Difference between revisions of "Team:XMU-China/Model"

Revision as of 08:47, 20 October 2019

The description of the dynamic form of such negative control system has been widely accepted: $$n*C_{in}+R{{k_{+4}\atop\rightharpoonup}\atop{\leftharpoondown\atop k_{-4}}}RC_{{in}_n}\eqno{[4]}$$ $$O+R{{k_{+5}\atop\rightharpoonup}\atop{\leftharpoondown\atop k_{-5}}}OR\eqno{[5]}$$ $R$:the repressor protein. $O$: the operating region. $C_{in}$: the intracellular inducers. $R{C_{in}}_n$: the repressor which binds to n intracellular inducers. $OR$: a operating region binds to a repressor protein. This form ignores the possibility of $C_{in}$ directly competing with (Hypothesis 7). In the cell, the total expression of repressor protein $[R_{tot}]$ is constant,and the total concentration of the operating region $[O_{tot}]$ is also constant, according to the law of conservation of materials: $$[R_{tot}]=[R]+[RC_{{in}_n}]+[OR]\eqno{(2.1.1)}$$ $$[O_{tot}]=[O]+[OR]\eqno{(2.1.2)}$$ As shown in the figure 3 above, this process can be broken down into two processes: first, the extracellular inducer binds to the repressor in the cell, causing a change in the concentration of the repressor, and then followed by a change in the concentration of the repressor resulting in a change in the concentration of the operating region. Firstly, analyze the first process. Since the change of repressors is related ton both chemical equation [4] and [5], it is necessary to analyze the two equations, equation (2.1.1) and (2.1.2) to obtain the transfer function of repressor concentration with respect to intracellular inducers. $$\frac{\mathrm{d}[R]}{\mathrm{d}t}=-k_{+4}*[C_{in}]^n*[R]+k_{-4}*[RC_{{in}_n}]-k_{+5}*[O]*[R]+k_{-5}*[OR]\eqno{(2.1.3)}$$ Substitute equation (2.1.1) and (2.1.2) into equation (2.1.3), the following equation can be obtained: \begin{equation} \left\ \frac{\mathrm{d}[R]}{\mathrm{d}t}= \begin{array} -k_{+4}*[C_{in}]^n*[R]+k_{-4}*[R_{tot}]-k_{-4}*[O_{tot}]+k_{-4}*[O] \\ -k_{-4}*[R]-k_{+5}*[O]*[R]+{k_{-5}*[O_{tot}]-k_{-5}*[O]}\\ \end{array} \right.\eqno{(2.1.4)} \end{equation*} Obviously, equation (2.1.4) is a non-linear function, so it needed to be linearized. Firstly, solve the steady-state operating point of the non-linear function. \begin{equation} \begin{array} (k_{+4}*[C_{in}]^n+k_{+5}*[O])*[R]+(k_{-5}-k_{-4})*[O]+k_{-4}*[R]\\ =k_{-4}*[R_{tot}]+(k_{-5}-k_{-4})*[O_{tot}]\\ \end{array} \right.\eqno{(2.1.5)} \end{equation*} According to the analysis, there is no inducer in the cell at the beginning, and the intracellular state at this time is the steady-state working point. $$[C_{in}]_s=0\eqno{(2.1.6)}$$ $$[R_{tot}]=(1+\frac{k_(+5)}{k_(-5)}*[O]_s)*[R]_s\eqno{(2.1.7)}$$ $$[O_{tot}]=(1+\frac{k_(+5)}{k_(-5)}*[R]_s)*[O]_s\eqno{(2.1.8)}$$ The steady-state operating point is obtained as: $$[C_{in}]_s=0$$ $$[O]_s=\frac{-b_1+\sqrt{{b_1}^2+4*\frac{k_{+5}}{k_{-5}}*[O_{tot}]}}{2*\frac{k_{+5}}{k_{-5}}}\qquad