GATE 2020 BT Q.45: solution and answer

Official answer: 526.01 to 529.99Source: Official answer key, GATE 2020 BT (IIT Delhi)

1. Let $X$ be the biomass concentration and $\mu$ the specific growth rate in the log phase, $\mu =\frac{1}{X}\frac{dX}{dt}=0.00417{\text{min}}^{-1}$. Split the total time into two parts. During the lag phase, $t_{lag}=30\text{min}$, the cells adapt but do not multiply, so the biomass stays at $X_0=1{\text{g L}}^{-1}$. After that, the cells grow exponentially at the constant rate $\mu$.

2. Exponential (log-phase) growth: $\frac{dX}{dt}=\mu X$. Integrating from $X_0$ at the end of the lag phase gives $X=X_0{e}^{\mu t_{exp}}$, so $t_{exp}=\frac{1}{\mu }\ln \frac{X}{X_0}$.

3. Substitute $X=8{\text{g L}}^{-1}$ and $X_0=1{\text{g L}}^{-1}$: $$t_{exp}=\frac{\ln 8}{0.00417{\text{min}}^{-1}}=\frac{2.07944}{0.00417}=498.67\text{min}$$

4. Check: $8=2^3$, so the biomass doubles three times. The doubling time is $t_d=\frac{\ln 2}{\mu }=\frac{0.693147}{0.00417}=166.22\text{min}$, and $3t_d=498.67\text{min}$, which agrees.

5. Add the lag phase: $$t_{total}=t_{lag}+t_{exp}=30+498.67=528.67\text{min}$$