Question 1

Nucleic acid fundamentals: what do the N-glycosidic bonds in DNA and RNA actually connect and how do they behave? Choose the most accurate statement about the glycosidic linkage in nucleotides (ribose/deoxyribose to the base).

Accepted Answer

Introduction / Context: Nucleic acids are polymers built from nucleotides. Each nucleotide contains a base linked to a (deoxy)ribose sugar and at least one phosphate. The bond that connects the base to the sugar is the N-glycosidic bond. Understanding this bond clarifies how nucleotide structure, base orientation, and overall DNA/RNA stability arise. Given Data / Assumptions: The question asks specifically about the N-glycosidic linkage in DNA and RNA. Backbone linkages (phosphodiester bonds) are separate from the glycosidic bond. Watson–Crick pairing occurs between complementary bases across strands, not via the glycosidic bond. Concept / Approach: The N-glycosidic bond forms between the anomeric carbon (C1′) of the sugar and a ring nitrogen of the base (N1 for pyrimidines; N9 for purines). This bond defines base orientation (anti/syn) but is distinct from the phosphodiester backbone and from the interstrand hydrogen bonds. It is relatively stable in base and more labile under acidic conditions (depurination). Step-by-Step Solution: Identify what is being linked: C1′ of sugar to a base nitrogen → N-glycosidic bond.Separate concepts: phosphodiester bonds link sugar–phosphate–sugar; glycosidic bonds do not.Assess stability: alkaline conditions attack RNA phosphodiester bonds (due to 2′-OH), not primarily the glycosidic bond; acid promotes depurination.Assess rotation: anti/syn conformations exist, but rotation is restricted by sterics and stacking; it is not freely rotatable 180° without consequence. Verification / Alternative check: Standard nucleic acid textbooks diagram the N-glycosidic link as sugar C1′–N9 (purines) or C1′–N1 (pyrimidines). Backbone linkages are shown separately as 3′–5′ phosphodiesters. Why Other Options Are Wrong: Hydroxide mainly promotes RNA backbone cleavage, not routine glycosidic hydrolysis. The glycosidic bond does not form or stabilize interstrand hydrogen bonds. Rotation is conformationally constrained, not freely 180°. Phosphate–sugar linkage is the phosphodiester, not the glycosidic bond. Common Pitfalls: Confusing glycosidic (base–sugar) with phosphodiester (sugar–phosphate–sugar) and assuming base-induced hydrolysis targets the glycosidic bond. Final Answer: Connect the sugar to the base.

Question 2

Thymidine (the thymine nucleoside) in nucleic acids: properties and pairing behavior Which combination of statements best describes thymidine in DNA and its relation to RNA?

Accepted Answer

Introduction / Context: Thymidine is the nucleoside composed of the base thymine attached to deoxyribose. Its chemical features determine base pairing, stacking, and differences between DNA and RNA. Recognizing these attributes helps explain DNA stability and the substitution of thymine with uracil in RNA. Given Data / Assumptions: Thymine bears a methyl group at the 5-position (5-methyluracil). RNA uses uracil instead of thymine in most biological contexts. Canonical base pairing is A•T with two hydrogen bonds in DNA. Concept / Approach: The 5-methyl group on thymine increases hydrophobic character and modulates stacking interactions, contributing to duplex stability. During transcription and in RNA genomes, uracil replaces thymine, providing comparable pairing with adenine while altering repair recognition pathways. Thymine (in DNA) and uracil (in RNA) pair with adenine via two hydrogen bonds. Step-by-Step Solution: Evaluate hydrophobic contribution: the 5-methyl substituent enhances base stacking; statement (a) is correct.Compare DNA vs RNA: RNA uses uracil in place of thymine; statement (b) is correct.Check pairing: A•T pairs via two hydrogen bonds; statement (c) is correct.Therefore, the aggregate choice that includes all correct statements is “All of the above.” Verification / Alternative check: Thermodynamic measurements show sequence-dependent contributions from methylated bases to stacking; structural analyses (X-ray, NMR) confirm two H-bonds for A•T and replacement of T by U in RNA. Why Other Options Are Wrong: “Forms three hydrogen bonds with guanine” is incorrect; three H-bonds are characteristic of G•C, not T with any base. Common Pitfalls: Confusing nucleoside vs nucleotide terminology (thymidine vs thymidylate) and mixing DNA vs RNA base identities. Final Answer: All of the above.

