Secondary 3 Biology Human Physiology Quiz

Secondary 3 Biology Quiz - Human Physiology (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions (10 marks)

1. A — Nose → Trachea → Bronchi → Bronchioles → Alveoli [1]
Explanation: Air enters through the nose, passes down the trachea, which branches into two bronchi (one for each lung), these further divide into smaller bronchioles, ending in tiny air sacs called alveoli where gas exchange occurs.

2. B — Large surface area and thin walls [1]
Explanation: Alveoli are adapted for efficient gas exchange by having a huge total surface area (millions of alveoli) and walls that are only one cell thick (simple squamous epithelium), minimising diffusion distance for O₂ and CO₂. Thick walls (A) would hinder diffusion. Cilia (C) and mucus (D) are features of the trachea/bronchi for trapping particles, not alveoli.

3. C — Glucose is partially broken down in the absence of sufficient oxygen [1]
Explanation: During vigorous exercise, oxygen supply to muscles may be insufficient for aerobic respiration. Glucose undergoes anaerobic respiration (glycolysis followed by lactic acid fermentation), producing lactic acid and a small amount of ATP. This is partial breakdown; complete oxidation (B) requires oxygen.

4. B — Small intestine [1]
Explanation: The small intestine (duodenum, jejunum, ileum) is the primary site for absorption of digested nutrients (glucose, amino acids, fatty acids, glycerol, vitamins, minerals) due to its large surface area from villi and microvilli. The stomach (A) absorbs only water, alcohol, and some drugs. The large intestine (C) absorbs water and salts. The oesophagus (D) has no absorptive function.

5. B — Pepsin [1]
Explanation: Enzyme X has an optimum pH of 2 (highly acidic), which matches pepsin, a protease active in the stomach (pH 1.5–3). Amylase (A) works at near-neutral pH (salivary amylase ~pH 7, pancreatic amylase ~pH 7–8). Trypsin (C) is a pancreatic protease with optimum ~pH 8. Lipase (D) also works at alkaline pH (~pH 8).

6. C — The left ventricle has a thicker muscular wall than the right ventricle [1]
Explanation: The left ventricle pumps blood at high pressure to the entire body (systemic circulation), requiring a thick muscular wall to generate this pressure. The right ventricle pumps blood only to the lungs (pulmonary circulation) at lower pressure. A is incorrect: right ventricle pumps deoxygenated blood to lungs. B is incorrect: left atrium receives oxygenated blood from lungs. D is incorrect: pulmonary vein carries oxygenated blood.

7. B — Gaseous exchange in the alveoli [1]
Explanation: In the alveoli, O₂ diffuses from alveolar air into blood (decreasing from 21% to 16%), and CO₂ diffuses from blood into alveolar air (increasing from 0.04% to 4%). Nitrogen is inert and unchanged. Photosynthesis (A) occurs in plants. Combustion (D) is burning. Fermentation (D) is anaerobic respiration in microorganisms/muscles.

8. B — Increasing the surface area for absorption [1]
Explanation: Villi are finger-like projections of the intestinal mucosa, covered in microvilli (brush border), massively increasing surface area for absorption. They contain blood capillaries and a lacteal for transport. Enzymes (A) are secreted by intestinal mucosa and pancreas. Bile (C) is from the liver. Peristalsis (D) moves food.

9. C — Red blood cells [1]
Explanation: Red blood cells (erythrocytes) are by far the most numerous blood cells (~4–6 million per µL), compared to white blood cells (~4,000–11,000 per µL) and platelets (~150,000–400,000 per µL). Plasma proteins are dissolved in plasma, not cells.

10. C — Glomerulus / Bowman's capsule [1]
Explanation: Ultrafiltration occurs at the glomerulus, a capillary network inside Bowman's capsule. High pressure forces water, glucose, amino acids, urea, and ions out of the blood into the capsule, forming glomerular filtrate. Large proteins and cells remain in blood. Reabsorption occurs in the proximal convoluted tubule (A), loop of Henle (B), and collecting duct (D).

