# Intro to Droplet Digital PCR

The utility of droplet digital PCR in gene quantification/copy number determination is becoming increasingly prevalent in industry. Its immediate benefits over traditional absolute quantitative PCR methods is hard to deny as it obviates the need to either generate rigorous standard curves or measure PCR efficiency of every gene pair to be measured. ddPCR delivers the same sensitivity as with conventional approaches with the added benefit of smaller errors. ddPCR relies on the Poisson algorithm and is essentially an end point measurement where the sample to be measured is randomly distributed into discrete partitions containing either none, one or more nucleic acid template copies. These individual partitions are then thermally cycled as in conventional PCR to their end point and then read to determine the fraction of partitions that are positive for the amplification.

Poisson distribution, essentially counting statistics, gives the probability Pr(n) that a droplet will contain n copies of target if the mean number of target copies per droplet is C. This can be written as: $Pr(n) = \frac{C^n . e^{-C}}{n!}$

So, the probability of getting a droplet with zero copies of the target gene is given by: $Pr(0) = e^{-C}$

In practice, C is defined as Copies Per Droplet (CPD) which essentially means the average number of copies per droplet. Take an instance where a typical 20 µl PCR reaction volume is divided up into 20,000 droplets each of 0.001 µl volume, then a CPD can be defined as either:

1. $CPD = \frac {total number of molecules}{total number of droplets}$ or
2. $CPD = (\frac {total number of molecules}{total reaction volume}) droplet volume (\mu l)$

So, assuming we had 100,000 molecules in a $20 \mu l$ reaction volume then by equation 1, we would have a CPD of 100,000 copies/20,000 droplets = 5 copies per droplet. Alternatively, using eq. 2, you would calculate (100,000 copies per $20 \mu l$) x 0.001 µl per droplet = 5 copies per droplet.

In other words, CPD can be thought of as the ‘expected value’ of the Poisson distribution. 100,000 copies (corresponding to a CPD of 5), as it so happens, is the limit of the dynamic range for this technique as CPD values beyond 5 lead to saturation.

So, why exactly do we need Poisson distribution anyway? Imagine, we had only 6 copies of a gene in a total reaction volume of $20 \mu l$. If the $20 \mu l$ were divided into 20,000 droplets, then it would be quite straightforward to find six droplets out of the 20,000 droplets that respond with a positive signal. In that case, we could simply count the number of positive droplets (6) out of 20,000 and know our concentration. But what happens when the concentration of the gene copy number is high enough to give droplets that might contain one or more copies of the gene. Here simply counting a positive signal as corresponding to a single copy of a gene would result in underestimation of the mean concentration. Some droplets might contain 2, 3, 4 or more copies and likewise, there might be more droplets that don’t contain any copies at all. Because some droplets can have more than one copy of the gene, Poisson statistics essentially tells us the true fraction of negative droplets which would otherwise be underestimated by simply taking an average value of positive droplets.

In these instances, the fraction of positive droplets is then used to calculate the concentration using the following equation: $Concentration (copies per volume) = \frac {-ln (\frac{N_{neg}}{N})}{V_{droplets}}$

Where $-ln (\frac{N_{neg}}{N})$ is the fraction of negative droplets and $V_{droplets}$ is the volume of droplets. This equation can likewise be rewritten as: $Concentration = \frac{-ln(1-p)}{V_{droplets}}$, where p is the fraction of positive reactions.

So, let’s take an example where a sample contains 5000 droplets in a $20 \mu l$ reaction volume partitioned into 20,000 droplets. This corresponds to a CPD of 5000/20000 = 0.25 copies per droplet. If the gene copies were evenly distributed amongst the 20,000 droplets, one might be tempted into thinking that 25% of the droplets contain one copy of the gene and that 75% of the droplets contain zero copies of the gene. However, Poisson statistics tells us to expect 78% of the droplets containing zero copies of the gene as follows: $Pr(0) = e^{-C} = e^{-0.25} = 0.7788 = 77.88%$, alternatively, for $Pr(1) = \frac {0.25^1 . e^{-0.25}}{1!} = 0.1947 or 19.5\%$ positive droplets with 1 copy of the gene rather than the expected 25%.

The mean and the variance of a Poisson distribution are the same $(\mu = \lambda and \mu^2 = \lambda)$. CV is expressed as standard deviation divided by the mean $(\frac{\sqrt\lambda}{\lambda})$.

The above concepts can be easily visualized using this interactive graphic below or view it here!