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BIOL253: Genetics > Transcription and control of gene expression > Flashcards

Transcription and control of gene expression Flashcards

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1
Q

List the basic steps involved in transcription

A
  1. Promoter recognition = RNA polymerase recognises and binds to promoter sequence
  2. Promoter opening = RNA polymerases separates the two strands of DNA to form a 14 bp bubble
  3. Initiation = RNA polymerase begins synthesising the first few nucleotides de novo
  4. Promoter clearance = once the RNA polymerase passes the promoter sequence it undergoes a conformational change that stabilises its interaction with the DNA
  5. Elongation = RNA polymerase continues elongating the complex in a 5’ to 3’ direction
  6. Termination = RNA polymerase reaches a termination sequence causing release from template
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2
Q

the transcribes pre-mRNA sequence has the same sequence as which strand of the DNA?

A

sense/non-template strand

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3
Q

Describe the structure of the simplest RNA polymerase?

A

Found in bacteria:
Has 5 subunits:
- beta and beta’ are the large subunits that bind to the DNA and form the active site at the point of intersection
- alpha I and alpha II subunits have a N-terminal domain closely bound to the beta and beta’ subunits. The N-terminal domain is linked by a flexible joint to the C-terminal domain (alpha CTD) that binds to an upstream promoter element (UP) of the DNA
- omega subunit

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4
Q

How many subunits does the eukaryotic RNA polymerase II have and what are they called?

A

12 subunits

called:
- RPB 1 to 12

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5
Q

How are the bacterial, eukaryotic, and archaeal RNA polymerases structurally similar?

A
  • over structure of the core RNA polymerase is conserved, particularly in the active site (which conserves the catalytic mechanism of transcription)
  • -> common evolutionary source.
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6
Q

What is the difference between the prokaryotic core polymerase enzyme and the holoenzyme?

A

core enzyme = the core polymerase enzyme (5 subunit polymerase as previously described)

holoenzyme = the core polymerase enzyme with the sigma factor

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7
Q

What is the role of the sigma factor? how does it carry out this function?

A

locates the transcriptional start site by recognising the promoter.

  • it identifies specific conserved sequences within the promoter at the -10 element (by domain 2 of sigma factor) and the -35 element (by domain 4 of sigma factor) upstream of start site.
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8
Q

What is the primary sigma factor in E. coli?

A

sigma 70

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9
Q

What is the role of a primary sigma factor?

A

transcription of housekeeping genes

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10
Q

Give an example of an alternative sigma factor in E. coli

A

sigma 32 (sigma H - heatshock)

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11
Q

What is the role of alternative sigma factors?

A

transcription of genes required to allow bacteria to respond to environmental changes by upregulating all genes with the consensus sequence that can be recognised by that sigma factor

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12
Q

How can the sigma factors expressed by an organism reflect its range of environment?

A
  • if a bacterium lives in a radically changing environment (such as free-living instead of in the gut), it will express more sigma factors
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13
Q

How does the similarity of the promoter consensus correlate with strength of the promoter?

A
  • frequently transcribed genes will have a promoter sequence similar to the consensus sequence to allow tight interaction between the sigma factor and the promoter - RNA polymerase has higher affinity
  • infrequently transcribed genes may have a promoter sequence that is more different to the consensus sequence so sigma factor binds less tightly - RNA polymerase has lower affinity
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14
Q

What is the location and what is transcribed by the eukaryotic RNA polymerase I?

A

Location: nucleolus

Transcribes: rRNA (28S, 5.8S large ribosomal subunit, and the 18s small ribosomal subunit)

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15
Q

What is the location and what is transcribed by the eukaryotic RNA polymerase II?

A

Location: nucleoplasm

Transcribes: mRNA, snRNAs (small nuclear RNAs found in spliceosomes), miRNAs (regulate gene expression)

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16
Q

What is the location and what is transcribed by the eukaryotic RNA polymerase III?

A

Location: nucleoplasm

Transcribes: rRNA (final 5S component of large ribosomal subunit), tRNAs, snRNAs (small nuclear RNAs found in spliceosomes)

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17
Q

What are general transcription factors? give examples?

