Secretion in eukaryotic cells is a fundamental process involving regulated release of a substance out of the cell. Cell secretion investigated in vitro in response to stimuli can help us to uncover complex intracellular microenvironment in biosystems.
A typical in vitro secretion assay begins with seeding a number of cell with a culture medium according to a protocol. After a while the medium containing secreted products is collected to quantify the exact amount of the products.
Neglecting cell proliferation or death (\(C=const\)) during in vitro experiment, cell product secretion is determined by the following equation:
\begin{equation} \frac{dP}{dt}=k \cdot C \label{eq:eq1} \end{equation}
In the cell secretion process, we also assume that the initial cell product concentration in media after stimulation is zero. Therefore, the initial condition is \(P=0\) at \(t=0\).
Integrating ODE (Equation \eqref{eq:eq1}) we can obtain an equation for \(k\):
\begin{equation} k = \frac{1}{C_{t_0}} \cdot \frac{P_{t_2}-P_{t_1}}{t_2-t_1} \label{eq:eq2} \end{equation}
So, having experimental timeseries in vitro and initial cell concentration we can estimate secretion rate constant.
This rate constant is a black box hiding the effects of different invironmental factors including media substances. Under controlled conditions we can pull out of the black box some influencers.
Consider non-obligatory effector when in in vitro experiment secretion is occur without investigated effector but it just accelerate or inhibit baseline secretion.
Based on InSysBio's approach to describe multiple effects of cytokines on cell dynamics processes we can obtain two equations \eqref{eq:eq3} and \eqref{eq:eq4} for \(k\) for obligatory effector (\(k_{obl}\)) or non-obligatory effector \(A\) (\(k_{nob}\)). In both equations effector \(A\) is affected by a modulator \(M\).
\begin{equation} k_{nob}=k_{base}\cdot \frac{1+E_{max}^A\cdot \frac{1+E_M^A\cdot M/EC_{50}^M}{1+M/EC_{50}^M}\cdot \frac{A}{EC_{50}^A}/\frac{1+EC_M^A\cdot M/EC_{50}^M}{1+M/EC_{50}^M}}{1+\frac{A}{EC_{50}^A}/\frac{1+EC_M^A\cdot M/EC_{50}^M}{1+M/EC_{50}^M}} \label{eq:eq3} \end{equation} \begin{equation} k_{obl}=k_{max}^A\cdot \frac{1+E_M^A\cdot \frac{M}{EC_{50}^M}}{1+\frac{M}{EC_{50}^M}}\cdot \frac{\frac{A}{EC_{50}^A}/\frac{1+EC_M^A\cdot M/EC_{50}^M}{1+M/EC_{50}^M}}{1+\frac{A}{EC_{50}^A}/\frac{1+EC_M^A\cdot M/EC_{50}^M}{1+M/EC_{50}^M}} \label{eq:eq4} \end{equation}
Above equations contain unknown parameters depending on experimental conditions. How can we estimate these parameters? Let’s analyse them.
Consider several experimental cases to estimate parameters.
An effector is absent, that is \(A = 0\) a modifier
\begin{align}k_{nob}^{A_0}&= k_{base}\\k_{obl}^{A_0}&= 0\end{align}
As expected, without an effector the product doesn’t secreted when an effector is obligatory but one is secreted with baseline rate constant when an effector is optional. Both cases are modulator-independent.
Remember, we assumed in this case constant cell concentration and absence of any effector. In this conditions we can estimate baseline secretion rate constant (for non-obligatory effectors):
\begin{equation} k_{base}=\frac{1}{C_{t_0}}\cdot\frac{P_{t_2}^{A_0}-P_{t_1}^{A_0}}{t_2-t_1} \label{eq:eq7} \end{equation}
Recommendation: due to linear assumption we need two measurements in linear region of the experimental curve.
