 |  |  CaM |
|  |  Molecule |
| | Reaction |
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Reaction List for pathway CaM (Pathway Number 373) | | Kd is calculated only for second order reactions, like nA+nB <->nC or nA<->nC+nD, where n is number and A,B,C,D are molecules, where as for first order reactions Keq is calculated. Kd for higher order reactions is not considered. |
| | Name | Kf | Kb | Kd | tau | Substrate | Product | | 1 |
CaM-bind-Ca | 8.4848
(uM^-1 s^-1) | 8.4853
(s^-1) | Kd(bf) = 1.0001(uM) | - | CaM
Ca
| CaM-Ca
| | | Lets use the fast rate consts here. Since the rates are so different, I am not sure whether the order is relevant. These correspond to the TR2C fragment. We use the Martin et al rates here, plus the Drabicowski binding consts. All are scaled by 3X to cell temp. kf = 2e-10 kb = 72 Stemmer & Klee: K1=.9, K2=1.1. Assume 1.0uM for both. kb/kf=3.6e11. If kb=72, kf = 2e-10 (Exactly the same !) 19 May 2006. Splitting the old CaM-TR2-bind-Ca reaction into two steps, each binding 1 Ca. This improves numerical stability and is conceptually better too. Overall rates are the same, so each kf and kb is the square root of the earlier ones. So kf = 1.125e-4, kb = 8.4853 | | 2 |
CaM-Ca-bind-Ca | 8.4848
(uM^-1 s^-1) | 8.4853
(s^-1) | Kd(bf) = 1.0001(uM) | - | CaM-Ca
Ca
| CaM-Ca2
| | | Lets use the fast rate consts here. Since the rates are so different, I am not sure whether the order is relevant. These correspond to the TR2C fragment. We use the Martin et al rates here, plus the Drabicowski binding consts. All are scaled by 3X to cell temp. kf = 2e-10 kb = 72 Stemmer & Klee: K1=.9, K2=1.1. Assume 1.0uM for both. kb/kf=3.6e11. If kb=72, kf = 2e-10 (Exactly the same !) 19 May 2006. Splitting the old CaM-TR2-bind-Ca reaction into two steps, each binding 1 Ca. This improves numerical stability and is conceptually better too. Overall rates are the same, so each kf and kb is the square root of the earlier ones. So kf = 1.125e-4, kb = 8.4853 | | 3 |
CaM-Ca2-bind-Ca | 3.6
(uM^-1 s^-1) | 10
(s^-1) | Kd(bf) = 2.7778(uM) | - | CaM-Ca2
Ca
| CaM-Ca3
| | | K3 = 21.5, K4 = 2.8. Assuming that the K4 step happens first, we get kb/kf = 2.8 uM = 1.68e6 so kf =6e-6 assuming kb = 10 | | 4 |
CaM-Ca3-bind-Ca | 0.465
(uM^-1 s^-1) | 10
(s^-1) | Kd(bf) = 21.5053(uM) | - | CaM-Ca3
Ca
| CaM-Ca4
| | | Use K3 = 21.5 uM here from Stemmer and Klee table 3. kb/kf =21.5 * 6e5 so kf = 7.75e-7, kb = 10 |
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