Gene Regulation Lac Operon

8 MCQs9-step worked example
Source: NCERT Structural Organisation in AnimalsPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The lac operon is one of the most NEET-tested examples of gene regulation in prokaryotes, and the trap that costs marks is confusing the roles of the structural genes, regulatory gene, and the two small-molecule signals — lactose and glucose.

The system. The lac operon in E. coli is an inducible operon that encodes three structural genes — lacZ (β-galactosidase), lacY (permease), and lacA (transacetylase) — arranged as a single transcription unit under the control of one promoter and one operator (NCERT Class 12 Biology Chapter 5, page 116). A separate regulatory gene (lacI), located upstream but not part of the operon itself, constitutively produces a repressor protein.

Default state (no lactose, glucose present). The repressor binds the operator, physically blocking RNA polymerase from transcribing the structural genes. The operon is OFF.

Induction (lactose present, glucose absent). Lactose (the inducer) is converted to allolactose, which binds the repressor and changes its shape so it can no longer bind the operator. RNA polymerase accesses the promoter and transcribes lacZ, lacY, and lacA as a polycistronic mRNA. The operon is ON.

The glucose layer (catabolite repression). Even when lactose is present, if glucose is also available, the operon stays largely OFF. High glucose lowers cyclic AMP (cAMP). Without cAMP, the CAP (catabolite activator protein) cannot bind upstream of the promoter to enhance transcription. Full expression requires lactose present AND glucose absent.

Watch-out for NEET. Questions frequently test whether you know (a) the repressor is the product of lacI, not of the structural genes, (b) lactose is the inducer but allolactose is the actual molecule that binds the repressor, and (c) the operon concept — a polycistronic unit with promoter + operator + structural genes — was proposed by Jacob and Monod.


Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

The lac operon of *E. coli* consists of which of the following structural genes?

MCQ 2Easy RecallPractice

The concept of the operon was proposed by:

MCQ 3Easy RecallPractice

Which molecule directly binds the lac repressor to inactivate it?

MCQ 4Direct ApplicationPractice

In *E. coli*, when both glucose and lactose are present in the medium, the lac operon is:

MCQ 5Direct ApplicationPractice

A mutation in the *lacI* gene produces a defective repressor that cannot bind the operator. In the absence of lactose, the structural genes of the lac operon will be:

MCQ 6Direct ApplicationPractice

The product of the *lacZ* gene in the lac operon is:

MCQ 7Concept TrapPractice

In the lac operon model, the operator region functions as:

MCQ 8Concept TrapPractice

Which of the following statements about the lac operon is INCORRECT?

Worked Example

  1. 1

    Given

    A wild-type *E. coli* culture is grown in a medium containing lactose as the sole carbon source (no glucose). The lac operon is fully induced. You are asked: Which proteins are being actively produced, and what would happen if glucose were suddenly added to the medium?

  2. 2

    Required

    (a) Identify the proteins produced when the lac operon is fully induced. (b) Predict the effect of adding glucose to the lactose-containing medium.

  3. 3

    Concept

    The lac operon is an inducible operon regulated at two levels: (i) negative regulation by the *lacI* repressor (removed by allolactose), and (ii) positive regulation by the cAMP–CAP complex (active when glucose is absent). Full transcription requires both conditions: repressor removed AND cAMP–CAP bound.

  4. 4

    Formula / Principle

    No mathematical formula applies. The regulatory logic is: - Lactose present → allolactose binds repressor → operator free → transcription possible. - Glucose absent → high cAMP → cAMP–CAP binds upstream → RNA polymerase strongly recruited → full transcription.

