Question

*** I need help with PART F of this problem. Hints are given in other parts of the problem and all information available is given on the worksheet. Please help, thank you!

The post-synaptic cell at a chemical synapse has the following distribution of ions between the cytosol and the extracellular environment. Consider that the relative permeabilities for Na, K and Cl' are 1, 100 and 50, and that there are no other significant permeabilities, lon Concentration (in mM) Extracellular Equilibrium potential mV) Intracellular 146 Na K 150 CI 15S 0.0001 145 Impermeable unions n/a a. What is the cell membrane potential? Hint: Use the GHK equation, b. Which ion(s) are at electrochemical equilibrium, if any? Explain. c. What happens to the membrane potential if a neurotransmitter causes opening of K channels, increasing the K permeability to 200 from the initial conditions? Hint: Calculate the membrane potential with Px = 200 as the anly change and then compare to the original membrane potential. d. What happens to the membrane potential if a neurotransmitter causes opening of Cl channels, increasing the C permeability to 200 (from the initial conditions)? See Hint for c) and make the change and for chloride. e. What happens to the membrane potential if a neurotransmitter causes opening of Na channels, increasing the Na permeability to 20 (from the initial conditions)? See Hint for (c) and make the change only for som f. If the Cl' permeability is increased to 200 from the initial conditions, what happens to the membrane potential when the Na permeability is increased to 20 at the same time? How does this compare to e above? What has happened to membrane excitability? That is, has the opening of the chloride channels been excitatory (increased chance to reach threshold) or inhibitory (decreased chance to reach threshold) based on their effect on the change in membrane potential due to opening the sodium channels? Explain.

Answer 1

For solving part F, we need to know about the answer to part A.

Upon solving part A, the membrane potential comes out to be -76.091 mV.

Now, solving part F by replacing Cl- permeability with 200 and Na+ permeability with 20, and then inserting relevant values into GHK equation :

So the new membrane potential comes out to be -55.156 mV.

Vm = RT in Px [k ]o + Pwa [Nalet Pale] F (PK [K]; + Pra [ Nat]; + Pe [u]. = 8,314x273 96485 ln (100x5) +(20x146) + 200x5) [(loox 150) + (20x5)+(200x155) = 0.023524 X ln 4420 46100 = 0.023524 X (-2.34461 I = – 0.055156 V. Vm = -55.156 mv.

Therefore difference between old and new membrane potential will be : (-76.091) - (-55.156) = -20.935 mV

Please do leave a thumbs up and an upvote if this answer really helped you :-)