Prokaryotic Transcription and rRna and tRna Questionnaire

NUCLEIC ACID FUNCTION AND PROTEINSYNTHESIS
CHEM 563
Lectures 8 and 9
Prokaryotic RNA transcription
Manal Swairjo, Ph.D.
• What is a gene? A nucleotide sequence in DNA carrying information.
• What defines a gene? Its start and end.
• The information has a direction: always 5’ to 3’. Understand the words
“downstream” and “upstream.”
• That information can be on either DNA strand, but not on both
simultaneously.
• Cells need to make an RNA copy of each gene, i.e. transcribe it.
• Therefore the sequence information in DNA that needs to be copied to
RNA can be either on the top strand or on the bottom strand. How does
the cell know which strand is the gene on and where it starts and ends?
Genes and operons
•
“Gene” is used loosely to refer to a DNA sequence that encodes a protein or
an RNA as the final product. A “gene” may also include the flanking
sequences that control its transcription and translation.
•
A gene that encodes a protein (i.e. copied to mRNA) is called a structural
gene.
•
Genes can be on either DNA strand throughout the chromosome.
•
Some genes occur next to each other in the chromosome, and are
transcribed as one unit so they are guaranteed to be expressed together. A
group of such consecutive genes is called an operon. E.g. lac operon (3
genes), trp operon (6 genes).
•
The RNA transcribed from an operon is called polycistronic RNA (i.e.
contains the sequence of several genes in a row).
This is the E. coli lac operon. It has 3 structural genes
that encode lactose metabolism enzymes.
We will discuss it in more detail later.
Different types of RNA in the cell
Structural gene
Nonstructural gene
coding RNA (mRNA)
protein
noncoding RNA.
1. Coding RNAs (mRNA) code for proteins. They get
translated to proteins.
2. Noncoding RNAs (next slide). They do NOT get
translated to proteins.
How transcription works
• This is on the next slide, but later watch
this short video:
https://dnalc.cshl.edu/resources/3d/13transcription-advanced.html
For a given gene, only one strand in the DNA double helix is
transcribed to RNA (unlike DNA replication). Here, the gene that’s being
transcribed lies in the blue strand. Like DNA before, RNA is synthesized in
the 5’ to 3’ direction.
Non-template strand
Template strand
•
The DNA strand carrying the sequence to be copied is called the sense (coding) strand.
•
The template DNA strand guiding the synthesis of RNA is called the antisense strand.
•
RNA being synthesized has the same sequence as the sense (coding) DNA strand, except U’s
instead of T’s.
•
The RNA being synthesized is complementary to the template (noncoding) DNA strand.
RNA chain growth always occurs in a 5′ to 3′ direction.
That is, incoming NTP’s are always added to the 3′ end
of the growing RNA chain.
1′
2′
3′
4′
5′
This is done by the processive enzyme RNA polymerase.
It polymerizes up to 1900 bases in one run.
RNA is synthesized by the enzyme RNA polymerase
(RNAP)
ω
• This is the bacterial RNAP for example.
• looks like an open crab claw.
β’
α
• 5 subunits, ααββ′ω. This is the core
enzyme that catalyzes RNA synthesis.
• Open complex, wide enough to
accommodate dsDNA.
• An additional sixth “σ subunit” joins this
core structure and acts as a cofactor
(not shown in figure).
α
β
1. How does RNA polymerase find where a
gene begins?
2. How does RNA polymerase recognize the
correct DNA strand that harbors the gene?
Stages of transcription in prokaryotes
RNAP holoenzyme
Sigma factor
How RNAP finds where the gene begins
• On the coding (sense) strand of the gene, on the 5’ side of a gene, i.e.
upstream of the gene, there is ~40 bp sequence that marks the initiation site
for transcription. It is called a “promoter” or a “Pribnow box” after David
Pribnow (Harvard) who first identified it. Also called the “-10 sequence”
because it is centered around 10 nt upstream of the gene.
• As RNAP moves along the dsDNA, it has low affinity to it (Kd ~ 1×10-7 M), so it
keeps moving, searching for the promoter sequence.
• When a promoter sequence comes along (on the sense strand), it is
recognized with high affinity (Kd ~ 1×10-12 M) by the sixth subunit of RNAP, the
“σ subunit” or “sigma factor” causing the whole complex to pause at that
site.
• RNAP complex closes over the template (antisense) DNA strand, separates
the strands over a distance of 15-20 base pairs, forming a transcription
bubble that moves downstream the DNA duplex.
• Each gene or operon has its own promoter site. There is a sigma factor
specific for each promoter site.
• Rates of transcription of different genes vary with rates of binding of RNAP to
their promoter sites.
An electron micrograph of E. coli RNAP bound to
various promoter sites on DNA. Each of those sites
marks the start of a gene.
12 of the 298 promoters found in the E. coli chromosome.
Each is on the sense (coding) strand of the gene it initiates.
+1
-1
+2
Bacterial RNAP with its sigma subunit recognizing a
promoter region on dsDNA. (Closed Complex)
• Note the -35 region (purple) and
-10 region (red) of the DNA sense
(coding) strand.
•
The template strand is in green.
•
The σ subunit is shown as Cα
trace in rainbow colors Note its
interaction with the -10 and -35
elements.
RNAP plus a σ factor =
RNAP holoenzyme.
