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DNA Flashcards

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1
Q

What does each diploid eukaryotic cell contain?

A

Diploid eukaryotic cells contain two copies of each chromosome (1 paternal, 1 maternal)

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1
Q

What does each diploid eukaryotic cell contain?

A

Diploid eukaryotic cells contain two copies of each chromosome

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2
Q

What differs between each chromosome pair?

A

Each chromosome pair differs in size and DNA sequence

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3
Q

What 2 aspects of a chromosome can staining highlight?

A

> This staining can be sued to show if there are any additions or defects.

> Could also see area where there’s cross over between chromosomes.

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4
Q

In dividing cells what state is DNA in?

A

DNA is in a condensed state known as chromatin, which is tightly packed to form chromosomes.

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5
Q

What is a karyotype?

A

The organized representation of all the chromosomes in a eukaryotic cell at metaphase

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6
Q

What is meant by a cell in interphase?

A

Not dividing

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7
Q

During interphase, where are uncondensed chromosomes found?

A

Individual chromosomes occupy distinct subnuclear territories even in interphase nuclei

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8
Q

What are nucleosomes and what is its function?

A

> Structures made up of 8 histone proteins, which chromatin is wound around

> Helps compact DNA into a condensed structure, and N terminal tails from all 8 core histones helps interact with other proteins to modify chromatin structure

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9
Q

What are linker Histones (H1) and their function?

A

> Type of histone (called H1) which interacts with DNA at the nucleosomes (not apart of the nucleosome structure itself, they bind to the phosphate backbone and further condense the DNA).

> Allowing for further folding and compaction of the chromatin fibre. This compaction helps to further restrict access to the DNA and can play a role in regulating gene expression (establishment of transcriptionally silent heterochromatin).

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10
Q

What is important about linker histone’s structure (H1)?

A

This area is rich in arginine and lysine So is very basic, so can bind to phosphate backbone of DNA to cause further condensing (not sequence specific binding, so can bind anywhere on DNA)

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11
Q

What enzyme uncondences DNA so transcription proteins can bind?

A

Chromatin can be remodelled by chemo remodelling enzymes These remove nucleosomes, opening DNA so transcription proteins can bind.

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12
Q

What is the Fractal globule model for chromatin organization and its function?

A

> The chromatin fibre is highly compacted and folded into a series of loops that are nested within each other, forming a fractal-like structure. Nuclear scale -> chromosome scale -> Megabase scale (like Russian dolls gets smaller and smaller)

> Allows chromatic to be reversibly condensed and recondensed without becoming knotted.

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13
Q

In an interphase cell, where is a) transcriptional active b) transcriptional inactive DNA found?

A

a) kept in the middle of nucleus

b) restricted to periphery of nucleus

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14
Q

What are 4 features which allow reliable replication and segregation of chromosomes?

A
  1. Telomeres on ends of chromosomes
  2. Replication origins
  3. Centromeres (centromeric DNA)
  4. Kinetochore
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15
Q

How do sister chromatids get pulled to opposite poles of the cell during anaphase of mitosis?

A

During mitosis centromeric DNA bind to kinetochore (inner plate) allowing microtubules to bind to one or more of the centromeres (microtubules bind to kinetochores outer plate), so spindles line up accurately. This then pulls chromatids to opposite poles of the cell (centromeric DNA found in middle of chromatids where kinetochore binds and microtubules pull).

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16
Q

What is the structure of Kinetochores and how do they interact with other proteins?

A

Made up of 2 plates:

  1. Outer plate proteins bind to protein components of mitotic spindle like microtubules
  2. Inner plate proteins bind to chromatin containing alpha-satellite DNA (centromeric DNA)
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17
Q

What is a centromere?

A

It is the site on the chromosome where the kinetochore forms (out plate binds here), contains alpha-satellite DNA repeats allowing formation of centric heterochromatin (highly condensed chromatin)

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18
Q

What is the role of Kinetochores?

A

A protein structure that attaches to spindle microtubules and facilitates chromosome movement during cell division.

