IGCSE Biology Paper 4 (Theory / Extended)

Topic 12: Respiration -- Challenge Exam 3

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0610

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

This paper covers the whole of Topic 12. Like a real Cambridge paper it ranges across every sub-topic — 12.1 respiration and the uses of energy, 12.2 aerobic respiration and 12.3 anaerobic respiration — and it mixes them inside single questions. All three Topic 12 papers do; they differ in the angle they come at it from, not in what they cover.

Question 1 — Woodlice, Soda Lime and a Drop of Liquid

Total: 12 marks

Fig. 1.1 shows a respirometer containing 3.0 g of woodlice. The whole apparatus was placed in a water bath. The distance moved by the drop of coloured liquid in 10 minutes was measured at three temperatures. The results are shown in the table.

Temperature / °C	10	20	35
Distance moved by the drop in 10 minutes / mm	15	30	8

(a) [3]

Calculate the rate of oxygen uptake by the woodlice at 20 °C, in mm³ per gram per hour. Show your working.

Model Answer — 1(a)

volume of oxygen used = 30 × 1.2 = 36 mm³ [1]

per gram = 36 ÷ 3.0 = 12 mm³ per gram in 10 minutes [1]

per hour = 12 × 6 = 72 mm³ per gram per hour [1]

⚠ If you missed marks here: Three conversions, done one at a time and each written down. If you try to do them in a single line and slip, you lose everything; shown working keeps the method marks. Sanity-check the last step — an hour is six times ten minutes, so the number must get bigger.

1 mark — volume of oxygen = 36 mm³ 1 mark — 12 mm³ per gram in 10 minutes 1 mark — 72 mm³ per gram per hour

(b) [2]

Explain the difference between the readings at 10 °C and 20 °C.

Model Answer — 1(b)

the rate has doubled, from 15 to 30 mm in the same time [1]

at the higher temperature the molecules have more kinetic energy, so enzyme and substrate collide more often and more successfully, and the enzyme-controlled reactions of respiration go faster [1]

⚠ If you missed marks here: Quote the numbers as well as explaining them — the first mark is for using the data, and it is free. Do not write that the woodlice “get more active”: that may be true but it is a consequence of the same thing, not an explanation of it.

1 mark — states the rate has doubled, quoting both figures 1 mark — explains in terms of kinetic energy and more frequent successful collisions

Suggest why the reading at 35 °C is much lower than the reading at 20 °C.

Model Answer — 1(c)

35 °C is above the optimum for a woodlouse, so its enzymes have begun to be denatured — the active sites have changed shape and the substrate no longer fits [1]

so the rate of respiration falls sharply; the woodlice may also have been killed by the high temperature [1]

⚠ If you missed marks here: A woodlouse is a small invertebrate that lives under damp logs, so 35 °C is dangerously hot for it — do not assume the optimum is 37 °C simply because that is the human value. Never write that enzymes are “killed”.

1 mark — enzymes are denatured above the optimum, active site changes shape 1 mark — so the rate of respiration falls sharply, and the woodlice may be killed

(d) [3]

Describe the control that should be set up alongside this apparatus, and explain why the whole respirometer must be kept in a water bath at a constant temperature.

Model Answer — 1(d)

a second, identical respirometer containing the same mass of dead woodlice (or glass beads of the same mass), with the same soda lime and capillary tube [1]

it shows that any movement of the drop in the first tube is caused by the woodlice respiring, and not by a leak or a physical change [1]

a constant temperature is needed because gases expand when warmed and contract when cooled, so a change in room temperature would move the drop and be mistaken for oxygen uptake [1]

⚠ If you missed marks here: The temperature mark is about physics, not biology: an unnoticed warm draught moves the drop just as convincingly as a respiring woodlouse. Note also that a control is not a repeat — it is a second experiment designed to rule out one specific alternative explanation.

1 mark — identical apparatus with dead woodlice or glass beads of the same mass 1 mark — shows the movement is caused by respiration rather than a leak or physical change 1 mark — constant temperature because gases expand and contract, which would move the drop

(e) [2]

Suggest two improvements to this investigation, other than adding a control.

