1Which enzyme is responsible for synthesizing RNA during transcription in prokaryotes?
Transcription in prokaryotes
Easy
A.DNA ligase
B.Primase
C.RNA polymerase
D.DNA polymerase
Correct Answer: RNA polymerase
Explanation:
In prokaryotes, a single RNA polymerase synthesizes all types of RNA by reading the DNA template strand.
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2Which subunit of prokaryotic RNA polymerase is required for recognizing the promoter sequence?
Transcription in prokaryotes
Easy
A.Omega () subunit
B.Beta () subunit
C.Sigma () factor
D.Alpha () subunit
Correct Answer: Sigma () factor
Explanation:
The sigma factor helps RNA polymerase recognize and bind to promoter regions, ensuring transcription begins at the correct site.
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3Which RNA polymerase transcribes protein-coding genes (mRNA) in eukaryotes?
Transcription in eukaryotes
Easy
A.RNA polymerase III
B.RNA polymerase I
C.RNA polymerase II
D.RNA polymerase IV
Correct Answer: RNA polymerase II
Explanation:
RNA polymerase II transcribes messenger RNA (mRNA) precursors in eukaryotic cells.
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4The TATA box in eukaryotic genes is a part of which region?
Transcription in eukaryotes
Easy
A.Enhancer far downstream
B.Terminator
C.Promoter
D.Coding sequence
Correct Answer: Promoter
Explanation:
The TATA box is a core promoter element located upstream of the transcription start site that helps position RNA polymerase II.
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5Which antibiotic inhibits bacterial transcription by binding to RNA polymerase?
antibiotic inhibitors of transcription
Easy
A.Rifampicin
B.Tetracycline
C.Streptomycin
D.Penicillin
Correct Answer: Rifampicin
Explanation:
Rifampicin binds to the beta subunit of bacterial RNA polymerase, blocking initiation of RNA synthesis.
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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
Correct Answer: RNA polymerase II
Explanation:
-Amanitin strongly inhibits eukaryotic RNA polymerase II, blocking mRNA synthesis.
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7The 5' cap added to eukaryotic mRNA is composed of which modified nucleotide?
Post-Transcriptional Modifications - capping
Easy
A.Adenosine monophosphate
B.5-methylcytosine
C.7-methylguanosine
D.Uridine triphosphate
Correct Answer: 7-methylguanosine
Explanation:
A 7-methylguanosine (m7G) cap is added to the 5' end of mRNA, protecting it and aiding in ribosome binding.
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8What is one major function of the 5' cap on eukaryotic mRNA?
Post-Transcriptional Modifications - capping
Easy
A.Removes introns from RNA
B.Signals the start of transcription
C.Adds amino acids to proteins
D.Protects mRNA from degradation
Correct Answer: Protects mRNA from degradation
Explanation:
The 5' cap protects mRNA from exonuclease degradation and assists in ribosome recognition during translation.
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9During RNA splicing, which segments are removed from the pre-mRNA?
RNA splicing
Easy
A.Introns
B.Promoters
C.Codons
D.Exons
Correct Answer: Introns
Explanation:
Splicing removes non-coding introns and joins the coding exons together to form mature mRNA.
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10Which cellular machine carries out the splicing of pre-mRNA in eukaryotes?
RNA splicing
Easy
A.Spliceosome
B.Nucleosome
C.Ribosome
D.Proteasome
Correct Answer: Spliceosome
Explanation:
The spliceosome, composed of snRNPs and proteins, recognizes splice sites and removes introns from pre-mRNA.
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11Polyadenylation adds a tail of which nucleotide to the 3' end of eukaryotic mRNA?
polyadenylation
Easy
A.Adenine
B.Guanine
C.Thymine
D.Cytosine
Correct Answer: Adenine
Explanation:
A poly-A tail consisting of many adenine nucleotides is added to the 3' end of mRNA to increase stability.
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12The poly-A tail added to mRNA primarily helps in which of the following?
polyadenylation
Easy
A.Removing exons
B.DNA replication
C.Amino acid activation
D.mRNA stability
Correct Answer: mRNA stability
Explanation:
The poly-A tail protects mRNA from degradation and enhances its stability and translation efficiency.
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13Which amino acid is carried by the initiator tRNA in prokaryotic translation?
