Microbiology

What is the primary function of the sigma (σ) factor in bacterial transcription?. What is the primary function of the sigma (σ) factor in bacterial transcription?. To terminate transcription by destabilizing the RNA-DNA hybrid. To recognize specific promoter sequences (-10 and -35 boxes). To unwind the DNA duplex ahead of the RNA polymerase during elongation.

DNA sequencing of a novel bacterial species has identified three genes encoding for various enzymes required to mount a response to increased heavy metal levels in soil. From your understanding of bacterial genome organisation, which of the following is the most likely organisation of these genes?. The genes are in random areas of the genome, under control of a constitutive promoter, and when transcribed produces a monocistronic mRNA. The genes are located together, under control of a constitutive promoter, and when transcribed produces a polycistronic mRNA. The genes are located together, under control of a regulated promoter, and when transcribed produces a monocistronic mRNA. The genes are located together, under control of a regulated promoter, and when transcribed produces a polycistronic mRNA.

Which initiator amino acid is specifically utilized during bacterial translation initiation?. Methionine (Met). N-formylmethionine (fMet). Valine (Val). D-formylmethionine (dMet).

What would be the consequences of Ef-Tu losing its activity?. impaired aa-tRNA delivery to the ribosomal A site. Impaired elongation due to lack of ribosome movement along mRNA. Termination will not occur due to impaired peptidyl-tRNA hydrolysis. No ribosome recycling due to lack of dissociation of 30s and 50s subunit.

What is the most energetically economical regulatory strategy to reprogram bacterial gene expression?. Transcriptional control. Translational control. Post-transcriptional contro. Post-translational control.

Which of the following is an example of post-transcriptional regulation in prokaryotes?. A repressor protein binds to the operator, blocking RNA polymerase progression and inhibiting transcription. An activator protein binds near the promoter and stimulates RNA polymerase recruitment, enhancing transcription. An sRNA binds near the Shine-Dalgarno sequence and occludes ribosome access, repressing translation and promoting mRNA degradation. An sRNA binds near the promoter and stimulates RNA polymerase recruitment, increasing translation efficiency.

Which methodology is currently considered the gold standard for targeted analysis of bacterial gene expression?. Northern blot. RT-qPCR. RNA-seq. lux operon fusion assays.

What is the correct RT-qPCR workflow for targeted bacterial gene expression analysis?. Total RNA extraction from bacterial samples – reverse transcription – RNase treatment – real time PCR – data analysis – results. Total RNA extraction from bacterial samples – reverse transcription – real-time PCR – data analysis –results. Total RNA extraction from bacterial samples – rRNA depletion -- library preparation – real-time PCR –data analysis – results. Total RNA extraction from bacterial samples – DNase treatment - reverse transcription – real-time PCR – data analysis – results.

Transcriptionally active regions of DNA adopt: An open euchromatin configuration. A densely packed euchromatin configuration. An open heterochromatin configuration. A densely packed heterochromatin configuration.

A gene contains six exons (E1–E6). Analysis of mature mRNA reveals that liver cells predominantly produce the transcript E1–E2–E3–E5–E6, whereas neuronal cells predominantly produce E1–E2–E3–E4–E5–E6. Both transcripts encode functional proteins with distinct biological properties. Which mechanism most plausibly explains this observation?. RNA Polymerase II uses different promoters in the two tissues, causing exon 4 to be deleted from the genomic DNA in liver cells. Tissue-specific splicing factors promote exclusion of exon 4 from the pre-mRNA in liver cells while favoring its inclusion in neuronal cells. Exon 4 is transcribed only in neuronal cells because RNA Polymerase III recognizes a neuron-specific promoter upstream of the exon. Liver cells selectively degrade exon 4 after translation, resulting in a shorter mature mRNA.

What is the main function of the poly(A) tail?. Initiate RNA translation. Initiate DNA replication. Promote the removal of introns. Increase the stability of mRNA and the efficiency of the translation process.

If a cell line has a mutation in the nuclear cap-binding complex (CBC), which of the following phenotypic defects will be immediately observed in these cells?. Retention of the poly(A) tail inside the nucleus. Inability of RNA Polymerase II to recognize the TATA box. Inability to protect the pre-mRNA from nuclear exonucleases and reduction in U1 snRNPdependent splicing efficiency. Total block in cytoplasmic translation initiation by elF4E.

