📚 StudyOS CBSE Class 5–12 AI Tutor

Molecular Basis of Inheritance

NCERT Class 12 · Biology Based on NCERT Class 12 Biology textbook · Free CBSE study kit

Chapter Notes

THE DNA

**DNA (deoxyribonucleic acid)** is a long polymer of deoxyribonucleotides that serves as the primary genetic material in most organisms. The length of DNA is measured in number of nucleotides or base pairs (bp) and is characteristic of each organism. For example:

  • Bacteriophage φ×174: 5,386 nucleotides
  • Bacteriophage lambda: 48,502 bp
  • Escherichia coli: 4.6 × 10⁶ bp
  • Haploid human DNA: 3.3 × 10⁹ bp
  • Structure of Polynucleotide Chain

    A **nucleotide** has three components:

  • **Nitrogenous base**: Purines (Adenine-A, Guanine-G) or Pyrimidines (Cytosine-C, Thymine-T in DNA; Uracil-U in RNA)
  • **Pentose sugar**: Deoxyribose in DNA, Ribose in RNA
  • **Phosphate group**
  • The structural organization follows this sequence:

    1. Nitrogenous base + OH of 1' carbon of pentose sugar = **Nucleoside** (N-glycosidic linkage)

    2. Nucleoside + phosphate group at 5' carbon = **Nucleotide** (phosphoester linkage)

    3. Two nucleotides linked via **3'-5' phosphodiester linkage** = **Dinucleotide**

    4. Multiple nucleotides form a **polynucleotide chain** with:

  • **5'-end**: Free phosphate moiety
  • **3'-end**: Free OH group of 3' carbon
  • **Backbone**: Sugar-phosphate groups
  • **Projecting bases**: Nitrogenous bases attached to sugar
  • **Key difference**: RNA has an additional **-OH group at 2' position** of ribose, making it more reactive and chemically labile than DNA.

    Double Helix Structure of DNA

    **Friedrich Meischer** first identified DNA as an acidic nuclear substance in 1869, naming it "nuclein." However, the structure remained elusive until **Watson and Crick (1953)** proposed the **Double Helix Model** based on X-ray diffraction data from **Maurice Wilkins** and **Rosalind Franklin**, and chemical observations by **Erwin Chargaff**.

    **Chargaff's Rules**: In double-stranded DNA, the ratio of Adenine to Thymine and Guanine to Cytosine each equals 1 (A=T, G=C).

    #### Salient Features of Double Helix Structure:

    1. **Two polynucleotide chains**: Sugar-phosphate backbone with bases projecting inward

    2. **Anti-parallel polarity**: One strand runs 5'→3', the other runs 3'→5'

    3. **Base pairing through hydrogen bonds**:

  • Adenine pairs with Thymine: **2 hydrogen bonds**
  • Guanine pairs with Cytosine: **3 hydrogen bonds**
  • Purine always pairs opposite pyrimidine, maintaining uniform distance between strands (~2.0 nm)
  • 4. **Right-handed helical coiling**:

  • Pitch of helix: 3.4 nm
  • Approximately 10 base pairs per turn
  • Distance between consecutive base pairs: 0.34 nm
  • 5. **Base stacking**: Planes of base pairs stack over each other, providing additional stability through hydrophobic interactions

    This complementary base pairing has profound genetic implications: if one strand sequence is known, the other can be predicted. If each parental strand acts as a template for new strand synthesis, two identical daughter DNA molecules result, explaining faithful genetic transmission.

    Central Dogma of Molecular Biology

    **Francis Crick proposed**: **DNA → RNA → Protein**

    Genetic information flows unidirectionally from DNA to RNA (transcription) to Protein (translation).

    *Exception*: Some viruses (e.g., retroviruses) show reverse information flow: **RNA → DNA** via **reverse transcriptase** (reverse transcription).

    Packaging of DNA Helix

    If human DNA has 6.6 × 10⁹ bp and the distance between consecutive bp is 0.34 × 10⁻⁹ m, the total length equals approximately **2.2 meters** — far exceeding the nucleus diameter (~10⁻⁶ m). Sophisticated packaging is essential.

    #### In Prokaryotes:

  • DNA is not scattered throughout the cell but organized in a **nucleoid** region
  • DNA (negatively charged) is held by positively charged proteins
  • DNA is organized in large **supercoiled loops** held by proteins, reducing volume significantly
  • #### In Eukaryotes:

  • **Histones**: Positively charged, basic proteins rich in lysine and arginine residues
  • **Histone octamer**: Unit of 8 histone molecules (2 copies each of H2A, H2B, H3, H4)
  • **Nucleosome**: DNA wrapped around histone octamer containing ~200 bp of DNA (Figure 5.4a)
  • **Chromatin**: Repeating nucleosome units; appears as "beads-on-string" structure under electron microscopy (Figure 5.4b)
  • **Non-histone Chromosomal (NHC) proteins**: Additional proteins for higher-level chromatin packaging
  • #### Chromatin Organization States:

  • **Euchromatin**: Loosely packed, stains light, **transcriptionally active**
  • **Heterochromatin**: Densely packed, stains dark, transcriptionally inactive
  • A mammalian cell nucleus contains approximately 30 million nucleosomes (calculated from 6.6 × 10⁹ bp ÷ 200 bp/nucleosome).

