CROSS-BRIDGE PROPERTIES DERIVED FROMMUSCLE ISOTONIC VELOCITY TRANSIENTS

BY R. J. PODOLSKY, A. C. NOLAN, AND S. A. ZAVELER

NATIONAL INSTITUTES OF HEALTH, BETHESDA, MARYLAND

Communicated by Terrell L. Hill, August 11, 1969

Abstract.-The rate constants for the turnover of cross-bridges during frogmuscle contraction were determined from an analysis of the motion that followsstep decreases in load. For a given projection from the myosin filament, thereappears to be a range of about 100 X along the length of the filament over whichthe projection can attach to the actin filament and form a cross-bridge. Thesite of attachment is then displaced by a distance of this same order before thelink is broken. The values of the rate constants also imply that a cross-bridge isformed each time an actin site comes within range of a myosin projection, so thatthe turnover of cross-bridges for a given contraction distance is independent ofthe speed of the motion.

There is considerable evidence that the contractile force of striated musclefibers is developed by cross-bridges that form between the interdigitating myo-filaments when the cell is activated. 1 The cross-bridges, which appear to consistof projections from the myosin-containing filaments that have attached to siteson the actin-containing filaments, break and reform when the myofilamentsslide past each other and the cell shortens. This type of cyclic contractionmechanism was analyzed by A. F. Huxley2 in terms of (1) rate constants f and gwhich characterize the making and breaking of cross-bridges, respectively, and(2) the force function for a cross-bridge, k; it was shown that the steady force-velocity and force-energy relations for frog muscle could be closely fitted whenthese functions were properly chosen and energy output was associated withcross-bridge turnover.

Additional constraints are imposed on the model parameters if non-steadyas well as steady motions are accounted for. In the present study we havedetermined values of f, g, and k that produce isotonic velocity transientsvery much like those seen in living muscle fibers after step changes in load fromPo to P < Po. 4 These functions provide quantitative information about theforce-generating process in muscle at the molecular level and thereby place reallimits on specific models of the contraction mechanism.'

Calculation technique.-The influence of f, g, and k on the response of themodel to step changes in load was found by simulating the operation of the model,with various values of the parameters, on a digital computer.' Following thetreatment of Huxley,2 actin sites were taken to be randomly distributed withrespect to the projections on the myosin filament.7 The state of the system at agiven time is then characterized by a distribution function, n(x), defined as thefraction of the actin sites at x that are attached to cross-bridges; the origin ofthe coordinate system is the position of the actin site at which the cross-bridgeexerts zero force (Fig. la). The contractile force developed by the cross-bridges is

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P = fk(x)n(x)dx. (1)

The distribution function, and therefore the force associated with the cross-bridges, can change due to both the passage of time and the relative movementof the myofilaments. The change of n with time is

On- = (1-n)f-ng, (2)at

where f and g are functions of x. The change of n with movement is On!Ox.

