Downstream formulas

In U/mg, where 1 U (enzyme unit) is the amount of enzyme that converts 1 μmol of substrate per minute under the stated assay conditions. It rises as contaminating protein is removed and is the working measure of purity for an enzyme or other active protein.

Yields use TOTAL activity (or total product mass), never specific activity. The overall yield is the product of the step yields.

Uses SPECIFIC activity; dimensionless, equal to 1 for the crude extract. The fold for one step alone is the ratio of specific activities after and before that step, and step folds multiply.

$R$ = protein released at time $t$, $R_m$ = maximum releasable protein, $k$ = first-order rate constant (for example min${}^{-1}$). Natural log. Equivalently $R/R_m=1-e^{-kt}$.

$N$ = number of passes, $P$ = operating pressure, $a$ = empirical pressure exponent that depends on the organism and its growth conditions, $k$ = rate constant (its numerical value depends on the units of $P$ and on temperature). Release is first order in the number of passes.

$d$ = particle diameter (m), ${\rho }_p$ and ${\rho }_f$ = particle and fluid densities (kg/m³), $\mu$ = fluid viscosity (Pa s), $g$ = 9.81 m/s². Valid for a rigid sphere settling freely in a dilute suspension (no hindered settling) when the particle Reynolds number $R{e}_p={\rho }_{f}vd/\mu$ is below about 1 (creeping flow). If ${\rho }_p<{\rho }_f$ the particle rises.

$r$ = radial position (m), $\omega$ = angular speed (rad/s); Stokes' regime assumed. The velocity grows with radius, so the particle speeds up as it moves outward.

$n$ in rpm, $\omega$ in rad/s, $r$ in m. $Z$ is quoted as 'times g': the factor by which the centrifuge speeds up settling compared with gravity. In practical units, $Z=1.118\times 10^{-5}r{n}^2$ with $r$ in cm and $n$ in rpm.

$\Sigma$ (m²) is the area of a gravity settling tank that would do the same separation; $Q$ is the feed rate at which particles of gravity settling velocity $v_g$ are just completely captured. For the same broth and the same separation, $Q/\Sigma$ is kept constant on scale-up between machines of the same type; comparing different types needs an efficiency correction.

$b$ = bowl length, $r_0$ = radius of the liquid surface (inner), $r_1$ = radius of the bowl wall (outer). Derived from Stokes' law, assuming uniform (plug) axial flow, for complete capture of a particle entering at the liquid surface. For a thin liquid layer ($r_0\to r_1=r$) it reduces to $2\pi rb\times {\omega }^{2}r/g$: the wall area times the G-number.

$N$ = number of discs ($N-1$ settling channels), $r_1$ and $r_0$ = outer and inner radii of the discs, $\theta$ = half-cone angle, measured between the disc surface and the axis of rotation. Same complete-capture basis as the tubular-bowl expression.

$v$ = sedimentation velocity at radius $r$: the velocity per unit centrifugal field. $s$ has units of time and is quoted in svedbergs, 1 S $=10^{-13}$ s. The second form is Stokes' law for a sphere. For a macromolecule, $s=M(1-\bar{v}{\rho }_f)/(N_{A}f)$, where $M$ = molar mass, $\bar{v}$ = partial specific volume, $f$ = frictional coefficient and $N_A$ = Avogadro's number, so $s$ depends on mass, density and shape.

$R$ = gas constant (8.314 J mol${}^{-1}$ K${}^{-1}$), $T$ = absolute temperature (K), $D$ = diffusion coefficient (m²/s), $\bar{v}$ = partial specific volume (m³/kg), ${\rho }_f$ = solvent density (kg/m³), $s$ in seconds; $M$ then comes out in kg/mol. It follows from $s=M(1-\bar{v}{\rho }_f)/(N_{A}f)$ and $D=RT/(N_{A}f)$, with $s$ and $D$ measured in the same solvent and at the same temperature.

