--%>

Sedimentation and Velocity

The first method begins with a well defined layer, or boundary, of solution near the center of rotation and tracks the movement of this layer to the outside of the cell as a function of time. Such a method is termed a sedimentary velocity experiment.


A particle of mass m at a distance x from center of rotation experiences a force given by 

ƒcentrif = m'xω2

Where w is the angular velocity in radiation per second m' is the distance effective mass of the solute particle, i.e. the actual mass corrected for the new effect of the solvent.

To express this buoyancy effect, we first recognize that v the specific volume of the solute, is the mass of 1 g of the solute. The volume of m g of solute is mv, and the mass of this volume of solvent is m of the solute is m - mvp = mj (1 - vp). We now can rewrite equation as:

Centrif = m (1 - vp) xω
2

Equating these two force expressions leads us to the constant drif velocity. A rearrangement of the equality:

M (1 - v) xω2 = 6∏r? dx/dt

Equating these two force expressions leads that collect the dynamic variables gives:

Dx/dt/xω2 = m (1 - vp)/6∏r?


The collection of dynamic terms on the left side of equation describes the results of sedimentation velocity experiments. This collection (dx/dt) xw2 can be looked on as the velocity with which the solute moves per unit centrifugal force. The sedimentation coefficient S is introduced as:

S = dx/dt/xω2

The experimentation results can therefore be tabulated as values of S. the value of S for many macromolecules is of the order of 10-13 has therefore been introduced, called a Svedberg, in honor of T. Svedberg, who did much of the early work with the ultracentrifuge.

Molar mass: s = dx/dt/xω2 = m )1 - vp)/ 6∏r?

Rearrangement and multiplication by Avogadro's number give:

M = Nm = 6∏r?NS/ 1- vp

Now the troublesome terms involving ? and r can be replaced by their effective values appear in the measurable values D of equation, to give the desired result:

M = RTS/ D (1 - vp)

Thus measurements of the substances of the sedimentation and diffusion coefficients and of the solvent and solute allow the deduction of the molar mass for a few macromolecules. The necessary data for such calculations for a few macromolecular materials are included.

A particular advantage of the sedimentation velocity technique is that a macromolecular solution containing two or more types of macromolecules is separated according to the molecular masses of the components. The type of sedimentation diagrams obtained for a system containing a number of macromolecular species.

Density gradient: better resolution can be obtained by allowing the sedimentation to occur in a density gradient solution, prepared, for example, by filling the centrifuge tube layer by layer with solutions of decreasing sucrose concentration. As the macromolecular substance or mixture of substances is centrifuged, it moves through a solvent with gradually increasing density. The result is more stable macromolecular zones and a better "spectrum" of the components. The technique is thus a modification of the sedimentation velocity method.

   Related Questions in Chemistry

  • Q : Structure of a DNA molecule Elaborate

    Elaborate the structure of a DNA molecule?

  • Q : What is electrolysis? Explain with

    Passage of a current through a solution can produce an electrolysis reaction.Much additional information on the properties of the ions in an aqueous solution can be obtained from studies of the passage of a direct current (dc) through a cell containing a s

  • Q : What are ion selective electrodes? Ion

    Ion Selective Electrodes An ion selective membrane can be used to form an electrochemical cell whose emf depends on the concentration of that ion. Before we proceed to an important application of emf measurements, brie

  • Q : Concentration of Calcium carbonate Help

    Help me to go through this problem. 1000 gms aqueous solution of CaCO3 contains 10 gms of carbonate. Concentration of the solution is : (a)10 ppm (b)100 ppm (c)1000 ppm (d)10000 ppm

  • Q : Molarity Give me answer of this

    Give me answer of this question. If 20ml of 0.4N, NaoH solution completely neutralises 40ml of a dibasic acid. The molarity of the acid solution is:(a) 0.1M (b) 0.2M (c)0.3M (d)0.4M

  • Q : Strength of any solution Give me answer

    Give me answer of this question. A solution contains 1.2046 x 1024 hydrochloric acid molecules in one dm3 of the solution. The strength of the solution is: (a) 6 N (b) 2 N (c) 4 N (d) 8 N

  • Q : Unit of mole fraction Provide solution

    Provide solution of this question. Unit of mole fraction is: (a) Moles/litre (b) Moles/litre2 (c) Moles-litre (d) Dimensionless

  • Q : What are lattices and unit cells? The

    The repeating, atomic level structure of a crystal can be represented by a lattice and by the repeating unit of the lattice, the unit cell.It was apparent very early in the study of crystals that the shapes of crystals stem from an ordered array of smaller

  • Q : Chemistry brief discription of relative

    brief discription of relative lowering of vapour pressure

  • Q : Why medications include the hcl Why do

    Why do various medications include the hcl? Describe briefly?