NCERT Solutions Class 12 Biology Chapter 5 Molecular Basis of Inheritance

In Class 12 Biology Chapter 5, Molecular Basis of Inheritance, students explore how genetic information is stored, copied, and expressed at the molecular level. The chapter also explains the concepts concerning 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 meticulously drafted NCERT Solutions that simplify the complex molecular mechanisms into clear, manageable steps. Our expert faculty focuses on making dense theoretical processes easy to visualize through high-quality diagrams and logical explanations. By practicing with ALLEN’s guided solutions, you gain the conceptual clarity and academic rigor needed to excel in your CBSE finals and competitive entrance exams.

1.0Download NCERT Solutions Class 12 Biology Chapter 5 Molecular Basis of Inheritance : Free PDF

This chapter explains how DNA carries genetic information and how it is used to make proteins. The NCERT Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance help students revise key molecular processes and practise textbook questions easily. Download the free PDF now to study anytime and prepare confidently for exams.

2.0Class 12 Biology Chapter 5: Key Concepts

The chapter focuses on understanding heredity at the molecular level. Some of the key lessons covered in this chapter are given below:

• Understand the structure of DNA, namely, the double helix model and components of nucleotides.
• Different types of RNA including mRNA, tRNA, and rRNA and their roles in the synthesis of protein.
• Understand the concept of DNA replication.
• The process of transcription and translation, namely how genetic information is copied from DNA to RNA and how proteins are formed from amino acids using the genetic code.
• Understanding how nucleotide sequences determine amino acid sequences.
• Learning about basic control of gene activity in cells.

3.0NCERT Class 12 Biology Chapter 5 Molecular Basis of Inheritance : Detailed Solutions

1. Group the following as nitrogenous bases and nucleosides: 

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

Ans: Nitrogenous Bases: Adenine, Thymine,Uracil and Cytosine

Nucleosides: Guanosine, Cytidine

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

Ans: According to Chargaff’s rule,

%C=%G and %A=%T

Therefor, C=20% implies G = 20%

Now, A+G+T+C = 100

⇒A+T = 60

Since A=T ⇒ A = 60%/2 = 30%

Therefore, A= 30%

3. If the sequence of one strand of DNA is written as follows: 

5' -ATGCATGCATGCATGCATGCATGCATGC-3' Write down the sequence of complementary strand in 5'→3' direction. 

Ans: Given: 5' -ATGCATGCATGCATGCATGCATGCATGC-3'

According to Chargaff’s rule, A pairs with T and G pairs with C

Complementary sequence in 3’-5’ direction:

3' – TACGTACGTACGTACGTACGTACGTACG – 5'

Now, convert to 5’-3’ direction

= 5' – GCATGCATGCATGCATGCATGCATGCAT – 3'

Therefore, the The complementary DNA strand in the 5' → 3' direction is:

5' – GCATGCATGCATGCATGCATGCATGCAT – 3'

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

5' -ATGCATGCATGCATGCATGCATGCATGC-3' Write down the sequence of mRNA.

Ans: Given:

The DNA’s coding strand sequence -  5' -ATGCATGCATGCATGCATGCATGCATGC-3'

Since the mRNA sequence is complementary to template strand and therefore identical to the coding strand, except that thymine (T) is replaced by uracil (U) in RNA, replace T with U in the given strand.

Therefore the final answer is: 5' – AUGCAUGCAUGCAUGCAUGCAUGCAUGC – 3'

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

Ans: The property of the DNA double helix that led Watson and Crick to hypothesize the semi-conservative mode of replication was the specific complementary base pairing between the two strands.

Explanation:

• Complementary Base Pairing: Watson and Crick observed that Adenine always pairs with Thymine, and Guanine always pairs with Cytosine.
• Template Mechanism: This specific pairing suggested that the two strands could separate and each act as a template for the synthesis of a new complementary strand.
• Faithful Copying: Because the sequence of bases in one strand dictates the sequence of bases in the newly synthesized strand, the two resulting double-stranded DNA molecules would be identical to the original parental molecule.
• Semi-Conservative Nature: In this model, each "daughter" DNA molecule would conserve one "parental" strand and possess one newly synthesized "daughter" strand.

6. 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. 

Ans:

1. DNA-dependent DNA Polymerase

• Template: DNA
• Product formed: DNA
• Function: This enzyme synthesises a new DNA strand using an existing DNA strand as a template.
• Role: It is mainly involved in DNA replication.

2. DNA-dependent RNA Polymerase

• Template: DNA
• Product formed: RNA
• Function: This enzyme synthesises RNA from a DNA template.
• Role: It plays a key role in transcription, forming mRNA, tRNA, and rRNA.

3. RNA-dependent DNA Polymerase (Reverse Transcriptase)

• Template: RNA
• Product formed: DNA
• Function: This enzyme synthesises DNA using RNA as a template.
• Role: It is commonly found in retroviruses and is involved in the formation of DNA from viral RNA.

