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Molecular Basis of Inheritance

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Chapter Notes

Overview
This chapter explains the molecular nature of genetic material, focusing on DNA as the primary genetic material. It covers the structure of DNA, its replication, transcription, translation, the genetic code, and regulation of gene expression. The one big idea a student must walk away with is that DNA stores genetic information, which is expressed through RNA to produce proteins, and this flow of information is central to all life processes.
Key Concepts (37)
DNA
Deoxyribonucleic acid, a long polymer of deoxyribonucleotides that acts as the genetic material in most organisms.
e.g. Human DNA contains about 3.3 × 10⁹ base pairs in its haploid genome.
RNA
Ribonucleic acid, a polymer of ribonucleotides that acts as a messenger, adapter, structural, and catalytic molecule, and as genetic material in some viruses.
e.g. mRNA carries genetic information from DNA to ribosomes for protein synthesis.
Nucleotide
The monomer unit of nucleic acids, consisting of a nitrogenous base, a pentose sugar (ribose or deoxyribose), and a phosphate group.
e.g. Adenine nucleotide has adenine base, deoxyribose sugar, and a phosphate group.
Nucleoside
A nitrogenous base linked to the 1' carbon of a pentose sugar through a N-glycosidic linkage, without a phosphate group.
e.g. Adenosine is a nucleoside formed by adenine and ribose.
Phosphodiester linkage
The bond between the 3' carbon of one sugar and the 5' carbon of the next sugar in a polynucleotide chain, forming the backbone.
e.g. Two nucleotides are joined by a 3'-5' phosphodiester linkage to form a dinucleotide.
Double helix
The three-dimensional structure of DNA proposed by Watson and Crick, consisting of two antiparallel polynucleotide chains coiled around each other with complementary base pairing.
e.g. The two strands of DNA are held together by hydrogen bonds between A-T and G-C base pairs.
Base pairing
The specific hydrogen bonding between adenine and thymine (two bonds) and between guanine and cytosine (three bonds) in DNA.
e.g. In DNA, A always pairs with T, and G always pairs with C.
Antiparallel polarity
The orientation of the two DNA strands such that one runs 5' → 3' and the other runs 3' → 5'.
e.g. If one strand has 5'-ATGC-3', the complementary strand runs 3'-TACG-5'.
Nucleosome
The repeating structural unit of chromatin, consisting of DNA wrapped around a histone octamer (eight histone proteins).
e.g. A typical nucleosome contains about 200 bp of DNA.
Euchromatin
Loosely packed chromatin that stains light and is transcriptionally active.
e.g. Genes in euchromatin are expressed because the DNA is accessible to RNA polymerase.
Heterochromatin
Densely packed chromatin that stains dark and is transcriptionally inactive.
e.g. Centromeric regions are often heterochromatic.
Transforming principle
The substance (later identified as DNA) that can transform non-virulent bacteria into virulent ones, as shown by Griffith's experiment.
e.g. Heat-killed S strain bacteria transformed live R strain bacteria into S strain.
Bacteriophage
A virus that infects bacteria, used by Hershey and Chase to prove DNA is the genetic material.
e.g. T2 bacteriophage was used in the Hershey-Chase experiment.
Semiconservative replication
DNA replication where each daughter DNA molecule has one parental strand and one newly synthesized strand.
e.g. Meselson and Stahl's experiment proved semiconservative replication using ¹⁵N and ¹⁴N isotopes.
Replication fork
The Y-shaped region where DNA replication occurs, with the two strands separated.
e.g. DNA polymerase works at the replication fork to synthesize new strands.
DNA-dependent DNA polymerase
The enzyme that catalyzes the polymerization of deoxynucleotides using a DNA template, with 5' → 3' directionality.
e.g. In E. coli, DNA polymerase III is the main replicative enzyme.
Okazaki fragments
Short, discontinuous DNA fragments synthesized on the lagging strand during replication.
e.g. Okazaki fragments are later joined by DNA ligase.
Transcription
The process of copying genetic information from DNA into RNA, following the rule of complementarity (A-U, T-A, G-C, C-G).
e.g. RNA polymerase transcribes a gene to produce mRNA.
Template strand
The DNA strand with polarity 3' → 5' that serves as the template for RNA synthesis during transcription.
e.g. The template strand is complementary to the RNA transcript.
Coding strand
The DNA strand with polarity 5' → 3' that has the same sequence as the RNA (except T instead of U).
e.g. The coding strand is not transcribed but is used as a reference.
