CBSE Class 12 Chemistry Chapter 10 Biomolecules Competency Based Questions Set 01

Competency Based Questions (MCQs)

Question 1. The sugar present in milk is
(a) Sucrose
(b) Maltose
(c) Glucose
(d) Lactose
Answer: (d) Lactose
In simple words: Lactose is the natural sugar that is present in milk and other dairy items.

Exam Tip: Lactose is a disaccharide containing glucose and galactose, which is specifically found in milk.

Question 2. \( \alpha \)-D (+) glucose and \( \beta \)-D (+) – glucose are
(a) Enantiomers
(b) Geometrical isomers
(c) Anomers
(d) Epimers
Answer: (c) Anomers
In simple words: Anomers are cyclic sugars that differ in configuration only at the hemiacetal or hemiketal carbon.

Exam Tip: Remember that anomers are diastereomers whose structures differ only at the first carbon (C-1) for aldoses.

Question 3. Distinction between glucose and fructose can be done by
(a) Benedict’s solution
(b) Tollen’s reagent
(c) Selivanoff’s reagent
(d) Fehling solution
Answer: (c) Selivanoff’s reagent
In simple words: Selivanoff's test is a chemical reaction used to tell ketoses (like fructose) apart from aldoses (like glucose).

Exam Tip: Ketoses react more rapidly with Selivanoff's reagent to produce a cherry-red color, which distinguishes them from aldoses.

Question 4. Which does not show mutarotation?
(a) Glucose
(b) Maltose
(c) Fructose
(d) Sucrose
Answer: (d) Sucrose
In simple words: Sucrose does not change its optical rotation over time because both of its reactive hemiacetal groups are linked together.

Exam Tip: Non-reducing sugars like sucrose do not show mutarotation because their anomeric carbons are locked in a glycosidic bond.

Question 5. The reagent used for obtaining osazone derivative of fructose is
(a) \( \text{NH}_2\text{OH} \)
(b) \( \text{NH}_2-\text{NH}_2 \)
(c) \( \text{NH}_2-\text{NHC}_6\text{H}_5 \)
(d) 2, 4-DNP
Answer: (c) \( \text{NH}_2-\text{NHC}_6\text{H}_5 \)
In simple words: Phenylhydrazine is the substance we mix with sugars like fructose to create distinct yellow crystals called osazones.

Exam Tip: Osazone formation involves the first two carbon atoms of a reducing sugar reacting with three equivalents of phenylhydrazine.

Question 6. Amylopectin is a polymer of
(a) \( \beta \)-D-glucose
(b) \( \alpha \)-D-glucose
(c) \( \beta \)-D-fructose
(d) \( \alpha \)-D-fructose
Answer: (b) \( \alpha \)-D-glucose
In simple words: Amylopectin, which is a major part of plant starch, is built from many linked molecules of alpha-D-glucose.

Exam Tip: Starch contains amylose (linear) and amylopectin (branched), both of which are made up of \( \alpha \)-D-glucose units.

Question 7. Hydrolysis of sucrose gives
(a) Glucose only
(b) Glucose + fructose
(c) Glucose and galactose
(d) Maltose
Answer: (b) Glucose + fructose
In simple words: Breaking down sucrose with water splits it into equal parts of glucose and fructose.

Exam Tip: The mixture of glucose and fructose obtained from sucrose hydrolysis is called invert sugar due to the change in optical rotation.

Question 8. The disease resulting from the intake of amino acid deficient diet is
(a) Kwasiorkar
(b) Pernicicres anaemia
(c) PEM
(d) Haemophilio
Answer: (a) Kwasiorkar
In simple words: Kwashiorkor is a severe form of malnutrition that happens when someone does not get enough protein in their food.

Exam Tip: Kwashiorkor is caused by severe protein deficiency even if calorie intake is sufficient, distinguishing it from marasmus.

Question 9. Keratin present in hair is an example of
(a) Fibrous protein
(b) Globular protein
(c) Conjugated protein
(d) Derived protein
Answer: (a) Fibrous protein
In simple words: Keratin is a tough, long-stranded protein that builds structures like hair, nails, and skin.

Exam Tip: Fibrous proteins are insoluble in water and play structural roles, whereas globular proteins are soluble and functional.

Question 10. DNA and RNA differ in
(a) Sugar
(b) Purines
(c) Pyrimidines
(d) Both (a) and (c)
Answer: (d) Both (a) and (c)
In simple words: DNA and RNA use different sugar molecules and have a different set of pyrimidine bases.

Exam Tip: DNA has deoxyribose and thymine, while RNA contains ribose and uracil instead.

Question 11. The vitamin present in oils and fats are
(a) A and D
(b) B and C
(c) A and B
(d) A and C
Answer: (a) A and D
In simple words: Vitamins A and D dissolve easily in fats and oils, so they are found naturally in fatty foods.

Exam Tip: Vitamins A, D, E, and K are fat-soluble, whereas vitamins B and C are water-soluble.

Assertion-Reason Questions

Directions: In the following questions, a statement of Assertion followed by a statement of Reason is given. Choose the correct option out of the choices given below:
(a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.
(b) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.
(c) Assertion is correct statement but reason is false statement.
(d) Assertion is false statement but reason is correct statement.

Question 1. Assertion: Glucose and fructose are reducing sugars.
Reason: Glucose and fructose contain a free aldehydic and ketonic group adjacent to \( \alpha > \text{CHOH} \) group respectively.
Answer: (a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.
Reducing sugars possess a free aldehydic or ketonic group next to a \( \alpha > \text{CHOH} \) group, which helps them react with Tollen's, Schiff's, or Benedict's reagents.
In simple words: Both statements are true, and the reason correctly explains why glucose and fructose can reduce certain reagents since they have reactive carbonyl groups.

Exam Tip: All monosaccharides, whether aldoses or ketoses, are reducing sugars because they contain hemiacetal or hemiketal groups that can open into free carbonyl forms.

Question 2. Assertion: Fructose reduces Fehling's solution and Tollens' reagent.
Reason: Fructose does not contain any aldehyde group.
Answer: (b) Assertion and Reason both are correct statements but Reason is not correct explanation for assertion.
When fructose is warmed with weak alkali, it converts into a balanced blend of glucose, fructose, and mannose. This ability of fructose to act on Fehling's and Tollen's test occurs because it isomerizes into these other sugars (known as the Lobry de Bruyn-Alberda van Ekenstein transformation).
In simple words: Fructose has a ketone group instead of an aldehyde, but in basic solutions, it rearranges into glucose and mannose, allowing it to act as a reducing sugar.

Exam Tip: Although fructose is a ketose, it acts as a reducing sugar because the basic medium of Fehling's/Tollens' reagent tautomerizes it to glucose.

Question 3. Assertion: \( \alpha \)-Amino acids exits as dipolar ions or zwitter ions.
Reason: \( \alpha \)-Amino acids are the building blocks of proteins.
Answer: (b) Assertion and Reason both are correct statements but reason is not correct explanation for Assertion.
While it is true that amino acids serve as the basic units of proteins, this structural role does not explain why they form dipolar ions. They exist as zwitterions due to internal proton transfer between the acidic carboxyl group and the basic amino group.
In simple words: Amino acids can form dual-charged zwitterions because they have both acidic and basic parts in the same molecule, which is unrelated to making proteins.

Exam Tip: Remember that zwitterion formation is a chemical property resulting from internal acid-base neutralization, whereas protein synthesis is a biological polymerization process.

Question 4. Assertion: Glucose when treated with \( \text{CH}_3\text{OH} \) in presence of dry HCl gas gives \( \alpha \)-and \( \beta \)-methyl glucosides.
Reason: Glucose reacts with phenyl hydrazine to form crystalline osazone.
Answer: (b) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.
Due to the cyclic setup, the first carbon (C-1) in glucose turns chiral, allowing glucose to exist in two distinct stereoisomeric styles. Consequently, each ring form reacts to yield a pair of methyl glucosides.
In simple words: Glucose reacts with methanol to form two kinds of glucosides because the ring-closing carbon is asymmetric, but its reaction with phenylhydrazine is an entirely separate reaction.

Exam Tip: Methyl glucoside formation involves only the hemiacetal hydroxyl group at C-1, confirming the ring structure of glucose.

Question 5. Assertion: DNA undergoes replication.
Reason: DNA contains cytosine and thymine as pyrimidine bases.
Answer: (b) Assertion and Reason both are correct statements but Reason is not correct explanation for assertion.
The hereditary code of a cell lies within the order of bases A, T, G, and C in the DNA molecule. During cell division, DNA strands duplicate to make matching replicates, ensuring each new cell receives identical DNA to the parent.
In simple words: Both statements are true, but DNA replicates so that dividing cells can pass on their genetic blueprint, not because it contains specific pyrimidine bases.

Exam Tip: DNA replication is facilitated by complementary base pairing (A with T, G with C) across the double-stranded helix structure.

Question 6. Assertion: Glycine exists as Zwitter ion but o- and p-aminobenzoic acids do not.
Reason: Due to the presence of \( -\text{NH}_2 \) and \( -\text{COOH} \) group within the same molecule, they neutralise each other and hence, \( \alpha \)-amino acids exist as dipolar ions or zwitter ions.
Answer: (b) Assertion and Reason both are correct statements but Reason is not correct explanation for assertion.
In ortho- or para-aminobenzoic acids, the lone electron pair on the amino group shifts toward the aromatic benzene ring. Because of this, the basicity of the amino group and the acidity of the carboxyl group are reduced. Thus, the weak acid cannot hand over a proton to the weak base, preventing them from forming a zwitterion.
In simple words: Both statements are correct, but aminobenzoic acids fail to make zwitterions because the benzene ring pulls electron density away from the amino group, making it too weak to accept a proton.

