Genomics

Question 1

Gene expression

Answer 1

• Exons code for proteins
• Not all genes are active at the same time → diversity in cells →ensured by cell type specific gene expression
• Gene expressed = the RNA transcribed from the gene is actually produced

Question 2

Measuring RNA production: qPCR

Answer 2

• rtPCR (retro-transcribed) –oldest method
• quantitative PCR allows gene of interest to understand how much cDNA is present in cells
• RNA cannot be directly measured by PCR → synthesise cDNA → complementary to RNA → reaction w/reverse transcriptase enzyme
• cDNA only includes exons because RNA is spliced
• cDNA made with specific fluorophore incorporated into RNA
• When Taq polymerase is completing second strand → fluorophore released→ qPCR quantifies how much fluorophore in reaction → more light = more RNA = more gene expression & earlier signal showing up

Question 3

qPCR Normalisation

Answer 3

• Results need to be normalised to measure actual change in transcription level → need to compensate for initial variations in mRNA and technical differences w/sequencing
• Housekeeping genes included in qPCR → have stable expression → important for cell components → always expressed above certain threshold

Question 4

qPCR Limitations

Answer 4

• Quick, relatively accurate, cheap but…
• Limited as to how many genes can be tested at any one time ~5-10 (not possible for 1000’s genes)

Question 5

Microarray

Answer 5

• Revolutionised gene expression analysis
• Allows for detection/comparison of thousands of genes simultaneously
• Relies on base-pairing hybridization with probes for each gene to be measured
• More expensive than PCR, but still relatively cheap
• Can measure:
  o Differing expression of genes over time, between tissues and different states
  o Co-expression of genes
  o Identification of complex genetic diseases

Question 6

Affy Gene Chip & Array Experiment

Answer 6

• Each gene has 16-20 pairs of probes synthesized on the chip
  1. RNA extraction
  2. Make cDNA using biotin (important for binding to streptavidin which is probe-specific)
  3. If binding between gene of interest and probe  chip releases light  quantify it

Question 7

Affy Expression Measurements

Answer 7

• A = absent
• M = marginal
• P = present (P-value gives confidence)

Question 8

Microarray Limitations

Answer 8

• Data is very noisy
• Probes not available for all genes; Affy probes only for ~75-80% of human genes
• Cannot detect genes w/very low expression levels
• Data requires lots of statistics and analysis
• Assay does not distinguish expression from different isoforms of the same gene

Question 9

Next generation sequencing

Answer 9

• Based on getting fragments out of a genome →adding sequences (adaptors) at the edges →adaptors always have same sequences →adaptors bind on flow cell and bend fragment then second adaptor binds → bridge-like formation
• Flow cell amplifies PCR → fragments w/adaptor still bind on sequencing machine to get sequenced →have millions of fragments ~75-150 base pairs → map onto reference genome
• “Seq” principle

Question 10

RNA-seq

Answer 10

• Uses next generation sequencing to measure gene expression
• Can assume that every mRNA present will be sequenced the same number of times
• If experiment shows 2x mRNA for particular gene as control, then gene expression is 2x greater
• Gives accurate measure of gene expression, even for genes w/v. low expression levels
• Can identify exact transcript being expressed
• Can potentially identify unknown transcripts with novel splice sites
• Method:
  1. Extract all mRNA → convert to cDNA fragments
  2. Add sequencing adaptors → obtain short sequence using high-throughput sequencing
  3. Resulting sequence reads aligned w/reference genome or transcriptome
  4. Base count profile for each gene is created
• Same procedure for control and variant
• Read counts are proportional to gene expression level

Question 11

RNA-Seq Normalisation

Answer 11

• Important to normalise:
  o Sequencing depth = how many reads are sequenced by the machine
  o Length when dealing w/ different organisms (e.g. human vs mouse)
  o Amount of fragments in each sample
• 2 main methods to normalise data:
  o Raw read count normalisation
  o Reads/fragments per KiloBase per Million reads (RPKM -single end reads; FPKM – paired end reads)

RPKM = 10⁹C / N L

C = raw count of reads in transcript
N = number of mappable reads in experiment
L = transcript length (bp)
Normalizes for gene length (C and L) and library size (N)

