Lecture #9 - RNA Processing Flashcards

Question

Processing tRNA (Nucleotide modification)

Answer 1

Many rRNAs and tRNAs have many modified nucleotides + Happens at many positions in rRNA and tRNA - Up to 25% of nucleotides are modified in tRNA Ribonicleotides can be modified in >100 ways Modifications can happen at all possessions on bases Some modifications affect W/C face = affects the BP capacity BUT some can be modified on the other side and affect the way RNA folds

Answer 2

1. Methylation 2. Isomerization of U 3. Demaination (Ex. demiantino of adenosine makes an Inosine) - Remove the Amino group that is used to BP A-U - Insoine = behaves like a G and pairs with a C 4. Psudeouracil - Made by cutting the glycosidic bond --> rotates the base 5. Modify the surgar – methylate the 2’OH on the ribsone sugar

Answer 3

Nucleotide modifications affect stricture and the ability to interact with other RNAs and protein 1. Modulationes of base pairing and decoding 2. Structure stabilization 3. Bidning to specific readers 4. 2’ OH methylarion affescts sturcture and stability

Answer 4

Modification occurs throughout the tRNA body AND have some within or next to the anticodon

Answer 5

tRNA uses nucleotides at positions 34,35,36 for anticodon (Forms basepairs with the nucleotides in codon of the mRNA) - Bases positioned at 34 = tend to be modified

Answer 6

tRNA is able to read non-perfect codons (have wobble position in the 3rd position) - Modified base in position 34 = can read the wobble position imprecisely = enables the tRNA to read multiple codons - Can have different nucleotides in the 3rd position - Position 34 of the anticodon tRNA can have whole sugars added onto the base - Position 37 (NOT in the anticodon) = modified so that it fixes the position of the codon/anticodon interactions

Answer 7

1. Have modifications that affect RNA structure (important for tRNA) 2. Have modifications that can be recognized by proteins – provides specificity for readers that recognize RNA 3. 2’OH methylation = provides stability - RNA is unstable because 2’ OH is a nucleophile that can attach the phosphodietsrer bond of the adjacent base and cleave the RNA molecule from within BUT if the 2’OH is methylated then it can’t do that

Answer 8

Overall - Specificity of modifications = sequence based + affected by structure (because regions need to be single stranded) Specificity of rRNA modifications is determined by base pairing interactions with snoRNAs that guide enzymes - Most modifications are added as the RNA is being folded - Modifications are guided by snoRNPs

Answer 9

Most common modifcation on rRNA = 2’OH methylation and Pseudouracil

Answer 10

SnoRNAps = SnoRNAs + proteins - SnoRNAs = enzyme that preforms the modification - RNA compenent = gudies the enzymes by BP complenetary of 2 RNAs (Similar to cas9 finding the target using a gRNA ) Example 1 – fibulin us guided by snoRNA (BOXC and BOXD snoRNA) - BOX D and BOXC = BP with RNA and the enzyme will methylate positions within the base pairing Example 2 – Box H and Box ACA react with dyskerin/CbfP5 to make uradin to psuedouradin

Answer 11

Overall – Specificity of tRNA modifications relies on global and local structures - tRNA modifying enzymes use structure and sequence for specificity Examples: 1. Pseduradine Syntehases (PUS) 2. Methytransferases (TRMT, NSUN, METTL) 3. Adenosine Demainase (ADAR)

Answer 12

1. 5' Capping 2. Splcing (inreveting sequences need to be reomved) 3. 3' cleavage and polyadenylatio 3. Base modifications Need to have transition from PremRNA --> Mature mRNA Most modifications = occur co-transcrtionally Modifications occur for all Pol2 products --> Pol2 brings along machinery for these steps

Answer 13

Start - RNA polymerase begins transcription with a nucleotide tripospahte (ex start with GTP – have PPP at the 5’ end of RNA product) 1. 5' phopshate (gamma phosphate) = removed and a GMP is added to the remaining diphosphate (Step 2) - WHEN added the GP is added in a different link form --> creates a 5’-5’ linkage (5’-5’ linkages = prevents degradation by exonuclease) - Cap = protects the 5’ end of mRNA 3. Methylation of added to position 7 of guanine

Answer 14

Includes cleavage and Polyadenylation Poly A tail = on ALL products of Pol 2 transcription

Answer 15

The determinateness for termination are split sequence - Pol recognized the sequence using factors that it brings along (Uses CPSF) Termination signal: 1. Have cleavage sequence where nascant RNA is cut - Recognition sequence of where to cut is provided by 20 BP conceeus sequence upstream (AU rich sequene) 2. Need U or GU rich region 30 nucleotides down stream

