Download Honor Genetics DNA structure and replication

Chapter 16
The Molecular Basis of
Inheritance
PowerPoint Lectures for
Biology, Seventh Edition
Neil Campbell and Jane Reece
Lectures by Chris Romero
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
• Overview: Life’s Operating Instructions
• In 1953, James Watson and Francis Crick
shook the world
– With an elegant double-helical model for the
structure of deoxyribonucleic acid, or DNA
Figure 16.1
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• DNA, the substance of inheritance
– Is the most celebrated molecule of our time
• Hereditary information
– Is encoded in the chemical language of DNA
and reproduced in all the cells of your body
• It is the DNA program
– That directs the development of many different
types of traits
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• Concept 16.1: DNA is the genetic material
• Early in the 20th century
– The identification of the molecules of
inheritance loomed as a major challenge to
biologists
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The Search for the Genetic Material: Scientific Inquiry
• The role of DNA in heredity
– Was first worked out by studying bacteria and
the viruses that infect them
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Evidence That DNA Can Transform Bacteria
• Frederick Griffith was studying Streptococcus
pneumoniae
– A bacterium that causes pneumonia in
mammals
• He worked with two strains of the bacterium
– A pathogenic strain and a nonpathogenic
strain
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• Griffith found that when he mixed heat-killed
remains of the pathogenic strain
– With living cells of the nonpathogenic strain,
some of these living cells became pathogenic
EXPERIMENT Bacteria of the “S” (smooth) strain of Streptococcus pneumoniae are pathogenic because they
have a capsule that protects them from an animal’s defense system. Bacteria of the “R” (rough) strain lack a capsule
and are nonpathogenic. Frederick Griffith injected mice with the two strains as shown below:
Living S
(control) cells
Living R
Heat-killed
(control) cells (control) S cells
Mixture of heat-killed S cells
and living R cells
RESULTS
Mouse dies
Mouse healthy
Mouse healthy
Mouse dies
Living S cells
are found in
blood sample.
Figure 16.2
CONCLUSION Griffith concluded that the living R bacteria had been transformed into pathogenic S bacteria by an
unknown, heritable substance from the dead S cells.
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• Griffith called the phenomenon transformation
– Now defined as a change in genotype and
phenotype due to the assimilation of external
DNA by a cell
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Evidence That Viral DNA Can Program Cells
• Additional evidence for DNA as the genetic
material
– Came from studies of a virus that infects
bacteria
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• Viruses that infect bacteria, bacteriophages
– Are widely used as tools by researchers in
molecular genetics
Phage
head
Tail
Tail fiber
Figure 16.3
Bacterial
cell
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100 nm
DNA
• Alfred Hershey and Martha Chase
– Performed experiments showing that DNA is
the genetic material of a phage known as T2
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• The Hershey and Chase experiment
EXPERIMENT
In their famous 1952 experiment, Alfred Hershey and Martha Chase used radioactive sulfur
and phosphorus to trace the fates of the protein and DNA, respectively, of T2 phages that infected bacterial cells.
1 Mixed radioactively
labeled phages with
bacteria. The phages
infected the bacterial cells.
Phage
2 Agitated in a blender to 3 Centrifuged the mixture
separate phages outside
so that bacteria formed
the bacteria from the
a pellet at the bottom of
bacterial cells.
the test tube.
Radioactive Empty
protein
protein shell
Radioactivity
(phage protein)
in liquid
Bacterial cell
Batch 1: Phages were
grown with radioactive
sulfur (35S), which was
incorporated into phage
protein (pink).
Batch 2: Phages were
grown with radioactive
phosphorus (32P), which
was incorporated into
phage DNA (blue).
4 Measured the
radioactivity in
the pellet and
the liquid
DNA
Phage
DNA
Centrifuge
Radioactive
DNA
Pellet (bacterial
cells and contents)
Centrifuge
Radioactivity
(phage DNA)
Pellet
in pellet
RESULTS
Phage proteins remained outside the bacterial cells during infection, while phage DNA entered the cells.
When cultured, bacterial cells with radioactive phage DNA released new phages with some radioactive phosphorus.
