Team:SCU-China/Enzyme Kinetic


1. Introduction

Naringenin is a natural production which has been synthesized in E.coli, S.cerevisiae and other chassis microorganisms, the synthesis of naringenin is almost the symbol of a universally accepted chassis organism. Due to the lack basal knowledge of V.natriegens’ metabolic network, however, we need to find out the best relative expression level of each enzyme.

Fig 1.Naringenin synthesis pathway.

Here, we applied enzyme kinetic model to explore the best expression combination of four enzymes and choose parts from SSR collection to optimize the production of naringenin.

2. Traditional Michaelis-Menten Theory & Reversible Michaelis-Menten Theory

In biochemistry, Michaelis–Menten kinetics is one of the best-known models of enzyme kinetics[1][2]. It involves an enzyme, $E$, binding to a substrate, $S$, to form a complex, $ES$, which in turn releases a product, $P$, regenerating the original enzyme. This may be represented schematically as

$$E+S \mathop{\rightleftharpoons}\limits^{k_f}\limits_{k_r}ES \xrightarrow{k_{cat}} E+P \tag{2.1.1}$$

where $k_f$(forward rate constant), $k_{r}$(reverse rate constant), and $k_{cat}$(catalytic rate constant) denote the rate constants[3], the double arrows between $S$ (substrate) and $ES$ (enzyme-substrate complex) represent the fact that enzyme-substrate binding is a reversible process, and the single forward arrow represents the formation of $P$ (product).

Applying the law of mass action, gives a system of four non-linear ordinary differential equations that define the rate of change of reactants with time $t$[4],

$$\frac{d[E]}{dt}=-k_f[E][S]+k_r[ES]+k_{kat}[ES] \tag{2.1.2}$$ $$\frac{d[S]}{dt}=-k_f[E][S]+k_r[ES] \tag{2.1.3}$$ $$\frac{d[ES]}{dt}=k_f[E][S]-k_r[ES]-k_{cat}[ES] \tag{2.1.4}$$ $$\frac{d[P]}{dt}=k_{cat}[ES] \tag{2.1.5}$$

In this mechanism, the enzyme E is a catalyst, which only facilitates the reaction, so that its total concentration, free plus combined,

$$[E]+[ES]=[E_0] \tag{2.1.6}$$

is a constant (i.e. $[E_0]=[E_{total}]$)

However, in situations where the reaction is low energy and a substantial pool of product(s) exists, the Michaelis–Menten equation breaks down, and the enzyme reaction is more correctly described as

$$E+S \mathop{\rightleftharpoons}\limits^{k_{f1}}\limits_{k_{r1}}ES \mathop{\rightleftharpoons}\limits^{k_{f2}}\limits_{k_{r2}} E+P \tag{2.2.1}$$

In our Naringenin Production System, due to the presence of intermediate products, it is obvious that the Traditional Michaelis–Menten Equation is not applicable. And we should use equation (2.2.1) for modeling. The difference between equations (2.2.1) and (2.2.2) lies in the parameter $k_{-2}$. This parameter may be small, but it is essential to limit the reaction speed.

Therefore, based on equation (2.2.1) we can rewrite the equation system (2.1.2-2.1.5) as $$\frac{d[E]}{dt}=-k_{f1}[E][S]+k_{r1}[ES]+k_{f2}[ES]-k_{r2}[E][P] \tag{2.2.2}$$ $$\frac{d[S]}{dt}=-k_{f1}[E][S]+k_{r1}[ES] \tag{2.2.3}$$ $$\frac{d[ES]}{dt}=k_{f1}[E][S]-k_{r1}[ES]-k_{f2}[ES] +k_{r2}[E][P] \tag{2.2.4}$$ $$\frac{d[P]}{dt}=k_{f2}[ES] -k_{r2}[E][P] \tag{2.2.5}$$


$$[E]+[ES]=[E_0] \tag{2.2.6}$$

is a constant (i.e. $[E_0]=[E_{total}]$)

3. Reversible Michaelis-Menten Theory in Naringenin Production System

According to the Reversible Michaelis-Menten Theory, we can simplify the synthesis route of naringenin into the following figure.