What is the main function of miRNAs?. Increasing DNA replication. Binding to mRNAs and repressing translation. Transcribing protein-coding genes in the nucleus. Modifying histone acetylation to activate transcription.

How does the RNA-Induced Silencing Complex (RISC) identify target mRNAs?. By recognizing CpG islands in promoters. Mainly through complementarity between the miRNA and sequences in the 3' UTR of target mRNAs. By binding directly to the poly(A) tail. Through interaction with histone acetyltransferases.

What is the main purpose of an antibiotic resistance gene in a plasmid vector?. To increase the copy number of the plasmid inside the bacterial cell. To allow only bacteria carrying the plasmid to survive on the selective medium. To create compatible sticky ends for DNA insertion. To ensure plasmid compatibility.

A bacterial gene is toxic when overexpressed, but the researcher still needs to express it from a plasmid to study its function. Which vector design would best reduce the risk of toxicity-related artifacts?. A high-copy vector with a strong constitutive promoter to maximize expression. A low-copy vector combined with a tightly regulated inducible promoter. A vector containing only an MCS and no selectable marker. A shuttle vector with a broad host range but no control over transcription.

What is the specific function of the 5' to 3' exonuclease in the Gibson Assembly master mix during the isothermal reaction?. To fill in the single-stranded gaps after the fragments have successfully annealed. To seal the remaining nicks in the phosphodiester backbone of the newly assembled DNA. To chew back the 5' ends of linear DNA fragments to expose single-stranded overlapping overhangs. To degrade the methylated template plasmid contamination carried over from the PCR amplification.

A student sets up a multi-fragment Gibson Assembly but mistakenly skips the DpnI treatment after PCRamplifying the inserts from a template plasmid. After transforming the mixture into E. coli, what experimental outcome should they expect?. The 5' to 3' exonuclease will fail to chew back the inserts because the DNA ends remain heavily methylated. The high-fidelity polymerase will introduce excessive mutations at the 20–40 bp overlap junctions. A high background of false-positive colonies carrying the original, unmodified template plasmid instead of the newly assembled construct. The E. coli host machinery will fail to replicate the assembled plasmid because recA- and endAcloning strains require completely unmethylated DNA.

What is the primary purpose of genetic "complementation" in bacterial functional analysis?. To restore the wild-type phenotype in a mutant strain by introducing a wild-type gene copy, proving that the gene is sufficient to restore the wild-type phenotype, confirming its role in that function. To artificially increase the gene dosage using high-copy vectors to observe gain-of-function phenotypes or toxic effects. To systematically substitute specific amino acid residues with alanine to dissect protein structurefunction relationships. To introduce targeted nucleotide deletions using CRISPR-Cas9 to generate a starting loss-of-function mutant strain.

A bacterial mutant loses motility after deletion of gene X. Introducing gene X on a plasmid restores motility, but only when the gene is expressed from a weak promoter. Strong expression causes abnormal cell morphology. What is the best interpretation?. Gene X is unrelated to motility because only weak expression restores the phenotype. Gene X likely contributes to motility, but its expression level must be controlled to avoid overexpression artifacts. Complementation always requires the strongest possible promoter to be valid. The plasmid ori, not gene X, is responsible for restoring motility.

What is the purpose of heat shock in the LiAc/PEG transformation method of yeast?. Heat shock indirectly permeabilizes the cell, promoting the uptake of exogenous DNA, which may enter the cell through an endocytosis-like membrane invagination process. Heat shock neutralizes the negative charges of both DNA and the plasma membrane, allowing DNA entry. Heat shock promotes the function of certain specific membrane DNA transporters that import plasmid DNA into the cell. Yeasts cells don't need heat shock to uptake foreign DNA, they are highly competent naturally, although purification protocols may be needed to obtain a homogenous genotype. It induces replication of the transforming plasmid before it enters the yeast cell.