Model Answer — 1(e)

any two of: take readings at more temperatures, at closer intervals, so the optimum can be located; repeat at each temperature and take a mean; leave the apparatus in the water bath for several minutes before starting so it reaches the set temperature; use more woodlice, all of the same species and size; take readings at fixed intervals rather than only after 10 minutes [1 each, to a maximum of 2]

⚠ If you missed marks here: Only three temperatures were used and two of them are on the rising side, so “more readings, closer together” is the strongest single improvement here. Vague suggestions such as “be more accurate” are never credited.

1 mark — first valid improvement 1 mark — second valid improvement

Question 2 — Definitions, Equations and a Bird in the Cold

Total: 12 marks

(a) [2]

Define anaerobic respiration, and state how the energy it releases per glucose molecule compares with aerobic respiration.

Model Answer — 2(a)

the chemical reactions in cells that break down nutrient molecules to release energy without using oxygen [1]

it releases much less energy per glucose molecule than aerobic respiration [1]

⚠ If you missed marks here: “Much less” is not the same as “none”, and the difference matters: if anaerobic respiration released nothing a sprinter would stop mid-race. Keep the verb correct as well — energy is released, not produced.

1 mark — releases energy without using oxygen 1 mark — releases much less energy per glucose molecule than aerobic respiration

(b) [3]

Write the balanced chemical equation for aerobic respiration, and state the total number of oxygen atoms on each side of it.

Model Answer — 2(b)

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O — formulae correct [1]

correctly balanced, with 6 in front of the oxygen, carbon dioxide and water [1]

18 oxygen atoms on each side: left 6 in glucose + 12 in the oxygen molecules; right 12 in the carbon dioxide + 6 in the water [1]

⚠ If you missed marks here: The third mark is only earned by counting atoms, not molecules, and by remembering that glucose itself contains six oxygen atoms. Candidates who count only the 6O₂ on the left get 12 and are then puzzled that the equation does not balance.

1 mark — correct formulae in the balanced equation 1 mark — correctly balanced with the right coefficients 1 mark — 18 oxygen atoms on each side, with the counting shown

A small bird eats about 30 % of its own body mass in food each day. A lizard of the same mass, living in the same place, eats about 3 %. Explain this difference in terms of the uses of the energy released by respiration.

Model Answer — 2(c)

a bird maintains a constant body temperature, well above that of its surroundings [1]

this requires a great deal of energy, released by respiration, some of which is transferred to the surroundings as thermal energy and must be continually replaced [1]

a lizard does not maintain a constant body temperature, so it spends almost none of its energy on this and needs far less food; the bird also uses a great deal of energy in muscle contraction during flight [1]

⚠ If you missed marks here: This is the seven-item list applied to an unfamiliar comparison, and it works only if you remember that maintaining a constant body temperature is one of the seven. Answers that say the bird is “more active” get part of the way but miss the larger and more specific reason.

1 mark — bird maintains a constant body temperature 1 mark — this costs a great deal of energy that must be continually replaced 1 mark — lizard does not, so needs far less food; flight also costs the bird energy in muscle contraction

(d) [4]

The cells lining the gills of a freshwater fish contain unusually large numbers of mitochondria. These cells take up ions from the water, in which the ion concentration is much lower than inside the fish. Explain why these cells need so many mitochondria, and predict what happens to ion uptake if the oxygen concentration of the water falls.

Model Answer — 2(d)

uptake is from a lower to a higher concentration, so it is against the concentration gradient and must be active transport [1]

active transport requires energy released by respiration, and the mitochondria are the site of aerobic respiration, so many mitochondria means a high rate of energy release [1]

if the oxygen concentration falls, less aerobic respiration can take place and less energy is released [1]

so the rate of active transport falls and ion uptake decreases [1]

⚠ If you missed marks here: This is a four-link chain in an unfamiliar organism, and it is exactly the same chain as the root hair cell in waterlogged soil. Write the links out one at a time rather than leaping from “less oxygen” to “the fish absorbs fewer ions”, which scores one.

