Equivalent Cylinders

Recall that the input conductance for the semi-infinite cable is given by:

$$G_\infty =\frac{\pi d^{3/2}}{2\sqrt{R_MR_A}}$$

Recall also that the space constant is

$$\lambda =\sqrt{(d/4)R_M/R_A}$$

The electrotonic length of a cable of physical length, $\ell$ is $L\equiv \ell/\lambda.$ The input conductance at X=0 for a finite cable of electrotonic length, L with a sealed end at X=L is just

$$G_{in}=G_\infty \tanh(L).$$

There are many other possibilities, but the sealed end is the most common.

  

Figure 4: Dendritic tree

The idea that Rall discovered is that if the dendrites were related in a particular fashion, then the whole thing could be collapsed to a single cylindrical cable. This is called the equivalent cylinder. Consider the tree shown in the figure 4. Suppose that the branches, 0,1 and 2 have the same membrane resistivities, R_M and R_A. Assume that the daughter branches, 1 and 2, have the same electrotonic length, that is, their physical length divided by their space constants (which of course depend on their diameters) are all the same. (For example, if both have equal diameters and are the same physical length.) Also, assume that the two have the same end conditions, eg sealed. We want to know if it is possible to combine the branches of the dendrite into a single equivalent cylinder. The key is that we must avoid impedence mismatches. Thus, to combine the dendrites, 1 and 2 with 0, we require:

1.

All the ends are the same conditions, sealed.

2.

The electrotonic lengths of 1 and 2 are the same

3.

The diameters match as follows:

$$d_0^\frac{3}{2} = d_1^\frac{3}{2}+d_2^\frac{3}{2}$$

It is clear that this last condition is a consequence of the impedance of the parent branch equalling that of the sum of the daughters. If these conditions hold then 0,1,and 2 can be collapsed into a single cylinder with diameter equal to that of dendrite 0 and total electrotonic length as the sum of 1 and 0 (or equivalently 2 and 0.)

Example

  

Figure 5: Example and exercise

In the above figure, we depict a dendritic tree consisting of several branches with their lengths and diameters in microns. (a) Can they be reduced to an equivalent cylinder (b) What is the electrotonic length (c) What is the input conductance. Assume sealed ends for all terminal dendrites and assume that $R_M=2000 \Omega\hbox{cm}^2$ and that $R_A=60 \Omega\hbox{cm}.$

Answer.

d_a^3/2 + d_b^3/2+d_c^3/2 = 1+1+1 = 3 = 2.08^3/2=d_d^3/2

d_d^3/2+d_e^3/2 = 3+3 = 6 = 3.3^3/2 = d_f^3/2

so the 3/2 rule is obeyed. Clearly a,b,c are all the same electrotonic length. The space constants are:

$$\lambda_a=\lambda_b=\lambda_c=\sqrt{d_aR_M/4R_A}=289\mu$$

$$\lambda_d=\lambda_e=\sqrt{d_eR_M/4R_A}=416\mu$$

$$\lambda_f=\sqrt{d_fR_M/4R_A}=524\mu$$

Thus, the total electrotonic length of abc with d is

$$L_{abcd} = \frac{\ell_a}{\lambda_a} + \frac{\ell_d}{\lambda_d} = \frac{10}{289} + \frac{10}{416} = .0586$$

$$L_{e} = \frac{\ell_e}{\lambda_e} = \frac{24}{416}=.0576$$

which are close enough to be considered equal (2% difference). Thus, we can combine the whole thing into an equivalent cylinder. The total electrotonic length is then:

$$L = L_f+ L_e = L_f+L_{abcd} = \frac{\ell_f}{\lambda_f}+L_e = 0.096\approx 0.1$$

Finally, the input conductance is

$$G_{in} = G_\infty \tanh(L) = \frac{\pi d^{3/2}}{2\sqrt{R_MR_A}}\tanh(L)$$

which is

$$G_{in} = \frac{\tanh(0.1)(3.14159)(3.3\times10^{-4})^{3/2}}{2\sqrt{2000\times 60}} = 2.7\times10^{-9} \hbox{S}$$

Exercise

Apply the same analysis to the bottom dendrite in the figure.