CBSE Class 12 Biology: 2023 board paper, set 57/1/1

1. Recombinants (new combinations of parental genes) arise from independent assortment of homologous chromosomes in meiosis I and from crossing over between non-sister chromatids of bivalents in prophase I.

   1 mark

2. Mutation produces new alleles rather than recombining existing ones, and meiosis II only separates sister chromatids, so (i) and (iii) are excluded. Answer: (ii) and (iv).

Answer: (ii) and (iv), option (B)

1. Transcription begins when RNA polymerase binds the promoter (X). The polymerase core (Z) uses the sigma factor (Y) to start transcription; the rho factor acts only at termination.

   1 mark

Answer: X = Promoter, Y = Sigma factor, Z = RNA polymerase, option (C)

1. The pre-reproductive base of the pyramid is narrower than the reproductive group, so fewer young individuals are entering the population. This urn-shaped profile shows a declining population.

   1 mark

Answer: Declining population, option (A)

1. Untreated sewage adds organic matter, which raises BOD, so microbes deplete dissolved oxygen. It also adds many ions. At point (i), dissolved oxygen is lowest and mineral ions are high, so this marks the sewage entry point.

   1 mark

Answer: Point (i), option (A)

1. The correct matches are:

   • Ligase joins DNA fragments (P-iii).
   • Exonuclease removes nucleotides from the ends of DNA (Q-i).
   • Taq polymerase extends the primer (R-ii).
   • Endonuclease cuts DNA at a specific site (S-iv).
   1 mark

Answer: P-iii, Q-i, R-ii, S-iv, option (D)

1. A child inherits half of its DNA bands from each parent. Every band of child X is accounted for by bands present in individuals 1 and 3, so 1 and 3 are the likely parents.

   1 mark

Answer: 1 and 3, option (C)

1. The curve peaks at medium neck length, which means the average phenotype is favoured. That is stabilising selection with medium-necked giraffes selected, so option (D) is correct for the graph shown.

   1 mark

2. The scheme also accepts (C), directional selection, on the basis of the classic account of giraffe evolution.

Answer: Option (D): stabilising selection (scheme also accepts C)

1. Each microspore mother cell divides by meiosis to form a tetrad of 4 pollen grains.

2. Number of pollen grains $=325\times 4=1300$

   1 mark

Answer: 1300, option (C)

1. The correct matches are:

   • Lippes loop is a non-medicated IUD that increases phagocytosis of sperms (P-ii).
   • Multiload 375 releases Cu ions, which make the cervix hostile to sperms (Q-iv).
   • Norplant is a progestin implant that thickens cervical mucus (R-iii).
   • Saheli is a non-steroidal pill that inhibits ovulation (S-i).
   1 mark

Answer: P-ii, Q-iv, R-iii, S-i, option (A)

1. Leydig cells secrete androgens, which initiate spermatogenesis, so (C) is correct.

   1 mark

2. The other options are wrong: Sertoli cells do not secrete testosterone, the placenta's estrogen does not initiate milk secretion, and FSH comes from the pituitary, not the ovary.

Answer: Option (C)

1. Cohen and Boyer joined an antibiotic-resistance gene to a native plasmid of Salmonella typhimurium to make the first recombinant DNA.

   1 mark

Answer: Salmonella typhimurium, option (B)

1. CSIR, led by Dr. R.A. Mashelkar, successfully challenged the US turmeric patent by citing traditional knowledge.

   1 mark

Answer: Dr. R.A. Mashelkar, option (D)

1. Lymphocytes are not immortal, so the genetically engineered cells must be infused periodically. The Assertion is true and the Reason is false.

   1 mark

Answer: Option (C)

1. The egret benefits from insects flushed out by cattle, while the cattle are unaffected. This is commensalism, and R correctly explains why.

   1 mark

Answer: Option (A)

1. Pea and bean seeds are ex-albuminous, but castor seeds are albuminous because they retain endosperm. So R is false while A is true.

   1 mark

Answer: Option (C)

1. In birds, ZW females produce two kinds of eggs, so they are heterogametic, while ZZ males are homogametic. R explains A.

   1 mark

Answer: Option (A)

1. (a) The stage is the blastocyst, a hollow ball of cells formed from the morula.

   ½ mark

2. (b) In a normal pregnancy the blastocyst becomes embedded in the endometrium, the innermost layer of the uterine wall. This is called implantation.

   ½ mark

3. (c) 'X' is the trophoblast, the outer layer. It attaches to the endometrium and helps the blastocyst implant in the uterus.

