CBSE Class 12 Biology: 2025 board paper, set 57/1/1

1. The man has a single X. It came from his mother, not from his colour-blind father, so he is normal ($X^{c}Y$ is ruled out): $XY$. The woman got $X^c$ from her colour-blind mother and $X$ from her normal father, so she is a carrier: $X^{c}X$.

2. Sons receive their X only from the mother: $\frac{1}{2}{X}^{c}Y$ (colour-blind) and $\frac{1}{2}XY$ (normal). So 50% of the sons are colour-blind, option (C).

   1 mark

Answer: 50%, option (C)

1. GEAC is the Indian government body that judges whether GM research and the release of GM organisms are safe. Its full form is Genetic Engineering Approval Committee, option (D).

   1 mark

Answer: Genetic Engineering Approval Committee, option (D)

1. The correct pairs are:

   • Ladybird beetle feeds on aphids.
   • Glomus is a mycorrhizal fungus.
   • Trichoderma is a biocontrol agent.
   • Methanobacterium produces biogas.

   So the match is (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a), option (B).

   1 mark

Answer: Option (B)

1. Humus is degraded very slowly by microbes. During this mineralisation, inorganic nutrients are released from humus. The answer is option (A).

   1 mark

Answer: Option (A)

1. Graft rejection is brought about by T-lymphocytes, which recognise the foreign tissue. This is the cell-mediated immune response, option (D).

   1 mark

Answer: Option (D)

1. The correct pairs are:

   • Trophoblast is the outer layer of the blastocyst.
   • Implantation is the embedding of the blastocyst in the endometrium.
   • The inner cell mass gives rise to the embryo.
   • The morula has 8–16 cells.

   So the match is (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii), option (A).

   1 mark

Answer: Option (A)

1. P is the single-celled zygote and Q is the suspensor. R is the pair of top lobes, the cotyledons. S lies between the cotyledons on the axis, so it is the plumule. This gives option (B).

   1 mark

Answer: Option (B)

1. Repeated PCR cycles can amplify a DNA segment about 1 billion times, option (C).

   1 mark

Answer: Option (C)

1. The mRNA has the same sequence as the coding strand, with U in place of T: 5'-AAUGCUAGGCAC-3'. This is option (B).

   1 mark

Answer: Option (B)

1. Each microspore mother cell (MMC) undergoes meiosis to give 4 pollen grains, so $50\times 4=200$ pollen grains.

2. Each megaspore mother cell (MMC) gives 4 megaspores, but only 1 is functional and one MMC lies in each ovule, so 50 ovules. The answer is option (B).

   1 mark

Answer: 200, 50, option (B)

1. The chronological order is Australopithecines → Homo erectus → Neanderthal → Homo sapiens, option (B).

   1 mark

Answer: Option (B)

1. Double-stranded RNA binds to the nematode's specific mRNA and silences it. This prevents translation of that mRNA, so the parasite cannot survive. The answer is option (A).

   1 mark

Answer: Option (A)

1. Assertion (A) is true: Saheli is a once-a-week non-steroidal pill.

2. Reason (R) is false: Saheli was developed by CDRI, Lucknow, not by NII, New Delhi. The answer is option (C).

   1 mark

Answer: Option (C)

1. Degeneracy means that a single amino acid can be coded by more than one codon. So Reason (R) is the definition of Assertion (A), and both are true. The answer is option (A).

   1 mark

Answer: Option (A)

1. Both statements are true. However, the type of bioreactor most commonly used does not explain why bioreactors provide optimal conditions. The answer is option (B).

   1 mark

Answer: Option (B)

1. Assertion (A) is true: tightly linked genes give far more parental combinations than recombinants.

2. Reason (R) is false: crossing over produces the non-parental (recombinant) types. Parental types are high because linked genes tend to stay together. The answer is option (C).

   1 mark

Answer: Option (C)

1. Relation: Morphine and heroin are both opioids. Heroin (smack, chemically diacetylmorphine) is made by acetylation of morphine. Morphine itself is extracted from the latex of the poppy plant Papaver somniferum.

