NCERT Solutions for Class 12 Biology Chapter 5: Molecular Basis of Inheritance
All 14 exercise questions of the latest NCERT textbook (2026–27 reprint) answered with maximum descriptive detail — full explanation of every point, labelled DNA double-helix and semiconservative-replication diagrams linked to the exact NCERT pages, base-pairing calculations, the lac operon mechanism, comparison tables, complete revision notes and multiple exam tips per answer.
Chapter at a glance — complete revision notes
5.1 The DNA
- DNA is a long polymer of deoxyribonucleotides; its length = number of nucleotides (or base pairs). E. coli has 4.6 × 10⁶ bp; human haploid DNA has 3.3 × 10⁹ bp.
- A nucleotide = nitrogenous base + pentose sugar (deoxyribose) + phosphate. Bases: purines (Adenine, Guanine) and pyrimidines (Cytosine, Thymine; Uracil in RNA).
- Nucleoside = base + sugar (no phosphate), e.g., adenosine, guanosine, cytidine, thymidine, uridine.
5.1.1 Structure of the Polynucleotide Chain
- Nucleotides are joined by 3′–5′ phosphodiester bonds to form a polynucleotide chain with a free 5′-phosphate and 3′-OH end.
- Watson & Crick (1953) gave the double-helix model based on Wilkins & Franklin’s X-ray data.
- Two antiparallel strands (5′→3′ and 3′→5′) are held by hydrogen bonds: A=T (2 bonds), G≡C (3 bonds).
- Chargaff’s rule: A = T and G = C, so (A+G) = (T+C).
- The helix is right-handed; one turn = 3.4 nm with 10 bp; distance between adjacent bp = 0.34 nm.
5.1.2 Packaging of DNA Helix
- In prokaryotes, negatively charged DNA is held by positively charged proteins in the nucleoid.
- In eukaryotes, DNA + positively charged histone proteins (rich in lysine & arginine) form the nucleosome (DNA wound around a histone octamer). Nucleosomes coil to form chromatin → euchromatin (loose, active) and heterochromatin (condensed, inactive).
5.2 The Search for Genetic Material
- Griffith (1928) — transformation experiment with Streptococcus pneumoniae (S and R strains).
- Avery, MacLeod & McCarty — showed the transforming principle is DNA (DNase destroyed transformation).
5.2.1 The Genetic Material is DNA (Hershey–Chase)
- Hershey & Chase (1952) used bacteriophages with radioactive ³²P (labels DNA) and ³⁵S (labels protein).
- Only ³²P (DNA) entered the bacteria and appeared in progeny — proving DNA is the genetic material.
5.3 RNA World
- RNA was the first genetic material. RNA acts as a genetic material and a catalyst (ribozyme). DNA evolved from RNA as a more stable molecule for information storage.
5.4 Replication
- DNA replication is semiconservative — each daughter DNA has one parental and one new strand (Watson & Crick’s prediction).
5.4.1 Experimental Proof
- Meselson & Stahl (1958) grew E. coli in ¹⁵N, then shifted to ¹⁴N; density-gradient centrifugation showed hybrid DNA after one generation, proving semiconservative replication.
- Taylor et al. confirmed it in Vicia faba (broad bean) using radioactive thymidine.
5.4.2 The Machinery and Enzymes
- Main enzyme: DNA-dependent DNA polymerase (adds nucleotides only in the 5′→3′ direction).
- Replication starts at the origin of replication (ori); the leading strand is synthesised continuously, the lagging strand discontinuously (as Okazaki fragments joined by DNA ligase).
5.5 Transcription
- Copying of genetic information from DNA (template strand) into RNA by RNA polymerase (5′→3′). Only the 3′→5′ template strand is copied.
5.5.2 Transcription Unit and the Gene
- A transcription unit has a promoter (RNA polymerase binding site, upstream/5′), the structural gene and a terminator (3′).
- Template strand (3′→5′) is copied; the coding strand (5′→3′) has the same sequence as mRNA (except T→U).
- In eukaryotes genes are split: exons (coding, expressed) and introns (non-coding, removed by splicing).
5.5.3 Types of RNA & Transcription
- Bacteria: a single RNA polymerase makes all RNA. Eukaryotes: RNA Pol I (rRNA), RNA Pol II (mRNA/hnRNA), RNA Pol III (tRNA, 5S rRNA).
