Module 1: Enzymatic Degradation of Antibiotics
The first module of our project forms the basis of what we wish to achieve through our project. For this module we aim to model the enzyme kinetics of antibiotic degradation. For our simulations we have considered the sulfonamide sulfamethazine for degradation. We assume the kinetics of SulX as it is the rate limiting step of the reaction. We assume the concentration of SulR is abundant for complete degradation of the antibiotic
We earlier had used a mono substrate kinetics equation. As per Dr. Raghu’s instruction (chemistry instructor), we looked into bi-substrate enzyme kinetics equations. We found two mechanisms that are likely to be followed: ordered bi-bi and random bi-bi (refer annexure for more details).
The paper reported the kinetic parameters of only one substrate. He asked to include a parameter that takes care of the other substrate by taking the Michaelis Menten value of FMNH2 to be:
where α1 is the parameter and [FHMN2] is the concentration of FHMN2
Since our enzyme won’t be acting in a lab environment but in an excreta medium, it might not function very efficiently. Hence, he told us to include another parameter that α2 such that,
where P = concentration of products and D is the concentration of the drug.
Since the enzymatic mechanism for this reaction is not known we modelled the reaction kinetics using both ordered bi-bi and random bi-bi mechanisms. The values of α1 and α2 need to be experimentally determined and would change depending on the system of interest In our models we have assumed the whole range for α1 and a restricted range for α2.
0 ≤ α1 ≤ 1
0.7 ≤ α2 ≤1
This would finally give us a range of time intervals that our enzyme would take to degrade the enzyme below the Proposed No Effect Concentration (PNEC) levels. Since we do not have laboratory access this year, we can only work with the whole range of α1 and α2. The range of alpha values encompasses the broad range of systems that our model can be used to describe.
ORDERED BI BI MECHANISM:
X represents the degrading enzyme SulX.
In this mechanism, there is an order that the substrates follow while as they bind to their respective binding sites. First FMNH2 binds to its active site and is precessed and then the drug molecule (D in the figure) comes and bind to its active site to give rise to the final complex that in turn releases the products (P in the figure).
Derivation for ordered bi bi mechanism:
From enzyme conservation, we have,
From the assumed equilibrium conditions, we have,
From (3), we get,
From (2) we get,
From (4) substituting for [X-FMNH2] in (5) we get,
Now finally substituting (4) and (6) in (1), we get,
Factoring out [X-FMNH2-D] from (7),
Now,
Substituting value of [X-FMNH2-D] from (7) in (9),
[X]
0 K
cat = V
m. Thus, (10) becomes,
Equation (11) is the final differential equation for ordered bi bi mechanism.
From the two alpha values introduced above, we get
Integrating the differential equation, we get an equation for time, where,
S0=initial concentration of antibiotic
S=final concentration of antibiotic
RANDOM BISUBSTRATE MECHANISM:
In random bi bi mechanism, the substrates can bind in any order. Either FMNH2 can bind first or the drug forming the corresponding complexes and then the other substrate will bind to that complex giving the final complex which releases the products. Again, as in the ordered case, the reactions in the red circle are assumed to be in equilibrium.
By making similar assumptions as in the previous case and following similar lines, we get the following differential equation,
By considering the alpha values, the above equation transforms to,
Upon integration, we get the equation for time to be, where,
S0=initial concentration of antibiotic
S=final concentration of antibiotic
| Symbol | Description | Value |
|---|
| KD | Concentration of Drug at which the rate is half of Vm | 16.83 uM |
| Kcat | Rate constant for rate limiting step | 0.58 * 46,000 = 26 s |
Module 2: Reduction of Horizontal Gene Transfer
Transformation
Transformation is the process in which the bacteria picks up plasmid from the environment.
In this submodule, we aim to find the time to form the first transformation mutant due to the introduction of our engineered bacteria. We present the model for transformation for both single-gene system and the double-gene system.
SINGLE GENE SYSTEM:
Here we assume only one plasmid is required to confer resistance. We adapted a model already published in literature [1] and tweaked it to fit our situation.