Question 12

Raw Read Count Normalisation –DESeq2

Answer 12

• Aims to make normalized counts for non-differentially expressed genes similar between samples
• Does not aim to adjust count distributions between samples
• Assume that:
  o Most genes are not differentially expressed
  o Differentially expressed genes divided equally between up and down
• Relies on housekeeping genes
• Normalisation looks for set of important, highly expressed genes → assume that expression is uniform across samples → same shift for housekeeping genes performed on all genes

Question 13

RNA-Seq Limitation

Answer 13

• Cofounded by heterogeneity of the sample:
  o Different cell types
  o Mutations
  o Different cell cycle stage
  o Epigenetic modifications
  o Stochastic gene expression

Question 14

Single Cell RNA-Seq

Answer 14

• Allows analysis of single cells
  o Enables improvement in resolution of gene expression within samples
  o Enables identification of heterogeneity in cell populations i.e. different cell types
  o Enables gene expression within single cells/cell types to be categorised
• Tissue → dissociation of cells → isolation of cells → single cell → RNA extraction → cDNA synthesis → single-cell sequencing → expression profile → cell type identification
• Plots show differing gene expression in cell types clearly distinguished; even cells difficult to separated (e.g. podocytes) are effectively dissociated
• Results can be illustrated by heat maps or using dimension reduction analysis tools such as PCA or t-SNE

Question 15

Future

Answer 15

Profile gene expression in vivo w/o need of isolating cells

Question 16

DNA Methylation

Answer 16

• Reversible
• Symmetrical so maintained thorugh cell division
• Adding methyl group (CH3) to 5’C of cytosine by methyltransferases
• In mammals, mainly occurs at CpG sites – CpG islands
• CpG islans used for identification of potential promoter regions
• Methylation of CpG island = silencing of gene expression
• Represses gene expression by:
  o Preventing binding of transcription factors
  o Modifies chromatin structure to repress transcription
• Methylation is major factor in epigenetic modifications
• Methyltransferases in mammals: DNMT3a and 3b
• During mitosis, hemi-methylated DNA is created → copied strand is unmethylated → recognised by DNMT1 →methylates new strand to maintain methylation state
• Methylation of histone → chromatin repressed → cannot be transcribed

Question 17

DNA methylation and disease

Answer 17

• Methylation patterns in disease tissue ≠ from normal tissue → aids in identification of disease-causing genes
  o Specially in cancer and neurodegeneration → disease correlates with loss of methylation
  o E.g Alzheimer’s disease (NEP gene); Colorectal cancer (MGMT gene); breast cancer (PRLR)
• Abnormal methylation silences tumour suppressor genes

Question 18

Where does methylation occur

Answer 18

• Intergenic regions = usually methylated
  o Maintains genomic integrity
  o Methylated DNA forms compacted chromatin → less accessible for recombination and translocation
  o DNMT1 deficient cells display genomic instability
• Repetitive elements = usually methylated
  o Transposable elements are highly mutagenic if they can transpose within genome → methylation protects genome from TEs
  o Methylated C mutates to T over evolutionary time → prevent transposition
  o Methylation prevents recombination
• Gene upstream regions = usually unmethylated
• Promoter regions = usually unmethylated so create CpG islands
• Lack of methylation creates relatively higher density of CpG due to lower rate of mutation to T

Question 19

Avoiding methylation

Answer 19

• When region is methylated at all times, DNA tries to find evolutionary solutions for methylation to be avoided → not having cytosine anymore → tend to mutate to T overtime
• If transposons accumulates mutation it loses functionality; 50% of human genome is made of transposons → 99% of them have lost their ability to be a parasite → cannot move anymore

Question 20

Identifying DNA-Methylation

Answer 20

MeDIP-Seq
* Antibody recognizes methylated cytosine → binds meth DNA → immunoprecipitation → retain only antibody bound DNA →fragmented → next gen sequencing → sequences mapped back onto genome to identify methylated regions
Bisulphite sequencing
* Samples treated bisulphite → converts unmethylated C to U → sequence and compare samples to determine methylation e.g. cancer vs normal cells
* PCR only able to amplify U-containing DNA (non-methylated); with other primers can amplify all fragments that contain methylated DNA
Both
* Expensive
* Great resolution
* MeDIP-Seq requires antibody

Question 21

X inactivation

Answer 21

• The silencing of one of the X chromosomes in all female mammals
• Required for dosage compensation to avoid over expression of genes on X chromosome
• Inactivated X chromosome packaged as compacted heterochromatin
  o Compaction by chromosome wide histone methylation –H3K27M3
• Inactivation by Xist gene (long non-coding RNA)

