1. Biochemical Identification of the Genetic Material
Nucleic Acid Structure
An Overview of DNA Replication
Molecular Mechanism of DNA Replication
Molecular Structure of Eukaryotic Chromosomes
1
Key Concepts:
Nucleic Acid Structure,
DNA Replication,
and Chromosome Structure
Chap: 11
2. What is the genetic material?
Four criteria necessary for genetic material:
1. Information
2. Replication
3. Transmission
4. Variation
Late 1800s – biochemical basis of heredity postulated
Researchers became convinced that chromosomes carry
the genetic information
1920s to 1940s – scientists expected the protein portion of
chromosomes would turn out to be the genetic material
2
Biochemical Identification
of the Genetic Material
4. Smooth strains (S) with
capsule are fatal; rough
strains (R) without capsule
are not
If mice are injected with
heat-killed type S, they
survive (because bacteria
are dead)
However, mixing live R
with heat-killed S kills the
mouse
Blood is found to contain
living type S bacteria
Known as
transformation
4
5. How is this possible?
Genetic material had been transferred
from the heat-killed type S bacteria to the
living type R bacteria
This gave them the capsule-secreting trait
and was passed on to their offspring
What was the biochemical basis of this
transforming principle? At the time there was
no way to know
5
6. Levels of DNA Structure:
1. Nucleotides – the building blocks of DNA and RNA
2. Strand – a linear polymer strand of DNA or RNA
3. Double helix – the two strands of DNA
4. Chromosomes – DNA associated with an array of
different proteins into a complex structure
5. Genome – the complete complement of genetic
material in an organism
6
Nucleic Acid Structure
14. Solving the structure of DNA
1953, James Watson and
Francis Crick, proposed the
structure of the DNA double
helix
Watson and Crick used Linus
Pauling’s method
of working out protein
structures using simple
ball-and-stick models
Rosalind Franklin’s
X-ray diffraction results
were crucial evidence,
suggesting a helical structure
with uniform diameter
14
15. Erwin Chargoff analyzed base composition
of DNA from many different species
Results consistently showed
amount of adenine (A) = amount of thymine (T)
amount of cytosine (C) = amount of guanine (G)
15
Base-pairing
17. Watson and Crick
Put together these pieces of information
Found ball-and-stick model consistent with
data
Double-stranded helix
Base-pairing: A with T and G with C
James Watson, Francis Crick, and Maurice
Wilkins awarded Nobel Prize in 1962
Rosalind Franklin had died and the Nobel Prize
is not awarded posthumously
17
19. Chargoff’s rule
A pairs with T
G pairs with C
Keeps width consistent
Complementary DNA strands
5’ – GCGGATTT – 3’
3’ – CGCCTAAA – 5’
Antiparallel strands
One strand 5’ to 3’
Other stand 3’ to 5’
19
21. Late 1950s – three different
models were proposed for DNA
replication
Semiconservative Model
Conservative Model
Dispersive Model
Newly-made strands are
“daughter strands”
Original strands are “parental
strands”
21
An Overview of DNA Replication
23. 23
Conservative Mechanism
Second round
of replication
First round
of replication
Original
double helix
(b) Conservative mechanism. DNA replication produces 1 double
helix with both parental strands and the other with 2 new
daughter strands.
