Molecular Basis of Inheritance NEET Biology Revision Notes, DNA Replication Transcription Translation and MCQs

Chapter Snapshot - Molecular Basis of Inheritance

This chapter is the molecular core of NEET Biology, covering the entire journey from the discovery of nucleic acids to the Human Genome Project. It opens with nucleic acids (DNA and RNA), their chemical composition (pentose sugar, phosphoric acid, nitrogenous bases), and the distinction between nucleosides and nucleotides. The Watson-Crick double helix model of DNA is described in detail, including antiparallel strands, complementary base pairing (A=T with 2 H-bonds, G≡C with 3 H-bonds), dimensions (3.4 A between base pairs, 34 A per turn, 20 A diameter, 10 bp per turn), and Chargaff's base equivalence rules. Five forms of DNA (A, B, C, D, Z) are compared. Landmark experiments establishing DNA as genetic material are covered: Griffith's transformation (1928), Avery-MacLeod-McCarty (1944), and Hershey-Chase (1952) blender experiment with S35 and P32. DNA replication is explained step by step: helicase unwinding, SSB proteins, RNA priming, DNA polymerase III elongation, Okazaki fragments on the lagging strand, primer removal by DNA polymerase I, and joining by DNA ligase. The Meselson-Stahl experiment (1958) with N15/N14 in E. coli proves semiconservative replication. RNA types (mRNA by Jacob and Monod, rRNA constituting 80% of cellular RNA, tRNA with clover leaf structure by Holley) and their functions are detailed. The genetic code properties (triplet, universal, commaless, non-overlapping, degenerate) and the code dictionary with 64 codons (61 sense + 3 stop: UAA, UAG, UGA) are presented. The central dogma (DNA to RNA to protein) and reverse transcription (Temin and Baltimore, RNA-dependent DNA polymerase) are explained. Transcription covers RNA polymerase (core enzyme plus sigma factor), promoter and terminator regions, initiation, elongation, and termination (rho factor in prokaryotes, poly-A tail in eukaryotes), and post-transcriptional processing (splicing of introns/exons in eukaryotic hnRNA). Translation is detailed with amino acid activation (aminoacyl-tRNA synthetase), initiation factors (IF1-3 in prokaryotes, eIF1-6 in eukaryotes), elongation (peptidyl transferase, translocase), and termination (release factors at stop codons). Gene regulation in prokaryotes is covered through the lac operon model (Jacob and Monod 1961) with structural genes Z, Y, A, operator, promoter, regulator gene, repressor, and inducer (lactose/allolactose). The tryptophan operon as a repressible system is contrasted. Eukaryotic gene regulation at replication, transcription, processing, and translation levels is described, including Britten-Davidson model. Advanced topics include DNA fingerprinting (Alec Jeffreys 1985, VNTR/minisatellites), PCR (Kary Mullis, Taq polymerase from Thermus aquaticus), cDNA and gene libraries, and the Human Genome Project (Francis Collins, Craig Venter, 3 billion bp, ~30,000 genes).

✓ Use This To Plan Your First 2–3 Hours

Expected Questions (Typical)

5-7

Molecular Basis of Inheritance is a high-weightage NEET chapter with questions spanning DNA structure, replication, transcription, translation, genetic code, lac operon, and DNA fingerprinting. Expect 5-7 questions including at least one assertion-reason and one application-based problem.

Time Required (Practical)

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16-20 hrs

This chapter covers the full molecular biology spectrum from nucleic acid chemistry through gene regulation and applied genomics. Requires strong conceptual understanding of sequential processes (replication, transcription, translation) and memorisation of key experiments, scientists, dimensions, and code properties.

Difficulty Level

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High

The chapter demands understanding of multi-step molecular mechanisms (DNA replication fork, transcription machinery, translation with initiation/elongation/termination), combined with precise numerical recall (DNA dimensions, codon table) and conceptual clarity on gene regulation models. The interplay between template/coding strands and mRNA/tRNA directions creates frequent exam traps.

Most Asked Style: Direct factual recall on DNA dimensions (3.4 A rise, 34 A per turn, 20 A diameter), Chargaff's rules, antiparallel strand polarity, semiconservative replication proof. Application-based questions on mRNA sequence from a given DNA template strand. Match-the-column on RNA polymerase types (I, II, III) with their products. Assertion-reason on genetic code degeneracy vs ambiguity, lac operon regulation, and Okazaki fragments on the lagging strand. Diagram-based questions on nucleosome structure, clover leaf tRNA model, and lac operon switching.Biggest Trap: Confusing the template strand (3' to 5', antisense) with the coding strand (5' to 3', sense, same sequence as mRNA except T instead of U). Students write mRNA complementary to the coding strand instead of the template strand. Another major trap: lac operon regulation is negative and inducible (repressor prevents transcription; lactose/allolactose removes repressor), NOT positive regulation. Also, the distance between adjacent base pairs is 3.4 A (0.34 nm), while one full turn is 34 A (3.4 nm) - these are frequently interchanged in options.Fast Win: Memorise DNA dimensions: 3.4 A rise per bp, 34 A per turn, 20 A diameter, 10 bp per turn for B-DNA. Master Chargaff's rule: A=T, G=C, so A+G = T+C (purines = pyrimidines). Know the three stop codons (UAA ochre, UAG amber, UGA opal) and that AUG is the universal start codon coding for methionine. Learn that RNA polymerase I makes rRNA, RNA polymerase II makes mRNA (hnRNA), and RNA polymerase III makes tRNA. Remember: Okazaki fragments form on the lagging strand and are joined by DNA ligase.Revision-Friendly: Create a master comparison table: DNA vs RNA (sugar, bases, strands, stability). Draw the central dogma flowchart including reverse transcription. Make a lac operon diagram showing both repressed and induced states. Build a genetic code summary table highlighting degenerate codons, stop codons, and the sole codons for Met (AUG) and Trp (UGG). Prepare a timeline of key experiments: Miescher 1869, Griffith 1928, Avery 1944, Hershey-Chase 1952, Watson-Crick 1953, Meselson-Stahl 1958, Jacob-Monod 1961, Nirenberg-Khorana 1961-66, Holley 1965, Jeffreys 1985.

