1. DNA: The Genetic Material
Chapter 14 1

2. 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

3. Frederick Griffith – 1928
Studied Streptococcus pneumoniae, a pathogenic bacterium causing pneumonia
2 strains of Streptococcus
S strain is virulent
R strain is nonvirulent

4. Griffith’s Experiment
Griffith infected mice with these strains hoping to understand the difference between the strains

5. Griffith’s Experiment
Live Nonvirulent
Strain of
S. pneumoniae
Live Virulent
Strain of S. pneumoniae
Polysaccharide
coat
Mice die
Mice live a. b.
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6. 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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7. 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

8. 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

9. Avery, MacLeod, & McCarty – 1944
Repeated Griffith’s experiment using purified cell extracts

10. 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

11. 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

12. Hershey & Chase –1952
Bacteriophage DNA was labeled with radioactive phosphorus (32P)
Bacteriophage protein was labeled with radioactive sulfur (35S)
Radioactive molecules were tracked

13. Hershey & Chase Experiment+
Phage grown in radioactive 35S,
which is incorporated into phage coat
Virus infect
bacteria
Blender separates
phage coat from bacteria
Centrifuge forms
bacterial pellet
35S in supernatant
35S-Labeled Bacteriophages
Phage grown in radioactive 32P.
which is incorporated into phage DNA
32P in bacteria pellet
32P-Labeled Bacteriophages
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14. 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!

15. Question 16
Mixing a killed virulent strain of bacteria and a living strain of benign bacteria together produces virulent bacteria. What does this demonstrate?
Genes are inactivated when a cell dies
DNA can only be passed on during reproduction
Cells can pick up genes from the environment
All bacteria are virulent
None of the above
Bloom’s level: Application,

16. Learning Objectives 14.2 DNA Structure
Understand the contributions of the following people in elucidating DNA’s structure
Chargaff
Wilkins & Franklin
Watson & Crick

17. 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

18. Nucleotide Subunits of DNA and RNA
Nitrogenous Base
Nitrogenous base
NH2
NH2 O 7 N 6 5 N C N C C N C N H 1 H C H C
Phosphate group 8
Purines C N C H C N C
NH2 2 N N O N 4 N H H 9 3
Adenine
Guanine –O P O
CH2 5′ O– O
NH2 O O 1′ C C C 4′ H C N
H3C C N H H C N H H C C O H C C O H C C O 3′ 2′
OH in RNA
Pyrimidines N N N OH H H H
Sugar
H in DNA
Cytosine
(both DNA and RNA)
Thymine
(DNA only)
Uracil
(RNA only)
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19. The chain of nucleotides has a 5′-to-3′ orientation
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
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 5′
PO4
Base
CH2 O C O
hydroxyl group
Phosphodiester
bond –O P O
Phosphate group O
Base
CH2 O OH 3′

20. 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)

21. Representation of Chargaff’s Data Table (1952)
Organism %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
31.6
1.05
1.00
35.2
64.8
Chicken
28.0
22.0
21.6
28.4
1.02
43.7
56.4
Rat
28.6
21.4
20.5
1.01
42.9
57.0
Human
29.3
20.7
20.0
30.0
1.04
40.7
59.3
Grasshopper
41.2
58.6
Sea Urchin
32.8
17.7
17.3
32.1
35.0
64.9
Wheat
27.3
22.7
27.1
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
51.7
48.3

22. Courtesy of Cold Spring Harbor Laboratory Archives
Rosalind Franklin
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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

23. 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

24. Double helix 2 strands are polymers of nucleotides
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5´
Phosphate group
2 strands are polymers of nucleotides
Phosphodiester backbone – repeating sugar and phosphate units joined by phosphodiester bonds
Wrap around 1 axis
Antiparallel P
5  O
4 
1 
Phosphodiester bond
3 
2  P
5  O
4 
1 
3 
2  P
5  O
4 
1 
5-carbon sugar
3 
2 
Nitrogenous base P
5  O
4 
1 
3 
2  OH
3 

25. Antiparallel Nature of DNA

26. 2nm 5′ 3′ T A G C 3.4nm Minor groove T G A T 0.34nm G C Major groove G
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2nm 5′ 3′ T A G C
3.4nm
Minor
groove T G A T
0.34nm G C
Major
groove G A T G C
Major
groove
Minor
groove 3′ 5′

