Chapter 5 Molecular Basis of Inheritance

NCERT Solutions Class 12 Biology Chapter 5 Molecular Basis of Inheritance

In Class 12 Biology Chapter 5, Molecular Basis of Inheritance, students learn about the molecular mechanism of storage, replication and expression of genetic information. The chapter also explains the concepts related to the structure of DNA and RNA, the processes of replication, transcription and translation etc., and builds a strong foundation among students for modern genetics and biotechnology.

ALLEN provides perfectly prepared NCERT Solutions which explain complex molecular mechanisms in simple and easy to follow steps. Our expert faculty is dedicated to helping you visualise the dense theoretical processes easily, with high quality diagrams and logical explanations. With ALLEN’s guided solutions, you will be able to gain the conceptual clarity and academic rigour required to ace your CBSE finals and competitive entrance exams.

1.0Key Concepts of Class 12 Biology Chapter 5 Molecular Basis of Inheritance

At the molecular level, heredity is explained through DNA, RNA, genes, and the processes that control the transfer and expression of genetic information. The major concepts in this chapter include:

Structure of DNA: Learn about the double helix structure of DNA, nucleotides, nitrogenous bases, and the arrangement of the DNA molecule.
Types of RNA: Understand the functions of mRNA, tRNA, and rRNA and their roles in protein synthesis.
DNA Replication: Study how DNA makes an identical copy of itself and understand the main steps involved in the replication process.
Transcription and Translation: Learn how genetic information is transferred from DNA to RNA through transcription and how proteins are produced from amino acids during translation.
Genetic Code: Understand how the sequence of nucleotides in DNA and RNA determines the sequence of amino acids in a protein.
Gene Expression and Regulation: Learn how gene activity is controlled in cells and how this regulation influences the production of proteins.

2.0NCERT Class 12 Biology Chapter 5 Molecular Basis of Inheritance : Detailed Solutions

Group the following as nitrogenous bases and nucleosides : Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine. Ans. Nitrogenous bases-Adenine, Thymine, Uracil and Cytosine Nucleosides-Cytidine and Guanosine.
If a double stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA. Ans. Cytosine 20%, therefore Guanine = 20% According to Chargaff's rule,

$A + T = 100 - (G + C)$

$A + T = 100 - 40$ . Since both adenine and thymine are in equal amounts

$Thymine = Adenine = \frac{60}{2} = 30%$

If the sequence of one strand of DNA is written as follows: [IMP.] 5'-ATGCATGCATGCATGCATGCATGCATGC-3' Write down the sequence of complementary strand in 5'→3' direction.

Ans. In 3' → 5' direction,

3-TACGTACGTACGTACGTACGTACGTACG-5' In 5' → 3' direction, 5'-GCATGCATGGATGCATGGATGCATGCAT-3

If the sequence of the coding strand in a transcription unit is written as follows: [IMP.] 5'-ATGCATGCATGCATGCATGCATGCATG-3' Write down the sequence of mRNA.

Ans. 5'-AUGCAUGCAUGCAUGCAUGCAUGCAUG-3'

Which property of DNA double helix led Watson and Crick to hypothesis semiconservative mode of DNA replication? Explain.

Ans. Watson and crick observed that the two DNA strands are antiparallel, and have opposite polarity. This means that 5' phosphate of one strand faces that 3' hydroxyl group of other strand and that the 5' phosphate group of two strands are present in opposite position. The antiparallel arrangement of two helices allows hydrogen bonding between amino and carbonyl group of complementary base

pairs. This led them to the hypothesis of the semiconservative mode of DNA replication where in two strands of DNA first separate from each other followed by copying of each template strands to form DNA molecules each carrying one parental strand and newly synthesized strands.

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

Ans. DNA template

(i) DNA polymerase for DNA replication. (ii) RNA polymerase for RNA synthesis or transcription. RNA template (i) RNA-dependent RNA polymerase for synthesis of RNA in some RNA viruses. (ii) Reverse transcriptase to synthesise cDNA (complementary DNA) over RNA template.

