LECTURE 30 - FINAL

Question 1

What are some DNA replication facts?

Answer 1

– DNA is copied before the cell divides during the cell cycle

– DNA is copied during the S or synthesis phase of interphase

– DNA is copied in a semi-conservative fashion

– DNA replication takes place in the S phase

• • four phases to cell cycle:
    
    1. G1 phase (cell grows physically larger, copies organelles, and makes the molecular building blocks it will need in later steps)
    2. S phase (DNA is copied in a semi-conservative fashion
    3. G2 phase (cell grows more, makes proteins and organelles, and begins prepping for mitosis
    4. M phase (the cell divides its copied DNA and cytoplasm to make 2 new cells)

Question 2

Describe DNA replication in E. Coli.

Answer 2

– Replication starts at Origin of Rep., called OriC

– One single OriC site

– Two replication forks move in opposite directions, synthesizing DNA at 1000 bp/sec. Takes about 40’

– They meet at the opposite end of circle

– They are always synthesizing new DNA in a 5’ to 3’ direction

– chromosome is one circular DNA molecule and replication occurs in bidirectional fashion from one fixed origin

Question 3

Describe a brief overview of Eukaryotic DNA replication.

Answer 3

– a little different than bacterial replication, this is mainly due to the fact that eukaryotic DNA is much larger and have linear chromosomes

– have multiple origins of replication on each chromosome such that replication can be complete during S phase

– replicated more slowly

– two strands open forming Replication Forks (Y-shaped region)

– New strands grow at the forks

Question 4

What are some of the enzymes involved in DNA replication?

Answer 4

– Enzyme Helicase unwinds and separates the 2 DNA strands by breaking the weak hydrogen bonds

– Single-Stand Binding Proteins attach and keep the 2 DNA strands separated and untwisted (they stabilize the single strand of DNA)

– Enzyme Topoisomerase attaches to the 2 forks of the bubble to relieve stress on the DNA molecule as it separates

– Before new DNA strands can form, there must be RNA primers present to start the addition of new nucleotides

– Primase is the enzyme that synthesizes the RNA primer

– DNA polymerase can only add nucelotides to the 3’ end of the template DNA

– The enzyme Logase joins the Okazaki fragments together to make one strand

Question 5

Describe the DNA polymerase reaction.

Answer 5

– DNA polymerases carry out specialized functions (Pol I, II, and III) and require Mg2+

– In prokaryotes, Pol III carries out most synthesis, but Pol I completes lagging strand synthesis

– substrates are dATP, dCTP, dGTP, and dTTP –> Pol III adds these depending on template sequence

– must have a DNA template strand

– dNTPs are added to a primer

– chain elongation occurs 5’ to 3’

– new nucleotide addition are determined by template strand of DNA

– DNA polymerase reaction involves nucleophilic attack by the 3’ OH of the primer terminus on the alpha-phosphate of the dNTP substrate, leading to phosphodiester bond formation

Question 6

Describe the structure of DNA Polymerase I.

Answer 6

– resembles a right hand

– the structure will only close on the correct base-pair and catalyze the reaction conformation change

– the structure allows the enzyme to hold on to the DNA repeat the catalytic cycle

– in addition to polymerase activity it also has two nuclease activities (3’ exonuclease and 5’ exonuclease)

– 3’ exonuclease serves as a “proofreading” function to make sure the DNA template is copied correctly

– 5’ exonuclease activity functions both in replication and in DNA repair

Question 7

What is the mechanism of DNA synthesis?

Answer 7

– once DNA primer binds to DNA template, Pol III beings adding complementary dNTPs

– 3’ OH of the primer need forms a phosphodiester bond with the alpha phosphate of the dNTP

– Pyrophosphate and water exit the reaction

– Hydrolysis of pyrophosphate drives the reaction

Question 8

describe the DNA polymerase active site.

Answer 8

– metal ions, Mg2+, are used to shield the negative charges of the deoxynucleotide 5’ – triphosphate (dNTPS) and activate the reactive 3’ OH

—> this makes the phosphodiester bond possible (in order for a phosphodiester bond to be formed from the reactive 3’ OH of one dNTP and the alpha phosphate from another dNTP, the charges need to be shielded)

– Mg2+ stabilizes these transition states which then enables the daughter strand to be synthesized in the 5’ –> 3’ direction

Question 9

Describe the proofreading exonuclease activity in DNA Polymerase.