Answer 16

Used for 3' cleavage and Polyadenylation - System for polyadenylation = has recogintion and then stimulation CPSF = comes with RNA pol --> when RNA polymerase transcribes the AAUAAA sequence CPSF can bind BUT it can’t do anything until Pol transcribes the GU rich sequence THEN when have transcription of the GU rich sequence the CstF (cleave stimulation factor) binds = activates CPSF for cleavage --> Cleave occurs at CA dinucletides THEN PABP comes and adds a poly A tail tp the 3’ end

Answer 17

CPSF system uses recogintion and then stimulation Done so that enzymes can’t cleave all AAUAAA sequences INSTEAD only cleav ethe right one

Answer 18

PolyA tail – used for mRNA enrichmemt - Use polA tail when want to purify just mRNA by pulling down on polyA mRNA using beads or using a oligioDT primer that will hybridize to polA NOTE - ONLY 3% of cellular RNA is mRNA (40% is rRNA) = needs to be enriched for = need to pull down and isolate it

Answer 19

EXCEPTION of 3’ polyA tail = histones mRNA which lack polyA tail BUT have stem loop structure at the end bound by SLBP Stem loop at 3’ end = generated by U7 snRNP-mediated reaction - U7 snRNA cleaves at position and allow for bidning of SLBP - SLBP = holds onto the stem of the histone mRNA and protects it during tranlsation

Answer 20

Histone transcription = coupled to the cell cycle (ex. Need ore histones when replicating DNA) = histones are made fast and degraded fast = histone mRNA have unique properties that allow for that

Answer 21

1. Helps direct downstream maturation including splicing and transcription termination 2. Important for transport to the cytoplasm - Licensing after transcription and processing that allows fully processed mRNA to exit the nucleus to the cytosplam 3. Stabilize mRNA (Protects the end of the mRNA from exonuclease on both sides) 4. Important for translation initiation - Protein bind to polyA tail and CAP to promote translation

Answer 22

Found cells infected with adenovirus --> purified the genome of the adenovirus and purified the adenovirus mRNA by taking the mRA that was being translated by the ribosomes THEN allowed the DNA genome and the mRNA of the virus to hybridize --> looked at electron microscope In microscope = saw RNA hybrdized to one part of the genome THEN skipped a part then continues hybridizing further downstream ***Shows mRNA is not fully contiguous with the genome

Answer 23

FOUND the same results in chickens – mapped all introns by measuring lengths on electron microscopy

Answer 24

Intron content differs across species - Yeast = few introns – small fraction of genome and genes with introns usually only have 1 intron - Drosophila = more introns (most gene shave 2+ introns) - Mammals = many introns (Many genes have a lot of introns + Many introns in 1 gene) ; titin gene = 363 introns ; Dystrophin gene has 79 introns that account for 98% of the length of the 2.4 mb transcript Splicing evoloved rapidly (seen in the number and distributions of introns)

Answer 25

Exon and intron lengths differ across species Exons = short in most species BUT Intron = length varies - Number and length of introns vary across species Example - Fungi = short introns vs. Mammales = longer introns - Yeast = infrequent short introns - Human genes = short exons and frequnet long introns

Answer 26

There are concensus sequences that define the end of the exon and the begining of the next one (the intervening intron seqeunce needs to be removed) Pre- mRNA = Have a 5’ splice site (5’ relative to teh intron) and a 3’ splice site AND have a branching point in the middle - AGGU concecus sequence at the 5’ end (GU are the most important nucletdoes) - CAAG concecus sequence at the 3’ end (AG are the improtant nucleptides) - A n the middle of a poly pyridine track = branching point on the middle of intron Cleavage occurs between the GG on the 5’ end and the GG on the 3’ end

Answer 27

The concesus sequences = needed BUT are not sufficient because have many GU and AG in the genome

Answer 28

Overall - transferication reactions - Splicing = occurs co-transcriptionally Steps of reaction (rearranging 2 phoshodietser bonds): 1. The branching point A is attached b a 5’ Phosphate and a 3’OH BUT it has a free 2’OH that can do a nucleophilic attack --> 2’OH will attack the phophate bond between the end of the exon and the begining of the intron = forms a 2’-5’ - Makes a layreyet on the intron where intron is attached to itself 2. Upstream exon has a free 3’OH that will do a nucleophilic attack on the phosphodietser bond between the intron and exon 2 --> makes a new phosphate bond between the exon - Causes relase of the 3’OH of introns END - have ligated exons and a branched leriet 2’5’ intron

