In the axon of a giant squid neuron, (i) The resting membrane potential is -70 mV. (ii) At the equilibrium potential of Na ^+ ( E_ Na ^+ = +55 mV) there is no net movement of Na ^+ ions across the membrane. In an experiment, when the axon is stimulated, the voltage-gated Na ^+ channels open. The membrane potential peaks at +30 mV, where the resistance of the axonal membrane for Na ^+ flow is 1 10^6 . The net Na ^+ cu

Step-by-step solution

The net ionic current across the membrane is determined by the driving force and the resistance of the membrane to that ion. The driving force is the difference between the membrane potential ( V_m ) and the equilibrium potential of the ion ( E_ Na ^+ ). Given: V_m = +30 mV E_ Na ^+ = +55 mV R_ Na ^+ = 1 10^6 The magnitude of the driving force for Na ^+ is: V = |V_m - E_ Na ^+ | = |30 - 55| = 25 mV Using Ohm's law, the net Na ^+ current ( I_ Na ^+ ) is: I_ Na ^+ = V R_ Na ^+ = 25 10⁻³ V 1 10^6 = 25 10⁻⁹ A = 25 nA S