(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, 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

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Answer 1

Question

Definition Definition Organ system that helps in the digestion of food by mechanical and biochemical processes. The digestive system breaks down food into simpler components that enable the body to then absorb or transport nutrients. The human digestive system includes the gastrointestinal tract or alimentary canal and the associate glands. The organs of the gastrointestinal tract include the mouth, pharynx, esophagus, stomach, small intestine, and large intestine. The associated glands include salivary glands, liver, gallbladder, and pancreas.

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, 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

Expert Solution

Answer to Problem 27.53AP

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Explanation of Solution

The molecular mass of amino acid 1000 and it has the composition Ala2, Arg, Gly, Ile. When the given peptide is treated with Edman reagents, then CF3CO2H it does not cause any change in the peptide. This means arginine residue is present at 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 α− amino group. The reaction of peptide C with Edman reagent then CF3CO2H is shown below.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 1

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 Arg−Ala2, Gly, Ile and the partial structure of peptide C is found to be Ala2, Gly, Ile. Also, peptide D on three cycles of Edman degradation shows the partial structure of the peptide Ala−Ile−Gly. Since the peptide D contains four amino acid so there must be Ala residue at both sides of the peptide D.the peptide D structure is Ala−Ile−Gly-Ala. It is already mentioned that arginine is present at the N− terminal of the peptide. Therefore, the peptide C structure is Arg−Ala−Ile−Gly−Ala.

Conclusion

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Interpretation Introduction

(b)

Interpretation:

The b-type fragmentation for 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.

Expert Solution

Answer to Problem 27.53AP

The b-type fragmentation for M+1 ion of the peptide C is shown below.

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

The b-type fragmentation for M+1 ion of the peptide D is shown below.

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

where Arg is asparagine, Gly is glycine, Ala is alanine, 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.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 2

Figure 2

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

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

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

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

Conclusion

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

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Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100

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For each reaction shown below follow the curved arrows to complete each equationby showing the structure of the products. Identify the acid, the base, the conjugated acid andconjugated base. Consutl the pKa table and choose the direciton theequilibrium goes. However show the curved arrows. Please explain if possible.

Answer 2

Question

Definition Definition Organ system that helps in the digestion of food by mechanical and biochemical processes. The digestive system breaks down food into simpler components that enable the body to then absorb or transport nutrients. The human digestive system includes the gastrointestinal tract or alimentary canal and the associate glands. The organs of the gastrointestinal tract include the mouth, pharynx, esophagus, stomach, small intestine, and large intestine. The associated glands include salivary glands, liver, gallbladder, and pancreas.

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, 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

Expert Solution

Answer to Problem 27.53AP

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Explanation of Solution

The molecular mass of amino acid 1000 and it has the composition Ala2, Arg, Gly, Ile. When the given peptide is treated with Edman reagents, then CF3CO2H it does not cause any change in the peptide. This means arginine residue is present at 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 α− amino group. The reaction of peptide C with Edman reagent then CF3CO2H is shown below.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 1

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 Arg−Ala2, Gly, Ile and the partial structure of peptide C is found to be Ala2, Gly, Ile. Also, peptide D on three cycles of Edman degradation shows the partial structure of the peptide Ala−Ile−Gly. Since the peptide D contains four amino acid so there must be Ala residue at both sides of the peptide D.the peptide D structure is Ala−Ile−Gly-Ala. It is already mentioned that arginine is present at the N− terminal of the peptide. Therefore, the peptide C structure is Arg−Ala−Ile−Gly−Ala.

Conclusion

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Interpretation Introduction

(b)

Interpretation:

The b-type fragmentation for 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.

Expert Solution

Answer to Problem 27.53AP

The b-type fragmentation for M+1 ion of the peptide C is shown below.

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

The b-type fragmentation for M+1 ion of the peptide D is shown below.

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

where Arg is asparagine, Gly is glycine, Ala is alanine, 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.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 2

Figure 2

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

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

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

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

Conclusion

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

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Answer 3

Question

Definition Definition Organ system that helps in the digestion of food by mechanical and biochemical processes. The digestive system breaks down food into simpler components that enable the body to then absorb or transport nutrients. The human digestive system includes the gastrointestinal tract or alimentary canal and the associate glands. The organs of the gastrointestinal tract include the mouth, pharynx, esophagus, stomach, small intestine, and large intestine. The associated glands include salivary glands, liver, gallbladder, and pancreas.

