Introduction to Bioorganic Chemistry and Chemical Biology
Answers to Chapter 3
(in-text & asterisked problems)
Answer 3.1
OH
-O P
O
O
O
O
5'
O
3'
O
-O P
O
O
NH
N
N
O
O
N
NH
N
dash/wedge
drawing
O
-O P
O
O
NH2
O
O
-O P
O
O
5'
N
The labor of drawing a simple tetranucleotide helps
one to appreciate why we use a combination of
one-letter code and atom-numbering to discuss DNA.
N
O
N
O
3'
pTGCAp
NH2
NH2
N
N
N
O
-O P
O
O
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Answer
3.2
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In the gas phase, the charge separation costs 83 kcal mol–1, which is much more costly
than the maximum benefit of aromaticity (35 kcal mol–1). However, in water, where the
dielectric constant is about 78, charge separation costs only 1.1 kcal mol–1, which is a
small compared with the energetic benefits conferred by aromaticity.
Energy = ke
=
q1
q2
4.0 Å
ε distance
9.0 x 109 J•m
C2
= 5.8 x 10-19 J
(1.6 x 10 -19 C)(1.6 x 10 -19 C)
1
×
×
(4.0 Å)
1Å
10-10 m
R N
+
gas phase
water (25 ° C)
1 kJ 6.0 x 1023
1000 J
mol
O
ε=1
ε = 78
= 35 kJ mol–1
or 83 kcal mol–1 (1 cal = 4.18 J)
in gas phase ( ε = 1)
Energy = 1.1 kcal mol–1 in water at 25 °C ( ε = 78)
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It is important to recognize that the positive charge on nitrogen and negative charge on
oxygen is a Lewis structure formalism. In fact, the nitrogen atom has a partial negative
charge and most of the positive charge is distributed on the various hydrogen atoms.
Surprisingly, ab initio calculations HF/STO-3G predict a dipole moment of 4.8 Debye
units, not too different from the dipole moment predicted from a +1 charge and -1
charge separate by 4.0 Å (4.0 Debye units).
1
2
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 3
Answer 3.3
- ..
:O:
O
amide
N
N+
N
amidine
most basic and most nucleophilic
+N
N
.. N:
most basic and most nucleophilic
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Answer 3.4
O
HO
: OH2
N
: AH
O
+ H
O
HO
NH2
N
HO
HO
NH2
N
H +
N
N
N
NH
N
HO
O
HO
NH2
N
+
N
OH
HO
.. H
N
NH2
:N
+
O
HO
N
HO
NH
A H
..
O
HO
N
N
+
O
HO
NH2
N
HO
+
NH
H A
..
N
H N
N
: A-
H
N+
N
NH2
N
N
NH2
HN
N
N
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Answer 3.5
The major groove.
major
groove
H
N H
N
R
N
N
R
O
H N
N
R
N
O
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Answer 3.6
NH2
O
NH
NH2
NH2
N
NH2
N
N
N
H
NH
or
N
O
O
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Answer 3.7
Five possible 18-base probes (A–C) will have complete overlap with the 14-base
sequence. Probes A–C will have the highest GC content and therefore the highest Tm.
Probes D and E will have a slightly lower Tm.
target
probe
probe
probe
probe
probe
5'-CGGGGGTGGCGCAGTGAGGAGG-3'
A 3'-GCCCCCACCGCGTCACTC -5'
B 3'-CCCCCACCGCGTCACTCC-5'
C
3'-CCCCACCGCGTCACTCCC -5'
D
3'-CCCACCGCGTCACTCCCT -5'
E
3'-CCACCGCGTCACTCCCTC -5'
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Tm =
Tm =
Tm =
Tm =
Tm =
4x2
4x2
4x2
5x2
5x2
+
+
+
+
+
14x4 °C =
14x4 °C =
14x4 °C =
13x4 °C =
13x4 °C =
64 °C
64 °C
64 °C
62 °C
62 °C
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 3
3
Answer 3.8
415 =
1,073,741,824
416 =
4, 294,967,296
17
4 =
17,179,869,184
418 =
68,719,476,736
419 = 274,877,906,944
420 = 1,099,511,627,776
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There are 68 billion possible DNA sequences composed of 18 base pairs, over 20 times
more than the number of bases in the human genome. At most, a human-size genome
could contain only about 3 billion different 18 bp sequences. Thus, it is reasonably
unlikely that an 18-base sequence would be present in the human genome as a result
of random chance.