Answer 29

NO NET energy gain = splciing is net zero

Answer 30

Shows how people found order of events ON gel - Have precursor at time 0 with 2 exons and introns --> then see free 5’ OH appear at the same time that exon 2 is joined to intron appears (first step of reaction) THEN see exon 2 appear at the same time as the lareit is released

Answer 31

Catalysis = done by splisosome - Splicing = multi step process onvolving more than 80 proteins and 5 snRNAs

Answer 32

Splisasome = ribonuclear protein machine - Splisasome = protein directed ribozyme Splisaome = maed up f 5 snRNPs + 100s of proteins - U1,U2,U5,U6/U4 = control RNA polecules and proteins - Splisome also uses 2 protein complexes (NTC and NTR)

Answer 33

Splisome = needs to be assembled on every intron (new each time) Assembly: 1. Each time Pol 2 transcribes introns U1 recognize the 5’ splice site by base pairing with te 5' splice site concensus sequence + U2 recognizes the branch point A and part of the 3’ splice site (MARK the ends that will be spliced) 2. AFTER U 1 and U2 --> U4/U5/U6 joins At some point you need to remove the snRNPs form the RNA to allow things to attach (ALL brought together then brought apart)

Answer 34

HAVE lots of rearrangement of RNA paring and breaking apart in order to bring ends together and catalyze reaction

Answer 35

Actual splicing reaction does not require energy because not making new bonds (amount of bonds broken = amount of binds made) BUT ATP is needed for helices - Helicase = catalyzes the rearrangement --> helicases allow for large rearengments of RNA:RNA interactions - Helicase requires ATP Example Rearangment - Need RNA ends to bind for catalysis to occur BUT need U4 RNP that holds part of the RNA to leave --> removal of U4 is done by Helicase - Removal of U4 allows for U6 to hybridize with U2 to make the active site of the ribosome

Answer 36

Each step of splicing requires a lot of rearrangement of the structures RNA helicases and step-specific splicing factors assist with remodeling of protein and RNA network during different steps

Answer 37

Consensus sequences = read by U1 and U2 Concesus sequencs = needed BUT they are to short to determine where splicing happens - Have many Splicing enhancers/silencers

Answer 38

There are many sequences around introns ad exons that enhance or inhibit the activity of the splisasome (Sequences = splicing enhancers or splicing silencers) Example silencers/enhancers: 1. Exonic splicing enhancer = sequence in exon that enhance splciing of Introns - Recognized by SR proteins (Ser and arginine rich proteins) 2. Intronic splicing enhancer = sequence in intron that enhances splicing 3. Exonic or introns splicing silencers = have regulator roles ad help avoid use of crytpic sites - Often bound by hnRNPs

Answer 39

Consequence of splicing = EJC deposition EJC = exon junction complex --> mark of correct splicing - Found at the junction of the joining exons - Splisome = brings proteins of EJC to a place where it catalyzes the reaction = leaves mark on mRNA that has been properly spliced - EJC = deposited by the splisome itself Use of EJC: 1. EJC = needed for export 2. Used for quality control processing required for translation

Answer 40

Splisasome = reminenet of the RNA world Splicing = catalyzed by the Splisasome BUT splciing orginated before the Splisasome existed - Before splisasome = had self-splicing introns (Ex. group 1 and group 2 self splicing introns) - Self splicing introns = 1st ribozymes found Example - Group 2 = uses the same mechasnims as us BUT doesn’t require energy

Answer 41

Alternative processing– frequent and plays regulatory roles - Results on multiple isoforms per gene - Splicing = enables generation of transcript diversity Use of alternative process: 1. Altenative splicing – skipped exons 2. Alternative 5’ splice site 3. Alternative 3’ splice site 4. Alsternative transcrtional start site 5. Alternative polyadnylation site or cleavage site

Answer 42

Splice site selection requires positive and negative regulators in addition to necessary consensus sequences - Additional elements can be regulated to result in alternatively spliced mRNA

Answer 43

Way decisions are made for alternative splicing or alternative cleavage/polyA: 1. Depends on how fast polymerase is transcribing 2. Depends on what other proteins are bound to mRNA + depends on strength of binding site (based on RNA binding proteins) - proteins may mask a cleave site or polyA site OR maybe have a weak binding site

Answer 44

Depending on speed of polymerase might be in 1 splice site over anotehr WATCH VIDEO

Answer 45

Example 1 - Protein Dscam - detrmines nueronal adhesion - Have alternative splicing at different cassettes (4 parts of the protein are ended by alternative exons - Any 1 molecules is made of 1 green, 1 red, 1 blue, 1 purple (have 12 gteen exons + 38 red exons etc.) = have 38,000 splice isoforms that a cell can make - Ensures all nuerons make a diferemt Dscam Example 2 of splicing – Procadherins in humans – diversity is generated by using alternative initiation and alternative splicing - Have 3 exos (ex. 15 purple) --> have different initiation sites (each has its own promoter) = get alternaive intiation (depending on which you use you splice differentley)

