Class 12Chemistry · Organic ChemistryFull chapter

Biomolecules

The whole chapter in one place — read it, then test yourself. Clear notes, key facts, a practice quiz, and worked NCERT solutions & PYQs.

Carbohydrates: Classification and Structure of Glucose

Quick answer Carbohydrates are polyhydroxy aldehydes/ketones classified as monosaccharides, oligosaccharides and polysaccharides; glucose and fructose are the key monosaccharides whose open-chain and cyclic (Haworth) structures explain their reactions.

Carbohydrates (saccharides) are optically active polyhydroxy aldehydes or ketones, or compounds that yield such units on hydrolysis. Most have the general formula Cn(H2O)m, which is why they were once called "hydrates of carbon", though some carbohydrates (e.g. rhamnose, C6H12O5) do not fit this formula while some non-carbohydrates (e.g. acetic acid, CH3COOH) do — so the formula is only a rough guide, not a definition.

On the basis of their behaviour on hydrolysis, carbohydrates are classified into three groups:

  • Monosaccharides: cannot be hydrolysed further into simpler polyhydroxy units, e.g. glucose, fructose, ribose. Classified by the number of carbons (triose, tetrose, pentose, hexose ...) and by the carbonyl group present — aldose (aldehyde group) or ketose (ketone group).
  • Oligosaccharides: yield 2–10 monosaccharide units on hydrolysis; those giving two units are disaccharides (sucrose, maltose, lactose), the most common in nature.
  • Polysaccharides: yield a large number of monosaccharide units on hydrolysis (starch, cellulose, glycogen). They are not sweet and are called non-sugars.

Carbohydrates are also classed as reducing sugars (reduce Fehling's solution and Tollens' reagent because a free −CHO or potentially free −C=O group is available in solution) or non-reducing sugars. All monosaccharides, and disaccharides such as maltose and lactose, are reducing sugars; sucrose is the important non-reducing disaccharide.

The open-chain structure of D-(+)-glucose, the most abundant monosaccharide, was established from a series of experimental facts: it forms a monoxime with hydroxylamine and adds one molecule of HCN (confirming a carbonyl group); on mild oxidation with bromine water it gives a six-carbon monocarboxylic acid, gluconic acid, showing the carbonyl is an aldehyde (a ketone could not be oxidised so mildly); acetylation gives glucose pentaacetate, confirming five −OH groups; and vigorous oxidation with dilute HNO3 gives a dicarboxylic acid, saccharic acid, showing that C6 is a primary alcohol (−CH2OH). Its stereochemistry (established by Fischer) is:

CHO−CHOH−CHOH−CHOH−CHOH−CH2OH, with the −OH at C2, C4 and C5 on the right and at C3 on the left in the Fischer projection (C5 fixes the D-configuration).

This open-chain structure, however, fails to explain some observations: glucose does not give the 2,4-DNP test or Schiff's test readily, it does not react appreciably with NaHSO3, and, most tellingly, freshly prepared solutions of glucose show mutarotation — the specific rotation of α-D-glucose (+111°) and β-D-glucose (+19°) both drift on standing to an equilibrium value of +52.7°. This is explained by a cyclic hemiacetal structure: the −OH at C5 attacks the C1 aldehyde to form a six-membered (pyranose) ring, generating a new stereocentre at C1 — the anomeric carbon — so that two cyclic forms (α and β anomers) exist and interconvert in solution. Fructose is the corresponding ketohexose (carbonyl at C2), which forms a five-membered (furanose) ring using the C5−OH, and also shows mutarotation and gives anomers.

Worked example: How many stereoisomers are possible for the open-chain structure of an aldohexose such as glucose? In CHO−CHOH−CHOH−CHOH−CHOH−CH2OH, the four middle carbons C2, C3, C4 and C5 are each attached to four different groups, so each is a chiral (asymmetric) centre: n = 4. The number of possible stereoisomers is N = 2n = 24 = 16 (8 belonging to the D-series and 8 to the L-series); D-(+)-glucose is just one specific member of this set of 16.

