(a) Interpretation: The m / z value of M + 1 fragment ion from b-type fragmentation of the given peptide 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.

Question

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Answer 1

Question

Chapter 27, Problem 27.27P

Interpretation Introduction

(a)

Interpretation:

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide 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 27.27P

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown below.

H−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

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 STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 1

Figure 1

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide 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−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Conclusion

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown above.

Interpretation Introduction

(b)

Interpretation:

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion 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 27.27P

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

Explanation of Solution

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

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 2

Figure 2

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide with M+1 fragment ion by subsequent addition of their residues in y-type fragment manner. The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Conclusion

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown above.

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Answer 2

Question

Chapter 27, Problem 27.27P

Interpretation Introduction

(a)

Interpretation:

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide 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 27.27P

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown below.

H−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

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 STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 1

Figure 1

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide 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−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Conclusion

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown above.

Interpretation Introduction

(b)

Interpretation:

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion 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 27.27P

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

Explanation of Solution

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

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 2

Figure 2

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide with M+1 fragment ion by subsequent addition of their residues in y-type fragment manner. The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Conclusion

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown above.

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Answer 3

Question

Chapter 27, Problem 27.27P

Interpretation Introduction

(a)

Interpretation:

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide 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 27.27P

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown below.

H−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

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 STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 1

Figure 1

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide 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−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Conclusion

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown above.

Interpretation Introduction

(b)

Interpretation:

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion 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 27.27P

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

Explanation of Solution

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

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 2

Figure 2

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide with M+1 fragment ion by subsequent addition of their residues in y-type fragment manner. The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Conclusion

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown above.

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Answer 4

Question

Chapter 27, Problem 27.27P

Interpretation Introduction

(a)

Interpretation:

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide 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 27.27P

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown below.

H−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

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 STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 1

Figure 1

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide 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−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Conclusion

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown above.

Interpretation Introduction

(b)

Interpretation:

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion 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 27.27P

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

Explanation of Solution

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

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 2

Figure 2

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide with M+1 fragment ion by subsequent addition of their residues in y-type fragment manner. The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Conclusion

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown above.

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Answer 5

Chapter 27, Problem 27.27P

Interpretation Introduction

(a)

Interpretation:

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide 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 27.27P

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown below.

H−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

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 STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 1

Figure 1

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide 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−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Conclusion

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown above.

Interpretation Introduction

(b)

Interpretation:

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion 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 27.27P

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

Explanation of Solution

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

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 2

Figure 2

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide with M+1 fragment ion by subsequent addition of their residues in y-type fragment manner. The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Conclusion

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown above.

Answer 6

Chapter 27, Problem 27.27P

Interpretation Introduction

(a)

Interpretation:

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide 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 27.27P

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown below.

H−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

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 STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 1

Figure 1

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide 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−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Conclusion

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown above.

Answer 7

Chapter 27, Problem 27.27P

Interpretation Introduction

(a)

Interpretation:

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide 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 8

Expert Solution

Answer to Problem 27.27P

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown below.

H−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

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 STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 1

Figure 1

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide 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−N m/z=115H−N−F m/z=262.1H−N−F−E m/z=391.2H−N−F−E−S m/z=478.2H−N−F−E−S−G m/z=535.2H−N−F−E−S−G−K−OH m/z=680.3

Answer 13

Conclusion

The m/z value of M+1 fragment ion from b-type fragmentation of the given peptide is shown above.

Answer 14

Interpretation Introduction

(b)

Interpretation:

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion 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 27.27P

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

Explanation of Solution

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

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 2

Figure 2

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide with M+1 fragment ion by subsequent addition of their residues in y-type fragment manner. The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Conclusion

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown above.

Answer 15

Interpretation Introduction

(b)

Interpretation:

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion 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 27.27P

The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.

H2−K−OH m/z=147.1H2−G−K−OH m/z=204.1H2−S−G−K−OH m/z=291.2H2−E−S−G−K−OH m/z=420.2H2−F−E−S−G−K−OH m/z=567.28H2−N−F−E−S−G−K−OH m/z=681.3

Where N is asparagine, F is phenylalanine, E is glutamic acid, S is serine, G is glycine, K is lysine amino acid.

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

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

EBK ORGANIC CHEMISTRY STUDY GUIDE AND S, Chapter 27, Problem 27.27P , additional homework tip 2

Figure 2

The given peptide is N−F−E−S−G−K where N is asparagine with m/z=114, F is phenylalanine with m/z=147.1, E is glutamic acid with m/z=129, S is serine with m/z=87, G is glycine with m/z=57 and K is lysine with m/z=128.1. The formation of the peptide with M+1 fragment ion by subsequent addition of their residues in y-type fragment manner. The m/z value of M+1 fragment ion from y-type fragmentation of the peptide in part (a) containing protonated H3N+— ion is shown below.