A continuum model for the growth of dendritic actin networks
AA continuum model for thegrowth of dendritic actin networks
Rohan Abeyaratne and Prashant K. Purohit Department of Mechanical Engineering,Massachusetts Institute of Technology,Cambridge, Massachusetts, 02139, USA Department of Mechanical Engineering and Applied Mechanics,University of Pennsylvania,Philadelphia, Pennsylvania, 19104, USAJune 15, 2020
Abstract
Polymerization of dendritic actin networks underlies important mechanical processes incell biology such as the protrusion of lamellipodia, propulsion of growth cones in dendritesof neurons, intracellular transport of organelles and pathogens, among others. The forces re-quired for these mechanical functions have been deduced from mechano-chemical models ofactin polymerization; most models are focused on single growing filaments, and only a fewaddress polymerization of filament networks through simulations. Here we propose a contin-uum model of surface growth and filament nucleation to describe polymerization of dendriticactin networks. The model describes growth and elasticity in terms of macroscopic stresses,strains and filament density rather than focusing on individual filaments. The microscopic pro-cesses underlying polymerization are subsumed into kinetic laws characterizing the change offilament density and the propagation of growing surfaces. This continuum model can predictthe evolution of actin networks in disparate experiments. A key conclusion of the analysisis that existing laws relating force to polymerization speed of single filaments cannot predictthe response of growing networks. Therefore a new kinetic law, consistent with the dissipa-tion inequality, is proposed to capture the evolution of dendritic actin networks under differentloading conditions. This model may be extended to other settings involving a more complexinterplay between mechanical stresses and polymerization kinetics, such as the growth of net-works of microtubules, collagen filaments, intermediate filaments and carbon nanotubes.
Actin polymerization drives the protrusion of lamellipodia and pseudopodia that play importantroles in cell sensing and motility [6, 19, 31]. Actin polymerization is also the mechanism behindmotion of growth cones in dendrites of neurons, organelles in cells, as well as pathogens, such as,1 a r X i v : . [ phy s i c s . b i o - ph ] J u l isteria and Shigella [12]. As such actin based motility has been studied using experiment andtheory for at least the last three decades [19]. It is now understood that the propulsive force in actinpolymerization is a result of the difference in chemical potential of the actin monomers in theirbound (to the filament) and unbound (in solution) states [7, 12].Early Brownian ratchet models [30] for the force generated by actin polymerization focusedon a single filament polymerizing against a load represented by a bead moving through a viscousmedium, and the goal was to relate the force on the filament with its elongation-rate measured bythe velocity of the bead [19, 27]. In these models thermal fluctuations of the filament and the beadcreated gaps between the bead and the polymerizing tip and resulted in an exponential dependenceof the velocity V of the bead and the force f on the filament (taken to be positive in compression): V = V exp( − ζf /f stall ) − exp( − ζ )1 − exp( − ζ ) , (1)where V is the velocity at f = 0 and f stall is the force at which polymerization stalls [27]; ζ is related to the size of the monomers a through ζ = af stall / ( k B T ) where k B is the Boltzmannconstant and T the absolute temperature. The stall force for actin was estimated to be around − [19] and this was confirmed through experiments on growing bundles of a few actinfilaments with forces exerted using an optical trap [12]. However, dendritic actin and the actingel in Listeria ‘comet tails’ is not a bundle of parallel growing filaments; rather, it is branchedand cross-linked with filament lengths that are in the range of 1 µ m or less [7, 24, 29]. Thus, indendritic actin a large number of short filaments exert forces on the lamellipodial membrane or themoving object (pathogen, bead, organelle, etc.) resulting in an elongation rate-force curve thatis not necessarily of the form (1). This was shown in subsequent models which accounted for thetransient nature of the contact between the load surface and growing filaments and the elasticityof the filament network [16, 26]. These models and others [22] recognized that polymerization indendritic actin occurs in a narrow zone next to the load surface (not all over the network), so thatthe barbed ends of most of the growing actin filaments are pointed toward the load surface.Experiments to measure the elongation rate-force relation of dendritic actin are scarce. Oneexperiment that yielded intriguing results was that of Parekh et al. [29] who polymerized dendriticactin between two atomic force microscope (AFM) cantilevers. The deflection of the AFMs al-lowed them to accurately measure both forces and elongation-rates. They were also able to controlthe conditions under which polymerization occurred (e.g., constant force, a force proportional tonetwork height, etc.). They found that the elongation-rate was independent of the force over a rangeof force and this was followed by a convex curve in which polymerization stalled over a short rangeof force; this elongation rate-force relation is very different from the exponential form in equation(1). Parekh et al. also demonstrated that the elongation-rate was loading history dependent and inparticular that there could be two (or more) steady state elongation-rates at the same force. Someof these findings were confirmed in a later experiment by Brangbour et al. [7] who used magneticbeads to control the force resisting polymerization. The actin polymerized between two beads with We will soon distinguish elongation-rate from growth-rate, but we note that they are the same when the monomersdo not deform. et al. [7] and Parekh etal. [29] through a continuum model of growth. Continuum models of growth fall into two broadcategories: volumetric growth and surface growth. In the former, new material is added to existingmaterial points whereas in the latter, new material points are added to the body at its evolvingboundary. One way in which to track the newly added material points is through a continuouslyevolving reference configuration . The driving force for surface growth is essentially the configu-rational force associated with this evolving reference configuration and can be calculated in severaldifferent (roughly equivalent) ways. The bio-physical micro-mechanisms underlying the growthprocess are captured at the continuum scale by a kinetic law for growth, with the second law ofthermodynamics imposing certain restrictions on such a kinetic law.In our continuum model the body is treated as a one-dimensional nonlinearly elastic barwhose length at time t is (cid:96) ( t ) in physical space and (cid:96) R ( t ) in reference space. The body we areconcerned with is comprised of actin filaments (F-actin) – long polymer chains – formed by theassembly of actin monomers (G-actin). It is surrounded by a pool of free actin monomers that,under suitable conditions, can bind to the tips of the existing filaments leading to their growth.The length (cid:96) R ( t ) evolves due to the addition of new material points. In addition, the formation ofnew filaments leads to a time-dependent filament density ρ ( t ) [29]. At this juncture, it is worthemphasizing the distinction between the elongation-rate ˙ (cid:96) and the growth-rate ˙ (cid:96) R . The former isthe rate of change of the length of the specimen in physical space, a quantity that can be observedand measured. Figure 4 shows a plot of ˙ (cid:96) versus the force σA as predicted by our model in aparticular setting; note the commonly observed rapidly declining nature of this curve. On the otherhand, any change in the length (cid:96) R in reference space occurs solely due to growth. This is in contrastto (cid:96) which changes due to both growth and stress. The relation between ˙ (cid:96) R and force is an inputinto the model, the kinetic law for growth. The particular relation used to predict Figure 4 is showndotted in Figure 8, and the qualitative distinction between the curves in the two figures is worthnoting.A phenomenon that brings out the distinction between growth and elongation is ‘tread-milling’, commonly observed in growing actin filaments [6, 19]. During treadmilling, the filamentlength (cid:96) (in physical space) remains constant because of a precise balance between the rates atwhich monomers attach to one end (growth) of a filament and detach from the other. Thus, growthoccurs continuously, but elongation does not. A recent paper [2] analyzes the existence and sta- There is a vast literature on this topic which we do not attempt to review here. Some books and review articlesthat the interested reader can refer to include [3, 5, 8, 11, 15, 18, 20, 21, 25, 32, 35, 38] The seminal paper by Skalak et al. [34] appears to be the first to model the kinematics of this. For a completetreatment of the kinematics, mechanics, thermodynamics and kinetics of surface growth, see, e.g. [36]. See for example [4, 13, 14, 36, 37] stall which occurs when the compressive force is so large that it preventsthe addition of new monomers [6, 19].In the rest of this section we describe (with little justification) some key aspects of our model.These will be explained in detail in later sections. Let σ and λ denote the compressive stress andstretch in the body. Since the material stiffens with increasing compressive stress, we take a simplestress-stretch relation that captures this phenomenon: σ = E ( λ − − . The tangent modulus ofthe material, − ∂σ/∂λ , then increases with increasing σ . The elastic modulus E could also dependon the filament density ρ . We are concerned with two settings: in the first (pertaining to Brangbour et al. [7]), the number of actin filaments is relatively small and therefore do not form a cross-linkednetwork. Consequently, filament bending is unimportant and so we take the elastic modulus todepend linearly on the filament density: E ( ρ ) ∼ ρ . In the second setting (pertaining to Parekh et al. [29]) the filaments form a cross-linked network where filament bending plays an importantrole. Therefore in this setting we take the modulus to depend quadratically on the filament density: E ( ρ ) ∼ ρ . Moreover, in the first setting the filament density remains constant, ˙ ρ = 0 , whereasin the second, it changes over time which we model through a suitable evolution law ˙ ρ = R ( ρ ) .In both settings, new material points can be added to one end of the specimen leading to surfacegrowth through polymerization. The growing boundary propagates at a speed V given by a kineticlaw V = V ( f ) where f = σ/ρ is the filament force. Finally, in the first setting, both ends of thespecimen are attached to supports and the length or force on the specimen can be controlled. In thesecond setting, one end is attached to an AFM cantilever modeled as a Hookean spring so that, inaddition to controlling the force, it is also possible to allow the specimen to freely evolve againstthe spring. Because of the way in which the experiments are set-up, polymerization only occursat one of the end of the specimen. At the other end, neither polymerization nor depolymerizationtakes place.This paper is organized as follows: in Section 2 we describe the basic aspects of the modelthat are common to both settings studied in the subsequent sections. Section 3 is concerned withproblems where the filament density is time independent. Further details of the model are givenin Section 3.1; the response of the model to various loading programs are examined in Section 3.2and compared with the experimental measurements of Brangbour et al. [7]; and finally in Section3.3 we explain how our model describes the setting in [7] even though they might seem different atfirst glance. We then turn in Section 4 to the growth of a network of actin filaments. The detailedconstitutive model is presented in Section 4.1 and the values of the various parameters are given inSection 4.1.4. In Section 4.2 we study the response of the specimen when it is growing under theaction of the spring, and in Section 4.3 we examine its response to loading programs that involvea sudden change in the force. The results of both sections are compared with the experimentalmeasurements of Parekh et al. [29]. In the appendix we derive the dissipation inequality, identifythe driving force, and examine the kinetic laws considered in Sections 3 and 4 in the context of the4issipation inequality. The Supplementary Material provides details on how the parameter valueswere chosen, the calculations in Section 4.3 were carried out, and the conformity of the solutionsin Sections 3 and 4 with the dissipation inequality. Since we will examine two rather different experimental settings in this paper, here we simplywrite down those constituents of the mathematical model that are common to both. More detaileddescriptions will be given in Sections 3.1 and 4.1.Imagine a test specimen composed of actin filaments held between two supports. One of thesupports will be compliant in Section 4 as shown schematically in Figure 7. Each filament is apolymeric molecule, comprised of a linear assembly of monomers. The filaments are surroundedby a solvent containing free monomers that can attach to one end of each filament (called thebarbed end in actin).In physical space the specimen is identified with the interval [ y ( t ) , y ( t )] at time t and so itscorresponding length is (cid:96) ( t ) = y ( t ) − y ( t ) . In reference space it is identified (at the same instant t ) with the interval [ x ( t ) , x ] where x ispermitted to be a function of time because growth may occur at that end. No growth occurs at x .The length of the specimen in reference space at time t is (cid:96) R ( t ) = x − x ( t ) . (2)The length in physical space is affected by both stress and growth, whereas the length in referencespace is only affected by growth. When monomers are added to the specimen at its left-handboundary, the specimen grows through the leftward motion of that surface in reference space at aspeed V = − ˙ x . (3)Thus growth corresponds to ˙ x < (and therefore by (2) to ˙ (cid:96) R > ). If there is no growth, x ( t ) isconstant. In order not to confuse the two velocities ˙ (cid:96) and ˙ (cid:96) R , we shall refer to ˙ (cid:96) as the elongation-rate and ˙ (cid:96) R as the growth-rate . In the problems of interest to us the stress and stretch fields arespatially uniform and so the stretch λ of the specimen relates the reference and current lengths: (cid:96) = λ (cid:96) R . (4)Let N ( t ) denote the number of load bearing filaments in the specimen and let A be the(fixed) area over which they are distributed. The filament density ρ ( t ) is defined as the number offilaments per unit cross-sectional area: ρ := N/A. Growth only occurs at one end of the specimen in the experiments of Parekh et al. [29] because the support atthat end was functionalized with an actin nucleating agent.