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19
Q

What % of our genome contains a) exons b) introns c) transposons d) what is the remining DNA for?

A

a) 1% exons (coding regions)

b) 20% introns (unique non-coding sequences between genes)

c) 50% transposons

d) The rest is non-repetitive DNA that is neither in introns nor codons

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20
Q

What are transposons and where do most of these come from?

A

> Repeated DNA sequences that are mobile and “jump” around the genome.

> Most of these elements are copies of retrotransposons known as “parasitic DNA”

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21
Q

What do more complex organisms contain more of in their genome than less complex organisms and why?

A
  1. Increased number of protein-coding genes for more proteins
  2. Increasing amounts of non-protein-coding DNA - these regions are important for regulating transcription and organising access to protein-coding genes.
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22
Q

What is one known role of introns and other non-coding DNA?

A

Some of the non-protein-coding DNA encodes cis-regulatory information which determines where and when in the body adjacent protein-coding genes are transcribed.

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23
Q

What are the 3 different types of Transposons and how do they function?

A
  1. DNA Transposons
    >Encodes transposase enzyme, bind and cut out transposons sequence and re-enter the genome at a target gene.
  2. Retroviral retrotransposons
    >Use reverse transcriptase from a virus capsid to transcribe RNA into DNA which is integrated into the genome at a new location.
  3. Non-retroviral PolyA retrotransposons
    >Similar to viral but doesn’t use viral capsid. Synthesis RNA transcriptase to synthesise RNA with PolyA tail. This cuts open DNA and reverse transcriptase transcribes a DNA copy of the RNA into this cut gap in the DNA.
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24
Q

What is an example of Non-retroviral PolyA retrotransposons causing a disease?

A

L1 insertions can cause disrupt genes causing haemophilia (blood doesn’t clot properly).

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25
Q

What is 7SL RNA?

A

A gene which is a shared ancestor of non-retroviral retrotransposons.

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26
Q

What must take place so that semi-conservative replication of DNA can take place?

A

The two strands must first be separated

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27
Q

What is the overall process of semi-conservative replication and what does it produce?

A

> Each separated strand is used as a template to determine the order of nucleotides in the 2 synthesised strands

> Produces: 2 double helix strands, 1 template strand and 1 newly synthesised strand.

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28
Q

What direction does DNA synthesis occur in?

A

5’ -> 3’ direction.

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29
Q

What direction is the leading strand and the lagging strand of DNA?

A

Leading strand (5’ -> 3’)

Lagging strand (3’ -> 5’)

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30
Q

What is the orientation of the primer strand in regards to the template strand?

A

The primer strand is anti-parallel to the template strand

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31
Q

How is the newly synthesised strand of DNA produced?

A

The newly synthesised strand is produced by extending the primer strand, by polymerisation of a new nucleotide onto the 3’ hydroxyl end.

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32
Q

What is the chemical reaction involved in nucleotide binding?

A

The chemical reaction is a nucleophilic attack on the phosphate group of the newly incoming dNTP (deoxyribonucleotide triphosphate), this binds the new nucleotide onto place on the primer strand while releasing 2 phosphate atoms (has to occur at 3’ end)

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33
Q

What is a nucleophilic attack?

A

Nucleophilic attack= transfer of electron to form diester bond

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34
Q

How are the leading and lagging strands orientated in regards to each other?

A

Anti-parallel

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35
Q

What are 9 proteins that are highly conserved and involved in DNA replication.

A
  1. DNA polymerase
  2. DNA helicase
  3. DNA primase
  4. Clamp loader
  5. Sliding clamp
  6. Ribonuclease H
  7. DNA Ligase
  8. DNA Topoisomerases I and II
  9. Telomerase
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36
Q

What 2 proteins make up the primosome?

A

DNA helicase and DNA primase

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37
Q

What proteins do E.coli use to produce the replication fork?

A

The same as humans, showing how highly conserved DNA replication machinery is.

