Delete if easily shocked by maths
klenchin at REMOVE_TO_REPLY.facstaff.wisc.edu
Mon Sep 10 22:40:03 EST 2001
In article , klenchin at REMOVE_TO_REPLY.facstaff.wisc.edu (Dima Klenchin) wrote:
>Michael Witty <mw132 at mole.bio.cam.ac.uk> wrote:
>> one of the reasons I took up biology was that I would never have
>>to do any mathematics. Sadly my careers advisor was economical with the
>>truth and I have a really difficult one. BUT maybe one of you has been
>>here before. Here is the problem:
>>One set of chromatography beads has 0.1mm diameter beads.
>>The other has 1mm diameter beads (approximately).
>>What is the difference in surface area (which could bind protein) for,
>>let us say, 10ml of each of the beads?
>>All this is so I have a rough idea about how much bigger to make the
>>column with coarser beads.
>>Formulae about the surface area of spheres don't seem to help. I know
>>because I have tried and retired with a headache.
>Simple. There are two cases, depending on what type your beads are.
>In both: if N spheres of diameter A fit in a given column diameter, then
>N/10 beads of diameter 10A fit there. The column is 3D entity, therefore
>we are talking about volumes, meaning three dimensions. Therefore,
>where 1 big sphere fits, 1000 10X smaller spheres fit (e.g., 10e3).
>1. Beads are non-porous (e.g. resins). Then binding surface to be
>concerned with is strictly spheres area. Area ~ R^2 => 10X larger
>sphere means 100X larger surface area.
>Thus we have that the same
>column volume occupied by bigger beads will have 1000X beads
>with each 100X larger surface.
Umm, this is horribly written. The above sentense shuld read:
.. the same column volume occupied by 10X larger beads will have
1000X _less number of_ beads, with each having 100X larger surface.
>Thus, the column has to be 1000/100=10
>times bigger to have the same capacity.
>2. Beads are highly porous (e.g. conventional macromolecular matrix
>based on cellulose, dextrane, agarose or PAAG). Then the binding depends
>on total volume occupied by beads. Volume ~ R^3. Then you have 1000
>less beads, each 1000X smaller volume. Therefore, the the column
>capacity is independent on bead size!
> (This is, of cource, first approximation and idealization, but it
>gives you right idea - when dealing with macroporous sorbents,
>don't be bothered too much with particle size if capacity is your
>concern. Other things like resolution and back pressure, OTOH,
>crucially depend on size).
> - Dima
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