--%>

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 : Describe characteristics of halides and

    Halides characteristics

  • Q : Determining of normality of sodium

    Can someone please help me in getting through this problem. The normality of a solution of sodium hydroxide 100 ml of which includes 4 grams of NaOH is: (a) 0.1 (b) 40 (c) 1.0 (d) 0.4

  • Q : Excel assignment I want it before 8 am

    I want it before 8 am tomorow please. I am just wondering how much is going to be ?

  • Q : Question relatede to calculate molarity

    Select the right answer of the question. What is molarity of a solution of HCl that contains 49% by weight of solute and whose specific gravity is 1.41 : (a) 15.25 (b) 16.75 (c) 18.92 (d) 20.08

  • Q : Kinds of insulators Describe all the

    Describe all the kinds of insulators which are present?

  • Q : Molarity of cane sugar solution 171 g

    171 g of cane sugar (C12H22O11)  is dissolved in one litre of water. Find the molarity of the solution: (i) 2.0 M (ii) 1.0 M (iii) 0.5 M (iv) 0.25 M Choose the right answer from above.

  • Q : Molecular Properties Symmetry Molecular

    Molecular orbitals and molecular motions belong to certain symmetry species of the point group of the molecule.Examples of the special ways in which vectors or functions can be affected by symmetry operations are illustrated here. All wave functions soluti

  • Q : Liquid Vapour Free Energies The free

    The free energy of a component of a liquid solution is equal to its free energy in the equilibrium vapour.Partial molal free energies let us deal with the free energy of the components of a solution. We use these free energies, or simpler concentration ter

  • Q : Dipole attractions-London dispersion

    Describe how dipole attractions, London dispersion forces and the hydrogen bonding identical?

  • Q : Calculate molarity of a solution

    Provide solution of this question. Molarity of a solution prepared by dissolving 75.5 g of pure KOH in 540 ml solution is: (a) 3.05 M (b) 1.35 M (c) 2.50 M (d) 4.50 M