Download Chapter 25 Amino Acids, Peptides, and Proteins

Chapter 25
Amino Acids, Peptides, and Proteins
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 25. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Amino acids in which the two functional groups are separated by exactly one
carbon atom are called _______ amino acids.
Amino acids are coupled together by amide linkages called ____________ bonds.
Relatively short chains of amino acids are called ___________.
Only twenty amino acids are abundantly found in proteins, all of which are ___
amino acids, except for ____________ which lacks a chirality center.
Amino acids exist primarily as _______________ at physiological pH
The ___________________ of an amino acid is the pH at which the concentration
of the zwitterionic form reaches its maximum value.
Peptides are comprised of amino acid _________ joined by peptide bonds.
Peptide bonds experience restricted rotation, giving rise to two possible
conformations, called _______ and _______. The _______ conformation is more
stable.
Cysteine residues are uniquely capable of being joined to one another via
______________ bridges.
______ is commonly used to form peptide bonds.
In the Merrifield synthesis, a peptide chain is assembled while tethered to
__________________________.
The primary structure of a protein is the sequence of _____________________.
The secondary structure of a protein refers to the ________________________
_____________________ of localized regions of the protein. Two particularly
stable arrangements are the ___ helix and ____ pleated sheet.
The tertiary structure of a protein refers to its _________________________.
Under conditions of mild heating, a protein can unfold, a process called
_________________.
Quaternary structure arises when a protein consists of two or more folded
polypeptide chains, called _____________, that aggregate to form one protein
complex.
666
CHAPTER 25
Review of Skills
Fill in the blanks and empty boxes below. To verify that your answers are correct, look
in your textbook at the end of Chapter 25. The answers appear in the section entitled
SkillBuilder Review.
25.1 Determining the Predominant Form of an Amino Acid at a Specific pH
CONSIDER THE FOLLOWING AMINO ACID,
AND DRAW THE FORM THAT PREDOMINATES
AT PHYSIOLOGICAL pH.
O
H2N
OH
NH2
25.2 Using the Amidomalonate Synthesis
IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION:
1)
O
2)
O
O
OEt
EtO
H
OH
N
3)
O
4)
N
H
NH2
25.3 Drawing a Peptide
DRAW A
BOND-LINE
STRUCTURE FOR
THE TRIPEPTIDE
Phe-Val-Trp.
25.4 Sequencing a Peptide via Enzymatic Cleavage
IDENTIFY THE FRAGMENTS OBTAINED FROM ENZYMATIC CLEAVAGE OF THE FOLLOWING PEPTIDE:
Ala-Phe-Lys-Pro-Met-Tyr-Gly-Arg-Ser-Trp-Leu-Hist
trypsin
chymotrypsin
CHAPTER 25
667
25.5 Planning the Synthesis of a Dipeptide
IDENTIFY ALL REAGENTS NECESSARY TO PREPARE THE DIPEPTIDE Ala-Gly:
Ala
Boc
Gly
+
Ala
Gly
Boc
OCH3
Ala
Gly
OCH3
Ala
Gly
25.6 Preparing a Peptide using the Merrifield Synthesis
IDENTIFY ALL REAGENTS NECESSARY TO PREPARE THE PENTAPEPTIDE Ile-Gly-Leu-Ala-Phe:
POLYMER
Boc
Phe
Boc
POLYMER
Phe
Ile
Gly
Leu
Ala
Phe
Ile
Gly
Leu
Ala
Phe
Boc
POLYMER
Ala
Phe
POLYMER
POLYMER
Review of Reactions
Identify the reagents necessary to achieve each of the following transformations. To
verify that your answers are correct, look in your textbook at the end of Chapter 25. The
answers appear in the section entitled Review of Reactions.
