S6 Biology Practice Paper: Genetics, Respiration, Photosynthesis And Energy Flow

S6 Biology Practice Paper: Genetics, Respiration, Photosynthesis and Energy Flow

2026-11-12

S6 Biology – Genetics, Respiration and Ecology

S6 Biology paper on genetic ratios and probability, respiration quotient, limiting factors in photosynthesis, ecosystem energy flow and gas exchange, with model answers.

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Paper details　S6 Biology practice paper with model answers on genetics, respiration quotient, photosynthesis and energy flow.

Part 1: Genetic ratios and probability (20 marks)

Four questions. Give ratios in simplest whole numbers and state genotypes.

#	Question	Answer and marking notes
1	(a) Two pea plants both of genotype Tt are crossed, T dominant (tall) and t recessive (dwarf). Give the ratio of tall to dwarf offspring and the number of tall plants if 800 offspring are produced.	Answer: Tt x Tt gives TT : Tt : tt = 1 : 2 : 1, so tall : dwarf = 3 : 1. Tall plants = 800 x 3/4 = 600 (dwarf 200). (3 marks genotypes, 1 mark ratio, 1 mark number)
2	(b) A tall offspring from (a) is test-crossed with a homozygous dwarf tt, giving 400 offspring. Give the expected numbers of each phenotype and state the use of a test cross.	Answer: test cross gives Tt : tt = 1 : 1, so tall = 400 / 2 = 200 and dwarf = 200. A test cross reveals whether a dominant phenotype is Tt or TT. (2 marks ratio, 2 marks numbers, 1 mark use)
3	(c) A dihybrid cross AaBb x AaBb (genes on different chromosomes, complete dominance) produces 1600 seeds. Find the number of each of the four phenotypes.	Answer: phenotypic ratio 9 : 3 : 3 : 1, giving 900, 300, 300 and 100 seeds (total 1600). (3 marks ratio, 2 marks numbers)
4	(d) A father is blood group AO and the mother BO. Find the probability that a child is group AB, and the expected number of AB children among 200.	Answer: AO x BO gives AB, AO, BO and OO in equal parts, so P(AB) = 0.25 (25%). Expected number = 200 x 0.25 = 50. (3 marks cross, 1 mark probability, 1 mark number)

Part 2: Cell respiration and respiratory quotient (16 marks)

The respiratory quotient (RQ) is moles of carbon dioxide released divided by moles of oxygen used.

#	Question	Answer and marking notes
5	(a) 0.5 mol of glucose is fully oxidised in excess oxygen. Write the balanced equation and find the moles of oxygen used, carbon dioxide released and the RQ.	Answer: C6H12O6 + 6O2 -> 6CO2 + 6H2O. O2 = 3.0 mol, CO2 = 3.0 mol, RQ = 3.0 / 3.0 = 1.00. (2 marks equation, 1 mark moles, 1 mark RQ)
6	(b) During exercise an athlete releases 1.4 dm3 of carbon dioxide and uses 2.0 dm3 of oxygen per minute. Find the RQ and identify the main substrate being oxidised.	Answer: RQ = 1.4 / 2.0 = 0.70. A value near 0.7 indicates fat is the main substrate, because fat is more reduced and needs more oxygen. (2 marks calculation, 1 mark substrate, 1 mark explanation)
7	(c) Another subject releases 1.6 dm3 of carbon dioxide and uses 2.0 dm3 of oxygen per minute. Find the RQ and explain why protein has an RQ of about 0.8 to 0.9.	Answer: RQ = 1.6 / 2.0 = 0.80. Protein contains less oxygen than carbohydrate and its nitrogen must be removed, so its RQ lies between fat and carbohydrate, at about 0.8 to 0.9. (2 marks calculation, 2 marks explanation)
8	(d) Compare the net ATP yield per mole of glucose in anaerobic yeast and in aerobic muscle, and find the ratio between them.	Answer: anaerobic fermentation yields 2 ATP per mole of glucose, aerobic respiration yields 38, so the ratio is 38 / 2 = 19. The anaerobic route uses glycolysis only and the rest of the energy remains in ethanol. (2 marks yields, 1 mark ratio, 1 mark explanation)

Part 3: Photosynthesis and limiting factors (16 marks)

Four questions. Show units and significant figures.

