answers to chapter 3 - garland · pdf fileintroduction to bioorganic chemistry and chemical...
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Introduction to Bioorganic Chemistry and Chemical Biology 1
Answers to Chapter 3(in-text & asterisked problems)
Answer 3.1
Answer 3.2
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.
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).
Introduction to Bioorganic Chemistry and Chemical Biology | A3211Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
pTGCAp
OO
P- OO
OO
PO
- OO
OO
POH
- OO
O
O
OPO
- OO
OP- O
O
O
NN
N
N NH2
NHN
N
N O
NH2
NHN
O
O
NN
NH2
O
5'
3'
5'
3'
dash/wedgedrawing
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.
Introduction to Bioorganic Chemistry and Chemical Biology | A3212Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
=(1.6 x 10 -19 C)(1.6 x 10 -19 C)
1 (4.0 Å)×
1 Å
10-10 m
9.0 x 109 J•m
C2
NR O+ -
4.0 Å
= 1= 5.8 x 10-19 Jmol
6.0 x 1023
= 35 kJ mol–1
or 83 kcal mol–1
1000 J1 kJ
(1 cal = 4.18 J)
q1 q2
distanceke
C) = 78
= 1)
1.1 kcal mol–1 in water at 25 °C ( = 78)
×
ε
ε
ε
ε
ε
Energy =
Energy =
in gas phase (
gas phasewater (25 °
2 Introduction to Bioorganic Chemistry and Chemical Biology: AnSwerS TO CHApTer 3
Answer 3.3
Answer 3.4
Answer 3.5
The major groove.
Answer 3.6
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.
Introduction to Bioorganic Chemistry and Chemical Biology | A3213Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
N
N
O
N
O
N
::..
N
N..
+
:
-
-
+
amide
amidine
most basic and most nucleophilic
most basic and most nucleophilic
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O
HO
NN
N N
NH2HO
.. H A
O
HO
NNH
N N
NH2HO
+..
O
HO
HO
+
NNH
N N
NH2+
O
HO
HO
+
:OH2
O
HO
HOO
H
H
+
: A-O
HO
OHHO
NNH
N N
NH2:HA NN
N N
NH2H
H
+N
N
N N
NH2H
H
+.. : A-
HNN
N N
NH2
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NN
O
O R
RN
N
N
N N
R
HH
H
majorgroove
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NH
O
N
NH2
N
NH2
NH
N
NH2
O N
NH
NH2
O
or
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CGGGGGTGGCGCAGTGAGGAGG5'- -3'
GCCCCCACCGCGTCACTC
CCCCACCGCGTCACTCCCCCCACCGCGTCACTCCCTCCACCGCGTCACTCCCTC
3'- -5'
3'- -5'3'- -5'
3'- -5'
Tm = 4x2 + 14x4 °C = 64 °CCCCCCACCGCGTCACTCC3'- -5' Tm = 4x2 + 14x4 °C = 64 °C
Tm = 4x2 + 14x4 °C = 64 °CTm = 5x2 + 13x4 °C = 62 °CTm = 5x2 + 13x4 °C = 62 °C
targetprobe Aprobe Bprobe Cprobe Dprobe E
Introduction to Bioorganic Chemistry and Chemical Biology: AnSwerS TO CHApTer 3 3
Answer 3.8
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.
Answer 3.10
Answer 3.11
Answer 3.12
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.
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415 = 1,073,741,824416 = 4, 294,967,296417 = 17,179,869,184418 = 68,719,476,736419 = 274,877,906,944420 = 1,099,511,627,776
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CGACCAGCTGGT
5'-3'-
1
22
ACGTGTCGCC
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3'-CCC CTACC GTG5'-GGG GATGG CACT
G
TT
GG
G
AThairpin
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NS
O
-O2C
MeMe
H
HN
R NH
OHN
HN SO
Ph
MeMe
HO2C
HO
Ph
R NH2..N
S-O2C
MeMe
H
HN
O
PhO -
NH
HR +
B..
NSHO2C
MeMe
H
HN
O
PhO
HNR
-:
:HB NSHO2C
MeMe
H
HN
O
PhO
HNR
-:
H +
good nuc
good L.G.
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N N
NH H
H
HH
+
O
OPO
O-
N N
OH
HH
O
OPO
O-arginine citrulline
stable salt bridge with DNA less stable interaction DNA
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-3ʹ5ʹ-CCATGCCTATGTTCATCGTGAACACCAATGT……CTGGCCCCACTTACCTGCACCGCTGTTC-3ʹ3ʹ-GTGAATGGACGTGGCGACAAG–5ʹ
Answer 3.14
Answer 3.15
Answer 3.16
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.)
