Introduction to Bioorganic Chemistry and Chemical Biology
Answers to Chapter 5
(in-text & asterisked problems)
Answer 5.1
H27
PDB 3ZNF
C8
C5
H21
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Answer
5.2
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+H
2N
NH2
NH
IRGERA
N
H
O
H
N
N
H
O
2N
O
O
H
N
N
H
O
NH
+H
O
H
N
O
O+H
NH2
NH2
2N
NH
ent-IRGERA
N
H
O
H
N
N
H
O
2N
O
O
H
N
N
H
O
NH
+H
O
H
N
O
O+H
NH2
NH2
2N
NH
O
AREGRI
O
O
H
N
N
H
N
H
O
NH
+H
2N
NH2
O
H
N
O
O
O-
N
H
H
N
1
2
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 5
+H
NH2
2N
NH
O
ent-AREGRI
N
H
O
O
H
N
N
H
O
NH
this “retro-inverso“ peptide is the best
match of the natural sequence IRGERA
+H
2N
O
H
N
O
O
N
H
H
N
O-
NH2
Answer 5.3
A For NKDVLRRMKK:
Residue
Charge
Arg (R)
+1 × 2
Lys (K)
+1 × 3
+H N terminus
3
+1 × 1
Glu (E)
–1 × 0
Asp (D)
–1 × 1
–O
–1 × 1
2C– terminus
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Net charge
+4
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Similar calculations for RFRGDYFAK and KGNIDKFTEK yield net charges of +2 and
+1, respectively.
BThe peptide is expected to bind better to a negatively charged surface through
ionic interactions, because opposite charges attract.
Answer 5.4
Residues in the peptide are labeled in red; residues in MDM2 are labeled in blue.
L26
V99
M62
W23
L54
F19
I61
I99
F91
L57
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Answer
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H383, H387, and E411.
H383
E411
H387
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Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 5
Answer 5.6
+
O
O
O
O:
OH+
H A
OH
-A:
HO
H
HO
+
HO
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Answer
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HN
N
O
O
O
O
O
N
N
H
OH
Fmoc-PNA-T-OH
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Answer
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Pro and Gly prefer turns
Val, Ile, and Tyr prefer sheets
HO
O
NH
O
H
N
N
H
OEt
O
N
H
Et O
H
N
N
O
O
N
H
N
O
O
NH
H
N
O
OH
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Answer 5.9
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A A1,3 strain is most important for ϕ angles; A1,2 strain is most important for ψ angles.
B
C
CONH
side chain
N
H H
side chain
O
CONH
O
H
side chain
N
H H
N
H
= 0°
= 160°
CONH
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side chain
O
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H
N
H
= 0°
CONH
= 160°
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Answer 5.10
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- O:
+ S: + H A
OH
+S
:
+ A-
HS
.. R
-A :
+ OH
S:
S
2
R
+
S:
S
..
R SH
R
A H
OH
S:
S
H + R
R + S
R
S
H
-A:
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R
S
S
R
: OH
S:
S
R
3
4
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 5
Answer 5.11
Cys
R
+
Cys-S P
R
R
: PR 3
S S
Cys
R
Cys-S P O
RR H
R ..
Cys-S P O RR
:B
R
H
Cys-S P O +
RR H
: OH2
Cys-S:
:B
Cys-SH
H B
R
R
P O
R
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Answer 5.12
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There are seven WD domains in each half of the dimer, leading to a total of 14 WD
domains.
Answer 5.13
A
H A
..
HN
+H
2N
R
3N
N H
+
H
R
R
N
+ H
R
R
R
N
R
B
-A:
enamine
R
R
NH
..
R
+H
+ H
N
H
R
3N
: A-
R
2N
H
R
..NH2
H
R
NH
.. 2
+H
NH2+
-A:
A H
R
R
NH
imine
: A-
H A
..
H2N
R
: A-
R
N:
H
R
R
..
