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Molecular Basis of Inheritance Class 12 Notes: CBSE Biology Chapter 5

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Chapter 5 Molecular Basis of Inheritance Notes PDF - FREE Download

Molecular Basis of Inheritance Class 12 Notes are prepared to simplify the important processes involved in genetics for students. These notes focus on key topics such as the structure of DNA, replication, transcription, translation, gene expression, and regulation. They provide clear explanations, summaries, and diagrams to help students grasp complex concepts like the central dogma of molecular biology and the genetic code.  With concise content and illustrative examples, Class 12 Biology Notes help students understand the fundamentals and excel in their exams.

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Table of Content
1. Chapter 5 Molecular Basis of Inheritance Notes PDF - FREE Download
2. Access Revision Notes For Class 12 Biology Chapter 5 Molecular Basis of Inheritance
    2.1Section–A (1 Mark Questions)
    2.2Section–B (2 Marks Questions)
3. Important Concepts
    3.1The DNA 
    3.2Packaging of DNA helix 
    3.3DNA as a Genetic Material 
    3.4The Genetic Material is DNA 
    3.5The Central Dogma of Molecular Biology 
    3.6DNA Replication 
    3.7Transcription 
    3.8Translation 
    3.9Genetic code 
    3.10Regulation of Gene Expression 
    3.11Lac Operon or Lactose Operon 
    3.12Human Genome Project 
    3.13Applications and Future Challenges in Genetic Research
    3.14DNA Fingerprinting
4. 5 Important Topics of Biology Class 12 Chapter 5 you shouldn’t Miss!
5. Importance of Molecular Basis of Inheritance Class 12 Notes
6. Tips for Learning the Class 12 Biology Chapter 5 Molecular Basis of Inheritance 
7. Conclusion
8. Related Study Materials for Class 12 Biology Chapter 5 Molecular Basis of Inheritance
9. Biology Notes for Class 12 Chapter Wise PDF FREE Download
10. Important Study Materials for Class 12 Biology
FAQs


Download the FREE PDF of Class 12 Molecular Basis of Inheritance Notes from Vedantu, updated according to the latest CBSE Class 12 Biology Syllabus, to enhance your study sessions and reinforce your understanding of molecular genetics.

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Access Revision Notes For Class 12 Biology Chapter 5 Molecular Basis of Inheritance


Section–A (1 Mark Questions)

1. Name the process in which unwanted mRNA regions are removed & wanted regions are joined.

Ans. RNA splicing is the process in which unwanted mRNA regions (introns) are removed & wanted regions (exons) are joined.


2. Mention the dual functions of AUG?

Ans. AUG has dual functions. It codes for Methionine (met), and it also acts as initiator codon.


3. In which direction, the new strand of DNA synthesized during DNA replication.

Ans. The DNA-dependent DNA polymerases catalyse polymerisation only in one direction, that is 5'→3'.


4. List three components of the transcription unit.

Ans. A transcription unit in DNA is defined primarily by the three regions in the DNA: 

(i) A Promoter 

(ii) The Structural gene 

(iii) A Terminator


5. What is a replication fork?

Ans. For long DNA molecules, since the two strands of DNA cannot be separated in its entire length, the replication occurs within a small opening of the DNA helix, referred to as replication fork.


Section–B (2 Marks Questions)

6. Give two reasons why both the strands of DNA are not copied during transcription.

Ans. Two reasons why both the strands of DNA are not copied during transcription are:

  1. If both the strands code for RNA two different RNA molecules & two different proteins would be formed hence genetic machinery would become complicated. 

  2. Since the two RNA molecules would be complementary to each other, they would wind together to form dsRNA without carrying out translation which means the process of transcription would be futile.


7. What do you mean by “Central Dogma of Molecular Biology?”

Ans. The central dogma of molecular genetics is the flow of genetic information from DNA to DNA through replication, DNA to mRNA through transcription & mRNA to proteins through translation. It was proposed by Francis Crick.


