1. List proteins (enzymes and other factors) involved in the process of DNA
replication in E. coli. Explain the role of each of these proteins in replication.
i. DnaA-initiation. Breaks the hydrogen bonds. The enzyme binds to 4 DnaA
boxes (9bp each). 10-20 DnaA proteins and 4 DnaA boxes form an initial complex.
N.B: Binding and opening occur at the 13bp sequence and requires ATP. Initiation occurs
only if DNA is negatively supercoiled.
ii. DnaB- helicase. They move along the dsDNA and separate the strands.
(requires ATP as it forms +ve supercoils infront of the enzyme)
iii. DnaC- Clamp ladder. DnaB requires DnaC to escort it to DnaA forming the
pre-priming complex. This is an active process and requires ATP to join DnaB and DnaC.
iv. SSB- Single strand binding proteins
They are small proteins that bind to ssDNA preventing pairing of separated strands.
They show cooperative binding i.e. binding of one molecule promotes the binding of the
The binding prevents internal pairing and double helix reformation. ssDNA is coated with
SSB to force the DNA into its extended conformation thus facilitating the action of DNAP.
v. DnaG – Primase
Forms the primers. DNAP can’t start replication on its own. It requires a 3’ OH group.
This is supplied by the primase. DnaG functions with DnaB to form the next priming site
vi. RNase H – breakdown of RNA
Removes the RNA primers when they are no longer needed. (except the last one)
vii. DNA Polymerase (I and III)
The replicate the DNA. III is the major. I is minor but was discovered first. II is involved
viii. DNA Ligase
They link fragments of the lagging strand together by forming a phosphodiester bond
between the 3’ OH terminus and the 5’ phosphorous terminus of the adjacent
They recognize and regulate supercoils. They introduce –ve supercoils around OriC
which are necessary for initiation of replication by DnaA. They also relax the +ve
supercoils formed by helicases and convert them to –ve coils (TII- Actively)
x. Dam methylase
Replication can’t start without methylation of the origin. C or A can be methylated.
A is methylated in GmATC and C in CmC(A/T)GG
Also, (A,B & C are trans factors)
xi. DNAPIII adds NA
xii. DNAPII proofreads.
xiii. DNAPI removes the last primer at the 5’ end and fills in the gap. 2. What is meant by replication being bidirectional? Semiconservative?
Continuous and discontinuous?
Replication is bidirectional as the anti parallel nature of the strands cause the polymerases to move in
opposite directions, the leading strand to be continuous while the lagging is discontinuous (DNAP can
only add NA in 5’ – 3’ direction and anti-parallel characteristic cause bidirectional NA addition).
Semi-conservative: Helicase unwinds the dsDNA into two ssDNA, each of which is used
as a template to synthesize the new DNA strand. Therefore, each new dsDNA molecule
is 50% old and 50% newly replicated.
Continuous: Replication of the leading strand starts at the RNA primer and reads from
the 3’ to the 5’ end continuously.
Discontinuous: Replication of the lagging strand starts at multiple sites with the help of
multiple primers. This type of replication is discontinuous as gaps are made in the
process that are later ligated with ligase.
3. Contrast the role of DNA polymerase I and III in E. coli DNA replication.
DNA pol III is the main enzyme involved in formation of the leading and lagging strands
(5’ to 3’). DNA pol I has a ribonucleotide exonuclease activity (5’ to 3’) and therefore,
removes primers and replaces them with DNA fragments.
Both enzymes have a 3’ to 5’ proofreading activity.
Pol III is more processive (250-1000 nt/sec) than pol I (20-100 nt/binding)
4. Which subunit of DNA polymerase III provides processivity? Which protein
complex loads this subunit onto the DNA?
The Beta subunit provides processivity. It associates with core at 3’ end of the growing
strand leading to an increase in processivity. (the clamps that activate DNAP activity)
The Gamma complex loads/unloads the Beta subunit using ATP in the loading.
5. How can discontinuos synthesis of the lagging strand keep up with
continuous synthesis of the leading strand?
Discontinuous keeps up as the DNAP cores of both strands are connected via t branches of gamma
subunit, while leading strand is worked primarily by only Pol III lagging has multiple enzymes like Pol
I, ligase, RNase H and Pol III constantly working together to speed it up, beta clamps make sure that
the strands cannot go too fast.
6. Why is decatenation required after replication of circular DNAs?
When circular DNA molecules are replicated, the product (two rings) are linked together
and are said to be catenated. In the catenated form, part of the strands are yet to be
replicated. For that to occur, the strands are denatured then decatenated to form two
separate daughter chromosomes. DNA replication can be completed before or after the
7. Why do eukaryotes need telomeres but prokaryotes do not?
Prokaryotes have circular DNA, therefore, when the primer is removed from the lagging
strand, the 3’ OH of the adjacent nucleotide provides the requirements for Pol I to fill the gap. This is not true of Eukaryotes as they have linear DNA and therefore no 3’ OH
8. What would be the components necessary to make DNA in vitro by using
DNA polymerase I.
i. DNA fragments
iii. RNA primer
iv. Template strand
vi. Mg2+ as a coenzyme
9. What properties would you expect