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HORIZON Publication Std. XII Sci. Success Chemistry - II 14 BIOMOLECULES Contents 14.0 Prominent Scientists 14.1 Introduction 14.2 Carbohydrates 14.3 Proteins 14.4 Enzymes 14.5 Lipids 14.6 Hormones 14.7 Vitamins 14.8 Nucleic acids Short Test Multiple Choice Questions 14.0 Prominent Scientist Name 1. Andreas Marggraf 2. Hermann Fischer 3. Har Gobind Khorana 4. Francis Crick 5. Frederick Sanger Contributions i. Pioneer in analytical chemistry. ii. Discovered formic acid, phosphoric acid, isolated zinc by heating calamine. iii. Discovered sugar in beetroots. i. Discovered hydrazine base, phenyl hydrazine. ii. Established relation between glucose, fructose and mannose. iii. Synthesized glucose, fructose and mannose from glycerol. iv. Prepared polypeptide containing 18 amino acids. v. Synthesized dipeptides, tripeptides and polypeptides. v. Awarded Nobel prize. i. Took interest in both proteins and nucleic acid at Cambridge in UK ii. Shared Nobel prize for medicine and physiology for cracking the genetic code. i. Most noted for one of two co-discoveries of the structure of DNA molecule. ii. Awarded Nobel prize i. Two times awarded Nobel prize. ii. First on structure of protein. iii. Second for determination of base sequence in nucleic acids. 14.1 Introduction *1. What are biomolecules ? Give examples. OR Define biomolecule and give two examples. Ans. Biomolecules are the lifeless, molecules which combine in a specific manner to produce life or control biological reactions. Eg. : Carbohydrates, proteins, lipids (fats & oils), nucleic acids, enzymes, vitamins, mineral salts, etc. Biomolecules 14. 1 Std. XII Sci. Success Chemistry - II www.horizonpublication.com 2. What are the different types of biomolecules ? Give two examples of each type. Ans. Two types of biomolecules are i. Simple biomolecules and ii. Macromolecules. i. Simple biomolecules : Water, mineral salts, vitamins, etc. ii. Macromolecules : Carbohydrates, proteins, lipids (fats and oils), etc. 3. Write a note on importance of biomolecules ? Ans. Biomolecules are simple molecules or macromolecules which are derived from reactions of simple molecules. i. ii. Biomolecules Carbohydrates Proteins iii. iv. Lipids Nucleic acids Functions Major constituents of food and source of energy. i. Help in proper functioning of living beings. ii. Important constituents of skin, hair, muscles. iii. Enzymes that catalyse chemical reactions taking place in cells are proteins. Are the storehouses of energy. RNA (Ribonucleic acid) and DNA (Deoxyribonucleic acid) are responsible for genetic characteristics and synthesis of proteins. 14.2 Carbohydrates *4. What are carbohydrates ? Give their general formula and one example. Ans. Carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones or the compounds that can be hydrolysed to polyhydroxy aldehydes or polyhydroxy ketones,. Carbohydrates are also called saccharides. Eg. : Glucose C6H12O6 , Maltose C12H22O11 5. How is the name carbohydrates derived ? Ans. i. According to earlier chemists the carbohydrates had molecular formula as Cx(H2O)y. ii. They considered carbohydrates as the compounds in which carbon atom is surrounded by water molecules i.e. hydrates of carbon, hence the name was given as carbohydrates. iii. Eg. : Compounds like glucose (C6H12O6) was considered as C6(H2O)6 and sucrose C12H22O11 as C12(H2O)11. 6. Give examples of the compounds which do not justify the earlier formula of carbohydrates. Ans. Earlier formula suggested for carbohydrates was Cx(H2O)y. However the compound like acetic acid, CH3COOH can be represented as C2(H2O)2 but cannot be classified as carbohydrate, but it is an acid. The compound rhamnose, CH3(CHOH)4CHO cannot be represented as Cx(H2O)y but it is a carbohydrate. Hence, the carbohydrates now have a different meaning or definition. 14.2.1 Classification of Carbohydrates 7. Give classification of carbohydrates. OR How are carbohydrates classified? Ans. On the basis of their behaviour on hydrolysis, carbohydrates are mainly classified into two groups as i. Simple carbohydrates and ii. Complex carbohydrates Biomolecules 14. 2 HORIZON Publication Std. XII Sci. Success Chemistry - II i. Simple carbohydrates : a. Are known as monoscharides. b. A carbohydrates that cannot be further hydrolyzed to simple sugar is called monosaccharide. c. Monosaccharides are the basic units of all carbohydrates. d. Depending upon the type of carbonyl group present are further classified as i. Alodses (containing an aldehyde group) and ii. Ketoses (containing keto group) e. Depending upon the number of carbon atoms and the functional group present are further classified as i. Triose ii. Tetrose iii. Pentose iv. Hexose etc. Sr. No. Type of Monosaccharide 1. 2. 3. 4. Triose Tetrose Pentose Hexose Class Triose (C3H6O3) CHO Aldoses H OH CH2OH D (+) Glyceraldehyde (Aldotriose) Ketoses CH2OH | C=O | CH2OH Dihydroxyacetone (Ketotriose) ii. Number of ‘C’ atoms present 3 4 5 6 Tetrose (C4H8O4) CHO OH H H OH CH2OH D () Erythrose (Aldotetrose) CH2OH | C=O H OH CH2OH D Erythrulose (Ketotetrose) Pentose (C5H10O5) Hexose (C6H12O6) CHO OH CHO OH H H H OH HO H H OH H OH H OH CH2OH D () Ribose (Aldopentose) CH2OH D (+) Glucose (Aldohexose) CH2OH | C=O OH H HO CH2OH | C=O H OH H OH H OH H CH2OH D Ribulose (Ketopentose) CH2OH D ()Fructose (Ketohexose) Complex carbohydrates. : a. Two categories of complex carbohydrates are i. Oligosaccharides ii. Polysaccharides. b. Oligosaccharides : A carbohydrate that yield two to ten monosaccharide units is called Oligosaccharide. c. Depending on the number of monosaccharides derived, oligosaccharides are further classified as : Biomolecules 14. 3 Std. XII Sci. Success Chemistry - II d. e. 8. www.horizonpublication.com Sucrose Glucose + Fructose Hydrolysis Maltose Glucose + Glucose Hydrolysis Lactose Glucose + Galactose Hydrolysis Raffinose Glucose + Fructose + Galactose Hydrolysis i. Disaccharides (C12H22O11) : ii. iii. Trisaccharides (C18H32O16) : Tetrasaccharides (C24H42O21) : Hydrolysis Glucose + Fructos + Galactose + Galactose Stachyose Polysaccharides : A carbohydrate which on hydrolysis yields a large number of monosaccharide units is called Polysaccharides. Polysaccharides are tasteless, amorphous, insoluble in water. They are long chain, naturally occurring polymers of carbohydrates. Eg. Cellulose, starch, glycogen. On the basis of solubility carbohydrates are classified as : i. Sugars : Crystalline, sweet to taste and soluble in water. Monosaccharides and Oligosaccharides are sugars. ii. Non-sugars : Amorphous, tasteless and insoluble in water. Polysaccharides are nonsugars and further classified as : a. Reducing Sugars : They reduce Fehling’s solution and Tollen’s reagent. Eg. : Maltose b. Non-reducing Sugars : They do not reduce Fehling’s solution and Tollen’s reagent. Eg. : Sucrose Referring to the table give below answer the following questions. [Intext Question text book page no. 513] Ans. Class Triose (C3H6O3) CHO Aldoses H OH CH2OH D (+) Glyceraldehyde (Aldotriose) Ketoses CH2OH | C=O | CH2OH Dihydroxyacetone (Ketotriose) Biomolecules Tetrose (C4H8O4) Pentose (C5H10O5) Hexose (C6H12O6) CHO OH H CHO OH H CHO OH H H OH H OH HO H H OH H OH H OH CH2OH D () Erythrose (Aldotetrose) CH2OH | C=O H OH CH2OH D Erythrulose (Ketotetrose) CH2OH D () Ribose (Aldopentose) CH2OH D (+) Glucose (Aldohexose) CH2OH | C=O OH H HO CH2OH | C=O H OH H OH H OH H CH2OH D Ribulose (Ketopentose) CH2OH D ()Fructose (Ketohexose) 14. 4 HORIZON Publication Std. XII Sci. Success Chemistry - II i. Identify the monosaccharides that contain only two chiral carbon atoms. Ans. CH2OH CHO H OH H OH | C=O OH H CH2OH D () Erythrose OH H CH2OH D Ribulose ii. Write the Fischer projection formulae of (a) L – (+) – erythrose (b) L – (+) – ribulose. Ans. a. b. CHO CH OH 2 H HO OH | C=O OH H H CH2OH L (+) Erythrose HO H CH2OH L (+) Ribulose iii. Is the followings sugar, D - sugar or L – sugar ? HO CHO H HO H HO H CH2OH Ans. The compound is L – sugar. 9. Explain D and L configuration in sugars. OR Explain Fischer projection formula. Ans. i. The simplest carbohydrates glyceraldehyde is chosen as the standard, to assign D and L configuration to monosaccharides. Glyceraldehyde contains one asymmetric carbon atom and exist in two enantiomeric forms as, CHO H CHO OH CH2OH D-Glyceraldehyde (+) Glyceraldehyde ii. iii. iv. v. HO H CH2OH (L) Glyceraldehyde () Glyceraldehyde The dextro entantiomer is represented as (+) glyceraldehyde and it is referred as D-configuration i.e., D-glyceraldehyde. The laevo enantiomer of glyceraldehyde is represented as () glyceraldehyde and is corelated as L-configuration i.e., L-glyceraldehyde. In Fischer projection formula, a monosaccharide is assigned D-configuration if the ( OH) hydroxyl group at the last chiral carbon and lies towards right hand side. On the other hand it is assigned L-configuration if the OH group on the last chiral carbon atom and lies on the left hand side. In monosaccharides, the most oxidised carbon (i.e., CHO) is at the top. Biomolecules 14. 5 Std. XII Sci. Success Chemistry - II Eg. : www.horizonpublication.com OH H CH2OH D(+)Glyceraldehyde CHO CHO CHO CHO H OH H OH H OH CH2OH D (+) Ribose H OH HO H HO H H OH H OH HO H H OH CH2OH HO H D (+) Glucose CH2OH L () Glucose 10. Identify D & L configurations of following monosaccharides. Ans. Monosaccharides Configuration CHO i. H HO L CH2OH ii. HO CHO H HO H HO H L CH2OH CH3 iii. OH H D COOH COOH HO H H OH iv. D COOH CH3 v. HO H L CH2OH Biomolecules 14. 6 HORIZON Publication Std. XII Sci. Success Chemistry - II 11. What is monosaccharide ? Explain how are monosaccharides classified ? Ans. The basic unit of all carbohydrates which is a simple carbohydrate and cannot be hydrolysed further is known as monosaccharide. The monosaccharide is crystalline and soluble in water. Eg. Glucose, fructose, ribose. Monosaccharides are classified as : H group in their structure. Eg. : Glucose. i. Aldoses : Aldoses contain aldehydic | C O Aldoses are further classified depending upon the number of carbon atoms present in the monosaccharide. The number of carbon atoms present in the molecule is indicated by the prefix tri for 3-carbon, tetra for 4-carbon, etc. The aldose accordingly is called Aldotriose, Aldotetrose, Aldopentose, Aldohexose, etc. Eg. : Glyceraldehyde C3H6O3 Aldotriose Erythose C4H8O4 Aldotetrose Ribose, Arabinose, Xylose C5H10O5 Aldopentose Glucose, Mannose C6H12O6 Aldohexose ii. Ketoses : Ketoses contain ketonic, C = O group in their structures. Eg. : Fructose. Ketoses are further classified depending upon the number of carbon atoms present in the monosaccharide. The number of carbon atoms present in the molecule is indicated by the prefix, tri for 3-carbons, tetra for 4-carbons, etc. The Ketoses accordingly are called Ketotriose, Ketotetrose, Ketopentose, Ketohexose, etc. Eg. : Dihydroxy acetone C3H6O6 Ketotriose Erythrulose C4H8O4 Ketotetrose Ribulose C 5H10O5 Ketopentose Fructose C6H12O6 Ketohexose *12. Classify the following into monosaccharides, oligosaccharides and polysaccharides. i. Starch ii. Glucose iii. Stachyose iv. Maltose v. Raffinose vi. Cellulose vii. Sucrose viii. Lactose. Ans. Glucose Monosaccharides Stachyose, maltose, raffinose, sucrose, Oligosaccharides lactose Starch, cellulose Polysaccharides 14.2.2 Preparation of Glucose *13. How is glucose prepared in the laboratory ? OR Give laboratory preparation of glucose. Ans. Preparation of glucose from sucrose (cane sugar) : Laboratory method i. When cane sugar or sucrose is boiled with dilute hydrochloric acid or sulphuric acid it undergoes hydrolysis to give glucose and fructose in equal amounts. ii. While cooling, alcohol is added as glucose is insoluble in alcohol it crystallized out first. Fructose being soluble in alcohol it remains in the solution. iii. Crystals of glucose are separated by filtration. dil HCl or H 2SO 4 C12 H 22O11 H 2O C6 H12O 6 C6 H12O 6 Sucrose Glu cos e Fructose Biomolecules 14. 7 Std. XII Sci. Success Chemistry - II www.horizonpublication.com 14. Give reason: At the time of cooling in laboratory method used for glucose preparation, alcohol is added. Ans. i. Hydrolysis of cane sugar gives a mixture of glucose and fructose. ii. Glucose is insoluble in alcohol and crystallized out first while Fructose being soluble in alcohol it remains in the solution. iii. Hence, in order to separate glucose from fructose at the time of cooling alcohol is added in laboratory method. *15. How is glucose prepared on commercial scale ? Ans. Commercially, on a large scale, glucose is prepared by hydrolysis of starch with dilute sulphuric acid. Starchy material is mixed with water and dilute sulphuric acid and heated at 393 K under pressure. Starch is hydrolysed to give glucose. dil H 2SO 4 (C 6 H10 O 5 ) n nH 2 O nC 6 H12 O 6 393K , 2 3 atm Starch glucose glucose When hydrolysis is complete, calcium carbonate (chalk powder) is added. Chalk powder neutralizes excess of acid. To the solution, activated charcoal is added which removes coloured impurities by adsorption. The solution is then filtered to remove insoluble impurities, The solution is then concentrated, when glucose is obtained in the crystalline form. Pure crystals of glucose are obtained by recrystallization. 16. Give reason : i. Chalk powder is added when the hydrolysis of starch, in the manufacture of glucose, is complete. ii. Activated charcoal is added at the end of the hydrolysis in the manufacture of glucose in commercial method. Ans. i. a. Hydrolysis of starch with dil. H2SO4 at 393 K under pressure gives glucose. b. Chalk powder is calcium carbonate (CaCO3). Chalk powder reacts with excess of sulphuric acid forming precipitate of calcium sulphate which is removed by filtration. Neutralization H2SO4 CaCO3 CaSO4 H2O CO2 Thus to remove excess of H2SO4 chalk powder is added. ii. a. The coloured impurities are removed by adding activated charcoal. b. Coloured impurities are adsorbed on the charcoal. c. Thus activated charcoal is added at the end of the hydrolysis in the manufacture of glucose in commercial method. 14.2.3 Open Chain Structure of Glucose *17. What is the action of following reagents on glucose ? i. HI ii. Hydroxyl amine (NH2OH) iii. Hydrogen cyanide iv. Bromine water v. Nitric acid vi. Acetic anhydride. Ans. i. Action of HI : Glucose on prolonged heating with HI gives n-hexane, indicates that all the six carbon atoms are linked in straight chain. CHO | hot HI (CHOH) 4 CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 | CH 2 OH n- Hexane Glu cos e Biomolecules 14. 8 HORIZON Publication Std. XII Sci. Success Chemistry - II ii. Action of hydroxyl amine : Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime. CHO | (CHOH) 4 | CH 2 OH Glucose NH 2 OH hydroxyl amine CH NOH | (CHOH) 4 + H2O | CH 2 OH Glucoxime iii. Action of hydrogen cyanide : Glucose reacts with hydrogen cyanide to form glucose cyanohydrin. CHO CN | | HCN (CHOH) 4 CHOH | | CH 2OH (CHOH) 4 | CH 2OH Glucose Glucose cyanohydrin iv. Action of bromine water : Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid, which shows that the carbonyl group in glucose is aldehyde group. COOH CHO | | Bromine water (CHOH) (CHOH) 4 (O) 4 | | CH 2 OH CH 2 OH Glu cos e Gluconic acid v. Action of dil. nitric acid : Glucose on oxidation with dilute nitric acid forms dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic group ( CH2OH) in glucose. CHO COOH COOH COOH | | | | dil HNO 3 dil HNO 3 (CHOH) (CHOH) (CHOH) 4 (CHOH) 4 4 4 ; | | | | COOH COOH CH 2 OH CH 2 OH Saccharic acid Saccharic acid Glu cos e Gluconic acid vi. Action of acetic anhydride : When glucose is heated with acetic anhydride in the presence of catalyst pyridine, glucose penta acetate is formed. It indicates that glucose is a stable compound and contains five hydroxyl groups. CHO O CHO O O | || | || || Pyridine (CHOH) 4 5(CH 3 C) 2 O (CH O C CH 3 ) 4 5 CH 3 C OH | | CH 2 OH CH 2 O C CH 3 Glucose || O Glucose pentaacetate Biomolecules 14. 9 Std. XII Sci. Success Chemistry - II www.horizonpublication.com 18. A monosaccharide on acetylation gives triacetate. It is not oxidized by bromine water but on oxidation by dil. HNO3 gives dicarboxylic acid. Write the possible structure of the monosaccharide. [Intext Question text book page no. 515] Ans. The compound C4H8O4 i.e. D – Erythrulose on acylation gives triacetate and on treatment with dil. HNO3 gives dicarboxylic acid. Hence, the structure of monosaccharide is CH2OH C=O CHOH CH2OH D – Erythrulose 19. What are the expected products of hydrolysis of lactose? Ans. Lactose on hydrolysis forms an equimolar mixture of D-glucose and D-galactose. 20. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose ? Ans. The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free –CHO. 14.2.4 Cyclic Structure of Glucose 21. List down the facts and reactions that are not explained by open chain structure of glucose. Ans. Open chain structure of glucose does not explain the following reactions and facts: i. Glucose in spite of having aldehyde group, does not give condensation reaction with 2,4 dinitro phenyl hydrazine, addition reaction with sodium bisulphite and gives negative test with Schiff base. ii. Glucose penta – acetate does not condense with hydroxylamine, indicating the absence of free aldehyde group. iii. Glucose is found to exist in two different crystalline forms α and β, called anomers. On crystallization from hot and saturated aqueous solution, α-glucose (m.p. 423 K) is obtained at 303 K while -glucose (m.p. 423 K) is obtained at 371 K. CH2OH CH2OH O H H OH H OH H HO H OH OH H HO OH H O H H H OH ( –) ( –) Write the two cyclic structures of -D-( + )-Glucose pyranose and -D-( + )-Glucose pyranose exist in equilibrium with open chain structures. Ans. O 1 1 1 || H C OH HO C H HC 2 2 H C OH H C OH O H C OH O 3 3 3 HO C H HO C H HO C H 4 4 4 H C OH H C OH H C OH 22. 5 HC 6 CH2OH D(+)Glucose pyranose Biomolecules 5 5 H C OH 6 CH2OH HC 6 CH2OH D(+)Glucose pyranose 14. 10 HORIZON Publication Std. XII Sci. Success Chemistry - II *23. Write the structures of -D-(+)-glucopyranose and -D-(+)-glucopyranose. Ans. 1C OH 2 OH H H HO H C H 2 OH 3 H 4 OH 5 6 CH2OH H HO H H O 3 H 4 OH 5 6 CH2OH H 1 HO D (+) Glucopyranose O D (+) Glucopyranose 24. Show how is Fischer projection formula of glucose converted into Haworth projection formula. Ans. The cyclic structures of glucose is converted into Haworth projection formula as follows : 1 CHO 2 H H 4 4 CH2OH H H H C H1 1 O 2 3 OH H 6 OH OH HO 5 6 5 H ≡ OH H CH2OH OH 3 HO 6 H H 5 H OH 4 HO 3 H OH O H C1 1 H OH 2 + OH CH2OH H+ D(+)Glucose 6 CH2OH H 4 CH2OH 5 H O H H OH HO H O H OH 1 OH H H HO 2 OH 3 OH H D (+) Glucopyranose (cis) OH H D (+) Glucopyranose (trans) 25. H Following are the structures of four stereoisomeric erythrofuranoses. Name each isomer using proper D, L and , notations. [Intext Question text book page no. 517] i. H O H H ii. H OH OH Biomolecules OH OH O H iii. OH H OH OH H O H iv. OH OH H OH OH O H H H OH 14. 11 Std. XII Sci. Success Chemistry - II Ans. i. D – erythrofuranose ii. L – erythrofuranose iii. –D– (–) erythrofuranose iv. –D– (–) erythrofuranose www.horizonpublication.com 14.2.5 Structure of Fructose 26. Explain the structure of fructose. Ans. Fructose has molecular formula C6H12O6. It contains ketonic functional group at carbon number 2 and six carbon atoms in straight chain. It belongs to D-series and is a laevo rotatory compound. It is written as D-( )-fructose. CH2OH | C=O H HO H OH H OH CH2OH D () Fructose 27. Write the Haworth projection formulae for -D-( )-Fructofuranose (trans) and -D-() fructofuranose (cis). Ans. 1 HOH2C C H 2 OH O 5 HO H OH OH H D() Fructofuranose (trans) CHO HO OH ii. H HO H HO CH2OH Fructose CHO OH H OH H D() Fructofuranose (cis) *28. Draw mirror images of glucose and fructose. Ans. i. Glucose H O 2 H H HOH2C H OH HO CH2OH CH2OH C=O C=O H H OH H OH HO H H OH HO H H OH HO H H OH HO H CH2OH D(+)Glucose Biomolecules CH2OH L()Glucose CH2OH D(+)Fructose CH2OH L(–)Fructose 14. 12 HORIZON Publication Std. XII Sci. Success Chemistry - II 29. Write the conversion of Fischer projection formula into Haworth projection formula for fructose. Ans. CH2OH H C=O 6 OH O: HO H 2C HO H HOH 2 C :OH O 2 6 H 5 H HO C H HO C 5 H OH 2 4 3 H H 4 3 C H 2 OH CH 2OH H OH 1 1 OH H OH H CH2OH H+ 1 6 HOH 2 C O 5 HOH 2 C HO 4 3 OH H OH H D() Fructofuranose (trans) 30. OH O 5 2 H H 6 C H 2 OH 2 H HO 4 3 C H 2 OH 1 OH H D() Fructofuranose (cis) Refer to the Fischer projection formula of ribose. Furanose form of ribose is produced by addition of – OH group at C-4 to aldehyde group. Draw Haworth projection formulae of four stereoisomeric ribofuranoses using D, L and , notations. [Intext Question text book page no. 518] Ans. CH2OH O H H H OH OH OH O H OH -D-(-)-Ribofuranoses OH -D-(-)-Ribofuranoses OH O H CH2OH OH HOH2C H OH OH OH -L-(-)-Ribofuranoses OH H OH H OH OH -L-(-)-Ribofuranoses Adjacent is the Haworth projection formula of -D-ribulofuranose. Write its Fischer projection formula. [Intext Question text book page no. 518] 5 Ans. Fischer projection formula is as follows : H H H HOH2C 31. CH2OH | C=O O O 1 CH 2OH 2 H 3 OH 4 OH OH CH2OH Biomolecules 14. 13 Std. XII Sci. Success Chemistry - II 14.2.6 Disaccharides www.horizonpublication.com *32. Write the hydrolysis products of (i). lactose and (ii). sucrose. Ans. i. Lactose on hydrolysis in presence of an acid or enzyme lactose gives one molecule each of glucose and galactose C12 H 22 O11 H 3O H 2 O or lactose Lactose ii. C 6 H12 O 6 C 6 H12O 6 D ( ) Glucose D ( ) Galactose Sucrose on hydrolysis in the presence of dil. acid or the enzyme invertase gives one molecule each of glucose and fructose . C12 H 22 O11 H 3O H 2 O or invertase Sucrose C 6 H12 O 6 C 6 H12O 6 D ( ) Glucose D ( ) Fructose 33. Explain the structures of following molecules : i. Sucrose ii. Maltose iii. Cellobiose iv. Lactose Ans. i. Sucrose : a. It is known as cane sugar or common table sugar. b. In sucrose C – 1 of D – Glucopyranose is linked to C – 2 of – D – Frouctofuranose by glycosidic linkage. c. The reducing groups of glucose and fructose are involved in glycosidic bond formation and hence sucrose is non reducing sugar. 6 C H 2 OH 5 H O H OH 4 H 1 H Glycosidic linkage OH 2 3 H OH 6 O HOH2C 5 H 2 H 4 OH ii. O HO 3 CH2OH 1 H Maltose : a. In Maltose C – 1 of one D – Glucopyranose is linked to C – 4 of another D – Glucopyranose molecule by glycosidic linkage. b. Thus in maltose 1 4 glycoside bond is present. c. It is reducing sugar as it gives – CHO group at C – 1 in second glucose molecule. Biomolecules 14. 14 HORIZON Publication Std. XII Sci. Success Chemistry - II 6 6 CH2OH CH2OH H 5 O H OH 4 HO O 2 3 H OH - D - Glucopyranose O H OH 4 1 H 5 H H H 1 H OH 2 3 6H OH - D - Glucopyranose Maltose ( – anomer), 4 – O – ( – D – Glucopyranosyl) – D – glucopyranose iii. Cellobiose : a. In this C – 1 of the – D Glucopyranose is linked to C – 4 of another – D Glucopyranose molecule by glycosidic linkage. b. It contains 1 4 glycoside bond. c. It is a reducing sugar as – CHO group is produced at C – 1 in second glucose molecule. 