Unit 5: RNA and Protein Synthesis and Processing - Practice Quiz

BTY426 — Cell And Molecular Biology 60 Questions
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1 Which enzyme is responsible for synthesizing RNA during transcription in prokaryotes?

Transcription in prokaryotes Easy
A. DNA ligase
B. DNA polymerase
C. RNA polymerase
D. Primase

2 Which subunit of prokaryotic RNA polymerase is required for recognizing the promoter sequence?

Transcription in prokaryotes Easy
A. Alpha () subunit
B. Omega () subunit
C. Beta () subunit
D. Sigma () factor

3 Which RNA polymerase transcribes protein-coding genes (mRNA) in eukaryotes?

Transcription in eukaryotes Easy
A. RNA polymerase I
B. RNA polymerase II
C. RNA polymerase III
D. RNA polymerase IV

4 The TATA box in eukaryotic genes is a part of which region?

Transcription in eukaryotes Easy
A. Enhancer far downstream
B. Coding sequence
C. Terminator
D. Promoter

5 Which antibiotic inhibits bacterial transcription by binding to RNA polymerase?

antibiotic inhibitors of transcription Easy
A. Rifampicin
B. Tetracycline
C. Penicillin
D. Streptomycin

6 -Amanitin, a toxin from mushrooms, primarily inhibits which enzyme?

antibiotic inhibitors of transcription Easy
A. DNA polymerase
B. RNA polymerase II
C. Reverse transcriptase
D. RNA polymerase I

7 The 5' cap added to eukaryotic mRNA is composed of which modified nucleotide?

Post-Transcriptional Modifications - capping Easy
A. 7-methylguanosine
B. Adenosine monophosphate
C. Uridine triphosphate
D. 5-methylcytosine

8 What is one major function of the 5' cap on eukaryotic mRNA?

Post-Transcriptional Modifications - capping Easy
A. Removes introns from RNA
B. Protects mRNA from degradation
C. Signals the start of transcription
D. Adds amino acids to proteins

9 During RNA splicing, which segments are removed from the pre-mRNA?

RNA splicing Easy
A. Promoters
B. Introns
C. Codons
D. Exons

10 Which cellular machine carries out the splicing of pre-mRNA in eukaryotes?

RNA splicing Easy
A. Proteasome
B. Spliceosome
C. Nucleosome
D. Ribosome

11 Polyadenylation adds a tail of which nucleotide to the 3' end of eukaryotic mRNA?

polyadenylation Easy
A. Cytosine
B. Guanine
C. Adenine
D. Thymine

12 The poly-A tail added to mRNA primarily helps in which of the following?

polyadenylation Easy
A. Removing exons
B. Amino acid activation
C. mRNA stability
D. DNA replication

13 Which amino acid is carried by the initiator tRNA in prokaryotic translation?

Protein synthesis in prokaryotes Easy
A. Alanine
B. Methionine
C. N-formylmethionine
D. Glycine

14 Which ribosomal subunit sizes are found in prokaryotes?

Protein synthesis in prokaryotes Easy
A. 30S and 50S
B. 20S and 40S
C. 50S and 70S
D. 40S and 60S

15 What is the size of the complete eukaryotic ribosome?

protein synthesis in eukaryotes Easy
A. 50S
B. 60S
C. 70S
D. 80S

16 Which site on the ribosome holds the growing polypeptide chain during translation?

protein synthesis in eukaryotes Easy
A. A site
B. E site
C. P site
D. T site

17 Which antibiotic inhibits translation by binding to the 30S ribosomal subunit and causing misreading of mRNA?

inhibitors of translation Easy
A. Actinomycin D
B. Penicillin
C. Streptomycin
D. Rifampicin

18 Chloramphenicol inhibits protein synthesis by blocking which enzymatic activity?

inhibitors of translation Easy
A. Peptidyl transferase
B. Topoisomerase
C. Ligase
D. Helicase

19 Which of the following is a common chemical post-translational modification of proteins?

Post translational modifications (PTMs) - chemical modifications Easy
A. Replication
B. Splicing
C. Phosphorylation
D. Transcription

20 Proteolytic cleavage of a protein involves which of the following?

proteolytic cleavage Easy
A. Breaking of peptide bonds
B. Adding a poly-A tail
C. Joining exons
D. Adding phosphate groups

21 In E. coli, a mutation in the (sigma) subunit of RNA polymerase would most directly affect which step of transcription?

