Biomolecules class 12th notes- Free ncert class 12 Chemistry Chapter 14 Notes- Download PDF

How do simple sugars give you energy? What makes food nutritious? How do proteins build your body? Biomolecules uncover the chemistry behind carbohydrates, proteins, nucleic acids, and enzymes, which form the foundation of every living organism; they play a unique role in the functionality and health of organisms. Biomolecules, or biological molecules, are made by living organisms and are crucial for various life functions. All biomolecules are naturally occurring compounds. They include large macromolecules, such as proteins, carbohydrates, lipids, and nucleic acids, as well as smaller molecules, including vitamins and hormones. These organic molecules are key to all life forms and support many biological processes. From simply eating food to the occurrence of all the processes inside our body, Biomolecules play an important role.

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NCERT Notes for Class 12 Chapter 10 Biomolecules: Download PDF

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NCERT Notes for Class 12 Chapter 10 Biomolecules

This chapter explores the structure, classification, and functions of essential biomolecules like carbohydrates, proteins, nucleic acids, and vitamins. It shows how biomolecules are linked to our health and influence everything from what we eat to the medicines we take. Carbohydrates provide energy for everyday activities, Proteins are essential for growth, Lipids serve the purpose of energy storage, and Nucleic Acids, like DNA and RNA, are essential for genetic information transmission. Biomolecules class 12 notes PDF is the best resource for quick revision, also they help build a clear understanding of fundamental principles and their real-life applications.

Biomolecules

Biomolecules are organic substances that form a basis for the growth and maintenance of the human body.

Biomolecules discussed here are carbohydrates, proteins, enzymes, vitamins, nucleic acids, and hormones.

A) Carbohydrates

• Manufactured in plants by performing photosynthesis.
• Optically active polyhydroxy aldehydes/ polyhydroxy ketones.
• Food, clothing, and shelter.

Classification of carbohydrates:

1. Monosaccharides

Monosaccharides are the simplest carbohydrates that cannot be hydrolyzed into simpler compounds.

Examples are glucose, fructose, and ribose.

General formula- (CH₂O)₅

Glucose:

Present in honey and fruits.

Belongs to D-family.

• Preparation of glucose

a)From sucrose-

${\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}+{\mathrm{H}}_2\mathrm{O}\stackrel{{\mathrm{H}}^{+}}{\to }{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$

b)From starch-

${\left({\mathrm{C}}_6{\mathrm{H}}_{10}{\mathrm{O}}_5\right)}_{\mathrm{n}}+{\mathrm{nH}}_2\mathrm{O}\underset{393\mathrm{K},2-3\text{ bar }}{\overset{{\mathrm{H}}^{+}}{\to }}{\mathrm{nC}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$

Structure Of Glucose

Here are a few reactions to find the structure of glucose-

a)Straight chain structure-

b)Presence of five hydroxyl groups-

c)Presence of an aldehyde group-

d)Oxidation of glucose-

Oxidation of glucose indicates the presence of the primary alcoholic group.

e)Open chain structure of Glucose-

D-(+)-Glucose

The cyclic structure of glucose

Limitations of open chain structure of glucose-

• Glucose does not undergo some of the characteristic reactions of aldehydes.
• No reaction with ammonia
• No reaction with hydroxylamine.

(+)-Glucose exists in two stereoisomeric forms i.e., α-D-glucose and β-D-Glucose.

Mutarotation-the two forms of glucose, convert into each other at equilibrium when glucose is dissolved in water and allowed to stand.

Anomers

Cyclization in the structure of glucose has been observed by the formation of hemiacetal between -CHO group and -OH group. On C5 carbon.

Optical isomers exist when configuration around only one of the carbons C₁ takes place. These are known as anomers.

α-D-glucose β-D-Glucose

Fructose

Fructose is obtained when disaccharides are hydrolyzed. Fructose has a six-member hemiacetal ring structure.

Structure of fructose

α-D-fructose β-D-fructose

Haworth structure of fructose

α-D-fructose β-D-fructose

2. Disaccharides

These are the types of carbohydrates that give more than one (can be the same or not) monosaccharide.

Examples-

3. Polysaccharides

These are polymer compounds that are formed by many numbers of monosaccharide units which are joined together by glycosidic linkages.

a)Starch

• Found in plants.
• Found in wheat, rice, maize, potatoes, barley, etc.
• Starch reacts with iodine solution to give a blue color.
• It is a non-reducing saccharide.
• Does not reduce Fehling’s solution.

b)Amylose

• Linear polymer
• It contains 200-300 α-D-glucose units linked together by glycosidic linkages.
• Molecular mass (10,000-500,000)

c)Amylopectin

• It is a highly branched polymer.
• Do not react with iodine solution to give a blue color.

d)Cellulose

• It is the most abundant organic polymer found on earth.
• It is a straight-chain polymer joined by β-glycosidic linkages.
• Used in the manufacturing of paper, textiles, and plastic industries.

e)Glycogen

• It is a polysaccharide found in animal cells occurring in muscles, and the liver.
• It is known as animal starch.
• It is a polymer made up of a thousand glucose units.

Importance of carbohydrates

• Plays a major role for both plants as well as animals.
• Carbohydrates are the major source of energy (except cellulose)
• Carbohydrates store energy for the functioning of living organisms.
• Carbohydrates are used as raw materials in the production of textiles, papers, lacquers, breweries, etc.

B) Proteins:

All living cells are made up of biomolecules having a high molecular mass, known as amino acids.

1. Amino acids

Amino acids are the building block units of proteins. These are the organic compounds that contain amino as well as a carboxyl group.

The above unit may be attached to any other carbon atom other than that of the -COOH group.

2. Classification of Amino Acids

Amino acids can be broadly classified as – acidic, basic, or neutral.

Neutral amino acids- These are the amino acids that contain an equal number of amino and carboxyl groups. Examples- glycine, alanine, valine, etc.

Acidic neutral acids- These are amino acids that contain more carboxyl groups than amino groups. Examples- are aspartic acid, asparagine acid, and glutamic acid, which contain two -COOH groups and one -NH₃ group.

Basic amino acids- These contain more amino groups than carboxyl groups. Examples- lysine, arginine, and histidine.

3. Structure of Proteins

Amino acids exist as zwitterion which is dipolar.

The basic nature of zwitterion is due to -COO^- ion.

The acidic character of zwitterion is due to the -NH₃⁺ group.

Peptide linkage- When two or more amino acids condense, the resulting -CO-NH- link is called peptide linkage of the peptide bond.

The structure of proteins is classified into four hierarchical levels based on different structural parameters such as the folding patterns, the sequence of amino acids, 3D conformation, and functional arrangement.

• Primary structure of proteins- amino acids are linked in one or more polypeptide chains, which are known as the primary structure of proteins.
• Secondary structure of proteins- the folding and arrangement of polypeptide chains give the shape or conformation of the protein.
• The secondary structure can be of α-helix structure type or β-pleated sheet structure.
• Tertiary structure- the tertiary structure of the protein arises due to the folding, bending, and coiling, which results in three-dimensional structures.
• Quaternary structure- many of the proteins exist as a grouping of two or more polypeptide chains. These polypeptide chains are called the sub-units.

Proteins can be classified into two types based on their molecular shape.

(a) Fibrous proteins: When the polypeptide chains run parallel and are held together by hydrogen and disulfide bonds, then fiber-like structure is formed. Such proteins are generally insoluble in water. Some common examples are keratin (present in hair, wool, silk) and myosin (present in muscles), etc

4. Denaturation of Proteins

Proteins can be denatured (physical changes and biological changes) but there is no chemical change in the protein structure.

Denaturation can arise due to many factors such as changes in temperature, pH, or certain chemical agents.

Enzymes

Enzymes are biological catalysts that catalyze biochemical reactions in living organisms. For example- hydrolysis of maltose is catalyzed by maltase.

Mechanism of enzyme

The mechanism is given as

1. The enzyme (E) binds to the substrate(s)

E+S→ES

2. Product Formation

ES→EP

3. Products released from the above complex.

EP→E+P

C) Vitamins

These are the biomolecules that are not produced by the body and hence, need to be supplied in small amounts for necessary biological functions of the body.

There are A, B, C, D, E, and K vitamins.

• Classification of Vitamins

Water-soluble vitamins- water-soluble vitamins are vitamin B, vitamin C, etc.

These vitamins need to be supplied to the body from time to time.

Fat-soluble vitamins- vitamins that are only soluble in fat are called fat-soluble vitamins. A, D, E, and, K vitamins are soluble in fat.

Vitamin deficiency-associated diseases:

Vitamin A - Night blindness, Xeropthalmia

Vitamin (Thiamine) B₁ - Beriberi

Vitamin (Riboflavin) B₂ - Cheilosis

Vitamin (Niacin) B₃ - Pellagra

Vitamin (Pyridoxine) B₂ - Convulsions , Anaemia

Vitamin B₁₂ - Pernicious anaemia

Vitamin C (Ascorbic acid) - Scurvy

Vitamin D - Rickets ( in children)

Osteomalacia ( in adults )

Vitamin E - Increased RBC fragility, muscular weakness

Vitamin K - Poor blood clotting

D) Nucleic Acids

• Nucleic acids are polymers that are present in all human bodies.
• Nucleic acids play an important role in the development and reproduction of all life forms.
• Nucleic acids have nucleotides as their repeating units.

