MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5d Nucleotides Nucleic Acids

What this quiz covers

This quiz focuses on 5d Nucleotides Nucleic Acids, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. RNA's 2′-OH favors an A-form helix by constraining sugar pucker, whereas DNA lacking a 2′-OH more readily adopts B-form geometry. (correct answer)
2. RNA forms A-form helices because uracil makes three hydrogen bonds with adenine, increasing helix diameter relative to DNA.
3. DNA adopts B-form helices because thymine is a purine, which stacks more strongly than pyrimidines in RNA.
4. RNA adopts A-form helices because its phosphodiester linkage is 2′→5′, whereas DNA is 3′→5′.

Explanation: This question tests knowledge of how sugar structure influences nucleic acid helical geometry. RNA contains ribose with a 2'-OH group, while DNA contains deoxyribose lacking this hydroxyl. The 2'-OH in RNA constrains the sugar to adopt a C3'-endo pucker, which favors the A-form helix with its wider, shorter structure. DNA's deoxyribose allows more conformational flexibility, typically adopting a C2'-endo pucker that promotes B-form geometry with its narrower, longer helix. Option B incorrectly claims uracil forms three hydrogen bonds with adenine (it forms two, like thymine), and option C wrongly identifies thymine as a purine (it's a pyrimidine). The structural difference at the 2' position is the primary determinant of helical form. When comparing RNA and DNA structures, always consider how the presence or absence of the 2'-OH affects sugar pucker and resulting helical geometry.

Question 2

1. The RNA initially carried a 5′-phosphate that blocked PNK labeling until it was removed to generate a 5′-OH. (correct answer)
2. The RNA initially carried a 3′-phosphate that blocked PNK labeling until it was removed to generate a 3′-OH.
3. The RNA initially lacked bases, preventing PNK from recognizing the sugar-phosphate backbone.
4. The RNA initially contained deoxyribose, and phosphatase converted it to ribose to enable labeling.

Explanation: This question tests knowledge of nucleic acid end modifications and enzymatic labeling requirements. Polynucleotide kinase (PNK) transfers the γ-phosphate from ATP to a free 5′-OH group, but cannot act on a 5′-phosphorylated end due to the lack of a hydroxyl acceptor. The passage indicates that RNA could only be labeled after alkaline phosphatase treatment, which removes terminal phosphate groups, suggesting the original RNA carried a 5′-phosphate that blocked PNK activity. The correct answer A accurately identifies that the RNA initially carried a 5′-phosphate that prevented PNK labeling until phosphatase treatment generated a free 5′-OH. Answer B incorrectly focuses on the 3′ end (PNK acts at the 5′ end), while answers C and D propose implausible scenarios about missing bases or sugar conversions. In nucleotide labeling experiments, always verify the chemical requirements of the labeling enzyme, particularly whether it requires free hydroxyl groups or can act on phosphorylated termini, as this determines necessary pretreatment steps.

Question 3

1. The newly synthesized strand will contain 5′-ATG-3′ at that position instead of 5′-ATG-3′.
2. The newly synthesized strand will contain 5′-ACG-3′ at that position instead of 5′-ATG-3′. (correct answer)
3. The newly synthesized strand will contain 5′-UAC-3′ at that position because uracil replaces thymine during replication.
4. The newly synthesized strand will be unchanged because point mutations on the template are corrected by base pairing alone.

Explanation: This question assesses understanding of Watson-Crick base pairing rules and their consequences for DNA replication fidelity. DNA replication relies on complementary base pairing where adenine (A) pairs with thymine (T) and guanine (G) pairs with cytosine (C), with the polymerase incorporating nucleotides complementary to the template strand. The passage describes a mutation changing the template strand from 3′-TAC-5′ to 3′-TGC-5′, which means the middle base changed from A to G. Following Watson-Crick pairing rules, the newly synthesized strand will incorporate C opposite the mutated G position instead of T opposite the original A, resulting in 5′-ACG-3′ instead of 5′-ATG-3′. The correct answer B accurately predicts this outcome based on complementary base pairing. Answer A shows no change (incorrect), answer C incorrectly introduces uracil (found in RNA, not DNA), and answer D incorrectly suggests mutations are corrected by base pairing alone. When analyzing mutations and replication, always apply Watson-Crick base pairing rules systematically to predict the sequence of newly synthesized strands.