Protein synthesis in prokaryotes
Easy
A.Alanine
B.N-formylmethionine
C.Glycine
D.Methionine
Correct Answer: N-formylmethionine
Explanation:
Prokaryotic translation begins with N-formylmethionine (fMet), which is carried by the initiator tRNA.
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14Which ribosomal subunit sizes are found in prokaryotes?
Protein synthesis in prokaryotes
Easy
A.50S and 70S
B.20S and 40S
C.30S and 50S
D.40S and 60S
Correct Answer: 30S and 50S
Explanation:
Prokaryotic ribosomes (70S) consist of a small 30S subunit and a large 50S subunit.
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15What is the size of the complete eukaryotic ribosome?
protein synthesis in eukaryotes
Easy
A.80S
B.60S
C.50S
D.70S
Correct Answer: 80S
Explanation:
Eukaryotic ribosomes are 80S, made up of a 40S small subunit and a 60S large subunit.
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16Which site on the ribosome holds the growing polypeptide chain during translation?
protein synthesis in eukaryotes
Easy
A.E site
B.T site
C.A site
D.P site
Correct Answer: P site
Explanation:
The P (peptidyl) site holds the tRNA carrying the growing polypeptide chain during elongation.
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17Which antibiotic inhibits translation by binding to the 30S ribosomal subunit and causing misreading of mRNA?
inhibitors of translation
Easy
A.Actinomycin D
B.Streptomycin
C.Rifampicin
D.Penicillin
Correct Answer: Streptomycin
Explanation:
Streptomycin binds to the 30S subunit, causing misreading of the genetic code and inhibiting protein synthesis.
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18Chloramphenicol inhibits protein synthesis by blocking which enzymatic activity?
inhibitors of translation
Easy
A.Ligase
B.Helicase
C.Topoisomerase
D.Peptidyl transferase
Correct Answer: Peptidyl transferase
Explanation:
Chloramphenicol binds to the 50S subunit and inhibits the peptidyl transferase activity that forms peptide bonds.
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19Which of the following is a common chemical post-translational modification of proteins?
Post translational modifications (PTMs) - chemical modifications
Easy
A.Splicing
B.Phosphorylation
C.Transcription
D.Replication
Correct Answer: Phosphorylation
Explanation:
Phosphorylation is a common PTM in which a phosphate group is added to amino acids like serine, threonine, or tyrosine.
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20Proteolytic cleavage of a protein involves which of the following?
proteolytic cleavage
Easy
A.Adding phosphate groups
B.Joining exons
C.Breaking of peptide bonds
D.Adding a poly-A tail
Correct Answer: Breaking of peptide bonds
Explanation:
Proteolytic cleavage is a PTM in which specific peptide bonds are cut, often converting inactive precursors into active proteins.
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21In E. coli, a mutation in the (sigma) subunit of RNA polymerase would most directly affect which step of transcription?
Transcription in prokaryotes
Medium
A.Proofreading of misincorporated nucleotides
B.Elongation speed along the template
C.Release of the completed transcript at the terminator
D.Recognition of promoter sequences and initiation
Correct Answer: Recognition of promoter sequences and initiation
Explanation:
The factor directs the RNA polymerase holoenzyme to promoter sequences (e.g., and boxes) and is essential for initiation. It dissociates once elongation begins.
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22A 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?
A GC-rich hairpin followed by a poly-U stretch in the transcript (template poly-A) forms a stem-loop that destabilizes the RNA-DNA hybrid, causing intrinsic termination without the Rho protein.
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23Which 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 IV
C.RNA polymerase II
D.RNA polymerase III
Correct Answer: RNA polymerase II
Explanation:
RNA polymerase II transcribes protein-coding genes into mRNA (plus most snRNAs). Pol I makes rRNA and Pol III makes tRNA and 5S rRNA.
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24The general transcription factor TFIIH possesses helicase and kinase activities. Its kinase activity contributes to transcription by:
transcription in eukaryotes
Medium
A.Phosphorylating the CTD of RNA Pol II to promote promoter clearance
B.Unwinding the promoter to expose the template strand
C.Adding the 7-methylguanosine cap to the nascent transcript
D.Recruiting the TATA-binding protein to the core promoter
Correct Answer: Phosphorylating the CTD of RNA Pol II to promote promoter clearance
Explanation:
TFIIH phosphorylates serine residues in the C-terminal domain (CTD) of RNA Pol II, triggering the transition from initiation to elongation (promoter clearance).
