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

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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>
<p>In order to simulate our design, we defined the initial condition of our system, which consists in estimating the following variables:</p>
<p>In order to simulate our design, we defined the initial condition of our system, which consists in estimating the following variables:</p>
 +
<p>
<ul>
<ul>
-
<li><p><span class="math">[<em>P</em>]<sub>0</sub></span> - initial concentration of the Plug&Play plasmids inside the bacteria.</p></li>
+
<li><span class="math">[<em>P</em>]<sub>0</sub></span> - initial concentration of the Plug&Play plasmids inside the bacteria.</li>
-
<li><p><span class="math">[<em>M</em>]<sub>0</sub></span> - initial concentration of recombinase monomers.</p></li>
+
<li><span class="math">[<em>M</em>]<sub>0</sub></span> - initial concentration of recombinase monomers.</li>
-
<li><p><span class="math">[<em>S</em>]<sub>0</sub></span> - initial concentration of ORF inside the bacteria</p></li>
+
<li><span class="math">[<em>S</em>]<sub>0</sub></span> - initial concentration of ORF inside the bacteria</li>
</ul>
</ul>
-
<p>To estimate the concentration of the variables, we need the volume of <em>''E. coli''</em>. According to [2]<br /><span class="math"><em>V</em><sub><em>e</em><em>c</em></sub> = 0.7*10<sup>−15</sup><em>L</em></span><br /> Using this estimate, it is possible to calculate the concentration of one molecule inside the bacteria in molar concentration <br /><span class="math">1<em>M</em> = 1<em>m</em><em>o</em><em>l</em> / 1<em>L</em> = 6*10<sup>23</sup><em>m</em><em>o</em><em>l</em><em>e</em><em>c</em><em>u</em><em>l</em><em>e</em><em>s</em> / <em>L</em></span><br /></p>
+
</p>
-
<p><br /><span class="math">$[1 molec] = \frac{1}{0.7*10^{-15} L} = \frac{1}{6*10^{23}  
+
<p>To estimate the concentration of the variables, we need the volume of <em>''E. coli''</em>. According to [2],  $V_{ec} = 0.7 \hspace{0.2cm}(\mu m)^3 = 0.7$ $10^{-15} L$. Using this estimation, it is possible to calculate the concentration of one molecule inside the bacteria in molar concentration <br /><span class="math">1<em>M</em> = 1<em>m</em><em>o</em><em>l</em> / 1<em>L</em> = 6*10<sup>23</sup><em>m</em><em>o</em><em>l</em><em>e</em><em>c</em><em>u</em><em>l</em><em>e</em><em>s</em> / <em>L</em></span><br /></p>
-
0.7*10^{-15}}M \simeq 10^{-9} M = 1 nM$</span><br /></p>
+
 
 +
\begin{align}
 +
[1 molec] = \frac{1}{0.7*10^{-15} L} = \frac{1}{6*10^{23} 0.7*10^{-15}}M \simeq 1 nM
 +
\end{align}
<h3 id="plasmid-concentration">Plasmid concentration</h3>
<h3 id="plasmid-concentration">Plasmid concentration</h3>
<p>According to [3] it is expected approximately 100-300 plasmid inside the bacteria (high copy) and approximately 10 plasmids (low copy). So, using the equation we have:  
<p>According to [3] it is expected approximately 100-300 plasmid inside the bacteria (high copy) and approximately 10 plasmids (low copy). So, using the equation we have:  
</p>
</p>
 +
<p>
<ul>
<ul>
-
<li><p><span class="math">$[P]_0 \simeq 10 nM$</span> - high copy plasmid.</p></li>
+
<li><span class="math">$[P]_0 \simeq 10 nM$</span> - high copy plasmid.</li>
-
<li><p><span class="math">$[P]_0 \simeq 100 nM$</span> - low copy plasmid.</p></li>
+
<li><span class="math">$[P]_0 \simeq 100 nM$</span> - low copy plasmid.</li>
</ul>
</ul>
 +
</p>
<p>
<p>
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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>
The values of these constants, obtained in [2] and [3], are presented below:
The values of these constants, obtained in [2] and [3], are presented below:
</p>
</p>
 +
<p>
<ul>
<ul>
-
<li><p><span class="math">$k_{degRNA} = 1/350$</span> (1/sec).</p></li>
+
<li><span class="math">$k_{degRNA} = 1/350$</span> (1/sec).</li>
-
<li><p><span class="math">$k_{transc} = 40$</span> (bp/sec)  for T7 promoter.</p></li>
+
<li><span class="math">$k_{transc} = 40$</span> (bp/sec) - for T7 promoter.</li>
-
<li><p><span class="math">$k_{transl} = 15$</span> (aa/sec).</p></li>
+
<li><span class="math">$k_{transl} = 15$</span> (aa/sec).</li>
-
<li><p><span class="math">$k_{degProt} = 0.0167/60$</span> (Prot/sec) Average protein degradation.</p></li>
+
<li><span class="math">$k_{degProt} = 0.0167/60$</span> (Prot/sec) Average protein degradation.</li>
-
<li><p><span class="math">$n_{bp} = 1032$</span> for CRE.</p></li>
+
<li><span class="math">$n_{bp} = 1032 $</span> - for CRE.</li>
-
<li><p><span class="math">$n_{aa} = n_{bp}/3 = 344$</span> for CRE.</p></li>
+
<li><span class="math">$n_{aa} = n_{bp}/3 = 344 $</span> - for CRE.</li>
-
<li><p><span class="math">$n_{bp} = 1119$</span> for FLP.</p></li>
+
<li><span class="math">$n_{bp} = 1119 $</span> - for FLP.</li>
