Team:ShanghaiTech/Model Interaction
Interaction between Data and Model
Nomination
$16S$ The quantity (copy number) of orthogonal 16S-RNA
Review of our system
\figure{figures/Data/sys3-1.png}
\figure{BIGPICTURE}
Equations for transcription layer (NFBL)
$$ \frac{dI}{dt}=C_{Inducer \cdot R}\cdot \frac{Inducer^{n_{Inducer}}}{k_{Inducer}+Inducer^{n_{Inducer}}} \cdot \frac{1}{k_{R}+R^{n_R}}-d_I \cdot I\ \ \ (1)$$
$$ \frac{dO}{dt}=C_{I\cdot O} \cdot \frac{I^{n_I}}{k_I+I^{n_I}}-d_O \cdot O\ \ \ (2)$$
$$ \frac{dR}{dt}=C_O \cdot \frac{O^{n_O}}{k_O+O^{n_O}}-d_R \cdot R\ \ \ (3)$$
$$ \frac{d16S}{dt} = C_{I \cdot 16S} \cdot \frac{I^{n_I}}{k_I+I^{n_I}}-d_{16S} \cdot 16S\ \ \ (4)$$
Equations for translation layer (orthogonal ribosome)
$$ \frac{dProduct}{dt}=k_{16S}\cdot [16S]-d_{Product}\cdot Product \ \ \ (5)$$
Determine the parameters in model before lab work
Parts in the designed circuit
\figure{Figure of Parts}
Part Lux and pT181
Part Lux is the input node in the negative feedback loop. Part pT181 is the repressor in the system. Then, we rewrite equation (1) and equation (2).
$$ \frac{dpLux}{dt}=C_{AHL \cdot pT181}\cdot \frac{AHL^{n_{AHL}}}{k_{AHL}+AHL^{n_{AHL}}} \cdot \frac{1}{k_{pT181Antisense}+pT181Antisense^{n_{pT181Antisense}}}-d_{pLux} \cdot pLux\ \ \ (1')$$
$$ \frac{dOutput}{dt}=C_{LuxAHL}\cdot \frac{LuxAHL^{n_{LuxAHL}}}{k_{LuxAHL}+LuxAHL^{n_{LuxAHL}}}-d_{Output}\cdot Output \ \ \ (2')$$
In this equation, the most important parameters are the Hill coefficient $n_{AHL}$ and $n_{pT181}$. Because in the biochemical reaction, the Hill coefficient represents the mechanism of the reaction. The other parameters only tell us the strength of reaction. Because our model includes optimization method, it is not important to know the explicit value of every parameter, but Hill coefficients are still important to make sure the system not change.
According to the research of igem team ETH Zürich 2013$^{[1]}$, the value of $n_{AHL}$ is 1.7.
The mechanism of pT181 is the pT181-Antisense combines with pT181 to repress the expression of downstream. $^{[2]}$ The reaction can be written in chemical equation.
\figure{figures/Data/pT181.png}
\figure{figures/Data/pT181equation.png}
$$ \phi_{pT181}=\frac{pT181}{TotalpT181}= \frac{K_{pT181}[pT181]}{pT181+K_{pT181}[pT181][pT181Antisense]}$$
$$ \phi_{pT181} = \frac{1}{\frac{1}{K_{pT181}}+[AntipT181]} $$
$$ rate(downstream) = C\cdot \phi_{pT181} =C\cdot \frac{1}{\frac{1}{K_{pT181}}+[pT181Antisense]} $$
From the last equation, we can see that the value of Hill function $n_{pT181}$ is 1. In conclude, if the all the coefficients in the chemical equation are 1, then the Hill function will be 1. Thus, because all of the coefficients in the reaction of LuxAHL activating the downstream expression is 1, the Hill function $n_{LuxAHL}$ is 1.
Part STAR
STAR is the product of input node, and will activate the expression of output node. Then equation (2') and (3) can be rewritten.
$$ \frac{dSTAR}{dt}=C_{LuxAHL}\cdot \frac{LuxAHL}{k_{LuxAHL}+LuxAHL}-d_{STAR}\cdot STAR \ \ \ (2'') $$
$$ \frac{dpT181Antisense}{dt}=C_{STAR} \cdot \frac{STAR^{n_{STAR}}}{k_{STAR}+STAR^{n_{STAR}}}-d_{pT181Antisense} \cdot pT181Antisense \ \ \ (3')$$
The mechanism of part STAR is shown below.
We can see that STAR also satisfies the condition that Hill function $n_{STAR}$ is 1.
Reference
[1] team ETH Zürich 2013, https://2013.igem.org/Team:ETH_Zurich/Parameter
[2] James Chappell ; Melissa K Takahashi ; Julius B Lucks, Nature Chemical Biology, 2015
[3] Westbrook, Alexandra M ; Lucks, Julius B, Nucleic acids research, 19 May 2017, Vol.45(9), pp.5614-5624