Chapter 9

Question 1

General features of DNA replication
•Double helical model for DNA includes the concept that 2 strands are complementary
•Each strand can serve as __ for making its own partner
•Semiconservative model for DNA replication is correct
•Half-discontinuous (short pieces later stitched together)
•Requires __ primers
•Usually bidirectional

Answer 1

General features of DNA replication
•Double helical model for DNA includes the concept that 2 strands are complementary
•Each strand can serve as template for making its own partner
•Semiconservative model for DNA replication is correct
•Half-discontinuous (short pieces later stitched together)
•Requires RNA primers
•Usually bidirectional

Question 2

DNA synthesis requires deoxynucleoside triphosphates and a ___ dNTP: (dATP,dCTP, dATP and dGTP). It has three phosphate attached to the 5’hydroxyl of the 2’-doxyribose.
Template: provides the ssDNA that directs the addition of each complementary deoxynucleotides. The primer is __, but shorter than the template. It must have a free __ adjacent to the ssDNA region of the template.
Essential substrates for DNA synthesis: dNTPs and primer:template junction (a combination of ___ regions)

Answer 2

DNA synthesis requires deoxynucleoside triphosphates and a primer:template junctiondNTP: (dATP,dCTP, dATP and dGTP). It has three phosphate attached to the 5’hydroxyl of the 2’ - doxyribose.
Template: provide the ssDNA that directs the addition of each complementary deoxynucleotides.
The primer is complementary to,
but shorter than the template. It must have a free 3’-OH adjacent to the ssDNA region of the template.
Essential substrates for DNA synthesis: dNTPs and primer: template junction (a combination of dsDNA and ssDNA
regions)

Question 3

DNA is synthesized by extending the 3’ end of the primer. DNA synthesis is initiated when the 3’-OH of the primer mediates the ___ of the __-phosphate of the incoming dNTP. This results in the extension of the 3’ end of the primer by one nucleotide and release one molecule of pyrophosphate. ___ rapidly hydrolyzes the released pyrophosphate into phosphate molecules.The orientation of the template is the __ of the growing DNA strand

Answer 3

DNA is synthesized by extending the 3’ end of the primer. DNA synthesis is initiated when the 3’-OH of the primer mediates the nucleophilic attack of the α-phosphate of the incoming dNTP. This results in the extension of the 3’ end of the primer by one nucleotide and release one molecule of pyrophosphate.
Pyrophosphatase rapidly hydrolyzes the released pyrophosphate into phosphate molecules.The orientation of the template is the opposite of the growing DNA strand

Question 4

XTP + (XMP)n (XMP)n+1 + (P~P). ∆G=__kcal/mol (rather small)
P~P 2Pi ∆G=__ kcal/mol
XTP + (XMP)n (XMP)n+1 + 2Pi ∆G=-10.5 kcal/mol

Answer 4

XTP + (XMP)n (XMP)n+1 + (P~P). ∆G=-3.5kcal/mol (rather small)
P~P 2Pi ∆G=-7 kcal/mol
XTP + (XMP)n (XMP)n+1 + 2Pi ∆G=-10.5 kcal/mo

Question 5

DNA polymerases use a single active site to catalyze DNA synthesis. The DNA polymerase monitors the ability of the incoming nucleotide to form A:T or G:C pair, rather than detecting the__.
Only when a correct base pair is formed are the __ of the primer and the__ of the incoming nucleotide triphopshate in the optimum position for catalysis to occur. (example of ___).

Answer 5

DNA polymerases use a single active site to catalyze DNA synthesis The DNA polymerase monitors the ability of the incoming nucleotide to form A:T or G:C pair, rather than detecting the exact nucleotide that enters the active site.
Only when a correct base pair is formed are the 3’-OH of the primer and the α-phosphate of the incoming nucleotide triphopshate in the optimum position for catalysis to occur. (example of kinetic proofreading).

Question 6

Steric constraints prevent DNA polymerase from using rNTP precursors. Although rNTPs are present at approximately 10-fold higher concentration in the cells, they are incorporated at a rate that is more than 1000-fold lower than dNTPs. This discrimination is mediated by the ___ of the rNTP from the DNA polymerase active site. In DNA Polymerase, the nucleotide-binding pocket cannot accommodate the __on the incoming nucleotide.

Discriminator amino acids: make van de Waals contact with the sugar ring. Change these amino acids to other amino acids with smaller ___ (i.e., by changing a glutamate to alanine) results in a DNA polymerase with significantly reduced ___.

