Status: brief note after discussion with Mike and Charlie aiming to resolve confusion or disagreement about wastewater pooling.

Pooled concentration

Consider that we have two planes indexed by \(f = 1, 2\) (where \(f\) stands for “flight”). Let \(A_{I, f}\) be the abundance of target pathogen, \(A_{U, f}\) be the amount (or abundance) of other nucleic acid and \(V_{f}\) be the total volume of wastewater from each flight.

Imagine that all of the wastewater from the flights is pooled together. Then the pooled amounts1 of pathogen and other nucleic acid are \[\begin{align} A_{I} &= \sum_{f = 1, 2} A_{I, f}, \\ A_{U} &= \sum_{f = 1, 2} A_{U, f}, \end{align}\] and the total volume of wastewater is \[ V = \sum_{f = 1, 2} V_{f}. \] The pooled concentration of the pathogen and other nucleic acids in are \[ C_{I} = \frac{A_{I}}{V_{f}}, \\ C_{U} = \frac{A_{U}}{V_{f}}. \]

Equivalent sampling approaches

Suppose we sample a fraction \(f \in (0, 1)\) of the pooled wastewater. The expected amounts of pathogen and other nucleic acid are simply \(f A_{I}\) and \(f A_{U}\). The expected concentrations remain unchanged as we both amount and volume are multiplied by \(f\), which cancels.

Note that we expect the proportion of each wastewater source in the sample to be the same as it is in the whole pool. To see this: if you have a gin and tonic made up of 1/4 gin and 3/4 tonic, then every sip you expect there to be (about) 1/4 gin and 3/4 tonic

This means that there are (at least) two equivalent ways that we could arrive at these pooled equations physically:

  1. All the of the wastewater is deposited into a holding tank, we assume that it is mixed together, then we take a sample with fraction \(f\)
  2. The wastewater from each plane is sent into a pipe, which operates such that the pipe separates a fraction \(f\) of each plane’s wastewater into a separate tank

Similarly, if you wanted a sip of gin and tonic you could either make up a whole drink and take a sip, or extract a sips worth of gin and tonic separately and then mix them together – it’s the same thing.

Sequencing bias

The number of mapped reads from the pathogen of interest \(M_{I, f}\) for a given flight \(f\) is given by \[ M_{I, f} = \frac{A_{I, f} B_{I, f}}{A_{I, f} B_{I, f} + A_{U, f}} \times M_{\text{tot}, f}, \] where \(B_{I, f}\) is the differential sequencing efficiency the pathogen of interest as compared with everything else on flight \(f\), which we consider to be \[ B_{I, f} = \frac{E_{I, f}}{E_{U, f}}. \] If the wastewater is pooled together, then weighted sequencing efficiencies are given by \[ E_I = \frac{\sum_{f = 1, 2} A_{I, f} E_{I, f}}{\sum_{f = 1, 2} A_{I, f}}, \\ E_U = \frac{\sum_{f = 1, 2} A_{U, f} E_{U, f}}{\sum_{f = 1, 2} A_{U, f}}, \] and the weighted differential sequencing efficiency is given by \[ B_I = \frac{E_I}{E_U}. \]

The number of mapped reads from the pathogen of interest \(M_{I}\) for the pooled wastewater is then given by \[ M_{I} = \frac{A_{I} B_{I}}{A_{I} B_{I} + A_{U}} \times M_\text{tot}, \]

  1. For brevity I do not use any notation to signify “pooled”. In a later version of this, you might want to add such an indicator.↩︎