Type Here to Get Search Results !

NEET Biology Notes: Molecular Basis of Inheritance Chapter Revision with MCQs

0

⚡ Quick Summary: Molecular Basis of Inheritance for NEET UG

Mastering molecular inheritance notes for neet requires a strong conceptual grip on nucleic acid organization, central dogma mechanisms, gene expression control, and genetic mapping. This high-yield NCERT revision guide simplifies key mechanisms like semi-conservative DNA replication, transcription, translation, operon models, and DNA fingerprinting into actionable insights for scoring 360/360 in NEET Biology.

NEET Biology Notes on Molecular Basis of Inheritance featuring NCERT-based chapter revision, high-yield concepts, and chapter-wise MCQs by @indiabiologyneet.


Parameter Details
NCERT Class Class 12 Biology (Chapter 6)
NEET Weightage 5 to 7 Questions (~20–28 Marks)
Key Focus Areas Replication Fork Enzymes, Transcription Units, Genetic Code Properties, Lac Operon Mechanism, VNTR Profiling

🎯 Target 360/360 in NEET Biology!

Boost your preparation with Chapter-wise NCERT revision notes and targeted test papers designed for top scores.

Access Full Chapter Guide →

Detailed Concept Breakdown & High-Yield Notes

1. Structure of Nucleic Acids & Packaging of DNA

A nucleotide consists of a nitrogenous base, a pentose sugar, and a phosphate group linked via 3'-5' phosphodiester bonds. Nitrogenous bases comprise purines (Adenine, Guanine) and pyrimidines (Cytosine, Thymine, Uracil).

  • Chargaff’s Rule: Applicable strictly to double-stranded DNA. [A] + [G] = [T] + [C] or ([A]+[G]) / ([T]+[C]) = 1.
  • DNA Packaging in Eukaryotes: Negatively charged DNA is wrapped around a positively charged basic protein octamer called histones (rich in lysine and arginine) to form a nucleosome (~200 bp DNA). Higher packaging relies on Non-Histone Chromosomal (NHC) proteins.
  • Euchromatin vs Heterochromatin: Euchromatin is loosely packed, lightly stained, and transcriptionally active. Heterochromatin is densely packed, darkly stained, and transcriptionally inactive.
💡 NEET High-Yield Trick / Mnemonic: To remember histone chemical properties, think "LA-Histone": Histones are rich in Lysine and Arginine, making them basic and positively charged to tightly attract acidic, negatively charged DNA (PO43- backbone).

2. DNA Replication: The Semi-Conservative Mechanism

Proven experimentally by Meselson and Stahl using heavy isotope 15N in E. coli and by Taylor et al. using radioactive 3H-thymidine in Vicia faba root tips.

  • Origin of Replication (ori): Specific site where replication starts. DNA unwinding by Helicase forms a Replication Fork.
  • DNA Polymerase: Catalyzes polymerization strictly in the 5' → 3' direction.
  • Leading vs Lagging Strand: The 3' → 5' template strand yields a continuous leading strand. The 5' → 3' template forms a discontinuous lagging strand made of Okazaki fragments, which are joined by DNA Ligase.

🔥 Crack NEET 2027 with 100% NCERT-Based MCQs!

Test your conceptual clarity with 90 high-yield, exam-pattern practice questions curated directly from NCERT lines.

Practice 90 NCERT MCQs Now →

3. Transcription: Converting DNA into RNA

A transcription unit comprises three structural elements: a Promoter (5'-upstream), a Structural Gene, and a Terminator (3'-downstream).

  • In prokaryotes, a single RNA Polymerase synthesizes all RNA types with the help of a Sigma (σ) factor for initiation and a Rho (ρ) factor for termination.
  • Eukaryotes feature three distinct RNA polymerases:
    • RNA Pol I: Synthesizes rRNAs (28S, 18S, and 5.8S).
    • RNA Pol II: Synthesizes precursor mRNA (hnRNA).
    • RNA Pol III: Synthesizes tRNA, 5S rRNA, and snRNAs.
  • Post-Transcriptional Modifications in Eukaryotes:
    1. Splicing: Introns (non-coding sequences) are spliced out, and exons are joined together.
    2. Capping: Methyl guanosine triphosphate is added to the 5'-end.
    3. Tailing: Adenylate residues (~200–300) are added to the 3'-end in a template-independent manner.

4. Translation & The Lac Operon Model

Translation: Protein synthesis occurs on ribosomes. It involves charging of tRNA (aminoacylation), initiation, elongation, and termination governed by non-catalyzed release factors responding to stop codons (UAA, UAG, UGA).

Lac Operon (Francois Jacob & Jacques Monod): A polycistronic inducible system regulated coordinately.

  • Regulator gene (i): Codes for the repressor protein.
  • Structural genes:
    • lac Z: Codes for β-galactosidase (breaks lactose into glucose + galactose).
    • lac Y: Codes for Permease (increases cell permeability to β-galactosides).
    • lac A: Codes for Transacetylase.
  • In the Absence of Inducer (Lactose): Repressor binds operator, blocking RNA Polymerase → Switched OFF.
  • In the Presence of Inducer: Allolactose binds repressor, inactivating it → RNA Polymerase transcribes genes → Switched ON.

5. DNA Fingerprinting

Developed by Alec Jeffreys. Uses Variable Number of Tandem Repeats (VNTRs) belonging to satellite DNA as probe markers due to their high degree of polymorphism.

Key Differences: Prokaryotic vs Eukaryotic Transcription

Feature Prokaryotic Transcription Eukaryotic Transcription
Cellular Location Cytoplasm (coupled directly with translation) Nucleus (transcription separated from translation)
RNA Polymerase Single RNA Polymerase for all RNA types Three specialized polymerases (RNA Pol I, II, III)
Gene Structure Polycistronic genes (multiple proteins from one mRNA) Monocistronic genes (single protein per mRNA)
RNA Processing No processing required; transcript is functional mRNA Requires post-transcriptional processing (Splicing, Capping, Tailing)

Top 10 High-Yield Practice MCQs for NEET UG

Q1. If a double-stranded DNA molecule contains 30% Cytosine, what will be the percentage of Thymine in this DNA?

  • (A) 30%
  • (B) 20%
  • (C) 40%
  • (D) 60%
View Correct Answer & Explanation

Correct Answer: (B) 20%
Explanation: By Chargaff’s Rule, [G] = [C] and [A] = [T]. If Cytosine = 30%, Guanine = 30%. Together, G + C = 60%. The remaining A + T = 100% - 60% = 40%. Since Adenine equals Thymine, Thymine = 40% / 2 = 20%.

Q2. Which enzyme is responsible for synthesizing the short RNA primer required to initiate DNA replication?

  • (A) DNA Polymerase I
  • (B) DNA Ligase
  • (C) Primase
  • (D) Topoisomerase
View Correct Answer & Explanation

Correct Answer: (C) Primase
Explanation: DNA polymerases cannot initiate synthesis de novo; they require a free 3'-OH end. Primase (an RNA polymerase) synthesizes a short RNA primer to provide this starting point.

Q3. Match the RNA Polymerases with their respective products in eukaryotes:

a. RNA Polymerase I — (i) tRNA & 5S rRNA
b. RNA Polymerase II — (ii) 28S, 18S, 5.8S rRNA
c. RNA Polymerase III — (iii) hnRNA

  • (A) a-(ii), b-(iii), c-(i)
  • (B) a-(i), b-(ii), c-(iii)
  • (C) a-(iii), b-(i), c-(ii)
  • (D) a-(ii), b-(i), c-(iii)
View Correct Answer & Explanation

Correct Answer: (A) a-(ii), b-(iii), c-(i)
Explanation: RNA Polymerase I transcribes larger ribosomal RNAs (28S, 18S, 5.8S). RNA Polymerase II transcribes hnRNA (precursor mRNA). RNA Polymerase III transcribes tRNA, 5S rRNA, and snRNA.

Q4. Statement I: The genetic code is degenerate because one amino acid can be coded by more than one codon.
Statement II: AUG serves a dual function; it codes for Methionine and acts as an initiation codon.