You are working in a research laboratory with a limited budget and no access to specialised equipment such as an electroporator. You need to introduce a plasmid carrying a gene of interest into E. coli K-12, a standard laboratory strain. Which transformation method is the most appropriate in this context?. Most laboratory strains are naturally competent, they have been engineered to contain the most refined machinery that allow foreign DNA uptake with ease. Use chemical competence with CaCl₂ and heat shock. Use electroporation, since it provides the highest transformation efficiency of all available methods. Use conjugation with an RP4 helper plasmid, since E. coli K-12 is known to be refractory to all artificial transformation methods and can only acquire foreign DNA through direct cell-to-cell contact.

What is the main difference between forward and reverse genetics in bacteria?. Forward genetics starts with a known gene, while reverse genetics starts with a phenotype. Forward genetics starts with a phenotype and identifies the responsible gene, while reverse genetics starts with a known gene and tests its function. Both forward and reverse genetics always use CRISPR interference. Reverse genetics is used only for random mutagenesis.

You are studying a gene that is suspected to be essential for bacterial growth. A classical deletion mutant cannot be recovered, suggesting that complete loss of the gene may be lethal. Which method would be the best choice to test the gene's function without permanently modifying the genomic locus?. Allelic exchange using a suicide vector, because it always works even when deletion of the gene is lethal. Transposon mutagenesis, because random insertion into an essential gene will produce many viable mutants. CRISPRi, because dCas9 guided by an sgRNA can repress transcription by blocking RNA polymerase without cutting DNA or permanently changing the genome. Tn-seq, because it generates a saturating transposon insertion library that can directly identify which genes are essential for growth.

Which transformation method in genetic engineering of filamentous fungi is commonly used for high efficiency transformation upon enzymatic degradation of fungal cell wall: Protoplast transformation. Agrobacterium-mediated transformation. Biolistic (Gene Gun). Non Homologous End Joining.

In protoplast transformation of filamentous fungi, PEG and CaCl₂ are mainly used to: initiate NHEJ. stabilize exogenous DNA. Enzymatically degrade fungal cell wall. facilitate uptake of exogenous DNA into protoplasts.

After transformation of a filamentous fungus, why must candidate transformants undergo multiple rounds of single-spore isolation before phenotypic analysis?. To increase the copy number of the integrated transgene. To ensure all nuclei carry the same genotype, producing a stable homokaryotic strain. To allow homologous recombination to complete integration of the deletion cassette. To amplify the selectable marker for Southern blot verification.

Why does genetic manipulation of essential genes in fungi require conditional silencing strategies instead of their classical deletion?. The physical removal of an essential gene from the genome induces direct cell lethality. Essential genes are protected by heterochromatin inaccessible to endonucleases. The full deletion of these genes activates defence mechanisms based on RNA interference. Auxotrophic markers are incompatible with genomic loci of vital functions.

Researchers often use "humanized" yeast strains to investigate human genes and disease-associated variants. Which of the following best describes the genetic modification involved in this approach?. Introducing or replacing yeast genes with functionally related human genes. Introducing mammalian artificial chromosomes into a yeast expression strain. Modifying yeast cells to produce fully human glycosylation patterns. Converting a haploid yeast strain into a diploid laboratory strain.

In the design of repair fragments for mutagenesis assays with the CRISPR-Cas9 system, why are silent mutations introduced into the target?. Disrupt the recognition of the PAM motif or the sequence hybridized by the sgRNA, preventing continuous re-cleavages after repair. Ensure that the final recombinant protein contains only high molecular weight amino acids. Facilitate thermal deletion of colonies carrying the desired mutation on selection plates. Eliminate the bacterial replication origin motifs contained in the base plasmid.

In a "one-step knockout" procedure in S. cerevisiae, what is the typical length of the homology arms required on a linear DNA fragment for successful targeted integration?. 30–50 base pairs. 1–5 base pairs. 500–1,000 base pairs. Over 1,000 base pairs.

A researcher uses a URA3-based counter-selection strategy to generate a markerless deletion of a target gene in yeast. Several colonies survive on 5-FOA medium, suggesting loss of URA3 function. However, PCR analysis reveals that the URA3 sequence is still present in some colonies. What is the most likely explanation?. A spontaneous mutation occurred in URA3 without excising the marker. The URA3 cassette was duplicated elsewhere in the genome before excision. The URA3 marker was successfully excised but remained detectable by PCR. The URA3 cassette remained functional after the counter-selection step.