Class 12 Biology Chapter 5 Molecular Basis of Inheritance Solutions. Explore Class 12 Biology Chapter 5 Molecular Basis of Inheritance Solutions with easy-to-understand answers, important questions, and exam-focused explanations for effective revision and better scores. Continue reading: Molecular Basis of Inheritance Solutions

Class 12 Biology Chapter 5 Molecular Basis of Inheritance Solutions

A. Very Short Answer Questions (1 mark each)

1. What is a nucleotide?

Ans: A nucleotide is the structural unit of nucleic acids (DNA and RNA). Each nucleotide contains a nitrogenous base, a pentose sugar and a phosphate group.

2. What are nucleotides made of?

Ans: Nucleotides contain carbon, hydrogen, oxygen, nitrogen and phosphorus, arranged in a nitrogenous base, a pentose sugar and a phosphate group.

3. Who coined the term “nucleic acid”?

Ans: Friedrich Miescher discovered nuclein; the term “nucleic acid” was introduced later. The source appears to attribute the naming to Altmann.

4. Who coined the term “nuclein”?

Ans: Friedrich Miescher.

5. Which nitrogenous bases occur in RNA?

Ans: Adenine, guanine, cytosine and uracil.

6. Name the nucleotides found in nucleic acids.

Ans: The corresponding nucleotide residues are adenylate, guanylate, cytidylate, thymidylate and uridylate; thymidylate is characteristic of DNA, whereas uridylate is characteristic of RNA.

7. Define autoradiography.

Ans: Autoradiography is a technique in which radiation emitted by a radioactive substance is recorded on a photographic or detector medium to reveal the location of the labelled substance.

8. Name the pentose sugar present in DNA.

Ans: Deoxyribose.

9. What is an RNA molecule made of?

Ans: RNA is made of ribonucleotides, each containing a nitrogenous base, ribose sugar and phosphate group.

10. What is the function of rRNA?

Ans: Ribosomal RNA forms part of the structure of ribosomes and plays a catalytic role in protein synthesis.

11. Where is mRNA produced?

Ans: mRNA is transcribed from DNA; in eukaryotic cells, this usually occurs in the nucleus. It is then transported to ribosomes for translation.

12. What are introns?

Ans: Introns are intervening sequences in a gene that are transcribed into the primary RNA transcript but are generally removed during RNA splicing.

13. What is transformation?

Ans: Transformation is the process by which a cell takes up external DNA and, as a result, may acquire new hereditary characteristics.

14. What is DNA replication?

Ans: DNA replication is the process by which one DNA molecule produces two DNA molecules with the same genetic information.

15. What carries the instructions for protein synthesis?

Ans: DNA carries the genetic instructions for protein synthesis.

16. How many nitrogenous bases make up a codon for one amino acid?

Ans: Three nitrogenous bases.

17. What is the genetic code?

Ans: The genetic code is the set of rules by which nucleotide triplets in mRNA specify amino acids or stop signals during protein synthesis.

18. What is a triplet or genetic codon?

Ans: A codon is a sequence of three nucleotides in mRNA that specifies an amino acid or a stop signal during translation.

19. What is a bacteriophage made of?

Ans: A bacteriophage consists mainly of nucleic acid (usually DNA) enclosed in a protein coat.

20. What did the presence of phosphorus-32 (³²P) in the DNA of progeny phages prove?

Ans: It supported the conclusion that DNA, rather than protein, is the hereditary material transmitted during bacteriophage reproduction, as shown in the Hershey–Chase experiment.

21. Who first deciphered a genetic codon?

Ans: Marshall W. Nirenberg and Heinrich J. Matthaei identified the first codon, UUU, as coding for phenylalanine.

22. What are the main building blocks of proteins?

Ans: Amino acids.

23. Define codon and anticodon.

Ans: Codon: A sequence of three nucleotides in mRNA that specifies an amino acid or a stop signal. Anticodon: A sequence of three nucleotides in tRNA that pairs with a complementary codon in mRNA.

24. Define transcription and translation.

Ans: Transcription is the synthesis of RNA using a DNA strand as a template. Translation is the synthesis of a polypeptide at a ribosome according to the codon sequence of mRNA, with tRNA bringing the corresponding amino acids.

