Class 12 · Chemistry · Chapter 10
Biomolecules
Biomolecules covers the molecules living systems are built from: carbohydrates (glucose and its open-chain and ring structures, sucrose, maltose, lactose, starch, cellulose, glycogen), proteins (amino acids, the peptide bond, four levels of structure, denaturation, enzymes), vitamins, nucleic acids and hormones. It is a memory chapter with no calculations, and the board repeats the same facts every year, so an organised revision earns near-full marks.
Before you startWhat the board asks from this chapter
Biomolecules is Unit 10 of the CBSE syllabus and carries 7 marks in the 70-mark theory paper. Typically one or two 1-mark items, one 2- or 3-mark question on glucose reactions, sucrose, vitamins or nucleic acids, and sometimes a case-based question on carbohydrates or proteins.
| Question type | Marks | Usually asked |
|---|---|---|
| MCQ | 1 | Reducing versus non-reducing sugars, product of glucose with HI, Br2 water or HNO3, sugar in DNA, base found only in RNA, vitamin deficiency diseases, linkage in maltose or lactose. |
| Assertion–Reason | 1 | Sucrose being non-reducing, glucose failing the Schiff and NaHSO3 tests, amino acids as zwitter ions, water-soluble vitamins not being stored, complementary DNA strands. |
| Very short answer | 2 | Hydrolysis products of sucrose, maltose or lactose; two reactions of glucose the open chain cannot explain; nucleoside versus nucleotide; define glycosidic or peptide linkage; anomers. |
| Short answer | 3 | Three-part questions mixing a vitamin deficiency, a glucose reaction and a definition; starch versus cellulose; essential and non-essential amino acids; DNA versus RNA; denaturation. |
| Long answer | 5 | Rare: evidence for the open-chain and cyclic structures of glucose, or the four levels of protein structure with denaturation, or vitamins combined with nucleic acids. |
| Case-based | 4 | A passage on carbohydrates (invert sugar, reducing sugars, starch and cellulose) or on proteins and nucleic acids, followed by four short parts. |
Hot topics, in order
- Reactions of glucose: HI, Br2 water, HNO3, NH2OH, HCN, acetic anhydride, and what each proves
- Evidence for the cyclic (pyranose) structure of glucose; anomers α and β
- Sucrose, maltose and lactose: hydrolysis products, glycosidic linkage, reducing or not, invert sugar
- Starch (amylose and amylopectin) versus cellulose versus glycogen
- Amino acids: zwitter ion, essential and non-essential, acidic, basic, neutral
- Peptide linkage, the four levels of protein structure, denaturation
- Vitamins: fat-soluble and water-soluble, sources and deficiency diseases
- Nucleic acids: nucleoside versus nucleotide, DNA versus RNA, base pairing
The whole chapter, in orderNotes
01 10.1 Carbohydrates and 10.1.1 Classification of Carbohydrates PYQ 2010 · 2014
NCERT defines carbohydrates as optically active polyhydroxy aldehydes or ketones, or compounds that give these on hydrolysis. Most fit Cx(H2O)y, the origin of the old name 'hydrates of carbon', but the formula is not the definition: acetic acid C2H4O2 fits it and is not a carbohydrate, while rhamnose C6H12O5 is one and does not. Carbohydrates are also called saccharides; the sweet, water-soluble ones such as glucose and sucrose are sugars.
Two classifications matter. By hydrolysis: monosaccharides cannot be hydrolysed to anything simpler (glucose, fructose, ribose; about 20 are known); oligosaccharides give two to ten units, the important case being disaccharides such as sucrose, maltose and lactose; polysaccharides give a large number of units (starch, cellulose, glycogen), are not sweet and are called non-sugars.
By reducing behaviour: a reducing sugar reduces Fehling's solution and Tollens' reagent because it has a free aldehyde or ketone group (a free hemiacetal in the ring form). All monosaccharides are reducing. Among disaccharides, maltose and lactose are reducing while sucrose is not, because both its carbonyl carbons are tied up in the glycosidic bond.
02 10.1.2 Monosaccharides: preparation and open-chain structure of glucose PYQ 2012 · 2014
Monosaccharides are classified by the carbonyl group (aldose or ketose) and by carbon count (triose to heptose). Glucose is an aldohexose, fructose a ketohexose, ribose an aldopentose.
Preparation of glucose. From sucrose: boil with dilute HCl or H2SO4 in alcoholic solution to get glucose and fructose. From starch, the commercial method: boil with dilute H2SO4 at 393 K under 2–3 atm.
Evidence for the open-chain structure. Learn it as a numbered list; the examiner ticks each point. (1) Molecular formula C6H12O6. (2) Prolonged heating with HI gives n-hexane, so all six carbons are in a straight chain. (3) It forms an oxime with hydroxylamine and a cyanohydrin with HCN, so a carbonyl group is present. (4) Bromine water, a mild oxidant, oxidises it to gluconic acid, a six-carbon monocarboxylic acid, so the carbonyl is an aldehyde. (5) Acetic anhydride gives a stable pentaacetate, so five –OH groups are present on five different carbons. (6) Glucose and gluconic acid are both oxidised by nitric acid to saccharic acid, a dicarboxylic acid, showing a primary alcoholic –CH2OH group.
Fischer fixed the spatial arrangement. Glucose is D-(+)-glucose: (+) because it is dextrorotatory, D because the –OH on the lowest asymmetric carbon (C5) is on the right in the Fischer projection, as in (+)-glyceraldehyde. D and L say nothing about the sign of rotation.
03 10.1.2 (continued) Cyclic structure of glucose PYQ 2009 · 2012 · 2014 · 2016
The open chain cannot explain three facts, and 'two reactions of glucose not explained by its open-chain structure' is a favourite 2-mark question. (1) Despite the aldehyde group, glucose does not give the 2,4-DNP test or Schiff's test and does not form the hydrogensulphite addition product with NaHSO3. (2) The pentaacetate of glucose does not react with hydroxylamine, so it has no free –CHO group. (3) Glucose exists in two crystalline forms: the α-form crystallises from a concentrated solution at 303 K (melting point 419 K) and the β-form from a hot saturated aqueous solution at 371 K (melting point 423 K).