Answer 9

– When polymerase is synthesizing DNA, every once in a while, an incorrect base is improperly paired with the template strand. To correct this mismatch, Polymerase has proofreading exonuclease activity

– the incorrect base pair (still attached to the newly synthesized DNA) moves from the polymerase active site to a second enzymatic site on the protein that contains 3’ to 5’ exonuclease activity

– Once the mismatch base has been removed, the DNA can then move back to the polymerase site and DNA synthesis resumes

Question 10

Describe the process of DNA synthesis initiation.

Answer 10

– initiated by RNA primer

– primer synthesized by Primase

– Primase is an RNA polymerase, which unlike DNA polymerases can begin synthesis without a primer

– Primase beings the DNA replication process by synthesizing a short (5-10 base) RNA molecule complementary to the template

– DNA polymerase can use the 3’ OH of the RNA primer to continue DNA synthesis

– here we’re in the S phase

– topoisomerase helps to prevent entire strand from unwinding completely

– polymerase moving in 5 to 3’ direction

Question 11

Why would we need RNA polymerase?

Answer 11

– DNA polymerase can’t come in bc there’s no hydroxyl group for DNA polymerase to do its nucleophilic attack

Question 12

Describe DNA synthesis at the replication fork.

Answer 12

– New strand is synthesized 5’ to 3’

– left-hand template can be synthesized easily, called Leading Strand –> it will go in the direction of fork

– Right-hand side has a problem. DNA is opening at the 5’ end of the synthesized strand. The side is called Lagging Strand

– *Note: there are multiple replication forks in Euk

Question 13

Describe the concept of opposite directions of DNA synthesis at a replication fork.

Answer 13

– DNA unwinding creates a fork structure that “moves” as more DNA is unwound

– Both DNA strands act as templates for DNA synthesis

– Because the DNA strands are antiparallel and DNA synthesis always occurs 5’ to 3’, the two strands are replicated in opposite directions relative to the movement of the replication fork

Question 14

What doe the multiple DNA polymerases do in prokaryotes?

Answer 14

– Pol I is involved in synthesis of lagging strand and repair

• —–> sliding clamp holds DNA polymerase in place
• —–> DNA polymerase III in 5’ –> 3’ direction synthesizing leading strand

– Pol II, IV, and V are for repair under unique conditions

– Pol III is primarily responsible for new synthesis

– in prokaryotes, Pol III carries out most synthesis, but Pol I completes lagging strand synthesis

Question 15

How is the lagging strand problem solved?

Answer 15

– Primase lays down primers as DNA is unwound. These are extended by DNA polymerase III

– Primase lays down lagging strand primer. DNA Pol III synthesizes from primer, DNA Pol I replaces RNA primer with DNA (nick translation). –> end up replicating in sort of this fragmented way

– DNA ligase joins Okazaki fragments –> allow nucleophilic attack so that phosphodiester bond is created –> seals the gap

– only 1 primer on leading strand

Question 16

Describe the processing Okazaki fragments for continuous DNA.

Answer 16

– When Pol III reaches the next Okazaki fragment, it disengages, leaves a “nick,” a break in the phosphate backbone

– RNA primer is removed with 5’ to 3’ exonuclease activity of Pol I. Pol I removes last base and synthesizes DNA 5’ to 3’ to replace RNA primer with DNA. The final nick is sealed by DNA ligase.

– cuts then adds a base

– each Okazaki fragment requires its own RNA primer

– Okazaki fragments –> Prokaryotes 1000 - 2000bp; Eukaryotes 100 - 200 bp

Question 17

Describe the leading and lagging strand at DNA synthesis at a Replication Fork.

Answer 17

– the leading strand is replicated in the same direction as fork as one long continuous strand of nascent DNA

– the lagging strand is replicated in opposite direction, and is synthesized discontinuously as Okazaki fragments ( a few hundred to a few thousand nucleotides in length)

Question 18

Describe the mechanism of Okazaki fragment synthesis.