-
<li><p><span class="math">$n_{aa} = n_{bp}/3 = 373$</span> for FLP.</p></li>
+
<li><span class="math">$n_{aa} = n_{bp}/3 = 373 $</span> - for FLP.</li>
</ul>
</ul>
 +
</p>
<p>
<p>
Once we want the concentration of the protein in the equilibrium state, both equations are equal to zero:</p>
Once we want the concentration of the protein in the equilibrium state, both equations are equal to zero:</p>
-
<p><br /><span class="math">$\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>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>So, an estimative of protein concentration 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 2000 $nM$ 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 2000 $nM$. This might occur because there are plenty of recombinase monomers to perform the recombination for concentrations higher than 2000 $nM$. Because of this, the following results are presented using 2000 $nM$ of monomer concentration.</p>
<h3 id="orf-concentration">ORF concentration</h3>
<h3 id="orf-concentration">ORF concentration</h3>
<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">$[So] = \frac{m_{dna} 2*10^{-15}}{50*10^{-6}}  
+
For a 800 <span class="math"><em>b</em><em>p</em></span> gene and 50 <span class="math"><em>μ</em><em>L</em></span> of solution we have:  
-
\simeq m_{dna} 0.4*10^{-10} M = 0.04 m_{dna} nM$</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>
+
\begin{align}
 +
n_{mols} = \frac{m_{dna}}{800*6*10^{23}.10^{-12}} \simeq m_{dna} 2*10^{-15}
 +
\end{align}
 +
and  
 +
\begin{align}
 +
[So] = \frac{m_{dna} 2*10^{-15}}{50*10^{-6}} \simeq m_{dna} 0.4*10^{-10} M = m_{dna}*0.04 nM.
 +
\end{align}  
 +
This means, for example, that in order to obtain 10 nM of concentration 250 ''ng'' of DNA are needed in a solution of 50 $\mu L$.</p>
<h1 id="results">Results</h1>
<h1 id="results">Results</h1>
Line 97: Line 143:
<p>The variable we are interested in optimizing is the concentration of Plug&Play plasmids with the inserted ORF. This variable is presented as a function of the degradation rate <span class="math"><em>k</em><sub><em>d</em></sub></span> and ORF concentration in figure 3 and 4, for CRE and FLP, respectively. The value of RNA degradation rate is indicated by a red arrow.</p>
<p>The variable we are interested in optimizing is the concentration of Plug&Play plasmids with the inserted ORF. This variable is presented as a function of the degradation rate <span class="math"><em>k</em><sub><em>d</em></sub></span> and ORF concentration in figure 3 and 4, for CRE and FLP, respectively. The value of RNA degradation rate is indicated by a red arrow.</p>
-
{{:Team:USP-UNESP-Brazil/Templates/RImage | image=ORFxkd.jpg | caption=<p>Fig. 3. The plasmid concentration with the inserted ORF as a function of degradation rate and ORF concentration for CRE recombinase. The red arrow indicates the RNA degradation rate <span class="math"><em>k</em><sub><em>d</em><em>R</em><em>N</em><em>A</em></sub> = 0. 0023</span> <span class="math">1 / <em>s</em></span>.</p> | size=600px }}
+
{{:Team:USP-UNESP-Brazil/Templates/RImage | image=ORFxkd.jpg | caption=Fig. 3. The plasmid concentration with the inserted ORF as a function of degradation rate and ORF concentration for CRE recombinase. The red arrow indicates the RNA degradation rate <span class="math"><em>k</em><sub><em>d</em><em>R</em><em>N</em><em>A</em></sub> = 0. 0023</span> <span class="math">1 / <em>s</em></span>. | size=600px }}
-
{{:Team:USP-UNESP-Brazil/Templates/RImage | image=ORFxkd_flp.jpg | caption=Fig. 4. Same as figure \ref{fig:ORFxkd} but for FLP recombinase. | size=600px }}
+
{{:Team:USP-UNESP-Brazil/Templates/RImage | image=ORFxkd_flp.jpg | caption=Fig. 4. Same as figure 3 but for FLP recombinase. | size=600px }}
<p>For CRE recombinase, linear DNA degradation do not play a fundamental role in our system and it can even be disregarded, figure 3. This may occur because the circularization of linear DNA by recombinases is faster than the degradation of it. For FLP, however, linear DNA degradation is an important effect and must be taken in account, figure 4. This occurs because the association of the first and second monomers for CRE is significantly higher than for FLP.</p>
<p>For CRE recombinase, linear DNA degradation do not play a fundamental role in our system and it can even be disregarded, figure 3. This may occur because the circularization of linear DNA by recombinases is faster than the degradation of it. For FLP, however, linear DNA degradation is an important effect and must be taken in account, figure 4. This occurs because the association of the first and second monomers for CRE is significantly higher than for FLP.</p>
-
