Team:USP-UNESP-Brazil/Plasmid Plug n Play/Modeling

From 2012.igem.org

(Difference between revisions)
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<p>The rate of change at each state over time was approached by using kinetic equations. To illustrate this process, we described the kinetic equation which refers to the change of the state <span class="math"><em>S</em></span> over time:</p>
<p>The rate of change at each state over time was approached by using kinetic equations. To illustrate this process, we described the kinetic equation which refers to the change of the state <span class="math"><em>S</em></span> over time:</p>
-
<p><br /><span class="math">$\frac{d}{dt}[S] = k_{-1}[S_{a}] - k_{1}[S][M]$</span><br /></p>
+
 
 +
\begin{align}
 +
&\frac{d}{dt}[S] = k_{-1}[S_{a}] - k_{1}[M][S]  
 +
\end{align}
<p>where <span class="math"><em>M</em></span> represents the concentration of recombinase monomers, <span class="math"><em>k</em><sub>1</sub></span> and <span class="math"><em>k</em><sub>−1</sub></span> represents the association and dissociation rate constant, respectively. As described in the above equation, there is only two possibilities of changing the concentration of the state <span class="math"><em>S</em></span>: it can increase (positive sign) if a molecule in the state <span class="math"><em>S</em><sub><em>a</em></sub></span> loses the monomer or it can decrease (negative sign) if a monomer binds the DNA.</p>
<p>where <span class="math"><em>M</em></span> represents the concentration of recombinase monomers, <span class="math"><em>k</em><sub>1</sub></span> and <span class="math"><em>k</em><sub>−1</sub></span> represents the association and dissociation rate constant, respectively. As described in the above equation, there is only two possibilities of changing the concentration of the state <span class="math"><em>S</em></span>: it can increase (positive sign) if a molecule in the state <span class="math"><em>S</em><sub><em>a</em></sub></span> loses the monomer or it can decrease (negative sign) if a monomer binds the DNA.</p>
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{{:Team:USP-UNESP-Brazil/Templates/RImage | image=Pplay_fig.jpeg | caption=Fig. 2. ORF insertion in the Plug&Play plasmid. | size=600px }}
{{:Team:USP-UNESP-Brazil/Templates/RImage | image=Pplay_fig.jpeg | caption=Fig. 2. ORF insertion in the Plug&Play plasmid. | size=600px }}
-
 
<h2 id="estimation-of-the-variables">Estimation of the variables</h2>
<h2 id="estimation-of-the-variables">Estimation of the variables</h2>
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To estimate the concentration of recombinase we used a simple model:
To estimate the concentration of recombinase we used a simple model:
</p>
</p>
-
<p><br /><span class="math">$\frac{d}{dt}[mRNA] = \frac{k_{transc}}{V n_{bp}} - k_{dRNA} [mRNA]$</span><br />
+
\begin{align}
-
</p>
+
