# Overall Objective

We aim to develop models that are based on mathematical calculations and analysis in order to establish a bridge between our theoretical framework and our experimental process. Our models analysis can give great help and guidance on our experiment.

In our project, we developed 2 models about our production analysis:
1. TnaA gene mutation analysis
2. 6Br-IG production analysis

# Model 1: TnaA Gene Mutation Analysis

### Hypothesis(1):

1. Differences in the solubility of 6Br-Indole and 6Br-Trp medium
2. Stable endogenous expression of the TnaA gene in E. coli, but lower than the promoter expression intensity of the plasmid.
3. The precipitation of 6Br-indole wouldn't participate in the following reactions.
4. Negligible changes in endogenous growth and impact coefficients between unmutated and mutated E. coli strain

### Formula(1):

Fermentation reaction formula
Fermentation production Tyrian purple equation.

Amount of tryptophan change in the medium:
$$v_{0}=-\frac{d C_{0}}{d t}=\frac{V_{\max 1} C_{0}}{K_{m 1}+C_{0}}=\frac{k_{c a t 1} E_{0} C_{0}}{K_{m 1}+C_{0}}$$

6Br-Trp changes the amount in two bacteria：
$$v_{1}=\frac{d C_{1}}{d t}=\frac{V_{\max 1} C_{0}}{K_{m 1}+C_{0}}-\frac{V_{\max 2} C_{1}}{K_{m 2}+C_{1}}=\frac{k_{c a t 1} E_{0} C_{0}}{K_{m 1}+C_{0}}-\frac{k_{c a t 2} E_{0} C_{1}}{K_{m 2}+C_{1}}$$
(Changes in Fre-L3-SttH-transformed E. coli )
Amount of variation of 6Br-Indole in TnaA-MaFMO transformed E. coli :
$${v_{2}}=\frac{d C_{2}}{d t}=\frac{V_{max2} εC_{1}}{K_{m2}+εC_{1}}-\frac{V_{max3} C_{2}}{K_{m3}+C_{2}}=\frac{k_{cat2}E_{0}εC_{1}}{K_{m2}+εC_{1}}-\frac{k_{cat3}E_{0}C_{2}}{K_{m3}+C_{2}}$$
ε is the proportion of 6Br-Trp representing transfer into TnaA-MaFMO E. coli.
ε from $0.99$ to $0.9$, to $0.8$, to $0.5$
The amount of 6BrIG variation in the culture medium:
$$\frac{d C_{3}}{d t}=\frac{V_{max3} C_{2}}{K_{m3}+C_{2}}=\frac{k_{cat3}E_{0}C_{2}}{K_{m3}+C_{2}}$$

(pathway of 6BrIG production)

### Result Analaysis & Conclusion(1):

#### Result Analysis

(Figure shows the comparison of different yields of Tyrian purple)

The curve shows that compared with the theoretical value (red line) of 6BrIG, only part of the tryptophan is converted to 6-bromotryptophan through SttH enzyme, and part of the 6-bromotryptophan is converted into 6-bromoindole. Bromoindole does not dissolve in water and turn into precipitation, thus reducing the final product, resulting in a decline in the yield of 6,6-dibromoindigo.

#### Conclusion

Since 6Br-Indole and 6Br-Trp enter into E.coli with different rates, ultimately, E. coli without knockout of the TnaA gene will produce 6Br-Indole due to the production of 6Br-Trp at the same time, which results in less efficient production of 6Br-IG in E. coli without TnaA mutation than in E. coli with TnaA mutation gene due to the low solubility of 6Br-Indole itself in solution and the low efficiency of entering E. coli (TnaA+MaFMO) than 6Br-Trp. Therefore, in order to ensure the production of 6,6-dibromoindigo, we should knock out the TnaA gene. This was also verified in experiments in the original literature. Without the knockout, the 6BrIG yield in the experiment will be lower than the 6BrIG produced after knockout. We can use CRISPR-Cas9 gene editing technology to mutate E. coli

# Model 2 - 6BrIG Production Model

### Hypothesis(2):

1. E. coli does not differ in population growth and holding capacity due to different plasmid transformations, and the coefficient of interaction between the two strain is close
2. It is considered that the produced 6Br-Trp enters completely into the TnaA-MaFMO ΔTnaA strain and undergoes further reactions.

### Formula(2):

Fermentation reaction formula
Fermentation production Tyrian purple equation.

Amount of tryptophan change in the medium:
$$v_{0}=-\frac{d C_{0}}{d t}=\frac{V_{\max 1} C_{0}}{K_{m 1}+C_{0}}=\frac{k_{c a t 1} E_{0} C_{0}}{K_{m 1}+C_{0}}$$

6Br-Trp change amount in the medium:
$$v_{1}=\frac{d C_{1}}{d t}=\frac{V_{\max 1} C_{0}}{K_{m 1}+C_{0}}-\frac{V_{\max 2} C_{1}}{K_{m 2}+C_{1}}=\frac{k_{c a t 1} E_{0} C_{0}}{K_{m 1}+C_{0}}-\frac{k_{c a t 2} E_{0} C_{1}}{K_{m 2}+C_{1}}$$
6Br-Indole changes amount in the medium:
$${v_{2}}=\frac{d C_{2}}{d t}=\frac{V_{max2} C_{1}}{K_{m2}+C_{1}}-\frac{V_{max3} C_{2}}{K_{m3}+C_{2}}=\frac{k_{cat2}E_{0}C_{1}}{K_{m2}+C_{1}}-\frac{k_{cat3}E_{0}C_{2}}{K_{m3}+C_{2}}$$
The amount of 6BrIG variation in the culture medium:
$$\frac{d C_{3}}{d t}=\frac{V_{max3} C_{2}}{K_{m3}+C_{2}}=\frac{k_{cat3}E_{0}C_{2}}{K_{m3}+C_{2}}$$

