Simulations of the membrane potential for two ion species

Bioelectricity: Electric model of the cell membrane

Simulations of the membrane potential for two ion species

As a reminder we once more write down the Nernst equation and linearized GHK equation again.

The the equilibrium potential of an ion is given by $E_\mathrm{ion}= \frac{RT}{zF}\cdot \ln\left(\frac{C_{\mathrm{e}}}{C_{\mathrm{i}}}\right) = -\frac{RT}{zF}\cdot \ln\left(\frac{C_{\mathrm{i}}}{C_{\mathrm{e}}}\right)$ This electric potential represents the resting membrane potential in a model of the cell with a single ion species and a single ion channel type, and the value depends on the absolute temperature $T$ , the valence $z$ (the number of unit charges per ion), and the ion concentrations inside ( $C_{\mathrm{i}}$ ) and outside ( $C_{\mathrm{e}}$ ) the cell near the membrane. $R$ is the gas constant, and $F$ is Faraday's constant (equal to the charge of one mole of monovalent ions). At a temperature of $29.2{}^{\circ}\mathrm{C}$ you can also compute the Nernst potential in millivolt through the equation $E_\mathrm{ion}= \frac{60}{z} \log_{10}\left(\frac{C_{\mathrm{e}}}{C_{\mathrm{i}}}\right)$

The linearized GHK equation for two ion channels, potassium and sodium chanels, allows you to compute the resting membrane potential $V_m$ : $V_m=\frac{g_\mathrm{K}}{g_\mathrm{K}+g_\mathrm{Na}}\cdot E_\mathrm{K} + \frac{g_\mathrm{Na}}{g_\mathrm{K}+g_\mathrm{Na}}\cdot E_\mathrm{Na}$

Use the simulation below to practice with the membrane potential in an artificial cell with two ion species, namely $\mathrm{K}^{+}$ and $\mathrm{Na}^{+}$ , and to answer the following questions. First, think about all answers and check them later in the explanation.

The initial condition in the simulation represents a "resting state" in which there is a relatively high conductivity of potassium compared to that of sodium. We place "resting state" between quotes because there are still ionic currents through the membrane. The corresponding currents are on the right-hand side. There is a net zero current because ion currents by transport of sodium and potassium are opposed. A sodium-potassium pump ensures the maintenance of the ion concentration differences inside and outside the cell. This pump is turned o in the simulation, but can be turned off and then you see in a high-speed animation what will happen.

Click on the link Simulation of membrane potential when you want to use the simulation widescreen in a separate window.

Experiment

Calculate the formulas given the Nernst potential for potassium and sodium and the membrane potential and check your answers with the values shown in the simulation.
Calculate the flow rate of the potassium and sodium channels and update your responses with the values shown in the simulation.
What do you expect to happen when the Na / K pump stops working? Give a statement before the turn off to pump into the simulation by pressing the on / off button to click;
Reset the animation and increase the conductivity of the sodium channel, starting from the initial state, and enter the following effect on the membrane potential $V_\mathrm{m}$ and currents $I_\mathrm{K}$ and $I_\mathrm{Na}$ .
Increase the conductivity of the potassium channel and enter the following effect on the membrane potential $V_\mathrm{m}$ and currents $I_\mathrm{K}$ and $I_\mathrm{Na}$ .
Depolarization means that the membrane potential $V_\mathrm{m}$ increases (strong). Consider four ways to make a membrane to depolarize and verify that in the simulation.
Repolarization means that the membrane potential $V_\mathrm{m}$ decreases (strong). Consider four ways repolarising them to a membrane.
How can you change the membrane potential without having to change the ion concentrations and the conductivity of the membrane?

