BIOMG 3320 Group 3 (Lectures 6-7)

  1. Explain the RNA World Hypothesis
    • RNA is the only biomolecule that functions both as genetic material and as an enzyme
    • Self replicating RNA, based on this hypothesis, is the likely critical component of the evolution of life.
  2. What do differences in gene expression by controlling transcription underlie?
    • Different cell types
    • Disease (Ex. Cancer is often caused by mutations that mess up gene expression patterns. Essentially all cancers show a change in transcriptional regulation.)
    • Speciation (Closely related species. A lot of what is changing is gene expression - not genes themselves)
    • Differences in a population
  3. Explain the chemistry of transcription
    • RNA polymerase transcribes in a 5’ to 3’ direction, using NTP as substrates and one DNA strand as a template.
    • Unlike DNA polymerase, no primer is necessary to initiate synthesis.
    • Incoming NTP is attacked at alpha phosphate by the 3’ OH of the growing RNA chain.
    • Three Asp residues in the RNA polymerase active site (catalytic core) coordinate two Mg ions which facilitate the reaction. This is called two metal catalysis.
    • RNA polymerase forms a 17 nt region of unwound DNA termed the transcription bubble.
    • 8 bp of DNA-RNA hybrid forms as RNA polymerase moves along the DNA.
    • 5’ end of nascent RNA is extruded out of the complex.
    • DNA helix is unwound and rewound as RNA polymerase moves along at ~50-90 nt.
  4. List 5 ways bacterial transcription can be regulated?
    • Presence of sigma factors
    • Heat shock - binds to the promoter of heat shock gene to promote expression
    • Regulatory binding sites close to promoter
    • Activator and repressor proteins
  5. Explain the structure of RNA polymerase
    • Holoenzyme is the core + a sigma subunit
    • Core enzyme subunits:
    • Two alpha subunits: assembly, interacts with regulatory factors, catalysis
    • Beta subunit: polymerase activity
    • Beta’ subunit: DNA binding, locks on tightly to keep the complex associated with DNA throughout transcription
    • Omega subunit: Stabilize the protein subunits, but not currently known
    • Sigma subunit: Brings the polymerase to the promoter.
    • Crab-shaped complex: Two pincers (beta and beta’) that grip DNA. The structure is similar between bacteria and eukaryotes.
  6. Explain sigma factors in E.coli
    • Seven different sigma factors in E.coli
    • Sigma controls which promoter RNAP holoenzyme recognizes
    • Depending on which sigma factors is bound to the holoenzyme, the holoenzyme will be guided to a different group of genes promoted by the promoter corresponding to the sigma factor. This makes the holoenzyme “gene specific”
    • Sigma70 is responsible for synthesizing most genes for E.coli under normal conditions
    • In heat stress, Sigma32 is switched on, which turns on heat-shock genes that allow E.coli to survive heat shock.
  7. Explain phage sigma factors
    • Phage has early, middle and late genes, expressed at different points of phage infection.
    • Phage early genes are transcribed by bacterial RNA polymerase and bacterial alpha70.
    • One of these early genes encodes a phage-specific sigma-factor, this phage sigma factor hijacks the cell’s transcription machinery and causes the phage’s middle genes to be transcribed.
    • This allows for gene transcription to occur in three levels.
  8. Explain promoters and sigma70
    • -10 region and -35 region: directly recognized and bound to by sigma70.
    • -10 region must be 5-9 N from RNA start site in order to be recognized.
    • -10 and -35 must have N17 in between in order to be recognized.
    • Promoters that match the consensus perfectly are strong promoters. Promoters that don’t match the consensus really well have low basal levels.
    • Consensus sequences for different sigma factors are different, creating the specificity of different sigma factors for different promoters.
  9. Explain the points of contact between RNA polymerase and promoters
    • Extended -10 region of some promoters provides additional points of contact between sigma factor and -10 region of promoter.
