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| | <img class="top_image" src="https://static.igem.org/mediawiki/2019/a/ac/T--XMU-China--model.png"> | | <img class="top_image" src="https://static.igem.org/mediawiki/2019/a/ac/T--XMU-China--model.png"> |
| | </div> | | </div> |
| − | The description of the dynamic form of such negative control system has been widely accepted:
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| − |
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| − | $$n*C_{in}+R{{k_{+4}\atop\rightharpoonup}\atop{\leftharpoondown\atop k_{-4}}}RC_{{in}_n}\eqno{[4]}$$
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| − |
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| − | $$O+R{{k_{+5}\atop\rightharpoonup}\atop{\leftharpoondown\atop k_{-5}}}OR\eqno{[5]}$$
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| − |
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| − | $R$:the repressor protein.
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| − |
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| − | $O$: the operating region.
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| − |
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| − | $C_{in}$: the intracellular inducers.
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| − |
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| − | $R{C_{in}}_n$: the repressor which binds to n intracellular inducers.
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| − |
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| − | $OR$: a operating region binds to a repressor protein.
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| − |
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| − | 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:
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| − |
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| − | $$[R_{tot}]=[R]+[RC_{{in}_n}]+[OR]\eqno{(2.1.1)}$$
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| − |
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| − | $$[O_{tot}]=[O]+[OR]\eqno{(2.1.2)}$$
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| − |
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| − | 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.
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| − |
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| − | 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.
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| − |
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| − | $$\frac{\mathrm{d}[R]}{\mathrm{d}t}=-k_{+4}*[C_{in}]^n*[R]+k_{-4}*[RC_{{in}_n}]-k_{+5}*[O]*[R]+k_{-5}*[OR]\eqno{(2.1.3)}$$
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| − |
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| − | Substitute equation (2.1.1) and (2.1.2) into equation (2.1.3), the following equation can be obtained:
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| − |
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| − | \begin{equation}
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| − | \left\ \frac{\mathrm{d}[R]}{\mathrm{d}t}=
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| − | \begin{array}
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| − | -k_{+4}*[C_{in}]^n*[R]+k_{-4}*[R_{tot}]-k_{-4}*[O_{tot}]+k_{-4}*[O] \\
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| − | -k_{-4}*[R]-k_{+5}*[O]*[R]+{k_{-5}*[O_{tot}]-k_{-5}*[O]}\\
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| − | \end{array}
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| − | \right.\eqno{(2.1.4)}
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| − | \end{equation*}
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| − |
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| − | 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.
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| − |
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| − | \begin{equation}
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| − | \begin{array}
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| − | (k_{+4}*[C_{in}]^n+k_{+5}*[O])*[R]+(k_{-5}-k_{-4})*[O]+k_{-4}*[R]\\
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| − | =k_{-4}*[R_{tot}]+(k_{-5}-k_{-4})*[O_{tot}]\\
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| − | \end{array}
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| − | \right.\eqno{(2.1.5)}
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| − | \end{equation*}
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| − |
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| − | 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.
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| − |
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| − | $$[C_{in}]_s=0\eqno{(2.1.6)}$$
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| − |
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| − | $$[R_{tot}]=(1+\frac{k_(+5)}{k_(-5)}*[O]_s)*[R]_s\eqno{(2.1.7)}$$
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| − |
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| − | $$[O_{tot}]=(1+\frac{k_(+5)}{k_(-5)}*[R]_s)*[O]_s\eqno{(2.1.8)}$$
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| − |
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| − | The steady-state operating point is obtained as:
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| − |
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| − | $$[C_{in}]_s=0$$
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| − |
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| − | $$[O]_s=\frac{-b_1+\sqrt{{b_1}^2+4*\frac{k_{+5}}{k_{-5}}*[O_{tot}]}}{2*\frac{k_{+5}}{k_{-5}}}\qquad b_1=\frac{k_{+5}}{k_{-5}}*[R_{tot}]-\frac{k_{+5}}{k_{-5}}*[O_{tot}]+1$$
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| − |
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| − | $$[R]_s=\frac{-b_2+\sqrt{{b_2}^2+4*\frac{k_{+5}}{k_{-5}}*[R_{tot}]}}{2*\frac{k_{+5}}{k_{-5}}}\qquad b_2=\frac{k_{+5}}{k_{-5}}*[O_{tot}]-\frac{k_{+5}}{k_{-5}}*[R_{tot}]+1$$
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| − |
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| − | Then the equation (2.1.4) is changed into an incremental model.
