Transcription And Translation

Question 1

Similarities between dna replication and transcription

Answer 1

• Similarities:
• Fundamental chemical mechanism
• Addition of nucleotides to 3’ end of growing chain
• Polarity of polynucleotide growth is from 5’ to 3’
• Use of a dna template
• 3 phases: initiation, elongation, termination

Question 2

Differences between dna replication and transcription

Answer 2

• Does not require a primer – rna pol will bind to primer and synthesise starting with first nucleotide
• Not all of dna is transcribed
• Only one strand of a dna template is transcribed by rna pol
• Both polymerases can proofread but rna pol is more sloppy, makes more mistakes
• This is tolerated as millions of copies of rna molecule so doesn’t matter if 1 is mutated

Question 3

Requirements of rna pol for transcription

Answer 3

• Prokaryotic cells have one rna pol to synthesise all types of rna
• Eukaryotic cells have 3 kinds, each rna species made by a different polymerase
• 3D structure highly conserved
• Implies that catalytic basis is identical
• Requires dna for activity
• Also requires ribonucelotides and Mg2+
• Mg2+ chealates incoming nucleotides at active site
• Structure resembles a claw
• DNA drawn into claw
• B’ subunit causes strands to separate – no helicase needed

Question 4

Subunit structure of rna pol

Answer 4

• E.coli rna polymerases:
• Very big >400kD
• Five kinds of subunit: alpha, beta, beta prime, omega, sigma
• 2 alpha subunits
• Initiation form is called holoenzyme
• Sigma only appears in holoenzyme

Question 5

Role of sigma in rna pol

Answer 5

• Sigma only appears in holoenzyme
• Sigma helps enzyme to recognise specific dna sequences called promoters, initiate transcription then dissociate
• Leaves the core catalytic enzyme which carries out catalysis / chain elongation
• Sigma wants to stay bound to promoter so must dissociate from enzyme to allow elongation phase
• Sigma decreases ability of core enzyme to bind dna non-specifically
• Without sigma rna pol has high affinity for non specific binding
• Allows holoenzyme to bind promoters
• Binds -10 and -35 sequences
• Allows holoenzyme to migrate along dna until a promoter is encountered (random walk)
• Different sigma factors permit binding to different promoters
• Allows for specific, regulated gene expression
• Falls off if it doesn’t find promoter
• Can regulate which promoters are bound
• Different sigma subunits for different states of the cell
• Bacteria in stationary phase express different sigma subunits under anaerobic conditions

Question 6

Roles of rna pol subunits

Answer 6

• Alpha subunit binds regulatory subunits/ proteins
• Beta subunit forms phosphodiester bonds
• Beta prime binds dna template
• Sigma subunit does promoter recognition
• Omega does rnap assembly

Question 7

Properties of e.coli promoters

Answer 7

• Highly conserved sequences at -10 and -35
• -10 region is TA rich sequence
• Relative position very well conserved- number of bases between each conserved sequence
• -35 region quite well conserved
• Consensus sequence is a representation of a series of promoters
• There are also some upstream sequences with AT rich regions that are less highly conserved and are found in highly expressed genes and operons

Question 8

Key steps in bacterial transcription initiation

Answer 8

• Key step in transcription – decision to express a gene
• Holoenzyme binds to about 70bp before transcription start site
• 70bp of promoter region, immmediately before transcription start site
• DNA protein binding can be determined by dna footprint g experiments
• Critical, conserved sequences occur at -35bp and -10bp regions before start site
• Not transcribed but critical for gene expression
• A series of changes occur in both the dna and rna polymerase upon binding promoters
• Closed complex formation (dna intact)
• Conformational change
• Open complex (DNA partially separated)
• Expose template to allow first nucletide to be added to strand
• Initiation (add first ribonucleotide)
• Promoter clearance (loss of sigma and change in holoenzyme to core enzyme – the elongation form of rna polymerase)