25. What is the objective of the Human Genome Project?

Ans: Its objective was to determine the sequence of human DNA and identify and analyse human genes, including their organisation and functions.

Fill in the Blanks

26. Uracil is found in ______.

Ans: RNA. Note: The source answer “DNA” is incorrect; uracil is normally found in RNA, while DNA contains thymine.

27. Chromatin that is tightly coiled and stains darkly is called ______.

Ans: Heterochromatin.

28. The movement of a gene from one location to another is called ______.

Ans: Transposition; such mobile genetic elements are called transposable elements or transposons.

29. New DNA strands are synthesised in the ______ direction.

Ans: 5′ to 3′.

30. The molecule with a cloverleaf-like structure is ______.

Ans: tRNA.

31. DNA replication is described as ______ replication.

Ans: Semiconservative replication.

32. A gene is considered the ______ unit of inheritance.

Ans: Basic (fundamental) unit.

33. Regulation of the lac operon by a repressor is called ______ control.

Ans: Negative regulation.

34. The untranslated sequences that remain in mature mRNA are called ______ regions.

Ans: Untranslated regions (UTRs).

35. A DNA segment that encodes a polypeptide is called a ______.

Ans: Cistron.

36. During transcription, RNA polymerase binds to the ______.

Ans: Promoter region.

B. Short-Answer Questions (Molecular Basis of Inheritance Solutions)

37. How many types of nitrogenous bases are there? Name and briefly describe them.

Ans: There are two main classes: (i) Purines: double-ring, nitrogen-containing bases—adenine (A) and guanine (G). (ii) Pyrimidines: single-ring, nitrogen-containing bases—cytosine (C), thymine (T) and uracil (U). Cytosine occurs in both DNA and RNA; thymine is normally found in DNA, while uracil occurs in RNA.

38. Draw the structures of adenine and guanine.

Ans: Adenine and guanine are purine bases with a fused double-ring structure. Adenine is 6-aminopurine; guanine is 2-amino-6-oxopurine. (A labelled structural drawing should be added in the printed version.)

39. Describe the structure of a nucleosome.

Ans: A nucleosome consists of a segment of DNA wrapped around a core of eight histone proteins (two copies each of H2A, H2B, H3 and H4). About 147 base pairs of DNA wrap around the histone octamer; linker DNA connects neighbouring nucleosomes.

40. What is a pentose sugar? Describe the two types found in nucleic acids.

Ans: A pentose is a monosaccharide containing five carbon atoms.

(i) Ribose (C₅H₁₀O₅) occurs in RNA and has a hydroxyl group at the 2′ carbon.

(ii) Deoxyribose (C₅H₁₀O₄) occurs in DNA and has hydrogen instead of a hydroxyl group at the 2′ carbon, so it contains one fewer oxygen atom than ribose.

41. What is satellite DNA?

Ans: Satellite DNA consists of highly repetitive DNA sequences that differ in base composition or repeat length from the bulk of genomic DNA. In density-gradient centrifugation, these sequences may form distinct bands or “satellite” peaks. Such DNA is often found in regions such as centromeres and is useful in DNA profiling.

42. List the functions of DNA.

Ans: (i) DNA stores and transmits hereditary information from parents to offspring. (ii) It contains instructions for synthesising proteins and functional RNAs. (iii) It serves as a template for transcription to RNA. (iv) DNA replication before cell division helps ensure that daughter cells receive genetic information.

43. What new information about DNA became known after the Watson–Crick model was proposed?

Ans: (i) Not all DNA is double-helical in every biological context; some viruses have single-stranded DNA. (ii) In double-stranded DNA, adenine pairs with thymine through two hydrogen bonds, while guanine pairs with cytosine through three. (iii) DNA molecules vary greatly in length and nucleotide number among organisms. Note: The source’s specific claim that a DNA molecule may contain only up to 30,000 nucleotides is not generally valid.

44. Write the functions of RNA.

Ans: (i) mRNA carries genetic instructions from DNA to ribosomes. (ii) tRNA brings amino acids to the ribosome during translation. (iii) rRNA forms part of ribosomes and participates in peptide-bond formation. (iv) In some viruses, RNA serves as the genetic material. Other small RNAs help regulate gene expression and RNA processing.