Question 4

1. Blocking cytosine's hydrogen-bonding pattern would selectively weaken C·G pairing and reduce duplex formation. (correct answer)
2. Blocking cytosine's hydrogen-bonding pattern would selectively weaken A·T pairing and reduce duplex formation.
3. Blocking cytosine has no effect because duplex formation depends only on ionic interactions of phosphates.
4. Blocking cytosine increases duplex formation by preventing mismatches from occurring during annealing.

Explanation: This question evaluates understanding of base-specific modifications and their effects on duplex formation. Cytosine forms three hydrogen bonds with guanine through specific donor and acceptor positions, and chemical modification that blocks these positions prevents proper C·G pairing. Since C·G pairs contribute significantly to duplex stability (three hydrogen bonds versus two for A·T), disrupting cytosine's ability to pair reduces overall duplex formation efficiency. The correct answer A accurately explains that blocking cytosine's hydrogen-bonding pattern would selectively weaken C·G pairing and reduce duplex formation. Answer B incorrectly suggests effects on A·T pairing, answer C incorrectly dismisses the role of base pairing, and answer D incorrectly claims blocking increases duplex formation. When analyzing base modifications, always consider which specific base pairs are affected and how disrupting hydrogen bonding patterns impacts overall duplex stability, with modifications to bases involved in stronger pairing having greater effects.

Question 5

1. Higher fluorescence with dsDNA is consistent with intercalation into stacked base pairs that are more prevalent in duplex DNA. (correct answer)
2. Higher fluorescence with dsDNA is consistent with covalent bond formation between the dye and phosphates unique to dsDNA.
3. Higher fluorescence with dsDNA indicates dsDNA has more nucleotides than ssDNA at the same concentration.
4. Higher fluorescence with dsDNA is best explained by increased 2′-OH content in DNA relative to RNA.

Explanation: This question assesses understanding of DNA intercalation and its relationship to duplex structure. Intercalating agents insert between stacked base pairs in double-stranded DNA, with the regular base stacking of duplex DNA providing optimal binding sites that are absent in the more flexible single-stranded form. The enhanced fluorescence with dsDNA reflects the dye's preferential binding to the organized, stacked structure of the double helix. The correct answer A accurately explains that higher fluorescence with dsDNA is consistent with intercalation into stacked base pairs prevalent in duplex DNA. Answer B incorrectly proposes covalent bond formation, answer C incorrectly suggests different nucleotide content, and answer D incorrectly invokes 2′-OH differences between DNA and RNA. When analyzing DNA-binding molecules, always consider how the regular structure of duplex DNA, particularly base stacking, creates unique binding sites for intercalators that enhance their fluorescent properties.

Question 6

1. The linkage described matches the canonical nucleic acid backbone connectivity found in DNA and RNA. (correct answer)
2. The linkage described is a peptide bond, explaining nucleic acid backbone stability.
3. The linkage described is 2′→5′, which is the dominant linkage in genomic DNA.
4. The linkage described requires thymine's methyl group to form, so it occurs only in DNA.

Explanation: This question tests knowledge of phosphodiester bond connectivity in nucleic acids. Natural DNA and RNA backbones are formed by 3′→5′ phosphodiester linkages, where the 3′-OH of one nucleotide attacks the 5′-phosphate of the next, creating a directional polymer with distinct 5′ and 3′ ends. This specific connectivity is universal in biological nucleic acids and determines the directionality of synthesis and degradation. The correct answer A accurately identifies that the described linkage matches the canonical backbone connectivity found in DNA and RNA. Answer B incorrectly identifies it as a peptide bond, answer C incorrectly claims 2′→5′ linkages dominate in genomic DNA, and answer D incorrectly requires thymine's methyl group for bond formation. In nucleic acid structure analysis, always verify that synthetic constructs maintain the natural 3′→5′ phosphodiester linkage pattern, as alternative connectivities can dramatically alter biological recognition and function.

Question 7

1. A free 3′-OH on the primer is required to form the next phosphodiester bond during chain elongation. (correct answer)
2. A free 5′-OH on the primer is required because synthesis proceeds 3′→5′ on the primer strand.
3. Extension fails because 3′-deoxy primers cannot hydrogen-bond to the template bases.
4. Extension fails because polymerases incorporate ribonucleotides only when the primer lacks a 3′-OH.

