Table 1.
Equations expressing the rate constants of the forward transitions in the reaction scheme according to the direction of the reaction flow during steady shortening
| k1,m, k1′,m (m−1 s−1) | = 4 × 105 | |
| k2 (s−1) | = 25 | |
| k3 (s−1) | = 30 exp(−0.1(x + x0 − 1.5)6) | |
| k4,M1 (s−1), k4′,M1 (s−1) | = 60 | |
| k4,M2 (s−1), k4′,M2 (s−1) | = 100 | |
| k4,M3 (s−1), k4′,M3 (s−1) | = 1000 | |
| k4,M4 (s−1), k4′,M4 (s−1) | = 5000 | |
| k5,M1 (s−1), k5′,M1 (s−1) | = 0.5 | |
| k5,M2 (s−1), k5′,M2 (s−1) | = 10 | |
| k5,M3 (s−1), k5′,M3 (s−1) | = 15 | |
| k5,M4 (s−1) | = 150 + 1450exp((−3z(x + x0 + 3z + 4))/kBθ)/(1 + exp(−3z(x + x0 + 3z + 4)/kBθ)) | |
| k5′,M4 (s−1) | = 150 + 1450exp((−3z(x + x0 + 3z +9.5))/kBθ)/(1 + exp(−3z(x + x0 + 3z +9.5)/kBθ)) | |
| k6,M1 (s−1) | = 40 + 1460exp(−zε(x + x0 + 1.15)/kBθ)/(1 + exp(−zε(x + x0 + 1.15)/kBθ) | |
| k6,M2 (s−1) | = 1500exp(−zε(x + x0 −1.5 + z)/kBθ)/(1 + exp(−zε(x + x0 −1.5 + z)/kBθ) | |
| k6,M3 (s−1) | = 1500exp(−zε(x + x0 − 1 + 2z)/kBθ)/(1 + exp(−zε(x + x0 − 1 + 2z)/kBθ) | |
| k6,M4 (s−1) | = 1500exp(−zε(x + x0 + 6 + 3z)/kBθ)/(1 + exp(−zε(x + x0 + 6 + 3z)/kBθ) | |
| k6′,M1 (s−1) | = 40 + 1460exp(−zε(x + x0 +6.65)/kBθ)/(1 + exp(−zε(x + x0 + 6.65)/kBθ) | |
| k6′,M2 (s−1) | = 1500exp(−zε(x + x0 + 4 + z)/kBθ)/(1 + exp(−zε(x + x0 + 4 + z)/kBθ) | |
| k6′,M3 (s−1) | = 1500exp(−zε(x + x0 + 4.5 + 2z)/kBθ)/(1 + exp(−zε(x + x0 + 4.5 + 2z)/kBθ) | |
| k6′,M4 (s−1) | = 1500exp(−zε(x + x0 + 11.5 + 3z)/kBθ)/(1 + exp(−zε(x + x0 + 11.5 + 3z)/kBθ) | |
| k7 (s−1), k7′ (s−1) | = 2000 | |
| k8,M2 (s−1) | = 0.1exp(−zε(x + x0 + 0.5 + z)/kBθ)/(1 + exp(−zε(x + x0 + 0.5 + z)/kBθ) | x < −1 |
| = 0 | x ≥ −1 | |
| k8,M3 (s−1) | = 60exp(−zε(x + x0 − 1 + 2z)/kBθ)/(1 + exp(−zε(x + x0 − 1 + 2z)/kBθ) | x < −1 |
| = 0 | x ≥ −1 | |
| k8,M4 (s−1) | = 0.1exp(−zε(x + x0 − 1.3 + 3z)/kBθ)/(1 + exp(−zε(x + x0 − 1.3 + 3z)/kBθ) | x < −1 |
| = 0 | x ≥ −1 | |
| k9,M2 (s−1) | = 1000exp(−zε(x + x0 + 0.5 + z)/kBθ)/(1 + exp(−zε(x + x0 + 0.5 + z)/kBθ) | x < −1 |
| = 0 | x ≥ −1 | |
| k9,M3 (s−1) | = 1200exp(−zε(x + x0 − 1 + 2z)/kBθ)/(1 + exp(−zε(x + x0 − 1 + 2z)/kBθ) | x < −1 |
| = 0 | x ≥ −1 | |
| k9,M4 (s−1) | = 10exp(−zε(x + x0 − 1.3 + 3z)/kBθ)/(1 + exp(−zε(x + x0 − 1.3 + 3z)/kBθ) | x < −1 |
| = 0 | x ≥ −1 | |
| kw1 (s−1) | = 73500exp(−zε·(x + x0 + 0.5)/kBθ)/(1 + exp(−zε(x + x0 + 0.5)/kBθ) | |
| kw2 (s−1) | = 49000exp(−zε(x + x0 + 0.5 + z)/kBθ)/(1 + exp(−zε(x + x0 + 0.5 + z)/kBθ) | |
| kw3 (s−1) | = 24500exp(−zε(x + x0 + 0.5 + 2z)/kBθ)/(1 + exp(−zε(x + x0 + 0.5 + 2z)/kBθ) | |
| kw1′ (s−1) | = 73500exp(−zε(x + x0 + 6)/kBθ)/(1 + exp(−zε(x + x0 + 6)/kBθ) | |
| kw2′ (s−1) | = 49000exp(−zε(x + x0 + 6 + z)/kBθ)/(1 + exp(−zε(x + x0 + 6 + z)/kBθ) | |
| kw3′ (s−1) | = 24500exp(−zε(x + x0 + 6 + 2z)/kBθ)/(1 + exp(−zε(x + x0 + 6 + 2z)/kBθ) |
The rate constants of the transitions between different biochemical states or between different actin monomers are indicated as klm where l (from 1 to 9) identifies the transition according to the scheme of Fig. 1 and m (M1→M4) indicates the structural state involved in the transition. The rate constants of the structural transitions for the same biochemical state are indicated as kw1, kw2 and kw3 for the transitions M1→M2, M2→M3 and M3→M4, respectively. These ‘structural’ rate constants have the same x-dependence for the different biochemical states but their values are multiplied by 0.01 for AM and A′M states. x0 is the strain of the M1 state at x = 0, which has been set to 1.15 nm. The prime symbol (′) is added to the subscript if the transition occurs on the second actin.