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Revision as of 10:26, 9 September 2015
Internal Model
0.5 Test
This is a very difficult text.
0.5 Test
This is a very easy text.
En alle beelden op tv
Van bloed en oorlog om ons heen
Werken daar ook niet echt aan mee
1. Introduction
My name is James Bond
In designing new circuits and implementing them, it is important to model the effect of it on the bacteria: how sensitive is the system, how much will it produce and will it affect the growth? These questions can be answerred in the internal model. We will use two approaches. Flux Balance Analysis (FBA) will study the steady-state values for production and growth rate, while Simbiology will study the sensitivity and dynamic processes in the cell.
2. Simbiology
In the next section we will describe our Simbiology model. Simbiology allows us to calculate systems of ODE's and to visualize them in a nice diagram. It also has options to make scans for different parameters which allows us to study the effect of the specified parameter. We will focus on the production of leucine, Ag43 and AHL in cell A and the changing behavior of cell B due to changing AHL concentration. In this perspective, we will make two models in Simbiology: one for cell A and cell B.
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:
2.1 Cell B
We can repeat this assignment for Cell B: $$\frac{{\large d} m_{cI}}{d t} = \alpha_1 {\cdot} cI_{gene} - d_1 {\cdot} m_{cI}$$ $$\frac{{\large d}{cI}}{d t} = \beta_1 {\cdot} {cI} -2 {\cdot} {k_{cI,dim}} {\cdot} {cI}^2 + 2 {\cdot}{k_{-cI,dim}}{\cdot} {[cI]_2} - d_{cI} {\cdot} {cI}$$ $$\frac{{\large d}{[cI]_2}}{d t}= k_{cI,dim} {\cdot} {cI}^2 - {k_{-cI,dim}}{\cdot} {[cI]_2} $$ $$\frac{{\large d} m_{luxR}}{d t} = (L_{lambda} + {\frac{\alpha_{lambda}}{1 + ({\frac{cI}{K_{d1}}})^{n_{cI}}}}) {\cdot} luxR_{gene} - d_{mluxR} {\cdot} m_{luxR} $$ $$\frac{{\large d} luxR}{d t} = \beta_{luxR} {\cdot} {m_{luxR}} - d_{luxR} {\cdot}{luxR} $$ $$\frac{{\large d} AHL_{in}}{d t} = ( {D_{IN,AHL}} {\cdot} {AHL_{out}} - {D_{OUT,AHL}} {\cdot} {AHL_{in}} ) - k_{lux,as}{\cdot}{luxR}{\cdot}{AHL_{in}} + k_{lux,dis}{\cdot}{[luxR/AHL]} - d_{AHL,in} {\cdot} {AHL_{in}} $$ $$\frac{{\large d} AHL_{out}}{d t} = ( {D_{OUT,AHL}} {\cdot} {AHL_{in}} - {D_{IN,AHL}}{\cdot}{AHL_{out}} ) -d_{AHL,out}{\cdot}{AHL_{out}} $$ $$\frac{{\large d} [luxR/AHL]}{d t} = k_{lux,as} {\cdot}{luxR}{\cdot}{AHL_{in}} - k_{lux,dis}{\cdot}{[luxR/AHL]} - 2 {\cdot} k_{lux,dim} {\cdot}{[luxR/AHL]^2} + 2 {\cdot}{[luxR/AHL]_{2}} $$ $$\frac{{\large d} [luxR/AHL]_{2}}{d t} = k_{lux,dim} {\cdot}{[luxR/AHL]^2} - k_{- lux,dim} {[luxR/AHL]} $$ $$\frac{{\large d} m_{CheZ}}{d t} = (L_{lux} + \frac{\alpha_{lux}}{1+(\frac{K_{d2}}{[luxR/AHL]_{2}})^{n_{lux}}} ) {\cdot} CheZ_{gene} - d_{mCheZ} {\cdot} {m_{CheZ}} $$ $$\frac{{\large d} m_{PenI}}{d t} = (L_{lux} + \frac{\alpha_{lux}}{1+(\frac{K_{d2}}{luxR/AHL]_{2}})^{n_{lux}}} ) {\cdot} PenI_{gene} - d_{mPenI} {\cdot} {m_{PenI}} $$ $$\frac{{\large d} CheZ}{d t} = \beta_{CheZ} {\cdot} {m_{CheZ}} - d_{CheZ} {\cdot} {CheZ} $$ $$\frac{{\large d} PenI}{d t} = \beta_{PenI} {\cdot} {m_{PenI}} - d_{PenI} {\cdot} {PenI} $$ $$\frac{{\large d} m_{RFP}}{d t} = (L_{Pen} + {\frac{\alpha_{Pen}}{1 + ({\frac{PenI}{K_{d3}}})^{n_{Pen}}}}) {\cdot} RFP_{gene} - d_{mRFP} {\cdot} m_{RFP} $$ $$\frac{{\large d} RFP}{d t} = \beta_{RFP} {\cdot}{m_{RFP}} - d_{RFP} {\cdot}{RFP} $$ We visualize these ODE's in the Simbiology toolbox. This gives us the following diagrams:
Test test test test test ultra test
To the lab!
This is the best website i have ever seen!
°CEO Alphabet, founder Google
°Founder of the United States of America
Test test test test test.
To-Do List
- Break mechanical cab driver.
- Drive to abandoned factory
- Watch video of self
X. Parameters
| Parameter | Value | Unit | Source | Comment |
|---|---|---|---|---|
| $D_a$ | $0.072 \cdot 10^{-3}$ | $cm^2/h$ | following [1] | |
| $D_b$ | $2.376 \cdot 10^{-3}$ | $cm^2/h$ | following [1] | |
| $D_r$ | $26.46 \cdot 10^{-3}$ | $cm^2/h$ | as found in [6] | $298.2 K$ |
| $D_h$ | $50 \cdot 10^{-3}$ | $cm^2/h$ | from [3] | |
| $K_{c}$ | $8.5 \cdot 10^{-3}$ | $cm^2/h$ | guessed | |
| $\gamma$ | $10^{-5}$ | $h^{-1}$ | from [1] | |
| $k_p$ | $1.0 \cdot 10^2$ | $cl^{-1}$ | from [1] | |
| $k_h$ | $17.9 \cdot 10^{-4}$ | $fmol/h$ | computed from [4] and [8] | |
| $k_r$ | $5.4199\cdot 10^{-4}$ | $fmol/h$ | computed from [7] and [8] | |
| $k_{lossH}$ | $1/48$ | $h^{-1}$ | from [5] | $ ph = 7$ |
| $k_{lossR}$ | $1/80$ | $h^{-1}$ | guessed |
| Protein | Translation rate (a.u.) | Translation rate (1/s) |
|---|---|---|
| LuxI | 261.13 | 0.016 |
| LuxR | 279.61 | 0.016 |
| cI857 | 5695.79 | 0.35 |
| Transaminase B | 2195.1 | 0.13 |
| Ag43-YFP | 33.22 | 0.0020 |
| PenI | 6696.73 | 0.41 |
| RFP | 2604.51 | 0.16 |
| CheZ-GFP | 396.95 | 0.024 |
