601. Two biological preparations contain proteins and carbohydrates in different arrangements.
P contains a purified protein made exclusively from alpha-amino acid residues joined through peptide bonds.
Q contains a protein component to which a carbohydrate component is covalently attached.
R contains a protein of the same kind as P physically mixed with free glucose, without any covalent bond between the protein and glucose.
The investigator wishes to distinguish a simple protein from a carbohydrate-containing protein conjugate without confusing covalent attachment with ordinary physical mixing.
Which conclusion is correct?
ⓐ. P cannot be a simple protein because every protein must contain a covalently attached carbohydrate component.
ⓑ. Q is necessarily a simple protein because its protein portion contains peptide bonds, while P is a carbohydrate.
ⓒ. R must be one carbohydrate-containing conjugated protein molecule because glucose and protein occur in the same preparation.
ⓓ. P is a simple protein; Q contains a carbohydrate-associated protein conjugate; R remains a physical mixture containing a simple protein rather than becoming a covalent conjugate.
Correct Answer: P is a simple protein; Q contains a carbohydrate-associated protein conjugate; R remains a physical mixture containing a simple protein rather than becoming a covalent conjugate.
Explanation: NCERT describes proteins as polymers of alpha-amino acids connected through peptide bonds. It also identifies proteins containing only alpha-amino acid constituents as simple proteins. Therefore, P satisfies the stated description of a simple protein. Q possesses a protein component together with a covalently attached carbohydrate component. Its complete chemical constitution is not limited to alpha-amino acid residues alone. Consequently, Q represents a carbohydrate-associated protein conjugate rather than a protein composed exclusively of amino acid-derived constituents. R differs from Q because its glucose molecules are physically mixed with the protein. The problem explicitly states that no covalent linkage joins these substances. Therefore, physical coexistence does not convert the protein molecules in R into carbohydrate-containing conjugates. The protein portion can retain its simple-protein classification. Detecting carbohydrate and protein within the same sample is insufficient to establish their covalent connectivity. The classification requires examining which components belong to the individual molecules rather than merely which substances are present in the preparation.
602. An enzyme system contains two reaction mixtures.
| Mixture | Chemical change |
|---|
| P | Maltose reacts with water and forms two glucose molecules. |
| Q | One organic substrate loses hydrogen while another accepts hydrogen under the action of an enzyme. |
The enzyme in P is called E₁, while the enzyme in Q is called E₂.
The changes occur without consumption of either enzyme as a stoichiometric reactant.
The reaction in Q is explicitly defined as a coupled oxidation–reduction transformation rather than hydrolysis.
Which classification correctly identifies the catalytic roles?
ⓐ. E₁ is an oxidoreductase because glucose is formed, while E₂ is a hydrolase because hydrogen is transferred.
ⓑ. E₁ acts as a hydrolase such as maltase, whereas E₂ acts as an oxidoreductase because the coupled reaction involves oxidation of one substrate and reduction of another.
ⓒ. Both are oxidoreductases because every enzyme-catalysed process necessarily transfers hydrogen between two organic substrates.
ⓓ. Both are hydrolases because the presence of an enzyme proves that water participates in every catalytic reaction.
Correct Answer: E₁ acts as a hydrolase such as maltase, whereas E₂ acts as an oxidoreductase because the coupled reaction involves oxidation of one substrate and reduction of another.
Explanation: Mixture P involves hydrolysis of the glycosidic linkage in maltose. A water molecule participates in cleavage of the disaccharide into two glucose molecules. Maltase is a familiar enzyme that catalyses this particular hydrolysis reaction. Therefore, E₁ performs the function of a hydrolase. Mixture Q describes a different type of transformation. One substrate loses hydrogen, which corresponds to oxidation under the stated conditions. Another substrate accepts hydrogen, corresponding to reduction. These complementary changes occur together in an oxidation–reduction reaction. NCERT identifies oxidoreductases as enzymes catalysing reactions in which oxidation of one substrate is coupled with reduction of another. Consequently, E₂ belongs to the oxidoreductase class. The absence of water as a specified reactant distinguishes Q from the hydrolysis process in P. Conversely, the production of glucose does not make maltose hydrolysis a redox reaction. Both E₁ and E₂ function as catalysts, but they accelerate different chemical transformations. The correct classification depends on the reaction mechanism rather than merely on the fact that an enzyme is present.
603. An experiment examines the roles of three types of RNA during protein synthesis.
Preparation P contains a particular messenger RNA carrying information required for synthesis of a protein.
Preparation Q contains the corresponding transfer RNA components needed to bring amino acids into the protein-synthesising system.
Preparation R contains functional ribosomal RNA as part of the ribosomal machinery.
Three independent experiments selectively remove P, Q or R while leaving the other components and the original DNA unchanged.
The missing RNA component is subsequently restored in each experiment.
Which conclusion is most defensible?
ⓐ. Removing messenger RNA cannot affect protein synthesis because the DNA sequence remains intact.
ⓑ. Transfer RNA is the permanent storage material for hereditary information, while messenger RNA forms every peptide bond independently of ribosomes.
ⓒ. Different RNA types perform complementary roles in protein synthesis; loss of any indispensable component can impair the process without requiring a change in DNA nucleotide sequence.
ⓓ. Restoring a missing RNA component necessarily changes the primary structure of the original DNA molecule.
Correct Answer: Different RNA types perform complementary roles in protein synthesis; loss of any indispensable component can impair the process without requiring a change in DNA nucleotide sequence.
Explanation: DNA contains hereditary information, including the information associated with synthesis of particular proteins. RNA molecules participate in using this information for protein synthesis. Messenger RNA carries the message needed for the synthesis of a particular protein. Transfer RNA participates in bringing the relevant amino acids to the protein-synthesising system. Ribosomal RNA is an important constituent of the ribosomal machinery involved in protein synthesis. These RNA types therefore perform related but different functions. Removing the required messenger RNA can interrupt the availability of the relevant information in the synthesising system. Removing indispensable transfer RNA components can impair the supply of amino acids to that system. Disrupting the necessary ribosomal RNA component can interfere with the operation of the ribosomal machinery. None of these manipulations requires alteration of the original DNA base sequence. Restoring the missing RNA component can permit recovery of the corresponding function if the other necessary conditions remain satisfied. The experiment distinguishes storage of hereditary information from the coordinated molecular processes involved in expressing that information.
