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HN series Flashcards

1
Q

whats is a gene?

A

a stretch of DNA that encodes diffusible products (RNA/proteins), which in turn carry out functions in the cell

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2
Q

how human genes identified

A

by:

  1. similarity to known genes
  2. expressed sequence tags (ESTs)
  3. in silico prediction
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3
Q

what is in silico prediction

A

computation prediction using known feature of the genes

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4
Q

what is a promoter

A

~100 bp, contains dispersed sequences that bind basal transcription apparatus

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5
Q

what is enhancer

A

~100bp, contains several closely arranged binding sites for transcription factors

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6
Q

e.g of regulatory regions

A

promoters and enhancers.

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7
Q

what is the aim of regulatory regions such as promoters and enhancers used for

A

alternative promoters/enhancers maybe employed to achieve specific expression patterns.

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8
Q

what made up the rest of the genome besides the genes

A
  1. regulatory regions,
  2. introns,
  3. centromeres, telomeres, origins of replication
  4. pseudogenes
  5. short repeat sequences
  6. long repetitive elements (LINES,SINES, transposons etc)
  7. other intergenic DNA
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9
Q

what does introns contain

A

introns can contain regulatory regions, e.g. enhancers, alternative promoters

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10
Q

what does intron do

A

intron

  1. allow for exon splicing/protein isoforms (alternative splicing, therefore high protein encoding capacity)
  2. allow for non-deleterious integration of DNA (e.g. from viruses, transposons, etc)
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11
Q

what is the function of centromeres

A

required for chromosomal division

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12
Q

what is the function of telomeres

A

required for chromosomal stability

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13
Q

what is the function of origins of replication

A

required for replication

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14
Q

what is the centromeres, telomeres and origins of replication characterised by

A

by short, tandem repeats.

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15
Q

what is pseudogenes

A

copies of functional genes that are altered
such that they no longer have function of
parent gene

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16
Q

what r the types of pseudogenes

A
  1. nonprocessed pseudogenes

2. processed pseudogenes

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17
Q

what is nonprocessed pseudogenes

A

they derived from gene duplication followed by inactivating mutation or incomplete duplication

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18
Q

give an e.g. of nonprocessed pseudogenes

A

the beta-globin pseudogene .
Features of active gene: promoter, splicing junctions and open reading frames
Changes in pseudogene:
promoter mutations, splicing junctions lost, nonsense mutation and missense mutations

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19
Q

what is processed pseudogenes

A

reverse transcription + insertion = processed pseudogenes

they are continuous stretch of exons but non-functional, due to lack of promoter, or poly-A tail

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20
Q

what r the short repeat sequences

A
  1. microsatellites

2. minisatellites

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21
Q

what is common for the short repeat sequences

A

the microsatellite and minisatellite are all unstable

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22
Q

what is microsatellites

A

smaller than 10bp repeating unit

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23
Q

what is minisatellites

A

~10 to 100 bp repeating unit, greater number of repeats than microsatellites

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24
Q

what r the e.g of long repetitive elements

A

LINES, SINES, transposons

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25
Q

what is LTR retrotransposons code

A

LTR retrotransposons code for REVERSE TRANSCRIPTASE AND INTEGRASE activities. they are able to make DNA copies of themselves and subsequently integrate into the genome

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26
Q

e.g. of non-LTR retroposon

A

LINES

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27
Q

what is LINES

A

long interspersed repetitive elements

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28
Q

what does LINES encode for

A

they encode a reverse transcriptase.

NO INTEGRASE activity

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29
Q

what is SINES

A

short intersperse nuclear elements

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30
Q

what does SINES originates

A

SINES originates as rRNA, tRNA that has been reverse transcribed
DO NOT ENCODE THEIR OWN REVERSE TRANSCRIPTASE , may use LINE machinery for replication

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31
Q

what is transposons

A

Transposons (jumping genes) is DNA elements that encode a transposase.
the transposase excises the transposon and inserts it into a new site in the genome.

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32
Q

what is euchromatin

A

light staining

accessible DNA, “active” genes

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33
Q

what is heterochromatin

A

dark staining

inaccessible DNA, “inactive” genes

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34
Q

what composed of chromatin

A

Chromatin is composed of subunits called nucleosomes

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35
Q

what is proteins in the nucleosome

A

histone protein

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36
Q

what does histone protein contain

A

contain many lysine and arginine residues (+ charged, non-specific interaction with the DNA backbone)

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37
Q

what r the core histone proteins

A

Core proteins – H2A, H2B, H3, H4 (octamer)

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38
Q

what protein is in the nucleosome besine the histone octamer

A

linker protein-H1, it is histone protein but different function

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39
Q

what is the length for nucleosomal DNA (core and linker DNA length)

A

core DNA 146 bp

linker DNA 8bp-114 bp

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40
Q

what does neucleosomes appear under EM

A

Electronmicroscopy: nucleosomes appear as a 10 nm fiber (“string of beads”)

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41
Q

what does nucleosome do

A

they further coil for form 30 nm solenoid
Further compaction is thought to
give rise to heterochromatin

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42
Q

what r the characteristic of histone tails

A
  • unstructured
  • protrude from the nucleosome
  • subjectto modifications
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43
Q

what does acetylated histones mean

A

euchromatin =active gnes

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44
Q

what does deacetylated histones mean

A

heterochromatin=inactive genes

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45
Q

what is the action on CpG dinucleotide

A

Cytosines in CpGs can be methylated by DNA methyltransferases

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46
Q

CpG summary

A

• Cytosines in CpGs can be methylated
• Cytosine spontaneously deaminates to
uracil, while 5meC deaminates to thymine • Uracil is recognised as “foreign” and is
replaced by cytosine
• Thymine can persist i.e mutation arises
• CpGs are underrepresented in the genome

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47
Q

what is CpG islands

A
  • Areas of the genome where this is a relative abundance of CpGs
  • Often lie in promoter regions of genes (particularly housekeeping genes)
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48
Q

• How have these CpGs survived? –

A

– protected from methylation (important, there are some protein protect CpG from methylated by preventing DNMT from coming in)
• DNA methylation (atCpGs) is associated with inactive genes, while a lack of methylation is associated with active, transcribed genes

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49
Q

• How are CpG islands protected from methylation?

A

– not completely sure
– mediated by transcription factors/cofactors – e.g Sp1 binds 5’-CCG CGC CCG-3’
• CpG methylation as a defence mechanism (switch off)

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50
Q

chromatin conserved between archaea and eukarya

A
  • prokaryotes dont have chromatin
    the archaea’s DNA is digested with MNase (micrococcal nuclease, an endonucleasease cut unspecifically in the linker region)
    -the result showed that when the MNase conc. is high, only 1 band at 50bp
    -> because the enzyme cut every nucleosome
  • when MNase conc. is lower, there are 2 bands at 50 bp and 150 bp
    -> showed that the enzyme cut every 2 or 3 nucleosome

the result show archaea have smaller nucleosome, because the core histone protein in archaea is tetramers.