Subtopics - Molecular Basis of Inheritance (NEET)

Complete guide to DNA structure, replication, transcription, translation, genetic code, gene regulation, DNA fingerprinting, and the Human Genome Project for NEET

Revision tip: Focus on DNA dimensions (3.4 A, 34 A, 20 A), Chargaff's rules (A=T, G=C), semiconservative replication proof (Meselson-Stahl with N15), RNA polymerase types and their products, genetic code properties (triplet, degenerate, universal, commaless), lac operon components and regulation, and DNA fingerprinting with VNTR. Practice writing mRNA sequences from given DNA template strands.

NCERT Lines MCQs Quick Test

1) Nucleic Acids: DNA and RNA Structure

Nucleic acids were first isolated by Friedrich Miescher (1869) from pus cell nuclei and named nuclein; the term nucleic acid was given by Altman (1899). DNA (deoxyribonucleic acid) is found in all living cells except plant viruses. It is composed of three chemical components: deoxyribose sugar (pentose, identified by Levene 1910), phosphoric acid (H3PO4, makes DNA acidic), and nitrogenous bases (purines: adenine and guanine, both double-ring; pyrimidines: cytosine and thymine, both single-ring; discovered by Kossel, Nobel Prize 1910). Nucleosides are formed by base + sugar; nucleotides by base + sugar + phosphate. The Watson-Crick double helix model (1953) describes two antiparallel polynucleotide chains with sugar-phosphate backbone on the outside and bases directed inward. Complementary base pairing: A=T (2 hydrogen bonds), G≡C (3 hydrogen bonds). Chargaff's rule (1950): A=T, G=C, so (A+G)/(C+T) = 1, and (A+T)/(G+C) is species-specific. B-DNA has 10 bp per turn, 3.4 A rise per bp, 34 A per turn, 20 A diameter, right-handed helix. Five DNA forms exist: A-DNA (11 bp/turn, right-handed), B-DNA (10 bp/turn, most common), C-DNA (9.33 bp/turn), D-DNA (8 bp/turn), Z-DNA (12 bp/turn, left-handed, 18 A diameter). Satellite DNA comprises small highly repetitive sequences in eukaryotes. Promiscuous DNA moves between mitochondria, chloroplasts, and nucleus (discovered in maize 1983). Denaturation (melting at ~90 C) separates strands by breaking hydrogen bonds; renaturation (annealing at ~25 C) reforms the double helix.

Watson-Crick modelChargaff's ruleDNA dimensionsDNA forms

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Chemical composition and nucleotide structureDNA comprises deoxyribose sugar, phosphoric acid, and four nitrogenous bases (A, G, C, T). Nucleoside = base + sugar; nucleotide = base + sugar + phosphate. Nucleotides linked by 3'-5' phosphodiester bonds form the polynucleotide chain.

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Watson-Crick double helix and Chargaff's rulesTwo antiparallel strands with sugar-phosphate backbone outside. A=T (2 H-bonds), G≡C (3 H-bonds). B-DNA: 10 bp/turn, 3.4 A rise, 34 A/turn, 20 A diameter. Chargaff's rule: molar A = T, G = C, purines = pyrimidines.

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Forms of DNA and special DNA typesA-DNA (11 bp, right), B-DNA (10 bp, most common), C-DNA (9.33 bp), D-DNA (8 bp), Z-DNA (12 bp, left-handed). Satellite DNA is highly repetitive. Promiscuous DNA moves between organelles and nucleus. Denaturation at 90 C; renaturation at 25 C.

2) DNA as Genetic Material: Key Experiments

Griffith's experiment (1928): Injected mice with virulent smooth (S-type) and non-virulent rough (R-type) Diplococcus pneumoniae. Heat-killed S-type + live R-type caused death, demonstrating transformation. Avery, MacLeod, and McCarty (1944) fractionated heat-killed S-type into DNA, protein, and carbohydrate; only intact DNA (without DNase) could transform R-type to S-type, proving DNA is the transforming principle. Hershey and Chase (1952) used bacteriophages labelled with S35 (protein coat) and P32 (DNA). After infection and blending, only P32 was found inside bacterial cells, providing unequivocal proof that DNA is the genetic material. Bacterial conjugation by Lederberg and Tatum (1946) demonstrated genetic recombination through DNA transfer between two auxotrophic E. coli strains via a cytoplasmic bridge. These experiments collectively established that DNA, not protein, carries hereditary information.

Griffith's transformationAvery's proofHershey-Chase experimentBacterial conjugation

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Griffith's transformation experimentS-type (virulent, smooth capsule) kills mice; R-type (non-virulent, rough) and heat-killed S-type alone do not. Heat-killed S + live R causes death with live S-type recovered. Transforming principle later identified as DNA by Avery et al. (1944).

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Hershey-Chase blender experimentBacteriophages labelled with S35 (protein) and P32 (DNA) used separately to infect E. coli. After blending, only P32 found inside cells. Sulphur is absent from DNA; phosphorus is absent from protein. Provided final proof DNA is the genetic material.

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Lederberg-Tatum bacterial conjugationTwo auxotrophic E. coli strains (mutant A: Met- Bio- Thr+ Leu+ Thi+; mutant B: Met+ Bio+ Thr- Leu- Thi-) could not grow in minimal medium alone, but mixed culture produced colonies. DNA transfer via cytoplasmic bridge in conjugation confirmed.

3) DNA Replication

DNA replication is semiconservative (each daughter molecule retains one parental strand) and semi-discontinuous (leading strand continuous, lagging strand in Okazaki fragments). Replication occurs in S-phase of cell cycle. Steps: (1) Helicase unwinds DNA using ATP energy and breaks hydrogen bonds. (2) Topoisomerase/gyrase (in E. coli) relieves supercoiling ahead of the fork. (3) Single-stranded binding proteins (SSB) stabilise separated strands. (4) Primase synthesises RNA primer (50-100 nucleotides). (5) DNA polymerase III elongates new strands in 5' to 3' direction from the 3' to 5' template. The leading strand is synthesised continuously; the lagging strand is synthesised as Okazaki fragments. (6) DNA polymerase I removes RNA primers (5' to 3' exonuclease activity) and fills gaps. (7) DNA ligase seals nicks between adjacent Okazaki fragments. Three DNA polymerases in E. coli: Pol I (discovered by Kornberg 1955, repair and primer removal), Pol II (unknown role), Pol III (main replicative enzyme, discovered by T. Kornberg and Gefter 1972). DNA repair involves excision of UV-induced thymine dimers by endonuclease, gap filling by Pol I, and sealing by ligase (photoreactivation). Meselson and Stahl (1958) proved semiconservative replication using N15-labelled E. coli transferred to N14 medium and CsCl density gradient centrifugation. After one generation, hybrid (N15/N14) DNA; after two generations, equal hybrid and light DNA. Taylor's autoradiography experiment on Vicia faba with tritiated thymidine confirmed this in eukaryotes. Cairns demonstrated theta replication in prokaryotes.

