It is not widely appreciated that the intercellular permeability is strongly influenced by the distribution of gap junctions in the intercellular membrane, although it is a common observation in introductory biology texts that there is a similar relationship between the distribution of stomata on leaves and the rate of evaporation of water through the leaf surface. Individual gap junctions are usually found in aggregates forming larger junctional plaques, as individual gap-junction particles are not easily distinguished from other nonjunctional particles. However, numerical simulations show that the permeability of the intercellular membrane decreases as the gap junction particles aggregate in larger groupings. This raises the intriguing possibility that intercellular permeability may be lowest when the gap-junctional plaques are easiest to see. This in turn provides a possible explanation for the fact that it has been difficult to establish a direct link between the number of recognizable gap junctions and the intercellular permeability.

Chen and Meng (1995) constructed a cubic lattice model of a two-cell system with a common border. A number of gap-junction particles, with varying degrees of aggregation, were placed on the border lattice points. Marker particles were placed in one of the cubes and followed a random walk over the lattice points of the cube. When they encountered a gap-junction lattice point on the boundary, there was an assigned probability that the marker particle would move across to the other cell. By measuring the time required for a certain percentage of marker particles to cross from one cell to the other, Chen and Meng obtained a quantitative estimate of the efficiency of intercellular transport as a function of gap-junction aggregation. Their results are summarized in Fig. 7.14. When the gap junctions are clumped together in a single junctional plaque, 10,000 time steps were required for the transfer of about 10% of the marker particles. However, when the gap-junction particles were randomly scattered, only 1,000 time steps were required for the same transfer. The magnitude of this discrepancy emphasizes the fact that gap junction distribution can have a huge effect on the rate of intercellular transport.

To get an analytical understanding of how the distribution of gap junctions affects the diffusion coefficient, we solve a model problem, similar to the one-dimensional problem solved in Section 7.2.1. We consider cells to be two-dimensional rectangles, with a portion of their ends open for diffusive transport (the gap junctions) and the remainder closed (Fig. 7.15A). The dashed lines in this figure are lines of symmetry across which there is no flux in a steady-state problem, so we can reduce the cell configuration to that shown in Fig. 7.15B.

To study diffusion in the x-coordinate direction, we assume that the vertical walls are separated by length L and have regularly spaced openings of width 25 with centers separated by length 2l. The fraction of the vertical separator that is open between cells is A = 5/l. To study how the distribution of gap junctions affects the diffusion coefficient, we hold A fixed while varying l. When l is small, the gap junctions are small and uniformly distributed, while when l is large, the gap junctions are clumped together into larger aggregates; in either case the same fraction (A) of the intercellular membrane is occupied by gap junctions.

Suppose that there are a large number of cells (say N) each of length L connected end to end. We impose a fixed concentration gradient across the array and use the definition (7.53) to define the effective diffusion coefficient for this array.

To find the flux, we solve Laplace's equation subject to no-flux boundary conditions on the horizontal lines y = 0 and y = l and on the vertical lines 5 < y < l,x = pL,p = 0,...,N. We further divide this region into two subregions, one fory > 5 and one for y < 5.

Percentage of signal molecules in target cell

Figure 7.14 Simulation results of the cubic-lattice gap-junction model on a 50 x 50 x 50 lattice with 1000 signal molecules in the source cell at time 0. In A, 100 gap-junction particles are arranged in a compact junctional plaque, while in B they are scattered randomly on the intercellular interface. The random scattering of gap-junction particles results in a greatly increased intercellular transfer rate (note the different scales for the two panels). (Chen and Meng, 1995, Fig. 1.)

Figure 7.15 A: Sketch of a single rectangular cell with gap-junctional openings in the end faces. B: Sketch of cell array, reduced by symmetry to a single "half-channel."

Consider first the solution on the upper region. The solution for a single cell 0 < x < L can be found by separation of variables to be

^ /nnx\ /nn(y - l)\ (x,y) = 2an cos^-^J cosh I -l- )•

This solution satisfies no-flux boundary conditions at y = l and at x = 0,L. Notice also that this solution is periodic, so it is a contender for the solution for any cell.