Analysis of Amino Acids
O
O
R
O
H2O
COOH
OH
+
+
N
OH
NH2
O
O
O
CO2
RCHO
668
CHAPTER 25
Synthesis of Amino Acids
O
R
O
O
R
R
OH
OH
Br
O
O
O
OH
NH2
O
O
R
EtO
OEt
EtO
H
N
N
NH2
H
O
O
O
N
O
R
OH
OEt
R
H2 N
H
R
NaCN
H2 N
C
R
H
O
C
OH
H
O
OH
NHAc
O
OH
NHAc
OH
NH2
Analysis of Amino Acids
Ph
O
PEPTIDE
H
PEPTIDE
NH2
N
N
H
+
NH
O
H
R
S
N
R
Synthesis of Peptides
O
O
+
OH
H2 N
N
H
O
O
H2N
O
H
N
OH
OH
O
R
[H+]
ROH
O
H2N
O
H2N
OH
R
OR
R
NaOH
H2O
R
669
CHAPTER 25
Solutions
25.1. In each case, the chirality center has the R configuration.
HO
O
HO
C
NH2
H
O
C
NH2
H
CH3
CH(CH3)2
D-Alanine
D-Valine
25.2.
O
O
O
OH
OH
NH2
a)
OH
NH2
N
H
b)
S
O
NH2
c)
OH
d)
NH2
25.3.
a) Pro, Phe, Trp, Tyr, and His
b) Phe, Trp, Tyr, and His
c) Arg, His, and Lys
d) Met and Cys
e) Asp and Glu
f) Pro, Trp, Asn, Gln, Ser, Thr, Tyr, Cys, Asp, Glu, Arg, His, and Lys
25.4.
O
O
H3C
a)
H
O
O
N
O
N
b)
H
H
O
H
c)
O
H2N
d)
O
O
O
H
N
H
e)
NH
H
N
H
H
H
O
O
N
H
HO
N
H
O
HO
f)
H
O
H
N
H
H
25.5. Arginine has a basic side chain, while asparagine does not. At a pH of 11,
arginine exists predominantly in a form in which the side chain is protonated. Therefore,
it can serve as a proton donor.
25.6. Tyrosine possesses a phenolic proton which is more readily deprotonated because
deprotonation forms a resonance-stabilized phenolate ion. In contrast, deprotonation of
the OH group of serine gives an alkoxide ion that is not resonance-stabilized. As a result,
the OH group of tyrosine is more acidic than the OH group of serine.
670
CHAPTER 25
25.7.
a) 2.77
b) 5.98
c) 9.74
d) 6.30
25.8.
a) aspartic acid
b) glutamic acid
25.9. Leucine and isoleucine
25.10. The pI of Phe = 5.48, the pI of Trp = 6.11, and the pI of Leu = 6.00.
a) At pH = 6.0, Phe will travel the farthest distance.
b) At pH = 5.0, Trp will travel the farthest distance.
O
25.11.
H
25.12.
O
O
1) Br2 , PBr3
OH
OH
2) H2O
NH2
3) excess NH3
(racemic)
a)
O
O
1) Br2 , PBr3
OH
2) H2O
3) excess NH3
b)
O
OH
NH2
(racemic)
O
1) Br2 , PBr3
OH
2) H2O
3) excess NH3
c)
OH
NH2
(racemic)
CHAPTER 25
671
25.13.
COOH
NH2
NH2
NH2
Leucine
a)
COOH
COOH
b)
H2N
d) Glycine
c) Phenylalanine
Valine
COOH
25.14.
O
O
O
1) NaOEt
OEt
EtO
H
OH
2)
N
NH2
Br
O
a)
O
3) H3O+, heat
O
O
1) NaOEt
OEt
EtO
H
N
b)
OH
2) CH3Br
NH2
3) H3O+, heat
O
O
O
O
1) NaOEt
OEt
EtO
H
c)
OH
2)
N
NH2
Br
O
3) H3O+, heat
25.15.
O
O
OH
OH
NH2
a) alanine
O
OH
NH2
b)
valine
NH2
c)
leucine
25.16. Leucine can be prepared via the amidomalonate synthesis with higher yields
than isoleucine, because the former requires an SN2 reaction with a primary alkyl halide,
while the latter requires an SN2 reaction with a secondary (more hindered) alkyl halide.
