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Replication + Repair + Recombination Flashcards

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1
Q

Why does the DNA replication must always proceed in a 5’ –> 3’ direction?

A

The initial phosphodiester bond is between the 3’ oxygen of growing strand and the alpha phosphate of a deoxyribonucleoside triphosphate (dNTP).

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2
Q

What are the three requirements for DNA synthesis by DNA polymerase

A

1) a primer strand with free 3’ terminus
2) a template strand base paired to primer
3) a source of dNTPs

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3
Q

Lagging strand is synthesized discontinuously by multiple RNA primers, producing ________ fragments.

A

Okazaki

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4
Q

Leading strand is synthesized continuously by ___ RNA primer at its 5’ end.

A

Single

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5
Q

In which direction does the DNA polymerase add nucleotides to a growing
daughter strand?

A

5’ -> 3’

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6
Q

Match the following terms:
1) ORC
2) RPA

A) binds to separated parental strands at an origin
B) 6 subunit protein complex that binds to each replication origin and loads 2 helicases in opposite directions

A

ORC - 6 subunit protein complex that binds to each replication origin and loads 2 helicases in opposite directions

RPA - binds to separated parental strands at an origin

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7
Q

Match the following terms regarding DNA synthesis:

Step -1
Step 0
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Step 7

  1. PRIMASE - POL α complexes synthesize short primers
  2. PCNA-Rfc-Pol ε complexes extend the leading strand
  3. HELICASES use ATP hydrolysis energy to unwind parental DNA in opposite directions, forming a replication bubble.
  4. PCNAc – Rfc–Pol ε complexes replace PRIMASE - POL α complexes and extend the lagging strand okazaki fragments, which is stitched to the 5’ of the leading strand by DNA LIGASE.

5.Two hexameric HELICASES bind at replication origin at opposite orientation and are activated by KINASES

  1. TOPOISOMERASE I removes torsional stress introduced by unwinding of strands
  2. PCNAc – Rfc–Pol δ complexes replace PRIMASE - POL α complexes and extend the short primers, generating leading strands at each fork
  3. At the same time, PRIMASE - POL α complexes synthesize RNA primers for lagging strand
    • As HELICASES unwind the parent strands, RPA proteins bind to newly exposed ssDNA.
A

Step -1 - TOPOISOMERASE I removes torsional stress introduced by unwinding of strands

Step 0 - Two hexameric HELICASES bind at replication origin at opposite orientation and are activated by KINASES

Step 1 - HELICASES use ATP hydrolysis energy to unwind parental DNA in opposite directions, forming a replication bubble.

Step 2 - PRIMASE - POL α complexes synthesize short primers

Step 3 - PCNAc – Rfc–Pol ε complexes replace PRIMASE - POL α complexes and extend the short primers, generating leading strands at each fork

Step 4 - As HELICASES unwind the parent strands, RPA proteins bind to newly exposed ssDNA.

Step 5 - PCNA-Rfc-Pol ε complexes extend the leading strand

Step 6 - At the same time, PRIMASE - POL α complexes synthesize RNA primers for lagging strand

Step 7 - PCNAc – Rfc–Pol δ complexes replace PRIMASE - POL α complexes and extend the lagging strand okazaki fragments, which is stitched to the 5’ of the leading strand by DNA LIGASE.

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8
Q

Match the following terms:

  1. Primase-pol α complexes
  2. PCNA
  3. RFC
  4. ORC (Origin Recognition Complex)
  5. RPA

A. Proliferating Cell Nuclear Antigen stabilizes DNA polymerase by acting as a “sliding clamp”
B. Replication factor C
C. Contain RNA polymerase and synthesizes RNA primers
D. Replication Protein A binds to separated parent strands at an origin
E. A 6 subunit protein complex that binds to each replication origin and loads 2 helicases in opposite directions

A

Primase-pol α complexes - Contain RNA polymerase and synthesizes RNA primers

PCNA - Proliferating Cell Nuclear Antigen stabilizes DNA polymerase by acting as a “sliding clamp”

RFC - Replication factor C

ORC (Origin Recognition Complex) - A 6 subunit protein complex that binds to each replication origin and loads 2 helicases in opposite directions

RPA - Replication Protein A binds to separated parent strands at an origin

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9
Q

At the end of DNA replication, RNA is removed by _______ and converted to DNA by ______.

A

ribonuclease, pol δ

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10
Q

Match the following terms regarding eukaryotic DNA polymerases.

