Bio Molecules

CONTENTS S.NO.

TOPIC

PAGE NO

1.

INTRODUCTION

2

2.

CARBOHYDRATES

2

3.

CLASSIFICATION OF

4.

PREPARATION OF GLUCOSE

3

5.

DISACCHARIDES

6

6.

POLY SACCHARIDES

8

7.

-AMINO ACIDS

10

8.

CLASSIFICATION OF AMINO ACIDS

10

9.

PROTEINS

12

10.

STRUCTURE OF PROTEINS

12

11.

NUCLEIC ACIDS

14

12.

LIPIDS

16

CARBOHYDRATES

2

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BIOMOLECULES 1. INTRODUCTION : Biochemistry may be defined as the study of the chemical composition and structure of living matter and the chemical changes that take place in them during the life processes. Complex organic molecules (carbohydrates, lipids, proteins and nucleic acids) which govern the common activitites of a living organism are called biomolecules.

2. CARBOHYDRATES : Hydrates of carbon because the ratio of hydrogen and oxygen in these compounds is the same as present in water. Carbohydrates are now defined as optically active polyhydroxy aldehydes or polyhydroxy ketones and other large molecules, like starch, glycogen or cellulose, which produce these compounds on hydrolysis.

3. CLASSIFICATION OF CARBOHYDRATES : Carbohydrates are classified into two major groups on the basis of their physical properties. (a)

Sugars

(b)

Nonsugars or polysaccharides

(a)

Sugars Sugars are crystalline substances, sweet in taste, and readily soluble in water. These have a fixedmolecular

weight and thus have a sharp m.p. Examples are glucose, fructose, sucrose and lactose etc. Classification of carhohydrates :

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ON THE BASIS OF HYDROLYSIS : Monosaccharides : Acarbohydarte that cannot be hydrolysed further into simpler unit of polyhydroxy aldehyde or ketone are R/a on the basis of no. of C-atoms they are : 1. Triose – Glyceraldehyde 2. Tetrose - Crythrose 3. Pentose – Ramnose, Ribose Hexose - Cellulose, fructose 4. Hebtose – Rham 5. Oligosaccharides : Oligosaccharides are sugars which on hydrolysis give two or more molecules of monosaccharides. These are further classified as di, tri or tetra saccharides, etc. Disaccharides : Disaccharides are sugars which on hydrolysis produce two molecules of the same or different monosaccharides. These are sucrose, maltose and lactose. They have the same chemical formula C12H22O11. Invertase  C6 H12 O6  C6 H12 O6 C12 H 22 O11  H 2O  or H 

Sucrose

Glucose

Fructose

C12 H 22 O11  H 2O maltase 2 C6 H12 O 6 Maltose

or H 

Glucose

Lactase C6 H11O6  C6 H12 O 6 C12 H 22 O11  2H 2O  or H

Lactose

Glucose

Galactose

Trisaccharides : Sugars which yield three molecules of the same or different monosaccharides on hydrolysis are called tri-sacchardies. e.g., raffinose C18H32O16 C18 H 32 O16  2H 2O  C6 H12 O6  C6 H12 O 6  C6 H12 O 6 Raffinose

Glucose

Fructose

Galactose

Polysaccharides : Carbohydrates which yield large number of monosaccharides unit on hydrolysis are R/Ar poly saccharides.

4. PREPARATION OF GLUCOSE : 1.

From sucrose : When sucrose is boiled with dil. H2SO 4 is alcoholic solution, to form glucose and fructose.  C12 H 22O11  H 2O H  C6 H12 O6  C6 H12 O6 Glucose

2.

Fructose

From starch :

 C6 H10O5 n  nH 2O H  C6 H12 O6 Starch or cellulose

Glucose

Chemical properties of glucose : (i) Reaction with hydrogen Iodide, (HI).

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(ii)

Reaction with hydroxylamite

(iii)

Reaction with hydrogen cyanide

Cyanohydrin

(iv)

reaction with bromine water.

(v)

Reaction with acitic and hydride

(vi)

Oxidation of glucose

The rxn indicates the presence of primary alcoholic gp. on qlucose.

STRUCTURE OF GLUCOSE : Conjugation of monogacchazides : • Open chain structure :

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D & L has no relation with optical activity & the optical rotation in glucose molecules is given by + Or – siqn. (R.H.S) +  OH gp (L.H.S) –  OH gp Hemiacetal structure

The two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl gp at C1 called Anomeric carbon (the aldehyde carbon before cyclisation). Such isomers i.e., -form and -form, are called anomers. The six membered cyclic structure of glucose is called pyranose. Structure because it resembele pyron ring structure.

Fructose : Structure of fructose :

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It also exists in two cyclic forms which are obtained by the add of –OH at C5 to the carbonyl gp. The ring, thus formed, is a five membered ring and in named as furanose.

–d–(–) – fructo furanose

Cyclic stucture of anomers of fructose :

5. DISACCHARIDES :

1.

When a sugar molecule on peydrolysis with dilute acids, or enzymes cyield two moelcules of either same of different monosaccharides are called disccharides. The linkge is formed by loss of water between two monosacchocides and attached byoxygen atom are called glcosidic linkage. Sucrose : Sucrose on hydrolysis quves equimolar mixture of D – (+) – Glucose and D – (–) – fructose. Invertase C12 H 22 O11  H 2O   C6 H12 O6  C6 H12 O6 Sucrose

or H 

Glucose

Fructose

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2.

The levo rotation of fructose (–92.4°) is more than dextrorotations fo glucose (+ 52.5°), so the mixture is levorotatory. The hydrolysis of sucrose brings a chauqe in the siqn of rotation from dextro thus the hydrolysis of rotation from dextro to lawo and the product is called invert sugar. Maltose : maltase C12 H 22 O11  H 2O  Maltose

3.

 2 C6 H12 O 6

or H 

Glucose

In C1 and C4 there is glycosidic likange. It shows reducing property so it is a reducing sugar. Lactose : It is also R/a milk sugar as it is found in milk. It is composed of B–D–galactose and B–D–Glucose. Lactase C6 H11O6  C6 H12 O 6 C12 H 22 O11  2H 2O  Lactose

or H 

Glucose

Galactose

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Biomolecules

6. POLY SACCHARIDES : 1.

Starch : It is a main storage poly saccharides of having general formula (C6H10O5)n. The main source is maize, wheat, barley rice and potates. It is made ofAneylose &Aneylopectin Amylose – 200 – 1000 zalton – 10 – 20% Aneylopectin – 1000 – 500 zaltone 80–85% Amylose is made up of long - unbroanched chain of  – D – (+) – glucose linkage. Amylopectin is a bronched chain polymer of  – D – glucose units is which chain is formed by C1–C4 qlycosidie linkage whereas branching occurs by C1–C6 qlycosidic linkage. H

(C6 H10 O5 ) n  nH 2O  n C6 H12 O6 Starch

Glucose

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S.No.

2.

Amylose

Amylopectin

1.

It is water soluble fraction of starch

It is water insoluble fraction of starch.

2.

It is 20% of the quantity of starch

It is 80% of starch.

3.

It is a straight chain polymer of D-glucose units.

It is branched chain polymer of D-glucose units.

4.

In amylose, the glucose units are joined by -1, 4 glycosidic linkage

In amylopectin, the glucose units are joined by -1, 6 glycosidic linkage.

5.

Its molecular mass lies in the range of 10,000 – 50,000

Its molecular mass is in the range of 50,000 – 1,00,000.

Cellulose : It is found in cellwall of plant cells. It is a linear polymer of D – C– glycose in which qlucose units are link together by  – 1 – 4 glycosidic linkage. It is a non-reducing sugar.

Relative sweetness of sugar Sugar Relative Sugar Relative sweetness sweetness 100 (standard) Glucose 74 Sucrose Lactose 16 Invertsugar 130 Galactose 32 Fructose 173 Maltose 33 Cellulose is the most important natural polymer whose chemical treatment gives various useful derivatives e.g., a. Rayon : Cellulose acetate and cellulose xanthate are used as a fibre. b. Celluloid : Cellulose dinitrate also known as pyroxylin, mixed with plasticizer and alcohol, is used for the manufacturing of photographic film, spectacle frames, piano keys, etc. It is known as artificial ivory. Explosive : Cellulose trinitrate is used extensively as a blasting and propellant explosive. c. Lacquer: Collodion is used for manufacturing washable cellulose paints. d. e. Water proofing : Solution of cellulose acetate is used to provide antishrink property to textile fabric. f. Methyl cellulose is used in fabric sizing, paste and cosmetics. Ethyl cellulose is used for manufacturing of rain coats and plastic films.

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7. -AMINO ACIDS : -Amino acids are substituted carboxylic acids in which one -hydrogen atoms of alkyl group is substituted by amino (–NH2) group. These may be represented by the general formula.  R  CH  COOH

where R = H or alkyl group

NH2

Structure of -amino acids The amino acids containing one carboxylic group and one amino group behave like a neutral molecule. This is due to the fact that in aqueous solution the acidic carboxylic group and basic-amino group neutralise each other intramolecularly to form an internal salt structure, known as zwitter ion or dipolar ions. +

NH2

NH3

R  C  COOH

R  CH  COO– Zwitter ion

However, the neutral zwitter ion (dipolar ions) changes to cation in acidic solution and exist as anion in alkaline medium. Thus amino acids exhibit amphoteric character. H2N  CH  COO– R Anion

Alkali

H3N+  CH  COO– R Zwitter ion

Acid

H3N+  CH  COOH R Cation

Therefore, amino acid exist as zwitter ion when the solution is neutral or its pH ~ 7. The pH at which the structure of an amino acid has no net charge is called its isoelectric point.

8. CLASSIFICATION OF AMINO ACIDS : -Amino acids are broadly classified in three main groups based on the relative number of –NH2 and –COOH group Neutral amino acids : Amino acids containing one –NH2 group and one –COOH group (i) are called neutral amino acids e.g., glycine, valine, alanine, etc. Basic amino acids: Amino acids which contain one –COOH group and two –NH2 groups (ii) are called basic amino acids e.g., lysine and arginine. Acidic amino acids : Amino acids containing two –COOH groups and one –NH2 group are (iii) classified as acidic amino acids e.g., aspartic acid and glutamic acid, etc.

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List of 20 naturally occuring -amono acids. Abbreviation or code

Structure

1.

NEUTRAL Glycine

Gly

H2N – CH2 – COOH

2.

Alanine

Ala

H2N – CH – COOH

-Amino acid

CH3 3.

Valine

Val

H2 N – CH – COOH CH (CH3)2

4.

5.

Proline

Pro

ACIDIC

Asp

Serine

Ser

H2 N – CH – COOH CH2 – OH

6.

H2N – CH – COOH

Aspartic acid

CH2 – COOH 7.

Glutamic acid

Glu

H2N – CH – COOH CH2 – CH2 – COOH

BASIC 8.

Lysine*

Lys

9.

Arginine*

Arg

H2N – CH – COOH CH2CH2CH2 – CH2NH2 H2N – CH – COOH CH2 (CH2)2NH – C – NH2 NH

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9. PROTEINS : Proteins are complex, nitrogeneous organic substance which occur in all animals and plants. They are so named because proteins are the most vital chemical substances of primary importance neccessary for the normal growth and maintenance of life. Protein serves following functions in our body. a. To promote growth. b. To supply essential amino acids to blood. c. To maintain body tissues. d. To synthesize various enzymes. e. To protect body from infection. Protein content of food stuffs Food stuff % Protein Food stuff % Protein Milk 5 Meat 24 – 26 Wheat 14 Egg yolk 16 12.5 Peas 21 Egg (white) Maize 10 Cheese 33 Classification of proteins Proteins are classified by two different methods. According to first mode of classification proteins are of two types depending upon their shape and functions. (a) Fibrous proteins (b) Globular proteins (a) Fibrous proteins : These have thread like molecules which lie side by side to form fibres. The various molecules are held together by hydrogen bonds. These are insoluble in water but soluble in concentrated acids and alkalis e.g., hair, nails, wood, feathers and horn, etc, are made up of keratin. Muscles have myosin. Silk is composed of fibroin. Bones and cartilage have collagen. (b) Globular proteins : This type of protein has molecule folded into compact units which often acquire spheroidal shape. Such proteins are soluble in water, dilute acids and alkalis e.g., insulin, hemoglobin, albumin, etc.

10. STRUCTURE OF PROTEINS : The structure of proteins is quite complex. Study of its structure is carried out under the following headings. (a) Primary structure of protein : The primary structure of protein refers to its covalent structure, i.e., the sequence in which various -amino acid are arranged in protein or in the polypeptide structure of protein. The linkage –CO – NH – is known as peptide linkage. H2N – CH – COOH + H2N – CH – COOH R

R

H2N – CH – CONH  CH – COOH + H2O Peptide bond

R

R

The dipeptide still has free amino and carboxyl groups through which it an react with other molecules of amino acid resulting in polypeptide formation O C

H

R

O

N

CH

C

H N

R CH

CH

C

N

CH

C

N

R

O

R

R

O

H

C-terminal N-terminal Linear polypeptide chain

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Biomolecules

In polypeptide chain, the free amino end is termed as N-terminal and the free carboxyl end is said to be C-terminal end. Secondary structure of proteins Secondary structure of protein refers to the arrangement of polypeptide chains into a definite three dimensional structure which protein assumes as a result of hydrogen bonding. Depending upon the size of the R-group of the amino acids in polypeptides, two different types of secondary structure are possible as discussed below (a) (b) -helix structure -Pleated structure -Helix structure : This type of secondary structure is acquired when the R groups in amino acids are large and involve in coiling of the polypeptide chains. The helical pattern in right handed coil and the shape is stabilised by the intra molecu lar hydrogen bond between t he >C = O group of one amino acid and –NH group of the fourth amino acid.

-Pleated structure : This type of secondary structure is acquired when the R groups of amino acids are small. In this structure the linear polypepotide chains are arranged side by side and held together by intermolecular hydrogen bond between the C = O and –NH group. R

H C C O

N

H N H

H

C O

O

H

R

H

C H

N H O C

N C

C

N R

C

R

C

R

H

H

C C

H N

O C

O

H

R

7.2Å

Tertiary structure of proteins The tertiary structure of protein is the most stable shape that a protein assumes under the normal conditions of temperature and pH. During acquiring of tertiary structure, various types of attractive forces between the amino acid chains are involved. These attractive forces, like hydrogen bond, disulphide bonds, ionic, chemical and hydrophobic bonds, result in a complex and compact structure of protein. The two important teritary structures of proteins are fibrous structures and globular structure. Fribrous proteins have largely helical structure and are rigid molecules of rod like shape. Globular proteins, on the other hand, have a polypeptide chain which consist partly of helical sections and partly -pleated structure and remaining in random coil form. These different segment of secondary structure then fold up to give protein a spherical shape.

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Quaternary structure of proteins The quaternary structure of protein is developed when the polypeptide chains, which may or may not be identical are held together by hydrogen bonds. It results in the increase of molecular mass of protein greater than 50,000 amu. For example, haemoglobin contains four submits, two identical chains containing 141 amino acids each and the other two identical -chains containing 161 amino acids each. Denaturation of proteins When the proteins are subjected to the action of heat, mineral acids or alkali, the water soluble form of globular protein changes to water insoluble fibrous protein resulting in the precipitation or coagulation of protein. This is called denaturation of proteins.

11. NUCLEIC ACIDS : Nucleic acids are vital biomolecules which are present in the nuclei of all living cells in the form of nucleoproteins. These are long chain polymers with a high molecular mass. These are also known as biopolymer having nucleotide as a repeating structural unit (monomer). These play an essential role in transmission of the heredity characteristics from one generation to the next and also in the biosynthesis of proteins. Therefore, the genetic information coded in nucleic acid governs the structure of protein during its biosynthesis and hence controls the metabolism in the living system. The nucleic acids are of two types differing mainly in the nature of carbohydrate present in them. (i) DNA (Deoxyribonucleic acid) (ii) RNA (Ribonucleic acid) Structure of nucleic acids The nucleic acids are the posthetic component of the nucleoproteins. Nucleic acid on stepwise hydrolysis gives following products as shown in the chart Nucleic acid Hydrolysis Nucleotides Hydrolysis

Nucleosides Hydrolysis

Pentose sugar Ribose or Deoxyribose

Purine bases (Adenine and guanine)

Phosphoric acid

Pyrimidine bases (Thymine and cytosine) from DNA (Uracil and cytosine) from RNA

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Difference between DNA and RNA The main points of difference betwen the two types of nucleic acids are given in the table.] DNA

RNA

1.

The pentose sugar present in it is 2deoxy D(–) ribose.

It has D (–) ribose sugar.

2.

It contains cytosine and thymine as pyrimidine bases

It contains cytosine and uracil as pyrimidine bases.

3.

DNA is double strand and pairing of bases is present throughout the molecule. It occurs in the nucleus of the cell.

It is a single strand molecule looped back on itself. The pairing of bases is present only in helical part.

5.

It is a very large molecule and the molecular weight vary from 6 million to 16 million amu.

It is a much smaller molecule and its molecular weight ranges from 20 thousand to 40 thousand.

6.

It has a characteristic property of replication

It does not replicate.

7.

DNA controls the heredity character.

RNA only governs the biosynthesis of proteins.

4.

It mainly occurs in the cytoplasm of the cell.

Ribonucleic acids 1. The pentose sugar in RNA is ribose. 2. Purine bases of RNA are represented by adenine and guanine, the pyrimidine bases are uracil and cytosine. The thymine in DNAis replaced by uracil in RNA. 3. 4. RNAis single stranded, but double stranded RNAis present in Rheovirus and wound tumour virus. 5. There are three major classes of RNA, each with specific functions in protein synthesis mRNA 6. Messenger RNAis produced by DNA; the process is called transcription. 7. Messenger RNA encodes the amino acid sequence of a protein in their nucleotide base sequence. Atriplet of nitrogenous bases specifying an amino acid in mRNAis called codon. 8. tRNA tRNA is also known as soluble RNA(sRNA) it is soluble in 1 molar solution of sodium chloride. 9. tRNAidentifies amino acids in the cytoplasm and transports them to the ribosome. 10. Molecules of tRNA are single-stranded and relatively very small. 11. Anticodon is a three-base sequence in a tRNA molecule that forms complementary base pairs with a 12. codon of mRNA. 13. All transfer RNApossess the sequence CCAat their 3 ends; the amino acid is attached to the terminal Aresidue. rRNA 14. Ribosomal RNA is found in ribosomes of cells and is also called insoluble RNA.

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15.

The mainfunction of rRNAis to attract and provide large surface for spreading oo m-RNAover ribosomes during translocation process of protein synthesis. Type SedimentationMol. Wt. Number of % of total cell coefficient nucleiotide residues RNA 75-3000 2 mRNA 6 to 25S 25,000-1000,000 tRNA 4S 23,000-30,000 75-90 16 rRNA 5S 35,000 100 82 16S 550,000 1500 23S 1100,000 3100 Genetic Code 16. The relationship between the sequence of amino acids in polypeptide with base sequence of DNAor mRNA is genetic code. Genetic code determines the sequence of amino acids in a protein. 17. 18. A triplet would code for a given amino acid as long as three bases are present in a particular sequence. 19. Later in a cell-free system, Marshall Nirenberg and Philip Leder (1964) were able to show that GUU codes for the amino acid valine. 20. The spellings of further codons were discovered by R. Holley, H. Khorana and M. Nirenberg. They have been awarded the Nobel Prize in 1968 for researches in genetic code.

12. LIPIDS The term lipids represent a group of biomolecules which are insoluble in water but soluble in organic solvents of low polarity such as chloroform, toluene, ether, carbon tetrachloride,. They also serve as the energy reserve for living cells. Lipids are classified in three groups (i) Triglyceride esters of higher fatty acids or oils and fats Phospholipids (ii) Waxes (iii)

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Biomolecules

SOLVED EXAMPLE Ex.1 Sol.

Ex.2 Sol.

Glucose or sucrose are souble in water but cyclohexane or benzene (simple six membered ring compound) are insoluble in water. Explain. Glucose contains five – OH groups and sucrose contains eiht – OH groups form hydrogen-bonds with water. Because of this extensive intermolecuar Hydrogen-bonding, glucose and sucrose are solublke in water irrespective of the fact that their molecular masses are 180 amu and 342 amu respectively. On the other hand, benzene (molecular mass = 78) and cyclohexane (molecular mass = 84) are simple molecular having low molecular masses. Even then they are insoluble in water. The reason being that these compounds do not contain –OH groups and hence do not form hydrogen-bonds with water. What products are expected when lactose is hydrolysed ? On hydrolysis, loctose gives two molecules of monosaccharides, i.e., onle molecule of D-(+)- glucose and D-(+)-Galactose. 

H3O C12 H12 O11 H 2O   C6 H12 O6  C6 H12O6 Lactose

or Lactase

D( )Glu cos e

D()Galactose

Ex.3 Sol.

What the constituents of maltose. It is a polysaccharide. It is polymer of D-glucose. It stores food in animals and human beings.

Ex.4 Sol.

What are polysaccharides ? Give one example ? Polysaccharides are the carbohydrates which on hydrolysis give a large number of molecules of monosaccharides. For example, starch or cellulose.

Ex.5 Sol.

What is difference between amylose and amylopectin. Amylose is water soluble linear of a-glucose.Amylopectin is water insoluble brancehd chain polymer of -glucose.

Ex.6 Sol.

What are oligosaccharides ? Carbohydrates which on hydrolysis give 2-10 molecules of monosaccharides are called oligosaccharides. For example, sucrose, raffinose, stachyrose, etc.

Ex.7 Sol.

What is mutarotion ? The spontaneous change of specific rotation of an optically active substance with time is called mutarotation.

Ex.8 Sol.

What are amino acids ? Amino acids are biomolecules which contain amino (–NH ) group at -carbon and a carboxylic acid 2

group (–COOH). Ex.9 Sol.

How do you explain the absence of aldehyde group in pentaacetate of glucose ? The cyclic hemicetal form of glucose contains an OH group at C-1 which gets hydrolysed in the queous solution to produce the open chain aldehydic form which then reacts with NH2OH to form the corresponding oxime. Therefore, glucose contains an aldehydic group. On the other hand, when glucose is reacted with acetic anhydride, the OH group at C-1, along with the four other OH groups at C-2, C-3,

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Biomolecules

C-4 and C-6 form a pentaacetate. As the pentaacetate of glucose does not contain a free OH group at C-1, it cannot get hydrolysed in aqueous solution to produce the open chain aldehydic form and thus glucose pentaacetate does not react with NH2OH to form glucose oxime.

Hence, glucose pentaacetate does not contain the aldehyde groups. Ex.10 State two main functions of carbohydrates ? Sol. (i) Carbohydrates act as biofuel to provide energy for functioning of living organisms. (ii) They act as constituents of cell walls. Ex.11 What is the structural feature characterising reducing sugars ? Sol. The main structural feature of reducing sugar is the presence of an alodehyde group (– CHO) such asin glucose, mannose galactose, etc. or -ketol grouping (–CO–CH2OH) as present in fructose. Ex.12 What is invert sugar ? Sol. An equimolar mixture of glucose and fructose is called invert sugar. Ex.13 What are constituent units of cellulose ? Sol. Cellulose is a linear polymer made up of D(+) glucose molecules linked by-glycosidec bonds. Ex.14 Why is cellulose not digested in human body ? Sol. It is due to the fact that humajhn beings do not have enzyme to digest cellulose. Ex.15 What is meant by inversion of sugar ? Sol. Sucrose is dextrorotatory but on hydrolysis, it gives an euimolar mixture of D(+)-glucose and D (–)fructose which is levorotatroy. This cange of specific rotation from dextrorotation to laevorotation is known as inversion of sugar.

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Biomolecules

Ex.16 The meltaing points and slubility in water of amino acids are higher than those of the corresponding halo acids. Explain. Sol.



The amino acids exist as zwitter ions, H3 N – OHR – COO – . Because of this dipolar salt like charactor, they have strong dipole-dipole attractions. So, their melting points are higher than halo acids which do not have salt like character. Moreover, due to this salt like character, they interact strongly with H O.2 Thus solubility in water of amino acids is higher than that of the corresponding halo acids which do not have salt like character.

Ex.17 Where does the present in the egg go after boiling the egg ? Sol. On boiling the egg, the proteins under go denaturation and the water present in the egg gets absorbed or adosorbed in the denaturated proteins probably through H-bonding. Ex.18 Explain why vitamin C can not be stored in the body ? Sol. Vitamin C is soluble in water, hence, it is readily excreted in urine and thus cannot be stored in the body. Ex.19 What products would be formed when a nucleotide from DNA containing thymine is hydrolysed ? Sol. Besides thymine, the two products are 2-deoxy-D-ribose and phosphoric acid. Ex.20 When RNAis hydrolysed, there is no relationship among the quantities of different bases obtained ? Sol.

What does this fact suggest about the structure of RNA? ADNAmolecule has two strands in which the four complementary bases pair each other, viz. cytosine (C) always pairs with guanine (G) while thymine (T) always pairs with adenine (A). Therefore, when a DNA molecule is hydrolysed, the molar amounts of cytosine is always equal to that of guanine and that of adenine is always equal to that of thymine. RNAalso contains four bases, the first three are same as in DNA but the fourth one is uracil (U) As in RNAthere is no relations hip between the quantities of four bases (C, G, Aand U) obtained, therefore, tthe base-pairing principle, viz. a pairs with U and C pairs with G is not followed. So unlike DNA, RNA has a single strand.

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EXERCISE-I Q.1

What happens when D-glucose is treated with the following reagents? (i) HI (ii) Bromine water (iii) HNO3

Q.2

Define the following terms in relation to proteins (i) Peptide linkage (ii) Denaturation.

Q.3

Define the following as related to proteins (i) Peptide linkage (ii) Primarystructure (iii) Denaturation.

Q.18 (a) Give reasons for the following statements : (i)Amino acids are amphoteric in nature . (ii) Amino acids have comparatively higher melting po ints than the corr esponding haloacids. (b) What deficiency diseases are caused due to lack to lack of vitamins A, B1, B6 and K in human diet? Q.19 The two strands in DNA are not identical but are complementary. Explain.

Q.4

What are the common types of secondary structure of proteins?

Q.5

How do you explain the amphoteric behaviour of amino acids?

Q.6

What is the effect of denaturation on the structure of proteins?

Q.20 State difference between the following pair (i) -helix and -pleated structures. (ii) Primary and secondary structures of a protein. Q.21 What are nucleic acids? Mention their two important functions.

Q.7

Describe the following (i) Glycosidic linkage (ii) Peptide linkage

Q.22 What is the difference between a nucleoside and a nucleotide?

Q.8

Enumerate the reactions of D-glucose which Q.23 What are reducing and non-reducing sugars ? cannot be explained by its open chain What is the structural feature characterising structure. reducing sugars ? What are essential and non-essential amino Q.24 Distinguish between -glucose and -glucose. acids? Give two examples of each type. Q.25 What happens when D-glucose treated with the following reagents ? List any for vitamins. Mention the chief sources (iii) HNO3 (i) HI (ii) Bromine water and functions of two of them. (i) What are essential and non-essential amino Q.26 Write the important structural and functional differences between DNAand RNA. acids ? Give two examples of each. (ii) What is a denatured protein Q.27 What are the different types of RNA found in the cell ? How are vitamins classified? Name the vitamin responsible for the coagulation of blood. Q.28 Define the following and give one example of each Why are vitamin A and vitamin C essential to (a) Isoelectric point (b) Mutarotation us? Give their important sources. (c) Enzymes. Draw open chain structure of aldopentose and Q.29 Answer the following queries about proteins ? aldohexose. How many asymmetric carbons are (i) How are proteins related to amino acid ? present in each ? (ii) How are oligopeptides different from (a) Describe the following giving one example: polypeptides ? Nucleotides. (iii) When is a protein said to be denatured ? (b) List four functions of carbohydrates in living Q.30 (a) Name the three major classes of carbohyorganisms. drates and give the distinctive characteristic of What type of bonding helps in stabilising the each class. helix structure of proteins? (b) What are nucleotides ? Name two classes of nitrogen containing bases found amongst Differentiate between globular and fibrous nucleotides. proteins.

Q.9 Q.10 Q.11

Q.12 Q.13 Q.14

Q.15

Q.16 Q.17

20

Biomolecules

EXERCISE-II Q.1 Q.2 Q.3

Q.4 Q.5

Q.6 Q.7

Q.8

Q.9

Q.10 Q.11

Q.12

Q.13 Q.14

State a use for the enzyme streptokinase in medicine. Describe the following : (i) Denaturation of Proteins Differentiate between: Primary structure and secondary structure of proteins Why is cellulose in our diet not nourishing ? Write the major classes in which the carbohydrates are divided depending upon whether these undergo hydrolysis, and if so, on the number of products formed. Explain mutarotation taking D-glucose as an example. Enumerate the structural difference between DNA and RNA. Write down the structure of sugar present in DNA. (a) Answer the following queries about proteins (i) How are proteins related to amino acids ? (ii) How are oligopeptides different from polypeptides ? (iii) When is a protein said to be denatured ? (a) Define and classify vitamins. Give at least two examples of each type. (b) Define enzymes and comment on the specificity in action of an enzyme. Illustrate with an example. What are essential an non-essential amino acids ? Give two examples of each. (a) Define the following terms: (i) Co-enzymes (ii) Nucleotides (b) List four main functions of carbohydrates in organisms. (a)Answer the following questions briefly (i) What areanytwo good sources ofvitaminA? (ii) What are nucleotides ? (b) How are carbohydrates classified ? Write two main functions of carbohydrates in plants. What happens when D-glucose is treated with the following reagents ?

Q.16

Q.17 Q.18 Q.19

Q.20 Q.21

Q.15

Why is cellulose in our diet not nourishing ? Write the major classes in which the carbohydrates are divided depending upon whether these undergo

Q.22 Aspartame, an artificial sweetener, is a peptide and has the following structures : NH2

CH2 C6H5

HOOC – CH2CH – CONH – CH – COOCH3

(a) Identify the four functional groups. (b) Write the zwitterionic structure (c) Write the structures of the amino acids obtained from the hydrolysis of aspartame. (d) Which of the two amino acids is more hydrophobic ?

Q.23 Q.24 Q.25 Q.26

Q.27

Q.28

(i) alk.KMnO4 (ii) Br 2 + CS 2 (iii) H2SO4 Name the four bases present in DNA. Which one of these is not present in RNA ?

Name two fat soluble vitamins, that sources and diseases caused due to their deficiency in diet. State a use for the enzyme streptokinase in medicine. Describe the following : Denaturation of Proteins Differentiate between: Primary structure and secondary structure of proteins

Give the chemical name of vitamin B12 . What are the following substances ? (i) Invert sugar (ii) Polypeptides Which forces are responsile for the stability of -helix ? Why is it named as 3.613 helix ? What are complementary bases ? Draw structure to show hydrogen bonding between adenine and thymine and between guanine and cytosine. Give reasons for the following : (i) On electrolysis in acidic solution amino acids migrate towards cathode, while in alkaline solution these migrate towards anode. (ii) The monoamino monocarboxylic acids have two pK values.

Q.29

Glycine exists as a Zwitter ion but anthranilic acid does not. Comment. Write the difference between DNA and RNA ?

Q.30

Explain structure of protin ?

21

Biomolecules

EXERCISE-III Q.1

The chromophore in the dye HO3S–

–N = N–

(a) –N(CH3)2 (c) –C6H5 Q.2

(c) R–CH–COO– | NH2

Peptide bond is (a) – CO – NH – (b) NH2 – CO – NH – R (c) R – CO – NH – R (d) – CONH2

Q.10

During hydrogenation of oils, higher melting ‘vegetable ghee’ is formed because (a) Hydrogen is dissolved in the oil (b) Hydrogen combines with oxygen of the oil (c) Esters of unsaturated fatty acids are reduced to those of saturated acids (d) Hydrogen drives off the impurities from the oil

Q.11

Structurally a biodegrable detergent should contain a (a) Normal alkyl chain (b) Branched alkyl chain (c) Phenyl side chain (d) Cyclohexyl side chain

Q.12

Which of the following dye has a nitro group ? (a) Malachite (b) Indigo (c)Aniline yellow (d) Martius yellow

is

(b) –SO3H (d) –N = N–

At the isoelectric point for an amino acid the species present are (a) R–CH–COOH (b) R–CH–COOH | NH2

Q.3

–N(CH3 )2

Q.9

|

+NH

3

(d) R–CH–COO– |

+NH

3

Secondary structure of a protein refers to (a) Mainly denatured proteins and structures of prosthetic groups (b) Regular folding patterns of contiguous portions of the polypeptide chain (c) Linear sequence of amino acide residues in the polypeptide chain (d) None of these

Q.4

The general formula of carbohydrates is: (a) CnH2n+1O (b) CnH2nO (c) Cn(H2O)n or Cx(H2O) y (d) Cn(H2O)2n

Q.13

Glucose and Fructose are (a) Tautomers (b) Chain isomers (c) Functional isomers (d) Geometrical isomers

Q.5

Which of the following is a disaccharide? (a) Sucrose (b) Glucose (c) Fructose (d) Starch

Q.14

Glucose is (a)Aldopentose (c) Ketopentose

Q.6

The iron in haemoglobin is bound by (a) Hydrogen bonds (b) Chelation (c) Ionic bonds (d) Covalent bonds

Q.15

Q.7

Whichofthe following statement(s) is(are) true? (a)All amino acids contain one chiral center (b) Some amino acids contain one, while some contain more chiral center or even no chiral center (c) All amino acids found in proteins have L configuration (d) All amino acids found in proteins have 1° amino group

Asubstance which can act both as an antiseptic and disinfectant is (a)Aspirin (b) Chloroxylenol (c) Bithional (d) Phenol

Q.16

Thrust imparted to the rocket is governed by the (a) Third law of thermodynamics (b) Gravitational law (c) Newton’s third law (d) None of these

Q.17

Which of the following represent a bi-liquid propellant? (a) N2O4 + unsymmetrical dimethylhydrazine (b) N2O4 + acrylic rubber (c) Nitroglycerine + nitrocellulose (d) Polybutandiene + ammonium perchlorate

Q.8

Anomers have different (a) properties (b) melting points (c) specific rotation (d) all of these

(b)Aldohexose (d) Ketohexose

22

Biomolecules

Q.18

The reagent used in Ruff’s degradation is (a) Baeyer’s reagent (b) Tollen’s reagent (c) Fenton’s reagent (d) Benedict’s reagent

Q.19

If K a1 and K a 2 are the ionization constants of H3N+CHICOOH and H 3N+CHICOO–, respectively, the pH of the solution at the isoelectric point is

Q.26

Choose the correct relationship for -Dglucose (1) and -D-glucose(2) (a)Aand B are epimers (b)Aand B are crystal modification (c) A is a pyranose sugar and B is furanose sugar (d)Ais an aldose and B is a ketose.

Q.27

Natural rubber is a polymer of (a) Chloroprene (b) Isoprene (c) 1,3-Butadiene (d) None

Q.28

Hydrolysis of sucrose is called (a) Saponification (b) Inversion (c) Esterification (d) Hydration

Q.29

Glucose and fructose give the same osazone. One may, therefore, conclude that (a) Glucose and fructose have identical structures (b) Glucose and fructose are anomers (c) The structures of glucose and fructose have mirror-image relationship (d) The structures of glucose and fructose differ only in those carbon atoms which take part in osazone formation.

Q.30

In vulcanization of rubber (a) Sulphur reacts to form a new compound (b) Sulphur cross-links are introduced (c) Sulphur forms a very thin protective layer over rubber (d)All statements are correct

Q.31

The simplest amino acid is (a) glycine (b) alanine (c) guanine (d) all of the above

Q.32

Which of the following belong to the class of natural polymers? (a) Proteins (b) Cellulose (c) Rubber (d)All of the above

Q.33

Pick out correct statements. (a) In an electrolysis experiment, amino acids migrate at the isoelectric point towards electrodes (b) p-aminobenzenesulphonic acid is a dipolar ion: while p-aminobenzoic acid is not (c) Sulphanilic acid is soluble in base, but not in acid

(a) pH  pK a1  pK a 2 (b) pH  (pK a 1  pK a 2 )

1/ 2

(c) pH  (pK a 1  pK a 2 )

1/ 2

(d) pH 

(pK a 1  pK a 2 ) 2

Q.20

‘Placedo’ is often given to patients. It is (a) an antidepressant (b) a broad spectrum antibiotic (c) a sugar pill (d) a tonic

Q.21

Coordination polymerization was developed by (a) Zeigler and Natta (b) Linus Pauling (c) Beckmann (d) None of these

Q.22

Teflon, polystyrene and neoprene are all (a) Copolymers (b) Condensation polymers (c) Homopolymers (d) Monomers

Q.23

Carbohydrates which differ in configuration at the glycosidic carbon (i.e. C1 in aldose and C2 in ketoses) are called (a)Anomers (b) Epimers (d) Enantiomers (c) Diastereomers

Q.24

Q.25

An ald ohexose (e.g.,gluco se) and 2-oxohexose (e.g., fruct ose) can be distinguished with the help of (a) Tollen’s reagent (b) Fehling’s solution (c) Benedict solution (d) Br2 / H2O The open-chain glucose on oxidation with HIO4 gives (a) 5 HCOOH + H2C = O (b) 4 HCOOH + 2 H2C = O (c) 3 HCOOH + 3 H2C = O (d) 2 HCOOH + 4 H2C = O

+

(d) H3 N CH2COOH(pka = 2.4) is more acidic than RCH2COOH (pKa = 4 – 5)

23

Biomolecules

Q.34

Q.35

-D-Glucose and -D Glucose differ from each other due to difference in one of carbon with respect to its (a) Size of hemiacetal ring (b) Number of—OH groups (c) Configuration (d) Conformation

Which of the following statements are true? (A) Water solubility is maximum at a pH when concentration of anions and cations are equal. (B) They give ninhydrin test (C) On reacting with nitrous acid give off N2 (a)All (b) B and C (c) Aand B (d) None of these Q.36

Glucose gives the silver mirror test with ammoniacal solution ofsilver nitrate because it contains (aAldehydes group (b) Ester group (c) Ketone group (d)Amide group

Q.37

Oligosaccharides contain ………. Simple sugar units (a) 2 to 10 (b) 4 to 8 (c) 6 to 12 (d) 6 to 10

Q.38

Apair of diastereomers that differ only in the configuration about a single carbon atom are called (a)Anomers (b) Epimers (c) Conformers (d) Enantiomers

Q.40

Q.41

Glucose molecule reacts with X number of molecules of phenylhydrazine to yield osazone. The value of X is (a) Three (b) Two (c) One (d) Four

Q.43

Main structural unit of protein is (a) Ester linkage (b) Ether linkage (c) Peptide linkage (d)All the above

Q.44

Which of the following statements is true of proteins? (a) They catalyse the biochemical reactions (b) They act as antibodies (c) They act as hormones (d) They perform all these functions

Q.45

Which of the following is a polysaccharide ? (a) Glucose (b) Galactose (c) sucrose (d) Pectines.

Q.46

Starch can be used as an indicator for the detection of traces of (a) Glucose in aqueous solution (b) Proteins in blood (c) Iodine in aqueous solution (d) Urea in blood

Q.47

Which of the following statements about ribose in incorrect ? (a) It is a polyhydroxy compound (b) It is an aldehyde sugar (c) It has six carbon atoms (d) It exhibits optical activity

Q.48

The letter ‘D’in carbohydrates represents (a) It direct synthesis (b) Its dextrorotation (c) Its mutarotation (d) Its configuration

For amino acid having the structure R—CH—CO2H | NH2

Q.39

Q.42

Pick out the incorrect statement aboutATP. Q.49 (a) It is a nucleotide (b) It contains the purine, adenine (c) The enzyme-catalysed hydrolysis of ATP t o ADP and AMP is accompanied by absorption of energy (d) Energy is stored in the cell in the form of ATP. If the sequence of bases in one strand of DNA is ATGACTGTC, then the sequence of bases in its complementary strand is (a) TACTGACAG (b) TUCTGUCUG Q.50 (c) GUAGTUAUG (d) None of these Cellulose is a linear polymer of (a) -glucose (b) -glucose (d) none of these (c) -fructose

Which of the following is the structure of D-xylose?

(a)

(c)

CHO HO — — H H — — OH H — — OH CH2OH CHO H — — OH HO — — H H — — OH CH2OH

(b)

(d)

CHO HO — — H HO — — H H — — OH CH2OH CHO H — — OH H — — OH H — — OH CH2OH

Glucose gives the silver mirror test with ammoniacal solution ofsilver nitrate because it contains the group (a)Aldehyde (b) Ester (c) Ketone (d)Amide

24

Biomolecules

Comprehension : Dye is a natural or synthetic colouring matter which is used in solution to stain materials especially fibres. All the coloured substances are not dyes. A coloured substance is termed as a dye if it fulfills the following conditions. (a) It must have a suitable colour. (b) It can be fixed on the fibre directly or which the help of a mordant. (c) When fixed, it must be fast to light and washing. According to the modern theory. (a) It must have a chromophoric group. (b) It must have certain groups which themself donot produce colour but when present with chromophore, it intensifies the colour. Q.51 The groups which by themselves do not produce colour but intensify the colour when present with the chromophores are called. (a) substituents (b) auxochromes (c) cataphores (d) chromogens Q.52 Which of the following is an anthraquinone dye? (a) phenolphthalein (b) congo red (c) malachite green (d) alizarin Q.53 Alizarin gives violet colour with which of the following mordants? (a) Al3+ (b) Fe3+ (c) Ba2+ (d) Cr3+

Q.54

Q.55

Q.56

Q.57

(a) If both Aand R are true and R is the correct explanation of A. (b) If both A and R are true but R is not the correct explanation of A. (c) If Ais true but R is fals. (d) If both A and R are false. A. The enzyme amylase hydrolyses starch to maltose. R. Starch is polymer containing glycosidic linkages. A. During emergency, hormone adrenaline stimulates the conversion of liver glycogen into glucose. R. Adrenaline is an example of peptide hormone. A. A so lut io n o f su cr o se in wat er is dextrorotatory but on hydrolysis in the presence of small amount of dil. HCl, it becomes laevorotatory. R. Sucrose on hydrolysis gives unequal amounts of glucose and fructose as a result of which change in sign of rotation is observed. A. Each turn of the a-helix structure of protein forms  13 membered ring is containing 3.6 amino acids. R. -helix is secondary structure of protein which gets stabilised via hydrogen bonding and disulphide linkages. A. -glucose and b-glucose are the anomers of glucose having different specific optical rotation. R. Glucose and fructose both are monosaccharides.

Assertion and Reason : Q.58 Each of the questions given below consists of two statements, an assertion (A) and rea son (R). Select the number corresponding to the appropriate alternative as follows Match the following Q.59 Match list I with list II and select the correct answer using the codes given below the lists. List I List II A. Nucleic acids (i) D.N.A. B. Hormones Uracil (ii) C. Thymine (iii) Polynucleotides double-helix structure D. R.N.A. (iv) A-(iii), B-(iv), C-(i), D-(i) A-(iv), B-(iii), C-(i), D-(i) (a) (b) none of these A-(iii), B-(i), C-(iv), D-(i) (c) (d) List II Q.60 List I A. (i) Genetic material Pepsin (ii) Sex hormone B. Nucleic acid (iii) Vitamin C C. Ascorbic acid (iv) Antibiotic D. Testosterone (v) Digestive enzyme (a) A-(iv), B-(i), C-(iii), D-(ii) (b) A-(v), B-(i), C-(ii), D-(iii) (c) A-(v), B-(i), C-(iii), D-(ii) (d) none of these

25

Biomolecules

ANSWER KEY EXERCISE - III Q.1 Q.8 Q.15 Q.22 Q.29 Q.36 Q.43 Q.50 Q.57

d c d c d a c a a

Q.2 Q.9 Q.16 Q.23 Q.30 Q.37 Q.44 Q.51 Q.58

d a c a b a d b b

Q.3 Q.10 Q.17 Q.24 Q.31 Q.38 Q.45 Q.52 Q.59

c c a d a b d d b

Q.4 Q.11 Q.18 Q.25 Q.32 Q.39 Q.46 Q.53 Q.60

c d c a d c c b c

Q.5 Q.12 Q.19 Q.26 Q.33 Q.40 Q.47 Q.54

a d d a (b,c,d) a c a

Q.6 Q.13 Q.20 Q.27 Q.34 Q.41 Q.48 Q.55

b c c b c b d c

Q.7 Q.14 Q.21 Q.28 Q.35 Q.42 Q.49 Q.56

b,c b a b b a c c

26