5f the force in each filament is f , the total force in the specimen is f N and so the stress is relatedto the filament force and filament density by σ = f N/A = f ρ. (5)We are concerned exclusively with compressive stress and so take σ to be positive in compression.The key variables in the model are the stress σ ( t ) , stretch λ ( t ) , filament density ρ ( t ) , lengthof the specimen in physical space (cid:96) ( t ) and length of the specimen in reference space (cid:96) R ( t ) . Thestress depends on the stretch and the filament density through a constitutive relation σ = (cid:98) σ ( λ, ρ ) .This relation is assumed to be invertible so that we can write λ = Λ ( σ, ρ ) . (6)Growth occurs by two mechanisms, one leading to an increase in (cid:96) R , the other to an increasein ρ . Typically, the specimen involves “load-carrying filaments” that extend from one support to theother. Monomers from the surrounding monomer pool can be added to the tips of such filaments atthe functionalized support at x . This changes their reference length (cid:96) R . There may also be “freefilaments” in the specimen, for example with one end attached to the existing polymer network andthe other free. Monomers can be added to the free ends of such filaments. When a free filamentgrows, it may eventually touch one of the supports and turn into a load-carrying filament of thefirst type. Such growth changes the density ρ of load-carrying filaments.We assume that the rate of increase of the number of load-bearing filaments, ˙ ρ , is governedby a kinetic law of the form ˙ ρ = R ( ρ ) , (7)where the kinetic response function R is allowed to depend on ρ but not σ . Since a “free filament”remains stress-free while it grows by polymerization at its free tip, such growth is expected to beunaffected by stress and hence we have taken R to be independent of σ . As for surface growth dueto polymerization, we assume that the speed of the moving boundary in reference space is relatedto the filament force by a kinetic relation of the form V = V ( f ) = V ( σ, ρ ) . (8)Since the kinetic response function V is a function of the filament force, it depends on both stressand filament density. Similar to earlier work [26], we write V as a function of the (average) filamentforce f = σ/ρ to facilitate comparison with kinetic laws such as equation (1) that are given in theliterature in terms of f for single growing filaments [19].The 5 variables (cid:96) ( t ) , (cid:96) R ( t ) , σ ( t ) , λ ( t ) and ρ ( t ) are governed by the equations (4), (6), (7) and(8), together with (2), (3) and a loading condition such as the prescription of the stress history. In this section we model the test specimen as involving a fixed number of nearly parallel filamentsas in the experiments of Brangbour et al. [7]. The filament density therefore does not change and6he constitutive relation of the specimen can be deduced from that of a single filament. Moreover,there is no cross-linking and filament bending is not important.
We now make the following choices for the constitutive response functions Λ , V and R . The stress σ in the network is a function of the stretch λ that vanishes when λ = 1 and becomes unbounded atextreme compression when λ → + . Moreover the material stiffens with increasing compressivestress. A simple model capturing this is σ = E ( λ − − , λ = Λ ( σ, ρ ) = 11 + σ/E where E = E ( ρ ) := ρE (9)is the effective Young’s modulus of the specimen. Note that the network elastic modulus is chosento be (proportional to the number N of filaments and therefore) linear in the filament density ρ since the elasticity of the network in [7] is of entropic origin such that E ∝ k B T . The tangentmodulus E t – the slope of the stress-stretch curve – is given as a function of stress by E t ( σ ) = − dσdλ (cid:12)(cid:12)(cid:12)(cid:12) λ =[1+ σ/E ] − = E (1 + σ/E ) . (10)As shown in Figure 1, the graph of E t ( σ ) versus σ according to (10) rises monotonically andis concave upwards which agrees with the trends observed experimentally, e.g. see Figure 3 ofChaudhuri et al. [9].Figure 1: Tangent modulus E t ( σ ) /E versus stress σ/E according to (10). The horizontal axis hasbeen drawn on a logarithmic scale so as to facilitate comparison with Figure 3 of [9].In this section, the number N of load-bearing filaments, and therefore the filament density ρ ,is assumed to remain constant during each experiment. Thus ˙ ρ = R ( ρ ) = 0 . V provided the filament force f is less thata certain critical value f stall , or equivalently, when the stress is less than a certain value σ stall .Accordingly we take the kinetic law for surface growth (8) to be V = V ( f ) = (cid:40) V for f < f stall , f > f stall , V = (cid:40) V for σ < σ stall , σ > σ stall , (11)where σ = ρf, σ stall = ρf stall . This piecewise constant kinetic relation is depicted by the dottedline in Figure 8. Such kinetic relations arise in other mechanics settings where they have beenassociated with a notion of maximum dissipation, e.g. plasticity [23, 33], phase transitions [1] andkink band motion [10]. They are referred to as maximally dissipative kinetics . In this section we shall consider various predictions of the model described above. Most of theresults can be readily and conveniently described using nondimensional variables. However, weshall continue to use the (dimensional) variables introduced above since that allows us to makequantitative comparisons with the experimental measurements of Brangbour et al. [7]. A briefdescription of their experiments and a discussion of how their apparently different model relatesto ours, is postponed to Section 3.3. Details on how the various parameter values were chosen isdescribed in Section S1 of the Supplementary Material.By (4) and (9) the length (cid:96) of the specimen at any instant is related to the stress and unstressedreference length at that instant by (cid:96) ( t ) = (cid:96) R ( t )1 + σ ( t ) /E . (12)In the first set of calculations, the filaments are initially permitted to grow freely under zerostress for some time interval [0 , t ) . During this stage, the unstressed reference length of the spec-imen is (cid:96) R ( t ) = V t as required by (11), (2) and (3). At time t the force on the specimen isincreased rapidly, and the force σA and specimen length (cid:96) are measured. If the loading-rate ismuch faster than the rate of growth, the value of (cid:96) R can be assumed to remain constant at thevalue (cid:96) R = V t . Figure 2 shows a plot of the force σA versus the specimen length (cid:96) in such anexperiment as predicted by (12) for two different fixed values of (cid:96) R = V t . This figure may becompared with Figure 3 of Brangbour et al. [7].In the second set of calculations, the filaments are again permitted to grow freely under zerostress for some initial time interval [0 , t ) during which (cid:96) R ( t ) = V t . At time t a force is appliedon the specimen and held constant. Then from (11) and (12), provided σ < σ stall , (cid:96) ( t ) = V t σ/E , t > t . (13)8igure 2: Force σA versus length (cid:96) according to (12) when the loading-rate is much faster than thegrowth-rate allowing (cid:96) R to be treated as constant. The figure has been drawn for EA = 2 .
259 pN and two different values of (cid:96) R . The vertical axis is on a logarithmic scale. Compare with Figure 3of [7].Figure 3 shows a plot of the specimen length (cid:96) ( t ) versus time t for three different fixed values ofthe force σA . Observe that the slope decreases as the force increases. This figure may be comparedwith Figure 1c of Brangbour et al. [7]. Figure 3: The length of the specimen (cid:96) versus time t according to (13) at two different values of theforce σA . In order to compare this with Figure 1c of Brangbour et al. [7], we have plotted (cid:96) + 2 R on the vertical axis where R = 550 nm . The figure has been drawn using V = 0 .
42 nm / sec , t =1000 sec and EA = 3 .
764 pN . The three lines correspond to σA = 0 . ; σA = 3 pN ; and σA = 17 pN .The elongation rate-force response plays a central role in the study of actin filament growth.Thus consider a third set of calculations in which the force is again held constant but now exam-ine the elongation-rate ˙ (cid:96) as a function of force σA and in particular on how it depends on EA . frequently referred to as the force-velocity relation in the literature, t at constant σ gives ˙ (cid:96) = V σA/ ( EA ) , (14)and Figure 4 shows a plot of the elongation-rate ˙ (cid:96) versus the force σA according to (14). The twocurves correspond to EA = 3 .
387 pN and EA = 0 .
979 pN . This figure may be compared withFigure 5 of Brangbour et al. [7].
Figure 4: Elongation-rate ˙ (cid:96) versus force σA at two different values of EA according to (14). Thefigure has been drawn for V = 0 .
42 nm / sec . Compare with Figure 5 of [7]. Fourth, suppose that the specimen is initially growing under some constant stress σ andthat at some instant t the stress is suddenly increased to a much larger value σ (less than thestall stress). The stress is kept constant at this higher value until time t at which instant it issuddenly decreased back to its original value σ and held constant at that value from then on. Thisis described by the loading history σ ( t ) = σ for 0 < t < t ,σ for t < t < t ,σ for t > t , where 0 < σ < σ ( < σ stall ) . (15)The corresponding response is found by solving (12), (11), (2) and (3): (cid:96) ( t ) = (cid:96) R ( t )1 + σ ( t ) /E , ˙ (cid:96) R ( t ) = V . (16)In solving (16) it is important to keep in mind that the referential length of the filament (cid:96) R ( t ) changes only due to growth. Thus at an instant at which the stress changes discontinuously, thelength (cid:96) ( t ) of the filament will also change discontinuously due to the jump in σ but the referential10ength will remain continuous since a finite segment of new material cannot appear in infinitesimaltime. Thus from (15) and (16) (cid:96) ( t ) = V t σ /E for t < t < t , V t σ /E for t < t < t , V t σ /E for t > t , ˙ (cid:96) ( t ) = V σ /E for t < t < t , V σ /E for t < t < t , V σ /E for t > t . (17)In Figure 3 we observed that there is effectively a one-parameter family of straight lines on the t, (cid:96) -plane, the force σA = F being the parameter. During the loading (15), the point ( t, (cid:96) ( t )) firstmoves on the straight line corresponding to σA = F in Figure 5. At time t it jumps down to thestraight line associated with σA = F and traverses that line for t < t < t , and finally at time t it jumps back up to the straight line σA = F where F i is the force. This figure may be comparedwith Figure 2 of Brangbour et al. [7].
500 600 700 800 900 100011001150120012501300135014001450
Figure 5: The length of the specimen (cid:96) versus time t according to (17) with σ A = 0 . and σ A = 39 pN . In order to compare this with Figure 1c of Brangbour et al. [7], we have plotted (cid:96) + 2 R on the vertical axis where R = 550 nm . The figure is drawn for V = 0 .
42 nm / sec and EA = 3 .
764 pN . See also Figure 3. et al. [7].
