Question 1

Vector stability in yeast — which is most genetically stable in vivo?  Among common Saccharomyces cerevisiae vector systems, which type is generally the most stable because it integrates into the host genome?

Accepted Answer

Introduction / Context: Different yeast vector classes balance copy number, stability, and ease of selection. Stability refers to faithful maintenance of genetic information over many cell divisions without selection pressure. Given Data / Assumptions: YIp vectors integrate into the yeast genome via homologous recombination. YEp vectors replicate episomally using 2μm sequences and can be lost without selection. YRp vectors with ARS can be unstable due to poor segregation. YCp vectors behave like mini-chromosomes but can be lost at low frequency. Concept / Approach: Integration into the host genome provides maximal genetic stability because the inserted sequence segregates with chromosomes. Episomal or ARS-based plasmids can be diluted or mis-segregated over time, particularly without selective pressure. Therefore, YIp is typically the most stable choice for long-term maintenance. Step-by-Step Solution: Define stability as retention over many generations without selection.Recognize integrative vectors (YIp) recombine into chromosomal DNA.Contrast with episomal (YEp) and ARS-based (YRp) plasmids.Select YIp as most stable. Verification / Alternative check: Yeast genetics manuals recommend YIp for gene replacement or stable integration, reflecting empirical stability compared with plasmid-based systems. Why Other Options Are Wrong: YEp/YRp may be lost without selection; YCp is fairly stable but still episomal; YACs are large and can undergo rearrangements. Common Pitfalls: Equating high copy number with stability; higher copy often increases instability, not the reverse. Final Answer: YIp (yeast integrative plasmid)

Question 2

SV40 early-region replacement vectors — required complement for growth  SV40-based vectors in which the early region is replaced by foreign DNA will replicate/express efficiently only if which support is provided?

Accepted Answer

Introduction / Context: SV40 vectors exploit the SV40 origin of replication and early/late gene organization. When the early region (coding for large T antigen) is replaced, the vector loses essential functions needed for replication and must be complemented. Given Data / Assumptions: SV40 early region encodes large T antigen required for initiating replication at the SV40 origin. COS cells constitutively express SV40 large T antigen. The question asks which support rescues vectors with early-region replacement. Concept / Approach: Large T antigen binds the SV40 ori, unwinds DNA, and recruits host replication machinery. Early-region replacement vectors therefore depend on a cell line that supplies large T in trans. COS cells are designed for precisely this purpose, enabling robust replication and expression of SV40-ori plasmids. Step-by-Step Solution: Acknowledge early region deletion removes large T antigen.Identify COS cells as a source of large T in trans.Conclude that COS cells are required for efficient vector replication/expression.Exclude helper viruses or unrelated plasmids that cannot provide large T in mammalian nuclei. Verification / Alternative check: Protocols for SV40 vectors consistently recommend COS-1/COS-7 cells to complement early-region defects, confirming this requirement. Why Other Options Are Wrong: Helper viruses providing late genes (option b) do not restore early functions; pBR322 (option c) is bacterial; baculovirus systems (option e) are insect-specific; early region is not dispensable (option d). Common Pitfalls: Assuming any SV40-related product suffices; only large T antigen complementation enables ori-driven replication. Final Answer: COS cell line supplying SV40 large T antigen in trans

Question 3

Properties of yeast integrative plasmids (YIp) — mechanism of transformation  Which statement correctly describes YIp vectors and how they establish themselves in Saccharomyces cerevisiae?

