Conformational Studies on 8-Oxoguanine Residue in DNA-strand
Nihar Sarkar
Department of Chemistry, University of Pittsburgh
Background:
In higher organisms, where aerobic respiration is observed, various reactive oxygen
species (such as O2.-, H2O2, OH.) may be formed and may lead to the oxidation of
guanine residues (G) of DNA to 8-oxoguanine (oxoG)(equation 1).
O
N
NH
8
9N
1
DNA O
DNA
N
O
O
1'
O
9N
NH2
O
[O]
DNA
O
NH
N
1
DNA O
+
2'
Guanine in
DNA strand
H
N
8
O
NH2
1'
2
'
(1)
oxoG in
DNA strand
During DNA replication oxoG residues are misread by the replicative enzymes and
adenine is incorporated into the growing strand (instead of cyctosine). This is
mutagenesis event. In biological systems a repair enzyme (8-oxoguanine DNA
glycosylase, hOGG1) is known which helps to repair “oxoG” to “G” (Figure A).
Figure A. Schematic representation of DNA-G → oxoG → T conversion and the role of
hOGG1 to repair the initial damage (Figure copied from: Bruner, S. D.; Norman, D. P.
G.; Verdine, G. L. Nature 2000, 403, 859-866.)
Learning Objective:
The modeling experiments in this project are designed to provide you an opportunity to
examine the conformational forms of G-DNA and oxoG. You will be able to interpret
how oxidation at C-8 position causes change in the conformation from its anti to syn
(Figure B). These experiments will also help you understand why there is a change in the
mode of base-pairing after oxidation (i.e., why oxoG can not make a correct WatsonCrick pair with cytosine anymore). This will be achieved by performing potential energy
calculation on several base pairs. In these experiments you will obtain dihedral angles,
map surface electron densities, derive potential energy curves, and optimize geometries
of molecules using semi-empirical calculations through the CAChe software package.
1
O
N
NH
8
9N
N
1
DNA O
O
NH2
1'
'
DNA
O
O
H
N
8
1
O
1
DNA
N
H
N
HN
H2N
DNA O
NH2
N
1
O
DNA
O
1'
2'
O
oxoG= 7,8 dihydro
8-oxoGuanine
oxoG(Syn)
oxoG(Anti)
8
N9
'
2'
O
O
Oxidation
NH
9N
DNA O
2
O
Guanine in
DNA strand (G-DNA)
Figure B. Oxidation of Guanine base unit in the DNA damaging process
In this study instead of single DNA-strand, which consists of repeating units of
nucleotide (nucleotide is a combination of a sugar, a base and a phosphate unit), only a
single guanosine (guanosine belongs to nucleoside family, not the nucleotide, where a
sugar unit and a base unit are present; for more details consult any organic text book)
derivative(s) will be considered (Figure C).
Base
Phosphate
O
8
O
O P O
O
N
9N
O
N
NH
N
N
NH2
HO
O
NH
N
NH2
O
1'
OH
Sugar
2'
OH
Example of Nucleotide
Example of Nucleoside
Nucleotide = sugar + base + phosphate
Nucleoside = sugar + base
Figure C. Nucleotide and Nucleoside
Electronically, sterically and environmentally the model molecules are far different from
the DNA-strand in the biological system. You will perform these experiments on the
model molecules and based on the results, you will be able to draw conclusion on how
the conformations get changed due to oxidation and causes mutagenesis in the actual
system.
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Learning Activity:
CAChe comes with “Fragment Library” which can be found in C:\Program
Files\Fujitsu\CAChe\FragmentLibrary. In the library, the folder called “Nucleotides”
contains all fours nucleic acids and the folder called “Nitrogen_base” contains the four
bases (A, T, C and G).