Answer 3.9
This hairpin was selected from a library of oligonucleotides because it binds to the amino
acid arginine. Only the first six and last six nucleotides form contiguous Watson–Crick
base pairs. NMR studies of the hairpin•arginine complex (PDB 1DB6) revealed additional
interactions not shown in the answer below, a Watson–Crick base pair between G9 and
C16 as well as non-Watson–Crick interactions between other pairs of bases.
CG T
G
T
CG C
1
5'-CGACCA A
3'-GCTGGTC
22
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Answer 3.10
TT
T
G
CAC T
3'-CCCGCTACC GTG A
GG
hairpin 5'-GGG GATGG
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Answer 3.11
B
..
R
NH2
..
-O
Ph
H
N
O
HO
N
H OH N
+
R
N
S
2C
-O
Me Me
2C
R
HO
H
N
B H
S
good nuc
O:
Ph
H
N
H
N
HO
:N
S
HO2C
Ph
H
H
N
R
N
HO
H N
good L.G.
+
S
HO2C
Me Me
Me Me
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Answer
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The guanidine moiety of arginine is protonated at pH 7, and these protonated arginine
side chains confer a significant affinity for the phosphate backbone of DNA. The urea
group of citrulline, which is not protonated at neutral pH, would have a lower affinity for
an anionic phosphate diester. Deimination of histones would release the bound DNA,
making it accessible to transcription factors.
arginine
H
N
H
H
N+
N
H
H
O
O
- P O
O
stable salt bridge with DNA
citrulline
O
N
H
N
H
O
O
- P O
O
H
less stable interaction DNA
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O:
Me Me
Ph
O
R
N
H
HN
HO2C
H
N
HO
S
Me
Me
Ph
4
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 3
Answer 3.13
DNA polymerase can only add to the 3ʹ end of a growing strand. The primer will be
identical to the 5ʹ end and complementary to the 3ʹ end.
5ʹ-CCATGCCTATGTTCATCGTGA-3ʹ
5ʹ-CCATGCCTATGTTCATCGTGAACACCAATGT……CTGGCCCCACTTACCTGCACCGCTGTTC-3ʹ
3ʹ-GTGAATGGACGTGGCGACAAG–5ʹ
Answer 3.14
NHBz
..
CH3NH2
B:
N
N
R
N
R
O
Ph
O
H H
N NHBz
H3C +
- N:
H3C
H
N
NHCH3
NH
- N:
N
R
O
N
N
R
O
O
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Answer
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OR i-Pr
H
P N
H A
O
i-Pr
N
N
N N
NC
N N
H
i-Pr2N :
OR + i-Pr
H
P N H
O
i-Pr
N
N
N N
NC
OR
H +
P
i-Pr
N
O
i-Pr
..
N
N -
NC
..
OR
P
i-Pr
..
N
O
i-Pr
H
N
N
N N
NC
OR
H
P
O + N
N
N N
NC
OR
P
N
O
N
N N
NC
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Answer
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B:
H O
:O
OH
OH
OH
O
Gal
Gal
O
Gal
O
O
O
OSO3Na
OSO3Na
O
O
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Answer 3.17
The amount of inhibitor drug is increasing from lane 1 to lane 6. At the highest
concentration of drug, most of the plasmid DNA remains in the fully supercoiled state.
(Figure adapted from I. Larosche et al., J. Pharmacol. Exp. Ther. 321: 526–535, 2007.)
least
inhibitor
sample
1
most
inhibitor
2
3
4
5
6
relaxed
partially supercoiled
supercoiled
actual [inhibitor] (µM)
~0
5
10
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20
50
100
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 3
Answer 3.18
5'- AACTGAATTTCAGGGGGATCCGCATGGCGT
-3'
AACTGAATTTCAGGGGGATCCGCATGGCGT
5'-5'-3'-3'
AACTGAATTTCAGGGGGATCCGCATGGCGT
3'-TTGACTTAAAGTCCCCCTAGGCGTACCGCA
-5'
3'-3'-5'-5'
TTGACTTAAAGTCCCCCTAGGCGTACCGCA
TTGACTTAAAGTCCCCCTAGGCGTACCGCA
A
B Design and synthesize two oligonucleotides with BamHI overhangs on the ends.
