DNA replication

Question 1

How long does DNA replication take?

Answer 1

DNA replication in bacteria takes approx. 20 minutes, but in eukaryotes can vary from 1-4 hours in yeast to 24 hours in cultured animal cells

Question 2

What were the theorized models of DNA replication?

Answer 2

1. Dispersive- breakage and reunion
2. Semi-conservative - unwind and replicate each
3. Conservative- replication without unwinding

Question 3

Who found evidence of the semi-conservative fashion for DNA replication?

Answer 3

in 1958, Meselson(Harvard )and stahl( uni of Oregon) produced evidence that DNA replicates in a semi-conservative fashion

Only semi-conservative model for DNA replication predicts experimental results

Question 4

What was the experiment Meselson and stahl used to deduce the semi-conservative model?

Answer 4

15N is a non-radioactive heavy isotope of nitrogen , was used to tag DNA

One cell division showed equal amounts of N- 14 and N-15 while a second division showed higher amounts of N-14 than N-15

Question 5

Who studied the enzymes responsible for DNA replication?

Answer 5

In 1959, Arthur kornberg received the Nobel prize for his research studying the enzymes responsible for DNA replication- most importantly- the identification

Question 6

What is common in DNA replication across all species?

Answer 6

1. DNA synthesized in 5’ to 3’ direction
2. All four dNTPs (Deoxynucleoside triphosphate ) A C T G
   
   • building blocks of the DNA molecule
3. A fragment of DNA to act as a template
4. Magnesium ions (Mg2+)
   
   • required for DNA polymerase activity
5. A primer providing a free 3’ -OH grou, most often is RNA but some organisms require DNA primer

Question 7

Where does DNA synthesis begin?

Answer 7

At the origin of replication (ORI)

Question 8

Prokaryotic genomes are usually

Answer 8

Circular and contain a single ORI

Question 9

In E. Coli where is the only place that the replication process is regulated ?

Answer 9

E. Coli is only regulated at the point of initiation, therefore once the replication fork is established- replication proceeded until completion

Question 10

Eukaryotic chromosomes don’t…

Answer 10

Have many replication origins and no replication termini

Question 11

Why are there multiple origins of replications in eukaryotic genomes?

Answer 11

They are required to relocate the large genomes in a timely fashion, e.gl drisophili

• human haploid genome = ~3.4 million base pairs
• average size of a chromosome= 100, 000,000bp

Rate of replication= 2000 bp/minute

Therefore if there was a single origin of replication on each human chromosome it would take 830 hours 9f replication to complete

Question 12

The number of origins increases with

Answer 12

Organisms with larger genomes

Question 13

What is OriC?

Answer 13

The origin of replication in E. Coli has a length of 245 bp and contains two important sequences

• 13 nucleotide sequences (three in tandem array)
• 9-nucleotide sequences

Question 14

What is the significance of the DNA sequences at the OriC?

Answer 14

The 13-mer tandem sequences at the Ori are A-T rich. Bonding between A-T is weaker than G-C. This facilitates easier melting and strand separation of the DNA molecule at this region

At the 9-nucleotide regions, DnaA initiator proteins bind to begin the process of replication

Question 15

Explain the model of initiation of replication at E. coli OriC

Answer 15

1. Multiple copies of initiator proteins (DnaA) bind to the 9mers at the origin
2. Strand separation occurs at the region of the 13-mer sequences
3. Helicase inhibitor (DnaC) protein delivers helicase (DnaB) to the template
4. Helicase clamps around each single strand of DNA
5. Helicase proceed to unwind the DNA in opposite directions away from origin

Question 16

Once ORI is opened, prokaryotic elongation requires:

Answer 16

1. DNA primase: polymerase which is required for DNA polymerase to begin elongation of DNA daughter strands
2. DNA polymerase III: multi-subunit protein for elongation of DNA daughter strands, DNA synthesis of leading and lagging strands
3. DNA polymerase I: removes and replaces RNA primer
4. DNA ligase: ligates Okazaki fragments together

Question 17

What are the main complications of elongation phase of replication in prokaryotes?

Answer 17

1. Template dependent DNA polymerase canno5 initiate synthesis from single strand
2. Both strands DNA must be copied in the 5’ to 3’ direction
3. Single stranded DNA tends to re-anneal and form secondary structures
4. How is DNA polymerase loaded and maintained on the single stranded DNA template, and unloaded when it reaches double stranded DNA (e.g. the next Okazaki fragment)

Question 18

Why are primers needed?

Answer 18

DNA polymerase requires short double stranded region to provide a 3’ hydroxyl group to add nucleotides to

Question 19

Explain the need for DNA primase

Answer 19

• in bacteria, RNA primers are synthesized by primase, a special RNA polymerase
  
  • unrelated to transcribing DNA
• each primer 4-15 nucleotides in length and provides a free 3’ -OH group
• Once primer is completed, DNA pol III will continue synthesis of the strand
• NB —> a primer is synthesized ONCE on the leading strand, and many times on the lagging strand

Question 20

Explain how DNA is replicated in a 5’ to 3’ direction

Answer 20

• One strand will be continuously replicated (leading strand )
• The other will be discontinuous (lagging strand) resulting in a series of short segments which need to be lighted together

The synthesis of both strands requires DNA polymerase III

Question 21

Summarize the roles of DNA polymerase III

Answer 21

1. Phosphodiester bond formation using the template strand and primer (elongation)
2. Repair of base misincorporation by DNA polymerase III

Question 22

What is the role of DNA polymerase III in phosphodiester bond formation ?

