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==Fluxes and Currents==
* A '''flux''', in the context of chemistry and physics, is the rate of flow of some substance from one place to another per unit of area. Flux usually has units of moles/meter<sup>2</sup>/second and is often represented by the symbol <math>$$J</math>$$.* A '''current''', in the context of electronics, is the rate of flow of charge. In metals, charge is carried by electrons; in biological cells, current is usually carried by the movement of ions through fluid. Current has units of coulombs/second, or amperes, and is represented by the symbol <math>$$I</math>$$.
* Note that electrons are negatively charged, and most ions found in biological systems are positively charged. In neurophysiology, the ''direction'' of the current flow is in the direction of the movement of positive charge (similar to the flow of ''conventional current'', which focuses on the flow of positive current without regard to charge carriers and is, consequently, opposite to the flow of negatively charged electrons).
* Thus, ionic fluxes can be represented by currents; but fluxes refer to the transfer of specific amounts of particular kinds of ''matter'' (e.g., the number of potassium ions that are moving from inside the cell to the outside), whereas currents refer to the amount of ''charge'' that is moving from one place to another, and this may not be the same. For example, if potassium ions are moving out of a nerve cell, and more chloride ions are moving into it, the net effect is an increase of negative charges within the cell; but the current is composed of the fluxes of more than one ion. This is the reason that the analysis of ionic fluxes often requires specific ion channel blockers; with these tools, it becomes possible to interpret current flows as the fluxes of specific ions.
These two features of the membrane—the lipid bilayer and the ion channels—can be represented using equivalent electrical components: a '''capacitor''' and a '''resistor'''. Let us focus first on the resistor (the channel).
* As an ion attempts to flow through a channel, it may encounter a series of obstacles, which may be due to constrictions in the diameter of the channel, protrusions of proteins into the channel, charges on some of the amino acids that make up the channel, and so on. These obstacles will tend to reduce the rate at which the ion can flow through the channel. What impels the ions to go through the channel? The electrical gradient across the membrane generates an electromotive force, i.e., a potential to do work on a charged entity, and this affects the probability that a charged ion will cross the membrane through the channel. The electromotive force is measured in volts, <math>$$V</math>$$.* Thus, it should be clear that the magnitude of the voltage across the membrane will affect the amount of current that crosses through the channel, or <math>$$V \varpropto I</math>$$.* The constant of proportionality is the resistance, <math>$$R</math>$$, measured in ohms.* Thus, one can re-write the equation as <math>$$V = I R</math>$$, which is known as '''Ohm's Law''', after its discoverer. Let us think about this equation for a bit:
** It implies that, for a given voltage, as the resistance to flow increases (it becomes harder for ions to move through a channel), the current will decrease.
** Another way to think about it is this: if one applied a fixed current, then a measure of the resistance would be how much the voltage changed for that fixed current. If the resistance is low, there would be a small voltage change; if the resistance is high, there would be a larger voltage change.
[[Image: EquilibriumFluxes.png|250px|thumb|Figure 2: Fluxes of potassium are balanced when the membrane potential equals the Nernst potential for potassium]]
Another way to describe the ease with which an ion can move through a channel is to take the '''reciprocal''' of the resistance, which is known as the '''conductance'''. The conductance, usually represented by the symbol <math>$$g</math> $$ and measured in units of siemens, is thus equal to <math>$$1/R</math>$$.
* Qualitatively, the relationship between resistance and conductance is clear: a high conductance implies a low resistance, and a low conductance implies a high resistance.
* To use conductances with Ohm's law, divide both sides of the equation by <math>$$R</math> $$ to obtain <math>$$V/R = I</math>$$; since <math>$$1/R = g</math>$$, Ohm's law becomes <math>$$g V = I</math>$$.
[[image: IonChannelCircuit.png|250px|thumb|Figure 3: An electrical equivalent circuit for potassium ion channels. On the left is shown the symbol for a voltmeter, and on the right the symbols for a resistor and a battery]]
Channels that are permeable to a specific ion will be associated with a specific Nernst or equilibrium potential. That is, the current flow through a channel will not only depend on the voltage across the membrane, but will also depend on how different the voltage is from the Nernst or equilibrium potential.