#	Question	Answer and marking notes
9	(a) Write the balanced equation for photosynthesis. If a leaf fixes 6 mol of carbon dioxide, all used to make glucose, find the moles and mass of glucose formed (molar mass 180 g/mol).	Answer: 6CO2 + 6H2O -> C6H12O6 + 6O2. Moles = 6 / 6 = 1.0 mol; mass = 1.0 x 180 = 180 g. (2 marks equation, 1 mark moles, 1 mark mass)
10	(b) In an experiment with pondweed, the illuminated set collects 60 oxygen bubbles in 10 minutes and the dark control 40. Find the percentage increase and state the source of the bubbles.	Answer: increase = (60 – 40) / 40 x 100% = 50%. The bubbles are oxygen released by photolysis of water in photosynthesis. (2 marks calculation, 1 mark value, 1 mark source)
11	(c) When light intensity rises from 100 to 200 units the rate rises from 12.0 to 18.0 mm/min; at 400 units the rate stays at 18.0 mm/min. Find the first increase and identify the limiting factor at that point.	Answer: first increase = (18.0 – 12.0) / 12.0 x 100% = 50%. Further increase in light gives no change, so carbon dioxide concentration or temperature is the limiting factor. (2 marks calculation, 1 mark value, 1 mark factor)
12	(d) Leaf pigments are separated by paper chromatography. One pigment moves 4.5 cm from the origin while the solvent front moves 9.0 cm. Find its Rf value and identify the pigment.	Answer: Rf = distance moved by pigment / distance moved by solvent front = 4.5 / 9.0 = 0.50. A value near 0.5 suggests chlorophyll b, since chlorophyll a is about 0.6 to 0.7 and carotenes higher. (1 mark formula, 2 marks value, 1 mark identification)

Part 4: Energy flow in ecosystems (16 marks)

Much energy is lost as respiration and excretion as it passes between trophic levels.

#	Question	Answer and marking notes
13	(a) Producers in an ecosystem fix 50000 kJ and transfer efficiency is about 10%. Find the energy available to primary, secondary and tertiary consumers.	Answer: primary = 50000 x 0.10 = 5000 kJ; secondary = 500 kJ; tertiary = 50 kJ. (1 mark each level, 1 mark units)
14	(b) A cow ingests 20000 kJ of grass energy per year and excretes 8000 kJ in faeces. Find its assimilation efficiency and state where the rest of the energy goes.	Answer: assimilated energy = 20000 – 8000 = 12000 kJ; efficiency = 12000 / 20000 x 100% = 60%. The remaining 8000 kJ leaves in faeces and is never assimilated. (1 mark amount, 2 marks efficiency, 1 mark fate)
15	(c) Of the assimilated energy above, 9600 kJ is used in respiration. Find the net production and net production efficiency, and explain why net production feeds the next trophic level.	Answer: net production = 12000 – 9600 = 2400 kJ; efficiency = 2400 / 12000 x 100% = 20%. Only net production stored in body tissue can be eaten by the next trophic level; energy used in respiration is lost as heat. (2 marks net production, 1 mark efficiency, 1 mark explanation)
16	(d) What is the overall transfer efficiency from producers to tertiary consumers, and why do food chains usually have only four or five levels?	Answer: overall efficiency = 50 / 50000 x 100% = 0.1%. About nine tenths of the energy is lost at each step, so by the fourth or fifth level too little remains to support a viable population. (2 marks efficiency, 2 marks explanation)

Part 5: Gas exchange and transport (16 marks)

Four questions. Express ventilation rate in dm3/min.