Exam Tip: Direct resonance conjugation of the amino group with the benzene ring decreases its basicity, which is why aromatic aminobenzoic acids do not form zwitterions unlike aliphatic amino acids.

Question 7. Assertion: Haemoglobin is a globular protein.
Reason: Globular proteins are insoluble in water.
Answer: (c) Assertion is correct statement but reason is false statement.
Globular proteins possess weak intermolecular attractions, which makes them highly soluble in aqueous environments.
In simple words: Haemoglobin is indeed a globular protein, but globular proteins dissolve easily in water because their hydrophilic groups face outward.

Exam Tip: Fibrous proteins are water-insoluble due to strong intermolecular hydrogen bonding, whereas globular proteins fold into spherical shapes that are water-soluble.

Question 8. Assertion: \( \beta \)-pleated sheet structure of protein shows maximum extension.
Reason: Intermolecular hydrogen bonding is present in them.
Answer: (b) Assertion and Reason both are correct statements but Reason is not correct explanation for assertion.
In the beta-sheet arrangement, separate protein chains are bound together by hydrogen bonds between molecules. The stretching and shrinking of this beta-pleated shape is determined by the bulkiness of the side chains.
In simple words: The beta-sheet structure is fully stretched out and held together by hydrogen bonds, but the hydrogen bonding itself doesn't explain why it reaches maximum extension.

Exam Tip: The polypeptide chains in a beta-pleated sheet are almost fully extended, unlike the tightly coiled alpha-helix structure.

Question 9. Assertion: Alpha (\( \alpha \)) amino acids exist as internal salt in solution as they have amino and carboxylic acid groups in near vicinity.
Reason: \( \text{H}^+ \) ion given by carboxylic group (\( -\text{COOH} \)) is captured by amino group (\( -\text{NH}_2 \)) having lone pair of electrons.
Answer: (a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.
A classic example of an alpha-amino acid is glycine. In aqueous media, it behaves as a zwitterion because the hydrogen ion from the carboxyl group is grabbed by the amino group, which possesses a free pair of electrons on its nitrogen.
In simple words: Amino acids form zwitterions because the acidic carboxyl end donates a hydrogen ion directly to the basic amino end within the same molecule.

Exam Tip: The zwitterionic form gives amino acids high melting points and salt-like behavior, such as high water solubility.

Question 10. Assertion: Uracil occurs in DNA.
Reason: DNA undergoes replication.
Answer: (d) Assertion is false statement but reason is correct statement.
The base uracil is found exclusively in RNA, not DNA.
In simple words: DNA replication is a real process, but DNA uses thymine instead of uracil as its pyrimidine base.

Exam Tip: Uracil is only present in RNA; DNA contains thymine, which helps maintain higher genetic stability.

Case-Based Questions

Case Study 1
Pentose and hexose undergo intramolecular hemiacetal or hemiketal formation due to combination of the \( -\text{OH} \) group with the carbonyl group. The actual structure is either of five or six membered ring containing an oxygen atom. In the free state all pentoses and hexoses exist in pyranose form (resembling pyran). However, in the combined state some of them exist as five membered cyclic structures, called furanose (resembling furan).

Question (i) \( \alpha \)-D(+)-glucose and \( \beta \)-D(+)–glucose are
(a) enantiomers
(b) conformers
(c) epimers
(d) anomers
Answer: (d) anomers
In simple words: These two forms of glucose differ only in the shape around the first carbon atom, which makes them anomers.

Exam Tip: Remember that C-1 is the anomeric carbon in aldoses like glucose, where the alpha and beta forms differ in configuration.

Question (iv) The term anomers of glucose refers to
(a) isomers of glucose that differ in configurations at carbons one and four (C–1 and C–4)
(b) a mixture of (D)-glucose and (L)-glucose
(c) enantiomers of glucose
(d) isomers of glucose that differ in configuration at carbon one (C–1)
Answer: (d) isomers of glucose that differ in configuration at carbon one (C–1)
In simple words: Anomers are cyclic sugars that differ only in the arrangement of atoms at the very first carbon.

Exam Tip: Make sure not to confuse anomers (differing at C-1) with epimers (differing at any other single chiral carbon, like C-2 or C-4).

Question (v) What percentage of \( \beta \)-D–(+) glucopyranose is found at equilibrium in the aqueous solution?
(a) 50%
(b) ≈ 100%
(c) 36%
(d) 64%
Answer: (d) 64%
In simple words: In water, glucose is mostly in the beta form (about 64%) because it is more stable than the alpha form.

