--%>

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 : 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 : Molarity A solution has volume 200ml

    A solution has volume 200ml and molarity 0.1.if it is diluted 5times then calculate the molarity of reasulying solution and the amount of water added to it.

  • Q : What are halogen oxoacids? Fluorine

    Fluorine yields only one oxyacid, hypo

  • Q : Neutralization of benzoic acid Choose

    Choose the right answer from following. How many grams of NaOH will be required to neutralize 12.2 grams of benzoic acid : (a) 40gms (b) 4gms (c)16gms (d)12.2gms

  • Q : Osmotic Pressure The O.P. (Osmotic

    The O.P. (Osmotic Pressure) of equimolar solution of Urea, BaCl2 and AlCl3, will be in the order:(a) AlCl3 > BaCl2 > Urea  (b) BaCl2 > AlCl3 > Urea  (c) Urea > BaCl2<

  • Q : How to calculate solutions ionic

    Transference numbers and molar conductors can be used to calculate ionic mobilities. This tables under is giving the transference numbers for positive ions at 25 degree C and the values obtained by extrapolation to infinite dilution:

    Q : How reactive is Trimethylindium towards

    Illustrate the reason, how reactive is Trimethylindium towards oxygen and water?

  • Q : Crystals of covalent compounds Crystals

    Crystals of the covalent compounds always contain:(i) Atoms as their structural units  (ii) Molecules as structural units  (iii) Ions held altogether by electrostatic forces (iv) High melting pointsAnswer: (i)

  • Q : Dipole moment Elaborate a dipole moment

    Elaborate a dipole moment?

  • Q : What are biodegradable polymers?

      These are polymers that can be broken into small segments by enzyme-catalysed reactions. The required enzymes are produced by microorganism. It is a known fact that the carbon-carbon bonds of chain growth polymers are inert to enzyme-catalysed reactions, and hence they are non biod