MICHALE FEE: So today, we're
going to start a new topic. We're going to be talking about
the propagation of signals in dendrites and axons. So the model that
we've considered so far is just a soma. We basically had a kind of a
spherical shell of insulator that we've been modeling that
has different kinds of ion channels in it that allow the
cell to do things like generate an action potential. So the reason that
we've been doing that is because in most
vertebrate neurons, the soma is the sight in the
neuron at which the decision to make an action
potential is made. So all kinds of inputs
come in, and then the soma integrates those inputs,
accumulates charge, reaches some spiking
threshold, and then generates an action potential. And so that's where
the decision is about whether a neuron is
going to spike or not. Now, in real neurons,
relatively few of the inputs actually come onto the soma. Most of the synaptic
inputs, most of the inputs arrive onto the dendrites, which
are these branching cylinders of cell membrane. And most of the
synapses actually form onto the dendrite at
some distance from the soma. There are synapses that
form onto the soma. But the vast
majority of synapses form onto these dendrites. And sometimes,
those synapses can be as far away as 1 or
2 millimeters for very large neurons in cortex. So there's a population
of neurons in deep layer V some of you may have heard
about that have dendrites that reach all the
way up into layer I. And those cells can be-- those dendrites can be as long
as a couple of millimeters. So we really have to think
about what this means, how signals get from out here in
the dendrite down to the soma. And that's what we're
going to talk about today. So the most important thing
that we're going to do is to simplify
this-- by the way, anybody know what
kind of cell this is? AUDIENCE: [INAUDIBLE] MICHALE FEE: Good. It's a Purkinje cell. And it was one of the-- this is one of the cells
that Ramon Cajal drew back in the late 1800s. So the most important
thing we're going to do is to simplify this very
complex dendritic arborization. And we're going
to basically think of it as a single cylinder. Now mathematically, there
are reasons why this is not actually unreasonable. You can write down--
if you analyze the structure of
dendritic trees, there is something
about the way the ratio of the diameters of
the different dendrites as they converge to
form thicker branches as you get closer
and closer to closer to the soma that
mathematically makes this not a bad approximation
for an extended dendritic arbor like this. So we're going to think
about the problem of having a synapse out here on this
cylindrical approximation to a dendrite. And we're going to
imagine that we're measuring the voltage
down here at the soma or at different positions
along the dendrite. And we're going to ask, how
does synaptic input out here on this cylinder affect
the membrane potential in the dendrite and
down here at the soma? And the basic conceptual
picture that you should have is that those signals propagate
some distance down the soma but gradually leak out. And there's a very simple kind
of intuitive picture, which is that the dendrite you
can think of this as a leaky pipe or a leaky hose. So imagine you took a
piece of garden hose and you poked holes
in the side of it or so that they're
kind of close together. And when you hook this
up to the water faucet, you turn the water on,
that some of that water flows down the hose. But some of it also leaks
out through the holes that you drilled. And you can see that
eventually the water is all going to leak out
through the sides, and it's not going to go all
the way down to the other end to get to your
hydrangeas or whatever it is that you're watering. And so you can see
that that signal isn't going to get very far
if the holes you drilled are big enough. And the general
kind of analogy here is that current is like
the flow of the water. Electrical current here is
like water current flowing down the pipe. And voltage is like pressure. So the higher the pressure here,
the higher the current flow you'll get. And we're going to develop
an electrical circuit model for a dendrite
like this that's going to look like a set
of resistors going down the axis of the dendrite
and a set of resistors that go across the membrane. And you can see that each
little piece of membrane here, a little piece
of the dendrite is going to look like a
resistor divider, where you have a resistor along
the axial direction and a resistance
across the membrane. And as you make a longer and
longer piece of dendrite, you're going to get
additional voltage dividers. Each voltage divider
divides the voltage by some constant factor. And as you stack
those things up, the voltage drops by some
constant factor per unit length of the dendrite. And so you can
see-- anybody want to just take a guess of
what kind of functional form that would give
you if you divide the voltage by some
constant factor each unit length
of the dendrite? AUDIENCE: Exponential. MICHALE FEE: Exponential. That's right. And that's where this
exponential falloff comes from. So today, we're going to
do the following things. And we're going
to basically draw a circuit diagram, an
electrical equivalent circuit of a piece of dendrite. And I would like you to be able
to make that drawing if you're asked to. We're going to be able
to plot the voltage in a piece of dendrite
as a function of distance for the case of a dendrite
that has leaky walls and for the case of a dendrite
that has non-leaky walls. And we're going to describe
the concept of a length constant, which I'll tell
you right now is just the 1 over the distance at which
the voltage falls by 1 over e as a function of length. So it's some length over
which the voltage falls by some amount 1 over e. We're going to go over how
that length constant depends on the radius of the dendrite. It's a function of the size. And also, we're going
