Question

***I need help on PART E of this problem. All of the information is given on the worksheet along with the hints on each part. 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

Abswer a)

a) When ions of opposite charge are separated by á permeable membrane, there is a transmembrane electrochemical gradient, a membrane potential (um) expressed in millivolts (mv). in the membrane potential is generally established . a result of the relative contributions of several ions when more than one ion channel in present and open) in the plasma membrane, the al potential membrane can can be be calculated a calculated by using the Goldman - Hodgkin - Katz equation (GHK). RT in / Pr [kt]o + pra [nat Jo+ Pa [el-1; Pac [K+]; + Pna [Nat]i + Palu-]. Uma - MT in lastent R-> Gas constant 8.315 J/mol. K Ty temperature in Kelvin F Faraday's constant 96480J/N, mol Pk, Pna, Pe Membrane permeability of K, Na, el [kt]o, [Nat]o, [41]. → Concentration of K, Na lemente el in extracellulas [kt];, [ Nat];, [et]; → Concentration of ions in intra cellular fluid

AR 25°C, V 500 96480 = 3 3 15 x 218K .( 10p(5) . (146) 96480mn 1 100 (150) = 0.025 in (0.033 + 29.2 + 0.032) = 0.025 in (29.26) Vm = 13.376 v or 3376 mul

Answer b)

til it exactly b) Extra cellular and Intracellular concentrations of ions are given - charge separation across the membrane leads to the establishment of an electrochemical gradient that grows in magnitude until it exactly balances the chemical gradient. The Nernst eqreation allows us to calculate the potential that will be established across the membrane based on the Nalence and concentration gradient of ions. w e N = RT In [kt]. zf Ik+]; all the equation the potential . z y valence of ions for Na, 2477.87 in 141 61) 96480 1 = 0.025 en 29.2 10.084 -1784 mV]

for K, V = 0.025 in -0.085 =[- 85 m - For I, = 2477.87 155 0.025 TD 96480 - - 0.025 in (31) - 0.085 = 85ml

Since the membrane potential of indiviaual ions are either positive or negative (and not zero), none of the ions are at electrochemical equilibrium.

Answer c)

So, changing the ion permeability does not change membrane potential.

Pk = 200 c) Permeability of potassium changes to Using GHK ean - 1200 (5) + 29.2+ Vm = 0.025 en 200 (150) = 3.376 v OR 3370 m V + 29.2 + 0.032

Answer d)

So, changing the ion permeability does not change membrane potential.

d) Nm = 0.025 in (0.033 + 29.2 + 20015)) = 3.376 V OR 3376 mV