   ½ mark

4. 'Y' is the inner cell mass. It differentiates to form the embryo.

   ½ mark

Answer: (a) Blastocyst; (b) endometrium of the uterus; (c) X (trophoblast) – attachment/implantation; Y (inner cell mass) – differentiates into the embryo

1. (a) The pollen tube enters the ovule through the micropylar end (micropyle).

   ½ mark

2. It then enters one of the synergids. The filiform apparatus at the micropylar tip of the synergids guides the pollen tube into the embryo sac.

   ½ mark

3. (b) One male nucleus fuses with the egg nucleus to form the diploid zygote. This is syngamy.

   ½ mark

4. The other male nucleus fuses with the two polar nuclei to form the triploid primary endosperm nucleus (PEN). This is triple fusion.

   ½ mark

Answer: (a) Through the micropyle, entering a synergid guided by the filiform apparatus; (b) one male nucleus + egg → zygote (syngamy), the other + two polar nuclei → PEN (triple fusion)

1. (i) 'A' is a plasmid, an extrachromosomal, circular, double-stranded DNA found in bacteria.

   ½ mark

2. 'B' is a bacteriophage, a virus that infects bacteria.

   ½ mark

3. (ii) Plasmids are used as cloning vectors. They carry a foreign gene into the host cell and replicate independently of chromosomal DNA, often in high copy number. They can also carry selectable markers such as antibiotic-resistance genes.

   ½ mark

4. Bacteriophages are also used as cloning vectors. They can replicate inside bacterial cells independently of the chromosomal DNA and reach a high copy number per cell, so the linked foreign DNA is multiplied.

   ½ mark

Answer: A – plasmid; B – bacteriophage; both act as cloning vectors that carry and multiply foreign DNA in a host

1. DNA is hydrophilic and cannot pass through the cell membrane on its own, so bacterial cells must first be made competent.

2. The bacterial cells are treated with a specific concentration of a divalent cation such as **calcium ($C{a}^{2+}$)**.

   ½ mark

3. This treatment increases the efficiency with which DNA enters the bacterium through pores in its cell wall.

   ½ mark

4. The recombinant DNA is then forced into the cells by incubating the cells with the DNA on ice.

   ½ mark

5. The cells are then given a brief **heat shock at $42^{\circ }C$** and placed back on ice, which allows the cells to take up the DNA.

   ½ mark

Answer: Treat cells with Ca²⁺ (raises DNA entry through wall pores) → incubate with rDNA on ice → heat shock at 42 °C → back on ice

1. 1. Ecological pyramids do not take into account the same species belonging to two or more trophic levels.

   1 mark

2. 2. They assume a simple food chain, which almost never exists in nature, and so cannot accommodate a food web.

   (Also acceptable: saprophytes/decomposers are given no place in the pyramid, although they play a vital ecological role.)

   1 mark

Answer: No account of a species occupying more than one trophic level; assumes a simple food chain (ignores food webs); saprophytes not included – any two

1. India has only about 2.4 percent of the world's land area.

   1 mark

2. Yet India accounts for about 8.1 percent of global species diversity. This very high share of species diversity makes India one of the 12 mega diversity countries.

   1 mark

Answer: 2.4% of the world's land area; 8.1% share of global species diversity

1. Step 1 – Recognition (diagram): Draw the vector DNA (plasmid) and the foreign DNA. Mark on each the palindromic EcoRI site $5^{\prime }$–GAATTC–$3^{\prime }$ / $3^{\prime }$–CTTAAG–$5^{\prime }$. Label 'EcoRI recognition site'.

   1 mark

2. Step 2 – Cutting (diagram): Show EcoRI cutting each strand between G and A. This leaves single-stranded overhangs $5^{\prime }$–AATT–$3^{\prime }$ (sticky ends) on both the vector and the foreign DNA fragment. Label 'sticky ends'.

   1 mark

3. Step 3 – Joining (diagram): Show the complementary sticky ends of the foreign DNA pairing with those of the opened vector. DNA ligase seals the nicks (phosphodiester bonds). Label the product 'recombinant DNA'.

   1 mark

Answer: Schematic: EcoRI recognises GAATTC in vector and foreign DNA → cuts between G and A to give AATT sticky ends → sticky ends anneal and DNA ligase seals them → recombinant DNA

1. (a) The trait is recessive.