An effector is added in saturable concentration without addition of a modifier. In experiment saturable concentration means effect doesn’t change at effector concentration above saturation or even the effect can decrease. If effector is an activator, dose-response curve have often bell-like shape due to tixicity of huge concentration of the effector. But we consider only first half of bell-shaped curve (for activators) in phisiological range of concentrations. From mathematical point of view saturable concentration is great enough to apply the concept of “limit at infinity”. So, experimental conditions \eqref{eq:eq3} and \eqref{eq:eq4} in math mode we can denote: \(A\to\infty\) and \(M = 0\). Then, firstly, rewrite the equations at condition \(M = 0\):
\begin{align}k_{nob}^{M_0}&= k_{base}\cdot\frac{1+E_{max}^A\cdot\frac{ A }{EC_{50}^A}}{1+\frac{A}{EC_{50}^A}}\end{align} \begin{align}k_{obl}^{M_0}&= k_{max}^A\cdot \frac{\frac{A}{EC_{50}^A}}{1+\frac{A}{EC_{50}^A}}\end{align}
Second,
\begin{align} \lim_{A\to\infty}k_{nob}^{M_0}=k_{base}\cdot E_{max}^A \end{align} \begin{align} \lim_{A\to\infty}k_{obl}^{M_0}=k_{max}^A \end{align}
Finally,
\begin{align}k_{nob}^{A_\infty,M_0}&= k_{base}\cdot E_{max}^A \label{eq:eq12} \end{align}
\begin{align}k_{obl}^{A_\infty,M_0}&= k_{max}^A \label{eq:eq13} \end{align}
Using Equation \eqref{eq:eq2} and assumptions applied in this case, we can estimate \(k_{max}^A\) and \(E_{max}^A\) as follow (note, \(k_{base}\) is estimated in Equation \eqref{eq:eq7}):
\begin{align} E_{max}^A =\frac{1}{k_{base}}\cdot\frac{1}{C_{t_0}}\cdot\frac{P_{t_2}^{A_\infty}-P_{t_1}^{A_\infty}}{t_2-t_1} \label{eq:eq14} \end{align} \begin{align}k_{max}^A=\frac{1}{C_{t_0}}\cdot\frac{P_{t_2}^{A_\infty}-P_{t_1}^{A_\infty}}{t_2-t_1}\end{align}
If required experimental data are absent we still able to roughly estimate parameter assume some simplification.
If initial cell concentration is not provided, then \(k_{base}\) or \(k_{max}^A\) cannot be calculated but \(E_{max}^A\) does. Insert Equation \eqref{eq:eq4} into Equation \eqref{eq:eq14}
\begin{align}E_{max}^A =\frac{P_{t_2}^{A_\infty}-P_{t_1}^{A_\infty}}{P_{t_2}^{A_0}-P_{t_1}^{A_0}}\cdot\frac{t_2^{A_0}-t_1^{A_0}}{t_2^{A_\infty}-t_1^{A_\infty}}\end{align}
Here we just need to know four experimental measurements without effector and at saturable effector cancentration.
When the first time point measurement coincide with the beginning of the experiment \(t_1 = t_0 = 0\) when the product is absent and product measurements occur at the same time after the beginning of the experiment at different concentration of the effector \(t_2^{A_\infty} = t_2^{A_0}\). Then,
\begin{align} E_{max}^A =\frac{P^{A_\infty}}{P^{A_0}} \end{align}
in this case both an effector and a modifier are added in saturable concentration. At first step let’s find constant at concentration of the effector in saturation:
\begin{align}k_{nob}^{A_\infty}=\lim_{A\to\infty} k_{nob} =k_{base}\cdot E_{max}^A\cdot\frac{1+E_M^A\cdot\frac{M}{EC_{50}^M}}{1+\frac{M}{EC_{50}^M}} \label{eq:eq18} \end{align} \begin{align} k_{obl}^{A_\infty}=\lim_{A\to\infty}k_{obl} =k_{max}^A\cdot\frac{1+E_M^A\cdot\frac{M}{EC_{50}^M}}{1+\frac{M}{EC_{50}^M}} \label{eq:eq19} \end{align}
Then,
\begin{align} k_{nob}^{A_\infty,M_\infty}=\lim_{M\to\infty}k_{nob}^{A_\infty}=k_{base}\cdot E_{max}^A\cdot E_M^A \label{eq:eq20} \end{align} \begin{align} k_{obl}^{A_\infty,M_\infty}=\lim_{M\to\infty}k_{obl}^{A_\infty}=k_{max}^A\cdot E_M^A \label{eq:eq21} \end{align}
Using Equation \eqref{eq:eq2} and assumptions applied in this case, we can estimate \(E_M^A\). For an obligatory effector:
\begin{align} E_M^A =\frac{k_{obl}^{A_\infty,M_\infty}}{k_{obl}^{A_\infty,M_0}}=\frac{P_{t_2}^{A_\infty,M_\infty}-P_{t_1}^{A_\infty,M_\infty}}{P_{t_2}^{A_\infty,M_0}-P_{t_1}^{A_\infty,M_0}} \end{align}
For a non-obligatory effector:
\begin{align} E_M^A =\frac{k_{nob}^{A_\infty,M_\infty}}{k_{base}\cdot E_{max}^A}=\frac{P_{t_2}^{A_\infty,M_\infty}-P_{t_1}^{A_\infty,M_\infty}}{P_{t_2}^{A_\infty,M_0}-P_{t_1}^{A_\infty,M_0}} \end{align}
\(E_M^A\) for both types of effectors are equal.
