Chapter 6: Molecular Basis of Inheritance – Long Answer Type Questions
CBSE Class 12 Biology Long Answer Questions (NCERT): Molecular Basis of Inheritance
Course & Examination Details
- Course: CBSE Class 12 Biology
- Unit: Unit II – Genetics and Evolution
- Chapter: Chapter 6 – Molecular Basis of Inheritance
- Prescribed Textbook: NCERT Biology Class XII
- Examination: CBSE Class 12 Board Examination
- Question Type: Long Answer Type
- Answer Length: 120–150 words each
Section A: DNA as Genetic Material
Q1. Explain the criteria required for a molecule to function as genetic material.
Answer:
A molecule functioning as genetic material must satisfy four essential criteria. First, it must be stable to ensure faithful transmission of information across generations. Second, it must be capable of self-replication, allowing genetic continuity during cell division. Third, it should have the ability to undergo mutation, generating variation necessary for evolution. Finally, it must be able to express information in the form of phenotype through protein synthesis. DNA fulfills all these criteria efficiently due to its stable double-stranded structure, complementary base pairing, and precise replication mechanism. RNA, although capable of mutation and expression, is less stable and therefore serves as genetic material only in certain viruses.
Q2. Describe the experimental evidence that established DNA as the genetic material.
Answer:
Several experiments confirmed DNA as the genetic material. Griffith’s transformation experiment suggested the existence of a “transforming principle.” Avery, MacLeod, and McCarty later proved that DNA was responsible for transformation by selectively destroying proteins, RNA, and DNA. Only destruction of DNA prevented transformation. This conclusion was further supported by the Hershey–Chase experiment using bacteriophages labelled with radioactive isotopes. Radioactive DNA entered bacterial cells and was inherited by progeny, whereas protein did not. These experiments conclusively established DNA as the hereditary material in living organisms.
Q3. Why is DNA considered superior to RNA as genetic material?
Answer:
DNA is considered superior to RNA due to its greater chemical and structural stability. DNA lacks a hydroxyl group at the 2′ carbon of sugar, making it less reactive. Its double-stranded structure provides an additional advantage, as complementary base pairing allows error correction and repair. RNA is single-stranded and more prone to degradation and mutation. DNA can store genetic information for long periods without significant alteration, ensuring accurate inheritance. These features make DNA the primary genetic material in most organisms, whereas RNA functions mainly in information transfer and protein synthesis.
Section B: Structure of DNA and RNA
Q4. Describe the double helical structure of DNA.
Answer:
DNA consists of two polynucleotide strands coiled around each other to form a right-handed double helix. Each strand has a sugar-phosphate backbone with nitrogenous bases projecting inward. The strands run antiparallel, one in the 5′→3′ direction and the other in the 3′→5′ direction. Complementary base pairing occurs between adenine and thymine through two hydrogen bonds, and between guanine and cytosine through three hydrogen bonds. This arrangement provides stability and specificity. The uniform diameter and regular helical turns allow efficient packing and replication of genetic material.
Q5. Compare the structure and function of DNA and RNA.
Answer:
DNA is a double-stranded molecule composed of deoxyribose sugar, phosphate groups, and nitrogenous bases adenine, guanine, cytosine, and thymine. RNA is single-stranded, contains ribose sugar, and has uracil instead of thymine. DNA stores genetic information and serves as a template for replication and transcription. RNA functions in gene expression, including mRNA for information transfer, tRNA for amino acid transport, and rRNA as a structural and catalytic component of ribosomes. Thus, DNA acts as the information reservoir, while RNA plays an active role in protein synthesis.
Q6. Explain Chargaff’s rule and its significance.
Answer:
Chargaff’s rule states that in DNA, the amount of adenine equals thymine and the amount of guanine equals cytosine. This rule highlights the complementary nature of base pairing. It provided crucial evidence supporting the double helical model of DNA. The rule ensures uniform diameter of the DNA molecule and explains how genetic information is replicated accurately. Chargaff’s observations were fundamental in understanding DNA structure and function, forming the basis for explaining replication, transcription, and mutation mechanisms.
Section C: Replication of DNA
Q7. Explain the semi-conservative nature of DNA replication.
Answer:
DNA replication is semi-conservative, meaning each daughter DNA molecule contains one parental strand and one newly synthesised strand. During replication, the double helix unwinds and the two strands separate. Each parental strand acts as a template for synthesis of a complementary strand. This mechanism ensures genetic continuity and accuracy. The semi-conservative nature was experimentally demonstrated by the Meselson–Stahl experiment using nitrogen isotopes. This process preserves original information while allowing precise duplication before cell division.
Q8. Describe the mechanism of DNA replication.
Answer:
DNA replication begins at a specific origin of replication. Helicase unwinds the double helix, forming replication forks. Primase synthesises short RNA primers. DNA polymerase extends the new strand in the 5′→3′ direction by adding complementary nucleotides. The leading strand is synthesised continuously, while the lagging strand is synthesised discontinuously as Okazaki fragments. DNA ligase joins these fragments to form a continuous strand. The process continues until replication is complete, ensuring accurate duplication of genetic material.
Q9. Explain the role of enzymes involved in DNA replication.
Answer:
Several enzymes coordinate DNA replication. Helicase unwinds the DNA strands. Primase synthesises RNA primers required for initiation. DNA polymerase adds nucleotides and proofreads the new strand. DNA ligase joins Okazaki fragments on the lagging strand. Topoisomerase relieves supercoiling ahead of the replication fork. Together, these enzymes ensure accurate, efficient, and rapid replication of DNA.