Central Dogma of Molecular Biology



8. “The genetic material in the majority of organisms is DNA and not RNA.” Comment.

Ans. The following characteristics of DNA contributes to that fact that it is the genetic material in the majority of organisms:

  1. 2'-OH group present at every nucleotide in RNA is a reactive group and makes RNA labile and easily degradable.

  2. DNA chemically is less reactive and structurally more stable when compared to RNA. The presence of thymine at the place of uracil also confers additional stability to DNA.


9. Which three codons on mRNA are not recognized by tRNA? What is the general term used for them and what is their significance in protein synthesis?

Ans. UAA, UAG & UGA are the three codons that are not recognized by tRNA. These are known as stop codons or nonsense codons. Since these three codonsare not recognized by any tRNA they help in the termination of the protein chain during translation.


 10. State the criterias which a molecule must fulfill to act as genetic material.

Ans. A molecule that can act as a genetic material must fulfill the following criteria: 

(i) It should be able to generate its replica (Replication). 

(ii) It should be stable chemically and structurally. 

(iii) It should provide the scope for slow changes (mutation) that are required for evolution. 

(iv) It should be able to express itself in the form of 'Mendelian Characters’.

11. Mention any three applications of DNA fingerprinting.

Ans. Three applications of DNA fingerprinting are:

(i) It is used in forensic science to identify potential crime suspects.

(ii) It is used to determine the real or biological parents in case of disputes. 

(iii) It is used to find out the evolutionary history of an organism and trace out the linkages between various groups of organisms.


Important Concepts

The DNA 

DNA (deoxyribonucleic acid) is a double helix structure that was cracked by Watson and Crick based on the results of X-ray crystallography. Each strand of a DNA helix consists of repeating nucleotide units. The nucleotide consists of 3 components: ribose or deoxyribose sugar, nitrogen-based. which is either Purines or pyrimidines and phosphate.


Structure of Nucleotide


Structure of Nucleotide 


  • There are two types of purines which are known as adenine, and guanine. There are three types of pyrimidines: thymine, cytosine, and uracil. 

  • All nucleotides are common in DNA and RNA, but uracil is found in RNA, and thymine is only found in DNA. 

  • DNA is negatively charged due to the presence of negatively charged phosphate groups. A nitrogenous base binds to the pentose sugar via the glycosidic bond. 

  • Two nucleotides join by a 3'5 'phosphodiester bond to form a dinucleotide. A polymer so formed has a free phosphate group at the 5 'end of the ribose sugar known as the 5' end of the polynucleotide chain. The other end of the polymer has a free 3'OH group of ribose sugar. 

  • This is known as the 3 'end of the polynucleotide chain. The bond between sugars and phosphates forms the backbone of a polynucleotide chain. The nitrogenous bases are bound to the sugar content and protrude from the backbone.


The salient feature of the double helix structure of DNA is as follows- 


  • Two polynucleotides chains that are present, wrap around each other. Here, the backbone is constituted by sugar-phosphate, and bases project inside. 

  • The two DNA chains are always antiparallel to each other. Antiparallel means that if one chain has the polarity 5'-3', the other has 3'-5'. 

  • The bases which are present in the two strands are paired through hydrogen bonding forming base pairs. Adenine form two hydrogen bonds with thymine whereas cytosine forms three hydrogen bonds with guanine. 

  •  The two strands are coiled in the right-handed pattern.  

  • The plane of one of the base pair stacks over the other in a double helix. This, in addition to H-bonds, confers the stability of the helical structure. 


