Chapter 1: Life Processes

Solution

Chlorophyll is the green pigment found in chloroplasts that absorbs sunlight (primarily red and blue wavelengths) and drives photosynthesis. Haemoglobin is the oxygen-carrying pigment in blood, melanin provides skin colour, and while carotene is a plant pigment, it is an accessory pigment, not the primary one for photosynthesis.

Solution

Fungi like bread mould (Rhizopus) exhibit saprophytic nutrition. They secrete digestive enzymes onto dead and decaying organic matter, break it down externally, and then absorb the simple nutrients. They do not ingest whole food (holozoic) nor depend on a living host (parasitic).

Solution

Salivary amylase (also called ptyalin) is secreted by the salivary glands in the mouth. It acts on starch and breaks it down into maltose (a disaccharide sugar). This is why if you chew plain bread for a long time, it starts to taste sweet. Salivary amylase works best at a slightly acidic to neutral pH (~6.8). It stops working in the acidic stomach environment.

Solution

The organ that removes urea from blood is: Kidneys.

Solution

The organ that removes carbon dioxide from blood is: Lungs.

Solution

The organ that removes uric acid from blood is: Kidneys.

Solution

The organ that removes bile pigments from blood is: Liver.

Solution

The organ that removes excess water and salts (sweat) from blood is: Skin.

Solution

The oxygen released during photosynthesis comes from the photolysis of water (splitting of water molecules) during the light-dependent reactions. The equation 2H₂O → 4H⁺ + 4e^- + O₂ shows that water is split, and its oxygen atoms form O₂. This was confirmed by isotope tracing experiments using heavy oxygen (¹⁸O).

Solution

Stomata open when the guard cells absorb water by osmosis and become turgid. The inner wall of guard cells (facing the stomatal pore) is thicker than the outer wall. When guard cells become turgid, the thinner outer wall expands more, causing the cells to curve apart and the pore to open. When guard cells lose water and become flaccid, they straighten and the pore closes.

Solution

Amoeba ingests food using pseudopodia (false feet) — temporary projections of the cell membrane and cytoplasm. The pseudopodia surround the food particle from all sides and fuse to form a food vacuole. This process is called phagocytosis. Cilia and oral grooves are used by Paramecium, not Amoeba.

Solution

Bile is produced by the liver and stored in the gall bladder. It does not contain digestive enzymes. Its main function is to emulsify fats — breaking large fat globules into smaller droplets. This increases the surface area available for the enzyme lipase to act upon, making fat digestion more efficient. Bile also makes the medium alkaline for the intestinal enzymes to work.

Solution

Villi are tiny finger-like projections that line the inner wall of the small intestine. They greatly increase the surface area available for absorption of digested food into the blood. Each villus is richly supplied with blood capillaries (for absorbing amino acids, glucose, etc.) and a lacteal (lymph vessel for absorbing fatty acids and glycerol).

Solution

In yeast, anaerobic respiration (fermentation) produces ethanol (alcohol) and carbon dioxide: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ + energy. This is used commercially in bread-making (CO₂ makes bread rise) and alcohol production. Note: in human muscles, anaerobic respiration produces lactic acid (not ethanol).

Solution

Alveoli are highly efficient for gas exchange because: (1) There are about 300 million alveoli in each lung, providing an enormous total surface area (~80 m²). (2) Their walls are extremely thin (one cell thick), minimising diffusion distance. (3) They are surrounded by a dense network of capillaries. (4) Their inner surface is moist, allowing gases to dissolve before diffusing. Cartilage rings are in the trachea, not alveoli.

Solution

Haemoglobin, an iron-containing protein found in red blood cells (RBCs), binds with oxygen in the lungs to form oxyhaemoglobin and releases it in the tissues. Each haemoglobin molecule can carry up to 4 molecules of oxygen. Plasma transports dissolved CO₂, nutrients, and hormones but not oxygen efficiently (only a tiny amount dissolves in plasma directly).

Solution

The primary force driving the upward movement of water through xylem is transpiration pull (also called the transpiration-cohesion-tension mechanism). As water evaporates from leaf stomata during transpiration, it creates a suction (negative pressure) that pulls water upward from the roots through the xylem vessels. Root pressure contributes but is insufficient alone, especially in tall trees.

