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| | [https://2014.igem.org/Team:Paris_Saclay/Modeling/oxygen_diffusion oxygen_diffusion] | | [https://2014.igem.org/Team:Paris_Saclay/Modeling/oxygen_diffusion oxygen_diffusion] |
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| - | == Odor ==
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| - | To synthesize the smell of lemon, we have to mainly produce Limonene, Geraniol and Pinene. The combination of these products will allow us to get the fragrance we need. We propose here a study of the evolution of these concentrations over time by using the technique Michaelis-Menten proposed in 1913 by Leonor Michaelis and Maud Menten. This will allow us to obtain a description of the kinetics of reactions between an enzyme (biological catalyst) and a substrate for giving a product.
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| - | === Basic model ===
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| - | We consider the reations below between the enzyme $E$ and the substrate $S$. $ES$ is an complex foregoing the formation of the product P.
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| - | \[ E + S {\rightarrow}^{k_1} ES {\rightarrow}^{k_2} E + P \]
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| - | \[ E + S {\leftarrow}_{k_{-1}} ES {\leftarrow}_{k_{-2}} E + P \]
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| - | where
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| - | \[
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| - | \left\lbrace
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| - | \begin{array}{llll}
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| - | k_1 \text{ is the association constant of } E \text{ and } S \\
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| - | k_{-1} \text{ is the dissociation constant of } ES \\
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| - | k_2 \text{ is the reaction constant of } ES \text{ in } E + P \\
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| - | k_{-2}\text{ is the association constant of } E \text{ and } P \\
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| - | \end{array}
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| - | \right.
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| - | \]
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| - | We want to model the kinetics of the reaction catalyzed by enzymes Michaelis. We are interested in the concentration of molecules and various constants that we have described above.
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| - | The speed of the reaction $V$ is caracterised by the speed of apparition of the product. It is given by $V = \frac{d[P]}{dt}$ where $[P]$ is the concentration of the product.
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| - | Modeling phenomena are inextricably linked to the reality of the phenomenon modeled, so we must take into account a number of experimental facts.
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| - | * At the beginning, we don't have a product $P$. So the reaction $ES {\leftarrow}_{k_{-2}} E + P$ doesn't exist.
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| - | * The speed of the reaction is constant equal to $V_0$. It only depend to the initiale concentration of substrate $[S]_0$ and enzyme $[E]_T$.
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| - | * Over time, the concentration of the substrate decrease while the concentration of P increase. We have only two possibles cases:
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| - | ** We have total reaction and it exist a time where $[S] = 0$. It imply that $V = 0$.
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| - | ** We reach a steady state between the formation and the destruction of the product. We have also $V = 0$.
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| - | In fact we can consider that the speed of reaction decrease to 0. It will allow us to get an expression of $V_0$ in function of known parameters. But we must add others experimentals conditions. We consider the conditions set by Michaelis.
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| - | ''' Michaelis condition '''
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| - | To simplify the problem, Michaelis consider that:
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| - | * the concentration of substrate is very large relative to the concentration of enzyme: $[S] >> [E]_T $ and the concentration of the product is negligible: $P \approx 0$. With the equation $ES {\leftarrow}_{k_{-2}} E + P$, we know that the speed formation of $ES$ is gived by $v = k_{-2}[E][P] \approx 0$.
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| - | * We suppose that we reach the equilibrium between $[E], [S]$ and $[ES]$ fastly. In the case of equilibrium, $[ES]$ is constant until $[P] \approx 0$, and so
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| - | $\frac{d[ES]}{dt} = 0$.
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| - | * $[S]_0 >> [E]_T \rightarrow [ES]_{max} << [S]_0$ because the concentration of the complex $[ES]$ is limited by the concentration of enzyme.
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| - | Since $[S] = [S]_0 - [ES]$ and $[ES] \approx 0$, we have $[S] = [S]_0$.
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| - | In the following, we will stay in the conditions of Michaelis.
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| - | $\forall t, V = \frac{d[P]}{dt} = k_2 [ES] - k_{-2}[E][P]$. Since initially $[P] \approx 0$, we have $\boxed{V_i = k_2 [ES]}$. The problem is that $[ES]$ is unknow ! \newline
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| - | The variation of the concentration of the complexe $ES$ is given by the difference between it formation and it dissociation. This imply
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| - | \begin{align*}
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| - | %
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| - | & \frac{d[ES]}{dt} = k_1 [E][S ] - (k_{-1} + k_2) [ES] = 0 \text{ in initial condition }. \\
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| - | & \text{ So, } \frac{[E][S]}{[ES]} = \frac{k_{-1} + k_2}{k1} =: Km = \text{ Michaelis constant }. \\
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| - | & \text{ We deduce the Michaelis equation } \boxed{[ES] = \frac{[E][S]}{Km}}.
