Here we have developed a mathematical model which will link and will make us understand the theory of our biological work and the physical feasibility of our designed experiment for further optimisations and understanding the results.
As we know that there is an alarming increase in the level of reactive nitrogen in the environment. Even though nitrogen in molecular form is one of the most abundant and essential compound in earth's atmosphere, but there are various forms of nitrogen(reactive type) which can come in contact with water and can give rise to N-Nitrosanime which posed a potential threat to human health during drinking water disinfection.
Here in the project we have demonstrated the construction of E.coli which constitutively produces a protein molecule - ADA(Adaptive response protein). This protein is released in E.coli upon DNA damaged caused due to alkylation. Here we have used the reporter fluroscence which will give us the idea about the N-Nitrosanime present in the water.
Our Goals :-
1.To engineer E.coli cell that can detect the presence of N-Nitrosanime in the water
2. To take an account of rate and range of N-Nitrosanime detected.
ParametersThe following parameters were used to design and construct Ordinary Differential Equation:
|K1||Reaction kinetics for enzymes||1||mg/ml|
|B||K1(1+ K2 Rtot)|
|Ǭm||maximum transcription rate of DNA||80 nt sec-1||t^-1|
|ƀd||maximum level of transcription||0.75||mol ml-1|
|f(E)||unbound operators of enzymes||50 nt||t^-1|
|ϒm/ϒI||degradation rate [mRNA/Protein]||3.3 nucleotides/20-40%of total bacterial protein|
|βI||maximum level of translation of mRNA|
|K2||reaction kinetics of repressor||mg/ml|
|Z||Alk-B ADA promoter|
- The DNA damaging effect of alkylating agents like nitrosamines is characterized by the formation of O6-MeG or O4-MeG lesions.
- The Ada response protein of ADA regulon, one of the proteins involved in Adaptive response for DNA repair, scans the DNA and gets methylated when comes in contact with O6-MeG or O4-MeG lesion.
- The methylated Ada protein acts as a transcriptional activator of three operons of ADA regulon. One of the operon has gene for Ada protein which leads to increase in the number of mRNA coding for Ada response protein.
- First equation is to find the rate of mRNA (mRNA of Ada protein) formation in E.coli when Ada responds to alkylated DNA. As per equation 1, the rate of mRNA can be defined as difference between the rate production of mRNA and rate of degradation of mRNA.
- • The rate of Ada mRNA production, corresponding to the repair mechanism, is represented as ƀd *Ǭm *f(E) and the rate of mRNA degradation as ϒm*M
- Therefore, the final equation rate of mRNA production becomes dM/dt= ƀd Ǭm f(E)- ϒmM (Equation 1)
- In our system we have placed fluorescent reporter gene (GFP) downstream to a methylated Ada induced promoter. Therefore, the expression of the fluorescence reporter (GFP) will confirm the presence of alkylating agents like Nitrosamines.
- Equation 2 demonstrates the rate of GFP expression, (i.e. dI/dt), βI *M is the protein that is translated by DNA and the degradation of that protein is represented as ϒI *I
- Therefore, the final equation for rate of GFP expression becomes dI/dt= βIM- ϒiI (Equation 2)
- As per the mechanism of Ada response protein, the Ada can transcriptionally activate the operons of Ada regulon in methylated form (active form) only.
- When the unmethylated form (inactive form) of Ada comes in contact with the lesion, it gets converted to an active methylated form due to the methyl shift reaction taking place at the point of contact.
With the following equilibrium equation
K2=ZXYn/ZXYn , K2=K2^-/K2^+
So the fraction of ALK-B ADA promoter not bound by unmethylated ADA is
Assuming as before , that the amount of X bound to Z is compared to the amount of unmethylated ADA gives
f(Y)=1+K1Y^n / 1 + (K1+K1K2Xtotal)Y^n = 1+K1Y^n / 1+KY^n
(1+K1E^n)/(A+BE^n )= Ǭm f(E) (Equation 3)
Now let, 3000 = A
Solving Eq. 1, we get
Complimentary Function: C1e-3.3t
Particular Integral: A / 3.3
M = Complimentary Function + Particular Integral
M = C1e-3.3t + A / 3.3
And solving Eq. 2, we get
DI = 20M - 0.4I
(D + 0.4)I = 20C1e-3.3t + 20.4/3.3
Complimentary Function: C1e-0.4t
Particular Integral: -20C1e-3.3t/2.9 + 20A/1.32
T = 2 minutes Interval = 10 seconds A = 3000
Graph MATLAB code:
u=[3000 - 3.3*x(1) , 20*x(1)-0.4*x(2)]';
[tout, xout]=ode23('vdpol2', [t0, tf], x0);
hp=plot(tout, xout(:,1),tout, xout(:,2));
The following graph is a representation of concentration of mRNA and intermediate protein against time.
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