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 Sumit Saini | Updated on Apr 23, 2022 11:36 AM IST

NCERT Class 12 Chemistry chapter 14 notes help to establish a relationship between inorganic molecules that makes a living organism. Biomolecules Class 12 notes cover a brief outline of topics such as carbohydrates, proteins, vitamins, enzymes, nucleic acids, hormones, etc.

The NCERT Class 12 chapter 14 notes discuss all the main topics of NCERT chapter 14 of Class 12. Some of the topics discussed in biomolecules Class 12 notes consist of carbohydrates such as monosaccharides, disaccharides, polysaccharides, amino acids, mechanism of enzyme action, nucleic acids, and hormones. Biomolecules are organic substances that build up the living organism and the same biomolecules are also important for three-dimensional growth and maintenance. The branch of science which deals with the study of biomolecules is known as biochemistry. Many of the biomolecules are polymers. For example, protein is a polymer made up of amino acids as the monomer unit.

This Story also Contains
  1. NCERT Class 12 Chapter 14 Notes-
  2. Biomolecules Class 12 Notes – Topic 1:
  3. Biomolecules-
  4. 2. Classification of Amino Acids-
  5. 3. Structure of Proteins-
  6. 4. Denaturation of Proteins-
  7. Classification of Vitamins-
  8. NCERT Class 12 Notes Chapter-Wise
  9. NCERT Books and Syllabus

All the topics in cbse class 12 chemistry chapter 14 notes are discussed in depth. CBSE Class 12 Chemistry chapter 14 notes also include quick points to remember from the exam point of view. All the topics mentioned below can be easily downloaded by the student by Class 12 Chemistry chapter 14 notes pdf download.

Also, students can refer,

NCERT Class 12 Chapter 14 Notes-

Biomolecules Class 12 Notes – Topic 1:

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.

  • Carbohydrates-

-Manufactured in plants by performing photosynthesis.

-optically active polyhydroxy aldehydes/ polyhydroxy ketones.

- Food, clothing, and shelter.

Classification of carbohydrates

  1. Monosaccharides

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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-

From sucrose-

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From starch-

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  • Structure of glucose-

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

Straight chain structure-

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Presence of five hydroxyl groups-

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Presence of an aldehyde group-

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Oxidation of glucose-

Oxidation of glucose indicates the presence of primary alcoholic group.

1663594706668

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-

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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.

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.

Amylose-

Linear polymer

It contains 200-300 α-D-glucose units linked together by glycosidic linkages.

Molecular mass (10,000-500,000)

Amylopectin-

It is a highly branched polymer.

Do not react with iodine solution to give blue color.

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.

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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-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.

  • 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.

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

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

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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 14 Notes

Biomolecules Class 12 notes contain step vise step headings and subheadings as per mentioned in the chapter 14 of Class 12 NCERT. NCERT Class 12 Chemistry chapter 14 notes are very helpful for a quick revision for competitive exams like VITEEE, BITSAT, JEE Core, NEET, etc. Class 12 Chemistry chapter 3 notes pdf download can be used for offline revision.

NCERT Class 12 Notes Chapter-Wise

Subject Wise NCERT Exemplar Solutions

Subject Wise NCERT Solutions

NCERT Books and Syllabus

Frequently Asked Questions (FAQs)

1. What are the main topics covered in the biomolecules Class 12 notes?

Ans: All the biomolecules such as proteins, carbohydrates, nucleic acids, hormones, vitamins, and enzymes are explained in Class 12 Chemistry chapter 14 notes.

2. What are the structures of the ncert Class 12 Chemistry chapter 14 notes?

Ans: The main structures explained in these notes are glucose, nucleic acid, amino acids.

3. How should a student study Class 12 Chemistry chapter 3 notes?

Ans: These notes have all the headings and subheadings given in chapter 14 of NCERT Class 12.

4. Highlight some of the important sections to cover from notes for Class 12 Chemistry chapter 14 for exams?

Ans: From an exam point of view, the structure of glucose, structure of proteins, amino acids, the structure of nucleic acids, the difference between RNA, DNA are some important topics.

5. What are some topics to cover for the chemical equations section in NCERT notes for Class 12 Chemistry chapter 14?

Ans: A simple way to prepare for the chemical equation is by checking any link for electrochemistry Class 12 notes pdf download or biomolecules Class 12 notes pdf download. Practice all the reactions given by glucose, fructose which are used to determine the structure of glucose.

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

Ans: Student should go through Class 12 Chemistry chapter 14 notes containing all the headings and subheadings with their brief explanation and solve questions from each topic.

Articles

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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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