Nucleic Acids

• The three components of a nucleotide: nitrogenous base, pentose sugar, and phosphate group
• The difference between purines (A, G) and pyrimidines (C, T, U) and how to remember which is which
• How DNA and RNA differ in sugar (deoxyribose vs ribose) and bases (T vs U)
• The Watson-Crick double helix model (1953): antiparallel strands, base pairing rules, dimensions
• The phosphodiester bond and the directionality convention (5'→3')
• The three types of RNA (mRNA, tRNA, rRNA) and their specific roles
• The central dogma: DNA → RNA → Protein (replication, transcription, translation)
• The biological significance of nucleic acids in heredity and protein synthesis

Nucleotide — The Monomer Unit

A nucleotide has three components joined covalently:

  Nitrogenous base
       |
  Pentose sugar — Phosphate group

1. Nitrogenous base — a heterocyclic aromatic compound (purine or pyrimidine)
2. Pentose sugar — 5-carbon sugar in furanose (5-membered ring) form
3. Phosphate group — —OPO₃²⁻ at the 5' carbon of the sugar

Nucleoside = base + sugar (no phosphate) Nucleotide = base + sugar + phosphate (nucleoside phosphate)

Nitrogenous Bases

Purines — double-ring structure (pyrimidine fused with imidazole):

• Adenine (A) — found in DNA and RNA
• Guanine (G) — found in DNA and RNA

Pyrimidines — single-ring structure:

• Cytosine (C) — found in DNA and RNA
• Thymine (T) — found in DNA only (has an extra —CH₃ group)
• Uracil (U) — found in RNA only (replaces thymine; no —CH₃ group)

Mnemonic — "Purines are Pure Gold (PureAG):" Purines = Adenine + Guanine Mnemonic — "CUT the Py(rimidines):" Pyrimidines = Cytosine + Uracil + Thymine

Memory hook for DNA vs RNA bases:

• DNA has T (Thymine) — "DNA is Durable, has Thymine (T for Two rings... wait no — T for DNA)"
• RNA has U (Uracil) — "RNA Uses Uracil"

Pentose Sugars

The C2'—OH in ribose makes RNA more chemically reactive and less stable than DNA — appropriate for its transient roles in gene expression.

DNA vs RNA — Complete Comparison

Watson-Crick Double Helix Model (1953)

James Watson and Francis Crick proposed the double helix model in 1953, based on X-ray data from Rosalind Franklin and Erwin Chargaff's base-composition rules.

Key features:

1. Two antiparallel polynucleotide strands — one runs 5'→3', the other 3'→5'
2. Right-handed helix — strands wind clockwise when viewed along the helix axis
3. Dimensions:
   • 3.4 Å (0.34 nm) per base pair (rise per residue)
   • 34 Å (3.4 nm) per full turn
   • 10 base pairs per turn (34 ÷ 3.4 = 10)
   • 20 Å (2.0 nm) diameter
4. Bases on inside, sugar-phosphate backbone on outside
5. Base pairing (Chargaff's rules):
   • A pairs with T: 2 hydrogen bonds (A=T)
   • G pairs with C: 3 hydrogen bonds (G≡C)
6. Major and minor grooves — asymmetry in the helix creates two grooves where proteins can bind

Chargaff's rules (from experiments): In any DNA sample, [A] = [T] and [G] = [C], therefore:

• A + G = C + T (purines = pyrimidines)
• (A + T)/(G + C) = constant for a species but varies between species

Mnemonic for H-bond count: "AT (2 H-bonds) — At a Table with 2 chairs; GC (3 H-bonds) — Good Connection with 3 bonds"

Phosphodiester Bond

The backbone of each DNA/RNA strand is formed by phosphodiester bonds — a phosphate group bridges the 3' carbon of one nucleotide and the 5' carbon of the next:

5' end
   |
  [Base]
   |
 Sugar
   |
  3'-O—P(=O)(OH)—O-5'
                  |
               Sugar
                  |
               [Base]
                  |
                3' end

Each phosphodiester bond is formed with loss of water (pyrophosphate in biosynthesis). DNA is synthesised in the 5'→3' direction.

