GATE 2019 BT Q.43: solution and answer

Official answer: DSource: Official answer key, GATE 2019 BT (IIT Madras)

1. P (true). An ideal chemostat is a perfectly mixed continuous stirred-tank reactor, so the composition is the same at every point in the vessel and the outlet stream is a sample of the vessel contents. Therefore $S_{\text{inside}}=S_{\text{exit}}$.

2. Setting up Q and R. Take Monod kinetics, $\mu =\frac{{\mu }_{max}S}{K_s+S}$. Here $\mu$ is the specific growth rate, ${\mu }_{max}$ is its maximum value and $K_s$ is the saturation constant. At chemostat steady state $\mu =D$, where $D=F/V$ is the dilution rate. Solving for the substrate concentration gives $S=\frac{K_{s}D}{{\mu }_{max}-D}$, and the substrate balance gives $X=Y_{X/S}(S_0-S)$. In these, $S_0$ is the feed substrate concentration, $X$ is the biomass concentration and $Y_{X/S}$ is the biomass yield on substrate.

3. Critical (washout) dilution rate. Washout happens when no substrate is used up, i.e. $S\to S_0$. Putting $S=S_0$ into $S=\frac{K_{s}D}{{\mu }_{max}-D}$ gives $$D_{crit}=\frac{{\mu }_{max}{S}_0}{K_s+S_0}={\mu }_{max}(1-a),a=\frac{K_s}{K_s+S_0},0<a<1$$

4. Optimal dilution rate. This is the $D$ at which biomass productivity $DX=Y_{X/S}D\left(S_0-\frac{K_{s}D}{{\mu }_{max}-D}\right)$ is highest. Setting $\frac{d(DX)}{dD}=0$ gives $S_0=\frac{K_{s}D(2{\mu }_{max}-D)}{({\mu }_{max}-D{)}^2}$. Now put $u={\mu }_{max}-D$, so that $D(2{\mu }_{max}-D)={\mu }_{max}^2-u^2$. The condition becomes $u^2(S_0+K_s)=K_s{\mu }_{max}^2$, so $u={\mu }_{max}\sqrt{a}$ and $$D_{opt}={\mu }_{max}\left(1-\sqrt{a}\right)$$

5. Q (true). When $0<a<1$, $\sqrt{a}>a$, so $D_{opt}<D_{crit}$ for every choice of $K_s$, $S_0$ and ${\mu }_{max}$. There is also a direct argument: productivity $DX$ is zero at $D=0$ and zero again at washout ($X=0$), and positive in between. Its maximum must therefore come before $D_{crit}$.

6. R (false). $S=\frac{K_{s}D}{{\mu }_{max}-D}$ goes up as $D$ goes up. So $X=Y_{X/S}(S_0-S)$ goes down as $D$ increases and reaches $X=0$ at $D_{crit}$. The statement is not always correct.

7. S (true). Suppose a fraction $\alpha$ of the feed flow is recycled, carrying cells at $\beta X$ with concentration factor $\beta >1$. The steady-state biomass balance on the vessel is $\alpha F\beta X-(1+\alpha )FX+\mu XV=0$, which gives $\mu =D(1+\alpha -\alpha \beta )<D$. The cells therefore only need to grow more slowly than the dilution rate. Washout now occurs only when $D>\frac{D_{crit}}{1+\alpha -\alpha \beta }$, so a steady state can be kept above the simple chemostat's $D_{crit}$.

8. P, Q and S are always correct and R is not, so the answer is option (D).