When the first time point measurement coincide with the beginning of the experiment \(t_1=t_0=0\) when the product is absent and product measurements occur at the same time after the beginning of the experiment at different concentration of the effector \(t_2^{A_\infty}=t_2^{A_0}\). Then,
\begin{align}E_M^A=\frac{P_{t_2}^{A_\infty,M_\infty}}{P_{t_2}^{A_\infty,M_0}} \end{align}
In fourth case an effector is added at concentration caused a half of effect (\(A = A_{50}\)) without addition of a modifier (\(M = 0\)). Definition: \(A_{50}\) is an effector concentration corresponding to its half-maximal effect on reaction corrected to baseline. In other words, at \(A_{50}\) secretion rate is a half of that at saturable effector concentration. Or, in mathematical notation, \(V^{A_{50}}=1/2 \cdot (V^{A_\infty}-V^{A_0})\) at any time point. In these experimental conditions
\begin{align} k_{nob}^{M_0}(A_{50})=k_{base}\cdot \frac{1+E_{max}^A\cdot \frac{A_{50}}{EC_{50}^A}}{1+\frac{A_{50}}{EC_{50}^A}} \end{align} \begin{align} k_{obl}^{M_0}(A_{50})=k_{max}^A\cdot \frac{\frac{A_{50}}{EC_{50}^A}}{1+\frac{A_{50}}{EC_{50}^A}} \end{align}
On the other hand,
\begin{align} V(A_{50})=\frac{1}{2}\cdot \left(V(A_\infty)+V(A_0)\right) \end{align}
\begin{align} 2\cdot k(A_{50})\cdot C=k(A_\infty)\cdot C+k(A_0)\cdot C \end{align}
\begin{align} 2\cdot k(A_{50})=k(A_\infty)+k(A_0) \label{eq:eq29} \end{align}
For an obligatory effector Equation \eqref{eq:eq29} takes the form:
\begin{align}2\cdot k_{max}^A\cdot \frac{\frac{A_{50}}{EC_{50}^A}}{1+\frac{A_{50}}{EC_{50}^A}}=k_{max}^A+0\end{align}
\begin{align}2\cdot \frac{A_{50}}{EC_{50}^A}=1+\frac{A_{50}}{EC_{50}^A}\end{align}
\begin{align}EC_{50}^A=A_{50}\end{align}
For a non-obligatory effector Equation \eqref{eq:eq29} takes the form:
\begin{align}2\cdot k_{base}\cdot \frac{1+E_{max}^A \cdot \frac{A_{50}}{EC_{50}^A}}{1+\frac{A_{50}}{EC_{50}^A}}=k_{base}\cdot E_{max}^A+k_{base}\end{align}
\begin{align}2\cdot(1+E_{max}^A\cdot \frac{A_{50}}{EC_{50}^A})=(E_{max}^A+1)\cdot (1+\frac{A_{50}}{EC_{50}^A})\end{align}
\begin{align}\frac{A_{50}}{EC_{50}^A}\cdot (E_{max}^A-1)=E_{max}^A-1\end{align}
\begin{align}EC_{50}^A=A_{50}\end{align}
So, for both cases \(EC_{50}^A = A_{50}\).
