401. Two amino acid-containing preparations are compared.
P contains free glycine and free alanine in equal molar amounts.
Q contains the dipeptide glycylalanine, Gly–Ala, in an amount equal to the initial glycine amount in P.
The glycine and alanine units have the same configurations and side-chain identities in both preparations.
Both samples undergo complete hydrolysis under conditions that preserve the amino acids.
A separate analytical procedure is performed before hydrolysis to determine whether the amino acid units are joined by a peptide bond.
Which conclusion correctly compares the information available before and after hydrolysis?
ⓐ. P and Q must be identical before hydrolysis because they contain equal numbers of glycine and alanine units.
ⓑ. P contains separate amino acids and Q a dipeptide; complete hydrolysis can yield the same amino acids from both.
ⓒ. Q releases glycine only because the alanine residue is consumed during complete peptide-bond hydrolysis.
ⓓ. P must possess one peptide bond because equal amounts of two amino acids necessarily undergo condensation whenever they are mixed.
Correct Answer: P contains separate amino acids and Q a dipeptide; complete hydrolysis can yield the same amino acids from both.
Explanation: A mixture of free glycine and free alanine consists of separate amino acid molecules unless a suitable condensation reaction occurs. Equal molar amounts do not automatically imply formation of a peptide bond. Therefore, P contains the two amino acids without the specified covalent peptide connection. Q is glycylalanine, a dipeptide with the sequence Gly–Ala. Its glycine carboxyl group is connected to the alanine amino group through a peptide linkage. Consequently, Q is chemically different from a simple mixture of the two free amino acids. Complete hydrolysis of Q breaks the peptide bond and regenerates glycine and alanine. Under the stated conditions, hydrolysis of P does not require breaking a peptide bond because the amino acids are already free. The resulting constituent amino acids can therefore be the same in both preparations. However, analysis performed only after complete hydrolysis cannot reveal that the two amino acids were originally joined in Q. The separate pre-hydrolysis procedure is needed to establish that connectivity. The comparison demonstrates the distinction between constituent composition and covalent molecular organisation.
402. Two protein-derived samples P and Q contain the same total numbers of glycine, alanine, valine and serine residues.
P is one continuous polypeptide chain with a particular amino acid sequence and stable native conformation.
Q consists of two shorter polypeptide chains whose combined amino acid composition is identical to that of P.
The two chains in Q associate non-covalently in a functional assembly.
Both preparations are subjected to complete hydrolysis, and the released amino acids are analysed quantitatively.
A second experiment selectively separates the non-covalent subunits of Q without breaking peptide bonds.
Which conclusion is justified?
ⓐ. Identical amino acid compositions prove that P and Q have the same primary structure and identical quaternary organisation.
ⓑ. Complete hydrolysis of P must release different amino acids from Q because P contains one chain and Q contains two.
ⓒ. The hydrolysates may be identical, but selective subunit separation distinguishes the original protein assemblies.
ⓓ. Separation of the two subunits of Q necessarily changes every amino acid residue into its corresponding D-enantiomer.
Correct Answer: The hydrolysates may be identical, but selective subunit separation distinguishes the original protein assemblies.
Explanation: Complete protein hydrolysis breaks peptide bonds and releases constituent amino acids. The problem specifies that P and Q contain identical combined quantities of the four amino acid types. Their complete hydrolysates can consequently give identical quantitative amino acid compositions. However, an amino acid analysis of fully hydrolysed material does not retain the original sequence of residues along each chain. It also does not preserve information about the number of chains that were associated in the starting assembly. P consists of one continuous polypeptide chain. Q consists of two shorter chains joined through non-covalent interactions in a functional assembly. The association of separate subunits is a feature of quaternary organisation. The specified second experiment can separate these subunits while preserving each chain's peptide connectivity. This reveals a structural difference that cannot be deduced solely from the common hydrolysis products. Separation does not require conversion of L-amino acid residues into D-enantiomers. The correct conclusion therefore distinguishes amino acid composition, primary sequence and the association of separate polypeptide subunits.
403. Two protein samples P and Q are subjected to different structural disturbances.
P contains a regular right-handed alpha-helical segment. The backbone –NH and C=O groups participate in hydrogen bonds between different turns of the coiled segment.
Q contains two extended polypeptide segments lying alongside one another. Hydrogen bonds form between backbone groups on the neighbouring segments.
Treatment I selectively disrupts these backbone hydrogen bonds without hydrolysing peptide bonds.
Treatment II, performed separately, replaces one amino acid residue in each sample with another residue while preserving the number of peptide bonds.
Which statement correctly distinguishes the structural changes?
ⓐ. Treatment I changes primary structure in both proteins, whereas Treatment II affects only secondary structure.
ⓑ. Treatment I converts both proteins into free amino acids, whereas Treatment II preserves the original amino acid sequences.
ⓒ. The secondary structures of P and Q are identical because both are stabilised by backbone hydrogen bonds.
ⓓ. Treatment I can disturb the alpha-helix of P and beta-pleated arrangement of Q without changing their primary sequences, whereas Treatment II changes primary structure.
Correct Answer: Treatment I can disturb the alpha-helix of P and beta-pleated arrangement of Q without changing their primary sequences, whereas Treatment II changes primary structure.
