Biomolecules MCQs with Answers – Part 8 (NEET Chemistry)

Explanation: Statement I is correct because glucose and galactose are both aldohexoses with unbranched six-carbon skeletons. Formation of n-hexane through prolonged heating with HI is consistent with this common carbon-chain feature. However, the reaction does not preserve information about the original configurations of their asymmetric carbon atoms. Statement II is correct because both sugars possess an aldehydic carbonyl group in their open-chain structures. Hydroxylamine can therefore react with either sugar to form an oxime. The formation of an oxime establishes carbonyl functionality but does not independently reveal the configuration at C4. Statement III is correct because both molecules contain five hydroxyl groups available for complete acetylation under suitable conditions. Consequently, both can form corresponding pentaacetate derivatives. Statement IV is incorrect because these three observations have the same qualitative interpretation for both candidate sugars. They do not distinguish the particular C4 configuration that differentiates glucose from galactose. Statement V is correct because the reactions establish carbon-chain length, carbonyl functionality and hydroxyl-group count without providing the required stereochemical discrimination. The observations are therefore useful for identifying broad structural features but insufficient for assigning the two individual sugar identities. This illustrates why several chemically correct tests may still fail to identify a compound uniquely when the candidate structures differ only in stereochemical configuration.

Explanation: \(\textbf{Principle of the rate–time graph:}\) The area under a rate–time graph gives the total amount of a substance consumed when the vertical axis represents consumption rate. Since the rate is expressed in \(\mathrm{mol\,min^{-1}}\), multiplying the rate by the corresponding time interval gives the amount of maltose consumed in moles. \(\textbf{Maltose consumption in mixture P:}\) The rate of maltose consumption in P remains constant for the first three minutes. \[ r_P=0.020\ \mathrm{mol\,min^{-1}} \] The corresponding reaction duration is: \[ t_P=3\ \mathrm{min} \] Therefore: \[ n_{\mathrm{consumed,P}}=r_Pt_P \] \[ n_{\mathrm{consumed,P}}=0.020\times3 \] \[ n_{\mathrm{consumed,P}}=0.060\ \mathrm{mol} \] At exactly 3 min, maltase is completely inactivated in mixture P. Consequently, no additional maltose is consumed during the remaining two minutes. \(\textbf{Maltose remaining in mixture P:}\) The initial amount of maltose is: \[ n_{\mathrm{initial,P}}=0.100\ \mathrm{mol} \] Subtracting the consumed amount: \[ n_{\mathrm{remaining,P}}=0.100-0.060 \] \[ n_{\mathrm{remaining,P}}=0.040\ \mathrm{mol} \] Thus, mixture P retains \(0.040\ \mathrm{mol}\) of unreacted maltose. \(\textbf{Maltose consumption in mixture Q:}\) The specified idealised rate in Q is: \[ r_Q=0.030\ \mathrm{mol\,min^{-1}} \] This rate is maintained for two minutes. Therefore: \[ n_{\mathrm{consumed,Q}}=r_Qt_Q \] \[ n_{\mathrm{consumed,Q}}=0.030\times2 \] \[ n_{\mathrm{consumed,Q}}=0.060\ \mathrm{mol} \] \(\textbf{Maltose remaining in mixture Q:}\) Initially, Q contains \(0.060\ \mathrm{mol}\) maltose. Hence: \[ n_{\mathrm{remaining,Q}}=0.060-0.060 \] \[ n_{\mathrm{remaining,Q}}=0 \] Therefore, Q exhausts its maltose supply at 2 min. Its subsequent zero rate is explained by substrate exhaustion rather than deliberate enzyme inactivation. \(\textbf{Glucose formation from maltose:}\) Maltose undergoes hydrolysis to produce two glucose molecules per molecule of maltose. The balanced chemical reaction is: \[ \mathrm{C_{12}H_{22}O_{11}+H_2O\rightarrow2C_6H_{12}O_6} \] Therefore: \[ n_{\mathrm{glucose}}=2n_{\mathrm{maltose\ consumed}} \] For mixture P: \[ n_{\mathrm{glucose,P}}=2(0.060) \] \[ n_{\mathrm{glucose,P}}=0.120\ \mathrm{mol} \] For mixture Q: \[ n_{\mathrm{glucose,Q}}=2(0.060) \] \[ n_{\mathrm{glucose,Q}}=0.120\ \mathrm{mol} \] Thus, both mixtures produce the same total amount of glucose. \(\textbf{Interpretation of the zero-rate intervals:}\) In mixture P, maltose remains available after 3 min, but the experimenter deliberately stops the catalytic reaction by inactivating maltase. In mixture Q, no deliberate enzyme inactivation occurs. The reaction stops because all available maltose has been consumed. Therefore, a zero reaction rate does not necessarily imply complete substrate exhaustion. It can also result from termination of catalytic activity. \(\textbf{Final comparison:}\) Both mixtures consume \(0.060\ \mathrm{mol}\) of maltose and produce \(0.120\ \mathrm{mol}\) of glucose. However, P retains \(0.040\ \mathrm{mol}\) maltose, whereas Q contains no unreacted maltose. The specified experimental conditions account for the different reasons why the two reaction rates become zero. \(\textbf{Final answer:}\) \[ \boxed{\begin{gathered} n_{\mathrm{glucose,P}}=n_{\mathrm{glucose,Q}}=0.120\ \mathrm{mol}\\ n_{\mathrm{remaining,P}}=0.040\ \mathrm{mol}\\ n_{\mathrm{remaining,Q}}=0 \end{gathered}} \]

