ICSE Class 10 Biology Board Exam Question Paper 2009 with Solutions

SECTION-I (40 Marks)

(Attempt all questions from this Section.)

Question 1.

(a) Name the following:

(i) The statistical study of the human population of a region. [1 Mark]

Answer: Demography.

Teacher's Note:
a) Demography is the scientific study of human populations, including their size, structure, distribution, and changes over time.
b) It involves analyzing birth rates, death rates, migration, and aging.
c) This field is crucial for understanding societal trends and planning public services.
d) Students should remember this specific term for the statistical study of populations.

(ii) The biological term given to the protective membranes of the brain. [1 Mark]

Answer: Meninges.

Teacher's Note:
a) The meninges are three layers of protective tissue that surround the brain and spinal cord.
b) These layers are the dura mater, arachnoid mater, and pia mater.
c) Their primary function is to protect the central nervous system from physical impact and infection.
d) Knowing the specific biological term for these structures is important.

(iii) The photosensitive pigment present in the rod cells of the retina. [1 Mark]

Answer: Rhodopsin.

Teacher's Note:
a) Rhodopsin is a light-sensitive receptor protein found in the rod cells of the retina.
b) It is responsible for vision in low light conditions (scotopic vision).
c) Rod cells do not detect color but are crucial for peripheral vision and night vision.
d) Students should differentiate between rhodopsin in rods and photopsins in cone cells.

(iv) The cell organelle responsible for photosynthesis. [1 Mark]

Answer: Chloroplast.

Teacher's Note:
a) Chloroplasts are organelles found in plant cells and other eukaryotic organisms that conduct photosynthesis.
b) They contain chlorophyll, the green pigment that absorbs sunlight.
c) The light-dependent reactions occur in the thylakoid membranes, and the light-independent reactions (Calvin cycle) occur in the stroma.
d) This is a fundamental concept in plant biology.

(v) The internal layer of the eye which prevents reflection of light. [1 Mark]

Answer: Choroid.

Teacher's Note:
a) The choroid is a vascular layer of the eye, containing connective tissue, and lying between the retina and the sclera.
b) It is rich in blood vessels and contains dark pigments (melanin) that absorb stray light.
c) This absorption prevents internal reflection, ensuring clear vision.
d) Students should understand the role of the choroid in maintaining visual clarity.

(b) State whether the following statements are true or false. If false, write the correct form of the statement by changing the first or last word only.

(i) The resting stage in mitosis is called interphase. [1 Mark]

Answer: False, "The resting stage in cell division is called interphase."

Teacher's Note:
a) Interphase is the stage of the cell cycle where the cell grows and DNA is replicated, preparing for cell division.
b) Mitosis itself is the process of nuclear division, which follows interphase.
c) The statement is false because interphase is a stage of the overall cell cycle, not specifically mitosis.
d) Students should clearly distinguish between the cell cycle and mitosis.

(ii) Photosynthesis occurs in all the cells of the plant. [1 Mark]

Answer: False, "Photosynthesis occurs in all the green cells of the plant."

Teacher's Note:
a) Photosynthesis primarily occurs in cells containing chloroplasts, which are typically found in the green parts of a plant.
b) Cells in roots, woody stems, and other non-green parts do not perform photosynthesis.
c) These non-photosynthetic cells rely on the products of photosynthesis transported from green parts.
d) Emphasize that the presence of chlorophyll is key for photosynthesis.

(iii) The pituitary gland is both exocrine and endocrine in function. [1 Mark]

Answer: False, "The pancreas gland is both exocrine and endocrine in function."

Teacher's Note:
a) The pituitary gland is purely an endocrine gland, secreting hormones directly into the bloodstream.
b) The pancreas is a mixed gland, with endocrine functions (insulin, glucagon) and exocrine functions (digestive enzymes).
c) Students should know examples of both pure endocrine glands and mixed glands.
d) This question tests the understanding of gland types and their functions.

(iv) Chemicals applied to spots and places to kill harmful microorganism are called disinfectants. [1 Mark]

Answer: True.

Teacher's Note:
a) Disinfectants are antimicrobial agents applied to non-living objects to destroy microorganisms.
b) They are typically too harsh for living tissues.
c) Antiseptics, in contrast, are applied to living tissues.
d) Understanding the distinction between disinfectants and antiseptics is important.

(v) All voluntary actions are controlled by the cerebellum. [1 Mark]

Answer: False, "All voluntary actions are controlled by the cerebrum."

Teacher's Note:
a) The cerebrum is the largest part of the brain and is responsible for voluntary actions, thought, memory, and sensory processing.
b) The cerebellum primarily coordinates voluntary movements, balance, and posture, but does not initiate them.
c) Involuntary actions are controlled by the brainstem and spinal cord.
d) Students should clearly differentiate the functions of the cerebrum and cerebellum.

(c) Give the specific function of the following structures found in the body of plants/animals :

(i) Hydathodes [1 Mark]

Answer: Hydathodes bring about the loss of water in the form of water droplets (guttation).

Teacher's Note:
a) Hydathodes are specialized pores, typically found on the margins or tips of leaves.
b) They are involved in guttation, the exudation of xylem sap from the tips or edges of leaves.
c) Guttation occurs when transpiration is low and root pressure is high.
d) This process is distinct from transpiration, which involves water loss as vapor.

(ii) Centrosome [1 Mark]

Answer: Centrosome initiates cell division in animal cells by forming spindle fibers.

Teacher's Note:
a) The centrosome is a microtubule-organizing center in animal cells.
b) It duplicates during interphase and organizes the mitotic spindle during cell division.
c) Plant cells do not have centrosomes but organize their spindle fibers from other microtubule-organizing centers.
d) Its role in spindle formation is crucial for accurate chromosome segregation.

(iii) Xylem [1 Mark]

Answer: Xylem conducts water and mineral salts from the roots up to the highest parts of the plant (leaves).

Teacher's Note:
a) Xylem is one of the two transport tissues in vascular plants, the other being phloem.
b) It consists of tracheids, vessels, xylem parenchyma, and xylem fibers.
c) The upward movement of water and minerals is driven by transpiration pull.
d) Students should understand its role in water transport and structural support.

(iv) Corpus luteum [1 Mark]

Answer: Corpus luteum secretes hormones such as oestrogen, progesterone, and relaxin, which are essential for maintaining pregnancy.

Teacher's Note:
a) The corpus luteum forms in the ovary after ovulation from the ruptured follicle.
b) Progesterone is its main hormone, crucial for thickening the uterine lining and inhibiting uterine contractions.
c) If pregnancy does not occur, the corpus luteum degenerates.
d) Its function is vital for the early stages of pregnancy and the menstrual cycle.

