Molecular Events of the Crossbridge Cycle Reflected in the Force–Velocity Relationship of Activated Muscle

Figure 1. (A) Illustration of a step release in force and the ensuing length transient. The 4 phases of the length response and their underlying causes are described in the text. The isotonic load after the release is kept constant during the transient and the phase 4 contraction (Modified from Seow, 2013). (B) Records from an experiment showing length response to a force step. A skinned single fiber of rabbit psoas muscle was used in the experiment. Both the fiber length and the laser-diffracted sarcomere length signals were recorded. The initial slope (dotted line) of the phase 4 response in the sarcomere length record is taken as the shortening velocity corresponding to the isotonic load.

Figure 2. An example of force–velocity data fitted by the Hill equation. The maximal shortening velocity (Vmax) and maximal isometric force (Fmax) are extrapolated from the fitted curve. Because the muscle’s power output is the product of force and velocity, a force–power curve (dotted curve) can be derived from the force–velocity curve.

Figure 3. Force–velocity (A) and force–power (B) curves with the value of a/Fmax and b/Vmax increasing from 0.15, 0.2, 0.25, to 0.3, resulting in a decrease in the curvature of the force–velocity curves (i.e., less curved) and an increase in Pmax of the force–power curves. The force at which Pmax occurs (open symbols) also increases as the curvature decreases.

Figure 4. (A) Schematic representation of a sarcomere. (B) Three different arrangements of 4 identical sarcomeres in parallel and in series. (C)Force–velocity curves for a single sarcomere (a) and each of the sarcomere arrangements (bi, bii, biii). (D) Force–power curves for a single sarcomere (a) and each of the sarcomere arrangements (bi, bii, biii).

Figure 5. Illustration of force–velocity relationship in the presence of an internal load of 0.05 Fmax magnitude. The Hill constants a and b are asymptotes of the hyperbolic curve. An internal load effectively increases constant a so that the new asymptote a’ equals a + Fi. The internal load also reduces maximal isometric force so that F’max equals Fmax – Fi. The apparent force–velocity curvature (F’max/a’) decreases in the presence of an internal load, compared with the curvature in the absence of the internal load (Fmax/a), because F’max/a’ = (Fmax – Fi)/(a + Fi). The maximal velocity seen in the absence of both internal and external loads (Vmax) is not affected by the internal load. However, the internal load causes an apparent decrease in the maximal velocity (from Vmax to V’max). Inset: Illustration of how partial activation in the presence of an internal load causes an apparent decrease in the maximal velocity from V’max at full activation to V″max at half-activation. The changes in Vmax, Fmax, and Fmax/a due to an internal load are called “apparent” because the changes are artifacts caused by the internal load and not by changes in the kinetics of actomyosin interaction.

Figure 6. A two-state model of cyclic interaction of myosin crossbridges with actin filaments. A, attached state, D, detached state. fAPP and gAPP denote the apparent forward and reverse rate constants, respectively.