    ---

    THE SEARCH FOR GENETIC MATERIAL

    Transforming Principle — Griffith's Experiments (1928)

    **Frederick Griffith** worked with **Streptococcus pneumoniae** (pneumococcal bacteria), observing two phenotypic variants:

  • **S strain (Smooth)**: Mucous polysaccharide coat, virulent, forms smooth colonies, kills mice
  • **R strain (Rough)**: No polysaccharide coat, avirulent, forms rough colonies, mice survive
  • **Experimental Design**:

    1. Heat-killed S strain injected into mice → mice survived

    2. Live R strain injected into mice → mice survived

    3. Heat-killed S strain + live R strain injected into mice → **mice died**

    4. Living S bacteria recovered from dead mice

    **Conclusion**: A **"transforming principle"** from heat-killed S cells transformed R cells into S cells (virulent). This proved genetic material transfer, though its biochemical nature remained undefined.

    Biochemical Characterisation of Transforming Principle — Avery-MacLeod-McCarty Experiments (1933-44)

    **Oswald Avery**, **Colin MacLeod**, and **Maclyn McCarty** identified the biochemical nature of Griffith's transforming principle.

    **Method**: Purified biochemicals from heat-killed S cells and tested which could transform live R cells:

  • **Proteases (protein-digesting enzymes)**: Did not inhibit transformation
  • **RNases (RNA-digesting enzymes)**: Did not inhibit transformation
  • **DNase (DNA-digesting enzyme)**: **Inhibited transformation**
  • **Conclusion**: **DNA is the hereditary material**, not protein or RNA. However, many biologists remained unconvinced.

    Genetic Material is DNA — Hershey-Chase Experiments (1952)

    **Alfred Hershey** and **Martha Chase** provided unequivocal proof using **bacteriophages** (viruses infecting bacteria).

    **Principle**: Viruses contain either protein coat or DNA. By radioactive labeling, they could track which entered bacterial cells.

    **Radioactive Labeling Strategy**:

  • **Radioactive phosphorus (³²P)**: Labels DNA only (DNA contains P; protein does not)
  • **Radioactive sulfur (³⁵S)**: Labels protein only (protein contains S; DNA does not)
  • **Experimental Steps** (Figure 5.5):

    1. Grew bacteriophages in media with ³²P or ³⁵S to produce radioactive viruses

    2. Allowed radioactive phages to attach to *E. coli* bacteria

    3. Used a **blender** to agitate and remove empty viral coats from bacterial surface

    4. **Centrifuged** to separate bacteria from viral coats

    **Results**:

  • **³²P-labeled phages**: Bacteria became radioactive → **DNA entered bacteria**
  • **³⁵S-labeled phages**: Bacteria remained non-radioactive → **protein did not enter bacteria**
  • **Conclusion**: **DNA is the genetic material** passed from virus to bacteria, definitively proving DNA is the hereditary substance.

    ---

    PROPERTIES OF GENETIC MATERIAL (DNA VERSUS RNA)

    A molecule fulfilling criteria for genetic material must:

    1. **Generate its replica** (support replication)

    2. **Be chemically and structurally stable**

    3. **Allow slow changes** (mutations) for evolution

    4. **Express itself** as observable Mendelian characters

    Comparative Analysis:

    #### Replication Capability:

    Both DNA and RNA can direct their duplication via **base pairing and complementarity** rules. Proteins fail this criterion entirely.

    #### Chemical Stability:

  • **RNA**: Contains **2'-OH group** at every nucleotide, making it highly reactive, labile, and easily degradable. RNA is catalytic (ribozymes), increasing reactivity.
  • **DNA**: Lacks 2'-OH group, making it chemically less reactive and structurally more stable. **Thymine (5-methyluracil) replaces uracil**, conferring additional stability and facilitating repair mechanisms.
  • **Result**: DNA is more stable than RNA, ideal for long-term genetic storage.

