Two questions (5 points each), due Thursday 17 February at 8:00 AM
1. The figure below plots the time-dependence for de-inactivation of Na+ currents in squid axon obtained using two voltage pulses to -21 mV (similar, but not identical to Hille, Fig. 2.15). The first pulse evokes (iNa)0. The second pulse is preceded by an adjustable time interval (t) and evokes (iNa)V1, where V1 is an adjustable voltage command prior to the second test pulse, as illustrated in the figure. The graph shows four traces that demonstrate the effect of voltage-clamping at different potentials V1 during t, including: -96, -81, -57, and -36 mV. What does this protocol tell you about the percentage of Na+ conductance available at resting potential (assume Vrest is -60 mV)? What does this protocol tell you about the available Na+ conductance if membrane potential were held at -100 mV at steady-state?
The available Na+ conductance at the resting potential of -60 mV is about 59% of the total Na+ conductance. At steady-state, with V1 = -57 mV, the Na+ current (iNa)V1 is about 0.7 (70%) of the (iNa)0 evoked at Vrest. Also, the Na+ current evoked when V1 = -96 mV is about 1.7 (170%) of the (iNa)0 evoked at Vrest. If we assume that -96 mV de-inactivates nearly all of the Na+ channels, and that the fraction (iNa)V1/(iNa)0 = 1.0 at Vrest, then the fraction of the Na+ channels available for activation is 1.0/1.7 = 0.59, or about 59% of them are available (41% are steady-state inactivated). At -100 mV, all the availlable Na+ channels will be available, because the plot of (iNa)V1/(iNa)0 is reaching an asymptotic value of near 1.7, which means that no more Na+ channels can be made available even if Vm is made more negative during V1. At -100 mV, the Na+ current is 170% of the magnitude of the Na+ current evoked from Vrest.
2. Study the traces below showing Vm (red, left axis) and a current stimulus (black, right axis) in the top graph, and the corresponding activity of the Na+ and K+ gating variables in the lower graph (n, m, and h in black, red, and green). Explain why the termination of a hyperpolarizing current pulse evokes neuronal excitation in this example. Make sure your answer explains the phenomenon in terms of voltage-dependence and kinetics of the gating variables, and the effects of the inward and outward currents on Vm.
At the beginning of the protocol the cell is stable at its resting potential near -63 mV. The gating variables are also at steady-state values and no excitation is possible unless the membrane is perturbed or stimulated somehow. Normally spikes are evoked by depolarizing stimuli, but here an action potential is evoked by a negative current pulse. The hyperpolarizing pulse causes voltage-dependent gating variables to change their state of "permissiveness." For example, both m and n decrease slightly as Vm hyperpolarizes. But the biggest change occurs with h, which moves to a much higher (more permissive) state during hyperpolarization. Hence, when the negative pulse terminates and t=40, the h-gates are in a much more permissive state. As m-gates rise again following the pulse, they are able to open more Na+ current since the h-gates are in a much more permissive state than prior to the negative current pulse. This fraction of additional Na+ current increases in a positive-feedback process, causing depolarization, evoking more Na+ current, and so on, which leads to the action potential. When a spike is evoked upon termination of a negative current stimulus we refer to the phenomenon as anode break excitation since the negative current is known as "anodal" stimulation.
