ISC Class 12 Biology Sample Paper 2026 with Solutions

SECTION A - 20 MARKS

Question 1
Answer the following questions briefly.

(i) Anita is suffering from dry and scaly lesions on various parts of the body. What is the biological name of the causative fungus? [1 Mark]

Answer:
Epidermophyton / Trichophyton / Microsporum

Teacher's Note:
a) Ringworm is caused by fungi belonging to these three genera.
b) Students must write the scientific name correctly with proper genus and species capitalization.

(ii) If the number of chromosomes in an endosperm is 12, then what will be the number of chromosomes in megaspore mother cell? [1 Mark]

Answer:
8

Teacher's Note:
a) Endosperm is triploid (\(3n = 12\)), so the haploid number (\(n\)) is 4.
b) The megaspore mother cell is diploid (\(2n = 2 \times 4 = 8\)).

(iii) Renu was conducting an experiment. She started with a single DNA strand and ran six cycles of PCR to produce number of molecules of DNA. [1 Mark]

Answer:
64

Teacher's Note:
a) The number of DNA molecules produced after \(n\) cycles is given by \(2^n\).
b) For 6 cycles, \(2^6 = 64\).

(iv) At what stage of embryonic development, can the zygote be introduced in the Fallopian tube in ZIFT? [1 Mark]

Answer:
8 - 16 blastomeres

Teacher's Note:
a) ZIFT stands for Zygote Intrafallopian Transfer, where embryos up to 8 blastomeres are transferred into the Fallopian tube.
b) Beyond 8 blastomeres, transfer is typically into the uterus (IUT).

(v) When a cross is made between tall plants with yellow seeds (TtYy) and tall plants with green seeds (Ttyy), what proportions of phenotype in the offspring could be expected to be tall plants with green seeds? [1 Mark]

Answer:
3/8

Teacher's Note:
a) For height (\(Tt \times Tt\)), the probability of tall is \(3/4\). For seed color (\(Yy \times yy\)), the probability of green (\(yy\)) is \(1/2\).
b) Multiplying the independent probabilities gives \(3/4 \times 1/2 = 3/8\).

(vi) A research scholar isolated a new restriction enzyme from Thermus aquaticus strain YT3. It was the fourth restriction enzyme discovered from this strain. Propose a scientifically accurate name for this enzyme using the standard nomenclature rules. [1 Mark]

Answer:
TaqYIV

Teacher's Note:
a) The first letter comes from the genus name (T) and the next two from the species name (aq).
b) The strain is denoted by Y and the Roman numeral IV indicates the chronological order of discovery.

(vii) Construct a pyramid of numbers for an ecosystem in which a single large aquatic plant supports a population of small herbivorous fish, which are consumed by a larger population of predatory fish such as kingfish. [1 Mark]

Answer:

[Figure: Inverted pyramid of numbers with a single large aquatic plant at the base, a larger number of herbivorous fish in the middle, and a larger population of predatory fish at the top]

Teacher's Note:
a) This represents an inverted pyramid of numbers commonly found in a single-plant aquatic or tree ecosystem.
b) Students must clearly draw the pyramid shape and label the trophic levels correctly.

(viii) Observe the relation between the first two words and then complete the analogy.
Chikungunya: Aedes :: Malaria: ________. [1 Mark]

Answer:
Anopheles

Teacher's Note:
a) The analogy relates the disease to its specific vector insect.
b) Malaria is transmitted by the female Anopheles mosquito.

(ix) A young boy named Arjun has sickle-shaped red blood cells. He experiences frequent fatigue, and shows signs of damage to kidney. Genetic analysis reveals a single point mutation in the gene coding for beta-globin.
Which one of the following genetic phenomena is MOST LIKELY responsible for the multiple symptoms observed in Arjun? [1 Mark]
(a) Polygenic inheritance
(b) Codominance
(c) Pleiotropy
(d) Incomplete dominance

Answer: (c) Pleiotropy

A single gene mutation affects multiple phenotypic traits (sickle cells, fatigue, kidney damage).

Teacher's Note:
a) Pleiotropy occurs when a single gene influences multiple unrelated phenotypic traits.
b) Sickle-cell anemia is a classic textbook example of pleiotropic inheritance.