   1 mark

2. Effect of morphine: It is a very effective sedative and painkiller. It is useful to patients who have undergone surgery.

   ½ mark

3. Effect of heroin: It is a depressant and slows down body functions. It binds to opioid receptors in the CNS and the gastrointestinal tract.

   ½ mark

Answer: Both are opioids; heroin is acetylated morphine. Morphine is a sedative/painkiller; heroin is a depressant that slows body functions.

1. (i)(1) With distillation: whisky, brandy or rum. These are made by distilling the fermented broth produced by yeast (Saccharomyces cerevisiae).

   ½ mark

2. (i)(2) Without distillation: wine or beer.

   ½ mark

3. (ii) Cyanobacteria such as Anabaena, Nostoc and Oscillatoria are autotrophic microbes that fix atmospheric nitrogen in the soil and in paddy fields.

   ½ mark

4. They also add organic matter to the soil, which increases its fertility. So they serve as a natural biofertiliser.

   ½ mark

Answer: (i)(1) Whisky/brandy/rum; (2) wine/beer. (ii) Cyanobacteria fix atmospheric nitrogen and add organic matter to soil, increasing fertility.

1. The natural old forest will be the most productive.

   ½ mark

2. It has a large standing biomass of producers and high biodiversity, with a multi-layered canopy. It therefore traps more solar radiation and stores it as organic matter, giving high primary productivity.

   ½ mark

3. A young forest is still developing. It has fewer and smaller trees, so it captures less solar radiation and its productivity is lower.

   ½ mark

4. A shallow polluted lake has few producers, a lot of dead organic matter and poor light and oxygen conditions. Its productivity is therefore the least of the three.

   ½ mark

Answer: Natural old forest is the most productive (more biomass and biodiversity trap more solar energy); the young forest and the shallow polluted lake are less productive.

1. Net Primary Productivity (NPP)	Gross Primary Productivity (GPP)
   GPP minus the energy lost in respiration by plants; $NPP=GPP-R$	Rate of production of organic matter by producers during photosynthesis
   Biomass actually available for consumption by heterotrophs (herbivores, decomposers)	Part of it is used up by the plants themselves in respiration, so it is not all available to consumers

   2 marks

Answer: GPP is the total rate of organic matter production in photosynthesis; NPP = GPP − respiration loss, the biomass available to heterotrophs.

1. 1 = Klinefelter's syndrome (44 + XXY). It results from fusion of the abnormal 22 + XY sperm with a normal 22 + X egg.

   ½ mark

2. 2 = Turner's syndrome (44 + XO). It results from fusion of the 22 + O sperm with a 22 + X egg.

   ½ mark

3. Klinefelter's syndrome	Turner's syndrome
   Extra X chromosome; 47 chromosomes (trisomy of sex chromosomes)	One X chromosome missing; 45 chromosomes (monosomy)
   Sterile male with overall masculine development but feminine features such as gynaecomastia; tall stature	Sterile female with rudimentary ovaries, short stature and lack of secondary sexual characters

   1 mark

Answer: 1 = Klinefelter's syndrome (44+XXY, sterile male); 2 = Turner's syndrome (44+XO, sterile female).

1. (a) Individual 1 (blood group B): genotype $I^{B}i$. Their child 7 has blood group A. That child must get $I^A$ from mother 2 and therefore $i$ from father 1.

   ½ mark

2. Individual 2 (blood group A): genotype $I^{A}i$. Their child 3 has blood group B. That child gets $I^B$ from father 1 and must get $i$ from mother 2.

   ½ mark

3. (b) Individual 5 (AB): both antigen A and antigen B are present on the RBC surface, because $I^A$ and $I^B$ are co-dominant.

   ½ mark

4. **Individual 8 (O, $ii$): neither antigen A nor antigen B** is present on the RBCs.

   ½ mark

Answer: (a) 1 = $I^{B}i$, 2 = $I^{A}i$; (b) 5 has both A and B antigens; 8 has neither.

