BMS334 Epigenetics Flashcards

1
Q

What are epigenetic mechanisms?

A
  • Create molecular environment that shape accessibility of genes to transcription machinery
  • Regulate levels of gene transcription and sensitivity to change by extrinsic DNA binding transcription factors
  • Epigenetics do not change the underlying DNA sequence
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2
Q

What do epigenetic mechanisms regulate?

A

Epigenetic mechanisms regulate growth, development, and maintenance of physiological homeostasis across the life course

  • Development and tissue homeostasis
  • Adaptive and maladaptive responses to environmental factors
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3
Q

What molecules are involved in epigenetic mechanisms?

A
  • DNA methylation of CpG dinucleotides
  • Covalent Modifications of nucleosomal histones within chromatin
  • Non-histone proteins that generate, recognise or remove DNA methylation, histone modifications
  • Non-coding regulatory RNAs
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4
Q

Give an example of the epigenetic mechanism that directly acts on promotors or enhancers

A
  • Steroid hormones e.g, ecdysone (potent regulators of gene transcription) bind to its TF receptor (ecdysone receptor) which undergoes a conformational change that allows entry to nucleus. It can then bind to recognised DNA and recruit proteins which stabilise the interaction between the RNA polymerase and the promotor regions of the ecdyson receptor
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5
Q

How does metylation of CpG dinucleotides result in gene regulation?

A
  • Methyl group added to cytosine bases using DNA methyltransferases.
  • Changes the interior surface of the major groove of DNA which changes the ability for TF’s to recognise the DNA surface
  • It produces recruitment sites for DNA binding proteins to mask the major groove stopping direct contact to DNA. This blocks the ability of DNA to be recognised, read and decoded
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6
Q

How does the DNA sequence of promotor regions allow the regulation of gene expression?

A

Promotor sequences in genes are rich in these nucleotides. Whether a gene is active is often determined by the methylation state of these CpG nucleotides

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7
Q

How can the methylation of core histones regulate gene transcription?

A

In transcriptionally silent chromosomes, there are histones that are highly methylated. The combination of methylated histones and methylated cytosine residues shuts the locus down

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8
Q

How can the acetylation of core histones regulate gene transcription?

A

Transcriptionally active chromatin has a much more open structure (nucleosome free regions of DNA) and the nucleosomes are instead hyper acetylated. These negatively charged acetyl groups help the DNA to remain accessible to TFs by maintaining the chromatin in an uncondensed state

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9
Q

What part of the core histones are modified?

A

The N terminal tails of core histones are targets of a wide range of different histone modifications by enzymes that function as modification “writers”.

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10
Q

What is the role of histone Acetyltransferases?

A

Histone Acetyltransferases add acetyl groups to multiple lysines in the N-terminal tails of core histones

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11
Q

What is the role of Histone Methyltransferases?

A

Histone Methyltransferases add methyl groups to specific lysines or arginines in the N-terminal tails of core histones

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12
Q

What is the role of Histone kinases?

A

Histone kinases add phosphoryl groups to Serines / Threonines of core histones

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13
Q

How are histone modifications recognised?

A

Recognised by proteins with modification-specific binding domains – modification “readers”

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14
Q

What are acetylated histones recognised by?

A

Proteins with bromodomains

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15
Q

What are methylated histones recognised by?

A

Proteins with chromodomains

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16
Q

What are modification erasers?

A

If there are no proteins bound then these modifications can then be selectively removed
- For acetylation, the protein to remove it is histone deacetylases and for methylation, the enzyme is histone demethylases

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17
Q

How can long non coding RNAs regulate gene expression?

A
  • Can be several KB in length and full of complex tertiary structures e.g. hair pin loops. These loops can interact with chromatin regulatory proteins.
  • These RNA protein complexes can impact on the structure and function of the chromatin of which they are apart
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18
Q

Give an example of a long non coding RNA that regulates gene expression

A

For example, the X chromosome inactivation RNA called Xist. Has complicated hair pinned loops which allow interaction with proteins which allow selective shutting down of one of the two X chromosome’s in female cells

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19
Q

Give examples of short non coding RNAs

A

MicroRNA

PiRNA

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20
Q

How can Micro RNAs regulate gene expression?

A

Micro RNAs are present in the cytoplasm and interfere with the transaction of existing mRNA that is complementary to the micro RNA
- Can also lead to the destruction of the mRNA once hybridisation occurs

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21
Q

Where are piRNA found?

A

Nucleus

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22
Q

What is the role of piRNAs?

A

piRNAs have complementarity with nascent mRNAs. These RNAs encode transposable elements which controls the expression of transposable elements in the genome.

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23
Q

How do piRNAs control gene expression of transposable elements?

A
  • They suppress the expression of transposable elements RNA through by recruitment of PIWI proteins to transposon loci and promote formation of transcriptionally silent heterochromatin
  • They also act as nucleation signals for chromosome modification enzymes which shut the transposable mechanism down
  • piRNA that are exported from the nucleus can also act in the cytoplasm to promote inhibition or degradation of transposable mRNA
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24
Q

What is meant by epigenetics ensuring a robust phenotype?

A

Epigenetic mechanisms guide development and ensure robust, stable phenotypes are produced through reliable mechanisms that regulate expression of the genome according to a predictable schedule.

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25
Q

What is meant by epigenetics ensuring plasticity?

A

Epigenetic mechanisms are also flexible and sensitive to physiological signals, which may originate within an organism or as a consequence of changes in the external environment

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26
Q

What kind of changes can the phenotypic plasticity of epigenetic mechanisms lead to?

A

This can either be beneficial changes or create capacities for the emergence of maladaptive chronic disease states

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27
Q

What are the two types of phenotypic plasticity?

A

Reaction norms

Polyphenisms

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28
Q

What are reaction norms in relation to phenotypic plasticity?

A

Reaction norms are ranges of phenotypic characteristics that are continuous and proportional to the environmental stimulus

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29
Q

Give an example of reaction norms in relation to phenotypic plasticity

A

Phenotypic Plasticity engenders beneficial adaptations that can improve fitness. Used resistance training to develop muscle hypertrophy. This is driven by PGC1alpha TF. The more resistance training, the more PGC1alpha, the more muscle hypertrophy

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30
Q

What did Ruas et al, 2012 discover about PGC1-alpha?

A
  • PGC1-alpha has two regulatory elements with one being responsive to resistance training which leads to a form of PGC1 alpha that leads to hypertrophy.
  • If over express this form of PGC1 alpha in mice then muscle size increases.
  • If do different training e.g. running then there is a different form of PGC1alpha activated and increases the metabolic function of the muscle and allow it to function over long periods of time
  • Saw that gene expression changes with muscle loading and unloading and saw the change in DNA methylation of the regulatory agents around these genes
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31
Q

What are polyphenisms?

A

Discrete transformation of phenotypic characteristics from one type to another in response to a threshold quantity of environmental factors being exceeded
- Switching between discrete alternative forms in response to changes in environmental stimulus

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32
Q

How are daphnia an example of polyphenisms?

A
  • In response to a predator Daphnia change shape to give a spikey helmet to keep the predators away. The survival of these daphnia is much higher than those without.
  • However, to get these helmets the daphnia must be in a predatory environment that release chemical stimuli and induce the change
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33
Q

How are butterflies an example of polyphenisms?

A
  • When the animals develop in the dry season they are brown but if develop in the wet season then the animals have eye spots on their wings.
  • This is linked to the production of levels of ecdysone. These eye spots are driven in an ecdysone dependant way by Distal-less TF expression.
  • In wet season, there is much more Distal-less expression meaning that the eye spot phenotype is much stronger
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34
Q

How is epigeneitcs linked to healthy ageing in humans?

A

Human behaviours and environmental exposures modules the activities of epigenetic mechanisms and promote chronic diseases

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35
Q

What was the dutch famine?

A
  • Rationed food, as little as 500 calories a day.
  • The babies conceived during the last three months of the famine were born with a low birth weight and these people then had an increased chance of chronic diseases such as coronary heart disease and schizophrenia
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36
Q

How was the phenotype of children conceived in the critical period of the dutch famine explained by epigenetic mechanisms?

A
  • Set of epigenetic modification revealed through analysis of blood samples
  • Can see the pattern of methylation changes. If conceived in the critical period, there is an increased level of methylation in some loci and decreased in other loci in comparison to unexposed sibling. This shows that exposure to the famine induced long lasting epigenetic changes to the genome
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37
Q

How can the epigenetic phenotype of children conceived in the critical period of the dutch famine explained?

A
  • This can be explained by the thrifty phenotype hypothesis: if developing under adverse conditions, metabolism attacks any nutritional sources available – this includes in laying down excess fat.
  • The metabolism is set in the period of adversity and still does this even when adverse conditions have been removed
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38
Q

How are chromosomal regions organised in the nucleus?

A
  • Chromosomal regions are condensed into distinct functional regions located on the periphery of the nucleus.
  • When this is the case, the domains are full of transcriptionally silent genes
  • Towards the centre, there are other domains that are associated with morphological structures that are defined by TFs e.g. RNA polymerase II.
  • These are called chromosomal domains and are topologically organised
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39
Q

What kind of tools do we need in order to analyse the structure function relationships within the nucleus?

A
  • Biochemical methods for defining and analysing structures e.g. Protein and nucleic acid purification and sequencing techniques
  • Genetic techniques for perturbing gene function and linking biochemical changes to phenotypes – model organism studies. These allow us to define distinct functions of genes
  • Computational tools for analysing genomic / epigenomic data. This has allowed the above two tools to be brought together – analyse the complex molecular interactions
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40
Q

What are the three main methods of detection of methylated DNA sequences and histone modifications?

A
  • Treatment of genomic DNA with Restriction Enzymes that selectively recognise
    and cleave DNA containing methylated OR unmethylated CpG dinucleotides
  • Chromatin immunoprecipitation with antibodies recognising specific covalent histone
    modifications
  • Bisulfite sequencing analysis of genomic DNA to map and quantify
    methylated CpG sites
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41
Q

What was the limitation of using resitiction enzymes that selectively recognised methylated or unmethylated CpG dinucleotides?

A

Was very limited as often focused on a specific gene or nucleotide pair - not genome wide

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42
Q

How can restriction enzymes be used to investigate specific DNA methylation?

A
  • Restriction enzymes HpaII and MSP1 recognise CCGG
  • However, HpaII will only cleave if there is no methylation at either C. MSP1 cleavage is blocked if the outer C is methylated but not the inner C. In mammalian genomes, methylation is limited to CG sites so MSP1 is affectively unaffected by methylation.
  • Can therefore digest genomic DNA from different tissues with MSP1 and HpaII to create different fragments and create a southern blot
  • Use probe specific for promotor of gene of interest
  • If it is methylated, it will be cut by Msp1 (does not discriminate between modified CpGs) but it won’t be cut by HpaII (which does not cut methylated CpGs).
  • The probe will reveal the length of the fragment and therefore whether it has been cleaved
  • Using this approach allows mapping of distribution of methylated CpGs across a section of DNA providing there are specific probes
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43
Q

Give an experiment that used restriction enzymes to investigate the methylation of a certain gene

A

Lu and Davies, 1997

  • Focused on the promotor of the Tissue Transgutaminase gene
  • Msp1 digest of genomic DNA from the Hela cells gives a 6Kb fragment whereas HpaII gives a 9Kb fragment. Shows that the CCGG sequence is methylated in the hela cells as it is not being cleaved by HpaII
  • If treat Hela cells with 5-AzaCytosine (inhibitor of DNA methyltransferases – prevents methylation from occurring). The fragment produced by HpaII disappears and the MspI fragment appears - Methylation was decreasing
  • This method allows investigation of the abundance of methylation in specific locations
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44
Q

How can antibodies specific for covalent modifications be used to detect DNA and histone modifications?

A

MeDIP
- Used anti-5-methylcytosine antibody to select out fragments of nucleic acids that contain methylated CpG and leave behind the Unmethylated DNA.
Variations of this technique
- Chromatin immunoprecipitation with histone modification-specific antibody, e.g. anti-acetyl-histone antibody. Selectively isolate these modifications and then use large genome sequencing
- These can then be mapped back to the reference genome and the locations of the modified histones can be assigned to specific locations within the genome

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45
Q

How is MeDIP carried out?

A
  1. Purified genomic DNA is sheared randomly into fragments.
  2. Fragments are converted into a library with primer sequences attached for DNA sequencing
  3. DNA fragments are incubated with an antibody that binds specifically to 5-methylcytosine
  4. Antibody-DNA complexes are immunoprecipitated
  5. Each fragment of immunoprecipitated DNA is sequenced and the DNA sequences are mapped to the genome to identify the location of methylated CpGs
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46
Q

How can MeDIP be used to create a methylation map of the genome of that cell?

A

This technique is usually carried out on a tissue sample so will therefore contain multiple cells (each having two copies of each DNA sequence). Can count the number of times each fragment is sequenced to see how much a specific sequence is methylated in the tissue compared to another fragment. If a specific fragment is being sequenced a lot compared to others then can make a map of the methylation levels over the genome. The more times sequenced, the more methylation of that locus in that tissue sample

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47
Q

How can MeDip be used to investigate cancer?

A

MeDip gives a genome wide methylation profile

  • Can then compare the methylation across the genome in different tissues e.g cancer cells v non cancer cells
  • E.g Can see different methylation fluctuations between fibroblasts and colon cancer cell lines. Methylation is an indicator of silencing, applying that that cancer cells have genes that are active that aren’t active in the fibroblasts – these could be oncogenes
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48
Q

How can methylation profiles be use to predict gene transcription in different cell types?

A

e. g. Methylation profile of NF-kappa B
- Abundantly expressed in white blood cells and involved in regulation of inflammatory response and T cell activation
- Can compare the transcript abundance and methylation profile in different tissue types. Can see that there is similarities between the profiles. However, in memory T cells and B cells there is no methylation. This allows us to predict the transcription of NF–kappa B in different cell types

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49
Q

How is Bisulphite sequencing carried out?

A
  • Treat genomic DNA with bisulphite. Bisulphite deaminates unmethylated cytosine to uracil. If methylated, there is no effect with bisulphite treatment and leaves the cytosine in tact. Where there is methylation, the cytosine is protected and will become an uracil.
  • Use PCR to amplify these changes – during PCR uracil will be replaced with thymine.
  • Can use this to measure the amounts of methylation in tissues – how many cells in complex tissues show high methylation across the genome
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50
Q

Why is Bisulphite sequencing the main technique used?

A

Because it gives a full account of methylation across the genome in different tissues and is less vulnerable to error as does not use antibodies which could vary in quality

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51
Q

Give an example of how Bisulphite sequencing can be used?

A

E.g. Hox cluster genes

  • How the methylation patterns of the hox cluster changes during development
  • Can see that there is low methylation in sperm and high in eggs. After fertilisation, the methylation is very high. This persists to the 256-cell stage (the maternal epigenetic phenotype).
  • At the sphere stage and zygotic gene transcription occurs, the methylation decreases dramatically
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52
Q

What is Chromatin immunoprecipitation (ChiP)?

A
  • Chromatin Immunoprecipitation (ChIP) requires an antibody that is specific for the histone modification or protein whose genomic distribution is of interest
  • Uses the same approach as MeDip to instead carry out analysis of histone modifications
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53
Q

What is the chromocentre in drosophila?

A

Heterochromatin in Drosophila polytene chromosomes is gathered together around the chromocentre, which comprises salivary gland heterochromatin from each drosophila chromosome that is rich in the silent marker H3K9me2 and deficient in the active euchromatin marker H3K4me2

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54
Q

What has ChiP analysis showed about histone methylation in the chromocentre?

A

Can map the distribution of H3K4 and H3K9 methylation. Regions within the chromocentre, it is abundant in H3K9 methylation but at the boarders there is more H3K4 methylation. In the body of heterochromatin there is no H3K4 methylation

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55
Q

What is position effect variegation?

A

Position effect variegation is the variable silencing of a gene that is understood to be active in a particular cell type but becomes silent when the gene gets relocated by chromosomal relocation

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56
Q

Who discovered the position effect variegation phenomenon?

A

Muller 1930

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57
Q

How was the position effect variegation phenomenon discovered?

A
  • Observed an unusual phenotype in which the eye was variegating with some patches of red and some white.
  • Red pigment is encoded by gene called white. If mutated and silenced then the eye is white
  • Inversion of polytene chromosomes so that one break-point is adjacent to the Wt and the other adjacent to the heterochromatin. This results in the Wt being translocated to the heterochromatin from euchromatin on the other chromosome resulting in its silencing. The eye therefore appears white. This is called position effect variegation
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58
Q

How can position effect variegation be used to screen for genes that regulate the properties of heterochromatin?

A

Can look for mutations that restore the eye phenotype or completely wipe out wt. This allows identification of Su(Var) (suppressors of variegation) and E(Var)(enhancers of variegation)

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59
Q

What Su(Var) was discovered through drosophila eye pigment screening?

A

Su(Var)3-9

  • Encodes a histone H3K9 methyltransferase which defines the chromocentre. When it is mutated the white allele is no longer in heterochromatin
  • If take Wt and mutant Su(Var)3-9 genes and run a western blot and probe with anti-H3K9me2. Can see that the Su(Var) mutations have no H3K9 methylation whereas the Wt does
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60
Q

What is Heterochromatin Protein 1 (HP1)?

A

Recognizes Su(VAR)3-9 –dependent H3K9-methylation

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61
Q

Give evidence that HP1 binds H3K9 and not H3K4?

A
  • Create a biochemical column on which have peptides from Histone H3 which contain methylated lysine 9 or 4 and add radiolabelled HP1. Find that HP1 will bind to lysine 9 column and not lysine 4
  • Can conclude that HP1 specifically binds to H3K9
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62
Q

What are heterochromatin and euchromatin?

