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Transcript of Control of Cell Volume and Membrane Potential James Sneyd Auckland University, New Zealand Basic...
Control of Cell Volume and Membrane Potential
James Sneyd
Auckland University, New Zealand
Basic reference: Keener and Sneyd, Mathematical Physiology (Springer, 1998)
A nice cell picture
• The cell is full of stuff. Proteins, ions, fats, etc.
• Ordinarily, these would cause huge osmotic pressures, sucking water into the cell.
• The cell membrane has no structural strength, and the cell would burst.
Basic problem
• Cells carefully regulate their intracellular ionic concentrations, to ensure that no osmotic pressures arise
• As a consequence, the major ions Na+, K+, Cl- and Ca2+ have different concentrations in the extracellular and intracellular environments.
• And thus a voltage difference arises across the cell membrane.
• Essentially two different kinds of cells: excitable and nonexcitable.
• All cells have a resting membrane potential, but only excitable cells modulate it actively.
Basic solution
Typical ionic concentrations (in mM)
Squid Giant Axon Frog Sartorius Muscle
Human Red Blood Cell
Intracellular
Na+ 50 13 19
K+ 397 138 136
Cl- 40 3 78
Extracellular
Na+ 437 110 155
K+ 20 2.5 5
Cl- 556 90 112
The cell at steady state
3 Na+
2 K+
Cl-
Ca2+
We need to model
• pumps and exchangers
• ionic currents
• osmotic forces
OsmosisP1 P2
waterwater +Solvent(conc. c)
At equilibrium:
€
P1 + kcT = P2
Note: equilibrium only. No information about the flow.
The cell at steady state
3 Na+
2 K+
Cl-
Ca2+
We need to model
• pumps and exchangers
• ionic currents
• osmotic forces
I’ll talk about this a lot more in my next talk.
Na,K-ATPase
CalciumATPase
Active pumping
• Clearly, the action of the pumps is crucial for the maintenance of ionic concentration differences
• Many different kinds of pumps. Some use ATP as an energy source to pump against a gradient, others use a gradient of one ion to pump another ion against its gradient.
• A huge proportion of all the energy intake of a human is devoted to the operation of the ionic pumps.
• Not all that many pump models that I know of. It doesn't seem to be a popular modelling area. I have no idea why.
A Simple ATPase
E E•ATP E•ATP•L
Ee•ATP•LEe•ATPEe•ADP•P
Inside
Outside
Li
Lo
ATP
ADP + P
k1 k2
k3
k4k5
k6
k-1 k-2
k-3
k-4k-5
k-6
€
J = k1[ATP ][E] − k−1[E • ATP ]
=
[ATP ][L e ]
[ADP][P][L i]− K1K2K3K4K5K6
a nasty function of the rate constants
=
[ATP ][L e ]
[ADP][P][L i]−
[ATP ]eq[L e ]eq
[ADP]eq[P]eq[L i]eq
a nasty function of the rate constants
Note how the flux is driven by how far the concentrations are away from equilibrium
flux
Reducing this simple model
E E•ATP E•ATP•L
Ee•ATP•LEe•ATPEe•ADP•P
Li
Lo
ATP
ADP + P
k1 k2
k3
k4k5
k6
k-1 k-2
k-3
k-4k-5
k-6
X3
Y3Y2Y1
X1 X2
Na+-K+ ATPase (Post-Albers)
E(N3)
Ee•K2
E(K2)
internal
occluded
external
Ei•K2 Ei•N3Ei
Ee•N3Ee
ATP
P
2Ki+
3Nai+
ADP
2Ke+
3Nae+
1 2
3
4
56
7
8
X1
Y2Y1
Z1 Z2
X3X2
Y3
Simplified Na+-K+ ATPase
X
Z2
Y
Z1
internal
occluded
external
3Nai+ 2Ki
+
2Ke+
3Nae+
ATP
P
ADPk-3
� 3� -8
� 7
� -4
k8
k-7k4
The cell at steady state
3 Na+
2 K+
Cl-
Ca2+
We need to model
• pumps
• ionic currents
• osmotic forces
The Nernst equation
€
Vi −Ve =RT
Fln
[S]e
[S]i
⎛
⎝ ⎜
⎞
⎠ ⎟
Note: equilibrium only. Tells us nothing about the current. In addition, there is very little actual ion transfer from side to side.