Question 3

Gel electrophoresis principle: which DNA fragments migrate fastest through an agarose or polyacrylamide gel? Select the most accurate statement about fragment mobility.

Accepted Answer

Introduction / Context: Nucleic acid gel electrophoresis separates DNA or RNA by size as molecules migrate through a porous matrix under an electric field. Mobility depends primarily on length when conformation and buffer are controlled. Given Data / Assumptions: Agarose or polyacrylamide gels act as sieves with pore sizes dependent on gel concentration. DNA carries a near-uniform charge density due to the phosphate backbone. Comparison is among linear fragments under standard conditions. Concept / Approach: Because DNA’s charge density is roughly constant, the electrophoretic force per unit length is similar across sizes. Frictional resistance increases with length, so shorter fragments experience less hindrance and travel farther in a given time. Gel concentration tunes the resolution range (agarose for kb range; PAGE for small fragments). Step-by-Step Solution: Assume linear fragments in the same buffer and gel matrix.Recognize size-dependent sieving: smaller fragments navigate pores more easily.Conclude that small fragments migrate faster and farther than large ones in a fixed time. Verification / Alternative check: DNA ladders demonstrate that fragments of 100–500 bp run closer to the dye front than 3–10 kb fragments at the same voltage, corroborating the size–mobility relationship. Why Other Options Are Wrong: Large fragments move more slowly. An intact genome is enormous and migrates poorly without prior shearing. A–T content can slightly affect curvature or melting but is not the primary determinant of mobility under standard conditions. Common Pitfalls: Comparing supercoiled vs linear forms (conformation matters), and using inappropriate gel percentage for the size range. Final Answer: Small fragments.

Question 4

Identify the base: which purine bears an amino (NH2) group at the 6-position on the ring? Choose the correct heterocycle based on its substituent pattern.

Accepted Answer

Introduction / Context: Nucleobases are distinguished by ring system (purine vs pyrimidine) and functional groups. Correctly identifying substituents is essential for understanding base pairing, tautomerism, and chemical reactivity. Given Data / Assumptions: Purines: adenine and guanine (fused imidazole–pyrimidine ring). Pyrimidines: cytosine, thymine, uracil (single six-membered ring). Question targets an amino group at C6 of a purine. Concept / Approach: Adenine is 6-aminopurine (NH2 at C6). Guanine is 2-amino-6-oxo purine (amino at C2, carbonyl at C6). Cytosine, thymine, and uracil are pyrimidines and do not match the description. Step-by-Step Solution: Classify bases: identify purines among options (adenine, guanine).Check substituents: adenine → NH2 at C6; guanine → NH2 at C2 and =O at C6.Select adenine as the correct base bearing NH2 at the 6-position. Verification / Alternative check: Standard chemical structures in biochemistry references annotate adenine as 6-aminopurine. Why Other Options Are Wrong: Pyrimidines do not have a purine C6 position; guanine’s 6-position is carbonyl, not amino. Common Pitfalls: Confusing numbering schemes between rings and mixing up the 2-amino group of guanine with the 6-amino of adenine. Final Answer: Adenine.

Question 5

DNA thermal stability: how do ionic strength and composition influence denaturation? Select the true statement regarding the melting behavior of the DNA duplex.

Accepted Answer

Introduction / Context: DNA melting (denaturation) is the disruption of base pairing and stacking, producing single strands. The melting temperature (Tm) depends on sequence composition and solution conditions, especially ionic strength. Given Data / Assumptions: DNA is a polyanion due to its phosphate backbone. Cations in solution shield electrostatic repulsion between strands. G–C pairs form three hydrogen bonds and often stronger stacking than A–T pairs. Concept / Approach: Increasing salt concentration screens negative charges on the backbone, reducing interstrand repulsion and stabilizing the helix, which raises Tm. Higher G–C content typically increases Tm. Hydrophobic base stacking contributes significantly to stability; it cannot be ignored. Step-by-Step Solution: Consider electrostatics: more salt → better charge screening → higher duplex stability → higher Tm.Consider composition: more G–C content → higher Tm due to three H-bonds and stacking.Reject statements that deny stacking contributions or sequence dependence.Choose the statement consistent with ionic strength increasing stability. Verification / Alternative check: Empirical formulas (for example, Tm ≈ 81.5 + 16.6 log[Na+] + 0.41(%G+C) − terms for mismatches) reflect positive dependence on salt and G–C content. Why Other Options Are Wrong: Claiming no dependence on stacking or G–C content contradicts established thermodynamics. Saying stability decreases with salt is opposite of observed behavior (within typical laboratory ranges). Tm is affected by pH and organic solvents. Common Pitfalls: Extrapolating beyond physiological ranges; extremely high salt or denaturants can introduce different effects, but the canonical trend in standard buffers is as stated. Final Answer: Increases with increasing salt concentration (ionic strength).

