Molecular Basis of Inheritance: Important Questions & Answers for Class 12 & NEET

The Molecular Basis of Inheritance is one of the most significant chapters in high school biology and competitive medical entrance exams. This chapter deals with the nature of genetic material, how it replicates, and how the information stored in DNA is converted into proteins. Below is a comprehensive collection of high-yield questions and detailed answers covering the entire syllabus from the search for genetic material to DNA fingerprinting.

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Topic 1: Search for Genetic Material and DNA Structure

Q1. Describe the experiment that unequivocally proved that DNA is the genetic material.

Answer: The unequivocal proof that DNA is the genetic material came from the experiment conducted by Alfred Hershey and Martha Chase (1952). They worked with bacteriophages (viruses that infect bacteria).

• Method: They grew some viruses on a medium that contained radioactive phosphorus (³²P) and others on medium that contained radioactive sulfur (³⁵S).
  • Viruses grown in ³²P contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not.
  • Viruses grown in ³⁵S contained radioactive protein but not radioactive DNA because protein contains sulfur but DNA does not.
• Infection: These radioactive phages were allowed to attach to E. coli bacteria.
• Blending and Centrifugation: The viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge.
• Observation: Bacteria which were infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria. Bacteria infected with viruses that had radioactive proteins were not radioactive.
• Conclusion: This proved that DNA is the genetic material that is passed from virus to bacteria.

Q2. State the salient features of the Double Helix structure of DNA as proposed by Watson and Crick.

Answer: In 1953, James Watson and Francis Crick proposed the Double Helix model of DNA based on X-ray diffraction data. The salient features are:

1. It is made of two polynucleotide chains, where the backbone is constituted by sugar-phosphate, and the bases project inside.
2. The two chains have anti-parallel polarity. It means, if one chain has the polarity 5' -> 3', the other has 3' -> 5'.
3. The bases in two strands are paired through hydrogen bond (H-bonds) forming base pairs (bp). Adenine forms two hydrogen bonds with Thymine (A=T) and Guanine is bonded with Cytosine with three H-bonds (G≡C).
4. The two chains are coiled in a right-handed fashion. The pitch of the helix is 3.4 nm and there are roughly 10 bp in each turn. Consequently, the distance between a bp in a helix is approximately 0.34 nm.
5. The plane of one base pair stacks over the other in double helix. This confers stability to the helical structure.

Q3. Explain the "Central Dogma" of molecular biology.

Answer: Proposed by Francis Crick, the Central Dogma states that the genetic information flows in a specific direction:

DNA → (Transcription) → mRNA → (Translation) → Protein

However, in some viruses (retroviruses), the flow of information is in reverse direction, i.e., from RNA to DNA. This is called Reverse Transcription.

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Topic 2: DNA Packaging

Q4. How is the long polymer of DNA packaged inside a microscopic nucleus?

Answer: In eukaryotes, DNA packaging involves positively charged basic proteins called histones.

• Histones are rich in the basic amino acid residues lysine and arginine.
• Eight molecules of histones are organized to form a unit called a histone octamer.
• The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called a nucleosome. A typical nucleosome contains 200 bp of DNA helix.
• Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin, giving a "beads-on-string" appearance.
• The chromatin fibers are further coiled and condensed at metaphase stage of cell division to form chromosomes. This packaging requires Non-Histone Chromosomal (NHC) proteins.

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Topic 3: DNA Replication

Q5. Why is DNA replication termed as "semi-conservative"? Which experiment proved this?

Answer: DNA replication is termed semi-conservative because, after the completion of replication, each DNA molecule would have one parental strand (conserved) and one newly synthesized strand.

Proof: It was experimentally proved by Matthew Meselson and Franklin Stahl (1958) using E. coli.

• They grew E. coli in a medium containing ¹⁵NH₄Cl (heavy isotope of Nitrogen) for many generations. The result was that ¹⁵N was incorporated into newly synthesized DNA.
• Then they transferred the cells into a medium with normal ¹⁴NH₄Cl and took samples at various time intervals.
• Using CsCl density gradient centrifugation, they found that after one generation (20 mins), the DNA was hybrid (intermediate density). After two generations (40 mins), it was composed of equal amounts of this hybrid DNA and "light" DNA. This proved the semi-conservative mode.