45. Write four characteristics of genetic material.

Ans: (i) It must be capable of replication. (ii) It must be sufficiently stable to preserve information. (iii) It must be capable of mutation, allowing variation and evolution. (iv) It must be able to express its information and transmit it according to inheritance patterns.

46. Describe Griffith’s experiment in the study of heredity. HS ‘24

Ans: Frederick Griffith studied Streptococcus pneumoniae, which has a virulent smooth (S) strain with a capsule and a non-virulent rough (R) strain without a capsule. Live S bacteria killed mice, while live R bacteria did not. Heat-killed S bacteria did not kill mice. However, a mixture of heat-killed S bacteria and live R bacteria killed mice, and live S bacteria were recovered. Griffith concluded that a “transforming principle” from the dead S cells had transformed the R cells into virulent cells. The experiment demonstrated bacterial transformation but did not itself identify DNA as the transforming material.

47. What is semiconservative DNA replication?

Ans: In semiconservative replication, the two strands of the parental DNA separate and each serves as a template for synthesis of a complementary new strand. Each daughter DNA molecule therefore contains one parental strand and one newly synthesised strand.

48. Describe the experiments of Avery, MacLeod and McCarty on the hereditary material.

Ans: Avery, MacLeod and McCarty investigated extracts from heat-killed virulent S bacteria of Streptococcus pneumoniae. They treated extracts with enzymes that destroyed proteins, RNA or DNA. Transformation of non-virulent R bacteria continued when proteins or RNA were destroyed, but stopped when DNA was destroyed by DNase. They concluded that DNA was the transforming principle and therefore the hereditary material in this experiment.

49. What conditions must a substance satisfy to function as genetic material?

Ans: (i) It must replicate accurately. (ii) It must be chemically stable. (iii) It must be capable of mutation to provide variation for evolution. (iv) It must be able to express its information and pass it to the next generation.

50. Differentiate between DNA and RNA.

Ans: DNA: (i) Usually double-stranded and helical. (ii) Contains deoxyribose sugar. (iii) Contains A, T, G and C. (iv) Stores hereditary information in most organisms. RNA: (i) Usually single-stranded. (ii) Contains ribose sugar. (iii) Contains A, U, G and C. (iv) Functions in gene expression and protein synthesis; it is the genetic material in some viruses.

51. How does DNA express biological information?

Ans: Genes in DNA are transcribed into RNA. For protein-coding genes, the mRNA carries codons to the ribosome, where tRNA brings amino acids in the order specified by the codons. The amino acids are joined into a polypeptide, which folds and may be modified to become a functional protein. Proteins and functional RNAs contribute to the structure, activities and regulation of cells.

52. Briefly describe DNA replication.

Ans: Before cell division, each DNA molecule is copied. The two parental strands separate, and each acts as a template for a complementary strand. DNA polymerases synthesise new DNA in the 5′ to 3′ direction. The process is semiconservative because each resulting DNA molecule contains one parental strand and one new strand.

C. Descriptive Questions (Molecular Basis of Inheritance Solutions)

53. Discuss the properties of the genetic code.

Ans: (i) Triplet: Each codon consists of three nucleotides. (ii) Degenerate: Most amino acids are specified by more than one codon; for example, threonine is specified by ACU, ACC, ACA and ACG. (iii) Non-overlapping: Each nucleotide is read as part of one codon in a reading frame. (iv) Nearly universal: Most organisms use the same code, with a few exceptions. (v) Start codon: AUG usually initiates translation and codes for methionine. (vi) Stop codons: UAA, UAG and UGA signal termination and do not code for amino acids. (vii) Commaless: Codons are read consecutively without punctuation between them.

54. Explain the lac operon model of gene regulation.

Ans: The lac operon in Escherichia coli regulates genes involved in lactose utilisation. It includes the promoter, operator and structural genes lacZ, lacY and lacA, and is regulated by the lacI repressor gene. In the absence of lactose, the repressor binds to the operator and prevents RNA polymerase from transcribing the structural genes. When lactose is available, its derivative allolactose binds to the repressor and reduces its ability to bind the operator, allowing transcription. Expression is strongest when glucose is low because the cAMP–CAP system enhances transcription. When lactose is depleted, the repressor becomes active again and transcription is reduced.

