Question 1

In DNA, which covalent linkage joins adjacent nucleotides to produce the repeating sugar–phosphate backbone of each strand?

Accepted Answer

Introduction / Context: The structure of DNA depends on two kinds of bonds: covalent bonds within each strand and noncovalent hydrogen bonds between complementary bases. This question asks which covalent bond links nucleotides together along a single DNA strand. Given Data / Assumptions: Standard deoxyribonucleotides with deoxyribose sugars. Backbone alternates sugar and phosphate groups. Bases project inward and pair by hydrogen bonds, not covalent bonds. Concept / Approach: Each nucleotide contributes a 3' hydroxyl on its sugar and a phosphate attached to the 5' carbon of the incoming nucleotide. A dehydration reaction forms a phosphodiester linkage connecting the 3' hydroxyl and the 5' phosphate, yielding a 3'–5' phosphodiester bond and a directional (5'→3') polymer. Step-by-Step Solution: 1) Consider two adjacent nucleotides positioned for polymerization.2) The 3' OH of the existing strand attacks the alpha phosphate of the 5' triphosphate on the incoming nucleotide.3) A phosphodiester bond forms, joining 3' of one sugar to 5' of the next.4) Repeating this reaction yields the sugar–phosphate backbone with sequence-specific bases attached via glycosidic bonds to each sugar. Verification / Alternative check: Enzymatic synthesis by DNA polymerases proceeds strictly 5'→3', consistent with formation of 3'–5' phosphodiester linkages and requirement for a free 3' OH primer terminus. Why Other Options Are Wrong: 2', 5' or 2', 3': The 2' carbon lacks the required hydroxyl in deoxyribose and these linkages are not used for DNA backbones. 3', 4': No standard biochemical mechanism makes a 3'–4' backbone in nucleic acids. Glycosidic bonds between adjacent bases: Glycosidic bonds join a base to its own sugar, not to neighboring nucleotides. Common Pitfalls: Confusing glycosidic base–sugar bonds with phosphodiester bonds, or forgetting polarity (5' phosphate to 3' OH). Final Answer: 3', 5' phosphodiester bonds to form a repetitive sugar–phosphate chain.

Question 2

In nucleic acid thermodynamics, TM (melting temperature) refers to which specific condition for a DNA duplex?

Accepted Answer

Introduction / Context: TM, or melting temperature, is a fundamental parameter in DNA hybridization, PCR primer design, and nucleic acid biophysics. It captures the temperature at which a DNA duplex is half melted and half intact under specified ionic and sequence conditions. Given Data / Assumptions: We are discussing DNA (or RNA–DNA) duplex stability. Salt concentration and sequence composition affect TM. Equilibrium conditions are assumed for the definition. Concept / Approach: As temperature rises, hydrogen bonding and base-stacking interactions that stabilize the double helix weaken. The melting temperature is defined where the fraction of base pairs that are paired equals the fraction that are unpaired, i.e., 0.5 duplex fraction at equilibrium. Step-by-Step Solution: 1) Start with a fully double-stranded DNA of known length and sequence.2) Increase temperature gradually while monitoring absorbance at 260 nm or using probes.3) Identify the temperature where 50% of molecules are denatured and 50% remain duplexed.4) Report this temperature as TM for those exact buffer and sequence conditions. Verification / Alternative check: Hyperchromic shift at 260 nm shows a characteristic S-shaped melting curve; TM corresponds to its midpoint. Predictive formulas incorporate GC content, length, and salt corrections and should agree with measurements within reasonable error. Why Other Options Are Wrong: Membrane 50% fluid: TM can also denote membrane transition temperature in lipid biophysics, but in nucleic acid questions it specifically refers to DNA duplex melting. Protein 50% unfolded: That concept describes protein melting temperature, not DNA TM. All of the above: These definitions are domain-specific and not simultaneously applicable here. RNA polymerase binding maximum: TM is unrelated to transcription factor binding optima. Common Pitfalls: Ignoring ionic strength and sequence length when comparing TM values across experiments. Another pitfall is assuming TM is a single universal property independent of context. Final Answer: 50% of a DNA duplex is denatured (single-stranded) at equilibrium.

Question 3

In Escherichia coli DNA repair, which DNA polymerase primarily fills short gaps during nucleotide-excision repair and Okazaki fragment processing because it has 5'→3' exonuclease activity?