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NN N
NH
ORP
O NNC
i-Pr
i-Pr..
ORP
O NNC
i-Pr
i-PrH +
..
OR
PO
NNCH
i-Pr
i-Pr
OR
PO
NNCH
N i-Pr
i-PrH
+OR
PO N
NCH
+
i-Pr2NH
N N
NN -
N N
NN
.. H A
NN N
N NN
ORP
O NNC
N NN
:
Introduction to Bioorganic Chemistry and Chemical Biology | A3226Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
OO
O OH
OSO3Na
Gal
HB:
OO
O OH
OSO3Na
Gal
:-
OO
OOH
Gal
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~0 5 10 20 50 100
1 2 3 4 5 6relaxed
supercoiled
actual [inhibitor] (µM)
sample
leastinhibitor
mostinhibitor
partially supercoiled
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N
N
NHBz
OR
CH3NH2
.. N
N
NHBz
OR
NH3C
H H
+:- N
N
NH
OR
HN
H3C:-
PhO
N
N
NHCH3
OR
B:
Introduction to Bioorganic Chemistry and Chemical Biology: AnSwerS TO CHApTer 3 5
Answer 3.18
A
B Design and synthesize two oligonucleotides with BamHI overhangs on the ends. These will hybridize to any site cut with BamHI.
C The sequence contains two BamHI sites. Any new overhangs on the ends could hybridize within the oligonucleotide or between oligonucleotides.
Answer 3.19
Answer 3.20
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AACTGAATTTCAGGGGGATCCGCATGGCGTTTGACTTAAAGTCCCCCTAGGCGTACCGCA3'-
5'- -3'-5'
BamH1
AACTGAATTTCAGGGGTTGACTTAAAGTCCCCCTAG3'-
5'-
AACTGAATTTCAGGGGGATCCGCATGGCGTTTGACTTAAAGTCCCCCTAGGCGTACCGCA3'-
5'- -3'-5'
GATCCGCATGGCGTGCGTACCGCA
-3'-5'
GATCCGCATGGGTTTCAAATCGGCGTACCCTTTGTTTAGCCTAG
AACTGAATTTCAGGGGTTGACTTAAAGTCCCCCTAG3'-
5'- GATCCGCATGGCGTGCGTACCGCA
-3'-5'
pGATCCGCATGGGTTTCAAATCGGCGTACCCTTTGTTTAGCCTAGp
GATCCGCATGGGTTTCAAATCGGCGTACCCTTTGTTTAGCCTAG
AACTGAATTTCAGGGGTTGACTTAAAGTCCCCCTAG3'-
5'- GATCCGCATGGCGTGCGTACCGCA
-3'-5'
Mg•ATP
GATCCTTTCATAAGTGGTGGGATCCCCATTCAATTGGAAAGTATTCACCACCCTAGGGGTAAGTTAACCTAG
design andsynthesize
T4 DNA ligase
kinase
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AACTGAATTTCAGGGGGATCCGCATGGCGTTTGACTTAAAGTCCCCCTAGGCGTACCGCA3'-
5'- -3'-5'
BamH1
AACTGAATTTCAGGGGTTGACTTAAAGTCCCCCTAG3'-
5'-
AACTGAATTTCAGGGGGATCCGCATGGCGTTTGACTTAAAGTCCCCCTAGGCGTACCGCA3'-
5'- -3'-5'
GATCCGCATGGCGTGCGTACCGCA
-3'-5'
GATCCGCATGGGTTTCAAATCGGCGTACCCTTTGTTTAGCCTAG
AACTGAATTTCAGGGGTTGACTTAAAGTCCCCCTAG3'-
5'- GATCCGCATGGCGTGCGTACCGCA
-3'-5'
pGATCCGCATGGGTTTCAAATCGGCGTACCCTTTGTTTAGCCTAGp
GATCCGCATGGGTTTCAAATCGGCGTACCCTTTGTTTAGCCTAG
AACTGAATTTCAGGGGTTGACTTAAAGTCCCCCTAG3'-
5'- GATCCGCATGGCGTGCGTACCGCA
-3'-5'
Mg•ATP
GATCCTTTCATAAGTGGTGGGATCCCCATTCAATTGGAAAGTATTCACCACCCTAGGGGTAAGTTAACCTAG
design andsynthesize
T4 DNA ligase
kinase
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AACTGAATTTCAGGGGGATCCGCATGGCGTTTGACTTAAAGTCCCCCTAGGCGTACCGCA3'-
5'- -3'-5'
BamH1
AACTGAATTTCAGGGGTTGACTTAAAGTCCCCCTAG3'-
5'-
AACTGAATTTCAGGGGGATCCGCATGGCGTTTGACTTAAAGTCCCCCTAGGCGTACCGCA3'-
5'- -3'-5'
GATCCGCATGGCGTGCGTACCGCA
-3'-5'
GATCCGCATGGGTTTCAAATCGGCGTACCCTTTGTTTAGCCTAG
