6. Membrane potential and the passive electrical properties of the neuron

Question 1

What are the two types of ion channels?

Answer 1

resting and gated

Question 2

What are the roles of resting and gated ion channels?

Answer 2

Resting channels are primarily important in maintaining the resting membrane potential, the electrical potential across the membrane in the absence of signalling. Some types of resting channels are constitutively open and are not gated by changes in membrane voltage; other types are gated by voltage but can open at the negative resting potential of neurons. Most voltage-gated channels, in contrast, are closed when the membrane is at rest and require membrane depolarisation to open.

Question 3

What causes the membrane potential (at an atomic level)

Answer 3

The neuron’s cell membrane has thin clouds of positive and negative ions spread over its inner (-) and outer (+) surfaces. At rest the extracellular surface of the membrane has an excess of positive charge and the cytoplasmic surface an excess of negative charge

Question 4

How is the membrane potential (Vm) defined as?

Answer 4

Vm = Vin - Vout

where Vin is the potential on the inside of the cell and Vout the potential on the outside.

Question 5

What is the membrane potential at rest called? What previously mentioned variable is it equal to?

Answer 5

The membrane potential of a cell at rest is called the resting membrane potential (Vr). Since by convention the potential outside the cell is defined as zero, the resting potential is equal to Vin.

Question 6

What is the typical range of the membrane potential?

Answer 6

−60 mV to −70 mV

Question 7

How is the direction of current defined?

Answer 7

The direction of current is conventionally defined as the direction of net movement of positive charge.

Question 8

What does this (definition of direction of current) mean in terms of the cations and anions in the cell?

Answer 8

Thus, in an ionic solution cations move in the direction of the electric current and anions move in the opposite direction.

Question 9

Describe the net charge movement of the cell at rest

Answer 9

In the nerve cell at rest there is no net charge movement across the membrane. When there is a net flow of cations or anions into or out of the cell, the charge separation across the resting membrane is disturbed, altering the electrical potential of the membrane.

Question 10

What is meant by depolarisation and hyperpolarisation?

Answer 10

A reduction or reversal of charge separation, leading to a less negative membrane potential, is called depolarisation. An increase in charge separation, leading to a more negative membrane potential, is called hyperpolarisation.

Question 11

What are meant by electrotonic potentials?

Answer 11

Changes in membrane potential that do not lead to the opening of gated ion channels are passive responses of the membrane and are called electrotonic potentials.

Question 12

What kind of responses are typically electrotonic potentials?

Answer 12

Hyperpolarising responses are almost always passive, as are small depolarisations.

Question 13

When is an action potential triggered?

Answer 13

When depolarisation approaches a critical level, or threshold, the cell responds actively with the opening of voltage-gated ion channels, which produces an all-or-none action potential.

Question 14

Describe how the membrane potential can be recorded from a cell

Answer 14

Glass micropipettes filled with a concentrated salt solution serve as electrodes and are placed on either side of the cell membrane. Wires inserted into the back ends of the pipettes are connected via an amplifier to an oscilloscope, which displays the amplitude of the membrane potential in volts. Because the diameter of a microelectrode tip is small (< 1 μm), it can be inserted into a cell with relatively little damage to the cell membrane.

Question 15

How can you know you’re recording from inside the cell?

Answer 15

When both electrodes are outside the cell, no electrical potential difference is recorded. But as soon as one microelectrode is inserted into the cell, the oscilloscope shows a steady voltage, the resting membrane potential. In most nerve cells at rest the membrane potential is approximately −65 mV.

Question 16

How could causal claims be made about changes to the membrane potential?

Answer 16

The membrane potential can be experimentally changed using a current generator connected to a second pair of electrodes—one intracellular and one extracellular. When the intracellular electrode is made positive with respect to the extracellular one, a pulse of positive current from the current generator causes positive charge to flow into the neuron from the intracellular electrode. This current returns to the extracellular electrode by flowing outward across the membrane.

Question 17

How are differences seen between small depolarising currents and larger currents being injected into the cell?

