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I N N 0 V A T I 0 N S A N D I D E A S

HYDRAULIC ANALOGS AS TEACHING TOOLS FOR

BIOELECTRIC POTENTIALS

Joaquim Procopio

Departamento de Fisiologia e Biofisica, Institute de Ciencias Biomedicas da Universidade de Sao Paula, Sao Paula, Brazil

H ydraulic analogs of bioelectrical potentials are proposed as teaching tools in helping students with no formal background in physics and mathematics.

Membrane capacitance is simulated by a water reservoir, V, whose variable level is the membrane potential. Resting membrane potential is simulated by a large capacitance reservoir of fixed level, connected to reservoir V through a tube having the role of the electrical conductance of the membrane. Injection of electrical current

into the cell is simulated by injection of water into the membrane capacitance reservoir. Reversal potentials of the end-plate potential are simulated by another water reservoir, with fixed level (analogous to the reversal potential of the end-plate potential) connected to the membrane capacitance reservoir through a tube corresponding to the acetylcholine-activated ion channels. Different phases of the synaptic

potential are then described using the hydraulic analogs. Hydraulic analogs have proved, in our experience, to be an efficient tool in complementing the already established electrical equivalents. AM. J. PHYiSIOL. 267 (ALWe PHYSIOL. EDUC. 12): S65-S76, 1994.

Key words: cell membrane potential; equivalent electrical circuits; bioelectrogenesis;

cell electricity

The teaching of bioelectricity to students having no formal background in physics and mathematics inevi- tably encounters some barriers. For those students, the usually employed electrical analogs do not, in our experience, gather sufficient intuitive appeal to fully convey the physicochemical details of ion transport and consequent voltage generation across the cell membrane. One of the particularly difficult concepts for the beginner is that of membrane capacitance. In our experience, the use of an electrical equivalent circuit containing a capacitor does not provide significant additional help in the way of explaining the capacitive behavior of the cell membrane.

We

hyd

have .raulic

been employing, for the last six years, analogs in concert wi th el .ectrical equiva-

lents. Although we still lack a statistically based comparison, we have experienced in all courses immediate acceptance of the hydraulic analogs. Students failing to grasp either the biological potential proper or its corresponding electrical equivalent circuit readily understood and accepted the hydraulic representation. When later confronted with the electrical equivalent, students have invariably re- ported better acceptance.

In the following we describe our approach to teaching the genesis of membrane potentials, perturbations of the resting potential, and some electric

tentials involved in cell excitability as applications of the use of the hydraulic analogs. This text is designed both to introduce the above strategy and to be used as material for a self-tutorial. We envision

1043 - 4046 / 94 - $3.00 - COPYRIGHT o 1994 THE AMERICAN PHYSIOLOGICAL SOCIETY

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this material being used by teachers and students as well. As such, the language is kept as simple as possible, and understandability is given priority over precision.

For the medical course we usually present the electrical equivalents before the hydraulic ones. This order has helped us to compare the efficiencies of both strategies and to conclude that the hydraulic analogs are more efficient. We recommend this order of presentation.

References are kept to a minimum because most concepts discussed here are deeply established in the literature. The textbook Principles of Neural Science, edited by Kandel and Schwartz (4) constitutes our main source and is recommended as an adjunct text.

MEMBRANE POTENTIAL AND RESTING POTENTIAL: GENESIS OF THE MEMBRANE POTENTIAL

As model cell we employ a typical mammal neuron. We begin by assuming that an idealized electrically neutral sodium-potassium (Na-K) pump with stoichiometry of 1: 1 (actual pump is electrogenic with stoichiometry of 32) maintains an asymmetry of Na+ and K+ concentrations across the cell membrane, such that intracellular (ic) and extracellular (ec) [ Na+] and [K+] are typically: [ Na]i, = 8- 15 (variable), [Na],, = 140, [K]i, = 140, and [I&, = 5 meq/l. We also ignore contributions of the Donnan phenomenon (7) to the genesis of membrane potential. This is clearly an oversimplification, justifiable by what follows.

It is also assumed that the cell membrane has channels selective to Naf and K+ (4). Cl- channels and channels for other ions will not be considered in a first approach because their contribution to the resting membrane potential is smaller. To further simplify matters we begin by working with two hypothetical types of cells: the Naf cell, having only Na+ channels in its membrane, and the K+ cell, having only K+ channels in its membrane. Consider first the Na+ cell, where Na ions are distributed as

[N 1 a ic = 5 (this is lower than the actual values of S-15) and [ Nalec = 140 meq/l.

