Secondary 5 Biology Practice Paper: Cell Division And Genetics, Photosynthesis And Respiration, Enzymes And Metabolism, Plant Transport, Animal Transport, Ecology And Energy Flow (100 Marks)

Secondary 5 Biology Practice Paper: cell division and genetics, photosynthesis and respiration, enzymes and metabolism, plant transport, animal transport, ecology and energy flow (100 marks)

2026-11-14

Biology – Secondary 5

S5 Biology practice paper: 24 structured questions with worked answers on genetics, enzymes, transport and ecology. Original questions for school tests.

Paper details　S5 Biology practice paper with 24 structured questions and full worked answers on cell division, genetics, gas exchange in plants, enzymes, transport and ecology.

Section A – Cell division and genetics (17 marks)

Study the information given and answer all questions.

#	Question	Answer and marking notes
1	A mammalian somatic cell contains 46 chromosomes. State the chromosome number of a daughter cell formed by mitosis and of a gamete, and explain how meiosis keeps the chromosome number stable between generations.	Answer: The daughter cell after mitosis still has 46 chromosomes; a gamete has 23. Meiosis halves the chromosome number and fertilisation restores 46, so the number stays constant from generation to generation.
2	Two tall pea plants with genotype Tt are crossed. State the expected genotypic and phenotypic ratios, and calculate the approximate number of tall and short plants when 240 offspring are produced.	Answer: Genotypic ratio 1 TT : 2 Tt : 1 tt; phenotypic ratio 3 tall : 1 short. Of 240 offspring about 180 are tall and about 60 are short.
3	A tall pea plant is crossed with a short plant of genotype tt, and about half of the offspring are tall and half are short. Deduce the genotype of the tall parent and state the use of this type of cross.	Answer: The tall parent is Tt. Crossing with a homozygous recessive individual reveals the genotype of an individual showing the dominant phenotype; appearance of recessive offspring proves it carries the recessive allele. This is a test cross.
4	State when crossing over occurs in meiosis and explain its contribution to genetic variation.	Answer: Crossing over occurs in prophase I of meiosis when homologous chromosomes pair, and non-sister chromatids exchange segments. Linked alleles are recombined, producing new gene combinations and increasing genetic variation.

Section B – Photosynthesis and respiration (17 marks)

The following questions concern gas exchange and energy conversion.

#	Question	Answer and marking notes
5	Write the word equation for photosynthesis. If a leaf fixes 12 molecules of carbon dioxide in the light, how many molecules of glucose can theoretically be formed?	Answer: Carbon dioxide + water -> glucose + oxygen (in the presence of chlorophyll and light). Six molecules of carbon dioxide give one molecule of glucose, so 12 molecules give 2 molecules of glucose.
6	In the dark a student finds that germinating seeds release 400 cm3 of carbon dioxide and use 500 cm3 of oxygen in 20 minutes. Calculate the respiratory quotient and deduce the main substrate being respired.	Answer: RQ = carbon dioxide released / oxygen used = 400 / 500 = 0.8. A value near 0.7 indicates that fat is the main substrate.
7	Compare photosynthesis and respiration in terms of site, energy change and gas exchange.	Answer: Photosynthesis occurs in chloroplasts, needs light, stores light energy as chemical energy, takes in carbon dioxide and releases oxygen. Respiration occurs in the cytoplasm and mitochondria, runs day and night, releases energy, uses oxygen and releases carbon dioxide.
8	Describe the experiment that shows starch is produced in a leaf exposed to light, and explain why the leaf is decolourised in hot alcohol first.	Answer: Keep the plant in light for several hours, kill a leaf in boiling water, decolour it in hot alcohol over a water bath, rinse and add iodine solution; a blue-black colour shows starch. Decolourising removes chlorophyll so the colour change is not masked.

Section C – Enzymes and metabolism (16 marks)

The following questions concern enzyme-catalysed reactions.