The explanation: the –OH on C5 adds across the C1 aldehyde to form a six-membered cyclic hemiacetal. C1 becomes a new asymmetric carbon, the anomeric carbon, so two ring forms exist differing only in the configuration at C1: α-D-glucose (C1 –OH below the ring in the Haworth projection) and β-D-glucose (C1 –OH above). Such a pair is called anomers. The ring has five carbons and one oxygen like pyran, so the forms are α-D-(+)-glucopyranose and β-D-(+)-glucopyranose. In solution the anomers are in equilibrium through a little open-chain form, which is why glucose still reduces Fehling's solution yet fails tests needing a stable free aldehyde.
04 10.1.2 (continued) Structure of fructose
Fructose, C6H12O6, is a ketohexose found with glucose in fruit and honey and as one half of sucrose. Its carbonyl is a ketone at C2, with –CH2OH groups at both ends. It has the D-configuration because the –OH on the lowest asymmetric carbon (C5) is on the right in the Fischer projection, yet it is laevorotatory, so it is written D-(−)-fructose. This is the standard example that D/L (configuration) and (+)/(−) (direction of rotation) are independent labels.
Like glucose, fructose forms cyclic hemiketals. The –OH on C5 adds to the C2 ketone, and the ring has four carbons and one oxygen, five-membered like furan. So the cyclic forms are furanose forms, α-D-(−)-fructofuranose and β-D-(−)-fructofuranose, differing at the anomeric carbon, C2. In the Haworth drawing C2 carries both the anomeric –OH and the C1 –CH2OH. Fructose is a reducing sugar although it is a ketose, because its α-hydroxy ketone is oxidised in the alkaline medium of Fehling's and Tollens' tests.
05 10.1.3 Disaccharides PYQ 2009 · 2013 · 2014 · 2016
A disaccharide is two monosaccharide units joined by an oxide linkage formed by loss of water between two –OH groups; this is the glycosidic linkage. Dilute acid or an enzyme hydrolyses it back to the monosaccharides.
Sucrose (cane sugar) gives one D-(+)-glucose and one D-(−)-fructose. The glycosidic bond joins C1 of α-D-glucose to C2 of β-D-fructose. Both reducing groups are used, so sucrose is non-reducing. Sucrose is dextrorotatory, but the hydrolysis mixture is laevorotatory because the laevorotation of fructose (−92.4°) exceeds the dextrorotation of glucose (+52.5°). The sign inverts, so the product is called invert sugar and the enzyme invertase (sucrase).
Maltose gives two α-D-glucose molecules; C1 of one unit is linked to C4 of the other. C1 of the second unit is free and can open to the aldehyde in solution, so maltose is reducing.
Lactose (milk sugar) gives β-D-galactose and β-D-glucose; C1 of galactose is linked to C4 of glucose. The C1 of the glucose unit is free, so lactose is reducing.
Put the three products and three linkages in one table; this is the most repeated 1- and 2-mark material in the chapter.
06 10.1.4 Polysaccharides PYQ 2012 · 2014 · 2016
Polysaccharides are long chains of monosaccharides joined by glycosidic linkages, the storage and structural materials of living things.
Starch is the main storage polysaccharide of plants and our chief dietary carbohydrate (cereals, potatoes, pulses). It is a polymer of α-glucose with two components. Amylose, 15–20 per cent, is water-soluble, a long unbranched chain of 200–1000 α-D-(+)-glucose units held by C1–C4 glycosidic linkages. Amylopectin, 80–85 per cent, is water-insoluble and branched: C1–C4 linkages build the chain and C1–C6 linkages attach the branches.
Cellulose is the most abundant organic substance in the plant kingdom and the chief material of plant cell walls. It is a straight-chain polysaccharide of β-D-glucose units joined by C1–C4 glycosidic linkages. That one letter, β instead of α, is 'the basic structural difference between starch and cellulose' and is why we digest starch but not cellulose.
Glycogen is the stored carbohydrate of animals, hence 'animal starch'. It occurs in the liver, muscles and brain, and in yeast and fungi. Its structure resembles amylopectin but is even more branched. When the body needs glucose, enzymes break glycogen down to glucose.
07 10.1.5 Importance of Carbohydrates PYQ 2012
This short section is asked as 'write two main functions of carbohydrates in plants' or 'state two functions of carbohydrates'. Know the list and pick what the question wants.
- Carbohydrates form the major portion of our food; the starch of cereals, pulses and potatoes is our main energy source, and honey (mostly glucose and fructose) has been used for centuries as an instant source of energy.
- Storage: plants store carbohydrate as starch and animals as glycogen, both broken down to glucose when energy is needed.
- Structure: cellulose builds the cell walls of plants and bacteria; wood is mostly cellulose and cotton is nearly pure cellulose, the raw material for paper and rayon.
- Carbohydrates are needed for the biosynthesis of fats and proteins.
- Two pentoses, D-ribose and 2-deoxy-D-ribose, are constituents of RNA and DNA.
- Plants make carbohydrates by photosynthesis, so they carry the energy of sunlight into every food chain.
The two functions 'in plants' the examiner expects are starch as the stored form of food and cellulose as the structural material of the cell wall.
08 10.2 Proteins, 10.2.1 Amino Acids and 10.2.2 Classification of Amino Acids PYQ 2009 · 2010 · 2012 · 2014
Proteins are the most abundant biomolecules in living systems; the name comes from the Greek proteios, 'of prime importance'. Chemically they are polymers of α-amino acids.
