401. Consider the following statements about carbohydrate structure and function.
I. Starch and cellulose can both be glucose homopolymers yet differ in function and iodine response.
II. Glycogen is a branched glucose storage polymer associated with animals.
III. Inulin is identified as a fructose polymer.
IV. Chitin is classified as a protein because its modified sugar units may contain nitrogen.
ⓐ. 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: Starch and cellulose illustrate how the same broad monomer identity can support different structures and functions. Starch serves as a plant reserve and possesses the organisation responsible for the blue iodine response, while cellulose is structural in plant cell walls and lacks that response. Glycogen is the highly branched glucose storage polysaccharide associated with animals, and inulin is a fructose polymer. Statement IV is incorrect since chitin remains a polysaccharide despite containing modified amino-sugar units. Nitrogen presence alone does not establish a protein; proteins require amino-acid residues linked through peptide bonds. The valid statements integrate monomer type, chain architecture, biological role and chemical classification. When applied to these observations, starch and glycogen function mainly in storage, whereas cellulose and chitin provide structural support.
402. Assertion: After selective hydrolysis of starch in a starch–cellulose mixture, the remaining residue may be iodine-negative yet still contain a glucose polymer.
Reason: Cellulose is a glucose polymer but lacks the starch-like iodine-holding helical organisation.
ⓐ. 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 Assertion is true because selective hydrolysis can remove starch, the component that produces the blue iodine response, while leaving cellulose in the residue. An iodine-negative result therefore does not prove that every glucose polymer has disappeared. The Reason is also true. Cellulose is built from glucose but differs from starch in linkage, conformation and biological role, and it lacks the starch-like helical arrangement that holds iodine to produce the blue colour. This property directly explains how a cellulose-containing residue can remain iodine-negative after starch is removed. The inference also illustrates a limit of the iodine test: it detects the appropriate intact starch organisation rather than glucose units in every possible polymer. Both statements are true, and the Reason supplies the structural basis for the Assertion without treating iodine negativity as evidence for absence of all carbohydrates.
403. Match each polysaccharide pair with the distinction that separates its members most directly. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Starch and cellulose | \(1\). Both are storage glucose polymers, but organismal association and branching differ. |
| Q. Starch and glycogen | \(2\). Both are structural polysaccharides, but their building units and biological locations differ. |
| R. Cellulose and chitin | \(3\). One is an animal storage glucose polymer, whereas the other is a fructose polymer. |
| S. Glycogen and inulin | \(4\). Both can be glucose homopolymers, but linkage, conformation, role and iodine response differ. |
ⓐ. P-\(1\), Q-\(4\), R-\(3\), S-\(2\)
ⓑ. P-\(2\), Q-\(3\), R-\(4\), S-\(1\)
ⓒ. P-\(3\), Q-\(2\), R-\(1\), S-\(4\)
ⓓ. P-\(4\), Q-\(1\), R-\(2\), S-\(3\)
Correct Answer: P-\(4\), Q-\(1\), R-\(2\), S-\(3\)
Explanation: Starch and cellulose can both be glucose homopolymers, yet starch is a plant reserve with iodine-holding organisation while cellulose is structural and iodine-negative, giving P-\(4\). Starch and glycogen are storage glucose polymers associated mainly with plants and animals, respectively, and glycogen is strongly branched, so Q-\(1\). Cellulose and chitin both serve structural roles, but cellulose contains glucose in plant cell walls whereas chitin contains modified amino sugars in arthropod exoskeletons; R maps to \(2\). Glycogen is an animal storage glucose polymer, while inulin is identified by fructose building units, giving S-\(3\). The matching task compares relationships rather than asking for isolated names. Each pair shares one broad feature but is separated by monomer identity, architecture, organismal location, function or experimental response. No Column II description can be exchanged without losing the decisive distinction for at least one pair.
404. Use the four arrangements described below.
P. An unbranched glucose chain forms a major structural component of a plant cell wall.
Q. A glucose reserve polymer contains numerous branches and occurs in animal tissue.
R. A plant reserve glucose polymer possesses iodine-holding helical regions.
S. A structural chain of modified amino sugars occurs in an arthropod exoskeleton.
Which ordered identification is correct?
ⓐ. P-starch, Q-cellulose, R-chitin, S-glycogen
ⓑ. P-inulin, Q-starch, R-cellulose, S-glycogen
ⓒ. P-cellulose, Q-glycogen, R-starch, S-chitin
ⓓ. P-glycogen, Q-inulin, R-starch, S-cellulose
Correct Answer: P-cellulose, Q-glycogen, R-starch, S-chitin
Explanation: P combines glucose composition with the structural plant-cell-wall role, identifying cellulose. Q is a branched glucose reserve polymer in animals and represents glycogen. R is the plant reserve polymer whose helical regions can hold iodine, identifying starch. S contains modified amino-sugar units and occurs in arthropod exoskeletons, which identifies chitin. The arrangement requires simultaneous interpretation of monomer, branching, biological location and function. Glucose composition alone would not separate P, Q and R because all three can be glucose polymers. Their distinct organisations and roles provide the necessary discriminating evidence. The correct sequence is cellulose, glycogen, starch and chitin. Functional comparison reveals that starch and glycogen function mainly in storage, whereas cellulose and chitin provide structural support.
405. Four nitrogen-containing compounds are analysed.
| Compound | Base | Sugar | Phosphate |
|---|
| P | Purine | Absent | Absent |
| Q | Pyrimidine | Ribose | Absent |
| R | Purine | Ribose | Present |
| S | Pyrimidine | \(2'\)-deoxyribose | Present |
Which ordered classification is correct?
ⓐ. P is a base, Q is a nucleoside, R is an RNA-type nucleotide and S is a DNA-type nucleotide.
ⓑ. P is a nucleoside, Q is a base, R is a DNA-type nucleotide and S is an RNA-type nucleotide.
ⓒ. P is a nucleotide, Q is an RNA polymer, R is a nucleoside and S is a free base.
ⓓ. P is a base, Q is a nucleotide, R is a DNA-type nucleoside and S is an RNA-type nucleotide.
Correct Answer: P is a base, Q is a nucleoside, R is an RNA-type nucleotide and S is a DNA-type nucleotide.
Explanation: P contains only a nitrogenous base and remains a free base. Q contains a pyrimidine attached to ribose but lacks phosphate, making it a nucleoside. R contains a purine, ribose and phosphate. The three-component set establishes nucleotide status, while ribose identifies it as RNA-type. S also contains all three nucleotide components, but its \(2'\)-deoxyribose identifies it as DNA-type. The table requires three independent checks: whether sugar is attached, whether phosphate is present and which pentose occurs. Purine or pyrimidine identity describes the base but does not by itself determine whether the complete molecule belongs to DNA or RNA. Given the supplied observations, a nitrogenous base joined to a pentose forms a nucleoside, and addition of phosphate forms a nucleotide. Base class, pentose and phosphate must be read together to distinguish a free base, a nucleoside and DNA- or RNA-type nucleotides.
406. A purine ribonucleotide is treated in two stages. First, its phosphate is removed without breaking the base–sugar bond. Next, its ribose is removed from the remaining molecule. What are the successive products?
ⓐ. DNA nucleotide followed by pyrimidine base
ⓑ. Free purine base followed by purine nucleoside
ⓒ. Purine nucleoside followed by free purine base
ⓓ. RNA polymer followed by DNA polymer
Correct Answer: Purine nucleoside followed by free purine base
Explanation: The starting molecule contains a purine base, ribose and phosphate, so it is a purine ribonucleotide. In the first stage, phosphate is removed while the base–sugar bond remains intact. The remaining two-component unit is a purine nucleoside because a base joined to a pentose forms a nucleoside. In the second stage, ribose is removed from that nucleoside, leaving the free purine base. The base class remains purine throughout because the nitrogenous base itself is not replaced. The sequence is nucleotide \(\rightarrow\) nucleoside \(\rightarrow\) free base. No \(2'\)-deoxyribose is introduced, so a DNA-type product cannot be inferred, and removal of components cannot create a polymer. Following the actual components present after each treatment gives the successive products without relying on the starting name alone.
