Question 1

DNA replication — The Meselson–Stahl experiment demonstrated which model of DNA replication?

Accepted Answer

Introduction: The Meselson–Stahl experiment (E. coli grown in heavy nitrogen followed by a light nitrogen shift) is a landmark demonstration of how DNA duplicates during the cell cycle. The question asks which replication model their data supported. Given Data / Assumptions: Three proposed models existed: conservative, semiconservative, dispersive. Density-gradient centrifugation can separate DNA by buoyant density (heavy vs light). Parental strands serve as templates during replication. Concept / Approach: Semiconservative replication predicts that each daughter DNA molecule contains one parental (heavy) strand and one newly synthesized (light) strand after one generation in light nitrogen, yielding hybrid density DNA. Step-by-Step Solution: 1) Grow cells in 15N to label parental DNA heavy.2) Shift to 14N; allow one round of replication.3) Observe a single intermediate-density band (hybrid), not separate heavy and light bands.4) After two rounds, see both hybrid and light bands, consistent with semiconservative replication. Verification / Alternative check: Conservative replication would show distinct heavy and light bands after one generation; dispersive would maintain only intermediates without a pure light band even after several generations. Why Other Options Are Wrong: a) Predicts separate heavy and light duplexes after one replication; not observed.c) Predicts all hybrids each generation without pure light; data contradict.d) A chromosomal disorder, not a replication model.e) Rolling-circle occurs in some systems but was not the universal model demonstrated here. Common Pitfalls: Confusing labeling (heavy vs light) with strand composition; forgetting the generational pattern of bands. Final Answer: Semiconservative replication.

Question 2

DNA transactions — Which enzyme is responsible for unwinding (unraveling) short segments of duplex DNA ahead of the replication fork?

Accepted Answer

Introduction: DNA replication requires coordinated activities to open the duplex, synthesize primers, extend strands, and relieve torsional stress. This question targets the specific enzyme that separates the two strands at the fork. Given Data / Assumptions: Duplex DNA is stabilized by hydrogen bonds and base stacking. Replication forks move directionally along the template. Distinct enzymes perform unwinding, priming, polymerization, ligation, and supercoil relief. Concept / Approach: Match each replication enzyme to its primary function: helicase unwinds; primase synthesizes RNA primers; polymerase extends DNA; ligase seals nicks; topoisomerase relieves supercoils but does not drive strand separation at the fork. Step-by-Step Solution: 1) Identify the mechanical task: disrupt base pairing to create single-stranded templates.2) Helicase uses ATP to translocate and unwind duplex DNA.3) Other enzymes act on the unwound templates (primase/polymerase) or on backbone integrity (ligase) or topology (topoisomerase). Verification / Alternative check: Loss-of-function helicase mutants stall forks despite intact polymerase activity, confirming helicase’s essential unwinding role. Why Other Options Are Wrong: a) Polymerase synthesizes DNA but cannot unwind duplex DNA efficiently.c) Ligase seals nicks between fragments; no unwinding.d) Primase makes RNA primers; no unwinding.e) Topoisomerase manages supercoils; it does not continuously separate strands at the fork. Common Pitfalls: Confusing topological stress relief (topoisomerase) with actual strand separation (helicase). Final Answer: Helicase.

Question 3

DNA thermal stability — If a cell’s DNA has Tm = 40 °C at 20% GC content, what happens to Tm if GC% rises to 60% (qualitative trend)?

