Medium consumption model #iGEMgoesgreen

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 on 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 growth of the E. coli is not affected, instead for every new E. coli one E. coli is put to waste. $$\frac{\partial c_{E. coli}}{\partial t} = 0$$

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 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 are allowed to grow at their maximum speed under the given conditions. So their ability to produce phage as fast as possible is not impaired.

Here we can make statements on how much energy we save with a specific setup relative to another one, but we do not estimate the absolute amount of energy that is used.

Configure the plot:

"Resolution" controls how many datapoints are calculated in each dimension, therefore complexity of both memory and computation increased with the square of Resolution. Increase for a smoother heatmap and decrease for a faster result. Values between 20 and 500 are reasonable

"Flow trough Chemostat", \(\Phi_{chemostat}\), the amount of medium that is pumped trough the chemostat during one hour. Depends on turbidostat volume, lagoon volume and quantity, flow trough lagoon. A greater total lagoon volume with a higher flow requires greater flow through the chemostat. For turbidostats the flow is calculated using the volume of the turbidostat and the generation time of the E. coli.$$\Phi_{turbidostat} = V_{Turbidostat} * \frac{\ln{2}}{t_{E. coli}}$$

"Maximum Duration", \(t_{E. coli}\) controls the upper border of the y-axis.

"Generation Time E. coli" sets the time in which the E. coli population doubles itself when in exponential phase. Only affects values for turbidostats, because chemostats control the generation time themselves.

"Volume of Turbidostat", actually the volume of the culture in the turbidostat. Only affects Turbidostats.

Calculation of the medium used in a turbidostat: $$V_{medium} = t * \Phi_{turbidostat} = t * V_{turbidostat} * \frac{\ln{2}}{t_{E. coli}}$$

Calculation of the medium used in a chemostat: $$V_{medium} = t * \Phi_{chemostat}$$

Read the plot:

When working with a turbidostat the upper x axis is relevant, when working with a chemostat the lower is. Put the mouse pointer on the heatmap to show the exact values for that point. You can zoom in by drawing the rectangle that should be shown with pressed left mouse button. You can always save the current plot by clicking on the camera icon.

Caution

Submitting one of the forms recalculates all values and the heatmap based on the values in all the forms.

Your experiment

You can annotate a point in the heatmap by providing it's coordinates and it's name. If you have a turbidostat, the value for Flow is ignored, if you have a chemostat, the volume is ignored. For the calculation of the flow trough 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 $$V_{turbidostat} = b * V_{lagoon} * N_{lagoon} * \Phi_{lagoon}$$

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 %. For a turbidostat, the volume can be calculated from the flow using $$V_{turbidostat} = \Phi_{turbidostat} * \frac{t_{E. coli}}{\ln{2}}$$

The calculation is based on wether turbidostatat 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 coverd is insufficient to find variants that are better than previous ones.

There are a lot of possible ways to estimate the size of the lagoons, here we show one based on the sequence length and mutation rate.

The size of the phage population per lagoon is $$N_{phage} = c_{phage} * V_{lagoon},$$ the total sequence length is $$l_{total} = N_{phage} * l_{sequence},$$ the number of mutations that occur during one generation is $$\frac{N_{mutations}^{real}}{N_{generations}} = l_{total} * r_{mutation}$$ with \(N_{phage}\), the total number of phage in one lagoon, \(l_{total}\)total, the total sequence length and \(l_{sequence}\) the length of a single sequence. \(r_{mutation}\) is the mutation rate, \(N_{mutation}\) the number of mutations during one generation

The number of possible n-fold mutants of a sequence with length \(l_{sequence}\) can be calculated by $$N_{n-fold mutations}^{theoretical} = 3^{n} * r_{mutation}^n * \frac{l_{total}!}{(l_{total}-n)!},$$ 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 $$N_{n-fold mutations}^{real} = N_{phage} * (r_{mutation} * l_{total})^{n},$$ therefore the required lagoon volume is $$V_{lagoon} = a * \frac{N_{n-fold mutations}}{N_{n-fold mutations}^{real}}$$.