201. Compare cellulose and chitin using the information below.
| Feature | Cellulose | Chitin |
|---|
| Building units | Glucose | Modified amino sugars |
| Major stated location | Plant cell wall | Arthropod exoskeleton |
| General role | Structural | Structural |
Which conclusion follows?
ⓐ. Both polymers contain identical monomers but perform unrelated roles.
ⓑ. Both are structural polysaccharides with distinct monomers and sites.
ⓒ. Cellulose is a protein, while chitin is a membrane lipid.
ⓓ. Chitin is an animal-storage glucose polymer corresponding to starch.
Correct Answer: Both are structural polysaccharides with distinct monomers and sites.
Explanation: The table reveals one shared relation and two important distinctions. Both cellulose and chitin are polysaccharides used structurally rather than serving as the stated principal storage polymers. Cellulose consists of glucose units and occurs in plant cell walls. Chitin contains modified amino-sugar units and occurs in arthropod exoskeletons. Their common broad function does not require identical chemical composition or biological location. Chitin should not be confused with glycogen, which is the animal storage glucose polymer, and neither cellulose nor chitin belongs to the protein or lipid category. Integrating all rows gives the precise comparison: shared structural polysaccharide status with different monomers and locations. The molecular evidence shows that starch and glycogen function mainly in storage, whereas cellulose and chitin provide structural support. Cellulose and chitin are both structural polysaccharides, but glucose versus modified amino-sugar units and plant-wall versus arthropod-exoskeleton occurrence distinguish them.
202. Assertion: Chitin is a protein merely because its modified sugar units may contain nitrogen.
Reason: Chitin is a complex polysaccharide of modified sugar units and occurs in arthropod exoskeletons.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain Assertion
ⓒ. Assertion is true and Reason is false; therefore Reason cannot explain Assertion
ⓓ. Assertion is false and Reason is true; therefore Reason cannot explain Assertion
Correct Answer: Assertion is false and Reason is true; therefore Reason cannot explain Assertion
Explanation: The Assertion uses nitrogen presence as the sole basis for calling chitin a protein, which is not valid. Proteins are polymers of amino acids linked through peptide bonds. Chitin instead has a sugar-based polymeric organisation involving chemically modified or amino-sugar units. The Reason correctly identifies both its molecular class and its major stated biological location in arthropod exoskeletons. A nitrogen-containing sugar does not become an amino-acid residue simply through nitrogen presence. Molecular classification requires the complete building-unit and backbone relation. The Assertion is false, while the Reason accurately describes chitin as a complex structural polysaccharide. The measured pattern shows that nucleic acids are polynucleotides whose repeating units contain a nitrogenous base, pentose and phosphate; their monomers therefore differ fundamentally from amino acids and monosaccharides.
203. Consider the following statements about nucleic acids.
I. DNA and RNA are polynucleotides.
II. Their repeating units are nucleotides containing a base, pentose and phosphate.
III. They are true biomacromolecular polymers.
IV. A single nucleoside is itself a complete polynucleotide.
ⓐ. Statements I and IV are correct; II and III are incorrect.
ⓑ. Statements II and III are correct; I and IV are incorrect.
ⓒ. Statements I, II and III are correct; IV is incorrect.
ⓓ. Statements I, II, III and IV are correct without exception.
Correct Answer: Statements I, II and III are correct; IV is incorrect.
Explanation: DNA and RNA are long chains formed from nucleotide units and are classified as polynucleotides. Each nucleotide contains three component types: a nitrogenous heterocyclic base, a pentose sugar and phosphate. Repetition of these complete nucleotide units produces a true macromolecular polymer, supporting Statements I, II and III. A nucleoside contains only a base and sugar and lacks phosphate. One nucleoside is neither a nucleotide nor a chain of nucleotides. Statement IV confuses a small two-component molecule with the complete polymeric organisation of a nucleic acid. The accepted combination retains the distinction among nucleoside composition, nucleotide composition and polynucleotide structure. When applied to these observations, nucleic acids are polynucleotides whose repeating units contain a nitrogenous base, pentose and phosphate; their monomers differ fundamentally from amino acids and monosaccharides.