b_1=\frac{k_{+5}}{k_{-5}}*[R_{tot}]-\frac{k_{+5}}{k_{-5}}*[O_{tot}]+1$$ $$[R]_s=\frac{-b_2+\sqrt{{b_2}^2+4*\frac{k_{+5}}{k_{-5}}*[R_{tot}]}}{2*\frac{k_{+5}}{k_{-5}}}\qquad b_2=\frac{k_{+5}}{k_{-5}}*[O_{tot}]-\frac{k_{+5}}{k_{-5}}*[R_{tot}]+1$$ Then the equation (2.1.4) is changed into an incremental model. \begin{equation} \left\ f_1=\frac{\mathrm{d}\Delta[R]}{\mathrm{d}t}= \begin{array} -k_{+4}*\Delta([C_{in}]^n*[R])+(k_{-4}-k_{-5})*\Delta[O] \\ -k_{-4}*\Delta[R]-k_{+5}*\Delta([O]*[R])\\ \end{array} \right.\eqno{(2.1.9)} \end{equation*} Lineariz it near the steady-state operating point: $$\frac{\mathrm{d}\Delta[R]}{\mathrm{d}t}=[a_1\quad a_2\quad a_3]*\[ \left[\begin{array}{cccc} \Delta[R]\\ \Delta[C_{in}]^n\\ \Delta[O] \end{array}\right] \eqno{(2.1.10)}$$ $$a_1=\frac{\partial f}{\partial\Delta[R]}=-k_{+4}*{[C_{in}]_s}^n-k_{+5}*[O]_s-k_{-4}=-k_{+5}*[O]_s-k_{-4}$$ $$a_2=\frac{\partial f}{\partial\Delta[C_{in}]^n}=-k_{+4}*[R]_s$$ $$a_3=\frac{\partial f}{\partial\Delta[O]}=k_{-4}-k_{-5}-k_{+5}*[R]_s$$ Then, equation (2.1.10) can be expressed as: $$\frac{\mathrm{d}\Delta[R]}{\mathrm{d}t}=a_1*\Delta[R]+a_2*\Delta C_{in}[s]+a_3*\Delta[O]\eqno{(2.1.11)}$$ Perform Laplace Transform on equation (2.1.11): $$s*\Delta R[s]-\Delta R(0)=a_1*\Delta R[s]+a_2*\frac{n!}{s^{n-1}}*\Delta C_{in}[s]+a_3*\Delta O[s]\eqno{(2.1.12)}$$ As $\Delta R(0)=0$, so (2.1.12) becomes: $$(s-a_1)*\Delta R[s]=a_2*\frac{n!}{s^{n-1}}*\Delta C_{in}[s]+a_3*\Delta O[s]\eqno{(2.1.13)}$$ The transfer function of $\Delta R[s]$ with respect to $\Delta C_{in}[s]$ is obtained: $$\frac{\Delta R[s]}{\Delta C_{in}[s]}=\frac{a_2*n!}{(s-a_1)*s^{n-1}}\eqno{(2.1.14)}$$ Next, analyze the second process. Simultaneous chemical formula [5] and equation (2.1.2), the differential equation of $[O]$ can be obtained: $$\frac{\mathrm{d}[O]}{\mathrm{d}t}=-k_{+5}*[O]*[R]+k_{-5}*[OR]\eqno{(2.1.15)}$$ Substitute equation (2.1.2) into equation (2.1.15) and get: $$\frac{\mathrm{d}[O]}{\mathrm{d}t}=-k_{+5}*[O]*[R]+k_{-5}*([O_{tot}]-[O])\eqno{(2.1.16)}$$ Similarly, the incremental model can be obtained: $$f2=\frac{\mathrm{d}\Delta[O]}{\mathrm{d}t}=-k_{+5}*[O]*[R]-k_{-5}*\Delta[O]\eqno{(2.1.17)}$$ After being linearized: $$\frac{\mathrm{d}\Delta[O]}{\mathrm{d}t}=a_4*\Delta[O]+a_5*\Delta[R]\eqno{(2.1.18)}$$ $$a_4=\frac{\partial f_2}{\partial[O]}=-k_{+5}*[R]_s-k_{-5}$$ $$a_5=\frac{\partial f_2}{\partial[R]}=-k_{+5}*[O]_s$$ Perform Laplace Transform on equation (2.1.18): $$s*\Delta O[s]-\Delta O(0)=a_4*\Delta O[s]+a_5*\Delta R[s]\eqno{(2.1.19)}$$ Similarly, $\Delta O(0)=0$, equation (2.1.19) becomes: $$(s-a_4)*\Delta O[s]=a_5*\Delta R[s]\eqno{(2.1.20)}$$ The transfer function of $\Delta O[s]$ with respect to $\Delta R[s]$ is obtained: $$\frac{\Delta O[s]}{\Delta R[s]}=\frac{a_5}{s-a_4}\eqno{(2.1.21)}$$ At last, simultaneous chemical equation (2.1.4) and equation (2.1.21), the transform of $\Delta O[s]$ with respect to $\Delta C_{in}[s]$ is obtained: $$\frac{\Delta O[s]}{\Delta C_{in}[s]}=\frac{\Delta O[s]}{\Delta R[s]}*\frac{\Delta R[s]}{\Delta C_{in}[s]}=\frac{a_2*a_5*n!}{(s-a_1)*(s-a_4)*s^{n-1}}\eqno{(2.1.22)}$$ Moreover, the region of P and the region of $O$ overlap in most operons, so the free concentration of $O$ can be replaced the free concentration of $P$ (Hypothesis 8). Therefore: $$\frac{\Delta P[s]}{\Delta C_{in}[s]}=\frac{\Delta O[s]}{\Delta R[s]}*\frac{\Delta R[s]}{\Delta C_{in}[s]}=\frac{a_2*a_5*n!}{(s-a_1)*(s-a_4)*s^{n-1}}\eqno{(2.1.23)}$$