Fig 2.Schematic diagram of the naringenin enzyme kinetic system

The meanings of enzymes (E1, E2, E3, E4), substrates (S1, S2) and intermediate products (P1, P2, P3, P4) are as follows.

Table 1.Meanings of enzymes, substrates and intermediate products

Enzymes Substrates  

Intermediate Products
E1 RtPAL S1 L-tyrosine P1 ρ-Coumaric acid
E2 Pc4CL S2 Malonyl-CoA P2 Coumaroyl-CoA
E3 PhCHS     P3 Naringenin   chalcone
E4 MsCHI     P4 Naringenin

Applying the law of mass action to the Figure 2, we can easily list the system of equations.

$$\frac{d[S_1]}{dt}=k_{-1}[E_1S_1]-k_1[E_1][S_1] \tag{3.2.1}$$ $$\frac{d[S_2]}{dt}=k_{-5}[E_3S_3]-k_5[P_2][E_3][S_2] \tag{3.2.2}$$ $$\frac{d[P_1]}{dt}=k_2[E_1S_1]-k_{-2}[P_1][E_1]-k_{3}[P_1][E_2]+k_{-3}[E_2S_2] \tag{3.2.3}$$ $$\frac{d[P_2]}{dt}=k_4[E_2S_2]-k_{-4}[P_2][E_2]-k_5[P_2][E_3][S_2]+k_{-5}[E_3S_3] \tag{3.2.4}$$ $$\frac{d[P_3]}{dt}=k_6[E_3S_3]-k_{-6}[P_3][E_3]-k_7[P_3][E_4]+k_{-7}[E_4S_4] \tag{3.2.5}$$ $$\frac{d[P_4]}{dt}=k_8[E_4S_4]-k_{-8}[P_4][E_4] \tag{3.2.6}$$ $$\frac{d[E_1]}{dt}=k_{-1}[E_1S_1]-k_1[P_1][E_2]-k_{-2}[P_2][E_2]+k_2[E_2S_2] \tag{3.2.7}$$ $$\frac{d[E_2]}{dt}=k_{-3}[E_2S_2]-k_3[P_1][E_2]-k_{-4}[P_2][E_2]+k_4[E_2S_2] \tag{3.2.8}$$ $$\frac{d[E_3]}{dt}=k_{-5}[E_3S_3]-k_5[S_2][E_3][P_2]-k_{-6}[P_3][E_3]+k_6[E_3S_2] \tag{3.2.9}$$ $$\frac{d[E_4]}{dt}=k_{-7}[E_4S_4]-k_7[P_3][E_4]-k_{-8}[P_4][E_4]+k_8[E_4S_4] \tag{3.2.10}$$ $$\frac{d[E_1S_1]}{dt}=-k_{-1}[E_1S_1]+k_1[P_1][E_1]+k_{-2}[P_1][E_1]-k_2[E_1S_1] \tag{3.2.11}$$ $$\frac{d[E_2S_2]}{dt}=-k_{-3}[E_2S_2]+k_3[P_1][E_2]+k_{-6}[P_3][E_3]-k_6[E_2S_2] \tag{3.2.12}$$ $$\frac{d[E_3S_3]}{dt}=-k_{-5}[E_3S_3]+k_5[S_2][E_3][P_2]+k_{-6}[P_3][E_3]-k_6[E_3S_3] \tag{3.2.13}$$ $$\frac{d[E_4S_4]}{dt}=-k_{-7}[E_4S_4]+k_7[P_3][E_4]+k_{-8}[P_4][E_4]-k_8[E_4S_4] \tag{3.2.14}$$

From equation (2.2.6), we know $$[ES]=[E_0]-[E]$$

So let

$$[E_1S_1]=[E_{t1}]-[E_1]$$ $$[E_2S_2]=[E_{t2}]-[E_2]$$ $$[E_3S_3]=[E_{t3}]-[E_3]$$ $$[E_4S_4]=[E_{t4}]-[E_4]$$