Answer 17

Small depolarising current pulses evoke purely
electrotonic (passive) potentials in the cell—the size of the change in potential is proportional to the size of the current pulses. However, a sufficiently large depolarising current triggers the opening of voltage-gated ion channels. The opening of these channels leads to the action potential, which differs from electrotonic potentials in the way in which it is generated as well as in magnitude and duration.

Question 18

How are differences seen between small hyperpolarising currents and larger currents being injected into the cell?

Answer 18

The responses of the cell to hyperpolarisation are usually purely electrotonic—as the size of the current pulse increases, the hyperpolarisation increases proportionately. Hyperpolarisation does not trigger an active response in the cell.

Question 19

Comment on the concentrations of different ion species inside and outside the cell

Answer 19

Of the four most abundant ions found on either side of the cell membrane, Na+ and Cl− are concentrated outside the cell and K+ and organic anions (A -, primarily amino acids and proteins) inside.

Question 20

What is meant by the equilibrium potential?

Answer 20

The membrane potential at which there is no net flux of the ion species across the cell membrane

Question 21

Roughly describe the distribution of these ions inside and outside of one particularly well-studied nerve cell process, the giant axon of the squid, whose extracellular fluid has a salt concentration similar to that of seawater. Give the exact equilibrium potential for the following:
K+
Na+
Cl-
A-

Answer 21

Species; Conc Cyto (Mm); Conc Extr (Mm); Eq pot
K+; 400; 20; -75
Na+; 50; 440; +55
Cl-; 52; 560; -60
A-; 385; none ; none

Question 22

What is the permeability of a cell membrane to a particular ion species determined by?

Answer 22

The relative proportions of the various types of ion channels that are open.

Question 23

The simplest case is that of the glial cell to demonstrate this permeability. Why is that?

Answer 23

The simplest case is that of the glial cell, which has a resting potential of approximately −75 mV. Like most cells, a glial cell has high concentrations of K+ and negatively charged organic anions on the inside and high concentrations of Na+ and Cl− on the outside. However, most resting channels in the membrane are permeable only to K +.

Question 24

Why is the membrane potential of glial cells negative then?

Answer 24

Because K+ ions are present at a high concentration inside the cell, they tend to diffuse across the mem- brane from the inside to the outside of the cell down their chemical concentration gradient. As a result, the outside of the membrane accumulates a net positive charge (caused by the slight excess of K+) and the inside a net negative charge (because of the deficit of K+ and the resulting slight excess of anions).

Question 25

Why do these potassium ions line the membrane of the extracellular fluid?

Answer 25

Because opposite charges attract each other, the excess positive charges on the outside and the excess negative charges on the inside collect locally on either surface of the membrane

Question 26

What does it mean to say that the flux of K+ out of the cell is self-limiting?

Answer 26

The efflux of K+ gives rise to an electrical potential difference; positive outside, negative inside. The greater the flow of K+, the more charge is separated and the greater is the potential difference. Because K+ is positive, the negative potential inside the cell tends to oppose the further efflux of K+.

Question 27

Thus, how can ions be driven across a membrane?

Answer 27

Thus ions are subject to two forces driving them across the membrane: (1) a chemical driving force, a function of the concentration gradient across the membrane, and (2) an electrical driving force, a function of the electrical potential difference across the membrane.

Question 28

How do these two forces relate to the equilibrium potetial?

Answer 28

Once K+ diffusion has proceeded to a certain point, the electrical driving force on K+ exactly balances the chemical driving force. That is, the outward movement of K+ (driven by its concentration gradient) is equal to the inward movement of K+ (driven by the electrical potential difference across the membrane). This potential is called the K+ equilibrium potential, Ek

Question 29

How is the equilibrium potential related to the resting membrane potential?

Answer 29

In a cell permeable only to K+ ions, Ek determines the resting membrane potential, which in most glial cells is approximately −75 mV.

Question 30

How can the equilibrium potential for any ion X be calculated?

