Molecular Biology

Molecular Biology

 

Unit I: Organization of DNA

1. Nucleic Acids as Hereditary Material

  • DNA and RNA are the carriers of genetic information.
  • Experiment by Avery, MacLeod, and McCarty confirmed DNA as hereditary material.
  • Hershey-Chase experiment provided further evidence.

2. Structure and Forms of DNA

  • DNA is composed of nucleotides: phosphate group, sugar (deoxyribose), and nitrogenous bases (Adenine, Thymine, Guanine, Cytosine).
  • Primary structure: Linear sequence of nucleotides.
  • Secondary structure: Double helix (Watson-Crick model).
    • Antiparallel strands held together by hydrogen bonds.
    • A-T has 2 hydrogen bonds, G-C has 3 hydrogen bonds.

3. Forms of DNA

  • A-DNA: Right-handed, short and broad. Found in dehydrated conditions.
  • B-DNA: Right-handed, most common form in cells.
  • Z-DNA: Left-handed, zigzag appearance, found in regions of high GC content.

4. RNA Structure

  • RNA is single-stranded, contains ribose sugar.
  • Bases: Adenine, Uracil, Guanine, Cytosine.
  • Types: mRNA, tRNA, rRNA, siRNA, miRNA.

5. Supercoiling of DNA

  • Supercoiling refers to the overwinding or underwinding of DNA.
  • Enzymes involved: Topoisomerase I and II.
  • Helps in compact packing of DNA in the nucleus.

Unit II: DNA Replication and Repair

1. DNA Replication Models

  • Conservative: Entire DNA acts as a template, but new DNA remains separate.
  • Semi-Conservative: Each strand serves as a template (proved by Meselson-Stahl experiment).
  • Dispersive: DNA fragments replicated and recombined randomly.

2. Mechanism of DNA Replication

  • Occurs in the S-phase of the cell cycle.
  • Key Steps:
    • Initiation: Origin of replication (ori) site, helicase unwinds DNA.
    • Elongation: DNA polymerase adds nucleotides to the 3′ end of the template.
    • Termination: Replication ends when replication forks meet.

3. DNA Damage and Repair Mechanisms

  • Types of DNA Damage:
    • Point mutations (base substitutions).
    • Double-strand breaks.
    • Thymine dimers caused by UV radiation.
  • Repair Mechanisms:
    • Base Excision Repair (BER).
    • Nucleotide Excision Repair (NER).
    • Mismatch Repair (MMR).
    • Double-Strand Break Repair (Homologous Recombination and Non-Homologous End Joining).
  • Diseases Due to Repair Defects:
    • Xeroderma pigmentosum (NER defect).
    • Hereditary non-polyposis colorectal cancer (MMR defect).

Unit III: Transcription

1. Importance of DNA Binding Proteins

  • DNA-binding proteins like transcription factors regulate gene expression.
  • Examples: Zinc fingers, leucine zippers.

2. RNA Polymerase

  • Prokaryotes: Single RNA polymerase synthesizes all RNA types.
  • Eukaryotes:
    • RNA Polymerase I: rRNA synthesis.
    • RNA Polymerase II: mRNA synthesis.
    • RNA Polymerase III: tRNA and 5S rRNA synthesis.

3. Mechanism of Transcription

  • Prokaryotes:
    • Initiation: Sigma factor binds promoter.
    • Elongation: RNA polymerase synthesizes RNA.
    • Termination: Rho-dependent or independent.
  • Eukaryotes:
    • Pre-initiation complex forms at TATA box.
    • Transcription factors and RNA polymerase II are involved.

4. Genetic Code

  • Triplet codons specify amino acids.
  • Features: Degenerate, unambiguous, universal.
  • Cracking of code: Experiment by Nirenberg and Khorana.

Unit IV: Translation

1. Machinery of Translation

  • Components: mRNA, tRNA, ribosomes, aminoacyl-tRNA synthetase.
  • Ribosomes: 70S (prokaryotes), 80S (eukaryotes).