The species into consideration are:
| SI no. | Species | Description | Units |
|---|
| 1 | B | Engineered Bacteria | cell/mL |
| 2 | B1 | Transformed Mutant | cell/mL |
| 3 | C | Wild Population | cell/mL |
| 4 | G | Plasmid | plasmids/mL |
Working: The engineered cells that will be degrading the antibiotics can die due to numerous reasons. When they die, they release the plasmid containing the antibiotic resistance gene into the environment. The wild population can take up these plasmids and can gain resistance to the antibiotics that we are trying to degrade. The cells replicate at a rate 𝝀 and follow a logistic growth curve. There is an intrinsic death rate in the population that we take as 𝛿. We have not considered any degradation rate of DNA in the environment as this would be the worst case scenario. In the single gene system the cells B upon death release the plasmid G into the environment. This contains the gene that can degrade the antibiotic and confer resistance. When the wild population of cells C come in contact with G they can uptake this genetic material which would lead to the formation of an AMR mutant in the population called B1. In the working of the single gene system we keep track of the number of B1 cells that are formed in the environment. The differential equations follow from this.
We can make a system of chemical equations from the cartoon.
Converting these chemical equations into a system of differential equations
| PARAMETER | DESCRIPTION | VALUE | UNITS | REFERENCE |
|---|
| 𝜆 | Growth rate | 0.03 | hr | [3] |
| 𝛿 | Death rate | 0.005 | hr-1 | [4] |
| r | Transformation rate | 10-8 | hr-1 | [5] |
| Ncap | Population cap | 109 | cells/mL | Chosen for this |
DOUBLE GENE SYSTEM:
In the double gene system, the cell needs to have both the genes to gain resistance. We have the two genes on two different plasmids. This means that, for the cell to have complete resistance, it needs to gain both the plasmids.
| SPECIES | DESCRIPTION | UNITS |
|---|
| B | Engineered Cells | cells/mL |
| B1 | Cells with plasmid G1 | cells/mL |
| B2 | Cells with plasmid G2 | cells/mL |
| B3 | Final Transformation mutant | cells/mL |
| G1 | Plasmid 1 | plasmid/mL |
| G2 | Plasmid 2 | plasmid/ml |
| C | Wild population | cell/mL |
Working: The engineered cells that will be degrading the antibiotics can die due to numerous reasons. When they die, they release the plasmid containing the antibiotic resistance gene into the environment. The wild population can take up these plasmids and can gain resistance to the antibiotics that we are trying to degrade. The cells replicate at a rate 𝝀 and follow a logistic growth curve. There is an intrinsic death rate in the population that we take as 𝛿. We have not considered any degradation rate of DNA in the environment as this would be the worst case scenario. In the double gene system the cells B upon death release the plasmids G1 and G2 into the environment. These contain the genes that need to be inherited together to confer resistance. When the wild population of cells C come in contact with G1 they can uptake this genetic material which would lead to the formation of B1. When the wild population of cells C come in contact with G2 they can uptake this genetic material which would lead to the formation of B2. B1 can come in contact with G2 or B2 can come in contact with G1 to form an AMR mutant in the population called B3. In the working of the double gene system we keep track of the number of B3 cells that are formed in the environment. The differential equations follow from this.
Converting the cartoon to chemical equations we get the following system of chemical equations,
We then make a differential equation model out of these chemical equations.
Conjugation
Conjugation is the process of DNA transfer by direct contact. The donor cells, the one having F plasmid forms a mating pair with the recipient cell, the one not having F plasmid. The DNA transfer is facilitated by the pilus that is formed during the formation of mating pair.
In this submodule, we aim to find the time it takes to form first F+ colikaze i.e., time it takes for our engineered cell to gain an F plasmid from the environment. For the worst case scenario, we assume that the wild population cells are F+.
Species considered are as given in the table below.
| SPECIES | DESCRIPTION | UNITS |
|---|
| R | Recipients | cells/mL |
| T | Transconjugant | cells/mL |
| D | Donor | cells/mL |
| IRT | Mating Pair between Recipients and Transconjugant | Mating pair/mL |
| ITT | Mating pair between two transconjugant | Mating pair/mL |
| IRD | Mating pair between Recipients and Donor | Mating pair/mL |
| ITD | Mating pair between Transconjugant and Donor | Mating pair/mL |
For the model we consider our Coli Kaze as R. All the bacterial cells grow at a rate of 𝝀. When a conjugal donor(bacteria that is F+) denoted by D comes in contact with R, they can form a mating pair at a rate k1 to form the intermediate IRD. This intermediate gets converted to another intermediate IDT at a rate 𝜸. This is the rate of DNA transfer. Finally the detachment rate k2 breaks the mating pair to give the final products of this interaction the transconjugant T and gives back the donor D. The detachment rate constantly operates over the system and can break the mating pairs at any point of time.