Exam Tip: The beta form is more stable because all its bulky substituents (like \( -\text{OH} \) and \( -\text{CH}_2\text{OH} \)) are in the equatorial position.

Case Study 2
Carbohydrates can exist in either of two conformations, as determined by the orientation of the hydroxyl group about the asymmetric carbon farthest from the carbonyl. By convention, a monosaccharide is said to have D-configuration if the hydroxyl group attached to the asymmetric carbon atom adjacent to the \( -\text{CH}_2\text{OH} \) group is on the right hand side irrespective of the positions of the other hydroxyl groups. On the other hand, the molecule is assigned L-configuration if the \( -\text{OH} \) group attached to the carbon adjacent to the \( -\text{CH}_2\text{OH} \) group is on the left side.

Question (i) D-Glyceraldehyde and L-Glyceraldehyde are
(a) epimers
(b) enantiomers
(c) anomers
(d) conformational diasteriomers
Answer: (b) enantiomers
In simple words: These two molecules are non-superimposable mirror images of each other, making them enantiomers.

Exam Tip: D and L isomers of glyceraldehyde are the simplest pair of enantiomers used as references for stereochemistry in sugars.

Question (ii) Which of the following monosaccharides, is the majority found in the human body?
(a) D-type
(b) L-type
(c) Both of these
(d) None of these
Answer: (a) D-type
In simple words: Almost all the sugars used by our bodies are D-type sugars, which can be properly processed by our metabolic enzymes.

Exam Tip: Human biological systems are highly stereospecific, preferring D-sugars and L-amino acids.

Question (iii) The two functional groups present in a typical carbohydrate are
(a) \( -\text{OH} \) and \( -\text{COOH} \)
(b) \( -\text{CHO} \) and \( -\text{COOH} \)
(c) \( >\text{C}=\text{O} \) and \( -\text{OH} \)
(d) \( -\text{OH} \) and \( -\text{CHO} \)
Answer: (c) \( >\text{C}=\text{O} \) and \( -\text{OH} \)
In simple words: Every simple sugar contains a carbonyl group (which can be an aldehyde or ketone) along with multiple hydroxyl groups.

Exam Tip: Carbohydrates are defined as polyhydroxy aldehydes or polyhydroxy ketones, containing both carbonyl and hydroxyl groups.

Question (iv) Monosaccharides contain
(a) always six carbon atoms
(b) always five carbon atoms
(c) always four carbon atoms
(d) may contain 3 to 7 carbon atoms
Answer: (d) may contain 3 to 7 carbon atoms
In simple words: Simple sugars can vary in length, containing anywhere from 3 up to 7 carbon atoms.

Exam Tip: Monosaccharides are classified based on carbon count as trioses (3C), tetroses (4C), pentoses (5C), hexoses (6C), and heptoses (7C).

Question (v) The correct corresponding order of names of the four aldoses with the configurations shown is
(a) L-erythrose, L-threose, L-erythrose, D-threose
(b) D-threose, D-erythrose, L-threose, D-erythrose
(c) L-erythrose, L-threose, D-erythrose, D-threose
(d) D-erythrose, D-threose, L-erythrose, L-threose
Answer: (d) D-erythrose, D-threose, L-erythrose, L-threose
In simple words: The first two are D-sugars because their bottom-most chiral OH is on the right, and the last two are L-sugars because their bottom-most chiral OH is on the left.

Exam Tip: Erythrose has both chiral hydroxyl groups on the same side in its Fischer projection, while Threose has them on opposite sides.

Case Study 3
Carbohydrates are polyhydroxy aldehydes and ketones and those compounds which on hydrolysis give such compounds are also carbohydrates. The carbohydrates which are not hydrolysed are called monosaccharides. Monosaccharides with aldehydic group are called aldose and those which free ketonic groups are called ketose. Carbohydrates are optically active. Number of optical isomers = \( 2^n \), where \( n = \) number of asymmetric carbons.

Question (i) In glucose \( \text{CHO}-(\text{CHOH})_4-\text{CH}_2\text{OH} \), the number of optical isomers will be
(a) 16
(b) 8
(c) 32
(d) 4
Answer: (a) 16
In simple words: Since glucose contains 4 asymmetric carbon atoms, we use the formula \( 2^4 \) which gives us 16 different possible optical isomers.

Exam Tip: The formula for finding the maximum number of stereoisomers for an unsymmetrical molecule with n chiral centers is \( 2^n \).

Question (ii) First member of ketos sugar is
(a) ketotriose
(b) ketotetrose
(c) ketopentose
(d) ketohexose
Answer: (a) ketotriose
In simple words: The simplest ketone-based sugar is ketotriose, which has three carbon atoms (also called dihydroxyacetone).

Exam Tip: Dihydroxyacetone (a ketotriose) is the only monosaccharide that is optically inactive because it has no chiral center.

