Team:Technion-Israel/Model

Modeling

In science, models are a mentally visual way of linking theory with experiment. Obtaining a complete operational picture using solely experimental approaches is challenging. Mathematical modeling provides an alternative path for this key problem, offering new approaches that incorporate detailed dynamics of sets of biochemical interactions.

In our project, we have established a comprehensive model of the entire "Synthetic Honey Stomach" metabolic pathway, based on our experimental data and a thorough study of all the underlying reactions.

We have successfully modeled the dynamic interactions between our three primary enzymes, their secretion from our engineered bacteria and the “Honey Circuit” controlling the process. Our model also accounts for the changing pH and its influence on our system, which addresses a significant challenge that remained unsolved in previous works we have encountered.

Modeling was required to better understand, and predict, the interplay between the enzymes and their products, as well as determine ideal initial concentrations of our enzymes.

Finally, we have integrated the results of our model into our wet lab work by designing several experiments based on the information we have obtained using our model simulation, minimizing the number of unknown variables and allowing for more efficient creation of our "BeeFree" Honey.

Based on the belief that any good model should extend beyond its limits of the lab, we have answered important questions asked by experts during our Integrated Human Practice work. Moreover, we have used the knowledge we gained while overcoming our modeling challenges and difficulties, to help other teams overcome theirs, read more in our collaborations.

Our whole model is the result of separate studies for each component of the system:

Enzymatic Processing of Nectar

• Overview

  Our “Synthetic Honey Stomach" includes three main enzymes responsible for the honey production process: invertase, catalase and glucose oxidase. In this part, we aimed to understand better the interactions between the enzymes and the products of the enzymatic reaction. This part also estimates the change in pH and its influence on our system, addressing a significant challenge that remains unsolved in previous studies we read.
  We decided to focus on modeling the changes of sucrose, glucose, fructose, gluco-𝛿-lactone (GDL), and hydrogen peroxide concentrations during honey production since their concentrations in the final product determine whether it can be considered as "real honey”.
  We believe our model must answer real questions and solve real problems. Therefore, this modeling section was motivated by questions directly derived from our Human Practices work.These questions guided us to include the pH change as well as its effect on the enzymes during the process.
  To model this part we had to model two sub-parts:

  1. Enzymes kinetics
  2. pH change

• Enzymes Kinetics

  Overview

  In this part, we have only modeled the kinetics of the enzymes, without considering the pH influence. To find the mathematical equations that accurately depict the kinetics of our enzymes, we have read various studies in the field and consulted with Prof. Tomer Shlomi from the Technion. We have concluded that our enzymes act according to Michaelis-Menten’s equations^[1][2][3].

  Chemical reactions

  $$Sucrose\ +\ Invertase\ \rightleftharpoons{\ \left[Suc-Inv\right]}_{complex}\ \longrightarrow\ \ Glucose\ +Fructose\\ $$ $$Glucose\ +\ GOX\ \rightleftharpoons{\ \left[Glu-GOX\right]}_{complex}\ \longrightarrow\ \ GDL\ +H_2O_2\\ $$ $$H_2O_2\ +\ Catalase\ \rightleftharpoons{\ \left[H_2O_2-Cat\right]}_{complex}\ \longrightarrow\ \ H_2O\ +\frac{1}{2}\bullet O_2\\ $$

  Mathematical equations and parameters

  Considering Michaelis-Menten’s equations and the dependance of the reaction on the interaction between the enzymes, we got the following mathematical equations:

  \(\frac{dSucr}{dt}=-K_{1,inv}\bullet\ Inv\bullet\ Sucr+K_{-1,inv}\bullet[Inv_0-Inv]\)
  
  \(\frac{dInv}{dt}=-K_{1,inv}\bullet\ Inv\bullet\ Sucr+K_{-1,inv}\bullet[Inv_0-Inv]+K_{2,inv}∙[Inv_{0}-Inv]\)
  
  \({\frac{dFruc}{dt}=K}_{2,inv}\bullet\left[Inv_0-Inv\right]\)
  
  \(\frac{dGluc}{dt}=-K_{1,GOX}\bullet\ GOX_{prot}\bullet\ Gluc+K_{-1,GOX}\bullet[GOX_0-GOX]\)+\(K_{2,inv}∙[Inv_0-Inv]\)
  
  \(\frac{d\left[GOX_{prot}\right]}{dt}=-K_{1,GOX}\bullet GOX_{prot}\bullet Gluc+K_{-1,GOX}\bullet[GOX_0-GOX]+K_{2,GOX}∙[GOX_0-GOX]\)
  
  \(\frac{dGDL}{dt}=K_{2,GOX}\bullet[GOX_0-GOX]\)
  
  \(\frac{dH_2O_2}{dt}=-K_{1,cat}\bullet\ Cat\bullet\ H_2O_2+K_{-1,cat}\bullet\left[Cat_0-Cat\right]\)+\(K_{2,GOX}\bullet[GOX_0-GOX]\)
  
  \(\frac{dp_{tot,cata}}{dt}=K_{-1,cat}\bullet\left[Cat_0-Cat\right]\)
  
  \(\frac{dCat}{dt}=-K_{1,cat}\bullet\ Cat\bullet\ H_2O_2+K_{-1,cat}\bullet\left[Cat_0-Cat\right]+K_{2,cat}\bullet\left[Cat_0-Cat\right]\)
  
  

  As known:

  $$K_m=\frac{K_{-1}+K_2}{K_1}$$

  Assumption 1: we assume that the reaction proceeds to the direction of products.

  
  \(K_1\gg K_{-1}\longrightarrow\) \(K_{-1}\approx0\longrightarrow\) \(K_1=\frac{K_2}{K_m}\)
  

  Considering our experiments’ conditions, we found from the extensive literature research we have done, the following parameters:

  Enzymes	Parameters
  Glucose Oxidase	\(K_m=39 mM\) [4]
  Glucose Oxidase	\(K_{cat}=446 sec^{-1} \)[5]
  Invertase	\(K_m=44.38 mM \)[6]
  Invertase	\(K_{cat}= 21.7 sec^{-1}\) [7]
  Catalase	\(K_m = 40.1 mM\) [8]
  Catalase	\(K_{cat} =175000 sec^{-1} \)[9]

  Table 1 - Michaelis-Menten's enzymes' parameters

  Based on assumption 1: \(K_1=\frac{K_2}{K_m}\)

  Hence:

  Enzymes	\(K_{cat}\)	\(K_{1}\)	\(K_{-1}\)
  Glucose Oxidase	446	11.43	\(\approx0\)
  Invertase	21.7	0.48	\(\approx0\)
  Catalase	175000	4364	\(\approx0\)

  Table 2 - Final Michaelis-Menten's enzymes' parameters

  From these parameters, we can predict that catalase is the fastest enzyme followed by GOx.

