Molecular Basis of Inheritance NCERT Solutions Class 12 PDF

Concept used. A nitrogenous base is a single nitrogen-containing ring molecule that belongs to one of two families: the double-ring purines (adenine, guanine) and the single-ring pyrimidines (cytosine, thymine, uracil). When a nitrogenous base is attached through an N-glycosidic bond to a five-carbon sugar (ribose in RNA, deoxyribose in DNA), the base–sugar unit is called a nucleoside. If a phosphate group is further joined to the 5' carbon of the sugar, the unit becomes a nucleotide.

1. Step 1: classify each item by its ending.
   • Adenine (-ine) ⇒ free base.
   • Cytidine (-idine) ⇒ base + ribose ⇒ nucleoside (cytosine + ribose).
   • Thymine (-ine) ⇒ free base.
   • Guanosine (-osine) ⇒ base + ribose ⇒ nucleoside (guanine + ribose).
   • Uracil (-il/-ine) ⇒ free base (RNA pyrimidine).
   • Cytosine (-ine) ⇒ free base.
2. Step 2: assemble the two groups.
   • Nitrogenous bases: Adenine, Thymine, Uracil, Cytosine.
   • Nucleosides: Cytidine, Guanosine.
3. Step 3: sanity check. Free bases must have no sugar attached. Adenine, Thymine, Uracil and Cytosine are all just ring molecules. Cytidine = cytosine + ribose; Guanosine = guanine + ribose. Both have a sugar attached, so they are nucleosides. Check passes.

Nitrogenous bases: Adenine, Thymine, Uracil, Cytosine. Nucleosides: Cytidine, Guanosine.

Concept used. Chargaff's rule of base pairing states that in any double-stranded DNA molecule, adenine pairs only with thymine and guanine pairs only with cytosine. Because every A on one strand faces a T on the other, and every G faces a C, % A = % T % G = % C. All four bases together account for every nucleotide in the molecule, so % A + % T + % G + % C = 100%. This is the same equivalence rule that confirmed the double-helix hypothesis when Watson and Crick built their model in 1953.

1. Step 1: write down what is given. % C = 20%. We are asked to find % A.
2. Step 2: use % G = % C. From Chargaff's rule, % G = % C = 20%.
3. Step 3: add up the GC content. % G + % C = 20% + 20% = 40%. So the remaining bases (A and T together) must make up 100% - 40% = 60%.
4. Step 4: split the AT content equally. Because % A = % T, % A + % T = 60% ⇒ 2 × % A = 60% ⇒ % A = 60%2 = 30%.
5. Step 5: sanity check. % A + % T + % G + % C = 30% + 30% + 20% + 20% = 100%.

The percentage of adenine in the DNA is 30%.

Concept used. A DNA double helix is built from two antiparallel strands. ``Antiparallel'' means that if one strand runs 5'3' from left to right, the other strand runs 3'5' over the same region. The bases on the two strands are joined by hydrogen bonds following complementary base pairing:

To write the complementary strand of any given sequence:

1. Replace each base in the given strand with its partner (A→T, T→A, G→C, C→G). This gives the partner sequence in the antiparallel direction.
2. Reverse the resulting sequence so that we read it in 5'3' direction (because by convention every DNA sequence is written 5' end first).

1. Step 1: write the given strand with the 5' end clear. 5'-A T G C A T G C A T G C A T G C A T G C A T G C A T G C-3' That is 28 bases long (4 bases × 7 repeats of ATGC).
2. Step 2: write the partner base under each position. Using A→T, T→A, G→C, C→G: 3'-T A C G T A C G T A C G T A C G T A C G T A C G T A C G-5' Notice the partner strand is written 3'5' because the leftmost base of the original was at 5', so its partner is at 3'.
3. Step 3: reverse the partner strand to put it in 5'3' direction. Reading the line in Step 2 right-to-left, and writing it left-to-right gives: 5'-G C A T G C A T G C A T G C A T G C A T G C A T G C A T-3'
4. Step 4: sanity check. Length: 28 bases, matches the input. First base of new strand = G, last base = T. The first base (G) pairs with the last base of the original (C, at the 3' end), which is correct for antiparallel pairing.