Question 9

DNA footprinting

Answer 9

• Take dna molecule expected to contain promoter
• Radioactively label one end
• Incubate half with rna pol and the other half without
• Add nucleases to cut molecule to give set of differing lengths
• Run on denaturing SDS acrylamide gel to get a ladder
• Differences between bands will be 1bp
• Can Sanger sequence the gel
• On side with rna pol rna pol holds part of the dna molecule inside it to prevent it being degraded
• Gives shadow of where rna pol is binding
• Sequencing gel next to it shows the sequence rna pol binds to
• Most protected regions are between -10 and -35

Question 10

Differences between template and non-template dna strands

Answer 10

• The strand that serves as a template for rna polymerase is the template strand
• The strand complementary to the template is the non-template or coding strand
• It is identical in base sequence to the rna

Question 11

How is elongation accompanied by unwinding and winding of dna

Answer 11

• When the open complex between dna and rna pol forms approx 17bp of dna at the start site are unwound
• After promoter clearance, rna chain is extended and the polymerase moves along the dna
• Unwinds in front and rewinds behind
• Keeps approx 17bp as a bubble of unwound dna as it goes
• In pol 2 strands are forced apart and have to go in different directions
• DNA entering and leaving pol causes large kink
• RNA exits from different pathway to dna
• Strands are forced apart as they interact with different residues within pol
• Template dna strand selects appropriate ribonucleotides by Watson crick base pairing
• In the transcription bubble, about 8bp of dna and the newly synthesised rna chain are base paired in A DNA form
• Transcription bubble moves at 170Asec-1 or about 50 nucleotides sec-1
• Structure of rna pol forces rna to exit from the helix (unwinding the duplex)

Question 12

Rho dependent termination

Answer 12

• Requirement for additional protein factor Rho which is a six subunit protein
• Binds CA rich sequences in the RNA with a Rut segment upstream
• Once bound at rut site Rho travels towards 3’ end of transcript
• Uses atp driven helicase activity to unwind the dna/rna helix
• Factor moves up by rotating helical strand and causes rna pol to fall off by unwinding the RNA-DNA hybrid within the transcription bubble

Question 13

Rho independent termination

Answer 13

• Occurs after transcription of a gc rich stretch followed by an A-rich stretch in the template strand
• GC-rich sequence is self complementary and forms a hairpin
• Polymerase pauses after synthesising the Us and then backtracks as the RNA-DNA hybrid in the transcription bubble is unstable
• When the backtracking pol encounters the hairpin the rna and the pol are released from the template
• Presence of hairpin structures caused by inverted repeats in the RNA followed by UUU which causes pol to pause and release transcript due to weaker A-U bp that are more easily pulled apart

Question 14

Eukaryotic transcription fundamentals vs prokaryotes

Answer 14

• Eukaryotic transcription is spatially distinct from translation
• Occurs in the nucleus- transcripts are exported
• Must take place on dna that is packaged onto nucleosomes and higher-order forms of chromatin structure
• 3 types of polymerase which share some subunits
• Prokaryotes have a single biochemical space. Chromosomes and ribosomes are in same compartment so rna is immediately bound by ribosomes
• In eukaryotes mRNA is exported from nucleus to be captured by ribosomes

Question 15

Types of eukaryotic rna pol

Answer 15

• Polymerase I- in the nucleolus makes pre-ribosomal RNA
• Polymerase II – makes mRNAs and some small RNAs- most susceptible to alpha-aminitin toxin
• Polymerase III- makes tRNAs, 5S rRNA and other small RNAs

Question 16

Alpha amanitin

Answer 16

• Alpha-amanitin is produced by death cap mushroom and shuts down transcription. If an affect is shut down by the toxin it is transcription. Shows transcription is key regulatory step

Question 17

Eukaryotic promoters

Answer 17

• Eukaryotic promoters are much larger and more complex than prokaryotic promoters
• They bind many more proteins
• E.g. pol III promoter sequences downstream required for initiation of transcription and inside regions to be transcribed
• Pol II promoters are similar but more complex than prokaryotic promoters
• Core promoters may have TATA box at -25bp or an Inr sequence near +1
• Also may have upstream and downstream elements