Molecular formula of glucose/fructose C₆H₁₂O₆ M = 180 g/mol · Both are hexoses; glucose is an aldose, fructose is a ketose.
General carbohydrate formula Cₙ(H₂O)ₘ Approximate/historical formula; not obeyed by all carbohydrates (e.g. deoxy sugars) nor exclusive to them.
Number of stereoisomers N = 2ⁿ n = number of chiral carbon atoms in the open-chain form; gives N = 16 for glucose (n = 4).
Remember
  • Carbohydrates classify as monosaccharides, oligosaccharides (2-10 units) and polysaccharides based on hydrolysis products.
  • Glucose is an aldohexose; fructose is a ketohexose; both have molecular formula C6H12O6.
  • Glucose's open-chain structure was deduced from oxime/HCN addition, bromine-water oxidation to gluconic acid, pentaacetate formation and HNO3 oxidation to saccharic acid.
  • Mutarotation (change of optical rotation to an equilibrium value) proves glucose exists mainly as a cyclic hemiacetal (pyranose) with alpha and beta anomers differing at C1.
  • Fructose forms a five-membered furanose ring using its C5-OH.

Disaccharides and Polysaccharides

Quick answer Disaccharides form when two monosaccharides join through a glycosidic linkage; polysaccharides such as starch, cellulose and glycogen are giant polymers of glucose that differ chiefly in the type of glycosidic linkage and branching.

A disaccharide is formed when two monosaccharide units are joined by a glycosidic linkage — an acetal linkage formed between the anomeric carbon (the hemiacetal −OH) of one unit and an −OH group of the other, with elimination of one molecule of water.

Sucrose (cane sugar) is obtained from sugarcane and sugar beet. It is formed by linking C1 of α-D-glucose to C2 of β-D-fructose (an α,β-1,2-glycosidic linkage). Because both anomeric carbons take part in the linkage, sucrose has no free anomeric −OH and cannot reduce Fehling's/Tollens' reagent — it is a non-reducing sugar. On hydrolysis (by dilute acid or the enzyme invertase) it gives an equimolar mixture of glucose and fructose. Sucrose is dextrorotatory (+66.5°) but the hydrolysis mixture is laevorotatory (−39.9°) because the large negative rotation of fructose outweighs the positive rotation of glucose; the change of sign is called inversion and the product mixture is called invert sugar.

Maltose consists of two α-D-glucose units joined by a C1−C4 (α-1,4) glycosidic linkage; one anomeric carbon remains free, so maltose is a reducing sugar. Lactose (milk sugar) consists of β-D-galactose joined to β-D-glucose by a C1−C4 (β-1,4) linkage, and is also a reducing sugar.

Polysaccharides are polymers of many monosaccharide units joined by glycosidic linkages; they are not sweet and serve mainly as storage or structural material:

  • Starch — the storage polysaccharide of plants; a polymer of α-D-glucose with two components: amylose (15–20%, a long unbranched chain with only α-1,4 linkages, coiled into a helix, gives a blue colour with iodine) and amylopectin (80–85%, a branched chain with α-1,4 linkages along the chain and α-1,6 linkages at branch points, gives a purple/red colour with iodine).
  • Cellulose — a linear, unbranched polymer of β-D-glucose joined by β-1,4-glycosidic linkages; the long straight chains pack together via hydrogen bonding into strong fibres that make up plant cell walls. Humans lack the enzyme (cellulase) that can hydrolyse β-linkages, so cellulose cannot be digested by humans though it is an essential dietary fibre.
  • Glycogen — the storage polysaccharide of animals ("animal starch"), stored in the liver and muscles; structurally similar to amylopectin but even more highly branched, which allows rapid release of glucose when energy is needed.

Worked example: 34.2 g of sucrose (molar mass 342 g mol⁻¹) is completely hydrolysed. Find the total mass of glucose and fructose formed.
Moles of sucrose = 34.2 g ÷ 342 g mol⁻¹ = 0.1 mol.
C12H22O11 + H2O → C6H12O6 (glucose) + C6H12O6 (fructose)
0.1 mol sucrose consumes 0.1 mol (1.8 g) of water and gives 0.1 mol glucose (0.1 × 180 = 18.0 g) and 0.1 mol fructose (18.0 g).
Total product mass = 18.0 g + 18.0 g = 36.0 g, which correctly equals 34.2 g (sucrose) + 1.8 g (water) by conservation of mass.

