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Genetics and Genomics (GEN) > GEN 8: Protecting the Genome > Flashcards

GEN 8: Protecting the Genome Flashcards

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1
Q

Observe the learning outcomes of this session

A
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2
Q

What are the two basic sources of DNA damage?

A
  • reactive chemicals:
  • free radicals from our ‘basic metabolism’
  • radiation
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3
Q

What are free radicals and how do they damage DNA?

A
  • they are molecules with an unpaired electrons
  • The most common types in our cells are reactive oxygen species (ROS) generated most often by the incomplete reduction of oxygen during mitochondrial oxidative phosphorylation.
  • ROS oxidise the DNA
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4
Q

What are the forms of chemical endogenous DNA damage sources?

A
  • ROS (oxidation)
  • spontaneous hydrolysis
  • alkylating agents
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5
Q

Which UV rays are most damaging?

A
  • DNA absorbs UV light most efficiently at 260nm wavelength.
  • This, and shorter wavelengths (higher energy), are highly damaging to DNA but fortunately these are efficiently absorbed by the atmosphere.
  • Wavelengths in the range 295-320nm (termed UVB) do reach us, however, and damage DNA in the skin cells that absorb them.
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6
Q

What are some other types of radiation that aren’t UV?

A
  • X-rays
  • radioactive elements (natural or man-made)
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7
Q

What is ionizing radiation (IR)?

A
  • penetrating radiation that causes DNA damage is called ionizing radiation (IR)
  • It includes both electromagnetic waves (e.g. gamma-rays, X-rays, UVC rays) and atomic or subatomic particles with sufficient energy to dislodge electrons from the atoms with which they collide.
  • Most of the IR we experience is natural, such as cosmic radiation and unstable isotopes in rocks (e.g. Uranium-235).
  • Much less is manmade (e.g. hospital X-ray equipment).
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8
Q

How does IR damage DNA?

A
  • either directly or by generating ROS from water
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9
Q

What type of DNA damage are ionising and UVB radiation?

A
  • exogenous
  • physical
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10
Q

What other exogenous chemicals cause DNA damage?

A
  • Environmental pollutants (e.g. alkylating agents and other chemicals in car fumes, tobacco smoke, crop sprays etc).
  • Natural toxins (e.g. fungal aflotoxins[SAM1]).
  • Dietary chemicals (e.g. products of cooking or curing processes including nitrosamines)
  • Anti-cancer drugs (e.g. cisplatin)
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11
Q

What is a form of endogenous physical damage?

A
  • Though less obvious, mechanical damage can arise from errors in chromosome replication or segregation, causing chromosomes to be torn apart by the mitotic spindle apparatus.
  • For example, this can result from unscheduled replication of centromeric DNA, as illustrated in the next image.
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12
Q

Explain how mechanical DNA damage occurs

A
  • The blobs and rings represent centromeres and cohesin, respectively.
  • Thin lines represent spindles attached to centromeres at metaphase (centre) and anaphase (right) and pulling chromatids to opposite poles to the left or right.
  • Failure of controlled replication origin firing (see GEN6) may cause re-replication in the centromeric region resulting in one sister with two centromeres.
  • Depending on how the spindles attach, the outcome may be chromosome mis-segregation (top), or physical breakage of the chromosomal DNA (bottom).
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13
Q

Give examples of biological genome damage sources

A
  • viruses
  • transposons
  • DNA replication errors
  • chromosome segregation failures
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14
Q

Describe how viruses are a biological DNA damage source

A
  • Many viruses (e.g. retroviruses) insert their genomes into the genome of their host cell.
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15
Q

Describe how transposons are a biological DNA damage source

A
  • though rare in human cells, transposons can move from site to site within a genome causing insertional or excisional mutagenesis.
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16
Q

Describe how DNA replication errors are a biological DNA damage source

A
  • DNA replication errors are another source of biological DNA damage. These result from:
  • Nucleotide misincorporation by DNA Polymerase
  • Replication slippage during microsatellite DNA replication. The image illustrates slippage during replication of a stretch of DNA with a single nucleotide repeat (AAAA). Note how insertion-deletion loops (IDLs) of unpaired bases are formed.
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17
Q

Describe how chromosome segregation failures are a biological source of DNA damage

A
  • Two pairs of sister chromatids are depicted at metaphase (left) and the ensuing anaphase (right).
  • The top pair segregates normally, each being pulled to opposite poles of the cell by spindles (microtubules) attached to centromeric proteins (kinetochores).
  • Each daughter cell will therefore receive one copy of the top chromosome.
  • The bottom pair, however, has one sister that fails to attach to the spindle.
  • As a result, both sisters will move to a single pole, so one daughter cell will receive two copies and the other will receive none.
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18
Q

Complete the table of the sources of DNA damage

A
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19
Q

What are the main types of DNA damage?