407. A DNA-type pyrimidine nucleotide is divided into three samples.
| Sample | Treatment | Observed components after treatment |
|---|
| P | Phosphate removed | Pyrimidine and \(2'\)-deoxyribose remain joined |
| Q | Sugar removed | Pyrimidine and phosphate remain, but no sugar |
| R | No treatment | Pyrimidine, \(2'\)-deoxyribose and phosphate remain |
Which conclusion is most appropriate?
ⓐ. P remains a DNA nucleotide because its base is a pyrimidine.
ⓑ. Q is a standard nucleoside because phosphate is present.
ⓒ. R is an RNA nucleotide because all nucleotides contain ribose.
ⓓ. P is a deoxyribonucleoside; R remains a DNA-type nucleotide.
Correct Answer: P is a deoxyribonucleoside; R remains a DNA-type nucleotide.
Explanation: Sample P retains a pyrimidine attached to \(2'\)-deoxyribose but lacks phosphate. It is a deoxyribonucleoside rather than a nucleotide. Sample R retains base, deoxyribose and phosphate and remains a DNA-type nucleotide. Sample Q lacks sugar and does not fit the standard base–sugar definition of a nucleoside or the base–sugar–phosphate definition of a nucleotide. Pyrimidine identity alone cannot preserve nucleotide status after phosphate removal. The experiment shows that each classification depends on the complete component set. The sugar identifies DNA-type character, but phosphate presence is still required for nucleotide classification. The structural evidence shows that a phosphate group distinguishes a nucleotide from its corresponding nucleoside, while the base alone lacks the pentose. Removing phosphate changes a nucleotide to a nucleoside, whereas an untreated base–deoxyribose–phosphate unit remains a DNA-type nucleotide.
408. An RNA-type nucleotide undergoes two changes: ribose is replaced by \(2'\)-deoxyribose, and phosphate is subsequently removed. The nitrogenous base remains unchanged. What is the final product?
ⓐ. An RNA-type nucleotide containing the original base
ⓑ. A DNA-type nucleoside containing the original base
ⓒ. A free nitrogenous base with no sugar
ⓓ. A DNA polynucleotide containing several bases
Correct Answer: A DNA-type nucleoside containing the original base
Explanation: The starting molecule is an RNA-type nucleotide because it contains ribose together with a nitrogenous base and phosphate. Replacing ribose with \(2'\)-deoxyribose changes the sugar-based identity to DNA-type while the molecule is still a nucleotide. The second change removes phosphate but leaves the base attached to the deoxyribose. A base–sugar unit without phosphate is a nucleoside, so the final product is a DNA-type nucleoside containing the original base. It is not a free base because the sugar remains attached, and one modified unit cannot become a polynucleotide. The two operations affect independent classification layers: pentose identity distinguishes RNA-type from DNA-type, while phosphate presence distinguishes nucleotide from nucleoside. Applying the changes in the stated order prevents the intermediate and final categories from being confused.
