Supplementary Information For

Supplementary Information for Mechanisms for achieving high speed and efficiency in biomolecular machines Jason A. Wagoner and Ken A. Dill1 1To whom correspondence should be addressed. E-mail: dilllaufercenter.org This PDF file includes: Supplementary text Fig. S1 Tables S1 to S4 References for SI reference citations Jason A. Wagoner and Ken A. Dill1 1 of 13 www.pnas.org/cgi/doi/10.1073/pnas.1812149116 Supporting Information Text 1. Flux for the two-state model The steady-state flux (number of full cycles per unit time) can be derived from J = PBfm − PArm, where the transition rates are labelled in Figure 1 of the main text, and PA, PB are steady state probabilities that can be solved using the 2 × 2 rate matrix. Here, we give a derivation of the steady state flux that can be used for a more general network of states and for other kinetic properties, like higher moments of flux. The evolution of probability density along the periodic two-state model is ∂P (A , t) j = P (B , t) f + P (B , t) r ∂t j−1 m j c −P (Aj , t) (1 − fc − rm) , [S1] where P (Aj , t) is the probability density of state Aj at time t. The steady state flux can be calculated from the generating functions X −ijk PÃ (k, t) = e P (Aj , t) , [S2] j and similarly for P˜B (k, t). The time evolution of the generating function is ∂PÃ (k, t) X = M P˜ (k, t) , [S3] ∂t An n n∈{A,B} where −ik 1 − fc − rm rc + fme M = ik [S4] fc + rme 1 − fm − rc, and we are using lettered rather than numbered indices (MAB is M12, etc.). The flux is: ∂ϑ J = i 0 ∂k k=0 f f − r r = c m c m , [S5] fc + fm + rc + rm where ϑ0 is the largest eigenvalue of M (1). 2. Maximizing the machine speed A. It is better to lower forward barrier heights than to increase reverse barrier heights. Equation 3 of the main text constrains the ratio of forward to reverse rates of a particular transition by ∆µdiss, the total change in basic free energy across the cycle. Any particular change to the basic free energy change across a transition, expressed with the parameter λ in equation 6, may ‡ ‡ affect the forward barrier height gfm, the reverse barrier height grm, or some combination. In this work we focus solely on changes to the forward barrier heights because they have the greatest impact on machine speed. To see this, we can expand the absolute rates to express how changes in free energy will be split (1) between the non-mechanical and mechanical transitions (expressed with the coupling parameter λ as before), and (2) across forward vs. reverse rates (expressed with α): ‡ ‡ gfc = gc − α (1 − λ) ∆µ, ‡ ‡ grc = gc + (1 − α) (1 − λ) ∆µ, ‡ ‡ gfm = gm − αλ∆µ + wδ, ‡ ‡ grm = gm + (1 − α) λ∆µ − w (1 − δ) , [S6] ‡ ‡ where, as in the main text, gc and gm are intrinsic barriers common to both the forward and reverse transitions. We consider α ∈ [0, 1], λ ∈ [0, 1] and, for simplicity, we assume the number of substeps N = 1. The parameters α and λ are agnostic to exact mechanism. These changes in basic free energy may come directly from ∆µ (e.g., ATP hydrolysis), or from up- and down-hill free energy changes of the machine’s conformational cycle, as in the conformationally-driven mechanical step described in the main text. Using these equations for barrier heights, we ask: is it better for changes in basic free energy expressed through α to increase the forward rates (lower the free energy barriers to forward motion) or to decrease the reverse rates (increase the free energy barriers to reverse motion)? First, we write equation S5 as: J = Π2S/F2S, [S7] 2 of 13 Jason A. Wagoner and Ken A. Dill1 where Π2S = fcfm − rcrm, [S8] F2S = fc + fm + rc + rm. [S9] We find the derivative of J with respect to α at fixed efficiency: ∂J 1 ∂Π Π ∂F = 2S − 2S 2S 2 ∂α ∆µ,w F2S ∂α ∆µ,w F2S ∂α ∆µ,w 1 ∂F = J ∆µ − 2S F2S ∂α ∆µ,w ≥ 0. [S10] The last line follows from ∂F2S = (1 − λ) ∆µ (fc + rc) [S11] ∂α ∆µ,w +λ∆µ (fm + rm) . Because all rates are strictly positive and λ ∈ [0, 1], 1 ∂F 2S ≤ ∆µ. [S12] F2S ∂α ∆µ,w Inserting equation S12 into the middle line of equation S10 and recalling that ∆µ ≥ 0 shows that ∂J/∂α is strictly positive. Therefore, higher values of α optimize speed, and the unique maximum is found when α = 1: for our two-state model, it is always better for a change in free energy to lower the forward barrier than to increase the reverse barrier. This matches the principles of maximizing machine flux found by Brown and Sivak (2, 3), who also showed that more