Solution

Translocation — the transport of food (mainly sucrose) from leaves (where it is synthesised) to other parts of the plant — occurs through the phloem. This is an active process requiring ATP. Xylem transports water and minerals upward, not food. Stomata are involved in gas exchange, and root hairs in water absorption.

Solution

The nephron is the structural and functional unit of the kidney. Each kidney contains about one million nephrons. Each nephron consists of a Bowman's capsule (containing the glomerulus), proximal convoluted tubule, loop of Henle, distal convoluted tubule, and collecting duct. Do not confuse nephron (kidney unit) with neuron (nerve cell).

Solution

Dialysis (haemodialysis) is the artificial process used when kidneys fail. Blood is passed through a machine containing a dialyser with a semi-permeable membrane. Waste products (urea, excess salts) diffuse out of the blood into the dialysing fluid, while useful substances are retained. Patients typically need dialysis 2-3 times per week until a kidney transplant is available.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 8 glucose molecules: 6 x 8 = 48 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 12 glucose molecules: 6 x 12 = 72 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 1 glucose molecules: 6 x 1 = 6 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 2 glucose molecules: 6 x 2 = 12 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 4 glucose molecules: 6 x 4 = 24 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 10 glucose molecules: 6 x 10 = 60 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 20 glucose molecules: 6 x 20 = 120 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 5 glucose molecules: 6 x 5 = 30 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 19 glucose molecules: 6 x 19 = 114 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 17 glucose molecules: 6 x 17 = 102 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 6 glucose molecules: 6 x 6 = 36 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 15 glucose molecules: 6 x 15 = 90 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 11 glucose molecules: 6 x 11 = 66 CO2 molecules needed.

Solution

The Left atrium pumps blood to the left ventricle.

Remember: Right side = lungs (pulmonary), Left side = body (systemic).

Solution

The Right ventricle pumps blood to the pulmonary artery.

Remember: Right side = lungs (pulmonary), Left side = body (systemic).

Solution

The Left ventricle pumps blood to the aorta.

Remember: Right side = lungs (pulmonary), Left side = body (systemic).

Solution

The Right ventricle pumps blood to the lungs.

Remember: Right side = lungs (pulmonary), Left side = body (systemic).

Solution

The Left ventricle pumps blood to the body (systemic circulation).

Remember: Right side = lungs (pulmonary), Left side = body (systemic).

Solution

The Right atrium pumps blood to the right ventricle.

Remember: Right side = lungs (pulmonary), Left side = body (systemic).

Solution

Blood group A can donate to O.

Answer: 0

Solution

Blood group O can donate to O.

Answer: 7

Solution

Blood group B can donate to A.

Answer: 0

Solution

Blood group O can donate to A.

Answer: 14

Solution

Blood group O can donate to O.

Answer: 7

Solution

Blood group A can donate to B.

Answer: 0

Solution

Blood group O can donate to B.

Answer: 18

Solution

Blood group O can donate to B.

Answer: 8

Solution

Blood group O can donate to A.

Answer: 22

Solution

Blood group O can donate to B.

Answer: 13

Solution

Blood group A can donate to AB.

Answer: 6

Solution

Blood group O can donate to AB.

Answer: 10

Solution

Blood group A can donate to B.

Answer: 0

Solution

Blood group A can donate to B.

Answer: 0

Solution

Blood group AB can donate to AB.

Answer: 11

Solution

Blood group O can donate to O.

Answer: 24

Solution

Blood group O can donate to A.

Answer: 29

Solution

Blood group B can donate to A.

Answer: 0

Solution

Blood group B can donate to A.

Answer: 0

Solution

Blood group A can donate to B.

Answer: 0

Solution

Blood group AB can donate to A.

Answer: 0

Solution

Blood group B can donate to A.

Answer: 0

Solution

Blood group O can donate to O.

Answer: 9

Solution

Blood group O can donate to O.

Answer: 20

Solution

Blood group B can donate to A.

Answer: 0

Solution

Blood group O can donate to A.

Answer: 29

Solution

Blood group O can donate to AB.

Answer: 12

Solution

Blood group O can donate to B.

Answer: 12

Solution

Blood group O can donate to O.

Answer: 28

Solution

Blood group AB can donate to A.