409. Consider the following statements about nucleic-acid components.
I. Purine–pyrimidine classification belongs to the nitrogenous base.
II. Ribose and \(2'\)-deoxyribose distinguish RNA-type and DNA-type units.
III. Phosphate presence distinguishes a nucleotide from the corresponding nucleoside.
IV. Replacing a purine with a pyrimidine automatically removes the sugar and phosphate.
ⓐ. 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: The first three statements describe three independent but connected classification axes. Purine or pyrimidine identifies the structural class of the nitrogenous base. Ribose or \(2'\)-deoxyribose identifies whether a complete unit is RNA-type or DNA-type. Phosphate determines whether a base–sugar molecule is a nucleotide rather than a nucleoside. Statement IV incorrectly links base replacement with removal of other components. A purine can be replaced by a pyrimidine while the sugar and phosphate remain attached, leaving the molecule a nucleotide but changing its base class. The valid set is I, II and III. Accurate classification requires each component to be assessed separately before the evidence is combined. The interpretation is that a phosphate group distinguishes a nucleotide from its corresponding nucleoside, while the base alone lacks the pentose.
410. Assertion: A molecule containing cytosine, ribose and phosphate is an RNA-type pyrimidine nucleotide.
Reason: Cytosine is a pyrimidine, ribose gives RNA-type identity and phosphate completes nucleotide composition.
ⓐ. 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: Cytosine belongs to the pyrimidine class of nitrogenous bases. Attachment to ribose supplies the sugar associated with RNA-type units, and the additional presence of phosphate makes the molecule a nucleotide rather than a nucleoside. The Assertion is correct. The Reason identifies each component and links it to the relevant classification decision: base class, nucleic-acid type and phosphate-dependent molecular category. It directly explains the complete description in the Assertion. No information about a polynucleotide chain is required because the evidence concerns one nucleotide unit. Both statements are true, and the component-by-component analysis in the Reason fully accounts for the stated classification. All three pieces of evidence are necessary: base class, pentose identity and phosphate-containing nucleotide status.
411. Match each molecular description with the correct category. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Adenine only | \(1\). Purine base |
| Q. Adenine joined to ribose | \(2\). Purine nucleoside |
| R. Adenine, ribose and phosphate | \(3\). RNA-type purine nucleotide |
| S. Cytosine, \(2'\)-deoxyribose and phosphate | \(4\). DNA-type pyrimidine nucleotide |
ⓐ. P-\(2\), Q-\(1\), R-\(4\), S-\(3\)
ⓑ. P-\(1\), Q-\(2\), R-\(3\), S-\(4\)
ⓒ. P-\(4\), Q-\(3\), R-\(2\), S-\(1\)
ⓓ. P-\(1\), Q-\(4\), R-\(3\), S-\(2\)
Correct Answer: P-\(1\), Q-\(2\), R-\(3\), S-\(4\)
Explanation: Adenine alone is a purine nitrogenous base, so P matches \(1\). Adenine joined to ribose forms a purine nucleoside, giving Q-\(2\). Adding phosphate to the adenine–ribose unit produces an RNA-type purine nucleotide, so R-\(3\). Cytosine is a pyrimidine, and its combination with \(2'\)-deoxyribose and phosphate forms a DNA-type pyrimidine nucleotide, giving S-\(4\). The mapping progresses from base to nucleoside to nucleotide while also using base class and pentose identity. Each added component changes a specific part of the classification without erasing the identities already established. The experimental evidence shows that a nitrogenous base joined to a pentose forms a nucleoside, and addition of phosphate forms a nucleotide. The condition given here demonstrates that omitting or adding one component changes the chemical category even when the nitrogenous base remains the same.
412. Use the molecular arrangement described below. Region P is a two-ring nitrogenous base. Region Q is ribose attached to P. Region R is a phosphate group esterified to Q. Which complete interpretation is correct?
ⓐ. P is a pyrimidine, and the whole molecule is a DNA nucleoside.
ⓑ. P is a purine, and the whole molecule is an RNA-type nucleotide.
ⓒ. P is a purine, and the whole molecule is a free nitrogenous base.
ⓓ. P is a pyrimidine, and the whole molecule is an RNA polynucleotide.
Correct Answer: P is a purine, and the whole molecule is an RNA-type nucleotide.
