BBeat regulation in twisted axonemes
Pablo Sartor , Veikko F Geyer , Jonathon Howard and Frank J¨ulicher Max Planck Institute for the Physics of Complex Systems. Noethnitzer Strasse 38 , 01187, Dresden,Germany. Department of Molecular Biophysics and Biochemistry, Yale University, New Haven, Connecticut
Cilia and flagella are hairlike organelles that propel cells through fluid. The active motion ofthe axoneme, the motile structure inside cilia and flagella, is powered by molecular motors of thedynein family. These motors generate forces and torques that slide and bend the microtubuledoublets within the axoneme. To create regular waveforms the activities of the dyneins must becoordinated. It is thought that coordination is mediated by stresses due to radial, transverse,or sliding deformations, that build up within the moving axoneme. However, which particularcomponent of the stress regulates the motors to produce the observed flagellar waveforms remainsan open question. To address this question, we describe the axoneme as a three-dimensional bundleof filaments and characterize its mechanics. We show that regulation of the motors by radial andtransverse stresses can lead to a coordinated flagellar motion only in the presence of twist. Bycomparison, regulation by shear stress is possible without twist. We calculate emergent beatingpatterns in twisted axonemes resulting from regulation by transverse stresses. The waveforms aresimilar to those observed in flagella of
Chlamydomonas and sperm. Due to the twist, the waveformhas non-planar components, which result in swimming trajectories such as twisted ribbons andhelices that agree with observations.
PACS numbers:
Cilia and flagella are slender cell appending organellesthat contain a motile internal structure called the ax-oneme. The axoneme, in turn, contains nine microtubuledoublets, a central pair of microtubules, motor proteinsin the axonemal dynein family, and a large number ofadditional structural proteins [1, 2], see Fig. 1A. The ax-oneme undergoes regular oscillatory bending waves thatpropel cells through fluids and fluids along the surfaces ofcells. This beat is powered by the dyneins, which gener-ate sliding forces between adjacent doublets [3]. Bendingoriginates from an imbalance of antagonistic motors act-ing on doublets on opposing sides of the beating plane,see Fig. 1A [4, 5].It has been suggested that the switching of dyneinactivity on opposite sides of the axoneme is the resultof feedback and non-linearities [4, 6–8]. The axonemaldyneins generate forces deforming the axoneme; the de-formations or the corresponding stresses, in turn, regu-late the dyneins. Several components of the stress canbe distinguished: radial stress, that tends to increase theaxoneme radius; transverse stress, that tends to separatethe doublets; and shear stress, that tends to slide dou-blets with respect to each other. How these componentscan regulate dynein activity remains poorly understood.One hypothesis, referred to as “geometric clutch”, is thatdynein is regulated by transverse stress. Due to the cir-cular symmetry of the transverse stress, motors on op-posite sides of the axoneme experience the same trans-verse stress, so transverse stresses are not antagonisticand cannot be used as a signal for switching. As a conse-quence, regulation by transverse stress requires an addi-tional asymmetry [5, 9, 10], whose origin remains elusive.Other suggested control mechanisms, such as regulationby sliding [6, 7, 11, 12] or curvature [5, 13, 14], do not have this circular symmetry and thus do not require anadditional asymmetry.In addition to sliding forces, axonemal dyneins can alsogenerate torques, which rotate microtubules in in vivo as-says [15, 16]. In the axoneme, this rotation will lead totwist [17]. Because of the possibility that twist mightbe important for beat generation, we have developed athree-dimensional model of the axoneme. This model, in-spired by earlier work [18], is able to distinguish betweentransverse and radial stresses. We show that twist is ca-pable of breaking the circular symmetry of these stresses.In this case, transverse stress can be used as a signal forswitching and so can act as a control mechanism. Theemergent beating patterns are similar to those of
Chlamy-domonas axonemes; they are non-planar and result incomplex swimming trajectories such as twisted ribbonsand helices.
