Recovery from fast inactivation involves two components. A paired pulse protocol (top) was used to examine the time course of recovery from inactivation. From a holding potential of −80 mV, a 5-ms activation step to 0 mV was used to produce Nav inactivation. In a given trial, the cell was then repolarized to a voltage between −60 and −120 mV for durations from 1 ms to 3 s before the second test step to 0 mV. The example protocol (top) shows example paired-pulse intervals for 50, 100, 200, 350, 700, 1,000, 2,000, and 3,000 ms recovery intervals at −120 mV. (A) A family of traces showing recovery from inactivation at −60 mV. On the right, traces are shown for the response to the initial step to 0 mV (pulse 1) and then for the response to the second step to 0 mV (pulse 2) following the recovery step. Colored traces highlight recovery at the indicated intervals. (B) As in A, traces are shown for recovery at −120 mV. (C) Fractional recovery from 1 ms to 3 s is plotted for recovery at −60, −80, −100, and −120 mV for the cell shown in A and B using a linear scale on the left and a logarithmic scale on the right. Lines are fits of a single exponential function to the recovery time course: It=A[1-exp(-t/τ)], where A and τ are the amplitude and time constant of the recovery process, respectively. (D) The same fractional recoveries shown in C are replotted along with fits of a double exponential function to the recovery time course with It=Af1−exp(−t/τf)+As1−exp(−t/τs), where Af and As are amplitudes of fast and slow recovery components, respectively, and τf and τs are the respective time constants of fast and slow recovery. For −60 mV, Af = 0.19 ± 0.03, τf = 49.9 ± 7.9 ms, As = 0.38 ± 0.02, and τs = 373.7 ± 34.6 ms (fitted value and 90% confidence limit). For −80 mV, Af = 0.40 ± 0.01, τf = 14.8 ± 0.9 ms, As = 0.59 ± 0.01, and τs = 332.0 ± 19.7 ms. For −100 mV, Af = 0.44 ± 0.01, τf = 5.0 ± 0.4 ms, As = 0.58 ± 0.1, and τs = 134.7 ± 8.4 ms. For −120 mV, Af = 0.46 ± 0.01, τf = 2.1 ± 0.1 ms, As = 0.57 ± 0.01, and τs = 46.1 ± 2.4 ms.
Figure 2.

Recovery from fast inactivation involves two components. A paired pulse protocol (top) was used to examine the time course of recovery from inactivation. From a holding potential of −80 mV, a 5-ms activation step to 0 mV was used to produce Nav inactivation. In a given trial, the cell was then repolarized to a voltage between −60 and −120 mV for durations from 1 ms to 3 s before the second test step to 0 mV. The example protocol (top) shows example paired-pulse intervals for 50, 100, 200, 350, 700, 1,000, 2,000, and 3,000 ms recovery intervals at −120 mV. (A) A family of traces showing recovery from inactivation at −60 mV. On the right, traces are shown for the response to the initial step to 0 mV (pulse 1) and then for the response to the second step to 0 mV (pulse 2) following the recovery step. Colored traces highlight recovery at the indicated intervals. (B) As in A, traces are shown for recovery at −120 mV. (C) Fractional recovery from 1 ms to 3 s is plotted for recovery at −60, −80, −100, and −120 mV for the cell shown in A and B using a linear scale on the left and a logarithmic scale on the right. Lines are fits of a single exponential function to the recovery time course: It=A[1-exp(-t/τ)], where A and τ are the amplitude and time constant of the recovery process, respectively. (D) The same fractional recoveries shown in C are replotted along with fits of a double exponential function to the recovery time course with It=Af1−exp(−t/τf)+As1−exp(−t/τs), where Af and As are amplitudes of fast and slow recovery components, respectively, and τf and τs are the respective time constants of fast and slow recovery. For −60 mV, Af = 0.19 ± 0.03, τf = 49.9 ± 7.9 ms, As = 0.38 ± 0.02, and τs = 373.7 ± 34.6 ms (fitted value and 90% confidence limit). For −80 mV, Af = 0.40 ± 0.01, τf = 14.8 ± 0.9 ms, As = 0.59 ± 0.01, and τs = 332.0 ± 19.7 ms. For −100 mV, Af = 0.44 ± 0.01, τf = 5.0 ± 0.4 ms, As = 0.58 ± 0.1, and τs = 134.7 ± 8.4 ms. For −120 mV, Af = 0.46 ± 0.01, τf = 2.1 ± 0.1 ms, As = 0.57 ± 0.01, and τs = 46.1 ± 2.4 ms.

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