Q. 1. Ammonium sulfate fractionation. You are purifying protein A and B from a 1
ID: 197164 • Letter: Q
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Q. 1. Ammonium sulfate fractionation. You are purifying protein A and B from a 1.4 liters of bacterial cell extract using salting out technique. Protein A requires 25% saturation, whereas protein B requires 55% saturation of the salt to precipitate. Calculate the amount of ammonium sulfate required to achieve 25% and 55% saturation of the cell extract. Q. 2. Dialysis. A purifled protein is in a sodium phosphate buffer at pH 7.4 with 500mM NaCI. A sample (1mL) of the protein is placed in a dialysis tube and dialyzed against 1L of the same phosphate buffer with 0 mM NaCl. Small molecules and ions such as Na", Cl and phosphate can diffuse across the dialysis membrane, but the protein cannot. Once the dialysis has come to equilibrium, what is the concentration of NaCI in the protein sample? Assume no volume changes occur in the sample during the dialysis. a. b. If the original 1ml sample were dialyzed twice, successively, against 100mL of the same Na phosphate buffer with Q. 3. lon exchange chromatography. At pH 7.0, in what order the following three peptides you would expect to be Table 1 OmM NaCI, what would be the final NaCI concentration in the sample? eluted from a column filled with carboxymethyl sephacryl (CM-sephacryl)? Peptide C 10 Amino acid Ala, % Arg, % Asn, % Asp, % Cys, % Gln, % Glu, % Gly, % His, % ile, % Leu, % Lys, % Met, % Phe, % Pro, % Ser, % Thr, % Trp, % Tyr, % Val, % Peptide A 10 10 Peptide EB 0 10 0 0 10 10 10 0 5 0 10 10 10 0 0 10 0 10 0 5Explanation / Answer
Q 2
Dialysis:
Principile behind the dialysis is diffusion:
- Diffusion is defined as the movement of molecules from the area of high concentration to low concentration via semipermeable membrane to reach equilibrium .once equilibrium is reached there is no net movement of molecules.
a. From the given data 1 ml of sample in buffer with 500 mM Nacl was dialyzed against 1 l of buffer without Nacl. and also it is given that Nacl will pass into and outside the dialysis unit.
when 1ml of sample dialysed against 1000ml of buffer it will reduce the concentration of dialyzed molecule by a factor of 1000. i.e
500mM/1000ml= 0.5 mM (or)
Using the formula C1*V1=C2*V2
C1= concentration of sample 1
V1 = volume of sample 1
C2=concentration of sample 2
V2=volume of sample 2
Using the above formula
C1= 500mM
V1=1ml
C2=X
V2= 1000ml
substituting the above values
500mM *1ml = x * 1000ml
500*1/1000=0.5mM
Conc of Nacl in protein sasmple after equilibrium is 0.5 mM
b.if 1 ml of sample dialysed twice with 100 ml of buffer (0mM Nacl)
After two successive dialysis in 100 ml of buffer ,concentration of Nacl will be reduced to a factor of 100*100
500mM/100*100=0.05mM
(or)
Using formula
After first dialysis
C1*V1=C2*V2
C1= 500mM
V1=1ml
C2=X
V2= 100ml
500mM*1=X*100
C2 = 5mM
After second dialysis
C1= 5mM
V1=1ml
C2=X
V2= 100ml
5mM*1=X*100
C2=0.05mM
Concentation of Nacl After two successive dialysis is 0.05mM
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