Explanation: This question tests knowledge of DNA polymerase mechanism and the role of the 3′-OH in nucleic acid synthesis. DNA polymerase catalyzes phosphodiester bond formation by facilitating nucleophilic attack of the primer's 3′-OH on the α-phosphate of an incoming nucleotide triphosphate, releasing pyrophosphate and extending the chain. The passage shows that a primer with a free 3′-OH supports full extension while a 3′-deoxy primer (3′-H) shows no extension, demonstrating the absolute requirement for the 3′-OH nucleophile. The correct answer A accurately states that the free 3′-OH is required for phosphodiester bond formation during chain elongation. Answer B incorrectly suggests synthesis proceeds 3′→5′ on the primer strand (it actually proceeds 5′→3′), while answers C and D propose incorrect mechanisms unrelated to the actual requirement for a nucleophilic 3′-OH. In nucleotide polymerization reactions, always verify that the growing strand has a free 3′-OH, as this functional group is essential for the catalytic mechanism of all DNA and RNA polymerases.

Question 8

1. The polymerase requires a free 3'-OH to form a new phosphodiester bond during elongation (correct answer)
2. The polymerase initiates synthesis by adding nucleotides to the 5' end of the growing strand
3. The polymerase can only incorporate ribonucleotides because the template is DNA
4. The polymerase catalyzes strand extension by hydrolyzing the phosphodiester backbone of the template

Explanation: This question tests understanding of DNA polymerase mechanism and primer requirements. DNA polymerases cannot initiate synthesis de novo; they require a primer with a free 3'-hydroxyl group to catalyze phosphodiester bond formation. During elongation, the 3'-OH attacks the α-phosphate of an incoming dNTP, releasing pyrophosphate and extending the chain. The experimental results show synthesis only occurs with a primer present (Condition 1), confirming this fundamental requirement. Option B incorrectly states polymerase adds to the 5' end, but synthesis always proceeds 5'→3'. Option C wrongly suggests only ribonucleotides can be incorporated with a DNA template. Option D mischaracterizes the mechanism as hydrolyzing the template backbone. In nucleic acid synthesis problems, always verify that the 3'-OH requirement for chain extension is properly understood.

Question 9

1. ddATP lacks a 2′-OH, preventing formation of Watson–Crick base pairs with thymine.
2. ddATP lacks a 3′-OH, preventing nucleophilic attack needed to form the next phosphodiester bond. (correct answer)
3. ddATP contains uracil instead of adenine, causing mismatch and polymerase stalling.
4. ddATP has an extra phosphate, increasing charge repulsion and forcing chain termination.

Explanation: This question tests understanding of the 3'-OH requirement for DNA synthesis. DNA polymerase catalyzes phosphodiester bond formation by facilitating nucleophilic attack of the 3'-OH on the α-phosphate of an incoming dNTP. Dideoxynucleotides (ddNTPs) lack the 3'-OH group, preventing this nucleophilic attack and terminating chain extension. The correct answer B accurately explains that ddATP lacks a 3'-OH, preventing the next phosphodiester bond formation. Answer A incorrectly focuses on the 2'-OH and base pairing. Answer C wrongly claims ddATP contains uracil instead of adenine. Answer D incorrectly suggests ddATP has an extra phosphate group. In nucleotide polymerization, always verify the presence of a 3'-OH for chain extension and recognize that ddNTPs are chain terminators due to missing 3'-OH.

Question 10

1. A 2'-hydroxyl group that can intramolecularly attack the adjacent phosphate to promote backbone cleavage (correct answer)
2. A 5'-hydroxyl group that hydrolyzes the phosphodiester bond by acting as a leaving group in base
3. A pyrimidine base that undergoes deamination in base, destabilizing the glycosidic bond
4. A methyl group on thymine that increases electron density on the phosphate and accelerates hydrolysis

Explanation: This question assesses understanding of nucleotide structural differences and their chemical reactivity. RNA contains a 2'-hydroxyl group on the ribose sugar that DNA lacks, making RNA susceptible to base-catalyzed hydrolysis. In alkaline conditions, the 2'-OH can act as a nucleophile and attack the adjacent phosphodiester bond, forming a cyclic 2',3'-phosphate intermediate that leads to backbone cleavage. The correct answer (A) identifies this key structural feature that explains why RNA is cleaved by alkali while DNA remains stable. Option B incorrectly suggests the 5'-OH acts as a leaving group, but the 5'-OH is not involved in alkaline hydrolysis. Options C and D propose base modifications that don't explain the differential alkali sensitivity between RNA and DNA. When analyzing nucleic acid stability, always consider the presence of the 2'-OH group in RNA as a key factor in chemical reactivity.

