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DNA structure and repair Flashcards

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1
Q

when can individual chromosomes be distinguished?

A

Individual chromosomes can be easily distinguished at metaphase of mitosis:
- In metaphase the chromosomes condense so they are visible
- They form a metaphase spread which can be stained in different colours specific for each chromosome
- Can easily identify any defects by using this methods
- Can identify areas of crossover for each chromosome – sign of genetic instability

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2
Q

how is genetic information contained in diploid eukaryotic cells?

A
  • Diploid eukaryotic cells contain 2 copies of each chromosome
  • Each chromosome pair differs in size and DNA sequence
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3
Q

what is a karyotype?

A

the organised representation of all the chromosomes in a eukaryotic cell at metaphase is called the Karyotype of the parent organism

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4
Q

how are chromosomes organised in the nucleus?

A

Individual chromosomes occupy distinct nuclear territories in interphase nuclei
- In non-dividing cells (interphase), staining can be used to see the different regions that correspond to different chromosomes
- Can see where the chromosomes are in the nucleus

Chromosomes are highly organised and exist in distinct nuclear territories, even when the cell isn’t dividing

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5
Q

what is a chromosome?

A

A chromosome is an organised, highly coiled fibre of chromatin

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6
Q

what is chromatin?

A

Chromosomes are organised into a highly coiled fibre of chromatin:
- Chromatin can exist as a 30nm fibre, or as a beads on a string
- Under the electron microscope, interphase chromatin resembles beads on a string
- The beads are nucleosomes
- chromatin forms a 30nm fibre

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7
Q

what is the structure of chromatin?

A
  • it is a supercoiled array of nucleosomes
  • the DNA helix wraps around the nucleosomes
  • it forms a highly supercoiled 30nm fibre
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8
Q

what is the structure of a nucleosome?

A

There are 8 protein subunits of the nucleosome called core histones
- The N-terminal tails of the 8 core histone subunits project out from the nucleosome core and are free to interact with other proteins
- This interaction facilitates regulation of chromatin structure and function

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9
Q

what are histones?

A

Histones make up cylindrical shape which the DNA helix is wrapped around the outside

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10
Q

what is the role of linker histone H1?

A

Linker histone H1 straps DNA onto histone octamers and limit movement of DNA relative to the histone octamer
- H1 is rich in lysine and arginine, so it can readily bind to the negatively charged DNA phosphate backbone, but is not sequence specific

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11
Q

what is the function of histones?

A

they limit movement of DNA relative to the histone octamer:
- This facilitates the establishment of transcriptionally silent heterochromatin
- Keeps chromatin tightly packed in regions that aren’t needed for transcription/regulation

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12
Q

how is DNA packaged by histone?

A

DNA is packaged by histone octamers into a compact, flexible, 30nm chromatin scaffold

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13
Q

how can the 30nm chromatin scaffold be remodelled? why does this occur?

A
  • 30nm scaffold can be remodelled to accommodate protein complexes involved in gene transcription and DNA replication
  • DNA remodelling enzymes remove the nucleosomes so that the DNA is unwound so proteins for transcription/replication can bind
  • chromatin is engineered to permit flexible responses to altered transcription factor activity caused by changes in cell differentiation status and changes in signalling pathway activities
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14
Q

how is interphase chromatin organised in a higher order fashion?

A

Interphase chromatin comprises a set of dynamic, fractal globules that can reversible condense and decondense without becoming knotted
- At the nucleus level, there are distinct patterns/territories of chromatin
- if we zoom in further at a higher resolution we can see areas of open and closed regions forming these patterns
- if we zoom in again within those closed regions there are distinct patterns of chromatin and those are the fractal globules

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15
Q

what are fractal globules?

A

Fractal globules are made up of globules within globules

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16
Q

why are fractal globules important in the higher order organisation of interphase chromatin?

A

It is this organisation of these globules that are fundamental to stabilisng the regions of inactive chromatin but also allows flexibility so that cells can react to certain cues such as progenitor cells differentiating into specialised cells

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17
Q

where do transcriptionally active and inactive DNA occur in the interphase nuclei?

A
  • The nuclear periphery in interphase cells is composed of transcriptionally inactive DNA
  • RNA transcripts are excluded from the periphery
  • Inactive DNA is restricted to nucleus periphery, whereas transcriptionally active DNA is at the centre of the cell
  • The fractal globules open up in the middle to allow transcription to occur
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18
Q

what do the specialised DNA sequences in chromosomes facilitate?

A

Chromosomes contain specialised DNA sequences that facilitate:
- Reliable and complete DNA replication
- Segregation of duplicated chromosomes during cell division

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19
Q

what specialised DNA sequences do chromosomes contain?