204. Four acid-insoluble substances are characterised below.
| Substance | Building units | Organisation | Stated role |
|---|
| P | Amino acids | Polypeptide | Catalytic protein |
| Q | Glucose units | Branched polymer | Carbohydrate storage |
| R | Nucleotides | Polynucleotide | Genetic-material role |
| S | Fatty acids and glycerol | Non-polymeric lipid | Membrane association |
Which substance is a nucleic acid?
ⓐ. Substance R
ⓑ. Substance P
ⓒ. Substance S
ⓓ. Substance Q
Correct Answer: Substance R
Explanation: Substance R combines all the evidence expected for a nucleic acid. Its building units are nucleotides, those units are organised as a polynucleotide, and the stated function is associated with genetic material. Substance P is a protein since its chain consists of amino acids. Substance Q is a glucose polysaccharide, while Substance S is a lipid assembled from fatty-acid and glycerol components without a repeating nucleotide backbone. Acid-insoluble recovery places several major organic classes in the same experimental fraction, so fraction location alone cannot identify the molecule. The building-unit and polymer evidence must be integrated with the biological role, and that complete pattern uniquely identifies R. Nucleic acids are polynucleotides whose repeating units contain a nitrogenous base, pentose and phosphate; their monomers differ fundamentally from amino acids and monosaccharides. Its acid-insoluble recovery is therefore consistent with, but not by itself sufficient for, the final identification.
205. Assertion: A single nucleotide is itself a polynucleotide and can be classified as a complete nucleic-acid polymer.
Reason: A nucleic acid contains many nucleotide units arranged in a polymeric chain.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain Assertion
ⓒ. Assertion is true and Reason is false; therefore Reason cannot explain Assertion
ⓓ. Assertion is false and Reason is true; therefore Reason cannot explain Assertion
Correct Answer: Assertion is false and Reason is true; therefore Reason cannot explain Assertion
Explanation: A nucleotide is the building unit from which a nucleic acid is constructed, but one unit does not constitute a polymer. The prefix poly indicates a chain containing many linked units. The Assertion is false since it treats a monomeric nucleotide as though it were already a complete polynucleotide. The Reason correctly describes nucleic-acid organisation as a polymer containing repeated nucleotide units. This distinction is comparable to the difference between one amino acid and a polypeptide or between one monosaccharide and a polysaccharide. The identity of the building unit remains important, yet polymer status depends on the presence of an extended chain rather than on one isolated component. The result is consistent with the fact that the sequence of nucleotide units forms the primary chemical organisation of a nucleic acid, while complete hydrolysis destroys the polynucleotide chain.
206. Match each nitrogenous base with its class. Column II entries may be reused.
| Column I | Column II |
|---|
| P. Adenine | \(1\). Purine |
| Q. Guanine | \(2\). Pyrimidine |
| R. Uracil | |
| S. Thymine | |
ⓐ. P-\(1\), Q-\(2\), R-\(1\), S-\(2\)
ⓑ. P-\(1\), Q-\(1\), R-\(2\), S-\(2\)
ⓒ. P-\(2\), Q-\(1\), R-\(2\), S-\(1\)
ⓓ. P-\(2\), Q-\(2\), R-\(1\), S-\(1\)
Correct Answer: P-\(1\), Q-\(1\), R-\(2\), S-\(2\)
Explanation: Adenine and guanine are purines, whereas uracil and thymine are pyrimidines. Since two bases belong to each structural class, the Column II entries must be reused. The mapping is therefore P-purine, Q-purine, R-pyrimidine and S-pyrimidine. Purines possess a fused two-ring nitrogenous system, while pyrimidines possess a single nitrogen-containing ring. This classification belongs to the base itself and remains unchanged when the base becomes part of a larger molecule. Attachment of a pentose converts a nitrogenous base into a nucleoside, and addition of phosphate converts the nucleoside into a nucleotide, but adenine and guanine remain purine bases and uracil and thymine remain pyrimidine bases within those compounds. The complete and biologically consistent mapping is P-\(1\), Q-\(1\), R-\(2\), S-\(2\). Cytosine would also map to pyrimidine, but it is not one of the four entries and is not needed to establish the supplied answer.