Question 11

1. The polymerase selects incoming nucleotides primarily by complementary base pairing to the template base (correct answer)
2. The polymerase selects incoming nucleotides primarily by recognizing the sugar (ribose vs deoxyribose)
3. The polymerase incorporates uracil opposite adenine during DNA replication because uracil is more stable
4. The polymerase extends DNA by forming peptide bonds between nucleotides in a sequence-specific manner

Explanation: This question evaluates understanding of base pairing specificity during DNA replication. DNA polymerases achieve high fidelity primarily through Watson-Crick base pairing between the template base and incoming nucleotide. The experiment shows that with adenine in the template, only dTTP supports extension, demonstrating the A-T complementary pairing rule. This selectivity occurs because correct base pairs have optimal geometry for catalysis in the polymerase active site. Option B incorrectly suggests sugar recognition is primary, but polymerases mainly discriminate based on base pairing. Option C wrongly states uracil pairs with adenine in DNA replication and claims uracil is more stable. Option D completely mischaracterizes the mechanism as forming peptide bonds rather than phosphodiester bonds. When analyzing polymerase fidelity, remember that complementary base pairing geometry is the primary selection mechanism.

Question 12

1. A transition mutation in which the original G–C pair can be converted to an A–T pair in one daughter duplex (correct answer)
2. A transversion mutation in which the original G–C pair is converted to a G–G pair in one daughter duplex
3. No mutation because uracil pairs preferentially with guanine in DNA and preserves the original information
4. A frameshift mutation because uracil causes insertion of an extra nucleotide during phosphodiester bond formation

Explanation: This question tests understanding of DNA damage, base pairing, and mutagenesis. Cytosine deamination produces uracil, which pairs with adenine rather than guanine during replication. In the absence of repair, one daughter strand will have U-A pairing (later becoming T-A after another round), while the other maintains G-C, resulting in a C→T transition mutation. This converts the original G-C pair to an A-T pair in one daughter duplex. Option B incorrectly suggests a G-G pair, which would not form stable Watson-Crick pairing. Option C wrongly claims uracil pairs with guanine. Option D incorrectly describes a frameshift, but deamination causes substitution, not insertion/deletion. When analyzing mutagenesis, trace the base pairing consequences through replication cycles to identify the mutation type.

Question 13

1. RNA polymerase requires a uracil-containing ribonucleoside triphosphate to pair with adenine in the template (correct answer)
2. RNA polymerase requires thymine because thymine forms three hydrogen bonds with adenine
3. RNA polymerase cannot use triphosphates and instead incorporates nucleoside monophosphates directly
4. RNA polymerase extends RNA by adding nucleotides to the 5' end, which requires thymine triphosphate

Explanation: This question tests understanding of RNA polymerase substrate requirements. RNA polymerase incorporates ribonucleoside triphosphates (NTPs) to synthesize RNA, and specifically requires UTP rather than TTP. Uracil pairs with adenine in RNA through two hydrogen bonds, just as thymine does in DNA. The experiment shows RNA synthesis only occurs with UTP present, confirming this fundamental requirement. Option B incorrectly states thymine forms three hydrogen bonds with adenine (it forms two). Option C wrongly claims RNA polymerase uses monophosphates instead of triphosphates. Option D incorrectly describes 5' addition and requires thymine for RNA synthesis. When analyzing transcription, remember that RNA polymerase specifically uses ribonucleoside triphosphates with uracil replacing thymine.

Question 14

1. A single-stranded overhang can provide a loading site that increases helicase-mediated unwinding efficiency. (correct answer)
2. Overhangs reduce unwinding because helicases require fully base-paired duplex ends to bind.
3. Overhangs increase unwinding because they add covalent bonds that destabilize the phosphodiester backbone.
4. Overhangs increase unwinding because they convert DNA into RNA, which helicases unwind more readily.

