Quasi-steady-state reduction of a model for cytoplasmic transport of secretory vesicles in stimulated chromaffin cells

Abstract

Neurosecretory cells spatially redistribute their pool of secretory vesicles upon stimulation. Recent observations suggest that in chromaffin cells vesicles move either freely or in a directed fashion by what appears to be a conveyor belt mechanism. We suggest that this observation reflects the transient active transport through molecular motors along cytoskeleton fibres and quantify this effect using a 1D mathematical model that couples a diffusion equation to advection equations. In agreement with recent observations the model predicts that random motion dominates towards the cell centre whereas directed motion prevails in the region abutting the cortical membrane. Furthermore the model explains the observed bias of directed transport towards the periphery upon stimulation. Our model suggests that even if vesicle transport is indifferent with respect to direction, stimulation creates a gradient of free vesicles at first and this triggers the bias of transport in forward direction. Using matched asymptotic expansion we derive an approximate drift-diffusion type model that is capable of quantifying this effect. Based on this model we compute the characteristic time for the system to adapt to stimulation and we identify a Michaelis–Menten-type law describing the flux of vesicles entering the pathway to exocytosis.

Appendices

Inner problem

Equating terms of equal order in (6) we find the following equations relating terms in the expansions of c, f and g.

$O\left( \varepsilon ^{-2} \right)$: We find that

$$\begin{aligned}&f_0=\alpha c_0 \; , \end{aligned}$$

(36)

$$\begin{aligned}&g_0=0 \; , \end{aligned}$$

(37)

i.e. the ratio of the leading order terms, free vesicles versus transported vesicles, is fixed and at the leading order the differential term vanishes.

$O\left( \varepsilon ^{-1} \right)$: Equating terms of order $\varepsilon ^{-1}$, the differential term at first order is given by

$$\begin{aligned} g_1=\nu c_0-\gamma f_0-a f_0' \; , \end{aligned}$$

(38)

and the ratio of free vesicles versus transported ones is given by $\alpha$ even at first order since

$$\begin{aligned} \alpha c_1-f_1=a g_0'= 0 \; , \end{aligned}$$

(39)

where we used (37).

O(1): From the first two systems in (6) we obtain

$$\begin{aligned}&\partial _t c_0 = c_0'' - (\alpha c_2 - f_2) + \gamma g_1 \; , \end{aligned}$$

(40)

$$\begin{aligned}&\partial _t f_0 + a g_1'+ b f_0' = (\alpha c_2 - f_2) - \gamma g_1 \; , \end{aligned}$$

(41)

which implies, taking the sum and using (36) and (38), that

$$\begin{aligned} \partial _t c_0 +c_0' \frac{a \nu +\alpha (b-a \gamma )}{1+\alpha }= \frac{1+a^2 \alpha }{1+\alpha }c_0'' \; . \end{aligned}$$

(42)

From the third system in (6) we obtain

$$\begin{aligned} g_2=\nu c_1-\gamma f_1-a f_1' \; , \end{aligned}$$

(43)

where we used (37). Finally, (40) combined with (42) shows that

$$\begin{aligned} \alpha c_2 - f_2 = -\partial _t c_0+ c_0'' + \gamma g_1 =-\partial _t c_0+\frac{a \nu +\alpha (b-a \gamma )}{1+a^2 \alpha } c_0'+\gamma g_1 \; . \end{aligned}$$

(44)

$O(\varepsilon )$: Likewise, taking the sum of the $O(\varepsilon )$ terms in the equations for c and f in (6) and using (39) and (43) we obtain

$$\begin{aligned} \partial _t c_1 +c_1' \frac{a \nu +\alpha (b-a \gamma )}{1+\alpha }= \frac{1+a^2 \alpha }{1+\alpha }c_1'' \; . \end{aligned}$$

(45)

Left boundary layer

We focus on an expansion up to leading order only, i.e. $c \approx \bar{c}_0 + O(\varepsilon )$, $f \approx \bar{f}_0 + O(\varepsilon )$ and $g \approx \bar{g}_0 + O(\varepsilon )$, and obtain

$$\begin{aligned} 0&= \bar{c}_0'' - (\alpha \bar{c}_0 - \bar{f}_0) \; , \end{aligned}$$

(46)

$$\begin{aligned} a \bar{g}_0'&= ( \alpha \bar{c}_0 - \bar{f}_0) \; , \end{aligned}$$

(47)

$$\begin{aligned} a \bar{f}_0'&= - \bar{g}_0 \end{aligned}$$

(48)

with boundary conditions

$$\begin{aligned}&\bar{f}_0(t, 0)+\bar{g}_0(t, 0)=0 \; , \end{aligned}$$

(49)

$$\begin{aligned}&\bar{c}_0(t, 0)=1 \; . \end{aligned}$$

(50)

Taking a derivative of (48) and coupling with (47) and (46) respectively, we find that $-a^2 \bar{f}_0'' = a \bar{g}_0' = \alpha \bar{c}_0-\bar{f}_0=\bar{c}_0''$. Hence $-a^2 \bar{f}_0=\bar{c}_0 + A x+B$ for two constants A and B. Since in the far field ($\bar{x} \rightarrow \infty$) we expect that the following matching conditions hold, $\bar{c}_0(t, \bar{x}) \rightarrow c_0(t, x=0)$ and, as a consequence of (36), $\bar{f}_0 (t, \bar{x}) \rightarrow \alpha c_0(t, x=0)$ we obtain $A=0$ and $-a^2 \alpha c_0(0) =c_0(0) +B$, i.e. $B=- c_0(0) (1+\alpha a^2)$.

