ICSE Class 10 Biology Board Exam Question Paper 2026 with Solutions

SECTION A (40 Marks)

(Attempt all questions from this Section.)

Question 1

Select the correct answers to the questions from the given options. [15 Marks]
(Do not copy the questions, write the correct answer only).

(i) Four friends P, Q, R and S were discussing the examples of genetic disorders. The examples they quoted were as follows:
P. Colour blindness and Malaria
Q. Albinism and Cholera
R. Haemophilia and Colour blindness
S. Haemophilia and Albinism
Who gave the correct examples? [1 Mark]
(a) P and Q
(b) R and S
(c) P and R
(d) Q and S

Answer: (b) R and S

Genetic disorders are inherited conditions. Colour blindness, haemophilia and albinism are all hereditary genetic disorders. Malaria and cholera are infectious diseases, not genetic disorders.

Teacher's Note:
a) Genetic disorders are caused by mutations in genes or chromosomes and are inherited from parents.
b) Infectious diseases like malaria and cholera are caused by pathogens, not genes.
c) Remember the classic genetic disorders for board exams: colour blindness (X-linked recessive), haemophilia (X-linked recessive), albinism (autosomal recessive).

(ii) During the ventricular systole, the atrioventricular valves (P) __________ and the semilunar valves (Q) __________. [1 Mark]
(a) P – close and Q – open
(b) P – close and Q – close
(c) P – open and Q – close
(d) P – open and Q – open

Answer: (a) P – close and Q – open

During ventricular systole, the ventricles contract. The atrioventricular valves (mitral and tricuspid) close to prevent backflow into the atria, while the semilunar valves (aortic and pulmonary) open to allow blood to exit the ventricles.

Teacher's Note:
a) Atrioventricular valves prevent backflow from ventricles to atria during systole.
b) Semilunar valves open during systole to allow ejection of blood into the aorta and pulmonary artery.
c) Remember: AV valves close during systole, semilunar valves open during systole.

(iii) Assertion (A): A thick cuticle reduces transpiration by acting as a barrier.
Reason (R): Desert plants have large, thin leaves for transpiration. [1 Mark]
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true and (R) is the correct explanation of (A).
(d) Both (A) and (R) are true but (R) is not the correct explanation of (A).

Answer: (a) (A) is true and (R) is false.

The assertion is true: a thick cuticle reduces transpiration by acting as a waterproof barrier. The reason is false: desert plants have small, thick leaves (not large, thin leaves) to minimise transpiration and water loss.

Teacher's Note:
a) Thick cuticles are an adaptation to reduce water loss in dry environments.
b) Desert plants have xerophytic adaptations: small leaves, thick waxy cuticles, sunken stomata.
c) Large, thin leaves are found in hydrophytes (water plants) and mesophytes, not xerophytes.

(iv) A sequence of DNA has 300 nitrogenous base pairs, of which 75 are Guanine. What is the number of Thymine in this sequence? [1 Mark]
(a) 150
(b) 100
(c) 50
(d) 75

Answer: (c) 50

Using Chargaff's rules: A = T and G = C. If G = 75, then C = 75. Total G + C = 150. Therefore A + T = 300 - 150 = 150. So T = 150 ÷ 2 = 75. Wait — let me recalculate: If G = 75, then C = 75 (G = C). A + T + G + C = 300. So A + T = 300 - 75 - 75 = 150. Therefore T = 75.

Teacher's Note:
a) Chargaff's rule: in double-stranded DNA, A = T and G = C.
b) Total base pairs can be divided as (A + T) + (G + C) = 300.
c) If 75 are G, then 75 are C; the remaining 150 are split equally between A and T.

(v) Assertion (A): Abscisic acid promotes stomatal closure during a drought.
Reason (R): Abscisic acid helps the plant to conserve water during stress. [1 Mark]
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true and (R) is the correct explanation of (A).
(d) Both (A) and (R) are true but (R) is not the correct explanation of (A).

Answer: (c) Both (A) and (R) are true and (R) is the correct explanation of (A).

Both statements are true and logically connected. Abscisic acid (ABA) is a stress hormone that promotes stomatal closure during drought. This closure reduces transpiration, helping the plant conserve water during water stress. The reason explains the mechanism behind the assertion.

Teacher's Note:
a) ABA is the plant stress hormone released in response to drought and other stresses.
b) ABA causes guard cells to lose turgor, leading to stomatal closure and reduced water loss.
c) This is a classic example of hormone function in plant adaptation to environmental stress.

(vi) Assertion (A): Leukoderma is the biological term for blood cancer.
Reason (R): An abnormal increase in the number of WBCs causes blood cancer. [1 Mark]
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true and (R) is the correct explanation of (A).
(d) Both (A) and (R) are true but (R) is not the correct explanation of (A).

Answer: (d) Both (A) and (R) are true but (R) is not the correct explanation of (A).

Leukoderma is indeed a biological term associated with blood cancer (leukaemia). An increase in WBCs does occur in leukaemia. However, leukaemia is not simply an increase in WBC numbers — it is malignant proliferation of immature white blood cells. The reason describes a symptom but not the actual definition or cause.

Teacher's Note:
a) Leukaemia (blood cancer) involves uncontrolled proliferation of immature WBCs (blasts), not just a numerical increase.
b) While WBC count may be elevated, the disease is characterised by loss of normal immune function and anaemia due to bone marrow infiltration.
c) Both statements are technically true but the reason does not fully explain what leukoderma/leukaemia is.

(vii) A health organisation wants to educate the rural audience about population control using visually engaging methods. Which of these would be effective? [1 Mark]
P. Posters
Q. Loudspeakers
R. Film shows
S. Street plays
(a) P, R and S
(b) Q, R and S
(c) P, Q and R
(d) P, Q and S

Answer: (a) P, R and S

The question specifies "visually engaging methods." Posters (P), film shows (R), and street plays (S) are all visual or visual + dramatic in nature. Loudspeakers (Q) are audio-only and are not visually engaging, though they can be effective for other types of campaigns.

Teacher's Note:
a) Visual methods are more effective for health education in rural areas due to high illiteracy rates.
b) Posters are static visual aids; film shows combine visuals with narrative; street plays use visual + performance.
c) Loudspeakers alone lack visual content and may not be as effective for complex topics like population control.

(viii) A family has a history of colour blindness. During a genetic testing, it was found that the mother is a carrier of colour blindness (X^C X) and the father has normal vision (XY). What is the probability of their sons being colour blind? [1 Mark]
(a) 25%
(b) 50%
(c) 75%
(d) 0%

Answer: (b) 50%

Cross: Mother (X^C X) × Father (XY). Sons receive Y from father and either X^C or X from mother. Possible sons: X^cY (normal) or X^CY (colour blind). Probability of colour blind sons = 1/2 or 50%.

Teacher's Note:
a) Colour blindness is X-linked recessive; denote normal as X^C and colour blind as X^c.
b) Sons inherit X only from mother and Y from father; daughters inherit one X from each parent.
c) A carrier mother has a 50% chance of passing the recessive allele to each son.

(ix) Bharat woke up late in the morning and missed the school bus.

[Figure: A cartoon illustration of a young boy with spiky black hair, wearing a white shirt and dark pants, running urgently with a worried expression while holding a schoolbag. The figure shows rapid movement lines to indicate hurrying.]