AACTGAATTTCAGGGGTTGACTTAAAGTCCCCCTAG3'-
5'- GATCCGCATGGCGTGCGTACCGCA
-3'-5'
pGATCCGCATGGGTTTCAAATCGGCGTACCCTTTGTTTAGCCTAGp
GATCCGCATGGGTTTCAAATCGGCGTACCCTTTGTTTAGCCTAG
AACTGAATTTCAGGGGTTGACTTAAAGTCCCCCTAG3'-
5'- GATCCGCATGGCGTGCGTACCGCA
-3'-5'
Mg•ATP
GATCCTTTCATAAGTGGTGGGATCCCCATTCAATTGGAAAGTATTCACCACCCTAGGGGTAAGTTAACCTAG
design andsynthesize
T4 DNA ligase
kinase
Introduction to Bioorganic Chemistry and Chemical Biology | A3229Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
HN
N
O
OR
CF3
S
Enz
..
HN
N
O
OR
F2C
S
Enz
- HN
N
O
OR
CF2
HNEnz
-
F HN
N
O
OR
CF2
S
Enz
H2NEnz
..
HN
N
O
OR
CF2
S
Enz
-
HNEnz
.. ..
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NH
N
N
N
O
RN
N+
NH
N
N
N
O -
R N2
:
+
GuaNH2
..N O
.. ..+
GuaN
NO
HH
+
-A
..
GuaN
NO
H
GuaN
NO
H
..
- H A+
GuaN
NOH.. H A
GuaN
NOH2
+..Gua N N
+ ..
GuaNH
NOH
+
A
..
ON
OH2+
:ON
O ..NO
..
..+
-: H A
ON
OH
H A
H2N
HN
N
N
N
O
R..
NH
N
N
N
O
RNH
HN
N
N
N
O
R
NH
HN
N
N
N
O
R
+
-
6 Introduction to Bioorganic Chemistry and Chemical Biology: AnSwerS TO CHApTer 3
Answer 3.21
Answer 3.22
Answer 3.23
Answer 3.24
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NC6H13
S C N
NC6H13
+
Nu..
.. NC6H13
N
NN
NH
DNA
O
NH2
+
fascicularin aziridinium ion
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HH
H H
H
H
+
H
HHH
HH
•
•
•
• abstractH atom
abstractH atom
abstraction of the two H atoms is not concerted
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HNO
O OH
OR
OH
R = CH(OH)CH 3
HNHO
HO OH
OHOH
RHO
H2O
HNHO
HO OH
OR
OH
HNHO
HO OH
HOR
OHOHSN1
H
H
uncialamycin
reducingintracellular
environment
reducedin the cell
Bergmanrearrangement
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H
H
H
H
H
HH
H
H
H
H
H
•
•
H
HH
H
HH
HH
H
H
H
H
H
H
•
H
H
HH
H
H
HH
HH
H
H
H
H H
H
HH
H
HH
H
+•
Introduction to Bioorganic Chemistry and Chemical Biology: AnSwerS TO CHApTer 3 7
*Answer 3.26
*Answer 3.27
*Answer 3.30
A In PDB 1AIO, cisplatin forms a crosslink between the N7 atoms of guanine bases (rendered in yellow).
B The platinum atom sits in the major groove.
C The crosslink is intrastrand, between adjacent guanine bases.
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pTATA
O
OOP
O-O
O
HOOP
O-O
O
OO
Thy
O
OOP
O-O
P
OH
O-O
A
OO
P- OO
OO
PO
- OO
OO
POH
- OO
O
O
OPO
- OO
HO
Thy
Thy
O
OOP
O-O
O
HOOP
O-O
O
OHO
Gua
O
OOP
O-O
d(GACA)
B
O
OOP
O-O
O
HOOP
O-O
O
OHO
Gua
O
OOP
O-O
GUCU
OH
OH
OH
OH
C
Ura
Cyt
Ura
Ade
Thy
Ade
Ade
Ade
Ade
Ade
Cyt
presence of Thysuggests that this is DNA
"d" pre�x means 2-deoxy
presence of Urasuggests that this is RNA
The conformational depiction below and the dash/wedge depiction to the right are equally acceptable
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NN
N
NdR
NH
H N
H NN
O dR
NCH3
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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.
