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A peptide with the primary structure Lys-Arg-Pro-Leu-Ile-Asp-Gly-Ala is sequenced by the Edman procedure. If each Edman cycle is \(96 \%\) efficient, what percentage of the amino acids liberated in the fourth cycle will be leucine? Do the calculation a second time, but assume a \(99 \%\) efficiency for each cycle.

Short Answer

Expert verified
84.93% for 96% efficiency; 96.06% for 99% efficiency.

Step by step solution

01

Understand the Edman Degradation Process

The Edman degradation process sequences peptides by cleaving the N-terminal amino acid one at a time. Each cycle has an efficiency rating, meaning not all peptides will have their respective N-terminal amino acids removed in each cycle.
02

Determine the Impact of Each Edman Cycle

For an efficiency rate of 96%, each cycle leaves a certain percentage of unreacted peptides. This percentage is calculated as 1 - 0.96 = 0.04 (4% remain). This means after each cycle, 96% of the remaining peptide population is reduced further by the next cycle's efficiency.
03

Calculate Percentage of Leucine at the Fourth Cycle for 96% Efficiency

After three cycles of cleavage, the peptide population is reduced by: 1. First cycle: 96% efficiency 2. Second cycle: 96% 3. Third cycle: 96% The fraction of the original population reaching the fourth cycle is (0.96)^3. Calculate this as:\( (0.96)^3 = 0.884736 \).This is the fraction left to start the fourth cycle. Of the peptides present in the fourth cycle, any that are successfully cleaved will have leucine liberated.
04

Solve for Amount Liberated at 96% Efficiency

Because the fourth cycle also has an efficiency of 96%, the percentage of peptides cleaved to release leucine is 96% of those starting the fourth cycle:\( 0.96 \times 0.884736 = 0.84934656 \) or 84.93%.
05

Repeat the Calculation for 99% Efficiency

To find the percentage liberated under 99% cycle efficiency, repeat the calculation:1. First cycle: 99% efficiency 2. Second cycle: 99% 3. Third cycle: 99% The fraction of peptides starting the fourth cycle is:\( (0.99)^3 = 0.970299 \).Then calculate the percentage successfully cleaved:\( 0.99 \times 0.970299 = 0.96059601 \) or 96.06%.

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Key Concepts

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

Peptide Sequencing
Peptide sequencing is a process where the sequence of amino acids in a peptide is determined. One popular method for peptide sequencing is Edman degradation. This chemical technique identifies the N-terminal amino acid of a peptide, then cleaves it off for identification. This method can be repeated multiple times to sequence a stretch of amino acids in a peptide. Each cleavage, or cycle, yields the next amino acid in line. It's crucial for decoding complex molecules and developing medicines.
In the outlined exercise, we began with the peptide Lys-Arg-Pro-Leu-Ile-Asp-Gly-Ala. The Edman process helps in sequentially identifying each of these amino acids. While the method is powerful, its efficiency is limited; in each cycle, not all peptides get cleaved, meaning careful calculations are needed to determine the cycle outcome. Understanding the efficiency of each cycle is important to predict which amino acids will be liberated at different stages.
Amino Acid Analysis
Amino acid analysis refers to identifying and quantifying the amino acids in a peptide. In Edman degradation, the focus is on determining the N-terminal amino acid in each cycle. This analysis progresses until the complete sequence is mapped out.
When analyzing a peptide like Lys-Arg-Pro-Leu-Ile-Asp-Gly-Ala, we focus on the N-terminal (Lysine in the first cycle within this exercise). As cycles progress through each amino acid, the efficiency of each step determines how accurately we can identify and separate each residue. When moving from one amino acid to the next, unfractionated peptides often throw challenges, leading to overlaps or incomplete sequencing unless corrected through efficiency or other methods. This analysis remains invaluable for researchers working on peptide drug synthesis or protein structure studies.
Cycle Efficiency
Cycle efficiency indicates the success rate of each Edman degradation cycle. It measures what percentage of peptides accurately undergo cleavage in every step. For instance, a 96% efficiency means 96% of the peptides in each cycle effectively cleave at the N-terminal.
  • Each cycle begins with the remaining peptides post-cleavage from the previous cycle.
  • With each subsequent cycle, fewer peptides may be available due to inefficiency (non-cleaved peptides).
In our exercise, we calculate the percentage of leucine liberated. At 96% efficiency, by the fourth cycle, a significant amount (84.93%) was cleaved to release leucine.
If efficiency improved to 99%, a higher percentage (96.06%) of peptides release leucine during the fourth cycle. This shows how slight efficiency changes significantly impact the overall peptide sequencing outcome, emphasizing the enhancement by optimizing Edman cycle conditions.

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Most popular questions from this chapter

A protein has a molecular mass of 400 kDa when measured by size-exclusion chromatography. When subjected to gel electrophoresis in the presence of sodium dodecyl sulfate (SDS), the protein gives three bands with molecular masses of \(180,160,\) and 60 kDa. When electrophoresis is carried out in the presence of SDS and dithiothreitol, three bands are again formed, this time with molecular masses of 160,90 , and 60 kDa. Determine the subunit composition of the protein.

