A purified protein is in a HEPES (N-(2-hydroxyethyl)piperazine-N-(2-ethanesulfon

Question

A purified protein is in a HEPES (N-(2-hydroxyethyl)piperazine-N-(2-ethanesulfonic acid) buffer at pH 7 with 400 mM NaCl. A sample (1 mL) of the protein solution is placed in a tube made of dialysis membrane and dialyzed against 2.0 L of the same HEPES buffer with 0 mM NaCI. Small molecules and ions (such as Na, Cl, and HEPES) can diffuse across the dialysis membrane, but the protein cannot. a) Once the dialysis has come to equilibrium, what is the concentration of NaCl in the protein sample? Assume no volume changes occur in the sample during the dialysis. Number mM (b) If the original 1 mL sample were dialyzed twice, successively, against 100 mL of the same HEPES buffer with 0 mM NaCl, what would be the final NaCl concentration in the sample? Number NaCl|- mM

Explanation / Answer

Dialysis is the process of Selctive difusion for small ions/ molecuses across the membrane. In your experiments only samll molecules will difuse through membrane but not protein sample.

Now in protein (1mL) sample, 400mM NaCl and HEPEs buffer and sepearted by Dialysis membrane from 2 litre of 0.0 mM NaCl with hepes

Now, The Na+ and Cl- ions difuse towords low concentration side. after equilibrium, the ions on both side are in same concentration.

Now,

a) Answer

Apply dilution formula, M1V1=M2V2 (2001 ml)

Now, Vf=V1+V2 , Both 1 ml and 2L ( you can even ignore and consider it 2 L)

400mM(1mL)=Mf Vf

Mf= 400mM(1mL)/2001mL

Mf= 0.1999 mM

The final concentartion NaCl is 0.1999 mM

b) If we do dialysis twice by 100 ml volume with 0.0mM NaCl buffer

Similar calculation but twice

first 100 ml

Mf= 400mM(1mL)/101mL

Mf= 3.96 mM

Second 100ml

Now we have to take salt concentration as 3.96 mM

So, Mf= 3.96mM(1mL)/100 mL

Mf= 0.0396 mM

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