Mechanisms for achieving high speed and efficiency in biomolecular machines

Expressing Model Mechanisms with Basic Free-Energy Diagrams.

For illustrating mechanisms in this paper, we use diagrams of free-energy–like functions. But first, we describe what our diagrams mean. The observable quantities in Fig. 1 are the forward and reverse rates. We represent these rates in terms of the following free-energy barrier definitions,fc=k0e−βgfc‡rc=k0e−βgrc‡fm=k0e−βgfm‡rm=k0e−βgrm‡,[1]where k0 is a frequency factor and the free-energy barriers gfc‡, etc., are defined to be positive.

Now, if our machines were at equilibrium, we could illustrate their properties using standard free-energy diagrams based on expressions such asfcfmrcrm=e−βgc,eq+gm,eq=1,[2]where gc,eq, gm,eq are the equilibrium free-energy changes between the states. Since these are cyclic processes, the free energies must sum to zero around an equilibrium cycle; i.e., gc,eq+gm,eq=0.

However, here we are interested in steady-state processes, not at equilibrium. But we can capture insights using similar expressions such asfcfmrcrm=e−βgc+gm=eβΔμ−w=eβΔμdiss.[3]Here, Δμ is a chemical work that is done on the system and w is a work done by the system. Their difference, Δμ−w=Δμdiss≥0, is a dissipation of heat and due to differences in concentrations of reactants (ATP) and products (ADP and Pi), which constrains the forward and reverse rates in Eq. 3 (15⇓–17). So, the quantities gc and gm are not true equilibrium free-energy differences between states A and B. Instead, these quantities give the changes in basic free energy shown in Fig. 2. The basic free energy is an analog of an equilibrium free energy that is corrected for nonequilibrium effects, such as a difference in chemical potentials of species that are out of equilibrium (e.g., [ATP]), to give an equivalent kinetic and thermodynamic formalism (15). Eq. 3 shows that, across a full cycle, these quantities are constrained by the measured observables Δμ and w: gc+gm=−Δμ−w=−Δμdiss. We hope these basic free-energy landscapes are helpful for illustrating mechanisms below.

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Fig. 2.

The basic free-energy changes and barrier heights across a machine’s landscape. For this two-state model, gc is the basic free-energy difference from state Ai to Bi, while gm is the basic free-energy difference from state Bi to Ai+1. Across the full cycle, gc+gm=−Δμdiss. The free-energy barriers for forward transitions are labeled in green, and the barriers for reverse transitions are in red. We explore mechanisms that have different values of gc and gm along with corresponding changes to the barrier heights.

Now, in terms of the rate coefficients above, we can compute the cycle flux (number of full cycles per unit time) as (SI Appendix, section 1)J=fcfm−rcrmfc+fm+rc+rm.[4]For a cytoskeletal walker, the velocity is V=Jd for step size d. For an ion pump that transports n ions per cycle, the current I=nJ. We use “speed” to refer to these quantities (flux, velocity, or current). We define the thermodynamic efficiency of a machine asη=wΔμ.[5]

Defining Parameters for the Free-Energy Transduction Across the Machine Cycle.

The machine landscape in Fig. 2 is defined by the four barrier heights g‡ and the basic free-energy changes gc and gm. For any given pair of macroscopic observables Δμ and w, there are many possible ways to define these quantities. Here, we define three parameters λ, δ, and N that represent essential features of this landscape and of the machine’s free-energy transduction. We then explore machine optimization principles over these parameters.

First, let λ be the fraction of free energy from the input chemical work that is expended within the mechanical step. Thus, 1−λ is the fraction expended across the nonmechanical step:gc=gfc‡−grc‡=−1−λΔμgm=gfm‡−grm‡=−λΔμ+w.[6]This quantity λ describes an inherent feature of the machine cycle, independent of w, and implicitly includes two contributions, from the input chemical energy Δμ and from changes in conformational free energy, which must sum to zero across the full cycle. In short, when λ=0, the decrease in basic free energy that is available from input chemical energy is expended in the first, nonmechanical step (gc=−Δμ) and all of the work is performed in the mechanical step (gm=w). In contrast, when λ=1, there is no change in basic free energy across the first step (gc=0) and all of the energetic changes happen in the second step (gm=−Δμ+w=−Δμdiss) (Fig. 3A).

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Fig. 3.

The basic free-energy landscapes of two theoretical machines. The first machine (orange) has most of its basic free energy drop across the nonmechanical step, λ=0.2. The second machine (blue) has most of its basic free energy drop across the mechanical step, λ=0.8. Here, gc‡=gm‡. (A) When no work is performed, these machines have the same speed. Each machine has one small and one large barrier to traverse. (B) At larger values of work, the second machine (λ=0.8, blue) is much faster. The first machine (λ=0.2, orange) has one very small and one very large barrier to traverse, while the second machine has more evenly distributed barriers. Here, w=0.6Δμ.

