Nucleotides and Nucleic Acids (5D)
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MCAT Chemical and Physical Foundations of Biological Systems › Nucleotides and Nucleic Acids (5D)
In a structural analysis of a 16-nt single-stranded nucleic acid isolated from a viral particle, enzymatic digestion showed the polymer was resistant to RNase A but was degraded by DNase I. Acid hydrolysis released only deoxyribonucleosides. The strand contained 6 guanine, 4 cytosine, 3 adenine, and 3 thymine residues. The core concept tested is base pairing and structural implications of nucleotide composition. Which statement is most consistent with the nucleotide structure described?
The polymer contains uracil and is expected to form A–U base pairs in a double helix.
The polymer is RNA because DNase I selectively cleaves phosphodiester bonds adjacent to ribose 2′-OH groups.
Thymine in the polymer indicates a 2′,3′-cyclic phosphate backbone typical of RNA cleavage products.
The polymer contains deoxyribose and could form a duplex in which G pairs with C via three hydrogen bonds.
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).
A research group synthesized a 12-nt oligonucleotide using standard solid-phase phosphoramidite chemistry. They confirmed that each coupling step adds one nucleotide to the 5′-hydroxyl of the growing chain while the first nucleotide is anchored through its 3′-end to the solid support. The core concept tested is nucleic acid directionality during synthesis. Which statement is most consistent with the nucleotide structure described?
The oligonucleotide is synthesized in the 5′→3′ direction because the growing end presents a free 5′-hydroxyl.
The oligonucleotide is synthesized in the 3′→5′ direction because new nucleotides attach to the 5′-hydroxyl.
The oligonucleotide is synthesized in the 3′→5′ direction because phosphodiester bonds always form between two 5′ carbons.
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.
A DNA duplex segment was analyzed for susceptibility to UV-induced lesions. Under identical irradiation conditions, a sequence containing 5′-TT-3′ on one strand produced a higher yield of cyclobutane pyrimidine dimers than a sequence containing 5′-TC-3′. The core concept tested is nucleotide identity and photochemical reactivity of pyrimidines. Which conclusion about nucleic acid function is best supported by the observation?
UV irradiation primarily causes depurination, and TT sequences accelerate glycosidic bond hydrolysis.
TT sequences are more reactive because thymine contains a 2′-OH that participates in photochemical crosslinking.
Adjacent purines form UV-induced dimers, and thymine behaves as a purine in DNA.
Adjacent pyrimidines can form UV-induced dimers, and thymine–thymine steps are especially prone to this lesion type.
Explanation
This question tests understanding of UV-induced DNA damage and pyrimidine photochemistry. UV radiation causes adjacent pyrimidines (thymine and cytosine) to form cyclobutane pyrimidine dimers through [2+2] cycloaddition. Thymine-thymine (TT) sequences are particularly susceptible to this damage. The correct answer A accurately states that adjacent pyrimidines form UV-induced dimers, with TT steps being especially prone. Answer B incorrectly claims adjacent purines form dimers and misidentifies thymine as a purine. Answer C wrongly focuses on depurination, which is not the primary UV-induced lesion. Answer D incorrectly states thymine contains a 2'-OH (DNA bases don't have sugar hydroxyls). In DNA damage analysis, remember that pyrimidine dimers are the major UV photoproduct, with TT dimers being most common.
In a biochemical assay, a DNA polymerase was supplied with a primer-template junction and all four dNTPs. When pyrophosphate (PPi) was added to 10 mM, net DNA synthesis slowed markedly, despite unchanged enzyme concentration. The core concept tested is the thermodynamic role of PPi in nucleotide polymerization. Which process best explains the data observed?
High PPi converts dNTPs into ddNTPs by removing the 3′-OH, causing chain termination.
High PPi increases the rate of PPi hydrolysis, which directly cleaves DNA phosphodiester bonds.
High PPi drives the polymerization equilibrium backward by favoring the reverse reaction (pyrophosphorolysis).
High PPi competitively inhibits base pairing by binding to nucleobases in the major groove.
