Molecular Basis of Inheritance

The chapter opens by fixing what DNA is, how long it is and how much of it a cell has to manage. DNA is a long polymer of deoxyribonucleotides; RNA is a polymer of ribonucleotides. Nucleic acids were first isolated by Friedrich Meischer in 1869, who called the material nuclein. DNA in most organisms is the genetic material; in some viruses RNA does that job instead.

The length of a DNA molecule is stated as the number of nucleotides it holds, or the number of base pairs (bp) if it is double stranded. Learn the four NCERT figures, because 1-mark questions pick them out: bacteriophage φ×174 has 5386 nucleotides, bacteriophage lambda has 48,502 bp, Escherichia coli has 4.6 × 10⁶ bp, and the haploid human genome has 3.3 × 10⁹ bp.

The size problem becomes obvious once you convert base pairs into metres. Adjacent base pairs are 0.34 nm apart along the helix, so a diploid human cell, with 6.6 × 10⁹ bp, carries about 6.6 × 10⁹ × 0.34 × 10⁻⁹ m ≈ 2.2 m of DNA inside a nucleus around 10⁻⁶ m across. Even E. coli must fit about 1.36 mm of DNA (the figure NCERT quotes) in a cell far smaller than that. The packaging problem, and its solution (section 5.1.2), follows directly from this calculation, and examiners like a question that asks you to do it.

Everything else in the chapter builds on two questions: what is the chemical structure of this polymer, and how does a cell copy it, read it and control it. Keep the numbers above on your formula sheet; they cost nothing to learn and are asked as fill-ups and MCQs.

A nucleotide has three parts: a nitrogenous base, a pentose sugar (ribose in RNA, 2'-deoxyribose in DNA) and a phosphate group. The bases are of two kinds: purines (adenine A, guanine G, double-ringed) and pyrimidines (cytosine C, uracil U, thymine T, single-ringed). Cytosine is common to both nucleic acids; thymine is found in DNA and uracil in RNA. Thymine is 5-methyl uracil.

The linkages are asked by name, so learn them exactly. Base joined to the 1' carbon of the pentose through an N-glycosidic linkage gives a nucleoside (adenosine or deoxyadenosine, guanosine or deoxyguanosine, cytidine or deoxycytidine, uridine or deoxythymidine). A phosphate added to the 5'-OH of the nucleoside through a phosphoester linkage gives a nucleotide. Two nucleotides join through a 3'-5' phosphodiester linkage to form a dinucleotide, and many join to form a polynucleotide chain. Every chain therefore has a free phosphate at its 5' end and a free 3'-OH at its 3' end; the backbone is sugar-phosphate and the bases project out from it.

Watson and Crick (1953) proposed the double helix using the X-ray diffraction data of Maurice Wilkins and Rosalind Franklin and the base-ratio observation of Erwin Chargaff that in double-stranded DNA the ratios A : T and G : C are constant and equal to one. The salient features of the double helix, as the marking scheme lists them: (i) it is made of two polynucleotide chains whose backbone is sugar-phosphate and whose bases point inward; (ii) the two chains are antiparallel, one 5'→3' and the other 3'→5'; (iii) the bases of the two strands pair by hydrogen bonds, A with T through two hydrogen bonds and G with C through three, and a purine always faces a pyrimidine, which keeps the distance between the strands uniform; (iv) the chains coil in a right-handed fashion with a pitch of 3.4 nm and roughly 10 bp per turn, so adjacent base pairs are 0.34 nm apart; (v) the planes of stacked base pairs lie one over the other, and this stacking, along with hydrogen bonding, confers stability on the helix.

The double helix explained how genetic information could be copied, and from it Francis Crick proposed the central dogma: genetic information flows DNA → RNA → protein. In some viruses the flow runs RNA → DNA, called reverse transcription.

The 2.2 m of DNA in a human cell and the 1.36 mm in E. coli must be folded into a very small space. Prokaryotes have no defined nucleus, yet their DNA is not scattered: the negatively charged DNA is held with some positively charged proteins in a region called the nucleoid, where it is organised into large loops.

