https://github.com/davidcsterratt/KappaNEURON
Revision 6a88c934b3f78ff0a213b88a04e6e7e6efa6df3d authored by David C Sterratt on 06 January 2017, 14:37:31 UTC, committed by David C Sterratt on 06 January 2017, 14:37:31 UTC
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Tip revision: 6a88c934b3f78ff0a213b88a04e6e7e6efa6df3d authored by David C Sterratt on 06 January 2017, 14:37:31 UTC
Added pulling of aclocal
Added pulling of aclocal
Tip revision: 6a88c93
tests.tex
\documentclass{article}
\usepackage{amsmath}
\usepackage{xspace}
\usepackage{mhchem}
\usepackage[amssymb]{SIunits}
\addunit{\molar}{M}
\newcommand{\ICa}{\ensuremath{I_\mathrm{Ca}}\xspace}
\newcommand{\gCa}{\ensuremath{g_\mathrm{Ca}}\xspace}
\newcommand{\ECa}{\ensuremath{E_\mathrm{Ca}}\xspace}
\newcommand{\Cmem}{\ensuremath{C_\mathrm{m}}\xspace}
\newcommand{\dif}{\textrm{d}}
\newcommand{\cCa}{[\textrm{Ca}]}
\begin{document}
\begin{table}
\centering
\begin{tabular}{lll}
Quantity & Symbol & Units \\
\hline
Voltage & $V$ & mV \\
Time & $t$ & ms \\
Conductance & $g$ & \siemens\per\centi\square\meter \\
Concentration & [Ca] & mM \\
Faraday const. & $F$ & C\per\mole \\
Length & $l$ & \micro\meter \\
Diameter & $d$ & \micro\meter \\
Capacitance & $\Cmem$ & \micro\farad\per\centi\square\meter \\
\hline
\end{tabular}
\caption{Quantities, symbols and units}
\label{tab:quantities}
\end{table}
\clearpage
\section{Test 1 - Ca accumulation}
\label{tests:sec:test-1-ca}
1 compartment, length $L$, diameter $d$, membrane capacitance \Cmem,
membrane potential $V$. Calcium current $\ICa=\gCa(V-\ECa)$, where $g$ is
conductance and $\ECa$ is calcium reversal potential. $g$ is 0 apart
from when it is set to $\overline{g}$ from $t_1$ to $t_2$.
ODEs describing membrane potential and calcium concentration:
\begin{equation}
\label{tests:eq:1}
\Cmem\frac{\dif V}{\dif t} = \ICa
\end{equation}
This can be solved:
\begin{equation}
\label{tests:eq:2}
V(t) = V(t_0) + (\ECa - V(t_0))(1-\exp((t-t_0)\gCa/\Cmem))
\end{equation}
The equation describing the calcium concentration [Ca] is:
\begin{equation}
\frac{\dif [\mathrm{Ca}]}{\dif t} = \frac{\ICa a }{2Fv} = \frac{2\ICa }{Fd}
\end{equation}
where $a= \pi Ld$ is the surface area and $v = \pi Ld^2/4$ is the
volume.
Hence following should be true:
\begin{equation}
[\textrm{Ca}](t_1) - [\textrm{Ca}](t_0) = \frac{\Cmem a (V(t_1) -
V(t_0)) }{2Fv} = \frac{2\Cmem (V(t_1) -
V(t_0)) }{Fd}
\end{equation}
Hence the factor-label method gives:
\begin{equation}
\begin{split}
\frac{\micro\farad}{\centi\square\meter}\cdot\frac{\micro\meter\squared\cdot\milli\volt}%
{\coulomb\per\mole\cdot\micro\meter\cubed}
= \frac{10^{-6} \farad}{10^{-4}\square\meter}\cdot\frac{\mole\cdot10^{-3}\volt}%
{\coulomb\cdot10^{-6}\meter}
= \frac{10^{-6} \coulomb}{10^{-4}\volt\square\meter}\cdot\frac{\mole\cdot10^{-3}\volt}%
{\coulomb\cdot10^{-6}\meter}\\
= \frac{10^1\mole}{\meter\cubed} =
\frac{10^1\mole}{10^3\deci\meter\cubed} = 10^{-2}\molar = 10\milli\molar
\end{split}
\end{equation}
\clearpage
\section{Test 2 - Ca accumulation with linear pump}
\label{tests:sec:test-1-ca}
\begin{itemize}
\item 1 compartment, length $L$, diameter $d$, membrane capacitance
\Cmem, membrane potential $V$.