Let us focus on the surroundings of a single Na+ channel. Na ions inside the channel “feel” a diffusional force (FdiE) resulting from the Naf concentration gradient across the membrane (7). This force tends to drive them toward the cytoplasm because [Na],, > [Na]i,. In consequence, Na ions move in, leaving outside “unpaired” anions and separating positive from negative charges. The charge separa- tion generates an electric field (E) inside the channel. As more and more Na ions penetrate the cell, thermodynamic equilibrium is eventually attained, and then the diffusional force acting on Na+ [FdiH = dpNal& where PNa is the Naf chemical potential (7)] is balanced by an electrical force (& = #, where 4 is the ionic charge). In this situation E’diB = &. This reasoning can be extended to any number of Na+ channels. When equilibrium is reached, a stable potential difference (PD) is established across the membrane (Fig. IA).

Grounding the extracellular medium makes the intracellular voltage numerically equal to the PD across the membrane, and its algebraic value is defined as the membrane potential o/in).

The PD that balances the diffusional force acting on Na+ is known as the Na+ equilibrium potential (ENa) and is given by the Nernst equation (7)

RT LNalic 5 E Na = ---&np-

P 1 a - ec -0.026 x In= = +O.O86v

whereR is 8.31 J.mol- l- ‘K-l, T is 300°K, F is 96,500 C/mol, and z is the valency. In this case, z = + 1. The same reasoning can be applied to the K+ cell, whose membrane has only K+ channels and in which K ions distribute as [K]i, = 140 and [K],, = 5 meq/l. The potassium equilibrium potential is given by the Nernst equation as

lU LKlic 140 EK = -

GP - = -0.026 x In5 = -0.086V PI ec

In the Na+ cell, ENa can be represented by a battery with electromotive force (emf) = +86 mV, pointing

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No+ hsmi - cell K+ hemi - cell

B 9 9

No+ hemi - cell K+ hemi - cell

FIG. 1 A: hypothetical Na+ and K+ cells showing Na+ and K+ channels and corresponding acting forces, Fd (diffusional) and Fe (electrical), shown equilibrated in each hemicell. Concentration asymmetry of Na and K ions, indicated in top part of cells, is guaranteed by the Na-K pump (not shown). Na+ cell is positive, and K+ cell is negative. In real cells, [Na]i, is higher, - 8-10 meq/l. B: electrical equivalent of Na+ and K+ cells before “fusion.” Batteries ENa and EK represent Nernst potentials of Na and K ions. RNa and RK represent resistance of Na+ and K+ channels. Na and K ions are in electro- chemical equilibrium across corresponding cell mem- branes, and consequently Vm = Eion for each cell.

inward as shown in Fig. 1B. In the K+ cell, EK is represented by a battery with emf = -86 mV, pointing outward (Fig. 1B). This means that the Na+ cell has an excess of positive charge in its cytoplasm, whereas the K+ cell has a negative excess charge.

Because real cells have both NaS and K+ channels in their membrane (other channels will be mentioned later), we improve our model by “fusing” the Na+ cell to the K+ cell, as shown in Fig. 2A, forming a Na-K cell. When the cells fuse, part of their corre-

sponding cytoplasmic excess charges migrates to the opposite hemicell, decreasing the magnitude of both polarities. Part of the positive charges in excess in the Na+ hemicell migrates to the K+ hemicell, and vice versa. This happens until a new stationary value is reached.

As a consequence of the above charge redistribu- tion, the average charge excess in the Na-K cell levels off at a value that produces a cytoplasmic voltage between ENa = +86 mV and EK = -86 mV This means that, in the Na-K cell, Na and K ions in their respective channels now lose their previous equilibrium. In both the Na+ and K+ hemicells, the diffusional forces now overweight their electrical counterparts (Fig. 2A). Because equilibrium is lost, the system now tends to a stationary voltage known as the resting potential (E,), whose value we estimate next.

In order that a steady-state membrane potential be reached in our hypothetical Na-K cell, Na+ and KS currents must be equal and opposite. These ionic currents are given by

iNa = (E m - ENa)GNa and & = (Em - EK)G (3)

where GNa and GK are the NaS and K+ conductances (l/Rio*) given by Gion = gion X NOpen. The conductance gion is the single-channel or unitary conductance of a NaS or K+ channel, and NOpen is the number of open channels of a given type per unit membrane area. Because the rates of NaS entry and K+ exit are numerically equal, iNa = -iK. Equating the right-hand side (RHS) of both equations above, we obtain Em

E Em =

Na * GNa + EK * GK

GNa + GK (9

Using ENa = +86mV,E,= -86 mV, and assuming that GK = 20 GNa, Eq. 4 gives Em = -77.8 mV.

