Answers to Chapter 3 - Garland · PDF fileIntroduction to Bioorganic Chemistry and Chemical Biology 1 Answers to Chapter 3 (in-text & asterisked problems) Answer 3.1 Answer 3.2 In

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.


=(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.


N

N

O

N

O

N

::..

N

N..

+

:

-

-

+

amide

amidine

most basic and most nucleophilic

most basic and most nucleophilic


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


NN

O

O R

RN

N

N

N N

R

HH

H

majorgroove


NH

O

N

NH2

N

NH2

NH

N

NH2

O N

NH

NH2

O

or


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.


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


CGACCAGCTGGT

5'-3'-

1

22

ACGTGTCGCC


3'-CCC CTACC GTG5'-GGG GATGG CACT

G

TT

GG

G

AThairpin


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.


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


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.)


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

:


OO

O OH

OSO3Na

Gal

HB:

OO

O OH

OSO3Na

Gal

:-

OO

OOH

Gal


~0 5 10 20 50 100

1 2 3 4 5 6relaxed

supercoiled

actual [inhibitor] (µM)

sample

leastinhibitor

mostinhibitor

partially supercoiled


N

N

NHBz

OR

CH3NH2

.. N

N

NHBz

OR

NH3C

H H

+:- N

N

NH

OR

HN

H3C:-

PhO

N

N

NHCH3

OR

B:


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


AACTGAATTTCAGGGGGATCCGCATGGCGTTTGACTTAAAGTCCCCCTAGGCGTACCGCA3'-

5'- -3'-5'

BamH1

AACTGAATTTCAGGGGTTGACTTAAAGTCCCCCTAG3'-

5'-


5'- -3'-5'

GATCCGCATGGCGTGCGTACCGCA

-3'-5'

GATCCGCATGGGTTTCAAATCGGCGTACCCTTTGTTTAGCCTAG


5'- GATCCGCATGGCGTGCGTACCGCA

-3'-5'

pGATCCGCATGGGTTTCAAATCGGCGTACCCTTTGTTTAGCCTAGp




-3'-5'

Mg•ATP

GATCCTTTCATAAGTGGTGGGATCCCCATTCAATTGGAAAGTATTCACCACCCTAGGGGTAAGTTAACCTAG

design andsynthesize

T4 DNA ligase

kinase



5'- -3'-5'

BamH1


5'-


5'- -3'-5'


-3'-5'




-3'-5'





-3'-5'

Mg•ATP



T4 DNA ligase

kinase



5'- -3'-5'

BamH1


5'-


5'- -3'-5'


-3'-5'




-3'-5'





-3'-5'

Mg•ATP



T4 DNA ligase

kinase


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

.. ..


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

+

-


Answer 3.21

Answer 3.22

Answer 3.23

Answer 3.24


NC6H13

S C N

NC6H13

+

Nu..

.. NC6H13

N

NN

NH

DNA

O

NH2

+

fascicularin aziridinium ion


HH

H H

H

H

+

H

HHH

HH

•

•

•

• abstractH atom

abstractH atom

abstraction of the two H atoms is not concerted


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


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

+•


*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.


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


NN

N

NdR

NH

H N

H NN

O dR

NCH3



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


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


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...


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


*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


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


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

+

: :


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'


*Answer 3.44

*Answer 3.48

*Answer 3.51


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'


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


RHN N

O OH+

N

NN

NDNA

NH H

..

HN N

HOO

N

NN

NDNA

NH H

+

NH

HN

O

O


A

B


*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.


A

B


O OH

O O

OMeNHEt

O

N

HN

canintercalatebetweenbase pairs

OH

SMe

Me

SNHCO2Me

O

HO

O

aromaticby-product


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


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


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


OOH

NCO

HO

OOH

para-benzyneintermediate

enediyneprecursor

OOH

O

HO

OOH

NC

OOH

O

HO

OOH

NC

..

..

:

-

+ -

+

Cl-

:Cl-

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Answers to Chapter 3 - Garland · PDF fileIntroduction to Bioorganic Chemistry and Chemical Biology 1 Answers to Chapter 3 (in-text & asterisked problems) Answer 3.1 Answer 3.2 In