27. Complementarity of bases A forms 2 hydrogen bonds with T
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Hydrogen
bond H
Complementarity of bases
A forms 2 hydrogen bonds with T
G forms 3 hydrogen bonds with C
Gives consistent diameter H N O H N H N G N H N C H
Sugar N N N H
Sugar H
Hydrogen
bond H H N N H O
CH3 N A N H N T H
Sugar N N H
Sugar

28. Question 7 Chargaff’s rule states that
DNA strands are in antiparallel alignment
G matches with C, and T matches with A
The DNA molecule is a double helix
DNA transformation occurs when an organism incorporates DNA from the environment
The nuclei of cells are totipotent
Bloom’s level: Knowledge/Understanding,

29. Question 11 3’- CATGGT- 5’ 3’- TCAGGT- 5’ 3’- TCAAAU- 5’
If a strand of DNA had the sequence 5’- AGTCCA- 3’, which of the following would be the complementary DNA strand?
3’- CATGGT- 5’
3’- TCAGGT- 5’
3’- TCAAAU- 5’
3’- AGTCCA- 5’
3’- GGTTCA- 5’
Bloom’s level: Application,

30. Question 12
If a DNA molecule contains 40% thymine, how much guanine will it contain?
10%
20%
30%
40%
Bloom’s level: Application

31. Learning Objectives 14.3 Basic Characteristics of DNA Replication
How did Meselson and Stahl figure out DNA replication?
What is required for DNA replication?

32. DNA Replication 3 possible models Conservative model
Semiconservative model
Dispersive model

33. Copyright © The McGraw-Hill Companies, Inc
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Conservative

34. Conservative Semiconservative 34
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Conservative
Semiconservative

35. Conservative Semiconservative Dispersive 35
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Conservative
Semiconservative
Dispersive

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. 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

38. From M. Meselson and F.W. Stahl/PNAS 44(1958):671
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DNA
E. coli
E. coli cells grown
in 15N medium
15N medium
Cells shifted to
14N medium and
allowed to grow
14N medium
Samples taken at
three time points
and suspended in
cesium chloride
solution
0 min
0 rounds
20 min
1 round
40 min
2 rounds
Samples are centrifuged
0 rounds
1 round
2 rounds 1 2
Top
Bottom
Rounds of
replication
From M. Meselson and F.W. Stahl/PNAS 44(1958):671

39. 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

40. DNA Replication Requires 3 things Something to copy
Parental DNA molecule
Something to do the copying
Enzymes
Building blocks to make copy
Nucleotide triphosphates

41. DNA replication includes
Initiation – replication begins
Elongation – new strands of DNA are synthesized by DNA polymerase
Termination – replication is terminated

42. Sugar– phosphate backbone
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Template Strand
New Strand
Template Strand
New Strand HO 3′ 5′ HO 3′ 5′ C P C P G O G O O
Sugar–
phosphate
backbone O P P T P T P A O A O O O P P P P A
DNA polymerase III A T O T O O O P P P P C C G O G O O O P P 3′ OH P P P A A O
Pyrophosphate O O T T P P P P P O 3′ OH A A O O OH P P 5′ 5′

43. All have several common features
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5 
5 
3 
3 
5 
3 
5 
RNA polymerase makes primer
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

44. Question 5 Conservative Semi-conservative Disruptive Differentiated
The Meselson-Stahl experiment demonstrated that DNA replication is
Conservative
Semi-conservative
Disruptive
Differentiated
Bloom’s level: Knowledge/Understanding

45. 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?

46. 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

47. Replisome Termination Origin Replisome Origin Termination Termination
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Replisome
Origin
Termination
Termination
Replisome
Origin
Termination
Termination
Termination
Origin
Origin
Origin

48. 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

49. Unwinding DNA causes torsional strain
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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

50. 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

51. 5′ 3′ First RNA primer Lagging strand (discontinuous) 5′ 3′
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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′

52. Partial opening of helix forms replication fork
DNA primase – RNA polymerase that makes RNA primer
RNA will be removed and replaced with DNA

53. Leading-strand synthesis Single priming event
Strand extended by DNA pol III
Processivity –  subunit forms “sliding clamp” to keep it attached
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. a. b.
a-b: From Biochemistry by Stryer. © 1975, 1981, 1988, 1995 by Lupert Stryer. Used with permission of W.H. Freeman and Company

