1. Start

2. So far you have 5 important equations
Name
What are they used for?
Describes the current-voltage relationship for ohmic (linear or non-rectifying) channels/ions
Ohm’s Law
V = i*R or V = i*1/g
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. Ex =
[X]o
[X]i RT zF ln
Nernst Equation Vm = RT F ln
PK[K+]o + PNa[Na+]o + PCl[Cl-]i
PK[K+]i + PNa[Na+]i + PCl[Cl-]o
Goldman equation
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.
Time Constant
t= Rm * Cm
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).
Space Constant
l =
(𝑟𝑚/𝑟𝑎)

3. Ex = [X]o [X]i RT zF ln Ex = [X]o [X]i 58 z mV log At 25oC:
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.
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. Ex =
[X]o
[X]i RT zF ln
The Nernst equation describes the equilibrium potential for a given ion: Ex =
[X]o
[X]i 58 z mV
log
At 25oC:
ENa = (58/1 ) log (140/ 10) = 66 mV
EK = (58/1 ) log (4/ 140) = -89 mV
ECa = (58/2 ) log (1/ ) = 116 mV
Given these ionic concentrations, predict the direction of current flow for each ion at positive and negative membrane potentials.

4. K+ movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
Ohm’s law
EK = (58/1 ) log (4/ 140) = -89 mV
Nernst i V EK
(Vm – EK)
driving force C E
Net K+
IK = gK [ -120 – (-89)]
= gK [ ] = -31gK
If Vm = -120 mV
So you’ve calculated the Nernst or equilibrium potential for K+ as -89 mV. What we want to ask now is for a particular membrane potential what is the current flow (which direction and relatively how much). We said there was 4 mM K+ out and 140 mM K+ in. [CLICK] So the chemical gradient is outward. [CLICK] If the membrane potential is -120 mV (inside relative to 0 mV outside). [CLICK] Then when the K+ channels are open, they are driving the cell to the equilibrium potential of So the electrical gradient for K+ is inward to bring the membrane potential from -120 to -89 it has to bring the positively charged K ions inside the cell. So that is why the electrical gradient is inward. So which one wins? What is the net current? We can determine that using Ohm’s law. [CLICK] We are going to put the equation in the form of I (the current that we want to find) = g (conductance) x V (voltage). Here we need to use the difference in the voltage between where we currently are at -120 mV and where the equilibrium potential is driving us at -89 mV. [CLICK] So calculating that for K+, we get -31 times the unknown potassium conductance. Remember that this voltage difference is the driving force. [CLICK] Resting membrane potential minus the equilibrium potential is the driving force for that ion. So now we can place the equilibrium potential and this relative value on an IV plot. [CLICK] We know that the current will be negative (because the conductance cannot be negative). So the sign for the driving force will tell you the direction of current flow. The current is negative. [CLICK] A negative current will be an inward current. You can see on the IV plot that you are more negative than the equilibrium potential so you have to bring K+ ions into the cell to get it back to the equilibrium potential back to -89 mV. +
Outward (+ current)
Inward (- current)

5. K+ movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
Ohm’s law
EK = (58/1 ) log (4/ 140) = -89 mV
Nernst i V EK
slope = gK (conductance) C E
Net K+
IK = gK [ -120 – (-89)]
= gK [ ] = -31gK
Vm = -120 mV
If Vm = -60 mV C
So now if the membrane potential is -60 mV [CLICK] What happens? You still have the same concentrations of K+, so there is still a chemical drive to push K+ out of the cell. The inside of the cell is still more negative than the outside of the cell, so there is still an electrical drive into the cell. [CLICK] So what wins? What is the net current? Again we use Ohm’s law to calculate the current through the K+ channels. [CLICK] Now we get the value of +29 times the conductance. So this is a positive current, so we can add that to the relative graph. [CLICK] Now these resting or leakage channels are linear or ohmic. [CLICK]. If we had actually measured the current then realize that we could calculate the conductance based one the slope of this line. [CLICK] Because it is a positive value, we know that it will be a net outward current [CLICK]. Also if you look at our graph here, the current potential is more positive than the equilibrium potential. So which direction of positive ions will make the inside of the cell more negative? Only if K+ leaves the cell can it get more negative. So that has to be an outward current. +
Outward (+ current)
Inward (- current)
Net E
IK = gK [ -60 – (-89)]
= gK [-60+89] = +29gK

6. K+ movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
Ohm’s law
EK = (58/1 ) log (4/ 140) = -89 mV
Nernst i V EK
slope = gK (conductance) C E
Net K+
IK = gK [ -120 – (-89)]
= gK [ ] = -31gK
Vm = -120 mV
If Vm = -89 mV C E
IK = gK [-89 – (-89)]
= gK [ ] = 0 C
So now what happens if the membrane potential is -89? [CLICK] You still have more K+ inside than out so the chemical gradient is still pushing K+ out. You still have more negative in than out, so the electrical gradient is pushing K+ in. But what is the actual K+ current when the membrane potential is -89 mV? Again use Ohm’s law to calculate the K+ current. [CLICK] and since the membrane is at the equilibrium potential, there is no net current.
Vm = -60 mV +
Outward (+ current)
Inward (- current)
Net E
IK = gK [ -60 – (-89)]
= gK [-60+89] = +29gK