Accepted Answer

Introduction / Context: E. coli encodes several DNA polymerases with distinct roles. Understanding which enzyme participates in repair versus replication is crucial for interpreting mutant phenotypes and biochemical assays. Given Data / Assumptions: Bacterial system (E. coli), not eukaryotes. Short single-stranded gaps arise after primer removal and during excision repair. An enzyme with both polymerase activity and 5'→3' exonuclease is advantageous for removing RNA primers and damaged nucleotides ahead of synthesis. Concept / Approach: DNA polymerase I possesses 5'→3' exonuclease activity that can remove RNA primers and excised oligonucleotides while simultaneously filling the gap with DNA. Polymerase III is the main replicative polymerase but lacks prominent 5'→3' exonuclease; Pol II is involved in repair under stress but is not the primary enzyme for primer removal. Step-by-Step Solution: 1) Identify the task: remove a stretch of nucleotides ahead of the polymerase and replace them with DNA.2) Select the enzyme that combines 5'→3' exonuclease with polymerase activity.3) DNA polymerase I meets these criteria and is well known for nick translation and primer removal.4) Conclude that Pol I primarily fills the gaps in excision repair and Okazaki processing. Verification / Alternative check: Classic experiments with Pol I mutants (polA) and biochemical nick-translation assays demonstrate its 5'→3' exonuclease and repair roles, while Pol III functions as the high-processivity replicase. Why Other Options Are Wrong: DNA polymerase II: Participates in repair and restart but lacks the principal 5'→3' exonuclease-driven nick translation function. DNA polymerase III holoenzyme: Main replicative polymerase; high processivity but not specialized for primer removal. None of these: Incorrect because Pol I fits the described function. Telomerase: Eukaryotic enzyme for telomere extension, absent in E. coli. Common Pitfalls: Assuming the major replicative polymerase also performs most repair gap-filling, or confusing 3'→5' proofreading with 5'→3' exonuclease used in nick translation. Final Answer: DNA polymerase I.

Question 4

During initiation of DNA replication, which protein actively unwinds the parental double helix ahead of the replication fork?

Accepted Answer

Introduction / Context: DNA replication requires coordinated action of multiple enzymes and accessory proteins. A critical early step is separation of the two parental strands to allow templated synthesis of new DNA. Given Data / Assumptions: We are focusing on the enzyme that actively breaks hydrogen bonds between bases. The replication fork advances directionally along duplex DNA. Concept / Approach: Helicases are ATP-dependent motors that translocate along DNA and unwind the duplex by disrupting base pairing. Other proteins assist: SSB stabilizes exposed single strands, topoisomerase relieves torsional stress, ligase seals nicks, and primase lays down RNA primers—but only helicase performs the active unwinding. Step-by-Step Solution: 1) Identify the task: separate complementary strands ahead of the fork.2) Determine which protein uses ATP hydrolysis to accomplish this task.3) Select DNA helicase as the enzyme that directly unwinds DNA. Verification / Alternative check: In vitro reconstitution of replication forks shows that without helicase, fork progression stalls even if polymerases, primase, and ligase are present. Addition of helicase restores unwinding and synthesis. Why Other Options Are Wrong: SSB: Binds single-stranded DNA to prevent reannealing but does not separate strands. DNA ligase: Joins nicks after synthesis; no unwinding activity. Topoisomerase: Relieves supercoils by transient breaks; does not unwind base pairs directly. Primase: Synthesizes RNA primers; no helicase function. Common Pitfalls: Equating stabilization of single strands with active unwinding, or attributing helicase-like roles to topoisomerases. Final Answer: DNA helicase.

Question 5

In bacteria such as E. coli, what is the specific chromosomal site where bidirectional DNA replication initiates?

