HKDSE Biology Paper 1 Practice Paper with Answers | 30 questions, 60 minutes, 100 marks

2026-10-29

Practice paper · HKDSE Biology Paper 1

HKDSE Biology Paper 1 practice paper: 25 multiple-choice and 5 structured questions on cells, genetics, physiology and ecology, 60 minutes, 100 marks, full worked answers and PDFs.

Download　Question paper PDF　|　Answer key PDF

Paper details　Level: S6 (HKDSE) | Subject: Biology (Paper 1) | Time: 60 minutes | Total: 100 marks | Questions: 30

Section A: Cells, Genetics and Molecules (24 marks)

1. Two marks each. Choose the best answer.

#	Question	Answer and marking notes
1	A cell has an actual diameter of 0.05 mm and measures 30 mm on a micrograph. What is the magnification?	Magnification = image size ÷ actual size = 30 ÷ 0.05 = 600 (×600). (2 marks; 0 if the fraction is inverted.)
2	A cell cycle lasts 20 hours, of which mitosis occupies 1 hour. What percentage of the cycle is mitosis?	Percentage = 1 ÷ 20 × 100% = 5%. (2 marks; a short mitotic phase within a long interphase is typical of the cell cycle.)
3	Which structure is found in plant cells but not in animal cells?	A cellulose cell wall and chloroplasts occur only in plant cells; the cell membrane, ribosomes, mitochondria and nucleus occur in both. (2 marks)
4	Salivary amylase is boiled at 100 ℃ and then cooled to 37 ℃; it no longer digests starch because:	Heat denatures the protein, so the shape of the active site changes, the substrate no longer fits, and denaturation is irreversible — cooling does not restore activity. (2 marks; 0 for “the enzyme is digested into amino acids”.)
5	An animal cell is placed in pure (distilled) water. What is the most likely result?	Pure water has a higher water potential than the cytoplasm, so water enters by osmosis; without a cell wall the animal cell swells and bursts (haemolysis). (2 marks)
6	In a double-stranded DNA sample adenine (A) makes up 30% of all bases. What percentage is guanine (G)?	A pairs with T, so T is also 30%; the remaining 100% − 30% − 30% = 40% is shared equally by G and C, so G = 20%. (2 marks)
7	A DNA fragment has 1000 base pairs. If each base pair adds 0.34 nm to the helix, how long is the fragment?	Length = 1000 × 0.34 = 340 nm (= 0.34 µm). (2 marks)
8	Which feature is unique to meiosis and absent from mitosis?	Only meiosis pairs homologous chromosomes (synapsis), allows crossing over and has two divisions, giving four haploid cells; mitosis divides once and keeps the chromosome number the same. (2 marks)
9	Two pea plants of genotype Aa are crossed, A being dominant. How many of 400 seeds are expected to show the dominant phenotype?	Aa × Aa gives 1 AA : 2 Aa : 1 aa, so 3/4 show the dominant phenotype; 400 × 3/4 = 300 seeds. (2 marks)
10	Which is the best way to find whether a tall pea plant (dominant phenotype) is homozygous?	Carry out a test cross with a dwarf (homozygous recessive) plant: all-tall offspring means the parent is homozygous, while any dwarf offspring means it is heterozygous. (2 marks; 1 mark for selfing, which can be inconclusive.)
11	A father of blood group Iᴬi and a mother of Iᴮi have children. What fraction is group O?	The combinations are IᴬIᴮ, Iᴬi, Iᴮi and ii in equal chance; only ii is group O, so 1/4. (2 marks; group O needs two i alleles.)
12	Sickle-cell anaemia is caused by a single base substitution. What kind of mutation is this and what is its most direct consequence?	It is a gene (point) mutation: the codon changes, one amino acid in the β-chain of haemoglobin is replaced, the protein’s structure and function change, and red cells become sickle-shaped at low oxygen. (2 marks)

Section B: Physiology, Ecology and Health (26 marks)