The diffusional forces acting on Na+ and KS can more conveniently be expressed in units of voltage as, respectively, E;y$ = ENa = + 86 mV and FFiR = EK = -86 mV. Because [ Na],, > [ Na]ic, Fzg is directed toward the cytoplasm, and because [K], > [K],,, ~~~~ is directed outward (Fig. 2C).

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A

t Fe I

Fd


C Fd


FIG. 2 A: balance of electrical and diffusional forces after fusion of tive force values of batteries ENa and Ex, and its value is i = Na+ to the K+ hemicell. Diffusional forces now exceed (J!& + &)/(RNa + Rx). C: concept of driving forces (DF) in electrical counterparts in each hemicell. Intracellular poten- steady-state Na-K fused cell (see text). Driving forces are tial of each hemicell is smaller in absolute value than defined as DFion = Vm - Eion for each hemicell. Membrane corresponding Nernst potentials. Electrical current flows potential has attained stationary value of -78 mV, as calcu- from positive (Na+) to negative (K+) hemicell. B: equivalent lated from Eq. 4. Asymmetry of Na+ and K+ concentrations electric circuit of fused Na-K hemicells (see text). Circulating across the membrane is maintained by the Na-K pump. current is driven, counterclockwise, by summed electromo- Electric current flows continually from Na+ to K+ hemicell.

The corresponding driving forces (DF) acting on Na+ and K+ will be DF,, = Vm - ENa and DFK = Vm - EK (6). In the numerical example, DFN, = -78 -(+86) = - 164 mV (inward) and DFK = Vm - EK = -78 - (-86) = +8 mV (outward) (Fig. 2C).

The collection of Na+ and K+ channels with their respective asymmetrical distribution can now be reduced to a single battery with emf = A!?~, pointed outward from the cell, in series with a conductance G, given by G, = GNa + GK + GCl + Gother ions cFi$?* 3).

Because, in the resting condition, iNa (in) = iK (out), there is no net current across the circuit element e-n -R,. This implies that Vm = Em.

APPLICATION 1 HYDRAULIC ANALOG OF THE RESTING MEMBRANE POTENTIAL

We consider a cylindrical and vertical hydraulic reservoir, V, which can be alternately connected to two other reservoirs, designated by the symbols “Na” and “K”, as shown in Fig. 4. We assume that the level of reservoir Na is kept fixed at a value designated by ENa, which is 86 meters above sea level. (Sea level is assumed to be the zero reference level.) The level of reservoir K, EK, is kept fixed at 86 meters below sea level. Many students ask at this point how the levels of reservoirs Na and K could be kept fixed. Despite this not being relevant for the subsequent understanding, one might picture the reservoirs as being very wide. In the cell case it is the Na-K pump that maintains [ K]i, and [Na], constant and, consequently, ENa and EK. The level of reservoir V, having no fixation mechanism, depends totally on the particular connections it makes with its neigh- bor reservoirs. The variable level of V is designated by Vm. To obtain an intuitive idea of how level Vm is modulated, we can play with the section of the connecting tubes. For example, if tube GK is nar- rowed, the level of V will tend to ENa = +86 meters,


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P and I E m

FIG. 3 Electrical equivalent circuit of Fig. 24 showing battery E, in series with resistance R,. In absence of external electrical perturbations, there is no current flowing across the system J&-R,, and consequently Vm = Em.

the level of reservoir Na, whereas total closure of GK will make Vm = I&. On the other hand, if tube GNa narrows, level Vm will tend to & = -86 meters, the level of reservoir K.

Along the above reasoning we conclude that, when tubes GNa and GK are both open, level Vm will attain a steady-state value somewhere between +S6 m and -86 m. At this stationary level, the flow of water entering V from reservoir Na is equal to the flow leaving V for reservoir K. Our task is now to calculate the stationary level of reservoir V.

We can generalize the above observation by stating that level Vm is a function of levels I& and EK and also of the relative values of the hydraulic conductances GNa and GK. To quantitate the above state- ment, we impose the condition that level Vm is constant in time. Applying the law of conservation of water, we have

Flow of water entering reser- Flow of water leav-

c-7 -

voir V from reservoir Na - ing reservoir

V to reservoir K

Because

Flow of water = Pressure difference

x Hydraulic conductance

Pressure difference = Level difference

we have

(E Na - Vm) X GNa = (Vm - Ed X G,

Pressure Hydraulic Pressure Hydraulic

difference conductance difference conductance

from which Vm is given by

E vm =

Na ’ GNa + EK * GK

GNa + GK (5)

Note that Eq. 5 is equivalent to Eq. 4.

Because Vm is now stationary and unperturbed, we call it the resting level of reservoir V and designate it byEm. Note that Em is the geometric mean OfENa and EK, where the “weights” are the hydraulic conductances GNa and GK. That reservoir, having the largest conductance, dominates the level of V. Also, because GNa and GK appear both in the numerator and denominator of Eq. 5, only their relative values count.