6 H CH2OH H 5 O H OH 4 4 H 2 3 OH H 1 HO H H OH - D - Glucopyranose 2 3 O H OH H H 1 OH O 5 6 CH2OH - D - Glucopyranose Cellobiose ( – anomer) 4 – O – ( – D – Glucopyranosyl) – D – glucopyranose iv. Lactose : a. In lactose C – 1 of – D galactopyranose is linked to C – 4 of – D – glucopyranose by glycosidic linkage. b. Thus it contains 1 4 glycoside bond. c. It is a reducing sugar as free – CHO group is produced at C – 1 of – D – glucopyranose molecule. 6 C H 2 OH HO 5 O H OH 4 H O H OH H 1 H 2 H OH - D - Galactopyranose OH 2 H 4 H 3 3 H 5 H O 1 OH CH2OH - D - Glucopyranose 6 lactose ( anomer) 4 – O – ( – D – Galactopyranosyl) – D – glucopyranose Biomolecules 14. 15 Std. XII Sci. Success Chemistry - II 14.2.7 Polysaccharides www.horizonpublication.com 34. Define polysaccharides and give general formula and example. Ans. The molecules in which a large number of same or different monosaccharides are linked together by glycosidic linkages is called polysaccharide. General formula (C6H10O5)n Eg. : Starch, Cellulose 35. Explain the structure of cellulose. Ans. Cellulose mainly occurs in plants. Cell wall of plant cells is made up of cellulose. It is a long chain polymer. In cellulose, D glucopyranose molecules are linked by glycosidic linkage between C1of one unit of glucose and C4 of another glucose unit. Thus cellulose contains 1 4 glycosidic linkages like those in cellobiose. 6 CH2OH 6 H 5 H CH2OH 4 5 O 4 H O 1 H OH O 3 2 H OH O H OH H 3 2 H OH O 1 H - link H Cellulose (–1,4’ polymer of D–glucose) 36. Explain the structure of starch. Ans. Starch is found in cereal grains, roots, tubers, potatoes etc. It is a polymer of D glucopyranose and consists of two components, amylose and amylopectin. Amylose is water soluble component and constitutes about 20% of starch. Amylose contains 200 to 1000 D glucopyranose molecules linked together by glycosidic linkage between C1 of one unit and C4 of another unit. 6 H 5 4 H OH O 3 CH2OH CH2OH H H 5 H O 4 1 H OH O H 6 6 CH2OH 2 OH 3 O H 2 H 5 H 4 1 O H OH O 3 H H 1 H 2 O OH H OH links Amylose (D Glucopyranose) Amylopectin is insoluble in water and constitutes about 80% starch. It is a branched chain polymer. In amylopectin, D glucopyranose molecules are linked together by glycosidic linkage between C1 of one unit and C4 of another unit to form long chain and branching occurs by glycosidic linkage between C1 and C6 glycosidic linkage. Biomolecules 14. 16 HORIZON Publication Std. XII Sci. Success Chemistry - II 6 CH2OH H 4 O 6 5 O H OH H H H CH2OH 5 O H OH H 4 1 O 2 3 H H O 6 CH2 6 CH2OH 4 O Branch at C-6(-1,6 glycosidic linkage) OH link H 1 2 3 OH H 5 O H OH H 3 H H H 4 2 OH 1 6 CH2OH 5 O H H 5 O H OH H 1 4 H OH H O 3 H 2 OH O 3 H 2 H 1 O OH Amylopectin 37. What is the basic structural difference between starch and cellulose ? Ans. Starch is a polymer of -glucose and consists of two components-amylose and amylopectin. In amylose D (+) glucose units held by C1 C4 glycosidic linkage and in amylopectin, D glucose units held by C1 C4 glycosidic linkage whereas branching occurs by C1 C6 glycosidic linkage. Cellulose is a straight chain polysaccharide composed only of D glucose units held by C1C4 glycosidic linkage. 38. Write a note on structure of glycogen. Ans. i. Glucose is stored in the form of glycogen in animal body. ii. Glycogen is known as animal starch, as its structure is similar to amylopectin. iii. The branching in glycogen is more extensive than in amylopectin. iv. Many end groups present in highly branched structure of glycogen, are available for quick hydrolysis to provide the glucose needed for metabolism. v. So it is a source of instant energy. *39. State the importance of carbohydrates. OR Why are carbohydrates important in diet ? Ans. i. Carbohydrates are essential for life in both plants and animals. ii. They are the important constituent of supporting tissues in plants e.g. cellulose found in wood, cotton, etc. iii. They form a major part of our food and store chemical energy in plants. iv. They act as the major source of energy for animals and human beings. v. Animals store glucose in the form of glycogen which is the main source of energy. vi. Carbohydrates satisfy the three basic needs i.e. food, clothes and shelter. Biomolecules 14. 17 Std. XII Sci. Success Chemistry - II www.horizonpublication.com 40. How is glycogen different from starch ? Ans. Starch is the main storage molecule of plants whereas glycogen is the main storage molecule of animals. Starch is found in cereals, roots, tubers, etc. Glycogen is present in liver, muscles and brain. *41. Write the structures of cellobiose, lactose, cellulose and starch. Ans. Compound Structure i. Cellobiose 6 C H 2 OH H 5 H O O OH H 1 H H OH - D - Glucopyranose 1 OH O 5 2 3 H H H HO 2 H 4 H OH 4 OH 3 CH2OH - D - Glucopyranose 6 Cellobiose (-anomer) 4 O ( D Glucopyranosyl) D glucopyranose ii. Lactose 6 C H 2 OH 5 HO O H O OH H H H H OH - D - Galactopyranose 1 OH O 5 2 3 2 H 1 H H OH 3 4 H OH 4 H CH2OH - D - Galactopyranose 6 Lactose (-anomer) 4 O ( D galactopyranosyl D glucopyranose iii. Cellulose 6 C H 2 OH H 6 C H 2 OH H 4 5 4 O H OH O 3 H H OH H 1 O O 1 H H 3 H H O 5 2 OH -link 2 OH Cellulose ( 1, 4’ polymer of D-glucose) Biomolecules 14. 18 HORIZON Publication Std. XII Sci. Success Chemistry - II iv. Starch : It is a mixture of two compounds (i) a water soluble : amylase (15 20%) & (ii) water insoluble : amylopectin (80 85%) 6 6 CH2OH H 5 H 4 OH O 3 H O H 6 H CH2OH H 5 O 1 4 H OH H O 2 OH H H CH2OH H 5 O H 1 4 H OH H 1 3 H 2 OH O 2 OH 3 O - links Amylose (D – Glucopyranose) CH2OH CH2OH O H H OH H H H O H O H OH H H OH O H OH Branch at C6 (-1,6 glucosidic linkage) - links O 6 CH2OH 5 H H 4 OH O 3 H 6 O H CH2OH H 5 H 1 4 6 CH2 O H H 2 OH H 5 H OH 4 1 O 3 H 2 OH O O H H OH H 3 H 2 OH 1 O Amylopectin 14.3 Proteins 42. Give reason ‘Proteins are ranked first of all the organic compounds’. Ans. i. Proteins are the most abundant biomolecules of the living system. ii. Proteins are the substances of life. iii. They are found in living cell and also are present in skin, hair, muscle, nerves, enzymes, antibodies and hormones. iv. Hormone that are proteins, regulate metabolic processess. v. Without proteins life would not be possible. vi. Thus proteins are ranked first of all the organic compounds. How is the word ‘Protein’ derived ? and mention the basic unit or hydrolysis product of proteins. Ans. i. The name protein is derived from the Greek word ‘proteios’ meaning first. ii. Hydrolysis product of proteins is amino acid. R | H 2O Proteins H 2 N CH COOH Mixture of - amino acids 43. Biomolecules 14. 19 Std. XII Sci. Success Chemistry - II www.horizonpublication.com *44. Define protein. Ans. Proteins are the biopolymers of a large number of -amino acids. They are naturally occurring polymeric nitrogenous organic compounds containing 16% nitrogen and peptide linkages ( CO NH ). 45. What are the sources of proteins. Ans. The sources of proteins are milk, egg, fish, pulses, cereals, peanuts, cheese etc. *46. What are -amino acids ? Ans. i. -amino acids are the derivatives of carboxylic acids obtained by replacing -H by amino group. ii. They are bifunctional as they have 2 functional groups i.e. – NH2, – COOH. iii. Proteins on hydrolysis give -amino acids. Note : Name of amino acid 2. 3. 4. 5. Neutral amino-acids Glycine Alanine Valine Leucine Isoleucine 6. 7. 8. Phenylalanine Tyrosine Tryptophan 1. 