604. One strand of an ordinary double-stranded DNA fragment acts as a template for formation of a complementary RNA segment.
The DNA template strand is initially:
5′-AGCTAC-3′
The RNA product is synthesised antiparallel to the template, with complementary pairing A–U, T–A, G–C and C–G.
A second DNA template is identical except that its final C nucleotide is replaced by T.
The RNA products are written in the conventional 5′→3′ direction.
No other nucleotide changes or additional processing occurs.
Which statement correctly gives the initial RNA sequence and the change produced by the DNA substitution?
ⓐ. Initial RNA: 5′-UCGAUG-3′; the final DNA substitution changes its last G into A.
ⓑ. Initial RNA: 5′-GUAGCU-3′; the final DNA substitution changes the first G of the RNA into A.
ⓒ. Initial RNA: 5′-GUAGCT-3′; the final DNA substitution converts uracil into thymine.
ⓓ. Initial RNA: 5′-CGAUAG-3′; the DNA substitution changes the second RNA nucleotide from G to C.
Correct Answer: Initial RNA: 5′-GUAGCU-3′; the final DNA substitution changes the first G of the RNA into A.
Explanation: DNA and RNA both contain complementary nitrogenous bases, but ordinary RNA uses uracil rather than thymine. The RNA product is formed antiparallel to the specified DNA template. Therefore, the complementary sequence must first be written opposite the template in its 3′→5′ direction. For the initial DNA template, 5′-AGCTAC-3′, the complementary RNA arrangement is 3′-UCGAUG-5′. Reversing the written orientation gives the conventional 5′→3′ RNA sequence, 5′-GUAGCU-3′. The second DNA template differs only at its final position, where C is replaced by T. The complementary RNA base opposite the original DNA C is G. The complementary RNA base opposite the replacement DNA T is A. Because the altered DNA position is at the template's 3′ end, it corresponds to the first position of the RNA when the RNA is written from 5′ to 3′. Thus, the modified RNA becomes 5′-AUAGCU-3′. No other complementary positions change under the stated conditions. Correctly predicting the RNA sequence depends on both base complementarity and the antiparallel orientation of the chains. Merely replacing DNA thymine with RNA uracil without reversing the sequence direction would not give the correct result.
605. A nutrition investigator compares two dietary patterns with two characteristic vitamin-deficiency observations.
| Observation | Finding |
|---|
| P | Fissuring at the corners of the mouth and lips, associated with digestive disorders |
| Q | Convulsions associated with inadequate vitamin intake |
According to the NCERT vitamin table:
Vitamin B₂ is riboflavin, with sources including milk, egg white, liver and kidney.
Vitamin B₆ is pyridoxine, with sources including yeast, milk, egg yolk, cereals and grams.
The investigator does not assume that a listed food source exclusively contains one vitamin or that either symptom establishes a diagnosis.
The purpose is to identify the corresponding textbook deficiency associations and one listed dietary source for each vitamin.
Which matching is correct?
ⓐ. P—vitamin B₆—egg white; Q—vitamin B₂—egg yolk
ⓑ. P—vitamin B₂—egg yolk; Q—vitamin B₆—egg white
ⓒ. P—vitamin B₁₂—egg white; Q—vitamin B₁—egg yolk
ⓓ. P—vitamin B₂—egg white; Q—vitamin B₆—egg yolk
Correct Answer: P—vitamin B₂—egg white; Q—vitamin B₆—egg yolk
Explanation: Vitamin B₂ is riboflavin, a member of the water-soluble B-group vitamins. NCERT associates its deficiency with cheilosis, including fissuring at the corners of the mouth and lips. Digestive disorders and a burning sensation of the skin are also included among the characteristic associations. Therefore, Observation P corresponds to vitamin B₂. The NCERT source list for B₂ includes milk, egg white, liver and kidney. Egg white is consequently an appropriate listed dietary source for P. Vitamin B₆ is pyridoxine, another water-soluble B-group vitamin. NCERT associates its deficiency with convulsions, identifying Observation Q. Its listed dietary sources include yeast, milk, egg yolk, cereals and grams. Egg yolk is therefore an appropriate source for Q. The comparison distinguishes two vitamins belonging to the same solubility group through their specific deficiency associations and source lists. Their common water solubility does not imply identical physiological functions. Nor does the presence of a vitamin in a food source establish that the food contains no other vitamins. The observations are being interpreted as textbook associations rather than as sufficient evidence for a clinical diagnosis.
606. Two enzyme preparations, E₁ and E₂, are used to study different biochemical transformations.
E₁ catalyses hydrolysis of a disaccharide into its constituent monosaccharides.
E₂ catalyses a reaction in which one organic substrate is oxidised while a second substrate is reduced.
Both enzymes are present in catalytic quantities.
The products of each reaction are recovered without contamination by the other reaction mixture.
A student concludes that E₂ must be a hydrolase because an enzyme participates in the reaction, while E₁ must be an oxidoreductase because its substrate is a carbohydrate.
Which statement best evaluates the student's classifications?
ⓐ. Both classifications are correct because enzyme names depend only on whether the substrate contains carbon and hydrogen.
ⓑ. E₁ is a hydrolase and E₂ an oxidoreductase; reaction type identifies the enzyme class.
ⓒ. E₁ and E₂ are necessarily the same enzyme because all enzymes use the same catalytic mechanism.
ⓓ. E₁ is an oxidoreductase and E₂ is a hydrolase because removal of hydrogen always means hydrolysis.
Correct Answer: E₁ is a hydrolase and E₂ an oxidoreductase; reaction type identifies the enzyme class.
