lectures 10-21 Flashcards
genome sizes are stated as….
bp per haploid genome
benefits from HGP?
personalised medicine
the central dogma
dna makes rna makes protein
dna and rna replication transfers info
DNA replication, transcription, translation, RNA replication, reverse transcription
reverse transcription
info at RNA level is copied to DNA
RNA viruses
retroviruses
mostly in RNA level
insert RNA into our DNA
why did Mendel work with peas?
large numbers of offspring short generation time self-fertilisation and cross-fertilisation possible cheap convenient
the first law of inheritance: the law of segregatin
2 coexisting alleles of an individual for each trait, segregate during gamete formation so that each gamete gets only 1 of the 2 alleles
what acts like Mendelian factors?
sex chromosomes
what did T. H. Morgan propose?
X chromosomes carried genes other than sex determinants (eye-colour in fruit flies)
sex-linked
what did Walter Sutton propose?
chromosome theory of inheritance: sex was determined by chromosome based inheritance
alkaptonuria disease
secrete homogenistic acid into urine which goes black following exposure to air
what did Garrod propose?
a gene for a metabolic step was defective in albinism and alkaptonuria
Life cycle of Neurospora crassa (haploid organism)
haploid ascospores (4A, 4a), ascus germination growth of conidiospores (reproduces by vegetative haploid spores) germinating conidium vegetative mycelium of A and a
cells of opposite mating type (A and a) fuse to form binucleate heterokaryon (2 diff nuclei) nuclear fusion meiosis, mitosis to form diploid nucleus fruiting body separate to haploid ascospores
Beadle and Tatum’s question
what did they do?
is there a separate gene for each step?
made arginine auxotrophs of Neurospora crassa
mutated arginine biosynthesis
X-RAY MUTATION in A and a, mate them to form fruiting body, contained mutants in arginine biosynthesis
MATE wild and mutated type of both A and a types
DISSECT individual microscopic ascospores to individual test tubes, grow colonies
IDENTIFY MUTANTS by transferring to minimal medium so failure to grow identified nutritional requirements
IDENTIFY NUTRITIONAL REQUIREMENT by diff minimal mediums so mutation in AA biosynthesis
IDENTIFY ARGININE AUXOTROPHS by testing all AAs
auxotroph
mutant that requires a particular additional nutrient
prototroph
normal strain which does not require that nutritional supplement
evidence for multiple steps of pathway
if auxotrophs came from different asci, probably have different mutations
if defective in different parts of arginine pathway
mated, then mutants might complement each other so heterokaryon would grow in absence of arginine
provides evidence of multiple steps
how does complementation work
heterokaryon contains both nuclei so between them they can perform all steps in complementation
defects complement each other
Beadle and Tatum’s results
each step of metabolism requires individual gene
What did Friedrich Miescher find?
discovered nucleic acids
sticky substance in pus
like protein but rich in phosphorus with no detectable sulphur
Griffith’s experiment on the transforming principle/factor
Avery-MacLeod-McCarthy follow up experiment
R rough colonies are non-pathogenic
S smooth colonies are pathogenic
dead S cells don’t cause pneumonia, no cells left
mixture of dead S and living R caused pneumonia so S cells transformed the R cells into pathogens
living S killed by heat, with enzyme treatment, with living R, only DNase destroyed the transforming principle
also purified DNA from S cells added to R cells transformed R cells to S cells, therefore hereditary material
T2 ‘phage research
phage infects E.coli with attachment mediated by base plate and fibres
remain attached but heads appear empty
labelled their DNA radioactively infect, centrifuge some bacteria were now radio-labelled grow bacteria in fresh medium phages were radiolabelled - confirms DNA is genetic material
labelled phage protein, not transferred to E.coli so not genetic material
what sugar is used in DNA?
pentose deoxyribose (5 carbon)
purine bases
pyrimidine bases
adenine, guanine (2 rings)
cytosine, thymine (1 ring)
which bond joins a nitrogenous base to deoxyribose sugar?
glycosidic bond
between sugar C-1’ and N-9 (purine) or N-1 (pyrimidine)
what other bond is in DNA?
ester bonds are formed between sugar C-5’ and phosphate
phosphate in DNA makes it …
negative
DNA has a 5’ and 3’ end so it is…
polar
phosphodiester bond in DNA
links 3’ C of 1 nucleotide to 5’ C of next so has polarity
5’ end
3’ end
phosphate group
hydroxyl group
what was Levene’s model?