We'll discuss the multi-ion case later.
[S]e=[S’]e[S]i=[S’]i
Vi Ve
Permeable to S,not S’
(The Nernst potential)
Only very little ion transferspherical cell - radius 25 m
surface area - 8 x 10-5 cm2
total capacitance - 8 x 10-5 F (membrance capacitance is about 1 F/cm2)
If the potential difference is -70 mV, this gives a total excess charge on the cell
membrane of about 5 x 10-12 C.
Since Faraday's constant, F, is 9.649 x 104 C/mole, this charge is equivalent to
about 5 x 10-15 moles.
But, the cell volume is about 65 x 10-9 litres, which, with an internal K+
concentration of 100 mM, gives about 6.5 x 10-9 moles of K+.
So, the excess charge corresponds to about 1 millionth of the background K+
concentration.
Electrical circuit model of cell membrane
C
outside
inside
Iionic C dV/dt
€
CdV
dt+ Iionic = 0
Vi −Ve = V
How to model this is the crucial question
How to model Iionic
• Many different possible models of Iionic
• Constant field assumption gives the Goldman-Hodgkin-Katz model
• The PNP equations can derive expressions from first principles (Eisenberg and others)
• Barrier models, binding models, saturating models, etc etc.
• Hodgkin and Huxley in their famous paper used a simple linear model
• Ultimately, the best choice of model is determined by experimental measurements of the I-V curve.
Two common current models
€
INa = gNa (V −VNa )
INa = PNa
F 2
RTV
[Na+]i −[Na+]e exp −VF
RT( )
1− exp −VF
RT( )
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥
GHK model
Linear model
These are the two most common current models. Note how they both have the same reversal potential, as they must.
(Crucial fact: In electrically excitable cells gNa (or PNa) are not constant, but are functions of voltage and time. More on this later.)
Electrodiffusion: deriving current models
x=0 x=L
[S1+] = [S2
-] = ci[S1
+] = [S2-] = ce
S1
S2
Inside Outside
f (0) = V f ( ) = 0L
cell membrane
€
d2φ
dx 2= −λ2(c1 − c2), λ2 = stuff × L2
J1 = −D1
dc1
dx+
F
RTc1
dφ
dx
⎛
⎝ ⎜
⎞
⎠ ⎟
J2 = −D2
dc2
dx−
F
RTc2
dφ
dx
⎛
⎝ ⎜
⎞
⎠ ⎟
€
c1(0) = c i, c1(L) = ce
c2(0) = c i, c2(L) = ce
φ(0) = V , φ(L) = 0
Boundary conditions
Poisson equation andelectrodiffusion
Poisson-Nernst-Planck equations.
PNP equations.
The short-channel limit
If the channel is short, then L ~ 0 and so ~ 0.
€
Then d2φ
dx 2= 0, which implies that the electric field,
dφ
dx, is constant through the membrane.
dφ
dx= v ⇒
dc1
dx− vc1 = −J1
€
⇒ J1 = vc i − cee
−v
1− e−v
⇒ I1 =D1F
2
LRTV
c i − ce exp −VF
RT( )
1− exp −VF
RT( )
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
This is the Goldman-Hodgkin-Katz equation.
Note: a short channel implies independence of ion movement through thechannel.
The long-channel limit
If the channel is long, then 1/L ~ 0 and so 1/ ~ 0.
€
Then 1
λ2
d2φ
dx 2= c1 − c2, which implies that c1 ≈ c2 through the membrane.
c1 = c2 ⇒ 2dc1
dx= −J1 − J2
€
⇒ c1 = c i + (ce − c i)x
⇒ φ = −v
v1
lnc i
ce
+ 1−c i
ce
⎛
⎝ ⎜
⎞
⎠ ⎟x
⎡
⎣ ⎢
⎤
⎦ ⎥ v1 = nondimensional Nernst potential of ion 1
⇒ J1 =ce − c i
v1
(v − v1)
This is the linear I-V curve.The independence principle is not satisfied, so no independent movement ofions through the channel. Not surprising in a long channel.
A Model of Volume Control
Putting together the three components (pumps, currents and osmosis) gives.....