ω
DN
β’
A
m
ov
em
en
α
α
β
t
The two modes of transcription initiation
Both modes rely on the sigma factor
1) Regular initiation: The σ subunit
holds on to the promoter elements,
while the rest of the RNAP holds on to
the template strand and pushes it into
active site. This causes “scrunching” of
the template strand. The strain
provides energy to strip the promoter
from RNAP which then continues its
progress along the template strand.
RNAP plus a σ factor =
RNAP holoenzyme.
ω
DN
β’
A
m
ov
em
en
α
α
2) Abortive initiation: RNAP fails to
escape the promoter. The tension in
the template strand is relieved by
letting go of the newly synthesized
RNA fragment. RNAP then re-initiates
transcription from the +1 position.
β
t
Sigma factors also control transcription of
genes at different times and in response to
environmental conditions.
Examples:
•Phage when it infects bacteria, it uses the bacteria’s σ factor to
express a few of its own genes, then one of those gene products
is the phage’s own σ factor which it uses to express another set
of its genes, etc.
•E. coli sigma factor σ70 controls transcription of most E. coli
genes at normal growth temperature, another sigma factor, σ32,
kicks in when temperature is high to allow transcription of heat
shock genes (i.e. expression of heat shock proteins, the
chaperones that protect the cell’s other proteins from heat).
•
An electron micrograph of three adjacent genes on the E. coli chromosome
being transcribed each many times by many RNA polymerases bound to the
gene consecutively.
•
Notice the emerging RNA molecules are longer near the 5’ end of each
gene (near the termination site).
• For highly expressed genes, e.g. rRNA, 1 RNAP binds to same gene per second.
• Speed/rate:
RNA transcription 20-50 nt/s (slow)
DNA replication 1000 nt/s (fast!)
• Error rate:
RNA transcription 1 in 104 . (more error)
DNA replication 1 in 104 – 106 . (less error)
WHY is the cell more ok with errors in RNA transcription than in DNA replication?
•
The tight association of RNAP with DNA, and its multiple associations with it
explains its high processivity even when it bumps into the DNA replisome.
• RNAP processivity: >> 1600 nt. Important for the super long eukaryotic mRNA.
RNAP has a problem: as it moves along the DNA, it
positively supercoils the DNA on the downstream side of the
transcription bubble, and negatively supercoils (underwinds)
it on the upstream side. This causes abortive initiation
(producing small incomplete RNAs).
DNA template strand
Topoisomrase I
relieves here
DNA gyrase
relieves here
Transcription Termination in Prokaryotes
• E. coli possesses two mechanisms for transcription
termination.
– Intrinsic termination: requires no protein components
other than RNA polymerase itself (Rho-independent
termination). The signals for this mode of termination
are contained entirely in the DNA, and the RNA
transcript derived from it.
– Rho-dependent termination: requires a termination
protein called Rho.
5’-end of RNA
transcript
1) Intrinsic termination: termination
site in the DNA template is a G-C rich
region followed by poly T.
How intrinsic termination works
RNA hairpin formation de-stabilizes
the RNA-DNA hybrid with the
template strand.
The A-U rich hybrid remaining is
not very stable since A-U bp are
weak. The whole RNA polymeraseDNA assembly falls apart as a
result, causing transcription to stop
and the nascent RNA chain to be
released.
Rho factor is a helicase that binds to the emerging RNA
strand, and unwinds the RNA-DNA helix to release the RNA,
thus transcription is terminated.
Rho factor bound to transcribed RNA
Inhibitors of bacterial RNA polymerase
-rifamycin B
-rifampicin
Bind specifically to a
pocket in the β subunit of
RNAP near to the RNADNA channel (clog the exit
channel).
Allow the formation of the
first phosphodiester bond,
but sterically inhibit the
formation of subsequent
bonds.
NUCLEIC ACID FUNCTION AND PROTEIN
SYNTHESIS
CHEM 563
Lecture 10
Transcription in eukaryotes
Manal Swairjo, Ph.D.
A eukaryotic cell uses three RNA
polymerases
They are all in the nucleus. They all have a core homologous to
ααββ’ω subunits of E. coli RNAP, plus additional subunits.
• RNA polymerase I:
– synthesizes the precursors of most ribosomal RNAs
• RNA polymerase II:
– synthesizes the precursors of mRNAs (code for proteins)
• RNA polymerase III:
– synthesizes the precursors of tRNAs, 5S rRNA, some other small
stable RNAs (e.g., snRNA)
RNA Polymerase II from yeast looks similar to bacterial
RNAP, i.e. crab claw. But no σ-factor!
Rpb1 (β’-like)
ω-like
Active site also similar.
Two Mg2+ ions bound.
Two-metal mechanism
α-like
(like bacterial RNAP and DNA Pol).
Bridge helix
α-like
The clamp of Rpb2
Rpb2 (β-like)
The bacterial RNAP
for comparison.
We rotate 90°around the horizontal axis to get
the view in the next slide.
Rpb1 (β’-like)
ω-like
α-like
90°
α-like
The clamp of Rpb2
Rpb2 (β-like)
When dsDNA binds, the clamp from Rpb2 closes on it.
The bridge from Rpb1 controls translocation down the DNA
template strand.
Coding DNA strand
Template DNA strand
Synthesized RNA
downstream
For each strand,
can you figure out
the 3’ and 5’ ends?