A
  • base modification or loss
  • single strand breaks
  • bulky adducts:
  • intra-strand-X-links
  • CPDs
  • mismatches and IDLs
  • inter-strand X-links
  • double strand breaks
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20
Q

Describe the causes, consequences and repair pathways of base modifications or losses

A
  • causes:
  • oxidation (ROS/IR)
  • hydrolysis
  • alkylation
  • consequences:
  • point mutations
  • replication stalling
  • repair pathways:
  • base excision repair
  • direct repair
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21
Q

Describe the causes, consequences and repair pathways of SSB (single strand breaks)

A
  • causes:
  • ROS
  • IR
  • consequences:
  • converted to double strand breaks by DNA replication
  • repair pathway:
  • base excision repair
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22
Q

Describe the causes, consequences and repair pathways of bulky adducts and intrastrand crosslinks

and additional info

A
  • causes: UVB or alkylating agents
  • consequences: replication stalling
  • repair pathway: nucleotide excision repair
  • many cancer chemotherapies use cross-linking agents
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23
Q

Describe the causes, frequency, consequences and repair pathways of mismatches and IDLs

A
  • causes: replication errors
  • frequency: after proofreading the mammalian replisome generate only one mismatch per 108 nucleotides
  • consequences: replication stalling
  • repair pathway: mismatch repair (MMR)
24
Q

Describe the causes, consequences and repair pathways of inter-strand crosslinks

  • uses
A
  • causes: bi-functional alkylating agents
  • consequences: replication stalling, cell death
  • repair pathway: HR
  • many cancer chemotherapies use crosslinking agents
25
Q

Describe the causes, consequences and repair pathways of double-strand breaks

A
  • causes: ROS, IR, mechanical breaks, replications of SSB
  • consequences: if unrepaired may lead to small or large mutations including chromosomal deletions, inversions, translocations and chromosome loss
  • repair pathway: non homologous end joining (NHEJ), homologous-directed repair (HR)
26
Q

Describe the oxidation of guanine to 8-oxoguanine (base modification)

A
  • A. The base guanine can be oxidised by ROS to form 8-oxoguanine (8-OG), but this can be repaired by base excision repair.
  • B. If not repaired, 8-OG (boxed) pairs with A instead of C during DNA replication.
27
Q

Describe depurination: hydrolysis of guanine to form an abasic site (base modification)

A
  • Depurination is the removal of a purine base (Guanine or Adenine) from DNA by hydrolysis, leaving an abasic site (i.e. there is no base there).
  • This blocks replication and transcription and needs repairing by base excision repair.
28
Q

Describe the hydrolytic deamination of cytosine to uracil (base modification)

A
  • Deamination is the removal of an amino group. If this happens to cytosine, it becomes uracil.
  • Uracil in DNA is thought to block DNA replication and must be repaired by base excision repair.
29
Q

Describe the deamination of 5-methyl-cytosine to thymine (base modification)

A
  • If 5-methyl-cytosine is deaminated, it becomes thymine which, unless removed, becomes a fixed mutation in the DNA sequence.
30
Q

What are three mechanisms our cells have to minimise some types of endogenous DNA damage?

A
  • neutralising reactive oxygen species (ROS)
  • avoiding DNA replication errors
  • avoiding chromosome segregation errors
31
Q

Describe how reactive oxygen species (ROS) are neutralised

A
  • Naturally occurring antioxidants such vitamin C, beta-carotene and glutathione, provide some protection from ROS, but further protection is provided enzymatically.
  • As indicated in the following image, superoxide dismutase (SOD) catalyses the decomposition of superoxide radicals into hydrogen peroxide, which is then converted into water by catalase (CAT) or glutathione peroxidase (GPx).
  • In the latter reaction, the tripeptide glutathione (GSH) is oxidised to form a dimer (GSSG) which must then be reduced back to GSH by glutathione reductase (GH).
32
Q

Describe how DNA replication errors are avoided

A
  • The DNA polymerases of the replisome (DNA pol α, 𝛿 and ε) have inbuilt proof-reading activity.
  • During chain elongation they can sense nucleotide mis-incorporation and use their 3’-to-5’ exonuclease activity to remove the offending nucleotide before resuming synthesis, as illustrated in the next image.
33
Q

Describe how chromosome segregation errors are avoided

A
  • A mechanism called the Spindle Assembly Checkpoint (SAC) prevents aberrant chromosome segregation at anaphase and any resulting aneuploidy or chromosome breaks.
  • riefly, the SAC uses specialised proteins to sense the spindle tension. When this is too low because of incorrect attachment of spindles, a signal is sent that keeps APC/C in an inactive state.
  • Recall from GEN6 that APC/C is responsible for triggering anaphase by promoting the degradation of cohesin, the protein that holds sister chromatids together
  • In this way anaphase can only proceed when all chromosomes are correctly attached to spindles. The SAC mechanism is outlined on the following image (click to enlarge).
34
Q

What are the DNA repair pathways?