25. In 1958, Matthew Meselson and Franklin Stahl devised an
experiment to differentiate among the three proposed DNA
replication mechanisms
Nitrogen comes in a common light form (14N) and a rare
heavy form (15N)
Grew E. coli in medium with 15N to label, then switched to
medium with 14N, collecting samples after each generation
Original parental strands would be 15N while newly made
strands would be 14N
Conclusion: Semiconservative DNA replication
25
Meselson and Stahl experiment
26. Semiconservative replication
The two parental
strands separate and
serve as template
strands
New nucleotides must
obey the AT/GC rule
End result: two new
double helices with
same base sequence as
original
27. Question 1
Because one original strand of the double stranded
DNA helix is found in each daughter cell (after cell
division), the DNA replication process is—
a. Conservative
b. Derivative
c. Dispersive
d. Paralleled
e. Semiconservative
28. Question 1
Because one original strand of the double stranded
DNA helix is found in each daughter cell (after cell
division), the DNA replication process is—
a. Conservative
b. Derivative
c. Dispersive
d. Paralleled
e. Semiconservative
29. Question 2
All are part of a nucleotide except—
a. Pentose sugar
b. Nitrogenous base
c. Phosphate group
d. Fatty acid tail
e. All of the above are parts of a nucleotide
30. Question 2
All are part of a nucleotide except—
a. Pentose sugar
b. Nitrogenous base
c. Phosphate group
d. Fatty acid tail
e. All of the above are parts of a nucleotide
31. Question 3
The pyrimidine bases in DNA are—
a. Cytosine, thymine and uracil
b. Adenine and guanine
c. Cytosine and thymine
d. Thymine, guanine and cytosine
e. Adenine, uracil and guanine
32. Question 3
The pyrimidine bases in DNA are—
a. Cytosine, thymine and uracil
b. Adenine and guanine
c. Cytosine and thymine
d. Thymine, guanine and cytosine
e. Adenine, uracil and guanine
33. Origin of replication provides an opening
called a replication bubble that forms two
replication forks
DNA replication proceeds outward from forks
Bacteria have single origin of replication
Eukaryotes have multiple origins of replication
33
Molecular Mechanism
of DNA Replication
36. DNA helicase
Binds to DNA and travels
5’ to 3’ using ATP to
separate strand and
move fork forward
DNA topoisomerase
Relives additional coiling
ahead of replication fork
Single-strand binding
proteins
Keep parental strands
open to act as templates
36
https://www.youtube.com/watch?v=EYGrElVyHnU
37. DNA polymerase
Covalently links nucleotides
Deoxynucleoside triphosphates
37
(a) Action of DNA polymerase
5′
3′
Incoming
deoxynucleoside
triphosphates
DNA polymerase
catalytic site
38. Deoxynucleoside triphosphates
Free nucleotides with three phosphate groups
Breaking covalent bond to release pyrophosphate (two phosphates) provides
energy to connect nucleotides
H
H
H
H
H
O
O
O
O–
P
N
O
H
O
CH3
O–
O
N
H
N
N
N
H
N
H H
H
H
H
O
O
O
O–
P
CH2
O
O
H
N
N
N
H
N
H H
H
H
H
O
O
O
O–
P
CH2
O–
N
H
H
H
H N
N
H
H
H
H
N
N
H H
H
H
H
O
O
O
O–
P
CH2
+
O
H
H
N
N
H
H
H
H
H
O
O
O
P
O–
N
O
OH
O
H
H
H
H
H
O
O
O
O–
P
N
O H
H
O
CH3
O
H
N
N
N
H
N
H H
H
H
H
O
O
O
O–
P
O
O
H
N
N
N
H
N
H H
H
H
H
O
O
O
O–
P
CH2
O–
H N
N
H
N
N
H H
H
H
H
O
O
O
O–
P
CH2
O
H
H
N
N
H
H
H
H
H
O
O
O
P CH2
O–
O
O
H
H
N
N
H
H
H
H
H
O
O
O
P CH2
O–
O
O
H
T
A
C
C
G
G
T
C
G
C
G
A
T
C
G
C
