Difference between revisions of "Team:KU Leuven/Modeling/Internal/Ultra/Test"

0.5 Test

This is a very easy text.

2.1 Cell A

The designed circuit in Cell A is under control of a temperature sensitive cI repressor. Upon raising the temperature, the circuit is activated and production of LuxR and LuxI initiates. LuxI will consectively produce AHL, which binds with LuxR to activate the production of Leucine and Ag43. We can extract the following ODE's from this circuit:
${\alpha}$:transcription term, ${\beta}$:translation term, ${d}$:degradation term, ${D}$:diffusion term ${K_d}$:dissociation constant, ${n}$:hill coefficient, ${L}$:leak term $$\frac{{\large d} m_{cI}}{d t} = \alpha_1 {\cdot} cI_{gene} - d_{mCI} {\cdot} m_{cI}$$ $$\frac{{\large d}{cI}}{d t} = \beta_{cI} {\cdot} {cI} -2 {\cdot} {k_{cI,dim}} {\cdot} {cI}^2 + 2 {\cdot}{k_{-cI,dim}}{\cdot} {[cI]_2} - d_{cI} {\cdot} {cI}$$ $$\frac{{\large d}{[cI]_2}}{d t}= k_{cI,dim} {\cdot} {cI}^2 - {k_{-cI,dim}}{\cdot} {[cI]_2} $$ $$\frac{{\large d} m_{LuxI}}{d t} = (L_{lambda} + {\frac{\alpha_{lambda}}{1 + ({\frac{[cI]_2}{K_{d1}}})^{n_{cI}}}}) {\cdot} LuxI_{gene} - d_{mLuxI} {\cdot} m_{LuxI} $$ $$\frac{{\large d} m_{LuxR}}{d t} = (L_{lambda} + {\frac{\alpha_{lambda}}{1 + ({\frac{[cI]_2}{K_{d1}}})^{n_{cI}}}}) {\cdot} LuxR_{gene} - d_{mLuxR} {\cdot} m_{LuxR} $$ $$\frac{{\large d} LuxI}{d t} = \beta_{LuxI} {\cdot} {m_{LuxI}} - d_{LuxI} {\cdot}{LuxI} $$ $$\frac{{\large d} LuxR}{d t} = \beta_{LuxR} {\cdot} {m_{LuxR}} -k_{lux,as} {\cdot}{LuxR}{\cdot}{AHL_{in}} + k_{lux,dis}{\cdot}{[LuxR/AHL]} - d_{LuxR} {\cdot}{LuxR} $$ $$\frac{{\large d} AHL_{in}}{d t} = {k_{luxI}} {\cdot} {luxI} - k_{lux,as} {\cdot}{luxR}{\cdot}{AHL_{in}} + k_{lux,dis}{\cdot}{[luxR/AHL]}+ ( {D_{IN,AHL}} {\cdot} {AHL_{out}} - {D_{OUT,AHL}} {\cdot} {AHL_{in}} ) - d_{AHL,in} {\cdot} {AHL_{in}}$$ $$\frac{{\large d} AHL_{out}}{d t} = ( {D_{OUT,AHL}} {\cdot} {AHL_{in}} - {D_{IN,AHL}}{\cdot}{AHL_{out}} ) -d_{AHL,out}{\cdot}{AHL_{out}}$$ $$\frac{{\large d} [luxR/AHL]}{d t} = k_{lux,as} {\cdot}{luxR}{\cdot}{AHL_{in}} - k_{lux,dis}{\cdot}{[luxR/AHL]} - 2 {\cdot} k_{lux,dim} {\cdot}{[luxR/AHL]^2} + 2 {\cdot}{k_{-lux,dim}}{\cdot}{[luxR/AHL]_{2}}$$ $$\frac{{\large d} [luxR/AHL]_{2}}{d t} = k_{lux,dim} {\cdot}{[luxR/AHL]^2} - k_{- lux,dim} {[luxR/AHL]} $$ $$\frac{{\large d} m_{ilvE}}{d t} = (L_{lux} + \frac{\alpha_{lux}}{1+(\frac{K_{d2}}{[luxR/AHL]_{2}})^{n_{lux}}} ) {\cdot} ilvE_{gene} - d_{milvE} {\cdot} {m_{ilvE}} $$ $$\frac{{\large d} m_{Ag43}}{d t} = (L_{lux} + \frac{\alpha_{lux}}{1+(\frac{K_{d2}}{luxR/AHL]_{2}})^{n_{lux}}} ) {\cdot} Ag43_{gene} - d_{mAg43} {\cdot} {m_{Ag43}} $$ $$\frac{{\large d} Transaminase B}{d t} = \beta_{TB} {\cdot} {m_{ilvE}} - d_{TB} {\cdot} {Transaminase B} $$ $$\frac{{\large d} Leucine_{out}}{d t} = (D_{OUT,Leu} {\cdot}{leucine_{in}} - D_{IN,Leu} {\cdot}{leucine_{out}}) - d_{Leu,out} {\cdot} {Leucine_{out}} $$ $$\frac{{\large d} Ag43}{d t} = \beta_{Ag43} {\cdot} {m_{Ag43}} - d_{Ag43} {\cdot} {Ag43} $$ We visualize these ODE's in Simbiology which results in the following diagram: