Cell growth, substrate utilisation and product formation

Definition, valid in any phase; units ${\mathrm{h}}^{-1}$. $X$ is the biomass concentration (g/L). When death is negligible this is the growth rate itself; when death or endogenous decay is significant, the measured $(1/X)dX/dt$ is the net rate $\mu -k_d$ (see below). The number-based rate ${\mu }_N$ uses the cell number concentration $N$ instead of $X$.

Applies to the exponential phase only: constant $\mu$, no lag inside the interval, death neglected ($X=X_0$ at $t=t_0$). If death is significant, use the net rate $\mu -k_d$ in place of $\mu$. Use the natural log. A plot of $\ln X$ against $t$ is a straight line of slope $\mu$.

Time for the biomass to double at constant $\mu$ (use $\mu -k_d$ if death is significant). The units follow $\mu$: $\mu$ in ${\mathrm{h}}^{-1}$ gives $t_d$ in h.

Based on cell NUMBER $N$ (cells/mL): after $n$ generations, $N=N_0{2}^n$. The relation $t_g=\ln 2/{\mu }_N$ holds for exponential growth at constant ${\mu }_N$. The generation time $t_g$ equals the biomass doubling time $t_d$ only during balanced growth, when the mass per cell is constant.

In the biomass balance, $k_d$ (${\mathrm{h}}^{-1}$) is a first-order constant for loss of biomass by death or endogenous decay, and ${\mu }_{\text{net}}=\mu -k_d$ is the net specific growth rate; in the stationary phase ${\mu }_{\text{net}}=0$. In the death-phase expression, $N$ is the viable cell concentration, $N_0$ is its value when the death phase starts and $t$ is the time since then; there $k_d$ is the specific death rate of viable cells, which need not equal the biomass decay constant.

Assumes one growth-limiting substrate, all other nutrients in excess, no inhibition and negligible death. ${\mu }_{\max }$ is in ${\mathrm{h}}^{-1}$; $K_s$ and $S$ are in the same concentration units. At $S=K_s$, $\mu ={\mu }_{\max }/2$.

Use these to simplify batch and chemostat problems. In batch culture with $S_0\gg K_s$, growth stays near ${\mu }_{\max }$ until the substrate is almost exhausted.

The Monod equation solved for $S$; valid only for $\mu <{\mu }_{\max }$. In a chemostat at steady state with sterile feed and negligible death, $\mu =D=F/V$, so this gives the residual substrate concentration $S=K_{s}D/({\mu }_{\max }-D)$, which does not depend on the feed substrate concentration.

Double-reciprocal (Lineweaver-Burk) plot: $1/\mu$ against $1/S$, with intercept $1/{\mu }_{\max }$, slope $K_s/{\mu }_{\max }$ and x-intercept $-1/K_s$. Second form: $S/\mu$ against $S$, with slope $1/{\mu }_{\max }$ and intercept $K_s/{\mu }_{\max }$. Third form: $\mu$ against $\mu /S$, with slope $-K_s$ and intercept ${\mu }_{\max }$. In the double-reciprocal plot the low-$S$ points (small $\mu$, large $1/S$) lie far from the axis, and taking reciprocals magnifies their errors, so they dominate the fitted line.

An unstructured model that needs no substrate balance: growth slows as $X$ approaches the carrying capacity $X_{\infty }$. $X_0$ is the biomass at $t=0$, and $k$ (${\mathrm{h}}^{-1}$) is the specific growth rate when $X\ll X_{\infty }$. The volumetric growth rate $dX/dt$ is highest at $X=X_{\infty }/2$, so the curve is sigmoidal when $X_0<X_{\infty }/2$.

Found by setting $d\mu /dS=0$. Below $S^{\ast }$, raising $S$ speeds growth; above it, raising $S$ slows growth. $\mu (S^{\ast })$ is always below ${\mu }_{\max }$.

Batch culture from inoculation ($t=0$, $X=X_0$, $S=S_0$) to time $t$, assuming Monod kinetics, constant yield, no lag, and negligible maintenance and death, so that $S=S_0-(X-X_0)/Y_{X/S}$. When $Y_{X/S}{K}_s\ll X_0+Y_{X/S}{S}_0$ and $S$ has not fallen close to zero, both correction terms are small and it reduces to $\ln (X/X_0)={\mu }_{\max }t$.

Units are g/g, C-mol/C-mol or mol/mol; always state which. Here $[{\mathrm{O}}_2]$ stands for the oxygen still available to the cells per litre of culture, which falls as it is consumed; it is not the dissolved-oxygen concentration. So $Y_{X/O_2}$ is the biomass formed per unit of oxygen consumed. Yields taken over the same interval combine as ratios: $Y_{P/X}=Y_{P/S}/Y_{X/S}$ and $Y_{S/X}=1/Y_{X/S}$.