Q6. Differentiate between the Leading strand and Lagging strand during replication.

Answer:

• Leading Strand: On the template strand with polarity 3' -> 5', the replication is continuous. The DNA polymerase adds nucleotides continuously towards the replication fork.
• Lagging Strand: On the template strand with polarity 5' -> 3', replication is discontinuous. DNA is synthesized in short fragments known as Okazaki fragments. These fragments are later joined by the enzyme DNA Ligase.

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Topic 4: Transcription

Q7. Explain the transcriptional unit and the process of transcription in prokaryotes.

Answer: A transcriptional unit in DNA is defined primarily by three regions: a Promoter, the Structural gene, and a Terminator.

The process in bacteria involves three steps:

1. Initiation: RNA polymerase binds to the promoter and initiates transcription. It uses a transient factor called sigma factor (σ) to start the process.
2. Elongation: The RNA polymerase unzips the DNA and uses nucleoside triphosphates as substrate, polymerizing in a template-dependent fashion following the rule of complementarity.
3. Termination: Once the polymerase reaches the terminator region, the nascent RNA falls off, as does the RNA polymerase. This is facilitated by the rho factor (ρ).

Q8. What are Exons and Introns? How is hnRNA processed in eukaryotes?

Answer: In eukaryotes, the primary transcript (hnRNA) contains both the coding sequences (Exons) and non-coding sequences (Introns).

Processing (Splicing, Capping, Tailing):

• Splicing: The introns are removed and exons are joined in a defined order.
• Capping: An unusual nucleotide (methyl guanosine triphosphate) is added to the 5'-end of hnRNA.
• Tailing: Adenylate residues (200-300) are added at the 3'-end in a template-independent manner (Poly-A tail).

The fully processed hnRNA is now called mRNA and is transported out of the nucleus for translation.

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Topic 5: Genetic Code and Translation

Q9. List the salient features of the Genetic Code.

Answer:

• The codon is triplet. 61 codons code for amino acids and 3 codons do not code for any amino acids (Stop Codons).
• Unambiguous and Specific: One codon codes for only one amino acid.
• Degenerate: Some amino acids are coded by more than one codon.
• Contiguous: The codon is read in mRNA in a contiguous fashion. There are no punctuations.
• Universal: For example, from bacteria to humans UUU would code for Phenylalanine (with some exceptions in mitochondrial codons).
• AUG has dual functions. It codes for Methionine (Met), and it also acts as an initiator codon.

Q10. Briefly explain the process of Translation.

Answer: Translation is the process of polymerization of amino acids to form a polypeptide based on the sequence of codons in mRNA.

• Charging of tRNA: Amino acids are activated in the presence of ATP and linked to their specific tRNA (aminoacylation of tRNA).
• Initiation: The ribosome binds to the mRNA at the start codon (AUG) that is recognized by the initiator tRNA.
• Elongation: Complexes composed of an amino acid linked to tRNA bind to the appropriate codon in mRNA by forming complementary base pairs with the tRNA anticodon. The ribosome moves from codon to codon along the mRNA. Amino acids are added one by one, translated into polypeptide sequences.
• Termination: When a release factor binds to the stop codon, the translation terminates and the complete polypeptide is released from the ribosome.

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Topic 6: Gene Expression and Regulation (Lac Operon)

Q11. Explain the regulation of the Lac Operon in E. coli.

Answer: The Lac Operon consists of one regulatory gene (the i gene) and three structural genes (z, y, and a).

• The i gene: Codes for the repressor protein.
• Gene z: Codes for beta-galactosidase (hydrolyses lactose).
• Gene y: Codes for permease (increases permeability of the cell to beta-galactosides).
• Gene a: Codes for transacetylase.