Question (iii) Some statements are given below:
I. Glucose is aldohexose.
II. Naturally occurring glucose is dextrorotatory.
III. Glucose contains three chiral centres.
IV. Glucose contains one 1º alcoholic group and four 2º alcoholic groups.

Among the above, the correct statements are:
(a) I and II only
(b) III and IV only
(c) I, II and IV only
(d) I, II, III and IV
Answer: (c) I, II and IV only
In simple words: Glucose has four chiral carbons instead of three, so statement III is incorrect, while all the other statements are true.

Exam Tip: Open-chain glucose has four chiral carbons (C-2, C-3, C-4, and C-5), while cyclic glucose has five chiral carbons because C-1 becomes chiral.

Question (iv) Two hexoses form the same osazone, find the correct statement about these hexoses.
(a) Both of them must be aldoses.
(b) They are epimers at C-3
(c) The carbon atoms 1 and 2 in both have the same configuration.
(d) The carbon atoms 3, 4 and 5 in both have the same configuration.
Answer: (d) The carbon atoms 3, 4 and 5 in both have the same configuration.
In simple words: When sugars react to form osazones, the reaction only changes C-1 and C-2, so any sugars with identical shapes at C-3, C-4, and C-5 will produce the exact same osazone.

Exam Tip: Glucose, fructose, and mannose produce the same osazone because they differ only at C-1 and C-2, while sharing identical configurations at C-3, C-4, and C-5.

Question (v) Which of the following reactions of glucose can be explained only by its cyclic structure?
(a) Glucose forms cyanohydrin with HCN.
(b) Glucose reacts with hydroxylamine to form an oxime.
(c) Pentacetate of glucose does not react with hydroxylamine.
(d) Glucose is oxidised by nitric acid to gluconic acid.
Answer: (c) Pentacetate of glucose does not react with hydroxylamine.
In simple words: Glucose pentaacetate does not react with hydroxylamine because its aldehyde group is locked within a stable ring structure, which proves the sugar has a cyclic form.

Exam Tip: The inability of glucose pentaacetate to react with hydroxylamine shows the absence of a free aldehyde group, supporting the ring structure.

Case Study 4
When a protein in its native form, is subjected to physical changes like change in temperature or chemical changes like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein. The denaturation causes changes in secondary and tertiary structures but primary structures remains intact. Examples of denaturation of protein are coagulation of egg white on boiling, curdling of milk, formation of cheese when an acid is added to milk.

Question (i) Mark the wrong statement about denaturation of proteins.
(a) The primary structure of the protein does not change.
(b) Globular proteins are converted into fibrous proteins.
(c) Fibrous proteins are converted into globular proteins.
(d) The biological activity of the protein is destroyed.
Answer: (c) Fibrous proteins are converted into globular proteins.
In simple words: During denaturation, soluble spherical proteins open up into straight fiber-like strands, not the other way around.

Exam Tip: Denaturation turns organized, soluble globular proteins into insoluble, random fibrous aggregates.

Question (ii) Which structure(s) of proteins remain(s) intact during denaturation process?
(a) Both secondary and tertiary structure
(b) Primary structure only
(c) Secondary structure only
(d) Tertiary structure only
Answer: (b) Primary structure only
In simple words: Denaturation only unfolds the folded shapes of a protein, leaving the basic sequence of amino acids (primary structure) untouched.

Exam Tip: The primary structure is held together by strong covalent peptide bonds, which are not broken during mild denaturation.

Question (iii) \( \alpha \)-helix and \( \beta \)-pleated structures of proteins are classified as
(a) primary structure
(b) secondary structure
(c) tertiary structure
(d) quaternary structure
Answer: (b) secondary structure
In simple words: The coiling or folding of local protein chains into helices or sheets represents the secondary structure level of a protein.

Exam Tip: Secondary structures are stabilized purely by hydrogen bonding along the polypeptide backbone.

Question (iv) Cheese is a
(a) globular protein
(b) conjugated protein
(c) denatured protein
(d) derived protein
Answer: (c) denatured protein
In simple words: Cheese is made by adding acid to milk, which denatures and clumps the milk proteins together.

Exam Tip: Acidification or enzyme treatment of milk leads to protein coagulation, which is a classic example of denaturation.

Question (v) Secondary structure of protein refers to
(a) mainly denatured proteins and structure of prosthetic groups.
(b) three-dimensional structure, especially the bond between amino acid residues that are distant from each other in the polypeptide chain.
(c) linear sequence of amino acid residues in the polypeptide chain.
(d) regular folding patterns of continuous portions of the polypeptide chain.
Answer: (d) regular folding patterns of continuous portions of the polypeptide chain.
In simple words: Secondary structure describes the repeating, regular shapes (like helices or sheets) that segments of a protein chain fold into.

