Question 1

Cell–cell communication in bacteria Which statement best captures the essence of a quorum sensing system in prokaryotes?

Accepted Answer

Introduction / Context: Quorum sensing enables bacteria to coordinate group behaviors—such as bioluminescence, biofilm formation, virulence, and competence—by monitoring population density through diffusible signals called autoinducers. Understanding what is sensed and how responses are triggered is essential in microbiology and antimicrobial strategy design. Given Data / Assumptions: The system responds to molecules synthesized by the cells themselves. Signal concentration reflects local cell density (quorum). Mechanisms differ between Gram-negative and Gram-positive organisms. Concept / Approach: In Gram-negative bacteria, LuxI-family enzymes synthesize acyl-homoserine lactones (AHLs) that diffuse and bind LuxR-family regulators. In many Gram-positive bacteria, autoinducing peptides (AIPs) are detected by membrane histidine kinases in two-component systems. In all cases, the core principle is sensing a self-produced signal that accumulates with cell density, not generic environmental parameters. Step-by-Step Solution: Identify what is sensed: an autoinducer synthesized by the population itself.Note pathway diversity: transcription factors (LuxR) vs. two-component systems.Choose the option that universally fits quorum sensing: sensing self-produced signals. Verification / Alternative check: Classic examples include Vibrio fischeri bioluminescence (AHL/LuxR) and Staphylococcus aureus agr system (AIP/histidine kinase), both using self-produced molecules as the signal. Why Other Options Are Wrong: Generic environmental signals (a): That is broader signal transduction, not specifically quorum sensing. Always a sensor kinase (c): Many Gram-negative systems use cytosolic LuxR receptors instead. Direct ribosome control (d): Quorum sensing regulates transcription mostly; effects on translation are indirect. Only in sporulation (e): QS is widespread and not limited to sporulating species. Common Pitfalls: Confusing quorum sensing with generic two-component environmental sensing. Final Answer: It senses an autoinducer compound produced by the bacterium itself (and its neighbors).

Question 2

Milestone in genomics What was the first free-living bacterial genome to be completely sequenced and made public?

Accepted Answer

Introduction / Context: The completion of the first bacterial genome marked a turning point for genomics, transforming biology with whole-genome data. Recognizing this landmark organism helps anchor timelines and contextualize later advances in sequencing technologies and assembly methods. Given Data / Assumptions: We refer to free-living bacterial genomes (not organelles or viruses). Completion and publication in the mid-1990s via whole-genome shotgun sequencing. Comparing prominent model organisms of that era. Concept / Approach: In 1995, the genome of Haemophilus influenzae Rd (KW20) was published as the first complete genome of a free-living organism using whole-genome shotgun methods. While Mycoplasma genitalium has a smaller genome and was sequenced soon after, H. influenzae came first chronologically among free-living bacteria. Step-by-Step Solution: Recall sequencing milestones of the 1990s.Identify H. influenzae (1995) as the first complete free-living bacterial genome.Select the corresponding option. Verification / Alternative check: Historical reviews of genomics list H. influenzae as the inaugural free-living genome; E. coli and B. subtilis followed later. Why Other Options Are Wrong: E. coli and B. subtilis: Sequenced later. No bacterial genome sequenced: False since multiple bacterial genomes have been completed since 1995. Mycoplasma genitalium: Very small genome, sequenced soon after, but not the first free-living one published. Common Pitfalls: Confusing “smallest genome” with “first sequenced.” Final Answer: Haemophilus influenzae

Question 3

Decoding codons Which of the following codons encodes the amino acid serine (Ser) under the standard genetic code?

Accepted Answer

Introduction / Context: Accurate decoding of codons is essential in genetics, cloning, and interpreting variants. Serine is encoded by six codons in the standard genetic code. This question tests recognition of one of those serine codons among distractors that encode other amino acids or serve different roles. Given Data / Assumptions: Standard genetic code (not mitochondrial variants). Single-codon selection among multiple options. Focus on serine (Ser). Concept / Approach: Serine codons are UCU, UCC, UCA, UCG, AGU, and AGC. Among the given options, AGU matches serine. Other codons here specify arginine (CGA), histidine (CAU), methionine/start (AUG), or tryptophan (UGG). Step-by-Step Solution: Recall the six serine codons: UCU/UCC/UCA/UCG/AGU/AGC.Compare each option to the list.Identify AGU as serine and choose it. Verification / Alternative check: Consulting any codon table confirms AGU → Ser, while CGA → Arg, CAU → His, AUG → Met (start), UGG → Trp. Why Other Options Are Wrong: CGA: Arginine. CAU: Histidine. AUG: Methionine and start signal. UGG: Tryptophan. Common Pitfalls: Mistaking AGG (Arg) or AGC (Ser) for AGU; only AGU among provided choices encodes Ser. Final Answer: AGU

Question 4

Stop codon nomenclature In classic genetic nomenclature, which termination codon is referred to as “amber”?