An amino acid has an amino group and a carboxyl group. In the α-amino acids of proteins the –NH2 sits on the α-carbon, next to –COOH, giving R–CH(NH2)–COOH. Twenty amino acids occur in proteins, differing only in R, each with a trivial name and three-letter and one-letter codes, e.g. glycine (Gly, G, R = H) and alanine (Ala, A, R = CH3).
Classification. By the relative number of amino and carboxyl groups: neutral when equal (glycine, alanine), acidic when carboxyl groups are more (aspartic acid, glutamic acid), basic when amino groups are more (lysine, arginine). By origin: essential amino acids cannot be synthesised in the body and must come from the diet (valine, leucine, isoleucine, phenylalanine, tryptophan, methionine, threonine, lysine, histidine, arginine); non-essential ones are made in the body (glycine, alanine, serine and others).
Properties. Amino acids are colourless, crystalline, water-soluble, high-melting solids that behave like salts. In aqueous solution the –COOH loses a proton and the –NH2 accepts it, giving the dipolar zwitter ion H3N+–CHR–COO−. This makes them amphoteric: –COO− accepts a proton from acids, –NH3+ donates one to bases. Except glycine, all α-amino acids are optically active because the α-carbon is asymmetric; natural ones have the L-configuration, drawn with –NH2 on the left.
09 10.2.3 Structure of Proteins PYQ 2009 · 2012 · 2014 · 2015 · 2016
Amino acids link through the peptide bond, an amide –CO–NH– formed when the –COOH of one amino acid condenses with the –NH2 of the next, losing water. Glycine and alanine give the dipeptide glycylalanine, Gly-Ala. Two amino acids make a dipeptide, three a tripeptide, many a polypeptide. A polypeptide with more than a hundred amino acid residues and a molecular mass above 10,000 u is called a protein.
By shape, proteins are fibrous or globular. In fibrous proteins the chains run parallel, held by hydrogen and disulphide bonds; they are water-insoluble: keratin (hair, wool, silk) and myosin (muscle). In globular proteins the chains coil into a spherical shape; they are usually water-soluble: insulin and albumins.
Four levels of structure. Primary: the sequence of amino acids; any change gives a different protein. Secondary: the local shape of the chain, held by hydrogen bonds between –C=O and –N–H of peptide bonds; the α-helix is a right-handed screw in which the –NH of each residue is hydrogen-bonded to the C=O of the adjacent turn, and the β-pleated sheet has chains stretched nearly to full length and laid side by side, held by intermolecular hydrogen bonds. Tertiary: the overall folding of the chain (further folding of the secondary structure) giving the fibrous or globular shape, stabilised by hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces. Quaternary: for proteins of two or more polypeptide chains, the spatial arrangement of the subunits with respect to one another.
10 10.2.4 Denaturation of Proteins PYQ 2010 · 2012 · 2013 · 2014
A protein found in a biological system with its own unique three-dimensional structure and biological activity is a native protein. When a native protein is subjected to a physical change such as temperature or a chemical change such as pH, the hydrogen bonds are disturbed, the globules unfold and the helix uncoils, and the protein loses its biological activity. This is denaturation.
The key marking point is what changes. Denaturation destroys the secondary and tertiary structures but leaves the primary structure intact: the amino acid sequence and the peptide bonds survive, only the folding is lost. Since the folding fits an enzyme to its substrate, a denatured enzyme is inactive.
Two NCERT examples appear in the paper. Boiling an egg coagulates the egg white: the soluble globular albumin unfolds into an insoluble fibrous mass, and the water in the egg is absorbed by the coagulated protein through hydrogen bonding. Milk curdles when lactobacillus bacteria produce lactic acid, which lowers the pH and denatures the milk protein.
Contrast with hydrolysis if a question hints at it: hydrolysis breaks the peptide bonds and destroys the primary structure; denaturation does not.
11 10.3 Enzymes and 10.3.1 Mechanism of Enzyme Action PYQ 2014 · 2016
Enzymes are biological catalysts. Almost all are globular proteins, and each is specific for one reaction or one type of reaction. Reactions that would need high temperature and strong acid in the laboratory run in the body at about 310 K and near-neutral pH because an enzyme is present. Enzymes are named after the compound they act on or the reaction they catalyse, usually ending in -ase: maltase hydrolyses maltose to glucose, and an enzyme that oxidises one substrate while reducing another is an oxidoreductase.
Mechanism. Like any catalyst, an enzyme lowers the activation energy of the reaction. Quote NCERT's figures: the activation energy for the acid hydrolysis of sucrose is 6.22 kJ mol−1, but only 2.15 kJ mol−1 when the hydrolysis is catalysed by the enzyme sucrase. The enzyme binds the substrate at its active site and provides an alternative path through a lower-energy transition state.
Because enzymes are proteins, denaturation applies: heat, extreme pH or heavy-metal ions unfold them and stop the catalysis. Diseases due to a missing enzyme are asked occasionally; phenylketonuria (lack of phenylalanine hydroxylase) and albinism (lack of tyrosinase) are the usual examples.
12 10.4 Vitamins and 10.4.1 Classification of Vitamins PYQ 2009 · 2012 · 2013 · 2014 · 2015 · 2016
Vitamins are organic compounds required in the diet in small amounts to perform specific biological functions for the normal maintenance of optimum growth and health. Most cannot be made by the body (vitamin D, formed in the skin in sunlight, is the exception NCERT flags), and both deficiency and excess cause disease. They are named A, B, C, D and so on, with the B group split into B1, B2, B6, B12 and others.
Classification is by solubility. Fat-soluble vitamins A, D, E and K are stored in the liver and adipose tissue. Water-soluble vitamins, the B group and C, must be supplied regularly in the diet because they are readily excreted in urine and cannot be stored, except vitamin B12, which is stored in the liver.
