Question 1

Restriction Digests—Nature of Overhangs The unpaired single-stranded nucleotides left at DNA termini after certain restriction enzyme cuts are referred to as:

Accepted Answer

Introduction / Context: Restriction enzymes can generate either blunt ends or single-stranded overhangs. Overhangs provide complementarity that facilitates specific annealing and efficient ligation in recombinant DNA techniques. Given Data / Assumptions: Some enzymes (EcoRI) produce 5' overhangs; others (KpnI) produce 3' overhangs. Overhangs enable base-pairing with compatible ends. Terminology distinguishes overhangs (“sticky ends”) from fully double-stranded “blunt ends.” Concept / Approach: “Sticky ends” describes cohesive termini with unpaired nucleotides that can hydrogen-bond to complementary sequences. This property is exploited to directional-clone inserts into vectors cut with compatible enzymes. Step-by-Step Solution: Identify that the question refers to unpaired nucleotides at ends.Recall the standard term: sticky (cohesive) ends.Exclude unrelated terms (ligase is an enzyme; restriction fragments are the cut pieces regardless of end type). Verification / Alternative check: Cloning manuals define sticky ends as overhangs that anneal via base pairing prior to ligase sealing of the backbone. Why Other Options Are Wrong: Single strands: too generic; does not denote cohesive ends specifically. Restriction fragments: includes blunt- and sticky-ended pieces. Ligases: enzymes that seal nicks; not DNA ends. Polylinkers: engineered multiple-cloning sites in vectors. Common Pitfalls: Confusing overhang type (5' vs 3') with the general property “sticky.” Final Answer: Sticky ends

Question 2

Human Genetics—RP13 Gene on Chromosome 17 The RP13 gene located on chromosome 17 encodes a protein primarily:

Accepted Answer

Introduction / Context: Retinitis pigmentosa (RP) comprises a group of inherited retinal degenerations. Loci are often designated RP followed by a number (e.g., RP13), and map to specific chromosomal regions. Genes at these loci typically encode proteins required for retinal structure or phototransduction pathways, directly affecting eye development and function. Given Data / Assumptions: RP locus numbering reflects discovery order and mapping. Chromosome 17 contains multiple genes implicated in ocular biology. RP-associated genes affect photoreceptors or retinal pigment epithelium. Concept / Approach: Interpret the RP nomenclature: it refers to retinitis pigmentosa, a disorder involving retinal degeneration. Therefore, the encoded proteins are tied to eye development or retinal maintenance rather than generalized transporters, keratins, personality determinants, or coagulation factors. Step-by-Step Solution: Recognize RP13 as an eye-related genetic locus.Connect locus function to retinal development/maintenance pathways.Select “Involved in eye development.” Verification / Alternative check: Genetic databases and ophthalmic literature align RP loci with ocular phenotypes, supporting the eye-development association. Why Other Options Are Wrong: Glucose transport and coagulation: unrelated to RP loci designation. Hair/nail structural proteins (keratins): different tissue specialization. “Personality traits”: not a recognized single-gene protein effect. Common Pitfalls: Misreading RP as “ribosomal protein” rather than retinitis pigmentosa; overinterpreting chromosome location without phenotype context. Final Answer: Involved in eye development

Question 3

cDNA Library Construction—Order of Steps from mRNA Choose the correct sequential order for constructing a double-stranded cDNA from a eukaryotic mRNA template using oligo-dT priming and tailing methods.

