DNA: The Genetic Material
Chapter 14
1
Learning Objectives
14.1 The Nature of the Genetic Material
• Understand experiments of
– Griffith & Avery
– Avery, MacLeod, and McCarty
– Hershey and Chase
• For both experiments, know/understand
major findings
2
Frederick Griffith – 1928
• Studied Streptococcus pneumoniae, a
pathogenic bacterium causing pneumonia
• 2 strains of Streptococcus
– S strain is virulent
– R strain is nonvirulent
3
http://o.quizlet.com/i/GEJK81oHlTEYutTzmSQK6Q.jpg
Griffith’s Experiment
• Griffith infected mice with these strains
hoping to understand the difference
between the strains
4
Griffith’s Experiment
Live Nonvirulent
Strain of
S. pneumoniae
Live Virulent
Strain of S. pneumoniae
Polysaccharide
coat
Mice die
Mice live
5
a.
b.
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Griffith’s Experiment
Mixture of Heat-killed Virulent
and Live Nonvirulent
Strains of S. pneumoniae
Heat-killed Virulent
Strain of S. pneumoniae
+
Mice die
Their lungs contain live
pathogenic strain of
S. pneumoniae
Mice live
c.
d.
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6
• Griffith’s results
– Live S strain cells killed the mice
– Live R strain cells did not kill the mice
– Heat-killed S strain cells did not kill the
mice
– Heat-killed S strain + live R strain cells
killed the mice
7
• Transformation
– Information specifying virulence passed
from the dead S strain cells into the live
R strain cells
• Our modern interpretation is that genetic
material was actually transferred between
the cells
8
Avery, MacLeod, & McCarty – 1944
• Repeated Griffith’s experiment using
purified cell extracts
9
http://biology.kenyon.edu/courses/biol114/KH_lecture_images/How_DNA_works/FG11_02.JPG
Avery, MacLeod, & McCarty – 1944
• Removal of all protein from the transforming
material did not destroy its ability to
transform R strain cells
• DNA-digesting enzymes destroyed all
transforming ability
• Supported DNA as the genetic material
10
Hershey & Chase –1952
• Investigated bacteriophages
– Viruses that infect bacteria
• Bacteriophage was composed of only
DNA and protein
• Wanted to determine which of these
molecules is the genetic material that is
injected into the bacteria
11
Hershey & Chase –1952
• Bacteriophage DNA was labeled with
radioactive phosphorus (32P)
• Bacteriophage protein was labeled with
radioactive sulfur (35S)
• Radioactive molecules were tracked
12
Hershey & Chase Experiment
35S-Labeled
Bacteriophages
+
Phage grown in radioactive 35S,
which is incorporated into phage coat
Virus infect
bacteria
32P-Labeled
Blender separates
phage coat from bacteria
Centrifuge forms
bacterial pellet
35S
in supernatant
Bacteriophages
+
Phage grown in radioactive 32P.
which is incorporated into phage DNA
Virus infect
bacteria
Blender separates
Centrifuge forms
phage coat from bacteria bacterial pellet
32P
in bacteria pellet
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13
Hershey & Chase –1952
• Only the bacteriophage DNA (as indicated
by the 32P) entered the bacteria and was
used to produce more bacteriophage
• Conclusion: DNA is the genetic material!
14
Question 16
Mixing a killed virulent strain of bacteria and a living
strain of benign bacteria together produces virulent
bacteria. What does this demonstrate?