A
  • Telomeres – protect ends of chromosomes from multiple rounds of replication so that genetic info is not lost at these ends – prevents genetic instability
  • Replication origins
  • Centromeres – regions of repeated DNA sequences where the chromosomes are connected to the mitotic spindle
  • Kinetochore – centromeres bind to the kinetochore to allow stabilisation of the mitotic spindle so segregation can properly occur
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20
Q

what are telomeres?

A

Telomeres – specialised repetitive DNA sequences at chromosome ends
- Usually 10-100 nucleotides long
- Telomeres are replicated by a specialised DNA Polymerase called Telomerase
- Telomeres define chromosome ends and maintain chromosome integrity
- Single-stranded 3’ overhanging TTAGGG repeat arrays are synthesised by the Telomerase enzyme and can be several hundred nucleotides long

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21
Q

what is the function of telomeres?

A

the telomere region is non-coding, but it prevents the loss of genetic info as it ensures that the coding regions of the DNA stay in tact
- they maintain chromosome integrity

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22
Q

what does chromosome segregation require?

A

Chromosome segregation requires attachment of chromosomes to the mitotic/meiotic spindle
- During mitosis, the centromeres bind to the kinetochore, allowing the microtubules of the spindle to bind to chromosomes and separate them

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23
Q

what are centromeres? how are they involved in chromosome segregation?

A

Centromeres contain specialised proteins and DNA sequences that facilitate chromosome segregation during cell division
- Centromeres contain alpha-satellite DNA repeats that readily form condensed chromatin with histone octamers containing unusual subunits

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24
Q

how are the kinetochore proteins and centromeres involved in chromosome segregation?

A
  • Kinetochore Inner Plate proteins bind to chromatin containing alpha-satellite DNA repeats of centromeres
  • Kinetochore Outer Plate proteins bind to protein components of mitotic spindle i.e. microtubules

Part of the mechanism for ensuring faithful segregation of sister chromatids at cell division

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25
Q

what is the structure of the kinetochore?

A

In yeast, the kinetochore is a basket that links a single nucleosome of centromeric chromatin to a single microtubule:
- There is a single nucleosome-H3 specific to centromere which binds to the sequence specific kinetochore
- Outer plate binds to inner plate which forms a ring around the microtubules

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26
Q

what percentage of eukaryotic genomes are encoding?

A

Only 1.5% of our DNA is coding

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27
Q

what is 50% of the human genome made up of?

A

~50% of the human genome is made up of repeated DNA sequence elements - transposons
- Most of these repeated elements are copies of retrotransposons known as parasitic DNA

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28
Q

what percentage of the human genome is made up by introns?

A

20%

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29
Q

what occurs to the genome as biological complexity of an organism increases?

A
  1. Increasing numbers of protein-coding genes
  2. Increasing amounts of non-protein coding DNA - for regulating transcription/expression of genes and organising access to protein-coding genes
    - 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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30
Q

what 3 types of transposons make up almost half of the human genome?

A
  1. DNA transposons
  2. Retroviral retrotransposons
  3. Non-retroviral polyA retrotransposons
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31
Q

what are transposons?

A

Transposons are mobile genetic elements that jump around the genome – also called transposable elements
- they are repeated DNA sequences
- These can be 1000 nucleotides in length of repeated sequences
- Most of the copies of these elements in the genome are defective, ancient relics of formerly functional elements, with many mutations that prevent expression of functional proteins.

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32
Q

how do DNA transposons jump around the genome?

A

DNA transposons move by a cut-and-paste mechanism without self-duplication:
- DNA transposons require the transposon-encoded enzyme Transposase
- E.g. P-element (fly), activator-dissociator (maize), Tn3/Tn10 (E. coli)
- Transposase allows the DNA transposons to move around in a cut-and-paste mechanism
- Short inverted repeat sequences can cut out the transposon and move it to anywhere in the genome
- The empty gap will rejoin
- The transposon will integrate into the target chromosome

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33
Q

how can DNA transposons be mutagenic?

A

if chromosome isn’t put back together properly after inserting the transposon, this process can be mutagenic

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34
Q

how do retroviral retrotransposons behave like retroviruses?

A

They replicate via RNA intermediates, producing new DNA copies that integrate at new genomic locations, using self-encoded reverse transcriptase
- They do not encode fully active infectious viruses
- Retroviral retrotransposons use reverse transcriptase
- DNA that was transcribed to RNA is reverse transcribed to DNA in viral capsid which can be reintegrated into the chromosome at a random location

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35
Q

give 3 examples of retroviral retrotransposons:

A
  1. Ty1-copia
  2. Ty3-gypsy
  3. ERV elements
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36
Q

what are non-retroviral polyA retrotransposons?

