Team:Manchester/Model/Continuous Culture
Continuous Culture of Bacteria
Aim
Estimate the amount of bacteria required to clean up a given amount of phosphate, and the associated production cost to determine the profitability of our project using computational modelling.
Introduction
We envision our bacteria to be cultured in a chemostat for production. A chemostat is a bioreactor in which fresh medium is continually added while culture liquid containing leftover nutrients, microorganisms, and metabolic products are continuously removed at the same rate. This technique is called continuous culture and allows microbial growth to take place under steady-state conditions - growth that occurs at a constant rate and in a constant environment. Unlike a batch culture method where bacterial cells undergo the full bacterial cell cycle, a continuous culture keeps the bacteria growing on its exponential phase of the bacterial cell cycle, thus a continuous supply of bacteria can be produced.
Background Theory
The growth of bacteria in its exponential phase can be represented in the following exponential growth equation:
\begin{equation} \frac{1}{x} \frac{dx}{dt} = \mu = \frac{log_e 2}{t_d} \end{equation} where:
\(x\) is the bacteria concentration (dry weight mass/unit volume) at time \(t\)
\(μ\) is the specific growth rate
\(t_d\) is the doubling time (time required for the concentration of organism to double)
In 1942, Jacques Monod showed that there is a relationship between the specific growth rate and the concentration of a limiting growth substrate that can be represented in this equation:
\begin{equation} \mu = \mu_{max} \bigg(\frac{s}{K_s + s}\bigg) \end{equation} where:
\(s\) the concentration of a limiting growth substrate
\(μ_{max}\) is the maximum growth rate (growth rate when organism is placed in excess nutrients without any limiting factors)
\(K_s\) is the saturation constant – the value of \(s\) when: \(\frac{μ}{μ_{max}}\) = \(\frac{1}{2}\)
A relationship between growth and utilization of substrate has also been shown by Monod by the equation:
\begin{equation} \frac{dx}{dt} = −Y \frac{ds}{dt} \end{equation}\begin{equation*} Y = \frac{\textrm{weight of bacteria formed}}{\textrm{substrate utilized}} \end{equation*}
where \(Y\) is known as the yield constant
If the values of the three growth constants: \(μ_{max}\), \(K_s\) and \(Y\) are known, equation (1) to (3) provides a complete quantitative description of the ‘growth cycle’ of a batch culture.
A chemostat is a continuous flow system in which fresh growth medium is added into the vessel at a steady flow-rate (\(F\)) and culture liquid exits at the same rate. Contents within the vessel are stirred so that the growth medium is uniformly dispersed. The rate in which nutrient is exchanged in the vessel is expressed as the dilution rate (\(D\)):
\begin{equation} D = \frac{\textrm{medium flow rate}}{\textrm{culture volume}} = \frac{F}{V} \end{equation}If we assume that the bacteria within the vessel stops growing and dividing, with equal stirring and continuous flow of medium, every organism will have an equal probability of leaving the vessel within a given time. The wash-out rate (rate in which organism initially present in the vessel will be washed out) can be expressed as:
\begin{equation} - \frac{dx}{dt} = Dx \end{equation}where \(x\) is the concentration of organisms in the vessel
1. Changes in concentration of organism
In a continuous culture, bacteria are growing at a rate expressed in equation (1) and at the same time, it is being washed out at a rate expressed in equation (5). The net rate of increase is therefore:
\begin{equation} \begin{split} \textrm{increase} & = \textrm{growth} - \textrm{output} \\\frac{dx}{dt} & = \mu x - Dx \end{split} \end{equation}
If:
\(μ\) > \(D\), concentration of organism will increase
\(μ\) < \(D\), concentration of organism will decrease
\(μ\) = \(D\), concentration of organism is constant; a ‘steady state’
Substituting (2) into \(μ\):
\begin{equation} \frac{dx}{dt} = x \bigg\{\mu_{max} \bigg(\frac{s}{K_s + s}\bigg) - D\bigg\} \end{equation}2. Changes in substrate concentration
In a continuous culture, substrate enters the vessel at a concentration \(S_{in}\), consumed by the bacterial cell and exits the vessel at concentration \(S_{out}\). The net rate of change is therefore:
\begin{equation*} \begin{split} \textrm{increase} & = \textrm{input} - \textrm{output} - \textrm{consumption} \\& = \textrm{input} - \textrm{output} - \frac{\textrm{growth}}{\textrm{yield constant}} \end{split} \end{equation*}
\begin{equation} \frac{ds}{dt} = D S_{in} - D S_{out} - \frac{\mu x}{Y} \end{equation}
Substituting (2) into \(μ\):
\begin{equation} \frac{ds}{dt} = D (S_{in} - S_{out}) - \frac{\mu_{max} x}{Y} \bigg(\frac{s}{K_s + s}\bigg) \end{equation}From equation (7) and (9), if \(D\) and \(S_{in}\) are held constant, there is a unique solution for \(x\) and \(s\) for which \(\frac{dx}{dt}\) and \(\frac{ds}{dt}\) are 0. When \(\frac{dx}{dt}\) and \(\frac{ds}{dt}\) is 0, the system is said to be in a ‘steady state’ because the concentration of organism and substrate within the continuous culture is kept constant. The values of steady state \(x\) and \(s\), designated as \(\tilde{x}\) and \(\tilde{s}\) are expressed as:
\begin{equation} \tilde{s} = K_s \bigg(\frac{D}{\mu_{max} - D}\bigg) \end{equation}\begin{equation} \tilde{x} = Y (S_{in} - \tilde{s}) = Y \bigg\{ S_{in} - K_s \bigg(\frac{D}{\mu_{max} - D}\bigg) \bigg\} \end{equation}
The steady state values depends solely on the value of \(D\) and \(S_{in}\) which can be manipulated within the chemostat. The values of growth constant \(μ_{max}\), \(K_s\) and \(Y\) are constant for a specific organism within a growth medium.