The Pump-Leak Model
3 Na+
2 K+
X-
Cl-
Na+ is pumped out. K+ is pumped in. So cells have low [Na+] and high [K+] inside. For now we ignore Ca2+ (horrors!). Cl- just equilibrates passively.€
−d
dt(qwN i) = gNa V −
RT
Fln
Ne
N i
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥+ 3pq
−d
dt(qwK i) = gK V −
RT
Fln
Ke
K i
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥− 2pq
−d
dt(qwCi) = gCl V +
RT
Fln
Ce
Ci
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
cell volume
[Na]ipump rate
Note how this is a reallycrappy pump model
Charge and osmotic balance
€
qw(N i + K i − Ci) + zxqX = qwe (Ne + Ke − Ce )
N i + K i + Ci +X
w= Ne + Ke + Ce
charge balance
osmotic balance
€
≤−1
• The proteins (X) are negatively charged, with valence zx.• Both inside and outside are electrically neutral.• The same number of ions on each side.
• 5 equations, 5 unknowns (internal ionic concentrations, voltage, and volume). Just solve.
Steady-state solution
If the pump stops, the cell bursts, as expected.The minimal volume gives approximately the correct membrane potential.In a more complicated model, one would have to consider time dependence also. And the real story is far more complicated.
RVD and RVI Okada et al., J. Physiol. 532, 3, (2001)
Ion transport
• How can epithelial cells transport ions (and water) while maintaining a constant cell volume?
• Spatial separation of the leaks and the pumps is one option.
• But intricate control mechanisms are needed also.
• A fertile field for modelling. (Eg. A.Weinstein, Bull. Math. Biol. 54, 537, 1992.) The KJU model.
Koefoed-Johnsen and Ussing (1958).
Steady state equations
€
PNavN i − Nme−v
1− e−v
⎛
⎝ ⎜
⎞
⎠ ⎟+ 3qpN i = 0
PKvK i − Kse
−v
1− e−v
⎛
⎝ ⎜
⎞
⎠ ⎟− 2qpN i = 0
PClvCi − Cse
v
1− ev
⎛
⎝ ⎜
⎞
⎠ ⎟= 0
w(N i + K i − Ci) + zX = 0
N i + K i + Ci +X
w= Ns + Ks + Cs
Note the different current and pump models
electroneutrality
osmotic balance
Transport control
€
Nm
Ns
<1+
3pq
PNa
1+2pq
PK
Simple manipulations show that a solution exists if
Clearly, in order to handle the greatest range of mucosal to serosal concentrations, one would want to have the Na+ permeability a decreasing function of the mucosal concentration, and the K+ permeability an increasing function of the mucosal Na+ concentration.
As it happens, cells do both these things. For instance, as the cell swells (due to higher internal Na+ concentration), stretch-activated K+ channels open, thus increasing the K+ conductance.
Inner medullary collecting duct cells
A. Weinstein, Am. J. Physiol. 274 (Renal Physiol. 43): F841–F855, 1998.
IMCD cellsReal men deal with real cells, of course.
Note the large Na+ flux from left to right.
Active modulation of the membrane potential: electrically excitable cells
Hodgkin, Huxley, and squid
Don't believe people thattell you that this is a smallsquid
Hodgkin Huxley
The reality
Resting potential• No ions are at equilibrium, so there are continual background currents. At steady-state, the net current is zero, not the individual currents.• The pumps must work continually to maintain these concentration differences and the cell integrity.• The resting membrane potential depends on the model used for the ionic currents.
€
gNa (V −VNa ) + gK (V −VK ) = 0 ⇒ Vsteady =gNaVNa + gKVK
gNa + gK
€
PNa
F 2
RT
⎛
⎝ ⎜
⎞
⎠ ⎟V
[Na+]i −[Na+]e exp(−VF
RT)
1− exp(−VF
RT)
⎛
⎝ ⎜
⎞
⎠ ⎟+ PK
F 2
RT
⎛
⎝ ⎜
⎞
⎠ ⎟V
[K+]i −[K+]e exp(−VF
RT)
1− exp(−VF
RT)
⎛
⎝ ⎜
⎞
⎠ ⎟= 0
⇒ Vsteady =RT
Fln
PNa [Na+]e + PK [K+]e
PNa [Na+]i + PK [K+]i
⎛
⎝ ⎜
⎞
⎠ ⎟
linear current model (long channel limit)
GHK current model (short channel limit)
Simplifications
• In some cells (electrically excitable cells), the membrane potential is a far more complicated beast.
