Transcription & Translation - Microbiology

1. What is the most likely consequence of this single base deletion on the resulting polypeptide?
   Answer: 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. 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.
2. If this strain is subjected to nitrogen starvation, what is the most likely expression pattern for the gal operon genes (galE, galT, galK)?
   Answer: Only galT and galK will be expressed, as transcription will initiate from the internal σ⁵⁴-dependent promoter. 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.
3. During translation termination in E. coli, release factor 1 (RF1) recognizes the stop codons UAA and UAG. What is the direct molecular function of RF1 once it binds to a stop codon in the ribosomal A site?
   Answer: It promotes the hydrolysis of the ester bond linking the completed polypeptide chain to the tRNA in the P site. When you encounter questions about translation termination, focus on the specific molecular events that must occur to release the completed protein from the ribosome. The key insight is that the growing polypeptide chain remains covalently attached to the tRNA in the P site via an ester bond, and this bond must be broken to free the protein. Release factor 1 (RF1) acts as a molecular mimic of aminoacyl-tRNA, but instead of delivering an amino acid, it delivers a water molecule to the peptidyl transferase center. When RF1 recognizes UAA or UAG stop codons in the A site, it promotes hydrolysis of the ester bond connecting the completed polypeptide to the tRNA in the P site. This hydrolysis reaction releases the finished protein from the ribosome, which is the essential function of translation termination. Answer A incorrectly describes ribosome translocation, which occurs during elongation, not termination, and involves elongation factors, not release factors. Answer B is wrong because RF1 has no helicase activity—it's specifically a hydrolysis-promoting factor. Answer C confuses transcription termination (where Rho factor functions) with translation termination, and Rho doesn't interact with ribosomes. For microbiology exams, remember that each step of protein synthesis has distinct molecular players and mechanisms. Translation termination specifically requires breaking the peptidyl-tRNA bond through hydrolysis—this is fundamentally different from the GTP-dependent processes in elongation or the RNA unwinding mechanisms in transcription.
4. What is the most likely outcome of this mutation on transcription of the gene?
   Answer: RNA polymerase will fail to pause and dissociate efficiently, leading to transcriptional read-through into the downstream DNA region. 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.
5. What is the most likely immediate effect on the transcription of housekeeping genes, such as those for glycolysis, which are recognized by the primary sigma factor σ⁷⁰ (RpoD)?
   Answer: Transcription of housekeeping genes will decrease due to competition between σ³² and σ⁷⁰ for a limited pool of core RNA polymerase enzymes. Both the primary sigma factor (σ⁷⁰) and alternative sigma factors (like σ³²) must bind to the same core RNA polymerase enzyme to initiate transcription. Because the pool of core RNA polymerase is finite, a sudden increase in the concentration and activity of σ³² leads to increased competition for these core enzymes. As a result, fewer core enzymes are available to bind with σ⁷⁰, causing a decrease in the transcription of housekeeping genes.
6. Which of the following represents the corresponding mRNA codon being transcribed and the tRNA anticodon that would bind to it during translation?
   Answer: mRNA codon: 5'-CGA-3'; tRNA anticodon: 3'-GCU-5' 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.
7. 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?
   Answer: A mixture of short, abortive transcripts approximately 10 nucleotides in length. 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.
8. 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?
   Answer: Catalyzing the formation of phosphodiester bonds during RNA elongation. 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.
9. 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?
   Answer: 10,890 Da 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.
10. The antibiotic most likely inhibits which specific step of protein synthesis?
    Answer: Translocation of the peptidyl-tRNA from the A site to the P site after the first peptide bond is formed. 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.
11. 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?
    Answer: It recognizes the same AUG codon as the elongator tRNAMet but is uniquely blocked from entering the ribosomal A site. 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.
12. A bacterial promoter has the following -10 consensus sequence: TATAAT. A 'down' mutation changes this sequence to TGTAAT. What is the most direct consequence of this mutation on the process of transcription?
    Answer: The open complex will form more slowly, decreasing the rate of transcription initiation. The -10 region (Pribnow box) is rich in A-T base pairs, which are easier to melt than G-C pairs. The sigma factor recognizes the -10 and -35 boxes to form the closed complex. The transition to the transcriptionally active open complex requires melting the DNA at the -10 region. Changing an A-T pair to a G-C pair (as in TATAAT to TGTAAT) makes the DNA more difficult to unwind. This slows down the rate of open complex formation, which is a key rate-limiting step in transcription initiation, thus decreasing overall transcription levels.
13. 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?
    Answer: Protein expression will be significantly lower due to reduced efficiency of translation initiation. 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.
14. 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?
    Answer: Translation will exclusively initiate at the second AUG because the strength of the SD sequence interaction is the dominant factor for initiation site selection. 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.
15. What is the most likely molecular mechanism explaining this polar effect on the expression of lacY and lacA?
    Answer: Premature translation termination in lacZ exposes a Rho utilization (rut) site on the nascent mRNA, leading to Rho-dependent termination of transcription. 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.
16. 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?
    Answer: The RNA polymerase and the ribosome move at the same effective speed, maintaining a relatively constant distance between them. 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.
17. 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?
    Answer: The 23S rRNA component of the 50S ribosomal subunit. 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.
18. In prokaryotes, transcription and translation are tightly coupled. A mutation that significantly slows down the rate of translation of an mRNA, without preventing its initiation, is observed to also decrease the steady-state level of that mRNA. What is the best explanation for this observation?
    Answer: The lack of ribosome protection on the nascent mRNA transcript exposes it to degradation by cellular ribonucleases. The coupling of transcription and translation provides a key protective function for mRNA in prokaryotes. Ribosomes translating the mRNA protect it from degradation by ribonucleases. If translation is slowed, a larger gap forms between the transcribing RNA polymerase and the translating ribosomes. This expanse of 'naked' mRNA is susceptible to attack by endonucleases and exonucleases, leading to a lower steady-state concentration of the transcript.
19. The specificity of translation initiation in prokaryotes is primarily determined by the base-pairing interaction between which two molecules?
    Answer: The Shine-Dalgarno sequence on the mRNA and a complementary sequence on the 16S rRNA of the 30S ribosomal subunit. While the start codon is essential, the initial positioning of the 30S ribosomal subunit on the mRNA is the key determinant of initiation specificity. This is achieved through the binding of the Shine-Dalgarno sequence (a purine-rich region upstream of the start codon) on the mRNA to a complementary anti-Shine-Dalgarno sequence near the 3' end of the 16S rRNA within the 30S subunit. This interaction correctly positions the start codon in the P site.
20. The bacterial transcription termination factor Rho functions as an ATP-dependent helicase. How does this helicase activity lead to the termination of transcription?
    Answer: Rho unwinds the RNA-DNA hybrid duplex within the transcription bubble, causing the release of the nascent RNA transcript. Rho factor first binds to a C-rich, G-poor 'rut' site on the nascent mRNA. It then moves along the transcript in the 5' to 3' direction, using ATP hydrolysis for energy. When the RNA polymerase pauses at a termination site, Rho catches up and its helicase activity actively unwinds the hybrid duplex formed by the RNA transcript and the DNA template strand within the transcription bubble. This destabilization leads to the release of the RNA chain and the dissociation of RNA polymerase from the DNA.

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