Question 6

Hydrogen-bonding penalty: approximate free-energy cost of leaving a potential H-bond unsatisfied inside a DNA duplex Choose the closest value for the unfavorable free energy.

Accepted Answer

Introduction / Context: When macromolecules fold or associate, buried polar groups prefer to form hydrogen bonds. Leaving such a group unsatisfied is energetically costly and can destabilize structures like DNA duplexes and protein interiors. Given Data / Assumptions: We are estimating an unfavorable free-energy penalty for an unsatisfied hydrogen-bonding opportunity after helix formation. Typical hydrogen bond strengths in aqueous systems are on the order of 10–30 kJ/mol, depending on geometry and environment. Values are approximate and context dependent. Concept / Approach: An unsatisfied polar group buried from solvent cannot form compensating interactions with water. The penalty is roughly comparable to the stabilization ordinarily conferred by forming a good hydrogen bond. A middle-of-the-range estimate around 20 kJ/mol is commonly cited for this penalty in biophysical discussions. Step-by-Step Solution: Recall typical H-bond free energies: approximately 8–25 kJ/mol in water; stronger in well-organized, low-dielectric environments.Associate penalty magnitude with loss of a stabilizing H-bond: pick a representative mid-value.Select 20 kJ/mol as the closest standard estimate among the options. Verification / Alternative check: Protein design and nucleic-acid thermodynamic models penalize buried unsatisfied donors/acceptors by values near tens of kJ/mol, aligning with ≈20 kJ/mol as a rule-of-thumb. Why Other Options Are Wrong: 10 kJ/mol may underestimate typical penalties for buried unsatisfied H-bonds; 25–30 kJ/mol can occur but are often higher than the commonly quoted average; 5 kJ/mol is too small. Common Pitfalls: Confusing enthalpy with free energy and ignoring environmental modulation (dielectric constant, geometry, cooperativity). Final Answer: 20 kJ/mol.

Question 7

Spectroscopy check: at what wavelengths do nucleotide bases and aromatic amino acids absorb most strongly (respectively)? Choose the correct pairing of peak absorbance maxima.

Accepted Answer

Introduction / Context: Ultraviolet absorbance is routinely used to quantify biomolecules. Nucleic acids and proteins have distinct absorbance peaks due to their chromophores, enabling rapid purity checks (for example, A260/A280 ratios). Given Data / Assumptions: Nucleotide bases have conjugated ring systems with maxima near 260 nm. Aromatic amino acids (tryptophan, tyrosine, phenylalanine) dominate protein absorbance near 280 nm. “Respectively” means first value corresponds to nucleotides, second to aromatic amino acids. Concept / Approach: DNA/RNA quantification uses A260 because nucleobases absorb strongly there. Protein quantification and contamination checks often reference A280. The A260/A280 ratio guides assessments of nucleic acid purity. Step-by-Step Solution: Map chromophores to peaks: nucleotides → 260 nm; aromatics → 280 nm.Select the option listing 260 nm first and 280 nm second. Verification / Alternative check: Laboratory spectrophotometer protocols standardize nucleic acid measurements at 260 nm and protein measurements at 280 nm, confirming the assignment. Why Other Options Are Wrong: Reversed or shifted values do not reflect the canonical maxima used in routine assays. Common Pitfalls: For proteins lacking Trp/Tyr, A280 underestimates concentration; complementary dye-binding methods may be preferred. Final Answer: 260 nm and 280 nm.

Question 8

Define a nucleotide: which component set is minimally required? Select the composition that best describes a nucleotide building block of DNA or RNA.