There are two types of nucleic acids-

DNA (deoxyribonucleic acid) and RNA (ribonucleic acid)

1. Chemical composition of nucleic acids-

Nucleotides consist of three chemical components such as a heterocyclic base, a five-carbon sugar, and a phosphate group.

2. Structure of Nucleic acids-

a) Nitrogen-containing heterocyclic base- Purines and pyrimidines are two types of heterocyclic bases. Example- Adenine and guanine are purines. Cytosine, thymine, and uracil are pyrimidines.

b) Sugars- the two types of sugars are RNA and DNA.

c) phosphate group- nucleotides are joined by these linkages.

d)Nucleoside- When a nitrogen base is attached to a sugar molecule a nucleoside unit is produced.

e) Nucleotide-

Base+Sugar+phosphate →nucleotide

1. The biological function of nucleic acids-

Some of the biological functions of nucleic acids are-

Replication- It is the property of a biomolecule to synthesize another molecule.

For example, DNA has a unique property to replicate itself.

Protein synthesis- genetic information stored in DNA in a specific base sequence is expressed in the form of a specific base sequence.

E) Hormones

Hormones are molecules that act as intercellular messengers. These are produced by endocrine glands in the body and are poured directly in the blood stream, which transports them to the site of action.

In terms of chemical nature, some of these are steroids, e.g., estrogens and androgens; some are poly peptides for example insulin and endorphins; and some others are amino acid derivatives such as epinephrine and norepinephrine. Hormones have several functions in the body. They help to maintain the balance of biological activities in the body.

Biomolecules Previous Years' Questions With Answers

Some important previous years' questions from this chapter are given below:

Question 1. Explain what is meant by:

(i) Peptide linkage

(ii) Glycosidic linkage

Answer:

(i) Peptide linkage: In this type of linkage, an amide bond is formed between the –COOH group of one amino acid and the –NH₂ group of another, releasing a molecule of water.

(ii) Glycosidic linkage: By the elimination of water, an ether (–O–) bond is formed between two monosaccharide units.

Question 2. Name two water‑soluble vitamins, their sources, and deficiency diseases.

Answer:

Vitamin B₂: Found in milk, yeast, and leafy vegetables. Deficiency causes glossitis, dermatitis.

Vitamin C: Found in citrus fruits and green veggies. Deficiency leads to scurvy.

Question 3. Name the four bases in DNA. Which one is not present in RNA?

Answer:

Four bases of DNA: Adenine, Guanine, Cytosine, Thymine.

Thymine is not present in RNA; it is replaced by Uracil.

Approach to Solve Questions of Class 12 Chemistry Chapter 10 Biomolecules

To effectively solve questions in the chapter on Biomolecules, follow a systematic and structured approach that helps understand all the concepts and reactions.

1. Classify Biomolecules

Biomolecules are usually classified into:

• Carbohydrates

• Proteins

• Lipids

• Nucleic Acids

• Vitamins & Hormones (sometimes included)

Main Points to Remember:

• Monosaccharides (glucose, fructose), Disaccharides (sucrose, lactose), Polysaccharides (starch, cellulose).

• Amino acids (essential & non-essential), peptide bonds, and protein structures (primary, secondary, tertiary, quaternary).

• Lipids (fats, oils, waxes) and their saponification.

• DNA vs RNA (differences in structure and function).

2. Practice Important Structures

• Carbohydrates:

  • Glucose structure (open-chain & cyclic forms – α-D-glucopyranose, β-D-glucopyranose).

  • Reducing vs Non-reducing sugars (e.g., Fehling’s test, Tollen’s test).

  • Glycosidic linkage in disaccharides.

• Proteins:

  • Zwitterion formation (amino acids at the isoelectric point).

  • Denaturation (heat, pH change).

  • Biuret test for proteins.

• Nucleic Acids:

  • Phosphodiester bonds in DNA/RNA.

  • Base pairing rules (A-T/U, C-G).

3. Practice the Reaction Mechanisms

• Hydrolysis of sucrose (into glucose + fructose).

• Formation of peptide bonds (condensation reaction).

• Saponification of triglycerides (formation of soap).

4. Differentiate between:

• Starch vs Cellulose (α-linkage vs β-linkage).

• Fibrous vs Globular proteins.

• DNA vs RNA (thymine vs uracil, double vs single strand).

5. Solve NCERT Questions Thoroughly

• Main Focus Should Be:

  • Defining the terms like "anomers" and "essential amino acids” in short answer type questions.

  • Practice questions on Haworth projections of glucose and ribose

  • Mutarotation and denaturation

  • Compare between DNA/RNA and classify carbohydrates/proteins.

7. Avoid Common Mistakes

• α-helix (protein) and double helix (DNA) are often confused

• Do not miswrite Fischer/Haworth projections.

• Avoid mistakes in zwitter ion formation conditions.

8. Practice previous years' questions and make short notes.

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