Transcription in prokaryotes Medium
A. Recognition of promoter sequences and initiation
B. Elongation speed along the template
C. Proofreading of misincorporated nucleotides
D. Release of the completed transcript at the terminator

22 A bacterial gene ends in a GC-rich palindrome followed by a run of A residues on the template. What type of termination does this predict?

Transcription in prokaryotes Medium
A. Rho-independent (intrinsic) termination
B. Attenuation-based termination
C. Sigma-mediated termination
D. Rho-dependent termination

23 Which RNA polymerase would be inhibited if a drug specifically blocked transcription of the genes encoding mRNAs in a human cell?

transcription in eukaryotes Medium
A. RNA polymerase I
B. RNA polymerase III
C. RNA polymerase II
D. RNA polymerase IV

24 The general transcription factor TFIIH possesses helicase and kinase activities. Its kinase activity contributes to transcription by:

transcription in eukaryotes Medium
A. Unwinding the promoter to expose the template strand
B. Recruiting the TATA-binding protein to the core promoter
C. Phosphorylating the CTD of RNA Pol II to promote promoter clearance
D. Adding the 7-methylguanosine cap to the nascent transcript

25 Rifampicin is effective against bacterial infections because it:

antibiotic inhibitors of transcription Medium
A. Binds the subunit of bacterial RNA polymerase and blocks initiation
B. Inhibits eukaryotic RNA Pol II elongation
C. Intercalates into DNA to block all polymerases
D. Binds the ribosomal 30S subunit to stop translation

26 -Amanitin from the death cap mushroom is lethal in humans primarily because it:

antibiotic inhibitors of transcription Medium
A. Prevents ribosomal translocation
B. Strongly inhibits RNA polymerase II
C. Inhibits bacterial RNA polymerase only
D. Blocks mitochondrial DNA replication

27 The 5' cap of eukaryotic mRNA is joined to the first nucleotide through an unusual linkage. Which linkage is it?

Post-Transcriptional Modifications - capping Medium
A. A 5'-to-3' pyrophosphate bond
B. A 2'-to-5' phosphodiester bond
C. A 5'-to-5' triphosphate bridge
D. A standard 3'-to-5' phosphodiester bond

28 If the enzyme guanylyltransferase were non-functional in a eukaryotic cell, the most likely consequence for mRNA would be:

Post-Transcriptional Modifications - capping Medium
A. Inability to transcribe the gene at all
B. Reduced mRNA stability and impaired translation initiation
C. Failure to remove introns during splicing
D. Loss of the poly-A tail at the 3' end

29 During spliceosome-mediated splicing, the branch point adenosine plays what role in the first transesterification reaction?

RNA splicing Medium
A. Its 3'-OH attacks the downstream exon
B. Its 2'-OH attacks the 5' splice site to form a lariat
C. It provides the phosphate for the exon-exon junction
D. It base-pairs with the 3' splice site to align exons

30 A single pre-mRNA gives rise to several different protein isoforms in different tissues. This is best explained by:

RNA splicing Medium
A. Alternative splicing of exons
B. Variation in poly-A tail length
C. Multiple transcription start sites
D. Differential capping efficiency

31 Which snRNP is responsible for recognizing the 5' splice site of an intron at the start of spliceosome assembly?

RNA splicing Medium
A. U1 snRNP
B. U2 snRNP
C. U5 snRNP
D. U6 snRNP

32 The cleavage and polyadenylation of a eukaryotic pre-mRNA is triggered by recognition of which conserved signal sequence?

polyadenylation Medium
A. -rich 5' splice site
B. box upstream of the gene
C. start codon
D. in the 3' untranslated region

33 The poly-A tail added to eukaryotic mRNA is synthesized:

polyadenylation Medium
A. Without a template, by poly-A polymerase
B. Using the DNA template strand
C. By the spliceosome after intron removal
D. By reverse transcriptase from an RNA primer

34 In prokaryotes, the ribosome is positioned correctly on the mRNA start codon by base-pairing between the 16S rRNA and which mRNA element?