Explanation: The reaction catalysed by E₁ involves cleavage of a glycosidic linkage through addition of water. This is a hydrolysis reaction. An enzyme catalysing such a transformation belongs to the functional category of hydrolases. The second reaction involves simultaneous oxidation and reduction of two substrates. One substrate loses hydrogen under the stated conditions, while the other accepts it. Such coupled changes are characteristic of an oxidation–reduction reaction. Therefore, E₂ is an oxidoreductase. The presence of an enzyme does not independently establish that water is a reactant. Likewise, the fact that E₁ acts on a carbohydrate does not automatically make the reaction an oxidation–reduction process. Enzyme specificity concerns the substrate and transformation recognised by the catalyst. Classification requires recognising the actual chemical change rather than simply naming the biomolecular class of the substrate. Both enzymes can accelerate their respective reactions without being consumed as stoichiometric reactants. The student's classifications are therefore reversed because the underlying reaction mechanisms have been misidentified.
607. An investigation is designed to determine whether loss of catalytic activity after heating is associated with the enzyme or with irreversible damage to the substrate.
| Trial | Experimental arrangement | Observation |
|---|
| P | Native enzyme with fresh substrate at suitable temperature and pH | Rapid product formation |
| Q | Previously heated enzyme, cooled, then mixed with fresh substrate at the original suitable conditions | Negligible product formation |
| R | Fresh native enzyme added to the unreacted substrate recovered from Q | Rapid product formation |
| S | Previously heated enzyme examined for peptide-bond cleavage | Original peptide backbone remains intact |
Which conclusion is best supported by the combined evidence?
ⓐ. The failure of Q proves that the substrate cannot undergo hydrolysis under any circumstances.
ⓑ. The recovery in R proves that the heated enzyme regained its original catalytic conformation without replacement.
ⓒ. Loss of activity indicates heat-induced enzyme denaturation, not necessarily peptide-backbone destruction.
ⓓ. The loss of activity must be caused by complete peptide-bond hydrolysis because every inactive protein consists of free amino acids.
Correct Answer: Loss of activity indicates heat-induced enzyme denaturation, not necessarily peptide-backbone destruction.
Explanation: Trial P establishes that the native enzyme can catalyse the reaction under suitable experimental conditions. Trial Q shows that the previously heated enzyme has lost detectable catalytic effectiveness even after being cooled and returned to those conditions. However, that observation alone would not establish whether the problem lies with the enzyme or the substrate. Trial R provides an additional distinction because fresh native enzyme converts the unreacted substrate recovered from Q. Therefore, the substrate remains capable of participating in the reaction. The failure in Q is consequently associated with the previously heated enzyme preparation. Heating can disturb hydrogen bonds and other interactions maintaining a protein enzyme's native conformation. Such structural changes can impair the arrangement required for effective catalysis. Trial S shows that the original peptide backbone remains intact after heating. This is consistent with denaturation affecting higher-order protein structure rather than complete hydrolysis into free amino acids. The rapid product formation in R is caused by the fresh enzyme and does not prove restoration of the previously heated enzyme. The experiment illustrates how a changed-component comparison can distinguish loss of catalytic function from loss of substrate reactivity.
608. Two equal portions of pure D-glucose are subjected to different carbonyl-group reactions.
Each portion contains \(0.10\ \mathrm{mol}\) of glucose.
Portion P reacts completely with hydroxylamine to form glucose oxime, eliminating one water molecule per glucose molecule.
Portion Q reacts completely with hydrogen cyanide to form glucose cyanohydrin without eliminating water.
No side reactions occur.
Take:
\[
M_{\mathrm{glucose}}=180\ \mathrm{g\,mol^{-1}}
\]
\[
M_{\mathrm{NH_2OH}}=33\ \mathrm{g\,mol^{-1}}
\]
\[
M_{\mathrm{HCN}}=27\ \mathrm{g\,mol^{-1}}
\]
\[
M_{\mathrm{H_2O}}=18\ \mathrm{g\,mol^{-1}}
\]
Which statement correctly compares the product masses and carbon-atom inventories?
ⓐ. P produces \(19.5\ \mathrm{g}\) of a six-carbon oxime; Q produces \(20.7\ \mathrm{g}\) of a seven-carbon cyanohydrin.
ⓑ. P produces \(21.3\ \mathrm{g}\) of a seven-carbon oxime; Q produces \(20.7\ \mathrm{g}\) of a six-carbon cyanohydrin.
ⓒ. P produces \(19.5\ \mathrm{g}\) of a seven-carbon oxime; Q produces \(18.0\ \mathrm{g}\) of a six-carbon cyanohydrin.
ⓓ. P produces \(18.0\ \mathrm{g}\) of a six-carbon oxime; Q produces \(19.5\ \mathrm{g}\) of a seven-carbon cyanohydrin.
Correct Answer: P produces \(19.5\ \mathrm{g}\) of a six-carbon oxime; Q produces \(20.7\ \mathrm{g}\) of a seven-carbon cyanohydrin.
Explanation: \(\textbf{Carbonyl reaction in Portion P:}\)
Glucose contains an aldehydic group in its open-chain structure.
Hydroxylamine reacts with this carbonyl group to produce an oxime.
The reaction eliminates one molecule of water for each oxime formed.
\[
\mathrm{Glucose+NH_2OH\rightarrow Glucose\ oxime+H_2O}
\]
\(\textbf{Molar mass of glucose oxime:}\)
\[
M_{\mathrm{oxime}}=180+33-18
\]
\[
M_{\mathrm{oxime}}=195\ \mathrm{g\,mol^{-1}}
\]
\(\textbf{Mass obtained from Portion P:}\)
\[
n_{\mathrm{oxime}}=0.10\ \mathrm{mol}
\]
\[
m_{\mathrm{oxime}}=0.10(195)
\]
\[
m_{\mathrm{oxime}}=19.5\ \mathrm{g}
\]
Hydroxylamine introduces nitrogen but no additional carbon atom.
Therefore, the oxime retains the six-carbon skeleton of glucose.
\(\textbf{Carbonyl addition in Portion Q:}\)
Hydrogen cyanide adds to the aldehydic group to form a cyanohydrin.