As one would expect, the DNA
binding cleft is lined with
positively charged amino acids,
just like in bacterial RNAP.
Functionally relevant features of Eukaryotic RNAP II
•
•
•
•
•
•
•
•
•
•
Mg2+ ions in active site. Two-metal catalysis.
The growing RNA chain hybridized to the DNA template
strand.
The clamp from Rbp2 (β-like subunit) swings over and
closes on the dsDNA.
The “Wall”: acts as a roadblock to direct the DNA template
strand out of the cleft by forcing it to make a 90o turn.
The “Rudder”: separates the RNA and DNA strands after ~
1 turn of the hybrid helix is formed.
The “RNA exit channel.”
The “Funnel”: the nucleoside triphosphate (NTP) entry
channel. It selects for NTP against dNTP.
The “Bridge Helix”: responsible for the translocation of
the RNA polymerase down the DNA template.
– Located near the active site.
– Bridges the two pincers of the clam claw.
– Alternates between bent and straight
conformations, causing movement of the RNA
polymerase relative to the DNA.
The “Trigger Loop“ (not shown): triggers catalysis
(nucleotide addition) when the correct NTP binds in the A
(addition) site. Responsible for selection of correct NTP.
α-amanitin binding site: more on that in slide 10.
3′
5′
Coding DNA strand
Template DNA strand
Synthesized RNA
Get oriented on the 3’
and 5’ ends.
3′
5′
3′
5′
How the “Bridge” helix helps eukaryotic RNAP II
translocate (or “walk”) down the DNA template
Note
difference in
position
The proposed transcription cycle and translocation mechanism of RNAP.
The trigger loop is responsible for selection of the
complementary NTP.
Template DNA strand
Growing
RNA strand
Incoming NTP flips
to addition (A) site
Bridge helix
Incoming NTP enters
in entry (E) site
Trigger Loop
When the correct NTP binds in the A site, the trigger loop swings in beneath
the correctly base-paired NTP and “triggers” catalysis (nucleotide addition).
Inhibitor of human RNAP is an anticancer drug
Amatoxins are deadly toxic cyclic
octapeptides produced by the
poisonous mushroom Amanita
phalloides.
• Amatoxins inhibit RNAP II and III (Kd = 10-8 M).
Reduces the elongation rate of RNAP II from several
thousand nucleotides per minute to only a few
nucleotides per minute.
• Mechanism of inhibition: α-amanitin binds to the
Bridge helix, impeding the conformational change in
the helix that’s needed for the translocation step.
• α-amanitin kills after several days of ingestion. Why
so slow?
• Conjugating α-amanitin to antibodies specific to
tumor cells created new anti-cancer drugs.
Eukaryotic Transcriptional Promoters
• Eukaryotic RNA polymerases do not directly
recognize promoter sequences in the DNA.
• Instead, proteins called basal or general
transcription factors recognize the promoter
sequences in the DNA, and the RNA polymerases
recognize these proteins already assembled on the
DNA.
• In all cases, promoters are on the sense (coding)
DNA strand.
Promoters for RNA Polymerase I are the simplest.
Remember RNAP I transcribes most rRNAs, i.e. these
are the promoters that control transcription of rRNA
(60% of all RNA in the cell).
+1
-187
-107
Upstream
Promoter
Element. GC
rich.
-31
+6
Core
Promoter
Element
Promoters for RNA Polymerase II are complex and diverse.
Remember RNAP II transcribes all mRNAs (structural genes).
Their variety allows a wide range of levels of expression of
proteins based on cell type.
+1
“Y” is pyrimidine [C or T]
“R” is purine [A or G]
N is any nucleotide
Typically, any particular mRNA gene would have only two or
three of these different promoter elements. They may occur in
different combinations.
Promoters for RNA Polymerase III are usually
located downstream of the transcription start site,
within the coding region of the gene.
Remember RNAP III synthesizes the precursors of tRNAs, 5S
rRNA, some other small stable RNAs (e.g., snRNA)
+1
+40
+80
The Eukaryotic General Transcription Factors (GTFs)
for RNAP II.
These proteins assemble on DNA to form the Pre-Initiation Complex
Nomenclature:
TF =Transcription Factor
II = transcription factor for RNAP II. (There are other ones specific for RNAP I & RNAP III)
A,B,D,E,F,and H designate the specific GTF.
What general transcription factors recognize TATA box,
the BRE, and the Inr, MTE, vs. DPE?
Recognized by TATA-box
binding protein of the
TFIID complex
Recognized by the TFIID complex
Recognized by TFIIB
Transcription
start site (+1)
TFIID is comprised of
• TATA box binding protein
• 14 other subunits
Assembly of the RNAP II preinitiation complex (PIC) on a TATA
box-containing promoter.
TATA-box binding protein binds to the minor
groove of TATA-box DNA and bends it by 90o
TATA-box binding protein
Note: most DNA binding
proteins approach their
target DNA from the
major groove side
because it is wider. The
TATA-box binding protein
is a rare case of minor
groove recognition.
e
jor
a
M
Minor groove
ov
o
r
g
The TATA box binding protein is the only
known universal transcription factor.
