Revision Tutorial for Lectures 23-28 Flashcards
Transcription, Post-transcriptional processing, Translation, Translational regulation
A research team has determined the DNA sequence of a bacterial gene, which encodes for an industrially important protein. They observed that this gene was infrequently expressed in bacteria. Which of the following may increase transcription of the gene?
A. Remove the 5’ untranslated region of the gene.
B. Decrease the distance between the -35 and -10 elements to be less than 10 bp apart.
C. Introduce mutations to the promoter region.
D. Edit the -10 and -35 elements to resemble their consensus
sequences.
a) 5’ untranslated region is in translation, not transcription ( includes the ribosome binding site (RBS) or Shine-Dalgarno sequence in bacteria)
b) distance and the sequence of these promoter elements are very important, and optimal distance between the -35 and -10 elements in bacterial promoters is typically about 17 base pairs. decreasing the optimal distance will not increase transcription
c) mutations further away from consensus, less likely for RNA polymerase to recognise the sequence
d) YESS!!! correct!! Making these elements resemble their consensus sequences (typically TATAAT for the -10 region and TTGACA for the -35 region) can increase the affinity of RNA polymerase for the promoter, leading to increased transcription.
Scientists found that a mutant bacterial strain that synthesises tryptophan in both tryptophan-rich and tryptophan-absent media. Which of the following mutations will NOT cause this behaviour?
A. Mutation in the repressor protein that stops it from binding to the trp operator
B. Mutation in the repressor protein that stops it from binding to tryptophan
C. Mutation in the repressor protein that makes it bind to the trp operator even without tryptophan
D. Mutation in the trp repressor gene that blocks expression of the repressor protein
For the trp operon, repressor protein stops synthesis of tryptophan. Tryptophan activates the repressor protein to bind to the operon. With no tryptophan, repressor protein does not bind to operon so trypotphan is produced.
a) If the repressor cannot bind to the operator, the operon will be continuously transcribed so true
b) If the repressor cannot bind to tryptophan, it cannot be activated to bind to the operator and inhibit transcription so true
c) YESS!! CORRECT ANSWER !! If the repressor can bind to the operator without tryptophan, it will continuously block transcription of the operon, preventing tryptophan synthesis so false
d) If the repressor protein is not expressed at all, it cannot bind to the operator to inhibit transcription so true
A patient with severe fatigue was found to be expressing unusually low amounts of Gene X mRNA.
Which of the following is the most likely cause of this patient’s condition?
A. The promoter of Gene X was found to be downstream of the transcription start site.
B. The promoter of Gene X was unable to recruit RNA polymerase I to the transcription start site.
C. The levels of Gene X-associated transcription factors were unusually low.
D. The promoter of Gene X was only functioning as a cis-acting element.
a) Can be found downstream or upstrea, not a problem
b) RNA polymerase II is responsible for transcribing mRNA in eukaryotic cells. RNA polymerase I transcribes rRNA, and RNA polymerase III transcribes tRNA and other small RNAs.
c) YESS!! CORRECT ANSWER!! Low levels of transcription factors specific to Gene X could result in inefficient transcription, leading to reduced mRNA levels
d) All promoters are cis-acting elements. A cis-acting element is a region of DNA that regulates the transcription of a gene on the same molecule of DNA.
A new toxin found in a rare mushroom species is found to be an RNA polymerase II inhibitor.
If the toxin is injected into the nuclei of liver cells, what will most likely happen?
A. tRNA will no longer be made
B. mRNA will no longer be made
C. No effect on the cells, since RNA pol II is found in the cytoplasm
D. No effect on the cells, since RNA pol I and III can replace the inhibited RNA pol II
a) tRNA synthesis is carried out by RNA polymerase III, not RNA polymerase II
b) YES!! CORRECT RNA polymerase II is responsible for synthesizing mRNA
c) incorrect, RNA polymerase II is found in the nucleus
d) incorrect, the polyermases have distinct roles
Roles of RNA Polymerases in Eukaryotes
RNA Polymerase I: Synthesizes rRNA (ribosomal RNA), specifically the 28S, 18S, and 5.8S rRNA.
RNA Polymerase II: Synthesizes mRNA (messenger RNA) and some snRNA (small nuclear RNA).
RNA Polymerase III: Synthesizes tRNA (transfer RNA) and the 5S rRNA.
Below are the results of a 5’ capping assay in the presence of increasing concentrations of a drug, mizoribine monophosphate (MZP). [increase MZP, decrease capping ability]
What effect might MZP have on cellular events?
A. Splicing of introns will be inhibited
B. mRNA will be degraded more easily
C. Translation levels will increase
D. Transcription levels will decrease
a) not really related to splicing
b) YESS!! CORRECT!! less capping, more degradation. The 5’ cap protects mRNA from degradation by exonucleases
c) Translation levels will decrease with less capping ability. A lack of 5’ cap would hinder the binding of the ribosome to mRNA
d) this is post transcription regulation
5’ cap of eukaryotic mRNA protects degration and helps transltion to occur
Given the pre-mRNA below, how many different mRNA
isoforms can be produced from alternative splicing?
sequence:
exon intron alt ss intron intron alt ss intron alt ss exon
A. 2
B. 3
C. 6
D. 8
2n different mRNAs from one pre-mRNA with n alternative splice sites
there is 3 alt ss so 2^3 = 8 different combinations so D
L25, Slide 16
Which of the following correctly describes the role of the initiator tRNA during prokaryotic translation?
