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The change from software such as Simbiology and Simulink called for a more fundamental method of modelling cell death. Preliminary research showed kill switches producing the proteins “Lysozyme c” and “DNase 1” both had very similar mechanisms; as both are enzymes. Therefore, it was decided that the two models would use the same code to simulate mRNA and protein production.
Initially, following advice from biologists and biochemists on our team, the first code incorporated two step functions, the first of these modelled mRNA production by a single E. coli cell. To calculate protein made by each mRNA a secondary step function triggered each time an mRNA was produced, the sum of these functions gave the total amount of protein. This program was simple with the only input variables being the production time of mRNA and the respective protein. The initial model was presented to the rest of the team receiving plenty of feedback, the most prevalent point being that the code modelled a single cell system with a single plasmid whilst it should model a single cell system that duplicates and has multiple plasmids. It was suggested the following factors were added to the model:
The initial feedback called for a re-write of the code as a considerable amount of the suggestions came in stages before mRNA production. A second version of the code was written which incorporated all main steps between the splitting of E. coli to the degradation rate of protein.
It is important to outline the factors that were overlooked due to research finding their effect on the model would be negligible. Firstly, after researching the production time of plasmids in E. coli it was found that plasmids will reproduce at a variable rate to maintain a constant population determined by their copy number (Nordström and Dasgupta, 2006). To address this, the production rate of plasmids was overlooked, allowing for a constant value of plasmids to be maintained throughout the simulation. In experiments a pSB1C3 strain was used - a high copy number plasmid; therefore the copy number was set to 300. Secondly, both enzymatic models will assume that travel time of protein to the substrate is negligible. Protein diffusing through the cytoplasm of E. coli have a diffusion coefficient on the scale of $10 \mu \text{m}^2 \text{s}^{-1}$ (Elowitz et al., 1998). Considering the surface area of E. coli is approximately $10 \mu \text{m}^2$, the protein will reach the cell wall in a several seconds which is several magnitudes of order smaller than the simulation time. Lastly, the models will assume no mutations occur; the aim of the simulations is to determine whether kill switches are a plausible method of biosafety, if the models show a kill switch is not a reliable way to terminate GMO’s then accounting for mutations will only that enforce statement.
Research showed that there is an upper limit of mRNA in E. coli, therefore an upper limit of $4 \times 10^3$ mRNA per E. coli cell (Thermofisher.com, 2016) has been included in the model. The lifetime of mRNA can be found from the observed half life of approximately 5 minutes or $300\text{s}$ (Bernstein et al., 2002), resulting in an average lifetime of $430\text{s}$. The production rate of mRNA along with ribosomes per coding region will be worked out for both the lysozyme and DNase models independently. In addition to this, both enzymatic models use the well known duplication time of E. coli - 17 minutes, at which point both the mRNA and protein is assumed to split among the two cells equally,.
$T_{(mRNA)} = \frac{T_{(mRNA) \frac{1}{2}}}{ln(2)} = \frac{300\text{s}}{ln(2)} = 430\text{s (2sf)}$(Hyperphysics.phy-astr.gsu.edu, 2016)
$T_{(mRNA)}$: Lifetime of mRNA [$\text{s}$]
$T_{(mRNA) \frac{1}{2}}$: Half life of mRNA [$\text{s}$]
In the case of the “Lysozyme c” kill switch the mechanisms beyond producing mRNA need to be modelled separately from the DNase model. There are several assumptions that will be made, the first of these is that enzymatic reactions can be modelled by Michaelis-Menten kinetics. Secondly, the temperature of the constants taken imply that this model is running in the range of $37-40^o\text{C}$ at an optimal pH for E. coli growth. The model will assume that when E. coli splits, the contents of the cell and damage of the cell wall is shared equally among the two resulting E. coli. Lastly, it will assume that lysozyme does not degrade throughout the simulation.
The plasmid used to produce lysozyme has a PCR with a length of $507\text{bp}$ with the promoter, RBS and terminator totalling a further $164\text{bp}$. Therefore the production rates of mRNA and lysozyme can be calculated using translation and transcription rates of $V_{translation} = 8.4\text{aas}^{-1}$ (Siwiak and Zielenkiewicz, 2013) and $V_{transcription} = 40\text{bps}^{-1}$ (García and Molineux, 1995) respectively. The translation time is for E. coli at $37^o\text{C}$, other values have been taken at $40^o\text{C}$ as this was the closest temperature that could be found. All reaction rates have been rounded to the nearest second as this reduces calculation times.
$t_{lysozyme} = \frac{L_{protein}}{V_{translation}} = \frac{\frac{507\text{bp}}{3}}{8.4\text{aas}^{-1}} = 20\text{s (2sf)}$
$t_{mRNA} = \frac{L_{plasmid}}{V_{transcription}} = \frac{507\text{bp} + 164\text{bp}}{40\text{bps}^{-1}} = 17\text{s (2sf)}$
$t_{lysozyme}$: Time to produce one lysozyme protein [$\text{s}$]
$t_{mRNA}$: Time to produce one mRNA [$\text{s}$]
$L_{plasmid}\text{, }L_{protein}$: Length of plasmid and protein coding region [$\text{bp}$]
To supplement the production of lysozyme, the program will implement the affects of multiple ribosomes on the coding site of lysozyme. For E. coli it has been found there are 3.46 codons per $100\text{bp}$ (Siwiak and Zielenkiewicz, 2013). The PCR used for lysozyme production has a length of $507\text{bp}$, meaning there are approximately 5.8 codons which will be rounded down to 5 as not to overproduce lysozyme.