Protein synthesis in prokaryotes Medium
A. The Shine-Dalgarno sequence
B. The poly-A tail
C. The TATA box
D. The Kozak sequence

35 The first amino acid incorporated during prokaryotic translation is carried by a special initiator tRNA. This amino acid is:

Protein synthesis in prokaryotes Medium
A. Methionine (unmodified)
B. Formylglycine
C. N-formylmethionine (fMet)
D. N-acetylserine

36 In eukaryotic translation initiation, the small ribosomal subunit typically locates the start codon by:

protein synthesis in eukaryotes Medium
A. Recognizing the poly-A tail first
B. Binding directly to a Shine-Dalgarno sequence
C. Scanning from the 5' cap until it reaches an
D. Base-pairing with the branch point

37 During elongation, the movement of the ribosome by one codon along the mRNA (translocation) in eukaryotes is powered by:

protein synthesis in eukaryotes Medium
A. Release factor eRF1 binding
B. GTP hydrolysis via eEF2
C. Peptidyl transferase activity of the rRNA
D. ATP hydrolysis via eIF4A

38 Diphtheria toxin kills human cells by ADP-ribosylating and inactivating eEF2. The direct result is:

inhibitors of translation Medium
A. Premature release of the polypeptide chain
B. Blockage of ribosomal translocation and halted protein synthesis
C. Inability to form the initiation complex at the 5' cap
D. Misreading of the genetic code

39 Why can chloramphenicol be used as an antibacterial agent with relatively low toxicity to the cytoplasmic ribosomes of human cells?

inhibitors of translation Medium
A. It inhibits bacterial RNA polymerase instead of ribosomes
B. It blocks the eukaryotic 5' cap structure
C. It targets the 50S subunit peptidyl transferase of 70S ribosomes
D. It binds only the 40S subunit of 80S ribosomes

40 The addition of a phosphate group to serine, threonine, or tyrosine residues of a protein is best described as:

Post translational modifications (PTMs) - chemical modifications Medium
A. A modification that always targets the protein for degradation
B. An irreversible cleavage that activates the protein
C. A step required for translation to begin
D. A reversible modification that often regulates enzyme activity

41 In E. coli, a mutation in the rpoB gene alters the subunit of RNA polymerase such that the enzyme fails to respond to the intrinsic terminator hairpin but still terminates normally at Rho-dependent sites. Which mechanistic interpretation best explains this phenotype?

Transcription in prokaryotes Hard
A. The mutation prevents factor release during promoter escape
B. The mutation blocks recruitment of NusA to the elongation complex globally
C. The mutation abolishes the ATPase activity required to translocate the polymerase
D. The mutation destabilizes the RNA:DNA hybrid recognition needed for hairpin-induced pausing and dissociation

42 During RNA Pol II transcription, the C-terminal domain (CTD) undergoes a defined phosphorylation cycle. If a kinase inhibitor selectively blocks CDK7 (part of TFIIH) but not CDK9 (P-TEFb), what is the most direct consequence?

transcription in eukaryotes Hard
A. Constitutive termination at every intron-exon junction
B. Loss of Ser2 phosphorylation, preventing 3'-end processing factor recruitment
C. Impaired Ser5 phosphorylation, disrupting promoter clearance and 5' capping enzyme recruitment
D. Failure to load the initiator tRNA onto the pre-initiation complex

43 Rifampicin inhibits bacterial transcription but has essentially no effect once a transcript exceeds a few nucleotides. Which observation best accounts for this length-dependence?

antibiotic inhibitors of transcription Hard
A. Rifampicin binds the RNA exit channel and sterically blocks synthesis of the first phosphodiester bonds only
B. Rifampicin chelates the catalytic only in the closed promoter complex
C. Rifampicin covalently modifies the factor and is displaced during elongation
D. Rifampicin intercalates into DNA and is stripped off by the moving polymerase

44 The 5' cap is added co-transcriptionally after only ~25–30 nucleotides are synthesized. Which enzymatic sequence and requirement correctly describes cap formation?