  
  

  Results

  

  Figure 1 - Products concentrations over time - whole enzyme system without pH influence

  • Sucrose: Due to the presence of invertase, we can see that the concentration of sucrose simply decreases.
  • Fructose: Once the concentration of sucrose decreases, the concentration of fructose increases to sucrose’s initial concentration.
  • Glucose: Once the concentration of sucrose decreases, the concentration of glucose increases, at a slower rate than fructose’s increasing rate, and decreases at the same point in which sucrose is over. This is due to the presence of GOx enzyme.
  • GDL: Once the glucose decrease, the GDL increases to sucrose’s initial concentration. This is due to the presence of GOx Enzyme.
  • \(H_2O_2\): In Figure 1 we see that the concentration of \(H_2O_2\) is almost zero but looking at Figure 2 we can see that once the glucose increases, the \(H_2O_2\) increases for a very low concentration, but decreases at the same point in which glucose is over. This is due to the presence of catalase enzyme which, as seen from its parameters, is a very fast enzyme.

  

  Figure 2 - \(H_2O_2\) concentration over time - whole enzyme system without pH influence

• pH change

  Overview

  In this section of the modeling chapter, our goal is to predict the change in the pH values over time in our “Synthetic Honey Stomach”. This section is especially important as the pH change influences the activity of the enzymes.

  What influences the pH value?

  The following chemical equations demonstrate the enzymatic reaction done by the three main enzymes in our system: $$Sucrose\ +\ Invertase\ \rightleftharpoons{\ \left[Suc-Inv\right]}_{complex}\ \longrightarrow\ \ Glucose\ +Fructose\ $$ $$Glucose\ +\ GOX\ \rightleftharpoons{\ \left[Glu-GOX\right]}_{complex}\ \longrightarrow\ \ GDL\ +H_2O_2$$ $$H_2O_2\ +\ Catalase\ \rightleftharpoons{\ \left[H_2O_2-Cat\right]}_{complex}\ \longrightarrow\ \ H_2O\ +\frac{1}{2}\bullet O_2\ $$

  Among the compounds presented in these equations, the pH is mainly influenced by these acidic compounds ^[10]:

  • Gluconic acid: \(\ C_6H_{12}O_7\)
  • Hydrogen Peroxide: \(H_2O_2\)

  
  
  

  To understand which one has the most significant influence, we looked at the pKa of each of them:

  • \(\ {pK}_{a,GA}=3.65\)^[10]
  • \(pK_{a,H_2O_2}=11.6\)^[11]

  
  
  

  As can be inferred from the pKa values, \({pK}_{a,GA}\ll pK_{a,H_2O_2}\). Thus, we can safely assume that the pH will be influenced mainly by gluconic acid -\( \ C_6H_{12}O_7\).

  Therefore, we focus on the following equation:

  $$Glucose\ +\ GOX\ \rightleftharpoons{\ \left[Glu-GOX\right]}_{complex}\ \longrightarrow\ \ GDL+H_2O_2$$

  The reaction product is Glucono-δ-lactone, which will eventually turn into gluconic acid (GA) as can be seen in Figure 3.

  

  Figure 3 - Illustration of the reaction turning Glucono-δ-lactone into gluconic acid

  Assumption: The reaction is not in equilibrium, and the gluconic acid product is favorable, due to the following reasons:
  
  

  • Our system is an aqueous solution, therefore there is excess of water. According to Le Chatelier's principle, the reaction will proceed in the product direction.
  • The equilibrium constant of the reaction is very high, \(K_{eq}=1.56\bullet{10}^4\).^[12]

  
  
  

  Therefore, we will only focus on the following equation:

  $$C_6H_{12}O_7\leftrightharpoons H^++C_6H_{11}O_7^-$$

  Equations

  In the GA reaction, as we have mentioned above, the pH change is evaluated by the following equation:

  $$(1) \ \ \ pH=-\log{\left(K_a\right)}+\ log\frac{[C_6H_{11}O_7^-]}{[C_6H_{12}O_7]}$$

  To find the concentration of \({[C}_6H_{11}O_7^-]\) and \([C_6H_{12}O_7]\) we used the following known equations:

  $$(2)\ \ K_a=\frac{[H^+]{[C}_6H_{11}O_7^-]}{[C_6H_{12}O_7]}$$ $$(3) \ \ \left[C_6H_{12}O_7\right]=\left[p_{GOX}\right]-[C_6H_{11}O_7^-]$$ $$(4){[H}^+]=[C_6H_{11}O_7^-]$$

  \(p_{GOx}\): is the concentration of all the Glucono-\delta-lactone produced during the degradation of glucose, which is one of the variables of GOx kinetics’ equations.

  In addition to the equations of the kinetics of our enzymes, equations (1) to (4) lead us to the following mathematical equations:

  $$\left[{C}_\mathbf{6}{H}_{\mathbf{11}}{O}_\mathbf{7}^-\right]^\mathbf{2}+{K}_{a}\left[{C}_\mathbf{6}{H}_{\mathbf{11}}{O}_\mathbf{7}^-\right]-{K}_{a}\left[{p}_{{GOX}}\right]=\mathbf{0}$$ $$ \ \ \ pH=-\log{\left(K_a\right)}+\ log(\frac{[C_6H_{11}O_7^-]}{\left[p_{GOX}\right]-[C_6H_{11}O_7^-]})$$ $$\frac{dSucr}{dt}=-K_{1,inv}\bullet\ Inv\bullet\ Sucr+K_{-1,inv}\bullet[Inv_0-Inv]$$ $$\frac{dInv}{dt}=-K_{1,inv}\bullet\ Inv\bullet\ Sucr+K_{-1,inv}\bullet[Inv_0-Inv]+K_{2,inv}∙[Inv_{0}-Inv]$$ $${\frac{dFruc}{dt}=K}_{2,inv}\bullet\left[Inv_0-Inv\right]$$ $$\frac{dGluc}{dt}=-K_{1,GOX}\bullet\ GOX_{prot}\bullet\ Gluc+K_{-1,GOX}\bullet[GOX_0-GOX]+K_{2,inv}∙[Inv_{0}-Inv]$$ $$\frac{d\left[GOX_{prot}\right]}{dt}=-K_{1,GOX}\bullet GOX_{prot}\bullet Gluc+K_{-1,GOX}\bullet[GOX_0-GOX]+K_{2,GOX}∙[GOX0-GOX]$$ $$\frac{dp_{tot,GOX}}{dt}=K_{2,GOX}\bullet[GOX_0-GOX]$$