B On the basis of the structures, the susceptible bases seem to be guanidines that 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
*Answer 3.38
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CATGCGTTCCCGTGGTGCGCAAGGGCAC
CATGCGTTCCCGTG
AGTCTA TAGACTTCAGAT ATCTGA
G
G
AGTCTATCAGATG or
ACGTCAG TGGCATTGCAGTC CCCGTA
G
A
AGTCTATCAGAT
GGG
T
TAGTCTA TAGACTTCAGAT ATCTGA
TGGGT
TGGGT
5'- -3'
5'- -3'3'- -5'
5'-3'-
5'- -3'3'- -5'
5'- -3'3'- -5'
5'-3'-
-3'-5'
or5'-3'-
duplex formation
ssDNA
hairpin orduplex with mismatch bulge
duplex with mismatch bulge
hairpin orduplex with mismatch bulge
strand not analyzed
strand not analyzed
arrows highlightsites of oxidation
Introduction to Bioorganic Chemistry and Chemical Biology | A3242Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
hv
H2ON
Ca
OHOOO
N
O OHO O
MeO
MeO
NO2
NH
-O2C
-O2C
HR'
NMeO
MeO
R2N
O -
O
+
*
NO
NMeO
MeO
CO2-CO2
-
OCa2+
hv
R'
NMeO
MeO
R2N
O -
OH
NCa
OHOOO
N
O OHO O
MeO
MeO
N
O
OH
..
R'HO
N
O
OHMeO
MeO
or...
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PhNH2
..N O
.. ..+
PhN
NO
HH
+
-A
..
PhN
NO
H
PhN
NO
H
..
- H A+
PhN
NOH.. H A
PhN
NOH2
+..Ph N N
+ ..
PhNH
NOH
+
A -..
ON
OH2+
:ON
O ..NO
..
..+
-: H A
ON
OH
H A
N
N
NH2
ODNA
N
N
N2+
ODNA
OH2+..
N
N
+N2
ODNA
OH2
-
+
N
N
OH
ODNA
H+A
..
N
N
OH
ODNA
..
H ANH
N
O
ODNA
+
H
A -..
NH
N
O
ODNA
:
NH
NN
N
DNANH2
ONaNO2
HCl
H2O
NH
NH
N
N
DNAO
O
N
NN
N
DNA
NH2 NaNO2HCl
H2O
NH
NN
N
DNA
OC
B
A
as in part A.
adenine
guanine
Introduction to Bioorganic Chemistry and Chemical Biology: AnSwerS TO CHApTer 3 9
*Answer 3.41
*Answer 3.42
A with 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
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PhNH2
..N O
.. ..+
PhN
NO
HH
+
-A
..
PhN
NO
H
PhN
NO
H
..
- H A+
PhN
NOH.. H A
PhN
NOH2
+..Ph N N
+ ..
PhNH
NOH
+
A -..
ON
OH2+
:ON
O ..NO
..
..+
-: H A
ON
OH
H A
N
N
NH2
ODNA
N
N
N2+
ODNA
OH2+..
N
N
+N2
ODNA
OH2
-
+
N
N
OH
ODNA
H+A
..
N
N
OH
ODNA
..
H ANH
N
O
ODNA
+
H
A -..
NH
N
O
ODNA
:
NH
NN
N
DNANH2
ONaNO2
HCl
H2O
NH
NH
N
N
DNAO
O
N
NN
N
DNA
NH2 NaNO2HCl
H2O
NH
NN
N
DNA
OC
B
A
as in part A.