No water molecule is eliminated.
\[
\mathrm{Glucose+HCN\rightarrow Glucose\ cyanohydrin}
\]
\(\textbf{Molar mass of glucose cyanohydrin:}\)
\[
M_{\mathrm{cyanohydrin}}=180+27
\]
\[
M_{\mathrm{cyanohydrin}}=207\ \mathrm{g\,mol^{-1}}
\]
\(\textbf{Mass obtained from Portion Q:}\)
\[
m_{\mathrm{cyanohydrin}}=0.10(207)
\]
\[
m_{\mathrm{cyanohydrin}}=20.7\ \mathrm{g}
\]
The carbon atom of HCN becomes part of the cyanohydrin molecule.
Consequently, the product contains seven carbon atoms.
\(\textbf{Chemical consistency:}\)
Both reactions establish the presence of a carbonyl-derived reactive group in open-chain glucose.
However, oxime formation is a condensation reaction, whereas cyanohydrin formation involves addition of HCN.
The different reaction stoichiometries explain the differences in product masses and carbon counts.
\(\textbf{Final answer:}\)
\[
\boxed{\begin{gathered}
m_{\mathrm{oxime}}=19.5\ \mathrm{g},\quad N_C=6\\
m_{\mathrm{cyanohydrin}}=20.7\ \mathrm{g},\quad N_C=7
\end{gathered}}
\]
609. A hypothetical linear tripeptide has the fixed residue sequence Gly–Ala–Ile.
Glycine contains no asymmetric carbon atom.
Alanine contains one asymmetric carbon atom at its alpha position.
Isoleucine contains two asymmetric carbon atoms, including one within its side chain.
Assume that every asymmetric centre can independently possess either configuration and that no meso-form reduction occurs.
The peptide sequence, covalent connectivity and functional groups remain unchanged.
A student proposes that the number of configurational possibilities is determined by the number of amino acid residues rather than by the number of asymmetric carbon atoms.
Which statement correctly evaluates the proposal?
ⓐ. The peptide has six configurational possibilities because it contains three residues and two possible terminal orientations.
ⓑ. The peptide has four configurational possibilities because every amino acid contributes at most one asymmetric carbon.
ⓒ. The peptide has eight configurational possibilities because its three asymmetric carbon atoms can independently adopt two configurations.
ⓓ. The peptide has sixteen configurational possibilities because glycine contributes one additional stereogenic carbon after peptide formation.
Correct Answer: The peptide has eight configurational possibilities because its three asymmetric carbon atoms can independently adopt two configurations.
Explanation: \(\textbf{Asymmetric centres in the constituent residues:}\)
Glycine has hydrogen as its side chain.
Its alpha carbon consequently possesses two identical hydrogen substituents and is not asymmetric.
\[
n_{\mathrm{Gly}}=0
\]
Alanine possesses one asymmetric alpha carbon.
\[
n_{\mathrm{Ala}}=1
\]
Isoleucine has an asymmetric alpha carbon and another asymmetric carbon within its side chain.
\[
n_{\mathrm{Ile}}=2
\]
\(\textbf{Total stereogenic-centre inventory:}\)
The peptide-bond formation specified here does not remove any of these existing asymmetric centres.
\[
n_{\mathrm{total}}=0+1+2
\]
\[
n_{\mathrm{total}}=3
\]
\(\textbf{Configurational possibilities:}\)
For three independently configurable asymmetric centres without meso-form reductions:
\[
N=2^n
\]
\[
N=2^3
\]
\[
N=8
\]
\(\textbf{Why residue count is insufficient:}\)
The sequence contains three amino acid residues, but one residue is achiral and another contains two asymmetric centres.
Consequently, the number of residues differs from the number of stereogenic centres.
The fixed peptide sequence also prevents rearrangement of residue positions from being counted as additional configurational possibilities.
Changing residue order would change primary structure rather than merely stereochemical configuration.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{configurational\ possibilities}}=8}
\]
610. Two ordinary tetrapeptides P and Q have the sequences Ala–Pro–Gly–Ser and Ala–Gly–Pro–Ser, respectively.
Both contain the same four amino acid residues.
In each peptide, proline occupies an internal position.
A normal peptide bond involving the backbone nitrogen of an internal proline residue does not leave that nitrogen with an N–H bond.
The regular alpha-helical structure described in NCERT is stabilised by hydrogen bonding involving peptide-backbone C=O and N–H groups.
Assume that no unusual covalent modification occurs.
Consider the following statements.
Statement I: P and Q have different primary structures.
Statement II: Their identical amino acid compositions establish that their primary structures are identical.
Statement III: Internal proline has a distinctive backbone hydrogen-bonding limitation that may influence the formation of a regular alpha-helical region.
Statement IV: The absence of an ordinary backbone N–H group at internal proline means that no peptide bond joins proline to the preceding residue.
Statement V: Moving proline from one internal position to another changes the location at which its distinctive structural constraint operates.
Which statements are correct?
ⓐ. I, II and III only
ⓑ. I, IV and V only
ⓒ. II, III and IV only
ⓓ. I, III and V only
Correct Answer: I, III and V only
Explanation: Statement I is correct because primary structure depends on the order of amino acid residues rather than merely their identities. P contains the sequence Ala–Pro–Gly–Ser, whereas Q contains Ala–Gly–Pro–Ser. The two peptides therefore have different primary structures despite having identical compositions. Statement II is incorrect because equal amino acid composition does not establish identical residue order. Statement III is correct because proline has a cyclic side-chain arrangement involving its nitrogen atom. When proline occurs internally in an ordinary peptide, its backbone nitrogen lacks the N–H group available in many other residues. This distinguishes its contribution to the regular backbone hydrogen-bonding pattern associated with an alpha-helix. Consequently, proline can impose a structural constraint on formation of a regular alpha-helical region. Statement IV is incorrect because the absence of a backbone N–H bond does not mean that the preceding peptide bond is absent. Proline remains covalently connected to the preceding residue through an ordinary peptide linkage. Statement V is correct because the location of proline within the sequence differs between P and Q. The position at which its structural constraint is introduced therefore changes. The observations do not prove that either complete tetrapeptide must adopt one particular secondary structure under all conditions.