Packaging of DNA helix 

Positively charged basic proteins that surround the DNA are known as histones. Histones are found to be rich in basic amino acids such as lysine and arginine. Histones molecules are organized in a specific manner to form a unit of eight molecules called histone octamer. The DNA is negatively charged in nature and is packaged by wrapping around the positively charged histone octamer. This forms a structure called a nucleosome. A nucleosome was found to be containing 200 base pairs of DNA helix. Nucleosomes form a specific repeating unit of a structure which is called chromatin in the nucleus. Chromatin is known to be a thread-like stained body seen in the nucleus. The nucleosomes in chromatin appear as a ‘beads-on-string’ structure when viewed under an electron microscope (EM). The beads on string structure in chromatin are packaged to form chromatin fibers that are further coiled and condensed at the metaphase stage of cell division to form chromosomes. At the high levels, chromatin packaging requires additional proteins. These proteins are the Non-Histone Chromosomal (NHC) proteins. In a typical nucleus, a loosely packed region of chromatin stains light and is referred to as euchromatin. The dense chromatin stains dark is called Heterochromatin. The euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is inactive. 


Packaging of DNA Helix

 

Packaging of DNA Helix 


DNA as a Genetic Material 

Griffith performed an experiment which is commonly known as the transforming experiment. He used the two different strains of Pneumococcus. These two different strains were used to infect the mice. The two strains which were used are type III-S (smooth), that contains outer capsule made up of polysaccharide and type II-R (rough) strain do not contain capsule. The capsule protects the bacteria from the host immune system. 


Griffith Experiment


Griffith Experiment


The experiment of Griffith is explained below- 


  • The rough strain of Pneumococcus is injected into the mouse. The mouse is alive. 

  • The smooth strain of Pneumococcus is injected into the mouse. The mouse dies

  • When the heat-killed smooth strain of Pneumococcus is injected into the mouse, the mouse is alive.

  • In the last set of experiments, rough strain, and heat-killed smooth strain are injected into the mouse. The mouse dies. 

  • This proves that there is some transforming substance present in the heat-killed S strain that is converting or transforming the rough strain into a virulent strain that is responsible for the death of the mouse. The substance that was transformed. later found to be DNA.  


The Genetic Material is DNA 

Alfred Hershey and Martha Chase in 1952 performed an experiment to prove that DNA is the genetic material. They worked on bacteriophages, which are viruses that infect the bacteria. It was found that when the bacteriophage attaches to the bacteria, its genetic material enters into the bacterial cell. The viral genetic material uses the bacterial cell to synthesize more viral particles. Hershey and Chase grew some viruses on a medium that contained radioactive phosphorus and another medium that contained radioactive sulfur. When phosphorus which was radioactive was present in the medium the viruses contained radioactive DNA but not radioactive protein. This is due to the fact that DNA contains phosphorus, but protein does not. Similarly,  when the growth medium contained radioactive sulfur the viruses contained radioactive protein but not radioactive DNA.  This is because DNA does not contain sulfur.  


Hershey and Chase Experiment


Hershey and Chase Experiment


These bacteria were only found to be radioactive when infected with viruses that contained radioactive DNA. This indicates that DNA was the material that was transferred from the virus to the bacteria. Bacteria which were infected with viruses containing radioactive proteins were not radioactive. This indicates that the proteins did not get into the bacteria from the virus. So DNA is the genetic material that is transferred from viruses to bacteria. This experiment shows that DNA is the genetic material.


The Central Dogma of Molecular Biology 

It is an explanation of how genetic information transfer in a biological system. It tells us how the DNA replicates and then transcribes into messenger RNA (mRNA). This mRNA will now be translated to make proteins. 