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

Ans. Hershey and Chase (1952) conducted experiments in bacteriophage to prove that DNA is the genetic material. Procedure: Some bacteriophage virus were grown on a medium that contained radioactive phosphorus $(^{32} P)$ and some in another medium with radioactive sulphur $(^{35} S)$ .

(a) Viruses grown in the presence of radioactive phosphorus ( $^{32} P$ ) contained radioactive DNA. (b) Similar viruses grown in presence of radioactive sulphur $(^{35} S)$ contained radioactive problem. (c) Both the radioactive virus types were allowed to infect E.coli separately. (d) Soon after infection, the bacterial cells were gently agitated in blender to remove viral coats from the bacteria. (e) The culture was also centrifuged to separate the viral particle from the bacteria cell.

Observation and Conclusions:-

(a) Only radioactive $(^{32} P)$ was found to be associated with bacterial cell, whereas radioactive ( $^{35} S$ ) was only found in surrounding medium and not in the bacterial cell. (b) This indicates that only DNA and not the protein coat entered the bacterial cell.

(c) This proves that DNA is the genetic material which is passed from virus to bacteria and not protein.

Differentiate between the following : [IMP.] (a) Repetitive DNA and satellite DNA (b) mRNA and tRNA (c) Template strand and coding strand Ans. (a) Repetitive DNA and satellite DNA :-

Repetitive DNA	Satellite DNA
DNA in which certain base sequence are repeated many times are called repetitive DNA.	DNA in which portion of the gene is tandemly repeated is called satellite DNA.
Repetitive DNA sequence are transcribed	Satellite DNA sequences are not transcribed

(b) mRNA and tRNA:-

mRNA	tRNA
It is linear.	It is clover-leaf shaped.
It carries coded information.	It carries information for association with an amino acid and an anticodon for its incorporation in a polypeptide.
mRNA undergoes additional processing, capping, tailing and splicing.	It does not require any processing.
Nitrogen bases are unmodified.	Nitrogen bases may be modified.

(c)Template strand and coding strand :-

Template strand	Coding strand
It is the strand of DNA which taken part in transcription.	It is the strand that dose not take part in transcription.
The polarity is $3^{'} \to 5^{'}$	The polarity is $5^{'} \to 3^{'}$ .
Nucleotide sequence is complementary.	The nucleotide sequence is same as the one present in mRNA except for presence of thymine instead of uracil.

List two essential roles of ribosome during translation. Ans. Two essential roles of ribosome during translation are: (i) One of the rRNA (23S in prokaryotes) acts as a peptidyl transferase ribozyme for formation of peptide bonds (ii) Ribosome provides sites for attachment of mRNA and charged (RNAs for polypeptide synthesis).
In the medium where E.coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium? Ans. It is because the repressor protein binds to the operator region of the operon and prevent RNA polymerase from transcribing the operon.

Explain (in one or two lines) the function of the following : (a) Promoter (b) tRNA (c) Exons

Ans.

(a) Promoter : It is the segment of DNA which lies adjacent to the operator and functions as the binding site for RNA polymerase to carry transcription if allowed by operator. (b) tRNA : It acts as an adaptor molecule that picks up a particular amino acid from cellular poll and takes the same over to site A of mRNA for incorporation into polypeptide chain. (c) Exons : These are the coding segments present in primary transcript which after splicing joined to form functional mRNA.

Why is the Human Genome Project called a mega project? Ans. Human Genome Project is called a mega project because of following reasons:

Sequencing of more than $3 \times 1 0^{9} bp$ .
Identification of all the approximately 20,000 - 25,000 genes in human DNA.
High expenditure of more than 9 billion US dollars.
Identification of all the alleles of genes and their functions.
Storage of data for sequencing would require space equal to 3300 books of 1000 pages each if each page will consist of 1000 letters.

What is DNA fingerprinting? Mention its application. Ans. DNA fingerprinting is the technique to determine the relationship between by studying the similarity and dissimilarity of VNTR (variable number of tandem repeats). Its applications are:

It is used as a tool in forensic tests to identify criminals.
To settle paternity disputes.
To identify racial groups to study biological evolution.