A

Heterochromatin - Region of chromosomes that is densely packed - can’t be transcribed
Euchromatin - Region of chromosomes that is loosely packed - can be transcribed

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63
Q

Give evidence that HP1a is essential for the for restricted binding of Su(Var)3-9 to heterochromatin

A

If HP1a is not present then then Su(Var)3-9 is no longer primarily associated with heterochromatin but is also found along the chromosome arms in the euchromatin and causes ectopic H3K9 methylation

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64
Q

What did Riddle et al, 2011 show about HP1 and Su(Var)3-9 distribution?

A
  • Compared distribution of HP1, Su(Var)3-9 and H3K9 methylation in chromatin. Can see that they are all colocalised to the same region of chromatin to each other (transcriptionally silenced)
  • Activation markers (H3K4) and RNA polymerase II are excluded from this location and instead emerge at the boundaries of the heterochromatic region
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65
Q

How does HP1 form heterochromatin?

A

(Machida et al, 2018)
- HP1 exists as dimers
HP1 binds to methylated H3K9 residues in nearby nucleosomes using their chromodomain.
- This pulls two nucleosomes together and compacts them.
- This stops RNA polymerase II action and stops transcription and forms heterochromatin.

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66
Q

What is meant by chromatin accessibility?

A

The accessibility of chromatin to incoming factors such as DNA binding proteins

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67
Q

What do changes in chromatin accessibility result in?

A

Changes in chromatin accessibility are accompanied by dynamic changes in patterns of histone modifications, DNA binding protein interactions and transcriptional activity of genes within chromatin

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68
Q

What is DNAse1?

A

DNAse1 are non-specific endonucleases and will fragment the DNA, at DNAse hypersensitive sites, if they are relatively free and accessible

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69
Q

What is DNAse1 hypersensitive site mapping?

A

Uses DNAse1 to identify regions of DNA in chromosomes that are relatively accessible.

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70
Q

Outline DNAse1 hypersensitive site mapping

A
  • Permeablise set of cells and isolate the nuclei so that they can take up DNAse1 which will fragment free DNA that it finds. It won’t be able to access DNA that is bound tightly to the nucleosome.
  • Extract the fragmented DNA and digest with specific restriction enzymes for gene of interest. The two ends of specific DNA sequence has a BamH1 restriction enzyme cleavage site at each end
  • Use DNA probe to detect the restriction fragment and run Sothern blot. The band will then be detected at a certain position on the blot.
  • If the DNAse1 endonuclease has fragmented this gene fragment at a DNAse1 hypersensitive site then the band will now give a smaller reading. This occurs if accessible to DNAse1
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71
Q

Give an example of DNAse1 hypersensitive site mapping

A
  • Can do this in different cell types to see the accessibility of the same gene in different cell types.
  • For example, in fibroblasts, the beta globin cluster is not accessible as only the original band appears on the southern blot but in erythrocytes the beta globin cluster is accessible as a band of a lower molecular weight appears.
  • This is because the transcription activator responsible for activating beta globin transcription is erythroid-specific and is not present in fibroblasts
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72
Q

What is the locus control region (LCR)?

A

An enhancer region that direct the developmental and tissue specific transcription of human globin genes

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73
Q

How is the LCR been able to be investigated using DNAse1 hypersensitive site mapping?

A

It is rich in DNAse hypersensitive sites e.g. HS4.

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74
Q

How was the LCR discovered?

A

DNAse1 hypersensitive site mapping
- In erythrocytes, there is a truncated BamH1 site which identifies a specific region in the chromatin near the beta globin gene that is sensitive to DNAse1 (accessible).

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75
Q

What did Grosveld et al, 1987 to investigate the role of LCR?

A

Took LCR DNA region and produced transgenic animals

- Do a northern blot to study the levels of transcription in different transgenic lines.

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76
Q

What did Grosveld et al, 1987 discover about the role of LCR?

A

The human b-globin LCR is therefore a super enhancer, conferring position independent, transgene copy number dependant transcriptional activity to multiple, linked transcription units

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77
Q

What is meant by LCR being position independent?

A

Usually, transgenes are position dependant as if they insert in a heterochromatin rich region of the chromosome the transgene will be silenced. However, the LCS acts as an enhancer that is so powerful that it is capable of overcoming the silencing power of heterochromatin so is therefore position independent

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78
Q

What is meant by LCR being transgenic copy number dependant?

A

If measure the number of copies of the transgene in the genome and relate it to the expression of the transgene then you see a high correlation between these number

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79
Q

If LCR is a powerful enhancer, why in its home location (erythrocytes) does it only regulate beta global genes?

A

Due to the presence of boundary elements which focus its action on beta globin genes

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80
Q

What is Micrococcal nuclease?

A

Micrococcal nuclease is a smaller enzyme then DNAse1 and can access regions of chromosomes that DNAse 1 may find difficult to access

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81
Q

How did Riddle et al, 2011 use micrococcal nuclease to decent heterochromatin in drosophila?

A
  • Made a transgenic fly where the white gene has been added to a euchromatic location meaning that the eye is red and one where the white gene has been added to a heterochromatic region of chromatin so the eye is white (gene is inactive)
  • Isolated the nuclei from the eye and digest with Micrococcal nuclease and do a southern blot to identify the white sequences on the transgene.
  • The transgene from the euchromatic region shows a smear on the southern blot showing that the enzyme is breaking the fragment down into smaller pieces.
  • In the heterochromatic region, the southern blot reveals a ladder where each band is one nucleosome length of DNA separate from each other. The Micrococcal nuclease is getting in between each nucleosome and can be used to detect heterochromatin
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82
Q

What is ATAC sequencing used for?

A

To look at the chromatin accessibility across the genome and compare between different cell types

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83
Q

How does ATAC sequencing work?

A
  • It uses a transposon in bacteria that move from one location to another and wherever they insert themselves they create a duplicated sequence either side of the transposable element
  • Can inject these transposons with the sequences attached into developing embryos or chromatin and they will look for accessible chromatin and insert. This will mark all the available chromatin with these tags. The DNA can then be extracted and sequenced and map the location of the tags. This will map all of the accessible chromatin across the genome
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84
Q

Give an example of ATAC sequencing?

A

Liu et al, 2017

  • Chromatin accessibility in early zebrafish embryogenesis from 64 cell stage to post zygotic gene activation (ZGA) to look at the period of when gene transcription begins
  • The ATAC sequencing shows peaks of ATAC sequencing accessibility that increase as gene transcription begins. The genome starts to open up as development increases
  • DNA methylation changes accompany the increase in accessibility revealed by ATAC sequencing
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85
Q

How can we discovery which DNA sequences are close together in the 3-dimensional nucleus?

A

Chromatin conformation capture (3C)

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86
Q

How is Chromatin conformation capture (3C) carried out?

A
  • Take chromatin from isolate nuclei and cross link the DNA to chromosomal components using formaldehyde which is a non-specific cross linker that will create bridges between the DNA and proteins.
  • Then use nuclease digestion to fragment the DNA and will be left with doublets which are short pairs of DNA sequences held together by proteins.
  • These are regions of the DNA that were close to one another in the nucleus. Can make a library of these sequences and can map back to the reference genome. Then ask which sequence is associated with another sequence in every experiment – pair wise comparison of sequence content
  • Can use this information to create a 3D model of what a region should look like
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87
Q

What did 3C reveal about the structure of the beta global network

A

The model produced showed that there is a loop where the transcribed region is and at the neck of this region exists the locus control region. The stable the neck of the loop subsequently contacts each of the genes in the cluster in turn

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88
Q

Give evidence for the 3D model of beta globin

A
  • There are peaks of high contact frequency that indicate that when that fragment was sequences so was the other sequences. This showed that sequences within the LCR there are sequences in close contact to a DNA sequence (3’ HS1).
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89
Q

What is CTCF?

A

There is a specific zinc finger transcription factor called CTCF that binds to HS5 and 3’ HS1 and defines the neck of the loop in the beta globin structure.
- It is a major organising factor for 3D structure

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90
Q

How can we look at the proximity of DNA sequenced across the whole genome?

A

2D heat map of pairwise contact between different regions of the locus

  • The diagonal line in the middle shows that the blue sequences are always found with the blue sequences. Can use this to compare to look for regions that sometimes interact
  • Can use this to characterise the pair wise interactions across a whole chromosome
  • Look at diagram
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91
Q

What did Dixon et al, 2016 do?

A

Used 2D heat mapping to define interactions of a cluster of olfactory receptor genes in the mouse

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92
Q

What are Topologically Associating Domains (TADs)?

A
  • Higher order organisational units of chromatin function within the nucleus.
  • TADs identify the extent of a chromatin domain containing transcriptionally co-regulated genes. TAD ends identify boundary elements that insulate genes within TADs from the effects adjacent elements
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93
Q

How are chromatin loops made?

A

Hand-cuff model
- Two ends of a TAD are brought together in 3D space by CTCF proteins which bind to each boundary via the DNA motifs ad recruit cohesin.
- This encloses the promotor and enhancers of genes to insulate its effects
- This is supported as the cohesin complex generally co-localises with CTCF throughout the mammalian genome
- However, a problem is that the number of CTCF binding sites are much more abundant than the number of TAD boundaries. Therefore, why are TADs not more abundant?
See diagram

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94
Q

What did E.B Lewis (20th century) discover?

A

His studies lead to the identification of the genes within the Antennapedia and Bithorax complex which make up the genes within the Hox clusters.

  • These genes are expressed in a nested fashion – some in the anterior/posterior. The domains overlap.
  • There is a collinearity between the structural organisation of these complexes of these genes and their expression domains
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95
Q

What are sex combs?

A

Sex combs are a small appendage on the foreleg and used in mating

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96
Q

How did E.B Lewis discover polycomb genes?

A
  • Scr mutants do not have these or reduced sex combs.
  • Lewis found some other genes which were unlinked to the homeotic complex which gave the opposite phenotype. It resulted in sex combs in the first three legs rather than just the first pair.
  • He mapped the mutations and they were elsewhere – not Hox gene complex
  • This family became known as the Polycomb complex
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97
Q

What is the phenotype of a polycomb mutant fly?

A

The segmental character of each of the larval segments were transformed. They usually have a distinct pattern (denticle patterns) in the thoracic segments and abdominal. However, in the Polycomb mutants each segment resembles the most posterior segment (8th).

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98
Q

How can the phenotype of a polycomb mutant fly be explained?

A

The explanation of this was that each segment now expresses many Hox genes. Usually there are fewer Hox genes expressed the more anterior of the larva - global derepression of the Hox gene

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99
Q

What is the function of polycomb genes in flies usually?

A

To restrict the expression of the homeotic cluster anteriorly

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100
Q

How are polycomb proteins clustered?

A

Into two complexes

  • PRC1
  • PCR2
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101
Q

What is the function of the Polycomb complexes?

A

These complexes are involved in the establishment of transcriptionally repressed chromatin. When these complexes can’t work (due to one gene mutation) then the system falls apart and it results in derepression of the Hox gene cluster

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102
Q

Give evidence that hox genes are evolutionary conserved?

A

There is collinearity between Hox gene organisation and the expression domains in drosophila, mice and human embryos showing that they are evolutionary conserved

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103
Q

What is the function of Hox genes in vertebrates?

A

Pattern the vertebral column

- There are five domains of the vertebral column

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104
Q

What is the function of vertebrate polycomb genes?

A

Polycomb genes are involved in setting the expression of Hox genes in the vertebral column and spinal cord

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105
Q

What happens if mutate vertebrate polycomb genes?

A

If mutate vertebrate Polycomb genes then it leads to posterior transformations of vertebrate elements

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106
Q

Give evidence for the involvement of polycomb genes in vertebrates

A

Carried out study on two closely related homologues of drosophila gene polyhomeatic called Phc1 and Phc2 in the mouse.

  • The cervical vertebrae in the vertebral column of both a Phc1 and Phc2 mutant mouse now has ribs coming out of the cervical column.
  • The final cervical vertebrae has the characteristics of a thoracic one. There is also one less of pair of ribs in Phc1 mutants.
  • The final segment of the vertebral column has been converted to a more posterior which do not have ribs
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107
Q

What makes up PRC2?

A

The PRC2 includes Ecz and the enhancer of zeste proteins

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108
Q

What makes up PRC1?

A

PCR1 contains Polycomb protein at its core

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109
Q

What are the vertebrate and C. Elegant orthologues of the enhancer of zeste?

A

It has an extended region of sequence similarity at the end of the C terminus including C. elegans genes such as MES2 and the vertebrate orthologue Enx1

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110
Q

What does the enhancer of zeste share homology to?

A

Has homology to Trithorax and a sequence from Su(var)3-9 (encodes a methyl transferase specific for lysine 9 methylation

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111
Q

What is the domain with high homology to Su(var)3-9 and trithorax in the enhancer of zeste?

A

STET

- Encodes the histone methyltransferase activity of all of these enzymes

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112
Q

What do the STET domains of Su(var)3-9, trithorax and the enhancer of zeste methylate?

A

The enhancer of zeste (in vertebrates Enx1) methylates Histone H3lysine27. Su(Var) does lysine 9 where as Trithorax methylates lysine 4

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113
Q

What did Cao et al, 2002 investigate?

A

The relationship between the enhancer of zeste and histone methylation

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114
Q

How did Cao et al, 2002 show if there was a protein present in a fraction that had methyl transferse activity?

A
  • Drosophila cells were cultured and prepared a protein extract
  • Separated the proteins in the extract into 100 different fractions and collected
  • Incubated a sample of each fractions with H-labelled S-adenosyl methionine (H-SAM) and a purified histone H3 and then a ran a western blot with the incubated extracts. If the extracts contained a protein with histone methyltransferase activity then the purified histone H3 would be methylated with a radiolabelled methyl group from H-SAM.
  • After incubation, the samples were electrophoresed by SDS-PAGE and auto radiographed to identify the fractions of the cell extract that were responsible for transferring methyl groups from H-SAM to histone H3.
  • Some fractions (20-53) were able to transfer the methyl group to histone H3 showing they contain these proteins
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115
Q

How did Cao et al, 2002 compare the methylation ability of the fractions to the expression of the enhancer of zeste?

A
  • Electrophoresed the samples from each fragment to create a western blot
  • Then probed the western blot with antibodies for Polycomb proteins and then used a secondary antibody that was conjugated to Alkaline phosphatase (AP)
  • Exposed the blot to autoradiography to detect the primary antibodies bound to their protein target
  • Can see that the enhancer of zeste is colocalised to the methylated activity (the same fractions have methylation ability and that protein)
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116
Q

What did Cao et al, 2002 conclude?

A
  • There is a correlation between the presence of enhancer of zeste and the methylating activity
  • These two experiments suggest that PRC2 contains a histone methyltransferase activity
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117
Q

What did Bender et al, 2004 investigate?

A

if Mes2 (c. elegant orthologue of enhancer of zeste) is required for trimethylation of lysine 27

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118
Q

How did Bender et al, 2004 investigate the role of Mes2?

A
  • Germ cells stained for nuclei and for trimethyl H3K27 (methylation mark)
  • The fluorescence antibody for DNA was labelled with RFP and the methylation mark labelled with GFP. Can see that there is Colocalisation that appeared yellow signal. Indicates that the trimethyllysine is present where DNA is in the embryo
  • In mutant of Mes2, there is no Colocalisation showing and appeared red showing that there is limited methylation mark being detected when there is no Mes2 present
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119
Q

What can Bender et al, 2004 conclude about Mes2?

A

Mes2 is required for trimethylation on lysine 27

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120
Q

What is Esc?

A

Esc is a beta propeller protein that produces an aromatic cage that the trimethylated lysine27 fits in.

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121
Q

How does PRC2 work?

A
  • The enhancer of zeste is the catalytic core of the complex to create the trimethylation and Esc recognises the mark. This pair of activity allows chromatin to be progressively methylated.
  • Enhancer of zeste methylates histone, the Esc then binds to this methylation mark and positions enhancer of zeste in the correct location to methylate the next histone H3 in the next nucleosome
  • The enhancer of zeste protein is inactive without ESC showing the need for the whole PRC2 complex
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122
Q

How does PRC1 work?

A
  • Recognises the trimethyl mark through the chromodomain
  • Methylation on lysine 27 from PRC2, the Polycomb protein in PCR1 then sits in between these methylations and creates condensed chromatin
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123
Q

What are PREs?

A

Cis-regulatory elements called PREs (Polycomb response elements) which recruit PRCs to establish a domain of silenced chromatin around the Hox cluster.

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124
Q

Does PREs affect the transcription of adjacent genes?

A

Yes

  • Transgenically label with LacZ reporter under the control of Ubx promotor which has a PRE next to it
  • Compare transgene activity in wild type and Polycomb mutant backgrounds
  • See expression in anterior domains but this restriction is lost in mutants that lack Polycomb function and instead appears everywhere. PcG binding to PREs therefore affects transcription of adjacent genes.
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125
Q

How do PREs interact with PRCs?

A

PREs bind a DNA binding component of the Polycomb family and through a series of interactions it recruits PRC2 which methylates and PRC1 which reads the methylations and condenses the chromatin

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126
Q

What are trithorax genes?

A
  • Animals that have mutated Trithorax have reduced sex combs but are not linked to the Hox genes
  • Promote Hox gene expression
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127
Q

How was the relationship between trithorax genes and polycomb proteins shown?-

A
  • In the absence of Trithorax function, Polycomb is not needed. This was shown in a double mutant of Trithorax and Polycomb and the resultant fly looked wild type – normal sex comb expression.
  • There is also wild type expression of Hox genes in these mutants but in the Polycomb mutants there is ectopic expression of Hox genes in the imaginal discs.
  • Polycomb proteins must be blocking the activity of Trithorax in regions where Hox gene expression must be prevented
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128
Q

What is the relationship between trithorax and polycomb proteins?

A

Functionally antagonistic

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129
Q

What recognise H3K4 methylation by trithorax?

A

PHD fingers

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130
Q

How does sex determination lead to an imbalance of gene product?

A

Females and hermaphrodites have two X chromosomes whereas males have one X and one Y. In simpler organisms, males only have one X (X0). This creates an imbalance in the number of chromosomes and therefore gene expression

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131
Q

What are autosomes?

