HKDSE Biology Paper 1 (Compulsory Part) Mock Paper | 10 Questions, 100 Marks, 2.5 Hours

2026-11-01

HKDSE Biology

HKDSE Biology Paper 1 mock: 10 questions, 100 marks, 150 minutes, covering magnification and field of view, enzyme activity and denaturation, limiting factors in photosynthesis, monohybrid crosses, ABO inheritance, the Lincoln index and energy transfer, osmosis and the incipient point, cardiac output and ventilation during exercise, emulsification and surface area, and vaccination and herd immunity, with marking notes for every question.

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Paper details　Level: HKDSE (S6) | Subject: Biology | Time: 2 hours 30 minutes | Marks: 100 | Questions: 10

Section A (Questions 1 to 5, 50 marks)

Answer ALL questions. In calculations show the formula, the substitution and the unit, and give answers to the accuracy asked; in explanation questions use biological terms and refer to the data.

#	Question	Answer and marking notes
1	A student uses a x10 eyepiece and a x40 objective to observe onion epidermal cells. (a) Calculate the total magnification of this combination. (b) At that magnification a cell measures 60 mm on the image; calculate its actual length in micrometres. (c) With a x10 objective the field diameter is 1,800 micrometres. Calculate the field diameter with the x40 objective and explain why the slide must be moved.	Answer: (a) Total magnification = 10 x 40 = 400 times. (b) Actual length = 60 mm / 400 = 0.15 mm = 150 micrometres. (c) Field diameter = 1,800 x 10 / 40 = 450 micrometres. A higher magnification narrows the field, so part of the specimen falls outside it; the slide is moved to centre the target, otherwise counts and shapes mislead.
2	A student uses 5 cm3 of enzyme solution to catalyse a substrate and measures the total product after 5 minutes: 5.0 mg at 20 C, 12.5 mg at 37 C, 3.0 mg at 60 C and 0 mg at 80 C. (a) Calculate the mean rate at each temperature in mg per minute. (b) State the optimum temperature and calculate the rate multiple at 37 C relative to 20 C (1 decimal place). (c) Explain why the rate is zero at 80 C and describe the rate after cooling that solution back to 37 C.	Answer: (a) 1.0, 2.5, 0.6 and 0 mg per minute respectively. (b) The optimum is 37 C; the multiple is 2.5 / 1.0 = 2.5 times. (c) At 80 C the enzyme is denatured: heat breaks the bonds holding its three-dimensional shape, the active site changes shape and the substrate no longer fits, so the rate is zero. Denaturation is irreversible, so after cooling to 37 C the rate stays at zero, unlike reversible low-temperature inhibition.
3	An aquatic plant releases the following numbers of bubbles per minute at different light intensities: 12 at 2 klx, 28 at 4 klx, 40 at 6 klx and 44 at 8 klx. (a) Calculate the increase in bubbles per minute from 2 to 4 klx and from 6 to 8 klx. (b) Calculate the total number of bubbles produced in 45 minutes at 8 klx. (c) Explain why the gain shrinks after 6 klx and name another factor that may then be limiting.	Answer: (a) From 2 to 4 klx the gain is 28 – 12 = 16 bubbles; from 6 to 8 klx it is only 44 – 40 = 4. (b) 44 x 45 = 1,980 bubbles. (c) Beyond a certain intensity the light-dependent stage is nearly saturated, so light is no longer limiting; carbon dioxide concentration or temperature then limits the rate, so extra light raises photosynthesis only slightly, cutting the gain from 16 to 4 bubbles per minute.
4	In mice black coat (B) is dominant to white coat (b). Two black mice produce 64 offspring, of which 16 are white. (a) State the genotypes of the two parents and explain your answer. (b) Calculate the expected numbers of black and white offspring out of 64 and compare with the observation. (c) If the pair has 3 more offspring, calculate the probability that all three are white, as a fraction in lowest terms.	Answer: (a) Both parents are Bb. A white offspring must be bb and needs one b from each parent, while both parents look black and therefore each carries a B. (b) Bb x Bb gives 1 BB : 2 Bb : 1 bb, a 3 : 1 phenotypic ratio, so 48 black and 16 white are expected, matching the observed 48 and 16 exactly. (c) Each offspring is white with probability 1/4, so all three white = (1/4)^3 = 1/64.
5	A father is blood group A (genotype I^A i) and the mother is group B (genotype I^B i). (a) Use a genetic diagram to list the possible blood groups of their children and the ratio. (b) Calculate the probability that a child is group O. (c) If they have two children, calculate the probability that both are group O (as a percentage, 2 decimal places) and explain why blood groups alone cannot establish parentage.	Answer: (a) The four combinations are I^A I^B (AB), I^A i (A), I^B i (B) and ii (O), in the ratio 1:1:1:1. (b) Group O (ii) has probability 1/4. (c) Both children group O = (1/4)^2 = 1/16 = 6.25%. Blood group is set by only three alleles of the ABO system and the same group is common in the population, so many parent-child pairs give identical groups; blood typing can only exclude, and DNA analysis is needed to confirm parentage.

Section B (Questions 6 to 10, 50 marks)

Answer ALL questions. In calculations show the formula, the substitution and the unit, and give answers to the accuracy asked; in explanation questions use biological terms and refer to the data.