A. It prevents the 50S ribosomal subunit from prematurely binding to the 30S subunit.
B. It binds to the promoter region and recruits RNA polymerase
C. It is chaperoned by an initiation factor to bind to the AUG start codon of mRNA
D. It carries modified methionine (fMet) and binds to all AUG codons in mRNA
a) typically associated with initiation factors instead
b) RNA polymerase and promoter region is all related to translation
c) YESS!! CORRECT! The initiator tRNA is indeed chaperoned by initiation factors (e.g., IF-2) to bind to the start codon on the mRNA.
d) nearly, does carry fMet but not to ALL AUG CODONS, only the start codon (AUG) at the beginning
Roles of the Initiator tRNA in Prokaryotic Translation
The initiator tRNA == a crucial role in the initiation of translation.
Binding to the Start Codon (AUG): recognizes and binds to the start codon (AUG) on the mRNA for translation
The initiator tRNA is chaperoned by initiation factors (IFs), particularly IF-2, which helps it bind to the small ribosomal subunit (30S) and subsequently to the start codon.
Formation of the Initiation Complex:
Once the initiator tRNA is bound to the start codon, the large ribosomal subunit (50S) joins to form the complete initiation complex, allowing translation to proceed.
- A scientist wants to decrease the levels of an E. coli protein that is responsible for opportunistic virulence.
Which would NOT be a potential method to achieve this?
A. Mutating the start codon of the gene
B. Mutating the -10 and -35 promoter sequences of the gene
C. Inhibiting the enzyme that adds a 5’ cap to mRNA
D. Introducing a mutation in the Shine-Dalgarno sequence of the gene
a) true : The start codon (typically AUG) is crucial for initiating translation. Mutating the start codon would prevent the ribosome from starting translation, thereby reducing or eliminating protein synthesis.
b) true : The -10 (Pribnow box) and -35 promoter sequences are essential for the binding of RNA polymerase to initiate transcription
c) YESS!! CORRECT, In E. coli and other prokaryotes, mRNA does not receive a 5’ cap so it is NOT a method that will work
d) Shine Dalgarno sequence is a ribosome binding site in the mRNA of prokaryotes, which is essential for the initiation of translation/ impair the ribosome’s ability to bind to the mRNA
An antibiotic was recently discovered to inhibit prokaryotic protein synthesis. In the presence of the drug, protein synthesis can be initiated, but only short polypeptides of ~2 amino acids are formed.
What step of protein synthesis is likely to be blocked by this antibiotic?
A. Modification of the methionine attached to the initiator tRNA
B. Recruitment of amino-acyl tRNA to the ribosomal A site
C. Translocation of tRNA and mRNA relative to the ribosomes
D. Recruitment of release factors to the ribosomal complex
a) initiation is not blocked (protein synthesis starts)
b) If the recruitment of aminoacyl-tRNA to the A site were blocked, it would prevent the addition of new amino acids beyond the initiator methionine, leading to very short polypeptides
c) **yess!! correct !! **The presence of the antibiotic that allows initiation but results in the formation of very short polypeptides (~2 amino acids) suggests that the translocation step is blocked.
d) Release factors are involved in terminating translation. If this step were blocked, translation would continue, but the polypeptide would not be released. This would not result in short polypeptides, but rather in incomplete, elongated chains.
A patient has been diagnosed with low levels of cellular iron.
Which of the following could be an underlying cause of their disease?
A. Underexpression of the transferrin receptor
B. Increased susceptibility of the ferritin mRNA to nucleases
C. Inability of the IRE binding protein to bind to iron
D. Mutation in the ferritin gene that interferes with normal protein function
a) YESS CORRECT!! cannot transport iron from blood to cells
b) result in less iron storage so higher cellular iron
c) therefore binds to ferritin, so ferritin cannot store iron, higher cellular iron
d) will lead to less iron storage and higher cellular iron
Which of the following statements regarding nonsense-mediated mRNA decay is FALSE?
A. This process destroys transcripts with a premature stop codon
B. During splicing, the site of an excised intron is marked by an exon junction complex
C. Ribosomes remove exon junction complexes as they move along mRNA
D. When a premature stop codon is present, no exon junction complexes remain on the mRNA
mRNA splicing occurs before translation, not during translation
==> So basically, you have a sequence with exons and introns, and since you only want exons, you remove them and then make a separate sequence by joining them with exon junction complexes (EJCs)
then during translation, these EJCs are removed by the ribosome as it moves along the mRNA
if there is a premature stop codon, translation terminates early and the EJCs are left intact
these remaining EJCs signal to proteins involved in nonsense-mediated mRNA decay come and degrade the mRNA
a) TRUE
a) TRUE
a) TRUE
a) YESS!! CORRECT this is FALSE! there