Eukaryotes use histones. These are small basic proteins rich in the amino acids lysine and arginine, whose side chains are positively charged. Eight histone molecules (two each of H2A, H2B, H3 and H4) form a histone octamer. The negatively charged DNA winds around the positively charged octamer to make a nucleosome; a typical nucleosome carries about 200 bp of DNA helix. Histone H1 sits where DNA enters and leaves the core. Nucleosomes repeat along the DNA to give chromatin, the thread-like structure seen as beads-on-string under the electron microscope. The string is coiled further to form the chromatin fibre, which is condensed at metaphase into chromosomes. These higher levels of packaging need additional non-histone chromosomal (NHC) proteins.

In a typical nucleus some chromatin stains light and some dark. The loosely packed, lightly stained region is euchromatin; it is transcriptionally active. The densely packed, dark-staining region is heterochromatin; it is transcriptionally inactive. This distinction is a 2-mark regular.

The 2-mark questions on this section are almost formulaic: name the positively charged component (histone octamer) and the negatively charged one (DNA); name the two amino acids that give histones their charge (lysine, arginine); draw and label the nucleosome; explain why a prokaryote's DNA is not scattered although it has no nucleus.

By the 1920s chromosomes were known to carry genes, but chromosomes contain both DNA and protein, and protein looked like the better candidate. Two experiments settled the question, and the board asks both.

Griffith's experiment (1928). Frederick Griffith worked with Streptococcus pneumoniae, which causes pneumonia. On a culture plate some colonies are smooth (S) because the cells have a mucous polysaccharide coat, and some are rough (R) because they lack it. S cells are virulent and R cells are not. Injected into mice: live S → mouse dies; live R → mouse lives; heat-killed S → mouse lives; heat-killed S mixed with live R → mouse dies, and living S bacteria are recovered from the dead mouse. Griffith concluded that some transforming principle had passed from the heat-killed S cells to the R cells and enabled them to synthesise the smooth polysaccharide coat and become virulent. He did not know its chemical nature.

Biochemical characterisation: Avery, MacLeod and McCarty (1933–44). They purified biochemicals (proteins, DNA, RNA) from heat-killed S cells to see which could transform live R cells into S cells. DNA alone transformed R cells. Proteases and RNases did not affect transformation, so the transforming substance was neither protein nor RNA. DNase inhibited transformation, showing that DNA caused the transformation. They concluded that DNA is the hereditary material, although not all biologists were convinced.

In an exam answer, structure it as: organism and the two strains, the four injections and their results, the conclusion, then the three enzyme treatments and their outcomes. A flow-chart question on Griffith carries 3 marks and wants the outcomes of each step written next to the arrows.

The unequivocal proof came from Alfred Hershey and Martha Chase (1952), who worked with bacteriophages, viruses that infect bacteria. A phage attaches to a bacterium, its genetic material enters the cell, and the cell then makes more viruses. The question was whether it was DNA or protein that entered.

Labelling. Some phages were grown on a medium containing radioactive phosphorus (³²P) and some on a medium containing radioactive sulfur (³⁵S). DNA contains phosphorus but not sulfur, so the first batch had radioactive DNA. Protein contains sulfur (in cysteine and methionine) but not phosphorus, so the second batch had radioactive protein coats.

Infection, blending, centrifugation. The radioactive phages were allowed to attach to E. coli. As infection proceeded the viral coats were removed from the bacteria by agitating them in a blender. The mixture was then centrifuged to separate the heavier bacterial cells (pellet) from the lighter viral particles (supernatant).

Result. Bacteria infected with ³²P-labelled phages were radioactive, meaning DNA had entered the cells. Bacteria infected with ³⁵S-labelled phages were not radioactive; the protein had stayed outside in the supernatant. Conclusion: DNA is the genetic material that passes from virus to bacterium; the protein coat does not.

The four-part question on this experiment (why phages, why two media, why a blender and a centrifuge, what conclusion) has appeared several times. Answer each part in one or two lines and always link ³²P to DNA and ³⁵S to protein explicitly.