The equations (3.2.1-3.2.14) can be simplified to

$$\frac{d[S_1]}{dt}=k_{-1}([E_{t1}]-[E_1])-k_1[E_1][S_1] \tag{3.2.15}$$ $$\frac{d[S_2]}{dt}=k_{-5}([E_{t3}]-[E_3])-k_5[E_3][S_2][P_2] \tag{3.2.16}$$ $$\frac{d[P_1]}{dt}=k_2([E_{t1}]-[E_1])-k_{-2}[P_1][E_1]-k_{3}[P_1][E_2]+k_{-3}([E_{t2}]-[E_2]) \tag{3.2.17}$$ $$\frac{d[P_2]}{dt}=k_4([E_{t2}]-[E_2])-k_{-4}[P_2][E_2]-k_5[P_2][E_3][S_2]+k_{-5}([E_{t3}]-[E_3]) \tag{3.2.18}$$ $$\frac{d[P_3]}{dt}=k_6([E_{t3}]-[E_3])-k_{-6}[P_3][E_3]-k_7[P_3][E_4]+k_{-7}([E_{t4}]-[E_4])\tag{3.2.19}$$ $$\frac{d[P_4]}{dt}=k_8([E_{t4}]-[E_4])-k_{-8}[P_4][E_4] \tag{3.2.20} $$ $$\frac{d[E_1]}{dt}=k_{-1}([E_{t1}]-[E_1])-k_{1}[S_1][E_1]-k_{-2}[P_1][E_1]+k_{2}([E_{t1}]-[E_1]) \tag{3.2.21}$$ $$\frac{d[E_2]}{dt}=k_{-3}([E_{t2}]-[E_2])-k_{3}[P_1][E_2]-k_{-4}[P_2][E_2]+k_{4}([E_{t2}]-[E_2]) \tag{3.2.22}$$ $$\frac{d[E_3]}{dt}=k_{-5}([E_{t3}]-[E_3])-k_{5}[P_2][E_3][S_2]-k_{-6}[P_3][E_3]+k_{6}([E_{t3}]-[E_3]) \tag{3.2.23}$$ $$\frac{d[E_4]}{dt}=k_{-7}([E_{t4}]-[E_4])-k_{7}[P_3][E_4]-k_{-8}[P_4][E_4]+k_{8}([E_{t4}]-[E_4]) \tag{3.2.24}$$

We query $k_{cat1}$ , $k_M1$(forward reaction), $k_{cat2}$ and $k_M2$( reverse reaction) values of the enzymes in the brenda database.

Table 2.Enzyme Kinetic Parameters

$k_{cat1}$(/s) $k_{M1}$(mM) $k_{cat2}$(/s) $k_{M2}$(mM)
RtPAL[5] 1.7612 0.34 0.1 1
Pc4CL[6] 0.068167 0.01189 0.1 1
PhCHS[5] 0.111656 0.0136 0.1 1
MsCHI[5] 130.05 0.0085 0.1 1

* Let $k_{cat2}$=0.1,$k_{M2}$=1

For each reversible reaction $E+S \mathop{\rightleftharpoons}\limits^{k_{f1}}\limits_{k_{r1}}ES \mathop{\rightleftharpoons}\limits^{k_{f2}}\limits_{k_{r2}} E+P $, we can obtain that[7]:

$$k_{f1} = \frac{(k_{cat1}+k_{cat2})}{K_{M1}} \tag{3.3.1}$$ $$k_{r1} = k_{cat2} \tag{3.3.2}$$ $$k_{f2} = k_{cat1} \tag{3.3.3}$$ $$k_{r2} = \frac{(k_{cat1}+k_{cat2})}{K_{M2}}\tag{3.3.4}$$

Based on the above equation (3.3.1-3.3.4) using the data in Table 2 to calculate

Table 3.Calculation results

$k_{f1}$ $k_{r1}$ $k_{f2}$ $k_{r2}$
RtPAL 5.474 0.1 1.7612 1.8612
Pc4CL 14.1436 0.1 0.0682 0.1681
PhCHS 15.5629 0.1 0.1117 0.2117
MsCHI 15311.7647 0.1 130.05 130.15

* Four decimal places.

