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So far you have 5 important equations
Name
What are they used for?
Ohm’s Law
V = i*R or V = i*1/g
Nernst Equation
Goldman
equation
Describes the current-voltage relationship for ohmic (linear
or non-rectifying) channels/ions
Used to calculate the equilibrium potential for a particular
ion. Note this is also the reversal potential. Also note that in
o cells with only one type of channel open at rest, the resting
i membrane potential will be equivalent to the equilibrium
potential for that ion.
[X]
RT
Ex =
ln
zF
[X]
RT PK[K+]o + PNa[Na+]o + PCl[Cl-]i
Vm = F ln
PK[K+]i + PNa[Na+]i + PCl[Cl-]o
Allows you to calculate the resting membrane potential when more than one type of channel is open at rest.
Note that permeability (P) reflects the conductance AND the number of open channels per unit membrane.
Time
Constant
Space
Constant
t= Rm * C
This is the time it takes for a passive or electrotonic potential to
decay to 63% of its value. Note that increasing either the
m membrane resistance or the membrane capacitance will increase
the time constant.
l = (𝑟𝑚/𝑟𝑎)
This is the distance it takes the potential to decay to 37% of its
initial value. Note that the better the insulation of the
membrane (increased rm) and the better the conducting
properties of the core (decreased ra), the longer is the space
constant (potential travels further before decaying).
1) One use of concentration gradients of ions across cell membranes is to drive the flow of ions during action
potentials of excitable cells. A concentration gradient of ions across a membrane may be expressed in terms of
an electrical potential at equilibrium by use of the Nernst Equation.
a) The concentrations of some of the ions inside (i) and outside (o) of a particular muscle cell are as follows:
[Na+]o = 140 mM; [Na+]i = 10 mM
[K+]o = 4 mM; [K+]i = 140 mM
[Ca2+]o = 1 mM; [Ca2+]i = 10-4 mM
Calculate the equilibrium potential for each of the ions in the muscle cell, assuming 25oC for temperature.
The Nernst equation describes the equilibrium
potential for a given ion:
At 25oC:
Ex =
[X]o
RT
ln
zF
[X]i
[X]o
58 mV
Ex =
log
z
[X]i
ENa = (58/1 ) log (140/ 10) = 66 mV
EK = (58/1 ) log (4/ 140) = -89 mV
ECa = (58/2 ) log (1/ 0.0001) = 116 mV
Given these ionic concentrations, predict the direction of current flow for each
ion at positive and negative membrane potentials.
K+ movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
EK = (58/1 ) log (4/ 140) = -89 mV
Ohm’s law
Nernst
i
E
V
EK
(Vm – EK)
IK = gK [ -120 – (-89)]
= gK [-120+89] = -31gK
Net
C
If
Outward (+ current)
+
+
Inward (- current)
Vm = -120 mV
K+
driving force
K+
K+ movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
EK = (58/1 ) log (4/ 140) = -89 mV
Ohm’s law
Nernst
i
slope = gK (conductance)
E
V
EK
Net
IK = gK [ -120 – (-89)]
= gK [-120+89] = -31gK
C
Vm = -120 mV
K+
If Vm
Outward (+ current)
= -60 mV
C
Net
+
+
Inward (- current)
E
IK = gK [ -60 – (-89)]
= gK [-60+89] = +29gK
K+
K+ movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
EK = (58/1 ) log (4/ 140) = -89 mV
Ohm’s law
Nernst
i
slope = gK (conductance)
E
V
EK
Net
IK = gK [ -120 – (-89)]
= gK [-120+89] = -31gK
C
Vm = -120 mV
If Vm
E
IK = gK [-89 – (-89)]
= gK [-89 +89] = 0
= -89 mV
K+
C
C
Vm = -60 mV
Outward (+ current)
Net
+
+
Inward (- current)
E
IK = gK [ -60 – (-89)]
= gK [-60+89] = +29gK
K+
K+ movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
EK = (58/1 ) log (4/ 140) = -89 mV
Ohm’s law
Nernst
i
slope = gK (conductance)
E
V
EK
Net
C
IK = gK [ -120 – (-89)]
= gK [-120+89] = -31gK
Vm = -120 mV
E
IK = gK [-89 – (-89)]
= gK [-89 +89] = 0
K+
Vm = -89 mV
K+
C
Vm = 0 mV
C
C
V = -60 mV Net
Net
m
Outward (+ current)
+
+
Inward (- current)
E
IK = gK [ -60 – (-89)]
= gK [-60+89] = +29gK
IK = gK [ 0 – (-89)]
= gK [0+89] = +89gK
Na+ movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
ENa = (58/1 ) log (140/ 10) = 66 mV
Ohm’s law
Nernst
i
C
V
INa = gNa [ +120 – (+66)]
= gNa [+120-60] = +54gNa
ENa
Net
Vm = +120 mV
C
E
INa = gNa [+66 – (+66)]
E
Na+
Vm = +66 mV
Na+
Vm = 0 mV
= gNa [-66 +66] = 0
Vm = -60 mV
Outward (+ current)
+
+
C
