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

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

Edited By Shivani Poonia | Updated on Apr 03, 2025 09:34 AM IST

Biomolecules form the foundation of every living organism; they play a unique role in 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 big macromolecules like proteins, carbohydrates, lipids, and nucleic acids, as well as smaller ones like 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

This Story also Contains
  1. NCERT Class 12 Biomolecules Notes
  2. Biomolecules
  3. Glucose
  4. Structure Of Glucose
  5. Anomers
  6. Fructose
  7. Proteins
  8. 2. Classification of Amino Acids-
  9. 3. Structure of Proteins
  10. 4. Denaturation of Proteins
  11. Vitamins
  12. Nucleic acids
  13. NCERT Class 12 Notes Chapter-Wise
  14. Subject Wise NCERT Exemplar Solutions
  15. Subject Wise NCERT Solutions
  16. NCERT Books and Syllabus
Biomolecules class 12th notes- Free ncert class 12 Chemistry Chapter 14 Notes- Download PDF
Biomolecules class 12th notes- Free ncert class 12 Chemistry Chapter 14 Notes- Download PDF


Carbohydrates provide energy for everyday activities, Proteins are essential for growth, Lipids serve the purpose of energy storage, Nucleic acids like DNA and RNA are essential for genetic information transmission. This shows how biomolecules are linked to our health and influence everything from what we eat to the medicines we take. Learning about them boosts student's knowledge of biology and helps Students for exam preparation Like JEE , NEET .

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NCERT Class 12 Biomolecules is a very important chapter from the point of view of CBSE boards as well as competitive exams and this chapter consists of lots of theoretical and factual knowledge and it's very important to remember all the topics well. To make learning easy, notes are prepared by our subject experts in a very comprehensive and systematic way. Scroll down to get access to these notes.

Also, students can refer,


NCERT Class 12 Biomolecules Notes


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.
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Classification of carbohydrates

  1. Monosaccharides

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

Examples- glucose, fructose and, ribose.

General formula- (CH2O)5

  1. Glucose

Present in honey and fruits.

Belongs to D-family.

  • Preparation of glucose-

a)From sucrose-

C12H22O11+H2OH+C6H12O6+C6H12O6

b)From starch-

(C6H10O5)n+nH2O393 K,23 bar H+nC6H12O6

Structure Of Glucose

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

a)Straight chain structure-

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b)Presence of five hydroxyl groups-

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c)Presence of an aldehyde group-

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d)Oxidation of glucose-

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

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e)Open chain structure of Glucose-

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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.

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α-D-glucose β-D-Glucose

  1. Fructose

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

Structure of fructose

1663594787436

α-D-fructose β-D-fructose

Haworth structure of fructose

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α-D-fructose β-D-fructose

  1. Disaccharides

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

Examples-

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  1. 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 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 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 thousand glucose units.
  1. 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.

Proteins

All the living cells are made up of biomolecules having 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.

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The above unit may be attached to any other carbon atom other than that of -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- aspartic acid, asparagine acid, 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.

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Basic nature of zwitterion is due to -COO- ion.

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

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

  • 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.
  • Secondary structure can be of α- Helix structure type or β- pleated sheet structure.
  • Tertiary structure- the tertiary structure of the protein arise due to the folding, bending, and coiling which results in three- dimensional structures.
  • Quaternary structure- many of the proteins exist as the grouping of two or more polypeptide chains. These polypeptide chains are called the sub-units.

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 action-

The mechanism is given as-

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

E+S→ES

  1. Product Formation

ES→EP

  1. Products releases from the above complex.

EP→E+P

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 chilldrens )

Osteomalacia ( in adults )

Vitamin E - Increased RBCs fragility , muscular weakness

Vitamin K - Poor blood clotting

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.

  1. Structure of Nucleic acids-

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

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

1663595195761

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

1663595219249

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

1663595242260

1663595267641

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.

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.


Hormones

Hormones are chemical compounds that are produced in ductless glands in the body. Because of their function hormones are also known as chemical messengers.

Significance of NCERT Class 12 Chemistry Chapter 10 Notes

Biomolecules Class 12 notes contain step vise step headings and subheadings as per mentioned in the chapter 10 of Class 12 NCERT. These are very helpful for a quick revision for competitive exams like VITEEE, BITSAT, JEE Core, NEET, etc.

NCERT Class 12 Notes Chapter-Wise


Subject Wise NCERT Exemplar Solutions


Subject Wise NCERT Solutions


NCERT Books and Syllabus

Frequently Asked Questions (FAQs)

1. Why are biomolecules important to study in Chemistry?

Understanding biomolecules is crucial because they are the foundation of all biological processes. Studying their structure, properties, and reactions helps us understand how living organisms function at a molecular level. This knowledge is then used in medicine, agriculture, biotechnology, and other scientific fields.

2. What are the four main types of biomolecules?

 The four major types of biomolecules are:

  • Carbohydrates
  • Proteins
  • Nucleic acids
  • Lipids
3. How do you classify carbohydrates?

Carbohydrates are classified into three main categories based on the number of sugar units they contain:

  • Monosaccharides: These are the simplest sugars, consisting of a single sugar unit. Examples include glucose, fructose, and galactose.
  • Disaccharides: These are composed of two monosaccharides joined together by a glycosidic bond. Examples include sucrose (glucose + fructose), lactose (glucose + galactose), and maltose (glucose + glucose).
  • Polysaccharides: These are complex carbohydrates made up of many monosaccharide units linked together. Examples include starch, glycogen, and cellulose.
4. What is the difference between reducing and non-reducing sugars?
  • Reducing sugars possess a free aldehyde or ketone group that can be oxidized, allowing them to reduce other substances (like metal ions). All monosaccharides are reducing sugars. Many disaccharides, like maltose and lactose, are also reducing sugars.
  • Non-reducing sugars do not have a free aldehyde or ketone group available for oxidation. Sucrose is a common example of a non-reducing sugar because the anomeric carbons of both glucose and fructose are involved in the glycosidic bond.
5. What are proteins made up of?

Proteins are made up of amino acids. Amino acids are organic molecules that contain an amino group (-NH2), a carboxyl group (-COOH), and a side chain (R-group), all attached to a central carbon atom. There are 20 common amino acids that are used to build proteins.

6. How to study Class 12 biomolecules notes in the right manner?

Student should go through Class 12 Chemistry chapter 10 notes containing all the headings and subheadings with their brief explanation and solve questions from each topic.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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