Question 22

Long non-coding RNA sequences

Answer 22

• Longer than 200 nucleotides
• Thousands identified but function largely unknown
  o Target different aspects of gene transcription mechanism
  o Can function as co-regulators or transcription factors
• Act in ‘cis’ (same chromosome they are transcribed from) or ‘trans’ (different chromosome)
• ncRNA Evf-2 = a co-activator for homeobox transcription factor Dlx2, involved in forebrain development and neurogenesis

Question 23

Xist

Answer 23

• 17kb long; acts in cis
• Expressed from only one of 2 X chromosomes  first detectable event in X inactivation
• Xist contains many repeats → 6 identified so far
• Repeat A (RepA) silences function of Xist → binds to PRC2 (Polycomb repressive complex –a histone methyltransferase complex) → lays down histone methylation along chromosome at Lys27

Question 24

HOTAIR

Answer 24

• Long ncRNA expressed from HOXC locus on chromosome 12 → represses HOXD on chrom.2
• ‘HOX’ = important developmental genes
• Acts in trans
• Binds to PRC2 and LSD1 → PCR2 adds repressive H3K27me → LSD1 removes active H3K4me → combined function produce repressive chromatin structure
• In cancer, HOTAIR acts on regions other than HOXD

Question 25

Transcription factors

Answer 25

• Up- or down-regulate a gene
• Recognise specific DNA motifs ~5,6,7 sequences
• Use several mechanisms to regulate gene expression:
  o Stabilising or blocking the binding of RNA polymerase to DNA
  o Recruit coactivator or corepressor proteins to the transcription factor DNA complex
  o Catalyse acetylation/deacetylation of histone proteins:
  i) Histone acyltransferase (HAT) activity –weakens association of DNA w/histones making DNA more accessible to transcription (up-reg)
  ii) Histone deacetylase (HDAC) activity –strengthens association of DNA w/histones making DNA less accessible to transcription (down-reg)
• ~95% of Tf can only bind motifs when chromatin is in active methylated state (non-condensed)

Question 26

Variability of TF

Answer 26

• Actual sequence of particular Tf binding site →variable → harder to identify → so described by general motif not fixed sequence
• E.g. Neurod family recognise small sequences; HOX recognise dimers
• P53 recognises dimers → most important tumour suppressor gene in mammalian cells → if mutation → cancer
• Very specific motifs; even single mismatch would affect function

Question 27

Pioneer transcription factors

Answer 27

• Able to go to region of inactive repressed chromatin → bind it → signal for active machinery to arrive

Question 28

Experimental Method (Tf) –ChIP-Seq

Answer 28

• Chromatin immunoprecipitation followed by sequencing
• Used to identify binding regions if binding protein is known
• Based on antibodies
• ChIP-Seq directly sequences the bound DNA → can then be mapped back onto genome for precise localization
• Consensus/variability of binding sites can be determined from sequence:
  o Map reads back to reference genome
  o Most frequently sequenced fragments form coverage peaks at specific locations
• Same approach as used for DNA methylation

Question 29

ChIP-Seq Method

Answer 29

1. Chromatin in nucleus is cross-linked (Tf cannot detach) → fragment it
2. DNA fragments include those w/target protein bound → incubate w/antibody specific for Tf of interest
3. Immune precipitation → everything bound to Tf will precipitate (normally use magnetic beads that recognise the antibody to keep fragments bound to Tf stuck in tube)
4. Reverse cross-linking
5. Use specific proteases to degrade Tf → have DNA only
6. Sequence using next-gen sequencing
7. Mapped back onto genome → reads only map in regions where Tf was binding

Question 30

Identifying histone modifications

Answer 30

• Also use ChIP-Seq
• Fractionate DNA → use antibody that binds to modified histone being studied → separated by immunoprecipitation →sequence isolated fraction using next gen seq → map sequence back onto genome to identify regions w/modified histones ie. genes under regulation

Question 31

ENCODE

Answer 31

• Encyclopaedia of DNA Elements
• 400 scientists involved
• 5-year project completed in 2012 → identify all functional elements in human genome sequence
  o Goal: to identify and characterise everything in the genome that is non-coding →the DNA/RNA regions regulated and the factors regulating them
• 2003- human genome was sequenced → now have a reference genome allowing development of this project