    #### Mutation Rate:

  • Both DNA and RNA can mutate
  • **RNA mutates faster** due to instability
  • **RNA viruses evolve rapidly** (shorter generation time, higher mutation rate)
  • #### Expression of Characters:

  • **RNA**: Directly codes for protein synthesis; can immediately express phenotypic characters
  • **DNA**: Dependent on RNA intermediates for protein synthesis; indirectly expresses characters through the RNA→Protein pathway
  • #### Conclusion:

    **DNA is the predominant genetic material** because its superior stability suits storage of genetic information across generations. **RNA is dynamic** in transmission and expression of genetic information. Both can function as genetic material, but DNA evolved as the primary storage molecule, while RNA evolved to facilitate and regulate information transfer.

    ---

    RNA WORLD

    Origin of Genetic Material

    **RNA was the first genetic material** in prebiotic Earth and early life forms.

    #### Evidence for RNA World Hypothesis:

    1. **RNA as catalyst**: Certain biochemical reactions are catalyzed by **RNA catalysts (ribozymes)**, not just protein enzymes. Examples: RNase P, ribosomal RNA (rRNA)

    2. **Evolution of protein synthesis**: The entire protein-synthesizing machinery evolved around RNA (ribosomal RNA in ribosomes)

    3. **Essential life processes** (metabolism, translation, splicing) show RNA involvement

    #### Evolution from RNA to DNA:

  • **RNA was reactive and unstable** because of its 2'-OH group and catalytic nature
  • **DNA evolved from RNA** through chemical modifications increasing stability
  • **Double-stranded DNA** with complementary strands enabled repair mechanisms, providing further protection against mutations and degradation
  • DNA became the **storage form** of genetic information; RNA retained roles in **information transfer and expression**
  • This transition explains why DNA dominates modern genetics while RNA performs dynamic regulatory and catalytic functions.

    ---

    REPLICATION

    Watson-Crick Semi-Conservative Model

    **Watson and Crick (1953)** immediately proposed a replication scheme from their double helix structure:

    > "It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material."

    **Mechanism**: The two DNA strands separate and act as **templates** for synthesis of new complementary strands. After replication, each new DNA molecule contains **one parental strand (old) and one newly synthesized strand (new)** — termed **semi-conservative replication**.

    Experimental Proof — Meselson-Stahl Experiment (1958)

    **Matthew Meselson** and **Franklin Stahl** elegantly proved semi-conservative replication using isotopic labeling in *E. coli*.

    #### Experimental Design:

    **Step 1 — Isotopic Labeling**:

  • Grew *E. coli* for many generations in medium containing heavy nitrogen (¹⁵N)
  • All DNA was labeled with ¹⁵N, appearing **heavy**
  • Then transferred bacteria to medium with normal nitrogen (¹⁴N)
  • **Step 2 — Replication Tracking**:

    After one DNA replication in ¹⁴N medium:

  • Extracted DNA and separated by **density gradient centrifugation** (cesium chloride density gradient)
  • DNA of different densities separated at different positions
  • **Step 3 — Results**:

    | Generation | Expected (Conservative) | Expected (Semi-Conservative) | Expected (Dispersive) | Observed |

    |-----------|-------------------------|------------------------------|----------------------|----------|

    | P (¹⁵N only) | Heavy | Heavy | Heavy | Heavy |

    | F₁ (after 1 replication in ¹⁴N) | 1 Heavy + 1 Light | Hybrid (¹⁵N-¹⁴N) | Hybrid | **Hybrid** |

    | F₂ (after 2 replications in ¹⁴N) | All Light (50%) + All Heavy (50%) | 50% Hybrid + 50% Light | Hybrid | **50% Hybrid + 50% Light** |

    **Observation**:

  • **F₁ generation**: All DNA was **hybrid density** (¹⁵N-¹⁴N), ruling out conservative replication (which would show heavy and light DNAs)
  • **F₂ generation**: **50% hybrid + 50% light** DNA, confirming semi-conservative model (dispersive would show all hybrid)
  • **Conclusion**: Semi-conservative replication definitively proven. Each strand of original DNA acts as template for new complementary strand.

    Mechanism of DNA Replication

    #### Semi-Conservative Nature:

    After replication:

  • Strand 1 (parental): Pairs with newly synthesized complementary strand
  • Strand 2 (parental): Pairs with newly synthesized complementary strand
  • Result: Two identical DNA molecules, each 50% old and 50% new
  • #### Directionality and Key Enzymes:

    **DNA Polymerase**:

  • Synthesizes DNA in **5' to 3' direction** only
  • Requires **primer** (short RNA or DNA sequence) with 3'-OH group
  • Cannot initiate synthesis *de novo*
  • **Template strand** is read in **3' to 5' direction**
  • #### Leading and Lagging Strands:

    Since DNA strands are **anti-parallel** and polymerase works only 5'→3', replication mechanisms differ for each strand:

    **Leading Strand**:

  • Template runs 3'→5'
  • Synthesized continuously in 5'→3' direction
  • Requires **one RNA primer**
  • Forms **Okazaki fragments** of 1000-2000 nucleotides (prokaryotes) or 100-200 nucleotides (eukaryotes)
  • **Lagging Strand**:

  • Template runs 5'→3'
  • Synthesized **discontinuously** in short segments (Okazaki fragments)
  • **Multiple RNA primers** needed
  • Fragments are joined by **DNA ligase**
  • #### Key Replication Proteins and Processes:

    1. **Helicase**: Unwinds double helix by breaking hydrogen bonds between base pairs; requires ATP

    2. **Primase**: RNA polymerase that synthesizes short RNA primers (~10 nucleotides)

    3. **DNA Polymerase III** (prokaryotes): Main replicative enzyme; adds nucleotides to 3'-OH of primer

    4. **DNA Polymerase I** (prokaryotes): Removes RNA primers and fills gaps; has 5'→3' exonuclease activity

    5. **DNA Ligase**: Joins Okazaki fragments by forming phosphodiester bonds between 3'-OH and 5'-phosphate

    6. **Single-Strand Binding Proteins (SSB)**: Protect single-stranded regions from nuclease degradation

    #### Replication Fork:

    The **Y-shaped structure** formed during replication where:

  • Parental strands separate
  • New strands are synthesized
  • Moves along DNA in 5'→3' direction
  • #### Semi-Conservative Implications:

  • Each DNA strand serves as **template for exactly one new strand**
  • Ensures **accurate replication** and faithful genetic transmission
  • Explains how genetic information is preserved across generations
  • Mutations in template strand result in daughter molecule carrying mutation
  • ---

    TRANSCRIPTION

    **Transcription** is the process of synthesizing **RNA from DNA template**, transferring genetic information from DNA to RNA. It is the first step in the **Central Dogma** (DNA → RNA → Protein).

    Importance of Transcription:

  • **mRNA** carries genetic information from DNA to ribosomes for protein synthesis
  • **tRNA** and **rRNA** are essential for translation machinery
  • **Regulatory RNA** controls gene expression
  • Characteristics of Transcription:

    1. **Template Strand** (Sense Strand; Antisense Strand):

  • Strand of DNA that serves as template
  • Read in **3' to 5' direction** by RNA polymerase
  • Only one strand per gene typically acts as template
  • 2. **Coding Strand** (Non-template Strand; Sense Strand):

  • Complementary to template strand
  • Has same sequence as mRNA (except T replaced by U)
  • Not used as template
  • 3. **Transcription Direction**: RNA synthesized in **5' to 3' direction** only

    4. **Promoter**: DNA sequence where RNA polymerase binds to initiate transcription

  • **In prokaryotes**: -35 box and -10 box (Pribnow box)
  • **In eukaryotes**: TATA box (TATAAA sequence ~25-30 bp upstream of start)
  • 5. **Terminator**: DNA sequence signaling end of transcription

    RNA Polymerase

    **Prokaryotes**:

  • Single RNA polymerase synthesizes all RNA types (mRNA, tRNA, rRNA)
  • Promoter recognition by sigma factor
  • **Eukaryotes**:

  • **RNA Polymerase I**: Synthesizes most rRNAs (18S, 5.8S, 28S)
  • **RNA Polymerase II**: Synthesizes mRNA and regulatory RNAs
  • **RNA Polymerase III**: Synthesizes tRNA, 5S rRNA, and other small RNAs
  • Stages of Transcription:

    #### Initiation (in Prokaryotes):

    1. **Sigma factor** binds to core RNA polymerase, enabling specific promoter recognition

    2. RNA polymerase-sigma complex recognizes and binds to **-35 and -10 boxes** of promoter

    3. DNA strands separate at promoter region

    4. RNA polymerase positions **first two nucleotides** of template strand

    5. First phosphodiester bond formed between first two NTPs

    #### Elongation:

    1. RNA polymerase moves along template strand in 3'→5' direction

    2. Free NTPs (ATP, GTP, CTP, UTP) enter and pair with template bases

    3. **Phosphodiester bonds** formed between incoming NTPs and growing RNA chain

    4. RNA elongates in **5'→3' direction**

    5. RNA transcript remains base-paired with template DNA

    6. Sigma factor released after ~10 nucleotides

    #### Termination:

    1. **Intrinsic termination**: GC-rich region in RNA forms hairpin/loop structure followed by poly-U tail

    2. Hairpin formation causes RNA polymerase to pause and release RNA transcript

    3. Transcript released with ~3'-OH

    **In Eukaryotes**: Similar stages but requiring **transcription factors** for initiation and more complex termination signals.

