Download (Chapter 9): Molecular Structure of DNA and RNA

LECTURE 1
Introduction to
DNA and RNA
(Chapter 09)
Slides 1-25; 35-48; 54-57
On your own:
Slides 26-34; 49-53; 58-62
1
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INTRODUCTION
• Genetics: Study of the structure, function,
transmission of genes
• Only living organisms have genes
• To understand genetics, we will start with the
question: “What is Life?”
– Characteristics shared by all living forms but not by
non-living forms
2
What is the role of DNA in life?
• DNA carries information
• Information, entropy and Solla Sollew
– “Duhka (A wheel out of kilter)
– I Had Trouble in Getting to Solla Sollew (where there aren’t any
problems at least very few)
UNIVERSE
LIFE (GENES)
(entropy)
(negative entropy)
3
How does life fight entropy?
• Living organisms “suck energy” (directly or
indirectly) from the Sun
If no life, medium λ =
medium radiation energy,
medium entropy
Short mean λ = high radiation
energy, low entropy
If life, long λ = low radiation energy
high entropy
Very high potential
energy;
Very low entropy
4
• Fusion reactions in sun’s core
– Hydrogen to helium with release of photons
– Mass converted to energy (E = mc2)
• 600 million tons hydrogen fused to helium per second
• Creates 596 tons helium per second + 3.9 x 1026 Joules of energy
• Life borrows this energy to build complex, highly ordered structures
5
• The source of all life’s “problems” is entropy
• The “solution” is stored as information in DNA
– Information and negative entropy
– Analog vs digital
• Different organisms have different genes;
different strategies for the survival of the genes
and their transmission to the next generation
• Those genes that built the best “survival
machines” are still with us today
• The vast majority of genes were lost
• This is a blind, undirected process
6
MOLECULAR GENETICS
• Chapter 9 examines the genetic material in detail:
molecular genetics
– The study of DNA structure and function at the
molecular level
• Recent dramatic advances in techniques and
approaches have greatly expanded our
understanding of molecular genetics
– And also of transmission and population genetics
• To a large extent, our knowledge of genetics
comes from our knowledge of the molecular
structure of DNA and RNA
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7
9.1 IDENTIFICATION OF DNA AS
THE GENETIC MATERIAL
• To fulfill its role, the genetic material must meet
several criteria
– 1. Information: It must contain the information necessary
to make an entire organism
– 2. Transmission: It must be passed from parent to
offspring
– 3. Replication: It must be copied
• In order to be passed from parent to offspring
– 4. Variation: It must be capable of changes
• To account for the known phenotypic variation in each species
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8
9.1 IDENTIFICATION OF DNA AS
THE GENETIC MATERIAL
• The data of many geneticists, including
Mendel, were consistent with these four
properties
– However, the chemical nature of the genetic material
cannot be identified solely by genetic crosses
• Indeed, the identification of DNA as the genetic
material involved a series of outstanding
experimental approaches
– These will be examined next
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9
Frederick Griffith Experiments with
Streptococcus pneumoniae
• Griffith studied a bacterium (pneumococci) now
known as Streptococcus pneumoniae
• S. pneumoniae comes in two strains
– S  Smooth
• Secrete a polysaccharide capsule
– Protects bacterium from the immune system of animals
• Produce smooth colonies on solid media
– R  Rough
• Unable to secrete a capsule
• Produce colonies with a rough appearance
10
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
In 1928, Griffith conducted experiments using two
strains of S. pneumoniae: type IIIS and type IIR

1. Inject mouse with live type IIIS bacteria



2. Inject mouse with live type IIR bacteria



Mouse survived
No living bacteria isolated from the mouse’s blood
3. Inject mouse with heat-killed type IIIS bacteria



Mouse died
Type IIIS bacteria recovered from the mouse’s blood
Mouse survived
No living bacteria isolated from the mouse’s blood
4. Inject mouse with live type IIR + heat-killed type IIIS cells


