Biology (9-1) 0970 | Paper 3 Mock 01 | Theory (Core)
Question 1 Report
A monitoring station high on a remote mountain, far from towns and factories, has recorded information about the atmosphere for many years. Two measurements taken at the station are the concentration of carbon dioxide in the atmosphere and the mean global temperature anomaly, which is the difference between the temperature in a given year and the average temperature in 1960. Both measurements have been recorded every year since 1960. The graph shows how these two quantities have changed over that time.
(a) State the carbon dioxide concentration recorded in the year 2000. [1]
(b) Using the graph, calculate the increase in carbon dioxide concentration between 1960 and 2020. [2]
(c) Describe the relationship between the carbon dioxide concentration and the temperature anomaly. [2]
(d) A student concluded that the rising carbon dioxide concentration has caused the rising temperature. Suggest why this graph alone does not prove that conclusion. [1]
(e) Suggest two possible consequences of a rising global temperature for living organisms. [2]
This question tests reading two lines from a graph, a subtraction, and the difference between correlation and cause.
(a) Carbon dioxide concentration in 2000 [1] Reading the solid carbon dioxide line at the year 2000 gives about 369 ppm.
(b) Increase in carbon dioxide, 1960 to 2020 [2] Subtract the 1960 value from the 2020 value: \( 414 - 317 = 97\ \text{ppm} \) (1 mark working, 1 mark answer). Accept values read consistently from the graph.
(c) Relationship between the two quantities [2] As the carbon dioxide concentration increases, the temperature anomaly increases (1); this is a positive correlation (1) - both lines rise together over time.
(d) Why the graph alone does not prove cause [1] A correlation does not prove causation; another factor could be responsible for the rising temperature, so the graph shows association only.
(e) Two consequences of rising global temperature [2] Any two, for example: melting of ice caps and rising sea levels flooding habitats; and changed distribution or migration of species, loss of biodiversity, or changed breeding patterns.
Answer Details
This question tests reading two lines from a graph, a subtraction, and the difference between correlation and cause.
(a) Carbon dioxide concentration in 2000 [1] Reading the solid carbon dioxide line at the year 2000 gives about 369 ppm.
(b) Increase in carbon dioxide, 1960 to 2020 [2] Subtract the 1960 value from the 2020 value: \( 414 - 317 = 97\ \text{ppm} \) (1 mark working, 1 mark answer). Accept values read consistently from the graph.
(c) Relationship between the two quantities [2] As the carbon dioxide concentration increases, the temperature anomaly increases (1); this is a positive correlation (1) - both lines rise together over time.
(d) Why the graph alone does not prove cause [1] A correlation does not prove causation; another factor could be responsible for the rising temperature, so the graph shows association only.
(e) Two consequences of rising global temperature [2] Any two, for example: melting of ice caps and rising sea levels flooding habitats; and changed distribution or migration of species, loss of biodiversity, or changed breeding patterns.
Question 2 Report
A farmer grows a crop of maize on part of the farm. Some of the maize is fed to cattle, and the cattle are later eaten by people, forming a short food chain. The way that energy flows along this chain is shown in the diagram, and the amount of energy available in the maize and in the cattle over one year is given in the table. Only some of the energy stored in the maize is passed on to the cattle, and only some of the energy in the cattle then reaches the people.
| Stage in food chain | Energy / kJ/m2/year |
|---|---|
| Maize | 15000 |
| Cattle | 1200 |
(a) State the original source of the energy that the maize uses to make its food. [1]
(b) Using the table, calculate the percentage of the energy in the maize that is transferred to the cattle. Show your working. [2]
(c) Describe three ways in which energy is lost from the cattle and so does not pass on to people. [3]
(d) State what is meant by the term herbivore. [1]
(e) Name the process in living cells that releases energy from food and produces heat. [1]
This question is about the flow of energy along a food chain.
(a) [1] The Sun (sunlight) [1] is the original energy source, trapped by the maize during photosynthesis.
(b) [2] Divide the energy in the cattle by the energy in the maize and multiply by 100: \[ \frac{1200}{15000} \times 100 \] [1] \[ = 8\% \] [1].
(c) [3] Any three ways in which energy is lost from the cattle and so is not passed on to people: in respiration, much of it is released as heat that keeps the body warm; energy is used for movement; energy is lost in egestion (undigested material in faeces); and energy is lost in excretion (for example in urine) [3, 1 each].
(d) [1] A herbivore is an animal that eats only plants (producers) [1].
(e) [1] Respiration [1].
Answer Details
This question is about the flow of energy along a food chain.
(a) [1] The Sun (sunlight) [1] is the original energy source, trapped by the maize during photosynthesis.
(b) [2] Divide the energy in the cattle by the energy in the maize and multiply by 100: \[ \frac{1200}{15000} \times 100 \] [1] \[ = 8\% \] [1].
(c) [3] Any three ways in which energy is lost from the cattle and so is not passed on to people: in respiration, much of it is released as heat that keeps the body warm; energy is used for movement; energy is lost in egestion (undigested material in faeces); and energy is lost in excretion (for example in urine) [3, 1 each].
(d) [1] A herbivore is an animal that eats only plants (producers) [1].
(e) [1] Respiration [1].
