Abstract

It has been known that when the length of active muscle fiber is released suddenly by ∼15% and re-stretched, tension falls to 0, then it redevelops exponentially with the rate constant ktr. The rise of tension has been interpreted to represent the force generation step (partial cross-bridge cycle). Here we propose a model, in which cross-bridges cycle many times in this period by stretching series elastic elements. One assumption needed is either the step size (η) or the stepping rate (v) decreases linearly with tension(F): η=η0{1-F(t)/F0} (1), or ν = ν0{1-F(t)/F0} (2), where F0 is isometric tension; η0 is step size and ν0 is the number of steps/sec without load. Eqs. 1 and 2 are mathematical representations of the Fenn effect. Distance travelled by cross-bridges in time dt is ηvdt, which stretches series elastic with stiffness σ. Thus, the increase in tension is: dF=σvηdt. By using Eq. 1 or 2, we get: dF/dt+kF=kF0 (3), where k=ση0ν0/F0 (4). By solving Eq. 3 with F(0)=0, we get F(t)=F0{1-exp(-kt)} (5). Eq. 5 is a good approximation of tension time course which deduced kTR. From Eq. 4, it can be concluded that kTR is proportionately related to the turn over rate (ν0). Because ATPase=M0ν0 (M0 is myosin concentration), k=ATPase•ση0/F0M0 (6). Our available data on rabbit psoas fibers on kTR and ATPase at 10°C–25°C demonstrate that these are proportionately related, as predicted by Eq. 6. Our data further demonstrate that kTR and 2πa (slowest rate constant of tension transients) are linearly related by 2πa=0.71kTR+0.89s−1, demonstrating that 2πa is also limited by the turn over rate. In the two state model with attachment rate (f) and detachment rate (g), the turnover rate is v0=fg/(f+g), hence kTR=(ση0/F0)(fg)/(f+g) (limited by a slow reaction). Therefore, kTR≠ f+g (limited by a fast reaction).

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