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25Rifampicin is effective against bacterial infections because it:
antibiotic inhibitors of transcription
Medium
A.Intercalates into DNA to block all polymerases
B.Binds the ribosomal 30S subunit to stop translation
C.Binds the subunit of bacterial RNA polymerase and blocks initiation
D.Inhibits eukaryotic RNA Pol II elongation
Correct Answer: Binds the subunit of bacterial RNA polymerase and blocks initiation
Explanation:
Rifampicin binds the subunit of prokaryotic RNA polymerase, blocking the extension of RNA beyond a few nucleotides. It does not affect eukaryotic polymerases, giving it selective toxicity.
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26-Amanitin from the death cap mushroom is lethal in humans primarily because it:
antibiotic inhibitors of transcription
Medium
A.Prevents ribosomal translocation
B.Inhibits bacterial RNA polymerase only
C.Blocks mitochondrial DNA replication
D.Strongly inhibits RNA polymerase II
Correct Answer: Strongly inhibits RNA polymerase II
Explanation:
-Amanitin binds tightly to RNA Pol II (and weakly to Pol III), halting mRNA synthesis. Loss of mRNA production causes severe liver failure and death.
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27The 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 standard 3'-to-5' phosphodiester bond
C.A 2'-to-5' phosphodiester bond
D.A 5'-to-5' triphosphate bridge
Correct Answer: A 5'-to-5' triphosphate bridge
Explanation:
The 7-methylguanosine cap is attached via a distinctive 5'-to-5' triphosphate linkage, which protects the mRNA from exonucleases and aids in ribosome binding.
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28If the enzyme guanylyltransferase were non-functional in a eukaryotic cell, the most likely consequence for mRNA would be:
Post-Transcriptional Modifications - capping
Medium
A.Reduced mRNA stability and impaired translation initiation
B.Loss of the poly-A tail at the 3' end
C.Failure to remove introns during splicing
D.Inability to transcribe the gene at all
Correct Answer: Reduced mRNA stability and impaired translation initiation
Explanation:
Guanylyltransferase adds the GMP that forms the cap. Without a cap, mRNA is rapidly degraded by exonucleases and cannot efficiently recruit the translation initiation machinery.
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29During spliceosome-mediated splicing, the branch point adenosine plays what role in the first transesterification reaction?
RNA splicing
Medium
A.It provides the phosphate for the exon-exon junction
B.Its 3'-OH attacks the downstream exon
C.It base-pairs with the 3' splice site to align exons
D.Its 2'-OH attacks the 5' splice site to form a lariat
Correct Answer: Its 2'-OH attacks the 5' splice site to form a lariat
Explanation:
In the first step, the 2'-OH of the branch-point adenosine attacks the 5' splice site, creating the lariat intermediate. The freed 3'-OH of the upstream exon then attacks the 3' splice site in step two.
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30A 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.Differential capping efficiency
C.Variation in poly-A tail length
D.Multiple transcription start sites
Correct Answer: Alternative splicing of exons
Explanation:
Alternative splicing allows different combinations of exons to be included or excluded from the mature mRNA, producing distinct protein isoforms from one gene in a tissue-specific manner.
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31Which snRNP is responsible for recognizing the 5' splice site of an intron at the start of spliceosome assembly?
RNA splicing
Medium
A.U2 snRNP
B.U1 snRNP
C.U6 snRNP
D.U5 snRNP
Correct Answer: U1 snRNP
Explanation:
U1 snRNP base-pairs with the 5' splice site to initiate spliceosome assembly, while U2 recognizes the branch point. U5 and U6 join later during catalytic activation.
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32The cleavage and polyadenylation of a eukaryotic pre-mRNA is triggered by recognition of which conserved signal sequence?
polyadenylation
Medium
A. in the 3' untranslated region
B. start codon
C.-rich 5' splice site
D. box upstream of the gene
Correct Answer: in the 3' untranslated region
Explanation:
The polyadenylation signal, recognized by CPSF, directs cleavage of the pre-mRNA about – nucleotides downstream, after which poly-A polymerase adds the adenine tail.
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33The poly-A tail added to eukaryotic mRNA is synthesized:
polyadenylation
Medium
A.By the spliceosome after intron removal
B.By reverse transcriptase from an RNA primer
C.Using the DNA template strand
D.Without a template, by poly-A polymerase
Correct Answer: Without a template, by poly-A polymerase
Explanation:
Poly-A polymerase adds 200 adenine residues to the cleaved 3' end in a template-independent manner. The tail enhances mRNA stability and export.
