Difference between revisions of "Team:Manchester/Model/Continuous Culture"

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 output is modelled through the 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.

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:

Since output rate is simply the concentration of bacteria produced per unit time, the units are:

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:

To calculate the time required to produce 1kg of bacteria, we can simply rearrange the output equation from before:

Plugging it into the cost equation:

Inserting our values and crossing out the same variables:

Volume of Chemostat

The volume of the chemostat is independent to the production cost as it is cancelled in the equation above. The volume of the chemostat affects how fast it produces our desired product and therefore would reduce the running cost of the chemostat. We can graph the time equation to see the relationship between time and volume at an output rate of 0.522g/L per hour.

From the figure above, we can see that the time required to produce 1kg of bacteria at an output rate of 0.522g/L per hour decreases exponentially as the volume increases. It can be seen that we can produce 1kg of bacteria in less than 24 hours in a chemostat volume above 100L.

Estimating Cost to Clean Phosphate in Waste Water

Now that the basic cost of production has been calculated, the next step is to calculate the cost to take up phosphate in a certain amount of waste water.

We were able to get some phosphate data from our visit to Davyhulme Treatment Works. Since Davyhulme Treatment Works do not treat phosphate in their site, we can use the data as a case study to see how much it would cost to treat waste water that comes to the plant. The data below shows the concentration of phosphate in incoming waste water from the past 12 months:

Assuming that we are treating 1 ton of waste water, equivalent to 1000 liters, we would need to sequester 7323mg or ~7.3g of phosphate. To calculate the cost, we will have to find out how much bacterial preparation we need by knowing how much phosphate can be accumulated per gram of bacteria. Then, we can multiply this by the production cost that we have calculated earlier:

Inserting values that we know:

To find out how much phosphate a bacterium can accumulate, we went to literature that we based our project on. In the paper “Bacterial microcompartment-directed polyphosphate kinase promotes stable polyphosphate accumulation in E. coli” (Liang et al., 2017), E. coli with microcompartment-direct PPK was able to accumulate (180 μg polyphosphate)/(mg protein in whole cell) in 48 hours.

So we know that:

The equation is not complete, there are other variables that would have to be added, such as how many grams of proteins are in a bacterium. We were able to obtain protein data from New England Biolabs which shows that a liquid culture of E. coli has a total protein weight of 0.15g per liter at 10^9 cells per ml. Therefore:

The next variable that is missing is the wet weight of bacteria per number of cells. A study has shown that an OD600 of 1.0 will have a cell wet weight of 1.7g/L (Glazyrina et al., 2010). For bacterial cell cultures, an OD600 of 1.0 corresponds to 8 x 108 cells/ml. Thus:

Adding it into the previous equation:

Now we have all the variables needed to cross out the units, thus we can solve it as:

Thus:

Therefore, we can plug it into the cost equation to determine how much it cost to treat 1 ton of wastewater:

Therefore, we would be able to treat 1 ton of waste water in 48 hours at a production cost of $87.37.

In Davyhulme Treatment Works, flows of more than 30,000 litres per second are treated for 24 hours straight every single day. This is equivalent to 2 592 000 tons of water treated every single day or 946 080 000 tons of water every year (United Utilities, 2017).

We have previously calculated that it would cost $87.37 to treat 1 ton of waste water in 48 hours. To treat 946 080 000 tons of water would therefore cost $82 654 975 231 or roughly £61.6 billion every year. With an estimate of 1.17 million households in Greater Manchester in 2016 (New Economy Manchester, 2016), the cost of implementing this technology would be an additional £52 705 per household each year which is fabulously expensive and would make our technology completely unrealistic on the real-world market. We would need substantial improvements of our system to make it competitive – costs would have to be reduced by more than 1000-fold before the technology could begin to be considered as a real-world solution to the phosphate problems.

At the current output of 0.522g/L per hour, the treating cost of £61.6 billion is too expensive to be feasible. This value does not even take into account the running cost of the chemostat which would add more cost. Therefore, there is a need to improve the efficiency of the device by increasing the output rate. This would not only reduce cost, but it will also reduce the time to produce 1kg of bacteria in a given volume.

To increase the output rate, we’ll have to increase the substrate concentration. In the previous calculation, molasses with a concentration of 1.8g/L is used. We’re going to use molasses with concentration of 5g/L, 10g/L and 20g/L. In truth, they would probably have a higher price than $0.07/kg. However, we’ll use a price of $0.07/kg as a benchmark for this calculation to show how much the cost can be reduced when glucose concentration is increased, when everything else is held constant.

From figure 1, we know that the maximum output rate of each initial substrate concentration can be obtained through these dilution rates:

Using the production cost equation, we can plug in the values for dilution rate and maximum output rate to obtain the cost of producing 1kg of bacteria for each different output rate:

Using the phosphate treatment cost equation, we can plug in the production cost into the equation to obtain the cost of treating 1 ton of waste water:

To treat 946 080 000 tons of water would therefore cost:

If we divide the total cost of phosphate treatment for a year with the number of households in Greater Manchester, implementing this technology would add an additional cost per household of:

These are still prohibitively expensive – our synthetic biology approach would not be competitive at such a high cost. We therefore considered alternative strategies to drastically re-engineer our system to achieve a substantially better economic position.