dna: the genetic material chapter 14 1. learning objectives 14.1 the nature of the genetic material...
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DNA: The Genetic MaterialChapter 14
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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
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Frederick Griffith – 1928 • Studied Streptococcus pneumoniae, a
pathogenic bacterium causing pneumonia
• 2 strains of Streptococcus
– S strain is virulent
– R strain is nonvirulent
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Griffith’s Experiment
• Griffith infected mice with these strains hoping to understand the difference between the strains
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Live NonvirulentStrain of
S. pneumoniae
Mice live
b.
Live VirulentStrain of S. pneumoniae
Mice die
Polysaccharidecoat
a.Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Griffith’s Experiment
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Heat-killed VirulentStrain of S. pneumoniae
Mice live
c.
+
Mixture of Heat-killed Virulentand Live Nonvirulent
Strains of S. pneumoniae
Their lungs contain livepathogenic strain of
S. pneumoniae
Mice die
d.Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Griffith’s Experiment
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• 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
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• 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
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Avery, MacLeod, & McCarty – 1944
• Repeated Griffith’s experiment using purified cell extracts
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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
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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
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• Bacteriophage DNA was labeled with radioactive phosphorus (32P)
• Bacteriophage protein was labeled with radioactive sulfur (35S)
• Radioactive molecules were tracked
Hershey & Chase –1952
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+
+
Phage grown in radioactive 35S,which is incorporated into phage coat
Virus infectbacteria
Blender separatesphage coat from bacteria
Centrifuge formsbacterial pellet
35S in supernatant
35S-Labeled Bacteriophages
Phage grown in radioactive 32P.which is incorporated into phage DNA
Virus infectbacteria
Blender separatesphage coat from bacteria
Centrifuge formsbacterial pellet
32P in bacteria pellet
32P-Labeled Bacteriophages
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Hershey & Chase Experiment
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• 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!
Hershey & Chase –1952
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Question 16Mixing a killed virulent strain of bacteria and a living strain of benign bacteria together produces virulent bacteria. What does this demonstrate?
a. Genes are inactivated when a cell dies
b. DNA can only be passed on during reproduction
c. Cells can pick up genes from the environment
d. All bacteria are virulent
e. None of the above
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Learning Objectives
14.2 DNA Structure
•Understand the contributions of the following people in elucidating DNA’s structure
– Chargaff
– Wilkins & Franklin
– Watson & Crick
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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
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Nucleotide Subunits of DNA and RNA
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Pu
rin
esP
yrim
idin
es
Adenine Guanine
NH2CC
NN
N
C
H
N
C
CH
O
H
H
OC
NC
H
N
C
NH2
H
CH O
O
C
NC
H
N
CH3C
CH
H
O
O
C
NC
H
N
CH
CH
NH2
CC
NN
N
C
H
N
C
CHH
Nitrogenous Base
4′
5′
1′
3′ 2′
2
8
7 6
39
4
5
1Phosphate group
Sugar
Nitrogenous base
CH2
N N
O
NNH2
OH in RNA
Cytosine(both DNA and RNA)
Thymine(DNA only)
Uracil(RNA only)
OH
H in DNA
O
P
O–
–O O
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• 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
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BaseCH2
O
5′
3′
O
P
O
OH
CH2
–O O
C
Base
O
PO4
Phosphodiesterbond
Phosphate group
hydroxyl group
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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)
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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 17.6 31.6 1.05 1.00 35.2 64.8
Chicken 28.0 22.0 21.6 28.4 0.99 1.02 43.7 56.4
Rat 28.6 21.4 20.5 28.4 1.01 1.00 42.9 57.0
Human 29.3 20.7 20.0 30.0 0.98 1.04 40.7 59.3
Grasshopper
29.3 20.5 20.7 29.3 1.00 0.99 41.2 58.6
Sea Urchin
32.8 17.7 17.3 32.1 1.02 1.02 35.0 64.9
Wheat 27.3 22.7 22.8 27.1 1.01 1.00 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 1.05 1.01 51.7 48.3
http://en.wikipedia.org/wiki/Chargaff%27s_rules
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Rosalind Franklin
• 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.
