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Propose a structure for an octapeptide that shows the composition Asp, Gly \(_{2}\), Leu, Phe, Pro \(_{2}\), Val on amino acid analysis. Edman analysis shows a glycine N-terminal group, and leucine is the C-terminal group. Acidic hydrolysis gives the following fragments: Val-Pro-Leu, Gly, Gly-Asp-Phe-Pro, Phe-Pro-Val

Short Answer

Expert verified
Gly-Gly-Asp-Phe-Pro-Val-Pro-Leu.

Step by step solution

01

List the Given Components

The problem gives us the octapeptide composition as Asp, Gly (2 units), Leu, Phe, Pro (2 units), Val. Edman degradation analysis provides the starting and ending groups: Gly is at the N-terminal and Leu is at the C-terminal. Also, we know the sequence breaks into smaller fragments upon acidic hydrolysis.
02

Analyze Fragmentation Data

The fragments given are Val-Pro-Leu, Gly, Gly-Asp-Phe-Pro, and Phe-Pro-Val. These fragments must be pieces of the complete octapeptide, and they contain clues about how the amino acids are connected.
03

Use Known Terminals to Connect Fragments

Since Gly is the N-terminal, and Leu is the C-terminal, look for where these fragments logically connect: the fragment Val-Pro-Leu seems to represent a clear end with Leu as its last amino acid. Gly can only connect as the first in a fragment, like Gly-Asp-Phe-Pro or it can stand alone.
04

Identify Overlaps in Fragments

Consider overlaps such as Gly in Gly-Asp-Phe-Pro, and the standalone Gly, ensuring that these Glycine units fit the sequence. The Phe-Pro-Val and Gly-Asp-Phe-Pro fragments share the amino acids, which suggests an overlap in the middle sequences.
05

Assemble the Sequence

We begin with the N-terminal Gly. Combine the fragments by starting with Gly, moving into Gly-Asp-Phe-Pro, then overlap Phe-Pro-Val with the middle section, linking to Val-Pro-Leu at the end, which closes with Leu at the C-terminal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Peptide Fragmentation
Peptide fragmentation is an important method in understanding the structure of peptides. When a peptide is subjected to fragmentation, it breaks into smaller pieces, or fragments. These fragments provide insight into how the individual amino acids are arranged in the peptide. For the octapeptide in the exercise, we used fragments like Val-Pro-Leu and Gly-Asp-Phe-Pro to reconstruct the sequence.
Fragmentation can occur naturally, such as in acidic conditions, or through analytical techniques like mass spectrometry. In this exercise, chemical fragmentation helps to map out a possible sequence for the octapeptide, which is crucial when other methods cannot provide the complete structure.
Understanding peptide fragmentation is beneficial for identifying proteins and determining their sequences, as it reveals which amino acids are adjacent to each other and how they might overlap in different fragments.
Amino Acid Sequence
The amino acid sequence of a peptide is the precise order in which amino acids are linked together to form the peptide chain. This sequence determines the eventual shape and function of the peptide. In our octapeptide example, the goal is to identify the correct linear sequence of amino acids: Asp, Gly, Leu, Phe, Pro, and Val.
Knowing the sequence is critical as it dictates how a peptide interacts with other molecules. The sequence Gly-Asp-Phe-Pro, identified from fragmentation data, is part of this process. Each fragment helps in placing each amino acid at the right spot within the sequence. By layering the fragments together, we achieve the full peptide sequence with identified termini — Gly at the N-terminal and Leu at the C-terminal.
Correctly identifying sequences is key in understanding biochemical pathways, drug design, and protein synthesis.
Edman Degradation
Edman degradation is a method for sequencing amino acids in a peptide. It sequentially removes one amino acid at a time from the N-terminal end, allowing its identity to be determined without disrupting the overall peptide structure. In the given exercise, Edman degradation indicates that glycine is the N-terminal amino acid.
This technique is particularly useful for short peptides, and can be done repeatedly to determine each amino acid in the sequence. It systematically peels away amino acids, revealing the structure one at a time, which can be vital when piecing together an octapeptide.
Learning to use Edman degradation weaves into a broader skill of determining sequences, allowing scientists to uncover detailed information about peptides and proteins efficiently.
Chemical Hydrolysis
Chemical hydrolysis is a process that breaks down peptides by adding water to cleave peptide bonds. In the context of the octapeptide exercise, acidic hydrolysis was used to split the peptide into smaller fragments.
This method capitalizes on the susceptibility of certain peptide bonds to acidic or basic environments, leading to predictable breaks. The fragments generated, such as Val-Pro-Leu and Phe-Pro-Val, are analyzed to determine how the entire chain is organized.
Understanding hydrolysis is critical for peptide research as it enables scientists to deduce sequences from larger peptides. It gives insights into peptide stability and allows experiments that would otherwise be impossible due to the complexity of the whole chain.

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Most popular questions from this chapter

Proline has \(\mathrm{pK}_{\mathrm{a} 1}=1.99\) and \(\mathrm{p} K_{\mathrm{a} 2}=10.60 .\) Use the HendersonHasselbalch equation to calculate the ratio of protonated and neutral forms at \(\mathrm{pH}=2.50 .\) Calculate the ratio of neutral and deprotonated forms at \(\mathrm{pH}=9.70\)

Proteins can be cleaved specifically at the amide bond on the carboxyl side of methionine residues by reaction with cyanogen bromide, \(\mathrm{BrC}=\mathrm{N}\). The reaction occurs in several steps: (a) The first step is a nucleophilic substitution reaction of the sulfur on the methionine side chain with BrCN to give a cyanosulfonium ion, \(\left[\mathrm{R}_{2} \mathrm{SCN}\right]^{+} .\) Show the structure of the product, and propose a mechanism for the reaction. (b) The second step is an internal \(S_{N}^{2}\) reaction, with the carbonyl oxygen of the methionine residue displacing the positively charged sulfur leaving group and forming a five-membered ring product. Show the structure of the product and the mechanism of its formation. (c) The third step is a hydrolysis reaction to split the peptide chain. The carboxyl group of the former methionine residue is now part of a lactone (cyclic ester) ring. Show the structure of the lactone product and the mechanism of its formation. (d) The final step is a hydrolysis of the lactone to give the product shown. Show the mechanism of the reaction.

Show how you could prepare the following \(\alpha\) -amino acids from the appropriate carboxylic acids: (a) Phenylalanine (b) Valine

Propose two structures for a tripeptide that gives Leu, Ala, and Phe on hydrolysis but does not react with phenyl isothiocyanate.

Evidence for restricted rotation around amide CO-N bonds comes from NMR studies. At room temperature, the \({ }^{1} \mathrm{H}\) NMR spectrum of \(N, N\) dimethylformamide shows three peaks: \(2.9 \delta\) (singlet, \(3 \mathrm{H}\) ), \(3.0 \delta\) (singlet, \(3 \mathrm{H}\) ), and \(8.0 \delta\) (singlet, \(1 \mathrm{H}\) ). As the temperature is raised, however, the two singlets at \(2.9 \delta\) and \(3.0 \delta\) slowly merge. At \(180^{\circ} \mathrm{C},\) the \({ }^{1} \mathrm{H}\) NMR spectrum shows only two peaks: \(2.95 \delta\) (singlet, \(6 \mathrm{H}\) ) and \(8.0 \delta\) (singlet, \(1 \mathrm{H}\) ). Explain this temperature-dependent behavior.

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