4 nucleotides in tetranucleotide blocks with bases pointing out
so DNA simple and not repetitive so not genetic material
also meant that each base present in equal amounts (not true)
x-ray crystallography
cross is typical of helical structure
stretched out DNA and left hydrated, produced cross pattern because alligned helices effectively form a diffraction grid which produces a cross pattern
closer spots on diffraction pattern means larger actual distance
calculated distances in DNA
why was Linus Pauling’s DNA model wrong?
triple helix means that sugar-phosphate backbone faced inwards but negative charges would repel each other
B-DNA
Z-DNA
right-handed clockwise (if slid down it)
left-handed anti-clockwise
6 key features of Watson-Crick model of DNA
right-handed
anti-parallel strands (for bases to match)
bases inwards, sugar-phosphate outwards
complementary base pairing
base pair distance (0.34nm apart)
major and minor grooves (backbones not equally spaced, important with how proteins interact w/ DNA)
how many hydrogen bond when A pairs with T?
C-G?
2
3
how is eukaryotic DNA organised?
DNA duplex (coiled in double helix) wrapped around histone proteins (nucleosome) chromatin fibre coiled chromatin fibre coiled coil metaphase chromatid (folded)
conservative model for DNA replication
parent strand transfers info to intermediate and this gets copied
parent helix is conserved and daughter is completely new
dispersive model for DNA replication
parent helix broken into fragments, dispersed, copied, assembled into 2 new helices
new and old DNA completely dispersed
caesium chloride equilibrium density gradient centrifugation
separates molecules on basis of densities
get gradient of CsCl
macromolecules will be where match CsCl density
Meselson and Stahl testing of models of DNA replication
used centrifugation to separate on densities: E.coli grown in light (N-14) and heavy (N-15) so when incorporated will label DNA
mixed together, separate, created dark bands with UV light
E.coli grown in heavy, then added light, so t=0 all would be heavy, as grow and divide would incorporate light
semi-conservative:heavy, intermediate, light and more intermediate, slowly goes to light and loses intermediate because more of new strands, loss of heavy and light (old and new)
this is what experiment showed
predicted bands for other models didn’t occur
what function does DNA polymerase I (Pol I) have other than 5’ to 3’ polymerising
5’ to 3’ exonuclease (break down)
3’ to 5’ exonuclease activity
so editing mistakes and proof-reading function
nucleotide incorporation into DNA
release of pyrophosphate (2 phosphates) from nucleotide and hydrolysis to phosphate
so 1 phosphate remains on nucleotide which binds to other nucleotide
DNA polymerase action during incorporation of nucleotides
conformational change in DNA polymerase, closes round DNA when correct nucleotide binds
testing DNA strand polarity
radio labelled phosphate on nucleotides about to be incorporated
phosphate is 5’ of new nucleotide that’s inserted, when DNAse breaks ester bonds so strand into pieces, labelled phosphate is now attached 3’ of nearest neighbour
so see how often bases near each other (phosphate go from 1 known radiolabelled base to next base)
proportions of bases can only match if strands are anti-parallel (GpA same amount as TpC so has be be opposite directions if correct bases are to match)
editing by DNA polymerase
incorrect nucleotide sticks out with unpaired 3’OH so DNA polymerase 3’ to 5’ exonuclease removes mismatch
enzyme has 2 active sites, 1 for polymerisation, 1 for editing
why is DNA replication asymmetrical?
because on lagging strand DNA replicated away from the fork because always 5’ to 3’ which means that as fork opens, needs to prime again so not continuous
what are the fragments on the lagging strand called?
Okazaki
why is there no 3’ to 5’ synthesis of DNA?
in 5’ to 3’, the E used to incorporate the next nucleotide is from when the high energy bond is broken when pyrophosphate is removed
if 3’ to 5’, when editing is required and the wrong nucleotide is removed, it will leave 5’ end with only 1 phosphate (because pyrophosphate was removed when nucleotide was added)
so there is no high E bond available for next time a nucleotide needs to be added
what evidence shows that the primer for Okazaki fragments is RNA not DNA?
