Introduction
We really want to thank the team iGEM Kent 2015 for this modeling born from the collaboration between the two groups.
Proteorhodopsin is a light-powered proton pump. This protein has the property to use light energy to generate an outward proton flux that can subsequently power cellular processes, such as ATP synthesis.After having experimentally assessed that our engineered E. coli was able to produce more ATP thanks to the proteorhodopsin, we decided to quantify this effect by modeling the amount of ATP that can be produced in function of the proteorhodopsin’s activation.
What is the number of ATP molecules that can be produced per second as a function of light irradiance that hits the bacterial membrane?
Once a photon is absorbed by proteorhodopsin (PR), PR must complete its photocycle before it can absorb another photon [1]. At high light irradiance, this leads to saturation. For this we choose to exploit the Michaelis-Menten kinetics, where V_max is the maximum rate of the system and the Michaelis-Menten constant, K_m, is the substrate concentration at which the reaction rate is $\frac{1}{2}V\max$.
Walter et al. demonstrated that the system is analogous to a circuit (figure 1), in this circuit representation; the proteorhodopsin (PR) acts like a battery with internal resistance. [2][3]
Membrane as an electric circuitElectric circuit analogy for the membrane[2]
The current through the system is inversely related to the PR resistor and is dependent on light irradiance.
$R_{PR=}\left( \frac{V_{\max }*I}{K_{m}+I} \right)^{-1}$
Walter et al. determined that $V_{\max}$ is fixed by the boundary condition that $R_{PR≈}\frac{R_{\sin k}}{10}$ at the highest light irradiance $I=\frac{160mW}{cm^{2}}$. $\; R_{\sin k}≈R_{\mbox{re}s}≈10^{15}\; \Omega$ and $K_{m=}\frac{60mW}{cm^{2}}$. Where light irradiance of $\frac{20mW}{cm^{2}}\;$ is roughly equivalent to PR absorption from solar illumination at sea level. [2]
At the boundary condition:
$Rpr=\frac{R_{\sin k}}{10}=10^{14}\Omega =\left( \frac{V_{\max }*I}{K_{m}+I} \right)^{-1}$
Hence:
$V\max \; =\; \frac{K_{m}+I}{R_{PR}*I}\; =\; 1.375*10^{-14}\; \Omega ^{-1}$
The rate of reaction, $v$, has units of $\Omega ^{\left( -1 \right)}$; through dimensional analysis we can see that $\Omega ^{-1}\; =\; \frac{Amps}{Volts}\; =\frac{coulombs}{\left( \sec ond*voltage \right)}$. The voltage across the PR, $V_{PR=}0.2\; Volts\;$ [2] and the charge of a proton is $q=1.6*10^{\left( -19 \right)}\; \mbox{C}$.
Therefore we can work out the number of protons pumped by the PR per second as
$N_{Proton}\; =\; \frac{V_{\max }\; I}{K_{m}+I}*\frac{V_{PR}}{q}$