Mouse died
Type IIIS bacteria recovered from the mouse’s blood
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11
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Living type S bacteria were
injected into a mouse.
Living type R bacteria were
injected into a mouse.
Heat-killed type S bacteria
were injected into a mouse.
Living type R and heat-killed
type S bacteria were injected
into a mouse.
Live
type R
Dead
type S
After
several
days
Mouse died
After
several
days
Mouse survived
After
several
days
Mouse survived
After
several
days
Mouse died
Type S bacteria were isolated
from the dead mouse.
No living bacteria were isolated
from the mouse.
No living bacteria were isolated
from the mouse.
Type S bacteria were isolated
from the dead mouse.
(a) Live type S
(b) Live type R
(c) Dead type S
(d) Live type R + dead type S
Figure 9.1
12

Griffith concluded that something from the dead
type IIIS bacteria was transforming type IIR
bacteria into type IIIS


He called this process transformation
The substance that allowed this to happen was
termed the transformation principle



Griffith did not know what it was
Suggested that the genetic information molecule is
resistant to high temperature
 Most proteins degrade at high temperature
Connect Griffith’s experiment to Experiment 1 (lab)
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13

We now know that the formation of the capsule is
guided by the bacteria’s genetic material





Transformed R bacteria acquired a gene from the dead
lysed S bacterua that directs the cell how to make a
capsule polysachharide
Variation exists in ability to make capsule (R cells carry a
non-function allele for the gene)
The gene required to create a capsule is replicated and
transmitted from mother to daughter cells
Once R cells acquired it they passed the trait to their
“offspring”
The molecular nature of the transforming principle
was determined using experimental approaches
that incorporated various biochemical techniques
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14
The Experiments of
Avery, MacLeod and McCarty
• Avery, MacLeod and McCarty realized that Griffith’s
observations could be used to identify the genetic material
• They carried out their experiments in the 1940s
– At that time, it was known that DNA, RNA, proteins and
carbohydrates are the major constituents of living cells
• They prepared cell extracts from type IIIS cells and
purified each type of macromolecule
– Only the extract that contained purified DNA was able to convert
type IIR bacteria into type IIIS
– Treatment of the DNA extract with RNase or protease did not
eliminate transformation
– Treatment with DNase did
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15

Avery et al. conducted the following experiments
Figure 9.2

To further verify that DNA, and not a contaminant (RNA
or protein), is the genetic material
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1
Type R
cells
2
3
Type S
DNA
extract
Type R
cells
Mix
4
Type S
DNA
extract
+
DNase
Type R
cells
Mix
5
Type S
DNA
extract
+
RNase
Type R
cells
Mix
Type R
cells
Type S
DNA
extract
+
protease
Mix
Allow sufficient time for the DNA to be taken up by the type R bacteria. Only a small percentage of the type R bacteria will be transformed to type S.
Add an antibody that aggregates type R bacteria (that have not been transformed). The aggregated bacteria are removed by gentle centrifugation.
Plate the remaining bacteria on petri plates. Incubate overnight.
Transformed
Transformed
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Transformed
16
Experiment 9A Hershey and Chase
Experiment with Bacteriophage T2
• In 1952, Alfred Hershey and Marsha Chase
provided further evidence that DNA is the genetic
material
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
They studied the
bacteriophage T2

Figure 9.3
It is relatively simple
since its composed of
only two
macromolecules

DNA and protein
DNA
(inside the
capsid head)
Inside the
capsid
Head
Sheath
Made up
of protein
Tail fiber
Base plate
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17
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Phage binds to host cell.
Capsid
Genetic material
Bacterial
cell wall
Bacterial
chromosome
Phage injects its
DNA into host cell.
Figure 9.4
Phage coat (protein)
precariously attached
to bacterial cell
The expression of
phage genes leads
to the synthesis of
phage components.
Life cycle of
bacteriophage T2
DNA inside of cell
Phage components
are assembled.
Host cell lyses
and new phages
are released.
18
• The Hershey and Chase experiment can be
summarized as follows:
– Used radioisotopes to distinguish DNA from proteins
• 32P labels DNA specifically
• 35S labels protein specifically
– Radioactively-labeled phages were used to infect
non-radioactive Escherichia coli cells
– After allowing sufficient time for infection to proceed,
the residual phage particles were sheared off the cells
• => Phage ghosts and E. coli cells were separated
– Radioactivity was monitored using a scintillation
counter
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19
9-14
The Hypothesis
– Only the genetic material of the phage is injected
into the bacterium
• Isotope labeling will reveal if it is DNA or protein
Testing the Hypothesis