CONTINUUM MECHANICS OF THE AXONEME
The axoneme contains a regular arrangement of mi-crotubule doublets in a cylindrical geometry that is sta-bilized by additional structural elements such as radialspokes, a central pair of microtubules, dynein molecu-lar motors, and other elements such as nexin linkers [2].The doublets n = 1 and n = 2 can be identified by ahigher density in electron microscopy images referred toas cross-bridge, see Fig. 1A.We characterize the axonemal structure by a bundle offilaments corresponding to the microtubule doublets thatare arranged on a cylindrical sheet of radius a [18], seeFig.1. The cylindrical sheet R ( s, φ ) is parametrized by a r X i v : . [ q - b i o . S C ] N ov A B
912 3 4 5678
FIG. 1:
Geometry of the axoneme. A . Schematic of anaxonemal cross-section, with numbering for the 9 doublets, asseen from the basal end. Dyneins appear in blue and green,elastic linkers in purple, and radial spokes in orange. Thecross-bridges between doublets 1-2, thick purple lines, definethe beating plane, dashed line. The center-line triad is ori-ented such that e , in red, points between doublets 3 and 4,and the beating plane is spanned by e , in green, and e , inblue. Bending in the direction of e requires the motors infilaments 2 − − e requires activity of motors in filaments 6 − B . Thebundle of filaments is parametrized by the azimuthal coordi-nate φ and the arc-length (cid:96) which depends on the center-linearc-length s . These define tangent vectors e s and e φ . Theunit vector n is normal to e s , and e a is orthogonal to thetangent plane. The basal end is on the right. an angular coordinate φ and the distance variable s as R ( s, φ ) = r ( s ) + e ( s ) a ( s, φ ) cos( ϕ ( s, φ ))+ e ( s ) a ( s, φ ) sin( ϕ ( s, φ )) . (1)Here, s is the arc-length of the center-line r ( s ) of thecylindrical sheet measured from base to tip. The vectors e and e are unit vectors normal to the centerline withunit tangent e = ∂ s r . We choose e to point betweenfilaments n = 3 and n = 4, and orient e perpendicular toboth e and e , see Fig. 1A. The angular parameter φ isused to identify filaments which correspond to the values φ = φ n with φ n = 2 π ( n − / /
9. The function ϕ ( s, φ )describing the azimuthal angle of filaments indexed by φ .Similarly we allow the cylinder radius a ( s, φ ) to dependon s and φ .To characterize the geometry of the filaments we in-troduce the tangent vectors e s = ∂ s R / | ∂ s R | and e φ = ∂ φ R / | ∂ φ R | , which form a basis of the tangent space onthe cylindrical sheet. We also introduce the filament nor-mal n in the tangent plane. It obeys e s · n = 0 and n = 1, see Fig. 1B. The vector e a = n × e s is nor-mal to the cylinder pointing outwards. The filamentcurvatures tangential and perpendicular to the cylindri-cal surface are then given by C ( s, φ ) = e s · ∂ s n and C ( s, φ ) = e a · ∂ s e s , respectively.In order to discuss the mechanics of the axoneme weintroduce the relevant deformation variables. For sim-plicity, we impose the constraints ϕ = φ and a = a forwhich the cylinder radius a and the separation betweenneighboring filaments 2 πa / a ∇ n ( (cid:96) b + (cid:96) ) , (2)where (cid:96) denotes the arc-length of a filament correspond-ing to angle φ at center-line distance s , with (cid:96) ( s, φ ) = (cid:90) s | ∂ s R ( s (cid:48) , φ ) | d s (cid:48) . (3)The length offset at the base defined as the mismatchbetween the center line and filament φ is denoted (cid:96) b ( φ ).Correspondingly, the sliding displacement at the base is∆ b = a ∇ n (cid:96) b . Here, the normal derivative is defined by ∇ n f = ( n s / | R s | ) ∂ s f + ( n φ / | R φ | ) ∂ φ f , where n s and n φ are the components of n = n s e s + n φ e φ . The filamentssplay is defined as Γ = e a · ∂ φ e s , (4)and corresponds to the out of plane rotation of the fila-ment tangent vector when changing φ .Molecular motors can induce sliding displacement andfilament splay by generating active stress conjugate tothese strains, see Fig. 2B. These conjugate stresses arethe motor force f m , which tends to slide filaments apart,and the motor torque m m which tends to induce splay.The geometry of the filaments is fully characterizedby the axonemal curvatures, Ω = e · ∂ s e and Θ = − e · ∂ s e , the twist Π = e · ∂ s e , and the length offset (cid:96) b ( φ ). To linear order in the curvatures, we can expressthe sliding displacement and the splay as∆( s, φ ) ≈ ∆ b ( φ ) − a Π( s ) + a cos( φ ) (cid:90) s Ω( s (cid:48) )d s (cid:48) + a sin( φ ) (cid:90) s Θ( s (cid:48) )d s (cid:48) , Γ( s ) ≈ a Π( s ) . (5)An analogous expansion can also be made for C and C , see Appendix. Note that, to lowest order in the de-formations, the splay directly corresponds to axonemaltwist. WORK FUNCTIONAL