#	Question	Answer and marking notes
17	(a) A person has a tidal volume of 500 cm3 and a breathing rate of 12 per minute. Find the ventilation rate in dm3 per minute.	Answer: ventilation = tidal volume x rate = 500 x 12 = 6000 cm3/min = 6.0 dm3/min. (1 mark formula, 2 marks calculation, 1 mark conversion)
18	(b) Inhaled air contains 21% oxygen and exhaled air 16%. If ventilation is 6.0 dm3/min, find the volume of oxygen absorbed per minute.	Answer: oxygen fraction difference = 21% – 16% = 5% (0.05), so absorption = 6.0 x 0.05 = 0.30 dm3/min. (2 marks fraction, 2 marks volume)
19	(c) Human alveoli have a total surface area of about 70 m2 and walls about 1 micrometre thick. Explain how these features aid gas exchange in the lung.	Answer: the large area (about 70 m2) gives a big diffusion surface for exchanging much gas at once, and the thin wall (about 1 micrometre) makes the diffusion distance very short, so diffusion is fast; with a rich blood supply and ventilation keeping a concentration gradient, gases continue to diffuse. (2 marks each point)
20	(d) Carbon monoxide binds haemoglobin about 200 times more strongly than oxygen. Explain carbon monoxide poisoning and state one first-aid measure with its physiological basis.	Answer: carbon monoxide occupies haemoglobin binding sites and forms carboxyhaemoglobin, greatly reducing oxygen carriage, and shifts the oxygen dissociation curve left so tissues obtain less oxygen. First aid is removal to fresh air and giving high concentration oxygen, which raises alveolar oxygen partial pressure and gradually displaces the carbon monoxide. (2 marks mechanism, 1 mark measure, 1 mark basis)

Part 6: Coordination and homeostasis (16 marks)

Four questions. Answer using the concept of negative feedback.

#	Question	Answer and marking notes
21	(a) After a meal a person blood glucose rises to 7.0 mmol/dm3 and falls to 5.0 mmol/dm3 two hours later. Find the percentage fall and name the main hormone involved and its source.	Answer: fall = (7.0 – 5.0) / 7.0 x 100% = 28.6%. The main hormone is insulin, secreted by beta cells of the pancreatic islets, which promotes glucose uptake and glycogen synthesis. (2 marks calculation, 1 mark hormone, 1 mark source)
22	(b) Using blood glucose control, explain how negative feedback maintains homeostasis and name one disorder that results when it fails.	Answer: when glucose rises, insulin secretion increases, cells take up glucose and the liver stores glycogen, so glucose falls; when glucose is low, glucagon secretion increases and glycogen is broken down, so glucose rises. This correction of any deviation keeps glucose within a narrow range; if insulin secretion is inadequate or receptors respond poorly, diabetes mellitus develops with persistently high glucose. (3 marks mechanism, 1 mark disorder)
23	(c) A patient temperature rises to 39.0 C (normal 37.0 C). Find the deviation and describe two heat-loss responses with their mechanisms.	Answer: deviation = 39.0 – 37.0 = 2.0 C. First, skin arterioles dilate so more blood flows near the surface, increasing heat loss by radiation and convection; second, sweat glands secrete more sweat and its evaporation absorbs much latent heat. Once temperature falls, the hypothalamus reduces these responses. (1 mark deviation, 1.5 marks each response)
24	(d) Describe the nerve pathway of the knee jerk reflex and explain why a reflex can be completed quickly without conscious control by the cerebrum.	Answer: tapping the patellar tendon stretches the quadriceps, stimulating stretch receptors whose impulses travel along sensory neurones to the spinal cord, synapse directly with motor neurones there, and the motor neurones carry impulses back to the quadriceps, which contracts. The pathway involves only the spinal cord and few synapses, with no relay to the cerebrum, so the response is fast and protective. (3 marks pathway, 1 mark explanation)

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