Semiconservative replicationOkazaki fragmentsMeselson-Stahl experimentDNA polymerases

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Mechanism and enzymes of replicationHelicase unwinds, SSB stabilises, primase makes RNA primer, DNA Pol III elongates 5' to 3', leading strand is continuous, lagging strand forms Okazaki fragments. DNA Pol I removes primers. DNA ligase joins fragments.

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Semiconservative replication proofMeselson and Stahl (1958) grew E. coli in N15 medium, transferred to N14. CsCl density gradient showed hybrid DNA after one generation, equal hybrid + light after two. Taylor confirmed in Vicia faba using tritiated thymidine autoradiography.

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DNA repair mechanismsUV-induced thymine dimers repaired by photoreactivation: endonuclease excises damaged segment, DNA Pol I fills the gap, DNA ligase seals. DNA polymerase I has both 5' to 3' and 3' to 5' exonuclease activities for proofreading and repair.

4) RNA: Types and Structure

RNA is found in cytoplasm, nucleolus, ribosomes, mitochondria, and chloroplasts. It is single-stranded (except in reovirus and wound tumour virus where it is double-stranded) and contains ribose sugar, phosphate, and bases A, G, C, U (uracil replaces thymine). Ochoa received Nobel Prize for artificial RNA synthesis. Three major types: (1) mRNA (messenger RNA, named by Jacob and Monod 1961): 5% of total cellular RNA, carries coded information from DNA to ribosomes for protein synthesis, has 5' methylated cap (7-methylguanosine) and 3' poly-A tail, short-lived. Monocistronic mRNA codes for one polypeptide; polycistronic codes for multiple. (2) rRNA (ribosomal RNA): constitutes 70-80% of total cellular RNA. Eukaryotic forms: 28S, 18S, 5.8S, 5S; prokaryotic: 23S, 16S, 5S. Synthesised in nucleolus/SAT region. 23S rRNA acts as ribozyme (Altman and Cech). (3) tRNA (transfer RNA, clover leaf model by Robert Holley 1965, Nobel Prize 1968 shared with Khorana and Nirenberg): 10-15% of total RNA, 75-80 nucleotides, smallest RNA. Four functional sites: amino acid attachment at 3' CCA end, DHU loop (activating enzyme recognition), anticodon loop (complementary to mRNA codon), and TpsiC loop (ribosome recognition). Other RNA types include snRNA (splicing, rRNA processing), scRNA (signal recognition), and hnRNA (precursor of mRNA in eukaryotes, contains introns and exons).

mRNA functionrRNA 80% of total RNAtRNA clover leaf modelHolley 1965

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mRNA: structure and functionNamed by Jacob and Monod (1961). 5% of total RNA. Has 5' cap (7-methylguanosine) and 3' poly-A tail. Carries genetic code from DNA to ribosomes. Monocistronic (one polypeptide) or polycistronic (multiple polypeptides). Short-lived.

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rRNA and ribosome compositionMost abundant RNA (70-80%). Eukaryotic ribosomes (80S): 60S subunit has 28S, 5.8S, 5S rRNA; 40S subunit has 18S rRNA. Prokaryotic ribosomes (70S): 50S has 23S, 5S; 30S has 16S. 23S rRNA is a ribozyme with peptidyl transferase activity.

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tRNA: clover leaf model and functional sitesClover leaf model by Holley (1965). 75-80 nucleotides. 3' CCA end binds amino acid. DHU loop for enzyme recognition. Anticodon loop reads mRNA codon. TpsiC loop for ribosome attachment. About 60 types for 20 amino acids.

5) Genetic Code and Central Dogma

The genetic code is the sequence of nitrogen bases in mRNA that specifies amino acids. Discovered through frame-shift mutations by Crick. Nirenberg and Mathaei (1961) used poly-U mRNA to show UUU codes for phenylalanine (first codon deciphered). Khorana received Nobel Prize for synthesising artificial polynucleotides. Properties: (1) Triplet - 3 nucleotides per codon, giving 64 codons (4 cubed). (2) Universal - same codon specifies same amino acid in all organisms. (3) Commaless - read continuously without pauses. (4) Non-overlapping - each nucleotide belongs to only one codon. (5) Degenerate - multiple codons can specify one amino acid (e.g., leucine has 6 codons); degeneracy discovered by Bernfield and Nirenberg; mainly due to wobble at third position (wobble hypothesis by Crick). (6) Unambiguous - each codon specifies only one amino acid. Of 64 codons: 61 are sense codons (code for 20 amino acids), 3 are stop/nonsense codons: UAA (ochre), UAG (amber), UGA (opal). AUG is the universal start codon, coding for methionine (formyl-methionine in prokaryotes). GUG can also serve as alternate initiator. Only methionine (AUG) and tryptophan (UGG) have single codons. The central dogma (Watson and Crick): DNA to RNA (transcription) to protein (translation). Reverse transcription by Temin and Baltimore (1970, Nobel Prize 1975) in Rous sarcoma virus: RNA to DNA via RNA-dependent DNA polymerase (reverse transcriptase).

64 codonsDegeneracy and wobbleAUG start codonCentral dogma

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Properties of the genetic codeTriplet (3 bases per codon), universal, commaless, non-overlapping, degenerate (multiple codons per amino acid due to wobble), unambiguous (one amino acid per codon). 61 sense codons + 3 stop codons (UAA, UAG, UGA). AUG is start codon for methionine.

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Cracking the genetic codeNirenberg and Mathaei (1961) used cell-free system with poly-U to decode UUU = phenylalanine. Khorana synthesised defined polynucleotides. Wobble hypothesis by Crick explains degeneracy at third codon position. Gamow proposed triplet nature.

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Central dogma and reverse transcriptionCentral dogma: DNA to mRNA (transcription) to protein (translation). Reverse transcription in retroviruses: RNA to DNA by reverse transcriptase (RNA-dependent DNA polymerase), discovered by Temin and Baltimore (1970, Nobel 1975) in Rous sarcoma virus.

6) Transcription

Transcription is the synthesis of mRNA from a DNA template (heterocatalytic function of DNA). The template strand (sense/master strand, 3' to 5') is copied. The DNA segment involved is a cistron with a promoter region (initiation) and terminator region (end). In prokaryotes, RNA polymerase is a multi-subunit enzyme: core enzyme (alpha2, beta, beta-prime, omega) for elongation, plus sigma factor for promoter recognition. Three stages: (1) Initiation - sigma factor recognises promoter; holoenzyme binds to TATA box (Pribnow box at -10 position, TATAAT; -35 sequence TTGACA). In eukaryotes, TATA box (Hogness box) serves same function. 5' cap has 7-methylguanosine. (2) Elongation - core enzyme moves along template strand in 3' to 5' direction, synthesising RNA in 5' to 3' direction. (3) Termination - rho (rho) factor terminates transcription in prokaryotes; poly-A tail signals termination in eukaryotes. In eukaryotes, three RNA polymerases exist: RNA Pol I (synthesises rRNA: 28S, 18S, 5.8S), RNA Pol II (synthesises hnRNA/mRNA), RNA Pol III (synthesises tRNA and 5S rRNA). Post-transcriptional processing in eukaryotes: hnRNA contains both exons (coding) and introns (non-coding, intervening sequences), called split genes. Splicing removes introns (aided by snRNA U1, U2), 5' capping adds methylguanosine, and 3' polyadenylation adds poly-A tail. Only mature mRNA exits the nucleus. Actinomycin D inhibits transcription; glucocorticoids increase it.

RNA polymerase subunitsSigma factorhnRNA processingSplit genes

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Prokaryotic transcription machineryRNA polymerase holoenzyme = core enzyme (alpha2, beta, beta', omega) + sigma factor. Sigma recognises promoter (Pribnow box at -10: TATAAT). Core enzyme elongates in 5' to 3'. Rho factor terminates. Transcription unit: promoter, structural gene, terminator.

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Eukaryotic transcription and RNA processingThree RNA polymerases: Pol I (rRNA), Pol II (mRNA/hnRNA), Pol III (tRNA, 5S rRNA). hnRNA has exons and introns (split genes). Processing: 5' capping (7-methylguanosine), splicing (intron removal by snRNA), 3' polyadenylation. Mature mRNA exported to cytoplasm.

7) Translation (Protein Synthesis)

Translation is the synthesis of protein from mRNA on ribosomes. Ribosomes have two subunits: 40S + 60S = 80S in eukaryotes (30S + 50S = 70S in prokaryotes). The larger subunit has A-site (aminoacyl/acceptor), P-site (peptidyl/donor), and peptidyl transferase enzyme. The smaller subunit has mRNA binding site. Amino acid activation: amino acid + ATP + aminoacyl-tRNA synthetase forms aminoacyl-AMP-enzyme complex, then transfers to tRNA 3' CCA end, forming charged tRNA (aminoacyl-tRNA); requires one ATP per activation. Initiation: mRNA attaches to smaller ribosomal subunit (5' cap contacts 3' end of 18S/16S rRNA); requires IF1, IF2, IF3 in prokaryotes (eIF1-eIF6 in eukaryotes); AUG at P-site attracts initiator Met-tRNA (formylated in prokaryotes, non-formylated in eukaryotes); larger subunit joins forming complete ribosome; Mg2+ required for subunit union. Elongation: new aminoacyl-tRNA enters A-site (requires EF-Tu/Ts in prokaryotes, eEF1 in eukaryotes + GTP); peptide bond formed by peptidyl transferase between COOH of P-site amino acid and NH2 of A-site amino acid; translocation by translocase (EF-G/eEF2 + GTP) moves ribosome one codon along mRNA, shifting peptidyl-tRNA to P-site; each amino acid incorporation requires 1 ATP + 2 GTP. Termination: stop codon (UAA/UAG/UGA) at A-site; release factor (RF1 for UAG/UAA, RF2 for UAA/UGA in prokaryotes; eRF1 in eukaryotes) triggers polypeptide release; ribosomal subunits dissociate. Post-translational modification: deformylation/removal of initiator methionine, folding into secondary (alpha-helix), tertiary, and quaternary structures. Polyribosomes (polysome): multiple ribosomes translating same mRNA simultaneously; ribosome nearest 5' end has shortest polypeptide. Puromycin inhibits translation.

Initiation factorsPeptidyl transferase1 ATP + 2 GTP per amino acidPolyribosomes

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Amino acid activation and initiationAminoacyl-tRNA synthetase charges tRNA with amino acid using ATP. AUG start codon at P-site attracts Met-tRNA (formylated in prokaryotes). Initiation factors: IF1-3 (prokaryotes), eIF1-6 (eukaryotes). Large subunit joins; Mg2+ essential.

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Elongation, translocation, and terminationAminoacyl-tRNA enters A-site (EF-Tu + GTP). Peptide bond by peptidyl transferase. Translocation by translocase (EF-G + GTP). Each amino acid costs 1 ATP + 2 GTP. Stop codons attract release factors, not tRNA. Polypeptide released, subunits dissociate.

