Unit 3: DNA REPLICATION AND EXTRACHROMOSOMAL REPLICATION

Question 1

Topoisomerase

Answer 1

An enzyme that changes the number of times the two strands in a closed DNA molecule cross each other.
It does this by cutting the DNA, passing DNA through the break, and resealing the DNA
-it also Allows chromosome to be relaxed

Question 2

Replisome

Answer 2

A multi-protein structure that assembles at the bacterial replication fork to undertake synthesis of DNA Replisomes do not exist as independent units and contain DNA polymerase and other enzymes. As the replisome moves along DNA, the parental strands unwind and daughter strands synthesize. Replisome synthesizes the daughter strand

Proteins bind to origin –> separate DNA strand –> DNAPoly and other proteins go into replisome –>synthesize daughter DNA

Question 3

Conditional lethal

Answer 3

a mutation that is lethal under one set of conditions but is not lethal under a second set of conditions such as temperature. Conditional lethals are able to accomplish replication under permissive conditions typically provided by the normal temperature of incubation) but they are defective under non-permissive or restrictive conditions (provided by the higher temperature of 42C)

Question 4

What is in vitro complementation used for?

Answer 4

A functional assay used to identify components of a process (in this case replication apparatus)
The reaction is reconstructed using extracts from a mutant cell aka an in vitro system for replication is prepared from a dna mutant and is operated under conditions in which the mutant gene is inactive
Fractions from wild-type cells are then tested for restoration of activity

Question 5

Dna Mutants

Answer 5

Temperature-sensitive replication mutants that are defective in replication elongation during synthesis of DNA

Distinguish two stages of replication by their behavior when the temperature is raised

1) Quick-stop mutants
2) Slow-stop mutants

Question 6

Quick-stop mutants

Answer 6

cease replication immediately upon a temperature increase. Defective in the components of the replication apparatus, typically in the enzyme needed for ELONGATION

Question 7

Slow-stop mutants

Answer 7

temperature-sensitive replication mutants that are defective in INITIATION of replication. Slow-stop mutants complete the current round of replication but cannot INITIATE another round. Hence defective in initiation of new cycle of replication at the origin

Question 8

DnaA-ATP and the prepriming complex bind to the origin of replication and do what?

Answer 8

start separating DNA (initiation of replication)

Question 9

Synthesis of DNA occurs both during what two events?

Answer 9

-replication and DNA repair

Question 10

What occurs in repair reactions?

Answer 10

the damage nucleotide and other bases are removed from the damaged strand and DNA polymerase comes in and synthesizes complementary strand accurately getting rid of damaged base

Question 11

An enzyme that can synthesize a new DNA strand on a template strand is called what?

Answer 11

DNA polymerase

Question 12

an enzyme that undertakes semiconservative replication (not involved in repair) is called a?

Answer 12

DNA replicase

Question 13

Fill in the blank: DNA is synthesized by adding new nucleotides to the ____ end of the growing chain/strand

Answer 13

3’-OH

Question 14

The incoming nucleotide that is added to the primer 3’-OH is a ? and what is released when the nucleotide is added to the growing chain?

Answer 14

5’ triphosphate

-diphosphate is released when nucleotide is added

Question 15

DNA grows in what direction?

Answer 15

5’ to 3’ direction

Question 16

In bacteria, which polymerase is involved in replication?

Answer 16

DNA polymerase III; responsible for synthesis of new strands of DNA

Question 17

What is holoenzyme?

Answer 17

a large protein structure of replicases

Question 18

What is the role of DNA poly II?

Answer 18

required to restart replication fork when its progressed is blocked by damage in DNA

Question 19

DNA polymerase IV and DNA polymerase IV are both involved in what?

Answer 19

translesion replication. Involved in allowing replication to bypass certain types of damage and are called error-prone polymerases

Question 20

What is meant by semiconservative replication?

Answer 20

The two strand of parental duplex are separated and each serve as a template for synthesis of daughter strand. Thus, there will be two daughter duplexes, each of which has one parental strand and one newly synthesized strand

Question 21

What is a DNA repair reaction?