Hershey and Chase settled the question for DNA-viruses and cells, but some viruses (tobacco mosaic virus, QB bacteriophage) use RNA as genetic material. NCERT therefore lists what any molecule must do to act as genetic material: (i) it should be able to replicate; (ii) it should be chemically and structurally stable; (iii) it should provide scope for slow changes (mutation) that evolution requires; (iv) it should be able to express itself as Mendelian characters.

Replication. Both DNA and RNA have complementary base pairing and can direct their own copying; proteins cannot. Stability. DNA wins. Its two strands, being complementary, can be separated by heat and will come back together; the 2'-OH group present in every ribose of RNA is a reactive group that makes RNA labile and easily degradable, and RNA is also catalytic and therefore reactive. In DNA, thymine replaces uracil, which gives extra stability. Mutation. Both can mutate, but RNA, being unstable, mutates faster, which is why RNA viruses have short life spans and evolve rapidly. Expression. RNA can directly code for protein synthesis; DNA depends on RNA for it. So, in NCERT's words, DNA is the better material for storage of genetic information and RNA is better for its transmission (expression). The 2-mark question 'why is DNA a better genetic material than RNA' wants stability (2'-OH absent, thymine, double strand) and slower mutation.

RNA world. RNA was the first genetic material. Essential life processes such as metabolism, translation and splicing evolved around RNA, and RNA served both as genetic material and as a catalyst (ribozymes). But being reactive it was unstable, so DNA, with chemical modifications that made it more stable, evolved from RNA, and the double-stranded form gave it resistance to change, with the second strand allowing repair. The three reasons for 'RNA was the first genetic material' are: RNA can act as genetic material (RNA viruses), RNA can act as a catalyst, and key processes still depend on RNA.

Watson and Crick's helix suggested the copying mechanism at once: the two strands separate and each acts as a template for a new complementary strand. After one round each daughter molecule has one parental strand and one new strand, so replication is semi-conservative. This is the property (complementary base pairing) that NCERT exercise 5.5 asks about.

Meselson and Stahl (1958) proved it in E. coli. They grew the bacteria in a medium whose only nitrogen source was ¹⁵NH₄Cl (¹⁵N is a heavy isotope, not radioactive) for many generations, so both strands of every DNA molecule contained ¹⁵N and the DNA was heavy. They then transferred the cells to a medium with normal ¹⁴NH₄Cl and took samples at definite intervals as the cells multiplied (generation time about 20 minutes). DNA extracted from each sample was centrifuged in a caesium chloride (CsCl) density gradient, which separates DNA by density. The heavy and light DNA could be distinguished because ¹⁵N-DNA and ¹⁴N-DNA form bands at different positions.

Result. DNA extracted after 20 minutes (one generation) formed a single hybrid (intermediate) density band. DNA after 40 minutes (two generations) gave equal amounts of hybrid and light DNA. This is exactly what the semi-conservative model predicts; a conservative model would have given heavy plus light after one generation, and a dispersive model would never give a purely light band. Conclusion: DNA replicates semi-conservatively, each daughter having one parental and one new strand.

Taylor and colleagues (1958) confirmed the same with radioactive thymidine in the root cells of Vicia faba (faba bean), showing that chromosomes also replicate semi-conservatively.

The 4-part question on this experiment (the nitrogen source, why heavy and light DNA were made, how they were distinguished, the conclusion) is a favourite; give the isotopes, the CsCl gradient and the 20-minute and 40-minute bands.

Replication is a complex, energy-demanding process run by a set of enzymes. The main one is DNA-dependent DNA polymerase, which uses a DNA template to catalyse polymerisation of deoxynucleotides. It is fast: E. coli, with 4.6 × 10⁶ bp, finishes replication in about 18 minutes, an average rate of about 2000 bp per second, and it is also highly accurate, since an error would be a mutation.

Deoxyribonucleoside triphosphates (dATP, dGTP, dCTP, dTTP) play a dual role: they are the substrates for polymerisation, and their two terminal high-energy phosphates provide the energy for it. This is the 'dual purpose' the 2-mark question asks for.