That is

Table 4.Parameters of equations (3.2.15-3.2.24)

$k_{1}$ $k_{-1}$ $k_{2}$ $k_{-2}$ $k_{3}$ $k_{-3}$ $k_{4}$ $k_{-4}$
5.474 0.1 1.7612 1.8612 14.1436 0.1 0.0682 0.1681
$k_{5}$ $k_{-5}$ $k_{6}$ $k_{-6}$ $k_{7}$ $k_{-7}$ $k_{8}$ $k_{-8}$
15.5629 0.1 0.1117 0.2117 15311.7647 0.1 130.05 130.15

Under different initial conditions (cases 1-5), bring the parameters in Table 4 into the equations (3.2.15-3.2.24), use Matlab to solve the equations and draw the images of the concentration of intermediate product (ρ-Coumaric acid, Coumaroyl-CoA, Naringenin Chalcone) and the final product (Naringenin) over time t.

Case 1: Limited substrate, E1:E2:E3:E4=2.5:2.5:2.5:2.5

The purple curve represents the change of the concentration of naringenin over time, and it can be seen that the production of naringenin reaches 15.5995mM in 100s.

Fig 3.Case 1: E1:E2:E3:E4=2.5:2.5:2.5:2.5

Case 2: Limited substrate, E1:E2:E3:E4=1:2:3:4,E1:E2:E3:E4=1:2:4:3,E1:E2:E3:E4=1:3:2:4,E1:E2:E3:E4=1:3:4:2,E1:E2:E3:E4=1:4:2:3,E1:E2:E3:E4=1:4:3:2

In this case, when the ratio is E1:E2:E3:E4=1:4:2:3 or E1:E2:E3:E4=1:4:3:2, naringenin produced at 100s reaches the highest amount (both 15.5995mM).

Fig 4.Case 2: E1:E2:E3:E4=1:2:3:4,E1:E2:E3:E4=1:2:4:3,E1:E2:E3:E4=1:3:2:4,E1:E2:E3:E4=1:3:4:2,E1:E2:E3:E4=1:4:2:3,E1:E2:E3:E4=1:4:3:2

Case 3: Limited substrate, E1:E2:E3:E4=2:1:3:4,E1:E2:E3:E4=2:1:4:3,E1:E2:E3:E4=2:3:1:4,E1:E2:E3:E4=2:3:4:1,E1:E2:E3:E4=2:4:1:3,E1:E2:E3:E4=2:4:3:1

In this case, when the ratio is E1:E2:E3:E4=2:4:1:3 or E1:E2:E3:E4=2:4:3:1, naringenin produced at 100s reaches the highest amount (both 16.5662mM).

Fig 5.Case 3: E1:E2:E3:E4=2:1:3:4,E1:E2:E3:E4=2:1:4:3,E1:E2:E3:E4=2:3:1:4,E1:E2:E3:E4=2:3:4:1,E1:E2:E3:E4=2:4:1:3,E1:E2:E3:E4=2:4:3:1

Case 4: Limited substrate, E1:E2:E3:E4=3:1:2:4,E1:E2:E3:E4=3:1:4:2,E1:E2:E3:E4=3:2:1:4,E1:E2:E3:E4=3:2:4:1,E1:E2:E3:E4=3:4:1:2,E1:E2:E3:E4=3:4:2:1

In this case, when the ratio is E1:E2:E3:E4=3:4:1:2 or E1:E2:E3:E4=3:4:2:1, naringenin produced at 100s reaches the highest amount (both 17.5363mM).

Fig 6.Case 4: E1:E2:E3:E4=3:1:2:4,E1:E2:E3:E4=3:1:4:2,E1:E2:E3:E4=3:2:1:4,E1:E2:E3:E4=3:2:4:1,E1:E2:E3:E4=3:4:1:2,E1:E2:E3:E4=3:4:2:1

Case 5: Limited substrate, E1:E2:E3:E4=4:1:2:3,E1:E2:E3:E4=4:1:3:2,E1:E2:E3:E4=4:2:1:3,E1:E2:E3:E4=4:2:3:1,E1:E2:E3:E4=4:3:1:2,E1:E2:E3:E4=4:3:2:1

In this case, when the ratio is E1:E2:E3:E4=4:3:1:2 or E1:E2:E3:E4=4:3:2:1, naringenin produced at 100s reaches the highest amount (both 17.5363mM).

