Force transients and minimum cross-bridge models in muscular contraction

Abstract

Two- and three-state cross-bridge models are considered and examined with respect to their ability to predict three distinct phases of the force transients that occur in response to step change in muscle fiber length. Particular attention is paid to satisfying the Le Châtelier–Brown Principle. This analysis shows that the two-state model can account for phases 1 and 2 of a force transient, but is barely adequate to account for phase 3 (delayed force) unless a stretch results in a sudden increase in the number of cross-bridges in the detached state. The three-state model $({\mathbf{A \rightarrow B\rightarrow C\rightarrow A}})$ makes it possible to account for all three phases if we assume that the ${\mathbf{A\rightarrow B}}$ transition is fast (corresponding to phase 2), the ${\mathbf{B \rightarrow C}}$ transition is of intermediate speed (corresponding to phase 3), and the ${\mathbf{C \rightarrow A}}$ transition is slow; in such a scenario, states A and C can support or generate force (high force states) but state B cannot (detached, or low-force state). This model involves at least one ratchet mechanism. In this model, force can be generated by either of two transitions: ${\mathbf{B\rightarrow A}}$ or ${\mathbf{B\rightarrow C}}$. To determine which of these is the major force-generating step that consumes ATP and transduces energy, we examine the effects of ATP, ADP, and phosphate (Pi) on force transients. In doing so, we demonstrate that the fast transition (phase 2) is associated with the nucleotide-binding step, and that the intermediate-speed transition (phase 3) is associated with the Pi-release step. To account for all the effects of ligands, it is necessary to expand the three-state model into a six-state model that includes three ligand-bound states. The slowest phase of a force transient (phase 4) cannot be explained by any of the models described unless an additional mechanism is introduced. Here we suggest a role of series compliance to account for this phase, and propose a model that correlates the slowest step of the cross-bridge cycle (transition ${\mathbf{C\rightarrow A}}$) to: phase 4 of step analysis, the rate constant k _tr of the quick-release and restretch experiment, and the rate constant k _act for force development time course following Ca²⁺ activation.

Appendices

Appendix 1

The solution of the differential equation (A1) can be found in the following way.

$$ dY(t)/dt+\lambda Y(t)=h $$

(A1)

Now we substitute Y(t) with Z(t) so that

$$ Y(t)\equiv h/\lambda+Z(t)\hbox{ exp}(-\lambda t) $$

(A2)

is satisfied. From Eq. A2,

$$ dY(t)/dt=[-\lambda Z(t)+dZ(t)/dt]\hbox{ exp}(-\lambda t) $$

(A3)

By substituting Eqs. A2 and A3 to Eq. A1, we arrive at

$$ dZ(t)/dt=0 $$

(A4)

By integrating Eq. A4, we get

$$ Z(t)=Y_{0} $$

(A5)

where Y ₀ is the integration constant. From Eqs. A2 and A5, we get

$$ Y(t)=Y_{0}\hbox{ exp}(-\lambda t)+h/\lambda $$

(A6)

Therefore, Eq. A1 has an exponential process with the rate constant λ, amplitude Y ₀, and the steady state value h/λ (Eq. A6). Y ₀ is determined by the initial conditions.

Appendix 2

In Scheme 3, the following differential equations can be set up.

$$ dA/dt=-(\alpha+\gamma^{\prime})A+\alpha^{\prime}B+\gamma \hbox{C} $$

(A7)

$$ dB/dt=\alpha A-(\beta+\alpha^{\prime})B+\beta^{\prime}\hbox{C} $$

(A8)

$$ dC/dt=\gamma^{\prime}A+\beta B-(\gamma+\beta^{\prime})C $$

(A9)

Eqs. A7–A9 can be written in matrix form.

$$ \frac{dX}{dt}=-HX $$

(A10)

$$ \hbox{where }\quad X(t)\equiv \left( {{\begin{array}{l} {A(t)} \\ {B(t)} \\ {C(t)} \\ \end{array} }} \right) $$

(A11)

$$ \hbox{And }\quad H\equiv \left( {{\begin{array}{lll} {(\alpha +\gamma^{\prime})}& {-\alpha^{\prime}}& {-\gamma \;} \\ {-\alpha }& {(\beta +\alpha^{\prime})}& {-\beta^{\prime}} \\ {-\gamma^{\prime}}& {-\beta }& {(\gamma +\beta^{\prime})} \\ \end{array} }} \right) $$

(A12)

$$ \hbox{Also},\quad A + B + C = A_{\rm T} $$

(A13)

Note the similarity of Eqs. A1 and A10. Note also that |H| = det(H) = 0, which is consistent with Eq. A13. The eigen values (λ) of matrix H can be found by determining the roots of Eq. A14.

$$ \left| {H-I\lambda } \right|=\left| {{\begin{array}{lll} {(\alpha +\gamma^{\prime})-\lambda }& {-\alpha^{\prime}}& {-\gamma \;} \\ {-\alpha }& {(\beta +\alpha^{\prime})-\lambda }& {-\beta^{\prime}} \\ {-\gamma^{\prime}}& {-\beta }& {(\gamma +\beta^{\prime})-\lambda } \\ \end{array}}} \right|=0 $$

(A14)

where I is the identity matrix. From Eq. A14,

$$ \lambda^{3}-2Q \lambda^{2}+M\lambda=0 $$

(A15)

$$ \hbox{where }\quad Q \equiv(\alpha+\alpha^{\prime}+\beta+\beta^{\prime}+\gamma+\gamma^{\prime})/2 $$

$$ \hbox{and }\quad M\equiv \, \alpha\beta+\beta\gamma +\gamma \alpha+\alpha\beta^{\prime}+\beta\gamma^{\prime}+\gamma\alpha^{\prime}+ \alpha^{\prime}\beta^{\prime}+ \beta^{\prime}\gamma^{\prime}+\gamma^{\prime}\alpha^{\prime} $$