1 mark — uptake against the gradient is active transport 1 mark — active transport needs energy from aerobic respiration in the mitochondria 1 mark — less oxygen means less aerobic respiration and less energy released 1 mark — so the rate of active transport and ion uptake falls

Question 3 — Six Readings and a Peak

Total: 12 marks

Fig. 3.1 shows the rate of carbon dioxide production by a yeast suspension, measured at six temperatures. The same volume and concentration of yeast suspension and of glucose solution was used each time, and a thin layer of oil was floated on the surface of every tube.

(a) [3]

Describe the results shown in Fig. 3.1. Use figures in your answer.

Model Answer — 3(a)

the rate increases from about 3 bubbles per minute at 5 °C to a maximum of about 61 at 35 °C [1]

above 35 °C the rate falls, to about 38 at 45 °C and about 2 at 55 °C [1]

the increase is gradual at first and then steep, while the fall is steeper still, so the curve is not symmetrical about its peak [1]

⚠ If you missed marks here: A description without figures rarely scores more than one. The third mark is the observation almost nobody makes: the two sides of the curve have different gradients, and noticing it is what leads into the explanation in part (b).

1 mark — rise described with figures from 5 to 35 °C 1 mark — fall described with figures above 35 °C 1 mark — notes the curve is not symmetrical about the peak

(b) [3]

Explain the shape of the curve. Your answer should make clear that the rise and the fall have different causes.

Model Answer — 3(b)

the rise: molecules gain kinetic energy, so enzyme and substrate collide more often and with more energy, and more enzyme–substrate complexes form [1]

the fall: the enzymes are denatured, so the shape of the active site changes and the substrate no longer fits [1]

the rise is fully reversible on cooling whereas denaturation is permanent, which is why the fall is so much steeper than the rise [1]

⚠ If you missed marks here: Giving the same explanation for both halves of the curve is the way this question is usually lost. Two sides, two causes — and the third mark is for the difference between them, not for repeating either one.

1 mark — rise explained by kinetic energy and more successful collisions 1 mark — fall explained by denaturation and the active site changing shape 1 mark — notes that cooling is reversible whereas denaturation is permanent

A student states that the optimum temperature for respiration in this yeast is exactly 35 °C. Explain why this cannot be concluded from Fig. 3.1, and describe how the investigation could be changed to find the optimum more precisely.

Model Answer — 3(c)

the readings are 10 °C apart, so all that can be said is that the optimum lies somewhere between 25 and 45 °C — the true peak may fall between two readings [1]

repeat the investigation using much closer intervals, for example every 2 °C between 25 and 45 °C, with repeats and means at each temperature [1]

⚠ If you missed marks here: The improvement must name both the interval and the range to be specific enough. “Do more temperatures” is too vague, and testing more temperatures at the far ends of the scale would not help at all.

1 mark — explains that 10 °C intervals cannot locate the peak, which lies between 25 and 45 °C 1 mark — suggests closer intervals over that range, with repeats

(d) [2]

State why a layer of oil was floated on every tube, and name the two products of respiration in the yeast under these conditions.

Model Answer — 3(d)

the oil prevents oxygen dissolving in from the air, so the yeast respires anaerobically [1]

the products are alcohol (ethanol) and carbon dioxide [1]

⚠ If you missed marks here: Do not answer “to stop evaporation” — that is what oil does in a transpiration experiment. And under oil the products are alcohol and carbon dioxide: lactic acid belongs to muscle and never appears in yeast, whatever the conditions.

1 mark — oil excludes oxygen so the yeast respires anaerobically 1 mark — products are alcohol and carbon dioxide

(e) [2]

A tube from 55 °C and a tube from 5 °C are both moved to a water bath at 35 °C and left for an hour. Predict what happens to the rate of carbon dioxide production in each tube, and explain the difference.

