Molecules to Medicine- Cell Physiology
1. Composition of cells 18 Nov 10 AM
Bill Betz
[email protected] http://www.cuphys.net/cellphys
Reading (optional): Alberts et al, 4th ed: pp 61-62; 112-113; 118; 583-588; 593-596; 636.
Lehninger, 4th ed: pp 47-56; 369-411.
Key words: plasma, interstitial fluid, membrane channels, membrane transporters
1: Composition of cells
Introduction. Today we begin a study of human physiology, a discipline that relates to
function as anatomy does to structure. This series will deal with the most basic principles in cell
physiology – things like ions, diffusion, osmosis, and electricity. Next semester (March), in the
CVPR Block, you will use these principles over and over as you study the heart, lungs, and
kidneys. The same goes for next year, in the Neuro Block (August), and later in the Endocrine
and Metabolism Blocks. In fact, these are principles that you will use throughout your medical
training and professional lives. We will try to preview some of these things as we progress.
A populist's view of the body. Let's begin by examining the overall composition of the
body. We'll focus on the fluid parts, ignoring bone, skin, nails, and other solid bits. It turns out that
about 99% of all the chemical particles in the 45 liters of body fluids are water molecules. Of the
remaining 1%, about 5 out of every 6 are simple inorganic ions, principally Na+, K+, and Cl-. All of
the rest -- all carbohydrates, proteins, nucleic acids, lipids, and so forth -- constitute a very small
fraction of the total. Thus, in terms of numbers of particles we're really not much more than a dilute
salt solution. This means that when you study the biochemistry of macromolecules you are
primarily concerned with a population of molecules that accounts for a small fraction of 1% of the
total. Just on the basis of these numbers, you might expect that there are special mechanisms in the
body responsible for maintaining proper water and ion balances, and indeed, this is what much of
physiology, and the next few lectures in particular, are all about. By and large, water and simple
ions are not handled like other chemicals important for sustaining life. The majority, for example,
don’t participate in enzymatic reactions, and so their biochemistry isn't a very exciting story.
What then do these biochemical dropouts do? The answer is: lots of vital things, like make
action potentials, which we'll study in detail later. For the time being, however, we can picture ions
as playing a passive role, merely providing a proper microscopic environment for the normal
enzymatic reactions in the body. It seems a rather dull subject, sort of like wanting to fly jet planes
and having to study the properties of air. That may be, but fluid and electrolyte balance is no trivial
topic. It is estimated that fully half of the energy we expend at rest is used to pump ions across
membranes. If ion or water concentrations were to change, even by a few percent, the
consequences can be disastrous, because enzymes are not very tolerant of changes in the
composition of their microenvironment. Speaking medically, changes in water and electrolyte
concentrations can make one sick, and a good bit of the practice of internal medicine, for example,
is concerned with maintaining or restoring proper water and solute composition of body fluids.
How simple matters would be if all 45 liters of body fluids had the same water and ion
composition; our concern then would be simple nutrition: to stay healthy, one need only balance
intake with output. Unfortunately, the fluids in the various body compartments do not have the
same composition; the ion content of cells, for example, is vastly different from that of blood
plasma. The differences in composition are maintained by a host of mechanisms; perhaps most
important are the molecular "pumps" that concentrate various solutes in particular compartments.
As you might expect, if any of these mechanisms is disturbed, fluid composition in the affected
compartment can change, upsetting metabolism, leading to a clinic visit. The point is that all of this
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can happen without any change in overall nutritional status (intake/output).
In summary, body fluid compartments have different compositions, maintained by a variety
of mechanisms. These facts make for a vulnerable situation: there are numerous ways that
imbalances in composition can arise without any change in nutrition. For example, as we'll see
later, if as little as 2% of the potassium inside cells should suddenly leak out of the cells, the result
can be, and usually is, fatal. The next several lectures will deal with the normal mechanisms that
determine the composition of fluids in different body compartments.
The two major fluid compartments are the intracellular (~2/3 total = ~27 liters) and
extracellular (~1/3 total = ~ 13 liters + ~ 5 liters ‘special fluid’ sequestered from the rest in a ‘third
space’ described later). The boundary between the intra- and extracellular fluids is of course the
plasma membrane. Each of these two compartments can be subdivided: extracellular fluid (ECF)
comprises plasma (~3 liters – plasma is whole blood without the cells), lymph, and interstitial
fluid, all of which have nearly identical inorganic ion and water compositions. Strictly speaking,
when we refer to the ECF we will mean the interstitial fluid. Intracellular fluid (ICF) comprises
mitochondrial, vesicular, nuclear, and other subcompartments. In these lectures, we can safely
picture the cell as just a bag of uniform cytoplasm, and leave it to the cell biologists to describe the
beauty and complexity of the cytoskeleton and intracellular organelles.
Our concern, then, is pretty simple: we want to explain what happens at the plasma
membrane that enables cells to maintain internal compositions that are very different from the fluid
in which they are bathed.
In order to discuss some basic principles that illustrate how the differences between ICF
and ECF arise, it will be helpful to make a simplified table (Table 1). We will lump some things
Table 1. Simplified table of ICF and ECF composition
ICF (mM)
ECF (mM)
Membrane Permeability
14
140
(-)
K+
145
5
+
Cl-
5
145
+
A-n
126
~0
-
Na
+
H2O
~55,000
same as ICF
+
=
=
together (HCO3 is put with Cl ; also HPO4 and SO4 are put with proteins, and this group is
named A-n, where A stands for "big anions", which have an average net charge -n of about -1.2).
We will also ignore other things (H+,
Mg++, Ca++, organic acids). Thus we will
Why are cells rich in potassium but poor in
+
+
-n
sodium? One idea concerns how the oceans grew
have four solutes (Na , K , Cl , and A )
salty as cells evolved (about a billion years ago).
and of course, H2O. The simplified
The early oceans were relatively rich in potassium;
composition is given in Table 1. In
they have grown more salty with sodium over time.
addition, the membrane permeability is
The reason is that, while there are about equal parts
shown (+ means the membrane is
potassium and sodium in the earth’s crust, silicates
permeable to the substance, and - means
(soils) bind potassium more tightly than they bind
the membrane is impermeable.) One other
sodium. Thus, as the rains fell, they preferentially
+
thing to note: the permeability to Na is
leached sodium out of the soil, loading up the
special: we will assume for starters that
oceans with sodium. Evidently, it was easier for
membranes are impermeable to Na+; later
cells to evolve mechanisms allowing them to retain
their primitive potassium-rich intracellular milieu
than to evolve enzymes more tolerant of sodium.
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we will see that Na+ can and does permeate cell membranes, but it is pumped in the other direction,
making real cells 'functionally impermeable' to Na+.
[A note on the word permeable: membranes are permeable (or impermeable) to solute,
while a solute is permeant (or impermeant) or permeating (or nonpermeating).]
As you can see from Table 1, the concentration of K+ is about 30 times higher inside the
cell than outside, and this in spite of the fact that the membrane is permeable to K+. Why doesn't all
the K+ leak out of the cells? As we'll see, we can handle this curiosity by understanding a few rules
about diffusion and electrical forces.
The Structure of the Plasma Membrane
All of the control mechanisms we will
discuss reside within the plasma membrane, so it
will be helpful to discuss some basic features of its
structure (Fig. 1). Two aspects will be especially
important to us. First, membranes are composed of
lipids. These molecules create a thin hydrophobic
skin around the cell, making it virtually
impermeable to charged substances like ions, and
even to neutral molecules, if they have a dipole
Fig. 1. Membranes are mostly lipid molecules,
moment, like water.
which are impermeable to water and charged
Lipids are also 'strong', in an electric sense.
substance. Proteins provide pathways for water
That is, they can keep opposite electric charges
and solutes to get across.
separated, without collapsing. As we will see, the
cytoplasm of nearly all cells is electrically negative, compared to the ECF, because cells contain a
few more negative than positive ions. The excess anions create an electrical potential difference
between the inside and the outside of the cell, and this membrane potential, which governs some
vital cell processes, is wholly dependent on the integrity of the plasma membrane. The ability of
the membrane to withstand the imposed electric force (which has a whopping strength of about
100,000 volts/cm) is due to its lipid composition.
The second physiologically important point about membranes is that they are apparently
shot full of holes. That is, many charged and polar molecules can cross membranes, no thanks to
the lipids. Their ability to do so depends primarily on the presence of certain proteins inserted in
the lipid membrane; these proteins have special abilities to transport particular substances between
the ECF and the ICF. From a functional (i.e., our) standpoint, there are two kinds of proteins that
mediate the transmembrane movements of charged substances:
1. Channels. Some charged/polar substances cross membranes by passing through
channels, which behave as passive pores, or tunnels in the membrane. Channels have two
especially important properties.
First, most are selective for particular ions. For example, sodium channels will pass Na+
ions, but not K+, or Cl-, or other ions. Some channels permit just about any cation to pass, but not
anions; these are called 'non-selective cation channels'. In this course, you will study channels
selective for Na+, K+, Cl-, Ca++, and a few other substances (including water).
Second, some channels contain molecular gates. Substances can pass through the channel
only when the gate is open. As you will see, opening and closing of the gates can govern important
functions, like action potentials. You will study channels with gates controlled by a variety of
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forces. Some depend on the electric field (membrane potential) across the membrane. The gates in
many such voltage-gated channels swing open when the membrane is depolarized (the cell is made
less negative inside). Gates in other channels require mechanical stimulation (stretching of the
membrane) to open (e.g., hair cells in the cochlea; touch receptors in the skin). Still others require
the binding of a particular chemical (synaptic receptors for a neurotransmitter). Others open or
close depending on temperature (cutaneous thermal receptors). Some gates in channels depend on
more than one force, making for some pretty complicated behavior. For example, a channel gate
may require the binding of a particular chemical substance and, simultaneously, a particular value
of membrane potential in order to open up. Some channels contain no gates (e.g., water channels,
also called aquaporins) , while others contain two gates (e.g. voltage-gated sodium channels), and
both have to be open to allow ions to pass!
Ion channel function. The patch clamp technique (Fig. 2) allows one to see individual ion
channels at work. A glass pipette touches a cell, and gentle suction draws a membrane bleb into the
mouth of the pipette, forming a tight seal with the glass. Then, when a channel gate opens, it is
possible to record the electric current through the channel. As channels flicker open and closed,
one can describe their behavior in great quantitative detail. In the example in Fig.2, about 1000
ions per millisecond flow through the open channel.
Fig. 2. The Patch Clamp was invented in 1980. A micropipette is lowered onto a cell, but instead of
punching through the membrane, it sucks onto it, and a tight seal is formed between the membrane
and glass pipette. In this situation, it is possible to see the tiny electric currents produced by the
openings and closings of single ion channels. Examples are shown on the right (with the averaged
response at the bottom). It is astonishing to see a single protein molecule at work, in real time, in its
native membrane environment. The work won a Nobel Prize in 1991 for its German inventors, Bert
Sakmann and Erwin Neher.
Ion channel structure. In recent years, we have begun to acquire knowledge of the
molecular structure of ion channels. It is challenging to make crystals of membrane proteins, which
are necessary for x-ray crystallography, owing to the amphipathic nature of the proteins
(membrane-spanning parts are lipophilic, while parts facing the ICF and ECF are hydrophilic). Fig.
3 shows the structure of a potassium channel at atomic resolution. The left panel shows a view
from outside the cell, revealing a potassium ion in the pore, with the four peptide helices arranged
symmetrically around it. The right panel is a space-filling model, with a cutaway view of the pore,
which contains three potassium ions, passing single file through the pore.
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These are extraordinarily powerful investigative techniques, and when combined with
molecular biological tools, which allow individual amino acids to be mutated, really awesome
studies are possible. In clinical medicine, an increasing number of genetic diseases due to ion
channel mutations are being recognized. This area is one of the most promising for the kind of
‘bench-to-bedside’ translational research efforts being heralded by the NIH ‘Roadmap for Medical
Research.’
Fig. 3. Molecular structure of a
potassium channel. Left: View from
outside. Right: Space-filling view,
cut away to reveal three potassium
ions traveling single file through the
pore.
2. Transporters. Some substances cross membranes by binding to proteins and being
escorted (‘carried’) across. Such a mechanism is needed to transport big molecules, like glucose,
selectively. Transporters are also needed in order to pump molecules across a membrane, that is, to
concentrate them on one side against their energy gradient. Such pumping is called active
transport, to distinguish it from passive transport pathways, like channels. Pumping means
performing work, and the energy to do the work can come from different sources. If the energy
comes directly from metabolism (usually splitting ATP), the transport is called primary active
transport. An example is the Na+ pump, which extrudes Na+ from cells, and requires ATP. If the
energy comes from other sources, it is called secondary active transport. There are many examples
of secondary active transport, and most of them use the energy released when Na+ ions leak into
cells, and the energy released is captured and used to pump another ion across the membrane.
Ultimately, secondary active transport also depends on metabolism.
Transporters work thousands of times slower than channels, usually not exceeding several
hundred transport cycles per second.
We are only just beginning to acquire information about the molecular structure of
transporters, and that information suggests that they are very similar to channels. For example, by
mutating only a few amino acid residues, a transporter of chloride and hydrogen ions, which
pumps only about a hundred ions per second, can be turned into a chloride-selective ion channel
that allows the passage of tens of thousands of ions per second.
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PROBLEM SET
Inorganic chemistry refresher
Atomic Weights:
H=1 C=12 N=14 O=16 Na=23 Mg=24 S=32 Cl=35.5 K=39 Ca=40
1. Calculate the formula weights of the following: a. H2O b. NaCl c. CaCl2 d. K2SO4 e. urea
(NH2)2CO f. glucose C6H12O6
2. The molar concentration (moles/liter, molarity, M) of a solution is the number of
formula weights in grams per liter of solution. Calculate the weights required to make 1 liter of a 1
molar solution of (b), (c), (d), and (f) in problem 1, above.
3. How much of each is required for a liter of 100 millimolar (mM) solution?
4. A liter of water weighs 1000 grams. What is its concentration in moles/l?
5. Some solutes, when dissolved in water, dissociate into their component ions. Which of
the solutes listed in problem 1 dissociate, and into what ions?
ANSWERS
1.a.H2O: 2 + 16 = 18. b. NaCl: 23 + 35.5 = 58.5. c. CaCl2: 40 + 2x35.5= 111. d. K2SO4:
2x39 + 32 + 4x16 = 174. e. urea: 60. f. glucose: 180
2.Same as answers in #1, in grams.
3.Same as answers in #1, divided by 10.
4.1000 g/l divided by 18 g/mole = 55.5 moles/l.
5.a.H2O: none; b.NaCl: Na+ & Cl-; c.CaCl2: Ca++ & Cl- & Cl-; d.K2SO4: K+ & K+ & SO4=
e.urea: none; f.glucose: none
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Molecules to Medicine- Cell Physiology
2. Cell Volume Regulation 18 Nov 11AM
Bill Betz
[email protected] http://www.cuphys.net/cellphys
Reading (optional): Lehninger, 4th ed: pp 56-58
Key words: molarity, osmolarity, equivalents, tonicity, reflection coefficient, lysis
2. Cell Volume Regulation
Making a model cell
A simple model cell must control its volume. Now let's return to the question first posed:
How do cells maintain an internal solute composition that is very different from the external fluid?
In order to simplify things, we will start with a hypothetical cell with an extremely simple
composition, and then add additional parameters in a stepwise fashion until arriving at the more
complex composition given last time, and reproduced in the Table below.
Table 1. Simplified table of ICF and ECF composition
ICF (mM)
ECF (mM)
Membrane Permeability
14
140
(-)
K
145
5
+
Cl-
5
145
+
A-n
126
-0
-
H2O
~55,000
same as ICF
+
Na
+
+
We will begin by imagining a hypothetical 'primitive' cell, bounded by a plasma
membrane, and containing only water and those molecules (uncharged) vital to its existence and
replication. The plasma membrane is impermeable to these vital molecules and permeable to water.
For a start, imagine the cell immersed in pure water (Fig. 1). Immediately water will tend to
diffuse into the cell, because the water concentration inside is lower than outside (that is, the
internal solute particles dilute the cellular water). This movement of water will cause the cell to
swell, and unless some countering force is applied, the plasma membrane will be stretched and
eventually will rupture.
Fig. 1. A very primitive cell.
The elementary force underlying this water movement is the random thermally-agitated
movement of the molecules, that is, diffusion. This movement causes water molecules outside the
cell to collide with the membrane; some will enter aqueous pores and so enter the cell. Similarly,
some water molecules inside the cell will enter pores and so leave the cell. However, the presence
of the nonpermeating molecules inside the cell will reduce the probability that a water molecule
will leave the cell, and this means more water will enter than will leave. This net inward movement
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of water across a semi-permeable membrane (i.e., a membrane permeable to solvent, but not to
solute) is called osmosis. Why is it given a special name, and not just called "diffusion of water"?
The reason is explained in Appendix I for those interested, but in fact you can picture osmosis as
diffusion of water. (You'll always get the direction of water movement right, but you may greatly
underestimate the rate at which it moves.)
The most important consequence of the movement of water into or out of a cell is that it
produces a change in cell volume. This is because biological fluids are dilute solutions; most of
their volume is water; the solute doesn't take up much space. For example, removing all of the
solutes from a liter of plasma leaves over 900 ml of pure water. Or to say it another way, for every
solute particle in plasma, there are about 150 water molecules. Thus, to a first approximation, all
volume changes are produced by the movement of water, or, stated in reverse, only water
movement can change the volume of a cell.
Osmotic balance restored. In the course of evolution, three mechanisms have evolved to
solve the problem of volume control. They are illustrated in Fig. 2.
A simple solution to the problem would be to make the cell membrane impermeable to
water, as well as to the internal solute molecules (Fig. 2, left). However, this would create many
new problems for the cell; growing cells, for example, must have a way of accumulating water as
they grow (remember, cells are mostly water by volume), and therefore they must have surface
membranes permeable to water. There are some special kinds of cell membranes that do exhibit
extremely low water permeability (sweat glands, some cells in the kidney). In general, however,
plasma membranes are quite highly permeable to water (although diffusion of water, and all
substances we will consider, through even the most permeable membrane, is still thousands of
times slower than diffusion in free solution).
Another solution (Fig. 2, middle) is to build a strong wall around the cell, and so keep the
cell from swelling by brute force (that is, apply a hydrostatic force to counter
the osmotic force drawing water into the cell). Plant cells, fungal cells, and
bacteria do just this, by building tough cell walls outside their plasma
membranes. But it's an expensive way to go, in the sense that the osmotic
force is not trivial (as shown on the right, a small container of pure water
would support a column of a 1 M solution, from which it is separated by a
semi-permeable membrane, that is 900 feet high), so that cell walls require
considerable metabolic resources (building materials, energy for synthesis,
etc.), and greatly limit cell shape (try to imagine a neuron in the cerebral
cortex, with all of its delicate dendrites surrounded by a cell wall!).
[In the drawing, the solute essentially tries to suck water through the
membrane from left to right; only the weight of the 900' column keeps the water from moving. It's
unfortunate that we usually think of osmosis as a pressure; really it's a suction; we have to exert a
pressure in order to counter the osmotic suction.]
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The quantitative relationship between the osmotic suction and the pressure one has to exert
in order to balance it was given last century by a Dutch physicist named Van't Hoff. He found that
the amount of pressure you have to apply is proportional to the difference in solute concentration
on the two sides of the (semi-permeable) membrane:  = RT.C, where  = osmotic pressure. The
constants of proportionality are the gas constant (R) and the temperature (T, in degrees Kelvin).