Similarly when a transconjugant T comes in contact with R, they can form a mating pair at a rate k1 to form the intermediate IRT. This intermediate gets converted to another intermediate ITT at a rate 𝜸. This is the rate of DNA transfer. Finally the detachment rate k2 breaks the mating pair to give the final products of this interaction are the two transconjugants T. The detachment rate constantly operates over the system and can break the mating pairs at any point of time. We keep track of the number of transconjugants T that are formed and consider them to be AMR mutants.
Our gene circuits involve two genes traT that reduces rate of mating pair formation and traS that reduces the rate of DNA transfer. These two genes produce proteins TraT and TraS respectively that act as inhibitors of mating pair formation and DNA transfer. Literature reports that when these two genes work together they cause a 33,000 fold reduction in the number of transconjugants formed
The system of chemical equations corresponding to this model is,
Converting this system of chemical equations into differential equations, the system we get is,
| PARAMETER | DESCRIPTION | VALUE | UNITS | REFERENCE |
|---|
| 𝜆 | Growth rate | 0.03 | hr-1 | [3] |
| 𝛿 | Death rate | 0.005 | hr-1 | [4] |
| 𝛄 | DNA transfer rate | 15 | hr-1 | [5] |
| K1 | Mating pair formation rate | 1.0 * 10-8 | hr-1 | [6] |
| K2 | Mating pair dissociation rate | 200 | hr-1 | [6] |
| Ncap | Population cap | 1010 | cells/mL | Arbitrarily Chosen |
Transduction
Transduction is the process in which the bacterial genetic material is transferred via a virus. The virus that infects the cell can pick up the genetic material of the cell and transfer it to some other cell while infecting it.
The aim of this submodule was to find out the time taken for transduction AMR mutants to form per ml of culture. We only assume generalized transduction events as our plasmids are very small for specialized transduction events to occur. We assume logistic growth curves for bacterial growth. There was one copy of AMR gene on the plasmid of the bacteria for the single gene system which conferred resistance (AMR). For the double gene system, there were two copies of the AMR gene on two different plasmids. Both the genes have to be taken up by the bacterium to gain resistance. Since we are tracking the number of resistant mutants that are formed due to the addition of our bacteria to the slurry, the fraction of Coli Kaze in the bacterial population determines the fraction of AMR bacteria in the slurry.
For our model, we have split every population into 2 parts. The first population would consist of a wild population of susceptible bacteria and the second part would consist of Coli Kaze population. Our Coli Kaze is also susceptible to viral infection and infection by transduced particles. Bacteria exist in 5 different states:- Susceptible, Newly infected, Processing plasmid, Lysogenic and Lytic. Lysogens are further divided into immune lysogens and non-immune lysogens. Immune lysogens cannot be further infected by bacteriophages.
Lytic bacteria are further divided into direct lytic and indirect lytic bacteria. The indirect lytic bacteria are induced for lysis upon induction by a lysogen. The direct lytic cycle is directly induced upon infection(newly infected bacteria). Susceptible bacteria are first infected and move into the newly infected stage. Then the bacteria proceeds to process the plasmid if it is not lysed(direct lytic stage). It then becomes a lysogen(immune/not immune). These are lysed upon induction(indirect lytic). The lytic bacteria (both direct and indirect) proceed into an intermediate stage called the burst stage and then phages are released from the bacteria. The phages follow exponential growth whereas the bacteria follow logistic growth.
The population split enables us to track the number of AMR mutants formed. We track the number of transduced particles released from Coli Kaze bacteria that carry the AMR gene and infect the wild population. The table in the above diagram summarizes infection and death dynamics.