Answer 21

Repair synthesis replaces a short stretch of one strand of DNA containing a damaged base.

One strand of DNA is damaged. It is excised and new material is synthesized to replace it.

Question 22

All prokaryotic and eukaryotic DNA polymerases share the same fundamental type of synthesis activity. What is it?

Answer 22

• antiparallel synthesis from 5’ to 3’ from a template that is 3’ to 5’
• this means adding nucleotides one at a time to a 3’-hydroxyl (OH) growing end meaning chain grows in the 5’-3’ direction
• the choice of the nucleotide to add to the chain is dictated by base pairing with the complementary template strand

Question 23

What is the unique role of DNA polymerase I?

Answer 23

DNA poly I has a unique 5’-3’ exonuclease activity. The exonuclease activity is also used to excise bases that have been added to DNA incorrectly. This provides a proofreading error control system. This exonuclease activity can be combined with DNA synthesis to perform nick translation.

Question 24

What happens to DNA pol I when treated with protease? What are the subunits of Pol I?

Answer 24

DNA pol I cleaves into two subunits when treated w/a protease. The large subunit is called the cleanout fragment and the smallest subunit has the 5’-3’ exonuclease activity which removes about 10 bases at a time.

Question 25

What is a nick translation? when does it occur?

Answer 25

nick translation replaces part of a preexisting strand of duplex DNA w/newly synthesized material. Nick generates 3’-OH, 5’-P groups.

Nick translation occurs when there is a nick on one strand of the DNA

Question 26

What degrades the old strand replaced by nick translation when there is a nick on one strand of DNA?

Answer 26

The old strand gets degraded by the 5’-3’ exonuclease activity while the cleanout fragment synthesizes a new strand as a result the nick gets translated along the DNA strand.

Question 27

What is meant by high fidelity DNA polymerases?

Answer 27

These polymerases are involved in replication that have a precisely constrained active site that favors binding of Watson-Crick base pairs. These polymerases are sensitive to change of structure and they use the geometry of the nucleotide pair to determine if the pairing is accurate

-if the pairing is mismatched it is unlikely to fit into the active site.

Question 28

What is meant by low fidelity polymerases?

Answer 28

Low fidelity polymerases are most often used for repair. They have a more open active site and can tolerate not only damaged based pairs but also mismatch pairs

examples: e.coli DNA poly IV that is used for damage bypass replication.

poly IV have open active sites that accommodates damaged nucleotides, but also incorrect base pairs.

Question 29

What is meant by processivity?

Answer 29

Processivity of enzyme is the ability of the enzyme to catalyze multiple reaction cycles on a single template without falling off between cycles and then getting back on.

aka the tendency to remain on a single template rather than dissociate and reassociate

Question 30

How do DNA polymerases control the fidelity of replication?

Answer 30

bacterial polymerase have a 5’-3’ exonuclease activity that is used to excise incorrectly paired bases and proofread which controls the fidelity of replication. DNA polymerases will check the nucleotide added at the end of the growing chain and will remove it if the nucleotide is incorrect.

Proofreading by DNA poly improves fidelity 100 fold

Question 31

How does DNA polymerase know that an incorrect base has been incorporated?

Answer 31

If an incorrect base is added then DNA helical structure gets warped and this will cause the DNA polymerase to slow down. This slowing down is a sign that the DNA 5’-3’ exonuclease activity needs to be exercised

Question 32

What is the advantage of a holoenzyme as opposed to a single unit?

Answer 32

The advantage of holoenzyme is that you can have the exonuclease part of the enzyme in a different subunit and this increases fidelity of a repair enzyme

Question 33

Explain how polymerase proofread:

Answer 33

1) Enzyme adds base to growing strand (3’-OH growing strand)
2) Enzyme moves on if new base is correct
3) Base is hydrolyzed and expelled if incorrect

Question 34

What is frameshift error?

Answer 34

occurs when an extra nucleotide is inserted or omitted. Fidelity w/regard to frame-shifts is affected by the processivity of the enzyme

Question 35

What is the klenow fragment

Answer 35

DNA polymerase I, the first DNA polymerase for which the structure was determined. Characterized by 3 domains: palm, fingers, and thumb (resembles a hand)

Question 36

In the Klenow structure where does the DNA sit?

