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Molecule Parameter List for CaM-Ca3

The statistics table lists the distribution of a molecule acting either as a substrate, product, enzyme or as a molecule within the network.
The text color of a molecule is highlighted by color.
Statistics
CaM-Ca3 participated asMoleculeSum total ofEnzymeSubstrate of an enzymeProduct of an enzymeSubstrate in ReactionProduct in Reaction
No. of occurrences1000011

Accession and Pathway Details
Accession NameAccession No.Accession TypePathway Link
CaMKIII90Network
Shared_Object_CaMKIII CaM 
CaMKIII is activated by Ca2+ and inactivated by S6K. Since CaMKIII inhibits eEF2, the net effect of Ca2+ on elongation is inhibitory and of S6K is excitatory

CaM-Ca3 acting as a Molecule in  
CaMKIII Network
NameAccession NamePathway NameInitial Conc.
(uM)
Volume
(fL)
Buffered
CaM-Ca3CaMKIII
Accession No. : 90
CaM
Pathway No. : 1094
01000No

CaM-Ca3 acting as a Substrate in a reaction in  
CaMKIII Network
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 reaction are not consider.
NameAccession NamePathway NameKfKbKdtauReagents
CaM-Ca3-bind-CaCaMKIII
Accession No. : 90
CaM
Pathway No. : 1094
0.465
(uM^-1 s^-1)
10
(s^-1)
Kd(bf) = 21.5051(uM)-Substrate
Ca
CaM-Ca3

Product
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

CaM-Ca3 acting as a Product in a reaction in  
CaMKIII Network
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 reaction are not consider.
NameAccession NamePathway NameKfKbKdtauReagents
CaM-Ca2-bind-CaCaMKIII
Accession No. : 90
CaM
Pathway No. : 1094
3.6001
(uM^-1 s^-1)
10
(s^-1)
Kd(bf) = 2.7777(uM)-Substrate
Ca
CaM-Ca2

Product
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



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