CBSE Biomolecules Class 12 Notes PDF

Section-A (1 Mark Questions)

1. Which carbohydrate is the main constituent of plant cell wall?

Ans. Cellulose.

2. What are the common types of secondary structure of proteins?

Ans.  (i) α-helix

(ii) β-pleated sheet

3. Name two α-amino acids which form a dipeptide which is 100 times more sweet than cane sugar.

Ans. Aspartic acid and phenylalanine.

4. What is the basic structural difference between starch and cellulose?

Ans. Starch is polymer of α-D glucose units whereas cellulose is linear polymer of β-D glucose units.

5. Why are carbohydrates generally optically active?

Ans. Carbohydrates have chiral or asymmetric carbon atom.

6. During curdling of milk, what happens to sugar present in it?

Ans. When milk is curdled, its sugar get oxidize to form lactic acid.

7. If one strand of a DNA has the sequence −ATGCTTCA−, what is the sequence of the bases is the complementary strand. 

Ans. As we know that in DNA molecules, adenine (A) always pairs with thymine (T) and cytosine (C) always pairs with guanine (G). Thus,

Sequence of bases in one strand : ATGCTTCA

Sequence of bases in the complementary strand : TACGAAGT.

8. Which one of the following is a disaccharide: Starch, Maltose, Fructose, Glucose?

Ans. Maltose is a disaccharide as it consists of two α-D-glucose units.

9. Define the following term: Anomers

Ans. The pair of stereoisomers which differ only in the configuration of the hydroxyl group at C1 are called anomers.

10. Define the following term: Polysaccharides

Ans. Carbohydrates which yield a large number of monosaccharide units on hydrolysis are called polysaccharides.

Section-B (2 Mark Questions)

11. What are nucleic acids? Mention their two important functions.

Ans.  Nucleic acids are biomolecules which are found in the nuclei of all living cells. They are polymers composed of repeating units called nucleotides. Nucleic acids play a very vital role in the transmission of the hereditary characteristics and the biosynthesis of proteins.

12. What is the difference between a nucleoside and a nucleotide?

Ans. Nucleoside is formed by condensation of a base with pentose sugar at position 1. When nucleoside is linked to phosphoric acid at 5 position of sugar moiety, we get a nucleotide. Hence, a nucleotide has their units–phosphate group, pentose sugar and a base, whereas nucleoside has two units–pentose sugar and a base.

13. How do you explain the presence of an aldehydic group in a glucose molecule ?

Ans. Glucose reacts with hydroxylamine to form a monoxime and adds one molecule of hydrogen cyanide to give cyanohydrin so it contains a carbonyl group which can be an aldehyde or a ketone. On mild oxidation with bromine water, glucose gives gluconic acid which is a six carbon carboxylic acid. Thus indicates that carbonyl group present in glucose is an aldehydic group.

14. Amino acids behave like salts rather than simple amines or carboxylic acids. Explain.

Ans. Amino acids behave like salts rather than simple amines or carboxylic acids. This behavior is due to the presence of both acidic and basic groups in the same molecule. In aqueous solution, the carbonyl group loses a proton and amino group accepts a proton to form a zwitter ion. 

15. Name two fat soluble vitamins, their sources and the diseases caused due to their deficiency in diet.

Ans. Examples of fat-soluble vitamins are vitamin A and D.

16. List two characteristic features of enzymes.

Ans. (i) Enzymes are highly specific for a particular reaction and for a particular substrate.

(ii) Very small amount of enzyme is required for the process of a reaction.

17. Describe what you understand by primary structure and secondary structure of proteins?

Ans. Primary structure: The specific sequence in which the various amino acids present in a protein are linked to one another is called its primary structure. Any change in the primary structure creates a different protein.

Secondary structure: The conformation of the polypeptide chain is known as secondary structure. The two types of secondary structure are α-helix and β-pleated sheet structure.

In α-helix structure, the polypeptide chain forms all the possible hydrogen bonds by twisting into a right-handed screw (helix) with the –NH groups of each amino acid residue hydrogen bonded to the >C=0 group of an adjacent turn of the helix. In β-pleated sheet structure, all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds.

18. Explain what is meant by a peptide linkage.

Ans. Proteins are the polymers of α-amino acids linked by amide formation between carboxyl and amino group. This is called peptide linkage or peptide bond e.g.,

19. Give the preparation of glucose from starch.

Ans. Commercially glucose is obtained by hydrolysis of starch by boiling it with dilute $H_{2}SO_{4}$ at 393 K under pressure.

$\underset{Starch\;cellulose}{(C_{6}H_{10}O_{5})_{n}}\;+nH_{2}O\xrightarrow[393K;2-3atm]{H^{+}}\;\underset{Glucose}{nC_{6}H_{12}O_{6}}$

20. Explain what is meant by the following: pyranose structure of glucose?

Ans. The six membered cyclic structure of glucose is called pyranose structure (α- or β-), in analogy with heterocyclic compound pyran.

PDF Summary - Class 12 Chemistry Biomolecules Notes (Chapter 14)

1. Introduction

Biomolecules are complex organic compounds that govern the common activities of living organisms. Carbohydrates, proteins, nucleic acids, lipids, and other complex biomolecules make up living systems. Furthermore, simple molecules such as vitamins and mineral salts play an important role in the functions of organisms. 

2. Carbohydrates 

Plants are the primary producers of carbohydrates, which comprise a large group of naturally occurring organic compounds. Cane sugar, glucose, starch, and other common examples The majority of them have the general formula \[{{\text{C}}_x}{{\text{H}}_{2y}}{{\text{O}}_y}\] and were thought to be carbon hydrates, hence the name carbohydrate. The molecular formula of glucose \[\left( {{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}} \right){\text{,}}\] for example, fits into this general formula, \[{{\text{C}}_{\text{6}}}\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right){\text{6}}{\text{.}}\] However, not all of the compounds that fit into this formula are carbohydrates. Rhamnose, \[{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{5}}}{\text{,}}\] is a carbohydrate, but it does not fall under this definition. Carbohydrates can be defined chemically as optically active polyhydroxy aldehydes or ketones, or compounds that produce such units upon hydrolysis. Some carbohydrates with a sweet taste are also known as sugars. Sucrose is the most common sugar found in our homes, whereas lactose is the sugar found in milk. 

2.1 Classification of Carbohydrates 

Carbohydrates are classified based on how they react to hydrolysis. They have been broadly classified into the three groups listed below: 

2.1.1 Monosaccharides 

A monosaccharide is a carbohydrate that cannot be hydrolyzed further to produce simpler units of polyhydroxy aldehyde or ketone. Examples include glucose, fructose, ribose, and others.

Monosaccharides are further classified based on the number of carbon atoms and the functional group they contain. When a monosaccharide contains an aldehyde group, it is referred to as an aldose, and when it contains a keto group, it is referred to as a ketose. As shown by the examples, the number of carbon atoms constituting the monosaccharide is also introduced in the name.

Different Types of Monosaccharides

2.1.2 Oligosaccharides 

Oligosaccharides are carbohydrates that, when hydrolyzed, yield two to ten monosaccharide units. They are further classified as disaccharides, trisaccharides, tetrasaccharides, and so on, based on the number of monosaccharides produced during hydrolysis. Disaccharides are the most common of these. The two monosaccharide units formed by hydrolysis of a disaccharide may be identical or dissimilar. Sucrose, for example, yields one molecule of glucose and one molecule of fructose when hydrolyzed, whereas maltose yields only two molecules of glucose.

2.1.3 Polysaccharides 

Polysaccharides are carbohydrates that when hydrolyzed yield a large number of monosaccharide units. Starch, cellulose, glycogen, gums, and other common examples of Polysaccharides do not have a sweet taste, so they are also known as non-sugars. 

2.1.4 Reducing and Non-Reducing Sugars 

Carbohydrates are also classified as reducing or non-reducing sugars. Reducing sugars are all carbohydrates that reduce Fehling's solution and Tollens' reagent. Sugars are reduced by all monosaccharides, whether aldose or ketose.

Nonreducing sugars, such as sucrose, are formed when the reducing groups of monosaccharides, such as aldehyde or ketonic groups, are bonded in disaccharides. Reducing sugars, on the other hand, are sugars that have these functional groups free, such as maltose and lactose.

3. Glucose (Aldohexose) 

3.1 Preparation of Glucose 

1. From Sucrose (Cane Sugar): 

When sucrose is boiled in alcoholic solution with dilute \[{\text{HCl or }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{,}}\] glucose and fructose are obtained in equal amounts.

\[\mathop {{{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}}\limits_{{\text{Sucrose}}}  + {{\text{H}}_{\text{2}}}{\text{O}}\xrightarrow{{{{\text{H}}^ \oplus }}}\mathop {{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}}\limits_{{\text{Glucose}}}  + \mathop {{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}}\limits_{{\text{Fructose}}} \]

2. From Starch 

Glucose is commercially obtained by hydrolyzing starch with dilute \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\] at 393 K under pressure.

\[\mathop {{{\text{C}}_{\text{6}}}{{\text{H}}_{10}}{{\text{O}}_{\text{5}}}}\limits_{{\text{Starch or cellulose}}}  + n{{\text{H}}_{\text{2}}}{\text{O}}\xrightarrow[{393{\text{ K,2 - 3 atm}}}]{{{{\text{H}}^ \oplus }}}\mathop {{\text{n}}{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}}\limits_{{\text{Glucose}}} \]

3.2 Reactions 

1. Oxidation

Oxidation Reactions of Glucose

2. Reduction

Reduction Reactions of Glucose

Note: Reduction with HI yields n-hexane, demonstrating that all six carbons of glucose are arranged in a straight chain.

3. Oxime Formation 

Oxime Formation by Glucose

4. Cyanohydrin Formation

Formation by Glucose Cyanohydrin

Note: Since carbonyl C has become chiral, two diastereomers are formed.

5. Acetylation

Acetylation of Glucose

6. Reaction with phenylhydrazine (formation of osazone)

Reaction of Glucose with Phenylhydrazine

Note: To produce osazone, one mole consumes three moles of PhNHNH2. Two moles produce hydrazone, and one is used to convert a CHOH group to a carbonyl group.

3.3 Configuration in Monosaccharides 

D(+)-glucose is the correct name for glucose. The letter ‘D' before the name of glucose represents the configuration, whereas the letter ‘(+)' represents the molecule's dextrorotatory nature. It should be noted that the letters 'D' and 'L' have no bearing on the compound's optical activity. The following is a definition of D– and L– notations. The letters 'D' or 'L' before the name of any compound indicate the stereoisomer's relative configuration. This refers to their affinity for a specific isomer of glyceraldehyde. Glyceraldehyde has one asymmetric carbon atom and comes in two enantiomeric forms, as shown. 

Configuration in Glyceraldehyde

All compounds that can be chemically correlated to the (+) isomer of glyceraldehyde have D-configuration, whereas those that can be correlated to the (–) isomer of glyceraldehyde have L-configuration. The lowest asymmetric carbon atom (as shown below) is compared when assigning the configuration of monosaccharides. As with (+) glucose, –OH on the lowest asymmetric carbon is on the right side, similar to (+) glyceraldehyde, so it is assigned D-configuration. The structure is written in such a way that the most oxidised carbon is at the top for this comparison.

(+) Isomer of D-configuration in Glyceraldehyde and Glucose

3.4 Cyclic Structure of Glucose 

Glucose has been discovered to exist in two different crystalline forms, which are known as \[\alpha \]and \[\beta .\] The \[\alpha \]-form of glucose (m.p. 419 K) is obtained by crystallisation from a concentrated glucose solution at 303 K, whereas the \[\beta .\]-form (m.p. 423) is obtained by crystallisation from a hot and saturated aqueous solution at 371 K. In aqueous solution, both \[\alpha \]-D-glucose and \[\beta \]-D-glucose undergo mutarotation. Although the crystalline forms of \[\alpha \]- and \[\beta \]-D (+)-glucose are quite stable in aqueous solution, each form gradually transforms into an equilibrium mixture of the two. 

This is demonstrated by the fact that the specific rotation of a freshly prepared aqueous solution of \[\alpha \]-D(+)-glucose decreases with time from \[{{ + 111^\circ  to  + 52}}{{.5^\circ ,}}\] whereas that of \[\beta \]-D(+)-glucose increases from \[{\text{ + 19}}{{.2^\circ  to 52}}{{.5^\circ }}{\text{.}}\] Thus, mutarotation refers to the spontaneous change in specific rotation of an optically active compound with time to an equilibrium value. It was discovered that glucose forms a six-membered ring, with –OH at C-5 playing a role in ring formation. This explains the lack of a –CHO group as well as the presence of glucose in two forms, as shown below. These two cyclic forms coexist with an open chain structure.

Cyclic Structure of Glucose

The only difference between the two cyclic hemiacetal forms of glucose is the configuration of the hydroxyl group at C1, which is referred to as anomeric carbon (the aldehyde carbon before cyclisation). Such isomers, namely -form and -form, are referred to as anomers. In analogy with pyran, the six-membered cyclic structure of glucose is known as the pyranose structure (– or –). Pyran is a cyclic organic compound with a ring containing one oxygen atom and five carbon atoms. The Haworth structure more accurately represents the cyclic structure of glucose.

Cyclic Structure of Glucose

3.4.1 How to Draw a Haworth Projection 

A ring of 6 atoms (5 ‘C' and 1 ‘O') is drawn, with the ‘O' atom in the upper right hand corner, as shown below.

Numbering in the Haworth Projection

Number one is assigned to the carbon atom on the right side of oxygen. Then, in a clockwise fashion, other carbon atoms are assigned numbers 2, 3,........... . In Fischer projection, groups attached to a carbon on the right side are placed below the ring, while groups attached to the carbon on the left side are placed above the ring. However, by convention, the \[{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}\] group of carbon 5 is placed above the plane of the ring.

(image will be uploaded soon)

4. Fructose (Ketohexose) 

Fructose has the molecular formula \[{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}{\text{,}}\]and based on its reactions, it was discovered to have a ketonic functional group at carbon number 2 and six carbons in a straight chain, just like glucose. It is a levorotatory compound from the D-series. D-(–)-fructose is the correct spelling. Its open chain structure is depicted below.

Fructose

It also exists in two cyclic forms, which are obtained by adding –OH to the carbonyl group at C5. The resulting ring is a five-membered ring known as furanose, after the compound furan. Furan is a five-membered cyclic compound that contains one oxygen atom and four carbon atoms.

Anomers of Fructose

Haworth structures are used to represent the cyclic structures of two anomers of fructose.

Haworth Structures of Anomers of Fructose

5. Comparison of Glucose And Fructose

6. Disaccharides 

An oxide linkage formed by the loss of a water molecule connects the two monosaccharides. The linkage of two monosaccharide units via an oxygen atom is known as glycosidic linkage. 

6.1 Sucrose 

Sucrose is a common disaccharide that, when hydrolyzed, yields an equimolar mixture of D-(+)-glucose and D-(–)-fructose.

${\begin{matrix} C_{12}H_{22}O_{11} & + & H_{2}O \longrightarrow & C_{6}H_{12}O_{6} & + & C_{6}H_{12}O_{6} \\ Sucrose & & & D-(+)-Glucose & & D-(-)-Fructose \\ [\alpha]_D=+66.5^\circ & & & [\alpha]_D=+52.5^\circ & & [\alpha]_D=-92.4^\circ\end{matrix}} $

A glycosidic linkage between C1 of \[\alpha \]-glucose and C2 of \[\beta \]-fructose holds these two monosaccharides together. Sucrose is a non-reducing sugar because the reducing groups of glucose and fructose are involved in the formation of glycosidic bonds.

Sucrose

Sucrose is a dextrorotatory compound that when hydrolyzed yields an equimolar solution of glucose and fructose.

Because the laevo rotation of fructose is greater than the dextro rotation of glucose, this solution is laevorotatory.

Thus, hydrolysis of sucrose causes a change in the sign of rotation, from dextro (+) to laevo (–), and the product is known as invert sugar, and the phenomenon is known as sugar inversion.

6.2 Maltose

Maltose, another disaccharide, is made up of two -Dglucose units, with C1 of one glucose (I) linked to C4 of another glucose unit (II). Because the free aldehyde group can be produced at C1 of second glucose in solution and has reducing properties, it is classified as a reducing sugar.

Maltose

6.3 Lactose

It is more commonly known as milk sugar since this disaccharide is found in milk. It is composed of \[\alpha \]-D-galactose and \[\beta \]-D-glucose. Fischer projections of -D-Glucose and -D-Galactose are drawn below:

Lactose

Galactose

Except for C-4, the configurations of all the carbon atoms in \[\beta \]-D-Glucose and \[\beta \]-D-Galactose are the same. Such stereoisomers that differ in configuration at only one carbon other than anomeric carbon are referred to as epimers, and the C atom in question is referred to as an epimeric carbon atom. As a result, \[\beta \]-DGlucose and \[\beta \]-D-Galactose are epimers, and C-4 is an epimeric carbon atom. The linkage in lactose is between C1 of galactose and C4 of glucose. As a result, it also reduces sugar.

Glycosidic Linkage in Lactose is Between C1 of Galactose and C4 of Glucose

7. Polysaccharides 

Polysaccharides are composed of many monosaccharide units linked together by glycosidic linkages. They primarily serve as food storage or structural materials.

7.1 Starch 

Plants' primary storage polysaccharide is starch. It is the most important source of nutrition for humans. Cereals, roots, tubers, and some vegetables have a high starch content. It is a -glucose polymer composed of two components: Amylose and Amylopectin. Amylose is a water-soluble component that accounts for about 15-20% of starch. Amylose is a long unbranched chain containing 200-1000 \[\alpha \]-D-(+)-glucose units held together by a C1-C4 glycosidic linkage. Amylopectin is insoluble in water and accounts for approximately 80-85 percent of starch. It is a branched chain polymer of \[\alpha \]-D-glucose units with a chain formed by C1-C4 glycosidic linkage and branching caused by C1-C6 glycosidic linkage.

Amylose

Amylopectin

7.2 Cellulose 

Cellulose is found only in plants and is the most abundant organic substance in the plant kingdom. It is a major component of the cell wall of plant cells. Cellulose is a straight-chain polysaccharide made up entirely of \[\beta \]-D glucose units linked together by a glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit.

Cellulose

7.3 Glycogen 

Carbohydrates are stored as glycogen in the animal body. It is also known as animal starch due to its structure, which is similar to amylopectin but more highly branched. It can be found in the liver, muscles, and brain. When the body requires glucose, enzymes convert glycogen to glucose. Glycogen is found in yeast and fungi as well.

7.4 Summary

Note: 

1. Reducing sugars are all carbohydrates that contain the CHO group, the \[\alpha \]-Hydroxy ketonic group, or the hemiacetal group.

2. Mutarotation is a phenomenon that occurs in all reducing sugars.

8. Proteins

Protein is derived from the Greek word "proteios," which means "primary" or "of primary importance." Proteins are all polymers of \[\alpha \]-amino acids.

8.1 Amino Acids

Amino acids have both amino (–$NH_2$) and carboxyl (–COOH) functional groups. The amino acids are classified as \[\alpha ,\beta ,\gamma ,\delta \] and so on based on the relative position of the amino group with respect to the carboxyl group. When proteins are hydrolyzed, only \[\alpha \]-amino acids are produced. They may also contain other functional groups.

Amino Acids

All \[\alpha \]-amino acids have innocuous names that usually reflect the compound's or its source's property. Glycine is named after its sweet taste (glykos means sweet in Greek), and tyrosine was first obtained from cheese (tyros means cheese in Greek).

8.2 Classification of Amino Acids 

The relative number of amino and carboxyl groups in an amino acid molecule determines whether it is acidic, basic, or neutral.

Non-essential amino acids are amino acids that can be synthesized in the body. Essential amino acids, on the other hand, are those that cannot be synthesized in the body and must be obtained through diet.

8.3 Properties of Amino Acids 

Amino acids are crystalline, colorless solids. These are water-soluble, high-melting-point solids that act more like salts than simple amines or carboxylic acids. The presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule causes this behavior.

Amino Acids Exhibit Amphoteric Behavior in Zwitter Ionic Form

In aqueous solution, the carboxyl group can lose a proton and the amino group can accept a proton, resulting in zwitter ion, a dipolar ion. This is neutral, but it does have both positive and negative charges. Amino acids exhibit amphoteric behavior in zwitter ionic form, reacting with both acids and bases.

Except for glycine, all other naturally occurring \[\alpha \]-amino acids are optically active due to the asymmetry of the -carbon atom.

These are available in both 'D' and 'L' forms. The majority of naturally occurring amino acids have an L-configuration. The –NH2 group on the left side is used to represent L-amino acids.

8.4.1 Isoelectric Point 

An amino acid's isoelectric point (pI) is the pH at which it has no net charge. In other words, it is the pH at which the negative charge on an amino acid exactly balances the positive charge. pI (isoelectric point) = pH at which no net charge exists. The pI of an amino acid with no ionizable side chain, such as alanine, is halfway between its two pKa values. This is due to the fact that at pH = 2.34, half of the molecules have a negatively charged carboxyl group and half have an uncharged carboxyl group, and at pH = 9.69, half of the molecules have a positively charged amino group and half have an uncharged amino group. As the pH rises from 2.34, more molecules' carboxyl groups become negatively charged; as the pH falls from 9.69, more molecules' amino groups become positively charged.

As a result, the number of negatively charged groups equals the number of positively charged groups when the two pKa values are averaged.

Isoelectric Point of Alanine

If an amino acid has an ionizable side chain, its pI is calculated by averaging the pKa values of similarly ionizing groups (positive ionizing to uncharged or uncharged ionizing to positive). For example, the pI of lysine is the average of the pKa values of two groups that are positively charged in their acidic form but uncharged in their basic form. The pI of glutamate, on the other hand, is the average of the pKa values of the two groups, which are uncharged in their acidic form and negatively charged in their basic form.

Isoelectric Point of Lysine

Isoelectric Point of Glutamic Acid

8.5 Structure of Proteins – 

The Peptide Bond Proteins are polymers of \[\alpha \]-amino acids that are linked together by peptide bonds or peptide linkage. Peptide linkage is chemically defined as an amide formed between the – COOH group and the –NH2 group. 