Answer 22

A) DNA polymerase catalyzes the formation of a phosphodiester bond between the 3’ OH group of the deoxyribose on the last nucleotide and the 5’ - phosphate of the dNTP precursor

B) At each step in the lengthening in the new DNA chain, DNA polymerase finds the correct precursor dNTP that can form a complementary base pair with the nucleotide on the template strand of the DNA. Pyrophosphate is liberated as a result of this incorporating a single base

Question 23

Explain the ability of DNA polymerase III

Answer 23

A) Occassionally DNA polymerase will misincorporate a nucleotide into the growing DNA chain

B) These misincorporations need to be repaired prior to the next to be repaired prior to the next DNA replication or a mutation will occur

C) the 3’ to 5’ proofreading exonuclease activity of DNA polymerase clips off any unpaired residues at the primer terminus. It continues this activity until a base paired 3’-OH terminus is encountered

Question 24

explain DNA synthesis on the lagging strand being discontinuous

Answer 24

Daughter DNA strands are synthesized in the 5’ to 3’ direction, therefore the DNA synthesized on the lagging strand must be made initially as a series of short DNA molecules called Okazaki fragments

In bacteria each Okazaki fragment is 1000-2000 bp in length

Question 25

What are the first events to occur on the lagging strand?

Answer 25

RNA primers synthesized by primase

Then DNA polymerase III synthesizes the Okazaki fragment

DNA polymerase I removes the RNA primer (RNAseH and FEN-1 protein in eukaryotes)

DNA ligase joins the DNA fragments together by forming a phosphodiester bond between the two fragments

Question 26

What is the function of DNA ligase?

Answer 26

Sealing the gap between adjacent DNA fragments to form a longer, covalently continuous chain

Question 27

What are the functions of DNA polymerase I?

Answer 27

Implicated in DNA repair

• 5’ to 3’ exonuclease activity (RNA primer removal)
• 5’ to 3’ polymerase activity (replace primer with DNA)
• 3’ to 5’ exonuclease activity (proofreading)

Question 28

What are the functions of Polymerase II?

Answer 28

Involved in repairing damaged DNA

-has 3’ to 5’ exonuclease activity

Question 29

What are the functions of polymerase III?

Answer 29

The main polymerase in bacteria

• 5’ to 3’ (responsible for elongation of leading and lagging strands)
• has 3’ to 5’ exonuclease activity (proofreading)

Question 30

Single stranded DNA tends to re-anneal and form secondary structures, how is this problem fixed?

Answer 30

Single strand binding proteins (SSB proteins)

SSB proteins binds to single stranded DNA in the replication fork. The cooperative binding of these proteins prevent reannealing and straightens out the DNA template to facilitate the DNA polymerization process

Question 31

How is DNA polymerase loaded and maintained on the single stranded DNA template, and unloaded when it reaches double stranded DNA (e.g., the next Okazaki fragment)?

Answer 31

Clamp protein

Clamp protein tightly holds the DNA polymerase onto the the template for synthesis of long templates, and releases DNA polymerase when it stalls at a region of double stranded DNA

Question 32

For a bacterial replication fork moving at 500 bp per second, the parental DNA helix ahead of the fork must rotate at 50 revolutions per minute. What is the solution?

Answer 32

Swivel the DNA using topoisomerase enzymes

Question 33

What is topoisomerase I?

Answer 33

This enzyme transiently forms a single covalent bond with the DNA and breaking a phosphodiester bonds; this allows free rotation of the DNA around the covalent backbone bonds

Question 34

Explain what the function of topoisomerase I is for DNA replication

Answer 34

1. One end of the DNA double helix cannot rotate relative to the other end. Type I DNA topoisomerase with tyrosine at the active site
2. DNA topoisomerase covalently attaches to a DNA phosphate, thereby breaking a phosphodu3ster linkage in one DNA strand
3. The two ends of the DNA double helix can now rotate relative to each other, relieving accumulated strain
4. The two ends of the DNA double helix can now rotate relative to each other, relieving accumulated strain
5. The original phosphodiester bond energy is stored in the phosphotyrosine linkage, making the reaction reversible
6. Spontaneous re-formation of the phosphodiester bond regenerates both the DNA helix and the DNA topoisomerase

Question 35

Why are there gals at the telomeres?

Answer 35

At the telomeres there are gaps at the 5’ end of the new DNA resulting from RNA primer removal

DNA polymerase can only synthesize DNA is the 5’ to 3’ direction - therefore no new DNA synthesis can fill these gaps

If the gaps found at the telomeres aren’t filled, the length of the chromosome would progressively decrease with subsequent rounds of DNA replication

Question 36

How can you completely replicate a linear chromosome in eukaryotes?

Answer 36

Synthesis of telomeric DNA by telomerase

Telomerase is similar to reverse transcriptase in that it synthesizes DNA using an RNA template covalently bound to the protein

Question 37

Explain the functioning process of telomerase

Answer 37

• Telomerase binds to the 3’ flanking end of the telomere with complementary base pairing to the telomerase RNA
• telomere is elongated using RNA template
• Telomerase translocates
• Second step is repeated a number of times: elongation

Later, DNA polymerase complements the lagging strand