Accepted Answer

Introduction / Context: Bacterial chromosomes begin replication at a defined locus to ensure complete and timely duplication before cell division. In E. coli, this locus is well characterized and called oriC. Given Data / Assumptions: Single circular chromosome in typical bacteria. Initiator protein DnaA binds specific repeats to start replication. Replication proceeds bidirectionally from the origin. Concept / Approach: At oriC, DnaA boxes and AT-rich DNA unwinding elements facilitate local strand separation. Helicase loading and priming establish two forks that move in opposite directions until they converge near ter sites. Step-by-Step Solution: 1) Identify the process: initiation of chromosomal replication.2) Locate the specialized sequence that recruits DnaA and promotes melting.3) Recognize that this site is named oriC in E. coli. Verification / Alternative check: Genetic and biochemical mapping shows that deletion or mutation of oriC blocks initiation, whereas ectopic insertion can create new initiation sites under controlled contexts. Why Other Options Are Wrong: Promoter: Site for RNA polymerase initiation of transcription, not DNA replication. Start codon: Translation initiation on mRNA, not chromosomal replication. Terminator (ter): Replication termination, not initiation. Ribosome binding site: A translation feature, unrelated to DNA replication origin. Common Pitfalls: Confusing regulatory sites for transcription or translation with those for DNA replication, or using the generic term origin (ori) without the specific bacterial designation oriC. Final Answer: oriC (origin of replication).

Question 6

According to the classical Watson–Crick model observed under physiological conditions, which DNA helix form represents the standard right-handed duplex of most cellular DNA?

Accepted Answer

Introduction / Context: DNA can adopt multiple helical conformations depending on hydration, sequence, and ionic conditions. The historically famous Watson–Crick structure corresponds to a particular form predominant in vivo. Given Data / Assumptions: Physiological (aqueous) conditions. Canonical duplex with antiparallel strands and complementary base pairing. Concept / Approach: B-DNA is the right-handed helix with about 10 base pairs per turn, wide major groove and narrow minor groove, matching the Watson–Crick model. A-DNA is a more compact right-handed form favored in lower hydration; Z-DNA is a left-handed helix occurring transiently in certain GC-rich sequences under high salt or negative supercoiling. Step-by-Step Solution: 1) Identify the structure historically described by Watson and Crick.2) Compare helical parameters of A, B, and Z forms.3) Select B-DNA as the standard cellular conformation. Verification / Alternative check: X-ray fiber diffraction and crystallography support B-form parameters under physiological hydration. Biochemical footprinting also aligns with B-DNA groove geometry for protein binding. Why Other Options Are Wrong: A-DNA: Right-handed but dehydrated form; common in RNA–DNA hybrids. Z-DNA: Left-handed, rarer, transient regulatory contexts. X-DNA and P-DNA: Not the classical cellular helix described by Watson–Crick. Common Pitfalls: Assuming any right-handed helix equals the classical form, or conflating A-DNA conditions with in vivo chromosomal DNA. Final Answer: B-DNA.

Question 7

Which classic experiment first provided direct evidence that a heritable factor from dead bacteria could change the phenotype of live bacteria, thereby implying that DNA is the genetic material?

Accepted Answer

Introduction / Context: Before DNA was universally accepted as the genetic material, several experiments probed how traits are inherited. Frederick Griffith's 1928 work with Streptococcus pneumoniae revealed a process where a trait could be transferred from dead cells to living cells. Given Data / Assumptions: Smooth (S) strain with capsule is virulent; rough (R) strain without capsule is nonvirulent. Heat-killed S cells alone do not cause disease. Mixing heat-killed S with live R leads to virulent infections in mice. Concept / Approach: Griffith observed that some heritable substance from dead S cells transformed live R cells into S phenotype. Later, Avery, MacLeod, and McCarty showed the transforming principle is DNA, cementing DNA's role as genetic material. Step-by-Step Solution: 1) Inject mice with live R: no disease.2) Inject with heat-killed S: no disease.3) Inject with live R + heat-killed S: disease occurs and live S bacteria are recovered.4) Conclude that R cells acquired information to synthesize a capsule, i.e., transformation. Verification / Alternative check: DNase treatment in Avery’s follow-up abolished transforming activity, while protease or RNase did not, identifying DNA as the transforming principle. Why Other Options Are Wrong: Transduction: DNA transfer via bacteriophages, discovered later by Zinder and Lederberg. Replication: DNA copying within a cell, not phenotype transfer between cells. Translation: Protein synthesis from mRNA. Conjugation: Direct plasmid transfer via cell-to-cell contact, discovered by Lederberg and Tatum. Common Pitfalls: Attributing Griffith’s findings to phage-mediated transduction or conflating his discovery with Avery’s identification of DNA. Final Answer: Transformation.

Question 8

Which of the following is NOT a structural DNA-binding motif commonly found in transcription factors?