The solution in animal cells to the problem of volume control (Fig. 2, right) is to fight fire
with fire, and balance the osmotic force osmotically, by having solute molecules in the ECF, in
order to balance those in the ICF. Thus, the concentration (or, more accurately, the activity) of
water is exactly the same in the ICF and ECF. How simple physiology can be.
There are two further important points about osmotic balance: First, the chemistry doesn't
matter: the ECF solutes can be different from the ICF solutes (and they are: mainly NaCl in the
ECF, and K+ and big anions inside the cell). What matters is that the total concentration of solute
particles, that is, the osmolarity of the two solutions is the same. Second, the solutes in the ECF
must be nonpermeating (to match those in the ICF); putting a permeating solute, like glycerol, in
the ECF would do no good at all in the long run, as the example problems below illustrate.
EXAMPLE PROBLEMS: Volume control
Assume we have a cell
Table 2. Example problems
that contains only water and
ECF (memb.perm.)
TONICITY/OSM VOLUME
nonpermeating protein P (no
A.
300
mosM
sucrose
(-)
charge on P) at a concentration
B. 150 mosM sucrose (-)
hypo/hypo
2x normal
of 300 mM. Calculate the
C.
600
mosM
sucrose
(-)
volume of the cell at
D. 300 mosM glycerol (+)
equilibrium (i.e., when all net
E. pure water
(+)
fluxes have stopped), relative
F.
600
mosM
glycerol
(+)
to normal, when it is immersed
G. 300 mosM sucrose (-)
in the solutions given in Table
& 300 mosM glycerol (+)
2. Note: + means the substance
H. 300 mM NaCl
(-)
can cross the membrane, means it cannot. Water can cross in all cases. None of the substances is electrically charged except
NaCl.
To solve these, note the following rules:
1. Assume that the ECF composition does not change, regardless of any fluxes into or out
of the cell. It's as though we continuously flow fresh ECF past the anchored cell, so anything that
comes out of the cell is swept away, and anything that goes in is immediately replaced.
2. Only water movement can change cell volume. Solute movement has a negligible
effect on volume.
3. Osmolarity is the total concentration of solute particles: for example, a 1 M solution of
CaCl2 gives a 3 osM solution (3 solute particles/molecule dissolved).
4. To solve these problems, you need only one rule: If the whole system is at equilibrium,
then each individual species must be at equilibrium. This means that if a substance can cross the
membrane, then the concentration on the two sides must be the same at equilibrium. Thus, if
glycerol is present, [Gly]i must equal [Gly]o at equilibrium.
Because water can always cross the membrane, [H2O]i= [H2O]o at equilibrium. Now
ordinarily, we don't talk about the "concentration of water". An equivalent statement is that the
osmolarities are the same inside and out: osMi= osMo.
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Now solve the problems (volumes only for now). (Answers are below).
In some of these exercises we introduced something new: the idea of a permeating solute
(glycerol). The cell membrane is no longer perfectly "semi-permeable" (i.e., permeable only to the
solvent H2O). A point to note is that the permeating solutes contributed nothing to controlling the
equilibrium volume of the cell (D and E had the same result, and D is just E with some glycerol;
the same goes for A and G, and for F and E). In other words, the final volume of the cell is the
same regardless of the presence of glycerol. All that matters is the concentration of nonpermeating
solute.
To say it another way, how a solute affects cell volume depends on the biology of the
situation, that is, whether or not the cell is permeable to the solute. This has forced biologists to
define a new term: tonicity, which is not necessarily the same as osmolarity. In brief, to determine
osmolarity, you need know only what’s in the solution. But to determine tonicity, you have to do
an experiment on the cell, to see if it shrinks or swells (and the effect might differ between
different cell types). So, any solution (such as A or D) with the same solute concentration as the
cell is isosmotic (same osmolarity) with the cell (at the beginning of the experiment). But solutions
A and D had wildly different effects on cell volume; that is, they had different tonicities. Any
solution that makes a cell shrink is hypertonic (it behaves like a hyperosmotic solution of
nonpermeating solute); any solution that makes a cell swell is hypotonic. Note that a solution that
is isotonic for one cell may not be isotonic for another cell. It all depends on the permeability of the
cell membrane. So you see, biology isn't always conceptually easier than chemistry.
To be sure you understand, go back and label each solution in terms of its osmolarity and
tonicity. Write the osmolarity of the solution relative to the normal cell (300 mosM). Write the
tonicity by considering the effect of the solution on the final volume of the cell (relative to the
initial volume).
ANSWERS (tonicity/osmolarity; volume change):
A. iso/iso; no change
E. hypo/hypo; swell & burst
B. hypo/hypo; 2X
F. hypo/hyper; swell & burst
C. hyper/hyper; 0.5X
G. iso/hyper; no change
D. hypo/iso; swell & burst H. hyper/hyper; 0.5X
Time course of volume changes. In these exercises we were concerned only with the
beginning and final (equilibrium) states of the cell. Now let's briefly return to example F (600
mosM glycerol) and consider what happens to cell volume immediately after the cell is exposed to
the solution. The test solution is hyperosmotic to the cell, so initially water will leave the cell.
Glycerol of course will enter the cell. Cell volume depends only on the movement of water, so the
cell will shrink at first. But then, as glycerol continues to
enter, the internal osmolarity will rise, and at some point the
movement of water will reverse direction, and begin to enter
the cell. From then on, glycerol and water will both enter,
and the cell will swell, and eventually burst (real cells can't
swell beyond about 140% of their resting volume).
Graphically, the change of volume with time would
Time course of volume
be as shown in the figure to the right. What would the curve
change when immersed in
for solution G look like? (Answer: shrink initially, then
600 mosM glycerol
return to normal)
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The point is that the rate of change of volume can tell us something about the way solutes
cross the membrane. The exact time course of volume change will depend on how easily (relative
to water) the solute can cross the membrane. If it crosses slowly, the cell will stay shrunken for a
longer period, although eventually it must swell and burst.
Reflection coefficient. Another consequence of a permeating solute is that it will not exert
as large an osmotic force as a nonpermeating solute at the same concentration. How much the
osmotic pressure is diminished depends again on how easily the molecule can cross the membrane,
relative to water. For example, consider a mixture of H2O and tritiated water, THO (one H is
replaced with one T). The THO can be considered as a solute. Of course, it exerts no osmotic
pressure, because THO and H2O cross membranes with equal ease. A molecule that crosses half as
easily as water will exert half of the ideal osmotic pressure of a non-permeating solute. The Van't
Hoff equation, which relates osmotic pressure to hydrostatic pressure, is modified to account for
this non-ideal behavior by multiplying by a factor called the reflection coefficient, which has a
range from zero (for THO) to one (for nonpermeating solute). It's a measure of how well the
membrane 'reflects' the solute. The modified equation is:
 =  RT C,
where  = reflection coefficient, R=gas constant, T=temperature, and C = difference in solute
concentration across the membrane.
Consider this example: Suppose we bathe a cell (internal osmolarity= 300 mosM;
nonpermeating internal solute) in a 600 mosM solution of solute that can cross the membrane half
as easily as water ( = 0.5). When first exposed to the solution, water will not move across the
membrane, because there is no osmotic force (1x300= 0.5x600). Then, as solute enters the cell,
water will follow, and eventually the cell will lyse.
Well, so what? Is any of this relevant to clinical medicine? Indeed it is. Consider this
unfortunate situation: A newborn infant suffered from severe dehydration (it usually happens from
a GI obstruction or from sweating in summer’s heat). Today in modern hospitals, rehydration is
accomplished by delivering fluids intravenously (IV). However, sticking an infant’s vein is not
easy, and in decades past, before special IV kits were developed (and even today, in underserved
regions), the fluid (~300 mosM glucose in water) was give subcutaneously (‘clysis’; usually in the
lower back where skin is loose), whence it would shortly be absorbed into the blood via capillaries
located there. Surprisingly, the ‘blister’ of injected fluid at first increased in size. Fluid was being
sucked out of the dehydrated infant’s circulation, exactly the opposite of the intended treatment!
(and risking circulatory collapse and death) Eventually, however, the blister decreased in size and
disappeared as the injected fluid entered the circulation.
Why did fluid flow in the wrong direction at first? Hint: NaCl crosses capillaries more
easily than glucose does.
Answer: While the injected fluid (isotonic glucose) had the same osmolarity as plasma, the
solutes were very different – glucose on one side, NaCl on the other (blood) side. Because the
capillaries were more permeable to NaCl than to glucose, it (NaCl) rapidly diffused out of the
blood into the injected fluid (which contained no NaCl). This of course raised the osmolarity of the
fluid in the blister , which caused water to move into the blister from the blood. To say it another
way: the reflection coefficient across capillaries is higher for glucose than for NaCl. Thus a 300
mM solution of glucose exerts a higher osmotic suction than 300 mosM NaCl, so that water moves
into the glucose solution. Eventually, as the sucrose diffuses into the blood, the blister will shrink.
Here is another, more common clinical challenge. Often, when children first present with
diabetes mellitus (Type I, or childhood diabetes), their body fluids are hyperosmotic due to the
11
high concentration of glucose in the ECF (lack of insulin prevents glucose uptake by most cells).
As insulin is administered in the clinic, the glucose is taken up by cells, and cellular metabolism is
rescued. Glucose concentration in the ECF falls. The potential danger concerns the brain.
Capillaries in the brain have a much lower permeability to nearly all substances than do systemic
capillaries (you will study this ‘blood-brain barrier’ in detail later). Thus, if plasma osmolarity falls
too quickly (insulin given too quickly), the glucose stranded in the brain ECF will create an
osmotic gradient across the brain capillaries (the plasma will be hypo-osmotic to brain interstitial
fluid) and the glucose in the brain ECF will osmotically suck water out of the capillaries.
Anywhere else in the body, this would produce swelling (edema), but the brain is encased in rigid
bone, and cannot swell very much. Instead, the pressure in the brain rises, which can seriously
disrupt brain function, sometimes with fatal consequences. This condition is called cerebral
edema. The danger is minimized by giving insulin slowly. The take-home message is simple: avoid
rapid changes in plasma osmolarity.
ECF composition can change. Finally, in the exercises above, we assumed that the ECF
composition didn't change (i.e., that the cells were bathed in a very large volume of ECF).
However, in the body, this condition does not hold: fluxes into or out of cells can and do change
the ECF composition. This is simply because the ECF volume is not large compared to the total
ICF volume. Specifically, in a healthy adult:
Total body fluid volume = 45 liters
ICF = 27 liters
*
Exchangeable ECF = 13 liters
*
The rest of the ECF (about 5 liters), is relatively inexchangeable, being sequestered
from plasma and interstitial fluid in the GI tract, bladder, etc.
Consider this example, which shows that K+ efflux from cells can indeed change K+
concentration in the ECF: Suppose a patient has a severe infection, with serious weight loss. The
weight loss is due largely to loss of muscle mass (muscle comprises much of the body mass, and
the main cation in the ICF is K+, so a lot of K+ leaks into the blood). Suppose 2% of the K+ in the
ICF leaks out. To what level does [K+]o rise? Assume this happens very quickly, and that none is
excreted by the kidneys (unlikely, unless renal failure has set in).
Answer: Normally, [K]i = 145 mM and assume ECF volume = 13 L and ICF volume = 27
L. So total K+ in cells = 3915 millimoles. If 2% (= 78.3 mmoles) leaks out, [K+]o would rise by
78.3/13 L = 6.0 mM from its normal of 4 mM. So final [K+]o = 10 mM. Not a big change – from
4 mM to 10 mM. Curiously, this change would be extremely dangerous, perhaps fatal, due solely
to its effect on membrane potential, which we will get to shortly.
A note on osmosis across capillaries. Plasma contains more proteins (mostly albumin, at a
concentration of about 1 mM) than interstitial fluid does. Capillaries are impermeable to protein.
So why don’t the proteins in plasma osmotically suck water out of the interstitial fluid? The answer
is that the blood is being pumped under hydrostatic pressure, which balances the osmotic 'suction'
created by the proteins in the plasma. This is one case where an osmotic gradient is not balanced
osmotically.
PROBLEM SET
1. The osmolarity, or osmotic concentration (osmolar, osM; milliosmolar, mosM) is the summed
concentration in formula weights of all solute particles -- undissociated molecules, anions and
cations. What is the mosM of the following solutions:
a.100 mM NaCl
12
b.100 mM K2SO4
c.100 mM CaCl2
d.100 mM glucose
e.100 mM glucose + 100 mM NaCl
2. The equivalent concentration (equivalents per liter, or more often milliequivalents (meq) per
liter, or mEq) of an ion in solution is used widely in clinical medicine, and can be confusing at first
when you try to convert between mM and mEq. The equivalent concentration is the number of
‘combining weights’ of an ion per liter. Combining weight concerns acid-base titration (e.g.,
titrating 100 mM H2SO4 takes twice as much base as does 100 mM HCl, so SO4= has twice as
many ‘combining equivalents’ per mole as Cl-). Converting from mM to mEq is a two-step
process: first, for each ion, convert to mosM, and second, multiply by valence of the ion in
question. In short, for each ion, multiply its osmolarity by valence to go from mM to mEq. (If the
solute is uncharged, mEq = mM.)
Try these examples: determine the concentration in
milliequivalents/liter (mEq) of each of the ions:
a.100 mM NaCl: Na+ = ___ mEq Cl- = ___ mEq
b.100 mM K2SO4: K+ = ___ mEq SO4= = ___ mEq
c. 100 mM CaCl2: Ca++ = ___ mEq Cl- = ___ mEq
3. Given below are the compositions of three widely used parenteral (not given by mouth; e.g.,
intravenous) fluids. Calculate the concentration in mM of each component, the osmolarity in
mosM of each dissolved species, and the equivalents per liter of each ion, in mEq.
a. "Isotonic" glucose solution: 60 grams glucose, add water to 1 liter.
b. “Isotonic" saline: 9 grams NaCl, add water to 1 liter.
c. Lactate-Ringer's Solution (NaLactate=Na-CH3CHOHCOO-): 6 g NaCl, 3.05 g Na-Lactate, 0.4 g KCl, 0.1 g CaCl2, 0.1 g MgCl2, add water to 1 liter. (By the way, Sydney Ringer was a
British physician who in 1882, while making up an IV fluid, accidentally substituted London tap
water for distilled water. The patient did much better, because of all the salts in the drinking water,
and thus began a study of the physiology of body fluids.)
4. The diagram below shows two fluid compartments separated by a membrane having "typical"
cell membrane permeability (i.e., impermeable to NaCl and glucose, highly permeable to urea - in
fact, as permeable to urea as it is to water, so the reflection coefficient = zero, which means it
(urea) exerts NO osmotic force). Concentrations at the start are also given.
MEMBRANE
NaCl
100 mM | 100 mM
Urea and water cross
Urea
100 mM ←→ 0 mM
the membrane equally well
Glucose 100 mM |
0 mM
a In which direction will water move?
b. What is the magnitude of the osmotic driving force for water (in mosM)?
c. Which way will NaCl move?
d. Which way will urea move?
5. A man eats 5.85 grams of NaCl (that’s 100 millimoles). Assuming all of the salt is absorbed
from the GI tract, none is excreted, and none enters cells, describe the changes (qualitatively only)
in volume of the ECF and ICF. (Given: normal mosM = 310; ICF volume = 27 l; ECF volume =
13
13 l).
ANSWERS
1. a.200 mosM b. 300 mosM c. 300 mosM d. 100 mosM e. 300 mosM
2. a. NaCl: 100 mEq Na+ & 100 mEq Clb. K2SO4: 200 mEq K+ & 200 mEq SO4=
c. 200 mEq Ca++ & 200 mEq Cl3.a. glucose: 60 g/l divided by 180 g/mole = .333 M = 333 mM = 333 mosM = 333 mEq.
b. NaCl: 9 g/l divided by 58.5 g/mole = 0.154 M = 154 mM = 308 mosM = 308 mEq.
c. Composition of Lactate Ringer:
NaCl: 6.0 g/l; divide by 58.50 g/mole, gives 0.103 M = 103 mM
NaLac: 3.05 g/l; divide by 112, gives 0.027 M = 27 mM
KCl 0.4 g/l; divide by 74.5, gives 0.0054 M = 5.4 mM
CaCl2 : 0.1 g/l; divide by 111, gives 0.0009 M = 0.9 mM
MgCl2: 0.1 g/l; divide by 95.0, gives 0.0011 M = 1.1 mM
Now calculate mosM:
Na+= 103 + 27 = 130 mosM
Cl- = 103 + 5.4 + 0.9*2 + 1.1*2 = 112.4 mosM
K+ = 5 mosM
Lac- = 27 mosM
Ca++ 0.9 mosM
Mg++ 1.1 mosM
TOTAL: 276.4 mosM
Now calculate mEq of each species:
Na+: 130 * 1 (valence) = 130 mEq
Cl-: 112.4 * 1 = 112.4 mEq
K+: 5 * 1 = 5 mEq
Lac- = 27 *1 = 27 mEq
Ca++: 0.9 * 2 = 1.8 mEq
Mg++: 1.1 * 2 = 2.2 mEq
TOTAL: 278.4 mEq
4 .a. right to left
b. 100 mosM
c. no movement
d. left to right
5. The quantitative answer to this question involves a lot of tedious algebra. Just think about it
intuitively: the added NaCl in the ECF will osmotically suck water out of the cells, until the
osmolarities of the ECF and ICF are equal. At the outset, the ECF contains 0.310 * 13 = about 4
moles of solute. The guy ate 0.1 moles of NaCl, which will dissociate to give 0.2 osmoles of
solute. That will increase ECF osmolarity by about 5% (4.2/4.0). So water will move out of cells,
diluting the ECF (and concentrating the ICF), until ECF and ICF osmolarities are equal. If ICF and
ECF volumes were the same, then the cells would shrink by about 5/2 = 2.5%. But ICF volume is
about twice as great ECF volume, so the cell shrinkage will be proportionately less (ICF will
14
shrink by about 1.5%, and ECF volume will increase by about 3%). That’s not such a big change in
plasma osmolarity, but the guy gets thirsty, which is a testament to the exquisite sensitivity of
neurons in the hypothalamic ‘thirst center’, which monitor the osmolarity of the plasma.
APPENDIX: Osmosis
Osmosis is an amazing force. Consider this experiment: The diffusion of water is studied
by placing some tritiated water (THO) on the right side of a chamber divided by a membrane with
molecular-sized pores (see drawing below, left side). The initial rate of appearance of THO on the
left side provides a measure of the unidirectional diffusion rate of water across the membrane.
Now the experiment is repeated with a little bit of nonpermeating solute on the left side of
the chamber (right panel of drawing). What should happen? Well, maybe the presence of
nonpermeating solute in the left chamber will
reduce the diffusion of water from left to right
a little (the solute takes up space that was
previously occupied by water), so the THO can
move more easily through the pores. Instead,
we find that the THO moves about 100 times
faster than before! The solute on the left
magically sucks water through the pores, just
as if a vacuum had been applied to the left side. Because the rate is so much faster than predicted
by diffusion, we give it a special name: osmosis (from the Greek, meaning "impulse"). The
direction water moves osmotically is the same as you would predict from only a consideration of
diffusion. What are the solute molecules doing to cause this effect? They must strike the
membrane, and interact with the pores. Evidently, as a solute molecule bounces off the mouth of a
pore, it creates a nanoscale negative pressure, which sucks the water through the pore.
15
Molecules to Medicine- Cell Physiology
3. Membrane Potential I 19 Nov 9AM
Bill Betz
[email protected] http://www.cuphys.net/cellphys
Reading (optional): Alberts et al, 4th ed: pp 631-634
Key words: Nernst equation, electrochemical gradient
3. Membrane Potential I
The Electric Force
We ended last lecture by considering a simple cell that consisted of an aqueous solution of
uncharged macromolecules which could not cross the plasma membrane. The model cell illustrated
an important problem facing real cells, namely controlling their volume. The problem arises
because of the presence of nonpermeating intracellular solutes. Animal cells solve the problem by
surrounding themselves with an external solution that also contains nonpermeating molecules. So
now we’re done with water. We know that solute concentrations have to be the same on the two
sides of the membrane to keep volume in check.