Fig 7.Case 5: E1:E2:E3:E4=4:1:2:3,E1:E2:E3:E4=4:1:3:2,E1:E2:E3:E4=4:2:1:3,E1:E2:E3:E4=4:2:3:1,E1:E2:E3:E4=4:3:1:2,E1:E2:E3:E4=4:3:2:1

It can be seen from the above five cases that if the substrate is sufficient, when E1:E2:E3:E4=3:4:1:2, E1:E2:E3:E4=3:4:2:1, E1:E2: E3:E4=4:3:1:2 or E1:E2:E3:E4=4:3:2:1, the naringenin production reaches the highest amount (17.5363mM) in all cases at 100s.

At the same time, we found that when the ratio values of E1 and E2, E3 and E4 are exchanged respectively, the production of naringenin at 100s is almost the same! (For example, E1:E2:E3:E4=1:2:3:4 and E1:E2:E3:E4=1:2:4:3 in 100s, the production of naringenin is both 13.6785mM; E1:E2: E3:E4=1:3:2:4 and E1:E2:E3:E4=3:1:2:4 in 100s, the production of naringenin is both 14.6367mM) Therefore, it indicates that the importance of enzymes is E1=E2>E3=E4 (RtPAL=Pc4CL>PhCHS=MsCHI).

To obtain the highest naringenin production, we need to determine the optimal ratio of the four enzymes. Based on the conclusion in 3.5, we assume that the content of E1 and E2 are the same, and the content of E3 and E4 are the same and explore the best ratio by increasing the ratio between E1/E2 and E3/E4.

It can be seen from the figure below, as the ratio between E1/E2 and E3/E4 continues to increase, the production of naringenin is getting higher and higher at 100s, while the growth rate is getting slower and slower.

In general, we think that E1:E2:E3:E4=9:9:1:1 (E1:E2:E3:E4=4.5:4.5:0.5:0.5 doubled) is a appropriate ratio.

Fig 8. E1:E2:E3:E4=3:3:2:2,E1:E2:E3:E4=3.5:3.5:1.5:1.5,E1:E2:E3:E4=4:4:1:1,E1:E2:E3:E4=4.5:4.5:0.5:0.5,E1:E2:E3:E4=4.75:4.75:0.25:0.25,E1:E2:E3:E4=4.9:4.9:0.1:0.1

Here are core codes we used in our research.

 function [f] = dXdT(~, x)

 k1F = 5.474;
 k1R = 0.1;
 k2F = 1.7612;
 k2R = 1.8612;
 k3F = 14.1436;
 k3R = 0.1;
 k4F = 0.0682;
 k4R = 0.1681;
 k5F = 15.5629;
 k5R = 0.1;
 k6F = 0.1117;
 k6R = 0.2117;
 k7F = 15311.7647;
 k7R = 0.1;
 k8F = 130.05;
 k8R = 130.15;
 E01 = 10;
 E02 = 10;
 E03 = 10;
 E04 = 10;


 ds1dt = k1R*(E01-e1) - k1F*e1*s1;
 dp1dt = k2F*(E01-e1) - k2F*p1*e1 - k3F*p1*e2 + k3R*(E02-e2);
 dp2dt = k4F*(E02-e2) - k4R*p2*e2 - k5F*p2*s2*e3 + k5R*(E03-e3);
 ds2dt = k5R*(E03-e3) - k5F*p2*e3*s2;
 dp3dt = k6F*(E03-e3) - k6R*p3*e3 - k7F*p3*e4 + k7R*(E04-e4);
 dp4dt = k8F*(E04-e4) - k8R*p4*e4;
 de1dt = k1R*(E01-e1) - k1F*s1*e1 - k2R*p1*e1 + k2F*(E01-e1);
 de2dt = k3R*(E02-e2) - k3F*p1*e2 - k4R*p2*e2 + k4F*(E02-e2);
 de3dt = k5R*(E03-e3) - k5F*s2*e3*p2 - k6R*p3*e3 + k6F*(E03-e3);
 de4dt = k7R*(E04-e4) - k7F*p3*e4 - k8R*p4*e4 + k8F*(E04-e4);

 f = [ds1dt; dp1dt; dp2dt; ds2dt; dp3dt; dp4dt; de1dt; de2dt; de3dt; de4dt];


 tspan = 0:0.1:200;
 initial1 = [10,0,0,10,0,0,2.5,2.5,2.5,2.5];

 [t1,x1] = ode45( @dXdT, tspan, initial1);


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