| |
Name | Accession
Type | Initial
Conc.
(uM) | Volume
(fL) | Buffered | Sum Total Of |
| 1 | IP3R* | Network | 0 | 0 | No | - |
| | This is the ligand-bound form of the IP3R. |
| 2 | IP3R | Network | 0.0166 | 0 | No | - |
| | The number of the IP3Rs in the cell is present only implicitly in the model, and is lumped in with the total permeability of the IP3R pool. The latter is constrained by the height of the Ca transient. |
| 3 | IP3 | Network | 0 | 0 | No | IP3(145)
|
| | Inositol(145)trisphosphate |
| 4 | inact_cap_entry | Network | 0 | 0 | No | - |
| | represents the portion of the capacitative-Ca entry channel which is blocked when there is lots of Ca sequestered in the stores |
| 5 | CaTransp-2Ca | Network | 0 | 0 | No | - |
| | equivalent to the enzyme-substrate complex. 2 Ca are bound to the transporter. ATP is ignored. |
| 6 | CaTransp | Network | 0.24 | 0 | No | - |
| | The calcium transporter levels are constrained by the resting levels of Ca in the cell. The rate of Ca sequestration depends on the amount of this pool. |
| 7 | capacitive_Ca_
entry* | Network | 0.01 | 0 | No | - |
| | This mechanism has taken a while to be more tightly confirmed as probably being the TRP channel. In this model the channel is implemented to match experimental observations about capacitative Ca entry. Levels are set by two constraints: the resting Ca levels, and the height of the response to IP3. |
| 8 | Calseq | Network | 9.09 | 0 | No | - |
| | This is Calsequestrin or the calcium buffer in the ER. from Cala & Jones, JBC 258(19), 1983: 11932-36 Calseq is present as 4mg/g of membrane protein; membrane protein = 2% of cell mass = 0.02 * 1g/cc * (1e-9)cc = (2e-11) g Hence Calseq = 8e-14/55000 moles per (1.6e-13)l = 9.091uM As per Guidebook to the calcium-binding proteins by Celio; and Mitchell et al, JBC 263, 1988: 1376-81; 1 mol of Calsequestrin binds 40 mol of Ca. This is the stoichiometry we use. System limitations do not allow us to create a 40th order reaction. Hence Calseq binding to Ca has been modeled in eight consecutive steps of 5th order each. The affinity of Calsequestrin for Ca in our model is constrained by the levels of free Ca in the stores (Ca-sequester). We use a Kd such that Ca-sequester levels remain similar to levels in the CaRegulation model without Ca buffering. |
| 9 | CaEPump | Network | 0.005 | 0 | No | - |
| | The calcium electrogenic pump: Mc Burney and Neering, TINS 10(4), 1987, 164-169. We treat the pump as a simple Michaelis-Menten enzyme. Levels are constrained tightly by the need to keep resting Ca levels at 80 nM. |
| 10 | Ca5-Cal | Network | 0 | 0 | No | - |
| | Calsequestrin with 5 Ca molecules bound |
| 11 | Ca40-Cal | Network | 0 | 0 | No | - |
| | Calsequestrin with 40 Ca molecules bound |
| 12 | Ca35-Cal | Network | 0 | 0 | No | - |
| | Calsequestrin with 35 Ca molecules bound |
| 13 | Ca30-Cal | Network | 0 | 0 | No | - |
| | Calsequestrin with 30 Ca molecules bound |
| 14 | Ca25-Cal | Network | 0 | 0 | No | - |
| | Calsequestrin with 25 Ca molecules bound |
| 15 | Ca20-Cal | Network | 0 | 0 | No | - |
| | Calsequestrin with 20 Ca molecules bound |
| 16 | Ca15-Cal | Network | 0 | 0 | No | - |
| | Calsequestrin with 15 Ca molecules bound |
| 17 | Ca10-Cal | Network | 0 | 0 | No | - |
| | Calsequestrin with 10 Ca molecules bound |
| 18 | Ca-sequester | Network | 5.796 | 0 | No | - |
| | Sequestered Ca pool The vol is 0.16 * the vol of the cell as a whole. |
| 19 | Ca-leak-to-cytop
lasm | Network | 0.024 | 0 | Yes | - |
| | represents the channels that leak Ca into the cytoplasm. Effects of membrane potential are not considered. The amount and total flux are constrained by the need to balance Ca flux and keep basal Ca around 80 nM. |
| 20 | Ca-leak-from-ext
racell | Network | 0.0008 | 0 | No | - |
| | This represents the pool of Ca leak channels. The conc gradient is so large that this pool needs only a small number of molecules. For an equilibrium at 0.1 uM we need flow of 36e3/sec. With a permeability of 0.01 and a conc gradient of 4mM->0.1 uM (4e4) we get flux = N * perm * grad => N = 36e3 / (1e-2 * 4e3) = 900 if flux = 20e3, N =500, which is what we use. This works out to a concentration of 0.83 nM. |
and 
color.