Brangbour et al. [7] devised a novel method for measuring the force-velocity relation for a systeminvolving a relatively small number of growing actin filaments. They used a suspension containing(actin monomers as well as) magnetic colloidal particles that when subjected to a magnetic fieldassembled into a linear chain. The surfaces of the colloidal particles were functionalized so thata certain controlled number of actin filaments grew radially from the surface of each particle.The interaction between the filaments on two adjacent particles caused the particles to move apartagainst the magnetically applied force. In this way the authors measured the force, the separationbetween particles and their velocity. Several of their results were described above.11 a) (b)
Figure 6: (a) Case (cid:96) > (cid:96) R : The filaments are not in contact with the wall and so remain stress-free.(b) Case (cid:96) < (cid:96) R : each of N filaments carry a force f . Both N and f increase as (cid:96) decreases (atfixed (cid:96) R ).Brangbour et al. [7] arrive at (what is effectively) equation (9) using a more accurate versionof the following simplified micro-mechanical argument: suppose that actin filaments of stress-freelength (cid:96) R are attached to a rigid spherical particle of radius R that is in the vicinity of a wall. Thewall represents the mid-plane between a pair of adjacent particles. When the distance (cid:96) from theparticle surface to the wall is greater than (cid:96) R as in Figure 6(a), there is no contact between thefilaments and the wall and so the filaments carry no force. On the other hand if the distance to thewall is less than (cid:96) R as in Figure 6(b), some filaments will interact with the wall and carry force.The (compressive) force f in a filament increases as the distance (cid:96) to the wall decreases. Supposethe force is given by the classical entropic model f = c /(cid:96) for (cid:96) < (cid:96) R , where the parameter c is related to the elastic modulus of a filament. Moreover, as the distancebetween the particle and wall decreases, the number, N , of filaments interacting with the wall, andtherefore carrying force, increases from the value N = 0 at (cid:96) = (cid:96) R . Assume that the number offorce carrying filaments is given by the linear relation N = c ( (cid:96) R − (cid:96) ) for (cid:96) < (cid:96) R , where the parameter c is related to the density of filaments on the particle surface. The total forcebetween the particle and wall, f N , when calculated using the two preceding equations, leads toprecisely a constitutive relation of the form (9) with EA = c c . It is worth emphasizing that when the constitutive relation (9) is derived in this way, it accounts forboth the stress-stretch behavior of a filament and the changing number of load carrying filaments.
According to the entropic model c = k B T where k B is the Boltzmann constant and T isthe absolute temperature, and the linearization of the geometric relation derived by Brangbour et l. [7] gives c = N GS / (4 R ) where N GS is the fixed number of filaments on a colloidal particle.Therefore one finds EA = c k B T N GS R where c is a factor they introduce to better fit the data. Observe from this that the effective modulus EA can be varied by changing the density of filaments on the particle surface.Our equation (12) is identical to equation (4) in Brangbour et al. [7] provided we identifytheir variables X, V t and ck B T N GS / (4 R ) with our (cid:96), (cid:96) R and E respectively. Figure 7: Schematic figure of test specimen depicting cross-linking, load-bearing filaments andfree filaments. The specimen is identified with the interval [ y ( t ) , y ( t )] in physical space. It isattached to an AFM-spring at its left-hand end y ( t ) and to a (movable) support at its right-handend y ( t ) . In reference space, the specimen is identified with the interval [ x ( t ) , x ] . Surfacegrowth occurs at the left end of the specimen causing the boundary x = x ( t ) to move leftward.We continue to consider the test specimen described in Section 2 but now with its left endattached to an AFM cantilever and its right end to a (movable) support as depicted schematicallyin Figure 7. The AFM acts like a linear spring whose force-elongation relation is σA = − k c ( y − Y ) . (18)Here σ ( t ) is the average (compressive) stress in the specimen, the cross-sectional area of the spec-imen over which the filaments are distributed is A , Y is the position of the AFM cantilever whenit is undeflected, y ( t ) is the position of the left-hand end of the specimen at time t and k c is the(usual) spring stiffness in units of force/displacement. It is more convenient to set k = k c /A andwrite (18) as σ = − k ( y − Y ) . (19)13 .1 Constitutive response functions. First consider the constitutive function Λ describing the relation between the stress σ , stretch λ and filament density ρ . Suppose that the force f in a filament is related to its stretch by f = E f A f ( λ − − where E f is its Young’s modulus and A f its cross-sectional area. This, togetherwith (5), gives the constitutive relation for the specimen to be σ = f ρ = E ( λ − − where E = ρE f A f .The filaments in the experiments of Brangbour et al. [7] are relatively short and do notform a network. In contrast, the actin filaments in the experiments of Parekh et al. [29] form across-linked network where filament bending becomes important. To account for an analogousphenomenon in foams, Gibson and Ashby proposed the modification E = ρ A f E f to the effectiveYoung’s modulus, see [17]. We adopt their model and write the stress-stretch-filament densityrelation in the form λ = Λ ( σ, ρ ) where Λ ( σ, ρ ) = 11 + σ/E , E = E ( ρ ) = ρ A f E f . (20)For a three-dimensional continuum model of actin networks see for example [28].Keeping in mind that we are taking compressive stress to be positive, we are concerned with σ ≥ whence λ ≤ . Since the total cross-sectional area taken up by the filaments, N A f , cannotexceed the cross-sectional area A over which the filaments are distributed, it is necessary that ≤ ρ ≤ /A f . (21) Next consider the kinetic relation ˙ ρ = R ( ρ ) governing the filament density. As the number offilaments increases, the number of monomers in the surrounding monomer pool decreases, andso the rate at which new filaments develop is expected to decrease, i.e. we anticipate ˙ ρ to be adecreasing function of ρ . Moreover, since the total number of monomers in the system is finite, aswell as because of (21), ρ cannot increase indefinitely. Therefore for the kinetic law ˙ ρ = R ( ρ ) wetake τ ρ ˙ ρ = ρ ∞ − ρ, τ ρ > , ρ ∞ > , (22)where τ ρ and ρ ∞ are constant parameters. The linear dependence of ˙ ρ on ρ is similar to that in [26].Since one can solve (22) explicitly, the response predicted by this kinetic relation is ρ ( t ) = ρ ∞ + ( ρ − ρ ∞ ) e − ( t − t ) /τ ρ , (23)where ρ = ρ ( t ) < ρ ∞ is the filament density at some particular instant t . Note that ρ ( t ) increases monotonically and ρ ( t ) → ρ ∞ as t → ∞ .14ince ρ ( t ) < ρ ∞ , this, together with E = ρ A f E f and ρ ( t ) → ρ ∞ as t → ∞ , tell us that theeffective Young’s modulus obeys E ( t ) < E ∞ and E ( t ) → E ∞ as t → ∞ where E ∞ := ρ ∞ A f E f . Thus E ∞ is the Young’s modulus when the system reaches steady state. It will be convenient towrite (20) in terms of E ∞ as E = E ∞ ρ /ρ ∞ . (24) We now present two models for the kinetic relation V = V ( f ) = V ( σ, ρ ) describing surface growthat the left-hand boundary of the specimen where V = − ˙ x . According to the literature, e.g. [7,29],growth is expected to stall at some critical value f stall of the filament force. Therefore we take V ( f ) to be a monotonically decreasing function of f with V ( f ) → as f → f stall .We first present the kinetic law (of Arrhenius form) that is exponential in the filament force f : V = V (cid:2) e − ζ f/f stall − e − ζ (cid:3) [1 − e − ζ ] for 0 ≤ f ≤ f stall , (25)where f stall , V and ζ are constant parameters with V > , f stall > , ζ (cid:54) = 0 . For reasons that we will explain below, we shall not use this kinetic relation in our calculationsbut note it here because of its frequent use in this field. Howard [19] gives an expression for theparameter ζ in terms of the temperature T , the Boltzmann constant k B and the length of a stressfree monomer a : ζ = af stall k B T , (26)indicating that the parameter ζ should also be positive. Keeping in mind that the filament force f ispositive in compression, (25) says that growth occurs ( V > ) for compressive forces in the range ≤ f < f stall and that growth stalls when f → f stall . The dashed red curve in Figure 8 shows thevariation of V with f according to (25) for ζ = 5 . When ζ → this curve approaches the straightline V /V = 1 − f /f stall .Next, recall the maximum dissipation kinetic law for growth (11) used in Section 3: V = V for ≤ f < f stall , V = 0 for f > f stall . This is shown dotted in Figure 8. In this section we adoptthe following regularized (smoothed out) version of this kinetic law: V = V ( f ) = V (cid:20) − (cid:18) ff stall (cid:19) m (cid:21) for 0 ≤ f ≤ f stall , (27)where f stall , V and m are constant parameters such that V > , f stall > , m > . .2 0.4 0.6 0.8 1.00.20.40.60.81.0 Figure 8: Growth speed
V /V versus filament force f /f stall for surface growth. Dotted blue:maximum dissipation kinetic law (11) from Section 3. Dashed red: exponential kinetic law (25)with ζ = 5 . Solid black: power-law kinetics according to (27) with m = 1 , and .For m = 1 this model is linear in the filament force and for m → ∞ it approaches the maximumdissipation kinetic law used in Section 3. The solid curves in Figure 8 show the variation of ˙ (cid:96) R with f according to (27) for some different values of m .The experimental observations in [29] indicate that the process of growth under spring load-ing involves an intermediate stage where the elongation-rate is almost independent of the stress,and therefore a plot of ˙ (cid:96) versus σ involves a more-or-less horizontal segment prior to stall. Thisis ideally captured by the maximum dissipation kinetic relation (which, as we saw in Section 3,also did very well in modeling the experiments of Brangbour et al. [7]). Because the approach tostall is gradual and not sudden, a regularized version of that kinetic relation is desirable such asthe power-law model (27) with a moderate value of m . As can be seen from Figure 8, the expo-nential model (25) does not capture this behavior. The exponential kinetic relation does approachthe maximum dissipation kinetic law when ζ → −∞ , though negative values of ζ appear not to bereasonable, e.g. Howard’s model (26).Instead of the growth speed parameter V it will sometimes be more convenient to use thetime-scale for growth defined by τ R := (cid:96) /V , (28)where (cid:96) is the distance between the AFM cantilever and the other support when the AFM isundeflected. When growth commences, the filaments attached to the AFM do not extend all theway to the other support. Therefore they initially grow under zero stress. Their length when theyfirst touch the other support is (cid:96) .The implications of the dissipation inequality on the kinetic law for growth are discussed inthe Appendix and Section S4 of the Supplementary Material.16 .1.4 Model parameters. Parameter values. The constitutive models described in Sections 4.1.1, 4.1.2 and 4.1.3 involve the following parame-ters: the stress-stretch-filament density relation (20) , (24) involves the Young’s modulus E ∞ andthe maximum filament density ρ ∞ . The kinetic law (22) for the nucleation of new filaments in-volves ρ ∞ and the time-scale τ ρ . The kinetic law for surface growth (27), (28) involves the timescale τ R , the stall force f stall and the exponent m . Since stall occurs when f = f stall and ρ = ρ ∞ it is convenient to define σ stall := ρ ∞ f stall . In addition, the constitutive relation of the AFM spring involves its stiffness k . It is useful to definean associated stress σ by σ := k(cid:96) , (29)where k is the stiffness of the AFM spring in units of stress/displacement; see (19).While it is natural to work with nondimensional variables and parameters, we shall not doso here since we want to make quantitative comparisons with the experimental results of Parekh etal. [29].How we arrived at the following specific values of the various parameters, including thesources of the data, is described in Section S2 of the Supplemental Material. Here we simplyrecord the values we shall use: (cid:96) = 3 µ m , σ stall = 0 .