The table to reproduce: A (fish liver oil, carrots, butter, milk) – xerophthalmia and night blindness. B1 thiamine (yeast, milk, green vegetables, cereals) – beri beri. B2 riboflavin (milk, egg white, liver, kidney) – cheilosis, digestive disorders, burning sensation of the skin. B6 pyridoxine (yeast, milk, egg yolk, cereals, grams) – convulsions. B12 (meat, fish, egg, curd) – pernicious anaemia, red blood cells deficient in haemoglobin. C ascorbic acid (citrus fruits, amla, green leafy vegetables) – scurvy, bleeding gums. D (sunlight, fish, egg yolk) – rickets in children, osteomalacia in adults. E (vegetable oils such as wheat-germ and sunflower oil) – increased fragility of red blood cells and muscular weakness. K (green leafy vegetables) – increased blood clotting time; K is the vitamin responsible for coagulation of blood.
13 10.5 Nucleic Acids and 10.5.1 Chemical Composition PYQ 2009 · 2012 · 2013 · 2014 · 2016
Every cell nucleus has chromosomes made of proteins and nucleic acids. The nucleic acids carry heredity, and there are two: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Both are long-chain polymers of nucleotides, also called polynucleotides.
Composition. Complete hydrolysis of DNA or RNA gives a pentose sugar, phosphoric acid and nitrogen-containing heterocyclic bases. The sugar is β-D-2-deoxyribose in DNA and β-D-ribose in RNA; the difference is the –OH at C2, present in ribose and absent in deoxyribose. Purines, with two fused rings, are adenine (A) and guanine (G); pyrimidines, with one ring, are cytosine (C), thymine (T) and uracil (U). DNA contains A, G, C and T; RNA contains A, G, C and U. So thymine is found only in DNA, uracil only in RNA, and A, G, C are common to both.
Nucleoside and nucleotide. A base attached to C1' of the pentose is a nucleoside (sugar + base). A phosphoric acid unit joined at the 5'-OH of the sugar of a nucleoside gives a nucleotide (sugar + base + phosphate). Nucleotides join through phosphodiester linkages between the 5' carbon of one sugar and the 3' carbon of the next, so the backbone is sugar–phosphate–sugar–phosphate with a base on each sugar.
14 10.5.2 Structure of Nucleic Acids PYQ 2009 · 2012 · 2016
The primary structure of a nucleic acid is the sequence of nucleotides, the order of bases along the chain, which carries the genetic information.
The secondary structure of DNA is the double helix proposed by James Watson and Francis Crick in 1953. Two polynucleotide strands wind round each other in a right-handed helix, sugar–phosphate backbones outside and bases inside, held by hydrogen bonds between pairs of bases. The pairing is fixed: adenine with thymine through two hydrogen bonds (A=T) and guanine with cytosine through three (G≡C). So the two strands are not identical but complementary: wherever one has A the other has T, wherever one has G the other has C.
RNA is single-stranded. It sometimes folds back on itself so that complementary bases within one strand pair, but there is no rule linking the amounts of its bases; that is why hydrolysis of RNA gives no fixed ratio among A, G, C and U, while in DNA A equals T and G equals C. The three types are messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA), each with a different role in reading the DNA message and building proteins.
15 10.5.4 Biological Functions of Nucleic Acids PYQ 2012 · 2013
The two important functions of nucleic acids, which NCERT exercise 10.21 asks for, are heredity and protein synthesis.
DNA is the reserve of genetic information. It is the chemical basis of heredity and carries the information for the characteristics of an organism from one generation to the next. When a cell divides, the double helix unwinds and each strand acts as a template for a new complementary strand, so two identical DNA molecules result. This is replication, and it works only because the strands are complementary.
Protein synthesis. DNA does not make proteins directly. A message copied from a segment of DNA is carried by messenger RNA from the nucleus to the ribosome (transcription). At the ribosome, made largely of ribosomal RNA, the message is read three bases at a time and transfer RNA brings the amino acid matching each triplet; the amino acids are joined in that order (translation). So the base sequence of DNA fixes the amino acid sequence of every protein, and hence the enzymes and chemistry of the cell.
A sentence that earns marks: DNA is the same in every cell and stays constant through life, while RNA is made as needed and broken down after use.
16 10.6 Hormones PYQ 2012
Hormones are molecules that act as intermolecular messengers. They are produced by the endocrine glands and released directly into the blood stream, which carries them to the site of action. They control growth, metabolism, reproduction and the response to stress. The syllabus asks only an elementary idea, no structures.
By chemical nature there are three kinds. Steroids: estrogens, androgens, and the adrenal cortex hormones cortisol and aldosterone. Polypeptides: insulin and endorphins. Amino-acid derivatives: epinephrine, norepinephrine and thyroxine.
Insulin is a polypeptide made by the β-cells of the pancreas and released in response to a rapid rise in blood glucose; it lowers blood glucose, while glucagon raises it, and together they keep it steady. Epinephrine and norepinephrine mediate the response to external stimuli. Thyroxine, from the thyroid gland, is an iodinated derivative of tyrosine; too little iodine in the diet leads to hypothyroidism and enlargement of the thyroid gland (goitre), which is why table salt is iodised with sodium iodide.
Among steroids, the adrenal cortex makes glucocorticoids such as cortisol, which control carbohydrate metabolism, modulate inflammation and take part in the stress response, and mineralocorticoids such as aldosterone, which control the excretion of water and salt by the kidney. Sex hormones: testosterone (androgen, testes) governs male secondary characteristics; estradiol (estrogen, ovaries) governs female characteristics and the menstrual cycle; progesterone prepares the uterus for implantation of the fertilised egg.