A
  • These are abundant in vertebrate genomes
  • They replicate via an RNA intermediate using its own retrotransposon-encoded reverse-transcriptase
  • Simpler than retroviral retrotransposons as they do not require a viral capsid
  • Copy and paste
  • They transcribe DNA to RNA with a polyA tail, which is then reverse transcribed to DNA which can be reintegrated into the chromosome-
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37
Q

what are examples of non-retroviral polyA retrotransposons

A
  • Human L1 elements (LINE-1 elements)
  • Human Alu elements
  • Mouse B1 elements
  • Long Interspersed Elements “LINEs”
  • Short Interspersed Elements “SINEs”
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38
Q

what is the advantage of non-retroviral polyA retrotransposons?

A

Products of L1 reverse transcription are integrated directly into the genome at new locations without the need to be packaged into a virus-like particles

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39
Q

how may non-retroviral polyA transposons lead to disease?

A

Some L1 insertions are known to disrupt genes and cause human disease e.g. haemophilia

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40
Q

how have non-retroviral transposons increased with increasing complexity of genomes?

A

Non-retroviral transposons have expanded hugely in numbers during evolution of higher mammal genomes

E.g. human Alu and mouse B1 originally evolve from the single copy 7SL RNA gene in the common ancestor, around 90 million years ago
- Around 1 million copies of Alu in human genome, even though it is non-coding
-Alu and B1 are SINEs (Short Interspersed Elements)

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41
Q

what type of replication does DNA undergo?

A

semi-conservative

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42
Q

what is semi-conservative replication?

A
  • 2 strands of the DNA double helix are separated
  • the 2 strands can then be used as templates to determine the order of nucleotides that are to be synthesised in the complementary strands
  • this results in 2 newly replicated double helix DNA, each containing one strand of parental (original) DNA, and one newly synthesised strand
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43
Q

in what direction does DNA synthesis occur?

A

in a 5’ to 3’ direction

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44
Q

how are the DNA strands and primers orientated to one another?

A

Orientation of the primer strand is antiparallel to the template strand

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45
Q

how are newly synthesised strands produced by the primer strand? what is the name of this reaction?

A
  • Newly synthesised strand is produced by extending the primer strand, via polymerisation of a nucleotide onto the 3’-hydroxyl end of the primer strand
  • This chemical reaction is a nucleophilic attack on phosphate group of the incoming deoxyribonucleotide triphosphate (DNTP)
  • DNTP binds the nucleotide into place onto the primer strand and releases pyrophosphate in the process
  • Nucleophilic attack is the transfer of electrons to form a diester bond
  • phosphodiester bonds are formed
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46
Q

how do the replication machineries of E. coli and humans compare?

A

Key components of the E. coli and human replication forks are highly conserved
- Proteins of DNA replication in E. coli are the same as in humans
- Highly conserved
- Can use single-cell organisms to study this process

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47
Q

how is DNA replication initiated?

A

creation of the replication fork, where the DNA strands are separated:
- DNA helicase hydrolyses the hydrogen bonds between the double helix, resulting in the separation of the 2 strands

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48
Q

what is the process of DNA replication?

A
  1. creation of replication fork by DNA helicase separating the 2 strands
  2. binding of primers to the 3’-OH ends of the 2 strands
    - the primers act as templates for DNAP to bind to and synthesise new nucleotide chains
  3. for the leading strand, addition of nucleotides is simple and occurs naturally in a 5’ to 3’ direction
    - continuous replication
  4. for the lagging strand, Okazaki fragments are required to ensure the synthesis occurs in 5’ to 3’ direction
    - discontinuous replication
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49
Q

why does the antiparallel orientation of the strands make replication complex for the lagging strand?

A

The antiparallel orientation of parental strands and the unidirectional orientation of DNA synthesis (5’ to 3’) means that both new strands cannot be synthesised continuously
- the lagging strand must undergo discontinuous DNA replication

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50
Q

how does leading strand synthesis occur?

A

Leading strand synthesis is continuous and occurs 5’ to 3’

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51
Q

how does lagging strand synthesis occur?

A

Lagging strand synthesis is discontinuous and occurs 5’ to 3’, using Okazaki fragments

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52
Q

what are Okazaki fragments?

A

Okazaki fragments are short sequences of DNA nucleotides which are synthesized discontinuously and later linked together by the enzyme DNA ligase to create the lagging strand during DNA replication

53
Q

what enzyme is responsible for the synthesising of DNA chains?

A

DNA polymerase (DNAP)

54
Q

how do DNAPs begin synthesis?

A

DNA polymerase can’t start making a DNA chain from scratch, but requires a pre-existing chain or short stretch of nucleotides called a primer
- Primers act as a hook that allow the DNAP to bind to and begin synthesis of the new strand
- DNAP can bind to these hybrids and begin synthesis

55
Q

how are primers made?

