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INTRODUCTION
•  In this chapter we will shift our attention to
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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9-2
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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9-3
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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9-4
Frederick Griffith Experiments with
Streptococcus pneumoniae
•  Griffith studied a bacterium (pneumococci) now
known as Streptococcus pneumoniae
•  S. pneumoniae comes in two strains
–  S à Smooth
•  Secrete a polysaccharide capsule
–  Protects bacterium from the immune system of animals
•  Produce smooth colonies on solid media
–  R à Rough
•  Unable to secrete a capsule
•  Produce colonies with a rough appearance
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9-5
n
In 1928, Griffith conducted experiments using two
strains of S. pneumoniae: type IIIS and type IIR
n
1. Inject mouse with live type IIIS bacteria
n
n
n
2. Inject mouse with live type IIR bacteria
n
n
n
Mouse survived
No living bacteria isolated from the mouse s blood
3. Inject mouse with heat-killed type IIIS bacteria
n
n
n
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
n
n
Mouse died
Type IIIS bacteria recovered from the mouse s blood
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9-6
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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
After"
several"
days
After"
several"
days
Mouse survived
Mouse survived
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
Mouse died
9-7
n
Griffith concluded that something from the dead
type IIIS bacteria was transforming type IIR bacteria
into type IIIS
n
n
He called this process transformation
The substance that allowed this to happen was
termed the transformation principle
n
Griffith did not know what it was
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9-8
n
The formation of the capsule is guided by the
bacteria s genetic material
n
n
n
n
Transformed bacteria acquired information to make the
capsule
Variation exists in ability to make capsule
The information required to create a capsule is replicated
and transmitted from mother to daughter cells
The nature of the transforming principle was
determined using experimental approaches that
incorporated various biochemical techniques
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9-9
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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9-10
n
Figure 9.2
Avery et al. conducted the following experiments
n
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
Type R"
cells
5
Type S"
DNA"
extract
+
RNase
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
9-11
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
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
n
They studied the
bacteriophage T2
n
Figure 9.3
It is relatively simple
since its composed of
only two
macromolecules
n
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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9-12
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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.
9-13
•  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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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
n
Refer to Figure 9.5
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9-15
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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 withE. 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 35
inS-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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9-16
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.
n
These results suggest that DNA is injected into the bacterial cytoplasm
during infection
n  Data is not conclusive since less than 100% of the DNA or protein
ended up in the cell or supernatant
n  Data is consistent with the hypothesis that DNA is the genetic
material
9-18
RNA Functions as the Genetic Material
in Some Viruses
n
In 1956, A. Gierer and G. Schramm isolated RNA
from the tobacco mosaic virus (TMV), a plant virus
n
Purified RNA caused the same lesions as intact TMV
viruses
n
n
Therefore, the viral genome is composed of RNA
Since that time, many RNA viruses have been
found
n
Refer to Table 9.1
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9-19
9-20
9.2 NUCLEIC ACID
STRUCTURE
•  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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9-21
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C
Nucleotides
A
TC
G
C
AA
TC
G
C
AA
TC
G
C
AAT
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
9-22
Nucleotides
n
n
The nucleotide is the repeating structural unit of
DNA and RNA
It has three components
n
n
n
n
A phosphate group
A pentose sugar
A nitrogenous base
Refer to Figure 9.7
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9-23
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Bases
Phosphate group
Sugars
Purines"
(double ring)
NH2
5′
HOCH2
H
O
H
H
O
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)
5
7
H
6
8
N
9
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′
Pyrimidines"
(single ring)
Uracil (U) (in RNA)
NH 2
H
1N
H
5
2
3
N
H
6
NH2
H
4
1
3N
2
N
O
H
Guanine (G)
Cytosine (C)
9-24
n
These atoms are found within individual nucleotides
n
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′
O
O
1′
4′
H
H
3′
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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9-25
n
Base + sugar à nucleoside
n
n
Base + sugar + phosphate(s) à nucleotide
n
n
Example
n  Adenine + ribose = Adenosine
n  Adenine + deoxyribose = Deoxyadenosine
Example
n  Adenosine monophosphate (AMP)
n  Adenosine diphosphate (ADP)
n  Adenosine triphosphate (ATP)
Refer to Figure 9.9
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9-26