VocabularyKey terms
Keep this on one pageFormula sheet
| Name | Expression | Notes |
|---|---|---|
| Glucose from sucrose | C12H22O11 + H2O → (H+) C6H12O6 (glucose) + C6H12O6 (fructose) | Dilute HCl or H2SO4 in alcoholic solution; the same hydrolysis gives invert sugar |
| Glucose from starch | (C6H10O5)n + nH2O → (H+, 393 K, 2–3 atm) nC6H12O6 | Commercial preparation |
| Glucose + HI | C6H12O6 + HI (prolonged heating) → n-hexane | Six carbons in a straight chain |
| Glucose + bromine water | HOCH2–(CHOH)4–CHO → HOCH2–(CHOH)4–COOH (gluconic acid) | Mild oxidation; the carbonyl is an aldehyde |
| Glucose + nitric acid | HOCH2–(CHOH)4–CHO → HOOC–(CHOH)4–COOH (saccharic acid) | Both ends oxidised; proves a –CH2OH group |
| Glucose + acetic anhydride, NH<sub>2</sub>OH, HCN | Glucose → pentaacetate (5 –OH groups); → oxime and → cyanohydrin (carbonyl group) | The pentaacetate gives no oxime: no free –CHO in the ring form |
| Invert sugar | Sucrose (+66.5°) → glucose (+52.5°) + fructose (−92.4°); mixture laevorotatory | Fructose's laevorotation outweighs glucose's dextrorotation |
| Peptide bond | R–CH(NH2)–COOH + H2N–CH(R')–COOH → R–CH(NH2)–CO–NH–CH(R')–COOH + H2O | Condensation with loss of water |
| Zwitter ion | R–CH(NH2)–COOH ⇌ R–CH(NH3+)–COO− | High melting point, solubility and amphoteric nature of amino acids |
| Enzyme activation energy | Ea (acid hydrolysis of sucrose) = 6.22 kJ mol−1; with sucrase = 2.15 kJ mol−1 | NCERT's example of an enzyme lowering activation energy |
| Base pairing in DNA | A = T (2 hydrogen bonds); G ≡ C (3 hydrogen bonds) | In RNA uracil replaces thymine |
Practise with a pencilDiagrams that earn marks
Fischer projection of D-(+)-glucose
Draw a vertical chain of six carbons with –CHO at the top (C1) and –CH2OH at the bottom (C6). Put –OH on the right at C2, left at C3, right at C4 and right at C5, with H on the opposite side each time. Label C1 to C6 and note that the –OH on C5 on the right gives the D-configuration.
Haworth structures of α- and β-D-(+)-glucopyranose
Draw a flat hexagon with the ring oxygen at the back right corner and C1 at the right. Put –OH below the ring at C1 for α and above for β; then –OH below at C2, above at C3, below at C4, and –CH2OH above the ring at C5. Draw both rings side by side, label C1 as the anomeric carbon and write 'anomers' between them.
Structure of sucrose
Draw a six-membered α-D-glucopyranose ring on the left and a five-membered β-D-fructofuranose ring on the right. Join C1 of glucose to C2 of fructose through one oxygen atom labelled 'glycosidic linkage (C1–C2)'. Note that both reducing groups are used, so sucrose is non-reducing.
Double helix of DNA
Draw two ribbons twisting round each other in a right-handed helix, labelled 'sugar–phosphate backbone'. Between them draw rungs labelled A–T and G–C with two dotted lines for A=T and three for G≡C, labelled 'hydrogen bonds'. Mark the strands 'complementary' and write 'Watson and Crick, 1953'.
TextbookNCERT questions the board keeps asking
10.3 How do you explain the absence of an aldehyde group in the pentaacetate of D-glucose?2 m
In the pentaacetate all five –OH groups, including the –OH at C1 of the cyclic hemiacetal, are acetylated. The ring can no longer open to give the free aldehyde, because opening needs a free –OH on the anomeric carbon. So the pentaacetate does not react with hydroxylamine; this is evidence for the cyclic structure of glucose.
10.4 The melting points and solubility in water of amino acids are generally higher than those of the corresponding halo acids. Explain.2 m
Amino acids exist as zwitter ions, H3N+–CHR–COO−, so molecules are held by strong electrostatic attractions like those in a salt, while halo acids are held only by weaker dipole–dipole forces. The ionic attraction raises the melting point, and the charged groups interact strongly with polar water, raising the solubility.
10.5 Where does the water present in the egg go after boiling the egg?1 m
On boiling, the globular protein of the egg white is denatured and coagulates into a fibrous, insoluble mass. The water present in the egg is absorbed by the coagulated protein through hydrogen bonding; it does not leave the egg.
10.6 Why cannot vitamin C be stored in our body?1 m
Vitamin C (ascorbic acid) is water-soluble. Water-soluble vitamins dissolve in body fluids and are readily excreted in urine, so they cannot be stored (except B12) and must be supplied regularly in the diet.
10.7 What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?1 m
A nucleotide is made of a sugar, a base and phosphoric acid. Complete hydrolysis of this DNA nucleotide gives β-D-2-deoxyribose, thymine and phosphoric acid.
10.8 When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact indicate about the structure of RNA?2 m
In DNA the strands are complementary, so A equals T and G equals C on hydrolysis. RNA shows no such relationship among A, G, C and U, so its bases are not paired with a complementary strand. RNA is therefore single-stranded.
10.8 What is the basic structural difference between starch and cellulose?2 m
Starch is a polymer of α-D-glucose: amylose has C1–C4 glycosidic linkages and amylopectin has C1–C4 linkages with C1–C6 branching. Cellulose is a straight-chain polymer of β-D-glucose joined by C1–C4 glycosidic linkages. The basic difference is the configuration of the glucose units: α in starch, β in cellulose.
10.9 What happens when D-glucose is treated with the following reagents? (i) HI (ii) Bromine water (iii) HNO33 m
(i) On prolonged heating with HI, glucose is reduced to n-hexane, CH3(CH2)4CH3, showing all six carbons in a straight chain.
(ii) Bromine water, a mild oxidising agent, oxidises the aldehyde group to carboxyl, giving gluconic acid, HOCH2–(CHOH)4–COOH.