Sucrose hydrolysis (invert sugar formation) C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆ Sucrose (dextrorotatory) hydrolyses to equimolar glucose + fructose (net laevorotatory) = invert sugar.
Molar mass of sucrose M = 342 g/mol · Used to convert mass of sucrose to moles for hydrolysis calculations.
Remember
  • Glycosidic linkage = acetal linkage between the anomeric carbon of one monosaccharide and an -OH of another, formed with loss of water.
  • Sucrose (glucose + fructose, both anomeric carbons used) is non-reducing; maltose and lactose (one free anomeric carbon each) are reducing sugars.
  • Hydrolysis of sucrose reverses its optical rotation sign (+66.5 degree to -39.9 degree); the product mixture is called invert sugar.
  • Starch = amylose (unbranched, alpha-1,4) + amylopectin (branched, alpha-1,4 and alpha-1,6); cellulose uses beta-1,4 linkages and is indigestible by humans.
  • Glycogen is the animal storage polysaccharide, structurally like a more highly branched amylopectin.

Amino Acids and the Structure of Proteins

Quick answer Proteins are polymers of alpha-amino acids joined by peptide bonds; the way a polypeptide chain folds defines four structural levels, and disturbing this folding (denaturation) destroys biological activity without breaking the amino-acid sequence.

Amino acids are the building blocks of proteins. Each contains an amino (−NH2) group and a carboxyl (−COOH) group; in the amino acids that make up proteins, the amino group is on the carbon next to −COOH, so they are called α-amino acids, general formula R−CH(NH2)−COOH, where R varies from one amino acid to another.

Amino acids are classified in two useful ways. By the relative number of amino and carboxyl groups: acidic (extra −COOH, e.g. glutamic acid, aspartic acid), basic (extra −NH2, e.g. lysine, arginine) and neutral (equal numbers, e.g. glycine, alanine). By nutritional requirement: essential amino acids (cannot be synthesised by the body and must come from the diet, e.g. valine, leucine) and non-essential amino acids (the body can synthesise them, e.g. glycine, alanine).

In the solid state and in aqueous solution, an amino acid exists mainly as a zwitterion (dipolar ion): the −COOH group transfers a proton to the −NH2 group intramolecularly, giving H3N+−CHR−COO. Because it carries both an ionisable acidic group (−NH3+) and an ionisable basic group (−COO), an amino acid is amphoteric — it reacts with added acid (−COO accepts H+) and with added base (−NH3+ loses H+). The pH at which an amino acid carries no net charge (exists purely as the zwitterion) is its isoelectric point.

Two amino acids join through a peptide linkage (an amide bond, −CO−NH−), formed between the −COOH of one molecule and the −NH2 of another with loss of a water molecule. Chains of amino acid residues are called dipeptides, tripeptides, oligopeptides or polypeptides depending on length; a polypeptide with more than about 100 amino acid residues (molecular mass greater than 10,000) is called a protein.

The structure of a protein is described at four levels:

  • Primary structure — the exact sequence of amino acids in the polypeptide chain(s); even a single change here can alter the protein's function.
  • Secondary structure — the local shape the chain adopts through hydrogen bonding between the C=O and N−H of peptide linkages: the coiled α-helix or the extended, zig-zag β-pleated sheet.
  • Tertiary structure — the overall three-dimensional folding of the whole chain, giving two broad classes of protein: fibrous proteins (long thread-like molecules held by hydrogen/disulphide bonds running parallel to each other, insoluble in water, e.g. keratin of hair, myosin of muscle) and globular proteins (chains fold into a compact, roughly spherical shape, generally water-soluble, e.g. insulin, albumins, most enzymes).
  • Quaternary structure — the spatial arrangement of two or more polypeptide subunits relative to each other, seen in proteins made of more than one chain, e.g. haemoglobin, which is built from four polypeptide subunits.

Denaturation occurs when a protein's native (biologically active) conformation is disturbed by a change of temperature or pH: the hydrogen bonds and other weak interactions that hold the secondary and tertiary structure are broken, the globular protein unfolds into a random coil, and its biological activity (e.g. enzymic action) is lost — even though the primary structure (the peptide-bonded amino acid sequence) remains unchanged. Coagulation of egg white on boiling and curdling of milk are everyday examples of denaturation.