7. K+ movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
Ohm’s law
EK = (58/1 ) log (4/ 140) = -89 mV
Nernst i V EK
slope = gK (conductance) C E
Net K+
IK = gK [ -120 – (-89)]
= gK [ ] = -31gK
Vm = -120 mV C E
IK = gK [-89 – (-89)]
= gK [ ] = 0
Vm = -89 mV
Vm = 0 mV C C
So now what happens if the membrane potential is -0 mV? There is still more K+ in than out so chemical drive still pushes K+ out. [CLICK] But there is now NO electrical drive because there is no electrical difference between inside the cell and outside the cell. Calculate the current through the K+ channels. [CLICK] You get a large positive value and indeed as is obvious, the net current will be the same as the chemical force because there is no electrical force. [CLICK]
Net
Vm = -60 mV +
Outward (+ current)
Inward (- current)
Net E
IK = gK [ -60 – (-89)]
= gK [-60+89] = +29gK
IK = gK [ 0 – (-89)]
= gK [0+89] = +89gK

8. Na+ movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
Ohm’s law
ENa = (58/1 ) log (140/ 10) = 66 mV
Nernst i V E C
Net
ENa
INa = gNa [ +120 – (+66)]
= gNa [ ] = +54gNa
Vm = +120 mV
Na+ E
INa = gNa [+66 – (+66)]
= gNa [ ] = 0
Vm = +66 mV
Net
Vm = 0 mV
Net
So now let’s do the same thing for Na+. You got +66 mV for the equilibrium potential for Na+ with these concentrations (140 mM out and 10 mM in) So more (bigger) Na+ out than in. [CLICK] We will once again look at 4 different membrane potentials. [CLICK] So because there is more Na+ out than inside, the chemical gradient will always be pushing Na+ into the cell at any membrane potential.
At +120 mV, there will be a big electrical push to get + ions out of the cell. To get the net Na+ current, let’s calculate Ohm’s law for Na+ current. ETC.
Vm = -60 mV +
Outward (+ current)
Inward (- current) E
INa = gNa [ 0 – (+60)]
= gNa [0-60] = -60gNa
INa = gNa [ -60 – (+60)]
= gNa [-60-60] = -120gNa

9. Ca2+ movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
Ohm’s law
ECa = (58/2 ) log (1/ ) = 116 mV
Nernst i V C E
ECa
Net
ICa = gCa [ +150 – (+116)]
= gCa [ ] = +34gCa
Vm = +150 mV
Ca++ E
ICa = gCa [+116 – (+116)]
= gCa [ ] = 0
Vm = +116 mV
Net
Vm = 0 mV
Net
So now let’s do the same thing for Na+. You got +66 mV for the equilibrium potential for Na+ with these concentrations (140 mM out and 10 mM in) So more (bigger) Na+ out than in. [CLICK] We will once again look at 4 different membrane potentials. [CLICK] So because there is more Na+ out than inside, the chemical gradient will always be pushing Na+ into the cell at any membrane potential.
At +120 mV, there will be a big electrical push to get + ions out of the cell. To get the net Na+ current, let’s calculate Ohm’s law for Na+ current. ETC.
Vm = -60 mV +
Outward (+ current)
Inward (- current) E
ICa = gCa [ 0 – (+116)]
= gCa [0-116] = -116gCa
ICa = gCa [ -60 – (+116)]
= gCa [ ] = -176gCa

10. Cl- movement (current flow) in a given electrochemical gradient
V = I R => I = V/R or I = g V (where g = 1/R)
Ohm’s law
ECl = (58/-1 ) log (140/ 7) = -75 mV
Nernst i V
Net Chloride is outward, net current is inward. C
Net E
ECl
ICl = gCl [ -120 – (-75)]
= gCl [ ] = -45gCl
Vm = -120 mV
Cl- E
ICl = gCl [-75 – (-75)]
= gCl [ ] = 0
Vm = -75 mV
Net
Vm = 0 mV
NOTE that when you are talking about cations (ions with a positive charge) the direction of the ion is the same as the direction of the current. This is because by convention we call the direction of the current the direction of positive charge flow. BUT WHEN LOOKING AT ANIONS – these two things are not in the same direction. Whatever direction chloride is moving, the current is moving in the opposite direction! So when you have an INWARD current, Chloride ions are moving OUT!
Vm = -20 mV +
Outward (+ current)
Inward
(-current)
Net
Cl- E
ICl = gCl [ 0 – (-75)]
= gCl [0+75] = +75gCl
Cl-
ICl = gCl [ -20 – (-75)]
= gCl [-20+75] = +55gCl

11. 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.

12. 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 ions that contribute to membrane potentials.
EK = -89 mV
ENa = 66 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) / ( ) = -156 / 1
gK/gNa = 156

13. Assume that at rest a particular neuron is permeable to K+ and Na+
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 = mV

14. 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

15. 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

16. 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.

17. 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

18. 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

19. Neuronal 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?
Phospholipid bilayer with glycoproteins
Maintain separation of charge
Ohm’s Law V = iR
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.
Any of a class of proteins that have carbohydrate groups attached to the polypeptide chain. Also called glycopeptide

20. Why 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.

21. What are the 3 type of gene superfamilies?
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.