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Accession Type:
Network
CaMKII_noPKA_
model3
Shared_Object_
CaMKII_noPKA_
model3
CaMKII
CaM
 Molecule
 Reaction
PP1
PP2B
PP1_PSD

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Reaction List for pathway CaM (Pathway Number 259)

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 KfKbKdtauSubstrateProduct
1 CaM-Ca3-bind-Ca-
PSD
0.465
(uM^-1 s^-1)
10
(s^-1)
Kd(bf) = 21.5048(uM)-CaM-Ca3-PSD
Ca-PSD
CaM-Ca4-PSD
  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-TR2-bind-Ca71.999
(uM^-2 s^-1)
72
(s^-1)
Kd(af) = 1(uM)-CaM
Ca
Ca
CaM-TR2-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 !)....
3 CaM-TR2-bind-Ca-
PSD
72
(uM^-2 s^-1)
72
(s^-1)
Kd(af) = 1(uM)-CaM-PSD
Ca-PSD
Ca-PSD
CaM-TR2-Ca2-PSD
  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 !)....
4 CaM-TR2-Ca2-bind
-Ca
3.6
(uM^-1 s^-1)
10
(s^-1)
Kd(bf) = 2.7778(uM)-CaM-TR2-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
5 CaM-TR2-Ca2-bind
-Ca-PSD
3.6
(uM^-1 s^-1)
10
(s^-1)
Kd(bf) = 2.7778(uM)-CaM-TR2-Ca2-PSD
Ca-PSD
CaM-Ca3-PSD
  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


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