MODEL

MATH MODELS

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MUTANT MHETase:

Introduction:

The lac repressor is a protein encoded by the lacI gene. It binds to the operator region of the lac operon's DNA, specifically to the operator sequence, and blocks the transcription of the adjacent genes (lacZ, lacY, and lacA) by preventing RNA polymerase from binding to the promoter. This binding occurs by default when lactose is absent in the cell.

When lactose is present, it acts as an inducer. Lactose is converted into allolactose, which binds to the lac repressor. This binding causes a conformational change in the repressor, preventing it from binding to the operator. As a result, RNA polymerase (RNAP) can bind to the promoter, initiating the transcription of the genes in the lac operon. In this way, the lac repressor regulates the expression of the lac operon genes in response to the presence or absence of lactose or its derivative allolactose.

Given below is the model for gene expression for the mutated MHETase enzyme. We’ve used the hill function to predict the concentration of the enzyme that will be produced by the E.coli cell.

The Hill function is a mathematical formula used to describe the cooperative binding of ligands to receptors or enzymes. It is commonly used in biology and biochemistry to model the relationship between the concentration of a ligand (such as a transcription factor) and the response of a biological system (such as gene expression).[1]

Assumptions and Data:

1. [Lactose]=[Allolactose]
2. Quasi-Steady-State Approximation (QSSA): → [for mRNA]

QSSA is used to simplify the analysis of reaction networks involving gene transcription and translation processes. Gene expression involves multiple steps, including the transcription of DNA into RNA (mRNA) and the translation of mRNA into proteins. Each step of this process can involve various intermediate chemical species and reactions.

QSSA assumes that the concentrations of certain intermediate species (like mRNA in this case) reach a quasi-steady state rapidly compared to the other reactions in the system. In other words, the rates of formation and degradation of these intermediates are much faster than the rates of other reactions in the system. By making this assumption, the concentrations of these intermediates can be considered approximately constant, simplifying the overall reaction mechanism.

This approximation allows us to reduce a complex system of differential equations representing the reactions involved in gene expression to a smaller set of equations, making the analysis and understanding of the system easier. [2][3]

3. Equilibrium → [for Protein]
4. All the constants taken have been derived from general data available for E.coli. The mutant enzyme size (1809 aa) has been taken into consideration.

Transcription rate:

k1 = 85 (nucleotides/second) → RNA polymerase transcription rate of stable RNA in Bacteria Escherichia coli. [4]

Translation rate:

k2 = 0.004422 (mutant MHETase/second) → Average translation rate in Bacteria Escherichia coli is ~8 amino acids/sec. Since our enzyme is 1809 a.a. long, we should get 8/1809 enzyme molecules per second. [5]

Copy number:

Cn = 40 (copies)

mRNA degradation rate:

d1 = 2.4444 (nucleotides/second) → Average size of mRNA for E.coli is 1100 bases. Then, the Average mRNA degradation time for Bacteria Escherichia coli is 5 to 10 min. So, let us consider it to be 7.5 min. Thus, the rate of degradation will be 1100/(7.5*60) nucleotides/second.[6][7]

Mutant MHETase secretion rate:

d2 = 0.0039798 (mutant MHETase/second) → Assuming 90% secretion efficiency (almost 100% secretion efficiency exists, however, it requires a large range of optimization methods and their combinations). Thus, rate of secretion will be 0.9*0.004422.[8]

Concentration at which the binding of TF is half-saturated:

Kd = 3*10^(-9) (M) → Equilibrium dissociation constant for specific binding of RNAP to phage T7 promoter in Bacteria Escherichia coli. Thus, very high affinity as suggested by the relatively low dissociation constant for promoter.[9]

Max rate at which LRP binds to Operator:

vm = 0.002825 (molecules/second) → Time for a single lac repressor dimer in one cell to find a specific operator in Bacteria Escherichia coli is 354s. Thus, the rate is 1/354. [10]

Dissociation constant of Allolactose:

ki = 6*(10^(-6)) (M) → Affinity of Allolactose and LacI in Bacteria Escherichia coli. [11]

5. The inactivation of Lac Repressor Protein (LRP) by Lactose is assumed to behave like a Michaelis Menten reaction with an inhibitor involved.