(v) Eustachian tube. [1 Mark]

Answer: The Eustachian tube equalizes atmospheric pressure and pressure in the ear (middle ear).

Teacher's Note:
a) The Eustachian tube connects the middle ear to the nasopharynx.
b) It opens during swallowing, yawning, or chewing to allow air to enter or leave the middle ear.
c) This equalization of pressure on both sides of the eardrum is essential for proper hearing.
d) Blockage of this tube can lead to discomfort and temporary hearing loss.

(d) Identify and name the following processes/terms from the statements given below:

(i) Movement of molecules from a region of high concentration to a region of low concentration. [1 Mark]

Answer: Diffusion (or Passive transport).

Teacher's Note:
a) Diffusion is a passive process where molecules move from an area of higher concentration to an area of lower concentration.
b) This movement occurs down a concentration gradient and does not require energy.
c) It is fundamental for gas exchange in lungs and nutrient absorption.
d) Osmosis is a specific type of diffusion involving water across a semi-permeable membrane.

(ii) Mild chemical applied on the skin to kill germs. [1 Mark]

Answer: Antiseptic.

Teacher's Note:
a) Antiseptics are substances that prevent the growth of disease-causing microorganisms.
b) They are safe enough to be applied to living tissues, such as skin.
c) Common examples include alcohol, iodine, and hydrogen peroxide.
d) This term contrasts with disinfectants, which are used on inanimate objects.

(iii) Chromosomes appear thread like. [1 Mark]

Answer: Prophase.

Teacher's Note:
a) During prophase of mitosis and meiosis, chromatin condenses into visible, thread-like chromosomes.
b) Each chromosome consists of two sister chromatids joined at the centromere.
c) This condensation makes them visible under a light microscope.
d) This is the first stage where chromosomes become distinctly observable.

(iv) The loss of water from injured parts of a plant. [1 Mark]

Answer: Bleeding (or Exudation of sap).

Teacher's Note:
a) Bleeding in plants refers to the exudation of sap from cut or injured surfaces.
b) This phenomenon is often due to root pressure, which pushes water and dissolved substances upwards.
c) It is different from transpiration, which is water loss as vapor from intact surfaces.
d) This process can be observed when a plant stem is cut.

(v) A pair of chromosomes carrying dissimilar alleles for a particular character. [1 Mark]

Answer: Heterozygous.

Teacher's Note:
a) A heterozygous individual has two different alleles for a specific gene.
b) For example, if 'T' represents the allele for tallness and 't' for dwarfness, a heterozygous individual would be 'Tt'.
c) This contrasts with homozygous individuals, who have two identical alleles (TT or tt).
d) Understanding heterozygosity is fundamental to Mendelian genetics.

(e) Classify the following actions as simple reflex or conditioned reflex:

(i) Playing a guitar. [1 Mark]

Answer: Conditioned Reflex.

Teacher's Note:
a) A conditioned reflex is a learned response developed through experience and training.
b) Playing a guitar involves complex motor skills, coordination, and memory, all of which are learned.
c) Simple reflexes are innate, involuntary, and unlearned responses.
d) This action requires conscious effort and practice to master.

(ii) Removing your hand suddenly when pricked by a thorn. [1 Mark]

Answer: Simple Reflex.

Teacher's Note:
a) A simple reflex is an involuntary, rapid, and automatic response to a stimulus.
b) The withdrawal reflex is an innate protective mechanism that does not involve conscious thought.
c) The nerve pathway for a simple reflex is called a reflex arc.
d) This action is crucial for immediate protection from harm.

(iii) Applying sudden brakes when a dog crosses the path. [1 Mark]

Answer: Conditioned Reflex.

Teacher's Note:
a) This action is a learned response based on prior experience and training (driving lessons, understanding traffic rules).
b) It involves processing visual information, making a decision, and executing a motor response.
c) While rapid, it is not an innate, unlearned reflex like withdrawing a hand from pain.
d) Conditioned reflexes are often complex and involve higher brain centers.

(iv) Blinking of eyelids on exposure to light. [1 Mark]

Answer: Simple Reflex.

Teacher's Note:
a) The blink reflex is an involuntary action that protects the eye from foreign objects, bright light, and dryness.
b) It is an innate reflex present from birth.
c) This reflex helps to keep the eyes moist and clean.
d) It is a rapid, automatic response that does not require conscious control.

(v) Tying one's shoe lace. [1 Mark]

Answer: Conditioned Reflex.

Teacher's Note:
a) Tying shoelaces is a learned motor skill that requires practice and repetition.
b) Initially, it requires conscious effort, but with practice, it becomes almost automatic.
c) This is an example of a complex motor skill that becomes habitual.
d) It demonstrates how repeated actions can lead to the formation of conditioned responses.

(f) The diagram given below is that of a developing human foetus in the womb. Study the same and then answer the questions that follow : [5 Marks]

[Figure: Diagram of a human foetus in the womb. Label 1 points to the umbilical cord. Label 2 points to the placenta. Label 3 points to the amniotic fluid/sac.]

(i) Name the part labelled 1. [1 Mark]

Answer: Umbilical cord.

Teacher's Note:
a) The umbilical cord connects the foetus to the placenta.
b) It contains blood vessels that transport oxygen and nutrients to the foetus and remove waste products.
c) It is a vital link for exchange between mother and foetus.
d) Students should be able to identify key structures in a foetal diagram.

(ii) Mention any two functions of the part labelled 2. [1 Mark]

Answer:
1. Provides diffusion for respiratory gases (oxygen and carbon dioxide) between mother and foetus.
2. Removes urea and other metabolic waste products produced by the foetus.

Teacher's Note:
a) Part 2 is the placenta, a temporary organ that connects the mother and the developing foetus.
b) It facilitates nutrient and oxygen uptake from the mother to the foetus.
c) It also allows for the elimination of foetal waste products.
d) The placenta also produces hormones necessary for maintaining pregnancy.

(iii) Explain the role played by the part labelled 3. [1 Mark]

Answer: Part labelled 3 is amniotic fluid. Its function is to act as a shock absorber, protecting the developing embryo from mechanical injuries and providing a stable environment.

Teacher's Note:
a) The amniotic fluid fills the amniotic sac, which surrounds the foetus.
b) It cushions the foetus from external impacts and allows for foetal movement.
c) It also helps maintain a constant temperature around the foetus.
d) This fluid is crucial for the physical protection and development of the foetus.