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Courtesy of Cold Spring Harbor Laboratory Archives
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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
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Double helix
• 2 strands are polymers of nucleotides
• Phosphodiester backbone – repeating sugar and phosphate units joined by phosphodiester bonds
• Wrap around 1 axis• Antiparallel
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5´
3
O
O
O
O
4
5
1
3 2
4
5
1
3 2
4
5
1
3 2
4
5
1
3 2
5-carbon sugar
Nitrogenous base
Phosphodiester bond
Phosphate group
OH
P
P
P
P
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Antiparallel Nature of DNA
25http://academic.brooklyn.cuny.edu/biology/bio4fv/page/molecular%20biology/dsDNA.jpg
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C
C
C
G
G
G
G
G
T
T
T
T
A
A
A
2nm5′ 3′
3.4nm
0.34nm
Minorgroove
Majorgroove
5′3′
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Majorgroove
Minorgroove
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• Complementarity of bases
• A forms 2 hydrogen bonds with T
• G forms 3 hydrogen bonds with C
• Gives consistent diameter
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A
H
Sugar
Sugar
Sugar
Sugar
T
G C
N
H
N O
H
CH3
H
HN
N N H N
N
N
H
H
H
N O H
H
H N
N H
N
N HN N
Hydrogenbond
Hydrogenbond
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Question 7
Chargaff’s rule states that
a. DNA strands are in antiparallel alignment
b. G matches with C, and T matches with A
c. The DNA molecule is a double helix
d. DNA transformation occurs when an organism incorporates DNA from the environment
e. The nuclei of cells are totipotent
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Question 11
If a strand of DNA had the sequence 5’- AGTCCA- 3’, which of the following would be the complementary DNA strand?
a. 3’- CATGGT- 5’
b. 3’- TCAGGT- 5’
c. 3’- TCAAAU- 5’
d. 3’- AGTCCA- 5’
e. 3’- GGTTCA- 5’
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Question 12
If a DNA molecule contains 40% thymine, how much guanine will it contain?
a. 10%
b. 20%
c. 30%
d. 40%
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Learning Objectives
14.3 Basic Characteristics of DNA Replication
•How did Meselson and Stahl figure out DNA replication?
•What is required for DNA replication?
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DNA Replication
3 possible models1. Conservative model
2. Semiconservative model
3. Dispersive model
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Conservative
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Conservative Semiconservative
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Conservative Semiconservative Dispersive
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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
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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
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Samples are centrifuged
E. coli
0
1
2
0 rounds 1 round 2 rounds
Bottom
15N medium
14N medium
E. coli cells grownin 15N medium
Cells shifted to14N medium andallowed to grow
DNA
Samples taken atthree time pointsand suspended incesium chloridesolution
Rounds ofreplication
Top
0 min0 rounds
20 min1 round
40 min2 rounds
From M. Meselson and F.W. Stahl/PNAS 44(1958):671
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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
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DNA Replication
• Requires 3 things
– Something to copy
• Parental DNA molecule
– Something to do the copying
• Enzymes
– Building blocks to make copy
• Nucleotide triphosphates
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• DNA replication includes
– Initiation – replication begins
– Elongation – new strands of DNA are synthesized by DNA polymerase
– Termination – replication is terminated
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P
P
P
P
P
P
P
P
P
P
P
Pyrophosphate
3′
3′
5′
5′
New StrandTemplate Strand
O
HO
OH
O
O
O
O
O
O
O
O
O
O
C
C
T
T
T
A
A
A
G
G
A
P
P
P
P
P
PP P
P P
P
P
P
P
3′
3′
5′
5′
New StrandTemplate Strand
O
HO
OH
OH
O
O
O
O
O
O
O
O
O
C
C
T
T
A
A
A
G
G
A
Sugar–phosphatebackbone
DNA polymerase III
TO
P
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• 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
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5
3
5
5 5 3
3
RNA polymerase makes primer DNA polymerase extends primer
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Question 5
The Meselson-Stahl experiment demonstrated that DNA replication is
a. Conservative
b. Semi-conservative
c. Disruptive
d. Differentiated
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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?
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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
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Replisome
Replisome
TerminationOrigin
Termination
Origin
Origin Origin
Origin
Termination TerminationTermination
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• 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
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• 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 replication49
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Supercoiling
Replisomes
No Supercoiling
Replisomes
DNA gyrase
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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
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RNA primer
Open helixand replicate
First RNA primer
Open helix andreplicate further
Lagging strand(discontinuous)
Second RNA primer
Leading strand(continuous)
RNA primer
5′
3′
3′
5′
5′
3′
3′
5′
5′
3′
5′
3′
5′
3′
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• Partial opening of helix forms replication fork
• DNA primase – RNA polymerase that makes RNA primer
– RNA will be removed and replaced with DNA
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Leading-strand synthesis
– Single priming event
– Strand extended by DNA pol III
• Processivity – subunit forms “sliding clamp” to keep it attached
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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
a-b: From Biochemistry by Stryer. © 1975, 1981, 1988, 1995 by Lupert Stryer. Used with permission of W.H. Freeman and Company
a. b.