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| - | %
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| - | \end{align*}
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| - | For all time, the total enzyme $E$ is composed by free molecule of $E$ and molecule associated to the substrate to form $ES$. It is expressed by
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| - | $\forall t, [E]_T = [E] + [ES]$, where we deduce $[E] = [E]_T - [ES]$. Replacing this in the Michaelis equation, we have :
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| - | \begin{align*}
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| - | %
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| - | & [ES] = \frac{([E]_T - [ES])[S]}{Km} = \frac{[E]_T [S]}{Km} - \frac{[ES][S]}{Km}. \\
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| - | & [ES](1 + \frac{[S]}{Km}) = \frac{[E]_T [S]}{Km} \\
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| - | & [ES] = \frac{\frac{[E]_T [S]}{Km} }{ \frac{Km + [S]}{Km} } = [E]_T \times \frac{[S]}{Km + [S]}.
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| - | %
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| - | \end{align*}
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| - | Returning to the expression of $V_i = k_2 [ES]$ and replacing $[ES]$ by the last expression below, we have $V_i = k_2 [E]_T \times \frac{[S]}{Km + [S]} $
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| - | \[ \text{Since }
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| - | \left\lbrace
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| - | \begin{array}{ll}
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| - | [ES] \leq [E]_T \\
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| - | V_i = k_2 [ES]
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| - | \end{array}
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| - | \right.
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| - | \text{ we have } \boxed{ V_{max} = k_2 [E]_T } (k_2 \text{ is also noted } k_{cat}).
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| - | \]
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| - | Taking into account all these results, we obtain the following equation of Michaelis-Menten
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| - | \[ \boxed{ V_i = V_{max} \frac{[S]}{[S] + Km} } . \]
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| - | We can determine $V_{max}$ and $Km$ experimentally. $[S] \approx [S]_0$ is the concentration of substrate at the beginning.
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| - | In the following, we will work directly with this expression.
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| - | === Biosynthetic pathway of limonene ===
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| - | === Complete model ===
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| - | By analysing the different reaction in pathway above and using the equation of Michaelis-Menten, we obtain the system of nine equation below:
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| - | \[
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| - | \left\lbrace
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| - | \begin{array}{llllllllll}
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| - | \frac{d[acetyl-CoA]}{dt} &= - \frac{Vm1 [acetyl-CoA]}{Km1 + [acetyl-CoA]} \\
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| - | \frac{d[acetoacetyl-CoA]}{dt} &= \frac{Vm1 [acetyl-CoA]}{Km1 + [acetyl-CoA]} - \frac{Vm2 [acetoacetyl-CoA]}{Km2 + [acetoacetyl-CoA]} \\
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| - | \frac{d[HMG-CoA]}{dt} &= \frac{Vm2 [acetoacetyl-CoA]}{Km2 + [acetoacetyl-CoA]} - \frac{Vm3 [HMG-CoA]}{Km3 + [HMG-CoA]} \\
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| - | \frac{d[Mevalonate]}{dt} &= \frac{Vm3 [HMG-CoA]}{Km3 + [HMG-CoA]} - \frac{Vm4 [Mevalonate]}{Km4 + [Mevalonate]} \\
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| - | \frac{d[MevalonateP-5]}{dt} &= \frac{Vm4 [Mevalonate]}{Km4 + [Mevalonate]} - \frac{Vm5 [MevalonateP-5]}{Km5 + [MevalonateP-5]} \\
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| - | \frac{d[5-pyroP-MevalonateP]}{dt} &= \frac{Vm5 [MevalonateP-5]}{Km5 + [MevalonateP-5]} - \frac{Vm6 [5-pyroP-MevalonateP]}{Km6 + [5-pyroP-MevalonateP]} \\
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| - | \frac{d[IPP]}{dt} &= \frac{Vm6 \frac{[5-pyroP-MevalonateP]}{2}}{Km6 + \frac{[5-pyroP-MevalonateP]}{2}} - \frac{Vm7 \frac{[IPP]}{2}}{Km7 + \frac{[IPP]}{2}} - \frac{Vm8 \frac{[IPP]}{2}}{Km8 + \frac{[IPP]}{2}} \\
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| - | \frac{d[DMAPP]}{dt} &= \frac{Vm6 \frac{[5-pyroP-MevalonateP]}{2}}{Km6 + \frac{[5-pyroP-MevalonateP]}{2}} + \frac{Vm7 \frac{[IPP]}{2}}{Km7 + \frac{[IPP]}{2}} - \frac{Vm8 \frac{[DMAPP]}{2}}{Km8 + \frac{[DMAPP]}{2}} \\
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| - | \frac{d[GPP]}{dt} &= \frac{Vm7 \frac{[IPP]}{2}}{Km7 + \frac{[IPP]}{2}} + \frac{Vm8 \frac{[DMAPP]}{2}}{Km8 + \frac{[DMAPP]}{2}} - \frac{Vm_f [GPP]}{Km_f + [GPP]} \\
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| - | \frac{d[Fragrance]}{dt} &= \frac{Vm_f [GPP]}{Kmf + [GPP]}
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| - | \end{array}
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| - | \right.
