Mathematical modeling and multiscale simulations for vesicular release at neuronal synapses
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Transcript: English(auto-generated)
00:15
The work I'm going to present to you today is the first step of a big project which
00:21
consists of building a model and numerical simulations to study vesicular release at neuronal synapses. So the first step is about asymptotic analysis of the mean first passage time of a benign particle to a small hole. I will first present a short overview of synaptic transmission at chemical synapses,
00:46
very short, no problem, and then I will show you the asymptotic analysis of narrow escape time. So the synapse, maybe everybody knows what it is, is a structure not very well known today
01:02
in the functioning, that permit to transmit an electrical signal from one neuron, the presynaptic neuron, to another, the postsynaptic. So when an electrical signal arrives, action potential in red here, it triggers the opening of the voltage gate of calcium channels, which are the blue channels on this picture.
01:25
And this will trigger the entry of calcium ions within the terminal, the orange dot. And those calcium ions can go to bind to specific receptors localized on docked vesicles. So docked vesicles are the ones which are just opposed to the membrane.
01:44
And if they find those docked vesicles and the specific receptors on those docked vesicles, it will trigger the release of neurotransmitters within the synaptic cleft, and then it will, they can bind to the pink receptors, specific receptors for neurotransmitters, and the chemical cascade and everything will trigger the, will create an electrical signal in the postsynaptic neuron.
02:10
So it's a very complex phenomenon. I'm just today interested in what's happening in the presynaptic part when calcium ions enter through the calcium channel
02:21
and have to find the very small receptors located at the docked vesicle. OK, so just a zoom on this part. So the docked vesicles are really attached to the membrane through the SNR complex, which is a complex of protein. And on the SNR complex, you have the receptors for the calcium.
02:40
So calcium enters through the blue channels, then they diffuse. And if few calcium, like four to eight, find the receptors, then the synaptic vesicle will fuse with the membrane, the membrane of the vesicle with the membrane, and then everything will start for the transmission. So the model here that we did is calcium ions are bromine particles that diffuse freely in the terminal.
03:08
We know that particles, that calcium ions are charged particles, but we will neglect this at the first step. And the docked vesicle is just a sphere tangent to the plan. OK, here you have the sphere in green and the tangent plane in blue.
03:26
And the binding on the SNR complex is when the bromine particle will find the red cylinder, you can find here. So this is our model of these ions finding the complex.
03:41
Find the small cylinder here, which is the centre of the cylinder, is the tangential point between the vesicle and the plane. So we are now interested in the time and in the further step of the project, the probability that an ion starting from this channel will find this small cylinder.
04:05
So I will present now the asymptotic analysis we did for this narrow escape time. So the narrow escape problem we have here is the following. A bromine particle is described by a stochastic equation, x dot equals square root of 2d w dot.
04:26
d is the diffusion coefficient of the particle and w dot is white noise. And the first time to exit the main here, omega bar, which represents the domain we are interested in, through the small hole in red, the cylinder in red in this picture, is given by this.
04:43
So the first time the particle finds the domain. OK, the particle is reflected everywhere on the border of the domain, except at the red cylinder where it's absorbed. So we are interested in the mean first passage time to this cylinder, which is u of x.
05:04
And we know, it's very well known, it's Dinkin, that this mean first passage time follows the solution of the mixed boundary value problem, which is here d laplacian u equals minus 1, d is diffusion coefficient, du dn equals 0 is a classical reflection at the boundaries,
05:23
and u equals 0 is a classical absorption at the red cylinder here. So we are going to try to find the asymptotic analysis of this time.
05:40
OK, so the first stuff we can see when we look at this domain is that it's invariant per rotation around the axis delta here, which is very easy to see. According to the equation we have, we know that this problem is independent of theta,
06:03
if we consider the cyclical coordinates around delta. So we can just study an equivalent problem in domain omega, which is a 2D domain, the projection of the domain on theta equals 0, for example. So on this domain, you see the blue is going to the blue, the red sphere is here,
06:24
the red arc of circle, the small cylinder is now just a segment. And the equation becomes the following equation, where you recognize here the laplacian in
06:42
cylindrical coordinates very simply. So I'm now in a 2D domain with a 2D problem, which is equivalent to the problem we had in 3D. You notice as well that I put the diffusion coefficient on the right, and that we have
07:01
reflection on the blue, black, and green boundaries, and absorption at the red segment. We are interested in the asymptotics of u when epsilon, which is the length of this segment, goes to 0. Very small, goes to 0, is very small in front of the rest of the boundary.
07:22
So the problem we have here is that we have a singularity, because when epsilon goes to 0, the point here is not regular. So to desingularize this singularity, we will have to... Okay, the problem of this singularity is that it
07:41
prevents us to use a classical method, which are, for example, the Green function on this domain, because we don't know the order of the Green function in epsilon and the singularity, this depends on the domain, and we don't have access to this. The other method, which is very classical, is matching asymptotics.
08:00
So we try to find a solution near the hole, we try to find a solution in the domain, we try to match, then we can obtain something. But here, because of the geometry, it's not possible as well. So the idea is to desingularize the singularity. And we are using conformal maps. So here, because we have tangent circles, we will use the inverse
08:25
map, which comes very easily when you start to think about it, because it will transform these domains into the domain tilde omega we have here, because the inverse map will transform the sphere tangent in 0 into parallel straight lines.
08:42
You notice that the small hole here in red is going to a straight line here, and you know that inverse maps are putting spheres, a circle into straight lines and a straight line into a circle. So in reality, the hole here is not a straight line,
09:02
it's part of a circle, but the radius of this circle is 1 over 2 rA, which is here the abscissa of the hole. And if you maybe don't remember, because I didn't tell you this before, rA is in order of square root of epsilon. It's a simple computation.