Accepted Answer

Introduction / Context: Complementary DNA (cDNA) synthesis converts processed eukaryotic mRNA into DNA for cloning and expression. A classical method uses an oligo-dT primer to anneal to the mRNA poly(A) tail, followed by first-strand synthesis, RNA removal or nicking, and tailing to prime second-strand synthesis. Correct step order ensures successful, full-length cDNA generation. Given Data / Assumptions: mRNA has a poly(A) tail at its 3' end. Reverse transcriptase synthesizes first-strand cDNA using oligo-dT primer. RNase (often RNase H) partially degrades RNA in RNA–DNA hybrids. Terminal transferase can add homopolymeric dG tails; oligo-dC can then prime the second strand for DNA polymerase. Concept / Approach: Map the workflow: prime first strand with oligo-dT → synthesize cDNA with reverse transcriptase → remove RNA template → tail the cDNA with dG → anneal oligo-dC → extend second strand with DNA polymerase. This produces double-stranded cDNA ready for cloning adapters or linkers. Step-by-Step Solution: Step 1: Bind oligo-dT to the poly(A) tail.Step 2: Use reverse transcriptase to synthesize first-strand cDNA.Step 3: Digest the RNA in the hybrid with RNase H.Step 4: Add dG tails to the 3' end of cDNA with terminal transferase.Step 5: Anneal oligo-dC to the dG tail.Step 6: Extend with DNA polymerase to make the second strand. Verification / Alternative check: Classical cDNA library protocols and kit manuals mirror this sequence, with variations (e.g., second-strand synthesis by RNase H/DNA Pol I method) but the listed tailing order remains valid for homopolymer tailing approaches. Why Other Options Are Wrong: Starting without oligo-dT or misordering tailing and reverse transcription disrupts priming. Adding 5' tails or ligase at early steps is not part of this method. Binding oligo-dC before first-strand synthesis is illogical since no dG tail exists yet. Common Pitfalls: Confusing random-primed cDNA approaches with oligo-dT tailing workflows; mixing up dA/dG tailing directions. Final Answer: Bind oligo-dT, treat with reverse transcriptase, digest with RNase, add G residues to the 3' end, bind oligo-dC, treat with DNA polymerase

Question 4

Laboratory methods: DNA agarose gel electrophoresis and protein SDS–PAGE are similar in which fundamental ways during biomolecule separation by gels?

Accepted Answer

Introduction: Agarose gel electrophoresis for DNA and SDS–PAGE for proteins are cornerstone techniques for separating biomolecules by size. Despite analyzing different macromolecules with different gel matrices, they share several core physical principles. This question asks you to identify the commonalities that make these methods conceptually similar in the lab. Given Data / Assumptions: An applied electric field drives migration of charged macromolecules through a hydrated gel matrix. DNA carries a nearly uniform negative charge per unit mass due to the phosphate backbone. In SDS–PAGE, sodium dodecyl sulfate coats proteins, conferring a near-constant negative charge-to-mass ratio. Gels (agarose or polyacrylamide) act as molecular sieves that hinder larger molecules more than smaller ones. Concept / Approach: Both assays separate primarily by size because the charge-to-mass ratio is roughly constant across species being compared. DNA is inherently uniformly charged; proteins become uniformly charged after SDS binding and denaturation. Under standard electrophoresis setups, negatively charged samples migrate toward the positively charged electrode (the anode). The hydrated gel network provides steric hindrance, so smaller molecules negotiate the pores more readily and run farther for a given time and voltage. Step-by-Step Solution: Recognize that an electric field is applied in both assays to move charged molecules.Note the charge normalization: DNA by nature, proteins via SDS coating and denaturation.Acknowledge the sieving role of agarose (for nucleic acids) and polyacrylamide (for proteins).Infer that mobility correlates inversely with size since charge per mass is similar. Verification / Alternative check: Marker ladders (DNA size standards or protein prestained ladders) resolve into predictable bands whose migration is logarithmically related to size under constant buffer, gel concentration, and voltage—consistent with sieving plus constant charge-to-mass assumptions. Why Other Options Are Wrong: Claiming no electric field is used is incorrect; electrophoresis literally requires an electric field. Suggesting only one of the features is shared ignores that all listed similarities apply under standard conditions. Common Pitfalls: Confusing native PAGE (variable charge) with SDS–PAGE (normalized charge), or assuming DNA might migrate to the cathode; in typical TAE/TBE systems, DNA runs to the anode because it is negatively charged. Final Answer: All of the above

Question 5

Mutagenesis and radiation biology: X-rays primarily cause which cellular effect that initiates DNA damage indirectly?

