Is sedimentation coefficient constant and why?

Question

Biochemistry by Voet writes the sedimentation rate of a macromolecule in a centrifuge this way: $$s=\frac{v}{\omega^2r}=\frac{M(1-\bar{V}_\rho)}{Nf}$$ where $\omega$ is the angular velocity of the rotor, $r$ is the position of the macromolecule with respect to the rotational axis, $M$ is its molar mass, $\bar{V}_\rho$ its partial specific volume, $N$ the number of moles, and $f$ the frictional coefficient.

The book further discusses the uses of $s=M(1-\bar{V}_\rho)/(Nf)$ and the information it reveals. Since they haven’t discussed any changes to $s$, it seems to me that they are holding it constant, but why should that be? If $s$ is a function of $r$, and $r$ changes, then $s$ should change too. Is my reasoning wrong? Is $s$ considered to be constant and, if so, why?

Answer 1

Sedimentation due to Gravity

The sedimentation coefficient of a particle falling in a fluid due to gravity is easier to understand. It is the ratio of the settling velocity ($v_{t}$) to the acceleration causing the settling($a_{g}$).

$$s=\frac{v_{t}}{a_{g}}$$

In fluids on earth $a_{g}$ is acceleration due to gravity which is known to be constant.

$v_{t}$ , which is also referred to as terminal velocity, is constant. It is constant because the force driving the acceleration of the particle through a fluid ($F_{g}$) is equal in magnitude and in an opposite direction to the drag force ($F_{d}$). Since these two forces cancel out the net force ($F_{n}$) is 0.

$$F_{g} - F_{d} = F_{n} = 0$$

There is no net acceleration $a_{n}$ since $F_{n}=m_{p}a_{n}=0$. The particle mass ($m_{p}$) cannot be zero so there is no net acceleration of the particle, thus the terminal velocity is constant.

A ratio of two constants are constant so the sedimentation coefficient is constant.

Sedimentation due to Centrifugal Acceleration

Radial acceleration, $a_{r}$, equals the angular velocity $\omega$ squared times the radius, $r$.

$$a_{r} =\omega^{2}r$$

If the angular velocity is constant, which it is for most of centrifugation, and the radius is constant, then the radial acceleration, like the acceleration due to gravity, is constant.

The sedimentation velocity in centrifugation, just as in gravity settling, is constant. It is constant because the drag force of the particle in the fluid is balanced by the centrifugal forces generated by the centrifuge. A centrifuge can generate forces several times higher than gravity so settling occurs much faster during centrifugation.

Again, the ratio of two constants are constant so the sedimentation coefficient is constant.

The other equation looks like it is a derivative of Stoke's law, which is used to determine the drag force of a particle in a fluid.

Answer 2

When a particle of mass m moves through a fluid under the action of a force, such as gravity, then a steady or terminal velocity $dz/dt$ can be reached as the frictional effect of the fluid balances the effect of the applied force. An example could be dropping a ball-bearing into a tall column of thick oil. In the case of a molecule a far greater force that that due to gravity is needed and this can be found in a centrifuge.

The force on a particle is reduced from simply mass times acceleration as in a fluid the particle will have some buoyancy. The force due to acceleration $a$ becomes $F=(1-\rho/\rho_m)ma$ where $\rho$ is the fluid's density and $\rho_m = M/V$ is the density of the particle where M is its molar mass and V the partial molar volume.

This force is balanced by the frictional force $f(dz/dt)$ where $f = 6\pi\eta r_0$ is the (Stokes) friction for a sphere of radius $r_0$ in a fluid of viscosity $\eta$ and has units $\pu {Pa\cdot s\cdot m}$. The friction is related to the diffusion coefficient ($\pu{m^2s^{-1}}$) as $D=kT/f$.

Equating the two forces produces $$ f\frac{dz}{dt}=(1-\rho/\rho_m)ma $$

In a centrifuge the rotational motion produces a force far greater than that due to gravity alone and on a mass m produces $F=(1-\rho/\rho_m)m\omega^2 r$ where r is the distance from the centre of rotation.

$$ f\frac{dz}{dt}=(1-\rho/\rho_m)m\omega^2 r \tag 1$$

The sedimentation constant is defined as $$ s= \left(\frac{1}{\omega^2 r}\right )\frac{dz}{dt}$$

and has units of $\pu{10^{-13}s^{-1}}$ (Svedberg's). This is a constant because $dz/dt$ is the steady terminal velocity measure at positoin r. Typical values for proteins are $2$ to $200\cdot \pu{10^{-13}s^{-1}}$

Equation $(1)$ can be re-written as

$$f=(1-\rho/\rho_m)m/s$$ which in not particularly useful but substituting for the diffusion coefficient produces

$$ m=\frac{skT}{D(1-\rho/\rho_m)}$$

from which the mass can be obtained.

There is a second method used in centrifugation which is the sedimentation equilibrium method. When equilibrium is reached the diffusion inwards caused by the concentration gradient balances the centrifugal force pushing molecules outwards. If a solute of low molecular weight is centrifuged there will be at equilibrium a concentration gradient, and hence density gradient along the tube.

If a molecule of high molecular mass is now added it will move outwards until its buoyant density is matched by the density of the solution. If a mixture of molecules is added then several bands will appear as each reaches equilibrium.