55. Discuss DNA as a medico-legal tool.

Ans: DNA profiling is used in forensic science and legal investigations. (i) It can help assess biological relationships, including disputed paternity, when interpreted with appropriate statistical methods. (ii) DNA evidence can link a biological sample to or exclude a suspect. (iii) It can help identify unidentified human remains or victims of disasters. (iv) It can assist in reuniting missing children and family members. (v) Genetic testing may identify inherited variants, although it does not by itself predict with certainty whether a person will develop every hereditary disease.

56. Describe the method of DNA fingerprinting.

Ans: DNA is extracted from an appropriate biological sample, such as blood, saliva, semen or cells from a hair root. The DNA is analysed at variable genetic markers, commonly short tandem repeats (STRs), using PCR. The resulting DNA fragments are separated by capillary electrophoresis, and a DNA profile is produced. The profile is compared with reference samples using statistical methods. Traditional DNA fingerprinting used restriction enzymes, gel electrophoresis, Southern blotting and labelled probes to detect variable repetitive sequences. Proper sample handling and contamination control are essential.

D. Additional Questions and Answers: Molecular Basis of Inheritance Solutions

1. Name two viruses that have RNA as their genetic material.

Ans: Tobacco mosaic virus (TMV) and Q-beta bacteriophage.

2. State Chargaff’s rule.

Ans: In double-stranded DNA, the amount of adenine equals thymine and the amount of guanine equals cytosine: A = T and G = C.

3. What forms the backbone of DNA?

Ans: Alternating deoxyribose sugar and phosphate groups form the sugar–phosphate backbone of each DNA strand.

4. What is the characteristic of complementary base pairing?

Ans: The bases pair specifically: adenine with thymine in DNA (or uracil in RNA), and guanine with cytosine. Knowing the sequence of one strand allows the complementary sequence to be predicted.

5. Mention four properties that allow both DNA and RNA to function as genetic material.

Ans: (i) They can be replicated. (ii) They are chemically stable enough to preserve information. (iii) They can undergo mutations that generate variation. (iv) Their information can be expressed and transmitted to subsequent generations.

6. How was the triplet nature of the genetic code established?

Ans: Experiments using synthetic RNA and later studies with defined nucleotide sequences showed that codons consist of three nucleotides. The source’s brief explanation is incomplete; the conclusion was established through biochemical and genetic experiments.

7. Why is energy required during translation?

Ans: Energy is required to charge tRNA with amino acids, deliver aminoacyl-tRNAs to the ribosome, and drive steps of translation. GTP and ATP are used at different stages.

8. How is a polynucleotide chain formed?

Ans: Nucleotides are joined by 3′–5′ phosphodiester bonds. A strand has a 5′ end, typically bearing a phosphate group, and a 3′ end with a free hydroxyl group.

9. What was the conclusion of Griffith’s experiment?

Ans: Griffith concluded that a transforming principle from heat-killed virulent bacteria could change live non-virulent bacteria into virulent bacteria. The experiment did not identify the chemical nature of the transforming principle.

10. What is the wobble hypothesis? Explain it.

Ans: The wobble hypothesis, proposed by Francis Crick, explains why some tRNAs can recognise more than one codon. Pairing at the first two positions of the codon is usually strict, while the third codon position can allow less stringent pairing with the first anticodon position. This accounts for some redundancy in the genetic code.

11. Write two important roles of the ribosome during translation.

Ans: (i) The ribosome binds mRNA and positions tRNAs so that codons pair with the corresponding anticodons. (ii) Its rRNA-containing large subunit catalyses peptide-bond formation; in bacteria, 23S rRNA forms the catalytic centre.

12. Why can the lac operon be switched off some time after lactose is added to a growing E. coli culture?

Ans: While lactose or allolactose is present, it inactivates the repressor and permits transcription of lac genes. As lactose is used up, the inducer level falls; the repressor can bind the operator again and block transcription. Glucose availability also affects the level of lac operon expression.

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