GATE 2024 BT Q.60: solution and answer

Official answer: 0.1Source: Official answer key, GATE 2024 BT (IISc Bengaluru)

1. At quasi-steady state in a fed-batch culture, the total biomass keeps rising but the biomass concentration $X$ and the substrate concentration $S$ stay constant. Growth has to exactly offset the dilution caused by the incoming feed, so the specific growth rate $\mu$ equals the instantaneous dilution rate: $\mu =D=F/V$, where $F$ is the feed flow rate and $V$ is the culture volume.

2. Dilution rate at $t=2\text{h}$, where $V=500\text{L}$: $D=\frac{F}{V}=\frac{50{\text{L h}}^{-1}}{500\text{L}}=0.1{\text{h}}^{-1}$.

3. Monod kinetics, with ${\mu }_m$ the maximum specific growth rate and $K_S$ the saturation constant (the substrate concentration at which $\mu ={\mu }_m/2$): $\mu =\frac{{\mu }_{m}S}{K_S+S}$.

4. Set $\mu =D$ and solve for $S$: $D(K_S+S)={\mu }_{m}S$, so $$S=\frac{K_{S}D}{{\mu }_m-D}$$

5. Substitute: $$S=\frac{(0.1{\text{g L}}^{-1})(0.1{\text{h}}^{-1})}{0.2{\text{h}}^{-1}-0.1{\text{h}}^{-1}}=\frac{0.01}{0.1}{\text{g L}}^{-1}=0.1{\text{g L}}^{-1}$$

6. Quick check: here $D={\mu }_m/2$, so by the definition of $K_S$ the answer has to be $S=K_S=0.1{\text{g L}}^{-1}$. The feed glucose concentration $S_0$ is not given and is not needed. At quasi-steady state it sets the biomass level ($X\approx Y_{X/S}{S}_0$, with $Y_{X/S}$ the biomass yield on substrate), not $S$.