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, 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

Expert Solution

Answer to Problem 27.53AP

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Explanation of Solution

The molecular mass of amino acid 1000 and it has the composition Ala2, Arg, Gly, Ile. When the given peptide is treated with Edman reagents, then CF3CO2H it does not cause any change in the peptide. This means arginine residue is present at 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 α− amino group. The reaction of peptide C with Edman reagent then CF3CO2H is shown below.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 1

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 Arg−Ala2, Gly, Ile and the partial structure of peptide C is found to be Ala2, Gly, Ile. Also, peptide D on three cycles of Edman degradation shows the partial structure of the peptide Ala−Ile−Gly. Since the peptide D contains four amino acid so there must be Ala residue at both sides of the peptide D.the peptide D structure is Ala−Ile−Gly-Ala. It is already mentioned that arginine is present at the N− terminal of the peptide. Therefore, the peptide C structure is Arg−Ala−Ile−Gly−Ala.

Conclusion

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Interpretation Introduction

(b)

Interpretation:

The b-type fragmentation for 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.

Expert Solution

Answer to Problem 27.53AP

The b-type fragmentation for M+1 ion of the peptide C is shown below.

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

The b-type fragmentation for M+1 ion of the peptide D is shown below.

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

where Arg is asparagine, Gly is glycine, Ala is alanine, 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.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 2

Figure 2

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

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

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

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

Conclusion

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

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Answer 4

Question

Definition Definition Organ system that helps in the digestion of food by mechanical and biochemical processes. The digestive system breaks down food into simpler components that enable the body to then absorb or transport nutrients. The human digestive system includes the gastrointestinal tract or alimentary canal and the associate glands. The organs of the gastrointestinal tract include the mouth, pharynx, esophagus, stomach, small intestine, and large intestine. The associated glands include salivary glands, liver, gallbladder, and pancreas.

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, 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

Expert Solution

Answer to Problem 27.53AP

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Explanation of Solution

The molecular mass of amino acid 1000 and it has the composition Ala2, Arg, Gly, Ile. When the given peptide is treated with Edman reagents, then CF3CO2H it does not cause any change in the peptide. This means arginine residue is present at 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 α− amino group. The reaction of peptide C with Edman reagent then CF3CO2H is shown below.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 1

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 Arg−Ala2, Gly, Ile and the partial structure of peptide C is found to be Ala2, Gly, Ile. Also, peptide D on three cycles of Edman degradation shows the partial structure of the peptide Ala−Ile−Gly. Since the peptide D contains four amino acid so there must be Ala residue at both sides of the peptide D.the peptide D structure is Ala−Ile−Gly-Ala. It is already mentioned that arginine is present at the N− terminal of the peptide. Therefore, the peptide C structure is Arg−Ala−Ile−Gly−Ala.

Conclusion

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Interpretation Introduction

(b)

Interpretation:

The b-type fragmentation for 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.

Expert Solution

Answer to Problem 27.53AP

The b-type fragmentation for M+1 ion of the peptide C is shown below.

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

The b-type fragmentation for M+1 ion of the peptide D is shown below.

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

where Arg is asparagine, Gly is glycine, Ala is alanine, 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.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 2

Figure 2

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

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

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

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

Conclusion

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

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Answer 5

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, 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

Expert Solution

Answer to Problem 27.53AP

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Explanation of Solution

The molecular mass of amino acid 1000 and it has the composition Ala2, Arg, Gly, Ile. When the given peptide is treated with Edman reagents, then CF3CO2H it does not cause any change in the peptide. This means arginine residue is present at 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 α− amino group. The reaction of peptide C with Edman reagent then CF3CO2H is shown below.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 1

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 Arg−Ala2, Gly, Ile and the partial structure of peptide C is found to be Ala2, Gly, Ile. Also, peptide D on three cycles of Edman degradation shows the partial structure of the peptide Ala−Ile−Gly. Since the peptide D contains four amino acid so there must be Ala residue at both sides of the peptide D.the peptide D structure is Ala−Ile−Gly-Ala. It is already mentioned that arginine is present at the N− terminal of the peptide. Therefore, the peptide C structure is Arg−Ala−Ile−Gly−Ala.

