Chapter 6: Molecular Basis of Inheritance – Short Answer Type Questions

CBSE Class 12 Biology Short 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: Short Answer Type
Answer Length: 60–80 words each

Section A: DNA as Genetic Material

Q1. Explain the criteria for a molecule to act as genetic material.

Answer:
A genetic material must be stable, capable of self-replication, able to undergo mutations, and capable of expressing information as phenotype. Stability ensures inheritance across generations, replication ensures continuity, mutation allows variation, and expression links genotype to phenotype. DNA fulfills all these criteria most effectively, making it the primary genetic material in most organisms.

Q2. Why is DNA preferred over RNA as genetic material?

Answer:
DNA is chemically more stable due to the absence of a hydroxyl group in deoxyribose sugar and its double-stranded structure. Complementary base pairing allows repair mechanisms. RNA is less stable and prone to mutations. Hence, DNA is better suited for long-term storage and transmission of genetic information.

Q3. Describe the Hershey–Chase experiment.

Answer:
Hershey and Chase used bacteriophages labelled with radioactive phosphorus (DNA) and sulphur (protein). Only phosphorus entered bacterial cells and was passed to progeny, proving DNA is the genetic material. Proteins remained outside, confirming that DNA carries hereditary information.

Q4. In which organisms does RNA act as genetic material and why?

Answer:
RNA acts as genetic material in certain viruses like tobacco mosaic virus. These viruses lack DNA and use RNA for replication and expression. RNA’s ability to mutate rapidly helps viruses adapt quickly to host defenses.

Q5. What is meant by transforming principle?

Answer:
Transforming principle refers to a substance that transfers genetic information from one organism to another. It was identified as DNA through Avery–MacLeod–McCarty experiments, proving DNA carries genetic traits.

Section B: Structure of DNA and RNA

Q6. Describe the double helical structure of DNA.

Answer:
DNA consists of two antiparallel polynucleotide strands forming a right-handed double helix. Each strand has a sugar-phosphate backbone with nitrogenous bases projecting inward. Complementary base pairing occurs between adenine–thymine and guanine–cytosine. The structure was proposed by James Watson and Francis Crick.

Q7. What are nucleotides and nucleosides?

Answer:
A nucleoside consists of a nitrogenous base attached to a pentose sugar. When a phosphate group is added to a nucleoside, it becomes a nucleotide. Nucleotides are the basic building blocks of nucleic acids like DNA and RNA.

Q8. Explain Chargaff’s rule.

Answer:
Chargaff’s rule states that in DNA, the amount of adenine equals thymine and the amount of guanine equals cytosine. This equality supports complementary base pairing and was crucial in understanding DNA structure.

Q9. Differentiate between DNA and RNA.

Answer:
DNA is double-stranded, contains deoxyribose sugar, and has thymine as a base. RNA is single-stranded, contains ribose sugar, and has uracil instead of thymine. DNA stores genetic information, while RNA helps in its expression.

Q10. What is the significance of antiparallel strands in DNA?

Answer:
Antiparallel orientation allows complementary base pairing and proper functioning of replication and transcription enzymes. It ensures correct synthesis of new strands during DNA replication.

Section C: Replication of DNA

Q11. Explain the semi-conservative nature of DNA replication.

Answer:
In semi-conservative replication, each daughter DNA molecule contains one parental strand and one newly synthesized strand. This ensures genetic continuity and accuracy. It was experimentally proven by Meselson and Stahl.

Q12. Describe the role of DNA polymerase.

Answer:
DNA polymerase synthesises new DNA strands by adding nucleotides in the 5′ to 3′ direction. It also proofreads newly formed strands, ensuring accuracy during replication.

Q13. Why does DNA replication require a primer?

Answer:
DNA polymerase cannot initiate synthesis on its own. A short RNA primer provides a free 3′-OH group for elongation of the new DNA strand.

Q14. Explain the formation of Okazaki fragments.

Answer:
Okazaki fragments form on the lagging strand because DNA synthesis occurs discontinuously due to the opposite direction of strand orientation relative to replication fork movement.

Q15. What is the role of DNA ligase?

Answer:
DNA ligase joins Okazaki fragments by forming phosphodiester bonds, creating a continuous DNA strand during replication.

Section D: Transcription

Q16. Define transcription and its importance.

Answer:
Transcription is the synthesis of RNA from a DNA template. It is essential because DNA cannot directly guide protein synthesis. RNA acts as an intermediary.

Q17. Why is only one strand of DNA transcribed?

Answer:
Transcribing both strands would produce complementary RNAs that could form double-stranded RNA, interfering with translation. Hence, only the template strand is used.

Q18. Explain the role of promoter in transcription.

