Molecular Basis of Inheritance — Part B (NEET Biology Class 12): Transcription, Genetic Code, Lac Operon & HGP
The Transcription Unit, Gene Structure & Transcription
🎯 NEET priority: THE highest-yield chapter in NEET Biology. 8–10 questions a year, 32–40 marks, and the trend is rising. The lac operon appears in essentially EVERY paper — as MCQ, match-the-column or assertion-reason. Master this article and you bank the single biggest block of marks available.
The transcription unit — three regions
A transcription unit has a promoter, the structural gene, and a terminator.
Because RNA polymerase only polymerises 5'→3', the strand with 3'→5' polarity acts as the TEMPLATE strand. The other strand (5'→3', same sequence as the RNA except T for U) is displaced and — confusingly — is called the CODING strand, even though it codes for nothing. All reference points are defined with respect to the coding strand.
The promoter is at the 5'-end (UPSTREAM) of the structural gene — it provides the RNA polymerase binding site, and its position is what DEFINES which strand is template and which is coding. Swap the promoter and terminator positions and the definitions reverse. The terminator is at the 3'-end (DOWNSTREAM).
Gene structure — cistron, exon, intron
A cistron is a segment of DNA coding for a polypeptide. Structural genes are MONOcistronic (mostly EUKARYOTES) or POLYcistronic (mostly BACTERIA/prokaryotes).
Eukaryotic genes are SPLIT: exons are the expressed sequences that DO appear in mature RNA; introns (intervening sequences) do NOT appear in mature RNA.
Transcription — prokaryotes vs eukaryotes
Feature | Bacteria (prokaryotes) | Eukaryotes |
RNA polymerase | A SINGLE DNA-dependent RNA polymerase transcribes ALL three RNA types | At least THREE nuclear RNA polymerases with clear division of labour |
RNA polymerase I | — | Transcribes rRNAs (28S, 18S, 5.8S) |
RNA polymerase II | — | Transcribes hnRNA — the precursor of mRNA |
RNA polymerase III | — | Transcribes tRNA, 5S rRNA and snRNAs |
Initiation/termination | RNA polymerase can ONLY catalyse elongation. It associates transiently with the initiation-factor SIGMA (σ) and termination-factor RHO (ρ) | More complex; involves additional factors |
mRNA processing | NONE needed — mRNA is active as made | hnRNA must be SPLICED (introns removed, exons joined), then CAPPED and TAILED |
Transcription + translation | COUPLED — translation can begin before transcription finishes (no nucleus/cytosol separation) | Separated — transcription in nucleus, translation in cytoplasm |
RNA processing in eukaryotes: in capping, an unusual nucleotide (methyl guanosine triphosphate) is added to the 5'-end of hnRNA. In tailing, 200–300 adenylate residues are added at the 3'-end. Only the fully processed hnRNA — now mRNA — is transported out of the nucleus.
The Genetic Code, tRNA & Translation
The genetic code — who proved what
Har Gobind Khorana developed the chemical method to synthesise RNA with defined base combinations (homopolymers and copolymers). Marshall Nirenberg's cell-free system cracked the code. Severo Ochoa's enzyme (polynucleotide phosphorylase) was also key.
Property of the genetic code | What it means |
Triplet | 3 bases per codon. 4³ = 64 codons — 61 code for amino acids, 3 are STOP codons |
Degenerate | Some amino acids are coded by MORE THAN ONE codon |
Contiguous / non-overlapping | Read in mRNA continuously, with no punctuation |
Nearly universal | The same code works from bacteria to humans. Exceptions exist in MITOCHONDRIAL codons and some protozoans |
AUG has DUAL function | Codes for METHIONINE and also acts as the INITIATOR codon |
Stop codons | UAA, UAG, UGA — these have NO tRNA |
tRNA — the adapter molecule
Francis Crick postulated the adapter molecule — something that reads the code at one end and binds a specific amino acid at the other, since amino acids themselves cannot read the code. tRNA (then called sRNA, soluble RNA) was already known, but its adapter role was assigned much later.
tRNA has an anticodon loop (bases complementary to the codon) and an amino acid acceptor end. Its secondary structure looks like a clover-leaf, but the actual 3-D structure is a compact INVERTED L. There is a special initiator tRNA — and NO tRNAs exist for stop codons.
