| |
Reaction
Name | Pathway Name /
Pathway No. | Kf | Kb | Kd | tau | Reagents |
| 1 | Ca-bind-CaNAB | PP2B
Pathway No. 248 | 10008
(uM^-2 s^-1) | 1
(s^-1) | Kd(af) = 0.01(uM) | - | Substrate:
CaNAB
Ca
Ca
Products:
CaNAB-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 | | 2 | CaMCa4-bind-CaNA
B | PP2B
Pathway No. 248 | 599.994
(uM^-1 s^-1) | 1
(s^-1) | Kd(bf) = 0.0017(uM) | - | Substrate:
CaM-Ca4
CaNAB-Ca4
Products:
CaM_Ca_n-CaNAB
|
| 3 | Ca-bind-CaNAB-Ca
2 | PP2B
Pathway No. 248 | 3.6001
(uM^-2 s^-1) | 1
(s^-1) | Kd(af) = 0.527(uM) | - | Substrate:
Ca
Ca
CaNAB-Ca2
Products:
CaNAB-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 |
and 
color.
