This year, our project focuses on the synthesis of abscisic acid (ABA), which involves a total of four enzymes, BcABA1-4. Since we attempt to construct an ABA expression system in Saccharomyces cerevisiae for the first time, it is necessary to describe the enzyme system by enzyme kinetic method.
We have two main tasks for this part of modeling. First, due to insufficient literature resources and experimental data, we are unable to determine the enzyme kinetics parameters, so we want to predict the BcABA parameters with the kinetic parameters through enzymes with high homology. Second, we hope to obtain the optimal ratio of BcABA1-4 enzymes in biochemical reactions through comprehensive enzyme kinetic simulations to achieve the most efficient heterologous synthesis of abscisic acids.
Restatement of the Problem
Before the start of modeling, let's review the BcABA-mediated biochemical pathways. In Botrytis cinerea, the bcABA gene cluster is composed of four genes. Among them bcABA1,2 expresses cytochrome P450 monooxygenase, bcABA4 expresses short-chain dehydrogenase/reductase, and bcABA3 is a member of the sesquiterpene synthase family. Farnesyl diphosphate (FPP) is first converted to the α-ionylideneethane catalyzed by BcABA3 and subsequently converted to α-ionylideneacetic acid by the carboxylation of BcABA1. Then the latter is converted to 1',4'-trans-dihydroxy-α-ionylideneacetic acid by BcABA2. Eventually, α-ionylideneacetic acid is converted to ABA with the help of NAD+ and BcABA4 [1]. In fact, the synthesis of ABA has been reported for a long time in fungi [2], but a complete description of the synthetic pathway has only appeared recently. Compared to the ABA synthesis pathway in plants [3], the pathway in Botrytis cinerea is undoubtedly much simpler. Also, fermentation through BcABA in yeast must have a better affinity than in bacteria.
Fig 1. BcABA-catalyzed ABA synthesis pathway.
Besides, since the heterologous expression of enzymes causes additional energy and resource consumption, it is necessary to give the optimal ratio of the four enzymes for adjusting their order on the polycistron by modeling. Therefore, we consider the enzyme kinetics to achieve this goal. In the field of enzyme kinetics, the most classical model must be the Michaelis-Menten equation, while most of our modeling is based on this. So before we start modeling, let’s briefly review the traditional form of the Michaelis-Menten equation.
Michaelis-Menten equation
Michaelis-Menten equation was originally proposed in 1913 by Leonor Michaelis and Maud Menten [4]. It describes the relationship between substrate concentration and reaction rate. In general, the chemical relationship between substrate, enzyme and product can be expressed by the following equation:
$$E+S \mathop{\rightleftharpoons}\limits^{k_1}\limits_{k_{-1}}
ES \xrightarrow{k_2} E+P \tag{1.1}$$
In this equation, $E$, $S$ and represents free enzyme, substrate and product respectively, while $ES$ stands for enzyme-substrate complex. And $k_1$, $k_2$ and $k_{-1}$ are the constant rate of reaction.
Based on the mass action law, we can list the ordinary differential equations:
$$\frac{d[S]}{dt}=-k_1[E][S]+k_{-1}[ES] \tag{1.2}$$
$$\frac{d[E]}{dt}=-k_1[E][S]+k_{-1}[ES]+k_2[ES]\tag{1.3}$$
$$\frac{d[P]}{dt}=k_2[ES]\tag{1.4}$$
$$\frac{d[ES]}{dt}=k_1[E][S]-k_{-1}[ES]-k_2[ES]\tag{1.5}$$
Where $[S]$ represents the concentration of the substrate in the system. Here we set the total concentration of all enzymes in the system as a constant, as $[E_0] = [E] + [ES]$. Given that $d[ES]/dt = 0$, we get
$$k_1[E][S]=(k_{-1}+k_2)[ES]\tag{1.6}$$
According to the definition of $K_m$,
$$K_m= \frac{[E][S]}{[ES]} = \frac{k_{-1}+k_2}{k_1}\tag{1.7}$$
Base on the relationship between $[E_0]$, $[E]$ and $[ES]$, we can get the equations:
$$K_m=\frac{([E_0]-[ES])[S]}{[ES]}$$
$$(K_m+[S])[ES] = [S][E_0]$$
$$[ES] = \frac{[S][E_0]}{K_m+[S]}\tag{1.8}$$
Then equation(1.4) can be rewrite as:
$$\frac{d[P]}{dt} =\frac{k_2 [S][E_0]}{K_m+[S]}\tag{1.9}$$
Which is the traditional form of Michaelis- Menten equation. Usually, $k2 [E_{0}]$ is written as $V_{max}$, which represents the maximum value of the reaction rate when the substrate saturates all the enzymes. $k_2$ is usually written as $k_{cat}$, denotes the molecular weight of the substrate transformed by each enzyme per unit time [4,5,6].