Worked example: A tripeptide is formed by condensing three amino acid molecules. How many peptide bonds are formed and how many water molecules are eliminated?
Joining n amino acid residues into one chain requires (n − 1) peptide-bond-forming condensations, each releasing one H2O.
For n = 3: peptide bonds = 3 − 1 = 2, and water molecules eliminated = 2.

Zwitterion of an amino acid H₃N⁺−CHR−COO⁻ Dipolar form in which amino acids exist in the solid state and in solution near the isoelectric point.
Peptide bond formation R−COOH + H₂N−R′ → R−CO−NH−R′ + H₂O Condensation (loss of water) between the -COOH of one amino acid and the -NH2 of the next.
Peptide bonds in a chain of n residues bonds = n − 1 Each additional residue adds exactly one peptide bond.
Remember
  • Proteins are built from alpha-amino acids, R-CH(NH2)-COOH, classified as acidic, basic or neutral, and as essential or non-essential.
  • Amino acids exist as zwitterions (H3N+-CHR-COO-) and are amphoteric, reacting with both acids and bases.
  • The peptide (amide) bond -CO-NH- links amino acid residues; a chain of more than 100 residues is called a protein.
  • Primary structure = amino acid sequence; secondary = alpha-helix/beta-sheet; tertiary = overall 3-D folding (fibrous vs globular); quaternary = arrangement of multiple subunits.
  • Denaturation disrupts secondary/tertiary structure (loses biological activity) while leaving the primary sequence of peptide bonds intact.

Enzymes: Biological Catalysts

Quick answer Enzymes are globular proteins that catalyse biochemical reactions with remarkable specificity and efficiency by binding the substrate at a precisely shaped active site.

Enzymes are globular proteins that act as biocatalysts, speeding up the biochemical reactions occurring inside living cells without themselves being consumed. Compared with ordinary chemical catalysts, enzymes are extraordinarily efficient (they can increase reaction rates by many orders of magnitude under the mild conditions of temperature and pH found in the body) and extraordinarily specific — a given enzyme generally catalyses only one reaction, or a small set of closely related reactions, of one particular substrate.

This specificity is explained by the shape of the enzyme molecule. Each enzyme has a small three-dimensional pocket or groove, the active site, whose shape and arrangement of chemical groups exactly complements the shape of its substrate — much as a key fits a particular lock (the lock-and-key model). The substrate binds non-covalently at the active site to form a transient enzyme–substrate (ES) complex; within this complex the reaction is catalysed and proceeds far more readily than it would in free solution; the products are then released and the free enzyme is regenerated unchanged, ready to bind another substrate molecule.

The following simple sequence summarises enzyme action:

  1. Substrate molecule diffuses to and binds at the active site of the free enzyme, forming the enzyme–substrate complex.
  2. The bound substrate is converted to product(s) while still held at the active site (the specific chemical environment of the active site lowers the activation energy of the reaction).
  3. The product(s) are released from the active site.
  4. The unchanged enzyme is free to bind a fresh substrate molecule and repeat the cycle.

Enzyme activity is sensitive to conditions: every enzyme has an optimum temperature (about 37°C for most human enzymes) at which it works fastest — activity falls off at lower temperatures because molecular motion (and hence collision with substrate) slows down, and it falls sharply above the optimum because heat denatures the enzyme's tertiary structure and destroys its active site. Similarly, each enzyme has an optimum pH, since extremes of pH disturb the ionisation of groups at the active site (and can also denature the protein). Because enzymes are proteins, anything that denatures a protein — heat, extremes of pH, heavy-metal ions — will also abolish enzyme activity; even a very small change in the substrate's structure can prevent it from being recognised at the active site at all.

Remember
  • Enzymes are globular proteins acting as highly efficient, highly specific biocatalysts.
  • The lock-and-key model: substrate binds the enzyme's active site to form an enzyme-substrate (ES) complex before conversion to product.
  • Each enzyme has an optimum temperature (about 37 degree C in humans) and optimum pH; activity falls off away from these, and denaturation destroys activity irreversibly.
  • Because the active site has a precise shape, even small changes in substrate structure can prevent enzyme binding (high specificity).

