Double Headed Motors

Experimentally, the speed and the force on a loaded motor were calculated.

\[F_\text{load}=6.5 \ \text{pN} \\ v_\text{motor}\approx 150 \ \text{nm/s}\]

Unloaded Motor

In the case of an unloaded motor there is no force.

\[v \approx \frac{d}{\tau}, \ \ \tau=\frac{d^2}{2D} \\ \Rightarrow v\approx\frac{2D}{d}\]

The motor feels a step potential, with step size \(\mathcal{E}=\mathcal{E}_\text{ATP}\approx30kT\).

Einstein Relation

The Einstein relation reveals the relationship between energy and diffusion.

\[Db=k_BT\]

Stoke’s Law says that the drag coefficient for a sphere is \(b=6\pi\eta R\). As a crude estimate, lets approximate the motor as a sphere.

\[R \sim 30 \ \text{nm} \\ \eta\approx10^{-3} \ \text{kg/ms} \\ \Rightarrow v \sim 2 \ \text{mm/s}\]

Ignoring the force on the motor leads to an answer that is way off of experimental results. Lets do better by taking into account the force. So let’s take into account the force of the load on the motor in order to get a more accurate result.

Loaded Motor

First consider a particle in a potential and thermal bath. The particle experiences diffusion. In addition to thermal diffusion on we will add a drift velocity term to the particle’s motion. The drift velocity term is to take into account the modulated potential the motor experiences. So we can the time derivative of the probability distribution in the following way:

\[\dot{p}=Dp''-v_d p' \\ \dot{p}=-\frac{\partial}{\partial x}\left[-Dp'+v_d p\right] \\ j \equiv -Dp'+v_d p\]

We can think of \(j\) as the rightward flux of particles. If the probability distribution is stationary that is equivalent to saying that the flux is constant.

\[\Rightarrow j=-Dp'+v_d p=\text{constant}\]

What causes the drift velocity?

As mentioned earlier, the drift velocity is cause by the modulated potential. The potential in turn exerts a drag force on the motor which is equal to

\[F = -\frac{dU}{dx} \ . \\ F=bv_d=\frac{kT}{D}v_d \ \Rightarrow \ v_d=-\frac{FD}{kT} \\ \text{let} \ \lambda\equiv\frac{kT}{F} \\ \therefore \ j = -D\left[p'+\frac{p}{\lambda}\right]\]

The flux being constant restrains what the probability distribution \(p\) can be. Let’s try to guess the solution.

\[p = \frac{c}{\lambda}\left[e^{-(x-d)/\lambda}-1\right]\]

Plugging this back into our equation for the flux above we can solve for \(j\).

\[j = \frac{cD}{\lambda^2}\]

This is a good result!

1. The expression for \(j\) is definitely constant.
2. \(j\) is positive, which means the flux is a flow of particle to the right, as desired.

This shows that the asymmetry in the potential (provided by the ATP hydrolisis) overcomes drift, diffusion (provided by the load). Now lets calculate the mean velocity of the motor from our results.

\[v = \frac{d}{\tau} \\ \tau=\left<\frac{1}{j}\right> = \int_0^d dx \ \frac{1}{j} \ p(x) \\ \tau = \int_0^d dx \ \frac{\lambda^2}{cD}\frac{c}{\lambda}\left[e^{-(x-d)/\lambda}-1\right] \\ \text{let} \ \alpha\equiv\frac{d}{\lambda} \\ \tau = \frac{\lambda^2}{D}\left[e^\alpha-\alpha-1\right] \\ \Rightarrow v = \frac{D}{d}\left(\frac{1}{e^\alpha-\alpha-1}\right)\]

Let’s make sure that our result is reasonable by checking some limits.

\[f=0 \ \Rightarrow \ \alpha=0 \\ v\rightarrow\frac{2D}{d} \\ f>>0 \ \Rightarrow \ \alpha>>1 \\ v\rightarrow\frac{D}{d}\alpha^2e^{-\alpha} \ \xrightarrow{\alpha\rightarrow\infty} \ 0\]

Comparison with Experimental Results

\[\text{set} \ \alpha=\frac{Fd}{kT}=\frac{13kT}{kT}=13 \\ v \approx \frac{D}{d}\alpha^2e^{-\alpha} \approx 400\text{nm/s}\]

This is a lot better than our previous result! This simple model got us an answer on the same order of magnitude. Note: Our model assumed that \(\mathcal{E}>>Fd\) and so the velocity cannot be less than zero.

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