Question 1

DNA replication fidelity: Which statement correctly describes the proofreading activity that maintains the accuracy of DNA synthesis in cells?

Accepted Answer

Introduction / Context: High fidelity DNA replication is essential for genome stability. DNA polymerases not only synthesize new strands but also correct errors in real time. This question tests your understanding of the intrinsic proofreading mechanism associated with replication. Given Data / Assumptions: Replication requires both polymerization and error correction. We compare on-the-fly proofreading versus post-replicative repair. Mechanistic detail: exonuclease directionality. Concept / Approach: Most replicative DNA polymerases possess 3'–5' exonuclease activity that removes a misincorporated nucleotide immediately after it is added. This backtracking and excision reduce error rates dramatically before mismatch repair acts. Proofreading is intrinsic to the polymerase holoenzyme in both prokaryotes and eukaryotes (e.g., E. coli Pol III, eukaryotic Pol δ and Pol ε). Step-by-Step Solution: During elongation, an incorrect base pairing slows polymerase and destabilizes the 3' end.The polymerase shifts the nascent strand into the 3'–5' exonuclease site.The mismatched nucleotide is excised, restoring a correct 3' terminus.Polymerization resumes, greatly improving fidelity. Verification / Alternative check: Mutations disabling the 3'–5' exonuclease increase mutation rates, demonstrating its role in proofreading apart from postreplicative mismatch repair pathways. Why Other Options Are Wrong: “Occurs after completion”: that describes mismatch repair, not polymerase proofreading. “Requires separate enzyme”: proofreading is within the polymerase complex, not a separate nuclease. “Occurs in prokaryotes but not eukaryotes”: incorrect; eukaryotic replicative polymerases also proofread. Common Pitfalls: Confusing proofreading (intrinsic, 3'–5' exonuclease) with mismatch repair (postreplicative pathway) or nucleotide excision repair (damage removal). Final Answer: It is a function of the 3'–5' exonuclease activity of DNA polymerases

Question 2

In molecular genetics, which enzyme complex unwinds short stretches of the DNA double helix at the replication fork so that polymerases can copy the strands?

Accepted Answer

Introduction / Context: DNA replication requires coordinated enzymes that open, stabilize, and copy parental strands. A common point of confusion is distinguishing the enzyme that directly unwinds base pairs at the fork from enzymes that relax supercoils or stabilize single strands. This question checks precise roles at the replication fork. Given Data / Assumptions: The site of interest is the replication fork. We are focusing on the enzyme that separates the two DNA strands (breaks base pairing by ATP-driven translocation). Other fork-associated proteins (topoisomerase, SSB, primase) may be present but have different tasks. Concept / Approach: Helicases are motor proteins that hydrolyze ATP to move along nucleic acid and disrupt hydrogen bonds between complementary bases, creating single-stranded templates. Topoisomerases (including gyrase) relieve torsional strain ahead of the fork by transiently breaking and rejoining DNA but do not separate base pairs at the fork. SSB proteins bind exposed single strands to prevent re-annealing. Primase synthesizes short RNA primers for initiation but does not unwind DNA. Step-by-Step Solution: Identify the action: strand separation at the fork requires ATP-driven motor activity.Match function to enzyme: helicase performs strand separation.Differentiate from topoisomerase: relieves supercoils but does not unzip base pairs at the fork.Differentiate from SSB: stabilizes single strands after helicase action.Differentiate from primase: lays RNA primers on the exposed template. Verification / Alternative check: Biochemistry texts describe DnaB (prokaryotic replicative helicase) encircling one strand and advancing 5′→3′ (or 3′→5′ depending on system) to open the fork while topoisomerase acts further upstream to prevent overwinding. Why Other Options Are Wrong: DNA polymerases: synthesize DNA; they require an already unwound template. SSB proteins: bind single strands; they do not unwind duplex DNA. Topoisomerases: relax supercoils; they are not the primary strand-separating motor at the fork. Primase: makes RNA primers; no unwinding activity. Common Pitfalls: Equating “unwinding ahead of the fork” with topoisomerase activity. Topoisomerase prevents torsional buildup; helicase performs the base-pair separation at the fork itself. Final Answer: Helicases

Question 3

Why eukaryotic chromosome replication finishes in a relatively short time: choose the best reason

