Introduction

Medium consumption model

PACE usually consumes an extraordinary amount of medium per experiment. This is due to the need for a continuous supply of host E. coli with a constant cell density. This can be achieved either by using a turbidostat or a chemostat. Here we provide a tool to both the community and ourselves to calculate medium consumption based on different tunable parameters of PACE. We also want to gain an understanding of how we can reduce the amount of medium needed for an experiment. Medium consumption is critical when it comes to the energy needed for an experiment because autoclaving needs a lot of energy. In a turbidostat the cell density is held constant by adjusting the medium influx to the cell density. That means the growht of the E. coli is not affected, instead for every new E. coli one E. coli is put to wate. In a Chemostat, the cell density is controlled by adjusting the influx of an essential nutrient to the cell density, which limits the growth of the culture. This may cause the E. coli to be less efficient in producing the proteins needed for PACE and replicating the phage genome. Additionally, there is a constant eflux from the chemostat to the lagoons, that compensates the growth of the E. coli. Currently, we are using a turbidostat, because probably PACE works better when the E. coli is allowed to grow at their maximum speed under the given conditions. So their ability to produce phage as fast as possible is not impaired. Calculation of the flow through a turbidostat: $$ \frac{\partial V_{M} }{\partial t} = \Phi_{T} = \frac{log(2)}{t_{E} } \cdot V_{T} $$

Your experiment

You can annotate a point in the heatmap by providing its coordinates \(t_{max}\) and \(\Phi_{T}\) or \(V_{T}\) and it's name. If you have a turbidostat, the value for the flow rate \(\Phi_{T}\) is ignored, if you have a chemostat, the volume \(V_{T}\) is ignored. For the calculation of the flow through the turbidostat, the value for generation time from the form above is used. Enter your specifications:

Minmal Turbidostat Volume

As larger turbidostats or chemostats with a larger flow need more medium for the same duration than smaller ones, working with the minimal required volume or flow is a way to save medium and thus energy. The minimal flow that is required can be calculated using $$ V_{T} = b \cdot V_{L} \cdot N_{L} \cdot \Phi_{L} $$ In case of fluctuations in the generation time of the E. coli it is crucial to have a buffer so that the turbidostat is not diluted when the culture grows slower. We currently use a buffer of 50 %, so \(b\) is set to \(1.5\). For a turbidostat, the volume can be calculated from the flow using $$ V_{T} = \Phi_{T} \cdot \frac{t_{E} }{log(2)} $$ The calculation is based on whether turbidostat or chemostat is picked above.

Minimal Lagoon Volume

Obviously smaller lagoons require smaller turbidostats or chemostats with a lower flow and are therefore saving medium. However it there is a lower limit to lagoon size, if the phage population is too small, the sequence space that can be covered is insufficient to find variants that are better than previous ones. Lagoon sizes used by other vary from 15 mlRN158 over 40 mlRN31 to 100 mlRN63. There are a lot of possible ways to estimate the ideal size of the lagoons, here we show one based on the sequence length and mutation rate. Alternatively, to adjust the size of the lagoons, it is possible to adjust the total duration of the experiment. But as that increases energy consumption for heating and stirring in addition to medium consumption, we decided to focus on the lagoons size. The size of the phage population \(N_{P}\) per lagoon is $$ N_{P} = c_{P} \cdot V_{L} $$ The total sequence length \(L_{T}\) is $$ L_{T} = N_{P} \cdot L_{S} $$ when \(L{S}\) is the length of one sequence. The number of mutations that occur during one generation \(N_{M}\) is $$ N_{M} = L_{T} \cdot r_{M} $$ Here \(r_{M}\) is the mutation rate. It is reported to be \(5.3 \cdot 10^{-7}\), or when increased by an induced mutagenesis plasmids \(5 \cdot 10^{-5}\) RN63. The number of possible n-fold mutants of a sequence with length sl can be calculated by $$ N_{n} = 3^{n} \cdot r_{M} \cdot \frac{L_{S}!}{(L_{S} - n)!} $$ as there are three possibilities for each basepair to be exchanged to and with each additional mutation there is one possible position less. The number of n-fold mutants that can occur in a lagoon can be calculated using $$ M_{n} = N_{P} \cdot (r_{M} \cdot L_{S})^n $$ therefore the required lagoon volume is $$ V_{L} = \frac{N_{n} \cdot t}{M_{n} } $$ With a theoretical coverage factor of the n-fold mutants of \(t\).