DNA for Class 12 — Structure, Importance and Packaging Explained

What is DNA?

DNA (Deoxyribonucleic Acid) is a long polymer of deoxyribonucleotides that stores genetic information in all living organisms. In most organisms DNA exists as a double-stranded helix formed by two complementary polynucleotide chains coiled around each other.

Examples of DNA content in organisms

• Bacteriophage Phi X 174: 5,386 nucleotides.
• Bacteriophage λ (lambda): 48,502 base pairs.
• Escherichia coli (E. coli): ≈ 4.6 × 10⁶ base pairs.
• Human haploid genome: ≈ 3.3 × 10⁹ base pairs.

Structure of the polynucleotide chain

Components of a nucleotide

A nucleotide consists of three components:

1. A nitrogenous base (purine or pyrimidine)
2. A pentose sugar (deoxyribose in DNA)
3. A phosphate group

Types of nitrogenous bases

• Purines: Adenine (A) and Guanine (G)
• Pyrimidines: Cytosine (C) and Thymine (T) (Uracil (U) is found in RNA)

Cytosine occurs in both DNA and RNA; thymine is characteristic of DNA while uracil occurs in RNA.

Nucleosides and nucleotides

• A base + sugar = nucleoside (e.g. adenosine).
• Nucleoside + phosphate (attached at the 5′-OH) = nucleotide.
• Nucleotides join by 3′–5′ phosphodiester bonds to form a polynucleotide chain.

Each polynucleotide chain has directionality: a free phosphate at the 5′ end and a free hydroxyl at the 3′ end. The sugar–phosphate backbone gives structural support.

Base pairing and complementarity

Complementary base pairing is central to DNA function:

• A (adenine) pairs with T (thymine) via two hydrogen bonds.
• G (guanine) pairs with C (cytosine) via three hydrogen bonds.

Because of complementarity, the sequence of one strand determines the sequence of the other; this is vital for accurate replication.

Salient features of the double helix

• Two polynucleotide chains form the structure.
• The chains are antiparallel (one runs 5′→3′, the other 3′→5′).
• Base pairs are held by hydrogen bonds (A=T, C≡G).
• The helix is right-handed (B-DNA).
• One helical turn contains about 10 base pairs and the pitch (distance per turn) is ≈ 3.4 nm.
• Distance between adjacent base pairs ≈ 0.34 nm.

Packaging of the DNA helix

Although the DNA double helix in a typical mammalian cell is about 2.2 metres long, it fits into a nucleus only a few micrometres across because of hierarchical packaging.

In prokaryotes (e.g. E. coli)

• DNA is not membrane-bound; it is organised in a nucleoid with looped domains, aided by DNA-binding proteins and supercoiling.

In eukaryotes (e.g. humans)

• DNA wraps around histone proteins forming nucleosomes (“beads on a string”).
• Nucleosomes fold into higher-order structures to form chromatin, which condenses into chromosomes during cell division.

This organisation allows DNA to be compact yet accessible for replication and transcription.

Importance of DNA

• Genetics & inheritance: DNA stores hereditary information and directs protein synthesis that determines traits.
• Medicine: DNA analysis enables genetic testing, diagnosis, forensics (DNA fingerprinting), and gene therapy research.
• Biotechnology: Recombinant DNA technology produces medicines (e.g. insulin), vaccines and GM crops.
• Research & society: Projects like the Human Genome Project have advanced personalized medicine and our understanding of biology.

Exercises

Exercise A — Short problems (SET A adapted)

1. DNA duplication in coli takes 20 minutes. An E. coli cell with heavy DNA is allowed to grow in ¹⁴N-containing NH₄Cl for 100 minutes. Find the ratio of hybrid DNA molecules to light molecules after 100 minutes. Draw the result (explain reasoning).
2. In a linear dsDNA molecule of 100 base pairs, 30% is guanine. Find the total number of hydrogen bonds and phosphodiester bonds in that DNA.
3. If the antisense strand of DNA (3′→5′) has the following sequence: 3′–T A C C C G A C G A T C–5′, find the sequence of its cDNA and of the mRNA.
4. coli has 4.6 × 10⁶ bp in its DNA. Find the length (in metres) of E. coli DNA.
5. Human haploid genome has 3.3 × 10⁹ bp. Find the length (in metres) of human haploid DNA.
6. If the pitch of B-DNA is 3.4 nm per 10 bp, how many base pairs are found within 34 nm?
7. Given an mRNA strand 5′–AUG CCC AAA UUU UUA UAA–3′, find how many amino acids are formed from this sequence; which end of tRNA carries the amino acid, and how many tRNA molecules are required?
8. (Variation of Q1) Same as Q1 but growth period = 80 minutes. Compute the ratio of hybrid to light DNA molecules after 80 minutes.
9. In a linear dsDNA molecule of 100 base pairs, 20% is guanine. Find the total number of hydrogen bonds and phosphodiester bonds.
10. If the antisense strand (3′→5′) is 3′–T A C C C G A C G A T C C C A A A–5′, find the cDNA and mRNA sequences.