    Post-Transcriptional Modifications in Eukaryotes:

    **mRNA Processing** (before translation):

    1. **5' Capping**:

  • Addition of **7-methylguanosine cap** to 5' end
  • Added co-transcriptionally
  • Protects mRNA from degradation
  • Enhances translation efficiency
  • Helps in mRNA export from nucleus
  • 2. **3' Polyadenylation**:

  • Addition of ~200-250 **adenine nucleotides (poly-A tail)** to 3' end
  • Cleavage signal (AAUAAA) in transcript triggers polyadenylation
  • Increases mRNA stability
  • Facilitates translation
  • 3. **Splicing**:

  • Removal of **introns** (non-coding sequences)
  • Retention of **exons** (coding sequences)
  • **Spliceosome** (snRNA + proteins) catalyzes splicing
  • 5' and 3' splice sites recognized by snRNPs
  • Alternative splicing increases protein diversity
  • **Mature mRNA Structure**:

  • 5' cap - 5' UTR - Exon 1 - Exon 2 - ... - Exon n - 3' UTR - Poly-A tail
  • **UTR** (Untranslated Regions): Not translated but important for stability and regulation.

    Differences Between Prokaryotic and Eukaryotic Transcription:

    | Feature | Prokaryotes | Eukaryotes |

    |---------|------------|-----------|

    | **RNA Polymerase** | Single | Three (I, II, III) |

    | **Promoter** | -35 and -10 boxes | TATA box and others |

    | **Transcription Factors** | Sigma factor | TFIID, TFIIB, etc. |

    | **mRNA Processing** | None; translation during transcription | 5' cap, 3' poly-A tail, splicing |

    | **Location** | Cytoplasm (no nucleus) | Nucleus |

    | **Coupling** | Transcription and translation coupled | Coupled but separate locations |

    | **Introns** | Absent | Present |

    ---

    GENETIC CODE

    **Genetic code** is the **"language" by which mRNA sequence is translated into amino acid sequence** of proteins. It is the set of rules governing conversion of nucleotide information to amino acid information.

    Discovery and Properties of Genetic Code:

    #### Codon Concept:

  • **Codon**: Sequence of **3 consecutive nucleotides** (triplet) on mRNA
  • Each codon specifies **one amino acid**
  • Total codons possible: 4³ = **64 codons**
  • **61 codons code for amino acids**
  • **3 codons are stop/termination codons** (UAA, UAG, UGA): "Ochre," "Amber," "Opal"
  • #### Evidence for Triplet Code:

    1. **Nirenberg and Matthaei Experiment**: Used synthetic mRNA (poly-U) to direct synthesis of polyphenylalanine

    2. **Crick's Frameshift Experiments**: Showed that genetic code is read in non-overlapping triplets with a specific reading frame

    #### Reading Frame:

    The mRNA sequence is read sequentially in non-overlapping triplets starting from:

  • **Start codon**: **AUG** (codes for methionine; initiates translation)
  • Example: 5'-... **AUG|GGA|UUC|UAA**... -3'
  • Stop codons (UAA, UAG, UGA) terminate translation
  • Properties of Genetic Code:

    #### 1. **Universal Code**:

  • Genetic code is essentially **universal** across all organisms (with minor exceptions in mitochondria and some microorganisms)
  • Same codons specify same amino acids in bacteria, plants, animals, fungi
  • #### 2. **Non-Overlapping**:

  • **Codons do not overlap**; each nucleotide is part of only one codon
  • Reading in triplets ensures no ambiguity: AUG|UAA|CGU (not AUG|UGA|ACG|U)
  • #### 3. **Comma-less Code**:

  • No punctuation marks between codons
  • **Correct reading frame established by start codon** (AUG)
  • Mutations shifting reading frame (frameshift mutations) cause complete loss of function downstream
  • #### 4. **Degenerate/Redundant Code**:

  • **Multiple codons code for same amino acid** (except methionine and tryptophan)
  • Example: Leucine coded by UUA, UUG, CUU, CUC, CUA, CUG (6 codons)
  • **Wobble hypothesis** (Crick): 3rd position of codon has less stringent base pairing
  • **Synonymous codons**: Codons specifying same amino acid often differ in 3rd position (usually A-G or U-C)
  • #### 5. **Specific Code**:

  • Each codon specifies **exactly one amino acid** (or stop)
  • No codon codes for two different amino acids
  • Ensures accuracy in translation
  • #### 6. **Initiation and Termination**:

  • **Start codon**: **AUG** (first codon; codes for methionine in prokaryotes as N-formylmethionine, in eukaryotes as methionine)
  • **Stop codons**: **UAA, UAG, UGA** (not recognized by tRNAs; recognized by release factors)
  • Standard Genetic Code Table:

    The 64 codons organized by position:

  • **1st position** (most important for specificity)
  • **2nd position** (determines amino acid family)
  • **3rd position** (wobble; variable)
  • Example patterns:

  • **UU_ codons**: Phenylalanine (UUU, UUC), Leucine (UUA, UUG)
  • **AU_ codons**: Isoleucine (AUU, AUC), Methionine (AUG - also start), Threonine
  • **GU_ codons**: Valine
  • **GC_ codons**: Alanine
  • *Note: Complete 64-codon table typically provided in textbooks and exams.*

    Genetic Code Variations (Exceptions):

    While essentially universal, exceptions exist:

    1. **Mitochondrial DNA**:

  • AUA codes for methionine (not isoleucine)
  • AGA and AGG are stop codons (not arginine)
  • UGA codes for tryptophan (not stop)
  • 2. **Some Bacterial Genes** (rare):

  • CTG codes for serine in some *Mycoplasma* species (instead of leucine)
  • 3. **Ciliate Nuclear Codes**:

  • UAA and UAG code for glutamine (not stop)
  • These variations support the concept of code evolution and demonstrate that while genetic code is universal, organisms can evolve variations.

    ---

    TRANSLATION

    **Translation** is the **process of protein synthesis**, where **mRNA codon sequence is translated into amino acid sequence** of a polypeptide chain. It occurs on **ribosomes** and requires **mRNA, tRNA, and rRNA**.

    Components of Translation Machinery:

    #### 1. **Ribosome**:

  • **rRNA + ribosomal proteins**
  • **Prokaryotic ribosome**: 70S (S = Svedberg unit of sedimentation)
  • Large subunit: 50S (23S + 5S rRNA + ~31 proteins)
  • Small subunit: 30S (16S rRNA + ~21 proteins)
  • **Eukaryotic ribosome**: 80S
  • Large subunit: 60S (28S + 5.8S + 5S rRNA + ~50 proteins)
  • Small subunit: 40S (18S rRNA + ~33 proteins)
  • **Catalytic activity**: rRNA (ribozyme) catalyzes peptide bond formation
  • **Ribosomal Sites**:

  • **A site** (Aminoacyl): Accepts incoming aminoacyl-tRNA
  • **P site** (Peptidyl): Holds tRNA with growing polypeptide
  • **E site** (Exit): Releases deacylated tRNA
  • #### 2. **tRNA (Transfer RNA)**:

  • **Adapter molecule** linking mRNA codons to amino acids
  • **Cloverleaf structure** (2D) folds into **L-shaped structure** (3D)
  • **Anticodon**: 3-nucleotide sequence complementary to mRNA codon (base-paired)
  • **Amino acid attachment site**: 3' CCA sequence where amino acid attaches
  • **Recognition by aminoacyl-tRNA synthetase**: Specific enzyme charges tRNA with correct amino acid
  • **tRNA Charging**:

  • **Aminoacyl-tRNA synthetase** catalyzes: Amino acid + tRNA + ATP → Aminoacyl-tRNA + AMP + PPi
  • ~20 different synthetases for 20 amino acids
  • Ensures accuracy of translation
  • #### 3. **mRNA**:

  • **Carries genetic information** from DNA to ribosome
  • **5' cap and 3' poly-A tail** in eukaryotes (prokaryotes lack)
  • **Codons** specify amino acids
  • **Ribosome binding site** (Shine-Dalgarno sequence in prokaryotes; Kozak sequence in eukaryotes) positions ribosome
  • #### 4. **Initiation Factors**:

  • **Prokaryotes**: IF1, IF2, IF3
  • **Eukaryotes**: eIF1, eIF2, eIF3, eIF4, eIF5, etc.
  • Facilitate ribosome assembly and initiator tRNA selection
  • #### 5. **Elongation Factors**:

  • **Prokaryotes**: EF-Tu, EF-Ts, EF-G
  • **Eukaryotes**: eEF1α (EF-Tu equivalent), eEF2 (EF-G equivalent)
  • Facilitate tRNA delivery and translocation
  • #### 6. **Release Factors**:

  • **Prokaryotes**: RF1 (recognizes UAA/UAG), RF2 (recognizes UAA/UGA), RF3
  • **Eukaryotes**: eRF1 (recognizes all three stop codons)
  • Terminate translation upon stop codon recognition
  • Stages of Translation:

    #### **Initiation**:

    **In Prokaryotes**:

    1. **30S subunit** binds to **Shine-Dalgarno sequence** on mRNA (~8 bp upstream of start codon)

    2. **IF3** helps position 30S on mRNA

    3. **Initiator tRNA** (fMet-tRNA; carrying N-formylmethionine) binds at **P site**

  • Special initiator tRNA recognizes **AUG start codon**
  • Different from elongator Met-tRNA used for internal methionines
  • 4. **IF2-GTP** facilitates fMet-tRNA binding at P site