Question 32

What they did in ENCODE

Answer 32

• Looked at how many different genes are expressed in different cell types
• Profiled binding of Tf across ~56 different cell types using ChIP-Seq
• Profile DNA methylation across ~56 different cell types
• Looked at chromatin conformation
  o Promoters & enhancers need to be in contact w/each other → contacts can be profiled by different techniques → look at 3D genome & chromatin
• Profile all accessible regions of chromatin (euchromatin state)
  o Euchromatin = less condensed, gene-rich, more easily transcribed; nucleosomes are depleted; DNA is accessible for binding of Tf
  o Heterochromatin = highly condensed, gene poor, transcriptionally silent

Question 33

Complexity of gene expression

Answer 33

• Regulating gene expression → complex → involves many regulatory factors → Tf, enhancers, silencers, methylation patterns
• Transcripts regulated by splicing factors e.g. exon and intron splicing enhancers and silencers
• Functionality often cell/tissue and time specific

Question 34

DNA Hypersensitive Sites (HS)

Answer 34

• Hypersensitive sites = regions of chromatin highly sensitive to DNase 1
  o Nucleose→ less compact→ enables DNA to bind to proteins e.g. Tf
• DNase-Seq = technique where DNase 1 cuts DNA only when accessible → isolating fragments → amplify through next gen seq → add adaptors w/ligation → cluster into flow cell→ cluster → reads mapped onto genome → they pile up in regions where chromatin was open and accessible
• Mapping HS sites → identify location of genetic regulatory elements e.g. promoters, enhancers, silencers, locus control regions →(ChIP-Seq can only identify Tfs)

Question 35

DNase-Seq Overview

Answer 35

• Open chromatin regions sequenced and mapped to reference genome
• Nuclear extraction → DNase 1 digestion → library preparation → PCR amplification → high-throughput sequencing

Question 36

DNase-Seq Footprinting

Answer 36

• Number of fragments that map to a sequence is a measure of regulatory activity
• Sites bound by some Tfs show highly specific patterns of DNase I cleavage = ‘DNase footprints’
• Footprints used to identify binding of specific Tfs; advantage over ChIP-seq
  o w/ChIP_seq need to know the transcription factor for immunoprecipitation
  o DNase-seq identifies the binding sites de novo
• DNAse-seq is a genome wide version of DNA footprinting method
• Footprint = prediction of what Tf could bind

Question 37

FAIRE-Seq

Answer 37

• Similar method to DNase-Seq + addition of formaldehyde for cross-linking
  o More efficient in nucleosome-bound DNA than in nucleosome-depleted regions of genome
• Phenol chloroform extraction of DNA → treatment to isolate nucleic acid from solutions
  o Cross-linked chromatin will go to bottom of tube (organic phase)
  o Condensed chromatin at bottom; active chromatin at top of tube
  o These increase resolution and reduce noise
• DNA extracted and mapped to reference genome to identify open DNA regions
• FAIRE-Seq higher coverage at enhancer regions over promoter regions
• DNase-Seq higher sensitivity towards promoter regions

Question 38

ATAC-Seq

Answer 38

• Uses mutated hyperactive transposase Tn5 instead of DNase1
• Tn5 enzyme derived from transposons → attacks and chews open active euchromatin efficienctly
• DNA fragments then isolated, sequenced, and mapped
• Advantages:
  o Requires smaller sample than DNase-seq and FAIRE-seq (requires 1000x more cells)
  o V. fast → completed in 3 hours
• Disadvantages:
  o V. expensive because monopoly

Question 39

MNase-Seq

Answer 39

• Uses micrococcal nuclease
• Cuts v. near to nucleosomes

Question 40

ChIP-Seq vs all others

Answer 40

• ChIP-seq requires antibody
  o Profiling something specific e.g. one Tf of interest
• All others don’t need antibody
  o Less specific → profile all active chromatin regions