Answer 36

across the palm, (groove cleft), between the finger and the thumb.

The palm domain has important conserved sequence motif that provide the catalytic site

Question 37

In the Klenow structure what is the role of the finger?

Answer 37

involved in positioning the DNA template correctly at the active site.

Question 38

In the klenow structure what is the role of the thumb?

Answer 38

the thumb binds the DNA as it leaves the enzyme and is key for processivity of the enzyme.

Question 39

What is semidiscontinous replication?

Answer 39

The mode of replication in which one strand is synthesized continuously (leading strand 5’-3’), while the other is synthesized discontinuously (lagging strand is synthesized by making short okazaki fragments that are later joined together)

Question 40

What elements are needed to convert DNA from double to single strand?

Answer 40

1) helicase
2) single-strand binding protein (SSB)

Replication requires a helicase to separate the strands of DNA using energy provided by hydrolysis of ATP

A single stranded DNA binding protein is required to maintain the separated strand

Question 41

What is the structure of a helixase?

Answer 41

multi-meric wi/hexamer being the most common form.

Question 42

What is the structure of SSB?

Answer 42

a tetramer

Question 43

How does SSB bind with the DNA?

Answer 43

cooperatively preventing the duplex from reforming

Question 44

Why does the antiparallel structure of the two strands of duplex DNA pose a problem for replication?

Answer 44

• As the replication fork advances, daughter strands must be synthesized on both of the exposed parental strands
• The fork template strands moves in the direction from 5’-3’ on one strand and in 3’-5’ on the other strand.

YET!!! DNA is synthesized only from a 5’ end toward a 3’ end (by adding a new nucleotide to the growing 3’ end)

Question 45

How is the antiparallel structure of DNA issue resolved when replicating DNA?

Answer 45

The problem is solved by synthesizing the new strand on the 5’ to 3’ template in a series of short fragments, each synthesized in the ‘backward’ direction; That is with the customary 5’-3’ polarity

Question 46

What is required for DNA polymerase to start DNA synthesis?

Answer 46

pre-existing 3’-OH end (PRIMER) to initiate synthesis. The template strand must also be converted to a single strand.

Question 47

TRUE OR FALSE: DNA polymerase can start DNA synthesis on their own

Answer 47

FALSE: DNA polymerases cannot start synthesizing the DNA strand on their own. They need a preexisting 3’-OH hydroxyl and it can then elongate by adding nucleotides to the 3’-hydroxyl group

Question 48

How can the priming end or primer be provided to a DNA strand?

Answer 48

There are several methods for providing the free 3’-OH end that DNA polymerases require to initiate DNA synthesis

1) RNA primer: is synthesized by a primase (or provided by base pairing RNA → DNA)
2) A nick in DNA: Duplex DNA is nicked to provide free end for DNA polymerase
3) Priming protein: A priming nucleotide is provided by a protein that binds to DNA

Question 49

How many primers are required on the leading and lagging strand for DNA polymerase to initiate synthesis?

Answer 49

The leading strand only requires one such primer while the lagging requires several such primers since each okazaki fragment requires its own primer.

Each okazaki fragment begins with a RNA primer of about 10 bases long

Question 50

Each okazaki fragment begins with a RNA primer of about 10 bases long. Where does the RNA primer come from?

Answer 50

from a special RNA polymerase called a primase.

e.coli has two types of priming reactions which occur at the bacterial origin (oriC) and the phageX174

Question 51

What is the differences between the two priming reactions for e.coli (oriC vs. phageX174)?

Answer 51

ori used DnaG primase to make its primer while phageX173 requires primosome to synthesize a primer and this is used when damage causes the replication fork to collapse and needs to be restarted

Question 52

Different enzymes units are required to synthesize the leading and lagging strands. Why?