Accepted Answer

Introduction / Context: Yeast cloning tools come in several flavors—integrative (YIp), replicative (YRp), episomal (YEp), centromeric (YCp), and artificial chromosomes (YAC). Understanding each class is crucial for choosing the right vector for stability versus copy number. Given Data / Assumptions: YIp lacks ARS and 2μm replication elements. Integration relies on homologous sequences (for example, selectable markers or gene fragments) to recombine into the yeast genome. Stable maintenance follows chromosomal segregation. Concept / Approach: Because YIp vectors do not carry yeast replication origins, they cannot persist as free plasmids. Instead, linearized or appropriately targeted circular YIp constructs integrate into chromosomal loci via single or double crossover events, conferring high genetic stability. Step-by-Step Solution: Confirm lack of ARS/2μm sequences on YIp.Recognize that integration requires homologous regions for recombination.Note that integrated sequences segregate with chromosomes, ensuring stability.Select the statement describing integration by crossing over. Verification / Alternative check: Yeast transformation protocols for gene replacement employ YIp-based constructs to knock in or disrupt genes, demonstrating their integrative mechanism. Why Other Options Are Wrong: High-copy episomal maintenance (option b) describes YEp; ARS-based replication (option c) describes YRp; centromere/telomere artificial chromosome behavior (option d) describes YAC; option e is incorrect since YIp can transform yeast via recombination. Common Pitfalls: Assuming all plasmids carry ARS; YIp specifically does not, which is the defining feature. Final Answer: They integrate into the genome by homologous recombination (crossing over) rather than autonomous replication

Question 4

Yeast centromeric plasmids (YCp) — defining features  Which statements are correct about YCp vectors used in Saccharomyces cerevisiae?

Accepted Answer

Introduction / Context: YCp vectors were designed to mimic chromosomal behavior by including a yeast centromere (CEN) sequence. This design improves segregation fidelity compared with ARS-only plasmids and reduces copy number to near-chromosomal levels. Given Data / Assumptions: YCp includes a CEN element and typically an ARS for replication. CEN confers attachment to the mitotic spindle via kinetochores. Faithful segregation occurs in mitosis and, in many cases, meiosis. Concept / Approach: Adding a centromere allows the plasmid to assemble a kinetochore and be partitioned like a natural chromosome. As a result, YCp vectors behave as mini-chromosomes: low copy number, relatively stable, and more predictable inheritance than ARS-only plasmids. Step-by-Step Solution: Identify presence of centromere: true for YCp.Infer segregation behavior: kinetochore-mediated, high fidelity.Recognize mini-chromosome behavior as a functional description.Therefore, all listed statements are correct. Verification / Alternative check: Genetic studies show YCp plasmids segregate 1:1 under selection, supporting their chromosome-like behavior. Why Other Options Are Wrong: Options a–c are each accurate; rejecting them (option e) contradicts established vector biology. Common Pitfalls: Confusing YCp (CEN-containing) with YRp (ARS-only) or YEp (2μm-based), which differ in stability and copy number. Final Answer: All of the above

Question 5

Reducing insert size when introns are present — vector system choice  If a eukaryotic gene contains multiple introns, which vector system is most naturally aligned with using cDNA (introns removed) and thereby reduces the effective insert size for c…

Accepted Answer

Introduction / Context: Many eukaryotic genes are interrupted by introns, which increase genomic DNA size and complicate expression in systems that cannot process introns. A common strategy is to use complementary DNA (cDNA) synthesized from spliced mRNA, yielding intron-free coding sequences. Given Data / Assumptions: Retroviral replication involves reverse transcription of spliced viral RNAs into DNA. cDNA lacks introns, making inserts smaller and more suitable for expression in non-splicing hosts or for size-limited vectors. The question asks which platform aligns with using cDNA to reduce insert size. Concept / Approach: Retroviral vectors are typically built from cDNA, mirroring the biology of retroviruses that package RNA and reverse transcribe it. Using cDNA removes introns, compressing large genomic loci into compact ORFs that fit within vector size constraints and express correctly without host splicing. Step-by-Step Solution: Note that genomic DNA with introns is large; spliced mRNA produces smaller coding sequences.Retroviral vectors natively use cDNA, mirroring the RNA-to-DNA life cycle.Conclude that retrovirus-based systems effectively reduce insert size versus intron-bearing genomic clones.Exclude vectors that tolerate or expect large genomic fragments with introns. Verification / Alternative check: Standard gene therapy practice uses retroviral or lentiviral vectors with cDNA inserts to stay within packaging limits and ensure correct expression in target cells. Why Other Options Are Wrong: Vaccinia (option b) and baculovirus (option c) accommodate large inserts, often including introns; YACs (option e) expand size rather than reduce it; option d is false because cDNA cloning removes introns routinely. Common Pitfalls: Assuming intron-containing genomic clones are required for expression; many systems prefer intron-free cDNA for efficiency. Final Answer: Retrovirus-based vector (cDNA derived from spliced mRNA)