A

DNA primase creates primers, by binding to a short RNA sequence and forming RNA/DNA hybrids

56
Q

how are primers used in leading strand synthesis?

A

On the leading strand, the short RNA primer is synthesised using template and NTPs by DNA Primase:
- The RNA primer is complementary to the template strand
- Once the RNA primer is in place, it provides the 3’-OH end for DNA polymerase to bind to and extend via addition of complementary nucleotides

57
Q

how does lagging strand synthesis occur?

A
  1. DNA Primase makes RNA primer (RNA/DNA hybrid) at a point on the lagging strand
  2. The primase dissociates and moves up the strand to bind to another point on the lagging strand
  3. DNA Polymerase extends the RNA primer into the gap – requires PRIMER-TEMPLATE junction
  4. Ribonuclease H removes the RNA primer, leaving a small gap
  5. DNA Polymerase extends across this small gap
  6. DNA Ligase seals the nick to join the new Okazaki fragment to the growing chain
58
Q

what enzymes are involved in lagging strand synthesis?

A

Lagging Strand synthesis requires:
- DNA Primase
- DNA Polymerase
- Ribonuclease H
- DNA Ligase

all to convert Okazaki fragments into a continuous strand of DNA

59
Q

what is the role of DNA helicase?

A

DNA Helicase uses ATP to separate parental DNA strands at the replication fork and move the Replication Fork forward
- It does this by breaking the hydrogen bonds between the 2 strands
- It forms a ring like structure around the DNA, and uses ATP hydrolysis to drive strand separation and form the replication fork

60
Q

what diseases can occur if DNA helicase is mutated?

A
  1. Werner’s syndrome
  2. Bloom syndrome
61
Q

what is Werner’s syndrome?

A
  1. Werner Syndrome: a progeria (premature ageing)
    - Werner Syndrome mutations are autosomal recessive, occurring in RECQ helicase gene WRN
    - Mutated helicase means that replication is inefficient or not occurring at all, leading to cell senescence and permanent arresting – hallmark of ageing
62
Q

what is Bloom syndrome?

A
  1. Bloom Syndrome: a rare cancer syndrome caused by loss-of-function mutations in a RecQ-family DNA helicase which maintains genome integrity
    - Bloom encodes an ATP-dependent DNA helicase, and is involved in removing blocks that may cause disruption to the replication fork – important in replication and repair
    - Related to cancerous genetic instability
    - Patients are sensitive to sun exposure
63
Q

what is the meaning of processivity?

A

Processivity: an enzymes ability to catalyse several reactions without releasing its substrate
– a processive enzyme always stays bound to its substrate
- non-processive enzyme will bind to and then release its substrate

64
Q

what is the degree of processivity of DNAP?

A

Degree of processivity of DNAP refers to the average number of nucleotides added each time it binds to a template:
- Average DNAP requires ~1 second to locate and bind to a primer-template junction
- If a non-processive DNAP is bound, it would add nucleotides at a rate of 1 nucleotide per second
- A processive DNAP adds multiple nucleotides per second by sliding along the template strand – increases rate of DNA synthesis

65
Q

how is the processivity of DNAP enhanced?

A

The processivity of DNA Polymerases is greatly enhanced by their association with a Sliding Clamp protein

66
Q

how does the sliding clamp protein interact with DNAP?

A
  • Once the first step of DNA synthesis has been accomplished, interaction of DNAP with the Primer-Template junction is maintained and addition of further nucleotides is very rapid
  • The Sliding Clamp (ATP-dependent) is positioned close to the Primer-Template Junction by a Clamp Loader
  • Sliding clamp encircles the DNA and hydrolyses ATP to release the clamp loader and help to lock DNAP onto the template strand and move forwards for synthesis
67
Q

what is the name of the human sliding clamp protein?

A

Proliferating Cell Nuclear Antigen (PCNA)
- near-identical 3D structure of the E. coli sliding clamp

68
Q

what is the name of the human sliding clamp protein?

A

Proliferating Cell Nuclear Antigen (PCNA)
- near-identical 3D structure of the E. coli sliding clamp

69
Q

what are single-stranded DNA binding proteins (SSBPs)?

A

SSBPs expose single-stranded DNA in the replication fork, making it available for template synthesis of the new DNA strand and easing replication fork progression

70
Q

what is the role of SSBPs?

A
  • As the helicase unwinds the double strand DNA, it creates sections of ssDNA in both leading and laggings strands
  • If the ssDNA are near regions that complementary to themselves, they can start to bind and form DNA hairpins, which may stop the processivity of DNAP
  • SSBPs bind to the ssDNA and prevent them from forming hairpins
71
Q

what are DNA topoisomerases?