Accepted Answer

Introduction / Context: Nucleotides are the monomeric units of nucleic acids and the currency of many metabolic reactions (for example, ATP). Distinguishing nucleotides from nucleosides clarifies structure and function. Given Data / Assumptions: Nucleoside = base + sugar. Nucleotide = nucleoside + phosphate(s). Number of phosphates may vary (mono-, di-, triphosphate). Concept / Approach: The minimal definition of a nucleotide requires a pentose sugar ((deoxy)ribose), a nitrogenous base, and at least one phosphate group. While triphosphates are common as substrates for polymerases, the term “nucleotide” does not require three phosphates. Step-by-Step Solution: Identify components: base + sugar + ≥1 phosphate → nucleotide.Reject definitions lacking a base (that would be a sugar phosphate) or insisting on three phosphates (too restrictive). Verification / Alternative check: Biochemistry texts define AMP, ADP, and ATP as nucleotides differing only in phosphate count; all share base, sugar, and phosphate. Why Other Options Are Wrong: “Sugar + phosphate” omits the base (not a nucleotide). “Paired bases only” lacks sugar and phosphate. “Three phosphates always” is incorrect because monophosphates are also nucleotides. Common Pitfalls: Confusing nucleoside (no phosphate) with nucleotide (has phosphate). Final Answer: A sugar, a base, and a phosphate.

Question 9

Purine–pyrimidine identification: which of the following bases is a purine? Choose the correct classification.

Accepted Answer

Introduction / Context: Nucleotide bases are grouped into purines and pyrimidines. This classification underlies Watson–Crick pairing rules and influences chemical behavior. Given Data / Assumptions: Purines: two fused rings (adenine, guanine). Pyrimidines: single six-membered ring (cytosine, thymine, uracil). Options include one purine among pyrimidines. Concept / Approach: Identify ring systems: only adenine among the listed choices is a purine. The others are pyrimidines used in DNA (C, T) and RNA (U). Step-by-Step Solution: Scan options for purines: adenine qualifies; cytosine, thymine, uracil do not.Select adenine as the correct answer. Verification / Alternative check: Chemical structures confirm that adenine is a bicyclic purine, whereas the others are monocyclic. Why Other Options Are Wrong: Cytosine, thymine, and uracil are pyrimidines; “all of these are pyrimidines” is false because adenine is included. Common Pitfalls: Memorization errors that swap guanine/adenine with pyrimidines; remembering “Cut the Py” (C, U, T are pyrimidines) helps. Final Answer: Adenine.

Question 10

Ribose vs deoxyribose: a five-carbon sugar lacking a 2′-hydroxyl group (2′-OH) is found in which nucleic acid? Select the molecule that contains 2′-deoxyribose as its sugar.

Accepted Answer

Introduction / Context: The key chemical difference between DNA and RNA sugars is the presence or absence of a hydroxyl group at the 2′ carbon. This small change affects stability, structure, and susceptibility to base-catalyzed hydrolysis. Given Data / Assumptions: DNA contains 2′-deoxyribose, which lacks a hydroxyl group at the 2′ position (it has hydrogen instead). RNA contains ribose, which bears a 2′-OH group. mRNA and tRNA are forms of RNA and thus contain ribose. Concept / Approach: The absence of the 2′-OH in DNA increases chemical stability, making the phosphodiester backbone resistant to alkaline cleavage compared with RNA, where the 2′-OH can attack the adjacent phosphate. Therefore, the nucleic acid featuring 2′-deoxyribose is DNA. Step-by-Step Solution: Identify the sugar: 2′-deoxyribose → no 2′-OH → DNA.Exclude RNA subclasses (mRNA, tRNA) which possess ribose with 2′-OH.Note that “rDNA” (recombinant DNA) is still DNA, but the standard, general answer is DNA itself. Verification / Alternative check: Textbook structures of deoxyribose vs ribose show H vs OH at the 2′ position, aligning with DNA vs RNA. Why Other Options Are Wrong: RNA (including mRNA and tRNA) carries the 2′-OH and is susceptible to alkaline hydrolysis; “rDNA only” needlessly restricts the correct category, as all DNA uses deoxyribose. Common Pitfalls: Misreading “lacking a hydroxyl” as “lacking oxygen entirely”; deoxyribose still contains oxygen atoms but not at the 2′ position. Final Answer: DNA.

Question 11

Foundational molecular biology: Which chemical component is present in ribonucleic acid (RNA) but absent from deoxyribonucleic acid (DNA)? Choose the most specific constituent that distinguishes RNA from DNA in standard cells.