  5. 5

    Substitution / Application

    (a) With lactose present and glucose absent: repressor is inactivated by allolactose, and cAMP–CAP enhances transcription. All three structural genes (*lacZ*, *lacY*, *lacA*) are transcribed as a polycistronic mRNA. The proteins produced are β-galactosidase, permease, and transacetylase. (b) Adding glucose: glucose lowers intracellular cAMP. Without cAMP, CAP cannot bind upstream of the promoter. Even though the repressor remains inactivated (lactose is still present), transcription drops to a low basal level because CAP-mediated enhancement is lost. This is catabolite repression — the cell preferentially uses glucose.

  6. 6

    Calculation

    Not applicable (conceptual question — no numerical computation).

  7. 7

    Final answer

    (a) β-galactosidase, permease, and transacetylase are actively produced. (b) Adding glucose causes catabolite repression: transcription drops to basal level despite lactose still being present, because the cAMP–CAP positive regulatory mechanism is disabled.

  8. 8

    Common trap

    Students often say "the operon switches completely OFF when glucose is added." This is incorrect — the repressor is still inactivated by allolactose, so the operator remains unblocked. What is lost is the CAP enhancement. Transcription continues at a low basal rate, not zero.

  9. 9

    Similar NEET-style question

    "In an *E. coli* mutant where the CAP-binding site is deleted, what would be the expression status of the lac operon when only lactose is present in the medium?" (Answer: The operon would be transcribed only at a basal level, because without the CAP-binding site, the cAMP–CAP complex cannot enhance transcription even when glucose is absent.) ---

Before solving, remember these

Key Fact

Lac operon

Jacob-Monod model. Lac repressor binds operator → blocks transcription. Lactose → allolactose → binds repressor → induces expression. Three structural genes: lacZ (β-galactosidase), lacY (permease), lacA (transacetylase). Negative regulation; CAP-cAMP positive regulation.

-- NCERT Class 12 Biology, Ch. 5, p. 116

Formulas

Mendel's monohybrid ratio

F2 ratio in monohybrid cross — products of independent assortment of two alleles per locus.

SymbolQuantitySI Unit
ratioF2 progeny ratio-

Valid when

  • Single gene with complete dominance
  • Pure-bred parents

Mendel's dihybrid ratio

F2 ratio in dihybrid cross with two independently-segregating loci, complete dominance, no linkage.

SymbolQuantitySI Unit
ratioF2 phenotype ratio-

Valid when

  • Two unlinked loci
  • Complete dominance

Hardy-Weinberg equation

In an idealised population (no mutation, drift, selection, gene flow, random mating), allele and genotype frequencies remain constant.

SymbolQuantitySI Unit
pfreq of dominant allele A-
qfreq of recessive allele a-

Valid when

  • Idealised population
  • All five conditions met
  • Diploid, autosomal, biallelic locus

Recombination frequency (genetic mapping)

Proportion of recombinant offspring measures genetic distance between linked loci. Capped at 50% (independent assortment).

SymbolQuantitySI Unit
RFrecombination frequency%

Valid when

  • Linked loci on same chromosome

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Category: Sign Convention

DNA polymerase synthesises only 5'→3'. Leading strand: continuous, same direction as fork. Lagging: discontinuous (Okazaki), opposite to fork.

When it triggers

Question on Okazaki, leading vs lagging, primer direction.

How to avoid

Reading template 3'→5'; synthesising 5'→3'. Lagging strand needs short fragments because it can't run continuously against fork direction.

Category: Negative Marking

Five forces disturb HW: mutation, gene flow, drift, selection, non-random mating. ANY of these violates equilibrium.

When it triggers

Question asks which factor maintains/disturbs HW.

How to avoid

Random mating + no other forces → equilibrium. ANY of mutation/migration/drift/selection/assortative mating → disequilibrium.

Category: Similar Terms

Monohybrid: genotype 1:2:1 (AA:Aa:aa); phenotype 3:1.

When it triggers

Question asks for one ratio while presenting cross details.

How to avoid

Always note dominance: phenotype merges Aa + AA; genotype keeps them separate.

Category: Similar Terms

X-linked recessive (haemophilia, colour-blindness): affects males predominantly; carrier mother → 50% sons affected; affected father → all daughters carriers but not affected.