Accepted Answer

Introduction / Context: Eukaryotic genomes are large, yet S phase completes in hours. This is made possible not by super-fast polymerases but by orchestrating replication initiation at many sites simultaneously. Understanding “replicons” and “origins” is key. Given Data / Assumptions: Eukaryotic chromosomes are linear and much larger than bacterial chromosomes. Replication speed per fork is limited by enzyme kinetics and chromatin context. Cells can fire many replication origins in parallel during S phase. Concept / Approach: A “replicon” is the DNA segment replicated from a single origin. Eukaryotic chromosomes have thousands of licensed origins; many fire to create numerous bidirectional forks that replicate different regions concurrently, dramatically shortening total time. Polymerase rates (nucleotides per second) are not orders of magnitude faster than bacteria; parallelism is the solution. Step-by-Step Solution: Identify constraints: polymerase speed is similar order of magnitude across life.Recognize strategy: increase fork number via multiple origins.Conclude: many replicons operating simultaneously compress overall replication time. Verification / Alternative check: DNA fiber assays show multiple initiation events along eukaryotic DNA; origin licensing factors (for example, ORC, Cdc6, MCM) support this mechanism. Why Other Options Are Wrong: More DNA (A) is a challenge, not a reason replication is fast. 1000× faster enzymes (B) is false; eukaryotic polymerases are not that much faster. Always single-stranded DNA (D) is incorrect; genomic DNA is double-stranded. Nucleosomes (E) actually impose structural hurdles; primers are still required. Common Pitfalls: Confusing polymerase processivity with fork number; speedups primarily come from simultaneous initiation, not extreme catalytic acceleration. Final Answer: Each eukaryotic chromosome contains multiple replicons (many origins of replication)

Question 4

During which fundamental process is an exact new copy of a DNA molecule synthesized?

Accepted Answer

Introduction / Context: Central dogma steps—replication, transcription, and translation—have distinct information flows. Only one process duplicates the genetic material itself to pass to daughter cells: replication. Given Data / Assumptions: The question asks about producing a precise, complementary copy of a DNA molecule. Other listed processes relate to RNA synthesis, protein synthesis, or DNA exchange. Concept / Approach: Replication is semi-conservative: each daughter duplex contains one parental and one newly synthesized strand. DNA polymerases copy templates in the 5′→3′ direction using primers. Transcription produces RNA, translation produces polypeptides, transformation and recombination alter DNA content but do not copy the entire DNA molecule as a dedicated process. Step-by-Step Solution: Define “exact new copy” as duplication of the DNA duplex.Identify the enzymatic process that accomplishes this: DNA replication.Exclude transcription (DNA→RNA), translation (RNA→protein), and transformation/recombination (DNA movement/reshuffling). Verification / Alternative check: Meselson–Stahl experiments demonstrated semi-conservative replication using isotope labeling of DNA strands, directly confirming the mechanism. Why Other Options Are Wrong: Transformation: introduces external DNA but does not copy the entire genome. Transcription: creates RNA, not DNA. Translation: uses mRNA to build proteins. Recombination: rearranges DNA segments, not wholesale duplication. Common Pitfalls: Equating “trans” words (transcription/translation/transformation) with duplication; only replication duplicates DNA. Final Answer: Replication (DNA-directed DNA synthesis)

Question 5

Name the enzyme that catalyzes template-directed synthesis of DNA during replication

Accepted Answer

Introduction / Context: Multiple enzymes collaborate at the replication fork. The central catalytic activity that elongates DNA strands belongs to the DNA polymerases, which add deoxyribonucleotides to a primer in a template-directed manner. Given Data / Assumptions: Reaction: polymerization of deoxynucleoside triphosphates into DNA. Directionality: 5′→3′ chain growth with a primer and template. Other fork proteins have different supporting roles. Concept / Approach: DNA polymerases catalyze phosphodiester bond formation, requiring a 3′-OH and a base-paired template. Helicase unwinds duplex DNA; gyrase relieves supercoils; ligase seals nicks between Okazaki fragments; primase lays RNA primers that polymerase extends. Step-by-Step Solution: Identify the reaction: DNA synthesis.Map enzyme to reaction: DNA polymerase performs chain elongation.Exclude accessory enzymes that enable but do not catalyze DNA polymerization. Verification / Alternative check: In E. coli, DNA polymerase III holoenzyme is the main replicase; DNA polymerase I removes RNA primers and fills gaps; eukaryotes use Pol α (primase), Pol δ/ε for elongation. Why Other Options Are Wrong: Gyrase: topological management, not polymerization. Ligase: joins nicks, no base addition. Helicase: unwinds DNA, no polymerization. Primase: synthesizes short RNA primers, not DNA extension beyond its brief role in eukaryotic Pol α complex. Common Pitfalls: Attributing the core polymerization to “ligase” or “helicase” due to their high visibility at the fork. Final Answer: DNA polymerase

Question 6

In prokaryotes with circular chromosomes, where does replication begin?