to describe the concept of an electrotonic length. And then finally,
we're going to go to some sort of extreme
simplifications, even beyond taking that
very complex dendrite, simplifying it as a cylinder. We're going to go to
an even simpler case where we can just treat the
cell as a soma connected to a resistor
connected by a resistor to a separate compartment. And that's sort of the
most extreme simplification of a dendrite. But, in fact, it's an
extremely powerful one from which you can
get a lot of intuition about how signals are
integrated in dendrites. So we're going to analyze
a piece of dendrite using a technique called
finite element analysis. We're going to
imagine-- we're going to approximate our piece
of dendrite as a cylinder of constant radius a, an
axial dimension that we're going to label x. We're going to break up this
cylinder into little slices. So imagine we just
took a little knife, and we cut little
slices of this dendrite. And they're going to
be very small slices. And we're going to model
each one of those slices with a separate little circuit. And then we're going to
connect them together. And we're going to
let the length of that slice be delta x. And then eventually, we're
going to let delta x go to 0. We're going to get some
differential equations that describe that relationship
between the voltage and the current in
this piece of dendrite. So let's start with a model for
the inside of this cylinder. So remember, in a
cell, we had the inside of the cell modeled by a wire. In a dendrite, we
can't just use a wire. And the reason is that
current is going to flow along the inside of the dendrite. It's going to flow,
and it's going to experience voltage drops. So we have to actually
model the resistance of the inside of the dendrite. And we're going to model
it like the resistance between each one of those slices
with a resistor value little r. We're going to model the outside
of the axon or the dendrite as a wire. And the reason we're going to
put resistors inside and just a wire outside is because the
resistance-- remember the axon or dendrite is very small. In the brain, dendrites might
be about 2 microns across. So the current is constrained
to a very small space. When currents then
flow outside, they're flowing in a much larger volume,
and so the effective resistance is much smaller. And we're going to essentially
ignore that resistance and treat the outside
as just a wire. Now we have to
model the membrane. Anybody want to take
a guess how we're going to model the membrane? AUDIENCE: [INAUDIBLE] MICHALE FEE: What's that? I heard two correct answers. What did you say Jasmine? AUDIENCE: The capacitor. MICHALE FEE: Capacitor. And? AUDIENCE: [INAUDIBLE] MICHALE FEE: Excellent. Whoops. I wasn't quite there. Let's put that up. Good. So we're going to
have a capacitance. We're going to imagine
that this membrane might have an ion selective ion
channel with some conductance G sub l and an equilibrium or
reversal potential E sub l. Now coming back to these terms
here, we're going to model. We're going to write
down the voltage in each one of our little
slices of the dendrite. So let's do that. Let's just pick one of
them as V, the voltage, at position, x, and time t. The voltage in the
next slides over is going to be V at
x plus delta x of t. And the voltage in this slice
over here is V of x minus delta x and t. So now, we can also write down
the current that goes axially through that piece of-- that slice of our dendrite. We're going to call
that I of x and t. And we can write down also the
current in every other time-- in every other slice
of the dendrite, I of x minus delta x and t. And we're going to model
this piece of membrane in each one of those
slices as well. Any questions about that? That's the basic setup. That's the basic finite
element model of a dendrite. No questions? Now we also have to model the
current through the membrane. That's going to be I
sub m, m for membrane. And it's going to be
a current per unit length of the dendrite. We're going to imagine
that there's current flowing from the inside to the
outside through the membrane. And there's going to be
some current per unit length of the dendrite. And we can also imagine that
we have current being injected, let's say, through a synapse
or through an electrode that we can also model as
coming in at any position x. And this is, again, current
per unit length times delta x. Does that make sense? So the first thing
we're going to do is we're going to write
down the relation between V in each node and the current
going through that node. So let's do that. We're going to use Ohm's law. So the voltage difference
between here and here is just going to be the
current times that resistance. Does that make sense? We're just going
to use Ohm's law-- very simple. So V of x and t minus
V of x plus delta x, t is just equal to little
r times that current. And now we're going
to rewrite this. Let's divide both sides of
this equation by delta x. So you see 1 over delta x V
of x minus V of x plus delta x is equal to little r over
delta x times the current. And can anyone tell me what that
thing is as delta x goes to 0? AUDIENCE: [INAUDIBLE] MICHALE FEE: Good. It's the derivative of--
it's the spatial derivative of the voltage. That's just the
definition of derivative when delta x goes to 0. So let's write that out. Notice that it's the
negative of the derivative because the derivative
would have V of x plus delta x minus V of x. So it's a negative
of the derivative. So negative dv/dx is
equal to some resistance times the current. And notice that
this capital R sub a is called the axial
resistance per unit length. It's this resistance per
unit length of the dendrite. Now notice that if you pass
current down that dendrite, the voltage drop is
going to keep increasing. The resistance is going to
keep increasing the longer that piece of dendrite is. So you can think