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34In 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 Kozak sequence
C.The TATA box
D.The poly-A tail
Correct Answer: The Shine-Dalgarno sequence
Explanation:
The Shine-Dalgarno sequence upstream of the start codon base-pairs with the 3' end of the 16S rRNA of the 30S subunit, positioning the ribosome for initiation at the correct .
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35The first amino acid incorporated during prokaryotic translation is carried by a special initiator tRNA. This amino acid is:
Protein synthesis in prokaryotes
Medium
A.N-acetylserine
B.N-formylmethionine (fMet)
C.Methionine (unmodified)
D.Formylglycine
Correct Answer: N-formylmethionine (fMet)
Explanation:
Prokaryotic translation initiates with N-formylmethionine carried by . The formyl group is often later removed, distinguishing bacterial initiation from eukaryotic (which uses unmodified Met).
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36In eukaryotic translation initiation, the small ribosomal subunit typically locates the start codon by:
protein synthesis in eukaryotes
Medium
A.Scanning from the 5' cap until it reaches an
B.Base-pairing with the branch point
C.Binding directly to a Shine-Dalgarno sequence
D.Recognizing the poly-A tail first
Correct Answer: Scanning from the 5' cap until it reaches an
Explanation:
The 40S subunit with initiation factors binds the 5' cap and scans along the mRNA until it finds the first favorable (usually in a Kozak context), where the 60S subunit then joins.
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37During 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.ATP hydrolysis via eIF4A
C.Peptidyl transferase activity of the rRNA
D.GTP hydrolysis via eEF2
Correct Answer: GTP hydrolysis via eEF2
Explanation:
eEF2 (the eukaryotic counterpart of prokaryotic EF-G) uses GTP hydrolysis to drive translocation, shifting the ribosome by one codon and moving tRNAs from A to P to E sites.
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38Diphtheria toxin kills human cells by ADP-ribosylating and inactivating eEF2. The direct result is:
inhibitors of translation
Medium
A.Misreading of the genetic code
B.Blockage of ribosomal translocation and halted protein synthesis
C.Premature release of the polypeptide chain
D.Inability to form the initiation complex at the 5' cap
Correct Answer: Blockage of ribosomal translocation and halted protein synthesis
Explanation:
By inactivating eEF2, diphtheria toxin prevents the translocation step of elongation, stopping protein synthesis and killing the affected cell.
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39Why 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 blocks the eukaryotic 5' cap structure
B.It binds only the 40S subunit of 80S ribosomes
C.It inhibits bacterial RNA polymerase instead of ribosomes
D.It targets the 50S subunit peptidyl transferase of 70S ribosomes
Correct Answer: It targets the 50S subunit peptidyl transferase of 70S ribosomes
Explanation:
Chloramphenicol inhibits the peptidyl transferase activity of the bacterial 50S subunit. Because human cytoplasmic ribosomes are 80S, they are largely unaffected (though mitochondrial ribosomes can be a toxicity concern).
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40The 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 step required for translation to begin
B.An irreversible cleavage that activates the protein
C.A modification that always targets the protein for degradation
D.A reversible modification that often regulates enzyme activity
Correct Answer: A reversible modification that often regulates enzyme activity
Explanation:
Phosphorylation by kinases (reversed by phosphatases) is a reversible PTM that acts as a molecular switch, commonly altering enzyme activity, localization, or protein interactions.
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41In 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 blocks recruitment of NusA to the elongation complex globally
B.The mutation prevents factor release during promoter escape
C.The mutation destabilizes the RNA:DNA hybrid recognition needed for hairpin-induced pausing and dissociation
D.The mutation abolishes the ATPase activity required to translocate the polymerase
Correct Answer: The mutation destabilizes the RNA:DNA hybrid recognition needed for hairpin-induced pausing and dissociation
Explanation:
Intrinsic (Rho-independent) termination requires the polymerase to pause at the hairpin and destabilize the weak hybrid. A subunit change affecting hairpin-induced conformational sensing selectively abolishes intrinsic termination while leaving Rho-dependent (helicase-driven) termination intact.