Accepted Answer

Introduction / Context: Bacterial chromosomes are typically circular with a single defined origin of replication (oriC). Accurate initiation ensures complete replication and proper cell cycle control. Given Data / Assumptions: Organism: a generic prokaryote like Escherichia coli. Chromosome: circular, with one well-characterized origin. Initiator proteins (for example, DnaA) recognize specific origin sequences. Concept / Approach: Replication initiates at oriC, where initiator proteins assemble, melt AT-rich elements, and recruit the helicase–primase–polymerase machinery. Bidirectional forks proceed from oriC to the terminus. Promoters are for transcription initiation, not replication. Step-by-Step Solution: Identify specialized DNA element: oriC.Recognize fixed sequence features: DnaA boxes, AT-rich DNA unwinding element.Conclude replication starts at the same locus each cycle under normal conditions. Verification / Alternative check: Genetic deletion or mutation of oriC prevents chromosomal replication; ectopic origins on plasmids demonstrate sequence-specific initiation. Why Other Options Are Wrong: AT-rich stretches (B) are common but are not sufficient; initiation is sequence- and protein-controlled. Random initiation (C) would cause catastrophic genome instability. Promoters (D) and ribosome-binding sites (E) control transcription/translation, not replication. Common Pitfalls: Confusing origin melting at AT-rich regions with the idea that any AT-rich region could serve as an origin. Final Answer: Always at the same DNA locus (the unique origin of replication, oriC)

Question 7

Which DNA polymerase possesses both 5′→3′ and 3′→5′ exonuclease activities?

Accepted Answer

Introduction / Context: Exonuclease activities provide proofreading and primer removal during replication and repair. Different polymerases carry distinct nuclease domains that determine their functional niches at the fork. Given Data / Assumptions: 5′→3′ exonuclease removes RNA primers/nicked DNA ahead of the polymerase. 3′→5′ exonuclease provides proofreading by excising misincorporated bases. We compare E. coli Pol I (Kornberg enzyme), Pol III, and thermostable Taq polymerase. Concept / Approach: E. coli DNA polymerase I has both 5′→3′ and 3′→5′ exonuclease activities, enabling nick translation and proofreading. Pol III, the main replicase, has strong 3′→5′ proofreading but lacks an intrinsic 5′→3′ exonuclease. Taq polymerase lacks 3′→5′ proofreading and therefore has relatively higher error rates; it also does not provide the robust 5′→3′ exonuclease used for primer removal in vivo. Step-by-Step Solution: List activities: Pol I = polymerase + 3′→5′ exonuclease + 5′→3′ exonuclease.Pol III = polymerase + 3′→5′ exonuclease; no 5′→3′ exonuclease.Taq = polymerase largely without 3′→5′ proofreading; not used for nick translation.Therefore, the enzyme with both exonuclease directions is Pol I (Kornberg enzyme). Verification / Alternative check: Biochemical assays of nick translation and primer removal in E. coli require Pol I’s 5′→3′ exonuclease; mutants in this domain abolish the activity. Why Other Options Are Wrong: Pol III: lacks 5′→3′ exonuclease domain. Taq Pol: lacks 3′→5′ proofreading and is error-prone. None of these: incorrect because Pol I has both activities. Ligase: joins nicks; no exonuclease activity. Common Pitfalls: Assuming the main replicase must have all nuclease functions; in reality, tasks are partitioned among enzymes. Final Answer: Kornberg enzyme (E. coli DNA polymerase I)

Question 8

Approximate rate of DNA replication in prokaryotes (nucleotides added per second per fork)