Accepted Answer

Introduction: Different types of radiation damage biomolecules via distinct mechanisms. Understanding which lesions are characteristic of X-rays versus ultraviolet (UV) light is essential for interpreting mutagenesis and DNA repair pathways. This question asks you to identify the hallmark primary effect of X-rays in cells. Given Data / Assumptions: X-rays are ionizing radiation with sufficient energy to remove electrons from molecules. Cells are mostly water; radiolysis of water yields reactive oxygen species (ROS). DNA damage can be direct (energy deposition in DNA) or indirect (ROS-mediated). Concept / Approach: In aqueous biological systems, the predominant effect of X-rays is ionization of water, generating hydroxyl radicals, hydrogen radicals, and other ROS. These highly reactive species attack DNA, creating single- and double-strand breaks, base modifications, and abasic sites. In contrast, UV (especially UV-B/UV-C) directly induces photochemical lesions such as cyclobutane thymine dimers and 6-4 photoproducts. Thermal heating is generally minimal at diagnostic or experimental doses of X-rays and is not the proximate cause of mutagenesis. Step-by-Step Solution: Identify X-rays as ionizing radiation capable of ejecting electrons.Acknowledge that water is the principal target; radiolysis forms ROS.ROS diffuse short distances and attack DNA, proteins, and lipids.Resulting DNA damage triggers repair pathways (BER, NHEJ, HR) and, if unrepaired, mutations. Verification / Alternative check: Scavengers of hydroxyl radicals (e.g., DMSO) reduce X-ray–induced DNA breaks in vitro, confirming an indirect ROS-mediated pathway. Conversely, UV-induced thymine dimers are repaired by nucleotide excision repair and photolyases—different lesion fingerprints. Why Other Options Are Wrong: Thymine dimers: characteristic of UV, not X-rays. Heat: not the primary mutagenic mechanism at typical X-ray exposures. Immediate crosslinking without intermediates: can occur but is not the principal pathway for X-ray damage. Common Pitfalls: Conflating UV photochemistry with ionizing radiation chemistry or assuming all radiation causes identical DNA lesions. Final Answer: Ionization of water within the cell producing reactive radicals

Question 6

Population genetics (Hardy–Weinberg): Under which condition below are allele (gene) frequencies most likely to remain unchanged across generations?

Accepted Answer

Introduction: The Hardy–Weinberg principle outlines conditions under which allele and genotype frequencies remain constant from generation to generation. Recognizing which forces disturb equilibrium helps interpret real population data and design genetic studies. Given Data / Assumptions: Equilibrium requires: random mating, infinitely large population, no selection, no mutation, and no migration. Any violation of these assumptions can shift allele frequencies. The options listed represent common departures from equilibrium. Concept / Approach: Random mating (panmixia) is one of the necessary conditions for Hardy–Weinberg equilibrium. While random mating alone does not guarantee equilibrium if other assumptions are violated, it is the only option among those given that is compatible with stability of allele frequencies. The other options explicitly introduce forces (selection, drift via small size, migration, mutation) that change allele frequencies. Step-by-Step Solution: List Hardy–Weinberg assumptions: no selection, no mutation, no migration, large population, random mating.Evaluate each option for conformity or violation.Identify random mating as consistent with equilibrium, given other assumptions hold.Conclude that random mating best fits the condition for unchanging allele frequencies among the choices. Verification / Alternative check: Under random mating with all other assumptions satisfied, genotype frequencies stabilize at p^2, 2pq, q^2, and allele frequencies p and q remain constant across generations, confirming the principle. Why Other Options Are Wrong: Selection for homozygotes shifts allele frequencies toward favored alleles. Small population size increases genetic drift, causing random fluctuations and fixation/loss. Active migration introduces new alleles and changes frequencies. Directional mutation changes allele proportions over time. Common Pitfalls: Thinking that random mating alone ensures equilibrium without considering the other assumptions; equilibrium is multifactorial, but among the listed options, it is the only compatible condition. Final Answer: Random mating (panmixia) with other equilibrium assumptions

Question 7

When population size is small, allele frequencies can fluctuate between generations and some alleles may become fixed or lost. This stochastic process is called __________.