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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–
Phosphate groups
Phosphates are
attached here
Figure 9.9
CH2
5′
4′
O
H
H
N
H
3
1′
H
Base always
attached here
2′
HO
OH
Ribose
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9-27
n
Nucleotides are covalently linked together by
phosphodiester bonds
n
n
Therefore the strand has directionality
n
n
n
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
n
The bases project from the backbone
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9-28
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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′
NH2
H
N
H
O
O
N
P
O
–
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
N
NH2
O
1′
H
2′
H
H
Sugar (deoxyribose)
3′
9-29
A Few Key Events Led to the Discovery of
the Structure of DNA
n
n
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
n
n
n
Linus Pauling
Rosalind Franklin and Maurice Wilkins
Erwin Chargaff
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9-30
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Linus Pauling
C
H
n
In the early 1950s, he
proposed that regions of
protein can fold into a
secondary structure
n
n
α-helix
H
N C C
C
C
H
To elucidate this structure,
he built ball-and-stick
models
C
H
C
C
N
C
O
C
N
C
H
C
O
Carbonyl "
oxygen
H
Amide "
hydrogen
C
N
O
C
C
N
O
Hydrogen "
bond
HO
O N
H
N
C
O
O N
N
H
H
C
C
H
C
C
O
N
N
C
C
O
O
Figure 9.11
(a) An α helix in a protein
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9-31
Rosalind Franklin
n
n
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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9-32
Rosalind Franklin
n
n
She made marked advances in X-ray diffraction
techniques with DNA
The diffraction pattern she obtained suggested
several structural features of DNA
n
n
n
Helical
More than one strand
10 base pairs per complete turn
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9-33
Erwin Chargaff s Experiment
n
n
n
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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9-34
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
n
Refer to Figure 9.13
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9-35
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y
Conceptual level
Experimental level
Solution of
1. For each type of cell, extract the
chromosomal
chromosomal material. This can be
extract
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.
2. Remove the protein. This can be done in
several ways, including treament with
protease.
3. Hydrolyze the DNA to release the bases
from the DNA strands. A common way
to do this is by strong acid treatment.
DNA +
proteins
Protease
DNA
Acid
G
A
C
C
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.)
5. Extract bands from paper into solutions Origin
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.)
Figure 9.13
C
G
T
G
AA A
A A
C
A
C C C CC C
G
G G GG G
TT
Individual
bases
T
A
T
T
T
G
T
6. Compare the base content in the DNA
from different organisms.
9-36
The Data
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9-37
Interpreting the Data
n
n
The data shown in Figure 9.13 are only a small
sampling of Chargaff s results
The compelling observation was that
n
n
n
Percent of adenine = percent of thymine
Percent of cytosine = percent of guanine
This observation became known as Chargaff s rule
n
It was a crucial piece of evidence that Watson and Crick
used to elucidate the structure of DNA
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9-38
Watson and Crick
n
Familiar with all of these key observations, Watson
and Crick set out to solve the structure of DNA
n
n
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
n
n
An early hypothesis proposed that DNA strands interact
through phosphate-Mg++ cross-links
Refer to Figure 9.14
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9-39
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Base
Base
Sugar
Sugar
O
Base
O
P
O
Sugar
–O
Mg2+
O–
O
Base
P
O
O
Sugar
Figure 9.14
n
This hypothesis was, of course, incorrect!
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9-40
Watson and Crick
n
n
They went back to the ball-and-stick units
They then built models with the
n
n
n
They first considered a structure in which bases form
H bonds with identical bases in the opposite strand
n
n
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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9-41
Watson and Crick
n
They then realized that the hydrogen bonding of A to
T was structurally similar to that of C to G
n
So they built ball-and-stick models with AT and CG
interactions between the two DNA strands
n
n
n
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
n
Rosalind Franklin died in 1958, and Nobel prizes are not
awarded posthumously
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9-42
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© Barrington Brown/Photo Researchers
(a) Watson and Crick
Figure 9.15
© Hulton|Archive by Getty Images
(b) Original model of the DNA double helix
9-43
The DNA Double Helix
n
General structural features (Figures 9.16 & 9.17)
Two strands are twisted together around a
common axis
n  There are 10 bases and 3.4 nm per complete turn
of the helix
n  The two strands are antiparallel
n
n
n
One runs in the 5 to 3 direction and the other 3 to 5
The helix is right-handed
n
As it spirals away from you, the helix turns in a
clockwise direction
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9-44
The DNA Double Helix
n
General structural features (Figures 9.16 & 9.17)
n
The double-bonded structure is stabilized by
n
1. Hydrogen bonding between complementary bases
n
n
n
A bonded to T by two hydrogen bonds
C bonded to G by three hydrogen bonds
2. Base stacking
n
Within the DNA, the bases are oriented so that the flattened
regions are facing each other
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9-45
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2 nm
Key Features
5ʹ′ P 3ʹ′
S P
P
A S
• Two strands of DNA form a "
right-handed double helix.