Answer: 0

Solution

The enzyme that acts on starch in the small intestine (from pancreas) is: Pancreatic amylase.

Solution

The enzyme that acts on fats in the small intestine is: Lipase.

Solution

The enzyme that acts on proteins in the stomach is: Pepsin.

Solution

The enzyme that acts on proteins in the small intestine (from pancreas) is: Trypsin.

Solution

The enzyme that acts on starch in the mouth is: Salivary amylase.

Solution

The enzyme that acts on milk protein (casein) in the stomach is: Rennin (in infants).

Solution

Minute ventilation = breathing rate x tidal volume

= 24 x 518 = 12432 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 16 x 575 = 9200 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 13 x 494 = 6422 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 17 x 579 = 9843 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 13 x 578 = 7514 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 24 x 478 = 11472 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 14 x 415 = 5810 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 22 x 474 = 10428 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 22 x 449 = 9878 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 16 x 457 = 7312 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 21 x 469 = 9849 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 19 x 404 = 7676 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 21 x 499 = 10479 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 25 x 582 = 14550 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 17 x 526 = 8942 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 22 x 543 = 11946 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 25 x 541 = 13525 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 24 x 586 = 14064 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 12 x 560 = 6720 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 21 x 514 = 10794 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 15 x 447 = 6705 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 14 x 529 = 7406 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 25 x 447 = 11175 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 12 x 402 = 4824 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 13 x 515 = 6695 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 25 x 575 = 14375 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 25 x 474 = 11850 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 17 x 464 = 7888 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 22 x 576 = 12672 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 21 x 529 = 11109 mL/min

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 3 glucose molecules: 6 x 3 = 18 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 18 glucose molecules: 6 x 18 = 108 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 7 glucose molecules: 6 x 7 = 42 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 16 glucose molecules: 6 x 16 = 96 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 13 glucose molecules: 6 x 13 = 78 CO2 molecules needed.

Solution

The balanced equation shows 6 CO2 -> 1 glucose.

For 9 glucose molecules: 6 x 9 = 54 CO2 molecules needed.

Solution

Blood group O can donate to B.

Answer: 26

Solution

Blood group O can donate to O.

Answer: 19

Solution

Blood group O can donate to A.

Answer: 24

Solution

Blood group A can donate to B.

Answer: 0

Solution

Blood group O can donate to O.

Answer: 9

Solution

Blood group A can donate to O.

Answer: 0

Solution

Blood group B can donate to B.

Answer: 10

Solution

Blood group A can donate to A.

Answer: 24

Solution

Blood group A can donate to AB.

Answer: 19

Solution

Blood group B can donate to A.

Answer: 0

Solution

Blood group A can donate to A.

Answer: 21

Solution

Blood group O can donate to O.

Answer: 16

Solution

Blood group A can donate to A.

Answer: 20

Solution

Blood group A can donate to A.

Answer: 6

Solution

Blood group AB can donate to AB.

Answer: 29

Solution

Blood group B can donate to B.

Answer: 25

Solution

Blood group AB can donate to A.

Answer: 0

Solution

Blood group B can donate to AB.

Answer: 13

Solution

Blood group B can donate to B.

Answer: 13

Solution

Blood group A can donate to O.

Answer: 0

Solution

Blood group O can donate to A.

Answer: 18

Solution

Blood group O can donate to O.

Answer: 24

Solution

Blood group O can donate to B.

Answer: 26

Solution

Blood group B can donate to AB.

Answer: 16

Solution

Blood group A can donate to B.

Answer: 0

Solution

Blood group A can donate to B.

Answer: 0

Solution

Blood group O can donate to A.

Answer: 27

Solution

Blood group AB can donate to AB.

Answer: 28

Solution

Blood group O can donate to O.

Answer: 16

Solution

Blood group A can donate to AB.

Answer: 18

Solution

Blood group AB can donate to A.

Answer: 0

Solution

Blood group B can donate to A.

Answer: 0

Solution

Blood group B can donate to AB.

Answer: 30

Solution

Blood group B can donate to A.

Answer: 0

Solution

Blood group A can donate to AB.

Answer: 18

Solution

Blood group B can donate to AB.

Answer: 26

Solution

Blood group AB can donate to A.