672
CHAPTER 25
25.17.
O
S
O
1) NH4Cl, NaCN
H
S
2) H3O+
OH
NH2
a)
N
b)
O
O
N
H
1) NH4Cl, NaCN
H
OH
2) H3O+
N
NH2
NH
O
O
1) NH4Cl, NaCN
H
OH
2) H3O+
NH2
c)
O
O
1) NH4Cl, NaCN
H
2) H3O+
OH
NH2
d)
25.18.
O
H3C
O
1) NH4Cl, NaCN
H3C
2) H3O+
H
OH
NH2
alanine
a)
O
O
1) NH4Cl, NaCN
H
2) H3O+
OH
NH2
leucine
b)
O
O
1) NH4Cl, NaCN
2) H3O+
H
OH
NH2
valine
c)
25.19.
O
O
H2 C
a)
OH
NHAc
OH
b)
NHAc
O
O
OH
OH
c)
NHAc
d)
HO
NHAc
CHAPTER 25
25.20. Glycine does not possess a chirality center, so the use of a chiral catalyst is
unnecessary. Also, there is no alkene that would lead to glycine upon hydrogenation.
25.21.
HS
H2N
a)
H
N
O
H
N
OH
N
H
O
O
H
N
H2N
O
COOH
OH
O
O
N
H
H
N
O
O
N
H
NH2
c)
OH
25.22.
Leu-Ala-Phe-Cys-Asp or L-A-F-C-D.
25.23.
Cys-Tyr-Leu
25.24. Constitutional isomers
25.25.
H
N
H2N
N terminus
O
C terminus
OH
O
25.26. Steric hindrance results from the phenyl groups:
O
OH
NH
H2N
O
N
H
N
H
N
N
H
O
b)
O
HN
S
H2N
O
OH
O
H
N
O
O
OH
673
674
CHAPTER 25
25.27.
O
H2N
OH
N
H
O
S
S
O
H
N
H2N
OH
O
25.28.
HOOC
H
N
H 2N
O
O
O
a)
HOOC
H
N
H2N
HOOC
O
O
H2N
O
HOOC
O
H
N
O
H2N
O
O
b)
25.29.
D-Glutamic
acid
L-Leucine
Not naturally occurring
L-Isoleucine
L-Lysine
O
O
H2 N
H
N
N
H
N
NH2
O
H
N
S
O
N
H
O
N
H
O
HN
O
L-Isoleucine
OH
HN
O
Not naturally occurring
O
O
NH
NH
H2N
Bacitr acin A
HN
O
H
N
O
L-asparagine
H
N
HO
D-Phenylalanine
O
NH
O
D-Aspartic
acid
N
L-Histidine
O
O
675
CHAPTER 25
25.30. An Edman degradation will remove the amino acid residue at the N terminus, and
Ala is the N terminus in Ala-Phe-Val. Therefore, alanine is removed, giving the
following PTH derivative:
Ph
S
N
O
NH
CH3
25.31.
Met-Phe-Val-Ala-Tyr-Lys-Pro-Val-Ile-Leu-Arg-Trp-His-Phe-Met-Cys-Arg-Gly-ProPhe-Ala-Val
25.32.
Ala-Phe-Val-Lys
25.33. Cleavage with trypsin will produce Phe-Arg, while cleavage with chymotrypsin
will produce Arg-Phe. These dipeptides are not the same. They are constitutional
isomers.
25.34.
a)
(Boc)2O
Trp
Boc
Trp
DCC
Boc
Trp
Met
OCH3
Boc
Ala
Ile
OCH3
1) CF3COOH
2) NaOH, H2O
Trp
Met
Ala
Ile
[H+]
Met
Met
OCH3
Boc
Ala
CH3OH
b)
(Boc)2O
Ala
DCC
[H+]
Ile
CH3OH
Ile
OCH3
1) CF3COOH
2) NaOH, H2O
676
CHAPTER 25
c)
(Boc)2O
Leu
Boc
Leu
DCC
Boc
Leu Val
1) CF3COOH
OCH3
2) NaOH, H2O
[H+]
Val
OCH3
Val
CH3OH
25.35.