Briefly describe the following: (a) Transcription (b) Polymorphism (c) Translation (d) Bioinformatics

Ans.

(a) Transcription: It is the formation of RNA over the template of DNA. It forms single-stranded RNA which has a coded information similar to the sense or coding strand of DNA with the exception that thymine is replaced by uracil. (b) Polymorphism: Genetic polymorphism means occurrence of genetic material in more than one form. It is of two major types, i.e allelic polymorphism and SNP. Allelic polymorphism: Allelic polymorphism occurs due to multiple alleles of a gene Allele possess different mutations which alter the structure and function of a protein formed by as a result, change phenotype may occur. SNPs or single nucleotide polymorphism: Over 1.4 million single-base DNA differences have been observed in human beings. According to SNP every human being is unique. SNP is very useful for locating alleles, identifying disease -associated sequences and tracing human history. (c) Translation: It is the process during which the genetic information which is stored in the sequence of nucleotides in a mRNA molecules is converted following direction of the genetic code into the sequence of amino acids in the polypeptide. It takes place in cytoplasm in both eukaryotes and prokaryotes. (d) Bioinformatics: The science which deals with handling, storing of huge information of genomics as databases, analysing, modeling and providing various aspects of biological information especially the molecules connected with genomics and proteomics is called bioinformatics.

3.0NCERT Solutions for Class 12 Biology | Other Chapter-wise Links

Access chapter-wise NCERT Solutions for Class 12 Biology with detailed explanations for all other chapters through the links below.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sexual Reproduction in Flowering Plants
Chapter 2	Human Reproduction
Chapter 3	Reproductive Health
Chapter 4	Principles of Inheritance and Variation
Chapter 6	Evolution
Chapter 7	Human Health and Disease
Chapter 8	Microbes in Human Welfare
Chapter 9	Biotechnology: Principles and Processes
Chapter 10	Biotechnology and Its Applications
Chapter 11	Organisms and Populations
Chapter 12	Ecosystem
Chapter 13	Biodiversity and Conservation

4.0Key Features and Benefits of Class 12 Biology Chapter 5 Molecular Basis of Inheritance

Clear Explanation of Molecular Processes: Understand DNA replication, transcription, translation, and other molecular processes through simple, step-by-step explanations.
Better Understanding of Genetic Code: Strengthen your knowledge of the genetic code, gene expression, and the flow of genetic information from DNA to RNA and proteins.
NCERT-Based Practice: Work through NCERT exercise questions to understand important concepts and improve your ability to answer chapter-based questions.
Effective Revision: Revise important terms, molecular processes, and biological sequences to improve understanding and recall.
Useful for NEET Preparation: Build a strong foundation in molecular genetics and related concepts that are important for NEET and other biology-based competitive exams.

NCERT Solutions Class 12 Biology Chapter 5 Molecular Basis of Inheritance

In Class 12 Biology Chapter 5, Molecular Basis of Inheritance, students learn about the molecular mechanism of storage, replication and expression of genetic information. The chapter also explains the concepts related to the structure of DNA and RNA, the processes of replication, transcription and translation etc., and builds a strong foundation among students for modern genetics and biotechnology.

ALLEN provides perfectly prepared NCERT Solutions which explain complex molecular mechanisms in simple and easy to follow steps. Our expert faculty is dedicated to helping you visualise the dense theoretical processes easily, with high quality diagrams and logical explanations. With ALLEN’s guided solutions, you will be able to gain the conceptual clarity and academic rigour required to ace your CBSE finals and competitive entrance exams.

