Continuous inoculation of slower organisms

Computer simulation of a chemostat shows that slower growing organisms can persist and even dominate if inoculated continuously. The fraction of slower organisms to be fed to the chemostat for dominance depends on the Monod coefficients and on the dilution rate. Practical reasons for controlling domination in a chemostat are:
  • minimizing the hazard of contamination,
  • controlling proportions in a synergistic fermentation,
  • creating a possibility of non-aseptic operation.

    Consider a two-stage system with a pure culture in the first stage and a mixture of two organisms in the second stage. We start with the same equations developed earlier for competition and add an input term for the organisms from the first stage. If we let µa = D, the equations can be solved for A and S at any specified D. Note that the A equation will not have any additional terms when a second organism, B, is also in the continuous culture, thus µa will still equal D whenever steady state is reached with A present. This is a simple but profound concept that make analysis of the mixed culture quite easy. When B is added continuously, the B equation becomes:

    dB/dt = D Ib - D B + µb B
    and a Monod equation for B is:

    When D is specified, S can be calculated from the Monod equation for organism A. Substituting this value for S in Monod equation defines µb, and the dB/dt equation can be solved knowing Ib to give:

    B = Ib / ( D - µb )
    With S and B known, A can be calculated using a rearranged form of the S equation:
    A = Y So - D S Y - Y µb B / (D Yb)
    In other words, µa and S are fixed by the property that steady state is reached with A present, and B and A are calculated from additional mass balance equations.

    Let us consider the conditions at which A approaches zero. The above equation set to zero and with Y factored out gives:

    So - D S - µb / (D Yb = 0
    After substituting the known terms, simplifying, and collecting terms, we get an equation in S squared, S, and a constant. The quadratic formula is used to find S. We now have the system defined, and solutions are shown by the following applet:
     

    Note that feeding B causes A to reach washout before D equals the maximum µ. When there is a small difference in specific growth rates, A is washed out at relatively low dilution rates when any appreciable amount of B is fed. When A is much the faster, B can predominate only if large amounts are fed.

    Endogenous metabolism was not considered, but the graphs of B and A versus D would be affected only at low dilution rates when reasonable values of maintenance energy coefficients are tested.

    Another practical situation of importance is when a slower-growing organism can persist because it adheres to surfaces in the vessel. The mathematical analysis is almost the same as above except for area terms and growth equations based on films.

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  • Competition in a pH Auxostat
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