Translation
Amino acids are first activated using ATP and linked to their cognate tRNA — called charging of tRNA or aminoacylation.
The ribosome is the cellular protein factory — structural rRNAs plus about 80 different proteins, existing as a large and a small subunit. The 23S rRNA in bacteria is the RIBOZYME that catalyses peptide bond formation — the ribosome itself is the catalyst, not a protein enzyme.
A translational unit runs from the start codon (AUG) to the stop codon. mRNA also has untranslated regions (UTRs) at both the 5'-end (before start) and 3'-end (after stop), needed for efficient translation.
Initiation: ribosome binds mRNA at AUG, recognised only by the initiator tRNA. Elongation: charged tRNAs bind codons via complementary anticodons; the ribosome moves codon to codon. Termination: a release factor binds the stop codon, freeing the polypeptide.
The Lac Operon — NEET's Most-Repeated Topic
The lac operon — NEET's most-repeated single topic
In eukaryotes, gene expression can be regulated at four levels: transcriptional (primary transcript formation), processing (splicing regulation), transport of mRNA out of the nucleus, and translational.
In E. coli, beta-galactosidase hydrolyses lactose → galactose + glucose. If no lactose is around, making the enzyme is wasteful — so the operon is switched off.
Component | What it does |
i gene | Codes for the REPRESSOR protein — expressed CONSTITUTIVELY (all the time) |
Promoter | RNA polymerase binding site |
Operator | Where the repressor binds to block transcription |
z gene | Codes for beta-galactosidase (hydrolyses lactose) |
y gene | Codes for PERMEASE (transports lactose into the cell) |
a gene | Codes for transacetylase |
INDUCER | LACTOSE (or allolactose) — it is the SUBSTRATE that switches the operon ON |
Condition | What happens |
Lactose ABSENT (operon OFF) | Repressor from the i gene binds the OPERATOR → RNA polymerase is blocked → no transcription |
Lactose PRESENT (operon ON) | Lactose enters via PERMEASE → acts as INDUCER → binds and INACTIVATES the repressor → repressor leaves the operator → RNA polymerase transcribes the operon |
Critical NEET points: regulation by a repressor is NEGATIVE regulation (the lac operon also has positive regulation, beyond NCERT's scope). Glucose and galactose CANNOT act as inducers — only lactose/allolactose can. And a very low level of lac operon expression must exist at all times, otherwise permease wouldn't exist and lactose could never get in to trigger induction. This can be neatly summarised as regulation of enzyme synthesis by its own substrate.
Human Genome Project, DNA Fingerprinting & Why This Matters for NEET
Human Genome Project (HGP)
Launched in 1990, a 13-year mega project coordinated by the US Department of Energy and the National Institute of Health, with the Wellcome Trust (UK) as a major partner. Completed in 2003. Closely tied to the rise of Bioinformatics.
Two methodologies: Expressed Sequence Tags (ESTs) — identify only genes expressed as RNA; and Sequence Annotation — blindly sequence the whole genome (coding + non-coding) and assign functions later.
Salient feature of the human genome | The number |
Total base pairs | 3164.7 million bp |
Average gene size | 3000 bases — but the LARGEST known human gene is DYSTROPHIN at 2.4 million bases |
Estimated total genes | About 30,000 — far lower than earlier estimates of 80,000–1,40,000 |
Bases identical across all humans | 99.9 per cent |
Genes of UNKNOWN function | Over 50 per cent |
Genome that codes for protein | LESS THAN 2 per cent |
Chromosome with MOST genes | Chromosome 1 — 2968 genes |
Chromosome with FEWEST genes | The Y chromosome — 231 genes |
SNP locations identified | About 1.4 million single-nucleotide polymorphism sites |
Model organisms also sequenced: bacteria, yeast, Caenorhabditis elegans (a free-living non-pathogenic nematode), Drosophila, and plants (rice and Arabidopsis).
DNA fingerprinting
Since 99.9% of human base sequence is identical, sequencing everyone would be absurdly expensive. DNA fingerprinting instead compares repetitive DNA — short stretches repeated many times.