<p>In the following analysis we evaluated the concentration of plasmids with the inserted ORF as a function of the DNA mass  in the solution during eletroporation and the variable <span class="math"><em>c</em></span> (the fraction of ORF concentration that enters in the bacteria), Figs 5 and 6. We are interested in concentrations of Plug&Play plasmids with the ORF inserted higher than <span class="math">1</span> <span class="math"><em>n</em><em>M</em></span> which means that, in average, there will be at least one plasmid with the ORF in the bacteria, represented by the red region on the Figs. 5 and 6. According to our results an amount of <span class="math">10000</span> <span class="math"><em>n</em><em>g</em></span> of DNA might be satisfactory when using CRE. Nevertheless, when using FLP this amount might not be enough and the amount needed is highly dependent of the linear DNA degradation rate.</p>
+
<p>In the following analysis we evaluated the concentration of plasmids with the inserted ORF as a function of the DNA mass  in the solution during eletroporation and the variable <span class="math"><em>c</em></span> (the fraction of ORF concentration that enters in the bacteria), Figs 5 and 6. We are interested in concentrations of Plug&Play plasmids with the ORF inserted higher than 1 $nM$ which means that, in average, there will be at least one plasmid with the ORF in the bacteria, represented by the red region on the Figs. 5 and 6. According to our results an amount of 10000 $ng$ of DNA might be satisfactory when using CRE. Nevertheless, when using FLP this amount might not be enough and the amount needed is highly dependent of the linear DNA degradation rate.</p>
-
<p>One possible strategy to improve the recombination without increasing this amount of DNA is to reduce the volume of the solution before eletroporation, which increase the ORF concentration in the solution. Values lower than <span class="math">10000</span> <span class="math"><em>n</em><em>g</em></span> of DNA may also be satisfactory since the ORF has a antibiotics resistance gene and once the ORF had been inserted the bacteria tend to keep and replicate the plasmid.</p>
+
<p>One possible strategy to improve the recombination without increasing this amount of DNA is to reduce the volume of the solution before eletroporation, which increase the ORF concentration in the solution. Values lower than 10000 $ng$ of DNA may also be satisfactory since the ORF has a antibiotics resistance gene and once the ORF had been inserted the bacteria tend to keep and replicate the plasmid.</p>
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{{:Team:USP-UNESP-Brazil/Templates/RImage | image=ORFxc.jpg | caption=<p>Fig. 5. The concentration of plasmids with the ORF inserted as a function of ORF mass in concentration and <span class="math"><em>c</em></span> (the fraction of ORF concentration that enters in the bacteria) for CRE recombinase. We suppose that eletroporation was done in a solution of 50 <span class="math"><em>μ</em><em>L</em></span>.</p>| size=600px }}
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{{:Team:USP-UNESP-Brazil/Templates/RImage | image=ORFxc.jpg | caption=Fig. 5. The concentration of plasmids with the ORF inserted as a function of ORF mass in concentration and <span class="math"><em>c</em></span> (the fraction of ORF concentration that enters in the bacteria) for CRE recombinase. We suppose that eletroporation was done in a solution of 50 <span class="math"><em>μ</em><em>L</em></span>.| size=600px }}
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+
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{{:Team:USP-UNESP-Brazil/Templates/RImage | image=ORFxc_flp.jpg | caption=Fig. 6. Same as figure \ref{fig:ORFxc} but for FLP recombinase. | size=600px }}
+
 +
{{:Team:USP-UNESP-Brazil/Templates/RImage | image=ORFxc_flp.jpg | caption=Fig. 6. Same as figure 5 but for FLP recombinase. | size=600px }}
<h1 id="discussion">Discussion</h1>
<h1 id="discussion">Discussion</h1>
<p>In order to identify differences between FLP and CRE, we compared the two enzymes using two analyses. Our results point to an obvious choice for the CRE-lox recombination system since it is less affected by DNA degradation and improves the insertion of the ORF compared with FLP-FRT system.</p>
<p>In order to identify differences between FLP and CRE, we compared the two enzymes using two analyses. Our results point to an obvious choice for the CRE-lox recombination system since it is less affected by DNA degradation and improves the insertion of the ORF compared with FLP-FRT system.</p>