Answer 52

Because leading and lagging are moving in different directions. The unit moving along the leading strand is moving in the same direction as the replication fork. The unit in the lagging strand is moving in a sense ‘backwards’

Each okazaki fragment has to be completely synthesized before the next fragment can begin synthesis at the replication fork. This means the DNA polymerase has to disengage from the template, move to the new location, and reconnect to the template and using the next primer start the next okazaki fragment

In e.coli the catalytic subunit is DnaE

In other like eukaryotes there are different catalytic subunits for each strand

Question 53

DnaE is?

Answer 53

the catalytic subunit in e.coli required to synthesize the leading and lagging strand.

Question 54

The DNA polymerase holoenzyme (pol III) consists of subcomplexes, what are they?

Answer 54

The replisome in e.coli consists of DNA pol III holoenzyme complex in association w/a primase and a helicase. The DNA pol III core contains 3 subunits including the catalytic subunit and proofreading subunit

Question 55

What are the catalytic cores of DNA polymerase III holoenzyme?

Answer 55

The holoenzyme has at least two catalytic cores: the processivity clamp and dimerization clamp loader complex

Question 56

It is proposed that DNA pol III has 3 polymerases cores, what are they?

Answer 56

two catalytic cores responsible for synthesis of the lagging strand and one catalytic core responsible for the leading strand
Each okazaki strand is synthesized by the alternating core polymerase

Question 57

What is the role of a clamp loader?

Answer 57

The clamp loader is a complex of seven proteins and its function is to place the processivity units (the beta subunits) on DNA to form the beta ring. The clamp is responsible for holding the catalytic core on to the template strands

-forms a circular clamp around the nucleic acid

Question 58

What is the catalytic core made up of?

Answer 58

The catalytic core is made up of the alpha subunit, which has the polymerase activity, epsilon - which has the 3’-5’ proofreading activity, and the theta subunit which stimulates the exonuclease

Question 59

How does DNA polymerase III assemble? (short explanation)

Answer 59

in stages, generating an enzyme complex that synthesizes

1) Clamp loader cleaves ATP to load clamp on DNA
2) Core enzyme joins
3) Tau + second core joins to give a symmetric dimmer
4) Tau subunits maintain dimeric structure

Question 60

long explanation for assembly of DNA poly III

Answer 60

The clamp loader uses the energy from the hydrolysis of ATP to bind the beta units to the template primer complex. Once the beta subunits bind to the DNA it is thought that the beta subunits have now increased affinity for the core complex through a conformational change, thus recruiting the core complex to the replication site

The tau dimer which is part of the clamp loader links to a second core complex and helps to maintain the dimeric structure

Each of the catalytic cores synthesizes one of the new strands of DNA namely the leading strand and the lagging strand

Question 61

Is the clamp on leading strand processive? What about lagging?

Answer 61

Yes, the clamp on leading is processive b/c its clamp keeps it on the DNA

On the lagging strand, the clamps is not processive b/c it dissociates at the end of each okazaki fragment and reassembles for the next fragment

Question 62

Why does the lagging strand create a loop, which gets bigger as the undwining of DNA advances?

Answer 62

as the holoenzyme moves continuously to form the leading strand the lagging strand gets pulled through creating a loop which gets bigger as the unwinding advances
The lagging strand core complex pulls the second single stranded template through the strand while synthesizing the new strand
The template should extend the length of at least one okazaki fragment before the polymerase on the lagging strand can start synthesizing the next fragment

Question 63

What is the role of helicase DnaB? okazaki

Answer 63

responsible for interacting w/the primase DNAG to initiate each Okazaki fragment.. Forward movement of the fork

Question 64

What happens to the clamp once an okazaki fragment is completely synthesized

Answer 64

Once okazaki is completely synthesized it will require the complete dissociation of the clamp from core poly and the reassociate again. The clamp loader causes the beta clamp to open up by changing its conformation
A new beta clamp is then recruited by the clamp loader to start the synthesis of the next okazaki fragment

STEPS

1) initiation of okazaki fragment
2) termination of okazaki fragment
3) beta clamp dissociates
4) beta clamp associates

Question 65

TRUE OR FALSE: Each okazaki fragment starts w/a primer and stops before the next fragment

Answer 65

TRUE; DNA polymerase I removes the primer and replaces it with DNA

Question 66

What is the role of polymerase I in okazaki fragments?

