The whole Fundamentals of Biological Engineering formula sheet, to print

Constant-volume batch. $k$ does not depend on concentration; it varies with temperature. Its units are $(\text{mol/L}{)}^{1-n}{\text{time}}^{-1}$.

Constant volume. $C_{A0}$ is the initial concentration of reactant A.

The rate does not depend on concentration. Valid only while $C_A>0$: A is used up completely at $t=C_{A0}/k$ and stays at zero after that.

Equivalently $C_A=C_{A0}{e}^{-kt}$, or $-\ln (1-X_A)=kt$ in terms of conversion. The half-life does not depend on $C_{A0}$, and A never quite reaches zero.

Applies when $-r_A=k{C}_A^2$, for example $2A\to P$ with $k$ defined on A, or $A+B\to P$ with $-r_A=k{C}_A{C}_B$ and $C_{A0}=C_{B0}$ (so $C_B=C_A$ throughout). The half-life is inversely proportional to $C_{A0}$.

Any $n\ne 1$. Gives the zero-order result for $n=0$ and the second-order result for $n=2$. For $n<1$ it holds only until A is used up, at $t=\frac{C_{A0}^{1-n}}{(1-n)k}$.

Any $n\ne 1$. The proportionality is the basis of the half-life method for finding the order.

$T$ in kelvin, $R=8.314{\text{J mol}}^{-1}{\text{K}}^{-1}$, $E_a$ in J/mol. A plot of $\ln k$ against $1/T$ is a straight line of slope $-E_a/R$.

Eliminates $A$. When $T_2>T_1$ and $E_a>0$, $k_2>k_1$.

$K_m$ has concentration units, the same as [S] (mM, mol/L or g/L). $v$ and $V_{max}$ are concentration per time (e.g. µmol L${}^{-1}$ min${}^{-1}$).

From $d[ES]/dt\approx 0$. Because $k_2>0$, $K_m$ is always larger than the dissociation constant $k_{-1}/k_1$.

$k_{cat}$ (s${}^{-1}$) is the turnover number: substrate molecules converted per active site per unit time when the enzyme is saturated. $[E{]}_0$ is the molar concentration of active sites.

Units L mol${}^{-1}$ s${}^{-1}$, a second-order rate constant. It compares substrates for the same enzyme. For the one-intermediate mechanism, $k_{cat}/K_m=k_1{k}_2/(k_{-1}+k_2)$, which is less than $k_1$, so it cannot exceed the rate constant for enzyme and substrate meeting by diffusion, of the order of $10^8$ to $10^9$ L mol${}^{-1}$ s${}^{-1}$.

At low substrate the reaction is first order in S with rate constant $V_{max}/K_m$. At high substrate it is zero order.

At $[S]=K_m$, $v=V_{max}/2$. This is the working definition of $K_m$ under either assumption.

Plot $1/v$ against $1/[S]$: slope $K_m/V_{max}$, $y$-intercept $1/V_{max}$, $x$-intercept $-1/K_m$.

Plot $v$ against $v/[S]$: slope $-K_m$, $y$-intercept $V_{max}$, $x$-intercept $V_{max}/K_m$.

Plot $[S]/v$ against [S]: slope $1/V_{max}$, $y$-intercept $K_m/V_{max}$, $x$-intercept $-K_m$.

Competitive: ${\alpha }^{\prime }=1$. Uncompetitive: $\alpha =1$. Pure non-competitive: $K_i=K_i^{\prime }$, so $\alpha ={\alpha }^{\prime }$.

The inhibited reaction still follows the Michaelis–Menten form, with these apparent constants at a fixed [I].

I binds only free E, competing with S for the active site. $K_{m,app}=K_m(1+[I]/K_i)$ increases and $V_{max}$ is unchanged, so enough substrate overcomes the inhibition.

I binds only ES. $V_{max,app}$ and $K_{m,app}$ both fall by the factor $1+[I]/K_i^{\prime }$, so their ratio $V_{max}/K_m$ is unchanged. The inhibition grows as [S] rises.

I binds E and ES with equal affinity, at a site other than the active site. $V_{max,app}=V_{max}/(1+[I]/K_i)$ and $K_m$ is unchanged. The fractional inhibition is the same at every [S].

A second substrate molecule binds ES to form an inactive $E{S}_2$, with $K_{SI}=[ES][S]/[E{S}_2]$. At low [S] this reduces to Michaelis–Menten. At very high [S], $v\approx V_{max}{K}_{SI}/[S]$, so the rate falls. On a Lineweaver–Burk plot the points curve upward near the $1/v$ axis.

From $dv/d[S]=0$. $v_{peak}$ is the highest rate the enzyme can reach and is always below $V_{max}$.