RNA Types and Functions

tRNA details:

• ~70–90 nucleotides long
• 3' end: CCA-OH — where the amino acid is attached (aminoacylation)
• Anticodon loop: 3-nucleotide sequence complementary to the mRNA codon
• Each tRNA is specific for one amino acid (charged by aminoacyl-tRNA synthetase)

The Central Dogma

Proposed by Francis Crick (1958):

DNA → RNA → Protein
 ↑
 (DNA → DNA: Replication)

Three processes:

1. Replication (DNA → DNA)

• DNA unwinds; each strand serves as template
• DNA polymerase adds complementary nucleotides in 5'→3' direction
• Semi-conservative: each daughter DNA has one old strand + one new strand

2. Transcription (DNA → mRNA)

• RNA polymerase reads the template (antisense) strand 3'→5'
• Synthesises mRNA in the 5'→3' direction
• Only a portion of DNA (a gene) is transcribed at a time
• Occurs in the nucleus (in eukaryotes)

3. Translation (mRNA → Protein)

• Ribosome reads mRNA codons (3-nucleotide units) in 5'→3' direction
• tRNA brings the corresponding amino acid
• Peptide bonds form between amino acids
• Occurs in the cytoplasm

Genetic code: 4 bases → 3-letter codons → 4³ = 64 codons for 20 amino acids

• Code is degenerate (multiple codons for one amino acid), non-overlapping, universal (same in all organisms)
• Start codon: AUG (codes for methionine)
• Stop codons: UAA, UAG, UGA (no amino acid; terminate translation)

Anti-parallel Orientation — Why It Matters

The two strands of DNA run antiparallel. If one strand reads 5'-ATCG-3', the complementary strand reads 3'-TAGC-5' (or 5'-GCTA-3'). This is fundamental to:

• DNA replication (leading strand synthesised continuously; lagging strand in Okazaki fragments)
• Transcription (only one strand is read)
• Hybridisation-based techniques (PCR, Southern blot)

In JEE problems: given one strand, always write the complementary strand antiparallel, with correct base pairs (A-T, G-C for DNA; A-U, G-C for RNA).

Stability of DNA — Role of Base Stacking and H-Bonds

The double helix is stabilised by two forces:

1. Hydrogen bonds between complementary bases (A=T: 2; G≡C: 3)
2. Base stacking interactions (hydrophobic + van der Waals between adjacent base pairs) — actually contribute MORE to stability than H-bonds

DNA with higher G+C content has higher melting temperature (Tm) because G≡C has 3 H-bonds vs A=T with 2. This is used to calculate Tm:

Tm (°C) ≈ 2 × (A+T count) + 4 × (G+C count)

Nucleotide Derivatives Beyond DNA/RNA

Nucleotides serve other biological roles:

• ATP (adenosine triphosphate): Energy currency of the cell (AMP + 2 phosphates)
• NAD⁺/NADH: Coenzyme in redox reactions (nicotinamide adenine dinucleotide)
• cAMP: Second messenger in cell signalling
• CoA (Coenzyme A): Contains adenosine; involved in acetyl group transfer

Differences Between Prokaryotic and Eukaryotic DNA

Example 1: Base Pairing, Strand Complement, and G+C Content

Problem: A DNA strand has the sequence 5'-ATGCGCAAT-3'.
(a) Write the complementary strand with correct polarity.
(b) Calculate the %G+C content.
(c) Predict which strand would have a higher Tm contribution.