3 Letter symbol (abbreviation) Side chain R of amino acids H– CH3 – (CH3)2CH – (CH3)2–CH – CH2 – CH 3 CH 2 CH | CH 3 C6H5 – CH2 – –HO–C6H4–CH2– Gly Ala Val Leu Ile Phe Tyr Try CH2 – N H 9. Serine 10. Proline 11. Glutamine HO–CH2– –CH2–CH2–CH2– O || H 2 N C CH 2 CH 2 12. Cysteine 13. Cystine SH – CH2 – 14. Methionine CH3–S–CH2–CH2– Met 15. Threonine CH 3 – CH – | OH Thr 16. Asparagine H2N–CO–CH2– Asn Biomolecules Ser Pro Gln O OC CH CH 2S CH 2S CH CO O | | NH 3 NH 3 Cys Cys.cys 14. 20 HORIZON Publication Std. XII Sci. Success Chemistry - II Acidic amino acids 17. Glutamic acid HOOC–CH2–CH2– 18. Aspartic acid HOOC–CH2– Basic amino acids H N C NH (CH 2 )3 2 19. Arginine | NH 2 20. Lysine H2N–(CH2)4 – 21. Histidine CH = C – CH2 – | | N NH Glu Asp Arg Lys His C | H *47. What is zwitter ion. Ans. i. Amino acids due to presence of both acidic and basic groups in the same molecule behave like salts. ii. H 2 N CH COOH H 3 N CH COO zwitter ion (dipolar ion) | | R R Such doubly charged ion is known as zwitter ion. *48. Write the classification of amino acids, giving examples. Ans. Amino acids are of three types : i. Acidic amino acids : If carboxyl groups are more in number than amino groups then amino acids are acidic in nature. COOH Eg. : Glutamic acid : HOOC CH2 CH2 – CH NH2 Aspartic acid : HOOC CH2 CH COOH NH2 ii. Basic amino acids : If amino groups are more in number than carboxyl groups then amino acids are basic in nature. COOH + Eg. : Arginine : H 2 N C NH (CH 2 ) 3 CH | NH 2 NH2 COOH Lysine : H2N – (CH2)4 – CH NH2 iii. Neutral amino acids : The amino acids having equal number of amino and carboxyl groups are called neutral amino acids. COOH Eg. : Alanine : CH3 – CH NH2 Biomolecules COOH Valine : (CH3)2 – CH – CH NH2 14. 21 Std. XII Sci. Success Chemistry - II www.horizonpublication.com *49. What are essential and non-essential amino adds ? Give two examples of each. Ans. i. The amino acids, which cannot be synthesised in the body and are supplied through diet are called essential amino acids. Eg. : Lysine, Valine ii. The amino acids which are synthesized in the body are called non-essential amino acids. Eg. : Glutamic acid, Serine *50. Explain the amphoteric behaviour of amino acids. Ans. i. Amino acids are bifunctional compounds containing acidic carboxyl group (COOH) and basic amino group ( NH2) within the same molecule. In aqueous solution, the carboxyl group loses a proton while the amino group accepts it. as a result, a dipolar or zwitter ion is formed. H 2 N CH C | R ii. O + H 3 N CH C | R Zwitter ion (dipolar ion) OH O O– -amino acids show amphoteric behaviour as they react with both acids and bases. In the acidic medium, COO ion of the zwitter ion accepts a proton to form the cation (I) while in the basic medium, N H 3 ion loses a proton to form the anion (II). + + H H H 3 N CH COOH H 3 N CH COO H 2 N CH COO | | | R R R (I) Zwitter ion (II) Thus, N H 3 group acts as the acid while COO groups acts as the base. 51. Define peptide linkage. Ans. Peptide linkage is an amide formed between – COOH and – NH2 group by elimination of water O molecule. || is peptide linkage. C NH *52. Write a note on peptide linkage. OR How is peptide linkage formed ? Ans. i. Peptide linkage is an amide formed between – COOH and – NH2 group by elimination of water O molecule. || is peptide linkage. C NH ii. Same or different amino acid molecules loose water molecule and get linked through peptide linkage. O H O O H O || | || || | || H 2 N CH C OH H N CH C OH H 2 N CH C N CH C OH H 2O | | | | R R R R Peptide bond iii. Many amino acid molecules combine together through peptide linkages and results in dipeptide, tripeptide, tetrapeptide and finally a protein molecule. Eg. Insulin contains 51 amino acids. Biomolecules 14. 22 HORIZON Publication Std. XII Sci. Success Chemistry - II *53. Write the structures of all possible dipeptides which can be obtained from glycine and alanine. Ans. i. Dipeptide from glycine : Carboxylic group of glycine reacts with amino group of another molecule of glycine to form dipeptide. O O || || H 2O H 2 N CH 2 C OH H N CH 2 COOH H 2 N CH 2 C NH CH 2 COOH | Glycine Glycine Dipeptide H ii. Dipeptide from alanine Carboxylic group of alanine reacts with amino group of another molecule of alamine to form dipeptide CH 3 O CH 3 CH 3 O CH 3 | || | | || | H 2O H 2 N CH C OH H N CH COOH H 2 N CH C NH CH COOH | H Alanine Alanine Dipeptide iii. Dipeptide from glycine and alanine. Carboxylic group of glycine reacts with amino group of another molecule of alanine to form dipeptide CH 3 CH 3 O O || | || | H 2O H 2 N CH 2 C OH H N CH COOH H 2 N CH 2 C NH CH COOH | H Alanine Glycine Dipeptide *54. How are proteins classified ? Ans. On the basis of their molecular shape proteins are classified into two types as follows : i. Fibrous proteins ii. Globular proteins i. Fibrous proteins : a. Long thread like and lie side by side to form fibres. b. Insoluble in water. c. Polypeptide chains are held together by hydrogen bonds. Eg. Collagen in tendons, keratin in hair, skin, nail etc. ii. Globular proteins : a. They are folded to form spherical shape and have intramolecular hydrogen bonding. b. Soluble in water and aqueous solutions of bases, acids and salts. c. Hydrogen bonding of these proteins in weak as compared to fibrous proteins. Eg. Haemoglobin (in blood), albumin (in egg) *55. Distinguish between globular and fibrous proteins. Ans. Globular proteins i. The proteins are folded to form i. spherical structure, are globular proteins. ii. Globular proteins are soluble in water. ii. iii. They are sensitive to small changes of iii. temperature and pH. iv. They possess biological activity. iv. Biomolecules Fibrous proteins The proteins in which the polypeptide chains lie parallel to form fibre like structure, are fibrous proteins. Fibrous proteins are insoluble in water. They are stable to moderate changes of temperature and pH. They do not possess biological activity. 14. 23 Std. XII Sci. Success Chemistry - II www.horizonpublication.com *56. Explain the structure of proteins. Ans. The structure and shape of proteins can be studied at four different levels called primary, secondary, tertiary and quaternary. Each level is more complex than the previous one. i. Primary structure : In primary structure, proteins contain polypeptide chain. Each protein has amino acids linked with each other in a specific sequence and this sequence of amino acids is termed as primary structure. O O || || H CH C NH CH C NH CH | | | R1 R2 R3 Any change in the primary structures creates different protein molecule. ii. Secondary structure : When long amino acid chain is coiled, looped or folded, it gives particular shape to a protein molecule, it is referred as secondary structure. There are two different types of structures -helix and -pleated sheets. a. -Helix structure : In -Helix structure, a poly peptide chain gets coiled by twisting into right handed spiral known as -helix. | The hydrogen bonding between C O and NH groups occurs in different parts of the same chain resulting in folding of polypeptide chain. Right Handed -helix -helix structure -helix b. -pleated structure : The polypeptide chain lie side by side and are held together by intermolecular hydrogen bonding. The poly peptide chains are stretched out resulting in a flat sheet. The contraction results in a pleated sheets called -pleated sheet. HRO H H R O N C C N H H R C O H C N O C C H R Less than 7.2 A Contracted peptide chain Biomolecules -pleated structure Pleated sheet structure 14. 24 HORIZON Publication Std. XII Sci. Success Chemistry - II iii. Tertiary structure : The secondary structure on folding gives rise to molecular shapes i.e., fibrous and globular. The polypeptide in tertiary structure held by disulphide or hydrogen bonds or Van Der Waals forces or electrostatic forces of attraction. iv. Quaternary structure : Two or more amino acid chains or polypeptide chains forms complex protein. The spatial arrangement of these polypeptide chains with respect to each other is known as quaternary structure. 