- hnRNA is processed by capping (methyl guanosine at 5′), tailing (poly-A at 3′) and splicing to form mature mRNA.
5.6 Genetic Code
- The code is a triplet (codon) — 64 codons: 61 code for amino acids, AUG = start (methionine), UAA, UAG, UGA = stop.
- Features: triplet, degenerate (one amino acid, several codons), non-overlapping, comma-less, universal, unambiguous. tRNA is the adapter molecule with an anticodon.
5.6.1 Mutations & Genetic Code
- Point mutation (e.g., sickle-cell anaemia) and frameshift mutation (insertion/deletion) show the code is a triplet read continuously.
5.7 Translation
- Synthesis of protein from mRNA on ribosomes. Amino acids are joined by peptide bonds. Stages: initiation, elongation, termination. The ribosome has a peptidyl transferase activity (ribozyme) and provides sites for tRNA binding.
5.8 Regulation of Gene Expression / 5.8.1 The Lac Operon
- Gene expression is regulated (in bacteria) mainly at the transcription level.
- Lac operon (Jacob & Monod) = regulatory gene (i) + promoter + operator + 3 structural genes z, y, a. z→β-galactosidase (splits lactose into glucose + galactose), y→permease, a→transacetylase.
- Lactose = inducer. No lactose → repressor (from i gene) binds the operator → operon OFF. With lactose → lactose binds/inactivates the repressor → RNA polymerase transcribes → operon ON. When lactose is used up, the repressor binds again → operon OFF.
5.9 Human Genome Project (HGP)
- A mega project (1990–2003) to sequence all 3 × 10⁹ bp of the human genome. Coordinated by the US Dept. of Energy & NIH, with the Wellcome Trust; Bioinformatics was developed to handle the data.
5.9.1 Salient Features of the Human Genome
- ~3164.7 million bp; ~20,000–25,000 genes; average gene ~3000 bp; <2% codes for proteins; over 99.9% of bases are the same in all people; chromosome 1 has the most genes (2968), the Y the fewest (231).
5.10 DNA Fingerprinting
- A technique to identify differences in specific regions of DNA (VNTRs — variable number of tandem repeats / satellite DNA) that are unique to each individual (except identical twins).
- Developed by Alec Jeffreys; in India by the CCMB, Hyderabad. Uses very small amounts of DNA. Applications: forensic (crime) investigation, paternity/parentage testing, and studying population/evolutionary genetics.
Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
To correctly classify these six terms, we must recall the building-block hierarchy of nucleic acids: a nitrogenous base is a nitrogen-containing ring compound on its own (either a purine — adenine or guanine — or a pyrimidine — cytosine, thymine or uracil). When a nitrogenous base is chemically joined to a pentose sugar (ribose or deoxyribose) but without any phosphate group, the resulting combination is called a nucleoside. (Adding a phosphate group to a nucleoside then produces a full nucleotide, the true monomer unit of DNA/RNA — but that is a step beyond what this question asks.)
| Nitrogenous bases | Nucleosides |
|---|---|
| Adenine, Thymine, Uracil, Cytosine | Cytidine, Guanosine |
Here, Cytidine is the nucleoside formed by joining cytosine to a ribose sugar, and Guanosine is the nucleoside formed by joining guanine to a ribose sugar — both therefore qualify as nucleosides because they already include the sugar component. The remaining four terms — Adenine, Thymine, Uracil and Cytosine — are simply the bare nitrogenous bases, with no attached sugar.
Conclusion: Adenine, Thymine, Uracil and Cytosine are nitrogenous bases on their own, while Cytidine (cytosine + ribose) and Guanosine (guanine + ribose) are nucleosides, since each already includes a sugar component.
If a double-stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
This calculation is based directly on Chargaff’s rule (Chargaff’s equivalence rule), an empirical finding (later explained structurally by Watson and Crick’s base-pairing model) which states that in any double-stranded DNA molecule, the total amount of adenine always equals the total amount of thymine (A = T), and the total amount of guanine always equals the total amount of cytosine (G = C). It follows directly from this that the four bases together must add up to the whole molecule: A + T + G + C = 100%.
Given information: Cytosine (C) = 20% of the total bases.
Step-by-step calculation:
- Step 1: Since Chargaff’s rule requires G = C, and we are given C = 20%, it follows that Guanine (G) = 20% as well.
