HKDSE Biology Paper 2 Elective Practice Paper | 2 questions, 40 marks, 1 hour

2026-10-31

HKDSE Biology Paper 2 Electives

HKDSE Biology Paper 2 elective practice paper: 2 questions, 40 marks, 60 minutes. Section A (E1 Human physiology) covers glomerular filtration and water reabsorption, the renal threshold and glycosuria, cardiac output and renal plasma flow, ADH and diuretics, and polyuria in diabetes; Section B (E2 Applied ecology) covers net primary production and energy transfer, mercury bioconcentration and biomagnification, carbon fixation by reforestation and the species-area relationship, with full worked solutions.

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Section A: E1 Human physiology (20 marks)

Elective questions: this paper covers two of the Paper 2 electives (E1 and E2), 20 marks each; show all working, give units, and use correct biological terminology in explanation questions. Total 40 marks.

#	Question	Answer and marking notes
1	An adult undergoes a renal function test: glomerular filtration rate (GFR) 125 cm³ min⁻¹, daily urine volume 1.5 dm³, plasma glucose concentration 5.5 mmol dm⁻³ (molar mass of glucose = 180 g mol⁻¹). In an exercise test the same person has a stroke volume of 70 cm³ and a heart rate of 72 min⁻¹. (a) Assuming the filtration rate is constant over 24 hours, calculate the volume of filtrate produced per day, in dm³. (2 marks) (b) Hence calculate the percentage of water reabsorbed by the tubules and collecting ducts, to 1 decimal place. (2 marks) (c) Calculate the amount of glucose filtered per day, in mmol. (2 marks) (d) If the plasma glucose concentration rises to 12.0 mmol dm⁻³ while the maximum rate of tubular glucose reabsorption is 990 mmol day⁻¹, calculate the amount of glucose lost in the urine per day, in mmol and in grams. (3 marks) (e) Use the exercise data to calculate the cardiac output in dm³ min⁻¹; if the kidneys receive 20% of the cardiac output, calculate the volume of plasma flowing through the kidneys per minute. (3 marks) (f) Explain how antidiuretic hormone (ADH) alters urine volume during water shortage, and name the osmoreceptors and the site of hormone secretion. (3 marks) (g) The diuretic furosemide inhibits the Na⁺–K⁺–2Cl⁻ cotransporter in the ascending limb of the loop of Henle. Explain why it increases urine output, and why blood potassium must be monitored during treatment. (3 marks) (h) Explain why a diabetic with poorly controlled blood glucose develops polyuria. (2 marks)	(a) Daily filtrate = 125 × 1440 = 180,000 cm³ = 180 dm³. (b) Water reabsorbed = (180 − 1.5) ÷ 180 × 100% = 99.2%, so about 99.2% of the filtrate is returned to the blood and only 1.5 dm³ is excreted as urine. (c) Filtered glucose = 5.5 mmol dm⁻³ × 180 dm³ = 990 mmol. (d) Filtered = 12.0 × 180 = 2160 mmol; excreted = 2160 − 990 = 1170 mmol = 1.17 mol; mass = 1.17 × 180 = 210.6 g. (e) Cardiac output = 70 × 72 = 5,040 cm³ min⁻¹ = 5.04 dm³ min⁻¹; renal plasma flow = 5.04 × 20% = 1.008 dm³ min⁻¹ (about 1.01 dm³ min⁻¹, i.e. 1008 cm³ per minute). (f) During water shortage blood osmotic pressure rises; osmoreceptors in the hypothalamus are stimulated and the posterior pituitary releases more ADH, which inserts more water channels (AQP2) in the collecting-duct membrane, so more water is reabsorbed and a smaller volume of concentrated urine is produced; the opposite happens after drinking. (g) Blocking the Na⁺–K⁺–2Cl⁻ cotransporter prevents the build-up of the medullary salt gradient, so the countercurrent multiplier fails and water cannot be reabsorbed osmotically from the collecting duct; large amounts of Na⁺, Cl⁻ and water remain in the lumen and are excreted, increasing urine volume. More Na⁺ reaching the distal tubule drives Na⁺–K⁺ exchange, so K⁺ is lost in urine and hypokalaemia (arrhythmia, muscle weakness) may develop, hence blood K⁺ is monitored. (h) Once blood glucose exceeds the renal threshold, the filtered glucose cannot all be reabsorbed; the glucose remaining in the lumen raises the osmotic pressure of the filtrate and opposes water reabsorption (osmotic diuresis), so urine volume and frequency increase.