2. Mechanism of Translation

  • Prokaryotes:
    • Initiation: Shine-Dalgarno sequence binds ribosome.
    • Elongation: Peptide bond formation by peptidyl transferase.
    • Termination: Stop codon recognized by release factors.
  • Eukaryotes: Similar steps but involve 5′ cap and poly-A tail.

3. Post-Translational Modifications (PTMs)

  • PTMs enhance protein stability and function.
  • Types:
    • Phosphorylation: Addition of phosphate groups.
    • Adenylation: Addition of adenine nucleotide.
    • Acylation: Addition of acyl groups.
    • Glycosylation: Attachment of sugar molecules.

Unit V: Regulation of Gene Expression

1. Regulation in Prokaryotes

  • Positive Control: Activators enhance transcription.
  • Negative Control: Repressors inhibit transcription.
  • Operons:
    • Lac Operon: Inducible system regulated by lactose.
    • Trp Operon: Repressible system regulated by tryptophan.

2. Regulation in Eukaryotes

  • DNA Level: Chromatin remodeling, DNA methylation.
  • Transcription Level: Activators, repressors, enhancers.
  • Translation Level: RNA interference, microRNAs.
  • Post-Translation: PTMs, protein degradation.

3. Antisense Technology

  • Molecular Mechanism: Antisense oligonucleotides bind mRNA to block translation.
  • Applications: Gene silencing, therapeutic use (e.g., treating genetic disorders).

 

Unit I: Organization of DNA – Questions and Answers


Question 1: What evidence supports that DNA is the hereditary material?
Answer:

  1. Griffith’s Experiment (1928):
    • Griffith demonstrated the phenomenon of bacterial transformation.
    • Heat-killed virulent bacteria (S-strain) transferred their genetic material to non-virulent bacteria (R-strain), making them virulent.
    • This suggested that a “transforming principle” exists, later identified as DNA.
  2. Avery, MacLeod, and McCarty (1944):
    • Used purified extracts of DNA, RNA, and protein from heat-killed S-strain.
    • Only DNA could transform R-strain into S-strain, confirming DNA as the hereditary material.
  3. Hershey-Chase Experiment (1952):
    • Used bacteriophages labeled with radioactive isotopes (^32P for DNA and ^35S for protein).
    • After infection, only ^32P (DNA) was found in host cells, proving that DNA is responsible for heredity.

Question 2: Describe the Watson-Crick model of the DNA double helix.
Answer:

  1. Structure:
    • DNA is a double helix made of two antiparallel strands.
    • Each strand consists of a sugar-phosphate backbone with nitrogenous bases attached.
  2. Base Pairing:
    • Adenine (A) pairs with Thymine (T) via 2 hydrogen bonds.
    • Guanine (G) pairs with Cytosine (C) via 3 hydrogen bonds.
    • This pairing maintains a uniform diameter of ~2 nm.
  3. Helix Features:
    • Helix has a right-handed twist.
    • 10 base pairs per turn (~3.4 nm per turn).
    • Major and minor grooves provide binding sites for regulatory proteins.
  4. Stability:
    • Stabilized by hydrogen bonding and base stacking interactions.
  5. Significance:
    • Complementary base pairing allows accurate DNA replication.

Question 3: Explain the different forms of DNA and their significance.
Answer:

  1. A-DNA:
    • Right-handed helix, shorter and thicker.
    • Occurs under dehydrated conditions or high salt concentration.
    • Plays a role in crystallographic studies.
  2. B-DNA:
    • Most common form in physiological conditions.
    • Right-handed helix with ~10.5 base pairs per turn.
    • Represents the native state of DNA in cells.
  3. Z-DNA:
    • Left-handed helix with a zigzag sugar-phosphate backbone.
    • Found in regions with high GC content or during transcription.
    • Believed to play a role in gene regulation.
  4. Significance:
    • DNA can adopt these forms to adapt to environmental conditions and functional requirements.