Figure 16.4
CONCLUSION
Hershey and Chase concluded that DNA, not protein, functions as the T2 phage’s genetic material.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Additional Evidence That DNA Is the Genetic Materia
• Prior to the 1950s, it was already known that DNA
– Is a polymer of nucleotides, each consisting of
three components: a nitrogenous base, a
sugar, and a phosphate group
Sugar-phosphate
backbone
5 end
O– 5
O P O CH2
O 1
O– 4 H
H
H
H
2
3
H
O
O P O CH2 O
O–
H
H
H
H
H
O
O P O CH2 O
O–
H
H
H
H
H
Figure 16.5
O 5
O P O CH2
O 1
O– 4 H
H
PhosphateH
H
3 2
OH H
Sugar (deoxyribose)
3 end
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Nitrogenous
bases
CH3
O
H
N
N
H
O
Thymine (T)
H
N
H
N
N
N
N
H
H
Adenine (A)
H H
H
N H
N
N
O
Cytosine (C)
H
N
N
N
O
N H
N H
H
Guanine (G)
DNA nucleotide
• Erwin Chargaff analyzed the base composition of DNA
– From a number of different organisms
• In 1947, Chargaff reported
– That DNA composition varies from one species to
the next
• This evidence of molecular diversity among species
– Made DNA a more credible candidate for the genetic
material
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Building a Structural Model of DNA: Scientific Inquiry
• Once most biologists were convinced that DNA
was the genetic material
– The challenge was to determine how the
structure of DNA could account for its role in
inheritance
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• Maurice Wilkins and Rosalind Franklin
– Were using a technique called X-ray
crystallography to study molecular structure
• Rosalind Franklin
– Produced a picture of the DNA molecule using
this technique
Figure 16.6 a, b
(a) Rosalind Franklin
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
(b) Franklin’s X-ray diffraction
Photograph of DNA
• Watson and Crick deduced that DNA was a
double helix
– Through observations of the X-ray
crystallographic images of DNA
G
C
A
T
T
A
1 nm
C
G
C
A
T
G
C
T
A
T
A
A
T
T
A
G
A
Figure 16.7a, c
3.4 nm
G
C
0.34 nm
T
(a) Key features of DNA structure
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(c) Space-filling model
• Franklin had concluded that DNA
– Was composed of two antiparallel sugarphosphate backbones, with the nitrogenous
bases paired in the molecule’s interior
• The nitrogenous bases
– Are paired in specific combinations: adenine
with thymine, and cytosine with guanine
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5 end
O
OH
Hydrogen bond
P
–O
3 end
OH
O
O
A
T
O
O
O
CH2
P
–O
O
H2C
O
–O
P
O
O
G
O
C
O
O
CH2
P
O
O–
O
P
H2C
O
O
C
O
G
O
O
O
CH2
P
–O
O–
O
O
O–
O
P
H2C
O
O
A
O
T
O
CH2
OH
3 end
O
O–
P
O
Figure 16.7b
(b) Partial chemical structure
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O
5 end
DNA
• Stands for Deoxyribonucleic acid
• Made up of subunits called
nucleotides
• Nucleotide made of:
1. Phosphate group
2. 5-carbon sugar
3. Nitrogenous base
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20
DNA Nucleotide
Phosphate
Group
O
5
CH2
O=P-O
O
O
N
C1 Nitrogenous ba
C4
Sugar
(deoxyribose)
C3
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(A, G, C, or T
C2
21
Pentose Sugar
• Carbons are numbered clockwise 1’
to 5’
5
CH2
O
C1
C4
Sugar
(deoxyribose)
C3
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C2
22
DNA
5
O
3
3
P
5
O
O
C
G
1
P
5
3
2
4
4
P
5
P
2
3
1
5
O
O
T
A
3
O
3
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
23
5
P
P
• Watson and Crick reasoned that there must be
additional specificity of pairing
– Dictated by the structure of the bases
• Each base pair forms a different number of
hydrogen bonds
– Adenine and thymine form two bonds,
cytosine and guanine form three bonds
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H
N
N
N
N
Sugar
O
H
H
CH3
N
N
N
O
Sugar
Thymine (T)
Adenine (A)
H
O
N
N
Sugar
N
H
N
N
N
N
N
Figure 16.8
H
H
Guanine (G)
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H
O
Sugar
Cytosine (C)
• Concept 16.2: Many proteins work together in
DNA replication and repair
• The relationship between structure and
function
– Is manifest in the double helix
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The Basic Principle: Base Pairing to a Template Strand
• Since the two strands of DNA are
complementary
– Each strand acts as a template for building a
new strand in replication
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• In DNA replication
– The parent molecule unwinds, and two new
daughter strands are built based on basepairing rules
T
A
T
A
T
A
C
G
C
G
C
T
A
T
A
T
A
A
T
A
T
A
T
G
C
G
C
G
C
G
A
T
A
T
A
T
C
G
C
G
C
G
T
A
T
A
T
A
T
A
T
A
T
C
G
C
G
C
A
G
(a) The parent molecule has two
complementary strands of DNA.