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51
Q

Why are we interested in regulatory elements in DNA?

A

• identifying regulatory DNA elements allows us to predict/identify their cognate DNA-binding proteins
• regulatory DNA elements + DNA-binding proteins =
“molecular switches”
– turn genes on/off(orup/down)
• understanding these molecular switches can allow us to artificially regulate gene expression
– drugs
– gene therapy
– designer proteins

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52
Q

regulation in globin clusters

A
  • alpha cluster have 4 functional gene
  • beta cluster have 5 functional gene
  • at different developing time, the globin cluster express different functional genes
  • > this is because the embryonic and fetal haemoglobin have higher affinity for oxygen, they need to get oxygen from mother’s blood.
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53
Q

Case study: haemoglobinopathies

A

• Mutated adult β-globin in beta cluster-> sickle cell anaemia-> this mutation cause protein to aggregate, clogged up in blood vessel

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54
Q

what is current research area for sickel cell anaemia

A

• Individuals with ↑ foetal γ-globin display less severe symptoms
• Can we reactivate γ-globin expression to reduce symptoms? – hydroxyurea
- study use zinc-finger transcriptional activator designed to interact with the gamma globin gene promoters enhances fetal hemoglobin producton in primary human adult erythroblasts
- in adult mice, they trying to correct sickel cell disease by interference with detal hemoglobin silencing

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55
Q

what r the Regulatory elements in DNA

A
  • Promoters
  • Enhancers (and silencers) • Locus Control Regions
  • Insulators
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56
Q

what is promoter

A
  • Region of DNA around the transcriptional start site of a gene
  • Comprises short motifs that are bound by transcription factors
  • “Promotes” transcription by recruiting RNA polymerase (when appropriate)
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57
Q

what does promoter define?

A

• Define the start site and direction of transcription

  • > the start site have INITIATOR ELEMENT A surrounded by pyramidine Y
  • > the recruitment of RNA pol by using TATA box ( in downstream elelment), the TATA box interact with TATA binding protein and recruit RNA polymerase
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58
Q

what is the positioning of RNA pol II

A
  • Positioning factors are required for RNA pol to find promoters
  • RNA pol II positioning factoris TFIID, which isTBP associated with up to 14 other subunits called TAFs
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59
Q

what does enhancers contain

A

• contain motifs bound by transcription factors

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60
Q

what does enhancer do

A
  • LOOP TO PROMOTERS to influence transcriptionrate
  • somewhat position and orientation-independent-> can be downstream of upstream, they have to be in the same double stranded DNA
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61
Q

what is

A

Locus Control Regions (LCRs)
• Enhancers that regulate gene clusters
e.g. the β-globin LCR
– allows developmentally appropriate expression of the different β-globin genes
- these LCR contain 5’ and 3’ hypersensitive sites which is hypersensitive to digestion of Nase
- LCR loops to the promoter sequence during different developmental stages-> allow different expression

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62
Q

what is insulators

A

insulators
• Block effects of enhancers, silencers and LCRs on unintended, neighbouring target genes
-> insulator prevent inappropriate activation of genes

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63
Q

what is the method used for detect physical interaction between regions of chromatin in vivo? i.e. looping of DNA

A
  1. folmaldehyde is used for crosslinking, enable join 2 proteins or protein+DNA force holding together
    e. g. enhancer and promoter
  2. restriction enzyme digestion cut the linker DNA
  3. ligase allow intramolecular ligation to join the DNA bound around enhancer and promoter
  4. reversal crosslinks and PCR
  5. sequence the DNA to find part of the DNA before binding
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64
Q

How do we check the activity of putative promoters and enhancers?
-> to check these promoter and enhancer actually working

A

• Clone the putative promoter
• Insert into an expression vector (upstream of a reporter gene)
• Why do we use reporter genes?
-> they are easy to visualise

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65
Q

How do we get the reporter construct into cells?

A
  • Genegun
  • Lipofection
  • Electroporation
  • Calcium phosphate precipitation
  • Microinjection (introduce DNA into organism)
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66
Q

e.g of reporter genes

A

GFP ( green fluorescent protein )

e.g. if the cell expresss lacz-> green
if the cell express firefly luciferase-> green

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67
Q

how do you Identifying the minimal promoter and important motifs

A

it monitor the effect of putative promoter and enhancer on the expression of the reporter gene

  1. empty plasmid as control-> should be no expression of reporter gene
  2. the activity of the Bklf promoter is activated by Eklf, adding Eklf protein, the expression is increased
  3. using the segments to identify which part of the Bklf is important. the mission Bklf parts maybe important for expression then
  4. identify the sequence for expression P23L3
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68
Q

How do we identify the proteins that bind to promoters?

A

• Largely by prediction (observing motifs that fit known consensus binding sites of transcription factors)
• Test by means of in vitro gel shift assays (a.k.a band retardation assays, electrophoretic mobility shift assays) or chromatin immunoprecipitation (in vivo)
BAND RETARDATION ASSAYS: if the protein bind to the gene of interested which labelled with radioactive label, their band is retardated, stay near the well. if the protein doesn’t bind to the gene of interest, the DNA sequence will migrate furthest

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69
Q

how do you know if the protein is important for regulatory sequence in disease

A

HAEMOPHILIA B- mutation in clotting factor IX (9)
this disease can be caused by the mutation in the coding sequence or mutation in the regulatory region e.g. -20 region
the HNF4 (hepatocyte nuclear factor 4) protein bind to the -26 and -20 position of the regulatory region, the clotting factor IX will be expressed
. found if the mutation is at -26 position, there is no recovery after puberty
. found that if the mutation is at -20 position of the regulatory region, haemophilia B is recovered after puberty
-> because another transcriptional factor AR (androgen binding, T is produce during puberty, which is a ligand for AR) protein bind to -26 position.
- as long as -26 position is intact, the haemophilia B can be recovered after puberty
Androgen receptor is activated at puberty and drives the clotting factor IX gene

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70
Q

how does DNA regulatory element act

A

DNA regulatory elements act in cis

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71
Q

what does regulatory proteins bind? and in which way?

A

• regulatory proteins that bind DNA act in trans

– “transcription factors”

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72
Q

how did regulatory protein produced?

A
  1. the regulator gene is anywhere in the genome
  2. it is transcribed into mRNA
  3. and it is translated into regulatory protein, which is diffusible, they can go anywhere for regulation of a specific gene
  4. the regulatory protein bind to the target site on the structural gene in a sequence specific manner
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73
Q

where is the transcriptional factor normally find and give an e.g.

A

transcriptional factor normally bind near the promoter region. e.g. a particular TF called NF1 (nuclear factor 1) bind relatively close to the promoter
NF1 binds to 5’ CCAAT-3’ at
around -100

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74
Q

Why are we interested in transcription factors ?

