that makes perfect sense and is actually much easier to understand than what I was originally thinking.
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