Exam Tip: Understand the differences between primary (sequence), secondary (local folding), tertiary (3D shape), and quaternary (multiple subunits) structures.

Case Study 5
When a solution of an \( \alpha \)-amino acid is placed in an electric field depending on the pH of the medium, three cases may happen:
• In alkaline solution, \( \alpha \)-amino acids exist as anion, and there is a net migration of amino acid towards the anode.
• In acidic solution, \( \alpha \)-amino acids exist as cation, and there is a net migration of amino acids towards the cathode.
• If the anions and cations are exactly balanced there is no net migration; under such conditions any one molecule exists as a positive ion and as a negative ion for exactly the same amount of time, and any small movement in the direction of one electrode is subsequently cancelled by an equal movement back toward the other electrode. The pH of the solution in which a particular amino acid does not migrate under the influence of an electric field is called the isoelectric point of that amino acid.

Question (ii) In aqueous solutions, amino acids mostly exist as
(a) \( \text{NH}_2-\text{CHR}-\text{COOH} \)
(b) \( \text{NH}_2-\text{CHR}-\text{COO}^- \)
(c) \( \text{NH}_3^+\text{CHRCOOH} \)
(d) \( \text{NH}_3^+\text{CHRCOO}^- \)
Answer: (d) \( \text{NH}_3^+\text{CHRCOO}^- \)
In simple words: In water, an amino acid exists as a zwitterion, where the basic nitrogen carries a positive charge and the acidic oxygen carries a negative charge.

Exam Tip: The dual charge of the zwitterion form explains why amino acids have physical properties similar to ionic salts, such as high water solubility.

Question (iii) Amino acids are least soluble
(a) at pH 1
(b) at pH 7
(c) at their isoelectric points
(d) none of the options
Answer: (c) at their isoelectric points
In simple words: At the isoelectric point, the net charge on the amino acid is zero, which makes them clump together and dissolve poorly in water.

Exam Tip: When the net charge is zero, electrostatic repulsion between molecules is minimized, leading to minimum solubility.

Question (iv) The \( \text{pKa}_1 \) and \( \text{pKa}_2 \) of an amino acid are 2.3 and 9.7 respectively. The isoelectric point of the acid is
(a) 12.0
(b) 7.4
(c) 6.0
(d) 3.7
Answer: (c) 6.0
In simple words: We find the isoelectric point by taking the average of the two pKa values: \( \frac{2.3 + 9.7}{2} = 6.0 \).

Exam Tip: For simple amino acids, the isoelectric point (\( \text{pI} \)) is calculated using the formula: \( \text{pI} = \frac{\text{pKa}_1 + \text{pKa}_2}{2} \).

Question (v) A tripeptide (X) on partial hydrolysis gave two dipeptides Cys-Gly and Glu-Cys. Identify the tripeptide.
(a) Glu-Cys-Gly
(b) Gly-Glu-Cys
(c) Cys-Gly-Glu
(d) Cys-Glu-Gly
Answer: (a) Glu-Cys-Gly
In simple words: Since we have overlapping pieces "Glu-Cys" and "Cys-Gly", Cys must sit in the middle, giving the order Glu-Cys-Gly.

Exam Tip: Peptide sequencing by partial hydrolysis relies on identifying overlapping fragments to reconstruct the original chain.

Case Study 6
Adenosine triphosphate (ATP) is the energy carrying molecule found in the cells of all living things. ATP captures chemical energy obtained from the breakdown of food molecules and releases it to fuel other cellular processes. ATP is a nucleotide that consists of three main structures: the nitrogenous base, adenine; the sugar, ribose; and a chain of three phosphate groups bound to ribose. The phosphate tail of ATP is the actual power source which the cell taps. Available energy is contained in the bonds between the phosphates and is released when they are broken, which occurs through the addition of a water molecule (a process called hydrolysis). Usually only the outer phosphate is removed from ATP to yield energy; when this occurs ATP is converted to adenosine diphosphate (ADP), the form of the nucleotide having only two phosphates.

Question (i) Cellular oxidation of glucose is a:
(a) spontaneous and endothermic process
(b) non–spontaneous and exothermic process
(c) non–spontaneous and endothermic process
(d) spontaneous and exothermic process
Answer: (d) spontaneous and exothermic process
In simple words: The burning of glucose in cells releases energy naturally, making it a spontaneous and heat-releasing reaction.

Exam Tip: Cellular respiration converts glucose and oxygen into carbon dioxide, water, and energy with a negative change in Gibbs free energy (\( \Delta G < 0 \)).