Accepted Answer

Introduction: The melting temperature (Tm) of DNA is the point at which half of the duplex molecules are denatured. It depends strongly on base composition because GC pairs have three hydrogen bonds and stronger stacking than AT pairs. This question asks for the qualitative direction of Tm change when GC% increases markedly. Given Data / Assumptions: Initial Tm is 40 °C at 20% GC. GC% is increased to 60% (substantially higher). We are asked for the trend, not an exact recalculated temperature. Concept / Approach: Higher GC content typically raises Tm. A common approximation for short DNAs is Tm ≈ 2 °C * (A + T) + 4 °C * (G + C). While exact values depend on length, salt, and sequence context, the direction is unambiguous: more GC → higher Tm. Step-by-Step Solution: 1) Recognize that GC base pairs contribute greater stability via three hydrogen bonds and stacking.2) Increasing GC from 20% to 60% increases the stabilizing fraction.3) Therefore, the temperature needed to melt half the duplex (Tm) increases. Verification / Alternative check: Empirical nearest-neighbor models show positive correlation between GC fraction and Tm across ionic strengths; calibration curves confirm the trend. Why Other Options Are Wrong: a) Stability changes with composition; Tm is not constant.c) More GC does not lower Tm.d) Although exact values require conditions, the trend is clear.e) Tm is a thermodynamic property, not a cyclic variable. Common Pitfalls: Ignoring salt concentration and length when computing exact Tm; here only the qualitative direction is required. Final Answer: Increases.

Question 4

Biopolymer structure — Why is the higher-order structure of most proteins far less regular (symmetrical) than the repeating, symmetric structure of DNA?

Accepted Answer

Introduction: DNA’s double helix is a repetitive polymer of only four bases attached to a uniform deoxyribose-phosphate backbone, producing a highly regular structure. Proteins, built from twenty amino acids with distinct side chains, show far greater structural diversity. This question explores why protein shapes are generally less symmetric than DNA. Given Data / Assumptions: Proteins perform many biochemical roles requiring diverse architectures. Amino acid side chains vary in size, charge, hydrophobicity, and hydrogen-bonding potential. Secondary structures (helices/sheets) combine into complex tertiary/quaternary folds. Concept / Approach: Evaluate how polymer chemistry influences shape. DNA’s repeating units encourage a uniform helix; proteins incorporate heterogeneous residues that drive different local conformations and interactions, reducing global symmetry and increasing functional adaptability. Step-by-Step Solution: 1) Protein function diversity (enzymes, scaffolds, motors, receptors) necessitates varied folds rather than a single repeating structure.2) Side-chain diversity (R groups) promotes local preferences for helix, sheet, turns, or loops, yielding complex 3D topologies.3) Certain residues (e.g., proline) introduce kinks; others (e.g., glycine) add flexibility; charged and hydrophobic residues drive packing and surface patterns. Verification / Alternative check: X-ray and cryo-EM structures show proteins adopting unique folds aligned to function, while DNA maintains B-form or related helices under physiological conditions. Why Other Options Are Wrong: e) Protein backbones do form extensive hydrogen bonds in helices and sheets; the claim is incorrect. Common Pitfalls: Assuming one canonical protein shape; underestimating how side-chain chemistry drives folding and asymmetry. Final Answer: All of the above.

Question 5

Nucleic acid chemistry — The 5′ and 3′ designations of a DNA (or RNA) strand refer specifically to what?

Accepted Answer

Introduction: Nucleic acids are directional polymers. The terms 5′ and 3′ indicate chemical polarity, which is essential for replication, transcription, and many enzymatic processes. This question asks what these numbers refer to at the molecular level. Given Data / Assumptions: DNA/RNA sugars are pentoses with labeled carbons 1′–5′. Backbone linkages connect the 3′-OH of one sugar to the 5′-phosphate of the next. Polymerases synthesize in the 5′→3′ direction. Concept / Approach: Associate strand polarity with sugar-carbon numbering: 5′ corresponds to the phosphate-bearing carbon; 3′ corresponds to the hydroxyl that accepts the next nucleotide. Orientation underlies antiparallel duplex formation and enzyme directionality. Step-by-Step Solution: 1) Identify the pentose sugar ring and its numbered carbons.2) Recognize phosphodiester bonds: 3′-OH to 5′-phosphate.3) Conclude that 5′ and 3′ denote specific sugar carbons, not strand length or base rules. Verification / Alternative check: Antiparallel duplexes are described as 5′→3′ and 3′→5′, consistent with sugar-carbon references, not base counts. Why Other Options Are Wrong: a,c,d,e) These do not define strand polarity; only sugar-carbon numbering does. Common Pitfalls: Confusing 5′/3′ with base-pairing or with the number of phosphates in a nucleotide triphosphate. Final Answer: They refer to the numbered carbons (5′ and 3′) of the pentose sugar.