During density gradient centrifugation, bulk DNA forms a major peak and the small extra peaks are satellite DNA. Classified by base composition, segment length and repeat number into micro-satellites, mini-satellites, etc.
These sequences do NOT code for proteins but form a large portion of the genome and show high polymorphism — the basis of fingerprinting. Because DNA from ANY tissue (blood, hair follicle, skin, bone, saliva, sperm) shows the same polymorphism, it is powerful in forensics; and because polymorphisms are inherited, it is the basis of paternity testing.
Polymorphism arises from MUTATION. A mutation in a germ cell can spread through the population by sexual reproduction; a somatic mutation cannot.
Why this matters for NEET
High-value one-liners: template strand = 3'→5'; RNA polymerase II → hnRNA; σ = initiation, ρ = termination; capping at 5', tailing at 3' with 200–300 adenylates; 61 coding + 3 stop codons; AUG = Met + initiator; tRNA is clover-leaf in 2D but inverted L in 3D; 23S rRNA = ribozyme; lac operon inducer = lactose only; HGP 1990–2003; dystrophin = largest gene; chromosome 1 most genes, Y fewest.
Trap: the CODING strand codes for nothing — it's the TEMPLATE strand that's actually transcribed. The naming is counter-intuitive by design and NEET exploits it constantly. Also, glucose is NOT a lac operon inducer — only lactose. And exons appear in mature RNA, introns do not — remember "EX = EXpressed, EXits into mRNA."
Test Yourself: MCQs, PYQs & Active Recall
Answer these, then close the article and do an Active Recall. Reveal each answer only after you commit to one.
Practice Questions
Q1. In a transcription unit, the strand with 3' to 5' polarity is called the:
(a) Antisense-free strand
(b) Template strand
(c) Coding strand
(d) Sense strand
Show answer
Answer: (b) — RNA polymerase polymerises only 5'→3', so the 3'→5' strand acts as the template; the other is called the coding strand.
Q2. The promoter in a transcription unit is located:
(a) Only in eukaryotes
(b) Downstream, at the 3'-end of the coding strand
(c) Upstream, at the 5'-end of the structural gene
(d) In the middle of the structural gene
Show answer
Answer: (c) — The promoter lies at the 5'-end (upstream), and its position is what defines which strand is template and which is coding.
Q3. Sequences that DO appear in mature or processed RNA are called:
(a) Exons
(b) Promoters
(c) Terminators
(d) Introns
Show answer
Answer: (a) — Exons are the expressed sequences that appear in mature RNA; introns (intervening sequences) do not.
Q4. Polycistronic structural genes are found mostly in:
(a) Eukaryotes
(b) Viruses only
(c) Plants only
(d) Bacteria (prokaryotes)
Show answer
Answer: (d) — Structural genes are polycistronic mostly in bacteria; eukaryotic genes are mostly monocistronic and split.
Q5. In eukaryotes, RNA polymerase II transcribes:
(a) All RNA types
(b) hnRNA, the precursor of mRNA
(c) rRNAs (28S, 18S, 5.8S)
(d) tRNA and snRNAs
Show answer
Answer: (b) — RNA polymerase I → rRNA; RNA polymerase II → hnRNA (mRNA precursor); RNA polymerase III → tRNA, 5S rRNA and snRNAs.
Q6. The initiation factor and termination factor of bacterial RNA polymerase are respectively:
(a) Rho and sigma
(b) Alpha and beta
(c) Sigma and rho
(d) Both sigma
Show answer
Answer: (c) — RNA polymerase can only catalyse elongation; it associates with sigma to initiate and rho to terminate.
Q7. In eukaryotic RNA processing, tailing involves adding:
(a) Introns at the 5'-end
(b) Uracil residues throughout
(c) 200-300 adenylate residues at the 3'-end
(d) Methyl guanosine triphosphate at the 5'-end
Show answer
Answer: (c) — Capping adds methyl guanosine triphosphate to the 5'-end; tailing adds 200-300 adenylate residues at the 3'-end.
Q8. Of the 64 codons, how many code for amino acids?