adenine
guanine
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NHC
O2N
N
NH
N
N
CH3
NO
O
B
A
2N
H B:
NH
N
N
CH3
NO
O2N
:-
NHN
C
CH3
NO
O2N
N : BHNC
N
H HH
+NC
NOH
:
HHH
B:
N +C
N -
H H
+
: :
Introduction to Bioorganic Chemistry and Chemical Biology | A3248Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
5'-GGGATCGAA3'-CCCTAGCTT
C GCGATGCA
C
N
N
dC
O
NH2N
N
Arg
B
C
D
NH
H
H
H
H+
N
N
dC
O
NH2N
NH
NH
H
H
H+
N
N
dC
O
NH2N
NH
NH
H
H
H+
Arg
Arg
or or
5'-GGGAGAATTCCCAGACCNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNCTGAGGGAAATTCTCCC-3'5'-GGGAGAATTCCCAGACC-3'
3'-GACTCCCTTTAAGAGGG-5'
5'-GGGAGAATTTCCCTCAG-3'
10 Introduction to Bioorganic Chemistry and Chemical Biology: AnSwerS TO CHApTer 3
*Answer 3.44
*Answer 3.48
*Answer 3.51
Introduction to Bioorganic Chemistry and Chemical Biology | A3248Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
5'-GGGATCGAA3'-CCCTAGCTT
C GCGATGCA
C
N
N
dC
O
NH2N
N
Arg
B
C
D
NH
H
H
H
H+
N
N
dC
O
NH2N
NH
NH
H
H
H+
N
N
dC
O
NH2N
NH
NH
H
H
H+
Arg
Arg
or or
5'-GGGAGAATTCCCAGACCNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNCTGAGGGAAATTCTCCC-3'5'-GGGAGAATTCCCAGACC-3'
3'-GACTCCCTTTAAGAGGG-5'
5'-GGGAGAATTTCCCTCAG-3'
Introduction to Bioorganic Chemistry and Chemical Biology | A3251Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
Me OH
Me
O
MeHO OH
SR:-
Me OH
Me
OH
MeHO OH
SR..
H A
Me OH
Me
OH
Me+H2O OH
SR
Me OH
Me
OHMe OH
SR+
:DNA Me OH
Me
OHMe
SR
DNA
OH
Introduction to Bioorganic Chemistry and Chemical Biology | A3255Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
RHN N
O OH+
N
NN
NDNA
NH H
..
HN N
HOO
N
NN
NDNA
NH H
+
NH
HN
O
O
Introduction to Bioorganic Chemistry and Chemical Biology | A3258Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
A
B
Introduction to Bioorganic Chemistry and Chemical Biology: AnSwerS TO CHApTer 3 11
*Answer 3.54
*Answer 3.57
5ʹ-TCCTNNAGGA-3ʹA The oligosaccharide is asymmetric with a head (H), a tail (T), a left-hand side and a
right-hand side.
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.
Introduction to Bioorganic Chemistry and Chemical Biology | A3258Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
A
B
Introduction to Bioorganic Chemistry and Chemical Biology | A3261Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
O OH
O O
OMeNHEt
O
N
HN
canintercalatebetweenbase pairs
OH
SMe
Me
SNHCO2Me
O
HO
O
aromaticby-product
Introduction to Bioorganic Chemistry and Chemical Biology | A3264Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
O
Me
O
OHMeO
HO
MeO
MeO Me
I
S O
O
Me
Me
HNO
O
HO
HO
OO
O
OMeEtHN
left-handside
right-handside
head (H)
tail (T)
TCCT
AH
T
H
T
H
T
GGA
TCCTNNAGGA
AGGANNTCCT
5'|
3'|
3'|
5'|
5'|
3'|
3'|
5'|
O
Introduction to Bioorganic Chemistry and Chemical Biology | A3264Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
O
Me
O
OHMeO
HO
MeO
MeO Me
I
S O
O
Me
Me
HNO
O
HO
HO
OO
O
OMeEtHN
left-handside
right-handside
head (H)
tail (T)
TCCT
AH
T
H
T
H
T
GGA
TCCTNNAGGA
AGGANNTCCT
5'|
3'|
3'|
5'|
5'|
3'|
3'|
5'|
O
12 Introduction to Bioorganic Chemistry and Chemical Biology: AnSwerS TO CHApTer 3
C If you link the oligosaccharide tail to tail, each half still wants to bind with the origi-nal orientation.
*Answer 3.58Introduction to Bioorganic Chemistry and Chemical Biology | A3264Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
O
Me
O
OHMeO
HO
MeO
MeO Me
I
S O
O
Me
Me
HNO
O
HO
HO
OO
O
OMeEtHN
left-handside
right-handside
head (H)
tail (T)
TCCT
AH
T
H
T
H
T
GGA
TCCTNNAGGA
AGGANNTCCT
5'|
3'|
3'|
5'|
5'|
3'|
3'|
5'|
O
Introduction to Bioorganic Chemistry and Chemical Biology | A3265Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz
OOH
NCO
HO
OOH
para-benzyneintermediate
enediyneprecursor
OOH
O
HO
OOH
NC
OOH
O
HO
OOH
NC
..
..
:
-
+ -
+
Cl-
:Cl-