Force-velocity relation and load-sharing in the linear polymerization ratchet revisited: the effects of barrier diffusion

Abstract

We study the velocity-force (V-F) relation for a Brownian ratchet consisting of a linear rigid polymer growing against a diffusing barrier, acted upon by a opposing constant force (F). Using a careful mathematical analysis, we derive the V-F relations in the extreme limits of fast and slow barrier diffusion. In the first case, V depends exponentially on the load F, in agreement with the well-known formula proposed by Peskin, Odell and Oster (1993), while the relationship becomes linear in the second case. For a bundle of two filaments growing against a common barrier, equal sharing of load in the corresponding V-F relation is predicted by a mean-field argument in both limits. However, the scaling behaviour of velocity with the number of filaments is different for the two cases. In the limit of large D, the validity of the mean-field approach is tested, and partially supported by a detailed and rigorous analysis. Our principal predictions are also verified in numerical simulations.

Graphic abstract

Appendix A: On load-sharing properties of a two-filament bundle

Here, we explore the question of load sharing in a two-filament bundle, outside the MFA. Consider the reflecting boundary condition $J(x_1=0,x_2)=0$ imposed at the first barrier, which implies the relation

$$\begin{aligned} \begin{aligned} -D\partial _{x_1}\theta (x_1,x_2)|_{x_1=0}=D\partial _{x_2}\theta (0,x_2)+f\theta (0,x_2). \end{aligned} \end{aligned}$$

(A.1)

After differentiating Eq. (A.1) with respect to $x_2$, and putting $x_2=0$, we arrive at

$$\begin{aligned} -D\frac{\partial ^2\theta }{\partial x_2\partial x_1}\bigg |_{x_1,x_2=0}\!\!\!=\!D\frac{\partial ^2}{\partial x_2^2}\theta (0,x_2)\bigg |_{x_2=0}\!+\!f\frac{\partial \theta (0,x_2) }{\partial x_2}\bigg |_{x_2=0}. \nonumber \\ \end{aligned}$$

(A.2)

Let us define the function $\psi (x)=\theta (x,0)=\theta (0,x)$, where the second equality follows from symmetry of $\theta (x_1,x_2)$ with respect to exchange of $x_1$ and $x_2$. This definition, when used in Eq. A.2, gives

$$\begin{aligned} D\frac{\partial ^2 \theta }{\partial x_2^2}\bigg |_{x_1,x_2=0}\!\!\!=\!-D\frac{\partial }{\partial x_1}\bigg (\frac{\partial \theta }{\partial x_2}\bigg )\bigg |_ {x_1=0,x_2=0}\!-\!f\bigg (\frac{\partial \psi }{\partial x_2}\bigg )_{x_2=0}. \nonumber \\ \end{aligned}$$

(A.3)

Next, use the reflecting boundary condition at the second barrier, i.e. $J(x_1,x_2=0)$, which provides the relation

$$\begin{aligned} \begin{aligned} D\frac{\partial \theta (x_1,x_2)}{\partial x_2}\bigg |_{x_2=0}=D\frac{\partial \psi (x_1)}{\partial x_1}-f\psi (x_1), \end{aligned} \end{aligned}$$

(A.4)

which, when used in Eq. A.3, yields the following useful relation between the second derivatives:

$$\begin{aligned} \begin{aligned} \frac{\partial ^2\theta (0,x_2)}{\partial x_2^2}\bigg |_{x_2=0}=\frac{\partial ^2\psi (x_1)}{\partial x_1^2}\bigg |_{x_1=0}. \end{aligned} \end{aligned}$$

(A.5)

Let us now revisit Eq. 31 for the two-filament bundle, and put $x_2=0$. It then follows that the function $\psi (x)$ satisfies the equation

$$\begin{aligned} \begin{aligned} D\frac{\partial ^2\psi }{\partial x^2}+2f\frac{\partial \psi }{\partial x}+\frac{f^2}{D}\psi =D\psi ^{\prime \prime }(0) \end{aligned} \end{aligned}$$

(A.6)

where we have used the exact result $T(x,0)=0$ for any x (see Eq. 32). In Laplace space, Eq. A.6 has the solution

$$\begin{aligned} \begin{aligned} \tilde{\psi }(s)=\frac{D\psi ''(0)}{s(s^2+2fs+\frac{f^2}{D})}+\frac{(\frac{3}{2}f+Ds)\psi (0)}{(s^2+2fs+\frac{f^2}{D})}, \end{aligned} \end{aligned}$$

(A.7)

where we have used the reflecting boundary condition $2D\psi ^{\prime }(0) +f\psi (0)=0$ for $\psi (x)$, which follows from $J(x_1=0,x_2=0)=0$. The inverse Laplace transform of Eq. A.7 gives

$$\begin{aligned} \psi (x)= & {} \frac{\psi ''(0)}{\alpha _1}\bigg [\frac{1}{\alpha _1}(1-e^{-\alpha _1 x})-xe^{-\alpha _1 x}\bigg ]\nonumber \\&+\psi (0)\bigg (e^{-\alpha _1 x}+\frac{f}{2D}xe^{-\alpha _1 x}\bigg ), \end{aligned}$$

(A.8)

where $\alpha _1=f/D$. Taylor expansion of Eq. A.8 around $x=0$ yields the expression

$$\begin{aligned} \begin{aligned} \psi (x)\equiv \psi (0)\bigg (1-\frac{f}{2D}x\bigg )+\frac{\psi ''(0)}{2}x^2. \end{aligned} \end{aligned}$$

(A.9)

It is confirmed readily by observation that Eq. 38 and Eq. 45, derived under the MFA, are consistent with Eq. A.9 up to ${{\mathcal {O}}}(x)$.