Question

8. Consider a neuron whose resting potential is -60 mV, and the E_K = -100 mV and the E_Na = +40 mV. (a) What is the driving force for Na+, and what is the driving force for K+ at the resting potential? Just give a number, don’t describe what a driving force is in words. (b) What is the relative resistance of the cell membrane for sodium and potassium (i.e. how many times greater or less than the R_K must the R_Na be? N.B. resistance not permeability!) Just give a number, no words. (c) A neurotransmitter-gated channel on this cell when opened has a synaptic reversal potential of -120 mV. Is it possible that the ion that flows through this channel is positively-charged? Give a yes or no answer, then briefly explain.

Answer 1

8

a. VDF = Vm − Veq

Driving force for Na+ : Vm= -100mV and Veq= -60mV hence, Driving force = -100-(-60) = -40mV

Driving force for K+   : Vm= +40mV and Veq= -60mV hence, Driving force =    40-(-60) = 100mV

b. Resistance of K+ is 2.5 times Resistance of Na+.

c. rm is the membrane resistance (the force that impedes the flow of electric current from the outside of the membrane to the inside, and vice versa), ri is the axial resistance (the force that impedes current flow through the axoplasm, parallel to the membrane), and ro is the extracellular resistance (the force that impedes current flow through the extracellular fluid, parallel to the membrane). In calculation, the effects of ro are negligible. So here the synaptic reversal potential of -120 mV. so the positivily charged ions have a tendancy to move across.