Inward (- current)
Net
C Net
INa = gNa [ 0 – (+60)]
= gNa [0-60] = -60gNa
E INa = gNa [ -60 – (+60)]
= gNa [-60-60] = -120gNa
Ca2+ movement (current flow) in a given electrochemical gradient
ECa = (58/2 ) log (1/ 0.0001) = 116 mV
V = I R => I = V/R or I = g V (where g = 1/R)
Ohm’s law
Nernst
i
C
V
ICa = gCa [ +150 – (+116)]
= gCa [+150-116] = +34gCa
ECa
Net
Vm = +150 mV
E
C
Ca++
E
ICa = gCa [+116 – (+116)]
Vm = +116 mV
Ca++
Vm = 0 mV
= gCa [+116-116] = 0
Vm = -60 mV
Outward (+ current)
+
+
C
Inward (- current)
Net
E
ICa = gCa [ -60 – (+116)]
= gCa [-60-116] = -176gCa
C Net
ICa = gCa [ 0 – (+116)]
= gCa [0-116] = -116gCa
Cl- movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
ECl = (58/-1 ) log (140/ 7) = -75 mV
Ohm’s law
Nernst
i
Net Chloride is outward,
net current is inward.
C
V
ICl = gCl [ -120 – (-75)]
= gCl [-120+75] = -45gCl
ECl
Net
E
Vm = -120 mV
C
ICl = gCl [-75 – (-75)]
Cl-
E
Vm = -75 mV
Cl-
Vm = 0 mV
= gCl [-75 +75] = 0
Vm = -20 mV
Outward (+ current)
Cl+
+
Cl-
Inward C Net
(-current)
E
ICl = gCl [ -20 – (-75)]
= gCl [-20+75] = +55gCl
C Net
ICl = gCl [ 0 – (-75)]
= gCl [0+75] = +75gCl
During the action potential, is there a net influx or efflux of Cl- ions during the
action potential?
When Vm > ECl expected ~ +25 mV at action potential peak, then chloride
ions are trying to drive the membrane to the chloride equilibrium potential at
-75 mV. So the answer is a net influx – you have to move the negative ions
into the cell to drive the membrane to a more negative potential. So that is
the flow of ions – but what about the direction of the current?? ALWAYS
opposite to anions. So the net current is outward.
i
V
If the experimenter holds the membrane potential of
the cell at -90 mV, what happens when GABA is
released on the membrane?
The GABA receptor is associated with a chloride channel. We
have found the chloride reversal potential to be -75 mV. When
GABA is released it binds to the receptor, causing the chloride
channel to open. Once open they are attempting to drive the
membrane to their equilibrium (-75 mV). Since this is more
positive than the membrane, a depolarizing (excitatory) current
will be produced. Chloride will move out of the cell and the net
current will be inward.
The actual measured membrane potential for the muscle cell
was -90 millivolts. From this information, what conclusion can
you draw concerning the relative conductances of sodium and
potassium in these cells at rest (i.e. in the absence of action
potentials) assuming that sodium and potassium are the only
ENa = 66 mV
ions that contribute to membrane potentials.
EK = -89 mV
At rest the sodium current is balanced with that of the potassium current.
From Ohm’s law I = gV:
gNa(Em-ENa) = - gK(Em – EK)
-gK/gNa = (Em-ENa)/(Em-EK)
-gK/gNa = (-90 – 66) / (-90 - -89) = -156 / 1
gK/gNa = 156
Assume that at rest a particular neuron is permeable to K+ and Na+. If EK = -89 mV and
ENa = +60 mV, AND gK = 5*gNa, then calculate the resting membrane potential.
At rest, INa = -IK (eq1)
and gK = 5* gNa (eq2)
Use Ohm’s law to substitute for current values in eq 1: gNa(Em – ENa) = -gK(Em-EK)
Now replace gK = 5*gNa :
gNa(Em – ENa) = -5gNa(Em-EK)
gNa is on both sides of equation – so can remove it: (Em – ENa) = -5 (Em-EK)
multiply the 5 through to separate Em: Em – ENa = -5Em+5EK
Add 5Em to both sides:
6Em – ENa = 5EK
Add ENa to both sides:
6Em = ENa + 5EK
Put in actual values:
Em = [60 + (5*-89)]/6 = -64.16 mV
If a few positive charges were moved from the inside to the outside of a cell, the
unbalanced negative charges would
a. distribute uniformly in the cytoplasm
b. end up at the membrane boundary
c. diffuse from the place they were left unbalanced to their final destination
d. violate the principle of electroneutrality
Answer: d
The Nernst potential describes
a. The potential at which only K+ channels are open in a neuron
b. The resting potential of a cell
c. The potential at which the electrical gradient is in balance with chemical
gradient (for a specific ion)
d The potential at which Na+/K+ ion pump is most active
answer: c
When you calculate the Nernst potential for K+, what else can this be
called?
