What this chapter is about
This chapter explains how genetic information is stored, copied and expressed at the molecular level. You will learn that DNA (deoxyribonucleic acid) is the genetic material in most organisms, understand how its double-helix structure allows faithful replication, and see how the information in DNA flows to RNA and then to proteins. This sequence—DNA → RNA → Protein—is often called the central dogma of molecular biology.
A Class 12 student meets this chapter after studying Mendelian inheritance and the chromosomal theory. Now you move from patterns of heredity to the chemistry behind them. By the end you should be able to describe DNA and RNA structure, outline the enzymes and steps in replication, transcription and translation, explain the genetic code, and appreciate how the Human Genome Project mapped our entire genetic blueprint.
Mastering these ideas prepares you for later study of biotechnology, genetic engineering and medical genetics, all of which rely on manipulating nucleic acids and understanding gene expression.
Key ideas
- DNA is the genetic material: Experiments by Griffith (transformation in bacteria) and by Hershey–Chase (using radioactive isotopes to label protein and DNA of bacteriophages) established that DNA, not protein, carries hereditary information.
- Double-helix structure of DNA: Two antiparallel polynucleotide strands wind around each other; adenine pairs with thymine (A–T, two hydrogen bonds) and guanine pairs with cytosine (G–C, three hydrogen bonds). The sugar-phosphate backbone lies outside; nitrogenous bases stack inside.
- Semi-conservative replication: Each original strand serves as a template; two daughter molecules each contain one parental strand and one newly synthesised strand. Key enzymes include helicase (unwinding), DNA polymerase (synthesising the new strand 5′→3′), primase (making RNA primers) and ligase (joining Okazaki fragments on the lagging strand).
- Transcription: RNA polymerase synthesises a single-stranded mRNA copy of the template (antisense) strand of a gene. In eukaryotes, the primary transcript undergoes capping, polyadenylation and splicing (removal of introns) before leaving the nucleus.
- Genetic code: A sequence of three bases (codon) in mRNA specifies one amino acid. The code is nearly universal, degenerate (most amino acids have more than one codon), non-overlapping and has start (AUG) and stop (UAA, UAG, UGA) codons.
- Translation: Ribosomes read mRNA codons; tRNA molecules carry specific amino acids and have anticodons complementary to mRNA codons. Peptide bonds form between successive amino acids, producing a polypeptide.
- Regulation of gene expression: The lac operon model in E. coli shows how genes can be switched on or off depending on environmental signals (presence or absence of lactose).
- Human Genome Project: This international effort sequenced approximately 3 × 10⁹ base pairs of human DNA, identified about 20 000–25 000 protein-coding genes, and revealed that coding sequences make up only about 2 % of the genome.
Formulas and facts to remember
- Chargaff's rules: In double-stranded DNA, percentage of A equals percentage of T; percentage of G equals percentage of C; hence (A + G) = (T + C).
- Base-pairing rule: A pairs with T (or U in RNA) by two hydrogen bonds; G pairs with C by three hydrogen bonds.
- DNA replication is semi-conservative: Each daughter DNA molecule has one old strand and one new strand.
- Direction of synthesis: DNA and RNA polymerases add nucleotides only in the 5′ → 3′ direction.
- Number of codons: 4³ = 64 possible triplet codons; 61 code for amino acids, 3 are stop codons.
- Start codon: AUG codes for methionine and signals the beginning of translation.
- Wobble position: The third base of a codon often shows flexibility in pairing, explaining degeneracy.
- Human genome facts: About 3.2 × 10⁹ bp; roughly 20 000–25 000 genes; most DNA is non-coding (introns, repetitive sequences).
Worked examples
Example 1: Calculating base composition
A double-stranded DNA molecule has 18 % guanine. Find the percentages of the other three bases.
Solution
By Chargaff's rule, G = C, so cytosine = 18 %.
Total G + C = 18 + 18 = 36 %.
Remaining bases: A + T = 100 − 36 = 64 %.
Since A = T, each = 64 / 2 = 32 %.
Answer: A = 32 %, T = 32 %, G = 18 %, C = 18 %.
Example 2: Determining mRNA and amino-acid sequence
A segment of the template (antisense) strand of a gene reads:
3′ – TAC GGA CTT AAC – 5′
Write the mRNA sequence and predict how many amino acids could be coded.
Solution
mRNA is synthesised 5′ → 3′, complementary to the template strand, with U replacing T.
Template: 3′ – T A C G G A C T T A A C – 5′
mRNA: 5′ – A U G C C U G A A U U G – 3′
Number of codons = 4 (AUG, CCU, GAA, UUG).
AUG is the start codon (Met). If we assume translation continues until a stop codon (not shown here), this mRNA segment codes for 4 amino acids.
Example 3: Predicting replication products
A bacterial chromosome undergoes two rounds of semi-conservative replication starting from one original double-stranded DNA molecule labelled with heavy nitrogen (¹⁵N). After two rounds, how many molecules contain only light nitrogen (¹⁴N)?
Solution
Round 1: 1 molecule → 2 molecules, each with one ¹⁵N strand and one ¹⁴N strand (hybrid).
Round 2: Each hybrid replicates → one hybrid daughter, one light-only daughter.
Total molecules = 4.
Hybrid molecules = 2; light-only molecules = 2.
Answer: 2 molecules contain only ¹⁴N.
Common mistakes
- Confusing template strand with coding strand → Remember: RNA polymerase reads the template (antisense) strand 3′→5′ and synthesises mRNA 5′→3′.
- Writing DNA base pairs as A–G or T–C → Only A–T and G–C pairs are allowed; mismatched pairs destabilise the helix.
- Forgetting that DNA polymerase needs a primer → Replication cannot begin de novo; primase must first lay down a short RNA primer.
- Assuming introns stay in mature mRNA → Introns are removed by splicing; only exons appear in the final mRNA that is translated.
- Treating the genetic code as organism-specific → The code is nearly universal; the same codons specify the same amino acids in bacteria, plants and humans (with minor exceptions).
Quick revision
- DNA is made of two antiparallel strands; A pairs with T, G pairs with C.
- Replication is semi-conservative and proceeds 5′ → 3′.
- Transcription produces mRNA from the template strand; eukaryotic mRNA is processed (cap, poly-A tail, splicing).
- Translation occurs on ribosomes; tRNA anticodons match mRNA codons to assemble amino acids into polypeptides.
- The lac operon illustrates inducible gene regulation in prokaryotes.
- The Human Genome Project revealed roughly 20 000–25 000 genes in about 3 × 10⁹ base pairs.