- Step 2: Adding these together, C + G = 20% + 20% = 40% of the total DNA is accounted for by cytosine and guanine combined.
- Step 3: Since all four bases must sum to 100%, the remaining two bases (A and T) together must make up A + T = 100% − 40% = 60%.
- Step 4: Since Chargaff’s rule also requires A = T, this remaining 60% must be split equally between them: Adenine (A) = 60% ÷ 2 = 30%.
∴ Percentage of Adenine in the DNA = 30%
As a check, the complete base composition works out to A = 30%, T = 30%, G = 20%, C = 20%, which correctly sums to 100%, and correctly satisfies both A = T and G = C.
Conclusion: applying Chargaff’s rule — since G = C = 20%, the remaining A + T must equal 60%, and since A = T, the percentage of adenine in the DNA works out to 30%.
If the sequence of one strand of DNA is written as 5′-ATGCATGCATGCATGCATGCATGCATGC-3′, write down the sequence of the complementary strand in 5′→3′ direction.
To solve this we must apply two key structural rules of the DNA double helix together: (1) complementary base pairing, where every A on one strand pairs with a T on the other, and every G pairs with a C; and (2) antiparallel orientation, meaning the two strands of DNA run in opposite directions to each other — if one strand is read 5′→3′ left to right, its complementary partner strand runs 3′→5′ left to right (or equivalently, 5′→3′ right to left).
The correct method is therefore: first pair every base with its complementary base directly underneath (giving the complementary strand in its natural 3′→5′ orientation, aligned against the given strand), and then reverse the entire sequence to express it conventionally in the standard 5′→3′ direction.
Direct complement (3′→5′): 3′-TACGTACGTACGTACGTACGTACGTACG-5′
Answer, reversed to 5′→3′: 5′-GCATGCATGCATGCATGCATGCATGCAT-3′
Conclusion: base-pairing each nucleotide (A↔T, G↔C) and then reading the antiparallel complementary strand in the conventional 5′→3′ direction gives the sequence 5′-GCATGCATGCATGCATGCATGCATGCAT-3′.
If the sequence of the coding strand in a transcription unit is written as 5′-ATGCATGCATGCATGCATGCATGCATGC-3′, write down the sequence of the mRNA.
This question tests the important distinction between the template strand and the coding strand of a gene, and their respective relationships to the mRNA transcript produced. During transcription, RNA polymerase actually reads and copies only the template strand (oriented 3′→5′), not the coding strand. However, because the template strand is itself complementary to the coding strand, the mRNA synthesised from the template strand ends up having exactly the same sequence as the coding strand — with just one small but important chemical substitution: wherever the coding strand (being DNA) has thymine (T), the mRNA (being RNA) has uracil (U) instead, since RNA does not use thymine.
mRNA (5′→3′) : 5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′
Notice that the polarity (5′→3′) and every base position remain identical between the coding strand and the mRNA — only every “T” has become a “U” — which is exactly why the coding strand is named as it is: it directly represents (or “codes for,” in sequence terms) the mRNA that will eventually be translated into protein.
Conclusion: because the mRNA has the same sequence as the coding strand (with U replacing T), the mRNA transcribed here is 5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′.
Which property of the DNA double helix led Watson and Crick to hypothesise the semiconservative mode of DNA replication? Explain.
The single structural property that led James Watson and Francis Crick to propose semiconservative replication was the complementary base pairing between the two antiparallel strands of the DNA double helix — the fixed rule that adenine always pairs specifically with thymine (via two hydrogen bonds), and guanine always pairs specifically with cytosine (via three hydrogen bonds), with no other pairing combinations occurring in normal double-stranded DNA.
Detailed explanation of the logical reasoning:
- Because base pairing is so strictly complementary, the sequence of bases on one strand completely and uniquely determines the sequence of bases that must be present on the other strand — there is no ambiguity. If you know one strand’s sequence, you automatically know its partner’s sequence.
- This meant that Watson and Crick immediately recognised that each individual strand, on its own, contains enough information to specify and reconstruct its missing partner strand — in other words, each strand could act as a template (mould) for building an entirely new complementary strand.
