3. • Replication has to be extremely accurate:
– One error/million bp leads to 6400 mistakes
every time a cell divides, which would be
catastrophic.
• Replication also takes place at high speed.
– E. coli replicates its DNA at a rate of 1000
nucleotides/second.
12.1 Genetic Information Must Be
Accurately Copied Every Time a Cell
Divides
4. 12.2 All DNA Replication Takes Place in a
Semiconservative Manner
• Proposed DNA Replication Models:
– Conservative replication model
– Dispersive replication model
– Semiconservative replication
5.
6. • Meselson and Stahl’s Experiment:
– Two isotopes of nitrogen:
• 14N common form; 15N rare heavy form
• E.coli were grown in a 15N media first, then
transferred to 14N media
• Cultured E.coli were subjected to equilibrium
density gradient centrifugation
12.2 All DNA Replication Takes Place in a
Semiconservative Manner
7.
8.
9. Concept Check 1
How many bands of DNA would be expected in
Meselson and Stahl’s experiment after two rounds
of conservative replication?
10. Concept Check 1
How many bands of DNA would be expected in
Meselson and Stahl’s experiment after two rounds
of conservative replication?
Two bands
11. • Modes of Replication
– Replicons: Units of replication.
• Replication origin
– Theta replication: circular DNA, E. coli; single
origin of replication forming a replication fork,
and it is usually a bidirectional replication.
– Rolling-circle replication: virus, F factor of
E.coli; single origin of replication.
12.2 All DNA Replication Takes Place in a
Semiconservative Manner
12.
13.
14. Linear eukaryotic replication
– Eukaryotic cells
– Thousands of origins
– A typical replicon: ~ 200,000-300,000 bp in
length.
– Fig. 12.6
12.2 All DNA Replication Takes Place in a
Semiconservative Manner
15.
16. Linear eukaryotic replication:
• Requirements of replication
– A template strand
– Raw material: nucleotides
– Enzymes and other proteins
12.2 All DNA Replication Takes Place in a
Semiconservative Manner
17.
18. Linear eukaryotic replication:
• Direction of Replication:
– DNA polymerase add nucleotides only to the 3 end of
growing strand.
– The replication can only go from 5 3
– Continuous and discontinuous replication
• Figs. 12.8 and 12.9
12.2 All DNA Replication Takes Place in a
Semiconservative Manner
19.
20. Linear eukaryotic replication:
• Direction of replication
• Leading strand: undergoes continuous replication.
• Lagging strand: undergoes discontinuous
replication.
• Okazaki fragment: the discontinuously synthesized
short DNA fragments forming the lagging strand.
12.2 All DNA Replication Takes Place in a
Semiconservative Manner
21.
22.
23. Concept Check 2
Discontinuous replication is a result of
which property of DNA?
a. Complementary bases
b. Antiparallel nucleotide strands
c. A charged phosphate group
d. Five-carbon sugar
24. Concept Check 2
Discontinuous replication is a result of
which property of DNA?
a. Complementary bases
b. Antiparallel nucleotide strands
c. A charged phosphate group
d. Five-carbon sugar
25. 12.3 Bacterial Replication Requires a
Large Number of Enzymes and Proteins
• Bacterial DNA Replication
– Initiation:
• 245 bp in the oriC. (single origin replicon)
• an initiation protein (DnaA in E.coli)
– Unwinding:
• Initiator protein
• DNA helicase
• Single-strand-binding proteins (SSBs)
• DNA gyrase (topoisomerase)
26.
27.
28. • Elongation:
– Primers: an existing group of RNA nucleotides
with a 3-OH group to which a new nucleotide
can be added. It is usually 10- 12 nucleotides
long.
– Primase: RNA polymerase
12.3 Bacterial Replication Requires a
Large Number of Enzymes and Proteins
29.
30. • Elongation: carried out by DNA polymerase III
• Removing RNA primer: DNA polymerase I
‒ Connecting nicks after RNA primers are removed:
DNA ligase
• Termination: when replication fork meets or
by termination protein.
12.3 Bacterial Replication Requires a
Large Number of Enzymes and Proteins
31.
32.
33.
34. The fidelity of DNA Replication
• Proofreading: DNA polymerase I: 3 5
exonuclease activity removes the incorrectly paired
nucleotide.
• Mismatch repair: corrects errors after replication is
complete.
12.3 Bacterial Replication Requires a
Large Number of Enzymes and Proteins
35. Concept Check 3
Which mechanism requires the ability to
distinguish between newly synthesized and
template strands of DNA?
a. Nucleotide selection
b. DNA proofreading
c. Mismatch repair
d. All of the above
36. Concept Check 3
Which mechanism requires the ability to
distinguish between newly synthesized and
template strands of DNA?
a. Nucleotide selection
b. DNA proofreading
c. Mismatch repair
d. All of the above
37. • Eukaryotic DNA Replication
– Autonomously replicating sequences (ARSs)
100–120 bps
– Origin-recognition complex (ORC) binds to
ARSs to initiate DNA replication.
– The licensing of DNA replication by the
replication licensing factor
• MCM: Minichromosome maintenance
– Eukaryotic DNA polymerase
12.4 Eukaryotic DNA Replication Is
Similar to Bacterial Replication but
Differs in Several Aspects
38.
39. • Eukaryotic DNA complexed to histone
proteins in nucleosomes
• Nucleosomes reassembled quickly following
replication
• Creation of nucleosomes requires:
‒ Disruption of original nucleosomes on the parental
DNA
‒ Redistribution of preexisting histones on the new
DNA
‒ The addition of newly synthesized histones to
complete the formation of new nucleosomes
12.4 Eukaryotic DNA Replication Is
Similar to Bacterial Replication but
Differs in Several Aspects
40. • The location of DNA replication within the
nucleus: DNA polymerase is fixed in location
and template RNA is threaded through it.
• Replication at the ends of chromosomes:
‒ Telomeres and telomerase.
‒ Fig. 12.18
‒ Fig. 12.19
12.4 Eukaryotic DNA Replication Is
Similar to Bacterial Replication but
Differs in Several Aspects
41.
42.
43. Concept Check 4
What would be the result if an organism’s
telomerase were mutated and
nonfunctional?
a. No DNA replication would take place.
b. The DNA polymerase enzyme would stall the
telomerase.
c. Chromosomes would shorten each generation.
d. RNA primers could not be removed.
44. Concept Check 4
What would be the result if an organism’s
telomerase were mutated and
nonfunctional?
a. No DNA replication would take place.
b. The DNA polymerase enzyme would stall the
telomerase.
c. Chromosomes would shorten each generation.
d. RNA primers could not be removed.
45. 12.5 Recombination Takes Place Through
the Breakage, Alignment, and Repair of
DNA Strands
• Homologous recombination: exchange is
between homologous DNA molecules
during crossing over.
• Holliday junction and single-strand break
• The double-strand break model of
recombination
46.
47.
48. 12.5 Recombination Takes Place Through
the Breakage, Alignment, and Repair of
DNA Strands
• Gene conversion:
– process of nonreciprocal genetic exchange
– produces abnormal ratios of gametes
• Gene conversion arises through
heteroduplex formation
