Biochemistry 441 Lecture 12 Ted Young March 6, 2000

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Arts and science are not cast into a mould, but
are formed and perfected by degrees, by often
handling and polishing, as bears leisurely lick
their cubs into form. Michel de Montaigne
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Biochemistry 441 Lecture 7Ted YoungJanuary 20,
2006

• Topic DNA replication enzymes

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Six enzymes to know-substrates, products,
co-factors, and function in DNA metabolism

• DNA polymerase
• b-clamp/processivity factor
• DNA ligase
• DNA helicase
• Single-strand binding protein
• Telomerase

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5- to 3-polymerization of dNTPs by DNA
polymerase
Template
4. Hydrolysis of dNTP and PPi are the driving
forces for polymerization
1. Specific base-pairing
5. Anti- parallel chain growth
2. 3-OH on primer
3. 5-dNTP

• All DNA and RNA polymerases synthesize polymers
  in this manner only the substrates and enzymes
  differ.

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DNA pol I
E. coli

• Genetic evidence that pol I is not the DNA
  replicating enzyme in bacteria

mutagenesis
make crude cell-free extracts from
2000 survivors (most have a mutation in some
gene)
Measure DNA poly- merase activity in crude cell
extracts
Colony 1856 no DNA polymerase activity! But it
grows normally.
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Pol I is important for repair of damage to DNA
and for primer excision

• Pol I mutants are more sensitive to DNA damaging
  agents than wild-type cells.

Wild-type
survival
Pol I-
UV dose
Pol I 5-3 exonuclease activity removes RNA
primers from Okazaki fragments, an essential
function that is present in the mutant.
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Comparison of Pol I and Pol III

• Pol I Pol IIIa
• Mass (kDa) 103 130
• Molecules/cell 400 10-20
• Turnover number (ntds/min/molecule) 600 9000
  
• Essential gene Yes Yes
• Polymerization 5-3
• Exonuclease 3-5 -
• Exonuclease 5-3 -

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Pol III holoenzyme

• Subunit Mass (kDa) Function
• a 130 polymerization
• e 27.5 3-5 exonuclease (editing)
• q 10 unknown
• t 71 unknown
• g, d, d 45.5, 35, g complex clamp loader
• c,y 33, 15, 12 with ATP
• b 40.5 clamp (processivity factor)
• Total mass 420 kDa

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Finding genes/proteins involved in DNA replication
Biochemical approach 1. Devise an assay (what
function are you looking for?) 2. Break open
cells and apply assay 3. Demonstrate that
the protein identified plays an important role in
vivo.

• Genetic approach-
• 1. isolate temperature-sensitive (ts) mutants.
• 2. screen for DNA synthesis at the restrictive
  temperature
• 3. characterize mutants defective in DNA
  synthesis
• 4. Identify biochemical defect

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The b-clamp-a processivity factor for polIII
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DNA ligase

ligase

• DNA ligase becomes activated during the reaction,
  conserving the energy of ATP or NAD as a high
  energy enzyme-adenylate intermediate.

Nicked DNA Substrate.

DNA-nick-adenylate.
The reaction requires an activated- phosphorylated
-intermediate, an enzyme adenylate complex, and a
5phosphate on the DNA. RNA ligases also exist.

ligase
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DNA helicase
DNA helicases are directional

• DNA helicases unwind duplex DNA in the presence
  of ATP, producing single stranded DNA, AMP, and
  pyrophosphate.

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ATP
5
AMP PPi
What, if anything, keeps the two complementary
strands from reforming the duplex?
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DNA single-strand binding proteins

• DNA single-strand binding proteins keep
  complementary single-strands of DNA from
  re-annealing. No energy is required.

denature
ss binding protein-
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Cooperative binding of ss DNA binding proteins
(SSBs)
Sigmoidal shape of the binding curve
indicates that binding of one molecule makes it
easier to bind the next, and so on...

• ss DNA binding proteins have a high degree of
  cooperativity binding of one protein makes it
  easier to bind a second protein.

Binding
protein
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Cooperative binding (cont)

• Cooperativity requires a conformational change in
  the shape of the protein that only occurs upon
  DNA binding.

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Helicase and SSB at the replication fork

• What type of helicase is this? 5 to 3 or 3 to
  5?

(Helicase)
DNA gyrase
Is this cartoon consistent with the EM picture we
saw earlier? It showed asymmetric
single-stranded regions at the replication forks.
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Initiation of DNA replication

• Model for initiation of leading strand synthesis
  requires special origin binding proteins and RNA
  polymerase. This process occurs only once per
  chromosomal replication in bacteria, but multiple
  times in eukaryotes.

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Enzymes at the replication fork
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Termination of replication

• Circular chromosomes decatenation by
  topoisomerase(s) is required.
• Linear chromosomes a special enzyme called
  telomerase replicates the linear ends. Why is
  this necessary? Because RNA primers are at the 5
  ends. After their removal, no DNA polymerase can
  add dNMP to these ends.

The end problem in replication of linear
chromosomes
replication
5 3
Rt
RNA
Chromosomal termini, or telomeres
Left
RNA
5 ends of last Okazaki fragments
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Replication of chromosome ends by telomerase
Note that the 3 end of the chromosome is not
base-paired to DNA

• Telomerase consists of an RNA template-telomerase
  RNA-and several proteins. The RNA acts as a
  template for the polymerase activity. The
  polymerase copies RNA into DNA, as does viral
  reverse transcriptase, extending the G-rich
  strand. The protein component of telomerase has
  homology to viral reverse transcriptases

Ch 26.3 Fig 26.35
Telomerase extends the single-stranded 3 end,
allowing Primase and DNA pol. to copy the
extended 3 end. The 3 end has redundant
sequence information so even if some is lost at
each replication, unique genetic information
isnt lost until all redundant sequence
information is lost.
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Summary

• DNA replication is very complex!
• Four distinct operations must occur
• initiation of the leading strand
• initiation of the lagging strand
• elongation
• termination
• Enzyme complexes are involved at each step of the
  process
• Termination of replication by telomerase requires
  a novel type of enzyme composed of both RNA and
  protein