Question 6

DNA replication enzymes — Which enzyme adds complementary bases to the growing DNA strand during replication?

Accepted Answer

Introduction: Replication requires a template, primers, and enzymes that catalyze phosphodiester bond formation. This question focuses on the enzyme that actually polymerizes deoxyribonucleotides into DNA by base pairing with the template strand. Given Data / Assumptions: Complementary base pairing guides nucleotide selection. DNA polymerases require a primer with a free 3′-OH. Synthesis proceeds in the 5′→3′ direction only. Concept / Approach: Identify the role of each enzyme: helicase unwinds, primase lays RNA primers, DNA polymerase extends DNA, ligase seals nicks, and RNA polymerase transcribes RNA, not DNA. Therefore, DNA polymerase is responsible for adding complementary bases to DNA during replication. Step-by-Step Solution: 1) Generate primers (by primase) to provide a 3′-OH.2) DNA polymerase catalyzes addition of dNTPs to the 3′ end, matching A with T and G with C.3) Proofreading by many polymerases enhances fidelity. Verification / Alternative check: Biochemical assays show polymerase activity as template- and primer-dependent incorporation of labeled nucleotides into acid-insoluble DNA. Why Other Options Are Wrong: a) Helicase unwinds DNA; it does not polymerize nucleotides.b,c) Terms are nonspecific and not standard for the polymerizing enzyme.e) RNA polymerase synthesizes RNA, not DNA. Common Pitfalls: Using “replicase” generically; in eukaryotes, multiple DNA polymerases (e.g., δ, ε) share roles. Final Answer: DNA polymerase.

Question 7

DNA double helix — What interactions hold the two antiparallel DNA strands together in the duplex?

Accepted Answer

Introduction: The stability and specificity of the DNA double helix depend on defined interactions between bases on opposing strands. This question asks which interaction is responsible for holding the two strands together in the Watson–Crick model. Given Data / Assumptions: Strands are antiparallel and complementary (A pairs with T, G pairs with C). Backbone consists of covalently linked sugar-phosphate units. Base pairing provides sequence-specific recognition. Concept / Approach: Differentiate the types of bonds: covalent bonds form the backbone within each strand; hydrogen bonds and base stacking stabilize pairing and helix formation between strands. The direct strand-to-strand connection comes from hydrogen bonds between bases. Step-by-Step Solution: 1) Identify complementary bases (A–T, G–C) facing each other inside the helix.2) Hydrogen bonds form between specific donor/acceptor groups on these bases.3) Base stacking further stabilizes the helix but is not the inter-strand bond listed in the options. Verification / Alternative check: Thermal denaturation disrupts hydrogen bonds, separating strands; the process is reversible upon cooling, consistent with non-covalent pairing. Why Other Options Are Wrong: a) There are no covalent bonds directly between paired bases across the two strands.c) The backbone carries negative charge; ionic bonds of this type do not hold the strands together.d) Association is specific, not random.e) Metal coordination does not link the two strands in normal physiological DNA. Common Pitfalls: Confusing backbone phosphodiester bonds (within a strand) with base pairing (between strands); overlooking the contribution of stacking as auxiliary stability. Final Answer: Hydrogen bonds between complementary base pairs.

Question 8

Molecular Genetics—Chromatin Structure Chromosomal DNA in eukaryotes complexes with core and linker histones to form nucleosomes. Which complete set correctly lists the canonical histone types?