In fifth case a modifier is added at concentration caused a half of effect (\(M = M_{50}\)) and an effector is added at saturable concentration (\(A\to\infty\)). This case and derivation are similar to previous fourth case. First, consider saturable effector concentration Equations \eqref{eq:eq18} and \eqref{eq:eq19}:
\begin{align} k_{nob}^{A_\infty}&=\lim_{A\to\infty}k_{nob}=k_{base}\cdot E_{max}^A\cdot \frac{1+E_M^A\cdot \frac{M}{EC_{50}^M}}{1+\frac{M}{EC_{50}^M}} \end{align} \begin{align} k_{obl}^{A_\infty}&=\lim_{A\to\infty}k_{obl}=k_{max}^A\cdot \frac{1+E_M^A\cdot \frac{M}{EC_{50}^M}}{1+\frac{M}{EC_{50}^M}} \end{align}
Now, rewrite these equations for \(M_{50}\):
\begin{align} k_{nob}^{A_\infty}(M_{50})=k_{base} \cdot E_{max}^A \cdot \frac{1+E_M^A\cdot \frac{M_{50}}{EC_{50}^M}}{1+\frac{M_{50}}{EC_{50}^M}} \label{eq:eq39} \end{align} \begin{align} k_{obl}^{A_\infty}(M_{50})=k_{max}^A\cdot \frac{1+E_M^A\cdot \frac{M_{50}}{EC_{50}^M}}{1+\frac{M_{50}}{EC_{50}^M}} \label{eq:eq40} \end{align}
On the other hand, by definition
\begin{align}V_{A_\infty}(M_{50})=\frac{1}{2}\cdot \left(V_{A_\infty}(M_\infty)+V_{A_\infty}(M_0)\right)\end{align}
\begin{align}2\cdot k_{A_\infty}(M_{50})\cdot C=k_{A_\infty}(M_\infty)\cdot C+k_{A_\infty}(M_0)\cdot C\end{align}
\begin{align}2\cdot k_{A_\infty}(M_{50})=k_{A_\infty}(M_\infty)+k_{A_\infty}(M_0) \label{eq:eq43} \end{align}
For an obligatory effector Equation \eqref{eq:eq43} takes the form (using Equations \eqref{eq:eq12},\eqref{eq:eq20},\eqref{eq:eq39}):
\begin{align} 2\cdot k_{max}^A\cdot \frac{1+E_M^A\cdot \frac{M_{50}}{EC_{50}^M}}{1+\frac{M_{50}}{EC_{50}^M}}=k_{max}^A\cdot E_M^A+k_{max}^A \end{align}
\begin{align} 2\cdot \left(1+E_M^A\cdot \frac{M_{50}}{EC_{50}^M}\right)=(E_M^A+1)\cdot \left(1+\frac{M_{50}}{EC_{50}^M}\right) \end{align}
\begin{align} \frac{M_{50}}{EC_{50}^M}\cdot (E_M^A-1)=E_M^A-1\end{align}
\begin{align}EC_{50}^M=M_{50} \end{align}
For a non-obligatory effector Equation \eqref{eq:eq43} takes the form (using Equations \eqref{eq:eq13},\eqref{eq:eq21},\eqref{eq:eq40}):
\begin{align}2\cdot k_{base}\cdot E_{max}^A\cdot \frac{1+E_M^A\cdot \frac{M_{50}}{EC_{50}^M}}{1+\frac{M_{50}}{EC_{50}^M}}=k_{base}\cdot E_{max}^A\cdot E_M^A + k_{base}\cdot E_{max}^A\end{align}
\begin{align}2+2\cdot E_M^A\cdot \frac{M_{50}}{EC_{50}^M}=(E_M^A+1)\cdot (1+\frac{M_{50}}{EC_{50}^M})\end{align}
\begin{align}\frac{M_{50}}{EC_{50}^M}\cdot (E_M^A-1)=E_M^A-1\end{align}
\begin{align}EC_{50}^M=M_{50}\end{align}
What if the effector is given at not saturable concentration? Consider a case number 6.