8) Gene Regulation, DNA Fingerprinting, and Human Genome Project

Gene regulation in prokaryotes follows the operon model by Jacob and Monod (1961). An operon consists of structural genes, operator gene, promoter gene, and is controlled by a regulator gene. The lac operon of E. coli has structural genes Z (beta-galactosidase, splits lactose into glucose and galactose), Y (galactoside permease), and A (transacetylase). The regulator gene (i gene) produces a repressor protein (MW 160,000, 4 subunits of 40,000 each) that binds to the operator (27 bp), blocking RNA polymerase passage. Lactose (or allolactose) acts as inducer: binds repressor, changes its conformation, freeing the operator. This is negative inducible regulation. cAMP is required for RNA polymerase to function. The system produces a polycistronic mRNA encoding all three enzymes. The tryptophan operon is a repressible system: 5 structural genes (E, D, C, B, A), normally ON. The regulator gene produces an aporepressor, which alone cannot block the operator. Tryptophan acts as corepressor; aporepressor + tryptophan = active repressor that blocks operator (feedback inhibition). Eukaryotic gene regulation occurs at four levels: replication (gene amplification), transcription (differential gene transcription), processing (80% nuclear RNA destroyed; splicing controls), and translation (regulated by histones as repressors per Frenster's model 1965, or Britten-Davidson gene battery model 1969 with integrator, sensor, producer, and receptor genes). DNA fingerprinting was developed by Alec Jeffreys (1985). Based on VNTR (variable number tandem repeats), 15-nucleotide minisatellites unique to each individual. Applications: forensic identification, paternity disputes, immigration verification, racial group identification. Southern blotting separates DNA fragments; gel electrophoresis and autoradiography employed. Human Genome Project: led by Francis Collins (HGP) and Craig Venter (Celera Genomics). Human genome has approximately 3 billion base pairs and ~30,000 genes. Chromosome 22 was the first fully sequenced (December 1999). Model organisms sequenced: E. coli (4.7 million bp, 4000 genes), yeast (12 million bp, 6000 genes), C. elegans (97 million bp, 18,000 genes), Drosophila (180 million bp, 13,000 genes). Prospects include designer drugs, genetically modified diets, and cancer gene therapy.

Lac operon regulationTryptophan operonDNA fingerprinting VNTRHuman Genome Project

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Lac operon: inducible systemJacob and Monod (1961). Structural genes Z, Y, A produce polycistronic mRNA. Repressor (from i gene) blocks operator. Lactose/allolactose induces by removing repressor. Negative inducible regulation. Operator is 27 bp. cAMP required for transcription.

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Tryptophan operon: repressible systemNormally ON with 5 structural genes (E, D, C, B, A). Aporepressor alone cannot bind operator. Tryptophan (corepressor) + aporepressor = active repressor that blocks operator. Anabolic pathway feedback inhibition. Opposite to lac operon logic.

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DNA fingerprinting and applicationsDeveloped by Alec Jeffreys (1985). Based on VNTR/minisatellites (15-nt repeats). Individual-specific DNA profile. Uses: forensic identification, paternity testing, immigration disputes. Technique involves Southern blotting, restriction enzymes, gel electrophoresis, autoradiography.

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Human Genome Project and genomicsFrancis Collins (HGP) and Craig Venter (Celera). 3 billion bp, ~30,000 genes. Chromosome 22 first fully sequenced (1999). Frederick Sanger developed DNA sequencing method. Promises designer drugs, gene therapy, genetic identity databases.

Molecular Basis of Inheritance Download Notes & Weightage Plan

For each topic in the Molecular Basis of Inheritance chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.

2 Downloads

Nucleic Acids: DNA and RNA Structure

Chemical composition of DNA and RNA, nucleosides vs nucleotides, Watson-Crick model, Chargaff's rules, DNA forms (A, B, C, D, Z), and special DNA types.

Watson-Crick modelChargaff's ruleDNA dimensionsDNA forms

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

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Topic Notes (Condensed)Make a DNA dimensions card: 3.4 A rise, 34 A per turn, 20 A diameter, 10 bp/turn for B-DNA. Table comparing all 5 DNA forms (bp per turn, handedness, diameter). Chargaff's rule equations: A=T, G=C, (A+G)/(T+C) = 1.

Download Notes Printable PDF

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NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.

NCERT Lines Flashcards

Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."

MCQ Set PYQs

How to revise: Draw the double helix with labelled dimensions. Practice Chargaff's rule calculations: given %A, find %T, %G, %C. Compare nucleoside vs nucleotide structures as a quick diagram.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2Expect 1-2 questions on DNA dimensions, Chargaff's rule calculations, or comparison of DNA forms.

Time Required3-4 hrsCovers fundamental structural chemistry with numerical problem practice.

DifficultyMediumConcepts are straightforward but require precise memorisation of numerical values and careful distinction between similar terms.

Scoring Focus: DNA dimensions are tested as direct MCQs almost every year. Chargaff's rule calculations appear as numerical problems. DNA forms comparison (especially B-DNA vs Z-DNA handedness) is a recurring match-the-column question.
High-risk Area: Mixing up 3.4 A (distance between base pairs) with 34 A (one complete turn). Confusing nucleoside (base + sugar) with nucleotide (base + sugar + phosphate). Forgetting Z-DNA is the only left-handed form.
Best Practice Style: Flashcards for dimensions. Practice numerical problems on Chargaff's rule. Draw and label the double helix from memory.

Priority rule: High - direct questions on DNA structure appear in every NEET paper.

DNA Replication and Key Experiments

Semiconservative replication mechanism, DNA polymerases I-III, Okazaki fragments, Meselson-Stahl experiment, Griffith's transformation, Hershey-Chase experiment.

Meselson-Stahl proofOkazaki fragmentsHershey-ChaseDNA polymerases

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

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Topic Notes (Condensed)Flowchart of replication: helicase -> SSB -> primase -> Pol III -> Okazaki fragments -> Pol I removes primer -> ligase joins. Table of DNA polymerases (I, II, III) with discoverers and functions. Meselson-Stahl generation-wise density results.

Download Notes Printable PDF

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NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.

NCERT Lines Flashcards

Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."

MCQ Set PYQs

How to revise: Draw the replication fork with all enzymes labelled. Practise Meselson-Stahl density ratios for generations 1-4. Timeline of experiments: Griffith (1928) -> Avery (1944) -> Hershey-Chase (1952) -> Meselson-Stahl (1958).

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2Expect questions on semiconservative replication proof, enzyme functions at the replication fork, and landmark experiment details.

Time Required3-4 hrsMulti-step mechanism with several experiments to master.

DifficultyHighUnderstanding the replication fork requires spatial reasoning. Meselson-Stahl ratios need mathematical clarity.

