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MBLG1001 Lecture 7
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COURSE: MBLG1001
Lecturer: Dale Hancock
Lecture 7
COMMONWEALTH OF AUSTRALIA
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MBLG1001 Lecture 7
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MBLG Lecture 7:
The Essentials of Molecular Biology: Chapters 3 & 6
DNA
Historically Nucleic acid, named because it was first isolated in the nucleus and it was acidic, was
not thought to be the genetic material. That role was thought to be fulfilled by proteins as they
contained more potential for variation (20 different amino acids rather than 4 different
nucleotides). The person credited with first discovering DNA was a Swiss biochemist Fredrich
Miescher circa 1860 who isolated nucleic acid from pus on bandages. Would have had a great job
description for the tax office!! The new polymer contained the usual suspects C, O, H and N and
also Phosphorus (this is not in protein).
The role of DNA
Work by Griffith then MacLeod, McCartney and Avery established to role of DNA as the genetic
material. Using Pneumococcus they were able to transform the bacterium from the rough R nonpathogenic form to the S (smooth) pathogenic form by adding a killed suspension of S bacteria. By
a process of elimination the active ingredient in the killed S bacterial lysate was identified as DNA
(page 101 – 105).
This was further confirmed by the Waring Blender experiment (Hershey and Chase), which put the
kitchen appliance on the map (how many blenders are instrumental in winning a Nobel prize)
(pages 106 – 109). The bacteriophage T2 which infects certain bacteria was labeled with 35S
(which labels protein only) and 32P (which only labels DNA) in separate experiments. The bacteria
and the T2 bacteriophage were mixed and quickly centrifuged (which separates the phage:bacteria
complexes from unattached phage and fragments). The pellet, containing the bacteria with the
attached phage, was then suspended and blended to remove the attached phage. The mix was then
centrifuged. Eighty % of the 35S was found in the supernatant, and only 20% in the pellet. Hence
the protein part of the phage was not transferred to the bacteria. The 20% was accounted for by
phage associated tail fragments which were too tightly bound to the bacterial cell surface to be
removed by blending.
When the 32P labeling experiment was performed, 70% of the 32P was found in the pellet,
indicating the DNA was indeed transferred to the bacteria. Most of the 32P found in the supernatant
was the result of lysed bacteria during the blending process and defective phage which, although
they could attach to the bacteria, were unable to inject their DNA. Reverse experiments were also
performed.
MBLG1001 Lecture 7
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The structure of DNA
Work by many chemists and biochemists between the 1870s and 1950s revealed this nucleic acid
to be made up of a sugar phosphate backbone with bases attached. There were four bases; two
based on the structure of the heterocyclic compound, purine and two based on pyrimidine. For a
long time the prevailing opinion was that there were equal amounts of each of the four nucleotides.
It wasn’t until a famous organic chemist Edwin Chargaff, by meticulous experimentation, showed
the base relationships, the rules of which bear his name (Chargaff’s rules):
•
A = T; G = C
•
•
The amount of purine = pyrimidine.
The #sugars (deoxyribose) = # phosphates = # bases
Then came Watson and Crick (not entirely alone) with the double helix. This model explained
Chargaff’s rules and also the X-ray crystallography pictures produced by Rosalind Franklin (and
less than scrupulously obtained by Maurice Wilkins, Watson and Crick).
Properties of DNA explained by the Watson and Crick Model
•
•
•
Ability to store genetic information
Ability to transfer a faithful copy of this information to daughter cells
Physical and chemical stability so that information can be stored for long periods of time
•
There is potential for small changes, which could be inherited, without the loss of parental
information
THE STRUCTURE!!
We will start with the sugar. DNA and RNA are very similar so we will consider them together to
begin with. DNA contains deoxyribose while RNA contains ribose. The missing –OH does alter
the properties of the 2 polymers (we will cover that later). Ribose is a 5 membered sugar, which is
very water soluble. Each of the 5 carbons in ribose has significant attachments.
MBLG1001 Lecture 7
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The 5' phosphate is attached here via a
phosphoester bond, forming a nucleotide.
The nucleotide is linked here to the 3' hydroxyl
of the ajoining nucleotide in DNA and RNA
by a second phosphoester bond (phosphodiester).
