Study Questions Set 8

Question 1

Describe the role of histone acetylation/deacetylation in regulation of transcription.

Answer 1

• histones are tightly bound to DNA due to the positive charges on the lysine groups, which bind to the negatively charged backbone
• when histones are acetylated, the positive charges on the lysine groups are neutralized
  o eliminates the tight bond with DNA
  o chromatin gets less condensed and DNA promoters become accessible
• when histones are deacetylated, DNA will more tightly bind to the histone
  o associated with repression of gene activity
• histone acetyltransferases are present in co-activators, while histone deacetyltransferases are present in co-repressors

Question 2

Describe the role of chromatin remodeling complexes in regulation of transcription.

Answer 2

• histone acetylation is essential but not sufficient for activation; nucleosomes are still intact, they have to be repositioned to expose the promoter elements
• histone octamers may slide along DNA; spacing between histone octamers may be altered producing nucleosome-free gap
• requires ATP and remodeling complex (e.g., SWI/SNF complex of yeast)
• (1) SwI/SnF proteins alter nucleosome core structure; (2) Swi proteins move nucleosomes on DNA so that enhancers are in nucleosome-free regions
• SwI/SnF are not considered transcription factors because they do not actually bind to specific DNA sequences, but are recruited by activators and repressors

Question 3

Describe an influence of activators and repressors on assembly of initiation complexes.

Answer 3

• Certain TFs need co-activator proteins that do not contact DNA but connect TF with general transcription factors to assemble the initiation complex
• Repressors can:
  o bind to sites that overlap activator binding sites
  o bind at site adjacent to activator site and interfere with activator
  o bind upstream, interact through mediator with GTFs to inhibit interaction
  o recruit co-repressors that alter nucleosomes in such a way to inhibit transcription

Question 4

Explain the role of enchancesomes and architectural proteins in regulation of transcription initiation?

Answer 4

• Enhancesome: allows for specific gene expression by interaction at the enhancer; complexes of tissue-specific transcription factors, non-specific transcription factors and co-activators with general transcription factors allows for regulation of various genes across many cell types
• Architectural Proteins: bind to specific binding sites in order to change the shape of a DNA in a control region so other proteins can interact with each other and/or GTFs and stimulate transcription; it is difficult for DNA to bend/loop if distances between core promoter and enhancer are short, therefore sometimes more than architectural TF is necessary

Question 5

Explain the role of mediators in regulation of transcription initiation.

Answer 5

mediators: proteins that are not in direct contact with DNA, but links transcription factors with general transcription factors; necessary for the assembly of pre-initiation complex
o without them, transcription factors would have no effect on transcription regulation

Question 6

Explain the role of insulators in regulation of transcription initiation

Answer 6

• insulators: set up boundaries between DNA domains
  o prevents activation/repression of genes by close but unrelated activators and repressors
  o also prevents gene silencing (spreads chromatin modifications – control boundaries between hetero and euchromatin – different on different tissues)

Question 7

What is the role of DNA methylation in regulation of transcription initiation?

Answer 7

• methylation causes genes to become inactive; methyl groups are attached to C in some CG doublets
  o i.e. CpG islands in promoters of housekeeping genes (genes constantly transcribed)
  o always unmethylated means they are always active
• methylation prohibits transcription
  o transcription factors can’t bind to promoter if they’re methylated
  o some binding proteins recognize methylated DNA and compete with transcription factors
  o methylation also affects chromatin structure
• cancer has been linked to global hypomethylation and gene specific hypermethylation
• during replication of transcriptionally active DNA, associated proteins shield the promoters of active genes from methylation

Question 8

Describe how steroid hormones regulate transcription.

Answer 8

regulate concentration and activity of TFs; important during development
o transcription of transcription factors is regulated many times with extracellular signal hormones (small, lipid soluble molecules, membrane permeable)
o e.g., the steroid hormone Dex is a dependent transport of the steroid hormone receptor that transcribes antibodies to β-Gal

Question 9

What is the role of transcription factors in development?