#	Question	Answer and marking notes
6	Snails in a grassland are estimated by capture-mark-recapture: 40 are caught, marked and released, and a week later 60 are recaught, of which 15 carry marks. (a) Estimate the population size using the Lincoln index. (b) State two assumptions this method requires. (c) If producers fix 50,000 kJ and primary consumers assimilate 5,000 kJ, calculate the ecological efficiency and the maximum energy available to secondary and tertiary consumers in kJ.	Answer: (a) Population = (40 x 60) / 15 = 160 snails. (b) The population is closed, with no immigration, emigration, births or deaths during the study; marked animals mix randomly, the mark does not affect survival or rub off, and every animal is equally catchable. (c) Efficiency = 5,000 / 50,000 x 100% = 10%; secondary consumers obtain at most 500 kJ and tertiary consumers at most 50 kJ, the rest being lost in respiration, excretion and uneaten tissue.
7	Five identical potato strips, each 5.0 g, are placed for 30 minutes in sucrose solutions of 0.0, 0.2, 0.4, 0.6 and 0.8 mol per dm3; the changes in mass are +8.0%, +2.0%, -4.0%, -10.0% and -16.0%. (a) Calculate the final mass of each strip in g to 2 decimal places. (b) Estimate the sucrose concentration at the incipient point where there is no net water movement, and state how you estimated it. (c) Explain why the mass change becomes more negative at higher concentration, and predict what happens when the strip from 0.8 mol per dm3 is returned to distilled water for 30 minutes.	Answer: (a) The final masses are 5.40, 5.10, 4.80, 4.50 and 4.20 g. (b) The change turns from +2.0% at 0.2 to -4.0% at 0.4 mol per dm3, so by linear interpolation the incipient point is 0.2 + (2.0 / 6.0) x 0.2 = 0.267, about 0.27 mol per dm3. (c) A stronger sucrose solution has lower water potential, so water leaves the vacuole by osmosis; the cells lose turgor and shrink (plasmolysis), so the mass falls further. In distilled water water re-enters and the mass rises again, though badly plasmolysed cells may not recover fully.
8	At rest a person has a heart rate of 70 per minute, stroke volume 70 cm3, breathing rate 12 per minute, tidal volume 500 cm3 and oxygen uptake 0.3 dm3 per minute. During vigorous exercise the values are 150, 110 cm3, 30, 1,800 cm3 and 2.4 dm3 per minute respectively. (a) Calculate the cardiac output at rest and during exercise in dm3 per minute, and the multiple (2 decimal places). (b) Calculate the pulmonary ventilation at rest and during exercise and the multiple. (c) Oxygen supply still falls short, so muscles respire anaerobically and lactate accumulates, creating an oxygen debt of 4.2 dm3. If recovery oxygen uptake exceeds the resting value by 0.7 dm3 per minute, calculate the time needed to repay the debt.	Answer: (a) Resting output = 70 x 70 / 1,000 = 4.9 dm3 per minute; exercise = 150 x 110 / 1,000 = 16.5 dm3 per minute, a multiple of 3.37. (b) Resting ventilation = 12 x 500 / 1,000 = 6.0 dm3 per minute; exercise = 30 x 1,800 / 1,000 = 54.0 dm3 per minute, a multiple of 9.0. (c) Time = 4.2 / 0.7 = 6 minutes.
9	Bile emulsifies a fat globule of diameter 2.0 mm into 1,000 droplets of diameter 0.2 mm (treat them as spheres with surface area 4 pi r squared, pi = 3.14). (a) Calculate the total surface area before and after emulsification in mm2. (b) Calculate the fold increase in total surface area. (c) Explain how emulsification speeds up fat digestion and state the difference in optimum pH between pepsin and trypsin and its significance.	Answer: (a) Before: r = 1.0 mm, area = 4 x 3.14 x 1.0 squared = 12.56 mm2. After: r = 0.1 mm, each = 0.1256 mm2, so 1,000 droplets give 125.6 mm2. (b) The increase is 125.6 / 12.56 = 10 times. (c) Lipase acts only at the water-fat interface, so a tenfold larger area lets lipase contact far more fat and hydrolyse it to fatty acids and glycerol faster. Pepsin works best near pH 2 in the acid stomach while trypsin works best near pH 8 in the alkaline small intestine, each matching the site where it acts.
10	A district has 500 residents, a vaccination rate of 80% and a vaccine efficacy of 90%; the basic reproduction number R0 of the disease is 5. (a) Calculate the number vaccinated, the number immune and the number still susceptible. (b) Calculate the proportion of immune people needed for herd immunity and judge whether the district reaches it. (c) Explain why an individual who is vaccinated is still less likely to become severely ill, and suggest one practical measure to raise the vaccination rate.	Answer: (a) Vaccinated = 500 x 80% = 400; immune = 400 x 90% = 360; still susceptible = 500 – 360 = 140, namely 100 unvaccinated people and 40 for whom the vaccine failed. (b) The required proportion is 1 – 1/5 = 0.80, or 80%, while actual immunity is 360/500 = 72%, so the district falls short of herd immunity. (c) Vaccination induces memory B and T cells and antibodies, so on infection the immune system recognises and clears the pathogen faster, lowering viral load and the risk of severe illness and death. Coverage can be raised by community health education and by offering free or school-based vaccination.

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