Post-Transcriptional Modifications - capping Hard
A. Methyltransferase first methylates the terminal nucleotide, then guanylyltransferase adds three phosphates
B. Guanylyltransferase adds GTP directly to the 3' end before phosphatase trimming and 2'-O methylation
C. RNA triphosphatase removes -phosphate, guanylyltransferase adds GMP via 5'-5' linkage, then methyltransferase adds a methyl to N7 of guanine
D. Poly(A) polymerase transfers a capped guanosine to the 5' end using ATP

45 In the two transesterification reactions of pre-mRNA splicing, what is the nucleophile and product of the FIRST step?

RNA splicing Hard
A. The 5'-phosphate of the intron attacks the branch point, releasing free exon
B. The 2'-OH of the branch-point adenosine attacks the 5' splice site, forming a lariat intermediate
C. The 3'-OH of the upstream exon attacks the 3' splice site, joining the exons
D. A spliceosomal snRNA 3'-OH attacks the 5' splice site to form a covalent adduct

46 Cleavage and polyadenylation of a pre-mRNA depends on cis-elements flanking the cleavage site. Which combination of signals and their bound factors correctly positions the cut?

polyadenylation Hard
A. Upstream AAUAAA bound by CPSF and downstream GU/U-rich element bound by CstF, with cleavage between them
B. Upstream GU-rich element bound by CstF and downstream AAUAAA bound by CPSF, with cleavage upstream
C. Upstream TATA box bound by CstF and downstream AAUAAA bound by poly(A) polymerase
D. Upstream Kozak sequence bound by CPSF and downstream poly(A) tract bound by PABP

47 A bacterial mRNA has a Shine-Dalgarno sequence spaced abnormally far (15 nt) from the start codon. What is the most likely translational consequence?

Protein synthesis in prokaryotes Hard
A. Enhanced initiation because the 30S subunit binds more mRNA
B. Reduced initiation efficiency due to poor alignment of the P-site codon with the initiator tRNA
C. Complete failure of elongation because EF-Tu cannot bind
D. Premature termination because RF1 recognizes the start codon

48 During prokaryotic elongation, GTP hydrolysis occurs at two distinct steps. Which pairing of factor and GTP-dependent role is correct?

Protein synthesis in prokaryotes Hard
A. EF-Tu hydrolyzes GTP after codon-anticodon proofreading; EF-G hydrolyzes GTP to drive translocation
B. EF-Tu hydrolyzes GTP for peptide bond formation; EF-G hydrolyzes GTP for tRNA delivery
C. EF-G hydrolyzes GTP to deliver aminoacyl-tRNA; EF-Tu hydrolyzes GTP for translocation
D. IF2 hydrolyzes GTP during elongation; EF-Ts hydrolyzes GTP for translocation

49 In cap-dependent eukaryotic initiation, eIF4E binds the m7G cap while the 43S complex scans for the start codon. If eIF4E is sequestered by hypophosphorylated 4E-BP, what happens?

protein synthesis in eukaryotes Hard
A. The 60S subunit fails to join, but scanning continues normally
B. The poly(A) tail is removed, triggering mRNA decay
C. Elongation halts because eEF2 cannot bind the ribosome
D. Cap-dependent initiation is repressed while IRES-driven translation can continue

50 Diphtheria toxin catalyzes ADP-ribosylation of a modified histidine residue (diphthamide) in eukaryotic elongation factor 2 (eEF2). What is the direct functional outcome?

inhibitors of translation Hard
A. eEF2 can no longer promote ribosomal translocation, arresting elongation
B. The 40S subunit cannot recruit initiator tRNA, blocking initiation
C. Release factors are prevented from recognizing stop codons
D. Peptidyl transferase activity of the 60S subunit is abolished

51 Puromycin causes premature chain release in both prokaryotes and eukaryotes. Which structural feature explains its mechanism?

inhibitors of translation Hard
A. It inhibits translocation by locking EF-G onto the ribosome
B. It mimics the 3' aminoacyl-adenosine end of aminoacyl-tRNA and accepts the peptide, then dissociates
C. It binds the 30S A site and causes codon misreading
D. It blocks the peptidyl transferase center by binding the P site tRNA

52 A secreted protein contains the sequon N-X-S where X is proline. Despite the presence of asparagine, N-linked glycosylation does not occur at this site. Why?