Let us look into the total lysogenic population of the population split. They are composed of immune lysogens and non-immune lysogens. Non-immune lysogens are transferred back to the susceptible population as per the model. The susceptible population is composed of the wild population and Coli Kaze. When the wild population is infected with a transduced particle it becomes an AMR susceptible bacteria or goes back to the susceptible bacterial population. If infected by a virion it goes back to the total lysogenic population. The total lysogenic population is just a placeholder for depiction and does not appear in the equations.
The above picture shows the different ways the virus activates the lytic cycle. The virus upon infecting the bacteria can directly activate the lytic cycle. Otherwise the virus waits for a while after infection until the conditions are ideal and then activates the lytic cycle. This is called the induced lytic cycle. All these populations go into a transient phase called the burst phase. This leads to the death of the bacterium and the release of virions and transduced particles. The transduced particle could contain the AMR gene or not depending on whether the AMR gene was packaged into the protein coat. A fraction of the virion population (N13) constitute the transduced particles.
The above figure summarizes infection dynamics of the system. Wild population can be infected by a transduced particle or a virion. If the transduced particle contains AMR gene it becomes an AMR susceptible population that we keep track of to count the AMR mutants We then find out the time it takes to form an AMR mutant per ml of culture in both the single gene and double gene systems for our purposes. Our enzyme must degrade the antibiotic before any transduction mutants are formed for maximum efficiency in the whole cell system.
| PARAMETER | DESCRIPTION | VALUE | UNIT |
|---|
| ph_br_n | Burst size per lysed bacterium | 250 | - |
| ph_fr_tp | Fraction of tp in burst | 0.02 | - |
| ph_d | Fractional extracellular deactivation rate of phage virions/tp | 0.0745 | h-1 |
| ph_cir_n | No. of circulating virions/tp | 107 | - |
| MOI | No. of circulating virions/tp per bacterium | ph_cir_n/Ncap | - |
| pr_inf_hit | Probability of virion/tp injecting the carried genetic material into the bacterium given absorption | 0.8 | - |
| pr_inf | Probability of infection for a growing susceptible or lysogen not immune to other enteric phages | 1 - exp(MOl*pr_inf_hit) | - |
| b_fr_inf_lys | Fraction of new infections going into lysogenic cycle | 0.9 | - |
| b_fr_limm | Fraction of lysogens cross-immune to other enteric phages | 1 | - |
| b_fr_lys_in | Fraction of lysogens in inflowing bacteria | 0.05 | - |
| b_fr_lys_en | Fraction of lysogens in enteric bacteria (start) | 0.013 | - |
| 𝜆 | growth rate of bacteria, | 0.03 | h-1 |
| 𝛿 | death rate of bacteria, | 0.005 | h-1 |
| Ncap | Population cap | 1010 | - |
| g_n | No. of AMR gene copies per bacterium | 1 | - |
| pr_gpick_gntr _pl | Probability tp picks plasmid bearing AMR gene, per lytic cycle per bacterium | 0.02 | - |
| pr_tp_pl_est | Probability tp picks plasmid bearing AMR gene, per lytic cycle per bacterium | 1 | - |
| ph _frAMRtp | Fraction of to with AMR gene of | 0.02 | - |
| t_newi | Newly infected bacteria | 1 | h |
| t_tp | Bacteria process tp | 1 | h |
| t_Itcl_v | Direct lytic cycle | 0.3833 | h |
| t_Itcl_t | Lytic cycle after prophage induction | 0.9583 | h |
| t_lscl | Lysogenic cycle | 100 | h |
A transduced particle (tp) of E. coli was assumed to carry 100 kbp of the accidentally packaged bacterial DNA (0.02 of the total bacterial genome, based on available estimates for coliphages). The target site for phage replication was nonspecific; hence, the probability that a tp randomly acquired the genome segment containing the AMR gene copy, if present in the lysed bacterium, assuming the gene copy number was 1, depending on the size of the segment picked (100 kbp) and the total genome size about (5,100 kbp). Hence, the probability of a tp acquiring the segment containing the AMR gene was approximately 1/51; the segment could be chromosomal or the plasmid assumed to be present in E. coli. This probability was squared for the double gene system for the formation of antimicrobial-resistant mutants.