Explanation: The D/L designation of a carbohydrate describes relative configuration according to its Fischer-projection relationship with the reference configuration. It does not directly indicate whether the substance is dextrorotatory or laevorotatory. P is D-(−)-fructose, showing that a D-series carbohydrate can rotate plane-polarised light towards the negative direction. Q is defined as the complete mirror-image stereoisomer of P. Therefore, all corresponding asymmetric centres have opposite configurations. Its configuration at the lowest asymmetric carbon is inverted, so Q belongs to the L-series. A pair of pure enantiomers has equal-magnitude but opposite-sign optical rotations under identical measurement conditions. Consequently, Q is dextrorotatory with a rotation opposite to that of P. R differs from P only at C3. Because its configuration at C5 remains unchanged, R retains the D-designation. However, R is a diastereomer rather than the complete enantiomer of P. Diastereomeric optical rotations do not follow the universal equal-and-opposite relationship characteristic of enantiomers. Therefore, the sign of R's optical rotation cannot be established solely from the supplied C3 configurational change. The correct answer requires separate application of the D/L assignment rule, the enantiomeric rotation relationship and the distinction between enantiomers and diastereomers.

Explanation: Primary protein structure describes the ordered amino acid sequence and its covalent peptide connectivity. Secondary structure describes regular arrangements of the polypeptide backbone, including α-helices and β-pleated sheets. The specified β-pleated sheet is initially formed by two extended regions belonging to the same polypeptide chain. Hydrogen bonds between suitable backbone groups maintain their relative arrangement. The first treatment hydrolyses one peptide bond in the connecting region R. Consequently, the original single polypeptide becomes two covalently independent chains. However, the question explicitly states that the hydrogen bonds between P and Q remain intact. The two extended segments can therefore retain their local β-pleated-sheet arrangement despite the loss of their original continuous covalent connection. A β-pleated sheet does not require every interacting segment to belong to one uninterrupted polypeptide chain. The second treatment instead disrupts the hydrogen bonds maintaining the sheet. It can therefore disturb that secondary-structure arrangement without cleaving additional peptide bonds. The two treatments act on different chemical connections and need not produce identical structural consequences. The experiment demonstrates that local secondary structure, covalent chain continuity and association between different polypeptide chains are related but distinguishable aspects of protein organisation.

Explanation: D-glucose and L-glucose are enantiomers because their corresponding asymmetric centres have opposite configurations. They have identical molecular formulas and the same types and numbers of ordinary functional groups. Their open-chain structures both contain an aldehydic group at C1. Consequently, a suitable non-stereoselective oxidising reagent can convert this group into a carboxyl group in either sample. Oxidation at C1 does not automatically reverse the configurations of the other asymmetric carbon atoms. Therefore, the corresponding products retain their enantiomeric relationship when no stereochemical change occurs elsewhere. The outcome with enzyme E must be interpreted differently. Enzymes exhibit specificity towards particular substrates and transformations. The problem explicitly establishes that E recognises D-glucose but not L-glucose under the experimental conditions. Thus, E can distinguish the two configurations even though X recognises the same aldehydic functionality in both. The difference does not contradict their shared molecular formula or functional-group inventory. It shows that recognising a chemical functional group and recognising the complete spatial arrangement of a substrate are different requirements. The experiment integrates carbohydrate stereochemistry, selective oxidation and enzyme specificity without assuming that enantiomers must behave identically in a biological catalytic environment.