Question 18

Transcription complex in eukaryotes

Answer 18

• Polymerase II does not contact these sequences directly and replies upon many other basal transcription factors to bind to promoters
• All require accessory protein factors to recruit polymerase
• In addition to core promoter sequences, enhancer sequences may lie several hundred or thousand base pairs away
• Unlike prokaryotic rna pol, pol II cannot initiate transcription independently
• Transcription initiation requires assembly of a very large transcription complex that includes core regulators and transcription factors
• Made up of:
• RNA pol II + basal transcription factors

Question 19

Roles of basal transcription factors

Answer 19

• TFIID binds TATA Binding protein (TBP)
• Also TFIIA, TFIIB, TFIIF, TFIIE, TFIIH - carry out functions similar to sigma subunit in bacteria
• All multi-subunit proteins
• TFIID recognises TATA box and TBP subunit binds it. TAF subunits recognise other dna sequences near transcription start point to regulate binding by TBP
• TFIIB accurately positions RNA pol at start site of transcription
• TFIIF stabilises rna pol interaction with TBP and TFIIB and helps attract TFIIE and TFIIH
• TFIIE attracts and regulates TFIIH
• TFIIH unwinds dna at transcription start point and releases rna pol from the promoter by phosphprylating ser5 of C terminal domain of rna pol, causing conformational change

Question 20

RNA processing

Answer 20

• Nearly all eukaryotic transcripts are processed
• Modification of bases
• (Some are chemical e.g. methylation to change nucleotides)
• Deletions and additions to 5’ and 3’ ends
• Removal of introns within primary transcript
• Pol II has C terminal domain that can be phosphorylated
• Responsible for closed to open conformational change and processing by splicing factors
• Eukaryotic mRNAs:
• Have methyl guanosine cap added to 5’ end (important for binding ribosome)
• Have polyadenyl tails (polyA) usually 100s of residiues, added to 3’ end after cleavage
• Have introns removed by splicing

Question 21

How was genetic code discovered

Answer 21

• Translation is the process of protein synthesis, based upon the dna code of four deoxyribonuceltoides
• Each protein consists of 20 types of amino acids
• Therefore, the code must be at least triplet as 4 to the power of 3 is 64 and 4 squared is 16
• Crick and Brenner showed it was triplet by mutational analysis
• Nirenberg added synthetic polyribonuceotides to bacterial extracts and showed they could make polypeptides e.g.polyU made polyPhe so UUU= Phe
• PolyA made polyLys
• Each codon = 1 amino acid

Question 22

What is redundancy

Answer 22

• Redundancy: more than one codon codes for an amino acid (as there are 20aa but 64 possible combinations)
• Tryptophan and methionine only have one codon
• Redundancy is usually 3rd amino acid in a codon
• E.g. arginine only 3rd nucleotide differs
• Each aa has a specific tRNA

Question 23

What is wobble base pairing

Answer 23

• Third position +tRNA – wobble base pairing (non-Watson Crick)
• TRNA binds to codon antiparallel (1st position of anticodon pairs with 3rd position of codon)
• 1st position – 5 possible nucleotides
• Inosine in anticodon = modified form of adenine, can base pairing with U, C or a in codon
• One tRNA can bind to more than one codon= flexibility
• There is some steric freedom in pairing of the 3rd base of the codon
• 1st and 2nd bases pair in a standard way
• Codons that differ in first 2 bases are recognised by different tRNA
• Inosine maximises number of codons that can be read by a tRNA molecule
• Interactions of tRNA + ribosome check whether Watson-crick base pairs are present in 1st 2 positions of codon-anticodon duplex but not third
• Ribosomeplays active role in decoding codon-anticodon interactions