A

• Each promoter is unique and is regulated by a
combination of specific transcription factors
• Transcription factors represent “molecular
switches” that can turn genes on/off (or up/down)
• Understanding these molecular switches can
allow us to artificially regulate gene expression
– drugs
– gene therapy
– designer transcription factors

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75
Q

How were mammalian transcription

factors discovered?

A

• In humans, are genes with similar functions
coordinately regulated by specific transcription
factors?
– if so, they might be expected to have similar promoters

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76
Q

“blood’ erythroid genes

A

Globin (alpha and beta) and haem biosynthesis
genes
• Alignment of promoters was difficult, but
small motifs stood out e.g GATA sites
• Perhaps there is a transcription factor that
binds to these sites to coordinately
regulate erythroid genes
• Check this by a band retardation assay

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77
Q

The hunt for the GATA-binding protein

A

• Lyse erythroid cells to get the protein
• High salt to disrupt DNA-protein interactions
• Obtain nuclear extracts to get full nuclear protein
• Affinity purification/chromatography using
IMMOBILISED DNA containing GATA sequences
– wash, wash, wash
– elute with higher and higher salt
• Test fractions for GATA-binding activity by band
retardation assays
• Run positive fractions on SDS-PAGE and sequence
• This is how the GATA-binding protein
GATA1 was discovered
– similarly, HNF4 (binds GTTAAT in liver genes
e.g clotting factor IX)

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78
Q

Which part of GATA1 binds DNA?

A

• Test a GATA1 deletion series in band retardation assays
• A central portion of GATA1 binds DNA
- the central portion contain 2 zinc fingers that binds to DNA
- the majority of mammalian TF use ZnFs to bind DNA

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79
Q

what r the other DNA-binding domains beside zinc fingers

A

• Other DNA-binding domains: basic region-leucine
zippers, basic helix loop helices, homeodomains,
ETS domains and MADS boxes

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80
Q

Once GATA1 has bound DNA, how might it regulate transcription?

A

• Which part of GATA1 is important for this?
– test truncated forms of GATA1 protein in
reporter assays
1. use full length GATA1,
2. use C-terminal deletion GATA1
3. use N-terminal deletion GATA1
the reporter gene GFP is used.
it was found that the N-terminal deletion GATA1 cause no expression of the reporter gene GFP, therefore The N-terminus of GATA1 contains an activation domain

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81
Q

what is activation domain

A

they are not folded domain, they are the short sequence of amino acid (short motifs) which r important in protein: protein interaction to influence the recruitment of RNA pol II

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82
Q

transcription factors like GATA1 are“modular”

A

– they generally have a DNA-binding domain and
one or more activation or repression domains
- contain DNA-binding domain, connecting domain and activating domain

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83
Q

what does activator do

A 
• activators interact
with the basal
transcription
apparatus 
1. activator contacts TAF in TFIID
2. activator contacts TFIIB
3. activator contact with coactivator
therefore most of the activity happening by protein and protein interactions
84
Q

what is activation/repression domains

A

• Activation and repression domains are
difficult to predict
– often composed of short motifs

85
Q

how many activation/repression domain does TF have

A 
• Transcription factors have one or more
activation/repression domains which
promote/inhibit transcription
– may recruit (or block) RNA polymerase and/or
DNA- and histone-modifying enzymes,
chromatin remodeling factors
86
Q

How do transcription factors bind to DNA?

A

DNA is 20A in diameter, it has major groove (22A across) and minor groove (12A across). it is the angle of the basepair makes the backbone causes asymmetry.
-most TF binds to major groove
because the major groove have more room for amino acid chin to interact, it has more chemical info. more H bonding. elelctostatic force. the backbone is -ve and the protein is +. allow non-specific interaction.
some TF also find to minor groove e.g. TATA binding protein

87
Q

How do transcription factors bind to DNA?

A 
– Classical zinc finger (Sp1)e.g. GATA1 TF
– Steroid receptor (AR)
– Leucine zipper (C/EBP)
– Helix-turn-helix
– Helix-loop-helix
88
Q

Classical zinc finger

A

• many transcription
factors contain zinc
fingers
• classical zinc fingers
- 2 beta sheet and 1 alpha helix
-TFIIIA
- ~23 amino acids (small domain)
- Cys-X2-4-Cys-X3-Phe-X5-Leu-X2-His-X3-His
-Cys2/His2 finger (Cys and His holding the zn finger together)
- the alpha helix is the recognition for bind to specific DNA sequence, the alpha helix fit into the major groove, it can accommodate many sequence, therefore a lot of variation and many specificity in classical zn finger

89
Q

how many bases can each zn finger recognize?

A

each finger recognises 3 bases

e.g. • Sp1 has 3 fingers and binds 5’-CCG CGC CCG-3’-> recognize 9bp, have 3 zn finger

90
Q

Steroid receptors

A

• steroid receptors contain zinc fingers (not classical)
• Cys-X2-Cys-X13-Cys-X2-Cys
• Cys2/Cys2 finger
• receptor activated by binding a particular
steroid (then the receptor can change its conformation and ble to interact with DNA)
• bind DNA as dimers

91
Q

describe steroid receptor

A

the steroid receptor have P box and D box.
- D box zn finger involve in dimerization, NO DNA interaction
- P box involve in DNA interaction. the alpha helix of the P-box interact with DNA
the steroid receptor need dimerise using its D-box. then it change the kind of the site it will recognize and P-box find the DNA sequence in the promoter region they control
- they recognize half site of the response elelment and separate by several nucleotides.
-palindromic

92
Q

Leucine zippers

A 
• amphipathic (hydrophillic and hydrophobic) alpha
helix with leucine
residue at every 7th
position ( allow it to zip together)
• dimerise through
zipper domain 
- the basic region recognize and interact with DNA
• major groove
• C/EBP binds
ATTGC(/)GCAAT
-palindromic
93
Q

how does helix turn helix TF bind to DNA

A
  • helices 1 and 2 lie above the DNA, no DNA interaction
  • helix 3 lies the major groove interact with DNA
  • N-terminal arm lies in minor groove
    • C-terminal helix lies in major groove
    • N-terminal arm lies in minor groove
94
Q

how do helix-loop-helix TF bind to DNA

A 
• amphipathic helices
• loop of 12-28 amino
acids
• dimer
• major groove
95
Q

how did TF achive specificity

A 
Specificity is achieved through
COMBINATIONS of short motifs
• Multiple short DNA motifs
combine to effectively
represent a longer motif
• These short motifs are
bound by multiple
transcription factors
• The different possible combinations of these transcription
factors determine which specific genes are expressed in different cells
96
Q

Once transcription factors have bound to a
DNA regulatory element, how do they
promote or inhibit transcription?

A

• Transcription factors recruit chromatin-modifying
enzymes (often via cofactors)
- the cofactor can recruit histone modifiers, DNA modifiers and chromatin remodellers

97
Q

How were histone modifying enzymes

discovered?