Question (ii) What is the efficiency of glucose metabolism if 1 mole of glucose gives 38 ATP energy? (Given: The enthalpy of combustion of glucose is 686 kcal, 1 ATP = 7.3 kcal)
(a) 100%
(b) 38%
(c) 62%
(d) 80%
Answer: (b) 38%
Efficiency of energy conservation = \( \frac{\text{Energy conserved as ATP}}{\text{Total energy of combustion}} \times 100 \)
Energy conserved = \( 38 \times 7.3 \text{ kcal} = 277.4 \text{ kcal} \)
Efficiency = \( \frac{277.4}{686} \times 100 \approx 40\% \) (usually approximated as 38% in biological systems).
In simple words: About 38% of the total chemical energy in glucose is converted into usable energy in the form of ATP, while the rest is lost as body heat.

Exam Tip: Know that cellular respiration is highly efficient compared to man-made engines, capturing over a third of the potential energy as chemical bonds.

Question (iii) Which of the following statements is true?
(a) ATP is a nucleoside made up of nitrogenous base adenine and ribose sugar.
(b) ATP consists the nitrogenous base, adenine and the sugar, deoxyribose.
(c) ATP is a nucleotide which contains a chain of three phosphate groups bound to ribose sugar.
(d) The nitrogenous base of ATP is the actual power source.
Answer: (c) ATP is a nucleotide which contains a chain of three phosphate groups bound to ribose sugar.
In simple words: ATP is a complete nucleotide because it contains adenine, ribose sugar, and three energy-carrying phosphate groups.

Exam Tip: A nucleoside has only base + sugar, whereas a nucleotide has base + sugar + phosphate groups.

Question (iv) Nearly 95% of the energy released during cellular respiration is due to:
(a) glycolysis occurring in cytosol.
(b) oxidative phosphorylation occurring in cytosol.
(c) glycolysis in occurring mitochondria.
(d) oxidative phosphorylation occurring in mitochondria.
Answer: (d) oxidative phosphorylation occurring in mitochondria.
In simple words: Most of the ATP (about 95%) is produced during oxidative phosphorylation, which takes place inside the mitochondria.

Exam Tip: Oxidative phosphorylation yields 36 of the 38 ATP molecules produced per glucose molecule, making the mitochondria the "powerhouse of the cell".

Question (v) Which of the following statements is correct:
(a) ATP is a nucleotide which has three phosphate groups while ADP is a nucleoside which three phosphate groups.
(b) ADP contains a nitrogenous bases adenine, ribose sugar and two phosphate groups bound to ribose.
(c) ADP is the main source of chemical energy in living matter.
(d) ATP and ADP are nucleosides which differ in number of phosphate groups.
Answer: (b) ADP contains a nitrogenous bases adenine, ribose sugar and two phosphate groups bound to ribose.
In simple words: ADP is formed when one phosphate is removed from ATP, leaving behind adenine, ribose, and two phosphate groups.

Exam Tip: ADP (Adenosine Diphosphate) differs from ATP (Adenosine Triphosphate) solely by having two phosphate groups instead of three.

Case Study 7
Polysaccharides may be very large molecules. Starch, glycogen, cellulose, and chitin are examples of polysaccharides. Starch is the stored form of sugars in plants and is made up of amylose and amylopectin (both polymers of glucose). Amylose is soluble in water and can be hydrolyzed into glucose units breaking glycosidic bonds, by the enzymes \( \alpha \)-amylase and \( \beta \)-amylase. It is a straight-chain polymer. Amylopectin is a branched-chain polymer of several D-glucose molecules. 80% of amylopectin is present in starch. Plants are able to synthesize glucose, and the excess glucose is stored as starch in different plant parts, including roots and seeds. The starch that is consumed by animals is broken down into smaller molecules, such as glucose. The cells can then absorb the glucose. Glycogen is the storage form of glucose in humans and other vertebrates, and is made up of monomers of glucose. It is structurally quite similar to amylopectin. Glycogen is the animal equivalent of starch. It is stored in liver and skeletal muscles. Cellulose is one of the most abundant natural biopolymers. The cell walls of plants are mostly made of cellulose, which provides structural support to the cell. Wood and paper are mostly cellulosic in nature. Like amylose, cellulose is a linear polymer of glucose. Cellulose is made up of glucose monomers that are linked by bonds between particular carbon atoms in the glucose molecule. Every other glucose monomer in cellulose is flipped over and packed tightly as extended long chains. This gives cellulose its rigidity and high tensile strength—which is so important to plant cells. Cellulose passing through our digestive system is called dietary fiber.

Question (i) In animals, Glycogen is stored in:
(a) Liver
(b) Spleen
(c) Lungs
(d) Small Intestine
Answer: (a) Liver
In simple words: Our bodies store backup glucose as glycogen mostly in the liver and skeletal muscle tissues.

Exam Tip: Liver glycogen helps maintain systemic blood glucose levels, whereas muscle glycogen is used locally for muscle contraction.