207. In a nucleotide, adenine is replaced by cytosine while the pentose and phosphate remain unchanged. What change has occurred?
ⓐ. A pyrimidine has been replaced by a purine, and the product is a nucleoside.
ⓑ. A purine has been replaced by another purine, and the product is a free base.
ⓒ. A purine has been replaced by a pyrimidine, while the product remains a nucleotide.
ⓓ. A pyrimidine has been replaced by another pyrimidine, while phosphate is lost.
Correct Answer: A purine has been replaced by a pyrimidine, while the product remains a nucleotide.
Explanation: Adenine belongs to the purine class, whereas cytosine is a pyrimidine. Replacing adenine with cytosine changes the base class from purine to pyrimidine. The pentose and phosphate are explicitly retained, so the molecule still contains all three components required for nucleotide status. It does not become a nucleoside, which would require loss of phosphate, or a free base, which would require removal of both sugar and phosphate. The case separates two simultaneous classification questions: the structural class of the nitrogenous base changes, but the overall base–sugar–phosphate category remains unchanged. The modified molecule is still a nucleotide containing a pyrimidine base.
208. Evaluate the following statements about purines and pyrimidines.
I. Adenine and guanine are purines.
II. Cytosine, uracil and thymine are pyrimidines.
III. The attached pentose determines whether a base is purine or pyrimidine.
IV. All five named compounds are nitrogenous heterocyclic bases.
ⓐ. I and III only
ⓑ. II and IV only
ⓒ. I, III and IV only
ⓓ. I, II and IV only
Correct Answer: I, II and IV only
Explanation: Statements I and II correctly divide the five named bases into two structural classes. Adenine and guanine are purines, while cytosine, uracil and thymine are pyrimidines. All five are nitrogen-containing heterocyclic compounds, making Statement IV valid. Statement III mixes base classification with sugar identity. Ribose and \(2'\)-deoxyribose distinguish RNA-type and DNA-type nucleotide units in the stated comparison, but they do not convert a purine into a pyrimidine or the reverse. The base retains its class when it becomes part of a nucleoside or nucleotide. The valid combination is I, II and IV, preserving the independence of base class and pentose type. The result is interpreted by noting that purines contain two rings and pyrimidines one, but ring classification alone does not identify whether a complete nucleotide belongs to DNA or RNA.
209. A classification table contains one incorrect entry.
| Row | Base | Recorded class |
|---|
| P | Adenine | Purine |
| Q | Guanine | Pyrimidine |
| R | Uracil | Pyrimidine |
| S | Cytosine | Pyrimidine |
Which correction is required?
ⓐ. Row Q should record guanine as a purine.
ⓑ. Row P should record adenine as a pyrimidine.
ⓒ. Row R should record uracil as a purine.
ⓓ. Row S should record cytosine as a purine.
Correct Answer: Row Q should record guanine as a purine.
Explanation: Adenine and guanine form the stated purine pair, while cytosine, uracil and thymine form the pyrimidine set. Row P correctly identifies adenine as a purine. Rows R and S correctly place uracil and cytosine among pyrimidines. Row Q alone conflicts with the classification, since guanine has been entered as a pyrimidine. Changing that entry to purine restores the complete table. The correction does not depend on whether the base later occurs in DNA, RNA, a nucleoside or a nucleotide. Purine–pyrimidine classification belongs to the nitrogenous base itself and remains valid when other molecular components are attached. The relevant structural distinction is the ring class of each free base: adenine and guanine are purines, whereas cytosine, uracil and thymine are pyrimidines. No sugar or phosphate information is needed to correct this table.
210. A nucleotide is described as containing a purine base, ribose and phosphate. What can be concluded from the supplied components?
ⓐ. It is a DNA nucleotide containing a pyrimidine base.
ⓑ. It is an RNA nucleotide containing a purine base.
ⓒ. It is a nucleoside containing a purine base.
