Is the Uptake Equation Different When $k_2 = 0$?

[Dustinsfl]

The question I have is: when $k_2 = 0$ sketch the uptake $r(u)$ as a function of $u$ and compare it with the Michaelis-Menten uptake.
If $k_2 = 0$, wouldn't that change everything significantly? Does that mean re-do everything with $k_2 = 0$.
Is the Michaelis-Menten uptake just Michaelis constant?$$
S + E \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} C_1 \xrightarrow{k_2} E + P,
$$
$$
S + C_1 \underset{k_{-3}}{\overset{k_3}{\rightleftharpoons}} C_2\xrightarrow{k_4} C_1 + P
$$
Let $s = $, $e = [E]$, $c_1 = C_1$, $c_2 = [C_2]$, and $p = [P]$.
Then by the Law of Mass action, we can write down the differential equation model as:
\begin{align}
\frac{ds}{dt} =& -k_1es + (k_{-1} - k_3s)c_1 + k_{-3}c_2,\\
\frac{dc_1}{dt} =& k_1se - (k_{-1} + k_2 + k_3s)c_1 + (k_{-3} + k_4)c_2,\\
\frac{dc_2}{dt} =& k_3sc_1 - (k_{-3} + k_4)c_2,\\
\frac{de}{dt} =& -k_1se + (k_{-1} + k_2)c_1,\\
\frac{dp}{dt} =& k_2c_2 + k_4c_2
\end{align}
The initial conditions for the differential equation model are
$$
s(0) = s_0,\quad e(0) = e_0,\quad c_1(0) = c_2(0) = p(0) = 0.
$$
Since the enzyme is the catalyst, the enzyme is conserved by adding equations 2, 3, and 4.
$$
\frac{de}{dt} + \frac{dc_1}{dt} + \frac{dc_2}{dt} = 0\Rightarrow e(t) + c_1(t) + c_2(t) = e_0
$$
Then $e(t) = e_0 - c_1(t) - c_2(t)$.
Since $dp/dt$ is uncoupled, we can directly solve for $p(t)$.
$$
\int dp = \int (k_2c_1 + k_4c_2)dt\Rightarrow p(t) = k_2\int c_1dt + k_4\int c_2dt
$$
The only equations left to solve for are equations 1, 2, and 3 which are
\begin{align}
\frac{ds}{dt} =& -k_1e_0s + (k_{-1} + k_1s - k_3s)c_1 + (k_1s + k_{-3})c_2,\notag\\
\frac{dc_1}{dt} =& k_1e_0s - (k_{-1} + k_2 +k_1s +k_3s)c_1 + (k_{-3} + k_4 - k_1s)c_2,\notag\\
\frac{dc_2}{dt} =& k_3sc_1 - (k_{-3} + k_4)c_2,\notag
\end{align}
after substituting $e(t) = e_0 - c_1(t) - c_2(t)$.
By making the substitutions
\begin{alignat*}{4}
\tau &= k_1e_0t, & u &= \frac{s}{s_0}, & v_1 &= \frac{c_1}{e_0}, & v_2 &= \frac{c_2}{e_0},\notag\\
a_1 &= \frac{k_{-1}}{k_1s_0}, &\quad a_2 &= \frac{k_2}{k_1s_0}, &\quad a_3 &= \frac{k_3}{k_1}, &\quad a_4 &= \frac{k_{-3}}{k_1s_0},\notag\\
a_5 &= \frac{k_4}{k_1s_0}, & \epsilon &= \frac{e_0}{s_0}
\end{alignat*}
we can non-dimensionalize then model as
\begin{align}
\frac{du}{d\tau} =& -u + (u -a_3u + a_1)v_1 + (a_4 + u)v_2 = f(u,v_1,v_2),\notag\\
\epsilon\frac{dv_1}{d\tau} =& u - (u + a_3u + a_1 + a_2)v_1 +(a_4 + a_5 - u)v_2 = g_1(u,v_1,v_2),\notag\\
\epsilon\frac{dv_2}{d\tau} =& a_3uv_1 - (a_4 + a_5)v_2 = g_2(u,v_1,v_2).\notag
\end{align}
The initial conditions for the dimensionless model are
$$
u(0) = 1,\quad v_1(0) = v_2(0) = 0.
$$
Just as in class this model is a single perturbation for $0 < \epsilon\ll 1$.
Solving for $v_1$ and $v_2$, we obtain
$$
v_1 = \frac{u}{a_1 + a_2 + u + a_3u^2(a_4 + a_5)^{-1}} \quad\text{and}\quad v_2 = \frac{a_3uv_1}{a_4 + a_5}.
$$
Then
$$
f(u,v_1(u),v_2(u)) = \frac{du}{d\tau} = -u\frac{a_2 + a_3a_5u(a_4 + a_5)^-1}{a_1 + a_2 + u + a_3u^2(a_4 + a_5)^{-1}} = -r(u)
$$
is the uptake equation for $u$.
Let$A = a_2$, $B = a_3a_5(a_4 + a_5)^{-1}$, $C = a_1 + a_2$, and $D = a_3(a_4 + a_5)^{-1}$.
Then
$$
\frac{du}{d\tau} = -r(u) = -u\frac{A + Bu}{C + u + Du^2}.
$$

 

[Dustinsfl]

Would it be a simply substitution or would the non-dimensionalization have to be re-worked?

