Important Questions Class 12 Biology Chapter 5 Molecular Basis of Inheritance With Answers

Molecular inheritance explains how genetic information is stored, copied, expressed, and transferred in living organisms. DNA, RNA, genes, chromosomes, codons, ribosomes, and regulatory sequences control inheritance at the molecular level.

Molecular Basis of Inheritance becomes scoring when students link DNA structure with replication, transcription, translation, and gene regulation. Important Questions Class 12 Biology Chapter 5 help students revise the chapter for CBSE 2026-27 board exams, school tests, and pre-board papers. NCERT Chapter 5 covers DNA, RNA, genetic material experiments, semiconservative replication, genetic code, tRNA, translation, lac operon, Human Genome Project, and DNA fingerprinting. These topics often appear as diagrams, short answers, reasoning questions, numerical questions, and process-based 5-mark answers.

Key Takeaways

• DNA Structure: DNA is a double helix with antiparallel strands and complementary base pairing.
• Genetic Material: Hershey and Chase proved that DNA, not protein, enters bacteria during bacteriophage infection.
• Gene Expression: Transcription forms RNA from DNA, while translation forms protein from mRNA.
• Class 12 Biology Focus: Lac operon, Human Genome Project, and DNA fingerprinting are repeated board-exam concepts.

Important Questions Class 12 Biology Chapter 5 Structure 2026-27

Important Questions Class 12 Biology Chapter 5 for CBSE 2026-27

Students usually lose marks in this chapter when they memorise terms without linking processes. CBSE 2026-27 questions can ask the same concept as a diagram, reason, sequence, experiment, or numerical.

1. What is DNA?

DNA is a long polymer of deoxyribonucleotides that acts as genetic material in most organisms. It stores hereditary information.

1. DNA stands for deoxyribonucleic acid.
2. It contains deoxyribose sugar.
3. It contains adenine, guanine, cytosine, and thymine.
4. Its length is usually expressed as base pairs.

Final answer:
DNA is the main genetic material in most living organisms.

2. What is RNA?

RNA is a nucleic acid that acts as genetic material in some viruses and mostly functions in gene expression. It works as messenger, adapter, structural, and catalytic molecule.

1. RNA contains ribose sugar.
2. RNA contains uracil instead of thymine.
3. mRNA carries genetic information.
4. tRNA acts as adapter.
5. rRNA forms ribosomes and catalyses peptide bond formation.

Final fact:
RNA performs several active roles in protein synthesis.

3. What are the components of a nucleotide?

A nucleotide has a nitrogenous base, a pentose sugar, and a phosphate group.

Components:

1. Nitrogenous base.
2. Pentose sugar.
3. Phosphate group.

Types of bases:

1. Purines: adenine and guanine.
2. Pyrimidines: cytosine, thymine, and uracil.

Final fact:
Thymine occurs in DNA, while uracil occurs in RNA.

Molecular Basis of Inheritance Class 12 Biology Important Questions on DNA Structure

DNA structure questions are highly searched because they connect Chargaff’s rule, Watson and Crick model, base pairing, and numerical base composition. Students should write exact bond names and polarity.

4. What is a polynucleotide chain?

A polynucleotide chain is a polymer formed by joining many nucleotides through 3'-5' phosphodiester linkages.

1. A nitrogenous base links to 1' carbon of sugar by N-glycosidic linkage.
2. A phosphate group links to 5' carbon by phosphoester linkage.
3. Two nucleotides join by 3'-5' phosphodiester linkage.
4. Sugar and phosphate form the backbone.
5. Bases project from the backbone.

Final fact:
A polynucleotide has a 5' end and a 3' end.

5. Who proposed the double helix model of DNA?

James Watson and Francis Crick proposed the double helix model of DNA in 1953. Their model used X-ray diffraction data from Maurice Wilkins and Rosalind Franklin.

1. Friedrich Miescher identified DNA as nuclein in 1869.
2. Watson and Crick proposed the DNA double helix.
3. Erwin Chargaff’s base ratio helped explain base pairing.

Final answer:
Watson and Crick proposed the DNA double helix model.

6. What are the salient features of Watson and Crick model Class 12 Biology?

The Watson and Crick model describes DNA as a right-handed double helix with antiparallel strands and complementary base pairing.