Accepted Answer

Introduction / Context: Eukaryotic chromosomal DNA is packaged into chromatin, whose fundamental unit is the nucleosome. Proper identification of histone types is essential to understand gene regulation, DNA replication, and epigenetics. Given Data / Assumptions: Nucleosome core contains an octamer: (H2A-H2B)2 and (H3-H4)2. Linker histone H1 associates with DNA at the entry/exit site to stabilize higher-order structure. Question asks for the complete canonical set of histones in eukaryotic chromatin. Concept / Approach: Recall the core histones (H2A, H2B, H3, H4) and the linker histone (H1). While variant histones exist (e.g., H2A.Z, H3.3), the canonical list remains five types. Histone H5 is specific to avian erythrocytes and is not a canonical eukaryotic histone across species. Step-by-Step Solution: Step 1: Identify core histones: H2A, H2B, H3, H4.Step 2: Add the linker histone: H1.Step 3: Assemble the full canonical set: H1, H2A, H2B, H3, H4. Verification / Alternative check: Chromatin reconstitution experiments and crystal structures of nucleosome cores confirm the histone composition and the role of H1 in chromatin compaction. Why Other Options Are Wrong: Three or two histones: incomplete; missing core members. Four with H1 but missing H2B: still incomplete. Including H5: not a ubiquitous canonical histone; mainly avian erythrocytes. Common Pitfalls: Confusing histone variants with canonical types; forgetting the linker histone H1. Final Answer: Five histones: H1, H2A, H2B, H3, and H4

Question 9

DNA Replication—Opening the Replication Bubble Which protein class is primarily responsible for opening a replication bubble by separating parental DNA strands?

Accepted Answer

Introduction / Context: DNA replication begins at origins where a replication bubble forms as parental strands are separated. Understanding which enzymes initiate strand separation versus those that stabilize or process DNA is fundamental to replication biology. Given Data / Assumptions: Replication requires helicase, SSB, topoisomerase, primase, polymerase, clamp, and ligase. Question asks for the protein class that actively opens (unwinds) the duplex. Concept / Approach: Helicases use NTP hydrolysis to translocate along DNA and separate complementary strands, generating single-stranded templates. SSB binds and stabilizes exposed single strands; topoisomerase relieves torsional stress; ligase seals nicks after synthesis; sliding clamp enhances polymerase processivity. Only helicase directly unwinds to create the bubble. Step-by-Step Solution: Step 1: Identify the motor protein that breaks base pairing: helicase.Step 2: Distinguish supportive roles: SSB prevents re-annealing; topoisomerase prevents supercoil buildup; ligase joins Okazaki fragments.Step 3: Conclude helicase is the principal opener of the bubble. Verification / Alternative check: Biochemical assays show helicase-dependent DNA unwinding; inhibition of helicase blocks bubble formation despite presence of SSB or topoisomerase. Why Other Options Are Wrong: SSB stabilizes but does not unwind. Ligase seals nicks; no unwinding function. Topoisomerase resolves supercoils but does not separate base pairs. Sliding clamp increases polymerase processivity; not an unwinding factor. Common Pitfalls: Attributing the opening to SSB because it binds single-stranded DNA; overlooking that SSB acts after helicase action. Final Answer: DNA helicases

Question 10

Base Pairing Rules During DNA Synthesis When DNA polymerase encounters guanine (G) on the template (parental) strand, which nucleotide does it add to the growing daughter strand?

Accepted Answer

Introduction / Context: High-fidelity DNA replication relies on specific base pairing (Watson–Crick rules). Knowing which nucleotide is incorporated opposite each template base is fundamental to molecular genetics and PCR design. Given Data / Assumptions: Template base is guanine (G). DNA polymerase synthesizes 5′→3′ by complementarity. Nucleotides are deoxyribonucleotides (no uracil in DNA). Concept / Approach: According to base-pairing rules, G pairs with C via three hydrogen bonds in B-form DNA. Therefore, the correct incoming deoxyribonucleotide is dCTP, which becomes cytosine (C) in the daughter strand after incorporation and pyrophosphate release. Step-by-Step Solution: Step 1: Identify complementary partner of G: C.Step 2: Determine the substrate used by polymerase: dCTP.Step 3: Incorporation yields a phosphodiester bond and adds C opposite template G. Verification / Alternative check: Melting temperature correlations show G≡C pairs (three H-bonds) contribute greater stability than A=T pairs (two H-bonds), aligning with the rule used here. Why Other Options Are Wrong: Phosphate is not a nucleotide base. Uracil is in RNA, not DNA. Guanine opposite G would be mismatched. Adenine pairs with thymine, not guanine. Common Pitfalls: Confusing RNA and DNA bases; forgetting that polymerase requires triphosphate forms (dNTPs) but incorporates the base + sugar + one phosphate into DNA. Final Answer: Cytosine

Question 11

Telomere Composition Human telomeric DNA repeats are typically rich in which nucleotide base?

