Replication, Repair and Recombination L4-6

Question 1

In E.coli what is the main polymerase used for replication?

Answer 1

DNA Polymerase III holoenzyme
DNA synthesis is catalysed by DNA pol. The reaction is driven by a large, favourable, free energy change caused by the release of pyrophosphate and its subsequent hydrolysis to 2 molecules of inorganic phosphate

Question 2

What is the core polymerase made out of?

Answer 2

alpha-the polymerase
epsilon- 3’-5’ EXOnuclease activity (proofreading)
sigma-stimulates epsilon

Question 3

What is the holoenzyme comprised of?

Answer 3

2/3 core pol III
2/3 flexible linker
clamp loader (3 gamma, 1 delta, 1 delta’ subunit) form PENTAMERIC ring shaped complex)
sliding clamp (2 beta subunits)
accessory proteins (dimer of psi and chi-important for the total activity of this complex)

Question 4

What DNA polymerase joins okazaki fragments?

Answer 4

DNA pol I

Question 5

How long are okazaki fragments in EUKs and PROKs

Answer 5

EUK-100/200

PROK 1000/2000

Question 6

What direction is dsDNA copied in?

Answer 6

Form the 3’ end
Add NTs to 3’ end
DNA pol therefore synthesises in a 5’-3’ direction

Question 7

What enzyme is needed to making the lagging strand continuous?

Answer 7

Repair enzyme RNAse H removes the RNA primer and DNA pol I replaces it with DNA
DNA ligase then joins the 3’ end of the new fragment to the 5’ end of the previous one

Question 8

Rates of copying DNA in E.coli and Euks

Answer 8

Ecoli=700bps per sec

EUKs=1000bps per sec

Question 9

Differences in Euks of the replisome

Answer 9

Different polymerases used for leading (pol epsilon) and lagging strand (pol delta)
A number of additional proteins are associated with the helicase in EUKs

Question 10

How many polymerases are there in E.coli and Euks?

Answer 10

E.coli=5
EUKs=12
The more complex the organism the more pols you need to carry out the various processes

Question 11

Differences b/w DNA pol III and DNA pol I

Answer 11

In Pol III the core enzyme (alpha) and the 3’-5’ exonuclease activity are separate subunits that interact via a protein-protein interface yet in pol I they are connected in a single polypeptide.

Question 12

What are the two roles of sliding clamps?

Answer 12

1) to increase processivity
2) act as a hub for interactions with multiple proteins at the replication fork or during repair (an assembly point)

In solution the sliding clamps are circular

Question 13

What is the difference between sliding clamps in E.coli and Euks?
Pentameric complex AAA+ ATPase

Answer 13

E.coli-homodimer (beta subunit of pol III)

Euks-trimer

Question 14

How many ATPS required for the clamp loader to load on the sliding clamp?

Answer 14

3 ATPs
The clamp loader has 5 subunits

The clamp loader complex is part of the sigma holoenzyme in E.coli

Question 15

What causes base pair specificity?

Answer 15

For the fingers of the DNA pol to close the enzyme needs to recognise the SHAPE of the base pair (not the H bonds)
The fingers move 40degrees to close over the bp. There are 3 catalytic residues, tyr lys and arg. Lys and arg coordinate the phosphates of the triphosphates.
The polymerases are dependent on dimeric cations like Mg to help stabilise the transition state and help the nuc attack of the 3’OH-also helps stabilise the triphosphate of the incoming dNTP.

The active site has evolved to avoid rNTPs (even tho there are 10x the conc of rNTPs than dNTPs) through having several amino acids present that fill a hole in the protein that prevents hydroxyl (OH) from entering. rNTP won’t fit correctly into the active site and you won’t get catalysis

Question 16

For what reason may an incorrect base pair be added by DNA polymerase?

Answer 16

If there is unusual chemistry in the bases such as an imino version or enol tautomer. imino cytosine base pairs with adenine
1 in 10^5 chance you get incorrect bases.

Question 17

What exonuclease activity does DNA pol I have?

Answer 17

5’-3’ and 3’ to 5’
This enzyme is less protective
Repairs okazaki fragments
Separate domain in the protein removes RNA primers ahead of the pol

Question 18

How do thymidine dimers arise in DNA?

Answer 18

UV light (DNA damage) cross-links b/w two ADJACENT thymidines

Question 19

How does replication deal with thymidine dimers?