Rewrite the system of Equation \eqref{eq:eq3} and \eqref{eq:eq4} for \(M_{50}\):
\begin{align} k_{nob}(M_{50})=k_{base}\cdot \frac{1+E_{max}^A\cdot \frac{1+E_M^A\cdot M_{50}/EC_{50}^M}{1+M_{50}/EC_{50}^M}\cdot\frac{A}{EC_{50}^A}/\frac{1+EC_M^A\cdot M_{50}/EC_{50}^M}{1+M_{50}/EC_{50}^M}}{1+\frac{A}{EC_{50}^A}/\frac{1+EC_M^A\cdot M_{50}/EC_{50}^M}{1+M_{50}/EC_{50}^M}} \end{align}
\begin{align} k_{obl}(M_{50})=k_{max}^A\cdot \frac{1+E_M^A\cdot \frac{M_{50}}{EC_{50}^M}}{1+\frac{M_{50}}{EC_{50}^M}}\cdot \frac{\frac{A}{EC_{50}^A}/\frac{1+EC_M^A\cdot M_{50}/EC_{50}^M}{1+M_{50}/EC_{50}^M}}{1+\frac{A}{EC_{50}^A}/\frac{1+EC_M^A\cdot M_{50}/EC_{50}^M}{1+M_{50}/EC_{50}^M}} \end{align}
On the other hand, by definition
\begin{align}V(M_{50})=\frac{1}{2}\cdot \left(V(M_\infty)+V(M_0)\right)\end{align}
\begin{align}2\cdot k(M_{50})\cdot C=k(M_\infty)\cdot C+k(M_0)\cdot C\end{align}
\begin{align}2\cdot k(M_{50})=k(M_\infty)+k(M_0)\end{align}
\begin{align}k_{nob}^{A,M_\infty}&=\lim_{M\to\infty}k_{nob}^{A}=k_{base}\cdot \frac{E_{max}^A\cdot E_M^A\cdot \frac{A}{EC_{50}^A}+EC_{50}^M}{\frac{A}{EC_{50}^A}+EC_{50}^M}\end{align}
\begin{align}k_{obl}^{A,M_\infty}&=\lim_{M\to\infty}k_{obl}^{A}=k_{max}^A\cdot \frac{E_M^A\cdot A}{EC_M^A\cdot EC_{50}^A+A}\end{align}
Define new variables: \(\boldsymbol A=A/EC_{50}^A\) and \(\boldsymbol M=M/EC_{50}^M\).
For an obligatory effector
\[k_{base}\cdot \frac{1+E_{max}^A\cdot \frac{1+E_M^A\cdot M_{50}/EC_{50}^M}{1+M_{50}/EC_{50}^M}\cdot \frac{A}{EC_{50}^A}/\frac{1+EC_M^A\cdot M_{50}/EC_{50}^M}{1+M_{50}/EC_{50}^M}}{1+\frac{A}{EC_{50}^A}/\frac{1+EC_M^A\cdot M_{50}/EC_{50}^M}{1+M_{50}/EC_{50}^M}}=k_{base}\cdot \frac{E_{max}^A\cdot E_M^A\cdot \frac{A}{EC_{50}^A}+EC_{50}^M}{\frac{A}{EC_{50}^A}+EC_{50}^M}+k_{base}\cdot \frac{1+E_{max}^A\cdot \frac{A}{EC_{50}^A}}{1+\frac{A}{EC_{50}^A}}\]
\[\frac{1+E_{max}^A\cdot \frac{1+E_M^A\cdot M_{50}/EC_{50}^M}{1+M_{50}/EC_{50}^M}\cdot \frac{A}{EC_{50}^A}/\frac{1+EC_M^A\cdot M_{50}/EC_{50}^M}{1+M_{50}/EC_{50}^M}}{1+\frac{A}{EC_{50}^A}/\frac{1+EC_M^A\cdot M_{50}/EC_{50}^M}{1+M_{50}/EC_{50}^M}}=\frac{E_{max}^A\cdot E_M^A\cdot \frac{A}{EC_{50}^A}+EC_{50}^M}{\frac{A}{EC_{50}^A}+EC_{50}^M}+\frac{1+E_{max}^A\cdot \frac{A}{EC_{50}^A}}{1+\frac{A}{EC_{50}^A}}\]