- Based on this insight, they proposed that DNA replication must proceed by the two strands of the parental double helix first separating (unwinding) from each other, breaking the hydrogen bonds between them, after which each of the two now-separated old strands independently directs the synthesis of a brand-new complementary strand alongside it, following the same strict base-pairing rules.
- The direct consequence of this mechanism is that each of the two resulting daughter DNA molecules ends up containing one original (parental, “old”) strand and one newly synthesised (“new”) strand — the original parental double helix is thus never fully conserved as a single unit, but is also never fully lost either; it is “half-conserved” in each daughter molecule. This is precisely why the mode of replication was named semiconservative.
Conclusion: the strictly complementary, antiparallel base-pairing structure of the double helix meant each strand could act as a template for building its partner, which is exactly the property that led Watson and Crick to hypothesise that DNA replication is semiconservative — each daughter molecule retaining one parental strand and gaining one newly made strand.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of the nucleic acid synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
A nucleic acid polymerase is broadly any enzyme that synthesises a new strand of a nucleic acid (DNA or RNA) by reading and copying an existing nucleic acid strand used as a template. Since both the template and the product being synthesised can independently be either DNA or RNA, there are logically four possible combinations, and biology makes use of all four, each associated with a specifically named enzyme and a specific biological process:
| Template read | Product synthesised | Name of the polymerase enzyme | Biological process / where found |
|---|---|---|---|
| DNA | DNA | DNA-dependent DNA polymerase | DNA replication — the routine copying of a cell’s genome before cell division, in essentially all cellular organisms. |
| DNA | RNA | DNA-dependent RNA polymerase | Transcription — the copying of a gene’s DNA sequence into a messenger, ribosomal or transfer RNA molecule. |
| RNA | DNA | RNA-dependent DNA polymerase (commonly called reverse transcriptase) | Reverse transcription — used by retroviruses (e.g., HIV) to convert their RNA genome into DNA, which then integrates into the host cell’s chromosome. |
| RNA | RNA | RNA-dependent RNA polymerase (also called RNA replicase) | RNA replication — used by many RNA viruses to copy their own RNA genome without ever going through a DNA intermediate. |
Conclusion: based on all four logical combinations of template and product, the four types of nucleic acid polymerases are DNA-dependent DNA polymerase (replication), DNA-dependent RNA polymerase (transcription), RNA-dependent DNA polymerase or reverse transcriptase (used by retroviruses), and RNA-dependent RNA polymerase (used by RNA viruses).
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Alfred Hershey and Martha Chase performed a now-classic experiment in 1952 that provided the first truly unequivocal, direct proof that DNA — not protein — is the genetic material, settling a debate that had persisted for decades among biologists (many of whom had earlier assumed, incorrectly, that proteins were too structurally complex and DNA too “simple” for DNA to be the carrier of genetic information). Their experimental system used bacteriophages — viruses that specifically infect bacteria — which are structurally simple, being made of essentially just DNA enclosed in a protein coat. This simplicity was the key to the experiment’s elegant design.
The central chemical trick that allowed DNA and protein to be distinguished was a difference in their elemental composition: DNA contains phosphorus but no sulphur, whereas protein contains sulphur (in the amino acids cysteine and methionine) but essentially no phosphorus. This meant that if the two molecules could be separately radioactively labelled with different labels, scientists could track which one physically entered the bacterial cell during infection.
Step-by-step procedure:
- One batch of bacteriophages was grown in a culture medium supplemented with radioactive phosphorus (³²P). Since only DNA (not protein) contains phosphorus, this specifically and exclusively labelled the phages’ DNA with radioactivity.
- A separate, second batch of phages was grown in a medium supplemented instead with radioactive sulphur (³⁵S). Since only protein (not DNA) contains sulphur, this specifically and exclusively labelled the phages’ protein coat with radioactivity.
- Each radioactively-labelled batch of phages was then allowed to infect fresh, non-radioactive E. coli bacteria separately.
- Shortly after the infection process had begun, the mixtures were agitated vigorously in a kitchen blender, a step specifically designed to physically shake loose and detach any remaining empty viral protein coats that were still clinging to the outside of the bacterial cells (since only the material that had actually entered the cell would remain firmly associated with it).
- The mixtures were then spun in a centrifuge, which separates the denser bacterial cells (which pellet at the bottom) from the lighter, detached viral coat fragments (which remain in the surrounding liquid).