1.0Key Concepts of Class 12 Biology Chapter 5 Molecular Basis of Inheritance

At the molecular level, heredity is explained through DNA, RNA, genes, and the processes that control the transfer and expression of genetic information. The major concepts in this chapter include:

Structure of DNA: Learn about the double helix structure of DNA, nucleotides, nitrogenous bases, and the arrangement of the DNA molecule.
Types of RNA: Understand the functions of mRNA, tRNA, and rRNA and their roles in protein synthesis.
DNA Replication: Study how DNA makes an identical copy of itself and understand the main steps involved in the replication process.
Transcription and Translation: Learn how genetic information is transferred from DNA to RNA through transcription and how proteins are produced from amino acids during translation.
Genetic Code: Understand how the sequence of nucleotides in DNA and RNA determines the sequence of amino acids in a protein.
Gene Expression and Regulation: Learn how gene activity is controlled in cells and how this regulation influences the production of proteins.

2.0NCERT Class 12 Biology Chapter 5 Molecular Basis of Inheritance : Detailed Solutions

Group the following as nitrogenous bases and nucleosides : Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine. Ans. Nitrogenous bases-Adenine, Thymine, Uracil and Cytosine Nucleosides-Cytidine and Guanosine.
If a double stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA. Ans. Cytosine 20%, therefore Guanine = 20% According to Chargaff's rule,

$A + T = 100 - (G + C)$

$A + T = 100 - 40$ . Since both adenine and thymine are in equal amounts

$Thymine = Adenine = \frac{60}{2} = 30%$

If the sequence of one strand of DNA is written as follows: [IMP.] 5'-ATGCATGCATGCATGCATGCATGCATGC-3' Write down the sequence of complementary strand in 5'→3' direction.

Ans. In 3' → 5' direction,

3-TACGTACGTACGTACGTACGTACGTACG-5' In 5' → 3' direction, 5'-GCATGCATGGATGCATGGATGCATGCAT-3

If the sequence of the coding strand in a transcription unit is written as follows: [IMP.] 5'-ATGCATGCATGCATGCATGCATGCATG-3' Write down the sequence of mRNA.

Ans. 5'-AUGCAUGCAUGCAUGCAUGCAUGCAUG-3'

Which property of DNA double helix led Watson and Crick to hypothesis semiconservative mode of DNA replication? Explain.

Ans. Watson and crick observed that the two DNA strands are antiparallel, and have opposite polarity. This means that 5' phosphate of one strand faces that 3' hydroxyl group of other strand and that the 5' phosphate group of two strands are present in opposite position. The antiparallel arrangement of two helices allows hydrogen bonding between amino and carbonyl group of complementary base

pairs. This led them to the hypothesis of the semiconservative mode of DNA replication where in two strands of DNA first separate from each other followed by copying of each template strands to form DNA molecules each carrying one parental strand and newly synthesized strands.

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

Ans. DNA template

(i) DNA polymerase for DNA replication. (ii) RNA polymerase for RNA synthesis or transcription. RNA template (i) RNA-dependent RNA polymerase for synthesis of RNA in some RNA viruses. (ii) Reverse transcriptase to synthesise cDNA (complementary DNA) over RNA template.

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

Ans. Hershey and Chase (1952) conducted experiments in bacteriophage to prove that DNA is the genetic material. Procedure: Some bacteriophage virus were grown on a medium that contained radioactive phosphorus $(^{32} P)$ and some in another medium with radioactive sulphur $(^{35} S)$ .

(a) Viruses grown in the presence of radioactive phosphorus ( $^{32} P$ ) contained radioactive DNA. (b) Similar viruses grown in presence of radioactive sulphur $(^{35} S)$ contained radioactive problem. (c) Both the radioactive virus types were allowed to infect E.coli separately. (d) Soon after infection, the bacterial cells were gently agitated in blender to remove viral coats from the bacteria. (e) The culture was also centrifuged to separate the viral particle from the bacteria cell.

Observation and Conclusions:-

(a) Only radioactive $(^{32} P)$ was found to be associated with bacterial cell, whereas radioactive ( $^{35} S$ ) was only found in surrounding medium and not in the bacterial cell. (b) This indicates that only DNA and not the protein coat entered the bacterial cell.

(c) This proves that DNA is the genetic material which is passed from virus to bacteria and not protein.