Section D: Transcription
Q10. Explain the process of transcription in prokaryotes.
Answer:
Transcription in prokaryotes involves synthesis of RNA from a DNA template. RNA polymerase binds to the promoter region, initiating transcription. The enzyme unwinds DNA locally and synthesises RNA complementary to the template strand in the 5′→3′ direction. Elongation continues until RNA polymerase reaches a terminator sequence, leading to release of the RNA transcript. The process is rapid and efficient, producing mRNA, tRNA, and rRNA required for protein synthesis.
Q11. Why is transcription described as asymmetric?
Answer:
Transcription is asymmetric because only one strand of DNA serves as the template for RNA synthesis. If both strands were transcribed, complementary RNA molecules would form double-stranded RNA, preventing translation. The coding strand remains untranscribed but has the same sequence as mRNA except for thymine replacing uracil. This asymmetry ensures accurate transfer of genetic information.
Section E: Genetic Code
Q12. Describe the salient features of the genetic code.
Answer:
The genetic code is a triplet code where three nucleotides form one codon. It is universal, with minor exceptions, and degenerate, meaning multiple codons code for the same amino acid. The code is non-overlapping and comma-less, ensuring continuous reading. AUG acts as the start codon, while UAA, UAG, and UGA are stop codons. These features ensure accurate and efficient translation of genetic information into proteins.
Q13. Explain the significance of degeneracy of genetic code.
Answer:
Degeneracy of the genetic code provides protection against mutations. Since multiple codons code for the same amino acid, a change in one nucleotide may not alter the amino acid sequence, resulting in a silent mutation. This reduces harmful effects of mutations and contributes to genetic stability while allowing variation and evolution.
Section F: Translation
Q14. Describe the process of translation.
Answer:
Translation is the synthesis of proteins from mRNA. Amino acids are first activated and attached to specific tRNAs. Ribosomes bind to mRNA at the start codon, initiating translation. During elongation, tRNA anticodons pair with mRNA codons, and peptide bonds form between amino acids. Translation ends when a stop codon is reached, releasing the completed polypeptide chain. This process converts genetic information into functional proteins.
Q15. Explain the role of tRNA and ribosomes in translation.
Answer:
tRNA acts as an adaptor molecule, carrying specific amino acids and matching its anticodon with mRNA codons. Ribosomes provide the platform for translation, ensuring proper alignment of mRNA and tRNA. rRNA within ribosomes catalyses peptide bond formation, making ribosomes essential for protein synthesis.
Section G: Regulation of Gene Expression
Q16. Explain the operon concept of gene regulation.
Answer:
An operon is a functional unit of gene regulation in prokaryotes. It consists of structural genes, a promoter, and an operator. The regulator gene produces a repressor protein that controls transcription. Operons allow coordinated expression of genes involved in a common pathway, ensuring efficiency and adaptability.
Q17. Describe the working of lac operon.
Answer:
The lac operon controls lactose metabolism in bacteria. In absence of lactose, the repressor binds to the operator, blocking transcription. When lactose is present, it binds to the repressor, inactivating it. This allows RNA polymerase to transcribe structural genes, enabling lactose utilisation. The lac operon is an inducible system.
Section H: Human Genome Project
Q18. Explain the objectives and significance of the Human Genome Project.
Answer:
The Human Genome Project aimed to identify and sequence all human genes, map the genome, and store data in databases. It revealed that humans have about 30,000 genes and that most DNA is non-coding. The project advanced medical research, improved disease diagnosis, and laid foundations for personalised medicine and gene therapy.
Q19. Mention major findings of the Human Genome Project.
Answer:
Major findings include identification of approximately 3.2 billion base pairs, discovery that only a small fraction codes for proteins, and confirmation that 99.9% of human DNA is identical. These insights enhanced understanding of genetic variation and disease mechanisms.
Section I: DNA Fingerprinting
Q20. Explain the principle and steps of DNA fingerprinting.
Answer:
DNA fingerprinting is based on polymorphism in repetitive DNA sequences such as VNTRs. DNA is isolated, digested with restriction enzymes, separated by gel electrophoresis, transferred to a membrane, and hybridised with probes. The resulting pattern is unique for each individual and used for identification.
Q21. Describe applications of DNA fingerprinting.
Answer:
DNA fingerprinting is used in forensic investigations, paternity testing, criminal identification, and wildlife conservation. It helps establish biological relationships and identify individuals with high accuracy.
Section J: Integrated Concepts
Q22. Explain the central dogma of molecular biology.
Answer:
The central dogma describes the flow of genetic information from DNA to RNA through transcription and from RNA to protein through translation. It explains how genetic information is expressed as phenotype and forms the basis of molecular biology.
Q23. Why is regulation of gene expression essential?
Answer:
Gene regulation ensures proteins are synthesised only when required, conserving energy and resources. It allows cells to respond to environmental changes and maintain proper development and function.
Q24. How do mutations influence molecular inheritance?
Answer:
Mutations alter DNA sequence, leading to changes in protein structure or function. They create genetic variation, which can be harmful, neutral, or beneficial, contributing to evolution and genetic diversity.
Q25. Why is the study of molecular basis of inheritance important?
Answer:
Understanding molecular inheritance explains how genetic information is stored, expressed, regulated, and transmitted. It forms the foundation of genetics, biotechnology, medicine, and evolutionary biology.