DNAse cannot completely destroy Okazaki fragments, it left little pieces of RNA, therefore they are the primers
DNA primase
type of RNA polymerase that is rifampicin-sensitive
catalyses the synthesis of a short RNA segment called a primer to inititate DNA synthesis on the lagging strand
makes a primer but doesn’t need a primer itself
lagging strand synthesis
DNA primase makes RNA primers
DNA Pol III extends primer to make Okazaki fragments
old primers erased by 5’ - 3’ exonuclease Pol I activity
replaced with new DNA
gap sealed by DNA ligase
Okazaki fragment joining
DNA ligase releases pyrophosphate from ATP and attaches resulting AMP to 5’ phosphate of downstream fragment
AMP released and phosphodiester bond forms between upstream 3’ OH and downstream 5’ phosphate
Processivity
enzyme’s ability to catalyse consecutive reactions without releasing its substrate
Leading strand synthesis
DNA helicase unwinds helix, with ATP
DNA primase makes RNA primer
primed duplex captured by Pol III
clamp holder transfers 2 halves of β clamp to Pol III
(new clamp halves maintain clamp holder in state of readiness)
clamping converts Pol III to high processivity so can replicate long stretches of DNA
helicase continues to unwind and Pol replicates leading strand
lagging strand synthesis in terms of DNA Pol
DNA primers makes primers
primed duplex captured by Pol III, clamped
lagging strand into loop so that 5’ end the right way away from the fork (same way as leading strand)
until old Okazaki fragments pulled back to Pol III so triggers unclamping so goes back to low processivity (has to had low to release fragments easily)
DNA Pol I and DNA ligase repair gaps
process restarts by clamping new lagging strand prier
DNA Pol I doesn’t require high processivity because works short period
DNA polymerase structure
clamp loader with spare clamp halves
arms with Pol III core, β clamp on it
DNA primase and helicase on end of clamp loader
separately has DNA Pol I and DNA ligase, and single-stranded DNA binding protein
single-stranded binding protein (SSB)
prevents inappropriate bsae pairing in ssDNA during replication
binds to ssDNA so keeps straight and stops forming strange structures
similar to when stem-loop structures can form if DNA is denatured and fails to re-anneal properly
constantly knocked off by DNA Pol III and replaced as helix unwinds
positive supercoiling
from overwinding DNA so can occur upstream of replication fork
makes strands difficult to separate
negative supercoiling
from unwinding/underwinding
easier to replicate so topoisomerases used to regulate degree and type of supercoiling - convert positive to negative
type II topoisomerases
convert overwound positively supercoiled DNA into underwound negatively supercoiled DNA
binds positive supercoil, makes nick in both strands, passes loop through break and religates it to negative supercoil
type I topoisomerase
relax negatively supercoiled DNA
bind to supercoil, nick in 1 strand, unwinds DNA and re-ligates it, releases tension
Bacterial DNA polymerisation
bi-directional
2 Pol III complexes enter DNA, replication in both directions at same time, 2 forks meet
2 chromosomes linked (catenated)
Topo IV separates catenated chromosomes by double stranded break and religation
telomeres
ends of chromosomes
problem with telomeres in eukaryotes
lagging strand template primed at telomere, DNA polymerase falls off
primers erased
gaps filled and repaired
gap at end cannot be filled by DNA polymerase because no primer
so cells get shorter after every cell division
telomerase
reverse transcriptase
provides primer that extends telomeres
contains bound RNA template
provides RNA template and uses that to synthesis DNA copy of template at 3’ end of parental lagging strand template
dideoxynucleotide (ddNTP)
no 3’ hydroxyl so both H not OH (2 deoxy)
so replication cannot contine when added
chain terminators
Sanger’s dideoxy sequencing method
add ssDNA, primer, DNA polymerase, deoxynucleoties, small proportion of radioactively-labelled ddATP/ddGTP etc.
produce diff sized strands that terminate in same base
separate with electrophoresis
read from bottom up
automated dideoxy sequencing
fluorescent ddNTPs in 1 tube
electrophoretic system, read with laser
What is PCR?
what material are needed?
amplify DNA sequence
template dsDNA 2 SPECIFIC oligodeoxynucleotide primers dNTPs buffer and MgCl2 Taq polymerase
how does PCR work?
heat to denature target DNA to ssDNA
hybridisation of primer to each strand (cool to allow to anneal)
extension of primers by Taq polymerase (72C)
products act as substrates for next cycle
PCR has both ______ and _______
and what do these mean?