Accepted Answer

Introduction / Context: DNA-binding proteins often use recurring structural motifs to contact specific sequences in promoters and enhancers. Recognizing these motifs helps identify likely transcription factors from domain annotations. Given Data / Assumptions: We are comparing structural motifs, not entire multiprotein complexes. Motifs are short, recurring 3D arrangements such as helix–turn–helix or zinc finger. Concept / Approach: Helix–turn–helix, zinc finger, bZIP, and helix–loop–helix are classical DNA-binding motifs. TFIID, however, is a large multi-subunit general transcription factor complex (containing TBP and TAFs), not a single structural motif. Step-by-Step Solution: 1) List common motifs used by sequence-specific DNA-binding proteins.2) Identify entries that are complexes or factors rather than motifs.3) TFIID stands out as a complex, so it is not a motif. Verification / Alternative check: Protein domain databases categorize helix–turn–helix, zinc finger, bZIP, and HLH as motifs, while TFIID is cataloged as a complex assembling at core promoters to recruit RNA polymerase II. Why Other Options Are Wrong: Helix–turn–helix: Classic motif in many bacterial repressors and eukaryotic factors. Zinc finger: Cys2His2 and other variants bind DNA via coordinated zinc ions. bZIP: Dimeric motif with basic region for DNA contact and leucine zipper for dimerization. Helix–loop–helix: Found in developmental regulators; facilitates dimerization and DNA binding. Common Pitfalls: Confusing a multiprotein complex name (TFIID) with a structural motif, or assuming all transcription-related terms are motifs. Final Answer: TFIID.

Question 9

Which enzyme synthesizes the short RNA primer required to initiate DNA synthesis on the lagging and leading strands at the replication fork?

Accepted Answer

Introduction / Context: DNA polymerases cannot start synthesis de novo; they require a preexisting 3' OH to extend. Cells therefore use a specialized RNA polymerase, primase, to lay down short RNA primers that DNA polymerases can elongate. Given Data / Assumptions: Replication forks operate on both leading and lagging strands. Primers are short RNA segments. DNA polymerase extends from the primer's 3' OH. Concept / Approach: Primase synthesizes RNA primers using ribonucleotides. On the leading strand, a single primer often suffices; on the lagging strand, each Okazaki fragment needs a new primer. Step-by-Step Solution: 1) Identify the biochemical limitation of DNA polymerase (cannot initiate).2) Provide a solution via an RNA primer created by primase.3) Allow DNA polymerase to extend and ligase to seal nicks after primer removal and replacement. Verification / Alternative check: Reconstituted replication systems without primase fail to initiate synthesis; adding primase restores initiation of both strands. In vivo, primase is part of the primosome with helicase. Why Other Options Are Wrong: DNA helicase: Unwinds DNA but does not synthesize RNA. DNA ligase: Seals nicks after DNA synthesis; cannot start chains. DNA topoisomerase: Manages supercoiling; not involved in primer synthesis. Exonuclease I: Degrades single-stranded DNA; not a priming enzyme. Common Pitfalls: Assuming that because primase is an RNA polymerase it only works on the lagging strand; in fact it primes both strands as needed. Final Answer: Primase.

Question 10

In molecular biology, what is the specific name for the Y-shaped points on a DNA molecule where the double helix is actively unwound and new daughter strands are synthesized during replication?

Accepted Answer

Introduction / Context: DNA replication proceeds from defined starting sites, and as enzymes open the helix, characteristic Y-shaped structures appear. These structures are central to understanding how new strands are built and how bidirectional replication progresses in both prokaryotes and eukaryotes. Given Data / Assumptions: DNA is a double helix that must be unwound for copying. Replication typically moves bidirectionally from an origin. Specialized enzymes (helicase, primase, polymerases, ligase) coordinate synthesis. Concept / Approach: The Y-shaped junction where parental strands separate and nascent strands elongate is called a replication fork. Helicase opens the helix, single-strand binding proteins stabilize exposed templates, and DNA polymerases extend primers on both leading and lagging templates. Step-by-Step Solution: Identify the structure formed where DNA strands separate during replication: this is the fork region.At each fork, one strand is synthesized continuously (leading) while the other is discontinuous (lagging) as Okazaki fragments.Because replication moves away from the origin, two forks often form and progress in opposite directions.Therefore, the proper term for the unwinding/synthesis junction is "replication fork." Verification / Alternative check: Origins are fixed sites where replication begins; forks are the moving fronts. Microscopy and replication bubble maps show two forks emanating from one origin, confirming the terminology. Why Other Options Are Wrong: Replication origins: these are start sites, not the moving Y-junctions. Leading strands: product strands, not the junction. Okazaki fragments: short pieces on the lagging strand, not the junction. Primosomes: primase–helicase complex, not the structure itself. Common Pitfalls: Confusing origin (a locus) with fork (a dynamic structure), or thinking Okazaki fragments name the junction. Final Answer: Replication forks

Question 11

In bacteria, negatively supercoiled chromosomal DNA is organized by nucleoid-associated proteins. Which set best represents such organizing proteins?

