Assessing the Impact of Electrostatic Drag on Processive Molecular Motor Transport

Abstract

The bidirectional movement of intracellular cargo is usually described as a tug-of-war among opposite-directed families of molecular motors. While tug-of-war models have enjoyed some success, recent evidence suggests underlying motor interactions are more complex than previously understood. For example, these tug-of-war models fail to predict the counterintuitive phenomenon that inhibiting one family of motors can decrease the functionality of opposite-directed transport. In this paper, we use a stochastic differential equations modeling framework to explore one proposed physical mechanism, called microtubule tethering, that could play a role in this “co-dependence” among antagonistic motors. This hypothesis includes the possibility of a trade-off: weakly bound trailing molecular motors can serve as tethers for cargoes and processing motors, thereby enhancing motor–cargo run lengths along microtubules; however, this introduces a cost of processing at a lower mean velocity. By computing the small- and large-time mean-squared displacement of our theoretical model and comparing our results to experimental observations of dynein and its “helper protein” dynactin, we find some supporting evidence for microtubule tethering interactions. We extrapolate these findings to predict how dynein–dynactin might interact with the opposite-directed kinesin motors and introduce a criterion for when the trade-off is beneficial in simple systems.

Appendices

Appendix

A Solution for Pinned Cargo System

The SDE used for pinned diffusion is the Ornstein–Uhlenbeck equation with nonzero mean. For notational efficiency, we write it as follows

$$\begin{aligned} \mathrm {d}X(t) = - \alpha (X(t) - \mu ) \mathrm {d}t + \sigma \mathrm {d}W(t), \end{aligned}$$

(31)

with $X(0) = x$. Because this SDE is linear, with additive noise, we can write the solution in convolution form using Duhamel’s formula

$$\begin{aligned} \begin{aligned} X(t)&= e^{-\alpha t} x + \int _0^t e^{-\alpha (t-u)} (-\alpha \mu ) \mathrm {d}u + \int _0^t e^{-\alpha (t-u)} \sigma \mathrm {d}W(u) \\&= e^{-\alpha t} x + (1 - e^{-\alpha t}) \, \mu \, + \sigma \int _0^t e^{-\alpha (t - u)} \mathrm {d}W(u). \end{aligned} \end{aligned}$$

Allowing for the possibility that the initial condition x and the center $\mu$ can be random variables, two quantities of interest follow immediately:

$$\begin{aligned} \mathbb {E}[X(t)]&= e^{-\alpha t} \mathbb {E}[x] + (1- e^{-\alpha t}) \, \mathbb {E}[\mu ] \end{aligned}$$

(32)

$$\begin{aligned} \mathbb {E}[X(t) \mu ]&= e^{-\alpha t} \mathbb {E}[x \mu ] + (1- e^{-\alpha t}) \, \mathbb {E}[\mu ^2]. \end{aligned}$$

(33)

To compute the MSD of X(t), we apply Itô’s formula to (31) using the function $f(x) = x^2$ to find that

$$\begin{aligned} \mathrm {d}X^2(t) = - 2 \alpha X(t) (X(t) - \mu ) + \sigma ^2 \mathrm {d}t + 2 \sigma X(t) \mathrm {d}W(t). \end{aligned}$$

It follows that

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t} \mathbb {E}\big [X^2(t)\big ] = -2 \alpha \mathbb {E}\big [X^2(t)\big ] + 2 \alpha \mathbb {E}[X(t) \mu ] + \sigma ^2, \end{aligned}$$

which, again by Duhamel’s formula has the solution

$$\begin{aligned} \begin{aligned} \mathbb {E}\big [X^2(t)\big ]&= e^{-2\alpha t} \mathbb {E}[x^2] + \int _0^t e^{-2 \alpha (t - u)} (2 \alpha \mathbb {E}[X(u) \mu ] + \sigma ^2) \mathrm {d}u \\&= e^{-2\alpha t} \mathbb {E}[x^2] + 2 \alpha \int _0^t e^{-2 \alpha (t - u)} \mathbb {E}[X(u) \mu ] \mathrm {d}u + \big (1 - e^{-2 \alpha t}\big ) \frac{\sigma ^2}{2 \alpha }. \end{aligned} \end{aligned}$$

