Microbiology Quiz: Transcription And Translation

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Transcription And TranslationQuestion 1 of 20

An in vitro coupled transcription-translation system is prepared using E. coli components and a DNA template for a 300-amino-acid protein. The experiment is run in the presence of a novel antibiotic. Analysis reveals that full-length mRNA transcripts are produced, but the only polypeptide products detected are dipeptides consisting of fMet linked to the second amino acid.

The antibiotic most likely inhibits which specific step of protein synthesis?

Binding of the 50S ribosomal subunit to the 30S initiation complex, thereby preventing the formation of a functional 70S ribosome.

Peptidyl transferase activity, preventing the formation of the first peptide bond between fMet and the second amino acid.

Binding of the initiator fMet-tRNA to the P site of the 30S ribosomal subunit.

Translocation of the peptidyl-tRNA from the A site to the P site after the first peptide bond is formed.

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Microbiology Quiz

Microbiology Quiz: Transcription And Translation

Practice Transcription And Translation in Microbiology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Transcription And Translation, giving you a quick way to practice the rules, question types, and explanations that matter most for Microbiology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The antibiotic most likely inhibits which specific step of protein synthesis?

Binding of the 50S ribosomal subunit to the 30S initiation complex, thereby preventing the formation of a functional 70S ribosome.
Peptidyl transferase activity, preventing the formation of the first peptide bond between fMet and the second amino acid.
Binding of the initiator fMet-tRNA to the P site of the 30S ribosomal subunit.
Translocation of the peptidyl-tRNA from the A site to the P site after the first peptide bond is formed. (correct answer)

Explanation: The production of full-length mRNA indicates that transcription is not inhibited. The formation of a dipeptide (fMet-aa2) demonstrates that initiation, codon recognition for the second amino acid, and the first peptide bond formation are all successful. The process must be stalling immediately after this step. The next step is translocation, where the ribosome moves one codon down the mRNA, shifting the dipeptidyl-tRNA from the A site to the P site. Inhibition of translocation would result in the accumulation of dipeptides, which matches the experimental observation.

Question 2

A bacterial gene contains a coding sequence of 1200 base pairs, beginning with ATG and ending with TAA. A nonsense mutation converts the 100th codon from CAG (Glutamine) to UAG (Stop). Assuming the average molecular weight of an amino acid is 110 Daltons, what is the approximate molecular weight of the resulting polypeptide?

44,000 Da
11,000 Da
43,890 Da
10,890 Da (correct answer)

Explanation: When you encounter questions about nonsense mutations and polypeptide molecular weight, you need to understand how premature stop codons affect translation and then calculate the resulting protein size. A nonsense mutation introduces a premature stop codon, causing translation to terminate early. In this case, the mutation converts the 100th codon from CAG (glutamine) to UAG (a stop codon). This means translation will stop at position 100 instead of continuing to the normal stop codon at the end of the gene. To find the molecular weight: The polypeptide will contain 99 amino acids (positions 1-99, since position 100 now contains the stop codon). With an average molecular weight of 110 Daltons per amino acid:

99 \times 110 = 10,890

Daltons. Looking at the wrong answers: Option A (44,000 Da) represents the approximate weight if the full-length protein were made (400 amino acids × 110 Da), ignoring the nonsense mutation entirely. Option B (11,000 Da) incorrectly includes the 100th position in the calculation (100 × 110 = 11,000), but the 100th codon is now a stop codon and doesn't contribute an amino acid to the polypeptide. Option C (43,890 Da) appears to subtract one amino acid from the full-length protein but uses the wrong total length. Remember: nonsense mutations create truncated proteins. Always count the amino acids that are actually incorporated before the premature stop codon hits. The stop codon itself doesn't contribute to the polypeptide's molecular weight.

Question 3

The formation of a peptide bond during translation is catalyzed by peptidyl transferase activity. A mutation in which of the following components would most directly abolish this specific activity in a bacterium?

The 16S rRNA component of the 30S ribosomal subunit.
The elongation factor G (EF-G) protein.
The 23S rRNA component of the 50S ribosomal subunit. (correct answer)
The aminoacyl-tRNA synthetase for methionine.