Promoter
A DNA sequence located upstream of the structural gene that provides the binding site for RNA polymerase.
e.g. The lac operon has a promoter where RNA polymerase binds to initiate transcription.
Terminator
A DNA sequence located downstream of the structural gene that defines the end of transcription.
e.g. The terminator region causes RNA polymerase to detach and release the RNA transcript.
Intron
Intervening sequences in eukaryotic genes that are removed during RNA splicing and do not appear in the mature mRNA.
e.g. In the beta-globin gene, introns are spliced out to form functional mRNA.
Exon
Coding sequences in eukaryotic genes that appear in the mature or processed RNA after splicing.
e.g. Exons are joined together to form the final mRNA sequence.
Genetic code
The set of rules by which the sequence of nucleotides in mRNA is translated into the sequence of amino acids in a polypeptide.
e.g. The codon AUG codes for methionine and also acts as the start codon.
Codon
A triplet of nucleotides in mRNA that specifies a particular amino acid or a stop signal.
e.g. UUU codes for phenylalanine.
Degeneracy of genetic code
The property that some amino acids are coded by more than one codon.
e.g. Leucine is coded by six different codons (UUA, UUG, CUU, CUC, CUA, CUG).
tRNA (transfer RNA)
An adapter molecule that carries specific amino acids to the ribosome and has an anticodon loop complementary to the mRNA codon.
e.g. tRNA with anticodon UAC carries methionine and binds to AUG on mRNA.
Anticodon
A triplet of nucleotides on tRNA that is complementary to the codon on mRNA.
e.g. The anticodon UAC pairs with the start codon AUG.
Translation
The process of polymerizing amino acids into a polypeptide chain based on the sequence of codons in mRNA.
e.g. Ribosomes translate mRNA to synthesize proteins like beta-galactosidase.
Ribosome
The cellular factory for protein synthesis, consisting of a large and a small subunit, and acting as a catalyst (ribozyme) for peptide bond formation.
e.g. In bacteria, 23S rRNA in the large subunit catalyzes peptide bond formation.
Operon
A cluster of genes transcribed as a single mRNA from a common promoter, regulated by a common operator, common in bacteria.
e.g. The lac operon includes z, y, and a genes for lactose metabolism.
Lac operon
An inducible operon that controls the metabolism of lactose in E. coli, consisting of a regulatory gene (i), promoter, operator, and structural genes (z, y, a).
e.g. In the presence of lactose, the repressor is inactivated, allowing transcription.
Repressor protein
A protein that binds to the operator region and prevents RNA polymerase from transcribing the operon.
e.g. The lac repressor binds to the lac operator in the absence of lactose.
Inducer
A molecule that inactivates the repressor, allowing transcription of the operon.
e.g. Lactose or allolactose acts as an inducer for the lac operon.
Human Genome Project (HGP)
A mega project launched in 1990 to determine the complete DNA sequence of the human genome, identify all genes, and develop tools for data analysis.
e.g. HGP identified approximately 20,000-25,000 genes in human DNA.
DNA fingerprinting
A technique used to identify individuals based on differences in their DNA sequences, often using variable number tandem repeats (VNTRs).
e.g. DNA fingerprinting is used in forensic science to solve crimes.
Formulas & Equations (7)
Number of base pairs = Length of DNA (nm) / 0.34 nm per bp
Reads as: Number of base pairs equals length of DNA in nanometres divided by 0.34 nanometres per base pair.
Variables: Length of DNA (nm): the total length of the DNA double helix in nanometres; 0.34 nm: distance between consecutive base pairs.
Use when: Calculating the number of base pairs from the length of DNA, or vice versa.
⚠ Forgetting to convert units (e.g., using metres instead of nanometres).
Pitch of DNA helix = 3.4 nm per turn
Reads as: The pitch of the DNA helix is 3.4 nanometres per turn.
Variables: Pitch: the distance for one complete turn of the helix (3.4 nm); approximately 10 base pairs per turn.
Use when: Describing the structural parameters of the DNA double helix.
⚠ Confusing pitch with the distance between base pairs (0.34 nm).
Number of nucleosomes = Total base pairs / 200 bp per nucleosome
Reads as: Number of nucleosomes equals total base pairs divided by 200 base pairs per nucleosome.
Variables: Total base pairs: the length of DNA in base pairs; 200 bp: typical DNA length wrapped around one nucleosome.
Use when: Estimating the number of nucleosomes in a given DNA molecule.
⚠ Assuming all DNA is packaged into nucleosomes (some regions are not).
Number of codons = Number of nucleotides in mRNA / 3