Substituting in 32, the second term on the right-hand side evaluates to

$$\begin{aligned} 2 \alpha \int _0^t e^{-2 \alpha (t - u)} \mathbb {E}[X(u) \mu ] \mathrm {d}u = 2 e^{-\alpha t} (1 - e^{-2 \alpha t}) \mathbb {E}[x\mu ] + (1 - e^{-\alpha t})^2 \mathbb {E}[\mu ^2]. \end{aligned}$$

(34)

We conclude that

$$\begin{aligned} \mathbb {E}[X^2(t)] = \mathbb {E}\left[ \left( e^{-\alpha t} x + (1 - e^{-\alpha t}) \mu \right) ^2\right] + (1 - e^{-2\alpha t}) \frac{\sigma ^2}{2 \alpha }. \end{aligned}$$

(35)

Expressing this in terms of the parameters used in the main text

$$\begin{aligned} \alpha = \frac{\kappa }{\gamma }, \, \sigma = \sqrt{\frac{2 k_B T}{\gamma }}, \, x = 0, \mu \sim N\left( 0, \frac{k_B T}{\kappa }\right) , \end{aligned}$$

we have

$$\begin{aligned} \mathbb {E}[X^2(t)] = 2 (1 - e^{-\kappa t /\gamma }) \frac{k_B T}{\kappa }. \end{aligned}$$

(36)

B Proof Omitted from the Main Text

Proof

(of Lemma 1) Since the rows of M sum to zero, $\lambda _0 = 0$ is an eigenvalue of M with right eigenvector $\mathbf {v}_0 = \left( 1,1,\ldots ,1\right) ^\top$. Let $\mathbf {u}_0$ be the left eigenvector associated with 0 for M.

Since the diagonal of M is negative, there exists some positive real number r such that $M' = M + r\mathbb {I}$ is a nonnegative matrix. Note that $\lambda$, $\mathbf {v}_\lambda$ are an eigenvalue-eigenvector pair for M if and only if $\lambda +r$ and $\mathbf {v}_{\lambda }$ are an eigenvalue-eigenvector pair for $M'$.

Now we show that $M'$ is irreducible. $M'$ differs from M only on the diagonal. Therefore, the associated directed graph of $M'$ will have the same edge values between distinct vertices as that of M. Since the associated graph for M is strongly connected, the one for $M'$ is strongly connected as well and consequently $M'$ is irreducible.

Since $M'$ is a nonnegative, irreducible matrix, the Perron–Frobenius theorem holds (Berman and Plemmons 1994). Thus, there exists a simple real eigenvalue $\lambda _p>0$ for $M'$ such that the spectral radius of $M'$ is $\lambda _p$, all other eigenvalues of $M'$ must have real part strictly less that $\lambda _p$, and the only eigenvectors whose components are all real and positive are those eigenvectors associated with $\lambda _p$. Note that

$$\begin{aligned} M'\mathbf {v}_0 = M\mathbf {v}_0 + r\mathbf {v}_0 = r\mathbf {v}_0. \end{aligned}$$

Ergo, since $\mathbf {v}_0$ is a eigenvector whose components are all real and positive, we must have that $\lambda _p =r$. Further,

$$\begin{aligned} \mathbf {u}_0 M' = \mathbf {u}_0M + \mathbf {u}_0r = \mathbf {u}_0r. \end{aligned}$$

Since $\mathbf {u}_0$ is an eigenvector associated with $\lambda _p = r$, its components are all real and positive.

As r is a simple eigenvalue for $M'$ and all other eigenvalues of $M'$ have real part strictly less than r, it now follows that 0 is a simple eigenvalue for M and all other eigenvalues of M have real part strictly less than 0.