The mechanical properties of the axoneme are charac-terized by the elasticity of filaments and linkers, and theactive forces and torques generated by molecular motors.We introduce the work functional G that describes themechanical work performed to induce axonemal deforma- BA relaxed deformed FIG. 2:
Deformation of the axoneme A . Relaxed and de-formed geometries of the axoneme with two filaments markedin green and blue, and a dynein represented as two green cir-cles. In the deformed state the tangent plane (in red) spannedby e s and e φ does not contain the tangent vector e (cid:48) s of thecontiguous doublet, which is displaced out of the plane by adistance Γ in the direction of e a . The sliding ∆ correspondsto the length mismatch of the two successive doublets mea-sured by projecting in the direction of the normal vector n . B . A dynein motor creates a pair of forces f m that producessliding ∆, and also torques m m that produce out of plane dis-placement Γ. The deformed states are shown from differentperspectives. tions: G = (cid:90) L (cid:90) π (cid:26) κ C + κ C + k s + k r (cid:96) b + (cid:96) − s ) + f m ∆+ m m Γ + σ φ a ( ∂ φ ϕ −
1) + σ a a ( a − a ) + Λ2 | ∂ s r | (cid:27) d s d φ + (cid:90) π (cid:26) K s + K r (cid:96) (cid:27) d φ . (6)Here κ and κ are bending rigidities corresponding todeformation of the filaments tangent and perpendicularto the cylindrical sheet. The sliding stiffness of elementsthat link neighboring doublets is denoted k s . Similarly k r denotes the radial stiffness of sliding linkers betweenfilaments and the central pair. These elastic constantsrelate to those of an individual filament by a geometricfactor of 9 / π , for example κ = 9 κ , db / π , with κ , db the doublet bending rigidity. The tangential stress σ φ and the radial stress σ a are Lagrange multipliers to im-pose the constraints ϕ = φ and a = a . The Lagrangemultiplier Λ imposes the constraint | ∂ s r | = 1. Finally, K s and K r are basal stiffnesses between neighboring fil-aments and between filaments and the central pair, re-spectively. The mechanical work performed by motors isgiven by f m ∆ and m m Γ.In the following, Fourier modes in φ will play an im-portant role. Therefore we express the φ dependence ofvariables in Eq. 6 by Fourier series. For instance, themotor force and sliding displacement can be written as f m = f (0)m + f (1)m cos( φ ) + f (2)m sin( φ ) + . . . ∆ = ∆ (0) + ∆ (1) cos( φ ) + ∆ (2) sin( φ ) + . . . , (7) In this analysis we truncate the series after the first modefor simplicity. Higher modes can be systematically takeninto account if they become relevant. Correspondingly,we also expand f , σ φ and σ a . Upper indices in paren-thesis in the following always denote azimuthal Fouriermodes. RADIAL AND TRANSVERSE STRESS IN THEAXONEME
A bent and twisted axoneme exhibits radial stress σ a and transverse stress σ φ , which can be key in regulatingthe motor activity. To determine their values, we firstestablish a force balance for the twist of the axoneme.In the case where twist relaxes quickly, the correspond-ing force balance is quasi-static. The torque balance δG/δ Π = 0 then implies a k s Π − a κ ∂ s Π = a m m − a f (0)m . (8)This equation shows that the cilium is twisted by motortorques m m and also by the zeroth harmonic of motorforce f (0)m . The corresponding twist stiffness is providedby the doublet sliding stiffness k s . Since twisting the cil-ium involves bending of the doublets, their bending stiff-ness κ couples to twist. The competition of sliding andbending of the filaments during twist is characterized bythe length-scale d = (cid:112) κ /a k s , at which twist deforma-tions decay along the axoneme in response to spatiallylocalized motor forces or torques.The transverse and radial stresses can be calculatedfrom the force balances δG/δϕ = 0 and δG/δa = 0 asLagrange multipliers to impose the constraints ϕ = φ and a = a . The full expressions of σ a and σ φ are givenin the Appendix. In the simple case of almost planardeformations with Θ = 0 and Π small we have σ (1) φ = M m Ω /a (9) σ (0) a = ( F (1) + F (2)r )Ω2 a (10) σ (1) a = 2 f (1) Π + 3 κ Ω ∂ s Π /a + Ω M m , (11)where we have introduced the integrated torque M m = − (cid:82) Ls m m d s (cid:48) and the integrated net sliding force F (1) = − (cid:82) Ls f (1) d s (cid:48) , with f (1) = f (1)m + k s ∆ (1) , as well as theradial force F (1)r = − (cid:82) Ls k r ( (cid:96) b , + (cid:96) − s )d s (cid:48) . Note that σ (2) φ and σ (0) φ vanish, while the first angular mode σ (1) φ ofthe transverse stress is proportional to curvature Ω. Thezeroth mode of the radial stress, in Eq. 10, is a gener-alization of the normal stress in two dimensional models[9, 10, 21]. Figure 3 depicts the angular profiles of radialand transverse stresses. BA C