Natural log. $V_{max}$ in concentration per time; $K_m$ and [S] in the same concentration unit.

Same assumptions. Gives the batch time for a target conversion $X$.

The fraction of fed A that has reacted. The concentration form holds only at constant density (liquid-phase systems).

The time needed to process one reactor volume of feed, measured at feed conditions (h). The space velocity $s$ is its reciprocal (h${}^{-1}$).

The bioprocess name for the reciprocal space time of a continuous stirred vessel. $F$ is the volumetric feed rate (L h${}^{-1}$, the same quantity as $v_0$) and $D$ is in h${}^{-1}$.

Steady-state chemostat with Monod growth $\mu ={\mu }_{\max }S/(K_s+S)$, a sterile feed, no cell death or maintenance, and a constant yield $Y_{X/S}=-dx/dS$ (g cells per g substrate). $x$ is the cell concentration (g L${}^{-1}$), written in lower case here so that it is not confused with conversion. $S$, $S_0$ and $K_s$ are in g L${}^{-1}$; ${\mu }_{\max }$ and $D$ are in h${}^{-1}$. The residual substrate $S$ does not depend on $S_0$, and the cell productivity is $Dx$ (g L${}^{-1}$ h${}^{-1}$). These results hold for $D<D_{\text{crit}}$; at $D\ge D_{\text{crit}}$ the only steady state is washout, $x=0$ and $S=S_0$.

$N_{A0}$ is the moles of A charged. This is the reaction time needed to reach $X_A$. It does not include the time to charge, empty or clean the vessel.

Steady state, with a feed that contains no converted A. The rate is evaluated at exit (f) conditions, which equal the reactor contents. The last form holds only at constant density.

Steady state, with no mixing along the length and a feed that contains no converted A. The rate varies along the reactor, so it stays inside the integral. The concentration form holds only at constant density.

For $-r_A=k{C}_A$ with $k$ in h${}^{-1}$. For a batch reactor, replace $\tau$ with $t$. The dimensionless group $k\tau$ is the first-order Damköhler number. The result does not depend on $C_{A0}$.

For $-r_A=k{C}_A$ at constant density and steady state. The result does not depend on $C_{A0}$.

Also covers $A+B\to$ products with $-r_A=k{C}_A{C}_B$ and equimolar feed ($C_{A0}=C_{B0}$), because the 1 : 1 stoichiometry then keeps $C_B=C_A$. $k$ is in L mol${}^{-1}$ h${}^{-1}$.

$V_{\max }=k_{cat}{E}_0$ stays constant only if the enzyme concentration is constant and the enzyme does not deactivate.

The integrated Michaelis–Menten equation. For a batch reactor, replace $\tau$ with $t$. The first term is the zero-order part and the second is the first-order part.

Steady state. The rate is evaluated at the exit substrate concentration $S$.

Same feed and same conversion, constant density. Neither ratio depends on $k$, and the second-order ratio does not depend on $C_{A0}$ either.

First order, constant density, with $N$ equal tanks each of space time ${\tau }_i$. The total space time is $\tau =N{\tau }_i$. For unequal tanks, multiply the stage factors: $C_{AN}/C_{A0}=1/[(1+k{\tau }_1)(1+k{\tau }_2)\cdots (1+k{\tau }_N)]$.

First order, with $N$ equal CSTRs in series reaching an overall conversion $X_A$.

$\tau$ is based on the fresh feed $v_0$, which contains no converted A. $X_{A1}$ is the conversion at the reactor inlet after mixing with the recycle stream.

First order, constant density, with $\tau =V/v_0$ based on the fresh feed. $R=0$ gives the PFR result, and $R\to \infty$ gives $k\tau =X_A/(1-X_A)$.

$z$ = signed charge number of the substrate, $\psi$ = signed electrostatic potential of the support (V), $F=96485$ C/mol, $R_g=8.314$ J mol${}^{-1}$ K${}^{-1}$, $T$ in K. Assumes a uniform potential in the enzyme's microenvironment and Boltzmann partitioning; the effect is largest at low ionic strength, because added salt screens the charge. Same signs ($z\psi >0$): substrate repelled, $K_{m,app}>K_m$. Opposite signs: substrate attracted, $K_{m,app}<K_m$. Partitioning alone leaves $V_{max}$ unchanged. More generally, any partition coefficient $K_p=S_{local}/S_b$ (electrostatic or hydrophobic) gives $K_{m,app}=K_m/K_p$ with $V_{max}$ unchanged.

Steady state; enzyme on the external surface only, no internal diffusion. Solve for $S_s$; the observed rate per unit area is either side of the equation.

Dimensionless. The maximum mass-transfer rate is $k_L{S}_b$, reached when $S_s=0$. Large $\mathrm{Da}$ favours film control, small $\mathrm{Da}$ favours reaction control (see the criteria below).