Scoring Focus: Meselson-Stahl density gradient results are frequently tested. Okazaki fragments and the role of DNA ligase are standard NEET questions. Hershey-Chase experiment with S35/P32 labelling appears regularly.
High-risk Area: Confusing Pol I (repair, primer removal) with Pol III (main replicative enzyme). Not understanding that the leading strand is continuous while lagging strand is discontinuous. Mixing up S35 (protein) and P32 (DNA) in Hershey-Chase.
Best Practice Style: Draw diagrams repeatedly. Practice MCQs on enzyme functions. Trace the Meselson-Stahl logic step by step.

Priority rule: High - replication mechanism and proof questions appear every year.

Genetic Code and Central Dogma

Properties of the genetic code, codon table, wobble hypothesis, central dogma, reverse transcription.

64 codonsDegeneracyStop codonsCentral dogma

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Topic Notes (Condensed)Genetic code properties mnemonic: TU-CNN-DU (Triplet, Universal, Commaless, Non-overlapping, Non-ambiguous, Degenerate, Unambiguous). Three stop codons: UAA (ochre), UAG (amber), UGA (opal). Single-codon amino acids: Met (AUG) and Trp (UGG). Central dogma with reverse transcription arrow.

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NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.

NCERT Lines Flashcards

Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."

MCQ Set PYQs

How to revise: Memorise the codon table by grouping: 6 codons (Leu, Ser, Arg), 4 codons (Val, Pro, Thr, Ala, Gly), 3 codons (Ile), 2 codons (Phe, Tyr, His, Gln, Asn, Lys, Asp, Glu, Cys), 1 codon (Met, Trp). Practice writing mRNA from DNA templates.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2Direct questions on code properties, mRNA sequence problems, and identification of start/stop codons.

Time Required2-3 hrsRequires thorough memorisation of the codon table and practice with sequence conversion.

DifficultyMediumMemorisation-heavy but conceptually straightforward once the code table is internalised.

Scoring Focus: Identifying stop codons, start codon, and degeneracy are direct NEET questions. Writing mRNA sequence from a given DNA strand is a common application question.
High-risk Area: Reading the wrong DNA strand (coding instead of template) when writing mRNA. Confusing degenerate (many codons per amino acid) with ambiguous (many amino acids per codon, which is false). Forgetting GUG can also serve as alternate start codon.
Best Practice Style: Practice codon-to-amino acid conversion problems. Write mRNA sequences from various DNA template strands daily.

Priority rule: High - genetic code questions appear in nearly every NEET paper.

Transcription and Translation

RNA polymerase structure and types, transcription steps, hnRNA processing, translation machinery, initiation-elongation-termination, energy cost per amino acid.

RNA polymerase typesSplicing and cappingPeptidyl transferase1 ATP + 2 GTP

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Topic Notes (Condensed)RNA polymerase types table: Pol I (rRNA), Pol II (mRNA/hnRNA), Pol III (tRNA, 5S rRNA). Translation energy budget: 1 ATP for activation + 2 GTP for elongation = 3 high-energy bonds per amino acid. Prokaryotic vs eukaryotic initiation factors comparison. Post-transcriptional processing: capping, splicing, polyadenylation.

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NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.

NCERT Lines Flashcards

Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."

MCQ Set PYQs

How to revise: Draw the complete transcription-translation flow in prokaryotes (coupled) vs eukaryotes (separated by nuclear membrane). Practise matching RNA Pol type to product. Enumerate initiation factors for both systems.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2Questions on RNA polymerase types, post-transcriptional modifications, translation machinery, and energy requirements.

Time Required4-5 hrsTwo complex multi-step processes with extensive molecular machinery to learn.

DifficultyHighBoth processes involve numerous factors, enzymes, and sequential steps that must be understood mechanistically.

Scoring Focus: RNA polymerase type matching is a standard NEET question. Energy cost per amino acid incorporation is frequently tested. The sequence of post-transcriptional processing (capping first, then splicing, then polyadenylation) is tested.
High-risk Area: Confusing RNA Pol I (rRNA) with Pol III (tRNA, 5S rRNA). Forgetting that 23S rRNA acts as ribozyme (peptidyl transferase activity). Not knowing that translation requires 1 ATP + 2 GTP per amino acid.
Best Practice Style: Create comparative tables. Practice energy cost calculations for protein synthesis. Draw coupled transcription-translation in prokaryotes.

Priority rule: High - transcription-translation questions are NEET staples.

Gene Regulation: Operon Model

Lac operon (inducible), tryptophan operon (repressible), eukaryotic gene regulation levels.

Lac operon negative regulationTryptophan operonHistones and regulation

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Topic Notes (Condensed)Lac operon components: i gene (repressor), P (promoter), O (operator, 27 bp), Z (beta-galactosidase), Y (permease), A (transacetylase). Repressor MW 160,000, 4 subunits. Lactose/allolactose = inducer. Tryptophan operon: aporepressor + tryptophan = active repressor. Inducible vs repressible table.

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NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.

NCERT Lines Flashcards

Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."

MCQ Set PYQs

How to revise: Draw lac operon in both OFF (repressor on operator) and ON (inducer removes repressor) states. Compare induction vs repression in a table. Learn the four levels of eukaryotic gene regulation.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1At least one question on lac operon regulation, gene functions, or inducible vs repressible distinction.

Time Required2-3 hrsFocused conceptual topic with clear diagrams to master.

DifficultyMediumConceptually elegant once the logic of negative regulation is understood, but tricky in exam MCQ format.

Scoring Focus: Lac operon regulation type (negative inducible) is frequently tested as a trap question. Functions of Z, Y, A genes are asked in match-the-column. Jacob and Monod (1961) credited for the model.
High-risk Area: Calling lac operon regulation positive (it is negative because repressor prevents transcription). Confusing Z gene product (beta-galactosidase) with Y gene product (permease). Mixing up inducible (lac) with repressible (trp) systems.
Best Practice Style: Draw operon diagrams from memory. Practice assertion-reason questions on operon regulation.

Priority rule: High - lac operon is tested almost every year in NEET.

DNA Fingerprinting and Human Genome Project

DNA fingerprinting technique, VNTR, applications, Human Genome Project milestones, PCR, cDNA.