FIG. 3:
Azimuthal stress profile. A . Components of thesliding force. The zeroth mode is homogeneous and createstwist. The first mode, positive above the dashed line (beatingplane) and negative below, is responsible for in-plane bending. B and C . The transverse stress has a first mode, and the radialstress has a zeroth and also a first mode. These first modes,possible only in the presence of twist, can regulate the firstharmonic of the sliding force. AXONEMAL DYNAMICS
The dynamics of the axoneme is governed by a balanceof fluid friction forces and axonemal forces ∂ t r = − (cid:16) ξ − ⊥ ( e e + e e ) + ξ − (cid:107) e e (cid:17) · δGδ r . (12)Here δG/δ r is the axonemal force, and ξ ⊥ and ξ (cid:107) arethe friction coefficients per unit length perpendicular andtangential to the axonemal axis. The explicit expressionfor δG/δ r is given in the Appendix. From Eq. 12 we canobtain dynamic equations for the axonemal curvatures Ωand Θ given by ∂ t Ω = − e · ∂ t ∂ s r and ∂ t Θ = e · ∂ t ∂ s r .In the following we will focus on the case of almostplanar and weakly twisted axonemal shapes. Physicallythis corresponds to enforcing the constraint Θ = 0. Thenon-linear equation of motion of Ω is to linear order inΠ given by ξ ⊥ ∂ t Ω = − ¯ κ∂ s Ω + ¯ a∂ s f (1) (13)+ ∂ s (Ω τ ) + ( ξ ⊥ /ξ (cid:107) ) ∂ s ((Ω (¯ κ∂ s Ω − ¯ af (1) ) + Ω ∂ s τ ))where ¯ κ = π ( κ + κ ) is an effective bending rigidityand ¯ a = πa . The center-line tension τ is related tothe Lagrange multiplier Λ introduced in Eq. 6 [8], seeAppendix.The dynamic shape equation 13 for the curvature Ωis a generalization of the previously introduced shapeequation for two dimensional beats [8, 13, 14]. Notethat Eq. 13 not only describes beats in a plane, but forΠ (cid:54) = 0 it describes beats on a twisted two-dimensionalmanifold. The resulting three dimensional shapes r ( s )can be determined from Ω and Π, which are solutions to Eqns. 13 and 8, by integrating ∂ s e = − Ω e and ∂ s e = Ω e − Π e ( s = 0). Thus, although we constrainedΘ = 0, the twist causes out of plane bending. To imposethe constraint of Θ = 0 in the presence of twist, thecomponent f (2) of the sliding force becomes a Lagrangemultiplier that corresponds to the force introduced bystructural elements. SELF-ORGANIZED BEATING BY MOTORCONTROL FEEDBACK
We now focus our attention on self-organized beatingpatterns with small amplitude waveforms and angularbeat frequency ω . In this case, the periodic flagellar beatis powered by oscillating sliding forces, which we write infrequency representation as f = ˜ f + ˜ f e iωt + ˜ f − e − iωt + . . . (14)where ˜ f is time independent, ˜ f = ˜ f ∗− is the amplitudeof the fundamental Fourier mode, and higher frequencyharmonics have been omitted for simplicity. We also de-fine time Fourier modes of the azimuthal force compo-nents denoted ˜ f ( n ) k , as well as of the transverse stress˜ σ ( n ) φ,k , sliding ˜∆ ( n ) k , and torque ˜ m m ,k , where k = − , , n = 0 , , f (1)1 ,but not the oscillating torque amplitude ˜ m m , . We pro-pose that motor regulation occurs through the sensitivityof the motor function to local axonemal deformations andstresses. For simplicity we illustrate our ideas by focus-ing on regulation by sliding displacement [6, 8, 18] and bymolecular deformations induced by the transversal stress[9, 10, 21]. We therefore write the oscillating motor forceto linear order as˜ f (1)1 = χ ( ω ) ˜∆ (1)1 + ζ ( ω )˜ σ (1) φ, , (15)where χ ( ω ) and ζ ( ω ) are complex frequency-dependentlinear response coefficients describing effects of slidingdisplacement and transverse stress, respectively [8]. SYMMETRIC AND ASYMMETRIC TWISTEDBEATING PATTERNS
We now discuss beating patterns that arise by an os-cillating instability from an initially nonoscillating statevia a Hopf bifurcation. This instability can be drivenby the mechanical feedbacks mediated by the regulationof motors by sliding displacements or tangential stressesdescribed by Eq (15). The nonoscillating state is char-acterized by a time independent curvature Ω = ˜Ω and FIG. 4:
Twisted beating patterns and swimming trajectories. A , B and C . Above, top, side, and front view of self-organized beating pattern regulated by transverse stress (A and B) or sliding (C). Due to the twist, all beating patterns arenon-planar, however only the waveforms in B and C are asymmetric. Below, three different views of the swimming path with alab-frame for reference. The gray line shows the path of the basal point, the yellow twisted surface the average trajectory, andthe blue arrow indicates the swimming direction. For the symmetric beating pattern (A) the swimming surface forms a twistedribbon, while for the asymmetric cases it forms a helix (B and C). Note that due to the presence of the head the precession ofthe basal end in C is much smaller than in A and B. a time independent twist Π = ˜Π . We again considerthe simple case Θ = 0. Starting from this non-oscillatingstate, an oscillating mode that represents the flagellarbeat emerges at the Hopf bifurcation. This mode can becharacterized by the Fourier amplitude of the fundamen-tal frequency component ˜Ω , which becomes non-zero be-yond the bifurcation point.We now discuss the symmetry properties of the emerg-ing beating patterns. We first consider the simple caseof vanishing static twist Π = 0, for which beats are con-fined to a plane. In this case we can distinguish sym-metric beats with vanishing average curvature ˜Ω = 0and asymmetric beats. Symmetric beats are mirror sym-metric within the beat plane [8] and the swimming tra-jectories are straight lines within the plane. This mirrorsymmetry is broken in the asymmetric case, for whichswimming trajectories are circles in the plane [23, 24].Generally, the static twist does not vanish. The beat-ing pattern then is confined to a twisted two dimensionalmanifold, see Fig. 4, bottom. Again we can distinguishsymmetric and asymmetric twisted beats. In the sym-metric case with ˜Ω = 0 the beat is now symmetric withrespect to π rotations with a rotation axis tangential tothe manifold, see Fig. 4A, and swimming trajectories are straight lines on the manifold. This symmetry is brokenin the case of asymmetric twisted beats, which exhibithelical swimming paths, see Fig. 4B.The static twist is determined from Eq. 8. The staticcurvature of asymmetric beats follows from the force bal-ance ¯ κ ˜Ω = ¯ a ˜ F (1)0 . For simplicity we illustrate the mainideas where the static force and torques act at the distalend, in which case F (1)0 and M m , are s independent. Asa consequence the static curvature is constant and theaverage shape is a twisted circular arc. We can now ex-press the dynamic equation for the beat shape. For thecase of a symmetric twisted beat, we find iωξ ⊥ ˜Ω = − ¯ κ∂ s ˜Ω + πa χ∂ s ˜Ω + πa β∂ s ˜Ω . (16)Here β ( ω ) = ( M m , /a ) ζ ( ω ) plays the role of an effectivemotor control feedback by curvature. The term propor-tional to χ describes a motor control feedback by slidingdisplacements. Solving Eq. 16 provides the time depen-dence of the curvature Ω from which we determine theaxonemal shape by integration along the arc-length. Anexample of such a beat is shown in Fig. 4A in the casewhere oscillations are generated by motors that are reg-ulated via transverse stresses. The resulting waveformis non-planar and the swimming path corresponds to atwisted ribbon. This beating patterns is similar to thatof the Chlamydomonas mutant mbo2 [23, 25].Asymmetric beating patterns can be studied in a sim-ilar way, by expanding Eq. 13 near a state with constantstatic curvature ˜Ω . This leads to a generalization ofEq. 16 describing asymmetric beats, see Appendix. Anexample of such a beating pattern for the case of mo-tor regulation by transverse stresses is shown in Fig. 4B.This beating pattern is non-planar and asymmetric, andthe resulting swimming path is a helical ribbon. Thiswaveform is similar to the one observed for the wild-type Chlamydomonas axoneme [23, 26].Alternatively we can also discuss beats by motor reg-ulation via sliding. It has been suggested that slidingcontrol likely governs the beat shape of bull sperm [7].In Fig. 4C we show such a beating pattern of asymmetri-cally beating axoneme with motors regulated via sliding.The result is a waveform similar to that of a freely swim-ming bull sperm. Due to the friction of the sperm head(green sphere in Fig. 4C) the resulting helical ribbon isnarrow as compared to Fig. 4B, see also [27, 28].