Section B: Structured Questions (20 marks)

11. (a) A: Trachea (or windpipe) [1]
B: Diaphragm [1]
Marking note: Accept "windpipe" for trachea. Must be specific structures visible on a standard respiratory system diagram.

(b) Inhalation mechanism: [3]

• The diaphragm contracts and flattens (moves downwards) [1]
• The external intercostal muscles contract, pulling the ribs upwards and outwards [1]
• This increases the volume of the thoracic cavity, decreasing the pressure inside the lungs below atmospheric pressure, so air flows in [1]
  Teaching note: Emphasise the sequence: muscle contraction → volume increase → pressure decrease → air inflow. "Negative pressure breathing."

(c) Function of C-shaped cartilage rings: [2]

• Prevent the trachea from collapsing during inhalation when intrathoracic pressure drops [1]
• Keep the airway open (patent) for continuous airflow [1]
• The C-shape (incomplete rings) allows the oesophagus behind to expand during swallowing [1 — bonus/detail, not required for 2 marks]
  Marking: Any two valid points. Common mistake: saying "protect the trachea" without explaining collapse prevention.

12. (a) Graph plotting: [3]

• Axes correctly labelled with units: x-axis "Temperature (°C)", y-axis "Time taken for starch digestion (minutes)" [1]
• Appropriate scales covering all data points (e.g., x: 0–80°C, y: 0–30 min) [1]
• All 5 points plotted accurately; smooth curve drawn through points (not dot-to-dot) [1]
  Marking note: Point at 70°C should be plotted at >30 min (off scale or marked "no digestion"). Curve should show a clear minimum at 37°C.

(b) Optimum temperature: 37°C [1]
Explanation: Shortest digestion time (4 minutes) indicates highest enzyme activity. This matches human body temperature.

(c) Explanation for decreased rate at 50°C and no digestion at 70°C: [3]

• At 50°C, the enzyme begins to denature — the high temperature breaks hydrogen bonds and other weak interactions maintaining its tertiary structure, altering the active site shape so fewer substrate molecules can bind [1]
• At 70°C, the enzyme is completely denatured — the active site is permanently destroyed, so no enzyme-substrate complexes can form and no digestion occurs [1]
• Denaturation is irreversible [1]
  Teaching note: Link temperature → kinetic energy → bond breaking → shape change → loss of function. Avoid "enzyme dies" — enzymes are proteins, not living.

(d) Improve reliability: [1]

• Repeat the experiment at each temperature and calculate the mean time / identify anomalies
  OR Use a larger sample size / more replicates
  OR Use a colorimeter/data logger to detect endpoint objectively instead of visual iodine test
  Marking: Any one valid suggestion. "Repeat" alone is sufficient.

13. (a) X: Lacteal (or lymphatic capillary) [1]
Y: Blood capillary (or capillary network) [1]
Marking note: Must distinguish between the two — lacteal carries absorbed fats (chylomicrons) as lymph; blood capillary carries glucose, amino acids, water-soluble vitamins, minerals.

(b) Adaptations of villus for absorption: [4]

• Large surface area: Villi are finger-like projections; epithelial cells have microvilli (brush border), greatly increasing surface area for absorption [1]
• Thin epithelium: Wall is one cell thick (simple columnar epithelium), providing a short diffusion distance for nutrients into blood/lacteal [1]
• Rich blood supply: Dense capillary network maintains a steep concentration gradient by rapidly carrying away absorbed nutrients (glucose, amino acids) [1]
• Lacteal: Central lymphatic capillary (lacteal) absorbs fatty acids and glycerol (re-esterified to triglycerides, packaged as chylomicrons) [1]
• Microvilli: Further increase surface area and contain membrane-bound enzymes (e.g., disaccharidases, peptidases) for final digestion [1]
• Smooth muscle fibres: Contract rhythmically to move villi, maintaining concentration gradients and moving lymph in lacteal [1]
  Marking: Any 4 distinct points. Must link structure → function. "Large surface area" and "microvilli" can be separate points if explained differently (villi vs microvilli level).