about resistance in a piece of dendrite
more appropriately as resistance per unit length. So there's Ohm's law-- minus dv/dx equals axial
resistance per unit length times the current. Any questions? And notice that
according to this, current flow to the
right, positive I is defined as current
to the right here produces a negative
gradient in the voltage. So the voltage is high on this
side and low on that side. So the slope is negative. So now let's take
this, and let's analyze this for some
simple cases where we have no membrane current. So we're going to
just ignore those. And we're just going to include
these axial resistances. And we're going to analyze
what this equation tells us about the voltage
inside of the dendrite. Does that makes sense? So let's do that. So if we take that equation,
we can write down the current at, let's say, these
two different nodes-- I of x minus delta x and I of x. And because there are
no membrane currents, you can see that
those two currents have to be equal to each other. Kirchoff's Current Law says
that the current into this node has to equal the current
out of that node. So if there are no
membrane currents, there's nothing
leaking out here, then those two currents
have to equal each other. And we can call that I0. So now, dv/dx is minus
axial resistance times I0. And what does that tell us
about how the voltage changes in a piece of dendrite if
there's no membrane current, if there's no leaky membrane? There's no leakage
in the membrane. If dv/dx is a constant,
what does it tell us? AUDIENCE: [INAUDIBLE]. MICHALE FEE: Yeah. But decreases how? What functional form? AUDIENCE: [INAUDIBLE] MICHALE FEE: Good. It changes linearly. So if there are no
membrane conductances, then the membrane
potential changes linearly. So you can see that the voltage
as a function of position-- sorry I forgot to
label that voltage-- just changes linearly
from some initial voltage to some final voltage
over some length l. We're considering a case of a
piece of dendrite of length l. Yes? AUDIENCE: [INAUDIBLE] MICHALE FEE: Yep. So I just rewrote this equation. Sorry, I just rewrote this
equation moving the minus sign to that side. Yep. Good. Now you can see that the delta
V that the voltage difference from the left side
to the right side is just the total resistance
times the current-- just Ohm's law again. And the total resistance
is the axial resistance per unit length
times the length. Really simple. Voltage changes linearly. If you don't have any
membrane conductances, and you can just write
down the relation between the voltage
difference on the two sides and the current. So, in general, let's
think a little bit more about this problem of being able
to what you need to write down the solution to this equation. It's a very simple equation. If you integrate
this over x, you can see that the voltage
as a function of position is some initial voltage
minus a resistance times the current times x. And that, again,
just looks like this. That's where that
solution came from. It's just integrating
this over x. And you can see that
in order to write down the solution to this equation,
we need a couple of things. We need to either know the
voltages at the beginning and end, or we need
to know the current. We need to know some combination
of those three things. So let's write down
the voltage here. Let's call it V0. Let's write down the
voltage there, V sub l, and plug those in. And you can see that-- there's V0. There's V sub l. You can see that if you know
any two of those quantities-- V0, V sub l, or Io-- you can calculate the third. So if you know V0 and Vl, you
can calculate the current. If you know V0 and the current,
you can calculate V sub l. That is the concept of
boundary conditions. You can write down the
voltages or the currents at some positions
on the dendrite and figure out
the total solution to the voltage
[AUDIO OUT] of position. Does that make sense? If you don't know some
of those quantities, you can't write down the
solution to the equation. It's just the simple
idea that when you integrate a
differential equation, you need to have an
initial condition in order to actually solve the equation. Any questions about that? So let's think about a couple
of different kinds of boundary conditions that you
might encounter. So this boundary condition
right here-- so let's say that we inject a
x amount of current I0 into a piece of dendrite. And we take that
piece of dendrite and we inject
current on one end, and we cut the other
end so that it's open. What does that produce
at the other end? So we have a wire that describes
the inside of the dendrite. We have a wire that describes
the outside of the dendrite. And if you cut the end of the
dendrite off so that they're-- it's leaky-- so
it's an open end-- what does that look
like electrically? Like what's the word for-- like those two wires
are touching each other. What's that called? They're shorts. If you cut the end
of a dendrite off, you've created a short circuit. The inside is connected
to the outside. So that's called an open
end boundary condition. And what can you say about
the voltage at this end? If the outside is
[AUDIO OUT] what can you say about the voltage
inside the dendrite at that end? AUDIENCE: [INAUDIBLE] MICHALE FEE: It's 0. Good. So we have injected current. We have V0, the
voltage at this end. And we know, if we
have an open end, that the voltage here is 0. Now we can write down. We know that the
initial voltage is V0. The voltage at position L is 0. And now you can-- you know that the current here
is equal to the current there, and you can write down
the equation and solve V0. So V0 is just the resistance,
the total resistance of the dendrite times
the injected current. And that Rin is known
as input impedance. It's just the resistance
of the dendrite. It tells you how
much voltage change you will get if you inject