Complementary strand in 5'3' direction:
5' -GCATGCATGCATGCATGCATGCATGCAT-3'

Concept used. In a transcription unit, the two DNA strands play different roles. The template strand (running 3'5') is actually read by RNA polymerase to build the mRNA. The other strand, running 5'3' with a sequence identical to the mRNA (apart from T → U), is called the coding strand (or ``sense'' strand). Therefore the mRNA can be written directly from the coding strand using one rule only: copy the coding strand as-is, but replace every T with U. The 5'3' direction is preserved because mRNA grows in the 5'3' direction, the same direction in which the coding strand is conventionally written.

1. Step 1: identify what we already have. coding strand = 5'-ATGCATGCATGCATGCATGCATGCATGC-3'. This is 28 nucleotides (4 bases × 7 repeats).
2. Step 2: apply the T → U substitution while keeping every other base unchanged. Going base by base: A stays A, T → U, G stays G, C stays C. So ATGC becomes AUGC, and the whole strand becomes: 5'-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'.
3. Step 3: write the answer in standard mRNA direction. mRNA is always reported 5'3' (the direction in which a ribosome reads it). Our sequence is already in that direction, so no reversal is needed.
4. Step 4: optional cross-check via the template strand. Template strand = complement of coding strand, written antiparallel: 3'-TACGTACGTACGTACGTACGTACGTACG-5'. Reading this template 3'5' and writing the complementary RNA base for each: T → A, A → U, C → G, G → C, ... gives exactly 5'-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'.

mRNA sequence:   5' -AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'

Concept used. Semi-conservative replication is the model in which each daughter DNA molecule consists of one parental strand and one newly synthesised strand. The clue that pointed Watson and Crick towards this model lies in the geometry of the helix itself: the two strands of DNA are complementary (every A pairs with T, every G with C) and they are joined only by weak hydrogen bonds. The complementarity means that if the two strands are gently pulled apart, each strand already carries the full information needed to rebuild its partner. The weak hydrogen bonds (in contrast to covalent backbone bonds) allow the strands to come apart without the molecule being destroyed.

1. Step 1: the key property – complementary base pairing. Within the helix, the order of bases on one strand determines the order of bases on the other. There is no independent information on the two strands; they are mirror images of each other through the A–T and G–C rule.
2. Step 2: physical separability via hydrogen bonds. The two strands are held together by 2 hydrogen bonds (A–T) or 3 (G–C). These are weaker than the covalent phosphodiester bonds along each backbone, so the strands can unwind without breaking either backbone. This makes strand separation a feasible step in replication.
3. Step 3: the implication Watson and Crick drew. Because each strand carries the complete information for its partner, the molecule can be copied by: (i) unwinding the helix, (ii) using each parental strand as a template, (iii) laying down a new strand whose bases are complementary to the parental template. After one round of replication, each daughter molecule has one old strand and one new strand – hence ``semi-conservative''.
4. Step 4: experimental confirmation. This hypothesis was later proved in 1958 by Meselson and Stahl using density-gradient centrifugation of E. coli DNA labelled with ¹⁵N, confirming the prediction beyond doubt.

The complementarity of the two strands, joined by weak hydrogen bonds, allowed each strand to act as a template for a new partner, leading directly to the semi-conservative model.

Concept used. A nucleic acid polymerase is an enzyme that synthesises a nucleic-acid strand by adding nucleotides one at a time, using a pre-existing nucleic acid as a template. Two choices define the family of the enzyme:

1. What kind of template does the enzyme read – DNA or RNA?
2. What kind of product does the enzyme build – DNA or RNA?

With two choices per axis, we get 2 × 2 = 4 possible types, each with a distinct biological role.

1. Step 1: list all four template-product combinations.

   tabularc c p4.2cm p4.0cm

   Template & Product & Enzyme & Biological role
   

   DNA & DNA & DNA-dependent DNA polymerase & DNA replication
   DNA & RNA & DNA-dependent RNA polymerase & Transcription
   RNA & DNA & RNA-dependent DNA polymerase & Reverse transcription (retroviruses)
   RNA & RNA & RNA-dependent RNA polymerase & Replication of RNA viruses
   

   tabular

2. Step 2: describe each enzyme briefly.
   • DNA-dependent DNA polymerase: copies DNA into DNA during cell division so each daughter cell gets a full genome.
   • DNA-dependent RNA polymerase: copies a DNA template into mRNA, rRNA or tRNA during transcription. In eukaryotes there are three sub-types (RNA pol I, II and III) for different RNA classes.
   • RNA-dependent DNA polymerase (reverse transcriptase): used by retroviruses such as HIV to write their RNA genome back into DNA so it can integrate into the host genome.
   • RNA-dependent RNA polymerase (RdRp): used by RNA viruses such as influenza, polio and SARS-CoV-2 to copy their RNA genome into more RNA.
3. Step 3: sanity check. The central dogma adds the reverse-transcriptase arrow only after Temin and Baltimore's 1970 discovery; the RdRp arrow was added for RNA viruses. Together the four enzymes cover every information-transfer route between DNA and RNA actually seen in nature.