Accepted Answer

Introduction / Context: This question checks core knowledge of nucleic acid chemistry by asking for the unique component that differentiates RNA from DNA. Recognizing these molecular signatures is essential in genetics, biotechnology, and clinical diagnostics. Given Data / Assumptions: We compare standard cellular RNA and DNA in bacteria, plants, and animals. Options include bases and backbone components common to nucleic acids. No unusual modifications (e.g., viral exceptions) are considered. Concept / Approach: Two main differences distinguish RNA from DNA: the sugar (ribose vs deoxyribose) and one of the bases (uracil vs thymine). The prompt asks for what is found on RNA but not DNA; among the choices, uracil fits uniquely. Step-by-Step Solution: Identify bases common to both: adenine is present in both RNA and DNA.Identify backbone components: phosphate is shared by both polymers.Deoxyribose is the DNA sugar, not an RNA component.Uracil replaces thymine in RNA and is not found in standard DNA → correct selection. Verification / Alternative check: Textbook base sets: RNA uses A, U, G, C; DNA uses A, T, G, C. Thus U is diagnostic for RNA. Why Other Options Are Wrong: Deoxyribose: specific to DNA, not RNA. Phosphate: shared in both nucleic acids. Adenine: present in both RNA and DNA. Common Pitfalls: Confusing uracil with thymine, or assuming phosphate is unique to one polymer. Final Answer: Uracil

Question 12

Nucleobase functional groups: Which pyrimidine base possesses an amino (–NH2) group at the carbon-4 position on the ring?

Accepted Answer

Introduction / Context: Knowing the functional groups on nucleobases is central to understanding base pairing, mutagenesis, and recognition by enzymes such as polymerases and deaminases. Given Data / Assumptions: Pyrimidines include cytosine, thymine, and uracil. We focus on substituents at the C4 ring position. Standard canonical structures (not tautomers) are considered. Concept / Approach: Cytosine is a pyrimidine with an amino group at C4 and a keto at C2. Thymine and uracil are 2,4-diketo pyrimidines lacking an amino at C4. Adenine is a purine, not a pyrimidine, and thus is irrelevant to the asked position. Step-by-Step Solution: List pyrimidines: C, T, U.Recall structural motifs: cytosine has –NH2 at C4; thymine and uracil have carbonyl (keto) groups at C4.Therefore, the base with an amino at C4 is cytosine. Verification / Alternative check: Base-pairing logic: cytosine pairs with guanine via hydrogen bonds partly supported by its C4 amino. Why Other Options Are Wrong: Thymine/Uracil: both have carbonyl at C4, not amino. Adenine: a purine with different ring numbering and groups. Common Pitfalls: Mixing up ring numbering or confusing purines with pyrimidines. Final Answer: Cytosine

Question 13

Experimental analysis of nucleic acids: Which routine property is most commonly measured to analyze and quantify DNA or RNA in the laboratory?

Accepted Answer

Introduction / Context: Every molecular biology lab relies on rapid methods to assess nucleic acids. Spectrophotometric absorbance is the standard first-line tool for concentration and purity assessment of DNA/RNA samples. Given Data / Assumptions: Nucleic acids contain conjugated double bonds in bases that absorb ultraviolet light. Typical instruments measure absorbance at 260 nm and 280 nm. We seek the most common/straightforward analytical readout. Concept / Approach: Nucleic acids exhibit strong absorbance at 260 nm. The A260 value is used for concentration (using standard conversion factors), while A260/A280 and A260/A230 ratios estimate purity. Visible-light absorbance is minimal for pure nucleic acids, and determining molecular weight usually requires gels or advanced instrumentation rather than quick spectrophotometry. Step-by-Step Solution: Recognize that nucleic acid bases absorb UV light at ~260 nm.Use A260 for concentration; use A260/A280 for protein contamination checks.Therefore, “absorption of uv light” best matches routine analysis. Verification / Alternative check: Nanodrop and cuvette spectrophotometers standardize A260 reads for DNA/RNA quantification. Why Other Options Are Wrong: Molecular weight: commonly determined by electrophoresis, not the quickest routine metric. Visible light absorption: nucleic acids have little absorbance in the visible range. None of these: incorrect since UV absorbance is widely used. Common Pitfalls: Interfering contaminants (phenol, guanidine) can skew A260/A280; interpret ratios with care. Final Answer: absorption of uv light

Question 14

Forces stabilizing nucleic acid structure: Which interaction provides the greatest stabilizing contribution to double-stranded nucleic acids under physiological conditions?