Question 41

Chromatin Interaction

Answer 41

• Within each chromosome TAD (Transcriptional active domain)- portions of chromosome isolated from each other
  o One TAD does not interact w/other TADs → enhancers/promoters can only regulate genes in same TAD (v. few expection)
• Lowest possible level = looping
• Knowing interactions is essential for understanding mechanisms of gene regulation in health and disease
• All possible interactions can be profiled by different techniques (e.g ChIA-PET, 3C, 4C, HiSEQ, etc)
  o All based on same approach w/one variation
  o Use restriction enzymes: fragmenting genome → religation→ profile what is ligated
• ChIA-PET → studies genome wide long range chromatin interactions involving protein factors
  o Involves additional step: antibody precipitation → to identify chromatin interactions that are regulated by a specific transcription factor, between distal and proximal regulatory sites and their associated promoters
• Diff-linker, PETs, used to identify non-specific ligation noise; identify ligations between different ChIP complexes

Question 42

RIP-seq and CLIP-seq

Answer 42

• V. accurate; can be reproduced
• Methods similar to DNA-protein interaction identification
• RIP-seq involves immunoprecipitation of RNA-binding protein (RBP) of interest → has to be done non-stringently
  o Low stringency = low specificity
• Developed to CLIP-seq → includes cross-linking step using UV light (irreversible)
• Final steps:
  o Digestion with proteinase K leaving peptide at binding site that modifies nucleotides to create cross-linked induced mutation sites (CIMS)
  o Reverse transcription to make cDNA → identify RNA-binding sites
  o Sequenced and mapped to transcript

Question 43

CLIP-seq variants

Answer 43

• PAR-CLIP = improves crosslinking w/photoreactive RNA nucleotides
• iCLIP = uses reverse transcriptase stalling to map individual nucleotide-protein interactions
• miCLIP = modifies RNA methylase to map its binding sites

Question 44

-seq key features

Answer 44

• All ‘seq’ methods use similar approaches and same final step to identify regions of interest (protein binding site, open chromatin, etc)
• Method:
  o Isolate sequence e.g. fragmentation and immunoprecipitation, phenol/chloroform etc
  o Sequence fragment and map back to genome

Question 45

ENCODE controversy

Answer 45

• Most of the genome is “functional” → controversial statement from ENCODE
  o ENCODE considers anything transcribed must be functional → but many transcripts are non-functional e.g. pseudogenes
• ENCODE emphasized sensitivity over specificity → lead to false positives
• Criticism: arbitrary choice of cell lines and transcription factors; lack of appropriate control experiments

Question 46

ENCODE limitations

Answer 46

• Conducted in immortalised cell lines (derived from human cancers → v. easy to manipulate but v. unstable)
  o Want something closer to real healthy cells
  o E.g HeLa cells sometimes have 3/4 pairs of chromosomes → not diploid
• Roadmap consortium
  o Profiled histone modifications across 25 human primary tissues (mark different chromatin states)

Question 47

What is a SNP

Answer 47

• DNA sequence variations → occur when single nucleotide (A, T, C, or G) in genome sequence is altered; must occur in at least 1% of the population
• SNPs make ~90% of all human genetic variation; occur approx every ~1000 bases; ~4-5 million SNPs in an individual human genome

Question 48

Why are SNPs important

Answer 48

• Can affect how humans develop diseases
• Can affect how an individual respond to pathogens
• Can affect how an individual respond to chemicals
• Can affect how an individual respond to drugs, etc
• Potentially their greatest importance in biomedical research is for comparing regions of the genome between cohorts
• Comparing cohorts with and without a disease – GWAS

Question 49

SNP location

Answer 49

• Intergenic region → possibly transcription enhancer/regulatory region
• Within promoter or transcription factor binding region
• Within exon → Could affect protein coding
• Within intron → Possibly regulatory region e.g. affecting splicing

Question 50

Disease SNPs

Answer 50

• Can be v. dangerous or neutral
• SNPs may be direct cause of disease or signal for increased likelihood of disease
• Disease associated SNPs:
  o Monogenic → one nucleotide change leads to disease; relatively easy to detect/analyze; simple traits
  o Polygenic → many nucleotide changes affect probability of disease; hard to detect/analyze; complex traits

Question 51

Coding SNPs

Answer 51

• Coding SNPs = potentially disease causing as they can affect the protein
• Types: Synonymous (silent) & Non- synonymous
• Synonymous mutation: change base but AA is the same
  o May still affect Exon Splicing Enhancers (ESE) or Exon Splicing Silencers (ESS) site so cannot always be ignored
• Non-synonymous – change in base changes AA → mutation could be detrimental