Four types: DNA-dependent DNA polymerase, DNA-dependent RNA polymerase, RNA-dependent DNA polymerase (reverse transcriptase), and RNA-dependent RNA polymerase.

Concept used. The Hershey–Chase experiment (1952) used radioactive isotope labelling to tell apart the two main components of a virus – DNA and protein – as they entered a host bacterium. The trick relies on the chemistry of these two molecules:

• DNA contains phosphorus but no sulphur (phosphate is part of the backbone, PO₄^3-).
• Protein contains sulphur but no phosphorus (sulphur sits inside the amino acids cysteine and methionine).

So if you grow viruses in a medium containing radioactive sulphur (³⁵S), only their proteins become radioactive. If you grow viruses in a medium containing radioactive phosphorus (³²P), only their DNA becomes radioactive. By following the radioactivity, you follow either the protein or the DNA, but never both at once.

1. Step 1: prepare two separate batches of bacteriophage T2.
   • Batch A: grow viruses in medium with ³⁵S. Outcome: virus protein coat is labelled, DNA is not.
   • Batch B: grow viruses in medium with ³²P. Outcome: virus DNA is labelled, protein coat is not.
2. Step 2: infect E. coli with each batch separately. Allow viruses time to attach to the bacterial wall and inject their genetic material.
3. Step 3: shear the empty viral coats off the bacteria using a blender. Whatever is inside the cell stays inside; whatever is outside (the empty coats) gets stripped from the bacterial surface.
4. Step 4: centrifuge to separate.
   • Heavy bacteria sink to the pellet.
   • Lighter, empty viral coats stay in the supernatant.
5. Step 5: measure radioactivity in pellet vs. supernatant.
   • Batch A (³⁵S, protein): radioactivity stays in the supernatant (the empty coats). Protein never entered the bacterium.
   • Batch B (³²P, DNA): radioactivity ends up in the pellet (inside the bacteria). DNA did enter the bacterium.
6. Step 6: interpret. Only the material that enters the host can be the genetic material (because the new viruses inside the bacteria came from that material). Since only DNA entered, DNA is the genetic material.

Hershey and Chase used ³⁵S to tag protein and ³²P to tag DNA; only the ³²P entered the bacteria, proving DNA, not protein, is the genetic material.

Concept used. Each pair contrasts two related-but-distinct molecules or sequences in molecular biology. To differentiate them cleanly we list, for each pair, the property that is unique to each member and the property they share, in a side-by-side table.

(a) Repetitive DNA vs. Satellite DNA.

(b) mRNA vs. tRNA.

(c) Template strand vs. Coding strand.

1. Step 1: spot the relationship in each pair. (a) Satellite DNA is a type of repetitive DNA. (b) mRNA and tRNA are two separate RNA classes with non-overlapping roles in translation. (c) Template and coding are the two strands of one transcription unit, only one of which is actually read.
2. Step 2: highlight the discriminator. For each pair, the single most useful discriminator is: density-gradient behaviour (a), structural shape (b), and polarity/role in transcription (c).

See tables above. Key discriminators:
(a) satellite DNA forms a separate density band; repetitive DNA is the broader category.
(b) mRNA carries codons; tRNA carries an anticodon and an amino acid.
(c) template is read 3'5' to make mRNA; coding strand is the same as mRNA except T/U.

Concept used. The ribosome is the molecular machine that builds a polypeptide chain from an mRNA template. It is made of rRNA (ribosomal RNA) and ribosomal proteins, and it has two subunits – a small subunit (30S in bacteria, 40S in eukaryotes) and a large subunit (50S/60S). Inside the ribosome there are three tRNA-binding pockets called the A, P and E sites. During translation the ribosome performs two distinct, essential tasks.