    • UP-element is found in some promoters. Interacts with C-terminal domain fo alpha subunit, increasing recruitment efficiency and strength hof promoter.
  10. What are the channels of RNAP?
    • DNA enter/exit channel
    • RNA exit channel
    • Template/non-template channel
    • Nucleotides channel
  11. What are the three steps of the transcription cycle
    • Initiation
    • Elongation
    • Termination
  12. Explain initiation of transcription
    • Promoter recognition and binding
    • Formation of closed complex (DNA still double stranded)
    • Transition to open complex (DNA is unwound). This is a regulatory step.
    • Multiple abortive rounds of transcription often occur - short transcription of small fragments of RNA, which are then lost.
    • Promoter clearance/escape - allows transition to elongation phase.
  13. Explain the 3 models of initiation mechanisms
    • Before the elongation mode, there are often many abortive RNA transcription rounds. This is one of the ways transcription of genes is controlled. Stronger promoters have fewer abortive rounds.
    • Very intensively studied, we still don’t know which model is definitively correct. Current consensus leans towards scrunching.
    • Transient excursions: The whole complex moves down the RNA. If failed, it backtracks after releasing abortive RNA.
    • Inchworming: Complex has some flexibility. Part of it moves down with the RNA that is transcribed, like an accordion.
    • Scrunching: The polymerase stays where it is, the DNA itself is scrunched.
  14. Explain elongation
    • Sigma subunit is displaced.
    • Other proteins (like NusA) join elongating RNA polymerase, aiding efficient elongation.
    • After termination, pro-elongation factors will dissociate.
    • Sigma subunit and RNAP core enzymes are recycled to start new rounds of transcription.
  15. Explain RNA polymerase correcting errors
    • Pyrophosphorolytic editing: same active site as polymerization, backward reaction. Removes one nt
    • Hydrolytic editing: Enzyme backtracks, cleaves several nts containing errors. GRE factors clip off the piece of RNA that contains the wrong (mispaired) nucleotide.
  16. Why does RNAP have a higher error rate than DNAP
    • DNA polymerase has exonucleolytic proofreading. RNA polymerases do not have this.
    • Error in DNA is permanent, error in RNA will change an amino acid in one protein.
  17. What are the roles of sigma factors?
    • Lead RNAP to promoter, starting initiation.
    • Lead to specific promoters, leading to different gene expression.
    • Assist going from closed complex to open complex.
  18. Explain termination pathways
    • Intrinsic, rho-independent termination: Symmetric sequences in the nascent RNA fold to form a termination hairpin with a poly(U) sequence. This induces the RNA polymerase to release the nascent RNA and dissociating from the DNA
    • Rho-dependent termination: ATP-dependent helicase recognizes rut elements (rho utilization) which are C-rich sequences. Rho binds nascent transcript and migrates towards polymerase; interactions between RNAP and rho result in termination.
  19. Give an overview of general types of regulation of transcription in bacteria
    • Basal level of transcription involves no additional factors: This regulation is governed entirely by the promoter – ability to recruit polymerase and proceed through initiation, i.e., the amount of transcription that the -35 and -10 regions are capable of initiating on their own.
    • Repression: repressor protein binds to operator sequence and prevents transcription by physically blocking the polymerase (sigma specifically) from binding to the promoter.
    • Activation: activator protein binds to activating binding site and increases or permits transcription. Facilitates process of transcription, for example by accelerating recruitment of polymerase or helping transformation from closed to open complex during initiation.
  20. Explain basal level of transcription
    How optimal the -10 and -35 regions are and to what extent they can facilitate transcription of the gene.
  21. Explain repressor proteins:
    • In-between the -10 and -35 regions (operator sequence).
    • Will stop the protein from latching on to the promoter.
  22. Explain Activator proteins:
    • Binds close to the promoter, interacts with the polymerase.
    • Often promotes the switch from to the open complex.