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38
Q

What are the 5 stages of DNA replication in the leading strand? (maybe add to this)

A
  1. Creation of replication fork by DNA helicase separating double helix.
  2. Binding of primers to 3’ hydroxyl ends on the template strand by DNA primase to act as template for DNA polymerase (gives 3’ overhang for it to bind to)
  3. Clamp loader loads sliding clamp onto DNA close to the primer template
  4. DNA polymerase is clamped to the template strand and the clamp loader is released
  5. DNA polymerase adds nucleotides continuously
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39
Q

Why can both strands of DNA be synthesised at the same time without DNA polymerases hitting into each other?

A

As leading and lagging strands are anti-parallel.

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40
Q

How is DNA synthesis described for a) leading b) lagging strands?

A

a) continuous

b) discontinuous

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41
Q

What are Okazaki fragments?

A

Okazaki fragments= short pieces of DNA that allow synthesis of DNA to occur to the next Okazaki fragment (formed during synthesis of new lagging strand)

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42
Q

What is the role of DNA primase?

A

Primase binds to short RNA sequences and creates RNA/DNA hybrids (RNA primer bound to DNA template strand) providing a 3’ overhang for DNA polymerase to bind and extend the primer to form the synthesised strand.

43
Q

What are the 5 steps in discontinuous DNA synthesis which converts Okazaki fragments into a continuous strand of DNA and which enzymes perform each part?

A
  1. DNA Primase – makes RNA primer
    >Then dissociates and makes another primer further up the strand in the 3’ direction.
  2. DNA Polymerase – extends RNA primer - requires PRIMER-TEMPLATE junction
    >Polymerase extends from the primer until reaching the lagging strand newly synthesised strand, forming an Okazaki fragment
  3. Ribonuclease H – removes RNA primer
    >When polymerase reaches the old primer left on the lagging synthesised strand, Ribonuclease H removes this old primer, leaving a gap between the Okazaki fragments
  4. DNA Polymerase – extends across gap
    >Gap between the Okazaki fragment and the lagging synthesised lagging strand (left behind after primer removal) is filled in.
  5. DNA Ligase seals the nick
    >DNA ligase joins the Okazaki fragment to the growing chain, to seal the nick.
44
Q

How does DNA helicase separate the leading and lagging strands to form the replication fork?

A

Helicases form a ring-like structure around the DNA, using ATP hydrolysis it drives strand separation by breaking hydrogen bonds between the double strands which intern drives progression of replication fork.

45
Q

What is Werner syndrome and how is it caused?

A

> Werner syndrome is a genetic disease causing premature ageing.

> Werner Syndrome mutations are autosomal recessive, occurring in RECQ helicase encoded for by gene WRN. Replication is not as efficient or do not replicate at all

46
Q

What is Bloom syndrome and what is it caused by?

A

> A rare autosomal recessive disorder causing mutation in the Bloom gene which encodes for the Bloom protein, an ATP dependent Helicase (part of RecQ-family).

> Bloom can no longer remove roadblocks which cause destruction to the replication fork, so DNA replication and repair cannot occur correctly.

47
Q

Why does DNA replication occur quickly?

A

Due to processivity of DNA Polymerase

48
Q

What is processivity of an enzyme?

A

> Processivity: Enzymes ability to catalyse consecutive reactions without releasing substrate

> Highly Processive enzyme: always stays bound to substrate

49
Q

What does DNA polymerase processivity refer to?

A

Polymerase processivity refers to the average number of nucleotides added each time it binds to a template

50
Q

What would DNA polymerase with a) high processivity b) low processivity do?

A

a) Adds multiple nucleotides a second for long stretches.

b) Adds only 1 nucleotide per second, as after each reaction has to dissociate from substrate.

51
Q

What helps increase processivity of DNA polymerase and how?

A

> Processivity of DNA polymerase is helped by sliding clamps

> Sliding clamps allow DNA polymerase to be locked onto the template strand (not release DNA substrate) increasing its processivity so long stretches of DNA can be replicated at a quick rate.

