Answer 10
a) Distance = 0.4 m
Conduction velocity = 60 m/sec
Conduction time = Distance/conduction velocity = 0.4/60 = 0.00667 seconds or 6.67 milliseconds
b) Three neurons ; so distance = 3 x 0.4 = 1.2m
Conduction time for one neuron = 6.67 milliseconds
Conduction time for a synapse = 1 millisecond
Total conduction time = 3 x conduction time for a neuron + 2 x conduction time for synapse
= 3 x 6.67 + 2 x 1
= 20.01 +2 = 22.01 milliseconds or 0.022 seconds
So, conduction velocity = Total distance/total conduction time = 1.2 / 0.022 = 54.54 m/sec
Ques 2 From given data,
a) INa+ = GNa+ (Vm - ENa+) ----------------(Ohm's Law)
For ENa+, applying nernst equation:
ENa+ (Equilibrium potential) = [(RT/zF] x {log [Naextracellular]/ [Naintracellular]}
ENa+ (Equilibrium potential) = [(60.15 mV)/z] x {log [Naextracellular]/ [Naintracellular]}
ENa+ = [(60.15 mV)/1] x {log [150 mM]/ [10 mM]}
ENa+ = [(60.15 mV)] x {log15}
ENa+ = [(60.15 mV)] x 1.1761
ENa+ = 70.74 mV
Using ENa+ in Ohm's law equation
INa+ = GNa+ (Vm - ENa+)
INa+ = 1nS (-70 mV -(+ 70.74 mV))
INa+ = 1nS (0.74 mV)
INa+ = 0.74 pA
b)As INa+ = 0.74 pA i.e. positive value, so it is leaving the cell
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