These will hybridize to any site cut with BamHI.
5'- AACTGAATTTCAGGGGGATCCGCATGGCGT
-3'
AACTGAATTTCAGGGGGATCCGCATGGCGT
5'-5'-3'-3'
AACTGAATTTCAGGGGGATCCGCATGGCGT
3'-TTGACTTAAAGTCCCCCTAGGCGTACCGCA
-5'
3'-3'-5'-5'
TTGACTTAAAGTCCCCCTAGGCGTACCGCA
TTGACTTAAAGTCCCCCTAGGCGTACCGCA
BamH1 BamH1
BamH1
GATCCGCATGGCGT
-3'
GATCCGCATGGCGT
GATCCGCATGGCGT
-3'-3'
GCGTACCGCA
-5'
GCGTACCGCA
design and
GCGTACCGCA
-5'-5' design
design
and
and
5'- AACTGAATTTCAGGGG
AACTGAATTTCAGGGG
5'-5'AACTGAATTTCAGGGG
synthesizesynthesize
synthesize
3'TTGACTTAAAGTCCCCCTAG
3'-3'TTGACTTAAAGTCCCCCTAG
TTGACTTAAAGTCCCCCTAG
kinase
kinase
kinase
pGATCCGCATGGGTTTCAAATCG
pGATCCGCATGGGTTTCAAATCG
pGATCCGCATGGGTTTCAAATCG
GCGTACCCTTTGTTTAGCCTAGp
GCGTACCCTTTGTTTAGCCTAGp
GCGTACCCTTTGTTTAGCCTAGp
5'- AACTGAATTTCAGGGG
GATCCGCATGGGTTTCAAATCG
GATCCGCATGGCGT
-3'
AACTGAATTTCAGGGG
GATCCGCATGGGTTTCAAATCG
5'-5'GATCCGCATGGCGT
AACTGAATTTCAGGGG
GATCCGCATGGGTTTCAAATCG
GATCCGCATGGCGT
-3'-3'
3'-TTGACTTAAAGTCCCCCTAG
GCGTACCGCA
-5'
GCGTACCCTTTGTTTAGCCTAG
GCGTACCGCA
GCGTACCCTTTGTTTAGCCTAG
3'-3'TTGACTTAAAGTCCCCCTAG
GCGTACCGCA
-5'-5'
GCGTACCCTTTGTTTAGCCTAG
TTGACTTAAAGTCCCCCTAG
T4 DNA
ligase
Mg•ATP
DNA
ligase
T4T4
DNA
ligase
Mg•ATP
Mg•ATP
5'- AACTGAATTTCAGGGG
GATCCGCATGGGTTTCAAATCG
-3'
GATCCGCATGGCGT
AACTGAATTTCAGGGG
GATCCGCATGGGTTTCAAATCG
5'-5'GATCCGCATGGCGT
AACTGAATTTCAGGGG
GATCCGCATGGGTTTCAAATCG
-3'-3'
GATCCGCATGGCGT
3'-TTGACTTAAAGTCCCCCTAG
GCGTACCGCA
GCGTACCCTTTGTTTAGCCTAG
-5'
GCGTACCGCA
GCGTACCCTTTGTTTAGCCTAG
3'-3'TTGACTTAAAGTCCCCCTAG
GCGTACCGCA
GCGTACCCTTTGTTTAGCCTAG
-5'-5'
TTGACTTAAAGTCCCCCTAG
C The sequence contains two BamHI sites. Any new overhangs on the ends could
hybridize within the oligonucleotide or between oligonucleotides.
GATCCTTTCATAAGTGGTGGGATCCCCATTCAATTG
GATCCTTTCATAAGTGGTGGGATCCCCATTCAATTG
GATCCTTTCATAAGTGGTGGGATCCCCATTCAATTG
GAAAGTATTCACCACCCTAGGGGTAAGTTAACCTAG
GAAAGTATTCACCACCCTAGGGGTAAGTTAACCTAG
GAAAGTATTCACCACCCTAGGGGTAAGTTAACCTAG
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Answer 3.19
O
O
CF 3
HN
O
S-
..