Mechanism:

• In the absence of inducer (Lactose): The repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon. Hence, the genes are "switched off".
• In the presence of inducer (Lactose or Allolactose): The repressor interacts with the inducer and becomes inactivated. As a result, the repressor cannot bind to the operator region. The RNA polymerase binds to the promoter and transcription proceeds. The genes are "switched on".

This is an example of negative regulation.

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Topic 7: Human and Rice Genome Projects

Q12. What are the major goals and findings of the Human Genome Project (HGP)?

Answer:

Goals: Identify all the genes in human DNA (approx. 20,000-25,000), determine the sequences of the 3 billion chemical base pairs, store this information in databases, and improve tools for data analysis.

Salient Features/Findings:

• The human genome contains 3164.7 million nucleotide bases.
• The average gene consists of 3000 bases, but sizes vary greatly. The largest known human gene is dystrophin (2.4 million bases).
• Less than 2% of the genome codes for proteins.
• Repeated sequences make up a very large portion of the human genome.
• Chromosome 1 has most genes (2968), and the Y has the fewest (231).
• Scientists have identified about 1.4 million locations where single base DNA differences (SNPs - Single Nucleotide Polymorphisms) occur in humans.

Q13. Write a brief note on the Rice Genome Project.

Answer: The International Rice Genome Sequencing Project (IRGSP) was a consortium of public laboratories. Rice was chosen because it has a relatively small genome compared to other cereals (approx. 400-430 million base pairs) and it serves as a model system for monocots. Sequencing the rice genome helps in understanding the genetics of crop yield, disease resistance, and adaptability, which is crucial for global food security.

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Topic 8: DNA Fingerprinting

Q14. Explain the principle and steps involved in DNA Fingerprinting.

Answer: DNA fingerprinting is a technique used to identify individuals based on their DNA profiles. It was developed by Alec Jeffreys.

Principle: It involves identifying differences in some specific regions in DNA sequence called repetitive DNA. A small stretch of DNA is repeated many times (Satellite DNA). These sequences show a high degree of polymorphism and form the basis of DNA fingerprinting. The specific type used is VNTR (Variable Number of Tandem Repeats).

Steps:

1. Isolation: Isolation of DNA from the sample (blood, hair follicle, skin, etc.).
2. Digestion: Digestion of DNA by restriction endonucleases.
3. Separation: Separation of DNA fragments by electrophoresis.
4. Blotting: Transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon (Southern Blotting).
5. Hybridization: Hybridization using labeled VNTR probe.
6. Detection: Detection of hybridized DNA fragments by autoradiography.

The autoradiogram gives a characteristic pattern of bands which is unique for every individual (except monozygotic twins).

Q15. What are the applications of DNA Fingerprinting?

Answer:

• Forensic Science: To identify criminals in rape and murder cases.
• Paternity Tests: To settle paternity disputes by comparing the DNA of the child with the parents.
• Evolutionary Biology: To determine population and genetic diversities and evolutionary history.

Note: These questions are curated based on the most frequently asked topics in board exams and competitive entrance tests. Ensure you practice the diagrams for the Double Helix, Nucleosome, Replication Fork, and Lac Operon.

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Very Short Answer Type Questions (VSAQ) - 1 Mark

Q1. Name the linkage joining two nucleotides in a polynucleotide chain.
Ans. 3'-5' Phosphodiester linkage.

Q2. Which nitrogenous base is present in DNA but absent in RNA?
Ans. Thymine.

Q3. Who discovered the nucleic acid and named it 'Nuclein'?
Ans. Friedrich Miescher (1869).

Q4. What is the net charge of histone proteins?
Ans. Positive.

Q5. Which two amino acids are abundant in histones?
Ans. Lysine and Arginine.

Q6. Name the enzyme responsible for DNA replication in E. coli.
Ans. DNA-dependent DNA polymerase.

Q7. What are Okazaki fragments?
Ans. Short DNA fragments synthesized on the lagging strand during DNA replication.

Q8. Which enzyme joins the Okazaki fragments?
Ans. DNA Ligase.

Q9. In which direction does DNA polymerization occur?
Ans. 5' to 3' direction.