Question 24

Open reading frame

Answer 24

• 1st codon is never at 5’ end of mRNA- always downstream of it
• The presence of a start (AUG) and stop (UAA, UGA and UAG) codons determines a sequence of codons called an open reading frame
• Before ORF 5’ untranslated regions are still in mRNA but not translated
• By chance have a stop codon = short open reading frames made of 20 codons, but real ORFs are much longer
• AUG= methionine or start (all proteins start with methionine)
• Can be predicted by computers in genome sequence
• In any mRNA sequence, can have 3 possible reading frames
• DNA can have 6 reading frames
• Usually find 2 out of 3 have more stop codons
• Start reading at 1st, 2nd, 3rd nucleotides gives different codons, 3 possible ORFs in any nucleotide sequence
• Correct ORF will be the longest sequence, the others will have many stop codons
• Different reading frames will have different codons
• Code for different proteins
• Generally only one reading frame is coding

Question 25

Structure of tRNA

Answer 25

• Serves as the adapter to recognise the triplet codon and link it to a particular amino acid • Specific base pairing between a triplet codon in the mRNA and three bases in tRNA – the anticodon • Specific amino-acid covalently linked to 3’ end • Structures of different tRNAs tend to be similar • L shaped: anticodon at one end, amino acid at the other Directed by intramolecular base pairing and base stacking • 5’ end next to 3’ end —> rna base pairs with itself and folds back on itself • All tRNAs can be arranged in a clover leaf pattern in which about half the residues are base paired

Question 26

Modified bases in tRNA

Answer 26

• TRNA molecules have many common structural features • Expected as tRNA molecules must all be able to interact in nearly the same way with ribosomes • Each is a single chain between 73 and 93 ribonucleotides • Contain 7-15 modified bases • Usually methylated or demethylated derivatives of A U C and G • Enzymatic modification • Some methylations prevent formation in base pairs, leaving bases accessible for interaction with other bases • Methylation imparts hydrophobic character to some regions of tRNA • About half of the bases are dsRNA • 5’ end is phosphorylated, usually pG • Activated amino acid attached to -OH group of adenosine residue of the invariant 3’ end : CCA , this single stranded region can change conformation in the course of amino acid activation and protein synthesis • Always U in same position in anticodon loop

Question 27

How are tRNAs charged with amino acids

Answer 27

• Covalent linkage of amino acids to tRNA • A crucial step in protein synthesis • Does not reply upon Watson-crick base pairing • Instead relies on activity of enzymes to maintain accuracy of code • Specificity comes from the exquisite ability of enzymes (aminoacetyl tRNA synthetases) to recognise even subtle differences in amino acid structure

Question 28

Mechanism of tRNA synthetases

Answer 28

Involves formation of an AMP-amino acid intermediate • Charge amino acids with ATP, pyrophosphate released • Form ester bond between alpha phosphate group and amino acid • 3’ hydroxyl group acts as a Nucleophile • Carbonyl susceptible to nucelophilic attack (electropositive) • Bond formed with 3’ carbon of adenine • 2’ hydroxyl can also act as Nucleophile • Transesterification reaction to shift ester to 3’ carbon • Some tRNA synthetases proof read • Chemically link tRNA Thr with Ser instead, (“mischarged” tRNA) • Incubate with Threonyl tRNA synthetases • Results in rapid hydrolysis of mischarged tRNA to Ser and free tRNA • Suggests if wrong amino acid incorporated, editing function activated and amino Acyl -tRNA bond hydrolysed • Like the proof reading by dna polymerase • Type 1 synthetases binds 1 tRNA • Type 2 synthetases is dimeric and can bind 2 tRNAs

Question 29

Structural principles of a ribosome

Answer 29

• Enormous structures (18nm diameter) consisting of rRNA (65% in e.coli) and protein (35%) • Sizes usually expressed in terms of sedimentation rates • Two subunits – large (50S bacteria, 60S eukaryotes), small (30S bacteria, 40S eukaryotes)

Question 30

Structure of ribosome (talk about sedimentation)

Answer 30

• Factory for translation • About ¼ dry weight of e.coli • 15000 per cell • Enormous structures, Mda • 18nm in diameter • Consists of rRNA (65%) and protein (35%) • Sizes usually expressed in terms of sedimentation rates • How quickly they sink in various mediums – measures density • Too large to purify by chromatographic means • Two subunits • Large (50S bacteria, 60S eukaryotes) • Small (30S bacteria, 40S eukaryotes) • Purified by sedimentation centrifugation • Sink according to density • More dense sink faster • Fractionates contents according to density • Can separate subunits • Can also sediment nucleic acids found in ribosome (3 types of rRNA) • Can put hole in bottom of tube to collect individual fractions