  23. Given an overview of lac operon
    • E. coli preferentially use glucose as energy (i.e., food) , but can use lactose sugar instead if necessary. This requires the proteins encoded in the lac operon.
    • To be most efficient, E. coli will not transcribe the lac operon when glucose is available, and will only transcribe the lac operon at high levels when lactose is available and glucose is not available.
  24. Explain the structure of lac operon
    • Lac operon consists of three regulatory sequences (CAP site, promoter, operator sequence) and three protein-coding genes (lacZ, lacY, and lacA)
    • CAP site is the DNA binding site for the activator protein called CAP (catabolite activator protein)
    • Promoter is the region of DNA that the RNA polymerase holoenzyme binds to
    • Operator sequence is the DNA binding site for the repressor protein
    • lacZ gene encodes galactosidase protein – critical enzyme in the metabolism of lactose sugar
    • lacY gene encodes permease protein – transmembrane protein that pumps lactose into the interior of the cell
    • lacA
  25. Explain mechanics of lac operon regulation: lactose is available, glucose is not available
    • Low glucose levels increase cAMP levels.
    • cAMP binds CAP.
    • When cAMP is bound to CAP site, CAP (catabolite activator protein) binds to the CAP site, enhancing transcription of the lac operon.
    • This is done by enhancing the recruitment of RNA polymerase holoenzyme by interacting with the C-terminal domain of the alpha subunit.
    • When cAMP is not present, CAP is unable to bind DNA.
    • Lactose is converted into allolactose.
    • Allolactose binds lac repressor. The repressor is unable to bind DNA.
    • When allolactose is not present, the lac repressor protein binds to the operator sequence and prevents transcription of lac operon.
  26. What is Helix-turn-helix?
    • Protein domain that binds DNA, found in CAP and lac repressor.
    • Both proteins form a dimer and bind to major grooves of DNA.
    • Both specific amino acids differ, allowing binding to specific DNA sequences.
  27. Mechanics of lac operon: If lactose is available and glucose is available
    • Basal level of transcription at the lac operon occurs.
    • Low cAMP means that CAP protein is not able to bind DNA, so there is no activation.
    • Some lactose is metabolized, so allolactose is produced, preventing the lac repressor protein from binding to DNA, so there is no repression.
  28. What does lac operon demonstrate?
    • Regulatory binding sites are very close to the promoter
    • Activity of regulatory proteins (activators and repressors) is coupled to the biochemical pathway the operon corresponds to.
  29. Mechanics of araBAD operon
    • Expression allows E.coli to metabolize arabinose.
    • There are two activators on the operon:
    • When there is no arabinose in the environment, AraC binds the DNA upstream of the promoter and causes the DNA to bend. This prevents the activation of RNA polymerase.
    • With arabinose present, arabinose binds to AraC, activating transcription.
    • With low glucose, which results in high cAMP levels, there is binding to CAP, further increasing transcription.
  30. Mechanics of glnA operon
    • Involved in helping the bacterium survive in environments with different levels of nitrogen.
    • Some NtrC binding sites function up to 2kb from promoter (unusual for bacteria)
    • NtrC is an ATPase that triggers polymerase to initiate closed to open transition.
    • Some promoters regulated by NtrC also have binding sites for IHF (integration host factor), which bends DNA, helping to bring NtrC close to polymerase.
  31. Mechanics of Trp operon
    • With high tryptophan, the trytophan binds to the Trp repressor, which binds to DNA and lowers transcription. (70 fold)
    • With intermediate tryptophan levels, transcription begins in the absence of a Trp repressor. However, transcription terminates at the attenuator sequence. The attenuator sequence forms an intrinsic terminator, which stops transcription. (10 fold)
    • With low tryptophan levels, ribosome gets stalled where the next amino acid is tryptophan. In this case, the attenuator does not form, and the polymerase gets past the intrinsic terminator.
Author
pelinpoyraz
ID
357975
Card Set
BIOMG 3320 Group 3 (Lectures 6-7)
Description
Updated