Accepted Answer

Introduction / Context: Bacterial chromosomes are compacted without canonical eukaryotic histones. Instead, nucleoid-associated proteins (NAPs) shape and organize negatively supercoiled DNA to enable replication, transcription, and segregation in a crowded cytoplasm. Given Data / Assumptions: We are considering bacterial chromosomal DNA (prokaryotic nucleoid). DNA is negatively supercoiled in vivo. Proteins that bend, bridge, or wrap DNA contribute to nucleoid architecture. Concept / Approach: Among the best-studied NAPs are HU (a small, abundant DNA-bending protein) and H-NS (a global gene-silencing/bridging protein that prefers AT-rich DNA). Together, they compact DNA, stabilize supercoils, and help regulate transcriptional programs in response to environmental cues. Step-by-Step Solution: Identify which proteins are bona fide bacterial NAPs: HU and H-NS qualify.Exclude typical eukaryotic histones (e.g., histone H1) that form nucleosomes in eukaryotes, not bacteria.Exclude heat-shock chaperones (e.g., HSP-family), which assist protein folding rather than directly organizing DNA.Therefore, the set that best represents organizing proteins is HU and H-NS together. Verification / Alternative check: Genetic deletions and biochemical reconstitution show HU induces flexible bends, while H-NS oligomerizes to bridge duplexes; both strongly influence nucleoid structure and gene expression. Why Other Options Are Wrong: HU only: incomplete; other major NAPs like H-NS also organize DNA. H-NS only: incomplete for the same reason. Histone H1: eukaryotic linker histone; not a bacterial NAP. HSP-1: chaperone, not a DNA-organizing NAP. Common Pitfalls: Equating bacterial DNA proteins with eukaryotic histones, or assuming any DNA-binding protein is a structural organizer. Final Answer: Both HU and H-NS

Question 12

Which statement captures a key difference between eukaryotic and prokaryotic DNA replication?

Accepted Answer

Introduction / Context: Genome size and chromatin organization impose distinct constraints on replication strategies in eukaryotes versus prokaryotes. Understanding origin usage and replisome dynamics is fundamental to cell biology and biotechnology. Given Data / Assumptions: Eukaryotic genomes are large, linear, and packaged with histones. Prokaryotic genomes are typically a single circular chromosome. Both systems rely on priming, polymerization, and ligation. Concept / Approach: The hallmark difference is origin usage: eukaryotic chromosomes fire many origins to complete S phase efficiently, while most bacteria use one (or very few) origins per chromosome, commonly oriC, with two forks moving bidirectionally. Step-by-Step Solution: Consider whether speed differences define the systems: prokaryotic polymerases are generally faster in elongation.Count polymerase varieties: eukaryotes possess many specialized polymerases; thus “fewer in eukaryotes” is incorrect.Assess primer requirement: both require RNA primers for DNA polymerases to initiate.Therefore, the accurate single best difference is multiple origins in eukaryotes. Verification / Alternative check: DNA combing and replication timing maps in eukaryotes reveal origin clusters; bacterial replication proceeds bidirectionally from oriC, supporting the stated difference. Why Other Options Are Wrong: Eukaryotic polymerases faster: typically not; bacterial elongation is faster. Fewer polymerases in eukaryotes: opposite of reality. No RNA primers in eukaryotes: incorrect; primase generates RNA primers in both. Prokaryotes unidirectional: most are bidirectional. Common Pitfalls: Assuming primer independence for eukaryotes or conflating number of polymerases with speed. Final Answer: Eukaryotic chromosomes have multiple origins of replication

Question 13

During DNA replication, what is the primary role of DNA ligase?

