Your protein, with a molecular weight of 18 kDa is suspended in 1mL of a 0.1 M T

Question

Your protein, with a molecular weight of 18 kDa is suspended in 1mL of a 0.1 M Tris

buffer that contains 6 M guanidine HCl. In order to fold this protein you decide to

remove the majority of guanidine HCl via dialysis in a 1000 molecular weight cut off

membrane (this means that the membrane is permeable to dissolved molecules <1000 Da

and is not permeable to dissolved molecules >1000Da). You can either dialyze the

protein into

(i) a beaker with 3L of 0.1M Tris buffer without guanidine HCl or

(ii) a beaker with 1L of 0.1 M Tris buffer followed by dialysis into another beaker

with 1L of 0.1 M Tris buffer.

Which method of dialysis would give you the lowest concentration of guanidine HCl?

Use approximate concentrations of guanidine HCl to back your argument.

Explanation / Answer

Second method will give better result. This is because, the dialysis works on the principel of diffussion of solute form higher concentration to a lower concentration.The first dialysis will reduce the salt concentraion approcibly and the amont remaining will undergo dialysis again. Total guanidine HCl present is 0.6 milimoles. In the first case it will be diluted to 3L. So the final concentration will be 2x10-4 M.Then the dialysis will stop. But in second case, in the first instance it will be diluted to 6x10-4 M. This one will undergo the second dialysis. where, 6x10-4 M will be diluted again to 6x10-7 M.    

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