(x) Kiwi is a dioecious species. Which of the following methods can be definitely ruled out as a possible mode of pollination in that case? [1 Mark]
(P) Cleistogamous autogamy
(Q) Chasmogamous autogamy
(R) Geitonogamy
(S) Xenogamy
(a) Only (P) and (R)
(b) Only (P) and (Q)
(c) Only (Q) and (S)
(d) Only (P), (Q) and (R)

Answer: (d) Only (P), (Q) and (R)

Dioecious plants bear male and female flowers on different plants, completely preventing autogamy and geitonogamy.

Teacher's Note:
a) Dioecious condition prevents self-pollination mechanisms like autogamy and geitonogamy.
b) Only cross-pollination (xenogamy) is possible in dioecious species.

(xi) Given below are two statements marked Assertion and Reason. Read both the statements carefully and choose the correct option. [1 Mark]
Assertion: An amino acid in polypeptide chain is not always altered even due to change in the third nitrogenous base of codon.
Reason: The amino acid does not change due to degeneracy of genetic code.
(a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Both Assertion and Reason are false.

Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.

Degeneracy means multiple codons can code for the same amino acid, especially when the third base changes (wobble hypothesis).

Teacher's Note:
a) The genetic code is degenerate, meaning most amino acids are specified by more than one codon.
b) Changes at the third position (wobble position) often result in synonymous codons.

(xii) Given below are two statements marked Assertion and Reason. Read both the statements carefully and choose the correct option. [1 Mark]
Assertion: 'Bt' toxin gene has been cloned from bacteria E.coli and expressed in plants to provide resistance from insect without the need of insecticides.
Reason: 'Bt' toxin is produced in a crystalline state by the above-mentioned bacterium.
(a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Both Assertion and Reason are false.

Answer: (d) Both Assertion and Reason are false.

Bt toxin gene is cloned from Bacillus thuringiensis, not E. coli, making the assertion and reason factually incorrect in their premise regarding the source bacterium.

Teacher's Note:
a) The bacterium that naturally produces Bt toxin is Bacillus thuringiensis, not Escherichia coli.
b) Although the gene is cloned into expression vectors using E. coli as a host, the native source and organism producing the crystal is B. thuringiensis.

(xiii) Why are tendrils of vine (Vitis) and pea (Pisum) considered to be analogous organs? [1 Mark]

Answer:
Tendrils of vine are modified stems and of pea are modified leaves or leaflets. They perform the same function - climbing and supporting the plants.

Teacher's Note:
a) Analogous organs have different embryonic origins and underlying anatomical structures.
b) They perform similar physiological functions due to convergent evolution.

(xiv) Ravi wanted to grow rice in his field. He was very concerned about environment degradation, so he did not want to use chemical fertilisers. Suggest a suitable biological method to Ravi. [1 Mark]

Answer:
He should use cyanobacteria (Anabaena, Nostoc, Aulosira) or Nitrogen-fixing bacteria (Azospirillum) or Azolla (fern plant that harbours Anabaena).

Teacher's Note:
a) Biofertilizers enrich the soil with nutrients and organic matter without causing chemical pollution.
b) Cyanobacteria and Azolla fix atmospheric nitrogen in paddy fields.

(xv) Answer the following: [2 Marks]
(a) Expand the abbreviation NPP. [1 Mark]
(b) Rule of Equivalence states that in DNA, adenine equals thymine and guanine equals cytosine. This is due to base pairing in double helix. thus, purines equal pyrimidines. Which scientist offered this concept? [1 Mark]

Answer:
(a) Net Primary Productivity
(b) Chargaff

Teacher's Note:
a) NPP is the available biomass for consumption to heterotrophs.
b) Erwin Chargaff formulated the base-pairing rules for DNA composition.

(xvi) A bilobed dithecous anther has 100 microspore mother cells per microsporangium. How many male gametophytes can this anther produce? [1 Mark]

Answer:
1600

Teacher's Note:
a) A bilobed dithecous anther has 4 microsporangia (\(4 \times 100 = 400\) microspore mother cells in total).
b) Each MMC produces 4 microspores/pollen grains (male gametophytes), giving \(400 \times 4 = 1600\).

(xvii) The diagram given below represents a specific stage of gestation period. Identify the structure labelled - B. [1 Mark]

[Figure: Diagram of a fetus inside the uterus with label B pointing to the placenta / chorionic villi region]

Answer:
Placenta / Chorionic villi

Teacher's Note:
a) The placenta acts as the structural and functional unit between the developing embryo and maternal body.
b) It facilitates the supply of oxygen and nutrients and removal of wastes.