Answer 30

The equilibrium potential for any ion X can be calculated from an equation derived in 1888 from basic thermodynamic principles by the German physical chemist Walter Nernst: Ex = (RT/ zF) (ln [X]o / [X]i) where R is the gas constant, T the temperature (in degrees Kelvin), z the valence of the ion, F the Faraday constant, and [X]o and [X]i the concentrations of the ion outside and inside the cell. (To be precise, chemical activities rather than concentrations should be used.)

Question 31

What is the value of the first part of the equation at 25C? (RT/F)

Answer 31

RT/F is 25 mV at 25°C (room temperature)

Question 32

Describe how you would get the value of -75mV for K+ in a glial cell given that Ko = 20 and Ki = 400

Answer 32

Since RT/F is 25 mV at 25°C (room temperature), and the constant for converting from natural logarithms to base 10 logarithms is 2.3, the Nernst equation can also be written as follows: Ex = (58mV / z) log(Xo/Xi) Thus, for K+, since z = +1 and given the concentrations inside and outside the squid axon: Ex = (58mV / 1) log(20/400) = -75mV

Question 33

How are nerve cells different to glial cells in their permeability?

Answer 33

Unlike glial cells, nerve cells at rest are permeable to Na+ and Cl− ions in addition to K+ ions. Of the abundant ion species in nerve cells, only the large organic anions (A−) are unable to permeate the cell membrane.

Question 34

Describe what would happen with the aforementioned K+ permeable cell with concentration gradients for K+, Na+, Cl-, and A- if a few resting Na+ channels are added to the membrane, making it slightly permeable to Na+.

Answer 34

Two forces drive Na into the cell: Na+ tends to flow into the cell down its chemical concentration gradient, and it is driven into the cell by the negative electrical potential difference across the membrane. The influx of Na+ depolarises the cell, but only slightly from the K+ equilibrium potential (−75 mV). The new membrane potential does not come close to the Na+ equilibrium potential of +55 mV because there are many more resting K+ channels than Na+ channels in the membrane. As soon as the membrane potential begins to depolarise from the value of the K+ equilibrium potential, K+ flux is no longer in equilibrium across the membrane. The reduction in the negative electrical force driving K+ into the cell means that there is now a net flow of K+ out of the cell, tending to counteract the Na+ influx. The more the membrane potential is depolarised and driven away from the K equilibrium potential, the greater is the net electrochemical force driving K+ out of the cell and consequently the greater the net K+ efflux. Eventually, the membrane potential reaches a new resting level at which the increased outward movement of K+ just balances the inward movement of Na+. This balance point (usually approximately −65 mV) is far from the Na+ equilibrium potential (+55 mV) and is only slightly more positive than the K+ equilibrium potential (−75 mV).

Question 35

What is the magnitude of the flux of an ion the product of?

Answer 35

The magnitude of the flux of an ion across a cell membrane is the product of its electrochemical driving force (the sum of the electrical and chemical driving forces) and the conductance of the membrane to the ion ion flux = (electrical driving force + chemical driving force) × membrane conductance.

Question 36

How does this understanding of the ion flux help us understand the balance point of Na+ and K+?

Answer 36

In a resting nerve cell relatively few Na+ channels are open, so the membrane conductance of Na+ is quite low. Thus, despite the large chemical and electrical forces driving Na+ into the cell, the influx of Na+ is small. In contrast, many K+ channels are open in the membrane of a resting cell so that the membrane conductance of K is relatively large. Because of the high relative conductance of Na+ to K+ in the cell at rest, the small net outward force acting on K is enough to produce a K efflux equal to the Na+ influx.

Question 37

What is required of the fluxes in order to have a steady membrane potential?

Answer 37

For a cell to have a steady resting membrane potential, the charge separation across the membrane must be constant over time. That is, at every instant the influx of positive charge must be balanced by the efflux of positive charge. If these fluxes were not equal, the charge separation across the membrane, and thus the membrane potential, would vary continually.

Question 38

Why can this steady leakage of ions cannot be allowed to continue unopposed for any appreciable length of time ?