Answer 66

Primer is RNA primer and needs to be removed and replaced with DNA, this is done by polymerase I using nick translation

Question 67

What steps are involved in synthesis of okazaki fragments? prokarotes

Answer 67

Require priming, extension, removal of RNA primer, gap filling, and nick ligation

1) Primase synthesizes RNA
2) DNA Pol III extends RNA primer into okazaki fragments
3) Next Okazaki fragment is synthesized
4) DNA polymerase I uses nick translation to replace RNA primer w/DNA
5) Ligase seals the nick

Question 68

How are okazaki fragments linked?

Answer 68

by ligase;

DNA ligase makes the bond that connects the 3’ end of one Okazaki fragment to the 5’ beginning of the next fragment

DNA ligase seals nick between adjacent nucleotides by employing an enzyme-AMP intermediate

Question 69

What happens when there’s damage to DNA?

Answer 69

If it’s not repaired it can be lethal so it bypasses lesion.

The replication fork stalls when it arrives at the damaged site and may collapse. This has to be replaced by a specialized DNA polymerase that can replicate through the lesion.

In e.coli two error prone polymerases IV and V and pol II are used for the translesion synthesis

Question 70

What happens after damage DNA has been bypassed by Poly IV and V

Answer 70

Primosome reinserts the replication complex after polymerase IV and V have repaired damage
-It reloads the DnaB so that the helicase activity can resume

THUS, the primosome is required to reinitiate replication or restart stalled replication fork by reinserting the replication complex

Question 71

How does replication terminate?

Answer 71

The two replication forks usually meet halfway around the circle, but there are ter sites that cause termination if the replication forks go too far.

These sites are situated near termination region
Each set of ter fork allow the replication fork to the termination region but does not allow it out the other side
So replication fork one can pass through ter B and ter C but cannot go to ter A, D or E
This serves as a replication fork trap. Hence if one replication fork is moving slower than the other the replication fork trapp will effectively trap the faster moving fork at the ter site. Until fork two catches up

Question 72

Plasmids

Answer 72

circular, extrachromosomal DNA. It is autonomous and can replicate itself

Question 73

Lysogeny

Answer 73

the ability of a phage to survive in bacteria as a stable prophage component of the bacterial genome

Question 74

Episome

Answer 74

if plasmid can integrate into the bacterial DNA/chromosome it is called an episome

Question 75

Immunity

Answer 75

plasmids and lysogenic phages do not allow another element of the same type to become established in the bacteria or cell -this ability is called immunity

Question 76

Why are the ends of linear DNA a in bacteria a problem for replication?

Answer 76

if we have linear DNA then the DNA polymerase needs to initiate replication at the 5’ end since the progression of the newly formed strand is in the 5’ to 3’ direction. There has to be some way for replication machinery to start at the very end of the template strand. it could do that by either converting the linear form into a circular or multimeric form as in T4, or the Lambda phage, or it could form long hairpin structures in the end as seen in paramecium. or it could use proteins which bind to the 5’ end as seen in adenoviruses

Question 77

TRUE OR FALSE:

Terminal proteins enable initiation at the ends of viral DNAs

Answer 77

TRUE; example adenovirus

Question 78

What is strand displacement?

Answer 78

Strand displacement - a mode of replication of some viruses in which a new DNA strand grows by displacing the previous (homologous) strand of the duplex.

Adenovirus DNA replication is initiated separately at the two ends of the molecule and proceeds by strand displacement

Question 79

In virus that use strand displacement, a terminal protein binds to the 5’ end of DNA, what is the role of this?

Answer 79

1) provides cytidine nucleotide with a 3’-OH group, which serves to prime replication.
2) terminal protein also associates DNA polymerase. the complex of the terminal protein and DNA polymerase bind to the5’ end of the DNA. elongation of the newly synthesized strand by the polymerase will result in a new strand that is bound to the initiating C nucleotide. the old terminal protein gets displaced

Question 80

What is the rolling circle method?

