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CMMB 411- Doreen > Topic 5 > Flashcards

Topic 5 Flashcards

1
Q

It is crucial for eukaryotic chromosomes to replicate at the S phase once and only once during each cell cycle. Why is ensuring this challenging?

A

It i challenging because of multi-origin of replications

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2
Q

Why does the formation of the replication bubble have to be well-coordinated with spindle fiber formation?

A

Chromosomal damage can occur if segregation occurs before replication.

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3
Q

Describe the 3 steps and results from the Meselson and Stahl experiment

A
  • Labelled DNA with heavy nitrogen isotope
  • Transferred heavy DNA to light nitrogen isotope medium and allowed DNA to replicate
  • Bands appeared halfway between heavy and light nitrogen bands which indicated semi-conservative replication
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4
Q

(DNA/RNA) primer is used in PCR, while (DNA/RNA) primer is used in vivo

A

DNA primer is used in PCR, while RNA primer is used in vivo

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5
Q

What are the 2 main steps in adding a nucleotide to a growing strand?

A
  1. Recognition of proper dNTP and attacking dNTP with a hydroxyl group
  2. dNTP has to be able to base pair with the base on the template strand
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6
Q

Processivity

A

An enzyme’s ability to catalyze consecutive reactions without releasing its substrate
- e.g. the average number of nucleotides added each time when an enzyme binds a primer:template junction

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7
Q

True or false: DNA polymerase has separate active sites for each of the 4 dNTPs

A

False
- DNA polymerase is a processive enzyme that catalyzes the DNA synthesis using a single active site for any of the 4 dNTPs

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7
Q

What 3 things is DNA polymerase able to do that makes it a processive enzyme??

A
  1. Sterically distinguish between dNTPs and its rNTPs precursors
  2. Show kinetic selectivity for adding correct base-paired dNTP (not just recognizing the dNTP but also recognizing the nucleotide on the template)
  3. Initiate synthesis using either RNA or DNA primer annealed to the template
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8
Q

ddNTPs serve as…

A

Terminators

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9
Q

What component of the DNA polymerase recognizes dNTPs over rNTPs?

A

There are discriminator amino acids that recognize the proper structure of the incoming nucleotide
- rNTP can’t fit into the polymerase due to its 2’OH that isn’t recognized by the discriminator amino acids

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10
Q

DNA polymerase forms a…

A

Pocket that fits the 3 phosphate groups of the incoming nucleotide

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11
Q

How does the DNA polymerase grip the primer-template junction?

A

Like a palm

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12
Q

Function of thumb domain in DNA polymerase (2)

A
  1. Maintaining the correct position of the primer and the active site
  2. Maintaining a strong association between DNA polymerase and its substrate
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13
Q

Function of palm domain in DNA polymerase

A

Catalytic site and monitor base pairing of the most recently added nucleotides
- Metal ions are present in the palm and are important in making the 3’OH nucleophilic

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14
Q

Function of fingers domain in DNA polymerase

A

Bind and enclose dNTP with the catalytic metal ions in the palm

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15
Q

What component of the DNA polymerase proofreads newly synthesized DNA?

A

3’ to 5’ exonuclease

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16
Q

Difference between endonuclease and exonuclease

A

Endonuclease: cuts DNA in the middle
Exonuclease: cuts DNA at the ends and can chow through the strands towards the middle of the DNA molecule

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17
Q

Helicase

A

Unwinds DNA at the replication form to create ssDNA template for primase

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18
Q

Primase

A

A specialized RNA polymerase that makes short RNA primers used to anneal to an ssDNA template to serve as a starting point during DNA replication

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19
Q

RNase H

A

Removes RNA primers from RNA:DNA hybrid to complete DNA synthesis
- DNA polymerase fills the gap and DNA ligase repairs the nick

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20
Q

What are the three essential DNA pol in eukaryotes?

