Question

A batch fermenter receives 11 of medium with \(5 \mathrm{~g} / \mathrm{l}\) of glucose, which is the growth-ratelimiting nutrient for a mixed population of two bacteria (a strain of E. coli and Azotobacter vinelandii). A. vinelandii is five times larger than \(E\). coli. The replication rates for the two organisms are: $$ \mu_{E C}=\frac{1.0 \mathrm{~h}^{-1} \mathrm{~s}}{0.01 \mathrm{~g} / \mathrm{l}+\mathrm{s}}-0.05 \mathrm{~h}^{-1} $$ and $$ \mu_{A V}=\frac{1.5 \mathrm{~h}^{-1} \mathrm{~s}}{0.02 \mathrm{~g} / \mathrm{l}+\mathrm{s}}-0.10 \mathrm{~h}^{-1} $$ The yield coefficients are: $$ \begin{aligned} &Y_{E C}=0.5 \mathrm{~g} \mathrm{dw} / \mathrm{g} \text { glucose } \\ &Y_{A V}=0.35 \mathrm{~g} \mathrm{dw} / \mathrm{g} \text { glucose } \end{aligned} $$ The inoculum for the fermenter is \(0.03 \mathrm{~g} \mathrm{dw} / \mathrm{l}\) of \(E\). coli \(\left(1 \times 10^{8} \mathrm{cells} / \mathrm{ml}\right)\) and \(0.15 \mathrm{~g} \mathrm{dw} / \mathrm{l}\) of A. vinelandii \(\left(1 \times 10^{8}\right.\) cells \(\left./ \mathrm{ml}\right)\). What will be the ratio of \(A\). vinelandii to \(E\). coli at the time when all of the glucose is consumed?

Short answer

The ratio of A. vinelandii to E. coli at the time when all glucose is consumed is calculated based on their final biomass developed from glucose consumption.

Step-by-step solution

Determine Initial Amounts of Glucose

The batch fermenter receives 11 L of medium with 5 g/L of glucose. Calculate the total amount of glucose in the medium.Total glucose = 11 L × 5 g/L = 55 g

Calculate Initial Biomass Concentrations

Given initial concentrations are 0.03 g dw/L for E. coli and 0.15 g dw/L for A. vinelandii. Calculate the total initial biomass for each bacteria in the 11 L medium.E. coli initial biomass = 0.03 g/L × 11 L = 0.33 gA. vinelandii initial biomass = 0.15 g/L × 11 L = 1.65 g

Calculate Glucose Allocation Based on Yield Coefficients

The yield coefficients are given as 0.5 g dw/g glucose for E. coli and 0.35 g dw/g glucose for A. vinelandii. Determine how the glucose gets converted to biomass.Let X_EC and X_AV be the amounts of biomass produced from glucose for E. coli and A. vinelandii respectively.Overall biomass: X_EC = 0.5 × glucose consumed by E. coliglucose consumed by E. coli = Biomass/X_EC = 0.33 g/Overall biomass: X_AV = 0.35 × glucose consumed by A. vinelandiiglucose consumed by A. vinelandii = Biomass/X_AV = 1.65/0.35 = 4.71 g

Calculate Replication Rates

E. coli's replication rate (µ_EC) is given by: \[\mu_{E C} = \dfrac{1.0 \mathrm{~h}^{-1} \cdot s}{0.01 \mathrm{~g} / \mathrm{l} + s} - 0.05 \mathrm{~h}^{-1}\]A. vinelandii's replication rate (µ_AV) is given by: \[\mu_{A V} = \dfrac{1.5 \mathrm{~h}^{-1} \cdot s}{0.02 \mathrm{~g} / \mathrm{l} + s} - 0.10 \mathrm{~h}^{-1}\]

Calculate Final Biomass

By the time all glucose is consumed, E. coli and A. vinelandii will have consumed their proportional share of glucose based on their yield coefficients. Thus, we calculate their final biomass using the glucose yield.Final biomass of E. coli = initial biomass + produced biomassFinal biomass of A. vinelandii = initial biomass + produced biomass

Calculate Ratio of Final Biomasses

Determine the ratio of the final biomass of A. vinelandii to E. coli when all glucose is consumed: \[\text{Ratio} = \dfrac{\text{Final biomass of A. vinelandii}}{\text{Final biomass of E. coli}}\]

Key concepts

Batch Fermenter

A batch fermenter is a type of bioreactor where the fermentation process occurs in a closed system. The medium, nutrients, and microorganisms are all placed inside the fermenter at the start, and the system is sealed.
The fermentation process runs until nutrients are depleted or desired by-products are formed.

• It is crucial for controlling contamination since no materials are added or removed during the process.
• Commonly used in the production of antibiotics, alcohol, and other biochemical products.

Understanding the operation of a batch fermenter is fundamental to grasping how different variables affect fermentation. For instance, factors like nutrient concentration and microorganism growth rates play a significant role in the final yield of the process.

Biomass Yield Coefficient

The biomass yield coefficient (Y) is a crucial parameter in fermentation. It denotes the efficiency with which microorganisms convert substrates (like glucose) into cellular biomass.
This coefficient is expressed as a ratio: the amount of biomass (g) produced per amount of substrate (g) consumed. For example, in the exercise, we have:

• For E. coli, the yield coefficient is \(Y_{\text{EC}} = 0.5 \, \text{g} \, \text{dw} / \, \text{g} \text{glucose}\).\supp,
• For A. vinelandii, it is \(Y_{\text{AV}} = 0.35 \, \text{g} \, \text{dw} / \, \text{g} \, \text{glucose}\).\suppli>

Higher yield coefficients mean more efficient biomass production. For example, E. coli has a higher biomass yield coefficient compared to A. vinelandii, making it more efficient in converting glucose to biomass.
However, it is crucial to balance yield coefficients with growth rates for an optimal fermentation process. An efficient biomass producer might not always have the fastest growth rate, and vice versa.

Growth-Limiting Nutrient

A growth-limiting nutrient is the essential nutrient present in the lowest concentration relative to the needs of the microorganisms, thereby limiting their growth and replication.
In this exercise, glucose is identified as the growth-limiting nutrient.

• It determines the maximum biomass that the mixed culture can achieve.
• Even if other nutrients are abundant, the availability of glucose dictates the extent of microbial growth and replication rates.

Understanding which nutrient is limiting helps in optimizing the fermentation process. For example, ensuring that the growth-limiting nutrient is adequately supplied can maximize biomass yield and product formation.
This concept is crucial for creating efficient and economical fermentation processes, especially in industrial applications where maximizing yield is key.