Let us assume, as a numerical example, that ENa = +86 meters, EK = -86 meters, and GK = 20 GNa.

Reservoir Na

Reservoir v

FIG. 4

Reservoir K

Hydraulic analog of membrane potential showing reservoirs Na, K, and V connected by tubes GNa and Gx. Note that conductance of tube GK is appreciably larger than that of tube GNa. Fixed levels of reservoirs Na and K are, respectively, EK = +86 meters and EK = -86 meters in relation to “sea level.” Level Vm, which is not fixed, is maintained between ENa and Ex, but much closer to level EK (see text).


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Applying the above figures to Eq. 5, we obtain

vm = E, = 86 x 1 + (-86 x 20)

1 + 20 = -77.8 meters

The consistency of Eq. 5 can now be checked in three situations. Making GK = 0 to simulate closure of tube GK and leaving open tube GNa, we find that Eq. 5 gives Vm = ENa. Making GNa = 0 to simulate closure of tube GNa and leaving open tube GK, we obtain Vm = EK. Finally, making GNa = GK, we get Vm = (ENa + &/2 = 0, which is the arithmetic mean of ENa and EK.

The system consisting of reservoirs Na+ and K+ and tubes GNa and GK can be reduced to a much simpler equivalent system consisting of a single reservoir E, connected to V by a single tube of conductance G,. This new equivalent reservoir will then have a level Em given by Eq. 5 and will be connected to V by a tube whose conductance is G, = GNa + GK + Gel + G other ions (Fig. 5). The following recipe can be used to make the substitutions: 1) Close the connections of V with reservoirs NaS and K+. 2) Exchange reservoirs Naf and K+ for a reservoir E, with fixed level Em given by Eq. 5.3) Connect reservoir E to reservoir V using a tube of conductance G, = GNa + Gk + Gel + G other ions* If the above recipe is followed, reservoir V connections to reservoirs Na and K by tubes GNa and

zero level

I Em I “rn I

Reservoir E

Reservoir V

FIG. 5 Reduced hydraulic equivalent of reservoirs Na and K and respective connecting tubes (shown in Fig. 4). Here, a single reservoir E, of ftxed level Em, is connected to V by tube G,. This is equivalent to set of reservoirs Na and K and their respective connecting tubes. In absence of external perturbations, levels Em and Vm are equal.

GK or to reservoir E by tube G, will not be distin- guishable. In the absence of external perturbations (to be defined below), the level of reservoir V stays constant and equal to the level of E. This unperturbed steady-state level is called the resting level of reservoir V.

PERTURBA’I’IONS OF THE CELL RESTING POTENTIAL

We now come back to the cell system, where the resting potential is represented by a battery Em in series with a resistance R, (Fig. 3). Electrical perturbations of this stationary state can now be easily studied. A perturbation will be defined as an injection of positive electrical charges into or aspiration of positive charges from the cytoplasm. This is usually performed by means of a micropipette impaling the cell membrane, as shown in Fig. 6. If a positive current is injected into the cell, part of newly entering charges accumulates into the cytoplasm, decreasing its previous negativity and partially discharging the membrane capacitance. How- ever, as soon as Vm becomes less negative than the fixed Em, charges begin to leak from the cell through the circuit element Em - R,. The fraction of the injected current that accumulates into the cell constitutes the capacitive current, whereas that fraction leaking from the cell through Em - R, is the

P IE m

FIG. 6 Micropipette impaling a cell, as an example of an electrical perturbation of cell resting potential. The electrical current introduced by pipette splits into 2 portions, capacitive (iC) and resistive (il- or &) currents. At any instant T, Vm = Em - R, X i, = E, - 1 i-r CJ 6 i,dt.


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resistive current (4)

Capacitive Resistive Injected

current current current

r dQ E, - vm

inj = dt + R,

(6)

where dQ/dt = Cm(dV,/dt).