Answer 80

-a rolling circle generate single-stranded multimers of the original sequence.

in the rolling circle method only one strand is used to generate several copies of the original sequence

if the template is a circular duplex DNA., one of the strands undergoes a Nick so providing a free 3’ hydroxyl group for the DNA polymerase to come in and synthesizes the complementary strand. as the new strand elongates it displaces the old strand generating a single stranded tail. now the polymerase can keep going around the circle indefinitely performing a multimeric single stranded tail with the original unit at the 5’ end.

Question 81

What determines the types of products generated by rolling circles?

Answer 81

The fate of the displaced tail

The displaced tails could get cleaved at unit lengths generating monomers or it could be cleaned as multiples which can remain linear or it could form a duplex by the synthesis of the complementary strand it could circularize to form single strand and duplex regular forms

Question 82

What is the phage. A protein?

Answer 82

a cis-acting relaxase that generates single-stranded circles from the tail produced by rolling circle replication.

Question 83

phageX174 DNA is a template for synthesizing single-stranded viral circles. What are the steps?

Answer 83

we see rolling circles when phage genomes are replicating.

1) PhageX174 genome is a single stranded circular DNA molecule and is referred to as the plus strand. for replication first the strand has to be converted into a duplex circle by synthesizing a complementary strand referred to as the minus strand.
2) the replication now proceeds in the rolling circle format.
3) the A protein nicks the positive strand and stays bound to the 5’ end of the strand, while the 3’ end gets extended by the polymerase. as it completes the circle it displaces the original strand
4) the A protein which is still bound to the 5’ end is still associated w/the rolling circle and will recognize the origin of the next rolling circle and nick it again, attaching itself to the 5’ end of the newly synthesized circle so that the next round of the low rolling circle can begin

Question 84

What is congjuation?

Answer 84

A process in which two cells come in contact and transfer genetic material. This process requires the presence of F plasmid

Question 85

What is an F plasmid?

Answer 85

The free F plasmid is a large circular DNA and a replicon that is maintained at the level of one plasmid per bacterial chromosome
-An F plasmid is an episome and can exist as a free plasmid or can integrate into the bacterial chromosome, in which case its own replication system is suppressed. Since it has its own origin of replication, which it utilizes if it is in the free form, it is maintained as a single copy per cell

Question 86

What is the transfer region on f plasmid

Answer 86

A segment of an F plasmid that is required for bacterial conjugation

Question 87

What happens to the ori on an F plasmid, if the F plasmid is integrated into the bacterial genome?

Answer 87

then the F-plasmid will not use its own Ori for replication but rather gets replicated as part of the bacterial genome.

Question 88

What is found at the regulatory region of plasmid?

Answer 88

a TraJ gene is found in this region

Question 89

What is the role of TraJ?

Answer 89

it turns on both TraM and TreI/Y

Question 90

What is the role of the regulator FinP?

Answer 90

It turns off TraJ

Question 91

How many Tra and Trb genes are involved directly with transfer of DNA?

Answer 91

only four

Question 92

Pili

Answer 92

form on the surface of the bacterium and are encoded by F plasmids

Question 93

Pilin

Answer 93

coded by gene TraA and polymerizes to form the pilus which is a hollow cylinder

Question 94

What role do pilus play on F positive cells?

Answer 94

F positive have 2-3 pilus and enables an F-positive bacterium to contact/connect to an F-negative bacterium and initiate conjugation

Question 95

TraS and TraT code for what on the surface of proteins?

Answer 95

code for surface exclusion proteins that prevent an F-positive cell from meeting w/another F-positive cell. After initial pairing of these cells, the F pili retract and disassmeble

Question 96

Transfer of DNA between F positive and F negative is provided by what ?

Answer 96

through a channel provided by the T4SS which is a membrane spanning protein complex

Question 97

Addiction system

Answer 97

a survival mechanisms produced by plasmids where the mechanism kills bacterium upon loss of plasmid.

Plasmid produces long lived killer substance and short lived antidote substance. plasmid provides antidote. If loss of plasmid occurs antidote will stop being produced and killer will kill bacterium