  1. DNA pol ε
  2. DNA pol α
  3. DNA pol δ

A. responsible for leading strand synthesis, with a 3’ -> 5’ exonuclease activity
B. responsible for lagging strand synthesis, with a 3’ -> 5’ exonuclease activity
C. synthesizes RNA primers, extends okazaki primers by 10 NTs but no 3’ -> 5’ exonuclease activity.

A

DNA pol ε - responsible for leading strand synthesis, with a 3’ -> 5’ exonuclease activity

DNA pol α - synthesizes RNA primers, extends okazaki primers by 10 NTs but no 3’ -> 5’ exonuclease activity.

DNA pol δ - responsible for lagging strand synthesis, with a 3’ -> 5’ exonuclease activity

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11
Q

Endogenous and Exogenous DNA damage can cause DNA mutation, which ultimately changes the DNA sequence. (T/F)

A

True!

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12
Q

What are the three excision repair systems?

A
  1. Base excision repair
  2. Non-homologous end joining
  3. Homologous recombination
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13
Q

DNA repair defects associated with _______.

A

Cancer

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14
Q

Match the following terms to their definition regarding mutations:

  1. Point mutations
  2. Transition
  3. Transversions
  4. Insertions/deletions
  5. Silent/synonomous
  6. Missense
  7. Nonsense

A. pyrimidines substituted for purines or vice versa
B. change of a single base/base pair
C. change in protein amino acid sequence
D. leads to a frameshift
E. gives rise to a stop codon
F. does not change amino acid sequence; usually in 3rd codon position
G. pyrimidines substituted for another pyrimidine or same with purines

A
  1. Point mutations - change of a single base/base pair
  2. Transition - pyrimidines substituted for another pyrimidine or same with purines
  3. Transversions - pyrimidines substituted for purines or vice versa
  4. Insertions/deletions - leads to a frameshift
  5. Silent/synonymous - does not change amino acid sequence; usually in 3rd codon position
  6. Missense - change in protein amino acid sequence
  7. Nonsense - gives rise to a stop codon
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15
Q

T/F Mutations can be in coding and/or non coding sequences; a gene regulatory region point mutation can’t change the gene sequence.

A

False. A gene regulatory region point mutation can change the gene sequence.

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16
Q

What are the endogenous sources of DNA damage that can cause DNA mutations? What does endogenous mean?

A

Endogenous; is a spontaneous cleavage of bonds produced within the cell

  1. Deamination
  2. Depurination
  3. Genotoxic chemical byproducts of metabolism
  4. Copy errors
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17
Q

What are the exogenous sources of DNA damage that can cause DNA mutations? What does exogenous mean?

A

Exogenous; applied by cell’s environment

  1. Carcinogens
  2. Ultraviolet + Ionizing radiations
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18
Q

What deamination is a frequent cause of point mutations? If it is not restored, what can it become during replication?

A

5-Methylcytosine to Thymine.

The C-G bond can become T-A during replication.

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19
Q

Deamination is the removal of an ______ group from a molecule. If the sugar base is ______, there is a higher chance to undergo deamination. There’s about ___ cytosine deaminations per day.

A

Amino

Methylated

100

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20
Q

What system with a high-fidelity recognizes and repairs deamination? What does it do?

A

DNA excision repair system. It removes the mismatched thymine and replaces it with the cytosine (the original NT).

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21
Q

Match the following terms regarding the DNA excision repair system to repair a deamination.

1) Step 1
2) Step 2
3) Step 3
4) Step 4

A) APURINIC ENDONUCLEASE I (APE 1) cuts DNA backbone at the abasic site

B) DNA POL β inserts the single base & DNA LIGASE links it to the backbone.

C) DNA GLYCOSYLASE specific for T-G mismatches flips + cuts the thymine out of the helix.

D) APURINIC LYASE associated with DNA POLYMERASE β (specialized for repair) removes the deoxyribose phosphate

A

Step 1: DNA GLYCOSYLASE specific for T-G mismatches flips + cuts the thymine out of the helix.

Step 2: APURINIC ENDONUCLEASE I (APE 1) cuts DNA backbone at the abasic site

Step 3: APURINIC LYASE associated with DNA POLYMERASE β (specialized for repair) removes the deoxyribose phosphate

Step 4: DNA POL β inserts the single base & DNA LIGASE links it to the backbone.

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22
Q

What are the five enzymes involved in a DNA excision repair system for deamination?

A

DNA glycosylase, Apurinic endonuclease I, Apurinic lyase, DNA polymerase β, and DNA ligase.

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23
Q

What is a depurination? What kind of mutations can it lead to? How many events per day?

A

Depurination (an endonuclease DNA damage) is the hydrolysis of N-glycosidic bond; it separates the base (adenine or guanine) from the rest of the nucleotide structure.