611. A purified polypeptide contains two cysteine residues, one asparagine residue and several alanine residues.
Initially, both cysteine residues possess free –SH groups, while asparagine contains an unmodified side-chain carboxamide group.
Treatment I oxidises the two cysteine thiol groups to one disulphide linkage.
Treatment II subsequently hydrolyses only the asparagine side-chain carboxamide group, converting that residue into an aspartic-acid-derived residue and releasing ammonia.
Treatment III reduces the disulphide linkage but does not reverse the side-chain deamidation.
No peptide bond is cleaved during any treatment.
Which conclusion correctly describes the final polypeptide?
ⓐ. It has two restored cysteine thiol groups, one aspartic-acid-derived residue and the original peptide-bond connectivity.
ⓑ. It has one permanent disulphide linkage, one unchanged asparagine residue and one fewer peptide bond.
ⓒ. It contains two new peptide bonds formed through oxidation of cysteine and one additional asparagine residue.
ⓓ. It has one remaining cysteine thiol group, one new glutamic-acid-derived residue and a completely hydrolysed backbone.
Correct Answer: It has two restored cysteine thiol groups, one aspartic-acid-derived residue and the original peptide-bond connectivity.
Explanation: Cysteine contains a thiol group in its side chain. Two suitable cysteine thiol groups can undergo oxidation to form a disulphide linkage. Therefore, Treatment I removes the two free –SH groups and creates one sulphur–sulphur bond. Asparagine contains a side-chain carboxamide group. Hydrolysis of this group converts the side chain into the carboxyl-containing structure associated with aspartic acid. This transformation releases ammonia and changes the chemical identity of that residue. It does not require cleavage of the polypeptide backbone. Consequently, Treatment II leaves the disulphide linkage intact while converting the asparagine-derived side chain into an aspartic-acid-derived side chain. Treatment III reduces the disulphide linkage. This regenerates two cysteine thiol groups. However, reduction of a sulphur–sulphur bond does not automatically reverse the independently performed deamidation. The final chain therefore retains the aspartic-acid-derived residue produced during Treatment II. Since every treatment explicitly preserves peptide bonds, the original residue connectivity remains intact, although the residue at the deamidated position has undergone a chemical change. The result demonstrates why successive functional-group reactions must be tracked separately instead of treating all reversible and irreversible changes as equivalent.
612. Two individuals P and Q show different deficiency manifestations despite receiving adequate quantities of carbohydrates and proteins.
P is a growing child whose vitamin deficiency produces poor bone mineralisation and rickets.
Q shows an abnormally prolonged blood-clotting time associated with a different vitamin deficiency.
Both deficient vitamins belong to the fat-soluble group.
A student suggests that supplying either fat-soluble vitamin should correct both conditions equally well.
Which conclusion correctly identifies the vitamins and evaluates the student's suggestion?
ⓐ. P lacks vitamin K and Q lacks vitamin D; both deficiencies produce identical changes in bone mineralisation.
ⓑ. P lacks vitamin E and Q lacks vitamin A; both vitamins directly replace every factor involved in blood clotting.
ⓒ. P lacks vitamin D and Q lacks vitamin K; the vitamins share a solubility class but perform different essential biological functions.
ⓓ. P lacks vitamin C and Q lacks vitamin B₁₂; both must be fat-soluble because their deficiencies cause visible physiological changes.
Correct Answer: P lacks vitamin D and Q lacks vitamin K; the vitamins share a solubility class but perform different essential biological functions.
Explanation: \(\textbf{Identification of the first deficiency:}\)
Vitamin D is important for normal calcium and phosphorus utilisation and bone mineralisation. Its deficiency in growing children is associated with rickets, a condition involving defective mineralisation of developing bones.
Therefore, the deficiency described for P is consistent with inadequate vitamin D.
\(\textbf{Identification of the second deficiency:}\)
Vitamin K is required for normal blood clotting. Inadequate vitamin K can impair the formation of functional clotting factors, resulting in an increased blood-clotting time.
Consequently, the deficiency described for Q is associated with vitamin K.
\(\textbf{Comparison of solubility classes:}\)
Both vitamin D and vitamin K belong to the fat-soluble group, which also contains vitamins A and E.
However, classification by solubility identifies a shared physical characteristic rather than identical biochemical functions.
\(\textbf{Evaluation of the proposed interchangeability:}\)
Supplying vitamin K does not replace the specific role of vitamin D in normal bone mineralisation. Similarly, providing vitamin D does not substitute for vitamin K's function in normal blood clotting.
The fact that both vitamins can be associated with fat-containing tissues does not make them functionally interchangeable.
\(\textbf{Integrated conclusion:}\)
The symptoms must be matched to the particular biological functions of the deficient vitamins. P requires correction of vitamin D deficiency, whereas Q requires correction of vitamin K deficiency.
Thus, common fat solubility is insufficient to establish functional equivalence between these vitamins.
613. Three molecules of uridine monophosphate are joined to form one ordinary linear RNA trinucleotide containing three uracil bases.
Each starting uridine monophosphate molecule contains one uracil base, one ribose sugar and one phosphate group.
Take the molar mass of uridine monophosphate as \(324\ \mathrm{g\,mol^{-1}}\).
Two ordinary 3′–5′ phosphodiester linkages are formed by condensation.
For this hypothetical chemical preparation, formation of each internucleotide linkage eliminates exactly one water molecule.
The original three phosphate groups remain incorporated into the trinucleotide, including its terminal phosphate group.
Take:
\[
M_{\mathrm{H_2O}}=18\ \mathrm{g\,mol^{-1}}
\]
No additional terminal-group modification or side reaction occurs.
Which statement correctly gives the trinucleotide molar mass and its number of internucleotide phosphodiester linkages?