Refer to Figure 9.5
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20
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Experimental level
Conceptual level
1. Grow bacterial cells. Divide into
two flasks.
Suspension of
E. coli cells
35S-labeled
protein capsid
32P-labeled
DNA
2. Into one flask, add 35S-labeled phage; in
the second flask, add 32P-labeled phage.
35S-labeled
32P-labeled
T2 phage
T2 phage
3. Allow infection to occur.
After blending
Bacterial cell
4. Agitate solutions in blenders for
different lengths of time to shear the
empty phages off the bacterial cells.
Suspension of
Suspension of
E. coli infected with E. coli infected with
35S-labeled phage
32P-labeled phage
5. Centrifuge at 10,000 rpm.
Supernatant
6. The heavy bacterial cells sediment to the with
pellet, while the lighter phages remain in 35S-labeled
the supernatant. (See Appendix for
empty phage
explanation of centrifugation.)
7. Count the amount of radioisotope in
the supernatant with a Geiger counter.
Compare it with the starting amount.
Figure 9.5
Pellet with
unlabeled
DNA in
infected
E. coli cells
Supernatant
with
unlabeled
empty phage
Pellet with
32P-labeled
DNA in
infected
E. coli cells
Viral genetic
material
Sheared empty
phage (labeled)
Bacterial
cell
Viral
genetic
material
(labeled)
Sheared
empty
phage
Sheared empty
phages (labeled)
Sheared
empty
phages
(unlabeled)
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21
The Data
Total isotope in supernatant (%)
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100
80
Extracellular 35S
80%
Blending removes 80%
of 35S from E. coli cells.
Extracellular 32P
35%
Most of the 32P (65%)
remains with intact
E. coli cells.
60
40
20
0
0
1
2
3
4
5
6
7
8
Agitation time in blender (min)
Data from A. D. Hershey and Martha Chase (1952) Independent Functions of Viral Protein and Nucleic Acid in Growth of
Bacteriophage. Journal of General Physiology 36, 39–56.
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22
Interpreting the Data
Most of the 35S was found
in the supernatant
But only a small
percentage of 32P
Total isotope in supernatant (%)
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100
80
Extracellular 35S
80%
Blending removes 80%
of 35S from E. coli cells.
Extracellular 32P
35%
Most of the 32P (65%)
remains with intact
E. coli cells.
60
40
20
0
0
1
2
3
4
5
6
7
8
Agitation time in blender (min)
Data from A. D. Hershey and Martha Chase (1952) Independent Functions of Viral Protein and Nucleic Acid in Growth of
Bacteriophage. Journal of General Physiology 36, 39–56.

These results suggest that DNA is injected into the bacterial cytoplasm
during infection
 Data is not conclusive since less than 100% of the DNA or protein
ended up in the cell or supernatant
 Data is consistent with the hypothesis that DNA is the genetic
material
23
9-18
RNA Functions as the Genetic Material
in Some Viruses

In 1956, A. Gierer and G. Schramm isolated RNA
from the tobacco mosaic virus (TMV), a plant virus

Purified RNA caused the same lesions as intact TMV
viruses


Therefore, the viral genome is composed of RNA
Since that time, many RNA viruses have been
found

