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Molecule Parameter List for Ca

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
Ca participated asMoleculeSum total ofEnzymeSubstrate of an enzymeProduct of an enzymeSubstrate in ReactionProduct in Reaction
No. of occurrences1000050

Accession and Pathway Details
Accession NameAccession No.Accession TypePathway Link
CaMKII_200349Network
Shared_Object_CaMKII_2003,  CaMKII,  CaM,  
PP1,  PP2B 
Model of regulation of CaMKII by Calcium, including parallel excitatory input from CaM and inhibitory input from PP1 as regulated by Calcineurin and PKA. Cell type: neuronal.
Bhalla US. Biophys J. 2004 Aug;87(2):733-44.

Ca acting as a Molecule in  CaMKII_2003 Network
NameAccession NamePathway NameInitial Conc.
(uM)		Volume
(fL)		Buffered
CaCaMKII_2003
Accession No. : 49
  • Shared_Object_
    CaMKII_2003
    Pathway No. : 201
    		• 0.081000Yes

    Ca acting as a Substrate in a reaction in      CaMKII_2003 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
    1CaM-TR2-bind-CaCaMKII_2003
    Accession No. : 49    		CaM
    Pathway No. : 203    		72
    (uM^-2 s^-1)    		72
    (s^-1)    		Kd(af) = 1(uM)-Substrate
    Ca
    Ca
    CaM

    Product
    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 !)....
    2
  • CaM-TR2-Ca2-bind
    -Ca
    		• CaMKII_2003
    Accession No. : 49    		CaM
    Pathway No. : 203    		3.6
    (uM^-1 s^-1)    		10
    (s^-1)    		Kd(bf) = 2.7778(uM)-Substrate
    Ca
    CaM-TR2-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
    3CaM-Ca3-bind-CaCaMKII_2003
    Accession No. : 49    		CaM
    Pathway No. : 203    		0.465
    (uM^-1 s^-1)    		10
    (s^-1)    		Kd(bf) = 21.5054(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
    4
  • Ca-bind-CaNAB-Ca
    2
    		• CaMKII_2003
    Accession No. : 49    		PP2B
    Pathway No. : 205    		3.6
    (uM^-2 s^-1)    		1
    (s^-1)    		Kd(af) = 0.527(uM)-Substrate
    Ca
    Ca
    CaNAB-Ca2

    Product
    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
    5Ca-bind-CaNABCaMKII_2003
    Accession No. : 49    		PP2B
    Pathway No. : 205    		10008
    (uM^-2 s^-1)    		1
    (s^-1)    		Kd(af) = 0.01(uM)-Substrate
    Ca
    Ca
    CaNAB

    Product
    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