GATE 2024 BT Q.53: solution and answer

Official answer: CSource: Official answer key, GATE 2024 BT (IISc Bengaluru)

1. Biomass balance on a chemostat with sterile feed ($X_0=0$). Here $\mu$ is the specific growth rate (${\text{h}}^{-1}$), $D=F/V$ is the dilution rate ($F$ = volumetric feed rate, $V$ = culture volume) and $X$ is the biomass concentration: $$\frac{dX}{dt}=\mu X-DX=(\mu -D)X$$

2. Monod kinetics, with ${\mu }_{max}$ the maximum specific growth rate and $K_s$ the saturation constant: $\mu =\frac{{\mu }_{max}S}{K_s+S}$. With sterile feed the reactor substrate $S$ cannot exceed the feed concentration $S_0$, so the highest growth rate the culture can reach is at $S=S_0$. Washout therefore happens when $D\ge D_{crit}=\frac{{\mu }_{max}{S}_0}{K_s+S_0}$.

3. When $D>D_{crit}$, $\mu <D$ for every $S\le S_0$, so $dX/dt<0$. Cells leave faster than they grow, and the only steady state is $X=0$ (the washout steady state).

4. Substrate balance, with $Y_{X/S}$ the biomass yield on substrate (g biomass per g substrate): $$\frac{dS}{dt}=D(S_0-S)-\frac{\mu X}{Y_{X/S}}$$ With $X=0$ nothing consumes substrate, so at steady state $D(S_0-S)=0$, which gives $S=S_0$.

5. (A) is false: at complete washout the biomass in the reactor is zero, its minimum. (Biomass productivity $DX$ peaks at an optimum dilution rate below $D_{crit}$, not at washout.)

6. (B) is false: with no cells no substrate is used, so the exit concentration is not lower than the inlet. (D) is false: $S=0$ would mean all the substrate was consumed, the opposite of washout.

7. (C) is true: the exit substrate concentration equals the inlet value, $S=S_0$.