GATE 2026 BT Q.55: solution and answer

Official answer: 2.2 to 2.4Source: Official answer key, GATE 2026 BT (IIT Guwahati)

1. Monod model: $\mu =\frac{{\mu }_{max}S}{K_s+S}$, where $\mu$ is the specific growth rate, ${\mu }_{max}=0.1{\text{h}}^{-1}$ is the maximum specific growth rate, $K_s=1{\text{mg L}}^{-1}$ is the Monod constant and $S$ is the limiting substrate concentration.

2. Put $S$ and $K_s$ in the same units: $S=23{\text{g L}}^{-1}=23000{\text{mg L}}^{-1}$.

3. $$\mu =0.1\times \frac{23000}{1+23000}=0.1\times 0.99996=0.099996{\text{h}}^{-1}$$ Because $S\gg K_s$, the culture grows at essentially ${\mu }_{max}$.

4. Convert to per day so it can be combined with the decay rate, which is given per day: $\mu =0.099996\times 24=2.3999{\text{day}}^{-1}$.

5. With endogenous decay, the net specific growth rate is ${\mu }_{net}=\mu -k_d$, where $k_d=0.1{\text{day}}^{-1}$ is the endogenous decay rate.

6. $${\mu }_{net}=2.3999-0.1=2.2999\approx 2.3{\text{day}}^{-1}$$