Explanation: The alpha-helix and beta-pleated sheet are two important forms of protein secondary structure. In the alpha-helix, the polypeptide backbone adopts a right-handed helical arrangement in the usual NCERT representation. Hydrogen bonds between backbone groups on neighbouring turns help stabilise this arrangement. In a beta-pleated sheet, extended polypeptide segments lie alongside one another and are stabilised by hydrogen bonds between the neighbouring segments. Thus, the same general types of backbone functional groups can participate in different geometrical arrangements. Treatment I is specified to disrupt hydrogen bonding without breaking peptide bonds. It can consequently disturb secondary structure while leaving the original sequence of amino acid residues unchanged. Treatment II instead replaces an amino acid residue with another residue. Such a replacement changes the ordered residue identity and therefore changes the primary structure. Preserving the number of peptide bonds does not guarantee preservation of the original amino acid sequence. The correct distinction depends on recognising that hydrogen-bond disruption and residue replacement act on different features of protein structure.
404. A sample of milk undergoes curdling after the activity of lactic-acid-producing bacteria. A separate sample of egg white coagulates on boiling.
In both experiments, a change in protein appearance and physical behaviour is observed.
Analysis of the protein components establishes that their peptide backbones have not undergone complete hydrolysis.
A student argues that the two phenomena cannot both be examples of protein denaturation because only one involves an increase in temperature.
Consider the following statements.
Statement I: Protein denaturation may be initiated by changes in temperature or by changes in pH.
Statement II: Lactic acid production can alter the environment of milk proteins and contribute to their denaturation.
Statement III: Coagulation proves that the proteins have been converted completely into free amino acids.
Statement IV: Loss of native secondary or tertiary organisation can accompany denaturation while the primary structure remains intact.
Which statements are correct?
ⓐ. I, II and IV only
ⓑ. I and III only
ⓒ. II, III and IV only
ⓓ. I, II and III only
Correct Answer: I, II and IV only
Explanation: Statement I is correct because NCERT describes denaturation as a consequence of physical changes such as heating or chemical changes such as alterations in pH. A protein's native conformation depends on interactions that stabilise its secondary and tertiary structures. Heating can disturb these interactions, causing loss of the native folded arrangement. Statement II is correct because NCERT identifies milk curdling caused by lactic acid produced by bacteria as an example of denaturation. The acid changes the chemical environment surrounding the milk proteins and can disturb their native organisation. Statement III is incorrect because visible coagulation does not prove complete cleavage of peptide bonds. Complete hydrolysis would release the constituent amino acids by breaking peptide connectivity. Denaturation instead involves disruption of the native structure and may cause loss of biological activity. Statement IV is therefore correct because the primary amino acid sequence can remain intact even when secondary and tertiary structures are disturbed. The fact that heating and acidification are different initiating conditions does not prevent them from producing the same broad class of protein structural change. The experiment distinguishes physical observations of coagulation from the specific chemical process of peptide-bond hydrolysis.
405. Three purified vitamin preparations are examined to distinguish solubility from storage behaviour.
Deficiency associated with P produces increased fragility of red blood cells and muscular weakness.
Deficiency associated with Q produces pernicious anaemia.
P partitions into a fat-like phase, while Q partitions into an aqueous phase under the specified separation conditions.
A researcher argues that Q cannot be retained in the body because it is water-soluble, whereas P must be the only preparation capable of appreciable storage.
Use the vitamin classification and explicit storage qualification given in NCERT.
Which evaluation of the researcher's argument is most appropriate?
ⓐ. Q must be fat-soluble if it can be retained, while P is water-soluble because its deficiency affects blood cells.
ⓑ. P and Q must enter the same phase because their deficiency manifestations involve red blood cells.
ⓒ. P is consistent with fat-soluble vitamin E; Q is water-soluble vitamin B₁₂, which is an exception to the usual limited-storage rule.
ⓓ. Q cannot belong to the B-vitamin group because every vitamin in that group is necessarily absent from body stores.
Correct Answer: P is consistent with fat-soluble vitamin E; Q is water-soluble vitamin B₁₂, which is an exception to the usual limited-storage rule.
Explanation: The first observation is associated in NCERT with deficiency of vitamin E, which can produce increased fragility of red blood cells and muscular weakness. Thus, P is consistent with vitamin E. The second observation is characteristic of deficiency of vitamin B₁₂, which identifies Q. Vitamins A, D, E and K belong to the fat-soluble group, whereas the B vitamins and vitamin C are water-soluble. The specified separation is consistent with these classifications, placing vitamin E in the fat-like phase and vitamin B₁₂ in the aqueous phase. However, solubility and biological storage are related but not identical classification questions. NCERT states that water-soluble vitamins generally need regular dietary supply because they are readily excreted and cannot be stored extensively. It explicitly gives vitamin B₁₂ as an exception to this general storage statement. Therefore, Q can be water-soluble without satisfying the researcher's claim of an absolute lack of storage. The researcher cannot convert a usual trend into a universal rule in the presence of a named exception. Likewise, the different blood-related deficiency manifestations do not imply that P and Q have the same chemical identity or solubility. The defensible conclusion combines the specific vitamin identities with the distinction between extraction behaviour and the qualified storage rule.
406. A single-stranded RNA molecule contains two short regions separated by a flexible unpaired segment.
The first region has the sequence 5′-GCA-3′.
A second region of the same RNA molecule can fold back alongside the first in an antiparallel orientation.
The two regions are to form three ordinary complementary base pairs without any mismatch.
Only A–U and G–C pairing is considered.
All candidate sequences for the second region are written in their own 5′→3′ direction.
Which sequence is compatible with the required paired arrangement?
ⓐ. 5′-CGU-3′
ⓑ. 5′-UCG-3′
ⓒ. 5′-UGC-3′
ⓓ. 5′-GCA-3′
Correct Answer: 5′-UGC-3′
Explanation: RNA is generally single-stranded, but suitable portions of the same molecule may fold back and form locally complementary regions. The nucleotide sequences must be examined with their strand directions in mind. Guanine pairs ordinarily with cytosine, while adenine pairs with uracil in RNA. The direct complementary sequence opposite 5′-GCA-3′ is therefore 3′-CGU-5′ when the two regions are shown alongside each other. However, every answer option is written in its own 5′→3′ direction. Reversing the displayed direction of the complementary region gives 5′-UGC-3′. Option A presents the complementary letters without the required reversal of orientation. The other candidates do not satisfy all three positional pairing requirements in the specified antiparallel arrangement. Thus, identifying the necessary bases is not sufficient unless their order and orientation are also correct. The sequence 5′-UGC-3′ can form the stated three complementary pairs with the first region. This local RNA secondary structure does not require converting the RNA molecule into double-stranded DNA.
407. Three purified hormones are placed in separate vessels.
| Vessel | Hormone | Chemical nature |
|---|
| P | Insulin | Polypeptide |
| Q | Estradiol | Steroid |
| R | Epinephrine | Amino acid derivative |
Each vessel receives a hypothetical reagent that quantitatively hydrolyses peptide bonds but leaves steroid frameworks, ordinary amino acid-derived hormone molecules and non-peptide covalent bonds intact.
Which outcome is consistent with the specified chemical selectivity?
ⓐ. P can undergo peptide-backbone fragmentation, while Q and R need not be fragmented by the reagent.
ⓑ. P and Q must fragment because hormones are always polymers of amino acids.
ⓒ. Q and R must fragment, whereas P remains unchanged because polypeptide hormones lack peptide bonds.
ⓓ. All three must generate identical free amino acids because they perform hormonal functions.
Correct Answer: P can undergo peptide-backbone fragmentation, while Q and R need not be fragmented by the reagent.
Explanation: Insulin is a polypeptide hormone and contains amino acid residues joined through peptide bonds. A reagent capable of hydrolysing these bonds can therefore disrupt insulin's polypeptide connectivity. Estradiol belongs to the steroid class of hormones. Its chemical constitution is not an ordinary chain of amino acid residues joined by peptide linkages. Therefore, the specified peptide-bond-hydrolysing reagent need not fragment estradiol. Epinephrine belongs to the amino acid-derived hormone class. Although its chemical origin is related to an amino acid, an epinephrine molecule is not itself an ordinary polypeptide chain. It consequently lacks the repeated peptide-bond framework targeted by the reagent. The shared function of insulin, estradiol and epinephrine as chemical messengers does not imply shared covalent structures. It is therefore incorrect to assume that all three yield amino acids under peptide hydrolysis. This experiment distinguishes classification by biological function from classification by chemical constitution. The expected outcome depends on the presence of the reagent's target bond rather than the common designation hormone.
408. A student uses the historical name “vitamine” to make several deductions about vitamins.
Statement I: The name originally arose from the idea of a vital substance containing an amine group.
Statement II: Subsequent evidence showed that vitamins do not all possess amino groups.
Statement III: Vitamins are required in relatively small dietary quantities, but excessive intake is not necessarily harmless.
Statement IV: Every vitamin must possess the same basic carbon–nitrogen structural framework because vitamins perform biological functions.
Statement V: Classification of vitamins into fat-soluble and water-soluble groups depends on solubility rather than on whether a molecule contains an amino group.
Which statements are correct?
ⓐ. I, II 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 NCERT explains that the historical term “vitamine” was derived from the words vital and amine. The name reflected an early belief that such essential substances contained amino groups. Statement II is correct because later discoveries showed that most vitamins did not possess the originally assumed amino-group feature. The final letter was consequently dropped, giving the modern term vitamin. Statement III is correct because vitamins are organic compounds required in small quantities for normal growth, maintenance and biological functioning. NCERT also cautions that excessive vitamin intake may be harmful. Statement IV is incorrect because vitamins belong to different chemical classes and cannot be defined by one universal molecular structure. Their shared biological importance does not require identical functional groups or carbon–nitrogen frameworks. Statement V is correct because the established broad classification depends on solubility in fats and oils or in water. Vitamins A, D, E and K are fat-soluble, whereas B-group vitamins and vitamin C are water-soluble. Thus, the correct combination separates historical nomenclature from actual chemical constitution and the modern classification criterion.
409. Three components involved in the biological use of hereditary information are present in a hypothetical cell-free preparation.
Component P is DNA containing the hereditary nucleotide sequence.
Component Q is messenger RNA carrying information associated with protein synthesis.
Component R consists of ribosomal RNA required for the normal functioning of the ribosomal component of the system.
Transfer RNA and the other necessary constituents are initially present and functional.
Treatment I selectively destroys R without changing P or Q.
Treatment II, applied to a separate preparation, selectively breaks interstrand hydrogen bonds in P without cleaving its phosphodiester backbones or destroying any RNA component.
Which conclusion is justified?
ⓐ. Treatment I must destroy the hereditary nucleotide sequence of DNA because RNA and DNA possess identical backbones.
ⓑ. I damages the ribosomal contribution to translation without necessarily altering DNA or mRNA; II need not hydrolyse DNA.
ⓒ. Treatment II necessarily converts the separated DNA strands into transfer RNA because their hydrogen bonds are broken.
ⓓ. Neither treatment can affect any biomolecular function because neither is specified to hydrolyse peptide bonds.