It can lead to deletions.

There are about 18,000 events per day.

24
Q

Another spontaneous DNA damage are replication errors. What happens in them? How are they corrected?

A

Incorrect base is added, which affects base pairing during next DNA replication.

They are corrected by 3’ -> 5’ exonuclease activity (proof reading) of DNA polymerase.

25
Q

DNA polymerase resembles a half opened ____ _____. The fingers bind the ss segment of the ________ strand, and the polymerase _________ ________ site lies in the junction between the fingers and the palm.

A

Right hand
Template
Catalytic site

26
Q

Briefly describe how the proofreading of the DNA polymerase works.

A

Incorrect base addition at the 3’ of the growing strand causes the newly formed end of the duplex melt and the polymerase pausing. This strand is transferred to the 5’ –> 3’ exonuclease site where the mispaired base + others removed. Then, the 3’ end flips back into the polymerase site and elongation continues.

27
Q

What is the product of the oxidation (another spontaneous DNA damage) of guanine? What can it bind to? What kind of mutations can this lead to?

A

8-oxoG, Adenine, Transversions

28
Q

What is caused by UV radiation (an exogenous source of DNA damage)?

A

Pyrimidine dimers

29
Q

Pyrimidine dimers generate a ____ in the helix, halting cell ______ (death).

A

Halt, replication.

30
Q

Which pyrimidine dimers caused by UV radiation get corrected and which ones lead to deamination?

A

T-T dimers get corrected because they have a kink but C-C often leads to deamination.

31
Q

What is the difference between the mismatch excision repair and the nucleotide excision repair?

A

Mismatch excision repair mechanisms correct errors during replication, while the nucleotide excision repair mechanisms fix DNA regions with chemically modified bases that distort DNA’s shape locally.

32
Q

Match the following terms regarding the mismatch excision repair system:

Step 1
Step 2
Step 3

A) DNA polymerase (pol δ) fills the gap and LIGASE connects.

B) An MSH2-MSH6 protein complex binds to a mispaired segment of DNA & distinguishes between the template and new strand

C) Binding triggers MLH1 and PMS2, which bind an ENDONUCLEASE to cut daughter strand. DNA HELICASE unwinds helix and an EXONUCLEASE removes the cut nucleotides with the mismatched base.

A

Step 1 - An MSH2-MSH6 protein complex binds to a mispaired segment of DNA & distinguishes between the template and new strand

Step 2 - Binding triggers MLH1 and PMS2, which bind an ENDONUCLEASE to cut daughter strand. DNA HELICASE unwinds helix and an EXONUCLEASE removes the cut nucleotides with the mismatched base.

Step 3 - DNA polymerase (pol δ) fills the gap and LIGASE connects.

33
Q

How does the MSH2-MSH6 protein complex (involved in the mismatch excision repair) distinguish the daughter and the parent strand in prokaryotes and in eukaryotes?

A

Prokaryotes daughter strand uniquely methylated

Unclear about eukaryotes

34
Q

Inheritable loss of function MSH2 or MLH1 (proteins involved in the mismatch excision repair) mutations cause predisposition to
__________ _______ _______.

A

Non-polyposis colorectal cancer

35
Q

What are the similarities and the differences between endonucleases and exonucleases?

A

Similarities:
- both cleave nucleic acids
- both act on DNA and RNA
- both involved in the DNA repair inside cell

Differences:
- endonuclease cleaves nucleic acids at the middle of the nucleic acids
- exonuclease cleaves nucleic acids at the ends (one by one)

36
Q

How were some of the proteins involved in the nucleotide excision repair mechanisms identified?

A

Study of defects in DNA repair in cultured cells from people with XERODERMA PIGMENTOSUM, which can lead to skin cancer if exposed to UV rays from sunlight.

37
Q

How do people with Xeroderma Pigmentosum become sensitive to UV rays?

A

Mutations in any of at least the 7 genes (XP-A through XP-G) can inactive the components of the nucleotide excision repair system

38
Q

Match the following terms regarding the nucleotide excision repair in human cells:

Step 1
Step 2
Step 3
Step 4

A. Endonucleases XP-F and XP-G (now acts as one) cut out the damaged strand at points 24-32 bases apart, which turns into mononucleotides.
B. An XP-C-23B protein complex recognizes double helix distortions such as the kink caused by the T-T dimer.
C. Gap filled by DNA polymerase and joined by DNA ligase.
D. The XP-C-23B protein complex recruits transcription factor TFIIH which unwinds the DNA through its ATP-powered helicase. XPG and RPA furhter bind n unwind n stablize helix, forming a bubble of about 25 bases.