ⓓ. It is a free purine base without sugar or phosphate.
Correct Answer: It is an RNA nucleotide containing a purine base.
Explanation: Phosphate is present together with a nitrogenous base and pentose, so the molecule is a nucleotide rather than a nucleoside or free base. Ribose identifies the unit as RNA-type in the stated DNA–RNA sugar comparison. The base is explicitly classified as a purine, although the description does not distinguish whether it is adenine or guanine. The strongest classification must retain all three observations: nucleotide status from the component set, RNA-type identity from the sugar and purine character from the base. No evidence supports a pyrimidine base or \(2'\)-deoxyribose. The complete molecular description is an RNA nucleotide carrying a purine base.
211. Assertion: Adenine and guanine are both purine bases.
Reason: A nitrogenous base is classified as a purine only when it is attached to \(2'\)-deoxyribose.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain Assertion
ⓒ. Assertion is true and Reason is false; therefore Reason cannot explain Assertion
ⓓ. Assertion is false and Reason is true; therefore Reason cannot explain Assertion
Correct Answer: Assertion is true and Reason is false; therefore Reason cannot explain Assertion
Explanation: Adenine and guanine are the two stated purines, making the Assertion true. Their purine identity belongs to the structure of the nitrogenous base and is not created by attachment to a particular sugar. A purine can occur in a nucleotide containing ribose or in one containing \(2'\)-deoxyribose. The Reason is false since it makes deoxyribose attachment a requirement for purine classification. Sugar identity instead helps distinguish RNA-type and DNA-type nucleotide units. The two classification systems operate at different component levels: purine or pyrimidine describes the base, while ribose or deoxyribose describes the pentose associated with the nucleic-acid type. Purine classification belongs to the nitrogenous base itself and does not depend on attachment to a particular pentose.
212. Two nucleotides contain the same nitrogenous base and phosphate. Nucleotide X contains ribose, whereas Nucleotide Y contains \(2'\)-deoxyribose. How should they be classified?
ⓐ. X is RNA-type, while Y is DNA-type.
ⓑ. X is DNA-type, while Y is RNA-type.
ⓒ. Both are nucleosides since their bases are identical.
ⓓ. Both are the same nucleic-acid type since phosphate is present.
Correct Answer: X is RNA-type, while Y is DNA-type.
Explanation: Both molecules contain a base, sugar and phosphate, so each remains a nucleotide. Their difference lies specifically in the pentose. Ribose is the sugar found in RNA nucleotide units, while \(2'\)-deoxyribose is the sugar found in DNA nucleotide units. X is RNA-type and Y is DNA-type. Sharing the same base does not erase the sugar distinction, and phosphate presence establishes nucleotide status rather than nucleic-acid type. The comparison shows how one component can be held constant while another changes the molecular classification. Base class and sugar class provide separate pieces of evidence that must be interpreted independently.
213. Match each molecular category with its pentose. Column II entries may be reused.
| Column I | Column II |
|---|
| P. RNA | \(1\). Ribose |
| Q. DNA | \(2\). \(2'\)-deoxyribose |
| R. RNA nucleotide | |
| S. DNA nucleotide | |
ⓐ. P-\(2\), Q-\(1\), R-\(2\), S-\(1\)
ⓑ. P-\(1\), Q-\(1\), R-\(2\), S-\(2\)
ⓒ. P-\(2\), Q-\(2\), R-\(1\), S-\(1\)
ⓓ. P-\(1\), Q-\(2\), R-\(1\), S-\(2\)
Correct Answer: P-\(1\), Q-\(2\), R-\(1\), S-\(2\)
Explanation: RNA contains ribose, and its nucleotide units also contain ribose. DNA contains \(2'\)-deoxyribose, so the corresponding DNA nucleotide units carry that pentose. Reuse of each Column II entry is required since the polymer and its building unit share the same sugar type. The mapping is P-ribose, Q-\(2'\)-deoxyribose, R-ribose and S-\(2'\)-deoxyribose. The distinction concerns the sugar component rather than the presence of phosphate, which is common to both nucleotide types. It also does not by itself identify the attached nitrogenous base. The pentose relation remains consistent from each nucleotide unit to the nucleic acid formed from those units. The sugar identity is carried from each nucleotide into the finished polymer, so the RNA–DNA distinction remains consistent across both levels. This continuity ensures that hydrolysis of each polymer releases nucleotide units carrying the corresponding pentose.