Observed results and interpretation:
- Bacteria that had been infected with the ³²P-labelled (DNA-labelled) phages were found to have become radioactive themselves — directly demonstrating that the radioactive DNA had physically entered the bacterial cell during infection.
- Bacteria that had been infected with the ³⁵S-labelled (protein-labelled) phages, by contrast, showed essentially no radioactivity — demonstrating that the radioactive protein coat had remained outside the bacterial cell and had never actually entered it.
Since it is well established that the genetic material of a virus must be the component that enters the host cell and subsequently directs the host’s machinery to manufacture new copies of the virus, and since only the DNA (³²P) was shown to enter — while the protein (³⁵S) was shown to stay outside — this experiment provided decisive proof that DNA, and not protein, is the genetic material.
Conclusion: Hershey and Chase distinguished DNA from protein by radioactively labelling DNA with ³²P (phosphorus, present only in DNA) and protein with ³⁵S (sulphur, present only in protein); since only the ³²P label was found to enter the infected bacteria, they proved that DNA — not protein — is the genetic material.
Differentiate between: (a) Repetitive DNA and Satellite DNA (b) mRNA and tRNA (c) Template strand and Coding strand.
(a) Repetitive DNA vs Satellite DNA
Not all of an organism’s DNA codes for proteins or is even unique in sequence — large portions of eukaryotic genomes consist of sequences that are repeated many times over, and satellite DNA is a specific, important sub-category of this repetitive DNA.
| Repetitive DNA | Satellite DNA |
|---|---|
| A broad category of short DNA sequences that are repeated many hundreds to many thousands of times throughout the genome, at various locations. | A specific sub-type of repetitive DNA, distinguished by having a base composition (proportion of A-T versus G-C) so different from the “bulk” genomic DNA that it separates out as a distinct, small “satellite” peak during density-gradient centrifugation — which is precisely how it got its name. |
| May be coding or non-coding, and constitutes a large fraction of the genome by mass. | Highly repetitive and non-coding, typically found in and around centromeric and other structural regions of chromosomes; it shows a high degree of individual-to-individual variation (polymorphism) in the number of repeats, which is precisely what makes it the molecular basis of DNA fingerprinting. |
(b) mRNA vs tRNA
Both are types of RNA essential to protein synthesis, but they play completely different, complementary roles in the process of translation.
| mRNA (messenger RNA) | tRNA (transfer RNA) |
|---|---|
| Carries the genetic message, encoded as a linear sequence of codons, copied from a gene on the DNA, all the way to the ribosome; it serves as the direct template that specifies the order in which amino acids must be joined during protein synthesis. | Acts as the “adapter” or “translator” molecule that physically reads each codon on the mRNA (via its own complementary anticodon) and delivers the correct, corresponding amino acid to the ribosome for incorporation into the growing polypeptide chain. |
| A relatively long, linear, largely unfolded single-stranded molecule, whose length varies depending on the size of the protein it codes for. | A comparatively short molecule (about 75–90 nucleotides) that folds back on itself through internal base-pairing into a characteristic clover-leaf secondary structure, with the anticodon exposed at one loop. |
(c) Template strand vs Coding strand
Within any gene’s double-stranded DNA, the two strands play asymmetric roles during transcription, and this distinction is a frequently tested concept.
| Template strand | Coding strand (also called the sense strand) |
|---|---|
| Runs with polarity 3′→5′ (relative to the direction transcription proceeds); this is the strand that RNA polymerase actually binds to and copies to synthesise the mRNA. | Runs with polarity 5′→3′; this strand is not directly transcribed by RNA polymerase — it simply sits as the “passive” partner strand during transcription. |
| Its base sequence is complementary to the mRNA produced (since the mRNA is built by base-pairing against it). | Its base sequence is identical to the mRNA sequence produced, except that every T in the DNA coding strand corresponds to a U in the mRNA. |
Conclusion: satellite DNA is a highly repetitive, base-composition-distinct, polymorphic sub-type of repetitive DNA (the basis of DNA fingerprinting); mRNA carries the coded genetic message to the ribosome while tRNA is the adapter molecule that supplies the matching amino acid; and within a gene, the template strand (3′→5′) is the one actually transcribed, while the coding strand (5′→3′) matches the resulting mRNA sequence.
List two essential roles of the ribosome during translation.
