0:29

So in this first part we're going

Â to discuss something that's called the reaction-diffusion equation.

Â In the last lecture on action potentials we derived this so

Â called cable equation for voltage in excitable cells such as neurons.

Â And we said this is an example of a reaction-diffusion equation.

Â Now we're going to give a second example, which

Â is how you describe ionic concentrations within a cell.

Â How does ionic concentrations vary from one region

Â of the cell to another region in the cell?

Â We're going to derive this and then we're going to illustrate how this

Â corresponds very very nicely with the

Â cable equation that we encountered previously.

Â 1:08

This slide illustrates where we left off last time, in

Â the last lecture on the Hodgkin-Huxon model of the action potential.

Â We derive this equation here, which we called the cable

Â equation, and we discussed how this describes the movement in,

Â with, it's a function of time and space of voltage

Â in an excitable cell such as a, such as a neuron.

Â A partial derivative of voltage with respect to time is equal to some constants

Â here, second derivative of voltage with respect

Â to location minus the the ionic current.

Â And we said that this is a reaction-diffusion equation and what

Â I promised you is that these appear in, in other contexts.

Â We also just briefly noted how partial differential equations could be

Â solved numerically by converting into discrete form in both space and time.

Â That's point number two down here.

Â We're going to get to that in the third lecture on PDEs.

Â What we're going to focus on today, on the, in the, in this first lecture

Â on partial differential equations, are other examples

Â of reaction-diffusion that are encountered in biological systems.

Â 2:13

The specific example that we're going to consider in

Â this lecture is diffusion across an epithelial cell.

Â We're not going to get in depth of the biology here,

Â but the, the basics of the biological background here are

Â that your kidneys regulate your pH and your body, in

Â part by transporting bicarbonate across the epithelium in your kidney.

Â If you were to lose all your bicarbonate in the

Â urine, then your body would be less able to regulate pH.

Â It's for acid-base balance and so you're kidney needs to make

Â sure that that it doesn't lose, doesn't lose all your bicarbonate.

Â It needs to take it back up, and it takes

Â it back up by transporting the bicarbonate across an epithelial cells.

Â And so the example we're going to discuss is bicarbonate

Â ion concentration, that's HCO3 minus in the renal proximal tubule.

Â This is a particular region of the kidneys

Â where most of this bicarbonate transport takes place.

Â 3:08

And so the sort of simplest way to

Â illustrate this is that your epithelial cell has an

Â apex at the top of the cell and it has, the bottom has a base of the cell.

Â And the bicarbonate is going to move from

Â this region over here, which is the limit of

Â your, of your nephron, to this region over here,

Â which is your, the interstitial space of your body.

Â And from here it goes into the blood.

Â And so there's going to be transport of bicarbonate across the apical membrane.

Â Then there's going to be diffusion of bicarbonate from left to right here,

Â from the apex of epithelial cell to the base of the epithelial cells.

Â Then there's going to be transport across the basal membrane.

Â 4:09

And the question becomes, in this case, how do

Â we describe the movement of bicarbonate from apex to base?

Â When we're both in the cell here, how do

Â we discuss, how do we describe mathematically the way

Â that bicarbonate is moving form the left part of

Â your diagram here, to the right part of the diagram?

Â We're going to go through that derivation and we're going

Â to see how we end up with a reaction-diffusion equation.

Â It's very similar to what we saw in the last slideshow on action potentials.

Â It's very similar to the cable equation

Â for, for memory potential, for memory voltage.

Â The way we're going to go through this derivation of, of

Â diffusion of bicarbonate across an epithelial cell, is similar to the

Â way that we went through the derivation of how voltage will

Â vary as a function of time and location in our neuron.

Â remember, with the neuron derivation we took a long

Â axon and we chopped it up into discrete parts.

Â We're going to do the same thing here.

Â We're going to represent the cell, you epithelial

Â cell, as a series of discrete segments.

Â 5:08

So we can think of the epithelial cell like this.

Â We've got a cube here on the apical side.

Â And then you've got another cube next to it

Â and another cube next to it, et cetera, until

Â you get to your nth cube, which is the one that's on the basal end of the cell.

Â So you're taking your epithelium, which goes from the, from the apex down

Â to the base and you're chopping it up into end pieces and cubes.

Â And each cube will have a length of delta x in this case.

Â 5:37

And the terminology we're going to use the

Â the way we're going to define our variables

Â is we're going to say HCO3 sub i,

Â is a concentration of bicarbonate in sub-cube i.

Â So i in this case refers to the to the number that we're in, that we're at.