A

Not the sex chromosomes

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132
Q

What are Monosomies?

A

When only one chromosome form the pair is present

- embryonic lethal

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133
Q

What are trisomes?

A

When there is three chromosomes in a pair instead of two

- Also embryonic lethal with the exception of Down syndrome (trisomy 21)

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134
Q

What is dosage compensation?

A

The mechanism to equalize gene expression from X-chromosome in males and females (hermaphrodites) to stop lethality

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135
Q

What strategy do drosophila use for dosage compensation?

A

Make the only X-chromosome in males twice more active.

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136
Q

What strategy do mammals use for dosage compensation?

A

Make one X-chromosome in females (hermaphrodites) totally inactive.

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137
Q

What strategy do C. elegans use for dosage compensation?

A

Make both X-chromosomes in females (hermaphrodites) twice less active.

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138
Q

What is Sxl?

A

Dosage compensation factor in drosophila

- expressed in a sex specific manner

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139
Q

What are the promoters for Sxl?

A

Sxl has two promotors: SxlPe – the establishment promotor and SxlPm – the maintenance promotor

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140
Q

Explain what happens in dosage compensation of female drosophila

A
  • Expression of Sxl splicing factor from SxlPe (establishment promoter) is activated early in development. Feedbacks to maintain its own expression.
  • The present Sxl protein then promotes splicing and binds to PolyU in RNA so that translation terminating exon 3 is excluded. This results in production of functional Sxl protein which maintains its own expression and prevents translation of MSL2 protein by binding its 5’-UTR and 3’-UTR of its RNA.
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141
Q

What is MSL2?

A

Male-Specific Lethal – 2

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142
Q

What is the MSL complex?

A
  • A group of proteins which bind together and is essential for dosage compensation and viability.
  • It consists of MSL1 which is a scaffold protein, MSL2 which is a ubiquitin ligase, MSL3 which binds trimethylated lysine 36 on histone 3 (mark of active genes), MLE which is RNA DNA helicase and MOF which is acetylates lysine 16 on histone 4. roX are two long non-coding RNAs which are required to bring the complex together.
  • Present on bodies of active genes on the X chromosomes in males
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143
Q

What are Polytene chromosomes?

A

The chromosomes found in salivary glands of drosophila larvae. This is very active tissue and produces a lot of protein. To allow this it undergoes multiple rounds of endoreplication without mitosis and sister chromatid separation. This allows them to be seen in light microscopy

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144
Q

Why were Polytene chromosomes useful when investigating MSL complex?

A

The condensed and non-condensed chromatin are easily see due to the dark and light bands. Can use antibody staining to see where proteins are localised – are they associated with condensed chromatin or active chromatin. Using this technique, can see that MSL3 is localised to the active X chromosome in males.

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145
Q

Give evidence for the involvement of MSL3 in the spreading of MSL complex

A

When MSL3 protein is not present, the complex forms and fails to spread. It localised in 150 distinct sites on the X chromosome instead of the whole chromosomes

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146
Q

How did Alekseyenko et al, 2008 investigate the recruitment of MSL complex to the X chromosome?

A
  • Used MSL2 chromatin immunoprecipitation to see what MSL complex bound to
  • Found that there was a motif enriched in the precipitated DNA suggesting that this is what the MSL complex bound to. When put this motif on other chromosomes it too recruited MSL complex showing it is sufficient for MSL recruitment
  • Named this motif MRE (MSL recognition element). This is enriched in X chromosomes in the drosophila and includes known recruitment sites of roX and roX2 genes
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147
Q

Outline dosage compensation in male drosophila

A
  • If there is no Sxl protein maintained from the SxlPe promotor then the terminating exon 3 is not spliced out leading to premature truncation of the protein.
  • This means that the prevention of MSL2 translation does not occur so MSL2 is expressed and forms the MSL complex
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148
Q

How does the MSL complex spread along the X chromosome in male drosophila?

A

This is achieved by MSL3 binding trimethylated lysine 36 on Histone 3. roX also helps this an unknown mechanism

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149
Q

How does the MSL complex exert its action in male drosophila?

A

Increases gene activities by two-fold.

  • MOF acetylates H4K16, which weakens a repressive internucleosomal structure to open the chromosome: DNA is more accessible for transcription
  • Topoisomerase II is recruited by MSL to relax torsional stress of X-linked genes making more accessible
  • MSL2 ubiquitinates H2B (only shown in vitro so far) and facilitates methylation H3K4 and H3K79 (human orthologue). These modifications important for transcription elongation
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150
Q

Give an example of when random X inactivation can be seen in mammals

A

X inactivation is a random event and is then inherited by all progeny of the cells. In female cats, there are patches of fur colour as the colour determinate gene is X linked so the X has been inactivated in different cells and then passed on to progeny.

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151
Q

What is Xic?

A

Xic is a single cis-acting master switch locus essential for silencing of X chromosome in cis (silences chromosomes where it is active) ensuring initiation of random inactivation.

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152
Q

What are Xist and Tsix?

A

Xist and Tsix encode non-coding RNAs which are complementary to each other, and the expression of one inactivates the other.

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153
Q

How is the expression of Xist and Tsix controlled?

A

Rnf12 encodes a ubiquitin E3 ligase and is a known Xist activator. It ubiquitinates and thus leads to degradation of the protein Rex1 which is found at promoter regions of both Xist and Tsix. This thought to simultaneously activate Xist and inactivate Tsix.

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154
Q

Outline the stochastic model of mama X chromosome inactivation

A
  • Autosomes produce a signal which results in Tsix expression whereas the X chromosomes express X linked Rnf12 from Xic which promotes Xist expression. These two proteins compete against each other
  • Stochastically, there is a probability that when there is a lot of Rnf12 produced (both X chromosomes producing it) then it will bind to Xist promotor and lead to inactivation of the one X chromosome.
  • This will lead to the reduction of Rnf12 as only one X chromosome producing it.
  • This means that the relative expression of Tsix from autosome will be higher than the Rnf12 so will reinforce the expression of Tsix in the other X chromosome.
  • This stabilises the active X chromosome and cannot express Xist
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155
Q

Because X activation occurs stochastically what can occur in real life?

A

There are some cases where both X chromosomes remain inactive or become inactive in some cells. This results in the cells eventually dying or outcompeted by cells with one active chromosome

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156
Q

How is the X inactivation spread across the entire X chromosome and maintained?

A
  • Xist encodes non-coding RNA which associates with inactivated X-chromosome and associates with mediating proteins, e.g. hnRNPU/SAFA and YY1.
  • Spreading must involve booster elements that are enriched on X-chromosome as XIC translocation to autosomes results in poor spreading unlike on the X chromosome
  • These could be L1 long-interspersed repeats (LINE-1). They are enriched on X, and there is a burst of LINE-1 expression from inactivated chromosome at the time of inactivation.
  • Locally produced short RNAs might also facilitate spreading.
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157
Q

How is thought that the inactivated X chromosome represses gene expression?

A
  • Inactivated X-chromosome has organization similar fashion to constitutive heterochromatin: depletion of active chromatin marks (acetylation, H3K4me2/me3), and enrichment of inactive chromatin marks (H3K27me3, H3K9me3, and methylated DNA including on housekeeping genes).
  • It also recruits PRC2 (places H3K27me3 and H2AK119u1) which is Xist-dependent,
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158
Q

How is X chromosome inactivation ensured that it is passed on to the cells progeny?

A

Protein, Xist RNA, and chromatin marks are the same on interphase and metaphase chromosomes: maintaining the same chromosome inactivated once inactivation was established. This contributes to inheritance to ensure that the same X chromosome remains inactive

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159
Q

Is X inactivation a continuous process?

A

Yes
- Some events happen much later than other e.g. DNA methylation. It is a continuous process and doesn’t all occur at once

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160
Q

What are the sex differences in C. elegans?

A

Males
- X0
Hermaphrodites (no females)
- XX

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161
Q

What signalling elements are released from the X chromosome in C. elegans?

A

XSEs are signalling elements from the X chromosome

  • sex-1 (signal element on X)
  • fox-1 (feminizing gene on X)
  • ceh-39
  • sex-2
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162
Q

What signalling elements are released from autosomes in C. elegans?

A

ASE is the signalling element from the autosome

- sea-1

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163
Q

Outline dosage compensation in C. elegans hermaphrodites

A
  • Four genes on XSEs cooperatively repress expression of xol-1.
  • sea-1 on ASE encodes a transcription factor, which activates xol-1 expression.
  • In hermaphrodites there is twice more products of XSEs, thus xol-1 is repressed.
  • XOL-1 protein is a kinase and inhibits SDC-2, which is then present only in hermaphrodites.
  • SDC-2 is a key regulator of Dosage Compensation Complex (DCC) assembly.
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164
Q

How does the Dosage Compensation Complex (DCC) lead to gene repression in C. elegans?

A
  • All the genes in DCC are maternally supplied, and form a complex like condensin.
  • The condensin complex is conserved in all eukaryotes and is essential for proper chromosome compaction and segregation during mitosis and meiosis.
  • Because of the similarity between these complexes, it is also involved in partial condensation of the X chromosome therefore reducing its activity
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165
Q

How is DCC recruited to the X chromosome in C. elegans?

A

There are X chromosomal regions which recruit DCC suggesting containing DCC recruitment sites

  • Csankovszki et al., 2004: Translocation of X-chromosome regions on autosomes. Was sufficient for DCC recruitment
  • Further mapping using ChIP identified smaller regions, called rex sites, enriched in a 12-bp DNA sequence motif, which are necessary (though not sufficient) for DCC recruitment on X chromosome
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166
Q

How does DCC spread along the X chromosome in C. elegans?

A

It is not known how DCC spreads from initial recruitment sites. It accumulates along X chromosome, especially at the promoters of actively transcribed genes. DCC spreading is not dependent on any particular property of X-linked DNA sequences

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167
Q

How much does the DCC complex reduce the activity of both X chromosomes in C. elegan hermaphrodites?

A

By about 2-fold

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168
Q

Give evidence for DCC causing reduced X chromosome activity in C. elegans?

A

In XX DCC mutants 40% of expressed genes show an increase in transcript levels

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169
Q

How is it thought that DCC causes reduced gene expression in c. elegans?

A

Histone 4 lysine 20 methylation, associated with gene repression, is enriched on X chromosome in hermaphrodites and is dependent on functional DCC, leading to reduction of histone 4 lysine 16 acetylation on X chromosome. Histone 4 lysine 20 methylation is greatly increased on condensed chromosomes during mitosis suggesting it affects the chromosomal structure

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170
Q

Does DCC have any autosomal targets?

A

There is one autosomal target of DCC – the her-1 gene, repression of which promotes hermaphrodites sexual development
- However, DCC recruitment to her-1 differs from that to X-chromosome: it requires SDC-3 instead of SDC-2 and uses a different DNA sequence motif.

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171
Q

What is the general principle of dosage compensation across species?

A
  • Express dosage compensation factors in sex-specific manner
  • Recruitment to the X-chromosome
  • Spreading along X-chromosome
  • Modifying chromatin to alter gene transcription
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172
Q

What did Csankowszki et al, 2004 identify?

A

A 793 base pair fragment that functions in vivo as an X recognition element to recruit DCC

  • The idea of there being a single recruitment site was eliminated by analysis of mutant hermaphrodites. DCC localised at both truncated X chromosomes showing there must be multiple recognition elements
  • Found a 4.5kb fragment that strongly recruits DCC and can distinguish the X chromosome from autosomes
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173
Q

What are through to be mammalian PREs?

A

Hypomethalyated CGIs (where the C and G content exceeds 50%) represent PRE like sequences that can recruit Pcg and TrxG

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174
Q

What happens when PcG complexes bind to CGIs?

A

These PcG bound CGIs correspond to repressed promotors and inhibition of transcription induces recruitment of PcG proteins to newly silenced CGIs

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175
Q

What is the chromatin sampling model?

A

Chromatin sampling model says that PcG proteins weakly interact with all potential binding sites but their stable binding is blocked by active transcription or the presence of activating TFs

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176
Q

Why was it though that long non coding RNAs can also regulate the recruitment of PcG complexes?

A

Long non coding RNAs also regulate recruitment of PcG complexes. Xist (long non coding RNAs) target PcG to the inactive chromosome. This lead to research in other long non coding RNAs (e.g. HOTAIR which associates with PRC2) that could target PcG to HOX

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177
Q

How have PcG complexes been linked to cancers?

A

PcG can play an oncogenic role by modulating cell proliferation and senescence:

  • PcG can cooperate with c-MYC to generate mouse lymphomas via direct silecning of the CDKN2a.
  • Also found tht elevated enhancer of zeste in prostate tumours correlates with poor prognosis
  • Missense mutation in EZhZ stops mono or di methylation of H3K27 but enhances trimethylation. This hypermethylation of H3K27 can act as a driver of many human cancers and can initate cell invasio and metastasis
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178
Q

How much of the genome encodes for proteins?

A

2% of genome encodes for proteins

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179
Q

What are the two types of non coding RNA?

A

Short (22-28 nt): miRNA, siRNA, piRNA;

Long (70nt – 118 kb): lncRNA, snRNA, tRNA, rRNA.

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180
Q

What is the function of short non coding RNAs?

A
Silencing expression (RNA interference)
- RNAi is a natural cellular process that silences gene expression and plays important roles in gene regulation and innate defence against invading viruses.
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181
Q

What are the three types of short non coding RNA?

A
SiRNA 
- cleaves mRNA
miRNA 
- inhibits mRNA translation 
piRNA 
- cleaves products of transposons in germ cells
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182
Q

How do siRNA and miRNA lead to gene silencing?

A
  • miRNAs are transcribed as longer primary pri-miRNAs which can include other miRNAs and even protein encoding exons, in addition to spacer sequences.
  • Pri-miRNAs are cleaved by the RNase III enzyme Drosha to generate a 60nt pre-miRNA that then exits the cytoplasm
  • They are then processed by DICER to create two mature miRNAs that are differentially assembled into RISC
  • siRNA would then assembles with Argonaute (AGO2) protein and other polypeptides to form the RISC complex. AGO2 has endonuclease function and then cleaves the complementary mRNA molecule to the siRNA and silences its expression.
  • However, miRNA is not perfectly complementary to the mRNA so instead prevents translation of the mRNA through the GW182 component of the RISC complex. The exact mechanism that this occur by are still unclear.
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183
Q

Outline the differences between siRNA and miRNA?

A
  • They differ in length and structure. siRNA are shorter consisting of 30-100nt with a 2bt 3’ overhang whereas miRNA have 70-100nt with interspersed mismatches and hairpin structure
  • siRNA are complementary to mRNA whereas miRNA are only partially complementary to mRNA typically the 3’ UTR
  • siRNA can only target one mRNA whereas miRNA can target over 100
  • siRNA cleaves the target and miRNA transcriptionally represses or degrades the mRNA (rare as requires good complementarity)
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184
Q

What developmental process are miRNA involved in?

A

miRNA is involved in many developmental processes including metabolism, cell proliferation, apoptosis, neuronal cell fate, ect

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185
Q

How does the amount of processes miRNA is involved make it susceptible to cause diseases?

A

If the developmental processes are misregulated they would therefore be involved in multiple diseases e.g. cancer, cardiovascular disease, skin disease ect.

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186
Q

What is Fragile X syndrome?

A
  • FXS is a genetic disorder linked to expansion of CGG trinucleotide repeats on X chromosome, which causes suppression of the fragile X mental retardation 1 (FXMR1) gene, and loss of fragile X mental retardation protein (FMRP). FMRP regulates neuronal connectivity and plasticity
  • One of the major causes of autism and mental retardation in humans
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187
Q

How did Jin et al, 2004 show that miRNA is inclined with fragile X protein?

A

Immunoprecipitation of fragile X syndrome protein. Shows that drosophila and human protein precipitates with both miRNAs and miRNA precursors

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188
Q

Why is the drosophila eye good to investigate changing phenotypes?

A

Drosophila eye consist of ommatidia with each having 8 photoreceptors. This regular pattern in the eye is easy to see when not normal so can look for mutations

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189
Q

How did Jin et al, 2004 show that FMR1 (fragile X protein) and Ago1 (involved in miRNA silencing) act in the same pathway?

A
  • Overexpression of drosophila Fmr1 protein lead to severe effects in retinal development. The same overexpression with also insertion of P element (drosophila natural transposon) which reduces Ago1 and the eye phenotype was normal.
  • The reduction of Ago1 therefore rescues the phenotype of overexpression of Fmr1.
  • They therefore interact within the same pathway.
  • The rescue was shown to be due to reducing the amount of Ago1 by removing the insertion, using a transposon, to see if the Fm1 would lead to the severe phenotype. This lead to the restoration of Ago1 and overexpression of Fmr1 and the disordered eye phenotype (control)
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190
Q

What is the current model of Fragile X syndrome protein action?

A

Pre-mRNA precursor binds to RISC complex and is processed. It also associates to Fragile X syndrome protein and in a phosphorylation dependant manner helps with the binding and silencing of mRNA and gene translation.

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191
Q

Why does the current model of Fragile X syndrome protein action allow for a quick switch between mRNA expression?

A

When it FMR1 is phosphorylated, it is bound to mRNA targets and when it dephosphorylates it is unbound and can be immediately translated again.
- This provides a quick switch between mRNA expression without the need for change in gene expression and waiting for other processes

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192
Q

What are long non coding RNAs?

A

They can be very long (200nt-thousands nt)

Due to their length, they can have much more diverse functions than just gene silencing.

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193
Q

What are the functions of long non coding RNAs?

A
  • Gene regulation in cis. Either by Transcription through a promoter blocks its function: transcriptional interference or recruiting histone-modifying enzymes in cis;
  • Scaffolding: Provide scaffold to recruit proteins and organise functional complexes
  • Decoy
  • Silencing through complementation (similar to miRNA)
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194
Q

Why are long non coding RNAs used as a scaffold instead of proteins?