Usually, we obtain the $K_m$ and $k_2$ through the literature. However, only through the above formula, we can not get the values of $k_1$ and $k_{-1}$, so it’s impossible to judge what kind of enzyme proportion is most conducive to the accumulation of products.
For the formula above, the value of $\frac{d[P]}{dt}$ is the opposite of $\frac{d[S]}{dt}$ (supposed that the $[ES]$ change rate is 0), as
$$\frac{-d[S]}{dt}=\frac{k_2[E_0][S]}{K_m+[S]}\tag{1.10}$$
Considering that $[S]$ is a function about $t$, we deform the equation and obtain the integral on both sides,
$$-\int^{[S_1]}_{[S_0]}\frac{K_m+[S]}{[S]}d[S]=\int^{t_1}_{t_0}k_2[E_0]dt$$
$$-(K_m ln[S]+[S]) \vert^{[S_1]}_{[S_0]} = k2[E_0]t|^{t_1}_{t_0}$$
$$-\Delta[S]+K_mln\frac{[S_0]}{[S_1]}=k_2[E_0]\Delta t \tag{1.11}$$
$$[S_1]+K_mln[S_1]=[S_0]+K_mln[S_0]-k_2[E_0]\Delta t\tag{1.12}$$
So we can deduce the concentration $[S_1]$ after unit time base on $[S_0]$. In the case of a finite time step, we can obtain the change rule of substrate concentration.
Based on the above equations, we can restate the enzyme system kinetics of BcABA, as the following figure shows:
Fig 2. Schematic diagram of the BcABA enzyme kinetic system.
Where $E1-E4$ represents BcABA1-4. $S$ represents FPP, and $P_1,P_2,P_3$ and $P_4$ represent α- ionylideneethane, α-ionylideneacetic acid, 1ʹ,4ʹ-trans-dihydroxy-α-ionylideneaceticacid and ABA, respectively. Different from (1.1), the above formula considers all the reactions as reversible ones, so the $k_{-2}$ is added, which indicates that the products and enzymes also have the trend of forming complexes. Based on the mass action law, we can list the following formulas according to the figure above:
$$\frac{d[S]}{dt}=k_{-1}[E_3S_1]-k_1[E_3][S] \tag{1.13}$$
$$\frac{d[P_1]}{dt}=k_2[E_3S_1]-k_{-2}[P_1][E_3]-k_{3}[P_1][E_1]+k_{-3}[E_1S_2]\tag{1.14}$$
$$\frac{d[P_2]}{dt}=k_4[E_1S_2]-k_{-4}[P_2][E_1]-k_5[P_2][E_2]+k_{-5}[E_2S_3]\tag{1.15}$$
$$\frac{d[P_3]}{dt}=k_6[E_2S_3]-k_{-6}[P_3][E_2]-k_7[P_3][E_4]+k_{-7}[E_4S_4]\tag{1.16}$$
$$\frac{d[P_4]}{dt}=k_8[E_4S_4]-k_{-8}[P_4][E_4] \tag{1.17}$$
$$\frac{d[E_1]}{dt}=k_{-3}[E_1S_2]-k_{3}[P_1][E_1]-k_{-4}[P_2][E_1]+k_{4}[E_1S_2]\tag{1.18}$$
$$\frac{d[E_2]}{dt}=k_{-5}[E_2S_3]-k_{5}[P_2][E_2]-k_{-6}[P_3][E_2]+k_{6}[E_2S_3]\tag{1.19}$$
$$\frac{d[E_3]}{dt}=k_{-1}[E_3S_1]-k_{1}[S][E_3]-k_{-2}[P_1][E_3]+k_{2}[E_3S_1]\tag{1.20}$$