Now recall that in the one-dimensional case, the solution is piecewise linear, with jumps at the cell boundaries, and that the slope of the solution within each cell is the same (Section 7.2.1). This suggests that the derivative of the solution in the two-dimensional case should be the same in each cell, or equivalently, the solution in each cell should be the same up to an additive constant. Thus, w(x'y)=n=a ^—l)gcos (L(x—pL))+ap (7.80)

for pL < x < (p + 1)L, 8 < y < l, and p = 0,...,N — 1. We have scaled the unknown constants by cosh(nn(8 — l)/L) for convenience.

On the lower region, a similar process gives

u(x,—) = (Ui - Uo)n-L + U0 + £ coshLnL) C sin ~ (7.81)

for 0 < x < NL, 0 <y < &■ Notice that this solution satisfies a no-flux boundary condition at y = 0 and has the correct overall concentration gradient.

Now, to make these into a smooth solution of Laplace's equation we require that u(x,y) and uy(x,y) be continuous at y = S. This gives two conditions,

2_^An cos ^L(x — pL)) = £ [Cn sin —— > + (U1 — Uo)— + Uo — ap (7.82)

2_^nAn tanh y—(8 — l)jcosy—(x — pL)J = ^ 2n tanh^-— Cn sin——— (7.83)

We now determine ap by averaging (7.82) over cell p. Integrating (7.82) from x = (p — 1)L to x = pL gives

since all the trigonometric terms integrate to zero. Hence, v — 1/2

Finally, for convenience, we choose U0 = N/2 and U1 = —N/2, which gives ap = p + (1 + N)/2. Since this is a linear problem, the values chosen for U0 and U1 have no effect on the effective diffusion coefficient.

To obtain equations for the coefficients, we project each of these onto cos kp by multiplying by cos -Lx and integrating from 0 to L. We find that

and where

2 -=i kA- tanh ( l(5 - l) j f = 2n tanh -f- (C„I2„,^), (7.87)

In- = j sin '-^p- cos'^¿x = — ( -—^—^— ) . (7.89)

There is an immediate simplification. Notice that I2nkk = 0 and Fk = 0 when k is even. Thus, Ak = 0 for all even k. Now we eliminate the coefficients Ak from (7.86) and (7.87) to obtain

In these terms, the average flux is

It follows that the effective diffusion coefficient is

L ^ , 2nnlA \ De = DljJ2Cn tanh —l— + A j . (7.92)

Since k can take on any odd positive integer value, (7.90) is an infinite set of equations for the coefficients Cn. Since the solution of the differential equation converges, we can truncate this system of equations and solve the resulting finite linear system numerically. Typical results are shown in Fig. 7.16A, where the ratio De/D is shown plotted as a function of l/L for different values of fixed A = S/l, and in Fig. 7.16B, where De/D is shown plotted as a function of A for fixed l/L.

There are a number of important observations that can be made. First, notice that in the limit A ^ 1, or l/L ^ to, f-Cn tanh 2nfA ^ 0. Thus, lim De = D, (7.93)

Figure 7.16 A: Effective diffusion ratio De/D as a function of the distribution ratio l/L for fixed gap-junction fraction A. B: Effective diffusion ratio De/D as a function of the gap-junction fraction A for fixed distribution ratio l/L.

Finally, from the numerical solution, it can be seen that De is a decreasing function of l/L. Thus, clumping of gap junctions lowers the effective diffusion coefficient compared with spreading them out uniformly.

From Fig. 7.16B we see that when gap junctions are small but uniformly spread, there is little decrease in the effective diffusion coefficient, unless A is quite small. Thus, for example, with A = 0.01 (so that gap junctions comprise 1% of the membrane surface area) and l = 0, the effective diffusion coefficient is about 86% of the cytoplasmic diffusion. On the other hand, with only one large gap junction with A = 0.01 in the end membrane of a square cell (l = 0.5), the effective diffusion coefficient is reduced to about 27% of the original.

It is interesting to relate these results to the one-dimensional solution. This two-dimensional problem becomes effectively one-dimensional in the limit L ^ 0, with a piecewise linear profile in the interior of the cells and small boundary or corner layers at the end faces. In this limit, the effective diffusion coefficient satisfies

De A

with x = 0.0016. This formula was found by plotting the curve A(DD — 1) against A, which, remarkably, is numerically indistinguishable from the straight line x(1 — A). Comparing this with the one-dimensional result, we find that the end-face permeability can be related to the fraction of gap junctions A through

7.3 EXERCiSES

1. (a) Verify the last step of (7.2). Hint: Use Stirling's formula n! & n"+1e "y ", and show thatlim^ ln[(1 - ")"] =-x.

