ICSE Class 10 Biology Board Exam Question Paper 2014 with Solutions

ICSE Class X Biology Board Exam Question Paper with Solutions

SECTION I (40 Marks)

Attempt all questions from this section.

(1-5) Name the following:

1. The part of the brain associated with memory. [1 Mark]

Answer: Cerebrum.

Teacher's Note:
a) The cerebrum is the largest part of the brain and is responsible for higher functions like memory, learning, reasoning, and voluntary actions.
b) Students often confuse the cerebrum with the cerebellum, which is primarily involved in coordinating muscle movements and maintaining balance.

2. The ear ossicle which is attached to the tympanum. [1 Mark]

Answer: Malleus.

Teacher's Note:
a) The malleus is the outermost of the three ossicles (malleus, incus, stapes) in the middle ear and is directly connected to the tympanic membrane (eardrum).
b) Its primary function is to transmit vibrations from the tympanum to the incus, initiating the mechanical amplification of sound waves.

3. The type of gene which is not expressed in the presence of a contrasting allele. [1 Mark]

Answer: Recessive gene.

Teacher's Note:
a) A recessive gene expresses its trait only when two copies of the allele are present (homozygous recessive condition).
b) In the presence of a dominant allele, the recessive allele's trait is masked, and the dominant trait is expressed.

4. The hormone secreted by islets of langerhans. [1 Mark]

Answer: Insulin (and Glucagon).

Teacher's Note:
a) The islets of Langerhans in the pancreas secrete two main hormones: insulin (from beta cells) and glucagon (from alpha cells).
b) Insulin lowers blood glucose levels, while glucagon raises them, working antagonistically to maintain blood sugar homeostasis.

5. The process of conversion of ADP into ATP during photosynthesis. [1 Mark]

Answer: Photophosphorylation.

Teacher's Note:
a) Photophosphorylation is the process by which light energy is used to convert ADP and inorganic phosphate into ATP during the light-dependent reactions of photosynthesis.
b) This ATP provides the energy required for the synthesis of glucose in the light-independent reactions (Calvin cycle).

(6-10) State the main function of the following:

6. Cerebrospinal fluid [1 Mark]

Answer: Protects the brain and spinal cord from mechanical shocks, supplies nutrients, and removes waste products.

Teacher's Note:
a) Cerebrospinal fluid (CSF) acts as a cushion, absorbing shocks and preventing injury to the delicate nervous tissue.
b) It also plays a vital role in maintaining the chemical stability of the central nervous system by transporting nutrients and removing metabolic waste.

7. Eustachian tube [1 Mark]

Answer: Equalises air pressure on both sides of the tympanum (eardrum).

Teacher's Note:
a) The Eustachian tube connects the middle ear to the nasopharynx, allowing air to enter or leave the middle ear cavity.
b) This equalisation of pressure is crucial for the proper vibration of the eardrum and clear hearing, especially during changes in altitude.

8. Suspensory ligament of the eye [1 Mark]

Answer: Holds the eye lens in position and helps in changing its curvature during accommodation.

Teacher's Note:
a) The suspensory ligaments connect the ciliary body to the lens, transmitting the tension from the ciliary muscles to the lens.
b) Relaxation or contraction of the ciliary muscles alters the tension in these ligaments, thereby changing the shape and focal length of the lens for focusing on objects at different distances.

9. Sperm duct [1 Mark]

Answer: Transports sperms from the epididymis to the urethra.

Teacher's Note:
a) The sperm duct, also known as vas deferens, is a muscular tube that propels mature sperm from the epididymis towards the ejaculatory duct.
b) It is a key component of the male reproductive system, facilitating the movement of sperm during ejaculation.

10. Lenticels [1 Mark]

Answer: Facilitate gaseous exchange between the atmosphere and the internal tissues of the stem.

Teacher's Note:
a) Lenticels are porous areas on the bark of woody stems and roots, composed of loosely arranged cells.
b) They serve a similar function to stomata in leaves, allowing for the diffusion of gases like oxygen, carbon dioxide, and water vapour.

11. Copy and complete the following by filling in the blanks 1 to 5 with appropriate words: [5 Marks]
The human female gonads are ovaries. A maturing egg in the ovary is present in a sac of cells called ________ (1). As the egg grows larger, the follicle enlarges and gets filled with a fluid and is now called the ________ (2) follicle. The process of releasing the egg from the ovary is called ________ (3). The ovum is picked up by the oviducal funnel and fertilisation takes place in the ________ (4). In about a week, the blastocyst gets fixed in the endometrium of the uterus and this process is called ________ (5).

Answer:
The human female gonads are ovaries. A maturing egg in the ovary is present in a sac of cells called follicle (1). As the egg grows larger, the follicle enlarges and gets filled with a fluid and is now called the Graafian (2) follicle. The process of releasing the egg from the ovary is called ovulation (3). The ovum is picked up by the oviducal funnel and fertilisation takes place in the fallopian tube/oviduct (4). In about a week, the blastocyst gets fixed in the endometrium of the uterus and this process is called implantation (5).

Teacher's Note:
a) This question tests knowledge of the female reproductive system and the process of human reproduction, from gamete maturation to implantation.
b) Key terms like 'follicle', 'Graafian follicle', 'ovulation', 'fallopian tube', and 'implantation' are essential for understanding the stages of reproduction.

(12-16) Given below are six sets with four terms each. In each set one term is odd and cannot be grouped in the same category to which the other three belong. Identify the odd one in each set and name the category to which the remaining three belong. The first one has been done as an example.
Example: Calyx, Corolla, Stamens, Midrib
Odd term: midrib
Category: Parts of a flower

12. Haemoglobin, Glucagon, Iodopsin, Rhodopsin [1 Mark]

Answer:
Odd term: Glucagon
Category: Pigments (or Photoreceptor pigments)

Teacher's Note:
a) Haemoglobin is a blood pigment, while iodopsin and rhodopsin are photoreceptor pigments found in the retina of the eye.
b) Glucagon is a hormone, not a pigment, and its function is to regulate blood glucose levels.

