NH4 Asp Glu Tyr Arg Met Pro Lys Ser Phe The following facts

NH^+_4, Asp, Glu, Tyr, Arg, Met, Pro, Lys, Ser, Phe The following facts were observed: Neither carboxylpeptidase A nor B treatment had any effect. Trypsin treatment yielded two tetrapeptides and free Lys. Clostripain treatment yielded a tetrapeptide and a hexapeptide. Cyanogen bromide treatment yielded an octapeptide and a dipeptide of sequence NP (using one letter codes). Chymotrypsin treatment yielded two tripeptides and a tetrapeptide. The N-terminal chymotryptic peptide had a net charge of -1 at neutral pH, and a net charge of -3 at pH 12. One cycle of the Edman degradation gave the PTH derivative: What is the amino acid sequence of this decapeptide? (Use the three letter codes for amino acids in your answer and if there is more than one possible sequence just enter one. Separate amino acids using spaces.)

Solution

Answer:Carboxypeptidase A will cleave at the C-terminus, exceptfor Pro, Lys, and Arg whereas Carboxypeptidase B cleavage occurs at C-terminal Arg and Lys. No action by either indicates either a circularpeptide or proline as C-terminus. One cycle of Edman degradation identifies serine as as the N-terminal amino acid.

Thus,

S_ _ _ _ _ _ _ _P

Cyanogen bromide treatment cleaves after M and releases two products that account for 10 amino acids residues. Thus, there must be only one M and it must be located near the C-terminus to release the dipeptide.

S_ _ _ _ _ _M N P

Tyrpsin cleavage yielded two tetrapeptides and free lysine. looking at the sequence it was conclude that so far the C-terminus must be in one of the tetrapeptides and so either an R or K must be at position 6. Positions 2 and 3 cannot be either R or K because the trysin products do not include a di- or tripeptide. So, there must be an R or K at position 4. Since 4 and 6 both contain R & K, the free lysin must come from position 5.

Thus,

S_ _ (R/K) K (R/K) _ M N P

Since the peptide contains both R and K the possibilities for the sequence of positions 4, 5 and 6 are RKR RKK and KKR and only RKK would generate fre lysine and two moles of it. This would put R at position 4, which is consistent with clostripain treatment that yielded a tetrapeptide and a hexapeptide.

So,

S _ _R K K _ M N P

F or Y must be located at position 7 to insure that chymotrypsin cleavage produces a tripeptide from the C-terminus and nothing longer. In addition an aromatic must also be located at position 3. The only amino acid unaccounted for is either Glu (E) or Gln (Q), which is located at position 2.

This gives: (1) S (E/Q) (F/Y) R K K (F/Y) M N P

For the N-terminal chymotryptic peptide, S (E/Q) (F/Y), to have a net charge of -1 at neutral pH and -3 at pH = 12 it must contain glumatic acid (E) and tyrosine (Y) to account for the charge.

S E Y R K K F M N P

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