Answer 6

Steric constraints preventing DNA polymerase from using rNTPprecursors. Although rNTPs are present at approximately 10-fold higher concentration in the cells, they are incorporated at a rate that is more than 1000-fold lower than dNTPs.This discrimination is mediated by the steric exclusion of the rNTP from the DNA polymerase active site. In DNA Polymerase, the nucleotide-binding pocket cannot accommodate the 2’-OH on the incoming nucleotide.
Discriminator amino acids: make van de Waals contact with the sugar ring. Change these amino acids to other amino acids with smaller side chains (i.e., by changing a glutamate to alanine) results in a DNA polymerase with significantly reduced discrimination between dNTPs and rNTPs

Question 7

Chemotherapeutic reagents that target DNA replication: Nucleotides that meet some but not all of the requirements for use by DNA Polymerase can inhibit DNA synthesis by terminating elongation. They can be used as anti-cancer or anti-viral drugs.

(a) 5-fluorouracil (pyromidineanalog)
(b) 6-mercaptopurine (purine analog)
(c) cytosine arabinoside (deoxycytidine analog): termination of elongation because of the difference between __ and __.
(d) Cisplatin (intrastrandDNA cross-link): the platinum atom binds covalently to the __ of __, forming interstrand and intrastrand ___.
(e) bis-chloroethylnitrosourea (intrastrandDNA cross-link)
(f) Azidothymidine (thymidine analog)that inhibits specialized DNA Polymerase (___). Used as anti-viral agents.
(g) Acyclovia(guanine analog): lack the __ group and therefore lacks the 3’-OH required for further nucleotide addition

Answer 7

Chemotherapeutic reagents that target DNA replication Nucleotides that meet some but not all of the requirements for use by DNA Polymerase can inhibit DNA synthesis by terminating elongation. They can be used as anti-cancer or anti-viral drugs.

(a) 5-fluorouracil (pyromidineanalog)
(b) 6-mercaptopurine (purine analog)
(c) cytosine arabinoside (deoxycytidine analog): termination of elongation because of the difference between deoxyribose and arabinose.
(d) Cisplatin (intrastrandDNA cross-link): the platinum atom binds covalently to the N7 of purines, forming inerstrand and intrastrand crosslinking.
(e) bis-chloroethylnitrosourea (intrastrandDNA cross-link)
(f) Azidothymidine (thymidine analog)that inhibits specialized DNA Polymerase (reverse transcriptase). Used as anti-viral agents
(g) Acyclovia(guanine analog): lack the ribose group and therefore the 3’-OH required for further nucleotide addition

Question 8

DNA Polymerase resembles a hand that grips the primer:template junction.
The palm domain is composed of a __and contains the primary elements of the __.
In particular, this region of DNA polymerase binds to two divalent ___ (typically __ or __). It also monitors the base-pairing of the most recently added nucleotides.
The thumb domain is not intimately involved in catalysis. It interacts with the ___

Answer 8

DNA Polymerase resembles a hand that grips the primer:template junction.
The palm domain is composed of a β-sheet and contains the primary elements of the catalytic site. In particular, this region of DNA polymerase binds to two divalent metal ions (typically Mg++ or Zn++). It also monitors the base-pairing of the most recently added nucleotides.
The thumb domain is not intimately involved in catalysis. It interacts with the DNA that have been recently synthesized

Question 9

Two metal ions bound to the DNA Polymerase catalyze
nucleotide addition. Metal ion A primarily interacts with the 3’-OH, resulting in ___. This leaves a nucleophilic 3’-O-. Metal ion B interacts with the ___to neutralize their negative charge. After catalysis, the pyrophosphate product is stabilized through interactions with ___. (the metal ions are usually Mg++ or Zn++)

Answer 9

Two metal ions bound to theDNA Polymerase catalyze
nucleotide addition
Metal ion A primarily interacts with the 3’-
OH, resulting in reduced association between the O and the H. This leaves a nucleophilic 3’-O-.
Metal ion B interacts with the triphosphate of the incoming dNTP to neutralize their negative charge. After catalysis, the pyrophosphate product is stabilized through interactions with metal ion B.

Question 10

DNA Polymerase “grips” the template and the incoming nucleotide when a correct base pair is made. Fingers are composed of __. Several residues located within the fingers bind to the incoming dNTP. More importantly, once a correct base pair is formed between the incoming dNTP and the template, the finger domain moves to ___. , therefore stimulating catalysis.
i.e. __ makes stacking interaction with the base of dNTP, and the two charged residue (Lys and Arg) associates with the ___.