Accepted Answer

Introduction: Not all evolutionary change is driven by natural selection. In small populations, random sampling effects can strongly influence allele frequencies, even in the absence of fitness differences. This question asks you to name that stochastic force—genetic drift. Given Data / Assumptions: Population size is small, so sampling error is large. No explicit selective advantage is assumed for the alleles discussed. Fixation (frequency = 1) or loss (frequency = 0) can occur by chance. Concept / Approach: Genetic drift refers to random fluctuations in allele frequencies due to chance events in reproduction and survival. The magnitude of drift is inversely related to effective population size. In extreme cases (founder effects, bottlenecks), drift can rapidly fix alleles, reduce genetic diversity, and increase the influence of deleterious alleles, independent of selection. Step-by-Step Solution: Recognize small population size as the key driver of strong stochastic effects.Connect random sampling of gametes/offspring to shifts in allele frequencies.Acknowledge that repeated random sampling can cause fixation or loss across generations.Identify the term for this phenomenon: genetic drift. Verification / Alternative check: Wright–Fisher and Moran models mathematically show variance in allele frequency per generation proportional to p*(1−p)/(2N) or related forms, demonstrating stronger drift at smaller N and eventual fixation probabilities equal to initial allele frequency. Why Other Options Are Wrong: Assortative mating: nonrandom mating pattern affecting genotype frequencies, not inherently causing fixation by chance. Inbreeding: increases homozygosity but does not by itself randomly change allele frequencies. Heterosis: increased fitness of heterozygotes; not a stochastic force. Balancing selection: maintains polymorphism via fitness differences, opposite to drift-driven fixation. Common Pitfalls: Equating increased homozygosity (from inbreeding) with drift; while often correlated in small isolated groups, the mechanisms are distinct. Final Answer: Genetic drift

Question 8

Recombinant DNA technology: Which enzyme catalyzes formation of covalent phosphodiester bonds to join compatible restriction fragments?

Accepted Answer

Introduction: After restriction enzymes generate compatible DNA ends, the fragments must be covalently joined to create a continuous backbone. The enzyme responsible for sealing the nick between adjacent nucleotides is a mainstay of cloning workflows. This question asks you to identify that enzyme—DNA ligase. Given Data / Assumptions: Restriction digestion produces either sticky (cohesive) or blunt ends. Annealing of sticky ends relies on base pairing but remains a nicked backbone until sealed. Cloning requires stable covalent joins for propagation in cells. Concept / Approach: DNA ligase catalyzes formation of a phosphodiester bond between adjacent 3′-OH and 5′-phosphate termini. T4 DNA ligase uses ATP, whereas E. coli DNA ligase uses NAD+. Ligation can join both cohesive and blunt ends (the latter typically at lower efficiency), enabling insert–vector assembly in plasmids and other constructs. Step-by-Step Solution: Generate compatible ends via restriction digestion or end-repair.Allow complementary overhangs to anneal if present.Add DNA ligase with appropriate buffer and cofactors (ATP or NAD+).Ligase seals the nicks, creating a continuous phosphodiester backbone. Verification / Alternative check: Successful ligation yields transformants with intact plasmids; control reactions lacking ligase fail to produce stable constructs, demonstrating the necessity of covalent sealing. Why Other Options Are Wrong: Primase synthesizes RNA primers; helicase unwinds duplexes; polymerase extends DNA but does not seal nicks; topoisomerase alters supercoiling and can transiently break/rejoin strands but is not used for cloning ligation. Common Pitfalls: Assuming annealed sticky ends are permanently joined without ligase; base pairing alone is insufficient for stable propagation. Final Answer: DNA ligase

Question 9

Restriction endonucleases: Which statements accurately describe what these enzymes do in bacteria and in vitro?

