Molecular Basis of Inheritance Class 12 Notes

DNA Structure — The Watson-Crick Model

• Chargaff's rule: In any species, A = T and G = C; therefore (A + G)/(T + C) = 1. Also, (A + T)/(G + C) varies between species but is constant within a species.
• Each turn of the helix spans 3.4 nm and contains 10 base pairs; the distance between adjacent base pairs is 0.34 nm.
• The diameter of the helix is 2 nm. The two strands are covalently bonded within each strand by phosphodiester bonds.
• Minor and major grooves in the helix serve as binding sites for regulatory proteins.

DNA as the Genetic Material

• Griffith's experiment (1928): Injected mice with live non-virulent (R) and heat-killed virulent (S) Streptococcus pneumoniae. Mice died — the R-strain was 'transformed' by some 'transforming principle' from the dead S-strain.
• Avery, MacLeod and McCarty (1944): Showed that the transforming principle was DNA — they purified components and treated with DNase, RNase and protease; only DNase destroyed transforming activity.
• Hershey-Chase experiment (1952): Used T2 bacteriophage labelled with ³⁵S (protein) and ³²P (DNA). After infection, only ³²P entered the bacteria and produced new phages, confirming DNA is the genetic material.
• Tobacco Mosaic Virus (TMV) — RNA is the genetic material in some viruses. In most organisms, DNA is the genetic material.

Semi-Conservative Replication

• DNA polymerase III catalyzes the addition of nucleotides in the 5'→3' direction, using the template strand (read 3'→5').
• The leading strand is synthesized continuously; the lagging strand is synthesized in short Okazaki fragments (each with an RNA primer) that are later joined by DNA ligase.
• DNA polymerase has proofreading (3'→5' exonuclease) activity, giving an error rate of about 10⁻⁹ per base pair per replication.
• Replication begins at the origin of replication (oriC in E. coli); eukaryotes have multiple origins per chromosome.

Transcription and mRNA Processing

Post-Transcriptional Processing (Eukaryotes)

• 5' capping: A methylated guanine (m⁷G) is added to the 5' end — protects mRNA and helps ribosome binding.
• 3' polyadenylation: A poly-A tail (100–200 adenines) is added to the 3' end — stabilizes mRNA and aids export from the nucleus.
• Splicing: Introns (non-coding) are removed and exons (coding) are joined by the spliceosome. Alternative splicing allows one gene to code for multiple proteins.

The Genetic Code

• Degenerate (redundant): Most amino acids are coded by more than one codon (e.g. leucine has 6 codons). The third base ('wobble' position) often varies without changing the amino acid.
• Universal: The same codons code for the same amino acids in almost all organisms — from bacteria to humans (with minor exceptions in mitochondria).
• Non-overlapping and comma-free: The code is read sequentially from a fixed starting point without gaps or overlaps.

Translation and the Lac Operon

Lac Operon Model (Jacob and Monod, 1961)

The lac operon is a model for gene regulation in prokaryotes (E. coli) that controls the metabolism of lactose. It consists of regulatory genes (lacI), an operator (lacO), a promoter (lacP), and three structural genes (lacZ, lacY, lacA) that code for β-galactosidase, permease and transacetylase.

• In the absence of lactose: the repressor protein (from lacI) binds to the operator and blocks RNA polymerase → structural genes are NOT transcribed (operon is OFF).
• In the presence of lactose (inducer): lactose (allolactose) binds to the repressor, inactivating it → repressor cannot bind the operator → RNA polymerase transcribes lacZYA → lactose is metabolized (operon is ON).
• When glucose is present and lactose is absent: the operon is OFF regardless. When glucose is absent and lactose is present: the operon is ON at high levels (cAMP-CAP complex enhances transcription).

Human Genome Project and DNA Fingerprinting

Human Genome Project (HGP)

• A mega-project launched in 1990 and completed in 2003, coordinated by the US Department of Energy and NIH (also called 'Moonshot of Biology').
• Sequenced approximately 3.2 × 10⁹ base pairs of the human genome.
• Key findings: humans have approximately 20,500 protein-coding genes (far fewer than the initially estimated 100,000); less than 2% of the genome codes for proteins.
• Applications: understanding genetic diseases, pharmacogenomics, forensic identification, evolutionary biology.
• Milestone technologies: chain-termination (Sanger) method, sequence tagged sites (STS), expressed sequence tags (ESTs).

DNA Fingerprinting

DNA fingerprinting (developed by Alec Jeffreys, 1984) is a technique to identify individuals based on their unique DNA profile. It uses VNTRs (Variable Number Tandem Repeats) or STRs (Short Tandem Repeats) — regions where the number of repeated sequences varies between individuals. These are non-coding, highly polymorphic regions.

• Process: DNA extracted → cut with restriction enzymes (RFLP method) → separated by gel electrophoresis → transferred to nylon membrane (Southern blotting) → hybridized with radioactive VNTR probes → autoradiography produces a unique banding pattern.
• Applications: forensic identification, paternity testing, immigration cases, identifying bodies in mass disasters, plant and animal breeding.
• VNTR analysis is so specific that it can distinguish between identical twins in some cases when somatic mutations differ.

Solved Examples

Example: In the Hershey-Chase experiment, why was ³²P used to label DNA and ³⁵S to label protein? Could the isotopes be swapped? Explain.

Solution: ³²P labels DNA because DNA contains phosphorus in its sugar-phosphate backbone but protein does not. ³⁵S labels protein because protein contains sulfur in amino acids like cysteine and methionine but DNA does not. The isotopes cannot be swapped — ³⁵P would label both DNA and protein (since both contain phosphorus in some amino acids), and ³²S would label only protein but is less specific. The key was choosing isotopes unique to each molecule.

Example: Explain why the lac operon is ON only when lactose is present AND glucose is absent. What would happen if both were present?

Solution: The lac operon requires two conditions: (1) Lactose must be present to inactivate the repressor (allolactose binds the repressor → operator is free → RNA polymerase can bind). (2) Glucose must be absent so that cAMP levels rise, and the cAMP-CAP complex binds to the promoter to enhance transcription. If both glucose and lactose are present, the repressor is removed, but without cAMP-CAP activation, transcription occurs only at very low (basal) levels. E. coli preferentially uses glucose (a more efficient energy source) via catabolite repression.