Exam 2 Study Guide

Question 1

What are tumor suppressor genes and proto-oncogenes? (Ch 9)

Answer 1

Proto-oncogenes: components of signal transduction pathways
–when mutated, can result in cancer

Tumor suppressor genes: genes whose products prevent the activation of CdK
enzymes when criteria have not been met (Rb and p53)

Question 1

Describe the general steps of G1/S checkpoint passage in eukaryotes. How is this passage related to
cyclins and Cdks? (Ch 9)

Answer 1

Steps of checkpoint passage:
1. Cell must express the correct growth factor receptor
2. Cell must have growth factor present to bind to the receptor
–Signal from cell membrane receptor to the nucleus is mediated by signal transduction pathways
3. Cell must receive continued growth factor signal long enough to trigger the production of active cyclin proteins
4. Cyclin proteins must activate a set of enzymes called cyclin-dependent kinases (Cdks)

Question 2

Describe the steps of Rb-mediated and p53-mediated control of cell cycle passage. How are the actions of these two proteins interrelated with respect to the control of cell cycle passage? (Ch 9)

Answer 2

Rb:
1. With continued growth factor signal, G1-Cdk (Cyclin D-Cdk) activity is stimulated
2. Active G1-Cdk will phosphorylate Rb (usually found in an active, unphosphorylated form)
3. Phosphorylated Rb will be inactivated and release the transcription factor E2F
4. E2F will activate the transcription of genes required for entry into S phase
5. Downstream components participate in positive feedback of E2F

p53:
– Prevents passage through checkpoint if DNA damage has occurred
– Transcription of p53 is upregulated by growth factor signal in preparation for restriction point decision
– Activity is under negative regulation by Mdm2
–Transcription of Mdm2 is upregulated by p53
–Binds to p53 and targets it to degradation pathways
–Also binds to p53 and inhibits its function as a transcriptional activator

p53 after DNA damage:
– p53 and Mdm2 are
phosphorylated
– Mdm dissociates from p53
– p53 degradation is decreased
– p53 translocates to the nucleus and activates the expression of genes that repress cell cycle progression
–p21 protein will bind to and inactivate G1/S-Cdk and S- Cdk complexes needed for passage through the restriction point and G1/S transition

Rb and p53
– Rb and p53 are negative effectors of cell cycle progression
–Tumor suppressor genes
– The action of these effectors (and others) must be integrated to facilitate rate of overall progression
– p21
–Expressed as a result of DNA damage and p53 activity
–Binds to cyclin/CdK complexes needed for Rb inactivation

Question 3

Explain how Dam methylase is associated with DnaA, oriC, and the inhibition of reinitiation of
replication. (Ch 10)

Answer 3

oriC: the replication origin of E. coli

DnaA: Bacterial replication initiator protein

OriC contains 11 copies of the palindromic sequence GATC
–The adenine in this palindrome is a target for methylation by Dam methylase enzymes

• Dam methylation sites are scattered throughout the genome, but several overlap DnaA boxes
• Before replication begins, the adenines of the palindromic sequence are methylated on both parental strands by Dam
• Replication incorporates nonmethylated adenines into the daughter strand
  –Hemimethylated DNA
• Only fully methylated origins can initiate replication
  –Hemimethylated origins inhibit initiation

Question 4

How is the methylation status of oriC related to its ability to initiate replication? (Ch 10)

Answer 4

Initiation at oriC begins with formation of a large six- membered protein complex
–DnaA, DnaB helicase, DnaC, HU, gyrase, SSB

DnaA binds ATP
–Only DnaA-ATP will bind
a fully methylated oriC
–oriC must be associated with the cell membrane for DnaA-ATP binding to occur

Question 5

Review the steps involved in bacterial initiation beginning with DnaA-ATP binding to oriC. (Ch 10)

