Difference between revisions of "Team:SCUT-China A/Model"

The procedure of 4S Pathway:

As the picture above illustrates, there are totally 4 enzymes participating in the 4S pathway. Through the pathway the C-S bonds in the DBTs are cleaved. After 4 steps of oxidation, the sulfur elements are converted into sulfate or sulfite, thereby dissolving in water and being removed. Meanwhile DBT is oxidized to HBP and remains in the oil phase, without degrading the hydrocarbons, which means the calorific value won’t be reduced. This is the process of 4s pathway biodesulfurization.

However, the lacking in durability of catalytic activity in the 4S pathway has become a major obstacle to BDS. Studies have found that HBP and HBPS in 4S pathway have majot inhibition effect on Dsz enzymes. HBP strongly inhibits the activity of enzyme DszA DszB and DszC; HBPS has certain inhibition effect on enzyme DszC, but only in the early stages in the BDS. The reason is that, when the concentration of HBP increases, the inhibition of HBP will become the major one. And some researchers have found that enzyme DszC is inhibited by HBP and HBPS noncompetitively, and the substrate DBT also has certain inhibition effect on enzyme DszC. Due to the limited high performance liquid chromatography (HPLC) detector resolution, the mechanisms of how the effects are brought by HBP to DszA, DszB are still unable to be determined.

2.2 Determination of related parameters

1).The catalytic pathway of DszA，DszB，DszD can be described as enzymatic reaction based on traditional Michaelis kinetic theory. The kinetic data can be fitted by Michaelis equation:

In the equation, v is the catalytic rate of enzymes, kcat is conversion number, [E0] is the concentration of enzymes, [S] is the concentration of substrates, Km is Michaelis constant, and Vmax is the maximum catalytic rate of enzymes.

2)Because DszC is inhibited by substrate DBTs, the Michaellis equation is no longer fit. We could only use substrate inhibition equantion to fit the gained data:

In the equation, KSI is substrate inhibition constant.

Using the fitting statistics of MATLAB, we obtain many kinetic parameters of different Dsz-enzymes.

(2) The determination of related parameters in four major inhibition reactions:

1).The determination of IC50：

The degree of inhibition can be reflected by the dosage-reaction; the dosage-reaction describes the influence brought by concentration variations of inhibitor to the activity of enzymes:

In the expression, vi is the catalytic rate of enzyme when concentration of inhibitor is [I], and v0 is that without enzymes，IC50 is the inhibitor’s concentration which makes the enzyme’s activity reduce to 50%, comparing to the original one.

2).The determination of enzyme DszC Ki and α:

The degree of inhibition can be reflected by the dosage-reaction; the dosage-reaction describes the influence brought by concentration variations of inhibitor to the activity of enzymes:

In the equation, Vmax is the maximum catalytic rate of enzymes, [S] is the concentration of enzymes’ substrates, Km is the Michaellis constant of enzymes, [I] is the concentration of inhibitors, KI is the inhibition constant towards enzymes DszC brought by HBP, and   shows the influence on enzymes and substrate affinity generated by inhibition or the one on enzymes and inhibition affinity generated by substrates.

Because substrate S has inhibition effect on DszC enzymes, we modify the model into:

In the equation, KSI is substrate inhibition constant.

Taking astringency into consideration, using MATLAB for models and experimental data fitting will be infeasible, so we choose graphical method to obtain K1 and  .

First we use double-reciprocal plot to analyze the 1/S changes over 1/v:

①The slope of double-reciprocal plot is a function of [I]：

②The intercept of 1/v –axis in the double-reciprocal plot is a function of [I]：

The intercept of I-axis is -αKI.

Based on models above, the value of KI,αparameters are derived.

Table 2: Conclusion of parameters related to four major inhibition reactions:

3. Determination of associated parameters of biological catalytic activity loss, without considering the effects of HBP.

Cell suspension is cultured in DBT-free liquid medium--30℃, 250rpm, 24h. Sample is detected for biological catalytic activity of BDS over time. The biological catalytic activity of BDS accords with an exponential attenuation, which is shown as below:

Hypothesis:

2.3 Modeling Part

1).Suppose the two intermediates--DBTO2 and HBPS—are only present in cytoplasm.