(A16)

Equation A15 has three roots λ₁, λ₂, and λ₃.

$$ \lambda _1 =0 $$

$$ \lambda _2 =Q+\sqrt{R} $$

(A17)

$$ \lambda _3 =Q-\sqrt{R}=\frac{M}{Q+\sqrt{R}} $$

(A18)

$$ \hbox{where }\quad R\equiv Q^{2}-M $$

With Eqs. A17 and A18, Eq. A10 can be solved to result in:

$$ X(t)=U \, E(t) $$

(A19)

$$ \hbox{where }\quad U\equiv \left( {{\begin{array}{lll} {A_1 }& {U_{12} }& {U_{13} } \\ {B_1 }& {U_{22} }& {U_{23} } \\ {C_1 }& {U_{32} }& {U_{33} } \\ \end{array}}}\right), \quad \hbox{ and }\quad E(t)\equiv \left( {{\begin{array}{l} 1 \\ {\exp (-\lambda _2 t)} \\ {\exp (-\lambda _3 t)} \\ \end{array} }} \right) $$

(A20)

The elements U _i2 and U _i3 (i = 1, 2, 3) are 2 eigen vectors (column vectors) of matrix H (Eq. A12) corresponding to λ₂ and λ₃, respectively, and they are time-independent variables. Their size (length) is determined by the initial conditions (2 independent variables). A ₁, B ₁ and C ₁ are the steady-state concentrations (see below). The correctness of Eq. A19 can be examined by substituting it into Eq. A10 keeping in mind that:

$$ U^{-1}HU=\left( {{\begin{array}{lll} {\lambda _1 }& 0& 0 \\ 0& {\lambda _2 }& 0 \\ 0& 0& {\lambda _3 } \\ \end{array} }} \right) $$

From Eqs. A19 and A20, the individual solutions are:

$$ A(t)=A_{1}+U_{12 }\hbox{exp}(-\lambda_{2}t)+U_{13 }\hbox{exp}(-\lambda_{3}t) $$

(A21)

$$ B(t)=B_{1 }+U_{22 }\hbox{exp}(-\lambda_{2}t)+U_{23}\hbox{exp}(-\lambda_{3}t) $$

(A22)

$$ C(t)=C_{1 }+U_{32 }\exp({-}\lambda_{2}t)+U_{33 }\exp({-}\lambda_{3}t) $$

(A23)

Therefore, according to Eq. 34, a force transient has two exponential processes that correspond to phases 2 and 3, with the rate constants λ₂ and λ₃, respectively.

If R < 0, then λ₂, λ₃, U _i2 and U _i3 are complex numbers, and λ₃ = λ₂* (Eqs. A17 and A18), and U _i3 = U _i2*, where * indicates the complex conjugate. In this case, Eq. A19 can be rearranged to result:

$$ X(t)=\left( {{\begin{array}{lll} {A_1 }& {2\Re (U_{12} )}& {2\Im (U_{12} )} \\ {B_1 }& {2\Re (U_{22} )}& {2\Im (U_{22} )} \\ {C_1 }& {2\Re (U_{32} )}& {2\Im (U_{32} )} \\ \end{array} }} \right)\left( {{\begin{array}{l} 1 \\ {\exp (-Qt)\cos \sqrt{-R}\cdot t} \\ {\exp (-Qt)\sin \sqrt{-R}\cdot t} \\ \end{array} }} \right) $$

(A24)

where ${\Re}(U_{ij})$refers to the real, and ${\Im}(U_{ij})$ to the imaginary part of the complex number U _ij . Equation A24 shows that this system has a damped oscillation. However, in literature dealing with muscle fibers, it is rare to find force transients with a damped oscillation in response to a sudden change in an experimental condition. If a force transient should oscillate, a resonance of the force transducer should be suspected before a conclusion is drawn. The fact that an oscillation is absent implies that the intrinsic rate constants among the three steps in Scheme 3 differ significantly.

Steady state

By setting dA/dt = 0 and dB/dt = 0 in Eqs. A7 and A8, and with the constraints of Eq. A13, we can solve the steady-state concentrations of A ₁, B ₁ and C ₁. These three values constitute an eigen vector (column vector) which belongs to the eigen value λ₁. M is defined in Eq. A16.

$$ A_{1}=A_{T}(\beta\gamma +\gamma\alpha^{\prime}+\alpha^{\prime}\beta^{\prime})/M $$

(A25)

$$ B_{1}=A_{T}(\gamma\alpha+\alpha\beta^{\prime}+\beta^{\prime}\gamma^{\prime}) /M $$

(A26)

$$ C_{1}=A_{\rm T}(\alpha\beta+\beta\gamma^{\prime}+\gamma^{\prime}\alpha^{\prime})/M $$

(A27)

The turnover (ATP hydrolysis) rate is:

$$ J=\alpha {A_{1}}-\alpha^{\prime}{B_{1}}= \beta {B_{1}}-\beta^{\prime}{C_{1}}=\gamma {C_{1}}-\gamma^{\prime}{A_{1}}=A_{T}(\alpha\beta\gamma-\alpha^{\prime}\beta^{\prime}\gamma^{\prime})/M $$

(A28)

A similar analysis based on Scheme 8 is found in the Appendix of Kawai (2003).