Accepted Answer

Introduction: Restriction endonucleases recognize specific DNA sequences and cleave phosphodiester bonds, forming the basis for many cloning and mapping techniques. In bacteria, they function as part of restriction–modification systems to defend against phage. This question evaluates your understanding of their biological role and cutting patterns. Given Data / Assumptions: Type II restriction enzymes recognize palindromic sites and cut at or near those sites. Cleavage can generate either cohesive (overhang) or blunt ends. Host DNA is typically protected by methylation at recognition sites. Concept / Approach: Restriction enzymes curtail phage infection by degrading unmodified foreign DNA while sparing methylated host DNA. In vitro, they are chosen for the type of ends they generate: staggered cuts with 5′ or 3′ overhangs facilitate directional cloning; blunt cutters are useful for end-agnostic ligations though often less efficient. Many different enzymes exist, each with defined recognition sites and cut patterns. Step-by-Step Solution: Identify the protective role in bacteria via restriction–modification systems.Differentiate cut types: sticky versus blunt ends.Connect cut type to downstream ligation strategies in cloning.Conclude that all three statements accurately describe restriction enzymes. Verification / Alternative check: Enzyme datasheets (e.g., EcoRI, HindIII for sticky ends; EcoRV, SmaI for blunt ends) explicitly state overhangs versus blunt termini, confirming diverse cut modes and uses. Why Other Options Are Wrong: “Only nick one strand” is false for standard restriction endonucleases; specialized nicking variants exist but classic enzymes cleave both strands to generate double-strand breaks. Common Pitfalls: Assuming all enzymes make sticky ends; selection of enzyme determines end type and cloning strategy. Final Answer: All of the above

Question 10

Small, inbred populations: Which statement is NOT generally true about the genetic consequences of inbreeding in small populations?

Accepted Answer

Introduction: Small, inbred populations experience characteristic genetic changes. Distinguishing which outcomes are driven by inbreeding and drift versus those that reflect altered mutation processes is important in conservation genetics and medical genetics. This question asks you to identify the statement that is not generally true. Given Data / Assumptions: Inbreeding increases the probability that alleles are identical by descent. Small population size magnifies genetic drift. Mutation rate is primarily a molecular property and not directly altered by mating structure. Concept / Approach: Inbreeding raises homozygosity, exposing deleterious recessive alleles and increasing recessive disease incidence. In small populations, drift can randomly fix or lose alleles, reducing overall genetic diversity. However, the per-base mutation rate is not inherently increased by inbreeding; new mutations arise largely independently of mating patterns, although their fate in the population is influenced by drift and selection. Step-by-Step Solution: Link inbreeding to homozygosity: F (inbreeding coefficient) > 0 increases homozygote probability.Relate small N to drift: increased variance in allele frequency changes, leading to fixation/loss.Identify disease risk: recessive conditions appear more often due to homozygosity.Recognize mutation rate as largely unaffected by inbreeding per se. Verification / Alternative check: Pedigree analyses and population-genetic models show increased homozygosity (and disease) with inbreeding without necessitating any change to the underlying mutation rate μ; instead, μ influences the supply of new alleles, not their immediate frequency under inbreeding. Why Other Options Are Wrong: Loss of diversity and allele fixation: classic outcomes of drift in small populations. Increased recessive disease: consequence of homozygosity revealing deleterious alleles. Elevated homozygosity: the defining genetic signature of inbreeding. Common Pitfalls: Attributing increased disease to a supposed rise in mutation rate; the driver is exposure of existing recessive alleles through homozygosity. Final Answer: Mutation rate is increased as a direct result of inbreeding

Question 11

Plasmid fundamentals: Which statements correctly describe typical plasmids used in molecular cloning?

