Team:UANL Mty-Mexico/MathModel/Protein

From 2014.igem.org

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<td>
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<p>These are all the equations we used in our model; they represent the mRNA and protein concentration of each respective
 
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part of our system that are the cI,Vip3Ca3 and GFP; and the maximums in <i>E.Coli</i>.These represent our deterministic model
 
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beacuse it doesn`t take account other variables or noises.</p>
 
 +
<p align="justify">The transcription of genes begins with the binding of transcription elements onto regulatory sites. The transcription rate is heavily dependant on how strong or weak these sites are. In our project, protein feedback will not be taken into account as there is no regulatory region capable of interacting with the products. We assume that both mRNA and protein synthesis have a constant rate of production, which does not vary throughout the process. Each component degrades randomly and is recycled within the cell. </p>
 +
 +
<ul>
 +
<li><b>Transcription</b></li>
 +
<p align="justify">The average transcription speed in E.coli is 70nt/s. Assuming all the transcription in our circuit works in such speed, we can calculate the maximum transcription rate for each transcription equation by using this formula: Maximum Transcription Rate = Transcription Speed(nt/min)/Gene Length(bp)=4200/Gene Length (nM/min)</p>
 +
<li><b>Translation</b></li>
 +
<p align="justify">The average translation speed in E.coli is 40Aa/s. Also assuming all the translation in our circuit works in the same speed, we can calculate the theoretical transcription rate. However, in wetlab, we can use different rbs to regulate the translation process, thus, the translation rate can be written as: Translation Rate = RBS * Translation Speed(Aa/min)/Protein Length(Aa) = 2400RBS/Protein Length (min^-1) This transformation does not change the degree of freedom of our system. However, this does limit the range of parameters since the strength of RBS can not be too extreme.</p>
 +
<li><b>Cell division rate</b></li>
 +
<p align="justify">We have assume the period of cell division is 30 mins, which means the "degradation rate" in our model is actually the sum of degradation rate of the substance(1/half life) and cell division rate(1/30 mins).</p>
 +
<li><b>RNA degradation rate</b></li>
 +
<p align="justify">From Ref1(Belasco 1993) and Ref2(Genome Biology 2006, 7:R99), we have decided that all the mRNA in our system have a half life of 4.4 mins.</p>
 +
</ul>
<br>
<br>
\begin{equation}
\begin{equation}
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\large \frac{d[mC]}{dt} = \alpha_{1} - \mu_{1}[mC]
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\frac{d[GFP]}{dt}= k_{translation}[mRNA] - k_{degradation}  
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\end{equation} </br></p>
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\end{equation} </br>
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</br>
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<br><p align="left"> Describe transcription of cI (C0051) mRNA over the change in time; that it is equal to the transcription rate of cI less of the degradation rate of cI(mRNA). Said in other words alpha(a) is how fast mRNA is produced by the mRNA polymerase minus the degradation speed of the mRNA produced (µ[mC]).</p></br>
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<br>
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\begin{equation}
\begin{equation}
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\large \frac{d[C]}{dt} = \alpha_{2} \cdot f_{RBS} \cdot [mC] - \mu_{2}[C]
+
\frac{d[TALEN]}{dt}= k_{translation}[mRNA] - k_{degradation}  
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\end{equation} </br></p>
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\end{equation}
 +
 
</br>
</br>
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<br><p align="justify"> Describe the translation of cI (C0051) over the change in time, that it is equal to the translation rate of cI(RNA polymerase velocity to make mRNA) by the function RBS of <i>E.Coli</i> ribosomes (expressed in equation 5) multiplied by the transcription of cI(equation 1). This part is the amount of cI(c0051)mRNA present in our bacteria, if this is zero (no mRNA produced) we would get no translation results; no material, no product. We need to reduce to the protein translated the degradation rate of cI(protein), because in all organic material they tend to decompose and we need to show this in the equation.</p></br>
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<br><p align="left"> These equations would describe the change of concentration of both GFP (our reporter) and the TALEN. As previous teams have described the <a href="http://parts.igem.org/Part:BBa_J23110:Experience">promoter activity</a>, we expect our system to behave similarly as previously reported. </p></br>
<br>
<br>
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\begin{equation}
 