Ac/rn f/ormenlto

<-Force 4o-Actrn 5/fe

~~~ X~~~05b ~~~~~~~~~~1.0

0.5

c 0 a h b

C In.

0ab

FIG. 1.-(a) Diagram of the coordinate system. x is distance between a site on the actinfilament and the position at which an attached cross-bridge exerts zero force. The Z line towhich the actin filament is connected is off the diagram to the right.

(b) Dependence of f and g on x. For f, a and b are the left- and right-hand limits, and his the midpoint. For g the origin, 0, is the right-hand limit, and c locates a discontinuityin value. The unit on the ordinate is the value of f at x = h.

(c) Displacement of n associated with sudden change in force from Po to P < Po. The dis-tribution function for P = Po is the rectangle (solid line) between x a and x = b. Whenthe load is reduced from Po to P, the actin filament moves an amount Azo relative to the myosinfilament, and the distribution of cross-bridges is displaced by the same amount; the dis-placed rectangle (shaded area) is the distribution function at the start of isotonic contraction.a is the additional displacement that brings the leading edge of n to the step increase in g at c.

In isometric contraction, there is no myofflament movement in the steadystate and the force developed by the cross-bridges reaches Po. In this case,an/at = 0 and, from equation 2,

npo = f/f + g. (3)In isotopic contraction, the change ill force due to the change ill n with time isjust balanced by the change in force due to the change in n with movement.The simulation starts when, because of a change ill load, the contraction

suddenly changes from isometric to isotonic and the contractile force changesfrom Po to P < Po. This corresponds to an initial translation of ntp, (eq. 3) by

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an amount Azo such thatco

P = fk(x)n(x + Azo)dx.

This displaced distribution function is then taken as the initial value of n(x) inan iterative procedure for calculating Az,, the displacement increment after theith time increment (At), and nf(x), the corresponding distribution function. Thesteps in the procedure are:

(I) Compute nf*(x) = ni-,(x) + {f(x) - nj_(x) [f(x) + g(x)]} At,co

(II) Find Azi such that P = fk(x)n,*(x + Azi)dx,(III) Let ni(x) = n*(x + Az,) and return to I.

Step I changes the distribution function according to equation (2). The forceassociated with the new (starred) distribution will generally be greater than P,so that muscle force exceeds the force of the load.8 Mechanical balance is re-stored by displacing the actin filament by an amount Az1, calculated in step IIby a search procedures This is equivalent to the translation of the distributionfunction by the same amount, as indicated in step III. The time course of themotion is then, z(t) = z(iAt) = NAzi.Two approximations were made to simplify the calculations. The first is to

make use of f and g functions that do not overlap.'0 Then np, = 0 where f = 0and nop = 1 where f 0 (Eq. 3). The second is to work with linear force func-tions, k = KX, where K is evaluated by normalizing the isometric force to unity,using equations (1) and (3). Under these conditions, the displacement of nfl,associated with the force step AP = Po - P is

Azo = hAP/Po (4)

where h = (a + b)/2, and a and b denote the left- and right-hand limits of the ffunction. For the f and g functions illustrated in Figure lb, the function np,and its displacement by Azo, are shown in Figure ic.

Isotonic contraction.-The upper of each pair of traces in Figure 2 shows theresponse of a frog muscle-fiber preparation to the force step given on the right.Except for the motion after the smallest force step, where shortening is practi-cally steady at the outset, the contraction velocity is first greater than, and thenless than, the steady value. The non-steady motion can be characterized by itsduration (the null time)" and its extent, 5, the contraction distance to the nulltime.The time course of the transient suggests that cross-bridges formed on the

right of the origin (Fig. la), where they exert positive force, move some distanceto the left of the origin, where they exert negative force and tend to retard themotion, before they have a very high probability of breaking. For this reason,the main part of the g function used in the present calculations was separatedfrom the f function by a "gap" (a-c in Fig. lb).The motion can be thought of as the continuous movement of n past the g

function (Fig. 1c). The leading edge of n is the residue of the displaced initial

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APFig. 2.-Conitractioni following step chaiiges iii load. PO

For each pair of displacement traces, the upper is the 0.06motion of a frog muscle fiber preparation for the inidi-cated force step, and the lower is the computed motionfor the same force step. The functions used in the calcu-lation are: Ado)

x<-60 f =Oandg = 260 \-60 < x< 0 f =O and g = 50 0.22

0 < x < 60 f = 5.5x + 145 and g = 060<x< 120 f = -5.5x+805andg= 0

where x is given in Angstrom units, and f and g are insec-. The experimental records are from Fig. 3 ofCivan and Podolsky,3 scaled to give the motion for ahalf sarcomere. Double arrows locate the first null \ 36times in the actual and the computed motion. Dotted 0.36lines above the transients in the motion join the start \of isotonic contraction to the first null time. 5 is thedistance between the start of isotonic contraction and 50\the displacement at the first null time. .4

I0msec