(xviii) Give a reason for each of the following statements: [2 Marks]
(a) Sameer drew a diagram of human sperm and showed it to his teacher. The teacher rejected the diagram. [1 Mark]
(b) Even under similar environmental conditions, decomposition of the exoskeleton of millipedes occurs more slowly than the leaves. [1 Mark]

Answer:
(a) Tail part has been labelled in the middle piece of sperm. Head is labelled as middle piece.
(b) Millipedes' chitinous exoskeleton decomposes more slowly.

Teacher's Note:
a) Correct anatomical labelling is critical in biological diagrams.
b) Chitin is a complex, highly resistant polysaccharide compared to plant leaf materials like cellulose.

SECTION B - 14 MARKS

Question 2 [2 Marks]
(i) Construct an ideal pyramid of energy when 1000,000 joules of sunlight is available. Label all its trophic levels.

Answer:

[Figure: Upright pyramid of energy showing sunlight at the base with 1,000,000 J, followed by plants with 100,000 J, herbivores with 10,000 J, and carnivores with 1,000 J according to the 10 percent law]

Teacher's Note:
a) Energy pyramids are always upright following the 10% law of energy transfer.
b) Only 10% of energy is transferred from one trophic level to the next.

OR

(ii) The graph given below is based on the data collected from a survey conducted on species richness of a group of mammals in three different climatic regions of the world: Brazil, France and Norway. Brazil has nearly 540 species of mammals, France has nearly 303 species of mammals and Norway has 65 species of mammals.
(a) Based on the species richness, identify the location of these countries in the respective climatic regions shown in the graph. [1 Mark]
(b) In which climatic region, will you place India? [1 Mark]

[Figure: Bar graph comparing species richness of mammals in Polar region, Temperate region, and Tropical region]

Answer:
(a) Brazil - Tropical region, France - Temperate region, Norway - Polar region.
(b) India - Tropical region.

Teacher's Note:
a) Species diversity increases as we move from polar regions towards the equator (tropics).
b) India, being largely a tropical country, exhibits high species richness.

Question 3 [2 Marks]
Given below is the relationship between the HIV levels in the blood and helper T-cell count in a person detected with AIDS. Study the relationship and answer the questions that follow.
(i) Describe the trend observed between viral load and immune response following initial infection. [1 Mark]
(ii) Is the virus permanently eliminated from the body? Justify your answer. [1 Mark]

[Figure: Line graph showing viral load and T-cell count in blood over months and years following HIV infection]

Answer:
(i) Initial viral spike -> T-cell drop -> partial recovery -> eventual immune suppression.
(ii) No; virus integrates/latent in host cells.

Teacher's Note:
a) HIV targets helper T-cells (\(CD4^+\)), drastically weakening cell-mediated immunity over time.
b) The virus persists in a latent proviral state within host DNA, preventing complete clearance.

Question 4 [2 Marks]
As a volunteer for an awareness programme in rural areas, you have been asked to design a poster on sexual health, highlighting birth control options available to males. What two contraceptive methods for males will you highlight in the poster that have negligible chances of failure? Mention the working principle for each method.

Answer:
Condom - Physical barrier preventing sperm entry into the female reproductive tract.
Vasectomy - Surgical sterilization blocking transport of sperms by cutting/ligating the vas deferens.

Teacher's Note:
a) Permanent methods like vasectomy have almost negligible failure rates.
b) Barrier methods prevent physical meeting of gametes.

Question 5 [2 Marks]
Study the diagram given below and answer the following questions:
(i) Identify the anode end in the diagram. [1 Mark]
(ii) How are these DNA fragments visualised? [1 Mark]

[Figure: Diagram of an agarose gel electrophoresis apparatus showing wells, DNA bands separated across a gel plate, and electrodes a and b]

Answer:
(i) Anode end is towards 'b'.
(ii) Agarose gel, containing DNA fragments is stained with ethidium bromide and exposed to UV radiation. Orange color bands of DNA become visible.

Teacher's Note:
a) DNA is negatively charged, so it migrates towards the positive electrode (anode).
b) Ethidium bromide intercalates with DNA and fluoresces under UV light.