a given amount of current. All right. Any questions about that? Let's consider another case. Rather than having an open end,
let's leave [? the end of ?] the dendrite closed so
that it's sealed closed. So we're going to consider
a piece of dendrite that, one end, we're
injecting current in, and the other end is closed. So what do you think
that's going to look like? It's called a closed end. What does that look like here? It's an open circuit. Those two wires are not
connected to each other. There's no resistance
between them. Let's say we define
the voltage here as V0. What can you say--
well, what you can say about the current
there is that the current is 0, because it's an open circuit. There's no current flowing. And so the current flowing
through this at this end is 0. Does that make sense? So what can you say about
the current everywhere? AUDIENCE: 0. MICHALE FEE: It's 0. And what can you say about
the voltage everywhere? It's V0. Exactly. So the voltage
everywhere becomes V0. And the input impedance? Anybody want to guess what
the input impedance is? How much-- what's the ratio
of the voltage at this end and the current at this end? AUDIENCE: Infinite? MICHALE FEE: It's infinite. That's right. So we're just trying to build
some intuition about how voltage looks
[AUDIO OUT] of distance for one special case, which
is a piece of dendrite of some finite length for which
you have no membrane currents. And you can see that the voltage
profile you get is linear, and the slope of it depends
on the boundary conditions, depends on whether
the piece of dendrite has a sealed end,
whether it's open. All right. So now we're going to
come back to the case where we have membrane
currents, and we're going to derive the
general solution to the voltage in a piece
of dendrite for the case where we have membrane
capacitance and membrane currents. All right. And I don't expect you to
be able to reproduce this, but we're going to derive
what's called the cable equation, which is the general
mathematical description, the most general mathematical
description for the voltage in a cylindrical
tube, of which-- that's what dendrites look like. So we're going to write down
that differential equation, and I want you to just
see what it looks like and where it comes
from, but I don't expect you to be able to derive it. All right. So let's come back to this
simple model that we started. We're going to put our model
for the membrane back in. Remember, that's a capacitor
and a conductance in parallel. We're going to-- we can write
down the membrane current, and we're going to have an
injected current per unit length. So Kirchoff's
current law tells us the sum of all of those currents
into each node has to be 0. So let's just write down-- let's just write
down an equation that sums together all of
those and sets them to 0. So the membrane current
leaking out minus that injected current coming in. They have positive signs because
one is defined as positive going into the dendrite,
and the other one is defined as
positive going out. So those two, the
membrane currents, plus the current going out this
way minus the current coming in that way is 0. So we're going to do the
same trick we did last time. We're going to
divide by delta x. So, again, membrane current per
unit length times the length of this finite element. We're going to
divide by delta x. So this thing right
here, i membrane minus i electrode, I guess, equals minus
1 over delta x I of x minus I of x minus delta x. So what is this? You've seen something
like that before. It's just a derivative. First derivative of I
with respect to position. So now what you see is
that the membrane current minus the injected current is
just the first derivative of I. So hang in there. We're going to substitute
that with something that depends on voltage. So how do we do that? We're going to take Ohm's law. There's Ohm's law. Let's take the derivative of
that with respect to position. So now we get the second
derivative of voltage with respect to position
is just equal to minus Ra times the first
derivative of current. And you can see we can just take
this and substitute it there. So here's what we get, that the
second derivative of voltage with respect to position is
just equal to the membrane or injected current coming into
the dendrite at any position. So the curvature of the
voltage, how curved it is, just depends on what's coming
in through the membrane. Remember, in the case where
we had no membrane current and no injected current,
the curvature was 0, d2V dx squared is 0, which,
if the curvature is 0, then what do you have? A straight line. Now, we're going to plug
in the right equation for our membrane current. What is that? That we know. It's just a sum of two terms. What is it? It's the sum of-- remember, this is going to be
the same as our soma model. What was that? We had two terms. What were they? The current through the
membrane in the model, in the Hodgkin-Huxley model is? What's that? AUDIENCE: [INAUDIBLE]. MICHALE FEE: Good. It's a capacitive current
and a membrane ionic current. So let's just plug that in. We're just going to substitute
into here the current through the capacitor
and the current through this conductance. That's just C dV dt
G times V minus EL. It's a capacitive part
and a resistive part. Now, the capacitance
is a little funny. It's capacitance per unit length
times the length of the element plus-- and the [AUDIO OUT] is
conductance per unit length times the length of
our finite element. Capacitance per unit length
and ionic conductance per unit length. And we're going to
plug that into there. We're first going to notice that
this E leak is just an offset, so we can just ignore it. We can just set it to 0. We can always add it
back later if we want. We divide both sides by the
membrane conductance per unit length to get this equation. And that's called
the cable equation. It's got a term with the
second derivative of voltage with respect to