$$\frac{d[E_4]}{dt}=k_{-7}[E_4S_4]-k_{7}[P_3][E_4]-k_{-8}[P_4][E_4]+k_{8}[E_4S_4]\tag{1.21}$$
$$\frac{d[E_1S_2]}{dt}=-k_{-3}[E_1S_2]+k_{3}[P_1][E_1]+k_{-4}[P_2][E_1]-k_{4}[E_1S_2]\tag{1.22}$$
$$\frac{d[E_2E_3]}{dt}=-k_{-5}[E_2S_3]+k_{5}[P_2][E_2]+k_{-6}[P_3][E_2]-k_{6}[E_2S_3]\tag{1.23}$$
$$\frac{d[E_3S_1]}{dt}=-k_{-1}[E_3S_1]+k_{1}[S][E_3]+k_{-2}[P_1][E_3]-k_{2}[E_3S_1]\tag{1.24}$$
$$\frac{d[E_4S_4]}{dt}=-k_{-7}[E_4S_4]-k_{7}[P_3][E_4]+k_{-8}[P_4][E_4]-k_{4}[E_4S_4]\tag{1.25}$$
The whole equation is somewhat more complicated because we have taken into account the reversibility of each reaction. For each reversible reaction:
$$E+S \mathop{\rightleftharpoons}\limits_{k_{-1}}\limits^{k_1}
ES \mathop{\rightleftharpoons}\limits_{k_{-2}}\limits^{k_2} E+P\tag{1.26}$$
We refer to the work of our team in 2017[6] to facilitate solving the parameters. For both forward and inverse reactions, we can reverse the role of $S$ and $P$, obtaining that:
$$K_{m1} = \frac{k_{-1}+k_2}{k_1}$$
$$k_{cat1} = k_2$$
$$K_{m2} = \frac{k_{-1}+k_2}{k_{-2}}$$
$$k_{cat2} = k_{-1} \tag{1.27}$$
Thus, we can express the four parameters as:
$$k_1 = \frac{(k_{cat1}+k_{cat2})}{K_{m1}} \tag{1.28}$$
$$k_2 = k_{cat1} \tag{1.29}$$
$$k_{-1} = k_{cat2} \tag{1.30}$$
$$k_{-2} = \frac{(k_{cat1}+k_{cat2})}{K_{m2}}\tag{1.31}$$
So if we acquire $K_m$ and $k_{cat}$ of the enzymes for the forward and reverse reactions, we can obtain the full differential equation parameters. However, we did not find any kinetic parameters of BcABA, and even the registered EC number of BcABA did not exist. Therefore, we wanted to start with the known enzyme kinetic parameters to prediction the enzyme kinetic parameters of BcABA.
Prediction of Enzyme Kinetic Parameters
In order to obtain the optimal ratio of the four enzymes of BcABA more precisely, we want to predict the enzyme kinetic parameters of BcABA1-4 from known parameters. Based on [1], we know that BcABA1 and 2 are P450 monooxygenases. Through BLAST, we also know that BcABA4 has 99% similarity with a variety of 3-oxoacyl-[acyl-carrier-protein] reductase, so we can use the mean value of the kinetic parameters of similar enzymes to approximate the parameters of BcABA1, 2, and 4. But for BcABA3 there’s no known homologous enzyme, so we decide to start with enzymes of the same substrate for parameter prediction.