A

DNA topoisomerases prevent DNA from becoming tangled during DNA replication and enhance processivity of DNA polymerase

72
Q

what is the role of DNA topoisomerases?

A
  • Helicase unwinding of parental DNA strands at the Replication Fork introduces superhelical tension into the DNA Helix.
  • Tension is relaxed by DNA Topoisomerases, which nick and reseal the backbone of the parental helix
73
Q

what is the role of Type 1 topoisomerases?

A

Type I Topoisomerases nick and reseal one of the 2 DNA strands, no ATP required

74
Q

what is the role of Type 2 topoisomerases?

A

Type II Topoisomerases nick and reseal both DNA strands, ATP required

75
Q

where does DNA replication begin?

A

DNA replication starts at a single point called the origin of replication
- at these origins are genetically defined replication sites with DNA sequences
- These specific DNA sequences recruit replication initiator proteins which bind to the origin and promote separation of the dsDNA
- This provides access for replication machinery and creates replication forks

76
Q

what is the origin of replication in E. coli, yeast and humans?

A

In E. coli, replication begins at genetically defined replication sites called OriC
- has only 1 origin of replication

In yeast, replication begins at Autonomously Replicating Sequences (ARS), and there are 600-700 of these origins

IN humans, replication is poorly understood, but begins at sequences near LMNB2 and MYC
- there are +100,000 of these origins
- origins are also defined by chromatin structure e.g. nucleosome-free

77
Q

how does initiation of replication in eukaryotes occur?

A

Initiation of replication in eukaryotes is biphasic
- To preserve the integrity of the genome, each origin of replication must fire only once per cell cycle – biphasic

  1. Replicator selection occurs in G1 phase - formation of pre-replicative complex (pre-RC)
  2. Origin activation occurs in S-phase – unwinding of DNA and recruitment of DNA polymerase
78
Q

why are the 2 phases of initiation of replication separated

A

Temporal separation of these 2 events ensures that each origin is used and each chromosome is only replicated exactly once per cell cycle

79
Q

what is the process of replicator selection in G1?

A

Eukaryotic replicator selection occurs in G1 and leads to the formation of a Pre-Replicative Complex (pre-RC):
- Origin Recognition Complex (ORC) binds to Replicator sequence e.g. ARS sequence in yeast
- Helicase-loading proteins Cdc6 and Cdt1 bind to ORC and recruit helicase Mcm2-7
- The Helicase Mcm2-7 binds to the origin to complete formation of pre-RC
- in G1, the pre-RC is INACTIVE

80
Q

what is the process of pre-RC activation in S-phase?

A

High levels of Cyclin-dependent kinase (Cdk) activity in S-phase activates existing pre-RC but prevents formation of new pre-RCs
- In G1 Cdk levels are low due to Cdk inhibitors, and this allows the pre-RC complex to form, but inhibits activation of pre-RC
- In S-phase Cdk activity increases
- Cdc6 and Cdt1 are substrates of Cdk and are phosphorylated, triggering activation of the helicase of the pre-RC
- Once Cdk activity is high, replication can start, and formation of any new pre-RCs is inhibited – no new origins of replications can be fired in any other phase of the cell cycle

81
Q

how is it ensured that chromosomes are replicated exactly once per cycle?

A

Close relationships between pre-RC function, Cdk levels and cell cycle ensures that chromosomes are replicated exactly once per cycle
- In G1, Cdk levels are low to allow assembly of inactive pre-RC
- In S-phase, Cdk levels increase, allowing activation of pre-RC to initiate replication
- Cdk levels remain high so that no other pre-RCs can form in G1
- Cdk levels must decrease to allow formation of new pre-RCs in G1

82
Q

when finishing DNA replication, why is there a problem at chromosome ends?

A
  • On the lagging strand, when an RNA primer at the end of the chromosome is removed by Ribo H, there is a gap that is left that cannot be synthesised as there is nowhere further upstream for DNAP to bind
  • This leaves 3’ overhangs at the end of the DNA strands
  • There are overhangs at the end of the lagging strand, and at the start of the leading strand where the initial DNA primer was bound
83
Q

what is the role of Ribonuclease H?

A

Ribonuclease H removes RNA primers, further shortening the newly synthesised DNA strands at 5’ ends of chromosomes, creating 3’ overhangs

84
Q

how may unregulated activity of Ribo H cause problems?

A

Chromosome shortening risks loss of valuable coding information, which may lead to premature ageing and mutation

85
Q

how can loss of genetic information by chromosome shortening be prevented?