2. Two marks each. Choose the best answer.

#	Question	Answer and marking notes
13	The overall equation of photosynthesis is 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂. If a leaf takes in 0.60 mol of CO₂ and converts it all to glucose (Mr = 180), what mass of glucose is formed?	The ratio CO₂ : glucose = 6 : 1, so n(glucose) = 0.60 ÷ 6 = 0.10 mol; m = 0.10 × 180 = 18 g. (2 marks)
14	At low light intensity the rate of photosynthesis rises with light intensity, but beyond a certain intensity it levels off. What does this show?	Light intensity is the limiting factor at first; once the rate levels off, some other factor such as carbon dioxide concentration or temperature becomes limiting. (2 marks; 0 for “light is no longer used”.)
15	In aerobic respiration 1 mol of glucose theoretically yields about 38 mol ATP. How much ATP is released from 10 mol of glucose oxidised completely?	It scales directly: 38 × 10 = 380 mol ATP. (2 marks; a real cell obtains about 30–32 mol, the rest being lost as heat.)
16	At rest a person breathes 15 times per minute, 500 cm³ of air per breath. What is the pulmonary ventilation rate?	Ventilation rate = breathing rate × tidal volume = 15 × 500 = 7500 cm³ min⁻¹ (7.5 dm³ min⁻¹). (2 marks)
17	A person has a heart rate of 70 beats min⁻¹ and a stroke volume of 70 cm³. What is the cardiac output?	Cardiac output = heart rate × stroke volume = 70 × 70 = 4900 cm³ min⁻¹ = 4.9 dm³ min⁻¹. (2 marks)
18	Which change most directly increases the rate of transpiration from a leaf?	Open stomata (as in light), together with air movement and low humidity, raise the transpiration rate; a thicker cuticle, closed stomata and high humidity all lower it. (2 marks)
19	In the kidney, where is glucose mainly reabsorbed, and where does ultrafiltration occur?	Glucose is reabsorbed completely by active transport in the proximal convoluted tubule; ultrafiltration occurs in the glomerulus (Bowman’s capsule), where pressure forces water and small molecules into the capsule. (2 marks)
20	Body temperature rises during vigorous exercise. The immediate heat-losing response is:	Increased sweating and vasodilation of skin arterioles, raising heat loss by evaporation and radiation; vasoconstriction and shivering conserve heat and are the opposite response. (2 marks)
21	In a reflex arc, the correct sequence of transmission is:	Receptor → sensory neurone → spinal cord (interneurone) → motor neurone → effector (muscle or gland). (2 marks)
22	After a meal blood glucose rises. Which hormone is secreted more and lowers it?	Insulin, from the β cells of the pancreatic islets, increases glucose uptake by cells and glycogen formation in liver and muscle, lowering blood glucose; glucagon acts in the opposite direction. (2 marks)
23	A 0.5 m × 0.5 m quadrat is used on a lawn and one quadrat contains 12 plants of a species. Estimate the population on 500 m² of lawn.	Quadrat area = 0.5 × 0.5 = 0.25 m²; density = 12 ÷ 0.25 = 48 plants m⁻²; total ≈ 48 × 500 = 24000 plants. (2 marks)
24	Producers fix 20000 kJ of energy. If the transfer efficiency is 10%, how much energy reaches the primary consumers?	20000 × 10% = 2000 kJ; the rest is lost through respiration, excretion and uneaten parts. (2 marks)
25	Which statement best explains why biodiversity should be conserved?	A wide range of species and genes makes an ecosystem more resistant to disease, climate change and other disturbance and keeps food webs stable; a single-crop system is easily wiped out. (2 marks)

Section C: Structured questions (50 marks)

