CIE A-Level Biology 9700 Paper 2 Structured Questions Practice Paper | 10 questions, 60 marks, 1 hour 15 minutes

2026-10-31

CIE A-Level Biology 9700 Paper 2

CIE A-Level Biology 9700 Paper 2 (AS structured) practice paper: 10 questions, 60 marks, 1 hour 15 minutes, with full worked solutions and marks on enzymes, osmosis, gas exchange, circulation, genetics and ecology.

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Paper details　Level: AS (A Level) | Subject: Biology | Time: 1 hour 15 minutes | Marks: 60 | Questions: 10

Section A (20 marks)

Structured questions: show your working and give correct units. Total 60 marks.

#	Question	Answer and marking notes
1	A student measures catalase activity in potato tissue: 1.6 cm³ of oxygen is collected in 30 s. (a) Calculate the rate of reaction in cm³ min⁻¹. (b) If the rate stays constant, how long would it take to collect 12 cm³ of oxygen? (c) The rate falls sharply when the temperature rises from 30 °C to 60 °C. Explain why, and explain why the rate is highest in the first few seconds.	(a) 3.2 cm³ min⁻¹ (1.6 ÷ 0.5 min); 2 marks for working, unit error −1. (b) 3.75 min (12 ÷ 3.2); 2 marks. (c) Above the optimum, hydrogen and ionic bonds break, so the tertiary structure and the active site are distorted and the enzyme is denatured (2 marks); denaturation is irreversible so the rate does not recover (1 mark). The initial rate is highest because substrate concentration is highest, giving the most collisions with active sites (2 marks).
2	A chloroplast in an electron micrograph measures 45 mm and its actual length is 150 µm. (a) Calculate the magnification. (b) In another micrograph a cell measures 12 mm at a magnification of ×400; find its actual length in µm. (c) Name two structures visible only with an electron microscope, giving one reason for each.	(a) 45 000 µm ÷ 150 µm = ×300 (2 marks). (b) 12 000 µm ÷ 400 = 30 µm (= 0.030 mm) (2 marks). (c) Ribosomes (about 20 nm across, smaller than the 200 nm resolving limit of a light microscope); cristae of mitochondria or detail of the endoplasmic reticulum (1 mark each, 2 marks).
3	Three equal potato strips (initial mass 2.40 g) are placed in distilled water, 0.5 mol dm⁻³ sucrose solution and a sucrose solution isotonic to the cell sap. After 30 minutes their masses are 2.58 g, 2.22 g and 2.40 g. (a) Calculate the percentage change in mass for the first two strips. (b) State which solution has the same water potential as the potato cell sap. (c) Explain the gain in mass of the first strip and why the third strip does not change.	(a) Distilled water: +7.5 %; sucrose solution: -7.5 % (1 mark each; wrong sign −1). (b) The isotonic sucrose solution (1 mark). (c) Distilled water has a higher water potential than the cell sap, so water moves into the cell by osmosis through the partially permeable membrane (2 marks); the vacuole swells and the cell becomes turgid (1 mark). In the third strip there is no water potential gradient, so the net movement of water is zero (2 marks).

Section B (20 marks)

Structured questions: show your working and give correct units. Total 60 marks.