If the injected current is kept constant for some time, the capacitive current will gradually decrease while the resistive current will gradually increase, until the resistive current equals the current being injected by the pipette. At this moment the membrane potential attains a stationary value defined by dVJdt = 0 and

I. vm - E,

inj = i, 1

Rm (7)

where i, is the resistive component of the current. Between the beginning of current injection and the steady state there is a time interval in which the membrane potential changes with a rate given by

dvm dQ 1 1 ----= dt - dt Cm (I inj - im> c (8)

m

where dQ/dt is the rate at which positive charges accumulate in the membrane capacitance. From Eq. 8, it is clear that as im increases with time, the term dQ/dt decreases, until eventually dQ/dt = dVm/dt = zero. At any time we have

im = (vm dvm

- E,)G, and i, = Cm dt = Iinj - i, (9)

At any instant (2) after the beginning of the current injection, the membrane potential is given by

vm AQ

=Em-c m

where AQ = JoT(Iinj - im)dt is the net amount of charge delivered to the cytoplasm. Alternatively Vm can be expressed as Vm = Em - R, x i,, where the term Rm X im is the ohmic voltage drop across resistance R,. When the steady state is reached, then

im = li,i and Vm = Em - R, X Iinj. This last expression shows that the membrane potential can be maintained indefinitely at any desired value (within limits) by means of the continuous injection of current: if the current is injected into the cell by the pipette, it exits the cytoplasm through Rm, and the voltage drop im X Rm subtracts voltage from the battery E,. In this case we say the current depolar- izes the membrane. When current is aspirated from the cytoplasm by the pipette, charges enter the cell through Rm, and the voltage drop Rm X im hyperpo- larizes the membrane.

APPLICATION 2 HYDRAULIC ANALOG OF THE MEMBRANE POTENTIAL PERTURBATIONS

As we saw in Fig. 5, the steady-state level of reservoir V in the absence of external perturbations is numerically equal t0 Ema A perturbation of Vm can be pictured as an injection of water (by opening a faucet, for example) directly into reservoir V, as shown in Fig. 7. This is analogous to injecting an electrical current directly into the cytoplasm. To simplify things, we assume the faucet is either closed or fully open, with constant flow I. Immediately after the faucet opens, the water injected into reservoir V

I I I rn Faucet

zero level I III

I I

Vm

Reservoir E

Gm

FIG. 7

Reservoir V

Hydraulic model of perturbation of resting potential produced by water injection into reservoir V. Constant water flow, I, injected by faucet into reservoir V, has 2 fates. Part of the water leaks to reservoir E (flow im) and part accumulates in reservoir V, increasing its level at a rate i,.


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splits into two portions. A first part of the injected water stays in V, increasing its level. However, as soon as Vm increases above Em, water begins to leak from reservoir V to reservoir E through tube G,. We can then write that the total flow being injected, Itotal (or I>, is made up of

I = wm total

Adt

+ Y-n - 4-n K-n

w

Total flow Capacitive flow Resistive flow

The term A(dVJdt) in Eq. 10 is equal to the rate of water accumulation in V. Observe that this term is equal to the rate of change of total water volume in V and is given by the base area of the reservoir (A) multiplied by the rate of change of Vm with time. This component is called the capacitive flow, because it measures that part of the water flow accumu- lating in the capacitance of the reservoir.

Note also that the capacitive flow does not go through any tube but simply accumulates into V. The second term on the RHS ofEq. 10 is that portion of the water leaking from V to E through G,. This portion is called resistive flow, to convey the idea that it goes through a hydraulic resistance (or conductance). Equation 10 is a differential equation whose solution is

(analog to the depolarizing action of the electric current in the cell membrane).

Equation 12 shows also that Vm can be kept indefinitely at a level lower than Em, so long as we continuously aspirate water from V. Such perturbation makes Vm depart from zero and corresponds to a hyperleveling action. This corresponds in the cell to a hyperpolarization of the membrane.

Between the two extreme instants, zero and infinite, after an injection of water in reservoir V, the level Vm follows an exponential curve given by Eq 10 and shown in Fig. 8. When t = RA, Vm = Em + Ia

R, (1-W) = Em + Ia R, (0.63). Remembering that vm = Em in time 0, then in t = RA, Vm attains 63% of its final value. The product R x A has dimension of time and is the time constant of the system (reservoir V + tube Gm).

From the above analysis, it is clear that no matter how intense the water flow is into reservoir V, its level does not rise immediately but has to follow an exponential curve. Also, when the faucet is suddenly closed after being open for some time, level Vm does not decay instantly but also follows an exponential.

It is illustrative to check the consistency of Eq. IO in two instants after the perturbation initiates. At time 0, we get Vm = Em, as expected. On the other hand, at t = m, we get

This last result indicates that, long after the faucet opens over V, Vm levels off at (I X R,) meters above Em. As long as the faucet stays open, Vm will stay constant at the value Em + IX,. If the injected flow1 increases, so does the product I-R,, and the level Vm will stay higher above Em. This is a useful conclusion because it says that, by keeping the flow1 at different values, the steady-state value of Vm can be kept at any desired level. The larger 1 is, the closer to zero will be the steady-state level of Vm, and consequently the more intense will be the deleveling action of 1

Time in ZOO msec intervals

FIG. 8 Exponential curve, drawn according to Eq. 22, for a hypothetical cell having R, = 1 x lo6 fi and C, = 1 x 10e6 F and a time constant of 1 s. Cell received a current injection of 2 X 10e5 A, beginning at time 0.