NH2
R
+H
H2N
R
NH2+
R
H2N
A H
R
NH
R
NH2
R
H2..N
aldol cross-link
R
*Answer 5.14
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NH3+
NH3+
spinorphin: LVVYPWT
NH3+
O
H2+ O
N
O
H
N
N
H
H
N
N
H
O
O
H
N
N
H
O
+H
O
3N
O
H
N
N
H
O-
OH
O
N
N
H
O
NH
N
H
O
O
H
N
O-
O
HO
O
NH
NH3+
NH2+
H2N
SV40 NLS sequence: PKKKRKV
N
H
OH
NH2
O
HN
O
+H
3N
N
H
S
H
N
O
O
N
H
H
N
O
O
O
N
H
O
NH2
N
O
N
H
H
N
O
O
O
O
O-
H
N
N
H
O
O
N
S
HN
H
O
HN
O
S
N
OO
H
N
NH
O
O
oxytocin: CYIQNCPLG
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O
O
N
H H
HO
S
phakellistatin 13:
cyclo-[FGPTLWP]
H
N
N
malformin A: cyclo-[D C D CV D LI]
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 5
*Answer 5.15
A substance P
RPKPQQFFGLM
B antimicrobial peptide
cyclo-[RRWWRF]
or cyclo-[RWWRFR]
or cyclo-[WWRFRR]
or cyclo-[WRFRRW]
or cyclo-[RFRRWW]
or cyclo-[FRRWWR]
or cyclo-[RRWWRF]
C synthetic integrin antagonist
CWLDVC
*Answer 5.17
A
Residue
Charge
Arg (R)
+1 × 0
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+1 ×and
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+H N terminus
+1 × 1
3
Glu (E)
–1 × 0
Asp (D)
–1 × 5
–O
–1 × 1
2C terminus
Net charge
–5
Residue
Charge
Arg (R)
+1 × 6
Lys (K)
+1 × 2
+H
+1 × 1
B
3N terminus
Glu (E)
–1 × 0
Asp (D)
–1 × 0
–O
–1 × 0
2C terminus
Net charge
+9
5
6
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 5
*Answer 5.19
:B
H
S
O
N
H
O
O
N
H
HN
..
-O S
O
N +
H2
..
SO
HN
H B+
O
HN
..
O
S
O
O
H2N
HN
B H
O
-O S
..
N
H
O
HN
..
-S
HS
O
H2N
O
H2N
HN
HN
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*Answer
5.23
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O:
Asp
H A
CF 3
O
O
O
O
Asp
OH+
R
CF 3
+
O
O
O
H
CF 3
OH
A H
NO : O
Asp
+
O
O:
HO
: A-
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*Answer
5.24
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Trp
O
O:
Asp
H A
OH+
Asp
OH
Asp
Trp
-A:
O
O
+
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N
H
H
Trp
+
N
H
N
H
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 5
7
*Answer 5.26
A Initial protonation occurs on the amide carbonyl, not the amide nitrogen.
A Initial protonation occurs on the amide carbonyl, not the amide nitrogen.
H A
..
H A
..
O
Ph Ph
O Ph Ph
N
Ph
R
R
N
Ph
H
H
H
H
Si i-Pr
i-Pr Si i-Pr
i-Pr
i-Pr
.. i-Pr
.. O
O
O
O-
OH++ Ph
OH Ph
R
R
CF 3
CF 3
Ph
Ph
Ph
Ph
N
N
H
H
H
H
i-Pr - Si
i-Pr Si
CF 3COO
CF 3COO
OH
OH
+
NH +
NH
R
R
Ph + Ph
Ph + Ph
Ph
i-Pr
Ph
i-Pr
i-Pr
i-Pr
H
O H
O
OH
OH
+CPh3
+CPh3
R
R
i-Pr
i-Pr
i-Pr
i-Pr Si i-Pr
Si i-Pr
CF 3COO
CF 3COO
NH
..
NH
..
H A
H A
R
R
NH2++
NH2
: A: A-
O
O
R
R
NH2
NH2
Ph
Ph
Ph
Ph
Ph H Ph
H
B Protonation of N would disrupt the aromaticity of the imidazole ring.
B Protonation of N would disrupt the aromaticity of the imidazole ring.
Therefore initial protonation occurs on the N not N .
Therefore initial protonation occurs on the N not N .
Ph Ph
Ph Ph
Ph Ph
Ph Ph
H A
Ph
Ph
H A
..
Ph
Ph
..
N
N
N
N
N
N
R
R
N
..
N
..
C
C
R
R
D
D
R
R N
N
H
H
R
R
H A
H A
H A
H A
..
..
O
O
R
R
R
R N
N
H
H
O
O
A H
OH
A H
OH
R ..
..
R N
O
N
O
H
H
H
H
i-Pr Si
i-Pr Si
CF 3COO
CF 3COO
i-Pr
i-Pr
i-Pr
i-Pr
R
R
N
N
H
H
H
H
+O
+O
H A
H A
..
..
O
O
R
R
N+
N
H+
H
R
R
OH++
OH
OH
OH
R
R N
N
H
H
O
O
H
H
i-Pr - Si
i-Pr Si
CF 3COO
CF 3COO
H+
H
N+
N
+
+
OH
OH
O
O
N
N
H
H
H
H
i-Pr - Si
i-Pr Si
CF 3COO
CF 3COO
+
+
+
+
+
+
H
O H
O
: A- O::
H
: A-O
H
R
R
R ..
..
R N
R N
R NH2
O
O
N
N
NH2
H2++ O
H2++ O
H2
H2
a concerted, one-step E2 mechanism is also plausible
a concerted, one-step E2 mechanism is also plausible
A
A
+ CO2
+ CO2
+
+
i-Pr3SiO 2CCF 3
i-Pr3SiO 2CCF 3
+ Me3C-H
+ Me3C-H
E Note that protonation of thioethers is extremely unfavorable (pKa' < -8).