A

• Cyclic AMP (cAMP)-responsive genes have
cAMP-response element(s) (CREs) in their
promoters
– CRE = 5’-TGACGTCA-3’
• CREB (CRE-binding protein) was found to
bind to CREs
– CREB is a bZIP transcription factor

98
Q

How does CREB respond to cAMP and drive transcription?

A

• cAMP activates protein kinase A
• protein kinase A phosphorylates CREB at
Ser133 (in its activation domain)
• CREB subsequently drives expression of target genes (CREB bind to CRF in an inactive state, once CREB phosphoralated by PKA, it will become activated)

99
Q

How does phosphorylation of CREB cause it

to drive gene expression?

A

• Looked for proteins that bind
phosphorylated CREB
– discovered CREB-binding protein (CBP)
• CBP is required by CREB to drive transcription
– investigated by mutating Ser133 (using reporter gene)
• P300 also discovered, similar to CBP (2 protein CBP and P300 bind to CREB)

100
Q

• What are CBP and p300?

A

– homology to the yeast regulatory protein Gcn5
– Gcn5 has homology to acetyltransferase
enzymes
• CBP and p300 acetylate histones and are
known as histone acetyltransferases
(HATs)

101
Q

what happen when histone acetylation

A

histone acetylation lead to recruitment of other proteins. the gene is active.
CREB is phosphorylated in response to cAMP, binds CBP
and activates transcription through histone acetylation

102
Q

Histone acetylation

A

– recruit histone deacetylases (HDACs)=inactive gene

103
Q

The histone code

A

• many transcriptional coregulators (such as
CBP) can read the histone code
• e.g CBP binds acetylated lysines on histones via its bromodomain (small protein domain recognize acetylated lysine and interpret the modification on histone protein. for CBP, it is a reader and also a writer of histone code)
• positive feedback loop (spread in local area)

104
Q

What are some of the specific
modifications and what are their
effects?

A

Histone tail have many sites of modification. E.g. sites of modification in H3, sites of modification in H4

105
Q

Acetylation of H3K9

A
  • Lysine9 on histone3(H3K9)is acetylated by histone acetyltransferases such as CBP
  • Major mark of active genes
  • Proteins with bromodomains (such as CBP) can bind acetylated H3K9
  • This further promotes histone acetylation in neighbouring nucleosomes （can then modify adjacent OR recruit CBP,enable fb, spread, EXPANSION!)
106
Q

Deacetylation of H3K9

A
  • Transcriptional repressors recruit histone deacetylases (HDACs)
  • HDACs remove the acetyl group on H3K9 and restore lysine’s positive charge-> inactive genes
107
Q

what stops H3K9 from being acetylated again?

Methylation of H3K9

A
  • Transcriptionalr epressors may recruit histone methyltransferases (HMTs)
  • HMTsaddamethylgrouptoH3K9
  • Lysine’spositivechargeisretained
  • Methylated H3K9 is protected from acetylation
  • METHYLATED H3K9 IS A MOJOR MARK OF INACTIVE GENES
108
Q

Methylated H3K9

A

Proteins with chromodomains bind to methylated H3K9
• eg heterochromatin protein 1 (HP1) -> it self cant make modification, need rely on CHROMSHADOW domain. chromoshadow domain allow protein to protein interactions-dimerisation and with Su(Var)3-9 (interact with HMT)
- HP1 recruits further HMTs-> this assist in further compacting mediate by dimerise the heterochromatin

109
Q

what r the feature of active gene

A

H3K9 Ace
histone acetyltransferases e.g. CBP
bromodomain

110
Q

what r the feature of inactive gene

A

H3K9 methylation (recognise by chromodomain)
HDAC
HMT
chromodomain

111
Q

Heterochromatin spreading

A 
• Su(Var)3-9 also contains a chromodomain
• It is recruited to methylated
H3K9 (the mark that it lays down!)
• This can lead to
heterochromatin spreading
112
Q

H3K9 summary

A

H3K9acetylation=“active”state,bromodomains


H3K9methylation=“inactive”state,chromodomains

113
Q

H3K4 methylation: an exception

A

• Methylated H3K4 is associated with active chromatin
• How is this the case?
– some transcriptional coactivators have chromodomains
• The methyl group can be removed by histone demethylases

114
Q

what determines the packaging of chromatin?

A

Chromatin remodeling
• carried out by remodeling complexes that use energy from ATP to move nucleosomes
• classified according tofamily that ATPase belongs to
• 4 major families
– SWI/SNF, +ve acting chromatin remodellling complexes, shuffle nucleosome CLEAR THE REGION AROUND PROMOTER
– ISWI,- CHD , -ve acting chromatin remodelling complexes, move into the promoter, repress gene expression
– INO80 histone exchange

115
Q

Chromatin-remodeling complexes

A

• SWI/SNF proteins
– ATPase domain
– shuffle nucleosomes along chromatin
• HowareSWI/SNF complexes recruited?
– by transcriptional activators (e.g GATA1) and HATs
– SWI/SNF proteins often have a bromodomain (which connext with histone modification)

116
Q

INO80 histone exchange using histone variants

A

histone variants change the interaction between nucleosome and DNA, Protein and protein interaction and gene expression
the H2A, H2B, H3, H4 called Canonical core histone

117
Q

INO80 histone exchange using histone variants

A

histone variants change the interaction between nucleosome and DNA, Protein and protein interaction and gene expression
the H2A, H2B, H3, H4 called Canonical core histone
e.g.H3.3. 4 amino acid longer, involve in trancriptional activation
macro H2A its much bigger than H2A, it contain leucine zipper TF, can dimerise between nuclease, increase packaging, heterochrotin and inactive gene, so the function isX chromosome inactivtion, transcriptional repression

118
Q

What determines the packaging of chromatin?

A

histone modification is on amino acid. DNA modification is on nucleotides.

  1. DNA methylation
  2. DNA methytransferases
  3. CpG and gene expression
119
Q

DNA methylation

A
  • Associated with gene silencing
  • Cytosinesin CpGs are methylated by DNA methyltransferases DNMT
  • Not readily reversible (normally)->robust gene repression, however it was found recent year it can be reversible, some nezyme can facilitate demethylation, not direct was. e.g. TAT enzyme
120
Q

DNA methyltransferases

A

• de novo methyltransferases – methylate cytosines in previously unmethylated CpG
– DNMT3a and DNMT3b (near CpG and add methyl)
• maintenance methylase- perpetuates existing methylation patterns (through replication)
– DNMT1

121
Q

maintain methylation patterns

A

DNMT1, not de nova
1. full methylated sites
2. replication become hemimethylated
3. methylation by DNMT1 in the daughter duplex
this is important in mainteining gene expression during cell division

122
Q

DNMT1 mechanism and meaintainence

A

this machinery also talk to histone modification machinery

  • DNMT1 makes 2 important contact
    1. contact with the HP1(heterochromatin protein1), HP1 have chromodomain that can recognise methylated H3K9
    2. also HP1 interact with H3K9 methytransferase, so that HP1 recognise the methylated H3K9 and recruit H3K9 methyltransferases. and because HP1 is recruited by the DNMT1, end up getting with a region of DNA methylation has occur, and around that also H3K9 modification. they all involve in gene silenceing
123
Q

CpG and gene regulation

A

CpG island at the promoter of houkeeping gene
these region are unmethylated
the way they denote this, on the DNA, the TSS(transcription start site ) the CpG are in this region are uniformly unmethylated. It is not always that an active gene is unmethylated. its true in promoter region, they need to be unmethylated, but in the actual region, CpG in the gene body is methylated.

