Kinesin Rectifies Noise: Why a Molecular Motor Is Not a Tiny Engine — Epoche B2
A common textbook image presents molecular motors such as kinesin — the protein that hauls cargo along the microtubule filaments inside our cells — as tiny engines: they burn fuel (the molecule ATP) and push themselves forward along a track, overcoming resistance step by step. This picture is intuitive because it copies the machines we know. It is also wrong in an important physical sense. At the scale of a protein, the rules that govern a car engine simply do not apply. This essay compares the two regimes quantitatively and argues that ATP hydrolysis in kinesin buys not propulsion but rectification — a biasing of random thermal motion that the surrounding water supplies for free. Two regimes: what the Reynolds number counts Any object moving through a fluid is subject to two quite different resistances. One is the cost of shoving fluid out of the way and setting it in motion, which grows with the fluid's density $\rho$ and with the square of the speed; this is the inertial part. The other is the internal stickiness of the fluid, its viscosity $\eta$, which resists shearing one layer of fluid past the next and grows only in proportion to speed. The Reynolds number is the ratio of the first to the second, $$Re = \frac{\rho v L}{\eta},$$ with $v$ the speed and $L$ the object's size. It has no units — the units of $\rho$, $v$, $L$ and $\eta$ cancel — so it is a pure number that says which resistance is in charge. Put in numbers. For a car, air has $\rho \approx 1.2\,\text{kg}\,\text{m}^{-3}$ and $\eta \approx 1.8 \times 10^{-5}\,\text{Pa}\,\text{s}$; take $v \approx 30\,\text{m}\,\text{s}^{-1}$ and $L \approx 4\,\text{m}$. Then $Re \approx (1.2)(30)(4)/(1.8\times10^{-5}) \approx 8 \times 10^{6}$. Inertia dominates by seven orders of magnitude: release the accelerator and the car coasts. For kinesin, water has $\rho \approx 10^{3}\,\text{kg}\,\text{m}^{-3}$ and $\eta \approx 10^{-3}\,\text{Pa}\,\text{s}$, while the motor has $L \sim 10^{-8}\,\text{m}$ and travels at $v \sim 10^{-6}\,\text{m}\,\text{s}^{-1}$. Then $Re \approx (10^{3})(10^{-6})(10^{-8})/(10^{-3}) = 10^{-8}$. Viscosity dominates by eight orders of magnitude — a swing of fifteen powers of ten between the two cases, which is why the analogy fails so completely. How far a protein coasts The equation of motion is Newton's second law with the two forces written out: a drag force proportional to velocity, a systematic force $F(x)$ from the molecule's own structure, and a rapidly fluctuating force $\xi(t)$ from water molecules hammering the protein from all sides: $$m\ddot{x} = -\gamma\dot{x} + F(x) + \xi(t).$$ Here $\gamma$ is the drag coefficient, the constant of proportionality between speed and drag. For a sphere of radius $a$ moving slowly through a fluid, Stokes' law gives $\gamma = 6\pi\eta a$ — and this formula is exactly the one to use here, because Stokes derived it by discarding the inertial term in the fluid equations, which is legitimate precisely when $Re \ll 1$. With $a \approx 5\,\text{nm}$ and $\eta = 10^{-3}\,\text{Pa}\,\text{s}$, $\gamma \approx 6\pi (10^{-3})(5\times10^{-9}) \approx 9 \times 10^{-11}\,\text{kg}\,\text{s}^{-1}$. Now ask how long inertia survives. Dropping the other forces, $m\ddot{x} = -\gamma\dot{x}$ has a decaying solution with time constant $\tau = m/\gamma$. Conventional kinesin has a mass of roughly $380$ kilodaltons, and one dalton is $1.66\times10^{-27}\,\text{kg}$, so $m \approx 6\times10^{-22}\,\text{kg}$ and $$\tau = \frac{m}{\gamma} \approx \frac{6\times10^{-22}}{9\times10^{-11}} \approx 7\times10^{-12}\,\text{s}.$$ In that time, moving at $10^{-6}\,\text{m}\,\text{s}^{-1}$, the protein coasts about $6\times10^{-18}\,\text{m}$ — a billionth of the $8\,\text{nm}$ it must travel to reach the next binding site, and far smaller than an atomic nucleus. Setting $m\ddot{x} \approx 0$ is therefore not an approximation of convenience but a statement about a term that has vanished, leaving $$\gamma\dot{x} = F(x) + \xi(t).$$ One clarification is needed here, because the original phrasing of this point invites a misreading. Saying that the protein "stops the instant the force stops" is true only of directed motion. The molecule does not come to rest; it is still being struck from every side by $\xi(t)$ and immediately begins to wander. What ceases within a few picoseconds is any memory of where it was going. The scale of the shaking How violent is that wandering? The natural energy unit is $k_{\mathrm{B}}T$, where $k_{\mathrm{B}} = 1.38\times10^{-23}\,\text{J}\,\text{K}^{-1}$ is Boltzmann's constant, the conversion factor between temperature and energy per degree of freedom. At body-like temperature $T = 300\,\text{K}$, $k_{\mathrm{B}}T = 4.14\times10^{-21}\,\text{J}$. Since $1\,\text{pN}\cdot\text{nm} = (10^{-12}\,\text{N})(10^{-9}\,\text{m}) = 10^{-21}\,\text{J}$, this is $4.1\,\text{pN}\cdot\text{nm}$ — a convenient unit, because forces on proteins are measured in piconewtons and their displacements in nanometres. Optical-trap experiments supply the comparison. An optical trap is a laser beam focused so tightly that its intensity gradient pulls a small transparent bead towards the focus; for small displacements the restoring force is proportional to the offset, so the bead behaves as a calibrated spring and its position reads out force directly. Svoboda and colleagues (1993) attached single kinesin molecules to such beads and watched them advance along microtubules in discrete steps of $8\,\text{nm}$. That number is not a property of the motor but of the track: the microtubule is built from tubulin dimers repeating every $8\,\text{nm}$ along each protofilament, so the motor's step is the lattice spacing of its binding sites. Loading the trap more heavily slows and finally halts the motor at a stall force of about $5$–$7\,\text{pN}$, depending on assay conditions; taking $6\,\text{pN}$ as a representative value, the mechanical work per step is $$W = F\,\Delta x \approx 6\,\text{pN} \times 8\,\text{