| Enzyme Molecule / Enzyme Activity | Accession Name | Pathway Name | Km (uM) | kcat (s^-1) | Ratio | Enzyme Type | Reagents |
1 | CycB / k20_lambdaB
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 100.002 | 5000 | 4 | explicit E-S complex | Substrate Rb
Product 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 |
2 | CycB / k21_phiB
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 10 | 200 | 4 | explicit E-S complex | Substrate PP1A
Product PP1
|
| phiB = 2. See calculation for k21_phiE |
3 | CycB / k31
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 0.00999989 | 0.7 | 4 | explicit E-S complex | Substrate IE
Product IEP
|
| Represented as k31.[IE].[CycB]/(J31 + [IE]) k31 = 0.7 J31 = 0.01 |
4 | CycB / k11
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 1 | 3 | 4 | explicit E-S complex | Substrate AminoAcids
Product Cdc20notA
|
| Represented simply as [CycB]*k11, where k11 is 1.5. As AAs are at 1, we get rate = [AAs].[CycB].kcat / (Km + [AAs]) So if we set Km = [AAs] = 1, then kcat = 3 gives our desired equation. |
5 | CycB / Cdh1_CycB
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 0.0099998 | 40 | 4 | explicit E-S complex | Substrate Cdh1
Product Cdh1_i
|
| Eqn 12. J4 = Km = 0.01 k4 = 40 GammaB = 1 kcat = k4 * GammaB = 40 |
6 | CycB / B_phosph_E2FA
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 9.99992 | 10 | 4 | explicit E-S complex | Substrate E2FA
Product E2FAP
|
| See A_phosph_E2F. Same rate of k23 = 1 applies. |
7 | CycB / B_phosph_ E2FA.Rb
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 9.99992 | 10 | 4 | explicit E-S complex | Substrate E2FA.Rb
Product E2FAP.Rb
|
| See A_phosph_E2F. Same rate of k23 = 1 applies. |
8 | CycB / Ak6_etaB
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 9.99992 | 1000 | 4 | explicit E-S complex | Substrate CycA_Kip1
Product CycA degraded
|
| See Ak6_etaE |
9 | CycB / k8_CycB
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 0.0999992 | 0.1 | 4 | explicit E-S complex | Substrate CycE
Product degraded
|
10 | CycB / k6_E_etaB
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 9.99992 | 1000 | 4 | explicit E-S complex | Substrate CycE_Kip1
Product CycE degraded
|
| See notes for k6_E_etaE. Here etaB = 1 so kcat = 1000, Km as before is 10 |
11 | CycB / k6_D_etaB
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 9.99992 | 1000 | 4 | explicit E-S complex | Substrate CycD_Kip1
Product CycD degraded
|
| 6 Apr 2005. Earlier used explicit E.S complex form with k1 = 1000, k2 = 10, k3 = 1. This gave low Km and lots of E.S. complex. So shift to MM form: k6 = 100, etaB = 1. Let Km = 10 >> substrate. Then kcat = Km * k6 * etaB = 1000 |
12 | CycB / k8_CycB_Kip1
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 0.0999992 | 0.1 | 4 | explicit E-S complex | Substrate CycE_Kip1
Product Kip1 degraded
|
| k8 = 0.2, psiB = 0.05, so kcat = 0.1. J8 = 0.1 |
13 | CycB / k6_kip1_B
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 9.99992 | 1000 | 4 | explicit E-S complex | Substrate Kip1
Product degraded_kip
|
| 6 Apr 2005. Using MM form: k6 = 100 Let Km = 10 >> substrate. Then kcat = Km * k6 * eta_B = 1000 |
14 | CycB / k20_lambdaB[1]
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 100.002 | 5000 | 4 | explicit E-S complex | Substrate 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 |
15 | CycB / k20_lambdaB[2]
| Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 100.002 | 5000 | 4 | explicit E-S complex | Substrate E2FA.Rb
Product E2FA 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 |
| Enzyme Molecule / Enzyme Activity | Accession Name | Pathway Name | Km (uM) | kcat (s^-1) | Ratio | Enzyme Type | Reagents |
1 | Cdc20 / k2_prime_prime | Mammalian_cell_ cycle Accession No. : 85 | CELLDIV Pathway No. : 1070 | 99.9992 | 100 | 4 | explicit E-S complex | Substrate CycB
Product degraded
|
| k2_prime_prime = 1. rate = k2_prime_prime * Cdc20 * CycB Using MM: rate = kcat * Cdc20 * CycB / (CycB + Km) Let Km >> CycB, ie, around 100. Then kcat = k2_prime_prime * Km = 100. |
2 | Cdh1_i / Cdh1_i_k2_prime | Mammalian_cell_ cycle Accession No. : 85 | Cdh1_grp Pathway No. : 1075 | 100.002 | 5 | 4 | explicit E-S complex | Substrate CycB
Product degraded
|
| k2_prime = 0.05. so actually this reaction is pretty negligible. rate = k2_prime * Cdh1_i * CycB From MM kinetics, rate = kcat * Cdh1_i * CycB / (CycB + Km). Let Km >>CycB, so Km = 10. Then kcat = k2_prime * Km = 0.5 |
3 | Cdh1 / Cdh1_k2 | Mammalian_cell_ cycle Accession No. : 85 | Cdh1_grp Pathway No. : 1075 | 100 | 2000 | 4 | explicit E-S complex | Substrate CycB
Product degraded
|
| k2 = 20 Let Km = 100, so it is >> substrate. Then kcat = Km * k2 = 2000 |