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42During 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
CDK7 within TFIIH phosphorylates Ser5 of the CTD, which drives promoter escape and recruits the capping enzyme. CDK9/P-TEFb phosphorylates Ser2 for elongation and 3'-processing. Blocking CDK7 specifically impairs Ser5-dependent early events.
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43Rifampicin 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 intercalates into DNA and is stripped off by the moving polymerase
B.Rifampicin chelates the catalytic only in the closed promoter complex
C.Rifampicin binds the RNA exit channel and sterically blocks synthesis of the first phosphodiester bonds only
D.Rifampicin covalently modifies the factor and is displaced during elongation
Correct Answer: Rifampicin binds the RNA exit channel and sterically blocks synthesis of the first phosphodiester bonds only
Explanation:
Rifampicin binds within the subunit near the RNA exit channel, physically obstructing extension of nascent RNA beyond 2–3 nucleotides. Once a longer transcript has formed, the RNA occupies the channel and rifampicin can no longer bind, so elongating complexes are resistant.
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44The 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.Poly(A) polymerase transfers a capped guanosine to the 5' end using ATP
C.RNA triphosphatase removes -phosphate, guanylyltransferase adds GMP via 5'-5' linkage, then methyltransferase adds a methyl to N7 of guanine
D.Guanylyltransferase adds GTP directly to the 3' end before phosphatase trimming and 2'-O methylation
Correct Answer: RNA triphosphatase removes -phosphate, guanylyltransferase adds GMP via 5'-5' linkage, then methyltransferase adds a methyl to N7 of guanine
Explanation:
Capping proceeds in three steps: (1) RNA triphosphatase removes the terminal -phosphate leaving a diphosphate; (2) guanylyltransferase transfers GMP from GTP forming the unusual 5'-5' triphosphate linkage; (3) methyltransferase adds a methyl group to N7 of the added guanine, yielding the m7G cap.
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45In 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 3'-OH of the upstream exon attacks the 3' splice site, joining the exons
C.A spliceosomal snRNA 3'-OH attacks the 5' splice site to form a covalent adduct
D.The 2'-OH of the branch-point adenosine attacks the 5' splice site, forming a lariat intermediate
Correct Answer: The 2'-OH of the branch-point adenosine attacks the 5' splice site, forming a lariat intermediate
Explanation:
In step one, the 2'-OH of the conserved branch-point adenosine performs a nucleophilic attack on the 5' splice site phosphate, creating the 2'-5' phosphodiester bond of the lariat. Step two then uses the freed 3'-OH of the upstream exon to attack the 3' splice site, ligating the exons.
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46Cleavage 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 Kozak sequence bound by CPSF and downstream poly(A) tract bound by PABP
B.Upstream AAUAAA bound by CPSF and downstream GU/U-rich element bound by CstF, with cleavage between them
C.Upstream GU-rich element bound by CstF and downstream AAUAAA bound by CPSF, with cleavage upstream
D.Upstream TATA box bound by CstF and downstream AAUAAA bound by poly(A) polymerase
Correct Answer: Upstream AAUAAA bound by CPSF and downstream GU/U-rich element bound by CstF, with cleavage between them
Explanation:
CPSF recognizes the highly conserved AAUAAA polyadenylation signal upstream, while CstF binds the downstream GU/U-rich element. These factors define the cleavage site located between them; after cleavage, poly(A) polymerase adds the untemplated adenine tail.
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47A 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
Correct Answer: Reduced initiation efficiency due to poor alignment of the P-site codon with the initiator tRNA
Explanation:
The Shine-Dalgarno sequence base-pairs with the anti-SD region of 16S rRNA to position the start codon in the P site. Optimal spacing is ~7 nt; abnormal spacing misaligns the AUG relative to the ribosome, lowering initiation efficiency rather than affecting later steps.
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48During 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 for peptide bond formation; EF-G hydrolyzes GTP for tRNA delivery
B.EF-G hydrolyzes GTP to deliver aminoacyl-tRNA; EF-Tu hydrolyzes GTP for translocation
C.IF2 hydrolyzes GTP during elongation; EF-Ts hydrolyzes GTP for translocation
D.EF-Tu hydrolyzes GTP after codon-anticodon proofreading; EF-G hydrolyzes GTP to drive translocation
Correct Answer: EF-Tu hydrolyzes GTP after codon-anticodon proofreading; EF-G hydrolyzes GTP to drive translocation
Explanation:
EF-Tu delivers aminoacyl-tRNA to the A site and hydrolyzes GTP once correct codon-anticodon pairing is verified (kinetic proofreading). EF-G then uses GTP hydrolysis to power translocation of the ribosome by one codon. Peptide bond formation itself is catalyzed by the ribozyme and needs no GTP.