Vitamins: Classification and Deficiency Diseases

Quick answer Vitamins are organic micronutrients the body cannot make in sufficient amounts; they are grouped as fat-soluble or water-soluble, and each has a characteristic deficiency disease.

Vitamins are organic compounds required in the diet in small amounts to maintain normal growth, health and metabolic function. Except for small amounts of a few (such as vitamin D, made in the skin on exposure to sunlight, and vitamin K and some B vitamins made by intestinal bacteria), the body cannot synthesise them, so they must be supplied through food. A deficiency of a specific vitamin (avitaminosis) produces a specific, recognisable disease, and both deficiency and (for some vitamins) excess can be harmful.

Vitamins are classified by solubility into two groups, which also governs how the body handles them:

  • Fat-soluble vitamins (A, D, E, K) dissolve in fats and oils, are absorbed along with dietary fat, and can be stored in appreciable amounts in the liver and adipose (fatty) tissue.
  • Water-soluble vitamins (the B-group and vitamin C) dissolve in the aqueous fluids of the body; because they are not stored to any significant extent and are continuously lost in urine, they must be supplied regularly through the diet.

Important vitamins, dietary sources and their deficiency diseases:

  • Vitamin A (retinol) — fish liver oil, carrots, milk, butter — deficiency causes xerophthalmia (hardening of the cornea) and night blindness.
  • Vitamin B1 (thiamine) — yeast, cereals, milk — deficiency causes beriberi (loss of appetite, weak muscles, enlarged heart).
  • Vitamin B2 (riboflavin) — milk, eggs, green vegetables — deficiency causes cheilosis (cracking at the corners of the mouth), digestive disorders and a burning sensation of the skin.
  • Vitamin B6 (pyridoxine) — meat, fish, egg yolk — deficiency can cause convulsions.
  • Vitamin B12 (cyanocobalamin) — meat, fish, curd — deficiency causes pernicious anaemia (the only water-soluble vitamin stored in the body, mainly in the liver, to a useful extent).
  • Vitamin C (ascorbic acid) — citrus fruits, amla, green leafy vegetables — deficiency causes scurvy (bleeding and swollen gums, slow wound healing).
  • Vitamin D (calciferol) — sunlight on skin, fish, egg yolk, milk — deficiency causes rickets in children (soft, deformed bones) and osteomalacia in adults.
  • Vitamin E (tocopherol) — vegetable oils, seeds — deficiency increases fragility of red blood cells and can cause muscular weakness; it also functions as a natural antioxidant.
  • Vitamin K — green leafy vegetables — deficiency increases the time taken for blood to clot (increased blood-clotting time).

Because water-soluble vitamins are excreted rather than stockpiled, deficiencies of B-group vitamins and vitamin C tend to appear relatively quickly if intake is inadequate, whereas fat-soluble vitamin deficiencies typically take longer to develop because of the body's stores — but excess fat-soluble vitamin intake can accumulate to toxic levels, which is far less likely with water-soluble vitamins.

Remember
  • Vitamins are essential organic micronutrients that the body largely cannot synthesise; deficiency causes a specific disease.
  • Fat-soluble vitamins (A, D, E, K) can be stored in liver/fat tissue; water-soluble vitamins (B-group, C) are not stored and are needed regularly.
  • Vitamin A deficiency: night blindness/xerophthalmia; Vitamin C deficiency: scurvy; Vitamin D deficiency: rickets/osteomalacia.
  • Vitamin B1 deficiency: beriberi; Vitamin B12 deficiency: pernicious anaemia; Vitamin K deficiency: increased blood-clotting time.
  • Excess of fat-soluble vitamins can accumulate to toxic levels; this is much less common with water-soluble vitamins.

Nucleic Acids: DNA and RNA

Quick answer Nucleic acids are polymers of nucleotides that store and transmit genetic information; DNA's double helix relies on complementary base pairing while RNA carries out protein synthesis.

Nucleic acids are polymers found mainly in the nucleus of the cell that carry genetic information and transmit it from one generation to the next; the two types are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).