Accepted Answer

Introduction: Plasmids are the workhorses of recombinant DNA technology. Understanding their key features explains how they replicate, how inserts are cloned, and how selection is enforced in bacterial hosts. This question tests recognition of the essential components and properties of standard plasmid vectors. Given Data / Assumptions: Cloning plasmids are typically small, circular double-stranded DNA molecules. They carry an origin of replication (ori) that determines copy number and host range. They include selectable markers and a multiple cloning site (MCS) with unique restriction sites. Concept / Approach: The circular topology makes plasmids stable and easy to propagate. The ori is indispensable for autonomous replication; without it, a plasmid cannot be maintained in dividing cells. The MCS provides flexibility in cloning by presenting multiple enzyme sites that are unique within the vector, facilitating directional insertion and screening. Additional features (promoters, tags) may be present depending on expression needs. Step-by-Step Solution: Confirm plasmid nature: circular dsDNA, extrachromosomal.Identify the ori as mandatory for replication and copy control.Recognize the presence of unique restriction sites in an MCS for cloning.Conclude that all listed statements are accurate for standard cloning plasmids. Verification / Alternative check: Popular vectors (e.g., pUC, pBR, pET, pGEM) all include an ori, a selectable marker, and an MCS; plasmid maps illustrate these conserved features. Why Other Options Are Wrong: Linear RNA molecules replicating without an origin: describes neither plasmids nor standard cloning vectors. Common Pitfalls: Confusing viral vectors (which may be linear DNA or RNA) with plasmids, or overlooking that some specialized plasmids can be linear yet still carry an ori; however, standard cloning plasmids are circular DNA with a defined ori. Final Answer: All of the above

Question 12

Gene targeting in mice: Knockout mice (loss-of-function for a chosen gene) are most commonly generated by which approach?

Accepted Answer

Introduction: Knockout mice enable precise functional studies by disrupting specific genes. The classic pre-CRISPR approach uses homologous recombination in embryonic stem (ES) cells to replace or disrupt the target locus. This question asks you to pick the method that defines traditional knockout generation. Given Data / Assumptions: ES cells can contribute to the germline when introduced into blastocysts. Gene targeting constructs contain homology arms flanking a selectable cassette. Chimeric founders transmit the targeted allele to progeny. Concept / Approach: Transfect ES cells with a targeting vector designed for homologous recombination. Select for correctly targeted clones using positive/negative selection (e.g., neomycin resistance and ganciclovir counterselection). Inject targeted ES cells into blastocysts to create chimeric mice, breed to germline transmission, and intercross to obtain homozygous knockouts. Although CRISPR/Cas systems now streamline this, the canonical method remains ES cell gene targeting for many historical and some current applications. Step-by-Step Solution: Design targeting vector with homology arms and a selection marker.Introduce vector into ES cells; select homologous recombinants.Inject ES cells into host blastocysts to produce chimeric founders.Breed for germline transmission and derive knockout lines. Verification / Alternative check: Genotyping by Southern blot or PCR across junctions confirms proper homologous recombination; Mendelian inheritance in offspring validates germline transmission of the knockout allele. Why Other Options Are Wrong: Random mutagenesis or retroviral insertion lacks locus specificity. Cell fusion chimeras without targeting do not guarantee gene disruption. Somatic nuclear transfer is cloning, not gene targeting per se. Common Pitfalls: Confusing general chimera creation with targeted gene disruption; chimerism is a step in ES-based knockouts but targeting is the defining feature. Final Answer: Transfecting embryonic stem (ES) cells with an engineered gene construct and selecting homologous recombinants

Question 13

Restriction enzyme end types: Which of the following enzymes generates blunt-ended DNA fragments upon cleavage?

