Team:Rice/Model/Metabolism

Metabolism

Overview

Cellular energy is generated via two co-expressed enzymes: a transporter and a metabolic enzyme. We assume that the cells grow in a chemostat with a constant concentration of nutrients. The cells import the nutrients and then convert them to cellular energy.

Details

Note: if the equations below seem messed up, please follow this link to a simplified version of this page.

To convert nutrients into energy, first the cell must import the nutrients from the outside source ($s_e$) to the inside ($s_i$). The transporter protein ($p_T$) accomplishes this via Michaelis-Menten kinetics, with a maximum speed of $v_T$ and a import threshold of $k_T$. Conversion to cellular energy is accomplished in a similar manner with the metabolic enzyme ($p_E$) and with constants $v_E$ and $k_E$. However, conversion is not perfect, and the efficiency of the conversion is given by the factor $\phi_E$. Energy is used during translation at a rate of $\gamma \sum c_i$, where the sum runs over all translation complexes. In addition, the cell slowly grows at a rate of $\lambda = \gamma \frac{\sum c_i}{M_p}$, where $M_p$ is the total mass (in amino acids) of the proteome. As a result, all species are diluted with a rate constant of $\lambda$.

Equations

\begin{equation*} \begin{aligned} \frac{ds_i}{dt} &= v_T p_T \frac{s_e}{k_T+s_e} - v_E p_E \frac{s_i}{k_E+s_i} - \lambda s_i \\ \frac{dE_c}{dt} &= \phi_E v_E p_E \frac{s_i}{k_E+s_i} - \gamma \sum c_i - \lambda E_c \\ \end{aligned} \end{equation*}