13. Urethra, Uterus, Urinary bladder, Ureter [1 Mark]

Answer:
Odd term: Uterus
Category: Parts of the excretory system (or Urinary system)

Teacher's Note:
a) The urethra, urinary bladder, and ureter are all components of the human excretory (urinary) system, involved in the formation and elimination of urine.
b) The uterus is a part of the female reproductive system, responsible for housing and nourishing the developing fetus.

14. Transpiration, Photosynthesis, Phagocytosis, Guttation [1 Mark]

Answer:
Odd term: Phagocytosis
Category: Plant processes

Teacher's Note:
a) Transpiration, photosynthesis, and guttation are all physiological processes occurring in plants.
b) Phagocytosis is a cellular process where a cell engulfs solid particles, primarily observed in animal cells (e.g., white blood cells) for defense or feeding.

15. Cyton, Photon, Axon, Dendron [1 Mark]

Answer:
Odd term: Photon
Category: Parts of a neuron

Teacher's Note:
a) Cyton (cell body), axon, and dendron are the main structural components of a neuron, which is the basic unit of the nervous system.
b) A photon is a quantum of light energy and is unrelated to the structure of a neuron.

16. Oxytocin, Insulin, Prolactin, Progesterone [1 Mark]

Answer:
Odd term: Insulin
Category: Hormones related to reproduction/pregnancy

Teacher's Note:
a) Oxytocin and prolactin are pituitary hormones involved in childbirth and lactation, respectively, while progesterone is a female sex hormone crucial for maintaining pregnancy.
b) Insulin is a pancreatic hormone primarily involved in glucose metabolism and is not directly related to reproduction.

(17-20) The figure given below represents an experimental setup with a weighing machine to demonstrate a particular process in plants. The experimental setup was placed in bright sunlight. Study the diagram and answer the following questions:

[Figure: An experimental setup showing two test tubes, A and B, placed on a weighing machine. Test tube A contains a potted plant with its roots immersed in water, and the top of the test tube is covered with oil. Test tube B contains only water with an oil layer on top, acting as a control. Both are exposed to sunlight.]

17. Name the process intended for study. [1 Mark]

Answer: Transpiration.

Teacher's Note:
a) The setup aims to measure the loss of water from the plant, which is the definition of transpiration.
b) The oil layer prevents evaporation from the water surface, ensuring that any weight loss is due to the plant's activity.

18. Define the above mentioned process. [1 Mark]

Answer: Transpiration is the process of loss of water in the form of water vapour from the aerial parts of a plant, primarily through stomata in leaves.

Teacher's Note:
a) It is a physiological process driven by the difference in water potential between the plant and the atmosphere.
b) Transpiration creates a 'transpirational pull' that helps in the ascent of sap and also cools the plant.

19. When the weight of the test tube (A & B) is taken before and after the experiment, what is observed? Give reasons to justify your observation in A & B. [2 Marks]

Answer:
Observation:
Test tube A (with plant) will show a significant decrease in weight.
Test tube B (control) will show a negligible or no change in weight.
Reasons:
In test tube A, the plant loses water through transpiration, leading to a reduction in the total weight of the setup.
In test tube B, there is no plant to transpire, and the oil layer prevents evaporation from the water surface, so no significant water loss occurs.

Teacher's Note:
a) This experiment demonstrates that plants actively release water vapour into the atmosphere, which can be quantified by measuring weight loss.
b) The control setup (Test tube B) is crucial to confirm that the observed weight loss in Test tube A is indeed due to the plant's activity and not other factors like evaporation from the water surface.

20. What is the purpose of keeping the test tube B in the experimental setup? [1 Mark]

Answer: Test tube B serves as a control to prove that water loss in test tube A is due to transpiration by the plant and not due to evaporation from the water surface.

Teacher's Note:
a) A control setup isolates the variable being tested (the plant's transpiration) by keeping all other conditions the same.
b) Without a control, it would be impossible to definitively conclude that the observed weight change is solely attributable to the plant's biological process.

21. Match the items given in Column A with the most appropriate ones in Column B and rewrite the correct matching pairs from Column A and Column B: [5 Marks]

Answer:

Teacher's Note:
a) This question assesses the student's ability to associate biological structures, substances, or processes with their related functions or components.
b) Understanding the specific roles of glands, environmental pollutants, reproductive structures, blood components, and plant processes is key to correctly matching the pairs.

(22-26) Choose the correct answer from the options given below:

22. Cretinism and Myxoedema are due to [1 Mark]
A. Hypersecretion of thyroxin
B. Hypersecretion of growth hormone
C. Hyposecretion of thyroxin
D. Hyposecretion of growth hormone

Answer: (C) Hyposecretion of thyroxin

Teacher's Note:
a) Cretinism is caused by hyposecretion of thyroxin in children, leading to stunted physical and mental development. Myxoedema is caused by hyposecretion of thyroxin in adults, resulting in lethargy, weight gain, and puffiness.
b) Hypersecretion of thyroxin leads to conditions like exophthalmic goitre (Graves' disease), while growth hormone imbalances cause gigantism or dwarfism.

23. Which of the following is not a natural reflex action? [1 Mark]
A. Knee-jerk
B. Blinking of eyes due to strong light
C. Salivation at the sight of food
D. Sneezing when any irritant enters the nose

Answer: (C) Salivation at the sight of food

Teacher's Note:
a) Knee-jerk, blinking, and sneezing are unconditioned (natural) reflex actions, meaning they are inborn and do not require prior learning.
b) Salivation at the sight of food is a conditioned reflex, which is learned through experience (e.g., Pavlov's experiment) and can be modified or extinguished.

24. After mitotic cell division, a female human cell will have [1 Mark]
A. 44 + XX chromosomes
B. 44 + XY chromosomes
C. 22 + X chromosomes
D. 22 + Y chromosomes

Answer: (A) 44 + XX chromosomes

Teacher's Note:
a) Mitotic cell division produces two daughter cells that are genetically identical to the parent cell.
b) A normal female human somatic cell has 46 chromosomes (diploid, 2n), consisting of 44 autosomes and two X sex chromosomes (XX).