Reads as: Number of codons equals number of nucleotides in mRNA divided by 3.
Variables: Number of nucleotides: total bases in the coding region of mRNA; 3: triplet nature of the genetic code.
Use when: Calculating the number of amino acids in a polypeptide from mRNA length.
⚠ Forgetting to exclude stop codons when counting amino acids.
Number of possible codons = 4³ = 64
Reads as: Number of possible codons equals 4 raised to the power 3, which is 64.
Variables: 4: number of different bases (A, U, G, C); 3: number of bases per codon.
Use when: Explaining why the genetic code has 64 codons for 20 amino acids.
⚠ Thinking that all 64 codons code for amino acids (3 are stop codons).
Chargaff's rule: A = T and G = C (for double-stranded DNA)
Reads as: In double-stranded DNA, the amount of adenine equals thymine, and guanine equals cytosine.
Variables: A: adenine; T: thymine; G: guanine; C: cytosine.
Use when: Predicting base composition or verifying DNA structure.
⚠ Applying Chargaff's rule to single-stranded DNA or RNA.
Length of DNA (m) = Number of base pairs × 0.34 × 10⁻⁹ m/bp
Reads as: Length of DNA in metres equals number of base pairs multiplied by 0.34 times 10 to the power minus 9 metres per base pair.
Variables: Number of base pairs: total bp; 0.34 × 10⁻⁹ m: distance between consecutive bp in metres.
Use when: Calculating the physical length of a DNA molecule from its base pair count.
⚠ Using 0.34 nm directly without converting to metres.
How It Works
DNA Replication (Semiconservative)
  1. Step 1: The two strands of the parental DNA separate at the replication fork, exposing the bases.
  2. Step 2: Each strand acts as a template for the synthesis of a new complementary strand.
  3. Step 3: DNA-dependent DNA polymerase adds deoxyribonucleotides in the 5' → 3' direction, using deoxyribonucleoside triphosphates as substrates and energy source.
  4. Step 4: On the leading strand (template 3' → 5'), replication is continuous; on the lagging strand (template 5' → 3'), replication is discontinuous, forming Okazaki fragments.
  5. Step 5: DNA ligase joins the Okazaki fragments to form a continuous strand.
  6. Step 6: The result is two daughter DNA molecules, each with one parental and one newly synthesized strand.
💡 Replication is semiconservative, not conservative or dispersive. The Meselson-Stahl experiment proved this using ¹⁵N and ¹⁴N isotopes.
Transcription in Bacteria
  1. Step 1: RNA polymerase binds to the promoter region of the DNA with the help of initiation factor (σ).
  2. Step 2: The enzyme opens the DNA helix and begins synthesizing RNA in the 5' → 3' direction, using the template strand (3' → 5').
  3. Step 3: Elongation continues as RNA polymerase moves along the template, adding ribonucleotides complementary to the template.
  4. Step 4: When RNA polymerase reaches the terminator region, termination factor (ρ) helps release the nascent RNA and the enzyme.
  5. Step 5: The RNA transcript is released and is functional without further processing in bacteria.
💡 Only one strand (template strand) is transcribed; the coding strand is not used. Transcription and translation can be coupled in bacteria.
Transcription in Eukaryotes
  1. Step 1: RNA polymerase II binds to the promoter of a gene and initiates transcription of hnRNA (pre-mRNA).
  2. Step 2: Elongation proceeds, producing a primary transcript containing both exons and introns.
  3. Step 3: Capping: A methyl guanosine triphosphate is added to the 5' end of the hnRNA.
  4. Step 4: Tailing: Adenylate residues (200-300) are added to the 3' end in a template-independent manner.
  5. Step 5: Splicing: Introns are removed and exons are joined together to form mature mRNA.
  6. Step 6: The mature mRNA is transported out of the nucleus to the cytoplasm for translation.
💡 Eukaryotic transcription involves three RNA polymerases (I, II, III) with distinct roles. RNA polymerase II transcribes mRNA precursors.
Translation (Protein Synthesis)
  1. Step 1: Activation of amino acids: Amino acids are activated in the presence of ATP and linked to their specific tRNA (charging/aminoacylation).
  2. Step 2: Initiation: The small ribosomal subunit binds to mRNA at the start codon (AUG). The initiator tRNA (carrying methionine) binds to the start codon.
  3. Step 3: The large ribosomal subunit joins, forming a functional ribosome with two sites (A and P sites).
  4. Step 4: Elongation: tRNAs with amino acids sequentially bind to the A site via codon-anticodon pairing. A peptide bond forms between adjacent amino acids, catalyzed by the ribosome (23S rRNA in bacteria).
  5. Step 5: The ribosome moves along the mRNA (translocation), and the process repeats, adding amino acids one by one.