Answer 10

DNA Polymerase “grips” the template and the incoming nucleotide when a correct base pair is made. Fingers are composed of α-helics. Several residues located within the fingers bind to the incoming dNTP. More importantly, once a correct base pair is formed between the incoming dNTP and the template, the finger domain moves to enclose the dNTP, therefore stimulating catalysis.
i.e. Tyr makes stacking interaction with the base of dNTP, and the two charged residue (Lys and Arg) associates with the triphosphate

Question 11

The Path of the template DNA through the DNA
Polymerase: The finger domain also associates with the ___, leading to a nearly 90° turn of the phosphodiester backbone between the first and the
second base of the template. (making sure only the first template base is ___)

Answer 11

The Path of the template DNA through the DNA
Polymerase: The finger domain also associates with the template region, leading to a nearly 90° turn of the phosphodiester backbone between the first and the
second base of the template. (making sure only the first template base is exposed to catalytic site)

Question 12

DNA Polymerases are processive enzymes. Catalysis by DNA Polymerase is rapid. DNA Poymerases can add as many as __ nucleotides/sec to a primer strand. This is due to the processive nature of the DNA polymerase.The degree of processivity of DNA Polymerase is defined as the ___. Processivity is facilitated by sliding of DNA Polymerase along the DNA template. Once bound to the primer:tamplate junction, DNA Polymerase interacts tightly with much of the double-stranded portion of the DNA in a ___ manner. The interactions include electrostatic interactions between the __ and __ interaction, as well as between __ and __ . Each time a nucleotide is added to the primer strand, the DNA partially ___(the minor groove are broken, but the electrostatic interactions with the thumb are maintained). The DNA then rapidly ___ in a position that is shifted by ___ using the same sequence-nonspecific mechanism.

Answer 12

DNA Poymerases can add as many as 1000 nucleotides/secto a primer strand. The degree of processivity of DNA Polymerase is defined as the average number of nucleotides added each time the enzyme binds a primer:template junction. Processivity is facilitated by sliding of DNA Polymerase along the DNA template. Once bound to the primer:tamplate junction, DNA Polymerase interacts tightly with much of the double-stranded portion of the DNA in a sequence-independent manner. The interactions include electrostatic interactions between the thumb domain and phosphate backbone interaction, as well as between palm domain and minor groove of DNA (as described in slide 16). Each time a nucleotide is added to the primer strand, the DNA partially releases from the polymerase (The hydrogen bonds with the minor groove are broken, but the electrostatic interactions with the thumb are maintained). The DNA then rapidly rebinds to the polymerase in a position that is shifted by 1bp using the same sequence-nonspecific mechanism

Question 13

___ proofread newly synthesized DNA.
DNA. replication error rate: __
DNA Polymerase error rate: __.
Proofreading exonuclease: originally identified in the same peptide as DNA Polymerase, degrade DNA starting from a ___ end.
Exonuclease: degrade DNA from a DNA end. Endonuclease: cut within a DNA strand

Answer 13

Exonucleases proofread newly synthesized DNA
DNA. replication error rate: 1 in 10~10
DNA Polymerase error rate: 1 in 10~5.
Proofreading exonuclease: originally identified in the same peptide as DNA Polymerase, degrade DNA starting from a 3’ DNA end.
Exonuclease: degrade DNA from a DNA end. Endonuclease: cut within a DNA strand

Question 14

Both strand of DNA are synthesized together at the replication fork In the cells, both strands of the DNA duplex are replicated. This requires the separation of the two strands of the double helix to create two template DNAs.
Replication fork: the junction between replicated DNA and the unreplicated DNA. DNA replication in E. coli (and in other organisms) is ____. One strand (the __ strand) is replicated continuously in the direction of the movement of the replicating fork. The other strand (the lagging strand) is replicated discontinuously as __kb (in bacteria) and __ bp in eukaryotes (both are called Okazaki fragments) in the opposite direction.This allows for both strands to be ___

Answer 14

Both strand of DNA are synthesized together at the replication fork.In the cells, both strands of the DNA duplex are replicated. This requires the separation of the two strands of the double helix to create two template DNAs.Replication fork: the junction between replicated DNA and the unreplicated DNA. DNA replication in E. coli (and in other organisms) is semi-discontinuous. One strand (the leading strand) is replicated continuously in the direction of the movement of the replicating fork.The other strand (the lagging strand) is replicated discontinuously as 1-2 kb (in bacteria) and 100-400 bp in eukaryotes (both are called Okazaki fragments) in the opposite direction.This allows both strands to be replicated in the 5’ to 3’-direction.