Although ribosomal RNA genes (transcribed by RNAP I)
and tRNA genes (transcribed by RNAP III) do not have
TATA boxes, and they use different sets of General
Transcription Factors,
The TATA-binding protein is required for transcription
by all 3 eukaryotic RNA polymerases!
RNAP II is also activated by enhancers
• Enhancers are DNA sequence elements that can be upstream
or downstream of transcription start site, on either DNA strand
(i.e. in either orientation), up to thousands of base pairs
away from a promoter. Unlike promoters, enhancers
locations and orientations vary.
• They are recognized by specific transcription factors that
activate RNAP II bound to a specific remote promoter.
• The DNA loops to bring the enhancer element near the remote
promoter.
• Enhancers are responsible for most selective gene
expression.
Simplified view of enhancer-mediated
activation of transcription in eukaryotes
A short RNA transcript is made
before elongation starts.
How elongation ensues
C-terminal domain (CTD) of the
largest RNAP II subunit.
The CTD gets phosphorylated at the
time transcription initiates.
Tyr-Ser-Pro-Thr-Ser-Pro-Ser(26)-COOH
Rpb1
α-like
β’-like
α-like
Rpb2
β-like
RNAP II enters the PIC in a non-phosphorylated state.
Elongation of transcription by RNAP II requires phosphorylation of the CTD of
its largest subunit (Rpb1, the β’ homolog). Rpb1 has an unusual amino acid
sequence at its C-terminus: The sequence Tyr-Ser-Pro-Thr- Ser-Pro-Ser – is
repeated 52 times in mammals (26 times in yeast). All of these serines (and
threonines and tyrosines!) can potentially be phosphorylated. This phosphorylation
occurs very soon after the PIC is formed.
= elongator (a protein factor that binds to the
phosphorylated CTD and stimulates elongation by RNAP II)
P P
P
P
P
P
P P
GTFs
RNAP
RNAPIIII
TFIIH is one of the
factors that
phosphorylates
the RNAP II CTD.
PP
P
P
P
P
P P
PP
P
P
P
P
P P
RNAP II
RNAP II
PP
P
P
P
P
P P
RNAP
RNAPII II
Phosphatase activity
RNAP II
Termination of transcription in eukaryotes
• The termination process in eukaryotes is imprecise and we
do not know if there are sequences signaling termination.
• It does not matter, because the resulting mRNA gets further
processed after transcription (e.g. polyadenylation which will
be discussed later).
What goes on inside your cells..

NUCLEIC ACID FUNCTION AND PROTEIN
SYNTHESIS
CHEM 563
Lecture 11
Posttranscriptional processing of RNA
Manal Swairjo, Ph.D.
Overview
• RNAPs produce primary RNA transcripts of mRNA, rRNA
and tRNA (pre-mRNA, pre-rRNA, pre-tRNA).
• These transcripts must be processed to maturity before use as
follows:
– Exo- and endonucleolytic removal of polynucleotide segments.
– Appending nucleotide sequences to their 3’- and 5’-ends.
– Modification of specific nucleotide residues.
• mRNA: no processing in prokaryotes. A lot of processing in
eukaryotes.
• rRNA and tRNA: get processed in both prokaryotes and
eukaryotes.
Post-transcriptional Processing of
Eukaryotic mRNAs
1. Addition of a cap to the 5’ end
• Protects mRNA from 5’ exonucleolytic degradation in the cytoplasm.
2. Addition of poly(A) tail to 3’ end
• Protects mRNA from 3’ exonucleolytic degradation in the cytoplasm
and aids in transcription termination, export of the mRNA from the
nucleus, and translation.
3. mRNA splicing
• Gathers the scattered coding regions of mRNA to generate a final
translatable transcript.
• Alternative splicing generates different proteins from same gene.
How the 5’-cap is added to
pre-mRNA: three steps
(know the structure of the cap)
1)
2)
3)
4)
5)
Removal of leading phosphate from 5’ end
of pre-mRNA (by RNA triphosphatase).
Adding a GMP to the new 5’-end to make
unusual 5’-5’ triphosphate bridge
(guanylyltransferase or capping enzyme).
Methylation of the added G at its N7 atom
(guanine-7-methyltransferase, SAM
cofactor).
Optional: Methylation of O2’ atom of
residues 1 & 2 (2’-O-methyltransferase).
Optional: methylation of N6 of residues 1
if it is an A.
When does 5’-capping happen?
1) Soon after transcription is initiated by RNA polymerase II (after about
30 nucleotides of the RNA chain have been synthesized).
2) The capping enzyme associates with RNA polymerase II’s
phosphorylated CTD (carboxyl terminal domain). This ensures that
only mRNAs are capped (recall only RNAP II has the CTD
phosphorylation feature).
3) Cap-binding complex then binds to the 5’-cap, protects mRNA from
exonucleolytic degradation.
Cap
binding 5’
complex
~30 nt of RNA
Capping enzyme P P
5’
P P
P P
GTFs
RNAP II
P
P
PP
P
P
P
P
P P
GTFs
RNAP II
Some viruses snatch the host cell’s mRNA caps!
This leads to novel viral proteins
From Russell, Cell 181, 2020
Polyadenylation of 3’ end
Eukaryotic mRNAs have heterogeneous 3’ sequences due
to imprecise RNAP II termination.
A poly(A) “tail” of about ~250 adenylate residues is
enzymatically added to the 3′ end.