52
Q

Where does DNA go through the sliding clamp complex?

A

DNA goes through the middle of sliding clamp

53
Q

Which 3 proteins aid DNA polymerase during DNA replication and their function?

A
  1. Sliding clamps
    >Increase processivity by locking DNA polymerase to template strand, so can can replicate long stretches of DNA.
  2. Single-stranded DNA binding proteins (SSBs)
    >Increase processivity by binding to single stranded DNA to keep it from forming DNA hair pins, stopping DNA replication blocks.
  3. Topoisomerases
    >Superhelical tension is relaxed, by nicking and resealing the backbone of DNA, prevents DNA from getting tangled during replication.
54
Q

Why does single stranded DNA in the replication fork form DNA hair pins and what issue do they cause?

A

If regions of single stranded DNA are complimentary to themselves, will rebind to each other causing hair pins (like mRNA self-interactions) which can block DNA replication.

55
Q

How is superhelical tension created when Helicase unwinds DNA?

A

When unwinding, DNA can become tangled at the replication fork, causing tension to build.

56
Q

What are the two types of Topoisomerases and their function?

A
  1. Type I Topoisomerases nick and reseal one of the 2 DNA strands, no ATP required
  2. Type II Topoisomerases nick and reseal both DNA strands, ATP required
57
Q

What is an origin of replication, what is the name of this in E.coli?

A

> Specific DNA sequences that attract DNA initiating proteins to the site which bind to the origin and promote melting of DNA double strand promoting access for DNA replication machinery and starting the replication force

> oriC in E.coli

58
Q

How many origins of replication are there in a) E.coli b) Humans?

A

a) 1

b) +100,000

59
Q

Why is initiation of DNA replication in eukaryotes described as biphasic?

A

Because it involves two separate phases.

60
Q

Describe the 2 phases of initiation of DNA replication in eukaryotes?

A
  1. Replicator Selection occurs in G1 phase
    - formation of a pre-Replicative Complex (Helicase)
  2. Origin Activation occurs in S phase
    - unwinding of DNA and recruitment of DNA Polymerase
61
Q

What is temporal selection in terms of initiation of DNA replication in eukaryotes and what is the importance of this?

A

> Temporal separation means replicator selection is separate from the activation of the origins of replication.

> This means each origin is only used once and each chromosome is only replicated exactly once per cell cycle.

62
Q

What occurs during the replicator selection phase of initiation of DNA replication in yeast?

A

> All occurs wihtin G1 of cell cycle:

  1. Origin Recognition Complex (ORC) binds to Replicator sequence, e.g. ARS sequence in yeast
  2. Helicase-loading proteins Cdc6 and Cdt1 bind to ORC
  3. The Helicase Mcm2-7 binds to complete formation of pre-RC
  4. It stays INACTIVE in G1, the machinery is just set up (due to low levels of Cdk).
63
Q

How does Cyclin-dependent kinase (Cdk) activity regulate activity of the Replicative Complex in G1 and S phases?

A
  1. In G1: low levels of Cdk (inhibited by Cdk inhibitors) allowing pre-replicative complex to form.
  2. In S phase: Cdk activity increases, as Cdc6 and Cdt1 (in yeast) are substrates for this enzyme becoming phosphorylated and activate helicase. Inhibiting creation of new pre-RC while activating the already made machinery.
64
Q

What is advantageous for Cdk to inhibit new pre-RC formation in S phase?

A

So no new origins of replication can be fired to start replication in other phases.

65
Q

Describe the role of Cdk’s role in the origin of replication complex in 3 steps.

A
  1. In G1 Cdk levels are low allowing assembly of pre-RC
  2. Remains inactive but loaded on DNA until S phase where Cdk activity increases and initiates replication
  3. Levels remain high so no more pre-RC are made, until levels become low again and can start to load onto DNA ready for S phase again
66
Q

Why is there a 3’ overhang left on the end of the lagging strand and the start of the leading strand

A

> Removal of last primer by Ribonuclease H

> Removal of initial primer by RNA H

(Primers are placed at 5’ end of new strand/ opposite to 3’ end of template strand)

67
Q

How does DNA primase synthesise RNA primers?