  6. Step 6: Termination: When a stop codon (UAA, UAG, UGA) is reached, a release factor binds, and the complete polypeptide is released from the ribosome.
💡 The ribosome acts as a ribozyme (catalytic RNA) for peptide bond formation. UTRs (untranslated regions) are present at both ends of mRNA for efficient translation.
Regulation of Lac Operon (Negative Regulation)
  1. Step 1: In the absence of lactose, the repressor protein (synthesized from the i gene) binds to the operator region.
  2. Step 2: The repressor-operator complex prevents RNA polymerase from binding to the promoter, so transcription is switched off.
  3. Step 3: When lactose is present, it is converted to allolactose, which binds to the repressor and inactivates it.
  4. Step 4: The inactive repressor cannot bind to the operator, allowing RNA polymerase to access the promoter and transcribe the structural genes (z, y, a).
  5. Step 5: The enzymes beta-galactosidase, permease, and transacetylase are synthesized to metabolize lactose.
💡 Lactose acts as an inducer, not a repressor. Glucose or galactose cannot induce the lac operon. This is negative regulation because the repressor blocks transcription.
Hershey-Chase Experiment
  1. Step 1: Bacteriophages were grown in media containing radioactive ³²P (to label DNA) or ³⁵S (to label protein).
  2. Step 2: The radioactive phages were allowed to infect E. coli bacteria.
  3. Step 3: After infection, the viral coats were removed from the bacteria using a blender.
  4. Step 4: The virus particles and bacteria were separated by centrifugation.
  5. Step 5: Bacteria infected with ³²P-labeled phages were radioactive, indicating DNA entered the bacteria.
  6. Step 6: Bacteria infected with ³⁵S-labeled phages were not radioactive, indicating protein did not enter.
  7. Step 7: Conclusion: DNA is the genetic material that is passed from virus to bacteria.
💡 DNA contains phosphorus (³²P labels DNA), protein contains sulfur (³⁵S labels protein). This experiment provided unequivocal proof that DNA is the genetic material.
Meselson-Stahl Experiment
  1. Step 1: E. coli were grown in a medium containing ¹⁵NH₄Cl (heavy nitrogen) for many generations, so all DNA became heavy (¹⁵N-labeled).
  2. Step 2: The cells were transferred to a medium with normal ¹⁴NH₄Cl.
  3. Step 3: DNA was extracted at various time intervals (0, 20, 40 minutes) and centrifuged in a CsCl density gradient.
  4. Step 4: After one generation (20 min), DNA had hybrid density (one ¹⁵N strand and one ¹⁴N strand).
  5. Step 5: After two generations (40 min), equal amounts of hybrid DNA and light DNA (both strands ¹⁴N) were observed.
  6. Step 6: This pattern matched semiconservative replication, not conservative or dispersive.
💡 The experiment used density gradient centrifugation, not radioactivity. The hybrid band after one generation proved semiconservative replication.
Exam Traps — Don't Make These Mistakes
⚠ Confusing template strand and coding strand
✗ The coding strand is used as a template for transcription.
→ The template strand (3' → 5') is actually used for transcription; the coding strand (5' → 3') has the same sequence as the RNA (except T/U) and is not transcribed.
✓ The template strand (3' → 5') serves as the template for RNA synthesis, while the coding strand (5' → 3') is not transcribed and has the same sequence as the RNA (except thymine replaced by uracil).
⚠ Thinking both DNA strands are transcribed
✗ Both strands of DNA are transcribed to produce two different RNA molecules.
→ Only one strand (the template strand) is transcribed for a given gene. Transcribing both strands would produce complementary RNAs that would form double-stranded RNA, preventing translation.
✓ Only one strand of DNA, the template strand, is transcribed into RNA. The other strand, the coding strand, is not transcribed.
⚠ Confusing introns and exons
✗ Introns are coding sequences that appear in mature mRNA.
→ Introns are non-coding intervening sequences that are removed during splicing and do not appear in mature mRNA. Exons are the coding sequences that appear in mature mRNA.
✓ Introns are intervening sequences that are removed during RNA splicing, while exons are expressed sequences that are joined together to form the mature mRNA.
⚠ Misunderstanding the role of the repressor in lac operon
✗ The repressor binds to the promoter to block transcription.
→ The repressor binds to the operator region, not the promoter. The operator is adjacent to the promoter, and when the repressor binds, it physically blocks RNA polymerase from accessing the promoter.