Until now, we have considered solute molecules to be uncharged, in order to illustrate the
principle of osmotic balance. In reality, however, most of the solutes in biological solutions are
electrically charged, and most are simple, inorganic ions. Today we will open a new can of worms
by considering the effect of electrical charge on the movement of solutes across membranes.
Consider the model cell in Fig. 1, which contains 300 mM nonpermeating, uncharged
solute (A), immersed in a solution of 150 mM NaCl. Note that the cell is in osmotic balance.
If the membrane is impermeable to both
Na+ and Cl-, nothing will happen and the cell will
survive. If the membrane is permeable to both
Na+ and Cl-, they both will diffuse into the cell
down their respective concentration gradients.
This will increase internal osmolarity, water will
enter the cell, and the cell will swell and burst.
Now what will happen if the membrane is
permeable to only one ion, say Cl-, and
impermeable to the counterion (Na+)? The Cl- ion
Fig. 1. Electrical gradient across membrane.
will continue to diffuse into the cell, as before.
But ions have more complex personalities than
uncharged solutes, owing to their extra baggage (an electrical charge). Thus, every time a Cl- ion
crosses the membrane, the inside of the cell acquires a net negative charge, because Na+ cannot
accompany the Cl-. This brings us to consideration of the electric force.
As you know, opposite charges attract and like charges repel each other. These forces are
very strong, much, much stronger in fact than osmotic forces. Recall that we said a beaker of H2O
will osmotically support a column of 1.0 M glucose 900 feet high. We can roughly compare this to
electrical forces by considering a mole of cations (6.02x1023 positive charges) confined to a one
liter box, with no anions inside to cancel the positive charges. Because cations repel each other we
can ask: How much weight do we need to keep the lid on the box? The answer is about 1022 lbs.,
which is about one-quarter the weight of the earth. This electric force is about 1018 times more
powerful than the osmotic force! Just try to intuit that.
As Cl- diffuses into the cell making it more negative, this electrical potential will tend to
16
repel further inward diffusion of the negative Cl- ions. And because the electric force is so
powerful, it doesn't take very many Cl- ions to produce a significant electrical potential inside the
cell. Eventually, a state will be reached wherein every time a Cl- ion diffuses into the cell down its
chemical (concentration) gradient, another Cl- ion diffuses out of the cell down its electrical
gradient, repelled by the negative interior. At this point the ion is in a state of equilibrium, and
requires no energy input to remain in this state, even though the concentration of Cl- is not equal on
the two sides.
The paragraph above contains qualitative descriptions of all the important points in this
lecture. It might be worthwhile to restate them:
1. The passive movement of an ion across the membrane is governed by two forces: its
concentration difference and the electrical potential difference across the membrane (the
membrane potential). The two forces, combined, make an electrochemical gradient.
2. Membrane potentials are produced by only one thing, namely an imbalance in the
number of cations and anions inside a cell.
3. The electric force is very much more powerful than the diffusional force produced by a
concentration difference, which means that relatively few excess ions are needed to counter large
concentration differences. In other words, as we will see, it is very safe and appropriate to consider
that the concentration of chloride in the cell does not change, or more generally, bulk solutions are
always electrically neutral. More on this later.
The Nernst equation. Returning to the model cell, we can ask: When will equilibrium be
reached, that is, when will the membrane potential be sufficiently negative to block the inward Cldiffusion? This is a quantitative question, and the equation that gives us the answer is called the
Nernst Equation. It is the most used equation in biology. It's most general form (applicable to any
ion) is: E = (RT/zF) lne (Co/Ci), where E= equilibrium potential; Co= outside concentration of the
ion; Ci= inside concentration; R= gas constant; T= temperature; z= valence of the ion in question;
F= Faraday constant.
The Nernst
Converting to base 10 logarithm, and at body temperature:
Equation
E is expressed in millivolts, mV. Please note that we
arbitrarily define E as the potential of the inside of the cell with respect to the outside. Thus, if E=
-40 mV, we say that the inside must be 40 mV negative to the outside in order for the ion in
question to be at equilibrium.
What is E? It is called the equilibrium potential for the ion in question. It is not an easy
concept, because it relates a concentration gradient to an electrical force (How can volts balance
molarity?) Specifically, it is the electrical potential difference across the membrane that must exist
if the ion is to be at equilibrium at the given concentrations. The Nernst equation can be applied to
any ion, and thus there are as many equilibrium potentials as there are ion species. The equilibrium
potential for chloride is ECl; for sodium it’s ENa; and so on – EK, ECa, EHCO3, EH, etc. If we know the
inside and outside concentrations of an ion, then we can calculate the value of E for the ion. Could
you calculate the equilibrium potential for sodium ions in Lake Michigan and the Pacific Ocean?
Yes, but it wouldn't mean very much (it’s about 120 mV, fresh water positive).
You can figure out the sign of an equilibrium potential easily. Suppose [K+]i = 140 mM and
+
[K ]o = 4 mM. Will EK be positive or negative? Well, to keep potassium from diffusing down its
concentration gradient out of the cell, we must make the inside of the cell negative to hold on the
positively-charged potassium ions. Nernst tells us that EK = -92.6 mV. Suppose an ion is at equal
concentrations across a membrane. What will be its equilibrium potential? (Answer: zero).
Equilibrium potentials are not real voltages - that’s why they’re written with an E (for
17
“electromotive force”) not a V. The real voltage is called the membrane potential, Vm, and can be
measured by impaling the cell with a microelectrode and measuring the real, honest-to-goodness
voltage difference between the tip of microelectrode inside the cell and a reference electrode
outside (Vm is also expressed in mV, and the sign is always written as the potential of the inside of
the cell). You will measure membrane potentials next August in the Synaptic Transmission Lab.
What does the value of E tell us? It's perhaps simplest to illustrate with an example. First
we calculate E for an ion, say Cl-, from the Nernst equation, knowing the chloride ion
concentration on each side of the membrane. Then we measure the real membrane potential with a
microelectrode. Then we compare the measured Vm with the calculated ECl. If they are the same,
then we can say that Cl- is distributed at its electrochemical equilibrium. We do not have to
postulate the existence of a Cl- "pump" to maintain it in a state away from equilibrium.
Consider a simple example. Assume [Cl-]o= 130 mM and [Cl-]i= 13 mM. Then ECl is
calculated from the Nernst equation to be -60 mV. Now we measure Vm by impaling the cell with a
microelectrode. Suppose it (Vm) is -60 mV. Well then, Vm= ECl, and we need look no further to
explain the distribution of Cl-; its concentration is ten times higher outside, but the internal
negativity (only a paltry 60 thousandths of a volt) is sufficient to keep Cl- from diffusing into the
cell. The electric force is indeed strong.
What if Vm is not the same as ECl? Then we can say that either a) the membrane is
impermeable to Cl-, or b) Cl- must be pumped across the membrane, because it is not distributed at
equilibrium. In this case, Cl- ions will perpetually diffuse in one direction, 'down' their
electrochemical energy gradient, and an exactly equal number of Cl- ions will be pumped in the
opposite direction. So the Cl- concentration in the cell won't change so long as the pump works.
Try these problems:
1. A cell has a membrane potential of -40 mV. It is permeable to Cl-. [Cl-]o= 130 mM and
[Cl-]i= 13 mM. Must there be a pump for Cl-, and if so, in which direction is Cl- pumped?
Answer: From the Nernst equation, ECl= -60 mV, which is different than Vm. Because the
membrane is permeable to Cl-, there must be a Cl- pump, because Cl- is not at equilibrium. Now the
hard part: in which direction does it pump the Cl-? There are different ways to think about it. Here's
one way (see drawing, next page): the Nernst equation says that to balance a ten fold concentration
gradient (130 mM outside, 13 mM inside) we need a 60 mV membrane potential (inside negative,
because it's an anion). In the drawing below, the arrows are vectors; their lengths reflect the
strength of the respective concentration and electrical gradients calculated from the Nernst
Equation. The electrical arrow says, "I need -60 mV to counter exactly the concentration force".
Ah, but the real membrane potential, Vm, is only -40 mV; the real arrow is shorter by one-third
than the one drawn. Thus, the net force on the ion is into the cell (the net force is obtained by
adding the two arrows). So the pump must pump Cl- out of the cell.
Here's another way to solve it: anthropomorphize, and give the Cl- ions a will. Their will is
to get to equilibrium, that is, to make Vm equal to ECl. To make Vm (-40 mV) move towards ECl (60 mV), the inside of the cell must become more negative, so the Cl- ions will leak into the cell.
18
Consequently, they must be pumped out of the cell.
[It's simplest to think about changing Vm, rather than changing ECl. Why? Because it takes
very few ions to change Vm (the electric force is very strong). What determines ECl? It's the
concentrations of Cl- inside and out, and it takes many more ions to change concentration than it
does to change Vm.]
Now try this one: A cell is permeable to Na+. Vm= -40 mV, [Na+]o= 140 mM, and [Na+]i=
14 mM. Is there a Na+ pump? In which direction does it pump Na+? Answer: Na+ is pumped out of
the cell.
One more problem: A cell is permeable to H+ ions. [H+]o= 50 nM, [H+]i= 100 nM, and Vm=
-80. In which direction is H+ pumped? (Note: LOG10 (0.5)= -0.3.) Answer: H+ is pumped out.
19
Molecules to Medicine- Cell Physiology
4. Membrane Potential II 19 Nov 10AM
Bill Betz
[email protected]
http://www.cuphys.net/cellphys
Reading (optional): see previous lecture
Key words: Principle of Electrical Neutrality, Donnan relation, electrogenic, steady state, ion
exchange resin, dialysis
4. Membrane Potential II
We have now covered all of the hard stuff. From here on, there really is nothing
conceptually new. We will be using the same rules that we have been discussing (diffusion,
osmosis, equilibrium and membrane potentials) to make the model cell more complex, more like
a real cell.
The NaCl cell. Let's go back to the model cell we were discussing earlier (reproduced in
Fig. 1, left panel), and consider a sample calculation. With the model in Fig. 1 (left panel) it is
impossible to calculate a realistic value for the equilibrium potential for Cl-, ECl (because [Cl]i= 0,
ECl= -infinity). Let's modify things slightly by putting a little NaCl inside the cell, as well as
outside (below, right). As before, internal solute A0 and Na+ do not permeate, Cl- and H2O do
permeate. Assume [A]i = 100 mOsm and [Na]i = 50 mM. Calculate the values of [Na]o, [Cl]i, and
ECl at equilibrium.
Fig. 1. A simple cell (left) acquires some NaCl inside (right).
Electrical neutrality. In order to solve this problem, we have to consider one more rule,
which we have already discussed in qualitative terms, and is pretty elementary and obvious. The
rule is called The Principle of Electrical Neutrality and it says simply that bulk solutions (inside
and out) have to be electrically neutral: the total cation concentration in the external solution must
equal the total anion concentration in the external solution. The same holds for the internal
solution: [cations]i = [anions]i. It's obvious in one sense, in that if you dip a glass of water from the
ocean, you get the same number of anions as you get cations, every time. However, this clearly
contradicts something said earlier, namely that membrane potentials arise from an excess of one
type of ion in the cell; a cell with Vm= -50 mV contains more anions than cations. Well, both are
correct. The key is once again to note that the electric force is so awesomely powerful that, in a
cell, the excess number of anions is very, very small compared to the total number of ions present
in the cell. For example, in a typical cell with a resting membrane potential of -80 mV, for every
100,000 cations in the cell there will be approximately 100,001 anions (see Appendix for
derivation). So electrical neutrality is only an approximation, but it is a very good one; it is
20
impossible to measure directly an ion's concentration to an accuracy of one part in 100,000, but it's
easy to measure a membrane potential of -80 mV (in fact, you will do it yourself next August when
you impale cells in the Synaptic Transmission Laboratory in the Neuro Block).
Now we can solve the problem in Fig. 1 (right side). First, if [Na+]i = 50 mM, then [Cl-]i
must also = 50 mM according to the principle of electrical neutrality. Second, total internal
osmolarity = 200 mosM, which must equal total external osmolarity to keep the cell from swelling
or shrinking. In other words, [Na+]o + [Cl-]o= 200 mosM. By electrical neutrality, [Na+]o= [Cl-]o.
Thus, [Na+]o= 100 mM = [Cl-]o . Third, from the Nernst Equation, we calculate the chloride
equilibrium potential, ECl= -18mV.
The Donnan rule. Continuing with the model cell, let's add some potassium to the system.
Specifically, let's replace the sodium inside the cell with potassium, and make the membrane
permeable to K+, as well as Cl- (Fig. 2).
First, note that the membrane must be kept
impermeable to Na+ or else the cell will swell and burst. What
about potassium? Clearly it will start to diffuse out of the cell,
down its concentration gradient. But as K+ diffuses out, the
inside becomes more negative, and this electrical gradient will
tend to retard further outward movement of the
positively-charged potassium ions. Thus, qualitatively, we can
imagine a situation for K+ that is like the one for Cl-, with
Fig. 2. Internal Na+ replaced by K+.
electrical and concentration forces cancelling each other,
allowing the cell to exist at equilibrium.
The problem is, can both permeating ions, K+ and Cl-, be at equilibrium simultaneously?
Well, if they can work things out, then there must be an arrangement whereby ECl = EK = Vm,
because a cell can have only one value of membrane potential. Then we can write: EK = 60 log
([K+]o/[K+]i)= ECl= -60 log ([Cl-]o/[Cl-]i). This simplifies to the Donnan Rule:
[K+]o.[Cl-]o = [K+]i.[Cl-]i
You'll note that there really is nothing new here. We have just equated the Nernst Equation
for two ion species. By equating them, we require that both ions be distributed at equilibrium.
The KCl cell. Consider an example of a cell permeable to both K+ and Cl-. First, no realistic
value of EK can be calculated if [K]o= 0 (EK= -infinity). So we will replace a little of the
extracellular Na+ with potassium (Fig. 3).
In solving this, we now have four rules we can use: 1) osmotic balance; 2) charge
neutrality; 3) Nernst Equation and 4) Donnan Equilibrium.
21
Here's how it's done: Because [K]i = 25mM, [Cl]i = 25mM (charge neutrality). Total
internal osmolarity = [K]i + [Cl]i + [A] = 308 mOsm = total external osmolarity. [Cl]o = total
external osmolarity/2 = 154 mM (charge neutrality). Now we need [Na]o and [K]o. Using the
Donnan Rule: [K]i.[Cl]i = 625 = [K]o.[Cl]o. Because [Cl]o = 154, [K]o = 625/154 = 4 mM. Finally,
[Na]o = [Cl]o - [K]o = 150 mM (charge neutrality). The equilibrium potentials, EK and ECl, are
calculated from the Nernst Equation: ECl= -60 log (154/25)= -45.5mV, and EK= 60 log (4/25)=
-45.5mV.
Fig. 3. The KCl cell.
Intracellular "big anions" are negatively charged.
One final change in the model is necessary. In real cells, the
internal impermeant species (mostly proteins, SO4=, and
HPO4=) carry a net negative charge. Assume that each A
molecule carries n negative charges (in real cells, n = -1 to 2). We can write down these simple equations:
1. Charge balance: [K]i + [Na]i = [Cl]i +
n[A] and [K]o + [Na]o = [Cl]o
2. Donnan Rule: [K]i.[Cl]i = [K]o.[Cl]o
3. Osmotic balance: [K]i + [Na]i + [Cl]i +
[A] = [K]o + [Cl]o + [Na]o
These are exactly like the previous example, except now n[A] appears in the charge
balance equation. Use these rules to solve the example in Fig. 4. (Answers: [K]i = 145 (osmotic
balance); EK =-87.7 mV; ECl = EK; n = -1.22)
So now we have fully accounted for
the cell whose composition was given at the
beginning of the first lecture. The model cell
could not be more efficient; it requires no
energy to remain in this state, even though the
membrane is permeable to K+ and Cl-, which
are not distributed evenly across the
membrane. Living cells, however, don't
behave quite so efficiently, and next we will
abandon model cells and get into real biology.
Fig. 4. The complete model cell. (Solve
for [K+]i, ECl, EK, and n)
22
Real cells
Finally, biology!
Sodium crosses the membrane. Until about 1940, the model cell discussed above was
generally accepted as an accurate representation of the way real cells maintain their composition.
Based on simple physical laws, infinitely efficient, the model attested to nature's ingenuity and
economy. The keystone of the model is the membrane's impermeability to sodium. Repeatedly, we
saw that if Na+ could cross the membrane, the cell would swell and burst. Thus, about 1940, it was
a major discovery to find that real cells are permeable to Na+.
If red blood cells are incubated in a medium containing radioactive 24Na+, and then washed
and counted for radioactivity, it is found that the cells have become 'hot' with radioactive Na+. If
the hot cells are then incubated in a medium without 24Na+, they lose their counts. This
fundamental discovery was made by a medical student (Leon Heppel), working part time in a lab at
the University of Rochester Medical School.
Consider the electrochemical energy gradient for Na+: Because the concentration of Na+
outside is higher than inside, the concentration gradient is directed into the cell. Because the inside
of the cell is electrically negative, the electrical gradient is also directed into the cell. Thus, given
the opportunity, Na+ is going to leak into the cell. So why don't real living cells fill up with Na+?
The answer is that Na+ is pumped back out of the cell. This means performing work, and the
energy to do the work comes from metabolism, by splitting ATP. The Na+ pump is found in all
animal cell surface membranes. If this pump is blocked, Na+ enters the cell, water follows, and the
cell swells. The pump can be blocked by interfering with ATP production (e.g., low temperature,
cyanide, hypoxia) or by specific drugs (mainly a group of plant alkaloids called cardiac glycosides,
which includes digitalis, ouabain, strophanthidin, and other clinically useful drugs) that selectively
block the Na+ pump by binding to a site on the outside surface of the membrane.
Potassium is pumped too. The Na+ pump also pumps potassium into cells. In fact, both Na+
+
and K ions must be present simultaneously, or the pump won't work. For example, the pump can't
extrude Na+ from the cell very well if the potassium concentration in the ECF is low. Also, cells
with a poisoned pump lose K+ as they gain Na+ (they still swell, because they also gain Cl- as Na+
leaks in). Thus, the sodium pump is really an obligatorily coupled sodium-potassium exchange
pump.
Pump kinetics.. The Na/K pump is "saturable", that is, it has an easily demonstrable
maximum rate of activity of only about 100 cycles per second. In general, such saturation is
characteristic of carrier mediated transport, but not of most fluxes through ion channels, which can
transport millions of ions per second.
Pump stoichiometry. Another property of the pump (not illustrated here) is that it is not a
1:1 pump; rather, the Na/K ratio is 3 Na+ for 2 K+. Thus, the pump is electrogenic, and not
electro-neutral. This has the effect of making Vm a little bit more negative than it would otherwise
be. The effect in most cases is very small, so that for our purposes we can ignore its
electrogenicity. It does, however, become important in certain neurons, which you will study next
year.
Pump mechanism. A simple way to envision the Na/K pump is like a channel with 2 gates,
one at the outside surface of the membrane, and one at the inside (see cartoon below). Only one
gate can be open at a time; both gates are never open simultaneously. In addition, the ion binding
affinity of the channel switches between Na and K, depending on which gate is open. ATP
provides the energy for the transitions (gates swinging and affinity changing).
Pump structure. The pump contains a large alpha subunit, and a smaller beta subunit, but
23
the exact (atomic scale) structure is unknown (it has not yet been possible to make crystals suitable
for x-ray diffraction studies).