Â And we're going to have it introduce a, a

Â number here, a constant called capital D, sub HCO3.

Â This is the intracellular diffusion constant for bicarbonate.

Â And delta x, we just defined, this is the distance between adjacent sub-cubes.

Â And the, and again the question that we're

Â trying to get at here and what we're trying

Â to derive is this, what are the equations

Â that describe movement of bicarbonate from apex to base.

Â As we derive the equations to describe movement of

Â bicarbonate from the apex to the base, we're going to

Â do that same thing that we did when we

Â were deriving the cable equation for, for memory voltage.

Â We're going to first consider what happens within three representative

Â sub-cubes that are somewhere in the middle of the cell.

Â So we're going to consider the ith cube here HCO3th of i.

Â 6:47

The, i minus 1 cube over here, and i plus 1 cube over here.

Â And we're going to define fluxes that describe the

Â transport of bicarbonate from one cube to the next.

Â So the flux that goes from i minus 1 to i is going to be J, from i minus 1 to i.

Â And this flux is going to be J from i to i, i plus 1.

Â And I just, I separated the cubes here just for,

Â for clarity just so I could draw the arrows here.

Â But, of course, in a real cell they're going to to be right next to

Â one another like we saw on the, on the last slide on the last diagram.

Â We're going through just one more variable here.

Â Capital a refers to the cross-sectional area along, along this dimension.

Â 7:42

In the first two equations that we can write down are J from i

Â minus 1 to i is the the fusion constant here D HCO3 times

Â the concentration difference, bicarbonate, i minus 1,

Â minus bicarbonate at i, divided by the distance delta x.

Â And then we can write a completely analogous equation for J from i to i plus

Â 1 and these these equations describes Fick's first law of diffusion.

Â But tou probably haven't heard of Fick's first law of diffusion before, but

Â these equations make intuitive sense, if we think about things a little bit.

Â First of all, what happens if the concentration in

Â i minus 1 and the concentration in i are equal?

Â Well, you know intuitively that if two concentrations are the

Â same, there's not going to be any diffusion from one

Â to the other and that they're the same and this

Â difference will be zero and therefore the flux will be zero.

Â So it makes sense in that regard.

Â And intuitively what would happen if you had to

Â travel a great distance from i minus one to i?

Â Well, if it had to travel a great

Â distance, then you could probably deduce that there

Â wouldn't be that much flux and sure enough,

Â that's why delta x is in the denominator here.

Â So as the distance becomes greater as the ions have

Â to travel a greater distance, the, the flux gets smaller.

Â And then the diffusion constant here, you probably you know, haven't seen before.

Â But this just sort of refers to how easy it is for, for the ions to move.

Â If it's very, very difficult for them to move from, from one

Â cube to another cube, then the diffusion constant will be very small.

Â And if it's easy for the ions to move, then the diffusion constant will be big.

Â And therefore, it makes sense that the diffusion

Â constant comes in in the numerator of this equation.

Â As the diffusion constant gets bigger the flux gets bigger.

Â And this is the equation for flux from i minus 1 to i and then

Â this is the equation here from flux from i to i plus 1.

Â And what we want to do now is we

Â want to relate these fluxes to changes in bicarbonate concentration.

Â This is what we care about, after all, right?

Â How much does bicarbonate concentration change as a, as a function of time?

Â What is the derivative, the time derivative of bicarbonate concentration?

Â And how does that relate to the, to the

Â transport of bicarbonate from one cube to the other cube?

Â 10:07

Intuitively, we can say that this derivative

Â should depend on the inflow versus the outflow.

Â If we're looking at bicarbonate concentration in this cube

Â here, in cube i, then if you have a lot

Â coming in and you have none or just a tiny

Â bit coming out, then its concentration is going to go up.

Â On the other hand, if you don't have much coming in and

Â you have a lot going out, then concentration is going to go down.

Â So somehow or another, you're going to have to subtract these two.

Â But in order to really look at this in quantitative terms, we're going to

Â need to consider units in order to be able to express this precisely.

Â So now I'll introduce some of the units of

Â these terms, as they, as they come into this equation.

Â 10:49

Delta x is a distance, so that's in, in some sort of distance units.

Â In this case, we'll just use centimeters for the sake of argument.

Â But it could easily be, you could use meters.

Â You could use micrometers, whatever.

Â Bicarbonate concentration, because we're talking about biological ion

Â concentrations, we usually use millimolar for this one.

Â And millimolar means you know, 10 to the minus 3 moles per liter.

Â And that's equivalent to a micromole per centimeter cubed.

