91Ó°ÊÓ

In the human nervous system, signals are transmitted along neurons as action potentials that travel at speeds of up to \(100 \mathrm{m} / \mathrm{s} .\) (An action potential is a traveling influx of sodium ions through the membrane of a neuron.) The signal is passed from one neuron to another by the release of neurotransmitters in the synapse. Suppose someone steps on your toe. The pain signal travels along a 1.0 -m-long sensory neuron to the spinal column, across a synapse to a second 1.0 -m-long neuron, and across a second synapse to the brain. Suppose that the synapses are each \(100 \mathrm{nm}\) wide, that it takes \(0.10 \mathrm{ms}\) for the signal to cross each synapse, and that the action potentials travel at \(100 \mathrm{m} / \mathrm{s} .\) (a) At what average speed does the signal cross a synapse? (b) How long does it take the signal to reach the brain? (c) What is the average speed of propagation of the signal?

Step by step solution

01

(a) Determine the synapse's average speed

To find the average speed across a synapse, we will use the formula: \(v = \frac{d}{t}\), where \(v\) is the average speed, \(d\) is the distance and \(t\) is the time. The distance is given as \(100 \mathrm{nm}\) (nanometers) and the time given is \(0.10 \mathrm{ms}\) (milliseconds). Let's first convert the given units into a standard unit (meter and second): Distance: \(100 \mathrm{nm} = 100 \times 10^{-9} \mathrm{m}\) Time: \(0.10 \mathrm{ms} = 0.10 \times 10^{-3} \mathrm{s}\) Now we can calculate the average speed: \(v = \frac{100 \times 10^{-9} \mathrm{m}}{0.1 \times 10^{-3} \mathrm{s}} = 10^{-6} \mathrm{m/s}\)

02

(b) Calculate the time it takes for the signal to reach the brain

To find the time it takes for the signal to reach the brain, we need to calculate the time it takes for the signal to travel through each of the neurons and synapses. We know that the action potentials travel at \(100 \mathrm{m/s}\) across neurons with a length of \(1.0 \mathrm{m}\). Time for a neuron: \(t = \frac{d}{v} = \frac{1.0 \mathrm{m}}{100 \mathrm{m/s}} = 0.01 \mathrm{s}\) There are two neurons, so the total time for neurons is: \(0.01 \mathrm{s} \times 2 = 0.02 \mathrm{s}\) We also know that it takes \(0.10 \mathrm{ms}\) for the signal to cross each synapse. There are two synapses, so the total time for synapses is: \(0.10 \mathrm{ms} \times 2 = 0.20 \mathrm{ms} = 0.20 \times 10^{-3} \mathrm{s} = 2 \times 10^{-4} \mathrm{s}\) To find the total time for the signal to reach the brain, we sum the time for neurons and synapses: \(0.02 \mathrm{s} + 2 \times 10^{-4} \mathrm{s} = 0.0202 \mathrm{s}\)

03

(c) Find the average speed of propagation

To find the average speed of propagation, we will use the formula: \(v = \frac{d}{t}\), where \(d\) is the total distance traveled by the signal and \(t\) is the total time taken. The total distance is the sum of distances traveled across neurons and synapses: \(distance_{neurons} = 2 \times 1.0 \mathrm{m} = 2.0 \mathrm{m}\) \(distance_{synapses} = 2 \times 100 \mathrm{nm} = 2 \times 100 \times 10^{-9} \mathrm{m} = 2 \times 10^{-7} \mathrm{m}\) \(total\_distance = 2.0 \mathrm{m} + 2 \times 10^{-7} \mathrm{m} = 2.0000002 \mathrm{m}\) We already found the total time taken in (b) which is \(0.0202 \mathrm{s}\). Now we can find the average speed of propagation: \(v = \frac{2.0000002 \mathrm{m}}{0.0202 \mathrm{s}} \approx 99.1 \mathrm{m/s}\)

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