Explanation: This question assesses understanding of helicase mechanism and substrate requirements. Many DNA helicases require a single-stranded DNA loading site to bind and begin translocation, after which they can unwind adjacent duplex regions using ATP hydrolysis. A single-stranded overhang provides an ideal entry point for helicase loading, eliminating the need for spontaneous duplex breathing to create an initial binding site. The correct answer A accurately explains that a single-stranded overhang can provide a loading site that increases helicase-mediated unwinding efficiency. Answer B incorrectly claims helicases require fully paired ends, answer C incorrectly invokes covalent bond effects, and answer D incorrectly suggests DNA is converted to RNA. In helicase activity analysis, always consider the enzyme's requirement for initial single-stranded DNA binding, as providing pre-existing single-stranded regions like overhangs can dramatically enhance unwinding efficiency.

Question 15

1. The polymer contains uracil and is expected to form A–U base pairs in a double helix.
2. The polymer contains deoxyribose and could form a duplex in which G pairs with C via three hydrogen bonds. (correct answer)
3. The polymer is RNA because DNase I selectively cleaves phosphodiester bonds adjacent to ribose 2′-OH groups.
4. Thymine in the polymer indicates a 2′,3′-cyclic phosphate backbone typical of RNA cleavage products.

Explanation: This question assesses understanding of nucleotide composition and base pairing in nucleic acids. The key evidence shows the polymer is DNA (resistant to RNase A, degraded by DNase I, contains deoxyribonucleosides and thymine). In DNA double helices, guanine pairs with cytosine via three hydrogen bonds, while adenine pairs with thymine via two hydrogen bonds. The correct answer B accurately states the polymer contains deoxyribose and could form G-C base pairs with three hydrogen bonds. Answer A is incorrect because the polymer contains thymine (not uracil), confirming it's DNA not RNA. Answer C incorrectly claims DNase I cleaves near ribose 2'-OH groups, but DNase I actually cleaves DNA phosphodiester bonds. When analyzing nucleic acid structure, always verify the sugar type (ribose vs deoxyribose) and characteristic bases (uracil in RNA, thymine in DNA).

Question 16

1. The oligonucleotide is synthesized in the 3′→5′ direction because new nucleotides attach to the 5′-hydroxyl. (correct answer)
2. The oligonucleotide is synthesized in the 5′→3′ direction because the growing end presents a free 5′-hydroxyl.
3. The oligonucleotide is synthesized in the 3′→5′ direction because phosphodiester bonds always form between two 5′ carbons.
4. The oligonucleotide is synthesized in the 3′→5′ direction because the first nucleotide is anchored via its 3′ end.

Explanation: This question tests understanding of nucleic acid synthesis directionality. In solid-phase oligonucleotide synthesis, the first nucleotide is anchored via its 3' end, and each new nucleotide is added to the 5'-hydroxyl of the growing chain. This means synthesis proceeds in the 3'→5' direction, opposite to biological DNA synthesis. The correct answer A accurately describes this: synthesis is 3'→5' because new nucleotides attach to the 5'-hydroxyl. Answer B incorrectly states 5'→3' direction. Answer C wrongly claims phosphodiester bonds form between two 5' carbons, when they actually form between 3'-OH and 5'-phosphate. Answer D has the right direction but wrong reasoning about the attachment point. In nucleotide synthesis problems, always track which end is fixed and where new units attach to determine directionality.

Question 17

1. Formation of a 2′,3′-cyclic phosphate intermediate enabled by the ribose 2′-OH, promoting backbone cleavage in RNA. (correct answer)
2. Protonation of thymine at physiological pH destabilized RNA base pairing, leading to spontaneous cleavage.
3. Hydrolysis of N-glycosidic bonds at purines generated abasic sites that are uniquely unstable in DNA.
4. Oxidation of deoxyribose at the 1′ carbon created strand breaks that occur only in RNA.