VNTR/minisatellitesAlec Jeffreys 1985HGP 30,000 genesPCR Kary Mullis

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Topic Notes (Condensed)DNA fingerprinting: Jeffreys 1985, based on VNTR (15-nt minisatellites), Southern blotting, restriction enzymes, gel electrophoresis. Applications: forensics, paternity, immigration. HGP: 3 billion bp, ~30,000 genes, chromosome 22 first sequenced. Model organisms genome sizes table. PCR: Kary Mullis, Taq polymerase from Thermus aquaticus, works at 72C. cDNA made by reverse transcriptase, RT-PCR technique.

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NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.

NCERT Lines Flashcards

Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."

MCQ Set PYQs

How to revise: Memorise the HGP model organism table (organism, bp count, gene count). Learn PCR temperature cycle (denaturation ~94C, annealing ~55C, extension ~72C). DNA fingerprinting flowchart: isolate DNA -> cut with restriction enzymes -> electrophoresis -> Southern blot -> probe hybridisation.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1Expect one question on DNA fingerprinting basis, PCR enzyme source, or HGP statistics.

Time Required2-3 hrsPrimarily memorisation-based with clear factual targets.

DifficultyLow-MediumFactual recall with minimal conceptual complexity; requires accurate memorisation of names, numbers, and techniques.

Scoring Focus: VNTR as basis of DNA fingerprinting is a direct question. Taq polymerase source (Thermus aquaticus) is commonly asked. HGP gene count (~30,000) has appeared in recent NEET papers.
High-risk Area: Confusing VNTR (minisatellites for fingerprinting) with microsatellites (SSR, shorter repeats). Forgetting that PCR extension uses Taq polymerase because it is thermostable. Getting model organism gene counts wrong.
Best Practice Style: Create flashcards for key numbers. Practice match-the-column: technique to scientist to year.

Priority rule: Medium - appears less frequently than structural/functional topics but is a quick-scoring memorisation target.

Molecular Basis of Inheritance Chapter NEET Traps & Common Mistakes (Topic-Wise)

Each subtopic below is of the Molecular Basis of Inheritance chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.

! Avoid Easy Negatives

DNA Dimensions Confusion

DNA structureB-DNAWatson-Crick model

Mistake Snapshot (What Students Do Wrong)

Swapping rise and pitch values: The distance between adjacent base pairs is 3.4 A (0.34 nm), and one complete turn is 34 A (3.4 nm). NEET options deliberately swap these values. The diameter of B-DNA is 20 A (2 nm), not 10 A.
Confusing B-DNA with Z-DNA handedness: B-DNA is right-handed with 10 bp/turn. Z-DNA is the only left-handed form with 12 bp/turn and 18 A diameter. Options may list A-DNA as left-handed - it is right-handed with 11 bp/turn.

2–3 Line Example (Typical Error)

NEET 2006 asked: One turn of B-form DNA helix is approximately (a) 3.4 nm (b) 2 nm (c) 20 nm (d) 0.34 nm. The answer is (a) 3.4 nm = 34 A. Students choosing (d) 0.34 nm confuse single base pair distance with full turn.

How NEET Frames The Trap

Options place 3.4 A (base pair distance) and 34 A (turn length) side by side, testing whether you know which is which.

NEET-Style Trap Question Format

Q. In B-form DNA, the distance between two adjacent base pairs and one complete helical turn are, respectively:
A. 0.34 nm and 3.4 nm B. 3.4 nm and 0.34 nm C. 2.0 nm and 3.4 nm D. 0.34 nm and 20 nm
Trick: Option (b) reverses the values. Option (c) substitutes diameter for base pair distance. The correct answer is (a): 0.34 nm (3.4 A) between base pairs and 3.4 nm (34 A) per turn.

Quick rule: 3.4 A per pair, 34 A per turn, 20 A diameter. Multiply: 3.4 x 10 bp = 34 A per turn. This multiplication check prevents the swap error.

Template vs Coding Strand in Transcription

TranscriptionmRNA sequenceTemplate strand

Mistake Snapshot (What Students Do Wrong)

Writing mRNA complementary to the wrong strand: The template strand (antisense, 3' to 5') is used for transcription. mRNA is complementary to the template and has the same sequence as the coding strand (sense, 5' to 3') with U replacing T. Students often write mRNA complementary to the coding strand, producing the wrong sequence.
Ignoring strand polarity direction: When a DNA sequence is given as 5'-ATACG-3', this is the coding strand. The template strand is 3'-TATGC-5'. The mRNA from this template is 5'-AUACG-3' (same as coding strand with U for T). Questions may specify which strand without clear labelling.

2–3 Line Example (Typical Error)

NEET repeatedly asks: if the DNA coding strand is ATACG, the mRNA sequence is UAUGC. This is because the template strand (3'-TATGC-5') is read, giving mRNA 5'-AUACG-3'. Wait - the textbook gives the answer as UAUGC, meaning the given strand IS the template and mRNA is its complement with U for T.

How NEET Frames The Trap

Questions may label a strand as template or coding, or give a strand without labelling and ask for the mRNA. Read the question stem carefully for 3' and 5' labels.

NEET-Style Trap Question Format

Q. During transcription, if the nucleotide sequence of the DNA template strand being coded is 3'-ATACG-5', the nucleotide sequence in the mRNA would be:
A. 5'-UAUGC-3' B. 5'-AUACG-3' C. 5'-TATGC-3' D. 5'-UACGU-3'
Trick: Option (b) gives the complement with U for T but reads the coding strand instead. The template is 3'-ATACG-5', so mRNA is 5'-UAUGC-3' (complement with U). Option (a) is correct.

Quick rule: mRNA = same as coding strand (T replaced by U). mRNA = complement of template strand. Always identify which strand is given before writing the answer.

Lac Operon Regulation Type

Gene regulationLac operonNegative regulation

Mistake Snapshot (What Students Do Wrong)

Calling lac operon positive regulation: Lac operon uses negative inducible regulation: the repressor protein prevents transcription by binding operator. Lactose (inducer) removes the repressor. It is NOT positive regulation. The distinction: negative = repressor blocks; positive = activator enables.
Confusing inducible with repressible: Lac operon is inducible (normally OFF, turned ON by inducer lactose). Tryptophan operon is repressible (normally ON, turned OFF when tryptophan accumulates as corepressor). NEET 2015 directly tested this distinction.