DISCUSSION
The role of the three dimensional architecture of theaxoneme for motor regulation and beat generation ispoorly understood. Here, using a three dimensional con-tinuum mechanical model of the axoneme, we showedthat both shear stresses (associated with sliding forces)and transverse stresses (associated with torques) can beused to regulate motors and generate periodic beatingpatterns. In particular we showed that in the presenceof twist transverse stresses are proportional to curvature,and motor regulation by transverse stresses effectively re-sults in control motor activity by axonemal curvature.Dynein generated shear forces induce relative sliding ofmicrotubule doublets. We show here that motor torquesinduce splay deformations of microtubule doublets, seeFig. 2. Because microtubule splay deformations lead toaxonemal twist, we conclude that motor torques twistthe axoneme even in the absence of shear forces. As aconsequence of the twist beating patterns are in generalchiral.Axonemal twist is governed by torsional stiffness. Ourwork shows that sliding elastic linkers between neighbor-ing microtubule doublets provide an effective torsionalstiffness a k s of the axoneme, see Eq 8. In additionthere could be a contribution to the torsional stiffnessof the axoneme from the torsional stiffness κ of in-dividual microtubule doublets, which for simplicity wehave not included in our discussion. This contributionbecomes relevant when doublets are constrained not torotate around their axis, and results in a net torsionalstiffness κ + a k s . Estimating κ ∼ κ , we have that κ / ( a k s ) ∼ ( d/a ) ∼ . In this case the torque nec- essary to twist the axoneme would be larger, and thetwist would decay faster along the axonemal length, seeAppendix.Our work shows that feedback control of motors bothby shear or by transverse stress can induce time peri-odic bending waves in three dimensions via a dynamicinstability. We show that motor regulation by transversestresses requires motor torques that twist the axoneme.This mechanism effectively gives rise to motor controlby axonemal curvature, which is very reliable and evenworks for short axonemes for which sliding control doesnot induce traveling waves [29]. These waveforms arethree dimensional because of the chirality of the axonemeand the resulting axonemal twist, see Fig. 4. Thesebeating patterns in general generate swimming trajec-tories that are either helical paths or can be representedas twisted ribbons. Such swimming paths have indeedbeen observed experimentally for different sperm cells[27, 28, 30]. If such swimmers are observed near surfacesthe chirality of the beat leads to circular trajectories [31]which have been reported for various systems [24, 32].The beating patterns of Chlamydomonas and of spermoften exhibit periodic deformations around an averagecurvature. In our work we introduced this average cur-vature via an asymmetry of static forces at the distal end.However the molecular origin of these beat asymmetriesremains unclear. We think that our continuum mechan-ical model of the axoneme will be an important tool tounderstand the mechanical origin of beat asymmetriesand the selection of the beat plane of flagellar beats.
APPENDIXCurvatures of filaments for small deformations
The filaments on the cylindrical surface have curva-tures C ( s, φ ) and C ( s, φ ). These are functions of theprincipal curvatures of the center-line, Ω( s ) and Θ( s ), aswell as the twist, Π( s ). For the case of small deforma-tions, we have C ( s, φ ) ≈ Ω( s ) cos( φ ) + Θ( s ) sin( φ ) − a ∂ s Π C ( s, φ ) ≈ Θ( s ) cos( φ ) − Ω( s ) sin( φ ) , where only terms linear in the axoneme curvature arekept, see also [19]. Note that the rate of twist, and notthe twist, affects the curvature on the tangent plane C .This is because, as shown in Eq. 5, a constant twist corre-sponds to a constant sliding. Constant twist contributesas a higher order correction to C . Radial and transverse stress