$$
S + E \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} C_1,
$$
$$
S + C_1 \underset{k_{-3}}{\overset{k_3}{\rightleftharpoons}} C_2\xrightarrow{k_4} C_1 + P
$$
Let $s = $, $e = [E]$, $c_1 = C_1$, $c_2 = [C_2]$, and $p = [P]$.
Then by the Law of Mass action, we can write down the differential equation model as:
\begin{align}
\frac{ds}{dt} =& -k_1es + (k_{-1} - k_3s)c_1 + k_{-3}c_2,\\
\frac{dc_1}{dt} =& k_1se - (k_{-1} + 0 + k_3s)c_1 + (k_{-3} + k_4)c_2,\\
\frac{dc_2}{dt} =& k_3sc_1 - (k_{-3} + k_4)c_2,\\
\frac{de}{dt} =& -k_1se + (k_{-1} + 0)c_1,\\
\frac{dp}{dt} =& 0 + k_4c_2
\end{align}
The initial conditions for the differential equation model are
$$
s(0) = s_0,\quad e(0) = e_0,\quad c_1(0) = c_2(0) = p(0) = 0.
$$
Since the enzyme is the catalyst, the enzyme is conserved by adding equations 2, 3, and 4.
$$
\frac{de}{dt} + \frac{dc_1}{dt} + \frac{dc_2}{dt} = 0\Rightarrow e(t) + c_1(t) + c_2(t) = e_0
$$
Then $e(t) = e_0 - c_1(t) - c_2(t)$.
Since $dp/dt$ is uncoupled, we can directly solve for $p(t)$.
$$
\int dp = \int (0 + k_4c_2)dt\Rightarrow p(t) = 0 + k_4\int c_2dt
$$
The only equations left to solve for are equations 1, 2, and 3 which are
\begin{align}
\frac{ds}{dt} =& -k_1e_0s + (k_{-1} + k_1s - k_3s)c_1 + (k_1s + k_{-3})c_2,\notag\\
\frac{dc_1}{dt} =& k_1e_0s - (k_{-1} + 0 +k_1s +k_3s)c_1 + (k_{-3} + k_4 - k_1s)c_2,\notag\\
\frac{dc_2}{dt} =& k_3sc_1 - (k_{-3} + k_4)c_2,\notag
\end{align}
after substituting $e(t) = e_0 - c_1(t) - c_2(t)$.
By making the substitutions
\begin{alignat*}{4}
\tau &= k_1e_0t, & u &= \frac{s}{s_0}, & v_1 &= \frac{c_1}{e_0}, & v_2 &= \frac{c_2}{e_0},\notag\\
a_1 &= \frac{k_{-1}}{k_1s_0}, &\quad a_2 &= \frac{0}{k_1s_0}=0, &\quad a_3 &= \frac{k_3}{k_1}, &\quad a_4 &= \frac{k_{-3}}{k_1s_0},\notag\\
a_5 &= \frac{k_4}{k_1s_0}, & \epsilon &= \frac{e_0}{s_0}
\end{alignat*}
we can non-dimensionalize then model as
\begin{align}
\frac{du}{d\tau} =& -u + (u -a_3u + a_1)v_1 + (a_4 + u)v_2 = f(u,v_1,v_2),\notag\\
\epsilon\frac{dv_1}{d\tau} =& u - (u + a_3u + a_1 + 0)v_1 +(a_4 + a_5 - u)v_2 = g_1(u,v_1,v_2),\notag\\
\epsilon\frac{dv_2}{d\tau} =& a_3uv_1 - (a_4 + a_5)v_2 = g_2(u,v_1,v_2).\notag
\end{align}
The initial conditions for the dimensionless model are
$$
u(0) = 1,\quad v_1(0) = v_2(0) = 0.
$$
Just as in class this model is a single perturbation for $0 < \epsilon\ll 1$.
Solving for $v_1$ and $v_2$, we obtain
$$
v_1 = \frac{u}{a_1 + 0 + u + a_3u^2(a_4 + a_5)^{-1}} \quad\text{and}\quad v_2 = \frac{a_3uv_1}{a_4 + a_5}.
$$
Then
$$
f(u,v_1(u),v_2(u)) = \frac{du}{d\tau} = -u\frac{0 + a_3a_5u(a_4 + a_5)^-1}{a_1 + 0 + u + a_3u^2(a_4 + a_5)^{-1}} = -r(u)
$$
is the uptake equation for $u$.
Let$A = a_2=0$, $B = a_3a_5(a_4 + a_5)^{-1}$, $C = a_1 + a_2$, and $D = a_3(a_4 + a_5)^{-1}$.
Then
$$
\frac{du}{d\tau} = -r(u) = -u\frac{Bu}{C + u + Du^2}.
$$

 

[Dustinsfl]

Michaelis-Menten uptake

So the Michaelis-Menten rate of uptake is
$$
R_0 = \frac{Q_{s_0}}{K_m+s_0}
$$
where $K_m=\dfrac{k_{-1}+k_2}{k_1}$ and $Q_{s_0}=k_2e_0s_0$

Q is the max velocity K_m is the Michaelis constant.

How can I plot this in Mathematica?

 

[Dustinsfl]

Re: Michaelis-Menten uptake

​Solved

 

[CellCoree]

When $k_2 = 0$, $A = 0$, and
$$
r(u) = u\frac{Bu}{C + u + Du^2} = \frac{u^2}{\frac{C}{B} + \frac{u}{B} + u^2D}.
$$
The Michaelis-Menten uptake equation is
$$
r(u) = \frac{u}{u + \frac{1}{D}}.
$$
The Michaelis constant is $K_M = \frac{1}{D}$.
When $k_2 = 0$, the uptake equation is not linear and does not have a Michaelis constant.