(Boc) 2O
Ile
Boc
Ile
DCC
Phe
Boc
Ile
[H+]
Phe
CH3OH
Phe
NaOH, H2O
OCH3
Boc
Ile
Phe
DCC
Ile
Phe Gly
OCH3
1) CF3 COOH
2) NaOH, H2O
Boc
Ile
Phe Gly
Gly
OCH3
OCH3
Leu
Val
CHAPTER 25
25.36.
677
678
CHAPTER 25
25.37.
a)
1) Boc
Cl
H
N
O
O
Ph
POLYMER
Phe-Leu-Val-Phe
2) CF3COOH
3) Boc
H
N
O
OH , DCC
4) CF3COOH
5) Boc
H
N
O
OH , DCC
6) CF3COOH
7) Boc
H
N
O
OH , DCC
Ph
8) CF3COOH
9) HF
b)
1) Boc
Cl
H
N
O
O
POLYMER
Ala-Val-Leu-Ile
2) CF3COOH
3) Boc
H
N
O
OH , DCC
4) CF3COOH
5) Boc
H
N
O
OH , DCC
6) CF3COOH
7) Boc
H
N
8) CF3COOH
9) HF
O
OH , DCC
679
CHAPTER 25
25.38.
(N terminus)
Val-Ala-Phe
(C terminus)
25.39. The regions that contain repeating glycine and/or alanine units are the most likely
regions to form β sheets:
Trp-His-Pro-Ala-Gly-Gly-Ala-Val-His-Cyst-Asp-Ser-Arg-Arg-Ala-Gly-Ala-Phe
25.40.
O
O
O
a)
H
N
H
O
O
H
b)
N
H
N
H
H2N
H
H
O
O
O
H
c)
N
H
O
N
H
d)
H
H
25.41. When applying the Cahn-Ingold-Prelog convention for assigning the
configuration of a chirality center, the amino group generally receives the highest
priority (1), followed by the carboxylic acid moiety (2), followed the side chain
(3), and finally the H (4). Accordingly, the S configuration is assigned to L amino
acids. Cysteine is the one exception because the side chain has a higher priority
than the carboxylic acid moiety. As a result, the R configuration is assigned.
25.42.
HO
C
H2N
O
HO
H
H
OH
H2N
CH3
a)
C
O
HO
H
H2N
CH2OH
b)
c)
C
O
HO
H
H2N
CH2Ph
d)
C
O
H
CH2CONH2
25.43.
a) Isoleucine and threonine
b) Isoleucine = 2S,3S. Threonine = 2S,3R
25.44.
O
S
S
O
R
OH
NH2
S
O
R
S
OH
NH2
O
R
OH
NH2
R
OH
NH2
25.45. The protonated form below is highly stabilized by resonance, which spreads the
positive charge over all three nitrogen atoms.
NH
H2N
H
O
N
H
Acid
OH
NH2
H2N
N
H
N
H
O
OH
NH2
680
CHAPTER 25
25.46. The protonated form below is aromatic. In contrast, protonation of the other
nitrogen atom in the ring would result in loss of aromatic stabilization.
O
O
Acid
OH
N
OH
H
NH2
NH
N
NH
NH2
25.47.
O
O
HO
O
OH
H
a)
N
H
O
HO
H
H
O
H
H
N
H
O
O
c)
O
b)
O
N
O
H
O
O
H
O
d)
H
N
H
25.48.
O
O
O
H
a)
N
H
O
H
N
H
b)
O
O
O
c)
N
H
N
H
O
H2N
H
H
H
H
O
O
O
H
d)
N
H
H
25.49.
a) 6.02
b) 5.41
c) 7.58
d) 3.22
25.50. Lysozyme is likely to be comprised primarily of amino acid residues that contain
basic side chains (arginine, histindine, and lysine), while pepsin is comprised
primarily of amino acid residues that contain acidic side chains (aspartic acid and
glutamic acid).