214. Consider the following statements about pentose sugars in nucleic acids.
I. RNA contains ribose.
II. DNA contains \(2'\)-deoxyribose.
III. Pentose identity can distinguish an RNA-type nucleotide from a DNA-type nucleotide.
IV. Pentose identity alone determines whether the attached base is purine or pyrimidine.
ⓐ. I and IV only
ⓑ. I, II and III only
ⓒ. II, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Ribose and \(2'\)-deoxyribose provide the stated pentose distinction between RNA and DNA. When a complete nucleotide is examined, finding ribose supports RNA-type identity, while finding \(2'\)-deoxyribose supports DNA-type identity. Statements I, II and III are valid. Statement IV incorrectly uses sugar evidence to classify the nitrogenous base. Purine–pyrimidine identity is a property of the base itself. Adenine or guanine remains a purine regardless of which of the two sugars is attached, and a pyrimidine remains a pyrimidine under the same comparison. The accepted combination keeps nucleotide-type classification separate from base-class classification. For the present comparison, purines contain two rings and pyrimidines one, but ring classification alone does not identify whether a complete nucleotide belongs to DNA or RNA.
215. Examine the molecular records below.
| Record | Base class | Sugar | Phosphate | Recorded identity |
|---|
| P | Purine | Ribose | Present | RNA-type purine nucleotide |
| Q | Pyrimidine | Ribose | Present | DNA-type pyrimidine nucleotide |
| R | Purine | \(2'\)-deoxyribose | Present | DNA-type purine nucleotide |
| S | Pyrimidine | \(2'\)-deoxyribose | Absent | DNA nucleotide |
Which records are internally consistent?
ⓐ. Q and S only
ⓑ. P and Q only
ⓒ. P and R only
ⓓ. R and S only
Correct Answer: P and R only
Explanation: Record P contains a purine base, ribose and phosphate. That combination is correctly described as an RNA-type purine nucleotide. Record R contains a purine, \(2'\)-deoxyribose and phosphate and is correctly identified as a DNA-type purine nucleotide. Record Q contains ribose, so its DNA-type label conflicts with the sugar evidence. Record S lacks phosphate and is a nucleoside rather than a nucleotide, even though its sugar is \(2'\)-deoxyribose. The consistent records are P and R. Solving the table requires three separate checks: base class, pentose identity and phosphate-dependent nucleotide status. No single column is sufficient. For this classification, DNA contains deoxyribose and commonly thymine, whereas RNA contains ribose and commonly uracil. The conclusion rests on the observation that phosphate is part of the repeating nucleotide unit, so a base-plus-sugar molecule remains a nucleoside until phosphate is added.
216. Assertion: Replacing ribose with \(2'\)-deoxyribose in a base–sugar–phosphate unit changes it from an RNA-type nucleotide to a DNA-type nucleotide.
Reason: Pentose identity distinguishes the stated DNA and RNA nucleotide types, while retained phosphate preserves nucleotide status.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain Assertion
ⓒ. Assertion is true and Reason is false; therefore Reason cannot explain Assertion
ⓓ. Assertion is false and Reason is true; therefore Reason cannot explain Assertion
Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion
Explanation: The starting molecule contains a base, ribose and phosphate and is an RNA-type nucleotide. Substitution of \(2'\)-deoxyribose for ribose changes the sugar criterion to the one associated with DNA nucleotide units. The base and phosphate remain present, so the molecule does not become a free base or nucleoside. The Reason identifies both parts of the classification: the pentose changes the DNA–RNA type, while phosphate retention maintains the three-component nucleotide category. The Assertion and Reason are true, and the Reason directly accounts for the stated change. This example shows how modifying one component can alter one classification layer without changing another. Changing the sugar changes DNA–RNA type while leaving the base class and nucleotide status intact.