25. The antibiotic penicillin is obtained from [1 Mark]
A. Protozoan
B. Bacteria
C. Virus
D. Fungus

Answer: (D) Fungus

Teacher's Note:
a) Penicillin, the first widely used antibiotic, is produced by the fungus Penicillium chrysogenum (formerly Penicillium notatum).
b) This discovery by Alexander Fleming revolutionised medicine by providing an effective treatment for bacterial infections.

26. The site of maturation of human sperms is the [1 Mark]
A. Seminiferous tubule
B. Interstitial cells
C. Epididymis
D. Prostate gland

Answer: (C) Epididymis

Teacher's Note:
a) Sperms are produced in the seminiferous tubules of the testes but undergo maturation and acquire motility in the epididymis.
b) The epididymis also serves as a storage site for mature sperms before ejaculation.

(27-31) State the exact location of the following:

27. Tricuspid valve [1 Mark]

Answer: Between the right atrium and the right ventricle of the heart.

Teacher's Note:
a) The tricuspid valve is one of the atrioventricular valves, preventing the backflow of blood from the right ventricle into the right atrium during ventricular systole.
b) Its name comes from having three cusps or flaps.

28. Amnion [1 Mark]

Answer: Innermost membrane enclosing the embryo/fetus, forming the amniotic sac filled with amniotic fluid.

Teacher's Note:
a) The amnion is one of the extraembryonic membranes that develops during pregnancy.
b) The amniotic fluid within the amnion protects the embryo from mechanical shocks, desiccation, and temperature fluctuations.

29. Yellow spot [1 Mark]

Answer: In the retina of the eye, slightly above the blind spot, containing the fovea centralis.

Teacher's Note:
a) The yellow spot, also known as the macula lutea, is the area of the retina responsible for sharp, detailed central vision.
b) The fovea centralis, located at the center of the yellow spot, has the highest concentration of cone cells, making it the point of clearest vision.

30. Seminal vesicle [1 Mark]

Answer: Posterior to the urinary bladder, at the base of the vas deferens, in males.

Teacher's Note:
a) The seminal vesicles are a pair of glands that secrete a viscous, alkaline fluid rich in fructose, prostaglandins, and clotting proteins.
b) This fluid constitutes a significant portion of semen, providing nourishment to sperm and aiding in their motility and survival.

31. Adrenal gland [1 Mark]

Answer: On top of each kidney.

Teacher's Note:
a) The adrenal glands are endocrine glands, each composed of an outer cortex and an inner medulla.
b) They produce a variety of hormones, including adrenaline, noradrenaline, cortisol, and aldosterone, which regulate stress response, metabolism, and electrolyte balance.

SECTION II [40 Marks]

Attempt any four questions from this section.

(32-36) Differentiate between the following pairs on the basis of what is mentioned within brackets:

32. Spinal nerves and Cranial nerves (number of nerves) [1 Mark]

Answer:

Teacher's Note:
a) Spinal nerves originate from the spinal cord and are typically mixed nerves, carrying both sensory and motor information to and from the body.
b) Cranial nerves originate directly from the brain and can be sensory, motor, or mixed, serving primarily the head and neck region.

33. Near vision and Distant vision (shape of the eye lens) [1 Mark]

Answer:

Teacher's Note:
a) For near vision, the ciliary muscles contract, relaxing the suspensory ligaments, which allows the elastic lens to become more convex to increase its converging power.
b) For distant vision, the ciliary muscles relax, increasing tension in the suspensory ligaments, which pulls the lens, making it thinner and less convex to decrease its converging power.

34. Corpus callosum and Corpus luteum (function) [1 Mark]

Answer:

Teacher's Note:
a) The corpus callosum is a large bundle of nerve fibers in the brain, essential for integrating sensory, motor, and cognitive functions between the left and right hemispheres.
b) The corpus luteum is a temporary endocrine structure in the ovary, formed after ovulation, critical for establishing and maintaining pregnancy by producing progesterone.

35. Turgor pressure and Wall pressure (explain) [1 Mark]

Answer:

Teacher's Note:
a) Turgor pressure is crucial for maintaining the rigidity and shape of plant cells, supporting non-woody plants, and driving cell expansion.
b) Wall pressure is an equal and opposite force to turgor pressure, ensuring that the cell does not burst due to excessive water intake.

36. Disinfectant and Antiseptic (definition) [1 Mark]

Answer:

Teacher's Note:
a) Disinfectants are generally stronger and more toxic, hence unsuitable for living tissues (e.g., bleach, phenol).
b) Antiseptics are milder and safe for topical application on the body (e.g., Dettol, Savlon, iodine solution).

(37-41) The diagram below represents the simplified pathway of the circulation of blood. Study the same and answer the questions which follow:

[Figure: A simplified diagram showing blood circulation. Arrows indicate blood flow.
- Head is connected to Heart via vessel 1 (going up from heart) and vessel 8 (coming down to heart).
- Lungs are connected to Heart via vessel 3 (going up from heart) and vessel 4 (coming down to heart).
- Heart is connected to Liver via vessel 2 (going from heart to liver) and vessel 5 (coming from liver to heart).
- Liver is connected to Small intestine via vessel 7 (going from small intestine to liver) and vessel 6 (going from liver to small intestine).
- Vessel 2 also connects to Small intestine.
- Vessel 8 also connects to Small intestine.
- Vessel 1 is an artery to the head. Vessel 8 is a vein from the head.
- Vessel 3 is pulmonary artery. Vessel 4 is pulmonary vein.
- Vessel 2 is hepatic artery (branch of aorta). Vessel 5 is hepatic vein.
- Vessel 7 is hepatic portal vein. Vessel 6 is an artery to the small intestine (branch of aorta).
- The diagram shows a double circulation pattern.]

37. Name the blood vessels labelled 1 and 2. [1 Mark]

Answer:
1: Aorta (or Systemic Artery leading to head)
2: Hepatic Artery

Teacher's Note:
a) Vessel 1 represents the main artery (aorta) carrying oxygenated blood from the heart to the upper body, including the head.
b) Vessel 2 is a branch of the aorta, the hepatic artery, supplying oxygenated blood to the liver.

38. State the function of blood vessels labelled 5 and 8. [1 Mark]

Answer:
5 (Hepatic Vein): Carries deoxygenated blood from the liver to the heart (inferior vena cava).
8 (Systemic Vein/Vena Cava): Carries deoxygenated blood from the head (and upper body) to the heart (superior vena cava).

Teacher's Note:
a) Veins generally carry deoxygenated blood towards the heart, with the exception of pulmonary veins.
b) The hepatic vein (5) is unique as it carries blood that has already passed through two capillary beds (intestine and liver) before returning to the heart.

39. What is the importance of the blood vessel labelled 6? [1 Mark]

Answer: Blood vessel 6 (Artery to small intestine) supplies oxygenated blood and nutrients to the small intestine for its metabolic activities.

Teacher's Note:
a) The small intestine is a highly active organ involved in digestion and absorption, requiring a rich supply of oxygen and nutrients.
b) Arteries are responsible for delivering this oxygenated blood from the heart to various organs.

40. Which blood vessel will contain a high amount of glucose and amino acids after a meal? [1 Mark]

Answer: Blood vessel 7 (Hepatic Portal Vein).

Teacher's Note:
a) After a meal, digested carbohydrates are absorbed as glucose and proteins as amino acids in the small intestine.
b) The hepatic portal vein (7) collects this nutrient-rich blood from the small intestine and transports it directly to the liver for processing before it enters the general circulation.

41. Draw a diagram of the different blood cells as seen in a smear of human blood. [1 Mark]

Answer:
[Figure: diagram to be added]
The diagram should show:
1. Red Blood Cells (Erythrocytes): Biconcave, disc-shaped, anucleated, numerous.
2. White Blood Cells (Leukocytes): Larger than RBCs, nucleated, less numerous, various types (e.g., Neutrophil with multi-lobed nucleus, Lymphocyte with large round nucleus, Monocyte with kidney-shaped nucleus, Eosinophil with bi-lobed nucleus and red granules, Basophil with bi-lobed nucleus and dark granules).
3. Platelets (Thrombocytes): Small, irregular fragments, anucleated, involved in clotting.

Teacher's Note:
a) Students should be able to identify and draw the characteristic shapes and features of each type of blood cell, especially the nucleus of WBCs and the absence of a nucleus in mature RBCs and platelets.
b) Emphasise the relative sizes and abundance of each cell type as seen in a blood smear.

(42-46) A candidate in order to study the process of osmosis has taken 3 potato cubes and put them in 3 different beakers containing 3 different solutions. After 24 hours, in the first beaker the potato cube increased in size, in the second beaker the potato cube decreased in size and in the third beaker there was no change in the size of the potato cube. The following diagram shows the result of the same experiment:

[Figure: Three beakers, each containing a potato cube.
- Beaker 1: Potato cube has increased in size.
- Beaker 2: Potato cube has decreased in size.
- Beaker 3: Potato cube size remains the same.]

42. Give the technical terms of the solutions used in beakers, 1, 2 and 3. [1 Mark]

Answer:
1: Hypotonic solution
2: Hypertonic solution
3: Isotonic solution

Teacher's Note:
a) A hypotonic solution has a lower solute concentration than the cell sap, causing water to enter the cell and make it swell.
b) A hypertonic solution has a higher solute concentration, causing water to leave the cell and make it shrink; an isotonic solution has the same solute concentration, resulting in no net movement of water.

43. In beaker 3, the size of the potato cube remains the same. Explain the reason in brief. [1 Mark]

Answer: In beaker 3, the solution is isotonic to the potato cells, meaning the concentration of solutes inside the potato cells is equal to that of the external solution. Therefore, there is no net movement of water across the cell membrane, and the potato cube's size remains unchanged.

Teacher's Note:
a) Isotonic conditions represent a state of equilibrium where water molecules move in and out of the cell at equal rates.
b) This balance prevents any significant change in cell volume or turgidity.

44. Write the specific feature of the cell sap of root hairs which helps in absorption of water. [1 Mark]

Answer: The cell sap of root hairs has a higher concentration of solutes (e.g., sugars, salts) than the surrounding soil water, creating a lower water potential.

Teacher's Note:
a) This difference in water potential establishes a concentration gradient, driving water into the root hair cells by osmosis.
b) The active transport of mineral ions into root hair cells contributes significantly to maintaining this high solute concentration.

45. What is osmosis? [1 Mark]

Answer: Osmosis is the movement of water molecules from a region of higher water concentration (lower solute concentration) to a region of lower water concentration (higher solute concentration) across a semi-permeable membrane.

Teacher's Note:
a) It is a passive process, meaning it does not require metabolic energy.
b) The semi-permeable membrane is crucial, allowing water molecules to pass through but restricting the movement of larger solute molecules.

46. How does a cell wall and a cell membrane differ in their permeability? [1 Mark]

Answer:
Cell wall: Fully permeable (allows all substances to pass through).
Cell membrane: Selectively permeable (allows only certain substances to pass through, while restricting others).

Teacher's Note:
a) The cell wall, being rigid and porous, provides structural support and protection but does not regulate the entry or exit of substances.
b) The cell membrane, composed of a lipid bilayer with embedded proteins, actively controls the movement of molecules, maintaining the cell's internal environment.

(47-52) A potted plant was taken in order to prove a factor necessary for photosynthesis. The
potted plant was kept in the dark for 24 hours. One of the leaves was covered with black paper in the centre. The potted plant was then placed in sunlight for a few hours.

47. What aspect of photosynthesis was being tested? [1 Mark]

Answer: The necessity of light for photosynthesis.

Teacher's Note:
a) By covering a part of the leaf with black paper, light is excluded from that specific area, allowing for a direct comparison with the exposed parts.
b) The subsequent starch test will reveal whether photosynthesis (and thus starch production) occurred in the presence or absence of light.

48. Why was the plant placed in the dark before beginning the experiment? [1 Mark]

Answer: To destarch the leaves, ensuring that any starch detected at the end of the experiment is newly formed during the experiment and not pre-existing.

Teacher's Note:
a) Keeping the plant in the dark for 24 hours forces it to use up any stored starch in its leaves for respiration.
b) This step is crucial for a valid experiment, as it provides a baseline (starch-free leaves) to accurately observe the effect of the experimental variable (light).

49. During the starch test, why was the leaf boiled in water [1 Mark]

Answer: To kill the cells and make the leaf permeable to iodine solution.

Teacher's Note:
a) Boiling in water denatures enzymes and breaks down cell membranes, allowing for easier penetration of subsequent reagents.
b) This step ensures that the iodine solution can reach the starch granules inside the cells for a clear colour change.

50. During the starch test, why was the leaf boiled in methylated spirit [1 Mark]

Answer: To remove chlorophyll (decolourise the leaf), as chlorophyll's green colour would mask the blue-black colour change of the starch test.

Teacher's Note:
a) Methylated spirit (ethanol) is a solvent for chlorophyll, extracting the pigment from the leaf.
b) Decolourisation makes the observation of the iodine test results much clearer and more accurate.

51. Write a balanced chemical equation to represent the process of photosynthesis. [1 Mark]

Answer:
6CO₂ + 6H₂O \(\xrightarrow{\text{Light energy}}\) C₆H₁₂O₆ + 6O₂

Teacher's Note:
a) This equation summarises the overall process where carbon dioxide and water are converted into glucose and oxygen in the presence of light energy and chlorophyll.
b) Students must ensure the equation is balanced and includes light energy and chlorophyll as conditions for the reaction.

52. Draw a neat diagram of a chloroplast and label its parts. [1 Mark]

Answer:
[Figure: diagram to be added]
The diagram should show:
1. Outer membrane
2. Inner membrane
3. Stroma (fluid-filled space)
4. Thylakoid (disc-shaped sac)
5. Granum (stack of thylakoids)
6. Stroma lamellae (intergranal thylakoid)
7. Starch granule (optional)
8. Lipid droplet (optional)

Teacher's Note:
a) A chloroplast is a double-membraned organelle, with the inner membrane enclosing the stroma and a system of flattened sacs called thylakoids.
b) The thylakoids are stacked into grana, which are the sites of light-dependent reactions, while the stroma is where light-independent reactions (Calvin cycle) occur.

(53-57) The diagram given below is a representation of a certain phenomenon pertaining to the nervous system. Study the diagram and answer the following questions:

[Figure: A diagram showing a reflex arc.
- Part 1: Receptor (e.g., skin)
- Part 2: Sensory neuron (afferent neuron) carrying impulse towards the spinal cord.
- Part 3: Relay neuron (interneuron) within the spinal cord.
- Part 4: Motor neuron (efferent neuron) carrying impulse away from the spinal cord.
- Part 5: Effector (e.g., muscle)
- Part 6: Nerve ending/dendrite of sensory neuron.]

53. Name the phenomenon which is being depicted. [1 Mark]

Answer: Reflex action (or Reflex arc).

Teacher's Note:
a) A reflex action is an involuntary, rapid, and automatic response to a stimulus, without conscious thought.
b) The pathway taken by the nerve impulse during a reflex action is called a reflex arc.

54. Give the technical term for the point of contact between the two nerve cells. [1 Mark]

Answer: Synapse.

Teacher's Note:
a) A synapse is the junction between two neurons where nerve impulses are transmitted from one neuron to another, typically via neurotransmitters.
b) It ensures unidirectional flow of nerve impulses.

55. Name the parts 1, 2, 3 and 4. [1 Mark]

Answer:
1: Receptor
2: Sensory neuron (Afferent neuron)
3: Relay neuron (Interneuron/Association neuron)
4: Motor neuron (Efferent neuron)

Teacher's Note:
a) The receptor detects the stimulus, the sensory neuron carries the impulse to the CNS, the relay neuron processes it, and the motor neuron carries the response to the effector.
b) Understanding the sequence of components in a reflex arc is fundamental to understanding nervous system function.

56. Write the functions of parts 5 and 6 [1 Mark]

Answer:
5 (Effector): Responds to the stimulus (e.g., muscle contracts, gland secretes).
6 (Dendrite/Nerve ending of sensory neuron): Receives the stimulus from the receptor.

Teacher's Note:
a) The effector is the final component of the reflex arc, carrying out the appropriate action in response to the stimulus.
b) The dendrites are specialised to receive incoming signals and transmit them towards the cell body.

57. How does the arrangement of neurons in the spinal cord differ from that of the brain? [1 Mark]

Answer:
Spinal cord: Grey matter (cell bodies) is internal (H-shaped), and white matter (axons) is external.
Brain: Grey matter (cerebral cortex) is external, and white matter is internal.

Teacher's Note:
a) Grey matter consists mainly of neuronal cell bodies, dendrites, unmyelinated axons, and synapses, responsible for processing information.
b) White matter consists primarily of myelinated axons, which transmit signals rapidly between different parts of the brain and spinal cord.

(58-62) Give scientific reasons for the following statements:

58. Use of CFC is banned in many countries. [1 Mark]

Answer: CFCs (Chlorofluorocarbons) are banned because they deplete the ozone layer in the stratosphere, which protects Earth from harmful ultraviolet (UV) radiation, leading to increased risks of skin cancer, cataracts, and damage to ecosystems.

Teacher's Note:
a) CFCs are very stable and can persist in the atmosphere for decades, eventually reaching the stratosphere where they release chlorine atoms.
b) These chlorine atoms act as catalysts, breaking down thousands of ozone molecules, thus thinning the ozone layer.

59. We cannot distinguish colours in moonlight. [1 Mark]

Answer: We cannot distinguish colours in moonlight because moonlight is very dim, and only the rod cells in our retina are sensitive enough to function in low light conditions. Rod cells are responsible for black-and-white vision, while cone cells (responsible for colour vision) require brighter light to be stimulated.

Teacher's Note:
a) The human retina contains two types of photoreceptor cells: rods and cones. Rods are highly sensitive to light and function in dim light (scotopic vision), but do not detect colour.
b) Cones are less sensitive to light but are responsible for high-acuity colour vision (photopic vision) in bright light.

60. Balsam plants wilt during mid-day even if the soil is well watered. [1 Mark]

Answer: Balsam plants wilt during mid-day because the rate of transpiration (water loss from leaves) exceeds the rate of water absorption by the roots, even if the soil is well watered. This leads to a temporary loss of turgor pressure in the cells.

Teacher's Note:
a) At mid-day, high temperatures, intense sunlight, and low humidity increase the rate of transpiration significantly.
b) While water is available in the soil, the plant's vascular system cannot transport it quickly enough to compensate for the rapid loss, causing temporary wilting.

61. Carbon monoxide is highly dangerous when inhaled. [1 Mark]

Answer: Carbon monoxide (CO) is highly dangerous because it has a much higher affinity (about 200-250 times) for haemoglobin than oxygen. When inhaled, it readily binds to haemoglobin to form carboxyhaemoglobin, which is a stable compound, thus reducing the oxygen-carrying capacity of the blood and leading to oxygen starvation in tissues.

Teacher's Note:
a) Carboxyhaemoglobin is stable and does not readily release oxygen, effectively rendering the haemoglobin unavailable for oxygen transport.
b) Even small concentrations of CO can be fatal because it displaces oxygen from haemoglobin, leading to hypoxia and cellular damage.

62. A person walks clumsily after consuming alcohol. [1 Mark]

Answer: Alcohol primarily affects the cerebellum, which is the part of the brain responsible for coordinating muscle movements, maintaining balance, and regulating posture. When the cerebellum is impaired by alcohol, a person loses coordination, leading to clumsy movements and an unsteady gait.

Teacher's Note:
a) Alcohol is a depressant that affects various parts of the brain, but its impact on the cerebellum is particularly noticeable in motor control.
b) Other effects of alcohol include impaired judgment, slowed reaction time, and slurred speech, which are due to its impact on other brain regions.

(63-67) Given below is a diagram representing a stage during mitotic cell division. Study it carefully and answer the questions which follow:

[Figure: A diagram showing a cell undergoing division. Chromosomes are aligned at the metaphase plate (equatorial plane). Each chromosome consists of two sister chromatids. There are no centrioles shown, and a cell wall is not explicitly depicted, but the overall shape is somewhat rectangular, suggesting a plant cell. Spindle fibers are attached to the centromeres.]

63. Is it a plant cell or an animal cell? Give a reason to support your answer. [1 Mark]

Answer: It is a plant cell.
Reason: The absence of centrioles (asters) at the poles and the somewhat rectangular shape of the cell indicate it is a plant cell.

Teacher's Note:
a) In animal cells, centrioles are present at the poles and form asters, which are absent in higher plant cells.
b) The presence of a rigid cell wall in plant cells often gives them a more fixed, angular shape compared to the typically rounded animal cells.

64. Identify the stage shown. [1 Mark]

Answer: Metaphase.

Teacher's Note:
a) The defining characteristic of metaphase is the alignment of all chromosomes at the equatorial plate (metaphase plate) of the cell.
b) Each chromosome is still composed of two sister chromatids, and spindle fibers are fully formed and attached to the kinetochores of the centromeres.

65. Name the stage which follows the one shown here. How is that stage identified? [1 Mark]

Answer: The stage that follows is Anaphase.
It is identified by the separation of sister chromatids, which move towards opposite poles of the cell, becoming individual chromosomes.

Teacher's Note:
a) During anaphase, the centromeres divide, and the sister chromatids are pulled apart by the shortening of spindle fibers.
b) This ensures that each daughter cell receives an identical set of chromosomes.

66. How will you differentiate between mitosis and meiosis on the basis of the chromosome number in the daughter cells? [1 Mark]

Answer:
Mitosis: Daughter cells have the same number of chromosomes as the parent cell (diploid, 2n).
Meiosis: Daughter cells have half the number of chromosomes as the parent cell (haploid, n).

Teacher's Note:
a) Mitosis is involved in growth, repair, and asexual reproduction, producing genetically identical cells.
b) Meiosis is involved in sexual reproduction, producing gametes with half the chromosome number, ensuring genetic diversity.

67. Draw a duplicated chromosome and label its parts. [1 Mark]

Answer:
[Figure: diagram to be added]
The diagram should show:
1. Two sister chromatids (each a replicated DNA molecule).
2. Centromere (the constricted region where sister chromatids are joined).
3. Kinetochore (protein structure on the centromere where spindle fibers attach, optional but good to include).

Teacher's Note:
a) A duplicated chromosome consists of two identical sister chromatids, formed during the S phase of the cell cycle.
b) The centromere is crucial for the proper segregation of chromosomes during cell division.

(68-69) Name the disease for which the following types of vaccines are given:

68. Salk's vaccine [1 Mark]

Answer: Polio (Poliomyelitis).

Teacher's Note:
a) Salk's vaccine is an inactivated polio vaccine (IPV), administered by injection.
b) It was developed by Jonas Salk and was one of the first effective vaccines against polio, a viral disease that can cause paralysis.

69. BCG [1 Mark]

Answer: Tuberculosis (TB).

Teacher's Note:
a) BCG stands for Bacillus Calmette-Guérin, and it is a vaccine primarily used against tuberculosis.
b) It is particularly effective in preventing severe forms of TB in children, such as TB meningitis and disseminated TB.

(70-73) Give one example of each of the following:

70. A water pollutant [1 Mark]

Answer: Industrial effluents (or sewage, pesticides, heavy metals, plastics, detergents).

Teacher's Note:
a) Water pollutants are substances that contaminate water bodies, making them harmful to humans, animals, and aquatic life.
b) Industrial effluents often contain toxic chemicals, heavy metals, and organic pollutants that can severely degrade water quality.

71. An aquatic plant used in the laboratory to demonstrate O₂ liberation during photosynthesis [1 Mark]

Answer: Hydrilla (or Elodea).

Teacher's Note:
a) Hydrilla is commonly used in experiments to demonstrate oxygen liberation because it is an aquatic plant that photosynthesises efficiently underwater.
b) The oxygen bubbles produced during photosynthesis can be easily observed and collected, providing direct evidence of the process.

72. An antibiotic [1 Mark]

Answer: Penicillin (or Streptomycin, Tetracycline, Amoxicillin).

Teacher's Note:
a) Antibiotics are medicines that kill or inhibit the growth of bacteria, used to treat bacterial infections.
b) They work by targeting specific bacterial structures or processes, such as cell wall synthesis or protein production.

73. A nitrogenous base in DNA [1 Mark]

Answer: Adenine (or Guanine, Cytosine, Thymine).

Teacher's Note:
a) DNA contains four nitrogenous bases: Adenine (A), Guanine (G), Cytosine (C), and Thymine (T).
b) These bases pair specifically (A with T, G with C) to form the rungs of the DNA double helix, carrying genetic information.

(74-77) Expand the following biological abbreviations:

74. ATP [1 Mark]

Answer: Adenosine Triphosphate.

Teacher's Note:
a) ATP is the primary energy currency of the cell, used to power most cellular processes.
b) Energy is released when the terminal phosphate bond is hydrolysed to form ADP (Adenosine Diphosphate).

75. TSH [1 Mark]

Answer: Thyroid Stimulating Hormone.

Teacher's Note:
a) TSH is secreted by the anterior pituitary gland.
b) Its main function is to stimulate the thyroid gland to produce and release thyroid hormones (thyroxine and triiodothyronine).

76. DPT [1 Mark]

Answer: Diphtheria, Pertussis (Whooping Cough), Tetanus.

Teacher's Note:
a) DPT is a combination vaccine that protects against three serious bacterial diseases.
b) It is typically given to infants and young children as part of routine immunisation schedules.

77. DNA [1 Mark]

Answer: Deoxyribonucleic Acid.

Teacher's Note:
a) DNA is the genetic material found in all living organisms, carrying the instructions for development, functioning, growth, and reproduction.
b) It is a double helix structure composed of nucleotides, each containing a deoxyribose sugar, a phosphate group, and a nitrogenous base.

(78-82) The given diagram represents a nephron and its blood supply. Study the diagram and answer the following questions:

[Figure: A diagram of a nephron and its associated blood vessels.
- Vessel 5: Afferent arteriole (entering glomerulus)
- Vessel 6: Efferent arteriole (leaving glomerulus)
- Structure 4: Glomerulus (capillary network)
- Structure 1: Bowman's capsule (cup-shaped structure enclosing glomerulus)
- Structure 2: Proximal Convoluted Tubule (PCT)
- Structure 3: Loop of Henle (descending and ascending limbs)
- The diagram shows the renal artery branching into afferent arteriole (5), forming glomerulus (4), then efferent arteriole (6), which then forms a capillary network around the tubules. Bowman's capsule (1) leads to PCT (2), then Loop of Henle (3), and then distal convoluted tubule (not numbered but implied before collecting duct).]

78. Label parts 1, 2, 3 and 4. [1 Mark]

Answer:
1: Bowman's capsule
2: Proximal Convoluted Tubule (PCT)
3: Loop of Henle
4: Glomerulus

Teacher's Note:
a) These four parts constitute the renal corpuscle (Bowman's capsule and glomerulus) and the renal tubule (PCT and Loop of Henle), which are essential for urine formation.
b) Correct identification of these structures is fundamental to understanding kidney function.

79. State the reason for the high hydrostatic pressure in the glomerulus. [1 Mark]

Answer: The afferent arteriole (vessel 5) entering the glomerulus is wider than the efferent arteriole (vessel 6) leaving it. This difference in diameter creates resistance to blood flow, leading to a build-up of high hydrostatic pressure within the glomerulus.

Teacher's Note:
a) This high hydrostatic pressure is crucial for ultrafiltration, forcing water and small solutes from the blood into Bowman's capsule.
b) The efferent arteriole's narrower lumen acts like a bottleneck, maintaining the pressure gradient necessary for filtration.

80. Name the blood vessel which contains the least amount of urea in this diagram. [1 Mark]

Answer: The blood vessel leaving the kidney (Renal Vein, not explicitly numbered but implied after filtration and reabsorption).

Teacher's Note:
a) The primary function of the kidney is to filter waste products, including urea, from the blood.
b) Therefore, blood that has passed through the nephrons and is leaving the kidney via the renal vein will have the lowest concentration of urea.

81. Name the two main stages of urine formation. [1 Mark]

Answer:
1. Ultrafiltration
2. Selective Reabsorption

Teacher's Note:
a) Ultrafiltration occurs in the glomerulus and Bowman's capsule, where blood plasma is filtered to form glomerular filtrate.
b) Selective reabsorption occurs along the renal tubule, where useful substances like glucose, amino acids, and most water are reabsorbed back into the blood, while waste products remain in the filtrate.

82. Name the part of the nephron which lies in the renal medulla. [1 Mark]

Answer: Loop of Henle (and collecting duct, though not explicitly numbered as part of the nephron in the diagram).

Teacher's Note:
a) The Loop of Henle is a U-shaped tubule that extends from the renal cortex into the renal medulla and back.
b) Its primary role is to create a concentration gradient in the medulla, which is essential for the reabsorption of water and the production of concentrated urine.

(83-87) Briefly explain the following terms.

83. Monohybrid cross [1 Mark]

Answer: A monohybrid cross is a genetic cross between two individuals that differ in only one specific trait or characteristic, typically involving alleles of a single gene.

Teacher's Note:
a) It is used to study the inheritance pattern of a single gene and to determine the dominance or recessiveness of alleles.
b) Mendel's experiments with pea plants, such as crossing tall and dwarf plants, are classic examples of monohybrid crosses.

84. Biomedical waste [1 Mark]

Answer: Biomedical waste (or hospital waste) refers to any waste generated during the diagnosis, treatment, or immunisation of human beings or animals, or in research activities pertaining thereto, or in the production or testing of biologicals.

Teacher's Note:
a) This waste includes items like discarded needles, syringes, bandages, human tissues, and laboratory cultures, which can be infectious or hazardous.
b) Proper segregation, treatment, and disposal of biomedical waste are crucial to prevent the spread of diseases and protect public health.

85. Innate immunity [1 Mark]

Answer: Innate immunity is the non-specific, inborn defense system that provides immediate protection against a wide range of pathogens without prior exposure. It includes physical barriers (skin), chemical barriers (stomach acid), and cellular defenses (phagocytes).

Teacher's Note:
a) It is the first line of defense and does not involve memory or specific recognition of pathogens.
b) Components of innate immunity are present from birth and act rapidly to prevent infection.

86. Diapedesis [1 Mark]

Answer: Diapedesis is the process by which white blood cells (leukocytes) squeeze through the intact walls of capillaries and venules to reach sites of infection or inflammation in the tissues.

Teacher's Note:
a) This ability of WBCs is crucial for the immune response, allowing them to migrate from the bloodstream to the affected areas where they can engulf pathogens or initiate other immune reactions.
b) It is a key mechanism for the body's defense against infection.

87. Hormones [1 Mark]

Answer: Hormones are chemical messengers secreted by endocrine glands directly into the bloodstream, which travel to target organs or cells to regulate specific physiological processes.

Teacher's Note:
a) Hormones act in very small concentrations and have specific effects on target cells that possess appropriate receptors.
b) They play vital roles in growth, metabolism, reproduction, mood, and many other bodily functions.

(88-91)

88. State any two harmful effects of noise pollution on human health. [1 Mark]

Answer:
1. Hearing loss (temporary or permanent)
2. Increased stress levels and anxiety (leading to hypertension, sleep disturbances)

Teacher's Note:
a) Prolonged exposure to high levels of noise can damage the delicate hair cells in the inner ear, leading to noise-induced hearing loss.
b) Noise pollution can also trigger physiological stress responses, affecting cardiovascular health and mental well-being.

89. Categorise the following activities according to the functions of the Red Cross Society and the WHO:
(a) To suggest quarantine measures to prevent spread of disease
(b) Humanitarian services to victims of war
(c) To educate people in accident prevention
(d) To promote projects for research on disease [1 Mark]

Answer:
Red Cross Society: (b) Humanitarian services to victims of war, (c) To educate people in accident prevention
WHO (World Health Organization): (a) To suggest quarantine measures to prevent spread of disease, (d) To promote projects for research on disease

Teacher's Note:
a) The Red Cross Society focuses on humanitarian aid, disaster relief, and health education at a community level.
b) The WHO is a global health agency that sets international health standards, coordinates disease prevention efforts, and promotes health research.

90. Write any two major reasons for the population explosion in India. [1 Mark]

Answer:
1. Decline in death rate (due to improved healthcare, sanitation, and control of diseases).
2. High birth rate (due to illiteracy, early marriage, religious beliefs, lack of family planning awareness).

Teacher's Note:
a) The population explosion is primarily a result of a significant gap between birth rates and death rates.
b) Social factors like poverty, traditional values favouring large families, and limited access to contraception also contribute to high birth rates.

91. State Mendel's Law of segregation. [1 Mark]

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

Teacher's Note:
a) This law explains why offspring inherit one allele from each parent and why recessive traits can reappear in later generations.
b) It is also known as the Law of Purity of Gametes because each gamete carries only one allele for a given trait.

(92-101) Give technical terms for the following:

92. A method of contraception in which the sperm duct is cut and ligated [1 Mark]

Answer: Vasectomy.

Teacher's Note:
a) Vasectomy is a permanent method of male sterilisation, where the vas deferens (sperm duct) is cut and sealed to prevent sperm from reaching the urethra.
b) It is a highly effective and safe contraceptive procedure.

93. Statistical study of human population [1 Mark]

Answer: Demography.

Teacher's Note:
a) Demography involves the study of population dynamics, including birth rates, death rates, migration, and population growth.
b) It helps in understanding population trends and planning for future resource allocation and social services.

94. The protective covering of the heart [1 Mark]

Answer: Pericardium.

Teacher's Note:
a) The pericardium is a double-layered sac that encloses the heart, providing protection and anchoring it within the chest cavity.
b) The fluid between its layers reduces friction during heartbeats.

95. A sudden heritable change in the gene [1 Mark]

Answer: Mutation.

Teacher's Note:
a) Mutations are random changes in the DNA sequence that can lead to new alleles and are the ultimate source of genetic variation.
b) They can be spontaneous or induced by mutagens like radiation or certain chemicals.

96. Repeated units of DNA molecule [1 Mark]

Answer: Nucleotides.

Teacher's Note:
a) DNA is a polymer made up of repeating monomer units called nucleotides.
b) Each nucleotide consists of a deoxyribose sugar, a phosphate group, and one of four nitrogenous bases (Adenine, Guanine, Cytosine, or Thymine).

97. The fluid portion of blood [1 Mark]

Answer: Plasma.

Teacher's Note:
a) Plasma is the yellowish liquid component of blood in which blood cells are suspended.
b) It constitutes about 55% of total blood volume and contains water, proteins, salts, hormones, nutrients, and waste products.

98. The nerve which transmits impulses from the ear to the brain [1 Mark]

Answer: Auditory nerve (or Vestibulocochlear nerve).

Teacher's Note:
a) The auditory nerve is a cranial nerve responsible for transmitting sound and balance information from the inner ear to the brain.
b) It comprises two main parts: the cochlear nerve (for hearing) and the vestibular nerve (for balance).

99. Group of hormones which influence other endocrine glands to produce hormones [1 Mark]

Answer: Tropic hormones (or Trophic hormones).

Teacher's Note:
a) Tropic hormones are secreted by the anterior pituitary gland and regulate the function of other endocrine glands.
b) Examples include TSH (Thyroid Stimulating Hormone), ACTH (Adrenocorticotropic Hormone), and FSH (Follicle-Stimulating Hormone).

100. Thin walled sac of skin which covers the testes [1 Mark]

Answer: Scrotum.

Teacher's Note:
a) The scrotum is an external pouch of skin that houses the testes outside the abdominal cavity.
b) Its primary function is to maintain the testes at a temperature slightly lower than body temperature, which is essential for viable sperm production.

101. The permanent stoppage of the menstrual cycle in a woman aged 50 years [1 Mark]

Answer: Menopause.

Teacher's Note:
a) Menopause is a natural biological process that marks the end of a woman's reproductive years, typically occurring around the age of 45-55.
b) It is characterised by the cessation of menstruation and is due to the ovaries producing fewer reproductive hormones.