Answer:
The promoter is a DNA sequence where RNA polymerase binds to initiate transcription. It determines the start point and direction of RNA synthesis.

Q19. How is transcription terminated?

Answer:
Transcription terminates when RNA polymerase reaches a terminator sequence, causing release of the RNA transcript.

Q20. Name the types of RNA and their functions.

Answer:
mRNA carries genetic information, tRNA transports amino acids, and rRNA forms ribosomes and catalyses peptide bond formation.

Section E: Genetic Code

Q21. What is genetic code?

Answer:
Genetic code is the sequence of codons on mRNA that determines the order of amino acids in a protein.

Q22. Why is the genetic code described as degenerate?

Answer:
Multiple codons code for the same amino acid, providing protection against mutations.

Q23. What is the significance of start and stop codons?

Answer:
Start codon AUG initiates translation, while stop codons terminate protein synthesis, ensuring correct protein length.

Q24. Explain why genetic code is universal.

Answer:
The same codons specify the same amino acids in almost all organisms, indicating a common evolutionary origin.

Q25. What happens when a codon mutates?

Answer:
Mutation may change the amino acid, remain silent, or introduce a stop codon, affecting protein structure and function.

Section F: Translation

Q26. Describe the process of translation briefly.

Answer:
Translation converts mRNA sequence into a polypeptide chain using ribosomes, tRNA, amino acids, and energy.

Q27. Explain the role of tRNA in translation.

Answer:
tRNA acts as an adaptor molecule, carrying specific amino acids and pairing its anticodon with mRNA codons.

Q28. What is the significance of ribosomes?

Answer:
Ribosomes are sites of protein synthesis and facilitate proper alignment of mRNA and tRNA.

Q29. What is peptide bond formation?

Answer:
Peptide bond formation links amino acids together during elongation, forming a polypeptide chain.

Q30. Why is ATP required during translation?

Answer:
ATP provides energy for amino acid activation and movement of ribosomes along mRNA.

Section G: Regulation of Gene Expression

Q31. What is gene regulation?

Answer:
Gene regulation controls when and how much a gene is expressed, preventing wasteful protein synthesis.

Q32. Explain the operon concept.

Answer:
An operon is a cluster of genes regulated together, common in prokaryotes, ensuring coordinated gene expression.

Q33. Describe the lac operon mechanism.

Answer:
In absence of lactose, repressor blocks transcription. Presence of lactose inactivates repressor, allowing gene expression.

Q34. Why is lac operon called inducible?

Answer:
It is activated only in presence of lactose, which induces transcription of lactose-metabolising enzymes.

Q35. What is the role of regulator gene?

Answer:
The regulator gene produces repressor protein that controls transcription of structural genes.

Section H: Human Genome Project

Q36. What was the aim of the Human Genome Project?

Answer:
The project aimed to sequence and map all human genes and understand genome structure and function.

Q37. State two major findings of HGP.

Answer:
Humans have about 30,000 genes and 99.9% of DNA sequence is identical among individuals.

Q38. Mention two applications of HGP.

Answer:
It aids in disease diagnosis, personalised medicine, and understanding genetic disorders.

Q39. What is bioinformatics?

Answer:
Bioinformatics uses computational tools to store, analyse, and interpret large biological datasets.

Q40. Why is non-coding DNA significant?

Answer:
Non-coding DNA regulates gene expression and contributes to genome organisation and evolution.

Section I: DNA Fingerprinting

Q41. Explain the principle of DNA fingerprinting.

Answer:
DNA fingerprinting is based on polymorphism in repetitive DNA sequences that vary among individuals.

Q42. Name the scientist who developed DNA fingerprinting.

Answer:
DNA fingerprinting was developed by Alec Jeffreys.

Q43. What are VNTRs?

Answer:
VNTRs are variable number tandem repeats that differ in length among individuals, forming unique DNA profiles.

Q44. Mention two applications of DNA fingerprinting.

Answer:
It is used in forensic investigations and paternity testing.

Q45. Why is DNA fingerprinting reliable?

Answer:
DNA patterns are unique to individuals, except identical twins, making identification highly accurate.

Section J: Integrated Concepts

Q46. Explain central dogma of molecular biology.

Answer:
Central dogma describes the flow of genetic information from DNA to RNA to protein.

Q47. Why is regulation of gene expression essential?

Answer:
It ensures proteins are produced only when needed, maintaining cellular efficiency.

Q48. What is the role of primase in replication?

Answer:
Primase synthesises RNA primers required to initiate DNA synthesis.

Q49. How does transcription differ from replication?

Answer:
Replication copies entire DNA, while transcription synthesises RNA from one DNA strand.

Q50. Why is Molecular Basis of Inheritance important in biology?

Answer:
It explains how genetic information is stored, expressed, regulated, and inherited at molecular level.