Question

Formation of lactic acid from glucose is realized in a continuous culture by Streptococcus lactis. The following information was obtained from experimental studies. $$ \begin{aligned} &S_{0}=5 \mathrm{~g} / 1, \mu_{m}=0.2 \mathrm{~h}^{-1}, K_{S}=200 \mathrm{mg} / 1, k_{d}=0.002 \mathrm{~h}^{-1}, Y_{X S}^{M}=0.4 \mathrm{~g} X / \mathrm{g} S, Y_{P / S}=0.2 \mathrm{~g} P / \mathrm{g} S, \\ &q_{P}=0.1 \mathrm{~g} P / \mathrm{g} X-\mathrm{h} . \end{aligned} $$ a. Plot the variations of \(S, X, P, \mathrm{DX}\), and DP with dilution rate. b. Determine (graphically) the optimum dilution rate maximizing the productivities of biomass (DX) and the product (DP).

Short answer

Use all rate relevant data, combine using cultivated formula’s graphical techniques, then determine optimal dilution around trial intervals.

Step-by-step solution

Review Given Information

First, gather and understand the given data:

Establish Formulae

List relevant formulae:- Monod equation: \[ \mu = \frac{\mu_mS}{K_S + S} \] - Specific growth rate with decay: \[ \text{Effective} \mu = \mu -k_d \] - Biomass productivity: \[ DX = Y_{X/S}^{M} \cdot (S_0 - S) \cdot D \] -Productivity of product: \[ DP = Y_{P/S} \cdot (S_0 - S ) \cdot D \] -Alternate product yield: \[ q_P \cdot X \] Where S-Substrate concentration, X-Biomass concentration and P-Product concentration

Define Dilution Rate

Dilution rate is defined as \[ D = \frac{F} {(V} \] Flowrate over Volume. Also equilibrium occurs when \mu_eff = D

Create Variations Equations

Write differential form equations: -Substrate \frac{dS}{dt} : \[ S_o - S -\frac {-\mu_s DX} \cdot X \]

Replacement and Calculation PM and PB

substitute in (3),(4) with data given and create necessary plots.

Obtain Other Values and Assemble Graphs

From previous separate single variable calculate ablative figurine DX and corresponding DP based rest give formaula at dilution values

:Optimalewline dilution Rate

By Graphical Inspect intervals at action limit and corresponding output.(tile 2 incls

Key concepts

Continuous Culture

Continuous culture refers to a method where fresh nutrient medium is continuously added to a bioreactor, while culture liquid is continuously removed to keep the culture volume constant. This method is used to grow microorganisms or cells at a steady state.

• Allows sustained growth and constant biomass production.
• Enables researchers to study microbial growth and product formation under steady-state conditions.
• Commonly used in industrial fermentation processes.

In the context of the given exercise, the continuous culture facilitates the conversion of glucose into lactic acid using *Streptococcus lactis*. To maintain a continuous culture, it is crucial to balance the rate of nutrient addition and the rate at which culture is removed. This balance is quantified using the concept of dilution rate, which helps in maintaining the desired concentration of cells and substrate.

Monod Equation

The Monod equation helps in understanding the relationship between microbial growth rate and substrate concentration. It's an essential tool in microbial kinetics:

• Equation: \( \frac{\text{mu}}{{K_s + S}} \), where \mu\ is the specific growth rate, \(S\) is the substrate concentration, and \(K_S\) is the half-saturation constant.
• Analogous to the Michaelis-Menten equation in enzyme kinetics.
• It characterizes how growth rate depends on the concentration of a limiting nutrient.

In this exercise, we use the Monod equation to determine the effective growth rate of *Streptococcus lactis*. By knowing \(\mu_m\), \(S\), and \(K_S\), we can assess the growth dynamics and adjust the parameters in the continuous culture to optimize biomass and product formation.

Biomass Productivity

Biomass productivity, expressed as DX, measures the amount of biomass produced per unit time in a continuous culture system:

• Formula: \( DX = Y_{X/S}^{M} \cdot\ (S_0 - S) \cdot\ D \), where \(Y_{X/S}^{M}\) is the yield coefficient of biomass from the substrate.
• Represents the conversion efficiency of substrate into new cell mass.
• Key parameter in optimizing continuous fermentation processes.

In the context of the given exercise, we need to plot the variations of DX with dilution rate. This helps in identifying the dilution rate at which biomass productivity is maximized. A high biomass productivity indicates efficient substrate conversion and effective microbial growth.

Dilution Rate

Dilution rate (D) is a pivotal concept in continuous culture, representing the rate at which fresh medium is added to the bioreactor relative to the culture volume:

• Formula: \( \frac{F}{V} \), where F is the flow rate of the incoming medium and V is the culture volume.
• Affects the substrate concentration and the specific growth rate of microorganisms.
• Critical in maintaining steady-state conditions.

In our exercise, the dilution rate directly impacts biomass and product productivity (DX and DP, respectively). By plotting DX and DP against various dilution rates, we can identify the optimal dilution rate that maximizes productivity. Understanding how dilution rate influences culture parameters is essential for designing efficient continuous bioprocesses.