In a chemostat (continuous stirred tank bioreactor), when washout has occurred (cells removed with the effluent so that the culture cannot sustain growth), what happens to the concentrations of biomass (X), growth-limiting substrate (S), and product (P) at steady state?

Introduction / Context:
A chemostat (continuous stirred tank bioreactor) operates at a fixed dilution rate with continuous inflow of fresh medium and outflow of culture. “Washout” occurs when the dilution rate exceeds the culture’s net specific growth rate so that the reactor cannot retain cells. Understanding the fate of biomass (X), limiting substrate (S), and product (P) at washout is fundamental to bioprocess control and design.

Given Data / Assumptions:

• Perfect mixing (reactor contents equal effluent composition).
• Limiting substrate concentration in the feed is Sf.
• At washout steady state, cells cannot be maintained (X → 0).
• P is a growth-associated product unless otherwise stated.

Concept / Approach:
At steady state, mass balances link production/consumption to dilution. If cells are washed out, there is essentially no biomass to consume substrate or to form product. Therefore, S rises to the feed value Sf, X tends to zero, and P tends to zero for a growth-associated product (or to a low value if it is non-growth associated with negligible formation at X ≈ 0).

Step-by-Step Solution:

Verification / Alternative check:
Steady-state chemostat equations give X = Yx/s (Sf − S). At washout, X = 0 implies S = Sf. For growth-associated product P = Yp/x X (or Yp/s (Sf − S)); with X = 0 and S = Sf, P → 0.

Why Other Options Are Wrong:

• X, S, and P all zero: mixing ensures S equals Sf, not zero.
• Cannot be predicted: standard chemostat theory predicts these values.
• None of the above or Only P accumulates: contradict mass balance logic at washout.

Common Pitfalls:
Assuming a non-growth-associated product necessarily accumulates; without biomass, its formation ceases and dilution flushes it out. Confusing transient behavior with steady state is another error.

Final Answer:
X and P will be ~0 while S equals the substrate concentration in the feed