  Results

  Based on the equations we plot the Figure 4:

  

  Figure 4 - Model of pH decrease over time

  As can be seen on Figure 5, the modeling results presented on Figure 4 match the results of the experiments performed in the wet-lab:

  

  Figure 5 - pH decrease over time in our wet lab experiment

• Enzyme kinetics considering the pH influence

  Overview

  This section is an integration of the previous two sections. To know how the pH change influences the kinetics of our enzymes, we have conducted a lab experiment, in which we have tested the activity of each enzyme (Invertase, GOx, and Catalase) separately. The experiment was conducted in the following pH values: 7, 6, 5 and 4.5. These values are selected since the initial pH of the honey production process is above 6, and at the end of the process, it reaches pH<4.5.

  Based on the experiment results, we have found the activity parameters in different pH values by normalizing the activity of the enzyme by the maximum activity, as shown in the following equation:

  $$k_{enzyme,pH,\ spicefic\ pH}=\frac{Activity\ in\ specific\ pH\ value}{Maximum\ activity}$$

  Wet-lab experiment

  The experiments had been conducted in appropriate buffers, made from monosodium-phosphate and diasodium-phosphate in varying amounts, in the total concentration of 0.1M, as specified in the protocol.

  The activity experiments based on the same reactions and absorbance-activity relations written in the enzymatic activity experiments above. To avoid the effects of other parameters, we have tested each pH level sample compared to its negative control (blank) solution that had the same pH level.

  Invertase

  We have performed an activity test on the invertase. The results demonstrate some influence of the pH level on the enzymatic activity, and an optimal pH level of 5 (Figure 6), which matches the data found in previous studies, see design page. However, the activity is not dramatically eliminated, indicate that the sucrose cleavage continues.

  

  Figure 6 - Absorbance at 492 nm, resulted from invertase activity in different pH levels

  pH	Absorbance at 492 nm	\(K_{enzyme,pH} \)
  4.5	0.05	0.608
  5	0.082	1
  6	0.0508	0.622
  7	0.01044	0.12

  Table 3 - Invertase activity parameters

  Glucose Oxidase

  The assay of both glucose oxidase and catalase relied on another enzyme present in the solution, horseradish peroxidase (HRP). Although the HRP was added to the solution in excess, we wanted to verify that the only variant is the glucose oxidase activity, i.e. that the other parameters of the experiment– HRP, ABTS, and hydrogen peroxide - are not affected by the pH level. Therefore, we had conducted a test using constant amounts of hydrogen peroxide, HRP and ABTS, and pH level as the sole variable. After we measured the absorbance, no significant difference observed between the different PH levels, implies that the only parameter afflicted by the pH is the enzyme teste (Figure 7).

  

  Figure 7 - verification of the method reliability by measuring HRP absorbance at 416nm for different PH levels.

  After we had verified the reliability of the method, we have performed the activity test. The results (Figure 8) indicate that glucose oxidase is nearly unaffected by the pH decrease.

  

  Figure 8 - Glucose oxidase activity in various pH levels by measuring the absorbance at 416nm.

  pH	Absorbance at 416 nm	\(K_{enzyme,pH} \)
  4.5	0.737	0.94
  5	0.75	1
  6	0.714	0.89
  7	0.73	0.93

  Table 4 - Glucose Oxidase activity parameters

  Catalase

  We have performed an activity test on the catalase enzyme. The results (Figure 9) demonstrate some influence of the pH level on the enzymatic activity, and an optimal pH level of 5, as documented in previous studies, see design page. However, the activity is not dramatically eliminated, indicates that the decomposition of hydrogen peroxide continues.

  

  Figure 9 - Absorbance difference at 416 nm, resulted from catalase activity in different pH levels

  pH	Absorbance difference at 416 nm	\(K_{enzyme,pH} \)
  4.5	0.064	0.64
  5	0.1	1
  6	0.072	0.72
  7	0.076	0.76

  Table 5 - Catalase activity parameters

  Mathematical equations

  Using the parameters found in our lab experiments, we have separated our mathematical equation into different ones for each pH range, as follow:

  $$\frac{dSucr}{dt}=-{k_{inv,pH,specific}\bullet K}_{1,inv}\bullet\ Inv\bullet\ Sucr+{k_{inv,pH,specific}\bullet K}_{-1,inv}\bullet[Inv_0-Inv]$$ $$\frac{dInv}{dt}=-k_{inv,pH,specific}\bullet\ K_{1,inv}\bullet\ Inv\bullet\ Sucr+{k_{inv,pH,specific}\bullet K}_{-1,inv}\bullet[Inv_0-Inv]+kinv,pH,7∙K_{2,inv}∙[Inv_{0}-Inv]$$ $${\frac{dFruc}{dt}=k_{inv,pH,specific}\bullet K}_{2,inv}\bullet\left[Inv_0-Inv\right]$$ $$\frac{dGluc}{dt}=-k_{GOX,pH,specific}\bullet\ K_{1,GOX}\bullet\ GOX_{prot}\bullet\ Gluc+k_{GOX,pH,specific}\bullet\ K_{-1,GOX}\bullet[GOX_0-GOX]+kinv,pH,7∙K_{2,inv}∙[Inv_{0}-Inv]$$ $$\frac{d\left[GOX_{prot}\right]}{dt}=-k_{GOX,pH,specific}\bullet K_{1,GOX}\bullet GOX_{prot}\bullet Gluc+k_{GOX,pH,specific}\bullet K_{-1,GOX}\bullet[GOX_0-GOX]+kGOX,pH,7∙K_{2,GOX}∙[GOX0-GOX]$$ $$\frac{dGDL}{dt}=k_{GOX,pH,specific}\bullet\ K_{2,GOX}\bullet[GOX_0-GOX]$$ $$\frac{dH_2O_2}{dt}=-{k_{Cat,pH,specific}\bullet K}_{1,cat}\bullet\ Cat\bullet\ H_2O_2+{k_{Cat,pH,specific}\bullet K}_{-1,cat}\bullet\left[Cat_0-Cat\right]+k_{GOX,pH,specific}\bullet K_{2,GOX}\bullet[GOX_0-GOX]$$ $$\frac{dp_{tot,cata}}{dt}=k_{Cat,pH,specific}\bullet\ K_{-1,cat}\bullet\left[Cat_0-Cat\right]$$ $$\frac{dCat}{dt}=-{k_{Cat,pH,specific}\bullet K}_{1,cat}\bullet\ Cat\bullet\ H_2O_2+k_{Cat,pH,specific}\bullet\ K_{-1,cat}\bullet\left[Cat_0-Cat\right]+{k_{Cat,pH,specific}\bullet K}_{2,cat}\bullet\left[Cat_0-Cat\right]$$