Answer: 0

Solution

Blood group O can donate to AB.

Answer: 11

Solution

Blood group A can donate to A.

Answer: 17

Solution

Blood group AB can donate to AB.

Answer: 26

Solution

Blood group O can donate to A.

Answer: 12

Solution

Blood group AB can donate to AB.

Answer: 22

Solution

Blood group A can donate to AB.

Answer: 29

Solution

Blood group O can donate to O.

Answer: 20

Solution

Blood group AB can donate to A.

Answer: 0

Solution

Blood group B can donate to B.

Answer: 10

Solution

Blood group O can donate to O.

Answer: 29

Solution

Blood group AB can donate to AB.

Answer: 24

Solution

Blood group O can donate to A.

Answer: 13

Solution

Blood group O can donate to A.

Answer: 23

Solution

Blood group B can donate to A.

Answer: 0

Solution

Blood group A can donate to A.

Answer: 5

Solution

Blood group B can donate to A.

Answer: 0

Solution

Blood group O can donate to O.

Answer: 5

Solution

Blood group B can donate to B.

Answer: 12

Solution

Blood group O can donate to A.

Answer: 22

Solution

Blood group AB can donate to A.

Answer: 0

Solution

Blood group AB can donate to AB.

Answer: 6

Solution

Blood group A can donate to O.

Answer: 0

Solution

Blood group O can donate to O.

Answer: 16

Solution

Minute ventilation = breathing rate x tidal volume

= 17 x 498 = 8466 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 14 x 404 = 5656 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 16 x 574 = 9184 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 18 x 541 = 9738 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 23 x 440 = 10120 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 25 x 517 = 12925 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 18 x 494 = 8892 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 23 x 483 = 11109 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 24 x 567 = 13608 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 14 x 414 = 5796 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 17 x 523 = 8891 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 16 x 429 = 6864 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 25 x 475 = 11875 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 17 x 444 = 7548 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 22 x 554 = 12188 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 23 x 510 = 11730 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 21 x 567 = 11907 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 20 x 482 = 9640 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 21 x 406 = 8526 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 20 x 511 = 10220 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 17 x 482 = 8194 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 17 x 499 = 8483 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 22 x 570 = 12540 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 18 x 412 = 7416 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 21 x 502 = 10542 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 14 x 599 = 8386 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 14 x 409 = 5726 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 24 x 432 = 10368 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 24 x 501 = 12024 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 15 x 431 = 6465 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 16 x 551 = 8816 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 17 x 540 = 9180 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 12 x 486 = 5832 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 19 x 503 = 9557 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 13 x 424 = 5512 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 19 x 463 = 8797 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 19 x 563 = 10697 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 22 x 537 = 11814 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 24 x 522 = 12528 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 17 x 433 = 7361 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 23 x 463 = 10649 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 16 x 540 = 8640 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 14 x 526 = 7364 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 25 x 500 = 12500 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 19 x 595 = 11305 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 25 x 589 = 14725 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 17 x 558 = 9486 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 18 x 527 = 9486 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 22 x 421 = 9262 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 19 x 435 = 8265 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 18 x 508 = 9144 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 20 x 537 = 10740 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 25 x 422 = 10550 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 23 x 503 = 11569 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 20 x 560 = 11200 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 25 x 472 = 11800 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 19 x 437 = 8303 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 24 x 566 = 13584 mL/min

Solution

Minute ventilation = breathing rate x tidal volume

= 14 x 584 = 8176 mL/min

Solution

By covering the leaf with black paper, the student blocked light from reaching that part. Since starch (detected by iodine turning blue-black) was not produced in the covered region, this demonstrates that light is necessary for photosynthesis. The uncovered part of the same leaf would have produced starch, serving as a control. This is the classic NCERT experiment on photosynthesis requirements.

Solution

HCl in gastric juice serves multiple functions: (1) It creates an acidic medium (pH ~2) which is optimal for the action of the enzyme pepsin. (2) It activates the inactive enzyme pepsinogen into its active form pepsin. (3) It kills most bacteria that enter with food. HCl itself does not digest any nutrient — it creates the right conditions for pepsin to break down proteins.