Explanation: Primary structure specifies the exact order of amino acid residues in a polypeptide. The experiment identifies fragments rather than merely the total amino acid composition. Experiment I establishes that glycine is followed directly by serine and then valine. Therefore, the original chain must contain the ordered segment Gly–Ser–Val. Experiment II establishes that alanine is followed directly by glycine. These two observations overlap at glycine. Joining them while preserving their individual orders gives Ala–Gly–Ser–Val. The original pentapeptide must also contain one lysine residue. The terminal-group experiment establishes that lysine occupies the C-terminal position. Consequently, lysine must follow valine in the reconstructed sequence. The complete sequence is therefore Ala–Gly–Ser–Val–Lys. Option B places alanine after the Gly–Ser–Val segment and contradicts the observed Ala–Gly fragment. Option C reverses the required Ser–Val order. Option D does not preserve either complete overlapping fragment in its required orientation. The reconstruction demonstrates why ordered partial-hydrolysis fragments and terminal information provide stronger evidence of primary structure than amino acid composition alone.

Explanation: Vitamin C is ascorbic acid and belongs to the water-soluble vitamin group. NCERT associates its deficiency with scurvy, including characteristic bleeding of the gums. Consequently, Observation P is consistent with inadequate vitamin C intake. Vitamin K belongs to the fat-soluble vitamin group. Its characteristic biological association concerns normal blood clotting, and deficiency can produce increased blood clotting time. Therefore, Observation Q is consistent with inadequate vitamin K intake. The two observations involve blood-related manifestations but do not describe identical biological functions. Their different responses to the controlled dietary corrections provide additional evidence for distinguishing the vitamin requirements. Improvement of P following correction of vitamin C intake supports the association between P and vitamin C. Improvement of Q following correction of vitamin K intake independently supports the corresponding association. The experiment does not require the two vitamins to possess the same chemical structure or solubility classification. Indeed, their classifications differ. The student's reasoning is incorrect because a superficially similar symptom category does not independently establish one common molecular cause. The appropriate interpretation combines the specific NCERT deficiency associations with the results of the controlled dietary comparison.

Explanation: Statement I is correct because the active preparation contains two polypeptide subunits associated in a defined spatial arrangement. Such an arrangement is characteristic of quaternary protein structure. Statement II is incorrect because separating associated subunits does not necessarily hydrolyse their peptide backbones. The question explicitly states that peptide bonds and the individual folded conformations remain intact. Consequently, the original primary structures of P and Q are preserved. Statement III is correct because neither isolated subunit displays appreciable catalytic activity under the tested conditions, whereas reassociation restores activity. The observation supports the importance of their correct combined organisation for the native catalytic function. Statement IV is incorrect because restoration of activity in the assembled protein does not establish activity in either isolated component. Indeed, the independent measurements indicate negligible activity in both separated preparations. Statement V is correct because the original activity disappears following disruption of subunit association despite preservation of the individual peptide backbones. The loss of activity is therefore not evidence that the proteins have been completely hydrolysed. The experiment demonstrates that catalytic function may depend on organisation at the quaternary level rather than on the mere presence of the correct amino acid residues in separate chains.

Explanation: \(\textbf{Original DNA backbone connectivity:}\) The DNA fragment initially contains two antiparallel polynucleotide strands. Each strand is covalently continuous through its phosphodiester backbone. \[ N_{\mathrm{covalent\ fragments,initial}}=2 \] \(\textbf{Effect of the single phosphodiester nick:}\) The reagent cleaves one internal phosphodiester bond in the first strand. This divides that strand into two covalently independent segments. The complementary second strand remains covalently continuous. \[ N_{\mathrm{fragments,after\ nick}}=2+1 \] \[ N_{\mathrm{fragments,after\ nick}}=3 \] \(\textbf{Hydrogen-bond preservation:}\) The nick does not break the hydrogen bonds between complementary nitrogenous bases. Consequently, the three covalent fragments can remain associated within one base-paired DNA assembly immediately after nick formation. Covalent connectivity and physical association through hydrogen bonds are therefore different properties. \(\textbf{Effect of heating:}\) The subsequent heating disrupts the interstrand hydrogen bonds. This separates the previously associated polynucleotide segments. However, heating does not introduce any additional phosphodiester cleavage. \[ N_{\mathrm{fragments,after\ heating}}=3 \] \(\textbf{G–C hydrogen bonds:}\) Each ordinary G–C base pair contributes three hydrogen bonds. \[ N_{\mathrm{H,G-C}}=5(3) \] \[ N_{\mathrm{H,G-C}}=15 \] \(\textbf{A–T hydrogen bonds:}\) Each ordinary A–T pair contributes two hydrogen bonds. \[ N_{\mathrm{H,A-T}}=3(2) \] \[ N_{\mathrm{H,A-T}}=6 \] \(\textbf{Original total hydrogen bonds:}\) \[ N_{\mathrm{H,total}}=15+6 \] \[ N_{\mathrm{H,total}}=21 \] \(\textbf{Consistency check:}\) The nick increases the number of covalently independent chain segments without reducing the initial interstrand hydrogen-bond count. Heating subsequently separates those segments by disrupting hydrogen bonding but does not change the number of covalent fragments. Therefore, the two treatments have distinguishable effects on the DNA structure. \(\textbf{Final answer:}\) \[ \boxed{\begin{gathered} N_{\mathrm{fragments,after\ nick}}=3\\ N_{\mathrm{fragments,after\ heating}}=3\\ N_{\mathrm{H,initial}}=21 \end{gathered}} \]

Explanation: \(\textbf{Terminal functionalities of the three substances:}\) Open-chain glucose contains an aldehydic group at C1 and a primary alcoholic group at C6. Its remaining four alcoholic hydroxyl groups occur at C2 through C5. Oxidation of C1 produces gluconic acid, whereas oxidation of both terminal groups produces saccharic acid. \(\textbf{Carboxyl-group inventory:}\) Glucose P has no carboxylic acid group. Gluconic acid Q contains one carboxylic acid group at C1. Saccharic acid R contains two carboxylic acid groups, at C1 and C6. \(\textbf{Carbon dioxide from Q:}\) Each carboxylic acid group reacts with sodium hydrogencarbonate to liberate one molecule of carbon dioxide. \[ n_{\mathrm{CO_2,Q}}=(0.050\ \mathrm{mol})(1) \] \[ n_{\mathrm{CO_2,Q}}=0.050\ \mathrm{mol} \] \(\textbf{Carbon dioxide from R:}\) Each molecule of R possesses two free carboxylic acid groups. \[ n_{\mathrm{CO_2,R}}=(0.050\ \mathrm{mol})(2) \] \[ n_{\mathrm{CO_2,R}}=0.100\ \mathrm{mol} \] \(\textbf{Alcoholic hydroxyl inventory of R:}\) Both terminal carbon atoms have been oxidised into carboxylic acid groups. The four secondary alcoholic hydroxyl groups at C2, C3, C4 and C5 remain. Therefore: \[ N_{\mathrm{alcoholic\ OH,R}}=4 \] \(\textbf{Acetylation requirement:}\) One molecule of acetic anhydride is consumed for each alcoholic hydroxyl group acetylated. The carboxylic acid hydroxyl groups are excluded by the specified reagent selectivity. \[ n_{\mathrm{anhydride,R}}=(0.050\ \mathrm{mol})(4) \] \[ n_{\mathrm{anhydride,R}}=0.200\ \mathrm{mol} \] \(\textbf{Independent chemical check:}\) Q retains the original C6 primary alcoholic group and consequently has five alcoholic hydroxyl groups. R has lost that alcoholic functionality through oxidation, but gained a second carboxylic acid group. Thus, R evolves more carbon dioxide while requiring less acetic anhydride than an equimolar sample of Q. The two tests distinguish acidic and alcoholic hydroxyl functionalities rather than merely counting every oxygen-containing group together. \(\textbf{Final answer:}\) \[ \boxed{\begin{gathered} n_{\mathrm{CO_2,Q}}=0.050\ \mathrm{mol}\\ n_{\mathrm{CO_2,R}}=0.100\ \mathrm{mol}\\ n_{\mathrm{anhydride,R}}=0.200\ \mathrm{mol} \end{gathered}} \]