(a) 20
(b) 60
(c) 64
(d) 61
Show answer
Answer: (d) — 61 codons code for amino acids and 3 (UAA, UAG, UGA) function as stop codons.
Q9. AUG is unique because it:
(a) Codes for methionine AND acts as the initiator codon
(b) Codes for two different amino acids
(c) Is a stop codon
(d) Has no corresponding tRNA
Show answer
Answer: (a) — AUG has a dual function — it codes for methionine and also serves as the initiator codon.
Q10. The adapter molecule hypothesis in protein synthesis was postulated by:
(a) Khorana
(b) Nirenberg
(c) Severo Ochoa
(d) Francis Crick
Show answer
Answer: (d) — Crick postulated an adapter molecule that reads the code at one end and binds a specific amino acid at the other — later identified as tRNA.
Q11. The actual three-dimensional structure of tRNA resembles:
(a) An inverted L
(b) A clover leaf
(c) A double helix
(d) A sphere
Show answer
Answer: (a) — The clover-leaf is the secondary structure representation; the actual 3-D tRNA is a compact inverted L.
Q12. The enzyme that catalyses peptide bond formation in bacterial ribosomes is:
(a) Aminoacyl synthetase
(b) DNA polymerase
(c) A protein enzyme in the small subunit
(d) The 23S rRNA — a ribozyme
Show answer
Answer: (d) — The ribosome itself acts as the catalyst; in bacteria the 23S rRNA is the ribozyme that forms peptide bonds.
Q13. Charging of tRNA, or aminoacylation, refers to:
(a) Adding a phosphate to tRNA
(b) Binding tRNA to the ribosome
(c) Activating an amino acid with ATP and linking it to its cognate tRNA
(d) Removing the anticodon
Show answer
Answer: (c) — Amino acids are activated in the presence of ATP and linked to their cognate tRNA — charging or aminoacylation.
Q14. In the lac operon, the y gene codes for:
(a) Transacetylase
(b) Beta-galactosidase
(c) Permease
(d) The repressor
Show answer
Answer: (c) — z → beta-galactosidase, y → permease (transports lactose in), a → transacetylase; the i gene codes for the repressor.
Q15. In the lac operon, the repressor protein binds to the:
(a) Promoter
(b) Terminator
(c) Operator
(d) Structural gene z
Show answer
Answer: (c) — The repressor binds the operator region, preventing RNA polymerase from transcribing the operon.
Q16. Which of the following CANNOT act as an inducer of the lac operon?
(a) Allolactose
(b) Glucose
(c) Lactose
(d) Both lactose and allolactose
Show answer
Answer: (b) — Only lactose or allolactose can induce the lac operon; glucose and galactose cannot act as inducers.
Q17. The Human Genome Project was completed in the year:
(a) 1995
(b) 2003
(c) 2010
(d) 1990
Show answer
Answer: (b) — HGP was launched in 1990 and completed in 2003 — a 13-year project.
Q18. The largest known human gene is:
(a) Dystrophin, at 2.4 million bases
(b) Haemoglobin beta
(c) Insulin
(d) Collagen
Show answer
Answer: (a) — The average human gene is about 3000 bases, but dystrophin, the largest known, spans 2.4 million bases.
Q19. What percentage of the human genome codes for proteins?
(a) Less than 2 per cent
(b) Over 90 per cent
(c) About 25 per cent
(d) About 50 per cent
Show answer
Answer: (a) — Less than 2 per cent of the human genome codes for proteins; repeated sequences make up a very large portion.
Q20. Satellite DNA used in DNA fingerprinting is separated from bulk DNA by:
(a) PCR
(b) Density gradient centrifugation
(c) Restriction digestion
(d) Gel electrophoresis
Show answer
Answer: (b) — During density gradient centrifugation, bulk DNA forms a major peak and the smaller extra peaks are satellite DNA.
NEET Previous Year Questions (PYQs)
Real NEET previous-year questions on this chapter — the highest-yielding chapter in NEET Biology, so this bank is deliberately larger. Explanations in our own words.
Q21. In the lac operon, in the ABSENCE of lactose: (NEET PYQ)
(a) Permease is overproduced
(b) The repressor is inactivated
(c) The repressor binds the operator and transcription is blocked
(d) RNA polymerase transcribes freely
Show answer
Answer: (c) — Without an inducer, the constitutively-made repressor binds the operator and blocks RNA polymerase — the operon is OFF.
