Ls4201_midsem_2016 2. A hexapeptide right handed helix has two Aib residues in t
ID: 196720 • Letter: L
Question
Ls4201_midsem_2016
2. A hexapeptide right handed helix has two Aib residues in the sequence and its crystal structure reveals the formation of Schellman motif at the C-terminus. please indicate the conformational parameter of two Aib residues and their location at the Ramachandran plot.
Boc is the N-terminus protecting group and OMe is c-terminus protecting group.The peptide sequence is Boc-pro-Aib-Gly-Leu-Aib-Leu-OMe.
Aib1 is in between pro- and Gly and Aib2 is in between two Leu residues. (3marks)
please answer only question no.2. (3marks)
1. A 20-residue helix is 23.2 Å long and has total 4.54 no. of turns and residues per turn is 4.4. Calculate the translation per residue and pitch value of this helix. 11.5+1.5-3 14 igh handecd 2. A hexapeptide helix has two Aib residues in the sequence and its crystal structure reveals the formation of Schellman motif at the C-terminus. Please indicate the conformational parameter of two Aib residues and their location at the Ramachandran plot. Boc is the N-terminus protecting group and OMe is C-terminus protecting group. The peptide sequence is Boc-Pro-Aib-Gly-Leu-Aib-Leu-OMe Aibl is in between Pro- and Gly and Aib2 is in between two Leu residues. [3]Explanation / Answer
Aib (- aminoisobutyrate) is an achiral residue, having a pair of methyl groups at C alpha carbon. This residue viewed as a combination of L-Alanine and D-Alanine. The steriochemically allowed regions of Aib residue in the Ramachandran space resides in the overlapping regions of L-Ala and D-Ala - maps . Achiral residue Aib at position 2 and 5 of the given heptapeptide, nucleates which of the secondary structure conformation, depends on the configuration of the flanking aminoacid residues. As in the given heptapeptide Aib residue at position 2 has flanked by the proline and glycine amino acid residue and it is known that proline tends to disrupt the helices because it lacks an -NH group and due to its ring structure its value get restricted in the ramachandran plot. Proline is a constrained residue and predominately found in turns and which inturn serve as a nuclei for antiparallel sheet. Aib residue present at 5th position flanked by the two Leucine residues ( Leucine is one of the aminoacid which show the higher tendency to form helices). We can conclude from above points that Aib1 is present at position 2nd permits strand due to the presence of Proline as flanking residue and Aib2 which is present at 5th postion permit helices conformation due to the presence of Leucine as flanking residue. The permitted values for and are usually indicated on 2-D map of the - plane, also known as Ramachandran plot where is the torsion angle between N-C bondand its value ranges from -180 to +180 and is the torsion angle between C-C bond and its values also ranges from -180 to +180. Ramachandran plot shows amino acid allowed conformations of - angles and have 4 quadrants upper left where strand conformation reside so Aib1 present in this quadrant and the two lower quadrants in which lower left quadrant where right handed alpha conformation reside so Aib2 present in this quadrant.
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