Refer to Table 9.1
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24
25
9.2 NUCLEIC ACID
STRUCTURE – On Your Own
• DNA and RNA are large macromolecules with
several levels of complexity
– 1. Nucleotides form the repeating unit of nucleic acids
– 2. Nucleotides are linked to form a linear strand of RNA
or DNA
– 3. Two strands can interact to form a double helix
– 4. The 3-D structure of DNA results from folding and
bending of the double helix. Interaction of DNA with
proteins produces chromosomes within living cells
– Refer to fig. 9.6
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26
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C
Nucleotides
A
TC
G
C
AA
TC
G
C
AA
TC
G
C
A AT
Single strand
C
G
C A
T GT
T A
A
G T
A C
GA
T
C A
T GT
T A
A
G
Double helix
Figure 9.6
Three-dimensional structure
27
Nucleotides

The nucleotide is the repeating structural unit of
DNA and RNA

It has three components




A phosphate group
A pentose sugar
A nitrogenous base
Refer to Figure 9.7
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28
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Bases
Phosphate group
Sugars
Purines
(double ring)
NH2
5′
HOCH2
H
O
H
H
P
H
5
7
H
N
9
4
O
CH3
1N
5
2
6
3
H
N
D-Deoxyribose (in DNA)
1′
4′
H
H
3′
HO
N
OH
O
H
H
2′
OH
D-Ribose (in RNA)
6
5
7
H
8
9
N
5
2
6
O
4
H
4
1
N
3N
2
O
H
Thymine (T) (in DNA)
O
HOCH2
1
H
3N
H
Adenine (A)
5′
4
O
N
H
O–
Figure 9.7
6
8
H
HO
O
N
2′
3′
O
OH
1′
4′
O–
Pyrimidines
(single ring)
Uracil (U) (in RNA)
NH 2
H
1N
H
5
2
3
N
H
Guanine (G)
6
NH2
H
4
1
3N
2
N
O
H
Cytosine (C)
29
9-24

These atoms are found within individual nucleotides

However, they are removed when nucleotides join together to make
strands of DNA or RNA
A, G, C or T
A, G, C or U
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O
O P
O
Base
O
O–
Phosphate
CH2
5′
4′
H
O
O
H
3′
1′
H
H
2′
OH
H
Deoxyribose
(a) Repeating unit of
deoxyribonucleic
acid (DNA)
Figure 9.8
P
Base
O
O–
Phosphate
CH2
5′
4′
H
H
3′
O
H
1′
H
2′
OH
OH
Ribose
(b) Repeating unit of
ribonucleic acid (RNA)
The structure of nucleotides found in (a) DNA and (b) RNA
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30

Base + sugar  nucleoside


Base + sugar + phosphate(s)  nucleotide


Example
 Adenine + ribose = Adenosine
 Adenine + deoxyribose = Deoxyadenosine
Example
 Adenosine monophosphate (AMP)
 Adenosine diphosphate (ADP)
 Adenosine triphosphate (ATP)
Refer to Figure 9.9
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31
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Adenosine triphosphate
Adenosine diphosphate
Adenosine monophosphate
Adenosine
Adenine
Phosphoester bond
NH2
N
N
H
O
–O
P
O–
O
O
P
O
O
O–
P
N
O
O–
CH2
5′
4′
Phosphate groups
Phosphates are
attached here
Figure 9.9
O
1′
H
H
N
H
H
Base always
attached here
2′
3
HO
OH
Ribose
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32

Nucleotides are covalently linked together by
phosphodiester bonds


Therefore the strand has directionality



A phosphate connects the 5’ carbon of one nucleotide to
the 3’ carbon of another
5’ to 3’
In a strand, all sugar molecules are oriented in the same
direction
The phosphates and sugar molecules form the
backbone of the nucleic acid strand

The bases project from the backbone
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33
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Backbone
Bases
O
5′
CH3
N
Thymine (T)
H
–
O
O
P
O
–
O
CH2
5′
4′
H
H
O
N
O
1′
H
2′
H
H
3′
NH2
N
Phosphodiester
linkage
N
Adenine (A)
H
N
O
O
P
O–
O
CH2
5′
4′
H
H
O
1′
H
2′
H
H
3′
P
O–
NH2
H
N
H
O
O
N
O
CH2
5′
4′
H
H
Cytosine (C)
O
N
O
1′
H
2′
H
H
3′
Guanine (G)
O
H
N
N
H
N
O
Single
nucleotide
O
P
O
CH2
5′
O–
4′
H
Phosphate H
3′
OH
Figure 9.10
3′
N
NH2
O
1′
H
2′
H
H
Sugar (deoxyribose)
34
A Few Key Events Led to the Discovery of
the Structure of DNA