1) How is the polyadenylation site recognized?
i.e., what are the cis-acting signals in the RNA that specify the
polyadenylation site?
2) What is the enzymatic machinery (the trans-acting factors)
that recognizes the polyadenylation site and carries out the
polyadenylation reaction?
The polyadenylation site is specified by two sequences.
Cleavage happens between them to rid of the extra 5’ sequence.
RNAP II
5’
5’
5’
5’
AAUAAA
G/U-rich
Cleavage by an unknown endonuclease
AAUAAA
AAUAAA
3’
5’
G/U-rich
A A A A A A A etc. 3’
The cleaved (extra) downstream 5’-sequences are degraded
RNAP II
A100-250
A100-250
gets
degraded
•
Poly(A) polymerase (or PAP) adds the poly(A) tail, from ATP, with the help of
specificity proteins. They all form a complex (polyadenylation machinery).
•
This is done in a template-independent manner.
•
The polyadenylation machinery associates with the phosphorylated CTD of
RNAP II and scans the mRNA as it is being synthesized, looking for the
polyadenylation signals in the mRNA.
•
When the trans-acting polyadenylation machinery senses the polyadenylation
signals, it cleaves the mRNA transcript and then add the poly(A) tail.
5’
polyadenylation
machinery
RNA
5’
GTFs
P
P P
P
P
P
P P
RNAP II
cleavage
PP
P
P
P
P
P P
RNAP II
PP
P
P
P
P
P P
RNAP II
Eukaryotic mRNA polyadenylation
summary
• AAUAAA and G/U-rich signals (cleavage occurs between them)
• poly(A) polymerase (PAP) is the enzyme that adds the poly(A)
tail (but several other factors are required for recognition of the
polyadenylation site and to cleave the RNA).
• this complex of factors is associated with the phosphorylated
CTD of RNAP II prior to recognizing the polyadenylation signals
in the RNA.
• ATP is the substrate used for polyadenylation.
• no template used.
• Polyadenylation of mRNA protects it from degradation and
increases efficiency of translation.
mRNA splicing (again, this occurs only in Eukaryotes)
• Introns are non-expressed intervening sequences of 2000-20,000 nt
encoded in the DNA and transcribed as part of the pre-mRNA. Introns
must be removed to produce a mature mRNA.
• Introns exist only in eukaryotes (none in prokaryotes).
• The shortest introns are about 65 nt in length.
• Introns are 4-10 times bigger than exons.
• The average length for introns in mammals is around 3500 nt.
• An intron in the gene for the muscle protein dystrophin has a length of
about 2,400,000 nt.
• The gene that has the greatest number of known introns is the gene for
the muscle protein titin. The titin gene contains 234 introns.
The production of a mature eukaryotic mRNA (shown here for
the chicken ovalbumin gene).
Polyadenylation Site
DNA
pre-mRNA
mRNA
Eukaryotic mRNA Splicing
Points of interest:
• cis-acting signals in the pre-mRNA
• trans-acting factors (the molecular machinery that
carries out the splicing process)
• mechanisms
– chemistry
– molecular recognition processes
• The evolution of the splicing process and its role in
evolution.
Four important cis-acting signals for
splicing pre-mRNAs:
branch-point sequence
(the “A” residue)
…YNCURAY…
• 5′ splice site junction sequence (conserved consensus GU)
• 3′ splice site junction sequence (conserved sequence AG)
• pyrimidine rich region about 5-15 bases upstream of the 3′
splice site.
• Branch-point sequence: an important A residue located
between the two splice site junctions.
Two transesterification reactions splice together the exons
of eukaryotic mRNAs.
• Eukaryotic mRNA splicing occurs
via two transesterification
reactions. Notice the unusual 2’,5’
linkage created in the first
reaction.
• The intron is removed as a lariat
structure (a loop with a branch
attached at the branch site).
• The transesterification reactions
preserve the free energy of each
cleaved phosphodiester bond
through the concomitant
formation of a new one.
What are the trans-acting factors?
A large macromolecular assembly called the Spliceosome is
recruited to the splice sites.
The spliceosome is made of:
• small nuclear ribonucleoprotein particles
(snRNPs) that contain
– Five snRNAs (U1, U2, U4, U5, and U6)
– associated proteins (~10 proteins per snRNA)
• other protein splicing factors (~65 total proteins
in the spliceosome).
The spliceosome is responsible for carrying
out the splicing reaction.
Recall small nuclear RNAs?
They are synthesized by RNAP II
Just like the situation with the
5’-cap, recruitment of the
splicing machinery to the premRNA is dependent upon
the phosphorylated CTD of
the large subunit of RNAP
II.
An electron micrograph of
spliceosomes in action.
Recognition of the cis-acting splicing signals by
specific components of the spliceosome.
U1 snRNA
3′
U C C
A U U C A U A 5′
Brach
Binding
Protein
U2AF
…YNCURAY…
5’ splice
site
U2 snRNP
U1 snRNA base pairs with the 5′ junction sequence.
U2 snRNA base pairs to the branch point sequence.
•Branch point binding protein also binds to the branch point sequence.
•U2 accessory factor (U2AF) binds to the polypyrimidine tract.
Structure of the U1 snRNP
(just to appreciate)
U1 snRNA base-pairing
with pre-mRNA 5’-splice
site sequence
5’
Sm protein
Why do genes in higher organisms have introns?