a. Genes are inactivated when a cell dies
b. DNA can only be passed on during
reproduction
c. Cells can pick up genes from the
environment
d. All bacteria are virulent
e. None of the above
Learning Objectives
14.2 DNA Structure
• Understand the contributions of the
following people in elucidating DNA’s
structure
– Chargaff
– Wilkins & Franklin
– Watson & Crick
16
DNA Structure
• DNA is a nucleic acid
• Composed of nucleotides
– 5-carbon sugar called deoxyribose
– Phosphate group (PO43-)
• Attached to 5′ carbon of sugar
– Nitrogenous base
• Adenine, thymine, cytosine, guanine
– Free hydroxyl group (—OH)
• Attached at the 3′ carbon of sugar
17
Nucleotide Subunits of DNA and
RNA
Nitrogenous Base
Nitrogenous base
NH2
1
8
Phosphate group
2
O
N
9
P
O
4
N
N C C
H
N
C
H
N C
C
N C C
N
H
N C
C
NH2
C
H
N
H
3
H
Adenine
CH2
N
Guanine
5′
O–
1′
4′
3′
2′
OH in RNA
OH
Sugar
O
NH2
O
H in DNA
Pyrimidines
–O
5
O
NH2
6
Purines
7N
H
C
H
C
C
N
N
C
O
H
Cytosine
(both DNA and RNA)
H3C
C
H
C
C
N
O
N
H
H
C
C
O
H
C
H
Thymine
(DNA only)
C
N
N
H
C
O
H
Uracil
(RNA only)
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18
• Phosphodiester
bond
– Bond between
adjacent nucleotides
– Formed between the
phosphate group of
one nucleotide and
the 3′ —OH of the
next nucleotide
• The chain of
nucleotides has a 5′to-3′ orientation
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5′
PO4
Base
CH2
O
C
O
Phosphodiester
bond
–O
hydroxyl group
O
P
Phosphate group
O
Base
CH2
O
OH 3′
19
Chargaff’s Rules
• Erwin Chargaff determined that
– Amount of adenine = amount of thymine
– Amount of cytosine = amount of guanine
– Always an equal proportion of purines
(A and G) and pyrimidines (C and T)
20
Representation of Chargaff’s Data
Table (1952)
Organis
m
%A
%G
%C
%T
A/T
G/C
%GC
%AT
φX174
24.0
23.3
21.5
31.2
0.77
1.08
44.8
55.2
Maize
26.8
22.8
23.2
27.2
0.99
0.98
46.1
54.0
Octopus
33.2
17.6
17.6
31.6
1.05
1.00
35.2
64.8
Chicken
28.0
22.0
21.6
28.4
0.99
1.02
43.7
56.4
Rat
28.6
21.4
20.5
28.4
1.01
1.00
42.9
57.0
Human
29.3
20.7
20.0
30.0
0.98
1.04
40.7
59.3
Grassho
pper
29.3
20.5
20.7
29.3
1.00
0.99
41.2
58.6
Sea
Urchin
32.8
17.7
17.3
32.1
1.02
1.02
35.0
64.9
Wheat
27.3
22.7
22.8
27.1
1.01
1.00
45.5
54.4
Yeast
31.3
18.7
17.1
32.9
0.95
1.09
35.8
64.4
E. coli
24.7
26.0
25.7
23.6
1.05
1.01
51.7
48.3
http://en.wikipedia.org/wiki/Chargaff%27s_rules
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Rosalind Franklin
• Performed X-ray diffraction
studies to identify the 3-D
structure
– Discovered that DNA is helical
– Using Maurice Wilkins’ DNA
fibers, discovered that the
molecule has a diameter of
2 nm and makes a complete
turn of the helix every 3.4 nm
a.
b.
Courtesy of Cold Spring Harbor Laboratory Archives
22
James Watson and Francis
Crick – 1953
• Deduced the structure
of DNA using
evidence from
Chargaff, Franklin,
and others
• Did not perform a
single experiment
themselves related to
DNA
• Proposed a double
helix structure
23
Double helix
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5´
Phosphate group
P
• 2 strands are polymers
of nucleotides
• Phosphodiester
backbone – repeating
sugar and phosphate
units joined by
phosphodiester bonds
• Wrap around 1 axis
• Antiparallel
5
O
1
4
3
Phosphodiester bond
2
P
5
O
4
1
3
P
2
5
O
1
4
5-carbon sugar
3
2
Nitrogenous base
P
5
O
1
4
3
2
OH
3
24
Antiparallel Nature of DNA
http://academic.brooklyn.cuny.edu/biology/bio4fv/page/molecular%20biology/dsDNA.jpg
25
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2nm
5′
A
T
G
Minor
groove
3′
3.4nm
C
T
G
A
T
0.34nm
C
G
Major
groove
G
A
G
T
C
Major
groove
Minor
groove
26
3′
5′
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Hydrogen
bond
• Complementarity of
bases
• A forms 2 hydrogen
bonds with T
• G forms 3 hydrogen
bonds with C
• Gives consistent
diameter
O
N
H
N
G
N
H
H
H
N
N
H
N
Sugar
H
C
N
N
H
Sugar
H
Hydrogen
bond
H
N
H
N
N
Sugar
A
N
CH3
O
H
H
N
H
T
N
N
H
Sugar
27
Question 7
Chargaff’s rule states that
a. DNA strands are in antiparallel alignment
b. G matches with C, and T matches with A
c. The DNA molecule is a double helix
d. DNA transformation occurs when an organism
incorporates DNA from the environment
e. The nuclei of cells are totipotent
Question 11
If a strand of DNA had the sequence 5’- AGTCCA3’, which of the following would be the
complementary DNA strand?