K+ equilibrium potential
or
K+ reversal potential
A particular mammalian cell displays a chloride equilibrium potential of -60 mV. Due
to a high resting potassium conductance (K+ equilibrium is at -90 mV) the cell’s resting
potential is at -80 mV. Please check the correct answer.
What would the direction of the chloride and potassium currents be at rest?
Chloride: ___inward
___outward
___no net current
Potassium:___inward
___outward
___no net current
What would the direction of the chloride and potassium ion movements be at rest?
Chloride: ___inward
___outward
___no net movement
Potassium:___inward
___outward
___no net movement
Answers:
What would the direction of the chloride and potassium currents be at rest?
Chloride: __X_inward
___outward
___no net current
Potassium:___inward
__X_outward
___no net current
What would the direction of the chloride and potassium ion movements be at rest?
Chloride: ___inward
__X_outward
___no net movement
Potassium:___inward
__X_outward
___no net movement
Remember that the direction of current is in the direction of positive charge movement.
So if Cl- ions are moving out of the cell, then the direction of current is into the cell.
Assume that a particular cell is equally permeable to chloride and potassium ions
and that gK = gCl. If the equilibrium potential for chloride is -60 mV, while for
potassium is -90 mV, what would you predict the resting membrane potential to be?
Show all work.
At rest IK = -ICl (inward and outward currents are equal and opposite in
direction so there is no net current).
Thus
gK (Vr - EK) = -gCl (Vr - ECl)
gK (Vr - (-90 mV)) = -gCl (Vr - (-60 mV))
gK (Vr +90 mV) = -gCl (Vr +60 mV)
(Vr +90 mV) = -(Vr +60 mV)
2Vr = -150 mV
Vr = -75 mV
Two microelectrodes are inserted into a cell: one is connected to a voltmeter to
measure the transmembrane potential; the second is hooked up to a tunable
current source (battery of variable output), which allows us to inject current
into the cell. These electrodes are then connected to a feedback circuit that
compares the measured voltage across the membrane with the voltage desired
by the experimenter. If these two values differ, then current is injected into the
cell to compensate for this difference. Thus the amount of current we provide
equals the amount of current passing through the cell membrane.
What is this mode of recording called?
Answer: Voltage Clamp
Can voltage clamp mode of recording be applied to whole cell recordings?
Can it be applied to inside-out patches of membrane?
Can it be applied to outside-out patches of membrane?
Answer: yes to all
Neuronal Membranes, Ion Channels and Pumps
What is a neuronal membrane made of?
Phospholipid bilayer with glycoproteins
What is one purpose of the membrane?
Maintain separation of charge
Artificial gramicidin –induced channels in
membranes behave according to what function?
Ohm’s Law V = iR
For a linear or Ohmic channel, when recording a
single membrane potential, does the amplitude of
the current vary?
No. The opening of the channel is all or none.
For a linear or Ohmic channel, how would you measure the conductance of the
channel?
Using a patch clamp recording, hold the cell at different
membrane potentials and measure the amplitude of the current
through the channel at each membrane potential. Plot this and
then measure the slope to obtain the conductance.
V = iR, y = mx + b, in this case you are measuring the current, i,
so you need to solve for i. Also we want conductance.
Remember that conductance g = 1/R. V= i/g, now solve for i.
i = gV. So the ‘m’ or slope is g.
Why might a channel not be Ohmic, and how could you tell?
Measure the current at different membrane potentials. If there is a change in slope at
some part of the graph, it is not Ohmic. Can be due to blocking particles.
When a channel passes more current at certain potentials than at others, what is this
called?
Rectification
What is happening to the ion channel protein when the channel opens?
A conformational (structural) change
How does gating of the channel work?
Conformation change in one part, along the whole channel length, or with a blocking
particle swinging in.
What are the ways to open a channel?
Ligand, voltage, phosphorylation, stretch/pressure
What is desensitization, and what causes it and relieves it?
A reduced second channel response after first opening. Typically caused by ligands
after exposure. In some cases a particular voltage must be reached to alleviate it.
Why might a channel respond differently early as opposed to late in development?
Channels are made of subunits that can have varying expression patterns over
development. AND changing the subunit composition of a receptor/channel can
change its function – including conductance, channel open time, desensitization, etc.
What are the 3 type of gene superfamilies?
Ligand-gated, gap junction, and Voltage-gated
How do ion pumps differ from ion channels?
Ion pumps move some ions against their electrochemical gradient;
Ion pumps require energy (ATP)
Ion pumps are very slow
Ion pumps have 2 gates not one.