GATE 2025 BT Q.62: solution and answer

Official answer: 3.4 to 3.8Source: Official answer key, GATE 2025 BT (IIT Roorkee)

1. Dilution rate, where $F$ is the volumetric flow rate and $V$ the working volume: $D=\frac{F}{V}=\frac{0.8{\text{L h}}^{-1}}{2\text{L}}=0.4{\text{h}}^{-1}$.

2. At steady state in a chemostat with sterile feed ($X_0=0$), and with no cell death or maintenance, the biomass balance $0=-DX+\mu X$ with $X\ne 0$ gives specific growth rate = dilution rate: $\mu =D=0.4{\text{h}}^{-1}$.

3. Monod kinetics: $\mu =\frac{{\mu }_{max}S}{K_s+S}$, where ${\mu }_{max}=0.6{\text{h}}^{-1}$ is the maximum specific growth rate, $K_s=0.5{\text{g L}}^{-1}$ is the Monod constant and $S$ is the substrate concentration in the reactor (which equals the exit concentration in a well-mixed chemostat). Setting $\mu =D$ and solving for $S$: $$S=\frac{K_{s}D}{{\mu }_{max}-D}=\frac{0.5\times 0.4}{0.6-0.4}=1.0{\text{g L}}^{-1}$$

4. Washout check, with $S_0=10{\text{g L}}^{-1}$ the feed substrate concentration: $D_{crit}=\frac{{\mu }_{max}{S}_0}{K_s+S_0}=\frac{0.6\times 10}{0.5+10}=0.571{\text{h}}^{-1}$. Since $D=0.4{\text{h}}^{-1}<D_{crit}$, a steady state without washout exists.

5. Steady-state substrate balance, with $Y_{X/S}=0.4{\text{g g}}^{-1}$ the biomass yield on substrate (grams of cells formed per gram of substrate consumed): $$X=Y_{X/S}(S_0-S)=0.4\times (10-1.0)=3.6{\text{g L}}^{-1}$$

6. Exit biomass concentration $=3.6{\text{g L}}^{-1}$.