Question

7. A peptide was isolated from sheep hypothalamus and subjected to sequence analysis. The experimental data are as follows. Deduce the amino acid sequence from this data. a. Composition (2Ala, 2Gly, Thr, Phe, Lys, His, Ser, Trp, Arg, Val) b. Reaction of dinitrofluorobenzene with the intact peptide yielded DNP-Thr. A brief treatment with carboxypeptidase a yielded glycine. c. Trypsin digestion yielded three fragments: i. Phe, Thr, Ala, Lys ii. Gly, Val iii. Trp, His, Ser, Gly, Arg, Ala Treatment of fragment I with phenylisothiocyanate yielded PTH-Val. Treatment of Fragment iii With phenylisothiocyanate yielded PTH-His. d. Chymotrypsin digestion yielded three fragments: i. Arg, Val, 2Gly, Ala ii. Thr, Phe iii. Lys, Trp, Ser, His, Ala Treatment of fragment i with phenylisothiocyanate yielded PTH-Ala. Treatment of Fragment lii with phenylisothiocyanate yielded PTH-Ala.

Answer 1

Sequence is N-Thr, Phen, Ala, Lys, His, Ser, Trp, Ala, Gly, Arg, Val, Gly-C (N to C)

(If the amino acid at a position is unknown, it is denoted by X)

• Since the reaction with dinitrofluorobenzene (Reacts with N terminal aminoacid) yields, thr, it is present at the N terminal. The reaction with carboxypeptidase gives glycine. So it is present at C terminal end.

• Trypsin clevage (Cleaves at C end of Lys and Arg) gives: (i) Phe, Thr, Ala, Lys, (ii) Gly, Val (iii) Trp, His, Ser, Gly, Arg, Ala.

So, in the first fragment, should be Thr, X, X, Lys. Since the treatment of one fragment with phenylisothiocyanate (Cleaves at C side) gives valine, it is bound to the terminal glycine. Third fragment has Trp, His, Ser, Gly, Arg, Ala in between the first two fragments, with Arg C end of the fragment. So sequence of this fragment is: X,X,X,X,X,Arg. Treatment of this fragment with phenylisothiocyanate gives histidine, it is present at the N side of the fragment. So the sequence after this step is:

Fragment I Fragment II    Fragment III

Thr, X, X, Lys His, X,X,X,X, Arg    Val, Gly

• Chymotrypsin cleavage (Cleaves at C end of aromatic amino acids: Phe, Trp, and Tyr) gives: (i) Arg, Val, 2Gly, Ala. (ii) Thr, Phe. (iii) Lys, Trp, Ser, His, Ala

From the second fragment, It is clear that Phe is adjacent to N terminal Thr.

In the third fragment, it is clear that the sequence should be in this order: Ala, Lys, His, Ser, Trp. Along with the previous interpretation, the order will be : Thr, Phe, Ala, Lys, His, Ser, Trp.

The first fragment has Arg, Val, 2Gly, Ala, out of which Arg, val and Gly are at the C terminal. The treatment of this fragment with phenylisothiocyanate gives Ala. So, it is present on the either side of this fragment. So the sequence of this fragment is Ala, Gly, Arg, Val, Gly.

So, the final sequence would be :

N- Thr, Phe, Ala, Lys, His, Ser, Trp, Ala, Gly, Arg, Val, Gly-C