A

Using NTPs makes complimentary RNA molecule to the template strand.

68
Q

Without telomeres why would DNA shorten during DNA replication and what is a risk caused by this?

A

> Ribonuclease H removes RNA primers, further shortening newly synthesised DNA strands at the 5’ ends of the chromosomes (leaving 3’ overhang of template strand)

> Chromosomes shortening risks loss of valuable coding information

69
Q

What are Telomeres and their function?

A

> Non-coding regions that occur at the ends of chromosomes to stop genetic info being lost.

> Stops Chromosome shortening during replication.

70
Q

What enzyme creates Telomeres and how doe this work?

A

> Telomerase

> Telomerase contains the sequence AAUCCCAAC which can base pair with TTAGGG telomere repeats, by providing own RNA template primer, telomerase can use reverse transcriptase activity to polymerase DNA from its RNA template via the Telomerase Shuffle (so keeps adding TTAGGG at the end of a chromosome).

71
Q

How do Telomeres stop loss of DNA during DNA replication in 4 steps?

A
  1. Telomerase RNA pairs up with the existing telomere repeat with a 3’ overhang at the end.
  2. Telomerase uses this as a template to reverse transcribe these 3 bases )adds a TTA onto the end
  3. Shuffles along and uses the CCCAAU region to add GGGTTA (Telomerase shuffle)
  4. Then shuffles along binding to TTA to redo this cycle (each cycle synthesizes 6 nucleotides (TTAGGG) and this can happen 100,000s of times so no genetic material is lost.)
72
Q

What is the only biological macromolecule that can self-repair in humans?

A

DNA

73
Q

What is our most important defence against cancer?

A

Stability of DNA

74
Q

What is the consequence of DNA damage (and not repairing) in a) actively proliferating cells b) non-dividing cells?

A

a) Cancer: Errors in Chromosome replication and repair leading to Mutations, gives advantage to cells which are highly replicative like cancer cells

b) Ageing: Blocking transcription, reducing gene expression leading to decline in tissues.

75
Q

What are the 2 types of source for DNA damage and examples of each?

A
  1. Endogenous sources of DNA damage
    >Happen spontaneously in cell
    >Reactions with molecules within the cell (Hydrolysis, oxygen species, by-products of metabolism)
  2. Exogenous sources of DNA damage
    >Environmental factors.
    >Reactions with molecules from outside the cell (UV, X-rays, carcinogens, chemotherapeutics)
76
Q

What are the 4 types of endogenous DNA damage (spontaneously occurring within cell) and how many strands of DNA does it effect?

A
  1. Deamination
    >Effects one strand of DNA helix
  2. Methylation
    >Effects one strand of DNA helix
  3. Replication errors
    >Effects one strand of DNA helix
  4. Depurination
    >Effects one strand of DNA helix
77
Q

What are the 3 types of exogenous DNA damage and how many strands of DNA does it effect?

A
  1. Pyrimidine dimers
    >Effects one strand of DNA helix
  2. Double strand breaks
    >Effects both strands of the DNA helix

3.Interstrand crosslinks
>Effects both strands of the DNA helix

78
Q

What type of DNA damage is a) more common b) more lethal and why?

A

a) Single stranded breaks (especially depurination)

b) Double stranded breaks are more lethal as are harder to repair than single stranded breaks.

79
Q

How does deamination work and what is the most common form?

A

> Removal of the amino group by hydrolysis, results in changes to DNA bases.

> Most common, is removal of amino group from cytosine, releasing ammonia creating uracil, so the original G-C is changed to A-U

80
Q

What are the 2 types of Point mutations in DNA?

A

Transitions and Transversions

81
Q

What occurs in a) Transversion b) Transition mutations and which is more common?

