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Molecule Parameter List for E2FAP

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
E2FAP participated asMoleculeSum total ofEnzymeSubstrate of an enzymeProduct of an enzymeSubstrate in ReactionProduct in Reaction
No. of occurrences1000620

Accession and Pathway Details
Accession NameAccession No.Accession TypePathway Link
  • Mammalian_cell_
    cycle
  • 85Network
    Growth CELLDIV Rb_grp 
    IE_GRP CycB_Grp Cdc20_Grp 
    Cdh1_grp E2F CycA_Grp 
    CycE_grp Early_Response_Genes Delayed_Response_Genes 
    CycD_Grp 
    This is a fairly complete mass-action reimplementation of the Novak and Tyson mammalian cell cycle model. It is inexact on two counts. First, it replaces many rather abstracted equations with mass action and Michaelis-Menten forms of enzymes. Second, it does not handle the halving of cellular volume at the division point. Within these limitations, the model does most of what the original paper shows including oscillation of the relevant molecules.

    E2FAP acting as a Molecule in  
    Mammalian_cell_cycle Network
    NameAccession NamePathway NameInitial Conc.
    (uM)
    Volume
    (fL)
    Buffered
    E2FAP
  • Mammalian_cell_
    cycle

    Accession No. : 85
  • E2F
    Pathway No. : 1076
    0200No

    E2FAP acting as a Product of an Enzyme in  
    Mammalian_cell_cycle Network
     Enzyme Molecule /
    Enzyme Activity
    Accession NamePathway NameKm (uM)kcat (s^-1)RatioEnzyme TypeReagents
    1CycD  /
    k20_lambdaD[1]
  • Mammalian_cell_
    cycle

    Accession No. : 85
  • CELLDIV
    Pathway No. : 1070
    10033004explicit E-S complexSubstrate
    E2FAP.Rb

    Product
    E2FAP
    Rb_P
        With a low Km, rate ~ kcat. Here we have rate = k20 * lambda_d = 10 * 3.3 = 33. 7 Apr 2005. Actually should have the substrate term in here. Use the form Km >> substrate, so rate = kcat * sub * enz / Km so kcat = Km * k20 * lambda_d = 10 * 10 * 3.3 = 330 The idea here is that these reactions phosphorylate the Rb protein attached to E2FAP, so that Rb_P is released and E2FAP is left.
    2CycE  /
    k20_lambdaE[1]
  • Mammalian_cell_
    cycle

    Accession No. : 85
  • CELLDIV
    Pathway No. : 1070
    100.00250004explicit E-S complexSubstrate
    E2FAP.Rb

    Product
    E2FAP
    Rb_P
        For Km ~ 0, rate ~ kcat. rate = k20 * lambdaE = 10 * 5 7 Apr 2005. Actually need to put in substrate term too. Let Km = 10 >> sub. Then, rate ~ kcat * sub * prd /Km so kcat = Km * k20 * lambdaE = 10 * 10 * 5 = 500
    3CycA  /
    A_phosph_E2F
  • Mammalian_cell_
    cycle

    Accession No. : 85
  • CELLDIV
    Pathway No. : 1070
    9.99992104explicit E-S complexSubstrate
    E2FA

    Product
    E2FAP
        Rate equn has form [CycA].[E2F].k23 k23 = 1 MM equn has form [CycA].[E2F].kcat/(Km + E2F) So, we set kcat = Km * k23 where Km >> E2F 25 Mar. Better: Use explicit enz form. rate = k3.k1/k2 if k3 << k2. Let k3 = 1, k2 = 10, so we get k1 = k23 * 10 = 10. 6 Apr. Problem with explicit form is that the enz-substrate complex may affect the levels of the CycA, B etc. Back to MM.
    4CycA  /
    k20_lambdaA[1]
  • Mammalian_cell_
    cycle

    Accession No. : 85
  • CELLDIV
    Pathway No. : 1070
    10030004explicit E-S complexSubstrate
    E2FAP.Rb

    Product
    E2FAP
    Rb_P
        Km ~ 0, so rate ~ kcat. Here rate = k20 * lambdaA = 10 * 3 7 Apr 2005: Fix it: rate should have substrate term in it. Set Km = 10 >> substrate. Then, kcat = Km * k20 * lambdaA = 10 * 10 * 3 = 300
    5CycB  /
    B_phosph_E2FA
  • Mammalian_cell_
    cycle

    Accession No. : 85
  • CELLDIV
    Pathway No. : 1070
    9.99992104explicit E-S complexSubstrate
    E2FA

    Product
    E2FAP
        See A_phosph_E2F. Same rate of k23 = 1 applies.
    6CycB  /
    k20_lambdaB[1]
  • Mammalian_cell_
    cycle

    Accession No. : 85
  • CELLDIV
    Pathway No. : 1070
    100.00250004explicit E-S complexSubstrate
    E2FAP.Rb

    Product
    E2FAP
    Rb_P
        With Km ~ 0, rate ~ kcat. Here rate = k20 * lambdaB = 10 * 5 7 Apr 2005. Changed to include substrate term. Use Km = 10 >> sub, so kcat = Km * k20 * lambdaB = 10 * 10 * 5 = 500

    E2FAP acting as a Substrate in a reaction in  
    Mammalian_cell_cycle 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
    1k22_23_prime
  • Mammalian_cell_
    cycle

    Accession No. : 85
  • E2F
    Pathway No. : 1076
    1
    (s^-1)
    0.005
    (s^-1)
    Keq = 0.005(uM)0.995secSubstrate
    E2FAP

    Product
    E2FA
      k22 is the forward rate of 1 k23_prime is the backward rate of 0.005
    2k26_P
  • Mammalian_cell_
    cycle

    Accession No. : 85
  • E2F
    Pathway No. : 1076
    10
    (uM^-1 s^-1)
    200
    (s^-1)
    Kd(bf) = 20.0001(uM)-Substrate
    E2FAP
    Rb

    Product
    E2FAP.Rb
      Same as k26, but acting on the phosph form.



    Database compilation and code copyright (C) 2022, Upinder S. Bhalla and NCBS/TIFR
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