25.51.
O
O
H2N
a)
O
O
H
N
H
H
O
O
b)
H
N
H
H
OH
O
O
N
c)
H
H
O
d)
H
N
H
H
CHAPTER 25
681
25.52.
O
C
H2N
O
O
OH
C
O
O
H
H
H2N
R
O
H
C
H2N
R
O
O
+
H
C
H
O
NH2
R
R
(planar)
Racemic mixture
25.53.
The pI of Gly = 5.97, the pI of Gln = 5.65, and the pI of Asn = 5.41.
a) At pH = 6.0, Asn will travel the farthest distance.
b) At pH = 5.0, Gly will travel the farthest distance.
25.54.
O
H
H
H
a)
b)
O
O
c)
O
H
d) no reaction
25.55.
a) Methionine, valine, and glycine.
O
O
N
O
O
b)
c) The compound is highly conjugated and has a λmax that is greater than 400 nm
(see Section 17.12)
25.56.
O
O
1) NH4Cl, NaCN
H
2) H3O+
OH
NH2
25.57. Alanine can be prepared via the amidomalonate synthesis with higher yields than
valine, because the former requires an SN2 reaction with a primary alkyl halide,
while the latter requires an SN2 reaction with a secondary (more hindered) alkyl
halide.
25.58. The side chain (R) of glycine is a hydrogen atom (H). Therefore, no alkyl group
needs to be installed at the α position.
O
O
H3O+, heat
O
OEt
EtO
H
N
OH
NH2
O
682
CHAPTER 25
25.59.
O
O
1) Br2 , PBr3
OH
OH
NH2
2) H2O
3) excess NH3
a)
O
H
1) NH4Cl, NaCN
b)
O
O
OEt
H
NH2
O
EtO
1) NaOEt
OH
2) CH3I
N
c)
OH
2) H3O+
O
3) H3O+, heat
O
NH2
25.60.
O
O
1) Br2 , PBr3
OH
OH
2) H2O
3) excess NH3
NH2
(racemic)
a)
O
O
O
1) NaOEt
OEt
EtO
H
NH2
Br
O
b)
O
3) H3O , heat
O
2) H3O+
OH
NH2
c)
25.61. 205 = 3,200,000
25.62.
O
OH
NH
+
1) NH4Cl, NaCN
H
N
OH
2)
N
NHAc
CHAPTER 25
25.63.
1) Leu-Met-Val,
4) Met-Leu-Val,
2) Leu-Val-Met,
5) Val-Met-Leu,
3) Met-Val-Leu,
6) Val-Leu-Met
25.64.
O
O
O
H
N
H3N
N
H
O
O
O
NH3
25.65.
HN
H2 N
O
H
N
N
H
O
N Terminus
OH
O
H
N
H
N
N
H
O
S
O
H2N
H
N
N
H
O
O
25.67.
O
O
H
N
H3N
O
O
O
25.68.
H2N
O
SH
OH
cysteine
H2N
O
OH
valine
O
N
H
C Terminus
COOH
25.66.
H
N
OH
O
S
HO
O
OH
O
683
684
CHAPTER 25
25.69.
HO
HO
tyrosine
O
H2N
H
OH
H
N
O
H2N
H2N
OH
HO
O
HO
OH
O
N
N
O
glycine
serine
25.70. It does not react with phenyl isothiocyanate so it must not have a free N terminus.
It must be a cyclic tripeptide:
Gly
Ala
Gly
Phe
Phe
Ala
25.71.
a) Arg
+
Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg
b) Arg-Pro-Pro-Gly-Phe
+
Ser-Pro-Phe
+
25.72. Phenylalanine
O
N
R=
HN
S
25.73. Val-Ala-Gly:
H
N
H2N
O
O
N
H
OH
O
Arg
685
CHAPTER 25
25.74. There cannot be any disulfide bridges in this peptide, because it has no cysteine
residues, and only cysteine residues form disulfide bridges.
