Molecular Basis of Inheritance Class 12 NCERT Notes & Solutions

Structure of DNA and the Polynucleotide Chain

DNA (Deoxyribonucleic acid) is a long polymer of deoxyribonucleotides. A nucleotide has three components: a nitrogenous base (Purines: Adenine, Guanine; Pyrimidines: Cytosine, Thymine), a pentose sugar, and a phosphate group. Nucleotides are linked by 3'-5' phosphodiester linkages to form a polynucleotide chain. According to Watson and Crick's Double Helix model: 1. DNA is composed of two polynucleotide chains running antiparallel (one 5' to 3', the other 3' to 5'). 2. The backbone is formed by sugar-phosphate, and bases project inside. 3. Adenine pairs with Thymine via two hydrogen bonds (A=T), while Guanine pairs with Cytosine via three hydrogen bonds (G=C). 4. The pitch of the helix is 3.4 nm with roughly 10 base pairs per turn. In eukaryotes, negatively charged DNA wraps around positively charged histone octamer proteins (rich in lysine and arginine residues) to form 'nucleosomes', which further pack to form chromatin fibers.

Mechanism of DNA Replication

1. Activation and Origin — Replication begins at a specific site called the Origin of Replication (ori). Deoxyribonucleoside triphosphates (dNTPs) serve a dual purpose: acting as substrates and providing energy via high-energy phosphate bonds.
2. Unwinding of Helix — The enzyme DNA Helicase unwinds the double helix, creating a Y-shaped replication fork. SSBP (Single-Stranded Binding Proteins) stabilize the unwound strands, and Topoisomerase relieves torsional strain.
3. Primer Synthesis — RNA Primase synthesizes a short RNA primer sequence because DNA polymerase cannot initiate DNA synthesis on its own; it can only add nucleotides to a pre-existing 3'-OH end.
4. Elongation (Continuous & Discontinuous) — DNA Polymerase III synthesizes new DNA only in the 5' to 3' direction. On the template strand with 3' to 5' polarity, synthesis is continuous (leading strand). On the template strand with 5' to 3' polarity, synthesis is discontinuous, forming short Okazaki fragments (lagging strand).
5. Ligation — The RNA primers are replaced with DNA by DNA Polymerase I, and DNA Ligase seals the nicks between Okazaki fragments to complete the continuous double-stranded molecule.

Key Concepts of the Central Dogma

Central Dogma

Proposed by Francis Crick, it states that genetic information flows from DNA -> RNA -> Protein.

Transcription

The process of copying genetic information from a template strand of DNA into a complementary RNA molecule by RNA Polymerase.

Genetic Code

The sequence of nucleotides in DNA or RNA that determines the amino acid sequence in protein synthesis. It is triplet, universal, degenerate, and non-overlapping.

Translation

The process by which ribosomes polymerize amino acids in a sequence determined by the codons on mRNA to form a polypeptide.

Lac Operon

A transcriptionally regulated system in E. coli where a polycistronic structural gene is regulated by a common promoter and a regulatory gene.

Critical Exam Tips & Board Pitfalls

Pay close attention to these high-yield zones for CBSE exams:

1. Polarity Confusion: Remember that DNA Polymerase can ONLY polymerize in the 5' -> 3' direction. Consequently, the template strand with 3' -> 5' polarity produces the leading strand, while the 5' -> 3' template strand produces the lagging strand (Okazaki fragments).

1. Coding vs Template Strand: During transcription, the template strand (3' -> 5') is read by RNA Polymerase. However, the sequence of the newly synthesized mRNA is identical to the coding strand (5' -> 3'), except Thymine (T) is replaced by Uracil (U). Always write your answers with correct polarities.

1. Chargaff's Rule Applicability: $A+G = T+C$ or $\frac{A+T}{G+C}$ is only constant for double-stranded DNA. If a question mentions single-stranded DNA or RNA, Chargaff's rule does not apply.

Practice Questions with Solutions

• Q: A double-stranded DNA has 20% Cytosine. Calculate the percentage of Adenine in this DNA. A: Step 1: Recall Chargaff's Rule, which states that for double-stranded DNA, the concentration of Cytosine (C) equals Guanine (G), and Adenine (A) equals Thymine (T). Step 2: Given Cytosine (C) = 20%, therefore Guanine (G) must also be 20%. Step 3: The total percentage of Guanine and Cytosine combined is 20% + 20% = 40%. Step 4: The remaining percentage of bases (A + T) is 100% - 40% = 60%. Step 5: Since A = T, the percentage of Adenine is half of the remaining percentage: 60% / 2 = 30%. Final answer: The percentage of Adenine in the DNA is 30%.
• Q: Explain the structure and transcriptional role of a Transcription Unit in DNA. A: Step 1: Identify the components. A transcription unit in DNA consists of three main regions: a Promoter, the Structural Gene, and a Terminator. Step 2: Define the Promoter. It is located at the 5' end (upstream) of the structural gene (with reference to the coding strand) and provides the binding site for RNA polymerase. Step 3: Define the Structural Gene. It contains the template strand (3' to 5' direction) from which transcription occurs, and the coding strand (5' to 3' direction) whose sequence matches the RNA transcript (with U instead of T). Step 4: Define the Terminator. It is located at the 3' end (downstream) of the coding strand and defines the end of the transcription process. Final answer: A transcription unit contains a promoter (RNA polymerase binding site), structural gene (template for mRNA), and terminator (signals process end).
• Q: Differentiate between repetitive DNA and satellite DNA in genomic analysis. A: Step 1: Define repetitive DNA. These are DNA sequences that contain highly repeated nitrogenous base sequences, occurring hundreds to thousands of times in the genome. Step 2: Define satellite DNA. This is a specific class of repetitive DNA that separates as a distinct satellite peak during density gradient centrifugation due to a different base composition (either AT-rich or GC-rich). Step 3: State their function and utility. While repetitive DNA forms a major bulk of the eukaryotic genome, satellite DNA does not code for proteins but shows a high degree of polymorphism, forming the basis of DNA fingerprinting. Final answer: Repetitive DNA refers to any sequence repeated multiple times; satellite DNA is a subcategory of highly repetitive DNA that separates as distinct peaks during density gradient centrifugation.
• Q: Explain the regulatory mechanism of the Lac Operon when lactose is present in the medium. A: Step 1: Identify the role of lactose. Lactose acts as an 'inducer' in the Lac Operon system. Step 2: Describe the action of the regulatory gene. The regulatory 'i' gene continuously transcribes and translates an active Repressor protein. Step 3: Explain the binding of the inducer. When lactose is present, it binds to the repressor protein, causing a conformational change that inactivates the repressor. Step 4: Explain transcription initiation. The inactive repressor can no longer bind to the operator region. This allows RNA Polymerase free access to bind to the promoter and transcribe the structural genes (lacZ, lacY, lacA). Step 5: Identify the enzymes produced. Transcription produces polycistronic mRNA, which translates into Beta-galactosidase (z), Permease (y), and Transacetylase (a) to metabolize lactose. Final answer: In the presence of lactose, the inducer inactivates the repressor, allowing RNA polymerase to transcribe structural genes (z, y, a) for lactose metabolism.

Frequently Asked Questions