Chemostat calculation: Lactococcus lactis has a maximum specific growth rate of 1.23 h^-1 in a glucose–yeast extract medium. What is its specific growth rate at steady state in a 4 L reactor fed at 2 L h^-1 (assume steady state is feasible)?

Introduction / Context:
In a chemostat at steady state, the specific growth rate equals the dilution rate. This simple relationship allows rapid checks that an operating point is below washout and consistent with organism kinetics.

Given Data / Assumptions:

• Reactor working volume V = 4 L.
• Feed volumetric flow F = 2 L h^-1.
• Maximum specific growth rate mu_max = 1.23 h^-1.
• Steady state is achieved (i.e., D < mu_max).

Concept / Approach:
Dilution rate D = F / V. At steady state (no washout), μ = D. If D > mu_max, washout occurs and steady state with biomass cannot be maintained; otherwise μ tracks D exactly.

Step-by-Step Solution:
Compute D: D = 2 / 4 = 0.5 h^-1.Compare with mu_max: 0.5 h^-1 < 1.23 h^-1, so steady state with cells is feasible.Therefore, μ at steady state equals D = 0.5 h^-1.

Verification / Alternative check:
Steady-state mass balance on biomass: d(XV)/dt = (μ − D) X V = 0 ⇒ μ = D. Independent of Monod parameters as long as the reactor can supply that μ with available substrate.

Why Other Options Are Wrong:
1.2 h^-1: close to mu_max, but μ at steady state equals D, not mu_max.2.4 or 4.0 h^-1: exceed mu_max and the computed D; physically impossible under given conditions.

Common Pitfalls:
Mixing up μ with μ_max; forgetting to divide F by V; ignoring the washout constraint μ_max > D.

Final Answer:
0.5 h^-1