A

It is very cost efficient
- RNA scaffolds can be very long: a typical RNA “arm” of 50 nt extends for 13 nm, whereas a 50 aa alpha helix extends for 7.5 nm (and require 150 nt to be encoded).
- This is longer and uses less nucleotides and doesn’t need protein production
More versatile
- A 100 nt RNA could easily bind multiple proteins, whereas a 100 aa domain might bind a single protein partner.

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195
Q

Give an example of a long non coding RNA used as a scaffold?

A

HOTAIR

- More than 2000nt in length

196
Q

How does HOTAIR interact with with proteins?

A
  • 5’ of HOTAIR interacts with PRC2

- 3’ of HOTAIR interacts with the histone-demethylase LSD1 (demethylates H3K4).

197
Q

What is the function of HOTAIR?

A
  • Necessary for PRC2 occupancy and histone H3 lysine-27 trimethylation of different genes on different chromosomes.
  • Activates Wnt signaling by recruiting PRC2 to the promoter region of WIF-1 (Wnt inhibitory factor 1). Involved in proliferation, cell migration and EMT.
  • Regulation of metalloproteases (MMP1 and MMP6). They change the environment of the cells and responsible for epithelial to mesenchymal transitions
198
Q

How can change in HOTAIR function lead to cancer?

A
  • Responsible for Wnt signalling activation which is involved in cell proliferation - misregualtion can lead to cancer
  • Responsible for metalloproteases which esponsible for epithelial to mesenchymal transitions. This is a critical step in cancer progression so misregulation can lead to cancer
199
Q

How are long non coding RNAs used as decoys?

A

They bind to proteins to inhibit their action

  • substrate can’t bound
  • acts a decoy
200
Q

Give an example of a long non coding RNA used as a decoy?

A

GAS5

201
Q

Outline the Glucocorticoid signaling (stress pathway)

A
  • Cortisol binds to glucocorticoid receptor which is then translocated to the nucleus
  • In the nucleus, it binds to DNA sequences (GRE) and regulates gene transcription. Can also lead to DNA methylation and methylation and acetylation of histones
202
Q

How did Kino et al, 2010 identify GAS5?

A
  • Immunoprecipitation with glucocorticoid receptors or control to see what is bound
  • Used PCR to identify the GAS5 product which was present in all pulldowns apart from the control
  • GAS5 binds glucocorticoid receptor
203
Q

Outline a luciferase assay?

A
  • When luciferase gene is expressed it produces a protein which upon substrate addition admits light.
  • The amount of protein produced is based on gene expression of the luciferase.
  • Can use this to see gene expression
204
Q

How did Kino et al, 2010 show that Gas5 directly bind to the glucocorticoid receptor?

A

Luciferase Assay

  • The GRE element was fused to luciferase. If glucocorticoid receptor is active then it will bind and activate transcription.
  • If there is no GAS5 then light emitted. If gas5 is added then glucocorticoid receptor was not able to activate luciferase well. Gas5 therefore inhibit the function of the glucocorticoid receptor.
  • A different promotor, gal4 (inducible), was fused to luciferase.
  • Normally this promotor is not recognised by glucocorticoid receptor. If gas5 acts specifically through glucocorticoid receptor then it would not inhibit transcription of luciferase – control.
205
Q

What did Kino et al, 2010 show?

A

That GAS5 was a long non coding mRNA that acts as a decoy for the glucocorticoid receptor

206
Q

What happens when GAS5 is not present?

A
  • If no Gas5 then GR actives claP2 and SGK1 expression which inhibits apoptosis. Glucocorticoid receptor activation therefore stops cells from entering apoptosis
207
Q

What happens when GAS5 is present?

A

Gas5 is expressed in starved cells and allows them to undergo apoptosis

208
Q

What is genomic imprinting?

A
  • Some mutations are only inherited when the mutation comes from one of the parent.
  • Some genes are only expressed on the maternal or paternal chromosome
209
Q

Give an example of genomic imprinting?

A

In mice, if the mutant allele the heterozygous male is crossed the wild type female so that the mutation is inherited from the paternal genome. The progeny of the mice is viable with a shorter tail. However, if cross a heterozygous female with a wild type male then it is lethal with more defects than just a short tale (polydactyl)

210
Q

Give evidence that Botha. maternal and paternal genome are needed for mammalian reproduction

A
  • Gynogenic embryos - Transplanted to maternal genomes into oocyte. They were lethal in early embryogenesis due to defects in extraembryonic tissue
  • Androgenetic embryos - Two paternal genomes also resulted in death due to defects in embryonic tissues
  • There is no overall bias in function of maternally and paternally imprinted genes, so this difference is due to a dominant effect of one or a few imprinted genes.
211
Q

Describe the genes imprinted in the mammalian genome

A

150 imprinted genes with 80% of them clustered in 16 genomic regions that contain two or more genes.
- Only 7 of these clusters are well characterized and contain 3-12 imprinted genes spread over 80-3700 kb of DNA.

212
Q

What mechanism is responsible for genomic imprinting?

A
  • DNA methylation is the only modification that was shown to function as the gametic imprint. Histone modifications and modifying enzymes play a minor role in genomic imprinting. Most other modifications require both DNA and histone modifications
213
Q

What is the DMR?

A

Differentially DNA-methylated region

  • DNA sequence carrying a gametic methylation imprint.
  • Many more maternally methylated DMRs than paternally are known.
214
Q

What is the ICE?

A

ICE (imprint control element): region which controls expression of the whole or part of imprinted cluster. DMR is ICE in 7 well-characterized clusters.

215
Q

Describe the 7 well characterised imprinting gene clusters

A
  • The range of size is large between 80 and 3700
  • Gene number also varies from 3 to 12
  • In most clusters, most protein encoding genes are expressed from one of the maternal or paternal chromosome. There is a non-coding gene which is then expressed from the other chromosome
  • Only one of the clusters there is no non-coding RNA genes
216
Q

When are genomic imprints acquired?

A

Acquired in the gametes when maternal and paternal genomes are separated. This makes sense because after the genomes are together in the nucleus, there is no way to differentiate between them

217
Q

How are DMR sequences not chosen for methylation in gametes?

A
  • Sequence comparison of known DMRs shows no sequence similarity (so cannot be simply due to recognition of a particular sequence by methyltransferases),
218
Q

How are DMR sequences chosen for methylation in gametes?

A
  • They do contain series of tandem repeats that may adopt a secondary structure that attracts DNA methylation
  • Maternal DMRs are CpG dinucleotides reach compared to the rest of genome, with CpG pairs spaced by 8-10 base pairs (optimal for recognition by some DNA methyl transferases: DNMT3A and DNMT3L).
  • Maternal DMRs de novo methylation is promoted by transcription across DMR - open chromatin domains
  • The combination of at least several factors provide a starting point for the acquisition of DNA methylation imprints.
219
Q

How are genomic imprints inherited?

A

Inherited through cell division but escapes the genome-wide reprogramming after fertilization (a combination of factors protects imprints from demethylation).

220
Q

When is the genomic imprint erased?

A

The methylation is then erased in primordial germ cells to set new imprints depending on progeny’s sex.

221
Q

How is strand specific gene expression silencing achieved?

A

There are two types of cis-acting silencing of imprinted genes

  • Insulator model (lgf2 cluster)
  • lncRNA-mediated silencing (lgf2r cluster)
222
Q

What did Hark et al, 2000 discover?

A

CTCF is a zinc finer protein implicated in chromatin boundary formation
- CTCF bind unmethylated ICE.

223
Q

How did Hark et al, 2000 find that CTCF can bind unmethylated ICE?

A
  • They used unmethylated or methylated DNA probes containing the ICE and incubated them with recombinant CTCF protein. If this protein can bind to the ICE after pulldown then it will be visualised on gel.
  • If it can’t bind then the protein will not be visualised in the western blot. The protein is only present when the DNA probe was not methylated
224
Q

Outline the structure of the IGF2 gene cluster?

A

IGF2 is upstream and the DMR is followed by a H19 gene and then two enhancers which drive expression of IGF2 in a tissue specific manne

225
Q

What did Kaffer et al, 2001 do to test how the IGF2 was expressed by the maternal and paternal chromosome?

A

They created transgenic mice.

  • One had a deletion where DMR usually was and another where there was an inserted DMR between the two enhancers.
  • They then looked at how H19 and IGF2 were expressed in the mice depending on whether they inherited the mutation allele from the mother or father using SNuPE (single nucleotide primer extension)
226
Q

What can SNuPE (single nucleotide primer extension) be used for?

A

Can be used to quantify allele specific expression

227
Q

Outline how Kaffer et al, 2001 used SNuPE to investigate the differental expression of genes in the IGF2 gene cluster?

A
  • The total RNA is isolated and converted to cDNA.
  • Used PCR with specific primers to amplify the gene product
  • Products were eluted from a gel and separated into two batches. Two different alleles of the same gene have a SNP and can be used to quantify the two different alleles. A specific primer for the changed nucleotide and anneal to the strand in both batches.
  • Instead of adding dNTPs, add dCTP (as one allele has G) or dTTP (other allele has A) labelled by radioactive phosphate.
  • When dCTP added, only the allele with G can be elongated by the labelled one nucleotide. When dTTP is added, only the allele with A can be elongated.
  • How many molecules labelled will depend on how much product was amplified by the maternal or paternal allele.
228
Q

What did Kaffer et al, 2001 find using SNuPE for the DMR deleted transgenic mouse?

A
  • In the control, H19 is expressed only when inherited from maternal allele and IGF2 is expressed only when mutation expressed from paternal allele.
  • When the DMR was removed, both H19 and IGF2 were expressed from both maternal and paternal allele.
  • This proved that methylation of DMR is required for imprinting and strand specific expression.
229
Q

What did Kaffer et al, 2001 find using SNuPE for the DMR inserted transgenic mouse?

A
  • When the DMR was introduced between the two enhancers, H19 expression in liver which is driven by one of the promotor and muscle the other promotor.
  • In control, H19 is expressed in liver and muscle when inherited from maternal allele. When the additional DMR is introduced, the liver expression was normal but no expression in muscle (this promotor was separated from H19 by DMR).
  • This shows that DMR acts as an insulator from action of an enhancer.
230
Q

What did Kaffer et al, 2001 find?

A

That methylation of the DMR is required for strand specific expression
That the unmethylated DMR acts as an insulator from the action of an enhancer

231
Q

What is the current model of IGF2 cluster silencing?

A
  • On paternal chromosome, DMR is methylated which silences it. This prevents expression of H19 (lncRNA) from the paternal allele and prevents CTCF binding. Enhancers can drive expression of IGF2 gene and other protein encoding genes. Paternal chromosome therefore expresses IGF2 but not H19.
  • On maternal chromosome, the DMR is not methylated which allows binding CTCF binding (forms boundary between chromatin domains) and allows to work as an insulator. This prevents the enhancers acting on the protein encoding genes. H19, however, is not methylated so can be expressed. Therefore, maternal chromosome expresses H19 but not IGF2.
232
Q

What is Beckwith-Wiedemann syndrome?

A

Uniparental disomy of Igf2 cluster (maternal copy is replaced by an extra paternal) results in too much IGF2 expression.
Leads to
- Microsomia
- Microcephaly
- Abdominal wall defects
- Neonatal hypoglycaemia (untreated, it can lead to permanent brain damage)
- Increased chance to develop neoplasms during childhood

233
Q

How did Stoger et al, 1997 show that the maternal IGF2R cluster was heavily methylated?

A
  • Methyl sensitive sites were used for restriction by enzymes so that when the site is methylated the site cannot be cut.
  • On the maternal locus, the sites cannot be cut showing that it is heavily methylated
  • The control showed that the paternal inherited locus is completely accessible to restriction enzymes so is not methylated
234
Q

Outline how lncRNAs silence the IGF2R cluster?

A
  • Maternal DMR is methylated. DMR overlaps with promotor of a lncRNA preventing its transcription. This allows the expression of protein encoding regions. IGF2R is expressed
  • On the paternal, DMR is not methylated and lncRNA is expressed which can propagate and prevent expression of protein encoding genes by knocking out Pol II. IG2R is not expressed
235
Q

What is meant by uni-parental progeny?

A

Two maternal or two paternal genomes used for reproduction

236
Q

How did Li et al, 2018 show that uni-maternal progeny is possible?

A
  • The embryonic stem cells were obtained from female organisms and made haploid
  • Used CRISP-Cas9 to delete two or three of the imprinted gene loci: H19 and DLK1 or H19, DLK1 and Rasgrf1, respectively
  • They injected into oocytes and implanted into host mother, In the case of two deletions, the progeny died early in development. With three imprinted loci deletion, the bi-maternal mice were produced with normal behaviour and produce viable progeny. The litter size was normal
237
Q

How did Li et al, 2018 attempt uni-paternal progeny?

A
  • The process was more complex: removed 6 imprinted loci (only 7 well characterised). They were then injected with sperm into an inoculated oocyte. If implanted into host mother it did not develop so they instead produced genetic diploid embryonic stem cells from this and implanted into a tetraploid zygote. Produced offspring but it resulted in many developmental abnormalities as well as increase birth weight
  • Then removed 7 imprinted loci, the offspring appeared normal but they died within the first 48 hours of birth for an unknown reason as no obvious defects
238
Q

Why did the mechanism of genomic imprinting evolve?

A

Hypothesis

  • There is a maternal and paternal conflict
  • Paternal genes promote growth of the embryo whereas maternal genes protect the mother and enables them to grow several embryos with different paternal origins. Therefore, reduces the size to give them a similar chance of developing
239
Q

What is meant by developmental plasticity?

A

Development is extremely plastic. The same genotype can result in many different phenotypes

240
Q

Give an example of how the same genotype can result in different phenotypes?

A
  • Shire and Shetland ponies. They can interbreed as they are the same species but have very different sizes. When Shetland mother breeds with shire male, the progeny is the size of the Shetland. The same vice versa. The size of the new-born is determined by the size of the mother even though the genotype is the same.
241
Q

What did Francis et al, 2003 show about the causes of behaviour?

A

Show that genotype and environment can also change behaviour
- Used two strains of mice which differ in colour and behaviour: B6 and BALB

242
Q

How did Francis et al, 2003 test the behaviour of the two strains of mice?

A

Then tested their behaviour using open field test. Rodents naturally have an aversion to open areas due to predators but also like to explore new areas. This conflict allows us to explore their anxiety levels. The amount of time spent on the outer edge or in the middle can determine their anxiety levels.

243
Q

How did Francis et al, 2003 show that the mice behavioural pattern was decided by genotype and environment?

A
  • The white (BALB) mice are generally more anxious then the black (B6) mice which are more exploratory. If BALB mice were prenatally transferred into B6 mice or if B6 mice were prenatally transferred into BALB mice but then spent with B6 mice then they have B6 behaviour
  • Despite genotype, both pre and post-natal periods affected their behaviour. The B6 mice who spent time with white mice were much more anxious then would naturally be
244
Q

What did Barker and Clark, 1997 show about diabetes and maternal diet?

A

Prevalence of diabetes (59-70 age) correlated with birth weight. This was also shown in systolic pressure in men and women. There is clear correlation with a smaller birth weight and higher systolic pressure. The birth weight can have effects on the health many years after birth

245
Q

What is the link between glomeruli and birth weight?

A
  • Low protein diet of the mother activates glucocorticoid response (regulates apoptosis) which causes cell death in developing kidney cells
  • This is important in evolution as when limited nutrition it goes to important tissues like heart and brain
246
Q

What is Developmental Origins of Health and Disease (DOHaD) hypothesis?

A
  • Since original discovery, prenatal environment was linked to a wide range of adult conditions, including lung disease, cognitive development abnormalities, breast cancer, prostate cancer, and leukemia leading to formulation of Developmental Origins of Health and Disease (DOHaD) hypothesis, in which both pre- and postnatal environments influence adult health.
247
Q

What is the Predictive Adaptive Response (PAR) model?

A

The early environmental clues shift the developmental pathway to modify the phenotype in expectation of the later environment.

248
Q

What is benefit of altering development in response to fetal environment?

A

Low nutrition in prenatal development, there is a high likely hood that there would be a low nutrition postnatally. Prepares the body to increase chances of survival

249
Q

Give an example of Predictive Adaptive Response (PAR)?

A

Butterflies

  • Colour of the wings depending on if the pupae is in the summer or spring.
  • Darker colour provides a functional advantage in cooler spring temperatures: dark pigment absorbs sunlight, raising body temperature more rapidly.
250
Q

How does PAR give an organism a survival advantage if the response matches the predicted environment?

A
  • Fetal malnutrition (e.g. low protein condition) sets the metabolism to be prepared for an environment where food is scarce and use every calorie and downregulate essential organs to increase chance of reproduction
  • Until recently, such predictions were mostly true enabling the individual to survive to reproductive age while staying small and thin due to nutrition in later life matching the nutrition of the mothers
251
Q

Why is PAR now more of a problem then an advantage?

A

With increasingly available food supplies within a single generation, the prediction often become false, and the individual is more likely to be obese and diabetic and to have high risk of heart disease due to their lower set metabolism

252
Q

What allows for the long term effects on metabolism?

A

Changes in gene expression

- DNA methylation is an epigenetic modification which enables such changes

253
Q

How is DNA methylation affected by diet?

A

The one carbon cycle

  • The product of this is DNMT which is the source of methylation
  • Many components are supplied by the diet including Folate, vitamins B6 and B12 and methionine
  • This process, which occurs in every single cell, is dependent on the diet and in development the mothers diet
254
Q

What two proteins were shown to be hypermethylated in response to a protein restricted diet by Gluckman et al, 2008?