Post translational modifications (PTMs) - chemical modifications Hard
A. Serine cannot serve as the third residue in an N-glycosylation sequon
B. N-linked glycosylation requires the sequon to be O-X-S instead
C. Proline at position X disrupts the -turn conformation required by oligosaccharyltransferase
D. Proline sterically blocks the transfer of GPI anchors, not glycans

53 -carboxylation of glutamate residues in clotting factors requires vitamin K as a cofactor. Warfarin inhibits this modification. What is the biochemical basis?

Post translational modifications (PTMs) - chemical modifications Hard
A. Warfarin degrades the mRNA encoding clotting factors
B. Warfarin blocks vitamin K epoxide reductase, depleting reduced vitamin K needed by the carboxylase
C. Warfarin chelates the calcium required for carboxylation
D. Warfarin directly inhibits the -glutamyl carboxylase active site

54 Proinsulin is converted to mature insulin by proteolytic processing. Which statement correctly describes the events?

proteolytic cleavage Hard
A. The signal peptide is removed to directly yield active insulin without further cleavage
B. Prohormone convertases excise the C-peptide, leaving A and B chains joined by disulfide bonds
C. Trypsin cleaves the A chain into two active fragments
D. The C-peptide remains attached and is required for receptor binding

55 During protein splicing, an internal intein is excised and the flanking exteins are ligated. Which feature is essential at the intein's C-terminal splice junction for excision to proceed?

protein splicing Hard
A. A conserved asparagine that cyclizes to form a succinimide, cleaving the intein from the C-extein
B. A glycine that undergoes ADP-ribosylation to trigger cleavage
C. A phosphorylated serine that is transferred to the N-extein
D. A disulfide bridge linking the two exteins before cleavage

56 -amanitin from death cap mushrooms shows differential sensitivity among the three eukaryotic RNA polymerases. Which ranking of sensitivity is correct?

transcription in eukaryotes Hard
A. Pol II highly sensitive > Pol III intermediate > Pol I resistant
B. Pol III highly sensitive > Pol I intermediate > Pol II resistant
C. Pol I highly sensitive > Pol II intermediate > Pol III resistant
D. All three polymerases equally and maximally sensitive

57 The factor recognizes -10 and -35 promoter elements. A promoter with an extended -10 motif (TGn upstream of the -10 box) can function even with a weak or absent -35 element. What is the mechanistic reason?

Transcription in prokaryotes Hard
A. The extended -10 replaces the requirement for the catalytic ion
B. The extended -10 recruits Rho to stabilize initiation
C. The extended -10 provides additional contacts with region 3 of , compensating for lost -35 recognition
D. The extended -10 acts as a Shine-Dalgarno equivalent for the polymerase

58 A point mutation converts the invariant GU at a 5' splice site to GC. Alternative splicing analysis shows partial exon skipping. Which explanation best fits?

RNA splicing Hard
A. Weakened U1 snRNA base-pairing reduces splice site recognition, favoring use of the next downstream site
B. The polypyrimidine tract is lengthened, enhancing U2AF binding
C. The branch point adenosine can no longer form the lariat
D. The 3' splice site AG is destroyed, blocking the second transesterification

59 Histone mRNAs in metazoans are unusual among protein-coding transcripts because they typically lack a poly(A) tail. How is their 3' end formed instead?

polyadenylation Hard
A. A stem-loop bound by SLBP and a downstream histone element guide endonucleolytic cleavage without polyadenylation
B. A ribozyme within the transcript self-cleaves to generate a 3'-OH
C. Poly(A) polymerase adds a short oligo-U tail recognized by exonucleases
D. The cap-binding complex loops to the 3' end and cleaves it internally

60 Phosphorylation of eIF2 by kinases such as PKR or PERK causes a global reduction in translation initiation. What is the molecular basis of this repression?

protein synthesis in eukaryotes Hard
A. Phosphorylated eIF2 cleaves initiator tRNA, blocking start codon selection
B. Phosphorylated eIF2 recruits release factors to initiation codons
C. Phosphorylated eIF2 destroys the m7G cap on target mRNAs
D. Phosphorylated eIF2 sequesters eIF2B, preventing GDP-to-GTP exchange needed to recycle the ternary complex