The Central Dogma of Molecular Biology

  

The Central Dogma of Molecular Biology 


DNA Replication 

DNA replication is the process by which two identical copies of DNA are made from a single DNA molecule. As we know, DNA is a double helix in which two strands are complementary to each other. These two strands of a helix separate at replication time to form two new DNA molecules. Of the two DNA strands formed, one is identical to one of the strands and the other is complementary to the original strand. This form of replication is defined as semi-conservative replication. goes into mitosis, DNA replicates in the S phase of the interface DNA polymerase is the most important enzyme involved in DNA replication is an energy-dependent process During the replication process, the two DNA chains do not separate completely, replication occurs within the small opening in the DNA helix known as the replication fork.DNA polymerase catalyzes the reaction in 5 'to 3' so that in one strand (the template with 3'5 'polarity) the replication is continuous, while in the other (the template with 5'3' polarity) the enzyme DNA ligase then binds to the discontinuously synthesized fragments. The continuously synthesized strand is referred to as the main strand, while the discontinuously synthesized strand is referred to as the lag strand. Replication begins at a specific location in DNA known as the origin of replication.


DNA Replication


DNA Replication


Transcription 

It is a process of making RNA, like messenger RNA, from DNA before gene expression or protein synthesis takes place. During transcription, one of the DNA strands acts as a template for mRNA formation. The synthesis of mRNA is carried out by the enzyme RNA polymerase. It generally occurs for a specific stretch of DNA that is most needed for gene expression. In addition to messenger RNA, other forms of RNA such as ribosomal RNA, microRNA, small nuclear RNA can be transcribed in a similar way. It consists of the following three regions: a promoter, a structural gene, and a terminator DNA-dependent RNA polymerase catalyzes polymerase in the 5'3 'direction. The promoter is the region to which the RNA polymerase binds. The terminator defines the end of the transcription.


Process of Transcription


Process of Transcription


  • Transcription consists of a total of three steps- initiation, elongation, and termination.  

  • Initiation is the step that involves the binding of RNA polymerase to the promoter. A single type of DNA-dependent RNA polymerase catalyzes the transcription of all types of RNA in bacteria. 

  • Elongation is the process of adding nucleotides to form RNA. 

  • The termination factor helps in the termination of the transcription. The RNA synthesized after transcription is called the primary transcript. The primary transcript undergoes modifications such as splicing, coating, tailing, etc.

  • The primary transcript consists of the introns and exons. Introns are known as splicing. The addition of the polyA tail to the 3 'end of the RNA is referred to as the tail. When capping, an unusual nucleotide (methylguanosine triphosphate) is attached to the 5 'end of the RNA. 

  • Some viruses have the property of reverse transcription. You can convert RNA templates into DNA. The enzyme used is known as reverse transcriptase. 

  • For example, the human immunodeficiency virus that causes AIDS."


Translation 

This is the process of gene expression or protein synthesis, that occurs in the cytosol. Ribosomes are known to be the cellular organelles that participate in protein synthesis. Messenger RNA, made by the transcription process, is deciphered by ribosomes to form a composite polypeptide. Of amino acids. Messenger RNA is made up of a polymer of nucleotides, or codons. Each codon consists of 3 nucleotides that code for a single amino acid. There are several important components involved in the synthesis of ribosome proteins, messenger RNA, and transfer RNA (tRNA). Transfer RNA is involved in the physical binding of mRNA and the amino acid sequence of proteins.


Translation


Translation


It involves 4 main steps- 


  • Activation of amino acids- amino acids bind to specific tRNA molecules. 

  • Initiation of the polypeptide synthesis- In the process of capping an unusual nucleotide (methyl guanosine triphosphate) is added to  the 5'-end of RNA 

  • Elongation of polypeptide synthesis- It involves the addition of amino acids to the growing polypeptide chains 

  • Termination of polypeptide synthesis- It involves the end of the translation of protein synthesis. 


Genetic code 

The set of different rules according to which the information encoded in genetic material is translated into proteins in living cells. The outstanding features of the genetic code are: 


  • The codon consists of 3 nucleotides. 61 codons code for 20 different types of amino acids. 

  • 1 codon codes for a single amino acid. 

  • 1 amino acid can be encoded by more than one codon.


Genetic Code


Genetic Code


Regulation of Gene Expression 

All the genes in the living cells are not active all the time. They become active when needed. Expression is controlled by genes are known as regulatory genes. Regulation in the eukaryotes can occur at the following different levels- 


  • Transcriptional level. 

  • Processing level. 

  • Transportation of mRNA from the nucleus to the cytoplasm. 

  • Translational level. 


Lac Operon or Lactose Operon 

An operon consists of structural genes, operator genes, promoter genes, promoter genes, regulator genes, and repressors.  Lac operon consist of lac Z, lac Y and lac A genes. Lac Z codes for galactosidase, lac Y codes for permease, and lac A codes for transacetylase. When repressor molecules are bound to the operator, genes are not transcribed. When the repressor does not bind the operator and instead inducer binds, transcription is switched on. In the case of the lac operon, lactose is an inducer. So, binding lactose to the operator, switch on the transcription. 


Lactose Operon


Lactose Operon


Human Genome Project 

The most important and basic features of the Human Genome Project are: 


  • The human genome contains 3,164.7 million nucleotide bases. 

  • The average gene is 3000 bases, but the size varies. 

  • Humans are said to have around 30,000 genes. 

  • The functions of more than 50 percent of the discovered genes are unknown. 

  • Less than 2 percent of the genome code for proteins. 

  • The human genome consists to a large extent of repetitive sequences. 

  • Chromosome 1 has the most genes (2968) and Y has the fewest (231).


Applications and Future Challenges in Genetic Research

  • Meaningful Knowledge from DNA Sequences:

    • Research in the coming decades will focus on deriving significant insights from DNA sequences.

    • Understanding biological systems will be a major goal.


  • Collaboration Across Disciplines:

    • The task will involve the expertise and creativity of scientists from various fields.

    • Collaboration between public and private sectors worldwide will be crucial.


  • Impact of Whole-Genome Sequencing:

    • Whole-genome sequences enable a new approach to biological research.

    • Researchers can study entire genomes rather than focusing on individual genes.


  • High-Throughput Technologies:

    • New technologies allow for systematic and large-scale investigations.

    • Examples include studying all genes in a genome, all transcripts in specific tissues or tumours, and analysing interconnected networks of genes and proteins.


  • Systematic Approach to Research:

    • Researchers can now address questions more comprehensively and broadly.

    • This approach facilitates a deeper understanding of the complex interactions within biological systems.


DNA Fingerprinting

  • Human Genetic Similarity and Differences:

    • 99.9% of base sequences are identical among humans.

    • With a genome of 3 × 10^9 base pairs, there are differences in a small fraction of sequences, making each individual's DNA unique.


  • Challenges of Sequencing:

    • Sequencing entire genomes to compare genetic differences is expensive and impractical.

    • DNA fingerprinting offers a quicker, cost-effective method to compare DNA sequences.


  • Repetitive DNA and Satellite DNA:

    • DNA fingerprinting focuses on repetitive DNA regions, where short DNA sequences are repeated multiple times.

    • These repetitive regions, known as satellite DNA, are separated from bulk DNA during density gradient centrifugation.

    • Satellite DNA includes microsatellites and minisatellites, which are highly variable and form the basis of DNA fingerprinting.


  • DNA Polymorphism:

    • Polymorphisms are genetic variations arising from mutations in DNA.

    • High-frequency variations in non-coding DNA sequences are key to DNA fingerprinting and genetic mapping.

    • Mutations in non-coding regions accumulate over generations, contributing to genetic diversity.


  • DNA Fingerprinting Technique:

    • Developed by Alec Jeffreys, initially used Variable Number of Tandem Repeats (VNTRs) as probes.

  • Steps:

    1. Isolation of DNA

    2. Digestion with restriction endonucleases

    3. Separation by electrophoresis

    4. Blotting onto synthetic membranes

    5. Hybridization with labelled VNTR probes

    6. Detection by autoradiography


  • VNTRs and Mini-Satellites:

    • VNTRs are a type of mini-satellite DNA with variable repeat numbers.

    • The size of VNTRs varies, producing distinctive patterns of bands after hybridization.

