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    NCERT Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance

    NCERT Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance

    Irshad AnwarUpdated on 06 Feb 2026, 11:14 AM IST

    The NCERT Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance explains the different concepts like DNA replication, transcription, and translation in an easy-to-understand manner. It also describes the structure and function of DNA and RNA with proper diagrams. Studying through the NCERT Solutions allows students to score well in school exams as well as in competitive entrance exams like NEET. These solutions also boost confidence and save a lot of time during the exams.

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    The overall CBSE class 12th Maths exam 2026 can be considered balanced and moderately difficult. Students with a clear conceptual understanding and regular practice of NCERT and exemplar problems would have found the paper manageable. The question paper was a fair mix of knowledge, application, and analytical questions, reflecting the ongoing shift towards competency-based assessment in school education.

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    This Story also Contains

    1. Download Molecular Basis of Inheritance NCERT Solutions PDF
    2. Access to Class 12 Biology Chapter 5 Molecular Basis of Inheritance Solutions (Exercise Questions)
    3. Approach to Solve the Molecular Basis of Inheritance Class 12 Question Answer
    4. Important Questions of NCERT Class 12 Biology Chapter 5
    5. What Students Learn from the Molecular Basis of Inheritance: NCERT Solutions
    6. What Extra Should Students Study Beyond the NCERT for NEET?
    7. NCERT Solutions for Class 12 Biology- Chapter-wise
    NCERT Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance
    NCERT Solutions for Class 12 Biology Chapter 5 - Molecular Basis of Inheritance

    The Molecular Basis of Inheritance Class 12 question answers explain how genetic information is stored, copied, and passed on to living organisms. These solutions also highlight important topics like the genetic code, mutations, and regulation of gene expression, which are frequently asked in exams. The NCERT Solutions Class 12 Biology help students understand important topics effectively without any confusion. Practising these questions regularly helps students improve conceptual clarity and strengthen their problem-solving skills.

    Download Molecular Basis of Inheritance NCERT Solutions PDF

    Students can get free PDFs of detailed answers to Chapter 5 from here. The Class 12 Biology Chapter 5 Molecular Basis of Inheritance question answers explain important topics like DNA replication, the genetic code, and how proteins are formed. These are clearly explained with diagrams and examples in the NCERT Solutions for Class 12.

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    Access to Class 12 Biology Chapter 5 Molecular Basis of Inheritance Solutions (Exercise Questions)

    Molecular basis of inheritance solutions explain complex topics like transcription and translation in a simple and step-by-step manner. The Class 12 Biology Chapter 5 Molecular Basis of Inheritance question answer includes detailed explanations that follow the NCERT guidelines, making it easier for students to revise and practice all important topics before exams.

    Q1. Group the following as nitrogenous bases and nucleosides:

    Adenine, Cytidine, Thymine, Guanosine, Uracil, and Cytosine.

    Answer:

    Nitrogenous bases- Adenine, thymine, uracil, and cytosine

    Nucleosides- Cytidine and guanosine

    Q2. If a double-stranded DNA has 20 per cent of cytosine, calculate the percentage of adenine in the DNA.

    Answer:

    According to Chargaff's rule, A=T and G=C. As there are 20% cytosines, the guanine percentage will also be 20%. The G+C count is 20+20= 40%. Hence, the remaining 60% will be shared equally between adenine and thymine. So, the percentage of adenine and thymine is 30% each.

    Q3. If the sequence of one strand of DNA is written as follows:
    5'-ATGCATGCATGCATGCATGCATGCATGC-3'

    Write down the sequence of the complementary strands in 5′→3′ direction.

    Answer:

    About base pairs, the DNA strands are complementary to each other. So, if the sequence of DNA is

    5'-ATGCATGCATGCATGCATGCATGCATGC-3'

    The sequence of the complementary strands will be

    3'- TACGTACGTACGTACGTACGTACGTACG - 5'.

    In the 5'→3' strand, it can be written as

    5'- GCATGCATGCATGCATGCATGCATGCAT-3'

    Q4. If the sequence of the coding strand in a transcription unit is written as follows:

    5'-ATGCATGCATGCATGCATGCATGCATGC-3'

    Write down the sequence of mRNA.

    Answer:

    The sequence of mRNA is the same as the coding strand of DNA, except that thymine is replaced by uracil. Thus, the sequence of mRNA will be

    5' - AUGCAUGCAUGCAUGCAUGCAUGCAUGC - 3'.

    Q5. Which property of the DNA double helix led Watson and Crick to hypothesise a semi-conservative mode of DNA replication? Explain.

    Answer:

    The characteristic of the DNA double helix that prompted Watson and Crick to speculate a semi-conservative mode of DNA replication is that the two DNA strands are antiparallel and complementary to one another in the sense of their base sequences. It follows from this organisation that DNA replication is semiconservative. When replication occurs, the two strands separate, and each strand serves as a template for the production of a new strand. By the completion of replication, parental types and recombinant types of DNA are produced.

    Q6. Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.

    Answer:

    Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), two types of nucleic acid polymerases are found

    1. DNA-dependent DNA polymerase - It uses DNA as a template to synthesise new strands of DNA.

    2. DNA-dependent RNA polymerase - It uses DNA as a template to synthesise RNA.

    Q7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

    Answer:

    Alfred Hershey and Martha Chase (1952) used a virus that infects bacteria, bacteriophages. They labelled DNA and protein with various radioactive isotopes. In their experiment, in one preparation, the protein component was rendered radioactive, and in the other, nucleic acid (DNA) was rendered radioactive. These two phage preparations were permitted to infect the culture of E. coli.

    As soon as after infection, before cell lysis, the E. coli cells were mildly shaken in a blender, loosening the attached phage particles, and the culture was centrifuged. The denser infected bacterial cells were pelleted to the bottom, and lighter viral particles were in the supernatant. When a bacteriophage with radioactive DNA was used to infect E. coli, radioactivity was found in the pellet. If a bacteriophage with a radioactive protein coat was employed to infect E. coli, the radioactivity was present in most of the supernatant. They demonstrated in their experiment that protein does not enter the bacterial cell, whereas DNA gets transformed. Thus, they established that DNA is the genetic material.

    Q8. Differentiate between the following:

    (a) Repetitive DNA and Satellite DNA

    Answer:

    Repetitive DNASatellite DNA
    Repetitive DNA refers to DNA sequences containing small segments that are repeated many times.DNA sequences containing highly repetitive DNA.


    (b) mRNA and tRNA

    Answer:

    mRNA or messenger RNAtRNA or transfer RNA
    It acts as the template for the translation of proteins.tRNA acts as the adapter molecule that carries specific amino acids to mRNA for the synthesis of a polypeptide.


    (c) Template strand and Coding strand

    Answer:

    Template strandCoding strand
    It is the template for the synthesis of mRNA during transcriptionStrand of DNA having the same sequence as mRNA (thymine in DNA is replaced by uracil in RNA
    Its polarity is 3' to 5'Its polarity is 5' to 3'


    Q9. List two essential roles of the ribosome during translation.

    Answer:

    The ribosome is a complex structure made up of ribonucleoproteins. It consists of two subunits, i.e., a larger subunit and a smaller subunit. The essential roles of ribosomes during translation are as follows:

    1. The ribosome acts as the site of protein synthesis. The larger subunit of the ribosome acts as an amino acid binding site, while the small subunit attaches to the mRNA.

    2. The ribosome acts as a catalyst for forming peptide bonds.

    Q10. In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does the lac operon shut down sometime after the addition of lactose in the medium?

    Answer:

    The lac operon is a set of genes in bacteria that help them use lactose for energy.

    • It has three main genes that make enzymes to bring lactose into the cell and break it down.

    • A regulatory gene makes a repressor protein that blocks these genes when lactose is absent.

    • When lactose is present, it binds to the repressor, stopping it from blocking the genes.

    • This allows the enzymes to be made and lactose to be used.

    • When lactose is gone, the repressor blocks the genes again to save energy.

    The operon is only active when lactose is available, and glucose (a preferred sugar) is not.

    lac operon

    Q11. Explain (in one or two lines) the function of the following:

    (a) Promoter

    Answer:

    Function of promoter

    The promoter gene refers to the site where the RNA polymerase enzyme binds and transcription of mRNA starts.

    (b) tRNA

    Answer:

    Function of tRNA

    tRNA plays a major role in the process of protein synthesis. It reads the genetic code present on mRNA.