specificity - provided by primers (complementary to opposite strands, with 3’ pointing towards each other)
sensitivity - 1 target molecule amplified a lot in short time because product of 1 cycle becomes template for next
mutations by PCR
introduced into amplified DNA by engineering mismatches in primer close to or at 5’ end
so next generations have mutation because use mutated as template
PCR in forensics
RFLP (restriction fragment length polymorphism)
requires large amounts of DNA so PCR allows small samples to be processed
VNTRs
variable number tandem repeats
act like allele
inherit different variant of each VNTR locus from mother and father
so can be used for identification
What is a possible reason for why uracil is not found in DNA?
cytosine can spontaneously deaminate to uracil
so a C-G pairing could be replaced with a U-A leading to a mutation
so U is removed by uracil-DNA glycosylase and repaired by DNA polymerase
3 classes of RNA
mRNA encodes proteins
rRNA makes up ribosomes and role in protein synthesis
tRNA are adaptors between mRNA and amino acids, role in protein synthesis
features of prokaryotic transcription
operon: groups of related genes transcribed from same promoter, controlled by operator
polycistronic RNA: multiple genes transcribed as 1 transcript
no nucleus so transcription same place as translation
prokaryotic transcription unit
5’ promoter binds RNA polymerase
transcribed sequence or RNA coding sequence
3’ terminator signals transcription stop
bacterial RNA polymerase
α β β’ ω subunits in 2:1:1:1 ratio
σ subunit converts enzyme to holoenzyme (active enzyme with coenzyme) - needs this to start and make RNA, directs polymerase to promoter
how to find out the promoter sequence
bind RNA polymerase holoenzyme to DNA in vitro
add nuclease so DNA degraded except for stretch bound to polymerase which is protected so this is the promoter
find sequence by electrophoresis
the ______ of the promoter sequence provides _________
asymmetry
directionality
3 stages of prokaryotic transcription
initiation: RNA Pol holoenzyme binds promoter, opens DNA helix and starts to transcribe
elongation: the σ disengages from holoenzyme and core enzyme continues to make RNA
termination: RNA polymerase core enzyme dissociates from DNA and transcription stops
initiation
core RNA polymerase binds DNA
σ subunit binds to core polymerase and directs to promoter
opens helix close to promoter (no ATP used unlike DNA helicase)
transcription start site exposed, RNA copy, doesn’t require primer
elongation proofreading
if misincorporates ribonucleotide, hesitates, then back-tracks, removes, continues
termination
Rho-independent: terminator sequence in RNA is recognised (palindromic GC rich followed by T rich)
stem loop structure with U tail after it, RNA Pol pauses and stem loop forms, RNA dissociates because only weak 4 A:U holding
Rho-dependent: requires rho protein to break RNa:DNA duplex in transcription bubble
protein (hexameric helicase) binds to C-rich G-poor sequences, uses helicase activity to chase RNA Pol, disrupts hybrid helix releasing RNA
what does RNA Pol do that allows the duplex to be opened more easily?
bends DNA so opens helix
Rifampicin
inhibits prokaryotic transcription because inhibits RNA polymerase because binds to exit channel so affects initiation
doesn’t affect if already in elongation because channel blocked so rifampicin doesn’t work
RNA Pol II
c-terminal extension on L’ subunit
RNA Pol II promoters
TATA box upstream or DPE downstream
Inr : initiator element (2 pyrimidine, C,A, 6 pyrimidine), where transcription starts
enhancer
may be CAAT box or GC box (on either strand)
TATA, CAAT, GC recognised by other proteins not Pol
TFIID
transcription factor
large
comprises 1 TAF (TBP associated factors) and TBP (TATA-box binding protein)
TFIIH
bi-functional
helicase that open helix so exposes initiator element
kinase that phosphorylates (phosphate added) C’-terminal domain (CTD) of RNA Pol II L’ subunit so triggers elongation
α-amanitin
binds to active site of RNA Pol II so reduce RNA production
constrains flexibility of RNA Pol II
enhancers
increase transcription levels of genes
DNA looping model
DNA bending proteins create loop of DNA so distant enhancer interacts with components of transcription initiation complex and basal transcription complex (enhancer binds to activator binds to mediator interacts with CTD of L’ of RNA Pol II)
DNA relaxes so easy to transcribe
histone acetylation
acetylation or methylation affects packing of chromatin and makes loose structure that boosts transcription
Processing of eukaryotic mRNA
5’ cap
3’ poly-A tail
introns removed by splicing
addition of a 5’ cap
5’ triphosphate modified (3 phosphates at 5’ can be degraded by nucleases)
1 phosphate removed and disphosphate left (2 phosphates) attacks phosphate of GTP to form 5’-5’ triphosphate linkage
mimics 3’ OH end
addition of a poly-A tail
3’ end (AAUAAA) cleaved by specific endonuclease
adenylate residues from ATP added by polyA polymerase
how do capping and tailing enzymes find the transcript?