2–3 Line Example (Typical Error)

AIPMT 2015 asked about lac operon regulation: the correct answer was 'negative and inducible because repressor protein prevents transcription.' Students selecting 'positive and inducible' confuse the removal of repressor (which enables transcription) with positive regulation.

How NEET Frames The Trap

Options pair regulation type (positive/negative) with operon type (inducible/repressible) in all combinations to test precise understanding.

NEET-Style Trap Question Format

Q. Gene regulation governing the lactose operon of E. coli involving the lac i gene product is:
A. Negative and inducible because repressor protein prevents transcription B. Positive and inducible because it can be induced by lactose C. Negative and repressible because repressor protein prevents transcription D. Feedback inhibition because excess beta-galactosidase switches off transcription
Trick: Option (b) is the most common wrong answer. Students reason that since lactose turns ON the operon, it must be positive regulation. But the mechanism works by removing a repressor (negative control), not by activating transcription directly.

Quick rule: If a REPRESSOR blocks transcription, the regulation is NEGATIVE. If an ACTIVATOR is needed to start transcription, the regulation is POSITIVE. Lac operon has a repressor, so it is negative. The inducer removes the repressor but does not directly activate transcription.

Chargaff's Rule Application Errors

DNA compositionChargaff's ruleNumerical problems

Mistake Snapshot (What Students Do Wrong)

Applying Chargaff's rule to single-stranded nucleic acids: Chargaff's rule (A=T, G=C) applies only to double-stranded DNA. It does NOT apply to RNA or single-stranded DNA. NCERT exemplar explicitly tests this: if A is not equal to T in a given nucleic acid, it must be single-stranded.
Calculation errors with percentage bases: If cytosine is 18%, then guanine is also 18% (Chargaff's rule). Remaining: 100 - 36 = 64% for A+T. So adenine = thymine = 32% each. Students sometimes add incorrectly or forget that A=T and G=C are separate equalities.

2–3 Line Example (Typical Error)

NCERT Exemplar: DNA analysis shows A=29%, G=17%, C=32%, T=17%. Since A is not equal to T and G is not equal to C, this violates Chargaff's rule, so the DNA must be single-stranded.

How NEET Frames The Trap

Provide base composition that violates or follows Chargaff's rule and ask whether the nucleic acid is double-stranded, single-stranded, RNA, or DNA. Alternatively, give one base percentage and ask for the others.

NEET-Style Trap Question Format

Q. In sea urchin DNA, 17% of the bases are cytosine. The percentages of the other three bases are:
A. G 17%, A 33%, T 33% B. G 17%, A 16.5%, T 32.5% C. G 34%, A 24.5%, T 24.5% D. G 8.5%, A 50%, T 24.5%
Trick: G=C=17%, so A+T = 100-34 = 66%, giving A=T=33% each. Option (b) incorrectly halves guanine. Option (c) doubles guanine. Option (a) is correct.

Quick rule: Always check: does A=T and G=C? If yes, double-stranded DNA. If not, single-stranded DNA or RNA. For percentage problems: %G = %C; %A = %T; %A + %G + %C + %T = 100%.

RNA Polymerase Type Confusion

TranscriptionRNA polymeraseEukaryotes

Mistake Snapshot (What Students Do Wrong)

Swapping RNA Pol I and RNA Pol III products: In eukaryotes: RNA Pol I synthesises rRNA (28S, 18S, 5.8S). RNA Pol II synthesises mRNA (hnRNA). RNA Pol III synthesises tRNA and 5S rRNA. Students frequently swap Pol I (rRNA) with Pol III (tRNA), or forget that Pol III also makes 5S rRNA.
Assuming prokaryotes have multiple RNA polymerases: Prokaryotes have a single RNA polymerase (with sigma factor for initiation). Multiple RNA polymerase types (I, II, III) exist only in eukaryotes. Questions may test whether a student incorrectly assigns polymerase types to prokaryotes.

2–3 Line Example (Typical Error)

NEET asks: Removal of RNA polymerase III from nucleoplasm will affect synthesis of - correct answer is tRNA (and 5S rRNA). Students selecting rRNA confuse Pol III with Pol I.

How NEET Frames The Trap

Match-the-column or assertion-reason pairing RNA polymerase types with their products. Also tested via removal/inhibition scenarios.

NEET-Style Trap Question Format

Q. In eukaryotes, removal of RNA polymerase III from the nucleoplasm would specifically affect the synthesis of:
A. mRNA B. rRNA (28S, 18S, 5.8S) C. tRNA and 5S rRNA D. hnRNA
Trick: Option (b) lists the rRNAs made by Pol I, not Pol III. Option (d) is hnRNA, which is the precursor of mRNA made by Pol II. The correct answer is (c): Pol III makes tRNA and 5S rRNA.

Quick rule: Remember I-r-R-m-III-t: Pol I = rRNA, Pol II = mRNA, Pol III = tRNA (+ 5S rRNA). Prokaryotes have only ONE RNA polymerase.

Molecular Basis of Inheritance

Chapter Snapshot - Molecular Basis of Inheritance

Subtopics - Molecular Basis of Inheritance (NEET)

1) Nucleic Acids: DNA and RNA Structure

2) DNA as Genetic Material: Key Experiments

3) DNA Replication

4) RNA: Types and Structure

5) Genetic Code and Central Dogma

6) Transcription

7) Translation (Protein Synthesis)

8) Gene Regulation, DNA Fingerprinting, and Human Genome Project

Molecular Basis of Inheritance Download Notes & Weightage Plan

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

Molecular Basis of Inheritance Chapter NEET Traps & Common Mistakes (Topic-Wise)

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

NEET > Biology > Genetics And Evolution Chapters

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Molecular Basis of Inheritance

Overview content

Chapter Snapshot - Molecular Basis of Inheritance

Subtopics - Molecular Basis of Inheritance (NEET)

Molecular Basis of Inheritance Download Notes & Weightage Plan

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

Molecular Basis of Inheritance Chapter NEET Traps & Common Mistakes (Topic-Wise)

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

NEET > Biology > Genetics And Evolution Chapters

Comments

Leave a comment

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