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BMSC320 > Gene expression and information transfer > Flashcards

Gene expression and information transfer Flashcards

1
Q

How do genes respond differently and uniquely to the same stimuli?

A
  • vary in # of response elements
  • vary in sequence of response element
  • engage an accessory element (s) to modulate its activity
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2
Q

response element vs composite response element

A

response element: having only one component/sequence such as TATA box
composite response element: having several components/two or more sequences like most promoters

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3
Q

How does reporter assay work to look at response elements?

A
  • fuses gene of interest to a gene whose product is visible (reporter gene)
  • where reporter gene product is present, the gene of interest’s product is present
  • can examine glucocorticoid responsiveness of Tyrosine aminotransferase (TAT) promoter for CAT; distal upstream glucocorticoid response element (GRE)
  • reporter gene codes for chloramphenicolacetyl-transferase (CAT) which acetylates chloramphenicol
  • if CAT isn’t expressed, GRE isn’t expressed
  1. fuse promoter element (TAT) with reporter gene (CAT)
  2. add 1-3 GREs upstream of TAT
  3. add dexamethasone (drives expression from GRE)
    - if no GRE: basal activity of CAT
    - if 1 GRE: no increase in CAT activity from basal activity
    - if 2 GRE: max increase in CAT activity
    - if 3 GRE: slight decrease from max CAT activity
  • CCAAT box (another promoter) increases CAT expression from basal rate when added upstream of GRE but only 1/4 of the max CAT expression
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4
Q

what are the 3 phases of transcriptional elongation?

A
  1. promoter escape
  2. proximal pausing
  3. productive elongation
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5
Q

Describe proximal pausing in transcriptional elongation

A
  1. PIC complex and initiation of transcription, RNA pol II begins transcribing approx. 20-60 nt
  2. NELF (negative elongation factor) and DSIF (DRB sensitivity inducing factor) are recruited and cause RNA pol II to pause to ensure it is transcribing the correct sequence
  3. If everything looks good, P-TEFb phosphorylates NELF and DSIF, RNA pol II is released and can continue transcribing
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6
Q

what can pausing be caused by?

A
  • bulky DNA lesion (cyclobutane pyrimidine dimers: CPDs) result: damage needs to be removed before RNA pol can continue transcribing, triggers TC-NER
  • less bulky DNA lesion such as mismatched base pairs. result: DNA damage tolerance may bypass mutation and keep transcribing, may produce mutations in mRNA
  • transcriptional pause site. result: continuation of productive elongation
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7
Q

Describe productive elongation in transcriptional elongation

A

SAGA complex acts on chromatin to initiate and elongate transcripts: main role to acetylate and sometimes ubiquinate histones (considered a HAT)
- many other histone acetyltransferases (HATs) modify proteins

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8
Q

what is SAGA?

A

Spt-Ada-Gcn5-Acetyltransferase
- recruited to chromatin via interaction with transcription factors such as Gal4
- interactions strengthened through binding to modified histones (need premodification of histones to keep SAGA around)
- interaction with pol II and TBP are facilitated through 19S RP

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9
Q

how does SAGA facilitate elongation?

A
  • acetylating and ubiquinating histones along coding region: drives transcription forward (goes along with RNA pol II and acetylates histones ahead of pol II to loosen up interactions and make DNA more accessible)
  • nucleosome eviction: removes histones from DNA then replaces it once RNA II has passed
  • coupling transcription to mRNA export: helps shuttle mRNA out of nucleus while transcription is ongoing
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10
Q

what signalling pathways regulate SAGA?

A
  • MAPK (mitogen activated protein kinase) pathway
  • PKC (protein kinase C) pathway
  • ER function
  • p53 (TF that inhibits proliferation)
  • nuclear receptors
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11
Q

What are 2 main types of epigenetic control?

A
  1. DNA methylation
  2. Histone modification

Epigenetics refers to genetic regulation by factors other than DNA sequence
An epigenetic trait is a stably inherited phenotype resulting from changes in a chromosome without alterations in DNA sequence

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12
Q

what is the role of chromatin in regulation of transcription?

A
  • euchromatin: loosely packed coding DNA
  • heterochromatin: tightly packed non-coding DNA
  • difference between the two types is based on modification of histone tails
  • control is important for aging of cell
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13
Q

what is a nucleosome?