✓ The repressor protein binds to the operator region of the lac operon, preventing RNA polymerase from transcribing the structural genes.
⚠ Thinking DNA replication is conservative
✗ In DNA replication, the parental DNA remains intact and an entirely new copy is made.
→ DNA replication is semiconservative, meaning each daughter DNA molecule has one parental strand and one newly synthesized strand. The Meselson-Stahl experiment proved this.
✓ DNA replication is semiconservative: each daughter DNA molecule consists of one parental strand and one newly synthesized strand.
⚠ Confusing the roles of RNA polymerases in eukaryotes
✗ RNA polymerase II transcribes rRNA.
→ RNA polymerase I transcribes rRNA (28S, 18S, 5.8S), while RNA polymerase II transcribes hnRNA (pre-mRNA). RNA polymerase III transcribes tRNA, 5S rRNA, and snRNA.
✓ In eukaryotes, RNA polymerase I transcribes rRNA, RNA polymerase II transcribes hnRNA (pre-mRNA), and RNA polymerase III transcribes tRNA, 5S rRNA, and snRNA.
Compare & Contrast
Chemical structure
DNA
RNA
DNA has deoxyribose sugar (no 2'-OH group) and thymine; RNA has ribose sugar (with 2'-OH group) and uracil instead of thymine.
Replication synthesis
Leading strand
Lagging strand
The leading strand is synthesized continuously in the 5' → 3' direction from a single primer; the lagging strand is synthesized discontinuously as Okazaki fragments, which are later joined by DNA ligase.
Packaging and activity
Euchromatin
Heterochromatin
Euchromatin is loosely packed, stains light, and is transcriptionally active; heterochromatin is densely packed, stains dark, and is transcriptionally inactive.
Presence in mature mRNA
Intron
Exon
Introns are non-coding sequences that are removed during splicing and do not appear in mature mRNA; exons are coding sequences that are joined and appear in mature mRNA.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA (3 marks), 1 LA (5 marks) — total 10 marks typically
Most asked: Structure of DNA (double helix, base pairing, antiparallel polarity) · Semiconservative replication (Meselson-Stahl experiment) · Transcription (template vs coding strand, promoters, terminators) · Genetic code (triplet nature, degeneracy, start and stop codons) · Translation (role of tRNA, ribosome, steps) · Lac operon (regulation, repressor, inducer) · Hershey-Chase experiment (proof that DNA is genetic material) · DNA fingerprinting and Human Genome Project
Don't skip: Explain the Hershey-Chase experiment and how it proved DNA is the genetic material.; Describe the process of DNA replication with the help of a diagram.; Explain the lac operon model of gene regulation.; Write the salient features of the genetic code.; Differentiate between transcription in prokaryotes and eukaryotes.
One-Line Revision (read aloud before the exam)
  1. DNA is a double helix with antiparallel strands held by hydrogen bonds between A-T (2 bonds) and G-C (3 bonds).
  2. The pitch of DNA is 3.4 nm with 10 base pairs per turn, and distance between base pairs is 0.34 nm.
  3. Chargaff's rule states that in double-stranded DNA, A = T and G = C.
  4. Nucleosomes are formed by DNA wrapped around a histone octamer, containing about 200 bp of DNA.
  5. Euchromatin is transcriptionally active; heterochromatin is transcriptionally inactive.
  6. Griffith's experiment showed the transforming principle, later identified as DNA by Avery, MacLeod, and McCarty.
  7. Hershey and Chase used radioactive ³²P and ³⁵S to prove that DNA, not protein, is the genetic material.
  8. DNA replication is semiconservative, as proved by Meselson and Stahl using ¹⁵N and ¹⁴N isotopes.
  9. DNA polymerase synthesizes DNA in the 5' → 3' direction, leading to continuous synthesis on the leading strand and discontinuous on the lagging strand.
  10. Transcription uses the template strand (3' → 5') to produce RNA, and the coding strand (5' → 3') has the same sequence as RNA (except T/U).
  11. In eukaryotes, RNA polymerase II transcribes hnRNA, which undergoes capping, tailing, and splicing to form mature mRNA.
  12. The genetic code is triplet, degenerate, non-overlapping, and nearly universal, with AUG as start codon and UAA, UAG, UGA as stop codons.
  13. tRNA acts as an adapter molecule with an anticodon loop and an amino acid acceptor end.
  14. Translation involves activation of amino acids, initiation, elongation, and termination on ribosomes.
  15. The lac operon is negatively regulated by a repressor that binds to the operator; lactose acts as an inducer.
  16. The Human Genome Project aimed to sequence the entire human genome and identified about 20,000-25,000 genes.
  17. DNA fingerprinting uses VNTRs to identify individuals based on their unique DNA sequences.