Why bother with the very elaborate pre-mRNA splicing process?
Is there some sort of evolutionary advantage?
1. Introns are a means to allow more rapid protein evolution
(rapid meaning on an evolutionary time scale of thousands
or millions of years).
2. Alternative splicing allows for the synthesis of many
different variant forms of a protein from just a single gene.
In evolution, new introns and exons are constantly being introduced
into genes via DNA recombination, leading to rapid protein evolution
intron
intron
intron
How alternative splicing sites are selected
• Alternative splicing sites are selected by
numerous RNA binding proteins and intronic
sequences that either enhance or silence
splicing, thus resulting in different mature
mRNAs, different proteins are translated.
• 15% of human genetic diseases are associated
with defects in mRNA splicing.
NUCLEIC ACID FUNCTION AND PROTEIN SYNTHESIS
CHEM 563
Lecture 12
Posttranscriptional processing of rRNA and tRNA
Manal Swairjo, Ph.D.
Electron micrograph of
amphibian ribosomal RNA genes being
actively transcribed by RNA polymerase I.
The posttranscriptional processing of prokaryotic rRNA is carried
out by a series of ribonuclease enzymes.
This is the E. coli example.
Notice the tRNA
transcripts!
Components of Ribosomes
Similarly, processing of eukaryotic rRNA primary transcript is
done via two cuts by specific ribonucleases.
processing by endoribonucleases
18S rRNA
5.8S rRNA
28S rRNA
RNA polymerase I synthesizes a 45S rRNA primary transcript, which then gets cut up into the three
rRNAs by specific ribonucleases.
To mature, eukaryotic rRNAs get further modified:
1. Methylation of nucleotides at certain positions.
2. Conversion of some uridines to pseudouridines.
Conversion of uridine to pseudouridine in rRNA, a
necessary step to produce mature rRNA in eukaryotes.
Pseudouridine
synthase
Note: this also
occurs in tRNA ☺
(U)
(ψ)
Methylation of rRNA in eukaryotes
The sites where these modifications occur in the rRNA are determined by small nucleolar RNAs
(snoRNAs) which guide (by base pairing with the rRNA) certain enzyme complexes (e.g.,
methyltransferases) to carry out the modifications.
In some eukaryotes, rRNAs are self-spliced!
• Remember that eukaryotic mRNAs require trans-acting snRNPs for their splicing.
• However, in fact, some rRNAs do not need the help of trans-acting RNPs. Instead
they are self-splicing RNAs. They fold up on themselves, identify splice junctions
within themselves, and then splice themselves without the help of any trans-acting
factors. Introns removed this way are called self-splicing introns.
•
The precursor RNA, by itself, catalyzes transesterification reactions that result in the
removal of an intron from within itself, and the ligation of the exons together.
•
There are two types of self-splicing introns: Group I and Group II
Example of a Group I self-splicing RNA
Self-splicing of the Tetrahymena thermophila
pre-rRNA.
This is an example of a Group I self-splicing
intron. It uses a free guanine nucleotide to
initiate the transesterification reactions, and one
of the products is a cyclized intron.
Step 1
Phosphor transfer
1st transesterification
Step 2
2nd transesterification
Step 3
To be able to do all that chemistry, the pre-rRNA Group I self-splicing
intron has extensive secondary and tertiary structure.
Secondary structure of the
Tetrahymena thermophila
Group I self-splicing intron.
Tertiary structure of the
Tetrahymena
thermophile Group I selfsplicing intron.
Group II Self-Splicing Introns
For a Group II self-splicing intron, the mechanism follows
exactly the same pathway as eukaryotic pre-mRNA splicing
(i.e., nucleophilic attack by the 2’ OH of a branch point “A” of
the 5’ splice junction, followed by excision of the intron as a
lariat.
However, no snRNAs or proteins are required to catalyze the
reaction in this case. The intron catalyzes its own removal
and the ligation of the exons.
Insight into evolution: The chemical similarities of the pre-mRNA and group II intron splicing
reactions suggest that spliceosomes are ribozymal systems whose RNA components (U1, U2, U4,
U5, and U6 snRNAs) have evolved from primordial self-splicing RNAs (that contained Group II
introns) and that the spliceosomal protein components serve mainly to fine-tune the ribozymal
structure and function. (This is consistent with the RNA World hypothesis for the origin of life.)
In prokaryotes and eukaryotes, tRNA is first synthesized as a raw transcript (a precursor or pretRNA), then it is processed in four steps to become mature tRNA.
3) CCA trinucleotide is
added to the 3’end.
1) 5′ end is trimmed here by
RNase P (another ribozyme!)
4) Some nucleotides are
modified (7 on average).
2) Intron removed by specific
endonucleases. Exons
ligated by specific ligases.
Summary of post-transcriptional tRNA processing
1. Removal of extra nucleotides at the 5′ end by RNase P. RNase P is another
example of a ribozyme. It consists of both an RNA component and a protein
component, but the catalytic activity resides in the RNA component.
2. Some eukaryotic tRNA primary transcripts have short introns (~10-15 nucleotides)
that have to be removed. Removal of tRNA introns occurs by a process
completely unrelated to mRNA splicing. No transesterification reactions occur.