Question 15

The initiation of a new strand of DNA requires an RNA primer. DNA Polymerase cannot initiate a new DNA strand ____.. Take advantage of the RNA Polymerase. Primase: is a specialized RNA Polymerase dedicated to make short RNA primers (5-10 nucleotides long) on an ssDNA template. Primase activity is dramatically increased when it is associated with ___. The requirement of ssDNA and DNA helicase association ensures that primase ____

Answer 15

The initiation of a new strand of DNA requires an RNA primer. DNA Polymerase cannot initiate a new DNA strand de novo.Take advantage of the RNA Polymerase. Primase: is a specialized RNA Polymerase dedicated to make short RNA primers (5-10 nucleotides long) on an ssDNA template. Primase activity is dramatically increased when it is associated with another protein that acts at the replication fork called DNA helicase.The requirement of ssDNA and DNA helicase association ensures that primase is only active at the replication fork

Question 16

RNA primers must be removed to complete DNA
Replication.
RNaseH (hybrid): specfically degrades __ that is __ (only cleaves bonds between __). The final ribonucleotide is removed by ___.
DNA Polymerase fills in the gap. DNA ligase seal the “nick”. DNA ligase uses __to create a phosphodiester bond between an adjacent __.

Answer 16

RNA primers must be removed to complete DNA
Replication.
RNaseH (hybrid): specfically degrades RNA that is base-paired with DNA (only cleaves bonds between ribonucleotides). The final ribonucleotide is removed by 5’ exonuclease.
DNA Polymerase to fill the gap.DNA ligase to seal the “nick”.
DNA ligase uses high energy co-factor (such as ATP) to create a phosphodiester bond between an adjacent 5’-phosphate and 3’-OH.

Question 17

Single-strand DNA—binding proteins stabilize ssDNA before replication.
SSBs: ssDNA-binding proteins, binds to the s__.
Cooperative binding: binding of one SSB promotes the binding of ___. That ’s because SSB molecules bind to each other.
SSBs interacts with ssDNAin a __manner. SSBs contacts ssDNA through electrostatic interactions with the __and stacking interaction with the __. When coated with SSBs, the ssDNA assumes a __ that inhibits ____.

Answer 17

Single-strand DNA—binding proteins stabilizes ssDNA before replication. SSBs: ssDNA-binding proteins, binds to the separated strands.
Cooperative binding: binding of one SSB promotes the binding of another SSB to the immediately adjacent ssDNA. That ’s because SSB molecules bind to each other.SSBs interacts with ssDNA in a sequence-independent manner.
SSBs contacts ssDNAthrough electrostatic interactions with the phosphate backbone and stacking interaction with the DNA bases.
When coated with SSBs, the ssDNAassumes a more extended conformation that inhibits the formation of intramolecular base pairs.

Question 18

Topoisomerases remove supercoils produced by DNA unwinding at the replication fork.
As __ supercoils accumulate in front of the replication fork, topoisomerases rapidly remove them. The enzymes break one or both strands of DNA without ___and pass the same number of DNAs strands through the break.It reduces ___, thus would only have to occur every __bp replicated.

Answer 18

Topoisomerases remove supercoils produced by DNA unwinding at the replication fork.
As positive supercoils accumulate in front of the replication fork, topoisomerases rapidly remove them. The enzymes break one or both strands of DNA without letting go of the DNA and pass the same number of DNAs strands through the break.It reduces the linking number by two, thus would only have to occur every 20bp replicated

Question 19

DNA helicase and DNA topoisomerase perform their function without __ or synthesizing new molecules.
DNA helicase only breaks ___. Topoisomerase breaks one or two of the target DNA’s __, each bond broken is precisely reformed. These proteins act at the replication fork in a sequence independent manner.

Answer 19

DNA helicase and DNA topoisomerase perform their function without altering DNA chemical structure, or synthesizing new molecules.
DNA helicase only breaks the hydrogen bonds that hold two strands of DNA together. Topoisomerase breaks one or two of the target DNA’s covalent bonds, each bond broken is precisely reformed.These protein act at the replication fork in a sequence independent manner.

