w5 slides fc Flashcards
2 types of genes that when transcribed…
- the resulting rna encodes a protein
- the resulting rna functions as rna and may NOT be translated into protein
rna polymerase function
opens up helix as it transcribes and adds ribonucleoside triphosphate
transcription cycle
steps:
-sigma factor binds to RNA polym. and finds PROMOTER sequence (signals start of transcription)
-localized unwinding of dna, a few short RNAs synthesized initially and then RNA polym. clamps down (sigma factor released and NO primer involved)
elongation
termination and release of RNA
-after terminator sequence is transcribed, sigma factor rebinds and process restarts
on the 5’ strand, on promoter sequence, what’s the sequence at -35 called
upstream
tata box at -10
true or false: you can use different sigma factors to recognize different promoter sequences
true
in a terminator sequence, the GC rich regions…
fold into hairpin loop
-pull messenger rna away from dna template
-H bonds b/w dna and rna (2 base pairs_
—>causes the loop
what happens during termination
rna polymerase releases and binds another sigma factor
rna synthesis vs elongation mode of rna polymerase
rna synthesis = relatively inefficient
elongation mode = highly processive
what are some of the characteristics of rna termination signals?
-hairpin structure formed as a result of GC rich sequences
-AT rich DNA sequences following hairpin sequences
how do termination signals help dissociate rna transcript from polymerase
-disrupts H bonding of new mrna transcript with dna template
prokaryotic gene expression
no nucleus= why bacteria replicate so quickly
-their transcription and translation are NOT separated
mRNAs
messenger rnas, code for proteins
rRNA
ribosomal rna, form the basic structure of the ribosome and catalyze protein synthesis
tRNA
trnsfer rna, central to protein synthesis as the adaptors between mRNA and AA
function of rna polymerse ll
ALL protein-coding genes, miRNA geenes, plus genes for other noncoding RNAs (e.g. those of the spliceosome)
eukaryotic rnap ll vs. bacterial rnap structure
bacterial rnap has 5 subunits, eukaryotic rna pol ll has 12
rna pol ll has special carboxyl terminal domain (CTD) not found in bacterial or other eukaryotic rnaps
eukaryotic vs bacterial rna polymerases
eukaryotic rna polymerases require proteins to help position them at the promoter called: transcription factors
these factors fulfill a similar role to the SIGMA subunit of the bacterial rna polymerase
eukaryotic rna polymerases need to deal with chromosomal structures
tata box function
the sequence tata is highly conserved, found 30 bp UPSTREAM from start site for transcription
-helps position rnap ll
steps in the initiation of transcription
binding of tata binding protein subunit of TFIID to tata box sequence
-mobilizes the binding of TFIIB complex adjacent to tata box
other transcription factors bind
-rnap ll together with other TFs, will be able to bind in correct orientation at transcription start site
-helicase activity and phosphorylation of C-terminal domain (CTD) of RNAP ll
–both jobs performed by TFIIH
transcription steps simplified
-tbp and tfiid bind tg onto helix and helix bends
-acts as helicase nd unwinds the double strnded helix = helicase domin
then kinase domain allows phosphorylation
mediator
assembles nd coordinates activator protein
steps in the initiation of transcription
- TBP a subunit of TFIID binds to TATA box promoter in the minor groove, bending and distorting the DNA
- this attracts other transcription factors, which help to orient and bind RNAP ll to the DNA
- the helicase activity of TFIIH uses atp to pry apart DNA strands at transcription start site
- TFIIH also phosphorylates the C-terminal domain of RNA polymerase ll, activating it so that transcription can begin
carboxyl terminal domain
found on largest subunit of rna polymerase ll
-tandem repeats of 7 AAs
how is rna polym ll activated
phosphorylation
what is phosphorylation
adding phosphate groups on S located on CTD
mRNA processing steps
- addition of 5’ cap
- removal of introns (splicing) while transcribing
- processing and polyadenylation of 3’ tails
phosphorylation of C-terminal tail of rna polym. ll results in binding of:
rna processing proteins
additional phosphorylation of ctd, including ser 2
5’ pre-mrna capping
helps to protect rna from exonucleases (dont want exonucleases cutting off nucleotides)
completed BEFORE mRNA fully transcribed
removal of introns from pre-mRNA
2 step process:
- branch point A attacks the 5’ splice site (cleave= breaks backbone)
- 3’ of one exon reacts with 5’ of next exon to RELEASE intron
catalytic mechanism
rna dependent
-2’ OH group of the ribose sugar is not present in deoxyribose
-this group is necessary for the formation of the LARIAT structure in intron splicing
splicing rxn
-pre-mRNA’s not able to self-splice
-spliceosomes contain snRNPs bound to protein (snRNPs) plus other associated protein
–assemble on mRNA to remove introns
splicing complete, EXON junction added
3’ end processing
3’ end processing proteins move from CTD to mRNA
-cleavage and addition of a poly-A-3’ tail along with Poly A-binding proteins result in the mature mRNA
C terminal domain will be de-phosphorylated = trigger rna polymerase to FALL off
What is the most comprehensive molecular definition of a gene?
(A) A sequence of DNA that codes for a protein
(B) A sequence of DNA that is transcribed into RNA, which may or may not encode a protein
(C) A region of DNA bound by RNA polymerase
(D) A section of DNA containing only exons
b
How do regulatory sequences contribute to gene function?
(A) They are transcribed into mRNA but not translated into protein
(B) They directly encode the active protein
(C) They control when and where transcription occurs
(D) They serve as primers for DNA replication
c
What is the role of the sigma factor in bacterial transcription?