Question 52

Transition and transversion

Answer 52

• Transition (Ti) - most common substitution
  o Replacing purine by purine i.e. A → G or pyrimidine by pyrimidine i.e. T → C
• Transversion (Tv) - less common
  o Replacing purine by pyrimidine or vice versa i.e. A → C
• Ti/Tv ratio – varies within genome; used to assess GWAS data quality
  o Across entire genome averages around 2
  o In protein coding regions typically higher, often above 3 due to transversions in third base of codon being more likely to change the encoded amino acid

Question 53

Sickle Cell Anaemia

Answer 53

• Inherited blood disorder due to mutations in beta globin HBB
• Found primarily in African and related populations
• Fragile, sickle-shaped cells deliver less oxygen to the body’s tissues
• Get stuck more easily in small blood vessels; break into pieces that interrupt healthy blood flow
• Symptoms: shortness of breath; infections (bone, gall bladder etc); joint pain
• Causes:
  o Mutation of β-globin gene at AA position 6 (HbS) - GAG → GTG: Glutamic acid → Valine
  o Only individuals homozygous for allele (T:T genotype) have sickle cell anaemia
  o Autosomal recessive mutation

Question 54

Alzheimer’s Disease

Answer 54

• Early onset familial
  o Hereditary; ~40yo
  o V. rare ~5% of all cases
  o Caused by mutation in amyloid precursor protein (APP) or presenilin-1 (PS1)
• Sporadic late onset
  o ~70yo
  o Associated w/many genes → e.g. Alipoprotein E (ApoE)
• ApoE contains 2 SNPs resulting in 3 possible alleles for the gene: E2, E3, E4
  o Protein product of each gene differs by one amino acid
  o E3 no effect regarding Alzheimer’s
  o 1x E4 allele → greater chance of developing Alzheimer’s
  o 1x E2 allele → person is less likely to develop Alzheimer’s
  o 2x E4 alleles → may never develop Alzheimer’s
  o 2x E2 alleles → may develop Alzheimer’s

Question 55

Non-coding SNPs

Answer 55

• Studies report: disease associated SNPs are enriched in regulatory DNA regions
  o Enhancers (~+1 million enhancers in human genome)
  o Silencers
  o Locus control regions
  o Promoters
  o Long non-coding RNAs maintaining higher order structure of 3D genome
• 98% of T2 diabetes associated SNPs were non-coding
• Changing 1 nucleotide in motif → Tf does not bind anymore → enhancer is not activated

Question 56

SNPs disrupt splice sites

Answer 56

• Causes ~10% of all mutations causing human inherited disease
• Splice sites found in proximity to exon → provide signal for proteins to cut RNA
• If SNP in splice site → cannot cut anymore
• SNP most likely causes total loss of associated exon; or introduces cryptic splice site
• OAS1 gene –associated w/T1 diabetes
• Can have synonymous mutation in coding DNA but affects splicing machinery

Question 57

Insertion or deletion

Answer 57

• Can cause:
  o Disrupted start codon
  o Disrupted stop codon
  o Disrupted splice site
  o Frame shift
• In a frameshift mutation, base is inserted/deleted, altering codon in which insertion or deletion took place, but also changing the reading frame so that all codons downstream are read out of frame → produces string of amino acid substitutions before a stop codon is reached (stop codons are frequent in coding sequences read out of frame)

Question 58

Genome Wide Association Studies (GWAS)

Answer 58

• GWAS = If 500 people with the same disease all share a half dozen SNPs in common, but a group of 500 healthy people don’t share those SNPs, the mutations behind the disease is probably around those SNPs (now more like 10,000 people)
  o GWAS has identified genetic variations that contribute to risk of: T2 diabetes, Parkinson’s, Heart disorders, Obesity, Crohn’s disease, Prostate cancer…
• GWAS is a very strong area of research → look at common SNPs → statistics
• The difficulty is accurately identifying the SNPs

Question 59

100,000 Genomes Project

Answer 59

• Genomics England project in collaboration w/NHS
• Aim: to sequence the genomes from approximately 70,000 people
  o Longer term aim → research on new & more effective treatments.
• Participants are NHS patients w/cancer or rare disease
• Genomes of families of patients w/rare diseases also sequenced → identify variants associated w/different conditions
• Objective: to create a new genomic medicine service for the NHS
• Patients may be offered a diagnosis where this wasn’t possible beforelonger term aim is research on new and more effective treatments.