Answer 9

• regulation of transcription of transcription factors is highly/strictly regulated
  o there is a critical point at which transcription of genes that produce transcription factors begins
  o dependent on where/when needed; both accessible binding sites and respective binding proteins are necessary for transcription to occur
  o e.g., in Drosophila development, maternal mRNA is already unevenly distributed in embryo
  o when fertilized and cell division occurs, daughter cells inherit different maternal mRNAs
  o certain cells will have different sets of active proteins (some transcription factors – cascade of regulation interactions starts at the very beginning of cell division)

Question 10

Describe the role of TBP during transcription (think about promoters for all three eukaryotic RNAPs and TATA-less RNAPII promoters – how do RNAPs bind to them; also, think about coordination of activities of all three polymerases).

Answer 10

• TBP is a component of the positioning factor, regardless of the RNAP transcribing; allows each type of RNAP to bind to its promoter
  o coordinates activities of all three polymerases by binding to other polymerase-specific factors
  o is the #1 commitment factor:
   within SL1 complex (RNAP I)
   within TFIID (RNAP II)
   also in TATA-less promoters
   within TFIIIB (RNAP III)

Question 11

What is the role of Sp1 protein? What is the role of SL1 protein? What do they have in common?

Answer 11

• Both increase the rate of transcription and bind directly to DNA
• Sp1: transcription factor that binds directly to DNA via a C2H2 zinc finger, enhancing transcription
• SL1: core-binding factor assembled by UBF-TBP + 3 TBP-associated factors specifically for RNAPI
  o responsible for ensuring RNAP is properly positioned at start point
  o required for high-frequency initiation

Question 12

What is unusual about type 1 and 2 promoters for RNAP III polymerase?

Answer 12

• They are internal (in reality, changes in regions upstream of start point alter efficiency of transcription)
• Only when transcription factors bind properly can RNAP III be recruited
• The details of interaction differ:
  o Type 1: found in genes for 5S rRNA; TFIIIA binds to box A and TFIIIC binds to boxC, enabling TFIIIB to bind at start point, recruiting RNAPIII
  o Type 2: TFIIIC binds to both box A and box B downstream of start point; enables TFIIIB to bind at start point, recruiting RNAPIII

Question 13

Explain the mechanism of attenuation of the Trp operon. What is the importance of this mechanism for a bacterium?

Answer 13

• attenuation is the regulated, premature termination of transcription
  o does not involve Rho, which is required for transcription termination in 50% of E. coli terminators
  o it is at specific site: the attenuator site (DNA sequence where RNAP either terminates/continues transcription)
  o differs from normal termination, which occurs at the t-site
• a feedback mechanism that is more directly controlled by molecule of interest (enables the bacteria to conserve resources more strictly)
• mRNA of trp operon contains 4 regions: region 1 which synthesizes “leader peptide” which is Trp rich, region 2, region 3, and region 4 which is followed by a polyU region; the region 3 and 4 together are called the attenuator
• chain of events:
  o attenuation (yes or no) depends on formation of the particular stem loop structure in leader sequence
  o formation of the particular stem loop structure depends on rate of ribosomal translation of leader sequence mRNA (connection between transcription and translation – they are going at the same time)
  o rate of ribosomal translation of leader sequence mRNA depends on the supply of tRNA for Trp
  o supply of tRNA for Trp depends on amount of Trp present in cell

Question 14

Could you imagine a mechanism similar to the mechanism of attenuation of the Trp operon in Eukaryotes? Explain your reasoning.

Answer 14

• no, because the attenuation mechanism involves a formation of a stem loop structure that is creates by DNA in transition between transcription and translation
  o only in prokaryotes can transcription and translation work at the same time on the same mRNA
  o in eukaryotes, transcription and translation work separately (nucleus vs. cytoplasm)

Question 15

Describe two distinctly different ways in which the trp operon is controlled by the overall availability of tryptophan.

