Competition in a Chemostat — Two species compete for a single limiting substrate: which organism avoids washout at steady state?

Introduction:
In a chemostat with a single limiting substrate, species compete through resource depletion. Classic theory predicts competitive exclusion: the species capable of reducing the substrate to the lowest steady-state concentration wins, provided both can grow at the imposed dilution rate.

Given Data / Assumptions:

• One limiting substrate in the feed; constant D.
• Both organisms follow Monod-type kinetics with different μm and Ks values.
• No other interactions (toxins, inhibition, syntrophy) dominate.

Concept / Approach:

At steady state, each species requires a critical substrate concentration Sc such that μ(Sc) = D. The species with the lower Sc (higher effective affinity μ/Ks combination at the given D) drives the bulk substrate toward that lower level, disadvantaging competitors that require higher substrate to sustain μ ≥ D. Those competitors wash out because μ < D in the environment created by the winner.

Step-by-Step Solution:

Verification / Alternative check:

Experimental chemostat studies show single-species dominance on a single limiting carbon source unless special trade-offs or spatial niches exist.

Why Other Options Are Wrong:

A and D: Higher residual S implies weaker competitive performance. C: Coexistence on a single resource is not generic. E: Cell size is irrelevant without kinetic advantage.

Common Pitfalls:

Confusing μm alone with competitive ability; at modest D, low Ks can dominate even if μm is lower.

Final Answer:

The species that can maintain the lowest residual substrate concentration (i.e., has the higher substrate affinity under the given D).