&\frac{d}{dt}[mRNA] = \frac{k_{transc}}{V n_{bp}} - k_{dRNA}[mRNA]
-
<p><br /><span class="math">$\frac{d}{dt}[Prot] = \frac{k_{transl}[mRNA]}{n_{aa}} - k_{dProt} [Prot] $</span><br />
+
\end{align}
-
</p>
+
\begin{align}
 +
&\frac{d}{dt}[Prot] = \frac{k_{transl}[mRNA]}{n_{aa}} - k_{dProt}[Prot]  
 +
\end{align}
<p>where the constant <span class="math"><em>k</em><sub><em>t</em><em>r</em><em>a</em><em>n</em><em>s</em><em>l</em></sub></span> represents the translation rate, <span class="math"><em>V</em></span> refers to the volume of bacterium, <span class="math"><em>n</em><sub><em>b</em><em>p</em></sub></span> refers to the number of base pairs of the protein, <span class="math"><em>k</em><sub><em>d</em><em>R</em><em>N</em><em>A</em></sub></span> represents the mRNA degradation rate, <span class="math"><em>k</em><sub><em>t</em><em>r</em><em>a</em><em>n</em><em>s</em><em>l</em></sub></span> represents the translation rate, <span class="math"><em>n</em><sub><em>a</em><em>a</em></sub></span> the number of amino acids of the protein and <span class="math"><em>k</em><sub><em>d</em><em>P</em><em>r</em><em>o</em><em>t</em></sub></span> the degradation rate of the protein.</p>
<p>where the constant <span class="math"><em>k</em><sub><em>t</em><em>r</em><em>a</em><em>n</em><em>s</em><em>l</em></sub></span> represents the translation rate, <span class="math"><em>V</em></span> refers to the volume of bacterium, <span class="math"><em>n</em><sub><em>b</em><em>p</em></sub></span> refers to the number of base pairs of the protein, <span class="math"><em>k</em><sub><em>d</em><em>R</em><em>N</em><em>A</em></sub></span> represents the mRNA degradation rate, <span class="math"><em>k</em><sub><em>t</em><em>r</em><em>a</em><em>n</em><em>s</em><em>l</em></sub></span> represents the translation rate, <span class="math"><em>n</em><sub><em>a</em><em>a</em></sub></span> the number of amino acids of the protein and <span class="math"><em>k</em><sub><em>d</em><em>P</em><em>r</em><em>o</em><em>t</em></sub></span> the degradation rate of the protein.</p>
<p>
<p>
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Once we want the concentration of the protein in the equilibrium state, both equations are equal to zero:</p>
Once we want the concentration of the protein in the equilibrium state, both equations are equal to zero:</p>
-
<p><br /><span class="math">$\frac{d}{dt}[mRNA] = 0$</span><br /></p>
+
\begin{align}
-
<p><br /><span class="math">$\frac{d}{dt}[Prot] = 0$</span><br /></p>
+
\frac{d}{dt}[mRNA] = 0
 +
\end{align}
 +
\begin{align}
 +
\frac{d}{dt}[Prot] = 0
 +
\end{align}
<p>and as a consequence:</p>
<p>and as a consequence:</p>
-
<p><br /><span class="math">$\frac{k_{transc}}{V.n_{bp}} - k_{dRNA} .[mRNA]  = 0$</span><br /></p>
+
\begin{align}\frac{k_{transc}}{V.n_{bp}} - k_{dRNA} .[mRNA]  = 0
-
<p><br /><span class="math">$\frac{k_{transl}[mRNA]}{n_{aa}} - k_{dProt} [Prot] = 0$</span><br /></p>