Population growth equation：

$$\frac{dA_1}{dt}=r_1A_1(1-(\frac{A_1+\alpha_{12}A_2}{K_1}))$$
(transformation of E. coli with SttH plasmid)
$$\frac{dA_2}{dt}=r_2A_2(1-(\frac{A_2+\alpha_{21}A_1}{K_2}))$$
(transformation of E. coli with TnaA-MaFMO plasmid)

### Result Analaysis & Conclusion(2):

#### Model 2.1 6BrIG Production

(The image shows the changes of the amount of Trp, 6Br-Trp, 6Br-Indole and 6BrIG over time.)

As time progresses. 6Br-Trp and 6Br-Indole reach the limit fast, at the same time 6BrIG increases with time, until the concentration of Trp reaches 0 where 6BrIG reaches its maximum value.

In the ideal situation, the Trp put in will be entirely turned into 6BrIG.

#### Model 2.2 - Bacteria Growth

(a) The ratio of the two E. coli was 1:1. With the increase of time, the concentration of A1(SttH) and A2( TnaA+MaFMO) increased; the concentration continued to increase rapidly until 6 min, and reached saturation after 8 min.

(b) The ratio of the two E. coli was 1:2. A1A2 both increase with time, reaching a maximum at around 6-7 min, leveling off and no longer increasing. A2(TnaA+MaFMO) starts at a higher concentration than A1Stth and eventually reaches equilibrium.

(c) The ratio of the two E. coli was 2:1. A1, A2 both increase with time, reaching a maximum at around 6-7 min, leveling off and no longer increasing. The starting concentration of A1 SttH is higher than that of A2 TnaA+MaFMO and finally reaches equilibrium.

(d) The ratio of the two E. coli was 1:100. A2(TnaA+MaFMO) always showed a rising trend, with a slow trend at the beginning and a large increase after 2 min, gradually leveling off after reaching 15 min. A1(SttH) kept increasing rapidly and reached six minutes with a slow decrease. In the case that SttH is much higher than TnaA+MaFMO, the two can still reach equilibrium eventually.

(e) The ratio of the two E. coli was 100:1. A1(Stth) always showed an upward trend, with a slow trend at the beginning, and a large increase after 2 min, which gradually leveled off after reaching 15 min. A2 (TnaA+MaFMO) kept increasing rapidly and reached six minutes with a slow decrease. In the case that TnaA+MaFMO is much higher than SttH, the two can still reach equilibrium eventually.

#### Bacterial Growth Summary

In the case of approximately the same content of E.coli, they can eventually reach the maximum capacity of the entire medium in a relatively short period of time, and the population is stable. While in a certain E. coli strain is dominant, in the early growth, the dominant strain will grow immediately, reaching the maximum capacity, while the minor strain will grow in a slow rate initially, but with time, it will grow fastly with population increase, and finally both E. coli strain will reach the maximum capacity of the medium Therefore, when co-culturing two E. coli strain , even if one of the strains is added too much, the two strains can still reach equilibrium(maximum population capacity) after a certain period of incubation time. So in operation, incubation time is more important than the amount of E. coli strain added. We need to ensure both strains are in the rapid growth state when the inducer is added.

# Parameters&References

#### Parameters

Below is a table that contains the data and sources used for all the variable in our modeling section

| Name | Value |
| ----------- | ----------- |
| $K_m$(SttH)$^1$ | $0.021$mM |
| $K_{cat}$(SttH)$^1$ | $0.2835$ s$^{-1}$ |
| $K_m$(TnaA)$^2$ | $0.39$ mM |
| $K_{cat}$(TnaA)$^2$ | $11.7$s$^{-1}$ |
| $K_m$(MaFMO)$^3$ | $0.09$ mM |
| $K_{cat}$(MaFMo)$^3$| $162$ s$^{-1}$ |
| Intrinsical Growth Value($\textit{E.coli}$ )$^4$ | $r_1=r_2=0.82$ |
| Population Capacity ($\textit{E.coli}$ )$^4$ | $K_1=K_2=0.62$ |
| $\textit{E.coli}$ Impact Coefficient$^5$ | $\alpha_{12}$=$\alpha_{21}$=$0.1$ |

#### References

1.Zeng, J., Zhan, J. Characterization of a tryptophan 6-halogenase from Streptomyces toxytricini . Biotechnol Lett 33, 1607–1613 (2011).

2.Phillips, R. S., Buisman, A. A., Choi, S., Hussaini, A., & Wood, Z. A. (2018). The crystal structure of Proteus vulgaris tryptophan indole‐lyase complexed with oxindolyl‐l‐alanine: implications for the reaction mechanism. Acta Crystallographica. Section D, Structural Biology, 74(8), 748–759.

3.Alfieri, A., Malito, E., Orru, R., Fraaije, M. W., & Mattevi, A. (2008). Revealing the Moonlighting Role of NADP in the Structure of a Flavin-Containing Monooxygenase. Proceedings of the National Academy of Sciences - PNAS, 105(18), 6572–6577.

4.Newbold, C. J. (n.d.). Microbial feed additives for ruminants. Biotechnology in Animal Feeds and Animal Feeding, 259–278.

5.https://2019.igem.org/Team:NCKU_Tainan/Model