Transcript 090711membrane potentials_11

Membrane potentials
Equivalent electrical circuit:
Parallel conductance model
Thermodynamic treatment:
Nernst-Planck flux equation
other useful stuff:
junction potential
Nernst equation
Constant field equation
Nernst Equation: how it works
= Cl• A membrane separates solutions of
100 mM and 10 mM KCl:
• The membrane is permeable to K+, but
not to Cl• K+ ions diffuse across the
membrane from left to right
= K+
IN = 100 mM KCl
• The movement of
makes the inside
negative with respect to outside - this
membrane potential prevents further net
movement of K+
K+
• Equilibrium is quickly established; the magnitude
of  is computed from the Nernst equation
 
+
`
-
+
+
OUT = 10 mM KCl
``
-
(-)
[K ]
60
10
log out  60 * log
 60 * (1)  60mv
z
[ K in ]
100
(+)
Electrochemical potential
ui is the electrochemical potential in Joules/mole.
Transport of an ion occurs from region of higher to lower electrochemical potential
For membrane transport:
ui = uio + RTlnCi + ziFV + …
uio is the standard electrochemical potential at C= 1M,
V=0
ziFV is the contribution of the electrical potential; ie the work
required to bring a mole of ions with charge z from zero potential
to potential V (F is Faraday constant = 96,500 coulombs/mole)
Equilibrium
At equilibrium the electrochemical potential
is the same on each side of membrane;
uio + RTlnC1 + ziFV1 = uio + RTlnC2 + ziFV2
V2-V1 = E = RT ln C1
ziF
C2
Nernst potential
Z = valence
R = gas constant: 8.314 (volts x coulombs/KO x mol)
T = absolute temperature
F = Faraday constant: 96,500 coulombs/mol
Diffusion
For a real cell
K+
140 mM K+
5 mM Na+
145 mM Cl-
5 mM K+
140 mM Na+
145 mM ClArea of cell membrane = 1000 µm2 (4p2)
Radius = 8.92 µm
Volume = 2974 x 10-15 liters (4/3pr3)
Vm = 58.17 log [5]/[140] = -87 mV
Questions: 1) How many K+ ions did the cell lose?
2) Did the K+ concentration inside the cell change?
How many K+ ions did the cell lose?
Use Q= CV
Know Vm = -87 mV
Calculate C
A = 1000 µm2 = 10-5 cm2 (1 µm = 10-4 cm)
Cm = 1 µF/cm2 x A (cm2) = 10-6 F/cm2 x 10-5 cm2 = 10-11 F = 10 pF
Q = 0.087 V x 10-11 F = 8.7 x 10-13 coulombs
Using Faraday’s constant:
8.7 x 10-13 C / 96,500 C/mol = 8.7 x 10-18 moles
Using Avogadro’s #:
8.7 x 10-18 moles x 6 x 1023 molecules/mole =
5 x 106 K+ ions
A single K channel ~2 pA current passes 107 ions/sec
Did the K+ concentration inside the cell change?
140 mM K+ = 0.14 moles/liter
0.14 moles/l x 6 x 1023 molecules/mole
= 0.84 x 1023 molecules/liter
Cell volume is 2974 x 10-15 liters
0.84 x 1023 molecules/liters x 2974 x 10-15 liters
= 2.3 x 1011 K+ ions in the cell
So losing 5 x 106 ions is not much
Ki does not change (in some cases Ko can!)
Ion equilibrium potentials in some cells
Ions are not in equilibrium across membranes
So Vmembrane does not equal Eion
Nernst-Planck Flux Equation
Need to calculate the flux of each individual ion
J = L dµ/dx
Flux (J) proportional to gradient of
electrochemical potential
J = moles/cm2s
J = L d [RTlnC + zFV]
dx
= L [RT dC + zF dV]
C dx
dx
need to find L
(Remember: d ln C = d ln C . dC )
dx
dC
dx
For a neutral diffusing substance (z=0), flux given by Fick equation
J = -D dC
dx
= L RT dC
C dx
so L = -DC/RT
Substituting and rearranging:
J = -D dC + C zF dV
dx
RT dx
Nernst-Planck Equation
µo
V
V(x)
Constant field dV = V
approximation: dx
d
d
Integrate Nernst-Planck eqn from 0 to d across membrane:
J = -zF D V C(d)ezFV/RT - C(0)
RT d
ezFV/RT - 1
The concentration of an ion just inside the membrane is related to
Its concentration in the bulk solution by a partition coefficient:
C (membrane) = k
C (bulk solution)
Permeability coefficient P= Dk/d
J = -zFV P C(1)ezFV/RT - C(2)
RT
ezFV/RT - 1
J=mol/cm2s
D= cm2/s
P= cm/s
What is the permeability constant?
Comes from flux equation: J = -DA dc/dx
J = flux (moles/second)
A= area
dc/dx = concentration gradient
k Is partition coefficient
dc/dx ~ k (Co-Ci)/ x
 big
Co
So J = -DA k (Co-Ci)/ x
small
Ci
x
or
J = -Dk/x A (Co-Ci)
= -P A (Co-Ci)
Permeability constant
(can be measured)
For membrane permeable to Na+, K+, and Cl-:
INa+ = FzJNa+
IK+ = FzJK+
Icl-= FzJCl-
J = mol/cm2s
F = 9.65 coul/mol
I = current density (coul/cm2 x sec-1)
Itotal = INa+ + IK+ + Icl-
When Itotal = 0
Vo = RT PNaNao + PKKo + PClCli
F PNaNai + PKKi + PClClo
Vo = zero current potential
Goldman-Hodgkin-Katz equation
Fluxes depend on ion concentrations
J = -zFV P C(1)ezFV/RT - C(2)
RT
ezFV/RT - 1
Limiting cases:
V∞, JzP C(1)FV
RT
V-∞, J-z P C(2)FV
RT
J=0
When C(1)ezFV/RT = C(2) or
V=RTlnC(2)/C(1)
zF
Rectification when
C(1) < or > C(2)
Resting Potential using the Goldman-Hodgkin-Katz eqn
•In real cells there are many more open
K+ channels than Na+ channels at rest
Vm = -82 mv
• Vm is somewhat more positive than EK
[Na]o = 145 mM
-
[Na]i = ~12 mM
-
[K]o =
-
[K]i = 140 mM
-
-
EK = -92.6 mv
-
-
ENa= +64.9 mv
-
4
• The Goldman-Hodgkin-Katz equation
(constant field equation)
PK [ K ]o  PNa [ Na]o
Vm  60 log
PK [ K ]i  PNa [ Na]i
K+
Na+
Where PK and PNa are the membrane
permeability to K and Na, respectively
-
-
-
-
K+-Dependence of Resting Potential
• membrane potentials measured using microelectrodes
• depolarization – Vm becomes more positive (less negative) than resting
potential
• hyperpolarization – Vm becomes more negative than resting potential
• increasing the external [K+] depolarizes the membrane
 [ K  ]O   [ Na  ]O 
RT