Accepted Answer

Introduction / Context: DNA duplex stability reflects multiple interactions. Distinguishing the dominant contributions helps explain melting behavior, GC-content effects, and sequence-dependent stability in genomics and PCR design. Given Data / Assumptions: Physiological ionic strength where electrostatic repulsion is screened. Covalent backbone remains intact; we consider noncovalent interactions. Standard B-form DNA considerations. Concept / Approach: While hydrogen bonds define base pairing specificity, the major stabilizing force for the helix is base stacking, which is dominated by Van der Waals interactions and hydrophobic effects between adjacent bases. Electrostatics mainly contributes repulsion between phosphates (reduced by cations). Entropy generally disfavors ordered duplex formation. Step-by-Step Solution: Acknowledge H-bonds: essential for pairing but not the primary stability driver.Recognize stacking: aromatic bases stack via Van der Waals and hydrophobic interactions.Therefore, Van der Waals (stacking) provides the largest stabilizing contribution. Verification / Alternative check: Sequence-dependent melting temperatures correlate strongly with stacking energies (nearest-neighbor parameters), not just number of hydrogen bonds. Why Other Options Are Wrong: Hydrogen bonds: necessary for specificity but contribute less to net stability than stacking. Electrostatic bond: phosphate–phosphate interactions are repulsive; cations screen them. Conformational entropy: ordering reduces entropy, opposing stability. Common Pitfalls: Equating more hydrogen bonds (GC vs AT) with sole stability; the real driver is stacking, though GC stacking is typically stronger. Final Answer: Van der Waals

Question 15

Sugar identity in nucleic acids: Fill in the blanks—The pentose sugar in RNA is _____, whereas the pentose sugar in DNA is _____.

Accepted Answer

Introduction / Context: The identity of the pentose sugar differentiates RNA and DNA chemistry, influencing stability, susceptibility to hydrolysis, and structural conformations like A-form vs B-form helices. Given Data / Assumptions: RNA contains a 2'-hydroxyl group on ribose; DNA lacks it (deoxyribose). Options include sugars and unrelated moieties (phosphate, uracil). We match RNA → ribose; DNA → deoxyribose. Concept / Approach: Ribose has OH at the 2' position, enabling greater reactivity and promoting A-form helices in RNA. Deoxyribose lacks this OH, aiding DNA’s chemical stability and preference for B-form under physiological conditions. Step-by-Step Solution: Identify RNA sugar: ribose.Identify DNA sugar: 2'-deoxyribose.Select the option “ribose, deoxyribose.” Verification / Alternative check: Chemical hydrolysis of RNA under alkaline conditions is facilitated by the 2'-OH; DNA is resistant—consistent with ribose vs deoxyribose. Why Other Options Are Wrong: Deoxyribose first is reversed. Phosphate and uracil are not sugars. Common Pitfalls: Confusing the sugar difference (ribose vs deoxyribose) with base difference (uracil vs thymine). Final Answer: ribose, deoxyribose

Question 16

Definition check: A nucleoside is best described as a purine or pyrimidine base that is __________ to a pentose sugar.

Accepted Answer

Introduction / Context: Precise terminology—nucleoside versus nucleotide—is foundational in biochemistry. Mislabeling can cause confusion in pathways like salvage, phosphorylation, and nucleic acid synthesis. Given Data / Assumptions: Nucleoside = base + sugar. Nucleotide = nucleoside + phosphate group(s). Bond type between base and sugar is asked. Concept / Approach: A nucleoside has a nitrogenous base linked to a pentose sugar via a covalent N-glycosidic bond (N9 of purines or N1 of pyrimidines to C1' of the sugar). Ionic or hydrogen bonds do not define the core nucleoside linkage. Step-by-Step Solution: Identify components: base + sugar.Recall linkage: N-glycosidic covalent bond.Therefore, “covalently bonded” is correct. Verification / Alternative check: Hydrolysis under acidic conditions cleaves the glycosidic bond—consistent with a covalent linkage. Why Other Options Are Wrong: Ionic and hydrogen bonds are noncovalent and do not define nucleosides. “None of the above” is incorrect because the covalent option is valid. Common Pitfalls: Confusing nucleosides with nucleotides (which include phosphate). Final Answer: covalently bonded

Nucleic Acids Questions