A
  1. Pol α
  2. Pol δ
  3. Pol ε
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21
Q

Pol α function

A

Primer synthesis during DNA replication

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22
Q

Pol δ function

A

Extend lagging strand

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23
Q

Pol ε function

A

Extend leading strand

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24
Q

DNA pol __ has relatively low processivity in comparison to DNA pols _______

A

DNA pol α has relatively low processivity in comparison to DNA pols δ and ε

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25
Q

Polymerase switching

A

Replacement of DNA pol α with DNA pol ε at the leading strand and pol δ at the lagging strand

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26
Q

Sliding clamp protein

A

Keeps the polymerase in close proximity to the DNA template which increases the processivity of the DNA polymerase

27
Q

Origin of replication

A

Sites at which DNA unwinding to initiate synthesis

28
Q

Replicator

A

The DNA sequence that direct the initiation of DNA replication
- Binding sites for the initiator protein
- AT-rich DNA sequence that unwinds easily

29
Q

Initiator protein

A

Sequence-specific DNA binding protein involved in the initiation of replication

30
Q

Replicon

A

The DNA replicated from a particular origin of replication

31
Q

Origin Recognition Complex (ORC)

A

6 protein complex that recognizes replicators
- Required to recruit other replication proteins, such as DNA helicase, to the replicator

32
Q

True or false: ORC:DNA interactions require ATP hydrolysis

A

True

33
Q

True or false: ORC binding leads to strand separation

A

False

34
Q

On S. cerevisiae, B1 and A make up the…

A

ORC binding site

35
Q

On S. cerevisiae, B2…

A

Facilitates DNA unwinding when the ORC binds

36
Q

Describe how replicators are activated to initiate replication, and then how they prevent multiple replications (4 steps)

A
  1. Activation of replicator 3 and 5
  2. Replicator 2 and 4 are passively activated by extension of origin 3 and 5
  3. Replicator 1 activates independently
  4. Each origin becomes inactive after it is replicated to prevent multiple replications.
37
Q

Describe the temporal control of the initiation of replication in eukaryotic cells (2 things)

A
  1. Replicator selection (G1): process to identify sequences that will direct the initiation of replication
  2. Origin activation (S): Triggers the replicator associated complex to initiate DNA unwinding and DNA pol recruitment
38
Q

When does helicase loading occur? What about replicator/origin activation?

A

Helicase loading: G1
Receptor/origin activation: S

39
Q

What are the 7 steps involved in helicase loading?

A
  1. Recognition of the replicator by the eukaryotic initiator, ORC.
  2. ORC recruits helicase loading proteins (Cdc6 and Cdt1) and Mcm2-7 helicase in the expense of ATP (head-to-head binding of Mcm2-7
    - The assembly of this pre-replicative complex does not unwind the DNA or initiate replication after cells enter the S phase.
  3. 2 kinases, CDK and DDK, phosphorylate Sld2/3 and Mcm2-7, respectively
  4. Formation of a complex with phosphorylated sld2/3 as well, but these molecules leave and result in the formation of CMG (Cdc45/Mcm2-7/GINS) complex
  5. The leading strand DNA pol ε is recruited before DNA unwinding
  6. DNA begins to unwind (Mcm2/7 subunits start separating and going opposite directions)
  7. DNA pol α/primase and DNA pol δ are recruited after DNA unwinding
40
Q

Helicase structure before activation

A

Helicase encircle dsDNA as head-to-head double hexamer

41
Q

Helicase structure after activation

A

Helicase encircle ssDNA as a single hexamer (and travel in opposite directions)

42
Q

How does CDK activity regulate eukaryotic helicase activation?

A
  • Low CDK activity is required for the loading of eukaryotic DNA helicase
  • High CDK activity is required to initiate DNA replication (activation of helicase)
  • After S phase, all loaded Mcm2-7 complexes will be removed from the DNA
43
Q

G1 in terms of helicase activity

A

Loading phase

44
Q

S in terms of helicase activity

A

Activation phase

45
Q

CDK levels in G1 and its impact

A

In G1, CDK activity is low and helicase is loaded, BUT cannot be activated

46
Q

CDK levels in S and its impact

A

In S, CDK activity is high and the helicase is activated to initiate DNA replication

47
Q

What happens to used helicase after DNA replication?