LAC REPRESSOR MODEL:

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Equations for the Hill function of this repressor:

Transcription rate:

k1 = 85 (nucleotides/second) → RNA polymerase transcription rate of stable RNA in Bacteria Escherichia coli. [4]

Hill coefficient:

n = 2 (ideal)

Concentration at which the binding of Lac Repressor Protein is half-saturated:

K = 10*(-13) (M). [13]

Designing a MATLAB code to simulate the equation:

% Parameters
k1 = 85; % transcription rate (nucleotides/second)
k2 = 0.004422; % translation rate (mutant MHETase/second)
Cn = 40; % copy number (copies)
d1 = 2.4444; % mRNA degradation rate (nucleotides/second)
d2 = 0.0039798; % mutant MHETase secretion rate (mutant MHETase/second)
Kd = 3*10^(-9); % concentration at which the binding of RNAP is half-saturated (M)
bo = m; % Basal expression (without repressor)
K = 10*(-13); % concentration at which the binding of Lac Repressor Protein is half-saturated (M)
n = 2; % Hill coefficient
% Calculate m
m = (k1/d1) * Cn * (RNAP./(Kd + RNAP));
% Define a range of R values
R = linspace(0, 500, 100); % Number of molecules of repressor
% Calculate Hill function using the given equations
H = k1*(K.^n)./((K.^n)+(R.^n));
% Plot the graphs
figure;
plot(R, H, 'g', 'LineWidth', 2);
xlabel('R');
ylabel('H');
legend('H');
title('Graph of H vs R');
grid on;
hold off;

The graph suggests that the hill function decreases as the concentration of LRP (R) goes on increasing. This indicates that gene expression will indeed get repressed in the presence of LRP.

WITH LACTOSE:

Designing a MATLAB code to simulate the equation:

% Define the parameters and range for [LRP]
vm = 0.002825; % per second
km = 10^(-13); % M
L = 1; % Concentration of Lactose in M
ki = 6*(10^(-6)); % M
LRP = linspace(0, 0.0000005, 100); % Generating 100 points from 0 to 10 for [LRP]
% Calculate Operator bound for each value of [LRP]
Operator_bound = (vm * LRP) ./ ((km * (1 + (L / ki))) + LRP);
% Plotting the graph
plot(LRP, Operator_bound, 'LineWidth', 2);
xlabel('[LRP]');
ylabel('Operator bound');
title('Operator bound vs [LRP] Plot');
grid on;

Thus, binding of Lac Repressor Protein [LRP] to the operator will eventually become negligible.

MUTANT MHETASE GENE EXPRESSION:

Modelling equations for the above reactions:

HTML- https://pastebin.com/aPYaT2k8

Equation 1

m= Rate of formation of mRNA
k1 = Rate of transcription
CN = Copy number
[TF] = Concentration/amount of transcription factor (RNAP)
Kd = Concentration at which the binding of TF is half-saturated (dissociation constant)
d1 = Rate of degradation of mRNA
m = Concentration/amount of mRNA

HTML- https://pastebin.com/ZjsF0nMk

Equation 2

p= Rate of formation of Mutant MHETase enzyme
k2 = Rate of translation
d2 = Rate of secretion of enzyme
p = Concentration/amount of protein

Thus, from equation 1

HTML- https://pastebin.com/14vLKqVx

QSSA = Quasi Steady State Approximation

HTML- https://pastebin.com/sBZ49n22

Equation 3 → plot → concentration of mRNA with respect to the amount of RNAP

HTML- https://pastebin.com/jVZjayLf

Substitute value of m in equation 2

HTML- https://pastebin.com/qJAFhFfp
HTML- https://pastebin.com/JetYGA5d

Equation 4

Designing a MATLAB code to simulate the equations:

% Parameters
k1 = 85; % transcription rate (nucleotides/second)
k2 = 0.004422; % translation rate (mutant MHETase/second)
Cn = 40; % copy number (copies)
d1 = 2.4444; % mRNA degradation rate (nucleotides/second)
d2 = 0.0039798; % mutant MHETase secretion rate (mutant MHETase/second)
Kd = 3*10^(-9); % concentration at which the binding of TF is half-saturated (M)
% Define the range of [TF] values
TF = linspace(0, 0.00000001, 100); % concentration of RNAP (M)
% Calculate alpha
alpha = k1 * k2 * (Cn / d1);
% Calculate m and p using the given equations
m = (k1/d1) * Cn * (TF./(Kd + TF));
p = (alpha / d2) * (TF ./ (Kd + TF));
% Plotting
figure;
plot(TF, m, 'LineWidth', 2, 'DisplayName', 'm');
hold on;
plot(TF, m, 'LineWidth', 2, 'DisplayName', 'p');
xlabel('RNAP');
ylabel('m and p');
title('Plot of m and p vs RNAP');
legend('show');
grid on;

CONCLUSION

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Hence, we can visualise how our mutant MHETase enzyme will be produced, while being regulated by the LAC operon system.

Enzyme kinetics Equation:

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The following model is based on Michaelis-Menten kinetics.For further introduction refer to [15] Ingalls.
The reaction between PET and PETase gives MHET, BHET, and TPA as the products(in our case we have used the more efficient mutant of PETase that is the FAST PETase, for further information refer to [19] Lu et al. However, as it can be observed from figure 2.B in Yoshida et al. [17], the amount of produced MHET is significantly greater than the produced BHET and TPA.Thus, they are taken as negligible products, since MHET is our main interest for the second reaction MHET + Mutant MHETase .
In the Ordinary differential equations given below the k values represent the rate of reaction, where {1} is the PET and FAST PETase reaction and C1 is the complex formed by PET and FAST PETase.
The rection {2} is the reaction between MHET and Mutant MHETase and C2 is the complex formed by MHET and Mutant MHETase.We keep track of the seven chemical species in (1) and (2) Here, we are assuming TPA and EG as a single product for keeping the equation simple and not increasing the complexity). This leads to seven ODEs describing the concentrations of the species over time, namely:
K1,k2,k3 are the reaction rates from equation {1},
K4,k5,k6 are the reaction rates from equation {2}

Where S1 is the concentration of PET,E1 is the concentration of FAST PETase,C1 is the concentration of the complex PET+ FAST PETase,S2 is the concentration of MHET, E2 is the concentration of Mutant MHETase, C2 is the concentration of the complex MHET+ Mutant MHETase, and
P is the final product TPA and EG.The main goal of this model is to know the concentration of TPA and EG produced after time t has passed.The references [18,19,20,21,22] were used for the assumption of the parameters

S1 = Substrate 1 (PET)
E1 = Enzyme 1 (FAST PETase)
C1 = Complex 1 ( formed by PET and FAST PETase)
S2 = Substrate 2 ( MHET)
E2 = Enzyme 2 ( Mutant MHETase)
C2 = Complex 2 (formed by MHET and Mutant MHETase)

P1 & P2 are negligible (BHET and TPA formed are in negligible quantity as compared the MHET formed)
P = TPA and EG formed along with the MHETase left

dS1/dt = -k1S1E1 + K2C1 -1
dE1/dt = -k1S1P1+ (K2+K3)C1 -2
dC1/dt = k1P1E1 -(K2+K3)C2 -3
dS2/dt = k3C1 - k4P2E2 + k5C2 -4
dE2/dt = k5P2E2 + (k5 + k6)C2 -5
dC2/dt = k4P2E2 - (k5 + k6)C2 -6
dP/dt = k6C2 -7

Figure 1. shows the chemical equations governing the reactions under the Michaelis-Menten assumption.

References

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3. Shi, X. (2021). A Hill type equation can predict target gene expression driven by p53 pulsing. FEBS Open Bio, 11(6), 1799–1808. https://doi.org/10.1002/2211-5463.13179
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