The output rate is the quantity of bacteria produced in unit time. The total output from a continuous culture in a steady state is equal to the product of flow-rate and concentration of organism. The output per unit volume of culture is therefore \(D\tilde{x}\):
\begin{equation} \textrm{Output} = D\tilde{x} = D * Y \bigg\{ S_{in} - K_s \bigg(\frac{D}{\mu_{max} - D}\bigg) \bigg\} \end{equation}We can measure the maximum output rate of organism by differentiating the above equation with respect to \(D\) and equating to zero. This will give us the dilution rate that gives the maximum output organism in unit time, \(D_{M}\):
\begin{equation} D_M = \mu_{max} \Bigg\{ 1 - \sqrt{\frac{K_s}{K_s + S_{in}}} \Bigg\} \end{equation}The steady-state concentration of organisms at this dilution rate can be obtained by substituting \(D_M\) in equation (11):
\begin{equation} \tilde{x}_M = Y \Bigg\{ (S_{in} + K_s) - \sqrt{K_s (S_{in} + K_s)} \Bigg\} \end{equation}The maximum output rate, \(D_{M}\tilde{x}_M\) is the product of the two equations:
\begin{equation} D_M \tilde{x}_M = \mu_{max} * Y * S_{in} \Bigg\{ \sqrt{\frac{K_s + S_{in}}{K_s}} - \sqrt{\frac{K_s}{S_{in}}} \Bigg\}^2 \end{equation}Parameters
| Parameter Name | Description | Value | Unit | Source |
|---|---|---|---|---|
| Growth Yield \(Y\) | A typical value of the observed growth yield of E. coli at rapid growth rates | 0.44 | grams dry weight / grams glucose | Majewski RA, Domach MM. Simple constrained-optimization view of acetate overflow in E. coli. Biotechnol Bioeng. 1990 Mar 25 35(7):732-8. p.736 left column paragraph above bottom |
| Saturation constant \(K_s\) | Monod substrate affinity constants (Ks) for E. coli growing on glucose as the only source of carbon and energy | 0.099 | grams / liter | Senn H, Lendenmann U, Snozzi M, Hamer G, Egli T. The growth of Escherichia coli in glucose-limited chemostat cultures: a re-examination of the kinetics. Biochim Biophys Acta. 1994 Dec 15 1201(3):424-36. P.425 left column top paragraph & P.433 table 4 |
| Maximum growth rate \(μ_{max}\) | Maximum growth rate for E. coli growing on glucose as the only source of carbon and energy | 1.05 | hour -1 | Senn H, Lendenmann U, Snozzi M, Hamer G, Egli T. The growth of Escherichia coli in glucose-limited chemostat cultures: a re-examination of the kinetics. Biochim Biophys Acta. 1994 Dec 15 1201(3):424-36. P.425 left column top paragraph & P.433 table 4 |
Modelling Output
The output is modelled through the equation:
\begin{equation*} \textrm{Output} = D\tilde{x} = D * Y \bigg\{ S_{in} - K_s \bigg(\frac{D}{\mu_{max} - D}\bigg) \bigg\} \end{equation*}There are two variables that can be changed according to the design of the chemostat: \(D\) (dilution rate) and \(S_{in}\) (initial substrate concentration).
We modelled the output vs. dilution rate on five different initial substrate concentration (1, 5, 10, 20 and 40g/L).
