Electrophysiology: Action Potentials and Membrane Potentials

A few more questions to help your understanding of these concepts:

1.  On slide 21 of Action Potential, you mentioned high Pk in absolute and relative refractory periods. I was just wondering what "high Pk" means? 

Pk is permeability to potassium, so high Pk means that during those two times there is high permeability to potassium.

2.  I was having a little trouble when working #56 on the CTL practice questions for Week 6.  When using the nernst equation to calculate the difference in voltage across the membrane, I keep getting (approx.) 60mV, not 30mV. Why is it 30?

CTL Question 56) Solutions A and B are separated by a semi-permeable membrane that is permeable to Ca2+ and impermeable to Cl-. Solution A contains 10 mM CaCl2, and solution B contains 1 mM CaCl2. At physiological temperatures, Ca2+ will be at electrochemical equilibrium when which of the following is achieved?
A. Solution A is +60 mV
B. Solution A is +30 mV
C. Solution A is -60 mV
D. Solution A is -30 mV
E. Solution A is +120 mV
F. Solution A is -120 mV
G. The Ca2+ concentrations of the two solutions are equal
H. The Cl- concentrations of the two solutions are equal

Similarly:

Earlier today I did the questions you posted for the membrane potential lecture and got number 14 incorrect. I used the Nernst equation to derive my answer, but your answer states that it is -30 instead of 30.

I believe I know the material well, so maybe I am just not understanding what the question is asking?? After doing the Nernst equation, I calculated Eca= +30mv.........(61/2)(log10/1)=30. So I dont understand why the answer is negative. The question is below.

MP Practice question 14. Solutions A and B are separated by a semi-permeable membrane that is permeable to Ca²⁺ and impermeable to Cl^-.  Solution A contains 10 mM CaCl₂, and solution B contains 1 mM CaCl₂.  At physiological temperatures, Ca²⁺ will be at electrochemical equilibrium when which of the following is achieved?

                A. Solution A is +60 mV

                B. Solution A is +30 mV

                C. Solution A is -60 mV

                D. Solution A is -30 mV

                E. Solution A is +120 mV

                F. Solution A is -120 mV

                G. The Ca²⁺ concentrations of the two solutions are equal

                H. The Cl^- concentrations of the two solutions are equal

And another: ( I tried using the Nernst Equation for this so I calculated 60 log (10/1) = +60 and I chose C. Could you explain to me how -30 mv was calculated?) 

You DO need to use the Nernst equation for this question.  There are two separate ways, however, that this can be achieved.  The most basic way would be to just plug in numbers (10mM and 1mM) and calculate out what the difference in voltage across the membrane is.  If you do that, you discover that there is a 30mV difference across the membrane.  

 

60/2 (Ca2+ has a valance of 2) * log (10/1) = 30mV

 

From there you would like to determine which side would have a particular charge.  Since all of the answer choices reference solution A, you need to determine what is the charge of solution A.  If the membrane is made permeable to Ca2+, then Ca2+ will move down its concentration gradient from solution A (where there is higher concentration) to solution B.  This will bring Ca2+ to solution B causing solution B to be positively charged and solution A to be negatively charged.  Therefore, solution A will be -30 mV.

Alternatively, you could utilize the Nernst equation directly.  The equation is written to calculate the concentration on the outside of the cell with respect to the inside to determine the membrane potential of the inside of the cell with respect to the outside.  Therefore, since all of the answer choices are solution A (with respect to solution B), therefore Solution A becomes the intracellular and solution B becomes the extracellular.  Then, utilizing the equation directly you can put 10 mM in for the concentration of intracellular solution and 1 mM for the concentration of the extracellular solution.  If you do that, you get -30mV of Solution A.

 

3.  MP Practice Question: 4.  A basic tenet of physics is that solutions of charged ions are electroneutral: there must be equal numbers of cations and anions.  When a resting membrane potential develops, this principle appears to be violated since there is a net migration of cations out of the cell.  For this to happen, there must exist a force that is greater than the attractive force between cations and anions in solution.  This force arises most directly from what?

