RNA Structure, Synthesis, and Processing (Euk); Control of gene expression (Lectures 15, 16, 18))

Question 1

coding strand

Answer 1

sense strand, identical to mRNA (except T for U)

Question 2

template strand

Answer 2

complementary to mRNA convention is to write template strand on the bottom, so the coding strand can be read 5’—-> 3’

Question 3

bacterial RNA polymerase

Answer 3

in bacteria, only one polymerase

• alpha subunits (2): has a structural role, may interact with regulatory proteins
• beta subunit: catalytic site
• beta prime subunit: binds DNA template
• sigma subunit: recognizes promoter, needed for transcription initiation
• core enzyme does not include the two alpha subunits, holoenzyme does. The holoenzyme is required for transcription initiation; elongation afterwards is carried out by the core enzyme

Question 4

classes of prok (and euk) RNA

Answer 4

• rRNA
• tRNA
• mRNA
• only euk: snRNA + miRNA

Question 5

prokaryotic promoter regions

Answer 5

• upstream of the initiation site (+1)
• written on the sense strand (-35 box and -10 box)
• specifically recognized by the sigma subunit
• the distance between the 2 sequence elements is critical; they need to be on the same face of the DNA helix to be recognized by RNA pol sigma subunit

Question 6

Differences between prok and euk transcription

Answer 6

1. Three polymerases
2. No sigma subunit* (recruitment of the polymerase to the appropriate promoter regions is mediated by transcription factors with regulatory elements)

*this difference enables genes to be expressed in different developmental stages, in different tissues and in response to different environmental stimuli

Eukaryotic RNA has to be processed before it can leave the nucleus –> cytoplasm. Prokaryotic RNA is made and processed in the cytoplasm, so no additional processing is necessary.

Eukaryotic processing (occurs in nucleus, needed for transport to cytoplasm):

• 5’ cap
• 3’ polyadenylation
• splicing to remove introns

Question 7

3 Eukaryotic RNA Polymerases

Answer 7

• RNA pol I: (nucleolus) transcribes precursor of large rRNAs (28S, 18S, 5.8S)
• RNA pol II: (nucleus) transcribes precursors of mRNAs (hnRNAs), miRNA and snRNAs
• RNA pol III: (nucleus) transcribes precursors of small RNAs (tRNAs and 5S rRNAs)

Question 8

RNA pol II promoter/enhancer regions

+ initiation step

Answer 8

1. Basal: site of initiation
2. Constitutive: regulates rate of transcription
3. Inducible: On or off, mediates transcription in response to environmental signals (tissue or response-specific)

Initiation:

• Formation of the open promoter complex involves ATP hydrolysis
• Basal apparatus is necessary but not sufficient to start initiation (RNA pol binding is stabilized by conformational changes in DNA caused by regulatory proteins binding to constitutive inducible promoter elements, activators, enhancers)
• TFIIH: kinase which phosphorylates Pol II

Elongation phase similar to that in prokaryotes, dephosphoryation is involved in termination

Question 9

Rho independent transcription termination (prokaryotes)

Answer 9

• most common method of termination
• DNA template contains inverted repeats (palindromic G-C regions + region of ~8 As on temp strand
• GC rich regions are hard to unwind, RNA pol slows (results in stem-loop structure ending with U residues)
• Only bp connecting RNA + DNA template are A-U, which are easily broken
• RNA strand then dissociates from the DNA template

Question 10

Rho-dependent termination of transcription

Answer 10

1. Rho (large hexameric protein-helicase, ATPase) binds to 5’ end of the new transcript
2. Polymerase slows at C-rich pre-termination sequence, rho moves to the 3’ end , hydrolyzing ATP; uses helicase to unwind DNA-RNA hybrid
3. Rho and RNA pol dissociate from the RNA

Question 11

Mechanism of action of rifampicin

Answer 11

• Binds to B subunit of RNA pol
• Prevents 1st phosphodiester bond from forming
• Blocks initiation of prokaryotic transcription, ongoing transcription isn’t affected

Question 12

actinomycin D

Answer 12

• (inhibits prok + euk transcription)
• intercalates between DNA strands (into major groove) and prevents DNA template strands from unwinding
• inhibits both initiation of transcription + elongation

Question 14

Addition of the 5’ cap

Answer 14

• GTP added “backwards” to 5’ end of the mRNA precursor to form 5’ - 5’ triphosphate linkage
  
  • Guanylyl transferase: release of gamma phosphate from RNA, release of PPi from GTP
• A methyl group is then added to the N7 of guanine
• The proteins that bind the cap (CAP binding proteins) enhance the efficiency of translation initiation and stabilize the mRNAs
  
  • cap also forms scaffold for protein binding
  • protects mRNA from nuclease digestion

Question 15

3’ Poly A tail

Answer 15

• added to the 3’ end of mRNA precursor by poly (A) polymerase
• AAUAA (memorize?) polyadenylation signal
• Related to transcription termination
• Protects 3’ end from degradation
• Stabilization of mRNA