Features:

1. DNA has two polynucleotide chains.
2. Sugar-phosphate backbone lies outside.
3. Bases project inside.
4. Strands run antiparallel.
5. A pairs with T through two hydrogen bonds.
6. G pairs with C through three hydrogen bonds.
7. The pitch is 3.4 nm.
8. Each turn has about 10 base pairs.
9. Distance between base pairs is 0.34 nm.

Final fact:
Base stacking and hydrogen bonds stabilise the double helix.

7. Why is DNA called antiparallel?

DNA is called antiparallel because its two strands run in opposite directions. One strand has 5' to 3' polarity, while the other has 3' to 5' polarity.

1. One chain runs from 5' end to 3' end.
2. The opposite chain runs from 3' end to 5' end.
3. This arrangement supports complementary base pairing.

Final fact:
Antiparallel polarity is essential for DNA replication and transcription.

8. What is Chargaff’s rule?

Chargaff’s rule states that adenine equals thymine and guanine equals cytosine in double-stranded DNA.

Rule:
A = T
G = C

Meaning:

1. Purine always pairs with pyrimidine.
2. A pairs with T.
3. G pairs with C.
4. Base ratios remain constant in double-stranded DNA.

Final fact:
Chargaff’s rule helped support the DNA double helix model.

9. If double-stranded DNA has 20 percent cytosine, calculate adenine percentage.

Adenine will be 30 percent. In double-stranded DNA, cytosine pairs with guanine.

Given data:
C = 20 percent

Using Chargaff’s rule:
G = 20 percent
C + G = 40 percent
A + T = 60 percent
A = T

Calculation:
A = 60/2
A = 30 percent

Final result:
Adenine = 30 percent

10. If one DNA strand is 5'-ATGCATGC-3', write the complementary strand in 5' to 3' direction.

The complementary strand in 5' to 3' direction is 5'-GCATGCAT-3'.

Given strand:
5'-ATGCATGC-3'

Complementary strand first forms as:
3'-TACGTACG-5'

Writing in 5' to 3' direction:
5'-GCATGCAT-3'

Final answer:
5'-GCATGCAT-3'

DNA Packaging Class 12 Biology Questions

DNA packaging explains how long DNA fits inside a tiny nucleus. CBSE questions often ask nucleosome, histone octamer, euchromatin, and heterochromatin.

11. What is a nucleosome?

A nucleosome is the basic repeating unit of chromatin in eukaryotes. It contains DNA wrapped around a histone octamer.

1. Histones are positively charged proteins.
2. DNA is negatively charged.
3. Eight histone molecules form a histone octamer.
4. DNA wraps around the octamer.
5. A typical nucleosome contains about 200 bp of DNA.

Final fact:
Nucleosomes appear as beads-on-string under an electron microscope.

12. Why are histones positively charged?

Histones are positively charged because they are rich in lysine and arginine. These amino acids carry positive charges in their side chains.

1. DNA has negatively charged phosphate groups.
2. Histones have positive charges.
3. DNA wraps around histone octamers.

Final fact:
Positive histones help package negatively charged DNA.

13. Differentiate between euchromatin and heterochromatin.

Euchromatin is loosely packed and transcriptionally active, while heterochromatin is densely packed and transcriptionally inactive.

Final fact:
Chromatin packing affects gene expression.

Hershey Chase Experiment Class 12 Biology and Search for Genetic Material

Students often search this section because the experiments look similar. The key difference is simple: Griffith showed transformation, Avery identified DNA chemically, and Hershey-Chase proved DNA enters bacteria.

14. What was Griffith’s transforming principle experiment?

Griffith’s experiment showed transformation in Streptococcus pneumoniae but did not identify the biochemical nature of genetic material.

Steps:

1. S strain bacteria were virulent and killed mice.
2. R strain bacteria were non-virulent and did not kill mice.
3. Heat-killed S strain did not kill mice.
4. Live R strain plus heat-killed S strain killed mice.
5. Living S strain bacteria were recovered from dead mice.

Final conclusion:
Some transforming principle changed R strain into virulent S strain.