The Na/K pump cycle.
A. Both gates are closed, 2 K+ ions
are inside.
B. ATP binds, the inner gate opens,
and affinity changes from K+ to Na+.
So K+ leaves and Na+ enters.
C. `ATP is split, leaving the pump
phosphorylated. The inner gate closes.
D. Spontaneously, the outer gate
opens and affinity changes from Na+
to K+.
D to A. The pump loses its phosphate
group, and the outer gate closes,
completing the cycle.
Let's return for the last time to the model cell we discussed earlier, and give it
honest-to-goodness biological significance by making it permeable to Na+ and adding a Na+/K+
pump (Fig. 1).
Fig. 1. A real cell, permeable to Na, with a
Na pump
Now that it is permeable to Na+, we need to examine how Na+ moves across the membrane.
We do this, of course, by calculating ENa, the sodium equilibrium potential, from the Nernst
equation: ENa= 60 log (140/14)= +60 mV. EK hasn't changed; it is still -85 mV. Now the question
is: What will determine the value of the measured membrane potential, Vm?
We have made a very fundamental change in going from the model cell to real biology.
The cell is no longer in a state of equilibrium; it requires a constant input of energy (from
metabolism) to keep its ICF composition from changing. It is in a steady state, which means that,
like the model cell at equilibrium, the ion concentrations aren't changing over time, but unlike the
model cell, a constant input of energy is needed in the real cell (in the form of ATP, to drive the
Na/K pump).
Thus, in real cells, Na+ is constantly leaking into the cell, trying to get to equilibrium. What
is equilibrium for Na+? It simply means getting Vm to equal ENa. For potassium, equilibrium is
24
achieved when Vm is equal to EK. So a continuous struggle ensues between the two ions - Na+
leaking in, trying to pull Vm up to ENa, and K+ leaking out, trying to pull Vm down to EK.
So, which one wins, Na or K? The answer is: it depends. It’s different for different cells. If
we were to take a micropipette and impale different cells all over the body, we would find a large
range in resting membrane potentials (Fig. 2). They are all negative, but range from nearly zero to
nearly EK.
What are the differences due to? The answer is:
relative permeability. A cell with many more K+
channels than Na+ channels will have a membrane
potential close to EK. Nerve and muscle cells are
examples; Vm is -70 to -90 mV. Conversely, a cell with
relatively more Na+ channels will have a membrane
potential closer to ENa. Glial cells are nearly perfect
‘potassium electrodes’ (permeable only to K), while red
blood cells are about equally permeable to Na and K.
A second determinant of Vm is ion concentration.
But in normal resting cells, this isn’t important, because
all cells have similar intracellular compositions (rich in
K) and are bathed in similar ECF (rich in Na). The most
important exception to this, of great medical importance,
is potassium. A small change in ECF potassium
concentration has a big effect on EK and (in nerve,
muscle, and other cells highly permeable to potassium) a
big effect on Vm.
Quantifying this intuitive explanation is not
Fig. 2. A sampling of resting
difficult. It involves nothing more than Ohm’s Law,
membrane potentials in different
which you may remember in its most familiar form:
cell types.
V=I.R. Here, V is the ‘driving force’ on the ion, I is the
current carried by the ion, and R is the membrane resistance to the ion. It's easier to think of
conductance, G, which is the reciprocal of resistance (G=1/R), which is close to membrane
permeability (the higher the permeability, the higher the conductance). Thus, Ohm’s law says,
I=(Driving Force) x G.
This says simply that the current carried by an ion (say, Na+) through the membrane is
given approximately by the product of [driving force on the Na+ ion] times [sodium conductance]
(conductance is proportional to the number of sodium channels that are conducting). It makes
sense: more channels, more current, and more driving force, more current.
What is the driving force on sodium? It is not just the value of membrane potential, Vm.
Think of it this way: At what membrane potential will the net movement of Na+ be zero? That will
occur when Vm= ENa (Na+ will then be at equilibrium). In other words, when Vm= ENa, the driving
force is zero. Thus, the driving force at any instant is the difference between Vm and ENa.
So the way we write Ohm's Law for sodium is as follows:
INa= GNa.(Vm-ENa).
The same holds for K+: IK= GK.(Vm-EK).
When these two currents are equal and opposite (INa = -IK) there will be no net movement
of charge across the membrane (we are ignoring other ions), so Vm will be at its resting value.
If we set INa = -IK and substitute, we get GNa.(Vm-ENa) = - GK.(Vm-EK).
25
Solving this for Vm gives: Vm = (GKEK + GNaENa) / (GK + GNa).
To illustrate that what matters is relative conductance, let GR=GNa/GK.. GR is the ratio of
sodium to potassium conductance, which gives
Notice that besides EK and ENa (which differ little from cell to cell), the only variable is GR, the
relative conductance of the membrane to sodium and potassium.
In summary, for all practical purposes (that is, under normal physiological conditions),
membrane potential is determined by relative conductance.
We can illustrate this with a simple example.
Consider two cells (Fig. 3). One (leaky membrane) has
100 K channels and 10 Na channels. The other (tight
membrane) has only a tenth as many channels (10 K and 1
Na). Their membrane potentials will be the same (in both
cases, GR = 0.1). But the cells are clearly different – how?
The leaky cell will have ten times more sodium leaking in,
Fig. 3. A leaky cell (left) and a tight
and ten times more potassium leaking out, compared to the cell (right) may have the same
tight cell, and so will have to expend 10 times more energy membrane potential.
to keep the Na/K pump working. Thus, the tight cell is ten
times more efficient, energetically speaking.
Here is an aside: You may read that the Goldman Equation gives the value of Vm. This is
correct. The Goldman Equation is very similar to Equation 1 (above):
where PR=PNa / PK (as before), the relative permeability of the membrane to sodium and
potassium. The little dots (...) mean that we should include all ions that can cross the membrane
(e.g., H+, Cl-, Ca++), but Na+ and K+ are the most important ones because they can carry the biggest
currents and both are pumped. The differences between the Goldman Equation and Equation 1
(above) are subtle, and you can ignore them.
The sodium/potassium pump and membrane potential. While the Na/K pump is absolutely
necessary over the long run, in many cells it plays only a negligible short term role. For example, if
the Na/K pump is completely blocked by drugs, nothing much happens initially (in many cells).
Gradually, of course., the cell will fill up with Na, and lose its K. As these changes occur, both ENa
and EK move towards zero, and cells will depolarize. How long it takes depends simply on how big
and how leaky the cell is. In a large, spherical cell (which has a low surface-to-volume ratio), with
few ion channels, it might take many hours for these changes to occur. On the other hand, in a cell
with a relatively large surface area (like many neurons in the brain, which have long, slender
processes), and a high density of channels, blocking the sodium pump can cause them to fill up
with Na and lose K in a matter of seconds.
This finishes our consideration of the basis of the resting membrane potential. We will now
turn to a few examples that illustrate how cells can change their membrane potentials from the
26
resting value, and how some clinical situations can be understood in these fundamental terms.
Given this cell: ICF (mM) ECF (mM)
Na+
20
150
+
150
5
K
Assume that the cell contains sufficient Cl- and A-n to be in a steady state, and in osmotic
balance.
From the Nernst equation: EK= -89 mV; ENa= +53 mV.
Assume PK= 70.PNa, which is typical for nerve and muscle. Calculate Vm from the Goldman
equation (answer: Vm= -79 mV).
OK. Now, what happens to Vm, EK, and ENa if the following things occur (Fig. 4): A. PNa
increases; B) [Na+]o decreases a lot, and C) [K+]o increases a little bit?
Fig. 4. A: PNa increases. B: [Na+]o decreases. C: [K+]o increases.
A. PNa undergoes a huge (700 fold) increase (this happens during an action potential), such
that PNa= 10.PK (Fig. 4A). We can say right away that EK and ENa will not change (no changes in
concentrations, at least in the short term). Using the Goldman equation,we calculate Vm= +38 mV,
a huge depolarization.
To say it in words, when PNa increases, Na+ ions are able to move into the cell much more
easily; the inward movement of positive charge depolarizes the membrane. As Vm moves closer to
ENa, the driving force on the Na+ ions is reduced, while the driving force on K+ is increased as Vm
moves away from EK (remember, the driving force on any ion is just the difference between Vm
and the ion's equilibrium potential). The new steady potential (+38 mV) is reached when the total
amount of Na+ entering equals the total amount of K+ leaving the cell. In other words, when the
Na+ and K+ currents are equal and opposite, Vm doesn't change.
2. External Na+ is reduced to 20 mM (that is, most Na+ is replaced with a nonpermeating
cation, such as choline+). This is illustrated in Fig. 4B. First of all, if we reduce [Na+]o, won’t that
make the extracellular fluid very negative, and so depolarize the cell? No! The solution is always
neutral. To reduce the sodium, we have to reduce an anion (Cl- probably) by exactly the same
amount. (we also would need to add solute (sucrose, for example) to keep the osmolarity constant).
The real answer is that ENa will move closer to zero, because the change in [Na+]o has made the
Na+ concentrations in the ICF and ECF more nearly equal (when [Na+]o= [Na+]i, ENa= 0, because
the log of 1 is zero). From the Nernst equation, ENa= 0 mV. From the Goldman Equation1, Vm=
27
-87 mV.
Thus, a very large change in [Na+]o (from 150 to 20 mM) produced an 8 mV
hyperpolarization (Vm changed from -79 to -87 mV). Clinically, one never sees such large
reductions in the plasma Na+ level. This is because Na+ is the major cation in the ECF, and so is
one of the major osmotic particles. Thus, Na+ loss would be accompanied by water loss, which
would reduce blood volume, without such a large change in Na+ concentration. The point for now
is simply that changes in ECF levels of Na+ have little effect on the membrane potential of this cell,
simply because the membrane is relatively impermeable to Na+.
3. External K+ is raised to 15 mM (Fig. 4C). Why is the following reasoning incorrect?
Because extracellular potassium ion concentration increased, and because it’s a cation, the
external solution became positively charged with the extra K+, so the cell membrane
hyperpolarized. (Answer: The previous statement is incorrect. Owing to the enormous strength of
the electric force, bulk solutions are always neutral. So, to raise [K+]o, we must either reduce
another cation (Na+) by the same amount, or add more anions (Cl-) with the K+.)
Intuitively, we can see that an increase in [K+]o will reduce the efflux of K+ from the cell.
Because less K+ leaves, the cell will depolarize. EK will move closer to zero (we have made the
internal and external concentrations more nearly equal); from the Nernst equation: EK= -60 mV.
That's a big change (almost 30 mV), and reflects the fact that the ratio of [K+]o to [K+]i determines
EK, and we have tripled the ratio with only a 10 mM change in [K+]o.
Because Vm must fall between EK and ENa, we predict a substantial depolarization, and
calculate from the Goldman equation: Vm= -57 mV. So you see, a rather modest change in [K+]o
(from 5 to 15 mM) has produced a large change in Vm (from -79 to -57 mV). Clinically, this can be
very important. First, it's easy to imagine how blood potassium could increase if a little bit of the
K+ leaked out of cells (remember, over 98% of the total K+ in the body is in the ICF). And second,
a 20-30 mV depolarization of cells in the heart can quickly lead to cardiac arrest.
In summary, external Na+ has little effect, and external K+ has a marked effect on
membrane potential in nerve and muscle cells. This is simply because the plasma membrane of
nerve cells and muscle cells is much more permeable to K+ than to any other ion: wherever EK
goes, Vm follows in these cells.
Clinical notes on acute hyperkalemia
As we saw above, a rather modest rise in ECF potassium ion concentration has a big effect
on Vm. This can have serious clinical consequences. The reason that such a small change (only 4
mM) has such a big effect goes back to the Nernst equation: the potassium equilibrium potential,
EK, depends on the ratio of external to internal potassium ion concentrations, and the ECF
potassium concentration has doubled. So we expect a big change in EK. In cells that have a
relatively high potassium permeability, Vm will slavishly follow EK, and so the membrane potential
will depolarize a lot. A large depolarization of cells can be life threatening.
In acute hyperkalemia, the main danger concerns the reliable conduction of electircal
signals (action potentials) in the heart. As you will study a little later, the heart relies on a special
electrical conduction system (intrinsic to the heart) to coordinate the contraction of its muscle
fibers each heartbeat. These synchronized electrical signals can become disrupted during acute
hyperkalemia, causing cardiac arrhythmias as conduction blocks occur and maverick pacemakers
arise in various locations of the conduction system.
The causes of hyperkalemia, as you might expect (knowing that 98% of all of the
potassium in the body fluids is in the ICF) mostly concern loss of potassium from cells. Crush
injuries, burns, and other trauma that disrupt cell membranes can do it. So can immunological
28
attack of red blood cells (causing hemolysis). One of the most important determinants of the
clinical course of the hyperkalemia is the status of the kidney, whose normal job it is to excrete
excess potassium. If kidney function is compromised, hyperkalemia can be much more serious
than if the kidney is functioning normally.
The diagnosis of hyperkalemia usually is via an electrocardiogram (EKG) to detect cardiac
arrhythmias, followed by measuring plasma potassium ion concentration.
CBIGK: The treatment of acute hyperkalemia involves attempts first to relieve any cardiac
arrhythmias, usually by giving intravenous calcium ions (you will study the mechanism by which
calcium ions quiet the conduction system later). Next, efforts are made to reduce the concentration
of potassium in the plasma. This can be done by encouraging cells to take up potassium by
alkalinizing the blood by giving sodium bicarbonate (Alka Seltzer), or by juicing up the energy
supply (ATP) for the sodium-potassium pump by giving insulin and glucose. Finally, potassium
can be removed from the body by administering an ion exchanger (oral or enema) like Kayexalate
(the exchanger is a big anion that is given as the sodium salt, but it has a higher affinity for
potassium than it does for sodium, so the it selectively binds up potassium ions).
A nice mnemonic always helps: when you C BIG K in a patient, what should you give, in
order? (answer: Calcium, Bicarbonate, Insulin + Glucose, Kayexalate). (In actual practice,
bicarbonate is used less often.)
Finally, a more drastic way to cleanse the blood of potassium is dialysis by an ‘artificial
kidney’. Blood is taken out of the body and passed through plastic capillary tubing, which allows
free exchange of ions and small molecules between the blood and the dialysate that bathes the
tubing. Because the dialysate contains a lower concentration of potassium than the plasma,
potassium leaves the blood, which is then returned to the patient, cleansed artificially of its
excess potassium ions. (Of course, the dialysate also contains normal concentrations of other
salts.)
APPENDIX: Excess anions
How many excess anions are there in a typical cell? Is the number big, compared to the
total number of ions in the cell?
Given: A spherical cell (radius= 15 µm = 0.0015 cm) with a membrane potential of -80 mV
and an internal cation concentration of 150 mM.
1) Excess anions. Consider the membrane as a charged capacitor (Like the dielectric in a
capacitor, the lipid bilayer can hold charges apart. The ICF and ECF are analogous to the metal
plates in a capacitor). The voltage V on a capacitor is given by: V= q/c, where q= coulombs of
charge, and c= capacitance. We are given that Vm= -80 mV. For all biological membranes, c=
about 10-6 farads/cm2. Thus q= 8.10-8 coulombs/cm2. Cell surface area A= 4.π.r2= 2.8.10-5 cm2. So
total cell q= 2.26.10-12 coulombs. There are about 6.2.1018 monovalent ions in one coulomb (i.e.
Avogadro's number/Faraday constant= 6.02.1023/96,500), so total excess anions = 1.4.107 ions. So
there is an excess of about fourteen million anions in the cell. That's a pretty sizable number.
2) But is 14 million significantly different from the total number of cations in the cell? The
total number of cations in the cell= (6.02.1023).Volume.Concentration. Cell volume= 4/3.π.r3=
1.4.10-11 liters. This gives the total number of cations as 1.28.1012, or 1.28 trillion cations. In other
words, there is about one excess anion for every 105 cations. In conclusion, the Principle of
Electrical Neutrality is an excellent approximation: bulk solutions are electrically neutral.
29
Molecules to Medicine- Cell Physiology
5. Membrane Transporters 20 Nov 9AM
Bill Betz
http://www.cuphys.net/cellphys
[email protected]
Reading: Alberts et al, 4th ed: pp 615-631.
Lehninger, 4th ed: pp 389-411.
Key words: primary active transport, secondary active transport, cotransporter, exchanger,
electrogenic
5. Membrane Transporters
Earlier we talked about the famous sodium-potassium pump, the ubiquitous transporter
that renders cells ‘functionally impermeable’ to sodium ions, and keeps the cells from swelling
uncontrollably. Today we will consider other types of carrier mechanisms.
The first thing to note is that none of them, including the Na/K pump, actually ‘carries’
solute cargo across the membrane. There is an antibiotic (valinomycin) that actually diffuses back
and forth across the membrane (shuttling potassium ions), but no biological molecule is known to
act this way, so transporter is a better word than carrier. How do protein transporters work? For
most, we don’t really know, but the general model of the Na/K pump (a tube with a gate on each
end and dynamic (changeable) binding affinity) is probably a good model for the others as well
(see Handout #4, page 5).
Conceptually, matters are pretty simple. Some transporters act like ion channels, shuttling a
single solute species in either direction. This is called facilitated diffusion (a historic term applied
to molecules that shouldn’t be able to diffuse across lipid membranes
because of their large size or charge, but do get across). The best known
example of this is the glucose transporter (cartoon at right). The glucose
transporter will transport glucose in either direction, and burns no energy
in the process. Thus, it is not a pump. You might then wonder how cells
accumulate glucose. The answer is that as soon as a glucose molecule gets
into the cell, it is phosphorylated to Glucose-6-Phosphate, which doesn't
fit on the transporter, and so is "trapped" inside.
Glucose uptake by cells is regulated by insulin, a hormone secreted by specialized cells in
the pancreas when plasma glucose levels rise. How does insulin turn on the transporter? It turns out
that in the absence of glucose, the transporter is not even present in the plasma membrane; it is
sequestered inside the cell, in the membrane of intracellular vesicles. Insulin triggers a biochemical
cascade that causes the vesicle membranes to fuse with the surface membrane (exocytosis),
exposing the glucose transporter to the ECF. The transporter then gets busy and ‘carries’ glucose
inside. When insulin subsides, the transporter molecules are reinternalized (endocytosis).
Pumps. Now we turn to transport proteins that can cause a substance to be accumulated or
expelled against its electrochemical gradient. Dozens have been identified, pumping nearly every
substance of importance in the body in one cell or another. The only evident exceptions are water
and urea (the end product of nitrogen metabolism in mammals). Isn’t it odd that no water pumps
exist anywhere in nature, given its importance? Water always moves down its concentration
(osmotic) gradient. By definition, pumps move solute against an energy gradient, and so require an
energy source. We subdivide pumps into two categories, depending on the source of that energy.
Primary active transporters, like the Na/K pump, derive their energy directly from the
splitting of ATP. There are no other ubiquitous primary active transporters in the plasma
30
membrane of cells. Some specialized cells have them, and you will study them later. For example,
the cells in the stomach that secrete acid, and certain cells in the kidney geared for excreting
protons from the body possess proton pumps in their plasma membranes that rely directly on ATP
for their energy source.
Inside cells (as opposed to the surface membrane) there are other primary active
transporters. One pumps protons into intracellular membrane-bound organelles (endosomes,
vesicles, lysosomes). Another pumps calcium ions into membrane-bound compartments. All
proton and calcium pumps are oriented to pump these ions out of the cytoplasm.
Inside mitochondria is a very special proton pump (the F1-ATPase, or ATP synthase) that,
when running backwards, lets protons leak across a membrane and synthesizes, rather than
hydrolyzes ATP.
