L3 · Chapter 6

How much entropy a fuel-driven enzyme produces each second

Sandbox Physics · 2026-10-29

The first five chapters followed one bead in a continuous trap. Many molecular machines in a cell instead hop between a few discrete states: an enzyme binds ATP, hydrolyses it and lets go of the phosphate, then releases ADP and is empty again. Each step happens at a random time, and the fuel makes forward steps more frequent than backward ones. This chapter writes the master equation of that three-step cycle, finds how many turns it makes per second, proves that the entropy it produces per second is the turn rate times the fuel per turn, and asks how much of that an observer who only counts turns can prove from the scatter of the count.

Three states and six rates

Number the enzyme states 0, 1 and 2: empty E, ATP-bound ET and ADP-bound ED. The rate of a forward hop from state $i$ to $i+1$ is $w^+_i$ and the reverse rate is $w^-_i$ . Labels wrap around modulo 3, so a forward hop from 2 lands on 0. One full forward turn uses one ATP and releases one ADP and one phosphate.

Energies are in units of $k_BT$ . The fuel releases $\Delta\mu=\mu_{\mathrm{ATP}}-\mu_{\mathrm{ADP}}-\mu_{\mathrm{P_i}}$ per turn, which we spread evenly over the three steps. State $i$ has its own free energy $E_i$ , with $E_0=0$ . Local detailed balance says that each forward-to-reverse rate ratio equals the drop in free energy along that step, as below. Thermodynamics fixes only this ratio; how fast the rates are depends on the enzyme, and here they are schematic.

Multiply the three ratios and every $E_i$ cancels, leaving $e^{\Delta\mu}$ . With no fuel, the products of forward and reverse rates are equal and the enzyme has no preferred direction.

\frac{w^+_i}{w^-_i}=e^{E_i-E_{i+1}+\Delta\mu/3},\qquad \prod_{i=0}^{2}\frac{w^+_i}{w^-_i}=e^{\Delta\mu}

Local detailed balance for one step, and its product over a full turn.

Source: A. C. Barato and U. Seifert

The master equation and its steady state

Let $p_i(t)$ be the probability that the enzyme is in state $i$ , and $J_i$ the net probability flow from $i$ to $i+1$ . The probability of state $i$ grows by what flows in minus what flows out. That is the master equation.

In the steady state no $p_i$ changes, so the net flow is the same on all three edges. Call it $J$ , in turns per second. Together with $p_0+p_1+p_2=1$ , three linear equations give the three probabilities. The enzyme lingers in whichever state precedes a slow step.

\frac{dp_i}{dt}=J_{i-1}-J_i,\qquad J_i=p_i\,w^+_i-p_{i+1}\,w^-_i

In the steady state the three net flows are equal to the cycle current.

Entropy production is turn rate times fuel

Chapter 4 measured dissipation by the ratio of forward to reverse path probabilities. For a jump process the same idea gives Eq. (128) of the Seifert review: for every pair of states, take the forward flow times the logarithm of the forward-to-reverse flow ratio, and add them up.

Pair each term with its reverse and every edge contributes $J_i\ln[p_iw^+_i/(p_{i+1}w^-_i)]$ . In the steady state every $J_i$ equals $J$ . Pull it out: the probability ratios cancel around the ring and the rate ratios multiply to $e^{\Delta\mu}$ . The entropy production is $J\Delta\mu$ . Each turn converts the fuel free energy $\Delta\mu$ entirely into heat in the surrounding water.

\sigma=\sum_{m\ne n}p_m w_{mn}\ln\frac{p_m w_{mn}}{p_n w_{nm}}=J\,\Delta\mu

k_BT

per second. The state energies

E_i

change

J

, but each turn still dissipates

\Delta\mu

Source: U. Seifert

What counting turns alone can prove

Suppose the observer cannot tell which state the enzyme is in and only counts net turns, for example ADP released minus ADP rebound. The net count $X_t$ after time $t$ has mean $Jt$ and a variance that grows linearly, written $2Dt$ . $D$ is the diffusion constant of the count.

Barato and Seifert proposed the thermodynamic uncertainty relation in 2015: a more precise count costs more dissipation. In terms of their product $\mathcal{Q}=2\sigma D/J^2$ , they proved $\mathcal{Q}\ge2$ for unicyclic networks and gave the sharper Eq. (20) for a ring of $N$ states. Read backwards, $J^2/D$ is a lower bound on entropy production that needs nothing but the count.

Check it on a ring whose states have equal energies and whose three steps share the same rates, $k^+$ forward and $k^-$ back. The enzyme then performs a biased random walk on sites, three sites per turn, so $J=(k^+-k^-)/3$ , $D=(k^++k^-)/18$ and $\ln(k^+/k^-)=\Delta\mu/3$ . Substituting gives the second formula below, which makes Eq. (20) an equality: for a given fuel and number of states, the uniform ring gives the most precise count for the entropy it produces.

With little fuel the fraction is close to 1, and counting proves almost all the entropy production. With a lot of fuel it falls to about $6/\Delta\mu$ , and counting proves only a small part. The relation also prices precision: a relative error $\epsilon$ in the count costs at least $2k_BT/\epsilon^2$ .