Question 20

There are 3 DNA polymerases, the enzymes that make DNA, found in E. coli:
–pol I
–pol II
–pol III
•E. coli DNA polymerase __ was the first polymerase identified.
•It was discovered in 1958 by ___
•Pol III is the primary enzyme involved in the __

Answer 20

There are 3 DNA polymerases, the enzymes that make DNA, found in E. coli:
–pol I
–pol II
–pol III
•E. coli DNA polymerase I was the first polymerase identified.
•It was discovered in 1958 by Arthur Kornberg
•Pol III is the primary enzyme involved in the replication of thechromosome

Question 21

E. coli DNA Polymerase I is specialized for the ___that are used to initiate DNA synthesis.
• Pol I is not highly ___, adding only 20-100 nucleotides per binding event.
•DNA polymerase I (pol I) is a versatile enzyme with 3 distinct activities:
1.__
2.___
3. ____ which removes the RNA-DNA linkage that is resistant to ___.
–Mild proteolytic treatment results in 2 polypeptides: the Klenow fragment (the large domain) and smaller fragment

Answer 21

E. coli DNA Polymerase I is specialized for the removal of the RNA primers that are used to initiate DNA synthesis.
• Pol I is not highly processive, adding only 20-100 nucleotides per binding event.
•DNA polymerase I (pol I) is a versatile enzyme with 3 distinct activities
1. DNA polymerase
2. 3’→5’exonucleas
3. 5’→3’exonuclease: remove the RNA-DNA linkage that is resistant to RNaseH.
–Mild proteolytic treatment results in 2 polypeptides: the Klenow fragment (the large domain) and smaller fragment

Question 22

Klenow Fragment Contains both___ and __ exonuclease activity, which serves as proofreading:
–If pol I added wrong nt, won’t base pair properly
–Pol I pauses, exonuclease removes mispaired nt
–Allows replication to continue
–Increases__ of replication

Answer 22

Klenow Fragment Contains both: Polymerase and 3’→5’ exonuclease activity, which serves as proofreading
–If pol I added wrong nt, won’t base pair properly
–Pol I pauses, exonuclease removes mispaired nt
–Allows replication to continue
–Increases fidelity of replication

Question 23

Eukaryotic DNA Polymerases. More than 15 Polymerases. Three of them are essential: DNA Pol δ, DNA Pol ε, and DNA Pol α/primase.
DNA Pol α/primase consists of ___. It is specifically involved in initiating new DNA strands.
DNA Pol δ specializes in synthesizing the __,
DNA Pol ε: specializes in synthesizes the ___.
he majority of the rest of the DNA polymerases are involved in ___.

Answer 23

Eukaryotic DNA Polymerases
More than 15 Polymerases. Three of them are essential: DNA Pol δ, DNA Pol ε, and DNA Pol α/primase. DNA Pol α/primaseconsists of a two-subunit DNA Pol αand a two-subunit primase. It is specifically involved in initiating new DNA strands.
DNA Pol δis specializing synthesizing the lagging strand, DNA Pol ε: leading strands.
The majority of the rest of the DNA polymerases are involved in DNA repair.

Question 24

In euks, DNA Polymerase switching occurs during eukaryotic DNA replication. Because of its low processivity, DNA Pol α/primase is rapidly replaced by __ and ___.

Answer 24

DNA Polymerase switching during eukaryotic DNA replication. Because of its low processivity, DNA Pol
α/primase is rapidly replaced by DNA Pol δ and DNA Pol ε.

Question 25

Sliding clamps dramatically increase DNA polymerase processivity. The proteins of the clamp are composed of multiple identical subunits assembled like a doughnut. The hole in the center is about 34A, larger than the width of ___ (20A). This provide enough space for___.
This is thought to allow the clamp to slide along the DNA easily. Sliding DNA clamp interacts with the part of the DNA polymerase that is ___as it emerges from the DNA Polymerase.

Answer 25

Sliding clamps dramatically increase DNA polymerase processivity. The proteins of the clamp are composed of multiple identical subunits assembled like a doughnut. The hole in the center is about 34A, larger than the width of DNA helix (20A).
This provide enough space for a thin layer of one or two water molecules between the sliding clamp and the DNA. This is thought to allow the clamp to slide along the DNA easily.
Sliding DNA clamp interacts with the part of the DNA polymerase that is closest to the newly synthesized DNA as it emerges from the DNA Polymerase.