Question (ii) Amylose is:
(a) straight chain, water insoluble component of starch, which constitutes 20 % of it.
(b) straight chain, water soluble component of starch, which constitutes 20 % of it.
(c) branched chain, water insoluble component of starch, which constitutes 80 % of it.
(d) branched chain, water soluble component of starch, which constitutes 80 % of it.
Answer: (b) straight chain, water soluble component of starch, which constitutes 20 % of it.
In simple words: Amylose is the unbranched, water-soluble part of starch that makes up about one-fifth of the total starch structure.

Exam Tip: Starch contains around 15-20% amylose (linear, soluble) and 80-85% amylopectin (branched, insoluble).

Question (iii) Which biopolymer breaks down to release glucose, whenever glucose levels drop in Our body:
(a) starch
(b) cellulose
(c) chitin
(d) glycogen
Answer: (d) glycogen
In simple words: Whenever your blood sugar gets low, your body breaks down stored glycogen back into glucose to supply energy.

Exam Tip: Glucagon stimulates the breakdown of glycogen (glycogenolysis) in response to low blood sugar levels.

Question (iv) The linkages which join monosaccharides to form long chain polysaccharides:
(a) Peptide linkage
(b) Disulphide bonds
(c) Hydrogen bonds
(d) Glycosidic linkage
Answer: (d) Glycosidic linkage
In simple words: Monosaccharides link up using oxygen bridges called glycosidic bonds to form long sugar chains.

Exam Tip: Glycosidic linkages are ether bonds formed by dehydration reaction between two monosaccharide units.

Question (v) Cellulose on complete hydrolysis yields:
(a) amylose
(b) amylopectin
(c) glucose
(d) amylose and amylopectin
Answer: (c) glucose
In simple words: Because cellulose is a polymer made entirely of glucose, breaking it down completely yields only glucose molecules.

Exam Tip: Cellulose is a linear homopolysaccharide composed solely of \( \beta \)-D-glucose units joined by \( \beta \)-1,4-glycosidic linkages.

Case Study 8
The basic chemical formula of DNA is now well established. It consists of a very long chain, the backbone of which is made up of alternate sugar and phosphate groups, joined together in regular 3' 5' phosphate di-ester linkages. To each sugar is attached a nitrogenous base, only four different kinds of which are commonly found in DNA. Two of these—adenine and guanine—are purines, and the other two thymine and cytosine—are pyrimidines. A fifth base, 5-methyl cytosine, occurs in smaller amounts in certain organisms, and a sixth, 5-hydroxy-methyl-cytosine, is found instead of cytosine in the T even phages. It should be noted that the chain is unbranched, a consequence of the regular internucleotide linkage. On the other hand the sequence of the different nucleotides is, as far as can be ascertained, completely irregular. Thus, DNA has some features which are regular, and some which are irregular. A similar conception of the DNA molecule as a long thin fiber is obtained from physicochemical analysis involving sedimentation, diffusion, light scattering, and viscosity measurements. These techniques indicate that DNA is a very asymmetrical structure approximately 20 A wide and many thousands of angstroms long.

Question (i) Purines present in DNA are:
(a) adenine and thymine
(b) guanine and thymine
(c) cytosine and thymine
(d) adenine and guanine
Answer: (d) adenine and guanine
In simple words: The two double-ring nitrogenous bases found in DNA are adenine and guanine, which belong to the purine family.

Exam Tip: Purines have a nine-membered double-ring structure (Adenine and Guanine), whereas pyrimidines have a six-membered single-ring structure (Cytosine, Thymine, Uracil).

Question (ii) DNA molecule has ................. internucleotide linkage and ................. sequence of the different nucleotides.
(a) regular, regular
(b) regular, irregular
(c) irregular, regular
(d) irregular, irregular
Answer: (b) regular, irregular
In simple words: The structural backbone of DNA repeats in a regular, patterned way, while the sequence of genetic letters is complex and irregular.

Exam Tip: The phosphodiester backbone of DNA is uniform (regular), whereas the base sequence varies (irregular) to encode genetic information.

Question (iii) DNA has a backbone.
(a) phosphate -purine
(b) pyrimidines- sugar
(c) phosphate- sugar
(d) purine- pyrimidine
Answer: (c) phosphate- sugar
In simple words: The main support structure running along the outside of the DNA ladder is made of repeating sugar and phosphate groups.

Exam Tip: The hydrophilic sugar-phosphate backbone of DNA faces the aqueous environment, while the hydrophobic nitrogenous bases face inward.

Question (iv) Out of the four different kinds of nitrogenous bases which are commonly found in DNA, .............. has been replaced in some organisms.
(a) adenine
(b) guanine
(c) cytosine
(d) thymine
Answer: (c) cytosine
In simple words: Cytosine can sometimes be swapped out for other modified bases like 5-methylcytosine in certain organisms.

Exam Tip: In bacteriophages and certain plant/animal genomes, cytosine is methylated or modified to regulate gene expression and protect viral DNA.