15. What did Avery, MacLeod and McCarty prove?

Avery, MacLeod and McCarty proved that DNA was the transforming principle. Their work identified the biochemical nature of genetic material.

1. Proteases did not stop transformation.
2. RNases did not stop transformation.
3. DNase stopped transformation.
4. DNA alone transformed R cells into S cells.

Final answer:
DNA caused bacterial transformation.

16. How did Hershey and Chase prove that DNA is genetic material?

Hershey and Chase proved that DNA enters bacteria during bacteriophage infection, while protein does not.

Steps:

1. They labelled phage DNA with radioactive phosphorus.
2. They labelled phage protein with radioactive sulphur.
3. Phages infected E. coli.
4. Blender removed viral coats.
5. Centrifugation separated bacteria from viral coats.
6. Radioactive DNA entered bacteria.
7. Radioactive protein did not enter bacteria.

Final conclusion:
DNA is the genetic material.

17. Why were radioactive phosphorus and sulphur used in Hershey Chase experiment?

Radioactive phosphorus labelled DNA, while radioactive sulphur labelled protein. DNA contains phosphorus but no sulphur.

1. DNA contains phosphorus in its phosphate backbone.
2. Protein contains sulphur in some amino acids.
3. This difference helped track DNA and protein separately.

Final fact:
Radioactive DNA entered bacteria, but radioactive protein stayed outside.

DNA Replication Class 12 Biology Questions With Answers

DNA replication questions commonly test semiconservative replication, Meselson-Stahl experiment, replication fork, DNA polymerase, leading strand, lagging strand, and DNA ligase.

18. What is semiconservative DNA replication?

Semiconservative DNA replication produces two DNA molecules, each with one parental strand and one newly synthesised strand.

1. DNA strands separate.
2. Each old strand acts as a template.
3. New complementary strands form.
4. Each daughter DNA has one old and one new strand.

Final fact:
Watson and Crick proposed this replication mechanism.

19. Which experiment proved semiconservative DNA replication?

Meselson and Stahl’s experiment proved semiconservative DNA replication in E. coli. They used heavy nitrogen and light nitrogen.

Steps:

1. E. coli was grown in 15NH4Cl medium.
2. Heavy nitrogen entered DNA.
3. Cells were transferred to 14NH4Cl medium.
4. DNA was extracted after generations.
5. CsCl density gradient centrifugation separated DNA.

Final result:
First generation showed hybrid DNA, proving semiconservative replication.

20. Why did DNA after one generation show intermediate density in Meselson Stahl experiment?

DNA showed intermediate density because each DNA molecule had one old heavy strand and one new light strand.

1. Old DNA contained 15N.
2. New DNA formed in 14N medium.
3. Each molecule had one heavy and one light strand.
4. The density became intermediate.

Final fact:
Intermediate density supports semiconservative replication.

21. What is a replication fork?

A replication fork is the small opened region of the DNA helix where replication occurs.

1. DNA strands separate at the fork.
2. Each strand acts as template.
3. DNA polymerase adds nucleotides.
4. Replication proceeds in 5' to 3' direction.

Final fact:
Long DNA cannot open completely due to high energy requirement.

22. What is the role of DNA-dependent DNA polymerase?

DNA-dependent DNA polymerase catalyses polymerisation of deoxynucleotides using DNA as template. It synthesises DNA only in 5' to 3' direction.

1. It uses deoxyribonucleoside triphosphates.
2. It follows base complementarity.
3. It works with high speed.
4. It works with high accuracy.

Final fact:
E. coli completes DNA replication in about 18 minutes.

23. Why is DNA replication continuous on one strand and discontinuous on the other?

DNA replication differs on the two strands because DNA polymerase works only in 5' to 3' direction.

1. The 3' to 5' template supports continuous synthesis.
2. The 5' to 3' template supports discontinuous synthesis.
3. Short fragments form on the discontinuous strand.
4. DNA ligase joins these fragments.

Final answer:
Antiparallel DNA strands cause continuous and discontinuous replication.

24. What is the role of DNA ligase?

DNA ligase joins discontinuously synthesised DNA fragments during replication. It completes the lagging strand.

1. DNA polymerase makes short fragments.
2. These fragments remain separated initially.
3. DNA ligase seals the gaps.

Final fact:
DNA ligase is essential for discontinuous strand synthesis.