S
P
• The bases in opposite strands "
hydrogen bond according to the"
AT/GC rule.
P
• The 2 strands are antiparallel with "
regard to their 5′ to 3′ directionality.
S
G
G
S
P
C
S
P
C
C
C
O P
P
P
S
S
S
P
A
T
P
G
C
S
P
Figure 9.16
H
O
H
C
-
NH2
CH2
C
H N
N
O
O
H
O P
-
P
H
CH2
H
O
CH3
T
N
H2N
O
O
H
H
N
H
H H
N
A
N
H H
H
O
H
H
H
-
O
CH2 O P O
O
H
P
S
O P
-
O
CH2
O
One nucleotide
0.34 nm
N
H
O
P S
-
O
N
O
O
CH2 O P
H
G S
P
O
H
H2N
N
O
H H
H
N
O
S
S
H
N
O
H
H
G
N
H
H
H2N
N H
N
NH2
O
C
H
H
N
OH
H
H H
H
H
O
-
O
CH2 O P
O
3ʹ′ end
G
-
H H
N
G
N
O
H
N
O
H
C P
A
S
O
CH2 O P
H
S
P S
PS
P
S C
O
H
H H
H
O
H
O
H
G
C
T
G
H
O
S
P S
S
N
HO
N
O
P
P
One complete "
turn 3.4 nm
A
H
T
A
P
G
S
P
S
H
N
N
N
O
T
NH2
H
H
G
P
-
CH2
H
• There are ~10.0 nucleotides in each "
P S
P
strand per complete 360° turn of"
S P
the helix.
S
A T S
S
O
N
H
-
O P O
S
P
3ʹ′ end
H
CH3
O
P S
S
5ʹ′ end
-
O
5ʹ′ end
P
S
5ʹ′
3ʹ′
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9-46
The DNA Double Helix
n
General structural features (Figures 9.16 & 9.17)
n
There are two asymmetrical grooves on the
outside of the helix
n
1. Major groove
n
2. Minor groove
n
Certain proteins can bind within these grooves
n
They can thus interact with a particular sequence of bases
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9-47
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
9-48
DNA Can Form Alternative Types of
Double Helices
n
The DNA double helix can form different
types of secondary structure
n
n
The predominant form found in living cells is
DNA
B-
However, under certain in vitro conditions,
A-DNA and Z-DNA double helices can form
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9-49
n
A-DNA
n
n
n
n
n
Right-handed helix
11 bp per turn
Occurs under conditions of low humidity
Little evidence to suggest that it is biologically important
Z-DNA
n
n
n
Left-handed helix
12 bp per turn
Its formation is favored by
n
n
n
Alternating purine/pyrimidine sequences, at high salt
concentrations (e.g. GCGCGCGCGC)
Cytosine methylation, at low salt concentrations
Evidence from yeast suggests that it may play a role in
transcription and recombination
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9-50
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C
G
C
G
C
G
G
C
C
C
G
G
G
G
C
C
G
C
C
G
T
A
G
A
T
Bases
substantially
tilted relative
to the central
axis
T
A
A
Bases
substantially
tilted relative
to the central
axis
C
C
T
G
G
G
C
C
C
C
G
G
G
C
C
G
C
G
G
C
C
C
G
A DNA
G
B DNA
Z DNA
(a) Molecular structures
Backbone
Minor"
groove
Bases
Major "
groove
Figure 9.18
B DNA
(b) Space-filling models
Bases relatively
perpendicular to the
central axis
Sugar-phosphate
backbone follows a
zigzag pattern
Z DNA
9-51
DNA Can Form a Triple Helix
n
In the late 1950s, Alexander Rich et al. discovered
triplex DNA
n
n
It was formed in vitro using DNA pieces that were made
synthetically
In the 1980s, it was discovered that natural doublestranded DNA can join with a synthetic strand of
DNA to form triplex DNA
n
The synthetic strand binds to the major groove of the
naturally-occurring double-stranded DNA
n
Refer to Figure 9.19
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9-52
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5'
3'
3'
5'
3' 5'
(a) Ribbon model