a.
b.
c.
d.
e.
3’- CATGGT- 5’
3’- TCAGGT- 5’
3’- TCAAAU- 5’
3’- AGTCCA- 5’
3’- GGTTCA- 5’
Question 12
If a DNA molecule contains 40% thymine, how much
guanine will it contain?
a. 10%
b. 20%
c. 30%
d. 40%
Learning Objectives
14.3 Basic Characteristics of DNA
Replication
• How did Meselson and Stahl figure out
DNA replication?
• What is required for DNA replication?
31
DNA Replication
3 possible models
1. Conservative model
2. Semiconservative model
3. Dispersive model
32
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Conservative
33
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Conservative
Semiconservative
34
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Conservative
Semiconservative
Dispersive
35
Meselson and Stahl – 1958
• Bacterial cells were grown in a heavy
isotope of nitrogen, 15N
• All the DNA incorporated 15N
• Cells were switched to media containing
lighter 14N
• DNA was extracted from the cells at
various time intervals
36
Meselson and Stahl – 1958
• Bacterial cells were grown in a heavy
isotope of nitrogen, 15N
• All the DNA incorporated 15N
• Cells were switched to media containing
lighter 14N
• DNA was extracted from the cells at
various time intervals
37
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DNA
E. coli
15N
medium
14N
medium
0 min
0 rounds
E. coli cells grown
in 15N medium
Cells shifted to
14N medium and
allowed to grow
20 min
1 round
40 min
2 rounds
Samples taken at
three time points
and suspended in
cesium chloride
solution
Samples are centrifuged
0 rounds
1 round
2 rounds
0
1
2
38
Top
Bottom
Rounds of
replication
From M. Meselson and F.W. Stahl/PNAS 44(1958):671
Meselson and Stahl’s Results
• Conservative model = rejected
– 2 densities were not observed after round 1
• Semiconservative model = supported
– Consistent with all observations
– 1 band after round 1
– 2 bands after round 2
• Dispersive model = rejected
– 1st round results consistent
– 2nd round – did not observe 1 band
39
DNA Replication
• Requires 3 things
– Something to copy
• Parental DNA molecule
– Something to do the copying
• Enzymes
– Building blocks to make copy
• Nucleotide triphosphates
40
• DNA replication includes
– Initiation – replication begins
– Elongation – new strands of DNA are
synthesized by DNA polymerase
– Termination – replication is terminated
41
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Template Strand
HO
New Strand
3′
Template Strand
HO
5′
G
5′
P
C
G
O
O
Sugar–
phosphate
backbone
3′
P
C
New Strand
O
O
P
P
P
T
A
P
T
A
O
O
O
O
P
P
P
A
DNA polymerase III
O
T
O
P
A
P
P
P
C
O
G
P
C
O
G
O
O
P
P
3′
P
OH
A
A
O
P
T
P
P
O
P
O
T
A
OH
O
OH
P
P
Pyrophosphate
A
O
5′
P
P
O
3′
P
O
T
O
5′
42
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5
3
5
RNA polymerase makes primer
5
3
3
5
DNA polymerase extends primer
• DNA polymerase
– Matches existing DNA bases with
complementary nucleotides and links them
– All have several common features
• Add new bases to 3′ end of existing strands
• Synthesize in 5′-to-3′ direction
• Requires a primer of RNA
43
Question 5
The Meselson-Stahl experiment demonstrated that DNA
replication is
a.
b.
c.
d.
Conservative
Semi-conservative
Disruptive
Differentiated
Learning Objectives
14.4 Prokaryotic Replication
• How many DNA pol’s does E. coli have
and what are their functions?
