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Objective # 25
Module 3E – DNA Structure
and Replication
Summarize the evidence
from the 1920s through the
early 1950s that convinced
scientists that DNA is the
genetic material.
 In
this module, we will examine:
 the molecular structure of the
genetic material
 how the genetic material replicates
 how damage to the genetic material
is repaired
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Objective 25
the late 19th and early 20th
centuries, scientists studying patterns
of inheritance concluded that inherited
traits are controlled by “hereditary
factors” or alleles which are located on
chromosomes.
 However, a key question remained
unanswered: what are these “hereditary
factors” made of?
Objective 25
 Scientists
 During
3
knew the genetic material
must carry out 2 basic functions:
code information
replicate itself
 They also knew coding could be done
by varying the sequence of monomers
that make up a polymer. [Similar to the
way we code information in books by
varying the sequence of 26 letters.]
Objective 25
lipids are not true polymers, and
since most polysaccharides are made of
repeating glucose units, this left 2 main
candidates: proteins and nucleic acids.
 Proteins were considered the more likely
candidate because coding information
would be more efficient using molecules
made of 20 different monomers rather
than just 4.
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Objective 25
 Since
 An
important breakthrough came in
1928, when Frederick Griffith
determined that something can pass
from one cell to another and alter the
characteristics of the recipient cell.
 Griffith called this process
transformation::
transformation
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Griffths Experiment
Live Virulent
Strain of
S. pneumoniae
Polysaccharide
coat
Mice die
Live Nonvirulent
Strain of
S. pneumoniae
Objective 25
Mixture of Heat-Killed
Virulent and Live
Heat-killed
Virulent Strain Nonvirulent Strains
of S. pneumoniae of S. pneumoniae
+
No
Polysaccharide
coat
Mice live
Mice die
Mice live
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Their Lungs contain
live pathogenic
strain of S.
pneumoniae
 Since
the agent that passed from one
cell to another in Griffith’s experiment
actually alters the characteristics of the
recipient cell, many scientists believed
it must be the genetic material.
 In 1944, Avery, MacLeod, and
McCarty determined that the
transforming agent was DNA.
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Objective 25
 In
spite of this evidence, many
scientists continued to believe that
proteins, rather than DNA, function as
the hereditary material.
 The question was finally settled in
1952 by Hershey and Chase who
carried out a classic experiment with
bacteriophages:
Objective # 26
Objective 26
 Once
scientists determined that the
genetic material was composed of
DNA, there was intense competition to
determine the structure of this
molecule
molecule.
 Scientists hoped that the structure of
DNA would provide important clues to
understanding how it works to replicate
itself and control genetic traits.
Name the scientists who
originally discovered the
structure off DNA.
DNA
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Objective 26
Chargaff’s Rules
 The
structure of DNA was finally
discovered at Cambridge University in
1953 by James Watson and Francis
Crick.
 Their discovery was based primarily on
x-ray crystallography data collected by
Maurice Wilkins and Rosalind Franklin
as well as on the chemical analysis of the
base composition of DNA carried out by
Irwin Chargaff.
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 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)
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James Watson and Francis Crick – 1953
Rosalind Franklin

 Watson
and Crick deduced the
structure of DNA using evidence from
Chargaff, Franklin, and others
 Did not p
perform a single
g experiment
p
themselves related to DNA
 Proposed a double helix structure
Performed XX-ray diffraction studies
to identify the 33--D structure
 Discovered that DNA is helical
 Us
Usingg Maurice
au ce Wilkins’
W
s DNA
DN
fibers, discovered that the
molecule has a diameter of 2 nm
and makes a complete turn of the
helix every 3.4 nm
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Objective # 27
Objective 27
 All
Describe the structure of a
DNA nucleotide and explain
h iit diff
how
differs from
f
the
h
structure of an RNA
nucleotide.
nucleotides have 3 parts:
A
pentose (5(5-carbon) sugar
A nitrogenous
g
base attached to the
1 prime carbon of the sugar
A phosphate group (PO4) attached
to the 5 prime carbon of the sugar:
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3
P
O–
NH2
7N
5
1
8
Phosphate group
O
4
N
9
–O
O CH2
5’
P
2
N
3
2´
OH in RNA
OH
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required for reproduction or display.