To make the predictions more accurate, we refer to BRENDA’s enzyme classification [7], assigning BcABA1-4 the basic classification base on E.C. number to facilitate finding related enzymes for parameter prediction. Reviewing the reaction process in Figure 1, we can see that BcABA4 removes hydrogen from hydroxyl group under the presence of NAD+, forming a carbon-oxygen double bond, which can be localized under EC1.1.1 ("Acting on the CH-OH group of donors, With NAD+ or NADP+ as acceptor"); BcABA1 adds a carboxyl group to the substrate, and BcABA2 adds two hydroxyl groups, which are consistent with the description of " Acting on paired donors, with incorporation or reduction of molecular oxygen " in EC1.14. However, BcABA3's role is more complicated. Through searching on the description of its function in [8] and on the enzyme which treats FPP as substrate, we found that EC 4.2.3 "Acting on The classification of "phosphates" fits its description. It’s also confirmed by search for related enzymes on SABIO-RK and BRENDA [7,9].
We first predict the enzyme kinetic parameters of BcABA4. Since the BLAST results indicate that the enzyme is highly homologous to EC 1.1.1.100 and its enzymatic reaction type is also consistent with 1.1.1, we decided to replace the parameters of BcABA4 with the enzymes’ kinetic parameters which catalyze similar reactions in 1.1.1. Setting the dehydrogenation reaction as forwarding reaction, we collected the following enzyme kinetic parameters [7]:
Table 1. BcABA4 Isozyme Kinetics Parameters
| EC | Reference | $K_{m1}(mM)$ | $k_{cat1}(/s)$ | $K_{m2}(mM)$ | $k_{cat2}(/s)$ |
| 1.1.1.8 | [10] | 0.052 | 240 | 0.052 | 240 |
| 1.1.1.37 | [11,12] | 0.269 | 48.6 | 0.022 | - |
| 1.1.1.82 | [13] | 0.2 | 51 | 0.05 | 78 |
| 1.1.1.145 | [14] | 0.021 | 0.1 | 0.004 | 1.03 |
| 1.1.1.169 | [15] | 0.13 | 3.8 | 0.006 | 23.1 |
| 1.1.1.195 | [16] | 0.009 | 16 | 0.003 | 32.2 |
| 1.1.1.252 | [17] | 0.025 | 0.38 | 0.021 | 13 |
| 1.1.1.282 | [18] | 0.26 | 31.1 | 0.2 | 329 |
| 1.1.1.286 | [19] | 0.027 | 37 | 0.011 | 61 |
| 1.1.357 | [20] | 0.003 | 230 | 0.001 | 500 |
BcABA1-2 have been certified as two cytochrome P450 monooxygenases. Therefore, we used the cytochrome P450 enzyme parameters as reference and considered hydroxylation and carboxylation as forwarding reactions. The obtained parameters are shown in Table 2:
Table 2. BcABA1,2 Isozyme Kinetics Parameters
| EC | Reference | $K_{m1}(mM)$ | $k_{cat1}(/s)$ | $K_{m2}(mM)$ | $k_{cat2}(/s)$ |
| 1.14.12.18 | [21] | 0.029 | 4.6 | - | - |
| 1.14.13.2 | [22] | 0.195 | 7.8 | - | - |
| 1.14.13.6 | [23] | 0.13 | 26 | - | - |
| 1.14.13.24 | [24] | 0.046 | 24 | 0.14 | 6 |
| 1.14.13.25 | [25] | 0.035 | 7.6 | - | - |
| 1.14.14.1 | [26] | 0.54 | 37.83 | - | - |
| 1.14.14.20 | [27] | 0.002 | 147 | 0.006 | 43 |
| 1.14.14.81 | [28] | 0.001 | 10.6 | - | - |
| 1.14.14.92 | [29] | 0.083 | 4.5 | - | - |
| 1.14.14.95 | [30] | 0.005 | 4.95 | - | - |
Unlike 1.1.1 enzymes, the P450 monooxygenase reaction seems to have few measured reverse reaction parameters, probably because the reverse reaction is weaker, which can also be seen from the data of 1.14.13.24 and 1.14.14.20.