(iii) Nitric acid, a strong oxidising agent, oxidises both –CHO and –CH2OH, giving saccharic acid, HOOC–(CHOH)4–COOH.
10.10 Enumerate the reactions of D-glucose which cannot be explained by its open-chain structure.3 m
(i) Despite having an aldehyde group, glucose does not give the 2,4-DNP test or Schiff's test and does not form an addition product with NaHSO3.
(ii) The pentaacetate of glucose does not react with hydroxylamine, showing the absence of a free –CHO group.
(iii) Glucose exists in two crystalline forms, α (melting point 419 K) and β (melting point 423 K).
All three are explained by the cyclic hemiacetal formed when the C5 –OH adds to the C1 aldehyde.
10.12 Define the following as related to proteins: (i) Peptide linkage (ii) Primary structure (iii) Denaturation.3 m
(i) Peptide linkage: the amide bond –CO–NH– formed between the –COOH group of one α-amino acid and the –NH2 group of another with loss of a water molecule.
(ii) Primary structure: the specific sequence in which the amino acids are linked in the polypeptide chain.
(iii) Denaturation: loss of the biological activity of a native protein when a physical change (temperature) or chemical change (pH) disturbs the hydrogen bonds, so the globule unfolds and the helix uncoils; secondary and tertiary structures are destroyed, the primary structure remains.
10.16 How do you explain the amphoteric behaviour of amino acids?2 m
In aqueous solution the –COOH group loses a proton and the –NH2 group gains it, so the amino acid exists as a zwitter ion, H3N+–CHR–COO−. The –COO− accepts a proton from an acid (acts as a base) and the –NH3+ donates a proton to a base (acts as an acid). Reacting with both, the amino acid is amphoteric.
10.22 What is the difference between a nucleoside and a nucleotide?2 m
A nucleoside has two units: a pentose sugar and a nitrogen base attached at its C1'. A nucleotide has three: the sugar, the base and a phosphoric acid group attached at the C5' –OH of the sugar. Nucleotide = nucleoside + phosphate; nucleotides are the monomers of nucleic acids.
10.23 The two strands of DNA are not identical but are complementary. Explain.2 m
The strands are held by hydrogen bonds between specific base pairs: adenine of one strand always pairs with thymine of the other (two hydrogen bonds) and guanine with cytosine (three). So the base sequence of one strand fixes that of the other: opposite A there is T, opposite G there is C. The sequences differ, but each is the complement of the other.
10.24 Write the important structural and functional differences between DNA and RNA.3 m
Structural: (i) DNA contains β-D-2-deoxyribose, RNA contains β-D-ribose. (ii) DNA has the bases A, G, C, T; RNA has A, G, C, U. (iii) DNA is double-stranded; RNA is single-stranded.
Functional: (i) DNA stores and transmits hereditary information and replicates itself. (ii) RNA (m-RNA, t-RNA, r-RNA) carries out protein synthesis on the instructions of DNA.
Previous years and frequently askedBoard questions with model answers
MCQ Which of the following is a non-reducing sugar?- Glucose
- Maltose
- Lactose
- Sucrose
Frequently asked1 m
MCQ The sugar present in DNA is:- β-D-ribose
- β-D-2-deoxyribose
- α-D-glucose
- D-fructose
Frequently asked1 m
MCQ On prolonged heating with HI, glucose gives:- n-hexane
- Gluconic acid
- Saccharic acid
- Glucose pentaacetate
Frequently asked1 m
MCQ The glycosidic linkage in maltose is between:- C1 of α-glucose and C2 of β-fructose
- C1 of one α-D-glucose and C4 of another α-D-glucose
- C1 of β-galactose and C4 of β-glucose
- C1 of α-glucose and C6 of another glucose
Frequently asked1 m
MCQ Deficiency of which vitamin causes pernicious anaemia?- Vitamin B1
- Vitamin B6
- Vitamin B12
- Vitamin K
Frequently asked1 m
MCQ Which of the following amino acids is optically inactive?- Alanine
- Glycine
- Valine
- Leucine
Frequently asked1 m
Assertion–Reason Assertion (A): Sucrose is a non-reducing sugar. Reason (R): The glycosidic linkage in sucrose is between C1 of α-D-glucose and C2 of β-D-fructose, so both reducing groups are involved in the bond.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): Glucose does not give Schiff's test and does not form an addition product with NaHSO3. Reason (R): Glucose contains a free aldehyde group in its stable form.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): Amino acids have high melting points and are soluble in water. Reason (R): Amino acids exist as zwitter ions.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Very short answer Name the products of hydrolysis of sucrose. Why is sucrose not a reducing sugar?CBSE 20092 m
Hydrolysis of sucrose gives one molecule of D-(+)-glucose and one of D-(−)-fructose. Sucrose is non-reducing because the glycosidic linkage is between C1 of α-D-glucose and C2 of β-D-fructose; both reducing carbons are involved in the bond, so no free aldehyde or ketone group is available to reduce Fehling's solution or Tollens' reagent.
Very short answer Write any two reactions of glucose which cannot be explained by the open-chain structure of the glucose molecule.CBSE 20122 m
(i) Glucose does not give the 2,4-DNP test or Schiff's test and does not form an addition product with NaHSO3, although it has an aldehyde group.
(ii) The pentaacetate of glucose does not react with hydroxylamine, showing there is no free –CHO group.
(Also acceptable: glucose exists in two crystalline forms, α and β, melting points 419 K and 423 K.)
Very short answer What is essentially the difference between the α-form of glucose and the β-form of glucose? Explain.CBSE 20092 m
Both are cyclic (pyranose) hemiacetal forms of D-glucose formed when the C5 –OH adds to the C1 aldehyde. They differ only in the configuration of the –OH at C1, the anomeric carbon: in α-D-glucose the C1 –OH is below the ring in the Haworth projection, in β-D-glucose it is above. Such isomers are called anomers. Their melting points differ: α 419 K, β 423 K.
