A1  ^0.50 Consider a protein for which the only two possible states are $A$ and $B$. Suppose that a direct transition from state $A$ to state $B$ occurs with a known characteristic time $\tau_{AB}$, and the reverse transition does not occur (actually it could be described with $\tau_{BA}\to{\infty}$). Consider a sample containing $N$ of such proteins. Let them all be in state $A$ at the initial moment. Find the dependence of the number of proteins in each state as a function of time $N_A(t)$ and $N_B(t)$. Express your answers in terms of $\tau_{AB}$ and $N$.

We know the probability of the transition per the unit of time, i.e. the probability $dp$ of the transition during $dt$ is $dp = dt/ \tau_{AB}$, Let $N_A(t)$ be the number of proteins in the state $A$ at the given time $t$. During $dt$ each of these proteins changes its state $B$ with the probability $dp$, so on average there are $N_A(t) dp$ transitions. We can write a following differential equation for $N_A(t)$: \[ d N_a = - N_a dp = - N_A \frac{dt}{\tau_{AB}}.\] This equaition can be easly integrated, so with $N_A(0) = N$ we have \[ \ln \frac{N_A(t)}{N} = - \frac{t}{\tau_{AB}}, \quad \Rightarrow \quad N_A(t) = N \cdot e^{-t/\tau_{AB}}.\] Also $d N_B = - dN_A$ and $N_B(t) = N - N_A(t) = N ( 1 - e^{-t/\tau_{AB}})$.

A2  ^0.50 Indicate which ones belong to which channel? The height of the step is the same everywhere and is equal to $I=1.4~\text{pA}$.

On the graph "C" we can see recognizable shift of the first peak in the scale of $10~\text{ms}$. so $\tau_{CO} \approx 10~\text{ms}$ and the first type of channels corresponds to the letter "C". The difference between graph "B" and graph "C" is that on the scale of $100~\text{ms}$ on graph "B" we don't see any secondary peaks, so for protein at "B" $\tau_{IC} \gg 100~\text{ms}$.

Graph $A$	Graph $B$	Graph $C$
2nd type	3rd type	1st type

A3  ^1.00 Let's consider 3 cells, the membranes of which have built-in channels described in the previous paragraph. In the membrane of each cell there are many ($\gg{1}$) channels, all of the same type. Measurements are made using the Whole Cell Patch Clamp method. The light turns on and in all three cases they wait until the current readings are established, after which the light turns off. Draw qualitatively the current versus time dependence for all three cells. What is the steady current for each of the cells if the voltage applied between the electrodes and the solutions in the pipette and in the external liquid were left unchanged after the experiment in point A2. Express your answers in terms of the number of channels in the cell $N=1000$, $\tau_{CO}$, $\tau_{OI}$ and $\tau_{IC}$.