The repeating unit of a nucleic acid is the nucleotide, built from three components: a pentose sugar (2-deoxy-D-ribose in DNA, D-ribose in RNA), a nitrogenous base, and a phosphate group. A base joined to the sugar alone (no phosphate) is called a nucleoside; adding a phosphate group gives a nucleotide. The nitrogenous bases are of two kinds: purines — adenine (A) and guanine (G) — and pyrimidines — cytosine (C), and either thymine (T, found only in DNA) or uracil (U, found only in RNA, replacing thymine).

Within a nucleotide, the base is attached to C1′ of the sugar by a β-N-glycosidic linkage, and the phosphate is attached to C5′ of the sugar by a phosphoester linkage. Successive nucleotides are then joined by a phosphodiester linkage between the 3′-OH of one sugar and the 5′-phosphate of the next, building up the alternating sugar–phosphate backbone of a polynucleotide chain, with the nitrogenous bases projecting from this backbone.

DNA normally exists as the famous double helix (Watson–Crick model): two polynucleotide chains coiled around a common axis, running in opposite (antiparallel) directions and held together by hydrogen bonds between specific, complementary pairs of bases on the two strands — adenine always pairs with thymine (2 hydrogen bonds) and guanine always pairs with cytosine (3 hydrogen bonds). This complementary base pairing means the two strands are not identical but exactly complementary, which is what allows DNA to be faithfully copied (replicated) and is the molecular basis of heredity.

RNA, in contrast, is generally single-stranded and occurs in three functional types, all involved in protein synthesis: messenger RNA (m-RNA) carries the genetic code from DNA to the site of protein synthesis; ribosomal RNA (r-RNA) forms part of the ribosome; and transfer RNA (t-RNA) carries specific amino acids to the ribosome for assembly into the growing protein chain.

Worked example (Chargaff's rule): A sample of double-stranded DNA contains 20% cytosine. Find the percentage of adenine, thymine and guanine.
By the base-pairing rule, %C = %G, so %G = 20% as well; hence %C + %G = 40%.
The remaining bases, adenine and thymine, must together make up 100% − 40% = 60%, and since %A = %T, each is 60% ÷ 2 = 30%.
So: %A = 30%, %T = 30%, %G = 20%, %C = 20% (total = 100%).

Nucleotide composition Nucleotide = Base + Sugar + Phosphate A nucleoside is just Base + Sugar (no phosphate).
Base pairing (Chargaff's rule) %A = %T ; %G = %C A pairs with T via 2 H-bonds; G pairs with C via 3 H-bonds, in double-stranded DNA.
Remember
  • A nucleotide = nitrogenous base + pentose sugar + phosphate group; a nucleoside is base + sugar only (no phosphate).
  • DNA sugar is 2-deoxy-D-ribose and its pyrimidine base is thymine; RNA sugar is D-ribose and its pyrimidine base is uracil (replacing thymine).
  • Successive nucleotides are linked by 3'-5' phosphodiester bonds forming the sugar-phosphate backbone.
  • DNA is an antiparallel double helix held by base pairing: A=T (2 H-bonds), G=C (3 H-bonds) - Chargaff's rule.
  • RNA is usually single-stranded; its three types (m-RNA, r-RNA, t-RNA) carry out protein synthesis, while DNA stores and transmits genetic information.

Key facts & terms

Every formula in this chapter, in one place — screenshot it before your exam.

C₆H₁₂O₆
Molecular formula of glucose/fructoseM = 180 g/mol
Cₙ(H₂O)ₘ
General carbohydrate formula
N = 2ⁿ
Number of stereoisomers
C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆
Sucrose hydrolysis (invert sugar formation)
M = 342
Molar mass of sucroseg/mol
H₃N⁺−CHR−COO⁻
Zwitterion of an amino acid
R−COOH + H₂N−R′ → R−CO−NH−R′ + H₂O
Peptide bond formation
bonds = n − 1
Peptide bonds in a chain of n residues
Nucleotide = Base + Sugar + Phosphate
Nucleotide composition
%A = %T ; %G = %C
Base pairing (Chargaff's rule)

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Carbohydrates easy

Glucose belongs to which class of carbohydrate based on its carbonyl group and carbon count?

Q2 Carbohydrates easy

Which of the following is a non-reducing sugar?

Q3 Vitamins easy

Deficiency of which vitamin causes scurvy?

Q4 Proteins medium

The bond that links two amino acid residues together in a protein chain is called a

Q5 Carbohydrates medium

The alpha and beta anomers of D-glucose differ in configuration at which carbon atom?