Correct Answer: I damages the ribosomal contribution to translation without necessarily altering DNA or mRNA; II need not hydrolyse DNA.
Explanation: DNA is the principal chemical store of hereditary information in the NCERT description. Its nucleotide sequence represents primary structural information that is retained within covalently connected polynucleotide strands. Messenger RNA performs a different role associated with carrying information for protein synthesis. Ribosomal RNA is another functional class of RNA associated with the ribosomal component of the protein-synthesising system. Treatment I selectively destroys that ribosomal RNA component. Therefore, the system can lose an important component required for normal protein synthesis even while its DNA and messenger RNA remain chemically unchanged. This illustrates why possession of hereditary information alone does not guarantee that every step involved in its biological use can proceed. Treatment II instead acts on hydrogen bonds between complementary DNA strands. The question states that the phosphodiester backbones remain intact. Consequently, strand separation does not itself remove the ordered nucleotide sequence from either individual DNA strand. Nor does separation convert DNA into a particular type of RNA. The correct conclusion distinguishes stored hereditary information, the functions of different RNA classes and the chemical nature of the bonds disrupted by each treatment.
410. Two purified biomolecules are examined using a hypothetical bond-selective reagent.
P is maltose, consisting of two glucose residues joined through an oxygen-containing glycosidic linkage.
Q is adenosine, consisting of adenine attached to the C1′ position of ribose through a nitrogen-containing base–sugar glycosidic linkage.
The reagent cleaves only the specified oxygen-containing linkage between monosaccharide residues.
It does not cleave the nitrogen-containing base–sugar linkage of Q.
No other reaction takes place.
Which conclusion correctly predicts the products?
ⓐ. Both P and Q release two glucose molecules because every glycosidic linkage joins carbohydrate monosaccharides.
ⓑ. P releases glucose and fructose, whereas Q releases adenine and 2-deoxyribose.
ⓒ. P can release two glucose molecules, while Q remains adenosine because its base–sugar linkage is excluded from the reagent's action.
ⓓ. P remains intact, while Q releases adenine and ribose because nitrogen-containing glycosidic linkages are oxygen-containing monosaccharide linkages.
Correct Answer: P can release two glucose molecules, while Q remains adenosine because its base–sugar linkage is excluded from the reagent's action.
Explanation: Maltose is a disaccharide containing two glucose units connected through a glycosidic linkage. Cleavage of this connection through hydrolysis releases the two constituent glucose molecules. The specified reagent is designed to act on this oxygen-containing connection between monosaccharide units. Therefore, P can undergo the stated hydrolysis. Adenosine is a nucleoside rather than a disaccharide. It contains adenine joined to ribose through a base–sugar linkage involving the sugar's anomeric C1′ position. Adenine is a nitrogenous base and not another glucose residue. The linkage in adenosine is consequently chemically different from the oxygen-containing connection between two glucose units in maltose. The reagent is explicitly stated not to hydrolyse that base–sugar connection. Therefore, Q remains adenosine under the specified conditions. Ribose, not 2-deoxyribose, is the sugar constituent of adenosine. The question demonstrates that a broad term such as glycosidic linkage does not establish identical bond connectivity or identical hydrolysis products in every biomolecular class.
411. A laboratory attempts to determine whether the enzyme sucrase remains active in a strongly acidic solution.
| Trial | Experimental conditions | Observation |
|---|
| P | Sucrose and native sucrase at suitable pH | Rapid glucose and fructose formation |
| Q | Sucrose in strongly acidic solution without enzyme | Glucose and fructose formation |
| R | Sucrose and sucrase in the same strongly acidic solution | Glucose and fructose formation |
| S | Sucrase recovered after acidic exposure, then tested with fresh sucrose at suitable pH | No detectable catalytic activity |
All trials are performed with appropriate controls for contamination and product identification.
Which conclusion is best supported?
ⓐ. Trial R proves that sucrase remains active in strong acid because glucose and fructose are detected.
ⓑ. Trials Q and R prove that sucrose is non-hydrolysable without an enzyme.
ⓒ. Trial S proves that every peptide bond of sucrase was hydrolysed during the acidic exposure.
ⓓ. Acid can hydrolyse sucrose in R independently of sucrase; S indicates loss of the enzyme's active conformation.
Correct Answer: Acid can hydrolyse sucrose in R independently of sucrase; S indicates loss of the enzyme's active conformation.
Explanation: Sucrase catalyses the hydrolysis of sucrose into glucose and fructose. Trial P confirms that the native enzyme is active under suitable conditions. However, sucrose can also undergo acid-catalysed hydrolysis. Trial Q demonstrates that the strongly acidic solution produces the same monosaccharides even when no enzyme is present. Therefore, product formation in Trial R cannot independently establish that sucrase remains active in that acidic medium. The observed products may arise from the acid-catalysed pathway. Trial S provides an additional distinction because the acid-exposed enzyme fails to catalyse hydrolysis when subsequently tested under otherwise suitable conditions. This is consistent with damage to its native functional conformation through an adverse pH change. Protein denaturation can cause loss of biological activity without complete hydrolysis of peptide bonds. Therefore, Trial S does not prove destruction of the primary amino acid sequence. The decisive experimental reasoning is to compare reactions with and without enzyme under the same acidic conditions, then separately test whether the exposed enzyme retains activity. A positive product test and a positive enzyme-activity test are not equivalent observations.