Secondary active transport is the mechanism by which most substances are pumped. In this
case, the energy to do the direct work of pumping comes not from metabolism (ATP), but from a
secondary source. Usually this energy source is the 'downhill leak' of Na+ into the cell. For
example, cells can accumulate amino acids against their energy gradients. This active uptake is
dependent on external Na+; if external Na+ is removed, amino acid uptake is abolished.
Conversely, removing the amino acid reduces the entry of Na+. The carrier ingeniously captures
the energy released by the inward leak of Na+ and instead of letting it escape as heat, uses it to
pump the amino acid into the cell.
There are two basic types of secondary active transporters, those that move different solute
species in the same direction (cotransport), and those that move solute in opposite directions
(antiport, or exchange). The cartoons below illustrate these two types (arrows pointing up mean the
substance is pumped; down means leak).
Secondary active transporters do not necessarily always run in the same direction. They
will always tap the bigger leak to drive the smaller pump. Consequently, they can reverse direction
sometimes. One of the most important examples of this is the sodium-calcium exchanger, which
reverses direction in heart muscle cells every time the heart beats (more on this later).
All secondary transport mechanisms depend ultimately on the Na+/K+ pump (and therefore
on ATP). For example, if the Na+/K+ pump is blocked, cells fill up with Na+, and thus the Na+
electrochemical gradient is reduced. Because this is the energy source for secondary active
transport, all of these transport mechanisms suffer.
Some secondary active transporters are electrogenic, in that one cycle produces a net
charge transfer across the membrane. For example, Na/amino acid transporters are electrogenic,
because one cycle transfers a net positive charge (Na+) into the cell. Other secondary active
transporters are not electrogenic; an example is the Na/K/2Cl cotransporter, which each cycle
moves one sodium ion, one potassium ion, and two chloride ions into the cell. The main feature of
electrogenic secondary active transporters is that their activity is governed by the membrane
potential (as described in the Appendix). Electrically silent transporters could not care less about
31
membrane potential.
Here are some important examples of secondary active transporters driven by the inward
sodium leak:
Calcium transport. There is a huge electrochemical gradient for calcium ions across cell
membranes. In fact, no other ion is further from equilibrium than calcium. The extracellular
(ionized) calcium concentration (about 1 mM) is nearly 10,000 times greater than the intracellular
concentration (about 0.0002 mM, or 200 nM); thus its concentration gradient is inward. And the
electrical gradient is also inward, of course, because the ICF is electrically negative and calcium is
positively charged. From the Nernst equation, ECa is calculated to be about +111 mV (ECa =
(60/2)*log(1/0.0002) = +111 mV). Thus, given the opportunity (i.e., an open calcium channel),
Ca++ ions will always leak into cells, so there must be a pump to extrude them. The Na/Ca
exchanger’s main job is to pump calcium ions out of the cell. The inward leak of sodium ions
provides the energy source.
The Na/Ca exchange pump takes on a special significance in the heart, where it actually
switches direction during each heartbeat. (You will study muscle contraction shortly, and will learn
that calcium ions are necessary for contraction.) At rest (while the ventricles are refilling with
blood during diastole), the exchanger runs forward, pumping Ca++ out (keeping the muscle
relaxed) as Na+ leaks in. When the ventricles contract (systole), the exchanger switches direction,
letting Ca++ leak into the cell, where it strengthens the force of contraction. The direction of Ca++
movement is controlled by the value of the membrane potential; when Vm is more negative than
about -60 mV, the sodium leak rules, and drives the outward pumping of calcium; when Vm is
more positive than -60 (during the action potential), the pump reverses direction, and calcium leaks
in (pumping sodium out).
Digitalis. For centuries it has been known that an extract of the beautiful purple foxglove
can help a weak heart beat stronger. Digitalis (and related drugs) exert their action by acting
directly on, not the Na/Ca exchanger, but the Na/K pump! In fact, digitalis blocks the Na/K pump.
How does this lead to an increase in the strength of contraction of heart muscle? Blocking the
Na/K pump of course allows intracellular sodium ion concentration to increase. In turn, this
reduces the energy available to all sodium-driven secondary active transporters, including the
Na/Ca exchanger. Thus, digitalis indirectly inhibits the Na/Ca exchanger, allowing intracellular
calcium ion concentration to rise, which increases cardiac contractility.
Hydrogen ions (protons) are also pumped out of most cells by a Na+/H+ exchange carrier,
which operates under the same principles as the Na/Ca exchanger. Protons are harder to study than
UFO's, because they capriciously vanish and reappear as they bind to and unbind from various
buffers. Typically, only about 1 in a million protons is free, as H+; the rest are hiding, bound to
buffers. The free concentration of protons in the ICF is about 100 nM (pH=7.0); in the ECF it’s
even lower, about 40 nM (pH=7.4). From the Nernst equation, EH = -24 mV. That means in cells
with membrane potentials more negative than -24 mV (most cells), H+ must be pumped out of the
cell. The mechanism is a secondary active transport system, in which the inward leak of Na+ drives
the outward pumping of H+.
Chloride ions are pumped into some cells by a secondary active transport process
(Na/K/2Cl cotransporter). As a result, ECl moves in a positive direction, away from the resting
membrane potential.
The (non-existent) ‘H+/K+ exchanger’. There are several clinical situations that suggest the
presence of a system that will exchange K+ for H+, and vice versa. For example, infusing K+ causes
32
acidemia (the K+ is taken up by cells ‘in exchange’ for H+), and infusing acid causes hyperkalemia
(elevated (hyper-) potassium (-kal- for Latin kalium) in the blood (-emia)). While it is conceptually
simple (and useful) to think in terms of an H/K exchanger, the reality is that such a transporter
probably does not exist. Rather, the process evidently involves different transporters, perhaps
working in pairs in parallel, the upshot being hydrogen/potassium exchange. For example,
hyperkalemia will cause extra K+ uptake via the Na/K pump. Hyperkalemia also will depolarize
cells (by shifting EK in a positive direction), and the change in membrane potential can affect the
rate of activity of electrogenic transporters. One such transporter, which transports 3 bicarbonate
ions and one sodium ion from the ICF to the ECF, is inhibited by depolarization. Thus, the
reduction in activity will reduce bicarbonate extrusion. Because bicarbonate is a base, its slower
extrusion will cause acidemia. So, which is easier to remember, the idea of a (non-existent) H/K
exchanger, or trying to keep track of several different cotransporters and exchangers that might be
playing a role?
Below are 10 cartoons of membrane transporters. One of them has been mislabeled. Which one?
answer: the last one (it is a sodium-bicarbonate
cotransporter, not an exchanger)
33
Molecules to Medicine- Cell Physiology
6. Epithelial Transport 23 Nov 9AM
Bill Betz
[email protected]
http://www.cuphys.net/cellphys
Reading (optional): Alberts et al, 4th ed: 1066-1070
Key words: apical, mucosal, lumenal, basolateral, serosal, peritubular, shunt pathway
6: Epithelial Transport
In previous lectures, we have assumed that the ECF composition is unvarying over time. In
health, this is quite true. However, a large and rapid turnover of water and solutes occurs in the
ECF. Water and nutrients are continually being added to the ECF by the G.I. tract (and O2 by the
lungs), and metabolic wastes, water and salts are continually being removed, mainly by the kidneys
(and CO2 by the lungs). Thus, alterations in the transport properties of any of these organs can alter
ECF (and, secondarily, ICF) composition. You will study the functions of these organs in detail
next spring. The main intent here is to introduce the cellular mechanisms by which these materials
are transported into and out of the ECF by epithelia.
There are many different epithelial tissues in the body, including the G.I. tract, kidneys, all
exocrine glands, gall bladder, choroid plexus, ciliary body, corneal epithelium, and mucous
membranes (but not the lungs – see box). Epithelial cells are organized into sheets of cells, often
forming tubular structures, like the epithelium lining
The main task of the lungs is to transport O2 into
the G.I. tract. The apical surface faces the ‘special’
and CO2 out of the blood. There are no
fluid (e.g., food in the gut, urine in the kidney, saliva
membrane transporters (active or otherwise) for
in the parotid duct) and usually contains the special
O2 and CO2; they are lipid soluble molecules
transporters that endow the epithelium with its
that diffuse freely through all membranes. Thus,
specialized transport properties. The basolateral
the endothelial cells in the lung that separate
surface is exposed to the interstitial fluid and usually
blood from air in the alveoli of the lungs do not
has generic transport properties like the plasma
possess special transporters like those in
membranes of non-epithelial cells (neurons, for
epithelia to transport O2 or CO2.
example). Fig. 1 shows a typical example. Synonyms
for apical include mucosal and lumenal; synonyms
for basolateral include serosal and peritubular.
The cells that compose the epithelial sheet are glued to their neighbors by tight junctions.
The glue is indeed tight is some epithelia, such as sweat glands and the distal parts of kidney
tubules, preventing virtually any substance from passing from one side to the other by passing in
between the cells. In most epithelia, however, the tight junctions are not very tight. For example,
tight junctions of the small and large intestine, gall bladder, and proximal part of the kidney tubules
are relatively leaky, and provide a pericellular shunt pathway for the movement of water and
solutes. Leaky tight junctions are somewhat selective in their leakiness; some are relatively more
permeable to cations, others to anions.
In general, epithelia engaged in massive transport of substances are leaky, while those
epithelia doing the finishing work (‘fine tuning’) are tight. Leaky epithelia cannot maintain
energy gradients as large as those produced by tight epithelia, because solutes and water leak
back across the epithelium through the pericellular shunt.
Thus, to get across an epithelium, a substance must follow one of two possible routes: it
may either cross two membranes by entering the epithelial cell on one side and leaving on the
other, or it may cross no membranes at all by passing in between cells through the pericellular
shunt pathway.
The driving force for nearly all transport – water, salts, nutrients, non-volatile metabolic
34
Fig. 1. A generic tight
epithelium for the
transport of NaCl and
water from the apical
to the basolateral
solution. Dashed lines
show leaks (passive
flux); solid lines show
pumps (active
transport). The size of
the letters indicates
relative concentrations
of the salts.
wastes – is the Na/K pump (always located in the basolateral membrane). By keeping intracellular
sodium ion concentration low, the Na/K pump provides the energy to drive a host of secondary
transporters. Protons are the main exception to the otherwise universal dependence of epithelial
pumping on the Na/K pump, because primary active transporters have evolved for protons, most
notably in the stomach (to secrete acid into the lumen of the stomach) and kidney (to excrete
protons, which are a metabolic waste product.)
Absorption of salt. Fig. 1 shows a cartoon of a generic tight epithelium that transports
sodium, chloride, and water from the apical to the basolateral solution, thus absorbing them into
the blood. Note that the basolateral membrane is just like many other cells, containing relatively
low sodium permeability and high potassium permeability. What value of membrane potential
would you expect to record across the basolateral membrane? Answer: because it’s like the plasma
membrane of a neuron, Vm would be about -70 mV. The basolateral membrane also contains some
chloride channels, and the Na/K pump.
The apical membrane is totally different. It is relatively highly permeable to sodium, not
potassium. What would be the potential across this membrane? (Answer: Vm would be more
positive, perhaps +10 mV.) In addition, there is no Na/K pump in the apical membrane.
Here is how salt and water are transported from apical to basolateral solution: Assume that
identical NaCl solutions are placed on either side of the epithelium. Sodium ions leak into the cell
across the apical membrane, down their electrochemical gradient. They are then pumped out of the
other side of the cell by the Na/K pump, across the basolateral membrane. This results in the net
transport across the epithelium of a positive charge, and chloride follows passively, drawn by the
electrical force. The net transport of NaCl produces an osmotic gradient, which in turn draws water
along.
What would happen to the transport if the Na/K pump were blocked? As the cell filled with
sodium (leaking in across the apical membrane), the driving force for further sodium entry across
the apical membrane would be reduced, and net transport of sodium, chloride, and water would
decrease.
If one were to measure the voltage across this epithelium, what would be the result? By
putting electrodes into apical and basolateral solutions and connecting them to a recording device,
one would find that the apical solution is negative with respect to the basolateral solution. This
makes perfect sense, because the movement of sodium out of this solution creates the electrical
35
gradient that draws the chloride along with it.
What would happen to this transepithelial voltage if the chloride ions in the apical solution
were replaced with a larger anion, say SO4=, which could not fit through the chloride channel? In
this case, the Na/K pump would continue to operate. Because no anion can follow, the transepithelial voltage will increase. Before long, the apical solution would become so negative that net
sodium transport would stop.
It can be confusing to put together the profile of voltage
as one moves across the apical membrane, through the cell, and
out again across the basolateral membrane (Fig. 2). Suppose the
basolateral membrane potential (Vm (B)) = -50 mV, and the
apical membrane potential (Vm (A)) = +10 mV. What is the
transepithelial potential difference (transPD)?
The short answer is:
TransPD = Vm (Basolateral) – Vm (Apical)
The explanation follows. Keep in mind a couple of rules: i) all
membrane potentials are written as the potential of the inside of
the cell with respect to the outside (i.e., outside = zero); ii) the
Fig. 2. Voltage profile across transepithelial potential is written as the potential of the apical
a generic tight epithelium
solution with respect to the basolateral (i.e., basolateral = zero);
iii) the cell is isopotential (all voltage drops are at membranes,
so all lines showing electric potential in Fig. 2 are horizontal – no change over distance, except
across membranes). Start at the basolateral side. The basolateral solution is defined as zero. You
are given that Vm (B) = -50 mV, so in crossing the basolateral membrane from outside to inside,
you must drop down 50 mV. The cell is isopotential (this is only an approximation, but a good
one), so the line across the cell is horizontal. In leaving the cell by crossing the apical membrane,
you must change by 10 mV, because Vm (A) = +10 mV. Do you move up or down? Because the
inside is 10 mV more positive than the outside, you must move down, emerging at a potential of 60 mV, compared to the basolateral solution.
Fig. 3 shows the same situation as
Fig. 1, except the epithelium is leaky, and
allows significant fluxes through the
paracellular shunt pathway. This would be
characteristic of an epithelium like the one
lining the G.I. tract, where relatively large
fluxes occur. The shunt pathway in this case
is relatively more permeable to chloride than
to sodium, but is by no means totally
impermeable to sodium (the molecular
mechanisms by which tight junctions
achieve selective permeabilities are not well
understood). Qualitatively, this epithelium
works just like the one in Fig. 1. The major
Fig. 3. Same as Fig. 1, except chloride and water
difference is that the leaky epithelium is
follow the shunt pathway
usually capable of moving a larger amount
of material. In addition, the leaky tight
junctions partially ‘short out’ the transepithelial potential difference, so that a lower transepithelial
36
voltage would be measured.
Fig. 4 illustrates another variation on the same
theme. Here, solute transport across the apical
membrane is by way of an electroneutral cotransporter
that carries one potassium, two chloride, and one
sodium ions into the cell each cycle (it sounds more like
pinocytosis than membrane transport!). This transporter
is found in several important epithelia, including the
kidney loop of Henle and the respiratory tract.
Qualitatively, though, the end result is no different than
Figs. 1 or 3.
In summary, we have seen three ways in which
Fig. 4. Apical sodium channel replaced
epithelia achieve the same result, namely the absorption
with Na/K/2Cl cotransporter.
of NaCl and water into the blood. They just use
different transport mechanisms in the apical membrane and different ‘tightness’ of the paracellular
shunt pathway.
Secretion of fluid by epithelia. Not all cells in a given epithelium are identical. On the
contrary, individual cells may be highly specialized, some for absorption, some for secretion.
The mechanisms described so far concern absorption by epithelia – moving solutes from lumen
to blood. Things can travel the other direction, too. One secretion pathway is of special clinical
importance, for it underlies important diseases, including some types of diarrhea (including the
worst of all, cholera), and also cystic fibrosis. (Clinical presentations on cholera (11/25 8 am)
and cystic fibrosis (12/3 8 am) will be given a little later.)
The key to understanding epithelial secretion in general and the main defect in those
diseases is a chloride channel in the apical membrane (especially in the GI tract and lungs). A
relatively simple way to think about it is as follows (see Fig. 5): Normally, in a resting cell, this
Cl- channel is closed. However, when this Cl- channel is activated, the cell begins to secrete
electrolytes and water into the lumen.
Fig. 5. Secretion of fluid by
an epithelial cell. The key
player here is Cl-. It is
pumped into the cell across
the basolateral membrane,
and leaks out the other side
through a Cl- channel. The
channel is closed in the
resting state, but opened
during digestion, and
during some types of
diarrhea, like cholera.
The secretion is driven by Cl- leaking out of the cell into the lumen. The Cl- concentration
in the cell is high, thanks to a Cl- pump in the basolateral membrane. The pump is a Na-K-2Cl
37
cotransporter that uses the downhill leakage of Na+ into the cell to drive the uptake of Cl- from
the interstitial fluid (it is just like the transporter in Fig. 4, but it is located in the basolateral
membrane, not in the apical membrane).
As the Cl- leaks out of the cell into the lumen, the electrical negativity that it creates
draws Na+ along passively. The Na+ flows mostly through the intercellular shunt pathway. The
resulting osmotic gradient draws water along, too, giving a net secretion of an isotonic solution
of NaCl.
Turning on the Cl- channel. So eptithelia have mechanisms both to absorb and to secrete.
Which one wins? As noted above, at rest, the apical Cl- channels are closed, so absorption wins.
A variety of stimuli can open the Cl- channel and turn on secretion. In the GI tract, it happens
physiologically during digestion (parasympathetic nerve stimulation, hormones in the blood), as
chemicals activate receptors in the basolateral membrane (the signal is carried across the inside
of the cell from the basolateral receptors to the apical membrane by ‘cell signaling’ mechanisms
(Ca++ ions, activated protein kinases, cyclic AMP, etc.) that you will study a little later.
Pathogens also can activate the Cl- channels. Cholera toxin, for example, acts by locking
open this channel, causing a massive efflux of fluid from the cell, leading to profound diarrhea
and dehydration. (The toxin, a protein, is secreted by a bacterium, Vibrio cholerae, which lives in
water. Don’t drink that water.)
While there are probably multiple types of Cl- channels (different gene products)
involved in fluid secretion by epithelia, one of the most important is the ‘Cystic Fibrosis
Transmembrane Conductance Regulator.’ (CFTCR, or more commonly CFTR). Cystic fibrosis
is a genetic disease; it is this Cl- channel that is mutated, which reduces the ability of epithelia to
secrete ‘serous’ (watery) fluid, leading to thickened mucous secretions, infections, and other lifeshortening complications. You will hear a clinical vignette about cystic fibrosis a little later.
Absorption of nutrients. In the G.I. tract, enzymes secreted by digestive glands hydrolyze
ingested proteins and polysaccharides, and the resulting amino acids and sugars are pumped from
the G.I. lumen into the blood by the G.I. epithelium. Virtually identical mechanisms operate in the
kidney, where glucose and amino acids are reabsorbed by the epithelial cells of the proximal
tubule after being filtered from the plasma in the glomerulus. As shown in Fig. 6, each nutrient is
pumped across the apical membrane, and then passively moves out of the cell into the interstitial
fluid.
The sugar and amino acid pumps are examples of sodium-dependent secondary active
transport systems. If [Na+] in the mucosal solution is removed (replaced by non-permeating
choline+), sugar and amino acid pumping stops. Conversely, removing sugars and amino acids
reduces the movement of Na+ from mucosal to serosal fluid. The transporter captures some of the
energy released as Na+ moves down its electrochemical gradient into the cell, and uses this energy
to pump the sugar or amino acid against its gradient into the cell. Note that the transport of glucose
by epithelial apical membranes is different from the transport across the plasma membranes of
nonepithelial cells (e.g., muscle), where glucose is transported by "facilitated diffusion", which is
not a pump. The important features of nutrient transport by epithelia are shown in Fig. 6.