2).The inhibition to DszA, DszB and DszC of HBP is considered, while the inhibitory effect of HBPS on DszC which is mainly exists in the early stages of BDS is neglected, with the purpose of simplifying our modeling.

3).The DszD’s activity is considered non-inhibited, therefore we took the enzyme out of our modeling.

4).The loss of biological catalytic activity without the effects of HBP is considered.

5).Assume that the concentrations of DBT and HBP stay the same in all the three different phrases as the stirring rate of bioreactors reaches certain degree.

Hypothesis:

1).Because DszA and DszB are both inhibited by HBP, the catalytic rates of these two enzymes fit the models as below respectively:

In the equation, kAcat,kBcat are the conversion coefficients of enzymes A and B respectively, and Km,A,Km,B are the Michaellis constant of enzymes A and B respectively, and IC50,A,IC50,B are the concentrations of inhibitors that lead to a 50% reduction in the enzymes’ activities respectively.

2).Because enzyme DszC is inhibited noncompetitively by HBP, and it is also inhibited by catalytic substrates HBP, the catalytic rate of enzyme DszC is:

In the equation，kCcat is the conversion coefficient of enzyme C ,and Km,C are the Michaelis constant of the enzyme C, respectively, and KI is the inhibition constant of HBPs to DzsC enzymes,  shows the influence on enzymes and substrate affinity generated by inhibition or the one on enzymes and inhibition affinity generated by substrates.

3).Estimate the variations of different components in 4S pathway, using differential equations:

2.4 Result Analysis

1).The optimal proportion of DszA, DszB and DszC..

Using MATLAB, we simulate different components’ concentration change with different DszA, DszB and DszC’s concentrations.

Using Matlab2015a, we found that:

when the proportion of three enzymes is A:B:C=1:2:4 or A:B:C=2:1:4, the decreasing rate of DBT is well suited, showing that DBT is well degraded. However, with the proportion of A:B:C=1:4:2 or A:B:C=4:1:2, the effect is not so obvious as the previous one.

In addition, taking HBP into consideration, methods with the proportions A:B:C=1:4:2 and A:B:C=1:2:4 will achieve better results;

However, intermediates are produced with greater amount in the conditions of A:B:C=1:2:4 and A:B:C=2:1:4. Because of the molecular movement effect, the reaction is harder to control.

In conclusions, we decided to multiple the proportion of enzymes, then we have:

The graghs above show that when multiple enzyme B and enzyme C respectively, the productive rate of HBP and the consumption rate of DBT almost stay the same. But regarding the quantity of the intermediate, with more expression of enzyme DszB, the reaction becomes more stable.

We can get a linear inequality equation based on the level of importance of three enzymes: DszB>DszC>DszA.

Under real conditions, we could regulate the intensity of the promoters in order to regulate the enzymes’ proportion.

The process of fermentation is shown as below:

3.1 Theories

Recently, the studies found that the number of DBP and HBP stored in the cytoplasm is determined by the cytoplasm and the distribution coefficient of oil phase and water phase. We managed to derive the best proportion of water and oil in the reactor to optimize our system.

We collected information and listed the distribution coefficient of DBT and HBP in different phrases by searching the literature:

3.2 Related parameters

Distribution coefficient refers to the proportion of the components in two phrases, in the condition of equilibrium and certain temperature.

3.3 Modeling

Hypotheses:

1).The volume percentage of the cell constantly maintains at 3%.

2).The stirring rate is fast enough to approach the chemical equilibrium and materials transfer balance at a fairly quick speed.

Construction:

In order to meet the actual need of making the remaining DBT least while maximizing the product HBP in oil phase, we establish the following models in order to eliminate the remaining DBT in the oil and preserve the HBP as much as possible:

In the initial state, the amount of DBT is a constant:

In the final state, the amount of HBP becomes a constant:

Assume that the volume is 1, then:

Substitute the known quantity into the equation, we get:

Which are equal to:

According to the actual demands, we define:

Which are:

Substitute the known quantity:

We can derive: xH2O≤0.08818349 or xH2O≥0.470590405

Because we can not calculate a result to satisfy both two optimal solutions, we take aonther new evaluation method.

Definition:

M1 represents the level of importance of DBP: 

M2 represents the level of importance of HBP: 

Because the toxicity of organic solvent has direct impact on the living of bacteria, then:

M is a parameter that has a positive correlation with the optimization of the whole system.