Differentiate between the following : [IMP.] (a) Repetitive DNA and satellite DNA (b) mRNA and tRNA (c) Template strand and coding strand Ans. (a) Repetitive DNA and satellite DNA :-

Repetitive DNA	Satellite DNA
DNA in which certain base sequence are repeated many times are called repetitive DNA.	DNA in which portion of the gene is tandemly repeated is called satellite DNA.
Repetitive DNA sequence are transcribed	Satellite DNA sequences are not transcribed

(b) mRNA and tRNA:-

mRNA	tRNA
It is linear.	It is clover-leaf shaped.
It carries coded information.	It carries information for association with an amino acid and an anticodon for its incorporation in a polypeptide.
mRNA undergoes additional processing, capping, tailing and splicing.	It does not require any processing.
Nitrogen bases are unmodified.	Nitrogen bases may be modified.

(c)Template strand and coding strand :-

Template strand	Coding strand
It is the strand of DNA which taken part in transcription.	It is the strand that dose not take part in transcription.
The polarity is $3^{'} \to 5^{'}$	The polarity is $5^{'} \to 3^{'}$ .
Nucleotide sequence is complementary.	The nucleotide sequence is same as the one present in mRNA except for presence of thymine instead of uracil.

List two essential roles of ribosome during translation. Ans. Two essential roles of ribosome during translation are: (i) One of the rRNA (23S in prokaryotes) acts as a peptidyl transferase ribozyme for formation of peptide bonds (ii) Ribosome provides sites for attachment of mRNA and charged (RNAs for polypeptide synthesis).
In the medium where E.coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium? Ans. It is because the repressor protein binds to the operator region of the operon and prevent RNA polymerase from transcribing the operon.

Explain (in one or two lines) the function of the following : (a) Promoter (b) tRNA (c) Exons

Ans.

(a) Promoter : It is the segment of DNA which lies adjacent to the operator and functions as the binding site for RNA polymerase to carry transcription if allowed by operator. (b) tRNA : It acts as an adaptor molecule that picks up a particular amino acid from cellular poll and takes the same over to site A of mRNA for incorporation into polypeptide chain. (c) Exons : These are the coding segments present in primary transcript which after splicing joined to form functional mRNA.

Why is the Human Genome Project called a mega project? Ans. Human Genome Project is called a mega project because of following reasons:

Sequencing of more than $3 \times 1 0^{9} bp$ .
Identification of all the approximately 20,000 - 25,000 genes in human DNA.
High expenditure of more than 9 billion US dollars.
Identification of all the alleles of genes and their functions.
Storage of data for sequencing would require space equal to 3300 books of 1000 pages each if each page will consist of 1000 letters.

What is DNA fingerprinting? Mention its application. Ans. DNA fingerprinting is the technique to determine the relationship between by studying the similarity and dissimilarity of VNTR (variable number of tandem repeats). Its applications are:

It is used as a tool in forensic tests to identify criminals.
To settle paternity disputes.
To identify racial groups to study biological evolution.

Briefly describe the following: (a) Transcription (b) Polymorphism (c) Translation (d) Bioinformatics

Ans.

(a) Transcription: It is the formation of RNA over the template of DNA. It forms single-stranded RNA which has a coded information similar to the sense or coding strand of DNA with the exception that thymine is replaced by uracil. (b) Polymorphism: Genetic polymorphism means occurrence of genetic material in more than one form. It is of two major types, i.e allelic polymorphism and SNP. Allelic polymorphism: Allelic polymorphism occurs due to multiple alleles of a gene Allele possess different mutations which alter the structure and function of a protein formed by as a result, change phenotype may occur. SNPs or single nucleotide polymorphism: Over 1.4 million single-base DNA differences have been observed in human beings. According to SNP every human being is unique. SNP is very useful for locating alleles, identifying disease -associated sequences and tracing human history. (c) Translation: It is the process during which the genetic information which is stored in the sequence of nucleotides in a mRNA molecules is converted following direction of the genetic code into the sequence of amino acids in the polypeptide. It takes place in cytoplasm in both eukaryotes and prokaryotes. (d) Bioinformatics: The science which deals with handling, storing of huge information of genomics as databases, analysing, modeling and providing various aspects of biological information especially the molecules connected with genomics and proteomics is called bioinformatics.