Model Answer — 3(e)

the tube from 5 °C produces gas rapidly again, because cooling only slowed the molecules down and the effect is fully reversible [1]

the tube from 55 °C produces little or no gas, because its enzymes have been permanently denatured and warming cannot restore the shape of the active site [1]

⚠ If you missed marks here: This is the reversibility test, and it is the single cleanest way to show that the two ends of the curve are not mirror images. If you have written that the cold tube was “denatured by the cold”, this question exposes it.

1 mark — tube from 5 °C recovers because cooling is reversible 1 mark — tube from 55 °C does not, because denaturation is permanent

Question 4 — A Sprinter and a Marathon Runner

Total: 12 marks

Two athletes were tested. Athlete J is a sprinter and athlete K is a marathon runner. Each ran as hard as possible for 60 seconds on a treadmill. Their oxygen uptake, peak blood lactic acid and recovery times were recorded.

Athlete	Oxygen required during the run / dm³ per minute	Oxygen actually taken in during the run / dm³ per minute	Peak blood lactic acid / arbitrary units	Time to return to resting lactate / minutes
J (sprinter)	6.2	3.4	13.5	62
K (marathon runner)	4.8	4.1	6.2	27

(a) [3]

Calculate the oxygen debt built up by each athlete during the 60-second run, and state which is the larger.

Model Answer — 4(a)

athlete J: shortfall = 6.2 − 3.4 = 2.8 dm³ per minute, and the run lasted 1 minute, so the debt is 2.8 dm³ [1]

athlete K: shortfall = 4.8 − 4.1 = 0.7 dm³ [1]

J built up the larger debt, four times that of K [1]

⚠ If you missed marks here: The debt is the shortfall, never the whole requirement — using 6.2 gives an answer that describes the oxygen J needed altogether. Notice that the run lasted exactly one minute, which is a detail the stem gives you so that no multiplication is needed.

1 mark — J: debt of 2.8 dm³ from the shortfall 1 mark — K: debt of 0.7 dm³ 1 mark — identifies J as the larger, ideally with the ratio

(b) [3]

Explain, using the data, why athlete J has a much higher peak blood lactic acid than athlete K.

Model Answer — 4(b)

J could not take in enough oxygen to meet the demand — a shortfall of 2.8 against only 0.7 for K [1]

so J’s muscles respired anaerobically to a much greater extent, in addition to respiring aerobically [1]

anaerobic respiration in muscle produces lactic acid, so more of it accumulated: 13.5 units against 6.2 [1]

⚠ If you missed marks here: Quote figures from two columns rather than describing the pattern in words, and make sure the word “anaerobic” is present with its product. Saying only “J is a sprinter so he produces more lactic acid” restates the result without using the data at all.

1 mark — quotes the difference in shortfall between the two athletes 1 mark — explains that J respired anaerobically to a much greater extent, as well as aerobically 1 mark — links this to lactic acid production, quoting the two peak values

Explain how the lactic acid in athlete J’s blood is removed during the 62 minutes after the run.

Model Answer — 4(c)

J’s heart rate remains fast after the run, transporting lactic acid in the blood from the muscles to the liver [1]

J’s breathing remains deeper and faster, so extra oxygen continues to be taken in — this is the oxygen debt being repaid [1]

the lactic acid is respired aerobically in the liver, so its concentration in the blood falls back to the resting value [1]

⚠ If you missed marks here: Three continuing changes, three marks, and the word that must be there is liver. Note that the debt is repaid over the whole 62 minutes rather than the first few — a long recovery is exactly what a large debt looks like.

1 mark — fast heart rate carries lactic acid in the blood to the liver 1 mark — deeper and faster breathing supplies the extra oxygen, repaying the debt 1 mark — lactic acid is respired aerobically in the liver

(d) [3]

A coach concludes from this table that athlete K is fitter than athlete J. Evaluate this conclusion.

