DNA Replication

Question 1

Aicardi-Goutieres syndrome

Answer 1

Cerebrospinal fluid has elevated white cells (lymphocytes) and high levels of interferon-alpha (INF) activity
Atrophy in frontal and temporal regions of brain
Followed by loss of almost all motor skills

Question 2

Initiation- prokaryotic replication

Answer 2

One start site for DNA replication –> initiation site
In ecoli: oriC

Initiator protein + ATP + histone like proteins attach to oriC causing DNA to wrap around the complex creating negative superhelical tension in the adjacent DNA

Results in the opening of the double helix at 3 x 13 AT rich repeats left to the protein complex

Further opening the DNA creates a replication bubble

Question 3

oriC

Answer 3

oriC has 4 sequences 9 bp in length that are recognized by the initiator protein

Adjacent to oriC are 3 AT rich repeats

Question 4

Why are AT rich repeats important?

Answer 4

Provides a site for DNA to open easily (more unstable than GC rich DNA)

Question 5

How many replication forks per replication bubble?

Answer 5

2

Question 6

What are the limits of DNA polymerase?

Answer 6

DNA polymerase cannot initiate DNA synthesis (new strand synthesis)

Requires a 3’ hydroxyl group at the end of a base paired strand to add dNTPs

Question 7

RNA primers

Answer 7

synthesized 5’ –> 3’ and are laid down antiparallel and complementary to the DNA template strands (both stands of the parent DNA)

Reaction is catalyzed by primase

Question 8

primosome

Answer 8

primase+ helicase + protein

binds both strands of the replication bubble and makes a short RNA primer

Question 9

DNA polymerase 1

Answer 9

RNA primer excision
DNA repair
3’ and 5’ exonuclease activity

not the main synthetic enzyme in E coli

Processivity: 3-200

Question 10

3’ exonuclease activity

Answer 10

when an incorrect base is incorporated the 3’ exonuclease removes it, correct base can be inserted

Question 11

5’ exonuclease activity

Answer 11

removal of RNA primers in Ecoli

Question 12

DNA polymerase 3

Answer 12

synthesizing enzyme in replication- replicative chain elongation

processivity- 500,000

functions as a holoenzyme, with a core of 3 polypeptides

Question 13

Core of DNA polymerase 3

Answer 13

alpha- dimer polymerase
epsilon- dimer 3’ exonuclease for proofreading
beta- dimer that forms sliding clamp around DNA

dimer so that it can replicate both strands of DNA at the same time

Question 14

beta dimer in pol III

Answer 14

internal diameter of 3.5 (just enough to fit 2 nm diameter for DNA) that has a low affinity in its inner surface to DNA so that the DNA ca slide smoothly along it

Question 15

Gamma complex of DNA polymerase 3

Answer 15

made up of various holoenzymes that help the beta subunit unload onto the DNA –> clamp loading

Question 16

Chi and Psi subunits of DNA pol 3

Answer 16

Also bind to the g complex

Chi subunit mediates transition from synthesizing RNA primers to DNA

Question 17

T complex DNA pol 3

Answer 17

Ensures that the core enzyme is a dimer, allowing both strands of the replication fork to be synthesized simultaneously

also binds to helicase

Question 18

Lagging strand synthesis

Answer 18

Has to go from 5 –> 3

the fork has to open up BEFORE new synthesis can be initiated

Question 19

What is the function of the loop?

Answer 19

Allows DNA synthesis to run smoothly and in the same direction for both strands

lagging strand folds to make a small loop, this allows the lagging strand to be in the same orientation as the leading strands and now they can be copied together by the holoenzyme dimers.

Question 20

What is formed at the end of DNA replication for bacteria?

Answer 20

2 lagging strands and 2 leading strands of new DNA

Question 21

Termination- prokaryotic replication

Answer 21

termination region is 180 degrees from the oriC (ter region)–> 20 bp segment

when replication fork meets TUS-ter complex, it halts…ensuring that each of the two forks initiated at oriC will travel no farther than ter

Question 22

Proofreading

Answer 22

DNA Polymerase III can recognize mismatched base pairing (mainly AG and CT)

The epsilon subunit removes the mismatched base pain with its 3’ to 5’ exonuclease activity

The chain is then extended by DNA polymerase III

Question 23

Nick Translation

Answer 23

DNA polymerase I 5’ to 3’ exonuclease hydrolyses the RNA primers

Simultaneously, the 3’ end of the Okazaki fragment (DNA) is extended by incorporation of deoxyrobonucleotides

DNA pol 1 moves to replace the RNA fragments in the previous Okazaki fragment with DNA, the “nick” keeps moving over one space until the end where ligase seals it together

Question 24

Final step in termination of DNA synthesis in E coli

Answer 24

type II topoisomerase separates the 2 interlinking strands of the replicated circles and refold the DNA into supercoils

Question 25

DNA polymerase α

Answer 25

Found in nucleus
Has primase activity –> can make RNA primer in lagging strand replication
Moderate inherent processivity
Moderate processivity with PCNA (like a clamp)

Question 26

DNA polymerase δ

Answer 26

Found in nucleus 
Lagging strand synthesis 
Low inherent processivity 
High processivity with PCNA
3' exonuclease activity

Question 27

DNA polymerase ε

Answer 27

Found in nucleus 
Leading strand synthesis 
High inherent processivity 
High processivity with PCNA
3' exonuclease activity

Question 28

DNA polymerase γ

Answer 28

Found in mitochondria  
mitochondrial DNA replication  
High inherent processivity 
High processivity with PCNA
3' exonuclease activity  

*not sensitive to aphidicolin

Question 29

Sensitivity to aphidicolin (fungal steroid that inhibits replication)

Answer 29

To see which specific polymerases were associated with replication

Question 30

3 prime structure in replication

Answer 30

Nucleosomes are dissociated ahead of the replication fork, and reassembled on newly synthesized DNA

Question 31

Okazaki fragments in eukaryotic DNA replication

Answer 31

Size is much smaller than in prokaryotic synthesis

Question 32

Telomeres

Answer 32

Highly repetitive sequences at the 3’ ends of linear chromosomes

Question 33

What is added by telomeres?

Answer 33

Protein component enzyme- reverse transcriptase- hTERT (human telomerase reverse transcriptase”

RNA component- hTR (human telomerase RNA)
Template

Question 34

Why do we need telomeres?

Answer 34

When you come to the last primer in the lagging stranding, polymerase 1 cannot complete gap at the end –> need OH group for DNA pol to work

Question 35

How does telomerase work?

Answer 35

Telomerase uses RNA component to add repeats

Telomerase repositions itself to add each repeat (moves 3’ –> 5’ on daughter strand)

The daughter strand is synthesized using complementary base pairing by DNA polymerase alpha

This process is repeated ultimately forming the long telomere

Question 36

3’ overhang at telomere end

Answer 36

A single stranded 3’ overhang left when telomerase moves away

Question 37

D-loop T-loop

Answer 37

Telomere ends are protects by loops that are stabilized by binding of telomere-binding proteins called TTAGGG repeat binding factors