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NCERT Exemplar Solution for CBSE Class 10 Science Chapter 6: Life Processes (Part-III)

May 22, 2017 12:13 IST

    Life Processes Class 9, Class 9 Science NCERT Exemplar Problems and SolutionsHere you get the CBSE Class 10 Science chapter 6, Life Processes: NCERT Exemplar Problems and Solutions (Part-III). This part of the chapter includes solutions for Question No.73 to 82 from the NCERT Exemplar Problems for Class 10 Science Chapter: Life Processes. These questions include only the Long Answer Type Questions framed from various important topics of the chapter.

    NCERT Exemplar Solution for CBSE Class 10 Science Chapter 6: Life Processes (Part-I)

    NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Science Board Examination 2017-2018 as well as other competitive exams.

    NCERT Exemplar Solution for CBSE Class 10 Science Chapter 6: Life Processes (Part-II)

    Find below the NCERT Exemplar problems and their solutions for Class 10 Science Chapter, Life Processes:

    Long Answer Type Questions

    Question. 73 Explain the process of nutrition in Amoeba.

    Answer.

    Amoeba follows the holozoic type nutrition. Holozoic refers to engulfment of whole/part of plant/animal in liquid/solid state. The heterotrophic Amoeba engulf the particulate food. The process of obtaining food by Amoeba is called phagocytosis. It is carried out as per the following steps:

    (i) Ingestion: Engulfment of food particle by forming temporary projections called pseudopodia is called as ingestion. The engulfed food is taken inside the body by  pinching off the food vacuole of pseudopodia.

    (ii) Digestion: Breakdown of complex substances into simpler ones by digestive enzymes takes place in food vacuole only.

    (iii) Absorption: The digested food present diffuses from food vacuole into the cytoplasm causing disappearance of food vacuole.

    (iv) Assimilation: Food absorbed is used by Amoeba for various metabolic processes.

    (v) Egestion: The waste products of Amoeba are disposed off in surrounding water by the process of simple diffusion through its cell membrane/ body surface.

    Nutrition in Amoeba

    Question. 74 Describe the alimentary canal of man.

    Answer.

    Human alimentary canal consists of organs of digestion which are namely, in sequence, mouth, oesophagus, stomach, small intestine, large intestine and anus.

    Mouth cavity: It is the oral cavity inside mouth, containing teeth, tongue and three pairs of salivary glands.

    Pharynx: The oral cavity opens into the pharynx. It is the part where food passage and air passage cross each other.

    Oesophagus: It is a long muscular, tube that passes through thoracic cavity and diaphragm into abdominal cavity and carries the food down to the stomach by peristalsis.

    Stomach: The stomach is a thick-walled, J-shaped organ present on the left side of the abdominal cavity and is continuous with the esophagus above and the duodenum of the small intestine below.

    Small intestine: The pyrolic valve of Stomach leads to a tubular structure called as small intestine. It has three divisions, i.e., duodenum, jejunum and ileum.

    Large intestine: The ileoceacal valve of small intestine leads to large intestine, which can be divided into two parts as anterior colon and posterior rectum.

    The rectum opens at the anus, which in turn serves as site of defecation, the expulsion of feces.  

    Human Alimentary Canal

    Question. 75 Explain the process of breathing in man.

    Answer.

    Breathing in humans involves inspiration and expiration that are explained below:

    (i) Inspiration: Inspiration, the active phase of ventilation, includes contraction of diaphragm and the external intercostals muscles and movement of rib cage upward and outward. Deep inspiration involves contraction and lowering down of diaphragm to facilitate expansion of lungs by increasing volume of thoracic cavity. The expansion of thoracic cavity and expansion of lungs create the air pressure difference outside the lung and alveoli to facilitate flow of air into alveoli.

    (ii) Expiration: It is passive phase of breathing that involves relaxed stage of diaphragm (now dome-shaped) and intercostals muscles causing downward and inward movement of rib cage. This causes decreased volume of thoracic cavity and recoiling of lungs. The resultant increased air pressure in alveoli as compared to atmospheric pressure causes air to move outside the body.

    CBSE Class 10 Science Syllabus 2017-2018

    Question. 76 Explain the importance of soil for plant growth.

    Answer.

    Soil is important for plant growth due to the reasons given below:

    (i) Anchorage: Soil holds the plants and provides the base on which-all the terrestrial and some aquatic plants grow.

    (ii) Source of water and mineral: Soil serve as reservoir of water and minerals absorbed by plant roots to support growth and development.

    (iii) The roots of a plant also take the oxygen required for respiration from the soil particles by the process of diffusion.

    (iv) Symbiotic association of plants with soil microbes help plant absorb otherwise unavailable nutrients such as nitrogen.

    Question. 77 Draw the diagram of alimentary canal of man and label the following parts.

    Mouth, Oesophagus, Stomach, Intestine

    Answer.

    The diagram of alimentary canal of man and its parts are shown below:

    Alimentary Canal of Man and its parts

    Question. 78 How do carbohydrates, proteins and fats get digested in human beings?

    Answer.

    Digestion of carbohydrates:

    (i) Oral cavity: Salivary amylase present in saliva breaks down starch into maltose.

    (ii) Small intestine:

    (a) Pancreatic amylase presents in pancreatic juice breakdown the remaining starch into maltose.

    (b) Maltase of intestinal juice breakdown maltose into glucose units.

    Digestion of Proteins

    (i) Mouth does not have any digesting enzyme for proteins and hence proteins digestion does not start in oral cavity.

    (ii) Stomach Pepsin is a protein digesting enzyme secreted by chief cells of stomach wall in its inactive form called as pepsinogen which in turn is converted into active pepsin upon exposure to HCl. Pepsin starts protein digestion into small peptides.