77 nN /µ m , τ ρ = 40 min , m = 5 ,σ = 0 . /µ m ( Figures 2 and 3a of [29] ) ,σ = 0 . /µ m ( Figure 3b of [29] ) . (30)The value of τ ρ was chosen arbitrarily, while that of m was chosen to ensure that the kinetic law V = V ( f ) was reasonably close to the maximum dissipation kinetic law as depicted in Figure 8.This was necessary in order to get qualitative agreement between the theoretical predictions and theexperiments. Two different AFM springs were used in [29] leading to the two values of σ = k(cid:96) above. The value of ρ ∞ turns out not to be needed but its value is of the order of filaments per µ m , significantly smaller than the maximum filament density /A f = 53 , µ m − .The two main sets of experiments carried out by Parekh et al. [29] pertain to their Figures 2and 3. It can be seen from those figures that the elongation-rate for growth under spring-loadingis ˙ (cid:96) = 72 nm / min in Figure 2 and ˙ (cid:96) = 129 nm / min in Figure 3. Therefore the conditions underwhich the two sets of experiments were carried out had to be different, and so it is not unreasonablefor the values of certain parameters in the model to also be different. The elongation-rate ˙ (cid:96) dependssensitively on the growth-speed V which in turn depends on the monomer concentration in thesurrounding solvent. We assume that the different values of ˙ (cid:96) observed is likely due to differentmonomer concentrations and so take different values for V (equivalently τ R = (cid:96) /V ) dependingon which experiment we are modeling. Furthermore, it can be readily shown from (36) and (37)below that the elongation-rate at the initial instant, ˙ (cid:96) (0) , depends sensitively on (the time scale for17rowth, τ R , and) the Young’s modulus at steady-state, E ∞ . Therefore we also take the value of E ∞ to be different in the two analyses. When concerned with the experiments focused on growthunder the action of the AFM spring (Figure 2 of [29]), we shall take E ∞ = 3 . /µ m , τ R = 34 min , (31)whereas when modeling the experiments focused on stress jumps (Figure 3 of [29]), we take E ∞ = 0 . /µ m , τ R = 10 min (Figure 3a) , τ R = 13 min (Figure 3b) . (32)As can be seen from the Supplementary Material, the values in (31) and (32) are within the rangeof experimentally measured data. First consider the problem where the specimen grows under spring-loading. Our main interest is incalculating the force, specimen length and elongation-rate as functions of time, and then lookingat a plot of elongation-rate versus force.In these calculations the position y of the support on the right-hand side is held fixed andthe velocity ˙ y of the AFM cantilever is measured. Thus the unstressed length of the specimenat the initial instant is (cid:96) := y − Y which represents the distance between the support and AFMcantilever when the AFM is not deflected. It then follows because (cid:96) = y − y that (cid:96) − (cid:96) = y − Y ,and so the spring loading equation σ = − k ( y − Y ) can be written as σ = k ( (cid:96) − (cid:96) ) : (cid:96) = (cid:96) + σ/k. The system of 4 equations to be solved to find (cid:96) ( t ) , (cid:96) R ( t ) , σ ( t ) , ρ ( t ) are (cid:96) = Λ ( σ, ρ ) (cid:96) R , ˙ (cid:96) R = V ( σ, ρ ) , ˙ ρ = R ( ρ ) , (cid:96) = (cid:96) + σ/k, (33)having used ˙ (cid:96) R = − ˙ x = V . From (33) , ˙ σ = k ˙ (cid:96). (34)Differentiating (33) with respect to time and using (34) leads to ˙ σ/k = Λ ρ (cid:96) R ˙ ρ + Λ ˙ (cid:96) R − k Λ σ (cid:96) R , (35)where we have set Λ σ = ∂ Λ /∂σ and Λ ρ = ∂ Λ /∂ρ . However from (33) and (33) (cid:96) R = (cid:96) Λ = k(cid:96) + σk Λ = σ + σk Λ . (cid:96) R from (35) yields ˙ σσ = Λ ˙ (cid:96) R /(cid:96) + (1 + σ/σ ) ˙ ρ Λ ρ / Λ − ( σ + σ ) Λ σ / Λ . (36)Equation (33) can be solved for ρ ( t ) . When this expression for ρ ( t ) , together with Λ = Λ ( σ, ρ ) and ˙ (cid:96) R = V ( σ, ρ ) , are substituted into (36), the resulting equation has the form ˙ σ = F ( σ, t ) .This can be solved for σ ( t ) . Thereafter one can calculate the associated value of ˙ (cid:96) ( t ) from ˙ (cid:96)/(cid:96) o (34) = ˙ σ/ ( k(cid:96) ) (29) = ˙ σ/σ , (37)and thus one can construct a parametric plot of ( σ ( t ) , ˙ (cid:96) ( t )) on the σ, ˙ (cid:96) -plane with time t as theparameter. The resulting figure can be compared with Figures 2B and 2C of Parekh et al. [29].Before proceeding to do this, since it will be useful when we discuss the dissipation in-equality later on, we now turn briefly to the ρ, σ -plane. Each radial line σ = f ρ on this planecorresponds to a constant filament force f ; see Figure 9. We are interested in the range ≤ f ≤ f stall , ≤ ρ ≤ ρ ∞ , corresponding to the wedge-shaped shaded region in the figure. The kineticrelation for surface growth has the form V = V ( f ) and so each radial line also corresponds to aconstant growth speed V . Since V ( f stall ) = 0 and V ( f stall ) > for ≤ f < f stall , the growthspeed vanishes on the bold red line σ = f stall ρ and is positive below it. Now consider a genericinitial-value problem for a spring-loaded specimen. In this case one solves the differential equa-tions ˙ σ = S ( σ, ρ ) , ˙ ρ = R ( ρ ) subject to initial conditions, say σ ( t ) = 0 , ρ ( t ) = ρ , where S isgiven by (36), (33) , and R by (22). The solution σ = σ ( t ) , ρ = ρ ( t ) , t ≥ t , of this problemdescribes a trajectory in the ρ, σ -plane shown schematically by the dashed blue curve in Figure 9.It starts at ( ρ, σ ) = ( ρ , and terminates at ( ρ, σ ) = ( ρ ∞ , σ stall ) corresponding to stall. Figure 9: The radial straight lines σ = f ρ are lines of constant filament force. The kinetic relation V = V ( f ) yields V = 0 on the bold red line σ = f stall ρ , and V > on the shaded region belowit. The trajectory defined by a solution ( ρ ( t ) , σ ( t )) , t ≥ t , of a generic initial-value problem isdepicted schematically by the blue dashed curve. It starts at the initial point ( ρ , and terminatesat stall corresponding to ( ρ ∞ , σ stall ) . 19 .2.1 Simple ρ -independent model. Before presenting the results of the preceding analysis, it is illuminating to first consider a simplermodel in which the number of filaments is fixed and does not evolve. Then we drop ρ from themodel so that equations (36) and (37) governing (cid:96) ( t ) , σ ( t ) specialize to ˙ (cid:96) = Λ ( σ ) V ( σ )1 − ( σ + σ ) Λ (cid:48) ( σ ) / Λ ( σ ) , ˙ σσ = ˙ (cid:96)(cid:96) , (38)where we have set V = V ( σ ) . Equation (38) gives ˙ (cid:96) as a function of stress σ , which can be plottedon the σ, ˙ (cid:96) -plane. Moreover, integrating (38) from σ = 0 at t = t to σ = σ stall at t = t stall givesthe time t stall at which growth stalls.We now specialize (38) to the particular choices (20), (27), Λ ( σ ) = 1 / (1 + σ/E ) , V ( σ ) = V (cid:20) − (cid:18) σσ stall (cid:19) m (cid:21) , (39)with the various parameters having the values given in (30) and (31). Figure 10 shows the plot of ˙ (cid:96) versus σ according to this model. Observe that the elongation-rate starts at ˙ (cid:96) = 83 nm / min andeventually stalls when the force reaches the value σ stall A = 293 nN . The time taken for the stressto reach of σ stall is
235 min . This figure may be compared with Figures 2B and 2C of Parekh et al. [29] where the elongation-rate starts at about
72 nm / min and growth stalls at a force of about
300 nN in a little over
200 min . Figure 10: Elongation-rate ˙ (cid:96) versus stress σ according to the simple model (38), (39). ρ -dependent model. Now consider the more detailed model (36), (37), specialized to the constitutive descriptions(20), (22), (27). The parameters have the same values used in the preceding subsection together It should be noted that this is not the same model we used in Section 3 since here we are using the regularizedmaximum dissipation kinetic relation for surface growth. This is needed in order to approach stall gradually. τ ρ taking the value given in (30) . The initial condition for ρ is chosen (arbitrarily) to be ρ ( t ) = 0 . ρ ∞ . Figure 11 shows a parametric plot of ( σ ( t ) , ˙ (cid:96) ( t )) on the stress-elongation-rateplane with time being the parameter. The elongation-rate starts at the value
70 nm / min and risesto a maximum value of
75 nm / min . The elongation-rate remains at about
70 nm / min during amore-or-less load-independent intermediate stage after which ˙ (cid:96) begins to decrease more rapidly asthe system approaches stall. The stress increases monotonically throughout this calculation untilit reaches of the stall force in
200 min , the stall force being
293 nN . As noted above, thecorresponding experimental values from [29] are an initial elongation-rate of about
72 nm / min ,growth stalling at a force of about
300 nN in a little over
200 min . Figure 11: Elongation-rate ˙ (cid:96) versus stress σ according to the general model (36), (37). We now turn to two calculations motivated by Figures 3a and 3b of Parekh et al. [29] wherethe force on the specimen is suddenly decreased while it is growing. Our primary interest is incalculating the resulting jump in the elongation-rate ˙ (cid:96) which, the experiments indicate, undergoesa striking sudden increase.It is worth noting at the outset that the kinetic law ˙ ρ = R ( ρ ) is independent of stress and so isunaffected by the details of how the stress varies. Thus we take ρ ( t ) and ˙ ρ ( t ) to vary continuouslythroughout. Consequently the effective Young’s modulus, E = ρ E ∞ /ρ ∞ , is also a continuousfunction of time. Moreover, keeping in mind that the referential length of the specimen (cid:96) R ( t ) changes only due to growth, and assuming that a finite segment of new material cannot appear inan infinitesimal instant of time, we require (cid:96) R ( t ) to be a continuous function of time. On the otherhand the specimen length (cid:96) ( t ) and elongation-rate ˙ (cid:96) ( t ) will be discontinuous when the stress isdiscontinuous.In the first calculation the specimen grows under spring loading conditions during an initialperiod t < t < t ; see left-hand figure in Figure 13. At time t , the force is suddenly decreased to According to the kinetic law (22) it takes infinite time for ρ ( t ) to reach the value ρ ∞ which is why we calculatethe time to reach of stall.