Question 4: What is DNA supercoiling, and why is it important?
Answer:

  1. Definition:
    • DNA supercoiling refers to the overwinding (positive supercoiling) or underwinding (negative supercoiling) of the double helix.
  2. Enzymes Involved:
    • Topoisomerase I: Relieves supercoils by cutting one strand.
    • Topoisomerase II (DNA gyrase): Introduces or removes supercoils by cutting both strands.
  3. Functions of Supercoiling:
    • DNA Packaging: Helps compact DNA to fit into the nucleus.
    • Replication and Transcription: Negative supercoiling facilitates unwinding of DNA strands.
    • Regulation: Influences gene expression by altering DNA accessibility.
  4. Clinical Relevance:
    • Antibiotics like ciprofloxacin target bacterial topoisomerases to prevent supercoiling, inhibiting bacterial replication.

Question 5: Differentiate between DNA and RNA in terms of structure and function.
Answer:

Feature DNA RNA
Strand Type Double-stranded Single-stranded
Sugar Deoxyribose Ribose
Bases A, T, G, C A, U, G, C
Stability More stable (due to absence of 2′-OH group) Less stable (due to 2′-OH group)
Forms A-DNA, B-DNA, Z-DNA mRNA, tRNA, rRNA, miRNA, siRNA
Function Long-term storage of genetic information Involved in protein synthesis and regulation
Examples Chromosomal DNA in eukaryotes mRNA for transcription, tRNA for translation
  1. Significance:
    • DNA acts as the genetic blueprint.
    • RNA serves as a functional molecule in gene expression and regulation.

 

Unit II: DNA Replication and Repair – Questions and Answers


Question 1: What are the models of DNA replication? Explain the semi-conservative model in detail.
Answer:
DNA replication is the process by which DNA is copied to ensure genetic continuity. There are three proposed models:

  1. Conservative Model: The parental DNA remains intact, and a completely new molecule is synthesized.
  2. Semi-Conservative Model: Each parental strand serves as a template for the new strand, resulting in two DNA molecules, each with one old and one new strand.
  3. Dispersive Model: Parental DNA is broken into fragments, and new DNA is synthesized in segments, resulting in a mix of old and new DNA in each strand.

Semi-Conservative Model:

  • Proposed by Watson and Crick, confirmed by the Meselson-Stahl experiment.
  • Process:
    • Parental strands separate with the help of helicase.
    • Each strand acts as a template for complementary base pairing.
    • DNA polymerase synthesizes the new strands by adding nucleotides.
    • The result is two DNA molecules, each with one parental and one newly synthesized strand.

Significance: Ensures genetic fidelity by preserving half of the original DNA in every replication cycle.


Question 2: What are the key enzymes involved in DNA replication, and what are their functions?
Answer:
Several enzymes coordinate DNA replication to ensure accuracy and efficiency:

  1. Helicase: Unwinds the DNA double helix by breaking hydrogen bonds between base pairs.
  2. Single-Strand Binding Proteins (SSBPs): Stabilize the unwound DNA strands to prevent them from reannealing.
  3. Primase: Synthesizes RNA primers to provide a starting point for DNA polymerase.
  4. DNA Polymerase:
    • Prokaryotes: DNA Polymerase III synthesizes the leading and lagging strands. DNA Polymerase I replaces RNA primers with DNA.
    • Eukaryotes: DNA Polymerase α (primer synthesis), δ (lagging strand synthesis), and ε (leading strand synthesis).
  5. Ligase: Seals nicks in the sugar-phosphate backbone, joining Okazaki fragments.
  6. Topoisomerase: Relieves supercoiling caused by DNA unwinding.

Importance: These enzymes work in a coordinated manner to ensure the replication process is accurate and error-free.