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| − |
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| − |
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| − | \begin{equation}
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| − | \left\ f_1=\frac{\mathrm{d}\Delta[R]}{\mathrm{d}t}=
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| − | \begin{array}
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| − | -k_{+4}*\Delta([C_{in}]^n*[R])+(k_{-4}-k_{-5})*\Delta[O] \\
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| − | -k_{-4}*\Delta[R]-k_{+5}*\Delta([O]*[R])\\
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| − | \end{array}
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| − | \right.\eqno{(2.1.9)}
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| − | \end{equation*}
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| − |
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| − | Lineariz it near the steady-state operating point:
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| − |
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| | $$\frac{\mathrm{d}\Delta[R]}{\mathrm{d}t}=[a_1\quad a_2\quad a_3]*\[ | | $$\frac{\mathrm{d}\Delta[R]}{\mathrm{d}t}=[a_1\quad a_2\quad a_3]*\[ |
| | \left[\begin{array}{cccc} | | \left[\begin{array}{cccc} |
| Line 96: |
Line 17: |
| | \eqno{(2.1.10)}$$ | | \eqno{(2.1.10)}$$ |
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| − | $$a_1=\frac{\partial f}{\partial\Delta[R]}=-k_{+4}*{[C_{in}]_s}^n-k_{+5}*[O]_s-k_{-4}=-k_{+5}*[O]_s-k_{-4}$$
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| − |
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| − | $$a_2=\frac{\partial f}{\partial\Delta[C_{in}]^n}=-k_{+4}*[R]_s$$
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| − |
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| − | $$a_3=\frac{\partial f}{\partial\Delta[O]}=k_{-4}-k_{-5}-k_{+5}*[R]_s$$
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| − |
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| − | Then, equation (2.1.10) can be expressed as:
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| − |
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| − | $$\frac{\mathrm{d}\Delta[R]}{\mathrm{d}t}=a_1*\Delta[R]+a_2*\Delta C_{in}[s]+a_3*\Delta[O]\eqno{(2.1.11)}$$
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| − |
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| − | Perform Laplace Transform on equation (2.1.11):
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| − |
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| − | $$s*\Delta R[s]-\Delta R(0)=a_1*\Delta R[s]+a_2*\frac{n!}{s^{n-1}}*\Delta C_{in}[s]+a_3*\Delta O[s]\eqno{(2.1.12)}$$
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| − |
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| − | As $\Delta R(0)=0$, so (2.1.12) becomes:
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| − |
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| − | $$(s-a_1)*\Delta R[s]=a_2*\frac{n!}{s^{n-1}}*\Delta C_{in}[s]+a_3*\Delta O[s]\eqno{(2.1.13)}$$
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| − |
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| − | The transfer function of $\Delta R[s]$ with respect to $\Delta C_{in}[s]$ is obtained:
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| − |
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| − | $$\frac{\Delta R[s]}{\Delta C_{in}[s]}=\frac{a_2*n!}{(s-a_1)*s^{n-1}}\eqno{(2.1.14)}$$
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| − |
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| − | Next, analyze the second process. Simultaneous chemical formula [5] and equation (2.1.2), the differential equation of $[O]$ can be obtained:
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| − |
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| − | $$\frac{\mathrm{d}[O]}{\mathrm{d}t}=-k_{+5}*[O]*[R]+k_{-5}*[OR]\eqno{(2.1.15)}$$
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| − |
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| − | Substitute equation (2.1.2) into equation (2.1.15) and get:
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| − |
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| − | $$\frac{\mathrm{d}[O]}{\mathrm{d}t}=-k_{+5}*[O]*[R]+k_{-5}*([O_{tot}]-[O])\eqno{(2.1.16)}$$
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| − |
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| − | Similarly, the incremental model can be obtained:
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| − |
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| − | $$f2=\frac{\mathrm{d}\Delta[O]}{\mathrm{d}t}=-k_{+5}*[O]*[R]-k_{-5}*\Delta[O]\eqno{(2.1.17)}$$
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| − |
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| − | After being linearized:
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| − |
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| − | $$\frac{\mathrm{d}\Delta[O]}{\mathrm{d}t}=a_4*\Delta[O]+a_5*\Delta[R]\eqno{(2.1.18)}$$
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| − |
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| − | $$a_4=\frac{\partial f_2}{\partial[O]}=-k_{+5}*[R]_s-k_{-5}$$
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| − |
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| − | $$a_5=\frac{\partial f_2}{\partial[R]}=-k_{+5}*[O]_s$$
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| − |
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| − | Perform Laplace Transform on equation (2.1.18):
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| − |
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| − | $$s*\Delta O[s]-\Delta O(0)=a_4*\Delta O[s]+a_5*\Delta R[s]\eqno{(2.1.19)}$$
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| − |
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| − | Similarly, $\Delta O(0)=0$, equation (2.1.19) becomes:
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| − |
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| − | $$(s-a_4)*\Delta O[s]=a_5*\Delta R[s]\eqno{(2.1.20)}$$
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| − |
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| − | The transfer function of $\Delta O[s]$ with respect to $\Delta R[s]$ is obtained:
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| − |
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| − | $$\frac{\Delta O[s]}{\Delta R[s]}=\frac{a_5}{s-a_4}\eqno{(2.1.21)}$$
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| − |
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| − | 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:
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| − |
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| − | $$\frac{\Delta O[s]}{\Delta C_{in}[s]}=\frac{\Delta O[s]}{\Delta R[s]}*\frac{\Delta R[s]}{\Delta C_{in}[s]}=\frac{a_2*a_5*n!}{(s-a_1)*(s-a_4)*s^{n-1}}\eqno{(2.1.22)}$$
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| − |
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| − | 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).
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| − |
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| − | Therefore:
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| − |
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| − | $$\frac{\Delta P[s]}{\Delta C_{in}[s]}=\frac{\Delta O[s]}{\Delta R[s]}*\frac{\Delta R[s]}{\Delta C_{in}[s]}=\frac{a_2*a_5*n!}{(s-a_1)*(s-a_4)*s^{n-1}}\eqno{(2.1.23)}$$
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| | | | |
| | <script type="text/x-mathjax-config">MathJax.Hub.Config({tex2jax: {inlineMath: [ ['$','$'], ["\\(","\\)"] ],displayMath: [ ['$$','$$'], ["\\[","\\]"] ]}});</script> | | <script type="text/x-mathjax-config">MathJax.Hub.Config({tex2jax: {inlineMath: [ ['$','$'], ["\\(","\\)"] ],displayMath: [ ['$$','$$'], ["\\[","\\]"] ]}});</script> |
$$\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)}$$