To obtain the radial stress σ a and the transverse stress σ φ we use the force balances given by δG/δa = 0 and δG/δϕ = 0. The procedure is straightforward, see [8, 18,21] for similar calculations, and for σ a directly gives σ a = 2( f − m m /a )Π + F (Ω cos( φ ) + Θ sin( φ )) /a + F r (Ω sin( φ ) − Θ cos( φ )) /a − κ (( ∂ s Π) − ∂ s Π(Ω cos( φ ) + Θ sin( φ )) /a ) . For the case Θ = 0 and to linear order in Π, the zerothand first azimuthal modes are those in Eqs. 10 and 11,where we have used the integral of Eq. 8. In the case ofthe transverse stress, the force balance results in a ∂ φ σ φ = κ (Ω cos( φ ) + Θ sin( φ ) − a ∂ s Π)(Θ cos( φ ) − Ω sin( φ )) − κ (Θ cos( φ ) − Ω sin( φ ))(Θ sin( φ ) + Ω cos( φ )) − a F (Θ cos( φ ) − Ω sin( φ )) − a F r (Ω cos( φ ) + Θ sin( φ )) . For the case Θ = 0 we have a ∂ φ σ φ = ( a F (0) + a κ ∂ s Π)Ω sin( φ )+ . . . . Using Eq. 8 and integrating givesthe harmonic in Eq. 9. General equations of motion
In the case in which the twist Π relaxes fast, it is cal-culated from Eq. 8, see [18] for a characterization of twistdynamics. The force balance describing the dynamics ofthe center-line (cid:126) r is given in Eq. 12, where δGδ(cid:126)r = − ∂ s (cid:110) ˆ e (cid:16) − ¯ a ( F (1)t Π + f (2) ) + ¯ κ ΩΠ + ¯ κ∂ s Θ (cid:17) + ˆ e (cid:16) − ¯ a ( F (2)t Π − f (1) ) + ¯ κ ΘΠ − ¯ κ∂ s Ω − ˆ e τ (cid:17) (cid:111) , where we have introduced the total force modes F (1)t = F (1) + F (2)r and F (2)t = F (2) + F (1)r , and the tension τ = 2 π Λ − ¯ a ( F (1) Ω + F (2) Θ) − ¯ κ (Ω + Θ ). Taking thetime derivative of the constraint ∂ s (cid:126) r = 1 we arrive atˆ e · ∂ s ∂ t (cid:126) r = 0, which together with Eq. 12 provides theequation for the tension. Finally, to calculate the modesof the basal length (cid:96) ( n )b we use the sliding force balances δG/δ(cid:96) ( n )b = 0. For n = 1 and n = 2 the equationsare K t (cid:96) (1)b = F (2)t (0) and K t (cid:96) (2)b = F (1)t (0), while doing δG/δ(cid:96) (0)b = 0 results in (cid:96) (0)b = 0. Here, we have definedthe total basal stiffness K t = K s + K r .The boundary forces and torques exerted by the fil-ament correspond to the boundary terms of δG/δ(cid:126)r and δG/δ Π, see [8, 18, 21]. Balancing these by external forces (cid:126)F ext and torques (cid:126)T ext we have at the base s = 0: (cid:126)F ext = − (cid:110) ˆ e (cid:16) − ¯ a ( F (1)t Π + f (2) ) + ¯ κ ΩΠ + ¯ κ∂ s Θ (cid:17) + ˆ e (cid:16) − ¯ a ( F (2)t Π − f (1) ) + ¯ κ ΘΠ − ¯ κ∂ s Ω (cid:17) − ˆ e τ (cid:111) (cid:126)T ext = (¯ aF (2) − ¯ κ Θ − πκ a Ω ∂ s Π)ˆ e + ( − ¯ aF (1) + ¯ κ Ω + πκ a Θ ∂ s Π)ˆ e M ext = κ a ∂ s Π , The lack of a third component in the torque balancecomes from neglecting the twist dynamics [18]. The thirdmoment balance comes from the contribution of filamentbending, and involves an additional external moment M ext . At the tip s = L the boundary conditions are anal-ogous. In this work we considered two types of boundaryconditions. For a freely swimming flagellum the externalforces and torques are null. For a flagellum attached to ahead the torques and forces at the base are (cid:126)F ext = ξ trans (cid:126)v and (cid:126)T ext = ξ rot (cid:126)ω , where (cid:126)ω = (ˆ e · ∂ t ˆ e , ˆ e · ∂ t ˆ e , ˆ e · ∂ t ˆ e ) and (cid:126)v = ∂ t (cid:126)r are the head’s rotational and translationalvelocities, with the subindex 0 indicating s = 0, and ξ trans and ξ rot the corresponding friction coefficients ofthe head. Asymmetric dynamics