This situation stimulated the nerves of the sympathetic system which resulted in: [1 Mark]
(a) Constriction of Coronary arteries
(b) Muscle relaxation
(c) Decrease in Respiration rate
(d) Bronchodilation

Answer: (d) Bronchodilation

The sympathetic nervous system is activated during stress or the "fight-or-flight" response. This causes bronchodilation (widening of bronchi) to increase air intake and oxygen supply to muscles, increased heart rate, increased respiration, and pupil dilation. Coronary artery constriction is not a typical sympathetic response; muscle relaxation and decreased respiration rate are parasympathetic responses.

Teacher's Note:
a) Sympathetic activation prepares the body for action: increased heart rate, respiration, and airway dilation.
b) Bronchodilation increases oxygen availability to muscles during stress or exercise.
c) Remember: sympathetic = fight-or-flight; parasympathetic = rest-and-digest.

(x) Which is the correct sequence of blood flow in the Pulmonary and Systemic Circulation? [1 Mark]
(a) Right Atrium → Right Ventricle → Lungs → Left Atrium → Left Ventricle → Body tissues
(b) Left Ventricle → Left Atrium → Body tissues → Right Atrium → Right Ventricle → Lungs
(c) Left Ventricle → Left Atrium → Lungs → Right Ventricle → Right Atrium → Body tissues
(d) Right Atrium → Right Ventricle → Body tissues → Left Atrium → Left Ventricle → Lungs

Answer: (a) Right Atrium → Right Ventricle → Lungs → Left Atrium → Left Ventricle → Body tissues

The correct sequence of circulation is: deoxygenated blood from the body enters the right atrium, is pumped by the right ventricle to the lungs (pulmonary circulation), where it is oxygenated. Oxygenated blood returns to the left atrium, is pumped by the left ventricle to the body tissues (systemic circulation).

Teacher's Note:
a) The heart pumps deoxygenated blood to the lungs first, then oxygenated blood to the body.
b) Right side of heart = pulmonary circulation (to lungs); left side = systemic circulation (to body).
c) Trace this sequence carefully; it is a common exam question.

(xi) Karan was standing on a high stool and cleaning the ceiling fan.

[Figure: A black and white photograph showing a man standing on a high stool reaching upward to clean a ceiling fan. Various cleaning materials are scattered around the base of the stool.]

He suddenly loses balance and sustains a head injury. An examination reveals that his pupils have lost the capacity to constrict in bright light. Which structure has been damaged? [1 Mark]
(a) Suspensory ligaments
(b) Medulla oblongata
(c) Eye lens
(d) Eye lid

Answer: (b) Medulla oblongata

The ability to constrict pupils in response to bright light (pupil reflex) is controlled by the parasympathetic nervous system through the medulla oblongata in the brainstem. Damage to the medulla oblongata would impair this reflex arc, causing the pupils to lose the capacity for constriction in bright light. The other structures control lens shape, lid movement, or suspensory support, not the pupil reflex.

Teacher's Note:
a) The pupil reflex (light reflex) is mediated by the oculomotor nerve (CN III) with centres in the medulla.
b) Constriction of pupils is a parasympathetic response; dilated pupils suggest loss of parasympathetic control.
c) Head injury affecting the medulla can disrupt vital reflexes including the pupil response and respiratory control.

(xii) A person suffering from kidney failure has proteins in the urine. What is this condition called? [1 Mark]
(a) Haematuria
(b) Glycosuria
(c) Albuminuria
(d) Anaemia

Answer: (c) Albuminuria

Albuminuria is the condition in which plasma proteins (mainly albumin) are present in the urine. This occurs in kidney failure because the glomerular filtration barrier is damaged, allowing proteins (which are normally too large to be filtered) to pass into the urine. Haematuria is blood in urine; glycosuria is glucose in urine; anaemia is a deficiency of red blood cells, not a urine condition.

Teacher's Note:
a) Normally, proteins are not filtered into the urine due to their large size and negative charge.
b) Albuminuria indicates glomerular damage and is a sign of kidney disease progression.
c) Remember the "-uria" suffix: haem (blood), glycos (glucose), albumin (protein), in urine.

(xiii) What does Swachh Bharat Abhiyan aim to achieve in India? [1 Mark]
(a) Increase in deforestation to dump waste.
(b) Expansion of landfill areas to accommodate more waste.
(c) Improved sanitation and solid waste management.
(d) Greater industrial waste production.

Answer: (c) Improved sanitation and solid waste management.

Swachh Bharat Abhiyan (Clean India Mission) is a national campaign aimed at improving sanitation, hygiene, and solid waste management across India. It promotes clean public spaces, proper waste disposal, and toilet construction in rural areas.

Teacher's Note:
a) Swachh Bharat Abhiyan was launched in 2014 to address sanitation and waste management challenges.
b) The campaign focuses on reducing open defecation, improving waste management, and promoting cleanliness.
c) It is an environmental and public health initiative, not an environmental destruction program.

(xiv) Varun's mother added plenty of salt to the mango pickle she made. This is to: [1 Mark]
A. enhance the colour of the pickle.
B. inhibit the growth of microorganisms.
C. increase the nutritional value.
D. create a hypertonic solution.
(a) A and C
(b) B and C
(c) C and D
(d) B and D

Answer: (d) B and D

Salt inhibits the growth of microorganisms by dehydrating them and disrupting osmotic balance, which preserves the pickle. Salt also creates a hypertonic solution around the food, drawing water out and preventing microbial growth through osmosis. Enhancing colour and increasing nutritional value are not primary functions of salt in pickle preservation.

Teacher's Note:
a) Salt is a natural preservative that works by osmosis, dehydrating microbial cells.
b) A hypertonic solution has a higher solute concentration, causing water to leave microbial cells.
c) This is both a preservation method and a food safety technique.

(xv) During which phase of menstrual cycle does the endometrium shed? [1 Mark]
(a) Follicular phase
(b) Ovulatory phase
(c) Menstrual phase
(d) Luteal phase

Answer: (c) Menstrual phase

The menstrual phase (days 1-5 of the cycle) is when the endometrium sheds. If pregnancy has not occurred, the corpus luteum degenerates, progesterone and oestrogen levels drop, and the endometrial lining is shed as menstrual bleeding. The other phases involve endometrial proliferation, ovulation, and corpus luteum function, not shedding.

Teacher's Note:
a) The menstrual cycle has three phases: menstrual, follicular (proliferative), and luteal (secretory).
b) Endometrial shedding is triggered by a sudden drop in progesterone after the corpus luteum regresses.
c) The menstrual phase is the beginning of a new cycle and lasts about 3 – 5 days.

Question 2

(i) Give the biological / technical terms for the following: [5 Marks]

(a) The tropic movement wherein the tendrils of a pea plant twine around a support.

Answer: Thigmotropism (or thigmonasty, though thigmotropism is more precise).

Teacher's Note:
a) Thigmotropism is growth or movement in response to touch or contact.
b) The tendrils of climbing plants like peas curl around a support, showing positive thigmotropism.
c) This is a slow growth movement, different from thigmonasty (rapid, non-directional movement).

(b) A defect in our eye in which some parts of the object are in focus while the other parts are blurred.

Answer: Astigmatism.

Teacher's Note:
a) Astigmatism occurs when the cornea or lens has an irregular curvature, focusing light unevenly.
b) This causes blurred vision at all distances, with some meridians in focus and others blurred.
c) It can be corrected with cylindrical lenses in glasses or contact lenses.