For generalized transduction of a plasmid, the probability that tp leaving the lysed bacteria had packaged the plasmid, if present, was pr_gpick_gntr_pl, and the probability of plasmid establishment in the next bacterium receiving the tp was pr_tp_pl_est. In the relevant modelled scenarios, the AMR gene copy, if present in the bacterium, was always on the plasmid. Further, the assumed fraction of enteric E. coli bacteria carrying the plasmid, b_fr_AMR_en, was the fraction of Coli Kaze in the population. Therefore, the probability that a tp received by a bacterium transduced the plasmid with the AMR gene was as follows: (ph_frAMRtp) x (b_fr_AMR_en) x (pr_gpick_gntr_pl) x (pr_tp_pl_est).
We also assumed that the plasmid always established in the recipient bacterium and enumerated the E. coli bacteria after the internalized genetic material of tp was processed (assuming a 1-hour processing time). The plasmids were approximately 5 kbp in size, both of them need to be inherited together to confer total resistance.
Differential Equation model for transduction:
Species and their descriptions used for the transduction model are as given in the table below
| SPECIES | DESCRIPTION | UNITS |
|---|
| N1 | Newly Infected bacteria of subpopulation 1 (Wild Population) | cells/mL |
| N2 | Newly Infected bacteria of subpopulation 2 (Coli Kaze) | cells/mL |
| N3 | Susceptible bacteria of subpopulation 1 (Wild Population) | cells/mL |
| N4 | Susceptible bacteria of subpopulation 2 (Coli Kaze) | cells/mL |
| N5 | Immune Lysogen of subpopulation 1 (Wild Population) | cells/mL |
| N6 | Immune Lysogen of subpopulation 2 (Coli Kaze) | cells/mL |
| N7 | Bacteria processing tp in subpopulation 1 (Wild Population) | cells/mL |
| N8 | Bacteria processing tp in subpopulation 2 (Coli Kaze) | cells/mL |
| N9 | Bacteria of subpopulation 1 (Wild Population) in Direct Lytic cycle | cells/mL |
| N10 | Bacteria of subpopulation 2 (Coli Kaze) in Direct Lytic cycle | cells/mL |
| N11 | Bacteria of subpopulation 1 (Wild Population) in Induced Lytic cycle | cells/mL |
| N12 | Bacteria of subpopulation 2 (Coli Kaze) in Induced Lytic cycle | cells/mL |
| N13 | Virions | virions/ml |
Using the species mentioned in the table we make a differential equation model as per the schematic mentioned above.
The concentration of AMR mutants would be calculated by the following equation:
For single gene system:
For double gene system, the term b_fr_AMR_en is squared as there are two plasmids that the virus has to pick up.
Therefore the double gene system,
References:
- 1. Kim, Dae-Wi, et al. "A novel sulfonamide resistance mechanism by two-component flavin-dependent monooxygenase system in sulfonamide-degrading actinobacteria." Environment international 127 (2019): 206-215
- 2. Mao, J., & Lu, T. (2016). Population-dynamic modeling of bacterial horizontal gene transfer by natural transformation. Biophysical journal, 110(1), 258-268.
- 3. Hendricks, C. W. (1972). Enteric bacterial growth rates in river water. Applied microbiology, 24(2), 168-174.
- 4. Menon, P., Billen, G., & Servais, P. (2003). Mortality rates of autochthonous and fecal bacteria in natural aquatic ecosystems. Water Research, 37(17), 4151-4158.
- 5. Lu, N., Massoudieh, A., Liang, X., Kamai, T., Zilles, J. L., Nguyen, T. H., & Ginn, T. R. (2015). A kinetic model of gene transfer via natural transformation of Azotobacter vinelandii. Environmental Science: Water Research & Technology, 1(3), 363-374
- 6. Zhong, X., Kro, J. E., Top, E. M., & Krone, S. M. (2010). Accounting for mating pair formation in plasmid population dynamics. Journal of theoretical biology, 262(4), 711-719.