Each base is paired by hydrogen
bonding with its specific partner,
A with T and G with C.
(b) The first step in replication is
separation of the two DNA
strands.
Figure 16.9 a–d
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(c) Each parental strand now
serves as a template that
determines the order of
nucleotides along a new,
complementary strand.
(d) The nucleotides are connected
to form the sugar-phosphate
backbones of the new strands.
Each “daughter” DNA
molecule consists of one parental
strand and one new strand.
• DNA replication is semiconservative
– Each of the two new daughter molecules will
have one old strand, derived from the parent
molecule, and one newly made strand
First
Second
Parent cell replication replication
(a) Conservative
model. The two
parental strands
reassociate
after acting as
templates for
new strands,
thus restoring
the parental
double helix.
(b) Semiconservative
model. The two
strands of the
parental molecule
separate,
and each functions
as a template
for synthesis of
a new, complementary strand.
Figure 16.10 a–c
(c) Dispersive
model. Each
strand of both
daughter molecules contains
a mixture of
old and newly
synthesized
DNA.
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• Experiments performed by Meselson and Stahl
– Supported the semiconservative model of DNA
replication
EXPERIMENT
Matthew Meselson and Franklin Stahl cultured E. coli bacteria for several generations
on a medium containing nucleotide precursors labeled with a heavy isotope of nitrogen, 15N. The bacteria
incorporated the heavy nitrogen into their DNA. The scientists then transferred the bacteria to a medium with
only 14N, the lighter, more common isotope of nitrogen. Any new DNA that the bacteria synthesized would be
lighter than the parental DNA made in the 15N medium. Meselson and Stahl could distinguish DNA of different
densities by centrifuging DNA extracted from the bacteria.
1
Bacteria
cultured in
medium
containing
15N
2
Bacteria
transferred to
medium
containing
14N
RESULTS
3
DNA sample
centrifuged
after 20 min
(after first
replication)
4
DNA sample
centrifuged
after 40 min
(after second
replication)
Less
dense
More
dense
The bands in these two centrifuge tubes represent the results of centrifuging two DNA samples from the flask
Figure 16.11 in step 2, one sample taken after 20 minutes and one after 40 minutes.
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CONCLUSION Meselson and Stahl concluded that DNA replication follows the semiconservative
model by comparing their result to the results predicted by each of the three models in Figure 16.10.
The first replication in the 14N medium produced a band of hybrid (15N–14N) DNA. This result eliminated
the conservative model. A second replication produced both light and hybrid DNA, a result that eliminated
the dispersive model and supported the semiconservative model.
First replication
Conservative
model
Semiconservative
model
Dispersive
model
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Second replication
DNA Replication: A Closer Look
• The copying of DNA
– Is remarkable in its speed and accuracy
• More than a dozen enzymes and other proteins
– Participate in DNA replication
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Getting Started: Origins of Replication
• The replication of a DNA molecule
– Begins at special sites called origins of
replication, where the two strands are
separated
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• A eukaryotic chromosome
– May have hundreds or even thousands of
replication origins
Origin of replication
1 Replication begins at specific sites
where the two parental strands
separate and form replication
bubbles.