   ½ mark

2. Reason: both parents in generation I are unaffected, yet their son (II-2) is affected. So the parents must be carriers of a hidden (recessive) allele.

   ½ mark

3. (b) The trait is autosomal.

   ½ mark

4. Reason: both males and females are affected (an affected female in generation II, affected males in generations II and III), so both sexes have an equal chance of showing it.

   (Also accepted by the scheme: sex-linked (X-linked recessive), because more males than females are affected and the affected son in generation III is born to an unaffected carrier mother.)

   ½ mark

5. (c) If autosomal: Child '1' (unaffected male) is AA or Aa.

   ½ mark

6. Child '3' (unaffected female with an affected son) must be Aa (carrier).

   (If X-linked: Child '1' is $XY$; Child '3' is $X^{h}X$, a carrier.)

   ½ mark

Answer: (a) Recessive; (b) autosomal (sex-linked also accepted); (c) Child 1 – AA/Aa (or XY), Child 3 – Aa (or XʰX carrier)

1. (a) Stem cells have the ability of self-renewal: they can divide repeatedly to produce more stem cells.

   ½ mark

2. They can also differentiate into any kind of specialised cell, tissue or organ.

   ½ mark

3. (b) Stem cells can be obtained from the inner cell mass of the blastocyst (embryonic stem cells), umbilical cord blood or bone marrow (any one).

   1 mark

4. (c) Application 1: treating diabetes by forming insulin-producing islets of Langerhans cells.

   ½ mark

5. Application 2: repairing spinal cord injuries or heart tissue damage, or restoring vision by injecting stem cells (any other correct application, such as rheumatoid arthritis or cystic fibrosis).

   ½ mark

Answer: (a) Self-renewal and ability to differentiate into any cell type; (b) inner cell mass of blastocyst/umbilical cord/bone marrow; (c) e.g. diabetes (islet cells), spinal cord injury/heart disease/vision restoration

1. (a)

   Malignant tumour	Benign tumour
   A mass of proliferating cells that grow very rapidly, invading and damaging surrounding normal tissue	Grows slowly and stays confined to its original location
   Shows metastasis	Does not show metastasis

   1 mark

2. (b) The most feared property of a malignant tumour is metastasis.

   1 mark

3. In metastasis, cells slough off from the tumour and travel through the blood to distant sites. Wherever they lodge, they start new tumours.

   1 mark

Answer: (a) Malignant – rapid invasive growth, shows metastasis; benign – slow, confined, no metastasis; (b) metastasis – tumour cells spread via blood and start new tumours at distant sites

1. (a) Primary sludge: in primary treatment, floating debris is removed by filtration and grit (soil, pebbles) by sedimentation.

   ½ mark

2. All the solids that settle down during this primary treatment form the primary sludge; the liquid left above is the primary effluent.

   ½ mark

3. (b) Activated sludge: in secondary (biological) treatment, the primary effluent is aerated so that aerobic microbes grow into flocs (masses of bacteria with fungal filaments). These microbes consume organic matter and greatly reduce the BOD.

   ½ mark

4. The effluent is then passed to a settling tank, where the bacterial flocs settle as activated sludge. Part of it is pumped back into the aeration tank as inoculum.

   ½ mark

5. (c) Anaerobic sludge digesters: the remaining activated sludge is pumped into large tanks called anaerobic sludge digesters.

   ½ mark

6. Inside, anaerobic bacteria digest the bacteria and fungi of the sludge and produce a mixture of gases such as $C{H}_4$, $H_{2}S$ and $C{O}_2$, which form biogas.

   ½ mark

Answer: (a) Solids settling in primary treatment; (b) sediment of aerobic microbial flocs after secondary treatment, BOD reduced; (c) tanks where anaerobic bacteria digest activated sludge, producing biogas (CH₄, H₂S, CO₂)

1. (a) The two primates are Dryopithecus and Ramapithecus.

   1 mark

2. (b) Time: about 2 million years ago (mya).

   ½ mark

3. Place: the East African grasslands.

   ½ mark

4. (c)

   Homo habilis	Homo erectus
   Brain capacity about 650–800 cc	Brain capacity about 900 cc
   Probably did not eat meat	Probably ate meat

   1 mark

Answer: (a) Dryopithecus and Ramapithecus; (b) ~2 mya in East African grasslands; (c) brain capacity 650–800 cc vs 900 cc; H. habilis probably did not eat meat, H. erectus probably did

1. (i) (1) ZIFT – Zygote Intra Fallopian Transfer

   ½ mark

2. (2) ICSI – Intra Cytoplasmic Sperm Injection

   ½ mark

3. (3) IUT – Intra Uterine Transfer

   ½ mark

4. (4) GIFT – Gamete Intra Fallopian Transfer

   ½ mark

5. (ii) GIFT cannot be considered an IVF procedure.