    (iii) Small intestine:

    (a) Pancreatic juice secretes Trypsinogen, chymotripsinogen (inactive from of trypsin). Trypsin and chymotrypsin digest proteins into peptides. 

    (b) Peptidases of intestinal juice digest the peptides into amino acids.

    Digestion of Fats

    (i) Small intestine:

    (a) Emulsification of large globules of fats into smaller droplets by bile salt.

    (b) Pancreatic lipase digests the emulsified fat globules into glycerol and fatty acids.

    (c) Glycerol and fatty acids enter the villi cells where they are rejoined and packaged as lipoprotein droplets before entering the lacteals.

    Question. 79 Explain the mechanism of photosynthesis.

    Answer.

    Photosynthesis is the process involving production of carbohydrates by using carbon dioxide as a chief carbon source and water as electron donor. It can be represented as.

    reaction of photosynthesis process

    Photosynthesis can be divided into two phases: light dependent phase and light independent phase (dark phase).

    (a) Light reaction: Since these reactions cannot take place in absence of light, they are called as light dependent reactions or light reactions. It includes:

    • Absorption of sunlight energy by main and accessory photosynthetic pigments  (chl a, b, c and carotenoids)
    • Oxidation of water into oxygen and hydrogen driven by light energy and preservation of released energy in form of ATP and NADPH

    (b) Dark reaction: Since these reactions do not require Sunlight and can occur in presence or absence of Sunlight, they are termed as dark reaction or light independent reactions.

    • Reduction of CO2 to produce carbohydrates using energy of ATP and NADPH
    • The extra glucose is stored as starch in the leaves of the plant.

    Question. 80 Explain the three pathways of breakdown in living organisms.

    Answer.

    Cellular respiration is the process of breakdown of food substances into simpler ones to release the stored energy. On the basis of presence or absence of oxygen, it can occur through aerobic or anaerobic pathway.

    (i) Aerobic respiration: Glucose in the presence of oxygen (air), breaks down to carbon dioxide and water with release of large amount of energy. It occurs in three steps, glycolysis , citric acid cycle and ETC.

    Glycolysis is breakdown of glucose into pyruvate. It is oxygen independent pathway occurs in cytoplasm. Pyruvate is decarboxylated into acetyl coA in cytoplasm only.

    Citric acid cycle is the second phase of cellular respiration in which acetyl Co are enzymatically oxidized into carbon dioxide and the released energy is stored in NADH and FADH2.

    Electron transport chain includes downhill flow of electrons to final electron acceptor (Oxygen, which is then reduced into water) through a chain of membrane-bound carriers to facilitate the uphill transport of protons across a proton-impermeable membrane and ATP synthesis.

    Enzymes of citric acid cycle are present in mitochondrial matrix while the electron transport chain is present in inner mitochondrial membrane, hence, the last two phases of aerobic respiration takes place in mitochondria only.

    (ii) Anaerobic respiration: Glycolysis is breakdown of glucose into pyruvate. It is oxygen independent pathway and is common to both aerobic and anaerobic respiration. Fate of pyruvate depends on oxygen availability. Anaerobic respiration can occur via lactic acid fermentation or alcohol fermentation.

    Alcohol fermentation Due to limited oxygen availability, pyruvate remains in cytoplasm where pyruvate decarboxylase and alcohol dehydrogenase enzymes carry out the second phase of anaerobic respiration and produce ethanol and carbon dioxide with limited amount of energy being released.

    Lactic acid fermentation Due to limited oxygen availability, pyruvate remains in cytoplasm where it is broken down into lactic acid with limited amount of energy being released.

    breakdown of glucose

    Question. 81 Describe the flow of blood through the heart of human beings.

    Answer.

    Human heart has four completely separated chambers which allow two completely separated circuits of oxygenated and deoxygenated blood; hence the name, double circulation.

    The pulmonary circuit includes flow of blood through heart-lung-heart while systemic circuit includes heart-rest of the body-heart. This ensures that left side of heart receives and pumps only oxygenated blood while the right side receives and pumps only deoxygenated blood.

    Two pulmonary veins come from each lung and pass O2-rich blood to left atrium. The left ventricle pumps blood into the aorta which in turn branches and delivers blood to the major body regions and organs. The anterior vena cava collects deoxygenated blood from the head, chest, and arms and enters the right atrium while the inferior vena cava collects blood from the lower body regions. Both venae cavae pass the deoxygenated blood to the right atrium. Blood from right atrium enters right ventricle and pulmonary arteries carry deoxygenated blood from right ventricle to lungs for oxygenation.

    Exchange of gases in human heart

    Question. 82 Describe the process of urine formation in kidneys.

    Answer.

    Urine formation occurs in the kidneys involves the following steps:

    (i) Glomerular Filtration: Glomerular filtration is passive filtration of the blood that starts with entry of blood through glomerulus, a special ball of capillaries. The overall process leads to removal of 15-25% water and solutes from the plasma entering the glomerulus. Filtration of any substance through glomerular filtration depends on molecular size, protein binding, ionization, polarity, and working conditions of kidney in general.

    (ii) Tubular reabsorption:  Glomerular filtrate enters the proximal convoluted tubule wherein nutrient and salt molecules are actively reabsorbed and pushed into the peritubular capillary network while water flows passively. The filtrate is now composed of urea, some salts and water.

    (iii) Tubular secretion: It includes active secretion of certain substances (H+, K+, ammonia, drug etc) from the peritubular capillary network into the distal convoluted tubule.

    The urine formed is collected in the urinary bladder.

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