phosphorylation of CTD of RNA Pol II recruits proteins for processing (capping/tail)
RNA editing by deamination
CAA deaminated to UAA (stop signal) so smaller protein and allows more than 1 protein from 1 gene
prokaryotic mRNA are often _______ which means….
while eukaryotic..
polycistronic
they contain amino acid coding info for more than 1 gene
mRNAs are monocistronic so they encode just 1 protein
in higher eukaryotes, more DNA is devoted to…… than to……
introns
exons
_______________________ don’t have introns
histone proteins
R-loop analysis
RNA-DNA hybridisation can be monitored by electron microscopy allowing qualitative and quantitative analysis of gene organisation, position, extension of homology regions and characterisation of transcription
can map specific DNA region
Prokaryotic R-loop
mRNA and corresponding DNA mixed
heated to separate DNA strands, cooled to allow hybridisation (mRNA binds to DNA in loop)
single displaced loop of ssDNA inidicates that gene is continuous and joined
Eukaryotic R-loop
2 loops of ssDNA and loop of dsDNA seen if gene contains intron
more introns = more loops
4 classes of intron
group I : self-splicing found in organelles and nuclear rRNA genes of some unicellular eukaryotes
group II : self-splicing in organelles in fungi, protists and plants
spliceosome-dependent introns : require spliceosome which removes introns, in nuclear mRNA
nuclear tRNA
ciliates
unicellular eukaryotes
conserved features of introns
5' splice site (with AG upstream) 3' splice site (with GU downstream) branch site - has specific adenine residue (in class II and spliceosomal introns)
self-splicing of group I introns
2 sequential transesterification reactions
exchnage organic R of ester with organic R of alcohol to produce different alcohol and ester
3’ OH of co-factor breaks phosphodiester bond at 5’ splice site
5’ exon end is now a G 3’ OH which breaks phosphodiester bond at 3’ splice site so exons join and introns removed
intron folding for splicing
tertiary structure so 5’ and 3’ splice sites close to allow transesterification
inside tertiary structure is G binding pocket which binds G nucleotide/nucleoside (co-factor) and presents 3’ OH at the right place
splicing of group II introns
no co-factor but internal nucleophile (2’ OH of branch site adenine) attacks phosphate at 5’ splice site to form lariat structure and freeing upstream exon
free exon attacks 3’ splice site so fuse and remove introns
ORF
open reading frame: bit that is translated, protein coded gene
ORFs in introns
some group II introns encode a splicing protein (maturase) in intronic ORF because require it for splicing
maturase stabilises structure of intron
intron-early hypothesis
since all 3 domains of life have introns, must be ancient
and play valuable role
intron-late hypothesis
some group I introns encode a homing endonuclease HEG which catalyses intron mobility
so may move intron from 1 organism to another
introns may be parasitic nucleic acids that encode protein that allows spread
spliceosome dependent intron splicing
U1 and U2 snRNP dimer binds to 5’ and branch site respectively so bring together
nucleation point for U4 U5 U6 trimer
U4 binding to U6 inactivates U6
dissociation of U4 activates U6 which displaces U1
active spliceosome (U2, U5, U6)
then same splicing as group II intron splicing
snRNP
small nuclear ribonucleic particle
comprises a snRNA and at least 7 protein subunits
what is the value of a complicated spliceosome?
bring 5’ splice site and branch site closer so more efficient transesterification
spliceosome recruited by phosphorylated C-terminal domain of L’ subunit of RNA Pol II so splicing coordinated with transcription
nuclear tRNA introns splicing
requires ATP and splicing endonuclease and ligase
alternative splicing
generates protein diversity
different introns removed so make more proteins than have genes
exon shuffling
recombination with breaks and rejoining of DNA
sometimes alignment of chromosomes goes wrong so non related genes line up and exons from unrelated genes end up in 1 protein
diamond hypothesis
amino acid fit into diamond pockets formed within grooves of DNA where 4 sides of each pocket defined by 4 bases
20 pockets and 20 amino acids so overlapping degenerate triplet code
why is the code non-overlapping?