Instead, tRNA splicing requires protein enzymes that are site-specific
endonucleases and RNA ligases.
3. Enzymatic addition of C-C-A residues to the 3′ end by CCA-adding polymerase.
(All tRNAs must end up with C-C-A at their 3′ end.)
4. Many and various nucleotide modifications.
Crystal structure of the T. maritima RNase P ribozyme in complex
with a pre-tRNA substrate.
Specificity domain
•
Notice the huge RNA component (dark and
light blue), and the small protein component
(green) of this ribozyme.
•
How does this ribozyme recognize only
tRNA to act on and not mRNA or rRNA?
– It recognizes the shape of tRNA (shape
complementarity).
– It recognizes the TψC loop (specific to
tRNAs) via base pairing interactions
between the ribozyme RNA and the
bases in the TψC loop.
Catalytic domain
NUCLEIC ACID FUNCTION AND PROTEIN SYNTHESIS
CHEM 563
Lecture 13
Reverse transcription and other examples of viral
replication
Manal Swairjo, Ph.D.
Retroviruses and Reverse Transcriptase
•
The retroviruses are a type of eukaryotic viruses. They have a single-stranded positive-sense
RNA genomes instead of a DNA genome.
•
Upon infecting a human cell, their genomic RNA is reverse-transcribed to DNA which is then
integrated into the host cell’s genome.
•
Examples:
–
Some tumor viruses that can cause cancer, like human T-cell leukemia-lymphoma virus (HTLV).
–
human immunodeficiency virus (HIV), the causative virus of AIDS.
•
Retroviruses contain an RNA-directed DNA polymerase, a reverse transcriptase. Reverse
transcriptase synthesizes DNA in the 5′—>3′ direction, guided by genomic viral RNA as a
template.
•
Reverse transcriptase is essential for the retroviral life cycle.
•
It is also an extremely important laboratory tool for making complementary DNA (cDNA) from mRNA.
Human Immunodeficiency Virus (HIV-AIDS) is a retrovirus
Two copies of plus-strand RNA
Reverse transcription of the HIV RNA genome in the host cell cytoplasm involves
multiple activities of the viral RT
A) Viral genomic RNA (positive-sense or plus-strand) has specific
sequence on both ends (R region). tRNA from human cell binds to
pbs sequence on 5’ end.
B) Reverse transcriptase synthesizes short negative-sense DNA
strand (or minus-strand) using the bound tRNA as a primer. The viral
RNA in that region gets degraded (dashed line) by RT’s RNase
activity.
5’
3’
3’
Human tRNA
3’
3’
C & D) The reverse transcriptase hops to the R sequence at the 3’
end & continues making minus-strand DNA. The rest of the viral
RNA strand is degraded, except for a short purine-rich sequence
(ppt) which is resistant to RNAse cleavage.
5’
3’
E) The remaining ppt segment acts as a primer for the synthesis of
plus-strand DNA. The tRNA primer is copied, so now the 3’ end of
plus-strand DNA contains the pbs sequence.
F) The tRNA primer is degraded. The pbs region in the newly
synthesized plus-strand DNA hops over to complement the 3’ end of
the minus strand. This is the second strand transfer event.
G) Extension of the plus and minus strands leads to the synthesis of
the complete double-stranded linear viral DNA.
3’
pbs
3’
3’
The first drugs to be clinically
approved to treat AIDS were
inhibitors of the HIV reverse
transcriptase.
Examples:
AZT (azidothymidine)
Beta-Thujaplicinol.
AZT binds in pol site
The Integration step in the HIV
life cycle is also a target of
successful anti-HIV drugs.
The HIV intasome
SARS-CoV2 Life Cycle
▪ Entry by attachment to ACE2
▪ Translation of viral genomic RNA gives one long
polyprotein.
▪ Viral polyprotein is cleaved into individual
functional proteins.
▪ Viral RNA (plus-strand) is replicated, i.e. plusstrand RNA used as template to make minusstrand RNA by the viral RNA-dependent RNApolymerase (RdRp).
▪ Same RdRp then makes a new plus-strand RNA
using the just synthesized minus-strand RNA as
template. This is now the final mRNA (plusstrand) that gets translated to viral proteins.
▪ The newly synthesized plus-strand viral RNA and
proteins are packaged into new virions.
SARS-CoV-2 is a positive-sense RNA virus
• The genome of SARS-CoV-2 is constituted of positive-sense RNA, i.e. it is the
coding mRNA and can be directly accessed by the host cell ribosomes to be
translated.
-ssRNA
RdRp
RdRp
Start here
+ssRNA
Host ribosomes
Viral proteins
RdRp : viral RNA-dependent RNA polymerase
+ssRNA = viral RNA positive-sense single strand
-ssRNA = viral RNA negative-sense single strand
Inhibition of SARS-CoV-2 RNA-dependent RNA
polymerase (RdRp) by Remdesivir
• Remdesivir (Gilead Biosc.) is the only FDAapproved drug currently used to treat Covid19
patients.
• Remdesivir is a adenosine nucleoside analog.
• It was originally intended for Hepatitis C virus and
tested against Ebola virus (two other +ve sense
RNA viruses) but it did not work.