Transcription Class 12 Biology Questions With Answers

Transcription is a high-intent search topic because students confuse template strand, coding strand, promoter, terminator, exons, introns, and hnRNA processing.

25. What is transcription?

Transcription is the process of copying genetic information from one DNA strand into RNA. It follows complementarity.

1. Only one DNA strand acts as template.
2. A pairs with U in RNA.
3. G pairs with C.
4. Only a DNA segment gets transcribed.

Final answer:
Transcription forms RNA from DNA.

26. Why are both DNA strands not copied during transcription?

Both DNA strands are not copied because they would produce two different RNA molecules and may form double-stranded RNA.

Reasons:

1. Two strands have different sequences.
2. They would code for different RNAs.
3. Complementary RNAs could pair with each other.
4. Double-stranded RNA cannot be translated properly.

Final fact:
Only one strand acts as template in a transcription unit.

27. What is a transcription unit?

A transcription unit is a DNA segment defined by promoter, structural gene, and terminator.

Parts:

1. Promoter.
2. Structural gene.
3. Terminator.

Final fact:
Promoter defines the template and coding strands.

28. Differentiate between template strand and coding strand.

Template strand forms RNA, while coding strand has the same sequence as RNA except thymine replaces uracil.

Final fact:
RNA sequence is complementary to template strand.

29. If the coding strand is 5'-ATGCATGC-3', write the mRNA sequence.

The mRNA sequence is 5'-AUGCAUGC-3'. It matches the coding strand except uracil replaces thymine.

Coding strand:
5'-ATGCATGC-3'

mRNA:
5'-AUGCAUGC-3'

Final answer:
5'-AUGCAUGC-3'

30. What is RNA polymerase?

RNA polymerase is the enzyme that catalyses RNA formation from a DNA template. It polymerises ribonucleotides in 5' to 3' direction.

1. It binds to the promoter.
2. It initiates transcription.
3. It elongates RNA.
4. It stops at the terminator.

Final fact:
Bacteria have one RNA polymerase for all major RNA types.

31. What are exons and introns?

Exons are expressed sequences that appear in mature RNA, while introns are intervening sequences removed during processing.

Final fact:
Eukaryotic genes are often split genes.

32. How is hnRNA processed into mRNA?

hnRNA becomes mature mRNA through splicing, capping, and tailing.

Steps:

1. Splicing removes introns and joins exons.
2. Capping adds methyl guanosine triphosphate at 5' end.
3. Tailing adds 200-300 adenylate residues at 3' end.
4. Processed hnRNA becomes mRNA.

Final fact:
Mature mRNA moves out of the nucleus for translation.

Genetic Code Class 12 Biology Questions

Genetic code questions often ask features, start codon, stop codons, degeneracy, universality, and frameshift mutation. These answers should stay exact and compact.

33. What is genetic code?

Genetic code is the set of rules by which mRNA codons specify amino acids during protein synthesis.

1. Codons are triplets of nucleotides.
2. 61 codons code for amino acids.
3. 3 codons act as stop codons.
4. AUG codes for methionine and acts as start codon.

Final fact:
The genetic code connects nucleotide sequence with amino acid sequence.

34. Write the salient features of genetic code Class 12 Biology.

Genetic code is triplet, degenerate, nearly universal, comma-less, and non-overlapping.

Features:

1. Codon is triplet.
2. 61 codons code for amino acids.
3. UAA, UAG, and UGA are stop codons.
4. AUG is start codon and codes methionine.
5. Code is degenerate.
6. Code is nearly universal.
7. Codons are read continuously.

Final fact:
UUU codes for phenylalanine from bacteria to humans.

35. What are start and stop codons?

AUG is the start codon, while UAA, UAG, and UGA are stop codons.

Start codon:
AUG = Methionine

Stop codons:

1. UAA
2. UAG
3. UGA

Final fact:
Stop codons do not code for any amino acid.

36. What is a frameshift mutation?

Frameshift mutation occurs when one or two bases get inserted or deleted from a coding sequence. It changes the reading frame.