Answer 5

1. DnaA-ATP initially binds as an extended multimeric form at the 9 bp repeats
2. Multimeric extended DnaA- ATP forms a helical core around which oriC is wound
3. Multimeric extended DnaA-ATP binds to the 13 bp repeats
4. Multimeric extended DnaA-ATP converts to a compact form that twists open and stabilizes a ssDNA bubble
   –Transcription of flanking genes assists in opening bubble
5. DnaA then recruits two “pre-priming” complexes
   –One hexameric DnaB bound to six DnaC-ATP monomers
   –One complex per replication fork in bidirectional replication
   –DnaB is the helicase responsible for unzipping the replication fork along the parental duplex
   —Cannot initially denature dsDNA
   —Can only extend ssDNA that has already been opened
   –DnaC is the DnaB helicase loader
   —Serves as a chaperone to repress DnaB activity until it is needed
6. DnaB also catalyzes the loading of DnaG onto the DnaB-(DnaC-ATP) pre-priming complex at each fork
   –DnaG is the primase required for initiating polymerase action on the leading and lagging strands at each fork
7. Loading of DnaG onto the pre-priming complex
   stimulates the dissociation of ATP from DnaC-ATP
   –The dissociation of ATP from DnaC-ATP will cause DnaC to be released from the pre-priming complex
   –The release of DnaC from the pre-priming complex will activate DnaB
8. DnaB will displace DnaA from oriC as it extends the bubble
9. Gyrase is also recruited
   –Type II topoisomerase
   –Allows one DNA strand to rotate around the other
   –Reduces positive supercoiling ahead of each fork that would inhibit helicase action
10. As helicase opens the bubble, the ssDNA is immediately stabilized by single-strand binding proteins
    –Modulates helicase action
    11.Protein HU is also found at the replicon
    –Stimulates initiation
    –General DNA binding protein
    –Thought to bend DNA and catalyze bubble formation

Question 6

Describe the three ways an E. coli cell prevents premature reinitiation of replication. (Ch 10)

Answer 6

1. OriC is sequestered from Dam methylase
   –The SeqA protein binds tightly to GATC palindromes in hemimethylated DNA
   —Specificity of SeqA to oriC palindromes that are bound by DnaA?
   –SeqA protein remains at the origin after initiation and inhibits Dam methylase action
   —Prevents methylation of new daughter strand
2. The amount of DnaA available for binding to oriC depends upon competition for its binding to other sites
   –More than 300 DnaA binding sites available outside of oriC
   —DnaA acts as transcriptional regulator at the promoter for several genes
   –Number of binding sites doubles following replication
   –Control of binding many DnaA molecules at dat locus
   —Dat is a site away from oriC that has a large number of DnaA binding sites
   —Binds a larger number of DnaA than the origin
3. Influence of Hda protein on ATPase activity of DnaA
   –Hda is recruited to the origin by DNA polymerase
   –Hda enhances ATPase activity of DnaA
   —DnaA-ADP is unable to bind to oriC and initiate
   replication
   –Exchange of DnaA-ADP for DnaA-ATP is slow
   –Delays accumulation of active DnaA available for initiation

Question 7

What is a licensing factor and what is the licensing factor in prokaryotes and in eukaryotes? (Ch 10)

Answer 7

Licensing factor: A factor located in the nucleus and necessary for replication; it is inactivated or destroyed after one round of replication. New factors must be provided for further rounds of replication to occur.

In prokaryotes, DnaA is the licensing factor

In eukaryotes, MCM is the licensing factor (cdc6 and cdt1?????)

Question 8

What is the pre-RC? Describe how the ORC, Cdc6, Cdt1, and Mcm2-7 are involved in initiation of replication in eukaryotes. (Ch 10)

Answer 8

Pre-replicative complex (Pre-RC): Complex of four components that must form in an ordered manner at each replicator in order to initiate replication (ORC, Cdc6, Cdt1, Mcm)

1. Binding of the origin recognition complex to an origin
   –ORC is highly conserved in all eukaryotes and can bind origins anytime during the cell cycle
   –Only participates in initiation of replication when part of a pre-RC
2. Cdc6 binds to the replicator only in late M phase and early G1
   –Cdc6 is deactivated and degraded as S-Cdk levels increase in S phase
3. Active Cdt1 is only present in G1
   –The geminin protein inactivates free Cdt1
   –Geminin is specifically degraded in early G1
4. Cdt1 escorts Mcm to the origin
   –Mcm2-7 contains the eukaryotic helicase
5. The pre-RC has now been formed
6. Cdc6 and Cdt1 are released
7. Mcm is fully loaded onto the origin

Question 9

How is Cdk activity related to helicase loading and activation? How does this prevent premature reinitiation of replication at each origin in eukaryotes? How are the Cdc45 and GINS proteins involved in this process? (Ch 10)

