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

From 2012.igem.org

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<ul>
<ul>
<li><p><span class="math">$k_{degRNA} = 1/350$</span> (1/sec).</p></li>
<li><p><span class="math">$k_{degRNA} = 1/350$</span> (1/sec).</p></li>
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<li><p><span class="math">$k_{transcription} = 40$</span> (bp/sec)  for T7 promoter.</p></li>
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<li><p><span class="math">$k_{transc} = 40$</span> (bp/sec)  for T7 promoter.</p></li>
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<li><p><span class="math">$k_{translation} = 15$</span> (aa/sec).</p></li>
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<li><p><span class="math">$k_{transl} = 15$</span> (aa/sec).</p></li>
<li><p><span class="math">$k_{degProt} = 0.0167/60$</span> (Prot/sec) Average protein degradation.</p></li>
<li><p><span class="math">$k_{degProt} = 0.0167/60$</span> (Prot/sec) Average protein degradation.</p></li>
<li><p><span class="math">$n_{bp} = 1032$</span> for CRE.</p></li>
<li><p><span class="math">$n_{bp} = 1032$</span> for CRE.</p></li>
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<p>
<p>
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<p>Once we want the concentration of the protein in the equilibrium state, both equations are equal to zero:</p>
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Once we want the concentration of the protein in the equilibrium state, both equations are equal to zero:</p>
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<p><br /><span class="math">$\begin{aligned}
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<p><br /><span class="math">$\frac{d}{dt}[mRNA] = 0$</span><br /></p>
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    &amp;\frac{d}{dt}[mRNA] = 0 \nn
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<p><br /><span class="math">$\frac{d}{dt}[Prot] = 0$</span><br /></p>
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    &amp;\frac{d}{dt}[Prot] = 0 \nn\end{aligned}$</span><br /></p>
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<p>and as a consequence:</p>
<p>and as a consequence:</p>
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<p><br /><span class="math">$\begin{aligned}
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<p><br /><span class="math">$\frac{k_{transc}}{V.n_{bp}} - k_{dRNA} .[mRNA]  = 0$</span><br /></p>
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    &amp;\frac{k_{tran}}{V.n_{bp}} - k_{dRNA} .[mRNA]  = 0 \nn
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<p><br /><span class="math">$\frac{k_{transl}[mRNA]}{n_{aa}} - k_{dProt} [Prot] = 0$</span><br /></p>
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    &amp;\frac{k_{trad}[mRNA]}{n_{bp}/3} - k_{dProt} [Prot] = 0 \nn\end{aligned}$</span><br /></p>
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<p>So, we have for both CRE and FLP recombinases:</p>
<p>So, we have for both CRE and FLP recombinases:</p>
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<p><br /><span class="math">$\begin{aligned}
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<p><br /><span class="math">$[Prot] = \frac{k_{transl} k_{transc}}{k_{dProt} k_{dRNA} n_{bp}^2/3 } \simeq 2000 nM$</span> (for each plug and play plasmid)<br /></p>
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    [Prot] = \frac{k_{trad}.k_{tran}}{k_{dProt} .k_{dRNA} .n_{bp}^2/3 } \simeq 2000 \hspace{0.1cm}nM \hspace{0.3cm}(for \hspace{0.1cm}each  
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    \hspace{0.1cm}plug \hspace{0.1cm}and \hspace{0.1cm}play \hspace{0.1cm}plasmid)\end{aligned}$</span><br /></p>
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<p>This result is an estimate of the amount of protein (CRE or FLP) produced by each plug and play plasmid and consequently, the total concentration should be higher than <span class="math">$2000 \hspace{0.1cm}nM$</span> and dependent of the kind of plug and play plasmid (high or low copy). Therefore, there is no significant change in the results presented here for concentrations higher than 2000 <span class="math"><em>n</em><em>M</em></span>. This might occur because there are plenty of recombinase monomers to perform the recombination for concentrations higher than <span class="math">$2000 \hspace{0.1cm}nM$</span>. Because of that, the following results are presented using <span class="math">$2000 \hspace{0.1cm}nM$</span> of monomer concentration.</p>