Question 31

Interaction of tRNA and ribosome

Answer 31

• 3tRNAs in cleft of large subunit • TRNA interacts with rRNA • RRNA is most significant in catalytic function and binding • Bacterial Large subunit contains 5S rRNA, 23S rRNA • Bacterial small subunit contains 16S rRNA • Eukaryotic large subunit contains 5S rRNA, 28S rRNA, 5.8S rRNA • Eukaryotic small subunit contains 18S rRNA

Question 32

Ribozyme

Answer 32

• The ribosome is a ribozyme • Visualisation of the ribsome highlights ability of RNA to form complex 3D structures for chemical catalysis • Most RNA in the cell is NOT mRNA • RNA is like proteins – folded chains • Not like dna which has a constant (ish) structure • RNA containing structures like the ribosomes are called ribozymes to distinguish from protein only enzymes • Agues for the idea that biochemical life began in an ‘rna world’

Question 33

Translational initiation and start codon

Answer 33

• All organisms have two tRNAs for the AUG methionine codon • One for initiating translation • One for ‘internal’ methionine residues • In bacteria, initiator tRNA has formyl-methionine linked to it • Formed enzymatically after Met-tRNA synthetases links Met to tRNAfMet • Met residue is N-formylated • Initiator tRNA is charged then formylated • Formulation at N terminus means the residue already has a peptide bond so can only be the N terminus of the protein • Internal Met is not formylated • In eukaryotes both tRNA have methionine, not formylated after charging

Question 34

Bacterial translation initiation complex

Answer 34

• Initiation factor 3 (IF3) binds small subunit, blocking interaction with large subunit • 3 tRNA binding sites in small subunit, E, P, A • E-exit site- where uncharged tRNA are ejected from (uncharged tRNA is tRNA with no amino acid attached) • P-peptidyl site • A- amino-Acyl site. tRNA enters here • IF1 binds to A to stop another tRNA coming in • MRNA binds to P site • Specific binding with Watson crick base pairing between Shine- Dalgarno (conserved sequence) in mRNA and 16S rRNA

Question 35

Shine - dalgarno sequence

Answer 35

• Shine dalgarno sequence is a conserved sequence upstream of the start codon • AUG binds to P and marks beginning of ORF • Positioning of AUG is due to the binding of the shine- dalgarno sequence anchoring mRNA in position

Question 36

Start of bacterial translation initiation

Answer 36

• Initiator tRNA binds to AUG, chaperoned by IF2, GTPase protein • IF2-GTP binds to initiator tRNA. Allows tRNA to bind at P site • GTP hydrolysis by IF2 causes conformational change • IF1, IF2, IF3 all ejected • Large subunit can bind • A site empty, tRNA can enter and interact with next codon in ORF

Question 37

Summary of bacterial translational initiation

Answer 37

• IF-1 and IF-3 bind the small 30S subunit • IF-3 prevents premature assembly with 50S subunit • MRNA binds to 30S subunit using its Shine-Dalgarno sequence • Base pairs with 16S rRNA in 30S subunit • FMet-tRNAfMet brought into P site with IF-2GTP • Combines with 50S subunit after GTP hydrolysis and departure of initiating factors • Ribosome stimulates the GTPase activity • Correct incorporation of fMet-tRNAfMet from: • Interaction of mRNA and 16S rRNA • Interaction of fMet-tRNAfMet with AUG codon • Interaction of fMet-tRNAfMet with P site • (Only tRNA that can directly enter the P site) • Start site is critical to the fidelity of code

Question 38

Eukaryotic initiation of translation

Answer 38

• Similar mechanism but more protein initiation factors required • No shine-dalgarno sequence • Presence of shine-dalgarno sequence means you can have more than one ORF in an mRNA as you can have multiple shine-dalgarno sequences • Positioning of mRNA relies upon interaction with both ends – the 5’ cap and 3’ polyA tail • Both recognised by initiation factors