The helix cannot be opened along its whole length because that would need too much energy, so it opens over a small stretch to form a replication fork, and replication proceeds as the fork moves. DNA polymerase polymerises only in the 5'→3' direction, and the two template strands are antiparallel, so replication on the two strands differs. On the template with 3'→5' polarity the new strand is made continuously (the leading strand). On the template with 5'→3' polarity it is made discontinuously, as short fragments (Okazaki fragments) that are later joined by DNA ligase. Ligase is the second enzyme the examiner expects by name.

Replication does not start anywhere: it begins at a definite region called the origin of replication (ori), which is why a foreign piece of DNA that becomes part of a chromosome can replicate only if it has, or is joined to, an origin; that is also why vectors in recombinant DNA technology need an ori. In eukaryotes replication takes place in the S phase of the cell cycle, and replication and cell division are tightly coordinated; if replication is not followed by division the cell becomes polyploid.

The 5-mark version wants the labelled fork (parental strands, 3' and 5' ends, continuous strand, discontinuous fragments, ligase, polymerase) and the roles of dNTPs, DNA polymerase and DNA ligase in words.

Transcription is copying genetic information from one strand of DNA into RNA, again using complementarity, with one change: adenine now pairs with uracil. Unlike replication, only a segment of DNA and only one strand is copied. NCERT gives two reasons: if both strands were templates they would code for RNAs of different sequences, hence different proteins, complicating the transfer of information; and the two RNAs would be complementary and would pair to form double-stranded RNA, so no translation could occur.

Transcription unit. A transcription unit has three regions: a promoter, the structural gene and a terminator. Since DNA polymerase and RNA polymerase both work 5'→3', the strand with 3'→5' polarity acts as the template strand. The other strand, with 5'→3' polarity, has the same sequence as the RNA (with T in place of U) and is displaced during transcription; it is called the coding strand, and by convention all reference points on a transcription unit are given with respect to it. The promoter is a DNA sequence at the 5'-end (upstream) of the structural gene, where RNA polymerase binds; the terminator is at the 3'-end (downstream) and ends transcription. The presence of the promoter is what defines which strand is template and which is coding.

Transcription unit and the gene. A gene is the functional unit of inheritance, located on DNA. A cistron is a segment of DNA coding for a polypeptide. Eukaryotic structural genes are monocistronic (one polypeptide each) and prokaryotic ones are polycistronic. In eukaryotes the coding sequence is interrupted: the coding regions, exons, appear in the mature or processed RNA, and the intervening sequences, introns, do not. Such genes are split genes. Regulatory sequences (promoter, operator) are often loosely called regulatory genes even though they do not code for RNA or protein.

A frequent 3-marker gives you one strand and asks you to construct the transcription unit or write the mRNA. Rule: mRNA is complementary and antiparallel to the template, and identical to the coding strand with U replacing T. Always mark polarity.

Bacteria have three major RNAs, all needed for protein synthesis: mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNA plays structural and catalytic roles in the ribosome.

Transcription in bacteria. A single DNA-dependent RNA polymerase transcribes all three RNAs, in three steps. Initiation: RNA polymerase binds the promoter with the help of the initiation factor sigma (σ). Elongation: the enzyme uses nucleoside triphosphates as substrates and polymerises RNA in the 5'→3' direction, following complementarity, opening a short stretch of the helix and elongating the RNA chain; only a short stretch of RNA stays bound to the enzyme. Termination: when the polymerase reaches the terminator, the termination factor rho (ρ) causes the nascent RNA and the enzyme to fall off. RNA polymerase alone can only elongate; it is the transient association with sigma and rho that gives initiation and termination. In prokaryotes the mRNA needs no processing, and because there is no nucleus, transcription and translation take place in the same compartment and are coupled: translation can begin before transcription finishes.

