91Ó°ÊÓ

We can model some of the electrical properties of an unmyelinated axon as an electric cable covered with defective insulation so that current leaks out of the axon to the surrounding fluid. We assume the axon consists of a cylindrical membrane filled with conducting fluid. A current of ions can travel along the axon in this fluid and can also leak out through the membrane. The inner radius of the cylinder is \(5.0 \mu \mathrm{m} ;\) the membrane thickness is \(8.0 \mathrm{nm} .\) (a) If the resistivity of the axon fluid is \(2.0 \Omega \cdot \mathrm{m},\) calculate the resistance of a 1.0 -cm length of axon to current flow along its length. (b) If the resistivity of the porous membrane is \(2.5 \times 10^{7} \Omega\).m, calculate the resistance of the wall of a 1.0 -cm length of axon to current flow across the membrane. (c) Find the length of axon for which the two resistances are equal. This length is a rough measure of the distance a signal can travel without amplification.

Step by step solution

01

1. Calculate the resistance of a 1.0-cm length of axon to current flow along its length

To calculate the resistance of the axon to current flow along its length, we use the formula: $$ R = \rho \frac{l}{A} $$ where \(R\) is the resistance, \(\rho\) is the resistivity, \(l\) is the length, and \(A\) is the cross-sectional area of the axon. Here, the resistivity of the axon fluid is \(2.0 \Omega\cdot m\), the length of the axon is \(1.0\:cm = 0.01\:m\), and the cross-sectional area of the axon is given by \(A = \pi r^2\), where \(r\) is the inner radius of the cylinder given by \(5.0 \mu m = 5.0 × 10^{-6} m\). Now, we calculate the resistance: $$ R = 2.0 \Omega\cdot m \frac{0.01\:m}{\pi (5.0 × 10^{-6} m)^2} $$

02

2. Calculate the resistance of the wall of a 1.0-cm length of the axon to current flow across the membrane

To calculate the resistance to current flow across the membrane, we use the formula for resistance in a cylindrical shell: $$ R = \rho \frac{2\pi rl}{t} $$ where \(R\) is the resistance, \(\rho\) is the resistivity, \(r\) is the radius of the cylinder, \(l\) is the length, and \(t\) is the thickness of the membrane. Here, the resistivity of the porous membrane is \(2.5 \times 10^{7} \Omega\cdot m\), the radius is \(5.0 \mu m = 5.0 × 10^{-6} m\), the length of the axon is \(1.0\: cm = 0.01\:m\), and the membrane thickness is \(8.0\:nm = 8.0\times10^{-9}\:m\). Now, we calculate the resistance: $$ R = 2.5 \times 10^{7} \Omega\cdot m \frac{2\pi(5.0 × 10^{-6} m)(0.01\:m)}{8.0\times10^{-9}\:m} $$

03

3. Find the length of axon for which the resistances are equal

To find the length for which the two resistances are equal, we can equate the resistances from steps 1 and 2 and solve for the length \(l\): $$ 2.0 \Omega\cdot m \frac{l}{\pi (5.0 × 10^{-6} m)^2} = 2.5 \times 10^{7} \Omega\cdot m \frac{2\pi(5.0 × 10^{-6} m)l}{8.0\times10^{-9}\:m} $$ We can now solve for the length \(l\): $$ l = \frac{2.0 \Omega\cdot m \cdot \pi (5.0 × 10^{-6} m)^2 \cdot 8.0\times10^{-9}\:m}{2.5 \times 10^{7} \Omega\cdot m \cdot 2\pi(5.0 × 10^{-6} m)} $$

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