    5. **50S subunit** binds, forming **70S initiation complex**

    6. **IF1 and IF3** release; **GTP hydrolyzed**

    7. **A site** is positioned over second codon, ready for incoming aminoacyl-tRNA

    **In Eukaryotes**:

    1. **40S subunit** recruits **eIF2-GTP** and **initiator Met-tRNA**

    2. **eIF4** binds 5' cap structure

    3. Ribosome scans along **5' UTR** until **first AUG** (Kozak sequence context: GCCRCCAUGG)

    4. **60S subunit** joins; **eIF5B-GTP** facilitates

    5. Initiator Met-tRNA positioned at **P site**

    6. **A site** ready for second codon

    #### **Elongation**:

    **Cycle repeats for each codon**:

    **Step 1 — Aminoacyl-tRNA Selection**:

    1. **Ternary complex** (EF-Tu/eEF1α-GTP + aminoacyl-tRNA) enters **A site**

    2. **Anticodon-codon pairing** checked:

  • Correct pairing → **GTP hydrolyzed** → EF-Tu/eEF1α-GDP released
  • Incorrect pairing → tRNA released without GTP hydrolysis (proofreading)
  • 3. Accuracy: ~10⁻⁴ error rate (one error per 10,000 amino acids)

    **Step 2 — Peptide Bond Formation**:

    1. **Peptidyl transferase activity** of **rRNA** (23S in prokaryotes, 28S in eukaryotes)

    2. Catalyzes bond between:

  • **Carboxyl (-COOH) group** of amino acid in P site (on peptidyl-tRNA)
  • **Amino (-NH₂) group** of incoming amino acid in A site (on aminoacyl-tRNA)
  • 3. **Peptide bond energy** (~3 kJ/mol) energetically favorable

    4. Results:

  • **Peptidyl-tRNA** (now +1 amino acid) in A site
  • **Deacylated tRNA** (no amino acid) in P site
  • **Step

    MCQs — 10 Questions with Answers

    Q1. Which of the following correctly describes the composition of a nucleotide?

    • A. A nitrogenous base, a pentose sugar, and a phosphate group ✓
    • B. A nitrogenous base and a pentose sugar only
    • C. A nitrogenous base and a phosphate group only
    • D. A pentose sugar and a phosphate group only

    Answer: A — A nucleotide is defined as the combination of all three components: nitrogenous base, pentose sugar (deoxyribose or ribose), and phosphate group.

    Q2. In DNA, adenine always pairs with thymine through hydrogen bonds. How many hydrogen bonds form in this base pair?

    • A. One hydrogen bond
    • B. Two hydrogen bonds ✓
    • C. Three hydrogen bonds
    • D. Four hydrogen bonds

    Answer: B — Adenine (purine) forms exactly two hydrogen bonds with thymine (pyrimidine), whereas guanine–cytosine pairs form three hydrogen bonds.

    Q3. Which statement correctly explains why the distance between the two strands of DNA remains almost constant?

    • A. Purines always pair with pyrimidines, maintaining uniform purine–pyrimidine distance ✓
    • B. All four bases have the same size and shape
    • C. The number of hydrogen bonds is identical for all base pairs
    • D. The DNA helix always contains equal numbers of adenine and guanine

    Answer: A — Purines (A, G—larger) always pair with pyrimidines (T, C—smaller), so purine–pyrimidine = purine–pyrimidine across all positions, keeping strands equidistant.

    Q4. If one strand of DNA has the sequence 5′-ATGC-3′, what is the sequence of the complementary strand?

    • A. 5′-CGTA-3′
    • B. 5′-TACG-3′ ✓
    • C. 5′-ATGC-3′
    • D. 5′-GCTA-3′

    Answer: B — The complementary strand is antiparallel (3′ end aligns with 5′ end), and base-pairing rules (A–T, G–C) give the reverse complement: 5′-TACG-3′.

    Q5. Which of the following is NOT a correct statement about DNA structure?

    • A. The two strands are antiparallel, running in opposite 5′→3′ directions
    • B. The pitch of the double helix is 3.4 nm with approximately 10 base pairs per turn
    • C. Adenine and guanine are pyrimidines because they have two aromatic rings ✓
    • D. The sugar-phosphate backbone is located on the outside of the helix

    Answer: C — Adenine and guanine are purines (two rings), not pyrimidines (one ring); cytosine, thymine, and uracil are pyrimidines.

    Q6. The central dogma of molecular biology states DNA → RNA → Protein. In some viruses, this flow is reversed. What is this reverse process called?

    • A. Replication
    • B. Transcription
    • C. Reverse transcription ✓
    • D. Translation

    Answer: C — Reverse transcription is the synthesis of DNA from an RNA template, occurring in retroviruses like HIV.

    Q7. How does the structure of RNA differ from DNA? (Select all correct differences.)