MCQs — 10 Questions with Answers

Q1. Which of the following correctly describes the composition of a nucleotide?

  • A. A nitrogenous base, a pentose sugar, and a phosphate group ✓
  • B. A nitrogenous base and a pentose sugar only
  • C. A nitrogenous base and a phosphate group only
  • D. A pentose sugar and a phosphate group only

Answer: A — A nucleotide is defined as the combination of all three components: nitrogenous base, pentose sugar (deoxyribose or ribose), and phosphate group.

Q2. In DNA, adenine always pairs with thymine through hydrogen bonds. How many hydrogen bonds form in this base pair?

  • A. One hydrogen bond
  • B. Two hydrogen bonds ✓
  • C. Three hydrogen bonds
  • D. Four hydrogen bonds

Answer: B — Adenine (purine) forms exactly two hydrogen bonds with thymine (pyrimidine), whereas guanine–cytosine pairs form three hydrogen bonds.

Q3. Which statement correctly explains why the distance between the two strands of DNA remains almost constant?

  • A. Purines always pair with pyrimidines, maintaining uniform purine–pyrimidine distance ✓
  • B. All four bases have the same size and shape
  • C. The number of hydrogen bonds is identical for all base pairs
  • D. The DNA helix always contains equal numbers of adenine and guanine

Answer: A — Purines (A, G—larger) always pair with pyrimidines (T, C—smaller), so purine–pyrimidine = purine–pyrimidine across all positions, keeping strands equidistant.

Q4. If one strand of DNA has the sequence 5′-ATGC-3′, what is the sequence of the complementary strand?

  • A. 5′-CGTA-3′
  • B. 5′-TACG-3′ ✓
  • C. 5′-ATGC-3′
  • D. 5′-GCTA-3′

Answer: B — The complementary strand is antiparallel (3′ end aligns with 5′ end), and base-pairing rules (A–T, G–C) give the reverse complement: 5′-TACG-3′.

Q5. Which of the following is NOT a correct statement about DNA structure?

  • A. The two strands are antiparallel, running in opposite 5′→3′ directions
  • B. The pitch of the double helix is 3.4 nm with approximately 10 base pairs per turn
  • C. Adenine and guanine are pyrimidines because they have two aromatic rings ✓
  • D. The sugar-phosphate backbone is located on the outside of the helix

Answer: C — Adenine and guanine are purines (two rings), not pyrimidines (one ring); cytosine, thymine, and uracil are pyrimidines.

Q6. The central dogma of molecular biology states DNA → RNA → Protein. In some viruses, this flow is reversed. What is this reverse process called?

  • A. Replication
  • B. Transcription
  • C. Reverse transcription ✓
  • D. Translation

Answer: C — Reverse transcription is the synthesis of DNA from an RNA template, occurring in retroviruses like HIV.

Q7. How does the structure of RNA differ from DNA? (Select all correct differences.)

  • A. RNA contains ribose sugar; DNA contains deoxyribose sugar
  • B. RNA contains uracil; DNA contains thymine
  • C. RNA has an –OH group at the 2′ position of the sugar; DNA has only –H
  • D. All of the above ✓

Answer: D — All three differences are accurate: ribose vs deoxyribose, uracil vs thymine, and the 2′-OH group present only in RNA.

Q8. According to Chargaff's rules, if a double-stranded DNA sample contains 30% adenine, what is the expected percentage of thymine?

  • A. 15%
  • B. 20%
  • C. 30% ✓
  • D. 40%

Answer: C — Chargaff's rule states A = T; therefore, if adenine is 30%, thymine must also be 30%.

Q9. A researcher observes that a newly discovered viral genome contains both adenine and uracil. Based on this observation, which conclusion is most reasonable?

  • A. The virus possesses DNA as its genetic material
  • B. The virus possesses RNA as its genetic material ✓
  • C. The virus possesses both DNA and RNA
  • D. The genome is denatured and cannot be identified

Answer: B — Uracil is exclusive to RNA; the presence of uracil (not thymine) indicates the genetic material is RNA, as in retroviruses.

Q10. Two DNA strands are held together by hydrogen bonds between base pairs. If you were to denature (separate) these strands by heating, how would you re-anneal (bring them back together) at a lower temperature while maintaining their original pairing?

  • A. Simply cool the solution; complementary strands will automatically find and bind to each other
  • B. Add excess thymine nucleotides to reform bonds
  • C. Add a catalyst enzyme to speed up base pairing
  • D. Adjust the salt concentration and gradually cool to allow slow, specific base-pairing ✓

Answer: D — Annealing requires gradual cooling in appropriate ionic conditions to allow complementary strands sufficient time and thermal energy to find their partners and form specific base pairs.

Flashcards

What is the 3′-5′ phosphodiester linkage?

A covalent bond linking the 3′-OH group of one sugar to the 5′-phosphate of the next, forming the sugar-phosphate backbone of DNA.

State Chargaff's rule for double-stranded DNA.

The ratio of adenine to thymine equals 1, and the ratio of guanine to cytosine equals 1 (A=T and G=C).

Why is the distance between DNA strands almost constant?

Purines always pair with pyrimidines (A with T, G with C), ensuring uniform purine–pyrimidine distance across the double helix.

What does 'antiparallel' mean in DNA strands?

