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Molecular Genetics > The Replicon > Flashcards

The Replicon Flashcards

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1
Q

synonymous mutation

A

A mutation in the genetic code that changes the DNA nucleotide sequence and RNA nucleotide sequence but does not ultimately change the encoded amino acid

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2
Q

nonsynonymous mutation

A

A mutation in the genetic code that changes the DNA and RNA nucleotide sequences as well as the encoded amino acid

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3
Q

the replicon

A

A region of a chromosome that replicates as a single unit starting at a specific origin of replication
- a unit of DNA in which individual acts of replication occur; contains an origin and may have a terminus
- can be linear or circular

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4
Q

origin of replication

A

The start point of DNA replication

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5
Q

unidirectional replication

A

Form of DNA replication in which the replication fork proceeds in one direction only
- replication fork created at the origin

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6
Q

bidirectional replication

A

Form of DNA replication in which the replication fork proceeds in both directions
- when origin creates two replication forks that move in opposite directions

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7
Q

plasmid

A

Small extrachromosomal circular DNA molecule that can replicate independently of the main cellular chromosome
- autonomous circular DNA genome that constitutes a separate replicon

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8
Q

extrachromosomal

A

Nucleic acids present in a cell, that are not a part of the main chromosome

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9
Q

in vivo

A

Preforming an experiment within a living organism

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10
Q

in vitro

A

Preforming an experiment under highly controlled conditions outside of a living organism.

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11
Q

how many replicons to bacteria have? how many do humans have?

A
  • bacteria usually only has one replicon
  • humans typically have thousands of replicons in each chromosome
  • ours typically has 10,000 replicons
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12
Q

what is the bacterial genome?

A
  • the bacterial genome is the replicon
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13
Q

what are some examples of replicons?

A
  • bacterial chromosome
  • eukaryotic chromosomes
  • plasmids
  • mitochondrial genome
  • phage
  • virus
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14
Q

how do bacterial chromosomes terminate?

A
  • they may or may not have a terminus
  • most common way is for DNA polymerases to crash into one another (humans do this)
  • E. coli have a different mechanism
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15
Q

what is the termination mechanism E. Coli have?

A
  • there are sequences that designate the termination point
  • in bacteria you do not want DNA replication to go in the opposite direction because the opposite is for transcription - especially for ribosomes
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16
Q

what organelles contain origins of replications?

A
  • nucleus
  • chloroplasts
  • mitochondria
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17
Q

what does the initiation of DNA replication commit the cell to?

A
  • commits the cell to division
  • division must not occur until after replication is completed
  • replication is controlled at the stage of initiation and is highly regulated
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18
Q

what are the ways you can map replication?

A
  • autoradiography
  • fluorescent labelling
  • electrophoresis
  • electron microscopy
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19
Q

what are the types of replication control?

A

single copy and multicopy

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20
Q

single copy

A
  • replicates once alongside bacterial chromosome
  • Usually when the bacterial chromosome itself replicates
  • Have a mechanism to make sure each copy is segregated to each daughter cell
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21
Q

multicopy

A
  • present in more copies than bacterial chromosome
  • small plasmids do this
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22
Q

unidirectional

A
  • replication in only one direction
  • mitochondria
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23
Q

bidirectional

A
  • replication is two directions
  • bacterial and eukaryotic chromosomes
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24
Q

replication fork

A
  • separation of the two strands of DNA
  • at each end of the replication eye
  • also called a theta structure
  • initiated at the origin
  • moves sequentially along DNA
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25
Q

replication eye

A

opening of the two strands to enable the reading
- also called a theta structure in circular DNA

26
Q

prokaryotic replication

A
  • single and circular
  • competitive inhibition between DNA A and seq A
  • at the origin seq A binds to the hemimethylated state on the new strand and blocks the binding of DNA A and slows the rate of DNA replication, causing a 13 minute delay
  • when Seq A binds to DNA is has an affinity for the membrane and pulls the origin to the membrane to hide it from DAM methylase
  • this also stops DAM methylase from binding as well
  • Seq A recognizes the origin because it has an affinity for DNA A sites called high affinity binding sites - DNA A binds to these sites but there are not enough for actually replication
  • bacterial oriC contains 11 GATC/CTAG repeats that are methylated on adenine on both strands
  • DAM methylates the A on each strand on each strand of the GATC sequence
  • the new strand contains a non methylated strand
  • once methylated by DAM methylase, DNA A is able to replicate - it takes 20-40 DNA A proteins to binds at the origin to pull the strands apart
27
Q

what is the role of termination sites and Tus proteins in bacterial replication?

A
  • bacterial chromosome is a single replicon that normally terminates when the replication forks meet
  • Ter sites serve as a backup to prevent replication from going too far
  • transcription of major genes is in the same direction as replication
  • ter sites are pointed in opposite directions
  • ter sites are pointed in opposite directions and binds to the Tus proteins that blocks helicase thus stopping DNA polymerase
  • Tus proteins are bounded asymmetrically - one side is bounded alot tighter than the other - if the helicase is going the right way it’ll just bounce off but if it’s going the wrong way the Tus protein stops
28
Q

why do you need ter sites?

A
  • one side of the DNA pol may be faster than another
  • this is to prevent replication from crossing into transcription
29
Q

bacteria number of replicons, length, and movement

A

replicons: 1
length: 4,200 kb
movement: 50,000 bp/min

30
Q

yeast number of replicons, length, and movement

A

replicons: 500
length: 20 kb
movement: 3,600 bp/min

31
Q

fruit fly number of replicons, length, and movement

A

replicons: 3,500
length: 40 kb
movement: 2,600 bp/min

32
Q

toad number of replicons, length, and movement

A

replicons: 15,000
length: 200 kb
movement: 500 bp/min

33
Q

mouse number of replicons, length, and movement

A

replicons: 25,000
length: 150 kb
movement: 2,200 bp/min

34
Q

plant number of replicons, length, and movement

A

replicons: 35,000
length: 300 kb

35
Q

what are the length of eukaryotic replicons?