A

a) Transversions is between purine and pyrimidine base (e.g. A to C or G to T)

b) Transitions more common to occur, a purine is swapped with another purine base (e.g. A to G or T to C)
>More easily to replace a single ringed structure with a single ringed structure or a double ringed structure with a double ringed structure

82
Q

Why does the Wobble Base Theory describe Transversion mutations to be more likely to result in amino acid substitutions than transition mutations?

A

> The theory suggests that the third nucleotide in a codon can sometimes tolerate a mismatched base pairing (wobbly base pairing)

> So a transition mutation swapping a purine for a purine wouldn’t effect this wobbly pairing much, while a transversion swapping the purine for a pyrimidine or vice versa would effect this wobbly pairing and therefore have a higher chance of substituting the amino acid.

83
Q

How does Depurination (abasic site) work and which type of nucleotide does this occur in more often?

A

> The base is cleaved away from a nucleotide, breaking the connection with the sugar phosphate backbone. leaves a sugar phosphate backbone with no nucleotide.

> More frequently in purine bases.

84
Q

If DNA is not repairs after depurination, what could be the effect?

A

Due to a loss of a nucleotide, could cause a frameshift in base order; which could result in a substitution of many amino acids.

85
Q

What are the reading frames of proteins, what is the effect of a shift in a reading frame?

A

> Proteins are translated from 3 reading frames (codons)

> If reading frame shifts by one and the base sequence changes, leads to different amino acids being made.

86
Q

What is a frameshift mutation and what can they cause?

A

> When a mutation causes a base to be added or delated so a different base codon is transcribed (a different base sequence)

> Leads to missense proteins

87
Q

How can UV light damage DNA in 3 ways?

A
  1. Distorts DNA to produce pyrimidine dimers (adjacent pyrimidines bind) meaning one strand of DNA doesn’t have flexibility (can’t be condensed and uncondensed for transcription as easily)

2/3. Interstrand crosslinks (incorrect base pairs) as well protein crosslinks (proteins bound to backbone of one strand) making DNA not unwind, blocking replication and transcription.

88
Q

Give 3 examples of damage that affects the phosphate backbone that are a) double strand break b) single strand break inducers?

A

a) X-rays, Ionising radiation, Topoisomerase II inhibitors

b) Reactive oxygen species, Hydroxyurea, Camptothecin

89
Q

How does DNA repair from depurination (abasic) sites and deamination in 3 steps (and the name of the process)?

A

> Base excision repair:

  1. DNA Glycosylase recognises and removes unwanted base such as Uracil instead of A (if is a deamination)
  2. AP endonuclease and phosphodiesterase remove sugar phosphate of were the base was (deamination) or were the base was always missing (depurination)
  3. DNA polymerase fills in the strand with a single-nucleotide gap and DNA ligase seals the nick (forms sugar-phosphate backbone)
90
Q

What bonds form between the adjacent nucleotides to form the sugar phosphate backbone and between the opposite bases on each strand?

A

> Diester bonds to form sugar phosphate backbone

> Hydrogen bonds between bases

91
Q

How does DNA repair from damage to more than one base such as pyrimidine dimers (by UV) in 3 steps (and the name of the process)?

A

> Nucleotide excision repair (NER):

  1. Excision nuclease creates breaks in sugar phosphate backbone of one DNA strand, at a number of bases on each side of the damage
  2. DNA helicase unwinds DNA and removes the excised parts, leaving a gap of nucleotides
  3. Template strand used as a template for DNA polymerase to fill the gap with the correct complimentary nucleotides.
92
Q

What is translesion synthesis in 3 steps?

A
  1. When sliding clamp encounters DNA damage on template strand, disassociates with DNA Polymerase.
  2. Associates with Translesion DNA polymerase which fills in the new strand with random nucleotides (can’t use complimentary to template as is damaged).
  3. After past the damage on template strand, clamp disassociates with Translesion polymerase and re-associates with normal DNA P to carry on synthesis.
93
Q

What is an issue with Translesion DNA polymerase?

A

Lacks precision in template recognition and substrate base choice, leading to base substitutions and single nucleotide deletion mutations due to random nucleotide selection.