Explanation: The ribosome is a ribozyme. The peptidyl transferase center, which catalyzes the formation of peptide bonds, is located within the large ribosomal subunit (50S in prokaryotes). Specifically, this catalytic activity resides in the 23S rRNA, not in any of the ribosomal proteins. Therefore, a mutation in the catalytic domain of the 23S rRNA would directly abolish peptide bond formation.

Question 4

A researcher discovers a bacterial mRNA with two potential start codons (AUG). The first AUG is preceded by a very weak Shine-Dalgarno (SD) sequence. The second AUG, 30 nucleotides downstream, is preceded by a strong, consensus SD sequence. Which statement most accurately predicts how this mRNA will be translated?

Translation will exclusively initiate at the first AUG because ribosomes always scan from the 5' end and initiate at the first available start codon.
Translation will exclusively initiate at the second AUG because the strength of the SD sequence interaction is the dominant factor for initiation site selection. (correct answer)
Two different proteins will be produced in roughly equal amounts, one starting at the first AUG and one at the second AUG.
No protein will be produced because the presence of two start codons will cause the ribosome to stall and dissociate from the mRNA.

Explanation: Unlike eukaryotic ribosomes, which typically employ a scanning mechanism, prokaryotic ribosomes are directly recruited to the initiation site via the Shine-Dalgarno (SD) sequence's interaction with the 16S rRNA. The strength of this interaction is the most critical factor determining where initiation occurs. A strong, consensus SD sequence will efficiently recruit the 30S subunit, leading to robust initiation at the nearby start codon. A very weak SD sequence will be a very poor initiator, so little to no translation will begin at the first AUG.

Question 5

Rifampicin is an antibiotic that specifically inhibits bacterial RNA polymerase. An experiment is conducted to identify the subunit it targets. Purified core enzyme (α₂ββ'ω) and holoenzyme (α₂ββ'ωσ) are separated into their individual subunits. Each subunit is then mixed with the other four to reconstitute the enzyme, but using one subunit from a rifampicin-resistant mutant in each reconstitution.

If the reconstituted enzyme containing the β subunit from the resistant mutant is active in the presence of rifampicin, while all other combinations are inhibited, what is the primary function of the β subunit?

Recognizing and binding to the -10 and -35 promoter consensus sequences.
Catalyzing the formation of phosphodiester bonds during RNA elongation. (correct answer)
Unwinding the DNA double helix to form the transcription bubble.
Assembling the core enzyme by scaffolding the other subunits.

Explanation: Rifampicin inhibits transcription by binding within the DNA/RNA channel of RNA polymerase, physically blocking the path of the elongating RNA chain. The binding site is located on the β subunit. Therefore, a mutation in the β subunit can confer resistance. The β and β' subunits form the catalytic core of the enzyme, with the β subunit containing the active site for phosphodiester bond formation. The experiment pinpoints the β subunit as the site of drug action and, by extension, its central role in catalysis.

Question 6

The initiator tRNA in E. coli, tRNAfMet, is charged with methionine, which is then formylated to N-formylmethionine (fMet). Which of the following correctly describes a key feature of the fMet-tRNAfMet complex?

It recognizes the same AUG codon as the elongator tRNAMet but is uniquely blocked from entering the ribosomal A site. (correct answer)
Its formyl group is essential for forming the first peptide bond, catalyzed by the 23S rRNA.
It is the only tRNA capable of binding to the 30S subunit before the 50S subunit joins to form the complete 70S ribosome.
It has a unique anticodon that allows it to bind to the Shine-Dalgarno sequence in addition to the start codon.

Explanation: Both initiator tRNAfMet and elongator tRNAMet recognize the AUG codon. However, their structures are different. The formylated methionine and other structural features of fMet-tRNAfMet allow it to bind to initiation factor IF2 and enter the P site of the 30S initiation complex directly. Conversely, these same features prevent it from binding to elongation factor Tu (EF-Tu), which is required for all elongator tRNAs to enter the ribosomal A site during the elongation phase.

Question 7

In a rapidly growing E. coli cell, the rate of transcription elongation is approximately 45 nucleotides per second, and the rate of translation elongation is approximately 15 amino acids per second. Given these rates, what is the physical relationship between RNA polymerase and the lead ribosome on a typical, actively expressed gene?