complex optimization principles may emerge for more complicated models and those with different intrinsic barrier heights. In the main text we focus exclusively on this ideal case when α = 1, where changes in basic free energy affect forward barrier heights with no impact to the reverse barriers, because these are the mechanisms that will have the greatest impact on machine speed. Though we do not explore it here, it is reasonable to expect that some machine mechanisms will perturb the reverse barrier heights. B. How to optimize forward barriers between the non-mechanical and mechanical steps. We next ask, how should dissipation be spread between the non-mechanical and mechanical transitions? In the main text equations 7, we consider the optimal case (α = 1) where changes to the dissipation across a transition will perturb the forward rate. To optimally parse free energy changes between the non-mechanical and mechanical transitions, we find the value of λ that maximizes speed for the rates given in equations 7, identical to the rates of equations S6 with α = 1. First, we express: ∂J 1 ∂Π Π ∂F = 2S − 2S 2S . [S13] 2 ∂λ ∆µ,w F2S ∂λ ∆µ,w F2S ∂λ ∆µ,w Noting that ∂Π2S = 0, we multiply throughout by F 2 and solve ∂λ ∆µ,w 2S ∂F 2S = 0. [S14] ∂λ ∆µ,w The solution to equation S14 gives λopt in equation 9 of the main text. C. What is the optimal number of mechanical substeps Nopt for a molecular machine?. First, we briefly derive the flux equation for the substep model. We define our model for a machine with N identical substeps to be a machine with 2N total states of alternating chemical and mechanical steps. Because they are identical, every (2-state) substep takes ∆µ/N as input and performs work w/N. Because the model is linear and periodic, we can view this 2N-state cycle as a 2-state cycle with N repeating units. Therefore, the equation for flux through the 2N-state model must match the equation for flux through the 2-state model, divided by N because the full cycle contains N repeating units of the 2-state unit. Therefore J2N (∆µ, w), the 1 flux through a machine with N substeps, is equivalent to N J (∆µ/N, w/N), where J is the flux of the two-state model given in equation S5, and dividing by N accounts for the mean transit time across N repeating units of this two-state unit. ∂J2N We use the substep model just described and solve ∂N = 0 to find Nopt. The value of Nopt depends on the amount of work w and the extent to which the mechanical step is rate-limiting. For example, if the machine has a very large non-mechanical ‡ barrier (a large value of gc ), Nopt will be smaller because there is less value in dividing the (relatively smaller) mechanical barrier. Figure S1 illustrates these two effects and shows that, under most conditions, Nopt is at least 4 and may be much larger. But, there are other practical considerations not included in our theoretical model. First, our model does not include the Jason A. Wagoner and Ken A. Dill1 3 of 13 Fig. S1. The optimal number of mechanical substeps, Nopt, of a molecular machine. The number of mechanical substeps that maximize speed depends both on the ‡ ‡ amount of work performed and the relative barrier heights of the mechanical (gm) to the non-mechanical (gc ) steps. As shown, a machine with more substeps has the greatest benefit at high work values or when the non-mechanical step is not substantially rate-limiting. 4 of 13 Jason A. Wagoner and Ken A. Dill1 portion of the intrinsic barrier height that would also be split up by the mechanical substeps (e.g., from large conformational changes of the machine). Including this component would increase the benefit of the mechanical substeps. Second, what are the design constraints on reaching a large N? We have shown that N = 4 is evolutionarily accessible by splitting the machine’s cycle between the four components of ATP hydrolysis. It may be difficult for a machine to achieve N > 4, though it is possible that substeps could be obtained from not only the components of ∆µ but also from up- and down-hill transitions in the conformational free energy. Third, these substeps may be applied to not only the mechanical step, but also other important transitions not included in our model. For example, the cytoskeletal walkers kinesin, myosin, and dynein have motor heads that undergo cycles of binding and release from their respective cytoskeletal tracks.

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