Answer 46

SMA = based on LOF of SMN --> healthy people have 2 copies of SMN - SMN = encodes a chaperone needed for snRNPs - SMN1 = makes full length protein - SMN 2 = often incorrectly splices (90% skip exon 7) IF have mutation in SMN1 = need to rely on SMN2 = don’t get enough SMN Researchers looked at why exon 7 is always skipped in SMN2 – found an intron silencer of splicing at the beginning of intron 7 that flanks exon 7

Answer 47

Did experiment - Run SMN1 and SMN2 in RT-qPCR = SMN1 makes full length product and SMN2 makes trunctaed product THEN Run SMN2 with a mutation in the splicing silencer = get full form of SMN2 --> gave people idea to help SMA patients by blocking the silencer NOW = give antisense oligos that are complementary to sequence in intron 7 -> ca turn SMN2 gene to make a full length product

Answer 48

Eukayotic mRNA modified nucleotdes (Base modifctaions = common in mRNA ) Use of Modification = Mark mRNA as self to prevent immune activation - Modifcation of ADA and psedouracil = ways that cells know that mRNA was self made

Answer 49

1. M6A – methylate A at position 6 - M6A = deposited by specific enzymes that modify mRNA - Abundent modification 2. Pseudouracil - A lot of Psudouracils on mRNA comes from tRNA modifying enzymes 3. A to I editting - ADARs = deaminiate A to I = change the coding capacity of mRNA - ADAR 1- endogenous dsRNAs prevnts immune activation - ADAR2 = specific editting events to recode important mRNAs

Answer 50

Modifications include: 1. Endonuclease cleavge 2. Exonuclease trimming 3. 3’ end modifications

Answer 51

Part of RNA porcessing = RNA decay Rate of production and decay determines how much of RNA or protein is present in a given cell - mRNA AND proteins can both be decayed - RNA decay has different purposes and uses distinct mechanisms - RNA decay = tightly regulated

Answer 52

Purpose of RNA decay: 1. RNA turnover to maintain a homeostasis or has regulatory purposes - Regulatory process (ex. During development) - can decay something fast if no longer needed 2. Degredation of defective RNAs that was not been processed correctley (quality control pathways) - IF mRNA is not properly spliced or doesn't have a CAP or polyA tail = mRNA will be degraded by an endonuclease 3. Removal of foreign RNA (Ex. CRIPSR, RNAi etc.)

Answer 53

Half life of RNA is different in different types of RNA - tRNA and rRNA can be stable for days vs. mRNA can be stable for minutes-hours - Rate of decay is under strong conrtol

Answer 54

Bacterial mRNA decay = needs endonucleaic cleavage (Decay starts with enodnuclease cleavage) BActeia ahve 3’ end and 5’ triphophate structure

Answer 55

Overall - Begins with deadenylation then decapping then exonucleaic cleave from the ends by XRN1 and exosome - miRNA and other proetins intiate process Process: 1. To degrade Eukaryotic mRNA = polyA tail needs to be degarded = need to remove the polyA protein and the deadenylate 2. Once have a short polyA tail = get decapping - Decapping = cleaves the 5’5’ phosphate bond) = allows exonuclease to chew mRNA from ends

Answer 56

Rate of deanylation = regulated based on mRNA strcture or the mRNA seqnce at the 3’UTR Rate is based on: 1. Based on proteins that guide a comples to the 3’ UTR of mRNA and begin deanylyating 2. Rate of decapping can be regulated by 3’ UTR structres and sequences through binding to other factors

Answer 57

Consequence of RNA Processing : 1. Production of functional RNAs and RNPs 2. Opportunity for regulation 3. Generation of molecular diversity 4. Biogenesis quality control

Answer 58

ALL of the mechanism appeared before complexity --> Means that even though the steps generate diversity that can’t be why they evolved because they evolve before diversity and you can’t select for something because it will be helpful in the future Prokaryotes show these “embellishments” are not necessary for the central dogma

Answer 59

Hypothesis for evolution of complexity = arms race between nucleic acids trying to invade genomes and the body coming up with solutions to combat this - Arms race = led to additions to gene regulatory process --> generates diverse patterns of gene expression --> can evolve more complexity - Likely evolved as part of defense against genomic parasites Example – nuclear envelope and chromatin = protect DNA from transposons and viruses --> THEN once tranpsosons and viruses could get in needed to change again - Once have nuclear envelope = need t get mRNA out