  Results

  As expected, and shown in figure 9, due to the Invertase activity, the sucrose levels decrease over time, while the glucose and fructose increase. Along with the rise in glucose levels, the GOx enzyme breaks it down to GDL and hydrogen peroxide, so their levels raised as well, as shown in the graph. However, as a result of the catalase enzyme activity in the native B. Subtilis, hydrogen peroxide levels not only increase at a lower rate compared to GDL, but also at some point it starts to decrease.

  

  Figure 10 - Products concentrations over time - whole enzyme system with pH influence

  

  Figure 11 - Products concentrations over time - whole enzyme system without pH influence

  Comparing figure 10 to figure 11, we can see the pH influence on the enzymes’ activity. It does not only show different behavior but also slower reactions overall.
  For example, after 400 seconds, the glucose concentration start decreasing, in contrast to the behavior presented in Figure 11, which demonstrates the moderate increase and then decreasing just after 950 seconds approximately.

  Moreover, the fructose and GDL reach plateau simultaneously around 1300 seconds in figure 10, while without pH influence fructose reaches the plateau at 900 seconds and GDL at 1200 seconds.

  

  Figure 12 - \(H_2O_2\) concentration over time - whole enzyme system with pH influence

  

  Figure 13 - \(H_2O_2\) concentration over time - whole enzyme system without pH influence

  Figures 12 and 13 are the zoom-in versions of figures 10 and 11 respectively. As can be seen, with pH influence, there is a change in the behavior of the hydrogen peroxide curve between 0 to 1400 seconds. These results can be due to the effect of the hydrogen peroxide, and its degradation products: water and oxygen, on the pH of the total solution.

• Conclusion

  In light of the information obtained in this section, we have achieved the following conclusions:

  • - The pH in our system drastically decreases. Therefore, accounting its influence on the enzymes was essential.
  • - The “Honey Circuit” is crucial to obtain the desired composition profile along with controlling the pH change in our system to match the “real Honey”.

  
  
  

Regulated Enzymes Factory using The Honey Circuit

• Overview

  In our Synthetic Honey Stomach, we use B. subtilis to secrete the enzymes involved in the honey-making process. Our goal is to model the secretion system as a whole, considering the bacterial growth. First, we have modeled the case in which we have a single bacterium. [See our honey circuit]. Then, along with bacterial growth and the enzyme secretion model, we have successfully built a comprehensive model that can estimate the number of bacteria and enzymes present in the system.

• The Honey Circuit

  The Honey Circuit

  To have better control over our system, we have decided to include a circuit that senses the composition of the solution and adjusts the production of enzymes accordingly. The precision of the concentrations in our final product was one of the most important aspects of our project, which means our purpose was to make our system as stable as possible. Thus, based on various references^[13][14] we decided to apply a synthetic negative feedback mechanism on the concentration of \(H_2O_2\). [See our honey circuit]

  
  

  

  Figure 14 - Honey Circuit

  
  

  Negative Feedback

  As in signal theory, negative feedback is a system where the output acts as the repressor of his synthesis, making the system "auto-regulated". In a genetic circuit, negative feedback has many advantages that fit our purpose perfectly:

  • Negative Feedback is known to speed up the reaction and allow it to reach a steady-state rapidly^[15]

  • Negative Feedback is known to enhance robustness^[16], which means that the deviation from the mean in our bacteria population is smaller.

   

  Time scale considerations and assumption

  For a LacI protein to be synthesized, a hydrogen peroxide molecule needs to bind to a perR transcription factor. This transcription factor unbinds the katA promoter, resulting in the transcription of LacI RNA.

  The binding of \(H_2O_2\) to perR transcription factor reaches equilibrium on a sub-second timescale (around the millisecond), considerably faster than the active transcription factor binding to the DNA site. Therefore, we assume that this \(H_2O_2\) /PerR binding reaction is constantly at steady-state^[17].

  Steady State of PerR/\(H_2O_2\) binding

  To find the appropriate steady-state value of the active PerR transcription factor, we used the thermodynamics model. In this approach, we first find all the possible states of bound/unbound combinations in our system and assign a weight to each state, according to its concentration and the affinity between binding sites. We then derive from those weights the probability of each state by dividing each weight by the sum of all weights. This leads us to the steady-state concentration of our transcription factor.

  In our case:

  State	Weight	Probabilities	Output
  \(H_2O_2\) not bound to PerR	1	$$\left(\frac{1}{1+\frac{\left[H_2O_2\right]}{K_{D,pKatA}}}\right)$$	None
  \(H_2O_2\) bound to PerR	$${\frac{\left[H_2O_2\right]}{K_{D,pKatA}}}$$	$$\left(\frac{\frac{\left[H_2O_2\right]}{K_{D,pKatA}}}{1+\frac{\left[H_2O_2\right]}{K_{D,pKatA}}}\right)$$	LacI mRNA synthesis

  Table 6 Events probabilities in pKatA transcription factors

  And for the pLac promoter:

  State	Weight	Probabilities	Output
  LacI not bound to pLac	1	$$\left(\frac{1}{1+\frac{\left[LacI\right]}{K_{D,pLac}}}\right)$$	GOx mRNA
  LacI bound to pLac	$${\frac{\left[LacI\right]}{K_{D,pLac}}}$$	$$\left(\frac{\frac{\left[LacI\right]}{K_{D,pLac}}}{1+\frac{\left[LacI\right]}{K_{D,pLac}}}\right)$$	None

  Table 7 Events probabilities in pLac transcription factors

  Parameters

  Finding the correct parameters is a difficult yet paramount step to get an accurate description of our system. To find these parameters, we have searched the literature and also found a former iGEM team that worked with the same Lac promoter, which helped us a lot in our research [iGEM 2017 William and mary team]. We found the following parameters:

  Parameters	Value
  $$K_{lacI}^{RNA} = K_{GOX}^{RNA}$$	\(8.7*10^{-8} [mM/sec]\)
  $$K_{lacI}^{protein} = K_{GOX}^{protein}$$	\(0.053 sec^{-1}\)
  $$\gamma_{LacI}^{RNA} = \gamma_{GOX}^{RNA}$$	\(5e-4 sec^{-1}\)
  $$K_{D,pKatA}=K_{D,pLac}$$	\(10*10^{-10} [mM]\)
  $$\gamma_{LacI}^{protein}$$	\(6.08e-4 sec^{-1}\)
  $$\gamma_{GOX}^{protein}$$	\(4.01e-6 sec^{-1}\)

  Table 8 - Honey circuits parameters

  We did not find a value for \(K_{D,pKatA} \) in literature, so we assumed it has the same value as \(K_{D,pLac} \).
  To determine the degradation constant of the proteins, we first found their half-life:
  

  Enzyme	Value
  $$t_{GOX\ enzyme}^{\frac{1}{2}}$$	1140 [sec]
  $$t_{Lac\ protein}^{\frac{1}{2}}$$	172800 [sec]

  Table 9 - Half life parameters of the enzymes

  
  Based on the following formula we have calculated \(\gamma_{degradation}\ \) $$\gamma_{degradation}=\frac{\ln{\left(2\right)}}{t^\frac{1}{2}}$$.

  Simulation of isolated circuit

  To examine the behavior of our model alone, we have used GOx as the autoregulated element. This was an appropriate approximation as the concentrations of GOx and \(H_2O_2\) should be strongly correlated. Hence, we used those equations:

  
  \( \frac{d\left[LacI_{RNA}\right]}{dt} = K_{lacI}^{RNA}\bullet\left(\frac{\frac{\left[GOX_{prot}^{produced}\right]}{K_{D,pKatA}}}{1+\frac{\left[GOX_{prot}^{produced}\right]}{K_{D,pKatA}}}\right)-\gamma_{LacI}^{RNA}\bullet[LacI_{RNA}]\)
  
  \(\frac{d\left[LacI_{prot}\right]}{dt} = K_{lacI}^{protein}\bullet[LacI_{RNA}]-γ_{LacI}^{protein}∙[LacIprot]\)
  
  \(\frac{d\left[{GOX}_{RNA}\right]}{dt} = K_{GOX}^{RNa}\bullet\left(\frac{1}{1+\frac{\left[LacI_{prot}\right]}{K_{D,pLac}}}\right)-\gamma_{GOX}^{RNA}\bullet[{GOX}_{RNA}]\)
  
  \(\frac{d\left[GOX_{prot}^{produced}\right]}{dt} = K_{GOX}^{protein}\bullet {GOX}_{RNA} - \gamma_{GOX}^{protein}\bullet[GOX_{prot}^{produced}]\)
  
  
  

  Using the lsoda numerical method in the Scipy solver, we have received the following results:

  

  Figure 15 - Proteins concentrations over time in presence of the Honey circuit

  
  

  As we can see from Figure 2, an increase in GOx leads to the transcription of its repressor, LacI, which results in GOx decrease. Using this approximation, we got a characteristic damped oscillation behavior in our system. It characterizes the negative feedback system, in which the autoregulation leads the product to oscillate around its steady-state value.

  Stochastic modeling

  We decided to use another simulation technique to confirm the results received by the ODE method. We have used the Gillespie algorithm^[21]
  As part of the Gillepsie algorithm, we randomly picked two numbers: one of them determines the time of the next reaction and the second one determines which reaction will occur next from all the possible reactions in our system (RNA synthesis, protein degradation, substrate/enzyme binding, etc.)
  After each step, we update the state of our system (in terms of quantities of each element) depending on the reaction that has just taken place. We have also updated the probabilities of each reaction at each step, since the probability of each reaction depends on the concentrations of the reactants, according to Michaelis Menten model.

  Event	Probability
  LacI RNA synthesis	$$8.7*10^{-8}*\left(\frac{\frac{\left[H_2O_2\right]}{10*10^{-10}}}{1+\frac{\left[H_2O_2\right]}{10*10^{-10}}}\right)$$
  LacI RNA degradation	\(5*10^{-4}[LacI_{RNA}]\)
  LacI synthesis	\(0.053[LacI_{RNA}]\)
  LacI degradation	\(6.08*10^{-4} [LacI]\)
  GOx RNA synthesis	\(8.7*10^{-8} * \left(\frac{1}{1+\frac{\left[LacI\right]}{K_{D,pLac}}}\right) \)
  GOx RNA degradation	\(5*10^{-4}[LacI_{RNA}]\)
  GOx synthesis	\(0.053[GOX_{RNA}]\)
  GOx degradation	\(4.01*10^{-6} [GOX]\)

  Table 10 - Events probabilities used in Gillepsie stochastic simulation

  Using this technique we achieved the following results:

  

  Figure 16 - Stochastic simulation of our isolated Honey Circuit

  As can be seen in Figure 16, the signal is noisy due to the technique used.
  To get a more accurate result, we ran the same experiment 10 times and calculated the average results.
  This graph presents the same oscillatory behavior we recieved in the ODE method although less damped in this case.
  In conclusion, our two approaches led us to an oscillatory behavior of our system that according to our ODE model quickly stabilizes to its final values. We can thus ensure that our system is stable and robust due to our negative feedback. The next step was to implement these results to our enzymes equations.
  

• Bacterial growth and enzymes secretion

  Bacterial growth

  In order to better understand how our Bacillus subtilis grows, we have conducted a wet lab experiment and received the following results:

  

  Figure 17 - Experimental bacterial growth over time

  As we can see in figure 17, our Bacillus subtilis grows exponentially with growth parameter \(\mu=1.045\) [hour].

  We used the following equation: $$ \frac{dBacteria}{dt}=\mu\bullet\ Bacteria $$

  Based on the equation and the growth's parameter we received in Figure 18 which approximately matches Figure 17.

  

  Figure 18 - Bacterial growth over time

  Enzymes secretion

  Since we could not find conclusive, unambiguous data in the professional literature regarding our proteins' secretin, we have concluded the amount of the active proteins that had been secreted based on the activity measured. Thus, the percentage of the supernatant activity was used to indicate the amount of the secreted enzyme, i.e., we have calculated the activity based on the following formula:

  $$Secreted\ Protein\ Percentage=\frac{Activity\ in\ Supernatant\ }{Total\ Activity}$$

  The supernatant activity measured in the Invertase supernatant was 0.12U/ml, while the total activity measured was 0.18 U/ml. Therefore, we evaluate the secretion percentage to be 40%. The Supernatant activity measure in the Glucose Oxidase solution was 0.45 U/ml while the total activity measured was 0.91 U/ml, therefore, we conclude that the secretion percentage is 49.4%. As for the catalase- the activity of the supernatant was 12.27 U/ml while the total activity measured was 22.26 U/ml, therefore, we conclude that the secretion percentage is 35.55%.

