At rest, and at 37 degrees C, a neuron has an extracellular sodium concentration
ID: 144203 • Letter: A
Question
At rest, and at 37 degrees C, a neuron has an extracellular sodium concentration of 145mM, an intracellular sodium concentration of 15mM, and a membrane permeability to sodium of 0.05.
The extracellular potassium concentration is 4mM, intracellular is 140mM, and membrane permeability is 1.0.
The extracellular chloride concentration is 110mM, intracellular is 4mM, and membrane permeability is 0.1.
The extracellular calcium concentration is 5mM, intracellular is 0.0001mM, and membrane permeability is 0.0.
Given the Nernst equation of: E(ion) = [61/z] log [ion]out/[ion]in
61 over z, times the log of the outside ion concentration divided by the inside ion concentration
(z is the valence of the ion (+1 for sodium))
What is the approximate equilibrium potential for sodium?
3 mV
60 mV
132 mV
72 mV
-29 mV
5 points
QUESTION 50
At rest, and at 37 degrees C, a neuron has an extracellular sodium concentration of 145mM, an intracellular sodium concentration of 15mM, and a membrane permeability to sodium of 0.05.
The extracellular potassium concentration is 4mM, intracellular is 140mM, and membrane permeability is 1.0.
The extracellular chloride concentration is 110mM, intracellular is 4mM, and membrane permeability is 0.1.
The extracellular calcium concentration is 5mM, intracellular is 0.0001mM, and membrane permeability is 0.0.
Given the Goldman-Hodgkin-Katz equation of:
Vm = 61 log { ( PK*K-out + PNa*Na-out + PCl*Cl-in ) / (PK*K-in + PNa*Na-in + PCl*Cl-out ) }
What is the approximate membrane potential of this neuron?
can't be determined - a variable is missing
-68 mV
-90 mV
68 mV
40 mV
3 mV
60 mV
132 mV
72 mV
-29 mV
Explanation / Answer
1) The equilibrium potential for sodium ion is given by the formula,
E = 61/z x log (Na out)/(Na 8n)
= 61/1 x log (145/15)
= 61 × log (9.67)
= 61 × 0.985
= + 60.085 mV
Approximately 60 mV
2) The membrane potential of the neuron can be calculated according to the formula,
Vm = 61 log ((4+7.25+0.4)/(140+0.75+11)
= 61 log (11.65/151.75)
= 61 log(0.077)
= 61(-1.11)
= -67.71 mV
Approximately -68 mV
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