A
  • the basic unit of DNA packaging
  • 146 bps of DNA wrapped twice around histone octamer made of 2 copies each of H2A, H2B, H3, and H4
  • histone core is tight but the tail ends of DNA (where DNA enters and exits the nucleosome) are looser
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14
Q

how do histone tails influence the nucleosome?

A
  • each histone has a tail which can contain up to 30 amino acids, some of which can be modified
  • massive amount of exposed protein because of tails sticking out: DNA is very protected and not visible
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15
Q

what is the histone code?

A
  • describes the 4 main types of histone modification that have effect on gene expression: acetylation, methylation, phosphorylation, and ubiquitination
  • sites of modifications aren’t random/can be specific, acts as a code to provide predictability
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16
Q

what effect do histone modifications have?

A
  • can be activating or repressing
  • lysine methylation (can be mono, di, or tri methylated): repressive
  • lysine acetylation: activating
  • histone modifications can be used as markers: H2AX is recruited to broken DNA ends to initiate DNA repair
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17
Q

Histone Acetylation and deacetylation

A
  • activation of gene transcription often (but not always) involves histone acetylation
  • repression of gene transcription is often associated with deacetylation
  • histone acetyltransferases (HATs) and histone deacetyltransferases (HDACs) exist in the nucleus and the cytosol and include co-regulators and certain TAFs that have HAT activity
  • Gcn5 of SAGA is a HAT
  • chemical HDACs are used in anti cancer agents
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18
Q

Histone methylation (lysine and arginine)

A
  • protein arginine (R) methyl transferase (PRMT) has two types
    type I: asymmetrical methylation: a single amino acid gets 2 methyl groups
    type II: symmetrical methylation: 2 amino acids each get 1 methyl group
  • effects on genes can be positive or negative
    -methylation effects vary and depend on specific residue, degree of methylation, and which gene it occurs on
  • mono-methylation can be activating and tri-methylation can be repressing on the same residue
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19
Q

Histone phosphorylation (serine)

A
  • phosphorylation of ser10 on H3 can lead to activation or repression of gene transcription depending on other histone modifications and on the stage of the cell cycle
  • if phosphorylated at ser10 but no other Lys around are phosphorylated, the cell is going through mitosis
  • phosphorylation of ser10 at mitosis facilitates chromosome condensation
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20
Q

Histone ubiquitination (Lysine)

A
  • occurs primarily on H2A and H2B
  • required for gene activation
  • appears to affect other histone modifications
  • can by polyubiquitination (add multiple ubiquitins)
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21
Q

Histone modification nomenclature

A

1: which histone is modified (H2A, H2B, H3, H4)
2: one letter amino acid code (K = lysine, S = Serine, R = arginine)
3: which amino acid # is being modified
4: subscript type of modification (acetylation: Ac, phosphorylation: P, Ubiquitination: Ub, Mono methylation: me1, dimethylation: me2, trimethylation: me3

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22
Q

What does H3K27^me2 mean?

A

Lysine 27 on histone 3 gets methylated twice

23
Q

what does H2AK19^ub mean?

A

Lysine 19 on histone 2A gets ubiquitinated

24
Q

What is an example of histone code and chromatin remodelling

A

activation of interferon β gene through successive series of events driven by histone modifications that recruit 2 protein complexes

  1. Gcn5 within SAGA complex interacts with histones via histone tails
  2. SAGA acetylates the histones so the tails release Gcn5
  3. Histone tails are now free to bind protein kinase
  4. protein kinase phosphorylates tails on ser10, further promoting acetylation of Ser14
  5. SWI/SNF (remodelling complex) is recruited
  6. SWI/SNF pushes things around to expose the TATA box
  7. TFIID can bind TATA box and recruit RNA pol II
  8. Transcription can occur
25
Q

Examples of regulatory aspects of histone modification

A
  • Hr4K16ac: present in promoters of active genes
  • H3K23me3: observed in promoters of low expressed genes, associated with repression of transcription elongation (lost in aging cells/progeria)
  • H3K4me3: associated with transcription initiation
  • H3K36me3: associated with removal of histone acetylations in the wake of an elongating RNA pol II (when SAGA is modifying histones)
  • absence of H3K4me3 in leukemias, correlation between changes in H3K4me3 and H3K9me3 and heart failure
26
Q

C.H. Waddington’s definition of epigenetics

A

the study of how phenotype is affected by the interaction between genes and their products

27
Q

Epigenetics = epigenesis + genetics

A

describes morphogenesis and development of an organism, and changed in morphology/phenotype due to changes in gene expression/genotype that depends on mechanisms other than changes in DNA sequence