Question 60

Identifying Variants

Answer 60

• Sequence multiple genomes from a population at low coverage and pool the data
• Align to reference and identify variants
• Pooling works as most of the genome will be the same; some individuals will also share variants
• Variant prediction software identifies which variants are real and which sequencing errors

Question 61

Limitations of GWAS

Answer 61

• Mostly white people genomes (UK or US) → no diversity → no representative pool
• Not easy to do functional validation
  o SNP may have effect in one cell type not in the other
• Coding mutations alter AA sequence of protein → effect is clear
  o Non-coding mutations disrupt regulatory elements are less clear
• Regulation of gene expression is dependent on multiple factors:
  o Cell-type
  o Temporal patterns such as circadian clock
  o Cell-tissue development
• GWAS identifies candidate SNPs but confirmation requires additional work
• Biggest limitation: linkage disequilibrium and GWAS
  o Linkage disequilibrium = the association of alleles at two or more loci within a population → haplotypes don’t occur at expected frequencies - not random
  o Can be used to improve genetic association studies, such as cancer
  o Enables identification of genetic markers for the associated disease
  o E..g 6 SNPs found in people w/Alzheimer’s → 3 found in linkage disequilibrium → cannot figure out which SNP is causative of disease

Question 62

Expression Quantitative Trait Loci (eQTL)

Answer 62

• First DNA sequencing → then RNA sequencing to quantify gene expression → involves mapping variants which alter gene expression
• eQTL = non-coding SNPs known to affect expression of specific gene ; variants associate w/RNA levels
• eQTL mapping enables identification of regulated genes → unlikely to be close to disease associated SNP
• Cis eQTL = affect expression of nearby gene
• Trans = does not map close to gene; could be other chromosome

Question 63

GWAS and eQTL

Answer 63

• 1000 people → get DNA → find SNPs → get RNA → quantify gene expression → leads to identification of disease causing genes

Question 64

SNP genotyping

Answer 64

• To only find out if pre-defined SNPs are present:
  o Microarrays w/probes for specific SNP → DNA only binds to probe if there is SNP

Question 65

Epigenetics

Answer 65

• Epigenetics = ‘above’ genetics → external modifications to chromatin that turn genes on/off
• Modifications do not change DNA sequence → they affect how cells read genes
• Epigenetic changes alter physical structure of DNA
• E.g. DNA and histone methylation
• Epigenetic modifications can be inherited → “An epigenetic system should be heritable, self-perpetuating, and reversible (Bonasio et al. - Science 29 October 2010: 612-616)”
• Depends on environment, diet, smoking

Question 66

Nucleosomes

Answer 66

• Genome is condensed and compacted into nucleosomes
• 146bp of DNA wrapped around histone octamer (8 → 2x H2A, H2B, H3, H4)
• Space between nucleosomes = linker DNA
• Nucleosomes disassemble then reform during replication
• Nucleosomes = repeating units of chromatin
• Interaction DNA-histones = sequence independent (H-bonding & ionic interactions w/sugar-phosphate backbone)

Question 67

H1

Answer 67

• Sits outside each nucleosome
• Structural function to keep nucleosome together

Question 68

Histones

Answer 68

• Can be methylated and acetylated (the tail of the spheres) histone modifications
• Covalently modified

Question 69

Euchromatin vs heterochromatin

Question 70

2 main ways to repress chromatin

Answer 70

1. Constitutive heterochromatin
   - H3K9me2/me3
   - H3 = histone; K9 = lysine 9; di- or tri- methylation
   - Older → weaker methylation of histones → heterochromatic regions get aberrantly activated e.g in Alzheimer’s
   - Permament
2. Facultative heterochromatin
   - Regions can turn on/off when necessary (e.g. gene promoters that don’t need to be active at all times i.e. developmental genes)
   - H3K27me2/me3 → methylation of Lys27 by H3, tri- or di- methylation
   - Non-permanent → deposited by PRC2
   * Need heterochromatin → otherwise DNA too big
   * Histone modifications repress transposons

Question 71

PCR2

Answer 71

• EZH2 = catalytic subunit (enzyme)
• Other subunits in complex: EED + SUZ12

Question 72

Typical signature of active enhancers

Answer 72

• Always acetylated in Lys27 of H3; 1x methylated in Lys4
• Sometimes methylation is a marker of activity
• Signature used by Tf to understand where to go; if chromatin is active or not
• Active promoters have trimethylation of H3 Lys4