Bubble
Parental (template) strand
Daughter (new) strand
0.25 µm
Replication fork
2 The bubbles expand laterally, as
DNA replication proceeds in both
directions.
3 Eventually, the replication
bubbles fuse, and synthesis of
the daughter strands is
complete.
Two daughter DNA molecules
(a) In eukaryotes, DNA replication begins at many sites along the giant
DNA molecule of each chromosome.
Figure 16.12 a, b
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(b) In this micrograph, three replication
bubbles are visible along the DNA of
a cultured Chinese hamster cell (TEM).
DNA Replication
• Enzyme Helicase unwinds
and separates the 2 DNA
strands by breaking the
weak hydrogen bonds
• Single-Strand Binding
Proteins attach and keep the
2 DNA strands separated
and untwisted
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35
DNA Replication
• Enzyme Topoisomerase attaches to
the 2 forks of the bubble to relieve
stress on the DNA molecule as it
separates
Enzyme
Enzyme
DNA
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36
DNA Replication
•
Before new DNA strands can form,
there must be RNA primers present
to start the addition of new
nucleotides
•
Primase is the enzyme that
synthesizes the RNA Primer
•
DNA polymerase III can then add
the new nucleotides
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37
Elongating a New DNA Strand
• Elongation of new DNA at a replication fork
– Is catalyzed by enzymes called DNA
polymerases, which add nucleotides to the 3
end of a growing strand
New strand
5 end
Sugar
Template strand
3 end
A
Base
Phosphate
T
A
T
C
G
C
G
G
C
G
C
A
T
A
OH
Pyrophosphate 3 end
P
OH
Figure 16.13
3 end
5 end
Nucleoside
triphosphate
C
P
C
2 P
5 end
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5 end
Antiparallel Elongation
• How does the antiparallel structure of the
double helix affect replication?
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• DNA polymerases add nucleotides
– Only to the free 3end of a growing strand
• Along one template strand of DNA, the leading
strand
– DNA polymerase III can synthesize a
complementary strand continuously, moving
toward the replication fork
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• To elongate the other new strand of DNA, the
lagging strand
– DNA polymerase III must work in the direction
away from the replication fork
• The lagging strand
– Is synthesized as a series of segments called
Okazaki fragments, which are then joined
together by DNA ligase
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• Synthesis of leading and lagging strands during
DNA replication
1 DNA pol Ill elongates
DNA strands only in the
5
3 direction. 3
5
Parental DNA
5
3
Okazaki
fragments
2
1
3
5
DNA pol III
2 One new strand, the leading strand,
can elongate continuously 5
3
as the replication fork progresses.
3 The other new strand, the
lagging strand must grow in an overall
3
5 direction by addition of short
segments, Okazaki fragments, that grow
5
3 (numbered here in the order
they were made).
Template
strand
3
Leading strand
Lagging strand
2
Template
strand
Figure 16.14
1
DNA ligase
Overall direction of replication
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4 DNA ligase joins Okazaki
fragments by forming a bond between
their free ends. This results in a
continuous strand.
Priming DNA Synthesis
• DNA polymerases cannot initiate the synthesis
of a polynucleotide
– They can only add nucleotides to the 3 end
• The initial nucleotide strand
– Is an RNA or DNA primer
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• Only one primer is needed for synthesis of the
leading strand
– But for synthesis of the lagging strand, each
Okazaki fragment must be primed separately
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1
Primase joins RNA nucleotides
into a primer.
3
5
5
3
Template
strand
RNA primer
3
5
3
DNA pol III adds DNA nucleotides to the
primer, forming an Okazaki fragment.
2
5
3
1
After reaching the next
RNA primer (not shown),
DNA pol III falls off.
Okazaki
fragment
3
3
5
1
5
4
After the second fragment is
primed. DNA pol III adds DNA
nucleotides until it reaches the
first primer and falls off. 5
3
5
3
2
5
1
DNA pol 1 replaces the
RNA with DNA, adding to
the 3 end of fragment 2.