• Later it was found to inhibit replication of
coronaviruses (SARS-CoV-1, MERS-CoV-1, SARSCoV-2).
d
ze
i
s
e
th
n
y
y s and
l
w
Ne A str
RN
3’
5’
Viral genomic
RNA template
3’
Structure of
remdesivir
Adenosine
Direction of movement of RNA template
View of remdesivir bound in the active site of SARS-CoV-2
RdRp after incorporation of the inhibitor in the growing chain
Remdesivir is incorporated by RdRp in the growing
RNA strand and this terminates further RNA
synthesis.
Remdesivir acts as a chain terminator.
Red: 3’ nucleotide of growing
RNA strand.
Magenta: remdesivir.
O3’
Cyano group
5’ U
Cyan: 5’ nucleotide of
template RNA strand.
Green spheres: Mg2+ ions.
sdsu.instructure.com
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Quiz: transcription of two E. coli genes at
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The following region of the E. coli genome has two consecutive genes. Their corresponding transcription promoters
are highlighted in color. The transcription start site for each gene is indicated in bold text.
Pages
(Will keep the highest of all your
Gene 1:
scores)
Files
5′ CCCGCTAGGCTTGACACTTGATCGGGCGCCTGTATAATACTACTGATGGCAGATCCTGGTGATCGATT 3′
Syllabus
3′ GGGCGATCCGAACTGTGAACTAGCCCGCGGACATATTATGATGACTACCGTCTAGGACCACTAGCTAA 5′
| Quizzes
Gene 2:
Modules
5′ CAGTAGCACAGATTGCAGAAGCATGATGATTATACTTGATATTGCGCCTGTGTCAACTCAGGCAGTTTGGATC 3′
Collaborations
3′ GTCATCGTGTCTAACGTCTTCGTACTACTAATATGAACTATAACGCGGACACAGTTGAGTCCGTCAAACCTAG 5′
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sdsu.instructure.com
Quiz: transcription of two E. coli genes: CHEM563-01: Nucleic Acid Function an…
gene transciption from left to right – Google Search
SOLUTION: Biochem Exam – Chemistry Homework Help – Studypool
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TE
Incorrect
Question 1
0/1 pts
Based on what you learned in class about the sequences of promoters E. coli uses for
transcription of its genes, are the two genes above oriented in the same direction?
yes
Ono
Incorrect
0/ 1 pts
Question 2
If RNA polymerase is moving along the DNA from left to right, which gene would it
be transcribing?
Gene 2
Gene 1
Both genes
Question 3
1 / 1 pts
If RNA polymerase is transcribing Gene 1, which strand would it be using as the
A sdsu.instructure.com
Quiz: transcription of two E. coli genes: CHEM563-01: Nucleic Acid Function and Protein Synthesis
SOLUTION: Biochem Exam – Chemistry Homework Help – Studypool
TE
Question 3
1 / 1 pts
If RNA polymerase is transcribing Gene 1, which strand would it be using as the
coding strand?
the top strand
the bottom strand
Incorrect
Question 4
0 / 1 pts
For Gene 1, where is the promoter located relative to the start site of transcription?
to the right of the start site
to the left of the start site
Question 5
1/1 pts
If RNA polymerase is moving along the DNA from right to left, which gene would it
be transcribing and which DNA strand would it be using as a template for
transcription?
A sdsu.instructure.com
Quiz: transcription of two E. coli genes: CHEM563-01: Nucleic Acid Function and Protein Synthesis
SOLUTION: Biochem Exam – Chemistry Homework Help – Studypool
TE
Question 5
1 / 1 pts
If RNA polymerase is moving along the DNA from right to left, which gene would it
be transcribing and which DNA strand would it be using as a template for
transcription?
a
Gene 2, the bottom strand
Gene 2, the top strand
Gene 1, the bottom strand
Gene 1, the top strand
Incorrect
Question 6
0 / 1 pts
For Gene 2, where is the promoter located relative to the start site of transcription?
to the left of the start site
to the right of the start site
Incorrect
Question 7
0 / 1 pts
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Quiz: transcription of two E. coli genes: CHEM563-01: Nucleic Acid Function and Protein Synthesis
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ATE
Incorrect
Question 7
0 / 1 pts
For Gene 2, what is the sequence of the -10 element of the promoter?
d
5′ TATAAT 3′
5′ ATTATA 3
5′ TAATAT 3′
Question 8
1/1 pts
For Gene 1, what is the sequence of the -35 element of the promoter?
5′ AACTGT 3
5′ TTGACA 3′
Incorrect
Question 9
0 / 1 pts
What are the first 10 nucleotides of the mRNA resulting from transcribing Gene 2?
A sdsu.instructure.com
Quiz: transcription of two E. coli genes: CHEM563-01: Nucleic Acid Function and Protein Synthesis
SOLUTION: Biochem Exam – Chemistry Homework Help – Studypool
ATE
Incorrect
Question 9
0/1 pts
What are the first 10 nucleotides of the mRNA resulting from transcribing Gene 2?
05’UGCAGAAGCA 3′
d
5’ACGUCUUCGU 3′
5′ AUGAUGAUUA 3′
5’UGCUUCUGCA 3′
Incorrect
Question 10
0 / 1 pts
What are the first 10 nucleotides of the mRNA resulting from transcribing Gene 1?
5′ UGAUGGCAGA 3′
5′ UAAUACUACU 3′
5′ AUUAUGAUGA 3′
5′ ACUACCGUCU 3′