Write the remaining eigenvalues of M as

$$\begin{aligned} 0=\lambda _0 < \left| {\lambda _1}\right| \le \left| {\lambda _2}\right| \le \cdots \le \left| {\lambda _k}\right| , \end{aligned}$$

where we assume that $\lambda _i$ has multiplicity $m_i$. Write M in Jordan normal form. That is write $M = PJP^{-1}$ where

$$\begin{aligned} J = \begin{pmatrix} 0&{}\quad &{}\quad &{}\quad \\ &{}\quad J_1 &{}\quad &{}\quad \\ {} &{}\quad &{}\quad \ddots &{}\quad \\ &{}\quad &{}\quad &{}\quad J_k\end{pmatrix}, \end{aligned}$$

and $J_i$ is an $m_i\times m_i$ matrix of the form

$$\begin{aligned} J_i = \begin{pmatrix} \lambda _i &{}\quad 1 &{}\quad &{}\quad &{}\quad &{}\quad \\ &{}\quad \lambda _i &{}\quad 1&{}\quad &{}\quad &{}\quad \\ &{}\quad &{}\quad \ddots &{}\quad \ddots &{}\quad \\ &{}\quad &{}\quad &{}\quad \lambda _i&{}\quad 1\\ &{}\quad &{}\quad &{}\quad &{}\quad \lambda _i\end{pmatrix}. \end{aligned}$$

Since zero is a simple eigenvalue, the first column of P is $\mathbf {v}_0$ and the first row of $P^{-1}$ is $\mathbf {u}_0$.

The solution of Eq. 16 is given by

$$\begin{aligned} \begin{aligned} \mathbf {Y}(t)&= e^{Mt}Y_0 + \int _0^te^{M(t-s)}\,\mathrm {d}s \cdot F\\&= Pe^{J t} P^{-1}Y_0 + Pe^{J t}\int _0^te^{-J s}\,\mathrm {d}s\, P^{-1}F. \end{aligned} \end{aligned}$$

(37)

Consider $e^{Jt}$. Write $J = D + N$, where D is a diagonal matrix with diagonal coinciding with J. Then N is a matrix with zeros everywhere except for some 1’s on the superdiagonal. Specifically, N is the block diagonal matrix

$$\begin{aligned} N = \begin{pmatrix} 0 &{}\quad &{}\quad &{}\quad \\ {} &{}\quad N_1&{}\quad &{}\quad \\ {} &{}\quad &{}\quad \ddots &{}\quad \\ {} &{}\quad &{}\quad &{}\quad N_k\end{pmatrix}, \end{aligned}$$

where each $N_i$ is an $m_i\times m_i$ nilpotent matrix of the form

$$\begin{aligned} N_i = \begin{pmatrix}0&{}\quad 1 &{}\quad &{}\quad \\ {} &{}\quad 0&{}\quad 1&{}\quad \\ {} &{}\quad &{}\quad \ddots &{}\quad \ddots \\ {} &{}\quad &{}\quad &{}\quad 0\end{pmatrix}. \end{aligned}$$

Notably, N commutes with D. Hence

$$\begin{aligned} e^{Jt} = e^{(D+N)t} = e^{Dt}e^{Nt}. \end{aligned}$$

Further, $e^{Nt}$ has the form

$$\begin{aligned} e^{Nt} = \begin{pmatrix}1 &{}\quad &{}\quad &{}\quad \\ &{}\quad E_1(t) &{}\quad &{}\quad \\ {} &{}\quad &{}\quad \ddots &{}\quad \\ {} &{}\quad &{}\quad &{}\quad E_k(t)\end{pmatrix}, \end{aligned}$$

where

$$\begin{aligned} E_i(t) = \begin{pmatrix} 1 &{}\quad t &{}\quad t^2/2&{}\quad \cdots &{}\quad \frac{t^{m_i-1}}{(m_i-1)!}\\ &{}\quad 1&{}\quad t&{}\quad \cdots &{}\quad \frac{t^{m_i-2}}{(m_i-2)!}\\ {} &{}\quad &{}\quad 1&{}\quad \cdots &{}\quad \frac{t^{m_i-3}}{(m_i-3)!}\\ {} &{}\quad &{}\quad &{}\quad \ddots &{}\quad \vdots \\ {} &{}\quad &{}\quad &{}\quad &{}\quad 1\end{pmatrix}. \end{aligned}$$