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Whe n Vm is constant but different that Vm is stationary and perturbed.

APPLICATION 3 END-PLATE POTENTLAL

The hydraulic model can now be employed to analyze the interplay of currents and potentials during the end-plate potential (EPP) at the neuro- muscular junction. Modifying slightly our strategy, electrical and hydraulic events will from now on be discussed in parallel. The following is a brief and downright simplified description of the biology/ biophysics of the end-plate potential in a mammal.

from Em, we say

When an action potential arrives at the end-plate region, it triggers a sequence of events that culmi- nates in the opening of a great number of acetylcholine (ACh)-activated ionic channels in the postsynaptic membrane. These channels are selective to both Na and K ions and about equally selective to them (2). Thus we may say that the electrical conductance for Naf and KS is about the same in these channels, and we write GTi = Gr (although in actuality these conductances are slightly different).

The set of ACh-activated channels can be reduced, from an electrical standpoint, to a single battery with emf = Es or Eepp in series with a conductance G,. Because Na+ and K+ move in parallel inside each channel, Es is given by the geometric mean of ENa and EK, where the weights are the conductances Ggt and Gr (see Eq. 5)

(13)

The equivalent channel conductance is Gs = GT, + GF (Fig. 9). Because ENa and EK have about the same modulus (86 mv) and opposite polarities, and

Ga = Gk, this gives Es = 0 in Eq. 13. Therefore, from an electrical point of view, the set of ACh- activated channels in the postsynaptic membrane (PSM) will be taken as equivalent to a battery, with emf - - Es - - and Gs = GT, + Gr (Fig. 9A).

internal conductance

The value Es is sometimes called the reversal potential of the ACh-activated channels. It corresponds to the value of membrane potential at which, under

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“:‘ili.i:‘:::i:I.:.:.: .,.,. ..::i:i:~$iii;;$: ,.,. .:.:.: :.._:_: . . . . : : : : : : : . . . . . . . . . .._... . . . . . . . . . . . . . . .:_:,:,. :.::::i:i:i:iiiil:ii: ::::::: .._: : ; :.:.:.:.:.:.:.:.:.: .:.:.:.:.:,: :.:.::i::.. . . . . . . . . . . . . . . ..1...::i:::::::::::::i:ii~~~.:.’ ,...:.:::::::::i:,:::::::::::: ,...

zero level EEPSP

im

FIG. 9 A: electrical equivalent of excitatory synapse, applying to all phases. In phase 0 (see text), switch S is open and all currents are equal to zero. Inpbase 1 (shown), switch S closes, current i, is directed inward, and currents i, and ic are directed outward. Current i, exceeds im by an amount i,, equal to rate at which membrane capacitance discharges. Inphase 2 (peak), currents i, and im maintain their directions and are numerically equal. Current i, is zero. In phase 3 (repolarization), current im exceeds i, by an amount i,, which inverts sign, being inward. B: hydraulic model of excitatory synaptic potential, applicable to all phases of excitatory potential. In phase 2 (deleveling), register R opens, allowing water to enter V from reservoir E epsp at a rate is. When Vm exceeds Em, water leaks from V to E at a rate i,. Difference i, - im is rate i,, at which level Vm increases. Inpbase 2 (peak), i, = im and i, = 0. In phase 3 (releveling), currents im and i, maintain their direction, but im now exceeds i,, which causes i, to invert sign, directing downward.

voltage clamp, the opening of these channels leads to no current, or to the Vm value where the synaptic current reverts sign (from outward to inward, or vice versa).

The opening of a population of channels with the above characteristics is hydraulically equivalent to inserting into the system of Fig. 5 an additional


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reservoir, Es, connected directly to reservoir V by means of a tube with conductance Gs, as shown in Fig. s)B. The level of reservoir E, is fixed at zero because the value E, obtained for the electrical analog of the cell in&@ 13 is zero.

The synaptic potential (or EPP) can now be analyzed during four different arbitrary phases: phase 0 is rest, phase 1 is depolarization, phase 2 is peak, and phase 3 is repolarization (2).

In phase 0 there is no ACh in the synaptic cleft, and the synaptic channels are closed. In the electrical equivalent (Fig. PA), switch S is open, and there is no current through the system E, - R,. The capacitor Cm is charged to Vm = E,, which means that the membrane potential coincides with ~5’~.

In the hydraulic equivalent of phase 0 (Fig. 9B), register R is closed so that there is no communica- tion between reservoir E, and reservoir V. Because reservoirs V and E communicate freely through tube G,, levels Vm and E, are equal.