Answer 15

• control in translation:
  o supply of tRNA depends on amount of trp present for ribosomal translation
  o if ribosome stalls at region 1 (requires trp), then stem loop structure forms between 2&3, which allows RNAP to transcribe the subsequent structural genes for trp (because obviously the cell needs more!)
  o if there is plenty of trp, plenty to charge tRNA, ribosome doesn’t stall and instead stem loop forms between 3&4; poly U region follows directly after region 4; triggers RNAP to stop transcription; no structural genes for trp is made, because cell doesn’t need it
• control in transcription:
  o tryptophan will bind to repressor as a co-repressor
  o enables repressor to make conformation that shuts off transcription

Question 16

Describe the mechanism responsible for shutdown of the trp operon when a plentiful supply of free tryptophan is available.

Answer 16

o RNAP transcribes; nascent mRNA is formed quickly; ribosome will initiate translation at start codon near 5’ end
o regions 1&2 of mRNA may form loop transiently, but no polyU region follows  RNAP pauses, but continues transcribing
o Trp-enriched leader peptide is readily synthesized (stop codon after region 1) – lots of Trp available for ribosome to incorporate rapidly
o Loop involving regions 1 & 2 melts
o Regions 3 & 4 of mRNA form loop –> polyU follows region 4 –> RNAP will terminate transcription before reaching trp structural gene

Question 17

Describe the mechanism responsible for shutdown of the trp operon when there is little tryptophan available

Answer 17

o RNAP transcribes; nascent mRNA is formed quickly; ribosome will initiate translation at start codon near 5’ end
o ribosome will stall at Trp codons in region 1 – has to wait for an available Trp-charged tRNA (due to low Trp)
o region 1 of mRNA cannot base-pair with region 2 (is stuck in ribosome) –> region 2 forms loop with region 3 as soon as 3 is synthesized
o no polyU after region 3 –> RNAP continues transcribing –> RNAP elongates past attenuator all the way to “t site” –> structural genes of trp operon are transcribed

Question 18

Describe the mechanism by which the leader-attenuator region fine tunes the extent of transcription of the structural genes in the trp operon when tryptophan is available (but not to the point to activate the repressor).

Answer 18

• when trp is available, ribosome will not stall at region one (which requires tryp)
• while loop forms between 1&2, it melts quickly and a loop between 3&4 formed instead
• poly(u) follows region 4, transcription stops at structural genes for tryp
• this is more temporary shut off/on mechanism; more closely regulates the expression of trp structural genes

Question 19

Describe rho-dependant transcription termination

Answer 19

• rho is a hexameric RNA helices with RNA-binding domain (N-terminal) and ATPase activities (C-terminal)
• when rho binds to rut site (rho utilization site), it scans along mRNA until reaches RNAP
• note: rho has time to catch up with RNAP because RNAP has to wait for hairpin to be formed by mRNA for termination site to work properly; rho unwinds RNA and DNA hybrid to release: mRNA, Rho, RNAP

Question 20

Describe rho-independent transcription termination

Answer 20

• Rho-independent termination site has two characteristics:
  o GC rich self-complementary region with several intervening nucleotides that forms stem loop structure
  o Followed by a series of U residues (polyU)
• after being transcribed, forms loop structure
• polyU site interacts with RNAP and causes it to pause (it’s a sequence that tells it to slow down)
  o while the polyU region is weakly bonded with complementary A-region on the DNA template, bonds melt as RNAP pauses
  o RNA chain is released
  o DNA strands re-anneal together
  o RNA dissociates from DNA

Question 21

Which proteins are involved in mRNA cleavage and polyadenylation? Describe their order of association with pre-mRNA and their role (don’t forget the role of CTD tail).

Answer 21

• CTD Tail: recruit enzymes necessary for polymerization
• CPSF: involved in cleavage and polyadenylation specificity factor
• CStF: cleavage stimulatory factor; binds to G-U region
• CFI&CFII: cleavage factor I & II bending; mRNA is cleaved
• PAP-Poly(A) Polymerase binds
• Slow phase for the first 12 nucleotides
• Rapid phase requires nuclear polyA binding protein II (PABII)

Question 22

How is 5’ cap added to the nascent RNA?

Answer 22

• added by enzyme that associates with CTD tail of RNAP II soon after RNAPII starts making transcript
  o carried out by mRNA guanyltransferase
  o acts as a barrier to 5’ exonucleases

Question 23

What is the relationship between hnRNA and mRNA?