  • around the TSS, there are histone variance can be corporated.
  • CTCF stop certain CpG from being methylated if they are bound by some protein called insulator e.g. CTCF, blocking factor prevent methylation
124
Q

what factors are associated with transcriptional activation

A
  1. euchromatin
  2. transcriptional activators (GATA1, AR, CREB)
  3. histone acetyltransferases (e.g. CBP)
  4. histone acetylation (H3K9)
  5. H3K4 methylation
  6. bromodomains
  7. SWI/SNF chromatin remodeling
125
Q

what factors are involved in transcriptional repression

A
  1. heterochromatin
  2. transcriptional repressors (BKLF and CtBP, CtBP is a co-repressor protein, e.g. repressor bound to promoter, TF recruit CtBP, itself doesnt bind to promoter)
  3. histone deacetylases (HDACs)
  4. Histone methyltransferases (e.g. Su(Var)3-9)
  5. H3K9 methylation
  6. chromodomains
  7. DNA methylation (DNMTs)
126
Q

in the transcriptional repression, the paper title: coordinated histone modifications mediated by a CtBp co-repressor complex. what protein is find in the gel. EXAM

A
  • TF bound to the promoter
  • TF itself recruites the CtBP
  • CtBP itself doesnt bind to sequences
  • there are several TFs can recruit this CtBP
  • instead of recruiting GATA1 binding directly, BKLF recruit through the intermediate
  • the experiment put the tag on CtBP and HA tag-> enable to pull out the antibody
  • run the gell -> pull out all the protein associated with promoter to do its job -> use mass spec. to identify the protein
  • they found ZEB1&2
  • ZEB1&2 are TFs, in this example, ZEB1&2 recruited the CtBP
  • HDAC1 and HDAC2 -> deactylases enzyme, remove actyle group from H3K9
  • EuHMT, histone methyl transferases. after acetyl group is removed, locking in the deacetylated state, then add a methyl group to it
  • CDYL contain a chromodomain-> can recognise methylated histone proteins
  • NPAO-> histone methylase enzyme-> 1st descovery of histone demethylase, methylation can both associated with transcriptional activation and repression-> methylation doesnt always deactivae gene. e.g. the histone demethylase can remove the methyl group from the methylated H3K4, therefore deactivate it
127
Q

TFs in development

A

understanding biological roles of transcription factors
– effects of overexpression
– effects of loss of expression
• use of transgenic and knockout mice for studying in vivo functions

128
Q

understand the biological roles of TFs

A 
effects of overexpression and loss of expression of TFs:
Pax-6
MyoD
Sry
GATA-1
AR
129
Q

what is endogenous expression and ectopic expression

A

endogenous expression is in the normal tissue in WT animal

ectopic expression of the gene of interest=express in other place where it shounldnt be

130
Q

describe Pax-6

A

Pax-6 i=protein is a homeodomain TF, specific helix-turn-helix TF using alpha-helic inserted into the major groove, it recognise specific sequences

  • mutation of Pax-6 related to human disease called Aniridia-> no iris, proble with eye development
  • homologue in other animal
  • Pax-6 is conserved among species
  • Pax-6 in fruitfly, testing whether this gene is the critical one-> a master regulator
  • they put the gene into other tissue, will you drive the expression of an eye? can it slone drive all the transcription down stream?-> result in production of an eye
  • they force the fly to express Pax-6 protein in different spots, not in the eye->they found eye develop in the antenae position, in the posterior position
  • the place they force Pax-6 expression develop real eye tissue
  • > just expression Pax-6 protein in different position will develop eye tissue
  • > this Pax-6 protein is a master regulator of eye development
131
Q

describe MyoD

A

MyoD is a helix-loop-helix TF, it is a master regulator of muscle development

  • they took the fibroblast and culture them
  • force the expression of MyoD, ectopically expression of this MyoD TF
  • the cell form muscle like cell and start to express kind of protein that r normally find in muscle cell e.g. myosin.
  • if you force other cell to express MyoD, you can turn them into muscle-like cell
132
Q

describe Sry

A

Sry is a sex specific regulator on the Y chromosome, belong to HMG (high mobility group TF) class-> once its bound to DNA, it bends the DNA, nor common seen
- it is a regulator of male-specific organs
• Sry on Y chromosome
• deletion in XY humans results in female development-> deletion of Sry gene in male Y chromosome
• some people with XX develop as males ( something went wrong in meiosis, somehow intact Sry gene at the X chromosome)

133
Q

describe GATA 1

A

GATA1 is involved in erythroid regulation, mutation in GATA 1 result in different blood disorders

  • mutations give rise to thalassaemia (problem with globin) and thrombocytopenia (some certain group cell are aberrant, increase number of lineage in the blood cell)
  • they dound individual have thalassaemia but have no mutation in their globin gene, they map the X-chromosome, found region contain GATA1 gene in the X chromosome
  • GATA1 is important in driving expression of blood related genes
  • they sequence the GATA1 genes to find any mutation give rise to thelassaemia.
  • enormous mutation in GATA1 give rise to to blood diseases
  • reinforce the importance of DNA binding TF
  • GATA1 have 2 zn finder, 1 is critical in stabilisation role rather than the 1 provide specificity in bidning
  • mutation on Arg216, the DNA binding phase second domain-> which is only involved in stabilisation, you can still end up twith bad outcome.
  • reinforce the importance of these functional domain intact
  • although zn finger domain primarily is DNA binding domain, in fact they can also have other functionality on zn finger domain .
  • e.g.GATA1, this particular zn finger stabalise the interaction b/t the DNA, the other side of zn finger is called the “FOG face (friend of GATA1)”, this side of zn finger involve in the protein to protein interaction and interact with other factors to regulate gene expresion.
  • if you have mutation disrupt the capacity of this zn finger to interact with FOG, end up with disease
  • only 1 single amino acid changes that are going to give rise to those disorders
  • complete loss of GATA1 is incompatible with life.
134
Q

describe AR

A

AR is another zn finger containing TF, work slightly different

  • they have 1 zn finger domain, important for DNA recognition
  • another zn finger enable it to dimeraise
  • it interact with with DNA as a dimer and it have 2 half sites and 2 recognition sequence
  • in human, if you are XY and have mutation on AR, cannot respond to to androgen, therefore genetic male will develop as female
  • AR itself is very important for maleness
  • AR also have a polyglutamine expansion in it e.g. huntington’s disease, can also happen in AR but it leads to a different syndrome-> more neuronlogical syndrome. when AR have a string of glutamine, it interact aberrantly and mess the process and influence neuron clearing the aggregated proteins
135
Q

how do do we experimentally determine the biological roles of genes?