Question 3: What are the different types of DNA damage, and how do they occur?
Answer:
DNA is continuously exposed to damaging agents that can cause mutations or structural changes. The major types of DNA damage include:

  1. Point Mutations: Single-base changes caused by errors in replication or chemical mutagens.
  2. Thymine Dimers: Formation of covalent bonds between adjacent thymine bases due to UV radiation.
  3. Single-Strand Breaks (SSBs): Breaks in one strand of DNA, often caused by reactive oxygen species (ROS).
  4. Double-Strand Breaks (DSBs): Breaks in both DNA strands, typically caused by ionizing radiation or severe mechanical stress.
  5. Chemical Modifications:
    • Deamination: Conversion of cytosine to uracil.
    • Alkylation: Addition of alkyl groups to bases.

Significance: If left unrepaired, these damages can lead to mutations, cancer, or cell death.


Question 4: Describe the mechanisms of DNA repair and their importance.
Answer:
DNA repair mechanisms are essential to maintain genetic stability. The key repair pathways include:

  1. Base Excision Repair (BER):
    • Repairs small, non-helix-distorting base lesions caused by oxidation or deamination.
    • Enzymes: DNA glycosylase, AP endonuclease, DNA polymerase, and ligase.
  2. Nucleotide Excision Repair (NER):
    • Removes bulky lesions like thymine dimers.
    • Steps: Damage recognition, excision of a short DNA segment, synthesis by DNA polymerase, and ligation.
  3. Mismatch Repair (MMR):
    • Fixes base pair mismatches introduced during DNA replication.
    • Proteins: MutS, MutL, and MutH.
  4. Double-Strand Break Repair:
    • Homologous Recombination (HR): Accurate repair using a sister chromatid as a template.
    • Non-Homologous End Joining (NHEJ): Direct joining of broken ends but prone to errors.

Importance: Prevents mutations, genomic instability, and diseases like cancer.


Question 5: What diseases are caused by defects in DNA repair mechanisms?
Answer:
DNA repair deficiencies can lead to various genetic disorders and predispose individuals to cancer. Some examples include:

  1. Xeroderma Pigmentosum (XP):
    • Cause: Defects in NER.
    • Symptoms: Extreme sensitivity to UV light, high risk of skin cancer.
  2. Hereditary Non-Polyposis Colorectal Cancer (HNPCC):
    • Cause: Mutations in mismatch repair (MMR) genes.
    • Symptoms: Increased risk of colorectal and other cancers.
  3. Ataxia Telangiectasia (AT):
    • Cause: Mutation in the ATM gene, affecting DSB repair.
    • Symptoms: Neurological defects, increased cancer risk.
  4. Fanconi Anemia:
    • Cause: Defects in homologous recombination.
    • Symptoms: Bone marrow failure, leukemia, and congenital abnormalities.

Significance: Understanding these diseases highlights the critical role of DNA repair in preventing cancer and maintaining cellular health.

 

Unit III: Transcription – Questions and Answers


Q1. What is the role of DNA-binding proteins in transcription regulation?
Answer:
DNA-binding proteins play a critical role in regulating transcription by interacting with specific DNA sequences and influencing RNA polymerase activity.

  • Functions of DNA-binding proteins:
    • Facilitate the recruitment of RNA polymerase to promoter regions.
    • Act as activators or repressors of gene expression.
    • Modify chromatin structure to enable or block access to DNA.
  • Types of DNA-binding proteins:
    • Transcription factors: Bind to promoter or enhancer regions to regulate transcription.
    • Zinc finger proteins: Use zinc ions to stabilize their DNA-binding domains.
    • Helix-turn-helix proteins: Recognize specific DNA sequences.
  • Examples:
    • Lac repressor in prokaryotes: Inhibits transcription of the lac operon.
    • CREB in eukaryotes: Binds to cAMP response elements to activate transcription.