38
Suppose a person eats a chocolate bar. It will be broken down into its constituent sugars,
including glucose. But how can the G.I. tract absorb the glucose unless some NaCl is eaten at the
same time? What will drive the glucose cotransporter? The answer is that NaCl will leak from the
blood across the leaky G.I. epithelium through the shunt pathway into the lumen of the G.I. tract.
This sodium will then provide the necessary drive for the active uptake of glucose from the lumen.
Fig. 6. Epithelial transport of
sugars and amino acids (X). In the
G.I. tract and kidney, solutes are
pumped into cells via a secondary
active transporter, driven by the
inward sodium leak. The solutes
diffuse across the cell, and are
transported down their energy
gradient out of the cell across the
basolateral membrane by way of
facilitated diffusion.
Finally, a word about regulation of the transporters that drive absorption in the G.I. tract. In
one sense, they are not regulated, at least not by the ECF composition. Instead, they seem geared
for maximum transport of nutrients any time, day or night. Thus, if a person drinks a glass of
water, whether thirsty or not, all of the water will be absorbed by the G.I. epithelium and put into
the blood. If a person eats five grams (that’s a lot) of table salt, virtually all of it enters the blood
plasma. Same for glucose, and virtually all other common nutrients. (There is great chemical
specificity of the transporters, however. For example, L-amino acids and D-sugars are selectively
transported, but not their stereoisomers. Eating L-glucose or D-amino acids (which are not
absorbed) is likely to produce an osmotic diarrhea as the unabsorbable solutes osmotically suck
water into the GI lumen.) The G.I. tract works so avidly probably because the transport
mechanisms evolved in far less abundant environments than the one we live in, nutritionally
speaking, and it was adaptive to be able to absorb all available nutrients. By not regulating ECF
composition at the input end, it falls by necessity to the kidneys, at the output end, to regulate the
composition of the ECF. Not very efficient.
Transport of Water. It is curious, given the importance of water, that no water pumps
exist in the body (or anywhere else). This means that water always moves passively down osmotic
gradients. As described above, absorption (by active transport) of salts, sugars, amino acids, and
other solutes by the G.I. tract epithelium is accompanied by water absorption (as the solutes are
absorbed, the lumenal contents become slightly hypo-osmotic to plasma, and so water is absorbed,
always passively). In a few places (sweat gland ducts, some parts of tubules in the kidney),
epithelia are relatively impermeable to water. In these special cases, osmotic gradients can be
maintained by pumping solute across the water-impermeable epithelium.
Consider, for example, sweat glands. Their job is to ‘excrete heat’. To do that, they deliver
water to the body surface where its evaporation (an endothermic process) cools the body. Nascent
sweat forms deep in the gland (by a secretory process); it has a salt composition that is similar to
plasma (about 300 mosM, mainly NaCl). As the fluid travels along the sweat duct on its way to the
39
surface, NaCl is (re)absorbed by the epithelium lining the duct and returned to the blood. Because
the ducts are impermeable to water, water cannot follow the salt, and the fluid in the lumen of the
duct becomes more and more dilute (it can be as low as about 50 mosM). This is useful, because
the solute in sweat does not help at all in the process of heat loss, and just leaves an unsightly white
crust on your tee shirt.
Ridding the body of metabolic wastes. Each day the cells in an adult human produce about
15 moles of metabolic waste solute. That is enough to more than double the osmolarity of the body
fluids! (45 liters of body fluids * 0.3 osmoles/liter = 13.5 osmoles present in body fluids at any one
time). And yet, a rise in plasma osmolarity of a few percent is enough to make a person thirsty, and
severe dehydration raises plasma osmolarity by less than ten percent. In other words, there is a
stupendous amount of metabolic waste to get rid of each day. The problem is largely solved by an
extraordinarily simple chemical fact: the end product of carbon metabolism, CO2, which composes
14.5 of the 15 moles of waste, is volatile. Consequently, it is simply exhaled via the lungs, and
requires no special transporters at all (CO2 permeates all membranes easily, by dissolving in them).
A little water - less than a liter – is lost as exhaled water vapor with the CO2 .
What about the remaining 0.5 moles (500 mmoles) of metabolic waste? They are nonvolatile molecules, and of course cannot be expelled by the lungs. Even though they account for a
small fraction of the total, getting rid of these non-volatile metabolic waste products is a big
problem. It falls to the magnificent kidney to solve the problem, and while it is responsible for
many other tasks (such as regulating the concentration of just about everything in the ECF), the
most important function of the kidney is to get rid of these non-volatile metabolic wastes, because
no other organ can do it. When kidney function is lost, death from uremia can follow. Uremia
literally means “urine in the blood.”
What kinds of metabolic waste molecules are non-volatile? The majority (about 450 of the
500 mmoles) are the end product of nitrogen metabolism, urea. Most of the remainder (about 50
mmoles) are protons, H+.
How does the kidney do it? The fundamental anatomical arrangement of the kidney is just
like the lungs, which get rid of the volatile waste product, CO2. That is, in both the lungs and the
kidney, blood capillaries pass close to the ends of dead-end tubules (glomeruli in the kidney,
alveoli in the lungs), and various chemical substances move from the blood into the tubules,
eventually becoming urine in the kidney, and expired air in the lungs.
Functionally speaking, however, matters are entirely different in the two organs. In the
lungs, CO2 diffuses passively from blood to air. The equivalent arrangement in the kidney would
be to have molecular transporters at the blood-tubule interface, and have them pump metabolic
wastes (mostly urea) out of the blood, into the tubules, and letting them, together with some water,
pass on. However, urea transporters do not exist.
This simple fact – no urea transporters - seems to underlie the entire design of the kidney.
That is, rather than possessing transporters to pump urea and other exotic non-volatile waste
products out of the blood, the kidney takes the exact opposite approach. In essence, it says, “I
know what I like.” Consequently, rather than trying selectively to pump waste products out of the
plasma, it forms an ultrafiltrate of plasma in the glomerulus, which contains water, salts, sugars,
amino acids, and all other beneficial compounds, as well as the non-volatile metabolic waste
products. Then, as this plasma ultrafiltrate passes along the renal tubules, the epithelial cells lining
the tubules reabsorb (pump back into the blood) the things that it wants to keep (glucose, salts,
bicarbonate, etc), allowing the wastes to pass on. It’s incredibly expensive, energetically speaking,
40
to do it this way, requiring a great deal of ATP to drive the reabsorbing pumps.
That’s enough responsibility for any organ, but the kidney does so much more. Earlier we
noted that the kidney regulates the ECF composition (by adjusting the activity of the transporters
that do the reabsorbing). This additional chore arises because the undisciplined GI tract absorbs
just about everything presented to it (see box below), regardless of the needs of the ECF. So the list
of functions of the kidney includes excreting non-volatile metabolic wastes and regulating the
composition of virtually all ECF solutes – nutrients and electrolytes – as well as water.
A note on water balance. Naturally, the kidney regulates water (like everything else)
balance in the body. Getting rid of extra water is relatively easy: solutes in the lumen of the kidney
tubule are reabsorbed as usual, but the epithelium is made water-impermeable, so water cannot
follow the solutes osmotically. Thus, the extra water stays in the lumen, and passes into the urine.
(The epithelium is made impermeable to water when the hypothalamus, sensing extra water on
board as a drop in plasma osmolarity, stops secreting anti-diuretic hormone (vasopressin), which
causes certain kidney epithelial cells to remove water channels (aquaporins) from their apical
membranes, thereby reducing water reabsorption.)
Conserving water in times of dehydration, like when stranded in the desert, is another
matter altogether. Remember, there are no water pumps, so the kidney cannot just pump water
back into the blood from the tubules. This makes for a huge physiological problem. You can guess
what organ has had to solve the problem. Lacking water pumps, the renal tubules have had to
evolve a far more complicated mechanism to conserve water. It works pretty well: human urine
can be as concentrated as 1200 mosM. Its basic design is described in the Appendix.
Our friend, the GI tract. Perhaps I should not disparage so completely the G.I. tract. While its
output, feces, comprises mainly bacteria and substances eaten but not absorbable, about 30 mmoles
per day of non-volatile metabolic wastes are excreted via the G.I. tract (compared to 500 by the
kidney). These wastes are mostly breakdown products of red blood cells (delivered from the liver to
the GI lumen), and are highly toxic if not promptly eliminated.
Moreover, as noted earlier, the transporters in the GI tract are incredibly specific. For example,
the stereo isomer D-glucose in absorbed, but L-glucose is not.
On the other hand, while most big molecules (e.g., whole proteins) that we eat are of course
not absorbed intact, but are broken down (digested) to much smaller molecules (e.g., amino acids)
before being absorbed into the blood, the most deadly substance on earth, botulinum toxin, if ingested
in food-gone-bad, is absorbed as an intact protein (and it’s a big protein), and once in the blood it goes
about its fiendish business. Thanks a lot, GI tract.
APPENDIX: Conserving water
In times of dehydration, we need to conserve water. Mammals are pretty good at it.
Humans can excrete urine that is as much as four times hyperosmotic to the body fluids (that’s
about 1200 mosM, more salty even than sea water (about 1000 mosM)). You will study the
detailed mechanisms by which this is accomplished later. What is introduced here is the basic
physiological strategy.
Conceptually, the simplest way would be to pump water out of the ultrafiltrate in the renal
tubules, back into the blood. But water pumps do not exist in biology, so a more complex process
has evolved that accomplishes the same thing. It involves separating in space the removal of solute
and the removal of water from the ultrafiltrate in the tubule.
41
There are three steps to the process (see Fig. A1 below). First, NaCl is actively transported
out of the ascending limb of the loop of Henle as it rises through the renal medulla. The tubule here
has a very low permeability to water, so the fluid in the tubule lumen becomes hypo-osmotic to
plasma (it's about 50 mosM at the top of the loop). In addition, the surrounding interstitial fluid is
poorly vascularized in the medulla, so the salt accumulates in the ECF and is not washed away,
creating a hyperosmotic interstitium. This is a key point.
Second, the distal tubule is permeable to water, and the interstitium here (in the renal
cortex) is well vascularized. Consequently, water passively leaves the tubule and is returned to the
blood. Thus, the fluid arriving at the end of the distal tubule is isosmotic with plasma (about 300
mosM). The volume of fluid in the lumen has been greatly reduced though. In other words, fivesixths of the salt was removed by the ascending loop of Henle (remember: no volume change,
because no water movement), and five-sixths of the water (and volume) left the distal tubule and
returned to the blood.
Third, the tubule (now called the collecting duct) plunges back down into the renal
medulla, through the hyperosmotic interstitium. The fluid in the lumen is isosmotic with plasma as
it begins its passage. But the collecting duct tubule is permeable to water, and so water leaves the
lumen, thereby making the lumenal fluid (urine) hyperosmotic to plasma. (Of course, water leaving
the collecting duct will dilute the medullary interstitium, partially defeating the system. But
because most (about 5/6) of the water was previously removed, the amount of water entering the
collecting duct is greatly reduced.)
In summary, the keys to this ingenious device is that most salt and water are removed from
the tubule at separate locations, and the salt is kept around to provide a special, hyperosmotic
interstitium to draw a little extra water from the lumen of the collecting duct. All because there are
no water pumps.
Fig. A1. Conserving water. This is a cartoon of a kidney
tubule. There are about a million of them in each kidney.
What comes in at the left (at the glomerulus) is filtered
plasma. What goes out at the lower right is urine. This
shows the 3 basic steps in conserving water. 1. NaCl is
pumped out of the tubule; water cannot follow; the
interstitium is not well vascularized, so that the salt
raises the osmolarity there. 2. Water is removed
passively in the distal tubule. 3. More water is removed
in the collecting duct, producing a hyperosmotic output
(urine).
42
Molecules to Medicine- Cell Physiology
7. Action Potential: Mechanism 23 Nov 10AM
Bill Betz
[email protected] http://www.cuphys.net/cellphys
Reading (optional): Alberts et al, 4th ed: pp 637-645
Key words: threshold, all-or-none, action potential, activation, inactivation, refractory period,
accommodation
7. Action Potential: Mechanism
Up until now, we have pictured inorganic ions as playing a rather passive role
in the body, just keeping enzymes and other macromolecules happy. Now we are
going to look at a vital chore in which ions take center stage, namely the rapid
propagation of electric signals over long distances. Consider a human sensory neuron
innervating the sole of the foot. It's axon, which may be about 10 m in diameter (much thinner
than a hair) courses uninterrupted from the foot up to the spinal cord, about a meter away. When
the axon terminal in the foot is depolarized (for example, by stepping barefoot on an object), an
electrical signal called an action potential travels faster than 100 mph up to the spinal cord (from
there, the signal will be relayed at a synapse to another axon that carries the signal up to the brain,
where conscious perception occurs).
The action potential depends upon a single molecular specialization: a voltage-gated
sodium channel. To understand the need for these channels, and to appreciate the astonishing
beauty of the action potential mechanism, we first need to deal with some basic electrical
properties of cells. The details are in Appendix I. The take-home message is simply that the
‘passive’ electrical properties of axons (that is, their electrical properties without the special
sodium channels) are pathetically, laughably inadequate to do the job. Without the sodium
channels, electrical signals can’t spread more than a few millimeters from the site of the stimulus,
and some axons are more than a meter in length. Thus, axons without the special sodium channels
are wretched little cables. The reasons (details in Appendix) are that the salty fluid inside the cables
is a lousy conductor of electricity and the insulation around the cable is leaky. [It should be noted,
however, that over distances of less than a few millimeters, passive spread of voltage can be very
important -- for example, along dendrites in neurons.]
To illustrate, consider the matter from an engineer's perspective. We want to send a
telephone message over a very thin wire. An engineer naturally would choose to use a copper wire
rather than a squishy axon filled with a dilute salt solution, because copper is a much better
conductor than intracellular fluid (ICF). In fact, the resistance of copper is about ten million times
lower than the resistance of cytoplasm. That means that looking down the inside of an axon one
meter long is like looking down a copper wire (of the same diameter) 10 million meters long.
That’s 6250 miles! It would be utterly fruitless to try to send an electrical signal over a 6250 milelong copper wire thinner than a hair. The electric signal would decay long before reaching its
destination. The only solution would be to construct booster stations at regular intervals along the
wire. Booster stations do two things: they have a source of energy to boost the decaying signal, and
they have a detector to recognize the arrival of a decaying signal. This is, in fact, exactly how
trans-Atlantic telephone cables work . The action potential mechanism in axons is exactly
analogous to the engineer's booster station; the energy source is the sodium ‘battery’ (the
difference between ENa and Vm) , and the detector is a voltage-sensing gate in the sodium channel.
The figure below illustrates how an action potential is evoked experimentally. The axon is
impaled with two electrodes; one records membrane potential (mV) and the other injects
depolarizing current (I) into the axon. With small amounts of current (a, b), the axon is depolarized
43
a little bit, and the membrane potential recovers when the current is
turned off. But with a larger current (c), the axon suddenly produces
an action potential - the membrane potential flies up, passing zero, on
its way to ENa. If an even larger current is passed (not shown), the
action potential looks the same. Thus, the action potential is all-ornone (you can't get half an action potential), and there is a threshold
membrane potential that must be reached in order to produce an
action potential. We'll return to these observations shortly.
Most importantly, we ask: what about propagation along the
axon? If we were to move the recording electrode, say one centimeter
away from the stimulating electrode and repeat the experiment (not
shown), we would see no change at all in Vm with pulses a and b. But with pulse c, we would see,
after a delay, an action potential, and it would be the same size as the one recorded at the source of
the stimulus. So, there is no decrement at all in the size of the signal as it propagates along the
axon! Moreover, the propagation velocity is pretty fast, up to 100 m/sec (about 200 mph) in some
axons.
Problem solved. Electrical signals do spread quickly, without decrement, over long
distances.
So now we are faced with two key questions: What is the mechanism that underlies the
action potential? How is it propagated down the axon? We'll consider the second question in the
next lecture. For the remainder of this lecture, we will consider what happens in a tiny patch of
axon membrane to make an action potential. Then, in the next lecture, we will ‘connect the
patches’, so to speak, to see how the action potential actually travels along the axon.
Within each square micrometer of a typical axonal membrane are about 100 gated sodium
channels. At the resting potential, the gates (called activation gates) are all closed. The activation
gates open when the cell is depolarized. This of course drastically alters the permeability of the
membrane, from one that is largely permeable to potassium to one that is much, much more
permeable to sodium. As sodium ions rush into the cell, the inside becomes more positive.
Threshold. We noted earlier that a certain amount of depolarization is required to initiate an
AP. That is, there is a threshold membrane potential that must be reached in order to generate an
AP. The threshold is not the point at which all of the Na channel activation gates open, as one
might initially suppose. In fact, there is no such point; the
relation between PNa and Vm is continuously graded with
membrane potential, as shown in Fig. 1.
What then determines the threshold voltage? The
answer is a bit subtle; it’s the point at which sodium and
potassium currents are exactly equal and opposite. Suppose
we depolarize an axon exactly to threshold, and then turn
off the stimulus. At that instant, some Na+ channels will
have opened and Na+ ions will be entering, 'trying' to
depolarize the cell even more. At the same time, however,
Fig. 1. Activation gates in Na channels
potassium ions will be leaving the cell, 'trying' to pull Vm
do NOT all open simultaneously at
back towards EK. At the threshold, a very precarious
threshold for the action potential (dashed
balance is struck; if the K+ current should by chance
line). Instead, they follow the solid line,
slightly exceed the Na+ current, Vm will be pulled back
and the definition of threshold is thus
towards the resting potential, Na+ channels will close, and
more subtle (see text).
44
there will be no AP. Conversely, if the inward Na+ current momentarily exceeds the outward K+
current, it will produce a little bit more depolarization, which will open more Na+ channels and the
Na+ entry process will become 'explosive', producing an AP. Thus, while it is certainly easier to
think of threshold as the point at which all of the sodium channels open, in fact threshold is defined
as the point at which the Na+ and K+ currents are exactly equal and opposite.
All or none response. Once threshold is exceeded, a miniature ‘explosion’ occurs.
Remember, at threshold, not all sodium channels are conducting yet. But the sodium entering the
channels that are conducting will depolarize the membrane further, and that depolarization will
cause more sodium channels to start conducting, which will cause more sodium entry, which will
cause more channels to open.... This is called ‘positive feedback.’ In a few tens of microseconds,
the sodium channels are all open, and Vm is well on its way to ENa..
What is it like right at the peak of the AP? The 'struggle' between Na+ and K+ goes on, as at
rest, but PNa is now much greater than PK, and so Vm is very close to the sodium equilibrium
potential. It's nothing more than a reflection of the basic rule: membrane potential is determined by
relative permeability to different ions.
Repolarization is accounted for in two ways: a) PNa declines back to its resting level and b)
PK undergoes a transient increase.
a) PNa decreases. There exists a second gate in the sodium channel, which is open at rest,
and closes upon depolarization. However, this gate (the inactivation gate), swings more slowly
than the Na gate we discussed previously (the activation gate). Thus, at rest, the activation gate is
shut and the inactivation gate is open. When the membrane is depolarized, they reverse states: the
activation gate opens and the inactivation gate closes. (Sometimes the gates are called 'm'
(activation) and 'h' (inactivation)).