The ribosome is a complex, dynamic molecular machine made of ribosomal RNA (rRNA) and proteins, organised into a large and a small subunit, and it is the actual physical site where translation (protein synthesis) takes place. It performs several coordinated functions during this process, of which two are considered the most essential:
- Provides the structural platform and binding sites for mRNA and tRNA, enabling accurate codon–anticodon recognition: the ribosome has distinct, specifically-shaped binding sites — commonly referred to as the P (peptidyl) site, A (aminoacyl) site, and E (exit) site — that hold the mRNA strand and the incoming charged tRNA molecules in the precise physical alignment needed for the codon on the mRNA to correctly pair with the complementary anticodon on the tRNA. Without this precise physical positioning, amino acids could not be reliably added to the growing polypeptide chain in the correct order specified by the genetic code.
- Catalyses the formation of the peptide bond between successive amino acids (acts as a ribozyme): remarkably, it is not a protein component of the ribosome but its ribosomal RNA — specifically within the large subunit — that possesses the actual catalytic activity, known as peptidyl transferase activity, which chemically links the amino acid held in the P site to the new amino acid just delivered to the A site, forming a peptide bond. Since this catalytic function is carried out by RNA rather than protein, the ribosome is classified as a ribozyme (a catalytic RNA). After each peptide bond is formed, the ribosome also physically moves (translocates) one codon further along the mRNA, making way for the next round of amino acid addition.
Conclusion: the ribosome’s two most essential roles in translation are providing the precisely-shaped binding sites (P, A and E sites) that allow accurate codon–anticodon pairing between mRNA and tRNA, and catalysing the actual formation of peptide bonds between successive amino acids through its ribozyme (peptidyl transferase) activity.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does the lac operon shut down some time after the addition of lactose in the medium?
The lac operon, first elucidated by the geneticist François Jacob and the biochemist Jacques Monod, is a classic example of an inducible operon — a cluster of genes whose transcription is switched on only in the presence of a specific triggering molecule, here called the inducer. In this operon, lactose itself functions as the inducer, and understanding exactly how it switches the operon both ON and eventually back OFF requires tracing the full sequence of molecular events:
- Switching ON: when lactose is added to the growth medium, it enters the bacterial cell (helped by a low, constitutive baseline level of permease that must always be present) and binds to the repressor protein — a protein continuously synthesised from the operon’s own regulatory i gene. This binding inactivates the repressor, changing its shape so that it can no longer attach to the operator region of the operon. With the operator now free, RNA polymerase is able to bind the promoter and transcribe the three structural genes — z, y and a — switching the operon ON.
- Enzyme action consumes the inducer: the z gene product, the enzyme β-galactosidase, actively hydrolyses (breaks down) the very lactose molecules that had switched the operon on, splitting each lactose molecule into glucose and galactose for the bacterium’s metabolic use.
- Depletion of the inducer over time: because β-galactosidase is continuously breaking lactose down while no fresh lactose is being added externally, the concentration of available lactose (inducer) inside the cell steadily falls and is eventually exhausted.
- Switching back OFF: once there is no lactose left to bind and inactivate it, the repressor protein reverts to its active form and once again binds tightly to the operator region, physically blocking RNA polymerase from transcribing the structural genes any further — so the lac operon shuts down (is switched OFF).
This entire cycle is an elegant example of negative feedback regulation: the operon produces the very enzymes needed to consume its own inducer, and once that inducer is gone, the operon automatically switches itself off again — an efficient strategy that prevents the bacterium from wasting energy and resources continuing to make lactose-digesting enzymes once there is no lactose left to digest.
Conclusion: the lac operon shuts down after some time because the β-galactosidase enzyme it produces actively consumes the lactose (inducer) that originally switched it on; once the lactose is exhausted, the repressor protein becomes active again, re-binds the operator, and blocks further transcription — a self-limiting negative feedback loop.
Explain (in one or two lines) the function of the following: (a) Promoter (b) tRNA (c) Exons.
Although this question asks for brief answers, each term still deserves a properly explained function rather than a bare one-word definition:
- (a) Promoter: the promoter is a specific DNA sequence located just upstream (at the 5′ end) of a gene’s coding region within a transcription unit; its essential function is to serve as the binding site to which RNA polymerase attaches in order to correctly initiate transcription, and it also determines which of the two DNA strands will act as the template for that particular gene.