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<p>This result is an estimate of the amount of protein (CRE or FLP) produced by each plug and play plasmid and consequently, the total concentration should be higher than <span class="math">$2000 nM$</span> and dependent of the kind of plug and play plasmid (high or low copy). Therefore, there is no significant change in the results presented here for concentrations higher than 2000 <span class="math"><em>n</em><em>M</em></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 that, the following results are presented using <span class="math">$2000 nM$</span> of monomer concentration.</p>
<h3 id="orf-concentration">ORF concentration</h3>
<h3 id="orf-concentration">ORF concentration</h3>
<p>Estimating 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. Because of that, 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 genes inside the bacteria. The variable <span class="math">[<em>S</em><em>o</em>]</span> represents the concentration of the genes 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 genes inside the bacteria becomes the same as in the solution.</p>
<p>Estimating 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. Because of that, 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 genes inside the bacteria. The variable <span class="math">[<em>S</em><em>o</em>]</span> represents the concentration of the genes 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 genes inside the bacteria becomes the same as in the solution.</p>
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<p>We can relate 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">$n_{mols} = \frac{m_{dna}}{n_{bp}\hspace{0.1cm}m_{bp}\hspace{0.1cm}n_{av}}$</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 DNA, <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>
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<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">$m_{bp} = \frac{650\hspace{0.1cm}10^{9}}{n_{av}} = \frac{650\hspace{0.1cm}10^{9}}{6\hspace{0.1cm}10^{23}} \simeq 10^{-12} \hspace{0.1cm} ng$</span><br /></p>
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<p>We can relate 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 DNA, <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>
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<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">$n_{mols} = \frac{m_{dna}}{800\hspace{0.1cm}6 \hspace{0.1cm}10^{23}.10^{-12}} \simeq m_{dna} \hspace{0.1cm}2 \hspace{0.1cm}10^{-15}$</span><br /> and <br /><span class="math">$[So] = \frac{m_{dna}\hspace{0.1cm}2\hspace{0.1cm}10^{-15}}{50\hspace{0.1cm}10^{-6}}  
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<p>The variables <span class="math"><em>m</em><sub><em>d</em><em>n</em><em>a</em></sub></span> and <span class="math"><em>m</em><sub><em>p</em><em>b</em></sub></span> should have the same unit. For example, if <span class="math"><em>m</em><sub><em>d</em><em>n</em><em>a</em></sub></span> is given in <span class="math"><em>n</em><em>g</em></span> we have <br /><span class="math">$m_{bp} = \frac{650*10^{9}}{n_{av}} = \frac{650*10^{9}}{6*10^{23}} \simeq 10^{-12} ng$</span><br /></p>
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\simeq m_{dna}\hspace{0.1cm} 0.4\hspace{0.1cm} 10^{-10}\hspace{0.1cm} M = 0.04\hspace{0.1cm} m_{dna} \hspace{0.1cm}nM$</span><br /> This means, for example, that in order to obtain <span class="math">10</span> <span class="math"><em>n</em><em>M</em></span> of concentration <span class="math">250</span> <span class="math"><em>n</em><em>g</em></span> of DNA are needed in a solution of <span class="math">50<em>μ</em><em>L</em></span>.</p>
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<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">$[So] = \frac{m_{dna} 2*10^{-15}}{50*10^{-6}}  
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\simeq m_{dna} 0.4*10^{-10} M = 0.04 m_{dna} nM$</span><br /> This means, for example, that in order to obtain <span class="math">10</span> <span class="math"><em>n</em><em>M</em></span> of concentration <span class="math">250</span> <span class="math"><em>n</em><em>g</em></span> of DNA are needed in a solution of <span class="math">50<em>μ</em><em>L</em></span>.</p>
<h1 id="results">Results</h1>
<h1 id="results">Results</h1>

Revision as of 18:28, 23 September 2012