ⓐ. \(918\ \mathrm{g\,mol^{-1}}\) and three phosphodiester linkages
ⓑ. \(936\ \mathrm{g\,mol^{-1}}\) and two phosphodiester linkages
ⓒ. \(954\ \mathrm{g\,mol^{-1}}\) and two phosphodiester linkages
ⓓ. \(972\ \mathrm{g\,mol^{-1}}\) and three phosphodiester linkages
Correct Answer: \(936\ \mathrm{g\,mol^{-1}}\) and two phosphodiester linkages
Explanation: \(\textbf{Starting nucleotide inventory:}\)
The preparation contains three uridine monophosphate molecules.
Each contributes one ribose, one uracil and one phosphate component.
Before condensation, their combined molar mass is:
\[
M_{\mathrm{initial}}=3(324)
\]
\[
M_{\mathrm{initial}}=972\ \mathrm{g\,mol^{-1}}
\]
\(\textbf{Internucleotide linkage count:}\)
An ordinary linear polynucleotide containing \(n\) nucleotide units possesses \(n-1\) internucleotide connections.
For three nucleotide units:
\[
N_{\mathrm{linkages}}=3-1
\]
\[
N_{\mathrm{linkages}}=2
\]
\(\textbf{Condensation water eliminated:}\)
The stated construction eliminates one water molecule per new phosphodiester linkage.
\[
n_{\mathrm{H_2O\ eliminated}}=2
\]
The corresponding molar-mass decrease is:
\[
\Delta M_{\mathrm{water}}=2(18)
\]
\[
\Delta M_{\mathrm{water}}=36\ \mathrm{g\,mol^{-1}}
\]
\(\textbf{Molar mass of the RNA trinucleotide:}\)
\[
M_{\mathrm{trinucleotide}}=972-36
\]
\[
M_{\mathrm{trinucleotide}}=936\ \mathrm{g\,mol^{-1}}
\]
\(\textbf{Phosphate-group interpretation:}\)
The molecule contains three nucleotide-derived phosphate components under the explicitly specified terminal arrangement.
However, it possesses only two internucleotide phosphodiester linkages because three nucleotide units are connected in one linear chain.
The number of phosphate groups should therefore not be confused with the number of internucleotide connections.
\(\textbf{Chemical consistency:}\)
All three uracil bases and all three ribose sugars remain part of the final product.
The mass decrease arises solely from elimination of the two specified water molecules.
\(\textbf{Final answer:}\)
\[
\boxed{M_{\mathrm{trinucleotide}}=936\ \mathrm{g\,mol^{-1}};\quad N_{\mathrm{phosphodiester}}=2}
\]
614. A DNA fragment initially contains an ordinary G–C base pair at one specified position.
At the chosen position, the first DNA strand contains guanine and the complementary strand contains cytosine.
During a hypothetical alteration, the guanine on the first strand is replaced by adenine while the cytosine on the second strand is initially unchanged.
A subsequent corrective replacement changes that opposing cytosine into thymine, restoring ordinary complementary pairing.
All other base pairs remain unchanged.
A–T pairs contribute two hydrogen bonds, while G–C pairs contribute three.
The analysis distinguishes the intermediate mismatched state from the final complementary DNA structure.
Which statement is correct?
ⓐ. The intermediate A–C combination is an ordinary complementary DNA base pair, and the final A–T pair contains one more hydrogen bond than the original G–C pair.
ⓑ. The intermediate A–C combination is not one of the ordinary complementary DNA pairs; after correction, the original G–C pair has become A–T, reducing the standard interstrand hydrogen-bond count by one.
ⓒ. The intermediate A–C combination is an ordinary G–C equivalent, and the final state preserves three hydrogen bonds because adenine is a purine.
ⓓ. The corrective replacement must produce A–U because thymine cannot pair with adenine in DNA.
Correct Answer: The intermediate A–C combination is not one of the ordinary complementary DNA pairs; after correction, the original G–C pair has become A–T, reducing the standard interstrand hydrogen-bond count by one.
Explanation: Ordinary double-stranded DNA contains complementary A–T and G–C base pairs. Adenine pairs with thymine through two hydrogen bonds. Guanine pairs with cytosine through three hydrogen bonds. The original position therefore contributes three hydrogen bonds through its G–C pairing. Replacing guanine with adenine while leaving cytosine unchanged produces an A–C mismatch. This combination does not satisfy the ordinary complementary pairing rules specified for DNA. Therefore, the intermediate state must not be treated as a normal A–T or G–C pair when applying the standard hydrogen-bond counts. The corrective replacement subsequently converts the opposing cytosine into thymine. The final position is consequently an ordinary A–T pair. Its standard hydrogen-bond contribution is two rather than the original three. Thus, the completed complementary substitution reduces the standard interstrand hydrogen-bond count by one. The number of base-pair positions remains unchanged throughout. The reasoning requires distinguishing a temporary mismatch from a completed complementary replacement instead of assigning ordinary pairing properties to an incorrect intermediate combination.
615. A biological system contains functional DNA, messenger RNA and the other components needed for protein synthesis.
Experiment P removes a particular DNA segment before the corresponding messenger RNA has been produced.
Experiment Q allows the messenger RNA to be formed first and then removes the original DNA segment without destroying the already formed messenger RNA.
In Experiment Q, all other components required for the next stage of protein synthesis remain functional.
The experiment concerns the availability of a pre-existing message, not continued synthesis of new copies of that message.
Which conclusion is most defensible?
ⓐ. DNA removal before transcription may prevent messenger RNA formation; an already formed message can still direct synthesis.
ⓑ. Removing DNA after messenger RNA formation necessarily converts the messenger RNA into deoxyribonucleic acid.
ⓒ. Messenger RNA cannot contain information about protein synthesis because hereditary information exists only as hydrogen bonds between DNA strands.
ⓓ. DNA and messenger RNA possess identical pentose sugars, so either can replace the other in every biochemical role.
Correct Answer: DNA removal before transcription may prevent messenger RNA formation; an already formed message can still direct synthesis.
