Name V i R or V i 1g Ohms Law E x X o X i RT zF ln V m RT F ln P K K o P Na Na o P Cl Cl i P K K i ID: 760788
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Slide1
Start
Slide2So far you have 5 important equations
Name
V = i*R or V = i*1/g
Ohm’s Law
E
x
=
[X]
o
[X]
i
RT
zF
ln
V
m
=
RT
F
ln
P
K
[K
+
]o + PNa[Na+]o + PCl[Cl-]i
PK[K+]i + PNa[Na+]i + PCl[Cl-]o
t
= Rm * Cm
l =
Nernst Equation
Goldman equation
Time Constant
Space Constant
What are they used for?
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 cells with only one type of channel open at rest, the resting membrane potential will be equivalent to the equilibrium potential for that ion.
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.
This is the time it takes for a passive or electrotonic potential to decay to 63% of its value. Note that increasing either the membrane resistance or the membrane capacitance will increase the time constant.
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).
Slide31) 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.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 mMCalculate the equilibrium potential for each of the ions in the muscle cell, assuming 25oC for temperature.
EK = (58/1 ) log (4/ 140) = -89 mV
E
x
=
[X]
o
[X]
i
58
z
mV
log
At 25
o
C:
E
Na
= (58/1 ) log (140/ 10) = 66 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.
E
x
=
[X]
o
[X]
i
RT
zF
ln
The Nernst equation describes the equilibrium potential for a given ion:
Slide4K+ movement (current flow) in a given electrochemical gradient
EK = (58/1 ) log (4/ 140) = -89 mV
K
+
K
+
Net
C
E
If
V
m
=
-120
mV
IK = gK [ -120 – (-89)] = gK [-120+89] = -31gK
Nernst
V = I R => I = V/R or I = g V
(where g = 1/R)
Ohm’s law
i
V
E
K
(V
m
– E
K
)
driving force
+
+
Outward (+ current)
Inward (- current)
Slide5K+ movement (current flow) in a given electrochemical gradient
EK = (58/1 ) log (4/ 140) = -89 mV
K
+
K
+
Net
C
E
V
m
=
-120
mV
I
K = gK [ -120 – (-89)] = gK [-120+89] = -31gK
Nernst
V = I R => I = V/R or I = g V
(where g = 1/R)
Ohm’s law
i
V
E
K
slope =
g
K
(conductance)
If
V
m
=
-60
mV
+
+
Outward (+ current)
Inward (- current)
C
E
I
K
=
g
K
[ -60 – (-89)]
=
g
K
[-60+89] = +29g
K
Net
Slide6K+ movement (current flow) in a given electrochemical gradient
EK = (58/1 ) log (4/ 140) = -89 mV
K
+
K
+
Net
C
E
V
m
=
-120
mV
I
K = gK [ -120 – (-89)] = gK [-120+89] = -31gK
Nernst
V = I R => I = V/R or I = g V
(where g = 1/R)
Ohm’s law
i
V
E
K
slope =
g
K
(conductance)
V
m
=
-60
mV
+
+
Outward (+ current)
Inward (- current)
C
E
I
K
=
g
K
[ -60 – (-89)]
=
g
K
[-60+89] = +29g
K
Net
If
V
m
=
-89
mV
C
E
I
K
=
g
K
[-89 – (-89)]
=
g
K
[-89 +89] =
0
Slide7K+ movement (current flow) in a given electrochemical gradient
EK = (58/1 ) log (4/ 140) = -89 mV
K
+
K
+
Net
C
E
V
m
=
-120
mV
I
K = gK [ -120 – (-89)] = gK [-120+89] = -31gK
Nernst
V = I R => I = V/R or I = g V
(where g = 1/R)
Ohm’s law
i
V
E
K
slope =
g
K
(conductance)
V
m
=
-60
mV
+
+
Outward (+ current)
Inward (- current)
C
E
I
K
=
g
K
[ -60 – (-89)]
=
g
K
[-60+89] = +29g
K
Net
V
m
=
-89
mV
C
E
I
K
=
g
K
[-89 – (-89)]
=
gK [-89 +89] = 0
Vm = 0 mV
C
IK =
gK
[ 0 – (-89)] = gK [0+89] = +89gK
Net
Slide8C
C
C
C
Na
+
movement (current flow) in a given electrochemical gradient
V
m
=
+120
mV
Nernst
V = I R => I = V/R or I = g V
(where g = 1/R)
Ohm’s law
i
V
E
Na
V
m
=
-60
mV
+
+
Outward (+ current)
Inward (- current)
V
m
= +66
mV
E
V
m
= 0 mV
E
Na
=
(58/1 )
log
(140/ 10)
=
66
mV
Na
+
Na
+
E
I
Na
=
g
Na [ +120 – (+66)] = gNa [+120-60] = +54gNa
Net
E
Net
I
Na
=
g
Na
[ -60 – (+60)]
= gNa [-60-60] = -120gNa
INa = gNa [+66 – (+66)] = gNa [-66 +66] = 0
Net
I
Na
=
g
Na
[ 0 – (+60)]
=
g
Na
[0-60] = -
60g
Na
Slide9C
C
C
C
Ca
2+
movement (current flow) in a given electrochemical gradient
V
m
=
+150
mV
Nernst
V = I R => I = V/R or I = g V
(where g = 1/R)
Ohm’s law
i
V
E
C
a
V
m
=
-60
mV
+
+
Outward (+ current)
Inward (- current)
V
m
= +116
mV
E
V
m
= 0 mV
Ca
++
C
a
++
E
I
Ca
=
g
Ca
[ +150 – (+
116)]
=
g
Ca
[+
150-116] = +34gCa
Net
E
Net
I
Ca
=
g
Ca
[ -60 –