Explanation: D-ribose is an aldopentose whose ordinary open-chain form contains an aldehydic group. The free sugar can also form a cyclic hemiacetal through intramolecular addition. When the anomeric hemiacetal hydroxyl is free, ring–chain interconversion permits access to the open-chain aldehydic form. Consequently, P possesses reducing character under suitable conditions. Q is adenosine, a nucleoside containing adenine covalently attached to ribose. Its base–sugar linkage involves the anomeric C1′ position of the sugar. Therefore, Q does not possess the corresponding free ribose hemiacetal group while that linkage remains intact. R is a nucleotide obtained through additional phosphorylation of the nucleoside at C5′. Its anomeric C1′ position remains occupied by the same base–sugar linkage. Consequently, the additional phosphate group does not restore ordinary free-ribose reducing behaviour. Treatment I removes the C5′ phosphate group from R, converting it into adenosine without breaking its C1′ glycosidic linkage. Thus, phosphate removal alone is insufficient to regenerate free ribose. Treatment II instead hydrolyses the base–sugar linkages of the nucleoside species. This releases adenine and free ribose. The liberated ribose can now undergo ordinary hemiacetal ring–chain interconversion and display reducing behaviour. The experiment demonstrates that the reducing character of a sugar-derived constituent depends on its anomeric connectivity, not merely on the presence of a pentose skeleton within a larger biomolecule.

Explanation: \(\textbf{Charge inventory of the original tetrapeptide:}\) The peptide contains two positively charged amino groups and two negatively charged carboxyl groups under the specified conditions. The positive contributions arise from the N-terminal alpha-amino group and lysine side chain. The negative contributions arise from the C-terminal alpha-carboxyl group and aspartate side chain. \[ q_{\mathrm{original}}=(+1)+(+1)+(-1)+(-1) \] \[ q_{\mathrm{original}}=0 \] \(\textbf{N-terminal acetylation in P:}\) Acetylation converts the original positively charged terminal amino functionality into a neutral amide-derived group. The original \(+1\) contribution is therefore removed. All other specified charges remain unchanged. \[ q_P=0-(+1) \] \[ q_P=-1 \] \(\textbf{Additional C-terminal esterification in Q:}\) Q possesses the same N-terminal modification as P. Its C-terminal carboxylate is also converted into a neutral methyl ester. Removal of the original \(-1\) contribution increases the net charge by one. \[ q_Q=q_P-(-1) \] \[ q_Q=-1+1 \] \[ q_Q=0 \] \(\textbf{C-terminal esterification alone in R:}\) R retains the positively charged N-terminal amino group. Only its C-terminal carboxylate becomes neutral. Therefore: \[ q_R=q_{\mathrm{original}}-(-1) \] \[ q_R=+1 \] \(\textbf{Chemical interpretation:}\) The three derivatives contain identical ordered amino acid residues. Their different charges arise from modification of terminal functional groups rather than from replacement of residues or hydrolysis of peptide bonds. The lysine and aspartate side-chain charges remain unchanged throughout. Consequently, the original neutral peptide becomes negatively charged after N-terminal protection alone, but positively charged after C-terminal protection alone. Protection of both termini restores zero net charge under the stipulated conditions. \(\textbf{Final answer:}\) \[ \boxed{q_P=-1,\qquad q_Q=0,\qquad q_R=+1} \]