217. An unknown molecule contains ribose, phosphate and an unidentified nitrogenous base. No structural information about the base has been obtained. What is the strongest conclusion?
ⓐ. a DNA-type nucleotide with an unconfirmed pyrimidine base
ⓑ. an RNA-type nucleotide with an unclassified nitrogenous base
ⓒ. a ribose-containing nucleoside with an unconfirmed purine base
ⓓ. a free pyrimidine base lacking both ribose and phosphate
Correct Answer: an RNA-type nucleotide with an unclassified nitrogenous base
Explanation: The molecule contains a base, ribose and phosphate. Presence of all three component types establishes nucleotide status, and ribose identifies the unit as RNA-type. The base has not been structurally identified, so no evidence separates the purine possibilities from the pyrimidine possibilities. Assigning either base class would exceed the available observations. The molecule is not a nucleoside since phosphate is present, and it cannot be a free base while both sugar and phosphate remain attached. The conclusion must include what the evidence establishes and what it leaves unresolved: RNA-type nucleotide identity is supported, but purine–pyrimidine classification requires additional information about the nitrogenous base.
218. A polypeptide is represented only as a linear list of residues from Position \(1\) to Position \(6\): \(\mathrm{Ala-Gly-Ser-Val-Lys-Tyr}\). No folding or subunit association is shown. The representation describes:
ⓐ. the quaternary-structure level
ⓑ. the tertiary-structure level
ⓒ. the secondary-structure level
ⓓ. the primary-structure level
Correct Answer: the primary-structure level
Explanation: Primary structure is the exact positional sequence of amino-acid residues from the first residue to the last in a polypeptide. The representation provides precisely that information: six named residues in one definite linear order. It gives no local helix or sheet, no compact three-dimensional folding of one chain and no arrangement of multiple subunits. Those features correspond to secondary, tertiary and quaternary structure, respectively. The residue list is not merely a composition record, since position is retained. Changing the order while keeping the same six residue types would produce a different primary structure. The described chain represents the first level of protein structural organisation.
219. A linear polypeptide has four labelled residues. P is the first residue, S is the last residue, Q lies immediately after P, and R lies between Q and S. What is the sequence from the N-terminal end to the C-terminal end?
ⓐ. P → R → Q → S
ⓑ. S → R → Q → P
ⓒ. P → Q → R → S
ⓓ. Q → P → S → R
Correct Answer: P → Q → R → S
Explanation: The first residue defines the N-terminal end, so the sequence must begin with P. Q lies immediately after P, fixing the first two positions as P → Q. R is stated to lie between Q and the last residue, while S is explicitly the final residue at the C-terminal end. The only order satisfying all four positional conditions is P → Q → R → S. This order represents primary structure since it specifies the exact residue sequence along one polypeptide. Reversing the chain would exchange the first and last positions and create a different primary structure rather than an equivalent drawing of the same sequence.
220. Evaluate the following statements about protein primary structure.
I. It records the exact order of amino-acid residues.
II. It extends from the first residue to the last residue of a polypeptide.
III. Two chains can have the same amino-acid composition but different primary structures.
IV. It refers to the overall compact three-dimensional fold of one chain.
ⓐ. I, II and III only
ⓑ. I and IV only
ⓒ. II, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Amino-acid composition alone does not preserve positional information; primary structure records the order of the residues. It describes the sequence from the first residue to the last, supporting Statements I and II. Two chains may contain identical numbers of alanine, glycine, serine and valine yet arrange them differently. Such chains have the same composition but different primary structures, making Statement III valid. Statement IV describes tertiary structure, where one polypeptide folds into its overall three-dimensional form. The correct combination is I, II and III. The distinction between composition and sequence is central: knowing which residues are present does not reveal the order in which they occur. The evidence supports only the conclusion that primary structure is the amino-acid sequence, secondary structure is local regular folding, tertiary structure is the overall three-dimensional fold, and quaternary structure concerns association of multiple polypeptide subunits.