Section B: E2 Applied ecology (20 marks)

#	Question	Answer and marking notes
2	A freshwater pond fixes a gross primary production (GPP) of 48,600 kJ m⁻² yr⁻¹ and loses 36,200 kJ m⁻² yr⁻¹ in respiration. A nearby factory discharges mercury: the water contains 2.0e-05 mg dm⁻³, the bioconcentration factor (BCF) for algae is 10,000, and mercury then increases about 5-fold at each higher trophic level. The local food safety limit is 0.5 mg kg⁻¹ wet mass. (a) Calculate the net primary production (NPP); if the ecological efficiency is 10%, calculate the energy available to primary consumers. (3 marks) (b) If tertiary consumers require 1,000 kJ m⁻² yr⁻¹, calculate the minimum NPP needed, and explain why food chains usually have only four or five levels. (3 marks) (c) Calculate the mercury concentration in algae, small fish and large fish (mg kg⁻¹ wet mass), and state how many times the limit the large fish exceed. (4 marks) (d) A 60 kg adult eats 0.3 kg of large fish per week. Calculate the weekly intake per kilogram body mass (µg kg⁻¹); the WHO provisional tolerable weekly intake is 1.6 µg kg⁻¹ body mass — calculate the multiple. (3 marks) (e) 12 hectares of abandoned farmland are reforested, each hectare fixing 8.5 tonnes of CO₂ per year. Calculate the annual fixation, and how many people it offsets at 4.5 tonnes CO₂ per person per year. (3 marks) (f) If habitat area falls by 75% and z = 0.25, calculate how many of an original 100 species remain, to 1 decimal place. (2 marks) (g) Suggest two measures to reduce mercury entering the food chain, and explain why top predators carry the highest concentrations. (2 marks)	(a) NPP = GPP − respiration = 48,600 − 36,200 = 12,400 kJ m⁻² yr⁻¹; available to primary consumers = 12,400 × 10% = 1,240 kJ m⁻² yr⁻¹. (b) Tertiary consumers are three transfer steps above the producers, so NPP × 0.1³ = 1,000, giving NPP = 1,000 ÷ 0.001 = 1,000,000 kJ m⁻² yr⁻¹. Only 10% of the energy passes to the next level, so beyond the fourth or fifth level too little energy remains to support a viable population — hence short food chains. (c) Algae = 2.0e-05 × 10,000 = 0.20 mg kg⁻¹; small fish = 0.20 × 5 = 1.00 mg kg⁻¹; large fish = 1.00 × 5 = 5.00 mg kg⁻¹. The large fish exceed the limit by 5.00 ÷ 0.5 = 10 times. (d) Weekly intake = 5.00 × 0.3 = 1.50 mg = 1500 µg; per kg body mass = 1500 ÷ 60 = 25.0 µg kg⁻¹, i.e. 25.0 ÷ 1.6 = 15.6 times the provisional tolerable weekly intake, so such a diet is unsafe in the long term, especially for pregnant women and young children. (e) Fixation = 12 × 8.5 = 102 tonnes CO₂ per year; this offsets 102 ÷ 4.5 ≈ 22.7, i.e. about 23 people. (f) Remaining area fraction = 0.25; since S = cA^z the species ratio is 0.25^0.25 = 0.7071, so 100 × 0.7071 ≈ 70.7 species survive — about 29.3 species are lost. (g) Measures: treat factory effluent by precipitation, ion exchange or activated carbon before discharge; replace mercury-based processes and recycle mercury waste; monitor fish and restrict fishing in affected waters. Top predators live long, are large, excrete mercury slowly and accumulate it in muscle and fat, while each trophic step concentrates it further (biomagnification).

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