\[2\cdot \frac{1+E_{max}^A\cdot \frac{1+E_M^A\cdot \boldsymbol M}{\color{red}1+\boldsymbol M}\cdot \boldsymbol A/\frac{1+EC_M^A\cdot \boldsymbol M}{\color{red}1+\boldsymbol M}}{1+\boldsymbol A/\frac{1+EC_M^A\cdot \boldsymbol M}{1+\boldsymbol M}}=\frac{E_{max}^A\cdot E_M^A\cdot \boldsymbol A+EC_{50}^M}{\boldsymbol A+EC_{50}^M}+\frac{1+E_{max}^A\cdot \boldsymbol A}{1+\boldsymbol A}\]
\[2\cdot \frac{\\(1+E_{max}^A\cdot \boldsymbol A\cdot \frac{1+E_M^A\cdot \boldsymbol M}{\color{red}1+EC_M^A\cdot \boldsymbol M}\\)\cdot \\({\color{red}1+EC_M^A\cdot \boldsymbol M}\\)}{1+\boldsymbol A+\boldsymbol M\cdot \\(\boldsymbol A+EC_M^A\\)}=\frac{\\(1+\boldsymbol A\\)\cdot \\(E_{max}^A\cdot E_M^A\cdot \boldsymbol A+EC_{50}^M\\)\\+\\(1+E_{max}^A\cdot \boldsymbol A\\)\cdot \\(\boldsymbol A+EC_{50}^M\\)}{\\(\boldsymbol A+EC_{50}^M\\)\cdot \\(1+\boldsymbol A\\)}\]
\[2\cdot \frac{1+E_M^A\cdot \boldsymbol A+\boldsymbol M\cdot \\(EC_M^A+\boldsymbol A\cdot E_{max}^A\cdot E_M^A\\)}{1+\boldsymbol A+\boldsymbol M\cdot \\(\boldsymbol A+EC_M^A\\)}=\frac{\\(1+\boldsymbol A\\)\cdot \\(E_{max}^A\cdot E_M^A\cdot \boldsymbol A+EC_{50}^M\\)+\\(1+E_{max}^A\cdot \boldsymbol A\\)\cdot \\(\boldsymbol A+EC_{50}^M\\)}{\\(\boldsymbol A+EC_{50}^M\\)\cdot \\(1+\boldsymbol A\\)}\]
Finally,
\begin{align}EC_{50}^M=M_{50}\cdot \frac{EC_M^A+\frac{A}{EC_{50}^A}}{1+\frac{A}{EC_{50}^A}} \label{eq:eq59} \end{align}
How one can check the Equation \eqref{eq:eq59}?
\[\lim_{A\to\infty}EC_{50}^M=M_{50}\]
which is the same as obtained in case 5.
So, using time- and concentration-dependences obtained in in vitro experiments we can estimate the following parameters of secretion rate constant:
\begin{align}k_{base}=\frac{1}{C_{t_0}}\cdot \frac{P_{t_2}^{A_0}-P_{t_1}^{A_0}}{t_2-t_1}\end{align}
\begin{align}k_{max}^A=\frac{1}{C_{t_0}}\cdot \frac{P_{t_2}^{A_\infty}-P_{t_1}^{A_\infty}}{t_2-t_1}\end{align}
\begin{align}E_{max}^A=\frac{1}{k_{base}}\cdot \frac{1}{C_{t_0}} \cdot \frac{P_{t_2}^{A_\infty}-P_{t_1}^{A_\infty}}{t_2-t_1}\end{align}
\begin{align}E_M^A=\frac{P_{t_2}^{A_\infty,M_\infty}-P_{t_1}^{A_\infty,M_\infty}}{P_{t_2}^{A_\infty,M_0}-P_{t_1}^{A_\infty,M_0}}\end{align}
\begin{align}EC_{50}^A=A_{50}\end{align}
\begin{align}EC_{50}^M=M_{50}\end{align}
\begin{align}EC_{50}^M =M_{50}\cdot \frac{EC_M^A+\frac{A}{EC_{50}^A}}{1+\frac{A}{EC_{50}^A}}\end{align}
These equations with predefined the first experimental point ( \(P_{t_1}=0\) at \(t_1=0\) ) are implemented in fIVE DB.
Follow us to know how estimate parameters for other cell processes from in vitro experiments…
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