Let's write the dynamic equations for the numbers of cells in each state: \[ \begin{cases} \dot{N}_C = -\frac{N_C}{\tau_{CO}}+ \frac{N_I}{\tau_{IC}}\\ \dot{N}_O = -\frac{N_O}{\tau_{OI}} + \frac{N_C}{\tau_{CO}} \\ \dot{N}_I = -\frac{N_I}{\tau_{IC}} + \frac{N_O}{\tau_{OI}} \end{cases}. \] First, let's find the steady state number $N_O$. At equilibrium, the derivatives are zero, and we can express $N_C$ and $N_I$ in terms of $N_O$: \[N_C = N_O \frac{\tau_{CO}}{\tau_{OI}}, \quad N_I = N_O \frac{\tau_{IC}}{\tau_{OI}}.\] A total number $N$ is $N_C + N_O + N_I$, so \[ N_O = N \frac{\tau_{OI}}{\tau_{CO}+\tau_{OI}+\tau_{IC}}\] Let's introduce the vector $v=\begin{pmatrix} N_C \\ N_O \\ N_I \end{pmatrix}$. With this vector, the dynamic equation could be rewritten in matrix form: \[ \dot{v} = \begin{pmatrix} -\frac{1}{\tau_{CO}} & 0 & \frac{1}{\tau_{IC}}\\ \frac{1}{\tau_{CO}} & -\frac{1}{\tau_{OI}} & 0\\ 0 & \frac{1}{\tau_{OI}} & -\frac{1}{\tau_{IC}} \end{pmatrix} v = Av\] Let's find the eigenvalues $\lambda$ and their eigenvectors $v_\lambda$: $A v_\lambda = \lambda v_\lambda$. This equation could also be rewritten in matrix form $(A - \lambda E) v_\lambda = 0$ and the value of $\lambda$ could be found as the solution of $\det(A - \lambda E) = 0$. \[ \det (A - \lambda E) = -\left( \frac{1}{\tau_{CO}} + \lambda \right) \left( \frac{1}{\tau_{OI}} + \lambda \right) \left( \frac{1}{\tau_{IC}} + \lambda \right) + \frac{1}{\tau_{IC}} \frac{1}{\tau_{CO}} \frac{1}{\tau_{OI}} = 0 \] We have one root $\lambda=0$ - it's the equilibrium point, and two others which are the roots of the quadratic equation: \[ \lambda \left[ \left( \frac{1}{\tau_{OI}} \frac{1}{\tau_{IC}}+ \frac{1}{\tau_{IC}} \frac{1}{\tau_{CO}} + \frac{1}{\tau_{CO}}\frac{1}{\tau_{OI}} \right) + \lambda \left( \frac{1}{\tau_{CO}} + \frac{1}{\tau_{OI}} + \frac{1}{\tau_{IC}} \right) + \lambda^2 \right] = 0 \] Thus, we have two characteristic rates of convergence: \[\lambda_{1,2} = \frac{-\left( \frac{1}{\tau_{CO}} + \frac{1}{\tau_{OI}} + \frac{1}{\tau_{IC}} \right) \pm \sqrt{\left( \frac{1}{\tau_{CO}} + \frac{1}{\tau_{OI}} + \frac{1}{\tau_{IC}} \right)^2 - 4 \left( \frac{1}{\tau_{OI}} \frac{1}{\tau_{IC}}+ \frac{1}{\tau_{IC}} \frac{1}{\tau_{CO}} + \frac{1}{\tau_{CO}}\frac{1}{\tau_{OI}} \right)}}{2}\] To find $v_{\lambda_0}$, $v_{\lambda_1}$ and $v_{\lambda_2}$ we solve the equation $(A-\lambda E) v_\lambda$: \[ \begin{pmatrix} -\frac{1}{\tau_{CO}} - \lambda & 0 & \frac{1}{\tau_{IC}}\\ \frac{1}{\tau_{CO}} & -\frac{1}{\tau_{OI}}- \lambda & 0\\ 0 & \frac{1}{\tau_{OI}} & -\frac{1}{\tau_{IC}}- \lambda \end{pmatrix} \begin{pmatrix} v_C\\ v_O \\ v_I \end{pmatrix} = 0 \] and the resulting time dependence of $v$ is \[v(t) = A v_{\lambda_0} + B e^{\lambda_1 t} v_{\lambda_1} + C e^{\lambda_2 t} v_{\lambda_2},\] where the constants $A$, $B$, $C$ should be found such that they satisfy the initial condition $v(0) = \begin{pmatrix} N & 0 & 0 \end{pmatrix}^T$. For the first type cells $\lambda_1 = -\frac{1}{0.10~\text{ms}}$, $\lambda_2 = -\frac{1}{4.97~\text{ms}}$. $I_\infty=1.4~\text{pA} \cdot N_O\approx 0.70~\text{nA}$.

For the second type cells $\lambda_1 = -\frac{1}{7.29~\text{ms}}$, $\lambda_2 = -\frac{1}{1.00~\text{ms}}$. $I_\infty=1.4~\text{pA} \cdot N_O\approx 0.34~\text{nA}$.

For the third type cells $\lambda_1 = -\frac{1}{9.89~\text{ms}}$, $\lambda_2 = -\frac{1}{1.00~\text{ms}}$. $I_\infty=1.4~\text{pA} \cdot N_O\approx 0.014~\text{nA}$.

B1  ^2.00 Find the total flux $J$ of potassium ions through the selective potassium channel. Express your answer in terms of the diffusion coefficient $D$, the molar concentration $c$ of potassium outside the cell and its derivative $\frac{dc}{dx}$ along the channel, the mobility $\mu$ of the potassium ions, the elementary charge $e$ and the electric field $E$.