Accepted Answer

Introduction / Context: Telomeres cap chromosome ends with tandem repeats (e.g., in humans, the repeat is TTAGGG). Their nucleotide composition influences secondary structures (e.g., G-quadruplexes) and telomerase activity. Given Data / Assumptions: Human telomeric repeat sequence: TTAGGG (5′→3′). G-richness promotes unique higher-order DNA conformations. Concept / Approach: Examine the canonical repeat: TTAGGG contains two thymines, one adenine, and three guanines, making the sequence guanine-rich. This G-rich nature is conserved across many species and is functionally relevant for binding proteins and telomerase substrate recognition. Step-by-Step Solution: Step 1: Parse TTAGGG: count G bases (three of six).Step 2: Conclude enrichment of guanine compared to other bases.Step 3: Select the guanine-rich option. Verification / Alternative check: Biophysical studies show G-quadruplex formation in telomeric DNA, a phenomenon dependent on G-rich sequences. Why Other Options Are Wrong: Adenine and thymine are present but not dominant. Cytosine is part of the complementary strand, not the overhang; not the defining enrichment. Uracil is not present in DNA. Common Pitfalls: Confusing RNA telomerase template with DNA telomere sequences; assuming A/T-rich repeats. Final Answer: Guanine

Question 12

Function of DNA Helicase In DNA metabolism, what is the primary role of DNA helicase?

Accepted Answer

Introduction / Context: Helicases are essential nucleic acid motor proteins in replication, repair, and recombination. They use NTP hydrolysis to destabilize base pairing and unwind DNA for downstream enzymes. Given Data / Assumptions: Helicase acts ahead of replication forks. Strand separation is prerequisite for polymerase action. Concept / Approach: Differentiate unwinding (helicase) from nick sealing (ligase) and backbone cleavage (nucleases). Helicase does not cut phosphodiester bonds; it disrupts hydrogen bonding between bases through mechanical translocation and energy usage. Step-by-Step Solution: Step 1: Identify the process requiring strand separation.Step 2: Map to helicase function (unwinding).Step 3: Exclude nuclease and ligase activities. Verification / Alternative check: In vitro helicase assays measure duplex unwinding; ligase shows nick sealing without unwinding capability. Why Other Options Are Wrong: Tight interaction to prevent unwinding: opposite of helicase action. Breaking phosphodiester bonds: nuclease, not helicase. Joining fragments: ligase function. Common Pitfalls: Confusing unwinding (hydrogen bonds) with cutting (phosphodiester bonds). Final Answer: Unwind the double helix by separating complementary strands

Question 13

Electrophoresis of DNA Under an electric field in an agarose gel, toward which electrode does DNA migrate?

Accepted Answer

Introduction / Context: Gel electrophoresis separates nucleic acids by size based on their movement in an electric field through a porous matrix. Understanding charge and directionality is foundational for interpreting bands and designing experiments. Given Data / Assumptions: DNA backbone carries negative charges due to phosphate groups. Standard electrophoresis uses a buffer-filled gel with applied voltage. Concept / Approach: Negatively charged molecules migrate toward the positive electrode (anode). Smaller fragments move faster through the matrix pores. Therefore, DNA moves from the cathode side toward the anode, forming size-dependent bands. Step-by-Step Solution: Step 1: Determine net charge of DNA: negative.Step 2: Apply electric field logic: negative species move to positive pole.Step 3: Conclude migration toward the anode. Verification / Alternative check: Tracking dyes and ladder standards consistently move toward the anode in agarose/TBE or TAE gels. Why Other Options Are Wrong: Negative pole: opposite direction for negatively charged DNA. No migration: contradicted by routine lab practice. Both poles equally: impossible under a unidirectional field. Common Pitfalls: Confusing electrode naming conventions; remembering that the anode is positive in electrophoresis setups. Final Answer: Positive pole (anode)

Question 14

Lab Techniques—Typical Volumes in Molecular Biology In routine molecular biology workflows (PCR setup, cloning, enzyme digests), the most common liquid volumes are measured in which unit?

