Chapter 27, Problem 27.53AP

Interpretation Introduction

(a)

Interpretation:

The structure of C from the given illustration and the explanation for how it is obtained is to be stated.

Concept introduction:

Trypsin is an enzyme that is found in the digestive system where it hydrolyzes the protein. It is also used to cleaved the peptide chain at the carboxyl end of lysine and arginine. Edman degradation is a method of peptide sequencing. In this method, $\text{N}-$ terminal of the peptide chain is treated with phenyl isothiocyanate which is known as Edman reagent. It results in the separation of that amino-terminal residue without affecting the rest peptide chain

Answer to Problem 27.53AP

The structure of C from the given illustration is $\text{Arg}-\text{Ala}-\text{Ile}-\text{Gly}-\text{Ala}$.

Explanation of Solution

The molecular mass of amino acid $1000$ and it has the composition ${\text{Ala}}_2$, $\text{Arg}$, $\text{Gly}$, $\text{Ile}$. When the given peptide is treated with Edman reagents, then ${\text{CF}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}$ it does not cause any change in the peptide. This means arginine residue is present at $\text{N}-$ terminal of the peptide chain. The amino acid arginine is basic in nature due to the presence of two amine groups. Therefore, the Edman reagent, phenyl isothiocyanate reacts with the other amine group instead of $\alpha -$ amino group. The reaction of peptide C with Edman reagent then ${\text{CF}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}$ is shown below.

Figure 1

As a result, the phenyl hydantoin derivative is not formed. Due to this, the peptide C remain unchanged on Edman degradation. When the peptide C reacts with trypsin enzyme, it forms a single peptide D. Also, the trypsin enzyme cleaves at lysine and arginine residue. The peptide D gives the same amino acid analysis as peptide C. This confirms that the peptide D does not contain arginine residue, due to this the amino acid analysis of the peptide C is similar to that of the peptide D. The partial structure of peptide D is found to be $\text{Arg}-{\text{Ala}}_{\text{2}}$, $\text{Gly}$, $\text{Ile}$ and the partial structure of peptide C is found to be ${\text{Ala}}_{\text{2}}$, $\text{Gly}$, $\text{Ile}$. Also, peptide D on three cycles of Edman degradation shows the partial structure of the peptide $\text{Ala}-\text{Ile}-\text{Gly}$. Since the peptide D contains four amino acid so there must be $\text{Ala}$ residue at both sides of the peptide D.the peptide D structure is $\text{Ala}-\text{Ile}-\text{Gly-Ala}$. It is already mentioned that arginine is present at the $\text{N}-$ terminal of the peptide. Therefore, the peptide C structure is $\text{Arg}-\text{Ala}-\text{Ile}-\text{Gly}-\text{Ala}$.

Conclusion

The structure of C from the given illustration is $\text{Arg}-\text{Ala}-\text{Ile}-\text{Gly}-\text{Ala}$.

Interpretation Introduction

(b)

Interpretation:

The b-type fragmentation for $\text{M}+1$ ion of both the peptide C and D is to be stated.

Concept introduction:

In mass spectroscopy, compounds can be identified on the basis of the mass of the compound. When the compound breaks into fragment then they can be distinguished from the other compounds. This technique is also used to differentiate the isotopes of compounds. In amino acids, three types of fragments are observed in low energy collisions are a, b and y ions. It is known as tandem mass spectrometry.

Answer to Problem 27.53AP

The b-type fragmentation for $\text{M}+1$ ion of the peptide C is shown below.

$\begin{array}{lrrr}\text{H}-\text{Arg}& m/z& =& 157.1\\ \text{H}-\text{Arg}-\text{Ala}& m/z& =& 228.1\\ \text{H}-\text{Arg}-\text{Ala}-\text{Ile}& m/z& =& 341.2\\ \text{H}-\text{Arg}-\text{Ala}-\text{Ile}-\text{Gly}& m/z& =& 398.2\\ \text{H}-\text{Arg}-\text{Ala}-\text{Ile}-\text{Gly}-\text{Ala}& m/z& =& 469.2\end{array}$

The b-type fragmentation for $\text{M}+1$ ion of the peptide D is shown below.

$\begin{array}{llll}\text{H}-\text{Ala}& m/z& =& 72\\ \text{H}-\text{Ala}-\text{Ile}& m/z& =& 185.1\\ \text{H}-\text{Ala}-\text{Ile}-\text{Gly}& m/z& =& 242.1\\ \text{H}-\text{Ala}-\text{Ile}-\text{Gly}-\text{Ala}& m/z& =& 313.1\end{array}$

where $\text{Arg}$ is asparagine, $\text{Gly}$ is glycine, $\text{Ala}$ is alanine, $\text{Ile}$ is isoleucine.

Explanation of Solution

In amino acids, b-type fragments appear due to an amino group or in other words charge is being carried by N-terminal. That is why it is also known as the N-terminus amino acid fragment. The b-type fragment is shown below.

Figure 2

The peptide C is $\text{Arg}-\text{Ala}-\text{Ile}-\text{Gly}-\text{Ala}$ where $\text{Arg}$ is asparagine with $m/z=156.1$, $\text{Gly}$ is glycine with $m/z=57$, $\text{Ala}$ is alanine with $m/z=71$, $\text{Ile}$ is isoleucine with $m/z=113.1$. The formation of the peptide C with $\text{M}+1$ fragment ion by subsequent addition of their residues in b-type fragment manner. The $m/z$ value of $\text{M}+1$ fragment ion from b-type fragmentation of the given peptide is shown below.

$\begin{array}{llll}\text{H}-\text{Arg}& m/z& =& 157.1\\ \text{H}-\text{Arg}-\text{Ala}& m/z& =& 228.1\\ \text{H}-\text{Arg}-\text{Ala}-\text{Ile}& m/z& =& 341.2\\ \text{H}-\text{Arg}-\text{Ala}-\text{Ile}-\text{Gly}& m/z& =& 398.2\\ \text{H}-\text{Arg}-\text{Ala}-\text{Ile}-\text{Gly}-\text{Ala}& m/z& =& 469.2\end{array}$

The peptide D is $\text{Ala}-\text{Ile}-\text{Gly}-\text{Ala}$ where $\text{Gly}$ is glycine with $m/z=57$, $\text{Ala}$ is alanine with $m/z=71$, $\text{Ile}$ is isoleucine with $m/z=113.1$. The formation of the peptide C with $\text{M}+1$ fragment ion by subsequent addition of their residues in b-type fragment manner. The $m/z$ value of $\text{M}+1$ fragment ion from b-type fragmentation of the given peptide is shown below.

$\begin{array}{ll}\text{H}-\text{Ala}& m/z=72\\ \text{H}-\text{Ala}-\text{Ile}& m/z=185.1\\ \text{H}-\text{Ala}-\text{Ile}-\text{Gly}& m/z=242.1\\ \text{H}-\text{Ala}-\text{Ile}-\text{Gly}-\text{Ala}& m/z=313.1\end{array}$

Conclusion

The b-type fragmentation for $\text{M}+1$ ion of both the peptide C and D is shown above.