Accepted Answer

Introduction / Context: Beyond their sequences, stop codons have historical names used in classic genetics literature and exam questions: amber, ochre, and opal (or umber). Remembering which name maps to which triplet helps in reading older papers and understanding mutant designations (e.g., amber suppressors). Given Data / Assumptions: Standard genetic code terminology. Mapping codon triplets to historical names. Single-choice identification. Concept / Approach: The three termination codons are UAG (amber), UAA (ochre), and UGA (opal/umber). The “amber” designation originated from a researcher’s surname (H. J. Amber—apocryphal play on words), later complemented by ochre and opal for the other stops. These names persist in genetics contexts, especially suppressor tRNA studies. Step-by-Step Solution: List stop codons and their names.Associate UAG specifically with amber.Choose UAG (amber). Verification / Alternative check: Standard codon tables and genetics glossaries consistently map UAG → amber; UAA → ochre; UGA → opal. Why Other Options Are Wrong: UAA is ochre, not amber. UGA is opal (or umber), not amber. AUG is the start codon (Met), not a termination codon. UUA encodes leucine. Common Pitfalls: Swapping ochre and amber; a mnemonic is “AmbEr = UAG (both contain A and G).” Final Answer: UAG (amber)

Question 5

Ribosome composition in bacteria What are the Svedberg sedimentation values of the two subunits that compose the 70S ribosome of Escherichia coli?

Accepted Answer

Introduction / Context: Prokaryotic ribosomes are termed 70S, composed of two unequal subunits whose sizes are crucial to antibiotic targeting, translation initiation, and molecular biology techniques. Recognizing the subunit S values helps differentiate bacterial from eukaryotic ribosomes (80S with 60S/40S subunits). Given Data / Assumptions: Bacterial ribosome overall: 70S (sedimentation coefficient, not additive). Goal: identify the pair of subunit S values for E. coli. Standard growth conditions and wild-type ribosomes assumed. Concept / Approach: The prokaryotic ribosome comprises a large 50S subunit (containing 23S and 5S rRNAs with proteins) and a small 30S subunit (16S rRNA with proteins). The combined particle is 70S due to non-linear sedimentation properties; Svedberg units are not strictly additive. Step-by-Step Solution: Recall standard bacterial ribosomal subunits: 50S large, 30S small.Confirm that 50S + 30S assemble into the 70S ribosome.Select 50S and 30S. Verification / Alternative check: Textbook diagrams and antibiotic mechanisms (e.g., macrolides binding 50S, aminoglycosides targeting 30S) corroborate the subunit identities and sizes. Why Other Options Are Wrong: 40S/30S or 40S/20S: Mixed or nonstandard values; 40S is eukaryotic small subunit. 50S/20S: No 20S bacterial subunit exists. 60S/40S: Eukaryotic ribosome (80S), not bacterial. Common Pitfalls: Assuming Svedberg units add linearly; they do not due to shape/mass/friction effects. Final Answer: 50S and 30S

Question 6

Global regulation in bacteria What do global regulatory systems in prokaryotes primarily accomplish within the cell?

Accepted Answer

Introduction / Context: Bacteria rapidly reprogram gene expression to adapt to nutrients, stresses, and signals. This is coordinated by global regulatory systems—networks (sigma factors, two-component systems, small RNAs, alarmones) that tune transcription and sometimes translation across many operons. Understanding their purpose clarifies phenomena from pathogenicity to industrial fermentations. Given Data / Assumptions: Focus is on the role (function) rather than specific mechanisms. Environmental conditions vary (carbon source, oxygen, temperature, stress). Multiple genes and pathways may be co-regulated. Concept / Approach: Global regulators (e.g., CRP-cAMP for carbon catabolite repression, RpoS for stationary phase, Fur for iron homeostasis, Pho regulon, stringent response via ppGpp) alter transcription of large gene sets to match current conditions. They do not determine physical location or fix specific DNA replication sites; rather, they orchestrate flexible, condition-dependent expression programs. Step-by-Step Solution: Identify that “global” implies many genes/operons.Recognize the trigger: environmental and physiological signals.Select the option describing broad, condition-dependent regulation. Verification / Alternative check: Transcriptomics shows genome-wide changes controlled by sigma factors and regulators under stress/starvation; classic examples include lac/catabolite repression and heat-shock responses. Why Other Options Are Wrong: Geographical location: Nonsensical for cellular regulation. Aid DNA replication mechanics or fix initiation site: These are controlled by oriC and replication proteins, not “global regulation” per se. Permanent rRNA silencing: Opposite of adaptive regulation; rRNA transcription varies with growth rate. Common Pitfalls: Confusing specific housekeeping processes with adaptive, global regulatory networks. Final Answer: Regulate many genes and pathways in response to changing environmental conditions

Question 7

Molecular genetics — identify the correct statement: Choose the one accurate statement about cDNA, restriction enzymes, RFLP, and protein sequence inference from nucleotide sequence (assume standard molecular biology practice and definitions).