Question 6

Presence of ARS — which yeast vector lacks autonomous replication? Autonomously replicating sequence (ARS) elements enable plasmids to replicate independently in yeast. Which vector class does not carry an ARS and therefore integrates for …

Accepted Answer

Introduction / Context: ARS sequences in yeast function as replication origins. Vector classes are distinguished by whether they carry ARS (and other elements) or rely on chromosomal integration. Given Data / Assumptions: YRp: ARS-based replicative plasmids. YEp: 2μm-based episomal plasmids (functionally analogous to ARS-driven replication). YCp: usually include both CEN and ARS. YIp: lack ARS and integrate into chromosomes. Concept / Approach: Without an ARS (or 2μm origin), a plasmid cannot persist as a free replicon in yeast. YIp vectors thus use homologous recombination to integrate into chromosomal DNA and are inherited with the genome, providing high stability but low copy number. Step-by-Step Solution: List vector classes and their replication features.Identify that YIp lacks ARS.Conclude YIp integrates and does not replicate autonomously.Select YIp as the correct answer. Verification / Alternative check: Vector maps and manuals confirm absence of ARS in YIp, distinguishing them from YRp/YEp/YCp/YAC-based systems. Why Other Options Are Wrong: YRp, YEp, YCp, and YAC have replication elements; they do not require integration for maintenance. Common Pitfalls: Assuming all yeast plasmids are autonomously replicating; YIp is the exception by design. Final Answer: YIp

Question 7

Insert size capacity of SV40 vectors  What is the approximate maximum length of foreign DNA that can be accommodated by SV40-based vectors while still allowing proper packaging and propagation?

Accepted Answer

Introduction / Context: Vector capacity determines the size of transgenes that can be cloned and packaged. SV40 is a small DNA virus (~5.2 kb genome), and its vectors are constrained by strict capsid packaging limits. Given Data / Assumptions: SV40 particle packaging tolerates genomes close to wild-type size. Replacing nonessential regions allows insertion of foreign DNA, but only up to a limited size. Oversized genomes are inefficiently packaged or not packaged at all. Concept / Approach: Because SV40 capsids package DNA of approximately the native genome length, only a fraction of the genome can be replaced by foreign sequences. In practice, about 2.5 kb of heterologous DNA is the commonly cited upper limit for stable packaging and propagation of recombinant SV40 vectors. Step-by-Step Solution: Start with wild-type SV40 genome size (~5.2 kb).Recognize only nonessential segments can be replaced.Account for packaging constraints around native size.Select ~2.5 kb as the practical maximum insert size. Verification / Alternative check: Method papers and vector guides consistently note a ~2–2.5 kb limit; larger inserts lead to low-efficiency packaging and instability. Why Other Options Are Wrong: 1 kb (option d) is overly conservative; 5–10–20 kb (options b, c, e) exceed the SV40 capacity and are more in line with adenoviral/HSV/baculoviral systems. Common Pitfalls: Confusing SV40 capacity with that of adenovirus (~7–8 kb inserts in first-generation vectors) or baculovirus (much larger). Final Answer: 2.5 kb

Question 8

Yeast gene regulation — Which promoter is specifically repressed by the presence of glucose via catabolite repression (consider standard Saccharomyces cerevisiae laboratory conditions)?