His-Ser-Gln-Gly-Thr-Phe-Thr-Ser-Asp-Tyr-Ser-Lys-Tyr-Leu-Asp-Ser-Arg-Arg-Ala-Gln-Asp-Phe-ValGln-Trp-Leu-Met-Asn-Thr
25.75. Prior to acylation, the nitrogen atom of the amino group is sufficiently
nucleophilic to attack phenyl isothiocyanate. Acylation converts the amino group into an
amide moiety, and the lone pair of the nitrogen atom is delocalized via resonance,
rendering it much less nucleophilic.
25.76.
O
H3N
a)
Boc
b)
O
OH
N
H
O
H2N
c)
O
O
OH
H3N
d)
O
25.77.
(Boc)2O
Phe
Boc
Phe
DCC
Boc
Phe
Ala
OCH3
[H+]
Ala
Ala
CH3OH
OCH3
25.78.
H
N
H 2N
O
O
CH3
OH
H2N
CH3
H
N
H 2N
O
H
N
O
OH
O
O
CH3
OH
H2N
O
H
N
O
OH
CH3
1) CF3COOH
2) NaOH, H2O
Phe
Ala
686
CHAPTER 25
25.79.
(Boc)2O
Phe
Boc
Ile
Phe
Ile
DCC
Boc
Phe
Ile
Val
Leu
OCH3
[H+]
Val
Leu
OCH3
Leu
Val
CH3OH
1) CF3COOH
2) NaOH, H2O
Phe
Ile
Val
Leu
25.80.
1) Boc
Cl
H
N
O
O
POLYMER
Leu-Val-Ala
2) CF3COOH
3) Boc
H
N
O
OH , DCC
4) CF3COOH
5) Boc
H
N
O
OH , DCC
6) CF3COOH
7) HF
Boc
H
N
O
OH
25.81.
25.82. A proline residue cannot be part of an α helix, because it lacks an N-H proton and
does not participate in hydrogen bonding. (The amino acid proline does indeed
have an N-H group, but when incorporated into a peptide, the proline residue does
not have an N-H group)
687
CHAPTER 25
25.83.
O
O
H
O
H2N
O
O
O
O
O
OH
O
OH
O
O
H
N
O
O
O
H
N
H
O
O
OH
O
H
N
+
O
O
O
OH
O
O
Br
O
HO
O
25.84.
25.85. The stabilized enolate ion (formed in the first step) can function as a base, rather
than a nucleophile, giving an E2 reaction:
H
O
H
O
OEt
EtO
H
N
O
EtO
OEt
EtO
H
O
Br
O
N
O
25.86. The lone pair on that nitrogen atom is highly delocalized via resonance and is
participating in aromaticity. Accordingly, the lone pair is not free to function as a
base.
688
CHAPTER 25
25.87.
O
OH
a)
HO
O
O
OH
Br
Br2
HO
OH
HO
b)
Br
25.88. At low temperature, the barrier to rotation keeps the two methyl groups in
different electronic environments (one is cis to the C=O bond and the other is
trans to the C=O bond), and they therefore give rise to separate signals. At high
temperature, there is sufficient energy to overcome the energy barrier, and the
protons change electronic environments on a timescale that is faster than the
timescale of the NMR spectrometer. The result is an averaging effect which gives
rise to only one signal.
25.89.
a) The COOH group does not readily undergo nucleophilic acyl substitution
because the OH group is not a good leaving group. By converting the COOH
group into an activated ester, the compound can now undergo nucleophilic
acyl substitution because it has a good leaving group.
b) The nitro group stabilizes the leaving group via resonance. As described in
Chapter 19, the nitro group serves as a reservoir for electron density.
c) The nitro group must be in the ortho or para position in order to stabilize the
negative charge via resonance. If the nitro group is in the meta position, the
negative charge cannot be pushed onto the nitro group.
25.90.
O
H2N
O
O
HCl, H2O
heat
OH
H2N
O
OH
OH
OH
NH2
threonine