Explanation: \(\textbf{Optical contribution of the original glucose enantiomers:}\) D-glucose and L-glucose are enantiomers with equal-magnitude but opposite-sign specific rotations. They are initially present in equal molar amounts. Since their molar masses are identical, their masses are also equal. Their optical contributions therefore cancel exactly. \[ R_{\mathrm{D\text{-}glucose}}+R_{\mathrm{L\text{-}glucose}}=0 \] \(\textbf{Rotation before sucrose hydrolysis:}\) Sucrose is initially present and is dextrorotatory. The original D-glucose/L-glucose mixture makes no net contribution. Therefore, the initial observed optical rotation is positive. \[ R_{\mathrm{initial}}\gt0 \] \(\textbf{Monosaccharides released by sucrose hydrolysis:}\) Complete hydrolysis of one mole of sucrose yields one mole each of glucose and fructose. For the specified amount: \[ n_{\mathrm{D\text{-}glucose,formed}}=0.10\ \mathrm{mol} \] \[ n_{\mathrm{D\text{-}fructose,formed}}=0.10\ \mathrm{mol} \] \(\textbf{Rotation contributions of the hydrolysis products:}\) Glucose and fructose have equal molar masses of \(180\ \mathrm{g\,mol^{-1}}\). Their respective masses formed are: \[ m_{\mathrm{glucose}}=0.10(180)=18.0\ \mathrm{g} \] \[ m_{\mathrm{fructose}}=0.10(180)=18.0\ \mathrm{g} \] Their combined optical contribution is proportional to: \[ R_{\mathrm{products}}\propto18.0(52.5)+18.0(-92.4) \] \[ R_{\mathrm{products}}\propto945-1663.2 \] \[ R_{\mathrm{products}}\propto-718.2 \] The negative fructose contribution exceeds the positive glucose contribution. \(\textbf{Rotation after hydrolysis:}\) The original equal D-glucose/L-glucose contributions still cancel. No sucrose remains to contribute positive rotation. Therefore: \[ R_{\mathrm{final}}\lt0 \] \(\textbf{Initially reducing carbohydrate molecules:}\) Both free glucose enantiomers are reducing. Intact sucrose is non-reducing. \[ n_{\mathrm{reducing,initial}}=0.050+0.050 \] \[ n_{\mathrm{reducing,initial}}=0.10\ \mathrm{mol} \] \(\textbf{Reducing carbohydrate molecules after hydrolysis:}\) The initial glucose enantiomers remain available. Sucrose hydrolysis introduces an additional \(0.10\ \mathrm{mol}\) each of free glucose and fructose. \[ n_{\mathrm{reducing,final}}=0.050+0.050+0.10+0.10 \] \[ n_{\mathrm{reducing,final}}=0.30\ \mathrm{mol} \] \(\textbf{Consistency check:}\) Optical inactivity of an equimolar enantiomeric glucose mixture does not imply that its molecules are non-reducing. Optical rotation and reducing behaviour measure different chemical properties. Sucrose hydrolysis therefore changes both the net optical rotation and the number of available reducing molecules, but through different structural considerations. \(\textbf{Final answer:}\) \[ \boxed{\begin{gathered} R_{\mathrm{initial}}\gt0,\qquad R_{\mathrm{final}}\lt0\\ n_{\mathrm{reducing}}:\ 0.10\longrightarrow0.30\ \mathrm{mol} \end{gathered}} \]

Explanation: The primary structure of a peptide includes the ordered sequence of its amino acid residues. Stereochemical configuration describes the spatial arrangement around its asymmetric centres. These two structural features must be distinguished. P contains L-alanine followed by L-serine. Q has exactly the same residue order, but both asymmetric centres have the opposite configuration. Since no other stereogenic elements are considered, P and Q form an enantiomeric pair. S also retains the Ala–Ser residue order. However, only the serine asymmetric centre has been inverted relative to P. The alanine centre retains its original configuration. Therefore, P and S are diastereomers rather than complete mirror-image enantiomers. R differs from P in a more fundamental way because it has the sequence Ser–Ala instead of Ala–Ser. The identities occupying its amino-terminal and carboxyl-terminal positions have been exchanged. This changes its primary residue order and chemical connectivity pattern. Consequently, P and R cannot be classified merely as configurational stereoisomers of the same ordered dipeptide. None of these relationships involves cyclic hemiacetal or hemiketal anomeric centres. The correct distinction requires examining residue connectivity first and stereogenic-centre configurations second.