Q2. Compare the structure and functions of RNA polymerase in prokaryotes and eukaryotes.
Answer:
RNA polymerase is the enzyme responsible for synthesizing RNA from a DNA template, but its structure and function differ between prokaryotes and eukaryotes.

  • Prokaryotic RNA polymerase:
    • Composed of a core enzyme (α2ββ′ω) and a sigma (σ) factor.
    • Core enzyme performs RNA synthesis; the sigma factor recognizes promoter regions.
    • Produces all types of RNA (mRNA, rRNA, tRNA).
  • Eukaryotic RNA polymerase:
    • Three types of RNA polymerases:
      • RNA Polymerase I: Synthesizes rRNA (28S, 18S, 5.8S).
      • RNA Polymerase II: Synthesizes mRNA and some snRNAs.
      • RNA Polymerase III: Synthesizes tRNA and 5S rRNA.
    • Requires general transcription factors (e.g., TFIIA, TFIIB) to bind promoter regions.
    • Operates in a more complex regulatory environment due to chromatin structure.

Q3. Describe the mechanism of transcription in prokaryotes.
Answer:
Transcription in prokaryotes involves three main steps: initiation, elongation, and termination.

  1. Initiation:
    • Sigma factor guides RNA polymerase to the promoter region.
    • Promoter contains conserved sequences: -10 (Pribnow box) and -35 regions.
    • RNA polymerase unwinds the DNA to form an open complex.
  2. Elongation:
    • RNA polymerase synthesizes RNA in the 5′ to 3′ direction using ribonucleotides (ATP, UTP, GTP, CTP).
    • A temporary DNA-RNA hybrid is formed during synthesis.
  3. Termination:
    • Rho-dependent termination: Rho protein binds to the RNA and causes dissociation of RNA polymerase.
    • Rho-independent termination: Formation of a hairpin loop in RNA followed by a U-rich sequence causes termination.

Q4. What are the key features of the genetic code, and how was it deciphered?
Answer:
The genetic code is the set of rules by which the nucleotide sequence in mRNA is translated into amino acids.

  • Features of the genetic code:
    • Triplet code: Each codon consists of three nucleotides.
    • Degenerate: Multiple codons can code for the same amino acid (e.g., UCU, UCC, UCA all code for serine).
    • Universal: Nearly identical in all organisms.
    • Non-overlapping: Each nucleotide is part of only one codon.
  • Cracking the code:
    • Nirenberg and Matthaei experiment (1961): Used synthetic RNA (e.g., poly-U) to demonstrate that UUU codes for phenylalanine.
    • Khorana’s experiments: Developed synthetic RNA sequences to identify codons for specific amino acids.
    • Nirenberg and Leder: Used tRNA and ribosomes to confirm codon assignments.

Q5. How does transcription differ between prokaryotes and eukaryotes?
Answer:
Transcription in prokaryotes and eukaryotes differs in complexity, location, and regulatory mechanisms.

  • Prokaryotes:
    • Occurs in the cytoplasm.
    • A single RNA polymerase transcribes all RNA types.
    • Transcription and translation occur simultaneously (coupled).
    • Promoters are simple with -10 and -35 regions.
    • Sigma factor directly recognizes promoters.
  • Eukaryotes:
    • Occurs in the nucleus.
    • Three RNA polymerases transcribe specific RNA types.
    • Transcription is separated from translation (transcription in nucleus, translation in cytoplasm).
    • Promoters are more complex with TATA box, enhancers, and silencers.
    • Requires transcription factors (e.g., TFIID binds TATA box).
    • Post-transcriptional modifications: Addition of 5′ cap, poly-A tail, and splicing of introns.

By understanding these differences, researchers can develop insights into regulatory mechanisms and therapeutic strategies targeting gene expression.

 

Unit IV: Translation

Here are five detailed questions and answers from Unit IV: Translation, enriched with relevant keywords for high ranking:


Q1: What is the role of ribosomes in protein synthesis, and how do prokaryotic and eukaryotic ribosomes differ?