57. Write a note on (i) -helix (ii) -pleated. Ans. Ref. Q. 56 (ii) (a) and (b). *58. What is denaturation of proteins ? How is it brought about ? Ans. i. The irreversible process of precipitation of proteins which takes place very easily is known as denaturation of proteins. ii. This process is brought about by heating the protein with alcohol, concentrated inorganic acids or by salts of heavy metals. iii. Denaturation uncoils the protein and destroys the shape and thus the characteristic biological activity. iv. Eg. a. Conversion of milk to curd b. Boiling of egg coagulates egg white 14.4 Enzymes *59. Define enzymes. Ans. Enzymes are the biological catalysts for various biochemical reactions in living organisms. 60. Write a note on enzymes and give example. Ans. i. All biological or bio-catalysts which catalyse the reactions in living organisms are called enzymes. ii. Chemically all enzymes are proteins. iii. They are required in very small quantities as they are catalyst also they reduce the activation energy for a particular reaction. iv. Eg. : Enzyme maltase converts maltose to glucose. maltase C12 H 22 O11 Maltose C6 H12 O 6 Glucose *61. Betaine C5H11O2N, occurs in beet sugar molasses. It is water soluble solid that melts with decomposition at 300 °C. It is unaffected by a base but reacts with hydrochloric acid to form a crystalline product C5H12O2NCl. It can be made from glycine with methyl iodide or treatment of chloroacetic acid with trimethyl amine. Draw structure for betaine which will account for all properties given above. HCl Ans. i. C5H11O 2 N C5H12O 2 NCl Betaine present in beetroot Crystalline product ii. Preparation of Betaine : CH 3 CH 3 | | H 2 N CH 2 COOH 3CH3I H 3C N CH 2 COOH H 3C N CH 2 COO HI 2HI | | CH 3 CH 3I Glycine Methyl iodide Betaine Biomolecules 14. 25 Std. XII Sci. Success Chemistry - II CH 3 | H 3C N CH 2 COOH H 3C | | CH 3 Cl Trimethyl amine iii. www.horizonpublication.com CH 3 | H 3C N CH 2 COOH HI | CH 3 Cl Chloroaceticacid CH 3 | N CH 2 COO | CH 3 Betaine CH 3 | The structure of Betaine is H 3C N CH 2 COO | CH 3 14.5 Lipids 62. What are lipids? Ans. Lipids are naturally occurring biomolecules having limited solubility in water and isolated from organisms by extraction with non-polar solvents. *63. How are lipids classified ? Ans. Lipids are classified into two categories : i. Complex lipids ii. Simple lipids i. Complex lipids : They are the esters of long chain fatty acids. It can be hydrolysed. Complex lipids includes triglycerides, glycolipids, phospholipids and waxes. a. Triglycerides are triesters of glycerol with higher fatty acids. O CH 2 OH || | H2C – O – C – R1 CH OH O | || CH 2 OH HC – O – C – R2 Glycerol O || H2C – O – C – R3 Triacyl glycerol b. c. d. ii. Fats and oils are mixtures of triacyl glycerols. R1, R2 and R3 may be same or different and may be saturated or unsaturated. Glycolipids : When lipids are associated with sugars, it forms glycolipids. The simplest animal glycolipids are cerebrosides. In plant glycolipids, the sugar group is commonly galactose. In phospholipids, two of the hydroxyl groups in glycerol are esterified by fatty acids and one by phosphate group. Waxes are mainly esters of long chain carboxylic acids with long chain alcohols, usually they are secreted by plants and animals. Simple lipids : These lipids have ester linkages and cannot be hydrolysed. These include steroids, terpenes and prostaglandins. Biomolecules 14. 26 HORIZON Publication Std. XII Sci. Success Chemistry - II a. Steroids are derived from cyclopentanoperhydrophenanthrene, which has nucleus of four rings. OH HO Steroid nucleus b. c. d. e. f. O Cholesterol (sterol) Testosterone (androgen) Sterols exist as free sterols or esters of fatty acids. Animal sterols include cholesterol and lanosterol. Terpenes are unsaturated hydrocarbons. Terpenoids, which are derivatives of terpenes include geraniol, menthol and vitamin A. Prostaglandins are group of C20 lipids. It contains five members ring with two long side chains. It is detected in many body tissues. Testosterone and androsterone is a male sex hormone. Esrtone and estradiol are female sex hormones. Note : Number of Carbon atom Class Example 10 Monoterpenes -Phellandrene 15 20 Sesquiterpenes Diterpenes Abscisic acid Cembrene 30 Triterpenes Squalene 40 Tetraterpenes -Carotene 64. State the functions of lipids in living organism. Ans. i. Food energy is stored as fats and oils. ii. Components of cell membrane are glycolipids. In plants, glycolipids are the main lipid constituents of chloroplasts. iii. Phospholipids and sterols like cholesterol are major components of cell membrane. Lipo proteins are found in cell membrane. iv. Waxes are also lipids and form protective covering on leaves, fruits, feathers of birds. v. Steroids contain adrenal hormones, sex hormones and bile acids. Bile acid helps digestion of fat in intestine. vi. Terpenes include vitamin E and K and phytol. These are biological active compounds useful for the growth and maintenance of the structure in human beings. vii. Prostaglandin can lower blood pressure, gastric secretion and stimulate uterine contractions during child birth. *65. What are sterols and give their classification. Ans. i. Sterols are free esters of fatty acids. ii. Sterols are classified into animal sterols and plant sterols. iii. Animal sterols include cholesterol and lanosterol. iv. Plant sterol is sitosterol. Fungal sterols (mycosterols) include ergosterol. *66. What are triacyl glycerols ? Ans. Triacyl glycerols are the triesters of glycerol with higher fatty acids. Fats and oils are mixtures of triacyl glycerols. R1, R2 and R3 may be same or different and may be saturated or unsaturated. The fatty acids in triacyl glycerols contains an even number of carbon atoms and an unbranched carbon chain. Biomolecules 14. 27 Std. XII Sci. Success Chemistry - II www.horizonpublication.com *67. Mention the examples of terpenes. Ans. Terpenes are unsaturated hydrocarbons and consist of isoprene units. i.e. H2C = C(CH3)CH = CH2 Eg. : -phellandrene, abscisic acid, cembrene, squalene, -carotene. 14.6 Hormones *68. What are hormones and how is the name derived ? OR Write a note on hormones. Ans. i. Hormones are the secretions of endocrine glands. ii. Hormone word is derived from Greek word ‘Hormaein’. iii. The part where hormones secreted in the body are called effectors and the cells where they act on are called targets. iv. Hormones are easily diffusible, have low molecular weight of affect biological processes. v. Hormones are derived from amino acid derivatives and proteins or steroids. Eg. : Thyroid gland secrete Thyroxine. 69. State the various functions of hormones. OR Give different examples, body parts where the hormones are secreted in human body and their functions. Ans. Functions of hormones : 1. 2. 3. 4. Hormone Thyroxine Body part Function Thyroid glands in the i. To increase the rate of energy exchange. neck ii. To increase consumption of O2. Insulin Pancreas i. To control carbohydrate metabolism by (Peptide hormone) increasing glycogen in muscles. ii. Oxidation of glucose in tissue. iii. To lower the blood sugar. Sex hormones Ovaries (females) i. To help in reproduction development. estrogen, progesterone ii. To develop secondary sexual (Steroids) characteristics. iii. To differentiate males from females Androgen Testes (Males) i. To help in reproduction, development. (Steroids) ii. To differentiate males from females. 