  A.  The membrane potential, which provides an attractive force for cations to exit the cell, thus making the extracellular surface   more positive.

  B.  The action of ATPases.

  C.  The concentration gradient of K+ across the cell membrane.

  D.  The concentration gradient of Na+ across the cell membrane

My understanding of this question is that whatever force is increasing the drive for cations to move out of the cell must be a force that is making the cell more positive. Since there is more Na+ on the outside of the cell than inside the cell at rest wouldn't the concentration gradient alone bring more Na+ into the cell making it more positive and driving Cations out?

Remember that the membrane potential of a typical cell is negative, that is the whole point of the question, it is stating that this gradient needs to over come the pull for positive ions to move inwards.  Therefore, it is actually the movement of cations OUT of the cell that is the force we are trying to identify.  Remember, the membrane is most permeable to K+ at rest and therefore it is the concentration gradient of K+ out of the cell that is overcoming the electrical drive for positively-charged ions to move into the cell.

4. MP Practice question: 11.  The sodium gradient across the nerve cell membrane is:

  A.  A result of the Donnan equilibrium

  B.  Significantly changed during an action potential

  C.  Used as a source of energy for the transport of other ions

  D.  An important determinant of the resting membrane potential

  E.  Maintained by a Na/Ca exchanger

 
I chose option D because at the resting membrane potential the movement of Na+ out of the cell through a Na+/K+ ATPase is important for maintaining the negative charge inside the cell.After looking at the answer I thought maybe the question isn't referring to a cell at rest but a nerve cell may imply an active cell so therefore it must be referring to how the flux of sodium causes depolarization during an action potential will because of the change in Vm trigger the slow opening of K+ channels to repolarize the cell. I struggle with this explanation still though because the answer choice says a "source of energy" which I associate with secondary transporters. I thought that movement of ions of Na+/K+ Channels were passive. Rather its the flux of Na+ that increases the drive of K+ to move out of the cell not necessarily requiring energy. Am I interpreting this correctly? 

You are correct that the movement of Na+ out of the cell is important to maintain the resting membrane potential somewhat, but certainly not as important as the potassium gradient. The question is referring to a nerve cell, but could be any cell for that matter.  Remember from the Membrane Transport Mechanism lecture that Na+ IS used in MANY secondary active transporters and is therefore a very important energy source to move other ions across the membrane.  The question is therefore asking you to remember all of the information you have learned about physiology thus-far and know that indeed Na+ does not contribute to the resting membrane potential very much, but it DOES play a significant role in membrane transport of other molecules by secondary active transport.

5. AP Practice question: 6.  I_Na is greater than I_K (I_ion = current for a given ion):

  A.  At the resting membrane potential.

  B.  At the peak of the action potential.

  C.  During the rising phase of the action potential.

  D.  At the beginning of the hyperpolarization.  

I chose A resting membrane potential instead of C, because my understanding of this is that as the cell reaches the rising phase of the action potential the drive of Na+ to move into the cell gradually decreases. Question 5 I thought had a similar concept and I chose the correct answer E. Could you explain why Question 5 was E but Question 6 was C and whether I am interpreting these questions incorrectly?

As indicated in the cell, I = the current for the given ion.  At rest, there is slight current for K+ out of the cell through the leak K+ channels, but even less current for Na+.  This is why the resting membrane potential is closer to the equilibrium potential for K+, because there is more permeability for that ion at rest.  However, during the depolarization phase of the action potential, or rising phase of the action potential (after the membrane reaches threshold), the voltage-gated Na+ channels open causing a great increase in the Na+ current into the cell CAUSING the membrane to depolarize.

6.  AP Practice question 10, or CTL #60.  If a ligand-gated channel permeable to both sodium and potassium was briefly opened at a specific location on the membrane of a typical resting neuron, what would result?

