Team:Tianjin/Model
/* OVERRIDE IGEM SETTINGS */
Construction of Bistable Model
Bistability is a common phenomenon in single-cell microbes, that two types of cell phenotype coexist. Bistability is very important for many single-cell microbes adapting to environmental changes. Single-cell microbes can choose the appropriate form according to changes of the environment, and bistability is the basis for achieving this change.
the reason for the existence of bistability in single-cell microbes is complex, and is generally thought to be related to the positive feedback of the gene network. We simulate the bistability in single-celled microbes by establishing a simplified gene regulation model.
\(y\) represents the concentration of the bistable substance, \(x\) is the activator of \(y\), which promotes the expression of the \(y\) gene to increase the concentration of \(y\), and \(R\) is the intracellular inhibitor of \(y\), whose concentration is in accordance with Hill Function. Establish the following equation(1)：
\[ \begin{cases} \frac{R}{R_{T}}=\frac{1}{1+{(x/x_{0})}^{n}}; \\ \tau_{x} \frac{dx}{dt}=\beta y-x; \\ \tau_{y}\frac{dy}{dt}=\alpha \frac{1}{1+R/R_{0}}-y; \end{cases} \]
Here \(R\) represents the concentration of the active inhibitory factor, \(R_{T}\) representing the total concentration, \(n\) is the Hill coefficient, \(x_0\) is the concentration at which the activation rate reaches the half, and the generation of \(y\) is described by the Michaelis-Menten equation, \(\tau_x\) 、\(\tau_y\) is the average survive time.
Under normal circumstances, the cells will be in a steady state, then the derivative of time on the equation (1) is zero, we can get equation (2):
\[ y=\frac{1+{(\beta y)}^{n}}{1+RT/R_{0}+{(\beta y)}^{n}} \]
Note that all of these formulations of derivation of Hill function from mass action kinetics assume that the protein has n sites to which ligands can bind. In practice, however, the Hill Coefficient n rarely provides an accurate approximation of the number of ligand binding sites on a protein.[2] We assume that the Hill Coefficient is 2 according to experience.
We can get the following cubic equation form equation(2), we can get equation(3)
\[ \begin{cases} y^3-ay^2+(\rho /{\beta}^2)y-(\alpha /{\beta}^2)=0 \\ \rho=1+R_{T}/R_{0} \\ \end{cases} \]
For the cubic equation, there can be 1, 2, 3 positive real solutions, to achieve the bistability in this model, the function should have two solutions, we assume that the equation form is equation(4):
\[ (y-a)^{2}(y-ka)=y^3-(2+k)ay^2+(1+2k)a^2y-ka^3=0 \]
Comparing equations 3 and 4, we can get the following parametric equation(5):
\[ \begin{cases} \rho=(1+2k)(1+2/k) \\ \alpha \beta=(2+k)^{1.5}k^{-0.5} \\ \end{cases} \]
Draw the parameters equation, can be obtained under the range of parameters in bistability:
The two curves are the bistable curves we want, and between the two curves the cells can achieve bistability.
The above result is obtained by the equation (1) assuming that the cell is in a steady state, and the stability analysis is given below for equation (1)
Let \(x^{*}\) and \(y^{*}\) represent stability：
\[ x=x^{'}+x^{*};y=y^{'}+y^{*} \]
Combine equation (1), we can get equation(5):
\[ \begin{cases} \tau_{x}\frac{dx^{'}}{dt}=\beta y^{'}-x^{'} \\ \tau_{y}\frac{dy^{'}}{dt}=a x^{'}-y^{'} \\ \end{cases} \]
In this equation:
\[ a=\frac{\partial}{\partial x}(\alpha\frac{1}{1+R/R_{0}})|_{x=x^{*}}=\frac{\partial}{\partial x}(\alpha\frac{1}{{1+(x/x_{0})}^n})|_{x=x^{*}} \]
According to the steady state theory of ordinary differential equations, the equilibrium point is stable if the eigenvalues of the coefficient matrix of equation (5) contain negative real parts, we can find the eigenvalues of equation (5)
\[ \lambda_{1,2}=-1\pm\sqrt{\alpha \beta} \]
When \(\alpha \beta \)<1, the equations reach the steady state. Combine the equation with equation(2):
\[ g(x):=a\frac{\beta(1+x^n)}{\rho+x^n}-x=0 \]
When \(\alpha \beta \)<1, \(g(x)^{'}＜0\), thus we can judge the stability of the resulting roots of equation(2).
A
B