28
Q

waddington’s epigenetic landscape

A
  • development looks like a bead running down a hill with many options, as it rolls it comes to a trough and has to go either left or right
  • the further it goes, the more decisions have to be made
  • decision maker (bead at the top of the hill, can make anything) is totipotent
  • developmental decisions (forked trough) = no longer totipotent, now lineage dependent
  • can become plurpipotent (reversible) using induced pluripotent stem cells (iPS) which are made by expressing Oct14, Sox2, Myc, and Klf4
  • can also use Yamanaka factors to reverse progeria but they can cause cancer
29
Q

4 Developmental potentials and epigenetic statesof cells at different stages of development

A
  1. Totipotent (zygote): global DNA methylation (has the capacity to do anything)
  2. Pluripotent (ICM/ES cells, EG + EX cells, mGS cells, iPS cells): only active X chromosomes; global repression of differentiation genes by polycomb proteins; promoter methylation (start to turn of genes that are not needed)
  3. Multipotent (adult stem cells/?partially reprogrammed cells): X inactivation; repression of lineage-specific genes by polycomb proteins; promoter hypermethylation
  4. Unipotent (differentiated cell types such as B cells, macrophage, fibroblast, muscle, etc.): X inactivation; derepression of polycomb sliced lineage genes; promoter hypermethylation
30
Q

Describe DNA methylation

A
  • covalent attachment of methyl group onto cytosine nucleotides
  • occurs at CG dinucleotide within CpG islands (associated with 60% of promoters in humans)
  • involved in imprinting: gene from only one parent is methylated
  • there are 3 known DNA methyltransferases (DNMT1,2,3)
  • hyper and hypomethylation of genome can lead to disease
31
Q

what problems arise from improper methylation?

A

hyper methylation: tumor supressor genes (silencing)
hypo methylation: repetitive sequences (activating)

32
Q

what is CpG island?

A

200bp region with >50% GC content and is CpG rich (high GC content leads to enrichment of CpGs)
- 60% of genes have them
- occur at a frequency less than what one would expect due to spontaneous and high mutation rate of methylated cytosines to thymines

33
Q

distribution of CpG and GpC sequences

A

CpG: enriched in upstream promoter region, non random
GpC: randomly distributed throughout upstream region and body of gene

34
Q

role of cytosine methylation in gene transcription regulation

A
  • promoters of actively transcribed genes with CpG islands are usually unmethylated (and therefore active) and in open nucleosome configuration
  • silenced genes possess CpG islands that are methylated and associated with chromatin condensation
  • CpG island-containing promoters can be methylated at specific sites which prevents binding of transcription factors
  • genes without CpG islands can still contain single CpG sites which can undergo methylation with funcitonal consequences
35
Q

effects of DNA methylation on gene expression

A
  • no methylation means transcription factors can bind and increase gene expression
  • methylation can either prevent transcription by reducing binding of transcription factors (decrease gene expression) or recruit factors that repress transcription such as methyl-CpG binding proteins recruit HDACs and co-repressors which decreases expression
36
Q

relationship between DNA methylation, histone deacetylation, and histone methylation

A
  • DNA methylation leads to recruitment of HDAC and deacetylation of histones which represses transcription
  • DNA methylation leads to recruitment of histone methyltransferases with repressive effects on transcription
  • Histone methylation leads to recruitment of DNMTs leading to DNA methylation
  • H3K4me prevents methylation of promoters and is associated with gene transcription activation
37
Q

What is the process of inheritance?

A
  • involves histone chaperones, histone modifying proteins, and DNMTs
  • modifications from parental strand will be spread to both daughter strands and act as template for further modification
  • the of the modifying enzymes move along with the replication fork and are ready for when they need to act
38
Q

What is epigenetic control of sex-based dosage compensation?