* If the membrane voltage is equal to the Nernst or equilibrium potential, then there should be no net flux of those particular kinds of ions across the membrane. For example, suppose you have a neuron with a high internal concentration of potassium, and the membrane potential is at the equilibrium potential for potassium. Imagine that the flux of potassium ions caused by the concentration gradient (an ''outward flux'') became a bit larger due to a random fluctuation in the movement of the ions. This would cause the inside of the membrane to become more negative and the outside more positive, leading to a hyperpolarization of the cell. This would increase the strength of the electrical gradient; in turn, the flux of potassium ions caused by the electrical gradient (an ''inward flux'') would strengthen, and the random fluctuation would be undone. Thus, there would be a balance between the flux of potassium ions out of the nerve cell (flux due to the concentration gradient) and the flux of potassium ions into the nerve cell (flux due to the electrical gradient), as illustrated in Figure 2.
* To represent all the ion channels of a particular kind (e.g., all potassium ion channels) using an electrical equivalent, one could use a battery whose voltage was at the Nernst or the equilibrium potential, which was in series with the resistance (or conductance) of the channels. This is shown in Figure 3 — the wavy lines are used to represent a resistor (or conductor), which is labeled with the symbol for the potassium conductance, <math>$$g_{\text{K}^+}</math>$$. In series with the resistor is the battery, labeled with the symbol for the potassium Nernst or equilibrium potential, <math>$$E_{\text{K}^+}</math>$$.
* This simple representation of the potassium ion channels can already be useful for predicting relationships between the membrane voltage, the conductance to those channels, and the current that will flow through them (assuming that the linear approximations we are making by using these electronic components are reasonably accurate).
* '''Kirchoff's Voltage Law''' can be stated as follows: the potential drops in a closed circuit must sum to zero. Applying that principle to the circuit shown in Figure 3, we see that the membrane voltage must equal the voltage drop across the resistor <math>$$\left(I/g\right)</math> $$ plus the voltage change across the battery <math>$$\left(E\right)</math>$$, or <math display="block"> V_m = \frac{I_{\text{K}^+}}{g_{\text{K}^+}} + E_{\text{K}^+}. \quad\text{(Equation 1)}</math>$$
* We can solve this equation for the current through the potassium channels to obtain <math display="block">
I_{\text{K}^+} = g_{\text{K}^+} \left(V_m - E_{\text{K}^+}\right). \quad\text{(Equation 2)}</math>$$
* This is an extremely important equation for understanding and mathematically modeling the properties of ion channels. In words, it says that the '''current''' through all the potassium ion channels is equal to the '''conductance''' of all the potassium ion channels multiplied by the '''driving force''', i.e., the difference between the membrane potential and the equilibrium potential for the potassium ions.
* An important implication of this equation is that current through a set of ion channels could drop to zero for two different reasons:
* Adding charges to one side of a membrane repels charges from the other side, inducing a current.
* If the charges are not added or removed, then the separation between the positive and negative charges across the membrane gives rise to a voltage.
* In other words, the charge separated across the membrane (or across a capacitor) is proportional to the voltage across the membrane, or <math>$$V \varpropto Q</math>$$, where <math>$$Q</math> $$ is the charge separated across the membrane.* To turn this proportionality into an equation, we define a constant of proportionality, the capacitance, measured in Farads and represented by the symbol <math>$$C</math>$$; the equation for the capacitor becomes <math display="block"> Q = C V. \quad\text{(Equation 3)}</math>$$* As we have just pointed out, changes in the number of charges can lead to changes in voltage, or <math display="block">C \Delta V = \Delta Q.</math>$$* If the change occurs in a small time increment <math>$$\Delta t</math>$$, then the change in the membrane's voltage and charge during that time will be, on average, <math display="block">C \frac{\Delta V}{\Delta t} = \frac{\Delta Q}{\Delta t}.</math>$$* As the time interval gets shorter and shorter, the fractions on the left and right side of the equation become the instantaneous rates of change, or derivatives, of the voltage and charge, respectively: <math display="block">C \frac{dV}{dt} = \frac {dQ}{dt}.</math>$$
* Since the instantaneous rate of change of the charge is the definition of the current, this equation becomes <math display="block">
C \frac{dV}{dt} = I. \quad\text{(Equation 4)}</math>$$
* In circuit diagrams, capacitors are represented by a symbol that shows two plates of equal length with a gap between; charges can be stored on the plates, but cannot directly jump across them. However, as positive charges are added or removed to one plate, positive charges will be removed or added to the other plate (respectively), so as '''Equation 4''' states, changing amounts of charge, or current, will lead to changes in voltage.