A

Telomeres can help avoid this – telomeres are repeated non-coding DNA sequences found at the ends of chromosomes to stop genetic info being lost
- These can prevent the chromosome shortening

86
Q

what enzyme forms telomeres?

A

telomerase

87
Q

what is the role of telomeres and telomerase?

A
  • The addition of TTAGGG telomere repeats by telomerase compensates for the loss of telomere sequences caused by RNA primer removal and prevents chromosome shortening
  • Extended 3’ end DNA is now long enough to enable DNA Primase to bind and initiate new RNA primer synthesis which can then be extended as an extra Okazaki fragment by DNA polymerase
88
Q

what is the structure telomerase and how does it function?

A

Telomerase contains an RNA component that specifies telomere sequence
- Telomerase is a ribonucleoprotein (protein and RNA subunits)
- Intrinsic RNA component (sequence: AUCCCAAU) can base pair with the TTAGGG repeats and act as a template on which telomere repeat sequences are synthesised in a step-wise process – the Telomerase Shuffle
- (note: Telomerase RNA has U not Ts!)
- Can use reverse transcriptase activity to polymerise DNA from its RNA template
- Telomerase RNA allows addition of multiple TTAGGG repeats to the 3’-OH at each telomere

89
Q

what is the telomerase shuffle?

A
  1. Telomerase RNA pairs up with an existing telomere repeat, forming a 3’ overhang at the end
  2. The telomerase uses this as a template to reverse transcribe the first 3 bases AAU to TTA onto the 3’ end
  3. Telomerase then shuffles forward and uses the next 3 bases to synthesise 6 new nucleotides onto the 3’ end (uses CCCAAU to form GGGTTA nucleotides)
  4. It shuffles forward again and binds to the TTA to repeat the process and form 6 new nucleotides

Occurs 100s-1000s of times to ensure genetic info isn’t lost

90
Q

why is DNA stability and repair important?

A
  • DNA encodes the genetic instructions used in the development and functioning of all known living organisms
  • The stability of DNA is therefore essential for cell survival.
  • Only biological macromolecule to repair – all others are replaced!
  • Genetic Stability is our most robust defence against cancer
91
Q

what are the consequences of DNA damage in proliferating/dividing cells?

A
  • In proliferating/dividing cells, DNA damage can cause errors in replication/repair which may lead to mutations
  • Mutations can give advantage to highly replicative cells such as cancer cells, leading to cancer progression
92
Q

what are the consequences of DNA damage in non-dividing cells?

A

In non-dividing cells, DNA damage causes blockage of transcription and reduced gene expression, eventually leading to the functional decline of tissues and organs and ultimately ageing

93
Q

what are endogenous sources of DNA damage? give examples:

A

Endogenous sources: spontaneously within the cell
- Reactions with other molecules within the cell
- Hydrolysis, oxygen species, by-products of metabolism

94
Q

what are exogenous sources of DNA damage? give examples:

A

Exogenous sources: Reactions with molecules from outside the cell
- UV, X-rays, carcinogens, chemotherapeutics

95
Q

what are the types of endogenous DNA damage?

A
  • Depurination (Abasic sites)
  • Deamination
  • Methylation
  • Replication errors
96
Q

what are the types of exogenous DNA damage?

A
  • Pyrimidine dimers from UV
  • Double strand breaks
  • Interstrand crosslinks
97
Q

which DNA damage affects both strands of the DNA double helix?

A

Double-strand breaks and interstrand crosslinks affect both strands

Depurination (Abasic sites), Deamination, Methylation, Replication errors and pyrimidine dimers from UV only affect one strand of the DNA elix

98
Q

how do single-strand damage and double-strand damage compare?

A
  • Depurination have 10000 lesions per cell per day, but is single-strand damage or damage to a single nucleotide so is easy to repair
  • Damage from X-rays cause 0.008 lesions per cell per day, but is more lethal as they cause double-strand breaks

Double strand breaks are harder to repair

99
Q

what is deamination?

A

Deamination – removal of amino group (endogenous)
- Removal of the amino group by hydrolysis results in changes to DNA bases
- Deaminations are also known as point mutations and there are 2 types: transitions and transversions

100
Q

what is a transition deaminiation mutation?

A
  • Transition mutation is when a purine base is swapped for another purine base (adenine and guanine), or a pyrimidine is swapped for another pyrimidine (thymine and cytosine)
  • C-T transtions (pyrimidines)
  • A-G transitions (purines)
101
Q

what are transversion deamination mutations?

A

Transversion mutation is when a purine (A or G) is swapped for a pyrimidine (T or C) or vice versa

102
Q

what is the most common deamination event?