Answer 9

In the S, G2, and early M phases, S-Cdk levels are high, so helicase activation is allowed, but helicase loading is not
–Once helicases move beyond the origin, they must be reloaded to initiate another replication event

As S-Cdk levels rise to their maxima in the S phase, Cdc45 and GINS bind Mcm.
–Mcm and ORC are phosphorylated
—The helicase is activated
—ORC is inactivated
–The Cdc45-GINS-Mcm complex recruit additional proteins that form the active replicative helicase
—CMG helicase
–The CMG helicases are loaded onto single stranded DNA and move away from the origin
–ORC is displaced
–Any inactive origins with bound ORC that are passed by the replication fork will be inactivated

Question 10

What are the differences between high fidelity and error-prone polymerases? What general functions within the cell do these polymerases perform? In which of these two classes do each of the E. coli polymerases fall? (Ch 11)

Answer 10

High-fidelity DNA polymerases have a precisely constrained active site that favors binding of standard base pairs
–Rate of incorporation of incorrect nucleotides is 10,000X slower than correct nucleotides
–3’-OH and a-phosphate must be in optimum position

Error-prone polymerase: Polymerase on lesion strand is removed and replaced with an error-prone polymerase, and damage is bypassed
–High fidelity polymerase replaces error-prone polymerase

E. coli polymerases:
–Polymerase III: primary replicase
–Polymerase II: required to restart the replication fork after it is stopped by damage
–Polymerase I: involved in both error repair and replication
–Polymerases IV and V: error-prone polymerases that allow bypass around DNA damage

Question 11

What types of mutational errors do the high fidelity and increased processivity of DNA polymerase III
prevent? What are some of the mechanisms that DNA polymerase III uses to maintain this high fidelity? (Ch 11)

Answer 11

High-fidelity DNA polymerases have a precisely constrained active site that favors binding of standard base pairs

prevents transversions and transitions

3’-OH and alpha-phosphate must be in optimum position
–Kinetic proofreading

A sugar with a 2’-OH cannot be easily accommodated in the nucleotide-binding pocket

Question 12

Describe the general structure of DNA polymerases. What is the function of each of the three domains of this class of enzymes? (Ch 11)

Answer 12

Many DNA polymerases have a structure that resembles a right hand and contains three domains
–Palm = Primary elements of the polymerase catalytic site
–Fingers = Binds to incoming dNTPs and moves correct dNTPs into close contact with polymerase catalytic site
—“Tightens” upon dNTP binding but before addition for initial proofreading step
–Thumb = Maintains the correct position of the 3’-OH and also a strong
association between the polymerase and the parental strand to facilitate processivity

Question 13

What role do metal cations have in the enzymatic activities of DNA polymerases? (Ch 11)

Answer 13

The palm domain is composed of a β sheet and contains the primary elements of the catalytic site

Also contains two divalent metal cations that interact with the 3’-OH of the primer and the correct paired dNTP
–One metal ion reduces the affinity of 3’-OH for its hydrogen
—More nucleophilic
–Other metal ion stabilizes the negative charges of the β- and gamma-phosphates of the incoming dNTP and the departing pyrophosphate

Question 14

What is the general mechanism used by DNA polymerase III to remove incorrectly incorporated
nucleotides? (Ch 11)

Answer 14

The 3’ —> 5’ exonuclease domain is located in a different domain than the catalytic domain

If a newly added nucleotide is mismatched
–Disruption of nonspecific hydrogen bonding between palm and base pairs in minor groove of new duplex leads to reduced catalysis rate
–Fingers cannot rotate towards palm to bind and direct dNTP to active site

The unpaired 3’ region will move into the exonuclease domain
–The exonuclease domain has a 10X higher affinity for single stranded 3’ ends of polynucleotide chains

Question 15

Describe the structure of the E. coli replisome. (Ch 11)

Answer 15

The replisome in E. coli consists of DNA polymerase III holoenzyme, primase, and helicase

The holoenzyme contains three catalytic core polymerase subunits
–One on leading strand
–Two on lagging strand

Question 16

Describe the structure of a DNA polymerase III holoenzyme. How many copies of each component are present in the holoenzyme and what is the function of each component? (Ch 11)