Accepted Answer

Introduction / Context: Glucose repression (also called catabolite repression) is a classic regulatory circuit in yeast. When glucose is abundant, Saccharomyces cerevisiae downregulates genes required to metabolize alternative carbon sources such as galactose. Identifying which promoters are glucose-repressed versus phosphate- or growth-driven is essential in designing expression systems and interpreting transcriptional data in molecular biology labs. Given Data / Assumptions: Organism of interest: Saccharomyces cerevisiae. Glucose present in the medium triggers catabolite repression via Mig1/Snf1 pathways. GAL system (GAL1, GAL10, GAL7) is the canonical glucose-repressed, galactose-inducible system. PHO system (PHO5/PHOS) is primarily phosphate regulated, not glucose dependent. PGK1 and ADH1 are robust glycolytic/housekeeping promoters active in glucose media. Concept / Approach: In glucose, the regulator Mig1 localizes to the nucleus and represses GAL promoters, preventing unnecessary expression of galactose-utilization genes. PHO5 responds to inorganic phosphate scarcity; its induction requires low phosphate rather than glucose absence. PGK1 and ADH1 are typically active in glucose-rich conditions, supporting high glycolytic flux and fermentation. Step-by-Step Solution: Map each listed promoter to its primary regulatory input.Recognize GAL1 as strongly repressed by glucose and induced by galactose (with functional Gal4/Gal80 regulation and Snf1 relief).Note PHO5 is low-phosphate inducible; PGK1 and ADH1 are generally strong in glucose and are not glucose-repressed.Therefore select GAL1 as the glucose-repressed promoter. Verification / Alternative check: Yeast expression vectors often use GAL1 for tightly controlled, off-in-glucose/on-in-galactose induction, confirming glucose repression in practice. Why Other Options Are Wrong: b: PHO5 depends mainly on phosphate starvation, not on glucose levels. c: Both is incorrect because PHO5 is not characteristically glucose-repressed. d: PGK1 is highly active in glucose; it is not repressed by it. e: ADH1 is a strong constitutive promoter in glucose media. Common Pitfalls: Assuming any “inducible” promoter is also glucose-repressed; regulatory signals are pathway-specific (carbon vs. phosphate regulation). Final Answer: Galactokinase promoter (GAL1)

Question 9

Baculovirus expression system — Where is a foreign gene positioned to achieve very high-level expression in insect cells?

Accepted Answer

Introduction / Context: Baculovirus expression vectors (BEVs), especially those based on Autographa californica multiple nucleopolyhedrovirus (AcMNPV), are widely used to produce recombinant proteins in insect cells. The hallmark of the system is the use of extremely strong late promoters to drive high-level expression of a heterologous gene. Given Data / Assumptions: Polyhedrin (polh) and p10 are strong late promoters in AcMNPV. Foreign open reading frames can replace native structural genes to co-opt the strong promoter. High expression requires correct promoter–ORF orientation and proper transcription/termination elements. Concept / Approach: Replacing the polyhedrin coding region with the gene of interest positions the heterologous gene immediately downstream of the polyhedrin promoter. This harnesses the natural late-phase transcriptional surge, often yielding very high protein levels suitable for structural biology and vaccine production. Step-by-Step Solution: Identify which location confers the strongest expression: immediately downstream of polh.Exclude placements that lack promoter control (e.g., mere proximity to replication origins).Avoid random insertion which risks poor expression or gene disruption.Select the classical configuration: foreign ORF replacing polh beneath the polh promoter. Verification / Alternative check: Commercial BEV kits and protocols map insertion at the polyhedrin locus; titers and expression levels confirm this strategy. Why Other Options Are Wrong: b: Origins of replication do not drive transcription. c: Random insertion leads to unpredictable expression. d/e: Suboptimal or incorrect promoter context reduces expression dramatically. Common Pitfalls: Confusing replication control with transcriptional control; neglecting late-promoter timing for maximal yield. Final Answer: Inserted downstream of the strong polyhedrin promoter (replacing the polyhedrin coding region)

Question 10

Bovine papillomavirus (BPV) vector basics — Approximately what is the size (kb) of the circular, double-stranded DNA genome of BPV?

Accepted Answer

Introduction / Context: Bovine papillomavirus (BPV) has been employed as a eukaryotic cloning and expression vector because of its ability to replicate episomally in certain animal cells. Knowing the approximate genome size helps in understanding vector capacity and engineering strategies. Given Data / Assumptions: BPV genome is circular and double-stranded DNA. Standard reference size is about 7.9 to 8.0 kb. Vector derivatives often incorporate portions of the BPV genome to maintain replication and maintenance functions. Concept / Approach: Recall canonical genome sizes: SV40 (~5.2 kb), BPV (~7.9 kb), many retroviral genomes (~9–10 kb). Matching these benchmarks allows rapid identification in exam settings and practical vector selection in lab work. Step-by-Step Solution: List plausible genome sizes for small DNA tumor viruses.Associate SV40 with ~5.2 kb, BPV with ~7.9 kb.Choose 7.9 kb as the correct size for BPV. Verification / Alternative check: Vector maps and original genomic characterizations report ~7.9 kb for BPV, consistent across literature and plasmid repositories. Why Other Options Are Wrong: a: 2.5 kb is too small for a complete papillomavirus genome. b: 5.2 kb corresponds to SV40, not BPV. d/e: Larger than canonical BPV genome. Common Pitfalls: Confusing BPV with SV40 due to both being circular DNA viruses used as vectors. Final Answer: 7.9 kb