• What other enzymes are needed for DNA
replication (in E. coli)
• What occurs at the replication fork?
• How is DNA replication semidiscontinous?
• What are the leading and lagging strands?
45
Prokaryotic Replication
• E. coli model
• Single circular molecule of DNA
• Replication begins at one origin of
replication
• Proceeds in both directions around the
chromosome
• Replicon – DNA controlled by an origin
46
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Replisome
Origin
Termination
Termination
Replisome
Origin
Termination
Origin
Termination
Termination
Origin
Origin
47
• E. coli has 3 DNA polymerases
– DNA polymerase I (pol I)
• Acts on lagging strand to remove RNA
primers and replace them with DNA
– DNA polymerase II (pol II)
• Involved in DNA repair processes
– DNA polymerase III (pol III)
• Main replication enzyme
– All 3 have 3′-to-5′ exonuclease activity –
proofreading
– DNA pol I has 5′-to-3′ exonuclase activity
48
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Supercoiling
Replisomes
No Supercoiling
Replisomes
DNA gyrase
• Unwinding DNA causes torsional strain
– Helicases – use energy from ATP to unwind
DNA
– Single-strand-binding proteins (SSBs) coat
strands to keep them apart
– Topoisomerase prevent supercoiling
• DNA gyrase is used in replication
49
Semidiscontinous
• DNA polymerase can synthesize only in 1
direction!!!!!
• Leading strand synthesized continuously
from an initial primer
• Lagging strand synthesized
discontinuously with multiple priming
events
– Okazaki fragments
50
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5′
3′
First RNA
primer
Lagging strand
(discontinuous)
5′
3′
Open helix and
replicate further
Open helix
and replicate
Second RNA primer
5′
3′
3′
3′
5′
5′
RNA primer
5′
3′
Leading strand
(continuous)
RNA primer
5′
3′
51
• Partial opening of helix forms replication
fork
• DNA primase – RNA polymerase that
makes RNA primer
– RNA will be removed and replaced with
DNA
52
Leading-strand synthesis
– Single priming event
– Strand extended by DNA pol III
• Processivity –  subunit forms “sliding
clamp” to keep it attached
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a.
b.
a-b: From Biochemistry by Stryer. © 1975, 1981, 1988, 1995 by Lupert Stryer. Used with permission of W.H. Freeman and Company
53
Lagging-strand synthesis
– Discontinuous synthesis
• DNA pol III
– RNA primer made by primase for each
Okazaki fragment
– All RNA primers removed and replaced by
DNA
• DNA pol I
– Backbone sealed
• DNA ligase
• Termination occurs at specific site
– DNA gyrase unlinks 2 copies
54
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5′
DNA ligase
Lagging strand
(discontinuous)
RNA primer
DNA polymerase I
Okazaki fragment
made by DNA
polymerase III
Primase
Leading strand
(continuous)
3′
55
Replication fork
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New bases
β clamp (sliding clamp)
Leading strand
3
5
Single-strand binding
proteins (SSB)
Clamp loader
DNA gyrase
Open β clamp
5
3
Parent
DNA
Helicase
Primase
DNA
polymerase III
Lagging strand
Okazaki fragment
New bases
3
5
DNA
polymerase I
DNA ligase
RNA primer
56
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57
Question 3
Where are the Okazaki fragments found?
a.
b.
c.
d.
e.
On the lagging strand
On the leading strand
At the replication origin
In the cytoplasm
On both strands
Question 4
Name the enzyme that links Okazaki fragments
a. DNA polymerase I
b. RNA primase
c. DNA ligase
d. Helicase
e. ATP synthase
Question 6
During DNA replication, what enzyme is responsible for
untwisting the DNA helix?
a. DNA polymerase I
b. RNA primase
c. DNA ligase
d. DNA polymerase III
e. Helicase
Question 9
Why does replication proceed in opposite directions on
the leading and lagging strands?
a. The polymerase enzyme needs a primer
b. DNA polymerase III can only add to the 3´ end of
a strand
c. The Okazaki fragments are only on the leading
strands
d. The parent strands are oriented in the same
direction
e. Helicase only allows for replication of one strand
at a time
Question 15
What would be the immediate consequence of a
non-functional primase enzyme?
a. The strands would break due to the torsional
strain from rapid untwisting
b. The helix could be opened
c. The DNA polymerase III enzyme would have
nothing to bind to
d. The Okazaki fragments would not be linked
together
e. The single DNA strands could not be held
open
Learning Objectives
14.5 Eukaryotic Replication
• What are differences between Prok and
Euk replication?