94
Q

What are the 2 mechanisms to repair double strand breaks in DNA and when do they occur?

A
  1. Nonhomologous end joining (NHEJ)
    >Occurs in G1 (no DNA replication as occurred so no sister chromatids).
  2. Homologous recombination (HR)
    >Occurs in S phase or G2 (as DNA replication has occurred so sister chromatids present)
95
Q

Describe non-homologous end joining (NHEJ) in 5 steps

A
  1. Resections double strand break via MRN protein complex To produce a 3’ overhang
  2. This initiates the Heterodimer Ku70/Ku80 A long with DNA-PKcs will dissociate both ends of the DNA strands
  3. Forms synaptic complex, pulls the ends of the strands together
  4. End processing via an endonuclease, cuts off 3’overhang
  5. Ligase re-joins the strands together.
96
Q

What is an issue with non-homologous end joining?

A

Error prone, leading to loss of nucleotides at break site and therefore loss of genetic information.

97
Q

Describe Homologous recombination in 6 steps

A
  1. Double strand break, MRN complex resections the double strand break producing 3’ overhang
  2. RPA along with BRCA1 BRCA2 coats 3’ overhang with Rad51, forming a Rad51 nucleofilament.
  3. Initiates strand invasion of sister chromatid forming holiday junction
  4. Homologous searching leads to the same sequence being found in the sister chromatid template as is damaged in the original strand, will synthesise the undamaged sequence.
  5. Forms another holiday junction to form a double holiday junction, then a Crossing over event re-joins the original strand
  6. Leads to an accurately repaired double strand break. (genetic info is swapped with sister chromatic via crossing over event)
98
Q

What is an advantage to homologous recombination?

A

Accurately repairs double strand damage without genetic information lost, as uses the sister chromatid as a template.

99
Q

Why does homologous recombination only occur in S and G2 phases?

A

As it requires a sister chromatid for a crossing over event to use genetic info as a template for the damaged DNA (so requires DNA replication to have occurred).

100
Q

Where are the 3 places during the cell cycle where DNA damage is detected and acted upon to stop the cell cycle and repair DNA?

A
  1. G1 checkpoint (cell growth checkpoint)
  2. Entry to S-phase (DNA replication checkpoint)
  3. Entry into mitosis
101
Q

What is an example of repair done in a) G1 b) G2?

A

a) G1 repair = non-homologous end joining

b) G2 repair = homologous recombination

102
Q

How does DNA damage cause the cell cycle to stop in 4 steps?

A
  1. Damage causes ATM/ATR protein kinases to get activated and associate with the site of DNA damage
  2. This activates other kinases like Chk1/Chk2 which phosphorylate P53
  3. p53 phosphorylation stabilises and activates it which binds to p21 which inhibits Cdk (kinase enzymes) (which is needed for S-phase) stopping cell cycle progression
  4. DNA is then repaired or apoptosis (if repair is not possible) so mutations aren’t replicated and passed onto the next cells.
103
Q

What happens to a cell if the DNA damage cannot be repaird?

A

Apoptosis

104
Q

What are 2 common effects of diseases caused by failure in DNA repair systems?

A

A pre-disposition to cancer, and pre-mature ageing.

105
Q

What is a disease which effects Nucleotide excision repair (NER) and what does this make people venerable to?

A

> Xeroderma Pigmentosum, an Autosomal Recessive Disease

> Defect in repair of UV damage, so damaged by UV quickly causing skin cancer

106
Q

What is an example of a disease causing defects in homologous recombination (double strand break repair) and what does this make them venerable to?

A

> 80-80% of inherited breast cancers cause loss of function of BRCA1/2 meaning homologous recombination cannot repair double stranded breaks.

> Venerable to mutations causing double stranded breaks so have predisposition to cancer e.g. X-rays, ionising radiation, topoisomerase II inhibiters. As well as single stranded breaks which turn into double stranded breaks.

Advanced Molecular Biology (9 decks)

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