One strand runs 5′→3′ while the complementary strand runs 3′→5′ in the opposite direction.

Name the two types of nitrogenous bases and give one example of each.

Purines (adenine, guanine—two rings) and pyrimidines (cytosine, thymine, uracil—one ring).

What is the pitch of the DNA double helix and how many base pairs fit per turn?

The pitch is 3.4 nm, with approximately 10 base pairs per complete turn of the helix.

Distinguish between DNA and RNA based on sugar and base composition.

DNA contains deoxyribose sugar and thymine, while RNA contains ribose sugar and uracil; RNA also has an –OH group at the 2′ position.

What is the central dogma of molecular biology?

Genetic information flows from DNA → RNA → Protein, establishing the direction of molecular information transfer.

What is a nucleoside and how does it differ from a nucleotide?

A nucleoside is a nitrogenous base linked to a pentose sugar; a nucleotide is a nucleoside with a phosphate group attached.

Why does base stacking in the DNA helix provide additional stability?

Overlapping base pairs create van der Waals forces and hydrophobic interactions that reinforce hydrogen bonding between strands.

Important Board Questions

Define a nucleotide. How is a nucleotide formed from a nucleoside? [2 marks]

State the three components of a nucleotide (base, sugar, phosphate). Explain that a nucleoside becomes a nucleotide when a phosphate group attaches to the 5′-OH of the sugar via a phosphoester linkage.

Explain how the complementary nature of the two DNA strands relates to the principle of DNA replication. Why is this property important for inheritance? [5 marks]

Describe base-pairing rules (A–T, G–C) and antiparallel orientation. Explain that if each parental strand serves as a template, the new strands synthesised will be complementary, producing two identical double-helical DNA molecules identical to the parent—ensuring faithful transmission of genetic information to daughter cells.

Discuss the salient features of the DNA double-helix model proposed by Watson and Crick. How did this structure explain the genetic implications of DNA as the hereditary material? [6 marks]

Include: two antiparallel chains with sugar-phosphate backbone and bases inside; right-handed coil (pitch 3.4 nm, ~10 bp/turn); complementary base-pairing via H-bonds (A–T, G–C); base stacking stability. Explain that complementarity allows each strand to serve as a template for accurate replication, producing genetically identical copies and explaining how traits pass unchanged from parent to offspring.

Assertion-Reason Questions for Molecular Basis of Inheritance

Q1. Assertion (A):

DNA is considered the genetic material in most organisms because its double helix structure allows each strand to serve as a template for synthesizing a complementary strand.

Reason (R):

Base complementarity (A-T and G-C pairing) in the DNA double helix ensures that if one strand is known, the other strand's sequence can be precisely predicted.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains why the double helix structure makes DNA suitable as genetic material—complementarity enables accurate replication and faithful inheritance.

Q2. Assertion (A):

RNA contains ribose sugar, while DNA contains deoxyribose sugar.

Reason (R):

The presence of the 2'-OH group in ribose makes RNA more chemically stable than DNA.

Answer: C — A is true but R is false — The assertion is correct—RNA has ribose and DNA has deoxyribose. However, the reason is false. The 2'-OH group actually makes RNA more chemically reactive and LESS stable than DNA (due to nucleophilic attack), not more stable.

Q3. Assertion (A):

Chargaff's rules state that in double-stranded DNA, the amount of adenine equals thymine and guanine equals cytosine.

Reason (R):

This equality occurs because of the specific base-pairing rules where adenine forms two hydrogen bonds with thymine and guanine forms three hydrogen bonds with cytosine.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains the basis of Chargaff's observation—complementary base pairing through hydrogen bonds creates 1:1 ratios for A:T and G:C.

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Case-Based Questions (CBQ) for Molecular Basis of Inheritance

Case Study:

A molecular biologist in India used X-ray crystallography to study the structure of DNA extracted from a legume plant (Vigna radiata). The analysis revealed a helical structure with 10 base pairs per complete turn, a pitch of 3.4 nm, and consistent base pairing where adenine always paired with thymine through two hydrogen bonds and guanine always paired with cytosine through three hydrogen bonds. The DNA showed anti-parallel strand orientation with a uniform diameter of 2.0 nm throughout the molecule.

Q1. Based on the crystallographic data, what is the distance between consecutive base pairs in this DNA molecule?

  • A. 3.4 nm
  • B. 0.34 nm ✓
  • C. 0.20 nm
  • D. 20 nm

Answer: B

Q2. Why is the uniform diameter of 2.0 nm maintained throughout the DNA molecule despite different base-pairing combinations?