Peptide Linkage in Protein

The reaction between two molecules of similar or different amino acids is initiated by the combination of one molecule's amino group with the carboxyl group of the other. This causes the removal of a water molecule and the formation of a peptide bond –CO–NH–.

8.6 Classification of Proteins 

Proteins can be classified into two types on the basis of their molecular shape. 

1. Fibrous Proteins 

A fiber like structure is formed when polypeptide chains run parallel and are held together by hydrogen and disulphide bonds. In general, such proteins are insoluble in water.

Keratin (found in hair, wool, and silk) and myosin (found in muscles) are two common examples.

2. Globular Proteins 

This structure is formed when polypeptide chains coil around to form a spherical shape. These are usually water soluble. Insulin and albumin are two common globular proteins. Protein structure and shape can be studied at four different levels: primary, secondary, tertiary, and quaternary.

1. Primary Structure 

One or more polypeptide chains can be found in proteins. Each polypeptide in a protein contains amino acids that are linked together in a specific sequence, and this sequence of amino acids is referred to as the protein's primary structure. Any change in this primary structure, i.e., the amino acid sequence, results in a different protein.

2. Secondary Structure 

A protein's secondary structure is the shape in which a long polypeptide chain can exist. They have been discovered to exist in two different structures: \[\alpha \]-helix and \[\beta \]-pleated sheet. These structures form as a result of the regular folding of the polypeptide chain's backbone caused by hydrogen bonding between the peptide bond's C = O and –NH– groups.

The –NH group of each amino acid residue is hydrogen bonded to the C = O of an adjacent turn of the \[\alpha \]helix in one of the most common ways in which a polypeptide chain forms all possible hydrogen bonds by twisting into a right handed screw (helix). 

All peptide chains in \[\beta \]-structure are stretched to nearly maximum extension and then laid side by side, held together by intermolecular hydrogen bonds. 

3. Tertiary Structure 

Protein tertiary structure represents overall folding of polypeptide chains, i.e., further folding of the secondary structure. It produces two major molecular shapes: fibrous and globular. Hydrogen bonds, disulphide linkages, van der Waals forces, and electrostatic forces of attraction are the main forces that keep proteins' 2° and 3° structures stable. 

4. Quaternary Structure 

Some proteins are made up of two or more polypeptide chains known as sub-units. The relationship of subunits to one another is referred to as quaternary structure.

8.7 Denaturation of Proteins 

A native protein is a protein found in a biological system that has a distinct three-dimensional structure and biological activity. When a protein in its native form is subjected to physical change, such as temperature change, or chemical change, such as pH change, the hydrogen bonds are disrupted. As a result, globules unfold, helixes uncoil, and protein loses biological activity. This is referred to as protein denaturation. A common example of denaturation is the coagulation of egg white when heated. Another example is milk curdling, which is caused by the bacteria present in milk producing lactic acid.

9. Nucleic Acids 

Every generation of every species resembles its ancestors in a variety of ways. How are these traits passed down from one generation to the next? It has been discovered that the nucleus of a living cell is in charge of the transmission of inherent characteristics, also known as heredity. The particles in the nucleus of the cell that are responsible for heredity are known as chromosomes, and they are made up of proteins and another type of biomolecule known as nucleic acids. These are primarily of two kinds: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) (RNA). Because nucleic acids are long chain polymers of nucleotides, they are also referred to as polynucleotides. 

9.1 Chemical Composition of Nucleic Acids 

When DNA (or RNA) is completely hydrolyzed, it produces a pentose sugar, phosphoric acid, and nitrogen-containing heterocyclic compounds (called bases). The sugar moiety in DNA molecules is \[\beta \]-D-2-deoxyribose, whereas the sugar moiety in RNA molecules is \[\beta \]-D-ribose.

Structure of Beta -D-ribose and Beta -D-2-deoxyribose

Adenine (A), guanine (G), cytosine (C), and thymine are the four bases found in DNA (T). The first three bases of RNA are the same as those of DNA, but the fourth is uracil (U).

Structure of Adenine (A) and Guanine (G)

Structure of Cytosine(C), Thymine(T) and Uracil(C)

9.2 Structure of Nucleic Acids 

A nucleoside is a unit formed by the attachment of a base to the 1' position of a sugar. The sugar carbons in nucleosides are labeled as 1', 2', 3', and so on. To differentiate these from the bases. A nucleotide is formed when a nucleoside is linked to phosphoric acid at the 5'-position of the sugar moiety.