When it triggers

Pedigree question; carrier vs affected.

How to avoid

Sex chromosomes: XX vs XY. Recessive on X needs both copies (XaXa) in female, only one (XaY) in male.

Category: Similar Terms

Allopatric: geographic isolation. Sympatric: same area, no physical barrier (e.g. polyploidy in plants, host-shift).

When it triggers

Question gives speciation scenario and asks which type.

How to avoid

If geographic barrier mentioned → allopatric. If population overlaps → sympatric.

Past Year Questions

51 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

Identify the statement that is NOT correct.

1Constant region of heavy and light chains are located at C-terminus of antibody molecules
2Each antibody has two light and two heavy chains.
3The heavy and light chains are held together by disulfide bonds.
4Antigen binding site is located at C-terminal region of antibody molecules.
NTA Answer: Option 4(final)
NEET 2025

Given below are two statements : Statement I : In the RNA world, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a catalyst for some important biochemical reactions in living systems. Being reactive, RNA is unstable. Statement II : DNA evolved from RNA and is a more stable genetic material. Its double helical strands being complementary, resist changes by evolving repairing mechanism. In the light of the above statements, choose the most appropriate answer from the options given below :

1Statement I is incorrect but statement II is correct
2Both statement I and statement II are correct
3Both statement I and statement II are incorrect
4Statement I is correct but statement II is incorrect
NTA Answer: Option 2(final)
NEET 2024

Which one of the following can be explained on the basis of Mendel's Law of Dominance? A. Out of one pair of factors one is dominant and the other is recessive. B. Alleles do not show any expression and both the characters appear as such in F generation. 2 C. Factors occur in pairs in normal diploid plants. D. The discrete unit controlling a particular character is called factor. E. The expression of only one of the parental characters is found in a monohybrid cross. Choose the correct answer from the options given below:

1A, B and C only
2A, C, D and E only
3B, C and D only
4A, B, C, D and E
NTA Answer: Option 2(final)
NEET 2024

Which of the following statement is correct regarding the process of replication in E.coli?

1The DNA dependent DNA polymerase catalyses polymerization in one direction that is 3’ → 5’
2The DNA dependent RNA polymerase catalyses polymerization in one direction, that is 5’ → 3’
3The DNA dependent DNA polymerase catalyses polymerization in 5’ → 3’ as well as 3’ → 5’ direction
4The DNA dependent DNA polymerase catalyses polymerization in 5’ → 3’ direction
NTA Answer: Option 4(final)
NEET 2023

The phenomenon of pleiotropism refers to

1Presence of two alleles, each of the two genes controlling a single trait
2A single gene affecting multiple phenotypic expression
3More than two genes affecting a single character
4Presence of several alleles of a single gene controlling a single crossover
NTA Answer: Option 2(final)
NEET 2023

Expressed Sequence Tags (ESTs) refers to

1All genes that are expressed as proteins.
2All genes whether expressed or unexpressed.
3Certain important expressed genes.
4All genes that are expressed as RNA.
NTA Answer: Option 4(final)
NEET 2023

Which of the following statements are correct about Klinefelter’s Syndrome? A. This disorder was first described by Langdon Down (1866). B. Such an individual has overall masculine development. However, the feminine developement is also expressed. C. The affected individual is short statured. D. Physical, psychomotor and mental development is retarded. E. Such individuals are sterile. Choose the correct answer from the options given below:

1C and D only
2B and E only
3A and E only
4A and B only
NTA Answer: Option 2(final)
NEET 2022

Given below are two statements : Statement I : Mendel studied seven pairs of contrasting traits in pea plants and proposed the Laws of Inheritance. Statement II : Seven characters examined by Mendel in his experiment on pea plants were seed shape and colour, flower colour, pod shape and colour, flower position and stem height. In the light of the above statements, choose the correct answer from the options given below :