In phase 1 the ACh-activated channels in the PSM open, allowing flow of Na and K ions down their respective driving forces. Consequently, Na ions begin to move into the cell, and K ions begin to move out. Despite the conductances for Na+ and K+ being about the same in these channels, the driving force for Na+ (164 mV in) is - 20 times larger than the driving force for KS (8 mV out), the net result being an intense inward current of Na+ ions.

The opening ofACh-activated channels in the PSM is represented by closing of switch S of the electrical equivalent (Fig. VA), or opening of register R in the hydraulic equivalent (Fig. 9B).

When switch S closes in the electrical equivalent of phase 1 (Fig. 9A), current begins to move in the left leg of the circuit, in the counterclockwise direction, driven uniquely by battery E,. This fact seems paradoxical to many students, inasmuch as they correctly reason that what actually drives Na ions into the cell (through the synaptic channel) is both a gradient of [Na+] and the cell negativity. This appar- ent inconsistency (which constitutes an excellent exercise) will be partially clarified by the use of the

hydraulic equivalent below. As soon as the switch closes, current begins to flow into the cell through the synaptic leg of the circuit and out of the cell through the nonsynaptic leg. The difference between the current entering and that leaving the cell constitutes the capacitive current.

In the hydraulic analog ofphase 1 (Fig. W), register Rs fully and rapidly opens, mimicking opening of the synaptic channels. Because excitatory potential level Eepsp is higher than Vm,


postsynaptic water flows

from reservoir & into reservoir V, through Gs, at a rate given by

is = CEepsp - KJGs WO

However, as soon as Vm increases above E,, water begins to flow also from reservoir V to reservoir E, through tube G,, at a rate given by

4n = K-n - E,)Gln w>

During phase 1, i, is larger than i,, and the differ- lates in reservoir V, increasing ence i, - i, accumu

the level Vm at a rate

dv- i capacitive

=i,-i,=A~

dt (-I61

As level Vm increases, i, progressively increases, making icap decrease. Eventually, icap gets SO small that i, = i,, and all the water entering reservoir V from& leaks to reservoir E. At this moment we are at the threshold ofphase 2, the peak.

In the peak phase, the rate dVJdt = 0, and Eq. 16 tells US that icap = 0. At this moment it is possible to estimate the amplitude of the Vm level displacement (the amplitude of the synaptic potential in the cell) provided some parameters are known. In the peak phase, the water flow entering V from E, is equal (by definition of peak) to the water flowing from V to E. As such

ipeak = s (E ePsP - Vm)Gs = i, = (Vi - E&T, (17)

Equation 17 gives the value of Vm at the peak, as

vm = EepspGs + EfnGrn

Gs + 6-l (peak)

Suppose that G, = 10 G, (and remember that E ePsP = 0). Then, Eq. 18 gives Vm = Em/1 1, or a


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quasi-total depolarization. On the other hand, if G, were 10 times greater than G,, then Vm = E,(lO/ll) or is practically unchanged. This means that to cause a significant change of the level Vm, G, must be significantly greater than G,. Indeed, at the peak of the EPP, G, is - 5 times G, (2).

After reaching maximum activation due to the action of ACh, synaptic channels begin to close as ACh gradually disappears from the synaptic cleft (by diffusion and hydrolysis). This actually happens well before the peak phase is reached, so that by the time of the peak, the synaptic current has decreased considerably.

After the peak is reached, despite the fact that synaptic current still continues entering the cell, its magnitude is smaller than that of the current leaking from the cell (across nonsynaptic channels). This means the cell is now losing, across its nonsynaptic membrane, more positive charges than those being gained through the synaptic channels. As a result, the cell begins to regain its former negativity. This constitutes the repolarization phase orphase (Fig. VA).

Repolarization is easily pictured using the hydraulic analog (Fig. %?). After the level Vm reaches its uppermost value, the flow of water leaving V toward E surpasses the flow entering V from reservoir E,. This happens both as a consequence of the gradual closure of register R (equivalent to gradual closure of synaptic channels) and of the progressively smaller level difference between Eepsp and Vm. Because now i, is larger than i,, the level Vm descends toward its resting value. When eventually register R fully closes (synaptic channels close), the level Vm is still higher than E,, but slowly the two reservoirs equilibrate through the conduct R,, until I&., = E,. The synaptic event has ceased.