Q10. Which factor associates with RNA polymerase to initiate transcription?
Ans. Sigma factor (σ).

Q11. Name the enzyme responsible for transcription.
Ans. DNA-dependent RNA polymerase.

Q12. What is the coding strand?
Ans. The strand of DNA which does not code for anything but has a sequence same as mRNA (except T instead of U).

Q13. Which codon acts as the initiator codon?
Ans. AUG.

Q14. What are the three stop codons?
Ans. UAA, UAG, and UGA.

Q15. Name the amino acid coded by AUG.
Ans. Methionine.

Q16. Which RNA carries the amino acid to the ribosome?
Ans. tRNA (transfer RNA).

Q17. What is the shape of tRNA in 2D structure?
Ans. Clover-leaf shape.

Q18. What are UTRs?
Ans. Untranslated Regions present at both 5' end and 3' end of mRNA.

Q19. Name the inducer in the Lac Operon.
Ans. Lactose (or Allolactose).

Q20. Which gene codes for the repressor protein in Lac Operon?
Ans. The i gene.

Q21. What is the function of Beta-galactosidase?
Ans. It hydrolyses lactose into glucose and galactose.

Q22. Who proposed the 'Central Dogma'?
Ans. Francis Crick.

Q23. Which chromosome in humans has the fewest genes?
Ans. Chromosome Y (231 genes).

Q24. What is the full form of VNTR?
Ans. Variable Number of Tandem Repeats.

Q25. Which technique is used to separate DNA fragments?
Ans. Gel Electrophoresis.

Q26. Name the blotting technique used in DNA fingerprinting.
Ans. Southern Blotting.

Q27. What is a 'Probe' in DNA fingerprinting?
Ans. A radioactive labelled single-stranded DNA molecule used to detect specific sequences.

Q28. Name the organism used in the Meselson and Stahl experiment.
Ans. E. coli.

Q29. What is Heterochromatin?
Ans. The densely packed and transcriptionally inactive region of chromatin.

Q30. Which bond links the sugar and base in DNA?
Ans. N-glycosidic linkage.

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Short Answer Type Questions (SAQ) - 2 to 3 Marks

Q1. Differentiate between Nucleoside and Nucleotide.
Ans. A nucleoside consists of a nitrogenous base linked to a pentose sugar. A nucleotide consists of a nucleoside linked to a phosphate group.

Q2. State Chargaff's Rule.
Ans. For a double-stranded DNA, the ratios between Adenine and Thymine and Guanine and Cytosine are constant and equal to one. (A=T and G=C; A+G = T+C).

Q3. Why is RNA less stable than DNA?
Ans. The 2'-OH group present at every nucleotide in RNA is a reactive group and makes RNA labile and easily degradable. Also, the presence of Uracil instead of Thymine makes RNA less stable.

Q4. What is the function of DNA Ligase?
Ans. DNA Ligase acts as a molecular glue. It joins the discontinuous Okazaki fragments on the lagging strand during DNA replication by forming phosphodiester bonds.

Q5. Distinguish between Euchromatin and Heterochromatin.
Ans. Euchromatin is loosely packed, stains light, and is transcriptionally active. Heterochromatin is densely packed, stains dark, and is transcriptionally inactive.

Q6. What is the role of 'ori' (origin of replication)?
Ans. It is a specific sequence of DNA where replication initiates. Any piece of DNA linked to this sequence can be made to replicate within the host cells.

Q7. Why is both strands of DNA not copied during transcription?
Ans. If both strands act as templates, they would code for RNA molecules with different sequences, leading to two different proteins. Also, the two RNA molecules produced would be complementary to each other and form dsRNA, preventing translation.

Q8. What is a Polycistronic structural gene?
Ans. It is a gene that codes for more than one polypeptide chain. It is mostly found in prokaryotes (bacteria).

Q9. Explain the term 'Degenerate' in Genetic Code.
Ans. Some amino acids are coded by more than one codon. For example, Serine is coded by UCU, UCC, UCA, UCG, AGU, and AGC. This property is called degeneracy.

