Kinetic model of Central Metabolism

A kinetic model of glycolysis with serine activation is constructed from the literature data .

Description of the model

Schematic diagram of the model is given here. The dotted line represents activation(+) or inhibition(-) and the dashed arrow indicate Pentose Phosphate Pathway reactions not included in the model. Click on a reaction to have more information

Reactions

Details of the abbreviations for this model is listed here. Reactions of the model are listed below.

Glucose Transporter Hexokinase Pentose Phosphate Pathway Phosphoglucomutase UDPG-pyrophosphorylase Glycogen synthase Glycogen phosphorylase Hexose-6-phosphate isomerase Transketolase Phosphofructokinase type 1 Fructose 1,6 bisphosphate aldolase Adenylate kinase Dehydrogenase Triosephosphate isomerase Glyceraldehyde-3-phosphate dehydrogenase

3-phosphoglycerate kinase Phosphoglycerate dehydrogenase Phosphoserine amino-transferase Phosphoserine phosphatase 3-phosphoglycerate mutase Enolase Pyruvate kinase ATPase Lactate dehydrogenase Monocarboxylate (Lactate) transporter Mitocondrial pyruvate metabolism Glycine hydroxymethyltransferase Serine out Glycine out Nucleosid diphosphate kinase Pyrophosphate phosphohydrolase

Initial concentration of the metabolites can be found here

Model File

The SBML file of the model can be found here

Notes

Necessary observation during model building and simulation is noted here.

Global parameters

Calculating $V_{max}$

Some of the Vmax value in the paper "Modeling cancer glycolysis" is given in $U \times \text{(mg total cellular protein)}^{-1}$ unit . To homogenize the units it is then converted back to $\frac{mM}{min}$ by multiplying with 65 as the HeLa cell was incubated in $65 \frac{\text{mg protein}}{\text{mL cell volume}}$.

• Converting $U \times \text{(mg total cellular protein)}^{-1}$ to $mM \times min^{-1}$:

$U \times \text{(mg total cellular protein)}^{-1} = \frac{\mu mole}{min \cdot \text{(mg total cellular protein)}} = \frac{10^{-6} mole}{min \cdot \text{(mg total cellular protein)}}$
Also, $\frac{mM}{min} = \frac{10^{-3} mole}{min \cdot liter}$
We want to convert 1 $U \times \text{(mg total cellular protein)}^{-1}$ to $mM \times min^{-1}$ where HeLa cell was incubated in $65 \frac{\text{mg protein}}{\text{mL cell volume}}$
Therefore, $\frac{ 1 \times 10^{-6} mole}{ min \times \text{(mg total cellular protein)}} \times \frac{65 \text{(mg total cellular protein)}}{10^{-3} \text{(L cellular protein)}}$
=> $\frac{ 1 \times 10^{-3} \times 10^{-3} mole}{ min \times \text{(mg total cellular protein)}} \times \frac{65 \text{(mg total cellular protein)}}{10^{-3} \text{(L cellular protein)}}$
Striking out the necessary elements we get $\frac{65 \times 10^{-3} mole}{L \times min}$
=> $65 mM \times min^{-1}$

Quantifying the flux

We used the Kinetic Flux Profiling (KFP) method to quantify the flux values. The central idea of KFP is that larger metabolic fluxes cause faster transmission of isotopic label from added nutrient to downstream metabolites. Under pseudosteady state, if an external nutrient is instantaneously switched from natural to isotopically labeled, for a metabolite X directly downstream of nutrient assimilation, unlabeled $X(X^U)$ will be replaced over time by its labeled counterpart ($X^*$) and the fraction of unlabeled $X (X^U/X^T)$ will decay. The rate constant of this decay $(K_X)$ is determined by the ratio of the flux through $X (f_X)$ to the total pool size of $X (X^T)$.

• We used first order rate constant because, labeling decay is an exponential decay and the exponential decay follows first order rate constant.

$\frac{dN}{dt} = - \lambda N$

and for first order rate equation

$- \frac{d[A]}{dt} = k[A]$

• For exponential decay the slope is the height of the function.

• First we determine the unlabeled from for each metabolites

$\text{Fraction Unlabeled} = \frac{\text{Peak Height}_{\text{unlabeled form}}}{\text{Sum of peak height}_{\text{every isotopic form}}}$

• If we can calculate the apparent first order rate constant $K_X$ then the flux can be calculated as $f_X = K_X \times X^T$ [$X^T$ is the intracellular concentration of X ].

• With $X^T$ at steady-state, we calculate the change of unlabeled fraction with the introduction of label as

$\frac{dX^{U}}{dt} = -f_X (X^{U}/X^{T})$

The analytical solution is $\frac{X^U}{X^T} = exp(-f_Xt/X^T)$

setting $K_X = \frac{f_X}{X^T}$, we get $\frac{X^U}{X^T} = exp(-K_Xt)$

Example:
To explain the procedure we took the example of Pyruvate in the presence of Serine and Glycine. $\frac{X^U}{X^T} = \frac{5 \times 10^4}{ 1.8 \times 10^5}$

So,
$\frac{5 \times 10^4}{ 1.8 \times 10^5} = exp(-K_Xt)$
$=> 0.277 = exp(-K_Xt)$
$=> log_e(0.277) = -K_Xt$
$=> K_x = \frac{log_e(0.277)}{t}$
$K_X = 0.0017 min^{-1}$

Now we know, $f_X = K_X \times X^T$. The intracellular concentration of $X^T = 3 \text{mM or}\frac{mmol}{vol}$.

So, $f_x = K_X \times X^T$
$f_x = 0.0017 \times 3 \frac{mmol}{vol \times min}$
$f_x = 0.0051 \frac{mmol}{vol \times min}$