DISCUSSION AND CONCLUSIONS

In our experience, hydraulic models can comple- ment with high efficiency the electrical equivalent circuits of bioelectric potentials. We do not recommend, however, leaving out the electricals. The advantages of this approach can be summarized. 1) The average medical/biology student is refractory to electricity concepts but accepts well mechanical or hydraulic models. 2) The main and more objective

advantage is that the water level in a reservoir is something readily seen either in a diagram or in a transparent vessel in a laboratory. On the other hand, the voltage across a membrane or capacitor is not easily visualized. To demonstrate or measure a voltage difference in the laboratory the teacher needs a voltmeter. Students are then asked to believe an instrument they barely understand. 3) The phenomena occurring in real cells are already electrical, so a further electrical representation intro- duces no new insight.

The order of presentation seems to be of impor- tance: when the electrical equivalent is presented first, students usually are refractory and less recep- tive. The hydraulic model, when presented soon afterward, usually makes them interact with the teacher, either agreeing or disagreeing. Many parallel discussions rise at this moment, and the teacher has invariably, by now, ample material to deal with. When the inverse order is employed (I have done that in a few instances), the students see little purpose in the electrical equivalents because they have already grasped the idea. The sequence that seems to work best is to present the electrical equivalents, then to present the hydraulic analogs, and finally to show both models comparatively, making the analogies between them.

We have also employed the above strategies in laboratory classes. To maintain levels E, and Eepsp constant, reservoirs E and Eepsp were made of a wide-mouthed bucket connected to a plastic soda bottle representing reservoir V. To further help in keeping levels E, and/or Eepsp fixed, a student was assigned the role of “pump,” and he or she had the task of keeping constant the water level in the bucket. These laboratory exercises proved most valuable and were highly appreciated by the students.

APPENDIX: ANALOGIES BETWEEN ELECTRICAL AND HYDRAULIC CIRCUITS

The following analogies are assumed to hold between electrical and hydraulic circuits: 1) The flow of electric charges is analogous to the flow of water. 2) The voltage difference (AV) driving a current across a resistance is analogous to a hydrostatic pressure difference (AP) between the two ends of a tube. As such, Ohm’s first law (V = Z&I) can be


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readily extended to water flow, with the following analogies: AP = AV, water flow = charge flow; electrical resistance = hydraulic resistance (1). 3)

The electrical capacitance (Gel) of a condenser is analogous to the hydraulic capacitance (C,,,) of a reservoir. In the electrical capacitor, Gel = AQ/AV, where Q is charge and V is voltage, whereas the hydraulic capacitance of a vertical cylindrical reservoir is given by C,,, = A (water volume) /A (water level) = base area.

The author is grateful to postgraduate student M. Oliveira Souza for ad apting the hyd raulic analogs to laboratory classes.

de

This study was supported by Grants 91/2264-4 from Fundacao de Amparo a Pesquisa do Estado de Sao Paulo and 301640-79.3

from Conselho National de Desenvolvimento Cientifico e Tecno- logico.

Address for reprint requests: J. Procopio, Departamento de Fisiologia e Biofisica, Instituto de Ciencias Biomedicas da Univer- sidade de Sao Paulo, Av. Prof. Lineu Prestes 1524, CEP 05508, Sao Paulo, SP, Brazil.

Received 13 August 1993; accepted in final form 14 July 1994.

References

1. Kandel, E. R., and J. H. Schwartz. Directly gated transmission at central synapses. In: Principles of Neural Science, edited by E. R. Kandel and J. H. Schwartz. Amsterdam: Elsevier, 1991, chapt. 11, p. 153.

2. Kandel, E. R., and S. A. Siegelbaum. Directly gated transmission at the nerve-muscle synapse. In: Principles of Neural Science, edited by E. R. Kandel and J. H. Schwartz. Amster- dam: Elsevier, 1991, chapt. 10, p. 135.

3. Kandel, E. R., S. A. Siegelbaum, and J. H. Schwartz. Synaptic transmission. In: Principles of Neural Science, edited by E. R. Kandel and J. H. Schwartz. Amsterdam: Elsevier, 1991, chapt. 9, p. 123.

4. Koester, J, Membrane potential. In: Principles of Neural Science, edited by E. R. Kandel and J. H. Schwartz. Amster- dam: Elsevier, 199 1, chapt. 6, p. 8 1.

5. Koester, J. Passive membrane properties of the neuron. In: Principles of Neural Science, edited by E. R. Kandel and J. H. Schwartz. Amsterdam: Elsevier, 1991, chapt. 7, p. 95.

6. Kuffler, S. W., and J. G. Nicholls. From Neuron to Brain. Sunderland, MA: Sinauer Associates, 1977.

7. Sten-Knudsen, 0. Passive transport processes. In: Membrane Transport in Biology, edited by G. Giebish, D. C. Tosteson, and H. H. Ussing. Berlin: Springer-Verlag, 1978, vol. I, chapt. 5,p. 5.


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