Growth kinetics formulas

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 ${\mu }_{\max }$.

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.

$\alpha$ 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 $\beta$.

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.