• To simplify modelling of these types of cells, it is simplest just to assume that the internal and external ionic concentrations are constant.
• Justification: Firstly, it takes only small currents to get large voltage deflections, and thus only small numbers of ions cross the membrane. Secondly, the pumps work continuously to maintain steady concentrations inside the cell.
• So, in these simpler models the pump rate never appears explicitly, and all ionic concentrations are treated as known and fixed.
Steady-state vs instantaneous I-V curves
• The I-V curves of the previous slide applied to a single open channel
• But in a population of channels, the total current is a function of the single-channel current, and the number of open channels.
• When V changes, both the single-channel current changes, as well as the proportion of open channels. But the first change happens almost instantaneously, while the second change is a lot slower.
€
I = g(V , t)φ(V )
I-V curve of singleopen channel
Number of open channels
Example: Na+ and K+ channels
K+ channel gating
S0 S1 S2
2� �
2��
S00 S01
S10 S11
€
dx0
dt= βx1 − 2αx0
dx2
dt= αx1 − 2βx2
x0 + x1 + x2 =1
€
x0 = (1− n)2
x1 = 2n(1− n)
x2 = n2
dn
dt= α (1− n) − βn
Na+ channel gating
€
x21 = m2h
dm
dt= α (1− m) − βm
dh
dt= γ(1− h) −δh
2� �
2��
S00 S01
S10 S11
S02
S12
�2�
2��� � � � � �
S i j
inactivation activation
activation
inactivation
Experimental data: K+ conductanceIf voltage is stepped up and held fixed, gK
increases to a new steady level.
€
gK = g K n4
dn
dt= α (V )(1− n) − β (V )n
τ n (V )dn
dt= n∞(V ) − n
time constant
steady-state
four subunits
Now just fit to the data
rate of rise gives n
steady state gives n∞
Experimental data: Na+ conductanceIf voltage is stepped up and held fixed, gNa
increases and then decreases.
€
gNa = g Nam3h
τ h (V )dh
dt= h∞(V ) − h
τ m (V )dm
dt= m∞(V ) − m
time constant
steady-state
Four subunits.Three switch on.One switches off.
Fit to the data is a little more complicated now, but still easy in principle.
Hodgkin-Huxley equations
∞
∞
∞€
CdV
dt+ g K n4 (V −VK ) + g Nam3h(V −VNa ) + gL (V −VL ) + Iapp = 0
τ n (V )dn
dt= n∞(V ) − n
τ m (V )dm
dt= m∞(V ) − m, τ h (V )
dh
dt= h∞(V ) − h
generic leak
applied current
much smaller thanthe others
inactivation(decreases with V)
activation(increases with V)
An action potential
• gNa increases quickly, but then inactivation kicks in and it decreases again.
• gK increases more slowly, and only decreases once the voltage has decreased.
• The Na+ current is autocatalytic. An increase in V increases m, which increases the Na+ current, which increases V, etc.
• Hence, the threshold for action potential initiation is where the inward Na+ current exactly balances the outward K+ current.
Basic enzyme kinetics
Law of mass actionGiven a basic reaction
A + B Ck1
k-1
we assume that the rate of forward reaction is linearly proportional to the concentrations of A and B, and the back reaction is linearly proportional to the concentration of C.
€
d[A]
dt= k−1[C] − k1[A][B]
Equilibrium
Equilibrium is reached when the net rate of reaction is zero. Thus
€
k−1[C] − k1[A][B] = 0
€
K1[C] = [A][B], K1 =k−1
k1
= eΔG0 / RTor
This equilibrium constant tells us the extent of the reaction, NOT its speed.
change in Gibb’sfree energy
Enzymes
• Enzymes are catalysts, that speed up the rate of a reaction, without changing the extent of the reaction.
• They are (in general) large proteins and are highly specific, i.e., usually each enzyme speeds up only one single biochemical reaction.
• They are highly regulated by a pile of things. Phosphorylation, calcium, ATP, their own products, etc, resulting in extremely complex webs of intracellular biochemical reactions.
Basic problem of enzyme kinetics
Suppose an enzyme were to react with a substrate, giving a product.
S + E P + E
If we simply applied the law of mass action to this reaction, the rate of reaction would be a linearly increasing function of [S]. As [S] gets very big, so wouldthe reaction rate.
This doesn’t happen. In reality, the reaction rate saturates.