    • Patterns are unique to individuals, except for monozygotic twins.


  • Modern Advancements:

    • Polymerase Chain Reaction (PCR) enhances sensitivity, allowing analysis from a single cell.

    • DNA fingerprinting is used in forensic science, population genetics, and determining genetic diversity.


This technique has revolutionised forensic identification and genetic research by providing a reliable method to analyse genetic differences and similarities among individuals.


DNA Fingerprinting


5 Important Topics of Biology Class 12 Chapter 5 you shouldn’t Miss!

Topic

Key Points

Structure of DNA

- Double-helix model by Watson and Crick

- Components: nucleotides (phosphate, deoxyribose, nitrogenous bases)

- Base pairing rules (A-T, G-C)

DNA Replication

- Enzymes: DNA polymerase, helicase, ligase

- Leading and lagging strands

- Okazaki fragments formation

Transcription

- Process of RNA synthesis from DNA

- Role of RNA polymerase

- mRNA processing (capping, poly-A tail, splicing)

Translation

- Process of protein synthesis from mRNA

- Role of ribosomes, tRNA, and amino acids

- Stages: initiation, elongation, termination

Genetic Code and Mutations

- Characteristics of the genetic code (universal, degenerate)

- Types of mutations (point, insertion, deletion)

- Effects of mutations on proteins


Importance of Molecular Basis of Inheritance Class 12 Notes

  • Molecular Basis of Inheritance Notes provides a comprehensive and detailed overview of critical topics such as DNA structure, replication, transcription, translation, and genetic code, ensuring all essential areas are thoroughly covered.

  • Complex genetic processes are explained in easy-to-understand language, making them accessible to all students.

  • Summaries and key facts are highlighted for rapid revision, helping students grasp and retain crucial information efficiently.

  • Diagrams, flowcharts, and illustrations facilitate a better understanding of important processes and enhance retention.

  • Class 12 Biology Chapter 5 Notes are structured clearly and concisely, allowing students to focus on key areas and manage their study time effectively.

  • With organised content and visuals, the Molecular Basis of Inheritance Notes PDF supports effective study sessions and improves recall during exams.


Tips for Learning the Class 12 Biology Chapter 5 Molecular Basis of Inheritance 

  • Focus on the double-helix model, base pairing rules, and the components like nucleotides. Understanding the structure is fundamental to grasping replication and transcription processes.

  • Learn the steps of DNA replication in detail, such as initiation, elongation, and termination. Diagrams can help visualise this process better.

  • Grasp the concept of how genetic information flows from DNA to RNA to proteins (transcription and translation). This is key to understanding gene expression.

  • Study the codons and how they determine the sequence of amino acids in a protein. Practice decoding sequences.

  • Learn about different types of mutations and their impact on DNA sequences and protein synthesis.

  • Focus on operons like the lac operon as examples of gene regulation in prokaryotes.

  • The chapter involves many processes like transcription and replication that can be better understood through diagrams. Practice drawing and labelling these.

  • Go through the in-text and exercise questions in NCERT to solidify your understanding of the concepts.


Conclusion

Molecular Basis of Inheritance Class 12 Notes highlights the essential role of DNA in heredity. Understanding DNA's structure, replication, and function is crucial for grasping how genetic information is passed from one generation to the next. The mechanisms of transcription and translation are key to how genes are expressed as traits. Understanding these concepts not only explains the foundation of genetics but also provides insight into how variations and mutations occur. This knowledge forms the backbone of modern genetics and helps in understanding complex biological processes.


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FAQs on Molecular Basis of Inheritance Class 12 Notes: CBSE Biology Chapter 5

1. What is the process of translation?

The process of protein synthesis in which mRNA is used to synthesize protein is known as translation. It is the process of gene expression or protein synthesis that occurs in the cytosol. The mRNA sequence is decoded to specify the amino acid of a polypeptide. Transcription consists of 3 steps.