Accepted Answer

Introduction / Context: Molecular biology experiments usually handle small reaction volumes to conserve reagents and fit microtube formats. Accurate volume measurement is central to reproducibility and enzyme kinetics. Given Data / Assumptions: Typical PCR volumes: 10–50 μl. Restriction digests/ligation: 10–20 μl. Micro-pipettes are calibrated for μl ranges (e.g., P10, P20, P200, P1000). Concept / Approach: Relate standard protocol volumes to units. Although nanoliter stages exist (e.g., robotics, qPCR droplets), most hands-on benchtop methods use microliter volumes. Milliliter volumes are common for culture media but not for reaction mixes. Step-by-Step Solution: Step 1: Recall typical reaction sizes (10–50 μl).Step 2: Match to micro-pipette ranges (P10, P20, P200, P1000).Step 3: Conclude microliters are the most common units. Verification / Alternative check: Vendor protocols for polymerases and ligases specify reagent additions in μl. Why Other Options Are Wrong: ml/l/dl: too large for typical reactions. nl: used in specialized platforms, not the most common at the bench. Common Pitfalls: Confusing culture volumes (ml) with reaction volumes (μl). Final Answer: μl (microliters)

Question 15

DNA Composition—Nitrogenous Bases How many different bases occur in DNA, and which are they?

Accepted Answer

Introduction / Context: DNA stores genetic information using four nitrogenous bases attached to a deoxyribose-phosphate backbone. Recognizing the correct set distinguishes DNA from RNA and from amino acids (e.g., tryptophan is an amino acid, not a nucleobase). Given Data / Assumptions: Pyrimidines in DNA: thymine (T), cytosine (C). Purines in DNA: adenine (A), guanine (G). RNA replaces thymine with uracil (U). Concept / Approach: Identify the canonical DNA bases and exclude non-DNA bases and amino acids. Tryptophan is an amino acid; uracil belongs to RNA, not DNA. Step-by-Step Solution: Step 1: List DNA bases: A, G, T, C.Step 2: Exclude U (RNA-specific) and tryptophan (amino acid).Step 3: Select the option listing A, G, T, C. Verification / Alternative check: Chargaff’s observations and modern sequencing confirm the four-base composition of DNA. Why Other Options Are Wrong: Includes tryptophan: not a nucleobase. Includes uracil: RNA base. Two or three bases: incomplete. Common Pitfalls: Mixing RNA and DNA bases; confusing amino acids with nucleotides. Final Answer: Four: adenine, guanine, thymine, and cytosine

Question 16

Chargaff’s Rules Application A double-stranded DNA sample has AT content of 40% (i.e., A% + T% = 40). What is the percentage of guanine (G) residue?

Accepted Answer

Introduction / Context: Chargaff’s rules state that in double-stranded DNA, A% ≈ T% and G% ≈ C%, and the total of A + T + G + C = 100%. Applying these rules allows inference of base percentages from limited information. Given Data / Assumptions: A% + T% = 40%. Therefore, G% + C% = 60%. Chargaff pairing implies G% = C%. Concept / Approach: Let G% = C% = x. Then G% + C% = 2x = 60%, so x = 30%. Hence guanine = 30% and cytosine = 30%. Step-by-Step Solution: Step 1: Compute GC fraction: 100 − 40 = 60%.Step 2: Set G = C; thus 2G = 60 → G = 30%.Step 3: Answer: G = 30%. Verification / Alternative check: Check totals: A + T = 40; G + C = 60; sum = 100, consistent with a valid composition. Why Other Options Are Wrong: 60%: that is GC combined, not G alone. 15% or 20%: inconsistent with G = C and GC = 60%. Cannot be calculated: it can, using Chargaff’s rules. Common Pitfalls: Forgetting that G equals C; mixing up total GC with individual G or C percentages. Final Answer: 30%

Question 17

Classic Experiment—Hershey–Chase Conclusion What overall conclusion emerged from the Hershey–Chase bacteriophage experiment using 32P and 35S labeling?