1. Why periodic infusion: Lymphocytes are taken from the patient's blood and given a functional ADA cDNA using a retroviral vector, then returned to the patient. These genetically engineered lymphocytes are not immortal. They die after some time, so the patient needs fresh infusions periodically.

   1 mark

2. Permanent cure: The gene producing ADA, isolated from marrow cells, can be introduced into cells (lymphocytes or stem cells) at early embryonic stages. All cells arising from them would then carry the functional gene.

   1 mark

Answer: Engineered lymphocytes are not immortal, so periodic infusion is needed; a permanent cure is introducing the ADA gene from marrow cells into cells at early embryonic stages.

1. Microinjection: Recombinant DNA is injected directly into the nucleus of an animal cell through a fine micro-needle.

   1 mark

2. Biolistics (gene gun): Plant cells are bombarded with high-velocity microparticles of gold or tungsten coated with DNA.

   *(Alternative technique: heat shock. Cells are made competent with divalent cations such as $C{a}^{2+}$ and incubated with rDNA on ice. They are then given a brief heat shock at 42 °C and put back on ice, so the DNA enters.)*

   1 mark

Answer: Microinjection (rDNA injected into the animal cell nucleus) and biolistics/gene gun (plant cells bombarded with DNA-coated gold/tungsten particles); heat shock is an alternative.

1. (a) The complementary strand is $3^{\prime }$ – CTTAAG – $5^{\prime }$. Read $5^{\prime }\to 3^{\prime }$, both strands give GAATTC, so the sequence is a palindrome.

   ½ mark

2. (b) This sequence is recognised by EcoRI.

   ½ mark

3. (c) Production: The restriction enzyme cuts each strand a little away from the centre of the palindrome, but between the same two bases on the opposite strands (between G and A). This leaves short single-stranded overhanging stretches at both ends, called sticky ends.

   1 mark

4. Role: Sticky ends form hydrogen bonds with their complementary cut counterparts. When vector DNA and foreign DNA are cut with the same enzyme, their sticky ends pair up, and DNA ligase then joins them to form recombinant DNA.

   1 mark

Answer: (a) 3'–CTTAAG–5'; (b) EcoRI; (c) staggered cuts between the same two bases on opposite strands leave overhangs (sticky ends), which pair with complementary ends and help DNA ligase join vector and foreign DNA.

1. (a) Sporozoites.

   ½ mark

2. (b) Fertilisation and development take place in the gut (stomach) of the female Anopheles mosquito.

   ½ mark

3. (c) P = the salivary glands of the female Anopheles mosquito, where sporozoites are stored.

   ½ mark

4. Q = gametocytes.

   ½ mark

5. (d) Asexual phase: in the human (liver cells and RBCs).

   ½ mark

6. Sexual phase: in the female Anopheles mosquito.

   ½ mark

Answer: (a) Sporozoites; (b) gut of female Anopheles; (c) P = salivary glands, Q = gametocytes; (d) asexual in human, sexual in mosquito.

1. (a) Alien species invasion: Nile perch introduced into Lake Victoria (East Africa) caused the extinction of more than 200 species of cichlid fish. Other examples include Parthenium, Lantana and water hyacinth, and the African catfish Clarias gariepinus, which threatens native catfish.

   1 mark

2. Habitat loss and fragmentation: The Amazon rainforest, home to millions of species, is being cleared to grow soybeans and to make grasslands for cattle. When large habitats are broken into small fragments, mammals and birds that need large territories decline badly.

   1 mark

3. (b)(i) IUCN Red List: It documents species facing a high risk of extinction (threatened species). It therefore helps identify which species need protection and guides conservation planning.

   ½ mark

4. (ii) Hotspots: They are regions with very high species richness and a high degree of endemism. Protecting these small areas saves a large share of global biodiversity.