G
A
N
H
H
N
H
H
N
H
H
H
H
N
N
H
H
H
H
N
O O–
–O
O O
P P
–O O–
5′ end
CH2
CH2
3′ end
HO
5′ end
O–
CH2
Template
strand
3′ end 3′ end
5′ end
Phosphate
3′ end Pyrophosphate
New phosphoester
bond
5′ end
CH2
OH +
An incoming nucleotide
(a deoxynucleoside triphosphate)
(b) Chemistry of DNA replication
HO
39. Features of DNA polymerase
1. DNA polymerase cannot
begin synthesis on a bare
template strand
Requires a primer to
get started
DNA primase makes
the primer from RNA
The RNA primer is
removed and replaced
with DNA later
2. DNA polymerase only
works 5’ to 3’
39
40. Leading strand
DNA synthesized in as one long
molecule
DNA primase makes a single RNA
primer
DNA polymerase adds nucleotides in
a 5’ to 3’ direction as it slides
forward
Lagging strand
DNA synthesized 5’ to 3’ but as
Okazaki fragments
Okazaki fragments consist of RNA
primers plus DNA
In both strands
RNA primers are removed by DNA
polymerase and replaced with DNA
DNA ligase joins adjacent DNA
fragments
43. DNA replication is very accurate
Three mechanisms for accuracy
1. Hydrogen bonding between A and T,
and between G and C is more stable than
mismatched combinations
2. Active site of DNA polymerase is unlikely to form
bonds if pairs mismatched
3. DNA polymerase can proofread to remove
mismatched pairs
DNA polymerase backs up and digests linkages
Other DNA repair enzymes are involved as well
43
44. DNA Polymerases Are a Family of
Enzymes With Specialized Functions
Important issues for DNA polymerase are speed,
fidelity, and completeness
Nearly all living species have more than one type of
DNA polymerase
Genomes of most species have several DNA polymerase
genes due to gene duplication
Independent genetic changes produce enzymes with
specialized functions
45. E. coli has 5 DNA polymerases
DNA polymerase III – multiple subunits,
responsible for majority of replication
DNA polymerase I – a single subunit, rapidly
removes RNA primers and fills in DNA
DNA polymerases II, IV and V – DNA repair and
can replicate damaged DNA
DNA polymerases I and III stall at DNA damage
DNA polymerases II, IV and V don’t stall but go slower
and make sure replication is complete
46. Humans have 12 or more DNA polymerases
Designated with Greek letters
DNA polymerase α – has its own built in primase subunit
DNA polymerase δ and ε – extend DNA at a faster rate
DNA polymerase γ – replicates mitochondrial DNA
When DNA polymerases α, δ or ε encounter
abnormalities they may be unable to replicate
Lesion-replicating polymerases may be able to synthesize
complementary strands to the damaged area
47.
48. DNA polymerase cannot copy the tip of the strand with a 3’ end
No place for upstream primer to be made
If this replication problem were not solved, linear chromosomes would become
progressively shorter
Telomerase enzyme attaches many copies of DNA repeat sequence to the ends
of chromosomes
Shortening of telomeres is correlated with cellular senescence
Telomerase function is reduced as an organism ages
99% of all types of human cancers have high levels of telomerase 48
3’ Overhang and Telomerase
49. Telomeres
Series of short nucleotide sequences repeated at the ends of
chromosomes in eukaryotes
Specialized form of DNA replication only in eukaryotes in the telomeres
Telomere at 3’ does not have a complementary strand and is called a 3’
overhang
50.