   ½ mark

6. Reason: in GIFT, the ovum and sperm (gametes) are transferred into the fallopian tube, where fertilisation occurs naturally inside the body (in vivo). In IVF, fertilisation takes place outside the body and the zygote or embryo is then transferred.

   ½ mark

Answer: (i) Zygote Intra Fallopian Transfer; Intra Cytoplasmic Sperm Injection; Intra Uterine Transfer; Gamete Intra Fallopian Transfer; (ii) GIFT – fertilisation occurs in vivo in the fallopian tube

1. (i)

   Perisperm	Pericarp
   Residual, persistent nucellus present in some seeds (e.g. black pepper, beet)	Wall of the fruit, developed from the ovary wall

   1 mark

2. (ii)

   Syncarpous pistil	Apocarpous pistil
   Many pistils (carpels) fused together (e.g. Papaver)	Many pistils free from each other (e.g. Michelia)

   1 mark

3. (iii)

   Plumule	Radicle
   Shoot tip at the terminal end of the epicotyl in the embryonal axis; forms the future shoot/stem	Root tip at the terminal end of the hypocotyl; forms the future root

   1 mark

Answer: (i) Perisperm – persistent nucellus; pericarp – fruit wall from ovary wall; (ii) syncarpous – fused pistils; apocarpous – free pistils; (iii) plumule – future shoot (epicotyl end); radicle – future root (hypocotyl end)

1. (a) The lac operon is controlled at the level of transcription, that is, by whether RNA polymerase is allowed to transcribe the structural genes. In the absence of lactose, the repressor binds the operator and blocks RNA polymerase, so no transcription occurs. In the presence of lactose (inducer), the repressor is inactivated and cannot bind the operator, so RNA polymerase transcribes the operon.

   1 mark

2. (b) Lactose cannot enter the cell on its own; it needs permease, coded by the y gene of the operon. So the operon must always work at a very low level to keep some permease present. This lets lactose enter the cell and act as the inducer that switches the operon on.

   1 mark

3. (c) The regulator gene is called 'i' because it codes for the inhibitor (repressor) protein. The repressor binds the operator and switches off the operon.

   1 mark

4. (d) In absence of inducer (diagram described): draw DNA showing genes in order i – p – o – z – y – a. The i gene is transcribed to mRNA and translated into the repressor. The repressor binds the operator (o), so RNA polymerase at the promoter (p) is blocked. Label: 'no transcription of z, y, a'.

   1 mark

5. (d) In presence of inducer (alternative, diagram described): draw the same gene order i – p – o – z – y – a. The repressor made by i is bound by the inducer (lactose/allolactose), becomes inactive and cannot bind the operator. RNA polymerase transcribes z, y, a into a polycistronic mRNA, which is translated into β-galactosidase, permease and transacetylase.

Answer: (a) Regulated by repressor allowing/blocking RNA polymerase transcription; (b) basal permease (y gene) needed for lactose entry; (c) i codes for inhibitor/repressor; (d) diagrams: repressor on operator (no transcription) / inducer–repressor complex, z, y, a transcribed

1. (a) Observation: species diversity decreases as we move from region A (tropics, about 30° N to 30° S) to region B (higher southern latitudes towards the pole).

   1 mark

2. Reason 1: the tropics have a less seasonal, relatively constant and predictable environment. They have also remained undisturbed for millions of years (no frequent glaciations), giving a long time for species diversification.

   1 mark

3. Reason 2: the tropics receive more solar energy, which gives higher productivity and supports greater diversity. Region B has less solar energy, lower productivity and less habitable land area.

   1 mark

4. (b) India has recorded more than 1200 species of birds.

   ½ mark

5. Reason: the Indian landmass lies largely in the tropical latitudes, which favour high diversity.

   ½ mark

6. (b) (alternative) The Amazon rainforest in South America has the greatest biodiversity on Earth.

7. Reason: it lies in the tropical region, with warm, stable climate, high solar energy and productivity, and a long undisturbed history.