  • Initiation

  • Elongation

  • Termination 

2. What does the process of DNA replication involve?

DNA replication is the course of creating dual copies of DNA from a single DNA molecule. The original strand is known as parent strands, and the new strand is known as the daughter strands. It is a process of biological inheritance. The initial stage in DNA replication is to unzip the double helix assembly of the molecule.

3. What are the differences between the leading and lagging strand?

Strands which are synthesized continuously are known as leading strands, whereas the strands which are synthesized discontinuously are known as lagging strands.

  • Leading strand does not require DNA ligase for its growth.

  • In the case of lagging strand DNA ligase is vital for constructing the Okazaki fragment.

  • The direction of growth of a leading strand is 5’ – 3’.

  • The direction of growth of a lagging strand is 3’ – 5’.

  • Formation of a leading strand starts instantly at the start of replication.

  • Development of a lagging strand starts after that of the leading strand.

4. What is DNA?

DNA which stands for Deoxyribonucleic acid, is the genetic material of humans and other animals. The DNA was first found by Friedrich Meischer in 1869. James Watson and Francis Crick made the discovery of the double helix structure of DNA in 1953. DNA is a continuous polymer made from simple units called nucleotides that are kept together firm by sugar and phosphate groups. This backbone contains four types of molecules known as bases, which are Adenine, Thymine, Guanine and Cytosine.

5. How is DNA packaged in Eukaryotes?

Eukaryotes contain a well-defined nucleus made up of DNA, a negatively charged polymer. This negatively charged polymer is enfolded with a positively charged histone octamer to form a nucleosome. Histone octamer is basically formed by organizing eight molecules of histones (which are positively charged basic proteins). The histones attain positive charge as they are rich in amino acid residues such as lysine and arginine. The nucleosomes formed are then coiled even more, resulting in the development of chromatin fibers.

6. Where can I find the NCERT solution for Class 12 Biology Chapter 5?

You can obtain NCERT solutions for all the questions provided at the end of the Class 12 Biology Chapter 5 on Vedantu. You can avail the entire solutions in PDF format for FREE of cost. These solutions are written by the professional subject matter experts at Vedantu keeping the CBSE curriculum in mind. The solutions provided to you explains the concept in a brief and simple manner. The language used is quite simple to make it understandable for the students. 

7. Why is Class 12 biology Chapter 5 important for students?

Class 12 biology chapter 12 ‘molecular basis of inheritance’ is an essential chapter for the students. This chapter holds the majority of the marks weightage in the board exams. Therefore, students must make sure to analyze the entire chapter thoroughly. This chapter tells the basic concepts of human genetics including DNA, RNA, process of replication, transcription, translation etc. Learn all the important definitions and practice all the important diagrams carefully. It is important to practice all the questions provided in the NCERT textbook.

8. Why should I prefer Revision notes of Vedantu for Class 12 biology Chapter 5 ?

Vedantu, today has become one of the renowned online learning platforms that offers the students with exclusive study materials and study guide for their learning process. Vedantu also provides well-written revision notes that encloses all the significant topics of any chapter, including for Class 12 Biology Chapter 5 Revision Notes. These revision notes are curated by experienced tutors and are available in PDF format on Vedantu. You can refer to these notes to enhance your concepts and revise the whole chapter in brief at FREE of cost on the Vedantu.

9. Why is studying the Molecular Basis of Inheritance Notes important for Class 12 students?

Understanding the Class 12 Molecular Basis of Inheritance Notes is crucial for grasping fundamental concepts in genetics, biology, and biotechnology, and it lays the foundation for advanced studies in medicine and life sciences.

10. How do Vedantu's Class 12 Biology Chapter 5 Notes help in understanding DNA replication?

Vedantu’s Molecular Basis of Inheritance Notes PDF simplifies the DNA replication process by breaking down each step and providing clear diagrams, making it easier for students to understand the mechanism.