Question 11

Animal cell gene delivery — Which viruses are commonly used as vectors for transferring genes into animal cells?

Accepted Answer

Introduction / Context: Viral vectors exploit natural infection and replication pathways to deliver genetic cargo into animal cells. Different vectors provide distinct advantages in host range, genome integration, packaging capacity, and expression persistence. Given Data / Assumptions: SV40 is a well-studied small DNA tumor virus used historically for transient expression. BPV-based vectors can replicate episomally in certain mammalian cells. Retroviruses (including lentiviruses) integrate into host genomes, enabling stable expression in dividing (and, for lentivirus, non-dividing) cells. Concept / Approach: Select the option that correctly lists legitimate animal cell vectors. Each listed virus has documented use as a vector system. Therefore the inclusive answer is correct. Step-by-Step Solution: Verify each listed virus has a vector system.Recognize that many labs have used all three in different contexts (transient, episomal, integrating).Choose the comprehensive option. Verification / Alternative check: Textbooks and vector supplier catalogs enumerate SV40-, BPV-, and retroviral-based vectors for animal transduction. Why Other Options Are Wrong: a/b/c individually are incomplete; the question asks which viruses are used, and all listed are valid. e is false because several are indeed used. Common Pitfalls: Assuming only integrating vectors count; episomal and transient systems are widely used in protein production and functional assays. Final Answer: All of these

Question 12

Vector taxonomy — Moloney murine leukemia virus (MoMLV / Moloney–Mu–leukemia virus) is best classified as which type of virus?

Accepted Answer

Introduction / Context: MoMLV is a prototypical gamma-retrovirus widely used in early gene therapy vectors. Understanding its classification informs properties such as integration into host DNA and the requirement for cell division during transduction. Given Data / Assumptions: MoMLV is an RNA virus that uses reverse transcriptase. Its life cycle includes conversion of RNA to DNA and integration into the host genome (provirus). Classically transduces dividing cells efficiently. Concept / Approach: Retroviruses are defined by their reverse transcription step and proviral integration. Baculoviruses infect insects; vaccinia is a large DNA poxvirus; adenoviruses are non-enveloped DNA viruses; herpesviruses are enveloped dsDNA viruses with latent replication—none match MoMLV’s key traits. Step-by-Step Solution: Identify MoMLV hallmark: reverse transcription and integration.Map this hallmark to the retrovirus family.Select “Retrovirus.” Verification / Alternative check: Retroviral packaging systems (gag/pol/env) are derived from MoMLV, confirming its class. Why Other Options Are Wrong: a/b/d/e: These are different DNA virus families without obligatory reverse transcription integration cycles. Common Pitfalls: Confusing lentiviruses (a retrovirus subgroup that infects non-dividing cells) with oncoretroviruses like MoMLV that require cell division for nuclear entry. Final Answer: Retrovirus

Question 13

Advantages of cloning in yeast (S. cerevisiae) — Which statement(s) describe recognized benefits of using yeast as a eukaryotic host?