A

PPAR-alpha
- A transcription factor, which is a major regulator of lipid metabolism.
- The target genes include AOX (acyl-CoA oxidase), an important component of fatty acid metabolism pathway. Suggests goes beyond just DNA methylation.
- There is high expression of PPAR-alpha and therefore there is more expression of its targets.
HGR (human glucocorticoid receptor)
- A transcription factor involved in regulation of multiple processes including hepatic gluconeogenesis through regulation of PEPCK (one of the enzymes in the pathway). This can linked to a stress response and anxiety

255
Q

How can the expression of proteins be restored when there is a protein restricted diet?

A

Folate is at the heart of the one carbon cycle

  • Therefore tested if supplementation of folate substitute was enough to rescue the phenotype of a protein restricted diet
  • Found that supplementation reversed the methylation of PPAR-apha and HGR and restored the expression levels of the downstream genes
256
Q

What did Altobelli et al, 2013 show about folic acid supplementation of animals who had been dietary restricted?

A
  • Did whole genome analysis of DNA methylation of the animals of mothers who were on a restricted diet and those who had folic acid supplementation
  • Can see that there were many genes (618 genes) that were differentially expressed in the progeny of mothers with restricted diet and control group. These changes were mostly reversed with folic acid supplementation. There were still some genes that were differentially methylated. There was good reversal but not complete
257
Q

What other genes are affected by dietary restriction?

A
  • DNA methyltransferases themselves: Dnmt1, Dnmt3b and Dnmt3a.
  • Hat1 (histone acetylate) also affected by diet restriction.
  • This suggests that it goes beyond DNA methylation and may also affect histone acetylation
258
Q

Outline the long term effects of maternal diet on offspring

A
  • Limited maternal nutrition leads to limited fetal nutrition. These lead to altered glucocorticoid concentration which changed the fetal developmental plasticity response.
  • This leads to changes in liver metabolism, pancreas insulin secretion, insulin sensitivity of muscles, cortisol secretion, decreased kidney nephron number, decreased vasculature and decreased myocyte number in the heart.
  • These changes are not lethal and allow the offspring to reproduce but it will have long term effects on them in later life leading to obesity, diabetes and hypertension and stroke
259
Q

What can over nutrition lead to?

A

Maternal high fat diet leads to increased risks of obesity and diabetes.
- Increased acetylation of H3K14 which associated with GCN5 expression (deacetylase).

260
Q

What induces nonalcoholic fatty liver disease?

A
  • In nonhuman primates, maternal high-fat diet induces nonalcoholic fatty liver disease.
  • SIRT1 HDAC liver specific knock-out in mice increases weight gain and hepatic steatosis in response to high-fat high-cholesterol diet.
  • This suggests that reduced SIRT1 activity causes nonalcoholic fatty liver disease which is induced by a high fat diet
261
Q

What did Shi et al, 2017 show?

A

Supplementation of nicotinic acid with water to Haao and Kano null mice helped sustain embryogenesis

262
Q

How did Shi et al, 2017 show that nicotinic acid helps embryogenesis?

A

There are two pathways for NAD synthesis:

  • One requires dietary tryptophan, and the other requires dietary niacin(vitamin B3, which is found in marmite)
  • They produced two mutant mice in components of the metabolic pathway which converts tryptophan to NAD. If were null for these enzymes and didn’t have any medication, the embryos died
  • If supplemented medicine with the diet, there were some live embryos present. Heterozygous embryos were normal and homozygous embryos had multiple developmental defects
263
Q

Does folic acid only improve prenatal health?

A

Mikael et al, 2012

  • Folic acid repairs most of the damage of malnutrition. However, high folic acid intact could also have a negative impact
  • Mouse mothers were supplemented with 10 times dose of daily recommended. This was associated with high embryonic loss and heart defects compared to mothers on the control diet
264
Q

What is SAM (S-Adenosyl methionine)?

A

Main methyl donor required for DNA methylation

265
Q

Give evidence that diet affects the one carbon cycle?

A

Dominguez-Salas et al, 2013

  • In some countries, diet changes throughout the year depending on the seasonal food availability
  • Can see in Gambia, there are monthly and seasonal in protein intake which are correlated with folate, B2 and DMG (components of one carbon metabolism pathway)
266
Q

Apart from the one-carbon cycle, what other process does the diet affect?

A

Diet is also the source of glucose which is required for cellular respiration which produces ATP. Each reaction produces a total of 36-38 ATP. Many of the intermediates of this reaction are involved in epigenetic regulation

267
Q

What intermediates of cellular respiration are involved in epigenetic regulation?

A
  • AcCoA – the main donor of acetyl groups required for acetylation
  • ATP - drives phosphorylation and is required for SAM production (main methyl donor)
  • NAD+ – cofactor for SIRT HDAC (deacetylases)
  • FAD – cofactor of demethylase LSD1
  • Micronutrients, amino acids and glucose supplied by diet determine availability of multiple epigenome effectors which are evolved in methylation, demethylation, acetylation, deacetylation and phosphorylation
268
Q

Why does nutrient intake have long term effects on metabolism?

A

Nutrient intake is sensed by cells and affect multiple metabolic pathways including one-carbon metabolism and cellular respiration. The resulting balance of metabolites contribute to multiple epigenetic changes in the genome – normally to adjust metabolism to environment to increase chances of survival.

269
Q

What epigenetic adaptions occur when there is calorie restriction?

A

PGC-1alpha is deacetylated and therefore active by SIRT1 (deacetylase).
This can then regulate the expression of downstream target genes increasing mitochondrial function.

270
Q

What epigenetic adaptions occur when there is calorie excess?

A
  • TCA-derived citrate is converted into AcCoA by ACL providing a nuclear/cytosolic pool of AcCoA, which is used for histone acetylation by histone acetyltransferases, such as GCN5
  • PGC-1alpha is then acetylated and therefore inactive by GDN5
  • Such a regulation of PGC-1α by nutrients inhibits mitochondrial function in situations with high energy supplies, favouring fat accumulation
271
Q

What model organisms were used to research healthy ageing and why?

A

Research into healthy lifespan were done in worms, flies and mice sue to their small size, short lifespan, genetic tractability and controllable environment

272
Q

How can ageing be measured?

A

There is no method to measure ageing so measure the age at death. Can use large populations to produce a curve and find the median half span. Can use this to look at the effects of dietary restriction on lifespan

273
Q

Give a study into the effect of dietary restriction on lifespan?

A

Partridge et al, 2005

  • Starvation of drosophila reduces life span
  • Moderate dietary restriction increases the life span. Control group had a median life span of 35 days while dietary restriction group had a median life span of 55 days, showing a significant increase in life span
  • Excessive food availability also reduces life span
274
Q

What percentage of adults are overweight in England?

A

62%

275
Q

What suggests that obesity has an epigenetic component

A
  • Permanent weight-loss is not possible for the majority of people with obesity. 5-year re-lapse rate for weight re-gain is greater than 90%
  • Thus, obesity is a chronic disease. This suggests long-term changes beyond the metabolism and nutrient consumption alone. This suggests that there are epigenetic components
276
Q

What percentage of people have type 2 diabetes?

A
  • 6% of people aged 17 years or older had diagnosed diabetes in England in 2013
  • Prevalence of both obesity and diabetes is rising in England
277
Q

What is the relationship between obesity and type2 diabetes?

A
  • 90% of adults with type 2 diabetes aged 16-54 years are overweight or obese
  • In England, 12.4% of people aged 18 years and over with obesity have diagnosed diabetes, five times that of people with a healthy weight
278
Q

How is obesity thought to promote type 2 diabetes?

A

It is not clear how obesity promotes T2D, but proposed factors include:

  • Mitochondrial dysfunction,
  • Increased inflammation,
  • Changes in lipid metabolism,
  • Changes in the gastrointestinal microbiota
279
Q

What does the agouti gene encode?

A
  • Agouti gene encodes paracrine-signalling molecule ASIP (Agouti signalling peptide)
  • ASIP inhibits hypothalamic signalling making mice hyperphagic (eat a lot)
280
Q

Why are Agouti viable yellow mice useful when studying diet?

A
  • ASIP is activated when IAP promotor is unmethylated of agouti viable yellow gene and inhibits hypothalamic signalling making mice hyperphagic (eat a lot). These mice are yellow and obese
  • Morgan et al, 1999 – Mice were supplemented with vitamin B12, folic acid, choline and betaine during pregnancy results in thin, brown pups. Without supplementation, the pups were yellow and hyperphagic
281
Q

How did Koza et al, 2006 investigate the epigenetic role of weight gain?

A
  • Used genetically identical B6 mice
  • At week 3 postnatally, they were kept at low fat diet, week 8-14 high fat diet, 2 weeks low fat and then high fat again
282
Q

What did Koza et al, 2006 observe?

A
  • Even though the mice were kept in the same conditions and were genetically identical, there was huge variation on the body weight and referred to high gainers and low gainers
  • At the starting point, there was no correlation with the original weight of the pup showing that there is no original difference between them so could be epigenetic component
283
Q

How is the Wnt pathway involved inhibiting adipocyte production?

A
  • When the pathway is active, Wnt5b binds frizzled receptor which activates non-canonical wnt pathway leading to inhibition of target genes.
  • At the same time Wnt10b binds to frizzled co receptor LRP leading to phosphorylation of Dvl which inactivation Axin and allows translocation of beta catenin into the nucleus. This can then activate the transcription of a subset of genes.
  • This is anti-adipogenic meaning that the stem cells of the adipocytes are maintained and fat is not produced
284
Q

How is the Wnt pathway involved promoting adipocyte differentiation and fat accumulation?

A

When there is a production of the two inhibitors Dkk and sFRP, Wnt10b can’t bind to the frizzled receptor inactivating the pathway. Independently, Wnt5b activates a different pathway which is pro-adipogenic so adipose tissues is actively made and cells are differentiated. This is well characterised in development but is also involved in weight gain through diet.

285
Q

How did Koza et al, 2006 show a correlation between the Wnt pathway and weight gain?

A
  • mRNA from SFRP5 (Wnt inhibitor that is pro-adipogenic) extracted from high and low gaining mice.
  • Expression of SFRP5 (Wnt inhibitor) correlates with mice potential to gain weight, and correlates with expression of known imprinted gene MEST, suggesting epigenetic component.
  • When more Wnt is inhibited, mice were overweight.
286
Q

Why does Koza et al, 2006 not prove that the Wnt inhibitor SFRP causes weight gain?

A

This does not prove that SFRP5 and MEST are causing weight gain as this is just a correlation. Would need to produce mutants of these genes and then analyse the weight gain

287
Q

Why is it important to study type 2 diabetes?

A

Type 2 diabetes is linked to chronic inflammation which leads to cancer, abnormal sex hormone production (infertility), damages endothelial and vascular muscle (cardiovascular disease and hypertension) and defect in chemo reflex action (sleep apnoea and cognitive impairment)

288
Q

Why is it important to study obesity?

A

Excessive food intake causes long-term changes in the epigenome leading to obesity and T2D. Obesity and T2D in tern increases risks of various other diseases including cancer and cardiovascular diseases

289
Q

Give an example of a food that can affect the epigenome and possibly cause long term effects?

A
  • Garlic contains different components which produces metabolites when broken down such as Allyl mercaptan. Allyl mercaptan is a potent inhibitor of HDAC. This results in rapid and sustained histone hyper acetylation in model organisms.
  • In human colon cancer cells, increased histone acetylation of promoter region of P21WAF1 gene elevates p21 protein expression and leads to cell cycle arrest.
290
Q

Give three ways to detected DNA methylation on a given sequence

A

Bisulphite sequencing
Restrition enzymes
Affinity purification

291
Q

How is affinity purification used to detect DNA methylation?

A
  • Takes advantage of the methyl-binding domain (MBD) which binds methylated CG sites
  • Tagged MBD domain expressed in E. coli is affinity purified and the MBD column s subsequently used to purify methylated DNA. MBD will bind to the methylated DNA which can then be sequenced and identified
292
Q

What is the problem with using affinity purification to detect DNA methylation?

A

A problem with this technique is that it measures the density if methylation in a given region. If the methylation of the target is region is sparese then it may be difficult to differentiate from an unmethylated region. This is a bigger issue in mammalian gebomes where the CG densiity is geerally low and CG dense sequences are typically unmethylated

293
Q

Give two ways to detected DNA methylation across the genome

A
MeDIP
DNA microarrays (bead arrays)
294
Q

How are bead arrays used to detect DNA methylation across a genome?

A
  • Bisulphite-converted DNA is assayed with two primers, each labelled with a different fluorescent dye.
  • One primer is designed to hybridize if the cytosine is methylated (and unconverted to uracil), whereas the other will only hybridize to a converted sequence.
  • The two primers are used in a PCR reaction with a locus-specific methylation-insensitive primer. The ratio of the PCR products is ascertained using a bead array platform, which can assay up to 1536 sites in 96 samples in one experiment.
295
Q

Evaluate the hand-cuff model of TAD formation?

A
  • Supported as the cohesin complex generally co-localises with CTCF throughout the mammalian genome
  • However, a problem is that the number of CTCF binding sites are much more abundant than the number of TAD boundaries. Therefore, why are TADs not more abundant?
296
Q

What are the two ways the mammalian X chromosome can be inactivated?

A

There are two ways that the X chromosome is inactivated in different in animals. In Metatherian mammals use the imprinted mode of X inactivation where the paternally derived X chromosome is always inactivated. Other mammals, e.g. mice and humans, use random X inactivation where either the maternal or paternal X chromosome can be inactivated

297
Q

How have miRNAs been harnessed by viruses?

A
  • Certain viruses have acquired and manipulated host miRNA genes to enhance infection while other harness specific miRNAs for viral functions such as genome replication
  • Therapeutics that block function therefore hold promise for combating disease states
298
Q

Give a study showing the effect of diet on epigenetic ability?

A

Dominguez-Salas et al, 2013

  • Studied the natural variations in dietary intake of Gambian women and looked how this affected methyl donor pathways and their respective metabolic biomarkers
  • They monitored the diets of 30 women in reproductive age and assessed blood biomarkers monthly
299
Q

What did Dominguez-Salas et al, 2013 find?

A
  • Found that dietary intakes of folate, choline and betaine all varied significantly and the biomarkers showed significant seasonality with vitamin B6 and folate having the highest fluctuations
  • There was a higher SAM/SAH ratio during the rainy season
  • Shows that chances in food consumption patterns have a profound effect on methylation
300
Q

What is a microbiome?

A

Human gut is a complex ecosystem, which harbors approximately 100 trillion microbes (mostly prokaryotes) of about 5000 species – collectively this ecosystem is termed microbiome

301
Q

Why are the gut microorganisms important?

A
  • The collective number of these microbes is greater than the total number of host cells due to the small size of prokaryotic cells.
  • Gut microorganisms co-evolved with their hosts, and are important for normal gastrointestinal tract functions such as digestion.
302
Q

What are the important questions to be answered when studying the microbiome?

A
  • Which species are present? Does this change between health and disease?
  • How to study role of microbiome in health of the host?
303
Q

What is next generation sequencing?

A

Relies on generating library of DNA fragments and sequencing short reads from paired or single ends. Then use bioinformatics approaches to align the resulting reads to each other and a reference genome and can then compare the mutations and genomic rearrangements within that patient

304
Q

What are the problems with using next generation sequencing to study the microbiome?

A
  • There are over 5000 species in the micro biome so how do we know which reference the read is derived from?
  • There so many genomes, most of them not fully represented by reads so are unlikely to be correctly represented
  • If the reads overlap they do not necessary come from the same genome, it could be because it is conserved.
  • Identification and removal of contaminants is challenging because it is hard to know which is contamination and which is one of the species
305
Q

What are the three main approaches used to investigate the species within the microbiome?

A
  • Analyzing taxonomically informative marker genes, e.g. 16S RNA;
  • Grouping sequences into defined taxonomic groups (i.e. binning)
  • De nova assembling sequences into distinct genomes
306
Q

What is the problem with using taxonomically informative marker genes to group species in the microbiome?

A
  • A marker gene is used to assess the taxonomy of the population
  • Difficult to find a good marker because genes present in all species are very important for the function so will be similar between evolutionary close species but genes not important will likely not be present in many. Difficult to find the right depth
307
Q

How can grouping sequences define taxonomic groups in the microbiome?

A

Can use CG content to split into distinct groups

Can be computationally challenging

308
Q

Why is De nova assembling sequences into distinct genomes the best method used to investigate the microbiome?

A
  • Most reliable in understanding what species are present and in what amounts
  • Only one that allows identification of new species as the others rely on previous knowledge
  • Relies on deep sequencing to ensure underrepresented sequences are still seen and is very computationally challenging
309
Q

How can the micro biome be used to study the health of the host?

A

Model the micro biome using germ free mice

310
Q

How are germ free mice produced?

A
  • Late 19th century due to the rise of antibiotics
  • Created by aseptic C-sections. They were there kept in a germ-free environment and fed germ free food – totally sterile.
311
Q

How can germ free mice allow us to study the roles of the microbiome?

A
  • To understand the impact of specific species, can add the specific bacteria directly to the germ free mice
312
Q

What has been learnt about the microbiome using germ free mice?

A
  • Learnt that without germs, would not live healthy life as animals that are raised germ free have lots of health issues including behavioural differences (anxiety). Also have changes in neural function (synapse gene expression, neuronal activity). Defects in stress hormone signalling (altered glucocorticoid receptor signalling)
313
Q

What are the three main ways that the microbiome can alter the epigenome?

A
  • Members of gut microbiota, the Bifidobacterium and Lactobacillus genera, produce through their metabolism and therefore supply host cells with vitamins B2, B6, B12, and folate
  • Members of gut microbiota produce other important metabolites, which alter epigenetic code
  • Members of gut microbiota produce outer-membrane and outer-inner membrane vesicles which contain RNAs and might function as miRNAs
314
Q

How does the production of vitamins B2, 6 and 12 and folate by the microbiome alter the epigenome?

A

They are therefore a source of the micronutrients required for the one-carbon metabolism. The product of this, SAM, is the key donor for DNA methylation

315
Q

What other metabolites do the microbiota produce that alter the epigenetic code?