Answer 19

DNA pol III falls off as cannot get past this damage and the translesion pols bind (Pol IV/V) and can copy over this damage since they have no proofreading/exonuclease activity and are not very processive. (exchange of pol III and the sliding clamp) Then the sliding clamp is reloaded along with DNA pol III. The cell has evolved to deal with the damage later.
Do not want the replication fork to stop-more serious than the damage in the DNA.

Question 20

Translesion pols IV/V

Answer 20

low processivity and no editing/proofreading
These have a much more flexible and open conformation, the fingers and thumb close over the substrate, catalysis occurs, even when incorrect base pairing is made. The catalytic pocket is more open.
Can easily get past damaged DNA. Low catalytic selectivity.

Question 21

In DNA damage MutS recognises the distortion of DNA and locks. It recruits MutH and MutL. MutH nicks the hemimethylated DNA. Then DNA is then processed as to whether the nick is upstream or downstream. What exonucleases are recruited?

Answer 21

If nick is upstream: Exo VII or recJ (5’-3’)

If nick is downstream: Exo I (3’-5’)

Question 22

How many replicons does E.coli have?

Answer 22

1 replicator (site of initiation) so 1 replicon
In E.coli the genome only has a single origin of replication
EUKs there are multiple replicators on each ch. so there are multiple replicons

Question 23

What are the initiators for replication called in E.coli and EUKs?

AAA+ ATPases

Answer 23

E.coli-DnaA (9 bp recognition seq)
protein makes site specific recognition of seqs within the replicator
EUKs-ORC (origin recognition complexes) In the case of Euks it seems the role of ATP is NOT required

Question 24

What enzyme methylates DNA in E.coli?

Answer 24

Dam methylase
Methylation control can turn off origins of replication
methylation occurs around the origin
hemi-methylated origins are resistant to methylation
when origins are fully methylated they are then competent for initiation
SeqA binds to GATC sites when they’re hemimethylated and coats the DNA around then initiation region preventing DnaA from binding to those sites and initiating replication

Question 25

SeqA (binds hemi-methylated DNA and prevents replication)

Answer 25

Binds weakly so that at times it will come off. Dynamic equilibrium b/w free and bound state so when it occasionally comes off Dam methylase comes in and methylate the other strand so that seqA can no longer bind.

Question 26

Eukaryotic replication is tightly controlled

Answer 26

Incomplete replication can lead to chromosome breakage/unequal segregation of chromosomes between the daughter cells.
If a region of DNA is not replicated the pair of chs. will stay linked by the fact rep has not finished yet.

Question 27

Steps in the formation of the pre-replicative complex

Answer 27

ORC binds to replicator (hexomeric complex) ORC recruits cdc6 and cdt1. This is followed by the recruitment of replicative helicase (Mcm2-7 heterohexamer)
The activation step is by cdk. Phosphorylation of the proteins at replicator sequence control whether or not you get formation of a pre-RC and whether you get initiation.
2 helicases bind to create 2 rep forks. DNA is ds going through the helicase at this stage=Pre-RC.

Question 28

Does the Pre-replicative complex need high or low cdk levels to begin replication?

Answer 28

When cdk levels are LOW the pre-RC forms (G1 phase-loading phase)
Phosphatases remove phosphate groups off ORC and allows the other proteins to bind (cdc6/cdt1)
When cdk levels rise in S/G2 and M phase this is the activation phase(phosphorylation occurs). Existing pre-RC is activated and new pre-RC formation is inhibited-ORC binds but doesn’t recruit the other complexes.
CDK is linked to cyclins which alter the conc

Question 29

In replication what do some bacteria with linear chromosomes, some virus and some mitochondrial DNA do?

Answer 29

Supply the RNA primer with a “priming protein” which has the necessary hydroxyl.
A primer is not needed and DNA pol binds via protein-protein interactions rather than interactions via the template.
Can initiate transcription right at end of ch.
TWO LEADING STRAND SYNTHESIS

Question 30

What is the telomere repeated sequence?

Answer 30

5’…TTAGGGTTAGGGTTAGGG…3’
Goes on for many 1000s of NTs
At ends of ch.s you have extended 3’ end

Question 31

What proteins in yeast regulated the lengths of telomeres?

Answer 31

INHIBITION
Rap1 targets telomere and binds to the dsDNA. Rif1 and Rif2 bind indirectly to Rap1 - weak inhibitors of telomerase
As the DNA extends these continue to bind co-operateively to sufficiently inactivate the telomerase
ACTIVATION
cdc13 binds to single stranded DNA and recruits telomerase, allowing further extension of the 3’ end.