Very short answer Write the main structural difference between DNA and RNA. Of the two bases, thymine and uracil, which one is present in DNA?CBSE 20122 m
DNA contains the sugar β-D-2-deoxyribose and is double-stranded; RNA contains β-D-ribose and is single-stranded. Thymine is present in DNA; RNA has uracil in its place.
Short answer (i) Write the names of the two monosaccharides obtained on hydrolysis of lactose. (ii) Why cannot vitamin C be stored in our body? (iii) What is the difference between a nucleoside and a nucleotide?CBSE 20163 m
(i) Lactose gives β-D-galactose and β-D-glucose.
(ii) Vitamin C is water-soluble; water-soluble vitamins are readily excreted in urine, so they cannot be stored and must be supplied regularly in the diet.
(iii) A nucleoside is a pentose sugar with a nitrogen base attached at C1'. A nucleotide is a nucleoside with a phosphoric acid group attached at the C5' –OH of the sugar.
Short answer Define the following terms: (i) Nucleotide (ii) Anomers (iii) Essential amino acids.CBSE 20143 m
(i) Nucleotide: the monomer unit of a nucleic acid, made of a pentose sugar with a nitrogen base attached at C1' and a phosphoric acid group at C5'.
(ii) Anomers: a pair of cyclic isomers of a sugar that differ only in the configuration of the –OH at the anomeric carbon (C1 in glucose), e.g. α-D-glucose and β-D-glucose.
(iii) Essential amino acids: amino acids that cannot be synthesised in the body and must be obtained from the diet, e.g. valine, leucine, lysine, phenylalanine.
Short answer (i) Which one of the following is a polysaccharide: starch, maltose, fructose, glucose? (ii) Write one difference between the α-helix and the β-pleated sheet structure of proteins. (iii) Write the name of the disease caused by the deficiency of vitamin B12.CBSE 20153 m
(i) Starch; maltose is a disaccharide, fructose and glucose are monosaccharides.
(ii) In the α-helix the chain coils into a right-handed screw held by intramolecular hydrogen bonds between the –NH of one residue and the C=O of the adjacent turn; in the β-pleated sheet the chains are stretched nearly to full length and laid side by side, held by intermolecular hydrogen bonds.
(iii) Pernicious anaemia (red blood cells deficient in haemoglobin).
Short answer Amino acids may be acidic, alkaline or neutral. How does this happen? What are essential and non-essential amino acids? Name one of each type.CBSE 20103 m
The nature depends on the relative number of amino and carboxyl groups. Equal numbers make the amino acid neutral (glycine, alanine); more –NH2 groups than –COOH make it basic (lysine, arginine); more –COOH groups than –NH2 make it acidic (aspartic acid, glutamic acid).
Essential amino acids cannot be synthesised in the body and must come from the diet, e.g. valine. Non-essential amino acids are synthesised in the body, e.g. glycine.
Long answer (a) Give the evidence that led to the open-chain structure of glucose (any four points with reactions). (b) State two facts the open-chain structure cannot explain and show how the cyclic structure explains them.Frequently asked5 m
(a) (i) Molecular formula C6H12O6. (ii) Prolonged heating with HI gives n-hexane, so the six carbons form a straight chain. (iii) Glucose gives an oxime with hydroxylamine and a cyanohydrin with HCN, so a carbonyl group is present. (iv) Bromine water oxidises it to gluconic acid, HOCH2–(CHOH)4–COOH, so the carbonyl is an aldehyde. (v) Acetic anhydride gives a stable pentaacetate, so there are five –OH groups on different carbons. (vi) Glucose and gluconic acid are both oxidised by HNO3 to saccharic acid, HOOC–(CHOH)4–COOH, so a –CH2OH group is present. Hence HOCH2–(CHOH)4–CHO.
(b) The open chain cannot explain that glucose fails the 2,4-DNP, Schiff and NaHSO3 tests, that its pentaacetate does not react with hydroxylamine, and that it exists as two crystalline forms, α (mp 419 K) and β (mp 423 K). In the cyclic structure the C5 –OH adds to the C1 aldehyde to form a six-membered hemiacetal (pyranose) ring. The aldehyde is no longer free, so the aldehyde tests fail and the fully acetylated pentaacetate cannot open to give –CHO. C1 becomes a new asymmetric carbon, so two anomers, α and β, differing only in the position of the C1 –OH, exist. Draw the Haworth structures of α- and β-D-(+)-glucopyranose with the open-chain form between them.
Long answer (a) What is a peptide linkage? (b) Describe the primary, secondary, tertiary and quaternary structures of proteins. (c) What is denaturation of a protein? Give two examples.Frequently asked5 m
(a) The amide bond –CO–NH– formed when the –COOH of one α-amino acid condenses with the –NH2 of another with loss of water, e.g. glycine + alanine → glycylalanine.
(b) Primary structure: the sequence in which amino acids are linked; a change in sequence gives a different protein. Secondary structure: the local shape of the chain held by hydrogen bonds between –C=O and –N–H of peptide bonds; the right-handed α-helix (each –NH bonded to the C=O of the adjacent turn) and the β-pleated sheet (chains stretched out and laid side by side with intermolecular hydrogen bonds). Tertiary structure: the overall three-dimensional folding that gives the fibrous or globular shape, stabilised by hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces. Quaternary structure: in proteins of two or more polypeptide chains, the spatial arrangement of the subunits with respect to one another.
(c) Denaturation is the loss of biological activity of a native protein when a change of temperature or pH disturbs the hydrogen bonds, so the globule unfolds and the helix uncoils; secondary and tertiary structures are destroyed, the primary structure is retained. Examples: coagulation of egg white on boiling; curdling of milk by the lactic acid that lactobacillus bacteria produce.