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<p>In our model we have considered all lox sites as loxP. However, there are mutated loxP and a combination of them can improve the insertion of the target gene (ORF) [4]. We have chosen to use lox66 and lox71 in our experimental design. We did not introduced the lox66 and lox71 in the model for two main reasons: there are no references about the values of rate constants for altered loxP and we prefer to keep the simplicity and clarity of the model. In order to take these variables in consideration, it would be necessary to use more equations and extra hypothesis.</p>
+
<p>In our model we have considered all lox sites as loxP. However, there are mutated loxP and a combination of them can improve the insertion of the target gene (ORF) [4]. We have chosen to use lox66 and lox71 in our experimental design. Nevertheless, we did not introduce the lox66 and lox71 in the model for two main reasons: there are no references about the values of rate constants for altered loxP and we prefer to keep the simplicity and clarity of the model. In order to take these variables in consideration, it would be necessary to use more equations and extra hypothesis.</p>
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<p>Although we did not consider the mutated loxP, we have some considerations about it. The insertion reaction is favored over the excision reaction by roughly fivefold using mutated recombination, when using CRE recombinases <span class="citation"></span>. This occurs because the double mutated loxP has a very low affinity for the CRE monomers. So, an intuitive conclusion is that the combination we chose may optimize the insertion of the ORF in the Plug&Play plasmid. Nevertheless, this conclusion could be false because the altered loxP demands more time in the circularization step since it has a lower association constant for CRE recombinase. This extra amount of time could be such, that the degradation of linear DNA plays a fundamental role in the process. However, as it is illustrated, in the case of CRE recombinases the degradation of linear DNA is not a fundamental variable and it may not interfere. Because of this, the combination of mutated loxP must optimize the amount of ORF inserted in the plasmid.</p>
+
<p>Although we did not consider the mutated loxP, we have some considerations about it. The insertion reaction is favored over the excision reaction by roughly fivefold using mutated recombination, when using CRE recombinases [4]. This occurs because the double mutated loxP has a very low affinity for the CRE monomers. So, an intuitive conclusion is that the combination we chose may optimize the insertion of the ORF in the Plug&Play plasmid. Nevertheless, this conclusion could be false because the altered loxP demands more time in the circularization step since it has a lower association constant for CRE recombinase. This extra amount of time could be such that the degradation of linear DNA plays a fundamental role in the process. However, as it is illustrated in Fig. 3, in the case of CRE recombinases, the degradation of linear DNA is not a fundamental variable and it may not interfere. Because of that, we may conclude that the combination of mutated loxP must optimize the amount of ORF inserted in the plasmid.</p>
 +
<h1 id="Appendix">Appendix</h1>
 +
<h2 id="Equations">Equations</h2>
 +
 +
\begin{align}
 +
&\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} ) \nonumber \\
 +
&\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}) \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}) \nonumber \\
 +
&\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}) \nonumber \\
 +
&\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]) \nonumber \\
 +
&\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_{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}) \nonumber \\
 +
&\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]) \nonumber \\
 +
&\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}) \nonumber \\
 +
&\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}) \nonumber \\
 +
&\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}) \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}) \nonumber \\
 +
&\frac{d}{dt}[E] = k_{-1}[E_{a}] - k_{1}[M][E] \nonumber
 +
\end{align}
<h1 id="references">References</h1>
<h1 id="references">References</h1>
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<p>[3] http://partsregistry.org/</p>
<p>[3] http://partsregistry.org/</p>
<p>[4] Zuwen Zhang and Beat Lutz. <em>Cre recombinase-mediated inversion using lox66 and lox71: method to introduce conditional point mutations into the CREB-binding protein.</em> Nucl. Acids Res. (2002) 30 (17): e90.</p>
<p>[4] Zuwen Zhang and Beat Lutz. <em>Cre recombinase-mediated inversion using lox66 and lox71: method to introduce conditional point mutations into the CREB-binding protein.</em> Nucl. Acids Res. (2002) 30 (17): e90.</p>
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Latest revision as of 02:42, 27 September 2012