Model Answer — 4(d)

the data do support it in one sense: K took in almost all the oxygen required (4.1 of 4.8), had a much lower peak lactate and recovered in less than half the time [1]

but each athlete was tested once, with no repeats and no means, and only two people were tested [1]

the test also suited K rather than J: a 60-second maximal run measures endurance, and J is trained for short explosive efforts, so the comparison measures different kinds of fitness rather than showing that one athlete is fitter overall [1]

⚠ If you missed marks here: The third mark is the interesting one and it needs you to think about what the test actually measured. An evaluation that lists only sample size and repeats is generic; naming why this particular test favours one athlete is what earns full marks.

1 mark — agrees the data point to K, quoting figures 1 mark — notes the single test, no repeats and only two subjects 1 mark — argues that the test measures a kind of fitness suited to K, so the comparison is not fair overall

Question 5 — Where the Lactic Acid Goes

Total: 10 marks

After a hard swim, a student measured the concentration of lactic acid in blood taken from a vein leaving her leg muscles, and in blood taken from a vein leaving her liver. The results are shown in the table.

Time after the swim / minutes	0	10	20	40
Lactic acid in blood leaving the leg muscles / arbitrary units	10.4	7.8	5.1	2.0
Lactic acid in blood leaving the liver / arbitrary units	6.9	4.8	3.0	1.2

(a) [3]

Describe the difference between the two sets of readings and explain what it shows.

Model Answer — 5(a)

blood leaving the liver always contains less lactic acid than blood leaving the muscles — for example 6.9 against 10.4 at 0 minutes [1]

so lactic acid has been removed as the blood passed through the liver [1]

it is respired aerobically in the liver, using the extra oxygen supplied by the continued deep and rapid breathing [1]

⚠ If you missed marks here: Quote a pair of figures rather than saying “the liver values are lower”. The reasoning is the same as reading a capillary in Topic 9: a substance falling in concentration along a route has been taken out somewhere along that route.

1 mark — states the liver blood contains less, quoting a pair of figures 1 mark — concludes that lactic acid is removed as blood passes through the liver 1 mark — names aerobic respiration of lactic acid in the liver as the mechanism

(b) [2]

Explain why the student’s heart continued to beat quickly for some minutes after she had stopped swimming.

Model Answer — 5(b)

so that blood continues to be pumped rapidly, transporting lactic acid from the muscles to the liver [1]

and delivering the extra oxygen taken in by the continued deep breathing to the liver so the lactic acid can be respired aerobically [1]

⚠ If you missed marks here: The heart is doing a transport job here, not simply “recovering”. Answers that say the heart is beating fast “because she is out of breath” describe a symptom rather than giving a function.

1 mark — transports lactic acid in the blood from the muscles to the liver 1 mark — delivers the extra oxygen for the aerobic respiration of lactic acid

Calculate the percentage fall in the concentration of lactic acid in blood leaving the leg muscles between 0 and 40 minutes, and explain why the concentration falls.

Model Answer — 5(c)

fall = 10.4 − 2.0 = 8.4 units; 8.4 ÷ 10.4 × 100 = 81 % [1]

the muscles are no longer respiring anaerobically, so no more lactic acid is being made [1]

while the lactic acid already present continues to be carried away in the blood and removed by the liver, so the concentration falls [1]

⚠ If you missed marks here: A percentage change is taken of the starting value; dividing by 2.0 gives 420 % and answers nothing. In the explanation, both halves are needed: production has stopped and removal is continuing. Give only one and you score one.

1 mark — percentage fall of about 81 % calculated from the starting value 1 mark — no more lactic acid is being produced once the exercise has stopped 1 mark — the lactic acid already present continues to be removed by the liver

(d) [2]

Suggest two reasons why these results, from one student on one occasion, should be treated with caution.

Model Answer — 5(d)

any two of: only one person was tested, so the results may not apply generally; the test was carried out once, with no repeats and therefore no means; how hard she swam, and for how long, was not measured or controlled; her fitness, diet and how recently she had eaten were not recorded; blood samples taken at only four times may miss the peak or a sudden change [1 each, to a maximum of 2]

⚠ If you missed marks here: Reasons must be specific to this investigation. “The results might be wrong” earns nothing; “only one person was tested, so the pattern may not be general” earns the mark because it names the limitation and its consequence.