14. (a) Glucose and amino acids in filtrate but not urine: [3]

• They are small molecules freely filtered at the glomerulus into Bowman's capsule [1]
• They are selectively reabsorbed in the proximal convoluted tubule by active transport (co-transport with Na⁺) back into the blood capillaries [1]
• In a healthy person, all filtered glucose and amino acids are reabsorbed (up to transport maximum), so none appear in urine [1]
  Teaching note: "Selective reabsorption" is key. Glucose appears in urine only if blood glucose exceeds renal threshold (~180 mg/dL) — diabetes.

(b) High urea concentration in urine: [2]

• Water is reabsorbed from the filtrate (in proximal tubule, loop of Henle, collecting duct) by osmosis, concentrating the remaining urea [1]
• Urea is not reabsorbed (or only partially in some parts); it remains in the tubule and is excreted, so its concentration rises as water leaves [1]
  Alternative: Some urea is passively reabsorbed in collecting duct, but net effect is concentration due to massive water reabsorption.

(c) Selective reabsorption (or active transport) [1]
Marking: Accept "active transport" or "selective reabsorption". Process occurs in proximal convoluted tubule.

15. (a) Arrows on diagram: [2]

• Deoxygenated blood: Vena cava → Right atrium → Right ventricle → Pulmonary artery → Lungs [1 for correct path]
• Oxygenated blood: Pulmonary vein → Left atrium → Left ventricle → Aorta → Body [1 for correct path]
  Marking: Arrows must show correct direction through chambers and vessels. Colour coding (blue/red) helpful but not required.

(b) Bicuspid valve (or mitral valve / left atrioventricular valve) [1]
Marking: "Tricuspid" is incorrect (right side). "Semilunar" is incorrect (ventricle to artery).

(c) Left ventricle thicker wall: [2]

• Left ventricle pumps blood to the whole body (systemic circulation) which has high resistance and requires high pressure [1]
• Right ventricle pumps blood only to the lungs (pulmonary circulation) which has low resistance and requires lower pressure [1]
  Teaching note: Pressure difference: systemic ~120/80 mmHg, pulmonary ~25/10 mmHg. Wall thickness proportional to pressure generated.

Section C: Extended Response Questions (10 marks)

16. Digestion of starch, protein, and fat: [6]

Starch (carbohydrate): [2]

• Mouth: Salivary amylase (ptyalin) from salivary glands breaks starch → maltose (optimum pH ~7)
• Small intestine (duodenum): Pancreatic amylase continues starch → maltose; maltase, sucrase, isomaltase on microvilli break maltose → glucose

Protein: [2]

• Stomach: Pepsin (secreted as pepsinogen, activated by HCl, optimum pH 2) breaks proteins → polypeptides
• Small intestine (duodenum): Pancreatic trypsin (activated from trypsinogen by enterokinase), chymotrypsin break polypeptides → peptides; peptidases (dipeptidases, aminopeptidases) on microvilli break peptides → amino acids

Fat: [2]

• Small intestine (duodenum): Bile (from liver, stored in gall bladder) emulsifies fats → small droplets (increases surface area) — not digestion, physical process
• Pancreatic lipase breaks triglycerides (fats) → fatty acids + monoglycerides (optimum pH ~8, aided by bile)
• Products absorbed by epithelial cells, re-esterified to triglycerides, packaged as chylomicrons into lacteal

Marking breakdown: 2 marks per macronutrient (enzyme + site + product). Must mention bile emulsification for fat (not an enzyme). End products: glucose, amino acids, fatty acids + monoglycerides/glycerol.

17. (a) Two structural differences: [2]

1. Artery has a thicker wall / thicker tunica media (more smooth muscle and elastic fibres) than vein
2. Artery has a narrower lumen relative to its wall thickness; vein has a wider lumen
   OR Vein has valves (to prevent backflow); artery has no valves (except semilunar at heart)
   OR Artery has more elastic fibres in tunica media; vein has less
   Marking: Any two distinct differences. Must be comparative.

