Difference between revisions of "Team:Tianjin/Model"

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<p>Scatchard curve is</p>
 
<p>Scatchard curve is</p>
 
<p>\[\frac{{[MX]}}{{{M_e}}} = K({X_0} - [MX])\]</p>
 
<p>\[\frac{{[MX]}}{{{M_e}}} = K({X_0} - [MX])\]</p>
 
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<h5> Biomass adsorption kinetics</h5>
 
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<p>The biosorption process can be divided into two stages. The first stage occurs on the cell wall surface, and mainly is the physical adsorption and ion exchange process which is going very fast. The second stage, also known as active adsorption, mainly is chemical adsorption, and metal ions at this stage can be transported through the active into the cell. This stage consumes the energy generated by cell metabolism, which was carried out very slowly.<br>
  <p>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.</p>
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Puranik and Paknikar describe the adsorption kinetics quantitatively in mathematical models. This mathematical model is based on the following two assumptions:
        <p>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.</p>
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        <p>\(y\) 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.
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Establish the following equation(1):
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</p>
 
</p>
    <p>
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<p> 1.During the adsorption process, the concentration of organic matter changed significantly;<br>
\[ 
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 2.Ignore the adsorption process occurred in the desorption, that is, adsorption is irreversible.
\begin{cases}
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\frac{R}{R_{T}}=\frac{1}{1+{(x/x_{0})}^{n}}; \\
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\tau_{x} \frac{dx}{dt}=\beta y-x; \\
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\tau_{y}\frac{dy}{dt}=\alpha \frac{1}{1+R/R_{0}}-y;
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\end{cases}
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\]
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</p>
 
</p>
        <p>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.</p>
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<p> Puranik and Paknikar On the basis of the above assumptions, the residual concentration is considered to have the following relationship with the adsorption time:
        <p>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):</p>
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        <p>
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\[ y=\frac{1+{(\beta y)}^{n}}{1+RT/R_{0}+{(\beta y)}^{n}} \]
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</p>
 
</p>
       
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<p>\[\frac{{dc}}{{dt}} =  - {K_t}{c_t}[1 - \theta (t)]\]</p>
        <p>Note that all of these formulations of derivation of Hill function from mass action kinetics
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<p>
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.
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\({{c_i}}\): The initial concentration;
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\({c_t}\), : The remaining concentration at time t;
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\({{c_{eq}}}\) : Balance concentration;
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\({K_t}\) : Rate constant.
 
</p>
 
</p>
        <p>We can get the following cubic equation form equation(2), we can get equation(3)</p>
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<p>According to the expression of θ (t), the physical meaning of 1 - θ (t) is that the remaining adsorption sites account for the percentage of total adsorption active sites at time t.  <br>
        <p>
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Integral:
\[
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\begin{cases}
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y^3-ay^2+(\rho /{\beta}^2)y-(\alpha /{\beta}^2)=0 \\
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\rho=1+R_{T}/R_{0} \\
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\end{cases}
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\]
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</p>
 
</p>
 
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<p>\[\ln \frac{{{c_t}}}{{{c_t} - {c_{eq}}}} = \ln \frac{{{c_i}}}{{{c_i} - {c_{eq}}}} + \frac{{{c_t}}}{{{c_t} - {c_{eq}}}}{K_t}t\]</p>
        <p>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):</p>
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<p>
        <p>
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According to the above formula, drawing the curve with the t, rate constant can be obtained from the slope.<br>
\[
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In the adsorption column, the adsorption capacity and the adsorption rate constant of the immobilized biosorption particles can be calculated by the kinetic model. Thomas dynamics model is:
(y-a)^{2}(y-ka)=y^3-(2+k)ay^2+(1+2k)a^2y-ka^3=0
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\]
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</p>
 
</p>
        <p>
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<p>
Comparing equations 3 and 4, we can get the following parametric equation(5):
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\[\frac{{{c_e}}}{{{c_0}}} = \frac{1}{{1 + \exp [\frac{K}{Q}({q_0}M - {c_0}V)]}}\]
 
</p>
 
</p>
        <p>
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<p>
\[
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\({{c_e}}\): effluent concentration;<br>
\begin{cases}
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    \({{c_0}}\): influent concentration;<br>
\rho=(1+2k)(1+2/k) \\
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    K: Thomas rate constant;<br>
\alpha \beta=(2+k)^{1.5}k^{-0.5} \\
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    \({{q_0}}\): maximum adsorption capacity;<br>
\end{cases}
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    M: the quality of the biosorption particles;<br>
\]
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    V: reactor volume;<br>
 +
    Q: flow rate.<br>
 
</p>
 
</p>
        <p>Draw the parameters equation, can be obtained under the range of parameters in bistability:</p>
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<p>The constant K and q0 values obtained from the operating data of the adsorption column can be used to design large scale adsorbent beds.</p>
<img 6>
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<div class="row">
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                        <div class="col-md-2"></div>
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<div class="col-md-8">
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<img src="https://static.igem.org/mediawiki/2017/e/e9/Tianjin_parameter_equation_1.png" alt="desktop"></div>
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                        <div class="col-md-2"></div>
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<figcation>Fig.X. parameter equation of \(\alpha \beta \) and \(\rho \) </figcaption>
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                  </div>     
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        <p>The two curves are the bistable curves we want, and between the two curves the cells can achieve bistability.</p>
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<img 7>
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<div class="row">
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                        <div class="col-md-2"></div>
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<div class="col-md-8">
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<img src="https://static.igem.org/mediawiki/2017/b/b6/Tianjin_parameter_equation_2.png" alt="desktop"></div>
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                        <div class="col-md-2"></div>
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<figcation>Fig.X.parameter equation of \(\frac{\alpha \beta}{\rho} \) and \(\frac{1}{\rho} \) </figcaption>
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                  </div>   
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        <p>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)</p>
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        <p>Let \(x^{*}\) and \(y^{*}\) represent stability:</p>
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        <p>
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\[
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x=x^{'}+x^{*};y=y^{'}+y^{*}
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\]
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</p>
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        <p>Combine equation (1), we can get equation(5):</p>
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        <p>
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\[
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\begin{cases}
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\tau_{x}\frac{dx^{'}}{dt}=\beta y^{'}-x^{'} \\
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\tau_{y}\frac{dy^{'}}{dt}=a x^{'}-y^{'} \\
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\end{cases}
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\]
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</p>
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        <p>In this equation:</p>
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        <p>
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\[
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a=\frac{\partial}{\partial x}(\alpha\frac{1}{1+R/R_{0}})|_{x=x^{*}}=\frac{\partial}{\partial x}(\alpha\frac{1}{{1+(x/x_{0})}^n})|_{x=x^{*}}
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\]
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</p>
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        <p>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)</p>
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        <p>
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\[
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\lambda_{1,2}=-1\pm\sqrt{\alpha \beta}
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\]
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</p>
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        <p>When \(\alpha \beta \)<1, the equations reach the steady state. Combine the equation with equation(2):</p>
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        <p>
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\[
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g(x):=a\frac{\beta(1+x^n)}{\rho+x^n}-x=0
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\]
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</p>
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        <p>When \(\alpha \beta \)<1, \(g(x)^{'}<0\), thus we can judge the stability of the resulting roots of equation(2).</p>
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Revision as of 03:44, 27 October 2017

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