14.7 Vitamins 70. What are vitamins ? Ans. i. Vitamins are the organic substances that must be supplied to permit proportionate growth in living beings (humans) or for the maintenance of the structure. ii. Plants synthesize nearly all the vitamins but most of them cannot be synthesized by our body. iii. Our daily diet like milk, vegetables, fruits etc, contain all necessary vitamins. However excess or lack of vitamins result in certain diseases. These vitamins perform specific biological functions. iv. Vitamins are designated by alphabets A, B, C, D, H etc. Some of them have subgroups, they are designated as B1, B2, B3, B12, etc. Biomolecules 14. 28 HORIZON Publication Std. XII Sci. Success Chemistry - II *71. Give classification of vitamins with examples. Ans. Vitamins are classified into two groups depending upon their solubility in water or fat and their chemical structure. i.. Depending upon their solubility vitamins are classified into two types : a. Water soluble vitamins : Vitamins B and C are soluble in water. They are readily excreted in urine, have low toxicity and cannot be stored in body. b. Fat soluble vitamins : Vitamin, A, D, E, K and P are soluble in oils and fats are stored in the body in liver and in tissues. ii. Vitamins are also classified according to their chemical structures as follows : a. Aliphatic series : Vitamins of this series contain a long chain of aliphatic compounds e.g. Vitamin C b. Aromatic series : Vitamins of this series contain long chains of aromatic compounds e.g. Vitamin K c. Alicyclic series : Vitamins of this series contain alicyclic rings in their structure. e.g. Vitamin A d. Heterocyclic series : Vitamins of this series contain rings containing hetero atoms, e.g. Vitamin B complex. It is mixture of vitamins B1 + B2 + B3 + B5 + B6 + B12 + Mesoinsositol + Folic acid + -Biotins. 72. Mention different types of vitamins, their sources and diseases due to deficiencies. Ans. Vitamins Sources Diseases due to deficiencies i. Milk, fish liver oil, tomatoes, Night blindness, retardation of A carrots, sweet potatoes growth, dryness of skin and hair. ii. B1 Rice, wheat, meat, green Causes the disease called Beriberi. (Thiamine) vegetables. iii. Egg yolk, fist, yeast, liver Inflammation of tongue, drying of B2 lips and at corners of mouth, cheilosis (Riboflavin) (retarding the growth and digestion) iv. Barley, liver, maize, wheat, rice Pellagra-pigmentation of the skin, B5 degeneration of spinal cord, mental (Nicotinamide) confusion. v. B6 Wheat, fish, maize, liver, milk, Convulsions, failure to gain weight, (A mixture of cereals, yeast mental changes, derangement of Pyriodoxine, enzymes (which control carbohydrate Pyridoxal and metablosim) pyridoxamine) vi. B12 Egg, curd, fish, liver of pig, sheep Degradation of spinal cord, anaemia vii. Oranges, grapes, lemon, green Scurvy (bleeding, spongy, swollen C vegetables like cabbage, gums.) tomatoes, onion, all citrus fruits. viii. Rice, liver of cattle, seed oils, Weakness of muscles, abnormal E soya bean oil, palm oil, wheat growth and deposition of tissue germ oil, cotton seed oil. ix. All green leafy vegetables like Increases blood clotting time K spinach, cauliflower, fish, meat etc. x. Orange, grapes Haemorrhage, decrease in capillary P resistance. Biomolecules 14. 29 Std. XII Sci. Success Chemistry - II www.horizonpublication.com 73. Write the names of all the sources of Vitamin A and C. [Intext Question text book page no. 529] Ans. Vitamins Sources A Milk, fish liver oil, tomatoes, carrots, sweet potatoes C Oranges, grapes, lemon, green vegetables like cabbage, totatoes, all citrus fruits. 74. Write the names of vitamins present in fish and eggs. Ans. Source Vitamins Fish A, B6, B12, K Eggs B2, B12 [Intext Question text book page no. 529] 14.8 Nucleic acids 75. What are nucleic acids ? Ans. Nucleic acids are bimolecules which are found in the nuclei of all living cells in the form of nucleo proteins or chromosomes. (Nucleoproteins = Proteins + Nucleic acid) (prosthetic group) Nucleic acids are esters of phosphoric acid with sugar. Base | O | Base | O | Base | O | Sugar O P O Sugar O P O Sugar O P O | O– | O– | O– Polynucleotide chain DNA is Deoxyribose nucleic acid. RNA is Ribonucleic acid. 76. Explain the chemical composition of nucleic acids. Ans. The complete hydrolysis of DNA or RNA gives mixture of three different components : i. Pentose sugar ii. Phosphoric acid iii. Nitrogen containing heterocyclic compounds (nitrogeneous bases) iv. Nucleoside and nucleotide i. Pentose Sugar : In RNA the sugar is D-ribose and in DNA D-2-deoxyribose, the sugars are found in furanose form. 2-Deoxy means no-OH group at C2 position. Biomolecules 14. 30 HORIZON Publication Std. XII Sci. Success Chemistry - II ii. Phosphoric acid : The sugar units are joined to phosphate through C3 and C5 hydroxyl groups. iii. Nitrogen containing heterocyclic compounds : The heterocyclic bases in DNA are adenine and guanine containing purine ring, cystosine and thymine which contains pyrimidine ring. The heterocyclic bases in RNA are adenine, gaunine, cystosine and uracil. Purines : Pyrimidines : iv. Nucleoside and Nucleotide : A base-sugar unit is called nucleoside and base-sugar phosphoric acid unit is called nucleotide. *77. A nucleotide from DNA containing thymine is hydrolysed. What are the products formed ? Ans. When nucleotide from DNA containing thymine is hydrolysed, 2-deoxy-D-ribose, thymine and phosphoric acid is obtained. Biomolecules 14. 31 Std. XII Sci. Success Chemistry - II 78. Draw the structures of DNA and RNA. Ans. Structure of DNA : www.horizonpublication.com Structure of RNA : – *79. Write the structures of (i) nucleoside and (ii) nucleotide ? Ans. i. Structure of nucleoside : A nucleoside contains two basic components of nucleic acids i.e. a pentose sugar and a nitrogenous base. It is represented a sugar-base. A nucleoside is formed when 1-position of a pyrimidine (cytosine, thymine or uracil) or 9-position of guanine or adenine base is attached to C 1 of sugar by -linkage. ii. Structure of nucleotide : A nucleotide contains all three basic components of nucleic acids i.e. a pentose sugar, a phosphoric acid and a nitrogenous base. These are obtained by esterification of C5 OH group of the pentose sugar by phosphoric acid. Nucleotides are joined together through phosphate ester linkage. O *80. What are the types of RNA ? How do they function ? Ans. i. There are three different types of RNA found in the cell. a. The messenger RNA which carries the message to the ribosome b. Ribosomal RNA where synthesis of protein takes place c. The transport RNA. ii. The messenger RNA calls up the series of transport RNA molecules which contain particular amino acid. iii. The order in which the transport RNA are called up by messenger RNA depends upon the sequence of bases of messenger RNA chains. Biomolecules 14. 32 HORIZON Publication Std. XII Sci. Success Chemistry - II 81. Give differentiating points between DNA and RNA Ans. DNA RNA i. Sugar present in -D-2-deoxyribose. Sugar present in -D-Ribose. ii. Pyrimidine bases present in DNA are Pyrimidine bases present are cystosine cystosine, thyamine. and uracil. iii. In secondary structure helixes are In secondary structure helixes are double stranded. single stranded. iv. DNA is the repository of the heredity RNA usually do not replicate and play information and uses it by replication important role in protein preparation. and duplicate the identical DNA. v. There are no types of DNA present. RNA has three types : i. Messenger RNA ii. Ribosomal RNA iii. Transport RNA 82. RNA on hydrolysis give four bases but there is no relationship between their quantities, similar to that observed for the bases obtained from DNA. From this what can be suggested about the strucutre of RNA ? [Intext Question from Textbook Page 531] Ans. i. This suggests that there are different types of RNA molecules which contain different quantities of bases. ii. The secondary structure of RNA is single stranded. iii. All the three types of RNA function on co-ordination basis with each other. Higher Order 83. Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain. Ans. Glucose and sucrose have –OH groups while cyclohexane and benzene do not. Thus glucose and sucrose can form hydrogen bonds with water and are soluble in water. While cyclohexane and benzene do not form hydrogen bonds with water and thus insoluble in water. The and -glucose have different specific rotations. When either is dissolved in water their rotation changes until the same fixed value results. i. What is this phenomenon called ? ii. Explain the chemistry behind it. Ans. i. Mutarotation. ii. The change in specific rotation of an optically active compound in solution with time, to an equilibrium value, is called mutarotation. When -D-glucose solution has a specific rotation of +111o is allowed to stand, the specific rotation gradually falls to +52.7o and remains constant at this value. Similarly when -D glucose solution has a specific rotation of +19o is allowed to stand, its specific rotation gradually increases and remains constant at +52.7o. In mutarotation, when either forms of glucose ( or ) is dissolved in water, it is converted into other anomer and an equilibrium mixture is formed together with small amount of open chain form. 84. D glucose Open chain form D glucose 36% 0.02% 63% Biomolecules 14. 33 Std. XII Sci. Success Chemistry - II www.horizonpublication.com 85. List the reactions of glucose which cannot be explained by its open chain structure. Ans. Limitation of the open chain structure : Although the open chain structure of D – (+) – glucose explains most of its reactions, yet it fails to explain the following facts. i. D (+) – glucose does not undergo certain reactions of aldehydes. For example, glucose does not form NaHSO3 addition product, aldehyde – ammonia 2, 4 – DNP derivative and does not respond to Schiff’s reagent test. ii. Glucose reacts with NH2OH to form an oxime but glucose penta acetate does not. 86. The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain. Ans. – NH2 group present in amino acids has the ability to form intermolecular H-bonds. Thus melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. 87. Where does the water present in the egg go after boiling the egg ? Ans. About three-fourth of an egg is plain water in which the albumin protein and fats are suspended. At room temperature, the protein strands are tightly folded in a complex three dimensional shape. Because the individual proteins are able to move freely, both the egg white and yolk remain liquid. But when egg is boiled, the protein strands straighten out. The protein ends, normally protected within the folds, become exposed and the open ends join together in bridge like bonds forming a giant structure. These new bonds prevent the molecules to move thereby making the egg material solid and the water seems to be disappeared. 88. Why cannot vitamin C be stored in our body ? Ans. Vitamin C is water soluble and is thus readily excreted in urine and cannot be stored in our body. It is due to this reason that it must be supplied regularly in diet. Biomolecules 14. 34 HORIZON Publication Std. XII Sci. Success Chemistry - II Quick Review Carbohydates O || Optically active, Polyhydroxy and – CHO or – C – On the basis of hydrolysis 1. Simple carbohydrates known as Monosaccharides Complex carbohydrates Aldoses – CHO group Triose 3 ‘C’ Tetrose 4 ‘C’ Ketoses O || C group Pentose 5 ‘C’ Hexose 6 ‘C’ In all functional group is CHO Triose 3 ‘C’ Tetrose 4 ‘C’ Pentose 5 ‘C’ Hexose 6 ‘C’ In all functional group is Keto Oligosaccharides on hydrolysis give two to ten monosaccharides Disaccharide give 2 monosaccharide Trisaccharide give 3 monosaccharides Polysaccharides give large no. of monocasccharides on hydrolysis Tetrasaccharide give 4 monosaccharides On the basis of solubility Sugars Soluble in H2O, monosaccharides oligosaccharides Non-sugars Insoluble in H2O, Polysaccharides On the basis of reducing property Reducing sugars reduce Fehling’s solution, Tollen’s reagent Biomolecules Non-reducing sugars do not reduce Fehling solution and Tollen’s reagent 14. 35 Std. XII Sci. Success Chemistry - II 2. Glucose known as dextrose Preparation Laboratory method from cane sugar alcohol is added to remove fructose 3. www.horizonpublication.com Commercial method from starch, chalk powder is added to remove excess H2SO4, charcoal powder to remove coloured impurity n-Hexane Glucose + CH3COOH pentaacetate HI Gluconic acid Oxidation (CH3CO)2O NH2OH Glucoxime Reactions of glucose HCN Oxidation dil. HNO3 Bromine water Oxidation dil. HNO3 Saccharic acid Glucose cyanohydrin Haworth projection formulae structures of 4. Monosaccharides i. Glucose ii. Fructose 5. Disaccharides i. Sucrose ii. Maltose iii. Cellobiose iv. Lactose Polysaccharides i. Starch ii. Cellulose Proteins Basic units -amino acids Fibrous proteins Eg. Keratin Globular proteins Eg. Haemoglobin 6. Structures of proteins Primary Biomolecules Secondary Tertiary Quaternary 14. 36 Std. XII Sci. Success Chemistry - II 7. HORIZON Publication Classification of amino acids Acidic more – COOH groups 8. Basic more – NH2 groups Neutral equal – COOH and – NH2 groups Enzymes : eg. Maltase Behaves mainly as catalysts in different reactions. Lipids 9. Complex Eg. Triacyl glycerol Simple Eg. Cholesterol 10. Hormones : Secretions of endocrine glands regulate vital body functions. Eg. : Thyroxin, Insuline 11. Vitamines : Based on solubility Water soluble vit. B, C Fat soluble vit. A, D Based on chemical structures Vitamins of aliphatic series vit. C Aromatic series vit. K 12. Heterocyclic series vit. B-complex Nucleic acids DNA Double helix Replicate, duplicate identical DNA Biomolecules Alicyclic series vit. A RNA Single strand No replication and duplication Perform functions through 3 types of RNA 14. 37 Std. XII Sci. Success Chemistry - II www.horizonpublication.com SHORT TEST Time : 1 Hour Max. Marks : 20 1. Answer the following questions. i. What happens when glucose is treated with [3] a. Bromine water b. Dilute nitric acid c. Hydrogen cyanide ii. Define complex lipids and give two functions of lipids. [2] iii. What are hormones and write the structure of simple triglycerides. [2] iv. What is meant by peptide linkage and biocatalysts ? [2] v. What is glycogen ? How is it different from starch ? [2] vi. What is essentially the difference between -glucose and -glucose ? what is meant by pyranose structure of glucose and give structures. [3] vii. Write the structures of nucleotide and nucleoside. [2] viii. Write the full forms of DNA and RNA. [2] 2. Choose the correct option : [2] i. Inflammation of tongue is due to the deficiency of (a) vit. B1 (b) vit. B2 (c) vit. B5 (d) vit. B6 ii. How many moles of acetic anhydride will be required to form glucose penta-acetate from 2M of glucose ? (a) 2 (b) 5 (c) 10 (d) 2.5 ANSWER 1. i. a. Refer to Q.17 (iv) b. Refer to Q.17 (v) c. Refer to Q.17 (iii) ii. Refer to Q.63 and Q.64. iii. Refer to Q.63 and Q.68. iv. Refer to Q.51 and Q.59. v. Refer to Q.40. vi. Refer to Q.21. vii. Refer to fig. of Q.76 (iv) viii. Refer to Q.75. 2. i. (b) Refer to Q.72 (iii) ii. (c) Biomolecules 1 Glucose + 5 acetic anhydride 1 Glucose Pentacetate 14. 38