Remark: mobility is the coefficient of proportionality between the drift velocity $v$ of the particles and the force acting on them: $\mu F=v$.

Two effects contribute into the flow of ions: a diffusion $J_1 = - D \frac{dc}{dx}$ and a drift $J_2 = vc = \mu E e c$.

B2  ^2.00 Find at the voltage $U_\text{rev}$ on the cell membrane (membrane potential) that corresponds to zero flux through the open potassium channel at temperature $T=20^{\circ}\text{C}$. Express your answer in terms of $T$, $c_\text{out}=4~\text{mmol}/\text{l}$ and $c_\text{in}=155~\text{mmol}/\text{l }$ – potassium concentrations outside and inside the cell, and fundamental physical constants.

Remark: the membrane potential is considered positive if the plus is on the inner surface of the cell membrane.

A zero current is in corresponding with the zero flux of ions $J=0$. Let the $x$-axis direction be inward the cell: \[ -\mu k T \frac{dc}{dx} + \mu \frac{dU}{dx} e c = 0.\] We obtain such a differential equation for the voltage $U$ and the concentration $c$: \[ dU = \frac{kT}{e} \frac{dc}{c} \quad \Rightarrow \quad U = \frac{kT}{e} \ln \frac{c_\text{out}}{c_\text{in}}\]

C1  ^3.00 From the data shown in the plots, determine the electrical properties of the cell membrane: $R_M$, $R_S$, $C_M$. All three graphs reflect the dependence of the current through the cell membrane on time for the same point (this is all one experiment, different sections are shown in time). Different lines from bottom to top correspond to the voltages applied to the membrane: $\text{−120 mV}$, $\text{−80 mV}$, $\text{−40 mV}$, $\text{0 mV}$, $\text{40 mV}$, $\text{80 mV}$, $\text{120 mV}$. The green line at the top indicates the time the light is on.

After converging to the equilibrium current value and before turning on the light:

$U,~\text{mV}$	$I,~\text{pA}$
-120	-48
-80	-32
-40	-16
40	16
80	32
120	48

The slope of the graph $U$ vs $I$ is $2.5~\text{G}\Omega$, but from the equivalent scheme we know that the slope is $R_M + R_S$. At the initial moment of time, the dependence of $I$ on time and voltage $U$:

$U,~\text{mV}$	$I,~\text{nА}$
	$10.0~\text{ms}$	$10.1~\text{ms}$	$10.2~\text{ms}$	$10.3~\text{ms}$	$10.4~\text{ms}$
-120	-24	-16	-11	-7	-5
-80	-16	-11	-7	-5	-3
-40	-8	-5	-3.5	-2.5	-1.5
40	9	5	3.5	2	1.5
80	16	10	7	5	3
120	23	16	10	7	5

According to the data the mean slope of the graph $\ln\left(\frac{I(t)}{I(10.0~\text{ms})}\right)$ vs $t$ is $-4.0~\text{ms}^{-1}$, but from the equivalent scheme it is $-\frac{1}{R_MC_M}$. The area under the line $I(t)$ is equal to the charge, which have to flow through the membrane to charge the capacitor to the voltage $U$. From the data we have:

$U,~\text{mV}$	$q,~\text{pC}$
-120	-6,03
-80	-3,94
-40	-1,94
40	1,87
80	3,89
120	5,89

The slope of the graph $q$ vs $U$ is $49~\text{nF}$ and it is also equal to $C_M$. Then $R_M = \frac{1}{4.0~\text{ms}^{-1} \cdot C_M} =5.0~\text{M}\Omega$ и $R_S = 2.5~\text{G}\Omega$.

С2  ^1.00 Determine which ion freely passes through the channel built into the membrane of this cell? The concentrations of all ions in the pipette and external solutions are shown in Figure 7.

From the graph we see that when $U\approx 40~\text{mV}$ the current with turrned-on light stays approximattly zero. It allow us to calculate the value of $U_\text{rev}$

Ion	$U_\text{rev},~\text{mV}$
$\rm K^+$	-84,1
$\rm Na^+$	40,6
$\rm Cl^-$	-40,6
$\rm Mg^{2+}$	8,8