1. Role 1: holds the mRNA, the codon, and the matching tRNAs together in space. The small subunit binds the mRNA and slides along it codon by codon. The A site exposes the next codon to be read. Incoming aminoacyl-tRNAs base-pair their anticodon with that codon inside the ribosome's reading frame.
   [2pt] The P site holds the tRNA carrying the growing peptide chain.
   [2pt] Without the ribosome, the mRNA codon and the matching tRNA anticodon would diffuse in solution and would not stay paired long enough for chemistry to happen.
2. Role 2: catalyses the formation of peptide bonds between adjacent amino acids (the peptidyl transferase activity). The peptide bond between the amino acid on the P-site tRNA and the amino acid on the A-site tRNA is formed by a catalytic site in the large subunit of the ribosome. That catalytic site is made entirely of rRNA, so the ribosome is a true ribozyme – an RNA enzyme. After the bond is made, the ribosome translocates to expose the next codon, repeating the cycle until a stop codon is reached.

Role 1: hold the mRNA and two adjacent tRNAs together so that codon–anticodon pairing happens precisely.
Role 2: catalyse peptide-bond formation between adjacent amino acids (peptidyl transferase activity of the large subunit's rRNA).

Concept used. The lac operon is a cluster of three structural genes (z, y, a) controlled together by a single promoter and operator, and a regulatory gene i that codes for the repressor protein. When lactose is absent, the repressor sits on the operator and blocks RNA polymerase from transcribing the operon. When lactose is present, a small amount of lactose is converted by basal β-galactosidase to allolactose, which acts as the inducer. Allolactose binds the repressor, changes its shape, and pulls it off the operator – so transcription begins. This explains why lactose ``switches on'' the operon. The question is the opposite: once switched on, why does the operon switch off again after some time?

1. Step 1: identify what changes as the operon runs. With the operon expressed, the structural genes are translated into three enzymes:
   • z → β-galactosidase: cleaves lactose into glucose + galactose.
   • y → permease: pumps more lactose into the cell.
   • a → transacetylase (a transferase).
   So as the operon runs, the lactose pool in the cell is rapidly being broken down by β-galactosidase.
2. Step 2: trace the consequence for the inducer. Allolactose is itself derived from lactose (a side reaction of β-galactosidase). When lactose is consumed, the amount of allolactose also falls.
3. Step 3: link to the repressor. With less allolactose around, the equilibrium repressor + allolactose repressor-allolactose complex shifts to the left: free repressor reappears.
4. Step 4: shut-down step. The free repressor binds back to the operator, physically blocks RNA polymerase, and transcription of the structural genes stops. Existing mRNAs are degraded quickly (mRNA half-life in bacteria is ∼ 2 minutes), so within a few minutes the cell stops making the lac enzymes and the operon is back to its ``off'' state.
5. Step 5: name the kind of regulation. Because lactose / allolactose is both the substrate and the inducer, the operon is autoregulated by negative feedback: the operon's output (enzymes) destroys the input (inducer), which turns the operon off. This is the classic example of end-product (substrate) regulation.

Once β-galactosidase digests away the lactose, the inducer (allolactose) is no longer present in enough quantity to keep the repressor off the operator. The repressor rebinds, blocks transcription, and the operon switches off, a negative-feedback, substrate-depletion shut-down.

Concept used. Each of these three terms names a different piece of the transcription–translation pipeline. We give a precise one-or-two-line function for each, and follow up with a one-line expansion of why that function matters.

1. (a) Promoter. A promoter is a short DNA sequence located upstream of a gene (with consensus elements at positions -10, the Pribnow box, and -35 relative to the transcription start site in bacteria) that serves as the binding site for RNA polymerase. By providing this binding site, the promoter defines exactly where transcription will start and which of the two DNA strands will be used as the template.
   [2pt] Why this matters. Different promoters have different strengths, so they control how often the gene downstream of them is transcribed. Mutations in a promoter can therefore increase or decrease gene expression without altering the protein-coding sequence at all.
2. (b) tRNA (transfer RNA). A tRNA is a small (∼ 73–90 nt) clover-leaf RNA that brings a specific amino acid to the ribosome and reads the mRNA codon via its anticodon loop, so that the correct amino acid is added next in the growing polypeptide.
   [2pt] Why this matters. Each of the 20 amino acids has at least one dedicated tRNA. Without tRNAs the codon table on the mRNA would have no way of being translated into a protein sequence – tRNAs are the physical embodiment of the genetic code.
3. (c) Exons. Exons are the segments of a eukaryotic protein-coding gene whose sequences are kept in the mature mRNA after splicing. They contain the codons that ultimately get translated into the protein sequence.
   [2pt] Why this matters. Splicing alternative combinations of exons (alternative splicing) lets one gene produce multiple distinct proteins, which is one major reason why humans have only ∼ 20,000 genes but hundreds of thousands of distinct proteins.