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-\noindent\rule[0.25\baselineskip]{\textwidth}{1pt}
+The description of the dynamic form of such negative control system has been widely accepted:
-Firstly, to facilitate the analysis and calculation, all cells was described as spheres corresponding to their corresponding volume equivalent diameters de under the engineering concept (Hypothesis 1). The equivalent diameter$^{[2]}$ is expressed as follows:
+$$n*C_{in}+R{{k_{+4}\atop\rightharpoonup}\atop{\leftharpoondown\atop k_{-4}}}RC_{{in}_n}\eqno{[4]}$$
-$$d_e=\sqrt[3]{\frac{6*V_p}{\pi}} \tag{1.1.1}$$
+$$O+R{{k_{+5}\atop\rightharpoonup}\atop{\leftharpoondown\atop k_{-5}}}OR\eqno{[5]}$$
-Vp: the volume of cells.\\
+$R$:the repressor protein.
-Secondly, the simple diffusion will reach a equilibrium state as both intracellular and extracellular concentions of inducers are the same. Since the cell volume in the solution occupies a so small proportion that the diffusion of inducers into the cell does not greatly affect the concentration in the whole solution. Therefore, it can be considered that the concentration of extracellular inducers is nearly unchanged (Hypothesis 2).\\
+$O$: the operating region.
-Meanwhile, Fick’s second law$^{[3]}$ shows:
+$C_{in}$: the intracellular inducers.
-$$\frac{\partial c(z,t)}{\partial t}=D*\nabla^2{c(z,t)}\tag{1.1.2}$$
+$R{C_{in}}_n$: the repressor which binds to n intracellular inducers.
+$OR$: a operating region binds to a repressor protein.
+This form ignores the possibility of $C_{in}$ directly competing with (Hypothesis 7). In the cell, the total expression of repressor protein $[R_{tot}]$ is constant,and the total concentration of the operating region $[O_{tot}]$ is also constant, according to the law of conservation of materials:
+$$[R_{tot}]=[R]+[RC_{{in}_n}]+[OR]\eqno{(2.1.1)}$$
+$$[O_{tot}]=[O]+[OR]\eqno{(2.1.2)}$$
+As shown in the figure 3 above, this process can be broken down into two processes: first, the extracellular inducer binds to the repressor in the cell, causing a change in the concentration of the repressor, and then followed by a change in the concentration of the repressor resulting in a change in the concentration of the operating region.
+Firstly, analyze the first process. Since the change of repressors is related ton both chemical equation [4] and [5], it is necessary to analyze the two equations, equation (2.1.1) and (2.1.2) to obtain the transfer function of repressor concentration with respect to intracellular inducers.