Q22. Regulation of the lac operon by the repressor protein is an example of: (NEET PYQ)
(a) Feedback inhibition
(b) Negative regulation
(c) Post-translational control
(d) Positive regulation
Show answer
Answer: (b) — Repressor-mediated blocking of transcription is negative regulation; the lac operon also has positive regulation beyond NCERT scope.
Q23. Which property of the genetic code means that some amino acids are coded by more than one codon? (NEET PYQ)
(a) Non-overlapping nature
(b) Universality
(c) Ambiguity
(d) Degeneracy
Show answer
Answer: (d) — Degeneracy means one amino acid may be specified by several different codons.
Q24. Exceptions to the universality of the genetic code have been found in: (NEET PYQ)
(a) Mitochondrial codons and some protozoans
(b) There are no exceptions
(c) Plant codons only
(d) Bacterial codons
Show answer
Answer: (a) — The code is NEARLY universal — exceptions occur in mitochondrial codons and in some protozoans.
Q25. Chromosome 1 and the Y chromosome carry approximately how many genes respectively? (NEET PYQ)
(a) 30,000 and 231
(b) 1000 and 1000
(c) 231 and 2968
(d) 2968 and 231
Show answer
Answer: (d) — Chromosome 1 has the most genes (2968) while the Y chromosome has the fewest (231).
Q26. In eukaryotes, transcription and translation are: (NEET PYQ)
(a) Both in the nucleus
(b) Coupled, as in bacteria
(c) Both in the mitochondria
(d) Separated — transcription in the nucleus, translation in the cytoplasm
Show answer
Answer: (d) — In bacteria they are coupled (no nuclear separation); in eukaryotes they are compartmentally separated.
Q27. The percentage of nucleotide bases that are exactly the same in all humans is: (NEET PYQ)
(a) 50 per cent
(b) 99.9 per cent
(c) 75 per cent
(d) 100 per cent
Show answer
Answer: (b) — Almost all — 99.9 per cent — of nucleotide bases are identical across humans; the 0.1 per cent difference underpins DNA fingerprinting.
Q28. Which free-living non-pathogenic nematode was sequenced as a model organism in the HGP era? (NEET PYQ)
(a) Caenorhabditis elegans
(b) Wuchereria bancrofti
(c) Drosophila melanogaster
(d) Ascaris lumbricoides
Show answer
Answer: (a) — Caenorhabditis elegans, a free-living non-pathogenic nematode, was among the model organisms sequenced.
Q29. Stop codons differ from other codons in that they: (NEET PYQ)
(a) Are read in reverse
(b) Have no corresponding tRNA
(c) Code for two amino acids
(d) Occur only in eukaryotes
Show answer
Answer: (b) — UAA, UAG and UGA are stop codons; there are NO tRNAs for stop codons — a release factor binds instead.
Q30. DNA fingerprinting is useful in forensics because DNA from any tissue of an individual: (NEET PYQ)
(a) Differs between tissues
(b) Changes with age
(c) Cannot be extracted from hair
(d) Shows the same degree of polymorphism
Show answer
Answer: (d) — DNA from blood, hair follicle, skin, bone, saliva or sperm all show the same polymorphism, making it a reliable identification tool.
Active Recall Prompt
Write everything you can recall about Molecular Basis of Inheritance (Part B), naming each part first: the transcription unit (promoter, structural gene, terminator; template vs coding strand); gene structure (cistron, monocistronic vs polycistronic, exons vs introns); transcription in prokaryotes vs eukaryotes (the three RNA polymerases, sigma and rho factors, splicing, capping and tailing); the genetic code (who cracked it, and all its properties); tRNA as the adapter molecule; translation (charging, the ribosome as ribozyme, initiation/elongation/termination); the lac operon (every gene, the inducer, and what happens with and without lactose); the Human Genome Project (dates, goals, methodologies, and every salient feature number); and DNA fingerprinting (satellite DNA, polymorphism, applications). Begin each fact with its topic and end it with a full stop.