A. Electron density calculation on the bases will provide the most preferred site(s) to
form H-bonding. In a DNA double helix, the two DNA strands are held together by Hbonding between the base pairs Cytosine/Guanine and Adenine/Thymine, which are also
known as Watson-Crick base-pairs. The first experiment is the mapping of electron
density on the surface of the base pair GC (Figure D) to help you understand the basis of
pairing. Open the workspace program from CAChe in the Start menu. Open Guanine.csf
from Fragment Library and resave it as Guanine.csf in the My Documents folder. Choose
New under Experiment, Property of chemical sample, Property electrostatic potential
on electron density Using PM3 geometry with PM3 wavefunction. Click start. After the
completion of the calculation, close the “Experiment” and the “Experiment Status”
windows. Go to Analyze, Show Surfaces, choose Guanine.csf.EonD, and click OK. This
will generate the color coded surface (Red implies more negative and blue implies more
positive; you may find more on color via the Surface Legend under Analyze menu).
O
N
N
H
H NH
N
N H
N
G
N H
H
O
N
H
C
C= Cytosine
G= Guanine
Figure D. Watson-Crick base pair between G and C
Do the same with cytosine. Save this file as Cytosine.csf.
1. Look at both the surfaces from the CAChe output files. Hand draw the two bases
below and label the H-bonding sites on each molecule as negative or positive
after interpreting the color codes. Now compare your answer with Figure D and
predict whether Watson-Crick pair between C and G is valid or not.
Due to oxidation, oxoG stays in its syn conformation as shown in Figure B (you will see
later why so) and the oxygenated surface is now exposed to cytosine. We know
oxoG(syn) does not form H-bonds with cytosine. You will be able to confirm this by
performing the following experiment. Construct the oxygenated guanine residue 1a
(Figure E).
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This side is exposed for
H-bond formation in
oxoG(syn)
O
8
H
N
N
H
O
N
N
H
NH2
1a
Figure E. Model Molecule 1a
(Hint: Open guanine.csf and make the required changes, beautify and save it as
oxoguanine.csf). Do the same experiment as you did for guanine.csf.
2. Cut and paste the structures and surfaces from CAChe drawing and output
window of both cytosine and 1a. Look at the exposed side of 1a and compare
with cytosine.
(This can be done as: after resaving the cytosine.csf and oxoguanine.csf, open both of
these windows (if it is already not opened). From the menu bar, choose Window\Tile.
This will place the two structures side by side. Press “Shift”+ “Print Screen” keys from
keyboard. Open Paint from All Programs\Accessories\Paint and Paste it. Then from paint
window copy the molecule. Open a new Word document and paste it. Do the same after
generating the surfaces and paste in the same word document. Now you have those four
windows in one page. Print it)
3. Why does C make pair with G but not with oxoguanine?
B. Geometry optimization will allow us to have the structure with minimized energy.
We have chosen guanosine derivatives as our model molecule. This has two main parts: a
sugar unit and the guanine base unit. By performing this experiment we will generate the
energy minimized structure. Open the Guanosine.csf from library and construct the
molecule 2a (Figure F).
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Guanine residue
Five member
sugar ring
O
N
NH
8
9N
H3C O
H3C
O
O
N
NH2
These three atom will
make H-bond with its
complementary pair
cytosine in DNA double
helix
1'
2
'
Figure F. Model structure 2a
Beautify and save as 2a.csf. Under Experiment\New, choose Property of chemical
sample, Property optimized geometry, Using PM3 geometry. After the experiment is
done, record the Final Heat of Formation. Close both of the experiment windows. Select
C-8, N-9, C-1′ and the ring oxygen, go to Adjust, Dihedral Angle and note the dihedral
angle. In the energy minimized structure, sugar has adopted the puckered structure and C8 stays toward the sugar ring making the latter “anti” with respect to free amine so that it
can form H-bonding with its complementary pair (cytosine) in DNA double helix.
Therefore assign this form as 2a(anti).