7. 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 used bacteriophages (viruses that infect bacteria) to determine whether DNA or protein is the genetic material. They grew the phages in two separate media:

• The first medium contained radioactive sulphur (³⁵S), which gets incorporated into proteins because proteins contain sulphur (in amino acids like cysteine and methionine), but DNA does not.
• The other medium contained radioactive phosphorus (³²P), which gets incorporated into DNA because DNA contains phosphorus in its phosphate backbone, while proteins do not.
• The labelled phages were then allowed to infect E. coli. After infection, to separate the viral protein coats from the bacterial cells, the mixture was agitated in a blender and then centrifuged.
• On observation, the ³²P-labelled DNA was found inside the bacterial cells thatis in the pellet, showing that it was DNA that entered the bacteria.
• The ³⁵S-labelled protein remained in the supernatant, showing that protein did not enter the bacterial cells.

This experiment demonstrated that the genetic material that was transported from the virus to the bacteria was DNA, and not protein.

8. Differentiate between the followings: 

(a) Repetitive DNA and Satellite DNA 

(b) mRNA and tRNA 

(c) Template strand and Coding strand 

Ans: (a) Repetitive DNA and Satellite DNA 

(b) mRNA and tRNA 

(c) Template strand and Coding strand 

9. List two essential roles of ribosome during translation. 

Ans: The two essential roles of ribosome during translationare:

• Structural and Catalytic Site: The ribosome consists of structural RNAs and about 80 different proteins; it acts as the site where all molecules involved in protein synthesis (mRNA, tRNA, amino acids) come together.
• Catalysis of Peptide Bond: In bacteria, the 23S rRNA acts as an enzyme (ribozyme) for the formation of peptide bonds between amino acids.

10. 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: The lac operon is an inducible system, and it functions only when the inducer, i.e, lactose, is present. The lac operon will shut down due to the following reasons:

• Depletion of Inducer: As E. coli consumes the lactose added to the medium, the concentration of lactose decreases, causing the lac operon to shut down.
• Repressor Binding: Once all the lactose is metabolised, there is no inducer left to bind to the repressor protein. Consequently, the free repressor binds back to the operator region, preventing RNA polymerase from transcribing the structural genes and shutting down the operon.

11. Explain (in one or two lines) the function of the followings: 

(a) Promoter 

(b) tRNA 

(c) Exons 

Ans: (a) Promoter: It is a DNA sequence that provides a binding site for RNA polymerase and determines the template strand and the start site of transcription.

(b) tRNA: It acts as an adapter molecule that reads the genetic code on mRNA and brings the corresponding amino acids to the ribosome during protein synthesis.

(c) Exons: These are the coding sequences or expressed sequences in a split gene that appear in mature or processed RNA.

12. Why is the Human Genome project called a mega project? 

Ans: The Human Genome Project (HGP) was called a mega project because of its enormous scale, complexity and requirements:

• Scale: The human genome has approximately 3 \times 10^9 base pairs and identifying every sequence required massive coordination.
• Cost: The estimated cost for the project was $3 per base pair, bringing the total cost to around $9 Billion.
• Data Storage: To store the sequence’s data in printed form would require around 3,300 books, each with 1,000 pages and 1,000 letters per page.
• Interdisciplinary Effort: It required the development of high-speed computational devices for data storage, retrieval, and analysis, giving rise to Bioinformatics.

13. What is DNA fingerprinting? Mention its application. 

Ans: A laboratory technique used to identify and compare the DNA profiles of individuals based on variations in their genetic makeup at the molecular level is known as DNA fingerprinting.

Some applications of DNA fingerprinting include:

• Forensic science: Used to identify criminals by matching DNA from crime scenes with suspects.
• Paternity and maternity testing: Helps in establishing biological relationships during legal disputes involving the parentage of the child.
• Identification of individuals: Used in cases of missing persons and disaster victim identification.
• Medical research: Helps in studying genetic disorders and population genetics.

14. Briefly describe the following: 

(a) Transcription 

(b) Polymorphism 

(c) Translation 

(d) Bioinformatics

Ans: (a)Transcription:

• Transcription is the process by which genetic information from DNA is copied into RNA. 
• It occurs in the nucleus (in eukaryotes) and is carried out by the enzyme DNA-dependent RNA polymerase.
• During this process, one strand of DNA acts as a template strand, and complementary RNA nucleotides are assembled to form mRNA, which carries the genetic code to the ribosomes for protein synthesis.

(b) Polymorphism

• Polymorphism refers to the presence of two or more genetically determined variations (alleles or DNA sequences) at a particular locus in a population. 
• These variations occur with a frequency of more than 1% and contribute to genetic diversity among individuals. 
• DNA polymorphisms form the basis of techniques like DNA mapping and fingerprinting.

(c) Translation

• Translation is the process by which the genetic code present in mRNA is decoded to synthesise a polypeptide (protein) with ribosomes as catalyst.
• During translation, tRNA molecules bring specific amino acids to the ribosome, and peptide bonds are formed between them in the sequence specified by the codons on the mRNA.

(d) Bioinformatics

• Bioinformatics is an interdisciplinary field that involves the use of computers and information technology to store, analyse, and interpret biological data, especially large volumes of DNA and protein sequence information. 
• It played a crucial role in projects like the Human Genome Project, helping scientists manage and understand complex genetic data.

4.0Key Features of NCERT Solutions for Class 12 Biology Chapter 5

• The molecular steps are clearly explained in the solutions, helping students follow DNA and RNA processes in the right order.
• The practice questions at the end of the chapters can strengthen understanding of concepts including genetic code.
• Regular revision of the solutions improves recall of complex biological terms and molecular sequences.
• Concepts like genetics and biotechnology help in the preparation for competitive exams like NEET.