Accepted Answer

Introduction / Context: Replication speed influences cell cycle timing and genome duplication efficiency. Prokaryotic replicative polymerases are highly processive and coordinated with helicase, clamp loader, and sliding clamp, enabling rapid synthesis. Given Data / Assumptions: Organism: typical bacteria such as E. coli under optimal growth. Metric: nucleotides incorporated per second at an active fork. Forks are generally bidirectional from the origin. Concept / Approach: Experimental measurements (pulse-chase labeling, DNA fiber autoradiography) consistently place bacterial fork rates on the order of 500–1,000 nt/s per strand under good conditions. A value near 1,000 nt/s is a widely taught benchmark. Step-by-Step Solution: Consider polymerase catalytic rate and replisome coordination.Select the standard textbook figure: ~10^3 nt/s.Reject implausibly low (10, 100) and unrealistically high (10,000, 50,000) rates. Verification / Alternative check: Doubling times of fast-growing bacteria align with genome sizes when assuming ~1000 nt/s and two forks per origin. Why Other Options Are Wrong: 10 or 100 nt/s: too slow to explain observed replication and doubling times. 10,000 or 50,000 nt/s: exceed known enzymatic and replisome constraints. Common Pitfalls: Confusing eukaryotic fork rates (typically slower) with prokaryotic rates; values are organism- and condition-dependent but order of magnitude is consistent. Final Answer: About 1,000 bases per second

Question 9

Directionality of DNA chain elongation by DNA polymerases during replication

Accepted Answer

Introduction / Context: All known DNA polymerases add nucleotides to the 3′-OH of a growing strand, making synthesis intrinsically 5′→3′. Understanding this constraint explains leading and lagging strand strategies with Okazaki fragments. Given Data / Assumptions: Enzyme: DNA polymerase (replicative or repair). Substrate: dNTPs; primer with free 3′-OH; template strand. Direction: determined by chemistry of phosphodiester bond formation. Concept / Approach: The 3′-OH of the primer attacks the alpha-phosphate of an incoming dNTP, releasing pyrophosphate. This mechanism only permits 5′→3′ extension. On the lagging strand, discontinuous synthesis produces Okazaki fragments that are later joined by ligase; still, each fragment is synthesized 5′→3′. Step-by-Step Solution: Identify the nucleophile (primer 3′-OH) and electrophile (dNTP α-phosphate).Recognize that reversal would create unstable chemistry and impair proofreading.Conclude universal 5′→3′ chain elongation. Verification / Alternative check: No cellular DNA polymerase has been found that synthesizes 3′→5′; this is consistent across domains of life. Why Other Options Are Wrong: 3′→5′, 5′→5′, 3′→3′: chemically inconsistent with polymerase mechanism. Either direction: not supported by known biology; both strands are synthesized 5′→3′. Common Pitfalls: Assuming the leading strand is 3′→5′ because it is continuous—direction refers to synthesis, not template orientation. Final Answer: 5′ to 3′ direction

Question 10

Select the correct statement about chromosomal DNA replication in Escherichia coli

Accepted Answer

Introduction / Context: E. coli replication is a textbook model: a single origin (oriC) fires to produce two replication forks that move in opposite directions around the circular chromosome. Given Data / Assumptions: Organism: E. coli in normal growth (no specialized phage systems controlling replication). Genome: circular with one oriC and a defined terminus region. Fork movement: two forks in opposite directions. Concept / Approach: Bidirectional replication halves the time needed to copy the chromosome compared with unidirectional synthesis. Phage infections can introduce their own polymerases for phage DNA replication, but chromosomal replication still relies on the host replisome and oriC regulation. Step-by-Step Solution: Identify initiation site: oriC.Describe fork dynamics: two forks diverge and meet at terminus.Exclude phage-dependent or promoter-dependent misconceptions. Verification / Alternative check: Marker frequency analysis and visualization of replication intermediates confirm two forks moving bidirectionally. Why Other Options Are Wrong: Unidirectional (A): inconsistent with well-established bidirectionality. Use of T7 polymerase (C) pertains to phage genome replication, not host chromosome. Only during λ infection (D): false; cells replicate in normal cycles. Promoters (E) control transcription, not replication initiation. Common Pitfalls: Conflating transcriptional promoters with replication origins and phage replication systems with host chromosomal replication. Final Answer: Replication proceeds bidirectionally from the origin

Question 11

Identify the antibiotic class that inhibits bacterial DNA gyrase (a type II topoisomerase)