- (b) tRNA (transfer RNA): tRNA functions as the “adapter” molecule of protein synthesis — it possesses a three-base anticodon that specifically recognises and base-pairs with a complementary codon on the mRNA, and it carries, covalently attached at its opposite end, the one specific amino acid that corresponds to that codon, thereby physically translating the genetic code into the correct amino-acid sequence of a protein.
- (c) Exons: exons are the expressed, protein-coding segments of a split (interrupted) eukaryotic gene; they are the portions that are retained in the mature mRNA after the intervening non-coding introns have been physically removed by the process of splicing, and it is only the joined sequence of exons that ultimately gets translated into protein.
Conclusion: the promoter provides the binding site that lets RNA polymerase begin transcription at the right place; tRNA is the adapter molecule linking each mRNA codon to its specific amino acid; and exons are the expressed, protein-coding portions of a gene that remain in the mRNA after splicing removes the introns.
Why is the Human Genome Project called a mega project?
The Human Genome Project (HGP), formally launched in 1990 and completed in 2003, set out with an extraordinarily ambitious goal: to determine the complete sequence of every single base pair present in the entire human genome, and to identify all the genes contained within it — both from a physical and a functional mapping perspective. The project is universally described in biology textbooks as a “mega project”, and this description is justified on several distinct, quantifiable grounds:
- The sheer size of the genome to be sequenced: the human genome consists of approximately 3 × 10⁹ (3 billion) base pairs. Sequencing a number of this magnitude, base by base, with the technology available at the time, represented an undertaking on a scale never before attempted in biology.
- The enormous financial cost involved: at the estimated cost of sequencing prevalent when the project began — roughly US $3 per base pair — simple multiplication shows that the total projected cost of sequencing the full genome would run into several billion US dollars (the actual completed project cost was around US $9 billion), an expenditure on a scale typically reserved for the largest scientific or engineering endeavours.
- The staggering volume of data generated: to appreciate just how much raw sequence data this represents, consider that if the sequence obtained were printed out in ordinary books (each with 1000 pages, and each page holding roughly 1000 letters), the data from just one single set of human chromosomes would fill approximately 3300 such books — an almost unimaginable quantity of information to store, let alone analyse by hand.
- The scale of international scientific collaboration required: no single laboratory or even single country could realistically undertake a project of this magnitude alone. It was coordinated principally by the US Department of Energy and the National Institutes of Health (NIH), working together with numerous international partners including the Wellcome Trust (UK) and research centres across many other countries, making it one of the largest coordinated international scientific collaborations in history.
- The demand for entirely new computational infrastructure and expertise: processing, storing, comparing and interpreting a dataset of this size was simply impossible using the analytical tools available at the time; the project necessitated the development of powerful high-speed computing systems and, in the process, gave rise to an entirely new interdisciplinary field of science — Bioinformatics — dedicated to managing and extracting meaning from massive biological datasets.
Conclusion: the Human Genome Project is called a mega project because of the enormous size of the genome sequenced (3 billion base pairs), its multi-billion-dollar cost, the vast quantity of data generated, the scale of international collaboration it required, and the fact that it necessitated the creation of an entirely new field, Bioinformatics, to handle its data.
What is DNA fingerprinting? Mention its applications.
DNA fingerprinting (also called DNA profiling or genetic fingerprinting) is a powerful molecular biology technique used to identify and distinguish individuals from one another based on differences found in certain highly variable regions of their DNA — specifically, the VNTRs (Variable Number of Tandem Repeats), a class of satellite DNA in which a short DNA sequence is repeated a variable number of times at a given location, and this number of repeats differs from person to person. Because these VNTR regions show an exceptionally high degree of polymorphism (sequence variation) across the population, the specific pattern of VNTRs an individual carries is essentially unique to that individual — a true genetic “fingerprint” — with the sole exception of identical (monozygotic) twins, who share the same VNTR pattern because they arise from a single fertilised egg.
The technique was originally developed by the British geneticist Alec Jeffreys. In India, DNA fingerprinting analysis and research is prominently carried out at the Centre for Cellular and Molecular Biology (CCMB), Hyderabad. One of the great practical advantages of the technique is that it requires only a very small quantity of DNA, which can be extracted from minute biological samples such as a single drop of blood, a few strands of hair (with roots attached), a trace of semen, or even saliva.