N
R
O
O
F2
.. C
- F
HN
N
R
H2..
N
O
Enz
O
CF 2
HN
S
Enz
N
R
HN
S
Enz
HN
.. CF 2
-
N
R
O
Enz
S
O
HN
N
R
O
Enz
Enz
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Answer 3.20
O
N
O:
-
O
N
H A
O
..
H
N
H A
..
-A
..
Gua
N
N
..
OH
Gua
H
N
N
H A
Gua
O
O
N
N
R
NH
N
N
+ N
N
R
N
NH HN
N
N
R
O
O
N
N
H2N
..
N
H
N
:O -
N
HN
N
R
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N
N
R
O
..
N
Gua
N
H
+
N
+
OH2
- H A
O
+
Gua N N
O
HN
NH
+
N
N
N
H
N
2
+
N
..
AH
N
OH
Gua + N
H H
O
N
Gua + N
OH2+
..
..
..
Gua
..
+
NH2 N O
..
O+
N
..
:O
N
N
R
Enz
HN
CF 2
5
6
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 3
Answer 3.21
..N
C6H13
C6H13
N
+
C6H13
N
..
Nu
S C N
fascicularin
O
N+
aziridinium ion
NH
N
DNA
N
NH2
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Answer 3.22
abstraction of the two H atoms is not concerted
H
H
•
•
•
H H
•
H H
abstract
H atom
H
abstract
H atom
H
+
H H
H
H
Answer 3.23
H
uncialamycin
O
HN
reducing
intracellular
environment
R
O
OH
O
HO
OH
OH
H2O
R = CH(OH)CH 3
OH
H
R
HN
HO
HO
OH
reduced
in the cell
Bergman
rearrangement
HO
HN
R
OH
HO
HN
HO
HO
O
R
OH OH
S N1
OH
HO
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OH
Answer 3.24
H
H
H
H
•
H
H
H
H
H
H
H
H
H
H
H
H
•
H
H
H
H
H
•
H
H
H
H
H
H
H
•
H
H
H
H
H
H
H
H
H
H
H
H
H
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H
H
H
H
+
H
H
H
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 3
7
*Answer 3.26
A
OH
-O P
O
O
The conformational depiction
below and the dash/wedge
depiction to the right are
equally acceptable
B
O
O
HO
O
Thy
-O
O
-O P
O
O
Gua
O
O
P O
O
-O
O
d( GACA)
Ade
"d" prefix means 2-deoxy
OH
-O
O
O
P O
O
-O P
O
O
Thy
O
-O
O
O
P O
O
-O
O
pTATA
Ade
O
P O
O
presence of Thy
suggests that this is DNA
C
O
HO
Thy
-O
O
O
P O
Ade
HO
O
O
-O P
O
O
Thy
O
Ade
HO
Gua
O
-O P O
O
OH O
O
-O
O
GUCU
Ura
P O
OH O
O
-O
presence of Ura
suggests that this is RNA
Cyt
O
P O
OH O
HO
Ura
OH
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*Answer 3.27
H
N H
N
dR
N
N
N
N
CH3
H N
N
N
O
dR
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*Answer 3.30
A In PDB 1AIO, cisplatin forms a crosslink between the N7 atoms of guanine bases
(rendered in yellow).
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B The
platinum
atom sits in the major groove.
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C The crosslink is intrastrand, between adjacent guanine bases.
Ade
O
O
P O
Cyt
O
-O
O
P O
O
HO
Ade
8
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 3
Answer 3.32
*A 3ʹ-AGCTTACGTAATAAGCA-5ʹ or
5ʹ-ACGAATAATGCATTCGA-3ʹ
*Answer 3.33
A The following secondary structures are expected for each oligonucleotide or pair of
oligonucleotides. The sites of oxidation are shown with arrows.
