(Hint: for calculations round all values of atomic masses of elements to integers) Angiotensins (Ang — Physical Chemistry — Kinetics Chemistry Question
Biologically active peptides and their metabolic pathways
(Hint: for calculations round all values of atomic masses of elements to integers)
Angiotensins (Ang) form a class of biologically active oligopeptides with numerous significant effects on human organism. They play an important role in regulating blood pressure, maintaining water-saline balance and performing intellectual and amnestic functions.
Decapeptide angiotensin I (Ang I) is the initial oligopeptide, a precursor of all members of the class. Complete acidic hydrolysis of Ang I leads to the mixture of nine amino acids: aspartic acid, arginine, valine, histidine, isoleucine, leucine, proline, tyrosine and phenylalanine.
Asparagine is hydrolyzed to form aspartic acid under the conditions required for complete hydrolysis of peptides.
Enzymes of several groups are involved in the metabolism of angiotensins. The first group includes amino peptidases (AMA and AMN), which cut off amino acids or peptide fragments from N-terminus of oligopeptides. The second group is represented by carboxypeptidases (Angiotensin-converting enzyme, ACE and its homolog ACE2), which cut off amino acids or peptide fragments from C-terminus of oligopeptides. The third group includes peptidases (neutral endopeptidase (NEP) and prolyl endopeptidase (PEP)), which split peptide bonds formed by specific amino acids residues.
Ang I is metabolized in man according to the scheme below:
[VISUAL]
1 – 5 are peptidases catalyzing corresponding reactions. Each of these peptidases catalyzes hydrolysis of only one peptide bond. One and the same peptidase may be encoded by different numbers.
To name angiotensins, a special nomenclature has been developed. Amino acid residues of Ang I are enumerated from N- to C-termini. Since all angiotensins contain fragments of Ang I, the word «angiotensin» in their names is followed by Arabic numerals in parenthesis, indicating the positions of N- and C-terminal residues they occupied in Ang I. For instance, Ang I should be named according to the nomenclature as «angiotensin (1-10)».
Write down the equation of the acidic hydrolysis of asparagine.
Model Answer
The equation of the acidic hydrolysis of asparagine is:
[VISUAL]
Write down all possible variants of amino acids and/or oligopeptides, which can be cut off as a result of Ang II formation from Ang I.
Model Answer
X and Z are nonapeptides. To pass from Ang I to these substances one terminal amino acid should be cut off in each case. Ang I is an acyclic peptide having two ends, thus N- and C-terminal residues are affected in these reactions. Heptapeptide Y is formed from Ang II, which is definitely not a nonapeptide (only two nonapeptides are possible, and these are X and Z). Thus, Ang II is an octapeptide. Since ACE is a carboxypeptidase, Y can be either Ang (1 – 7) or Ang (2 – 8). The fact that Y is formed directly from Ang I through one step process allows attributing Y to Ang (1 – 7).
By the other reaction Y is formed directly from X. Thus, the latter comprises Y in its structure and has the same N-terminal amino acid as Ang I and Y. Then nonapeptide X is formed as a result of cleavage of C-terminal peptide bond in Ang I.
The molecular mass of the leaving amino acid is: 1295 – 1182 + 18 = 131, which corresponds to either leucine or isoleucine.
Ang II is formed from Ang I as a result of cutting off two C-terminal amino acids. The molecular mass of 9th (from the N-terminus) amino acid in Ang I is: 1182 – 1045 + 18 = 155, which corresponds to histidine.
Finally, two dipeptides are possible as leaving fragments: His-Leu and His-Ile.
Name oligopeptides X, Y and Z according to the Angiotensin nomenclature. Determine whether enzymes 1-3 are amino or carboxypeptidases.
Model Answer
X – Ang (1 – 9)
Y – Ang (1 – 7).
Z – Ang (2 – 10), since is being formed by cutting off N-terminal amino acid.
2 - Amino peptidase;
1 and 3 - Carboxypeptidase.
Determine the gross amino acid content of Ang I. Show evidence to prove that the answer is unambiguous.
Model Answer
Gross amino acid content of Ang I can be determined from its molecular mass using the following calculations:
M(Ang I) – sum of molar masses of amino acids formed as a result of hydrolysis + 9 M(H2O) = molar mass of the repeating amino acid (this is correct only if Ang I does not contain Asn).
If Ang I contains Asn, the calculated above value of molecular mass will be different from the molecular mass of the repeating amino acid by 1 g mol–1 (in case 1 residue of Asn present) or 2 g mol–1 (in case 2 residues of Asn present). This deviation is due to the difference of the molecular masses of Asn and Asp (132 and 133 g mol–1, respectively).
Calculations:
Mr (repeating-amino acid) = 1295 – (155 + 2⋅131 + 133 + 174 + 117 + 181 + 115 + 165 – 18⋅9) = 155.