Applications of DNA fingerprinting, explained:
- Forensic science and crime investigation: biological evidence left at a crime scene — such as bloodstains, hair, or bodily fluids — can be DNA fingerprinted and compared against samples taken from suspects, allowing investigators to conclusively identify or exclude a suspect, or to identify unknown victims, with a very high degree of scientific certainty.
- Paternity and parentage testing: since a child inherits exactly half of their VNTR pattern from each biological parent, DNA fingerprinting can be used to definitively confirm or rule out a disputed biological relationship between a child and an alleged parent.
- Study of genetic relationships, population and evolutionary genetics: by comparing DNA fingerprint patterns across individuals and populations, researchers can study patterns of genetic diversity, relatedness and evolutionary history within and between populations.
- Identification in mass disasters and other legal contexts: the technique is also routinely used to identify victims of accidents or disasters where bodies are otherwise difficult to identify, and in various other legal and immigration-related contexts requiring proof of biological relationship or identity.
Conclusion: DNA fingerprinting identifies individuals from the unique, highly polymorphic pattern of VNTR (satellite DNA) repeats in their genome, developed by Alec Jeffreys and carried out in India at CCMB Hyderabad, and it finds wide application in forensic crime investigation, paternity/parentage testing, and studies of population and evolutionary genetics.
Briefly describe the following: (a) Transcription (b) Polymorphism (c) Translation (d) Bioinformatics.
- (a) Transcription: transcription is the fundamental molecular process by which the genetic information encoded in a gene’s DNA sequence is copied into a complementary RNA molecule. The enzyme DNA-dependent RNA polymerase binds at the gene’s promoter and moves along the template strand of the DNA (read 3′→5′), synthesising a new RNA strand in the 5′→3′ direction, following the same base-pairing rules as replication (except that A pairs with U, not T, since the product is RNA). The process occurs in three defined stages — initiation (RNA polymerase binds the promoter and begins), elongation (the RNA chain is extended nucleotide by nucleotide), and termination (RNA polymerase reaches a terminator sequence and releases the finished RNA transcript) — and only the segment of DNA lying between the promoter and the terminator is actually transcribed, defining the “transcription unit.”
- (b) Polymorphism: polymorphism refers to the naturally occurring variation in DNA sequence found at a particular genetic locus among different individuals of the same species, formally defined as any variant present in the population at a frequency greater than 1% (0.01). Such variation arises through mutations accumulated over generations, and is especially pronounced in certain repetitive, non-coding regions such as VNTRs (satellite DNA). Because these inheritable DNA polymorphisms are largely unique to each individual, they form the essential scientific basis of techniques such as DNA fingerprinting, and are also widely used in constructing genetic maps and studying evolutionary relationships between populations.
- (c) Translation: translation is the process by which the genetic information carried by an mRNA molecule is converted into an actual protein (polypeptide chain), and it takes place on cellular structures called ribosomes. Each three-base codon on the mRNA is recognised by a specific tRNA molecule carrying the corresponding anticodon and its matching amino acid; as successive tRNAs deliver their amino acids to the ribosome, these amino acids are joined together in sequence by peptide bonds (catalysed by the ribosome’s own ribozyme activity), building up the polypeptide chain in the exact order specified by the mRNA, until a stop codon is reached and the completed protein is released.
- (d) Bioinformatics: bioinformatics is the interdisciplinary branch of science that combines biology with computer science, mathematics and statistics, using specialised software tools and large searchable databases to store, organise, retrieve, compare and interpret enormous quantities of biological data — most notably DNA sequences, RNA sequences and protein sequences. This field developed largely out of necessity during and after the Human Genome Project, whose data output (billions of base pairs of sequence) was far too vast to be analysed manually, and bioinformatics tools remain essential today for tasks such as comparing genomes across species, predicting protein structure and function, and identifying disease-associated genetic variants.
Conclusion: transcription copies a gene’s DNA into RNA via RNA polymerase in three stages (initiation, elongation, termination); polymorphism is inheritable variation in DNA sequence across a population, forming the basis of DNA fingerprinting; translation builds a protein from mRNA on the ribosome using tRNA adapters and peptide-bond formation; and bioinformatics is the computational discipline, born largely out of the Human Genome Project, that stores and analyses such vast biological sequence data.