Assumes a constant yield and negligible maintenance. $X_{\max }$ is reached when the limiting substrate is used up ($S=0$).

Rates per unit biomass, e.g. g substrate/(g cells·h). OUR is the volumetric oxygen uptake rate, e.g. mol ${\mathrm{O}}_2$/(L·h). The last relation uses the instantaneous observed yield and neglects death, since then $Y_{X/S}=(dX/dt)/(-dS/dt)=\mu /q_S$. A volumetric rate is the specific rate times $X$, e.g. $r_S=q_{S}X$.

$Y^M$ is the true (maximum) growth yield, the yield if no substrate went to maintenance. $m_S$ is the maintenance coefficient, in g substrate/(g cells·h), taken as constant; $m_{O_2}$ is the same for oxygen. Maintenance is energy spent on turnover of cell components, osmotic and ion gradients, and motility, not on making new cells.

Follows from $Y_{X/S}^{\text{obs}}=\mu /q_S$ (death neglected). A plot of $1/Y^{\text{obs}}$ against $1/\mu$ is a straight line with slope $m_S$ and intercept $1/Y_{X/S}^M$. In a chemostat at steady state with sterile feed and negligible death, $\mu =D$, so the plot is against $1/D$. Whenever $m_S>0$, $Y^{\text{obs}}<Y^M$, and the gap widens as $\mu$ falls.

The substrate is split three ways: new biomass, product and maintenance. Use this form when the product is made by a pathway separate from the cell's energy supply. If the product is itself the end product of energy metabolism (ethanol in anaerobic yeast culture), the substrate that forms it is the same substrate that supplies energy for growth and maintenance, so adding a separate product term counts that substrate twice.

$\alpha$ is the growth-associated coefficient (g P/g X). $\beta$ is the non-growth-associated coefficient (g P/(g X·h)). The two forms are equivalent when $dX/dt=\mu X$ (death neglected). To find both constants, plot $q_P$ against $\mu$: the slope is $\alpha$ and the intercept is $\beta$.

Death neglected. The exponential-phase form uses ${\int }_0^{t}Xdt=(X-X_0)/\mu$ for constant $\mu$. In the stationary phase, with $X=X_s$ constant from time $t_s$, when the product concentration is $P_s$, product rises linearly: $P=P_s+\beta X_s(t-t_s)$.

No extracellular product. $c$ is the biomass yield in C-mol cells per C-mol substrate. Here $m$, $n$, $\alpha$, $\beta$ and $\delta$ are formula subscripts, not the maintenance coefficient, the number of generations or the Luedeking-Piret constants. If a product is formed, add $f$ mol of product, with $j$ carbon atoms per molecule, to the right-hand side.

Four equations in five unknowns ($a$ to $e$), so one more measured quantity is needed, usually $c$ or the RQ. The electron balance below replaces the H and O balances and is quicker.

Available electrons per C-mol, counting C = +4, H = +1, O = -2 and N = -3 (ammonia as the nitrogen reference). The reference compounds $\mathrm{C}{\mathrm{O}}_2$, ${\mathrm{H}}_2\mathrm{O}$ and $\mathrm{N}{\mathrm{H}}_3$ all have $\gamma =0$. If nitrate is the nitrogen source, count N as +5; if ${\mathrm{N}}_2$ is, count it as 0. One mole of ${\mathrm{O}}_2$ accepts 4 electrons.

Per C-mol of substrate. ${\gamma }_S$, ${\gamma }_B$ and ${\gamma }_P$ are the degrees of reduction of substrate, biomass and product per C-mol, and $fj$ is the C-mol of product formed. Electrons in the substrate end up in biomass, product or oxygen; $\mathrm{C}{\mathrm{O}}_2$, ${\mathrm{H}}_2\mathrm{O}$ and the nitrogen source carry none, provided ${\gamma }_B$ is computed with the valence of N for that source.

$\eta$, $\zeta$ and $\epsilon$ are the fractions of the substrate's available electrons that go to biomass, product and oxygen. For a heterotroph growing on one organic carbon source (no $\mathrm{C}{\mathrm{O}}_2$ fixation), the biomass yield in C-mol/C-mol can never exceed the carbon limit of 1 or the electron limit ${\gamma }_S/{\gamma }_B$. The electron limit is reached only with no oxygen uptake and no product. The same reasoning bounds the product: $fj\le \min (1,{\gamma }_S/{\gamma }_P)$.

RQ is the mol of $\mathrm{C}{\mathrm{O}}_2$ formed per mol of ${\mathrm{O}}_2$ consumed. $M_X$ and $M_S$ are masses per C-mol (g/C-mol). To express the coefficients $a$ to $e$ per mole of substrate instead of per C-mol, multiply them by the number of carbon atoms in the substrate; the g/g yields and the RQ do not change.