Vm 
 ln 


F
 [ K ]i   [ Na ]i 
Dependence of the
resting membrane
potential on [K+]o and on
the PNa/PK ratio. The
blue line shows the case
with no Na+ permeability
(i.e., PNa/PK = 0). Orange
curves describe the Vm
predicted by the GHK
equation for  > 0. The
deviation of these curves
from linearity is greater
at low [K+]o, where the
Na+ contribution is
relatively larger.
Application to resting and active membrane in squid axon
(Hodgkin & Katz, 1949)
Neurons can have different PNa/PK
PNa
Na+
PNa
PK
K+
Na+
ENa
ENa
E
E
K
K
PK
K+
External K+ can modify K conductance of resting membrane
K+o increases K permeability
K+
PNa
Na+
PK
K+
PK
PNa
K+
K+K+
K+ K+
K+
K
K+
K+
K+
+
Ko = 4 mM
Ko = 10 mM
PNa/PK = 1
PNa/PK = 0.5
Vm is halfway between
Vna and VK
Vm closer to VK
Hyperkalemia in a neuron with a large resting K conductance
PNa
PK
K+
PNa
Na+
PK
K+
ENa
EK*
E
K
hyperkalemia
Hyperkalemia in a pacemaker cell
PNa
Na+
PK
K+
PNa
Na+
ENa
E
K
hyperkalemia
PK
K+
Removal of Mg2+ block accounts for effects of Ko on PK
K+
K+
K+
K+
Mg2+
K+
K+
K+
K+
Ko = 4 mM
K+
Mg2+
K+
K+
K+
K+
Inward K flux
clears Mg from
channel
K+
K+
K+
K+
Ko = 10 mM
K+
Resting potential
Active ion transport in nerve
(Hodgkin & Keynes, 1955)
24Na
24Na
A case of too much K+
A 35 year old male with history of bipolar disorder, taking no medications,
presented to the emergency department complaining of nausea, vomiting,
lethargy, and abdominal pain 5 hours after ingesting an unknown
number of digoxin tablets He was found partly conscious by a friend who
called for emergency medical help. Paramedics found him to be agitated
with a pulse of 30. Upon arrival to the ED, vital signs were blood pressure 162/87,
pulse of 30 beats per minute, respiratory rate of 22 per minute and temperature
of 97 oF. An ECG showed second degree AV block. The patient was intubated and
digoxin antibody administered. The next day his digoxin level was reduced
from 2.49 ng/ml to 0.69 ng/ml and he was transferred out of the ICU.
How does digoxin work?
Can the GHK eqn explain the electrical effects of digoxin?