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BB20023 MCQ 20% EXAM > DNA replication > Flashcards

DNA replication Flashcards

(96 cards)

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1
Q

Define cellular homeostasis?

A

finely tuned balance between minimising the generation and persistence of DNA mutations that affect genomic stability and those that do not affect genetic stability

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2
Q

Genetic diversity in the germ line is essential to allow ?

A

Genetic fitness (Darwin evolution)

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3
Q

chromosomal polyploidy or aneuploidy will cause what

A

diseases such as cancer

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4
Q

DNA is used to?

A

to encode hereditary information

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5
Q

Where does replication occur?

A

replisomes

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6
Q

Function of replisomes?

A

assemble at DNA loci start sites or ‘replication origin sequences’

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7
Q

Replication starts from ?

A

single site of origin along both leading and lagging strands

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8
Q

Once replication is complete?…..

A

replisomes dis-assemble

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9
Q

The replisome is composed of?

A

The core replication proteins (in all cell types) include a helicase, primase, DNA polymerases, sliding clamp, clamp loader, and single-strand binding (SSB) protein

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10
Q

Replication is divided into which three stages?

A

Initiation, elongation, termination

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11
Q

INITIATION?

A

Initiator proteins recognise and bind to site of origin on the DNA

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12
Q

ELONGATION?

A

Movement of replisomes and DNA synthesis on leading and lagging strands

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13
Q

Prokaryotic genomes are typically ? (3)

A

haploid, smaller in size and usually having a single circular chromosome found in the nucleoid

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14
Q

Eukaryotic genomes are (3) ?

A

diploid; bigger is size and organized into multiple densely compacted linear chromosomes found in the nucleus

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15
Q

Time taken for E.coli unwind and duplicate DNA, error rate?

A

1000 bp/s, one mutation for each 10− 6 to 10− 7 NT incorporated

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16
Q

In e.coli replication initiation involves which two structures? (2)

A
  • Replicator sequence on the DNA called oriC

- An oriC binding initiator protein called DnaA

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17
Q

Describe the process in which a complex of supercoiled DNA wrapped around a central core- Initiation. Dependence and proteins required? (PROS)

A

5 monomers of the DnaA protein (52kD each), which recognises and binds 9bp repeats in oriC (consensus sequence of 5’-TTATCCACA-3’).

This ATP dependent process is facilitated by histone like HU or IHC proteins to help DNA bending.

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18
Q

Describe the formation of the open complex in initiation? And what two structures are required
(PROS)

A

DnaA proteins melt 3 tandemly repeated 13bp AT rich segments (consensus sequence of 5’-GATCTNTTNTTTT-3’ to the left of the oriC. This 45bp segment is insensitive to P1 nuclease in the double stranded state but can be cut by the P1 nuclease in the single stranded state.

Requires: ATP and DnaA

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19
Q

Describe the formation of the prepriming complex?(PROS)

A

DnaA protein complex recruits the DnaB6 and DNAC6 hexametric complex into melted region to form the prepriming complex. This allows 5 additional DnaA monomers to bind to the bound DnaA dimers. DnaC6 is then released.

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20
Q

DnaC6

A

an ATPase that promotes loading

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21
Q

What is the next step which occurs to the the pripriming complex? - initiation of pros

A

DnaB6, a helicase, unwinds the DNA in the prepriming complex in both directions. SSB (single strand binding proteins) and gyrase proteins are necessary for DnaB6 activity

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22
Q

helicase/gyrase function in initiation of pros

A

further unwinds the helix to allow entry of the RNA polymerase, which activates the primase to begin RNA primer synthesis.

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23
Q

Bidirectional replication of PROS by what process?

A

by the recruitment of the replisome complex

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24
Q

Dif between the origins of replication in eus vs pros, why? (2)

A
  1. eukaryotes have multiple ori sites that can be activated (‘fired’) simultaneously
  2. Differences in the number and complexity of the proteins required in this process.

Why?
To enable replication of the larger and more complex genomes.