Naturally, if the gates were to swing at the same rate, the channel would never conduct,
because both gates have to be open in order for sodium ions to move through the channel. This is
avoided because the activation gate swings faster than the inactivation gate, so that when the axon
is first depolarized there is a brief instant when both gates are open and sodium can then rush into
Fig. 2. The behavior of
the voltage-sensitive
gates that control ion
fluxes during an action
potential. Sodium
channel activation gate
(m), inactivation gate (h),
and Potassium channel
gate (k) are shown. Thin
arrows point to the parts
of the AP in which gate
positions are as drawn in
the cartoon.
the cell. Then the inactivation (h) gates close (a delayed response to the depolarization), and PNa
promptly declines. Fig. 2 illustrates this behavior of the Na channels. What an awesome molecule one protein with two gates, exquisitely timed to allow sodium to enter for a brief instant only.
b) PK increases. The closing of the Na+ channels at the peak of the AP would indeed cause
the axon to repolarize. In addition, the permeability of the membrane to potassium ions increases.
45
This too of course promotes repolarization. The K+ channels open as the Na+ channel inactivation
gates are closing. Because Vm is very far removed from EK, K+ ions rush out of the cell, and Vm
plummets back towards its resting level.
What controls the behavior of the K+ channel gate? Again, the answer is membrane
potential. The K+ channel gates are closed at rest, and open in response to depolarization. However,
they respond with a delay. This delay is certainly a good thing, for if K+ channels and Na+ channels
opened simultaneously, the AP would be much reduced in height, or might not occur at all. The
delay in K+ channel opening is timed to allow both maximum height of the AP and rapid
repolarization. The behavior of the K gates is also illustrated in Fig. 2. Notice that Vm
"undershoots" the resting potential after repolarization. This is the result of the increased PK:
repolarization will cause the K+ channels to close, but they respond with a delay, so that when Vm
first gets back to resting level, the K+ channels are still open, and so Vm moves closer to EK. Then,
as the K+ channels close, Vm finally relaxes back to its resting level. Notice that while the explosive
opening of Na+ channels is an example of positive feedback, the K+ channel behavior is an
example of negative feedback: depolarization causes them to open, which causes repolarization,
which causes them to shut.
Role of potassium channels. What is gained by having voltage-gated potassium channels?
After all, with the closure of the inactivation gates in the sodium channels, Vm would return to its
resting level. The answer is: faster repolarization. The K channels provide additional pathways for
potassium ions to leave the cell, pulling the membrane potential back to its resting value faster than
in the absence of an increase in potassium ion permeability. Without the voltage-gated K channels,
the membrane would indeed repolarize, but the repolarization would be slower, determined only by
the passive (ungated) K+ channels responsible for the resting potential. Having lots of channels
provides lots of pathways for rapid potential changes. And the faster the potential change, the more
action potentials can be generated in a given period of time.
Refractory period. After producing an action potential, an axon cannot generate another
one for a few milliseconds. This is called the refractory period. It can be subdivided into an
absolute refractory period, during which time no stimulus, no matter how strong, can evoke
another AP, followed by a relative refractory period, during which time a stronger-than-normal
stimulus is required to evoke another AP. The refractory period results primarily from the fact that
the sodium channel inactivation (h) gates require time to reopen after repolarization. If a stimulus
is applied when some h gates are still closed, the sodium channel activation (m) gates may swing
open, but no Na+ can flow owing to the closed inactivation gates. Whenever any channel is blocked
by a closed inactivation gate, we say that the channel is inactivated.
The potassium channels also contribute to refractoriness. After the axon repolarizes, it takes
time (several msec) for the K+ channel gates to close again. The higher K+ conductance makes it
harder for a stimulus to depolarize the axon.
Accommodation. If an axon is depolarized slowly, it may fail to generate an action
potential, even if depolarized beyond what had been threshold for a rapid depolarization. In other
words, the axon accommodates to the slow, steady stimulus.
What is the mechanism of action potential accommodation? Normally, physiological
stimuli evoke fast membrane depolarization, so that sodium channel activation gates swing first,
and the inactivation gates, being slower to respond to a change in membrane potential, do not
snap shut until after the action potential rising phase has been generated. A slow depolarization,
however, provides time for the inactivation gates to close first, so that, when activation gates do
open, any channel in which the inactivation gate has already closed is useless; it cannot conduct
46
sodium ions. Susceptibility to accommodation varies widely among different neurons. Some fail
altogether to give an action potential when depolarized slowly, while others are barely affected.
Accommodation also manifests itself during hyperkalemia. That is, the steady membrane
depolarization produced by elevated plasma potassium ion concentration closes some
inactivation gates. Consequently, when a physiological stimulus arrives, producing a rapid
depolarization, inactivated channels are incapable of contributing to the action potential. This is
the mechanism that underlies the generation of cardiac arrthymias, described more fully next
lecture.
Energy and action potentials. Clearly, an axon must pay a price for producing an action
potential. During each AP, a few thousand Na+ ions enter through each Na+ channel, and about the
same number of K+ ions exit through each K+ channel. Ultimately the Na+/K+ pump must be called
upon to restore proper ion balances. If the pump is poisoned and an axon is stimulated repeatedly,
gradually it will fill up with Na+ and lose K+. How quickly that happens depends mainly on the
surface-to-volume ratio of the axon. Big axons have a relatively small surface area (compared to
their volume), and so run down relatively slowly, compared to small axons. The giant axon of the
squid (about 1000 microns (1 mm) in diameter) can give over 100,000 AP's without needing its
sodium pump. Tiny neurons in the human cerebral cortex (about 10 µm in diameter) are much
more sensitive to loss of their sodium pumps. In many cells, for one AP, intracellular [Na+] does
not increase very much - probably by less than 0.1 mM. Thus, while the Na/K pump is necessary
over the long run, most axons can give long bursts of AP's without requiring pump activity. It's an
elegant system: a large reserve of stored energy (the Na+ and K+ 'batteries') can be tapped rapidly
and repeatedly, and restored leisurely at a later time.
In summary, we have examined in some detail the behavior of the voltage-sensing gates in
sodium and potassium channels, to see how various phenomena – refractoriness, accommodation –
arise. But we have not yet considered how the action potential spreads along the length of the axon.
That's next.
APPENDIX: Axons are wretched electrical cables.
How do electrical signals travel along biological cables? We'll begin simply and work up to
more complex situations. Consider an experiment on a hypothetical spherical cell, one with a pure
lipid membrane that has no ion channels. The experiment is illustrated in Fig. 1. We impale the cell
with two microelectrodes. One is for recording membrane potential, and the other is for passing
electric current into the cell; it is connected to a battery through a switch. When the switch is
closed, positive charge is injected into the cell. The accumulation of charge in the cell of course
changes Vm. Because we are injecting constant current (i.e., injecting charges at a constant rate),
Vm is depolarized linearly with time. When the switch is opened, current stops flowing, and Vm
stays at its final level (because the injected charges, which are repelling each other and would love
to get out of the cell, can't get out, because there are no ion channels).
The membrane is behaving as a capacitor; the ICF and ECF are like the two parallel metal
Fig. A1. Injecting
current into a cell
that has no ion
channels. The cell
behaves like a
capacitor (right
side).
47
plates of an electronic capacitor, and the membrane is like the dielectric material between the
plates. We can model this cell's membrane properties with a capacitor, as shown in Fig. A1, right
side.
Now let's move a little closer to reality by putting some ion channels in the membrane and
repeating the same experiment (Fig. A2). When we inject current, Vm starts to change at the same
rate as before, but soon lags. Eventually a state is reached in which, even though charge continues
to be injected into the cell, Vm does not change any more. At this time (and beyond), every time a
charge is injected into the cell, another leaves (and so Vm doesn't change, because Vm is always
determined by the number of excess anions (or cations) in the cell). The injected charges, which
repel each other, now have a way to get out of the cell: through the ion channels. But clearly, the
channels do not allow the charges to move freely, because Vm has changed from its resting value
(i.e., not all injected charges have left the cell). The ion channels have a resistance to ion flow.
When the switch is opened and charges are no longer injected, Vm returns to its former
level as the remaining injected charges leave the cell. The peak value of the change in Vm (Vm)
depends on the membrane resistance; by Ohm's Law: Vfinal = I R, where I= current and R= the
lumped resistance of the ion channels (called the 'input resistance' of the spherical cell).
The electronic model of the cell can be updated by putting a resistor (whose resistance
represents the (parallel) resistance of all of the ion channels) in parallel with the capacitor, as
shown in Fig. A3. A battery represents the value of the resting membrane potential. This is a
parallel RC circuit. The shape of the fall of Vm when the current is turned off is exponential, and
the 'time constant' (which is the time for ΔV to decay to 1/e or about 37% of its plateau value) is
given simply as the product R.C, where C= the total capacitance of the spherical cell ('input'
capacitance). The rise and fall have the same shape.
Fig. A2. Current injection in
spherical cell
Fig. A3. Electronic analogue
of spherical cell
As shown in Fig. A4, when current is injected into this spherical cell, it follows two routes:
some current flows into the membrane capacitor and some through the ion channels. The
capacitative component (Ic) is transient, and the resistive component (IR) makes up the remainder.
48
Fig. A4. Capacitative and resistive components of the current.
Now that we have reviewed some basic electronics, we can directly address the question:
How are electrical signals transmitted in the body? Let us repeat the experiment, only now we will
perform it on a nerve axon, which is cylindrical in shape (Fig. 5, below). The electrical analogue of
a patch of membrane is essentially unchanged, there being a number of identical membrane
elements strung out along the length of the axon. The individual elements are connected together
inside the fiber by resistors (ri) representing the intracellular resistance of the cytoplasm. Outside
the cell, the elements are 'shorted' together, because the resistance of the bathing solution is very
low.
[This is not to say that axoplasm and ECF have fundamentally different electrical
properties. The 'specific' resistivity of cytoplasm is nearly the same as that of the ECF, but the
axon’s cross-sectional area is much smaller, so that, for a given length of axon, intracellular
longitudinal resistance, ri, is much greater than ECF resistance. For any simple conductor, the
resistance r is given by r = R.l/ A, where R= 'specific' resistivity, l= length, and A= cross sectional
area.]
If we insert both current
passing and recording electrodes
near each other (left side), pass
negative current, and record the
potential change, the results (Fig.
5, Vo) are very similar to the
previous experiment. If now we
remove the recording electrode
and re-impale the axon at a more
remote point and repeat the
experiment, the response is very
different (Fig. 5, Vx). First, Vx is
smaller than Vo because some
of the injected current leaks out of the fiber before reaching the recording site. Second, ΔVx is
slower because there is a lot of membrane capacitance between the injection and recording sites,
and it takes time to charge and discharge (longer than at the source because the current must flow
through the axoplasmic resistance). These are called cable properties of axons, because the
equations describing this kind of behavior were first solved in the 19th century by engineers
designing transatlantic telephone cables (the cable cores had a small but finite resistance, and the
49
insulation was a little leaky and capacitative). A leaky cable is a terrible design for a nerve axon:
electrical signals are slowed and diminished over distance, which makes rapid transmission of
signals over long distances virtually impossible if one relies on these passive properties.
Quantitatively, how bad is the
design? What we need to know to
answer this question is how the potential
change falls off with distance along the
axon. If we pass a long duration current
(in order to charge fully the membrane
capacitance and reach a steady state) and
measure Vm at different distances from
the current source, we find that ΔVm falls
off exponentially from the source, as
shown in the graph to the right.
Intuitively, one can see that, the bigger Rm is, the farther charges will spread down the inside of the
axon (there being fewer ion channels for the charges to leak out), and the smaller Ri, the farther
along the inside of the axon the charges will spread. Quantitatively, the length constant lambda (λ)
is the distance at which a steady applied voltage has fallen to 1/e (about 37%) of its maximum
value. The value of λ is given by:
The length constant of an axon
where rm is the total membrane resistance and ri is the total axoplasmic resistance of a 1 cm length
of the axon being studied.
In a typical nerve axon, λ has a value of less
The first trans-Atlantic cable was
than 1 millimeter. Over a distance of three length
completed in 1858. Queen Victoria
constants, the signal will have virtually disappeared
sent the first message, to President
(0.373 is about five per cent). To get the voltage to be Buchanan. Transmission was slow
(1 letter every 2 minutes), so they
transmitted over a one meter long axon seems a
turned up the voltage, and at about 2000 volts, the
hopeless prospect. In other words, the passive spread
cable fried and died. Four cables and 30 years later,
of electric signals over distances greater than a few
they got it to work reliably. The cable cores had
millimeters is utterly hopeless.
diameters of a several cm, and length constants
measured in miles.
50
Molecules to Medicine- Cell Physiology
8. Action Potential: Conduction 23 Nov 11 AM
Bill Betz
[email protected]
http://www.cuphys.net/cellphys
Key words: safety factor, compound action potential, salutatory, myelin, volume
conductor, fixed negative charges, hypocalcemia, hyperkalemia
8. Action Potential: Conduction
We have considered what goes on in a microscopic patch of membrane
during an action potential. In a nutshell, at threshold, (nearly) all of the sodium
channel activation gates open up, and at the peak of the AP, (nearly) all of the
sodium channel inactivation gates close and (nearly) all of the potassium channel
gates open. Characteristics that arise from this behavior include threshold, all-ornone behavior, refractory period, and accommodation. With this as background, we
will now consider the mechanisms involved in conduction of the action potential along a nerve
fiber. No new principles are involved.
Suppose we could 'freeze' an action potential in time and measure the membrane potential
at each point along the length of the axon. The spatial profile of potential at that instant would be
as shown in Fig. 1 (top). The flow of current is shown in Fig. 1 (bottom). At the ‘active locus’, Na+
ions are rushing into the fiber, making the region positive with respect to the rest of the fiber.
Current then flows through the axoplasm in both directions away from the active locus. This flow
of current ahead of the active locus depolarizes the membrane to threshold; Na+ channels there will
then open, so that this next stretch of membrane will itself become an active locus, and so on along
the membrane to the right. A millisecond or so before the AP reached the point shown in Fig. 1 the
membrane behind the locus (to the left) was itself an active locus. Now however, this patch of
membrane is undergoing repolarization; Na channels have stopped conducting and K channels
have opened. It is refractory.
The inward membrane current at the active locus is of course carried by Na+ ions. The
axoplasmic and outward membrane currents are carried principally by K+ ions. Chloride ions carry
little membrane current, because PCl is relatively low.
Fig. 1. Top: An action potential
propagating left to right along an
axon is frozen in time, and the spatial
profile of potential is shown at that
instant. Bottom: Currents are shown.
Inward current at the active locus is
carried by sodium. Notice that the
axon is not drawn to scale. The
diameter of the axon is only 10 µm
but the length of the active zone,
drawn the same size, is really a
thousand times bigger (1 cm). Thus,
when an action potential propagates,
at any instant the sodium channels are
open over a length of a centimeter or
more.
51
Now suppose we add a local anesthetic like novocaine to just the right side of the axon, just
ahead of the active locus. The anesthetic blocks sodium channels, keeping activation gates from
opening. The depolarization generated by the last active sodium channels might spread passively
farther along the axon, but it would become smaller and smaller, so that if the anesthetic was
applied to a long enough length of axon, any depolarization getting past the anesthetic block would
not reach threshold, and the sodium channels there would not open, and the action potential would
die. This is what happens when your dentist administers novocaine. Pain fibers in your tooth or
gum may be firing action potentials like crazy, but they don’t get past the novovaine block and
your brain remains blissfully ignorant of the licensed violence being committed in your mouth.
(How long would the length of block need to be? To answer this, you need to remember the length
constant of a typical axon – about 1 mm. The depolarization will fall to about one-third each length
constant. So, several millimeters of blocked axon should do the job.)
Safety factor. It should be clear that a crucial factor in normal conduction is the density of
Na channels in each imaginary patch of membrane. There must be enough Na channels to supply
enough current to depolarize the next bit of membrane to threshold. This is key to understanding
action potential conduction. It turns out that axons have more than enough Na channels to do the
job; their safety factor of transmission is 5-10 times the minimum required for successful
propagation. Is this excess just a needless waste of Na+ channels? After all, for every Na channel
that opens, the sodium-potassium pump must eventually pump several thousand ions. There are
several good reasons for this apparent extravagance. First, axons branch, sometimes profusely. At
each branch point, the safety factor for transmission is reduced: the membrane just before the
branch must deliver sufficient current to depolarize both branches to threshold. A high safety factor
insures that action potentials will spread down both branches of the axon.
Another good reason for having a high safety factor concerns sodium channel inactivation.
Immediately after an AP, the axon is refractory, in part because Na channel inactivation gates have
not yet reopened. The absolute refractory period ends when enough inactivation gates have
reopened to enable the axon to conduct again. This requires that a certain minimum number of
inactivation gates must be open in each ring of membrane. That number will be reached sooner if
there is an excess of Na channels in the membrane. The upshot of this is simply that the axon will
be able to transmit AP's at higher frequency than if it had only the minimum number of Na
channels required for propagation. Moreover, the velocity of propagation of the AP will be
increased, because the excess Na channels will supply extra current, and so drive the ring ahead of
the active locus to threshold quicker.
Mini-problem set
Action potential aerobics. Here are some questions to test your understanding. Answers
follow the last question.
1. In a clinical neurological test, a nerve trunk is stimulated with a brief shock delivered
transcutaneously in the forearm. In each axon that reaches threshold, how many action potentials
are generated?
2. Backfiring: Why doesn’t a propagating action potential reverse direction? After all, when
repolarization occurs (behind the active zone), Vm comes back to threshold. So, why isn’t a second
AP generated, which would then propagate in the opposite direction?
3. An axon is stimulated simultaneously at two points separated by about 10 cm. How
many action potentials are created?
4. Same as number 3. What happens when the two APs traveling towards each other
collide? Does one win and the other stop? Do both stop? Do both pass on through?
52
5. An axon is perfused intracellularly with ECF, and is bathed in ICF (believe it or not, this
experiment has been done, using the squid giant axon). The membrane potential, instead of being
-70 mV, now is about +70 mV. Negative current is injected to drop Vm to about +40 mV. Does the
axon fire an upside-down action potential?
ANSWERS:
1. When an axon is stimulated, unless the stimulus is given exactly at one end, then two
APs arise, and travel in opposite directions. There is no inherent directionality to the AP
mechanism.
2. An AP doesn’t reverse direction because, looking backwards (like riding in the last car
on a train and looking back), all one sees is refractory axon – the sodium channels are incapable of
firing an AP immediately after the active zone passes.
3. Two stimulating electrodes on an axon will generate four APs, two of them traveling
towards each other, and two traveling towards the ends.
4. When the two APs traveling towards each other collide, they annihilate each other.
Why? Pretend you’re one of the APs, traveling down the axon. As you collide with the oncoming
AP, all you see ahead of you is refractory membrane.
5. Reversing ICF and ECF, do you get an upside-down AP? If the membrane potential is
+70 mV, what is the position of the inactivation gates in the sodium channels? They are all tightly
closed, and they wouldn’t open unless Vm were made very much more negative (-70 mV or so). So
there will be no upside-down AP. [This might lead you to ask, suppose in addition to switching
solutions, we also inverted the membrane, like turning a sock inside-out? In that case, all
membrane potentials would be positive, and all APs would indeed be upside-down.]
Extracellular recording. Hopefully by now you are comfortable thinking about
experiments involving measurement of membrane potential with an intracellular microelectrode.
There are, however, no clinical tests involving the use of intracellular microelectrodes (they are
technically too difficult), although many do involve extracellular recording techniques (e.g., EKG,
EMG, EEG). Such records of course give less precise information than an intracellular recording,
but they report simultaneously the activity of many cells, which can provide useful diagnostic
information.
Like intracellular recordings, extracellular recordings involve nothing more than measuring
the potential difference between two electrodes. The ECF has a small but finite electrical resistance
(which we have ignored previously), and a current flowing will produce a voltage drop across this
resistance. That’s what extracellular recording electrodes detect. The currents are generated, of
course, by action potentials. Extracellular signals are small (a few mV) in amplitude. Because
extracellular recordings detect activity of many nearby axons, their amplitudes and shapes vary a
lot. Some details of extracellular recording and stimulation are described in Appendix Ia, Ib, and
Ic.