A
  • mismatch repair (MMR)
  • Base Excision Repair (BER)
  • Nucleotide Excision Repair (NER)
  • Homologous Recombination (HR)
  • Non-Homologous End-Joining (NHEJ)
35
Q

Place the correct pathway in the blue boxes

A
36
Q

Describe base excision repair

A
  • Base excision repair is required to remove different types of base damage.
  • This could be a removing uracil, 8-oxo-guanine (as shown here) or dealing with missing bases.
  • Four steps are involved in this process.
  • The key enzyme involved in this repair mechanism is DNA glycosylase.
  • DNA gyclosylase identifies the modified base and removes it from the double helix.
  • This results in an empty sugar phosphate region.
  • An endonuclease then cuts the DNA backbone.
  • The baseless sugar-phosphate is removed and DNA polymerase adds a new base, and DNA ligase seals the gap.
37
Q

Describe nucleotide excision repair (NER)

A
  • Nucleotide excision repair removes bulky damage such a pyrimidine dimers.
  • You can see here that these two Ts represent a pyrimidine dimer, the type of damage caused by UV radiation.
  • This type of damage will be recognised by RNA pol II, if it occurs in transcribed DNA.
  • However, if it occurs elsewhere in the genome, the protein XPC recognises it.
  • Multiple proteins are involved in DNA repair and you will see many starting with XP, after the disease Xeroderma pigmentosum (XP).
  • This is an autosomal recessive disease caused by mutations in an XP gene.
  • XP patients suffer from sensitivity to UV light resulting in corneal ulcerations and dry skin prone to blistering and cancer.
  • DNA around the damage is unwound by helicases.
  • Specifically, proteins XPD and XPB which are helicases. Endonucleases, in this case, XPF and XPG, snip the DNA up and downstream and 25-30 nucleotides are excised. \
  • New DNA is synthesised by DNA polymerase, and then sealed by DNA ligase.
38
Q

Describe mismatch repair (MMR)

A
  • So DNA is replicated and proofreading ensures most errors are rectified. However, this is not 100% efficient.
  • As such, we can be left with mismatched bases.
  • Mismatch repair can tackle this in a similar way to excision repair, by cutting out the damaged DNA and synthesising a new strand. However, in this case, there is no recognisable damage.
  • The bases are simply incorrect.
  • In this figure you can see guanine, which should pair with cytosine, is actually paired with thymine due to a replication error.
  • When new DNA is synthesised, it contains nicks, that you can see here.
  • This distinguishes the daughter strand, contained the mismatch, from the parent strand, and is therefore a way the cell can determine which strand has the error.
  • The protein MutS binds to the mismatched DNA and forms a complex with MutL.
  • These then slide along the daughter strand until they find one of these nicks.
  • An exonuclease degrades the mismatched region of DNA.
  • This gap is then filled once again by DNA polymerase in the 5’ to 3’ direction, and then sealed by DNA ligase.
  • The importance of the mismatch repair pathway is illustrated by Hereditary Nonpolyposis Colorectal Cancer (HNPCC), which accounts for 5-7% of colon cancers.

HNPCC is inherited in an autosomal dominant fashion as a result of mutations in any of several genes encoding MMR proteins, such as MutSα and MutLα.

39
Q

Why is BER significant?

A
  • base excision repair is very versatile because it can choose from multiple glycosylase enzymes, each recognising a different kind of damaged base.
  • Furthermore, the last two steps in BER contribute to the repair of IR-induced SSBs that have been recognised by a protein call polyADP ribose polymerase (PARP).
  • As we will see later, PARP turns out be an important therapeutic target.
40
Q

Recall the enzymes used in base excision repair

A
  • glycosylase
  • Endonuclease
  • DNA polymerase & ligase
41
Q

Recall the enzymes used in nucleotide excision repair

A
  • XPC or RNApolII
  • Helicase (XPB, XPD)
  • Endonuclease (XPF, XPG)
  • DNA polymerase & ligase
42
Q

What are the enzymes used in mismatch repair?

A
  • MutL & MutS
  • Exonuclease
  • DNA polymerase & ligase
43
Q

What processes repair double-stranded DNA damage?