Question 39

Initiation factors binding for eukaryotic translation

Answer 39

• IFs bind to tail by polyA binding proteins. Joins cap and tail to make a circle • Eukaryotic mRNAs are usually monocistronic (one ORF) : no need for internal initiation • Ribosomes can ‘scan’ the mRNA in a 5’ to 3’ direction until they encounter the first AUG codon • eIF1, eIF3 and eIFA bind small subunit • EIF1 binds E and eIFA binds A • Initiator tRNA brought in with eIF2-GTP GTPase • EIF4F complex binds mRNA and allows entry into ribosome • One protein in eIF4F complex is an RNA helicase (eIF4A) • Ribosome activates helicase • Helicase pulls RNA through • Interaction between codon and initiator tRNA anticodon causes pause • GTP hydrolysis causes conformational change • EIF4F still associated • Ribosome checks RNA is intact as polyA binding proteins interact with eIF4F complex which also activates helicase acitivity • No scanning without polyA tail

Question 40

Mechanism of translational elongation

Answer 40

• Start with large subunit bound • A site is empty, second codon exposed • TRNAs come in and out • A site has high affinity for tRNA • TRNA stays if there is interaction between codon and anticodon • TRNAs can only enter if bound to GTPase – Tu-GTP • Amino acids covalently linked at 3’ end • If recognition, ribsome activates GTPase activity causes conformational change • Aminoacyl end of tRNA swings so it is moved closer to first amino acid • Tu-GDP ejected • GEF in cytoplasm exchanges GDP for GTP • Nuelotide residues of 23S rRNA surround amino acids and catalyse peptide transferase reaction where peptide bond is formed • Uncharged tRNA now in P site

Question 41

Role of elongation factors in translational elongation

Answer 41

• Uncharged tRNA now in P site • Elongation factor EF-G enters A site • Ribosome translocates and moves RNA along • EF-G is in A site • GTP hydrolysis removes EF-G • A site now empty • EF-G has similar structure to EF-Tu bound to tRNA. Example of molecular mimicry as protein structure of EF-G mimics tRNA

Question 42

Summary of translational elongation

Answer 42

• Requires factors EF-Tu, EF-TS and EF-G • Aminoacyl-tRNA brought into A site by EF-TuGTP • GTP hydrolysed and EF-TuGDP released • EF-TuGTP regenerated using EFTs • Peptide bond formation catalysed by 23S rRNA from nucleophilic attack by alpha amino group of amino acid in A site on carbonyl group of peptide in P site • Translocation requiring EF-G and GTP hydrolysis • Moves ribsome one codon along mRNA (5’ to 3’) • Moves newly synthesised peptidyl-tRNA into P site • Moves ‘uncharged’ tRNA into E site

Question 43

Eukaryotic translational elongation

Answer 43

• Analogous elongation factors • eEF1alpha – EF-Tu • eEF1betagamma- EF-Ts • eEF2- EF-G • Nt interchangeable • Same mechanism as bacteria

Question 44

Mechanism of translational termination

Answer 44

• No tRNA for recognising UAA, UGA or UAG • Release factors enter A site: • RF1 recognises UAA and UAG • RF2 recognises UAA and UGA • Most bacterial codes end in UAA as more efficiently terminated bc two RFs recognise it • Encourage ribosome to transfer peptide to water rather than aminoacyl tRNA, causes release • Ribosome dissociation requires EF-G, RRF and IF-3, each one removes the preceding one • Eukaryotes have a single factor called eRF

Question 45

How do antibiotics inhibit protein synthesis

Answer 45

• Number of naturally occurring toxins and antibiotics inhibit translation implying structural differences between the eukaryotic and prokaryotic ribsome • Puromycin – similar to 3’ end of a tRNA, binds A site and causes premature termination • Tetracyclines- block A site • Chloramphenicol- Blocks peptidyltransferase • Cycloheximide- blocks eukaryotic peptidyltransferase • Streptomycin- causes misreading of codons