(b) Verify that the Poisson distribution P(k) = e mkTk has mean m, by verifying that {k) = E^g kP(k) = m. '

(c) Verify that the sum of k identical Gaussian distributions with mean / and variance a2 is a Gaussian distribution with mean k/ and variance ka2.

2. This question is based on the model presented by Peskin (1991). Motivated by the smallness of ci with respect to ce, simplify the Llinas model by setting ci & 0. How much difference does this make to the Ca2+ currents plotted in Fig. 7.5? Calculate the steady-state Ca2+ current as a function of V and show that it is bell-shaped. Solve for a general step in voltage from V1 to V2 and demonstrate synaptic suppression.

3. Calculate the analytic solution to (7.8) when V is a given function of t.

4. Construct a simple function F(t) with the same qualitative shape as the function W(t) used in the Magleby and Stevens model, and calculate the analytic solution to (7.43) for that F. Compare to the numerical solutions shown in Fig. 7.11.

5. In the Magleby and Stevens model, a simple choice for the release function f (t) results in end-plate conductances with considerable qualitative similarity with those in Fig. 7.11. Suppose there is a sudden release of ACh into the synaptic cleft at time t = 0. We take f (t) = yS(t), where S is the Dirac delta function. Show that the resulting differential equation is dc

for t > 0. Solve for c, and, assuming that y is small, substitute this expression for c into the differential equation for x, (7.39). Look for a solution for x of order y, and show that ypn

which is always positive. Sketch the solution.

6. Peskin (1991) presented a more complex version of the Magleby and Stevens model. His model is based on the reaction scheme

ki P

where E is some enzyme that degrades ACh in the synaptic cleft. (The assumption of enzymatic degradation of ACh is one of the ways in which the Peskin model differs from the Magleby and Stevens model. The other difference is that the Peskin model does not assume that the amount of ACh bound to its receptor is negligible.) Write down the equations for the 6 dependent variables. Use conservation laws to eliminate two of the equations.

Then assume that the reactions involving ACh with R, and ACh with E (with reaction rates ki,i = 1,4), are fast to obtain expressions for [R] and [E] in terms of the other variables. Substitute these expressions into the differential equations for [ACh • R*] and [ACh] + [R] — [E] to end up with two differential equations in [ACh • R*] and [ACh]. Solve these equations when the stimulus is a small sudden release of ACh (i.e., assume that rT = eS(t) and look for solutions of O(e)), and show that the solution has the same form as (7.98) but that the exponential coefficients are given by the roots of a quadratic polynomial. What is the rate of ACh degradation?

7. Solve the above exercise (and obtain the same solution!) by nondimensionalizing, finding a small parameter, and then solving in terms of an asymptotic expansion. Hint: One method is to nondimensionalize time by y and let y/k2 = e be the small parameter. To lowest order in e one gets only three equations for four unknowns, and so to solve the lowest-order problem completely it is necessary to go to the higher-order terms. The differential equations obtained at higher order must then be added in the appropriate manner (as in the previous question) so that unwanted terms cancel.

8. Solve (7.49) and plot the solution (Peskin, 1991). What is the solution as y ^ 0? What is the slope of the solution at t = 0? Compare to the curve d in Fig. 7.8.

9. By linking the output of the Llinas model to the input of the single-domain/bound-calcium model, and then linking this to the input of the Magleby and Stevens model (the rate of production of ACh) construct a unified model for the synaptic cleft that connects the presynaptic action potential to the postsynaptic voltage via the concentration of ACh in the synaptic cleft. Solve the model numerically and compare to the simpler model presented briefly in Fig. 7.8.

10. Incorporate the effects of nicotine into a model of ACh activation of receptors.

11. Calculate the effective diffusion coefficient De for a periodic medium with periodic diffusion coefficient D(x).

assuming that D is periodic with a basic spatial subunit ^ of total volume V. Show that u = U(x) + eW(x) • VU(x) + O(e2), where U(x) satisfies the averaged Poisson equation

Answer:

where the period of D(x) is P. 12. Use homogenization arguments to solve the Poisson equation

and where the effective diffusion coefficient is and where the effective diffusion coefficient is

CHAPTER 8

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