A

Step 1 - An XP-C-23B protein complex recognizes double helix distortions such as the kink caused by the T-T dimer.

Step 2 - The XP-C-23B protein complex recruits transcription factor TFIIH which unwinds the DNA through its ATP-powered helicase. XPG and RPA furhter bind n unwind n stablize helix, forming a bubble of about 25 bases.

Step 3 - Endonucleases XP-F and XP-G (now acts as one) cut out the damaged strand at points 24-32 bases apart, which turns into mononucleotides.

Step 4 - Gap filled by DNA polymerase and joined by DNA ligase.

39
Q

Match which proteins are involved in which step of the nucleotide excision repair:

Step 1
Step 2
Step 3
Step 4

A. XP-F and XP-G as endonucleases
B. XP-C-23B protein complex
C. DNA Polymerase, DNA ligase
D. XP-C-23B protein complex, TFIIH, Helicase, XP-G, and RPA

A

Step 1 - XP-C-23B protein complex
Step 2 - XP-C-23B protein complex, TFIIH, Helicase, XP-G, and RPA
Step 3 - XP-F and XP-G as endonucleases
Step 4 - DNA Polymerase, DNA ligase

40
Q

What are the two causes of DNA double-stand breaks?

A
  1. Ionizing radiation
  2. Anticancer drugs
41
Q

What are the two excision repair systems used to repair double-strand breaks?

A
  1. Non-homologous end joining
  2. Homologous end joining
42
Q

T/F Non-homologous end joining accounts for most double-strand break repairs outside of meiosis.

A

True!

43
Q

Match the following steps regarding non-homologous end-joining:

Step 1
Step 2
Step 3

  1. The two ds molecules are ligates together with several base pairs missing
  2. Nucleases remove bases from the DNA ends
  3. A ku-DNA-dependent protein kinase (DNAPK) complex binds to the ends of a double-strand break forming a “synapse”
A

Step 1 - A ku-DNA-dependent protein kinase (DNAPK) complex binds to the ends of a double-strand break forming a “synapse”

Step 2 - Nucleases remove bases from the DNA ends

Step 3 - The two ds molecules are ligates together with several base pairs missing

44
Q

How is non homologous end joining error prone?

A

1) Loss of good bases = gene mutations
2) Incorrect rejoining

45
Q

What does incorrect rejoining lead to?

A
  1. Can create a hybrid gene that codes for the N-terminal portion of one aa sequence fused to the C-terminal portion of a diff aa sequence.
  2. Chromosomal rearrangement mutations that can affect the expression of genes by bringing a promoter of one gene close to the coding region of another.
46
Q

T/F. Homologous recombination accounts for double strand breaks outside of meiosis.

A

False! Homologous recombination plays an important role during meiosis.

47
Q

What is homologous recombination? What are its two functions?

A

Exchange of DNA strands between two homologous DNA sequences.

  1. Is a repair mechanism; damaged DNA sequence is copied from an undamaged copy of same/ highly homologous sequence
    - of a collapsed replication fork
    - double-strand break
  2. Creates genetic diversity by exchanging large regions of chromosomes between maternal and paternal pair of homologous chromosomes during meiosis
48
Q

What happens if a break in the phosphodiester backbone of a DNA strand is not repaired before a replication fork passes?

A

The replicated portions of the daughter chromosomes become separated because there are no covalent bonds between the two fragments on either side of the nick.

49
Q

Match the following steps regarding the homologous recombinational repair of a collapsed replication fork

Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Step 7

A. RecA (bacteria) or Rad51 (eukaryotes) protein bind to the 3’ end of the ssDNA and catalyze STRAND INVASION
B. Cleavage of the phosphodiester bonds that cross over from one parent strand to the other.
C. Rebuild replication fork and continue replication
D. Ligation of the 5’ and 3’ ends base paired to the same parent strands generates a structure similar to replication fork
E. A nick in one strand causes replication fork collapse
F. Target DNA-invading strand is extended in the direction away from the break (BRANCH MIGRATION) forming a HOLIDAY STRUCTURE
G. 5’-EXONUCLEASE digests the strand with its 5’ end at the broken end of DNA, leaving the strand with its 3’ end at break to be ssDNA. Other daughter stand bp to unbroken parent strand is ligated to the unreplicated portion of the parent chromosome with the nick.

A

Step 1 - A nick in one strand causes replication fork collapse

Step 2 - 5’-EXONUCLEASE digests the strand with its 5’ end at the broken end of DNA, leaving the strand with its 3’ end at break to be ssDNA. Other daughter stand bp to unbroken parent strand is ligated to the unreplicated portion of the parent chromosome with the nick.