For BcABA3, on the other hand, since the reaction itself is more complex, we selected the parameters of the enzymatic reaction with the same substrate as BcABA3 (FPP), which removes the pyrophosphate group during the reaction and eventual form ring structure. The parameters were obtained as shown in Table 3, using the ring-forming reaction as the forward reaction:
Table 3. BcABA3 Isozyme Kinetics Parameters
| EC | Reference | $K_{m1}(mM)$ | $k_{cat1}(/s)$ | $K_{m2}(mM)$ | $k_{cat2}(/s)$ |
| 4.2.3.6 | [31] | 0.006 | 0.138 | - | - |
| 4.2.3.21 | [32] | 0.002 | 0.04 | - | - |
| 4.2.3.37 | [33] | 0.001 | 0.045 | - | - |
| 4.2.3.38 | [34] | 0.077 | 0.01 | - | - |
| 4.2.3.43 | [35] | 0.0001 | 0.027 | - | - |
| 4.2.3.48 | [36] | 0.001 | 0.24 | - | - |
| 4.2.3.64 | [37] | 0.0003 | 0.26 | - | - |
| 4.2.3.89 | [38] | 0.0211 | 0.455 | - | - |
| 4.2.3.133 | [39] | 0.0048 | 0.04 | - | - |
| 4.2.3.194 | [40] | 0.0096 | 0.025 | - | - |
In this section, we found these enzymatic reactions do not have the measured results of reverse reaction parameters. We speculated this is due to the large amount of energy provided by FPP in breaking the pyrophosphate bond, driving the ring-forming reaction, which makes the reaction irreversible. We therefore considered that $k_{-2}=0$.
After getting above parameters, we started calculating parameters of BcABA1-4. Here we take $k_{cat1}$ as the base, and then take the average value of it as the value of BcABA1-4 $\bar{k_{cat1}}$. For the column with missing values (such as $k_{cat2}$), we first remove the missing values and take the average of the remaining values, as $\bar{k_{cat2}}'$, then find the average value of $k_{cat1}$ removed the missing rows, called $\bar{k_{cat1}}'$. Calculate the ratio of $k_{cat1}$ to $\bar{k_{cat1}}'$, and then scale the average of column with missing values proportionally. i.e.
$$\bar{k_{cat2}} = \bar{k_{cat2}}'\frac{\bar{k_{cat1}}}{\bar{k_{cat1}}'} \tag{1.32}$$
As the same,
$$\bar{K_{m2}} = \bar{K_{m2}}'\frac{\bar{K_{m1}}}{\bar{K_{m1}}'} \tag{1.33}$$
Through the data in Tables 1-3 and Equations (1.28-1.33), we can predict the enzyme kinetic parameters of BcABA1-4, as follows:
Table 4. BcABA1-4 Kinetics Parameters (Units Ditto)
| Name | $K_{m1}$ | $k_{cat1}$ | $K_{m2}$ | $k_{cat2}$ | $k_{1}$ | $k_{2}$ | $k_{-1}$ | $k_{-2}$ |
| BcABA1 | 0.13 | 27.49 | 0.40 | 6.91 | 264.62 | 27.49 | 6.91 | 85.15 |
| BcABA2 | 0.13 | 27.49 | 0.40 | 6.91 | 264.62 | 27.49 | 6.91 | 85.15 |
| BcABA3 | 0.01 | 0.13 | - | - | 23.00 | 0.13 | 0.10 | 0 |
| BcABA4 | 0.10 | 65.80 | 0.04 | 139.02 | 2048.20 | 65.80 | 139.00 | 5120.50 |
At this point, we have all the parameters for solving the enzyme kinetic equation. Let's see which combination of enzyme ratios yields the most amounts of ABA products.