Answer 16

Composition of DNA polymerase III holoenzyme:
–Three copies of the polymerase III catalytic core, each containing
—One alpha subunit for DNA polymerase activity
—One epsilon subunit for 3’ —> 5’ proofreading exonuclease
—One theta subunit used to stimulate exonuclease activity when mispaired nucleotides are present
–Three copies of the linker protein tau
—tau joins the core enzymes to the sliding clamp loader
–Three copies of the clamp, each containing β2 ring that binds the DNA for processivity
–One copy of the clamp loader
—gamma complex of seven proteins that load the clamp on the DNA

Question 17

What are the steps involved in the assembly of the DNA polymerase III holoenzyme? (Ch 11)

Answer 17

ATP bind β2 in clamp
Clamp opens
Beta binds dna
Lose loader cause change affinity
Clamp binds polymerase
Tau links clamp to core

Question 18

Describe the structure of the clamp associated with DNA polymerase III. What aspects of its structure
contribute to the high processivity of the enzyme? (Ch 11)

Answer 18

The sliding clamp looks like a horseshoe, and DNA goes inside the hole, the core DNA polymerase binds on top.

β2 ring that binds the DNA for processivity

Question 19

How do interactions between DnaB and other components of the replisome regulate Okazaki fragment length and the rate of fork movement? (Ch 11)

Answer 19

As DnaB helicase moves along the lagging strand and denatures the parental duplex, SSB coat the single stranded DNA that is produced
–The core on the leading strand displaces the SSB and continues synthesis

Interactions between DnaB helicase and tau results in an increase in helicase and core enzyme activity by 10X
–Prevents helicase from “running away” from holoenzyme

DnaG primase is the only replisome component that is not tightly associated with the fork

The only association between DnaG and the other fork components occurs via weak interactions with DnaB
–After primer synthesis, DnaG is released back into solution

Interactions between DnaG and DnaB stimulate DnaG activity 1000X

A stronger association between DnaG and DnaB would result in more frequent periods of interaction between the components and more priming events
–Shorter Okazaki fragments

A weaker association would result in less frequent interactions and fewer priming events
–Longer Okazaki fragments

Question 20

What is the general structure of the eukaryotic replisome? (Ch 11)

Answer 20

One DNA polymerase alpha
–Initiates DNA synthesis
–Contains primase domains

One DNA polymerase epsilon
–Elongation of the leading strand

One DNA polymerase delta
–Elongation of the lagging strand

Question 21

Describe DNA polymerase action on the leading and lagging strands in eukaryotes. (Ch 11)

Answer 21

On the leading strand:
–CMG helicase moves along the strand and denatures parental
duplex
–Attached polymerase epsilon will continuously replicate

On the lagging strand:
–Polymerase alpha will prime each Okazaki fragment
–After polymerase switch, polymerase delta will complete fragment
–Eukaryotic Okazaki fragment length of approximately 200 nucleotides

Question 22

What are the general steps of nucleosome assembly? (Ch 11)

Answer 22

Nucleosome assembly:

H3(v2)-H4(v2) tetramer binds to DNA, followed by stepwise addition of H2A-H2B dimers
–Replication-coupled pathway

Process is facilitated by histone chaperones
–Chromatin remodeling complexes

Reverse process for disassembly

Question 23

Describe the steps of the replication-coupled nucleosome assembly pathway. How are “old” and “new” histones used in the pathway? Which histone chaperones are included in this pathway and how do they act? (Ch 11)

Answer 23

1. Replication fork displaces octamers
2. Octamers disassociate into H3(v2)-H4(v2) tetramers and H2A-H2B dimers
3. “Old” H3(v2)-H4(v2) tetramers are transferred randomly to one of the two daughter strands by FACT and NAP1
   –At approximately the same place in the genome
4. “Old” H2A-H2B dimers are released into the soluble histone pool
5. “New” H3(v2)-H4(v2) tetramers are assembled onto other daughter strand by CAF1
6. “Old” or “new” H2A-H2B dimer chosen from the soluble pool by NAP1
7. The H2A-H2B dimer is placed onto daughter strands by NAP1
   –Along with the already present tetramers
   –Forms complete octameric nucleosomes

Question 24

How is the pattern of histone modifications on the parental duplex accurately transferred to both two daughter duplexes following replication? (Ch 11)