412. Two proteins P and Q each contain an internal lysine residue and an internal arginine residue.
Lysine possesses an additional free primary aliphatic amino group in its side chain.
Arginine possesses a nitrogen-containing guanidino group in its side chain.
All the alpha-amino groups of the two residues are already involved in ordinary peptide linkages.
A hypothetical reagent selectively modifies free primary aliphatic side-chain amino groups but does not react with guanidino groups or peptide-bond amides.
The treatment leaves every peptide bond intact.
Which conclusion correctly combines chemical selectivity and amino acid classification?
ⓐ. The lysine side chain can be modified while the arginine guanidino group remains unchanged; both amino acids belong to the basic and essential categories in the NCERT classification.
ⓑ. Both side chains must be modified identically because the basic character of lysine and arginine proves that they have the same side-chain structure.
ⓒ. Only arginine is modified because lysine contains a carboxamide side chain rather than an amino group.
ⓓ. Neither side chain can react because peptide-bond formation consumes every nitrogen-containing group of both residues.
Correct Answer: The lysine side chain can be modified while the arginine guanidino group remains unchanged; both amino acids belong to the basic and essential categories in the NCERT classification.
Explanation: Lysine and arginine are both naturally occurring alpha-amino acids with nitrogen-containing side chains. Their additional basic groups make them examples of basic amino acids in the elementary NCERT classification. NCERT also marks both as essential amino acids. However, the two side-chain structures are not identical. Lysine contains a terminal primary aliphatic amino group in its side chain. Arginine contains a guanidino group, in which the nitrogen atoms belong to a different functional arrangement. The hypothetical reagent is explicitly selective for the free primary aliphatic amino group and excludes guanidino functionality. Therefore, it can modify the lysine side chain while leaving the specified arginine group unchanged. Incorporation into a peptide uses the alpha-amino and alpha-carboxyl groups involved in backbone formation. It does not automatically remove every nitrogen-containing functional group from the side chain. The specified treatment also preserves the peptide backbone. Thus, two amino acids can share both nutritional and acidic/basic classifications yet respond differently to a chemically selective reagent. The correct conclusion requires identifying their actual side-chain functional groups rather than relying on the label basic alone.
413. Two deficiency-related observations concern the same vitamin.
Case P involves a child with bone deformities associated with inadequate availability of the vitamin.
Case Q involves an adult with soft bones and joint pain associated with inadequate availability of the same vitamin.
The vitamin is obtainable from certain dietary sources, and suitable sunlight exposure is another recognised source-related factor.
A student concludes that the two cases must involve different vitamins because the associated conditions have different names.
Which statement correctly evaluates the conclusion?
ⓐ. P involves vitamin A deficiency and Q involves vitamin K deficiency because both vitamins are associated with bones.
ⓑ. P involves vitamin C deficiency and Q involves vitamin B₁ deficiency because both are water-soluble.
ⓒ. P involves vitamin E deficiency and Q involves vitamin A deficiency because each is stored in adipose tissue.
ⓓ. Both cases are consistent with vitamin D deficiency, which is associated with rickets in children and osteomalacia in adults.
Correct Answer: Both cases are consistent with vitamin D deficiency, which is associated with rickets in children and osteomalacia in adults.
Explanation: Vitamin D belongs to the fat-soluble group of vitamins. NCERT identifies exposure to sunlight, fish and egg yolk among its recognised sources. Its deficiency is associated with disorders involving bones. In children, vitamin D deficiency can produce rickets, which is characterised by bone deformities. In adults, the corresponding deficiency is associated with osteomalacia, including soft bones and joint pain. Consequently, the different names of the two conditions do not require different vitamin identities. The age-related descriptions distinguish two recognised manifestations associated with deficiency of vitamin D. Vitamin A deficiency is instead associated with night blindness and xerophthalmia. Vitamin K deficiency is associated with increased blood clotting time. Vitamin C deficiency causes scurvy, while vitamin B₁ deficiency is associated with beri-beri. The correct interpretation therefore combines vitamin identity, solubility, sources and the age-related qualification of its deficiency manifestations. The observations alone do not provide a basis for prescribing treatment or determining individual vitamin concentrations.
414. A simplified endocrine model contains three hormone-related functions.
| Component | Specified function |
|---|
| P | Regulates the excretion of water and salts by the kidneys |
| Q | Influences carbohydrate metabolism, inflammatory reactions and responses to stress |
| R | Participates in lowering an elevated blood glucose concentration |
The adrenal cortex is assumed to be functioning inadequately, while the hormone represented by R remains available.
A student argues that R can automatically replace both P and Q because all three influence the body's internal balance.
Which conclusion is most consistent with NCERT?
ⓐ. P and Q are adrenal-cortex hormone classes, but R is glucagon and therefore cannot lower elevated blood glucose.
ⓑ. P is mineralocorticoids, Q glucocorticoids and R insulin; insulin cannot replace their distinct adrenal-cortex functions.
ⓒ. P corresponds to insulin, Q to glucagon and R to testosterone; their functions are identical because all three are proteins.
ⓓ. P is mineralocorticoids, Q glucocorticoids and R insulin; therefore insulin can replace both because all maintain homeostasis.
Correct Answer: P is mineralocorticoids, Q glucocorticoids and R insulin; insulin cannot replace their distinct adrenal-cortex functions.