09:27
So sorry, this means that this, at the first order, we can consider that this part is a straight line. The first order is a straight line. So putting setting v of st equal u of
09:43
rdv, we can now consider the equation in the map domain, which is much simpler domain, because at the first order, it's like rectangular domain. The equation is much more complicated, because it captures the geometry of the problem in omega, but still the domain now is much easier,
10:01
and we get this equation. We have the reflection dv dn, again on the blue, black and green boundaries, and the absorption on the red boundary. So to solve this equation, we will now use a scaling method. So we put zeta equal s square root of 2r epsilon,
10:31
square root of tilde epsilon, which means that we are going, we are taking
10:40
our interest is in the region very close to the hole, so very close to the red boundary. So by setting this, using the scaling, we can put big y equals of psi t equals v of st, very simple, and then we can do a regular expansion of big y in power of epsilon tilde.
11:05
Injecting this regular expansion into the equation and expanding the equation now, we obtain here the asymptotic of the equation in power of epsilon, and we will now take care of the terms of leading of much interest, leading other terms of the
11:20
equation. The first one is in one over tilde epsilon square, then the second one is one over epsilon, et cetera, et cetera. So using this expansion now, we are looking, because we are looking for asymptotic analysis, we are looking of the leading order term, looking for the other term, sorry. We will first consider the first term, which is
11:41
the second derivative of y0 over t equals 0. So using this equation and the boundary condition we have on the blue and green boundaries, we can show very easily that y0 is independent of t.
12:02
This is what we get from the first term. The second term, the second order term gives us this equation here and integrating over t again, we can show that y1 is as well independent of t. It's simple computation. And using this,
12:21
we obtain an equation for y0 in zeta. And using the boundary condition now at the hole, which says that in zeta equal 1, the solution equals to 0, we obtain the form of the leading order term of the solution, big Y0, near the hole. Going back into our first parameters,
12:48
we have that V of st take this form. So it's equals to 0 near the hole. This solution is an approximation, not an approximation, it's the leader order term of the solution near the hole.
13:03
Our problem is we have to compute A. So usually you use the rest of the parameters, but this approximation of the solution is not good, far from the hole. We cannot use the reflection, the last boundary condition we get. So to compute A,
13:23
we will use the divergence theorem, which says that the integral of bar omega of Laplacian mu is equal to the integral of the border, a very classical equation. So this, the Laplacian mu equals minus the volume of tilde omega over d just comes from the equation, d Laplacian
13:44
mu equals minus 1. And the other part, because we have reflection everywhere except at the hole, says that this integral on the border is exactly the integral of the solution at the hole. It's good because we have a solution at the hole. So we can then, using
14:03
our first order term of the solution at the hole, give this equation and then find A. And we now have the first order term, the leading order term of the solution in epsilon.
14:22
For the whole domain. So to get tau, to get the mean first passage time, sorry, we just put the value of A in V of s t, and then consider when s is equal to zero, when we are far from the hole, s equals to one at the hole, s equals to zero for the hole,
14:42
and we get the mean first passage time for our Bernoulli particle to the small hole. In the boundary layer, we also have the leading order term of the mean first passage time, and we know as well the way it behaves. Okay, so we computed the leading order term of the mean first passage time to a small ribbon located between two tangents first. And we
15:05
could show that this mean first passage time is constant outside of the boundary layer, as we just noticed. And as well, this tells us that this mean first passage time is well approximated by a Poisson process. So this is a result we already know. So it means that we can
15:27
consider that the arrival of this particle to this small hole is Poissonian. And the first approximation, the law of arrival, is an exponential. This is very interesting because
15:44
using this analysis in further development of the project that I don't have the time to present here, we could compute as well the probability when the particles start inside the boundary layer, because here we have the mean first passage time to go to the hole when you are
16:04
far from the hole, because we set s equals to zero, so it means we are far. What's happening when you are close? The problem is still a problem. So using the same kind of analysis, what we have here, we could compute the probability to reach the cylinder before escaping
16:21
the domain. Okay, it's the same kind of computation. And using this, we could build a model of the active zone, which is the place where vesicle are located and the calcium channel are located as well, to investigate the influence of the position of the channels on the crowding of vesicles or not crowding of vesicles, etc, etc, on the release probability,
16:44
which is the probability to have diffusion. So the probability to have four ions that are coming to this vesicle or not. And using this, we could build a model of the presynaptic terminal, much bigger, with the whole stuff we have over there.
17:03
And the fact that this mean first passage time is Poissonian is really what is important here in our study, because if you want to model the presynaptic terminal, you have a Brownian particle. You can do Brownian simulations if you want, it's very simple. Not that simple,
17:22
because the domain is not that fun, but okay, you can do it. But you will wait for hours because of the very small hole you have here. And because of the very long time you have to find this small hole. The fact that it's a Poisson process allows you to use simply Gillespie algorithm to say, okay, it's Poissonian. I can at least use Markov chain for what's happening, because
17:45
it's constant. It doesn't depend on the position. So it doesn't depend on the time. And it's what we did in the following of this project. And I wanted to show you this because it's the very first step, but everything is going after that, because we could show that the mean first
18:07
pass time is independent of the position and that this time is Poissonian time. So this work was published in Siam political modeling simulations. So I want to thank just my PhD
18:22
supervisor David Altman and my lab mates and the organizer of this conference for giving me the opportunity to present my work today. Thank you.
18:41
I was just wondering what you mean by it's well approximated by Poisson process. To me, the first pass time is approximately exponentially distributed because of this. Because it doesn't depend on the position, in fact. So it doesn't depend when you start.
19:07
The mean and the variance is the same, and it's well approximated. Any other questions? Well, if not, let's thank the speaker again.
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