distribution. Calculations with various combinations of f and g show that thesteady contraction velocity is established very soon after the leading edge of nencounters the step rise in the g function at x - c. Since Azo, the initial displace-ment of the distribution function, increases in proportion to the force step (Eq. 4),the contraction distance to the first null time should decrease correspondingly ifa gap were present between f and the major part of g. The data in Figure 3,taken from an experiment published by Civan and Podolsky,3 show that this isthe case. The extrapolated value of 6 for AP/P0 = 0 gives the value of a-c, thedistance between the left-hand edge of f and the step rise in the g function; fromFigure 3 this is 60 A. When AP/Po = 1, Azo = h (Eq. 4), and the value of 6 atthis force step is the additional displacement, (a-c)-h, required to bring the leadingedge of n to the step rise in g. The data in Figure 3 show that this additionaldisplacement is close to zero, so that h = 60 A. The same analysis of data fromtwo other preparations (Civan and Podolsky, unpublished experiments) gavevalues of 45 A and 55 A for a-c, and 50 A and 60 A, respectively, for h.When f and g were systematically varied within these constraints, we found

that the transients in the model could be brought into reasonably good agree-ment with those in the actual motion by using the functions illustrated in Figurelb with a = 0, b = 120 !, and c = -60 X. The calculated contractions for arange of force steps are plotted in Figure 2 below the traces of the actual motion.It is clear that the null times and the steady speeds are about the same in the twomotions. For larger force steps, where the transients are less conspicuous, thesteady speed of the computed motion remains within 20% of the observedvelocity.When the width of the f function was increased above 120 A (which seems

unlikely since it implies that some of the cross-bridges exert negative force duringisometric contraction) the magnitude of the displacement deviation during thenon-steady motion was reduced. Decreasing the f function width below 80 A

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TVI\s Ftu. 3.-Relatwion l)etweellthe isotonic contraction dis-

40 - tance to the first null time(a) and the magnitude of the

30 force step (AP/Po). Thedata are taken from the ex-

20 periment shown in Fig. 3 of.Civan and Podolsky;3 verti-cal lines through points give

10 estimated limits of uncer-tainty. The line was fitted

0.2 0.4I 0.6 0.8 D~ by the method of leastAP/ Po0 squares.

produced nonsteady motions which clearly differed from the actual motion;in addition, the steady motion could not be made to satisfy the physiologicalforce-velocity relation. It appears, therefore, that the width of the f function isbracketed between 80 A and 120 A.

Force-energy relation.-Figure 4 shows the distribution functions at the begin-ning of isotonic contraction (shaded area) and in the steady state (solid line) forthe f and g functions used in Figure 2. For both large and small forces (i.e., slowand fast steady contraction speeds) the value of n at x = 0 is very close to 1.Since n(0) is the probability that an actin site will form a cross-bridge duringa transit of a myosin projection, this implies that cross-bridge turnover for agiven amount of shortening is independent of contraction speed. If cross-bridge turnover requires the hydrolysis of one ATP molecule, which is generallyassumed to be the case, the energy consumption per unit shortening due to turn-over would be independent of contraction speed.

Compliance.-Cross-bridges exerting force are characterized by a compliance,

y = - z/&P. Since P = fkndx and Oz/OP = -AxP,

y= I9fc dx I

For k KX,co

= [Kfndx]-l- Co

so that the compliance is inversely proportional to the area under the distribu-tion function.The quick displacement required to reduce the force of the cross-bridges from

0

Po to 0 is fydP. This length is equal to h (Eq. 4), which is close to 60 A/halfPo

sarcomere (Fig. 3), or about 0.5% of the half-sarcomere length. Since the quickdisplacement required to reduce the force of the living muscle fiber from Po to 0is about 1% of the fiber length, 12 half of the total appears to be due to strainof the cross-bridges.

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FIG. 4.-Change in n during isotonic contraction for three dif-ferent force steps. The value of n at the start of isotonic con-traction is shown by the shaded rectangle, and the steady statevalue is given by the solid line. Note that the area enclosedby the steady-state n is greater than that enclosed by the initialn. The rate constants are those given in the legend of Fig. 2.

Figure 5 shows the time course of the cross-bridge compliance of the modelduring isotonic contraction. The compliance decreases, nearly monotonically,with time. The magnitude of the compliance change is larger for the larger forcesteps; the time required for a steady value to be reached is close to the nulltime in the corresponding displacement trace.

This effect could be examined experimentally by comparing the complianceof a fiber that had been contracting isotonically for a given length of time withthat of the same fiber contracting isometrically. However, the magnitude of theeffect is relatively small: for a force step of 0.47, for example, the expected de-crease in cross-bridge compliance is about 20%, which amounts to a change of10% in the total fiber compliance. Nevertheless, data which gave only the signof the compliance change would be a useful check on the present analysis, since

F-G. 5.-Calculated changeof cross-bridge compliance _0.9during isotonic contraction ffor four different force steps. TThe ordinate is the ratio of athe cross-bridge compliance of 0e47after isotonic contraction to c0.8the initial compliance; thetabscissa is the contraction a 0.87time. The rate constantsused in the calculation are 0.7given in the legend of Fig. 2.

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the decrease in cross-bridge compliance of the model reflects the fact that thearea under the distribution function increases wlhien the steady state is established(Fig. 4). This increase in area is due to the gap betweeII f anid the step in g. blimodels where the gap is absent, the cross-bridge compliance generally becomesgreater, rather than smaller, in the steady state. This, for example, is the casewith the rate functions used by Huxley (Fig. 6 in ref. 2), since for all loads thearea under the distribution function decreases, relative to the isometric value,when the steady state is established (Fig. 7 in ref. 2).Discussion.-The center of the f function we have used is offset from the posi-