    • A. RNA contains ribose sugar; DNA contains deoxyribose sugar
    • B. RNA contains uracil; DNA contains thymine
    • C. RNA has an –OH group at the 2′ position of the sugar; DNA has only –H
    • D. All of the above ✓

    Answer: D — All three differences are accurate: ribose vs deoxyribose, uracil vs thymine, and the 2′-OH group present only in RNA.

    Q8. According to Chargaff's rules, if a double-stranded DNA sample contains 30% adenine, what is the expected percentage of thymine?

    • A. 15%
    • B. 20%
    • C. 30% ✓
    • D. 40%

    Answer: C — Chargaff's rule states A = T; therefore, if adenine is 30%, thymine must also be 30%.

    Q9. A researcher observes that a newly discovered viral genome contains both adenine and uracil. Based on this observation, which conclusion is most reasonable?

    • A. The virus possesses DNA as its genetic material
    • B. The virus possesses RNA as its genetic material ✓
    • C. The virus possesses both DNA and RNA
    • D. The genome is denatured and cannot be identified

    Answer: B — Uracil is exclusive to RNA; the presence of uracil (not thymine) indicates the genetic material is RNA, as in retroviruses.

    Q10. Two DNA strands are held together by hydrogen bonds between base pairs. If you were to denature (separate) these strands by heating, how would you re-anneal (bring them back together) at a lower temperature while maintaining their original pairing?

    • A. Simply cool the solution; complementary strands will automatically find and bind to each other
    • B. Add excess thymine nucleotides to reform bonds
    • C. Add a catalyst enzyme to speed up base pairing
    • D. Adjust the salt concentration and gradually cool to allow slow, specific base-pairing ✓

    Answer: D — Annealing requires gradual cooling in appropriate ionic conditions to allow complementary strands sufficient time and thermal energy to find their partners and form specific base pairs.

    Flashcards

    What is the 3′-5′ phosphodiester linkage?

    A covalent bond linking the 3′-OH group of one sugar to the 5′-phosphate of the next, forming the sugar-phosphate backbone of DNA.

    State Chargaff's rule for double-stranded DNA.

    The ratio of adenine to thymine equals 1, and the ratio of guanine to cytosine equals 1 (A=T and G=C).

    Why is the distance between DNA strands almost constant?

    Purines always pair with pyrimidines (A with T, G with C), ensuring uniform purine–pyrimidine distance across the double helix.

    What does 'antiparallel' mean in DNA strands?

    One strand runs 5′→3′ while the complementary strand runs 3′→5′ in the opposite direction.

    Name the two types of nitrogenous bases and give one example of each.

    Purines (adenine, guanine—two rings) and pyrimidines (cytosine, thymine, uracil—one ring).

    What is the pitch of the DNA double helix and how many base pairs fit per turn?

    The pitch is 3.4 nm, with approximately 10 base pairs per complete turn of the helix.

    Distinguish between DNA and RNA based on sugar and base composition.

    DNA contains deoxyribose sugar and thymine, while RNA contains ribose sugar and uracil; RNA also has an –OH group at the 2′ position.

    What is the central dogma of molecular biology?

    Genetic information flows from DNA → RNA → Protein, establishing the direction of molecular information transfer.

    What is a nucleoside and how does it differ from a nucleotide?

    A nucleoside is a nitrogenous base linked to a pentose sugar; a nucleotide is a nucleoside with a phosphate group attached.

    Why does base stacking in the DNA helix provide additional stability?

    Overlapping base pairs create van der Waals forces and hydrophobic interactions that reinforce hydrogen bonding between strands.

    Important Board Questions

    Define a nucleotide. How is a nucleotide formed from a nucleoside? [2 marks]

    State the three components of a nucleotide (base, sugar, phosphate). Explain that a nucleoside becomes a nucleotide when a phosphate group attaches to the 5′-OH of the sugar via a phosphoester linkage.

    Explain how the complementary nature of the two DNA strands relates to the principle of DNA replication. Why is this property important for inheritance? [5 marks]

    Describe base-pairing rules (A–T, G–C) and antiparallel orientation. Explain that if each parental strand serves as a template, the new strands synthesised will be complementary, producing two identical double-helical DNA molecules identical to the parent—ensuring faithful transmission of genetic information to daughter cells.

    Discuss the salient features of the DNA double-helix model proposed by Watson and Crick. How did this structure explain the genetic implications of DNA as the hereditary material? [6 marks]

    Include: two antiparallel chains with sugar-phosphate backbone and bases inside; right-handed coil (pitch 3.4 nm, ~10 bp/turn); complementary base-pairing via H-bonds (A–T, G–C); base stacking stability. Explain that complementarity allows each strand to serve as a template for accurate replication, producing genetically identical copies and explaining how traits pass unchanged from parent to offspring.

    Next chapterEvolution →

    Practice with interactive flashcards, mind maps, upload your own chapters and get AI study kits instantly

    Try StudyOS Free →