21 smaller value and clamped at that value from then on. (The specimen is not under spring-loadingconditions for t > t .)In the second calculation (see left-hand figure in Figure 14) the specimen grows with theforce clamped at a fixed value during an initial stage t < t < t . At the instant t = t the forceclamp is released and the specimen is allowed to grow under spring loading conditions for a period t < t < t . At the instant t = t the force is suddenly decreased back to the value it had duringthe original constant force stage and clamped at that value for t > t .In both of these calculations the load levels are always in the “load-independent” range ofFigure 11.In all processes, whether the stress is held constant or the specimen is spring-loaded, we have (cid:96) = Λ (cid:96) R , ˙ (cid:96) R = V ( σ, ρ ) , ˙ ρ = R ( ρ ) . (40)Differentiating (40) with respect to time, and then using (40) and (40) to eliminate (cid:96) R and ˙ (cid:96) R from the result, leads to ˙ (cid:96) = ΛV + (cid:96) Λ σ Λ ˙ σ + (cid:96) Λ ρ Λ ˙ ρ. In processes where σ ( t ) = constant , this reduces toConstant stress: ˙ (cid:96) = ΛV + (cid:96) Λ ρ Λ ˙ ρ, (41)whereas when the specimen is spring-loaded, so that σ = k ( (cid:96) − (cid:96) ) , it yieldsSpring loading: ˙ (cid:96) = ΛV + ˙ ρ (cid:96) (1 + σ/σ ) Λ ρ / Λ − ( σ + σ ) Λ σ / Λ , (42)where σ = k(cid:96) as before.Consider the instant t at which the stress changes discontinuously. For any time-dependentfunction g ( t ) that suffers a finite jump discontinuity at time t we write g + = g ( t +2 ) , g − = g ( t − ) , and if g ( t ) is continuous at t we simply write g = g ( t ) . In both of Parekh et al. ’s experiments,the specimen is spring-loaded just before the stress jump and has the force clamped just after. Thus(42) holds at t − , while (41) holds at t +2 . The elongation-rates just before and just after the stressjump are therefore ˙ (cid:96) − = Λ − V − + ˙ ρ (cid:96) (1 + σ − /σ ) Λ − ρ / Λ − − ( σ − + σ ) Λ − σ / Λ − , ˙ (cid:96) + = Λ + V + + ˙ ρ(cid:96) + Λ + ρ / Λ + , (43)where we have written Λ ± = Λ ( σ ± , ρ ) and V ± = V ( σ ± , ρ ) having used the fact that ρ ( t ) variescontinuously. 22onsider an instant at which the stress has the values σ ± . In order to calculate ˙ (cid:96) + using(43) we need the value of (cid:96) + . While we can use σ − = k ( (cid:96) − − (cid:96) ) to calculate (cid:96) − , we cannot use σ + = k ( (cid:96) + − (cid:96) ) to calculate (cid:96) + since the specimen is not spring-loaded at time t + . Instead, wecalculate (cid:96) + using (cid:96)/(cid:96) R = Λ , i.e. (cid:96) + = Λ ( σ + , ρ ) Λ ( σ − , ρ ) (cid:96) − . (44)In writing (44) we have used the fact noted at the beginning of this section that the filament density ρ ( t ) and referential length (cid:96) R ( t ) vary continuously. ρ -independent model. In order to get a sense for how the elongation-rate suffers a sudden increase when the stress issuddenly decreased, consider again the special case where the constitutive functions Λ and V areboth independent of the filament density ρ . Then equations (41) and (42) take the formsConstant stress: ˙ (cid:96) = Λ ( σ ) V ( σ ) , (45)Spring loading: ˙ (cid:96) = Λ ( σ ) V ( σ )1 − ( σ + σ ) Λ (cid:48) ( σ ) / Λ ( σ ) . (46)These equations tell us how the elongation-rate ˙ (cid:96) varies as a function of stress σ for the two typesof loading.Subject to mild assumptions on Λ and V , the curve on the σ, ˙ (cid:96) -plane defined by (46) liesbelow the one described by (45). For example, Figure 12 shows these curves for the particularchoice (39) with the parameters having the values given in (30) and (32).When σ = 0 we have Λ (0) = 1 , Λ (cid:48) (0) = − /E and V (0) = V , and therefore from (45) and(46) ˙ (cid:96) = V (constant force), V σ /E (spring loading), at σ = 0 . These are the values of ˙ (cid:96) at which the curves in Figure 12 intersect the vertical axis. Therefore theseparation between the two curves (at least at σ = 0 ) increases as σ /E increases. Therefore inorder to increase the separation at σ = 0 we should decrease E . In our quantitative calculationswe have therefore taken the smallest value of E from the range of possible values determinedexperimentally. The value of σ is determined by the stiffness of the AFM spring.Now consider the response of a spring-loaded specimen starting from point A in Figure 12where the stress is σ ( A ) . The point ( σ ( t ) , ˙ (cid:96) ( t )) evolves along the green curve starting from A andmoving to the right (towards stall). Suppose that when it reaches point B the stress is suddenlydecreased back to the value σ ( A ) and clamped at that value. Then ( σ ( t ) , ˙ (cid:96) ( t )) jumps from point B to point C (and remains there for subsequent time). Therefore, as the stress decreases suddenly23rom σ ( B ) to σ ( C )(= σ ( A )) the elongation-rate increases discontinuously from the value ˙ (cid:96) ( B ) to ˙ (cid:96) ( C ) . A BC
Figure 12: Elongation-rate ˙ (cid:96) versus stress σ for (a) loading at constant stress (equation (45), redcurve) and (b) spring-loading (equation (46), green curve). Suppose that the spring-loaded speci-men evolves from A to B, at which point the stress is suddenly decreased back to the value σ ( A ) and clamped at that value. Accordingly the system must jump from B to C and remain there fromthen on. Thus during the jump, the stress decreases discontinuously from σ ( B ) to σ ( C ) (= σ ( A )) while the elongation-rate increases discontinuously from ˙ (cid:96) ( B ) to ˙ (cid:96) ( C ) .In our first quantitative calculation we took the values of the force just before and after thejump from Figure 3a of [29] and determined the elongation-rates from ˙ (cid:96) − = Λ ( σ − ) V ( σ − )1 − ( σ − + σ ) Λ (cid:48) ( σ − ) / Λ ( σ − ) , ˙ (cid:96) + = Λ ( σ + ) V ( σ + ) , (47)which follow from (45) and (46). The functions Λ and V are given by (39) with the parametershaving the values in (30) and (32). Equation (47) then led to ˙ (cid:96) − = 120 nm / min , ˙ (cid:96) + = 204 nm / min . The corresponding experimentally measured elongation-rates were ˙ (cid:96) − = 129 nm / min and ˙ (cid:96) + =275 nm / min .In our second numerical calculation the values of the force just before and after the jumpwere taken from Figure 3b of [29]. The elongation-rates were calculated as above and this led to ˙ (cid:96) − = 162 nm / min , ˙ (cid:96) + = 239 nm / min , whereas the experimentally measured values were ˙ (cid:96) − = 129 nm / min and ˙ (cid:96) + = 270 nm / min . First and second refer to the two processes described at the very beginning of Section 44.3. .3.2 General ρ -dependent model. When the filament density ρ is taken into account, a simple graphical description based on (41)and (42) is no longer possible since the right-hand sides of those equations now depend on both σ and t (through ρ ( t ) ). Instead we solve the relevant initial-value problem based on (41) and (42)to calculate the response of the specimen and in particular to determine the elongation-rates justbefore and after the stress jump.
95 100 105 1100.300.320.340.360.380.40
95 100 105 110120140160180200220240
Figure 13:
Stress and elongation-rate versus time; compare with Figure 3a of [29]. The specimen grows underspring loading for t < t < t and with the stress fixed for t > t . In the first calculation we solve the differential equation (42) for t < t < t using suitableinitial conditions at t . The conditions at time t +2 are then determined using (44) and this infor-mation is used as initial conditions to solve (41) for t > t . The details of this calculation can befound in Section S3 of the Supplementary Material. Figure 13 shows plots of σ ( t ) and ˙ (cid:96) ( t ) versustime as predicted by our model which may be compared with Figure 3a of [29]. In particular wefind σ ( t − ) = 0 .
373 nN /µ m , ˙ (cid:96) ( t − ) = 112 nm / min and ˙ (cid:96) ( t +2 ) = 212 nm / min , the correspond-ing experimentally determined values being σ ( t − ) = 0 .
381 nN /µ m , ˙ (cid:96) ( t − ) = 129 nm / min and ˙ (cid:96) ( t +2 ) = 275 nm / min .
50 60 70 80 900.170.180.190.200.210.22
50 60 70 80 90120140160180200
Figure 14:
Stress and elongation-rate versus time; compare with Figure 3b of [29]. The specimen grows with thestress fixed at the same value for both t < t < t and t > t , and under spring loading for intermediate times t < t < t . In the second calculation we start by solving (41) for t < t < t using suitable initial25onditions at t . The conditions at time t +1 are then deduced from continuity and the results areused as initial conditions to solve (42) for t < t < t . The conditions at time t +2 are then calculatedusing (44). Finally we solve (41) for t > t using the information from t +2 as initial conditions. Thedetails of these calculations can be found in Section S3 of the Supplementary Material. Figure 14shows plots of σ ( t ) and ˙ (cid:96) ( t ) versus time as predicted by our model which may be compared withFigure 3b of [29]. In particular we find σ ( t − ) = 0 .
216 nN /µ m , ˙ (cid:96) ( t − ) = 119 nm / min and ˙ (cid:96) ( t +2 ) =187 nm / min , the corresponding experimentally determined values being σ ( t − ) = 0 .
218 nN /µ m , ˙ (cid:96) ( t − ) = 129 nm / min and ˙ (cid:96) ( t +2 ) = 270 nm / min .In this second calculation the stress is held fixed at the same value .
218 nN /µ m for both t < t < t and t > t . Therefore on both these time intervals the corresponding elongation-rate is given by (41) with σ = 0 .
218 nN /µ m . However the right-hand side of (41) also involvesthe filament density ρ ( t ) and therefore though σ has the same constant value, the elongation-rateevolves as a function of time due to the evolution of ρ ( t ) . In the specific calculation above, wefind (in particular) that ρ ( t ) = 0 . ρ ∞ and ρ ( t ) = 0 . ρ ∞ at t = 73 min and t = 79 min .This small difference in the filament densities leads to a small difference in the correspondingelongation-rates, viz. ˙ (cid:96) ( t − ) = 186 nm / min and ˙ (cid:96) ( t +2 ) = 187 nm / min . This is in contrast to thelarge difference, ˙ (cid:96) ( t − ) = 170 nm / min and ˙ (cid:96) ( t +2 ) = 270 nm / min , observed in the experiments. Inorder to capture this we will need to modify the kinetic relation ˙ ρ = R ( ρ ) and possibly ˙ (cid:96) R = V ( σ, ρ ) and λ = Λ ( σ, ρ ) as well.Finally it is worth noting the qualitative similarity between the right-hand figure in Figure 14and Figure 5 from our earlier calculation related to the experiments of Brangbour et al. [7]. In bothcases the force on the specimen is held constant at the same value for t < t < t and t > t . In thecase of Figure 5 the force was fixed at a smaller value during the intermediate interval t < t < t whereas the specimen grew under spring loading in Figure 14. In Figure 5, the curve (straight-line) pertaining to t > t is the continuation of the curve pertaining to t < t < t . Because of thedependency on the filament density ρ ( t ) , this is not true of the corresponding curves in Figure 14. In this paper we have shown that a continuum model of surface growth and filament nucleationcan quantitatively capture the evolution of growing dendritic actin networks. Such growing net-works provide the propulsive force for a variety of processes in live cells. A surface growth modelis appropriate because polymerization in dendritic actin occurs in a narrow zone next to the loadsurface, not all over the network. A distinguishing feature of our model is that it describes theactin network as a growing continuum subject to the balance laws of continuum thermomechanicstogether with constraints imposed by the dissipation inequality. The microscopic details of poly-merization of individual filaments are distilled into a kinetic law for the propagation speed of thegrowing surface. This continuum model is applied to two different experiments. The first is theset of experiments of Brangbour et al. [7] in which the density of filaments does not change andthe filaments are not cross-linked. Our model is in remarkable agreement with experiment on the26volution of filament lengths, forces (or stresses), etc., using a maximally dissipative kinetic law.The second is the set of experiments of Parekh et al. [29] in which the density of filaments changesand the filaments are cross-linked. Once again, a smoothed version of the maximally dissipativekinetic law is able to describe the evolution of variables in this experiment. Our kinetic law relatesgrowth rate (in the reference configuration) to an (average) force per filament to facilitate com-parison with well-known ‘force-velocity’ relations for single growing filaments. We find that thekinetic law that best describes the two experiments is different from the exponential force-velocityrelation that is often used in the context of single filaments. This is hardly surprising since the nar-row zone in which polymerization of dendritic actin occurs is a network of interacting filamentswhich are not parallel and are acted upon by different forces. This kinetic law is one of many thatsatisfy the dissipation inequality and there could yet be others that perform better; we use it herebecause of its simplicity and its ability to capture the convex approach to stall that is distinct fromthe well-established exponential force-velocity relation. The continuum framework proposed heremay be applicable to growing networks of other biological and non-biological filaments such asthose of collagen, microtubules, carbon-nanotubes, etc.