A

Most common event: removal of amino group from cytosine which releases ammonia and produces uracil
- Therefore uracil, usually found in RNA is now in DNA
- When uracil is bound in DNA during replication, it is recognised as a thymine so that it will base pair with an adenine on the new strand
- This leads to a change from a guanine to an adenine on the new strand
- The other strand is unchanged as the guanine was not deaminated, meaning one strand has mutated, and one strand is unchanged

Transition mutation from CG to TA

103
Q

which is more likely to occur, transition mutations or transversion mutations? why?

A

In general, transition mutations are more likely than transversions:
- Substituting a double ring structure for another double ring structure, or a single ring for a single ring, are more likely than substituting a double ring for a single ring and vice versa.
- C – T transitions
- A – G transitions

104
Q

are transitions or transversions more likely to result in amino acid substitutions?

A

Transitions are less likely to result in amino acid substitutions:
- Transitions have a smaller effect on amino acid sequences
- Amino acids have multiple codons that code for them, so a single nucleotide change may not alter the amino acid e.g. mutation from AAA lysine to AAG will still code for lysine

105
Q

what is depurination?

A

depurination (abasic site) - whole base is removed (endogenous)
- The N-glycosidic bond is a common substrate for hydrolysis = abasic site (AP site)
- Whole base is cleaved off the sugar-phosphate backbone, leaving an abasic site (no nucleotide is at that site)
- These occur ~18,000 per genome per day, and are more frequent at purine bases than pyrimidine bases

106
Q

if depurination is not repaired, what happens?

A

Failure to undertake DNA repair of depurination results in mutation:
- Depurination damage results in a frameshift mutation – potentially very damaging to the cell
- One new strand will completely miss out the base and be mutated, leading to the frameshift, and the other strand will be synthesised normally

107
Q

how do 3 reading frames significantly impact fraemshift mutations?

A
  • Proteins are translated by triplets of DNA sequence, so they have 3 reading frames
  • If the reading frame is shifted by one, a totally different amino acid is expressed, resulting in a totally different protein
108
Q

what to frameshift mutations generate?

A

Frameshift mutations generate missense proteins:
- In this example, the A has been depurinated and is missing
- In the original strand, the amino acid sequence would stop at TGA
- In the missense mutation, the missing A has lead to a frameshift mutation, so there is no longer a TGA stop codon, and the amino acid sequence is much longer, forming a missense protein that cannot function normally

109
Q

how does UV light result in the formation of pyrimidine dimers?

A

UV light induces the formation of pyrimidine dimers (exogenous)
- UV light produces a pyrimidine dimer via a photochemical reaction between bases forming cyclobutane rings (pyrimidine dimer)
- These rings distort the DNA structure and flexibility
- This has consequences for DNA replication and transcription

110
Q

how else may UV light distort DNA structure?

A

UV can also cause interstrand DNA crosslinks and DNA-protein crosslinks
- DNA cross-links are when the incorrect bases pair with each other e.g. G-T
- DNA-protein cross-links are when one strand of the DNA has a base which interacts with the amino acid sequence of a protein

Both are highly toxic to the cell as they block replication and transcription

111
Q

what may cause double-strand breaks?

A
  • X-rays
  • Ionising radiation
  • Topoisomerase II inhibitors
112
Q

what may cause single-strand breaks?

A
  • Reactive oxygen species
  • Hydroxyurea
  • Camptocthecin – used for chemotherapy
113
Q

what is the base excision repair (BER) pathway?

A

Repairs base damage e.g. abasic sites, deamination
- Uses base-flipping strategy to identify errors – if a mismatch is recognised during replication, the DNA will flip the base out e.g. uracil and the excision proteins cleave the incorrect base from the DNA strand
- The enzyme that flips uracil out is called uracil DNA glycosylase
- AP endonuclease enzyme and phosphodiesterase removes the sugar-phosphate backbone where that base was, and then joins the free ends together respectively
- DNA polymerase uses the other strand as a template to refill the correct base (cytosine)
- DNA ligase seals the nicks in the backbone

114
Q

what is the nucleotide excision repair (NER) pathway?

A

Repairs damage when more than one base is involves e.g. pyrimidine dimers caused by UV
- Involves the excision of short patches of single-stranded DNA to remove the affected bases (same as BER but excises larger fragment of DNA)
-Excision nuclease creates breaks in the sugar-phosphate backbone of the damaged DNA strand, affecting bases either side of the damage
- DNA helicase unwinds this section of the DNA and removes it
- This leaves the DNA helix with a 12 nucleotide gap in its sequence
- The undamaged template strand will aid DNA polymerase synthesise the strands together
- DNA ligase seals any nicks

115
Q

what is translesion synthesis repair?