Accepted Answer

Introduction: Different restriction enzymes leave different end chemistries, which dictate cloning strategy. Sticky (cohesive) ends enable directional ligation via complementary overhangs, while blunt ends are compatible in any orientation but ligate less efficiently. This question tests knowledge of a classic blunt-cutter used in many protocols. Given Data / Assumptions: EcoRV recognizes the sequence GAT^ATC and cleaves in the center to produce blunt ends. SalI, XhoI, HindIII, BamHI generate cohesive overhangs. End type selection affects ligation efficiency and insert orientation control. Concept / Approach: By knowing recognition sequences and cut positions, you can predict end types. EcoRV cuts both strands at the same position within its site, yielding no overhangs (blunt). In contrast, SalI (G^TCGAC), XhoI (C^TCGAG), HindIII (A^AGCTT), and BamHI (G^GATCC) leave 5′ overhangs that are complementary only to matching sticky ends. Step-by-Step Solution: Recall or consult enzyme cut patterns for common cutters.Identify EcoRV as the enzyme cleaving centrally to give blunt ends.Exclude enzymes known for 5′ overhangs (SalI, XhoI, HindIII, BamHI).Select EcoRV as the correct blunt-end cutter. Verification / Alternative check: Digesting a plasmid with EcoRV yields fragments that cannot anneal via sticky ends; ligation requires higher concentrations and T4 DNA ligase, often with dephosphorylated vector controls to reduce background. Why Other Options Are Wrong: SalI/XhoI/HindIII/BamHI: all produce cohesive 5′ overhangs, not blunt ends. Common Pitfalls: Confusing EcoRV (blunt) with EcoRI (sticky 5′ AATT overhang); the “V” versus “I” matters for end type. Final Answer: EcoRV

Question 14

Human cDNA libraries in genetic engineering  What best describes a human cDNA library used in recombinant DNA technology and functional genomics?

Accepted Answer

Introduction / Context: A complementary DNA (cDNA) library is a foundational resource in molecular biology. It is generated by reverse transcribing messenger RNAs (mRNAs) into DNA, capturing the protein coding repertoire that a particular human cell type expresses under the conditions sampled. This question checks whether you can distinguish a cDNA library from a genomic library and recognize its appropriate uses in cloning and expression. Given Data / Assumptions: cDNA is synthesized from mRNA via reverse transcriptase. Only expressed genes present as mRNA at the time of RNA isolation are represented. cDNA lacks introns and most regulatory/intergenic sequences. Libraries are cloned into vectors (e.g., plasmids) for propagation and screening. Concept / Approach: A cDNA library reflects the transcriptome of the source tissue, not the entire human genome. Because introns are spliced out of mRNA, cDNAs correspond to exons joined together, which is advantageous for expressing eukaryotic proteins in prokaryotes. Radioactivity is not intrinsic to cDNA libraries; radiolabeled probes may be used in screening, but the library itself is simply DNA cloned in vectors. Step-by-Step Solution: Identify what cDNA represents: DNA copies of mRNAs.Note tissue specificity: only genes expressed in that tissue/time are captured.Infer intron absence: suitable for bacterial expression.Select the description that matches these properties. Verification / Alternative check: Comparing to a genomic library clarifies differences: genomic libraries include introns, promoters, and intergenic DNA; cDNA libraries do not. Expression screening of cDNA libraries is routine for isolating coding sequences. Why Other Options Are Wrong: Option B overclaims universality; no single cDNA library covers all proteins. Option C confuses screening radioactivity with the library itself. Option D is false because a correct description exists. Option E describes a genomic library, not cDNA. Common Pitfalls: Assuming cDNA equals genome; forgetting tissue and condition dependence; thinking introns remain in cDNA. Final Answer: A collection of DNA copies corresponding to mRNAs for specific human proteins expressed in the sampled cells or tissues

Recombinant DNA Technology Questions