Step 3 - RecA or Rad5 protein bind to the 3’ end of the ssDNA and catalyze STRAND INVASION

Step 4 - Target DNA-invading strand is extended in the direction away from the break (BRANCH MIGRATION) forming a HOLIDAY STRUCTURE

Step 5 - Cleavage of the phosphodiester bonds that cross over from one parent strand to the other.

Step 6 - Ligation of the 5’ and 3’ ends base paired to the same parent strands generates a structure similar to replication fork

Step 7 - Rebuild replication fork and continue replication

50
Q

Match the following terms to their definition:

  1. Strand Invasion
  2. Branch migration
  3. Holiday structure
  4. RecA
  5. Rad51

A. Protein found in bacteria that mediates strand invasion
B. Protein found in eukaryotes that mediates strand invasion
C. A result of branch migration where all bases are base paired to complementary bases in the parents strands
D. The target DNA-invading strand region is extended away from the break
E. The broken single-strand DNA and the protein search and identify a region of homology in intact duplex DNA to base pair to

A

Strand invasion - The broken single-strand DNA and the protein search and identify a region of homology in intact duplex DNA to base pair to

Branch migration - The target DNA-invading strand region is extended away from the break

Holiday structure - A result of branch migration where all bases are base paired to complementary bases in the parents strands

RecA - Protein found in bacteria that mediates strand invasion

Rad 51 - Protein found in eukaryotes that mediates strand invasion

51
Q

Match the following steps regarding the repair of ds break by homologous recombination

Step 1
Step 2
Step 3
Step 4
Step 5
Step 6

A. DNA polymerase extends the 3’ end of the invading DNA strand (using the displaced ss loop of parent DNA as template) until the loop base-pairs with the other 3’ ss end.
B. Broken ends digested by 5’-exonuclease, leaving each with 3’ single-stranded ends.
C. Holiday structures resolved; cleavage of the strands, followed by ligation of the alternative 5’ and 3’ ends.
D. RecA or Rad51 mediate strand invasion of one of the 3’ end into the homologous region of the homologous chromosome
E. The 3’ ends are ligated to the exonuclease digested 5’ ends, generating two holiday structures.
F. DNA polymerase extends the other 3’ end sequence.

A

Step 1 - Broken ends digested by 5’-exonuclease, leaving each with 3’ single-stranded ends.

Step 2 - RecA or Rad51 mediate strand invasion of one of the 3’ end into the homologous region of the homologous chromosome

Step 3 - DNA polymerase extends the 3’ end of the invading DNA strand (using the displaced ss loop of parent DNA as template) until the loop base-pairs with the other 3’ ss end.

Step 4 - DNA polymerase extends the other 3’ end sequence.

Step 5 - The 3’ ends are ligated to the exonuclease digested 5’ ends, generating two holiday structures.

Step 6 - Holiday structures resolved; cleavage of the strands, followed by ligation of the alternative 5’ and 3’ ends.

52
Q

What are the products of repair of a ds break by homologous recombination?

A

Two recombinant molecules that contain the DNA of one parent on one side of the initial break point, and the other parent DNA molecule on the other side of the break point

53
Q

A holiday structure can be cleaved and relegated in two different ways (vertical/horizontal), horizontal cleavage leads to generation of ________ chromosomes, while vertical cleavage leads to the generation of ________ chromosomes.

A

Horizontal cleavage leads to generation of ORIGINAL chromosomes, while vertical cleavage leads to the generation of RECOMBINANT chromosomes.

54
Q

Match the following terms to their definitions:

Spo11
Mre11 exonuclease complex
Gene conversion

A) Digests the 5’ end of the ds breaks during meiotic recombination
B) A type of DNA recombination in which one DNA sequence is converted to the sequence of a second homologous DNA, leading to a loss of heterozygosity
C) Directs a programmed double strand break for meiotic recombination (similar to topoisomerase)

A

Spo11 - Directs a programmed double strand break for meiotic recombination (similar to topoisomerase)

Mre11 exonuclease complex - Digests the 5’ end of the ds breaks during meiotic recombination

Gene conversion - A type of DNA recombination in which one DNA sequence is converted to the sequence of a second homologous DNA, leading to a loss of heterozygosity

55
Q

T/F Base-pair mismatches between the two parent strands during meiotic recombination usually repaired to generate a complementary bp. In this process, the sequence differences between the two parents are not lost.

A

False, the sequence differences between the two parents are lost. AKA GENE CONVERSION –> loss of heterozygosity.

z Molecular Bio (9 decks)

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