    (c) Exon

    Answer:

    Exons are the coding sequences on DNA that are transcribed into proteins.

    Q12. Why is the Human Genome Project called a mega project?

    Answer:

    The Human Genome Project is called a mega project because its specific goal is to sequence every base pair in the human genome. HGP took approximately 13 years to accomplish in 2006. The main aim of HGP was to develop new technology and generate new information in genomic studies. Because of this project, many new areas have opened in genetics. Thus, HGP is a mega project.

    Q13. What is DNA fingerprinting? Mention its application.

    Answer:

    DNA fingerprinting is a very easy and quick way to compare the DNA sequence of any two individuals. It includes identifying differences in some specific regions in DNA sequences called repetitive DNA sequences. In these regions, a small stretch of DNA is repeated many times, and it is specific to every individual. The technique of fingerprinting was initially developed by Alec Jeffrey.

    Applications of DNA Fingerprinting

    1. It is used in forensic science to identify individuals.

    2. It can be used to establish paternity or maternity-related disputes.

    3. DNA fingerprinting is used to establish evolutionary relationships between organisms.

    Q14. Briefly describe the following:

    (a) Transcription

    Answer:

    It is the process of replication of genetic information from a strand of DNA into mRNA. Transcription involves only one strand that is replicated into RNA. During replication, the adenine position is occupied by uracil. DNA transcription contains a transcription unit. The transcription unit has a promoter, the structural gene, and a terminator. The 3'→5' polarity strands serve as a template and are referred to as template strands, whereas the other strand is referred to as the coding strand. A schematic organisation of a transcription unit is represented below.

    transcription

    The promoter is located at the 5’ end, and it binds the enzyme RNA polymerase to start transcription. The sigma factor also helps in the initiation of the process of transcription. The terminator is located at the 3’ end of the coding strand and usually defines the end of transcription, where the rho factor will bind to terminate transcription.

    14.(b) Polymorphism

    Answer:

    Polymorphism- It refers to a special kind of genetic variation in which nucleotide sequence can exist at a particular site in a DNA molecule. This heritable mutation is caused by a mutation in either the somatic cell or the germ cell. It ultimately results in the accumulation of various mutations at one site. Polymorphism brings revolution in the process of finding a location on the chromosome for disease-associated sequences and tracing human history.

    14.(c) Translation

    Answer:

    Translation- The process of amino acid polymerisation to form a polypeptide chain is called translation. The sequence and order of the amino acids within a polypeptide chain rely on the sequence of mRNA. Translation consists of three steps, i.e., initiation, elongation, and termination. In the process of initiation, the ribosome attaches at the beginning codon, which is AUG. Ribosomes then continue to travel codon by codon along the mRNA to add onto the polypeptide chain. Ultimately, the release factors recognise the stop codon, which results in the end of translation and release of the polypeptide from the ribosome.

    translation process

    14. (d) Bioinformatic

    Bioinformatics- It is a recent and very effective area in the field of biology. In this discipline, the knowledge from the DNA sequences is used to solve various doubts regarding organisms that can not be studied in real time. Therefore, we can say that we utilise the biological information stored in the DNA of an organism.

    Also, check NCERT Books and NCERT Syllabus here:

    Approach to Solve the Molecular Basis of Inheritance Class 12 Question Answer

    To solve the questions effectively, one needs to understand the concepts first. The chapter has certain key terms, such as replication, transcription, translation, and operon. Students can make notes from the Class 12 Biology Chapter 5 Molecular Basis of Inheritance question answer, which helps in highlighting all important points and definitions. Students should practice the questions given in the Class 12 Biology Chapter 5 Molecular Basis of Inheritance Solutions to have an overall idea of the chapter and build confidence before the exams. Easy explanations for all the questions present in the NCERT books help students build a solid foundation in molecular genetics.

    NCERT Solutions for Class 12- Subject-wise

    Important Questions of NCERT Class 12 Biology Chapter 5

    Given below are some of the solved practice questions from this chapter. To understand the complex processes, such as transcription, translation, the lac operon, etc., students need to refer to the NCERT Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance.