A 
• Loss of function studies
– RNA interference (“knockdown”), knockdown the function of specific TRANSCRIPT, NOT interfere with actual DNA sequence, it influence the amount of transcripts that translated into proteins/ it target specific gene and knock them down, reduce their function 
– gene ablation切除 (“knockout”)
• Gain of function studies
– expression/overexpression
136
Q 
Generating knockout (KO) mice
method
A 
• Clone a targeting construct
– includes a selectable marker (e.g neomycin resistance gene)
• Electroporation
• Selection with neomycin
(G418)
• Three possible outcomes
– no integration of
construct
– random integration of construct
– homologous integration
of construct
137
Q

how to generate Pax-6 knowck out mice

A
  1. disrupt Pax-6 gene by inserting something in it. in this case, Neomycin gene into the exon, destroy the sequence and knockout Pax-6 gene
  2. use selectable marker e.g. neo gene which is neomycin resistance gene and HSV TK gene
    -neo gene insert in exon, it is a +ve selection, select the one taking this DNA
    -HSV TK gene encode for thymidine kinase enzyme, it is a -ve selection, normal cell without TK gene expose to the chemical gancyclovir will survive, if the cell express TK gene, it will phosphoralate gancyclovir, then it will breakdown into toxic product which will kill the cell, therefore it select AGAINST the one take up the TK gene
    - 3 possible outcome:
  3. no integration of construct-> expose to G418 neomycin, since there is no incorporation of neo gene, they will die
  4. random integration of construct and 3. homologues integration of construct will have neo gene and survive the positive selection.
    - random integration of construct is not at Pax-6 gene
    - homologous integration of construct is at Pax-6 gene
    HOW TO TELL random integration or homologues integration?
    - in random integration, the whole construct go in,so there is gene + neo gene + TK gene in the exon , put gancyclovir, TK make the protein and phosphoralate them and the toxic product kill the cell
138
Q

HOW TO TELL random integration or homologues integration?

A
  • in random integration, the whole construct go in,so there is gene + neo gene + TK gene in the exon , put gancyclovir, TK make the protein and phosphoralate them and the toxic product kill the cell
139
Q

how do you prepare generation of chimaeric mice

A
  1. prepare DNA from mouse
  2. donor mouse has dark phenotype
  3. manipulate embryoic stem cell from blastocyst
  4. transfect cells with DNA from dark mouse,
  5. use gancyclovir and neomycin to screen/enrich for transfected cells,
  6. then the cell have disrupted Pax6 gene
    1a. take a host blastocyst from a light mouse, inject the ES cells to the donor cell (which contain the disrupted Pax-6 gene),
  • together they implant blastocyst into foster mother, them make progeny, this progeny is chimeric progeny
  • then test if it is a germline transmission by cross with light mouse, because we want the knowckout gene transmit this to next generation, the offspring should be black if it is germ line transmission
140
Q

how to generalising conditional knowouts

A
  • rather than insert neo gene into exon, where the gene is always disrupted.
  • they instead flanked their gene at interest by recombination site
  • the lox gene doesnt normally destroy the function of the target gene
  • they drive the recombinase enzyme acts on lox sites
  • Cre recombinase enzyme cuts the genome, everything b/t the 2 lox gene sites
  • the seqeunce b/t lox gene site is released
  • left with the gene is completely cut out, left with lox gene reminant site
  • they can control the cell by control which cell is going to express the recombinase= which cell have knockout gene
141
Q

Transcription factor KO phenotypes

A
  • p53 – susceptibility to cancer • Hox11 – no spleen
  • Ikaros – no B or T cells
  • Fos – no teeth
  • MyoD- knockout in MyoD in mice have a normal phenotype, why???
142
Q

why do some KOs display no phenotype

A
  1. COMPENSATION
    - when there is a smilar TF important for muscle development, if 1 is missing , the other one can compensate. then you can using double knockout, then they will fail to get muscle development
  2. NOT CHALLENGE THE MICE IN THE RIGHT WAY
    - if you knockout some important gene in immunity, but not give the pathogen, you cant see the affect of KO, you need to know the right way in order to reveal the phenotype of the gene of interest
143
Q

gain of function studies

A 
• Promoter selection for expression vector 
– endogenous promoter?
– other naturally occurring promoter?
– synthetic promoter?
– whole locus?
• Genomic integration 
– site? (integration into heterochromatin will silent the gene because of the heterochromatin spread)
– copy number?
  • if TF present at certain level, adding more of it, sometimes it wont make a difference, add more protein to a system is not going to reveal anything about that function of protein. but in some case e.g. Pax-6, it can be informative.
  • so key Q: how r u going to drive the expression of the gene of interest in different tissue to ask what it does?
  • sometimes using normal promoter of the gene of interest e.g Pax-6 gene promoter, they introduce another copy, normal promoter, normal gene sequence , therefore increase dosage, ask what affect it have on eye development
  • eg expression Pax-6 gene in a particular tissue see whether you can make an eye development in the heat, then use a promoter you know that drive the expression in heart, you have to use CHARACTERISE PROMOTER.
144
Q 
gain of function phenotype
MyoD
Sry
Myc
GATA1
A
  • MyoD: fibroblasts ->muscle
  • Sry: female->male
  • Myc: cells->immortalised
  • GATA1…
145
Q

Epigenetics in human health and disease

A
  • regulation of gene expression in the nervous system

* epigenetics and cancer

146
Q

Experience and the nervous system

A
  • memory = processes used by the brain for long-term storage of information
  • many paradigms used in memory research
  • contextual fear conditioning = animals learn to predict aversive events
  • associate neutral context with aversive stimulus such that the context alone elicits the fear response
147
Q

Contextual fear conditioning

A

involve 2 parts: training for contexual fear condititoning and then testing for memory formation.
during training, the mouse receive foot chock, and was return to its home cage. 24hr later, the trained mouse was brought back to the training chamber and tested for contexual memory.

contexual memory was determined by fear response, indicated by the percentage of freezing. during training, mouse shoed minimum freezing. when tested, only the shocked mice displayed significant freezing, indicating that they learned to associated the chamber with foot shock and remembered.

148
Q

What is the mechanism of memory
formation in the contextual fear
conditioning paradigm?

A

• hippocampus involved in long-term memory – long-term potentiation
• changes in gene expression observed (1 hr after conditioning)
•epigenetic modifications appear to play some
role
– histone acetylation
– histone methylation
– DNA methylation

149
Q

Histone acetylation

A

• recall that acetylation of lysine residues in histone tails is associated with active gene expression
• increased acetylation of histone H3 observed in cells of the hippocampus after animal trained in contextual fear conditioning paradigm
- using western blot, there should be no changes in total histone

150
Q

What mediates this acetylation?