3.4 Result Analysis

Using MATLAB, we can fit the image that indicates the variations of M over xH2O:

When xH2O =88%, M reaches the maximum equal to 2.45481884523198, which is our optimal result.

[1]Andres Abin-Fuentes, Magdy El-Said Mohamed, Daniel I. C. Wang, Kristala L. J. Prather. Explori-ng the Mechanism of Biocatalyst Inhibition in Microbial Desulfurization[J]. Applied and Environmental Microbiology.2013: p. 7807–7817.

[2]Brian Ingalls. Mathematical Modelling in Systems Biology: An Introduction[M]. University of Waterloo.2012:199-272.

[3]Lei Jinzhi. Systematic Biological Mathematic Basis[M]. Zhou Yuanpei mathematic application center of Tsinghua University. 2007:13-15

5.1 Source Code

Enzyme proportional：

enzyme.m

function dy=enzyme(t,y)

%y1=oDBT; y2=wDBT; y1=DBT; y2=DBTO2; y3=HBPS; y4=HBP; y7=wHBP; y8=oHBP;

global kC kA kB KA KB KC ksi ki kd aHBP ICA ICB;

dy=zeros(7,1);

dy(1)=-kC*y(7)*y(1)*exp(-kd*t)/(KC*(1+y(4)/ki)+y(1)*(1+y(4)/(aHBP*ki))+y(1)*y(1)/ksi);

dy(2)=kC*y(7)*y(1)*exp(-kd*t)/(KC*(1+y(4)/ki)+y(1)*(1+y(4)/(aHBP*ki))+y(1)*y(1)/ksi)-kA*y(5)*y(2)*exp(-kd*t)/((KA+y(2))*(1+y(4)/ICA));

dy(3)=kA*y(5)*y(2)*exp(-kd*t)/((KA+y(2))*(1+y(4)/ICA))-kB*y(6)*y(3)*exp(-kd*t)/((KB+y(3))*(1+y(4)/ICB));

dy(4)= kB*y(6)*y(3)*exp(-kd*t)/((KB+y(3))*(1+y(4)/ICB));

dy(5)=0;

dy(6)=0;

dy(7)=0;

progress1.m

global kC kA kB KA KB KC ksi ki kd aHBP ICA ICB Kawc Kaoc Kaow PWCD POCD POWD PWCH POCH POWH fc fw fo;

fo=0.1; fw=0.87; fc=0.03; PWCD=1; POWD=21000;

POCD=21000; PWCH=7; POCH=4.1; POWH=29; kA=11; kB=1.7; kC=1.6; KA=3.6;

KB=1.3; KC=1.4; ksi=1.8; ICA=60; ICB=110; aHBP=0.4; ki=40; kd=0.00055;

Kawc=1; Kaoc=1; Kaow=1;

y0=zeros(7,1);y0(1,1)=1;y0(5,1)=1;y0(6,1)=2;y0(7,1)=4;

y1=zeros(7,1);y1(1,1)=1;y1(5,1)=1;y1(6,1)=4y1(7,1)=2;

y2=zeros(7,1);y2(1,1)=1;y2(5,1)=4;y2(6,1)=1;y2(7,1)=2;

y3=zeros(7,1);y3(1,1)=1;y3(5,1)=2;y3(6,1)=1;y3(7,1)=4;

y4=zeros(7,1);y4(1,1)=1;y4(5,1)=2;y4(6,1)=4;y4(7,1)=1;

y5=zeros(7,1);y5(1,1)=1;y5(5,1)=4;y5(6,1)=2;y5(7,1)=1;

e=linspace(0,7,100);

[x0,z0]=ode45('enzyme',e,y0);

[x1,z1]=ode45('enzyme',e,y1);

[x2,z2]=ode45('enzyme',e,y2);

[x3,z3]=ode45('enzyme',e,y3);

[x4,z4]=ode45('enzyme',e,y4);

