--%>

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 : Symmetry Elements The symmetry of the

    The symmetry of the molecules can be described in terms of electrons of symmetry and the corresponding symmetry operations.Clearly some molecules, like H2O and CH4, are symmetric. Now w

  • Q : Molarity of solution Help me to go

    Help me to go through this problem. When 7.1gm Na2SO4 (molecular mass 142) dissolves in 100ml H2O , the molarity of the solution is: (a) 2.0 M (b) 1.0 M (c) 0.5 M (d) 0.05 M

  • Q : Donnan Membrane Equilibria The electric

    The electric charge acquired by macromolecules affects the equilibrium set up across a semipermeable membrane.Laboratory studies of macromolecule solutions as in osmotic pressure and dialysis studies confine the macromolecules to one compartment while allo

  • Q : Mole fraction of Carbon dioxide Choose

    Choose the right answer from following. If we take 44g of CO2 and 14g of N2 what will be mole fraction of CO2 in the mixture: (a) 1/5 (b) 1/3 (c) 2/3 (d) 1/4

  • Q : Determining concentration in ppm A 500

    A 500 gm tooth paste sample has 0.2g fluoride concentration. Determine the concentration of F in terms of ppm level: (a) 250 (b) 200 (c) 400 (d) 1000Answer: (c) F-ions in ppm = (0.2/500) x 106 = 400

  • Q : How to calculate solutions molar

    The contribution of an electrolyte, or an ion electrolyte, is reported as the molar of a conductance. The definition of the molar conductance is based on the following conductivity cell in which the electrodes are 1 m apart and of sufficient area that th

  • Q : Explain Rotational Vibrational Spectra

    The infrared spectrum of gas samples shows the effect of rotational-energy changes along with the vibrational energy change.As we know from the interpretations given to thermodynamic properties of gases, gas molecules are simultaneously rotating and vibrating. It follows that an absor

  • Q : Ionic radius of chloride ion The edge

    The edge length of the unit cell of Nacl crystal lattice is 552 pm. If ionic radius of sodium ion is 95. What is the ionic radius of chloride ion:(a) 190 pm  (b) 368 pm  (c) 181 pm  (d) 276 pm     <

  • 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 : Describe Point Groups. For any

    For any symmetric object there is a set of symmetry operations that, together, constitute a mathematical group, called a point group.It is clear from the examples that most molecules have several elements of symmetry. The H2O