Question

The resistor-capacitor network model we developed in Lecture (see figure below where it is shown only for the myelinated case) for the transmission of signals in nerve axons can actually be applied to both unmyelinated and myelinated nerve cells to estimate the propagation speed of a nerve impulse. In the unmyelinated case the 6-7 nm thick plasma membrane (the lipid bilayer) acts both as a (very thin) dielectric layer forming a capacitor between the inside of the axon and the outside, and as an insulating (high resistance) boundary between the inside and outside of the axon. The myelination does the same thing but now the dielectric/insulating layer layer is MUCH thicker, typically 2 µm instead of 6 nm.

Plasma membrane - high resistivity, but thin Unmyelinated Axoplasm: conducting medium Myelinated Axoplasm resistance V(t) Vt) Membrane (myelin) resistance Membrane (myelin) capacitance Consider a single short segment of a nerve axon of length Ar, as shown below, where the outer cylindrical insulating layer can be either the plasma membrane of an unmyelinated fiber or the combination of the plasma membrane plus myelin layer for the myelinated case. AX AX K, Pm ER Cm SR Myelinated Unmyelinated

(a) Taking the axon to be a cylinder of radius a and electrical resistivity ρa, calculate the resistance Ra for currents traveling along a length ∆x of the axon. Note that since ρa is much less than the membrane (or myelin) resistivity ρm (see below), you can assume for this calculation that the current stays inside the axon.

(b) Calculate the capacitance Cm of the segment between the interior of the axon (axoplasm) and the conducting medium outside the nerve cell. The insulating layer has dielectric constant κ. You may consider the thickness of the insulating layer to be thinner than the radius of the axon, i,.e, t < a and so you can treat the segment as a “rolled up” parallel plate capacitor.

(c) Calculate the resistance of the segment for currents flowing through the membrane. The resistivity of the membrane is ρm. Again, take t < a and use the simple form Rm = ρL/A for your resistance calculation.

(d) Compute the time constant τ = RmCm. Your answer should not depend on ∆x.

Answer 1

(a) We have the resistivity = $\rho_a$

current flowing along the length = $\Delta x$

cross-section area of the cylinder = $\pi a^2$

We know that the resistance of a resistor ( R ) = $\frac{\rho l}{A}$ ,

where, $\rho$ is the resistivity of the resistor,

l is the length of the resistor,

A is the area of the cross-section of the resistor,

So, required resistance ( $R_a$ ) = $\frac{\rho l}{A}$ ,

putting the values,    ......( 1 )

Ra = Pada

(b) We have the dielectic constant of the medium =

the distance between the parallel plate = t

  we can consider that the , so we can have the approx area of paralle plate capacitor =
2παλι

We know that the capacitance of a parallel plate capacitor ( C ) = $\frac{A\epsilon _o}{d}K$ ,

where, A is the area of the parallel plate capacitor,

  $\epsilon _o$ is the permittivity of the vaccum,

d is the distance between the parallel plates,

K is the dielectric constant of the medium.

Now, for the required capacitance ( $C_m$ ) = $\frac{A\epsilon _o}{d}K$ ,

putting the values,    ........( 2 )

;,(rܠ - ( 2roa c

(c) We have the resistivity = $\rho_m$

current flowing along the length =

we can consider that the , so we can have the approx area of the resistor =
2παλι

We know that the resistance of a resistor ( R ) = $\frac{\rho l}{A}$ ,

where, $\rho$ is the resistivity of the resistor,

l is the length of the resistor,

A is the area of the cross-section of the resistor,

So, required resistance ( $R_m$ ) = $\frac{\rho l}{A}$ ,

putting the values,    ......( 3 )

R = 2πα

(d) Time constant   ,

T=CRO

putting the values from equation ( 2 ) and ( 3 ),

  

(2παΔα)εο. Pmt 2παΔη

.

T = Ekt