A

Used helicase is disassembled

48
Q

What happens to CDK levels in S, G2 and M? Why?

A

CDK activity remains high is S, G2 and M to prevent loading new helicase; thereby restrict one DNA replication/cell cycle

49
Q

(Circular/Linear) DNA can be completely replicated

A

Circular

50
Q

After circular DNA replication, the new DNA and the parental DNA are linked together (catenation). What separates the two daughter DNA molecules?

A

Topo II

51
Q

Describe the “end replication problem” of linear chromosomes

A
  • Because a primer is needed for the lagging strand, a little bit of DNA with be lost from the end of the chromosome during each replication cycle.
  • Shortening of one of the two daughter DNA molecules is resulted from each round of DNA replication.
  • Slowly disrupt the complete propagation of the genetic material from generation to generation
52
Q

What are the 2 possible solutions for the end replication problem?

A
  1. To use a protein, instead of an RNA, as the primer for the last Okazaki fragment at each end of the chromosome.
  2. Use of telomere sequence
53
Q

Describe the use of a priming protein instead of a primer to solve the end replication problem

A
  • Used for some linear bacterial chromosomes, and some viruses
  • An OH of an amino acid substitutes for the 3’-OH of an RNA primer
  • The protein allows priming at the last nucleotide; therefore, no sequence is lost.
  • The protein binds to the 3’ end of the lagging template strand (5’ end of the lagging strand itself)
54
Q

Describe the use of a telomere sequence to solve the end replication problem

A
  • Head-to-tail repeats of a TG-rich DNA sequence (NOT TA-RICH)
  • Serves as an origin of replication for 3’ end of each chromosome
  • Use telomerase as the DNA polymerase
  • Telomeres shorten instead of DNA sequence
55
Q

Describe the structure of telomerase

A
  • A ribonucleoprotein complex that consists of:
    1. An RNA subunit: serves as a template and no exogenous template is needed
    2. Reverse transcriptase subunit: A DNA polymerase that uses the RNA subunit as a template.
56
Q

What are 4 similarities between telomerase and DNA polymerase

A
  1. Require a template to direct nucleotide addition
  2. Extend at 3’OH end
  3. Use the same nucleotide precursors
  4. Act in a processive manner
57
Q

What are 4 differences between telomerase and DNA polymerase?

A
  1. Telomerase has an RNA component
  2. Telomerase does not require an exogenous template (because the RNA itself is a template)
  3. Telomerase is able to use ssRNA substrate to make ssDNA (i.e. able to reverse transcribe)
  4. Telomerase has RNA:DNA helicase activity to displace its RNA template from the DNA for repeated rounds of synthesis
58
Q

Which solution to the end replication problem is used by most eukaryotes?

A

Telomerase

59
Q

True or false: repetitive telomeric DNA is protein coding

A

False
- Repetitive telomeric DNA is non-protein coding; thus, the extra sequence does not interfere with the cellular functions

60
Q

Functions of telomere-binding proteins

A

Longer telomeres inhibit telomerase activity by recruiting telomere-binding proteins
- Telomere-binding proteins also protect the chromosome ends from recombination and chromosome fusion (repair processes that can fix the break)

61
Q

Cdc13 function

A

Telomerase recruitment

62
Q

Rif1 and Rap1 function

A

Inhibits telomerase activity

63
Q

POT1 function

A

Inhibits telomerase activity

64
Q

Shelterin

A

Protein complex that protects telomeres from DNA repairing enzymes

65
Q

What proteins make up shelterin? (6)

A
  • POT1
  • Rap1
  • TPP1
  • TIN2
  • TRF1
  • TRF2
66
Q

Telomere structure looks very similar to the ends of chromosomes that have been accidentally broken. How to telomeres prevent inappropriate attempts to fuse or repair the telomere?

A

Human telomeres form t-loop structures to prevent DNA repairing.

CMMB 411- Doreen (9 decks)

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