Acknowledgements
RA gratefully acknowledges stimulating discussions with Eric Puntel andGiuseppe Tomassetti. This research was partially carried out while PKP was Visiting Professorat MIT in Fall 2019. PKP acknowledges generous support from the Department of MechanicalEngineering at MIT and an NIH grant R01 HL 135254.
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Appendix
In this section we examine the implications of the dissipation inequality (“second law of thermo-dynamics”) on the kinetic law for surface growth V = V ( f ) . Recall that the kinetic laws (11) and(27) imply that the growth speed V is > when the filament force is in the range ≤ f < f stall .As we shall see, the dissipation inequality requires f driv V ≥ where f driv ( ρ, σ ) is the driving forcefor growth (which will be identified below). Our task therefore is to examine the implications ofthe inequality f driv ≥ on the inequality ≤ f < f stall , or in terms of stress and filament density,the connection between f driv ( ρ, σ ) ≥ and ≤ σ < ρf stall , ≤ ρ ≤ ρ ∞ . As shown schematically in Figure 7, the specimen occupies the interval [ y ( t ) , y ( t )] in physicalspace and its associated length is (cid:96) = y − y . Its left end is attached to an AFM cantilever ofstiffness k c . The elastic energy stored in the cantilever (modeled as a Hookean spring) is k c ( y − Y ) where Y is the position of the cantilever when it is undeflected. In reference space thespecimen occupies the interval [ x ( t ) , x ] and its associated length is (cid:96) R = x − x . Let W be thefree energy of the specimen per unit reference volume. Then, the rate of increase of the energystored in the specimen and spring is ddt (cid:2) W A ( x − x ) (cid:3) + ddt (cid:104) k c ( y − Y ) (cid:105) = ˙ W A(cid:96) R − W A ˙ x − σA ˙ y , (A.1)having equated the compressive force σA in the specimen to the force − k c ( y − Y ) in the spring.The rate at which work is being done on the specimen by the compressive force σA at the right-hand end is = − σA ˙ y . (A.2)Next we model the inflow of chemical energy into the specimen. If a denotes the length ofa single stress-free monomer (i.e. its length in a stress-free reference configuration), the numberof monomers in a single filament is (cid:96) R /a and the total number of monomers in the specimen is N (cid:96) R /a . Thus the monomer concentration c , defined as the number of monomers per unit referencevolume, is c = N (cid:96) R /a(cid:96) R A = NaA = ρa , where ρ = N/A is the filament density. When the body grows, the left-hand boundary of thespecimen moves outwards in reference space at a speed − ˙ x , and so the rate at which monomersare added to a filament at that end is − ˙ x /a . Therefore the associated rate of intake of chemicalenergy is − µ ˙ x /a per filament. In addition, there is an intake of chemical energy due to theformation of new filaments. Since the number of monomers in a filament is (cid:96) R /a and its chemical30nergy µ(cid:96) R /a , the rate of intake of chemical energy due to the formation of new filaments is ( µ(cid:96) R /a ) ˙ N . Thus the total inflow of chemical energy into the specimen per unit time is − µ ( ˙ x /a ) N + ( µ(cid:96) R /a ) ˙ N = − µcA ˙ x + µ ˙ cA(cid:96) R . (A.3)Therefore the dissipation rate is given by (A.2) plus (A.3) less (A.1) : D = − σA ˙ y + (cid:2) − µcA ˙ x + µ ˙ cA(cid:96) R (cid:3) − (cid:2) ˙ W A(cid:96) R − W A ˙ x − σA ˙ y (cid:3) == ( − σλ + µc − W ) AV + (cid:16) − σ ˙ λ + µ ˙ c − ˙ W (cid:17) A(cid:96) R , (A.4)where λ = (cid:96)/(cid:96) R = ( y − y ) / ( x − x ) is the stretch and V = − ˙ x is the outward propagationspeed of the left-hand boundary of the specimen. Suppose that the material is described by theconstitutive characterization W = W ( λ, c ) together with − σ = ∂W∂λ , µ = ∂W∂c , (A.5)keeping in mind that σ is positive in compression. In view of (A.5), the dissipation rate (A.4)reduces to D = ( − σλ + µc − W ) AV, (A.6)and we therefore identify the driving force for growth to be f driv := − σλ + µc − W = − σλ + µρ/a − W. (A.7)Note that µ/a is the chemical potential per unit reference length. By (A.6) and (A.7), the dissipa-tion inequality D ≥ requires f driv V ≥ . Growth corresponds to V = − ˙ x > which therefore requires f driv ≥ . Next we calculate an explicit expression for the driving force associated with the stress-strain-filament density relation − σ = ∂W∂λ = E (cid:0) − λ − (cid:1) , E = E ( ρ ) . (A.8)Note from (A.8) and λ > that σ > − E. (A.9)Integrating (A.8) with respect to λ gives the free energy W ( λ, ρ ) = E ( λ − log λ −
1) + g ( ρ ) , (A.10)31nd the corresponding chemical potential is µ = ∂W∂c = a ∂W∂ρ = aE (cid:48) ( ρ )( λ − log λ −
1) + ag (cid:48) ( ρ ) . (A.11)The driving force is then given by (A.7), (A.8), (A.10) and (A.11) to be f driv = E ( ρ ) log λ + ρE (cid:48) ( ρ )( λ − − log λ ) + f ( ρ ) , (A.12)where we have set f ( ρ ) = ρg (cid:48) ( ρ ) − g ( ρ ) . This can be written in terms of the stress (and filamentdensity) using λ = (1 + σ/E ) − : f driv = f driv ( ρ, σ ) = − E log (cid:16) σE (cid:17) + ρE (cid:48) (cid:20)
11 + σ/E − (cid:16) σE (cid:17)(cid:21) + f . (A.13)The driving force is a function of both σ and ρ since both E and f depend on ρ . All functions of ρ including f ( ρ ) here and σ st ( ρ ) below are defined for ≤ ρ ≤ ρ ∞ .When σ = 0 (or equivalently λ = 1 ), the driving force reduces to f driv = f ( ρ ) . This suggeststhat f ( ρ ) is determined entirely by chemistry and so we refer to it as the chemical driving force .Recall that according to the kinetic relations (11) , (25) and (27), the growth speed V vanisheswhen the filament force f = f stall . Since f = σ/ρ , this means that the growth speed vanishesat the stress σ = f stall ρ . We now make the assumption that the driving force vanishes when thegrowth-rate vanishes : f driv ( ρ, σ st ( ρ )) = 0 where σ st ( ρ ) := f stall ρ for 0 ≤ ρ ≤ ρ ∞ . (A.14)Note that σ st ( ρ ∞ ) = σ stall . From (A.13) and (A.14) we now obtain the following expression forthe chemical driving force: f ( ρ ) = E log (cid:16) σ st E (cid:17) − ρE (cid:48) (cid:20)
11 + σ st E − σ st E ) (cid:21) , σ st = σ st ( ρ ) , (A.15)where E = E ( ρ ) as well. Equation (A.15) can be used to eliminate f ( ρ ) in favor of σ st ( ρ ) from(A.13) allowing the driving force to be written as f driv ( ρ, σ ) = E log (cid:18) σ st E σE (cid:19) + ρE (cid:48) (cid:20)
11 + σE −
11 + σ st E − log (cid:18) σ st E σE (cid:19)(cid:21) . (A.16)The driving force f driv ( ρ, σ ) is defined on a subdomain of the ρ, σ -plane. It, and the growthspeed V , vanish on the bold red line σ = σ st ( ρ ) = f stall ρ shown in Figure 9. The shaded regionbelow that line corresponds to ≤ f < f stall and therefore to V > by the kinetic relation. Itremains to examine the consequences of the dissipation inequality f driv ( ρ, σ ) ≥ on this region. but not necessarily the converse. Recall for example from (27) that V (0) = V > . .4 The driving force further specialized. It order to examine where on the shaded region of Figure 9 one has f driv ( ρ, σ ) ≥ we limit attentionto the particular forms of the elastic modulus used in this paper, specifically, E ( ρ ) = E ρ n where n = 0 , and . Case E ( ρ ) = E : This case is applicable to a specimen involving a few parallel actin filamentsand the expression (A.16) for the driving force specializes to f driv = E log (cid:18) σ st /E σ/E (cid:19) , σ st ( ρ ) = f stall ρ, E ( ρ ) = E . (A.17)Keeping (A.9) in mind, we conclude from (A.17) that if < σ < σ st then necessarily f driv > .Therefore the dissipation inequality requires V ≥ on the shaded region of Figure 9 and so allprocesses obeying the kinetic law V = V ( f ) are admissible. In addition, (A.17) shows that f driv decreases monotonically with increasing σ on this interval, or said differently, the driving forcedecreases as the stress becomes progressively more compressive. Case E ( ρ ) = ρE : This case pertains to our model of the Brangbour et al. [7] experiments andthe expression (A.16) for the driving force reduces to f driv = E (cid:20)
11 + σE −
11 + σ st E (cid:21) , σ st ( ρ ) = f stall ρ, E ( ρ ) = E ρ. (A.18)It follows from (A.18) in view of (A.9) that f driv > when < σ < σ st . Therefore in thiscase also the dissipation inequality requires V ≥ on the shaded region of Figure 9 and so allprocesses satisfying the kinetic law V = V ( f ) are admissible on this region. Moreover, f driv decreases monotonically with increasing σ (at each fixed ρ ) as can be seen from (A.18) , and sothe driving force for polymerization decreases as the stress becomes increasingly compressive. Case E ( ρ ) = ρ E : This case is applicable to the experiments of Parekh et al. [29] and accountsfor filament bending. Recall from (24) that in Section 4 we wrote E = E ∞ ρ /ρ ∞ , so that in termsof those parameters, E = E ∞ /ρ ∞ .In this case the expression (A.16) for the driving force specializes to f driv E = 21 + σE −
21 + σ st E − log (cid:18) σ st E σE (cid:19) , σ st ( ρ ) = f stall ρ, E ( ρ ) = E ∞ ρ /ρ ∞ . (A.19)The driving force f driv ( ρ, σ ) is defined on the wedge shaped region, ≤ ρ ≤ ρ ∞ , ≤ σ ≤ σ st = f stall ρ , of the ρ, σ -plane; see Figure A.1. We want to know where f driv is positive on this region.First consider two limiting cases. If σ is close to σ st at fixed ρ , one can approximate (A.19)to read f driv E = (cid:16) − σ stall E ∞ ρ ∞ ρ (cid:17)(cid:0) σ st E (cid:1) σ st − σE + higher order terms , ρ/ρ ∞ > σ stall /E ∞ and negative for ≤ ρ/ρ ∞ < σ stall /E ∞ . Note also that the driving force is proportional to f stall − f in this case.In the other limit where, at fixed ρ , σ is small, one gets f driv E = 2 −
21 + σ st E − log (cid:16) σ st E (cid:17) + higher order terms . It is not difficult to show that this expression for the driving force is positive provided σ st /E < /α or equivalently when ρ/ρ ∞ > α σ stall /E ∞ where α ≈ . is the unique positive root of theequation −