Question 26

In the absence of sliding clamp, a DNA polymerase dissociates and diffuses away from the template DNA on average once every ___bp synthesized. The association with the sliding clamp prevents it from diffusing away from the DNA. After DNA Polymerase has completed synthesis of the template, the absence of a ___ causes a change in the DNA Polymerase that releases it from the sliding clamp

Answer 26

In the absence of sliding clamp, a DNA polymerase dissociates and diffuses away from the template DNA on average once every 20-100bp synthesized. The association with the sliding clamp prevents it from diffusing away from the DNA. After DNA Polymerase has completed synthesis of the template, the absence of primer:template junction causes a change in the DNA Polymerase that releases it from the sliding clamp

Question 27

Sliding clamps are opened and placed on DNA by clamp loaders. Clamp loaders are 5-subunit protein complex. To catalyze Sliding clamp opening, the clamp loader must be __. Once bound to ATP, the clamp loader binds the clamp and ___at one of the ____. Now the complex can bind to DNA, preferentially to the primer: template junctions. dsDNA position to the “hole” of the clamp. Clamp loader releases by ___. Sliding clamp loaders and DNA Polymerase cannot interact with each other because they have ____ on the sliding clamp

Answer 27

Sliding clamps are opened and placed on DNA by clamp loaders Clamp loaders are 5-subunit protein complex. To catalyze Sliding clamp opening, the clamp loader must be bound to ATP.
Once bound to ATP, the clamp loader binds the clamp and opens the ring at one of the subunit:
subunit interfaces. Now the complex can bind to DNA, preferentially to the primer: template junctions.
dsDNA position to the “hole” of the clamp. Clamp loader release by ATP hydrolysis. Sliding clamp loaders and DNA Polymerase cannot interact with each other because they have overlapping binding sites on the sliding clamp

Question 28

The replicon model of replication initiation: DNA synthesis generally initiates at internal regions.
1963, Francois Jacob proposed a replicon model to explain replication initiation in bacteria.
Replicon: particular region of DNA required for initiation of DNA replication. The replicon model proposed two components that controlled the initiation of replication: the replicator and the initiator. Replicator is defined as the___ that are sufficient to direct the ___.
This is in contrast to the origins of replication, the specific sites at which DNA unwinding and initiation of replication occur. Origin of DNA replication is always part of the replicator.
Initiator: the protein specifically recognizes __ in the __ and __. Contains a common core ___, regulated by ATP binding and hydrolysis. The initiator protein is the only sequence-specific DNA-binding protein involved in the initiation of replication.
AAA or AAA+: ___associated with diverse cellular activity. (They share a conserved domain of about 23o residues

Answer 28

Replicator is defined as the cis-acting DNA sequence that are sufficient to direct the initiation of DNA replication.
This is in contrast to the origins of replication, the specific sites at which DNA unwinding and initiation of replication occur. Origin of DNA replication is always part of the replicator.
Initiator: the protein specifically recognizes a DNA element in the replicator and activates the initiation of replication. Contains a common core AAA+ ATP binding motif, regulated by ATP binding and hydrolysis. The initiator protein is the only sequence-specific DNA-binding protein involved in the initiation of replication.
AAA or AAA+: ATPase sassociated with diverse cellular Activity. (They share a conserved domain of about 23o residues

Question 29

A DNA fragment containing an origin of replication will take on a __ shape. A DNA fragment that doesn’t contain a origin of DNA replication will take on a variety of ____This method requires the investigator already know___.

The Y-shaped molecule that results from the ___ is similar in shape to a linear molecule two times the size of the unreplicated fragment. ___ will migrate most slowly of all such molecules derived from a particular DNA fragment.
In contrast, Y-shaped molecules with ___ or ___ will migrate similarly to the unreplicated version of the same DNA fragment. (bubble-to-arch transition is easily detected as a ___ and is highly indicative of origin of replication

Answer 29

A DNA fragment containing an origin of replication will take on a “bubble” shape. A DNA fragment that doesn’t contain a origin of DNA replication will take on a variety of “Y-shaped” shape. This method requires the investigator already knows the approximate location of the potential origin of replication.

The Y-shaped molecule that results from the almost completely replicated fragment is similar in shape to a linear molecule two times the size of the unreplicated fragment. Y-shaped molecule that has three equal length arms will migrate most slowly of all such molecules derived from a particular DNA fragment.
In contrast, Y-shaped molecules with very short replicated arms or large replicated regions will migrate similarly to the unreplicated version of the same DNA fragment. (bubble-to-arch transition is easily detected as a discontinuity in the arc and is highly indicative of origin of replication.

Question 30

E. Coli DNA replication is regulated by DnaA•ATP and SeqA
(a)Before DNA replication, GATC sequences
throughout the E. Coli genome are __
on both strands
(b) DNA replication converts these sites to a __ state.
(c)Hemimethylated DNA are rapidly bound by SeqA.
(d) Bound SeqA protein inhibits___and __.
(e)When seqA infrequently dissociates from the GATC sites, the sequences can become fully methylated by ___, preventing rebinding by SeqA (It adds methyl group to the A within every GATC sequence)
(f) When the GATC sites become fully methylated, DnaA
binds to the __ and direct a new round of replication form the daughter ___.