N C
C H
N
H
C
N C C
N
H
N C
C
NH2
H
Adenine
N
Guanine
1’
H in DNA
Sugar
OH in RNA
H
C
H
C
C
N
H3C
N
C
O
O
NH2
3´
2’
N
H
1´
O
OH
N C C
O
4´
3’
C
5´
O–
4’
H
CH2
O
O
NH2
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Purines
O
Nitrogenous Base
Nitrogenous base
Pyrimidines
Phosphate group
–O
There are 5 possible nitrogenous bases and 2
possible sugars:
Nitrogenous base
NH2
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7N 5
N1
8
2
9 N 4 N3
Structure of a
Nucleotide
O
H
Cytosine
(both DNA and RNA)
H
C
C
C
N
N H
H
C
C
H
C
H
Thymine
(DNA only)
O
C
N
N
H
C
O
H
Uracil
(RNA only)
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R
Sugar
H in DNA
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Objective 27
Objective # 28
 DNA
nucleotides contain the sugar
deoxyribose and the nitrogenous bases
A, G, C, T.
Describe the structure of a
DNA molecule and explain
how it differs from the
structure of a RNA
molecule.
 RNA
nucleotides contain the sugar
ribose and the nitrogenous bases A, G,
C, U.
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Objective 28
Double helix
 DNA:
 consists
of 2 unbranched chains of
DNA nucleotides twisted into a double
helix..
helix
 the 2 chains are held together by
hydrogen bonds between the
nitogenous bases.
 A always pairs with T, and G with C.
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5´
Phosphate group
P
5´
O
1´
4´
3´
Phosphodiester bond
2´
P
5´
O
4´
1´
3´
P
2´
5´
O
1´
4´
3´
5-carbon sugar
2´
Nitrogenous base
P
5´
O
1´
4´
3´
OH
• each strand is a polymer of nucleotides
• Phosphodiester backbone – repeating sugar and phosphate
sugar and phosphate units joined by phosphodiester bonds
• Nitrogenous bases project from the sugar‐
phosphate backbone
2´
3´
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4
Complementarity of bases:
Hydrogen
bond
N
H
• G forms 3 hydrogen bonds with C
• A forms 2 hydrogen bonds with T
bonds with T
• Gives consistent diameter
O
N
N
G
H
H
N
N
H
H
H
C
N
N
Sugar
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
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Objective 28
 RNA
is a polynucleotide composed
of a single, unbranched chain of
RNA nucleotides.
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5’ end
RNA
P
P
Objective # 29
Copyright © The McGraw-Hill Companies, Inc.
Permission required for reproduction or display.
In detail, describe the process of
DNA replication including the
names and functions of the
enzymes involved.
involved Be able to
explain why DNA replication is
discontinuous along one strand
and continuous along the other.
Phosphate group
O
Phosphodiester
bonds
O
P
O
5-carbon sugar (ribose)
P
O
OH
3’ end
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Nitrogenous base
(A,C,G or U)
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Objective 29
 In
1958, Meselson and Stahl carried out
an experiment at Cal Tech supporting
the hypothesis that the replication of
DNA is semiconservative
semiconservative..
 According to this hypothesis, during
replication, the 2 strands of DNA
separate and each one serves as a
template for a new strand.
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Objective 29
 The
following animation provides a
brief overview of the process of
semiconservative DNA replication.
 Note that the ability of one strand of
DNA to act as a template for the
matching strand is based on the fact
that A always pairs with T and G
always pairs with C.
Objective 29
 The
next animation provides a more
detailed description of the process of
DNA replication including the names and
functions of the major enzymes involved.