To facilitate the solution of the differential equation model, we should reduce the number of formulas and variables appropriately. Substitute $[Eo]=[E]+[ES]$ into Equations (1.13-1.25) and simplify it obtaining:
$$\frac{d[S]}{dt}=k_{-1}([E_{t3}]-[E_3])-k_1[E_3][S]$$
$$\frac{d[P_1]}{dt}=k_2([E_{t3}]-[E_3])-k_{-2}[P_1][E_3]-k_{3}[P_1][E_1]+k_{-3}([E_{t1}]-[E_1])$$
$$\frac{d[P_2]}{dt}=k_4([E_{t1}]-[E_1])-k_{-4}[P_2][E_1]-k_5[P_2][E_2]+k_{-5}([E_{t2}]-[E_2])$$
$$\frac{d[P_3]}{dt}=k_6([E_{t2}]-[E_2])-k_{-6}[P_3][E_2]-k_7[P_3][E_4]+k_{-7}([E_{t4}]-[E_4])$$
$$\frac{d[P_4]}{dt}=k_8([E_{t4}]-[E_4])-k_{-8}[P_4][E_4] $$
$$\frac{d[E_1]}{dt}=k_{-3}([E_{t1}]-[E_1])-k_{3}[P_1][E_1]-k_{-4}[P_2][E_1]+k_{4}([E_{t1}]-[E_1])$$
$$\frac{d[E_2]}{dt}=k_{-5}([E_{t2}]-[E_2])-k_{5}[P_2][E_2]-k_{-6}[P_3][E_2]+k_{6}([E_{t2}]-[E_2])$$
$$\frac{d[E_3]}{dt}=k_{-1}([E_{t3}]-[E_3])-k_{1}[S][E_3]-k_{-2}[P_1][E_3]+k_{2}([E_{t3}]-[E_3])$$
$$\frac{d[E_4]}{dt}=k_{-7}([E_{t4}]-[E_4])-k_{7}[P_3][E_4]-k_{-8}[P_4][E_4]+k_{8}([E_{t4}]-[E_4])\tag{1.34}$$
By solving the above differential equation, we get simulation of the substrate, final product and several intermediates concentration over time. Under the default that concentration of four enzymes is in equal proportions and resources is limited, we obtain the simulation in Figure 3 below:
Fig 3. The concentration of substrate and products with limited resources
Through figure 3, we can observe that the concentrations of intermediates 1 and 2 are very low throughout the process; while at the final steady-state, the concentration of ABA is only lower than that of intermediate 3, which provides that the final yield of ABA is high enough. At a beginning of the substrate concentration of 100 mM, ABA is about 30 mM in the steady-state. It should be noted that sum of products and substrate concentration is not equal to the initial product value, which is because some substrates or intermediates are bound to the enzyme as a complex, and concentration of the complex is not reflected in the figure.
Since we want to obtain the optimal ratio of the four enzymes for ABA production, we try a total of 24 ratio combinations of BcABA. That is, we explored different combinations of 10, 20, 30, and 40 mM concentrations of BcABA1-4 to observe the final concentration of products. Here we list the concentration plots of four of these combinations:
Fig 4. Concentration of substrate and products catalyzed by different ratios of enzymes under limited resources
From Figure 4, it can be seen that the steady-state is reached at different times with different enzyme ratios, but the final ABA concentrations are almost the same. Among the above four enzyme ratios, the ratio of 2:1:4:3 about BcABA1-4 is the first to steady-state, and also the concentration of the final product is the highest.
However, during fermentation, experimentalists continue to add nutrients, which allows maintaining a stable concentration of substrates. Therefore, we also need to know the concentration changes of substrates and products with sufficient substrate concentration:
Fig 5. Concentration of substrate and products with unlimited resources
The concentration of final products and intermediates is increasing due to sufficient resources. Similarly, we also explored the effect of different enzyme ratios on end-product concentrations in the presence of sufficient resources.
Fig 6. Concentration of substrate and products catalyzed by different ratios of enzymes under unlimited resources
The concentration of final products and intermediates is increasing due to sufficient resources. Similarly, we also explored the effect of different enzyme ratios on end-product concentrations in the presence of sufficient resources.
Unlike Figure 4, there is a large difference in the ABA concentration at final moment under sufficient resources for different enzyme ratios. For example, BcABA1-4 reached an ABA concentration of 210mM at a ratio of 2:1:4:3, while 3:2:1:4 was only 45mM.
To visualize the effect of the 24 enzyme ratio combinations on ABA, we plot Figures 7-8 by observing only the ABA concentration.