Extracts from the bacterium Bacillus brevis contain a peptide with antibiotic properties. This peptide forms complexes with metal ions and seems to disrupt ion transport across the cell membranes of other bacterial species, killing them. The structure of the peptide has been determined from the following observations. (a) Complete acid hydrolysis of the peptide followed by amino acid analysis yielded equimolar amounts of Leu, Orn, Phe, Pro, and Val. Orn is ornithine, an amino acid not present in proteins but present in some peptides. It has the structure (b) The molecular weight of the peptide was estimated as about 1,200. (c) The peptide failed to undergo hydrolysis when treated with the enzyme carboxypeptidase. This enzyme catalyzes the hydrolysis of the carboxyl- terminal residue of a polypeptide unless the residue is Pro or, for some reason, does not contain a free carboxyl group. (d) Treatment of the intact peptide with 1 -fluoro- 2,4 dinitrobenzene, followed by complete hydrolysis and chromatography, yielded only free amino acids and the following derivative: (Hint: The 2,4 -dinitrophenyl derivative involves the amino group of a side chain rather than the \(\alpha\) -amino group.) (e) Partial hydrolysis of the peptide followed by chromatographic separation and sequence analysis yielded the following di- and tripeptides (the amino- terminal amino acid is always at the left): Leu-Phe Phe-Pro Orn-Leu Val-Orn Val-Orn-Leu Phe-Pro-Val Pro-Val-Orn Given the above information, deduce the amino acid sequence of the peptide antibiotic. Show your reasoning. When you have arrived at a structure, demonstrate that it is consistent with each experimental observation.

A purified protein is in a Hepes (N-(2-hydroxyethyl)piperazine- \(N^{\prime}-(2 \text { -ethanesulfonic acid })\) ) buffer at pH 7 with \(500 \mathrm{mm}\) NaCl. A sample \((1 \mathrm{mL})\) of the protein solution is placed in a tube made of dialysis membrane and dialyzed against \(1 \mathrm{L}\) of the same Hepes buffer with \(0 \mathrm{m}\). NaCl. Small molecules and ions (such as \(\mathrm{Na}^{+}, \mathrm{Cl}^{-}\), and Hepes) can diffuse across the dialysis membrane, but the protein cannot. (a) Once the dialysis has come to equilibrium, what is the concentration of NaCl in the protein sample? Assume no volume changes occur in the sample during the dialysis. (b) If the original 1 mL sample were dialyzed twice, successively, against \(100 \mathrm{mL}\) of the same Hepes buffer with \(0 \mathrm{my} \mathrm{NaCl}\) what would be the final NaCl concentration in the sample?

One method for separating polypeptides makes use of their different solubilities. The solubility of large polypeptides in water depends on the relative polarity of their \(R\) groups, particularly on the number of ionized groups: the more ionized groups there are, the more soluble the polypeptide. Which of each pair of polypeptides that follow is more soluble at the indicated pH? (a) (Gly) \(_{20}\) or \((\mathrm{Glu})_{20}\) at \(\mathrm{pH} 7.0\) (b) (Lys-Ala) \(_{3}\) or (Phe-Met) \(_{3}\) at pH 7.0 (c) (Ala-Ser-Gly) \(_{5}\) or (Asn-Ser-His) at pH 6.0 (d) (Ala-Asp-Gly) or (Asn-Ser-His) at pH 3.0

At pH 7.0, in what order would the following three peptides be eluted from a column filled with a cation-exchange polymer? 'Their amino acid compositions are: Peptide A: Ala \(10 \%,\) Glu \(5 \%, \operatorname{Ser} 5 \%,\) Leu \(10 \%,\) Arg \(10 \%\) His \(5 \%,\) Ile \(10 \%,\) Phe \(5 \%,\) Tyr \(5 \%,\) Lys \(10 \%,\) Gly \(10 \%,\) Pro \(5 \%\) and Trp \(10 \%\). Peptide B: Ala \(5 \%,\) Val \(5 \%,\) Gly \(10 \%,\) Asp \(5 \%,\) Leu \(5 \%,\) Arg \(5 \%,\) Ile \(5 \%,\) Phe \(5 \%,\) Tyr \(5 \%,\) Lys \(5 \%,\) Trp \(5 \%,\) Ser \(5 \%,\) Thr \(5 \%\) Glu \(5 \%,\) Asn \(5 \%,\) Pro \(10 \%,\) Met \(5 \%,\) and Cys \(5 \%\). Peptide C: Ala \(10 \%,\) Glu \(10 \%,\) Gly \(5 \%,\) Leu \(5 \%,\) Asp \(10 \%\) Arg \(5 \%,\) Met \(5 \%,\) Cys \(5 \%,\) Tyr \(5 \%,\) Phe \(5 \%,\) His \(5 \%,\) Val \(5 \%,\) Pro \(5 \%,\) Thr \(5 \%,\) Ser \(5 \%,\) Asn \(5 \%,\) and \(\operatorname{Gln} 5 \%\).

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