We can see from the graph that for all five different concentrations of initial substrate, the output rate increase as dilution rate is increased. However, at a certain dilution rate, the output then starts to fall off dramatically. This is when the dilution rate value is higher than the specific growth rate of bacteria. At this point, the bacteria are washed out of the reactor at a faster rate than it can reproduce. Eventually, all the bacteria will be displaced out of the reactor.
The different points in the graph represent the dilution rate that gives the maximum output. As the initial substrate concentration increases, the maximum output rate is also shown to increase. In the figure below, we graphed the correlation between maximum output rate and initial substrate concentration from the values obtained from the graph above.
We can see that the relationship between initial substrate concentration and maximum output rate is a positive linear line. There is theoretically no limit to the output rate that we can achieve, as long as we can afford the initial substrate concentration. In this case, the limitations would be a financial limit and depends on how much we can afford to buy the substrate concentration.
However, it is important to note from figure 1 that as initial substrate concentration increases, the graph gets steeper. The dilution rate required to get maximum output rate gets closer and closer to a critical dilution rate where output then starts to fall off. Therefore, we can see that while initial substrate concentration increases the maximum output rate at a constant rate (equal to the Growth Yield), there is also a greater risk of reaching that dilution rate that would ultimately wash out all the bacteria from the reactor.
The main assumption is that all other factors for growth are in sufficient supply to ensure that glucose is kept as the limiting factor in the chemostat. In high concentrations of glucose, for example, there are other factors that may become limiting. Therefore, there would be an additional cost in supplying enough of these other factors, such as oxygen or nitrogen to ensure that it reaches the desired growth rate.
Cost Estimation
Estimating Cost of Production
Our interaction with John Liddell from the Centre Process of Innovation told us that sources of glucose usually come from cheap media such as molasses. To determine the concentration of glucose in molasses, we went to literature and found a study about high level ethanol production from sugar cane molasses that was applied in an Egyptian distillery factory. The microorganism in the study was fermented at a final concentration of 18% (w/v) or 1.8g/L sugar level (Fadel et al., 2013).
As seen in figure 3, at a dilution rate of 0.81 we can achieve maximum output rate of 0.522 g/L of bacteria liquid culture per hour. Now that we know the maximum output rate, the next step is to determine the cost of producing a certain amount of bacteria.
-Molasses cost $0.07/kg
-The density of molasses is roughly 1.4kg/L
-Therefore, 1L of molasses will cost $0.07 x 1.4 = $0.098
To calculate the cost of 1kg of bacteria, we’ll start by going back to the output equation:
\begin{equation*} \textrm{Output} = D\tilde{x} = D * Y \bigg\{ S_{in} - K_s \bigg(\frac{D}{\mu_{max} - D}\bigg) \bigg\} \end{equation*}Since output rate is simply the concentration of bacteria produced per unit time, the units are:
\begin{equation*} \textrm{Output} = \frac{g/L}{h} = \frac{Amount (g)}{Time (h)*Volume (L)} \end{equation*}The total cost of producing a 1kg of bacteria can be calculated by taking into account several variables and multiplying it with the cost of 1L of molasses.
1. The time it takes to produce 1kg. The faster we can produce 1kg, the cheaper it will be.
2. The dilution rate or turnover rate (the rate of nutrient exchange in the chemostat) which is equal to the flow rate divided by the volume of the chemostat. Higher dilution rate means a larger volume of substrate used which increases the output rate up to a certain point.
3. The volume of the chemostat. The bigger the volume, the more bacteria can be produced per time.
Therefore:
\begin{equation*} \textrm{Cost} = \textrm{Time*Dilution rate*Volume (L)}*\frac{$0.098}{L} \end{equation*}To calculate the time required to produce 1kg of bacteria, we can simply rearrange the output equation from before:
\begin{equation*} \textrm{Output} = \frac{\textrm{Amount (g)}}{\textrm{Time (h)*Volume (L)}} \end{equation*} \begin{equation*} \textrm{Time (h)} = \frac{1kg}{\textrm{Output rate (g/L/h)*Volume (L)}} \end{equation*}Plugging it into the cost equation:
\begin{equation*} \textrm{Cost} = \textrm{Time*Dilution rate*Volume (L)*$0.098} \end{equation*} \begin{equation*} \textrm{Cost} = \frac{1kg}{\textrm{Output rate*Volume}}*Dilution rate*Volume*$0.098 \end{equation*}Inserting our values and crossing out the same variables:
\begin{equation*} \textrm{Production cost} = \frac{1000g}{0.522g/L/h}*0.81h^(-1)*$0.098/L \end{equation*} \begin{equation*} \textrm{} = \frac{1000g*L*h}{0.522g}*\frac{0.81}{h}*\frac{$0.098}{$0.098} \end{equation*} \begin{equation*} \textrm{} = \textrm{$152.07 for 1kg} \end{equation*}