Answer 38

The passive movement of K+ out of the resting cell through open channels balances the passive movement of Na+ into the cell. However, this steady leakage of ions cannot be allowed to continue unopposed for any appreciable length of time because the Na+ and K+ gradients would eventually run down, reducing the resting membrane potential.

Question 39

How is this dissipation of the ionic gradients prevented?

Answer 39

By the sodium-potassium pump (Na+-K+ pump), which moves Na+ and K+ against their electrochemical gradients: It extrudes Na+ from the cell while taking in K+.

Question 40

How does the pump get the energy for this?

Answer 40

Hydrolisis of ATP

Question 41

Thus at the resting membrane potential the cell is not in equilibrium. How is it decribed?

Answer 41

In a steady state: There is a continuous passive influx of Na+ and efflux of K+ through resting channels that is exactly counterbalanced by the Na+-K+ pump.

Question 42

Loosely describe the structure of the Na+-K+ pump

Answer 42

The Na+-K+ pump is a large membrane-spanning protein with catalytic binding sites for Na+ and ATP on its intracellular surface and for K+ on its extracellular surface.

Question 43

What is the rate of ATP expenditure?

Answer 43

With each cycle the pump hydrolyses one molecule of ATP. (Because the Na+-K+ pump hydrolyses ATP, it is also referred to as the Na+-K+ ATPase.) It uses this energy of hydrolysis to extrude three Na+ ions from the cell and bring in two K+ ions.

Question 44

Why is the ion pump said to be electrogenic?

Answer 44

The unequal flux of Na+ and K+ ions causes the pump to generate a net outward ionic current. Thus, the pump is said to be electrogenic.

Question 45

What effect does the pump have on the membrane potential and what

Answer 45

This pump-driven efflux of positive charge tends to set the resting potential a few millivolts more negative than would be achieved by the passive diffusion mechanisms discussed above.

Question 46

What can change the rate of the ion pump and how?

Answer 46

During periods of intense neuronal activity the increased influx of Na+ leads to an increase in Na+-K+ pump activity that generates a prolonged outward current, leading to a pro- longed hyperpolarising after-potential that can last for several minutes, until the normal Na+ concentration is restored.

Question 47

How can the Na+-K+ pump be inhibited?

Answer 47

The Na+-K+ pump is inhibited by ouabain or digitalis plant alkaloids, an action that is important in the treatment of heart failure.

Question 48

What family does this pump belong to?

Answer 48

The Na+-K+ pump is a member of a large family of pumps known as P-type ATPases (because the phosphoryl group of ATP is temporarily transferred to the pump).

Question 49

What other ion pump in the neuron belongs to this P-type ATPases family? Comment on the extra-intracellular concentration of this ion

Answer 49

P-type ATPases include a Ca2+ pump that transports Ca2+ across cell membranes. All cells normally maintain a very low cytoplasmic Ca2+ concentration, between 50 and 100 nM. This concentration is more than four orders of magnitude lower than the external concentration, which is approximately 2 mM.

Question 50

Where do these calcium pumps pump the calcium? (2)

Answer 50

Calcium pumps in the plasma membrane transport Ca2+ out of the cell; other Ca2+ pumps located in internal membranes, such as the smooth endoplasmic reticulum, transport Ca2+ from the cytoplasm into these intracellular Ca2+ stores.

Question 51

What is the rate of the Ca2+ pump?

Answer 51

Calcium pumps are thought to transport two Ca2+ ions for each ATP molecule that is hydrolyzed, with two protons transported in the opposite direction.

Question 52

Describe the molecular structures of the two pumps

Answer 52

The Na+-K+ pump and Ca2+ pump have similar structures. They are formed from 110 kD α-subunits, whose large transmembrane domain contains 10 membrane- spanning α-helixes

Question 53

Most neurons have relatively few Ca2+ pumps in the plasma membrane. How then is Ca2+ typically transported from the cell?