From $dr/dt=s{\omega }^{2}r$ at constant $s$: the time for a particle to travel from $r_{\min }$ (top of the liquid) to $r_{\max }$ (bottom of the tube). The clearing factor $k$ of a rotor at speed $n$ (rpm) gives the pelleting time in hours when $s$ is in svedbergs. For the same particle, run times on two rotors scale as $t_1/t_2=k_1/k_2$.

$V$ = filtrate volume (m³), $A$ = filter area (m²), $\Delta P$ = pressure drop (Pa), $\mu$ = filtrate viscosity (Pa s), $\alpha$ = specific cake resistance (m/kg), $c$ = mass of dry cake solids per volume of filtrate (kg/m³), $R_m$ = filter-medium resistance (m${}^{-1}$). $\alpha$ is constant for an incompressible cake; for a compressible one see below.

Integrated form for constant $\Delta P$ with $V=0$ at $t=0$. A plot of $t/V$ against $V$ is a straight line: the slope gives $\alpha$ and the intercept gives $R_m$. With $R_m$ negligible, $t=\mu \alpha c{V}^2/(2A^2\Delta P)$, so $V\propto \sqrt{t}$.

At a constant filtrate rate $Q=V/t$ (m³/s) and constant $\alpha$ (incompressible cake), the pressure drop must rise linearly with time.

$s$ = compressibility index: $s=0$ for an incompressible cake, typically between 0 and 1 for the compressible cakes formed by cells, so raising the pressure gives less gain in rate than an incompressible cake would.

$J$ = permeate volume per membrane area per time (m³ m${}^{-2}$ s${}^{-1}$, often quoted in L m${}^{-2}$ h${}^{-1}$); $R_m$ = membrane resistance, $R_f$ = resistance of the fouling or polarised layer; $P_{\text{in}}$ and $P_{\text{out}}$ = feed-side pressures at the module inlet and outlet. When the retained solute exerts an osmotic pressure, the driving force becomes $\Delta P_{TM}-\sigma \Delta \pi$ ($\sigma$ = reflection coefficient).

$k$ = boundary-layer mass-transfer coefficient (m/s), $C_G$ = gel (wall) concentration, $C_B$ = bulk concentration. In this regime flux is independent of pressure. It is the film model $J=k\ln [(C_w-C_p)/(C_B-C_p)]$ with full rejection ($C_p=0$) and $C_w=C_G$.

$C_p$ = permeate concentration, $C_B$ = bulk (retentate) concentration. $R=1$: fully retained; $R=0$: passes freely. The intrinsic rejection uses the wall concentration $C_w$ in place of $C_B$.

Well-mixed retentate reduced in volume from $V_0$ to $V$ at constant observed rejection $R$. The fraction of solute still in the retentate is $CV/(C_0{V}_0)=(V/V_0{)}^{1-R}$.

Buffer is added as fast as permeate leaves, so the well-mixed retentate volume $V$ stays constant; $N_D$ = number of diavolumes. A freely permeating solute ($R=0$, such as a salt being exchanged out) falls to $e^{-N_D}$ of its starting level: about 5 % after 3 diavolumes and under 1 % after 5.

Equilibrium concentration in the extract (or top) phase divided by that in the raffinate (or bottom) phase. Treated as constant for dilute solutions.

$K_0$ = partition coefficient of the uncharged form, assuming only that form enters the organic phase; for the base, $\text{p}{K}_a$ is that of its conjugate acid ${\text{BH}}^{+}$. At pH = p$K_a$ the apparent coefficient is $K_0/2$; once the pH is more than about one unit on the ionising side of the p$K_a$, each further pH unit cuts it roughly tenfold.

The ratio of solute AMOUNTS in the two phases at equilibrium; $V_E$ and $V_R$ = extract and raffinate volumes (or volumetric flow rates in continuous operation).

For an ATPS the yield in the top phase is $Y_T=E/(1+E)$ with $E=K{V}_T/V_B$, where $V_T$ and $V_B$ are the top and bottom phase volumes.