(a) Promoter: DNA site upstream of a gene where RNA polymerase binds; sets the start point of transcription.
(b) tRNA: clover-leaf RNA that carries an amino acid to the ribosome and reads the mRNA codon via its anticodon.
(c) Exons: the coding segments of a eukaryotic gene that are retained in the mature mRNA and translated into protein.

Concept used. The Human Genome Project (HGP) was an international research programme (1990–2003) that aimed to determine the complete nucleotide sequence of the ∼ 3 × 10⁹ base pairs of the human genome and to identify every gene in it. It is called a ``mega project'' not because of one feature but because of the scale, cost, duration, organisational scope, technological demands and societal impact – all of which were unprecedented for any single biology project up to that point.

steps Step 1: scale of data. The human genome contains about 3.1 × 10⁹ base pairs. Sequencing each pair, storing it, checking it and annotating it meant handling a dataset that, in 1990 computing terms, was enormous. If printed at 1000 letters per page in standard books, the data would fill about 3300 books of 1000 pages each. Step 2: cost. The total budget was approximately US  9 billion. Such a sum, contributed by multiple governments (mainly the US Department of Energy and NIH, with UK, France, Germany, Japan and China), is in the same league as a space exploration programme. Step 3: duration. The project ran for about 13 years, with a working draft published in 2001 and the substantially complete sequence announced in 2003. A decade-plus commitment is unusual for any biology project; most biology grants run 3--5 years. Step 4: technology demands. HGP drove the development of automated DNA sequencers (capillary-based Sanger machines), bacterial artificial chromosome (BAC) cloning libraries, expressed sequence tag (EST) databases, and bioinformatics tools. The cost-per-base fell by orders of magnitude during the project. Step 5: international coordination. Twenty laboratories across six countries (US, UK, France, Germany, Japan, China) sequenced different chromosomes in parallel, with regular data deposition into the public GenBank database. Coordinating that effort -- with shared standards, shared software and free open access to data -- was itself a major achievement. Step 6: societal and ethical impact. For the first time, biological research had to be paired with formal ELSI (Ethical, Legal and Social Implications) programmes covering privacy, informed consent, gene patents and insurance discrimination. About 5% of the HGP budget was set aside for ELSI work. Step 7: legacy. The HGP made personalised medicine, comparative genomics, rapid cancer-genome sequencing, and the COVID-19 vaccine pipelines (which depend on fast viral-genome sequencing) possible. Its return-on-investment, conservatively estimated, runs into hundreds of billions of dollars. steps

center tikzpicture[font=, scale=0.9] % bars showing scale [->, thick, cdInk] (0,0) -- (8.4,0); [->, thick, cdInk] (0,0) -- (0,4.0); [rotate=90, font=] at (-0.6,2.0) Scale; x/h//in 1.0/3.5/cdAccent/3 10^9 bp, 2.8/3.0/cdMint/9 billion, 4.6/2.4/cdAmber/13 years, 6.4/2.0/cdViolet/20 labs / 6 countries [!70] (-0.4,0) rectangle (+0.4,); [!50!black, very thick] (-0.4,0) rectangle (+0.4,); [above, font=, color=!60!black, align=center] at (,) ;

[below, font=] at (1.0,-0.05) Data; [below, font=] at (2.8,-0.05) Cost; [below, font=] at (4.6,-0.05) Duration; [below, font=] at (6.4,-0.05) Teams; tikzpicture
[2pt] cdMutedHGP at a glance: vast data, very high cost, long duration and global teamwork. center

Because the HGP combined an unprecedented scale of data (3× 10⁹ bp), cost (US 9 billion), duration ( 13$ years), international coordination, technology development and societal/ethical impact, no single feature but the combination of all of them earned it the ``mega'' label.

Concept used. DNA fingerprinting is a laboratory technique that identifies an individual from a tiny amount of biological sample by comparing highly variable, person-specific regions of the genome. The variability comes from short DNA sequences called VNTRs (Variable Number Tandem Repeats) or minisatellites, which are part of the satellite DNA component of the genome. Each person has a unique combination of copy numbers at many VNTR loci – so unique that the chance of two unrelated people sharing the same pattern is vanishingly small. The technique was developed by Alec Jeffreys in 1984.