(+116)] = gCa [-60-116] = -176gCa
ICa = gCa [+116 – (+116)] = gCa [+116-116] = 0
Net
I
Ca
= gCa [ 0 – (+116)] = gCa [0-116] = -116gCa
E
Ca
=
(58/2 )
log
(1/ 0.0001)
=
116
mV
Slide10+
+
Outward (+ current)
Inward
(-current
)
C
C
C
C
Cl
-
movement (current flow) in a given electrochemical gradient
V
m
= -
120
mV
Nernst
V = I R => I = V/R or I = g V
(where g = 1/R)
Ohm’s law
i
V
E
Cl
V
m
=
-20
mV
V
m
= -75
mV
E
V
m
= 0 mV
E
Cl
=
(58/-1 )
log
(140/ 7)
=
-75 mV
Cl
-
Cl-
E
I
Cl
= gCl [ -120 – (-75)] = gCl [-120+75] = -45gCl
Net
E
Net
I
Cl
=
g
Cl
[ -20 –
(-75)] = gCl [-20+75] = +55gCl
ICl = gCl [-75 – (-75)] = gCl [-75 +75] = 0
Net
I
Cl
= gCl [ 0 – (-75)] = gCl [0+75] = +75gCl
Cl
-
Cl
-
Net Chloride is outward, net current is inward.
Slide11During 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.
Slide12The 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 ions that contribute to membrane potentials.
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
E
K = -89 mV
E
Na
= 66
mV
Slide13Assume 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,
I
Na
= -I
K
(eq1)
and
g
K
= 5*
g
Na
(eq2)
Use Ohm’s law to substitute for current values in
eq
1:
g
Na
(
E
m
– E
Na
) = -
g
K
(
E
m
-E
K
)
Now replace
g
K
= 5*
g
Na
:
g
Na
(
E
m
– E
Na
) =
-5g
Na
(
E
m
-E
K
)
gNa
is on both sides of equation – so can remove it:
(
E
m
– E
Na
) =
-5 (
E
m
-E
K
)
multiply the 5 through to separate
E
m
:
E
m
–
E
Na
=
-5E
m
+5E
K
Add 5E
m
to both sides: 6E
m
– E
Na
= 5E
K
Add E
Na
to both sides: 6E
m
= E
Na
+ 5E
K
Put in actual values:
E
m
= [60 + (5*-89)]/6 = -64.16 mV
Slide14If a few positive charges were moved from the inside to the outside of a cell, the unbalanced negative charges woulda. distribute uniformly in the cytoplasmb. end up at the membrane boundaryc. diffuse from the place they were left unbalanced to their final destinationd. violate the principle of electroneutrality
Answer: d
Slide15The Nernst potential describesa. The potential at which only K+ channels are open in a neuronb. The resting potential of a cellc. 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
Slide16A 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 currentPotassium:___inward ___outward ___no net currentWhat would the direction of the chloride and potassium ion movements be at rest?Chloride: ___inward ___outward ___no net movementPotassium:___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 currentPotassium:___inward __X_outward ___no net currentWhat would the direction of the chloride and potassium ion movements be at rest?Chloride: ___inward __X_outward ___no net movementPotassium:___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.
Slide17Assume 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 I
K
= -
I
Cl
(inward and outward currents are equal and opposite in direction so there is no net current).
Thus
g
K
(
V
r
- E
K
) = -
g
Cl
(
V
r
- E
Cl
)
g
K
(V
r
- (-90 mV)) = -g
Cl
(V
r
- (-60 mV))
g
K
(V
r
+90 mV) = -g
Cl
(V
r
+60 mV)
(V
r
+90 mV) = -(V
r
+60 mV)
2V
r
= -150 mV
V
r
= -75 mV
Slide18Two 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
Slide19Neuronal Membranes, Ion Channels and Pumps
What is a neuronal membrane made of?
What is one purpose of the membrane?Artificial gramicidin –induced channels in membranes behave according to what function?For a linear or Ohmic channel, when recording a single membrane potential, does the amplitude of the current vary?
For a linear or Ohmic channel, how would you measure the conductance of the channel?
Phospholipid bilayer with glycoproteins
Maintain separation of charge
Ohm’s Law V = iR
No. The opening of the channel is all or none.
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.
Slide20Why might a channel not be Ohmic, and how could you tell?When a channel passes more current at certain potentials than at others, what is this called?
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.
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.
Slide21Why 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.