#	Question	Answer and marking notes
4	A student breathes 12 times per minute with a tidal volume of 500 cm³. Inhaled air contains 21 % oxygen and exhaled air 16 %. (a) Calculate the pulmonary ventilation rate in dm³ min⁻¹. (b) Calculate the volume of oxygen absorbed per minute in cm³. (c) Explain how the structure of the alveolar wall speeds up diffusion of oxygen, and state two physiological responses that raise the rate of diffusion.	(a) 500 × 12 = 6 000 cm³ = 6.0 dm³ min⁻¹ (2 marks). (b) 6.0 × 5 % = 0.30 dm³ = 300 cm³ min⁻¹ (2 marks; 1 mark if the 5 % is not multiplied by the ventilation rate). (c) Alveolar and capillary walls are one flattened cell thick, so the diffusion distance is very short; the total alveolar surface area is large; a thin film of moisture lets gases dissolve first (3 marks). Responses: deeper or faster breathing, faster heart rate, dilation of pulmonary capillaries (2 marks).
5	A potometer records that the bubble moves the equivalent of 2.4 cm³ of water in 20 minutes; the leaf area is 60 cm². (a) Calculate the transpiration rate in cm³ min⁻¹. (b) Calculate the rate per cm² of leaf per minute. (c) Explain why the reading is much higher in strong wind.	(a) 0.12 cm³ min⁻¹ (2 marks). (b) 0.12 ÷ 60 = 0.0020 cm³ cm⁻² min⁻¹ (2 marks). (c) Wind removes water vapour from around the leaf, steepening the gradient between the sub-stomatal air spaces and the outside air and thinning the boundary layer, so evaporation and diffusion are faster (2 marks).
6	An athlete at rest has a stroke volume of 70 cm³ and a heart rate of 72 beats min⁻¹. During strenuous exercise the stroke volume is 110 cm³ and the rate is 150 beats min⁻¹. (a) Calculate the cardiac output in both states (dm³ min⁻¹). (b) Find how many times greater the cardiac output is during exercise. (c) Explain why output cannot rise by increasing heart rate alone, and describe the blood vessel changes during exercise.	(a) At rest: 70 × 72 = 5 040 cm³ = 5.04 dm³ min⁻¹; during exercise: 110 × 150 = 16 500 cm³ = 16.5 dm³ min⁻¹ (4 marks, 2 each). (b) 16 500 ÷ 5 040 = 3.27 times (2 marks). (c) A very high rate shortens diastole, so filling time and coronary blood supply fall and stroke volume drops; during exercise arterioles in skeletal muscle and skin dilate while those to the gut constrict to maintain blood pressure (1 mark).

Section C (20 marks)

Structured questions: show your working and give correct units. Total 60 marks.

#	Question	Answer and marking notes
7	Two heterozygous black mice (Bb) are crossed and produce 800 offspring. (a) Predict the number of black and brown offspring. (b) A test cross of a black mouse with a brown mouse gives 118 black and 122 brown offspring out of 240. State the genotype of the black parent and explain. (c) If two black parents produce only black offspring, state their genotypes.	(a) Black 600 (3/4), brown 200 (1/4) (2 marks). (b) The black parent is heterozygous Bb (2 marks): the near 1:1 ratio shows it produces B and b gametes while the brown parent contributes only b (2 marks). (c) Both are homozygous BB and pass on only B alleles.
8	A messenger RNA molecule has 300 bases, all of which are translated into one polypeptide. (a) How many amino acids does the polypeptide contain? (b) How long is the double-stranded DNA that codes for it, in nm (0.34 nm per base pair)? (c) How many peptide bonds does it contain?	(a) 300 ÷ 3 = 100 amino acids (2 marks). (b) 300 base pairs × 0.34 nm = 102 nm (= 0.102 µm) (2 marks). (c) 100 − 1 = 99 peptide bonds (1 mark).
9	Yeast ferments 5.4 g of glucose (M = 180) anaerobically. (a) How many moles of carbon dioxide are produced? (b) What volume does it occupy at room temperature (24 dm³ mol⁻¹)? (c) How much energy in kJ would complete aerobic respiration release (2 880 kJ per mol of glucose)?	(a) 0.03 mol glucose → 0.06 mol CO₂ (2 marks). (b) 0.06 × 24 = 1.44 dm³ (= 1440 cm³) (2 marks). (c) 0.03 × 2 880 = 86.4 kJ (1 mark); anaerobic fermentation yields only about 2 ATP, so most energy remains in the ethanol.
10	Researchers capture and mark 60 snails, then release them. A week later they catch 80 snails, 12 of which are marked. (a) Estimate the population size. (b) If primary consumers fix 25 000 kJ of energy, how much energy reaches secondary and tertiary consumers at 10 % transfer efficiency?	(a) (60 × 80) ÷ 12 = 400 snails (2 marks). (b) Secondary consumers receive 2500 kJ and tertiary consumers 250 kJ (1 mark each). Assumptions: marking does not affect survival or catchability and there is no immigration, emigration, birth or death between samples.

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