A
  • Sliding clamp keeps DNAP on the DNA strand
  • Sliding clamp moves along the DNA, and if it encounters damaged section of DNA template, it dissociates from the normal DNAP and associates with a Translesional DNAP
  • Translesional DNA polymerases are specialised and can replicate highly damaged DNA - They can skip over the DNA damage
  • Once skipped over, the sliding clamp will then swap out the Translesional DNAP for a normal DNAP
116
Q

what do translesional DNAPs lack? why is this a problem?

A
  • Precision in template recognition and substrate base choice
  • exonucleolytic proof-reading activity

Translesional DNAPs therefore may cause base substitution and single nucleotide deletion mutations

117
Q

what 2 mechanisms exist to repair double strand breaks?

A
  1. NHEJ (non-homologous end joining) – occurs in G1
  2. HR (homologous recombination) – occurs in S-phase and G2
118
Q

what is non-homologous end joining (NHEJ)?

A

NHEJ: Restricted to G1 phase
- Double-strand break is reset by protein complex MRN to produce a 3’-overhang
- This initiates the heterodimer Ku70/80 which, with DNA-PK, will dissociate to both ends of the DNA strands
- It forms a synaptic complex that pulls the ends together to close proximity
- End processing via an endonuclease cuts off the 3’-overhang
- Ligase reseals the strands together

119
Q

what are the disadvantages of NHEJ?

A
  • Error-prone as it often results in loss of nucleotides around the break site
  • Important genetic info may be lost, and may lead to mutations
  • Quick process
120
Q

what is homologous recombination?

A

Occurs only in S-phase and G2:
- Uses intact sister chromatid as a template
- MRN complex re-sections the double-strand break and produces a 3’-overhang
- RPA protein, BRCA1 and BRCA2 coat the 3’-overhang with Rad51 to form a Rad51 nucleofilament
- This initiates strand invasion of the sister chromatid
- A holiday junction is formed to search for homology between the break site and the sister chromatid template
- Once homology is found the DNA sequence is synthesised
- Another holiday junction is formed to form a double-holiday junction, and the DNA is reformed by a crossing-over event to transfer homologous DNA from the sister chromatid

121
Q

what is the advantage of HR compared to NHEJ?

A

HR has error-free repair - double-strand break is accurately repaired

NHEJ is error prone and can result in the loss of nucleotides, leading to mutations

122
Q

what 3 points in the cell cycle can DNA damage be detected and repaired?

A
  1. G1 checkpoint
  2. Entry to S-phase
  3. Entry into mitosis

Plus a check for chromosome non-disjunction

Examples:
* G1 repair = non-homologous end joining
* G2 repair = homologous recombination

123
Q

how is DNA damage detected?

A
  • ATM/ATR protein kinases get activated and associate with the site of DNA damage
  • This activates other kinases to block the cell cycle, such as Chk1 and Chk2, which phosphorylate p53
  • p53 is stabilised and activates p21
  • p21 renders the G1/S-CDK and S-CDK complexes INACTIVE., thus preventing cycle progression
  • DNA is then repaired
124
Q

what happens if DNA cannot be repaired?

A

If repair is not possible, apoptosis is activated

125
Q

what is an example of a disease associated with a defect in NER?

A

Xeroderma Pigmentosum – defect in repair of UV damage
- An Autosomal Recessive Disease
- 1 in 250000 in europe
- 1 in 40000 in Japan
- 2000-fold increased risk of skin cancer
- Skin cancer occurs 50 years earlier – mean onset age = 12 years

126
Q

what is an example of a disease associated with defects in double-strand break repair?

A

breast cancer:
- 10% breast cancer is inherited
- 80-90% of all inherited breast cancers are BRCA1/2 associated (also associated with ovarian and prostate cancer)
- BRCA1/2 carriers have a 80% lifetime risk (10x higher than normal)

127
Q

why are mutations associated with cancer?

A
  • BRCA2 deficient cells exhibit genomic instability, are sensitive to DNA damaging agents and are defective in homologous recombination (cannot repair double-strand breaks)
  • Leads to predisposition to cancer
128
Q

what methods can be used to study DNA damage?

A
  1. See if cells are sensitive to DNA damage via survival assays
    - E.g. see how sensitive cells are to ionising radiation if they cannot repair the damage
  2. Identify markers of double-strand breaks e.g. gamma-H2AX which is involved in detecting breaks within the cell
    - Can be detected using immunofluorescence
    - Can find differences in prolonged formation/lack of formation of the protein
  3. Comet assay – electrophoresis to see overall damage to the cell
    - If they have no damage, the fragments stay in tact
    - If there is damage, damage is pulled out of the cell and looks like a comet
  4. Western blots can identify markers of DNA damage e.g. BRCA2 to see if they are functioning properly

Molecular cellular biology (13 decks)

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