Explanation
This question assesses understanding of pyrophosphate's role in DNA synthesis thermodynamics. DNA polymerase catalyzes: (DNA)n + dNTP ⇌ (DNA)n+1 + PPi. High PPi concentration drives this equilibrium backward through mass action, favoring the reverse reaction (pyrophosphorolysis) where PPi attacks the terminal phosphodiester bond. The correct answer A accurately describes how high PPi favors the reverse reaction. Answer B incorrectly suggests PPi hydrolysis directly cleaves DNA bonds. Answer C wrongly claims PPi binds to nucleobases in the major groove. Answer D incorrectly proposes PPi converts dNTPs to ddNTPs by removing 3'-OH. When analyzing polymerization reactions, remember that product accumulation (like PPi) can reverse the reaction through Le Chatelier's principle.
A researcher prepared a 20-nt DNA oligonucleotide and quantified its UV absorbance at 260 nm before and after annealing to its fully complementary strand. Upon duplex formation, the absorbance at 260 nm decreased by 30% at constant concentration and path length. The core concept tested is UV hypochromicity and base stacking in nucleic acids. Which process best explains the data observed?
Duplex formation exposes bases to solvent, increasing UV absorbance via the hyperchromic effect.
Annealing converts deoxyribose to ribose, shifting the absorption maximum away from 260 nm.
Duplex formation increases base stacking interactions, decreasing UV absorbance via the hypochromic effect.
Annealing increases phosphodiester bond hydrolysis, reducing nucleotide concentration and lowering absorbance.
Explanation
This question tests understanding of UV hypochromicity in nucleic acids. When single-stranded DNA forms a duplex, bases stack on top of each other, reducing their exposure to UV light and decreasing absorbance at 260 nm - this is called the hypochromic effect. The correct answer A accurately describes how duplex formation increases base stacking, decreasing UV absorbance. Answer B incorrectly claims duplex formation exposes bases and increases absorbance (hyperchromic effect occurs during denaturation, not annealing). Answer C nonsensically suggests annealing converts deoxyribose to ribose. Answer D incorrectly proposes phosphodiester bond hydrolysis during annealing. In nucleic acid spectroscopy, always remember that base stacking in duplexes reduces UV absorbance, while denaturation increases it.
A lab engineered a point mutation in a DNA binding site such that one base pair in a duplex changed from G≡C to A=T, without altering the overall length of the duplex. Melting experiments were performed in 100 mM NaCl. The measured melting temperature $T_m$ decreased by 2.5 °C after the mutation. The core concept tested is hydrogen bonding and duplex stability. Based on the passage, which outcome is most likely following the mutation?
The duplex becomes more stable because A=T pairs increase base stacking relative to G≡C in all sequence contexts.
The duplex becomes less stable because adenine is a pyrimidine and disrupts helix geometry.
The duplex becomes less stable because replacing G≡C with A=T reduces the number of hydrogen bonds at that position.
The duplex becomes more stable because A=T pairs form three hydrogen bonds compared with two for G≡C.
Explanation
This question assesses understanding of hydrogen bonding and DNA duplex stability. G-C base pairs form three hydrogen bonds while A-T base pairs form only two hydrogen bonds. Replacing a G-C pair with an A-T pair reduces the total number of hydrogen bonds, decreasing duplex stability and lowering the melting temperature (Tm). The correct answer B accurately states the duplex becomes less stable due to reduced hydrogen bonding. Answer A incorrectly claims A-T pairs form three hydrogen bonds. Answer C wrongly identifies adenine as a pyrimidine (it's a purine). Answer D incorrectly suggests A-T pairs increase stability through enhanced base stacking. When analyzing duplex stability, remember that G-C pairs contribute more stability than A-T pairs due to the extra hydrogen bond.
A lab compared electrophoretic mobility of two 30-nt single-stranded nucleic acids under denaturing conditions (8 M urea) at pH 8.0. Strand X contained a standard phosphodiester backbone. Strand Y was identical in base sequence but contained one phosphorothioate substitution (a non-bridging oxygen replaced by sulfur) at each linkage. Strand Y migrated more slowly. The core concept tested is backbone chemistry and charge-to-mass effects on mobility. Which statement is most consistent with the nucleotide structure described?