Explanation: \(\textbf{Original glycosidic connectivity:}\) The four-residue main chain contains three α-(1→4) linkages. The three-residue side chain contains two more α-(1→4) linkages. One α-(1→6) linkage attaches the side chain to the main chain. \[ N_{\alpha(1\rightarrow4)}=3+2=5 \] \[ N_{\alpha(1\rightarrow6)}=1 \] \(\textbf{Selective cleavage during Treatment I:}\) Treatment I hydrolyses all five α-(1→4) linkages. Therefore, the original main-chain connections A–B, B–C and C–D are broken. The side-chain connections E–F and F–G are also broken. \(\textbf{Connection that survives:}\) The α-(1→6) linkage between B and E remains intact. Thus, B and E remain joined as one disaccharide. The other five residues, A, C, D, F and G, become independent glucose molecules. \(\textbf{Product inventory after Treatment I:}\) \[ N_{\mathrm{disaccharides}}=1 \] \[ N_{\mathrm{free\ glucose}}=5 \] The total number of glucose-derived units is conserved. \[ 2(1)+5=7 \] \(\textbf{Water consumed during Treatment I:}\) Each of the five hydrolysed α-(1→4) linkages consumes one water molecule. \[ n_{\mathrm{H_2O,I}}=5\ \mathrm{mol} \] \(\textbf{Hydrolysis during Treatment II:}\) The only remaining inter-residue connection is the α-(1→6) linkage in the B–E disaccharide. Treatment II hydrolyses this bond. \[ n_{\mathrm{H_2O,II}}=1\ \mathrm{mol} \] The B–E disaccharide consequently releases two additional free glucose molecules. \(\textbf{Final product and water balance:}\) \[ N_{\mathrm{free\ glucose,final}}=5+2=7 \] \[ n_{\mathrm{H_2O,total}}=5+1 \] \[ n_{\mathrm{H_2O,total}}=6\ \mathrm{mol} \] \(\textbf{Consistency check:}\) A single connected oligosaccharide containing seven glucose residues requires six inter-residue linkages. Complete hydrolysis of all six linkages must consume six water molecules per original molecule. The two-stage consumption agrees with that requirement. The key distinction is that Treatment I removes the linear α-(1→4) connections while retaining the one branch-forming α-(1→6) linkage. \(\textbf{Final answer:}\) \[ \boxed{\begin{gathered} \text{After I: }1\text{ disaccharide}+5\text{ glucose molecules}\\ n_{\mathrm{H_2O,II}}=1\ \mathrm{mol} \end{gathered}} \]

Explanation: Statement I is correct because vitamin B₁ and vitamin B₂ are both water-soluble vitamins, but their characteristic deficiency associations are different. Vitamin B₁, or thiamine, is associated with beri-beri, whereas vitamin B₂, or riboflavin, is associated with cheilosis and related symptoms. Statement II is correct because NCERT identifies vitamin B₁₂ as an exception to the general statement about the poor storage of water-soluble vitamins. The fact that B₁₂ can be stored does not change its classification as water-soluble. Statement III is correct because vitamin E belongs to the fat-soluble group and its deficiency is associated with increased fragility of red blood cells and muscular weakness. Statement IV is incorrect because a common solubility class does not imply identical molecular constitution or identical physiological functions. Individual B-group vitamins are chemically and biologically distinct substances. Statement V is incorrect because the numerical designation in vitamin B₁₂ is a naming convention rather than a count of peptide bonds. Vitamins are not universally polypeptides. The comparison shows that solubility, storage behaviour, molecular constitution and deficiency manifestations must be assessed independently. Correct interpretation requires preserving the explicit NCERT exception for B₁₂ without extending that exception to all water-soluble vitamins.

Explanation: Both P and Q contain glucose-derived residues, so complete hydrolysis of either can yield free glucose. However, the identity of the constituent monosaccharide does not fully describe the original polymeric structure. Amylose contains essentially unbranched α-D-glucose chains with α-(1→4) linkages. Cellulose contains β-D-glucose residues connected through β-(1→4) linkages. Therefore, P possesses amylose-like connectivity, while Q possesses cellulose-like connectivity. The difference concerns the configuration at the anomeric carbon participating in the glycosidic linkage. Enzyme E is specified to recognise the alpha linkage type and can therefore hydrolyse P. It cannot hydrolyse Q through that linkage-selective mechanism. Conversely, enzyme F can hydrolyse Q through its beta linkages but not P through its alpha linkages. The two polymers can consequently show different enzymatic responses despite producing the same monosaccharide after complete acid hydrolysis. The experiment distinguishes monomer composition from the stereochemistry of inter-unit connectivity. It also demonstrates why enzyme specificity can reveal structural information that is lost when a polymer is completely broken down into its constituent monosaccharides.