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 2(final)
NEET 2022

Statements related to human Insulin are given below. Which statement(s) is/are correct about genetically engineered Insulin? (a) Pro-hormone insulin contain extra stretch of C-peptide (b) A-peptide and B-peptide chains of insulin were produced separately in E.coli, extracted and combined by creating disulphide bond between them. (c) Insulin used for treating Diabetes was extracted from Cattles and Pigs. (d) Pro-hormone Insulin needs to be processed for converting into a mature and functional hormone. (e) Some patients develop allergic reactions to the foreign insulin. Choose the most appropria

1(c), (d) and (e) only
2(a), (b) and (d) only
3(b) only
4(c) and (d) only
NTA Answer: Option 3(final)
NEET 2021

Match List-I with List-II. (2) (a)-Replication; (b)-Transcription; (c)-Transduction; (d)-Protein (3) (a)-Translation; (b)-Replication; (c)-Transcription;(d)-Transduction (4) (a)-Replication; (b)-Transcription; (c)-Translation; (d)-Protein Answer (4) Choose the correct answer from the options given 127. Which of the following are not secondary below. metabolites in plants? (a) (b) (c) (d)

1Rubber, gums (1) (ii) (i) (iv) (iii)
2Morphine, codeine (2) (ii) (iv) (i) (iii)
3Amino acids, glucose (3) (iv) (iii) (ii) (i)
4(iii) (i) (iv) (ii) (4) Vinblastin, curcumin
NTA Answer: Option 2(final)
NEET 2020

Which of the following refer to correct example(s) of organisms which have evolved due to changes 97. ÁŸêŸ ◊ ∑§ÊÒŸ, ∞ ‚ ¡ËflÊ ∑ § ‚„Ë ©ŒÊ„⁄UáÊÊ ∑§Ê ‚ ŒÁ÷¸Ã ∑§⁄UÃÊ „Ò in environment brought about by anthropogenic ¡Ê ◊ÊŸfl ∑§Ë Á∑˝§ÿÊ•Ê mÊ⁄UÊ flÊÃÊfl⁄UáÊ ◊ ’Œ‹Êfl ∑ § ∑§Ê⁄UáÊ action ? Áfl∑§Á‚à „È∞ „Ò? (a) Darwin’s Finches of Galapagos islands. (a) ªÒ‹Ê¬ÒªÊ mˬ ◊ «UÊÁfl¸Ÿ ∑§Ë Á» §ø (b) Herbicide resistant weeds. (b) π⁄U¬ÃflÊ⁄UÊ ◊ ‡ÊÊ∑§ŸÊ‡ÊË ∑§Ê ¬˝ÁÃ⁄UÊ œ (c) Drug resistant eukaryotes. (c) ‚‚Ë◊∑ §ãŒ˝∑§Ê ◊ ŒflÊßÿÊ ∑§Ê ¬˝ÁÃ⁄UÊ œ (d) Man-created breeds of domesticated animal

1∑ §fl‹ (d) (1) only (d)
2∑ §fl‹ (a) (2) only (a)
3(a) ∞fl (c) (3) (a) and (c)
4(b), (c) ∞fl (d) (4) (b), (c) and (d) Hindi+English 23 H3
NTA Answer: Option 4(final)
NEET 2020

•ŸÈ‹ πŸ ∑ § ‚◊ÿ «UË.∞Ÿ.∞. ∑§Ë ∑È §«U‹Ë ∑§Ê πÊ ‹Ÿ ◊ ∑§ÊÒŸ‚Ê

1RNA polymerase ∞ ¡Êß◊ ◊ŒŒ ∑§⁄UÃÊ „Ò?
2DNA ligase (1) •Ê⁄U.∞Ÿ.∞. ¬ÊÚÁ‹◊⁄ U$¡
3DNA helicase
4DNA polymerase (2) «UË.∞Ÿ.∞. ‹Êߪ $¡ (3) «UË.∞Ÿ.∞. „Ò‹Ë∑ §$¡ 117. Snow-blindness in Antarctic region is due to : (4) «UË.∞Ÿ.∞. ¬ÊÚ‹Ë◊⁄ U$¡ (1) Damage to retina caused by infra-red rays (2) Freezing of fluids in the eye by low
NTA Answer: Option 1(final)
NEET 2020