Accepted Answer

Introduction / Context: Foundational concepts in molecular genetics include how cDNA relates to genes, what substrates restriction endonucleases recognize, where RFLPs commonly occur, and how protein sequences are inferred. This question checks precise understanding of these basics used across biotechnology and bioinformatics. Given Data / Assumptions: cDNA represents the reverse-transcribed, intron-free, mature mRNA sequence. Conventional restriction endonucleases recognize palindromic sites in double-stranded DNA. RFLPs reflect sequence variation at restriction sites, often in noncoding regions. Standard genetic code enables translation of open reading frames from cDNA. Concept / Approach: To find the correct statement, match each claim against standard definitions. The only universally correct statement among the options is that cDNA, which mirrors spliced mRNA, can be translated in silico to deduce the encoded amino acid sequence of the protein. Step-by-Step Solution: Confirm cDNA lacks introns; thus intron size/sequence cannot be deduced from cDNA alone.Recall that classical restriction endonucleases act on double-stranded DNA; single-stranded cleavage is atypical and not a general rule.Note that RFLPs are frequently found in noncoding regions; they are not “usually formed” in coding regions.Use the genetic code to translate cDNA open reading frames and obtain the protein’s amino acid sequence — this statement is correct. Verification / Alternative check: Practical workflows (e.g., cloning a cDNA and Sanger sequencing) routinely translate cDNA to predict protein sequence and motif content; proteomics then validates peptides experimentally. Why Other Options Are Wrong: Introns from cDNA: cDNA is post-splicing; introns are absent. Restriction enzymes on ssDNA: the canonical activity is on double-stranded DNA recognition sites. RFLP location: commonly noncoding; coding regions are not the usual source. Common Pitfalls: Confusing genomic DNA (with introns) and cDNA (without introns), or overgeneralizing enzyme capabilities beyond their standard substrates. Final Answer: The amino acid sequence of a protein can be deduced from its corresponding cDNA nucleotide sequence

Question 8

Translation initiation in prokaryotes: Which modified amino acid is used at the start of most prokaryotic proteins during translation initiation?

Accepted Answer

Introduction / Context: Protein synthesis in bacteria begins uniquely with a specialized initiator tRNA charged with a modified amino acid. Recognizing this initiator residue is essential for understanding translation initiation, ribosome assembly, and how bacterial proteins are distinguished from eukaryotic ones. Given Data / Assumptions: Bacterial initiation uses a dedicated tRNA (tRNAfMet). The first amino acid in most bacterial proteins is chemically modified before incorporation. Question asks for the specific modified residue’s name. Concept / Approach: In bacteria, methionine is formylated to N-formylmethionine (fMet) and delivered to the P-site by initiation factors along with the small ribosomal subunit at the start codon (usually AUG, occasionally GUG or UUG reinterpreted as start). Step-by-Step Solution: Identify the bacterial initiator: methionine on initiator tRNA is formylated.Match the correct chemical name: N-formylmethionine.Exclude other N-formyl–amino acids, which are not used physiologically as initiators.Select N-formylmethionine as the answer. Verification / Alternative check: Proteomic studies show N-terminal formyl groups are often deformylated post-translationally, but initiation still uses fMet. Eukaryotes instead start with methionine (not formylated), supporting the distinction. Why Other Options Are Wrong: N-formylserine/leucine/alanine are not the bacterial initiator residues. Common Pitfalls: Confusing bacterial fMet initiation with eukaryotic Met initiation or with mitochondrial translation, which resembles bacterial mechanisms but is context specific. Final Answer: N-formylmethionine

Question 9

Gene regulation (prokaryotic): Which mechanism describes negative regulation of protein synthesis (transcriptional control) in bacteria?

Accepted Answer

Introduction / Context: Negative regulation is a cornerstone of bacterial gene control, exemplified by the lac operon. Understanding how repressors modulate transcription distinguishes transcriptional control from enzymatic regulation and positive control mechanisms. Given Data / Assumptions: Negative regulation reduces or blocks transcription. Operons use operators where regulatory proteins bind. RNA polymerase promoter binding alone is not “negative regulation.” Concept / Approach: In negative control, a repressor binds a specific operator DNA sequence near the promoter, physically blocking RNA polymerase binding or elongation. Inducers or corepressors modulate repressor DNA affinity to switch genes on or off based on cellular conditions. Step-by-Step Solution: Define negative regulation: repression mediated by repressor binding to operator DNA.Assess options: only repressor–DNA binding at the operator fits the definition.Reject enzyme allosteric inhibition (post-translational) and polymerase binding (positive/neutral), and nonstandard “repressor to polymerase” as the primary model.Select repressor–operator interaction. Verification / Alternative check: Classic systems (lacI–lacO, trpR–trpO) illustrate repressor occupancy reducing transcription; inducers (allolactose) relieve repression by reducing operator affinity. Why Other Options Are Wrong: Allosteric inhibition: regulates enzyme activity, not transcription initiation. RNA polymerase promoter binding: required for transcription, not repression. Repressor binding to RNA polymerase: not the canonical mechanism for negative control in textbook operons. Common Pitfalls: Conflating transcriptional control with enzyme feedback inhibition, or mixing positive and negative control concepts. Final Answer: Binding of a repressor protein to DNA (operator)

Translation and Regulation Questions