Accepted Answer

Introduction / Context: DNA replication on the lagging strand proceeds discontinuously as short Okazaki fragments. Converting these fragments into a continuous strand requires a sealing step that completes the sugar-phosphate backbone. Given Data / Assumptions: Okazaki fragments are produced after priming and polymerization. RNA primers are removed and replaced with DNA. A final enzymatic step must close remaining nicks. Concept / Approach: DNA ligase catalyzes phosphodiester bond formation between a 3'-OH and a 5'-phosphate at single-strand breaks (nicks), using an energy cofactor (ATP in eukaryotes, NAD+ in many bacteria). This reaction yields a continuous DNA strand. Step-by-Step Solution: Identify the unresolved feature after polymerase action: unsealed nicks remain between fragments.Determine the enzyme that seals nicks by forming phosphodiester bonds: DNA ligase.Exclude enzymes with other jobs (helicase unwinds, polymerase synthesizes, RNase removes primers).Therefore, ligase’s role is to join adjacent DNA fragments. Verification / Alternative check: In vitro replication assays show loss of ligase yields discrete fragments; adding ligase restores high-molecular-weight DNA, confirming its function. Why Other Options Are Wrong: Synthesizes lagging strand from scratch: DNA polymerases synthesize; ligase does not polymerize de novo. Copies mRNA to DNA: that is reverse transcriptase, not ligase. Degrades mRNA: RNases perform this function. Unwinds helix: helicase performs unwinding. Common Pitfalls: Confusing sealing of nicks (ligase) with primer removal (RNase H/Pol I flap removal) or with helicase action. Final Answer: Seals nicks by joining adjacent DNA fragments

Question 14

Which nucleic acid sample would require a higher temperature to denature (melt) under comparable ionic conditions?

Accepted Answer

Introduction / Context: Thermal denaturation (melting) breaks hydrogen bonds and base-stacking interactions in nucleic acids. The base composition significantly influences the melting temperature (Tm). Given Data / Assumptions: We compare samples under similar salt and pH conditions. Double-stranded polymers exhibit cooperative melting behavior. G–C pairs have three hydrogen bonds; A–T (or A–U) pairs have two. Concept / Approach: Higher G–C content strengthens duplex stability through extra hydrogen bonds and stronger stacking, raising Tm. While RNA duplexes can be very stable, the question asks for the general sample that requires higher temperature among typical choices and conditions. Step-by-Step Solution: Relate base composition to Tm: more G–C increases Tm.Hold salt constant: higher ionic strength screens backbone repulsion and elevates Tm, but composition effect persists.Among listed options, GC-rich dsDNA is the canonical sample with highest Tm.Therefore, select dsDNA with high G–C content. Verification / Alternative check: Empirical formulas (e.g., Tm ≈ 69 + 0.41*%GC for long DNA at moderate salt) predict larger Tm for GC-rich DNA, matching experimental melting curves. Why Other Options Are Wrong: High A–T DNA: fewer hydrogen bonds and weaker stacking lower Tm. High G–C RNA: can be stable, but the canonical, straightforward answer in comparative DNA contexts is GC-rich DNA. High A–U RNA: generally lower Tm than GC-rich duplexes. Single-stranded DNA: lacks duplex to melt cooperatively. Common Pitfalls: Equating base-pair count only with hydrogen bonds and forgetting base stacking; ignoring salt effects that shift absolute Tm but not the ordering. Final Answer: Double-stranded DNA with high G–C content

Question 15

In eukaryotic cells, which phases together constitute the full cell cycle framework within which DNA replication occurs?

Accepted Answer

Introduction / Context: DNA replication is restricted to a defined window of the eukaryotic cell cycle, but understanding that window requires viewing the entire cycle framework that orchestrates growth, synthesis, checkpoint control, and division. Given Data / Assumptions: Interphase consists of G1, S, and G2. M phase encompasses mitosis and cytokinesis. DNA replication itself occurs during S phase. Concept / Approach: The complete canonical eukaryotic cell cycle is G1 (growth), S (DNA synthesis), G2 (preparation/checking), and M (mitosis). Many exam items ask for the set of phases constituting the cycle, not only the single phase where replication happens. Step-by-Step Solution: List all standard phases: G1, S, G2, M.Identify replication window: S phase.Select the option that includes the full framework (not a subset): G1, S, G2 and M.Confirm that other options omit necessary phases. Verification / Alternative check: Cell-cycle markers (cyclins/CDKs) and DNA content analysis (flow cytometry) define these stages, with 1C → 2C DNA increase in S phase. Why Other Options Are Wrong: G1, G2 and M: omits S phase. S, G2 and M: omits G1. G2 and M: incomplete. S only: names the synthesis phase, not the whole cycle. Common Pitfalls: Answering “S phase” when the question asks for the cell-cycle composition rather than the replication window. Final Answer: G1, S, G2 and M

Question 16

DNA segments that can move from one genomic location to another are known as what?