A
  • 40-200 kb in length
  • replicons only work once before cell division
  • a chromosome is divided into many replicons
36
Q

what portion of DNA replication did eukaryotes inherit from archaea?

A
  • Orc1 and Cdc6
37
Q

what are the differences between replication termination for prokaryotes and eukaryotes?

A
  • eukaryotes are terminated by the fusion of replication eyes
  • no ter sites or Tus proteins
38
Q

what is the cell cycle?

A
  • mother cells divides by mitosis
  • two daughter cells form and grow
  • interphase occurs
  • gap 1 in synthesis
  • synthesis
  • gap 2 in synthesis
39
Q

foci

A
  • active replicons are clustered (foci)
  • 300-500 active sites at any given time
  • eukaryotic nucleus contains 100-300 foci
  • one focus can contain more than 300 replicons
40
Q

ORC

A
  • origin recognition complex
  • a complex of sic proteins that binds to an ARS at the A and B1 sites
  • 6 protein
  • about 400 kD (roughly the size of DNA polymerase)
41
Q

how does licensing factor components work?

A
  • necessary for initiation at each origin
  • is present in the nucleus prior to replication but is activated or destroyed by replication
  • initiation of another replication cycle becomes possible only after licensing factor reenters the nucleus after mitosis
  • ORC is already there and bound there
  • ORC has to bind to ATP, then recruits cdc6 and cdc6 recruits cdt1
  • in turn cdt1 recruits MCM 2-7 to the origin site
  • then DNA synthesis can initiated
42
Q

cdc6

A
  • part of the licensing factor that binds to the orc
  • recruits cdt1
43
Q

cdt1

A
  • cannot arrive until cdc6 arrives
  • recruits the helicase
  • irreversibly committed to replication
44
Q

geminin

A
  • orc normally gets too many helicases, so geminin is recruited and bind to cdt1 so it no longer recruits the helicases
45
Q

MCM 2-7

A
  • a helicase
  • minichromosome maintenance
  • 6 subunits
  • moves in opposite polarity from the direction of the bacterial helicase
46
Q

what is the percentage of eukaryotic replicons active at any given time?

A

about 15%
- if all replicons were acrive, replication could be completed in 1 hour, but the S phase lasts about 6 hours which explains that only 15% of replicons are active at any given time

47
Q

what are the general rules for replication?

A
  • initiation of DNA replication commits cells to division
  • division must not occur until after replication is completed
48
Q

what does the bacterial origin support?

A
  • initiation of replication
  • control frequency
  • segregates replicated chromosomes - not in eukarotes
49
Q

what is controlling reinitiation?

A
  • physical sequestration of oriC
  • delay in remethylation
  • inhibition of DNAA binding
  • repression of DNAA transcription
50
Q

what can the beginning of eukaryotic DNA replication be traced back to?

A
  • archaeal chromosomes
  • orc1/cdc6 are the same proteins found in archaea and eukaryotes
51
Q

what is the system controlling reinitiation?

A
  • physical sequestration of OriC
  • delay in remethylation
  • inhibition of DnaA binding
  • repression of DnaA transcription
52
Q

oriC

A
  • 245 bp fragment
  • smallest fragment that can support replication
  • very AT rich
53
Q

how is eukaryotic replication terminated?

A
  • fusion of replication etes - no ter sites
54
Q

eukaryotic replication

A
  • The A domain is where ORC attaches and is very AT rich
  • the B domain is where transcription factors attach
  • once ATP binds to ORC initiation can begin
  • at the B domain ABF1 stimulates initiation by opening the two strands
  • then cdc6 binds to ORC which in turn recruits cdt1
  • cdt1 recruits MCM which is a helicase
  • when too many MCM binds, geminin binds to cdt1 to stop recruitment of MCM
  • now replication can begin
55
Q

mitotic cell cycle

A
  • G1 is the first growth stage
  • then synthesis occurs
  • then G2 in another growth phase
  • then mitosis can begin and the cell can divide
  • interphase is everything in the cycle except mitosis
56
Q

explain the connection between eukaryotic replication and the cell cycle

A
  • in early G1 ORC is bound
  • cdc6 and cdt1 are synthesized and binds
  • in late G1 the helicase and possibly geminin are recruited
  • then in synthesis, the DNA replication machinery is recruited and synthesizes the DNA to prepare for mitosis
  • cdc6 and cdt1 are degraded in S phase to prevent reinitiation
  • mcm is phosphorylated at the beginning of S phase and becomes active by leaving the origin
  • In the G2 phase all the proteins are degraded
  • finally mitosis occurs
57
Q

where are MCM proteins in metazoans? in budding yeast?

A
  • metazoans = MCM proteins are in the nucleus throughout the cycle
  • budding yeast = only after mitosis
58
Q

what does MCM stand for?

A

minichromosomal maintenance

59
Q

how can ORC, cdc6, cdt1, and mcm be downregulated?

A

by cdk activity
- cell cycle dependent kinase
- phosphorylates enzyme

60
Q

when is geminin degraded?

A

during late mitosis and early G1
- this allows cdt1 to bind which results in the recruitment of the MCM helicase in early G1

61
Q

which of the following proteins or complexes is not a component of licensing factor?

A

ORC

62
Q

which of the following proteins blocks binding of mcm to the origin

A

geminin

Molecular Genetics (28 decks)

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