- 7. Volkova, V. V., Lu, Z., Besser, T., & Gröhn, Y. T. (2014). Modeling the infection dynamics of bacteriophages in enteric Escherichia coli: estimating the contribution of transduction to antimicrobial gene spread. Applied and environmental microbiology, 80(14), 4350-4362
Module 3: DNA Degradation and Cell Death (‘kill switch’)
The aim of this module is to determine the time required for complete genome degradation after induction of the ‘kill switch.’ We also try to figure out the concentration of the sugars Arabinose and Glucose indirectly, so that the efficiency of our kill switch is increased.
Working: Following is the sequence of the genes we use for the ‘kill switch’. The AraC protein is constitutively produced by a medium anderson promoter and production of DNase 1 (bovine pancreatic DNase 1) is controlled by the arabinose promoter or the P-araBAD.
The AraC protein forms a dimer, each molecule binds to the araI1 and araO2 regulatory regions of the DNA and prevents the binding of RNA polymerase molecule to the P-araBAD and hence repressing the production of DNase 1 protein. But during this, AraC protein is still constitutively produced. We assume that every cell is by default in this stage only.
In the presence of Arabinose (and low amount of glucose sugar, will be explained along with equations), the Arabinose binds to AraC protein dimer, brings about a conformational change and binds to the regions araI1 and araI2 regions of the DNA releasing the P-araBAD. Now RNA polymerase is free to bind to the promoter and proceed with transcription.
The incorporation of RNA polymerase at the promoter is enhanced by the CAP-cAMP complex bound to the CAP binding site. As the concentration of cAMP is inversely proportional to the concentration of glucose, for a constant amount of CAP protein, lesser the concentration of glucose, more will be the concentration of cAMP molecules, hence enhanced transcription of DNase 1 gene.
The complete overview of Module 3
| SPECIES | DESCRIPTION | EXPLANATION |
|---|
| G1 | AraC genes | Number of AraC genes in a cell |
| R1 | AraC RNA | Number of AraC RNA's in a cell |
| P1 | AraC protein | Number of AraC molecules in a cell |
| A | Arabinose sugar | Number of molecules of arabinose sugar outside the cell |
| X1 | Arabinose sugar bound to AraC | Number of A-AraC complexes in a cell |
| θ1 | AraC protein bound to PBAD promoter | Number of actively repressed p-bad promoters in a cell |
| θ2 | A-AraC (arabinose sugar AraC complex) bound to 11 and 12 sites of DNA) near p-bad promoter (this actually releases the DNA loop to give free promoter) | Number of active PBAD promoters in a cell |
| θ3 | A-AraC-CC (arabinose sugar AraC complex bound to 11 and 12 sites and CAP-cAMP complex bound to the CAP binding site of DNA) near to P-bad promoter (this enhances the efficiency with which RNA polymerase binds to the promoter) | Number of promoters which have enhanced activity in a cell |
| X2 | c-AMP molecules | Number of c-AMP molecules in a cell |
| Eg | External glucose | Number of moles of glucose sugar outside the cell |
| X3 | Catabolite activator protein (CAP) | Number of CAP molecules inside a cell in free form |
| X4 | CAP bound to c-AMP | Number of CAP-c-AMP complex in a cell |
| G2 | DNase 1 gene | Number of DNase 1 gene in a cell |
| R2 | DNase 1 mRNA | Number of DNase 1 mRNA in a cell |
| P2 | DNase 1 protein | Number of DNase 1 molecules in a cell |
| S | Substrate concentration | Number of phosphodiester bonds in the DNA of a cell |
| Vc | Rate of c-AMP production (depending on external glucose concentration) | Number of c-AMP molecules produced in cell per second |
| GT | External genes added to cell | Total number of gene segments added to cell (both AraC and DNase 1) |
Production of AraC Protein:
In this part, we aim to model the dynamics of the repressor protein AraC. The AraC gene (G1) produces the AraC mRNA (R1) at the rate a1. The mRNA degrades at a rate d1. The AraC mRNA produces AraC protein (P1) at a rate a2 and the AraC protein (P1) degrades at a rate d2.
Assumptions: The cell has the capacity to produce sufficiently large amount of AraC mRNA and and protein (i.e in the interval we consider, production does not reach saturation).
Growth rate of bacteria is in the stationary phase.
Rate of degradation of protein and mRNA of AraC are same as that of other proteins and mRNAs (actually very negligible).