Question 6 [2 Marks]
Microbes especially yeasts have been used from time immemorial for the production of beverages like wine, beer, whisky brandy, or rum. Depending on the type of the raw materials used for fermentation and type of processing (with or without distillation) different types of alcoholic drinks are obtained.
(i) Mention the scientific name of the organism used to prepare fermented beverages. [1 Mark]
(ii) Name any one beverage obtained without distillation of fermented broth. [1 Mark]

Answer:
(i) Saccharomyces cerevisiae
(ii) Wine, Beer

Teacher's Note:
a) Brewer's yeast ferments sugars anaerobically to produce ethanol and carbon dioxide.
b) Beverages like wine and beer are produced without distillation, unlike whisky, brandy, and rum.

Question 7 [2 Marks]
(i) Ryan had developed a GM organism. Which government organisation will he approach to obtain the clearance for its mass production? [1 Mark]
(ii) Which bioactive product is used for the treatment of emphysema? [1 Mark]

Answer:
(i) GEAC (Genetic Engineering Appraisal Committee)
(ii) Alpha - 1 antitrypsin

Teacher's Note:
a) GEAC makes decisions regarding the validity of GM research and safety of introducing GM organisms for public services.
b) Alpha-1 antitrypsin is used to treat emphysema as a transgenic biological product.

Question 8 [2 Marks]
(i) In a barn, there were 30 rats. 5 more rats entered the barn and 6 of the rats were eaten by the cats in one week. If 8 rats were born during the same period, and during the same time, 7 rats left the barn, find the resultant rat population in the barn at the end of one week. [1 Mark]
(ii) Define carrying capacity. [1 Mark]

Answer:
(i) \(30 + [(5 + 8) - (6 + 7)] = 30$
(ii) Habitats are species specific and have resources up to a limit that can support maximum number of individuals to grow and reproduce, this limit of habitat to subsist a species is called carrying capacity.

Teacher's Note:
a) Population density is calculated by adding natality and immigration, then subtracting mortality and emigration.
b) Carrying capacity represents the maximum sustainable population size for a given habitat.

SECTION C - 21 MARKS

Question 9 [3 Marks]
Study the pedigree chart given below showing the pattern of blood group inheritance in a family.
(i) State the genotypes of the following: [2 Marks]
(a) Parents
(b) The individual 'X' in the second generation
(ii) State the possible blood group(s) of the individual 'Y' in \(3^{rd}\) generation. [1 Mark]

[Figure: Pedigree chart showing inheritance of ABO blood groups across three generations with individuals marked A, B, AB, X, Y]

Answer:
(i) (a) Father: \(I^A i\), Mother: \(I^B i$
(b) \(I^A I\), \(I^B i\), \(ii$
(ii) A or O

Teacher's Note:
a) ABO blood grouping exhibits codominance and multiple allelism.
b) Pedigree analysis helps trace the transmission of specific genetic traits through generations.

Question 10 [3 Marks]
(i) Darwin's finches are best example of adaptive radiation. Justify.

Answer:
Darwin on voyage on the ship H.M.S Beagle, Galapagos Island separated from mainland South America - 13 species of Ground finches found on different islands. Original finches were found on mainland - seed eating beak; different islands - different types of beaks - for e.g. Vegetarian tree finches, insectivorous eating, Cactus eating. Beak got modified according to the food available on that particular island. It supports adaptive radiation.

Teacher's Note:
a) Adaptive radiation refers to the evolutionary process where species evolve from a common ancestor into diverse forms adapted to new habitats.
b) Changes in beak morphology reflect dietary specializations.

OR

(ii) (a) The recessive allele 'b' occurs with a frequency of \(0 \cdot 8\) in a population of moths that is in Hardy Weinberg Equilibrium. What is the frequency of homozygous dominant individuals? [2 Marks]
(b) List any two differences between Homo habilis and Homo erectus. [1 Mark]

Answer:
(a) Frequency of homozygous dominant individuals = \(0 \cdot 04\) or \(4\%\).
(b)

Teacher's Note:
a) According to Hardy-Weinberg equilibrium, \(p + q = 1\). Given \(q = 0 \cdot 8\), \(p = 0 \cdot 2\). Homozygous dominant frequency is \(p^2 = (0 \cdot 2)^2 = 0 \cdot 04\).
b) Human evolution exhibits progressive increases in cranial capacity and stature.

Question 11 [3 Marks]
Medical interns Arun and Sam are undergoing training in a hospital. They were assigned three patient cases to review. While they successfully diagnosed the diseases in cases B and C, they could not recall the names of the causative agents. Additionally, they were unable to diagnose the disease of patient A.
The table given below shows their diagnosis.

Identify the disease affecting patient 'A'. Mention the biological name of the causative agents responsible for the diseases in patients 'B' and 'C'.

Answer:
(A) Plague
(B) Streptococcus pneumoniae
(C) Wuchereria bancrofti

Teacher's Note:
a) Pathogen identification is fundamental in clinical pathology and diagnostics.
b) Students must memorize common human infectious diseases and their respective causative agents.

Question 12 [3 Marks]
A twenty-year old boy, Reshu has become addicted to alcohol.
(i) Mention any two possible reasons for his alcohol addiction. [2 Marks]
(ii) Suggest any one measure by which addiction can be prevented. [1 Mark]

Answer:
(i) Consumption in social gathering, to relieve social or physical discomforts, desire for excitement, to escape from disappointments, family atmosphere. (Any two)
(ii) Avoid undue peer pressure, education, and counselling, seeking help from parents and peers, looking for danger sign, seeking professional and medical help. (Any one)

Teacher's Note:
a) Adolescent substance abuse often stems from peer pressure, curiosity, or stress.
b) Preventive measures require psychological support, awareness, and healthy coping mechanisms.

Question 13 [3 Marks]
A scientist is attempting to create a recombinant DNA molecule by combining a plasmid vector and a foreign DNA. A plasmid DNA and a linear DNA of the same size have a single site for the restriction enzyme EcoRI. When cut by the same RE and separated by gel electrophoresis, the plasmid shows one DNA band, and the linear DNA shows two bands.
(i) What causes the difference between the number of DNA bands generated from the plasmid and from the linear DNA? [1 Mark]
(ii) What is the advantage of using agarose in gel electrophoresis? [1 Mark]
(iii) How does EcoRI differ from an exonuclease? [1 Mark]

Answer:
(i) Plasmid is circular, therefore when cut by RE it produces only one fragment, whereas the linear DNA is cut to release two fragments.
(ii) Agarose is non-reactive and large pore size allows easier separation of fragments.
(iii) EcoRI is an endonuclease, it cuts the specific inner bonds while exonucleases cut outer bonds starting either from 3' - terminus or from 5' - terminus.

Teacher's Note:
a) Circular DNA molecules yield linear fragments equal to the number of restriction sites, while linear DNA produces one more fragment than the number of sites.
b) Endonucleases cut within specific recognition sequences, whereas exonucleases remove nucleotides from the ends.

Question 14 [3 Marks]
Study the diagram given below that shows the modes of pollination and answer the questions that follow.
(i) The given diagram shows three methods of pollination in plants. What are the technical terms used for pollen transfer methods labelled '2' and '3'? [2 Marks]
(ii) How does Salvia achieve pollination successfully? [1 Mark]

[Figure: Diagram showing floral pollination modes with arrows labeled 1, 2, and 3 representing autogamy, geitonogamy, and xenogamy]

Answer:
(i) 2 - Geitonogamy, 3 - Xenogamy / Allogamy
(ii) Pollinated by insects

Teacher's Note:
a) Geitonogamy is functionally cross-pollination but genetically similar to autogamy as pollen comes from the same plant.
b) Salvia employs a specialized lever mechanism adapted for insect pollination (entomophily).

Question 15 [3 Marks]
A population pyramid is a graphic representation of the distribution of a population by age groups. Diagrammatically represent the three kinds of age pyramids.

[Figure: Three population age pyramids showing Expanding population, Stable population, and Declining population with pre-reproductive, reproductive, and post-reproductive age classes]

Answer:
(Diagram representing expanding, stable, and declining population age pyramids with pre-reproductive, reproductive, and post-reproductive age classes).

Teacher's Note:
a) Expanding populations have a broad base with a high proportion of pre-reproductive individuals.
b) Declining populations show a constricted base due to low birth rates.

SECTION D - 15 MARKS

Question 16 [5 Marks]
During a field excursion, a group of class XII students visited Kaziranga National Park and observed efforts to protect the Indian rhinoceros in its natural habitat. They also toured a botanical garden and a gene bank, where they saw endangered plant species and preserved seeds.
Based on this context, answer the following questions:
(i) Name and define the two types of conservation methods mentioned above. [2 Marks]
(ii) What is cryopreservation? [1 Mark]
(iii) Is Captive breeding an ex-situ or an in-situ strategy. Justify your answer with a reason. [2 Marks]

Answer:
(i) In-situ and ex-situ conservation methods respectively. In-situ conservation is the conservation of biotic resources in their natural habitats. Ex-situ conservation is the conservation of threatened species outside their natural habitats.
(ii) Cryopreservation is in vitro conservation of tissues, organs, embryos, seeds etc. at low temperature of \(-196^{\circ}\text{C}$.
(iii) Captive breeding is ex-situ conservation method because threatened and endangered species are bred and reared in zoological parks under human supervision.

Teacher's Note:
a) In-situ conservation protects species in their natural ecosystem, whereas ex-situ involves artificial or managed environments.
b) Liquid nitrogen is used in cryopreservation to maintain biological samples in a suspended metabolic state.

Question 17 [5 Marks]
(i) (a) Write a brief note on Griffith's experiment. What was the conclusion drawn from this experiment? [3 Marks]
(b) Name the parts 'A' and 'B' of the transcription unit shown below. [2 Marks]

[Figure: Diagram of a transcription unit with a promoter region, structural gene, and terminator showing labels A and B on the DNA strands]

Answer:
(a) Two strains of Streptococcus pneumoniae - Smooth Virulent Strain (S-III) and Rough Avirulent Strain (R-II).
- R-II were injected into mice - mice remained healthy.
- S-III were injected into mice; mice developed pneumonia and died.
- S-III were heated to \(60^{\circ}\text{C}\) and killed. These heat-killed bacteria were injected to mice; mice remained healthy.
- A mixture of R-II and heat killed S-III bacteria were injected into mice; mice developed pneumonia and died.
Conclusion: Heat killed S-III bacteria introduced some transforming principle that caused transformation of R-II avirulent into virulent S-III bacteria.
(b) A - Promoter; B - Coding / Anti template / Sense strand

Teacher's Note:
a) Griffith's experiment laid the foundation for discovering DNA as the genetic material by demonstrating bacterial transformation.
b) The promoter acts as the binding site for RNA polymerase during transcription initiation.

OR

(ii) Study the diagram given below and answer the questions that follow.
(a) Name the molecule 'X' synthesised by 'I' gene. How does this molecule get inactivated? [2 Marks]
(b) Name the enzymes coded respectively by genes- z, y, and a. [2 Marks]
(c) Which enzyme binds to the operator to initiate transcription? [1 Mark]

[Figure: Diagram of the lac operon showing regulator gene i, promoter p, operator o, and structural genes z, y, a with mRNA transcription and translation steps]

Answer:
(a) X = Repressor protein. This molecule gets deactivated in the presence of inducer (Lactose).
(b) gene 'z' - beta-galactosidase; gene 'y' - permease; gene 'a' - transacetylase.
(c) RNA polymerase

Teacher's Note:
a) The lac operon is an inducible operon where lactose acts as the inducer.
b) Binding of the inducer to the repressor causes a conformational change that prevents it from binding to the operator.

Question 18 [5 Marks]
(i) The following graph represents the relative concentrations of the four hormones present in the blood plasma of a woman during her menstrual cycle. Identify the hormones A, B, C and D. [3 Marks]

[Figure: Line graph showing fluctuations of hormones FSH, Estrogen, LH, and Progesterone across days 1 to 28 of the menstrual cycle, labeled A, B, C, D]

Answer:
A - Follicle Stimulating Hormone (FSH)
B - Estrogen
C - Luteinizing Hormone (LH)
D - Progesterone

Teacher's Note:
a) The LH surge (C) triggers ovulation around the middle of the menstrual cycle.
b) Progesterone (D) levels rise during the luteal phase to maintain the uterine endometrium.

(ii) During a fertility consultation, a woman mentions her menstrual cycle lasts for 33 days regularly. To help her conceive, the doctor advised her to track ovulation. On which day of the menstrual cycle is she most likely to ovulate? [2 Marks]

Answer:
\(33 - 14 = 19^{\text{th}}\) day

Teacher's Note:
a) Ovulation always occurs 14 days prior to the next expected menses, regardless of cycle length.
b) For a 33-day cycle, ovulation happens on day 19 (\(33 - 14\)).