A

Short-chain fatty acids (butyrate, acetate, propionate, succinate, and lactate)
- These can be incorporated by intestinal epithelial cells or can diffuse across epithelium into the underlying intestinal lamina propria and enter blood stream.
Polyphenols
- Which are transformed in the gut into various derivatives of aromatic short-chain fatty acids, such as phenylacetate and phenylbutyrate.

316
Q

What affects is Butyrate (short chain fatty acid) thought to have on the health of the individual?

A
  • Anti-cancer activity
  • Anti-obesity
  • Anti-inflammatory
317
Q

How does Butyrate have an anti-cancer activity?

A
  • Activation of pro-apoptotic protein and cell-cycle inhibitor p21. This is epigenetically silenced in cancer cells (like allyl mercaptan). Its activation could therefore stop the proliferation of cancer cells and shrink the tumour
  • Modulation of Wnt signalling, which is constitutively activated in most colorectal cancers. Modulating Wnt could therefore prevent cancer
318
Q

How does Butyrate have an anti-obesity activity?

A
  • Normally, when calorie restriction, deacetylases activate PCG-1alpha which increases mitochondrial biogenesis and activity
  • Butyrate increases expression of PGC-1α and modulation of Wnt signalling.
  • Can speculate that there is more PGC-alpha at the same calorie intake when butyrate increases PGC-alpha expression.
  • This would therefore act like was in calorie restriction
319
Q

How does Butyrate have an anti-inflammatory activity?

A

Increased expression of FoxP3 gene in naïve peripheral CD4+ T cells resulting in an increase in circulating Tregs (regulatory T cells). Treg expansion reduces local inflammation.

320
Q

What is a BTBR mouse?

A
  • Model of autism
  • The genetic nature of this mice is unknown and is an inbred strain. There are multiple changes compared to normal mice and it is unclear which change causes the autism traits
321
Q

How did Kratsman et al, 2016 model the effect of short chain fatty acids?

A
  • Supplied different concentrations of butyrate to the BTBR mice and looked how the behaviour changed
  • Behavioural test: look to see how well the mice could bury marbles in different locations. Butyrate supplementation reduced time spent on these actions
  • See acetylation of histone H3. Affected genes involved in cell-cell signalling, behaviour, neuron differentiate
  • Butyrate supplementation therefore modulated excitation inhibition valance in the prefrontal correct which has a positive effect on social behaviour in an autism mouse model
322
Q

What affects do polyphenols have on the epigenome?

A
  • Inhibit histone deacetylates activity, but degree of inhibition varies from one compound to another. The resulting composition of the polyphenols is therefore difficult to produce and its effect on histones would vary
  • Additionally, at least some compounds inhibit DNMTs.
323
Q

Give evidence for how polyphenols modulate microbiome ecology?

A

Kang et al, 2016

  • Plant polyphenols modulate microbiome ecology: for example, EA supplementation decreases Firmicutes: Bacteriodetes ratio (the first cannot metabolite EA). One can’t metabolise the EA so the other species can benefit from it
  • Different types of microbes produce different metabolites of EA with different activities – so the final exact result of ellagic acid intake depends on the composition of microbiome.
324
Q

What effects are polyphenols thought to have on the health of an individual?

A

Like short-chain fatty acids such as butyrate, polyphenols are reported to have wide range of activities in the prevention and alleviation of various diseases and conditions, for example cancer, neuroinflammation, diabetes, and ageing.

325
Q

How are vesicle produced by microbiota thought to affect the hosts gene expression?

A

Very controversial – only two papers talking about it
- Bacteria produce vesicles and release them from their surface. Depending on gram negative or positive depends on whether outer vesicle or outer inner vesicles. The vesicles contain RNAs which could act as miRNAs and regulate the expression of the host genes by the same pathways of host miRNAs

326
Q

What are the structures of the miRNAs produced by microbiota?

A
  • Secreted RNAs are generally smaller then 16nt and are enriched in non-coding RNAs and are very different to bacterial intracellular RNA.
  • Aligned the RNAs to human genome and matched region was doing to match DNAse1 hypersensitivity clusters with elevated transcription levels
  • This is a very new theory so any functional relevance of this alignment is yet to be determined
327
Q

What applications to human health does the understanding of the importance of the microbiome?

A

Probiotics
- Efficiency is questionable
Stool transplant therapy
- Faecal samples from a healthy subject to an unhealthy subject

328
Q

How has stool transplant therapy been shown to aid immunotherapy cancer treatments?

A
  • In some patient’s immunotherapy works very well but in other patients it doesn’t work.
  • Faecal transplants from patients who responded well to immunotherapy improved the response in those who were not responsive
  • This is due to the microbiome providing the correct metabolites to alter the epigenetic code of the host
329
Q

What is cancer?

A

A collection of related diseases when cells start to divide uncontrollably and spread into surrounding tissues

330
Q

How does cancer cause death?

A

It causes death directly through the failure of organs to which is spreads and indirectly through cachexia. This is the uncontrollable growth of the cell mass which uses a lot of energy. This energy is taken from other tissues such as muscles causing these tissues to also fail

331
Q

What are the hallmarks of cancer?

A
  • Sustaining proliferative signaling
  • Genome instability and mutation
  • Resisting cell death
  • Activating invasion and metastasis
  • Evading growth suppressors
  • Avoiding immune destruction
  • Enabling replicative immortality
  • Inducing angiogenesis
332
Q

How can cancer and normal cells easily be distinguished between?

A

Chromatin organisation is severely affected in cancer. This can easily be distinguished using stained tissue samples.
Microscopic examination of nuclear architecture by a pathologist remains a gold standard in cancer diagnostics

333
Q

What are the differences between cancer and normal cells that can be seen through chromosomal staining?

A
  • Size of the nucleus
  • Nuclear outline
  • Hyperchromatic chromatin
  • High nuclear/cytoplasmic ratio (larger nucleus than usual
  • All these are suggestive of major chromatin alterations and an epigenetic component.
334
Q

How did Feinberg & Vogelstein, 1983 investigate the changes in DNA methylation in cancer cells?

A
  • Restricted genomic DNA with restriction enzymes that were sensitive to DNA methylation (HpaII or MSPI). Do not cut when DNA methylation is present
  • Stained with ethidium bromide and ran on a gel
  • Cancer cells resulted in lighter weight products than normal cells meaning restriction enzymes cut the DNA in more places due to less methylation
  • Looked at specific genes. larger products in normal cells (not restricted by enzymes) but smaller in cancer cells showing hypomethylation
  • Then investigated three specific genes and in all cancers they were hypermethylated instead and even more when metastatic.
335
Q

What patterns of DNA methylation are seen in cancers?

A
  • Hypermethylation of CpG islands in the 5’ regions of some genes
  • Global genome hypomethylation also occurs (40-60% of CpG methylation instead of normal 80%).
336
Q

What does CDKN2A encode?

A

Two products from alternative splicing
P16ink4A
P14ARF

337
Q

How can changes in P16ink4A expression lead to cancer progression?

A

It is a cyclin dependant kinase inhibitor that keeps Rb protein in its active state. Its inactivation leads to uncontrolled cell cycle progression (evading growth suppression).

338
Q

How can changes in P14ARF expression lead to cancer progression?

A

p14ARF initiates p53-dependant apoptosis allowing cancer cells to resist cell death if mutated

339
Q

Give evidence for CDKN2A methylation causing cancer

A

Tonnessen & Haass, 2015

  • Looked at different samples of carcinomas and normal DNA
  • In normal cells, there is little methylation of CDKN2A promotor but in cancer there are pretty much no unmethylated sites left
  • DNA methylation inhibits gene expression. In cancer cells the proteins are not expressed
  • When there is no P16, there is no inhibition so CDK4 and cyclin D will be active and Rb will be inactive meaning they evade cell cycle arrest
  • P14 inactivation results in the inhibition of p53 meaning there is no apoptosis
340
Q

What does hypermethylation of BRCA2 cause?

A

Breast cancer

341
Q

What was the first paper showing how breast cancer has a genetic component?

A

Smithers, Br J Cancer, 1948

  • First paper to show genetic component in cancer
  • They looked at different female patients with breast cancer and built a genetic inheritance tree and saw a strong genetic linkage between cases
  • DNA structure had not been discovered so unsure how
342
Q

What is the current model of BRCA1 function?

A
  • It acts at multiple regulatory pathways
  • When BRCA1 is hypermethylated (not functional) the normal DNA damage response pathway cannot be activated. This means that rearrangements cannot be repaired. This is the source of additional mutations – genomic instability
  • Also stops chromatin remodelling causing changes in epigenetic code affecting histone acetylation. DNA methylation and histone acetylation both are involved in cancer promotion
343
Q

What is CDH1?

A
  • Hypermethylated in all carcinomas
  • Encodes a protein called E-cadherin which is the key protein in cell-cell adhesion
  • It is a transmembrane protein which physically binds to E-cadherin on the neighbouring cell. It acts with the cytoskeleton inside the cell
344
Q

How can change in CDH1 expression cause cancer progression?

A

In humans, loss of E-cadherin is linked to numerous types of malignant tumors.
- This is because for a cell to escape from the tumour and invade the rest of the body it needs to detach from the primary tumour. To do this it needs to lose its cell adhesion – activates invasion and metastasis

345
Q

Give evidence for CDH1 methylation causing cancer

A

Yoshiura et al., 1995

  • Used different restriction sites
  • Different cancer samples were treated with different enzyme combinations. E-cadherin positive carcinoma lines have no products visible – cut in multiple places but in E-cadherin negative lines there are producers meaning restriction enzymes could not cut it. This shows that when E-cadherin is not present there is much hypermethylation in the E-cadherin promotor
346
Q

Why is most the genome hypomethylated in cancer cells?

A

This is due to transposable elements

347
Q

What are transposable elements/

A
  • Sequences that can move to a different place in the genome

- Up to 50% of the human genome is composed of transposable elements

348
Q

What type of transposable element is important in cancers?

A
TEs are classified based on their transposition mechanisms, organization, and distribution. 
- One class of TEs is particularly important in cancers: LINEs (long interspersed nuclear elements.
349
Q

Describe LINEs

A
  • LINEs retain their ability as active mobile elements (although many lost it due to truncations).
  • LINE1 sequence represents 17% of human genome showing how widely LINEs are represented in the whole genome
  • Normally LINEs are highly methylated stopping the expression of enzymes required for their movement.
350
Q

What happens when LINEs are hypomethylated?

A
  • Transcription and produce short RNA sequences which are complementary to normal RNA and act as microRNA so degrades of normally expressed genes.
  • Produce transposases allowing excision of the sequence which causes a double stranded break. Mutations/methylation in genes like BRCA1 stop the repair of these breaks resulting in genome instability - new mutations, promote cancer development
  • Transposable elements can then insert into the genome and change expression of these genes
351
Q

What causes the changes in DNA methylation in cancer cells?

A

IDH1/2

  • They modulate the products required for enzymes involved in methylation
  • It loses their catalytic function which usually acts as an antagonist to histone demethylases and DNA hydroyxlases which in turn affects the methylation profile of the genome
  • If their activity is altered then it will later the DNA methylation efficiency and will result in whole genome changes in DNA methylation patterns. This changes gene expression and promotes tumorigenesis
352
Q

How does histone modifications affect cancer prognosis?

A

Elsheikh et al., 2009

  • Looked at wide range of histone modifications and different samples
  • They found for each histone modification, different patients had different alterations. Some high and some low levels. There was some correlation with the severity of disease and prognosis. Stronger alterations of histone modification profile lead to more severe cancer. This cannot be said for all cancers and patients
353
Q

Why does most of our knowledge on polycomb complexes come from drosophila?

A
  • In drosophila, the system is relatively simple with only one gene homologue for each
  • Repressive complexes in vertebrates are much more varied
354
Q

Is PRC1 or 2 biologically more complex?

A

PRC1
- There are at least 100 different forms of PRC1 could functionally exist in each cell at a given developmental stage. The specific functional properties and regulation of these complexes are poorly understood.

355
Q

Describe the structure of the PRC2 core

A
  • EZH1 and EZH2 are mutually exclusive (PCR2 only has one type)
  • They methylate H3K27 and require stable associative with Eed and Suz12 and recruit RBBP4 or RBBP7. Finally, it recruits some proteins such as AEB2, JARID2 and one of PCL1,2 or 3.
356
Q

Describe the structure of the PRC1 core?

A

The core contains histone ubiquitinates: RING1A and RING1B which are mutually exclusive. They require a stable association with one of the polycomb group ring finger (PCGF) proteins. There are 6 different PCGF proteins. In case of PCG2/4 they further recruits either CBX (90% in ESC) or RYBP, which are mutually exclusive, and further components.

357
Q

What is the phenotype of EzH2 knockout mice?

A

Knockout of EZH2 in mice results in death shortly after implantation

358
Q

What is the phenotype of PRC2 knockout mice?

A

Knockout in PCR2 results in an increase in developmental gene expression, differentiation defects and premature lineage specification but this does not occur correctly as it is not accompanied by silenced of pluripotent genes. This means that pluripotency is not efficiently turned off and differentiation occurs prematurely causing a loss of balance between differentiation and proliferation

359
Q

How is PRC2 expression related to cancer progression

A

Both over expression and inactivation lead to poor prognosis

  • Low EZH2 expression in differentiated cells and over expressed in invasive breast carcinomas. This has also been shown in prostate cancer. It is linked to poor prognosis and more metastasis
  • Neoplasms that have PRC2 mutation resulted in a poorer prognosis. This was seen in different mutations in the PRC2 including copy number alterations and uniparental disomy
360
Q

What type of PRC2 mutations are widespread in cancer?

A

Both gain and loss of function alterations in genes encoding core components of PRC2 are widespread in cancer

361
Q

How does PRC2 have an oncogenic role?

A

Nakagawa et al, 2013
- Can promote cell cycle by inhibiting P16 (gene hypermethylated in cancers). Its inactivation helps the cells to uncontrollable proliferate
Cao et al, 2008
- Inhibits E-cadherin promoting metastasis and invasion
- EZH2 expression is mutually exclusive with E-cadherin. Can’t say which is cause and which is consequence. However, then mutated EZH2 and showed that is directly involved in mutating E-cadherin expression

362
Q

Outline notch signalling

A
  • NOTCH receptor bind to different ligands such as delta – transmembrane proteins. To sense to the signalling, it has to come from adjacent cell
  • Binding releases of notch intracellular domain and enters the nucleus and act as a TF to repress or activate target genes
363
Q

How is T-cell acute lymphoblastic leukaemia linked to Notch signalling?

A

T-cell acute lymphoblastic leukaemia is a hematologic malignancy characterized by activating mutations in NOTCH1 gene. NOTCH doesn’t need to bind to its ligand to act meaning the signalling is always active

364
Q

In a hyper activated notch pathway, what is the histone methylation profile?

A
  • Associated with EzH2 reduction

- Histone H3K27 methylation was depleted are the promotor region

365
Q

How does PRC2 act as a tumour suppressor gene?

A
  • Notch1 constitutively activation leads to the loss of histone H3K27 methylation by eviction of PRC2
  • The consequence of this is that the activity of notch signalling is even further elevated – positive feedback. Promotes its own function
  • If PRC2 is overexpressed it cannot be evicted by notch meaning PRC2 would not be effected and notch cannot upregulate itself
  • PRC2 therefore inhibits notch overexpression and prevents cancer progression
366
Q

What outcomes can changes in PRC2 expression lead to?

A

Different underlying mechanisms of PRC2 changes can result in different outcomes for the cancer: can either promote or inhibit the progression

367
Q

Why is PRC2 a good therapeutic target for cancer?

A

PRC2 changes are very frequent in cancer making it a good target. It is an enzyme so easier to find an inhibitor to modify its activity

368
Q

Give an example of the targeting of PRC2 expression as a treatment for cancer

A

Knutson et al, 2012

  • Mutated forms of EZH2 hypersensitive to a set of small molecules. These act as selective inhibitors of mutant EZH2 with little effect of WT.
  • Mutations in EZH2 are associated with malignant non-Hodgkin’s Lymphoma
  • They are hyperactive and only catalyse trimethylation of H3K27 (not able to cause mono- or di-methylation, unlike wild-type EZH2)
  • Small molecule EPZ005687 can specifically block H3K27 trimethylation in a dose dependant manner.
  • This is a cancer specific inhibitor of proliferation. This therefore will not have the lethal effects on normal cells
369
Q

Why is there a need for personalised medicine when treating cancer?

A
  • Inhibiting EzH2 would only help when PRC2 is overexpressed (oncogene).
  • It is difficult to know how patients will react to the different treatments as the cancers appear the same but have different underlying mechanisms causing it.
  • Using these inhibitors when PRC2 is acting as a tumour suppressor (notch1 overexpression) could make the cancer worse – need for personalised medicine.
370
Q

Where do embryonic stem cells arise?

A
  • First arise in the inner cell mass of the blastocyst in a preimplantation embryo.
  • The fertilised egg divides a few times to form a morula and the inner cell mass forms. This takes place at around embryonic day 2.5-3.5
  • . Post implantation (5.5 days) results in the formation of the epiblast. The ES cells persist into the post implantation embryo
371
Q

Outline the pattern of methylation of the genome of a developing embryo

A

As preimplantation embryo is created and develops, methylation is decreasing. Once is becomes the epiblast (post implantation), remethylation occurs. This demethylation wave in the preimplantation embryo is erasing the epigenetic imprints that the sperm and egg has

372
Q

What is a DMR?

A

Differentially methylation sites

- Present in the egg and sperm and are demethylated in the early embryo

373
Q

How does the methylation contribution of the egg and sperm differ?

A
  • The methylation regions of the genome in the egg become demethylated (50% of them)
  • In the sperm epigenome, the sperm specific methylation sites become demethylated in early embryogenesis and upon implantation methylation of these sites increases.
  • The embryo is requiring the set of methylation marks and reapplying them to the whole genome (sperm and egg).
  • This is an indicator that the methylonome is dynamically reprogrammed
374
Q

What enzymes are required for de novo DNA methylation in the embryo?