Â 11:16

Centimeter cubed, a lot of you probably know, is equal to a, a milliliter.

Â So, you're just taking the units here, milli- in front

Â of the molar becomes a micro in front of the molar.

Â And then, when you go from liters to milliliters, you're just taking

Â the numerator, numerator and the denominator

Â and dividing each one by a thousand.

Â So our bicarbonate concentration, our micromoles per centimeter cubed.

Â In the diffusion constant, diffusion constants in general

Â are in units of length squared per unit time

Â and the units we need here to get

Â everything to work out are centimeters squared per second.

Â So if we plug the units into this equation here, centimeters squared per second

Â micromole per centimeter cubed, and then centimeter in the denominator here.

Â Then we end up with units of micromole per centimeter squared per second.

Â So what the flux is, is it's saying, how much substance, how many moles

Â of something do you have, per unit time, and then also, per unit area.

Â 12:24

So if you have the same amount of stuff going across twice the area, then your

Â flux will decrease then you, you know, your

Â flux will be decreased be a factor of two.

Â So that's another normalization term, is, is a

Â cross-sectional area that your flux is travelling across.

Â So we have our fluxes in units of micromole per centimeter squared

Â per second, and we want to relate that to the derivative here.

Â And the derivative that we need to get in

Â units of micromoles per centimeters cubed per second, right.

Â Because this needs to be in units of concentration per times.

Â So this is the unit of concentration for time, right?

Â Micromole per centimeter cubed is a concentration and then

Â we have to have time in the denominator here.

Â So somehow or another we need to

Â get another centimeter in the denominator here.

Â 13:10

So what we're going to show next is how to relate that

Â to the, to the geometry of our cubes that define our epithelial cell.

Â As we just discussed, what we need to do is, we need to convert from micromole per

Â centimeter squared per second, to micromole per centimeter cubed per second.

Â So there's one more thing about how diffusion is

Â treated mathematically that we need to introduce now, which

Â is that if you want, we, we said that

Â a flux is normalized per centimeter square per per area.

Â So we need to multiply by the inter-cube cross-sectional area A.

Â Then that just tells us how much total stuff,

Â micromoles per second, is traveling into that, into that cube.

Â Then if we want to convert that into a concentration

Â change, what we do is we divide by the volumes.

Â And we need to divide by the volume of the ith cube Vi.

Â So the way that we relate our two fluxes to

Â the concentration change, the derivative of bicarbonate concentrations, we take

Â the difference between these two, we multiply them by the

Â cross sectional area A, and we divide by the volume b.

Â So remember that we need to get another centimeter in

Â the denominator in order to get our units to work out.

Â So we're taking something in the numerator that's going to be the nearest

Â centimeter squared and then in, in

Â the denominator we're going to have centimeter cubed.

Â And therefore we're going to have another centimeter in the denominator.

Â And our units are going to work out correctly.

Â But, remember, we're talking about a cube here.

Â Or not a cube, but, in general a, a, a parallel pipe guide.

Â It's you know, it's some sort of square, squar-ish or rectangl-ish shape.

Â So our volume is equal to our cross-sectional area times our distance.

Â That's would be true of any such that's how

Â you would calculate the volume of any such shape.

Â So because our volume is equal to A times delta x,

Â we're going to have A in the numerator here, and then

Â we're going to have A times delta x in the denominator, so

Â our cross-sectional area A is going to, is going to cancel out.

Â 15:33

Now if we plug in our equations for for the two fluxes.

Â Remember this is the concentration difference divided by

Â delta x, concentration difference divided by delta x.

Â Each one of these has diffusion constant in front of it.

Â We took the two diffusion, diffusion constants.

Â We factored that out, put that front here.

Â This difference, minus this difference each of those divided

Â by delta x, and then divided by delta x here.

Â In the next slide we're going to, we're going to

Â simplify this, and we're going to see how we can generalize

Â 16:10

What we want to do now with this equation is, we want to be able to generalize it.

Â What's the limit as delta x goes to 0?

Â We divided this into into discrete cubes, in

Â order to make it conceptually, you know, something that

Â we could wrap our heads around and in order

Â to be able to write down some straightforward equations.

Â But what we care about overall is what is true for all cubes on any arbitrary side.

Â What's the limit is delta X goes to 0?

Â Well, this term here.

Â One concentration minus another concentration

Â divided by the spatial difference.

Â The limit of delta x goes to 0 here is

Â our spatial derivative, the derivative of bicarbonate with respect to x.

Â What do we have in our, our equation that we were just looking at here?