3.0NCERT Solutions for Class 12 Biology | Other Chapter-wise Links

Access chapter-wise NCERT Solutions for Class 12 Biology with detailed explanations for all other chapters through the links below.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sexual Reproduction in Flowering Plants
Chapter 2	Human Reproduction
Chapter 3	Reproductive Health
Chapter 4	Principles of Inheritance and Variation
Chapter 6	Evolution
Chapter 7	Human Health and Disease
Chapter 8	Microbes in Human Welfare
Chapter 9	Biotechnology: Principles and Processes
Chapter 10	Biotechnology and Its Applications
Chapter 11	Organisms and Populations
Chapter 12	Ecosystem
Chapter 13	Biodiversity and Conservation

4.0Key Features and Benefits of Class 12 Biology Chapter 5 Molecular Basis of Inheritance

Clear Explanation of Molecular Processes: Understand DNA replication, transcription, translation, and other molecular processes through simple, step-by-step explanations.
Better Understanding of Genetic Code: Strengthen your knowledge of the genetic code, gene expression, and the flow of genetic information from DNA to RNA and proteins.
NCERT-Based Practice: Work through NCERT exercise questions to understand important concepts and improve your ability to answer chapter-based questions.
Effective Revision: Revise important terms, molecular processes, and biological sequences to improve understanding and recall.
Useful for NEET Preparation: Build a strong foundation in molecular genetics and related concepts that are important for NEET and other biology-based competitive exams.

Frequently Asked Questions

Are these NCERT Solutions for Class 12 Biology based on the latest NCERT textbook?

Do NCERT Solutions help improve answer presentation in Biology exams?

Are NCERT Solutions for Class 12 Biology helpful in preparing for competitive exams?

Is the PDF version of NCERT Solutions helpful for last-minute revision?

What is the difference between repetitive DNA and satellite DNA in NCERT Solutions?

What are the applications of DNA fingerprinting explained in NCERT Solutions?

How does the lac operon function according to NCERT Solutions for Class 12 Biology?

What are the stages of translation explained in NCERT Solutions for Molecular Basis of Inheritance?

Join ALLEN!

NCERT Solutions Class 12 Biology Chapter 5 Molecular Basis of Inheritance

1.0Key Concepts of Class 12 Biology Chapter 5 Molecular Basis of Inheritance

2.0NCERT Class 12 Biology Chapter 5 Molecular Basis of Inheritance : Detailed Solutions

3.0NCERT Solutions for Class 12 Biology | Other Chapter-wise Links

4.0Key Features and Benefits of Class 12 Biology Chapter 5 Molecular Basis of Inheritance

Table of Contents

Frequently Asked Questions

Are these NCERT Solutions for Class 12 Biology based on the latest NCERT textbook?

Do NCERT Solutions help improve answer presentation in Biology exams?

Are NCERT Solutions for Class 12 Biology helpful in preparing for competitive exams?

Is the PDF version of NCERT Solutions helpful for last-minute revision?

What is the difference between repetitive DNA and satellite DNA in NCERT Solutions?

What are the applications of DNA fingerprinting explained in NCERT Solutions?

How does the lac operon function according to NCERT Solutions for Class 12 Biology?

What are the stages of translation explained in NCERT Solutions for Molecular Basis of Inheritance?

Join ALLEN!

NCERT Solutions Class 12 Biology Chapter 5 Molecular Basis of Inheritance

1.0Key Concepts of Class 12 Biology Chapter 5 Molecular Basis of Inheritance

2.0NCERT Class 12 Biology Chapter 5 Molecular Basis of Inheritance : Detailed Solutions

3.0NCERT Solutions for Class 12 Biology | Other Chapter-wise Links

4.0Key Features and Benefits of Class 12 Biology Chapter 5 Molecular Basis of Inheritance

Table of Contents