For asymmetric beats we expand Eq. 13 around a staticshape given by a constant static curvature ˜Ω = ¯ a ˜ F (1)0 / ¯ κ .The dynamic mode then obeys iωξ ⊥ ˜Ω = − ¯ κ∂ s ˜Ω + ¯ a∂ s f (1)1 + (1 + ξ ⊥ /ξ (cid:107) ) ˜Ω ∂ s τ + ( ξ ⊥ /ξ (cid:107) ) ˜Ω (¯ κ∂ s ˜Ω − ¯ a∂ s f (1)1 ) , where τ is obtained from expanding the equation for thetension. Twist by tip-accumulated torques
If the source of twist is a tip accumulated torque,we have that m m = M δ ( s − L ). In the case in which f (0)m = 0, the solution to Eq. 8 is using the boundaryconditions ∂ s Π( s = 0) = 0 and a κ ∂ s Π( s = L ) = M is Π = ( M d/κ a ) sinh( s/d ) / cosh( L/ L (cid:28) d the twist created changes little along the length.Conversely, for L (cid:29) d the twist quickly decays away fromthe tip. Swimming trajectories
The force density exerted by the cilium in the fluid is (cid:126) f fl = (cid:0) ξ ⊥ (ˆ e ˆ e + ˆ e ˆ e ) + ξ (cid:107) ˆ e ˆ e (cid:1) · ∂ t (cid:126) r . Imposing that thesum of all forces and torques in the fluid must vanish wehave ξ trans (cid:126) v + (cid:82) L (cid:126) f fl d s = 0 and ξ rot (cid:126)ω + (cid:82) L (cid:126) r × (cid:126) f fl d s = 0,where the terms outside the integrals come from thehead’s drag [24, 33, 34]. Given a beating pattern we cancalculate (cid:126)f fl and use these equations to obtain the trans-lational and rotational velocities at each instant duringthe beat. Parameters used
The
Chlamydomonas cilium is L = 10 . µ m long andhas frequency ω/ π = 73 . L = 58 . µ m and ω/ π = 19 . = 0 . µ m − [7], for wild-type Chlamy-domonas we took ˜Ω = 0 . µ m − and for mbo2 ˜Ω = 0[23, 26]. The radius we used is a = 0 . µ m [2]. Toestimate M , note that a density of 500 µ m − dyneinswith a force of 0 . F (1)0 ≈
300 pN, compatiblewith ˜Ω for Chlamydomonas . If ∼
5% of this force re-sults in a torque over a distance of 0 . µ m, we obtain M ≈ .
25 pN µ m, used for Chlamydomonas . For bullsperm, with a smaller ˜Ω , we used M ≈ .
025 pN µ minstead. A doublet has 24 protofilaments compared to13 in a microtubule. The bending stiffness scales asarea squared, and so κ db ≈ (24 / κ mt ≈
80 pN µ m ,with κ mt ≈
23 pN µ m for microtubules [35]. For sim-plicity we take κ , db = κ , db = κ db , which results in κ ≈
115 pN µ m and ¯ κ = 9 κ db ≈
700 pN µ m , compara-ble to measurements of sea urchin sperm [36]. The slidingstiffness k s was determined in [37], and corresponds to d between 3 . µ m and 10 µ m, we chose d ≈ µ m. The re-sulting distal twist angle for bull sperm is 0 . .
25 rad obtained in [27]. The friction co-efficients are ξ (cid:107) ≈ πµ/ (ln(2 L/a ) − /
2) and ξ ⊥ ≈ ξ (cid:107) ,where µ is the viscosity [34, 38, 39]. For water at 22 Cwe have µ = 0 .
96 10 − pN s µ m − , which for L ≈ µ mresults in ξ (cid:107) ≈ . µ m − . For the head of bullsperm we use ξ trans = 6 πrα t µ and ξ rot = 8 πα r r µ [36],where r is the radius of the head and α t = 1 / (1 − / α r = 1 / (1 − /
8) are corrections due to the prox-imity to a wall [40]. In [7] it was observed that slidingcontrolled beating patterns require a large head friction.We take r ≈ µ m, large for bull sperm but adequatefor other species [41]. This results in ξ rot ≈
35 pN s µ mand ξ trans = 0 .
45 pN s µ m − . The values of the re-sponse coefficients for Chlamydomonas wild-type beatswere ζ = − i µ m and χ = − /µ m , for the mbo2 mutant ζ = − i µ m and χ = − /µ m ,and for bull sperm χ = ( − − i /µ m .We are thankful to M. Bock and E. M. Friedrich forinsightful discussions, and particularly to G. Klindt forsuggestions regarding the hydrodynamic aspects of thispaper. [1] Pazour GJ, Agrin N, Leszyk J, Witman GB (2005) Pro-teomic analysis of a eukaryotic cilium. Science
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