A

recognise the promoter and recruit the RNA polymerase to the transcriptional start site.

Examples:
- TFIIA, TFIIB, TFIID, TFIIE, TFIIF, TFIIH

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18
Q

Are general transcription factors composed of multiple subunits?

A

Yes, except TFIIB

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19
Q

What are DNA core promoter elements?

A

in eukaryotes, core promoter elements are DNA sequences upstream of the transcriptional start site that direct the loading of the transcription initiation complex

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20
Q

Describe how the general transcription factors nucleate the formation of the pre-initiation complex?

A

1) core promoter element TATA box is bound by TBP subunit of TFIID, this induces a kink deformation in the DNA as the A-T base pairs of the TATA box are easier to bend
2) this facilitates the binding of TFIIB to the BRE core promoter element

  • from here the other general transcription factors and RNA polymerase II are recruited to form the pre-initiation complex
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21
Q

in RNA polymerase II, how is promoter opening achieved?

A

TFIIH has a helicase subunit that separates the two strands requiring ATP

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22
Q

What is abortive initiation? why does this occur

A

the formation of short RNA transcripts released before the transcription complex passes the promoter (before promoter clearance)

Occurs because:

  • sigma factor (bacteria) or TFIIB (for eukaryotes) have a loop that extends into the active site region and blacks the elongating transcript from reaching the exit channel
  • RNA polymerase stalls and is unable to move from promoter
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23
Q

What is promoter clearance/how does this occur?

A
  • displacement of the sigma factor (bacteria) or TFIIB (eukaryotes) loop from the active site region aids in the breaking away of the polymerase from the promoter
  • polymerase undergoes conformational change:
  • -> increases association with DNA (stable)
  • -> decreases association with transcription factors
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24
Q

What is the equivalent of the bacterial polymerase alpha subunit in eukaryotes?

A

RPB1 - has CTD

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25
Q

Describe how eukaryotic RNA polymerase II couples elongation to mRNA processing

A

The 5th serine in heptad repeats of CTD region of RPB1 are phosphorylated by TFIIH kinase

This acts as signal for recruitment of negative elongation factors that bind to polymerase and temporarily arrest transcription

Phosphorylation also recruits RNA-processing enzymes that add the 5’ guanosine cap to the end of the mRNA

5; capping leads to phosphorylation of the 2nd serine in the heptad repeats of RPB1 CTD, causing dissociation of negative elongation factors and resumes elongation

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26
Q

What supercoiling occurs as a result of the unwinding of the DNA strands?

A
  • positive supercoiling ahead of polymerase

- negative supercoiling behind polymerase

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27
Q

How can the formation of supercoils affect transcription progression?

A
  • can stall RNA polymerases so tension must be relieved by topoisomerases
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28
Q

What are the topoisomerases in E. coli responsible for the resolution of supercoiling created by the polymerase?

A
  • DNA gyrase = removes positive supercoils

- DNA topoisomerase I removes negative supercoils

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29
Q

What are the roles of histone chaperones during eukaryotic transcription? Give some examples of histone chaperones

A

remove nucleosomes ahead of polymerase and reassemble them behind polymerase

Examples: FACT (facilitates chromatin transcription), Asf1, Spt6

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30
Q

How can RNA polymerase correct transcription mistakes?

A

1) RNA polymerase stalls
2) RNA polymerase reverses direction causing the most recently transcribes RNA to protrude from the complex
3) this protruding RNA is cleaved by RNA polymerases endonuclease activity (which may or may not require stimulation by extrinsic transcript cleavage factors)

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31
Q

Describe the two types of terminators in E. coli?