Explanation: Region P has two fused rings, which places the nitrogenous base in the purine class. Attachment of ribose at Q creates a purine ribonucleoside component. Region R adds an esterified phosphate group to the sugar, completing the three-part composition of a nucleotide. Because the pentose is ribose rather than \(2'\)-deoxyribose, the whole unit is RNA-type. The molecule cannot be a free base, since both sugar and phosphate are present, and one nucleotide is not a polynucleotide. A pyrimidine interpretation is also excluded by the two-ring description. The classification requires all three spatially described observations: ring number establishes the base class, the sugar establishes RNA-type identity, and phosphate establishes nucleotide status. Ignoring any one region would produce an incomplete or incorrect molecular category.
413. Four lipid samples are described below.
| Sample | Structural evidence | Observed relation |
|---|
| P | Glycerol esterified with three fatty acids lacking \(C=C\) bonds | Relatively high melting behaviour |
| Q | Glycerol esterified with three fatty acids containing \(C=C\) bonds | Liquid under the stated room condition |
| R | Lipid containing fatty-acid components and phosphate | Occurs in cell membranes |
| S | Low-mass membrane lipid recovered in acid-insoluble material | Not a repeating-unit polymer |
Which interpretation is correct?
ⓐ. P is an unsaturated monoglyceride; Q is a saturated triglyceride; R is a nucleotide; S is a protein polymer.
ⓑ. P is a saturated diglyceride; Q is an unsaturated monoglyceride; R is a glyceride; S is a nucleic-acid polymer.
ⓒ. P is a saturated monoglyceride; Q is an unsaturated diglyceride; R is phosphate-free; S is acid-soluble.
ⓓ. P is a saturated triglyceride; Q is an unsaturated triglyceride; R is a phospholipid; S is membrane-associated lipid.
Correct Answer: P is a saturated triglyceride; Q is an unsaturated triglyceride; R is a phospholipid; S is membrane-associated lipid.
Explanation: P and Q each contain three fatty acids esterified to glycerol, so both are triglycerides. Their difference lies in fatty-acid saturation: P lacks \(C=C\) bonds and is saturated, while Q contains such bonds and is unsaturated. The associated melting observations are consistent with the stated fat–oil distinction. R contains phosphate and occurs in cell membranes, supporting phospholipid classification. S has a low individual molecular mass and is non-polymeric but remains with acid-insoluble membrane material. It demonstrates that lipid recovery in that fraction results from membrane association and water insolubility rather than true macromolecular polymer status. The table integrates structure, physical behaviour, location and fractionation. Its acid-insoluble recovery reflects aggregation with membrane fragments, not a molecular mass comparable to that of true macromolecular polymers.
414. A lipid isolated from seeds contains glycerol esterified with three fatty acids. At least one fatty-acid chain contains a \(C=C\) bond, and the material is liquid under the stated room condition. Which description best fits the evidence?
ⓐ. It is an unsaturated triglyceride showing oil-like melting behaviour.
ⓑ. It is a saturated monoglyceride because only one glycerol is present.
ⓒ. It is a phospholipid because every seed lipid contains phosphate.
ⓓ. It is a true polysaccharide polymer because it is stored in living tissue.
Correct Answer: It is an unsaturated triglyceride showing oil-like melting behaviour.
Explanation: Three fatty acids esterified to one glycerol identify a triglyceride; the number of glycerol molecules does not determine the mono-, di- or triglyceride name. Presence of at least one \(C=C\) bond establishes unsaturation in the fatty-acid component. The liquid state under the stated room condition is consistent with oil-like melting behaviour. Phosphate has not been detected, so phospholipid classification is unsupported. Storage in seeds also does not make the molecule a polysaccharide or a true repeating-unit polymer. The evidence must be combined across three decision axes: esterification number identifies triglyceride status, double bonds identify unsaturation and melting behaviour supports its classification as an oil under the stated condition. All three observations converge on the same molecular description rather than supplying independent alternative classifications.