Long answer (a) How are vitamins classified? Name one vitamin of each class with its source and deficiency disease. (b) Write three structural differences between DNA and RNA. (c) Name the three types of RNA.Frequently asked5 m
(a) By solubility. Fat-soluble vitamins (A, D, E, K) are stored in the liver and adipose tissue; e.g. vitamin A, from fish liver oil, carrots, butter and milk, whose deficiency causes xerophthalmia and night blindness. Water-soluble vitamins (B group and C) are excreted in urine and must be supplied regularly, except B12; e.g. vitamin C, from citrus fruits, amla and green leafy vegetables, whose deficiency causes scurvy.
(b) (i) DNA has β-D-2-deoxyribose; RNA has β-D-ribose. (ii) DNA has thymine (bases A, G, C, T); RNA has uracil instead (A, G, C, U). (iii) DNA is a double-stranded helix with complementary base pairing; RNA is single-stranded.
(c) Messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA).
Case-based Read the passage and answer the questions. Sucrose, the common sugar, is a disaccharide of glucose and fructose. It is dextrorotatory, but when hydrolysed by dilute acid or by the enzyme invertase the product mixture rotates plane-polarised light to the left. Unlike sucrose, its hydrolysis products reduce Fehling's solution. Honey is largely this hydrolysed mixture, which is why it is a quick source of energy.
(i) Why does the sign of rotation change on hydrolysis of sucrose? What is the mixture called?
(ii) Why does sucrose not reduce Fehling's solution while its hydrolysis products do?
(iii) Name the glycosidic linkage in sucrose.
(iv) Fructose is a ketose, yet it is a reducing sugar. Comment in one line.Frequently asked4 m
(i) Hydrolysis gives equal amounts of glucose (+52.5°) and fructose (−92.4°); the laevorotation of fructose exceeds the dextrorotation of glucose, so the mixture is laevorotatory. It is called invert sugar.
(ii) In sucrose the glycosidic linkage uses C1 of glucose and C2 of fructose, so no free aldehyde or ketone group is available. After hydrolysis each monosaccharide has a free reducing group.
(iii) Between C1 of α-D-glucose and C2 of β-D-fructose (a C1–C2 glycosidic linkage).
(iv) The α-hydroxy ketone of fructose is oxidised in the alkaline medium of Fehling's or Tollens' reagent, so all monosaccharides, including ketoses, are reducing.
Case-based Read the passage and answer the questions. In 1953 Watson and Crick proposed that DNA consists of two polynucleotide strands wound round each other in a right-handed helix. The sugar–phosphate backbones lie outside and the bases point inward, where each base on one strand is hydrogen-bonded to a specific base on the other. This pairing explains how DNA copies itself and why the sequence of one strand fixes the sequence of the other.
(i) Name the sugar in DNA and the base present in DNA but not in RNA.
(ii) Which base pairs with adenine and which with guanine? How many hydrogen bonds does each pair have?
(iii) A strand of DNA has the sequence AGCTTA. Write the complementary strand.
(iv) How does complementary pairing help DNA replicate?Frequently asked4 m
(i) The sugar is β-D-2-deoxyribose. Thymine is present in DNA but not in RNA.
(ii) Adenine pairs with thymine through two hydrogen bonds; guanine pairs with cytosine through three.
(iii) TCGAAT (A opposite T, G opposite C, C opposite G, T opposite A, T opposite A, A opposite T).
(iv) During cell division the strands separate and each acts as a template on which a new complementary strand is built by base pairing, giving two identical double helices, so the genetic information is copied exactly.
Test yourselfPractice quiz
Marks lost every yearCommon mistakes
- Writing that sucrose is a reducing sugar, or that maltose is not. Sucrose is the only common non-reducing sugar; maltose and lactose keep a free C1.
- Swapping the products of bromine water and nitric acid. Bromine water is mild and gives gluconic acid; nitric acid oxidises both ends and gives saccharic acid.
- Saying denaturation breaks peptide bonds or destroys the primary structure. Only secondary and tertiary structures are lost; confusing denaturation with hydrolysis costs the whole mark.
- Confusing nucleoside with nucleotide. Nucleoside is sugar + base; nucleotide is sugar + base + phosphate.
- Treating D/L as the same as (+)/(−). D-(−)-fructose has the D configuration and is laevorotatory; assertion–reason items test exactly this.
- Writing 'starch has α-linkage, cellulose has β-linkage' without saying what is α or β. Say starch is a polymer of α-D-glucose and cellulose of β-D-glucose, both with C1–C4 glycosidic linkages.
- Mixing up vitamin deficiency diseases, such as scurvy for vitamin D or beri beri for B12. A wrong pairing earns zero, not partial credit.
- Answering 'where does the water in the egg go' with 'it evaporates'. It is absorbed by the coagulated protein through hydrogen bonding.
How to prepareStrategy for this chapter
This is the easiest 7 marks in the paper if you prepare it as facts rather than as chemistry to derive. Give it two focused sessions of about two hours, then ten minutes of revision every second day until the exam.
First session, carbohydrates: write the six evidences for the open-chain structure of glucose with their reactions, then the three facts the open chain cannot explain and the cyclic answer to each. Draw the Fischer projection and both Haworth rings from memory until they come out right. Make one table for sucrose, maltose and lactose (products, linkage, reducing or not) and one for amylose, amylopectin, cellulose and glycogen.
Second session, proteins, vitamins and nucleic acids: write the four levels of protein structure in a sentence each, define peptide linkage and zwitter ion, and write the denaturation paragraph with both examples. Copy the vitamin table in your own hand: name, source, disease. For nucleic acids learn the sugar, the bases, nucleoside versus nucleotide, base pairing with hydrogen-bond counts, and the DNA versus RNA differences. Finish with the definitions of enzyme and hormone.
In the exam, answer this chapter early; it is quick and settles you. In three-part questions answer each part on a labelled line. When a question says 'define', give the NCERT sentence; when it says 'explain', add the reason in one line. Do not spend more than eight minutes on the chapter.