Question

Show your work to get credit. 1. Draw the tryptic digestion fragments resulting from the polypeptide. N-ERGDEWKTHR-C Provide a scheme for how these fragments can be purified from one another at pH 5

Answer 1

Answer 1 : We have drawn the tryptic digestion fragments resulting from the polypeptide i.e. N'- ERGDEWKTHR - C'

which is shown as follows :

(1) The first fragment would be : N'- ER - C'

The structure of fragment 1 showing the net charge at pH 5 is shown as follows :

The structure of the first peptide fragment lie. N'-E-R-C) ie N-Glutamic acid - Arginine act along with its net charge at pH s is shown as follows: t-Ex-&-cooo dua CH2 And since we have two positive and two regative "charge on this fragment at pH 5: this net Charge at pH s vould be (+2-2=0 CH2 CH2 NH CH2 code c = NHI M N' terminal - Glutamic Acid - Arginine -c' terminal

(2) The second fragment would be : N'- GDEWK - C'

The structure of fragment 2 showing the net charge at pH 5 is shown as follows :

The structure of the fragment 2 il. N' JGDE WK-C le. N'- Glycine - Aspartic Acid - Glatonic Acid - Trylotofken-lysse along with its net charge at pH 5 is shown as treinal Jollan. И И ир H -cooo oave H си, CH I сил- H CH2 1 CH2 CH2 L си - cool dun N'terminal -Glycine - Asportic Acid - Glutamic Acid- -Trylotopohon-Lysine - c'terminal And the net charge on this focynent at 645 would be (+2-3 = -1)

(3) The third fragment would be : N'- THR - C'

The structure of fragment 3 showing the net charge at pH 5 is shown as follows :

structure of the third beptide fragmentie N'-THR-C ire N' terminal -Threonine - Histidine Arginine-c' tegnind and the net charge on it at pus which is shown as follows: Pya H H H CHE CH2 And the net charge on this fragment would offits be (+3-1=+2) CH2 HG-on CH HC CH2 ми N'terminal - Threonine Histidine- Arginine - c'terminal NHL

This is due to the fact that as we know that enzyme trypsin cleave peptides on the C-terminal side of lysine (K) and arginine (R) amino acid residues and thus frees its $\alpha$ carboxyl group which then generate two fragment at the site of action of the trypsin enzyme. The one fragment would be containing free $\alpha$ carboxyl group of lysine (K) and arginine (R) amino acid residues and the other fragment contain the free $\alpha$ amino group of amino acid residues which was present next to the lysine (K) and arginine (R) amino acid residues present in the polypeptides.

Thus, the fragment resulting from the the tryptic digestion of the polypeptide i.e. N'- ERGDEWKTHR - C' is shown as follows :

(1) The first fragment would be : N'- ER - C'

(2) The second fragment would be : N'- GDEWK - C'

(3) The third fragment would be : N'- THR - C'

We have provided the scheme for how these fragments would be purified from one another at pH 5 is shown as follows :

So for that now let us see the net charge on each fragment which is shown as follows :

(1) The first fragment i.e. N'- ER - C' net charge at pH 5 would be : zero

(2) The second fragment i.e. N'- GDEWK - C' net charge at pH 5 would be : - 1.

(3) The third fragment i.e. N'- THR - C' net charge at pH 5 would be : + 2

Thus since as we see above there is difference in the net charge of each fragment at pH 5. So, we can purify these fragment by use of the Ion exchange chromatography whose pattern is shown as follows :

(1) We can separate the fragment 3 i.e. N'- THR - C' by use of cation exchange chromatography i.e. a type of Ion exchange chromatography since the the fragment 3 i.e. N'- THR - C' carries net charge of +2 at pH 5.

(2) We can separate the fragment 2 i.e. N'- GDEWK - C' by use of anion exchange chromatography i.e. a type of Ion exchange chromatography since the the fragment 2 i.e. N'- GDEWK - C' carries net charge of - 1 at pH 5.

(3) And the solution left (eluted) of the peptide fragment after the cation exchange chromatography and anion exchange chromatography would be of the fragment 1 i.e. N'- ER - C' which carries the net charge 0 at pH 5; thus not purified by cation exchange chromatography and anion exchange chromatography.

Thus, in this way, we have provided the scheme for how these fragments would be purified from one another at pH 5.