Answer 30

(a)Before DNA replication, GATC sequences
throughout the E. Coli genome are methylated
on both strands (shown by red hexagons)
(b)DNA replication converts these sites to hemimethylated state.
(c)Hemimethylated DNA are rapidly bound by SeqA.
(d) Bound SeqA protein inhibits the full methylation of the sequences and binding of OriCby DnaA proteins
(e)When seqAinfrequently dissociates from the GATC sites, the sequences can become fully methylated by Dam DNA methytransferase, preventing rebinding by
SeqA (It adds methyl
group to the A within every GATC sequence)
(f) When the GATC sites become fully methylated, DnaA
binds to the 9-mer sequences and direct a new round of replication form the daughter oriC replicators.

Question 31

Helicase loading is the first step in the initiation of
replication in eukaryotes. Eukaryotic replication initiation occurs at distinct times in the cell cycle. Helicase loading occurs during __; Replicator activation occurs after cells enter ___. (different from prokaryotes);
A. Association of ATP-bound origin recognition complex (ORC) with the replicator;
B. In G1 phase, ORC bound to the origin recruits two ___ (ATP-bound cdc6 and Cdt1) and two copies of the Mcm2-7 helicase to the origin;
C.The assembly of proteins triggers ATP hydrolysis by __, resulting in the loading of___ of the __encircling double strand origin DNA and the release of cdc6 and cdt1 from the origin;
D.Subsequent ATP hydrolysis by ORC is required for ____

Answer 31

Helicase loading is the first step in the initiation of
replication in eukaryotes. Eukaryotic replication initiation occurs at distinct times in the cell cycle. Helicase loading occurs during G1 phase; Replicator activation occurs after enter S phase.
(different from prokaryotes);
A.Association of ATP-bound origin recognition complex (ORC) with the replicator;
B. In G1 phase, ORC bound to the origin recruits two helicase loading protein (ATP-bound cdc6 and Cdt1) and two copies of the Mcm2-7 helicase to the origin;
C.The assembly of proteins triggers ATP hydrolysis by cdc6, resulting the loading of head-to-head dimer of the Mcm2-7 complex encircling double strand origin DNA and the release of cdc6 and cdt1 from the origin;
D.Subsequent ATP hydrolysis by ORC is required for helicase release and reset of the process

Question 32

Similarity between eukaryotic and prokaryotic DNA replication initiation:
A.recognition of the replicator by the initiator protein.
B.The initiator protein in combination with one or more helicase loading proteins assembles the DNA helicase on the replicator.
C.The helicase generates a region of
ssDNA that can act as a template for RNA primer
synthesis.
D.Once the primers are synthesized, the remaining components of the replisome assemble through interactions with the resulting primer: template junction.
•Difference: Regulation.
A.In bacterial cells, rapidly dividing cells initiate replication
more than once per cell cycle.
B. Eukaryotic cells focus on regulation on the initial loading of the MCM helicase onto the origin DNA. Whereas bacterial cells focus regulation on the binding of the DnaA initiator protein to the DNA (replication is regulated by Dna•ATP levels and SeqA)

Answer 32

Similarity between eukaryotic and prokaryotic DNA replication initiation:
A.recognition of the replicator by the initiator protein.
B.The initiator protein in combination with one or more helicase loading proteins assembles the DNA helicase on the replicator.
C.The helicase generates a region of ssDNA that can act as a template for RNA primer synthesis.
D.Once the primers are synthesized, the remaining components of the replisome assemble through interactions with the resulting primer: template junction.
•Difference: Regulation.
A.In bacterial cells, rapidly dividing cells initiate replication more than once per cell cycle.
B. Eukaryotic cells focus on regulation on the initial loading of the MCM helicase onto the origin DNA. Whereas bacterial cells focus regulation on the binding of the DnaA initiator protein to the DNA (replication is regulated by Dna•ATPlevels and SeqA)

Question 33

The end replication problem
A. As the lagging-strand replication machinery reaches the end of the chromosome, at some point, ___ no longer has sufficient spaces to synthesize a new RNA primer. (even with the end primer present, once the RNA is removed, there will be short region of unreplicated ssDNA at the end of the chromosome)
B. This results in incomplete replication and a short ssDNA region at the 3’ end of the lagging-strand DNA product.
C. When the DNA product is replicated in the next round, one of the two products will be __and will ___

Answer 33

The end replication problem
A. As the lagging-strand replication machinery reaches the end of the chromosome, at some point, primase no longer has sufficient spaces to synthesize a new RNA primer. (even with the end primer present, once the RNA is removed, there will be short region of unreplicated ssDNA at the end of the chromosome)
B. This results in incomplete replication and a short ssDNA region at the 3’ end of the lagging-strand DNA product.
C. When the DNA product is replicated in the next round, one of the two products will be shortened and will lack the region that was not fully copied in the previous round of replication.

