--%>

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 : Which solution will have highest

    Which solution will have highest boiling point:(a) 1% solution of glucose in water  (b) 1% solution of sodium chloride in water  (c) 1% solution of zinc sulphate in water  (d) 1% solution of urea in waterAnswer: (b) Na

  • Q : Problem on making solutions The weight

    The weight of pure NaOH needed to made 250cm3 of 0.1 N solution is: (a) 4g  (b) 1g  (c) 2g  (d) 10g Choose the right answer from above.

  • Q : Help 1) Chromium(III) hydroxide is

    1) Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to support your answer. 2) Explain how dissolving t

  • Q : Some basic concepts of chemistry an

    an atom of an element is 10.1 times heavier than the mass of a carbon atom.What is its mass in amu?

  • Q : Mole fraction and Molality Select the

    Select the right answer of the following question.What does not change on changing temperature : (a) Mole fraction (b) Normality (c) Molality (d) None of these

  • Q : Molarity of sodium hydroxide Can

    Can someone please help me in getting through this problem. Determine the molarity of a solution having 5g of sodium hydroxide in 250ml  solution is: (i) 0.5  (ii) 1.0  (iii) 2.0   (d) 0.1Answer: The right answer i

  • Q : Gibberella fusarium in bioremediation

    in bioremediation gibberella fusarium is used to break down____?

  • Q : Production of alcoholic drinks give all

    give all physical aspects in the production of alcohol

  • Q : Explosions produce carbon dioxide

    Illustrate all the explosions produce carbon dioxide?

  • Q : 6. 20 gm of hydrogen is present in 5

    6. 20 gm of hydrogen is present in 5 litre vessel. The molar concentration of hydrogen is