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\large [mC]_{max} = \frac{\alpha _{1}}{\mu _{1}}
 
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\end{equation} </br></p>
 
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</br>
 
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<p align="justify">In order to get a maximum cI mRNA capacity we described the transcription of cI (C0051) when the transcription of cI (C0051) mRNA over the change in time is 0 because we supposed at that specific time the transcription will be in a maximum production. Then µ[mC] is isolated; getting that µ[mC] is the transcription rate of cI divided by the degradation rate of cI.</p>
 
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<br>
 
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\begin{equation}
 
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\large [C]_{max} = \frac{\alpha _{1} \cdot \alpha _{2}} {\mu _{1} \cdot \mu _{2}}
 
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\end{equation} </br></p>
 
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</br>
 
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<p align="justify">It describes the translation of cI (C0051) when the translation of cI (C0051) over the change in time is 0 because we supposed at that specific time the translation will be in the maximum production, such as the third equation. It is solved and the [C]max is equal to the transcription of cI by the translation of cI both divided by the degradation rate of cI(mRNA) multiplied by the degradation rate(protein)</p>
 
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<br>
 
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\begin{equation}
 
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\large f_{RBS} = \left\{ \begin{array}{rcl} 0 & \mbox{if}
 
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& t \geq ON \\ 1 & \mbox{if} & t < OFF
 
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\end{array} \right.
 
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\end{equation} </br></p>
 
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<p align="justify">The function of RBS of ribosome says that at the range of temperature between 32 ºC and 37 ºC, the production of Vip3Ca3 will be null such as GFP and that´s why we relation with the zero and the term of greater or equal than is ON (cI), and the term 1 express that at below temperature (32ºC) it will start the production of Vip3Ca3 and GFP (ON) and the cI off, the riboswitch. This equation express the primary and secondary structures the riboswitch k115017 can take depending on the temperature. When temperature is below 32ºC the riboswitch have a special structure named hairpin structure that avoids the sequence to entering the ribosome to translate, but at higher temperatures the mRNA is at a normal shape that can enter the ribosome to create a protein. </p>
 
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<br>
 
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\begin{equation}
 
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\large \frac{d[mV]}{dt} = \alpha_{3} \cdot \frac{K_{D}^h}{K_{D}^h + [C]^h} - \mu_{1}[mV]
 
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\end{equation} </br></p>
 
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</br>
 
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<p align="justify">It describes the transcription of Vip3Ca3 mRNA over the change in time; that it is equal to the transcription rate of Vip, regulated depending on the production of cI, less the degradation rate of Vip (mRNA)</p>
 
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<br>
 
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\begin{equation}
 
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\large \frac{d[V]}{dt} = \alpha_{4} \cdot [mV]  - \mu_{4}[V]
 
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\end{equation} </br></p>
 
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</br>
 
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<p align="justify">It describes the translation of Vip3Ca3 over the change in time, that it is equal to the translation rate of Vip multiplied by the transcription of Vip (equation 7) less of degradation rate of Vip (protein).</p>
 
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<br>
 
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\begin{equation}
 
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\large [mV]_{max} = \frac{\alpha _{3}}{\mu _{3}}
 
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\end{equation} </br></p>
 
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</br>
 
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<p align="justify">It describes the transcription of Vip3Ca3 when the time is 0 because we supposed at that specific time the transcription will be in the maximum production. Then it is solved the octave equation and is the transcription rate of Vip3Ca3 divided by the degradation rate of VIP.</p>
 
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<br>
 
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\begin{equation}
 
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\large [V]_{max} = \frac{\alpha _{3} \cdot \alpha _{4}} {\mu _{3} \cdot \mu _{4}}
 
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\end{equation} </br></p>
 
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</br>
 
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<p align="justify">It describes the translation of Vip3Ca3 when the time is 0 because we supposed at that specific time the translation will be in the maximum production, such as the ninth equation. It is solved and the equation is equal to the transcription of Vip by the translation of Vip is divided by the degradation rate of Vip mRNA by the degradation rate of Vip protein  </p>
 