   ½ mark

Answer: (a) Nile perch in Lake Victoria → extinction of 200+ cichlids; clearing of the Amazon rainforest/fragmentation → decline of species. (b)(i) Red List identifies threatened species for protection; (ii) hotspots have high species richness and endemism, so they are priority areas for conservation.

1. (a) Nucleosome.

   ½ mark

2. (b) P = DNA (the negatively charged helix wrapped around the core).

   ½ mark

3. Q = histone octamer (the core of eight histone molecules).

   ½ mark

4. (c) Histones are basic, positively charged proteins, rich in the amino acids lysine and arginine.

   ½ mark

5. (d)

   Euchromatin	Heterochromatin
   Loosely packed	Densely packed
   Stains lightly	Stains darkly
   Transcriptionally active	Transcriptionally inactive

   1 mark

Answer: (a) Nucleosome; (b) P = DNA, Q = histone octamer; (c) basic/positively charged; (d) euchromatin is loosely packed, light-staining and active; heterochromatin is dense, dark-staining and inactive.

1. (a) Amniocentesis is a prenatal diagnostic technique.

   ½ mark

2. A small amount of the amniotic fluid around the developing foetus is withdrawn. The foetal cells and dissolved substances in it are analysed to detect chromosomal or genetic disorders such as Down's syndrome or cystic fibrosis.

   ½ mark

3. (b) Medical Termination of Pregnancy (MTP).

   ½ mark

4. Yes, it is safe for Radha at present.

   ½ mark

5. She is in her second month, which falls in the first trimester. MTP is comparatively safe up to 12 weeks of pregnancy.

   ½ mark

6. (c) MTP is illegal when it is done by unqualified quacks, or when it is used after sex determination to abort a normal female foetus (female foeticide).

   ½ mark

Answer: (a) Amniocentesis: amniotic fluid is analysed for foetal genetic disorders; (b) MTP, safe now as it is within the first trimester (up to 12 weeks); (c) illegal when done by quacks or for female foeticide.

1. Basis (1): DNA is negatively charged. In an electric field, DNA fragments therefore move through the matrix (agarose gel) towards the anode.

   1 mark

2. Basis (2): The agarose gel acts as a sieve. Fragments separate according to their size: smaller fragments move faster and farther. The separated bands are seen after staining with ethidium bromide under UV light, and are then cut out and eluted.

   1 mark

3. Uses of the eluted fragments (any two):

   • constructing recombinant DNA by joining them with cloning vectors
   • DNA fingerprinting
   • amplification by PCR
   1 mark

Answer: Negatively charged DNA moves towards the anode and is separated by size through the sieving gel (smaller fragments move faster); the fragments are used in rDNA technology, DNA fingerprinting and PCR.

1. Difference 1: In divergent evolution, the same basic structure develops along different directions as an adaptation to different needs. In convergent evolution, different structures evolve to perform the same function and so come to look similar.

   1 mark

2. Difference 2: Divergent evolution gives homologous organs and indicates common ancestry (e.g. forelimbs of whale, bat, cheetah and human). Convergent evolution gives analogous organs and shows that a similar habitat selected similar adaptive features (e.g. wings of butterfly and bird, eyes of octopus and mammals).

   1 mark

3. Similarity: Both are driven by the environment through natural selection, and both contribute to the evolution of species (adaptive radiation).

   1 mark

Answer: Divergent evolution: same structure modified for different functions, giving homologous organs and indicating common ancestry. Convergent evolution: different structures adapted for the same function, giving analogous organs and indicating similar habitat. Both are shaped by the environment/natural selection and contribute to species evolution.

1. (a) The peak is caused by luteinising hormone (LH), called the LH surge.

   ½ mark

2. Function: it induces rupture of the Graafian follicle, which releases the ovum (ovulation).

   ½ mark

3. (b) Ovary: primary follicles grow into a fully mature Graafian follicle.

   1 mark

4. Uterus: the endometrium regenerates and proliferates (thickens).

   1 mark

5. (c) Q = progesterone.