+
\end{align}
-
<p>So, we have for both CRE and FLP recombinases:</p>
+
\begin{align}\frac{k_{transl}[mRNA]}{n_{aa}} - k_{dProt} [Prot] = 0
-
<p><br /><span class="math">$[Prot] = \frac{k_{transl} k_{transc}}{k_{dProt} k_{dRNA} n_{bp}^2/3 } \simeq 2000 nM$</span> (for each plug and play plasmid)<br /></p>
+
\end{align}
 +
 
 +
<p>So, the concentration of the protein per Plug&Play plasmid, for both CRE and FLP recombinases, is given by:</p>
 +
\begin{align}
 +
[Prot] = \frac{k_{transl} k_{transc}}{k_{dProt} k_{dRNA} n_{bp}^2/3 } \simeq 2000 nM
 +
\end{align}
<p>This result is an estimation of the amount of protein (CRE or FLP) produced by each Plug&Play plasmid and consequently, the total concentration should be higher than <span class="math">$2000 nM$</span> and dependent of the kind of Plug&Ply plasmid (high or low copy). Therefore, there is no significant change in the results presented here for concentrations higher than <span class="math">$2000 nM$</span>. This might occur because there are plenty of recombinase monomers to perform the recombination for concentrations higher than <span class="math">$2000 nM$</span>. Because of this, the following results are presented using <span class="math">$2000 nM$</span> of monomer concentration.</p>
<p>This result is an estimation of the amount of protein (CRE or FLP) produced by each Plug&Play plasmid and consequently, the total concentration should be higher than <span class="math">$2000 nM$</span> and dependent of the kind of Plug&Ply plasmid (high or low copy). Therefore, there is no significant change in the results presented here for concentrations higher than <span class="math">$2000 nM$</span>. This might occur because there are plenty of recombinase monomers to perform the recombination for concentrations higher than <span class="math">$2000 nM$</span>. Because of this, the following results are presented using <span class="math">$2000 nM$</span> of monomer concentration.</p>
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<p>Estimate the ORF concentration inside the bacteria (<span class="math">[<em>S</em>]<sub>0</sub></span>) is not simple because we do not know the amount of DNA that will get inside the bacteria during eletroporation. So we introduced a variable - lets call <span class="math"><em>c</em></span> - so that <span class="math">[<em>S</em>]<sub>0</sub> = <em>c</em>[<em>S</em><em>o</em>]</span> represents the concentration, in average, of ORF copies inside the bacteria. The variable <span class="math">[<em>S</em><em>o</em>]</span> represents the concentration of the DNA (ORF) in the solution before eletroporation and <span class="math"><em>c</em></span> is a constant such that <span class="math"><em>c</em> ≤ 1</span>. In the most optimist scenario we have <span class="math"><em>c</em> = 1</span> which means that during eletroporation the concentration of DNA (ORF) inside the bacteria becomes the same as in the solution.</p>