  • A. All nitrogenous bases have identical molecular sizes
  • B. Purine-pyrimidine pairing ensures a purine (larger) pairs with a pyrimidine (smaller), maintaining constant width ✓
  • C. The sugar-phosphate backbone independently determines helix width
  • D. Different numbers of hydrogen bonds adjust the spacing between bases

Answer: B

Q3. In a sample of this plant DNA, analysis showed 35% adenine content. What would be the expected percentage of guanine according to Chargaff's rules?

  • A. 35%
  • B. 15%
  • C. 30% ✓
  • D. 65%

Answer: C

Q4. The anti-parallel orientation of strands in this DNA means:

  • A. The two strands repel each other magnetically
  • B. The base pairs are arranged in opposite order along the helix
  • C. One strand runs 5' to 3' while the other runs 3' to 5' ✓
  • D. Each strand contains different nitrogenous bases

Answer: C

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Previous Year Questions (PYQ) for Molecular Basis of Inheritance

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 57/1/1 · 1 mark

Which of the following statements about plasmids is incorrect?

  • Plasmids have the ability to replicate within the bacterial cell.
  • Their replication is controlled by chromosomal DNA.
  • They are autonomously replicating circular extra-chromosomal DNA.
  • They often carry antibiotic resistant genes.

CBSE 2026 · Set 57/1/1 · 2 marks

Study the sequence of bases in a DNA molecule that is shown below and answer the questions. 5' – GAATTC – 3' / 3' – CTTAAG – 5' (a) What are such sequences called? Name the enzyme used that recognises such nucleotide sequences. (1 mark) (b) How do these enzymes function? (1 mark)

CBSE 2026 · Set 57/1/1 · 2 marks

(b) (i) Differentiate between a DNA and RNA nucleotide. (1 mark) (ii) Name the two types of bonds present within a single nucleotide. (1 mark)

CBSE 2026 · Set 57/1/1 · 5 marks

(a) (i) Watson and Crick's discovery of double helical structure was based on which two findings. Also mention the name of the scientists associated with these findings. (2 marks) (ii) Write down the salient features of double helix structure of DNA (any three points). (3 marks)

CBSE 2026 · Set 57/1/2 · 1 mark

Messelson and Stahl's experiment carried out centrifugation in CsCl2 density gradient to separate:

  • DNA from RNA
  • DNA from protein
  • RNA from protein
  • Normal DNA from heavy DNA

CBSE 2026 · Set 57/1/2 · 5 marks

(a) (i) Watson and Crick's discovery of double helical structure was based on which two findings. Also mention the name of the scientists associated with these findings. (2 marks) (ii) Write down the salient features of double helix structure of DNA (any three points). (3 marks) OR (b) Given below is a stretch of DNA showing the coding strand of structural gene of transcription unit. 5' - ATG ACC GTA TTT TCT GTA GTG CCC GTA CTT CAG GCA TTA 3' (i) Write the corresponding template strand and m-RNA strand that will be transcribed along with its polarity. (2 marks) (ii) If GUA of transcribed mRNA is an intron, then depict the sequence involved in formation of mRNA / mature / processed hnRNA strand: (1) In a bacterium (1 mark) (2) In humans (1 mark) (iii) How many amino acids the resulting polypeptide will have after the process of translation in humans? (1 mark)

CBSE 2026 · Set 57/1/3 · 1 mark

A DNA molecule is 140 base pairs long. It has 20% thymine. How many guanine bases are present in the molecule?

  • 64
  • 84
  • 32
  • 56

CBSE 2026 · Set 57/1/3 · 5 marks

Describe how the lac operon operates both in the presence and absence of an inducer in E.coli.

CBSE 2026 · Set 57/2/1 · 1 mark

To separate repetitive DNA from bulk genomic DNA, which of the following method is used ?

  • Elution
  • Southern Blotting
  • Polymerase Chain Reaction
  • Density Gradient Centrifugation

CBSE 2026 · Set 57/2/1 · 1 mark

A bacterium having radioactive thymidine in its DNA is allowed to multiply in a medium having non-radioactive thymidine for two generations. What percentage of bacteria will have radioactive thymidine in its DNA ?

  • 100
  • 50
  • 75
  • 25

CBSE 2026 · Set 57/2/1 · 2 marks

(a) The diagram below shows the sequence of amino acids in part of haemoglobin molecule. (i) If the base *T is substituted with A, in 'C*TT' of DNA strand how would it affect the haemoglobin chain ? (ii) Name the condition and the effects associated with the above substitution.

CBSE 2026 · Set 57/2/1 · 3 marks

(a) What is a genetic code ? (b) Why did scientists feel the need to propose a genetic code ? (c) Genetic code is said to be 'Universal'. Why ?

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Class 12 Biology Previous Year Questions, Chapter-wise

All 1074 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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