   ½ mark

6. Function: it maintains the endometrium and so maintains pregnancy.

   ½ mark

7. (d) (alternative to (c)) The corpus luteum remains functional in Case 2, because hCG keeps it going during pregnancy.

   ½ mark

8. It is formed from the remaining cells of the ruptured Graafian follicle after ovulation.

   ½ mark

Answer: (a) LH; it induces ovulation. (b) Ovary: follicles mature into a Graafian follicle; uterus: endometrium proliferates. (c) Progesterone; it maintains the endometrium/pregnancy. (d) Corpus luteum, formed from the ruptured Graafian follicle after ovulation.

1. (a) The butterfly acquires the chemical during its caterpillar stage, by feeding on a poisonous weed.

   1 mark

2. (b) Formula:

   $N_{t+1}=N_t+[(B+I)-(D+E)]$

   ½ mark

3. Substitution:

   $800=N_t+[(200+200)-(150+100)]$

   $800=N_t+150$

   $N_t=650$

   ½ mark

4. Comment: Natality plus immigration exceeds mortality plus emigration, so the population is growing. The age pyramid will therefore be expanding (triangular), with a broad base of pre-reproductive individuals.

   1 mark

5. (c) Counting individuals is not a meaningful measure here, because one huge banyan tree cannot be compared with 200 small plants. The banyan tree is measured by biomass or percent cover, while the carrot grass (Parthenium) is measured by percent cover.

   1 mark

6. (d) (alternative to (c)) Tiger density is estimated indirectly from pug marks and faecal pellets.

   1 mark

Answer: (a) From a poisonous weed eaten in its caterpillar stage; (b) $N_t$ = 650, expanding age pyramid; (c) banyan tree by biomass/percent cover, carrot grass by percent cover; (d) pug marks and faecal pellets.

1. (i) MALT stands for Mucosa-Associated Lymphoid Tissue. It makes up about 50% of the lymphoid tissue in the human body.

   ½ mark

2. It is located within the lining of the major tracts: the respiratory, digestive and urogenital tracts.

   ½ mark

3. (ii) Cytokine barriers: Virus-infected cells secrete proteins called interferons. These protect neighbouring non-infected cells from further viral infection.

   1 mark

4. (iii) The diagnostic test for AIDS is ELISA (Enzyme Linked Immuno-Sorbent Assay).

   1 mark

5. It works on the principle of antigen–antibody interaction, detecting the antibodies produced against HIV. (PCR, based on amplification of the viral nucleic acid, is also accepted.)

   1 mark

6. (iv) Bone marrow and thymus are primary lymphoid organs. Bone marrow produces all blood cells, including lymphocytes. Some lymphocytes migrate to the thymus, where they mature into T-lymphocytes. Together, these organs provide the micro-environment in which immature lymphocytes develop and differentiate into antigen-sensitive lymphocytes.

   1 mark

Answer: (i) Mucosa-Associated Lymphoid Tissue, in the lining of the respiratory, digestive and urogenital tracts; (ii) interferons from virus-infected cells protect uninfected cells; (iii) ELISA, based on antigen–antibody interaction; (iv) they produce lymphocytes and provide the micro-environment for their development and maturation into antigen-sensitive cells.

1. (i)(a) H = Clostridium butylicum

   ½ mark

2. I = Bacterium

   ½ mark

3. (b) J = Statins

   ½ mark

4. K = Yeast (fungus)

   ½ mark

5. (c) L = Trichoderma polysporum

   ½ mark

6. M = Immunosuppressive agent, used in patients who have undergone organ transplantation

   ½ mark

7. (ii) Reasons baculoviruses are used as biocontrol agents:
   1. They are species-specific and have narrow-spectrum insecticidal action, attacking only the target insects and arthropods (e.g. Nucleopolyhedrovirus).
   2. They have no negative impact on non-target organisms such as plants, mammals, birds, fish and beneficial insects, so they are ideal in IPM and ecologically sensitive areas.