Hence,

$$\begin{aligned} e^{Jt} = \begin{pmatrix}1 &{}\quad &{}\quad &{}\quad \\ &{}\quad e^{\lambda _1 t}E_1(t)&{}\quad &{}\quad \\ {} &{}\quad &{}\quad \ddots &{}\quad \\ {} &{}\quad &{}\quad &{}\quad e^{\lambda _kt}E_k(t)\end{pmatrix}. \end{aligned}$$

Since $\text {Re}\left( \lambda _i\right) < 0$ for all $i\ne 0$, we have

$$\begin{aligned} \lim _{t\rightarrow \infty }e^{J t} = \begin{pmatrix} 1&{}\quad &{}\quad &{}\quad \\ {} &{}\quad 0&{}\quad &{}\quad \\ {} &{}\quad &{}\quad \ddots &{}\quad \\ {} &{}\quad &{}\quad &{}\quad 0\end{pmatrix}. \end{aligned}$$

It follows that

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{1}{t}Pe^{J t}P^{-1}Y_0 = \mathbf {0}. \end{aligned}$$

(38)

We now turn our attention to the integral $\int _0^te^{J(t-s)}\,\mathrm {d}s$. Note that for any integer $a\ge 1$,

$$\begin{aligned} \int _0^t (t-s)^ae^{\lambda _i(t-s)}\,\mathrm {d}s = \frac{\varGamma \left( a+1,-\lambda _it\right) - a(a!)}{\lambda _i^{a+1}}, \end{aligned}$$

where $\varGamma (s,x)$ represents the incomplete gamma function. For large x, $\varGamma (s,x) \sim x^{s-1}e^{-x}$ (Abramowitz and Stegun 1964). So, for sufficiently large t,

$$\begin{aligned} \int _0^t(t-s)^ae^{\lambda _i(t-s)}\,\mathrm {d}s \sim \frac{\left( -\lambda _it\right) ^a e^{\lambda _i t}-a(a!)}{\lambda _i^{a+1}}. \end{aligned}$$

In particular, since Re$(\lambda _i)<0$,

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{1}{t}\int _0^t(t-s)^ae^{\lambda _i(t-s)} \mathrm {d}s= 0. \end{aligned}$$

(39)

Now,

$$\begin{aligned} \int _0^te^{J(t-s)}\,\mathrm {d}s = \begin{pmatrix}t &{}\quad &{}\quad &{}\quad \\ {} &{}\quad \tilde{E}_1(t) &{}\quad &{}\quad \\ {} &{}\quad &{}\quad \ddots &{}\quad \\ {} &{}\quad &{}\quad &{}\quad \tilde{E}_k(t)\end{pmatrix}, \end{aligned}$$

where $\tilde{E}_i(t)$ is an upper triangular matrix for which the entry on the jth row and $(j+k)$th column (where $k \ge 0$) is

$$\begin{aligned} \tilde{E}_i(t) = \int _0^t (t-s)^{k} e^{\lambda _i (t-s)} \mathrm {d}s \end{aligned}$$

It follows now from Eq. 39 that

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{1}{t}\int _0^te^{J(t-s)}\,\mathrm {d}s = \begin{pmatrix} 1 &{}\quad &{}\quad &{}\quad \\ {} &{}\quad 0 &{}\quad &{}\quad \\ &{}\quad &{}\quad \ddots &{}\quad \\ &{}\quad &{}\quad &{}\quad 0 \end{pmatrix}. \end{aligned}$$

(40)

Equations 38 and 40, together with the fact that $\mathbf {v}_0 = (1,1,\ldots ,1)^\top$ is the first column of P and $\mathbf {u}_0$ is the first row of $P^{-1}$, imply

$$\begin{aligned} \lim _{t\rightarrow \infty } \frac{1}{t}\left( Pe^{Jt}P^{-1}Y_0 + P\int _0^te^{J(t-s)}\,\mathrm {d}s\,P^{-1}F\right) = \mathbf {v}_0\mathbf {u}_0F. \end{aligned}$$

Hence, $\lim _{t\rightarrow \infty }\frac{1}{t}Y_j(t) = \mathbf {u}_0 F$ for any $j\in \{0,\ldots ,n-1\}$ as claimed. $\square$