From this we conclude $\bar{f}_0(t, \bar{x})= \frac{-1}{a^2} \left( \bar{c}_0(t, \bar{x}) - c_0(t, 0) (1+\alpha a^2) \right)$ which we substitute in (46) to obtain an equation for $\bar{c}_0$,

$$\begin{aligned} \bar{c}_0''&=\alpha \bar{c}_0 -\bar{f}_0\\&=\alpha \bar{c}_0 + \frac{1}{a^2} \left( \bar{c}_0 - c_0(t, 0) (1+\alpha a^2) \right) \\&= \left( \bar{c}_0 - c_0(t, 0)\right) \left( \alpha + \frac{1}{a^2} \right) \; . \end{aligned}$$

Coupling this with (50) we obtain

$$\bar{c}_0=c_0(t, 0) + (1-c_0(t, 0)) \exp \left( -x \sqrt{\alpha + \frac{1}{a^2}} \right) \; ,$$

$$\begin{aligned} \bar{f}_0&= \frac{-1}{a^2} \left( \bar{c}_0 - c_0(t, 0) (1+\alpha a^2) \right) \\&= \frac{-1}{a^2} \left( \bar{c}_0 -c_0(t, 0)-\alpha a^2 c_0(0) \right) \\&= \frac{-1}{a^2} \left( (1-c_0(t, 0)) \exp \left( -x \sqrt{\alpha + \frac{1}{a^2}} \right) -\alpha a^2 c_0(t, 0) \right) \\&= \frac{-1}{a^2} (1-c_0(0)) \exp \left( -x \sqrt{\alpha + \frac{1}{a^2}} \right) +\alpha c_0(t, 0) \; , \end{aligned}$$

and

$$\begin{aligned} \bar{g}_0&=-a \bar{f}_0' = -a \left( \frac{-1}{a^2} (1-c_0(t, 0)) \exp \left( -x \sqrt{\alpha + \frac{1}{a^2}} \right) + \alpha c_0(t, 0) \right) '\\&= - a \left( \frac{1}{a^2} (1-c_0(t, 0)) \sqrt{\alpha + \frac{1}{a^2}} \exp \left( -x \sqrt{\alpha + \frac{1}{a^2}} \right) \right) \\&= \frac{-1}{a} (1-c_0(t, 0)) \sqrt{\alpha + \frac{1}{a^2}} \exp \left( -x \sqrt{\alpha + \frac{1}{a^2}} \right) \; . \end{aligned}$$

We determine the far field $c_0(t, 0)$ so that (49) is satisfied, i.e. $0= (1-c_0(t, 0)) - c_0(t, 0) \alpha a^2 + (1-c_0(t, 0)) \sqrt{\alpha a^2 + 1}$ and conclude that

$$\begin{aligned} c_0(t, 0) = \frac{1 + \sqrt{\alpha a^2 + 1}}{ 1 +\alpha a^2 + \sqrt{\alpha a^2 + 1}} = \frac{1 + \sqrt{\alpha a^2 + 1}}{(1 + \sqrt{\alpha a^2 + 1}) \sqrt{\alpha a^2 + 1}} = \frac{1 }{ \sqrt{\alpha a^2 + 1}}=\frac{1}{K} \; , \end{aligned}$$

where K is defined in (14) and represents the ratio between $c_0(t, 0)$ and the actual boundary value $\bar{c}_0(t, 0)$ for every $t>0$.

Right boundary layer

The system (6) written with respect to the right boundary layer variables $\hat{c}$, $\hat{f}$, $\hat{g}$ and $\hat{x}$ is given by (15), coupled to the boundary conditions (16) and (17). The leading order terms in the expansion $\hat{c} \approx \hat{c}_0 + O(\varepsilon )$, $\hat{f} \approx \hat{f}_0 + O(\varepsilon )$ and $\hat{g} \approx \hat{g}_0 + O(\varepsilon )$ satisfy the system of equations

$$\begin{aligned} 0&= \hat{c}_0'' - (\alpha \hat{c}_0 - \hat{f}_0)\;, \end{aligned}$$

(51)

$$\begin{aligned} a \hat{g}_0'&= ( \alpha \hat{c}_0 - \hat{f}_0) \;, \end{aligned}$$