Fig 7. Concentration of ABA with limited resources Under different enzyme ratios
Fig 8. Concentration of ABA with unlimited resources Under different enzyme ratios
As can be seen, regardless of whether the resources are sufficient or not, the ABA concentration under the 24 enzyme ratios are concentrated in four clusters of curves. We can see that the ratio of BcABA3 has the greatest influence on the final concentration of ABA. In the case of sufficient resources, it directly determines the ABA yield. The combination of enzyme ratio with the highest ABA yield is 3:2:4:1 in the case of limited resources, and 1:3:4:2 in the case of sufficient resources, while the combination with the lowest ABA yield is 2:3:1:4 in both case of limited and sufficient resources.
To obtain a more precise effect of the enzyme ratio on ABA concentration, we performed linear regression analysis on the final ABA concentration and the BcABA1-4 concentration. With limited resources, we obtained ABA final concentration rules as follows：
$$[ABA_{Lim}] = -0.0504[E2] + 0.3384[E3] -0.0719[E4] +
33.0567\tag{1.35}$$
And with sufficient resources,
$$[ABA_{Unl}] = 5.6241[E3] -0.0058[E4]-11.4475 \tag{1.36}$$
Thus, we know that BcABA3 has a crucial effect on the final ABA concentration. For the same amount of total enzyme, an increase in the ratio of BcABA2 and 4 seems to be detrimental to ABA accumulation. Therefore, we believe that a better enzyme ratio would be 3:1:4:2 or 3:2:4:1, so we apply the ratio into our wet lab.
This year, our project focuses on the synthesis of abscisic acid (ABA), which involves a total of four enzymes, BcABA1-4. Since we attempt to construct an ABA expression system in Saccharomyces cerevisiae for the first time, it is necessary to describe the enzyme system by enzyme kinetic method.
We have two main tasks for this part of modeling. First, due to insufficient literature resources and experimental data, we are unable to determine the enzyme kinetics parameters, so we want to predict the BcABA parameters with the kinetic parameters through enzymes with high homology. Second, we hope to obtain the optimal ratio of BcABA1-4 enzymes in biochemical reactions through comprehensive enzyme kinetic simulations to achieve the most efficient heterologous synthesis of abscisic acids.
// Code:
function [f] = dXdT(t, x)
k1F = 23;
k1R = 0.1;
k2F = 0.13;
k2R = 0;
k3F = 264.62;
k3R = 6.91;
k4F = 27.49;
k4R = 85.15;
k5F = 264.62;
k5R = 6.91;
k6F = 27.49;
k6R = 85.15;
k7F = 2048.2;
k7R = 65.80;
k8F = 139.02;
k8R = 5120.5;
E01 = 10;
E02 = 10;
E03 = 10;
E04 = 10;
s=x(1);
p1=x(2);
p2=x(3);
p3=x(4);
p4=x(5);
e1=x(6);
e2=x(7);
e3=x(8);
e4=x(9);
dsdt = k1R*(E03-e3) - k1F*e3*s;
dp1dt = k2F*(E03-e3) - k3F*p1*e1 +k3R*(E01-e1);
dp2dt = k4F*(E01-e1) - k4R*p2*e1 - k5F*p2*e2 + k5R*(E02-e2);
dp3dt = k6F*(E02-e2) - k6R*p3*e2 - k7F*p3*e4 + k7R*(E04-e4);
dp4dt = k8F*(E04-e4) - k8R*p4*e4;
de1dt = k3R*(E01-e1) - k3F*p1*e1 - k4R*p2*e1 + k4F*(E01-e1);
de2dt = k5R*(E02-e2) - k5F*p2*e2 - k6R*p3*e2 + k6F*(E02-e2);
de3dt = k1R*(E03-e3) - k1F*s*e3 + k2F*(E03-e3);
de4dt = k7R*(E04-e4) - k7F*p3*e4 - k8R*p4*e4 + k8F*(E04-e4);
f = [dsdt; dp1dt; dp2dt; dp3dt; dp4dt; de1dt; de2dt; de3dt; de4dt];
end
tspan = 0:0.001:100;
initial = [100,0,0,0,0,10,10,10,10];
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2. Marumo, Shingo, et al. "Microbial Production of Abscisic Acid by Botrytis cinerea." Agricultural and biological chemistry 46.7 (1982): 1967-1968.
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