| |
Reaction
Name | Pathway Name /
Pathway No. | Kf | Kb | Kd | tau | Reagents |
| 1 | CaM-Ca3-bind-Ca | CaM
Pathway No. 814 | 0.465
(uM^-1 s^-1) | 10
(s^-1) | Kd(bf) = 21.5051(uM) | - | Substrate:
CaM-Ca3
Ca
Products:
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 | | 2 | CaM-bind-Ca | CaM
Pathway No. 814 | 8.4851
(uM^-1 s^-1) | 8.4853
(s^-1) | Kd(bf) = 1(uM) | - | Substrate:
CaM
Ca
Products:
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 | | 3 | CaM-Ca2-bind-Ca | CaM
Pathway No. 814 | 3.6
(uM^-1 s^-1) | 10
(s^-1) | Kd(bf) = 2.7778(uM) | - | Substrate:
CaM-Ca2
Ca
Products:
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-Ca-bind-Ca | CaM
Pathway No. 814 | 8.4851
(uM^-1 s^-1) | 8.4853
(s^-1) | Kd(bf) = 1(uM) | - | Substrate:
CaM-Ca
Ca
Products:
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 |
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