A
  • CBP(CREBbindingprotein)–histone acetyltransferase activity

* Interfering with CBP activity in hippocampus of impairs contextual fear conditioning p8L8

151
Q

CBP∆1 inhibits CRE-mediated transcription in tissue culture. p9L8

A

test trucated CBP, can it interfere with the function of CBP? the result show there is less activation of luciferase, therefore the truncated CBP intereferes.
the use of Forsk+IBMX activates CRE-mediated transcription

152
Q

CBP∆1 transgenic mice exhibit impaired long-term memory for contextual fear conditioning

A

CaMKIIα promoter drives expression in specific neurons including those of the hippocampus.
CaMKIIα promoter drive the mutated version of CBP to produce it in hippocampus. so they have normal CBP and mutated CBP in hippocampus. they showed that there is reduce level of freezing (because they still have normal CBP) compare to WT, therefore CBP activity is important for memory

153
Q

Histone methylation

A
  • recall that histone methylation can be associated with active gene expression or gene repression, depending on the amino acids that are modified
  • increased trimethylation of lysine 4 of histone H3 (active mark) found 1 hr after contextual fear conditioning. they found that the total histone level is the same in WT and stimulated mouse, but increased trimethylation of lysine 4 of histone H3 increased
154
Q

Determining location of histone

modifications

A
  1. cross link sample with formaldehyde
  2. extract crosslinked chromatin
  3. shear chromatin
  4. immunoprecipate target protein
  5. purify DNA from immunoprecipatate
155
Q

zif268

A

• H3K4 trimethylation found at zif268 promoter zif268
• expression of zif268 increased in hippocampus in contextual fear conditioning
• ZIF268 is a transcription factor
• regulates expression of
synaptobrevin

156
Q

Detection of DNA methylation

A

using bisulfite treatment
1. unmethylated (C)
after bisulfite treatment, the C become U
2. methylated, after bisulfite treatment, the C remain as a C, do not convert to U.

157
Q

DNA methylaton

A

DNA methylation
• DNMT3a and DNMT3b are upregulated within 30 min of contextual fear conditioning
• promoter of gene encoding protein phosphatase 1 (PP1) is methylated one hour after conditioning, and transcript decreased
• Decrease in PP1 leads to increase in CREB-dependent gene expression

158
Q

summary L8

A 
• histone acetylation – CBP
• histone methylation
– H3K4 in zif268 promoter
• DNA methylation
– PP1 promoter methylated
159
Q

Can epigenetic modifications be manipulated to influence memory?

A

• in CBP defective mice, HDAC inhibitor restored normal long-term memory (balance between HAT/HDAC activity is important)
HDAC inhibitor reduced acetylase and deacetylase activity
• mouse model of alzheimers, acetylation is defective, HDAC inhibitors improved memory to the level of normal mice

160
Q

Epigenetics and cancer

A

• deregulation of epigenetic modifiers • DNA methylation

161
Q

Deregulation of epigenetic modifiers in cancer

A

in cancer, deregulatation of these epigenetic modifiers:

  1. chromatin remodelling
  2. DNA methylation
  3. Histone acetylation
162
Q

Deregulation of epigenetic modifiers in cancer

A

mutation or changes in expression reported for at least 50 epigenetic modifiers

  1. DNMT3a/3b = overexpression
  2. CBP= mutations/translocations/deletions 3. HDAC3 = overexpression
163
Q

DNA methylation and cancer cancer P22L8 exam

A

two distinct changes to methylation status

1) CpG islands in promoters are hypermethylated
2) genomes of cancer cells are hypomethylated relative to normal cells

164
Q

DNA methylation and cancer cancer P22L8 exam

A

• Hypermethylation of promoters
– decreased expression of genes that play important roles in regulating cell cycle, apoptosis, DNA repair, differentiation, and cell adhesion
• Hypomethylation of repetitive sequences – decondensation of heterochromatin facilitates
recombination between repetitive sequences
– genomic instability

165
Q

Targeting cancer by epigenetic therapy

A

P24L8 look at it, try to think about why and its mechanism

166
Q

Epigenomics

A
  • epigenome – describes complete collection of epigenetic modifications to a genome
  • although DNA sequence of genome is invariant across tissues within an organism, epigenome shows tissue-specific variation
167
Q

An example of non-Mendelian inheritance

A 
Igf-2 = mouse gene that encodes a growth hormone called the insulin-like growth factor
Igf-2m = mutant allele of Igf-2 that encodes defective version of protein
168
Q

how did they find Igf-2 is non-mendelian inheritance?

A

they crossed the normal male (homozygous) with mutant female (homozygous), the offspring are heterozygous for the mutation and appear normal

  • however when they cross the mutant male with normal female, they found the offspring are heterozygous for the mutation and are dwarf.
  • same genotype, different phenotype. so is it the mother or father that the mutated gene inherited from?
  • they found that the maternal Igf-2 never express in any way, only the paternal gene is expressed.
169
Q

Monoallelic expression I

A
  • in somatic cells of the mouse (and humans) there are two copies of each gene – one inherited from mother and one from father
  • most genes expressed from both copies
  • some genes from one copy only = monoallelic expression
170
Q

Monoallelic expression II

A

• for some genes, the allele from which they are expressed is random
– some cells express the copy that came from the father (paternally-inherited)
– other cells express the copy that came from the mother (maternally-inherited)
• for other genes, the allele from which they are expressed is NOT random
– these genes are called “imprinted”, so wither paternal expressed in all cells or vice versa, just 1 from specific parent

171
Q

Monoallelic expression III

Igf-2

A

Monoallelic expression III
• Igf-2 is an example of an imprinted gene
• recall Igf-2 Igf-2m heterozygotes were only dwarf if the mutant allele had come from the father
• this is because Igf-2 is only expressed from the paternally-inherited allele

172
Q

Parental/genomic imprinting

A
  • the process by which “imprinted” genes are established is called parental or genomic imprinting
  • definition: the epigenetic modification of a specific parental chromosome in the gamete that leads to differential expression of the two alleles of a gene in the somatic cells of the offspring
173
Q

what is the definition os imprinting exam

A

• definition: the epigenetic modification of a specific parental chromosome in the gamete that leads to differential expression of the two alleles of a gene in the somatic cells of the offspring

174
Q

What is the imprint? H19 enhancer

A

imprinted genes are differentially methylated
• for example – H19 gene in mouse (neighbour of Igf-2) only expressed from maternal allele
• expression dependent on methylation status of a nearby segment of DNA
• segment called differentially methylated domain (DMD) or imprinting control region (ICR)

175
Q

how the epigenetic change in germ cell maintain in somatic cell?

