Endoplasmic reticulum ¶ The Endoplasmic reticulum fluxes include ryanodine receptor (RyR) flux (Jrel), SERCA flux (Jup), SR leakage (Jleak), and calcium diffusion from NSR to JSR (Jtr).
J C a S R = V N S R V s u b S R ( J l e a k − J u p ) + J r e l J r e l = k R y R ⋅ P O 1 R y R ⋅ ( c a J S R − c a s r ) J t r = k t r C a S R ( c a J S R − c a N S R ) J l e a k = 0.5 ( 1 + 5 R y R C K p ) k S R l e a k J u p = V m a x S R f S R − r S R 1 + f S R + r S R d d t P O 1 R y R = k a p o s R y R ⋅ H ( c a s r , K m R y R , 4 ) ⋅ P C 1 R y R − k a n e g R y R ⋅ P O 1 R y R d d t c a J S R = β S R ( − J r e l V s u b S R + J t r V N S R ) / V J S R d d t c a N S R = J u p − J l e a k − J t r β S R = ( c a J S R + K m c s q n ) 2 ( c a J S R + K m c s q n ) 2 + Σ C s q n K m c s q n f S R = ( c a s r K m f p ) 2 r S R = ( c a N S R K m r S R ) 2 K m R y R = 3.51 ⋅ e x p i t ( − c a J S R − 530 200 ) + 0.25 P C 1 R y R = 1 − P O 1 R y R K m f p = min ( f C K I I P L B , f P K A P L B ) f P K A P L B = ( 1 − 0.5531 ) 1 − f r a c P L B p f r a c P K A P L B o + 0.5531 f C K I I P L B = ( 1 − 0.5 ∗ f r a c P L B C K p ) f r a c P L B C K p = 0.2 C a M K A c t \begin{align}
J_{CaSR} &= \frac{V_{NSR}}{V_{subSR}} (J_{leak} - J_{up}) + J_{rel} \\
J_{rel} &= k_{RyR} \cdot PO1_{RyR} \cdot (ca_{JSR} - ca_{sr}) \\
J_{tr} &= ktrCa_{SR} (ca_{JSR} - ca_{NSR}) \\
J_{leak} &= 0.5 (1 + 5 RyR_{CKp}) kSR_{leak} \\
J_{up} &= Vmax_{SR} \frac{fSR - rSR}{1 + fSR + rSR} \\
\frac{d}{dt} PO1_{RyR} &= kapos_{RyR} \cdot H(ca_{sr}, Km_{RyR}, 4) \cdot PC1_{RyR} - kaneg_{RyR} \cdot PO1_{RyR} \\
\frac{d}{dt} ca_{JSR} &= \beta_{SR} (-J_{rel} V_{subSR} + J_{tr} V_{NSR}) / V_{JSR} \\
\frac{d}{dt} ca_{NSR} &= J_{up} - J_{leak} - J_{tr} \\
\beta_{SR} &= \frac{(ca_{JSR} + Km_{csqn})^2}{(ca_{JSR} + Km_{csqn})^2 + \Sigma Csqn Km_{csqn}} \\
fSR &= \left( \frac{ca_{sr}}{Kmfp} \right)^2 \\
rSR &= \left( \frac{ca_{NSR}}{Kmr_{SR}} \right)^2 \\
KmRyR &= 3.51 \cdot expit(-\frac{ca_{JSR} - 530}{200}) + 0.25 \\
PC1_{RyR} &= 1 - PO1_{RyR} \\
Kmfp &= \min(fCKII_{PLB}, fPKA_{PLB}) \\
fPKA_{PLB} &= (1 - 0.5531) \frac{1 - fracPLBp}{fracPKA_PLBo} + 0.5531 \\
fCKII_{PLB} &= (1 - 0.5 * fracPLB_{CKp}) \\
fracPLB_{CKp} &= 0.2CaMKAct \\
\end{align} J C a SR J re l J t r J l e ak J u p d t d PO 1 R y R d t d c a J SR d t d c a NSR β SR f SR r SR K m R y R PC 1 R y R K m f p f P K A P L B f C K I I P L B f r a c P L B C K p = V s u b SR V NSR ( J l e ak − J u p ) + J re l = k R y R ⋅ PO 1 R y R ⋅ ( c a J SR − c a sr ) = k t r C a SR ( c a J SR − c a NSR ) = 0.5 ( 1 + 5 R y R C K p ) k S R l e ak = Vma x SR 1 + f SR + r SR f SR − r SR = ka p o s R y R ⋅ H ( c a sr , K m R y R , 4 ) ⋅ PC 1 R y R − kan e g R y R ⋅ PO 1 R y R = β SR ( − J re l V s u b SR + J t r V NSR ) / V J SR = J u p − J l e ak − J t r = ( c a J SR + K m cs q n ) 2 + Σ C s q n K m cs q n ( c a J SR + K m cs q n ) 2 = ( K m f p c a sr ) 2 = ( K m r SR c a NSR ) 2 = 3.51 ⋅ e x p i t ( − 200 c a J SR − 530 ) + 0.25 = 1 − PO 1 R y R = min ( f C K I I P L B , f P K A P L B ) = ( 1 − 0.5531 ) f r a c P K A P L B o 1 − f r a c P L Bp + 0.5531 = ( 1 − 0.5 ∗ f r a c P L B C K p ) = 0.2 C a M K A c t