In 1953, James Watson and Francis Crick elucidated
the double helical structure of DNA

The scientific framework for their breakthrough was
provided by other scientists including



Linus Pauling
Rosalind Franklin and Maurice Wilkins
Erwin Chargaff
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35
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Linus Pauling
C
H

In the early 1950s, he
proposed that regions of
protein can fold into a
secondary structure

a-helix
N C C
C
C
H
C
To elucidate this structure,
he built ball-and-stick
models
C
H
C
C
C
N
O
Carbonyl
oxygen
H
Amide
hydrogen
C
N
O
C
C
N
O
Hydrogen
bond
HO
H
O N
H
N
N
O
C
H
N
C
O
O N
C

H
H
C
C
H
C
C
N
C
O
C
N
O
O
Figure 9.11
(a) An a helix in a protein
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36
Rosalind Franklin


She worked in the
same laboratory as
Maurice Wilkins
She used X-ray
diffraction to study
wet fibers of DNA
X rays diffracted
by DNA
Wet DNA fibers
X-ray beam
The pattern represents the
atomic array in wet fibers.
(b) X-ray diffraction of wet DNA fibers
Figure 9.12b
The diffraction pattern is interpreted
(using mathematical theory)
This can ultimately provide
information concerning the
structure of the molecule
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37
Rosalind Franklin

She made marked advances in X-ray diffraction
techniques with DNA

The diffraction pattern she obtained suggested
several structural features of DNA



Helical
More than one strand
10 base pairs per complete turn
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38
Erwin Chargaff’s Experiment

Chargaff pioneered many of the biochemical
techniques for the isolation, purification and
measurement of nucleic acids from living cells

It was known that DNA contained the four bases:
A, G, C and T

Chargaff analyzed the base composition of DNA
isolated from many different species
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39
The Hypothesis
– An analysis of the base composition of DNA in
different species may reveal important features
about the structure of DNA
Testing the Hypothesis

Refer to Figure 9.13
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40
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y
Conceptual level
Experimental level
1. For each type of cell, extract the
chromosomal material. This can be
done in a variety of ways, including the
use of high salt, detergent, or mild alkali
treatment. Note: The chromosomes
contain both DNA and protein.
Solution of
chromosomal
extract
DNA +
proteins
Protease
2. Remove the protein. This can be done in
several ways, including treament with
protease.
DNA
Acid
3. Hydrolyze the DNA to release the bases
from the DNA strands. A common way
to do this is by strong acid treatment.
G
A
Individual
bases
T
A
C
C
G
T
C
G
4. Separate the bases by chromatography.
Paper chromatography provides an easy
way to separate the four types of bases.
(The technique of chromatography is
described in the Appendix.)
AA
A A
C
C
C
A
C CC C
G
G G GG G
5. Extract bands from paper into solutions
and determine the amounts of each base
by spectroscopy. Each base will absorb
light at a particular wavelength. By
examining the absorption profile of a
sample of base, it is then possible to
calculate the amount of the base.
(Spectroscopy is described in the
Appendix.)
A
G
Origin
TT
T
T
T
T
6. Compare the base content in the DNA
from different organisms.
Figure 9.13
41
The Data
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42
Interpreting the Data

The data shown in Figure 9.13 are only a small
sampling of Chargaff’s results

The compelling observation was that



Percent of adenine = percent of thymine
Percent of cytosine = percent of guanine
This observation became known as Chargaff’s rule

It was a crucial piece of evidence that Watson and Crick
used to elucidate the structure of DNA
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43
Watson and Crick

Familiar with all of these key observations, Watson
and Crick set out to solve the structure of DNA