Question 16

Splicing steps

Answer 16

1. Involves sn (single nuclear) RNA + proteins = snRNPs
2. Mechanism involves 2’-5’ phosphodiester bond at branch point

REMEMBER: all introns begin with GU and end with AG; branch point is an a located in a pyrimidine-rich sequence approx 50 bases from the 3’ end of the intron

Steps:

1. U1 (splice site) and U2 (branch site) bind to mRNA
   * Intron begins to loop, branch point A is close to 5’ splice junction
2. Binding of U4, U5, and U6 completes conformational change
3. U6 replaces U1, U1/U4 complex is released
4. Catalytic center is formed by U2 (branch point) and U6 (5’ splice junction) U6 is a ribozyme
5. U5 holds two exons together during splicing
6. U6 catalyzes two trans-esterification reactions

• 1st 2’OH of A attacks 5’ splice site
• Newly formed 3’OH attacks 3’ splice site
• These steps result in the 5’ exon being joined to the 3’ exon, and the intron is removed as a lariat.
• DNA does not have 2’OH, preventing it from undergoing this rxn, even though both DNA + RNA processing are in the nucleus
• Abberrant splicing often forms a truncated protein, usually due to a frameshift mutation that results in an early stop codon

Question 17

snRNPs involved in splicing

Answer 17

sn (single nuclear RNA) + proteins = snRNPs

• U1: binds 5’ splice site
• U2: binds branch site, part of catalytic center
• U5: binds the 3’ splice site, loops over to the 5’ site
• U4: masks the catalytic activity of U6
• U6: catalyzes splicing

Question 18

Specificity of alternative RNA splicing

Answer 18

• RNA binding proteins facilitate alternative splicing
• (ESE/ISE) exon/intron splicing enhancer
  
  • Bound by SR (serine/arginine rich) proteins
• (ESS/ISS) exon/intron splicing silencer
  
  • bound by hnRNP (heterogeneous nuclear ribonucleoprotein)
    
    • prevents binding of U2 to the branch, which excludes the exon from the processed mRNA

Question 19

Lac operon mechanism

Answer 19

operon: function as single transcription unit and comprise 2 or more adjacent genes + regulatory regions; transcript contains polycistronic message (RNA transcript for more than 1 gene)

In the presence of glucose: repressor binds to operator, and RNA polymerase cannot bind to promoter

When lactose is present: allolactose binds to repressor + inactivates it, but glucose metabolism predominates

When glucose levels are low: increase in cAMP, which interacts with the CRP (cAMP receptor protein). This enhances transcription by recruiting RNA polymerase to the promoter

Question 20

Transcription factors &

4 types of transcription factor motifs

Answer 20

1. Helix-turn-helix
2. Zn finger
3. Leucine zipper
4. Basic helix-loop-helix

• Many weak contacts amount to strong specific interaction between protein and DNA
• Hydrogen bonding between amino acid side chains and DNA bases in major groove give recognition and specificity
• Heterodimers give large range of potential interactions with additional regulatory proteins
• Regulation of transcription factors:
  
  • protein synthesis
  • ligand binding
  • protein phosphorylation
  • addition of subunit
  • unmasking
  • stimulation of nuclear entry
  • release from membrane

Question 21

Helix - turn - helix motif

Answer 21

• Often form dimers
• Recognition alpha helix
• Unstructured turn
• Second alpha helix

Question 22

Zn finger

Question 23

Control of gene expression in eukaryotes

Answer 23

• genes aren’t organized into operons: each gene possess reg. sequences appropriate for its expression
  
  • sites of transcription factor interaction
• transcriptional repressors + activators contain DNA-binding motifs as part of the protein
  
  • recurrring structure is alpha helices that fit directly into the major groove
  • H bonding critical for recognition between aa + bp in major groove

Question 24

Zn finger motif

Answer 24

• Don’t form dimers
• Use one or more molecule of Zn
• DNA binding regions: alpha helices
  
  • repetitive motif of a pair of Cys and a pair of His (or Cys)
  • each repeat coordinates a zinc ion via 2 Cys and 2 His (or Cys)
  • 12 or so residues separating the coordination sites loop + form Zn finger, which binds the major groove of DNA

Question 25

Leucine zipper motif

Answer 25

* Form dimers * N terminal; DNA binding regions: alpha helices (known as basic region) * Zipper: alpha helix with periodic repetition of leucine * Leucine protrudes from the serine side of the helix have hydrophobic interations with another set of Leu side chains on a second polypeptide * forms stable noncovalent linkage, fostering dimerization of 2 polypeptides * bZIP: different amino acids replace leucine and form the coiled coil through hydrophobic attraction (heterodimers; expands the DNA recognition + regulator possibilities of these proteins)

Question 26

basic helix-loop-helix (bHLH) motif

Answer 26

* similar to leucine zipper * also involves alpha helix in one polypeptide interacting hydrophobically with alpha helix of 2nd polypeptide * each polypeptide has 2 alpha helices separated from each other by a non-helical stretch of aa that loop out * each polypeptide has a basic helical DNA binding domain with (+) aa to contact DNA. Like b-ZIP, homo + hetero dimers can form