Answer 18

– All DNA polymerases synthesize DNA in a 5’ –> 3’ direction, they attach incoming nucleotides to the 3’ end of the growing daughter strand

– lagging strand has to loop around in order to come back through the second polymerase in correct direction

– lagging strand is looped such that it is oriented with the same polarity as the leading strand synthesis

– loop gets re-formed and enlarged as new RNA primers get synthesized and elongated by DNA Pol

– The new strand will meet the previous Okazaki fragment at its 5’ end

– The Pol then detaches and a new loop forms

Question 19

What protein machinery is the replication fork? (Prokaryotic)

Answer 19

– Helicase unwinds (ATP), Primase is attached to it, makes primers, called Primasome

– Single-stranded binding protein keep DNA strands unwound

– In E. Coli, there are 2 Pol III molecules, one for leading strand and one for lagging strand

– The two sides have the same core enzyme, but lagging strand has extra subunits (not shown)

– Together called the holoenzyme (DNA polymerase + extra subunits)

– DNA Pol I – fills in the necessary nucleotides between the Okazaki fragments and proofread for any mistakes as it moves alongs

Question 20

How does the Pol III Holoenzyme replicate leading and lagging strands (Prokaryote)?

Answer 20

– Primary enzyme involved in prokaryotic DNA replication

– Tau proteins link Pol III subunits to the clamp loader complex

– The polymerases handle the replication of leading and lagging in concert

– Tau also links helicase to polymerases

Question 21

T or F, “sliding clamp” attaches to DNA Pol III to the template DNA

Answer 21

True
– Multi-subunit protein structure surrounds the DNA (in the middle)

– in Euk, it’s a trimer (PCNA), but in E. Coli it’s a dimer (Beta protein)

– the proteins are different and have little sequence homology

– makes the enzyme highly processive

Question 22

How do sliding clamps increase the processivity of DNA polymerases?

Answer 22

– component of the DNA pol III holoenzyme complex

– proteins that encircle the DNA (sliding clamps) bind to a polymerase and keep it associated with the DNA template strand

– Beta and PCNA are the prokaryotic and eukaryotic clamps, respectively

– Sliding clamps are multi-subunit proteins that are opened up by a “clamp-loader” protein complex between subunits to load them onto the DNA (requires ATP binding and hydrolysis to load the clamp)

Question 23

Describe the sliding clamp loading in E. coli.

Answer 23

– exchange of ADP for ATP, increases affinity of clamp loader for closed sliding clamp

– Binding causes opening of clamp. Binding to template-primer, causes ATP hydrolysis and clamp closing around DNA and ejection of clamp loader

– Pol III binds clamp and DNA synthesis begins. Only has to occur once on leading strand, but must occur many times on the lagging strand

Question 24

Describe how the single-stranded binding protein keeps the DNA unwound for DNA synthesis.

Answer 24

– For synthesis, DNA must be single-stranded (no intramolecular base-pairing)

– binds near the replication fork to stabilize the single strand

– SSB binds the unwound DNA (sequence-independent) immediately in a cooperative manner (binding of one protein facilitates the next, and so on)

– SSB binding prevents intramolecular base-pairing and keeps the DNA in an extended conformation with the bases exposed to facilitate synthesis

Question 25

Describe how the DNA Helicases unwind DNA for replication.

Answer 25

– it functions to separate double-stranded DNA into single strands allowing each strand to be copied

– DNA helicases use the energy of ATP (lots of it) to unwind DNA helix

– These enzymes typically are hexamers that form a donut shape that can encircle the DNA, keeping them bound, which makes them processive

– The well known helicase bind around one DNA strand and move in only one direction (they have polarity)

Question 26

Describe the coordination of leading and lagging strand synthesis at the replication fork.

Answer 26

– Helicase unwinds DNA

– SSB protein keep template strands apart

– Primase lays down primer on lagging strand (primer for leading strand not shown)

– Sliding clamp loaded onto 3’ end of primer

– DNA Pol III begins lagging strand synthesis. DNA Pol III releases when the next RNA primer is reached. Pol I will further synthesize from this 3’ end and replace next RNA primer

– DNA ligase will attach the two Okazaki fragments on lagging strand

Question 27

T or F, Eukaryotic DNA polymerase beta is most similar to E coli DNA Pol I

Answer 27

True

Question 28

T or F, Eukaryotic DNA polymerase omega (Pol omega) and DNA polymerase epsilon (Pol epsilon) are replicative DNA polymerases at the replication fork

Answer 28

True