 Note that the main enzyme involved in
strand elongation, DNA polymerase III,
cannot initiate synthesis of a new
nucleotide strand. It can only add
nucleotides to the 3’ end of an existing
strand.
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Objective # 30
Objective 30
 DNA
replication in prokaryotes and
eukaryotes is fundamentally the same.
 On the next slide we can see how the
single circular DNA molecule found in
prokaryotic organisms like E. coli is
replicated. (Note that the circular
DNA molecule is drawn as linear in
order to simplify the animation.)
Explain how DNA
replication in prokaryotes
differs from DNA
replication in eukaryotes.
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Objective 30
Objective 30
 In
order to speed up the replication of
such a large amount of DNA, the
linear DNA molecule found in each
eukaryotic chromosome may have
hundreds or even thousands of
replication origins, while the circular
DNA molecule in a prokaryotic cell
has just one.
 Although
DNA replication in
prokaryotes and eukaryotes is very
similar, there are a few differences.
 One difference is due to the fact that
eukaryotes have so much more DNA
than prokaryotes.
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Objective 30
Objective 30
 Because
DNA polymerase III can only
add new nucleotides to the 3’ end of
an existing strand, there is no way to
complete the final segment of the
lagging strand located at each end of a
linear DNA molecule.
 As a result, linear DNA molecules gets
shorter and shorter with each round of
replication:
 Another
difference in the replication
process is related to the fact that
eukaryotes
y
have linear rather than
circular DNA. This creates a unique
problem for eukaryotes.
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Objective 30
Original strands shown in green and newly synthesized strands in blue:
Eukaryotes have 2 ways to deal with
this problem:
1) The ends of eukaryotic chromosomes
have extra DNA in the form of
telomeres.. Telomeres consist of a short
telomeres
nucleotide sequence that is repeated
over and over again. As a result, when
eukaryotic chromosomes replicate the
telomeres shorten rather than the actual
protein--coding genes.
protein
Replication first
round

3
5
5
3
Leading strand (no problem) Lagging strand (problem
at the end)
3
5
5
3
Last primer
Origin
Leading
strand
Pi
Primer
removall
3
5
5
Lagging
strand
3
Replication second
round
Removed primer
cannot be replaced
5
3
3
5
3
5
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required for reproduction or display.
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5
3
Shortened template
Objective 30
Objective 30
 Scientists
2) Some cells, such as germ cells, have
an enzyme called telomerase.
telomerase. This
enzyme has an internal RNA
template that is used to lengthen the
telomeres so that chromosomes can
continue to replicate without
shortening the actual protein
protein--coding
genes.
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believe that normal shortening
of telomeres during DNA replication
may help protect against cancer by
limiting the number of divisions that
cells can undergo.
undergo
 Interestingly, telomerase activity has
been detected in cancer cells. By
lengthening the telomeres, this may
allow the cells to continue to divide
indefinitely.
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Objective # 31
Describe the importance
and the mechanisms of
DNA repair.
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Objective 31
 Many
DNA polymerases can
“proofread” added bases and correct
mistakes as DNA is being replicated.
This increases the accuracy of
replication but some errors still occur
replication,
occur.
 These mistakes or mutations are a
mixed blessing. They provide the
genetic variation that is essential for
evolution but, unfortunately, most are
harmful.
 In
Objective 31
addition to mistakes that occur during
replication, DNA is constantly exposed to
damaging agents such as UV light, XX-rays,
and chemicals.
 Mechanisms to repair this damage fall into
2 categories: specific and non
non--specific.
 Specific repair mechanism target a
particular type of damage. Some
examples are shown in the next animation.
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Objective 31
 Excision
repair is a nonnon-specific repair
mechanism that can be used if only
one strand of the DNA is damaged. It
involves 3 steps:
 Recognition of the damage
 Removal of the damaged strand
 Synthesis of a new strand using the
undamaged strand as a template
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