Answers & worked solutions

General notes for answers: E. coli DNA replication is semi-conservative. Each replication cycle (20 minutes) doubles the number of DNA molecules; after nnn cycles there are 2n2^n2n total DNA molecules. “Heavy” vs “light” refers to parental (heavy) vs newly synthesised (light) strands; hybrid molecules contain one heavy and one light strand.

1. After 100 min:
   Replication time = 20 min → cycles = 100/20=5100/20 = 5100/20=5 cycles. Total molecules = 25=322^5 = 3225=32. Semi-conservative replication: after 1 cycle all molecules are hybrid (1 heavy + 1 light strand). After 2 cycles some become light–light, and so on. The number of hybrid molecules after nnn cycles = 2n−12^{n-1}2n−1. So for n=5n=5n=5, hybrid = 24=162^{4} = 1624=16. Light (completely light–light) molecules = total − hybrid − parental (if parental remains) but parental heavy–heavy are 0 after first round (parental distributes). Standard semi-conservative result: after nnn cycles hybrid = 2n−12^{n-1}2n−1, light = 2n−1−2^{n-1}-2n−1− (but simpler: after 5 cycles hybrid = 16, light = 16). So ratio hybrid : light = 1 : 1 (16:16).
   (Graphic: 32 circles — 16 hybrid, 16 light.)
2. 100 bp, 30% G:
   G = 30 bp → C = 30 bp (complementary). Remaining bases = 40 bp → A = 20 bp, T = 20 bp. Hydrogen bonds: each A–T = 2 bonds; each G–C = 3 bonds. Total H bonds = (20×2)+(30×3)=40+90=130 (20\times2) + (30\times3) = 40 + 90 = 130(20×2)+(30×3)=40+90=130 hydrogen bonds (per double helix). Phosphodiester bonds: each strand of 100 bases has 99 phosphodiester bonds; two strands → 2×99=1982 \times 99 = 1982×99=198 phosphodiester bonds.
3. Antisense (given 3′→5′): 3′–T A C C C G A C G A T C–5′
   First write antisense 5′→3′: 5′–C T A G C A G C C C A T–3′ (reverse the order). But easier method: cDNA (complement of antisense) will be the sense/coding strand (5′→3′). The coding strand (cDNA) sequence is: 5′–A T G G G C T G C T A G–3′ (which corresponds to original). mRNA (transcribed from coding strand replacing T with U): 5′–A U G G G C U G C U A G–3′. (Translate accordingly: starts at AUG).
4. Length of E. coli DNA:
   Number of bp = 4.6 × 10⁶. Each bp rise ≈ 0.34 nm = 0.34 × 10⁻9 m. Length = 4.6×106×0.34×10−94.6\times10^6 \times 0.34\times10^{-9}4.6×106×0.34×10−9 m = 1.564×10−31.564\times10^{-3}1.564×10−3 m = 1.564 mm.
5. Length of human haploid DNA:
   3.3 × 10⁹ bp × 0.34 × 10⁻9 m = 1.122×1001.122\times10^01.122×100 m ≈ 1.12 metres (per haploid set).

**6. Pitch 3.4 nm per 10 bp ⇒ 34 nm contains (34 / 3.4) × 10 bp = 10 × 10 = 100 bp.

7. mRNA 5′–AUG CCC AAA UUU UUA UAA–3′
   Translate into codons: AUG | CCC | AAA | UUU | UUA | UAA

• AUG = Met (start, also codes Met).
• CCC = Pro
• AAA = Lys
• UUU = Phe
• UUA = Leu
• UAA = Stop (termination)
  So amino acids formed before stop: Met, Pro, Lys, Phe, Leu → 5 amino acids.
  tRNA carries amino acid on its 3′ end (the CCA acceptor end). One tRNA molecule is required for each amino acid added (so up to 5 tRNAs will be used during translation of these five codons, though the initiating Met tRNA is reused in other translations).

8. After 80 min: cycles = 80/20 = 4 → total molecules =24=16=2^4=16=24=16. Hybrid molecules =23=8=2^{3}=8=23=8. Light molecules =8=8=8. Ratio hybrid:light = 1:1.
9. 100 bp, 20% G → G = 20, C = 20, remaining 60 → A = 30, T = 30.
   Hydrogen bonds = (30×2)+(20×3)=60+60=120 (30\times2) + (20\times3) = 60 + 60 = 120(30×2)+(20×3)=60+60=120. Phosphodiester bonds = 2×99=1982\times99 = 1982×99=198.
10. Antisense 3′–T A C C C G A C G A T C C C A A A–5′
    cDNA (coding strand, 5′→3′) = 5′–A T G G G C T G C T A G G G T T T–3′
    mRNA = 5′–A U G G G C U G C U A G G G U U U–3′

Short revision checklist (quick facts)

• Base pair distance = 0.34 nm.
• 10 bp ≈ 3.4 nm (one helical turn).
• A–T = 2 H bonds, G–C = 3 H bonds.
• Direction: strands are antiparallel (5′ ↔ 3′).
• Human haploid DNA length ≈ 1.12 m.