Accepted Answer

Introduction / Context: Mobile DNA elements reshape genomes across evolution. They can insert, excise, or copy themselves, influencing gene regulation, genome size, and mutational landscapes in both prokaryotes and eukaryotes. Given Data / Assumptions: We are considering DNA sequences that change position within a genome. Movement may be cut-and-paste or copy-and-paste. Terminology distinguishes endogenous mobile elements from exogenous agents. Concept / Approach: Such sequences are called transposons (or transposable elements). DNA transposons use transposase for excision/integration, while retrotransposons use an RNA intermediate and reverse transcription. Step-by-Step Solution: Identify the defining feature: intragenomic mobility.Map to terminology: "transposons" is the accepted umbrella term.Exclude external agents (viruses) and non-mobile genomic features (introns, centromeres).Thus, the correct term is transposons. Verification / Alternative check: Classic McClintock maize elements and modern genomic surveys reveal abundant transposable element families that relocate or copy within genomes. Why Other Options Are Wrong: Retroviruses: infectious agents requiring packaging and cell-to-cell transmission. Introns: noncoding segments removed by splicing; not mobile by default. Moving elements (generic): imprecise; the precise term is transposons. Centromeres: structural chromosomal loci; not mobile elements. Common Pitfalls: Confusing retroviruses with endogenous retrotransposons; while related mechanisms exist, the term sought is transposons. Final Answer: Transposons

Question 17

A double-stranded DNA sample has 50% G–C content. Approximately at what temperature will it melt (denature) under standard laboratory ionic conditions?

Accepted Answer

Introduction / Context: Melting temperature (Tm) is the midpoint of duplex denaturation where half of the base pairs are disrupted. Approximate formulas relate Tm to base composition and salt concentration for long DNA molecules. Given Data / Assumptions: Double-stranded DNA with 50% G–C content. Moderate salt typical of standard buffers. Long polymer behavior (not short oligos which use nearest-neighbor rules). Concept / Approach: A commonly used empirical relationship for long DNA is Tm ≈ 69 + 0.41*(%GC) under moderate ionic strength. Substituting %GC = 50 gives a value close to 90 °C, consistent with many textbook approximations. Step-by-Step Solution: Use approximation: Tm ≈ 69 + 0.41*(GC%).Compute: 69 + 0.41*50 = 69 + 20.5 = 89.5 °C.Round to the nearest option: approximately 90 °C.Hence, choose 90 °C. Verification / Alternative check: UV absorbance melting curves for genomic DNA with 50% GC typically center around 88–92 °C depending on salt, aligning with 90 °C as an acceptable estimate. Why Other Options Are Wrong: 60 or 70 °C: too low for GC50% long DNA at normal salt. 100 °C: overly high; most DNA melts below boiling under lab conditions. 120 °C: exceeds water’s boiling point at 1 atm and is not a typical Tm. Common Pitfalls: Applying short-oligo rules to long polymers or ignoring salt effects that shift Tm modestly but do not overturn the estimate. Final Answer: 90 °C

Question 18

Eukaryotic DNA replication occurs during which specific cell-cycle phase?

Accepted Answer

Introduction / Context: Cell-cycle control restricts DNA synthesis to a defined window to preserve genome integrity and coordinate replication with division. Given Data / Assumptions: Eukaryotic cell cycle has G1, S, G2, and M phases. Checkpoints prevent re-replication and ensure fidelity. Concept / Approach: Licensing of replication origins occurs in late M/G1, but firing and bulk DNA synthesis are confined to S phase. This temporal separation prevents multiple rounds of replication in one cycle. Step-by-Step Solution: Identify the synthesis window: S phase.Confirm that other phases are preparatory (G1), checking (G2), or segregation (M).Thus, select S phase. Verification / Alternative check: DNA content by flow cytometry increases from 2C to 4C specifically during S phase, verifying synthesis timing. Why Other Options Are Wrong: G1: growth and licensing, not bulk synthesis. G2: post-replication gap; repair and preparation for mitosis. M: chromosome segregation, not synthesis. G0: quiescence outside the cycle. Common Pitfalls: Confusing licensing (G1) with replication (S) or assuming low-level synthesis outside S phase. Final Answer: S phase

Question 19

Whose X-ray diffraction work provided crucial evidence that helped Watson and Crick deduce the DNA double helix?