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25
What structure is formed in initiation of eus
two bi-directional replication forks
26
eukaryotic replicative helicase is ?
an 11-subunit complex referred to as CMG (an acronym of the three components required for helicase activity “C” = Cdc45, “M” = six Mcm subunits that form the ring shaped heterohexameric Mcm2–7 complex, “G” stands for the heterotetramer GINS complex ).
27
CMG stands for?
“C” = Cdc45, “M” = six Mcm subunits that form the ring shaped heterohexameric Mcm2–7 complex, “G” stands for the heterotetramer GINS complex.
28
activation of the ori sites triggers in both pros and eus
Replication
29
Activation is split into?- eu initiation
Two distinct phases: G1 AND S
30
what proteins are required when licensed in eus initiation
a process that requires the six-subunit origin recognition complex (ORC), along with Cdc6, Cdt1, and the Mcm2–7 complex.
31
The licensing process results in?- eus initiation
The licensing process results in two Mcm2–7 hexamers that encircle dsDNA and are positioned head-to-head (ie, the two hexamers bind via their N-terminal regions)
32
In proceeding to S phase what is required- eus initiation
several origin activation factors work together, along with two cell cycle kinases (CDK and DDK), to form the CMG complex
33
Result of proceeding to S phase- eus initiation
Cdc45 and GINS become tightly associated with Mcm2–7, and each CMG complex encircles only one strand of DNA
34
The CMG helicases function in initiation- eus
unwind DNA, forming the ssDNA needed for priming and replisome assembly
35
Following kinase activity?- eus initiation
Following kinase activity, replication factors assemble and unwind the DNA helix and culminate in the assembly of the eukaryotic DNA polymerases
36
DNA polymerises required for assembly- initiation of eus
Dpb11, Sld2, Sld3, and Sld7
37
What is and isn't associated with the mature replisome?- eus initiation
Dpb11, Sld2, Sld3, and Sld7 is required for assembly but are not thought to be associated with the mature replisome, whereas Cdc45, GINS, and Pol ε, are associated with it.
38
Function of DDK and CDK kinase activity?- initiation in eus
DDK and CDK kinase activity completes assembly and primes the complex for helicase activation that is accomplished by Mcm10 and RPA.
39
Where in DnaB found and what is its function in Pros
ELONGATION IN PROS DnaB is found at the apex of the replication fork on the lagging strand function: uses the energy of ATP hydrolysis to unwind dsDNA
40
How is the lagging strand template formed and protected?- elongation in pros
the lagging strand template is formed by helicase action and is protected by SSB (single strand binding proteins).
41
Function of DNA polymerase III holoenzyme?- elongation in pros
Pol III HE function: u Uses the single strands of DNA produced by DnaB as templates to synthesize new DNA on both the leading and lagging strands.
42
Function of the β2 sliding clamp- elongation in pros
confers high processivity on the DNA Pol III HE by tethering the Pol III αɛθ cores onto the DNA.
43
Clamp loader complex function in elongation of pros
CLC:uses ATP hydrolysis to assemble the β2 clamp onto RNA primer junctions on template DNA
44
Up to how many Pol III HE cores are coupled to what proteins?- DIRECTLY AND INDIRECTLY- elongation of pros
Up to three Pol III cores are coupled through the τ subunits of the CLC through their extreme C-terminal domains, and the τ subunits also interact with DnaB, thus organizing and coupling the DNA Pol III HE to DnaB
45
how is the lagging strand synthesised and why?- pro elongation
. Due to the opposite polarity of the two DNA strands, the lagging strand is synthesized in a series of short Okazaki fragments in the opposite direction to the leading strand.
46
How is the start of synthesis of an Okazaki fragment initiated?- pro elongation
DnaG primases interact with DnaB to synthesize RNA primers to initiate DNA synthesis on the lagging strand
47
When an Okazaki fragment is complete... to that previous frag ?-pro elongation
Pol I replaces RNA primers with DNA and DNA ligase A joins the fragments into a contiguous lagging-strand DNA chain.
48
Why does the complexity of replication increase in eus?
genomes are very large (over 4 billion base pairs in the human), linear and divided into multiple, densely compacted chromosomes.
49
Chromosomes in elongation of eus and what must interact
are dynamic complexes of DNA and protein, with which the replication machinery must interact
50
Key replisome proteins in eu elongation
an 11-subunit CMG helicase, DNA polymerase alpha-primase, leading strand DNA polymerase epsilon, lagging strand DNA polymerase delta, PCNA clamp, RFC clamp loader, and the RPA single strand binding (SSB) protein
51
DNA polymerase cannot initiate DNA synthesis in eu ....? They require?
de novo, they require a pre-exsisiting primed site
52
Initiation of replication requires what type of enzyme and does what fuction?- eus