Conclusion

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Interpretation Introduction

(b)

Interpretation:

The b-type fragmentation for 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.

Expert Solution

Answer to Problem 27.53AP

The b-type fragmentation for M+1 ion of the peptide C is shown below.

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

The b-type fragmentation for M+1 ion of the peptide D is shown below.

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

where Arg is asparagine, Gly is glycine, Ala is alanine, 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.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 2

Figure 2

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

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

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

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

Conclusion

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

Answer 6

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, 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

Expert Solution

Answer to Problem 27.53AP

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Explanation of Solution

The molecular mass of amino acid 1000 and it has the composition Ala2, Arg, Gly, Ile. When the given peptide is treated with Edman reagents, then CF3CO2H it does not cause any change in the peptide. This means arginine residue is present at 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 α− amino group. The reaction of peptide C with Edman reagent then CF3CO2H is shown below.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 1

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 Arg−Ala2, Gly, Ile and the partial structure of peptide C is found to be Ala2, Gly, Ile. Also, peptide D on three cycles of Edman degradation shows the partial structure of the peptide Ala−Ile−Gly. Since the peptide D contains four amino acid so there must be Ala residue at both sides of the peptide D.the peptide D structure is Ala−Ile−Gly-Ala. It is already mentioned that arginine is present at the N− terminal of the peptide. Therefore, the peptide C structure is Arg−Ala−Ile−Gly−Ala.

Conclusion

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Answer 7

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, 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 8

Expert Solution

Answer to Problem 27.53AP

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The molecular mass of amino acid 1000 and it has the composition Ala2, Arg, Gly, Ile. When the given peptide is treated with Edman reagents, then CF3CO2H it does not cause any change in the peptide. This means arginine residue is present at 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 α− amino group. The reaction of peptide C with Edman reagent then CF3CO2H is shown below.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 1

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 Arg−Ala2, Gly, Ile and the partial structure of peptide C is found to be Ala2, Gly, Ile. Also, peptide D on three cycles of Edman degradation shows the partial structure of the peptide Ala−Ile−Gly. Since the peptide D contains four amino acid so there must be Ala residue at both sides of the peptide D.the peptide D structure is Ala−Ile−Gly-Ala. It is already mentioned that arginine is present at the N− terminal of the peptide. Therefore, the peptide C structure is Arg−Ala−Ile−Gly−Ala.

Answer 13

Conclusion

The structure of C from the given illustration is Arg−Ala−Ile−Gly−Ala.

Answer 14

Interpretation Introduction

(b)

Interpretation:

The b-type fragmentation for 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.

Expert Solution

Answer to Problem 27.53AP

The b-type fragmentation for M+1 ion of the peptide C is shown below.

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

The b-type fragmentation for M+1 ion of the peptide D is shown below.

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

where Arg is asparagine, Gly is glycine, Ala is alanine, 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.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 2

Figure 2

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

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

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

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

Conclusion

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

Answer 15

Interpretation Introduction

(b)

Interpretation:

The b-type fragmentation for 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 16

Expert Solution

Answer to Problem 27.53AP

The b-type fragmentation for M+1 ion of the peptide C is shown below.

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

The b-type fragmentation for M+1 ion of the peptide D is shown below.

H−Alam/z=72H−Ala−Ilem/z=185.1H−Ala−Ile−Glym/z=242.1H−Ala−Ile−Gly−Alam/z=313.1

where Arg is asparagine, Gly is glycine, Ala is alanine, Ile is isoleucine.

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

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.

EBK ORGANIC CHEMISTRY, Chapter 27, Problem 27.53AP , additional homework tip 2

Figure 2

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

H−Argm/z=157.1H−Arg−Alam/z=228.1H−Arg−Ala−Ilem/z=341.2H−Arg−Ala−Ile−Glym/z=398.2H−Arg−Ala−Ile−Gly−Alam/z=469.2

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