Exam 2 Study Guide Flashcards
What are tumor suppressor genes and proto-oncogenes? (Ch 9)
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)
Describe the general steps of G1/S checkpoint passage in eukaryotes. How is this passage related to
cyclins and Cdks? (Ch 9)
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)
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)
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
Explain how Dam methylase is associated with DnaA, oriC, and the inhibition of reinitiation of
replication. (Ch 10)
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
How is the methylation status of oriC related to its ability to initiate replication? (Ch 10)
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
Review the steps involved in bacterial initiation beginning with DnaA-ATP binding to oriC. (Ch 10)
- DnaA-ATP initially binds as an extended multimeric form at the 9 bp repeats
- Multimeric extended DnaA- ATP forms a helical core around which oriC is wound
- Multimeric extended DnaA-ATP binds to the 13 bp repeats
- 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 - 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 - 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 - 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 - DnaB will displace DnaA from oriC as it extends the bubble
- 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 - 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
Describe the three ways an E. coli cell prevents premature reinitiation of replication. (Ch 10)
- 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 - 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 - 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
What is a licensing factor and what is the licensing factor in prokaryotes and in eukaryotes? (Ch 10)
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?????)
What is the pre-RC? Describe how the ORC, Cdc6, Cdt1, and Mcm2-7 are involved in initiation of replication in eukaryotes. (Ch 10)
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)
- 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 - 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 - Active Cdt1 is only present in G1
–The geminin protein inactivates free Cdt1
–Geminin is specifically degraded in early G1 - Cdt1 escorts Mcm to the origin
–Mcm2-7 contains the eukaryotic helicase - The pre-RC has now been formed
- Cdc6 and Cdt1 are released
- Mcm is fully loaded onto the origin
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)
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
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)
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
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)
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
Describe the general structure of DNA polymerases. What is the function of each of the three domains of this class of enzymes? (Ch 11)
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
What role do metal cations have in the enzymatic activities of DNA polymerases? (Ch 11)
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
What is the general mechanism used by DNA polymerase III to remove incorrectly incorporated
nucleotides? (Ch 11)
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
Describe the structure of the E. coli replisome. (Ch 11)
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
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)
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
What are the steps involved in the assembly of the DNA polymerase III holoenzyme? (Ch 11)
ATP bind β2 in clamp
Clamp opens
Beta binds dna
Lose loader cause change affinity
Clamp binds polymerase
Tau links clamp to core
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)
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
How do interactions between DnaB and other components of the replisome regulate Okazaki fragment length and the rate of fork movement? (Ch 11)
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
What is the general structure of the eukaryotic replisome? (Ch 11)
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
Describe DNA polymerase action on the leading and lagging strands in eukaryotes. (Ch 11)
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
What are the general steps of nucleosome assembly? (Ch 11)
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
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)
- Replication fork displaces octamers
- Octamers disassociate into H3(v2)-H4(v2) tetramers and H2A-H2B dimers
- “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 - “Old” H2A-H2B dimers are released into the soluble histone pool
- “New” H3(v2)-H4(v2) tetramers are assembled onto other daughter strand by CAF1
- “Old” or “new” H2A-H2B dimer chosen from the soluble pool by NAP1
- The H2A-H2B dimer is placed onto daughter strands by NAP1
–Along with the already present tetramers
–Forms complete octameric nucleosomes
How is the pattern of histone modifications on the parental duplex accurately transferred to both two daughter duplexes following replication? (Ch 11)
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
How is RecA involved in the strand invasion process of E. coli homologous recombination? (Ch 13)
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
How are Spo11, Dmc1, and Rad51 involved in eukaryotic homologous recombination and meiotic recombination? (Ch 13)
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
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)
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