Question 73

Acetylation and methylation enzymes

Answer 73

• Histone acetylases and deacetylases
• Methyltransferases and demethylases
• If these are impaired → histone code is aberrant → have inappropriate labels → e.g. gene that should be repressed is labelled w/active markers → leads to disease e.g cancer

Question 74

Epigenetics and Cancer

Answer 74

• Epigenetic change that silences tumour suppressor gene → lead to uncontrolled cellular growth
• Turn off genes that help repair damaged DNA → lead to increase in DNA damage → cancer risk
• Prostate cancer associated w/gene silencing by CpG island hypermethylation within promoter region of GSTP1 gene
• If issue upstream in epigenetics → leads to cascade of problems

Question 75

X inactivation

Answer 75

• Example of epigenetics
• Marsupials: paternal X chromosome always silenced
• Tortoiseshell cat:
  o All female
  o Black/orange alleles of fur coloration are in X chromosome
  o If heterozygous →resulting colour depends on which X is inactivated
  o Tortoiseshell pattern (phenotype) determined by X inactivation

Question 76

Agouti Mice

Answer 76

• Genetically identical mice
• Mother 1 →skinny-brown mouse→ methyl-rich diet → methylated agouti gene repressed
• Mother 2 →obese, yellow mouse prone to diabetes/cancer→ unmeth. agouti gene expressed
• Agouti gene common to all mammals

Question 77

Epigenetics and Twins

Answer 77

• Identical twins = identical genes → differences due to epigenetic changes → lead to different disease susceptibility for example
• Label different histone modifications w/fluorescent probes → chromosome pairs in twins digitally superimposed →one tag red other green  overlapping shown as yellow

Question 78

Difficulty identifying inheritance

Answer 78

• Histone modifications can be inherited  environmental factors impact it (e.g. smoking mother)
• Changes in epigenome inherited due to mechanisms (still under study) that allow cells of offspring to remember epigenome of parents
• Complicated to understand where inheritance is coming from → always present in family? → something environmentally driven?
• Need to show that epigenetic effect can pass through enough generations to rule out possibility of direct exposure → potentially 3 generations at once exposed to same environmental conditions → prove epigenetic inheritance requires epigenetic change in 4th generation
• WW2 in Netherlands → extreme lack of food  diet poor in methyl groups → kids born now show these effects → e.g. hypomethylation of IGF2 (insulin-like growth factor) involved in diabetes and cardiovascular diseases

Question 79

Erasing of methylation

Answer 79

• DNA of human sperm is highly methylated; eggs less
• Egg fertilized → methylation/acetylation in chromatin largely erased esp. from paternal genome
• As embryo develops, methylation marks continue to be lost from maternal genome up to blastocyst stage
• Not all methylation erased, hence inheritance possible
• After this → cells differentiate → DNA becomes methylated again for specialized cell types

Question 80

How is methylation inherited

Answer 80

• Methylation patterns normally erased in primordial germ cells
• Methylation marks converted to hydroxymethylation then progressively diluted as cells divide
  o V. efficient mechanism → resets methylation patterns of genes for each generation
• Research found: some rare methylation can escape this reprogramming process → can be passed on to offspring → enables inheritance of epigenetic traits

Question 81

Imprinting

Answer 81

• Most genes inherit 2 working alleles from mother & father
• Imprinted genes inherit only one working copy → other is epigenetically silenced by addition of methyl groups → repressed forever but reset during egg/sperm formation
• Expression of gene comes from only 1 allele → less protein or less variation
• Almost all imprinted genes associated w/growth

Question 82

Genetic conflict hypothesis

Answer 82

• Theory of why imprinting happens
• Males multiple offspring w/multiple partners simultaneously at low cost
• Females only one set of offspring at a time at high cost of personal resources
• Many imprinted genes involved in growth and metabolism
• Paternal imprinting favours production of larger offspring; maternal imprinting favours smaller offspring

Question 83

Epigenetics in plants

Answer 83

• Plants depend on epigenetic processes for proper function
• Flowering controlled by set of genes affected by environmental conditions through alteration in expression pattern → ensures production of flowers even when plants are growing under adverse conditions
• Epigenetic modifications include DNA methylation, histone modifications, and production of micro RNAs (miRNAs)