A

Two enzymes that will take an unmethylated CpG and de novo methylate it: DNMT3A and DNMT3B

375
Q

How is DNA methylation maintained across cell divisions?

A

Upon division, that methyl mark needs to be maintained. This maintenance methylation is carried out by the enzyme DNMT1 which recognises hemi-methylated DNA (only one strand is methylated from the parent cell).

376
Q

What would occur in the absence of maintenance methylation enzymes?

A

In the absence of maintenance methylation there would be progressive loss of methylation in the genome passively. Originally thought that the progressive loss of methylation in the embryo was passive but was active

377
Q

How can DNA methylation be removed actively?

A

DNA methylation can also be removed by TET enzymes in a multi-step process

  • Forms hydroxyl methylcytosine (sign of active demethylation)
  • Leads to full removal of the nucleotide to create a gap in the DNA
  • Repaired by the base excision repair mechanism
  • Is then repaired to unmethylated cytosine
378
Q

What are the two ways ES cells can be maintained in culture?

A

These two cultures are known as 2i and serum culture

379
Q

What are the two types of ES cells?

A
Naïve 
- Preimplantation embryo
- Culture indefinitely 
- Only these when reintroduced into host blastocyst (after culture) can function as full ES cells
Primed
- Post implantation embryo
- Reduced potency in comparison to naïve 
- Culture indefinitely
380
Q

What is 2i culturing?

A

Maintaining ES stem cells in the presence of FGF and Wnt inhibitors
- Makes naive ES cells

381
Q

How did Ficz et al, 2013 show the need for TETs in demethylation?

A
  • ES cells are maintained in 2i, 5methylcyytosine decreases at different CpG sites while hydroxymethylcytosine increases and then decreases (intermediate).
  • Cells that carry a mutation in TET and Wt were cultured in 2i and investigated the methylation of the CpG sites.
  • Wt methylation decreases and hydroxymethylcytosine increases but in the TET mutant most of the methylation is preserved and there is less hydroxymethylcytosine increases.
  • At a different CpG, the TET mutant cells can still be demethylated. Shows that TET is not always required for demethylation (passive)
  • There is also a reduction in the de novo methylases
382
Q

What genes are required for pluripotency?

A

Oct4, Nanog and Sox2

383
Q

How was it shown that the pluripotency genes were required?

A

Mitsui et al, 2003

  • Nanog and Oct4 knockout cells
  • Both begin to spontaneously differentiate into endoder and knockout embryos cannot form blastocysts
  • Knockout Sox2 and there is an increase in trophoectoderm markers (extraembryonic tissue) and a decrease in pluripotency markers
  • Therefore these three genes are essential in maintaining pluripotency
384
Q

How do the pluripotency factors work?

A

They work by binding to cis regulatory sequences in pluripotency maintaing genes and recruit enzyme histone acetyltransferase P300 which targets histone H3K27. P300 has a histone acetyltransferase domain to carry out histone acetylation

385
Q

How does P300 activate the gene expression of pluripotency genes?

A

Has a bromodomain so can recognise the acetylated hsitone marks. This allows P300 to both bind to acetylation marks and then make more in the vacinity. These genes also contain H3K4 methylation domains. Recruits methylation to sustain their expression.

386
Q

How does the epigenome of differentiated and ES cells differ?

A
  • ES have a bivalent pattern of modification. Contains both methylated lysine 4 and lysine 27 which have distinct functional characteristics. Lysine 4 is activating and lysine 27 is repressive
  • In differentiated cells some retain the lysine 4 or the lysine 27 methylation. They make the decision to remove one of the methylation marks allowing either transcription or repression. It loses bivalency.
  • Genes required for a specific lineage maintain their lysine 4 mark and remove their lysine 27 mark.
387
Q

What is meant by ES cells being in a poised state?

A
  • The vast majority of promoters are enriched in H3K27 and H3K4 methylation. They are bivalent and are all inactive genes but have an association with RNA polymerase II so are ready to differentiate (poised)
388
Q

Give evidence for H3K4 methylation requiring trithorax function

A

H3K4 methylation requires the activity of the Trithorax protein Mll2

  • In Wt ES cells there is a strong peak of H3K4 methylation in various genes
  • Knockout of Mll2 then the level of methylation has been completely abolished
  • However, H3K27 methylation is unaffected by Mll2 knockouts
389
Q

Give evidence for H3K27 methylation requiring polycomb function

A

H3K37 require polycomb function

  • Knockout components of the Polycomb complexes
  • Knockout of EED (binds to EzH2), or EzH2 then lose tri and di methylation of H3K27
  • Expression of pioneer transcription factors (nanog ect) in knockouts is gone. Polycomb function is required for pluripotency
  • Loss of PRC2 function causes spontaneous differentiation of ESCs into meso-endodermal tissue and lose pluripotency
390
Q

How do ES cells have the bivalent pattern of histone modification?

A

Polycomb and Trithorax Group proteins are functionally antagonistic

  • Come together on bivalent genes to create the state of poised transcriptional inactivity
  • Induction of differentiation disturbs this balance and breaks the transcriptional silence to allow for the removal of H3K27 marks and differentiation into a specific lineage.
391
Q

What are the functions of the brain?

A
  • Receives, stores, processes and integrates information from environmental sources and the rest of the body.
  • Transmits information to different parts of the body and regulates the body’s behavioural interactions with factors in the environment.
  • Adapts its own functions and enables other parts of the body to adapt in the light of information received, to improve physiological performance and optimise behaviour.
392
Q

What is long term potentiation and depression?

A
  • Neural activity itself elicits changes in synaptic strength. This can be increases or decreases
  • The changes are normally due to changes in neurotransmitter receptors number and modulation
  • Long term potentiation is an increase in synaptic strength but long term depression is a decrease
393
Q

What mechanisms cause an increased number of glutamate receptors in long term potentiation?

A
  • There are stores of glutamate receptors already synthesised in the neuron ready to move to the membrane
  • De novo synthesis of gene products also occurs as a result of induced gene transcription.
394
Q

How does de novo synthesis of glutamate receptors occur?

A
  • A critical transcription factor that responds to neural activity is CREB
  • CREB responds to Ca2+ influx triggered by glutamate binding. This causes its phosphorylation by CaMKII. It moves into the nucleus and drives the expression of target genes
395
Q

How did Lin et al, 2008 show that neuronal activity can stimulate gene expression?

A

Lin et al, 2008

  • Culture neurons of a mouse brain
  • Quiescent in culture but will spontaneously depolarisation upon addition of KCl. Leads to increase in phosphorylated CREB and CREB target genes expression e.g. Npas4 and C-fos
  • Shown in a western blot – occurs quickly upon addition of KCl and can see phosphorylated CREB
  • Can be induced by a variety of stimuli e.g. neurotrophins
396
Q

How did Ghosh et al, 1994 show that activity facilitates neurones remodelling?

A

Ghosh et al, 1994

  • Expand neurons in culture and treat with KCl. They will depolarise and interact with each other
  • Add an antibody that blocks BDNF then the synaptic interconnections fail to develop
  • The role of activity dependence is therefore to facilitate remodelling
397
Q

What is contextual fear conditioning?

A
  • Animals can be transferred from home cage to new cage which contains electrodes that give them an electric shock. The electric shock will then be associated with the unusual features of the cage e.g. if it had red lines on the bottom of the cage
  • If repeat this a few times and return the animal to the home cage and then reintroduce to the cage but without the shock, the mouse will remember it.
  • This can be measured by a behavioural shock response (will freeze in response to an electric shock). Being in the cage alone is enough to cause the freezing of the mouse. This is the formation of a memory
398
Q

What are the three stages of contextual fear conditioning?

A

Acquisition
- Associating the shock to the environment
Consolidation
- Reflecting in home environment
Retrieval
- The behavioural response without the shock

399
Q

What did Miller and Sweatt, 2007 investigate?

A

Are there any epigenetic changes in animals with the contextual fear conditioning?

400
Q

How did Miller and Sweatt, 2007 discover the need for DNA methyl transferases for memory formation?

A
  • Animals in which the conditioning had taken place showed an increase in the de novo methyl transferase DNT3A.
  • If treated with DNA methyltransferase inhibitor, then the ability of the animal to remember the shock was significantly reduced.
  • This implied that the DNA methyl transferase was required for full recall of the negative experience
401
Q

What target genes did Miller and Sweatt, 2007 find that DNA methyl transferases targeted?

A

Protein Phosphatase 1 (PP1)

  • Added a DNA methyltransferase inhibitor and caused PP1 gene expression increase.
  • This gene is a memory suppressor gene
  • Proposed that DNA methyl transferases are involved in the acquisition of memory by inhibiting the expression of memory suppressor genes
402
Q

Another target gene identified by Miller and Sweatt, 2007 was CaN, what is its role?

A

Involved in regulating the methylation of CREB

403
Q

What did Miller and Sweatt, 2007 see about CaN expression in mice who had contextual fear conditioning?

A

Over time can see that methylation of the CaN promoter is increased as the context and shock become co-administered. The level of transcription of CaN decreased. Therefore, CaN expression limits the recall of memory and it is blocked upon memory formation

404
Q

What did Rudenko et al, 2013 show about TET1?

A
  • Mutated TET1 in mice and developed with abnormal levels of methylation in synaptic activity regulated genes. This is seen in both the cortex and the hippocampus and is not seen in the Wt (only see small scattered methylation)
  • This is the case for a number of genes
  • Hypermethylation resulted in a reduction of transcription in the cortex and hippocampus in comparison to Wt animals
  • This is as a result of the loss of the active demethylase activity
405
Q

In Rudenko et al, 2013, what did contextual fear conditioning of TET1 mutants show?

A
  • The ability to form the memories is not impaired. Still froze in response to context without the shock
  • Put animals back into home cage
  • Wt animals dissociate the fear and the memory becomes extinguished overtime but TET1 knockout mice have persistently high levels of recall and are unable to extinguish the memory. They cannot demethylate and cannot lose the memory
  • This implies that the loss of methylation of these genes is associated with the loss of memory and the methylation patterns are associated with the retention of memories
406
Q

How is it through that DNA methylation affects memory formation?

A

Methylation represses the expression of memory suppressor genes
Failure to remove methylation results in failure to extinguish the memory

407
Q

What affect does histone acetylation have on memory formation?

A

Guan et al, 2009

  • Overexpression of histone deacetylate 2 results in a reduction in the power of recall. HDAC2 knockout enhance recall.
  • There is no affect with histone deacetylase 1. Suggests that HDAC2 suppresses conditioned memory so histone acetylation is required for memory formation
  • Conditioned fear also resulted in a change in neuronal dendritic spine formation. HDAC2 represses this change.
408
Q

What did ChiP analysis in Guan et al, 2009 show about the regulation of neuronal activity genes?

A
  • Found that the synaptic regulating genes such as neuronal activity genes, genes involved in synaptic formation are associated with HDAC2 protein. They all bind HDAC2 showing that HDAC2 is actively on the loci and antagonise acetylation
  • In HDAC2 knockout, there is an increase in acetyl marks on the activity regulated genes
409
Q

How did Guan et al, 2009 use a HDAC inhibitor to the role of HDAC2 in memory formation?

A
  • Administering the HDAC inhibitor SAHA increases the ability to recall the memory.
  • Overexpression of HDAC2 depressed the power of recall but SAHA addition rescues recall.
  • Can look at dendritic formation in the presence of SAHA. SAHA increases the numbers of dendrites in Wt and restores the numbers of dendrites when HDAC2 is overexpressed
410
Q

What is CBP?

A

CREB binding protein

CBP binds to CREB and locally acetylates target genes

411
Q

What disease do humans who are heterozygous for CBP mutants have?

A

A severe developmental abnormality called Rubinstein-Taybi syndrome

412
Q

What is the function of CBP?

A

It regulated the function of CREB and potentiates its activity.
CBP mutants show reduction in H2B acetylation.
The heterozygous animals show a reduced capacity to retain the memories

413
Q

What did ChiP analysis of C-fos gene (target of CREB) show about the action of CBP?

A
  • Enhancer elements upstream and downstream of the C-fos gene and the gene body itself becomes hyper acetylated. At the same positions, there is the accumulation of histone H3K4 methylation (made by Trithorax genes) but this is not activity induced. CBP is also localised to the enhancer elements
  • DNAse1 hypersensitive sits correspond to the enhancer elements that are activity modified
  • All activity regulated genes respond in the same way to KCl depolarisation culture by hyper acetylating regulator elements by recruiting CBP
414
Q

What epigenetic processes govern memory formation?

A
  • DNA methylation promotes memory formation
  • TET1 demethylases and promotes memory extension
  • Histone acetylation by CBP is required for memory formation
  • Histone deacetylation by HDAC2 inhibits memory formation
415
Q

Do the epigenetic processes that govern memory formation act together or separately?

A

Data suggests that there are some overlaps in gene targets but does not appear to have strong antagonist functions. Thought that they have evolved to work together in some unknown way

416
Q

Define sensitivity in terms of addiction

A

Inability to tolerate the adverse effects of a drug: start point

417
Q

Define tolerance in terms of addiction

A

Reduction in sensitivity when a drug is used repeatedly over time (higher amount is required to achieve the same level of response and adverse effects)

418
Q

How do different drugs exert their affects?

A
  • Opiates - µ opiate receptors
  • Nicotine - acetylcholine receptors
  • Cannabis - cannabinoid receptors
  • Alcohol - several; major being GABA (gamma-aminobutyric) and NMDA (N-methyl-d-aspartate) receptors
  • They then act through a secondarily system through dopamine reward system
419
Q

How does CAMP pathway exert its action?

A

Leads to the activation of PKA. IT can then translocate into the nucleus and phosphorylate CREB and bind to co factor CBP. Together they activate transcription of target genes

420
Q

What system did Korzus et al, 2004 use to look at the role of CDP in memory consolidation?

A
  • They use binary transgenic system which has two transgenes inserted into the genome. One encodes a tetracycline transactivator protein (tTA) which is under the regulation of CamKII promotor – in the frontal neurones. The second transgene has a tTA responsive promotor and the encoding region of CBP (wt or mutated)
  • Normally, in cells that express CamKII, tTA is expressed and will drive the expression of CBP protein in frontal neurons. If DOX is added then it can bind and prevent the function of tTA so cannot drive the expression of CBP.
421
Q

What did Korzus et al, 2004 find abut the role of CBP in memory consolidation?

A

Mutant CBP mice had impaired memory conservation compared to wt

422
Q

How did Pandey et al, 2008 investigate the anxiety of mice?

A

Used open maze test to check the anxiety of the mice. Compared the open branches of the maze and closed. More anxious mice hide in the closed branches

423
Q

What did acute injection of alcohol lead to in Pandey et al, 2008?

A
  • Increased the time mice spent in the open branches of the maze
  • Increased density of the dendritic spines in the amygdala
  • Amount of phospho CREB in the nuclei increased. At the same time, the expression of BDNF (downstream target) increased at mRNA and protein level
  • Increased acetylated histone 2 by CBP and also 3 and 4
  • Reduces HDAC activity
424
Q

Outline what happens in response to acute alcohol?

A

Acute exposure to ethanol activates CAMP pathway and activates CBP and inhibits HDAC. This leads to massive changes in histone acetylation and changes the expression of target genes such as DBNF and NPY (responsible for behavioural changes)

425
Q

What did chronic injection of alcohol lead to in Pandey et al, 2008?

A
  • Dendritic spine density was unaffected
  • Acetylation of histone H3 and CBP expression were the same as the control group
  • BDNF expression is not distinguishable from the control
426
Q

Why did chronic injection of alcohol have no affect in Pandey et al, 2008?

A

There has been a neuro adaptation which restores the density of the dendritic spine to normal

427
Q

What happened in the withdrawal group (after chronic injection) in Pandey et al, 2008?

A
  • The dendritic density drops significantly below the control
  • Acetylation of histone H3 and CBP expression were reduced
  • BDNF significantly reduced
428
Q

Outline the model of chromatin alterations during alcohol addiction

A
  • Acute exposure to alcohol opens chromatin to exposure from CBP and HDACs changing expression of proteins involved in behaviour and memory formation e.g. BDNF and NPY
  • Chronic exposure and the expression of proteins and chromatin structure is the same as normal due to neuroadaption
  • In the withdrawal, the chromatin are more impacted and reduced the chromatin acetylation and protein exposure. This results in behavioural changes when in withdrawal period and craves more drugs to open the chromatin up again to return to normal.
429
Q

What other epigenetic changes occur in response to alcohol exposure?

A

Epigenetic changes in the liver
- Decreased the levels of H3K9me2 and increased H3K4me2 giving an open chromatin state and increases alcohol dehydrogenase gene
Cortical neurons
- Decrease in site-specific DNA methylation in cortical neurons, which persisted following ethanol withdrawal: Could be why addicts often relapse

430
Q

How does the epigenetic changes in response to alcohol give a potential link to cancer?

A

Varela-Rey et al., 2013

  • Chronic ethanol consumption: abnormal regulation of the methylation pathway by inhibiting methionine adenosyl transferase, the enzyme that converts methionine to SAM, which leads to global DNA hypomethylation
  • Chronic alcohol abuse is linked to hypomethylation of DNA. This could link to cancers as hypomethylation of the genome is associated with genome instability
431
Q

What did Kumar et al, 2005 show about the effect of acute cocaine exposure?

A
  • Inject cocaine into mouse model and looked for acetylation on specific gene promotors – not genome wide: BDNF, C-fos and FosB
  • Half an hour after injection, there was massive increases in histone H4 acetylation at these genes promoters and decrease after 3 hours
  • Acute treatment affects acetylation of histone H4 but not 3
432
Q

What did Kumar et al, 2005 show about the effect of chronic cocaine exposure?