Answer 53

Ca2+ is transported out of the cell by the Na+-Ca2+ exchanger

Question 54

Describe how this Na+-Ca2+ exchanger is different to that of the other transporters

Answer 54

This membrane protein is not an ATPase but a different type of molecule called a co-transporter. Co-transporters move one type of ion against its electrochemical gradient by using the energy stored in the electrochemical gradient of a second ion.

Question 55

Describe how the Na+-Ca2+ exchanger works including its rate

Answer 55

The electrochemical gradient of Na+ drives the efflux of Ca2+ . The exchanger transports three or four Na+ ions into the cell (down the electrochemical gradient for Na+) for each Ca2+ ion it removes (against the electrochemical gradient of Ca2+).

Question 56

Why is the Na+-Ca2+ exchanger known as an antiporter?

Answer 56

Because Na+ and Ca2+ are transported in opposite directions, the exchanger is termed an antiporter.

Question 57

Why is ion flux driven by cotransporters often referred to as secondary active transport?

Answer 57

Ultimately, it is the hydrolysis of ATP by the Na+-K+ pump that provides the energy (stored in the Na+ gradient) to maintain the function of the Na+-Ca2+ exchanger. For this reason, ion flux driven by cotransporters is often referred to as secondary active transport, to distinguish it from the primary active transport driven directly by ATPases.

Question 58

So far, for simplicity, we have ignored the contribution of an ion to the resting potential. Which is this?

Answer 58

In most nerve cells the Cl− gradient is controlled by one or more active transport mechanisms so that ECl will differ from Vr . As a result, the opening of Cl− channels will bias the membrane potential toward its Nernst potential.

Question 59

How are chloride ions typically transported?

Answer 59

Chloride transporters typically use the energy stored in the gradients of other ions—they are cotransporters.

Question 60

Cell membranes contain a number of different types of Cl− cotransporters. Describe one type of cotransporter which is a symporter that transports Cl- into the cell and why it is a symporter

Answer 60

Some transporters increase intracellular Cl− to levels greater than those that would be passively reached if the Cl− Nernst potential was equal to the resting potential. In such cells ECl is positive to Vr so that the opening of Cl− channels depolarises the membrane. An example of this type of transporter is the Na+-K+-Cl− cotransporter. This protein transports two Cl− ions into the cell together with one Na+ and one K+ ion. As a result, the transporter is electroneutral. The Na+-K+-Cl− cotransporter differs from the Na+-Ca2+ exchanger in that the former transports all three ions in the same direction—it is a symporter.

Question 61

How is the Cl- gradient determined in most neurons?

Answer 61

In most neurons the Cl− gradient is determined by cotransporters that move Cl− out of the cell. This action lowers the intracellular concentration of Cl− so that ECl is typically more negative than the resting potential. As a result, the opening of Cl− channels leads to an influx of Cl− that hyperpolarises the membrane.

Question 62

Give an example of a cotransporter that moves calcium out of the cell

Answer 62

The K+-Cl− cotransporter is an example of such a transport mechanism; it moves one K+ ion out of the cell for each Cl− ion it exports.

Question 63

Describe the relationship between Na+-K+-Cl− and K+-Cl− transporters

Answer 63

Interestingly, in early neuronal development cells tend to express primarily the Na+-K+-Cl− cotransporter. At this stage the neurotransmitter γ-aminobutyric acid (GABA), which activates Cl− channels, typically has an excitatory (depolarising) effect. As neurons develop they begin to express the K+-Cl− cotransporter, such that in mature neurons GABA typically hyperpolarises the membrane and thus acts as an inhibitory neurotransmitter.

Question 64

How can pathologies affect these Cl- transporters?

Answer 64

In some pathological conditions in adults, such as certain types of epilepsy or chronic pain syndromes, the expression pattern of the Cl− cotransporters may revert to that of the immature nervous system. This will lead to aberrant depolarising responses to GABA that can produce abnormally high levels of excitation.

