/en-s6-biology-cell-structure-enzymes-photosynthesis-respiration-genetics-ecology-physiology-biotechnology-practice-paper/
/en-s6-biology-cell-structure-enzymes-photosynthesis-respiration-genetics-ecology-physiology-biotechnology-practice-paper/
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S6 Biology Practice Paper: Cell Structure and Enzymes, Photosynthesis and Respiration, Genetics, Ecology, Human Physiology, Biotechnology (with answers)

S6 Biology Practice Paper: Cell Structure and Enzymes, Photosynthesis and Respiration, Genetics, Ecology, Human Physiology, Biotechnology (with answers)

  • 2026-11-15

HKDSE Biology

S6 Biology practice paper: enzymes, photosynthesis, respiration, genetics, ecology, physiology, biotechnology; 24 parts, 100 marks, 60 minutes, full answers.

Paper details S6 Biology practice paper with full answers: 100 marks, 60 minutes.

Cell Structure and Enzymes

These questions cover surface-area-to-volume ratio, enzyme reaction rate, diffusion and microscope calculation.

# Question Answer and marking notes
1 For cube models of side 2 cm and side 1 cm, what is the surface-area-to-volume ratio (per cm) of each? Which is larger, and by what factor? Answer: the 2 cm cube gives 24 / 8 = 3.0 per cm, the 1 cm cube gives 6 / 1 = 6.0 per cm; the smaller cube has 2.0 times the ratio, so diffusion supplies it more easily.
2 An enzyme catalyses 120 reaction units per minute at 30 degrees and only 90 units at 50 degrees. What is the percentage drop, and why? Answer: a fall of 30 units, i.e. 25.0 per cent; heat denatures the enzyme so the active site no longer fits the substrate.
3 Diffusion time is proportional to the square of distance. If the distance becomes 3 times longer, by what factor does the time change? Answer: time goes as the square of distance, and 3 squared is 9, so the time becomes 9 times longer.
4 Under 500 times magnification, a cell of actual diameter 20 micrometres has an image of what diameter in millimetres? Answer: 20 micrometres times 500 gives 10000 micrometres, that is 1 cm or 10.0 millimetres.

Photosynthesis and Respiration

These questions cover the stoichiometry of photosynthesis, aerobic versus anaerobic yield, and transpiration loss.

# Question Answer and marking notes
5 In the photosynthesis equation, 12 moles of carbon dioxide produce how many moles of oxygen and how many moles of glucose? Answer: 6 carbon dioxide to 6 oxygen to 1 glucose, so 12 moles of carbon dioxide give 12 moles of oxygen and 2 moles of glucose.
6 To make 9 g of glucose (molar mass 180), how many moles of carbon dioxide are needed, and what mass is that (molar mass 44)? Answer: 9 / 180 = 0.05 mol glucose, needing 0.3 mol carbon dioxide, that is 0.3 x 44 = 13.2 g.
7 Aerobic respiration yields up to 38 ATP per glucose molecule, anaerobic only 2. How many times more is the aerobic yield? Answer: 38 / 2 = 19 times, because anaerobic respiration does not break glucose down completely.
8 A leaf of total area 50 square centimetres loses 0.4 cubic centimetres of water per square centimetre per hour. How much water is lost in 8 hours of sunlight? Answer: 50 x 0.4 = 20.0 cubic centimetres per hour, so 160.0 cubic centimetres in 8 hours.

Genetics and Molecular Biology

These questions cover monohybrid and dihybrid ratios, DNA base composition, and the genetic code.

# Question Answer and marking notes
9 Two pea plants of genotype Aa are crossed, giving 400 offspring. How many are expected to show the dominant and the recessive phenotype? Answer: the ratio is 3 to 1, so about 300 dominant and 100 recessive out of 400.
10 Two individuals of genotype AaBb are crossed, giving 640 offspring. With a 9 to 3 to 3 to 1 ratio, how many show each phenotype? Answer: 360, 120, 120 and 40, totalling 640.
11 In a double-stranded DNA sample adenine is 30 per cent. What percentage is thymine, guanine and cytosine? Answer: thymine 30 per cent; the remaining 40 per cent is shared equally by guanine and cytosine, 20 per cent each.
12 A messenger RNA has 90 bases used for coding. With 3 bases per amino acid, how many amino acids are translated? Answer: 90 / 3 = 30 amino acids.

Energy Flow in Ecosystems

These questions cover energy pyramids, net primary production, population growth and the role of decomposers.

# Question Answer and marking notes
13 Producers fix 50000 kJ. If only one tenth passes to the next trophic level, how much energy reaches the third-level consumer? Answer: 5000 kJ, then 500 kJ, and 50 kJ at the third level.
14 An ecosystem has gross primary production of 20000 kJ and plant respiration of 8000 kJ. What is net primary production, and what do primary consumers get at one tenth efficiency? Answer: net primary production is 12000 kJ, so primary consumers get 1200 kJ.
15 An island deer population is 500 and grows by one fifth each year. With no immigration or emigration, roughly how many after 2 years? Answer: 500 x 1.2 to the power 2 = 720 deer.
16 State two roles of decomposers in the carbon cycle, and the effect on nutrient cycling if decomposers decline sharply. Answer: decomposers break down organic carbon in dead bodies and waste into carbon dioxide for producers, and release nitrogen and phosphorus back to the soil. A sharp decline leaves organic matter accumulating and inorganic nutrients short, lowering productivity.

Human Physiology and Coordination

These questions cover cardiac output, kidney filtration, blood glucose and nerve conduction speed.

# Question Answer and marking notes
17 With a heart rate of 70 beats per minute and a stroke volume of 70 mL, what is the cardiac output in mL per minute, and in litres per minute? Answer: 70 x 70 = 4900 mL per minute, that is 4.9 litres per minute.
18 Glomerular filtration rate is 125 mL per minute. How many litres are filtered in a full day of 1440 minutes? Answer: 125 x 1440 = 180000 mL, that is 180.0 litres.
19 Blood glucose is 5.5 mmol per litre and plasma volume is 3 litres. What is the total amount of glucose in mmol? Answer: 5.5 x 3 = 16.5 mmol.
20 A nerve impulse travels at 120 metres per second over 1.2 metres. How long does it take in seconds, and in milliseconds? Answer: 1.2 / 120 = 0.01 s, that is 10.0 ms.

Biotechnology and Conservation

These questions cover PCR amplification, gel electrophoresis, reserve planning and yield comparison.

# Question Answer and marking notes
21 Each PCR cycle doubles the number of target DNA fragments. Starting from a single copy, how many copies are present after 30 cycles? Answer: 2 to the power 30 = 1073741824 copies.
22 In gel electrophoresis three fragments are 800, 400 and 200 base pairs long. Roughly how many times further does the shortest travel than the longest, and why? Answer: 800 / 200 = 4 times; shorter fragments meet less resistance in the gel mesh and move faster.
23 A reserve covers 2000 hectares, with the core zone taking 35 per cent. What is the core area in hectares? Answer: 2000 x 35 / 100 = 700 hectares.
24 A genetically modified variety yields 6.5 tonnes per hectare against 5.8 tonnes for the traditional one. Over 10 hectares, what is the difference in total yield and the percentage gain? Answer: 65.0 tonnes against 58.0, a difference of 7.0 tonnes, or a gain of 12.1 per cent.

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