Answer:

  • Ribosomes are the molecular machines responsible for translating mRNA into proteins.
  • They consist of two subunits:
    • Prokaryotic Ribosomes: 70S, composed of 50S (large) and 30S (small) subunits.
    • Eukaryotic Ribosomes: 80S, composed of 60S (large) and 40S (small) subunits.
  • Functions in Translation:
    • Facilitate binding of mRNA and tRNA.
    • Catalyze peptide bond formation via peptidyl transferase activity (rRNA acts as a ribozyme).
    • Coordinate the stepwise elongation of the polypeptide chain.

Q2: Explain the initiation process of translation in prokaryotes.

Answer:

  • Initiation is the first step in protein synthesis, where the ribosome assembles on the mRNA to begin translation.
  • Key Steps:
    1. Recognition of Shine-Dalgarno Sequence:
      • The 16S rRNA of the 30S ribosomal subunit binds to the Shine-Dalgarno sequence on mRNA.
    2. Binding of Initiator tRNA:
      • The initiator tRNA carrying formylmethionine (fMet) binds to the start codon (AUG) at the P-site.
    3. Assembly of the Ribosome:
      • The 50S subunit joins the 30S subunit to form the functional 70S ribosome.
  • Factors Involved:
    • Initiation factors (IF-1, IF-2, IF-3) assist in proper assembly.
  • Energy Requirement: GTP hydrolysis drives the process.

Q3: What is the elongation phase of translation, and how does it differ between prokaryotes and eukaryotes?

Answer:

  • Elongation is the stage where the polypeptide chain grows by the sequential addition of amino acids.
  • Steps:
    1. Codon Recognition:
      • The appropriate aminoacyl-tRNA enters the A-site of the ribosome based on codon-anticodon pairing.
    2. Peptide Bond Formation:
      • The amino acid at the A-site forms a peptide bond with the growing chain at the P-site.
      • Catalyzed by peptidyl transferase activity of the ribosome.
    3. Translocation:
      • The ribosome moves one codon forward, shifting the tRNA from the A-site to the P-site, and the empty tRNA exits from the E-site.
  • Differences Between Prokaryotes and Eukaryotes:
    • Prokaryotes: Elongation factors EF-Tu and EF-G are involved.
    • Eukaryotes: Corresponding factors are eEF-1 and eEF-2.

Q4: Describe the termination process of translation in detail.

Answer:

  • Termination occurs when a stop codon is reached on the mRNA, signaling the end of translation.
  • Stop Codons: UAA, UAG, UGA (do not code for any amino acid).
  • Steps:
    1. Recognition of Stop Codon:
      • Release Factors (RFs) bind to the stop codon in the A-site.
      • Prokaryotes: RF-1 (UAA, UAG) and RF-2 (UAA, UGA).
      • Eukaryotes: eRF1 recognizes all stop codons.
    2. Polypeptide Release:
      • Peptidyl transferase hydrolyzes the bond between the polypeptide and tRNA, releasing the newly formed protein.
    3. Ribosome Disassembly:
      • The ribosome dissociates into subunits with the help of GTP hydrolysis.
  • Energy Requirement: GTP hydrolysis by release factors.

Q5: What are post-translational modifications, and why are they important for protein function?

Answer:


molecular biology, DNA structure, RNA structure, double helix, nucleotides, genetic material, DNA replication, transcription, translation, gene expression, genetic code, DNA damage, DNA repair mechanisms, mutations, operon model, lac operon, trp operon, RNA polymerase, tRNA, mRNA, rRNA, DNA-binding proteins, ribosomes, chromatin remodeling, epigenetics, antisense technology, gene regulation, post-translational modifications, protein synthesis, helicase, topoisomerase, base excision repair, nucleotide excision repair, homologous recombination, genetic engineering, biotechnology, molecular cloning, PCR, CRISPR-Cas9, DNA sequencing, genome editing, proteomics, functional genomics.

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