5
3
6
5
1
DNA ligase forms a bond
between the newest DNA
and the adjacent DNA of
fragment 1.
5
3
Figure 16.15
3
2
7
The lagging strand
in this region is now
complete.
3
2
1
Overall direction of replication
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5
Other Proteins That Assist DNA Replication
• Helicase, topoisomerase, single-strand binding
protein
– Are all proteins that assist DNA replication
Table 16.1
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• A summary of DNA replication
Overall direction of replication
1 Helicase unwinds the
parental double helix.
2 Molecules of singlestrand binding protein
stabilize the unwound
template strands.
3 The leading strand is
synthesized continuously in the
5 3 direction by DNA pol III.
DNA pol III
Lagging
Leading
strand Origin of replication strand
Lagging
strand
OVERVIEW
Leading
strand
Leading
strand
5
3
Parental DNA
4 Primase begins synthesis
of RNA primer for fifth
Okazaki fragment.
5 DNA pol III is completing synthesis of
the fourth fragment, when it reaches the
RNA primer on the third fragment, it will
dissociate, move to the replication fork,
and add DNA nucleotides to the 3 end
of the fifth fragment primer.
Replication fork
Primase
DNA pol III
Primer
4
DNA ligase
DNA pol I
Lagging
strand
3
2
6 DNA pol I removes the primer from the 5 end
of the second fragment, replacing it with DNA
nucleotides that it adds one by one to the 3 end
of the third fragment. The replacement of the
last RNA nucleotide with DNA leaves the sugarphosphate backbone with a free 3 end.
Figure 16.16
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1
3
5
7 DNA ligase bonds
the 3 end of the
second fragment to
the 5 end of the first
fragment.
The DNA Replication Machine as a Stationary Complex
• The various proteins that participate in DNA
replication
– Form a single large complex, a DNA replication
“machine”
• The DNA replication machine
– Is probably stationary during the replication
process
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Proofreading and Repairing DNA
• DNA polymerases proofread newly made DNA
– Replacing any incorrect nucleotides
• In mismatch repair of DNA
– Repair enzymes correct errors in base pairing
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• In nucleotide excision repair
– Enzymes cut out and replace damaged
stretches of DNA
1 A thymine dimer
distorts the DNA molecule.
2 A nuclease enzyme cuts
the damaged DNA strand
at two points and the
damaged section is
removed.
Nuclease
DNA
polymerase
3 Repair synthesis by
a DNA polymerase II
fills in the missing
nucleotides.
DNA
ligase
Figure 16.17
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4 DNA ligase seals the
Free end of the new DNA
To the old DNA, making the
strand complete.
Proofreading New DNA
• DNA polymerase initially makes about 1
in 10,000 base pairing errors
• Enzymes (DNA polymerase II) proofread
and correct these mistakes
• The new error rate for DNA that has been
proofread is 1 in 1 billion base pairing
errors
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51
Crash Course Biology DNA Structure and
Replication
• https://www.youtube.com/watch?v=8kK2zwjRV
0M
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Replicating the Ends of DNA Molecules
• The ends of eukaryotic chromosomal DNA
– Get shorter with each round of replication
5
End of parental
DNA strands
Leading strand
Lagging strand
3
Last fragment
Previous fragment
RNA primer
Lagging strand
5
3
Primer removed but
cannot be replaced
with DNA because
no 3 end available
for DNA polymerase
Removal of primers and
replacement with DNA
where a 3 end is available
5
3
Second round
of replication
5
New leading strand 3
New lagging strand 5
3
Further rounds
of replication
Figure 16.18
Shorter and shorter
daughter molecules
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• Eukaryotic chromosomal DNA molecules
– Have at their ends nucleotide sequences,
called telomeres, that postpone the erosion of
genes near the ends of DNA molecules
Figure 16.19
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1 µm
• If the chromosomes of germ cells became
shorter in every cell cycle
– Essential genes would eventually be missing
from the gametes they produce
• An enzyme called telomerase
– Catalyzes the lengthening of telomeres in
germ cells
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• In eukaryotes, the DNA-protein complex, is called
chromatin
– Is ordered into higher structural levels than the
DNA-protein complex in prokaryotes
Figure 19.1
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• Both prokaryotes and eukaryotes