[x5,z5]=ode45('enzyme',e,y5);

subplot(321), plot(x0,z0(:,1),'x',x1,z1(:,1),x2,z2(:,1),'v',x3,z3(:,1),x4,z4(:,1),'diamond',x5,z5(:,1));

legend('A:B:C=1:2:4','A:B:C=1:4:2','A:B:C=4:1:2','A:B:C=2:1:4','A:B:C=2:4:1','A:B:C=4:2:1');

title('DBT');

xlabel('time');

ylabel('concentration');

axis([0,7,0,1])

subplot(312), plot(x0,z0(:,2)+z0(:,3),x1,z1(:,2)+z1(:,3),x2,z2(:,2)+z2(:,3),x3,z3(:,2)+z3(:,3),x4,z4(:,2)+z4(:,3),x5,z5(:,2)+z5(:,3));

title('inter production');

ylabel('concentration');

axis([0,7,0,1])

subplot(313), plot(x0,z0(:,4),x1,z1(:,4),x2,z2(:,4),x3,z3(:,4),x4,z4(:,4),x5,z5(:,4));

title('HBP');

xlabel('time');

ylabel('concentration');

axis([0,7,0,1]) end

progress2.m

global kC kA kB KA KB KC ksi ki kd aHBP ICA ICB Kawc Kaoc Kaow PWCD POCD POWD PWCH POCH POWH fc fw fo;

fo=0.1; fw=0.87; fc=0.03; PWCD=1; POWD=21000;

POCD=21000; PWCH=7; POCH=4.1; POWH=29; kA=11; kB=1.7; kC=1.6; KA=3.6;

KB=1.3; KC=1.4; ksi=1.8; ICA=60; ICB=110; aHBP=0.4; ki=40; kd=0.00055;

Kawc=1; Kaoc=1; Kaow=1;

y0=zeros(7,1);y0(1,1)=1;y0(5,1)=1;y0(6,1)=10;y0(7,1)=100;

y1=zeros(7,1);y1(1,1)=1;y1(5,1)=1;y1(6,1)=100;y1(7,1)=10;

y2=zeros(7,1);y2(1,1)=1;y2(5,1)=100;y2(6,1)=1;y2(7,1)=10;

y3=zeros(7,1);y3(1,1)=1;y3(5,1)=10;y3(6,1)=1;y3(7,1)=100;

y4=zeros(7,1);y4(1,1)=1;y4(5,1)=10;y4(6,1)=100;y4(7,1)=1;

y5=zeros(7,1);y5(1,1)=1;y5(5,1)=100;y5(6,1)=10;y5(7,1)=1;

e=linspace(0,7,100);

[x0,z0]=ode45('enzyme',e,y0);

[x1,z1]=ode45('enzyme',e,y1);

[x2,z2]=ode45('enzyme',e,y2);

[x3,z3]=ode45('enzyme',e,y3);

[x4,z4]=ode45('enzyme',e,y4);

[x5,z5]=ode45('enzyme',e,y5);

subplot(321), plot(x0,z0(:,1),'x',x1,z1(:,1),x2,z2(:,1),'v',x3,z3(:,1),x4,z4(:,1),'diamond',x5,z5(:,1));

legend('A:B:C=1:10:100','A:B:C=1:100:10','A:B:C=100:1:10','A:B:C=10:1:100','A:B:C=10:100:1','A:B:C=100:10:1');

title('DBT');

axis([0,7,0,1])

xlabel('time');

ylabel('concentration');

subplot(312), plot(x0,z0(:,2)+z0(:,3),x1,z1(:,2)+z1(:,3),x2,z2(:,2)+z2(:,3),x3,z3(:,2)+z3(:,3),x4,z4(:,2)+z4(:,3),x5,z5(:,2)+z5(:,3));

title('inter production');

axis([0,7,0,1])

subplot(313), plot(x0,z0(:,4),x1,z1(:,4),x2,z2(:,4),x3,z3(:,4),x4,z4(:,4),x5,z5(:,4));

title('HBP');

axis([0,7,0,1]);

xlabel('time');

ylabel('concentration');

end;

Oil-Water ratio：

t=0:0.01:0.97;

KD2=21000;

KD3=21000;

KH2=29;

KH3=4.1;

XC=0.03;  %XC is x(cell)

y=1./(1-t).*(t/KD3+XC/KD2+1-t-XC)./(t/KH3+XC/KH2+1-t-XC);

plot(t,y);

end;

Not for yet.

5.2 Related Experiments