Explanation: This question assesses understanding of structural differences between RNA and DNA. RNA contains ribose with a 2'-OH group, while DNA contains deoxyribose lacking this hydroxyl. The 2'-OH in RNA can act as a nucleophile to attack the adjacent phosphodiester bond, forming a 2',3'-cyclic phosphate intermediate and cleaving the backbone. The correct answer A accurately describes this mechanism unique to RNA. Answer B incorrectly focuses on thymine protonation, which doesn't explain RNA-specific cleavage. Answer C describes depurination creating abasic sites, but claims this is unstable only in DNA, which is backwards. Answer D incorrectly states oxidation at the 1' carbon causes breaks only in RNA. When analyzing nucleic acid stability, remember that the 2'-OH makes RNA chemically labile compared to DNA.

Question 18

1. The $T_m$ will decrease because replacing G–C with A–T reduces hydrogen bonding and base-stacking contributions at that site. (correct answer)
2. The $T_m$ will increase because A–T pairs contain an extra hydrogen bond compared with G–C pairs.
3. The $T_m$ will be unchanged because duplex stability depends only on phosphodiester bonds, not on base pairing.
4. The $T_m$ will increase because thymine is a purine and enhances stacking interactions relative to cytosine.

Explanation: This question tests understanding of how base pair identity affects DNA duplex stability and melting temperature. G-C base pairs form three hydrogen bonds while A-T base pairs form only two, making G-C pairs more stable. Additionally, G-C pairs contribute more to base stacking interactions due to their larger aromatic surface area. Replacing a G-C pair with an A-T pair reduces both hydrogen bonding and stacking contributions at that position, decreasing overall duplex stability and lowering the melting temperature (Tm). Option B incorrectly claims A-T pairs have more hydrogen bonds than G-C pairs, while option D wrongly identifies thymine as a purine (it's a pyrimidine). The relationship between GC content and Tm is fundamental to DNA analysis and PCR primer design. When predicting effects of mutations on duplex stability, consider both hydrogen bonding differences (3 vs 2) and base stacking contributions between different base pairs.

Question 19

1. The polymerase likely requires the 2b2-hydroxyl of ribose for efficient incorporation, so dUTP is a poorer substrate than UTP. (correct answer)
2. dUTP cannot form WatsonCrick base pairs with adenine, preventing templated incorporation.
3. dUTP hydrolyzes phosphodiester bonds during elongation, causing chain scission and reduced yield.
4. Replacing UTP with dUTP forces synthesis in the 3b215b2 direction, reducing processivity.

Explanation: This question assesses nucleotide structural requirements for polymerase catalysis in nucleic acid synthesis. Nucleotides for RNA synthesis are ribonucleoside triphosphates (NTPs) with a 2'-hydroxyl group on the ribose sugar, which may influence polymerase substrate recognition and incorporation efficiency. In the experiment, replacing UTP with dUTP (lacking the 2'-OH) reduces the yield of full-length RNA product, indicating the polymerase prefers substrates with the 2'-OH for optimal activity. This occurs because the enzyme's active site is adapted for ribonucleotides, making dUTP a poorer substrate and slowing incorporation. A distractor might claim dUTP cannot base-pair, but it can form Watson-Crick pairs with adenine, highlighting the error in overlooking sugar structure. In nucleotide analysis, always verify substrate specificity based on sugar and phosphate features. This principle applies to distinguishing DNA and RNA polymerases in synthesis assays.

Question 20

1. DNA polymerases require a free 3b2-OH on the primer to attack the incoming dNTP b1-phosphate and form a phosphodiester bond. (correct answer)
2. A blocked 3b2 end prevents WatsonCrick base pairing between the primer and template, eliminating hybridization.
3. 3b2-O-methylation removes the primers phosphate groups, preventing Mg$^{2+}$ binding needed for catalysis.
4. Blocking the 3b2 end forces polymerase to extend from the 5b2 end, which is slower but should still occur.

Explanation: This question assesses nucleotide structural requirements for phosphodiester bond formation in nucleic acid synthesis. Nucleotides are extended by DNA polymerases through nucleophilic attack of the primer's 3'-OH on the incoming dNTP's alpha-phosphate, forming a new bond and extending the chain. The 3'-O-methyl blocked primer prevents extension because it lacks the free 3'-OH needed for this attack, halting synthesis. This modification blocks the reactive group essential for catalysis, while the unmodified primer proceeds normally. A common distractor might claim it prevents base pairing, but the blockage affects elongation chemistry, not hybridization. In nucleotide analysis, always verify the role of 3'-OH in polymerase mechanisms. This principle applies to chain-terminating analogs in sequencing.