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49In 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.Elongation halts because eEF2 cannot bind the ribosome
B.The 60S subunit fails to join, but scanning continues normally
C.Cap-dependent initiation is repressed while IRES-driven translation can continue
D.The poly(A) tail is removed, triggering mRNA decay
Correct Answer: Cap-dependent initiation is repressed while IRES-driven translation can continue
Explanation:
Hypophosphorylated 4E-BP binds and sequesters eIF4E, preventing eIF4F assembly on the cap and blocking cap-dependent initiation. Internal ribosome entry site (IRES)-containing mRNAs bypass the cap requirement and can still be translated, which is why some viral and stress-response mRNAs stay active.
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50Diphtheria 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.Release factors are prevented from recognizing stop codons
C.eEF2 can no longer promote ribosomal translocation, arresting elongation
D.Peptidyl transferase activity of the 60S subunit is abolished
Correct Answer: eEF2 can no longer promote ribosomal translocation, arresting elongation
Explanation:
Diphtheria toxin ADP-ribosylates the diphthamide residue of eEF2, inactivating it. Since eEF2 is required for GTP-dependent translocation during elongation, the ribosome cannot advance, halting protein synthesis at the elongation stage.
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51Puromycin causes premature chain release in both prokaryotes and eukaryotes. Which structural feature explains its mechanism?
inhibitors of translation
Hard
A.It binds the 30S A site and causes codon misreading
B.It blocks the peptidyl transferase center by binding the P site tRNA
C.It inhibits translocation by locking EF-G onto the ribosome
D.It mimics the 3' aminoacyl-adenosine end of aminoacyl-tRNA and accepts the peptide, then dissociates
Correct Answer: It mimics the 3' aminoacyl-adenosine end of aminoacyl-tRNA and accepts the peptide, then dissociates
Explanation:
Puromycin structurally resembles the aminoacyl-tRNA 3' terminus (aminoacyl-adenosine). It enters the A site, accepts the growing peptide chain via peptidyl transferase, but its non-hydrolyzable amide bond and lack of tRNA body cause the peptidyl-puromycin to fall off, prematurely terminating synthesis.
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52A 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.Proline at position X disrupts the -turn conformation required by oligosaccharyltransferase
C.Proline sterically blocks the transfer of GPI anchors, not glycans
D.N-linked glycosylation requires the sequon to be O-X-S instead
Correct Answer: Proline at position X disrupts the -turn conformation required by oligosaccharyltransferase
Explanation:
The N-glycosylation consensus is N-X-S/T where X is any amino acid except proline. Proline at the X position forces a conformation incompatible with the -turn that oligosaccharyltransferase requires to present the asparagine, so glycosylation is blocked even though the sequon superficially appears valid.
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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 directly inhibits the -glutamyl carboxylase active site
B.Warfarin degrades the mRNA encoding clotting factors
C.Warfarin blocks vitamin K epoxide reductase, depleting reduced vitamin K needed by the carboxylase
D.Warfarin chelates the calcium required for carboxylation
Correct Answer: Warfarin blocks vitamin K epoxide reductase, depleting reduced vitamin K needed by the carboxylase
Explanation:
-glutamyl carboxylase uses reduced vitamin K (hydroquinone), converting it to the epoxide during each reaction. Vitamin K epoxide reductase regenerates the reduced form. Warfarin inhibits this reductase, so reduced vitamin K is depleted and carboxylation of clotting factors is impaired.
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54Proinsulin is converted to mature insulin by proteolytic processing. Which statement correctly describes the events?
proteolytic cleavage
Hard
A.Trypsin cleaves the A chain into two active fragments
B.Prohormone convertases excise the C-peptide, leaving A and B chains joined by disulfide bonds
C.The signal peptide is removed to directly yield active insulin without further cleavage
D.The C-peptide remains attached and is required for receptor binding
Correct Answer: Prohormone convertases excise the C-peptide, leaving A and B chains joined by disulfide bonds
Explanation:
After signal peptide removal (forming proinsulin) and disulfide bond formation, prohormone convertases (PC1/3 and PC2) plus carboxypeptidase E excise the connecting C-peptide. The resulting A and B chains remain held together by the previously formed disulfide bonds, yielding mature insulin.