51. Typical eukaryotic chromosome may be
hundreds of millions of base pairs long
Length would be 1 meter
But must fit in cell 10-100µm
Chromosome
Discrete unit of genetic material
Chromosomes composed of chromatin
DNA-protein complex
51
Molecular Structure of
Eukaryotic Chromosomes
52. Three levels of DNA compaction
1. DNA wrapping
DNA wrapped around histones to form nucleosome
Shortens length of DNA molecule 7-fold
2. 30-nm fiber
Current model suggests asymmetric, 3D zigzag of
nucleosomes
Shortens length another 7-fold
52
55. Cell division
When cells prepare to divide, chromosomes
become even more compacted
Euchromatin not as compact
Hetrochromatin much more compact
Metaphase chromosomes highly compacted
55
59. Question 1
Because one original strand of the double stranded
DNA helix is found in each daughter cell (after cell
division), the DNA replication process is—
a. Conservative
b. Derivative
c. Dispersive
d. Paralleled
e. Semiconservative
60. Question 1
Because one original strand of the double stranded
DNA helix is found in each daughter cell (after cell
division), the DNA replication process is—
a. Conservative
b. Derivative
c. Dispersive
d. Paralleled
e. Semiconservative
61. Question 2
All are part of a nucleotide except—
a. Pentose sugar
b. Nitrogenous base
c. Phosphate group
d. Fatty acid tail
e. All of the above are parts of a nucleotide
62. Question 2
All are part of a nucleotide except—
a. Pentose sugar
b. Nitrogenous base
c. Phosphate group
d. Fatty acid tail
e. All of the above are parts of a nucleotide
63. Question 3
The pyrimidine bases in DNA are—
a. Cytosine, thymine and uracil
b. Adenine and guanine
c. Cytosine and thymine
d. Thymine, guanine and cytosine
e. Adenine, uracil and guanine
64. Question 3
The pyrimidine bases in DNA are—
a. Cytosine, thymine and uracil
b. Adenine and guanine
c. Cytosine and thymine
d. Thymine, guanine and cytosine
e. Adenine, uracil and guanine
65. Question 4
The double helix structure of DNA was first described by
—
a. Chargoff
b. Watson, Crick and Wilkins
c. Griffith
d. Avery and Mcleod
e. Hershey and Chase
66. Question 4
The double helix structure of DNA was first described by
—
a. Chargoff
b. Watson, Crick and Wilkins
c. Griffith
d. Avery and Mcleod
e. Hershey and Chase
67. Question 5
The enzyme responsible for initiating the unwinding of
double-stranded DNA (alleviates coiling) by nicking a
single strand of the DNA molecule is—
a. Ligase
b. Helicase
c. Topoisomerase
d. Gyrase
e. DNA polymerase I
68. Question 5
The enzyme responsible for initiating the unwinding of
double-stranded DNA (alleviates coiling) by nicking a
single strand of the DNA molecule is—
a. Ligase
b. Helicase
c. Topoisomerase
d. Gyrase
e. DNA polymerase I
69. Question 6
The enzyme that proceeds along one of the strands of a
DNA molecule adding nucleotides to the other strand is
called—
a. DNA polymerase II
b. Primase
c. DNA polymerase I
d. DNA polymerase III
70. Question 6
The enzyme that proceeds along one of the strands of a
DNA molecule adding nucleotides to the other strand is
called—
a. DNA polymerase II
b. Primase
c. DNA polymerase I
d. DNA polymerase III
71. Question 7
The enzyme that creates a short RNA segment at the
initiation sites where replication is to be carried out is
called—
a. Primase
b. DNA ligase
c. DNA gyrase
d. Exonuclease
72. Question 7
The enzyme that creates a short RNA segment at the
initiation sites where replication is to be carried out is
called—
a. Primase
b. DNA ligase
c. DNA gyrase
d. Exonuclease
73. Question 8
The enzyme that stitches Okazaki fragments together on
the lagging strand is—
a. DNA polymerase II
b. DNA polymerase III
c. Topoisomerase
d. DNA ligase
e. DNA helicase
74. Question 8
The enzyme that stitches Okazaki fragments together on
the lagging strand is—
a. DNA polymerase II
b. DNA polymerase III
c. Topoisomerase
d. DNA ligase
e. DNA helicase
75. Question 9
Because DNA polymerase III can only act from 5' to
3', continuous strand growth can be achieved only
along the leading strand and strand growth along
the other strand must occur discontinuously
resulting in the production of a series of short
sections of new DNA called —
a. Replicon fragments
b. Okazaki fragments
c. Klenow fragments