1 mark — first valid limitation with its consequence 1 mark — second valid limitation with its consequence

Question 6 — Yeast, Alcohol and a Sealed Vessel

Total: 12 marks

A yeast suspension in glucose solution was sealed in a vessel at 30 °C. The concentration of alcohol was measured every two hours.

Time / hours	0	2	4	6	8	10
Alcohol concentration / %	0.0	2.6	5.4	8.1	9.6	9.7
Glucose remaining / g per dm³	50	38	27	17	12	12

(a) [3]

Write the word equation and the balanced chemical equation for the reaction taking place in the sealed vessel, and state the name of this type of respiration.

Model Answer — 6(a)

glucose → alcohol + carbon dioxide [1]

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂, correctly balanced [1]

this is anaerobic respiration (in yeast) [1]

⚠ If you missed marks here: Naming the process is a mark in its own right and takes two words — do not leave it out because the equations feel like the real answer. Never put oxygen on the left of an anaerobic equation.

1 mark — correct word equation 1 mark — correct balanced equation for yeast 1 mark — names the process as anaerobic respiration

(b) [3]

Calculate the mean rate of alcohol production between 0 and 6 hours, in percent per hour. Then describe and explain what happens between 8 and 10 hours.

Model Answer — 6(b)

(8.1 − 0.0) ÷ 6 = 1.35 % per hour [1]

between 8 and 10 hours the alcohol concentration is almost unchanged (9.6 to 9.7) and the glucose stops falling, staying at 12 g per dm³ — so respiration has effectively stopped [1]

even though glucose remains, the accumulated alcohol has reached a concentration that kills or inhibits the yeast, so the reaction cannot continue [1]

⚠ If you missed marks here: The table deliberately shows glucose still present at 10 hours, so any answer blaming a lack of food has ignored the evidence. The second mark needs both rows of the table quoted — the alcohol plateau and the glucose plateau together are what make the conclusion safe.

1 mark — rate of 1.35 % per hour calculated 1 mark — describes both the alcohol and the glucose levelling off, with figures 1 mark — explains that accumulated alcohol has killed or inhibited the yeast although glucose remains

Predict how the results would differ if air were bubbled continuously through the vessel, and explain your prediction.

Model Answer — 6(c)

much less alcohol would be produced, and the glucose would be used up faster or the yeast would grow more [1]

because with oxygen available the yeast would respire aerobically, producing carbon dioxide and water instead of alcohol, and releasing far more energy per glucose molecule [1]

⚠ If you missed marks here: Yeast is not restricted to anaerobic respiration — it uses whichever route the conditions allow, which is why the bung matters so much in this experiment. Do not predict that the yeast would produce lactic acid; it never does.

1 mark — predicts much less alcohol produced 1 mark — because with oxygen the yeast respires aerobically, giving carbon dioxide and water and far more energy

(d) [2]

Describe a control that should be set up alongside this investigation, and state what it would show.

Model Answer — 6(d)

an identical sealed vessel of glucose solution containing boiled yeast (or no yeast at all), kept at the same temperature [1]

no alcohol should be produced and the glucose concentration should not fall, showing that the changes in the first vessel were caused by the living yeast respiring rather than by the glucose breaking down on its own [1]

⚠ If you missed marks here: State what the control shows, not just what it is. A control that is described but not interpreted usually scores one, because the mark scheme is looking for the alternative explanation that has been ruled out.

1 mark — identical vessel with boiled yeast or no yeast, same conditions 1 mark — no alcohol and no fall in glucose, showing the changes were caused by living yeast

(e) [2]

A student says that human muscle would behave in the same way in a sealed vessel. Explain why this is wrong.