(52)

$$\begin{aligned} a \hat{f}_0'&= - \hat{g}_0 \;, \end{aligned}$$

(53)

as well as the boundary conditions

$$\begin{aligned}&\hat{f}_0(t, 1)=\hat{g}_0(t, 1) \;, \end{aligned}$$

(54)

$$\begin{aligned}&\hat{c}_0'(t, 1)=a \hat{f}_0(t, 1) \; . \end{aligned}$$

(55)

Using Eqs. (53), (52) and (51) we obtain $-a^2 \hat{f}_0'' = a \hat{g}_0' = \alpha \hat{c}_0-\hat{f}_0=\hat{c}_0''$. Hence, for two arbitrary constants A and B it holds that $-a^2 \hat{f}_0=\hat{c}_0 + A x+B$. It is our goal to match the solution of the right boundary layer problem to the solution to the inner problem, therefore we expect that in the far field $\hat{x} \rightarrow -\infty$ it holds that $\hat{f}_0(t, \hat{x})=f_0(t, 1) = \alpha c_0(t, 1)$ where we used (36). Therefore we find that $A=0$ and $B=- c_0(1) (1+\alpha a^2)$. As a consequence it holds that

$$\begin{aligned} \hat{f}_0(t, \hat{x})= \frac{-1}{a^2} \left( \hat{c}_0(t, \hat{x}) - c_0(t, 1) (1+\alpha a^2) \right) \; , \end{aligned}$$

(56)

which we substitute in (51) to obtain

$$\begin{aligned} \hat{c}_0''&=\alpha \hat{c}_0 + \frac{1}{a^2} \left( \hat{c}_0 - c_0(t, 1) (1+\alpha a^2) \right) \\&=\alpha \hat{c}_0 + \frac{1}{a^2} \left( \hat{c}_0 - c_0(t, 1)\right) -\alpha c_0(t, 1) \\&= \left( \hat{c}_0 - c_0(t, 1)\right) \left( \alpha + \frac{1}{a^2} \right) \;. \end{aligned}$$

Coupling this to the matching condition $\lim _{\hat{x} \rightarrow -\infty } \hat{c}_0(t, \hat{x})=c_0(t, 1)$ we obtain

$$\begin{aligned} \hat{c}_0(t, \hat{x})= c_0(t, 1) \left( 1+(\hat{K}-1) \exp \left( (\hat{x}-1) \sqrt{\alpha + \frac{1}{a^2}} \right) \right) \end{aligned}$$

(57)

for a constant $\hat{K}$. Substituting (57) in (56) we also find

$$\begin{aligned} \hat{f}_0 (t, \hat{x})&= c_0(t, 1) \left( \alpha - \frac{\hat{K}-1}{a^2} \exp \left( (\hat{x}-1) \sqrt{\alpha + \frac{1}{a^2}} \right) \right) \; . \end{aligned}$$

(58)

The constant $\hat{K}$ can be determined by coupling these results to either (54) or (55). Substituting (57) and (58) in (55) we find that

$$\begin{aligned} 0&=c_0(t, 1) (\hat{K}-1) \sqrt{\alpha + \frac{1}{a^2}}-a c_0(t, 1)\left( \alpha - \frac{\hat{K}-1}{a^2} \right) \\&= c_0(t, 1) \left( (\hat{K}-1) \left( \sqrt{\alpha + \frac{1}{a^2}} + \frac{1}{a} \right) - \alpha a \right) \end{aligned}$$

which implies

$$\begin{aligned} \hat{K}= \frac{ \alpha a }{ \sqrt{\alpha + \frac{1}{a^2}} + \frac{1}{a} }+1= \frac{ \alpha a^2 }{ \sqrt{\alpha a^2 + 1} + 1 }+1= \frac{ \alpha a^2 + \sqrt{\alpha a^2 + 1} + 1 }{ \sqrt{\alpha a^2 + 1} + 1 }= \sqrt{\alpha a^2 + 1} \end{aligned}$$

(59)

which equals the constant of proportionality K we found before for the left boundary layer. Note that coupling (57), (56) with (54) instead leads to the same result according to computations which mimic those to find K in the left boundary layer: First substitute (58) in (53) to find

$$\begin{aligned} \hat{g}_0&=-a \hat{f}_0' = c_0(t, 1) \sqrt{\alpha + \frac{1}{a^2}} \frac{\hat{K}-1}{a} \exp \left( (\hat{x}-1) \sqrt{\alpha + \frac{1}{a^2}} \right) \; , \end{aligned}$$

which, together with (58), we substitute in (54) to obtain

$$\begin{aligned} 0=c_0(t, 1) \left( \alpha - \frac{\hat{K}-1}{a^2} \right) - c_0(t, 1) \sqrt{\alpha +\frac{1}{a^2}} \frac{\hat{K}-1}{a} \; , \end{aligned}$$

which leads to (59).