HO
OH
5
CH2
O
4
1
C
H
H
3
2
C
Reacts with the aldehyde
from C1 forming an acetal
ring.
C
The base is attached here
by an N-glycosidic bond,
forming a nucleoside.
C
H
H
OH
OH
This is the 3' hydroxyl.
The 5' phosphate of the
ajoining nucleotide links
here by a phosphoester bond
The hydroxyl is absent
at this position in DNA
Terminology:
Base: the purine or pyrimidine like ring structure; adenine, guanine, cytosine, thymine, uracil
Nucleoside: base + sugar
Nucleotide: base + sugar + phosphate.
The bases
The two purine bases, adenine and guanine are based on the heterocyclic (a fancy chemical
name for a ring structure containing both carbon and nitrogen) compound, purine (see below).
Note the numbering conventions for the ring atoms.
6
C
1
7
N
5
N
C
8
C
2
4
C
C
9
3
N
N
H
attaches to sugar
MBLG1001 Lecture 7
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The oxygen is very electronegative (δ-ve)
and acts as an H-bond acceptor
O
The pKa of the ring N is ~9.
It will act as an H-bond donor
at physiological pH
The amino group (δ+ve) acts
as an H-bond donor
N
HN
H2N
N
N
H
Guanine
The amino group (δ+ve)
acts as an H-bond donor
The ring N (pKa ~4)
acts as an H-bond acceptor
at physiological pH
NH2
N
N
N
N
H
Adenine
The two pyrimidine bases found in RNA are Uracil and Cytosine. The pyrimidine bases are
based on the heterocyclic compound pyrimidine. Note the numbering of the ring atoms.
MBLG1001 Lecture 7
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4
C
3
5
2
6
N
C
C
C
1
N
attaches to sugar
The amino group (δ+ve)
is an H-bond donor
NH2
The ring N has a pKa~4.5.
It acts as an H-bond
acceptor at pH 7.
The keto group acts as
an H-bond acceptor
N
O
N
H
Cytosine
The keto group acts as
an H-bond acceptor
O
The ring N has a pKa~9.
It acts as an H-bond donor.
The keto group acts as
an H-bond acceptor
This carbon is
methylated in thymine
HN
O
N
H
Uracil
Differences between DNA and RNA:
Structural differences: the sugar is deoxyribose, which has no –OH at position 2’ and the uracil is
methylated at position 5 forming thymine (see below). These differences at first seem subtle but
are not accidental. We will discuss them further.
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Methyl added to
5-carbon
O
CH3
HN
O
O
-O
P
N
O
O
O-
H
H
OH
H
H
H
Thymine 5’ Monophosphate
DNA structure
The nucleotide residues are joined together via a phosphodiester bond in a 5’ 3’ direction.
One of the best known characteristics of DNA is its double helical structure. Any double helix
symbol is virtually synonymous with DNA.
What forces maintain the double helix?
The double helix is maintained by a number of weak forces:
•
Hydrophobic interactions resulting in base stacking
•
Hydrogen bonding
•
Ionic interactions
•
Van der Waal’s forces
Base Stacking. The bases are flat and aromatic in character quite hydrophobic. They tend to be
“buried” in the interior of the molecule. Because they are flat the bases can stack favourably on top
of each other at a distance of ~3.4 A. This interaction is actually more energetically important than
hydrogen bonding. Van der Waals forces become significant at this close proximity, with mutual
distortion of the electron clouds. The nature of this base stacking interaction is poorly understood.
It is not quite the same as the hydrophobic interactions that operate with proteins (the hydrophobic
interactionsd in proteins are entropic while the base stacking appears to be more enthalpic).
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Hydrogen bonding. Hydrogen bonding between the bases when they are in their correct
tautomeric form gives the double helix its specificity. Adenine pairs with thymine; the amine
group on adenine acts as a donor to the keto group of thymine (provided both bases are in these
forms) and the ring N of adenine acts as an acceptor; the ring N of thymine (and uracil) is
protonated at physiological pH and acts as a donor. Likewise the amine groups on guanine and
cytosine act as donors and the keto groups as acceptors. If the wrong tautomeric forms of the bases
exist then the base pairing is altered. This will be discussed in the mutation lecture.