Explanation: Mineralocorticoids are hormones produced by the adrenal cortex and are associated with regulation of water and salt excretion by the kidneys. Therefore, P represents mineralocorticoids. Glucocorticoids are another class of adrenal-cortex steroid hormones. NCERT associates them with carbohydrate metabolism, modulation of inflammatory reactions and responses to stress. Consequently, Q represents glucocorticoids. Insulin is a polypeptide hormone associated with lowering an elevated blood glucose concentration. Therefore, R represents insulin. These hormones all contribute to biological regulation, but their specified functions are not identical. The fact that insulin influences blood glucose does not establish that it performs the kidney-related regulatory role of mineralocorticoids. Similarly, its action cannot simply be equated with the broader adrenal-cortex functions described for glucocorticoids. NCERT associates inadequate adrenal-cortex function with Addison's disease and describes the importance of the affected hormonal classes. The question does not require detailed physiological mechanisms or treatment decisions. Its central distinction is that participation in maintaining biological balance does not make chemically and functionally different hormones interchangeable.
415. A purified chromosome-containing preparation is subjected to a treatment that selectively denatures associated proteins.
The treatment disturbs hydrogen bonding and other interactions responsible for the native folding of the proteins.
Every peptide bond remains intact.
The DNA retains its original nucleotide sequence and its covalent phosphodiester backbones.
No nucleotide base is chemically altered.
The preparation is subsequently examined for protein folding and the chemical information stored in DNA.
Which conclusion correctly distinguishes the structural and hereditary consequences?
ⓐ. The DNA sequence must change because every loss of protein secondary structure alters the order of DNA nucleotides.
ⓑ. Protein denaturation necessarily converts the DNA backbone into peptide bonds.
ⓒ. The proteins may lose native higher-order structures while retaining sequences; the DNA may retain its original nucleotide sequence.
ⓓ. The DNA must lose all hereditary information because its associated proteins no longer possess their original native conformations.
Correct Answer: The proteins may lose native higher-order structures while retaining sequences; the DNA may retain its original nucleotide sequence.
Explanation: Chromosomes contain proteins and nucleic acids, including DNA. The chemical structures of these biomolecular classes are fundamentally different. Protein primary structure is determined by the sequence of amino acid residues joined through peptide linkages. Its secondary and tertiary structures depend on folding interactions, including hydrogen bonding and other stabilising forces. The stated treatment disrupts those folding interactions without hydrolysing peptide bonds. Consequently, the associated proteins can become denatured while retaining their original amino acid sequences. DNA, in contrast, contains nucleotide units connected through phosphodiester linkages. The question explicitly states that these covalent connections and the nucleotide-base identities remain unchanged. Therefore, the stored nucleotide-sequence information can remain chemically intact. NCERT identifies DNA as the chemical basis of heredity and as a reserve of genetic information. Denaturation of associated proteins does not, by itself, establish destruction of that DNA information. However, preservation of the nucleotide sequence alone does not prove that every original chromosome-level biological function remains unaffected. The scientifically justified conclusion is limited to the specified structural changes and the preservation of the underlying covalent sequences.
416. The following observations are available for glucose and its derivatives:
Observation I: Glucose has the molecular formula \(\mathrm{C_6H_{12}O_6}\).
Observation II: It forms glucose pentaacetate on treatment with acetic anhydride.
Observation III: Glucose pentaacetate is a stable compound.
Observation IV: Glucose can form a cyclic hemiacetal through intramolecular reaction of its C5 hydroxyl group with the C1 aldehydic group.
A student proposes that Observation II alone establishes the presence of five hydroxyl groups, whereas Observations II and III together provide additional evidence about their distribution among carbon atoms.
Which conclusion best evaluates the proposal?
ⓐ. The proposal is incorrect because acetylation determines only the number of carbon atoms and cannot provide information about hydroxyl groups.
ⓑ. The proposal is correct because pentaacetate formation supports five hydroxyl groups, and the stability of the derivative supports their attachment to different carbon atoms.
ⓒ. The proposal is correct only if all five hydroxyl groups are attached to the same carbon atom of glucose.
ⓓ. The proposal is incorrect because cyclic hemiacetal formation consumes all five hydroxyl groups and makes pentaacetate formation impossible.
Correct Answer: The proposal is correct because pentaacetate formation supports five hydroxyl groups, and the stability of the derivative supports their attachment to different carbon atoms.
Explanation: Acetic anhydride reacts with suitable hydroxyl groups to form acetate ester groups. The formation of glucose pentaacetate consequently provides evidence for five hydroxyl groups in the glucose molecule. This is the first structural conclusion obtained from the acetylation experiment. NCERT makes an additional deduction from the stability of the pentaacetate derivative. It states that the five hydroxyl groups should be attached to different carbon atoms. Thus, the observations concern both the number and the distribution of hydroxyl functionalities. They do not imply that five hydroxyl groups are attached to one carbon atom. The cyclic hemiacetal structure of glucose is consistent with this evidence. During cyclisation, the C5 hydroxyl oxygen enters the ring, while the original C1 carbonyl oxygen becomes the anomeric hydroxyl oxygen. Consequently, the cyclic form also possesses five hydroxyl groups available for suitable acetylation. Intramolecular hemiacetal formation does not remove all hydroxyl groups from glucose. The experiment therefore illustrates how a reaction product and its stability can supply complementary structural evidence rather than merely repeating the same conclusion.
417. The usual cyclic forms of two monosaccharides are compared.
P is D-glucopyranose, formed when the C5 hydroxyl group of open-chain glucose adds to the aldehydic carbon at C1.
Q is D-fructofuranose, formed when the C5 hydroxyl group of open-chain fructose adds to the ketonic carbon at C2.
In P, C6 remains a –CH₂OH group.