(b) Artery structure adapted to function: [2]

• Thick tunica media with elastic fibres allows artery to stretch (expand) when high-pressure blood is ejected from heart (systole), then recoil (elastic recoil) to maintain pressure and smooth flow during diastole [1]
• Thick smooth muscle layer allows vasoconstriction/vasodilation to regulate blood pressure and distribution [1]
• Narrow lumen maintains high blood pressure for rapid delivery to tissues [1]
  Marking: Any two valid adaptations linked to function (transport at high pressure, pulse smoothing).

18. Thermoregulation during marathon (negative feedback): [5]

• Stimulus: Body temperature rises above set point (37°C) detected by thermoreceptors in hypothalamus (and skin) [1]
• Control centre: Hypothalamus (thermoregulatory centre) processes signal and sends nerve impulses to effectors [1]
• Effector responses:
  • Skin arterioles vasodilate (smooth muscle relaxes) → increased blood flow to skin capillaries → increased heat loss by radiation/convection/conduction [1]
  • Sweat glands secrete sweat → evaporation from skin surface removes latent heat → cooling [1]
  • Erector pili muscles relax → hairs lie flat → reduces insulation (trapped air layer) [1]
• Negative feedback: As temperature falls back to set point, hypothalamus reduces signals → responses decrease → temperature stabilises [1]
  Marking: Must mention hypothalamus as detector/control centre, skin arterioles vasodilation, sweat evaporation, and negative feedback loop. 5 distinct points for 5 marks.

19. (a) Pressure drop during inhalation: [3]

• Diaphragm contracts and flattens; external intercostal muscles contract → ribs move up and out [1]
• Volume of thoracic cavity increases → volume of lungs increases (pleural membranes adhere) [1]
• According to Boyle's Law ($P \propto 1/V$ at constant temperature), increase in volume causes decrease in intrapulmonary pressure below atmospheric pressure [1]
• Pressure gradient drives air into lungs [1 — implied]
  Marking: 3 marks for: muscle action → volume increase → pressure decrease (Boyle's Law).

(b) Boyle's Law calculation: [2]

• $P_1V_1 = P_2V_2$
• $P_1 = 101.3 \text{ kPa}$ (atmospheric pressure at rest)
• $V_1 = 500 \text{ cm}^3$
• $V_2 = 3000 \text{ cm}^3$
• $P_2 = \frac{P_1V_1}{V_2} = \frac{101.3 \times 500}{3000} = \frac{50650}{3000} = 16.88 \text{ kPa}$
• Answer: 16.9 kPa (or 16.88 kPa) [2]
  Marking: 1 mark for correct substitution/formula, 1 mark for correct answer with units. Common mistake: using 101.3 as $P_2$ or inverting volumes.

20. (a) Hormone: Insulin [1]
Organ: Pancreas (specifically beta cells of islets of Langerhans) [1]

(b) How insulin lowers blood glucose: [3]

• Binds to receptors on target cell membranes (liver, muscle, adipose) → triggers translocation of GLUT4 glucose transporters to cell membrane → increases glucose uptake into cells [1]
• Stimulates glycogenesis in liver and muscle: glucose → glycogen (storage polysaccharide) [1]
• Stimulates lipogenesis in adipose tissue: glucose → fatty acids → triglycerides (fat storage) [1]
• Inhibits glycogenolysis and gluconeogenesis in liver [1 — bonus]
  Marking: Any 3 valid mechanisms. Must mention cellular uptake and storage.

(c) Why injection > oral for insulin: [2]

• Insulin is a protein (polypeptide hormone) [1]
• If taken orally, it would be digested (hydrolysed) by proteases (pepsin, trypsin, peptidases) in the stomach and small intestine into amino acids, destroying its activity [1]
• Injection (subcutaneous/intravenous) delivers it directly into bloodstream, bypassing the digestive tract [1]
  Marking: 1 mark for "protein/digested", 1 mark for "bypasses digestion/bloodstream". "Stomach acid destroys it" is partial — proteases are main reason.

End of Answer Key