Fig. 7. Use-dependent block of tetrodotoxin-resistant Nav1.8 currents in the absence (control,  open symbols ) or presence (  closed symbols ) of isoflurane (0.56 ± 0.08 mm; 1.6 minimum alveolar concentration [MAC]). Whole-cell currents were evoked by 60-step depolarization commands (holding potential −70 mV; test potential +10 mV; pulse duration 10 ms) delivered at 1 Hz (  A ), 3 Hz (  B ) or 10 Hz (  C ). Peak-current amplitude values (mean ± SEM, n = 5) were normalized to that of the first response at each frequency and plotted against pulse number. The normalized first pulse amplitude was reduced to 0.54 ± 0.03 of control by isoflurane. (  D ) Representative recordings at 10 Hz in the absence (  left ) or presence (  right ) of isoflurane.  Arrows mark traces for pulse 1 and pulse 60 (  dashed line represents baseline). Data were fitted by a mono-exponential equation, and values for fractional block of the plateau of normalized  I Naare shown in  E . ***  P < 0.001, paired two-tailed Student  t test, n = 5. 

Fig. 7. Use-dependent block of tetrodotoxin-resistant Nav1.8 currents in the absence (control,  open symbols ) or presence (  closed symbols ) of isoflurane (0.56 ± 0.08 mm; 1.6 minimum alveolar concentration [MAC]). Whole-cell currents were evoked by 60-step depolarization commands (holding potential −70 mV; test potential +10 mV; pulse duration 10 ms) delivered at 1 Hz (  A ), 3 Hz (  B ) or 10 Hz (  C ). Peak-current amplitude values (mean ± SEM, n = 5) were normalized to that of the first response at each frequency and plotted against pulse number. The normalized first pulse amplitude was reduced to 0.54 ± 0.03 of control by isoflurane. (  D ) Representative recordings at 10 Hz in the absence (  left ) or presence (  right ) of isoflurane.  Arrows mark traces for pulse 1 and pulse 60 (  dashed line represents baseline). Data were fitted by a mono-exponential equation, and values for fractional block of the plateau of normalized  I Naare shown in  E . ***  P < 0.001, paired two-tailed Student  t test, n = 5. 

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