Q6 Nucleic acids medium

Which nitrogenous base is present in RNA but not found in DNA?

Q7 Proteins medium

During denaturation of a globular protein, which level(s) of structure are disrupted while the primary structure remains intact?

Q8 Carbohydrates medium

Amylopectin differs from amylose mainly because amylopectin

Q9 Carbohydrates hard

Complete hydrolysis of 17.1 g of sucrose (molar mass 342 g/mol) is carried out. What is the total combined mass of glucose and fructose produced?

Q10 Nucleic acids hard

A double-stranded DNA sample contains 24% adenine. What percentage of its bases is guanine?

Q11 Enzymes hard

In the lock-and-key model of enzyme action, the substrate binds to the enzyme at the

Q12 Carbohydrates hard

The open-chain form of an aldohexose has four chiral carbon atoms. How many stereoisomers (including D-glucose) are possible for this constitution?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 What are reducing sugars? Give an example each of a reducing and a non-reducing disaccharide.Carbohydrates

Carbohydrates that reduce Fehling's solution and Tollens' reagent (i.e. possess a free aldehyde group, or a free anomeric -OH that can open to give a free carbonyl group in solution) are called reducing sugars. All monosaccharides (both aldoses and ketoses) are reducing sugars.

Among disaccharides: maltose (two alpha-D-glucose units joined C1-C4, one anomeric carbon free) is a reducing disaccharide, while sucrose (glucose joined to fructose through both anomeric carbons, C1-C2) is a non-reducing disaccharide because no free anomeric -OH/-CHO remains.

2 What is the basic structural difference between starch and cellulose?Carbohydrates

Starch is a polymer of alpha-D-glucose units. It consists of amylose (an unbranched chain held together purely by alpha-1,4-glycosidic linkages, coiled into a helix) and amylopectin (a branched chain with alpha-1,4 linkages along the chain and alpha-1,6 linkages at the branch points).

Cellulose is a linear, completely unbranched polymer of beta-D-glucose units joined only by beta-1,4-glycosidic linkages.

So the basic structural difference is the configuration of the glycosidic linkage: alpha in starch versus beta in cellulose (starch may also be branched, while cellulose is always linear). This difference is why humans can digest starch (using alpha-amylase) but cannot digest cellulose, lacking an enzyme that hydrolyses beta-linkages.

3 Amino acids show amphoteric behaviour. Explain why.Proteins

Every amino acid contains an acidic group (-COOH) and a basic group (-NH2) in the same molecule. In aqueous solution, an intramolecular proton transfer occurs from the -COOH group to the -NH2 group, giving a dipolar ion (zwitterion): H3N+-CHR-COO-.

Because the zwitterion carries both a protonated amino group (-NH3+, which can donate H+ and thus act as an acid towards an added base) and a deprotonated carboxylate group (-COO-, which can accept H+ and thus act as a base towards an added acid), the amino acid can react with both acids and bases. This dual reactivity is called amphoteric behaviour.

4 What is meant by denaturation of a protein? Give an example.Proteins

Denaturation is the loss of a protein's biological activity that occurs when its native, folded (secondary and tertiary) structure is disrupted by a change in temperature or pH.

In the native state, a protein (especially a globular protein) is held in a specific three-dimensional shape by hydrogen bonds and other weak interactions. When subjected to heat or strong changes in pH, these weak interactions break, the polypeptide chain unfolds into a random coil, and the protein loses its characteristic biological activity — even though the primary structure (the sequence of amino acids joined by peptide bonds) remains unchanged.

Example: The coagulation of egg white (albumin) on boiling an egg is denaturation — the soluble, folded albumin unfolds and aggregates into an insoluble solid mass. Curdling of milk into curd is another example.

5 Why can vitamin C not be stored in our body?Vitamins

Vitamin C (ascorbic acid) is a water-soluble vitamin. Water-soluble vitamins dissolve in the aqueous fluids of the body rather than in fat, so they are not deposited in fatty tissue or the liver the way fat-soluble vitamins (A, D, E, K) are. Instead, any excess vitamin C is continuously excreted from the body in the urine.

