To maintain a succesful continuous PACE experiment persisting contaminations of the lagoons with foreign bacteria or funghi, which would harm or interfere with the host E. coli strain have to be avoided. Therefore the question arose, whether a unwanted microorganism would propagate inside the lagoons in a way that it would persist and displace the host strain. Since a cancelled PACE run due to suspected contamination can be very costly an simulation of contamination growth would be vital to save precious resources. If the constant dilution rate of the lagoons is sufficiently high that the overall change in population size of the contamination is negative, a contamination of suspended cells would likey wash out and wouldn't be a lasting problem.

Theory

Note: contamination with viruses such as bacteriophages is not accounted for, because only exponential growth of the contamination is assumed in this model. This is plausible for bacteria and funghi since the lagoons are constantly diluted so that there is always fresh medium and space available. The change in concentration of the contamination can be described as $$ \frac{\partial c_{X}(t)}{\partial t} = -\Phi_{L} \cdot c_{X}(t) + \frac{\ln{2} }{t_{X} } \cdot c_{X}(t) $$ The concentration of the contamination \(x_{X}\) is assumed to only depend on the time, since the constant dilution reduces effects such as spent resources or growth inhibition by waste products. The growth factor \(\frac{ln(2)}{t_{X} }\) is the factor by which the current concentration and its derivative are proportional. There is the dilution term containing the factor \(\Phi_{L}\) that is the flow rate trough the lagoon in volumes per hour and the growth term of the contamination based on the contaminations generation time \(t_{X}\). Solving this equation yields $$ c_{X}(t) = c_{X}(t_{0}) \cdot e^{\big(\frac{ln(2)}{t_{X} } - \Phi_{L}\big) \cdot t} $$ Here \(c_{X}(t_{0})\) is introduced, which is the initial concentration of the contamination.
When working with the model it is more practical to calculate relative changes in the concentration: $$\frac{c_{X}(t)}{c{X}(t_{0})}$$ That simplifies the above equation to $$ c_{X}(t) = e^{\big(-\Phi_{L} + \frac{\ln{2} }{t_{X} }\big) \cdot t} $$ Differentiation after \(t\) gives $$ \frac{\partial c_{X}(t)}{\partial t} = \Big(\frac{ln(2)}{t_{X} } - \Phi_{L}\big) e^{\big(\frac{ln(2)}{t_{X} } - \Phi_{L}\Big) \cdot t} $$ The following statements can be made:
The contamination expands if the flow rate \(\Phi_{L}\) is lower than the growth factor, $$ c_{X}(t) > 0, \quad if \quad \Phi_{L} < \frac{ln(2)}{t_{X} } $$ it remains constant if the flow rate is exactly the growth factor $$ c_{X}(t) = 0, \quad if \quad \Phi_{L} = \frac{ln(2)}{t_{X} } $$ and it diminishes, when the flow rate is higher than the growth factor. $$ c_{X}(t) < 0, \quad if \quad \Phi_{L} > \frac{ln(2)}{t_{X } } $$

Practice

All PACE experiments we planned used lagoon flow rates \(\Phi_{L}\) between \(0.8\) and \(1.5\) volumes per hour. We assumed that the contaminations have generation times greater than \(20\) minutes.

Conclusion

Contaminations that have a generation time below 40 min are theoretically able to grow in the lagoons. That does not apply to most funghi even if there are some with generation times below one hourtrinci1969, so the main concern remaining is bacteria. The host E. coli for PACE experiments carry two or three plasmids with different resistance genes, to which the corresponding antibiotics are added to the medium. The combination of multiple antibiotics makes bacterial contamination highly unlikely, since the contaminating bacteria would need a defined set of resistance genes in order to grow in one of the lagoons used.
Since bacteriophages are not covered by this model, they remain as the main contamination risk.