Answer 24

Old H3(v2)-H4(v2) tetramers (with their associated modifications) are passed to the daughter duplexes The specific modifications of these tetramers will result in recruitment of specific enzymes that are able to propagate modifications onto the H2A-H2B members of the nucleosome --Also able to propagate modifications onto neighboring nucleosomes

Question 25

How is RecA involved in the strand invasion process of E. coli homologous recombination? (Ch 13)

Answer 25

The RecA protein is a key component of the homologous recombination mechanism in E. coli --Similar Rad51 protein in eukaryotes RecA binds to the single stranded 3’ overhang produced after DbSB --Invading strand RecA forces ssDNA into a “double stranded” helical conformation The RecA/ssDNA complex can displace single strands from duplexes --“Samples” sequences of nearby duplexes If RecA/ssDNA finds a complimentary sequence of >15 nt --Complimentary strand will pair with ssDNA to form heteroduplex

Question 26

How are Spo11, Dmc1, and Rad51 involved in eukaryotic homologous recombination and meiotic recombination? (Ch 13)

Answer 26

Dmc1 and Rad51 coat ssDNA overhangs and promote strand invasion --Homologues of prokaryotic RecA Dmc1 only active in meiosis and has specificity for homologous chromosome --Prevents invasion of sister chromatid Rad51 is used throughout the cell cycle for all DsDB recombination in eukaryotes --Loaded onto the ssDNA by the Brca2 protein --Loss of Brca2 function leads to reduced DsDB repair Spo11 protein produces the DbSB needed for meiotic recombination --A tyrosine side chain of Spo11 attacks the phosphodiester backbone of DNA and creates a covalent enzyme-DNA complex --Energy of phosphodiester linkage is stored in enzyme-DNA linkage --DbSB is reversible and DNA strands can be easily resealed if recombination does not proceed

Question 27

What are the differences between the crossover (splice) and non-crossover (patch) products that result from resolution of a single Holliday junction? Describe the molecular processes that lead to these different outcomes beginning with the formation of the DbSB and ending with resolution. How are the different outcomes related to DbSB repair and meiotic recombination? (Ch 13)

Answer 27

Strand invasion generates a Holliday junction, where two homologous helices are held together by the exchange of two of the four strands 90% of the time, the DbSB are resolved as non-crossover (patch) products Original duplexes are reformed, except for small region (patch) where homologous sequence was as a template to fill area around DbSB Closely resembles products from general homologous recombination process used to repair spontaneous DNA breaks The other 10% of the time --The second strand is captured --Another Holliday junction is formed --The strands are cut and resolved as crossover (splice) products

Question 28

What is gene conversion and how does DbSB-coupled repair lead to a gene conversion event? (Ch 13)

Answer 28

Gene conversion: One allele converted to another. DbSB-coupled repair can lead to gene conversion two ways: 1. If a gene is located inside the region that is degraded during DbSB repair, then the original allele will be replaced by one from the homologous duplex --Heterozygous Dd genotype --After DbSB, DD/dd --> Dd/dd --One descendant allele has been converted from D to d 2. Regions of heteroduplex DNA resulting from branch migration will be randomly repaired by mismatch repair enzymes

Question 29

How do base excision, nucleotide excision, and mismatch excision repair differ from each other? What types of specific damage do they repair? (Ch 14)

Answer 29

Nucleotide Excision Repair: Recognizes bulky lesions in DNA --UV-induced pyrimidine dimers a) Global genome repair --Repairs damage anywhere in the genome b) Transcription-coupled repair --Repairs damage in the transcribed strand of active genes Base Excision Repair: Recognizes damage to individual nitrogenous bases --Deamination or alkylation Mismatch Excision Repair: A type of repair that corrects mispaired bases ---Usually immediately following replication --Preferentially corrects the sequence of the daughter strand --Distinguishes between the daughter strand and parental strand ---Sometimes based on methylation status --Also recognizes insertion and deletion hairpin loops

Question 30

Describe some of the specific mutational events that can lead to single base changes and structural distortions. (Ch 14)

Answer 30

Single Base Changes: Affect the DNA sequence but don’t greatly distort the structure --deamination or replication errors Structural Distortions: Provide a physical impediment to replication or transcription --Pyrimidine dimers (UV) --Alkylation --Nitrogenous base removal Depurination

Question 31

What are the two types of enzymes involved in base excision repair? How are the activities of these enzymes related to long-patch versus short-patch repair in eukaryotes? (Ch 14)

