Isotonic Responses

Thus far we have shown how the Huxley model can explain the Hill force-velocity curve using crossbridge dynamics. However, for the model to give an acceptable explanation of muscle dynamics, there is a great deal of additional experimental data with which it should agree. In particular, the model should explain the response of a muscle, first, to a step change in tension (isotonic response) and second, to a step change in length (isometric response). After all, the Hill model was rejected as a satisfactory explanation because of its inability to explain all such data.

It is instructive to consider how one calculates the response of the Huxley model to a step change in tension, as the procedure is not obvious. Suppose a muscle exerts its isometric tension, p0, at some length L. Then, the steady-state crossbridge distribution is f (x)

Now suppose the tension on the muscle is reduced to pi < p0 so suddenly that no cross-bridges are able to associate or dissociate during the reduction. In a typical experiment of Civan and Podolsky (1966), a muscle fiber of length 15,000 pm was subjected to a change in tension that changed the fiber length by less than 50 pm, a relative length change of 1/300. Hence, a typical sarcomere of length 2.5 pm changed in length by less than 10 nm, and so the length of each crossbridge was changed by less than 10 nm. A crossbridge is able to absorb such length changes without dissociating from the binding site. The extension of each crossbridge decreases by an unknown amount AL, and so the crossbridge distribution suddenly changes to ns (x + AL) and is no longer at steady state. The change in length is found by constraining the new tension to be p1, and hence AL satisfies

Although (18.28) cannot in general be solved analytically, AL can be determined numerically, since it is easy to determine p1 as a function of AL.

Following the sudden change in tension, the crossbridge population is not at steady state, so it must change according to the differential equation (18.18) with initial condition n(x, 0) = ns(x + AL) and subject to the constraint that the tension is constant at p = p 1. However, during this evolution, v is not constant, as there is some transient behavior before the muscle reaches its steady contraction velocity (cf. Fig. 18.9). However, we can determine an expression for v(t) in terms of n(x, t) that guarantees that the tension remains constant at p1.

Since pi is constant, it must be that d- = 0, or rœ rœ / du \

0 = k(x)ut(x,t)dx = k(x) v(t)--+ (1 — u)f (x) — ug(x) dx. (18.29)

We solve this for v(t) to get f—œ k(x) ((1 — u)f (x) — ug(x)) dx

Thus, for the tension to remain constant, the partial differential equation (18.18) must have the contraction velocity specified by (18.30).

Using a slightly different approach, Podolsky et al. (1969; Civan and Podolsky, 1966) showed that the Huxley model does not agree with experimental data in its response to a step change in tension. We saw in Fig. 18.9 that immediately after the tension reduction the muscle length changes also, and after an initial oscillatory period, the muscle contracts with a constant velocity. However, the Huxley model does not show any oscillatory behavior, the approach to constant velocity being monotonic.

Motivated by this discrepancy, Podolsky and Nolan (1972, 1973) and Podolsky et al. (1969) altered the form of the functions f and g to obtain the required oscillatory responses in the Huxley model. Of course, in Huxley's original model no physiological justification was given for the functions f and g, and modification of these functions is therefore an obvious place to start when fiddling with the model to fit the data. Julian (1969) also showed that the Huxley model can be adjusted to give the correct responses to a step change in length. The details of these analyses do not concern us greatly; the main point is that Huxley's crossbridge model has enough flexibility to explain a wide array of experimental data.

Was this article helpful?

0 0

Post a comment