Accepted Answer

Introduction / Context: Yeast bridges bacterial ease and eukaryotic processing, making it a prime platform for biotechnology. It supports industrial-scale production, eukaryotic protein maturation, and large-fragment cloning, all of which are crucial in modern genomics and biopharma. Given Data / Assumptions: S. cerevisiae and related yeasts are GRAS (generally recognized as safe) organisms in many applications. Secretory pathways in yeast can glycosylate and fold proteins. Yeast artificial chromosomes (YACs) can carry megabase-sized inserts. Concept / Approach: Match each benefit to known yeast capabilities. Yeast can overexpress proteins using strong promoters, carry out N-linked glycosylation, and maintain large genomic inserts with YAC technology. Hence the inclusive option is correct. Step-by-Step Solution: Confirm protein overproduction with inducible/constitutive promoters (e.g., GAL1, PGK1, ADH1).Acknowledge secretory signal peptides enabling glycosylation and export.Recognize YACs for large DNA cloning compared with plasmids/cosmids. Verification / Alternative check: Industrial enzyme and vaccine subunit production in yeast validates options a and b; genome projects historically used YACs for physical mapping (option c). Why Other Options Are Wrong: e: Incorrect because multiple advantages are real and documented. Common Pitfalls: Assuming yeast glycosylation is identical to mammalian; patterns differ but still provide key eukaryotic modifications. Final Answer: All of the above

Question 14

Yeast selectable markers — Growth is specifically inhibited by α-aminoadipate when which marker gene is present (i.e., functional)?

Accepted Answer

Introduction / Context: Genetic selections in yeast exploit toxic analogs that become lethal only in the presence of specific metabolic functions. α-Aminoadipate (α-AA) is a classic reagent used in LYS2-based selections to enrich for particular recombinants or counterselect for marker presence. Given Data / Assumptions: LYS2 encodes an enzyme in the lysine biosynthetic pathway (α-aminoadipate pathway). α-Aminoadipate can be toxic to LYS2+ cells under selection conditions. URA3, HIS3, TRP1 are unrelated markers with different counterselections (e.g., 5-FOA for URA3). Concept / Approach: Counterselection relies on a toxic intermediate produced or processed when the corresponding gene is active. In α-AA selection, cells with functional LYS2 are sensitive; loss-of-function lys2 mutants are resistant and can grow, allowing targeted genetic manipulations. Step-by-Step Solution: Identify which marker interacts with α-AA toxicity: LYS2.Relate others to their known systems (URA3 ↔ 5-FOA; HIS3, TRP1 ↔ amino acid prototrophy).Select LYS2 as the correct answer. Verification / Alternative check: Protocols for lys2 selection routinely specify α-aminoadipate plates to enrich lys2 mutants. Why Other Options Are Wrong: a/d/e: Not the cognate marker for α-AA toxicity. c: URA3 is counterselected with 5-FOA, not α-AA. Common Pitfalls: Confusing different counterselectable markers; memorize the specific analog: 5-FOA ↔ URA3; α-AA ↔ LYS2. Final Answer: LYS2

Question 15

Yeast counterselection — 5-Fluoroorotic acid (5-FOA) inhibits growth if which selectable marker gene is present (functional) in yeast?

Accepted Answer

Introduction / Context: 5-Fluoroorotic acid (5-FOA) is a cornerstone reagent for yeast genetics. It enables efficient counterselection against URA3+ cells, facilitating marker recycling and precise genome engineering via homologous recombination. Given Data / Assumptions: URA3 encodes orotidine-5′-phosphate decarboxylase in uracil biosynthesis. URA3 converts 5-FOA to a toxic product (5-fluorouracil), killing URA3+ cells. URA3- cells survive on 5-FOA medium when supplemented with uracil, enabling selection for URA3 loss. Concept / Approach: Counterselectable markers rely on gene products that metabolize harmless analogs into toxins. URA3’s activity on 5-FOA is the canonical example used to pop out integrated cassettes or select against background URA3 activity after cassette shuffling. Step-by-Step Solution: Match 5-FOA toxicity to URA3 enzymatic conversion.Eliminate HIS3, LYS2, TRP1, LEU2 since they lack this conversion.Choose URA3 as the correct marker for 5-FOA counterselection. Verification / Alternative check: Standard gene deletion cassettes (e.g., URA3 pop-in/pop-out) leverage 5-FOA plates to isolate URA3- recombinants. Why Other Options Are Wrong: a/b/d/e: These markers do not render cells sensitive to 5-FOA. Common Pitfalls: Forgetting that URA3+ is lethal on 5-FOA unless uracil synthesis is bypassed; always supplement appropriately during selections. Final Answer: URA3

Vectors Uses for Animal Cell Culture Questions