primase enzyme that synthesizes an RNA primer to initiate DNA synthesis
53
The leading strand is extended in which direction ?
direction of the fork movement
54
The leading strand most of the time ?...
May only need to be primed once
55
The lagging strands priming?
May need to be primed multiple times at each Okazaki fragment
56
Size of Okazaki fragments- eus
100–200 NT
57
What is the eu primase & its structure. Dif between pros and eus
Polymerase Alpha: four subunit enzyme, contains both RNA primase and DNA polymerase activities in separate subunits. In pros: single unit bacterial primase
58
# fill in the blanks Polymerase alpha generates ........ for the replicative polymerases .......... and ............ It also lacks ..... ............. unlike replicative polymerases
RNA/DNA primer epsilon and delta proof- reading
59
leading and lagging strand B-family polymerases
leading: pol epsilon lagging: pol delta
60
Both B-family polymerases contain?
a 3′-5′ exonuclease proofreader
61
A mutation in the exonuclease site.....
Cancer
62
Function of pol alpha
synthesis of the primed sites for both strands
63
Function of pol delta
bulk replication of the lagging strand is performed by a second DNA polymerase: pol delta
64
Pol delta requires?
accessory factors, PCNA (the clamp) and replication factor C - RFC (the clamp loader).
65
After the pre-RC has been activated by the kinases?....
replication factors unwind the DNA helix and culminate in the assembly of the eukaryotic DNA polymerases
66
Pol alpha interacts with?
CMG to prime both strands
67
Unwinding of DNA strand is performed by? This allows what to happen next?
3' helicase This allows the assembly of the primase DNA polymerase , which lays down a RNA primer
68
The polymerase switch is initiated by
Clamp loader and replicative factor factor (FRC)
69
# Fill in the gaps: In an ........... .................. ............. RFC dissociates ........... .................... .... and assembles the ......... near the primer terminus
ATP DEPENDENT PROCESS DNA pol alpha PCNA
70
PCNA stands for, structure, function?
PCNA: proliferating cell nuclear antigen PCNA is a homotrimeric ring like protein that encircles DNA
71
Pol epsilon is stabilised where and by what
on the leading strand, CMG helicase complex
72
CMG stabilises ..... and what is the impact of this?
Stabilises the replicative DNA polymerase delta and makes it highly processive adding thousands of NT without dissociation
73
Pol delta is stabilised by the ?
CGM helicase complex
74
laggin strand synthesis occurs in the ?....which is opposite in directional movement to ?
occurs in the 5' to 3' orientation (opposite to the direction of movement of the replication folk
75
encountering a downstream primer induces?
the dissociation of the polymerase complex
76
What protein replaces RNA primers with NT?
Rnase H1
77
Rnase H1 function?
Removes all but the last ribonucleotide of RNA primer
78
What is the last ribonucleotide removed by? what happens next
FEN1/RTH1 and then DNA ligase seals the nick by linking the last phosphodiester bind to the newly synthesised strand
79
Replicator activation thing in eus
not all replicators are activated simultaneously
80
Timing of origin activation is controlled by?
large regions in the genome
81
What part of the cell cycle are origins of replication fired?
S phase
82
Chromatin replication is largely dependent on?
state of compaction of chromatin
83
Transcriptionally active chromatin tends to be bound in what way, and therefore replicates late or early?
Loosely- euchromatin early
84
Transcriptionally inert chromatin tends to be bound in what way, and therefore replicates late or early?
Tightly- heterochromatin replicates later in the cell cycle
85
Late replication is associated with which two frequent NT sequence
AT
86
Replication licensing
How eu cells ensure each replicon is replicated once
87
Replication licensing involves which two factors
Replication licensing fator (RLF) and s phase promoting factor (SPF)
88
How do the 'factors' in replication licensing ensure the replicon is only replicated once
Controlling the availability and location of RLF and SPF
89
When is RFL activated?
metaphase-anaphase transition
90
When does RFL decline?
Near the end of G1
91
How does RFL licence replicators? (simple)
By putting them in an initiation competent state
92
Which components are involved in the RFL licensing
pre-RC (ORC. cdc6, csc6, cdt1 and Mcm proteins)
93
SPF function (harder) ...and causes cells to do what
induced licensed replicators to begin initiation and in doing so RLF is removed. This causes cells to move from G1 to s
94
Which kinases are involved in the SPF signal
cell-cycle dependent kinases CD/DDkinases
95
What is the OriC?
highly conserved, unique 250 bp segment, that supports bi-directional replication
96
Which three complexes are found within initiation?
first: initial complex second: open complex third: prepriming complex

BB20023 MCQ 20% EXAM (2 decks)

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