1. Step 1: isolate DNA from the sample. Sources can be blood, semen, skin, hair root, saliva, bone – any tissue with nucleated cells.
2. Step 2: amplify and cut. Either amplify the VNTR loci with the polymerase chain reaction (PCR), or digest the genomic DNA with a restriction enzyme that cuts outside the VNTR region. The fragment length then reflects the number of repeats in that region.
3. Step 3: separate by size. Run the fragments through an agarose gel by electrophoresis; shorter fragments migrate further.
4. Step 4: detect with a probe. Transfer the bands to a membrane (Southern blot) and probe with a radioactively labelled VNTR sequence to make the bands visible. The result is a ladder of bands – the fingerprint.
5. Step 5: compare. Compare the fingerprints of the unknown sample and one or more reference samples. Identical fingerprints across many loci means the samples are from the same individual (or identical twins). Sharing half the bands on average means a parent–child relationship.

Applications:

• Forensic science – matching biological evidence at a crime scene (blood, semen, hair) to a suspect.
• Paternity / maternity testing – a child shares half the bands with each biological parent.
• Identification of victims of mass disasters, plane crashes, fires.
• Pedigree analysis – tracing family trees, identifying smuggled wildlife, etc.
• Population and evolutionary studies – estimating genetic distance between populations, conservation genetics of endangered species.

DNA fingerprinting is the identification of an individual by analysing person-specific variable regions of DNA (VNTRs) using PCR/restriction digestion + electrophoresis + DNA hybridisation. Applications: forensic identification, paternity testing, victim identification, pedigree analysis, population genetics.

Concept used. Each of these four terms names one of the foundational ideas of molecular biology. We define each precisely, then briefly describe what happens in the process or why the phenomenon matters.

(a) Transcription. Transcription is the process by which a single-stranded mRNA molecule is synthesised on a DNA template, by the enzyme RNA polymerase. It has three phases:

• Initiation: RNA polymerase binds the promoter of a gene and unwinds the local DNA.
• Elongation: the polymerase reads the template strand 3'5' and builds a complementary mRNA 5'3', using ribonucleoside triphosphates as substrates and following the rule A→U, T→A, G→C, C→G.
• Termination: at a terminator sequence the polymerase, DNA and nascent mRNA dissociate.

In eukaryotes the mRNA is further processed (5' cap, 3' poly-A tail, splicing of introns) before leaving the nucleus.

(b) Polymorphism. Polymorphism is the occurrence of two or more variants (alleles or sequence forms) of a DNA sequence in a population at a frequency > 1% for the rarer variant. Most polymorphisms are single-nucleotide polymorphisms (SNPs): a single base position where different individuals carry different bases (e.g. A in some, G in others). VNTRs (used in DNA fingerprinting) are a length polymorphism. Polymorphism is the raw material of genetic diversity within a species and is the basis for studying disease susceptibility, ancestry and evolution.

(c) Translation. Translation is the process by which the codon sequence of mRNA is decoded into the amino acid sequence of a polypeptide, on the ribosome. Each tRNA brings one amino acid that matches the codon being read at the ribosome's A-site; the ribosome's large subunit catalyses peptide-bond formation; the ribosome then translocates one codon along the mRNA. The chain starts at the AUG start codon and ends at a stop codon (UAA, UAG, or UGA). The new polypeptide then folds (often with the help of chaperones) into a functional protein.

(d) Bioinformatics. Bioinformatics is the application of computational tools – algorithms, databases and statistical models – to store, search and analyse biological data, especially DNA and protein sequences. Typical tasks include genome assembly, sequence alignment, gene prediction, protein-structure prediction, phylogenetic-tree construction and gene-expression analysis. Bioinformatics became indispensable during the Human Genome Project, and today drives drug discovery, COVID variant tracking and personalised medicine.

(a) Transcription = DNA → mRNA by RNA polymerase (initiation → elongation → termination).
(b) Polymorphism = >1% frequency of two or more sequence variants (SNPs, VNTRs) at a locus in a population.
(c) Translation = mRNA codons decoded into amino acids on the ribosome with tRNA adapters.
(d) Bioinformatics = computational analysis of biological sequence and structural data.