Module 4: Section 1A

Question 1

Bacterial chromosome replication

Answer 1

• Replication is bidirectional and begins at oriC site
• oriC is AT-rich, making it easier to separate than GC-rich regions

Question 2

Replication fork and replisome activity - bacterial chromosome

Answer 2

• After the double-stranded DNA is opened, replication forks move in opposite directions, powered by the replisome, until they meet at the termination site
• In fast-growing cells, new replication rounds can start at oriC before previous ones finish, allowing multiple chromosome copies to form

Question 3

What does replication of circular chromosomes result in?

Answer 3

two interlocked (catenated) daughter chromosomes

Question 4

How is the catenate resolved?

Answer 4

• One chromosome is cleaved by topoisomerase
• Failure to separate these strands leads to cell death

Question 5

Circular vs linear chromosomes - free ends

Answer 5

• Circular chromosomes avoid replication issues caused by “free ends”
• Eukaryotes with linear chromosomes use telomeres to protect and manage these ends during replication

Question 6

Components of the bacterial replication fork - step 1

Answer 6

• Helices unwinds DNA
• Lagging strand is coated with single stranded DNA binding proteins
• RNA primer is synthesized by primate

Question 7

Components of the bacterial replication fork - step 2

Answer 7

• DNA polymerase III holoenzyme is tethered to DNA via a B clamp
• It is loaded into the complex by the t clamp loader

Question 8

Components of the bacterial replication fork - step 3

Answer 8

• One core enzyme of Pol III carries out leading strand synthesis
• Synthesizes DNA continuously

Question 9

Components of the bacterial replication fork - step 4

Answer 9

• Two or more Pol III enzymes carry out lagging strand synthesis
• Synthesizes pieces of DNA called Okazaki fragments

Question 10

Components of the bacterial replication fork - step 5

Answer 10

• When a core enzyme reaches a completed region it is released from the DNA and Okazaki fragments are joined together

Question 11

How does DNA polymerase I join Okazaki fragments together?

Answer 11

• Removes RNA primer
• Fills in gaps with DNA and DNA ligase joins the strands

Question 12

Components of the eukaryotic replication fork - step 1

Answer 12

• Helicase must be complexed with additional proteins to be activated to unwind DNA
• Single stranded DNA is protected by replication protein A

Question 13

Components of the eukaryotic replication fork - step 2

Answer 13

• helicase associates indirectly with a primate (Poly a)
• Synthesizes RNA primer

Question 14

Components of the eukaryotic replication fork - step 3

Answer 14

• Polymerase E engages in leading strand synthesis
• Polymerase S engages in lagging strand synthesis

Question 15

How are polymerase E and S loaded onto the DNA?

Answer 15

• By proliferating cell nuclear antigen
• PCNA is carried to the DNA by replication factor C

Question 16

Components of the eukaryotic replication fork - step 4

Answer 16

• Okazaki fragments are much shorter than prokaryotes
• 100-200 bp vs 1000-2000 bp
• RNA primer is removed by strand displacement and filled in by polymerase S
• Single strand gap is repaired by ligase

Question 17

Factor 1 that slows down eukaryotic replication

Answer 17

• Replisomes are subject to cell-cycle dependent regulation of initiation and termination
• Nucleosomes get displaced during passage of the replication fork

Question 18

Factor 2 that slows down eukaryotic replication

Answer 18

Many of the enzymes in the replisome are subject to post-translational modifications that activate or inactivate them

Question 19

Proofreading

Answer 19

• If an error is made while DNA polymerases are replicating DNA, enzymes can reverse, excise the incorrect base and insert correct base
• Also called 3’-5’ exonuclease activity

Question 20

Repair only

Answer 20

• Other polymerases only engage in repair
• Important at the replication fork in order to repair breaks in double stranded DNA if the fork becomes “stalled”

Question 21

Prokaryotic replication

Answer 21

• Simple bacterial chromosomes are synthesized fast
• Have mechanisms to correct mistakes
• Mutations can be beneficial and a bacterial population could gain the conferred advantage

Question 22

Eukaryotic replication

Answer 22

• Requires simultaneous synthesis of multiple chromosomes
• Accumulation of mutations could be deleterious to the function of complex organisms
• Multiple checks and balances are in place to ensure fidelity of DNA template

Question 23

Viral replication

Answer 23

• Relies on host cell machinery
• Small genome sizes help ensure the numerous copies of viral genome are produced before a cell’s resources are exhausted

Question 24

Retrovirus replication

Answer 24

• Genomes made of single stranded RNA
• Need to create DNA base template in order to synthesize a new RNA genome
• Have very high mutation rates since RNA polymerases are not good at proofreading and repair