• What are telomeres and how are they
replicated?
14.6 DNA Repair
• What are the three forms of DNA repair?
• Why is DNA repair important for the cell?
63
Eukaryotic Replication
• Complicated by
– Larger amount of DNA in multiple
chromosomes
– Linear structure
• Basic enzymology is similar
– Requires new enzymatic activity for
dealing with ends only
64
Telomeres
• Specialized structures found on the ends
of eukaryotic chromosomes
• Protect ends of chromosomes from
nucleases and maintain the integrity of
linear chromosomes
• Gradual shortening of chromosomes
with each round of cell division
65
http://www.scientificamerican.com/media/inline/telomeres-telomerase-and_1.jpg
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Replication first round
5´
3´
3´
5´
Leading strand (no problem)
Lagging strand (problem at the end)
5´
3´
3´
5´
Last primer
Origin
Leading
strand
Primer removal
3´
5´
5´
Lagging
strand
3´
Replication second round
Removed primer
cannot be replaced
5´
3´
3´
5´
5´
3´
3´
5´
Shortened template
66
• Telomeres composed of short repeated
sequences of DNA
• Telomerase – enzyme makes telomere section
of lagging strand using an internal RNA template
(not the DNA itself)
– Leading strand can be replicated to the end
• Telomerase developmentally regulated
– Relationship between senescence and telomere
length
• Cancer cells generally show activation of
telomerase
67
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5́
T
T
G
3́
Synthesis by telomerase
Telomerase
5́
3́
T
A
T
A
G
C
G
G
C
Telomere
extended
by telomerase
G
C
C
A
A
C
Template RNA is
part of enzyme
Telomerase moves and
continues to extend telomere
5́
3́
T
T
G
G
G
Now ready
to synthesize
next repeat
G
T
T
G
A
A
C
C
C
C
A
A
C
68
DNA Repair
• Errors due to replication
– DNA polymerases have proofreading ability
• Mutagens – any agent that increases the
number of mutations above background
level
– Radiation and chemicals
• Importance of DNA repair is indicated by
the multiplicity of repair systems that have
been discovered
69
DNA Repair
Falls into 2 general categories
1. Specific repair
– Targets a single kind of lesion in DNA and
repairs only that damage
2. Nonspecific
– Use a single mechanism to repair multiple
kinds of lesions in DNA
70
Photorepair
• Specific repair mechanism
• For one particular form of damage caused
by UV light
• Thymine dimers
– Covalent link of adjacent thymine bases in
DNA
• Photolyase
– Absorbs light in visible range
– Uses this energy to cleave thymine dimer
71
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DNA with adjacent thymines
T
A
T
A
UV light
Helix distorted by
thymine dimer
Thymine dimer
A
A
Photolyase binds
to damaged DNA
Photolyase
A
A
Visible light
Thymine dimer
cleaved
T
A
T
A
72
Excision repair
• Nonspecific repair
• Damaged region is removed and replaced
by DNA synthesis
• 3 steps
1. Recognition of damage
2. Removal of the damaged region
3. Resynthesis using the information on the
undamaged strand as a template
73
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Damaged or incorrect base
Excision repair enzymes recognize damaged DNA
Uvr A,B,C complex
binds damaged DNA
Excision of damaged strand
Resynthesis by DNA polymerase
DNA polymerase
74
Question 13
The enzyme telomerase attaches the last few bases
on the lagging strand. As cells age, telomerase
activity drops. What would happen to the
chromosomes in the absence of telomerase activity?
a. Chromosome replication would be terminated
b. Okazaki fragments would not be linked
together
c. Chromosomes would shorten during each
division
d. The leading strand would become the lagging
strand
e. The cells would become cancerous
Question 10
Mutations can be caused by copying mistakes, and by
exposure to chemicals or electromagnetic radiation.
a. This is True
b. This is False