+$$\frac{\mathrm{d}[R]}{\mathrm{d}t}=-k_{+4}*[C_{in}]^n*[R]+k_{-4}*[RC_{{in}_n}]-k_{+5}*[O]*[R]+k_{-5}*[OR]\eqno{(2.1.3)}$$
+Substitute equation (2.1.1) and (2.1.2) into equation (2.1.3), the following equation can be obtained:
+\begin{equation}
+\left\ \frac{\mathrm{d}[R]}{\mathrm{d}t}=
+\begin{array}
+-k_{+4}*[C_{in}]^n*[R]+k_{-4}*[R_{tot}]-k_{-4}*[O_{tot}]+k_{-4}*[O] \\
+-k_{-4}*[R]-k_{+5}*[O]*[R]+{k_{-5}*[O_{tot}]-k_{-5}*[O]}\\
+   \end{array}
+  \right.\eqno{(2.1.4)}
+\end{equation*}
+Obviously, equation (2.1.4) is a non-linear function, so it needed to be linearized. Firstly, solve the steady-state operating point of the non-linear function.
+\begin{equation}
+\begin{array}
+(k_{+4}*[C_{in}]^n+k_{+5}*[O])*[R]+(k_{-5}-k_{-4})*[O]+k_{-4}*[R]\\
+=k_{-4}*[R_{tot}]+(k_{-5}-k_{-4})*[O_{tot}]\\
+   \end{array}
+  \right.\eqno{(2.1.5)}
+\end{equation*}
+According to the analysis, there is no inducer in the cell at the beginning, and the intracellular state at this time is the steady-state working point.
+$$[C_{in}]_s=0\eqno{(2.1.6)}$$
+$$[R_{tot}]=(1+\frac{k_(+5)}{k_(-5)}*[O]_s)*[R]_s\eqno{(2.1.7)}$$
+$$[O_{tot}]=(1+\frac{k_(+5)}{k_(-5)}*[R]_s)*[O]_s\eqno{(2.1.8)}$$
+The steady-state operating point is obtained as:
+$$[C_{in}]_s=0$$
+$$[O]_s=\frac{-b_1+\sqrt{{b_1}^2+4*\frac{k_{+5}}{k_{-5}}*[O_{tot}]}}{2*\frac{k_{+5}}{k_{-5}}}\qquad b_1=\frac{k_{+5}}{k_{-5}}*[R_{tot}]-\frac{k_{+5}}{k_{-5}}*[O_{tot}]+1$$
+$$[R]_s=\frac{-b_2+\sqrt{{b_2}^2+4*\frac{k_{+5}}{k_{-5}}*[R_{tot}]}}{2*\frac{k_{+5}}{k_{-5}}}\qquad b_2=\frac{k_{+5}}{k_{-5}}*[O_{tot}]-\frac{k_{+5}}{k_{-5}}*[R_{tot}]+1$$
+Then the equation (2.1.4) is changed into an incremental model.
+\begin{equation}
+\left\ f_1=\frac{\mathrm{d}\Delta[R]}{\mathrm{d}t}=
+\begin{array}
+-k_{+4}*\Delta([C_{in}]^n*[R])+(k_{-4}-k_{-5})*\Delta[O] \\
+-k_{-4}*\Delta[R]-k_{+5}*\Delta([O]*[R])\\
+   \end{array}
+  \right.\eqno{(2.1.9)}
+\end{equation*}
+Lineariz it near the steady-state operating point:
+$$\frac{\mathrm{d}\Delta[R]}{\mathrm{d}t}=[a_1\quad a_2\quad a_3]*\[
+\left[\begin{array}{cccc}
+\Delta[R]\\
+\Delta[C_{in}]^n\\
+\Delta[O]
+\end{array}\right]
+\eqno{(2.1.10)}$$
+$$a_1=\frac{\partial f}{\partial\Delta[R]}=-k_{+4}*{[C_{in}]_s}^n-k_{+5}*[O]_s-k_{-4}=-k_{+5}*[O]_s-k_{-4}$$
+$$a_2=\frac{\partial f}{\partial\Delta[C_{in}]^n}=-k_{+4}*[R]_s$$
+$$a_3=\frac{\partial f}{\partial\Delta[O]}=k_{-4}-k_{-5}-k_{+5}*[R]_s$$
+Then, equation (2.1.10) can be expressed as:
+$$\frac{\mathrm{d}\Delta[R]}{\mathrm{d}t}=a_1*\Delta[R]+a_2*\Delta C_{in}[s]+a_3*\Delta[O]\eqno{(2.1.11)}$$
+Perform Laplace Transform on equation (2.1.11):
+$$s*\Delta R[s]-\Delta R(0)=a_1*\Delta R[s]+a_2*\frac{n!}{s^{n-1}}*\Delta C_{in}[s]+a_3*\Delta O[s]\eqno{(2.1.12)}$$
+As $\Delta R(0)=0$, so (2.1.12) becomes:
+$$(s-a_1)*\Delta R[s]=a_2*\frac{n!}{s^{n-1}}*\Delta C_{in}[s]+a_3*\Delta O[s]\eqno{(2.1.13)}$$
+The transfer function of $\Delta R[s]$ with respect to $\Delta C_{in}[s]$ is obtained:
+$$\frac{\Delta R[s]}{\Delta C_{in}[s]}=\frac{a_2*n!}{(s-a_1)*s^{n-1}}\eqno{(2.1.14)}$$
+Next, analyze the second process. Simultaneous chemical formula [5] and equation (2.1.2), the differential equation of $[O]$ can be obtained:
+$$\frac{\mathrm{d}[O]}{\mathrm{d}t}=-k_{+5}*[O]*[R]+k_{-5}*[OR]\eqno{(2.1.15)}$$
+Substitute equation (2.1.2) into equation (2.1.15) and get:
+$$\frac{\mathrm{d}[O]}{\mathrm{d}t}=-k_{+5}*[O]*[R]+k_{-5}*([O_{tot}]-[O])\eqno{(2.1.16)}$$
+Similarly, the incremental model can be obtained:
+$$f2=\frac{\mathrm{d}\Delta[O]}{\mathrm{d}t}=-k_{+5}*[O]*[R]-k_{-5}*\Delta[O]\eqno{(2.1.17)}$$
+After being linearized:
+$$\frac{\mathrm{d}\Delta[O]}{\mathrm{d}t}=a_4*\Delta[O]+a_5*\Delta[R]\eqno{(2.1.18)}$$
+$$a_4=\frac{\partial f_2}{\partial[O]}=-k_{+5}*[R]_s-k_{-5}$$
+$$a_5=\frac{\partial f_2}{\partial[R]}=-k_{+5}*[O]_s$$
+Perform Laplace Transform on equation (2.1.18):
+$$s*\Delta O[s]-\Delta O(0)=a_4*\Delta O[s]+a_5*\Delta R[s]\eqno{(2.1.19)}$$
+Similarly, $\Delta O(0)=0$, equation (2.1.19) becomes:
+$$(s-a_4)*\Delta O[s]=a_5*\Delta R[s]\eqno{(2.1.20)}$$
+The transfer function of $\Delta O[s]$ with respect to $\Delta R[s]$ is obtained:
+$$\frac{\Delta O[s]}{\Delta R[s]}=\frac{a_5}{s-a_4}\eqno{(2.1.21)}$$
+At last, simultaneous chemical equation (2.1.4) and equation (2.1.21), the transform of $\Delta O[s]$ with respect to $\Delta C_{in}[s]$ is obtained:
+$$\frac{\Delta O[s]}{\Delta C_{in}[s]}=\frac{\Delta O[s]}{\Delta R[s]}*\frac{\Delta R[s]}{\Delta C_{in}[s]}=\frac{a_2*a_5*n!}{(s-a_1)*(s-a_4)*s^{n-1}}\eqno{(2.1.22)}$$
+Moreover, the region of P and the region of $O$ overlap in most operons, so the free concentration of $O$ can be replaced the free concentration of $P$ (Hypothesis 8).
+Therefore:
+$$\frac{\Delta P[s]}{\Delta C_{in}[s]}=\frac{\Delta O[s]}{\Delta R[s]}*\frac{\Delta R[s]}{\Delta C_{in}[s]}=\frac{a_2*a_5*n!}{(s-a_1)*(s-a_4)*s^{n-1}}\eqno{(2.1.23)}$$
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