#	Question	Answer and marking notes
9	State three properties of enzymes and their chemical nature.	Answer: Enzymes are proteins. They are specific to particular substrates, can be used repeatedly, and their activity is affected by temperature and pH; high temperature denatures them.
10	In an experiment 40 cm3 of product is collected in 10 minutes. Calculate the mean rate of reaction and explain why the rate falls later in the reaction.	Answer: Mean rate = 40 / 10 = 4.0 cm3 per minute. Later the substrate concentration falls and the active sites become saturated, so the rate decreases.
11	Explain how the rate of an enzyme-catalysed reaction changes when the temperature rises from 37 degrees to 60 degrees, and state whether the change is reversible.	Answer: 37 degrees is close to the optimum for human enzymes, so the rate is highest. At 60 degrees the three-dimensional structure is disrupted, the active site loses its shape and the rate falls sharply; denaturation is irreversible.
12	Explain how a competitive inhibitor slows an enzyme-catalysed reaction and how its effect can be reduced.	Answer: A competitive inhibitor resembles the substrate and occupies the active site so the substrate cannot bind. Raising the substrate concentration increases the chance that substrate occupies the site, reducing the inhibition.

Section D – Transport in plants (17 marks)

The following questions concern the transport of water and nutrients.

#	Question	Answer and marking notes
13	A cut shoot stands in a measuring cylinder. After 24 hours the water level falls from 100.0 mL to 88.0 mL. Calculate the mean transpiration per day and per hour.	Answer: Water lost per day = 100.0 – 88.0 = 12.0 mL; per hour = 12.0 / 24 = 0.5 mL.
14	Explain how stronger wind, lower humidity and brighter light each affect the rate of transpiration.	Answer: Wind removes water vapour from the leaf surface and steepens the diffusion gradient, so transpiration speeds up. Lower humidity also steepens the gradient. Brighter light opens the stomata and warms the leaf, which also raises the rate.
15	Compare xylem and phloem in the materials carried and the direction of transport.	Answer: Xylem carries water and dissolved mineral salts mainly upwards from the roots to the leaves. Phloem carries organic solutes such as sucrose and amino acids in both directions between leaves and other organs.
16	Explain how guard cells control the opening and closing of stomata and why this matters to the plant.	Answer: When guard cells take up water they swell and bend outwards, opening the stoma so carbon dioxide can enter and transpiration can occur. When water is scarce they lose turgor, the stoma closes and water loss is reduced.

Section E – Transport in animals (17 marks)

The following questions concern the circulatory system.

#	Question	Answer and marking notes
17	At rest a person has a stroke volume of 70 mL and a heart rate of 72 beats per minute. Calculate the cardiac output in mL per minute and in litres per minute.	Answer: Cardiac output = stroke volume x heart rate = 70 x 72 = 5040 mL per minute, which is 5.04 litres per minute.
18	During exercise the stroke volume rises to 110 mL and the heart rate to 140 beats per minute. Calculate the cardiac output and how many times the resting value it is.	Answer: Cardiac output during exercise = 110 x 140 = 15400 mL per minute, about 3.06 times the resting value, supplying more oxygen and glucose to the muscles.
19	Explain how haemoglobin transports oxygen between the lungs and the tissues.	Answer: In the lungs the oxygen partial pressure is high, so haemoglobin binds oxygen to form oxyhaemoglobin. In the tissues the partial pressure is low, so oxygen is released for cellular respiration.
20	Compare arteries, veins and capillaries in wall structure and blood flow.	Answer: Arteries have thick elastic walls, high pressure and fast flow. Veins have thin walls with valves that prevent backflow, and low pressure. Capillaries have a wall one cell thick and the slowest flow, which favours exchange.

Section F – Ecology and energy flow (16 marks)

The following questions concern energy flow in ecosystems.

#	Question	Answer and marking notes
21	Producers in an ecosystem fix 100000 kJ of energy. If ten per cent of the energy is passed to the next trophic level, calculate the energy available to the primary, secondary and tertiary consumers.	Answer: Primary consumers 10000 kJ, secondary consumers 1000 kJ, tertiary consumers 100 kJ.
22	Explain why food chains usually have only four or five trophic levels.	Answer: Much energy is lost at each level through respiration, excretion and parts not eaten, so only about ten per cent passes on. The more levels there are, the less energy reaches the top consumer, too little to support a population.
23	Describe the carbon cycle and state how human activity affects the carbon dioxide content of the atmosphere.	Answer: Producers fix carbon dioxide by photosynthesis. Organisms release carbon dioxide by respiration and decomposers release it when breaking down organic matter. Burning fossil fuels and clearing forests raise atmospheric carbon dioxide.
24	If the top predator in a food chain is heavily hunted, predict the effect on the ecosystem.	Answer: Prey numbers would rise in the short term and may overgraze or over-hunt the next level, so those numbers fall sharply. The balance of the whole food web is disturbed and biodiversity declines.

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