In Q, both C1 and C6 are –CH₂OH groups.
All hydroxyl groups are unmodified, and no glycosidic linkage joins either monosaccharide to another molecule.
A reagent reacts selectively with free primary alcoholic hydroxyl groups and does not react with secondary alcohols, anomeric hydroxyls or ring ethers.
Which prediction is correct?
ⓐ. P and Q each contain one reactive primary alcoholic hydroxyl group because both possess five total hydroxyl groups.
ⓑ. P contains two reactive primary alcoholic hydroxyl groups, whereas Q contains one because its carbonyl group was originally ketonic.
ⓒ. P contains five reactive primary alcoholic hydroxyl groups, whereas Q contains four because fructofuranose has a smaller ring.
ⓓ. P contains one reactive primary alcoholic hydroxyl group, whereas Q contains two; their equal total hydroxyl counts do not imply identical hydroxyl types.
Correct Answer: P contains one reactive primary alcoholic hydroxyl group, whereas Q contains two; their equal total hydroxyl counts do not imply identical hydroxyl types.
Explanation: Glucose and fructose both have the molecular formula \(\mathrm{C_6H_{12}O_6}\), but their open-chain carbonyl positions differ. In glucose, the aldehydic group occurs at C1, while the primary alcoholic group occurs at C6. Formation of glucopyranose involves the C5 hydroxyl group and the C1 aldehydic group. The C6 primary alcoholic hydroxyl group remains free. Therefore, P possesses one primary alcoholic hydroxyl group satisfying the reagent's selectivity. Fructose has a ketonic group at C2 in its open-chain form. Its C1 and C6 positions are both hydroxymethyl groups. Formation of the usual fructofuranose structure involves the C5 hydroxyl group and the C2 carbonyl group. Neither the C1 nor the C6 hydroxymethyl group is consumed by that ring-forming reaction. Consequently, Q retains two free primary alcoholic hydroxyl groups. Both cyclic monosaccharides have five hydroxyl groups in total, but those groups occupy different structural environments. The number of primary alcoholic hydroxyl groups must therefore be determined from carbon connectivity rather than from the total hydroxyl count alone.
418. Two hypothetical glucose polymers P and Q each contain ten glucose residues in one connected, non-cyclic molecule.
P is an ordinary unbranched chain with nine α-(1→4) glycosidic linkages.
Q contains seven α-(1→4) linkages and two α-(1→6) branch-forming linkages.
Every free cyclic glucose molecule is taken to contain five hydroxyl groups before condensation.
Each glycosidic linkage is formed by condensation involving two hydroxyl groups, with elimination of one water molecule.
An α-(1→4) linkage consumes a hydroxyl group at C4 of the accepting glucose residue, whereas an α-(1→6) linkage consumes its C6 hydroxyl group.
Which comparison correctly gives the total free hydroxyl groups and the free primary alcoholic C6 hydroxyl groups in P and Q?
ⓐ. Both have 32 total free hydroxyl groups; P has 10 free C6 hydroxyl groups and Q has 8.
ⓑ. P has 32 total free hydroxyl groups and Q has 30; both have 10 free C6 hydroxyl groups.
ⓒ. Both have 41 total free hydroxyl groups; P has 9 free C6 hydroxyl groups and Q has 7.
ⓓ. P has 32 total free hydroxyl groups and Q has 34; P has 8 free C6 hydroxyl groups and Q has 10.
Correct Answer: Both have 32 total free hydroxyl groups; P has 10 free C6 hydroxyl groups and Q has 8.
Explanation: \(\textbf{Initial hydroxyl inventory:}\)
Each free cyclic glucose molecule contains five hydroxyl groups.
Both polymers are formed from ten glucose molecules.
Therefore:
\[
N_{\mathrm{OH,initial}}=10(5)
\]
\[
N_{\mathrm{OH,initial}}=50
\]
\(\textbf{Total glycosidic connectivity:}\)
P contains nine α-(1→4) linkages.
Q contains seven α-(1→4) linkages and two α-(1→6) linkages.
Thus:
\[
N_{\mathrm{linkages},P}=9
\]
\[
N_{\mathrm{linkages},Q}=7+2=9
\]
\(\textbf{Total free hydroxyl groups:}\)
Each specified glycosidic condensation consumes two hydroxyl groups.
Consequently:
\[
N_{\mathrm{OH,free},P}=50-2(9)
\]
\[
N_{\mathrm{OH,free},P}=32
\]
For Q:
\[
N_{\mathrm{OH,free},Q}=50-2(9)
\]
\[
N_{\mathrm{OH,free},Q}=32
\]
\(\textbf{Primary alcoholic hydroxyl groups in P:}\)
The α-(1→4) connections do not consume the C6 hydroxyl groups.
All ten C6 primary alcoholic groups therefore remain free.
\[
N_{\mathrm{C6-OH},P}=10
\]
\(\textbf{Primary alcoholic hydroxyl groups in Q:}\)
The two α-(1→6) branch linkages each consume one C6 hydroxyl group.
\[
N_{\mathrm{C6-OH},Q}=10-2
\]
\[
N_{\mathrm{C6-OH},Q}=8
\]
\(\textbf{Structural interpretation:}\)
Equal residue counts and equal total numbers of glycosidic linkages give equal overall free hydroxyl counts under the stated assumptions.
Nevertheless, branching changes which hydroxyl positions are involved in glycosidic bonding.