• Overall Enzymes Factory

  We added the bacterial growth equation to our set of Honey-Circuit ODE and multiplied each enzyme equation by the number of bacteria considering the derivative of multiplication. Finally, we get the following equations:

  \(\frac{d\left[LacI_{RNA}\right]}{dt}=K_{lacI}^{RNA}\bullet\left(\frac{\frac{\left[GOX_{prot}^{total\ prod}\right]}{K_{D,pKatA}}}{1+\frac{\left[GOX_{prot}^{total\ prod}\right]}{K_{D,pKatA}}}\right)-\gamma_{LacI}^{RNA}\bullet[LacI_{RNA}]\)
  
  \(\frac{d\left[LacI_{prot}\right]}{dt}=K_{lacI}^{protein}\bullet[LacI_{RNA}]-γ_{LacI}^{protein}∙[LacIprot]\)
  
  \(\frac{d\left[{GOX}_{RNA}\right]}{dt}=K_{GOX}^{RNA}\bullet\left(\frac{1}{1+\frac{\left[LacI_{prot}\right]}{K_{D,pLac}}}\right)-\gamma_{GOX}^{RNA}\bullet[{GOX}_{RNA}]\)
  
  \(\frac{d[GOX_{prot}^{produced}]}{dt}=K_{GOX}^{protein}∙[GOX_{RNA}]-γ_{GOX}^{protein}∙GOX_{prot}^{produced}\)
  
  \(\frac{dBacteria}{dt}=\mu\bullet\ Bacteria\)
  
  \(\frac{d[GOX_{prot}^{total\ prod}]}{dt}=K_{secretion}K_{GOX}^{protein}∙[GOX_{RNA}]-γ_{GOX}^{protein}∙GOX_{prot}^{produced}∙Bacteria+GOX_{prot}^{produced}∙μ∙Bacteria\)
  
  

  The solution of above equations gave us the results presented on Figure 19:

  

  Figure 19 - Stabilization of overall Glucose Oxidase concentration as a result of the Honey-Circuit

  From Figure 19, we can obtain information about the total GOx concentration in our system. We can see that when the concentration of GOx stabilizes by autoregulation and the bacterial growth reaches the stationary phase, our overall GOx concentration stabilizes as well. If we zoom in, we can still see damped oscillations in the overall GOx concentration (Figure 20). This is expected regarding the results we got for single bacteria. We can assume that in real conditions all the bacteria will not be in the same phase. Considering that, the concentration of GOx should display less oscillatory behavior.

  

  Figure 20 - Damped oscillations in overall GOx concentration

Synthetic Honey Stomach - Metabolic Pathway

• Overview

  
  In this section, we have modeled the integration of both previous sections- the "Enzymatic Processing of Nectar" and "Regulated Enzyme Factory". We aimed to understand the interplay between the enzymes and their enzymatic reaction products, in the presence of our “Honey Circuit”. To do that we have followed several steps, each introducing another enzyme. In the previous section, we have modeled the "Honey Circuit" and included in the Enzyme Factory, using GOx as the auto-regulated element. Subsequently, our first step in this section is modeling the Enzyme Factory using \(H_2O_2\) as the auto-regulated element (simulation of isolated honey circuit). Consequently, we have added the two other enzymes, Invertase and Catalase, and finally the pH influence on all enzymes

• Enzyme Factory With GOx Only

  Overview

  After we have seen that our synthetic circuit works as expected, we have modeled the Enzyme Factory using \(H_2O_2\) as the auto-regulated element. That means, using the Honey Circuit allows the bacteria to sense the concentration of \(H_2O_2\), and as a result, represses the transcription of GOx. Since GOx is responsible for hydrolyzing glucose to form \(H_2O_2\) and GDL, the concentration of \(H_2O_2\) will decrease once the transcription of GOx decreases. To model this negative feedback, we needed to account for the relation between \(H_2O_2\) and GOx along with regulation of the genes.

  Mathematical equations

  \(\frac{d\left[LacI_{RNA}\right]}{dt}=K_{lacI}^{RNA}\bullet\left(\frac{\frac{\left[H_2O_2\right]}{K_{D,pKatA}}}{1+\frac{\left[H_2O_2\right]}{K_{D,pKatA}}}\right)-\gamma_{LacI}^{RNA}\bullet[LacI_{RNA}]\)
  
  \(\frac{d\left[LacI_{prot}\right]}{dt}=K_{lacI}^{protein}\bullet[LacI_{RNA}]-γ_{LacI}^{protein}∙[LacIprot]\)
  
  \(\frac{d\left[{GOX}_{RNA}\right]}{dt}=K_{GOX}^{RNA}\bullet\left(\frac{1}{1+\frac{\left[LacI_{prot}\right]}{K_{D,pLac}}}\right)-\gamma_{GOX}^{RNA}\bullet[{GOX}_{RNA}]\)
  
  \(\frac{d[GOX_{prot}^{produced}]}{dt}=K_{GOX}^{protein}∙[GOX_{RNA}]-γ_{protein}^{GOx}∙GOX_{prot}^{produced}\)
  
  \(\frac{dBacteria}{dt}=\mu\bullet\ Bacteria\)
  
  \(\frac{d[GOX_{prot}^{total prod}]}{dt}=K_{secretion}\left(\left(K_{GOX}^{protein}\bullet\left[{GOX}_{RNA}\right]-\gamma_{GOX}^{protein}\bullet\left[GOX_{prot}^{produced}\right]\right)\bullet B a c t e r i a+GOX_{prot}^{produced}\bullet\mu\bullet B a c t e r i a\right)\)
  
  \(\frac{d\left[GOX_{prot}\right]}{dt}=-K_{1,GOX}\bullet GOX_{prot}\bullet Gluc+K_{-1,GOX}\bullet{Comp}_{GlucGOX}+K_{2,GOX}\bullet{Comp}_{GlucGOX}\)
  \(+K_{secretion}\left(\left(K_{GOX}^{protein}\bullet\left[{GOX}_{RNA}\right]-\gamma_{GOX}^{protein}\bullet\left[GOX_{prot}^{produced}\right]\right)\bullet B a c t e r i a+GOX_{prot}^{produced}\bullet\mu\bullet B a c t e r i a\right)\)
  