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| - | \]
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| - | where $Vm_f$ and $Km_f$ are constant relative to the Fragrance ie Limonene, Geraniol or Pinene.
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| - | To solve these equation, we use the logiciel R and we obtain these following results for the production of Limonene:
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| - | The differents constant are obtained via Brenda enzyme database: http://www.brenda-enzymes.org/
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| - | == Oxygen Diffusion ==
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| - | To realise our artwork, we use an agarose gel to obtain the shape of a lemon. To push the resemblance to the extreme, we wish to have a crust in the edge of the lemon when we seperate it. In fact, we build bacteria who produce yellow/green color in presence of oxygen. Thus we must evaluate the penetration of the oxygen in the gel !
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| - | Oxygen penetrates into the gel by diffusion phenomenon that we will study below. We first use the following phenomenological law suggest by Adolphe Fick in 1855:
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| - | ----
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| - | '' In an homogeneous and isotropic environment, containing particles distributed inhomogeneously,appears spontaneously a volumetric flow density vector particle $\overrightarrow{J}(M,t) $. In any point $M$ in space, this vector is proportional to the gradient of the particle density $n(M,t)$. Mathematicaly, this relationship take the form: \[ \overrightarrow{J}(M,t) = - D \times \nabla n(M,t) \qquad (1) \] where $D$ is the diffusion coefficient and $\nabla$ is the gradient vector.''
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| - | ----
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| - | In equation after Fick's law above, we need to know the parameter $D$. We then searched in scientific literature and article '''[1]''' describes a method to get it. Referring to this article, the diffusion coefficient of oxygen in agarose is $ D = 0{,}256 \times 10^{-8} m^2 s^{-1} $.
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| - | The phenomenon we are facing is quite difficult to tackle if taken as a whole and We must emit a number of intuitively acceptable hypothesis.
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| - | * To simplify the problem, we consider that the diffusion of oxygen particle occurs only in one direction. So $\overrightarrow{J}(M,t) = J(x,t) \overrightarrow{e}_x $. This hypothesis is credible because we seek the maximum penetrance of oxygen. This is why we consider that the diffusion occurs in the line of greatest slope.
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| - | * Spatial variations in the density of particles are connected to spatial variations of the vector $\overrightarrow{J}(M,t)$ by '''the material's equation of conservation''' in presence of volume distribution of particle source $\sigma (x,t)$ (device which injects or subtracted particles to the system) :
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| - | \[ \frac{\partial n}{\partial t} (x,t) = - \frac{\partial J}{\partial x} (x,t) + \sigma (x,t) \qquad (2) \]
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| - | By replacing $(2)$ in $(1)$, we obtain the following '''equation of diffusion''' :
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| - | \[ \forall t, \forall x, \bigg( \frac{\partial}{\partial t} - D \frac{\partial^2}{\partial x^2} \bigg) n(x,t) = \sigma (x,t) \qquad (3) .\]
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| - | As our lemon is exposed to the ambient air, we stay in steady state where the source $ \sigma (x,t) $ is equal to $N_0$ the quantity of $O_2$ in the air.
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| - | To solve this equation, we use Fourier's analysis and it is kwown (classical solution of the heat equation) that
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| - | \[ \forall x, \forall t>0, n(x,t) = \frac{N_0}{\sqrt{4 \pi D t}} exp \bigg(- \frac{x^2}{4 D t} \bigg) + \int_{0}^{t} \underbrace{N_0 * exp \bigg(- \frac{|x|^2}{4 D \tau} \bigg)}_{= 0 \text{ by symmetry of the gaussian distribution }} \frac{d\tau}{\sqrt{4 \pi D \tau} } \]
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| - | Whence, \[ \forall x, \forall t>0, n(x,t) = \frac{N_0}{\sqrt{4 \pi D t}} exp \bigg(- \frac{x^2}{4 D t} \bigg) \]
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| - | Thus we deduce easily that the average dispersion particle is given by the variance $\Delta x = \sqrt{2Dt}$. Using this formula, we deduct that, for example, oxygen will penetrate $3 \times 10^{-3} m$ in $1956.522 s = 32.6082 $ minutes.
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| - | The charts below give us a very clear idea of penetrance evaluated over time.
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| - | === Graphical visualization ===
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| - | [[File:Paris_Saclay_oxygenGraph.jpeg|600px]]
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| - | [[References:]]
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| - | '''[1]''' A.C. Hulst, H.J.H. Hens, R.M. Buitelaar and J. Tramper, ''Determination of the effective diffusion coefficient of oxygen in gel materials in relation to gel concentration'', Biotechnology Techniques Vol 3 No 3 199-204 (1989).
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| - | '''[2]''' Vincent Renvoizé, ''Physique PC-PC*'', Cap Prepas, Pearson Education, 2010.
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| - | '''[3]''' Gilles Camus, La cinétique des enzymes michaeliennes et l'équation de Michaelis-Menten, 22 novembre 2012.
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| | {{Team:Paris_Saclay/default_footer}} | | {{Team:Paris_Saclay/default_footer}} |
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