Consider a typical recording from a nerve trunk, which may contain thousands of
individual axons (Fig. 2). The external electrode will sample the activity from all of the fibers
simultaneously (something which certainly is not possible with an intracellular micropipette). The
result is called a compound action potential because it comprises many discrete units. Assume we
stimulate the ulnar nerve where it enters the upper arm, and record extracellularly in the arm near
the stimulating electrodes (Fig. 2, V1). The stimulating electrodes, like the recording electrodes, are
extracellular. During the pulse, current flows between the stimulating electrodes; some of it will
flow across the axonal membranes, and depolarize the axons. With pulses of low intensity, we
record a small negative deflection. With further increase in stimulus strength, the size of the
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recorded response grows; that is, the response is graded with stimulus intensity. This is certainly
different from an intracellular recording where the action potential is an all-or-none event. The
extracellular response is graded because the axons in the nerve trunk do not all reach threshold
together. Some axons require a stronger stimulus than others in order to initiate an action potential.
The more axons that fire, the bigger the response. We
will return to this point shortly.
Continuing the experiment, we move the
recording electrodes much farther from the stimulating
electrodes - say, down to the wrist - and stimulate again
with increasingly strong shocks. Now, of course, the
latency (i.e., the elapsed time between the stimulus
artifact and response) is increased because the action
potentials must cover a greater distance before reaching
the recording electrodes (V2 in Fig. 2). As before, we
find that the amplitude of the compound action potential
is graded with stimulus strength. With strong shocks,
however, an additional change in the response is
observed: now a second, smaller bump is observed on
Fig. 2. Superimposed extracellular recordings
the tail of the first. In other words, the compound action
from sciatic nerve. Stimuli were applied where
potential is dispersed in time. The dispersion means
nerve enters the thigh; three different stimulus
strengths (top). Recordings were made from the simply that some axons conduct at a lower velocity than
others and so reach the recording electrodes later than
sciatic nerve in the thigh (V1) and at the ankle
the fast action potentials. This dispersion wasn’t
(V2).
observed when the recording was made in the thigh
because the fast conducting APs had not had time to outrun the slow ones. And the fact that the
second peak (corresponding to the slowly conducting fibers) is seen only with relatively high
stimulus strengths means that the slow fibers require a stronger stimulus to reach threshold than do
fast fibers.
So slow conduction velocity and high threshold (to external stimulation) go together. If we
assume that the microscopic properties of all axon membranes are identical, we can explain this
behavior on the basis of one very simple parameter: axon diameter. As we will see, small diameter
fibers have a high threshold to (external) stimulation and conduct action potentials at a lower
velocity compared to large diameter axons. In addition, they have a low safety factor for
conduction. Let's take these three properties in turn.
1. Threshold to external stimulation. Small diameter fibers are harder to stimulate than
large diameter fibers. The basic mechanism is described in Appendix Id. This difference in
threshold to extracellular stimulation is useful in clinical tests. For example, the conduction
velocity in peripheral nerves is measured during regeneration of damaged nerves (velocity
increases as recovery proceeds). Such measurements are also useful in localizing compressive
tumors or inflammatory processes (conduction velocity is reduced where a nerve trunk is
compressed or inflamed). A potential practical problem arises because there are pain fibers in the
nerve trunk, and if they are stimulated, the testing procedure can be very unpleasant for the patient.
However, it turns out that pain fibers have small diameters. Thus the large diameter fibers (which
include motor axons) can be selectively stimulated, and the testing procedure is relatively painless.
2. Safety factor. Small diameter fibers have a low safety factor for conduction of AP's. The
quantitative analysis of the effect of axon diameter on safety factor is complex. However
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complicated the explanation is, though, the fact is easy to demonstrate. For example, small
diameter axons can be selectively blocked by a low dose of local anesthetic such as procaine. (The
anesthetic works by blocking some of the Na channels.) Large axons, however, are spared. If you
have ever had a laceration stitched up, you may have felt the skin being manipulated (large
diameter "touch" fibers were still conducting), but felt no pain (small diameter pain fibers blocked
by anesthetic).
3. Conduction velocity. Bigger is faster. Myelin is even faster. Through the course of
evolution, organisms that could react quickly to changes in their environment had a strong
selective advantage over their slower peers. Reaction time is limited, in the end, by the velocity of
action potential conduction along sensory and motor axons. Two mechanisms have evolved to
boost conduction velocity: axon diameter and myelin. We will consider these in turn.
a. Axon diameter. Axons with big diameters conduct faster than axons with small
diameters. The mechanism by which this arises is described in the Appendix.
The biggest human axons are about 20 µm in diameter. The world record is held by the squid,
which has an axon that may be over one millimeter (1000 µm) in diameter, and mediates the
squid's escape behavior. Now that is a lot of space to devote to one cell. The human olfactory nerve
is only a millimeter wide, yet contains more than 100,000 individual axons! Furthermore, the
squid's response to natural selection reaches a point of diminishing returns, because conduction
velocity depends on the square root of diameter, so that doubling diameter (which quadruples the
volume) increases conduction speed by only 41%.
b. Myelin. Vertebrates have evolved a novel specialization to increase conduction speed
still further, namely myelin. Glial cells create the multiple layers of tightly packed myelin
membrane that surround an axon, which increases the electrical resistance between the inside of
the axon and the extracellular fluid, thereby reducing the leaky cable properties of the underlying
axon membrane. The insulating sheath of myelin is interrupted every few millimeters, exposing the
bare axonal membrane. This naked region is called a Node of Ranvier, and is the site where all of
the voltage-gated sodium channels are localized (their tight packing may reach a density of several
thousand channels per square micrometer of membrane); there are no sodium channels beneath the
myelin. Thus, when a node of Ranvier is depolarized to threshold, Na+ channels in the nodal
membrane open and the node becomes an active locus. The current then spreads effectively inside
the axon through the internodal region to the next node, which becomes depolarized , and the
process repeats, 'jumping' from node to node, a process called saltatory conduction (L. saltare, to
jump).
Myelination greatly increases conduction velocity; if the squid giant axon were myelinated
its conduction velocity would increase by a factor of 100! Myelination also changes the
quantitative relation between fiber diameter and conduction velocity; the relationship now becomes
directly proportional so that doubling the diameter doubles the conduction velocity.
Fig. 3. Myelin Parameters
If fast conduction is so great, why aren’t all axons big? (There is a 100-fold range in human
axon diameters (from 0.2 to 20 µm).) The explanation for the huge range in diameters lies in the
55
fact that there is not an unlimited amount of space or precursor materials in the body for axons; a
compromise is struck between the need for speed and the need for content. The choice becomes
whether to make all axons the same size, or make some big at the expense of others - one axon that
can conduct at 90 m/sec takes up as much space as nine smaller axons that conduct at 30 m/sec. So
for the sacrifice of several milliseconds in conduction time, an enormous increase in information
content can be obtained - nine fibers can tell you a lot more than one fiber can. Furthermore,
processing by the central nervous system (CNS) of information of any complexity requires a fair
amount of time (tens of milliseconds), and supplying this information to the brain over 90 m/sec
axons would be gross mismatching, because the long central computation time precludes a very
rapid response. Thus, Nature has struck a balance between conflicting needs in the evolution of the
nervous system, and the outcome is an efficient and economical compromise between speed and
content. In the human nervous system, two of the most extreme examples of the speed vs content
struggle are olfactory axons and muscle spindle afferent axons. The latter, which you will study
later, convey feedback sensory information to the CNS from muscles, and play an important role in
the fine tuning of movements. Such information is needed in a hurry by the CNS while one is, say,
playing a Chopin scherzo on the piano, and muscle spindle afferents are the largest and fastest
conducting of all axons (about 20 microns diameter, 100 m/s in velocity). On the other hand,
responses to olfactory stimuli are not very rapid; the axons in the olfactory nerve are only about 0.2
µm in diameter, and so conduct slowly. However, there is enough space for about 100,000
olfactory axons in the olfactory nerve which, at about 1 mm diameter, is about the size of the squid
giant axon. The net result is a very rich input to the CNS, but you cannot savor an odor in fifty
milliseconds; you need more central computing time.
Finally, if myelin is so great for
increasing conduction velocity, why aren’t all
axons myelinated? You may recall that, for
nonmyelinated fibers, conduction velocity
increases as the square root of diameter,
whereas in myelinated axons, conduction
velocity is proportional to fiber diameter. A
graph of these relations is shown in Fig. 4.
Notice that below a diameter of about one
micrometer, the conduction velocity is faster in
the nonmyelinated fiber (see Appendix If for
details). Nature has discovered this by trial and
Fig. 4. Axons with diameters below about
error, and only the smallest axons (diameters less
1.0 µm are nonmyelinated.
than one micrometer) in mammals are
nonmyelinated.
Acute Hyperkalemia and CBIGK. Let’s review one more time the physiology of acute
hyperkalemia, and consider the role calcium ions as a treatment - the C in CBIGK. The causes are
usually due to K+ escaping rapidly from cells (hemolytic anemia, crush accident, electrocution,
severe infection). Acute hyperkalemia is most dangerous in the heart where the elevated
extracellular potassium depolarizes cells, disrupting the rhythm of the master clock (in the S-A
node) that triggers each beat of the heart. Normally, cells there, called pacemakers, undergo
spontaneous depolarizations (see figure below), and when they reach threshold, they fire APs,
which spread to the muscle fibers throughout the heart and trigger contraction. Hyperkalemia
depolarizes these cells, disrupting the exquisite timing of the clock. New pacemakers - mavericks -
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Normal pacemaker activity of the heart's master clock (S-A node). Cells there undergo spontaneous
depolarization and fire APs when they reach threshold (about once per second). Abnormal
depolarization of these cells by hyperkalemia can disrupt this pattern.
can arise anywhere in the conduction
system, disrupting the synchrony of muscle contraction, causing the EKG arrhythmias .
The maverick pacemakers can be silenced by administering calcium ions, thereby
restoring the normal synchronous pattern of excitation and contraction. How calcium ions exert
this quieting effect on sodium channels is described in detail in Appendix II. In brief, the
mechanism is unlike anything that we have discussed: calcium ions act by binding to fixed
negative charges on the outside surface of cells and trick the sodium channels into thinking that the
membrane has been hyperpolarized, thereby raising the threshold for action potential initiation.
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APPENDIX I: Details of extracellular recordings and threshold
a. Extracellular recording from a single axon. Consider a simple experiment: a single axon
is dissected from a nerve trunk, and is gently laid across two widely separated (ca. 10 cm)
electrodes (Fig. A1). An action potential is initiated with a stimulator (not shown) and propagates
from left to right. The recordings can look very different, depending on the electrode arrangement
(Fig.A3).
Figs. A1 and A2. Extracellular recordings
from a single axon. Action potential
propagates from left to right (arrows). In Fig.
A1 the nerve is in Ringers solution; in Fig.
A2, part of the nerve is lifted into an
insulating medium.
In the first panel (Fig. A3), the response is very small - the individual bumps can barely be
resolved above the background noise of the recording system. Matters are greatly improved if the
axon is lifted out of the saline into an insulating medium, such as paraffin oil or moist air (Fig. A2);
this gets rid of most of the short-circuiting effect of the saline; a thin film
of saline will, however, cling to the axon and provide an extracellular
current path).
So you see right away that the amplitude of externally recorded
signals depends a lot on the geometry of the external fluid. The negative
deflection occurs as the active locus of the action potential passes the first
electrode. When it passes the second electrode, the deflection is of course
reversed in sign. Ordinarily, it is impractical to dissect such a long length
(10 cm) of nerve; usually the recording electrodes are less than one cm
apart. In such a case, the second wave overlaps the first, and a diphasic
potential is recorded (Fig. A3, panel C). This adds a further complexity,
because it now is difficult to know exactly what happens at each individual
electrode because of the overlap of the two waveforms. So, in
experimental studies, the action potential is usually blocked between the
electrodes (by crushing the nerve, or by local cooling). Because the action
potential never reaches the second electrode, the response is now a pure
Fig. A3. Extracellular
monophasic potential (Fig. A3, panel D).
recordings of electrical
activity in a nerve trunk.
b. With extracellular stimulation, will depolarization occur
at the negative or positive electrode? Answer: The axon will be
depolarized at the negative electrode. As the negative electrode is
moved close to the axon, it makes the outside of the axon negative
at that point. Making the outside negative is equivalent to making
the inside positive (it's the potential difference that matters), so an
action potential is initiated at the cathode.
c. When recording compound APs, why is it better to have
the recording electrode far from the stimulating electrode? Answer:
to give time for slow and fast conducting APs to separate in time.
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d. Why do large axons have a lower threshold to extracellular stimulation than do small
axons? A large diameter and a small diameter axon are placed near stimulating electrodes (Fig.
A4). Current flows between the stimulating electrodes, mostly in the extracellular fluid (because of
the low resistance). Some of the current will enter the axons under the positive electrode, flow
through the axoplasm, and out of the axon at the negative electrode. The axon will be depolarized
at the negative electrode.
We can represent this situation schematically with two resistors: r1 (representing a ring of
membrane under each electrode) and r2 (representing the internal resistance of the axon). Now r1 is
inversely proportional to the axon diameter, but r2 varies inversely with the square of fiber
diameter: r2 is proportional to 1/d2. The applied voltage V will be dropped across the 3 resistors
arranged in series. We are interested in ΔV across the resistor, r1, under the cathode (when ΔV is
large enough, the membrane will be depolarized to threshold). This is just a voltage divider, and
the potential change across r1 is simply:
Fig. A4. Left: spread of current between
stimulating electrodes. Right:
Schematic diagram of axon. r1 =
resistance of a "ring" of membrane, r2 =
internal resistance between "rings".
Substituting the proportionate diameter values:
where k is a constant of proportionality. Simplifying:
Thus, as d increases, ΔVr1 will also increase. In other words, for a given stimulus, the potential
change will be greater in a large diameter axon.
e. What factors determine optimal myelin thickness, and optimal internodal distance? As
shown in Fig. 3 (reproduced below), assume we have a fiber of constant thickness T (note that T =
axon + myelin thickness). How much of the total diameter T should be devoted to axon, and how
much to myelin, in order to maximize conduction velocity? As we increase axon diameter
(keeping T constant), the internal resistance of the axoplasm of course decreases. This will
increase conduction velocity. However, as axon diameter increases, the myelin thickness must be
decreased in order to keep the overall diameter (T) constant. Thinner myelin will make the
internode leakier, resistively and capacitatively, and this will reduce conduction velocity. Thus
there will be a balance point between axon diameter and myelin thickness, and at this point
conduction velocity will be maximal (for the given fiber diameter T). Purely theoretical
considerations show that the axon diameter should be about two-thirds of the overall fiber diameter
in order to achieve maximum conduction velocity. In real axons, as you might expect, the value
falls between 60% and 80%. In other words, without recourse to theoretical considerations,
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evolution has selected the optimum design for maximizing conduction velocity.
Fig. 3 (reproduced).
Myelin Parameters
The optimum internodal distance is more complex, and depends on many factors. It turns
out that internodal length is not too critical for conduction velocity, although, in general, large
diameter fibers have longer internodal distances than do small diameter fibers. Usually, the length
of fiber between nodes is about 100 times the fiber diameter (in human nerve, most nodes are
separated by about 1-2 mm).
Appendix II: Calcium and excitability
To understand the basics of the cardiac arrhythmias, we need to know a little bit about how
heart beats are triggered (details come next spring). In brief, there is a master clock in the wall of
the right atrium that triggers each beat. The clock comprises a clump
of connected cells (the sino-atrial (SA) node) that depolarize
spontaneously (and synchronously). When they reach threshold, each
fires one action potential. Axon-like processes carry the APs out to all
of the muscle fibers, initiating muscle contraction (systole). In fact, a
kind of backup system exists, because all cells in the cardiac
conduction system depolarize spontaneously. It’s just that the ones in
the SA node depolarize fastest and so get to threshold first, setting the
overall tempo of beats. But the backups can make trouble during
hyperkalemia.
During hyperkalemia, the conduction cells are at risk. The depolarization moves Vm
closer to threshold, making it easier to fire an action potential, but it also inactivates some
sodium channels, making it harder to fire an action potential. Maverick pacemakers can arise for
either reason. In the latter case (block of conduction), other pacemaker cells distal to the block
will keep depolarizing spontaneously, and when they get to threshold, they will fire an AP. It’s
bad news either way, because now these mavericks are driving part of the heart at a rate that is
independent of the rest, and when heart muscle fibers don’t contract synchronously, the heart
cannot pump blood effectively. If other maverick pacemakers arise, the situation can deteriorate
rapidly, leading ultimately to ventricular fibrillation, a condition in which no blood is pumped.
Calcium ions and excitability. Calcium ions are one of the most important intracellular
signaling molecules in the body, participating in a bewildering variety of important cellular events,
such as muscle concentration, cellular motility, bone deposition, and exocytosis. You will study
these mechanisms in the months ahead. Ca++ is also important in governing the threshold for the
action potential, which is the rationale for giving calcium when cardiac arrhythmias emerge during
hyperkalemia.
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How does calcium affect AP threshold? The key features are illustrated in Fig. A5, which
shows the effects of extracellular Ca++ on the surface negativity of the membrane. By way of
Fig. A5. Calcium ions screen
fixed negative charges, increasing
the membrane electric field (Vm).
A. In the normal situation, some
charges are screened, and the
local Vm is -70 mV. B. If [Ca++]
in the ECF increases, more
charges are screened, and the
local Vm hyperpolarizes to -80
mV, thereby moving farther from
threshold for an AP. C. In
hypocalcemia, more fixed
negative charges are exposed,
reducing the local Vm to -50 mV,
leading to spontaneous AP
generation.
introduction to this arcane but important phenomenon, the outside surfaces of all plasma
membranes are studded with fixed negative charges (they are negative headgroups of phospholipid
molecules and also anionic carbohydrate residues bound to
membrane proteins, including sodium channels -see cartoon
to the right). These fixed negative charges make the outside of
the membrane a little bit more negative than it would
otherwise be, but the negativity is exceedingly localized in
space – a tiny nanodomain near the membrane surface. For
example, if a positive ion in the ECF approaches the
membrane, it would not even begin to detect the presence of
the fixed negative charges until it was within a few
nanometers of the surface (less than the thickness of the
membrane itself!). That's because all the other free ions and water molecules in the bulk ECF block
its electrical 'view' of the fixed charges beyond this short distance. Membrane proteins, however,
lie within the nanodomain, and thus ‘see’ that the outside of the membrane is made negative by the
fixed negative charges. Making the outside of the membrane negative is equivalent to a
depolarization. A microelectrode recording Vm would, however, never see this nanodomain. For
example, in Fig. A5A, a microelectrode would record a Vm of -80 mV, but the ‘local’ Vm
perceived by sodium channels and other membrane proteins is less, only -70 mV, owing to the
unscreened fixed negative charges on the outside surface of the membrane. In summary,
unscreened fixed negative charges reduce the electric field across the membrane, making it easier
for sodium channels to start conducting.
As shown in Fig. A5B, giving extra calcium ions screens naked fixed negative charges,
hyperpolarizing the local Vm, making it more difficult for the cell to fire an action potential. The
binding of calcium is entirely electrical; no special ‘receptor’ is involved. In summary, pacemaker
cells in the conduction system of the heart are depolarized by the high [K+]o, and maverick
pacemakers arise. Calcium ions administered intravenously bind to surface membranes, increasing
the threshold for AP generation. The clinical hope, of course, is that the normal pacemakers will
take over once again and restore the normal excitation of the heart muscle.
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