    Question 1. In a DNA strand, the nucleotides are linked together by:

    Options:

    1. Glycosidic bonds

    2. Phosphodiester bonds

    3. Peptide bonds

    4. Hydrogen bonds

    Answer:

    In a DNA molecule, nucleotides are joined through phosphodiester bonds, which are crucial covalent connections between the 3'-hydroxyl group of sugar in one nucleotide and the 5'-phosphate group of the sugar in the next.

    This bonding pattern generates the DNA's sugar-phosphate backbone, ensuring a clear directionality from 5' to 3' ends. The significance of these bonds lies in their contribution to DNA's structural integrity and facilitation of nucleotide chain elongation.

    Hence, the correct answer is option 2) Phosphodiester bonds.

    Question 2. In E.coli, the lac operon gets switched on when:

    Options:

    1. Lactose is present, and it binds to the repressor

    2. The repressor binds to the operator

    3. RNA polymerase binds to the operator

    4. Lactose is present, and it binds to RNA polymerase

    Answer:

    Lactose metabolism involves the regulation of gene expression that is determined by the presence or absence of the lac operon in bacteria. Here is how it works:
    Lactose Binding: Once lactose exists in the surrounding environment, it will bind to the attached repressor protein along the operator region of the lac operon.

    Hence, the correct answer is option 1) Lactose is present, and it binds to the repressor.

    Question 3. To initiate translation, the mRNA first binds to:

    Options:

    1. The smaller ribosomal subunit

    2. The larger ribosomal subunit

    3. The whole ribosome

    4. No such specificity exists.

    Answer:

    To initiate translation, the mRNA first attaches to the smaller ribosomal subunit at a specific binding site. Then, the initiator tRNA binds to the start codon (AUG) on mRNA.

    Hence, the correct answer is option 1) The smaller ribosomal subunit.

    NCERT Exemplar Class 12 Solutions

    CBSE Class 12th Syllabus: Subjects & Chapters
    Select your preferred subject to view the chapters

    What Students Learn from the Molecular Basis of Inheritance: NCERT Solutions

    Students gain a deeper understanding of how the genetic material carries information and directs the functioning of living organisms.

    1. Learners explore the detailed structure and features of DNA, including the double-helix model.

    2. They understand the mechanism of replication through the Molecular Basis of Inheritance NCERT Solutions.

    3. They also study the process of transcription, translation, and regulation of gene expression in both prokaryotes and eukaryotes, which are important for both school exams and the NEET exam.

    4. Gain knowledge about important experiments like Griffith’s, Hershey-Chase, and Meselson-Stahl through the Molecular Basis of Inheritance Class 12 question answer for conceptual clarity.

    5. Explore the concept of the Human Genome Project and its significance in understanding genetic material.

    What Extra Should Students Study Beyond the NCERT for NEET?

    Studying concepts beyond the NCERT will help in gaining conceptual clarity. It also allows students to answer application-based questions easily, especially for competitive exams or school exams. Some extra concepts that can be studied for NEET, along with the NCERT Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance, are given below.

    Frequently Asked Questions (FAQs)

    Q: What are the key topics covered in NCERT Class 12 Biology Chapter 5?
    A:

    The important topics which are covered in the NCERT Solutions for Class 12 Biology Chapter 5 Molecular basis of inheritance are given below:

    • DNA structure and replication

    • RNA and its types

    • Transcription, translation, and the genetic code

    • Regulation of gene expression

    • Human Genome Project and DNA fingerprinting

    Q: How does DNA replication occur as per NCERT Class 12 Biology?
    A:

    DNA replication is semi-conservative, where each strand serves as a template. Enzymes like DNA helicase, DNA polymerase, and ligase assist in unwinding, copying, and sealing the DNA strands. 

    Q: What are the differences between transcription and translation in Chapter 5?
    A:
    • Transcription: Formation of mRNA from DNA inside the nucleus.

    • Translation: Conversion of mRNA sequence into a protein at ribosomes.

    Q: How does gene expression work in prokaryotes and eukaryotes?
    A:
    • Prokaryotes: Operon model (e.g., lac operon) regulates gene expression.

    • Eukaryotes: Gene expression is regulated at multiple levels (transcription, RNA processing, translation).

    Q: What are the key differences between DNA and RNA mentioned in NCERT?
    A:
    • DNA: Double-stranded, contains deoxyribose sugar, has thymine (T).

    • RNA: Single-stranded, contains ribose sugar, has uracil (U) instead of thymine.

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