   2 marks

Answer: (i) H = Clostridium butylicum, I = bacterium, J = statins, K = yeast/fungus, L = Trichoderma polysporum, M = immunosuppressant in organ transplant patients; (ii) species-specific, narrow-spectrum action with no harm to non-target organisms.

1. (a) The two cells are the vegetative cell and the generative cell.

   ½ mark

2. Vegetative cell	Generative cell
   Large, with abundant food reserve and a large, irregularly shaped nucleus	Small, spindle-shaped, with dense cytoplasm; floats in the cytoplasm of the vegetative cell
   Forms the pollen tube	Divides mitotically to form two male gametes

   1 mark

3. (b)(i) 1 = Autogamy (within the same flower)

   ½ mark

4. 2 = Geitonogamy (between flowers of the same plant)

   ½ mark

5. 3 = Xenogamy (between flowers of different plants)

   ½ mark

6. (ii)(a) Water lily: the flowers emerge above the water surface and are pollinated by insects or wind.

   ½ mark

7. (b) Vallisneria: pollination is by water. The female flower reaches the surface on a long stalk. Male flowers or pollen grains released onto the water surface are carried passively by water currents and reach the stigma.

   ½ mark

8. (iii) Advantages of xenogamy (any two):

   • it brings genetically different pollen, giving genetic variation
   • it produces healthier, more vigorous offspring (heterosis) and avoids inbreeding depression
   • it improves disease resistance and helps evolution
   1 mark

Answer: (a) Vegetative cell is large with food reserve and forms the pollen tube; generative cell is small and forms two male gametes. (b)(i) 1 = autogamy, 2 = geitonogamy, 3 = xenogamy; (ii) water lily by insects/wind, Vallisneria by water; (iii) genetic variation, vigour/no inbreeding depression.

1. (i) Fate of P, Q and R: Sperm P has reached the zona pellucida and its acrosome reacts. It is the one that penetrates and fertilises the ovum. Sperms Q and R are still in or beyond the corona radiata. They cannot enter the ovum and degenerate, because only one sperm is allowed in.

   1 mark

2. (ii) Role of zona pellucida: When the first sperm contacts the zona pellucida, it changes the membrane and zona of the ovum (zona reaction). These changes block the entry of additional sperms and prevent polyspermy, so only one sperm fertilises the ovum.

   1 mark

3. (iii) Changes in the ovum after sperm entry: Entry of the sperm stimulates the secondary oocyte to complete meiosis II. This produces a large haploid ovum (ootid) and a small second polar body. The haploid nucleus of the sperm then fuses with that of the ovum, forming a diploid zygote.

   1 mark

4. (iv) Acrosome: The acrosome is a cap-like structure over the head of the sperm. It is filled with hydrolytic (lytic) enzymes that dissolve the corona radiata and zona pellucida, which lets the sperm enter the cytoplasm of the ovum.

   1 mark

5. (iv) Middle piece: The middle piece contains numerous mitochondria. They produce energy (ATP) for the movement of the tail, giving the sperm the motility it needs to reach the ovum and fertilise it.

   1 mark

Answer: (i) P fertilises the ovum; Q and R cannot enter. (ii) Zona pellucida changes on contact with the first sperm and blocks entry of other sperms. (iii) The secondary oocyte completes meiosis II, forming the ootid and second polar body; the nuclei fuse into a zygote. (iv) Acrosome enzymes help penetration; middle-piece mitochondria give energy for motility.

1. (i) Cross between two sickle-cell carriers: Parents are $H{b}^{A}H{b}^S\times H{b}^{A}H{b}^S$. Gametes from each parent are $H{b}^A$ and $H{b}^S$.

   	$H{b}^A$	$H{b}^S$
   $H{b}^A$	$H{b}^{A}H{b}^A$ (disease free)	$H{b}^{A}H{b}^S$ (carrier)
   $H{b}^S$	$H{b}^{A}H{b}^S$ (carrier)	$H{b}^{S}H{b}^S$ (diseased)

   ½ mark

2. Ratio in F1: Carrier : Disease free : Diseased = 2 : 1 : 1.

   ½ mark

3. Base substituted: In the β-globin gene, the codon GAG changes to GTG, so adenine is substituted by thymine (in mRNA, GAG → GUG, i.e. A replaced by U). As a result, glutamic acid is replaced by valine at the 6th position of the β-globin chain of haemoglobin.