Second, let δ represent the location of the transition state along the mechanical step, which dictates how a change in w will affect the forward gfm‡ and reverse grm‡ barrier heights (7⇓–9). Together, λ and δ define the barrier heightsgfc‡=gc‡−1−λΔμ  grc‡=gc‡gfm‡=gm‡−λΔμ+wδ grm‡=gm‡−w1−δ,[7]where gc‡ and gm‡ are intrinsic barriers common to both the forward and reverse transitions. We have assumed that changes in λ affect the forward and not reverse barrier heights because these are the changes that will have the greatest impact on machine speed (SI Appendix, section 2.A).

Finally, let N define the number of mechanical substeps used by the machine. When N=1, the machine has a single mechanical step (two total states illustrated in Fig. 1). For a machine with N substeps, the cycle has 2N total states of alternating chemical and mechanical steps. For simplicity, we assume each of these N substeps is identical. From Eqs. 4 and 7, JΔμ,w is the speed of a machine with a single mechanical step (two total states) and J2N=1NJΔμ/N,w/N is the speed of a machine with N substeps (2N total states of alternating chemical and mechanical steps) (SI Appendix, section 2.C).

Maximizing Machine Speed by Varying the Basic Free-Energy Drop Across the Mechanical Step.

We first explore how machine speed and efficiency depend on the λ parameter, which determines the drop in basic free energy across the mechanical step. Here, for simplicity, we assume the machine has a single mechanical step (N=1). Using Eq. 7 to define the barrier heights, we solve ∂J/∂λ=0, with all other model quantities (δ, gc‡,gm‡) fixed, to find λopt (SI Appendix, section 2.B):λoptΔμ,w=12+wδ2Δμ+gm‡−gc‡2Δμ.[9]Inserting λopt into the definitions of the barriers, Eq. 7, gives the result that the machine operates fastest when the two barrier heights are equal, gfc‡=gfm‡=0.5gc‡+gm‡−Δμ+wδ. This optimization principle for molecular machines is similar to the principle of Knowles, who showed for enzymes that turnover rates are maximal when the rates of binding reactants and releasing products are equal (12, 13), and the work of Brown and Sivak who, using a two-state cycle, showed that machine speed is optimized when the forward rates are equal (10).

Here is more intuition about Eq. 9. The first term on the right-hand side of Eq. 9 (=1/2) shows that the chemical input free energy should be distributed evenly across the cycle, all else being equal. The second term expresses that if the machine performs much work, there is a large barrier in the mechanical step, and hence more of the available basic free energy should be expended across the mechanical step. The third term shows that the optimal distribution of basic free-energy changes will depend on the intrinsic free-energy barriers of the steps gc‡ and gm‡. That is, if the mechanical step has a large intrinsic barrier [e.g., from the large entropic barrier that accompanies the conformational change in myosin V’s 36-nm step (18)], then more of the basic free energy should be expended in the mechanical step to reduce this intrinsic barrier. Alternatively, if the chemical barrier is intrinsically high, then less of the basic free energy should be expended in the mechanical step to achieve optimal speed.

Fig. 3 shows the landscapes of two theoretical machines. The first one (orange, λ=0.2) has most of the total basic free-energy drop across the first, nonmechanical step. The second one (blue, λ=0.8) has most of the basic free-energy drop across the mechanical step. When no work is performed, each machine has one small and one large barrier (Fig. 3A), and they have identical speeds. When work is performed, the second machine (λ=0.8) is faster and has more evenly distributed barrier heights (Fig. 3B). Fig. 4 shows the speed–efficiency tradeoffs. A machine with a large decrease in basic free energy across the mechanical step (large λ) is faster at high efficiency.

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Fig. 4.

The speed–efficiency tradeoffs for a machine with λ=1 (blue) and λ=0 (orange). The machine with a large basic free-energy drop across the mechanical step (λ=1) is faster at high efficiency. Here, gc‡=gm‡ and δ=1.

A Conformationally Driven Mechanical Step Optimizes Speed.

Here, we give a molecular interpretation of λopt from Eq. 9 in terms of a machine’s conformational free-energy landscape. The idea is that, through changes in the conformational free energy, the input free energy can be stored in the machine in the first, chemical step and then expended in the mechanical step. The conformational free energy contributes to the change in basic free energy across a transition. Because changes in conformational free energy must sum to zero across a full cycle, these contributions are implicitly included in the parameter λ.

We assume here that all of the chemical free energy Δμ is released across the first transition. We label the change in conformational free energy across this chemical step as ΔGconf. The conformational free energy across the mechanical step is then −ΔGconf. We expand the definition of basic free-energy changes from Eq. 6 to explicitly include these terms:gc=gfc‡−grc‡=−(1−λ)Δμ=−Δμ+ΔGconf,gm=gfm‡−grm‡=−λΔμ+w=−ΔGconf+w.[10]Eq. 10 shows that λ=ΔGconf/Δμ. So, using Eq. 9, we can now express what conformational free energy optimizes the machine speed:ΔGconf, optΔμ,w=Δμ2+wδ2+gm‡−gc‡2.[11]Fig. 5 shows the prediction of Eq. 11 that, under most conditions, the fastest machines will be those having a relatively large and specific value of conformational free-energy storage. The machine takes up and stores this free energy in the chemical step and expends that free energy in the mechanical step, like the cocking and releasing of a spring.

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Fig. 5.

To achieve maximum speed, a molecular machine has an optimal change in conformational free energy. Shown is the speed of a theoretical machine with respect to the conformational free-energy change across the mechanical step (−ΔGconf). Here, Δμ=20 kT, w=12 kT, gc‡=gm‡, and δ=1.

Fig. 5 shows that a machine having a suboptimal conformational change can run orders of magnitude slower. Conformational energy storage modulates the barrier heights across the machine cycle. A machine with small ΔGconf would have one small barrier and one large barrier (orange curve in Fig. 3B). Larger conformational free-energy storage will raise the barrier of the chemical step and lower the larger barrier of the mechanical step. This serves to more evenly distribute the barrier heights (blue curve in Fig. 3B).

Machines Are Faster That Have an “Early” Transition State.

Our model also allows for a parameter δ, which can be interpreted as the position of the transition state along the machine’s mechanical step (Eq. 7). It has been shown that machines are faster if they have a small value of δ (7, 8) and we have shown that this conclusion is general (9). That is, ∂J/∂δ<0 under all conditions and for arbitrarily complex machine models (which include futile hydrolysis cycles, ATP-driven backsteps, etc.). A smaller value of δ (i.e., a transition state close to the initial state) always corresponds to a machine that is faster, more powerful, and more efficient at constant velocity. And it corresponds to a favorable, concave speed–efficiency tradeoff as in the blue curve in Fig. 4.

We note that a machine’s power stroke, depending on how it is defined, may correspond to this optimal value of δ (8, 9, 19) or to the optimal value of λ (20, 21). We do not advocate any specific definition here, but by either definition, a power stroke may refer to the release of conformational strain across a machine’s mechanical step.

Speeding up Machines with Mechanical Substeps.

Machine speed can be optimized by breaking the large output step into smaller alternating chemical and mechanical steps, which we call mechanical substeps. A machine that has four 10-kT–sized barriers is faster than a machine with one 40-kT barrier. This principle applies broadly but is best illustrated for a machine where all of the chemical energy is taken up in the nonmechanical step and all of the work is performed in the forward mechanical step (λ=0 and δ=1). Consider a set of machines that have different numbers N of mechanical substeps but that are otherwise identical. As discussed above and in SI Appendix, section 2.C, if JΔμ,w is the speed of a machine with a single mechanical step (two total states), then J2N=1NJΔμ/N,w/N is the speed of a machine with N substeps (2N total states of alternating chemical and mechanical steps).

The mechanical substeps reduce and distribute the free-energy barriers of the machine, as shown in Fig. 6. Fig. 6, Inset shows that, when performing a large amount of work, a machine with more substeps is several orders of magnitude faster than an analogous machine with one large step. Biological machines can leverage the four separate downhill components of ATP hydrolysis (ATP binding, hydrolysis, ADP release, and Pi release) to split up their work steps. This is observed in the myosin family, where N=2 for the 6-nm step of myosin II (22), and N=3 for the 36-nm step of myosin V (18).

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Fig. 6.

Splitting one large work step (black) into two (red) or four (blue) mechanical substeps. ξ is the reaction coordinate. Modifying a molecular machine to have more substeps reduces the barrier height along the mechanical step. Inset shows that, when performing a large amount of work, a machine with several mechanical substeps can be several orders of magnitude faster than a machine with a single mechanical step. Speed is plotted on a log scale.