A

1) Type I terminators (aka Rho-independent terminators, or Intrinsic terminators)
- no additional factors required for termination
- formation of hairpin within region of self-complementarity followed by a run of uridines - destabilises interaction between RNA and DNA template

2) Type II terminators (aka Rho-dependent terminators)
- require Rho factor (uses ATP)
- Hexameric Rho ring structure assembled at particular RNA sequences around the RNA and facilitates dissociation

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32
Q

Describe the two models of termination by eukaryotic RNA polymerase II

A

Model 1: allosteric model

  • poly-A-tail added through adenylation
  • RNA polymerase continues to transcribe after polyadenylation signal and cleaves the 3’poly-A-tail causing conformational change that destabilises interaction of RNA with DNA template and RNA polymerase dissociates

Model 2: torpedo model

  • poly-A-tail added though adenylation
  • cleavage at poly-A-tail
  • RNA downstream of poly-A-tail is digested by 5’ to 3’ ribonuclease Rat1
  • disrupts polymerisation and causes polymerase dissociation from the DNA
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33
Q

Name the five possible methods of pre-mRNA processing to produce functional mRNa

A

1) 5’ guanine cap
2) 3’ Polyadenylation (Poly-A-tail)

3) Cleavage
- exonuclease (digestion from one end other other)
- endonuclease (cut specific sequence within DNA)

4) splicing
- removal of introns

5) editing
- base insertion
- base deletion
- base modification (e.g. deamination of adenosine produces inosine)

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34
Q

Describe how RNA cleavage is used to process precursor RNA

A

rRNA and tRNA usually synthesised as long precursor RNA

- processed to the correct length by ribonucleases

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35
Q

Describe the 5’ cap structure at the end of eukaryotic mRNA and its functions

A

guanine ribonucleotide attached to the mRNA via a 5’ to 5’ triphosphate link (so 5’ end of mRNA actually has an exposed 3’ from the guanine ribonucleotide

Functions:

  • protect from degradation
  • enhances translatability of mRNA
  • transport from nucleus to cytoplasm
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36
Q

Describe the steps involved in the 5’ capping process

A

1) an RNA triphosphatase removes the terminal 5 phosphate
2) guanylyl transferase uses GTP to attach a GMP
3) guanine of GMP is methylated by a methyltransferase

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37
Q

What is special about the RNA triphosphatase and guanylyl transferase enzymes?

A

They are part of a bifunctional enzyme complex

38
Q

Describe the steps involved in the 3’ polyadenylation of eukaryotic mRNAs (except those encoding metazoan histones)

A

polymerase reaches a polyadenylation signal (5’ hexameric AAUAAA sequence followed by downstream CA, followed by downstream U or GU rich region)

mRNA is cleaved immediately after CA between the AAUAAA sequence and the U/GU rich region.

poly(A)polymerase adds approximately 200 adenosines to the 3 end

39
Q

true or false: poly adenylation is added post-transcriptionally?

A

True

40
Q

What is the spliceosome? what is in comprised out?

A

In metazoa, it facilitates the removal of introns at intron boundaries to produce mature mRNA

comprised of 5 small nuclear ribonucleoproteins (snRNPs) each of which contains proteins a snRNAs (transcribed by RNA polymerase II and III)

41
Q

Describe the steps involved in splicing

A

1) transesterification reaction: branch point adenosine hydroxyl attacks phosphodiester bond between intron and exon and forming a new phosphodiester bend between the adenosine and the adenosine hydroxyl and the 5’ phosphate of the intron, leaving behind a 3’ OH of the exon
2) transesterification reaction: the 3’ OH of the exon attacks the phosphodiester bond between the intron (3’ end) and exon (5’ end of next exon) forming a phosphodiester bond between the 3’ OH of one exon and the 5’ phosphate of the next (an exon junction complex/EJC is deposited at splice junctions)
3) the intron is excised in a lariat structure

42
Q

What defines intron boundaries?

A

5’ GU

3’ AG

43
Q

How does the cell know that splicing has occurred?

A

The presence of the EJC marks the transcript as having been processed and now suitable for export from the nucleus

44
Q

What is alternative splicing?

A

generating different mature mRNA from the same genes by combining different exons and producing different proteins = increase diversity that can be produced by the cell

45
Q

What percentage of human genes have alternative splice forms?

A

90%

46
Q

How is the splicing machinery recruited to the RPB1 CTD of RNA polymerase II?

A

when the CTD is fully phosphorylated at serines 2 and 5 of the heptad repeats

47
Q

True or false: processing is intimately linked to transcription?

A

True

48
Q

What is RNA editing?