Answer: (a) Diversity decreases from A to B, because the tropics have a stable, undisturbed environment and more solar energy/productivity; (b) >1200 bird species in India, being largely tropical; OR Amazon rainforest, being tropical

1. (i) They used **normal (light) nitrogen, ${}^{14}N$** (as ${}^{14}N{H}_{4}Cl$)

   ½ mark

2. and **heavy nitrogen, ${}^{15}N$** (as ${}^{15}N{H}_{4}Cl$).

   ½ mark

3. Why: nitrogen is incorporated into DNA bases, so the two isotopes make DNA of two different densities, heavy (${}^{15}N$) and light (${}^{14}N$). Parental and newly made strands can then be told apart.

   1 mark

4. (ii) E. coli divides about every 20 minutes. The ${}^{15}N$-labelled cells were shifted to a ${}^{14}N$ medium and sampled at 20-minute intervals, so each sample showed the DNA after one more round of replication. This revealed the mode of replication generation by generation.

   1 mark

5. (iii) Caesium chloride density gradient centrifugation separates DNA molecules by density. It distinguishes heavy (${}^{15}N$), hybrid (${}^{15}N{/}^{14}N$) and light (${}^{14}N$) DNA as separate bands.

   1 mark

6. (iv) After one generation (20 min), all DNA was of hybrid/intermediate density. After two generations (40 min), DNA was equal amounts of hybrid and light. So they concluded that DNA replicates semiconservatively: each daughter DNA has one parental strand and one new strand.

   1 mark

Answer: (i) ¹⁴N and ¹⁵N, to make light and heavy DNA; (ii) generation time 20 min, so each sample = one replication round; (iii) separates DNA by density; (iv) DNA replication is semiconservative

1. (i) Parents: Tall round TTRR × Dwarf wrinkled ttrr

   ½ mark

2. Gametes: TR × tr

   ½ mark

3. F₁: TtRr, all tall with round seeds. F₁ phenotypic ratio: 100% tall round (1 : 0).

   ½ mark

4. Selfing of F₁: TtRr × TtRr

   ½ mark

5. Gametes from each parent: TR, Tr, tR, tr. Punnett square:

   ♂ \ ♀	TR	Tr	tR	tr
   TR	TTRR	TTRr	TtRR	TtRr
   Tr	TTRr	TTrr	TtRr	Ttrr
   tR	TtRR	TtRr	ttRR	ttRr
   tr	TtRr	Ttrr	ttRr	ttrr

   2 marks

6. F₂ phenotypes: Tall round 9 : Tall wrinkled 3 : Dwarf round 3 : Dwarf wrinkled 1, i.e. 9 : 3 : 3 : 1.

   ½ mark

7. (ii) Law of Independent Assortment: when two pairs of traits are combined in a hybrid, the segregation of one pair of characters is independent of the other pair.

   ½ mark

Answer: F₁: all tall round (TtRr); F₂: tall round : tall wrinkled : dwarf round : dwarf wrinkled = 9 : 3 : 3 : 1; Law of Independent Assortment

1. (i) A vaccine contains antigenic proteins of the pathogen, or inactivated/weakened pathogens.

   ½ mark

2. These antigens activate the immune system. B-lymphocytes produce antibodies against them, giving a primary (humoral) immune response, so antibody concentration rises (the first, smaller peak in the graph).

   1 mark

3. (ii) The vaccination also produced memory B and T cells.

   ½ mark

4. On a later infection by the same pathogen, these memory cells recognise it quickly and mount a rapid, intense secondary (anamnestic) response. This is why the antibody level rises very fast and very high.

   ½ mark

5. (iii) P = Yes (natural active immunity forms memory cells).

   ½ mark

6. Q = catching the infection (being naturally infected by the pathogen).

   ½ mark

7. R = No (passively received antibodies leave no memory cells).

   ½ mark

8. S = Yes (the vaccine makes the body produce its own antibodies).

   ½ mark

9. T = No (ready-made antibodies are injected; the body makes none).

   ½ mark

Answer: (i) Vaccine antigens stimulate B-cells to make antibodies (primary response); (ii) memory cells give a rapid secondary/anamnestic response; (iii) P – Yes, Q – catching an infection, R – No, S – Yes, T – No

1. (i) The chemical name of 'smack' is diacetylmorphine (heroin). It is a white, odourless, bitter crystalline compound obtained by acetylation of morphine.

   1 mark

2. Its consumption is treated as abuse because (1) it is highly addictive, so the user quickly becomes dependent on it, and (2) it is a depressant that slows down body functions. It is not taken for any medical purpose.