Answer 31

Base excision repair is triggered by direct removal of a damaged nitrogenous base from DNA The enzymes that remove bases from DNA are called glycosylases --Glycosylases sometimes also have lyase activity that cleaves the ring structure of the deoxyribose sugar The removal of a nitrogenous base triggers the removal and replacement of a stretch of polynucleotides Glycosylase action alone leads to the DNA Pol-delta-epsilon pathway that results in the replacement of a long stretch of nucleotides --Long-patch pathway Both glycosylase and lyase action leads to the DNA Pol-beta pathway that results in the replacement of a short stretch of nucleotides --Short-patch pathway

Question 32

Which DNA polymerases are associated with error-prone repair and how do their activities differ from those of high fidelity polymerases? (Ch 14)

Answer 32

Damaged DNA that has not been repaired causes DNA Polymerase III to stall during replication DNA polymerase V or DNA polymerase IV can temporarily replace polymerase III and synthesize a complement to the damaged strand --Polymerase V is the only E. coli enzyme that can bypass pyrimidine dimers The DNA synthesized by repair DNA polymerases often have errors in their sequence --Error-prone synthesis --Error-prone polymerases add random nucleotides to daughter strand at site of damage ---Lack specificity in active site that is found in high fidelity replicases

Question 33

How is methylation related to directionality of repair in prokaryotes and eukaryotes? (Ch 14)

Answer 33

Methylation and directionality of mismatch: The strand lacking methylation at a Me-A/T is usually replaced in prokaryotes --Repair must be completed before daughter strand is methylated Methylation status is not used by eukaryotic repair enzymes

Question 34

How does recombination repair differ in the G1 phase versus the G2/M phases? (Ch 14)

Answer 34

The recombination repair system functions when you have a gap or a double stranded break in a strand opposite a damaged sequence --DNA replication errors Nuclease enzymes that resect 5’ ends of broken strands become very active in S and G2 --When sister chromatids are present In G1 phase, no sister chromatids are present, so recombination repair would use homologous chromosome for repair --Loss of heterozygosity Similar mechanisms in prokaryotes and eukaryotes --Rec genes in E. coli --Mechanism of non-crossover (patch) formation from recombination --Mechanism used for Cas-CRISPR editing

Question 35

How is recombination repair involved in the Cas-CRISPR gene editing process? (Ch 14)

Answer 35

A Cas9 transgene can be transferred into a cell along with a guide RNA that targets a specific DNA sequence in the genome and the gene --Cas9 and guide RNA are usually included together on an expression vector The Cas9/guide RNA complex binds to the target sequence in the genome and creates a DbSB DbSB is repaired using recombination repair mechanism If homologous sequence is present, it will replace the damaged region --Gene editing --Homologous sequence is sometimes introduced into the cell on same vector as Cas9 and guide RNA

Question 36

What is NHEJ and how is its mechanism related to the relatively low accuracy rate we observe in NHEJ repair compared to that of recombination repair? (Ch 14)

Answer 36

If the homologous recombination DbSB repair mechanisms cannot find an identical or homologous sequence to use as a template for repair, then the cell will use the NonHomologous End-Joining (NHEJ) pathway --Primary pathway used to repair DsDB in mammalian somatic cells --Spend most of their time in G1 when identical or homologous sequence is less available The NHEJ pathway is error-prone and usually results in errors in the repaired DNA --Acceptable in somatic cells because most of the mammalian genome is not essential for life and can tolerate NHEJ errors The Brca1 protein regulates an early step in homologous recombination DbSB repair --Defective Brca1 will shunt cells to error-prone NHEJ pathway

Question 37

proto-oncogenes

Answer 37

Proto-oncogenes: components of signal transduction pathways --when mutated, can result in cancer

Question 38

tumor suppressor genes

Answer 38

Tumor suppressor genes: genes whose products prevent the activation of CdK enzymes when criteria have not been met (Rb and p53)

Question 39

Rb mediated control

Answer 39

1. With continued growth factor signal, G1-Cdk (Cyclin D-Cdk) activity is stimulated 2. Active G1-Cdk will phosphorylate Rb (usually found in an active, unphosphorylated form) 3. Phosphorylated Rb will be inactivated and release the transcription factor E2F 4. E2F will activate the transcription of genes required for entry into S phase 5. Downstream components participate in positive feedback of E2F

