Question

5) A new drug has been discovered that completely opens the voltage-gated potassium channels. If this drug is used experimentally on an isolated axon, the membrane potential can be easily measured. Assume that this experiment provided the following results. The extracellular Na+ is 145 mM and the extracellular K+ is 4mM, the measured membrane potential in the presence of the drug is -94mV. Using this information, calculate the intracellular K+ concentration.

To solve this equation, you will need to use the following concentrations inside (i) and outside (o) the cell and permeabilities (P) of sodium (Na+), potassium (K+) and chloride (Cl-) for the neuron: Ko = 5 (all) Ki = 150 Nao = 150 Nai = 15 Clo = 125 Cli = 9 PNa= 10-9, PK = 10-8 PCl = 10-9 (All concentrations in mM and permeabilities are in units of cm/s but see “hints” below)

Answer #1

Membrane Potential: An electrical potential generated due to the movement of ion from higher concentration to the lower concentration and vies versa during excited state is called membrane potential.

To calculate the membrane potential Goldman equation is used According to the equation

Em = RT/F ln X_{O}/X_{in} X = is the
ion

R = ideal gas constant

T = Temperature

F = Faraday constant

From these constant value

E_{m =} 61.5 * log X_{o} / X_{in}

_{then, for interacellular K+ is}

E_{m =.} 61.5 log 4 / K+

-94 = 61.5 *( log 4 - log K+)

-94 / 61.5 - .602 = log K+

Log K+ = 2.13

Divided by log both side

K+ = log^{-1}2.13

K+ = 10^{2.13}

^{K+ = 135mM}

^{The interacellular concentration of K+ is 135 mM.}

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