Accepted Answer

Introduction / Context: DNA gyrase introduces negative supercoils and relieves torsional stress during replication and transcription in bacteria. Quinolone antibiotics exploit this essential function by stabilizing DNA–enzyme cleavage complexes, blocking replication and causing lethal DNA damage. Given Data / Assumptions: Target enzyme: DNA gyrase (topoisomerase II) in Gram-negative bacteria; topoisomerase IV also targeted in some organisms. Drugs to compare: quinolones vs. inhibitors of protein synthesis or cell wall synthesis. Outcome: inhibition of DNA replication and bacterial death. Concept / Approach: Nalidixic acid is the prototypical quinolone; fluoroquinolones (ciprofloxacin, levofloxacin) are widely used derivatives. They bind to the DNA–gyrase complex, preventing re-ligation of cleaved DNA. Tetracyclines block aminoacyl-tRNA binding to the 30S ribosome; cephalosporins inhibit peptidoglycan cross-linking; macrolides bind 50S to impede translocation—none of which directly inhibit gyrase. Step-by-Step Solution: Match mechanism (gyrase inhibition) to drug class (quinolones).Identify nalidixic acid as a quinolone.Exclude protein synthesis and cell wall inhibitors. Verification / Alternative check: Genetic mutations in gyrA/gyrB confer quinolone resistance, confirming the target pathway. Why Other Options Are Wrong: Tetracycline: ribosomal inhibitor; no direct effect on gyrase. Both (C): incorrect because tetracycline does not inhibit gyrase. Cephalosporin: targets penicillin-binding proteins in cell wall synthesis. Macrolide: ribosomal 50S inhibitor; not a topoisomerase inhibitor. Common Pitfalls: Assuming all bactericidal antibiotics target DNA; mechanisms vary widely across classes. Final Answer: Nalidixic acid (a quinolone targeting DNA gyrase)

Question 12

In molecular genetics, which enzyme repairs a single-strand nick in DNA by sealing the backbone (forming a phosphodiester bond between adjacent nucleotides)?

Accepted Answer

Introduction / Context: DNA integrity depends on enzymes that maintain the sugar–phosphate backbone. A very common lesion is a single-strand nick, where the phosphodiester bond is missing between adjacent nucleotides. The cell must reseal this break to restore continuity before replication or recombination can proceed. Given Data / Assumptions: The break is a clean nick with correct base pairing already in place. Only the phosphodiester linkage between the 3'-OH and 5'-phosphate is absent. No unwinding or supercoil resolution is required at this step. Concept / Approach: DNA ligase catalyzes the formation of a phosphodiester bond joining a 3'-hydroxyl to a 5'-phosphate, typically using ATP (eukaryotes, many viruses) or NAD+ (many bacteria) as an energy source. This reaction is the final step in DNA replication fragments joining, base-excision repair, and many recombination events. Step-by-Step Solution: Identify the biochemical task: seal a backbone nick (3'-OH to 5'-phosphate).Recall the enzyme whose canonical role is ligation of DNA ends: DNA ligase.Exclude enzymes that unwind or relax DNA (helicase, gyrase, topoisomerases). Verification / Alternative check: In laboratory cloning, T4 DNA ligase is routinely used to join compatible DNA ends, directly demonstrating its role in phosphodiester bond formation. Why Other Options Are Wrong: Helicase: unwinds duplex DNA; does not seal nicks. DNA gyrase/topoisomerases: change DNA topology (supercoiling), not ligation of a simple nick. SSB protein: stabilizes single strands; no catalytic ligation activity. Common Pitfalls: Confusing topoisomerase-mediated strand passage (which involves transient breaks) with permanent sealing by ligase; only ligase leaves the backbone intact at the end of repair. Final Answer: DNA ligase

Question 13

In Escherichia coli DNA replication, which enzyme synthesizes the short RNA primers required to initiate Okazaki fragment synthesis on the lagging strand?

