Please show all work 10. Suppose point A is 0.4m from point B. a. What is the co

Question

Please show all work 10. Suppose point A is 0.4m from point B. a. What is the conduction time required for an action potential to travel this distance if a single neuron spans this distance and has a conduction velocity is 60m/sec? b. What is the conduction yelocity if the pathway consists of three neurons, two synapses and the synaptic delay is 1 msec? 11. Ohm's Law describes the movement of ions across a membrane: I-G(Vm-E). I is the current in amperes (A), G is the conductance of x (a measure of its permeability) in Siemens(S), Vm is the membrane voltage and E is the equilibrium potential of x. Assume 30C. Note: nS x mV pA a. Using Ohm's Law, G(VarE), and the data below, calculate the magnitude of the I Na+ (sodium current). Extracellular Na" 150 mM Intracellular Na+ = 10 mM b. Is Na* entering or leaving the cell'?

Explanation / Answer

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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