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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
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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
5′
3′
Primase
RNA primer
Okazaki fragmentmade by DNApolymerase III
Leading strand(continuous)
DNA polymerase I
Lagging strand(discontinuous)
DNA ligase
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Replication forkCopyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
5 3
New bases β clamp (sliding clamp) Leading strand
Single-strand bindingproteins (SSB)
DNA gyrase
ParentDNA
PrimaseHelicase
3 5
Clamp loader
Open β clamp
Lagging strandOkazaki fragment
5 3
DNA ligase
polymerase IDNA
RNA primer
New bases
polymerase IIIDNA
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Question 3
Where are the Okazaki fragments found?
a. On the lagging strand
b. On the leading strand
c. At the replication origin
d. In the cytoplasm
e. On both strands
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Question 4
Name the enzyme that links Okazaki fragments
a. DNA polymerase I
b. RNA primase
c. DNA ligase
d. Helicase
e. ATP synthase
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Question 6
During DNA replication, what enzyme is responsible for untwisting the DNA helix?
a. DNA polymerase I
b. RNA primase
c. DNA ligase
d. DNA polymerase III
e. Helicase
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Question 9
Why does replication proceed in opposite directions on the leading and lagging strands?
a. The polymerase enzyme needs a primer
b. DNA polymerase III can only add to the 3´ end of a strand
c. The Okazaki fragments are only on the leading strands
d. The parent strands are oriented in the same direction
e. Helicase only allows for replication of one strand at a time
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Question 15
What would be the immediate consequence of a non-functional primase enzyme?
a. The strands would break due to the torsional strain from rapid untwisting
b. The helix could be opened
c. The DNA polymerase III enzyme would have nothing to bind to
d. The Okazaki fragments would not be linked together
e. The single DNA strands could not be held open
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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?63
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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
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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
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Leading strand (no problem)Lagging strand (problem at the end)
Last primer
Replication first round
Shortened template
Origin
5´
3´
3´
5´
3´
5´
5´
3´
3´
5´5´
3´
5´
3´
5´
3´
3´
5´
3´
5´
Removed primercannot be replaced
Leadingstrand
Laggingstrand
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Primer removal
Replication second round
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• 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
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G
GGGGG
T T
TTT TG
TT
GGG
GG
T
TTT
CCCCC AAAA
CCCCC AAAA
Telomere extendedby telomerase
Template RNA ispart of enzyme
Telomerase
Now readyto synthesizenext repeat
5 ́
3 ́
5 ́
3 ́
5 ́
3 ́
Synthesis by telomerase
Telomerase moves andcontinues to extend telomere
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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
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DNA Repair
Falls into 2 general categories1. Specific repair
– Targets a single kind of lesion in DNA and repairs only that damage
2. Nonspecific– Use a single mechanism to repair multiple
kinds of lesions in DNA
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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
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T
A
T
A
A A
A A
TA
TA
TT
T T
Thymine dimercleaved
Photolyase
Helix distorted bythymine dimer
Thymine dimer
DNA with adjacent thymines
UV light
Visible light
Photolyase bindsto damaged DNA
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Excision repair
• Nonspecific repair
• Damaged region is removed and replaced by DNA synthesis
• 3 steps1. Recognition of damage
2. Removal of the damaged region
3. Resynthesis using the information on the undamaged strand as a template
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Damaged or incorrect base
Uvr A,B,C complexbinds damaged DNA
DNA polymerase
Excision of damaged strand
Resynthesis by DNA polymerase
Excision repair enzymes recognize damaged DNA
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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?
a. Chromosome replication would be terminated
b. Okazaki fragments would not be linked together
c. Chromosomes would shorten during each division
d. The leading strand would become the lagging strand
e. The cells would become cancerous
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Question 10
Mutations can be caused by copying mistakes, and by exposure to chemicals or electromagnetic radiation.
a. This is True
b. This is False