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55During 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 disulfide bridge linking the two exteins before cleavage
B.A conserved asparagine that cyclizes to form a succinimide, cleaving the intein from the C-extein
C.A phosphorylated serine that is transferred to the N-extein
D.A glycine that undergoes ADP-ribosylation to trigger cleavage
Correct Answer: A conserved asparagine that cyclizes to form a succinimide, cleaving the intein from the C-extein
Explanation:
Protein splicing involves an N-S/O acyl shift, transesterification, and then cyclization of the conserved C-terminal asparagine of the intein to form a succinimide, which cleaves the peptide bond releasing the intein. A final S/O-N acyl shift forms the native peptide bond between the exteins.
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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 III highly sensitive > Pol I intermediate > Pol II resistant
B.Pol II highly sensitive > Pol III intermediate > Pol I resistant
C.Pol I highly sensitive > Pol II intermediate > Pol III resistant
D.All three polymerases equally and maximally sensitive
Correct Answer: Pol II highly sensitive > Pol III intermediate > Pol I resistant
Explanation:
-amanitin binds the bridge helix of RNA Pol II and strongly inhibits it at very low concentrations. Pol III is inhibited only at higher concentrations (intermediate sensitivity), while Pol I is essentially resistant. This differential sensitivity is a classic tool for distinguishing polymerase activities.
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57The 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 provides additional contacts with region 3 of , compensating for lost -35 recognition
B.The extended -10 replaces the requirement for the catalytic ion
C.The extended -10 acts as a Shine-Dalgarno equivalent for the polymerase
D.The extended -10 recruits Rho to stabilize initiation
Correct Answer: The extended -10 provides additional contacts with region 3 of , compensating for lost -35 recognition
Explanation:
The extended -10 motif (a TGn sequence just upstream of the -10 hexamer) makes additional contacts with region 3 of . These extra interactions stabilize the polymerase-promoter complex, allowing efficient initiation even when the -35 element is weak or missing.
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58A 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.The branch point adenosine can no longer form the lariat
B.The 3' splice site AG is destroyed, blocking the second transesterification
C.Weakened U1 snRNA base-pairing reduces splice site recognition, favoring use of the next downstream site
D.The polypyrimidine tract is lengthened, enhancing U2AF binding
Correct Answer: Weakened U1 snRNA base-pairing reduces splice site recognition, favoring use of the next downstream site
Explanation:
The 5' splice site GU is recognized by U1 snRNA through base-pairing. A GU→GC change weakens but does not abolish this pairing, reducing recognition efficiency of that splice site. The spliceosome may then use a downstream 5' splice site, causing partial exon skipping.
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59Histone 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 ribozyme within the transcript self-cleaves to generate a 3'-OH
B.The cap-binding complex loops to the 3' end and cleaves it internally
C.Poly(A) polymerase adds a short oligo-U tail recognized by exonucleases
D.A stem-loop bound by SLBP and a downstream histone element guide endonucleolytic cleavage without polyadenylation
Correct Answer: A stem-loop bound by SLBP and a downstream histone element guide endonucleolytic cleavage without polyadenylation
Explanation:
Replication-dependent histone mRNAs end in a conserved stem-loop bound by the stem-loop binding protein (SLBP). Together with the U7 snRNP recognizing a downstream histone element, this directs a single endonucleolytic cleavage to form the 3' end, bypassing polyadenylation entirely.
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60Phosphorylation of eIF2 by kinases such as PKR or PERK causes a global reduction in translation initiation. What is the molecular basis of this repression?
B.Phosphorylated eIF2 sequesters eIF2B, preventing GDP-to-GTP exchange needed to recycle the ternary complex
C.Phosphorylated eIF2 destroys the m7G cap on target mRNAs
D.Phosphorylated eIF2 recruits release factors to initiation codons
Correct Answer: Phosphorylated eIF2 sequesters eIF2B, preventing GDP-to-GTP exchange needed to recycle the ternary complex
Explanation:
eIF2-GTP delivers initiator Met-tRNA to the ribosome; after initiation it is released as eIF2-GDP and must be recycled by the GEF eIF2B. Phosphorylation of eIF2 makes it bind eIF2B tightly and inhibit it. Since eIF2B is limiting, ternary complex regeneration stalls, globally repressing initiation.
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