Exact for Michaelis–Menten kinetics. It comes from the quadratic $x^2+(\mathrm{Da}+\kappa -1)x-\kappa =0$; take the positive root ($0<x<1$).

Referred to the bulk concentration. Always $\le 1$ for Michaelis–Menten kinetics, because $S_s\le S_b$.

$k_s=V_{max}^{\prime }/K_m$ (m/s) is the surface first-order rate constant. Film and reaction act as resistances in series. Note ${\mathrm{Da}}_1=\mathrm{Da}/\kappa$: it uses $K_m$ in place of $S_b$.

Referred to the surface concentration $S_s$, not the bulk. $0<\eta \le 1$ for first-order and Michaelis–Menten kinetics.

${\varphi }^2$ compares the characteristic reaction rate with the characteristic diffusion rate. For a slab, $L$ is the half-thickness when both faces are exposed (the full thickness when one face is sealed). $R$ is a radius, not a diameter.

$L$ = half-thickness with both faces exposed. Exact for isothermal first-order kinetics.

$R$ = particle radius. Exact for isothermal first-order kinetics.

Gives identical numbers; only the modulus is defined without the factor 1/3. In this convention the large-modulus limit is $\eta \approx 3/{\varphi }_s$.

Small $\varphi$: reaction is slow compared with diffusion and the whole particle sees $S_s$. Large $\varphi$: substrate is consumed in a thin shell near the surface and the core is starved. For a slab $\tanh \varphi /\varphi$ is within 1% of $1/\varphi$ once $\varphi \ge 2.65$ ($\tanh 3=0.995$). A sphere converges more slowly — at $\varphi =3$, $1/\varphi$ overestimates $\eta$ by about 12% — hence the correction term, which is within 1% of the exact value for $\varphi \ge 1$.

${\varphi }_1$ is the first-order modulus; $\ln$ is the natural logarithm. The factor multiplying ${\varphi }_1$ tends to 1 as $\beta \to 0$ and to $1/\sqrt{2\beta }$ as $\beta \to \infty$, and decreases steadily in between, so $\Phi$ falls as $S_s$ rises.

$v_{obs}$ = measured rate per unit particle volume. The identity ${\Phi }_W=\eta {\varphi }^2$ holds for first-order kinetics with $\varphi$ based on $V_p/A_p$.

Rate per unit particle volume, from $k_L{a}_p(S_b-S_s)=\eta k_1{S}_s$. For isothermal first-order kinetics $\eta$ does not depend on $S_s$, so the two resistances simply add.

First order, with $\varphi$ based on $V_p/A_p$ and ${\eta }_o$ referred to $S_b$. ${\mathrm{Bi}}_m$ compares film transport with internal diffusion; the identity $k_1/(k_L{a}_p)={\varphi }^2/{\mathrm{Bi}}_m$ turns the series-resistance result into this form. If a question defines $\mathrm{Bi}=k_{L}R/D_e$ for a sphere together with ${\varphi }_s=R\sqrt{k_1/D_e}$, the second term is ${\varphi }_s^2/(3\mathrm{Bi})$. Large ${\mathrm{Bi}}_m$: the film is negligible and ${\eta }_o\to \eta$.

Referred to the bulk concentration. It is the product of the external and internal factors because $v_{obs}=\eta v(S_s)$.

Valid for $\varphi \gg 1$. The observed rate scales with $\sqrt{k_1}$, hence with the square root of the enzyme loading (uniformly spread enzyme), and inversely with particle size.

Strong internal diffusion limitation ($\varphi \gg 1$) in an isothermal particle, where $v_{obs}\propto \sqrt{k_1{D}_e}$. $E$ = intrinsic activation energy of the enzyme reaction; $E_D$ = activation energy of diffusion, usually small. Under external film control the apparent value falls further, to the low value of liquid-film mass transfer.

Strong internal limitation ($\varphi \gg 1$), isothermal particle, intrinsic power-law order $n$ ($n>-1$): first order stays first order and zero order appears as half order. A Michaelis–Menten enzyme that is saturated at the surface ($\beta \gg 1$) under strong limitation therefore shows $v_{obs}\propto S_s^{1/2}$.

Surface-bound enzyme with Michaelis–Menten kinetics and a liquid film. The plateau is still $V_{max}^{\prime }$; half of it needs $S_s=K_m$ at the surface, and the film must then carry $V_{max}^{\prime }/2$, so $k_L(S_b-K_m)=V_{max}^{\prime }/2$. This bulk half-saturation concentration is what the measured curve shows as its apparent $K_m$. Because the whole curve is not exactly Michaelis–Menten, a $K_m$ fitted by linearisation depends on the substrate range used.

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.