Because the body cannot build up a reserve of it, vitamin C must be supplied regularly through the diet (citrus fruits, amla, green leafy vegetables); if intake is inadequate for even a short period, deficiency symptoms (scurvy) can appear.

6 What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?Nucleic acids

A nucleotide is made up of a nitrogenous base, a pentose sugar and a phosphate group linked together. On complete hydrolysis, a DNA nucleotide containing thymine breaks down into its three components:

  • 2-deoxy-D-ribose (the pentose sugar of DNA)
  • Thymine (the pyrimidine nitrogenous base)
  • Phosphoric acid (H3PO4)

(Partial hydrolysis, breaking only the phosphoester bond, would instead give the corresponding nucleoside — thymidine, i.e. thymine joined to 2-deoxy-D-ribose — plus phosphoric acid.)

Previous-year board questions 4

Q1 What is a peptide linkage? Write the equation for the formation of a peptide bond between two molecules of glycine. 2023 2 marks

A peptide linkage is an amide bond, -CO-NH-, formed between the -COOH group of one amino acid molecule and the -NH2 group of another amino acid molecule, with the elimination of one molecule of water. It is this linkage that joins amino acid residues together to build peptides and proteins.

Formation of the dipeptide glycylglycine from two molecules of glycine:

H2N-CH2-COOH + H2N-CH2-COOH → H2N-CH2-CO-NH-CH2-COOH + H2O

The product, glycylglycine, contains one peptide (-CO-NH-) linkage.

Q2 Define the term 'invert sugar'. Name the two monosaccharide products obtained on complete hydrolysis of sucrose and state the type of glycosidic linkage originally present between them. 2022 3 marks

Sucrose, on hydrolysis with dilute acid or the enzyme invertase, gives an equimolar mixture of D-glucose and D-fructose.

Sucrose itself is dextrorotatory (specific rotation +66.5°), but the resulting glucose-fructose mixture is laevorotatory (specific rotation −39.9°), because the large negative optical rotation of fructose outweighs the positive rotation of glucose. This reversal in the sign of optical rotation on hydrolysis is called inversion, and the resulting equimolar mixture of glucose and fructose is called invert sugar.

In sucrose, C1 of alpha-D-glucose is joined to C2 of beta-D-fructose by an alpha,beta-1,2-glycosidic linkage; since this linkage involves both anomeric carbons, sucrose has no free anomeric -OH and is a non-reducing sugar.

Q3 How many chiral (asymmetric) carbon atoms are present in the open-chain (Fischer projection) structure of D-(+)-glucose? Hence calculate the total number of stereoisomers possible for an aldohexose of this constitution. 2024 4 marks

The open-chain structure of D-(+)-glucose is:

CHO-CHOH-CHOH-CHOH-CHOH-CH2OH (C1 to C6)

Here C1 is the aldehyde carbon and C6 is -CH2OH; neither is attached to four different groups, so neither is chiral. The four middle carbons, C2, C3, C4 and C5, are each bonded to four different groups (H, OH, and two different chain fragments), so each is a chiral centre.

Number of chiral carbon atoms, n = 4.

Total number of possible stereoisomers, N = 2n = 24 = 16.

These 16 stereoisomers comprise 8 aldohexoses of the D-series and 8 of the L-series; D-(+)-glucose is one specific member of the D-series, distinguished by having the -OH groups at C2, C4 and C5 on the right and at C3 on the left in the Fischer projection.

Q4 Differentiate between DNA and RNA on the basis of (i) the pentose sugar present, (ii) the pyrimidine base(s) present, and (iii) their principal biological function. 2023 3 marks
  • Sugar: DNA contains 2-deoxy-D-ribose as its pentose sugar, while RNA contains D-ribose (which has an extra -OH at C2' compared to deoxyribose).
  • Pyrimidine base: DNA contains cytosine and thymine as its pyrimidine bases; RNA contains cytosine and uracil (uracil replaces thymine in RNA).
  • Function: DNA is the storehouse of genetic information and is responsible for transmitting hereditary characters from one generation to the next (via replication); RNA (as m-RNA, r-RNA and t-RNA) mainly carries out protein synthesis, using the genetic information copied from DNA.

(DNA also typically occurs as a double-stranded double helix, while RNA is generally single-stranded.)

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