“Lysozyme c” is an enzyme that hydrolyses bonds holding together peptidoglycan in the cell wall, this is explained in detail on the project page. The next task of the lysozyme model is to connect the amount of protein at each time to the degradation of the E. coli cell wall, to do this Michaelis-Menten kinetics were applied, which gives the reaction rate of one enzyme (Berg et al., 2002).
$k_{(cat)} = \frac{[S]k_{(cat)max}}{[S] + K_M}$
$k_{(cat)}$: Reaction rate of one lysozyme [$\text{s}^{-1}$]
$k_{(cat)max}$: Maximum reaction rate of one lysozyme [$\text{s}^{-1}$]
$[S]$: Substrate concentration [$\text{M}$]
$K_M$: Michaelis constant [$\text{M}$]
To use this model two constants are required, the Michaelis constant ($K_M$), the concentration of the substrate when the reaction rate is exactly one half of the maximum reaction rate. The average reaction rate assumed to be $k_{(cat)avg} = (k_{(cat)max}/2$). The logarithms of both of these values has been calculated at $40^oC$ to be $-log(K_M) = 5.18 \pm 0.3$M and $-log(k_{cat}^{obs}) = 0.15 \pm 0.005$s$^{-1}$ (Banerjee et al., 1975). Giving values of:
$K_M = 5.6\text{mM}$ (2sf)
$k_{(cat)avg} = \frac{k_{(cat)max}}{2} \approx \frac{k_{(cat)}^{obs}}{2} = \frac{0.86\text{s}^{-1}}{2} = 0.43\text{s}^{-1}$ (2sf)
Lastly, the initial concentration of the substrate peptidoglycan is calculated. An assumption is made that determines that all the peptidoglycan in E. coli is spread out over the entire volume of the cell. Peptidoglycan or murein amount has been calculated to be approximately $3.5\text{x}10^6$ molecules per cell in a strain of E. coli (Vollmer and Höltje, 2004). Using an approximate volume of E. coli of $0.7 \mu \text{m}^3$, the concentration of peptidoglycan is:
$[Pep]_{int} = \frac{\frac{N_{pep}}{N_A}}{V_{E. coli}} = \frac{\frac{3.5\text{x}10^6}{6.02\text{x}10^{23}}}{0.7 \mu \text{m}^3} = 8.3\text{mM}$ (2sf)
$[Pep]_{int}$: Initial concentration of peptidoglycan [$\text{mM}$]
$N_{pep}$: Amount of peptidoglycan [molecules]
$N_A$: Avogadro's constant [molecules/mole]
$V_{\textit{E. coli}}$: Volume of E. coli [$\mu\text{m}^3$]
This calculation gives a value in the same order of magnitude as the Michaelis constant, which represents the concentration of substrate when the reaction rate is at half of its maximum value.
Fig. 1. Using Michaelis-Menten kinetics the reaction rate of each lysozyme enzyme has
been plotted for each peptidoglycan substrate concentration. The
average reaction rate of $0.43\text{s}^{-1}$ occurs when the concentration is equal to $K_M = 0.0056\text{M}$.
The maximum or initial concentration $[Pep]_{int} = 0.0083\text{M}$ of the substrate causes a reaction rate of $0.51\text{s}^{-1}$.
Enzyme reaction rates have been modelled by the Michaelis-Menten kinetics model in Fig. 1, therefore the reaction rate decreases as the substrate concentration decreases. This graph shows that the reaction rate will be greatest at the beginning of the simulation and approach zero when the cell wall is most damaged.
Fig. 2. The percentage of peptidoglycan compared to the original concentration plotted against time.
Fig. 3. Plots a smaller range of times as Fig. 2. To show the rapid decrease in peptidoglycan concentration.
The model predicted the complete degradation of the cell wall to be within the first generation of E. coli, Fig. 2. The reaction rate is slow at first due to the cell having no initial lysozyme, this slowly increases, until the low concentration of the substrate casues the reaction rate of lysozyme to slow considerably. The peptidoglycan concentration in the cell is negligible until 17 minutes at which point the E. coli splits sharing the cell wall damage equally between the two child cells, hence cell damage drops from almost 100% to 50%. The immediate concern is that in this model cell death would occur far before it is able to duplicate, meaning that assuming no mutations the cell would terminate before 17 minutes.
The cell death threshold of peptidoglycan concentration in the cell wall is not well defined from research. Fig. 3 demonstrates that any threshold that is chosen is likely to fall in between 2 and 5 minutes of the simulation which is well before the reproduction rate of E. coli of 17 minutes. Therefore it is a reasonable assumption that given no mutations were to occur that the cell would be terminated before the E. coli could reproduce.