A

exam p11L9

176
Q

Igf-2 imprinting p12L9 exam

A

• some loci are more complicated - it is not always the case that
the methylated allele is silenced
• CCCTC-binding factor (CTCF) is an enhancer-blocking protein
• CTCF only binds the ICR if it is unmethylated
• on the paternal allele the enhancer has access to Igf-2, while on the maternal allele, absence of methylation results in CTCF binding and blocking of the enhancer

177
Q

how is the imprint established and maintained P13L9 exam recording too. 9:28 08/04

A

exam

178
Q

Human syndromes arising from defects at imprinted loci

A
  • Prader-Willi and Angelman syndromes

* Beckwith-Wiedermann and Silver Russell syndromes

179
Q

Prader-Willi and Angelman

syndrome

A

• human chromosome 15 region called PWS and AS domain
• name derived from syndromes
associated with defects in expression of
genes in this region
• different genes expressed from maternal vs paternal copy of the locus
• imprinting control region (ICR)
• methylation status of ICR affects expression of the 6 genes
p16L9

180
Q

What happens there is a
deletion/mutation in this region of
the genome?

A

• since the genes are imprinted, the
outcome depends on whether the mutation
was inherited from the mother or the father

181
Q

Prader-Willi syndrome (PWS) and Angelman syndrome (AS) are caused by epigenetic dysregulation of imprinting control regions in chromosomal region 15q11-q13.

A 
T 
p18L9
If maternal allele is defective...
• no expression of UBE3A
• mental retardation, speech impairment and behavioural abnormalities = Angelman syndrome
If paternal allele is defective...
• no expression of MAGEL2 or SNRPN
• hyperphagia and obesity during early childhood, mental retardation and behavioural abnormalities = Prader-Willi syndrome
182
Q

Beckwith-Wiedermann syndrome and Silver Russell syndrome

A

• BWS: pre- and/or postnatal growth greater
than the 90th centile
• SRS: pre- and postnatal growth retardation

183
Q

p22L9 exam

A

diagrams, listen to recording again

184
Q

Imprinted genes

A

• ~60imprintedgenelociidentifiedinhumans, many control growth

185
Q

what are the paternal gene that is important in promote growth

A 
Gene KO weight
Igf2    60% 
Peg1   82% 
Peg3  80%
undergrowth mean the gene is responsible for promote growth and vice versa
186
Q

what are the maternal gene that is important in promote growth

A 
Gene KO weight
Igf2r (receptor)  130% 
H19 130% 
Grb10 130%
overgrowth mean the gene is responsible for restrain growth
187
Q

what is X-inactivation

A

random “turning off” of one of the X
chromosomes inherited by a female from
her parents

188
Q

why does X-inactivation happen

A 
DOSAGE COMPENSATION 
human females XX, males XY
– to equalise expression of X-linked genes
between XX and XY, majority of genes
on one X chromosome are silenced in
XX females
189
Q

what is the methods of dosage compensation in drosophila

A 
Drosophila melanogaster
– female XX
– male XY
– increase level of expression from
male X chromosome 2-fold
190
Q

what is the methods of dosage compensation in C.elegans

A 
• Caenorhabditis elegans
– hermaphrodite XX
– male X0
– decrease level of expression from
both hermaphrodite X
chromosomes to 50% (cf male)
191
Q

what is Lyon hypothesis

A

1) X-inactivation occurs early in embryonic
life
2) inactivation is random (ie either paternal
or maternal X is inactivated)
3) entire X-chromosome is inactivated
4) X-inactivation is permanent and clonal

192
Q

X-inactivation in Tortoiseshell cats

A 
• coat-colour gene on X-chromosome in
cats that is called “0”
• two alleles
– O: codes for enzyme that converts black
pigment to orange
– o: codes for an inactive form of the enzyme
(ie black pigment can’t be converted to
orange)
female XOXO cat is orange
female XoXo cat is black
193
Q

what colour is heterozygote XOXo cat

A

mosaic coat = tortoiseshell

194
Q

Human X-linked skin disease

A
  • incontinentia pigmenti

- anhidrotic ectodermal dysplasia

195
Q

Counting and choice

A

• XX – 1 Barr body
• XY – no Barr bodies
• XO – Turner syndrome – no Barr bodies
• XXX – triple X syndrome – 2 Barr bodies
• counting ensures that one and only one X
remains active
• how is counting achieved? how is choice
made?

196
Q

X-X pairing

A

• in differentiating mouse embryonic stem cells, X
chromosomes begin at random positions
• after 2-4 days of differentiation, X-chromosomes
pair – this is time at which X-inactivation begins
• cross-talk enabled by pairing results in marking
of one X to become Xa and the other to become
Xi

197
Q

X inactivation center (Xic)

A 
Xic is 450 kb domain on X chromosome
that controls X-inactivation
Xist (X inactive specific transcript) –
large non-coding nuclear RNA
Tsix antisense Xist transcript
198
Q

Initiation of X-inactivation

A 
Xist is ncRNA
prior to X inactivation Xist is
unstable
Tsix acts as negative regulator of
Xist
antisense transcription
on the future Xi the Xist transcript is
stabilised – possibly by silencing of
Tsix on this chromosome
Xist transcript accumulates and,
starting from Xist gene, spreads in
both directions to coat the entire
chromosome
on the future Xa, Xist doesn’t coat
the chromosome
expression of Xist also dependent
on methylation status of promoter –
methylated on Xa
199
Q

Maintenance of X-inactivation

A 
• coating of X chromosome with Xist RNA is
followed by conversion of the DNA into
heterochromatin
• process involves many epigenetic
modifications – histone modifications,
chromatin remodelling, DNA methylation
Removal of euchromatinassociated
histone modifications
• modifications lost are
– acetylation of lysine 9 of histone H3
– methylation of lysine 4 of histone H3
200
Q

Appearance of repressive histone

modifications

A 
• includes methylation of lysine 27 of
histone H3
• Polycomb complexes PRC2 (ESC-E(Z))
and PRC1 are responsible for this and
other repressive histone modifications
201
Q

Inclusion of variant histone

A 
• inactive X chromosome contains variant
histone protein called macroH2A
• in vitro macroH2A can interfere with
transcription factor binding and SWI/SNF
nucleosome remodelling activity
202
Q

DNA methylation

A

• promoters of genes on inactive X

chromosome become methylated

203
Q

Overview of X-inactivation

A

• X chromosomes counted
• choice made to inactivate one or the other
• Xist RNA coats one X chromosome –
inactive
• coating by Xist is followed by histone
modifications etc that result in compaction
to Barr body

204
Q

Imprinted X-inactivation

A

• X-inactivation in marsupials is not random
• always inactivate paternal X chromosome
• parentally imprinted
• in mouse extra-embryonic cells (eg
placenta) – paternal X is inactivated

205
Q

Lyon hypothesis

A

1) X-inactivation occurs early in embryonic
life
2) inactivation is random (ie either paternal
or maternal X is inactivated)
3) entire X-chromosome is inactivated
4) X-inactivation is permanent and clonal

zBCHM (5 decks)

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