Question 65

Describe what happens to the Na+ and K+ channels as the threshold for an AP is approached

Answer 65

However, this balance changes when the membrane is depolarised toward the threshold for an action potential. As the membrane potential approaches this threshold, voltage-gated Na+ channels open rapidly. The resultant increase in membrane conductance to Na+ causes the Na+ influx to exceed the K+ efflux once threshold is exceeded, creating a net influx of positive charge that causes further depolarisation. The increase in depolarisation causes still more voltage-gated Na+ channels to open, resulting in a greater influx of Na+, which accelerates the depolarisation even further. This regenerative, positive feedback cycle develops explosively, driving the membrane potential toward the Na+ equilibrium potential of +55 mV

Question 66

Does Vm ever reach +55mV? Why/ why not?

Answer 66

The membrane potential never quite reaches ENa because K+ efflux continues throughout the depolarisation. A slight influx of Cl− into the cell also counteracts the depolarising effect of the Na+ influx.

Question 67

How close does Vm get to ENa?

Answer 67

So many voltage-gated Na+ channels open during the rising phase of the action potential that the cell membrane’s Na+ conductance is much greater than the conductance of either Cl or K . Thus, at the peak of the action potential the membrane potential approaches the Na+ equilibrium potential, just as at rest (when permeability to K+ is predominant) the membrane potential tends to approach the K+ equilibrium potential.

Question 68

Why does the membrane potential remain at this large positive value near ENa indefinitely? (2)

Answer 68

First, following the peak of the action potential the voltage- gated Na+ channels gradually close by the process of inactivation. Second, opening of the voltage-gated K+ channels causes the K+ efflux to gradually increase. The increase in K+ conductance is slower than the increase in Na+ conductance because of the slower rate of opening of the voltage-gated K+ channels. The slow increase of K+ efflux together with the decrease in Na+ influx produces a net efflux of positive charge from the cell, which continues until the cell has repolarised to its resting membrane potential.

Question 69

As a general rule, when Vm is determined by two or more species of ions, the contribution of one species is determined not only by the concentrations of the ion inside and outside the cell. What else should be taken into account

Answer 69

The ease in which the ion crosses the membrane

Question 70

What is a measure of the ease at which an ion crosses the membrane?

Answer 70

One convenient measure of how readily the ion crosses the membrane is the permeability (P) of the membrane to that ion, which has units of velocity (cm/s). This measure is similar to that of a diffusion constant, which determines the rate of solute movement in solution driven by a local concentration gradient.

Question 71

Therefore, how is Vm calculated?

Answer 71

The Goldman equation: Vm = ( RT / F ) ( ln ( Pk[K+]o +PNa[Na+]o +PCl[Cl–]i ) / ( Pk[K+]i+PNa [Na+]i+PCl [Cl–]o ) ) R = gas constant T = Temperature in Kelvin F = Faradays constant Pk = permeability of potassium [K+]o = amount of potassium outside the cell [K+]i = amount of potassium in the cell

Question 72

When does the Goldman equation apply?

Answer 72

This equation applies only when Vm is not changing.

Question 73

What does the Goldman equation state about the concentration of an ion species?

Answer 73

It states that the greater the concentration of an ion species and the greater its membrane permeability, the greater its contribution to determining the membrane potential.

Question 74

What is the relationship between the Nernst equation and the Goldman equation?

Answer 74

In the limit, when permeability to one ion is exceptionally high, the Goldman equation reduces to the Nernst equation for that ion. For example, if PK >> PCl or PNa, as in glial cells, the equation becomes as follows: Vm =~ RT / F ln [K+]o / [K+]i

Question 75

Alan Hodgkin and Bernard Katz used the Goldman equation to analyse changes in membrane potential in the squid giant axon. They measured the variations in membrane potential in response to systematic changes in the extracellular concentrations. What did they find?

Answer 75

They found that if Vm is measured shortly after the extracellular concentration is changed (before the internal ionic concentrations are altered), [K+]o has a strong effect on the resting potential, [Cl−]o has a moderate effect, and [Na+]o has little effect.

Question 76

With what ratios could the membrane at rest be fit accurately by the Goldman equation according to Hodgkin and Katz?