A
  • species with chromosome based sex determination must have a form of dosage compensation or risk massive developmental flaws
  • in drosophila, increase expression of single X chromosome in males is dependent on chromatin formation and histone modification
39
Q

Histone modifications in dose compensation in drosophila

A
  • polytene chromosomes from male drosophila salivary glands are stained for H4K16 acetylation which is a marker for active genes
  • found that male X chromosome is rich in H4K16ac which indicates more expression on X chromosome
  • levels of H4K5/8/12 are similar to autosomes (no difference in expression)
  • happens because H4K16ac co-localizes with members of the dosage compensation system (MLE, MSL1, MSL2 (only male-specific protein), MSL3)
  • MOF acetylates H4K16, compensation members are present even in the absence of H4K16
40
Q

Dosage compensation in male drosophila

A
  • SXL is absent/off and prevents assembly of the MSL complex (which is required for acetylation of H4K16)
  • MSL complex is free to form
  • MOF (part of MSL complex) acetylates H4K16 - opens up chromatin of the X chromosome
  • transcription occurs at a double rate because chromatin is open
41
Q

Dosage compensation in female drosophila

A

-SXL is expressed and prevents assemble of MSL complex
- no acetylation occurs
- transcription occurs at normal rate

42
Q

what is the X:A ratio?

A
  • how many Xs to how many sets of As is how dosage compensation is recognized
  • 1.0 = female : SXL expressed, TRA and DSXf on, MSL2 off
  • 0.5 = make : X- needs upregulation, MSL complex forms and binds X-linked genes which increases transcription
43
Q

Describe silencing

A

non-coding RNAs (ncRNAs)
- many small RNAs direct DNA methylation and histone modification because they play a role in imprinting (one gene in male or female is expressed when the other isn’t) and silencing of heterochromatin and genomic repeats
- Barr bodies are silent X chromosomes: female mammals only express one X chromosome early in development, but without compensation is lethal (Turner syndrome)
- random silencing occurs early in embryonic stage after cell differentiation (mosaic expression)

44
Q

describe Barr bodies

A

Xa = active
Xi = inactive (resembles heterochromatin)
- increased DNA methylation so genes are repressed
- low histone acetylation
- low H3K4 methylation (which marks active transcription)
- high H3K9 methylation (marks transcriptionally silent genes)
ncRNA Xist facilitates silencing of one X chromosome is only expressed by Xi (physically binds chromatin it’s on so we know there is a silent X chromosome where ever it is present)
- inactivation requires presence of X-inactivation center (XIC) on X chromosome

45
Q

what are monozygotic twins?

A

identical = same DNA = expect the same phenotype
- methylation between twins is similar at birth/when young, but differs as they get older = phenotype differs with age

46
Q

Aging and epigenetics

A
  • cancer and aging show alterations in chromatin formation: heterochromatin is activated, and euchromatin (and its important genes for cell survival, growth, and control) is silenced by DNA methylation and histone modifications
47
Q

Epigenetic clock

A
  • search for biological markers to predict biological age of a cell vs chronological age of the organism
    -DNA methylation changes with age and is predictive
  • highest when born and slowly peters out with age
  • DNAm is affected by diet, seasonal correlations, disease exposure, toxic chemicals, psychological state, exercise, etc.
48
Q

chronological age vs biological age

A

chronological: calendar time that’s passed since birth - negative numbers = prenatal ages, positive numbers = postnatal ages
biological/physiological age: organismal age or phenotypic age (subjective) - dependent on biological state of the organism

49
Q

explain epigenetic age

A
  • age estimated in years from mathematical algorithm based on methylation state of specific CpGs in genome
50
Q

Age deceleration/accerleration

A

deceleration: healthy food intake, exercise
acceleration: gender, cancer risk/incidence, neurodegenerative disease, BMI/metabolic syndrome markers, stress, infections, etc.

  • an individual’s DNA methylome is the perfect track record of past environmental exposures and read-out to monitor health status to estimate future disease risk or life expectancy
51
Q

Explain epigenetic drift

A
  • stochastic (random and inevitable loss of DNAm heratibility with DNA replication) and environmental (random and specific exposure to certain environments influences DNAm) effects on DNAm are random and increase with age
52
Q

why can estimating epigenetic age be useful?

A
  • investigating how epigenetic age differs from a group of people of same chronological age can help determine impact of endogenous and exogenous factors on biological aging
  • epigenetic changes are reversible which means that DNAm age estimates might be useful for identifying or validating anti-aging interventions
53
Q

What bioinformatic methods are needed to estimate DNAm age?

A

microarrays, pyrosequencing, quantitative PCR, and next-generation sequencing

DNAm age estimators are usually built by regressing transformed version of chronological age on set of CpGs using supervised machine learning methods

54
Q

How to measure DNA methylation?

A

bisulfite conversion
- bisulfite converts Cs to Us which causes a C/G to T/A conversion in sequence
- bisulfite will not convert methylated Cs
- number of converted C/G to T/A pairs indicates degree of methylation

BMSC320 (9 decks)

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