A

adds or deletes bases from pre-mRNA, or chemically alters bases
- result in mRNA with bases that don’t match the DNA sequence

49
Q

Describe RNA editing in Trypanosomes

A

blood borne parasite transmitted by tsetse fly between mammals - cause trypanosomiasis in humans/sleeping sickness
- DNA genome of single mitochondrion located in the kinetoplast at the basal body of the flagellum

When align the DNA sequences of cytochrome oxidase subunit III mitochondrial gene of T. brucei and other close relatives:
- see that T. brucei has missing nucleotides from the genomic DNA but not from the cDNA

RNA editing: addition of uridines that were not present in the DNA template

  • mitochondrial genome also encodes guide RNA (gRNA) molecules that can anneal to mRNA
  • some sites have imperfect base pairing, resulting in looping out of gRNA
  • endonuclease cuts mRNA at mismatch
  • uridylyl transferase adds uridines to 3’ end of mRNA at cut, guided by the gRNA
  • the nick is ligated by RNA ligase

RNA editing: removal of uridines not required in the final transcript

  • gRNA anneals to mRNA imprecisely creating a mismatch
  • endonuclease cleaves mRNA at mismatch
  • uridine is cleaved off by exonuclease
  • nick ligated by RNA ligase
50
Q

What type of RNA editing is seen frequently in eukaryotes? give an example of a gene that is edited in this way

A

substitution editing

example: editing of gene for human apolipoprotein B (APOB gene)

51
Q

Describe the process of substitution editing of APOB gene

A

codon CAA at 2153 position encodes glutamine (Gln)

Liver cells: produces full length mRNA that encodes APOB100 protein important for cholesterol transport

Intestinal cells: additional step of pre-mRNA processing where there is deamination of the C nucleotide in CAA codon 2153 to U (changing codon from CAA to UAA, which is a stop codon)
- produces smaller mRNA that encodes for a truncated APOB48 protein important for uptake of lipids in small intestine thorough chylomicrons.

52
Q

name one determinant of RNA stability in eukaryotes

A

poly-A-tail

53
Q

Describe the mechanisms of RNA degradation

A

exonuclease deadenylases recruited to 3’ end and remove the poly-A-tail causing mRNA to be destabilised

  • -> this can have have two effects:
    1) another exonuclease degrades the RNA in 3’ to 5’ direction
    2) decapping recruited to remove the 5’ cap and results in recruitment of exonuclease that degrades the RNA in 5’ to 3’ direction
54
Q

How is RNA degradation controlled, giving an example?

A
  • AREs (AU rich elements) are found in the 3’ untranslated region (3’ UTR) of some mRNAs (often seen in mRNA of proteins that only need to be present for a short period of time)
  • AREs recruit AUBPs (AU binding proteins) that either recruit a decapping enzyme and result in 5’ to 3’ exonuclease RNA degradation, or recruit a deadenylase enzyme resulting in 3’ to 5’ exonuclease degradation of RNA.

Example:

  • fos is a proto-oncogene that encodes a transcription factor that promotes proliferation, contains an ARE in mRNA so has a short half-life
  • viral fos contained by tumour-promoting viruses lacks ARE in mRNA so has a long half-life = mRNA remains for longer and promote uncontrolled proliferation.
55
Q

Describe the small RNAs that mediate targeted RNA degradation in eukaryotes

A

uses two types of sRNA:

1) siRNAs
- derived from processing longer dsRNAs so no processing in the nucleus (all occurs in cytoplasm)
- exonuclease activity by Dicer/TRBP2 complex breaks down into smaller functional siRNA duplexes

2) miRNAs
- derived from RNA polymerase II transcripts and has regions of self-complementarity that enables it to form a hairpin structure
- 3’ poly-A-tail and 5’ cap are cleaved off to produce pre-miRNA that is exported from nucleus
- exonuclease activity by Dicer/TRBP2 complex produces the functional miRNA duplex

Both functional miRNA and siRNA duplexes can be loaded onto argonaut (Ago), one strand in degraded to produce mature miRISC or siRISC

56
Q

Describe the process of RNAi

A
  • dsRNA cleaved by dicer into smaller fragments
  • smaller fragment loaded onto argonaut and one strand is degraded to form the RISC complex
  • guide strand directs the cleavage and degradation of complementary mRNA
57
Q

How do miRISCs silence genes?

