Part 1 Summary

The yield of bacterial cellulose is related to many culture conditions, such as culture temperature, pH value, fermentation time and inoculum size, etc., and these factors also influence each other. We use the relationship of inoculum size and fermentation time as an example to illustrate this problem.

When the inoculum size is small, the bacteria grow slowly and synthesize cellulose slowly. However, if the inoculum size is too large, the bacteria will quickly reach the growth peak, and the number of bacteria will reach saturation. At that time, the nutrients in the culture medium will also be quickly exhausted, which is not conducive to the production of bacterial cellulose. In order to acquire the maximum economic benefit, we tried to use mathematical model to find the optimal inoculation quantity and the optimal fermentation time, so as to acquire the maximum yield of bacterial cellulose.

The inoculum size was expressed as a percentage of the culture medium volume; the bacterial density was expressed as OD600 value, and the unit of bacterial cellulose yield was g/L.

Part 2 General assumption and parameter

2.1 Assumption

2.1.1 First, the bacteria are growing in an ideal environment (with a suitable growth temperature, humidity and pH value).

2.1.2 Additionally, since there is no formula for explaining the relationship between the fermentation time and the inoculum size with different initial value of OD600, which is well studied, we assume that they are in a polynomial relation.

2.2 Parameter

**//*: In the experiment in each times, we use different initial different inoculation amounts. Details are in here:

Part 3. Mathematic analysis

In this topic, the speed of growing is obviously not keeping increasing all the time, because of some limitation in real environment. Therefore, bn/t will no keep increasing. I try to use such a unique information as a breakthrough point to find a better situation for E.coli to grow.

Such is the diagram of bn/t vs t(day).

As is shown in the diagram, most of trend of each set of data has one turning-point, thus we can assume that quadric expression can represent the relation of them. Then, if bn/t and t has a quadric relation, the bn and t must have a cubic relation. According to this, we can do the cubic fitting for each set of b(g/L) with t(day).

Result of fitting:

However, it is not the eventual result that we want, because what we want is finding the best option after considering the speed of increasing in bn and the amount of OD600 we need to use instead of the value of b at each t their own.

The graph of them produced by MATLAB can intuitively reflect some useful information of day, OD600, BC and initial inoculation amounts.

Comparison table of the graph:

In terms of choosing a best environment for E.coil to grow, we need to see which plane can reach a higher BC amounts. The models show the 7% initial inoculation amount with day 7 is the best in all.

Part 4. Conclusion and self-reflection

4.1 Conclusion

In the experiment of our project, 7% initial inoculation amount is the best, because it is not only effective but also economized.

4.2 Disadvantages of the model

① The assumption that the relation of them can be represented as a polynomial is a little cursory.

② The result of the fitting still has a constant, though it is small enough for us to ignore them.

Team:JNFLS/Model