The equations corresponding to this processes are:
Where,
a1 is 0.289 mRNAs per gene per second (annexure 2- 1*),
a2 is 0.143 molecules per mRNA per cell (annexure 2- 2*),
d1 is 3.0X10-7 s-1 (annexure 2- 3*) and
d2 is 1.2X10-11 s-1 (annexure 2- 4*).
Effect of AraC and Arabinose on PBAD promoter:
Here we wish to model the dependence of AraC and Arabinose on the araBAD promoter. AraC protein (P1) represses the araBAD promoter (araBAD promoter, θ1), whereas Arabinose (A) binds with the AraC protein (P1) to release the araBAD promoter.
Assumptions:
Amount of arabinose in the system is in excess (i.e. at least 100 times more than the number of AraC proteins)
ka, being a first order rate constant, depends only in the number of AraC proteins
kd, being a 2nd order reaction’s equilibrium constant, depends on θ1 and θ2 alone
The reactions of this processes:
The corresponding equation are:
Where kd is 1.8X104 molecules per cell and ka is -0.26 s-1. [1]
Effect of Extracellular Glucose on araBAD promoter:
The external concentration of glucose affects the production of cAMP (𝛘2) molecules in the cell. There is a fixed amount of CAP proteins in a cell at any point of time.The Catabolite activator protein (𝛘3) binds to the Cellular cAMP (𝛘2) to form the CAP-cAMP complex (𝛘4) which enhances the transcription of DNase 1 gene (G2).
Assumptions:
1. Rate of production of cAMP inside the cell is affected by external glucose concentration
2. Amount of cAMP molecules inside a cell is constant.
3. Kinetics of CAP-cAMP complex binding to araBAD promoter is the same as that of its binding with the Lac-promoter as the genetic material (DNA) is same for both the cases.
How does the external concentration of glucose affect the production?
Where 0.035 s-1 is the rate constant of cAMP degradation.
These (given below) are the reactions of how cAMP enhances the activity of the free promoter.
And these are the equations:
The number of promoters bound to A-AraC-CAP will be [2]
This equation is obtained by solving the equations (10.i) to (10.viii)
Where,
References: [2] and Annexure 2- (12*,9*,5*,6*)
Production of DNase 1 protein:
DNase 1 gene (G2) produces DNase 1 mRNA (R2) at the rate b1, this DNase 1 mRNA (R2) degrades at a rate d1, DNase 1 mRNA (R2) produces protein at the rate b2, DNase 1 protein (P2) degrades at the rate d2.
Assumptions:
same as that for AraC production
The reactions are
The equations for this process are as follows:
Where,
Kelong 0.046 is mRNAs per gene per second (annexure 2- 5*),
b2 is 0.0006026 molecules per mRNA per cell (annexure 2- 7*),
d1 is 0.0000003 s-1 (annexure 2- 3*) and
d2 is 0.000000000012 s-1 (annexure 2- 4*).
DNA Degradation:
The DNase 1 protein degrades the DNA by digesting the phosphodiester bonds of DNA backbone and producing single stranded breaks at the rate:
Where kcat is 176 s-1 and km is 0.000978 molecules per cell, according to [3]
However, substrate concentration (or number of phosphodiester bonds) never reaches zero as the single strand breaks made by the DNase 1 are efficient (rate=r) only till the length of each strand reaches 2-3 bases (i.e number of [S] in each single strand is 2 or 1).
Let us assume that all strands are reduced to a size of 2 bases (i.e. 1 phosphodiester bond per final segment). This implies that the final [S] after which DNase 1 will be inefficient is “half of initial substrate concentration” that is 6143249.
But by the time this happens, the genes coding for AraC and DNase 1 protein also would have degraded. As the time taken for degradation is very less, the degradation of the genes would hardly matter. But to be accurate and develop a rigorous mathematical model, we do the following,
According to the [4]:
The mass distribution of DNA fragments at a given interval depending upon the amount and time of UV radiation it is exposed is given by:
We take M to be 5074 (number of base pairs on AraC and DNase 1 gene together) and r is dS/dt.
References:
- 1. https://2011.igem.org/Team:St_Andrews/modelling
- 2. Mathematical_Model_of_the_lac_Operon
- 3. Wiley Library
- 4. Researchgate