<p>Estimate the ORF concentration inside the bacteria (<span class="math">[<em>S</em>]<sub>0</sub></span>) is not simple because we do not know the amount of DNA that will get inside the bacteria during eletroporation. So we introduced a variable - lets call <span class="math"><em>c</em></span> - so that <span class="math">[<em>S</em>]<sub>0</sub> = <em>c</em>[<em>S</em><em>o</em>]</span> represents the concentration, in average, of ORF copies inside the bacteria. The variable <span class="math">[<em>S</em><em>o</em>]</span> represents the concentration of the DNA (ORF) in the solution before eletroporation and <span class="math"><em>c</em></span> is a constant such that <span class="math"><em>c</em> ≤ 1</span>. In the most optimist scenario we have <span class="math"><em>c</em> = 1</span> which means that during eletroporation the concentration of DNA (ORF) inside the bacteria becomes the same as in the solution.</p>
-
<p>We can correlate the variable <span class="math">[<em>S</em><em>o</em>]</span> with the amount of mass of DNA using the following relation: <br /><span class="math">$[So] = \frac{n_{mols}}{V}$</span><br /> where <br /><span class="math">$n_{mols} = \frac{m_{dna}}{n_{bp} m_{bp} n_{av}}$</span><br /> where <span class="math"><em>m</em><sub><em>d</em><em>n</em><em>a</em></sub></span> is the mass of DNA, <span class="math"><em>n</em><sub><em>b</em><em>p</em></sub></span> is the number of base pairs of the ORF, <span class="math"><em>m</em><sub><em>b</em><em>p</em></sub></span> is the mass of one base pair, <span class="math"><em>V</em></span> is the volume of the solution and <span class="math"><em>n</em><sub><em>a</em><em>v</em></sub></span> is the Avogadro’s number.</p>
+
<p>We can correlate the variable <span class="math">[<em>S</em><em>o</em>]</span> with the amount of mass of DNA using the following relation:  
-
<p>The variables <span class="math"><em>m</em><sub><em>d</em><em>n</em><em>a</em></sub></span> and <span class="math"><em>m</em><sub><em>p</em><em>b</em></sub></span> should have the same unit. For example, if <span class="math"><em>m</em><sub><em>d</em><em>n</em><em>a</em></sub></span> is given in <span class="math"><em>n</em><em>g</em></span> we have <br /><span class="math">$m_{bp} = \frac{650*10^{9}}{n_{av}} = \frac{650*10^{9}}{6*10^{23}} \simeq 10^{-12} ng$</span><br /></p>
+
\begin{align}
 +
[So] = \frac{n_{mols}}{V}
 +
\end{align}
 +
where  
 +
\begin{align}
 +
n_{mols} = \frac{m_{dna}}{n_{bp} m_{bp} n_{av}}
 +
\end{align}
 +
where <span class="math"><em>m</em><sub><em>d</em><em>n</em><em>a</em></sub></span> is the mass of DNA, <span class="math"><em>n</em><sub><em>b</em><em>p</em></sub></span> is the number of base pairs of the ORF, <span class="math"><em>m</em><sub><em>b</em><em>p</em></sub></span> is the mass of one base pair, <span class="math"><em>V</em></span> is the volume of the solution and <span class="math"><em>n</em><sub><em>a</em><em>v</em></sub></span> is the Avogadro’s number.</p>