The ribosome is stalled far behind the RNA polymerase, waiting for the entire mRNA to be synthesized before initiating translation.
The RNA polymerase and the ribosome move at the same effective speed, maintaining a relatively constant distance between them. (correct answer)
The ribosome moves significantly faster than the RNA polymerase, frequently catching up to it and pausing until more mRNA is available.
The ribosome moves significantly slower than the RNA polymerase, causing the distance between them to continuously increase over time.

Explanation: The rates are coupled. The rate of translation is 15 amino acids per second. Since each amino acid is coded by a 3-nucleotide codon, the ribosome covers the mRNA at a rate of 15 codons/sec * 3 nucleotides/codon = 45 nucleotides per second. This rate is identical to the rate of transcription elongation. This elegant coordination ensures that transcription and translation proceed in lockstep, maintaining a consistent distance between the polymerase and the lead ribosome and minimizing the exposure of naked mRNA.

Question 8

A mutation introduces a stop codon (UAG) early in the lacZ gene of the E. coli lac operon. This mutation not only abolishes β-galactosidase activity but also significantly reduces the expression of the downstream lacY (permease) and lacA (transacetylase) genes, even in the presence of an inducer.

What is the most likely molecular mechanism explaining this polar effect on the expression of lacY and lacA?

The nonsense codon in lacZ destabilizes the entire polycistronic mRNA molecule, leading to its rapid degradation by ribonucleases before downstream genes can be translated.
Ribosomes stalling at the premature stop codon physically obstruct the progression of RNA polymerase, thereby preventing the transcription of downstream genes.
Premature translation termination in lacZ exposes a Rho utilization (rut) site on the nascent mRNA, leading to Rho-dependent termination of transcription. (correct answer)
The premature stop codon prevents the correct folding of the nascent mRNA, which in turn blocks ribosome binding at the Shine-Dalgarno sequences of lacY and lacA.

Explanation: This phenomenon is known as a polar mutation. In prokaryotes, transcription and translation are coupled. When ribosomes terminate translation prematurely at the nonsense codon, a long stretch of nascent, ribosome-free mRNA is exposed. This can expose a cryptic Rho utilization (rut) site, allowing the Rho termination factor to bind and terminate transcription before RNA polymerase reaches the downstream lacY and lacA genes. This is a primary mechanism for polarity.

Question 9

Consider a bacterial gene being actively expressed. The RNA polymerase is transcribing the DNA template strand, which has the sequence 3'-...GCTAAC...-5', and the ribosome is closely following behind.

Which of the following represents the corresponding mRNA codon being transcribed and the tRNA anticodon that would bind to it during translation?

mRNA codon: 5'-CGAUUG-3'; tRNA anticodon: 3'-GCUAAC-5'
mRNA codon: 5'-AAGUUC-3'; tRNA anticodon: 3'-UUCAAG-5'
mRNA codon: 5'-GCU-3'; tRNA anticodon: 3'-CGA-5'
mRNA codon: 5'-CGA-3'; tRNA anticodon: 3'-GCU-5' (correct answer)

Explanation: When you encounter transcription and translation problems, you need to carefully track the direction and complementarity rules at each step. RNA polymerase reads the DNA template strand in the 3' to 5' direction and synthesizes mRNA in the 5' to 3' direction using complementary base pairing (A-U, T-A, G-C, C-G). Given the DNA template sequence 3'-GCTAAC-5', RNA polymerase will transcribe this to produce mRNA with the sequence 5'-CGAUUG-3'. However, since we're looking for individual codons, we need to consider that codons are read in triplets. Taking the first three bases, the mRNA codon would be 5'-CGA-3'. During translation, tRNA anticodons bind to mRNA codons through complementary base pairing, but in antiparallel orientation. The anticodon for CGA would be 3'-GCU-5'. Answer choice A incorrectly shows a six-nucleotide sequence instead of a three-nucleotide codon, and the anticodon matches the original DNA template rather than being complementary to the mRNA. Choice B appears to use incorrect base pairing rules and doesn't follow from the given DNA sequence. Choice C has the wrong mRNA codon (GCU instead of CGA) and consequently the wrong anticodon. Choice D correctly shows mRNA codon 5'-CGA-3' and its complementary tRNA anticodon 3'-GCU-5', following proper transcription and translation rules. Study tip: Always work step-by-step through transcription first (DNA template → mRNA), then translation (mRNA codon → tRNA anticodon). Remember that complementary sequences are always antiparallel, and double-check your base pairing at each step.