415. Two purified triglycerides contain the same glycerol backbone and the same number of esterified fatty acids. Lipid P contains no \(C=C\) bonds and melts at a higher temperature than Lipid Q, whose fatty acids contain several \(C=C\) bonds. Which conclusion is best supported?
ⓐ. P and Q differ in esterification number rather than saturation.
ⓑ. Q is a phosphate-containing phospholipid with lower melting.
ⓒ. Q is more unsaturated and melts lower at equal esterification.
ⓓ. P is polymeric, whereas Q is a non-polymeric micromolecule.
Correct Answer: Q is more unsaturated and melts lower at equal esterification.
Explanation: Both samples contain three fatty acids attached to glycerol, so both remain triglycerides. Their esterification number is controlled and cannot explain the melting difference. The structural variable is fatty-acid saturation. P has no \(C=C\) bonds and is saturated, while Q contains several double bonds and is more unsaturated. The observed lower melting behaviour of Q is associated with its greater unsaturation under the supplied comparison. No phosphate evidence is present, and melting temperature does not determine whether a molecule is a polymer. The experiment isolates saturation as the meaningful changed variable while holding the glycerol backbone and fatty-acid number constant. It prevents unsaturation from being confused with incomplete esterification. The lower melting point of Q is consistent with poorer packing of its unsaturated chains.
416. A membrane lipid loses its phosphate-containing component while retaining its fatty-acid-containing portion. Which prediction is most justified?
ⓐ. It remains a phospholipid because its fatty acids are retained.
ⓑ. Loss of phosphate removes a defining feature of phospholipids.
ⓒ. It becomes a triglyceride regardless of its esterification number.
ⓓ. It enters the acid-soluble pool solely because phosphate is absent.
Correct Answer: Loss of phosphate removes a defining feature of phospholipids.
Explanation: Phospholipids are identified by the presence of phosphorus-containing components and their membrane occurrence. Removing the phosphate-containing part eliminates a defining piece of evidence for that category. The retained material remains lipid-derived, but its exact new subclass cannot be determined without knowing the remaining components and esterification pattern. It cannot be called a triglyceride unless glycerol is shown to carry three fatty acids. Removal of phosphate also does not create a nucleotide, which requires a nitrogenous base, sugar and phosphate, or a repeating-unit polymer. The strongest prediction remains limited: the molecule loses the structural feature required for phospholipid classification, while a more precise replacement name needs additional evidence. The altered molecule may retain lipid character, but it no longer satisfies the stated phospholipid description.
417. Consider the following statements about integrated lipid classification.
I. Saturation depends on the presence or absence of \(C=C\) bonds in fatty-acid chains.
II. Mono-, di- and triglyceride names depend on the number of fatty acids esterified to glycerol.
III. Phosphate-containing lipids such as lecithin are associated with cell membranes.
IV. Acid-insoluble recovery proves that every lipid is a high-mass polymer.
ⓐ. 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: Statements I, II and III describe separate but connected lipid decisions. Double-bond evidence determines whether a fatty acid is saturated or unsaturated. The number of fatty acids attached to glycerol determines whether the product is a mono-, di- or triglyceride. Phosphate-containing lipids such as lecithin occur in cell membranes. Statement IV incorrectly interprets experimental fractionation as proof of polymer structure. Individual membrane lipids generally have relatively low molecular masses and are not repeating-unit polymers. They enter the acid-insoluble fraction because disrupted membranes remain as water-insoluble aggregates or fragments. The correct set integrates covalent structure, physical classification, membrane occurrence and the important fractionation exception without collapsing them into one rule. Statement IV fails because experimental fraction membership does not convert a lipid into a polymer.
418. Assertion: Membrane lipids may occur in the acid-insoluble fraction even though individual lipid molecules are not true macromolecular polymers.
Reason: Disrupted membrane lipids can remain associated in water-insoluble fragments or vesicle-like aggregates during fractionation.