Ionic or electrostatic interactions. Theses forces give the DNA its twist. The phosphate group in
each residue is negatively charged at physiological pH and hence they actually repel each other. So
you have some conflict here. The bases actually want to associate together but the phosphates want
to be as far as possible away from each other. The net result is the twist to accommodate both
forces. If the molecule twists the phosphates can be separated while still maintaining the base
pairing.
NH2
N
N
O
HN
N
deoxyribose
N
N
deoxyribose
O
Adenine base paired to Thymine
O
N
N
deoxyribose
NH
H2N
N
N
N
NH2
O
deoxyribose
Guanine base paired to Cytosine
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The general statistics:
Right handed double helix
10 residues per turn
diameter ~20 Å
vertical distance between turns 3.4 nm or 34 Å
contains a major and minor groove
The major and minor grooves. These are the result of the angle of the N-glycosydic bond. They
enable proteins to interact with the DNA in a base specific manner. The major groove offers a
“window” into bases. Without this life would not exist!
Major Groove
NH2
CH3
O
N
C
C
C
C
C
N
N
H
O
H
HN
C
N
C
C
N
H
O
H
O
H
OH
H
HO
H
H
Minor Groove
H
H
The significance of these forces experimentally:
Many of the techniques used in molecular biology involve creating an environment that promotes
or disrupts hydrogen bonding. The manipulation of this environment is done in one or more of the
following ways:
•
Increased temperature will disrupt the H bonds
MBLG1001 Lecture 7
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•
Increased pH will remove the ring N proton from Thymine and Guanine, thus preventing
the H-bonding
•
The addition of formamide. Formamide offers an alternative molecule for H bonding.
The addition of formaldehyde. Formaldehyde reacts with the amine groups of adenine,
guanine and cytosine forming a Schiff’s base and preventing H bonding.
Ethanol Precipitation. The ethanol is able to disrupt the base stacking and the bases slide
•
•
off each other (into a mess!!).
The Physical and chemical stability of DNA (pages 112 -114).
Information stored in DNA must be passed on from one generation to the next over millions of
years. To do this DNA molecules must be very stable. They have evolved over time to be just that.
Initially it was thought that life started as RNA. After all RNA can store and transfer information
like DNA and it can catalyse some reactions like proteins. DNA and proteins then evolved and
became more specialized with time.
Properties which make DNA ideal as a store of genetic information:
• The sugar phosphate backbone is very stable to chemical attack.
•
The absence of the –OH at position 2 in the sugar makes DNA very resistant to alkali
(unlike RNA). The –OH at position 2 in ribose enables the phosphate nearby to form a
cyclic compound when the proton is removed (at high pH). This cyclic compound rapidly
breaks the phosphodiester bond resulting in a 2’ or 3’ Phosphorylated nucleotide.
•
The N-glycosidic bond to the bases is very stable as the bases are hydrophobic (if they
were more hydrophilic the bond would not be so stable).
•
The tightly stacked bases exclude almost all water. This protects the information (in the
more polar section in the center of the double strand) from water soluble compounds which
may react with the charged groups.
•
The two strands, which gives a double copy and a template for repair. The information (in
the form of charged and polar groups on the bases) is found, “buried” at the very center of
the double stranded helix, protected by a hydrophilic sugar phosphate backbone and then
the hydrophobic bases. These two layers with distinctly different properties make it
difficult for possible mutagenic compounds to reach the information. With single stranded
RNA the polar groups on the bases are exposed to all the chemical compounds in the cell
(there are a lot).
•
Uracil is replaced by thymine. Cytosine is deaminated to uracil spontaneously by a process
called oxidative deamination. In a given day about 100 cytosines are deaminated to uracils
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in each cell. These uracils are a corruption of the code and need to be removed and
replaced. But if uracil was a normal constituent of DNA there would be no way of knowing
whether the uracils were supposed to be there or not. By converting uracils to thymines (by
a methylation) before incorporation into DNA, any uracils found in the DNA must have
come from cytosine and can thus be excised and replaced.