¡ËŸ ‘I’ ¡Ê ABO ⁄UÄà flª¸ ∑§Ê ÁŸÿ òÊáÊ ∑§⁄UÃÊ „Ò ©‚∑ § ‚ Œ÷¸ 121. Identify the wrong statement with reference to the gene ‘I’ that controls ABO blood groups. ◊ ª‹Ã ∑§ÕŸ ∑§Ê ¬„øÊÁŸ∞–

1Allele ‘i’ does not produce any sugar. (1) ‘i’ ∞ ‹Ë‹ ∑§Ê ߸ ÷Ë ‡Ê∑¸§⁄UÊ ©à¬ÛÊ Ÿ„Ë ∑§⁄UÃÊ–
2The gene (I) has three alleles. (2) ¡ËŸ (I) ∑ § ÃËŸ ∞ ‹Ë‹ „Ê Ã „Ò –
3A person will have only two of the three (3) ∞∑§ √ÿÁÄà ◊ ÃËŸ ◊ ‚ ∑ §fl‹ ŒÊ ∞ ‹Ë‹ „Ê ª – alleles.
4¡’ IA ∞fl IB ŒÊ ŸÊ ß∑§_ „Ê Ã „Ò , ÿ ∞∑§ ¬˝∑§Ê⁄U ∑§Ë (4) When IA and IB are present together, they ‡Ê∑¸§⁄UÊ •Á÷√ÿÄà ∑§⁄Uà „Ò – express same type of sugar.
NTA Answer: Option 4(final)
NEET 2020

≈˛UÊ ‚‹ ‡ÊŸ (•ŸÈflÊŒŸ/SÕÊŸÊ Ã⁄UáÊ) ∑§Ë ¬˝Õ◊ •flSÕÊ ∑§ÊÒŸ ‚Ë 123. The first phase of translation is : „Ê ÃË „Ò?

1Recognition of an anti-codon (1) ∞∑§ ∞ ≈UË-∑§Ê «UÊÚŸ ∑§Ë ¬„øÊŸ
2Binding of mRNA to ribosome (2) ⁄UÊß’Ê ‚Ê ◊ ‚ mRNA ∑§Ê ’㜟
3Recognition of DNA molecule (3) «UË.∞Ÿ.∞. •áÊÈ ∑§Ë ¬„øÊŸ
4Aminoacylation of tRNA (4) tRNA ∑§Ê ∞ ◊ËŸÊ ∞‚Ë‹ ‡ÊŸ
NTA Answer: Option 4(final)
NEET 2020

◊ «U‹ Ÿ Sflà òÊ M§¬ ‚ ¬˝¡ŸŸ ∑§⁄UŸ flÊ‹Ë ◊≈U⁄U ∑ § ¬ÊÒœ ∑§Ë 130. How many true breeding pea plant varieties did Mendel select as pairs, which were similar except Á∑§ÃŸË Á∑§S◊Ê ∑§Ê ÿÈÇ◊Ê ∑ § M§¬ ◊ øÈŸÊ ¡Ê Áfl¬⁄UËà Áfl‡Ê ∑§Ê in one character with contrasting traits ? flÊ‹ ∞∑§ ‹ˇÊáÊ ∑ § •‹ÊflÊ ∞∑§ ‚◊ÊŸ ÕË?

18 (1) 8
24 (2) 4
32 (3) 2
414 (4) 14 H3 30 Hindi+English
NTA Answer: Option 4(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Sources

NCERT refs: Class 12 Biology Chapter 5, p.116

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