 +
<p>The variables <span class="math"><em>m</em><sub><em>d</em><em>n</em><em>a</em></sub></span> and <span class="math"><em>m</em><sub><em>p</em><em>b</em></sub></span> should have the same unit. For example, if <span class="math"><em>m</em><sub><em>d</em><em>n</em><em>a</em></sub></span> is given in <span class="math"><em>n</em><em>g</em></span> we have  
 +
\begin{align}
 +
m_{bp} = \frac{650*10^{9}}{n_{av}} = \frac{650*10^{9}}{6*10^{23}} \simeq 10^{-12} ng
 +
\end{align}
<p>For a 800 <span class="math"><em>b</em><em>p</em></span> gene and 50 <span class="math"><em>μ</em><em>L</em></span> of solution we have: <br /><span class="math">$n_{mols} = \frac{m_{dna}}{800*6*10^{23}.10^{-12}} \simeq m_{dna} 2*10^{-15}$</span><br /> and <br /><span class="math">$[So] = \frac{m_{dna} 2*10^{-15}}{50*10^{-6}}  
<p>For a 800 <span class="math"><em>b</em><em>p</em></span> gene and 50 <span class="math"><em>μ</em><em>L</em></span> of solution we have: <br /><span class="math">$n_{mols} = \frac{m_{dna}}{800*6*10^{23}.10^{-12}} \simeq m_{dna} 2*10^{-15}$</span><br /> and <br /><span class="math">$[So] = \frac{m_{dna} 2*10^{-15}}{50*10^{-6}}  
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\begin{align}
\begin{align}
-
  \frac{d}{dt} [S]     &= k_{-1}[S_{a}] - [S] \left(k_{1}[M] + k_{d}\right) \\
+
&\frac{d}{dt}[S] = k_{-1}[S_{a}] - [S](k_{1}[M] + k_{d}) \nonumber \\
-
  \frac{d}{dt} [S_{a}] &= k_{1}[S][M] + k_{-1}[S_{aa}] + k_{-2}[S_{ab}] - [S_{a}]( k_{1}[M] + k_{-1} + k_{2}[M] + k_{d} ) \\
+
&\frac{d}{dt}[S_{a}] = k_{1}[S][M] + k_{-1}[S_{aa}] + k_{-2}[S_{ab}] - [S_{a}]( k_{1}[M] + k_{-1} + k_{2}[M] + k_{d} ) \nonumber \\
-
  \frac{d}{dt} [S_{aa}] &= k_{1}[S_{a}][M] + k_{-2}[S_{3}] - [S_{aa}](k_{2}[M] + k_{-1} + k_{d}) \\
+
&\frac{d}{dt}[S_{aa}] = k_{1}[S_{a}][M] + k_{-2}[S_{3}] - [S_{aa}](k_{2}[M] + k_{-1} + k_{d}) \nonumber \\
-
  \frac{d}{dt} [S_{ab}] &= k_{2}[S_{a}][M] + k_{-1}[S_{3}] - [S_{ab}](k_{-2} + k_{1}[M] + k_{d}) \\
+
&\frac{d}{dt}[S_{ab}] = k_{2}[S_{a}][M] + k_{-1}[S_{3}] - [S_{ab}](k_{-2} + k_{1}[M] + k_{d}) \nonumber \\
-
  \frac{d}{dt} [S_{3}] &= k_{1}[S_{ab}][M] + k_{2}[S_{aa}][M] + k_{-2}[S_{4}] - [S_{3}](k_{-1} + k_{-2} + k_{2}[M] + k_{d}) \\
+
&\frac{d}{dt}[S_{3}] = k_{1}[S_{ab}][M] + k_{2}[S_{aa}][M] + k_{-2}[S_{4}] - [S_{3}](k_{-1} + k_{-2} + k_{2}[M] + k_{d}) \nonumber \\
-
  \frac{d}{dt} [S_{4}] &= k_{2}[S_{3}][M] + k_{-34}[I_c] - [S_{4}](k_{-2} + k_{34} + k_{d}) \\
+
&\frac{d}{dt}[S_{4}] = k_{2}[S_{3}][M] + k_{-34}[I_c] - [S_{4}](k_{-2} + k_{34} + k_{d}) \nonumber \\
-
  \frac{d}{dt} [I_c]   &= k_{34}[S_{4}] + k_{-5}[L_{2}][C_{2}] - [I_c](k_{-34} + k_{-5}) \\
+
&\frac{d}{dt}[I_c] = k_{34}[S_{4}] + k_{-5}[L_{2}][C_{2}] - [I_c](k_{-34} + k_{-5}) \nonumber \\
-
  \frac{d}{dt} [C_{2}] &= k_{5}[I_c] + k_{2}[C_{1}][M] - [C_{2}](k_{-5}[P_{2}] + k_{-2}) \\