position, and it's got a term that's the
first derivative of voltage with respect to time. That's because of the capacitor. And then it's got a term that
just depends on [AUDIO OUT].. Now, that's the most
general equation. It describes how the voltage
changes in a dendrite if you inject a
pulse of current, how that current will propagate
down the dendrite or down an axon. We're going to take
a simplifying case. Next, we're going to study the
case just of the steady state solution to this. But I want you to
see this and to see how it was derived just using
finite element analysis, deriving Ohm's law in a
one-dimensional continuous medium. And by plugging in the
equation for the membrane that includes the capacitive
and resistive parts, you can derive
this full equation for how the voltage changes
in a piece of dendrite. Now, there are a couple of
interesting constants here that are important-- lambda and tau. So lambda has units of length. Notice that all of
the denominators here have units of voltage. So this is voltage
per distance squared. So in order to have
the right units, you have to multiply
by something that's distance squared. This is voltage per
unit time, so you have to multiply by something
that has units of time. So that is the length
constant right there, and that is a time constant. And the length constant
is defined as 1 over membrane conductance. That's the conductance of the
membrane, through the membrane, and this is the axial
resistance down the dendrite. So this is conductance
per unit length, and this is resistance
per unit length. And when you multiply
those things together, you get two per unit length
down in the denominator. So when you put those
in the numerator, you get length squared. And then you take
the square root, and that gives you
units of length. The time constant is
just the capacitance per unit length divided by the
conductance per unit length. And that is the
membrane time constant, and that's exactly the
same as the membrane time constant that we
had for our cell. It's a property of the
membrane, not the geometry. So any questions about that? It was-- it's a lot. I just wanted you to see it. Yes, [INAUDIBLE]. AUDIENCE: Like, two
slides ago [INAUDIBLE] MICHALE FEE: This one, or-- AUDIENCE: One more
slide [INAUDIBLE].. MICHALE FEE: Yes, here. AUDIENCE: So when you plug that
in for the derivative of V, were we not assuming that there
was no membrane [INAUDIBLE]?? MICHALE FEE: No. That equation is still correct. AUDIENCE: OK. MICHALE FEE: It's-- voltage
is the derivative with respect to position as a function
of the axial current. AUDIENCE: OK. MICHALE FEE: OK? Remember, going back
up to here, notice that when we derive this
equation right here, we didn't even have to
include these membrane. They don't change anything. It's just Ohm's law. It's the voltage here
minus the voltage there has to equal the current
flowing through that resistor. Doesn't matter what other
currents-- whether current is flowing in other
directions here. AUDIENCE: OK. MICHALE FEE: Does
that make sense? The current through
that resistor is just given by the
voltage difference on either side of it. That's Ohm's law. So now we're going to
take a simple example. We're going to
solve that equation for the case of steady state. How are we going to
take the steady state? How are we going to
find the steady state version of this equation? Any idea? AUDIENCE: [INAUDIBLE]. MICHALE FEE: Good. We just set dV dt
to 0, and we're left with this equals that. So we're going to take
a piece of our cable, and we're going
to imagine that we take a piece of dendrite
that's infinitely long in either direction. And somewhere here in the middle
of it, we're going to inject-- we're going to put an
electrode, and we're going to inject current
at one position. So it's injecting
current at position 0. How many of you have heard of
a delta function, a Dirac delta function? OK. So we're going to
define the current as a function of position
as just a current times a Dirac delta function
of x, that just says that all the current is
going in at position 0, and no current is
going in anywhere else. So the Dirac delta
function is just-- it's a peaky thing that is
very narrow and very tall, such that when you integrate
over it, you get a 1. So we're going to go to
the steady state solution. And now let's write down that. So there's the steady
state cable equation. And we're going to inject
current at a single point. So that's what it looks like. Does anyone know the
solution to this? Notice, what this says
is we have a function. It's equal to the second
derivative of that function. Anybody know? There's only one
function that does this. It's an exponential. That's right. So the solution to this
equation is an exponential. V of position is V0, some
voltage in the middle, e to the minus x over lambda. Why do I have an absolute value? What is the voltage
going to look like if I inject current right here? You're going to have
current flowing. Where's the current going to go? If I inject current into the
middle of a piece of dendrite, is it all going to go this way? No. What's it going to do? It's going to go both ways. And the current-- the voltage
is going to be high here, and it's going to fall as
you go in both directions. That's why we have an
absolute value here. So the voltage is going
to start at some V0 that depends on how much
current we're injecting, and it's going to drop
exponentially on both sides. And notice what's right here. The lambda tells us
the 1 over e point, how far away the 1 over e
point of the voltage is. What that means is
that the voltage is going to fall to 1 over e
of V0 at a distance lambda from the side at which
the current is injected. Does that make sense? That is the steady
state space constant. It has units of length. It's how far away do you have
to go so that the voltage falls to 1 over-- falls to 1 over 2.7 of