Similarly, FoF1-ATPase uses mechanical substeps. In a recent study, Anandakrishnan et al. (6) used a simple kinetic model to study the evolutionary optimization of this motor. Their work shows that the rotary mechanism of FoF1-ATPase optimizes speed with the same principle described here for mechanical substeps—the rotary mechanism reduces and distributes the kinetic barriers of the machine cycle. This motor takes, as input, the downhill transport of 8–15 protons, depending on the species (23). The rotary mechanism intersperses these downhill steps with the synthesis of three ATP molecules (output), each of which is further subdivided into separate work steps of reactant binding, synthesis, and product release (24, 25). This gives a more even application of torque across the angular coordinate of the motor (5).

Other biological machines also use mechanical substeps. In SI Appendix, section 2.C, we use Eq. 8 to calculate Nopt for different machines and to discuss various practical considerations.

Biological Machines Are Both Fast and Efficient.

Fig. 7 shows that biomolecular machines in vivo tend to have high efficiencies. For many machines, these in vivo values of input free energy Δμ and output work w used to calculate efficiency will vary, depending on cell conditions. But for the machines in Fig. 7, these values are known under typical conditions and appear to be fairly consistent (SI Appendix, section 3). These machines are as efficient as macroscopic electric motors and more efficient than heat engines; η>50% for 10 of the 12 machines shown.

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Fig. 7.

The in vivo efficiencies of molecular machines. Included are Na-K ATPase; SERCA; the proton PPi pump; the plasma membrane proton pump; PMCA; V-ATPase; myosin II; NCX; NCKX; and animal, Escherichia coli, and chloroplast FoF1-ATPase. The efficiency η=w/Δμ, where these values of Δμ and w are calculated from experimental data (SI Appendix, section 3).

Fig. 8 shows the speed–efficiency relationships for various biomolecular machines as measured from in vitro experiments. These machines have tradeoff curves that are much more favorable than would occur in a single Arrhenius barrier process, also shown for reference.

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Fig. 8.

The speed–efficiency relationships for six molecular machines. Speed is normalized with respect to the maximum speed of each machine. These machines maintain a high speed even while operating at high efficiency. The black line shows a reference model of a single-barrier, irreversible, Arrhenius process. Data are from refs. 30 and 32⇓⇓⇓–36. Fits to the data are described in SI Appendix, section 4.

We now rationalize some of the differences in the speed–efficiency behaviors shown in Fig. 8. (We do not discuss F1-ATPase because these speed–efficiency data, obtained for the isolated F1 domain running in the hydrolysis direction, do not match the motor’s in vivo function.) Consistent with the model, machines with the best speed–efficiency curves in Fig. 8 have more substeps, N. And these machines all have large values of λ, except for RNAP, which has the worst speed–efficiency tradeoff. Kinesin and dynein also have small values of δ, the transition-state location. See SI Appendix, section 4 for the values of N, λ, and δ (estimated from model fits to the in vitro single-molecule data from Fig. 8).

Na-K ATPase has the best speed–efficiency tradeoff of the machines in Fig. 8. This pump hydrolyzes ATP as input to export three Na⁺ ions from the cell and import two K⁺ ions uphill against their chemiosmotic gradients. This machine has such a good tradeoff because the output work is delocalized across the entire cycle and therefore the rate-limiting work step (the release of one of the Na⁺ ions) is only weakly dependent on the total work of the machine (26). This pump effectively has many substeps, although we do not report an exact value since it is difficult to estimate the work contribution over the various transitions (the occlusion, transport, and release of each of the five ions).

The cytoskeletal transport motors myosin V, kinesin, and dynein are fast at high efficiency, but they have fewer substeps than Na-K ATPase [N=3 for myosin V (18)]. The tradeoffs of these three motors may also be affected by the run length L, the average number of full cycles a motor completes before falling off its cytoskeletal track. At zero load, L≈25 for myosin V (27), L≈100 for kinesin (28), and L≈250 for dynein (29). For these three motors, the trend in L (dynein > kinesin > myosin V) is anticorrelated with the speed–efficiency behavior (myosin V > kinesin > dynein). It is possible that the motor’s run length introduces an extra tradeoff, where different biological mechanisms can use the available free energy to achieve a greater run length, but at the expense of speed or efficiency.

RNAP has the worst speed–efficiency behavior of the machines in Fig. 8. RNAP synthesizes RNA from a template DNA strand (30). As output work, RNAP generates force that melts the DNA template. This output work is physically separated from the machine itself—RNAP breaks hydrogen bonds that are several base pairs upstream from the motor (31). This physical separation may make it difficult or impossible for RNAP to use the speed-optimizing strategies we have described. The motor does not use mechanical substeps and it has a small value of λ=0.22, in agreement with the observation that it operates as a Brownian ratchet (30).