Structure of (a) a Nucleoide and (b) a Nucleotide

The phosphodiester linkage between the pentose sugar's 5' and 3' carbon atoms connects nucleotides. The following is a simplified version of the nucleic acid chain.

Simplified Version of the Nucleic Acid Chain

The primary structure of a nucleic acid is information about the sequence of nucleotides in its chain. Secondary structure is also present in nucleic acids. The double strand helix structure of DNA was proposed by James Watson and Francis Crick. Two nucleic acid chains are wrapped around each other and held together by hydrogen bonds formed by base pairs. Because hydrogen bonds form between specific pairs of bases, the two strands are complementary to one another. Adenine and thymine form hydrogen bonds, whereas cytosine and guanine form hydrogen bonds. There are helices in the secondary structure of RNA that are only single stranded. They can fold in on themselves to form a double helix structure. There are three types of RNA molecules, each of which serves a different purpose. They are known as messenger RNA (m-RNA), ribosomal RNA (r-RNA), and transfer RNA, respectively (t-RNA).

Double Helix Strand of DNA

9.3 DNA Vs RNA

9.4 Biological Functions of Nucleic Acids 

DNA is the chemical basis of heredity and can be thought of as a repository of genetic information. Over millions of years, DNA has been solely responsible for maintaining the identity of various species of organisms. During cell division, a DNA molecule can self-replicate, and identical DNA strands are transferred to daughter cells. Protein synthesis in the cell is another important function of nucleic acids. Actually, proteins are synthesised by various RNA molecules in the cell, but the message for a specific protein's synthesis is present in DNA. 

10. Enzymes 

Enzymes function as biological catalysts. Enzymes are all globular proteins chemically. The following are some important enzymes and their functions.

11. Vitamins

12. Hormones 

Hormones are biomolecules that are produced in the ductless (endocrine) glands and transported by the bloodstream to various parts of the body where they control various metabolic processes. These are required in minute amounts and, unlike fats and carbohydrates, are not stored in the body but are constantly produced. 

12.1 Steroidal Hormones

12.2 Peptide Hormones

12.3 Amine Hormones

13. Test For Biomolecules 

13.1 Test of Carbohydrates 

13.1.1 Molish Test 

The Molisch test detects all types of carbohydrates, including monosaccharides, disaccharides, and polysaccharides. Molisch reagent (1 percent alcoholic solution of \[\alpha \]-naphthol) is added to an aqueous carbohydrate solution, followed by conc. \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\] along the test tube's sides. At the intersection of the two layers, a violet ring forms.

13.2 Test for Proteins 

13.2.1 Biuret Test 

An alkaline solution of a protein when treated with a few drops of 1% \[{\text{CuS}}{{\text{O}}_{\text{4}}}\] solution, produces a violet colouration. The colour is due to the formation of a coordination complex of \[{\text{C}}{{\text{u}}^{{\text{ + 2}}}}\] with carbonyl group and –NH– groups of the peptide linkages.

13.2.2 Xanthoproteic Test 

When a protein is treated with concentrated \[{\text{HN}}{{\text{O}}_{\text{3}}}{\text{,}}\] it turns yellow. This test is performed by a protein composed of \[\alpha \]-amino acids containing a benzene ring, such as tyrosine, phenylalanine, and others, and the yellow color is caused by benzene ring nitration. An important example of this test is when concentrated \[{\text{HN}}{{\text{O}}_{\text{3}}}{\text{,}}\] is spilled on your hands, the skin turns yellow due to nitration of the benzene ring of the amino acids in your skin's proteins. 

13.2.3 Millon’s Test 

Millon's reagent is a nitric acid solution of mercurous nitrate and mercuric nitrate with little nitrous acid. When Millon's reagent is added to an aqueous protein solution, a white ppt. is formed. All proteins containing phenolic \[\alpha \]-amino acids, such as tyrosine, pass this test. As a result, gelatin that does not contain phenolic \[\alpha \]-amino acids fails this test.

13.2.4 Ninhydrin Test 

A blue-violet color is produced when proteins are boiled in a dilute aqueous solution of ninhydrin (2, 2-dihydroxyindane-1,3-dione). This test is provided by all \[\alpha \]-amino acids. Proteins and peptides both pass this test because they give \[\alpha \]-amino acids when hydrolyzed.

Sub-Topics Covered Under Biomolecules

Let us find out the sub-topics that fall under the topic, Biomolecules.