  • (A) Both Statement I and Statement II are incorrect.
  • (B) Statement I is correct, but Statement II is incorrect.
  • (C) Statement I is incorrect, but Statement II is correct.
  • (D) Both Statement I and Statement II are correct.
View Correct Answer & Explanation

Correct Answer: (D) Both Statement I and Statement II are correct.
Explanation: Degeneracy means multiple codons specify a single amino acid. AUG acts both as the initiation codon and codes for Methionine.

Q5. What will happen if a nonsense mutation occurs in the lac Z gene of the Lac Operon?

  • (A) Only Permease will be produced.
  • (B) All three enzymes will be synthesized normally.
  • (C) β-Galactosidase translation will be prematurely terminated, hindering downstream gene expression.
  • (D) The repressor protein will fail to bind to the operator region.
View Correct Answer & Explanation

Correct Answer: (C) β-Galactosidase translation will be prematurely terminated, hindering downstream gene expression.
Explanation: A nonsense mutation introduces a premature stop codon in the lac Z sequence, interrupting polycistronic translation.

Q6. Assertion (A): Taylor and colleagues proved the semi-conservative mode of DNA replication in eukaryotes.
Reason (R): They used radioactive tritiated thymidine (3H-thymidine) on root tip chromosomes of Vicia faba.

  • (A) Both (A) and (R) are true, and (R) is the correct explanation of (A).
  • (B) Both (A) and (R) are true, but (R) is NOT the correct explanation of (A).
  • (C) (A) is true, but (R) is false.
  • (D) (A) is false, but (R) is true.
View Correct Answer & Explanation

Correct Answer: (A) Both (A) and (R) are true, and (R) is the correct explanation of (A).
Explanation: Taylor and his team experimentally demonstrated eukaryotic chromosomal semi-conservative replication using radioactive thymidine on Vicia faba.

Q7. Which post-transcriptional step involves adding adenylate residues to the 3'-end of eukaryotic hnRNA?

  • (A) Splicing
  • (B) Capping
  • (C) Tailing (Polyadenylation)
  • (D) Aminoacylation
View Correct Answer & Explanation

Correct Answer: (C) Tailing (Polyadenylation)
Explanation: Polyadenylation adds ~200–300 adenylate residues to the 3'-end of hnRNA without a template.

Q8. During DNA fingerprinting, the hybridization probe belongs to which DNA category?

  • (A) Exonic coding sequences
  • (B) Variable Number of Tandem Repeats (VNTRs)
  • (C) Ribosomal DNA
  • (D) Non-polymorphic single-copy genes
View Correct Answer & Explanation

Correct Answer: (B) Variable Number of Tandem Repeats (VNTRs)
Explanation: VNTRs belong to mini-satellite DNA featuring high polymorphism, enabling accurate genetic identification.

Q9. The amino acid attaches to which specific site of the tRNA molecule during charging?

  • (A) Anticodon loop
  • (B) 5'-phosphate end
  • (C) 3'-OH CCA end
  • (D) TψC loop
View Correct Answer & Explanation

Correct Answer: (C) 3'-OH CCA end
Explanation: Amino acids bind covalently via ester linkage to the terminal adenosine residue at the 3'-OH end (CCA sequence) of tRNA.

Q10. Who unequivocally proved that DNA, and not protein, is the genetic material in organisms?

  • (A) Frederick Griffith (1928)
  • (B) Avery, MacLeod, and McCarty (1944)
  • (C) Alfred Hershey and Martha Chase (1952)
  • (D) Watson and Crick (1953)
View Correct Answer & Explanation

Correct Answer: (C) Alfred Hershey and Martha Chase (1952)
Explanation: Hershey and Chase used radioactively labeled bacteriophages (32P for DNA and 35S for protein) to conclusively establish DNA as the hereditary material.

📸 Join Our NEET Biology Community on Instagram!

Get daily NCERT revision reels, high-yield diagrams, memory tricks, and real-time NEET updates directly on your feed.

Follow @gujaratbiologyneetplus on Instagram 🚀

Post a Comment

0 Comments

Top Post Ad

Below Post Ad