3. Answer all questions. Show your working, units and reasoning.

#	Question	Answer and marking notes
26	A student observes onion epidermis under a compound microscope and measures it with an eyepiece graticule. (a) At 40× objective the image is 30 mm across and the real cell is 0.05 mm. Find the magnification. (2 marks) (b) Describe three structures found in plant cells but not in animal cells, and give one function of each. (3 marks) (c) Why is iodine solution used before observing? (2 marks) (d) If the slide is moved to the left, which way does the image move? Explain. (3 marks)	(a) Magnification = image ÷ actual = 30 ÷ 0.05 = 600. (2 marks) (b) Cell wall — support and fixed shape; chloroplast — photosynthesis; large vacuole — stores water and solutes and maintains turgor. (3 marks, 1 each) (c) Iodine stains the nucleus and starch dark, raising contrast so the cell wall and nucleus are easier to see. (2 marks) (d) The image moves to the right: a compound microscope inverts the image in both axes, so moving the slide left moves the image right. (3 marks; 1 mark for “opposite” alone.)
27	In peas, round (R) is dominant to wrinkled (r) and yellow (Y) to green (y); the two genes are on different chromosomes. Two RrYy plants are crossed, giving 320 seeds. (a) List the gametes of both parents with their ratio, and find the phenotypic ratio of the offspring by a Punnett square or probability. (4 marks) (b) How many round-yellow, round-green, wrinkled-yellow and wrinkled-green seeds are expected? (3 marks) (c) If 20 wrinkled-green seeds were actually obtained, is this reasonable? (1 mark) (d) A round-yellow seed grows into a plant that gives wrinkled-green offspring when test crossed with rr yy. What is the genotype of that seed? (2 marks)	(a) Each plant makes four gametes RY, Ry, rY and ry in 1 : 1 : 1 : 1; the genes assort independently, giving a 9 : 3 : 3 : 1 phenotypic ratio. (4 marks) (b) Round-yellow = 320 × 9/16 = 180; round-green = 60; wrinkled-yellow = 60; wrinkled-green = 20. (3 marks; each computed as 320 × its fraction) (c) Reasonable: 20 is close to the expected 20, and small samples fluctuate at random. (1 mark) (d) It must be able to pass on ry, so the seed was RrYy (double heterozygote). (2 marks)
28	A leaf takes in 0.60 mol of CO₂ in light and releases the same amount of O₂ (6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂, Mr(glucose) = 180). (a) Write the overall equation and state where the light reactions and carbon fixation (dark reactions) occur in the chloroplast. (3 marks) (b) Find the mass of glucose formed in g. (3 marks) (c) Explain why a plant does not photosynthesise at night but still respires. (2 marks) (d) If temperature rises from 20 ℃ to 35 ℃ and respiration rises more than photosynthesis, what happens to net photosynthesis? (2 marks)	(a) 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂; the light reactions occur on the thylakoid (grana) membranes and carbon fixation in the stroma. (3 marks) (b) n(glucose) = 0.60 ÷ 6 = 0.10 mol; m = 0.10 × 180 = 18 g. (3 marks) (c) Without light the light reactions stop, so no ATP or NADPH is available for carbon fixation; respiration needs no light and the mitochondria keep oxidising respiratory substrates, so it runs day and night. (2 marks) (d) Net photosynthesis = photosynthesis − respiration; if respiration rises more, net photosynthesis falls and the plant accumulates less. (2 marks)
29	A student uses 0.5 m × 0.5 m quadrats to survey a wild flower on 500 m² of grassland, placing 10 quadrats at random; one quadrat holds 12 plants. (a) Find the quadrat area and the plant density in plants m⁻². (3 marks) (b) Estimate the total number of plants on the grassland. (2 marks) (c) Explain why many random quadrats are used rather than a single one. (2 marks) (d) Producers fix 20000 kJ and the transfer efficiency is 10%. Find the energy available to primary and secondary consumers, and explain why an energy pyramid narrows upwards. (3 marks)	(a) Area = 0.5 × 0.5 = 0.25 m²; density = 12 ÷ 0.25 = 48 plants m⁻². (3 marks) (b) Total ≈ density × area = 48 × 500 = 24000 plants. (2 marks) (c) Plants are not evenly distributed, so one quadrat may fall on a dense clump or bare ground and misrepresent the site; many random quadrats reduce sampling error and give a more reliable mean. (2 marks) (d) Primary consumers = 20000 × 10% = 2000 kJ; secondary = 2000 × 10% = 200 kJ. Most energy at each level is lost in respiration (heat), excretion and uneaten parts, with only about one tenth passed on, so higher levels hold less energy and biomass. (3 marks)
30	At rest a person has a heart rate of 70 min⁻¹ and a stroke volume of 70 cm³; during vigorous exercise the rate is 140 min⁻¹ and the stroke volume 100 cm³. (a) Find the cardiac output at rest and during exercise in dm³ min⁻¹. (4 marks) (b) How many times greater is the exercising cardiac output? (1 mark) (c) Explain the physiological value of the rise in heart rate and stroke volume during exercise. (3 marks) (d) Breathing also deepens and quickens; explain how this matches the higher cardiac output. (2 marks)	(a) At rest: 70 × 70 = 4900 cm³ min⁻¹ = 4.9 dm³ min⁻¹; during exercise: 140 × 100 = 14000 cm³ min⁻¹ = 14.0 dm³ min⁻¹. (4 marks) (b) 14.0 ÷ 4.9 ≈ 2.9 times (about 3 times). (1 mark) (c) Working muscles need more oxygen and produce more carbon dioxide, so a higher cardiac output delivers oxygen and glucose faster and removes carbon dioxide and heat, keeping the internal environment stable. (3 marks) (d) A higher cardiac output sends more blood through the alveolar capillaries; the matching rise in ventilation keeps the concentration gradients of oxygen and carbon dioxide between air and blood, so gas exchange stays efficient. (2 marks)

Find a tutor — free

A TutorZone practice paper; original questions and passages, not publisher material.