A

bind to the 3’UTR of mRNA where they:

  • repress translation
  • activate de-adenylation, de-capping, exonuclease degradation

(no cleavage as they are not totally homologus to target)

58
Q

How do siRISCs silence genes?

A

perfectly base pair with the target mRNA resulting in cleavage of the mRNA by argonaut.

cleaved mRNA is then degraded.

59
Q

Why do cells need to regulate transcription?

A

1) may require different quantities of different proteins
2) may only need to produce some proteins at particular times
3) some cells only ever need to produce a subset of proteins encoded by the genome

60
Q

In eukaryotes: what are the two domains of a transcription factor?

A

DNA binding domain (recognises specific sequence)

Transactivating domain (interacts with other proteins to assemble RNA polymerase onto start site)

61
Q

What are the three types of DNA binding domains?

A

1) Helix-turn-helix motif (found in all types of organisms) - Two alpha helices joined by a tight turn

2) zinc-finger domain (very common in humans)
- alpha helix with beta sheet orientated by binding a zinc via Cys and His residues

3) coiled-coil (found in leucine zipper and helix-loop-helix motifs)
- extended alpha helices (disordered when not bound to DNA)

62
Q

What are the transcriptional regulatory mechanisms in prokaryotes?

A
  • promoter strength (similarity between promoter sequence and consensus sequence recognised by particular sigma factor)
  • Targeted gene regulation (most often at initiation)
  • Repressors and activators (regulatory proteins that bind to operators - proximal regulatory sequences)
63
Q

What are the transcriptional regulatory mechanisms in eukaryotes?

A
  • enhancers - distal regulatory sequences upstream or downstream of the gene (DNA folds so protein bound to enhancer can still interact with the RNA polymerase bound at the promoter)
64
Q

How does the control of gene expression of regulated genes differ in prokaryotes and eukaryotes?

A

prokaryotes = transcriptional regulation

eukaryotes = mostly post-transcriptional regulation

65
Q

What is an operon? give two examples

A

co-located genes that encode for proteins that work together to perform a particular cellular function
- the genes are transcribed into one polycistronic mRNA which is then translated into each protein

examples:

  • lac operon (metabolise lactose)
  • trp operon (synthesise tryptophan)
66
Q

What is the main carbon and energy source for E. coli?

A

glucose

67
Q

Describe the structure of the lac operon (and the lacI gene)

A

Lac operon encodes three proteins via genes lacZ (beta-galactosidase - breakdown lactose into galactose and glucose, and can isomerise lactose into allolactose), lacY (permease - allow lactose to enter cell) and lacA (transacetylase)

the lacI gene produces the lac repressor protein (LacI) is immediately upstream of the operon and has its own promoter and is constitutively expressed

68
Q

Describe the regulation of the lac operon in E. coli

A

no lactose present:
- LacI binds to the lac operator and no transcription occurs as RNA polymerase unable to bind.

lactose present:
- allolactose (produced by low levels of lac operon synthesis of beta-galactosidase) binds to LacI, conformational change so no longer able to bind to operator so polymerase now able to bind to promoter

no glucose present:

  • adenylate cyclase active and produces large amounts of cAMP.
  • cAMP binds to CAP (catabolite activator protein) causes CAP to bind as dimer to DNA CAP operator upstream of promoter and bends DNA by 90 degrees (kink) - RNA polymerase more able to bind (cAMP-CAP interacts with the alpha-CTD to facilitate RNA polymerase holoenzyme binding)

glucose present:

  • inactivation of adenylate cyclase so dramatic reduction in cAMP concentration
  • no cAMP bound to CAP, so CAP not bound to DNA CAP operator
69
Q

State the scenarios in which the lac operon is strongly on, weakly on and off

A

Strongly on:
- lactose present (RNA polymerase can bind at promoter), glucose absent (cAMP-CAP enhances RNA polymerase binding)

Weakly on:
- lactose present (RNA polymerase can bind), glucose present (cAMP-CAP not bound so transcription is not enhanced)

Off:
- lactose absent (RNA polymerase cannot bind) regardless of present or absence of glucose.