if was overlapping, would limit what amino acids can follow certain ones, so would be some impossible combinations
but found that any amino acid can follow any so code is non-overlapping
RF
recombination frequency: frequency with which crossover and rare recombination will take place
depends on how far apart genes are
further apart means more likely to recombine
cistron
gene that encodes a polypeptide
non-essential part of gene
tolerate mutations
evidence for triplet code
+3 or -3 restored reading frame and wild type of bacteriophage
evidence for degenerate code
more proline than expected so must be more than 1 triplet code for 1 amino acid
tRNA
half of molecule is base paired
3 conserved loop - D loop, T loop, anticodon loop
clover leaf pattern
variably sized extra loop between anticodon and T loop
specific amino acid attached at 3’ CCA motif
folds into L structure that exposes anticodon
the wobble hypothesis
3rd base less role in determining amino acid than 1st 2 bases
1st 2 bases form strong Watson-Crick base pairs with last 2 bases f anticodon
1st base of anticodon which pairs with 3rd base of codon ca wobble and determines number of codons that can be recognised by tRNA
U pairs with A or G so 2 codons
G pairs with C or U so 2 codons
I (inosine modified nucleotide) pairs with A U or C so 3 codons
uncharged tRNA
charged tRNA
doesn’t have amino acid attached, written as tRNA^aa
has AA attached written as aa-tRNA^aa
charging tRNA
aminoacyl tRNA synthetases attach AA to 3’ of corresponding tRNA
tRNA synthetase for each AA so very specific
synthetase can also edit and replace incorrect AA
requires energy
amino acid activated by energy derived from ATP hydrolysis, high energy bond created
so codon in mRNA selects tRNA charged with activated high energy amino acid corresponding to that codon
structure of prokaryotic ribosome
large subunit (50S) : catalyses formation of peptide bonds to link amino acids small subunit (30S) : provides framework where tRNAs can be matched to codons
assembled 70S ribosome has 3 sites:
E: tRNA exit site
P: peptidyl site
A: aminoacyl-tRNA site
how does a ribosome find a prokaryotic mRNA?
Shine-Dalgarno sequence (SD) dominated by purines
initiation of prokaryotic translation
IF-3 prevents assembly of ribosome until both mRNA and initiating tRNA are available
S subunit binds SD sequence
IF-1 prevents entry into A site
IF-2 guides initiating aminoacyl tRNA to the P site
L subunit recruited to 30S initiation complex
IF-1 and IF-2 released
prokaryotic translational elongation
EF (elongation factor) guides next tRNA to A site
peptide bond forms
EF-G (translocase) enters A site so ribosome moves 1 codon towards 3’ end so tRNA in E (exit) site and other tRNA enters P site
incoming tRNA displaces EF-G and other tRNA in E site leaves
prokaryotic translational termination
RF (release factor) binds STOP codon
peptidyl-tRNA bond is cleaved, releasing the protein
mRNA-ribosome complex disassembles
head polymerisation
high energy bond of incoming monomer is used for next polymerisation
structure of eukaryotic ribosome
60S subunit
40S subunit
80S ribosome with 2 sites
P: peptidyl site
A: aminoacyl-tRNA site
eukaryotic initiation of translation
cap binding protein bound to 5’ cap
poly A binding protein bind poly A tail
bridge between cap and poly A binding proteins so form circularised mRNA
5’ untranslated region form stem-loop structures which is scanned by mRNA (instead of SD)
multiple initiation factors assemble with the 40S subunit
allow binding of tRNA and 5’ cap of mRNA binds to S subunit
initiation complex scans mRNA using helicase activity of eIF4A to unwind stem loop structure
takes first AUG after Kozak as the start
60S subunit recruited, tRNA at P site
eukaryotic translational elongation
tRNA enters A site and peptide bond forms
translocase enters A site and ribosome translocates by 1 codon in 5’ to 3’ directions
etc.
eukaryotic translational termination
single release factor that recognises STOP codon
peptidyl-tRNA bond is cleaved, releasing the protein and ribosome disassociates