21 + 1 /α − log (1 + 1 /α ) = 0 . Therefore from these two limiting cases we conclude that the driving force is positive on the upperand lower boundaries of the shaded region in Figure A.1.Returning to the general expression (A.19) for the driving force, keep in mind that we areconcerned with the range ≤ σ/E ≤ σ st /E corresponding to filament force values in the range ≤ f ≤ f stall . In the next paragraph we will show that (at each fixed ρ ), the equation f driv ( ρ, σ ) =0 has two roots σ , the smaller of which we denote by Σ st . For σ stall /E ∞ ≤ ρ/ρ ∞ ≤ onehas Σ st = σ st , whereas Σ st < σ st for ≤ ρ/ρ ∞ < σ stall /E ∞ . Furthermore Σ st > provided ρ/ρ ∞ > α σ stall /E ∞ where α ≈ . . (One cannot write a closed form expression for Σ st onthis range.) The curve σ = Σ st ( ρ ) for α σ stall /E ∞ ≤ ρ/ρ ∞ ≤ is shown in Figure A.1. Finally,we find that f driv ( ρ, σ ) > on the shaded region of the figure between the ρ -axis and the curve σ = Σ st ( ρ ) . The figure has been drawn assuming σ stall /E ∞ < . If σ stall /E ∞ > the curve σ = Σ st ( ρ ) lies below the straight line σ = σ st ( ρ ) throughout the range of interest.In order to establish the results described in the preceding paragraph, let ρ be fixed at anyvalue in the interval (0 , ρ ∞ ) , in which event E ( ρ ) and σ st ( ρ ) are also fixed. Consider the graph of f driv /E versus σ/E . First observe that f driv /E → ∞ when both σ/E → − + and σ/E → ∞ .Second, the slope of this curve is negative for − < σ/E < , positive for σ/E > and vanishesat σ/E = 1 . Third, it intersects the horizontal axis at σ/E = σ st /E . It follows that this curvenecessarily intersects the horizontal axis at precisely two points , corresponding to two values of σ/E ( > − , one of them less than unity, the other greater. Let σ/E = Σ st /E ≤ correspondto the left-most intersection point. It then follows that f driv > for − < σ/E < Σ st /E , andthat f driv decreases monotonically with increasing σ on this interval. It is not difficult to show that Σ st = σ st if ρ/ρ ∞ > σ stall /E and that Σ st /E < < σ st /E for ≤ ρ/ρ ∞ < σ stall /E .The kinetic law V = V ( f ) gives V > on the wedge shaped region ≤ ρ ≤ ρ ∞ , ≤ σ < σ st = f stall ρ , in Figure A.1. The dissipation inequality requires f driv V ≥ which reducesto f driv ≥ when V > . Therefore a solution σ ( t ) , ρ ( t ) , t ≥ t , of an initial-value probleminvolving growth will conform to the dissipation inequality only if the trajectory defined by it liesin the shaded region of Figure A.1 corresponding to f driv ≥ . We have confirmed this to be true except if σ st /E = 1 in which case the two intersection points coalesce. Figure A.1: Case E = E ρ : The driving force vanishes on the curve σ = Σ st ( ρ ) for ασ stall /E ∞ ≤ ρ/ρ ∞ ≤ where α ≈ . . Note that Σ st ( ρ ) = σ st ( ρ ) for σ stall /E ∞ ≤ ρ/ρ ∞ ≤ . The drivingforce is positive on the shaded region between the curve σ = Σ st ( ρ ) and the ρ -axis. Thereforethe dissipation inequality implies that V ≥ indicating that growth is permitted on this region.Recall that the kinetic law V = V ( f ) also gives V > here. The trajectory defined by a solution ( ρ ( t ) , σ ( t )) , t ≥ t , of a generic initial-value problem is depicted schematically by the blue dashedcurve starting at the initial condition ( ρ , and terminating at stall corresponding to ( ρ ∞ , σ stall ) .Only solutions in the shaded region conform to the dissipation inequality.in the case of the particular solutions determined in Section 4 of this paper . The details of thiscan be found in Section S4 of the Supplementary Material.Finally we remark that if the kinetic relation for growth had the form V = V ( f driv ) with V ( f driv ) f driv ≥ (rather than V = V ( f ) with V ( f ) f ≥ ), all solutions of an initial-value problemwould automatically conform with the dissipation inequality. The solutions in Section 3 pertain to the case E = E ρ , and per the earlier discussion, obey the dissipationinequality. NLINE S UPPLEMENTARY M ATERIAL to accompany
A Continuum Model for the Growth of Dendritic Actin Networks byRohan Abeyaratne and Prashant K. Purohit Department of Mechanical Engineering, Massachusetts Institute of Technology, Cambridge,Massachusetts, 02139, USA Department of Mechanical Engineering and Applied Mechanics, University of Pennsylvania,Philadelphia, Pennsylvania, 19104, USAJuly 9, 2020
S1. Numerical values of the parameters used in modeling the experiments of Brangbour etal. [2]
From [2] we obtain V = 0 .
42 nm / sec , R = 1 ,
100 nm . (xx)– Figure 2 of this paper (Figure 3 of [2]): Brangbour et al. give N GS = 4000 and c = 0 . ± . .Thus we take c = 0 . so that EA = c k B T N GS R = 0 . × . × , .
258 pN . (xxi)They also give the filament length at the end of two stress-free growth periods to be
200 nm and
400 nm . Therefore we take (cid:96) R = 200 nm and (cid:96) R = 400 nm . (xxii)– Figure 3 of this paper (Figure 1c of [2]): Brangbour et al. give N GS = 10 , . Based on theitem above we take c = 0 . . Then EA = c k B T N GS R = 0 . × . × , .
764 pN . (xxiii)Next we want to calculate the time t at which the force was applied. Let d ( t ) = (cid:96) ( t ) + 2 R ;it represents the distance between the centers of two adjacent particles in the model in [2].Then from the formula (cid:96) ( t ) = v t σ ( t ) /E ⇒ d ( t ) = v t σ ( t ) /E + 2 R, (xxiv)36nd therefore at the instant t +0 just after the application of the force, one has d ( t +0 ) = v t σ ( t +0 ) /E + 2 R, (xxv)and therefore t = (cid:2) d ( t +0 ) − R (cid:3)(cid:2) σ ( t +0 ) A/EA (cid:3) v . (xxvi)From Figure 1c of [2], with EA = 3 .
764 pN , R = 1100 nm and V = 0 .
42 nm / sec , we find σ ( t +0 ) A d ( t +0 ) t using (xxvi) . Therefore we take t = 1000 sec .– Figure 4 of this paper (Figure 5 of [2]): Brangbour et al. give. N GS = 4000 , c = 0 . ± . and N GS = 10000 , c = 0 . ± . . So we take c = 0 . and c = 0 . and then obtain EA = c k B T N GS R = 0 . × . × , .
979 pN , (xxvii) EA = c k B T N GS R = 0 . × . × , .
387 pN . (xxviii)– Figure 5 of this paper(Figure 2 of [2]): Brangbour et al. give N GS = 10 , . Based on thefirst item above we take c = 0 . and then EA = c k B T N GS R = 0 . × . × , .
764 pN (xxix)They also give t = 650 sec ; t = 855 sec ; the smaller value of force to be . ; and thelarger value of force to be
39 pN . – CONTINUED –37
2. Numerical values of the parameters used in modeling the experiments of Parekh etal. [6]
Table 1:
Primitive Parameters. From the literature.
Quantity Description Value Source1 k B T Boltzmann constant times .
14 pN nm At o K absolute temperature k B = 1 . × − J / K E ∞ Young’s modulus . − . /µ m Marcy et al. [5]. Average . /µ m of polymer network Parekh et al. [6] refer to Marcy’s datain their supplement3 A Specimen cross-sectional area µ m Parekh et al. [6] supplementary material4 k c AFM stiffness
30 nN /µ m Parekh et al. [6] supplement: two cantilevers.(force/deflection)
20 nN /µ m k c = 0 .
03 nN / nm and k c = 0 .
02 nN / nm σ stall A Force in specimen at stall
294 nN
Parekh et al. [6] Figure 26 (cid:96) Length of unstressed specimen
Parekh et al. [6] Figure 2A. Value ofat initial instant t (cid:96) ( t ) at time t = 0 E f Filament Young’s modulus . Howard [3] Table 3.28 A f Filament cross-sectional area
19 nm [= π (2 . ] Howard [3] Table 7.1Boal page 24 π (4) = 50 nm f stall Stall force for one filament
Howard [3] page 170
Koˇsmrlj [4]10 a Length of a monomer (G-actin) . Koˇsmrlj [4]More or less Howard’s δ Derived Parameters
Quantity Description Value Sourcea (cid:96) R (0) initial length in ref space (cid:96) R (0) = (cid:96) (0) / Λ (0) = (cid:96) (0) b k AFM stiffness . /µ m k = k c /A = 30 / (stress/deflection) . /µ m k = k c /A = 20 / c σ = k(cid:96) . /µ m σ = k(cid:96) = 0 . × . /µ m σ = k(cid:96) = 0 . × d σ stall Stress in specimen at stall .
77 nN /µ m σ stall = 294 /A = 294 / e V V at f = 0 32 − / min Min value from Brangbour [2]Max value from Marcy et al. [5]See Remark 1f τ R Time scale for growth . −
94 min τ R = (cid:96) /V with V fromat tips row-eg τ ρ Time scale for development Arbitrary Unknownof new filaments – CONTINUED –39 emark 1: Estimating V in row-e of Table 2 from other people’s data. Brangbour et al. [2], Figure 5, gives V = 0 .
39 nm / s at f = 0 . . If V = V exp( − f a/kT ) ,then with f = 0 . , a = 2 . , kT = 4 .
14 pN nm , one gets V = 32 nm / min (and so τ R = (cid:96) /V = 94 min ).Marcy et al. [5], page 5995, right column, top paragraph gives values of V (which they call V F =0 ) in the range . ± . µ m / min = 1750 ±
600 nm / min . Therefore V can be as large as / min . This means τ R = (cid:96) /V can be as small as .
28 min .Therefore we have the ranges V = 32 − / min and τ R = 1 . −
94 min . Remark 2: Determining which springs were used in the experiments underlying Figure 3of [6]:
Recall from the supplementary material in Parekh et al. [6] that they used 2 springs in theirexperiments with stiffnesses k c = 20 nN /µ m and k c = 30 nN /µ m .From the data in Figure 3a of [6] : ˙ σ = 145 − × (100 − .
5) = 0 . − µ m − ,k = ˙ σ ˙ (cid:96) = 0 . . /µ m . By comparing this with the value in the top row of row-b, we infer that they would have used thespring with stiffness k c = 30 nN /µ m in the experiment related to their Figure 3a.From the data in Figure 3b of [6]: ˙ σ = 83 − × . − µ m − ,k = 0 . . /µ m . By comparing this with the value in the bottom row of row-b, we infer that they would have usedthe spring with stiffness k c = 20 nN /µ m in the experiment related to their Figure 3b. Remark 3: Value of ρ ∞ . Though we do not need the value of ρ ∞ since it gets nondimensionalizedout, it is still worth calculating it in two different ways, as a consistency check of the model. Thefirst estimate is ρ ∞ = σ stall f stall = 0 .