the initial voltage. Any questions? It's pretty simple. We took an unusually
complicated route to get there, but that's the-- the nice
thing about that is you've seen the most general
solution to how a cable-- a dendrite will behave when
you inject current into it. So now we can
calculate the current as a function of position. Any idea how to do that? What-- if you know
voltage, what do you use to calculate current? Which law? AUDIENCE: Ohm's. MICHALE FEE: Ohm's law. Anybody remember what
Ohm's law looks like here? AUDIENCE: [INAUDIBLE]. MICHALE FEE: Yes. And we have to do
something else. The-- remember, the
current is what? Ohm's law in a continuous
medium, the current is just going to be
what of the voltage, the blank of the voltage? AUDIENCE: Derivative? MICHALE FEE: The
derivative of the voltage. So we're just going to take
this and take the derivative. That's it. dV dx is just equal
to minus R times I. So the current is proportional
to the derivative of this. What's the derivative
of an exponential? Just another exponential. So there we go. The current, and
then there's some-- you have to bring
the lambda down when you take the derivative. So the current is now just minus
1 over the axial resistance per unit length times
minus V0 over lambda-- lambda comes down when
you take the derivative-- times e to the
minus x over lambda. Notice, the current is to
the right on this side, so the current is
positive it's flowing to the left on that side,
so the current is negative. So to do this properly,
you'd have-- this is the solution on the right side. You'd have to write another
version of this for the current on the left side, but I
haven't put that in there. And, again, the current
starts out at I0, and drops exponentially,
and it falls to 1 over e at a distance lambda. Why is that? Because the current
is leaking out through the holes
in our garden hose. So as you go further down,
less and less of the current is still going
down the dendrite. I don't expect you to
be able to derive this, but, again, just know
where it comes from. Comes from Ohm's law. So I want to show you one really
cool thing about the space constant. It has a really
important dependence on the size of the dendrite. And we're going
to learn something really interesting about why
the brain has action potentials. So let's take a closer
look at the space constant, and how you
calculate it, and how it depends on the
size, this diameter, this radius of the dendrite. So we're going to take a
little cylinder of dendrite of radius a length little l. G sub m is the membrane
conductance per unit length. Let's just derive what
that would look like. The total membrane conductance
of this little cylinder of dendrite, little
cylinder of cell membrane, is just the surface area of that
cylinder times the conductance per unit area. Remember, this is the
same idea that we've talked about when
we were talking about the area of our soma. We have a conductance
per unit area that just depends on the
number of ion channels and how open they are on
that piece of membrane. So the total conductance
is just going to be the conductance per
unit area times the area. And the area of that
cylinder is 2 pi a-- that gives us the
circumference-- times the length, 2 pi al. And the conductance
per unit length is just that total conductance
divided by the length. So it's 2 pi a times g sub l,
the conductance per unit area. So that's membrane
conductance per unit length. The axial resistance
per unit length along this piece, this
little cylinder of dendrite, we can calculate
in a similar way. The total axial resistance
along that dendrite is-- can be calculated using this
equation that we developed on the very first day,
the resistance of a wire in the brain, the resistance
of a chunk of extracellular or intracellular solution. The resistance is
just the resistivity times the length
divided by the area. The longer-- for a given
medium of some resistivity, the longer you have to
run your current through, the bigger the resistance
is going to be. And the bigger the area,
the lower the resistance is going to be. So that total resistance is--
it has units of ohm-millimeters. So it's the resistivity
times l divided by A. In intracellular space, that's
around 2,000 ohm-millimeters. And the cross-sectional area
is just pi times a squared. So now we can calculate
the axial resistance per unit length. That's the total
resistance divided by l. So that's just
resistivity divided by A, which is resistivity divided
by pi radius squared, just the cross-sectional
area, and that has units of ohms per millimeter. So now we can calculate the
steady state space constant. Conductance per unit
length and axial resistance for unit length-- the
space constant is just 1 over the product of
those two, square root. We're just going to
notice that that's siemens per millimeter, ohms
per millimeter, inverse ohms. So those cancel, and you're
left with millimeter squared, square root, which
is just millimeters. So, again, that has the
right units, units of length. But now let's plug
these two things into this equation
for the space constant and calculate how
it depends on a. So let's do that. Actually, the first
thing I wanted to do is just show you what
a typical lambda is for a piece of dendrite. So let's do that. Conductance per area is around
5 times 10 to the minus 7, typically. So the conductance
per unit length of a dendrite, 6
nanosiemens per millimeter. You don't have to remember that. We're just calculating
the length constant. Axial resistance is--
plugging in the numbers for a piece of dendrite that's
about 2 microns in radius, the axial resistance
per unit length is about 60 megaohms
per millimeter. And so when you plug those two
things to calculate lambda, you find that lambda for a
typical piece of dendrite is about a millimeter. So that's a number that I would
hope that you would remember. That's a typical space constant. So if you inject a signal into
