BCEM 393 - DNA Replication

Fill in the blanks: the key enzymes in DNA replication are called __________, which promote the formation of the ______________ joining units of the DNA backbone.

DNA polymerases, phosphodiester linkages

Fill in the blanks: E. coli has five DNA polymerases, but two are best understood; _____________ and ____________.

Polymerase I, polymerase III

Fill in the blanks: in E. coli, polymerase I is responsible for ___________ and _____________.

Primer removal, DNA repair

Fill in the blanks: in E. coli, polymerase III is responsible for ____________.

Replication

Fill in the blank: binding of the correct incoming dNTP induces a _______________ in DNA polymerase.

Conformational change (induced fit)

Fill in the blanks: DNA polymerase catalyzes the strand-elongation reaction in the following steps:

1. The reaction requires _____, _____, _____, and _____, as well as the _____ ion.

2. The new DNA strand is assembled directly on the _________________.

3. DNA polymerases require a ___________ with a free ________ group to begin synthesis. Elongation of the DNA strand proceeds in the ________ direction.

4. Many ___________ are able to correct mistakes in DNA by removing mismatched nucleotides.

dATP, dGTP, dCTP, TTP, Mg²⁺, pre-existing DNA template, primer, 3’-OH, 5’ to 3’, DNA polymerases

True or False: DNA elongation is in the 3’ to 5’ direction.

False; DNA elongation is in the 5’ to 3’ direction

Fill in the blank: DNA polymerases can correct mistakes using ___________ activity (proofreading).

3’ to 5’ exonuclease

Multiple Choice: you are setting up a PCR reaction, and you have added the DNA template, polymerase, primers, and buffer to the reaction tube. You realize after starting the reaction that you added ddNTPs instead of dNTPs. Is this a problem? Why?

a) It will still work fine, as they are very similar

b) ddNTPs will not base pair with the complementary base in the template

c) ddNTPs lack a 3’ hydroxyl so they will not be recognized by the polymerase and be incorporated into the growing strand, so synthesis will stop

d) ddNTPs lack a 3’ hydroxyl and will still be recognized by the polymerase and incorporated into the growing strand, terminating synthesis

d)

Fill in the blanks: when mistakes are made during DNA replication, the _______________ removes incorrect nucleotides from the ___ end of the growing strand by ___________. A mismatch results in a “stall” which gives additional time for the incorrect region to _____________ the ___________ where it is removed.

3’ to 5’ exonuclease domain, 3’, hydrolysis, flop into, exonuclease active site

Fill in the blank: in E. coli, the ____________ is a region of DNA with special features and is the start site of DNA replication.

Origin of replication (oriC)

Fill in the blanks: in DNA replication in E. coli the __________ protein binds to the origin of replication to initiate the prepriming complex. The _____________ protein, a helicase, separates DNA. ____________ then bind to the newly generated single-stranded regions, preventing the complementary strands from reforming the double helix. This process results in a structure called the _____________.

DnaA, DnaB, single strand binding (SSB) proteins, prepriming complex

Fill in the blanks: hydrolysis of ATP by ____________ causes “ratcheting” of the subunits of the hexamer, pulling __________ through the center.

Helicases, single stranded DNA

Fill in the blank: __________ is an RNA polymerase that makes a short primer.

Primase

Fill in the blanks: ___________ simultaneously synthesizes the leading and lagging strands at the replication fork. The ____________ improves processivity.

DNA polymerase III, sliding clamp (B₂)

Multiple Choice: a mutation impairs the 5’ to 3’ exonuclease activity of DNA polymerase I. What effect would this have on DNA replication in E. coli?

a) The polymerase wouldn’t be able to proofread so there could be errors in the replicated DNA

b) RNA primers will remain on the lagging strand after DNA replication

c) Okazaki fragments could not be synthesized

d) All of the above

b)

Fill in the blanks: in E. coli, the leading strand is made continuously by __________ in the ______ direction.

DNA polymerase III, 5’ to 3’

Fill in the blanks: in E. coli, the lagging strand is ___________, then starting from an RNA primer, ___________ adds ~1000 nucleotides in the ______ direction. The sliding clamp releases, a new loop is formed, and a sliding clamp is added. ____________ adds RNA primers and ___________ makes a new Okazaki fragment. ____________ fills gaps between fragments and removes RNA primer with _______ exonuclease activity. Finally, ____________ seals the fragments.

Looped out, DNA polymerase III, 5’ to 3’, primase, DNA polymerase III, DNA polymerase I, 5’ to 3’, ligase

Fill in the blanks: the mode of replication where the lagging strand is looped out, has about 1000 nucleotides added to it, and a new loop is formed with the continued synthesis of short stretches of RNA primer and Okazaki fragments is termed the ____________ because the size of the loop lengthens and shortens like a __________.

Trombone model, slide on a trombone

Fill in the blank: ________ plays a key role in replication by serving as a sliding DNA clamp.

B₂

Fill in the blanks: ligase catalyzes the formation of a _____________ between the __________ group and the __________ group of two fragments.

Phosphodiester linkage, 3’-hydroxyl, 5’-phosphate

Fill in the blanks: before it can bind fragments together, ligase must go through an initial charging step. In this step, catalytic __________ attacks the alpha-phosphate of ___________, generating ___________ adduct and releasing ____________. The adenylated ligase then transfers ___________ from _____________ to the 5’ phosphate at the nicked backbone. Now, the 3’OH at the nick site attacks adenylated 5’ phosphate, making a _____________ and releasing ____________.

Lysine, ATP, lysyl-AMP, pyrophosphate, AMP, lysine, phosphodiester bond, AMP

True or False: in E. coli, ligase, DNA polymerase I, and the sliding clamp are not required for continuous DNA synthesis (on the leading strand).

False; ligase and DNA polymerase I are not required for continuous synthesis, but the sliding clamp increases processivity on the leading strand