\(\textbf{Final answer:}\)
\[
\boxed{\begin{gathered}
P:32\ \mathrm{free\ OH},\ 10\ \mathrm{free\ C6-OH}\\
Q:32\ \mathrm{free\ OH},\ 8\ \mathrm{free\ C6-OH}
\end{gathered}}
\]
419. A freshly prepared aqueous solution of ordinary lactose contains its characteristic galactose–glucose glycosidic linkage.
The anomeric C1 of the β-D-galactose unit is connected through oxygen to C4 of the D-glucose unit.
The glucose residue retains a free hemiacetal group at C1.
During the experiment, the glucose reducing end can interconvert between its alpha and beta configurations through an open-chain form.
The galactose–glucose glycosidic linkage is not hydrolysed.
Which statement correctly describes the structural changes that can occur?
ⓐ. Mutarotation necessarily converts the β-D-galactose glycosidic linkage into an α-D-galactose linkage.
ⓑ. Mutarotation requires the lactose molecule to split into free galactose and glucose before any change in optical rotation is possible.
ⓒ. The configuration at the free glucose anomeric centre can change while the β-configured galactose glycosidic linkage remains intact.
ⓓ. The glucose reducing end cannot undergo anomeric interconversion because another residue of the same lactose molecule contains a glycosidic bond.
Correct Answer: The configuration at the free glucose anomeric centre can change while the β-configured galactose glycosidic linkage remains intact.
Explanation: Lactose contains a β-D-galactose unit joined to a D-glucose unit through a C1–C4 glycosidic linkage. The anomeric carbon of galactose is involved in forming this bond. Its glycosidic configuration is therefore fixed as long as the linkage remains intact. The glucose residue has a different structural role. Its C1 anomeric centre remains free and can behave as a hemiacetal. Ring opening at this centre gives access to an open-chain glucose-derived aldehydic form. Subsequent ring closure can regenerate either anomeric configuration at that glucose centre. Therefore, the intact lactose molecule can participate in anomeric interconversion at its reducing end. This does not require cleavage of the galactose–glucose glycosidic linkage. Nor does it automatically reverse the stereochemical configuration at the galactose anomeric centre already involved in bonding. The molecule can consequently exhibit reducing behaviour and anomeric equilibration while retaining its original two-unit constitution. This distinction separates stereochemical changes at a free reducing end from changes in the glycosidic configuration of a linked residue.
420. Two glycine molecules and one alanine molecule are used to prepare an ordinary linear tripeptide.
Exactly one residue of alanine and two residues of glycine must occur in every product molecule.
The amino acid sequence may vary, but the direction from the free amino terminus to the free carboxyl terminus is significant.
Each peptide bond is formed by condensation with elimination of one water molecule.
Assume no side-chain reactions or cyclic peptide formation.
Molar masses are: glycine \(=75\ \mathrm{g\,mol^{-1}}\), alanine \(=89\ \mathrm{g\,mol^{-1}}\), water \(=18\ \mathrm{g\,mol^{-1}}\).
Which statement correctly gives the number of different residue sequences and the molar mass of each tripeptide?
ⓐ. Six different sequences, each with molar mass \(203\ \mathrm{g\,mol^{-1}}\)
ⓑ. Three different sequences, each with molar mass \(203\ \mathrm{g\,mol^{-1}}\)
ⓒ. Three different sequences, each with molar mass \(221\ \mathrm{g\,mol^{-1}}\)
ⓓ. Two different sequences, each with molar mass \(239\ \mathrm{g\,mol^{-1}}\)
Correct Answer: Three different sequences, each with molar mass \(203\ \mathrm{g\,mol^{-1}}\)
Explanation: \(\textbf{Sequence possibilities:}\)
A tripeptide contains three residue positions.
Two positions must contain glycine, while the remaining position contains alanine.
The two glycine residues are chemically identical for this sequence-counting purpose.
Therefore, changing their positions relative to each other does not produce a new residue sequence.
\(\textbf{Distinct residue arrangements:}\)
Alanine can occupy the first, second or third position.
The resulting sequences are:
\[
\mathrm{Ala-Gly-Gly}
\]
\[
\mathrm{Gly-Ala-Gly}
\]
\[
\mathrm{Gly-Gly-Ala}
\]
Hence:
\[
N_{\mathrm{sequences}}=\frac{3!}{2!}
\]
\[
N_{\mathrm{sequences}}=3
\]
\(\textbf{Combined mass of free amino acids:}\)
\[
M_{\mathrm{free\ units}}=2(75)+89
\]
\[
M_{\mathrm{free\ units}}=239\ \mathrm{g\,mol^{-1}}
\]
\(\textbf{Peptide-bond formation:}\)
An ordinary linear tripeptide has two peptide bonds.
Each condensation eliminates one water molecule.
\[
N_{\mathrm{peptide\ bonds}}=3-1=2
\]
\(\textbf{Molar mass after condensation:}\)
\[
M_{\mathrm{tripeptide}}=239-2(18)
\]
\[
M_{\mathrm{tripeptide}}=239-36
\]
\[
M_{\mathrm{tripeptide}}=203\ \mathrm{g\,mol^{-1}}
\]
\(\textbf{Chemical interpretation:}\)
The three sequences have identical amino acid compositions and molar masses but different primary structures.
Complete hydrolysis would produce the same amino acid composition from all three.
Therefore, composition and molecular mass alone cannot determine which sequence was originally present.
\(\textbf{Final answer:}\)
\[
\boxed{3\ \text{distinct sequences};\quad M=203\ \mathrm{g\,mol^{-1}}}
\]