– Must alter their patterns of gene expression in
response to changes in environmental
conditions
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• Chromatin structure is based on successive
levels of DNA packing
• Eukaryotic DNA
– Is precisely combined with a large amount of
protein
• Eukaryotic chromosomes
– Contain an enormous amount of DNA relative
to their condensed length
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Nucleosomes, or “Beads on a String”
• Proteins called histones
– Are responsible for the first level of DNA
packing in chromatin
– Bind tightly to DNA
• The association of DNA and histones
– Seems to remain intact throughout the cell
cycle
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• In electron micrographs
– Unfolded chromatin has the appearance of beads
on a string
• Each “bead” is a nucleosome
– The basic unit of DNA packing
2 nm
DNA double helix
Histones
Histone
tails
Histone H1
Linker DNA
(“string”)
Nucleosome
(“bad”)
(a) Nucleosomes (10-nm fiber)
Figure 19.2 a
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10 nm
Higher Levels of DNA Packing
• The next level of packing
– Forms the 30-nm chromatin fiber
30 nm
Nucleosome
(b) 30-nm fiber
Figure 19.2 b
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• The 30-nm fiber, in turn
– Forms looped domains, making up a 300-nm
fiber
Protein scaffold
Loops
300 nm
(c) Looped domains (300-nm fiber)
Figure 19.2 c
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Scaffold
• In a mitotic chromosome
– The looped domains themselves coil and fold
forming the characteristic metaphase
chromosome
700 nm
1,400 nm
(d) Metaphase chromosome
Figure 19.2 d
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• In interphase cells
– Most chromatin is in the highly extended form
called euchromatin
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• Gene expression can be regulated at any
stage, but the key step is transcription
• All organisms
– Must regulate which genes are expressed at
any given time
• During development of a multicellular organism
– Its cells undergo a process of specialization in
form and function called cell differentiation
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Differential Gene Expression
• Each cell of a multicellular eukaryote
– Expresses only a fraction of its genes
• In each type of differentiated cell
– A unique subset of genes is expressed
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• Many key stages of gene expression
– Can be regulated in eukaryotic cells
Signal
NUCLEUS
Chromatin
Chromatin modification:
DNA unpacking involving
histone acetylation and
DNA demethlation
DNA
Gene available
for transcription
Gene
Transcription
RNA
Cap
Exon
Primary transcript
Intron
RNA processing
Tail
mRNA in nucleus
Transport to cytoplasm
CYTOPLASM
mRNA in cytoplasm
Degradation
of mRNA
Translation
Polypetide
Cleavage
Chemical modification
Transport to cellular
destination
Active protein
Degradation of protein
Figure 19.3
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Degraded protein
Regulation of Chromatin Structure
• Genes within highly packed heterochromatin
– Are usually not expressed
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Histone Modification
• Chemical modification of histone tails
– Can affect the configuration of chromatin and
thus gene expression
Chromatin changes
Transcription
RNA processing
mRNA
degradation
Translation
Protein processing
and degradation
Histone
tails
DNA
double helix
Amino acids
available
for chemical
modification
Figure 19.4a
(a) Histone tails protrude outward from a nucleosome
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• Histone acetylation (acetyl group COCH3)
– Seems to loosen chromatin structure and
thereby enhance transcription
Unacetylated histones
Figure 19.4 b
Acetylated histones
(b) Acetylation of histone tails promotes loose chromatin structure that
permits transcription
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DNA Methylation
• Addition of methyl (CH3) groups to certain
bases
in DNA
– Is associated with reduced transcription in
some species
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Epigenetic Inheritance
• Epigenetic inheritance
– Is the inheritance of traits transmitted by
mechanisms not directly involving the
nucleotide sequence
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Regulation of Transcription Initiation
• Chromatin-modifying enzymes provide initial
control of gene expression
– By making a region of DNA either more or less
able to bind the transcription machinery
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