\documentclass[a4paper,12pt,pdftex]{exam}

\newcommand{\ptitle}{Activation curve}
\input{../header.tex}
\firstpagefooter{Supervisor: Lukas Sonnenberg}{}%
{email: lukas.sonnenberg@student.uni-tuebingen.de}

\begin{document}

\input{../instructions.tex}


%%%%%%%%%%%%%% Questions %%%%%%%%%%%%%%%%%%%%%%%%%
\section{Estimation of the activation curve}
Mutations in genes, encoding for ion channels, can result in a variety of neurological diseases like epilepsy, autism and intellectual disability. One way to find a possible treatment is to compare the voltage dependent kinetics of the mutated channel with its corresponding wild-type. These kinetics are described in voltage-clamp experiments and the subsequent data analysis. 

In this task you will compute and compare the activation curves of the Nav1.6 wild-type channel and a variation named A1622D (the amino acid Alanine (A) at the 1622nd position is replaced by Aspartic acid (D)) that causes intellectual disability in humans.



\begin{questions}
    \question In the accompanying datasets you find recordings of cells with WT or A1622D transfections. The cells were all clamped to -70mV for some time to bring all ion channels in the same closed states. They are activated by a step change in the command voltage to a value described in the "steps" vector. The corresponding recorded current (in pA) and time (in ms) traces are also saved in the files.

\begin{parts}
    \part Plot the current traces of a WT and a A1622D cell. Because the number of transfected channels can vary the peak values have little value. Normalize the curves accordingly (what kind of normalization would be appropriate?). Can you already spot differences between the cells? 

    \part \textbf{IV curve}: Find the peak values for each voltage step and plot them against the steps.

    \part \textbf{Reversal potential}: Use the IV-curve to estimate the reversal potential of the sodium current. Consider a linear interpolation to increase the accuracy of your estimation.

    \part \textbf{Activation curve}: The activation curve is a representation of the voltage dependence of the sodium conductivity. It is computed with a variation of Ohm's law:
    \begin{equation}
        g_{Na}(V) = \frac{I_{peak}}{V - V_{reversal}}
    \end{equation}

    \part \textbf{Compare the two variants}: To compare WT and A1622D activation curves you should first parameterise your data. Fit a sigmoid curve
    \begin{equation}
        g_{Na}(V) = g_{max,Na} / ( 1 + e^{ - \frac{V-V_{1/2}}{k}} )
    \end{equation} 
    to the activation curves. With $g_{max,Na}$ being the maximum conductivity, $V_{1/2}$ the half activation voltage and $k$ a slope factor. Now you can compare the two variants with a few simple parameters. What do the differences mean?

    \part \textbf{BONUS question}: Take a good look at your raw data. What other differences can you see? How could you analyse these?

  \end{parts}
\end{questions}

\end{document}