Explanation: \(\textbf{Charge inventory at pH P:}\) There are two positively charged amino groups. One belongs to the N-terminal alpha-amino functionality. The other belongs to the lysine side chain. There are three negatively charged carboxylate groups. One is the C-terminal alpha-carboxylate, while the others belong to aspartate and glutamate side chains. \[ q_P=2(+1)+3(-1) \] \[ q_P=+2-3 \] \[ q_P=-1 \] \(\textbf{Change at pH Q:}\) Only the aspartate side-chain carboxylate is protonated. It changes from charge \(-1\) to zero. The total molecular charge therefore increases by one. \[ \Delta q_{P\rightarrow Q}=0-(-1) \] \[ \Delta q_{P\rightarrow Q}=+1 \] \[ q_Q=-1+1 \] \[ q_Q=0 \] \(\textbf{Independent change at pH R:}\) The experiment returns to the original pH P ionisation pattern before applying the change specified for R. Only the lysine side-chain amino group loses its positive charge. \[ \Delta q_{P\rightarrow R}=0-(+1) \] \[ \Delta q_{P\rightarrow R}=-1 \] \[ q_R=-1-1 \] \[ q_R=-2 \] \(\textbf{Chemical interpretation:}\) The peptide contains the same amino acid residues and the same number of peptide bonds under all three conditions. Only the specified protonation states change. Neutralising an originally negative group makes the total charge more positive, whereas neutralising an originally positive group makes it more negative. These changes do not require cleavage of the peptide backbone or replacement of an amino acid residue. \(\textbf{Final answer:}\) \[ \boxed{q_P=-1,\qquad q_Q=0,\qquad q_R=-2} \]

Explanation: An enzyme is a biological catalyst that increases the rate of a particular chemical transformation. Its activity depends on possessing an appropriate molecular structure. Mixture I contains active enzyme and its substrate under suitable conditions, so it can exhibit catalytic conversion. In Mixture II, the enzyme has undergone irreversible denaturation before being introduced into the reaction mixture. Preservation of peptide bonds does not guarantee preservation of the folded catalytic arrangement. Consequently, Mixture II need not show the acceleration produced by the native enzyme. The uncatalysed reaction is still described as very slow, so absence of catalytic activity should not be confused with absolute chemical impossibility. Mixture III contains active enzyme but lacks the specified substrate. The catalyst cannot convert an absent substrate into the designated product. Mixture IV differs because some product forms before the enzyme is denatured. Subsequent denaturation can eliminate further enzyme-mediated acceleration without automatically destroying product already formed. The problem explicitly excludes direct chemical alteration of S or P during denaturation. Therefore, the observations distinguish substrate availability, enzyme conformation, catalytic action and product preservation. The experiment also demonstrates why detecting product after an enzyme has been denatured does not prove that the denatured enzyme remains catalytically active.

Explanation: \(\textbf{Governing base-pair relationship:}\) Every fragment contains eight complementary base pairs. If the number of G–C pairs is \(g\), then the number of A–T pairs is \(8-g\). \[ N_{\mathrm{A-T}}=8-g \] The total hydrogen-bond count is: \[ H=3g+2(8-g) \] \[ H=16+g \] \(\textbf{Initial composition of R:}\) R initially has the same G–C count as P. P corresponds to \(g=2\). Therefore: \[ g_{\mathrm{R,initial}}=2 \] \[ H_{\mathrm{R,initial}}=16+2=18 \] \(\textbf{Effect of the two substitutions:}\) Each replacement removes one A–T pair and introduces one G–C pair. The number of G–C pairs therefore increases by two. \[ g_{\mathrm{R,final}}=2+2 \] \[ g_{\mathrm{R,final}}=4 \] \(\textbf{Recalculated hydrogen-bond count:}\) \[ H_{\mathrm{R,final}}=16+4 \] \[ H_{\mathrm{R,final}}=20 \] \(\textbf{Comparison with P and Q:}\) P has two G–C pairs and eighteen hydrogen bonds. Q has six G–C pairs and twenty-two hydrogen bonds. R after modification has four G–C pairs and twenty hydrogen bonds. \[ g_R=\frac{2+6}{2}=4 \] \[ H_R=\frac{18+22}{2}=20 \] \(\textbf{Interpretation of the graph:}\) The straight-line relationship arises because replacing an A–T pair with a G–C pair adds one hydrogen bond while preserving the total number of base pairs. The change is not two hydrogen bonds per substitution. Thus, R lies midway between P and Q for both quantities represented by the graph. \(\textbf{Final answer:}\) \[ \boxed{N_{\mathrm{G-C,R}}=4;\qquad H_R=20} \]