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?

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:

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).