Question 10

A researcher constructs two versions of a bacterial gene. Version A has the consensus Shine-Dalgarno sequence (5'-AGGAGGU-3') located 8 bases upstream of the start codon. Version B has a mutated sequence (5'-AGCAGGU-3') at the same position. Both genes are transcribed at identical rates. How will the protein expression from Version B most likely compare to Version A?

Protein expression will be significantly lower due to reduced efficiency of translation initiation. (correct answer)
Protein expression will be identical because the start codon, not the Shine-Dalgarno sequence, is the primary determinant of translation rate.
Protein expression will be higher because the mutation removes a repressive secondary structure in the mRNA.
Protein expression will be abolished because the mutated sequence recruits a translational repressor protein.

Explanation: The efficiency of translation initiation depends on the strength of the interaction between the Shine-Dalgarno (SD) sequence on the mRNA and the anti-SD sequence on the 16S rRNA. The consensus sequence provides the strongest binding. The mutation in Version B weakens this base-pairing interaction, reducing the efficiency with which the 30S ribosomal subunit is recruited to the mRNA. Since transcription rates are identical, the lower translation initiation efficiency will result in significantly less protein being produced from Version B.

Question 11

A bacterial gene is terminated by a Rho-independent (intrinsic) terminator. The DNA sequence encoding the 3' end of the mRNA transcript contains an inverted repeat followed by a stretch of eight adenine-thymine pairs. A mutation occurs that changes the template strand sequence in this region from 5'-TTTTTTTT-3' to 5'-TTGTTGTT-3'.

What is the most likely outcome of this mutation on transcription of the gene?

The stability of the mRNA hairpin structure will be compromised, causing the polymerase to stall indefinitely at the terminator.
Transcription will terminate prematurely because the altered sequence is recognized as a stop signal by the RNA polymerase core enzyme.
RNA polymerase will fail to pause and dissociate efficiently, leading to transcriptional read-through into the downstream DNA region. (correct answer)
The affinity of Rho factor for the nascent transcript will increase, causing termination to become Rho-dependent.

Explanation: Rho-independent termination relies on two key features: a stable hairpin loop in the nascent RNA that causes RNA polymerase to pause, and a weak rU-dA hybrid in the transcription bubble (transcribed from a poly-A/T tract in the DNA). The mutation disrupts the poly-T tract on the template strand, which means the transcribed RNA will no longer have a poly-U tract. This strengthens the RNA-DNA hybrid, preventing the dissociation of the RNA from the DNA template even if the polymerase pauses at the hairpin. The result is read-through transcription.

Question 12

Which of the following represents the corresponding mRNA codon being transcribed and the tRNA anticodon that would bind to it during translation?

mRNA codon: 5'-CGAUUG-3'; tRNA anticodon: 3'-GCUAAC-5'
mRNA codon: 5'-AAGUUC-3'; tRNA anticodon: 3'-UUCAAG-5'
mRNA codon: 5'-GCU-3'; tRNA anticodon: 3'-CGA-5'
mRNA codon: 5'-CGA-3'; tRNA anticodon: 3'-GCU-5' (correct answer)

Question 13

What is the most likely molecular mechanism explaining this polar effect on the expression of lacY and lacA?

The nonsense codon in lacZ destabilizes the entire polycistronic mRNA molecule, leading to its rapid degradation by ribonucleases before downstream genes can be translated.
Ribosomes stalling at the premature stop codon physically obstruct the progression of RNA polymerase, thereby preventing the transcription of downstream genes.
Premature translation termination in lacZ exposes a Rho utilization (rut) site on the nascent mRNA, leading to Rho-dependent termination of transcription. (correct answer)
The premature stop codon prevents the correct folding of the nascent mRNA, which in turn blocks ribosome binding at the Shine-Dalgarno sequences of lacY and lacA.

Question 14

A researcher identifies a single nucleotide deletion within the 5th codon of a bacterial gene's coding sequence. The original sequence of codons 4, 5, and 6 was 5'-UGG UAC GGU-3', coding for Trp-Tyr-Gly.

What is the most likely consequence of this single base deletion on the resulting polypeptide?