Answer 23

• hnRNA = heterogeneous nuclear RNA
• pre-mRNA = primary transcript
• snRNAs = small nuclear RNA
• hnRNA = pre-mRNA + snRNAs; transcript that is going to be processed to give mRNAs

Question 24

What are the general steps in processing of a pre-mRNA into an mRNA?

Answer 24

1) transcription
2) modification of ends: 5’ capping, addition of 3’ end
3) splicing, exon-intron junctions are broken
4) exons are joined

Question 25

What is the role of snRNAs in the spliceosome?

Answer 25

• snRNAs and their associated proteins form the snRNP which perform spiceosomal splicing
• the catalytically active components of snRNPs (ribozymes)
• associated with phosphorylated CTD tail, which:
  o assists snRNAs and Sm proteins in splice site recognition and binding to exons
  o spliceosomal protein-protein interactions
  o conformational changes during spliceosome assembly
  o i.e., U1, U2, U4, U5 and U6; rich in uridine

Question 26

What is the role of Sm proteins in the spliceosome? What other proteins (apart from Sm) are found to be associated with splicing?

Answer 26

• bind “smRNA motif” in snRNAs
• bind U1, which attaches to 5’ splice site of intron
  o complementarity between intron splice site and snRNA of U1
• other factors involved in splicing: helicases, proteins that could recognize single-stranded regions

Question 27

What is the difference between splicing of group I and group II introns? Between splicing of group II introns and spliceosomal splicing?

Answer 27

• group I introns: large, self-splicing ribozymes; catalyze their own excision from mRNA, tRNA and rRNA precursors; utilizes GTP
• group II introns: intron excision occurs in the absence of GTP
  o involves formation of a lariat
  o lariat strongly resembles lariat made during splicing of nuclear pre-mRNAs
• spliceosomal splicing necessitates the snRNPs subunits and associated protein factors come together; group II introns splice without this complex

Question 28

Describe the current model of spliceosomal splicing.

Answer 28

1. U1 attaches to 5’ splice site of intron (due to complementarity b/w sequences)
2. U2 binds to branch point A; requires ATP
3. U4/U6 (base paired with each other) & U5 bind in a tri-complex; binds to 3’ intron splice site; U4 inhibits U6
4. U4 and U6 change conformation due to the binding, forming B1 complex
5. U6 dissociates from U4; displaces U1; U1 & U4 are released (ATP dependent)
6. Once U1 and U4 are released, the spliceosome is activated; it contains U2, U5 and U6 snRNPs (B2 complex); U2-U6 represent active site of spliceosome
7. U5 is responsible for positioning and holding; U2-U6 complex has catalytic activity, mediates splicing at both sites; formation of C1 and C2 complexes; requires ATP; two trans-esterification reactions
8. Lariat structure (the pretzel shape) degrades, making splicing irreversible
9. snRNPs released as individual particles and released

Question 29

List the roles of 5’ methyl cap. List the roles of CTD tail (yes, again).

Answer 29

5’ methyl cap:
• recognized by transport machinery
• protection from endonucleases (degradation)
• important in splicing (1st U1 protein binds to 5’ cap)
• ribosome recognizes and binds to 5’ methyl cap for translation initiation

CTD tail:
•	5’ capping
•	polyadenylation: brings enzymes involved
•	splicing
•	initiation of transcription
•	elongation of transcription

Question 30

Describe transcriptional units in Eukaryotes. Use diagrams.

Answer 30

• simple transcription units: when there is only one mRNA product possible; constitutive splicing
• complex: when there is alternative 3’ exons or alternative internal exons; alternative splicing regulated by specific slicing factors

Question 31

List and explain two means of control of gene expression that could happen during pre-mRNA processing.

Answer 31

• inhibited polyadenylation
• alternative splicing
  • alternative spliceosome
  • tissue specific RNA splicing control
  • tissue specific alternative poly (A)
  • regulated splicing
• trans-splicing
• regulation of mRNA processing

Question 32

Describe one case of control of gene expression by means of splicing.

Answer 32

• alternative splicing where one product may be a functional repressor/inducer, while the other product may be the opposite; these molecules play a role in regulation of gene expression