• Carbohydrates - This refers to sugars, fibres, and starches which exist in different kinds of grains, fruits, vegetables, and milk products

• Monosaccharides - It refers to the simple sugar, and it happens undoubtedly to be the simplest form of sugar

• Disaccharides - The term disaccharide points to a sugar whose formation takes place when the joining of two monosaccharides takes place by the glycosidic linkage

• Polysaccharides - This happens to be a particular type of carbohydrate, whose molecules comprised to be the number of sugar molecules

• Structure of Proteins - The students can be able to learn about the structure of proteins in detail here

• Amino Acids - These refer to the organic compounds, whereas their combination indeed results in the formation of proteins

• Enzymes - These molecules happen to be the biological molecules that enhance up the rate of all chemical reactions which occurs inside the cells

• Vitamins - Vitamins means the organic compounds which are vital in small quantities for the sustenance of life

• Nucleic Acids - These refer to either biopolymers or small biomolecules whose presence is crucial to all forms of life

• Structure of Nucleic Acids - Here, the students will study the structure of nucleic acids in detail under this sub-unit

Importance of Biomolecules in Competitive Exams

Biomolecules and Polymers are very important chapters in the JEE Main Chemistry. In general, 1 question is asked from each chapter, Biomolecules and Polymers, which carries about 8 marks in total. Thus, their overall weightage in JEE Main is about 2-3 per cent, and also these chapters cannot be skipped. So, most of the questions will be asked from Biomolecules and Polymers, and it is relatively easy to score. 

Some of the topics tested under the Polymers can be listed as - About Polymerization, Different Types of Polymers, Lipids, Proteins, and Carbohydrates, are a few topics tested under Biomolecules. Reading to the study or revision notes for biomolecules and polymers will help you cover JEE Main Chemistry Syllabus.

Tricks to Solve the Questions from Biomolecules and Polymers

• As these questions are mostly theory-based, it is highly recommended you draw all the diagrams and list the equations on paper of every concept on multiple times

• Attempt the questions from within a specific text and those after going through the chapters

• Primarily focus on preparing from NCERT books. This is because most of the questions will be asked in JEE Main Chemistry belongs to these

• While attempting the MCQs (Multiple Choice Questions), take a moment to mentally visualize the concepts, equations, and diagrams that you have learnt pertaining to a specific question. Then, go for the correct answer

Benefits of Class 12 Chemistry Revision Notes on Chapter 14 - Biomolecules

The following are some of the prime benefits of Class 12 Chemistry Revision Notes for Chapter 14 - Biomolecules.

• The revision notes include detailed explanations of all the concepts of Chapter 14 Biomolecules of CBSE Class 12 Chemistry.

• Going through the revision notes on CBSE Class 12 Chemistry Chapter 14 Biomolecules will help improve students’ understanding of the topics. 

• These revision notes are prepared by our Chemistry experts having years of experience in this field to provide quality study material.

• Each and every concept is explained in a step-by-step manner, which will help the students to secure good marks in the Class 12 board exams.

• Tricks to solve the questions from Biomolecules and Polymers are also covered in these revision notes.

Why Choose Vedantu?

Because, Vedantu is a more significant online platform that provides all types of information including the revision notes, question papers of previous years, and many more. You can find the direct documents in the PDF format such as class 12 chemistry chapter 14 revision notes, revision notes class 12 chapter 14, revision notes class 12 chemistry chapter 14, class 12 notes biomolecules and more. These are some examples of the keywords where you can find many like these and can download them on the official website of Vedantu.

Related Topic Links for Biomolecules Class 12 Notes PDF

Explore a comprehensive collection of resources on Biomolecules Class 12 Notes PDF through our curated list of related topics. Dive deeper into the fascinating world of biomolecules, including carbohydrates, proteins, lipids, nucleic acids, and enzymes. Gain insights into their structures, functions, and biological significance, as well as their roles in living organisms. Whether you're a student seeking to enhance your understanding or an educator looking for supplementary materials, these links provide valuable resources to support your learning journey.

• Carbohydrates

• Proteins

• Lipids

• Nucleic Acids

• DNA and RNA structure

• Michaelis-Menten Kinetics

• Factors affecting enzyme activity

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Important Biomolecules Related Links

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Conclusion

If you are preparing for Class 12 Chemistry and got stuck in Chapter 14, you can always refer to Vedantu’s Revision Notes on Class 12 Chemistry for help. Biomolecules Class 12 Chemistry Notes has been the best study guide for the students for several years. Therefore, by going through the concepts of Vedantu’s revisions notes of Chapter 14 of Class 12 Chemistry, students will be well-prepared for the exam and can ace their board examinations with remarkable marks. You can also find the other study materials like Class 12 NCERT Solutions, important questions, syllabus, and previous years’ question papers from our website.