Transcription in eukaryotes has two additional complexities. First, there are three RNA polymerases in the nucleus: RNA polymerase I transcribes rRNAs (28S, 18S, 5.8S), RNA polymerase II transcribes the precursor of mRNA, heterogeneous nuclear RNA (hnRNA), and RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs. Second, the primary transcript contains both exons and introns and is non-functional until processed. Splicing removes the introns and joins the exons in a defined order. Capping adds an unusual nucleotide, methyl guanosine triphosphate, to the 5' end; tailing adds 200–300 adenylate residues (the poly-A tail) to the 3' end in a template-independent manner. The fully processed hnRNA, now mRNA, is transported out of the nucleus for translation. Splicing hints at the RNA world, since the split-gene arrangement is thought to be an ancient feature of the genome.

Replication and transcription rely on complementarity, but translation does not: the four bases of mRNA must direct the sequence of twenty amino acids. The genetic code is the set of rules that maps nucleotide triplets to amino acids. Learn the four names and their contributions, because a 3-marker asks for them. George Gamow, a physicist, argued that a code of three nucleotides is needed, since 4 × 4 × 4 = 64 triplets can cover 20 amino acids while 4² = 16 cannot. Har Gobind Khorana chemically synthesised RNA molecules of defined base combinations (homopolymers and copolymers). Marshall Nirenberg developed a cell-free system for protein synthesis. Severo Ochoa discovered the enzyme polynucleotide phosphorylase, which helped polymerise RNA with defined sequences without a template. Together these gave the checkerboard of 64 codons.

Salient features of the code. (i) It is a triplet code; 61 codons code for amino acids and 3 (UAA, UAG, UGA) are stop or termination codons. (ii) One codon codes for only one amino acid; the code is unambiguous and specific. (iii) Some amino acids are coded by more than one codon; the code is degenerate. (iv) The codons are read continuously in the mRNA, with no punctuation; the code is commaless. (v) The code is nearly universal: UUU codes for phenylalanine in bacteria and in humans, with a few exceptions in mitochondrial codons and in some protozoans. (vi) AUG has a dual function: it codes for methionine and it is the initiator codon. Only two amino acids, methionine (AUG) and tryptophan (UGG), have a single codon.

Mutations and the code. A point mutation changes a single base pair; the NCERT example is sickle-cell anaemia, where a change in one base in the β-globin gene (GAG → GUG) replaces glutamic acid with valine. Insertion or deletion of one or two bases shifts the reading frame from that point onward, a frame-shift mutation; inserting or deleting three bases (or multiples of three) adds or removes one codon, so one amino acid changes and the rest of the frame is unchanged. NCERT's sentence 'RAM HAS RED CAP' illustrates the point: delete one letter and every word after it becomes nonsense.

How does a codon on mRNA specify an amino acid when the two cannot recognise each other directly? Francis Crick postulated an adapter molecule that reads the code on one side and binds a specific amino acid on the other. That molecule is tRNA (once called sRNA, soluble RNA). A tRNA has an anticodon loop carrying three bases complementary to the codon, and an amino acid acceptor end to which the amino acid is attached. tRNAs are specific for each amino acid; a codon UAC for tyrosine is read by a tRNA with anticodon AUG (written 3'→5' opposite the codon's 5'→3'), and it carries tyrosine.

Shape. The secondary structure, as drawn in NCERT, looks like a clover leaf; the actual three-dimensional structure is a compact, inverted L. In a diagram show the anticodon loop at the bottom, the two side loops, the amino acid acceptor end at the 3' end on top, the 5' end beside it, and the hydrogen-bonded stem regions.

There are no tRNAs for stop codons, which is why translation stops there. There is a special initiator tRNA for methionine at the initiation codon AUG.

Charging of tRNA (aminoacylation). Before an amino acid can be used it is activated in the presence of ATP and linked to its cognate tRNA by the enzyme aminoacyl-tRNA synthetase; this is charging. It matters because only a charged tRNA can bring the amino acid to the ribosome, and because the high-energy bond made during charging drives peptide bond formation when two charged tRNAs are brought close together. The 2-mark question 'what is aminoacylation and why is it necessary' wants both the definition (with ATP) and the reason (peptide bond formation).

Translation is the polymerisation of amino acids to form a polypeptide, with the order of amino acids set by the sequence of bases in mRNA. Amino acids are joined by peptide bonds, and forming a peptide bond needs energy. So, in the first phase, each amino acid is activated in the presence of ATP and linked to its specific tRNA, the charging of tRNA or aminoacylation. When two charged tRNAs are brought close enough, a peptide bond forms between their amino acids, and the ribosome is the cellular factory that makes this happen efficiently.