Only the 5th amino acid will be changed, and the rest of the polypeptide will be synthesized normally, resulting in a full-length protein.
The polypeptide will be truncated at the 4th amino acid because the ribosome cannot read past the deletion.
The amino acid sequence will be altered starting from position 5, and the protein will likely be truncated due to a newly formed stop codon. (correct answer)
The polypeptide will be completely unchanged because the wobble hypothesis allows the ribosome to accommodate single base changes.

Explanation: A single nucleotide deletion causes a frameshift mutation. The reading frame is shifted by -1 for all codons downstream of the deletion. This will alter every amino acid from position 5 onwards. Frameshift mutations almost invariably lead to the creation of a premature stop codon within a short distance, resulting in a truncated and nonfunctional polypeptide. This is distinct from a point mutation (missense) which would only change one amino acid.

Question 15

Recognizing and binding to the -10 and -35 promoter consensus sequences.
Catalyzing the formation of phosphodiester bonds during RNA elongation. (correct answer)
Unwinding the DNA double helix to form the transcription bubble.
Assembling the core enzyme by scaffolding the other subunits.

Question 16

Protein expression will be significantly lower due to reduced efficiency of translation initiation. (correct answer)
Protein expression will be identical because the start codon, not the Shine-Dalgarno sequence, is the primary determinant of translation rate.
Protein expression will be higher because the mutation removes a repressive secondary structure in the mRNA.
Protein expression will be abolished because the mutated sequence recruits a translational repressor protein.

Question 17

An E. coli strain has a mutation rendering its primary sigma factor, σ⁷⁰, unable to recognize the promoter for the gal operon. However, a cryptic promoter recognized by the nitrogen-starvation sigma factor, σ⁵⁴, exists within the first gene of the operon, galE. Under normal growth conditions, the operon is not expressed.

If this strain is subjected to nitrogen starvation, what is the most likely expression pattern for the gal operon genes (galE, galT, galK)?

None of the genes will be expressed because the primary promoter is non-functional.
All three genes will be expressed from the polycistronic mRNA transcribed from the σ⁷⁰-dependent promoter.
Only galT and galK will be expressed, as transcription will initiate from the internal σ⁵⁴-dependent promoter. (correct answer)
All three genes will be expressed, but the resulting GalE protein will be a truncated, non-functional version.

Explanation: Under nitrogen starvation, σ⁵⁴ becomes active and directs RNA polymerase to its cognate promoters. Since there is a σ⁵⁴-dependent promoter within galE, transcription will initiate from this internal site. This means the polymerase will not transcribe the beginning of the galE gene but will transcribe the downstream genes, galT and galK, as part of a shorter polycistronic message. The galE gene itself will not be properly expressed.

Question 18

The ribosome is stalled far behind the RNA polymerase, waiting for the entire mRNA to be synthesized before initiating translation.
The RNA polymerase and the ribosome move at the same effective speed, maintaining a relatively constant distance between them. (correct answer)
The ribosome moves significantly faster than the RNA polymerase, frequently catching up to it and pausing until more mRNA is available.
The ribosome moves significantly slower than the RNA polymerase, causing the distance between them to continuously increase over time.

Question 19

44,000 Da
11,000 Da
43,890 Da
10,890 Da (correct answer)

99 \times 110 = 10,890

Question 20

A researcher performs an in vitro transcription assay with purified E. coli RNA polymerase holoenzyme and a linear DNA template containing a single strong promoter. In one reaction, transcription proceeds normally, producing a 500-nucleotide RNA. In a second reaction, the antibiotic streptolydigin is added after the first 10 nucleotides have been incorporated. What is the most likely product observed in the second reaction?

No RNA transcripts, as streptolydigin prevents the binding of RNA polymerase to the promoter.
Full-length 500-nucleotide RNA transcripts, but at a much lower concentration than the control.
A mixture of short, abortive transcripts approximately 10 nucleotides in length. (correct answer)
A single 500-nucleotide RNA transcript that remains permanently bound to the DNA template.

Explanation: Streptolydigin inhibits transcription elongation, but not initiation. It binds to RNA polymerase and prevents the conformational changes required for translocation after a phosphodiester bond has been formed. This is different from rifampicin which blocks the path of the growing chain. Because the drug was added after initiation and the first 10 bases were incorporated, the polymerase is already on the template. It will attempt to add the next nucleotide, but the antibiotic will prevent it from moving forward, leading to the release (abortion) of the short ~10 nt transcript. This cycle repeats, producing many short transcripts.