   1 mark

3. (ii)(1) Marijuana: source plant is Cannabis sativa (inflorescence). Effect: cannabinoids affect the cardiovascular system of the body.

   1 mark

4. (ii)(2) Cocaine: source plant is Erythroxylum coca (coca plant). Effect: it interferes with the transport of the neurotransmitter dopamine, which produces euphoria and a feeling of increased energy. Excessive doses cause hallucinations.

   1 mark

5. (ii)(3) Morphine: source plant is Papaver somniferum (poppy plant, from its latex). Effect: it acts as a depressant/sedative and painkiller that slows down body functions.

   1 mark

Answer: (i) Diacetylmorphine; it is highly addictive and a depressant that slows body functions. (ii) Marijuana – Cannabis sativa, affects cardiovascular system; Cocaine – Erythroxylum coca, interferes with dopamine transport (euphoria); Morphine – Papaver somniferum, depressant/sedative, painkiller.

1. (i) Method 1: transfer of pollen from the anther to the stigma of the same flower is autogamy.

   ½ mark

2. Method 2: transfer of pollen between different flowers of the same plant is geitonogamy.

   ½ mark

3. Method 3: transfer of pollen from a flower of one plant to the stigma of a flower on a different plant is xenogamy.

   ½ mark

4. (ii)(1) Water lily: its flowers emerge above the water surface. They are therefore pollinated by insects or wind, as in most land plants, and not by water.

   ½ mark

5. (ii)(2) Vallisneria: the female flowers reach the water surface on long stalks. The male flowers or pollen grains are released onto the water surface and are carried passively by water currents. Some of them reach the female flowers and their stigmas, which brings about pollination.

   1 mark

6. (iii) Hereditary (genetic) device – self-incompatibility: this is a genetic mechanism that stops pollen from the same flower, or from other flowers of the same plant, from fertilising the ovules. It works by inhibiting pollen germination or pollen-tube growth in the pistil.

   1 mark

7. Physiological device – non-synchronised maturation (dichogamy): pollen release and stigma receptivity do not happen at the same time. Either the pollen matures before the stigma becomes receptive, or the stigma becomes receptive much before pollen is released. This prevents self-pollination.

   1 mark

Answer: (i) 1 – Autogamy, 2 – Geitonogamy, 3 – Xenogamy. (ii) Water lily – by insects/wind; Vallisneria – female flowers reach the surface on long stalks and water currents carry the released male flowers/pollen to them. (iii) Hereditary: self-incompatibility; Physiological: pollen release and stigma receptivity not synchronised.

1. (i)(1) Days 1–7: menstrual phase (menstruation).

   ½ mark

2. (i)(2) Days 8–12: follicular phase, also called the proliferative phase.

   ½ mark

3. (i)(3) Days 16–28: luteal phase, also called the secretory phase.

   ½ mark

4. (i)(4) Days 13–15: ovulatory phase (ovulation occurs around day 14).

   ½ mark

5. (ii)(1) Days 8–12 – pituitary: rising levels of FSH (and LH) from the anterior pituitary stimulate the development of ovarian follicles and the secretion of estrogen by the growing follicles.

   ½ mark

6. Days 8–12 – ovarian: the estrogen secreted by the growing follicles causes proliferation (regeneration) of the endometrium.

   ½ mark

7. (ii)(2) Days 13–15 – pituitary: LH and FSH reach their peak, and the LH surge induces rupture of the Graafian follicle, releasing the ovum (ovulation).

   ½ mark

8. Days 13–15 – ovarian: high estrogen levels complete the maturation of the Graafian follicle and keep the endometrium thickened.

   ½ mark

9. (ii)(3) Days 16–28 – pituitary: LH acts on the remains of the ruptured follicle and maintains it as the corpus luteum.

   ½ mark

10. Days 16–28 – ovarian: the corpus luteum secretes large amounts of progesterone, which maintains the endometrium for implantation. If fertilisation does not occur, the corpus luteum degenerates, progesterone falls and the endometrium breaks down, leading to the next menstruation.

    ½ mark

Answer: (i) 1–7: Menstrual phase; 8–12: Follicular/proliferative phase; 16–28: Luteal/secretory phase; 13–15: Ovulatory phase. (ii) 8–12: FSH stimulates follicle growth, and estrogen proliferates the endometrium; 13–15: LH surge ruptures the Graafian follicle (ovulation), and high estrogen matures the follicle and thickens the endometrium; 16–28: LH maintains the corpus luteum, whose progesterone maintains the endometrium.