User talk:Rvfrolov/Sandbox

Estimations of the number of ions involved in generating the upstroke of the action potential
A question concerning a number of ions traversing the membrane during action potential and whether or not they can significantly alter the ionic gradients is addressed in this section.

If K+ is leaving the cell, won't this deplete the cell of intracellular K+, thus decreasing the concentration gradient? The answer to the latter question is "not much", the concentration gradient is not dissipated very much. So how can this be? To answer this question, one must understand is that the number of potassium ions that must cross the membrane is actually quite small. This is because the membrane potential occurs only right next to the membrane. The membrane potential is stored in the membrane capacitance, which physically is only the charges in direct contact with the membrane surfaces. The voltage across the capacitor is the membrane potential (see action potential for an example of a membrane RC circuit). Knowing that, one needs only to calculate how much current needs to flow to discharge the membrane capacitance to calculate the number of charges that must flow across the membrane to change the capacitor voltage (membrane potential) by, say 100 mV.

The capacitative current that flows into or out of a capacitor to effect a change in membrane potential of amplitude dV, is given by the equation:


 * $$c_m A \frac{\mbox{d}V} {\mbox{d}t} =-I_c $$

To make a back-of-the-envelope calculation, let’s assume the normal conditions for cylindrical cardiomyocytes during the action potential upstroke:
 * dV: 100 mV
 * dt: 2 ms
 * cm: 0.01 to 0.02 farads·m-2, the latter for cardiomyocytes due to the membrane invaginations of the T tubules.
 * A: cell membrane surface area
 * Length L: 100 μm
 * radius r: 5 μm
 * Thus, since the cell is a cylinder,
 * Area, A: 3.3 m2
 * Volume, V: 7.85 liters
 * Intracellular sodium concentration, [Na+]i: 15 mM


 * $$I_c = \left(0.02 \ \mathrm{F\cdot m}^{-2} \right) \left(3.3\times 10^{-9} \ \mathrm{m}^2 \right) \frac{0.1 \ \mathrm{V}} {0.002 \ \mathrm{s}} = 3.3\times 10^{-9} \ \mathrm{C\cdot s}^{-1}$$

This equals 3.3 nanoamps. Now, we wanted to know the number of ions, n, this corresponds to. dt is 2 ms, every ion has 1 elementary charge, e, each, and one coulomb is 6.24 e.


 * $$n = \left(3.3 \times 10^{-9} \ \mathrm{C\cdot s}^{-1} \right) \left( 0.002\ \mathrm{s} \right) \left( 6.24 \times 10^{18} \ e \cdot \mathrm{C}^{-1} \right) = 4.12 \times 10^7 \ e$$

Therefore, it would take the movement of 41 million charges (i.e. sodium ions) across the membrane to change the voltage of the cardiomyocyte by 100 mV in 2 ms. We want to know what that will do to the intracellular sodium concentration, which in cardiomyocytes is about 15 millimoles per liter. Avogadro's number, which relates numbers to moles, is 6.022·1023 mol-1:


 * $$\Delta [\mbox{Na} ^+]_{ \mbox{i}} = \frac{4.12 \times 10^7 \ e} {\left(6.022 \times 10^{23} \ e \cdot \mathrm{mol}^{-1} \right) \left( 7.85 \times 10^{-12} \ \mathrm{L} \right)} = 8.71 \times 10^{-6} \ \mathrm{mol \cdot L}^{-1}$$

Since the normal inside concentration of sodium is some 15 mM, 8.7 μM entering the cell during the action potential upstroke is only about 0.058% extra sodium ions

The bottom line for those who don’t want to sift through the calculation is that only about 40,000,000 sodium ions must cross the membrane to move the membrane potential by 100 mV in 2 ms, and that this constitutes only some 0.06% of the sodium already present in the cell. This is at the action potential's most extreme and in a very large cell. In other cells, calculations may look different, depending on surface-to-volume ratio, max dV/dt etc.