ⓐ. 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 Assertion states the central lipid exception correctly. Proteins, polysaccharides and nucleic acids are true polymeric biomacromolecules in the acid-insoluble fraction, whereas individual lipid molecules are generally much smaller and lack indefinite repeating-unit organisation. The Reason explains how lipids nevertheless enter the same operational fraction. During tissue disruption, membrane lipids remain together in fragments or vesicle-like aggregates that are water-insoluble and are retained with the acid-insoluble material. Their fraction location reflects physical association and solubility rather than polymer status. Both statements are true, and the Reason directly accounts for the apparent mismatch between low individual molecular mass and insoluble recovery. Membrane aggregation explains insoluble recovery without converting low-mass lipid molecules into polymers.
419. An unknown cellular molecule has a molecular mass of \(720\,\text{Da}\), contains fatty-acid components and phosphate, and is recovered mainly with disrupted membrane fragments in the acid-insoluble fraction. What is the strongest conclusion?
ⓐ. It is a nucleotide polymer because phosphate is present.
ⓑ. It is a membrane phospholipid recovered insolubly through membrane association.
ⓒ. It is a high-mass polysaccharide because it remains in the retentate.
ⓓ. It is an acid-soluble amino acid because its mass is below \(800\,\text{Da}\).
Correct Answer: It is a membrane phospholipid recovered insolubly through membrane association.
Explanation: Fatty-acid components together with phosphate and membrane association support phospholipid classification. The molecular mass of \(720\,\text{Da}\) lies within the general size range of individual lipids rather than that of high-mass biomacromolecular polymers. Its recovery in the acid-insoluble fraction is explained by continued association with membrane fragments after tissue disruption. Phosphate presence alone cannot establish nucleotide identity because a nucleotide also requires a nitrogenous base and sugar. Likewise, fraction location alone cannot convert the molecule into a polysaccharide. Although the mass lies within the usual acid-soluble range, actual fraction behaviour also depends on water solubility and physical association. The integrated evidence identifies a membrane phospholipid displaying the lipid fraction exception.
420. Four unknown molecules are compared.
| Molecule | Fatty acids attached to glycerol | \(C=C\) bonds | Phosphate | Fraction behaviour |
|---|
| P | \(1\) | Absent | Absent | Not specified |
| Q | \(3\) | Present | Absent | Oil-like under the stated condition |
| R | \(3\) | Absent | Absent | Fat-like under the stated condition |
| S | Fatty-acid-containing lipid | Not decisive | Present | Recovered with membrane fragments |
Which synthesis is correct?
ⓐ. P is a saturated triglyceride; Q is a saturated monoglyceride; R is an unsaturated diglyceride; S is a nucleotide.
ⓑ. P is a phosphate-containing lipid; Q is a monoglyceride; R is a diglyceride; S is a polysaccharide polymer.
ⓒ. P and Q differ only in phosphate; R and S are repeating-unit macromolecular polymers.
ⓓ. P is a monoglyceride; Q is an unsaturated triglyceride; R is a saturated triglyceride; S is a phospholipid.
Correct Answer: P is a monoglyceride; Q is an unsaturated triglyceride; R is a saturated triglyceride; S is a phospholipid.
Explanation: Molecule P has one fatty acid esterified to glycerol and is a monoglyceride. Q and R each contain three fatty acids, making both triglycerides. Q contains \(C=C\) bonds and displays oil-like behaviour, supporting unsaturated triglyceride classification. R lacks double bonds and displays fat-like behaviour, supporting saturated triglyceride classification. S contains phosphate and fatty-acid components and is recovered with membrane fragments, identifying a membrane phospholipid. Its acid-insoluble behaviour does not imply true repeating-unit polymer status. The table integrates three independent structural questions—esterification number, saturation and phosphate presence—with physical and fractionation evidence. Each molecule is classified through the complete pattern rather than through one isolated column. The combined evidence separates esterification number, saturation and phosphate content instead of treating all lipids as one class.