tion corresponding to zero force, which is the origin of the local coordinate sys-tem.'3 Equation (1) shows that this offset is necessary to produce isometricforce, since nflO = 1 wherever f differs from 0. The interval between the zero offorce and the center of the f function, derived from the influence of force step onthe contraction distance to the null time (Fig. 3), appears to be about 60 X.Another characteristic of the rate constants is the "gap" of 60 A between f

and the major part of g. This gap is responsible for the characteristic shape ofthe transients, for it results in cross-bridges which were formed on the right of theorigin being carried to the left of the origin where they exert negative force andslow down the motion. The influence of the g of small magnitude in the gap isfelt mainly when the motion is relatively slow (e.g., AP/Po = 0.06), in which caseit removes curvature otherwise present in the displacement trace.We have assumed that f is symmetrical with respect to h. This was largely

done for convenience since the influence of f to the left of its midpoint has prac-tically no effect on the motion. The reason for this is that the value of n is almost1 by the time an actin site reaches the midpoint of f, even at the faster contrac-tion velocities (Fig. 4). It appears, therefore, that the amplitude of f between aand h can be any value that drives the value of n in this region to 1 during iso-metric contraction, and that the contraction kinetics are affected almost entirelyby the amplitude of f between h and b.Although the calculated motion simulates the major features of the actual

motion, discrepancies still remain. These are: (1) the magnitude of the dis-placement deviation is about half of that seen in the actual motion, and (2)second null times3 for the larger force steps are not seen in the model motion.Nevertheless, the general agreement between the model and the actual motionstrongly suggests that the f and g functions used in the simulation representprocesses that actually occur within the cell.The values of these functions imply that the turnover of cross-bridges for a

given contraction distance is independent of the speed of the motion. This doesnot agree with the usual interpretation of the heat and work production in con-tractions with different loads, which is that the chemical change associated withshortening is a function of contraction velocity.14 It should be pointed out,however, that the assumptions' that must be made when energy output ofmuscle is directly related to chemical change have not been verified, and theymay not be correct. Additional experimentation is required to resolve thisquestion.The picture that emerges from our analysis is that a cross-bridge is made each

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time an actin site comes within range of a myosin projection (in contrast to theview that the probability of interaction is a function of the transit velocity) andthat the average reach of a cross-bridge between making and breaking is of theorder of 100 A. These properties of the force-generating process are both con-sequences of the relatively large gap between the left-hand limit of the f functionand the major part of the g function.

IThe evidence has recently been reviewed by Huxley, H. E., Science, 164, 1356 (1969).Also see Gordon, A. M., A. F. Huxley, and F. J. Julian, J. Physiol., 184, 143 (1966).

2 Huxley, A. F., Prog. Biophys., 7, 255 (1957).3Podoleky, R. J., Nature, 188, 666 (1960); Civan, M. M., and R. J. Podolsky, J. Physiol.,

184, 511 (1966).4Armstrong, C. F., A. F. Huxley, and F. J. Julian, J. Physiol., 186, 26P (1966).See, for example, Huxley, H. E., Proc. Roy. Soc. B, 160, 442 (1964); Hill, T. L., these

PROCEEDINGS, 64, 267 (1969).6Preliminary results with this technique were reported briefly by Podolsky, R. J., N. Z.

Shapiro, and S. A. Zaveler, Fed. Proc., 26, 553 (1967).7 Low angle X-ray diffraction studies of living muscle by Huxley, H. E., and J. Brown,

J. Mol. Biol., 30, 383 (1967) provide evidence for this assumption, since the periodicities on themyosin and actin filaments are different.

8 This was checked during each iteration because certain choices of f and g make it possiblefor the contractile force to fall below the force of the load in the course of the motion.

9 A search procedure was used to keep open the option of using arbitrary force functions.For a linear force function, the value of Azi is given directly by the expression

co co

[Kfxfni*(X)dX - P]1/ fni*(x)dx.co -c

10 The kinetic response of the model to step decreases in load would not be substantially af-fected if f and g did overlap, provided that g << f in the region of overlap. This seems likelysince otherwise the maintenance of tetanic tension would be very inefficient.

11 This refers to the first of the null times described by Civan and Podolsky. The displace-ment deviation between the first and second null time is relatively small and, in the presentdiscussion, will be taken as zero.

12Jewell, B. R., and D. R. Wilkie, J. Physiol., 143, 515 (1958).13 It is worth pointing out that the orientation of the myosin projection is not specified at the

zero of force and, therefore, the model gives no information about where the root of the cross-bridge is relative to the tip when positive force is being exerted.

14 Hill, A. V., Proc. Roy. Soc. B, 126, 136 (1938); ibid., 159, 297 (1964).15 These are that (1) the isometric heat rate is the base line for shortening heat, (2) shorten-

ing heat is due only to the net chemical reaction driving cross-bridge turnover, and (3) the driv-ing chemical reaction is the same for all contraction speeds. See reference 14, and also Podolsky,R. J., "Thermodynamics of Muscle," in The Structure and Function of Muscle, ed. G. B.Bourne, (New York: Academic Press, 1960), vol. 2, p. 359.

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