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Result: 1 - 4 of 4 rows are displayed

Reaction List for pathway CaM (Pathway Number 814) in Accession Ajay_Bhalla_2007_ReacDiff2 (Accession Number 83)

Entries are grouped according to Pathway Number and they are alternately color coded using  and  color.
Further ordering can be done to the table header.  indicates that ordering is done according to ascending or descending order.
Keq is calculated only for first order reactions.
Kd is calculated only for second order reactions. [nA+nB <->nC or nA<->nC+nD, where n is number and A,B,C,D are molecules]
  Reaction
Name
Pathway Name / 
Pathway No.
KfKbKdtauReagents
1 CaM-bind-CaCaM

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
2 CaM-Ca-bind-CaCaM

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
3 CaM-Ca2-bind-CaCaM

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-Ca3-bind-CaCaM

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

 
Result: 1 - 4 of 4 rows are displayed



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