Figure 9.19
T binds to an
AT pair in
biological DNA
C binds to a
CG pair in
biological DNA
3ʹ′ 5ʹ′"
T – A"
G – C"
C – G"
C – G"
T – A"
A – T"
G – C"
3ʹ′ G – C"
T • A – T"
T • A – T"
T • A – T"
T • A – T"
T • A – T"
C + G – C"
T • A – T"
T • A – T"
T • A – T"
C + G – C"
T • A – T"
T • A – T"
C + G – C"
T • A – T"
T • A – T"
T • A – T"
T • A – T"
C + G – C"
T • A – T"
T • A – T"
5ʹ′ G – C"
G – C"
C – G"
C – G"
C – G"
A – T"
G – C"
5ʹ′ 3ʹ′"
n
n
n
Triplex DNA formation
is sequence specific
The pairing rules are
Triplex DNA has been
implicated in several
cellular processes
n
n
Replication,
transcription,
recombination
Cellular proteins that
specifically recognize
triplex DNA have been
recently discovered
9-53
The Three-Dimensional Structure of
DNA
n
To fit within a living cell, the DNA double helix must
be extensively compacted into a 3-D conformation
n
This is aided by DNA-binding proteins
n
n
Refer to 9.20
This topic will be discussed in detail in Chapter 10
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9-54
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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
9-55
RNA Structure
n
The primary structure of an RNA strand is much
like that of a DNA strand
n
n
n
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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9-56
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Backbone
Bases
5′
O
H
H
N
Uracil (U)
O–
O
P
O–
O
CH2
5′
4′
H
H
H
O
N
O
1′
H
2′
OH
H
3′
NH2
N
Phosphodiester"
linkage
N
Adenine (A)
H
N
O
O
P
O–
O
CH2
5′
4′
H
H
1′
H
2′
OH
H
NH2
H
N
H
O
P
O
N
O
3′
O
H
O
–
CH2
5′
4′
H
H
Cytosine (C)
O
N
O
1′
H
2′
OH
Guanine"
(G)
H
3′
O
H
N
N
H
N
O
RNA"
nucleotide
Figure 9.21
O
P
CH2
O
5′
O
4′
H
Phosphate H
3′
OH
N
NH2
O
–
1′
H
2′
OH
H
Sugar (ribose)
3′
9-57
n
Although usually single-stranded, RNA molecules
can form short double-stranded regions
n
This secondary structure is due to complementary basepairing
n
n
n
This allows short regions to form a double helix
RNA double helices typically
n
n
n
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
n
Refer to Figure 9.22
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9-58
Complementary regions
Held together by
hydrogen bonds
Figure 9.22
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U
U A
A U
U A
A
U
C
G
C
C
U
C
C G
A
A U
(a) Bulge loop!
C
C G
U
G
U A
G C 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
Noncomplementary 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
A
G
U
G C
A U
A
A C G U
U G C A
G G C A
C C G U
(d) Stem-loop
Also called
hair-pin
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9-59
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Double helix
n
Many factors
contribute to the
tertiary structure of
RNA
n
5′ end
Molecule contains
single- and doublestranded regions
For example
n
n
3′ end"
(acceptor"
site)
Base-pairing and
base stacking within
the RNA itself
These spontaneously
fold and interact to
produce this 3-D
Double helix
structure
Interactions with
ions, small
molecules and large
proteins
Anticodon
(a) Ribbon model
n
Figure 9.23
Figure 9.23 depicts the tertiary structure of tRNAphe
n
The transfer RNA that carries phenylalanine
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9-60