A
  • homologous recombination (HR)
  • non-homologous end joining (NHEJ).
44
Q

Describe homologous recombination (HR) in more detail

A
  • Following a double-strand break in the DNA, exonuclease proteins bind to both broken ends and cleave to produce single stranded overhangs.
  • These overhangs are protected by the protein RPA.
  • RPA is then replaced by important repair proteins you have have heard of before: BRCA1, BRCA2 and RAD51.
  • The end bound by RAD51 then invades the homologous region of the sister chromatid and this forms a Holliday junction.
  • Other proteins are involved in this process, however it is just key for you to know here that DNA polymerase synthesises a new strand from this template, and DNA ligase seals the DNA.
  • You may have heard of BRCA1 and 2 gene mutations being linked to inherited or acquired cancers, which shows the importance of HR in repairing DNA.
45
Q

Describe non-homologous end joining (NHEJ) in more detail

A
  • NHEJ is an error prone process for repairing double-stranded DNA breaks.
  • When a break occurs, if the ends are blunt, DNA ligase can seal the gap right away.
  • However, if the ends are not blunt the need processing first. Protein Ku70/80 binds to both ends and recruits DNA protein kinase.
  • DNA-PK then recruits two more enzymes; an exonuclease called Artemis, and DNA polymerase.
  • These can trim and fill in the ends of the broken DNA respectively, creating blunt ends.
  • DNA ligase can then be recruited to seal the gap.
  • There are aberrant translocations of ends, derived from different chromosomes, associated with cancer.
46
Q

Which DNA repair pathways require which enzymes?

A
47
Q

Choose one or more pathways for each of the descriptions below

A
48
Q

Look at some of the proteins that are involved in the DNA damage response (DDR)

Describe their functions

A
  • includes:
  • sensors: detect DNA damage
  • mediators: recruit and activate transducers
  • signal transducing kinases: phosphorylate and activate effector proteins
  • effector kinases, effector proteins: lead to transcription, apoptosis and cell cycle arrest
49
Q

Why do you think it might be advantageous to arrest the cell cycle in response to DNA damage?

A
  • Cell cycle arrest allows DNA damage to be repaired before the DNA is replicated (G1/S checkpoint) or transmitted to daughter cells (G2/M checkpoint).
  • It also avoids complications such as stalling of DNA replisome at damaged DNA.
50
Q

What is the advantage of a cell undergoing apoptosis in response to DNA damage?

A
  • Apoptosis eliminates the high risk of tumorigenesis from cells where damage was too severe to be properly repaired.
51
Q

What is the paradox of cancer therapy?

A
  • Paradoxically, despite its cancer-causing effects, DNA damage can be used for cancer therapy.
52
Q

Describe radiotherapy and chemotherapy

A
  • In the treatment of many cancers, radiotherapy (ionising radiation) and chemotherapeutic drugs work by inducing DNA damage, particularly double strand breaks.
  • Because cancer cells grow and divide more frequently that normal cells, they are more susceptible to such agents.
  • These treatments have many side effects, such as killing the normally fast-growing cells in the bone marrow, digestive tract and hair follicles, leading to infections, sickness and hair loss.
  • More seriously, DNA damage in normal cells increases the risk of developing treatment-related cancers.
  • Chemotherapy for cancer is also hampered by cancers that develop resistance to the drug.
53
Q

Describe synthetic lethality

A
  • Synthetic lethality is relative new approach for targeting cancers with a known defect in DNA repair, such as a mutation in a BRCA gene. To treat such cancers without harming normal cells, drugs have been developed that inhibit PARP.
  • The drug Olaparib is a PARP inhibitor used to treat BRCA1 or BRCA2-deficient ovarian cancers. Other synthetic lethal treatments are under development.
54
Q

Can you predict the effect of a PARP inhibitor on DNA repair?

A
  • It will impair the ability of BER to repair single strand breaks.
55
Q

What happens to unrepaired SSBs?

A
  • They are converted into DSBs during DNA replication.
56
Q

So what will be the effect of PARP inhibition on normal cells and on BRCA1-deficient cancer cells?

A
  • Normal cells will be able to repair the DSB by HR, but the cancer cells will not because BRCA1 is required for DSBR by HR.
  • Normal cells will therefore survive while the cancer cells will prone to undergo apoptosis.
  • This is summarised in the figure attached showing the mechanism of cell death from synthetic lethality as induced by inhibition of Poly(Adenosine Diphosphate [ADP]–Ribose) Polymerase 1 (PARP1).

Genetics and Genomics (GEN) (10 decks)

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