Â We have a derivative here and a derivative here

Â divided again by the spatial distance, by delta x.

Â So we have a derivative minus a derivative, divided by distance.

Â We've got our second derivative here.

Â So this is, this equation here is is delta x goes to 0.

Â This term the limit there becomes the

Â second derivative of bicarbonate with respect to

Â x, and we can understand that when we consider just what's the first derivative.

Â Well, the first derivative is the

Â concentration difference divided by delta x.

Â So you've got one concentration difference, a

Â second concentration difference, again divided by delta x.

Â The limit as delta x goes to 0 of

Â this whole thing here, is going to be the second

Â 17:45

derivative of bicarbonate with respect to location.

Â Therefore, in the limit of small delta x, our overall equation becomes the

Â partial derivative of the bicarbonate with respect

Â to time, that's what we've previously had

Â on the left hand side of the equation, is equal to the diffusion constant,

Â times the second derivative of bicarbonate with

Â respect to x, with respect to location.

Â What we've just done here is we've derived a one-dimensional diffusion equation.

Â 18:42

What if we made this slightly more complicated where you could have

Â flux by carbonate into the cube, flux of bicarbonate out of the

Â cube, but then once the bicarbonate was within the cube, you could

Â have some rate constant k that

Â was consuming bicarbonate or degrading somehow.

Â This could be binding to a buffer.

Â In the case of bicarbonate, it's actually

Â convergent it's combining with hydrogen ions and

Â then being converted into water plus carbon

Â dioxide through an enzyme called carbonic anhydrase.

Â But we're not going to get into that level of biological detail.

Â Really what we care about here is what if you can have flux of this species

Â into the cube, flux out of the cube,

Â and some first order process that's consuming bicarbonate.

Â So, what we would have in that, in that case

Â is, are the derivative of bicarbonate with respect to time, is

Â this flux divided by delta x minus this flux divided delta x, minus

Â some term, rate constant k times the bicarbonate concentration.

Â 19:46

And again we want to look at this in a continuum limit.

Â Remember that as delta x goes to 0, this equation here simplifies to the

Â second partial derivative of bicarbonate with respect to location.

Â And then, again, minus k times HCO3.

Â And now we have a reaction-diffusion equation rather than just

Â the diffusion equation like we had on a previous slide.

Â And this should, this should look somewhat familiar to you.

Â So what we're going to conclude this first lecture

Â by comparing what we just derived here for

Â bicarbonate with respect to location, with respect to

Â time, with what we derived in the previous lecture.

Â 20:23

Let's look at our one dimensional cable

Â equation for voltage versus our epithelial reaction-diffusion equation

Â for the derivative, partial derivatives of bicarbonate with

Â respect to x and with respect to time.

Â Our cable equation looks like this,

Â capacitance times partial derivative of voltage with

Â respect to time is equal to these terms that are related to geometry.

Â Here are these constants times second derivative of voltage with

Â respect to x, minus i ion, minus the ionic current.

Â 20:53

Why do we call this the reaction-diffusion equation?

Â Well, this term here describes diffusion of

Â voltage or diffusion of charge in the neuron.

Â And this term here is a, is

Â a reaction that either increases or decreases voltage.

Â If you remember back to the lectures on the Hodgkin Huxley model,

Â when you have a negative ionic current going into the cell you're

Â going to increase voltage, when you have positive ionic current, which is

Â cation moving out of the cell, you're going to decrease the voltage.

Â So the ionic current can be thought of as

Â the reaction that will either increase voltage or decrease voltage.

Â And by analogy, what we have when we have diffusion of

Â bicarbonate across an epithelium, this is the equation that we just derived.

Â 21:40

This term here describes diffusion and this

Â term here describes a reaction that consumes bicarbonate.

Â So in either case, you have the same sort of general structure.

Â You have a term describing diffusion and

Â you have a term that described the reaction.

Â And in either case your, your diffusion term is

Â the second derivative of your species with respect to location.

Â Here we have d squared voltage over dx squared,

Â sorry, I left off the squared here, that's a mistake,

Â but we have d squared bicarbonate over dx squared, again

Â this is the second derivative of bicarbonate with respect to

Â 22:17

space.

Â So to summarize this lecture, what we've seen is that the

Â partial differential equation, the PDE that describes transport of an ion across

Â a cell, is analogous to the cable equation PDE that we encountered

Â in neurons and both of these PDEs are examples of reaction-diffusion equations.

Â What we're going to see next are some

Â more specific examples and we're going to talk about

Â some studies that have used these PDEs, use

Â these reaction-diffusion equations to gain new insight into biology.

Â