They tried to build ball-and-stick models that incorporated
all known experimental observations
A critical question was how the two (or more strands)
would interact


An early hypothesis proposed that DNA strands interact
through phosphate-Mg++ cross-links
Refer to Figure 9.14
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44
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Base
Base
Sugar
Sugar
O
Base
O
P
O
Sugar
–O
Mg 2+
O–
O
Base
P
O
O
Sugar
Figure 9.14

This hypothesis was, of course, incorrect!
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45
Watson and Crick


They went back to the ball-and-stick units
They then built models with the



They first considered a structure in which bases form
H bonds with identical bases in the opposite strand


Sugar-phosphate backbone on the outside
Bases projecting toward each other
ie., A to A, T to T, C to C, and G to G
Model building revealed that this also was incorrect
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46
Watson and Crick

They then realized that the hydrogen bonding of A to
T was structurally similar to that of C to G

So they built ball-and-stick models with AT and CG
interactions between the two DNA strands



These were consistent with all known data about DNA structure
Refer to Figure 9.15
Watson, Crick and Maurice Wilkins were awarded
the Nobel Prize in 1962

Rosalind Franklin died in 1958, and Nobel prizes are not
awarded posthumously
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47
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
© Barrington Brown/Photo Researchers
(a) Watson and Crick
© Hulton|Archive by Getty Images
Figure 9.15
(b) Original model of the DNA double helix
48
The DNA Double Helix
– On Your Own

General structural features (Figures 9.16 & 9.17)



Two strands are twisted together around a
common axis
There are 10 bases and 3.4 nm per complete turn
of the helix
The two strands are antiparallel


One runs in the 5’ to 3’ direction and the other 3’ to 5’
The helix is right-handed

As it spirals away from you, the helix turns in a
clockwise direction
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49
The DNA Double Helix

General structural features (Figures 9.16 & 9.17)

The double-bonded structure is stabilized by

1. Hydrogen bonding between complementary bases



A bonded to T by two hydrogen bonds
C bonded to G by three hydrogen bonds
2. Base stacking

Within the DNA, the bases are oriented so that the flattened
regions are facing each other
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50
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2 nm
Key Features
5 P 3
S P
P
A S
• Two strands of DNA form a
right-handed double helix.
• The bases in opposite strands
hydrogen bond according to the
AT/GC rule.
S
P
P
• The 2 strands are antiparallel with
regard to their 5′ to 3′ directionality.
5 end
G
G
S
S
P
C
C
O P O
C
O P
P
S
P
S
A
T
P
S
G
C
S
P
-
NH2
H
CH2
C
O
H
H N
N
H
-
O
P
H
H
T
O
O
H
CH3
H
N
H2N
N
H
N
-
O
O
H
H
H
H H
H
CH2 O P
N
H
S
O P
-
O
H
G
O
H
H
N
H
H
H2N
N
CH2
O
O
O
H
One nucleotide
0.34 nm
S
H H
N
A
O
P S
O
O
P
P
-
CH2 O P
H
CH2
O
O
O
H
H2N
N
O P
H H
H
N
O
G S
C
N H
N
NH2
O
C
H
H
N
OH
H
H H
H
H
O
-
O
CH2 O P
O
3 end
G
O
H H
N
G
N
O
S
A
S
-
CH2 O P
N
O
H
C P
P S
PS
P
S C
O
O
H
H
O
H
G
C
T
G
O
H
H H
H
O
H
S
S
S
N
N
HO
N
H
O
P S
P
H
A
O
P
P
One complete
turn 3.4 nm
N
O
H
T
A
P
G
S
P
S
-
CH2
T
NH2
H
H
G
P
O
N
H
H
P S
P
S P
S
A T S
S
H
CH3
O
S
P
3 end
P
C
P S
S
• There are ~10.0 nucleotides in each
strand per complete 360° turn of
the helix.
S
-
O
5 end
P
S
5
3
Figure 9.16
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51
The DNA Double Helix

General structural features (Figures 9.16 & 9.17)