Accepted Answer

Introduction / Context: The Hershey–Chase experiment (1952) used T2 bacteriophages labeled with radioactive isotopes to determine whether DNA or protein is the genetic material. This landmark study complemented Avery–MacLeod–McCarty’s work and helped establish DNA as the carrier of hereditary information. Given Data / Assumptions: Phage protein shells were labeled with 35S (sulfur in methionine/cysteine). Phage DNA was labeled with 32P (phosphate in nucleotides). After infection and blender treatment, radioactivity was tracked between pellet (cells) and supernatant (ghosts). Concept / Approach: If DNA is genetic, it should enter bacteria and be found in progeny phage. If protein is genetic, labeled protein should enter cells. Observations showed 32P (DNA) inside cells and in progeny, while 35S (protein) largely remained with phage coats outside the cells. Step-by-Step Solution: Step 1: Infect bacteria with 32P-DNA–labeled phage and 35S-protein–labeled phage in parallel.Step 2: Shear off phage coats (blender) and separate by centrifugation.Step 3: Detect radioactivity: 32P in the cell pellet; 35S in the supernatant.Step 4: Conclude DNA, not protein, enters cells and directs progeny production. Verification / Alternative check: Progeny phage produced from infected bacteria carried 32P labeling, confirming DNA transmission across generations. Why Other Options Are Wrong: Proteins as genetic material: contradicted by labeling results. A=T ratio statement: Chargaff’s rule, not the experiment’s main conclusion. “Phage DNA identical to bacterial DNA”: not the question addressed. Protein entering while DNA stays outside: opposite of observed outcome. Common Pitfalls: Confusing the roles of 32P and 35S; misinterpreting pellet vs supernatant radioactivity. Final Answer: DNA was responsible for heredity

Question 18

Molecular biology – In a cesium chloride (CsCl) gradient centrifugation experiment, DNA molecules are separated primarily based on which physical property?

Accepted Answer

Introduction: Cesium chloride (CsCl) density-gradient centrifugation is a classic analytical and preparative technique in molecular biology. It is famous for separating nucleic acids according to their buoyant density and for enabling the historic demonstration of semi-conservative DNA replication. This question tests the core principle behind how DNA bands form in a CsCl gradient. Given Data / Assumptions: Under ultracentrifugation, CsCl forms a continuous density gradient. DNA migrates until it reaches a position where its density equals the surrounding solution. Bouyant density depends on base composition (notably GC content) and isotopic labeling (e.g., 15N vs 14N). Concept / Approach: When spun at high speed, CsCl solutions stratify such that each position corresponds to a specific solution density. DNA does not separate by size in this setup; instead, it moves to the exact point in the gradient where net buoyancy is zero. Differences in GC content or isotopic labeling alter buoyant density slightly, allowing discrete bands for different DNA populations. Step-by-Step Solution: Prepare DNA sample and CsCl solution at appropriate concentration.Ultracentrifuge to establish a stable density gradient along the tube.DNA molecules migrate through the gradient.Each DNA species bands where DNA density equals local CsCl density (buoyant equilibrium).Visualize bands (e.g., with ethidium bromide) at distinct positions corresponding to different densities. Verification / Alternative check: In the Meselson–Stahl experiment, 15N-labeled DNA bands at a higher density than 14N DNA, proving separation is by density rather than length or charge. Mixing populations yields distinct or hybrid-density bands exactly where predicted by buoyant-density calculations. Why Other Options Are Wrong: Absorption/resorption: not the driving force for banding in CsCl. Adhesion to the tube: irrelevant to equilibrium band formation. Net electric charge: electrophoresis uses charge/size; CsCl gradients use buoyant density. Common Pitfalls: Confusing CsCl equilibrium density gradients with sucrose rate-zonal gradients (which separate primarily by size/shape over time, not by equilibrium density). Final Answer: Density (buoyant density in the CsCl gradient)

Question 19

DNA structure – Which statement about base pairing in the DNA double helix is incorrect?

