ch11_lecture_ppt REVISED.ppt

 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

 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

Griffith’s bacterial transformation
 Late 1920s – Frederick Griffith was working with
Streptococcus pneumoniae bacteria
 Two strains of S. pneumoniae:
 Strains that secrete capsules look smooth (S)
and infections are fatal in mice
 Strains that do not secrete capsules look rough (R) and
infections are not fatal in mice
 The capsule shields the bacteria from the immune
system, so they survive in the blood
3
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
1 Type S cells
are virulent.
Control:
Injected living
type R bacteria
into mouse.
2 Type R cells
are benign.
Control:
Injected living
type S bacteria
into mouse.

 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

 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

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

7
Single strand
Nucleotides
Double helix
DNA associates with
proteins to form a
chromosome.

DNA
 Formed from nucleotides (A, G, C, T)
 Nucleotides composed of
three components
 Phosphate group
 Pentose sugar
 Deoxyribose
 DNA = Deoxyribonucleic Acid
 Nitrogenous base
 Purines – Adenine (A), Guanine (G)
 Pyrimidines – Cytosine (C), Thymine (T)
8
Base
Phosphate
Deoxyribose

9
DNA nucleotides
(a) DNA nucleotide
Phosphate
Deoxyribose
Base
Thymine (T)
Adenine (A)
Cytosine (C)
Guanine (G)
NH2
O
H
H
N
N
NH2
N
H H
H H
N
H
N
N
O
NH2
H
N
N
N
H
N
H
H
OH
H
H
O
O
O–
CH2
O–
P
O
H
H
O
N
O
N
H
Purines
(double ring)
Pyrimidines
(single ring)
CH3

RNA
 Formed from nucleotides (A, G, C, U)
 Nucleotides composed of
three components
 Phosphate group
 Pentose sugar
 Ribose
 RNA = Ribonucleic Acid
 Nitrogenous base
 Purines – Adenine (A), Guanine (G)
 Pyrimidines – Cytosine (C), Uracil (U)
10
Base
Phosphate
Ribose
H
OH

11
RNA nucleotides
(b) RNA nucleotide
Ribose
Uracil (U)
Adenine (A)
Guanine (G)
H
Cytosine (C)
NH2
O
H
H
N
N
H
H H
H
H
H
NH2
N
N
H
N
N
O
NH2
H
N
N
N
H
N
Phosphate
Base
H
OH
OH
H
H
O
O
O–
CH2
O–
P
O
H

 Sugar carbons are 1’ to 5’
 Base attached to 1’ carbon on sugar
 Phosphate attached to 5’ carbon on sugar
12
Nucleotide numbering system
H
H
H
OH
H
H
O
O
O–
O–
5′
4′ 1′
2′
3′
3
2
1
6
5
4
P
O
O
H
H
N
O
N
Thymine
CH3
CH2
Phosphate
Deoxyribose

Strands
 Nucleotides are
covalently bonded
 Phosphodiester bond
– phosphate group links
two sugars
 Backbone – formed from
phosphates and sugars
 Bases project away from
backbone
 Written 5’ to 3’
 ex: 5’ – TACG – 3’
13
O
H
H
N
O
O–
N
N
H
N
N
H
H
H
3′
2′
3′
N
Thymine (T)
Adenine (A)
Bases
Backbone
CH3
Sugar (deoxyribose)
OH
H
Phosphate
NH2
H H
O
O P
O–
O
O
O
5′
4′
3′
P
5′ CH2
CH2
H
H
H
O
1′
2′
O
O
N
N
H
H
H
H
H
O
O
O P
O–
H
N
N
N
H
N
H
H
O
O
O
O
P
5′
4′ 1′
2′
3′
5′
4′ 1′
Cytosine (C)
Guanine (G)
Phosphodiester
linkage
Single
nucleotide
CH2
NH2
NH2
CH2
H
H
H
H
H
O
1′
2′
3′
O
H
H
O–
5′
4′
O–

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

 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

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

 Double stranded
 Antiparallel strands
 Right-handed helix
 Sugar-phosphate
backbone
 Bases on the inside
 Stabilized by H-bonding
 Specific base-pairing
 ~10 nts per helical turn
18
Features of DNA
Bases
Hydrogen bond
2 nm
(a) Double helix
5′ end
3′ end
Sugar-phosphate
backbone
One nucleotide
0.34 nm
3′ end
5′ end
Complete turn
of the helix
3.4 nm

 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

 Grooves are revealed in
the space-filling model
 Major groove
 Proteins bind to affect gene
expression
 Minor groove
 Narrower
20
Major groove
Minor groove
Major groove
Minor groove

 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

22
Semiconservative Mechanism
Second round
of replication
First round
of replication
Original
double helix
Parental strand
Daughter strand
(a) Semiconservative mechanism. DNA replication produces
DNA molecules with 1 parental strand and 1 newly made
daughter strand.

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.

