Team:Grenoble-EMSE-LSU/Project/Modelling/Building

From 2013.igem.org

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<h3> Third Model</h3>
<h3> Third Model</h3>
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<p> Until now, our equations describe the phototoxic effect of KillerRed as instantaneous: in the presence of light Killer Red produces ROS, which immediately kills a certain proportion of the bacteria but has no lasting effect. It is well-known, however, that cells can repair damages due to ROS, up to a certain level. We can thus consider that bacteria are unable to instantly repair all the damages caused by ROS, and with damages mounting up, they are more and more fragile and close to death.In other words, the effect of a certain amount of KR at a certain time $u$, $K(u)$, illuminated by a light intensity $I(u)$, will affect cell growth at time $t$ later than $u$, weighted by a factor $e^{-p(t-u)} that vanishes as $t$ increases. The effect of this ROS production at time $u$ will thus exponentially decrease with time. The term $– kI.K$  (photokilling) was thus replaced by the integral:</p>
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<p> Until now, our equations describe the phototoxic effect of KillerRed as instantaneous: in the presence of light Killer Red produces ROS, which immediately kills a certain proportion of the bacteria but has no lasting effect. It is well-known, however, that cells can repair damages due to ROS, up to a certain level. We can thus consider that bacteria are unable to instantly repair all the damages caused by ROS, and with damages mounting up, they are more and more fragile and close to death.In other words, the effect of a certain amount of KR at a certain time $u$, $K(u)$, illuminated by a light intensity $I(u)$, will affect cell growth at time $t$ later than $u$, weighted by a factor $e^{-p(t-u)}$ that vanishes as $t$ increases. The effect of this ROS production at time $u$ will thus exponentially decrease with time. The term $– kI.K$  (photokilling) was thus replaced by the integral:</p>
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The variable 'tox' is representative of the amount of damages inflicted to bacteria, and so of their probability of dying. During a time step (for us, a minute), bacteria cures part of their injuries ($l<1$) and suffers new damages ($k'I(t)K(t)$). At first, few bacteria die : $\mbox{tox}(t)\cong k'I(t)K(t)$, then, 'tox' increases until it reach $\mbox{tox}(t)\cong\frac{k'}{1-l}I(t)K(t)$
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The variable 'tox' is representative of the amount of damages inflicted to bacteria, and so of their probability of dying. During a time step (for us, a minute), bacteria cures part of their injuries ($l<1$) and suffers new damages ($k'I(t)K(t)$). At first, few bacteria die : $\mbox{tox}(t)\cong k'I(t)K(t)$, then, 'tox' increases until it reach $\mbox{tox}(t)\cong\frac{k'}{1-l}I(t)K(t)$. $k$ is the same in the continuous-time and the discrete-time forms, and $p$ and $l$ are directly related : $p=-\ln(l)$.
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Revision as of 11:02, 2 October 2013

Grenoble-EMSE-LSU, iGEM


Grenoble-EMSE-LSU, iGEM

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