Model Answer — 6(e)

human muscle respiring anaerobically produces lactic acid, not alcohol and carbon dioxide [1]

so no alcohol would be produced, and no gas would be released either — lactic acid is the only product of anaerobic respiration in muscle [1]

⚠ If you missed marks here: Two organisms, two equations, no overlap. The second mark rewards spotting the consequence rather than just the product: with no carbon dioxide there would be nothing to collect, which is why muscle cannot be used for the classic yeast experiment.

1 mark — muscle produces lactic acid, not alcohol and carbon dioxide 1 mark — so no alcohol and no gas would be produced, lactic acid being the only product

Question 7 — Words, Colours and a Headline

Total: 10 marks

(a) [4]

A student wrote: “Plants respire at night and photosynthesise during the day. Respiration in the leaf produces energy, which the plant uses for transpiration. At high temperatures the enzymes are killed, so respiration stops.” Identify four errors and give the correct version of each.

Model Answer — 7(a)

plants respire all the time, day and night; in daylight they photosynthesise as well, and faster [1]

respiration releases energy rather than producing it — the energy was already stored in the glucose [1]

transpiration is not a use of the energy released by respiration; it is evaporation driven by the sun. A correct use would be active transport, protein synthesis, cell division or growth [1]

enzymes are denatured, not killed — a molecule was never alive; the shape of the active site changes so the substrate no longer fits [1]

⚠ If you missed marks here: Every one of these four errors is a single word or phrase, which is how a short topic is examined. Build the habit of checking four things in any respiration answer: the verb, whether respiration is said to stop, whether the use of energy is really a use of energy, and the word denatured.

1 mark — plants respire continuously, day and night 1 mark — energy is released, not produced 1 mark — transpiration is not a use of energy from respiration; names a correct one 1 mark — enzymes are denatured, not killed, with the active site explanation

(b) [3]

Hydrogencarbonate indicator is red in ordinary air, yellow when carbon dioxide rises and purple when it falls. Three sealed tubes of indicator are prepared: A contains a woodlouse and is kept in the light; B contains a leaf and is kept in the light; C contains a leaf and is wrapped in foil. Predict the colour in each tube after two hours and explain each prediction.

Model Answer — 7(b)

A turns yellow: an animal only respires, so carbon dioxide can only rise [1]

B turns purple: the leaf photosynthesises as well as respiring, and photosynthesis uses carbon dioxide faster than respiration releases it [1]

C turns yellow: with no light there is no photosynthesis, so only respiration takes place and carbon dioxide accumulates [1]

⚠ If you missed marks here: The three tubes together test one idea: whether you believe respiration ever stops. It does not — the leaf in tube B is respiring throughout, and the indicator is reporting the net result of two processes rather than one.

1 mark — A yellow, because an animal only respires 1 mark — B purple, because photosynthesis exceeds respiration in the light 1 mark — C yellow, because without light only respiration occurs

A newspaper headline reads: “Cold storage stops vegetables going off — scientists prove respiration is switched off at 4 °C.” The evidence is that peas stored at 4 °C lost 0.6 % of their dry mass in a week, while peas at 20 °C lost 4.1 %. Evaluate this claim.

Model Answer — 7(c)

the data do show a large difference: 0.6 % against 4.1 %, so the loss at 4 °C is about seven times smaller [1]

but a loss of 0.6 % is not zero, so respiration has been slowed, not switched off — low temperature reduces the kinetic energy of the molecules but does not denature the enzymes, and the effect is reversible [1]

a fair judgement: cold storage clearly reduces the rate of respiration and so preserves the vegetables for longer, but the headline overstates the result, and no details of sample size or repeats are given [1]

⚠ If you missed marks here: The claim fails on the word “stops”, and the number in the data disproves it directly — using that number is the second mark. Notice that a good evaluation credits what the data really show before attacking what has been claimed for them.

1 mark — quotes both figures and the size of the difference 1 mark — uses the non-zero loss at 4 °C to show respiration is slowed, not stopped, and notes cooling is reversible 1 mark — reaches a judgement, noting the headline overstates the result and the evidence is thin

Self-Assessment

Tick marks earned, then click Calculate Grade.

Your Score

Out Of

Percentage