d. Chargoff’s fragments
e. None of the above
76. Question 9
Because DNA polymerase III can only act from 5' to
3', continuous strand growth can be achieved only
along the leading strand and strand growth along
the other strand must occur discontinuously
resulting in the production of a series of short
sections of new DNA called —
a. Replicon fragments
b. Okazaki fragments
c. Klenow fragments
d. Chargoff’s fragments
e. None of the above
77. Question 10
A repeating DNA sequence at the end of chromosomes
that prevents them from losing base pair sequences at
their ends and from fusing together is—
a. A telomere
b. A telomerase
c. A replicon
d. A primer
e. A promoter
78. Question 10
A repeating DNA sequence at the end of chromosomes
that prevents them from losing base pair sequences at
their ends and from fusing together is—
a. A telomere
b. A telomerase
c. A replicon
d. A primer
e. A promoter
79. Question 11
In the DNA double helix, complementary base
pairs are held together by—
a. Phosphodiester bonds
b. Peptide bonds
c. Hydrogen bonds
d. N-glycodisic bonds
e. Ionic bonds
80. Question 11
In the DNA double helix, complementary base
pairs are held together by—
a. Phosphodiester bonds
b. Peptide bonds
c. Hydrogen bonds
d. N-glycodisic bonds
e. Ionic bonds
81. Question 12
A major difference between DNA replication in
prokaryotes and eukaryotes is—
a. There is only one replication origin in prokaryotes
b. Replication is conservative in prokaryotes
c. DNA amylase performs the function of DNA helicase in
prokaryotes
d. Prokaryotes do not use topoisomerase in the replication
process
82. Question 12
A major difference between DNA replication in
prokaryotes and eukaryotes is—
a. There is only one replication origin in prokaryotes
b. Replication is conservative in prokaryotes
c. DNA amylase performs the function of DNA helicase in
prokaryotes
d. Prokaryotes do not use topoisomerase in the replication
process
83. Question 13
A cell in interphase would most likely have
heterochromatin present.
a. This is true
b. This is false
84. Question 13
A cell in interphase would most likely have
heterochromatin present.
a. This is true
b. This is false
85. Question 14
What does transformation involve in bacteria?
a. The infection of cells by a phage DNA molecule
b. The creation of a strand of RNA from DNA
c. The replication of DNA
d. Assimilation of external DNA into a cell
e. None of the above
86. Question 14
What does transformation involve in bacteria?
a. The infection of cells by a phage DNA molecule
b. The creation of a strand of RNA from DNA
c. The replication of DNA
d. Assimilation of external DNA into a cell
e. None of the above
89. Question 16
Genetic material must–
a. Relay information
b. Be able to be replicated
c. Be able to be passed on
d. Account for variation
e. All of the above
90. Question 16
Genetic material must–
a. Relay information
b. Be able to be replicated
c. Be able to be passed on
d. Account for variation
e. All of the above
91. Question 17
If a DNA strand with the base sequence TTGCAGG
were to be unzipped and be replicated, what would
be the base sequence of TTGCAGG old partner’s
NEW complementary strand–
a. AACGTCC
b. AATGCAA
c. TTGCAGG
d. GGACGTT
e. AACCACC
92. Question 17
If a DNA strand with the base sequence TTGCAGG
were to be unzipped and be replicated, what would
be the base sequence of TTGCAGG old partner’s
NEW complementary strand–
a. AACGTCC
b. AATGCAA
c. TTGCAGG
d. GGACGTT
e. AACCACC
93. Question 18
What might be the consequence of a deficiency of
DNA ligase–
a. Lack of a replication fork.
b. Lack of a replication bubble.
c. Failure of nucleotide addition to complementary bases
at the 3’ end of a molecule.
d. Failure of nucleotide addition to complementary bases
at the 5’ end of a molecule.
e. Failure of preformed fragments to link into a single
strand
94. Question 18
What might be the consequence of a deficiency of
DNA ligase–
a. Lack of a replication fork.
b. Lack of a replication bubble.
c. Failure of nucleotide addition to complementary bases
at the 3’ end of a molecule.
d. Failure of nucleotide addition to complementary bases
at the 5’ end of a molecule.
e. Failure of preformed fragments to link into a
single strand