77 nN /µ m µ m − , and the second follows from E ∞ = ρ ∞ A f E f : ρ ∞ = 1 A f (cid:115) E ∞ E f = 119 nm (cid:115) . /µ m . × nN /µ m = 67 µ m − . The upper bound on the number of filaments is ρ < /A f = 53 , µ m − . emark 4: Stiffness of the spring used in the experiments underlying Figure 2 of [6]: FromFigure 2a of [6] ˙ F = 175 − −
33 = 2 .
015 nN / min , ˙ (cid:96) = 8 . − −
33 = 0 . µ m / min , and therefore k c = ˙ F ˙ (cid:96) = 2 . .
067 = 30 nN /µ m . S3. Calculations underlying Figures 13 and 14.S3.1. Figure 13:
Figure 3a of Parekh et al. [6] shows that when the specimen grows under springloading the force increases linearly from the value
120 nN to
145 nN in . . Assuming thatthe force increased linearly from the start, and extrapolating backwards, we conclude that the forcewas zero at time t = 62 . . Thus we take as initial conditions σ ( t ) = 0 , ρ ( t ) = 0 . ρ ∞ , (cid:96) ( t ) = (cid:96) R ( t ) = 3000 nm at t = 62 . , where . ρ ∞ is the arbitrarily chosen initial condition for the filament density and is thedistance between the AFM cantilever and the support when the cantilever is not deflected. In thisexperiment the specimen grows under spring loading for t < t < t = 100 min ; at time t thestress is suddenly decreased to .
315 nN /µ m ; and thereafter it is held at σ ( t ) = 0 .
315 nN /µ m for t > t . The differential equation (4.25) with the preceding initial conditions and σ = k ( (cid:96) − (cid:96) ) can now be solved to find σ ( t ) , ˙ (cid:96) ( t ) and (cid:96) ( t ) for t < t < t . In particular one obtains σ ( t − ) = 0 .
373 nN /µ m , ˙ (cid:96) ( t − ) = 111 . / min , (cid:96) ( t − ) = 7742 nm . Next, the conditions at time t +2 can be found by first calculating (cid:96) ( t +2 ) from (4.27) (keeping in mindthat we are given σ ( t +2 ) = 0 .
315 nN /µ m and we know ρ ( t ) from (4.6)) and ˙ (cid:96) ( t +2 ) from (4.24): σ ( t +2 ) = 0 .
315 nN /µ m , ˙ (cid:96) ( t +2 ) = 212 nm / min , (cid:96) ( t +2 ) = 8329 nm . Finally, the differential equation (4.24) is solved for t > t using the known information at t +2 asinitial conditions. Figure 13 shows plots of σ ( t ) and ˙ (cid:96) ( t ) versus t resulting from these calculations.This figure is to be compared with Figure 3a of [6]. S3.2: Figure 14.
As seen in Figure 3b of [6], in their second experiment Parekh et al. kept thestress fixed at the value σ ( t ) = 0 .
178 nN /µ m for t < t < t = 73 min ; the force clamp wasthen released at time t and the specimen allowed to grow under spring loading conditions for Presumably prior to that, for < t < t , the actin filaments did not extend all the way from from the AFMcantilever to the other support and so the specimen was growing freely under stress-free conditions without engagingthe AFM spring. < t < t = 79 min ; at the instant t the value of the stress was suddenly decreased back to thevalue .
178 nN /µ m , and held there for t > t . In order to calculate the response predicted byour model, the first task is to estimate the time t at which the stress was initially applied on thespecimen (which we assume was done at the instant when the filaments extended all the way fromthe AFM cantilever to the other support).In order to determine t we proceed as follows: according to Figure 3b of [6] the stressvaries continuously at the instant t when the force clamp is released and therefore σ ( t +1 ) =0 .
178 nN /µ m . Thus from σ ( t +1 ) = k ( (cid:96) ( t +1 ) − (cid:96) ) we find (cid:96) ( t +1 ) = 6390 nm . When σ variescontinuously it follows from (4.27) that (cid:96) varies continuously (since ρ varies continuously). Thus (cid:96) ( t − ) = 6390 nm . Now focus on the time interval t < t < t . At the instant t we have theinitial conditions (cid:96) R ( t +0 ) = 3000 nm and ρ ( t ) = 0 . ρ ∞ where the initial value of the fila-ment density has been chosen to be consistent with the dissipation inequality; see Section S4.2.Thus integrating (4 . with respect to time from t to t and using (cid:96) R ( t +0 ) = 3000 nm and (cid:96) R ( t − ) = (cid:96) ( t − ) / Λ( σ ( t − ) , ρ ( t )) = 6390 / Λ(0 . , ρ ( t )) with ρ ( t ) given by (4.6) leads to a non-linear algebraic equation for t . This yields t = 48 .
23 min . Thus we have the initial conditions σ ( t ) = 0 .
178 nN /µ m , ρ ( t ) = 0 . ρ ∞ , (cid:96) R ( t ) = 3000 nm at t = 48 .
23 min . Determining σ ( t ) and ˙ (cid:96) ( t ) is now straightforward. We first determine ρ ( t ) for all time from(4.6) with ρ ( t ) = 0 . ρ ∞ . We then integrate (4 . with respect to t and use (4.23) and thepreceding initial conditions to find the stress and elongation-rate for t < t < t . In particular weobtain σ ( t − ) = 0 .
178 nN /µ m , ˙ (cid:96) ( t − ) = 186 nm / min , (cid:96) ( t − ) = 6390 nm . By the aforementioned continuity of the stress and elongation at time t and (4.25) we have σ ( t +1 ) = 0 .
178 nN /µ m , ˙ (cid:96) ( t +1 ) = 124 nm / min , (cid:96) ( t +1 ) = 6390 nm . Next we use these as initial conditions to solve (4.25) together with σ = k ( (cid:96) − (cid:96) ) to determine σ ( t ) and ˙ (cid:96) ( t ) for t < t < t . In particular we find σ ( t − ) = 0 .
216 nN /µ m , ˙ (cid:96) ( t − ) = 119 nm / min , (cid:96) ( t − ) = 7118 nm . Turning to the instant t +2 , we know the stress σ ( t +2 ) = 0 .
178 nN /µ m and so use (4.27) to find (cid:96) ( t +2 ) . This together with (4.24) gives σ ( t +2 ) = 0 .
178 nN /µ m , ˙ (cid:96) ( t +2 ) = 187 nm / min , (cid:96) ( t +2 ) = 7508 nm . Finally we solve (4.24) to find the stress and elongation-rate for t > t using the above informationat t +2 as initial conditions. Figure 14 shows plots of σ ( t ) and ˙ (cid:96) ( t ) versus t resulting from thesecalculations. This figure is to be compared with Figure 3b of [6]. Keep in mind that the length (cid:96) ( t +0 ) (cid:54) = 3000 nm since the specimen length will change suddenly as the stress isapplied.
4. The solutions in Sections 4.2 and 4.3 obey the dissipation inequality.S4.1. Spring-loaded solution.
The solution ρ ( t ) , σ ( t ) , t ≥ t , found in Section . . for the spring loaded specimen isshown on the ρ, σ -plane in Figure S.1. The trajectory starts from ( ρ ( t ) , σ ( t )) = (0 . ρ ∞ , and terminates at stall corresponding to ( ρ ∞ , σ stall ) . Since σ stall = 0 .
77 nN /µ m and E ∞ =3 . /µ m , the figure has been drawn with σ stall /E ∞ = 0 . . We are only concerned withthe region on and below the straight line σ = f stall ρ since the filament force then lies in therange ≤ f ≤ f stall . The driving force is positive in the lightly shaded region below this line,and therefore on this region both the dissipation inequality f driv V ≥ and the kinetic relation V = V ( f ) give V > . The solution associated with the trajectory shown therefore satisfies thedissipation inequality. Figure S.1:
The specimen is spring-loaded per Section 4.2. The dissipation inequality and kinetic relation both give
V > in the lightly shaded region. The reader is referred to the appendix for a definition of the stress Σ st ( ρ ) . S4.2. Loading programs involving discontinuous stress.
Next consider the loading programs studied in Section . involving a discontinuous changein stress at a certain instant t . Since σ stall = 0 .
77 nN /µ m and E ∞ = 0 . /µ m , in this casewe have σ stall /E ∞ = 1 . .In the first calculation, the specimen initially grows under spring loading conditions andfollows the trajectory shown in Figure S.2 for t < t < t . It starts from ( ρ ( t ) , σ ( t )) = (0 . ρ ∞ , . σ stall at time t , it issuddenly decreased to the value . σ stall and held constant at that value for t > t . The trajectorytherefore drops vertically at this instant and follows a rightward pointing horizontal line thereafter;this part of the trajectory falls within a very narrow sliver and is barely visible in Figure S.2. Again,the trajectory lies in the lightly shaded part of the figure where f driv > and V > and thereforethe solution satisfies the dissipation inequality. Figure S.2:
The specimen is spring-loaded for t < t < t . The stress is decreased suddenly at time t and thenheld constant for t > t . Here t = 0 and t = 100 min . The reader is referred to the appendix for the definition ofthe stress Σ st ( ρ ) . In the second calculation, initially, for a period t ≤ t ≤ t , the stress is held constant at thevalue . σ stall and so the trajectory of ( ρ ( t ) , σ ( t )) follows a rightward pointing horizontal line onthe ρ, σ -plane. In order for this trajectory to be admissible by the dissipation inequality it mustlie in the lightly shaded region in Figure S.3 and therefore ρ ( t ) must exceed the value ≈ . .We take ρ ( t ) = 0 . ρ ∞ . At the instant t the force clamp is released and the specimen now growsunder spring loading conditions. The trajectory for t < t < t is determined as in Section . with the initial condition σ ( t ) /σ stall = 0 . , ρ ( t ) /ρ ∞ = 0 . where the value of ρ ( t ) wasdetermined using equation (24). The solution now follows the curved trajectory shown in FigureS.3. At time t , when the stress has reached the value . σ stall , it is suddenly decreased back to . σ stall and held constant at that value from then on. The trajectory lies in the lightly shadedregion of the figure and therefore satisfies the dissipation inequality. References [1] Boal D.,
Mechanics of the Cell , Cambridge University Press, (2002). .0 0.2 0.4 0.6 0.8 1.00.00.20.40.60.81.0 Figure S.3:
The stress is kept constant at the value . σ stall for time t < t < t . The force clamp is releasedat time t and the specimen is spring-loaded for t < t < t . At time t the stress is decreased suddenly back to itsoriginal value, and kept constant for t > t . Here t = 0 , t = 73 min and t = 79 min . The reader is referred to theappendix for the definition of the stress Σ st ( ρ ) . [2] Brangbour, C., du Roure, O., Helfer, E., Demoulin, D., Mazurier, A., Fermigier, M., Carlier, M-F.,Bibette, J., Baudry, J., “Force-velocity measurements of a few growing actin filaments”, PLoS Biology (4), e1000613, (2011).[3] Howard J., Mechanics of motor proteins and the cytoskeleton , Sinauer Associates, (2001).[4] Koˇsmrlj, A, Online class notes – MAE 545: Lessons from Biology for Engineering Tiny Devices,Princeton University, Spring, 2017.[5] Marcy, Y., Prost, J., Carlier, M-F., Sykes, C., “Forces generated during actin-based propulsion: A directmeasurement by micromanipulation”,
Proceedings of the National Academy of Sciences (16), 5992-5997, (2004).[6] Parekh, S.H., Chaudhuri, O., Theriot, J.A., Fletcher, D.A., “Loading history determines the velocity ofactin-network growth”,
Nature Cell Biology (12), 1119-1123, (2005).(12), 1119-1123, (2005).