How ATP gives random transitions a direction

A visible step can contain hidden waiting

At lattice position $n$, call the states $A_n$ and $B_n$. A chemical edge connects them with forward rate $u_0$ and reverse rate $w_0$. A mechanical edge connects $B_n$ to $A_{n+1}$ with rates $u_1$ and $w_1$. The mechanical displacement is $d$.

These are coarse chemical–mechanical states, not assignments of every nucleotide-bound structure of a particular protein. A bead records displacement and can miss a chemical transition at fixed position.

Reading: U. Seifert · K. Svoboda & S. M. Block

The forward-to-backward cycle ratio is constrained

Define opposing load $F$ as positive. Advancing one step delivers work $Fd$ against it. One forward cycle releases chemical free energy $\Delta\mu$. Multiplying forward-to-reverse rate ratios around the cycle cancels internal-state free energies.

The remaining dimensionless quantity $\mathcal A$ is the cycle affinity. A positive value favors forward cycling; zero removes this preference. Local detailed balance constrains rate ratios. It leaves kinetic scales to be determined separately.

Reading: U. Seifert

Where ATP enters the teaching network

Let $c$ be ATP concentration divided by one micromolar. All rates below are in inverse seconds. The chemical forward rate grows with concentration; opposing force slows forward mechanical transitions and accelerates reverse ones. We choose $d=8\;\mathrm{nm}$ and $k_BT=4.114\;\mathrm{pN\,nm}$ as teaching scales.

These choices give $\Delta\mu=k_BT\ln(c/0.01)$. The reference $0.01\;\mathrm{\mu M}$ absorbs fixed ADP, phosphate and standard chemical potentials. It is not a universal cellular equilibrium concentration. A real assay must specify the products and solution conditions.

Reading: U. Seifert

From state occupancies to speed

At stationarity, incoming and outgoing probability balance. The total rate leaving $A$ is $u_0+w_1$; the total leaving $B$ is $w_0+u_1$. Equating probability flows and normalizing gives the occupancies. Mechanical forward flow is $p_Bu_1$ and reverse flow is $p_Aw_1$. Their difference times step length gives mean speed.

For $100\;\mathrm{\mu M}$ ATP and $1.50\;\mathrm{pN}$ load, the four rates are approximately $300$, $3.00$, $14.0$ and $2.58$ per second. Most time is spent in $B$. The net current is about $13.1$ cycles per second, giving $105\;\mathrm{nm\,s^{-1}}$. Individual four-second records fluctuate around this long-time mean.

Increasing load reduces the current to zero while transitions continue. At the same ATP concentration the chemical affinity is $\ln10000$, giving a stall load of $4.74\;\mathrm{pN}$. Real kinesin data can involve reaction pathways omitted by this minimal graph.

Reading: U. Seifert · K. Svoboda & S. M. Block

Efficiency and power behave differently near stall

During forward operation, net chemical input per cycle is $\Delta\mu$ and mechanical output is $Fd$. Their ratio is the tight-coupling efficiency. It approaches one near stall, but current approaches zero and output power vanishes.

A finite record still contains forward and backward fluctuations. Long-time or repeated-record averages reveal the small net speed. Testing equilibrium requires setting both zero load and the reference ATP concentration.

Reading: U. Seifert

One default trajectory can outpace its long-time mean

The synthetic default uses $100\;\mathrm{\mu M}$ ATP, $1.50\;\mathrm{pN}$ load and a $60\;\mathrm{Hz}$ digital detector cutoff. Seed $4919$ generates $2001$ samples over four seconds. Position and step counts in the table are hidden model truth. Fitted speed uses only detector positions, excluding the first $0.10\;\mathrm{s}$.

This particular trajectory advances faster than the long-time mean. Random molecular waiting changes the step count in a finite four-second record, so different seeds give different fitted speeds. Compare stationary theory with averages of independent records.

Reading: Sandbox Physics