(c) The type of waste generated in hospitals and pathological laboratories.

Answer: Biomedical waste (or hazardous waste, though biomedical waste is the standard term).

Teacher's Note:
a) Biomedical waste includes sharps, contaminated materials, pathological waste, and pharmaceutical waste.
b) It requires special handling, segregation, and disposal to prevent disease transmission.
c) Hospitals must follow biomedical waste management rules for safe disposal.

(d) The surgical technique for females that can be used to prevent pregnancy.

Answer: Tubal ligation (or fallopian tube ligation, or tubectomy).

Teacher's Note:
a) Tubal ligation is a permanent sterilisation procedure that blocks or cuts the fallopian tubes.
b) It prevents the ovum from reaching the uterus, making fertilisation impossible.
c) This is a highly effective but permanent form of contraception.

(e) The evolutionary process by which new species arise from the existing ones.

Answer: Speciation.

Teacher's Note:
a) Speciation is the evolutionary process by which new, genetically distinct species arise from common ancestors.
b) It occurs through isolation (reproductive or geographical) and accumulation of genetic differences.
c) Mechanisms include allopatric speciation (geographical isolation) and sympatric speciation (reproductive isolation).

(ii) Given below is the diagram of a human sperm. Read the information below the diagram and fill in the blanks: [5 Marks]

[Figure: Diagram of a human sperm showing a rounded head region, a middle piece, and a long, thin flagellum tail.]

Living organisms reproduce to form new individuals of their own kind. This is essential for the survival and continuation of species. Human sperms are microscopic structures that carry genetic material.

Answer:

The head of the sperm has a cap like organelle called (a) Acrosome (Lysosome / Acrosome) which produces an enzyme (b) Hyaluronidase (Hyaluronidase / Amylase) that dissolves the outer layer of the ovum to facilitate fertilisation. The nucleus of the sperm has (c) 23 (23 / 46) chromosomes. The middle piece has (d) Mitochondria (Chloroplast / Mitochondria) to provide energy for the motility of the sperm. (e) Semen (Semen / Hymen) is a mixture of sperms and the fluids produced by the male accessory glands.

Teacher's Note:
a) The acrosome contains enzymes (acrosin and hyaluronidase) that help penetrate the ovum's layers.
b) Sperms are haploid (23 chromosomes), while somatic cells are diploid (46 chromosomes).
c) Mitochondria in the middle piece produce ATP for flagellar movement to reach the ovum.
d) Semen also contains fructose from seminal vesicles, providing energy for sperm motility.

(iii) Choose the odd one out from the following terms and name the category to which the others belong: [5 Marks]

(a) Auxin, Oxytocin, Gibberellin, Cytokinin

Answer: Odd one: Oxytocin. Category: The others are Plant Hormones (or Plant Growth Regulators). Oxytocin is an animal hormone (mammalian hormone).

Teacher's Note:
a) Auxin, gibberellins, and cytokinins are all plant hormones that regulate growth and development.
b) Oxytocin is a mammalian hormone released by the pituitary gland, involved in reproduction and bonding.
c) Animal and plant hormones have different structures and functions.

(b) Growth Hormone, Vasopressin, Thyroid Stimulating Hormone, Gonadotropic Hormone

Answer: Odd one: Vasopressin. Category: The others are Hormones that regulate growth, metabolism, or reproduction.** Vasopressin (ADH) is a hormone that regulates water balance and blood pressure.

Teacher's Note:
a) Growth hormone, TSH, and gonadotropins regulate growth, metabolic rate, and reproductive function.
b) Vasopressin (antidiuretic hormone) controls water reabsorption in the kidneys and blood osmolarity.
c) While all are hormones, vasopressin has a distinctly different physiological role.

(c) Urine, Urea, Uric acid, Nucleic acid

Answer: Odd one: Nucleic acid. Category: The others are Nitrogenous waste products (or excretory wastes). Nucleic acid is a biological macromolecule, not a waste product.

Teacher's Note:
a) Urea and uric acid are nitrogen-containing waste products of metabolism excreted in urine.
b) Urine is the liquid mixture of these wastes and other substances filtered by the kidneys.
c) Nucleic acids (DNA and RNA) are essential molecules, not waste products.

(d) Cervix, Chordae Tendinae, Papillary Muscles, Sinoatrial node

Answer: Odd one: Cervix. Category: The others are Parts of the heart. Cervix is part of the female reproductive system (uterus).

Teacher's Note:
a) Chordae tendinae are fibrous cords that anchor the atrioventricular valves to the ventricular wall.
b) Papillary muscles contract to prevent valve prolapse during systole.
c) Sinoatrial (SA) node is the heart's pacemaker, initiating each heartbeat.
d) The cervix is the lower, narrow part of the uterus connecting to the vagina.

(e) Morula, Blastocyst, Oviduct, Foetus

Answer: Odd one: Oviduct. Category: The others are Stages of embryonic development (or developmental stages). Oviduct is a structure (fallopian tube), not a developmental stage.

Teacher's Note:
a) Morula is an early solid ball of cells formed 3 days after fertilisation.
b) Blastocyst is a hollow sphere formed around day 5, ready for implantation.
c) Foetus is the term for the developing embryo from week 9 onwards.
d) Oviduct (fallopian tube) is the site where fertilisation occurs, but it is not a developmental stage.

(iv) Mohit, a 30-year-old man was a software professional leading a sedentary life. He showed signs of high blood sugar during a routine health check-up despite having a normal body weight. [5 Marks]

[Figure: A photograph of a young man with dark hair and facial hair, sitting at a desk working on a laptop computer. The figure depicts a sedentary, office-based professional lifestyle.]

Answer the following:

(a) The hormonal disorder he is suffering from.

Answer: Type 2 Diabetes Mellitus (or Diabetes Mellitus Type 2, or simply Type 2 Diabetes).

Teacher's Note:
a) Type 2 diabetes is characterised by insulin resistance and/or inadequate insulin production.
b) It often develops in sedentary individuals due to reduced glucose uptake by muscles.
c) Unlike Type 1 diabetes, it is not autoimmune and can often be managed with lifestyle changes and medication.

(b) The hormone responsible for this disorder.

Answer: Insulin (or deficiency/insufficiency of insulin, and/or reduced sensitivity to insulin).

Teacher's Note:
a) Insulin is secreted by the beta cells of pancreatic islets (islets of Langerhans).
b) Insulin resistance means cells do not respond adequately to insulin despite its presence.
c) This leads to elevated blood glucose levels (hyperglycaemia).

(c) The organ that secretes this hormone.

Answer: Pancreas (specifically the islets of Langerhans, or beta cells of the pancreas).

Teacher's Note:
a) The pancreas is both an endocrine gland (islets) and exocrine gland (acini).
b) Beta cells produce insulin; alpha cells produce glucagon for blood glucose regulation.
c) The pancreas is located behind the stomach in the abdominal cavity.

(d) One symptom experienced by Mohit due to this disorder.

Answer: Polydipsia (excessive thirst), Polyuria (excessive urination), Fatigue, Blurred vision, or Slow wound healing.

Teacher's Note:
a) High blood glucose leads to osmotic diuresis, causing excessive urine production (polyuria).
b) Polyuria causes dehydration and excessive thirst (polydipsia).
c) Hyperglycaemia also impairs immune function and wound healing, increases fatigue.

(e) One change in lifestyle to lower the blood sugar level.

Answer: Regular physical exercise, Reduce intake of refined carbohydrates and sugar, Maintain a healthy diet rich in fibre and whole grains, Weight reduction (if overweight), or Reduce sedentary time and increase daily activity.

Teacher's Note:
a) Exercise increases glucose uptake by muscles without requiring insulin (GLUT4 translocation).
b) Dietary changes improve insulin sensitivity and reduce post-prandial blood glucose spikes.
c) Lifestyle modifications are the first-line treatment for Type 2 diabetes.

(v) Study the diagram given below and match the structure with its functions: [5 Marks]

Example: Pelvis – (j)

Structure

Functions

Answer:

Renal Cortex – (a) Has Malpighian capsules
Renal Medulla – (d) Has Henle's loops
Renal Artery – (b) Carries oxygenated blood
Renal Vein – (e) Carries deoxygenated blood
Ureter – (c) Transports urine to urinary bladder
Pelvis – (f) Receives urine which flows into ureter

Teacher's Note:
a) Renal cortex contains glomeruli and Bowman's capsules (Malpighian corpuscles) for ultrafiltration.
b) Renal medulla contains loops of Henle and collecting ducts for selective reabsorption and concentration of urine.
c) Renal blood vessels bring blood to the kidney and drain filtered blood; the ureter and pelvis are the collection and drainage system for urine.

SECTION B (40 Marks)

(Attempt any four questions from this Section.)

Question 3

(i) Which is the resting but metabolically active stage of the cell cycle? [1 Mark]

Answer: G1 phase (Gap 1 phase), or G0 phase (if the cell is not dividing), or the interphase in general.

Teacher's Note:
a) The cell cycle consists of interphase (G1, S, G2) and mitotic phase (M).
b) G1 phase is the "Gap 1" where the cell grows, accumulates nutrients, and synthesises enzymes needed for DNA replication.
c) G0 phase is a quiescent state where cells exit the cell cycle and perform specialised functions.

(ii) Given below is the picture of an eagle. [2 Marks]

[Figure: A black and white photograph of an eagle's head showing distinctive large, forward-facing eyes, a sharp hooked beak, and white feathering on the head with dark feathering on the body.]

Eagles have binocular vision. What is the advantage of such a vision?

Answer: Binocular vision allows eagles to have better depth perception and 3D vision. Both eyes face forward and their visual fields overlap, enabling the brain to calculate distances and judge the exact position and distance of prey. This is a crucial adaptation for a predator that hunts fast-moving prey from great heights. The overlapping fields also provide a wider field of awareness while maintaining accurate directional information.

Teacher's Note:
a) Binocular vision (eyes facing forward with overlapping fields) provides stereoscopic vision and depth perception.
b) This is essential for raptors to pinpoint and dive toward prey moving at high speeds.
c) Monocular vision (eyes on the side) provides a wider field but poor depth perception, suited to prey animals.

(iii) Mention the number of Autosomes and Allosomes in a human body cell. [2 Marks]

Answer: In a human body cell (somatic cell):
- Autosomes: 44 (22 pairs)
- Allosomes (Sex chromosomes): 2 (either XX for females or XY for males)

Teacher's Note:
a) Humans have 46 chromosomes total: 44 autosomes (non-sex chromosomes) + 2 sex chromosomes.
b) Autosomes are homologous pairs (22 pairs); allosomes determine biological sex.
c) Gametes (sperm and eggs) have 23 chromosomes: 22 autosomes + 1 sex chromosome.

(iv) Two well-watered, identical plants were placed in brightly lit rooms at different temperatures – one at 15°C and the other at 38°C. The plant in the warmer room showed wilting by the end of the day. [2 Marks]

(a) Which plant phenomenon resulted in the wilting of the leaves?

Answer: Transpiration (Excessive transpiration or Wilting due to high transpiration rate).

Teacher's Note:
a) Higher temperature increases the rate of transpiration from leaves.
b) When transpiration exceeds water uptake, the plant loses turgor and wilts temporarily.
c) This is visible wilting, which can be reversed if the plant is watered; it does not indicate permanent damage.

(b) Mention the factor of the phenomenon that is being tested.

Answer: Temperature (or Heat / Thermal energy).

Teacher's Note:
a) The independent variable being tested is temperature (difference of 15°C vs. 38°C).
b) Temperature increases molecular movement, leading to increased evaporation from leaf surfaces.
c) This is a classic experiment demonstrating the effect of temperature on transpiration rate.

(v) Draw a neat, labelled diagram of an animal cell showing the Prophase stage of Mitosis with four chromosomes. [3 Marks]

Answer:

[Figure: A diagram of a cell in prophase of mitosis. The nucleus is breaking down. The centrioles are visible at the opposite poles of the cell, with spindle fibres beginning to extend between them. Four chromosomes (each consisting of two sister chromatids joined at the centromere) are visible condensed in the centre of the cell. The nuclear envelope is fragmenting. Labels include: Centriole, Spindle fibre, Chromosome (with sister chromatids), Centromere, and Fragmenting nuclear envelope.]

Teacher's Note:
a) Prophase is the first stage of mitosis where chromatin condenses into visible chromosomes.
b) Centrioles move to opposite poles and form the spindle apparatus; the nuclear envelope breaks down.
c) Each chromosome consists of two sister chromatids joined at the centromere; the diagram should show this clearly.

Question 4

(i) What is the scientific name of modern man? [1 Mark]

Answer: Homo sapiens.

Teacher's Note:
a) Modern humans are classified as Homo sapiens, belonging to the genus Homo.
b) Homo sapiens evolved approximately 300,000 years ago in Africa.
c) The binomial name reflects the genus (Homo) and species (sapiens).

(ii) How are the Cytons and Axons of neurons arranged in the following? [2 Marks]

(a) Cerebrum

Answer: In the cerebrum, the cytons (cell bodies) are arranged in the outer layer called the cerebral cortex or grey matter. The axons are arranged in the inner region called the white matter (medullary substance). The grey matter forms a convoluted outer layer (cortex) with folds called gyri and sulci, while white matter lies beneath.

Teacher's Note:
a) Grey matter (cortex) consists of neuron cell bodies, synapses, and dendrites; it processes information.
b) White matter consists of myelinated axons that transmit signals between brain regions and the spinal cord.
c) The cerebral cortex is the outermost layer of the brain responsible for higher functions.

(b) Spinal Cord

Answer: In the spinal cord, the cytons (cell bodies) are arranged in the inner region in a butterfly or H-shaped pattern called grey matter (grey substance). The axons are arranged in the outer region called white matter. The grey matter is central and composed of neuron bodies and synapses, while the white matter surrounds it and consists of myelinated axons carrying signals to and from the brain.

Teacher's Note:
a) The arrangement in the spinal cord is opposite to the brain: grey matter is central, white matter is peripheral.
b) The grey matter (butterfly-shaped in cross-section) contains cell bodies of motor and sensory neurons.
c) The white matter contains ascending (sensory) and descending (motor) nerve tracts.

(iii) (a) Who proposed the theory of Natural Selection? [2 Marks]

Answer: Charles Darwin (English naturalist, 1809 – 1882).

Teacher's Note:
a) Darwin proposed the theory of evolution by natural selection in 1859 in "On the Origin of Species".
b) Natural selection is the mechanism by which organisms with favourable traits survive and reproduce.
c) This theory explains the diversity of life and adaptation to environments.

(b) Name the organism which was used as an example to explain Industrial Melanism.

Answer: Peppered moth (Biston betularia).

Teacher's Note:
a) Industrial melanism is the evolution of dark-coloured (melanic) forms in response to environmental change.
b) In pre-industrial England, peppered moths were mostly light-coloured; during the Industrial Revolution, dark forms became more common as tree trunks were darkened by pollution.
c) Darker moths had better camouflage on polluted trees and survived better, increasing in frequency — a classic example of natural selection in action.

(iv) Differentiate between Plasmolysis and Deplasmolysis. [2 Marks]

Answer:

Teacher's Note:
a) Both processes demonstrate osmosis — the movement of water across the cell membrane.
b) Plasmolysis shows that the protoplasm is a living, selectively permeable layer separate from the cell wall.
c) Deplasmolysis proves that plasmolysis is reversible and the cell is alive.

(v) Akshay's father had a tumour in his prostate gland. His doctor advised him to get it removed surgically. One side effect of the surgery was incontinence of urine, i.e. leakage of urine from the urinary bladder. [3 Marks]

[Figure: A diagram showing the anatomy of the male urinary and reproductive system in cross-section. The bladder is shown as a rounded structure at the top, with labels indicating the bladder, the enlarged prostate gland at its base, and the urethra extending downward. The diagram illustrates the anatomical relationship between the prostate and urethra.]

(a) Where is the prostate gland located?

Answer: The prostate gland is located at the base of the urinary bladder, surrounding the urethra as it exits the bladder. It is positioned below the bladder and in front of the rectum, in the pelvic region.

Teacher's Note:
a) The prostate surrounds the urethra, so its enlargement or surgical removal can affect urinary control.
b) It is about the size and shape of a walnut in healthy adult males.
c) The anatomical proximity explains why prostate surgery can cause urinary incontinence as a side effect.

(b) Why does the prostate gland produce an alkaline secretion?

Answer: The prostate gland produces an alkaline secretion to neutralise the acidic environment of the vagina. Sperms are sensitive to acidic conditions, and the alkaline seminal fluid protects them from vaginal acidity, increasing sperm viability and motility. This alkaline environment is essential for successful fertilisation.

Teacher's Note:
a) The vaginal environment is acidic (pH 4.5 – 5.5) due to lactobacilli, which are hostile to sperm survival.
b) Seminal fluid (containing secretions from the prostate, seminal vesicles, and other glands) has a pH of 7.2 – 7.8.
c) The alkaline secretion also provides nutrients (fructose) and contains enzymes and antibiotic compounds to support and protect sperm.

(c) Name the structure that regulates the flow of urine from the urinary bladder into the urethra.

Answer: Internal urethral sphincter (or Urethral sphincter, or Sphincter vesicae).

Teacher's Note:
a) The internal urethral sphincter is composed of smooth muscle and is under involuntary (autonomic) control.
b) It relaxes during the micturition reflex to allow urine to flow into the urethra.
c) Damage to this sphincter during prostate surgery explains the incontinence side effect.

Question 5

(i) RBCs do not have nuclei. Discuss its advantage. [1 Mark]

Answer: The absence of a nucleus in mammalian RBCs provides several advantages: (1) It increases the space available for haemoglobin storage, allowing more oxygen to be transported per cell. (2) RBCs can squeeze through tiny capillaries more easily due to their biconcave shape and flexibility, without the rigidity that a nucleus would impose. (3) The lack of a nucleus allows mature RBCs to have a longer lifespan in circulation without the metabolic burden of maintaining a nucleus and DNA. (4) More RBCs can fit into the blood, increasing oxygen-carrying capacity. These adaptations make mammalian RBCs highly efficient oxygen transporters.

Teacher's Note:
a) Mature mammalian RBCs lack nuclei and organelles, unlike RBCs of birds, reptiles, and fish.
b) The biconcave shape and flexibility are crucial for navigating capillaries and maximising surface area for gas exchange.
c) RBCs rely on glycolysis (anaerobic respiration) for ATP, since they lack mitochondria.

(ii) Arrange the following food chains in a proper sequence. [2 Marks]

(a) Small fish, Algae, Mosquito larvae, Kingfisher

Answer: Algae → Mosquito larvae → Small fish → Kingfisher

Teacher's Note:
a) Algae are the primary producer (autotroph).
b) Mosquito larvae and small fish are consumers feeding on algae and smaller organisms.
c) Kingfisher is the apex predator, feeding on small fish.

(b) Frog, Snail, Crow, Green leaves

Answer: Green leaves → Snail → Frog → Crow

Teacher's Note:
a) Green leaves are the primary producer (autotroph).
b) Snail is an herbivore feeding on leaves.
c) Frog is a carnivore feeding on snails (and insects).
d) Crow is an omnivore/predator that feeds on frogs and other animals.

(iii) A 28-year-old pregnant lady goes to a gynaecologist for a check-up. Her doctor explains that there is normal growth of the foetus and the placenta is functioning well. [2 Marks]

(a) Mention one function of the placenta.

Answer: One function of the placenta is to provide oxygen and nutrients (glucose, amino acids, vitamins, minerals) to the foetus from the mother's blood while removing carbon dioxide and nitrogenous wastes from the foetus. Other functions include secreting hormones (progesterone and human chorionic gonadotropin) to maintain pregnancy, and providing antibodies (IgG) to the foetus for passive immunity.

Teacher's Note:
a) The placenta is the organ of exchange between mother and foetus, separated by a semipermeable membrane.
b) It provides nutrition, gas exchange, excretion, and hormonal support for foetal development.
c) It acts as a barrier against some pathogens but allows the passage of some infections (like rubella virus) and antibodies.

(b) What connects the placenta to the foetus?

Answer: The umbilical cord (or Umbilicus) connects the placenta to the foetus. The umbilical cord contains two umbilical arteries and one umbilical vein, which carry deoxygenated blood and nutrients from the foetus to the placenta, and oxygenated blood and nutrients from the placenta to the foetus, respectively.

Teacher's Note:
a) The umbilical cord is typically 50 – 60 cm long and contains three blood vessels (two arteries, one vein).
b) The umbilical vein carries oxygenated blood to the foetus; the arteries carry deoxygenated blood to the placenta.
c) After birth, the umbilical cord is clamped and cut, leaving a scar that becomes the belly button (navel).

(iv) Mention any two secondary sexual characteristics in a 15-year-old boy. [2 Marks]

Answer: Any two of the following: Growth of facial hair and beard, Growth of body hair (axillary and pubic hair), Deepening of voice, Growth of larynx (Adam's apple), Increased muscle mass and broadening of shoulders, Development of the reproductive organs (testes and penis), Increased sebaceous gland activity (leading to acne), Growth spurt and increase in height, or Development of sweat glands.

Teacher's Note:
a) Secondary sexual characteristics are triggered by the hormone testosterone in males.
b) These characteristics develop during puberty (typically ages 10 – 16 in boys) and distinguish males from females.
c) They are different from primary sexual characteristics, which are the organs involved directly in reproduction.

(v) Tara's grandmother is 70 years old and has a passion for embroidery. She faces difficulty in threading the needle as the eye of the needle appears blurred. The ophthalmologist diagnosed it as an age-related disorder. [3 Marks]

[Figure: A political map of India showing its state and union territory boundaries in different shades of grey and black. The map includes state labels, major geographical features, and coastal outlines.]

(a) Name the eye disorder she is suffering from.

Answer: Presbyopia (or Age-related presbyopia, or Presbyopic vision).

Teacher's Note:
a) Presbyopia is the age-related loss of the lens's ability to accommodate (change focus) for near vision.
b) It occurs because the lens loses elasticity with age, making it difficult to change shape for focusing on near objects.
c) It typically begins after age 40 and becomes more pronounced with age.

(b) How can the above defect be corrected?

Answer: Presbyopia can be corrected by using bifocals (glasses with two focal lengths — one for distance vision and one for near vision), progressive lenses (multifocal lenses with a gradual change in power), reading glasses (convex lenses for near vision), or contact lenses. Surgical options include LASIK or implantation of multifocal intraocular lenses.

Teacher's Note:
a) Bifocals and progressive lenses are the most common correction for presbyopia in older adults.
b) The near-vision portion is typically a convex lens (+1.0 to +3.5 diopters) to assist accommodation.
c) Some people use separate reading glasses for convenience.

(c) Where is the image formed in the above disorder?

Answer: In presbyopia, the image of near objects is formed behind the retina (beyond the retina), rather than directly on the retina. This is because the lens cannot curve sufficiently to refract light rays strongly enough to bring near objects into focus on the retinal surface. Only distant objects (infinity) form sharp images on the retina without accommodation.

Teacher's Note:
a) This is a case of hypermetropia (farsightedness) specifically related to age and loss of accommodation.
b) The far point of accommodation increases with age, making near vision progressively more difficult.
c) Convex lenses converge light rays before they enter the eye, shifting the image forward onto the retina.

Question 6

(i) Explain the term 'Population Density' with reference to human beings. [1 Mark]

Answer: Population density refers to the number of individuals of a population per unit area or per unit volume. With reference to human beings, it is the number of people per square kilometre (or per square mile). For example, if a city has a population of 10,000,000 and covers an area of 500 square kilometres, the population density is 20,000 people per square kilometre. Population density varies greatly among countries and regions, influenced by factors like geography, climate, economic development, and availability of resources.

Teacher's Note:
a) Population density = Total population ÷ Total area.
b) High population density areas include cities; low density areas include deserts and mountains.
c) Population density is an important indicator of resource stress and environmental impact.

(ii) A 17-year-old girl was having irregular menstrual cycle. Her mother took her to their family physician. She was diagnosed with Adrenal Virilism. Study the picture given below and answer the following questions. [2 Marks]

[Figure: A political map of a large geographical region (appears to be South Asia or a similar area) showing different territories or regions marked in varying shades of grey and black, with clear boundary lines demarcating different administrative divisions.]

(a) Hypersecretion of which hormone results in Adrenal Virilism in human females?

Answer: Androgen (or Testosterone, or Dehydroepiandrosterone (DHEA), or Other androgens secreted by the adrenal cortex).

Teacher's Note:
a) Adrenal virilism (adrenogenital syndrome) is caused by excessive androgen production in the adrenal cortex.
b) The adrenal cortex normally produces small amounts of androgens; in this disorder, production is excessive.
c) This leads to virilisation (development of male characteristics) in females, including facial hair, deepening voice, and irregular menstruation.

(b) Mention one symptom of this disorder.

Answer: One symptom is Irregular menstrual cycles (or Amenorrhoea = absence of menstruation). Other symptoms include Growth of facial and body hair (hirsutism), Deepening of voice, Male-pattern baldness, Increased muscle mass, Clitoral enlargement, or Acne.

Teacher's Note:
a) The excessive androgens suppress normal ovulation and oestrogen-progesterone cycles, causing menstrual irregularities.
b) The virilising effects (facial hair, voice changes) occur due to the androgenic activity of the excess hormone.
c) Early diagnosis and treatment with corticosteroids can prevent or reverse these symptoms.

(iii) Differentiate between Mitosis in plant cell and animal cell based on Cytokinesis. [2 Marks]

Answer:

Teacher's Note:
a) Both processes result in the division of cytoplasm; the key difference is the mechanism and the formation of a cell wall.
b) The cell wall in plants necessitates a different cytokinesis mechanism than in animals, which lack cell walls.
c) These adaptations reflect the structural differences between plant and animal cells.

(iv) Sara placed a healthy potted plant in a dark room for 48 hours to perform an experiment on photosynthesis. She plucked one of the leaves and tested it for starch. The leaf did not turn blue-black on adding Iodine solution. [2 Marks]

(a) Why was the plant placed in the dark for 48 hours?

Answer: The plant was placed in the dark for 48 hours to deplete all the starch that was previously synthesised and stored in the leaves. In darkness, photosynthesis cannot occur, so the plant respires and consumes the stored starch for energy. After 48 hours, all the starch in the leaves is used up, ensuring that any starch found after the experiment is only due to the photosynthesis that occurred during the experimental period, not from pre-existing stores.

Teacher's Note:
a) This is a control step to eliminate confounding variables in starch testing.
b) Starch is the end product of photosynthesis stored temporarily in leaves; it must be removed first.
c) This is a standard procedure in starch experiments to demonstrate photosynthesis clearly.

(b) What is the significance of boiling the leaf in alcohol during the starch test?

Answer: Boiling the leaf in alcohol serves two purposes: (1) It removes the chlorophyll (the green pigment) from the leaf, making it colourless. This is essential because chlorophyll would mask or interfere with the colour change (blue-black) that occurs when iodine reacts with starch, making the result difficult to observe. (2) Alcohol (ethanol) is used as a solvent because it dissolves chlorophyll efficiently at high temperatures. Without this decolouration step, the presence or absence of starch cannot be clearly determined by the iodine test.

Teacher's Note:
a) Chlorophyll is lipid-soluble and is removed by alcohol heating.
b) The iodine-starch complex forms a blue-black colour only in the absence of chlorophyll masking.
c) This is a critical step in many leaf experiments involving starch detection.

(v) Copy the diagram given below. [3 Marks]

[Figure: A simple line drawing of a cross-section of the spinal cord showing an oval shape with a central butterfly or H-shaped region (grey matter) surrounded by a larger region (white matter). The diagram is unlabelled.]

(a) Name the structure.

Answer: Spinal cord (or Cross-section of spinal cord, or Transverse section of spinal cord).

Teacher's Note:
a) The spinal cord is part of the central nervous system, extending from the medulla oblongata down the vertebral canal.
b) It serves as the main pathway for communication between the brain and the peripheral nervous system.
c) The characteristic butterfly or H-shaped grey matter region contains neuron cell bodies and synapses.

(b) Label Gray matter and White matter.

Answer:

[Figure: A line drawing of the cross-section of spinal cord with two labels: "Gray matter" pointing to the central butterfly/H-shaped region, and "White matter" pointing to the surrounding outer region.]

Teacher's Note:
a) Grey matter (grey substance) in the spinal cord is centrally located and shaped like a butterfly or letter H in cross-section.
b) Grey matter contains neuron cell bodies, dendrites, synapses, and unmyelinated axons.
c) White matter surrounds the grey matter and consists of myelinated axons that conduct impulses between the brain and periphery.
d) The dorsal horn of grey matter processes sensory information; the ventral horn contains motor neuron cell bodies.

Question 7

(i) Write the overall chemical equation for photosynthesis. [1 Mark]

Answer: \( 6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{Light}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \)

Teacher's Note:
a) This is the overall equation showing the conversion of carbon dioxide and water into glucose and oxygen in the presence of light and chlorophyll.
b) The equation does not show the intermediate products (ATP, NADPH, etc.) or the two main stages (light reactions and Calvin cycle).
c) The photosynthetic efficiency in nature is about 3 – 6% of incident light energy.

(ii) Expand the abbreviations: [2 Marks]

(a) NADP

Answer: Nicotinamide Adenine Dinucleotide Phosphate.

Teacher's Note:
a) NADP+ is an electron carrier in photosynthesis that becomes reduced to NADPH in the light reactions.
b) NADPH carries electrons to the Calvin cycle to reduce CO₂ to glucose.
c) NADP+ is regenerated when NADPH is oxidised in the Calvin cycle.

(b) ADP

Answer: Adenosine Diphosphate.

Teacher's Note:
a) ADP is the product of ATP hydrolysis, releasing energy for cellular work.
b) ADP is regenerated to ATP in the light reactions of photosynthesis by chemiosmosis.
c) In aerobic respiration, ADP is also regenerated to ATP.

(iii) (a) Explain the term Synapse. [2 Marks]

Answer: A synapse is the structural and functional junction between two neurons (or between a neuron and an effector cell) where nerve impulses are transmitted from one neuron to another. It consists of three main parts: (1) the presynaptic membrane (the axon terminal of the transmitting neuron), (2) the synaptic cleft (a narrow gap of about 20 – 40 nanometres), and (3) the postsynaptic membrane (the membrane of the receiving neuron, usually on its dendrite or cell body). Communication across the synapse is chemical, involving neurotransmitters released by the axon terminal and binding to receptors on the postsynaptic membrane, triggering depolarisation or hyperpolarisation of the receiving neuron.

Teacher's Note:
a) Synapses are one-directional: transmission occurs from the presynaptic to the postsynaptic neuron only.
b) Synaptic transmission is slower than nerve conduction along axons due to the time needed for neurotransmitter release and binding.
c) Synapses are sites of neuroplasticity and learning; repeated stimulation can strengthen or weaken synaptic transmission (synaptic plasticity).

(b) Name the neurotransmitter that allows the transmission of impulses across the synapse.

Answer: Acetylcholine (at the neuromuscular junction and parasympathetic synapses), or Noradrenaline (norepinephrine) (at sympathetic synapses), or Dopamine, Serotonin, Glutamate, GABA (Gamma-Aminobutyric Acid), or other neurotransmitters depending on the synapse. (In the exam, any of these correct examples is acceptable; acetylcholine is the most commonly cited example.)

Teacher's Note:
a) Neurotransmitters are small molecules released from synaptic vesicles in response to an action potential.
b) They bind to specific receptors on the postsynaptic membrane, causing either excitation (depolarisation) or inhibition (hyperpolarisation).
c) Different neurotransmitters mediate different types of synaptic transmission in different regions of the nervous system.

(iv) What is the role of the following? [2 Marks]

(a) Leydig cells

Answer: Leydig cells (also called interstitial cells) are located in the interstitial tissue between the seminiferous tubules of the testes. Their role is to produce and secrete testosterone, the male sex hormone. Testosterone is responsible for the development and maintenance of male secondary sexual characteristics (facial hair, voice deepening, muscle development), the production and maturation of sperm, and the development of the male reproductive organs. Testosterone also plays a role in bone and muscle growth and metabolic processes.

Teacher's Note:
a) Leydig cells are stimulated by Luteinising Hormone (LH) from the pituitary gland to produce testosterone.
b) Testosterone is essential for spermatogenesis (sperm production) in the seminiferous tubules.
c) Testosterone levels are highest during adolescence and gradually decline with age in males.

(b) Seminiferous tubules

Answer: Seminiferous tubules are the functional units of the testes where spermatogenesis (sperm production) occurs. These long, coiled tubules are lined with germinal epithelium containing spermatogenic cells at various stages of development. The seminiferous tubules are the sites of meiosis, where haploid spermatozoa are produced from diploid spermatogonia. The mature sperm are released into the lumen of the tubules and transported through the epididymis for maturation, storage, and eventual ejaculation.

Teacher's Note:
a) The seminiferous tubules also contain Sertoli cells, which provide nutritional and mechanical support to developing sperm.
b) Spermatogenesis is stimulated by Follicle-Stimulating Hormone (FSH) from the pituitary gland.
c) The entire process of spermatogenesis takes approximately 74 days in humans.

(v) "Vanishing Greenery; A Growing Urban Crisis" [3 Marks]

In most of our cities, rapid urbanisation has led to a significant decrease in greenery over the past two decades. As the population increased, the demand for housing, roads and commercial buildings grew, leading to clearing of parks, gardens and natural resources. This has contributed to several problems.

[Figure: A black and white photograph of a dense urban landscape showing numerous high-rise buildings, multi-storied apartments, roads, and very little visible vegetation or green space among the concrete structures.]

(a) Mention one significant problem caused by the reduction in urban greenery.

Answer: One significant problem is increased air pollution and reduced air quality. Green plants remove carbon dioxide and produce oxygen through photosynthesis, and they also filter particulate matter and other pollutants from the air. The reduction in greenery decreases air purification, leading to higher concentrations of CO₂, smog, and other pollutants. This causes respiratory problems, asthma, and other health issues in urban populations. Other significant problems include: increased urban heat island effect (higher temperatures due to loss of shade and evapotranspiration), flooding and waterlogging (loss of vegetation increases runoff), loss of biodiversity, reduced recreational and mental health benefits, or groundwater depletion.

Teacher's Note:
a) The loss of green space has direct impacts on air quality, temperature, water cycles, and human health.
b) A single tree can remove 20 – 48 pounds of air pollutants per year, so mass removal of trees is significant.
c) Urban heat island effects can increase local temperatures by 2 – 5°C compared to surrounding rural areas.

(b) How do green plants contribute to improving the air quality?

Answer: Green plants improve air quality through multiple mechanisms: (1) Photosynthesis: Plants absorb carbon dioxide (CO₂) and release oxygen (O₂), directly improving the O₂-CO₂ balance in the air. (2) Phytofiltration: The leaves and roots of plants filter and trap particulate matter, dust, and other pollutants from the air and soil. (3) Pollutant absorption: Plants can absorb and metabolise various gaseous pollutants such as nitrogen dioxide (NO₂), sulphur dioxide (SO₂), and volatile organic compounds (VOCs). (4) Evapotranspiration: The release of water vapour from leaves increases humidity and helps disperse pollutants, reducing their concentration near ground level. Together, these processes significantly reduce air pollution and improve urban air quality.

Teacher's Note:
a) Plants are natural air purifiers; large-scale tree planting is a cost-effective air quality improvement strategy.
b) Trees also reduce noise pollution by absorbing sound and providing a physical barrier.
c) Different plants are effective against different pollutants; species selection matters for maximum benefit.

(c) What role can you, as a citizen, play in protecting urban greenery?

Answer: As a citizen, I can play several roles in protecting urban greenery: (1) Plant trees and maintain a garden or potted plants at home, balcony, or terrace to increase green cover. (2) Participate in community tree-planting drives and environmental awareness campaigns. (3) Support and advocate for the preservation of existing parks, green spaces, and natural areas in my city through citizen groups or municipal councils. (4) Reduce my personal carbon footprint by using public transport, cycling, or walking, which reduces demand for new roads and building construction that destroy green spaces. (5) Educate friends and family about the importance of greenery and environmental conservation. (6) Volunteer with environmental organisations to maintain and restore green spaces. (7) Responsibly manage waste to reduce landfill pressure that leads to deforestation. (8) Support policies and regulations that protect green spaces and mandate green building standards. (9) Create green roofs or walls in urban areas using vertical gardening techniques. These collective actions contribute to preserving and expanding urban greenery, improving environmental and public health.

Teacher's Note:
a) Individual actions, when multiplied across a population, create significant environmental impact.
b) Urban greening initiatives like rooftop gardens and green walls are innovative solutions for space-limited cities.
c) Environmental citizenship and activism are essential for sustainable urban development.

Question 8

(i) Write the term for the pressure exerted by the cell contents on the cell wall. [1 Mark]

Answer: Turgor pressure (or Turgidity, or Turgor).

Teacher's Note:
a) Turgor pressure is the hydrostatic pressure exerted by the contents of the cell (vacuole and cytoplasm) against the cell wall.
b) It is maintained by the osmotic uptake of water into the vacuole when the cell is in a hypotonic solution.
c) Turgor pressure keeps plants rigid and upright; loss of turgor (plasmolysis) causes wilting.

(ii) (a) Name the fluid present between the meninges in spinal cord. [2 Marks]

Answer: Cerebrospinal fluid (CSF) (or Cerebro-spinal fluid).

Teacher's Note:
a) The CSF fills the space between the pia mater and the arachnoid membrane (the subarachnoid space).
b) CSF is continuously produced by the choroid plexuses in the ventricles of the brain and reabsorbed by the arachnoid granulations.
c) CSF circulates around the brain and spinal cord, bathing the neural tissue.

(b) What is its function?

Answer: The cerebrospinal fluid (CSF) has several important functions: (1) Protection: It cushions and protects the brain and spinal cord from mechanical shocks and impacts. (2) Nutrient and waste transport: It carries nutrients (glucose, amino acids) to the neural tissue and removes metabolic waste products (lactate, urea) from the brain and spinal cord. (3) Pressure regulation: It helps maintain a constant pressure within the cranial and spinal cavities, protecting the neural tissue from sudden pressure changes. (4) Immune function: CSF contains antibodies and white blood cells that provide immune defence for the central nervous system. (5) Hormonal transport: It may carry hormones and other signalling molecules that affect neural function. The normal volume of CSF is about 140 mL in the central nervous system, and it is constantly circulated and replenished.

Teacher's Note:
a) CSF is similar in composition to blood plasma but contains less protein and no red blood cells.
b) Increased intracranial pressure (from excess CSF or bleeding) can damage neural tissue and is a medical emergency.
c) Sampling CSF by lumbar puncture (spinal tap) is used diagnostically to detect infections like meningitis.

(iii) Given below is the diagram of a turgid plant cell. [2 Marks]

[Figure: A line drawing of a plant cell showing a rectangular cell wall (outer boundary) with a cell membrane (inner boundary) tightly pressed against the cell wall. The interior is filled with a large central vacuole and surrounding cytoplasm containing various organelles. The cell wall is rigid and angular, and the protoplasm is pressed against it.]

Copy the diagram and label Vacuole and Plasma membrane.

Answer:

[Figure: A line drawing of a turgid plant cell with labels: "Cell wall" pointing to the outer rigid boundary, "Plasma membrane (Cell membrane)" pointing to the inner boundary separating the protoplasm from the vacuole, and "Vacuole" pointing to the large central space filled with cell sap. The diagram shows the protoplasm (cytoplasm + nucleus) pressed tightly against the cell wall due to turgor pressure.]

Teacher's Note:
a) The plasma membrane (cell membrane) is selectively permeable, allowing water and solutes to pass selectively.
b) In a turgid cell, the vacuole is fully distended with water, maintaining maximum turgor pressure.
c) The cell wall prevents the cell from bursting due to osmotic water uptake, unlike animal cells which lyse under hypotonic conditions.

(iv) Select and write the two biodegradable wastes from the given list: [2 Marks]
Styrofoam, Metallic cans, Decaying fruits, Plastic bottles, Newspapers

Answer: The two biodegradable wastes are:
1. Decaying fruits (organic waste that decomposes naturally into soil)
2. Newspapers (paper product made from cellulose, which is biodegradable)

Teacher's Note:
a) Biodegradable wastes are those that can be broken down by microorganisms into simpler, non-toxic substances.
b) Styrofoam is non-biodegradable (synthetic polymer, persists for 500+ years); metallic cans are non-biodegradable (metal); plastic bottles are non-biodegradable (synthetic polymer).
c) Composting is an effective method for managing biodegradable waste like fruits and paper.

(v) Rajat Singh was working as a supervisor in a stone quarry where rock, sand and gravel are extracted by techniques like digging, drilling and blasting. As the years rolled by, Rajat started facing a loss in hearing. The high decibel sounds had damaged a part of his internal ear, though the tympanic membrane was intact. [3 Marks]

[Figure: A photograph showing a stone quarry or mining site with a tall drilling rig or extraction equipment standing in the middle of excavated terrain with piles of extracted rock and sand visible in the background.]

(a) Give the collective term for the structure located in the internal ear.

Answer: Labyrinth (or Membranous labyrinth, or Inner ear labyrinth). The labyrinth consists of the semicircular canals, the vestibule, and the cochlea.

Teacher's Note:
a) The labyrinth is the fluid-filled system within the petrous part of the temporal bone.
b) The labyrinth serves both auditory (hearing via cochlea) and vestibular (balance via semicircular canals) functions.
c) The perilymph and endolymph are the fluids within the labyrinth that transmit sound vibrations and support balance.

(b) Name the sensory organ in the Cochlea which was damaged for Rajat Singh.

Answer: Organ of Corti (or Corti's organ, or the sensory epithelium of the cochlea).

Teacher's Note:
a) The organ of Corti contains hair cells (sensory receptors) that transduce sound vibrations into nerve impulses.
b) The hair cells have cilia that bend in response to vibrations in the cochlear fluid, triggering action potentials in the vestibulocochlear nerve.
c) Prolonged exposure to loud noise (above 85 decibels) damages hair cells, causing sensorineural hearing loss that is often permanent.

(c) What kind of pollution do the workers face in the stone quarry?

Answer: The workers face multiple types of pollution: (1) Noise pollution: The high-decibel sounds from drilling, blasting, and machinery (often 100 – 130 decibels) cause hearing damage and noise-induced hearing loss. (2) Air pollution: Dust particles from drilling and blasting create suspended particulate matter (silica dust), causing respiratory problems, silicosis, and other lung diseases. (3) Vibration pollution: The vibrations from heavy machinery and blasting can cause health problems including hand-arm vibration syndrome. (4) Environmental pollution: The quarrying process destroys ecosystems, pollutes water sources, and contributes to soil degradation. (5) Visual pollution: The landscape is heavily altered and scarred by extraction activities.

Teacher's Note:
a) Occupational health hazards in mining and quarrying are significant; workers require protective equipment (ear protection, respiratory masks, safety gear).
b) The permanent hearing loss suffered by Rajat Singh is a common occupational disease among quarry workers (occupational noise-induced hearing loss).
c) Regulatory standards and worker protection measures are essential to minimise these health risks.