A
  • Compared acetylation of histone H3 and 4 at specific gene promoters between control and chronic cocaine exposure
  • Mice could self-administer cocaine using a lever (behavioural trait to seek the drug)
  • Large increase in acetylation of Histone H3 and BDNF and FosB but not C-fos. There were no changes in acetylation of histone H4.
  • Chronic exposure partially reverts the phenotype of acute exposure but there are site specific changes in acetylation of histone H3
  • Don’t fully understand the mechanism
433
Q

How does chronic cocaine exposure effect HDAC5?

A
  • HDAC5 was hyper phosphorylated but returns after 24 hours
  • Normally, HDAC5 is localised to the nucleus colocalised with DAPI staining.
  • After 30 minutes after chronic exposure, HDAC5 was found in the cytoplasm – leads to extrusion from the nucleus meaning it cannot deacetylase Histone H3 which is why it is hyper acetylated after chronic exposure
  • Found that the same changes to HDAC5 happened after exposure to chronic stress – could possibly be modulated by glucocorticoid signalling
434
Q

How does chronic cocaine exposure effect CREB?

A
  • Chronic exposure but not acute increases the amount of phosphorylated CREB
  • This increase is associated with expression of two miRNAs: Mir-212 and 132. There is positive feedback loop between these and CREB. CREB upregulates their expression and the miRNAs increase phosphorylation of CREB
435
Q

What are the differences and similarities between alcohol and cocaine exposure?

A

Similarities
- Both act through cAMP signaling, CREB, and histone acetylation;
- Both induce multiple additional epigenetic changes including in DNA and histone methylation
Differences
- Chronic ethanol exposure neuroadaptation fully restores histone acetylation and leads to withdrawal but chronic cocaine does not alter histone acetylation back

436
Q

How did Pandley et al, 2008 suggest a treatment for withdrawal from alcohol?

A
  • TSA is a HDAC inhibitor
  • Withdrawal reduces histone acetylation and ethanol inhibits HDAC.
  • Mice were also injected with TSA (as well as ethanol and control)
  • This was completely restored histone acetylation in withdrawal but didn’t affect the CBP protein
  • Used Open maze test and during the withdrawal, mice were hiding and didn’t want to be in the open maze but this was restored by TSA injection
437
Q

How did Sakharkar et al, 2014 further consolidate TSA as a treatment for withdrawal from alcohol?

A
  • Two inbred mice strains: one prefers ethanol and one doesn’t
  • They could self-apply ethanol – choose between water and ethanol
  • Three days and then injected with saline or TSA.
  • If the mice were injected with saline they continued to drink ethanol but if injected with TSA then their preference for the ethanol significantly dropped
438
Q

How did Dunbar and Taylor, 2017 suggest a treatment for withdrawal from cocoaine?

A
  • Increase in histone acetylation with cocaine so inhibition of histone acetyl transferases could be a treatment
  • Used Garcinol which is a non-specific HAT inhibitor
  • Rats self-administer cocaine for 10 days and showed conditional stimuli to associate environment. Then enter withdrawal.
  • They then reactivated addiction by single dose of cocaine and showed one of the stimuli they associate with cocaine. Looked if the rats wanted to press the lever if shown the stimuli
  • Injected animals with Garcinol. The seeking behaviour was significantly reduced
439
Q

What are the two types of immunity?

A

Acquired
- Results from the creating of B and T lymphocytes.
- B cells clono-synthesis of immunoglobulins
- T cell receptor are rearranged to produce clonal antigen specific antiforms
Innate
- Myeloid cells like macrophages and dendritic cells
- These cells do not have clonally specific receptors for antigens but instead have pattern recognition receptors which interact with pathogen specific molecules (PAMPs which bind to PRRs)

440
Q

What parts of the immune system can adapt?

A

Both arms of the immune system can be modulated by experience to exhibit adaptive characteristics. Usually thing of acquired as adaptive but now known innate can too. This can also have maladaptive effects

441
Q

What must happen to B cells for them to mature?

A
  • Every B lymphocyte that emerges from the spleen carries an immunoglobulin on its surface. This has got there by a process of development in which a lymphoid progenitor cells has been lineage restricted to a pro-B cell and then undergoes immunoglobulin gene rearrangements
442
Q

What is Productive V(D)J recombination?

A
  • This generates the recombined immunoglobulin gene
  • Together with this, epigenetic and transcript machinery ensure that any given pro-B cell only rearranges one allele of any given immunoglobulin gene in any given cell so that the B cell is specific for one individual antigen
443
Q

What is the structure of an immunoglobulin?

A
  • Made up of two heavy chains and two light chains which are encoded by different genes
  • The polypeptides fold into globular domains which are held together by disulphide bridges
  • They both have variable domains which are created by VDJ recombination and are antigen specific
444
Q

How must the heavy chain of an immunoglobulin be structured?

A

The heavy chain has to be structured so that it has a variable region containing a diversity and joining segment attached to a constant portion

445
Q

How did Chowdhury and Sen, 2001 show that there was a variable region in the immunoglobulin locus which reorganises in pro-B cells?

A
  • When pro-B cells are specified they acquired a domain of DNAse1 hypersensitivity meaning that the chromatin are open.
  • Locus segments give rise to variable components: some constant and some variable (these domains are known as diversity and joining segments)
  • Probe the locus with DNAse1 and can test which regions are open.
  • Can see that in Pro-T cells this does not occur (negative control).
  • There are hypersensitive sites appearing in the pro-B cells in the middle section of the locus.
446
Q

What is the enhancer present in the immunoglobulin locus in pro-B cells?

A

Mu enhancer
- When reporter genes are transfected into B cells can see that it is transcribed meaning that this enhancers activate B cell specific genes

447
Q

Give evidence for the action of the Mu enhancer?

A

Malin et al, 2010

  • There are many regions on the locus which have variability and extends for several Kb- largest part of the locus
  • Using sequence based ChiP across the region to look at lysine acetylation and methylation. Can see there is low levels of acetylation and H3K4 methylation and high H3K9 acetylation. As approach the Mu enhancer there is a big peak in acetylation and H3K4 methylation
448
Q

What enzyme is required for Productive V(D)J recombination?

A

RAG1/RAG2 recombinase complex

449
Q

What are the functions of the enzymes in the RAG1/RAG2 recombinase complex?

A
  • RAG1 is the portion of the recombinase that binds to the recombination signal sequences and then cleaves it.
  • This can only occur if it is bound to RAG2 which is the key agent that recruits the complex to the locus due to its PHD finger that binds H3K4 methylatio
450
Q

Give evidence for the role of RAG2?

A

Liu et al, 2017

  • Immunoprecipitation from pro-B cells what antibody for H3K4 trimethylation and blotting with an antibody for RAG2. H3K4 methylation also contains RAG2
  • In vitro experiment which contains columns containing recombinant histones and pour on radiolabelled RAG2, it only binds to columns using H3K4 methylated
  • Rag2 is therefore the recruiting agent for this complex
451
Q

What does the RAG1/RAG2 recombinase complex bind to?

A

Recombination signal sequences
- Immediately either side of each D and J segment there are repeat structures which the RAG recombinases bind to cut and region in order to create the subsequent fusions.

452
Q

How does RAG1/RAG2 recombinase complex produce a the DJ fusion?

A
  • RAG1/RAG2 bind to the recombination signal sequences next to corresponding D and J region.
  • This creates a U-shaped structure and the RAG1 enzymes cleave the DNA to release the portion to create a circular recombination product which is discarded.
  • Left with ligation between the D and J portion which produces the fusion junction
453
Q

What is required for the DJ region of the immuglobulin locus to fuse with the variable region?

A

To do this it needs to bring the variable portions (which could be kb away) into proximity to the DJ region. The distal variable segments are moved towards the proximal region through locus contraction

454
Q

Give evidence for locus contraction?

A

3C was used to visualise the domains of chromatin
- In Pro-B cells when recombination has begun you can see that there is a large area of chromatin interaction
- Therefore, the variable domains are brought into proximity to the DJ fusion
Fluorescent in situ hybridisation
- Specific oligonucleotides specific for variable domains, diversity segments and joining segments and labelled with fluorescence
- Can see that all of these signals are all colocalised in the Pro-B cells but not in the non-B cells: these are all distinct because they are not condensed

455
Q

What promotes locus contraction?

A

Pax5

  • Contains trans activator and trans repressor domains.
  • It represses myeloid genes and promotes transcription of B lymphocytes specific genes. It also promotes locus contraction
456
Q

Give evidence for the role of Pax5 in locus contraction?

A

Wt B cells have colocalisation between variable and diverse segments. In mutant Pax5, this colocalisation is lost. This shows that Pax5 is required for locus contraction to cause the chromatin to overlap with each other

457
Q

What else is required for locus contraction?

A
  • Pax5 binds to a series of DNA sequences in the V clusters (15 binding sites at uniform intervals).
  • At the Pax5 binding site there is lots of Pax5 and CTCF. This is a loop forming structure that sits at the base of chromatin loops and organises the 3D structure. It is involved in promoting loop formation involved in the locus contraction.
  • Regions also rich in H3K4 methylation and H3K9 acetylation. If remove Pax5 function then you lose the methylation.
458
Q

How does trained immunity occur in the innate immune system?

A
  • Upon second infection, the immune response is potentiated.
  • This is dependent on the epigenetic reprogramming of the immune cells that are responsible for the immune response e.g. macrophages.
  • It produces more pattern recognition receptors so that the signals are stronger when transduced
  • This is a conditioning process and not necessarily pathogen specific. One pathogen can potentiate the response to other pathogens.
459
Q

Give examples of how trained immunity is not pathogen specific

A
  • Immunisation of a fungal pathogen and can protect against a bacterial infection
  • Oligonucleotides containing unmethylated CpGs (TLR9 ligands) protects against-E. coli-induced sepsis/meningitis
  • Immunization of mice with BCG vaccine (against Mycobacterium tuberculosis) protects against fungal protozoan infections and can now even be used as therapy against bladder cancer
460
Q

Give evidence for how trained immunity is not pathogen specific

A

Kleinnijenhuis et al, 2014

  • SCID mice lack B and T cells so depend upon innate immunity
  • Immunise half with BCG vaccine and half with nothing and expose to a pathogen like a fungus. Ones when have been immunised will survive
  • This is immune training. This party works through strong induction of cytokines such as TNF-alpha
461
Q

How did Foster et al, 2007 discover the genes involved in trained immunity?

A
  • Animals were immunised with bacterial cell wall component (LPS). There were four different groups: either a couple of hours or 24 hours after immunisation, the animals were sacrifice and the macrophages were analysed for gene expression or they were treated with LPS a second time just after collection
  • Look at immediate response to LPS and then see what are active after
462
Q

What classes of genes did Foster et al, 2007 identify as being involved in trained immunity?

A

Class T genes
- Induced by the first exposure to LPS but suppressed by second exposure to LPS
- Rapid response then decays and don’t respond at second time
- There is a rapid increase in H4 acetylation and H3K4 methylation and then rapid decay
Class NT genes
- Induced by first exposure at low levels to LPS and induction enhanced by second exposure to LPS
- These first acquire a small amount of histone modification which is then potentiated - localised with CBP expression

463
Q

What controls CBP expression and histone acetylation of inflammatory genes?

A

CBP is localised to regions of these genes that are associated with macrophages specific TF PU1. This is recruited to the genes independently of exposure to immune challenge and sets up the sites ready for induction

464
Q

How is thought that autoimmune diseases are caused?

A
  • Over expression if inflammatory genes in macrophages is part of the pathological phenotype of people with autoimmune disorders
  • The idea has started to grow that people with these diseases over express their inflammatory programmes because their immune system is over trained
465
Q

How can the idea of trained immunity causing autoimmune diseases provide a therapeutic treatment?

A

Nicodeme et al, 2010

  • I-BET is a small molecule mimic of acetylated lysine and blocks the function of CBP and histone acetylation enzymes.
  • When administered to macrophages, the level of histone acetylation in the inflammatory genes (e.g. TNFalpha) is reduced and the ability of proteins that bind to these acetylated lysine (bromodomains) is also blocked.
  • The presence of acetylation on inflammatory genes is induced by LPS
466
Q

What is a teratogen?

A

A chemical or biological agent to which an embryo or foetus is exposed during development that can cause non-heritable developmental malformations and other birth defects

467
Q

Give examples of teratogens

A
  • Cyclopamine: Impact on Shh signaling
  • Alcohol: causes dose dependent malformations of brain formation
  • Thalidomide: Prescribed to pregnant women and lead to malformations of children. Interfered with vascular development in the limbs
468
Q

What is a carcinogen?

A

A chemical or biological agent that can cause cancer

469
Q

What is a Endocrine Disruptor?

A

An exogenous substance that interferes with any aspect of hormonal action – by disrupting, modifying or mimicking endocrine signals in the body.

470
Q

Give examples of Endocrine Disruptors?

A
  • DDT: used to kill pests and had a devastating ecological impact – bird egg shells thinned so can’t contain embryo. It also increases the risk of breast cancer between 5 and 10-fold. Risk decreases if encountered in later life
  • Oestrogen mimics: synthetic small molecule that mimic natural oestrogen e.g. Bisphenol A – binds to oestrogen receptor. DES also does
471
Q

What is a tobacco smoke?

A

Tobacco Smoke is a teratogen and carcinogen, containing at least 70 compounds with known defect- and disease- causing activities
It substantially increases the risk of birth defects

472
Q

Name a class of disease causing compound in tobacco smoke

A

Polyaromatic compounds

  • e.g. Benzo pyrene, Dioxin (exongenous), there are also endogenous ligands
  • Has an aromatic ring
473
Q

What happens when the body encounters Polyaromatic compounds found in tobacco smoke?

A

Aryl Hydrocarbon Receptor Family (AHR)

  • Ligand binds to AHR and forms a heterodimer with ARNT which assembles on DNA sequences on target genes and promotes transcription
  • Targets cytochrome P450 (specifically CYP1A1) which are involved in metabolising the multi ring structures
  • This turns the Polyaromatic compounds into a powerful mutagen. They are converted to epoxides which are DNA damaging agents and covalently attach to guanine and damage its coding properties
474
Q

What is the role of AhRR?

A

Another member of the receptor family is AhRR which is a repressor of AHR and ARNT
- binds to AHR and stops it from promoting of CYPA1

475
Q

What did Shenker et al, 2015 find about the methylation pattern of AhRR in response to tobacco smoke?

A
  • Did an Epigenome wide association study
  • Reported that there was an association between methylation of the AhRR locus (4 CpGs) and smoking
  • In both alveolar macrophages and white blood cells there is a lower level of methylation in smokers v non-smokers – hypomethylated.
  • This correlated with increasing expression of AhRR.
476
Q

What is the global epigenetic effect of tobacco smoke?

A
  • Between 400 and 500 CpG sites have changes in methylation by smoking and stopping smoking reduces these differences apart from the AhRR sites.
  • The body still remembers of the exposure to tobacco smoke in the AhRR CpGs – even 40 years later
477
Q

Why was it thought that AhRR might be a tumour suppressor?

A
  • AhRR expression is increased in smokers
  • Maybe AhRR is being de-repressed to counteract the production of mutagenic derivatives of tobacco smoke
  • This would stop the activation of CYP1A1 activation which is cancer causing
478
Q

What is the original function of the AHR family?

A

Modulate endogenous chemicals such as tryptophan

479
Q

How does tryptophan act on AHR?

A
  • In day light, the endogenous tryptophan binds to FICZ which is an immune modulator. This binds to AHR. This induces CRYPA1 activation and breaks down the products into non-toxic products.
  • High levels of AhRR bind to AHR and limit the production of CYP1A1 allowing FICZ to accumulate and modulate the immune system
480
Q

Is AhRR a tumour supressor?

A

No

  • The hypomethylation of AhRR doesn’t actually decrease the risk of lung cancer, it increases the risk. It is not a tumour suppressor gene. However, it is responding in some way.
  • KO AHR mice have decreased survival. 20% of LOF AHR heterozygous died - showing that its repression by AhRR wouldn’t decrease likelyhood of tumour
  • When added a tumour inducing chemical, Wt survived and AHR mutant formed liver tumours
481
Q

What is the agouti mouse?

A

The fur for agouti mouse have a band of yellow in an otherwise black strand. The hair produces two types of pigment in different times in its development – defined period when black pigment is off and yellow is on

482
Q

What does the Agouti viable yellow mutant allele lead to?

A

Yellow fur

483
Q

How does the Agouti viable yellow mutant allele lead to yellow fur?

A

There is a IAP retro transposon upstream to the transcription start site of the agouti gene. The proximal promotor is constitutively active meaning agouti transcription is favoured causing yellow fur and the repression of black

484
Q

What is meant by the Agouti viable yellow mutant being variable and what causes this?

A
  • The phenotype is variable: Some carriers of the dominant mutation are yellow and others appear agouti. This is a bit like position effect variegation. Sometimes the animals have constitutively active expression and sometimes it is much lower so normal agouti phenotype
  • This is due to functional instability of the Agouti viable yellow allele
485
Q

What methylation changes of the IAP promoter occur when the phenotype of the agouti variable yellow mutant changes?

A
  • Correlation between methylation of the agouti locus in the IAP promotor.
  • When heavily methylated they are pseudo agouti but when demethylated they look yellow.
  • This methylation is not just in the coat but also in the brain testis and lung ect
486
Q

What is the effect of bisphenol A on agouti variable yellow mice?

A
  • If when developing the mother was fed bisphenol A, the degree of methylation of the promotor is reduced and increases the amount of yellow offspring
  • This exposure is shifting the methylation pattern. It is unclear exactly how this is happening but it is an endocrine disrupter so be mimicking signalling
487
Q

What is the effect of bisphenol A on global methylation patterns?

A

Much wider effect on methylation than just the agouti yellow locus. Genome wide study (Kim et al, 2014) in the liver

  • Progeny of pregnant mice that had even been fed control, small amounts of bisphenol A or large amounts of bisphenol A
  • Some genes show heavy unmethylation and some acquire methylation
  • Can see that the function of many of the genes that have affected methylation are involved in metabolism