Michaelis and Menten
In 1913, Michaelis and Menten proposed the following mechanism for a saturating reaction rate
S + E k1
k-1
C k2 P + E
Complex. product
• Easy to use mass action to derive the equations.• There are conservation constraints.
Equilibrium approximation
€
k−1c = k1se
And thus, since
€
c + e = e0
€
c =e0s
Ks + s
Thus
€
V = k2c =k2e0s
Ks + s=
Vmaxs
Ks + s
reaction velocity
Pseudo-steady state approximation
€
(k−1 + k2)c = k1se
And thus, since
€
c + e = e0
€
c =e0s
Km + s
Thus
€
V = k2c =k2e0s
Km + s=
Vmaxs
Km + s
reaction velocityLooks very similar to previous, but is actually quite different!
Basic saturating velocity
s
V
Vmax
Km
Vmax/2
Lineweaver-Burke plots
€
1
V=
1
Vmax
+Km
Vmax
1
s
Plot, and determine the slope and intercept to get the required constants.
Cooperativity
S + E k1
k-1
C1k2 P + E
S + C1 k3
k-3
C2k4 P + E
Enzyme can bind two substrates molecules at different binding sites.
or
E C1 C2
E E
S S
S S
P P
Pseudo-steady assumption
€
c1 =K2e0s
K1K2 + K2s + s2
c1 =e0s
2
K1K2 + K2s + s2
V = k2c1 + k4c2 =(k2K2 + k4s)e0s
K1K2 + K2s + s2
Note the quadraticbehaviour
Independent binding sites
€
k1 = 2k3 = k+
2k−1 = k−3 = k−
2k2 = k4
E C1 C2
E E
S S
S S
P P
2k+ k+
2k-k-
€
V = 2k2e0s
K + sJust twice the single binding rate, as expected
Positive/negative cooperativity
Usually, the binding of the first S changes the rate at which the second S binds.
• If the binding rate of the second S is increased, it’s called positive cooperativity
• If the binding rate of the second S is decreased, it’s called negative cooperativity.
Hill equation
In the limit as the binding of the second S becomes infinitely fast, we get a nice reduction.
€
Let k3 → ∞, and k1 → 0, while keeping k1k3 constant.
€
V =(k2K2 + k4s)e0s
K1K2 + K2s + s2→
Vmaxs2
Km2 + s2
Hill equation, withHill coefficient of 2.
This equation is used all the time to describe a cooperative reaction. Mostly use of this equation is just a heuristic kludge.
VERY special assumptions, note.
Another fast equilibrium model ofcooperativity
E C1 C2
E E
S S
S S
P P
Let C=C1+C2
€
V = k2c1 + k4c2 =k2K3 + k4s
s + K3
⎛
⎝ ⎜
⎞
⎠ ⎟c = ϕ (s)c
k-1
k1 k3
k-3
k2 k4
S + E k1
k-1
C s)P + E
Monod-Wyman-Changeux model
A more mechanistic realisation of cooperativity.
Equilibrium approximation
Don’t even think about a pseudo-steady approach. Waste of valuable time.
€
Y =r1 + 2r2 + t1 + 2t2
2(r0 + r1 + r2 + t0 + t1 + t2)
K1r1 = 2sr0,K
which gives
€
Y =sK1
−1(1+ sK1−1) + K2
−1[sK3−1(1+ sK3
−1)]
(1+ sK1−1)2 + K2
−1(1+ sK3−1)2
occupancy fraction
and so on for all the other states
Note the sigmoidal character of this curve
Reversible enzymes
Of course, all enzymes HAVE to be reversible, so it’s naughty to put no back reaction from P to C. Should use
S + E k1
k-1
Ck2
P + Ek-2
I leave it as an exercise to calculate that
€
V =e0(k1k2s − k−1k−2 p)
k1s + k−1p + k−1 + k2
Allosteric modulation
substrate binding
inhibitorbinding at adifferent site
this state canform no product
(Inhibition in this case, but it doesn’t have to be)
X
Y Z
Equilibrium approximation
€
(e0 − x − y − z)s − K1x = 0
(e0 − x − y − z)i − K3y = 0
ys − K1z = 0
and thus
x =e0K3
K3 + i
s
K1 + s
V = k2x =Vmax
1+ i /K3
s
K1 + s
X
Y Z
Could change these rate constants, also.
Inhibition decreases theVmax in this model