+
&\frac{d}{dt}[C_{2}] = k_{5}[I_c] + k_{2}[C_{1}][M] - [C_{2}](k_{-5}[P_{2}] + k_{-2}) \nonumber \\
-
  \frac{d}{dt} [C_{1}] &= k_{1}[C][M] + k_{-2}[C_{2}] - [C_{1}](k_{-1} + k_{2}[M]) \\
+
&\frac{d}{dt}[C_{1}] = k_{1}[C][M] + k_{-2}[C_{2}] - [C_{1}](k_{-1} + k_{2}[M]) \nonumber \\
-
  \frac{d}{dt} [C]     &= k_{-1}[C_{1}] - k_{1}[M][C] \\
+
&\frac{d}{dt}[C] = k_{-1}[C_{1}] - k_{1}[M][C] \nonumber \\
-
  \frac{d}{dt} [L_{2}] &= k_{5}[I_c] + k_{2}[L_{1}][M] - [L_{2}](k_{-5}[C_{2}] + k_{-2} + k_{d}) \\
+
&\frac{d}{dt}[L_{2}] = k_{5}[I_c] + k_{2}[L_{1}][M] - [L_{2}](k_{-5}[C_{2}] + k_{-2} + k_{d}) \\
-
  \frac{d}{dt} [L_{1}] &= k_{1}[L][M] + k_{-2}[L_{2}] - [L_{1}](k_{-1}+ k_{2}[M] + k_{d}) \\
+
&\frac{d}{dt}[L_{1}] = k_{1}[L][M] + k_{-2}[L_{2}] - [L_{1}](k_{-1}+ k_{2}[M] + k_{d}) \nonumber \\
-
  \frac{d}{dt} [L]     &= k_{-1}[L_{1}] - [L](k_{1}[M] + k_{d}) \\
+
&\frac{d}{dt}[L]     = k_{-1}[L_{1}] - [L](k_{1}[M] + k_{d}) \nonumber \\
-
  \frac{d}{dt} [P]     &= k_{-1}[P_{1}] -  k_{1}[M][P] \\
+
&\frac{d}{dt}[P]     = k_{-1}[P_{1}] -  k_{1}[M][P] \nonumber \\
-
  \frac{d}{dt} [P_{1}] &= k_{1}[P][M] + k_{-2}[P_{2}] - [P_{1}](k_{-1} + k_{2}[M]) \\
+
&\frac{d}{dt}[P_{1}] = k_{1}[P][M] + k_{-2}[P_{2}] - [P_{1}](k_{-1} + k_{2}[M]) \nonumber \\
-
  \frac{d}{dt} [P_{2}] &= k_{5}[I] + k_{2}[P_{1}][M] - [P_{2}](k_{-5}[C_{2}] + k_{-2}) \\
+
&\frac{d}{dt}[P_{2}] = k_{5}[I] + k_{2}[P_{1}][M] - [P_{2}](k_{-5}[C_{2}] + k_{-2}) \nonumber \\
-
  \frac{d}{dt} [I]     &= k_{34}[M_{4}] + k_{-5}[P_{2}][C_{2}] - [I](k_{-34} + k_{5}) \\
+
&\frac{d}{dt}[I]     = k_{34}[M_{4}] + k_{-5}[P_{2}][C_{2}] - [I](k_{-34} + k_{5}) \nonumber \\
-
  \frac{d}{dt} [E_{4}] &= k_{-34}[I] + k_{2}[E_{3}][M] - [E_{4}](k_{34}+ k_{-2}) \\
+
&\frac{d}{dt}[E_{4}] = k_{-34}[I] + k_{2}[E_{3}][M] - [E_{4}](k_{34}+ k_{-2}) \nonumber \\
-
  \frac{d}{dt} [E_{3}] &= k_{-2}[E_{4}] + k_{2}[E_{aa}][M] + k_{1}[E_{ab}][M] - [E_{3}](k_{2}[M] + k_{-2} + k_{-1}) \\
+
&\frac{d}{dt}[E_{3}] = k_{-2}[E_{4}] + k_{2}[E_{aa}][M] + k_{1}[E_{ab}][M] - [E_{3}](k_{2}[M] + k_{-2} + k_{-1}) \nonumber \\
-
  \frac{d}{dt} [E_{aa}] &= k_{-2}[E_{3}] + k_{1}[E_{a}][M] - [E_{aa}](k_{2}[M] + k_{-1}) \\
+
&\frac{d}{dt}[E_{aa}]= k_{-2}[E_{3}] + k_{1}[E_{a}][M] - [E_{aa}](k_{2}[M] + k_{-1}) \nonumber \\
-
  \frac{d}{dt} [E_{ab}] &= k_{-1}[E_{3}] + k_{2}[E_{a}][M] - [E_{ab}](k_{1}[M] + k_{-2}) \\
+
&\frac{d}{dt}[E_{ab}]= k_{-1}[E_{3}] + k_{2}[E_{a}][M] - [E_{ab}](k_{1}[M] + k_{-2}) \nonumber \\
-
  \frac{d}{dt} [E_{a}] &= k_{-1}[E_{aa}] + k_{-2}[E_{ab}] + k_{1}[E][M] - [E_{a}](k_{1}[M] + k_{2}[M] + k_{-1}) \\
+
&\frac{d}{dt}[E_{a}] = k_{-1}[E_{aa}] + k_{-2}[E_{ab}] + k_{1}[E][M] - [E_{a}](k_{1}[M] + k_{2}[M] + k_{-1}) \nonumber \\
-
  \frac{d}{dt} [E]     &= k_{-1}[E_{a}] - k_{1}[M][E] \\
+
&\frac{d}{dt}[E] = k_{-1}[E_{a}] - k_{1}[M][E] \nonumber
\end{align}  
\end{align}  

Revision as of 22:39, 25 September 2012