Question 40

p53 control

Answer 40

p53 after DNA damage: – p53 and Mdm2 are phosphorylated – Mdm dissociates from p53 – p53 degradation is decreased – p53 translocates to the nucleus and activates the expression of genes that repress cell cycle progression

Question 41

shorthand Review the steps involved in bacterial initiation beginning with DnaA-ATP binding to oriC. (Ch 10)

Answer 41

1- 9 2- helical core 3- 13 4- compact 5- preprime 6- B brings G 7- G says bye C and activate B 8- B displace A 9- gyrase 10 - helicase 11- HU

Question 42

shorthand Describe how the ORC, Cdc6, Cdt1, and Mcm2-7 are involved in initiation of replication in eukaryotes. (Ch 10)

Answer 42

1- orc binds origin 2 - cdc6 binds in M/G1 3 - cdt1 active in G1 4 - cdt1 brings Mcm2-7 5 - pre-rc formed 6 - cdc6 and cdt1 released 7-mcm loaded

Question 43

palm

Answer 43

Primary elements of the polymerase catalytic site

Question 44

fingers

Answer 44

Binds to incoming dNTPs and moves correct dNTPs into close contact with polymerase catalytic site ---“Tightens” upon dNTP binding but before addition for initial proofreading step

Question 45

thumb

Answer 45

Maintains the correct position of the 3’-OH and also a strong association between the polymerase and the parental strand to facilitate processivity

Question 46

alpha subunit

Answer 46

1 copy DNA polymerase activity

Question 47

epsilon subunit

Answer 47

one copy 3’ ---> 5’ proofreading exonuclease

Question 48

theta subunit

Answer 48

one copy used to stimulate exonuclease activity when mispaired nucleotides are present

Question 49

tau

Answer 49

3 copies joins the core enzymes to the sliding clamp loader

Question 50

clamp

Answer 50

3 copies containing β2 ring that binds the DNA for processivity

Question 51

clamp loader

Answer 51

one copy gamma complex of seven proteins that load the clamp on the DNA

Question 52

how many copies of polymerase catalytic core in holoenzyme

Answer 52

3

Question 53

short okazaki fragments

Answer 53

A stronger association between DnaG and DnaB would result in more frequent periods of interaction between the components and more priming events

Question 54

long okazaki fragments

Answer 54

weak association between DnaG and DnaB

Question 55

DNA polymerase alpha

Answer 55

--Initiates DNA synthesis --Contains primase domains primes lagging strand

Question 56

DNA polymerase epsilon

Answer 56

Elongation of the leading strand

Question 57

DNA polymerase delta

Answer 57

Elongation of the lagging strand

Question 58

structure of the eukaryotic replisome?

Answer 58

one polymerase alpha one polymerase epsilon one polymerase delta

Question 59

nucleotide excision repair

Answer 59

Recognizes bulky lesions in DNA a) Global genome repair --Repairs damage anywhere in the genome b) Transcription-coupled repair --Repairs damage in the transcribed strand of active genes

Question 60

base excision repair

Answer 60

Recognizes damage to individual nitrogenous bases

Question 61

mismatch excision repair

Answer 61

A type of repair that corrects mispaired bases ---Usually immediately following replication --Preferentially corrects the sequence of the daughter

Question 62

glycosylase action alone

Answer 62

leads to the DNA Pol-delta-epsilon pathway that results in the replacement of a long stretch of nucleotides --Long-patch pathway

Question 63

glycosylase and lyase action

Answer 63

leads to the DNA Pol-beta pathway that results in the replacement of a short stretch of nucleotides --Short-patch pathway

Question 64

Old h3h4 tetramer put on by

Answer 64

FACT CAF1

Question 65

New tetramer put on by

Answer 65

CAF1

Question 66

Old and new h2a h2b diners put in by

Answer 66

NAP1

Question 67

RecA

Answer 67

Binds to 3’ overhang and displaces dna to find compliment

Question 68

Dmc1 and rad51

Answer 68

Coat ssdna and promote strand invasion

Question 69

Dmc1

Answer 69

Active in meiosis Prevents invasion of sister

Question 70

Rad51

Answer 70

Throughout cell cycle All dbsb recombination

Question 71

Spo11

Answer 71

Produces dbsb for meiotic recombination