4. Heat of formation for 2a(anti) =
Dihedral angle =
Now calculate the final heat of formation and the dihedral angle for 2a(syn)(Hint: The
structure shown in Figure F is for 2a(anti). You need to draw the same structure as 2a,
then beautify. Select C-8, N-9, C-1′ and the ring oxygen, go to Adjust, Dihedral Angle,
and choose Angle =180, click Define Geometry Label, click Lock Geometry, click OK.
This will lock the geometry. Save as 2asyn.csf). Do the geometry optimization
calculation. Which has the higher energy?
5. Heat of formation for 2a(syn) =
C. Potential energy map calculation will generate the most stable to least stable
conformations. Conformational changes in the DNA strand can cause enormous effect in
terms of its property. An oxidation of G-DNA (Figure B) causes oxoG to adopt its “syn”
conformation rather than the “anti”, and this causes the mutagenesis. In this experiment,
the potential energy map calculation will help you to find out the most and least
conformations (i.e. syn and anti conformer) of the model molecule. Construct the model
molecule 3a as shown in Figure G.
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O
8
H
N
9N
H3C O
O
NH
N
NH2
O
1'
H3C
O
2
'
Figure G. Model structure 3a
Open Guanosine.csf from C:\CAChe\FragmentLibrary\Nucleotides in workspace. Modify
as it is shown in 3a, beautify and save it as 3a.csf. Beautify and save. Select C-8, N-9, C1′ and the ring oxygen. Then go to Adjust, Dihedral Angle, check Define Geometry
Label, Select Search from -180 to 180, using 24 steps (You may use more steps), and
click OK. Under Experiment/New, choose Property of conformations of small (<100
atoms) molecule, Property potential energy map, Using exhaustive search with
semiempirical PM3 (1 label). Click start. After its completion a potential map will appear
along with a window showing a structure. Click on the curve to view the corresponding
conformation of 3a. Identify the most and least stable conformations.
6. Cut and paste the most and least stable conformations of 3a along with the
corresponding potential energy maps from CAChe output. Assign the least
stable conformation as 3a(anti) and the most stable conformation as 3a(syn).
7. Record the dihedral angle for both the conformers. Which conformation has the
highest energy?
.
8. Now from the results of the model molecule 3a, predict which conformation of
oxoG will have higher energy?
From these experiments its clear that 2a prefers to stay in its anti form because in the syn
conformation, there is significant steric hindrance between the sugar ring and the guanine
residue. But in the case of 3a(anti) the C-8 oxygen and sugar ring oxygen are very close
to each other and the electronic repulsion between these two oxygen atoms make this
conformer unstable, forcing it to adopt syn conformation even though there is some steric
hindrance too. Actually in this example the electronic effect prevails over the steric
(Figure H).
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O
N
NH
8
9N
H3C O
H3C
O
N
NH2
O
N
HN
H2N
H3C O
N
2a(anti)
Significant electronic
repulsion between
H
these two oxygen
N
8
atoms
O
9N
O
1'
2'
O
H3C
2a(syn)
O
O
NH
N
NH2
HN
H2N
H3C O
H
N
O
2'
8
O
N
N9
O
1'
1'
H3C
8
N9
O
1'
2'
O
H3C O
Sterically
hindered
H3C
3a(anti)
O
2'
3a(syn)
Figure H Preferred conformation of 2a and 3a
Therefore, in the actual system G-DNA adopts its anti conformation and forms correct
Watson-Crick pair with cytosine during the replication process. But once the C-8 position
of guanine residues get oxygenated (i.e. oxoG), it is no longer able to stay in its anti form
due to the strong electronic repulsion and switches to its syn form. Consequently, oxoG
loses its capability to form Watson-Crick base pair with its complementary base,
cytosine. Instead, it forms pair with adenine. This is mutagenesis.
References:
1. Bruner, S. D.; Norman, D. P. G.; Verdine, G. L. Nature 2000, 403, 859-866.
2. Fromme, J. C.; Banerjee, A.; Huang, S. J.;Verdine, G. L. Narure 2004, 427, 652656.
3. Organic Chemistry, Vollhardt & Schore, 5th Edition.
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