Q10. What is a Point Mutation? Give an example.
Ans. Mutation arising due to a change in a single base pair of DNA is called point mutation. Example: Sickle cell anemia (GAG to GUG).

Q11. What is the function of aminoacyl-tRNA synthetase?
Ans. This enzyme catalyzes the activation of amino acids and attaches them to their specific tRNA molecules (charging of tRNA) using ATP.

Q12. What is a Ribozyme?
Ans. An RNA molecule that acts as an enzyme and catalyzes biochemical reactions is called a Ribozyme (e.g., 23S rRNA in bacteria acts as a ribozyme for peptide bond formation).

Q13. Briefly explain the function of the Promoter gene.
Ans. The promoter is a DNA sequence located upstream of the structural gene. It provides the binding site for RNA polymerase to initiate transcription.

Q14. What are Constitutive genes?
Ans. These are genes that are constantly expressed in a cell because their products are needed for basic cellular functions (housekeeping genes).

Q15. Why is the Lac Operon called an inducible operon?
Ans. It is normally "off" but can be turned "on" in the presence of an inducer (lactose). Hence, it is called an inducible operon.

Q16. What is the role of Permease in Lac Operon?
Ans. Permease facilitates the entry of lactose into the bacterial cell. Without permease, lactose cannot enter to act as an inducer.

Q17. Define SNPs (Single Nucleotide Polymorphisms).
Ans. SNPs are single base differences in the DNA sequence that occur in the population. They are useful in finding chromosomal locations for disease-associated sequences.

Q18. What are ESTs (Expressed Sequence Tags)?
Ans. It is an approach in genome sequencing where only those genes that are expressed as RNA are identified and sequenced.

Q19. What is Sequence Annotation?
Ans. It is the approach of sequencing the whole genome (coding and non-coding) and later assigning different regions in the sequence with functions.

Q20. What is Satellite DNA?
Ans. These are repetitive DNA sequences that do not code for proteins and form a large portion of the human genome. They show high polymorphism and are the basis of DNA fingerprinting.

Q21. Why is DNA replication discontinuous on the lagging strand?
Ans. DNA polymerase can only polymerize in the 5' -> 3' direction. On the template strand with 5' -> 3' polarity, replication must occur away from the fork in short segments to maintain the 5' -> 3' synthesis direction.

Q22. What happens during the 'Capping' of hnRNA?
Ans. An unusual nucleotide (methyl guanosine triphosphate) is added to the 5' end of the primary RNA transcript to protect it from degradation.

Q23. What is the significance of the Poly-A tail?
Ans. The addition of adenylate residues at the 3' end (Poly-A tail) protects the mRNA from enzymatic degradation and helps in its transport from the nucleus to the cytoplasm.

Q24. Why did Hershey and Chase use radioactive Sulfur (³⁵S)?
Ans. They used ³⁵S to label proteins because sulfur is present in proteins (amino acids cysteine and methionine) but not in DNA.

Q25. What is Reverse Transcription?
Ans. The process of synthesizing DNA from an RNA template using the enzyme Reverse Transcriptase. It is seen in retroviruses (like HIV).

Q26. What is a Frame-shift mutation?
Ans. Insertion or deletion of one or two bases changes the reading frame of the genetic code from the point of insertion or deletion, leading to a completely different protein.

Q27. Why are tRNA molecules called 'Adapter Molecules'?
Ans. Because they read the code on mRNA on one hand (anticodon loop) and bind to specific amino acids on the other hand (acceptor arm).

Q28. What is the difference between Template strand and Coding strand?
Ans. The template strand (3'->5') acts as a template for RNA synthesis. The coding strand (5'->3') does not code for RNA but its sequence matches the RNA sequence.

Q29. What is the role of the Rho factor?
Ans. The Rho factor binds to the RNA polymerase when it reaches the terminator region and facilitates the termination of transcription and release of the RNA.

Q30. Mention two applications of the Human Genome Project.
Ans. 1. Diagnosis and treatment of genetic disorders. 2. Understanding human evolution and migration patterns.