There are two asymmetrical grooves on the
outside of the helix

1. Major groove

2. Minor groove

Certain proteins can bind within these grooves

They can thus interact with a particular sequence of bases
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52
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Minor
groove
Minor
groove
Major
groove
Major
groove
© Laguna Design/Photo Researchers
(a) Ball-and-stick model of DNA
Figure 9.17
(b) Space-filling model of DNA
53
To calculate how often the enzyme will cut:
4(# bp in recognition site)
For Eco RI = 46 = cuts every 4,096 bp
Homodimerizes then binds in successive
major grooves to cut each DNA strand
To calculate number of cuts/genome:
(bp in genome)/4(# bp in recognition site)
Eco RI cuts the human genome:
(3.2 x 109 bp)/4096 cuts/bp = 780,000 cuts
54
The Three-Dimensional Structure of
DNA

To fit within a living cell, the DNA double helix must
be extensively compacted into a 3-D conformation

This is aided by DNA-binding proteins

Refer to 9.20
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55
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Radial loops
(300 nm in diameter)
Metaphase
chromosome
30 nm fiber
Nucleosomes
(11 nm in diameter)
DNA
(2 nm in diameter)
Histone
protein
DNA wound
around histone
proteins
Each chromatid
(700 nm in diameter)
Figure 9.20
56
57
RNA Structure – On Your Own

The primary structure of an RNA strand is much
like that of a DNA strand

Refer to Figure 9.21 vs. 9.10

RNA strands are typically several hundred to
several thousand nucleotides in length

In RNA synthesis, only one of the two strands of
DNA is used as a template
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58
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Backbone
Bases
5′
O
H
H
N
Uracil (U)
O–
O
P
O–
O
5′
4′
CH2
H
H
O
N
O
1′
H
2′
OH
H
H
3′
NH2
N
Phosphodiester
linkage
N
Adenine (A)
H
N
O
O
P
O–
O
5′
4′
CH2
H
1′
H
2′
OH
H
H
NH2
H
N
H
O
P
O
N
O
3′
O
H
O
–
5′
4′
CH2
O
N
O
H
H
Cytosine (C)
H
1′
H
Guanine
(G)
2′
OH
3′
O
H
N
N
H
N
O
RNA
nucleotide
O
P
O
–
O
Phosphate
5′
4′
CH2
H
N
NH2
O
H
3′
OH
H
1′
H
2′
OH
Sugar (ribose)
Figure 9.21
3′
59

Although usually single-stranded, RNA molecules
can form short double-stranded regions

This secondary structure is due to complementary basepairing



This allows short regions to form a double helix
RNA double helices typically



A to U and C to G
Are right-handed
Have the A form with 11 to 12 base pairs per turn
Different types of RNA secondary structures are
possible

Refer to Figure 9.22
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60
Complementary regions
Held together by
hydrogen bonds
Figure 9.22
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A
U
U A
A U
A U
U A
A
U
C
G
C
C
U
C
C G
A
A U
(a) Bulge loop
C G
U
C
G
U A
C
G
A A G
C G U U C
A
C
C C
G A A
G G C U U
G C
C G
A U
C G
A U
(c) Multibranched junction
Non-complementary regions
Have bases projecting away
from double stranded regions
C G
U
G C
A U
A U
C G
G
C G
(b) Internal loop
C
G
C G
G
C
U
G C
C G
C G
U
A U
U A
A
G
U
G C
A
C
U
A
G
G
U
C A
C
G G
A
C C G U
(d) Stem-loop
Also called
hair-pin
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61
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Double helix

Many factors
contribute to the
tertiary structure of
RNA

5′ end
Molecule contains
single- and doublestranded regions
For example


3′ end
(acceptor
site)
Base-pairing and
base stacking within
the RNA itself
Double helix
Interactions with
ions, small
molecules and large
proteins
These spontaneously
fold and interact to
produce this 3-D
structure
Anticodon
(a) Ribbon model

Figure 9.23
Figure 9.23 depicts the tertiary structure of tRNAphe

The transfer RNA that carries phenylalanine
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