  \(\frac{dH_2O_2}{dt}=K_{2,GOX}\bullet{Comp}_{GlucGOX}\)
  
  \(\frac{dGluc}{dt}=-K_{1,GOX}\bullet\ GOX_{prot}\bullet\ Gluc+K_{-1,GOX}\bullet{Comp}_{GlucGOX}\)
  \(+K_{2,inv}\bullet[Inv_0-Inv]\)
  
  \(\frac{d\left[{Comp}_{GlucGOX}\right]}{dt}=K_{1,GOX}\bullet GOX_{prot}\bullet Gluc-K_{-1,GOX}\bullet{Comp}_{GlucGOX}-K_{2,GOX}\bullet{Comp}_{GlucGOX}\)
  
  \(\frac{dGDL}{dt}=K_{2,GOX}\bullet{Comp}_{GlucGOX}\)
  
  

  Results and conclusions

  At the cell level, we saw repression of GOx as a result of LacI stabilization (Figure 21)

  

  Figure 21 - LacI and GOx concentrations per bacteria as a result of the gene regulation along with the enzymatic relation between \(H_2O_2\) and GOx

  At the system level, we received a stabilization of our \(H_2O_2\) and glucose concentrations

  

  Figure 22 - \(H_2O_2\) and glucose concentrations over time under our "Honey circuit"

  Conclusions:

  This is not only a proof that our Honey Circuit works, but also that our previous approximation, that we can model our Honey Circuit using GOx as the auto-regulated element,is valid.(See simulation of isolated honey circuit).

• Enzyme Factory With Enzyme Combination

  Overview

  In this section, we modeled the integration of our three enzymes and their kinetics, as well as the Enzyme Factory regulated by the “Honey Circuit”.

  Mathematical equations

  $$\frac{d\left[LacI_{RNA}\right]}{dt}=K_{lacI}^{RNA}\bullet\left(\frac{\frac{\left[H_2O_2\right]}{K_{D,pKatA}}}{1+\frac{\left[H_2O_2\right]}{K_{D,pKatA}}}\right)-\gamma_{LacI}^{RNA}\bullet[LacI_{RNA}]$$ $$\frac{d\left[LacI_{prot}\right]}{dt}=K_{lacI}^{protein}\bullet[LacI_{RNA}]-γ_{LacI}^{protein}∙[LacI_{prot}]$$ $$\frac{d\left[{GOX}_{RNA}\right]}{dt}=K_{GOX}^{RNA}\bullet\left(\frac{1}{1+\frac{\left[LacI_{prot}\right]}{K_{D,pLac}}}\right)-\gamma_{GOX}^{RNA}\bullet[{GOX}_{RNA}]$$ $$\frac{d[GOX_{prot}^{produced}]}{dt}=K_{GOX}^{protein}∙[GOX_{RNA}]-γ_{GOx}^{protein}∙GOX_{prot}^{produced}$$ $$\frac{dBacteria}{dt}=\mu\bullet\ Bacteria$$ $$\frac{d[GOX_{prot}^{total prod}]}{dt}=K_{secretion}((K_{GOX}^{protein}∙GOX_{RNA}-γ_{GOX}^{protein}∙GOX_{prot}^{produced})∙Bacteria+GOX_{prot}^{produced}∙μ∙Bacteria$$ $$\frac{d\left[GOX_{prot}\right]}{dt}=-K_{1,GOX}\bullet GOX_{prot}\bullet Gluc+K_{-1,GOX}\bullet{Comp}_{GlucGOX}+K_{2,GOX}\bullet{Comp}_{GlucGOX}$$$$\ \ \ +K_{secretion}\left(\left(K_{GOX}^{protein}\bullet\left[{GOX}_{RNA}\right]-\gamma_{GOX}^{protein}\bullet\left[GOX_{prot}^{produced}\right]\right)\bullet B a c t e r i a+GOX_{prot}^{produced}\bullet\mu\bullet B a c t e r i a\right)$$ $$\frac{dH_2O_2}{dt}=K_{2,GOX}\bullet{Comp}_{GlucGOX}$$ $$\frac{dGluc}{dt}=-K_{1,GOX}\bullet\ GOX_{prot}\bullet\ Gluc+K_{-1,GOX}\bullet{Comp}_{GlucGOX}$$$$\ \ \ +K_{2,inv}\bullet[Inv_0-Inv]$$ $$\frac{d\left[{Comp}_{GlucGOX}\right]}{dt}=K_{1,GOX}\bullet GOX_{prot}\bullet Gluc-K_{-1,GOX}\bullet{Comp}_{GlucGOX}-K_{2,GOX}\bullet{Comp}_{GlucGOX}$$ $$\frac{dGDL}{dt}=K_{2,GOX}\bullet{Comp}_{GlucGOX}$$ $$\frac{dSucr}{dt}=-K_{1,inv}\bullet\ Inv\bullet\ Sucr+K_{-1,inv}\bullet[Inv_0-Inv]$$ $$\frac{dInv}{dt}=-K_{1,inv}\bullet\ Inv\bullet\ Sucr+K_{-1,inv}\bullet[Inv_0-Inv]+K_{2,inv}∙[Inv_{0}-Inv]$$ $${\frac{dFruc}{dt}=K}_{2,inv}\bullet\left[Inv_0-Inv\right]$$

  Results

  As we can see in Figure 23, our “Honey Circuit” stabilizes the concentrations of glucose and GDL to a steady stat value, In contrary to the results we received without the “Honey Circuit”, in which glucose decreases to zero. [See Enzyme kinetics - results]

  

  Figure 23 - Sugars and products evolution over time in presence of the “Honey Circuit”, Invertase, GOx and catalase

  If we zoom in on our \(H_2O_2\) concentration, we can see that the concentration stabilizes before its final degradation by catalase (Figure 24).

  

  Figure 24 - Evolution of \(H_2O_2\) concentration over time in presence of the "Honey Circuit", Invertase, GOx and catalase

• Enzyme Factory With Enzyme Combination Under pH Influence

  Overview

  In this section, we integrated all the parts mentioned before, including the enzyme kinetic, secreted and regulated by our Enzyme Factory, and their effect from the pH change. This gives us a comprehensive model of the whole “Synthetic Honey Stomach” metabolic pathway.