The ribosome has two subunits, a larger and a smaller one. In bacteria the ribosome is 70S (50S + 30S). In its inactive state the subunits are separate; when the smaller subunit encounters an mRNA, translation begins. The larger subunit has two sites for subsequent amino acids to come close enough for a peptide bond to form. The rRNA of the large subunit acts as the catalyst for peptide-bond formation: a ribozyme. This answers NCERT exercise 5.9 on the two essential roles of the ribosome: it provides the sites for mRNA and tRNAs to bind, and its rRNA catalyses peptide-bond formation.

Translational unit. On mRNA the sequence of RNA that is translated is flanked by the start codon AUG and one of the stop codons. The mRNA also has additional sequences that are not translated, the untranslated regions (UTRs), present at both the 5' end (before AUG) and the 3' end (after the stop codon); they are required for efficient translation.

Steps. Initiation: the ribosome binds the mRNA at the start codon AUG, which is recognised only by the initiator tRNA carrying methionine. Elongation: complexes of an amino acid with its tRNA bind to the appropriate codon on mRNA by base pairing between codon and anticodon; the ribosome moves along the mRNA one codon at a time and the amino acids are added to the growing chain, peptide bond by peptide bond. Termination: when the ribosome reaches a stop codon, for which there is no tRNA, a release factor binds, the completed polypeptide is released and the ribosome dissociates from the mRNA.

A gene's expression is regulated at several levels. In eukaryotes NCERT lists four: (i) transcriptional level (formation of the primary transcript), (ii) processing level (splicing), (iii) transport of mRNA from nucleus to cytosol, and (iv) translational level. In prokaryotes control of the rate of transcriptional initiation is the predominant site of regulation. Genes are switched on or off in response to metabolic, physiological or environmental conditions; for example, E. coli makes the enzyme β-galactosidase to hydrolyse lactose only when lactose is present.

Transcription of a gene is regulated by accessory proteins that interact with the promoter. An operator is a sequence, usually next to the promoter, to which a repressor protein binds; each operon has its own specific operator and repressor. The operon concept was proposed by Jacob and Monod: a polycistronic structural gene regulated by a common promoter and regulatory genes. Examples are the lac, trp, ara, his and val operons.

The lac operon has a regulatory gene i (inhibitor), the promoter p, the operator o and three structural genes: z codes for β-galactosidase, which hydrolyses lactose into galactose and glucose; y codes for permease, which increases the permeability of the cell to β-galactosides; a codes for transacetylase. All three are needed to metabolise lactose. Lactose is the inducer: it is the substrate of β-galactosidase and it regulates switching the operon on and off. It enters the cell through permease, and a very low basal level of lac operon expression is always present so that some permease exists.

Absence of inducer (switched off): the i gene is transcribed constitutively and its repressor protein binds the operator, preventing RNA polymerase from transcribing the structural genes. Presence of inducer (switched on): lactose binds the repressor and inactivates it, so the repressor cannot bind the operator, RNA polymerase transcribes z, y and a, and the enzymes are made. Since the repressor acts by blocking transcription, this is negative regulation. (NCERT notes the lac operon is also under positive control, but that is outside the syllabus.)

NCERT exercise 5.10 asks why the operon shuts down some time after lactose is added: as β-galactosidase hydrolyses the lactose, the inducer is used up, the repressor is again free to bind the operator, and transcription stops.

The Human Genome Project (HGP) was a 13-year mega project launched in 1990 and completed in 2003, coordinated by the US Department of Energy and the National Institutes of Health, joined later by the Wellcome Trust (UK) and other countries. NCERT exercise 5.12 asks why it is called a mega project: the human genome has about 3 × 10⁹ bp; at about 3 US dollars per base the sequencing cost would be about 9 billion US dollars; stored as typed text with 1000 letters per page and 1000 pages per book, the sequence would need 3300 books; and handling this needed new high-speed computational tools, which grew into bioinformatics.