\item Calcium channel current $\ICa^\mathrm{chan}=\gCa(V-\ECa)$, where $g$ is conductance
and $\ECa$ is calcium reversal potential.
\item Calcium pump current $\ICa^\mathrm{pump}=- \frac{Fdk_1}{2}\cCa$,
where $k_1$ is a constant.
\item $g$ is 0 apart from when it is set to $\overline{g}$ from $t_1$
to $t_2$.
\end{itemize}
ODEs describing membrane potential and calcium concentration:
\begin{equation}
\begin{split}
\Cmem\frac{\dif V}{\dif t} = -\ICa^\mathrm{chan} - \ICa^\mathrm{pump} = \gCa(\ECa - V) - \frac{Fdk_1}{2}\cCa ) \\
\frac{\dif\cCa}{\dif t} = -\frac{2\ICa^\mathrm{chan}}{Fd} - k_1\cCa =
\frac{2}{Fd} \left(\gCa(\ECa - V) - \frac{Fdk_1}{2}\cCa \right)
\end{split}
\end{equation}
This can be solved:
\begin{equation}
\label{tests:eq:2}
V(t) = V(t_0) + (V_\infty - V(t_0))(1-\exp((t-t_0)/\tau))
\end{equation}
Where
\begin{equation}
\label{tests:eq:5}
V_\infty = \frac{g\ECa + \Cmem k_1 V(t_0)}{g + \Cmem k_1} \quad
\mbox{and} \quad \tau = \frac{\Cmem}{g + k_1\Cmem}
\end{equation}
For the dimensions of our quantities we need:
\begin{equation}
\label{tests:eq:5}
V_\infty = \frac{10^3g\ECa + \Cmem k_1 V(t_0)}{10^3g + \Cmem k_1} \quad
\mbox{and} \quad \tau = \frac{\Cmem}{10^3g + k_1\Cmem}
\end{equation}
\clearpage
\section{Test 3 - Ca accumulation with nonlinear pump}
\label{tests:sec:test-3-ca}
\begin{itemize}
\item 1 compartment, length $L$, diameter $d$, membrane capacitance
\Cmem, membrane potential $V$.
\item Calcium channel current $\ICa^\mathrm{chan}=\gCa(V-\ECa)$, where
$g$ is conductance and $\ECa$ is calcium reversal potential.
\item Pump is modelled using a pump molecule P with starting density
$[\ce{P}]_0$ and pump reactions:
\begin{equation}
\label{tests:eq:3}
\begin{aligned}
\textrm{Ca binding:}\quad & \ce{P + Ca ->[k_1] P.Ca} \\
\textrm{Ca release:}\quad & \ce{P.Ca ->[k_2] P}
\end{aligned}
\end{equation}
\item Thus the calcium pump current $\ICa^\mathrm{pump}=-
\frac{Fdk_2}{2}[\ce{P.Ca}]$, where $k_1$ is a constant.
\item $g$ is 0 apart from when it is set to $\overline{g}$ from $t_1$
to $t_2$.
\end{itemize}
ODEs describing membrane potential and calcium concentration:
\begin{equation}
\begin{aligned}
\Cmem\frac{\dif V}{\dif t} &
%= -\ICa^\mathrm{chan} - \ICa^\mathrm{pump}
= \gCa(\ECa - V) - \frac{Fdk_2}{2}[\ce{P.Ca}] )
= \gCa(\ECa - V) - \frac{Fdk_2}{2}([\ce{P}]_0 - [\ce{P}] ) \\
\frac{\dif\cCa}{\dif t} &
= -\frac{2\ICa^\mathrm{chan}}{Fd} - k_1[\ce{P}][\ce{Ca}]
= \frac{2}{Fd} \left(\gCa(\ECa - V) - k_1[\ce{P}][\ce{Ca}] \right) \\
\frac{\dif[\ce{P}]}{\dif t} &
= -k_1[\ce{P}][\ce{Ca}] + k_2[\ce{P.Ca}]
= -k_1[\ce{P}][\ce{Ca}] + k_2([\ce{P}]_0 - [\ce{P}])
\end{aligned}
\end{equation}
This pump is nonlinear, even with fixed $E_{\mathrm{Ca}}$. Thus an
analytical solution is harder, if not impossible. However, we can say
the following:
\begin{itemize}
\item If $k_2=0$, Ca and voltage will rise during pulse, then voltage
will remain steady, but Ca concentration will decline.
\end{itemize}
\end{document}
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