70
Q

What two regulatory mechanisms is the lac operon under control by?

A
  • negative inducible regulation (LacI repressor)

and

  • positive regulation (cAMP-CAP)
71
Q

What type of metabolic pathways are negatively regulated by repressible regulation? give an example

A

anabolic/biosynthetic pathways (turned off when product readily available = negative regulation by repression)

Example: the trp operon

72
Q

Describe the structure of the Trp operon

A
  • encodes 5 genes for tryptophan biosynthesis
  • transcribed as polycistronic mRNA
  • turned off by excess tryptophan
  • the regulatory gene (trpR) is located away from the operon
73
Q

Describe the two regulation of the trp operon

A

1) at level of initiation: repressible regulation (on/off switch)
trpR encodes an aporepressor
- no tryptophan = aporepressor doesn’t bind to operator = transcription occurs
- tryptophan present = aporepressor binds tryptophan (acts a co-repressor) allowing binding to the operator to prevent transcription.

2) At level of production of full length transcript: attenuation
the trp leader mRNA has four regions (1, 2, 3, and 4) that can form alternative secondary structures due to self-complementarity and has sequence containing trp codons that encodes the leader peptide
- tryptophan starvation = polymerase stalls at the trp codons (incomplete leader peptide), trp leader mRNA forms hairpin between regions 2 and 3 (antiterminator structure) that allows transcription to proceed to produce full length mRNA encoding tryptophan biosythesis.
- presence of tryptophan = no stalling at trp codons (complete leader peptide forms) and polymerase reaches region 2 of the leader mRNA. This results in formation of hairpin followed by row of uridines (attenuator structure) between regions 3 and 4, which terminates transcription in region 2 to prevent transcription of structural genes

74
Q

What is the structure and function of a riboswitch? give an example

A

Structure:
- portion of a transcript that can directly bind a small molecule that controls the RNA secondary structure
- two regions: - the aptamer (binds metabolite)
- expression platform (controls
transcription or translation

Function:
regulates transcription or translation

Example:
adenine riboswitch

75
Q

Describe the adenine riboswitch in B. subtilis

A

regulates adenine synthesis and transport

low adenine levels:
- regions 2 and 3 form an anti-terminator structure (aptamer region cannot form) and transcription proceeds

high adenine levels:
- self-complementation between regions 1 and 2 form the aptamer region that binds adenine, and regions 3 and 4 form a type I terminator structure to prevent transcription.

76
Q

What is a type one terminator/attenuator strucutre?

A

hairpin loop followed by a row of uridines

77
Q

What are the three general effects of regulatory protein/transcription factor binding to DNA sequences?

A

1) aid oligomerisation (recruitment of other proteins)
2) activate or repress transcription
3) interact with other regulators that do not have DNA-binding activity themselves (co-activators or co-repressors - Trp)

78
Q

How does histone acetylation regulate transcription in eukaryotes?

A

Histone acetylation: modification of Lys residues

  • Hypoacetylation = strong internucleosomal interactions between positively charged Lys residues and negative DNA (DNA wrapped around histones occludes binding sites so transcription factors cannot bind)
  • Hyperacetylation by HATs = weak internucleosomal interactions as Lys now neutrally charged - DNA more accessible to transcription factors
79
Q

What are nucleosome remodelling complexes and how are they recruited? give an example

A

use ATP to move the histone octamer to expose the DNA binding sites and allow transcription factors to bind.

Recruited:

  • sequence-specifically
  • histone modification

Example: SW1/SNF

80
Q

How does DNA methylation regulate transcription in eukaryotes?

A
  • in mammals cytosine residues are methylated at CpG sites by DNA methyltransferase

hypomethylation = transcriptionally active

hypermethylation = transcriptionally silenced

81
Q

How can DNA methylation be an issue with regards to mutation?

A

methylation of cytosine produces 5-methyl cytosine which makes the amino group more unstable and susceptible to hydrolytic deamination

–> deamination produces thymine result in a mismatch that can result in a fixed mutation in replicated

82
Q

changes in methylation pattern occur in what three scenarios?

A
  • fragile X syndrome
  • cancer
  • aging
83
Q

How can DNA methylation and histone modification affect each other?