1. One base insertion shifts codon reading.
2. Two base insertion also shifts codon reading.
3. Three base insertion adds one codon but preserves frame.
4. One or two base deletions shift the reading frame.

Final fact:
Frameshift mutations can change many amino acids after the mutation point.

37. What point mutation causes sickle cell anaemia?

Sickle cell anaemia occurs due to a single base-pair change in the beta-globin gene. It changes glutamate to valine.

1. A point mutation occurs in the beta-globin chain gene.
2. One codon changes.
3. Glutamate gets replaced by valine.
4. Haemoglobin structure changes.

Final fact:
Sickle cell anaemia is a classic point mutation example.

Translation Class 12 Biology Questions With Answers

Translation questions need mRNA, tRNA, ribosome, codon, anticodon, peptide bond, aminoacylation, and release factor. The flow should always move from codon reading to polypeptide formation.

38. What is translation?

Translation is the process of polymerising amino acids to form a polypeptide. The amino acid order is decided by mRNA codons.

1. mRNA provides codons.
2. tRNA brings amino acids.
3. Ribosome provides the translation site.
4. Peptide bonds join amino acids.

Final answer:
Translation forms protein from mRNA information.

39. Why is tRNA called an adapter molecule?

tRNA is called an adapter molecule because it reads mRNA codons and carries specific amino acids.

1. It has an anticodon loop.
2. The anticodon pairs with mRNA codon.
3. It has an amino acid acceptor end.
4. It binds a specific amino acid.

Final fact:
Francis Crick predicted the need for an adapter molecule.

40. What is charging of tRNA?

Charging of tRNA is the attachment of an amino acid to its specific tRNA using ATP. It is also called aminoacylation of tRNA.

1. Amino acid gets activated.
2. ATP provides energy.
3. Amino acid binds to its cognate tRNA.
4. Charged tRNA enters translation.

Final fact:
Charged tRNA supports peptide bond formation.

41. What are the two roles of ribosome during translation?

The ribosome provides a platform for translation and catalyses peptide bond formation.

Roles:

1. It binds mRNA and charged tRNAs.
2. It positions amino acids close together.
3. It moves from codon to codon.
4. 23S rRNA acts as a ribozyme in bacteria.

Final fact:
Ribosome acts as the protein-synthesis factory.

42. What is a translational unit?

A translational unit is the mRNA sequence flanked by a start codon and a stop codon. It codes for a polypeptide.

1. AUG starts translation.
2. Codons specify amino acids.
3. Stop codon terminates translation.
4. UTRs occur at 5' and 3' ends.

Final fact:
UTRs help efficient translation.

Lac Operon Class 12 Biology Questions

Lac operon is one of the most searched Class 12 Biology topics because students need the roles of i, z, y, a, promoter, operator, repressor, and inducer in one clear flow.

43. What is lac operon?

Lac operon is a transcriptionally regulated gene system in E. coli that controls lactose metabolism. It was explained by Jacob and Monod.

1. It has one regulatory gene.
2. It has three structural genes.
3. It is controlled by lactose.
4. It shows negative regulation.

Final fact:
Lac operon is a model operon in bacteria.

44. Name the genes of lac operon and their functions.

Lac operon contains i, z, y, and a genes. The i gene is regulatory, while z, y, and a are structural genes.

Final fact:
All three structural genes support lactose metabolism.

45. What is the role of lactose in lac operon?

Lactose acts as an inducer in lac operon. It inactivates the repressor and allows transcription.

1. Repressor binds operator in absence of lactose.
2. RNA polymerase cannot transcribe structural genes.
3. Lactose or allolactose binds repressor.
4. Repressor becomes inactive.
5. RNA polymerase transcribes the operon.

Final fact:
Lac operon switches on when lactose is available.

46. Why does lac operon shut down after some time?

Lac operon shuts down after lactose gets used up. Without inducer, the repressor binds the operator again.

1. Lactose enters the bacterial cell.
2. It induces lac operon.
3. Enzymes metabolise lactose.
4. Inducer level falls.
5. Repressor binds operator again.

Final answer:
Lac operon stops when lactose is no longer available.

Human Genome Project Class 12 Biology Questions

Human Genome Project questions test goals, methods, salient features, and bioinformatics. Use exact NCERT figures because they often appear in assertion-reason and one-mark questions.