[[image: Channels&Capacitor.png|200px|thumb|Figure 4: An electrical equivalent circuit for ungated potassium channels and the passive membrane]]
* We can now represent all of the ungated potassium channels in the membrane, and how they will respond to charges injected into the nerve cell, with an electrical equivalent circuit that shows the capacitor in parallel with the ion channels. The capacitor is in parallel because ions can either flow through ion channels, or affect charges on the other side of the membrane if they land on the membrane, and thus there are two parallel paths for the movement of charge. In Figure 4, the capacitance of the membrane is labeled <math>$$C_m</math>$$.
* An analogy, which can be shown to be mathematically equivalent, is to think about what happens to a hose with flexible sides when water pressure is applied to one end. The water pressure is equivalent to voltage, the center of the hose is equivalent to the resistor, and the flexible sides are equivalent to the capacitor.
* Initially, the hose center will fill, and the sides will begin to expand. If the hose was placed in water, this would cause water to be pushed away from the sides of the hose — a current of water would be generated at right angles to the hose, even though no water could flow through the sides of the hose!
* We can use the circuit diagram to derive an electrical equivalent equation for the resting potential, which can be as useful as the Goldman-Hodkgin-Katz constant field equation that we learned about in the previous unit, if the assumptions that allow the membrane to be represented using these electrical components are valid.
* The total current flowing across the membrane is the sum of the currents flowing through each of the branches of the equivalent circuit. In symbols, we can state this as <math display="block">
I_m = I_{\text{K}^+} + I_{\text{Na}^+} + I_{\text{Cl}^-}. \quad\text{(Equation 5)}</math>$$
[[image: RestingPotentialFluxes.png|200px|thumb|Figure 6: Balanced fluxes of sodium and potassium ions across the membrane at the resting potential]]
* At the resting potential, the net flux of currents across the membrane is zero, even though the net flux of each of the ions is not zero. Figure 6 shows a schematic diagram for two of the three ions, for simplicity.
* If the fluxes were not in balance, the membrane would depolarize or hyperpolarize, and the change in potential would increase the fluxes of ions further from their Nernst potential, and reduce the fluxes of ions closer to their Nernst potential; the net effect would be to change the membrane potential back to the resting potential.
* If the fluxes are in balance, then the current across the membrane is zero, or <math>$$I_m = 0</math>$$.
* Repeatedly applying '''Equation 2''' to the right-hand side of '''Equation 5''', we obtain <math display="block">
0 = g_{\text{K}^+} \left(V_m - E_{\text{K}^+}\right) + g_{\text{Na}^+} \left(V_m - E_{\text{Na}^+}\right) + g_{\text{Cl}^-} \left(V_m - E_{\text{Cl}^-}\right). \quad\text{(Equation 6)}</math>$$
* A little algebra (which you are encouraged to do yourself) leads to a solution for the membrane voltage in terms of the conductances to the three ions and their Nernst or equilibrium potentials: <math display="block">
V_m = \frac{g_{\text{K}^+} E_{\text{K}^+} + g_{\text{Na}^+} E_{\text{Na}^+} + g_{\text{Cl}^-} E_{\text{Cl}^-}}{g_{\text{K}^+} + g_{\text{Na}^+} + g_{\text{Cl}^-}}. \quad\text{(Equation 7)}</math>$$
* In words, the equation says that the membrane potential is a weighted sum of the equilibrium potentials, in which the weighting factors are the conductance to each of the ions.
* Note that if the conductance to any two ions were to go to zero, the equation would simplify to the Nernst or equilibrium potential for the remaining ion.