   1 mark

4. (ii) Garden pea (law of dominance): Violet flower colour (V) is completely dominant over white (v). The F1 shows only the dominant parental trait.

   ½ mark

5. Pea cross: Violet (VV) × White (vv) gives gametes V and v. F1 is Vv, all violet.

   F1 × F1 (Vv × Vv) gives F2 of VV : Vv : vv = 1 : 2 : 1. The phenotypic ratio is 3 violet : 1 white.

   1 mark

6. Snapdragon (incomplete dominance): Red (R) is not completely dominant over white (r). The F1 heterozygote shows an intermediate pink colour that resembles neither parent.

   ½ mark

7. Snapdragon cross: Red (RR) × White (rr) gives gametes R and r. F1 is Rr, all pink.

   F1 × F1 (Rr × Rr) gives F2 of RR (red) : Rr (pink) : rr (white) = 1 : 2 : 1. Here the phenotypic and genotypic ratios are the same, unlike the 3 : 1 seen in pea.

   1 mark

Answer: (i) Carrier : disease free : diseased = 2 : 1 : 1; adenine is substituted by thymine (GAG→GTG; A→U in mRNA), so glutamic acid is replaced by valine. (ii) Pea shows complete dominance (F1 all violet, F2 3 : 1). Snapdragon shows incomplete dominance (F1 all pink, F2 1 red : 2 pink : 1 white).

1. Structure of the lac operon: It has a regulator gene i (with its own promoter), a promoter (p), an operator (o) and three structural genes z, y, a. These code for β-galactosidase, permease and transacetylase respectively. The i gene constitutively produces a repressor protein.

   ½ mark

2. (i) Presence of inducer — step 1: Lactose, or its derivative allolactose, acts as the inducer. Permease lets it enter the cell, where it binds to the repressor and inactivates it.

   ½ mark

3. (i) Presence of inducer — step 2: The inactivated repressor cannot bind to the operator, so the operator is free.

   ½ mark

4. (i) Presence of inducer — step 3: RNA polymerase binds the promoter and transcribes z, y, a into a polycistronic mRNA. This is translated into β-galactosidase, permease and transacetylase. β-galactosidase hydrolyses lactose into glucose and galactose. The operon is switched ON (inducible operon).

   ½ mark

5. Diagram (i), operon ON: Draw DNA in order i – p – o – z – y – a.

   • Show repressor mRNA from the i gene giving the repressor protein.
   • Show the repressor bound to the inducer (lactose), forming an inactive complex that is not attached to the operator.
   • Show RNA polymerase moving from p through z, y, a, producing lac mRNA.
   • Show that mRNA translated into β-galactosidase, permease and transacetylase.

   Label every part.

   1 mark

6. (ii) Absence of inducer — step 1: The repressor produced by the i gene stays active and binds to the operator region.

   ½ mark

7. (ii) Absence of inducer — step 2: The bound repressor blocks RNA polymerase, so z, y, a are not transcribed and no enzymes are made. The operon is switched OFF (negative regulation).

   ½ mark

8. Diagram (ii), operon OFF: Draw the same DNA, i – p – o – z – y – a.

   • Show the repressor protein from the i gene attached to the operator.
   • Show RNA polymerase on the promoter but blocked, with a cross mark meaning no transcription.
   • Show no lac mRNA and no enzymes formed.

   Label every part.

   1 mark

Answer: (i) With an inducer: lactose/allolactose binds and inactivates the repressor, the operator is free, RNA polymerase transcribes z, y, a, and β-galactosidase, permease and transacetylase are made (operon ON). (ii) Without an inducer: the active repressor binds the operator, RNA polymerase is blocked, and there is no transcription (operon OFF).