Strand Y migrates more slowly because phosphorothioate substitutions force the strand to become double-stranded under denaturing conditions.
Strand Y migrates more slowly because sulfur converts the backbone into peptide bonds that do not carry charge.
Strand Y migrates more slowly because phosphorothioate linkages increase molecular mass while maintaining a similar net negative charge.
Strand Y migrates more slowly because phosphorothioate linkages neutralize the phosphate charge at pH 8.0.
Explanation
This question assesses understanding of phosphorothioate modifications and electrophoretic mobility. Phosphorothioate linkages replace a non-bridging oxygen with sulfur, increasing molecular mass while maintaining similar negative charge at pH 8.0. In gel electrophoresis, molecules migrate based on charge-to-mass ratio; increased mass with similar charge results in slower migration. The correct answer A accurately explains that phosphorothioates increase mass while maintaining similar charge. Answer B incorrectly claims phosphorothioates neutralize charge at pH 8.0. Answer C nonsensically suggests sulfur creates peptide bonds. Answer D wrongly proposes phosphorothioates force duplex formation under denaturing conditions. When analyzing modified nucleic acids, consider how chemical changes affect both charge and mass for predicting electrophoretic behavior.
A DNA oligonucleotide was treated with an enzyme that cleaves specifically at abasic (AP) sites generated by prior loss of a nucleobase. In a separate reaction, the same oligonucleotide was treated with a glycosylase that removes uracil from DNA, creating AP sites. The core concept tested is the distinction between nucleobase removal and backbone cleavage. Which process best explains the relationship between the two enzymes’ effects?
The glycosylase converts uracil to thymine by methylation, and the AP endonuclease recognizes thymine and excises it.
The glycosylase removes a nucleobase by cleaving an N-glycosidic bond, and the AP endonuclease cleaves the backbone at the resulting abasic site.
The glycosylase removes the ribose sugar, and the AP endonuclease restores the missing base by ligation.
The glycosylase directly hydrolyzes phosphodiester bonds, and the AP endonuclease then removes uracil bases.
Explanation
This question tests understanding of DNA repair enzyme mechanisms. DNA glycosylases remove damaged or incorrect bases by cleaving the N-glycosidic bond between the base and sugar, creating an abasic (AP) site. AP endonucleases then recognize these abasic sites and cleave the phosphodiester backbone to initiate repair. The correct answer B accurately describes this two-step process: glycosylase removes the base, AP endonuclease cleaves the backbone. Answer A incorrectly reverses the functions, claiming glycosylase cleaves phosphodiester bonds. Answer C wrongly suggests glycosylases methylate uracil to thymine. Answer D incorrectly states glycosylases remove ribose sugars. In DNA repair pathways, always distinguish between base removal (glycosylase activity) and backbone cleavage (nuclease activity).
A polymerase assay compared incorporation of dATP versus ddATP into a primer-template DNA duplex. Reaction conditions were identical except for nucleotide identity. When ddATP was present as the only adenine-containing substrate, extension halted immediately after a single incorporation event. The core concept tested is the role of the 3′-OH in phosphodiester bond formation. Which conclusion about nucleic acid function is best supported by the observation?
ddATP lacks a 2′-OH, preventing formation of Watson–Crick base pairs with thymine.
ddATP has an extra phosphate, increasing charge repulsion and forcing chain termination.
ddATP lacks a 3′-OH, preventing nucleophilic attack needed to form the next phosphodiester bond.
ddATP contains uracil instead of adenine, causing mismatch and polymerase stalling.
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.
An in vitro transcription reaction produced an RNA transcript that was then treated with a base that selectively cleaves the phosphodiester backbone at positions containing a 2′-hydroxyl group. Under the same conditions, an otherwise identical DNA strand showed no cleavage. The core concept tested is structural differences between ribose and deoxyribose. Which process best explains the data observed?
Formation of a 2′,3′-cyclic phosphate intermediate enabled by the ribose 2′-OH, promoting backbone cleavage in RNA.
Hydrolysis of N-glycosidic bonds at purines generated abasic sites that are uniquely unstable in DNA.
Protonation of thymine at physiological pH destabilized RNA base pairing, leading to spontaneous cleavage.
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.