arrows highlight
sites of oxidation
ssDNA
5'-CATGCGTTCCCGTG -3'
duplex formation
5'-CATGCGTTCCCGTG -3'
3'-GTGCGCAAGGGCAC -5'
hairpin or
duplex with mismatch bulge
5'- AGTCTA
3'- TCAGAT G
duplex with mismatch bulge
5'-ACGTCAG
3'-TGCAGTC
hairpin or
duplex with mismatch bulge
5'- AGTCTAT G
3'- TCAGATT G
strand not analyzed
G
5'- AGTCTA
3'- TCAGAT
TAGACT -3'
ATCTGA -5'
G
or
G
TGGCAT -3'
ACCCGTA -5'
G
strand not analyzed
TGGGT
TAGACT -3'
ATCTGA -5'
TGGGT
5'-AGTCTA
3'-TCAGAT
or
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B
On the basis
ofbythe
structures, the susceptible bases seem to be guanidines that
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are not protected by both base-pairing and π stacking. Therefore, Gs at the ends of
duplexes or in bulges or loops are susceptible.
*Answer 3.37
MeO
MeO
N
O
O
HO O
hv
O
O
O OH
-O
*
MeO
hv
R 2N
O
or...
HO
N
CO2-
2C
MeO
ON
OH
R'
MeO
N
MeO
N
N
Ca
O
O
HO O
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*Answer 3.38
A
O
N
O:
-
O
N
H A
O
..
H
N
H A
..
-A
N
N
..
OH
N
H A
Ph
O
..
N
Ph
N
+
OH2
H
+
N
N
- H A
O
+
Ph N N
..
Ph
Ph
H
N
OH2+
..
H H
O
N
Ph + N
..
AH
N
OH
Ph + N
..
..
Ph
..
+
NH2 N O
..
O+
N
..
:O
OH
..
O
R 2N
R'
O
R'
CO2-
MeO
+ ON
O
H
MeO
NH
2C
-O
OH
N
MeO
N
O
MeO
Ca2+
H2O
N
Ca
MeO
MeO
NO2
O
O
O OH
N
O:
-
N
H A
N
H A
N
..
OH2+
..
-A
H
O
..
H to Bioorganic Chemistry
H
H H Introduction
- H Aand Chemical Biology: Answers to Chapter 3
+
O
O
N
N
N
O
Ph + N
Ph
N
Ph
N
..
..
..
Ph
..
+
NH2 N O
..
B
Ph
N
N
..
OH
H A
Ph
..
N
N
+
OH2
+
Ph N N
..
AH
N
OH
Ph + N
as in part A.
NH2
N
N
DNA
O
..
N
DNA
+
OH2
:N-
N
DNA
O
C
N
H2O
N
O
guanine
N
N
DNA
N
NH2
H2O
N
DNA
O
H
O
+
NH
N
DNA
O
A-
O
NH
N
DNA
O
O
O
N
N
DNA
NH
N
O
NaNO 2
HCl
NH
H A
N
N
DNA
NaNO 2
HCl
N
N
DNA
OH
OH
+
N
O
NH2
adenine
H
A
..
N
2
..
+N
OH2+
..
N2+
N
N
DNA
NH
N
H
O
Introduction to Bioorganic Chemistry and Chemical Biology | A3243
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*Answer 3.41
A
:
B
N
O 2N
N
H
H
:
:B
:N
O
N
O 2N
N
CH3
N
H
O
N
N
CH3
O 2N
N
C
NH
O:
N
H B
N
CH3
O 2N
N
C
NH
:N
+
H
OH
N
C
H
H
Introduction to Bioorganic Chemistry and Chemical Biology | A3246
Van Vranken & Weiss | 978-0-8153-4214-4
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*Answer 3.42
AWith a calculator: 430 = 1,152,921,504,606,847,000 = 1.2 × 1018
Even better, work this without a calculator using the approximation 410 ≈ 106: [410]3 =
[106]3 = 1018
B
5'-GGGAGAATTCCCAGACC-3'
5'-GGGAGAATTCCCAGACCNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNCTGAGGGAAATTCTCCC-3'
3'-GACTCCCTTTAAGAGGG-5'
5'-GGGAGAATTTCCCTCAG-3'
C
AC GT
5'-GGGATCGAA
A
G
3'-CCCTAGCTT
CC GC
Arg
D
H
+
N
H
N
NH2
H
+
N
N
NH2
Arg
H
+
N
H
N
NH2
B:
N
N+
C
H
H
H
NN+
C
H
H
9
5'-GGGAGAATTTCCCTCAG-3'
C
AC GT
5'-GGGATCGAA
A Bioorganic Chemistry and Chemical Biology: Answers to Chapter 3
10
Introduction to
G
3'-CCCTAGCTT
CC GC
Arg
D
H
H
N
H
+
N
N
NH2
H
H
N
H
or
O
H
+
N
H
N
N
NH2
H
N
H
O
N
or
H
N
N
O
dC
*Answer 3.44
Me
OH
Me
HO
DNA:
Me
Me
HO
..