Explanation: DNA is the principal hereditary information-bearing molecule described in NCERT. The information required for synthesis of a particular protein is associated with the nucleotide sequence of DNA. Messenger RNA carries a corresponding message used during protein synthesis. In Experiment P, removal of the required DNA segment occurs before formation of the relevant messenger RNA. Consequently, the information needed to produce that particular message may no longer be available in the specified system. Experiment Q is different because the messenger RNA has already been formed. The question explicitly states that this RNA remains intact and that the other components required for the next stage are functional. Therefore, the existing RNA message can continue to provide the relevant information without immediate dependence on the removed DNA segment. This does not imply that messenger RNA and DNA have identical chemical constitutions. RNA contains ribose and ordinarily uses uracil, whereas DNA contains 2-deoxyribose and thymine. Nor does it establish that an existing messenger RNA can support unlimited protein synthesis indefinitely. The correct conclusion distinguishes the original hereditary information source from a previously produced molecular message that can be used subsequently.
616. Two endocrine preparations P and Q are subjected to a chemical test.
P contains insulin, a polypeptide hormone involved in lowering elevated blood glucose.
Q contains a steroid hormone belonging to the glucocorticoid group, which participates in carbohydrate metabolism and responses to stress.
A reagent selectively hydrolyses peptide bonds but does not react with the steroid framework.
The investigator separately examines the chemical identity of the products and the biological roles of the original hormones.
Which statement is correct?
ⓐ. The reagent must hydrolyse both hormones because participation in carbohydrate metabolism implies the presence of peptide bonds.
ⓑ. The reagent must leave insulin unchanged because a molecule involved in signalling cannot possess a polypeptide backbone.
ⓒ. Insulin has hydrolysable peptide bonds, but the glucocorticoid does not; related regulation does not imply shared structure.
ⓓ. Both hormones must possess comparable peptide backbones, so the difference can lie only in their rates of hydrolysis.
Correct Answer: Insulin has hydrolysable peptide bonds, but the glucocorticoid does not; related regulation does not imply shared structure.
Explanation: Hormones act as chemical messengers coordinating biological processes. However, NCERT explains that hormones belong to different chemical classes. Insulin is a polypeptide hormone and contains amino acid residues joined through peptide bonds. Therefore, a reagent that selectively hydrolyses peptide bonds can act on its backbone. Glucocorticoids are steroid hormones released by the adrenal cortex. They participate in carbohydrate metabolism, inflammatory responses and reactions to stress. Their steroid framework is chemically different from an ordinary polypeptide backbone. Consequently, the specified peptide-selective reagent does not cleave the glucocorticoid through peptide-bond hydrolysis. Both insulin and glucocorticoids can influence carbohydrate-related biological processes, but this shared context does not make their structures identical. Nor does it establish that either hormone directly produces glucose through hydrolysis of itself. The distinction between chemical structure and biological role is therefore central to the experiment. The correct conclusion recognises hormone diversity without confusing regulatory signalling with catalytic bond cleavage.
617. Three derivatives of D-glucose are examined for ordinary mutarotation.
P is free D-glucose with an unmodified anomeric hemiacetal hydroxyl group at C1.
Q is a glucose derivative in which only the hydroxyl groups at C2, C3, C4 and C6 have been converted into acetate esters; the C1 hemiacetal hydroxyl remains free.
R is glucose pentaacetate, in which all five original hydroxyl groups, including the C1 anomeric hydroxyl, are acetylated.
The experiment is performed under conditions that preserve the acetate groups and allow ordinary hemiacetal ring–chain interconversion whenever the required anomeric hydroxyl is free.
Which prediction is correct?
ⓐ. Only P can undergo ordinary mutarotation because acetylation of any non-anomeric hydroxyl must destroy the glucose carbon skeleton.
ⓑ. P and Q can access ordinary anomeric interconversion, whereas R lacks the free hemiacetal hydroxyl needed for that process under the stated conditions.
ⓒ. Q and R undergo ordinary mutarotation, whereas P cannot because its C1 hydroxyl is unprotected.
ⓓ. All three necessarily undergo identical mutarotation because they contain six carbon atoms and the same number of ring oxygens.
Correct Answer: P and Q can access ordinary anomeric interconversion, whereas R lacks the free hemiacetal hydroxyl needed for that process under the stated conditions.
Explanation: Mutarotation of glucose arises from interconversion between its alpha and beta anomers through an open-chain form. The process requires a free anomeric hemiacetal centre capable of ring opening. P is ordinary free glucose and therefore has the necessary C1 hemiacetal functionality. Consequently, P can undergo mutarotation. Q is different because four non-anomeric hydroxyl groups have been acetylated. However, its C1 anomeric hydroxyl remains free. Under the specified conditions, this free hemiacetal centre can still participate in ring–chain interconversion. Therefore, Q can also undergo anomeric interconversion, even though its other hydroxyl groups have been chemically protected. R is glucose pentaacetate, in which the C1 anomeric hydroxyl has also been acetylated. It lacks the required free hemiacetal group while the protecting groups remain intact. Consequently, ordinary mutarotation through the free-glucose hemiacetal pathway is blocked in R. The comparison shows that the chemical state of the anomeric hydroxyl is decisive. Acetylation of every hydroxyl group does not have the same mechanistic consequence as acetylation of only the non-anomeric hydroxyl groups.
618. Two samples of a polypeptide are examined under different conditions.
Both samples contain a lysine residue and an aspartic acid residue positioned near one another in the folded protein.
In native Sample P, lysine carries a positive side-chain charge while aspartic acid carries a negative side-chain charge.
Sample Q is exposed to strongly acidic conditions that neutralise the aspartate carboxylate while preserving the lysine positive charge.
No residue is replaced and no peptide bond is broken.
A third preparation is instead modified by replacing aspartic acid with asparagine while maintaining the same backbone connectivity.
Consider the following statements.
Statement I: The initial lysine–aspartate attraction is consistent with an electrostatic interaction stabilising tertiary structure.
Statement II: Strong acid can weaken the stated attraction by neutralising the aspartate side-chain charge without necessarily changing primary structure.
Statement III: Replacement of aspartic acid with asparagine changes the residue identity and therefore the primary amino acid sequence.
Statement IV: Neutralisation of the aspartate side chain and replacement by asparagine are chemically identical operations.
Statement V: The two modifications can affect the same initial electrostatic interaction through different chemical mechanisms.
