--%>

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 : Problems related to entropy change A)

    A) Two compartments each of 1 m3 capacity are joined by a valve and insulated from the surroundings and from one another. One compartment has saturated steam at 683.6 kPa and the other contains steam at the same temperature but at a pressure of 101.3 kPa. T

  • Q : Acid Solutions Choose the right answer

    Choose the right answer from following. Volume of water needed to mix with 10 ml 10N NHO3 to get 0.1 N HNO3: (a) 1000 ml (b) 990 ml (c) 1010 ml (d) 10 ml

  • Q : Concentration of Sodium chloride

    Provide solution of this question. If 25 ml of 0.25 M NaCl solution is diluted with water to a volume of 500ml the new concentration of the solution is : (a) 0.167 M (b) 0.0125 M (c) 0.833 M (d) 0.0167 M

  • Q : P- block why pentahalids are more

    why pentahalids are more covalent than tetrahalids

  • Q : Problem on Molar solution Can someone

    Can someone please help me in getting through this problem. 2.0 molar solution is acquired, when 0.5 mole solute is dissolved in: (i) 250 ml solvent (ii) 250 g solvent (iii) 250 ml solution (iv) 1000 ml solvent

  • Q : Whether HCl is a base or an acid

    Whether HCl is a base or an acid? Briefly state your comments?

  • Q : Explain various chemicals associated

    During processing of food, several chemicals are added to it to augment its shelf life and to make it more attractive as well. Main types of food addi

  • Q : Problem on decinormal strength Can

    Can someone please help me in getting through this problem. How many grams of dibasic acid (having mol. wt. 200) must be present in 100ml  of its aqueous solution to provide decinormal strength: (i) 1g  (ii)2g  (iii) 10g  (iv) 20g<

  • Q : Carnot cycle show how a mathematical

    show how a mathematical definition of entropy can be obtauined from a consideration of carnot cycle?

  • Q : Molarity of Sodium hydroxide Select the

    Select the right answer of the question. Molarity of 4% NaOH solution is : (a) 0.1M (b) 0.5M (c) 0.01M (d) 0.05M