| Name | Accession Name | Pathway Name | Kf | Kb | Kd | tau | Reagents |
1 | CaM_dash_Ca3_ dash_bind_dash_ Ca | mRNA synthesis Accession No. : 94 | kinetics Pathway No. : 1112 | 0.465 (uM^-1 s^-1) | 10 (s^-1) | Kd(bf) = 21.5051(uM) | - | Substrate Ca CaM_dash_Ca3
Product CaM_dash_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_dash_bind_ dash_Ca | mRNA synthesis Accession No. : 94 | kinetics Pathway No. : 1112 | 8.4846 (uM^-1 s^-1) | 8.4853 (s^-1) | Kd(bf) = 1.0001(uM) | - | Substrate Ca CaM
Product CaM_dash_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_dash_Ca2_ dash_bind_dash_ Ca | mRNA synthesis Accession No. : 94 | kinetics Pathway No. : 1112 | 3.6001 (uM^-1 s^-1) | 10 (s^-1) | Kd(bf) = 2.7777(uM) | - | Substrate Ca CaM_dash_Ca2
Product CaM_dash_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_dash_Ca_ dash_bind_dash_ Ca | mRNA synthesis Accession No. : 94 | kinetics Pathway No. : 1112 | 8.4846 (uM^-1 s^-1) | 8.4853 (s^-1) | Kd(bf) = 1.0001(uM) | - | Substrate Ca CaM_dash_Ca
Product CaM_dash_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 |
5 | Ca_dash_bind_ dash_CaNAB_ dash_Ca2 | mRNA synthesis Accession No. : 94 | kinetics Pathway No. : 1112 | 3599.928 (uM^-2 s^-1) | 1 (s^-1) | Kd(af) = 0.0167(uM) | - | Substrate Ca Ca CaNAB_dash_Ca2
Product CaNAB_dash_Ca4
|
| This process is probably much more complicated and involves CaM. However, as I can't find detailed info I am bundling this into a single step. Based on Steemer and Klee pg 6863, the Kact is 0.5 uM. kf/kb = 1/(0.5 * 6e5)^2 = 1.11e-11 |
6 | Ca_dash_bind_ dash_CaNAB | mRNA synthesis Accession No. : 94 | kinetics Pathway No. : 1112 | 10008360 (uM^-2 s^-1) | 1 (s^-1) | Kd(af) = 0.0003(uM) | - | Substrate Ca Ca CaNAB
Product CaNAB_dash_Ca2
|
| going on the experience with CaM, we put the fast (high affinity) sites first. We only know (Stemmer and Klee) that the affinity is < 70 nM. Assuming 10 nM at first, we get kf = 2.78e-8, kb = 1. Try 20 nM. kf = 7e-9, kb = 1 |