(iv) What is the normal gestation period (in days) of the developing foetus? [2 Marks]

Answer: 280 days (or approximately 9 months).

Teacher's Note:
a) The gestation period is the time from conception to birth.
b) For humans, it is typically around 40 weeks or 280 days when calculated from the last menstrual period.
c) This period allows for the complete development of the foetus.
d) Students should know the approximate duration of human pregnancy.

(g) Explain the following terms:

(i) Natality [1 Mark]

Answer: Natality is the number of children born per person in the population per unit time, also termed as birth rate. It is calculated as: Natality = (Total no. of children born (live) in a year / Total population in the middle of the year) × 1000.

Teacher's Note:
a) Natality refers to the birth rate within a population.
b) It is a key factor influencing population growth.
c) High natality rates can lead to rapid population increase.
d) Students should understand its definition and significance in population dynamics.

(ii) Photolysis in photosynthesis [1 Mark]

Answer: Photolysis in photosynthesis is the splitting of water molecules into its two component ions (H⁺ and OH^-) in the presence of light energy and chlorophyll.
H₂O \( \xrightarrow[\text{Chlorophyll}]{\text{Sunlight}} \) H⁺ + OH^-

Teacher's Note:
a) Photolysis is a crucial step in the light-dependent reactions of photosynthesis.
b) It provides electrons to replace those lost by chlorophyll and releases oxygen as a byproduct.
c) The H⁺ ions are used to form NADPH, and OH^- ions combine to form water and oxygen.
d) Students should remember the reactants and products of this reaction.

(iii) Antibiotic [1 Mark]

Answer: Antibiotics are chemical substances derived from microbes (like bacteria, yeast, moulds) which kill or prevent the growth of other microbes or pathogens. For example, Penicillin is an antibiotic.

Teacher's Note:
a) Antibiotics are a type of antimicrobial drug used in the treatment and prevention of bacterial infections.
b) They work by targeting specific bacterial processes, such as cell wall synthesis or protein synthesis.
c) Overuse or misuse of antibiotics can lead to antibiotic resistance.
d) Students should know the definition and a common example.

(iv) Root pressure [1 Mark]

Answer: Root pressure is the pressure generated by the cortical cells of the root, which pushes water and minerals into the xylem of the root for upward conduction.

Teacher's Note:
a) Root pressure is a positive pressure in the xylem sap of roots.
b) It is caused by the active transport of mineral ions into the root xylem, followed by osmotic movement of water.
c) Root pressure is responsible for guttation and contributes to the ascent of sap, especially when transpiration is low.
d) It is generally a minor force compared to transpiration pull in tall plants.

(v) Parturition. [1 Mark]

Answer: Parturition is the technical term for the delivery of the foetus or childbirth.

Teacher's Note:
a) Parturition is the physiological process of expelling the foetus and placenta from the uterus.
b) It is initiated by complex hormonal signals from both the foetus and the mother.
c) The process involves uterine contractions and cervical dilation.
d) Students should be familiar with this biological term for childbirth.

(h) Given below are five statements or questions followed by four choices. Select and rewrite the correct answer to the given statements from the four choices given below each statement:

(I) The cerebral hemispheres in mammals are connected by: [1 Mark]
(i) Corpus luteum
(ii) Hypothalamus
(iii) Pons varolli
(iv) Corpus callosum

Answer: (iv) Corpus callosum

Teacher's Note:
a) The corpus callosum is a large, C-shaped nerve fiber bundle found beneath the cerebral cortex.
b) It connects the two cerebral hemispheres, allowing communication between them.
c) This structure is crucial for integrating sensory, motor, and cognitive functions.
d) Students should know the major parts of the brain and their connections.

(II) Insulin is secreted by: [1 Mark]
(i) Beta cells of pancreas
(ii) Alpha cells of pancreas
(iii) Delta cells of pancreas
(iv) None of the above

Answer: (i) Beta cells of pancreas

Teacher's Note:
a) The pancreas contains clusters of endocrine cells called islets of Langerhans.
b) Beta cells within these islets are responsible for producing and secreting insulin.
c) Insulin lowers blood glucose levels by promoting glucose uptake by cells.
d) Alpha cells secrete glucagon, which raises blood glucose levels.

(III) A de-starched plant is one whose : [1 Mark]
(i) Leaves are free from chlorophyll
(ii) Aerial parts are free from starch
(iii) Leaves are free from starch
(iv) Plant is free from starch.

Answer: (iii) Leaves are free from starch

Teacher's Note:
a) A plant is de-starched by keeping it in darkness for 24-48 hours.
b) During this period, any starch already present in the leaves is converted to sugar and transported away.
c) This ensures that any starch detected after an experiment was newly synthesized during the experiment.
d) De-starching is a crucial preliminary step in experiments demonstrating photosynthesis.

(IV) The onset of menstruation in the female is termed : [1 Mark]
(i) Ovulation
(iii) Menopause
(ii) Menarche
(iv) Parthenogenesis

Answer: (ii) Menarche

Teacher's Note:
a) Menarche is the first menstrual period in a female, marking the onset of puberty.
b) It signifies the beginning of a female's reproductive years.
c) Menopause is the cessation of menstruation, marking the end of reproductive years.
d) Ovulation is the release of an egg from the ovary, which occurs during the menstrual cycle.

(V) BCG vaccine provides immunity against: [1 Mark]
(i) Tetanus
(iii) Tuberculosis
(ii) Cholera
(iv) AIDS

Answer: (iii) Tuberculosis

Teacher's Note:
a) BCG stands for Bacillus Calmette-Guérin, a vaccine primarily used against tuberculosis (TB).
b) It is particularly effective in preventing severe forms of TB in children.
c) Vaccines work by stimulating the immune system to produce antibodies against specific pathogens.
d) Students should be aware of common vaccines and the diseases they prevent.

SECTION-II (40 Marks)

(Attempt any four questions from this Section.)

Question 2.

(a) The diagram below represents a stage in cell division. Study the same and answer the questions that follow : [5 Marks]

[Figure: Diagram showing a cell undergoing division. Chromosomes are aligned at the metaphase plate. Spindle fibers (A) are attached to the centromeres (D) of the chromosomes (C). The cell boundary (B) is visible. The chromosomes are clearly visible and condensed.]

(i) Identify the stage of cell division. [1 Mark]

Answer: Metaphase.

Teacher's Note:
a) In metaphase, chromosomes align at the metaphase plate (equatorial plate) of the cell.
b) This alignment ensures that each daughter cell receives an identical set of chromosomes.
c) Spindle fibers are fully formed and attached to the centromeres.
d) This stage is characterized by the orderly arrangement of chromosomes in the middle of the cell.

(ii) Name the parts labelled A, B, C and D. [1 Mark]

Answer:
A = Spindle Fiber
B = Cell wall
C = Chromatid (or Chromosome)
D = Centromere

Teacher's Note:
a) Spindle fibers are microtubules that move chromosomes during cell division.
b) The cell wall is a rigid outer layer characteristic of plant cells.
c) A chromatid is one of the two identical halves of a replicated chromosome.
d) The centromere is the constricted region of a chromosome that holds sister chromatids together and where spindle fibers attach.

(iii) What is the unique feature observed in this stage ? [1 Mark]

Answer: Chromosomes are aligned at the equatorial plate (metaphase plate) of the cell.

Teacher's Note:
a) The alignment of chromosomes at the metaphase plate is the defining characteristic of metaphase.
b) This precise arrangement is critical for ensuring equal distribution of genetic material.
c) Each sister chromatid faces opposite poles, ready for separation.
d) This feature helps in identifying metaphase in microscopic observations.

(iv) Where does this type of cell division usually occur? [1 Mark]

Answer: Somatic cells (or growing regions like meristems in plants).

Teacher's Note:
a) Mitosis, the type of cell division shown, occurs in somatic cells for growth, repair, and asexual reproduction.
b) In plants, it is prominent in meristematic tissues such as root tips, shoot tips, and cambium.
c) This ensures that daughter cells are genetically identical to the parent cell.
d) Meiosis, in contrast, occurs in germ cells to produce gametes.

(v) How many daughter cells are formed from this type of cell division ? [1 Mark]

Answer: Two daughter cells.

Teacher's Note:
a) Mitosis results in the formation of two genetically identical daughter cells from a single parent cell.
b) Each daughter cell has the same number of chromosomes as the parent cell.
c) This process maintains the chromosome number across generations of somatic cells.
d) Meiosis, on the other hand, produces four haploid daughter cells.

(vi) Is the dividing cell shown a plant or an animal cell? Give a reason to support your answer. [1 Mark]

Answer: It is a plant cell because a cell wall (labelled B) is clearly visible, which is a unique feature of plant cells.

Teacher's Note:
a) The presence of a rigid cell wall is a distinguishing characteristic of plant cells.
b) Animal cells lack a cell wall and have a more flexible cell membrane.
c) Other indicators of plant cells include the absence of centrioles (though not explicitly shown here, centrosomes are absent in higher plants) and the presence of a large central vacuole (not visible in this stage).
d) Students should be able to differentiate between plant and animal cells based on structural features.

(b) The diagram below represents a structure found in a leaf. Study the same and answer the questions that follow : [5 Marks]

[Figure: Diagram of a stomatal apparatus. Label A points to the guard cells. Label B points to the stomatal pore.]

(i) Name the parts labelled A and B. [1 Mark]

Answer:
A = Guard Cells
B = Stoma (or Stomatal pore)

Teacher's Note:
a) Guard cells are specialized epidermal cells that surround the stomatal pore.
b) The stoma is the pore itself, through which gas exchange and transpiration occur.
c) Together, the guard cells and stomatal pore form the stomatal apparatus.
d) Accurate labeling of these parts is essential for understanding leaf anatomy.

(ii) What is the biological term for the above structure ? [1 Mark]

Answer: Stomata (plural) or Stoma (singular).

Teacher's Note:
a) Stomata are microscopic pores on the surface of leaves, primarily on the lower epidermis.
b) They regulate the exchange of gases (CO₂ and O₂) and water vapor between the plant and the atmosphere.
c) The opening and closing of stomata are controlled by the turgidity of the guard cells.
d) This structure is vital for photosynthesis and transpiration.

(iii) What is the function of the parts labelled A? [1 Mark]

Answer: Part labelled A, i.e., guard cells, control the opening and closing of stomata.

Teacher's Note:
a) Guard cells regulate the size of the stomatal pore.
b) Their turgor pressure changes due to the movement of water and potassium ions.
c) When turgid, they bow outwards, opening the stoma; when flaccid, they close the stoma.
d) This control is crucial for balancing water loss and carbon dioxide uptake.

(iv) Mention two structural features of A, which help in the function mentioned in (iii) above. [1 Mark]

Answer:
1. Guard cells are structurally bean-shaped (or kidney-shaped).
2. The cell wall of guard cells is differentially thick; the inner wall (facing the stomatal pore) is thick and inelastic, while the outer wall is thin and elastic.

Teacher's Note:
a) The bean shape allows them to bow outwards when turgid.
b) The differential thickness of the cell wall ensures that when turgor pressure increases, the thinner outer wall stretches more, causing the guard cells to curve and open the pore.
c) Guard cells also contain chloroplasts, enabling them to perform photosynthesis and produce sugars that contribute to turgor changes.
d) These features are adaptations for efficient stomatal regulation.

(v) Where is this structure likely to be found in a leaf? [1 Mark]

Answer: This structure is more likely to be found on the lower epidermis of the leaves.

Teacher's Note:
a) Stomata are generally more abundant on the lower epidermis to minimize water loss due to direct sunlight.
b) This arrangement helps in reducing the rate of transpiration.
c) Some plants, like aquatic plants, may have stomata on the upper epidermis or lack them entirely.
d) Students should understand the distribution of stomata and its ecological significance.

(vi) The above structure helps in the process of transpiration Explain the term transpiration. [1 Mark]

Answer: Transpiration is the process by which plants absorb water through the roots and then give off water vapor through pores (stomata) in their leaves, stems, and flowers. It is essentially the loss of water from the aerial parts of the plant in the form of water vapor.

Teacher's Note:
a) Transpiration is a vital process that drives the ascent of sap in plants.
b) It also helps in cooling the plant and transporting minerals.
c) The majority of water absorbed by plants is lost through transpiration.
d) Students should distinguish between transpiration and guttation.

(vii) How many other cells are found surrounding this structure as seen in the diagram. [1 Mark]

Answer: There are five (5) other cells surrounding this structure (subsidiary cells/epidermal cells).

Teacher's Note:
a) The cells surrounding the guard cells are called subsidiary cells or accessory cells.
b) These cells assist the guard cells in their function by providing a reservoir of water and ions.
c) The number and arrangement of subsidiary cells can vary among different plant species.
d) Students should be able to identify these accessory cells in a stomatal diagram.

Question 3.

(a) The diagram shows a section of the human brain. Answer the questions that follow : [5 Marks]

[Figure: Diagram of a human brain in sagittal section. Label A points to the Cerebrum. Label B points to the Cerebellum. Label C points to the Medulla Oblongata.]

(i) Name the parts labelled A, B and C. [1 Mark]

Answer:
A = Cerebrum
B = Cerebellum
C = Medulla Oblongata

Teacher's Note:
a) The cerebrum is the largest part of the brain, responsible for higher-level functions.
b) The cerebellum is located at the back of the brain, involved in motor control.
c) The medulla oblongata is part of the brainstem, controlling vital involuntary functions.
d) Correct identification of these major brain regions is fundamental.

(ii) Give the main function of each of the parts A, B, and C. [1 Mark]

Answer:
Cerebrum (A): It is the site of controlling memory, reasoning, thinking, perception, emotions, and speech.
Cerebellum (B): It maintains posture, equilibrium, and muscular coordination.
Medulla Oblongata (C): It contains centers for cardiac, respiratory, and vasomotor activities. It also coordinates reflexes for swallowing, coughing, sneezing, and vomiting.

Teacher's Note:
a) The cerebrum is the center for conscious thought and voluntary actions.
b) The cerebellum refines motor movements and ensures balance.
c) The medulla oblongata regulates essential life-sustaining functions.
d) Students should be able to associate specific functions with each major brain part.

(iii) Name the three protective membranes covering the brain. [1 Mark]

Answer: The three protective membranes covering the brain are: Dura mater, Pia mater, and Arachnoid mater.

Teacher's Note:
a) These three membranes collectively form the meninges.
b) They provide physical protection and support to the brain and spinal cord.
c) The cerebrospinal fluid circulates between the arachnoid and pia mater, providing further cushioning.
d) Knowing the names of these layers is important for understanding brain protection.

(iv) Name the basic unit of the brain. [2 Marks]

Answer: Neuron.

Teacher's Note:
a) The neuron is the fundamental structural and functional unit of the nervous system.
b) Neurons transmit electrical and chemical signals throughout the body.
c) They are specialized cells that process and transmit information.
d) Understanding neuron structure and function is key to neurobiology.

(b) (i) Mention any three adaptations found in plants to favour the process of photosynthesis. [2 Marks]

Answer:
1. Leaves are thin and flat, offering a large surface area for light absorption and gas exchange.
2. Internally, the loose arrangement of mesophyll tissue allows rapid diffusion of gases (CO₂).
3. The high concentration of chloroplasts in the mesophyll, especially the palisade cells of dorsiventral leaves, helps to trap solar energy.
4. Presence of stomata for efficient gas exchange.

Teacher's Note:
a) The broad, flat shape of leaves maximizes light capture.
b) Air spaces in the spongy mesophyll facilitate CO₂ diffusion to photosynthetic cells.
c) Chloroplasts are strategically located in palisade cells to receive maximum sunlight.
d) Stomata regulate the entry of CO₂ and release of O₂.

(ii) Why does one feel blinded for a short while on coming out of a dark room? [1 Mark]

Answer: When coming out of a dark room, one feels blinded for a short while due to light adaptation. In the dark, the pupils are dilated to allow maximum light entry, and rhodopsin (visual purple) accumulates. Upon sudden exposure to bright light, the pupils constrict, and rhodopsin rapidly breaks down, causing temporary overstimulation and dazzling until the eyes adapt to the new light intensity.

Teacher's Note:
a) This phenomenon is called light adaptation.
b) In darkness, pupils dilate and rhodopsin regenerates, increasing light sensitivity.
c) In bright light, pupils constrict, and rhodopsin bleaches, reducing sensitivity.
d) The temporary blindness is due to the time taken for the eye to adjust its sensitivity to the sudden change in light intensity.

(iii) Explain how the rate of transpiration is affected on:
1. a windy day
2. a foggy day. [2 Marks]

Answer:
1. On a windy day, the rate of transpiration increases because moving wind removes the water vapor from around the leaves, maintaining a steep water potential gradient between the leaf and the atmosphere.
2. On a foggy day, the rate of transpiration decreases because atmospheric moisture (humidity) is high, which reduces the water potential gradient between the leaf and the atmosphere, inhibiting the rate of water vapor diffusion.

Teacher's Note:
a) Wind increases transpiration by sweeping away the humid air layer (boundary layer) around the leaf.
b) High humidity (as in fog) reduces the water potential gradient, thus slowing down water evaporation.
c) Transpiration is directly influenced by environmental factors affecting the water potential gradient.
d) Students should understand the inverse relationship between humidity and transpiration rate, and the direct relationship with wind speed.

Question 4.

(a) The diagram given alongside represents the structure found in the inner ear. Study the same and then answer the questions that follow : [5 Marks]

[Figure: Diagram of the human inner ear. Label A points to the Semicircular canals. Label B points to the Utriculus. Label C points to the Sacculus. Label D points to the Cochlea. The Auditory Nerve is also visible.]

(i) Name the parts labelled A, B, C and D. [1 Mark]

Answer:
A = Semicircular canals
B = Utriculus
C = Sacculus
D = Cochlea

Teacher's Note:
a) The inner ear consists of the bony labyrinth, which houses the membranous labyrinth.
b) Semicircular canals, utricle, and saccule are part of the vestibular system, responsible for balance.
c) The cochlea is responsible for hearing.
d) Accurate identification of these structures is crucial for understanding ear physiology.

(ii) Name the part of the ear responsible for transmitting impulses to the brain. [1 Mark]

Answer: Auditory Nerve (or Vestibulocochlear nerve).

Teacher's Note:
a) The auditory nerve (specifically the cochlear nerve part of the vestibulocochlear nerve) transmits sound information.
b) The vestibular nerve (the other part) transmits balance information.
c) These nerves carry sensory impulses from the inner ear to the brain for processing.
d) Understanding the neural pathway for hearing and balance is important.

(iii) Name the part labelled above which is responsible for:
1. Static equilibrium.
2. Dynamic equilibrium.
3. Hearing. [1 Mark]

Answer:
1. Static equilibrium: Utriculus and Sacculus (B and C)
2. Dynamic equilibrium: Semicircular canals (A)
3. Hearing: Cochlea (D)

Teacher's Note:
a) The utricle and saccule detect linear acceleration and head position relative to gravity (static equilibrium).
b) The semicircular canals detect rotational movements of the head (dynamic equilibrium).
c) The cochlea converts sound vibrations into electrical signals.
d) Each part of the inner ear has a distinct role in maintaining balance and hearing.

(iv) Name the audio receptor cells which pick up vibrations. [1 Mark]

Answer: Organ of Corti (specifically hair cells within it).

Teacher's Note:
a) The Organ of Corti is located within the cochlea and contains the hair cells.
b) Hair cells are mechanoreceptors that convert mechanical vibrations into electrical signals.
c) These signals are then transmitted via the auditory nerve to the brain.
d) This structure is essential for the process of hearing.

(v) Name the fluid present in the inner ear. [1 Mark]

Answer: Endolymph (and Perilymph).

Teacher's Note:
a) The inner ear contains two types of fluid: perilymph and endolymph.
b) Perilymph fills the space between the bony and membranous labyrinths.
c) Endolymph fills the membranous labyrinth (cochlear duct, utricle, saccule, semicircular canals).
d) These fluids are crucial for transmitting vibrations and detecting head movements.

(b) Name the hormone responsible for the following functions :

(i) Increase in heart beat. [1 Mark]

Answer: Adrenalin (or Epinephrine).

Teacher's Note:
a) Adrenalin is a hormone secreted by the adrenal medulla.
b) It is part of the "fight or flight" response, increasing heart rate, blood pressure, and blood flow to muscles.
c) This prepares the body for immediate physical activity or stress.
d) Students should recognize its role in emergency situations.

(ii) Maintains glucose level in the blood. [1 Mark]

Answer: Insulin (lowers blood glucose) and Glucagon (raises blood glucose).

Teacher's Note:
a) Insulin, secreted by beta cells of the pancreas, lowers blood glucose by promoting uptake and storage.
b) Glucagon, secreted by alpha cells, raises blood glucose by stimulating glycogenolysis and gluconeogenesis.
c) These two hormones work antagonistically to maintain blood glucose homeostasis.
d) Both are crucial for regulating energy metabolism.

(iii) Converting Glycogen to Glucose. [1 Mark]

Answer: Glucagon.

Teacher's Note:
a) Glucagon stimulates glycogenolysis, the breakdown of glycogen into glucose, primarily in the liver.
b) This process releases glucose into the bloodstream, increasing blood sugar levels.
c) It is secreted when blood glucose levels are low.
d) This is a key mechanism for maintaining glucose supply during fasting.

(iv) Regulates basal metabolism. [1 Mark]

Answer: Thyroxine (Thyroid hormone).

Teacher's Note:
a) Thyroxine, produced by the thyroid gland, regulates the body's metabolic rate.
b) It affects almost every cell in the body, influencing energy production and heat generation.
c) Imbalances in thyroxine can lead to conditions like hyperthyroidism or hypothyroidism.
d) This hormone is essential for normal growth and development.

(v) Ossification of bones. [1 Mark]

Answer: Calcitonin (and Parathormone, Vitamin D).

Teacher's Note:
a) Calcitonin, secreted by the thyroid gland, lowers blood calcium levels by inhibiting osteoclast activity and promoting calcium deposition in bones.
b) Parathormone (PTH) and Vitamin D also play crucial roles in bone formation and calcium metabolism.
c) Ossification is the process of bone formation.
d) These hormones work together to maintain calcium homeostasis and bone health.

(vi) Prepares the body during emergency. [1 Mark]

Answer: Adrenalin (or Epinephrine).

Teacher's Note:
a) Adrenalin is the primary hormone involved in the "fight or flight" response.
b) It rapidly mobilizes energy reserves, increases cardiovascular activity, and diverts blood flow to essential organs.
c) This prepares the body to respond quickly to perceived threats.
d) It is a key component of the sympathetic nervous system's response to stress.

(vii) Responsible for normal growth of the whole body. [1 Mark]

Answer: Somatotropic Hormone (STH) or Growth Hormone.

Teacher's Note:
a) Growth hormone, secreted by the anterior pituitary gland, stimulates growth in almost all tissues.
b) It promotes protein synthesis, fat breakdown, and glucose conservation.
c) Deficiency leads to dwarfism, while excess leads to gigantism or acromegaly.
d) It is crucial for skeletal and muscular development during childhood and adolescence.

(viii) Regulates the functioning of the male and female reproductive organs. [1 Mark]

Answer: Gonadotropic Hormone (GTH) (or FSH and LH).

Teacher's Note:
a) Gonadotropins, Follicle-Stimulating Hormone (FSH) and Luteinizing Hormone (LH), are secreted by the anterior pituitary.
b) In males, they stimulate sperm production and testosterone secretion.
c) In females, they regulate ovarian function, including follicle development, ovulation, and hormone production.
d) These hormones are central to sexual development and reproduction.

(ix) Increased reabsorption of water in the kidneys [1 Mark]

Answer: Antidiuretic Hormone (ADH) or Vasopressin.

Teacher's Note:
a) ADH is secreted by the posterior pituitary gland in response to increased blood osmolarity or decreased blood volume.
b) It increases the permeability of the collecting ducts and distal convoluted tubules in the kidneys to water.
c) This leads to increased water reabsorption and production of concentrated urine.
d) ADH plays a critical role in maintaining body fluid balance.

(x) Increased blood supply to muscles. [1 Mark]

Answer: Adrenalin (or Epinephrine).

Teacher's Note:
a) Adrenalin causes vasodilation in the blood vessels supplying skeletal muscles.
b) This increases blood flow to muscles, providing more oxygen and nutrients for increased activity.
c) Simultaneously, it causes vasoconstriction in other areas like the digestive tract.
d) This redirection of blood flow is part of the body's emergency response.

Question 5.

(a) The diagram given alongside represents a surgical sterilization method in males. Study the same and then answer the questions that follow : [5 Marks]

[Figure: Diagram showing male reproductive organs and a surgical procedure. Label A points to the Seminal Vesicle. Label B points to the Vas Deferens (Sperm Duct). Label C points to the Urethra. Label D points to the Testes. Label E points to the Scrotum Sac. The vas deferens is shown cut and tied.]

(i) Name the parts marked A, B, C, D and E. [1 Mark]

Answer:
A = Seminal Vesicle
B = Sperm Duct (or Vas Deferens)
C = Urethra
D = Testes
E = Scrotum Sac

Teacher's Note:
a) The seminal vesicles produce a fluid that forms part of semen.
b) The vas deferens transports sperm from the epididymis to the ejaculatory ducts.
c) The urethra carries both urine and semen out of the body.
d) The testes produce sperm and male hormones, and the scrotum houses the testes.

(ii) Give the name of the surgical method represented in the diagram. [1 Mark]

Answer: Vasectomy.

Teacher's Note:
a) Vasectomy is a permanent method of male contraception.
b) It involves cutting or sealing the vas deferens.
c) This prevents sperm from mixing with semen during ejaculation.
d) It is a relatively simple and effective procedure.

(iii) Which part is ligated or cut? [1 Mark]

Answer: Sperm Duct (or Vas Deferens) (Part B).

Teacher's Note:
a) Ligation refers to the surgical tying off of a duct or blood vessel.
b) In vasectomy, the vas deferens is cut and tied to block the passage of sperm.
c) This prevents sperm from reaching the urethra.
d) The procedure does not affect hormone production or sexual function, only fertility.

(iv) Name the corresponding surgical method conducted on females. [1 Mark]

Answer: Tubectomy.

Teacher's Note:
a) Tubectomy is the female equivalent of vasectomy, a permanent birth control method.
b) It involves cutting, tying, or sealing the fallopian tubes.
c) This prevents the egg from reaching the uterus and sperm from reaching the egg.
d) It is also known as tubal ligation.

(v) Name the part which is ligated in females and why ? [1 Mark]

Answer: The Oviduct (or Fallopian tube) is ligated in females. This is done to prevent the egg from meeting with the sperms, thus preventing fertilization.

Teacher's Note:
a) The fallopian tubes are the site where fertilization typically occurs.
b) By blocking these tubes, the egg cannot travel from the ovary to the uterus, and sperm cannot reach the egg.
c) This effectively prevents pregnancy.
d) The procedure does not affect ovarian hormone production.

(b) (i) Explain the following terms:
1. Monohybrid cross.
2. Gene
3. Phenotype [3 Marks]

Answer:
1. Monohybrid Cross: A cross between two parents taking into consideration the alternative traits of one single character. For example, a cross between tall and dwarf pea plants.
2. Gene: Genes are the basic unit of heredity and are found on chromosomes. Each character or trait is controlled by one gene.
3. Phenotype: The observable character which is genetically controlled.

Teacher's Note:
a) A monohybrid cross helps study the inheritance pattern of a single trait.
b) Genes are segments of DNA that carry genetic information.
c) Phenotype is the physical expression of an organism's genotype.
d) Understanding these terms is fundamental to genetics.

(ii) Name the two sex-linked diseases in males. [1 Mark]

Answer: Two sex-linked diseases in males are: Haemophilia and Colour blindness.

Teacher's Note:
a) Sex-linked diseases are typically X-linked recessive disorders.
b) Males are more commonly affected because they have only one X chromosome.
c) If a male inherits a recessive allele on his X chromosome, he will express the trait.
d) Females, with two X chromosomes, usually need two copies of the recessive allele to express the trait.

(iii) State Mendel's law of segregation. [1 Mark]

Answer: Mendel's Law of Segregation states that during gamete formation, the two alleles for a heritable character separate (segregate) from each other such that each gamete receives only one allele. These alleles then randomly unite at fertilization.

Teacher's Note:
a) This law is also known as the Law of Purity of Gametes.
b) It explains why offspring can inherit traits from both parents, even if one parent expresses a dominant trait.
c) The segregation of alleles ensures genetic variation in offspring.
d) This is one of Mendel's fundamental laws of inheritance.

Question 6.

(a) The diagram below represents an experiment conducted to prove the importance of a factor in Photosynthesis. Study the same and then answer the questions that follow : [5 Marks]

[Figure: Diagram of an experiment to show the necessity of light for photosynthesis. A potted plant has one of its leaves partially covered with a black paper clip. The rest of the leaf and plant are exposed to sunlight.]

(i) Name the factor being studied in this experiment? [1 Mark]

Answer: Sunlight (or Light).

Teacher's Note:
a) The experiment uses a black paper clip to block light from a portion of the leaf.
b) This creates a control (covered part) and an experimental part (uncovered part) for light exposure.
c) The aim is to demonstrate that light is essential for photosynthesis.
d) This is a classic experiment to prove the necessity of light.

(ii) Why was the plant kept in a dark room before conducting the experiment? [1 Mark]

Answer: The plant was kept in a dark room before conducting the experiment to de-starch it, meaning to remove any pre-existing starch from its leaves.

Teacher's Note:
a) De-starching ensures that any starch detected after the experiment was newly synthesized during the experiment.
b) In darkness, the plant utilizes its stored starch, and no new starch is produced.
c) This step is crucial for obtaining reliable results in photosynthesis experiments.
d) It establishes a baseline for starch content in the leaves.

(iii) Why was the experimental leaf then kept in
(1) boiling water (2) methylated spirit? [1 Mark]

Answer:
1. In boiling water, the leaf was kept to kill the cells and make the leaf permeable to iodine solution.
2. In methylated spirit (alcohol), the leaf was kept to dissolve and remove the chlorophyll, making the color change with iodine visible.

Teacher's Note:
a) Boiling water denatures enzymes and breaks down cell membranes, allowing reagents to penetrate.
b) Chlorophyll is removed by alcohol because it would mask the color change of the iodine test.
c) Alcohol is heated indirectly (water bath) because it is flammable.
d) These steps are preparatory for the starch test.

(iv) Name the solution used to test for presence of starch in the leaf. [1 Mark]

Answer: Iodine solution.

Teacher's Note:
a) Iodine solution (potassium iodide solution) is a common indicator for starch.
b) Starch reacts with iodine to produce a blue-black color.
c) The absence of starch results in a yellowish-brown color (the color of iodine solution itself).
d) This test is fundamental for demonstrating starch production in photosynthesis.

(v) What will we observe in the experimental leaf at the end of the starch test? [1 Mark]

Answer: On treating the leaf with iodine solution, the portion of the leaf that was covered with the black paper strip will not give a positive test for starch (it will remain yellowish-brown), while the uncovered portion will turn blue-black, indicating the presence of starch. This proves that light is necessary for photosynthesis.

Teacher's Note:
a) The covered part of the leaf did not receive light, so no photosynthesis occurred, and no starch was produced.
b) The uncovered part received light, performed photosynthesis, and produced starch.
c) This observation directly supports the hypothesis that light is essential for starch formation (photosynthesis).
d) The color change is a clear visual indicator of starch presence or absence.

(vi) Give a balanced chemical equation to represent the process of photosynthesis. [1 Mark]

Answer: \( 6\text{CO}_2 + 12\text{H}_2\text{O} \xrightarrow[\text{Chlorophyll}]{\text{Sunlight}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{H}_2\text{O} + 6\text{O}_2 \)

Teacher's Note:
a) This equation represents the overall process of photosynthesis.
b) Carbon dioxide and water are reactants, and glucose, water, and oxygen are products.
c) Sunlight and chlorophyll are essential conditions for the reaction.
d) Students should ensure the equation is balanced and includes the conditions.

(b) (i) Mention three adaptations found in plants to reduce transpiration. [2 Marks]

Answer:
1. Leaves may be covered with a thick cuticle.
2. Stomata may be sunken in pits or grooves, especially in xerophytes.
3. Some plants contain latex, mucilage, resins, etc., to retain water in the plant body.
4. Reduced leaf surface area (e.g., spines), fewer stomata, or shedding leaves during dry periods.

Teacher's Note:
a) A thick cuticle reduces water loss from the leaf surface.
b) Sunken stomata create a humid microenvironment, reducing the water potential gradient.
c) Water-retaining substances help store water and reduce evaporation.
d) These adaptations are crucial for plants in arid or dry environments.

(ii) Name any three germ killing secretions of our body. [1 Mark]

Answer: Three germ-killing secretions of our body are: Tears, Saliva, and Hydrochloric Acid (HCl) in the stomach.

Teacher's Note:
a) Tears contain lysozyme, an enzyme that breaks down bacterial cell walls.
b) Saliva also contains lysozyme and other antimicrobial agents.
c) HCl in the stomach creates a highly acidic environment that kills most ingested pathogens.
d) These secretions are part of the body's innate immune defense system.

(iii) What are the age restrictions for marriage for boys and girls in India? [1 Mark]

Answer: The age restriction for marriage for boys in India is 21 years, and for girls, it is 18 years.

Teacher's Note:
a) These age limits are set by law to ensure physical and mental maturity before marriage.
b) Marrying below these ages is considered child marriage and is illegal.
c) These laws aim to protect individuals, especially girls, from early marriage and its associated health and social risks.
d) Students should be aware of these legal and social aspects related to human reproduction and health.

(iv) Mention two activities of the Red Cross. [1 Mark]

Answer: Two activities of the Red Cross are:
1. It assists the medical services of a country by providing ambulance services and first aid.
2. It provides training for medical and paramedical staff for dealing with war and natural calamities, and offers disaster relief and humanitarian aid.

Teacher's Note:
a) The Red Cross is an international humanitarian movement.
b) Its core mission includes protecting human life and health, ensuring respect for all human beings, and preventing and alleviating human suffering.
c) They are active in disaster response, health services, and promoting humanitarian values.
d) Students should know the general scope of work of such international organizations.

Question 7.

(a) Given below is a diagram depicting a defect of the human eye. Study the same and then answer the questions that follow : [5 Marks]

[Figure: Diagram showing an eye defect where parallel light rays converge in front of the retina. The eyeball appears elongated. Label 1 points to the cornea/lens, Label 2 points to the retina, Label 3 points to the optic nerve.]

(i) Identify the defect. [1 Mark]

Answer: Myopia (or Nearsightedness).

Teacher's Note:
a) Myopia is a common vision condition where close objects appear clear, but distant objects appear blurry.
b) This occurs because light rays focus in front of the retina, rather than directly on it.
c) The diagram clearly shows the focal point before the retina.
d) Students should be able to identify common eye defects from diagrams.

(ii) Name the parts labelled 1, 2, and 3. [1 Mark]

Answer:
Part labelled 1 = Vitreous Humor
Part labelled 2 = Yellow spot
Part labelled 3 = Optic Nerve

Teacher's Note:
a) The vitreous humor is the clear gel that fills the space between the lens and the retina.
b) The yellow spot (macula) is the area of the retina responsible for sharp, detailed central vision.
c) The optic nerve transmits visual information from the retina to the brain.
d) Precise anatomical labeling is important for understanding eye function.

(iii) Give two possible reasons for this eye defect. [1 Mark]

Answer: Two reasons for this defect are:
1. The eyeball becomes too long from front to back.
2. The lens becomes too curved/convex, increasing its converging power.

Teacher's Note:
a) An elongated eyeball causes light to focus before reaching the retina.
b) A lens that is too strong (too convex) also causes light to converge too quickly.
c) Both conditions result in blurry distant vision.
d) These are the primary anatomical causes of myopia.

(iv) Draw a labelled diagram to show how the above mentioned defect is rectified. [2 Marks]

Answer: This defect is rectified by using a concave lens (diverging lens). The concave lens diverges the incoming light rays before they enter the eye, causing them to focus correctly on the retina.
[Figure: Diagram showing the rectification of myopia. Parallel light rays pass through a concave lens, diverge slightly, and then enter the eye. Inside the eye, the rays converge precisely on the retina, resulting in clear vision. The concave lens is shown in front of the eye.]

Teacher's Note:
a) Concave lenses are diverging lenses, meaning they spread out parallel light rays.
b) By diverging the light before it enters the myopic eye, the focal point is shifted backward onto the retina.
c) This allows the image to be focused sharply on the retina, correcting distant vision.
d) Students should understand the principle of corrective lenses for refractive errors.

(b) Complete the following by filling in the blanks numbered 1 to 10 with the appropriate word/term: [5 Marks]

Photosynthesis involves light reaction and dark reaction. During light reaction, the chlorophyll present in the (1) ......... gets activated by absorbing light energy. This energy splits (2) ......... molecules to (3) ......... and oxygen and releases two electrons. This process is called (4) ......... . The (5) ......... ions are picked up by NADP to form (6) ......... . The ADP is converted to (7) ......... . This process is called (8) ......... . During the dark phase, the compound produced at the end of light reaction reacts with carbon dioxide to form (9) ......... . This product is converted to starch. The process is called (10) ......... .

Answer:
1. Grana
2. Water
3. H⁺ (Hydrogen ions)
4. Photolysis
5. H⁺ (Hydrogen ions)
6. NADPH
7. ATP
8. Photophosphorylation
9. Glucose
10. Carbon fixation (or Calvin cycle)

Teacher's Note:
a) The light reaction occurs in the grana (thylakoids) and involves light absorption, water splitting, and ATP/NADPH formation.
b) Photolysis of water provides electrons, protons (H⁺), and releases oxygen.
c) ATP and NADPH are energy carriers produced during the light reaction.
d) The dark reaction (Calvin cycle) uses ATP and NADPH to fix carbon dioxide into glucose, which is then converted to starch. The official key shows "Isomerisation" for blank (10); however, "Carbon fixation" or "Calvin cycle" is the correct term for the overall process of forming glucose from CO₂ in the dark phase. The conversion of glucose to starch is starch synthesis.