Chemostat washout: An E. coli strain has μ_max = 0.8 h^-1 on glucose. If the dilution rate D = 1.2 h^-1, what is the steady-state cell concentration in the reactor?
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AIt will increase above normal steady state
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BIt will be zero (washout occurs)
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CIt will change randomly around some mean
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DIt will decrease slightly but remain nonzero
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EIt will oscillate but not wash out
Answer
Correct Answer: It will be zero (washout occurs)
Explanation
Introduction / Context:In a chemostat, the dilution rate D (h^-1) sets the specific growth rate at steady state for a limiting substrate, provided D < μ_max. If D exceeds μ_max, cells cannot reproduce fast enough to balance outflow, leading to washout. This question evaluates recognition of the washout criterion.
Given Data / Assumptions:
- Monod kinetics; single limiting substrate.
- μ_max = 0.8 h^-1; dilution rate D = 1.2 h^-1.
- Well-mixed continuous stirred-tank reactor at steady input/output.
Concept / Approach:Steady state in a chemostat requires μ = D. Since μ ≤ μ_max, if D > μ_max, no positive biomass concentration can satisfy μ = D. The trajectory moves toward X → 0 as cells are washed out faster than they divide.
Step-by-Step Solution:
Compare D with μ_max: 1.2 h^-1 > 0.8 h^-1.Conclude μ cannot reach D at any feasible substrate concentration.Therefore, biomass decreases toward zero: washout.Verification / Alternative check:Phase-plane or mass-balance solutions show X* = 0 is the only steady state when D ≥ μ_max given standard Monod form.
Why Other Options Are Wrong:
- Increase/oscillate/random: contradict dynamical behavior under D > μ_max.
- Decrease but nonzero: impossible at steady state when D exceeds μ_max.
Common Pitfalls:Confusing transient overshoots with steady state; forgetting μ is bounded by μ_max.
Final Answer:It will be zero (washout occurs).