--%>

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 : Question on Mole fraction Mole fraction

    Mole fraction of any solution is equavalent to: (a) No. of moles of solute/ volume of solution in litter (b) no. of gram equivalent of solute/volume of solution in litters (c) no. of  moles of solute/ Mass of solvent in kg (d) no. of moles of any

  • Q : Vapour pressure Vapour pressure of

    Vapour pressure of methanol in water Give me answer of this question. An aqueous solution of methanol in water has vapour pressure: (a) Equal to that of water (b) Equal to that of methanol (c) More than that of water (d) Less than that of water

  • Q : Finding Active mass of urea Can someone

    Can someone please help me in getting through this problem. 10 litre solution of urea comprises of 240 gm urea. The active mass of urea is: (i) 0.04 (ii) 0.02 (iii) 0.4 (iv) 0.2

  • Q : Problem on MM equation How to obtain

    How to obtain relation between Vm and Km,given k(sec^-1) = Vmax/mg of enzyme x molecular weight x 1min/60 sec S* = 4.576(log K -10.753-logT+Ea/4.576T).

  • Q : Molecular Symmetry Types The number of

    The number of molecular orbitals and molecular motions of each symmetry type can be deduced. Let us continue to use the C2v point group and the H2O molecule to illustrate how the procedure develop

  • Q : Problem based on normality Choose the

    Choose the right answer from following. NaClO solution reacts with H2SO3 as,. NaClO + H2SO3→NaCl+ H2SO4. A solution of NaClO utilized in the above reaction contained 15g of NaClO per litre. The

  • Q : The three facts on the evaporation

    Describe briefly the three facts on the evaporation?

  • Q : Explosions produce carbon dioxide

    Illustrate all the explosions produce carbon dioxide?

  • Q : P block bif3 is ionic while other

    bif3 is ionic while other trihalides are covalent in nature

  • Q : Question based on relative lowering of

    Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre