Question 1

Question 1(e)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: [5](i) Name the process intended for study.(ii) Define the above mentioned process.(iii) 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.(iv) What is the purpose of keeping the test tube B in the experimental setup?

Accepted Answer

(i) Transpiration.(ii) Transpiration is the process in which water is lost in the form of vapour from the aerial parts of the plant.(iii) Weight of test tube A before the experiment was more than its weight after the experiment. This is because water from test tube A has evaporated due to transpiration.Weight of test tube B remains the same before and after the experiment, because no loss of water occurs in test tube B as there is no plant and the oil on the surface prevents direct evaporation of water.(iv) It is a control experiment where the purpose of using test tube B is to compare the level of water in both test tubes.

Question 2

Question 2(a)Differentiate between the following pairs on the basis of what is mentioned within brackets: [5](i) Spinal nerves and Cranial nerves (number of nerves)(ii) Near vision and Distant vision (shape of the eye lens)(iii) Corpus callosum and Corpus luteum (function)(iv) Turgor pressure and Wall pressure (explain)(v) Disinfectant and Antiseptic (definition)

Accepted Answer

(i) Difference between Spinal nerves and Cranial nerves (number of nerves)Spinal nervesCranial nervesSpinal nerves are composed of 31 nerve pairsCranial nerves are composed of 12 nerve pairs.(ii) Difference between near vision and Distant vision (shape of the eye lens)Near visionDistant visionIn near vision, the lens is too curved.In distant vision, the lens is too flat.(iii) Difference between Corpus callosum and Corpus luteum (function)Corpus callosumCorpus luteumConnect the left and right cerebral hemispheres.Secrete female sex hormones Oestrogen and Progesterone.(iv) Difference between Turgor pressure and Wall pressure (explain)Turgor pressureWall pressureThe pressure of cell contents on the cell wall is called Turgor pressure.The pressure exerted by the cell wall on the cell content is called Wall pressure.(v) Difference between Disinfectant and Antiseptic (definition)DisinfectantAntisepticThese are strong chemical substances that are applied on spots and places where germs thrive and multiply.These are mild chemical substances which, when applied on the body, kill germs. These substances cause no harm to the skin and body.

Question 3

Question 2(b)The diagram below represents the simplified pathway of the circulation of blood. Study the same and answer the questions that follow: [5](i) Name the blood vessels labelled 1 and 2.(ii) State the function of blood vessels labelled 5 and 8.(iii) What is the importance of the blood vessel labelled 6?(iv) Which blood vessel will contain a high amount of glucose and amino acids after a meal?(v) Draw a diagram of the different blood cells as seen in a smear of human blood.

Accepted Answer

(i) The blood vessels labelled 1 and 2 are:1 → Anterior / Superior venacava2 → Dorsal aorta(ii) Function of blood vessels —5 → Supply oxygenated blood to liver.8 → Carries deoxygenated blood from the posterior parts of the body to the right auricle of the heart.(iii) Blood vessel 6 is called Hepatic portal vein. It carries deoxygenated blood from intestine to liver. This blood contains excess glucose, some toxic substances etc. which are sent to liver where they are detoxified and the excess glucose is converted to glycogen and stored. This prevents these substances from directly entering and damaging the heart.(iv) Hepatic portal vein (i.e., blood vessel 6) contains a high amount of glucose and amino acids after a meal.(v) Below diagram shows the different blood cells as seen in a smear of human blood:

Question 4

Question 3(a)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: [5](i) Give the technical terms of the solutions used in beakers 1, 2 and 3.(ii) In beaker 3, the size of the potato cube remains the same. Explain the reason in brief.(iii) Write the specific feature of the cell sap of root hairs which helps in absorption of water.(iv) What is osmosis?(v) How does a cell wall and a cell membrane differ in their permeability?

Accepted Answer

(i) The technical terms of the solutions used in beakers 1, 2 and 3 are:Beaker 1 — Hypotonic solution.Beaker 2 — Hypertonic solution.Beaker 3 — Isotonic solution.(ii) In beaker 3, the solution present is an isotonic solution, i.e. the relative concentration of water molecules and solutes is the same in the solution as well as inside the cell. There is no movement of water molecules across the cell membrane. Hence, the size of potato cubes remains the same.(iii) The cell sap of root hair has a higher concentration of solute than the surrounding water. Due to this endosmosis takes place helping the root hairs in absorption of water.(iv) Osmosis is the movement of water molecules from their region of higher concentration (dilute solution or with a lower solute concentration) to their region of lower concentration (concentrated solution or with a higher solute concentration) through a semi-permeable membrane.(v) The cell wall is freely permeable to all substances, while the cell membrane is selectively permeable and allows only certain substances to enter or exit the cell.

Question 5

Question 3(b)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. [5](i) What aspect of photosynthesis was being tested?(ii) Why was the plant placed in the dark before beginning the experiment?(iii) During the starch test, why was the leaf (1) boiled in water (2) boiled in methylated spirit.(iv) Write a balanced chemical equation to represent the process of photosynthesis.(v) Draw a neat diagram of a chloroplast and label its parts.

Accepted Answer

(i) The above experiment is conducted to show that Sunlight is necessary for photosynthesis.(ii) The plant was placed in the dark before beginning the experiment to destarch the leaves.(iii) The leaf was boiled in water to kill the cells and boiled in methylated spirit to remove chlorophyll.(iv) Balanced chemical equation representing the process of photosynthesis is given below:6CO2+12H2O  ⟶chlorophylllight energy  C6H12O6+6H2O+6O2↑6CO_2 + 12H_2O \space \space \overset{	ext{light energy}}{\underset{	ext{chlorophyll}}{\longrightarrow}} \space \space C_6H_{12}O_6 + 6H_2O + 6O_2 \uparrow6CO2​+12H2​O  chlorophyll⟶​light energy​  C6​H12​O6​+6H2​O+6O2​↑(v) Below is the labelled diagram of a chloroplast:

Question 6

Question 4(b)Give scientific reasons for the following statements: [5](i) Use of CFC is banned in many countries.(ii) We cannot distinguish colours in moonlight.(iii) Balsam plants wilt during mid-day even if the soil is well watered.(iv) Carbon monoxide is highly dangerous when inhaled.(v) A person after consuming alcohol walks clumsily.

Accepted Answer

(i) Chlorofluorocarbons (CFC) are one of the major causes for ozone depletion. Hence they are banned in many countries.(ii) Cone cells in the retina are responsible for colour vision. They are functional only in bright light. Moon light is unable to stimulate the cone cells. However, the rod cells respond to dim light. Hence, we can see in moon light but cannot distinguish colours.(iii) In case of balsam plants, the rate of transpiration during mid-day exceeds the rate of absorption of water by the roots. The cells therefore lose turgidity. In the evening or during the night, the stomata are constricted and the temperature is not very high. Therefore, there is no loss of water through transpiration, and the turgidity of the leaves is re-acquired.(iv) Haemoglobin has very strong affinity for Carbon Monoxide forming a stable compound Carboxyhaemoglobin (HbCO). This cuts down the capacity of the blood of transporting Oxygen, sometimes resulting in death. Hence, Carbon monoxide is highly dangerous when inhaled.(v) Alcohol affects the cerebellum part of the brain which is responsible for muscular co-ordination. So after consuming alcohol the cerebellum is unable to coordinate muscular movements properly resulting in the person walking clumsily.

Question 7

Question 5(a)Given below is a diagram representing a stage during mitotic cell division. Study it carefully and answer the questions which follow: [5](i) Is it a plant cell or an animal cell? Give a reason to support your answer.(ii) Identify the stage shown.(iii) Name the stage which follows the one shown here. How is that stage identified?(iv) How will you differentiate between mitosis and meiosis on the basis of the chromosome number in the daughter cells?(v) Draw a duplicated chromosome and label its parts.

Accepted Answer

(i) Plant cell.Reason — Cell wall present and Centrioles absent.(ii) Prophase.(iii) Metaphase follows the prophase. The chromosomes will be arranged at the equator attached to the spindle fibre.(iv) In Mitosis — The chromosome number in the daughter cell remains the same as that of the parent cell.In Meiosis — Daughter cells receive only half the number of chromosomes from the parent cells.(v) Below labelled diagram shows a duplicated chromosome:

Question 8

Question 5(b)(i) Name the disease for which the following types of vaccines are given: [5]Salk's vaccineBCG(ii) Give one example of each of the following:A water pollutantAn aquatic plant used in the lab to demonstrate O2 liberation during photosynthesisAn antibioticA nitrogenous base in DNA(iii) Expand the following biological abbreviations:ATPTSHDPTDNA

Accepted Answer

(i) The vaccines are given for:Salk's vaccine → PoliomyelitisBCG → Tuberculosis(ii) One example of each are:A water pollutant → SewageAn aquatic plant used in the lab to demonstrate O2 liberation during photosynthesis → Hydrilla.An antibiotic → PenicillinA nitrogenous base in DNA → Adenine(iii) The full forms are:ATP → Adenosine Triphosphate.TSH → Thyroid Stimulating Hormone.DPT → Diphtheria Pertussis Tetanus.DNA → Deoxyribonucleic Acid.

Question 9

Question 6(a)The given diagram represents a nephron and its blood supply. Study the diagram and answer the following questions: [5](i) Label parts 1, 2, 3 and 4.(ii) State the reason for the high hydrostatic pressure in the glomerulus.(iii) Name the blood vessel which contains the least amount of urea in this diagram.(iv) Name the two main stages of urine formation.(v) Name the part of the nephron which lies in the renal medulla.

Accepted Answer

(i) The parts 1, 2, 3 and 4 are:1 → Collecting duct.2 → Distal convoluted tubule.3 → Loop of Henle.4 → Bowman's capsule.(ii) The efferent arteriole is narrower than the afferent arteriole. This creates a high hydrostatic pressure in the glomerulus.(iii) Efferent arteriole.(iv) Two main stages of urine formation are:Ultrafiltration.Selective Reabsorption.(v) Henle's loop and Collecting tubules

Question 10

Question 6(b)Briefly explain the following terms. [5](i) Monohybrid cross(ii) Biomedical waste(iii) Innate immunity(iv) Diapedesis(v) Hormones

Accepted Answer

(i) A cross between two parents taking the alternative traits of one single character. For example, a cross between tall and dwarf pea plants.(ii) Biomedical waste refers to the waste from medical services such as needles, syringes, dirty dressings, unused discarded medicinal tablets and powders, used surgical instruments, etc.(iii) Immunity which is by birth, i.e., acquired from parents, is called Innate immunity.(iv) The process in which WBCs (specially monocytes and neutrophils) produce pseudopodia and migrate through the walls of the blood vessels into the tissues to fight against disease causing germs is termed as Diapedesis.(v) Hormones are secretions from some glandular part of the body, that is poured directly into blood and that acts on the target organs or cells of the same individual, bringing about coordination between distant parts of the body.

Question 11

Question 7(a)(i) State any two harmful effects of noise pollution on human health. [5](ii) Categorize the following activities as per the functions of the Red Cross Society and the WHO:To suggest quarantine measures to prevent spread of disease.Humanitarian services to victim of war.To educate people in accident prevention.To promote projects for research on disease.(iii) Write any two major reasons for the population explosion in India.(iv) State Mendel's Law of segregationAnswer(i) Two harmful effects of noise pollution are:It lowers efficiency of work.It disturbs sleep and leads to nervous irritability.(ii) The activities are categorized below:To suggest quarantine measures to prevent spread of disease. → WHOHumanitarian services to victim of war. → Red CrossTo educate people in accident prevention. → Red CrossTo promote projects for research on disease. → WHO(iii) Two major reasons for the population explosion in India are —Desire for a male child.Illiteracy and Lack of awareness about birth control measures.(iv) Law of Segregation (also called the law of purity of gametes) — The two members of a pair of factors separate during the formation of gametes. They do not blend but segregate or separate into different gametes. The gametes combine together by random fusion at the time of zygote formation.

Accepted Answer

Answer: WHO, Red Cross, Red Cross, WHO

Question 12

Question 7(b)Give technical terms for the following: [5](i) A method of contraception in which the sperm duct is cut and ligated.(ii) Statistical study of human population.(iii) The protective covering of the heart.(iv) A sudden heritable change in the gene.(v) Repeated units of DNA molecule.(vi) The fluid portion of blood.(vii) The nerve that transmits impulses from the ear to the brain.(viii) Group of hormones which influence other endocrine glands to produce hormones.(ix) Thin walled sac of skin which covers the testes.(x) The permanent stoppage of the menstrual cycle in a woman aged 50 years.

Accepted Answer

(i) Vasectomy(ii) Demography(iii) Pericardium(iv) Mutation(v) Nucleotide(vi) Plasma(vii) Auditory nerve(viii) Tropic hormones(ix) Scrotum(x) Menopause.

Solved 2014 Question Paper ICSE Class 10 Biology

Section I 40 Marks

Section Ii 40 Marks