Step 1: Complement each base (A↔T, G↔C) and reverse (antiparallel):
  Given:         5'-A T G C G C A A T-3'
  Complement:    3'-T A C G C G T T A-5'
  Written 5'→3': 5'-A T T G C G C A T-3'

Step 2: Count G+C in the original strand:
  Bases: A,T,G,C,G,C,A,A,T → G=2, C=2, A=3, T=2
  Total = 9 bases; G+C = 4
  %G+C = 4/9 × 100 = 44.4%

Step 3: Since G≡C pairs have 3 H-bonds vs A=T with 2 H-bonds,
  regions with higher G+C content melt (denature) at higher temperature.
  The given strand and its complement are the SAME double helix,
  so Tm is a property of the double-stranded molecule.
  Tm ≈ 2(3) + 4(2) ×... 
  (Using simplified: Tm = 2×(A+T) + 4×(G+C) per strand)
  = 2×(3+2) + 4×(2+2) = 2×5 + 4×4 = 10 + 16 = 26 (relative units only)

Answer: Complementary strand = 5'-ATTGCGCAT-3'; %G+C = 44.4%

Example 2: Tracing the Central Dogma

Problem: A segment of the coding (sense) strand of DNA reads:
  5'-ATGAAACCCUAA-3'
Identify: (a) the mRNA sequence, (b) the amino acids coded, 
           (c) what happens at UAA.

Note: The given sequence contains U — this looks like mRNA directly.
      If this IS the mRNA sequence:

mRNA: 5'-AUG-AAA-CCC-UAA-3'
           ↓     ↓     ↓    ↓
Codons:  AUG   AAA   CCC  UAA

AUG → Methionine (START codon)
AAA → Lysine
CCC → Proline
UAA → STOP codon (no amino acid added; translation terminates)

Protein produced: Met — Lys — Pro  (tripeptide)
                  (then released from ribosome)

Key takeaways:
• AUG always initiates translation and codes for Met.
• UAA, UAG, UGA are stop codons — no tRNA recognises them.
• The protein has 3 amino acids (not 4 — stop codon adds nothing).

1. Name the three components of a nucleotide and state how a nucleotide differs from a nucleoside.
2. A DNA sample has 20% adenine. What is the percentage of each of the other three bases?
3. What is the anticodon of tRNA that recognises the mRNA codon 5'-GUC-3'?
4. Why is RNA less stable than DNA at physiological conditions?
5. (Stretch) A researcher finds a newly discovered organism in which the DNA has [A] ≠ [T] but [A] = [G]. What does this tell you about the structure of this organism's DNA? Is it single-stranded or double-stranded? How would you experimentally test this?

• NCERT Class 12 Chemistry, Chapter 14 — Biomolecules, Section 14.6 (Nucleic Acids)
• NCERT Class 12 Biology, Chapter 6 — Molecular Basis of Inheritance (for deeper treatment of central dogma)
• JEE Foundation: 1–2 questions on DNA structure dimensions, base pairing, central dogma steps
• Common MCQ formats: H-bond count between strands, complementary strand writing, identify DNA vs RNA, identify mRNA/tRNA/rRNA role

Study strategy: Memorise the DNA helix dimensions (3.4 Å/bp, 34 Å/turn, 10 bp/turn) as a set — they often appear together in MCQs. Practise writing complementary strands quickly with correct polarity. Know all 3 stop codons and the start codon.

Practice MCQs on DNA structure, base pairing, and central dogma in the Biomolecules — Nucleic Acids topic bank. Start with Easy (base pairing rules), then Medium (central dogma tracing).

Confused about antiparallel strands and how to write the complementary strand correctly? Ask the Drishti AI tutor to walk you through a step-by-step example with directionality arrows.

Complete the Quick Check questions and mark them in your Drishti progress tracker. Aim for 4/5 before attempting the Biomolecules mixed MCQ set.

• Read: Vitamins — water-soluble (B-complex, C) vs fat-soluble (A, D, E, K), deficiency diseases
• Then: Hormones — nature (protein, steroid), function
• Practice: Full Biomolecules Mixed MCQ Set (Medium + Hard for JEE Foundation)