A

methyl groups on either DNA or histone tails results in recruitment of methyl-binding proteins, which recruit other proteins (such as histone and/or DNA modifying enzymes) –> affects expression of genes

84
Q

What is the mediator complex?

A

the mediator complex consists of about 20 proteins and is required by RNA polymerase to bind to the promoter and initiate transcription

85
Q

Describe transcriptional regulation by ELK1 in eukaryotes

A

REGULATION AT LEVEL OF INITIATION

  • absence of mitogens = ELK1 binds to serum response factor (SRF) bound to DNA = no activation of transcription
  • mitogen binds to cell surface receptor = mitogen-activated signal transduction pathway = phosphorylation of ELK1 = recruits the mediator complex = promote transcription of proliferative genes
86
Q

Describe transcriptional regulation by GAL in yeast (eukaryotes)

A

REGULATION AT LEVEL OF INITIATION - inducible transcription

GAL1, GAL7 AND GAL10 are co-located (but not in operon) encode enzymes in galactose metabolic pathway

  • GAL4 produces regulatory protein Gal4 (a transcriptional activator) that binds to UASG (Upstream Activator Sequence - Galactose) and activates transcription
  • presence of glucose (preferred carbon and energy source) = prevents transcription as glucose exerts catabolite repression
  • absence of galactose = Gal80 binds to Gal4 transactivation domain = preventing transcription
  • presence of galactose = Gal3 binds to Gal80, allowing Gal4 transactivation domain to recruit SAGA through Tra1 , SAGA recruits the mediator complex and the mediator complex recruits RNA polymerase = rapid transcription
87
Q

Describe transcriptional regulation by Nuclear receptor proteins in eukaryotes

A

REGULATION AT LEVEL OF INITIATION

Nuclear receptor proteins are specialist intracellular transcription factors that bind to particular signalling molecules (ligands - E.g. hormones) which results in:

1) conformational change that allows then to drive transcription (E.g. Oestrogen receptor - ligand enters nucleus binding of ligand causes co-repressor to dissociate from NRP and co-activator to bind)
2) translocation to the nucleus (e.g. glucocorticoid receptor - NRP bound to chaperone protein in cytoplasm, ligand (cortisol) binds, conformational change, chaperone dissociates and the NRP can enter the nucleus, bind DNA and recruit a co-activator to drive transcription)

88
Q

Describe transcriptional regulation by Ume6 in yeast (eukaryotes)

A

REGULATION AT LEVEL OF INITIATION

Ume6 responds to nutritional cues and can activate or repress transcription

  • plenty of nutrients = Ume6 binds DNA and recruits co-repressors (Sin3, Rpd3 - histone deacetylase (off), Isw2 - nucleosome remodelling enzyme)
  • absence of nutrients = signal transduction pathway results in phosphorylation of Ume6 so co-repressors dissociate and co-activator Ime1 (a histone acetyltransferase - on) is recruited = transcription occurs
89
Q

What is promoter proximal pausing in eukaryotes?

A

stalling of RNA polymerase II after 35-50 bp have been transcribed

  • promoted by negative elongation factors (E.g. NELF and DSIF)
  • initiation has already occurred so relief of this pausing rapidly upregulates gene expression in response to environmental change
90
Q

Describe transcriptional regulation by Hsf in Drosophila (eukaryotes)

A

REGULATION AT THE ELONGATION STAGE

regulation of the Hsp70 protein

Absence of heat shock:

  • GAGA transcription factor binds to GAGA sequence and recruits NURF (nucleosome remodelling factor)
  • NURF changes the structure of the chromatin to expose control elements on the DNA (such as HSE sequence)
  • RNA binds but negative elongation factors NELF and DSIF bind and prevent phosphorylation of the Rbp1 CTD = promoter proximal pausing.

Presence of heat shock (sudden temperature rise):

  • Hsf protein rapidly forms trimer that binds to the HSE sequence, interacting with the mediator complex
  • together with mediator, Hsf trimer recruits a kinase to phosphorylate the CTD = dissociation of negative transcription factors and relief of pausing

BIOL253: Genetics (6 decks)

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