Question 34

Protein priming as a solution to the end replication problem. By binding to the ___ and to the __, a protein provides a priming hydroxyl group (typically tyrosine) to initiate DNA synthesis. In the example shown, the protein primers all DNA synthesis has is seen for many viruses. For longer DNA molecules, this methods combines with conventional origin function to replicate the chromosomes.(Bacterial and viruses)

Answer 34

Protein priming as a solution to the end replication problem. By binding to the DNA Polymerase and to the 3’ end of the template, a protein provides a priming hydroxyl group (typically tyrosine) to initiate DNA synthesis. In the example shown, the protein primers all DNA synthesis as is seen for many viruses. For longer DNA molecules, this methods combines with conventional origin function to replicate the chromosomes.(Bacterial and viruses)

Question 35

Replication of the telomeres by telomerase.
Telomere: ends of the eukaryotic chromosomes, generally are composed of head-to-tail repeats of ___ DNA sequences. In human, 5’-TTAGGG-3 ’.
Telomerase uses its ___ to anneal to the 3’ end of the ssDNA region of the telomere. Telomerase uses its ___ activity to synthesize DNA to the end of the RNA template. Telomerase then displaces the RNA from the DNA product and rebinds at the end of the telomere and repeats the process

Answer 35

Telomere: ends of the eukaryotic chromosomes, generally are composed of head-to-tail repeats of TG-rich DNA sequences. In human, 5’-TTAGGG-3 ’
Telomerase uses its RNA component to anneal to the 3’ end of the ssDNA region of the telomere. Telomerase uses its reverse transcription activity to synthesize DNA to the end of the RNA template. Telomerase then displaces the RNA from the DNA product and rebinds at the end of the telomere and repeats the process

Question 36

Extension of the 3’ end of the telomere by telomerase
solves the problem of end replication. When the 3’ end is extended by telomerase, how is the 5’ end extended? Using___. Even after the action of the lagging-strand machinery, there remains a short ssDNA region at the end of the chromosome. Indeed, the presence of a __ may be important for the end protection function of the telomerase. Nevertheless, telomerase and the___mechanism ensures the telomere is maintained at sufficient length to protect the end of the chromosome from shortening.

Answer 36

Extension of the 3’ end of the telomere by telomerase
solves the problem of end replication. When the 3’ end is extended by telomerase, how is the 5’ end extended? Use lagging-strand DNA replication machinery. Even after the action of the lagging-strand machinery, there remains a short ssDNA region at the end of the chromosome. Indeed, the presence of a 3’ overhang may be important for the end protection function of the telomerase. Nevertheless, telomerase and the lagging-strand replication mechanism ensures the telomere is maintained at sufficient length to protect the end of the chromosome from shortening.

Question 37

Telomere-binding proteins regulate telomerase activity and telomere length (a). S. cerevisiae cells. Rap1 directly
binds to the double-stranded telomere repeats DNA. Rif1, Rif2 binds to __. All these proteins have been implicated in the Inhibition of telomerase activity. Cdc13 binds to __ and is involved in ___.
(b) Human cells: TRF1 and TRF2 bind directly to the __. Human R1p1, TIN2, TPP1 and POT1 all associated with ___. Together these protein form a complex called __ (shelter the telomeres from the action of DNA repair enzymes). __ also binds directly to the single-stranded telomere repeat DNA and inhibits telomerase activity.

Answer 37

Telomere-binding proteins regulate telomerase activity and telomere length (a). S. cerevisiaecells. Rap1 directly
binds to the double-stranded telomere repeats DNA. Rif1, Rif2 binds to Rap1. All these proteins have been implicated in the Inhibition of telomerase activity. Cdc13 binds to the single-stranded telomere repeat DNA and is involved in telomerase recruitment.
(b) Human cells: TRF1 and TRF2 bind directly to the double-stranded telomere repeat DNA. Human R1p1, TIN2, TPP1 and POT1 all associated with TRF1 or TRF2. Together these protein form a complex called shelterin (shelter the telomeres from the action of DNA repair enzymes). POT1 also binds directly to the single-stranded telomere repeat DNA and inhibits telomerase activity.