a piece of dendrite, it's gone-- it's mostly gone or about
2/3 gone in a millimeter. And that's how you
can have dendrites that are up in the range
of close to a millimeter, and they still are
able to conduct a signal from synaptic inputs
out onto the dendrite down to the soma. So a millimeter is a
typical length scale for how far signals propagate. So now let's plug in those-- the expressions that we derived
for conductance per unit length and axial resistance
for unit length of a into this equation for
the space constant. And what you find is
that the space constant is a divided by 2
times the resistivity times the membrane
conductance per unit area, per area to the 1/2. It goes as the square root. The space constant, the
length, goes as the square root of the radius. And notice that
the space constant gets bigger as you increase
the size of the dendrite. As you make a dendrite
bigger, what happens is the resistance down
the middle gets smaller. And so the current can go
further down the dendrite before it leaks out. Does that make sense? But the resistance
[AUDIO OUT] is dropping as the
square of the area, but the surface area is
only increasing linearly. And so the resistance
down the middle is dropping as the square. The conductance out the
side is growing more slowly. And so the signal
can propagate further the bigger the dendrite is. So that's why-- it's very
closely related to why the squid giant axon is big. Because the current
has more access to propagate down the axon
the bigger the cylinder is. But there are limits to this. So you know that, in
our brains, neurons need to be able to send signals
from one side of our head to the other side of our
head, which is about how big? How far is that? Not in Homer, but in [AUDIO OUT]
seen the cartoon with little-- OK, never mind. How big across is the brain? How many millimeters, about? Yes. Order of magnitude,
let's call it 100. So a piece of dendrite
2 microns across has a length constant
of a millimeter. How-- what diameter dendrite
would we need if we needed to send a signal across the
brain passively through a piece of-- a cylindrical piece
of dendrite like this? So lambda scales with radius. 2 microns diameter, radius,
gives you 1 millimeter. Now you want to go to-- you
want to go 100 times further. How-- by what factor larger
does the radius have to be? AUDIENCE: [INAUDIBLE]. MICHALE FEE: 10,000. Good. And so how big does our 2-micron
radius piece of dendrite have to be to send a
signal 100 millimeters? 10,000 times 2
microns, what is that? Anybody? AUDIENCE: [INAUDIBLE]. MICHALE FEE: 2 centimeters. So if you want to make
a piece of dendrite that sends a signal from
one side of your brain to the other 100
millimeters away, you need 2 centimeters across. Actually, that's the radius. It needs to be 4
centimeters. across. Doesn't work, does it? So you can make things-- you can
make signals propagate further by making dendrites
bigger, but it only goes as the square root. It's like diffusion. it's only-- it
increases very slowly. So in order to get a signal
from one side of your brain to the other with the
same kind of membrane, your dendrite would have to
be 4 centimeters in diameter. So that's why the
brain doesn't use passive propagation of
signals to get from one place to the other. It uses action potentials that
actively propagate down axons. Pretty cool, right? All right. So I want to just introduce
you to the concept of electrotonic length. And the idea is very simple. If we have a piece of dendrite
that has some physical length l, you can see
that that length l might be very good at conducting
signals to the soma if what? If-- what aspects
of that dendrite would make it very good at
conducting signals to the soma? AUDIENCE: [INAUDIBLE]. MICHALE FEE: So it's big. Or what else? AUDIENCE: Short. MICHALE FEE: It's got a
fixed physical length l, so let's think of
something else. AUDIENCE: [INAUDIBLE]. MICHALE FEE: Less leaky. Right. OK. So depending on the
properties of that dendrite, that piece of dendrite
of physical length l might be very good at
sending signals to the soma, or it might be very bad if
it's really thin, really leaky. So we have to compare
the physical length to the space constant. So in this case, there's
very little decay. The signal is able to propagate
from the site of the synapse to the soma. In this case, a slightly
smaller piece of dendrite might have a shorter
lambda, and so there would be more decay by the
time you get to the soma. And in this case, the
lambda is really short, and so the signal
really decays away before you get to the soma. So people often refer
to a quantity as the-- referred to as the
electrotonic length, which is simply the ratio of
the physical length to the space constant. So you can see
that in this case, the physical length is
about the same as lambda, and so the electrotonic
length is 1. In this case, the
physical length is twice as long as lambda,
and so the electrotonic length of that piece of dendrite is 2. And in this case, it's 4. And you can see that the amount
of signal it gets from this end to that end will go like what? Will depend how on the
electrotonic length? It will depend something
like e to the minus L. So a piece of dendrite that
has a low electrotonic length means that the synapse out
here at the other end of it is effectively very
close to the soma. It's very effective at
transmitting that signal. If the electrotonic
length is large, it's telling you that some
input out here at the end of it is very far away. The signal can't
propagate to the soma. And the amount of signal
that gets to the soma goes as e to the minus L,
e to the minus [AUDIO OUT].. So if I told you
that a signal is at the end of a
piece of dendrite that has electrotonic length
2, how much of that signal arrives at the soma. The answer is e to the minus
2, about 10%, whatever that is. So I want to tell
you a little bit more about the way people model