Answers

The formulas lead to the following questions (with extra decimals in increments): $\begin{aligned} E_\mathrm{K}&=60\log_{10}\left(\frac{[\mathrm{K}^{+}]_\mathrm{e}}{[\mathrm{K}^{+}]_\mathrm{i}}\right) \\ &= 60\log_{10}\left(\frac{20}{400}\right) \\ &= -60\log_{10}(20) \\ &= -78.0618 \approx 78.1\,\mathrm{mV} \\ \\ E_\mathrm{Na} &=60\log_{10}\left(\frac{[\mathrm{Na}^{+}]_\mathrm{e}}{[\mathrm{Na}^{+}]_\mathrm{i}}\right) \\ &= 60\log_{10}\left(\frac{430}{50}\right) \\ &= -60\log_{10}(8.6) \\ &= 56.0699\approx 56.1\,\mathrm{mV} \\ \\ V_\mathrm{m} &=\frac{g_\mathrm{K}}{g_\mathrm{K}+g_\mathrm{Na}}\cdot E_\mathrm{K} + \frac{g_\mathrm{Na}}{g_\mathrm{K}+g_\mathrm{Na}}\cdot E_\mathrm{Na} \\ &= 0.9\cdot -78.0618+ 0.1\cdot 56.0699 = -64.6486\approx -64.6\,\mathrm{mV} \end{aligned}$
Ohm's law leads to the following results (with extra decimals in increments)
$\begin{aligned} I_\mathrm{K}&=g_\mathrm{K}(V_\mathrm{m} -E_\mathrm{K}) \\ &= 9.0\bigl(-64.6486-(-78.0618)\bigr) \\ &= 120.7188\approx 120.7 \mu\mathrm{A/cm}^2 \\ \\ I_\mathrm{Na}&=g_\mathrm{Na}(V_\mathrm{m} -E_\mathrm{Na}) \\ &= 1.0\bigl(-64.6486-56.0699) \\ &= 120.7185 \approx 120.7 \mu \mathrm{A/cm}^2 \end{aligned}$
If the Na / K-pump stops working potassium then flows from the inside to the outside of the cell, as there is a positive current for the potassium ion because the membrane potential is higher than the Nernstpotententiaal of potassium. Just as sodium flows from the outside to the inside. Tengevolgen of the ion flow, the Nertstpotentiaal of potassium to be higher and that of sodium lower. the membrane potential will also be less negative. Because the sum of the initial concentrations of the ions inside the cell is equal to the sum of the initial concentrations of the ions outside the cell (and that remains so), will eventually the potentials and ionic currents are equal to zero. You see accelerated simulation event
The linearized Goldman-Hodgkin-Katz equation can be written as a weighted sum of Nernstpotentialen $V_\mathrm{m}=\alpha\cdot E_\mathrm{K}+\beta\cdot E_\mathrm{Na}$ with $\alpha = \frac{g_\mathrm{K}}{g_\mathrm{K}+g_\mathrm{Na}}\quad\text{en}\quad \frac{g_\mathrm{Na}}{g_\mathrm{K}+g_\mathrm{Na}}$ If the conductivity $g_\mathrm{Na}$ increases, $\alpha$ smaller and $\beta$ larger. The membrane potential, as a weighted sum of Nernstpotentialen, slides in the direction of the Nernst potential of sodium on, and thus comes to lie higher. The potential difference $V_\mathrm{m}-E_\mathrm{K}$ is thus larger, and the current intensity $I_\mathrm{K}$ will be greater in absolute value since it is proportional to the potential difference. In other words, more potassium ions to flow from the inside to the outside of the cell. The ion currents of potassium and sodium remain cancel each other out, and thus there will also be more sodium ions from the outside to the inside start to flow. The current strength of the sodium channels will therefore also increase in absolute value. This is all to observe the simulation.
If the conductivity $g_\mathrm{K}$ increases, $\alpha$ larger and $\beta$ smaller. The membrane potential, as a weighted sum of Nernstpotentialen, slides in the direction of the Nernst potential of potassium on, and thus comes to lie lower. The potential difference $V_\mathrm{m}-E_\mathrm{Na}$ is thus larger, and the current intensity $I_\mathrm{Na}$ will be greater in absolute value since it is proportional to the potential difference. In other words, more sodium ions flow from outside to inside the cell. The ion currents of potassium and sodium remain cancel each other out, and thus there will also be more potassium ions from inside to outside to flow. The current strength of potassium channels will therefore also increase in absolute value. This is all to observe the simulation.
Four ways are to a membrane polarize
1. Clean the conductivity of sodium channels $g_\mathrm{Na}$ greater;
2. Clean the conductivity of potassium channels $g_\mathrm{K}$ smaller;
3. Make the quotient $[\mathrm{Na}]_\mathrm{e}/[\mathrm{Na}]_\mathrm{i}$ larger (for example by increasing the sodium concentration outside the cell because then $E_\mathrm{Na}$ up);
4. Make the quotient $[\mathrm{K}]_\mathrm{e}/[\mathrm{K}]_\mathrm{i}$ larger (for example by increasing the potassium concentration outside the cell because then $E_\mathrm{K}$ up).
You can verify all the way into the simulation.
Four ways are to a membrane repolarising honor
1. Clean the conductivity of sodium channels $g_\mathrm{Na}$ smaller;
2. Clean the conductivity of potassium channels $g_\mathrm{K}$ greater;
3. Make the quotient $[\mathrm{Na}]_\mathrm{e}/[\mathrm{Na}]_\mathrm{i}$ smaller (eg by reducing the sodium concentration outside the cell);
4. Make the quotient $[\mathrm{K}]_\mathrm{e}/[\mathrm{K}]_\mathrm{i}$ smaller (for example, by lowering the potassium concentration outside the cell).
You can verify all the way into the simulation.
You can change the membrane potential without having the ion concentrations and the conductivity of the membrane changes the conductivity of the potassium and sodium channels ( $g_\mathrm{K}$ and $g_\mathrm{Na}$ change) without changing the sum of the conductivities (ie, $g_\mathrm{K}+g_\mathrm{Na}$ kept constant). You can verify this in the simulation.