 

  A.  Local currents will flow from the outside to the inside of the membrane and away from that region

  B.  Local currents will flow from the inside to the outside of the membrane and away from that region

  C.  Local currents will travel without decrement all along the cell’s length

( Ruled out: Need substantial depolarization to produce action potential and propagate down cell length (axon). Process of action potential is highly dependent on sequence of movement of ions first Na+ flows into the cell and then K+ out)

  D.  A brief local hyperpolarization of the membrane would result

( Ruled out: Same explanation for option C)

  E.  Fluxes of sodium and potassium would be equal, so no local currents would flow

(Ruled out: Both Ions that are antagonist of each other being able to travel through the same channels at the same brief short period of time to achieve equilibrium flux seems unlikely.)

 
I didn't really know what to choose for this question since I have a limited understanding of ligand gated channels in Neuron cells. Instead I ruled out C, D, E with my explanations below. Can you explain why A and not B and whether I ruled out C-E correctly?

Similarly: For CTL #60, I was also confused. I thought the answer was e because Na and K flow in opposite directions so wouldn't the charge be the same on each side; so no local currents would need to flow? The answer was a for this one. I read the explanations; but still don't quite understand. 

   

Your explanation for C is correct in that you need to have an action potential for this to travel the entire length and this is just a small movement of ions, either Na+ or K+ or both.  D would mean that just K+ was effluxing causing a hyperpolarization and as I will explain below that does not occur.  If the equilibrium potentials for both Na+ and K+ were equal-distance away from the membrane potential then making the cell equally permeable to both of these ions would cause their driving forces to be equal to each other resulting in no local current movement.  However, if you take a cell that is at resting membrane potential and immediately make it permeable to both Na+ and K+, which you are doing here, you have an equilibrium potential for Na+ that is approximately 130mV away from the resting membrane potential (-70mV + +60mV) and an equilibrium potential for K+ that is approximately 20mV away from the resting membrane potential (-90mV + -70mV).  Therefore the driving force for Na+ to come into the cell is much greater allow for those depolarizing charges to enter the cell and diffuse away from the area.  

Alternatively, if you were to determine an approximate 'equilibrium potential' for this channel, it would be at a membrane potential that is exactly halfway between these two ions equilibrium potentials, so it would want to bring the membrane to a potential of approximately -15mV.  Now, in order for the membrane to move from its resting membrane potential of approximately -70mV to the potential that this channel 'desires', positively charged ions would have to enter the cell, thereby depolarizing the membrane and flowing away from the region.  In fact, this is EXACTLY what happens when a nicotinic ACh opens causing a graded potential in the postsynaptic cell.  You will cover this in more detail in synaptic transmission in a few weeks!

7.  On the CTL questions, I am confused retarding the correct answers: #54 I thought was a because wouldn't you need a positive charge (Na+) going inside to make the cell more positive (from -70 to -60)? The answer is c.

CTL #54. The figure below represents a solution system at 37^oC with a membrane separating the two compartments.  What ion or ions was the membrane permeable to in order to create a membrane potential of approximately -60mV on the right with respect to the left?

A) Na⁺            B) K⁺            C) Cl^-            D) Ca²⁺           E)  Equally permeable to K⁺ and Cl^- 

F)  Equally permeable to K⁺ and Na⁺

 

This question is just like the third turningpoint question from the lecture.  The difference of that question, however, was that it indicated that the right side of the system was negative.  In this question, the right side of the system is more than just negative, but it is at a specific membrane potential of -60 mV.  If the membrane was only permeable to K+, the right side of the system would in fact be negative, but the membrane potential would be equal to the equilibrium potential for K+, that given the concentration differences in this system is not -60 mV (calculate it out utilizing the Nernst equation).  The equilibrium potential for chloride, however, in this system IS -60 mV.  Therefore, if the membrane was solely permeable to Cl-, the membrane potential that would exist at equilibrium would be the equilibrium potential for Cl-, -60mV.

 

In essence, it is not asking you which way any ions would move, but that you MEASURE this system and it has a membrane potential of -60mV, the permeability of WHICH ion solely (therefore the equilibrium potential for which ion) would cause the right side of the system to be -60mV.