complex dendrites in-- sort of in real life. So most of the time, we're
not integrating or solving the cable equation. The cable equation
is really most powerful in terms
of giving intuition about how cables respond. So you can write
down exact solutions to things like pulses
of current input at some position, how the
voltage propagates down the dendrite, the functional
form of the voltage as a function of distance. But when you actually want
to sort of model a neuron, you're not usually integrating
the cable equation. And so people do
different approximations to a very complex dendritic
structure like this. And one common way
that that's done is called
multi-compartment model. So, basically,
what you can do is you can model the soma with this
capacitor-resistor combination. And then you can
model the connection to another part of the
dendrite through a resistor to another sort of
finite element slice, but we're gonna let the
slices go to 0 length. We're just going to model them
as, like, chunks of dendrites, that are going to be modeled
by a compartment like this. And then that can
branch to connect to other parts of the
dendrite, and that can branch to connect to
other parts of the model that model other
pieces [AUDIO OUT] So you can basically
take something like this and make it
arbitrarily complicated and arbitrarily close
to a representation of the physical structure
of a real dendrite. And so there are
labs that do this, that take a picture
of a neuron like this and break it up
into little chunks, and model each one of
those little chunks, and model the branching
structure of the real dendrite. And you can put in real ionic
conductances of different types out here in this model. And you get a gazillion
differential equations. And you can [AUDIO OUT]
those differential equations and actually compute,
sort of predict the behavior of a complex
piece of dendrite like this. Now, that's not my favorite
way of doing modeling. Any idea why that would be-- why there could be a better way
of modeling a complex dendrite? I mean, what's the--
one of the problems here is that, in a
sense, your model gets to be as complicated
as the real thing. So it would be-- it's a great
way to simulate some behavior, but it's not a great way of
getting an intuition about how something works. So people take simplified
versions of this, and they can take this
very complex model and simplify it even more by
doing something like this. So you take a soma
and a dendrite. You can basically just
break off the dendrite into a separate piece and
connect it to the soma through a resistor. Now, we can simplify
this even more by just turning it into
another little module, a little compartment, that's
kind of like the soma. It just has a capacitor,
and a membrane resistance, and whatever ion
channels in it you want. And it's a dendritic
compartment that's connected to the somatic
compartment through a resistor. And if you write that down,
it just looks like this. So you have a
somatic compartment that has a somatic
membrane capacitance, somatic membrane conductances,
a somatic voltage. You have a dendritic compartment
that has all the same things-- dendritic membrane capacitance,
conductances, and voltage, and they're just connected
through a coupling resistor. It turns out that
that very simple model can explain a lot of complicated
things about neurons. So there are some
really beautiful studies showing that this kind
of model can really explain very diverse kinds
of electrophysiological [AUDIO OUT] neurons. So you can take, for example,
a simple model of a layer 2/3 pyramidal cell that has a
simple, compact dendrite. And you can write
down a model like this where you have different
conduct [AUDIO OUT] dendrite. You have Hodgkin-Huxley
conductances in the soma. You connect them
through this resistor. And now, basically,
what you can do is you can model that
spiking behavior. And what you find is that if
you have the same conductances in the dendrite and in
the soma but you simply increase the area, the total
area of this compartment, just increase the total
capacitance and conductances, that you can see that-- and that would model
a layer 5 neuron that has one of these very
large dendrites-- you can see that the spiking
behavior of that neuron just totally changes. And that's exactly what the
spiking behavior of layer 5 neurons looks like. And so you could imagine
building a very complicated thousand-compartment
model to simulate this, but you wouldn't really
understand much more about why it behaves that way. Whereas [AUDIO OUT] a
simple two-compartment model and analyze it,
and really understand what are the
properties of a neuron that give this kind
of behavior as opposed to some other kind of behavior. It's very similar
to the approach that David Corey took in
modeling the effect of the T tubules on muscle fiber
spiking in the case of sodium-- failures of the sodium
channel to inactivate. That was also a
two-compartment model. So you can get a
lot of intuition about the properties
of neurons [AUDIO OUT] simple extensions of an
additional compartment onto the soma. And, next time,
on Thursday, we're going to extend
a model like this to include a model
of a [AUDIO OUT] So let me just remind you of
what we learned about today. So you should be able
to draw a circuit diagram of a dendrite, just that
kind of finite element picture, with maybe three or
four elements on it. Be able to plot the
voltage in a dendrite as a function of
distance in steady state for leaky and
non-leaky dendrites, and understand the concept
of a length constant. Know how the length constant
depends on dendritic radius. You should understand the
idea of an electrotonic length and be able to say
how much a signal will decay for a dendrite of a
given electrotonic length. And be able to draw
the circuit diagram of a two-compartment model. And we're going to spend more
time on that on Thursday.