Solution

During heavy exercise, the oxygen supply to muscles becomes insufficient, so muscle cells switch to anaerobic respiration, producing lactic acid. The accumulation of lactic acid in the muscles causes muscle fatigue, pain, and cramps. The cramp subsides when the person rests and oxygen supply is restored — the lactic acid is then broken down by oxygen (this extra oxygen needed is called the 'oxygen debt').

Solution

Glycolysis — the breakdown of glucose (6C) into two molecules of pyruvate (3C) — occurs in the cytoplasm of the cell. This is common to both aerobic and anaerobic respiration. If oxygen is available, pyruvate enters the mitochondria for further oxidation (aerobic pathway). If oxygen is absent, pyruvate is converted to ethanol or lactic acid in the cytoplasm itself.

Solution

The left ventricle pumps oxygenated blood to the entire body through the aorta — organs from head to toes. This requires generating much higher pressure than the right ventricle, which only pumps blood to the nearby lungs. Therefore, the left ventricle has a thicker muscular wall to generate the necessary force. This is a frequently asked question in CBSE board exams.

Solution

Double circulation (found in mammals and birds) means blood goes through the heart twice — once for pulmonary circulation (heart → lungs → heart) and once for systemic circulation (heart → body → heart). The key advantage is that oxygenated and deoxygenated blood do not mix, so tissues receive fully oxygenated blood. This is essential for maintaining a high metabolic rate and constant body temperature in warm-blooded animals.

Solution

Selective reabsorption occurs mainly in the proximal convoluted tubule. As the filtrate passes through the tubule, all glucose, most amino acids, necessary salts, and most water are actively reabsorbed back into the surrounding blood capillaries. Urea is NOT reabsorbed (it is waste). Blood cells and proteins are never filtered in the first place — they are too large to pass through the glomerular capillaries.

Solution

Xylem transport is largely passive — driven by transpiration pull (evaporation of water from leaves creating suction) and cohesion of water molecules. No metabolic energy is spent. Phloem transport (translocation) is active because sucrose must be loaded into the sieve tubes using ATP. This increases osmotic pressure in the phloem, drawing in water and creating a pressure flow. Phloem cells are living (not dead), which is necessary for active transport.

Solution

Both the Assertion and Reason are true, and the Reason correctly explains the Assertion. Plants lack a specialised excretory system because they have a lower metabolic rate compared to animals, generating fewer waste products. Additionally, plants can reuse CO₂ (waste from respiration) in photosynthesis and O₂ (waste from photosynthesis) in respiration. Excess wastes are stored in vacuoles, old leaves, or bark. These factors together eliminate the need for a dedicated excretory organ.

Solution

Both statements are true and the Reason correctly explains the Assertion. Veins carry blood back to the heart under relatively low pressure (since they are far from the heart's pumping action). Without valves, blood would flow backward due to gravity, especially in the legs. Valves ensure one-way flow toward the heart. Arteries do not need valves because blood flows through them under high pressure from the heart's contractions, which is sufficient to maintain forward flow.

Solution

The pancreas secretes pancreatic juice containing three key enzymes: trypsin (digests proteins), pancreatic lipase (digests fats), and pancreatic amylase (digests starch). If the pancreas fails, all three types of nutrients will be poorly digested in the small intestine. Salivary amylase in the mouth (option A) would still work. Bile from the liver (option D) would still emulsify fats. Water absorption (option C) in the large intestine is unrelated to the pancreas.

Solution

Both are true and the Reason correctly explains the Assertion. In aerobic respiration, glucose is completely oxidised (C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O), releasing all the stored energy — about 38 ATP molecules. In anaerobic respiration, glucose is only partially broken down (to ethanol or lactic acid), so much of the energy remains locked in the products, yielding only 2 ATP. The complete oxidation in the presence of oxygen is the reason for the much higher energy yield.

Solution

Large volumes of dilute urine indicate that water is not being adequately reabsorbed from the filtrate back into the blood. The Loop of Henle creates a concentration gradient in the kidney medulla, and the collecting duct (under the influence of ADH — antidiuretic hormone) reabsorbs water to concentrate urine. If these structures malfunction, water passes straight through and is excreted, producing copious dilute urine. This condition resembles diabetes insipidus. A glomerular problem would affect filtration rate, not dilution. Protein leakage (option C) and glucose loss (option D) cause different symptoms.