The value corresponds to histidine as the repeating amino acid and Asp. Thus, the gross amino acid content of Ang I is:
2 His : 1 Asp : 1 Arg : 1 Ile : 1 Leu : 1 Phe : 1 Pro : 1 Tyr.
Metabolic pathways of Ang I derivatives are summarized in the following scheme:
[VISUAL]
6 – 12 are peptidases catalyzing corresponding reactions. One and the same peptidase may be encoded by different numbers. Pancreatic proteinase trypsin catalyzes hydrolysis of peptide bonds formed by carboxyl groups of arginine or lysine. Z1 has the highest molecular mass among all peptides formed as a result of trypsin catalyzed proteolysis of Ang I. Determine which fragments are cut off as a result of the transformation from Ang II to Ang IV.
Model Answer
Z1 is formed in two ways: from Ang I in the trypsin catalyzed reaction and from nonapeptide Z (Ang (2-10)) in the AM-N (N-peptidase) catalyzed reaction. Thus, Z1 is Ang (3-10), whereas Arg is the 2nd amino acid residue in Ang I.
Studying the transformation of Ang II to Ang IV, we come to the conclusion that Ang III is a heptapeptide (pay attention to the reactions catalyzed by enzymes 7, 8, 10). Since Ang IV is formed from heptapeptide Ang III and further hydrolyzed to pentapeptide Y3, it is a hexapeptide. Taking into account that Ang IV is formed from both Ang (3-10) and Ang (1-8), we finally attribute Ang IV to Ang (3-8). Thus, on the way from Ang II to Ang IV the 1st and 2nd amino acids residues are consecutively cut off. The 2nd residue was earlier found to be Arg. The first residue can be easily determined from the difference of relative molecular masses of Ang II and Ang IV:
1045 – 774 – 174 + 2⋅18 = 133, which corresponds to Asp.
PEP selectively cleaves peptide bonds formed by carboxyl group of proline. Determine the C-terminal amino acid in Ang II and structure of the dipeptide released when heptapeptide Y is treated with ACE.
Model Answer
PEP cuts off the 8th amino acid residue from Ang (3-8), revealing that proline is the 7th residue in Ang I. Relative molecular mass of the 8th eighth amino acid in Ang I is: 774 – 627 + 18 = 165, which corresponds to Phe.
Heptapeptide Y is Ang (1 – 7). ACE catalyzed hydrolysis can lead only to one pentapeptide, Ang (1 – 5). Relative molecular mass of the 6th amino acid which is released from Y as a part of the dipeptide, is: 1045 – 664 – 165 – 115 + 3⋅18 = 155, which corresponds to His.
Thus, C-terminal amino acid of Ang II is Phe, and dipeptide released from Y is His-Pro.
Pancreatic proteinase chymotrypsin catalyzes hydrolysis of peptide bonds formed by carboxyl groups of aromatic amino acids phenylalanine, tyrosine or tryptophane. Quite often chymotrypsin also reveals specificity towards leucine, which is close to the mentioned above amino acids in hydrophobicity. Only two tetrapeptides are formed when Ang II is treated with chymotrypsin. Write down the finally established exact amino acid sequence of Ang I.
Model Answer
Only two tetrapeptides are formed when octapeptide Ang II is treated with chymotrypsin. This means that one the following amino acids: Tyr, Phe or Leu is among the first 7 amino acids and occupies the 4th position. Phe was earlier established as the 8th amino acid, and can be thus omitted from subsequent considerations. If the 4th position is occupied by Leu, Tyr should be either the 3rd or 5th residue (the 10th position is already occupied by either Leu or Ile, see answer to question 2), which will result in a complicated mixture of products of chymotrypsin catalyzed hydrolysis. Thus, the 4th amino acid is Tyr. For similar reasons, Leu can be placed in the 3rd or 5th position. So, it is Leu that occupies the 10th position.
There are only two positions (the 3rd and 5th) and two amino acids (Val and Ile) left. Exact assignment can be done by calculating possible molecular masses of tetrapeptides formed as a result of Ang II treatment with NEP.
Variant 1.
Val – 3, Ile – 5: Mr(angiotensin (1 – 4)) = 133 + 174 + 117 + 181 – 3⋅18 = 551;
M(angiotensin (5 – 8)) = 131 + 155 + 115 + 165 – 3⋅18 = 512;
Variant 2.
Val – 5, Ile – 3: Mr (angiotensin (1-4)) = 133 + 174 + 131 + 181 – 3⋅18 = 565;
Mr (angiotensin (5-8)) = 117 + 155 + 115 + 165 – 3⋅18 = 498.
It is seen that Variant 1 is in agreement with the task conditions. Finally, Ang I structure:
Asp-Arg-Val-Tyr-Ile-His-Pro-Phe-His-Leu
Name oligopeptides X1, Y1 and Z1 according to the Angiotensin nomenclature.
Model Answer
X1 – Ang (5 – 8);
Y1 – Ang (2 – 7);
Z1 – Ang (3 – 10)