Accepted Answer

Introduction / Context: DNA polymerases cannot start synthesis de novo; they extend an existing 3'-OH. Cells therefore use an RNA primase to lay down short RNA primers that provide the first 3'-OH for polymerase extension, especially important for discontinuous lagging-strand synthesis (Okazaki fragments). Given Data / Assumptions: Organism: E. coli. Task: create RNA primers on the lagging strand. Protein naming convention: DnaA, DnaB, DnaC, DnaG are replication factors with distinct roles. Concept / Approach: DnaG is the primase in E. coli. It binds the helicase (DnaB) at the replication fork to intermittently synthesize short RNA primers (about 10–12 nucleotides), allowing DNA polymerase III to extend these primers into Okazaki fragments. Step-by-Step Solution: Identify the needed activity: RNA primer synthesis.Map E. coli proteins to functions: DnaG = primase.Select DnaG as the correct enzyme. Verification / Alternative check: Mutations in dnaG abolish primer synthesis and block lagging-strand replication, confirming its essential role. Why Other Options Are Wrong: DnaA: initiator that melts oriC; does not synthesize RNA. DnaC: loads DnaB helicase; no primase activity. “All of these”: only DnaG has primase activity. DNA polymerase I: removes RNA primers (5'→3' exonuclease) and fills gaps; it does not create primers. Common Pitfalls: Conflating priming (RNA synthesis) with primer removal and gap filling (Pol I); these are separate steps carried out by different enzymes. Final Answer: DnaG (primase)

Question 14

During which DNA-related process do both strands of the parental DNA act as templates at the same time, yielding two daughter duplexes (leading and lagging synthesis coordinated at a replisome)?

Accepted Answer

Introduction / Context: Cells must duplicate their genomes before division. Semi-conservative replication uses each parental strand as a template to build a complementary daughter strand, producing two identical duplex molecules. At the replication fork, synthesis on both templates is coordinated by a multi-enzyme replisome. Given Data / Assumptions: Replication forks generate leading and lagging strands. Leading strand: continuous 5'→3' synthesis. Lagging strand: discontinuous 5'→3' synthesis (Okazaki fragments) with RNA priming. Concept / Approach: In replication, both templates are copied simultaneously: one continuously, one discontinuously. Repair pathways (excision, mismatch) typically excise and resynthesize short patches on one strand rather than duplicating the entire chromosome concurrently from both templates. Step-by-Step Solution: Identify the process that necessarily uses both strands as templates across long regions: replication.Contrast with repair pathways that replace limited tracts, not full duplex duplication.Select replication as correct. Verification / Alternative check: Meselson–Stahl experiments demonstrated semi-conservative replication: each daughter duplex contains one old and one new strand, proving both parental strands serve as templates. Why Other Options Are Wrong: Excision/mismatch repair: template choice is local and patch-based; not simultaneous full-strand copying. “None”: incorrect because replication precisely fits the description. Common Pitfalls: Assuming “both strands” implies bidirectional repair; in reality, only replication mandates long, coordinated synthesis on both templates at forks. Final Answer: Replication

Question 15

Which enzyme listed lacks 3'→5' exonuclease (proofreading) activity under standard conditions used in PCR and DNA synthesis workflows?

Accepted Answer

Introduction / Context: Proofreading improves fidelity by removing misincorporated nucleotides via 3'→5' exonuclease activity. Different polymerases vary in this capability, directly affecting error rates in amplification and synthesis applications such as PCR or cloning. Given Data / Assumptions: Klenow fragment retains 3'→5' exonuclease. Taq polymerase (from Thermus aquaticus) lacks 3'→5' proofreading. Ligases do not have exonuclease activity and are not polymerases. Concept / Approach: Taq DNA polymerase performs 5'→3' polymerization but lacks 3'→5' exonuclease. Consequently, its base substitution error rate is higher than that of proofreading polymerases (for example, Pfu, Phusion). Klenow and E. coli Pol III possess proofreading; ligase is irrelevant to exonuclease discussion. Step-by-Step Solution: Identify which enzyme is classically known as non-proofreading: Taq.Eliminate Klenow (has 3'→5' exonuclease) and Pol III (proofreading core subunits).Ignore ligase for exonuclease properties; it seals nicks, not edits bases. Verification / Alternative check: High-fidelity PCR protocols replace Taq with proofreading enzymes specifically to gain 3'→5' exonuclease activity and reduce errors. Why Other Options Are Wrong: Klenow: retains proofreading even though it lacks 5'→3' exonuclease of full Pol I. T4 DNA ligase: not a polymerase; exonuclease activity is not its function. “All of the above”: false because Klenow and Pol III proofread. Common Pitfalls: Assuming thermostable equals proofreading; thermostability and exonuclease capability are independent traits. Final Answer: Taq DNA polymerase

DNA Replication Questions