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<br>
 
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\begin{equation}
 
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\large \frac{d[mG]}{dt} = \alpha_{5} \cdot \frac{K_{D}^h}{K_{D}^h + [C]^h} - \mu_{5}[mG]
 
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\end{equation} </br></p>
 
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</br>
 
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<p align="justify">It describes the transcription of GFP mRNA over the change in time; that it is equal to the transcription rate of GFP, regulated depending on the production of cI, less the degradation rate of GFP (mRNA)</p>
 
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<br>
 
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\begin{equation}
 
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\large \frac{d[G]}{dt} = \alpha_{6}  \cdot [mG]  - \mu_{6}[G]
 
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\end{equation} </br></p>
 
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</br>
 
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<p align="justify">It describes the translation of GFP over the change in time, that it is equal to the translation rate of GFP multiplied by the transcription of GFP (equation 11) less of degradation rate of GFP (protein).</p>
 
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<br>
 
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\begin{equation}
 
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\large [mG]_{max} = \frac{\alpha _{5}}{\mu _{5}}
 
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\end{equation} </br></p>
 
-
</br>
 
-
<p align="justify">It describes the transcription of GFP when the time is 0 because we supposed at that specific time the transcription will be in the maximum production. Then it is solved the octave equation and is the transcription rate of GFP divided by the degradation rate of GFP.</p>
 
-
<br>
 
-
\begin{equation}
 
-
\large [G]_{max} = \frac{\alpha _{5} \cdot \alpha _{6}} {\mu _{5} \cdot \mu _{6}}
 
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\end{equation} </br></p>
 
-
</br>
 
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<p align="justify">It describes the translation of GFP when the time is 0 because we supposed at that specific time the translation will be in the maximum production, such as the 13 equation. It is solved and the equation is equal to the transcription of GFP by the translation of Vip is divided by the degradation rate of GFP mRNA by the degradation rate of Vip protein  </p>
 
-
<br>
 
-
<br>
 
</tr>
</tr>
</table>
</table>

Latest revision as of 23:11, 17 October 2014

Math Model
Protein Production Model

The transcription of genes begins with the binding of transcription elements onto regulatory sites. The transcription rate is heavily dependant on how strong or weak these sites are. In our project, protein feedback will not be taken into account as there is no regulatory region capable of interacting with the products. We assume that both mRNA and protein synthesis have a constant rate of production, which does not vary throughout the process. Each component degrades randomly and is recycled within the cell.

  • Transcription

    • The average transcription speed in E.coli is 70nt/s. Assuming all the transcription in our circuit works in such speed, we can calculate the maximum transcription rate for each transcription equation by using this formula: Maximum Transcription Rate = Transcription Speed(nt/min)/Gene Length(bp)=4200/Gene Length (nM/min)

  • Translation

    • The average translation speed in E.coli is 40Aa/s. Also assuming all the translation in our circuit works in the same speed, we can calculate the theoretical transcription rate. However, in wetlab, we can use different rbs to regulate the translation process, thus, the translation rate can be written as: Translation Rate = RBS * Translation Speed(Aa/min)/Protein Length(Aa) = 2400RBS/Protein Length (min^-1) This transformation does not change the degree of freedom of our system. However, this does limit the range of parameters since the strength of RBS can not be too extreme.

  • Cell division rate

    • We have assume the period of cell division is 30 mins, which means the "degradation rate" in our model is actually the sum of degradation rate of the substance(1/half life) and cell division rate(1/30 mins).

  • RNA degradation rate

    • From Ref1(Belasco 1993) and Ref2(Genome Biology 2006, 7:R99), we have decided that all the mRNA in our system have a half life of 4.4 mins.


\begin{equation} \frac{d[GFP]}{dt}= k_{translation}[mRNA] - k_{degradation} \end{equation}
\begin{equation} \frac{d[TALEN]}{dt}= k_{translation}[mRNA] - k_{degradation} \end{equation}

These equations would describe the change of concentration of both GFP (our reporter) and the TALEN. As previous teams have described the promoter activity, we expect our system to behave similarly as previously reported.



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