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Team:CIDEB-UANL Mexico/Math-Equations
From 2013hs.igem.org
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Equations
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\begin{equation} \large \frac{d[mC]}{dt} = \alpha_{1} - \mu_{1}[mC] \end{equation} 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]). \begin{equation} \large \frac{d[C]}{dt} = \alpha_{2} \cdot f_{RBS} \cdot [mC] - \mu_{2}[C] \end{equation} 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 E.coli 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. \begin{equation} \large [mC]_{max} = \frac{\alpha _{1}}{\mu _{1}} \end{equation} 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. \begin{equation} \large [C]_{max} = \frac{\alpha _{1} \cdot \alpha _{2}} {\mu _{1} \cdot \mu _{2}} \end{equation} 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) \begin{equation} \large f_{RBS} = \left\{ \begin{array}{rcl} 0 & \mbox{if} & t \geq ON \\ 1 & \mbox{if} & t < OFF \end{array} \right. \end{equation} 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. \begin{equation} \large \frac{d[mV]}{dt} = \alpha_{3} \cdot \frac{K_{D}^h}{K_{D}^h + [C]^h} - \mu_{1}[mV] \end{equation} 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) \begin{equation} \large \frac{d[V]}{dt} = \alpha_{4} \cdot [mV] - \mu_{4}[V] \end{equation} 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). \begin{equation} \large [mV]_{max} = \frac{\alpha _{3}}{\mu _{3}} \end{equation} 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. \begin{equation} \large [V]_{max} = \frac{\alpha _{3} \cdot \alpha _{4}} {\mu _{3} \cdot \mu _{4}} \end{equation} 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 \begin{equation} \large \frac{d[mG]}{dt} = \alpha_{5} \cdot \frac{K_{D}^h}{K_{D}^h + [C]^h} - \mu_{5}[mG] \end{equation} 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) \begin{equation} \large \frac{d[G]}{dt} = \alpha_{6} \cdot [mG] - \mu_{6}[G] \end{equation} 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). \begin{equation} \large [mG]_{max} = \frac{\alpha _{5}}{\mu _{5}} \end{equation} 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. \begin{equation} \large [G]_{max} = \frac{\alpha _{5} \cdot \alpha _{6}} {\mu _{5} \cdot \mu _{6}} \end{equation} 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 |
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