Answer 76

The data for the membrane at rest could be fit accurately by the Goldman equation using the following permeability ratios: PK : PNa : PCl = 1.0 : 0.04 : 0.45.

Question 77

With what ratios could the point in which Vm is not changing at the peak of an AP be fit accurately by the Goldman equation according to Hodgkin and Katz?

Answer 77

At that point the variation of Vm with external ionic concentrations is fit best if a quite different set of permeability ratios is assumed: PK : PNa : PCl = 1.0 : 20 : 0.45. For these values of permeability the Goldman equation approaches the Nernst equation for Na+

Question 78

Why is the use of the Goldman equation limited? (2)

Answer 78

The usefulness of the Goldman equation is limited because it cannot be used to determine how rapidly the membrane potential changes in response to a change in permeability. It is also inconvenient for determining the magnitude of the individual Na+, K+, and Cl− currents.

Question 79

How can this information (speed of Vm changes in response to P; magnitude of currents) be calculated?

Answer 79

This information can be obtained using a simple mathematical model derived from electrical circuits. The model, called an equivalent circuit, represents all of the important electrical properties of the neuron by a circuit consisting of conductors or resistors, batteries, and capacitors.

Question 80

What is the first step in developing an equivalent circuit?

Answer 80

To relate the membrane’s discrete physical properties to its electrical properties.

Question 81

What is denoted as Cm in a circuit model of a neuron?

Answer 81

The lipid bilayer, which endows the membrane with electrical capacitance, the ability of an electrical nonconductor (insulator) to separate electrical charges on either side of it. The nonconducting phospholipid bilayer of the membrane separates the cytoplasm and extracellular fluid, both of which are highly conductive environments. The presence of a thin layer of opposing charges on the inside and outside surfaces of the cell membrane, acting as a capacitor, gives rise to the electrical potential difference across the membrane.

Question 82

How is the electric potential difference or voltage across a capacitor calculated?

Answer 82

V = Q / C Where Q is the net excess positive or negative charge on each side of the capacitor and C is the capacitance.

Question 83

In what units are capacitance and charge measured?

Answer 83

Capacitance is measured in units of farads (F), and charge is measured in coulombs (where 96,500 coulombs of a univalent ion is equivalent to 1 mole of that ion)

Question 84

What does a charge separation of 1 coulomb across a capacitor of 1F produce?

Answer 84

A charge separation of 1 coulomb across a capacitor of 1 F produces a potential difference of 1 volt.

Question 85

What is the typical values of membrane capacitance of a nerve cell approximately? What does this mean for the charges required for a potential difference?

Answer 85

A typical value of membrane capacitance for a nerve cell is approximately 1 μF per cm2 of membrane area. Very few charges are required to produce a large potential difference across such a capacitance. For example, the excess of positive and negative charges separated by the membrane of a spherical cell body with a diameter of 50 μm and a resting potential of −60 mV is 29 × 10^6 ions. Although this number may seem large, it represents only a tiny fraction (1/200,000) of the total number of positive or negative charges in solution within the cytoplasm. The bulk of the cytoplasm and the bulk of the extracellular fluid are electroneutral.

Question 86

The membrane is a _____ capacitor. Why?

Answer 86

The membrane is a leaky capacitor because it is studded with ion channels that can conduct charge. Ion channels endow the membrane with conductance and with the ability to generate electromotive force (emf). The lipid bilayer itself has effectively zero conductance or infinite resistance. However, because ion channels are highly conductive, they provide pathways of finite electrical resistance for ions to cross the membrane.

Question 87

How can we represent K+ channels in an equivalent circuit?

Answer 87

In an equivalent circuit we can represent each K+ channel as a resistor or conductor of ionic current with a single-channel conductance γK (conductance = 1/resistance).

Question 88

How and why is Cl different to K and Na in the Goldman equation

Answer 88

The i (concentration inside the neuron) is on top of the equation while the o is on top. This is the reverse of the others because Cl is negatively charged so it has an opposite effect o Vm