Which statements are correct?
ⓐ. I, II, III and IV only
ⓑ. II, III, IV and V only
ⓒ. I, III and IV only
ⓓ. I, II, III and V only
Correct Answer: I, II, III and V only
Explanation: Statement I is correct because tertiary protein structure is stabilised partly through electrostatic attractions between appropriately charged side chains. Positively charged lysine and negatively charged aspartate can participate in such an interaction. Statement II is correct because lowering the pH can protonate a carboxylate group. Protonation removes the original negative charge from the aspartate side chain and can weaken its electrostatic attraction to lysine. This acid–base change does not require replacement of the amino acid residue or cleavage of a peptide bond. Statement III is correct because asparagine and aspartic acid possess different side-chain functional groups. Replacing one residue with the other changes the identity of the residue occupying that sequence position. Therefore, the primary amino acid sequence changes, even when the chain length and backbone connections remain the same. Statement IV is incorrect because protonation changes the ionisation state of an existing carboxyl group, whereas replacement with asparagine introduces a carboxamide side chain. These are chemically different operations. Statement V is correct because both operations can remove the original negative-charge contribution to the specified interaction. However, their mechanisms and possible consequences for other interactions need not be identical. The conclusion distinguishes changes in ionisation state from changes in residue identity.
619. A nutrition laboratory compares two substances obtained from ordinary dietary materials.
P is a water-soluble vitamin required in relatively small quantities.
Q is a protein enzyme that catalyses a chemical transformation involving a specific substrate.
The investigator detects nitrogen in both preparations.
The investigator then concludes that P and Q must both contain peptide bonds and that each must catalyse the same reaction.
Which evaluation is most appropriate?
ⓐ. The conclusion is correct because every nitrogen-containing biomolecule is a polypeptide.
ⓑ. The conclusion is correct only if the vitamin is water-soluble, because fat-soluble vitamins cannot contain nitrogen.
ⓒ. The conclusion is unsupported: nitrogen content alone establishes neither peptide linkage nor enzyme activity.
ⓓ. The conclusion is unsupported only because protein enzymes cannot contain nitrogen.
Correct Answer: The conclusion is unsupported: nitrogen content alone establishes neither peptide linkage nor enzyme activity.
Explanation: Proteins contain amino acid residues connected through peptide bonds. Protein enzymes possess such polypeptide structures and function as catalysts for particular biochemical transformations. Vitamins are chemically diverse organic substances required in relatively small quantities for normal biological functioning. Some vitamins contain nitrogen, but the presence of that element does not establish a repeating alpha-amino acid backbone. Therefore, a nitrogen-containing vitamin need not possess peptide bonds. The detection of nitrogen in Q is consistent with its protein constitution but does not independently establish which substrate it recognises. Likewise, detecting nitrogen in P does not prove that it acts as a catalyst for Q's reaction. Chemical composition, connectivity and biological function represent different kinds of information. Two substances may share an element without sharing the same structural class. Similarly, biological importance does not make every molecule an enzyme. The investigator has therefore inferred both molecular structure and catalytic function from insufficient evidence. The correct interpretation requires examining characteristic functional groups and biological action rather than relying solely on elemental analysis.
620. A mixture contains equal molar amounts of three different RNA nucleosides: adenosine, guanosine and cytidine.
Each nucleoside contains one ribose sugar joined to one nitrogenous base.
The corresponding nucleotides are formed by esterifying the C5′ hydroxyl group of each nucleoside with one phosphoric acid molecule.
For each esterification, one water molecule is eliminated.
The three resulting nucleotides are then joined to form one ordinary linear trinucleotide by two additional condensation reactions, each eliminating one water molecule.
Assume that all three initial nucleoside molecules are incorporated, that no further phosphate group is introduced and that no other reaction occurs.
How many water molecules are eliminated in total per trinucleotide formed, and how many phosphate groups remain incorporated in that trinucleotide?
ⓐ. Three water molecules and two phosphate groups
ⓑ. Five water molecules and three phosphate groups
ⓒ. Six water molecules and five phosphate groups
ⓓ. Four water molecules and three phosphate groups
Correct Answer: Five water molecules and three phosphate groups
Explanation: \(\textbf{Conversion of nucleosides into nucleotides:}\)
A nucleoside contains a nitrogenous base attached to a pentose sugar.
Addition of a phosphate component converts it into a nucleotide.
The reaction is specified to involve esterification at C5′ with loss of one water molecule.
\(\textbf{Water eliminated during phosphorylation:}\)
Three nucleosides are independently converted into their corresponding nucleotides.
\[
N_{\mathrm{phosphorylations}}=3
\]
Each phosphorylation eliminates one water molecule.
\[
N_{\mathrm{H_2O,phosphorylation}}=3
\]
\(\textbf{Linear trinucleotide formation:}\)
Three nucleotide units are connected into one linear polynucleotide segment.
An ordinary linear chain of three units requires two internucleotide phosphodiester linkages.
\[
N_{\mathrm{internucleotide\ links}}=3-1
\]
\[
N_{\mathrm{internucleotide\ links}}=2
\]
\(\textbf{Additional condensation water:}\)
The specified formation of each inter-unit linkage eliminates one water molecule.
\[
N_{\mathrm{H_2O,linkage}}=2
\]
\(\textbf{Total condensation water:}\)
\[
N_{\mathrm{H_2O,total}}=3+2
\]
\[
N_{\mathrm{H_2O,total}}=5
\]
\(\textbf{Phosphate-group inventory:}\)
Each of the three nucleotides contributes one phosphate group.
No phosphate group is removed during the stated condensation sequence.
\[
N_{\mathrm{phosphate}}=3
\]
\(\textbf{Chemical interpretation:}\)
The trinucleotide possesses three incorporated phosphate components but only two internucleotide phosphodiester connections.
The first three water molecules are eliminated during nucleoside phosphorylation, while the other two are eliminated during chain formation.
These are distinct reaction stages and must not be combined into a single unexamined linkage count.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{H_2O}}=5;\qquad N_{\mathrm{phosphate}}=3}
\]
