IGEM:IMPERIAL/2007/Projects/Biofilm Detector/Modelling/Construct1
Construct 1: Introduction
With the chemical pathway below for construct 1 we can make form some initial equations about the system.
Our aim in modelling Infector Detector is to determine the concentration of biofilm we can detect such that we report a visible signal. Therefore we want to relate the GFP concentration to the concentration of AHL. Starting that the rate of change in GFP concentration:
- [math]\displaystyle{ \frac{d[GFP]}{dt}=k_{6}[AP]-\delta_{GFP}[GFP]\cdots\cdots(1) }[/math]
Aussume : [AP] reaches a constant level
Reason : Treat 1st formation of AP complex as a black box - it just reaches steady state.
- [math]\displaystyle{ \therefore\frac{d[AP]}{dt}=0=k_{4}[A][P]-k_{5}[AP]\cdots\cdots(2) }[/math]
- [math]\displaystyle{ \Rightarrow[AP]=\frac{k_{4}}{k_{5}}[A][P]\cdots\cdots(3) }[/math]
Assume : Conservation of Plux Promoters
Reason : no damage to Promoters will occour eg. no DNA damage due to old age or cell defence mechanisms attacking the DNA
- [math]\displaystyle{ \displaystyle[P]_{0}=[P]+[AP]\cdots\cdots(4) }[/math]
- [math]\displaystyle{ \displaystyle[P]=[P]_{0}-[AP]\cdots\cdots(5) }[/math]
Substitute this into (3)
- [math]\displaystyle{ [AP]=\frac{k_{4}}{k_{5}}[A]\left([P]_{0}-[AP]\right)\cdots\cdots(6) }[/math]
Solving for [AP]
- [math]\displaystyle{ [AP]+\frac{k_{4}}{k_{5}}[A][AP]=\frac{k_{4}}{k_{5}}[A][P]_{0}\cdots\cdots(7) }[/math]
- [math]\displaystyle{ [AP]\left(1+\frac{k_{4}}{k_{5}}[A]\right)=\frac{k_{4}}{k_{5}}[A][P]_{0}\cdots\cdots(8) }[/math]
- [math]\displaystyle{ [AP]=\frac{k_4[A][P]_{0}}{k_{5}+k_{4}[A]}=\frac{K_{\beta}[A][P]_{0}}{1+K_{\beta}[A]}\cdots\cdots(9) }[/math]
- [math]\displaystyle{ K_{\beta}=\frac{k_{4}}{k_{5}} }[/math]
From 9 [math]\displaystyle{ \because }[/math] [AP] is a constant [A] is a constant
From Reaction pathways at top of page we can see that because 3 is in equilibrium 2 is in equilibrium
From these two statements we can say that at equilibrium:
- [math]\displaystyle{ [A]=K_{\alpha}[AHL][LuxR]\cdots\cdots(10) }[/math]
- [math]\displaystyle{ K_{\alpha}=\frac{k_{2}}{k_{3}} }[/math]
Substituting all of this into expression for GFP
- [math]\displaystyle{ \frac{d[GFP]}{dt}=k_{6}\left[\frac{K_{\alpha}K_{\beta}[AHL][LuxR][P]_{0}}{1+K_{\alpha}K_{\beta}[AHL][LuxR]}\right]-\delta_{GFP}[GFP]\cdots\cdots(11) }[/math]