SR
OH
OH
OH
SR
OH
OH
DNA
Me
Me
Me
+H O
2
OH
SR
OH
H A
Me
OH
Me +
OH
Me
OH
: SR
-
O
Me
OH
Me
Me
OH
Me
SR
OH
*Answer 3.48
Introduction
Chemistry and Chemical Biology | A3251
Hto Bioorganic
H
N | 978-0-8153-4214-4
Van Vranken & Weiss
© www.garlandscience.com design by www.blink.biz
N
N
DNA
N
H O
+
N
DNA
N
R
O
HN
HO
Introduction to Bioorganic Chemistry and Chemical Biology | A3255
Van Vranken & Weiss | 978-0-8153-4214-4
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*Answer 3.51
A
B
N
N
N
..
HN
H
N
dC
Introduction to Bioorganic Chemistry and Chemical Biology | A3248
Van Vranken & Weiss | 978-0-8153-4214-4
© www.garlandscience.com design by www.blink.biz
Me
NH2
H
H
N
dC
Arg
H
N
H
+
N
Arg
H
N
N
+
H
N
N
O
N
H
O
O
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 3
*Answer 3.54
O
Introduction to Bioorganic Chemistry and Chemical
Biology | A3258
OMe
NHCO2Me
Van Vranken & Weiss | 978-0-8153-4214-4
HO
NHEt
Sby www.blink.biz
design
© www.garlandscience.com
O
O
OH
N
O
aromatic
by-product
O
can
intercalate
between
base pairs
Me S
Me O
HN
OH
Introduction to Bioorganic Chemistry and Chemical Biology | A3261
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*Answer 3.57
5ʹ-TCCTNNAGGA-3ʹ
head (H)
A The oligosaccharide
is asymmetric with a head (H), a tail (T), a left-hand side and a
OH
right-hand
side.Me
MeO
head (H)
O
HO
OH
MeO
O
MeO
Me
I
O
HO
MeO
lefthand
side
HN
HO
HN
O
HO
Me
EtHN
O
tail (T)
hand
side
O
O
O
O
Me
Me
O
OMe
Me right-
HO
O
I
side
left- MeMeO
hand
O
S
side
HO
O
O
S
O
Me rightMeOhand
O
O
O
EtHN
OMe
tail (T)
B Because the minor groove is also asymmetric, the 3ʹ end is different from the 5ʹ end.
Therefore the oligosaccharide will bind in a preferred orientation.
5'
|
3'
|
T
A
3'
H 5'|
|
C
G
C
GT H A
G
T T AC
|
|
C
G
3'
5'
T T A
|
3'
|
5'
11
12
C
G
T T A
|
|
Introduction to Bioorganic Chemistry and Chemical
Biology:
Answers to Chapter 3
3'
5'
C If you link the oligosaccharide tail to tail, each half still wants to bind with the original orientation.
5'
3'
|
|
T
C
C
T
N
N
A
G
G
A|
A
G
G
T A
N
O
N
T T
C
C
H T
|
H
3'
5'
*Answer
3.58 Chemistry and Chemical Biology | A3264
Introduction to Bioorganic
Van Vranken & Weiss | 978-0-8153-4214-4
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- Cl :
NC
para-benzyne
intermediate
+
- ..
O
HO
NC
- Cl :
O
OH
OH
O
enediyne
precursor
- ..
+
O
HO
NC
O
OH
OH
O
Introduction to Bioorganic Chemistry and Chemical Biology | A3265
Van Vranken & Weiss | 978-0-8153-4214-4
© www.garlandscience.com design by www.blink.biz
O
O
OH
HO
OH
O