Goals of HGP. Identify all the approximately 20,000–25,000 genes in human DNA; determine the sequences of the 3 billion chemical base pairs; store the information in databases; improve tools for data analysis; transfer related technologies to other sectors such as industry; and address the ethical, legal and social issues (ELSI) that arise.

Methodologies. Two approaches were used: expressed sequence tags (ESTs), which identify all genes expressed as RNA, and sequence annotation, the blind approach of sequencing the whole genome, coding and non-coding, and later assigning function. Steps: isolate the total DNA and convert it into random fragments of manageable size; clone the fragments in suitable vectors, usually bacterial artificial chromosomes (BAC) and yeast artificial chromosomes (YAC), to amplify them; sequence the fragments with automated DNA sequencers working on the principle developed by Frederick Sanger; align the overlapping sequences by computer; and prepare genetic and physical maps using microsatellite polymorphisms and RFLP markers. Genomes of many non-human organisms were also sequenced: bacteria, yeast, Caenorhabditis elegans, Drosophila, rice and Arabidopsis.

Salient features of the human genome (asked as a 3-marker): it contains 3164.7 million bases; the average gene has 3000 bases, but sizes vary, the largest known being dystrophin at 2.4 million bases; the total number of genes is about 30,000, far fewer than expected, and 99.9% of nucleotide bases are identical in all people; functions are unknown for over 50% of discovered genes; less than 2% of the genome codes for proteins; repeated sequences make up a very large portion; chromosome 1 has the most genes (2968) and the Y chromosome the fewest (231); about 1.4 million locations of single nucleotide polymorphisms (SNPs) were identified, which promise to help find disease-associated sequences and trace human history.

Applications and challenges. Knowing the genome allows studies of how thousands of genes and proteins work together in the interconnected networks of a living organism, and helps in understanding disease, drug design and forensics.

Since 99.9% of the human genome is the same in everyone, telling individuals apart by sequencing would mean finding differences in the remaining 0.1%, a huge task. DNA fingerprinting is a quick way to compare DNA sequences of two individuals. It identifies differences in specific regions of the DNA called repetitive DNA, where a small stretch is repeated many times. These repeats separate as a distinct peak during density-gradient centrifugation, away from the bulk genomic DNA, and are therefore called satellite DNA. Satellite DNA is classified into micro-satellites and mini-satellites on the basis of base composition (A : T rich or G : C rich), length of the segment and number of repetitive units. These sequences do not code for proteins but form a large portion of the genome and show a high degree of polymorphism, which is the basis of fingerprinting.

Polymorphism (variation at the genetic level) arises from mutation. A mutation in a germ cell is passed to the next generation, so the variation is inheritable. Mutations in non-coding DNA accumulate over generations because they have no immediate effect on reproductive ability, which is why non-coding regions are so variable. By definition, an allele is polymorphic when it occurs in the population at a frequency of more than 0.01. DNA polymorphism is the basis of genetic mapping and of DNA fingerprinting.

The technique (Alec Jeffreys) uses a satellite DNA called VNTR (variable number of tandem repeats), a mini-satellite of 0.1 to 20 kb, whose copy number varies from person to person and from chromosome to chromosome. The steps, in NCERT order: (i) isolation of DNA; (ii) digestion of DNA by restriction endonucleases; (iii) separation of fragments by gel electrophoresis; (iv) transfer (blotting) of the separated fragments to a synthetic membrane such as nitrocellulose or nylon (Southern blotting); (v) hybridisation using a labelled VNTR probe; (vi) detection of hybridised fragments by autoradiography. The autoradiogram shows bands of different sizes, and this band pattern is unique to each individual except for identical twins. Since very little DNA is usually available, the sample is now amplified by PCR first.

Applications: forensic science (matching a suspect to a crime-scene sample), paternity disputes, identification of victims (as in a train accident where bodies are burnt beyond recognition), study of genetic diversity in populations, and conservation of species. Half of a child's bands match the mother's and the rest must match the biological father's; that is how a drop of blood or a tissue sample settles paternity.