Accepted Answer

Introduction / Context: The double-helix model of DNA proposed by James Watson and Francis Crick in 1953 relied on converging lines of evidence: base-pairing rules, fiber diffraction, and chemical constraints. Understanding whose data underpinned the helix geometry is part of the history of science. Given Data / Assumptions: X-ray diffraction patterns of DNA fibers constrain helical pitch and symmetry. Contemporaneous biochemical data (e.g., Chargaff’s ratios) limited base pairing possibilities. Multiple scientists contributed to the era’s insights. Concept / Approach: Rosalind Franklin’s high-resolution X-ray diffraction photographs (notably Photo 51, taken by her student Raymond Gosling) revealed helical features, spacing of stacked bases (~3.4 Å), and dimensions consistent with a two-strand helix, guiding model building. Step-by-Step Solution: Identify whose data directly constrained the helix: Franklin’s fiber diffraction was pivotal.Differentiate historical contributors: Bragg advanced crystallography methods; Pauling proposed alternative structures; Lederberg contributed to bacterial genetics; Chargaff provided base ratios.Select the scientist whose X-ray work was decisive for the helix parameters: Rosalind Franklin. Verification / Alternative check: Historical accounts and primary sources credit Franklin’s patterns with helical inference and correct helical parameters used by Watson and Crick. Why Other Options Are Wrong: W. H. Bragg: Nobel laureate in X-ray crystallography methods, but not the DNA fiber data. Linus Pauling: proposed a triple-helix model that was incorrect for DNA. Joshua Lederberg: genetics pioneer; not responsible for DNA X-ray fiber images. Erwin Chargaff: provided base ratio rules; crucial but not X-ray diffraction. Common Pitfalls: Attributing “X-ray evidence” generically to famous names without distinguishing the specific DNA fiber data. Final Answer: Rosalind Franklin

Question 20

In structural biochemistry, which description correctly characterizes the B-DNA form observed in aqueous solution, including strand orientation and approximate helical pitch per turn?

Accepted Answer

Introduction / Context: B-DNA is the predominant DNA conformation in physiological aqueous solution. This item tests recognition of its handedness, strand polarity, and approximate base pairs per turn—core facts for molecular biology and genetics. Given Data / Assumptions: Context: nucleic acid structure at near-neutral pH and moderate ionic strength. We must select the accurate structural description of B-DNA. Approximate helical repeat is acceptable (~10 bp/turn commonly cited; many texts use 10–10.5). Concept / Approach: Canonical B-DNA is a right-handed helix with two antiparallel polynucleotide strands (one 5'→3', the other 3'→5'). Under physiological conditions, its helical repeat is about 10–10.5 base pairs per turn, with bases roughly perpendicular to the helix axis and distinctive major/minor grooves. Step-by-Step Solution: 1) Handedness: B-DNA is right-handed (Z-DNA is left-handed; A-DNA is right-handed but is a dehydrated form).2) Strand orientation: DNA strands in duplex form are antiparallel, not parallel.3) Helical pitch: accept ~10 bp/turn as the standard simplified value.4) The description that matches all three properties is the right-handed antiparallel double helix with ~10 bp/turn. Verification / Alternative check: Fiber diffraction and high-resolution structures consistently show right-handedness and antiparallel polarity for B-DNA; only Z-DNA is left-handed and it has a zig-zag backbone but is not the predominant solution form. Why Other Options Are Wrong: Left-handed antiparallel: describes Z-DNA, not B-DNA.Left-handed zig-zag: again Z-DNA (special sequences/conditions), not the common solution form.Right-handed parallel: the strands in DNA are not parallel; they are antiparallel. Common Pitfalls: Mixing Z-DNA traits (left-handed, zig-zag backbone) with B-DNA, or assuming parallel strand orientation. Final Answer: Right-handed double helix of antiparallel chains (~10 bp/turn)

DNA Structure and Replication Questions