\sigma\ge\frac{J^2}{D},\qquad \mathcal{Q}=\frac{2\sigma D}{J^2}\ge\frac{\Delta\mu}{3}\coth\frac{\Delta\mu}{6}\ge 2

The first is the bound that counting proves; the second is Eq. (20) of Barato and Seifert with

N=3

\frac{J^2/D}{\sigma}=\frac{2(k^+-k^-)}{(k^++k^-)\ln(k^+/k^-)}=\frac{\tanh(\Delta\mu/6)}{\Delta\mu/6}

On a uniform three-state ring, the fraction of the entropy production that counting proves.

Source: A. C. Barato and U. Seifert

Worked example: an ATP-driven enzyme

Take the Lab defaults: $\Delta\mu=8$ , $E_1=-1$ , $E_2=1$ , a base rate of $20\,\mathrm{s^{-1}}$ for every step, and each free-energy drop split evenly between the forward and reverse rates. Steps 0 and 2 both drop by $1+8/3=3.67$ , giving rates of $125\,\mathrm{s^{-1}}$ and $3.20\,\mathrm{s^{-1}}$ . The hydrolysis step drops by only $0.667$ , giving $27.9\,\mathrm{s^{-1}}$ and $14.3\,\mathrm{s^{-1}}$ .

The steady state is $p=(0.157,\,0.699,\,0.144)$ : the enzyme spends 70% of its time in ET, waiting for the slow hydrolysis. Check two edges: $0.157\times125-0.699\times3.20=17.4$ and $0.699\times27.9-0.144\times14.3=17.4$ ; the exact value is $J=17.45\,\mathrm{s^{-1}}$ . The entropy production is $\sigma=17.45\times8=139.6\,k_BT/\mathrm{s}$ .

The diffusion constant has no short formula; the Lab computes $D=5.09\,\mathrm{s^{-1}}$ from the rate matrix. Then $J^2/D=17.45^2/5.09=59.8\,k_BT/\mathrm{s}$ , only 43% of $\sigma$ . $\mathcal{Q}=2\times139.6\times5.09/17.45^2=4.67$ , above the Eq. (20) value $(8/3)\coth(4/3)=3.06$ . With the same fuel a uniform ring would let counting prove 65%. Here the uneven energy levels make the count scatter more, so counting proves less.

Try it in the Lab

What counting proves

Predict first: if the fuel per cycle drops from 8 to 1, does the fraction that counting proves rise or fall? Check with “Near equilibrium, counts suffice”, then try “Strong fuel, counts fall short”.
Compute both fractions from $\tanh(\Delta\mu/6)/(\Delta\mu/6)$ and compare them with “Share proved by counting cycles” in the details.
Return to the default and drag the ↕ handle to raise ET. Watch the entropy production, the cycle rate and the counting bound, then explain why the entropy production is still the cycle rate times 8.

Exercises with explained answers

On a uniform three-state ring, at what fuel per cycle does counting prove only half of the entropy production?
Show the answer
Let $x=\Delta\mu/6$ and solve $\tanh x/x=1/2$ . The numerical root is $x=1.915$ , so $\Delta\mu\approx11.5\,k_BT$ .
A three-state enzyme ring burns $8\,k_BT$ per cycle and must count its cycles with 1% relative error. By Eq. (20), how much free energy must it use at least, and how many cycles is that?
Show the answer
Eq. (20) gives $\mathcal{Q}\ge3.06$ , and $\sigma t=\mathcal{Q}/\epsilon^2$ , so it must use at least $3.06\times10^4\,k_BT$ . At $8\,k_BT$ per cycle that is about 3830 cycles. With the default rates, $\mathcal{Q}=4.67$ and it needs about 5840 cycles.
Why does changing the ET and ED energy levels leave the entropy production equal to cycle current times fuel? What do the levels change instead?
Show the answer
Around one turn the free-energy drops add up to $\sum_i(E_i-E_{i+1})+\Delta\mu=\Delta\mu$ , because the levels cancel in pairs; in the steady state the logarithms of probability ratios also cancel around the ring. The levels change the step rates, and so $J$ and $D$ . That changes the entropy produced per second and the fraction that counting proves.

Sources and what we recomputed

U. Seifert, Stochastic thermodynamics, fluctuation theorems and molecular machines, arXiv:1205.4176v1; Reports on Progress in Physics 75, 126001 (2012)
Version 1, 18 May 2012. Section 2: overdamped Langevin dynamics, Eq. (14) for the heat released into the medium and Eqs. (15)–(19) for the first law along one trajectory. Section 4: Eqs. (58) and (61), the Jarzynski and Crooks relations. Section 6.2: entropy production for master equations, Eqs. (120)–(129). Section 7.2: relative-entropy bounds on dissipation, Eqs. (146)–(150). Section 7.3: measurement and feedback, Eqs. (151)–(163).
A. C. Barato and U. Seifert, Thermodynamic uncertainty relation for biomolecular processes, arXiv:1502.05944v2; Physical Review Letters 114, 158101 (2015)
Version 2, 30 April 2015. Eq. (4): for a biased random walk the dissipation times the squared relative uncertainty equals A coth(A/2), at least two thermal energy units. Eqs. (7) and (10): the diffusion constant and the product Q for a general network. Eq. (19): the three-state ATP hydrolysis cycle. Eq. (20): the sharper bound for a ring of N states. The enzyme rates and numbers in this course are our own.