Accepted Answer

Introduction: Watson–Crick base pairing underpins the stability, fidelity, and information storage capacity of DNA. Knowing which bases pair in standard B-form DNA is fundamental for understanding replication, transcription, and hybridization technologies. Given Data / Assumptions: DNA is a double helix with antiparallel strands. Complementary base pairing occurs via hydrogen bonds. A pairs with T; G pairs with C in canonical pairing. Concept / Approach: In standard B-DNA, adenine (A) forms specific hydrogen bonds with thymine (T), whereas guanine (G) pairs with cytosine (C). These pairings enforce a uniform helix geometry and enable predictable melting behavior. Any statement contradicting A–T or G–C pairing is incorrect for canonical DNA under physiological conditions. Step-by-Step Solution: Identify canonical pairs: A–T and G–C.Scan statements for mismatched pairs.Option e claims A pairs with C, which is false for standard B-DNA.Therefore, the incorrect statement is the one asserting A–C pairing. Verification / Alternative check: Thermodynamic parameters and structural data consistently show stability of A–T and G–C pairs, with mismatches (e.g., A–C) destabilizing duplexes and used diagnostically in SNP genotyping assays. Why Other Options Are Wrong: Strands held by base interactions: correct (hydrogen bonding plus stacking). A with T and G with C: both correct. “None of the above are incorrect”: incorrect because A–C is wrong. Common Pitfalls: Confusing RNA pairing (A–U) with DNA; in DNA, thymine replaces uracil. Final Answer: Adenine normally pairs with cytosine in standard B-DNA.

Question 20

DNA replication (prokaryotes) – Which DNA polymerase removes RNA primers during DNA synthesis in bacteria such as E. coli?

Accepted Answer

Introduction: During bacterial DNA replication, short RNA primers synthesized by primase must be removed and replaced with DNA. Recognizing the enzyme responsible for primer removal is essential for understanding lagging-strand processing and Okazaki fragment maturation. Given Data / Assumptions: Primase synthesizes RNA primers to initiate DNA synthesis. Okazaki fragments on the lagging strand are RNA–DNA hybrids. Enzymes differ in exonuclease activities and processivity. Concept / Approach: E. coli DNA polymerase I possesses a unique 5'→3' exonuclease activity that removes RNA primers while simultaneously filling the resulting gaps with DNA (nick translation). DNA polymerase III is the main replicative enzyme but lacks the 5'→3' exonuclease for primer removal. DNA ligase subsequently seals remaining nicks by forming phosphodiester bonds. Step-by-Step Solution: Primase lays down RNA primers on both strands (especially frequent on lagging strand).DNA pol III extends primers, creating Okazaki fragments.DNA pol I uses 5'→3' exonuclease to remove RNA primer nucleotides.DNA pol I fills in with DNA using its polymerase activity.DNA ligase joins adjacent fragments by sealing the nick. Verification / Alternative check: Mutants lacking DNA pol I 5'→3' exonuclease activity accumulate RNA primers and unprocessed Okazaki fragments, confirming the enzyme's specialized role in primer removal and gap filling. Why Other Options Are Wrong: DNA pol II: involved in DNA repair and restart; lacks primer removal role. DNA pol III: high-processivity replication polymerase but lacks 5'→3' exonuclease. Primase: synthesizes primers; does not remove them. None of these: incorrect because DNA pol I is correct. Common Pitfalls: Assuming the main replicative polymerase (pol III) also removes primers; in bacteria, primer removal is a distinct function of pol I. Final Answer: DNA polymerase I

DNA Structure and Replication Questions