Biology - Botany - How to Solve: Molecular Basis of Inheritance (NEET UG) – Complete Guide

How to Solve: Molecular Basis of Inheritance (NEET UG) – Complete Guide

Introduction

"Mastering DNA replication, transcription, translation, and the lac operon unlocks 8–10 direct questions in NEET UG—worth 32–40 marks. This is your golden ticket to a top biology score."

WHAT YOU NEED TO KNOW FIRST

1. Central Dogma of Molecular Biology – DNA → RNA → Protein.
2. Basic DNA structure – Double helix, antiparallel strands, 5’ → 3’ direction.
3. Enzyme basics – Polymerases, ligases, and their roles in synthesis.

KEY TERMS & FORMULAS

DNA Replication

MEMORIZE THIS: - DNA replication direction: Always 5’ → 3’. - Okazaki fragment length: ~100–200 nucleotides in prokaryotes, ~1000–2000 in eukaryotes.

Transcription

MEMORIZE THIS: - Transcription direction: 5’ → 3’ (RNA synthesis). - Prokaryotes: Single RNA polymerase. - Eukaryotes: RNA Polymerase I (rRNA), II (mRNA), III (tRNA).

Translation

MEMORIZE THIS: - Genetic code is: - Universal (same in all organisms). - Degenerate (multiple codons for one amino acid). - Non-overlapping (read in triplets). - Commaless (no gaps between codons).

Lac Operon

MEMORIZE THIS: - Lac operon is: - Inducible (turned on by lactose). - Repressible (turned off by glucose via CAP-cAMP). - No lactose → Repressor binds operator → No transcription. - Lactose present → Allolactose binds repressor → Transcription occurs. - Low glucose → High cAMP → CAP binds → Enhanced transcription.

STEP-BY-STEP METHOD

1. DNA Replication (Prokaryotes)

Step 1: Identify the origin of replication (ori). Step 2: Helicase unwinds DNA at the replication fork. Step 3: SSBPs stabilize single-stranded DNA. Step 4: Topoisomerase relieves supercoiling ahead of the fork. Step 5: Primase synthesizes a short RNA primer (5’ → 3’). Step 6: DNA Polymerase III adds nucleotides to the 3’ end of the primer (5’ → 3’). Step 7: Leading strand is synthesized continuously. Step 8: Lagging strand is synthesized discontinuously (Okazaki fragments). Step 9: DNA Polymerase I replaces RNA primers with DNA. Step 10: Ligase joins Okazaki fragments.

2. Transcription (Prokaryotes)

Step 1: RNA Polymerase binds to the promoter (e.g., -10 and -35 sequences in E. coli). Step 2: DNA unwinds, forming a transcription bubble. Step 3: RNA synthesis begins at the +1 site (5’ → 3’). Step 4: RNA Polymerase reads the template strand (3’ → 5’). Step 5: RNA is synthesized complementary to the template strand (U instead of T). Step 6: Transcription stops at the terminator (rho-dependent or rho-independent).

3. Translation (Prokaryotes)

Step 1: Small ribosomal subunit binds to mRNA at the Shine-Dalgarno sequence. Step 2: Initiator tRNA (fMet-tRNA) binds to AUG at the P-site. Step 3: Large ribosomal subunit joins, forming the 70S ribosome. Step 4: Next tRNA enters the A-site (anticodon matches mRNA codon). Step 5: Peptide bond forms between amino acids (P-site → A-site). Step 6: Ribosome translocates (moves 3 nucleotides 5’ → 3’). Step 7: Empty tRNA exits via the E-site. Step 8: Repeat until a stop codon (UAA, UAG, UGA) is reached. Step 9: Release factor binds, releasing the polypeptide.

4. Lac Operon Regulation

Step 1: No lactose → Repressor (Lac I) binds operator (O) → No transcription. Step 2: Lactose present → Allolactose binds repressor → Repressor dissociates → Transcription allowed. Step 3: Low glucose → High cAMP → CAP-cAMP binds promoter → Enhanced transcription. Step 4: High glucose → Low cAMP → CAP does not bind → Reduced transcription.

WORKED EXAMPLES

Example 1 – Basic: DNA Replication

Question: If the template strand of DNA is 3’-ATCGGTA-5’, what is the sequence of the newly synthesized strand?

Step 1: Identify the template strand (3’ → 5’). Step 2: New strand is synthesized 5’ → 3’, complementary to the template. Step 3: Write the complementary bases: - A → T - T → A - C → G - G → C Step 4: New strand: 5’-TAGCCAT-3’

What we did and why: - Used base-pairing rules (A-T, C-G). - Remembered 5’ → 3’ synthesis direction.

Example 2 – Medium: Transcription & Translation

Question: A DNA template strand has the sequence 3’-TACGGATTC-5’. a) What is the mRNA sequence? b) What is the amino acid sequence?

Part a) Transcription: Step 1: Identify the template strand (3’ → 5’). Step 2: mRNA is synthesized 5’ → 3’, complementary to the template (U instead of T). Step 3: Write the complementary bases: - T → A - A → U - C → G - G → C Step 4: mRNA sequence: 5’-AUGCCUAAG-3’

Part b) Translation: Step 1: Split mRNA into codons: AUG | CCU | AAG Step 2: Use the genetic code table: - AUG → Methionine (Met) - CCU → Proline (Pro) - AAG → Lysine (Lys) Step 3: Amino acid sequence: Met-Pro-Lys

What we did and why: - Transcription: Used complementary base pairing (U instead of T). - Translation: Split mRNA into triplets and matched to the genetic code.

Example 3 – Exam-Style: Lac Operon

Question: In E. coli, if lactose is present but glucose is absent, what happens to the lac operon? Options: A) Repressor binds operator, no transcription. B) Repressor is inactive, CAP-cAMP binds, high transcription. C) Repressor is active, CAP-cAMP does not bind, low transcription. D) Repressor is inactive, CAP-cAMP does not bind, low transcription.

Step 1: Lactose present → Allolactose binds repressor → Repressor inactive → Transcription allowed. Step 2: Glucose absent → High cAMP → CAP-cAMP binds promoter → Enhanced transcription. Step 3: Correct answer: B

What we did and why: - Applied lac operon regulation rules: - Lactose → Repressor inactive. - No glucose → CAP-cAMP active. - Eliminated wrong options by checking both conditions.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second crash course for NEET’s molecular biology questions.

1. DNA Replication:
2. Semi-conservative, 5’ → 3’ synthesis, Okazaki fragments on lagging strand.

3. Helicase unwinds, primase adds RNA primer, DNA Pol III extends, ligase seals gaps.

4. Transcription:

5. RNA Polymerase binds promoter, reads template strand (3’ → 5’), makes mRNA (5’ → 3’).

6. Prokaryotes: 1 RNA Pol. Eukaryotes: 3 (I, II, III).

7. Translation:

8. mRNA → Protein.
9. AUG = Start (Met), UAA/UAG/UGA = Stop.

10. tRNA brings amino acids (anticodon matches codon).

11. Lac Operon:

12. No lactose → Repressor binds → No transcription.
13. Lactose present → Repressor inactive → Transcription allowed.
14. No glucose → CAP-cAMP binds → High transcription.

Memorize the genetic code table, directionality, and lac operon conditions. You’ve got this!