47. What was the Human Genome Project?

Human Genome Project was a mega project launched in 1990 to sequence the complete human genome. It was completed in 2003.

1. It aimed to identify human genes.
2. It aimed to sequence 3 billion base pairs.
3. It stored data in databases.
4. It improved tools for data analysis.
5. It addressed ethical, legal, and social issues.

Final fact:
HGP supported the growth of bioinformatics.

48. Why was the Human Genome Project called a mega project?

HGP was called a mega project because it aimed to sequence about 3 × 10^9 base pairs of human DNA. It required huge cost, data storage, and global collaboration.

1. Estimated cost was about 9 billion US dollars initially.
2. Storing typed sequence data would need about 3300 books.
3. It required high-speed computational tools.
4. It involved several countries.

Final fact:
HGP generated massive biological data.

49. Write the main goals of Human Genome Project Class 12 Biology.

The main goals of HGP were gene identification, DNA sequencing, data storage, data analysis, technology transfer, and ELSI review.

Goals:

1. Identify about 20,000-25,000 genes.
2. Determine 3 billion base-pair sequence.
3. Store information in databases.
4. Improve tools for analysis.
5. Transfer technologies to other sectors.
6. Address ethical, legal, and social issues.

Final answer:
HGP mapped and analysed the human genome.

50. What are the salient features of human genome?

Human genome data showed that humans have 3164.7 million base pairs and less than 2 percent of the genome codes for proteins.

Features:

1. Human genome contains 3164.7 million bp.
2. Average gene has about 3000 bases.
3. Total genes are about 30,000.
4. 99.9 percent nucleotide bases are same in all humans.
5. Over 50 percent discovered genes have unknown functions.
6. Chromosome 1 has most genes.
7. Y chromosome has the fewest genes.
8. About 1.4 million SNP locations were identified.

Final fact:
Repeated sequences form a large part of the human genome.

DNA Fingerprinting Class 12 Biology Questions

DNA fingerprinting is a genuine high-search topic because it connects forensic science, paternity testing, VNTR, Southern blotting, autoradiography, and DNA polymorphism.

51. What is DNA fingerprinting?

DNA fingerprinting is a technique used to compare DNA sequences of individuals by studying polymorphic repetitive DNA regions.

1. It identifies differences in DNA sequence.
2. It uses repetitive DNA.
3. It produces individual-specific banding patterns.
4. Identical twins show the same pattern.

Final fact:
Alec Jeffreys developed DNA fingerprinting.

52. What is VNTR in DNA fingerprinting?

VNTR means Variable Number of Tandem Repeats. It is a minisatellite DNA sequence used as a probe in DNA fingerprinting.

1. VNTR sequences repeat many times.
2. Copy number varies among individuals.
3. This variation gives different banding patterns.
4. VNTR shows high polymorphism.

Final fact:
VNTR size can vary from 0.1 to 20 kb.

53. What are the steps of DNA fingerprinting?

DNA fingerprinting includes DNA isolation, digestion, electrophoresis, blotting, hybridisation, and autoradiography.

Steps:

1. Isolation of DNA.
2. Digestion by restriction endonucleases.
3. Separation by electrophoresis.
4. Blotting onto nitrocellulose or nylon membrane.
5. Hybridisation with labelled VNTR probe.
6. Detection by autoradiography.

Final fact:
PCR has increased the sensitivity of DNA fingerprinting.

54. What are the applications of DNA fingerprinting?

DNA fingerprinting is used in forensic science, paternity testing, genetic diversity studies, and evolutionary biology.

Applications:

1. Crime investigation.
2. Paternity disputes.
3. Population studies.
4. Genetic diversity analysis.
5. Evolutionary studies.

Final fact:
DNA from blood, hair follicle, skin, bone, saliva, or sperm can be used.

55. Why is DNA fingerprinting useful in forensic science?

DNA fingerprinting is useful because every individual has a characteristic DNA banding pattern except identical twins.

1. Crime-scene DNA can be compared with suspect DNA.
2. VNTR patterns differ among individuals.
3. The matching band pattern helps identification.

Final answer:
DNA fingerprinting helps identify individuals from biological samples.

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