24
Dispersive Mechanism
Second round
of replication
First round
of replication
Original
double helix
(c) Dispersive mechanism. DNA replication produces DNA
strands in which segments of new DNA are interspersed
with the parental DNA.

 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

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

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

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

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

 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

34
2
1
3
2
DNA strands unwind.
DNA replication begins outward
from two replication forks.
DNA replication
continues in both
directions.
2
Replication
forks
Replication
fork
Replication
fork
DNA replication
is completed.
Site where
DNA replication
ends
DNA strands unwind,
and DNA replication
begins.
DNA strands unwind,
and DNA replication
begins at multiple
origins of replication.
DNA replication
is completed.
Kinetochore proteins
at the centromere
(c) Multiple origins of replication in eukaryotes
(b) Single origin of replication in bacteria
(a) Bidirectional replication
Circular
bacterial
chromosome
Origin of
replication
Origin of
replication
Origin of
replication
1
1
https://www.youtube.com/watch?v=yyUNaSQf4zs

 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

 DNA polymerase
 Covalently links nucleotides
 Deoxynucleoside triphosphates
37
(a) Action of DNA polymerase
5′
3′
Incoming
deoxynucleoside
triphosphates
DNA polymerase
catalytic site

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

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

 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

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

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

 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

 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

 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

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

 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

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

53
Nucleosome:
8 histone proteins +
146 or 147 nucleotide
base pairs of DNA
DNA
Linker
region
Amino
terminal
tail of
histone
protein
H2B
H2B
H4
H4
H2A
H3
H1
11 nm
30 nm
(a) Micrograph of a 30-nm fiber
(b) Three-dimensional zigzag model
a: Photo courtesy of Dr. Barbara HamkaloZ

3. Radial loop domains
 Interaction between
30-nm fibers and
nuclear matrix
 Each chromosome
located in discrete
territory
 Level of compaction is
not uniform
 Heterochromatin
 Euchromatin
54
Protein fiber inside the nucleus
30-nm fiber
Radial loop
domain
Protein that attaches the base
of a DNA loop to a protein fiber

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

56
2 nm
30 nm
1
2
11 nm
Histone H1
Wrapping of DNA around
histone proteins
Formation of a 3-dimensional
zigzag structure via histone
H1 and other DNA-binding
proteins
Histones
Nucleosome
DNA double helix
(a) DNA double helix
(b) Nucleosomes (“beads on a string”)
(c) 30-nm fiber
a: © Dr. Gopal Murti/Visuals Unlimited; b: © Ada L. Olins and Donald E. Olins/Biological Photo Service; c: Courtesy Dr. Jerome B. Rattner,
Cell Biology and Anatomy, University of Calgary

57
4
3
5
Anchoring of radial loop
domains to the nuclear matrix
Further compaction of radial
loops to form heterochromatin
Metaphase chromosome with
2 copies of the DNA
1,400 nm
700 nm
300 nm
(d) Radial loop domains
(e) Heterochromatin
(f) Metaphase chromosome
d: Courtesy of Paulson, J.R. & Laemmli, U.K. James R. Paulson, U.K. Laemmli, “The structure of histonedepleted
metaphase chromosomes,” Cell, 12:817–28, Copyright Elsevier 1977; e-f: © Peter Engelhardt/
Department of Virology, Haartman Institute

58
2 nm
30 nm
1
4
3
2
5
11 nm
Histone H1
Wrapping of DNA around
histone proteins
Formation of a 3-dimensional
zigzag structure via histone
H1 and other DNA-binding
proteins
Anchoring of radial loop
domains to the nuclear matrix
Further compaction of radial
loops to form heterochromatin
Metaphase chromosome with
2 copies of the DNA
1,400 nm
700 nm
300 nm
Histones
Nucleosome
DNA double helix
(a) DNA double helix
(b) Nucleosomes (“beads on a string”)
(c) 30-nm fiber
(d) Radial loop domains
(e) Heterochromatin
(f) Metaphase chromosome
a: © Dr. Gopal Murti/Visuals Unlimited; b: © Ada L. Olins and Donald E. Olins/Biological Photo Service; c: Courtesy Dr. Jerome B. Rattner, Cell Biology and
Anatomy, University of Calgary; d: Courtesy of Paulson, J.R. & Laemmli, U.K. James R. Paulson, U.K. Laemmli, “The structure of histonedepleted
metaphase chromosomes,” Cell, 12:817–28, Copyright Elsevier 1977; e-f: © Peter Engelhardt/Department of Virology, Haartman Institute

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

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

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

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

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

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

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

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

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

Question 13
A cell in interphase would most likely have
heterochromatin present.
a. This is true
b. This is false

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

Question 15
A bacteria’s DNA is located in the nucleus.
a. This is true
b. This is false

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

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

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
e. Failure of preformed fragments to link into a single
strand

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
d. Failure of nucleotide addition to complementary bases
e. Failure of preformed fragments to link into a
single strand

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