Each stage contacts the raffinate with an equal volume of fresh solvent; $E$ is the extraction factor of one stage (based on the solvent used in that stage).

Feed enters stage 1 and solvent enters stage $n$; $E$ is based on the total solvent and feed flows, with constant $K$ and immiscible phases. For $n=1$ it reduces to $1/(1+E)$.

Assumes a monolayer on identical, independent sites. $q_{\max }$ = monolayer capacity; $K$ = association constant (for example L/g). Written equivalently $q=q_{\max }C/(K_d+C)$ with $K_d=1/K$, so $q=q_{\max }/2$ at $C=K_d$. At low $C$ it becomes linear, $q=q_{\max }KC$. Linearised: $1/q=1/q_{\max }+1/(q_{\max }KC)$.

Empirical, for heterogeneous surfaces; it has no saturation limit. $n>1$ indicates favourable adsorption. Linearised: ${\log }_{10}q={\log }_{10}{K}_F+(1/n){\log }_{10}C$, so a log-log plot has slope $1/n$.

$V$ = liquid volume, $m$ = mass of adsorbent (initially free of solute), $C_0$ = initial concentration. The equilibrium point $(C^{\ast },q^{\ast })$ lies on both this straight operating line and the isotherm.

$S$ = protein solubility; $\beta$ = intercept (depends on the protein, pH and temperature); $K_s$ = salting-out constant; $I$ = ionic strength (mol/L), with $c_i$ the molar concentration and $z_i$ the charge of ion $i$. Base-10 logarithm. Applies in the salting-out region (high $I$).

$t_0$ = elution time of an unretained species (dead time); $k^{\prime }$ = retention (capacity) factor; $K$ = distribution coefficient (stationary/mobile concentration); $V_s$ and $V_m$ = stationary- and mobile-phase volumes. Retention volumes obey the same relation, $V_R=V_m(1+k^{\prime })$.

Component 2 elutes later. $\alpha =1$ means no separation is possible, however efficient the column. (Not the specific cake resistance of filtration.)

For a Gaussian peak: $w_b$ = baseline width between the tangents ($=4\sigma$), $w_{1/2}$ = width at half height ($5.54=8\ln 2$). $t_R$ and $w$ in the same units; $L$ = column length. Smaller $H$ means a more efficient column.

$R_s=1.5$ gives baseline separation of two Gaussian peaks of similar size (about 99.7 % separation); $R_s=1.0$ still leaves a small overlap (about 98 % separation).

$N$ is the plate number for the later peak. Resolution grows only as $\sqrt{N}$, that is as $\sqrt{L}$ at fixed plate height.

$u$ = linear velocity of the mobile phase. $A$ = eddy diffusion (independent of $u$; smaller for small, uniformly packed particles), $B/u$ = longitudinal molecular diffusion, $Cu$ = resistance to mass transfer between the phases (also smaller for smaller particles). Large proteins diffuse slowly, so the $C$ term dominates and their optimum velocity is low.

$V_e$ = elution volume, $V_0$ = void volume (liquid outside the beads, found with a fully excluded marker such as blue dextran), $V_t$ = total bed volume. $K_{av}=0$ for a fully excluded molecule, approaching 1 for one that permeates all the pores. (A related coefficient, $K_d=(V_e-V_0)/V_i$, uses the internal pore volume $V_i$ and runs from 0 to 1 exactly.) Within the fractionation range, $K_{av}$ is approximately linear in log(molecular mass) for molecules of similar shape, such as globular proteins, which is the basis of the calibration curve.

$m_s$ = mass of dry solid, $A$ = drying area, $N_c$ = constant drying rate (kg water m${}^{-2}$ h${}^{-1}$), $X_0$ = initial moisture content (dry basis). Applies while $X>X_c$.

Assumes the rate falls linearly from $N_c$ at $X_c$ to zero at $X^{\ast }$; $X_f$ = final moisture content ($X_f>X^{\ast }$). Total drying time $=t_c+t_f$.