(A) It catalyzes phosphodiester bond formation
(B) It binds to RNA polymerase and directs it to the promoter
(C) It unwinds DNA at the replication fork
(D) It removes RNA primers
Answer: (B) (Sigma factors guide RNA polymerase to promoters and are released after initiation.)
How does bacterial RNA polymerase differ from eukaryotic RNA polymerases?
(A) Bacteria have three types of RNA polymerase, while eukaryotes have one
(B) Bacterial RNA polymerase requires general transcription factors for initiation
(C) Bacterial RNA polymerase contains a sigma factor instead of multiple general transcription factors
(D) Eukaryotic RNA polymerases function without protein cofactors
Answer: (C) (Bacteria use sigma factors, while eukaryotes require multiple transcription factors.)
What feature of bacterial promoters is critical for RNA polymerase recognition?
(A) The presence of TATA boxes
(B) The -10 and -35 consensus sequences
(C) A high GC content throughout the gene
(D) An intron-exon boundary
Answer: (B) (The -10 and -35 regions help RNA polymerase bind to bacterial promoters.)
How do bacterial transcription terminators function?
(A) They recruit RNA polymerase inhibitors
(B) They form RNA secondary structures or use rho-dependent mechanisms to halt transcription
(C) They add a poly-A tail to the transcript
(D) They trigger protein degradation
Answer: (B) (Termination occurs via hairpin loops (intrinsic) or rho proteins.)
What is a key difference between bacterial and eukaryotic transcription initiation?
(A) Eukaryotic transcription requires multiple transcription factors, while bacteria use a sigma factor
(B) Eukaryotic RNA polymerase binds directly to DNA without protein assistance
(C) Bacterial genes require a TATA box for transcription
(D) Only eukaryotic transcription occurs in the cytoplasm
Answer: (A) (Eukaryotic transcription involves transcription factors like TBP and TFIID.)
What is the function of the TATA box in eukaryotic promoters?
(A) It signals transcription termination
(B) It recruits transcription factors and RNA polymerase II
(C) It prevents splicing errors
(D) It enhances ribosome binding
Answer: (B) (The TATA box positions the transcription machinery at the start site.)
How does RNA polymerase II differ from RNA polymerase I and III?
(A) RNA polymerase II transcribes rRNA
(B) RNA polymerase II synthesizes mRNA
(C) RNA polymerase III transcribes mRNA
(D) RNA polymerase I transcribes tRNA
Answer: (B) (RNA Pol II is responsible for protein-coding mRNA transcripts.)
Why is RNA polymerase II’s C-terminal domain (CTD) important?
(A) It binds to DNA during initiation
(B) It recruits RNA processing enzymes
(C) It catalyzes the splicing reaction
(D) It forms the ribosome
Answer: (B) (CTD phosphorylation allows binding of capping, splicing, and polyadenylation factors.)
What is the function of the 5’ cap on eukaryotic mRNA?
(A) It promotes mRNA degradation
(B) It facilitates nuclear export and translation initiation
(C) It signals for splicing
(D) It terminates transcription
Answer: (B) (The 5’ cap protects mRNA and aids in translation.)
What sequence signals for polyadenylation in eukaryotic mRNA?
(A) The -35 sequence
(B) The TATA box
(C) The AAUAAA consensus sequence
(D) The Shine-Dalgarno sequence
Answer: (C) (The AAUAAA sequence recruits polyadenylation enzymes.)
How does alternative splicing contribute to proteome diversity?
(A) It removes DNA from genes
(B) It allows different exons to be included in the final mRNA
(C) It inserts introns into transcripts
(D) It changes RNA polymerase specificity
Answer: (B) (Alternative splicing creates multiple protein isoforms from a single gene.)
What is a consequence of a mutation in a splice donor site?
(A) The 5’ cap is removed
(B) An exon may be skipped or retained improperly
(C) The mRNA cannot exit the nucleus
(D) The poly-A tail is removed
Answer: (B) (Splice site mutations disrupt exon recognition and inclusion.)
How would a mutation in the TATA box likely affect transcription?
(A) Transcription initiation would be impaired
(B) RNA polymerase would elongate faster
(C) More introns would be retained
(D) The mRNA would lack a poly-A tail
Answer: (A) (A damaged TATA box disrupts RNA polymerase recruitment.)
How does histone acetylation affect transcription?
(A) It condenses chromatin, making genes less accessible
(B) It loosens chromatin, facilitating transcription factor binding
(C) It causes histone degradation
(D) It blocks enhancer function
b
Why is RNA splicing necessary in eukaryotic cells?
(A) It removes exons that contain mutations
(B) It prevents transcription errors
(C) It allows introns to be removed and exons to be joined into a functional mRNA
(D) It repairs DNA damage
c
How does transcription termination differ between bacteria and eukaryotes?
(A) Bacteria use poly-A tails for termination, while eukaryotes use rho-dependent termination
(B) Bacteria use either rho-dependent or rho-independent termination, while eukaryotic RNA Pol II termination is linked to polyadenylation
(C) Eukaryotic mRNAs form hairpin loops to stop transcription
(D) Only bacteria have a defined transcription stop site
Answer: (B) (Bacteria use rho-dependent or intrinsic mechanisms; eukaryotic RNA Pol II termination involves polyadenylation signals.)
How do microRNAs (miRNAs) regulate gene expression?
(A) They enhance ribosome binding
(B) They catalyze transcription
(C) They modify DNA bases
(D) They degrade target mRNAs or inhibit translation by binding complementary sequences
d