  Mathematical equations

  As in previous section [See Enzymes kinetics and pH influence], we have separated our equation to different pH ranges:

  $$\frac{d\left[LacI_{RNA}\right]}{dt}=K_{lacI}^{RNA}\bullet\left(\frac{\frac{\left[H_2O_2\right]}{K_{D,pKatA}}}{1+\frac{\left[H_2O_2\right]}{K_{D,pKatA}}}\right)-\gamma_{LacI}^{RNA}\bullet[LacI_{RNA}]$$ $$\frac{d\left[LacI_{prot}\right]}{dt}=K_{lacI}^{protein}\bullet[LacI_{RNA}]-γ_{LacI}^{protein}∙[LacI_{prot}]$$ $$\frac{d\left[{GOX}_{RNA}\right]}{dt}=K_{GOX}^{RNA}\bullet\left(\frac{1}{1+\frac{\left[LacI_{prot}\right]}{K_{D,pLac}}}\right)-\gamma_{GOX}^{RNA}\bullet[{GOX}_{RNA}]$$ $$\frac{d[GOX_{prot}^{produced}]}{dt}=K_{GOX}^{protein}∙[GOX_{RNA}]-γ_{GOx}^{protein}∙GOX_{prot}^{produced}$$ $$\frac{dBacteria}{dt}=\mu\bullet\ Bacteria$$ $$\frac{d[GOX_{prot}^{total prod}]}{dt}=K_{secretion}∙((K_{GOX}^{protein}∙GOX_{RNA}-γ_{GOX}^{protein}∙GOX_{prot}^{produced})∙Bacteria+GOX_{prot}^{produced}∙μ∙Bacteria)$$ $$\frac{d\left[GOX_{prot}\right]}{dt}=-k_{GOx, pH dependant}∙K_{1,GOX}\bullet GOX_{prot}\bullet Gluc+k_{GOx, pH dependant}∙K_{-1,GOX}\bullet{Comp}_{GlucGOX}+k_{GOx, pH dependant}∙K_{2,GOX}\bullet{Comp}_{GlucGOX}$$ $$+K_{secretion}\left(\left(K_{GOX}^{protein}\bullet\left[{GOX}_{RNA}\right]-\gamma_{GOX}^{protein}\bullet\left[GOX_{prot}^{produced}\right]\right)\bullet B a c t e r i a+GOX_{prot}^{produced}\bullet\mu\bullet B a c t e r i a\right)$$ $$\frac{dGluc}{dt}=-k_{GOx, pH dependant}∙K_{1,GOX}\bullet\ GOX_{prot}\bullet\ Gluc+k_{GOx, pH dependant}∙K_{-1,GOX}\bullet{Comp}_{GlucGOX}+k_{Inv, pH dependant}∙K_{2,inv}\bullet[Inv_0-Inv]$$ $$\frac{d\left[{Comp}_{GlucGOX}\right]}{dt}=k_{GOx, pH dependant}∙K_{1,GOX}\bullet GOX_{prot}\bullet Gluc-k_{GOx, pH dependant}∙K_{-1,GOX}\bullet{Comp}_{GlucGOX}-k_{GOx, pH dependant}∙K_{2,GOX}\bullet{Comp}_{GlucGOX}$$ $$\frac{dGDL}{dt}=k_{GOx, pH dependant}∙K_{2,GOX}\bullet{Comp}_{GlucGOX}$$ $$\frac{dH_2O_2}{dt}=-k_{Cat, pH dependant}∙K_{1,Cat}∙Cat∙H_2O_2+k_{Cat, pH dependant}∙K_{-1,Cat}∙(Cat_0-Cat)$$ $$+k_{GOx, pH dependant}∙K_{2,GOX}\bullet{Comp}_{GlucGOX}$$ $$\frac{dP_{tot,cat}}{dt}=k_{Cat, pH dependant}∙K_{-1,Cat}∙(Cat_0-Cat)$$ $$\frac{dCat}{dt}=-k_{Cat, pH dependant}∙K_{1,Cat}∙Cat∙H_2O_2+k_{Cat, pH dependant}∙K_{-1,Cat}∙(Cat_0-Cat)+k_{Cat, pH dependant}∙K_{2,Cat}∙(Cat_0-Cat)$$ $$\frac{dSuc}{dt}=-k_{Inv, pH dependant}∙K_{1,inv}\bullet\ Inv\bullet\ Sucr+k_{Inv, pH dependant}∙K_{-1,inv}\bullet\left[Inv_0-Inv\right]$$ $$\frac{dInv}{dt}=-k_{Inv, pH dependant}∙K_{1,inv}\bullet\ Inv\bullet\ Sucr+k_{Inv, pH dependant}∙K_{-1,Inv}\bullet[Inv_0-Inv]+k_{Inv, pH dependant}∙K_{2,inv}∙[Inv_{0}-Inv]$$ $$\frac{dFruc}{dt}=k_{Inv, pH dependant}∙K_{2,inv}\bullet\left[Inv_0-Inv\right]$$
• Results

  
  In figure 25, we can see that the final concentrations of part of our products changed. Since the pH change affects the Glucose oxidase, GDL concentration is stabilized at a lower value, approximately 30% lower than its value without pH consideration. For the same reason, glucose gets to a higher steady-state. Moreover, the overall reaction is slower when considering the pH influence.

  

  Figure 25 - Sugars and products evolution over time in presence of the "Honey Circuit" and all of enzymes while considering the pH influence

Integration Between Modeling and Wet Lab Work

To forecast the results of future experiments with commercial enzymes, we adapted our model to the experimental results previously obtained with the same enzymes.
In a previous experiment, in which we have tested the sugar levels, we received the following results:

Figure 26 Lab results - Degradation of sucrose

Based on Km, a Michaelis constant, mentioned in the commercial Invertase information sheet, along with the calculation of K1 from the results shown in figure 26, we have determined also K2, the turnover number of the enzyme. Moreover, relying on the activity ratio of GOx and Invertase giving in the product sheet, we have determined the GOx kinetics constants.
Overall, using the same algorithm we have developed to model the composition profile in the presence of the three enzymes, considering their influence by the pH change, we received the following results:

Figure 27 Adapted model of sucrose degradation

Comparing figure 26 to figure 27, we can see that our theoretical model matches the wet lab results, as the sucrose reaches its half initial value in 7 hours. However, afterward, the theoretical model no longer applies. In our opinion, this was for the reason that the invertase level reached zero through the reaction because it has a half-life of 10 hours^[22], whereas our model assumes that we have a constant value of Invertase.
Eventually, we have used this optimized Model in order to design future experiments.