54. 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

55. Leading strand (continuous)
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DNA ligase
Lagging strand
(discontinuous)
RNA primer
DNA polymerase I
Okazaki fragment
made by DNA
polymerase III
Primase
Leading strand
(continuous) 3′

56. Replication fork New bases β clamp (sliding clamp) Leading strand 3 
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New bases
β clamp (sliding clamp)
Leading strand
3 
Single-strand binding
5 
proteins (SSB)
Clamp loader
DNA gyrase
Open β clamp
5 
3 
Parent
DNA
Helicase
Lagging strand
DNA
Primase
Okazaki fragment
polymerase III
New bases
3 
DNA
5 
polymerase I
DNA ligase
RNA primer

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58. Question 3 On the lagging strand On the leading strand
Where are the Okazaki fragments found?
On the lagging strand
On the leading strand
At the replication origin
In the cytoplasm
On both strands
Bloom’s level: Knowledge/Understanding

59. Question 4 DNA polymerase I RNA primase DNA ligase Helicase
Name the enzyme that links Okazaki fragments
DNA polymerase I
RNA primase
DNA ligase
Helicase
ATP synthase
Bloom’s level: Knowledge/Understanding

60. Question 6 DNA polymerase I RNA primase DNA ligase DNA polymerase III
During DNA replication, what enzyme is responsible for untwisting the DNA helix?
DNA polymerase I
RNA primase
DNA ligase
DNA polymerase III
Helicase
Bloom’s level: Knowledge/Understanding

61. Question 9
Why does replication proceed in opposite directions on the leading and lagging strands?
The polymerase enzyme needs a primer
DNA polymerase III can only add to the 3´ end of a strand
The Okazaki fragments are only on the leading strands
The parent strands are oriented in the same direction
Helicase only allows for replication of one strand at a time
Bloom’s level: Knowledge/Understanding

62. Question 15
What would be the immediate consequence of a non-functional primase enzyme?
The strands would break due to the torsional strain from rapid untwisting
The helix could be opened
The DNA polymerase III enzyme would have nothing to bind to
The Okazaki fragments would not be linked together
The single DNA strands could not be held open
Bloom’s level: Application

63. 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?

64. 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

65. 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

66. Leading strand Lagging strand Removed primer cannot be replaced
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Replication first round 5´ 3´ 3´ 5´
Leading strand (no problem)
Lagging strand (problem at the end) 3´ 5´ 3´ 5´
Last primer
Origin
Primer removal
Leading
strand 3´ 5´ 5´
Lagging
strand 3´
Removed primer
cannot be replaced
Replication second round 5´ 3´ 3´ 5´ 5´ 3´ 3´ 5´
Shortened template

67. 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

68. continues to extend telomere
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5 ́ T T G
3 ́
Synthesis by telomerase
Telomerase G
Telomere
extended
by telomerase T
5 ́ G G T T T G G A A C C C C A A C
3 ́
Template RNA is
part of enzyme
Telomerase moves and
continues to extend telomere T T
5 ́ G T T G G G G T T G A A C C C C A A C
3 ́
Now ready
to synthesize
next repeat

69. 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

70. DNA Repair Falls into 2 general categories Specific repair Nonspecific
Targets a single kind of lesion in DNA and repairs only that damage
Nonspecific
Use a single mechanism to repair multiple kinds of lesions in DNA

71. 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

72. Photolyase binds to damaged DNA
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DNA with adjacent thymines T T A A
UV light
Helix distorted by
thymine dimer
Thymine dimer T T A A
Photolyase binds
to damaged DNA
Photolyase T T A A
Visible light
Thymine dimer
cleaved T T A A

73. Excision repair Nonspecific repair
Damaged region is removed and replaced by DNA synthesis
3 steps
Recognition of damage
Removal of the damaged region
Resynthesis using the information on the undamaged strand as a template

74. Damaged or incorrect base
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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

75. 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?
Chromosome replication would be terminated
Okazaki fragments would not be linked together
Chromosomes would shorten during each division
The leading strand would become the lagging strand
The cells would become cancerous
Bloom’s level: Application

76. Question 10 This is True This is False
Mutations can be caused by copying mistakes, and by exposure to chemicals or electromagnetic radiation.
This is True
This is False
Bloom’s level: Knowledge/Understanding