E Note that protonation of thioethers is extremely unfavorable (pKa' < -8).
Cys
Cys
H A
H A
Ph
Ph
Ph Ph
Ph Ph
..
..
S
S
H
H
i-Pr - Si
i-Pr Si
CF 3COO
CF 3COO
i-Pr
i-Pr
i-Pr
i-Pr
Cys
Cys
Ph + Ph
Ph + Ph
Ph
Ph
H
+H
+S
S
Ph
Ph
Ph
Ph Ph
Ph
i-Pr3SiO 2CCF 3
i-Pr3SiO 2CCF 3
Cys
Cys
SH
SH
+ Ph3C-H
+ Ph3C-H
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+
+
+CPh3
+CPh3
Ph + Ph
Ph + Ph
Ph
i-Pr
Ph
i-Pr
i-Pr
i-Pr
i-Pr3SiO 2CCF 3
i-Pr3SiO 2CCF 3
+ Ph3C-H
+ Ph3C-H
i-Pr3SiO 2CCF 3
i-Pr3SiO 2CCF 3
+ Ph3C-H
+ Ph3C-H
+
+
i-Pr
i-Pr
i-Pr
i-Pr
8
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 5
*Answer 5.28
A KVKVKVKVKVKVK, because β-branched amino acids have high β-sheet propensities
and low α-helix and turn propensities.
B GPPGPPGPPGPPGPPGPPGPP, because Pro has low α-helix and β-sheet propensities.
CNLEDKAEELLSKNYHLENEVARL, because it has a leucine zipper motif with Leu at
every seven residues.
D GPPGPPGPPGPPGPPGPPGPP, because (GPP)n forms a collagen triple helix.
E
GPPGPPGPPGPPGPPGPPGPP, because Pro has low α-helix and β-sheet propensities.
*Answer 5.31
AThe calcium-binding loops are underlined: MADQLTEEQIAEFKEAFSLFDKDGDGTITTKELGTVMRSLGQNPTEAELQDMINEVDADGNGTIDFPEFLTMMARKMKDTDSEEEIREAFRVFDKDGNGYISAAELRHVMTNLGEKLTDEEVDEMIREADIDGDGQVNYEEFVQMMTAK
BThe first calcium-binding loop is positions 21–32, -DKDGDGTITTKE-. The first and
last positions of the calcium-binding loop are the only positions conserved as anionic amino acids. Therefore Asp21 and Glu32 are most likely to bind as anionic carboxylates. In the crystal structure for PDB 1CFC, both of the side-chain carboxylate
oxygens of Glu32 bind to the calcium ion. Because asparagines are also tolerated
at the second and third ligand positions, Asp23 and Asp25 probably coordinate as
neutral ligands through the side-chain carbonyls. Because phenylalanine is also tolerated at the fourth ligand position, it is likely that the backbone carbonyl, and not
the side chain, is coordinating to calcium. Because serine is a common ligand for the
fifth position, Thr29 probably binds through the side-chain hydroxyl. As a result of
the high pKa of alcohols compared with carboxylic acids, Thr29 probably binds as a
neutral OH ligand.
Asp25
H
N
O
Thr27
HN
Asp23
O
O
OH
HN
OH
O
Ca
O
O
NH
OH
O O
Asp21
OH
NH
O
CH3
Thr29
O
O
HN
NH
Glu32
O
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C Van
Glycine
highly
flexible and can adopt tight turns. Isoleucine and valine are
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β-branched hydrophobic amino acids that limit the flexibility of the side chain and
loop.
*Answer 5.32
ATwo types of dimers are known. The dimer based on swapping the S-peptide (PDB
1BSR) has the S-peptide from the first molecule bound in the cleft of the second
molecule, and vice versa.
domain-swapped dimer
domain-swapped dimer
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 5
BThere may be many solutions to this problem. Any mutation to the loop that prevents
the helix from binding to the cleft in residues 1–113 will favor the formation of the
domain-swapped dimer. Investigators have induced dimerization by generating a
deletion variant Δ(114:119) that lacks the flexible loop.
wild type
protein
engineering
(truncate loop)
deletion
mutant
domain-swapped dimer
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*Answer
5.36
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bonds with limited rotation due to A1,3 strain
A 1,3 strain
R
R'
H
HO
R"
O
R
OH
O
O
OH
H
R"
R'
O
NH2
OH
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*Answer
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A Cys28 makes contacts with DNA (rendered as a brown surface).
BThe nearby cysteine residue is Cys120.
C CysteineS thiols
are Hefficient at conjugate addition reactions.
H
Cys
Cys
O
S
H
O
H
HO
HO
H O
O
SO
H
Cys
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S
Cys
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9
10
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 5
*Answer 5.41
resonance-stabilized
R
R
X•
H
+
X H
O
R
•
R
tautomerization
OH
R
H
•
O
O
R
O
etc
•
•O
O
R
R
H
O
keto tautomers
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R
HO
enol tautomers (aromatic)