Question 3 Report
A student investigated how the direction from which light comes affects the way a shoot bends. Bean seedlings were grown in identical pots and then lit in several ways: from the left side, from the right side, from directly above, with even light from all directions at once, and in complete darkness. After 48 hours the mean angle through which each shoot had bent was measured, and a shoot that grew straight up was recorded as a curvature of 0 degrees. The apparatus is shown in the diagram and the mean curvature for each lighting condition is given in the table.
| Direction of light | Mean curvature of shoot / degrees |
|---|---|
| from the left side | 42 |
| from the right side | 40 |
| from directly above | 3 |
| all-round even light | 2 |
| complete darkness | 1 |
(a) State the direction of light that produced the largest curvature. [1]
(b) Describe the effect of the direction of light on the curvature of the shoots. [2]
(c) Calculate the difference in mean curvature between light from the left side and complete darkness. [2]
(d) Explain why the shoots showed almost no curvature in all-round even light. [3]
(e) Suggest why the complete darkness result was included. [1]
This question tests phototropism - the growth of a shoot towards light - using results from different lighting directions.
(a) Direction giving the largest curvature [1] From the left side (42 degrees, the highest value in the table).
(b) Effect of light direction on curvature [2] Light from the side (left or right) gives a large curvature (1); while light from above, all-round even light and darkness give little or no curvature (1). Shoots only bend when light comes from one side.
(c) Difference between left light and darkness [2] Subtract the darkness value from the left-side value: \( 42 - 1 = 41 \) degrees (1 mark working, 1 mark answer).
(d) Why almost no curvature in all-round even light [3] The light reaches all sides of the shoot equally (1); so auxin is distributed evenly around the shoot (1); therefore all sides elongate equally and the shoot grows straight up (1). Bending only happens when auxin collects on the shaded side.
(e) Why the darkness result was included [1] As a control, to show the curvature is caused by the light and not by some other factor.
Answer Details
This question tests phototropism - the growth of a shoot towards light - using results from different lighting directions.
(a) Direction giving the largest curvature [1] From the left side (42 degrees, the highest value in the table).
(b) Effect of light direction on curvature [2] Light from the side (left or right) gives a large curvature (1); while light from above, all-round even light and darkness give little or no curvature (1). Shoots only bend when light comes from one side.
(c) Difference between left light and darkness [2] Subtract the darkness value from the left-side value: \( 42 - 1 = 41 \) degrees (1 mark working, 1 mark answer).
(d) Why almost no curvature in all-round even light [3] The light reaches all sides of the shoot equally (1); so auxin is distributed evenly around the shoot (1); therefore all sides elongate equally and the shoot grows straight up (1). Bending only happens when auxin collects on the shaded side.
(e) Why the darkness result was included [1] As a control, to show the curvature is caused by the light and not by some other factor.
Question 4 Report
A student set up four boiling tubes, labelled A to D, to find out which conditions are needed before seeds will germinate. Each tube held ten cress seeds resting on a piece of cotton wool. The tubes were treated in different ways so that in each one a single condition was changed. In tube C the water was first boiled to drive out any dissolved air and then covered with a thin layer of oil. Fig. 6.1 shows how the apparatus was arranged and the table lists the conditions inside each tube, together with whether the seeds had germinated. The tubes were left in the same place for one week before the results were recorded.
| Tube | Water | Oxygen | Temperature | Germinated? |
|---|---|---|---|---|
| A | present | present | 20 °C | yes |
| B | none (dry) | present | 20 °C | no |
| C | present | none | 20 °C | no |
| D | present | present | 2 °C | no |
(a) State which tube acts as the control and give a reason for your answer. [2]
(b) State the condition being tested in tube B and the condition being tested in tube D. [2]
(c) Explain why the seeds in tube C did not germinate. [3]
(d) Name the gas released by the germinating seeds during respiration. [1]
(e) Suggest one way of making the results of this investigation more reliable. [1]
This is a controlled experiment on germination. A control has all conditions favourable so it shows what normal germination looks like; each other tube removes one factor to test whether it is needed.
- (a) Tube A is the control [1], because all the conditions are present/favourable, so it shows normal germination for comparison with the others [1].
- (b) Tube B (dry) tests water [1]; tube D (2 \(^{\circ}\)C) tests temperature / warmth [1].
- (c) In tube C there is no oxygen (it was boiled out and the oil stops it re-dissolving) [1]; so the seeds cannot carry out aerobic respiration [1]; and no energy is released for the processes of germination [1].
- (d) The gas released in respiration is carbon dioxide [1].
- (e) Make it more reliable by using more seeds in each tube / repeating the whole investigation and comparing the results [1].
Answer Details
This is a controlled experiment on germination. A control has all conditions favourable so it shows what normal germination looks like; each other tube removes one factor to test whether it is needed.
- (a) Tube A is the control [1], because all the conditions are present/favourable, so it shows normal germination for comparison with the others [1].
- (b) Tube B (dry) tests water [1]; tube D (2 \(^{\circ}\)C) tests temperature / warmth [1].
- (c) In tube C there is no oxygen (it was boiled out and the oil stops it re-dissolving) [1]; so the seeds cannot carry out aerobic respiration [1]; and no energy is released for the processes of germination [1].
- (d) The gas released in respiration is carbon dioxide [1].
- (e) Make it more reliable by using more seeds in each tube / repeating the whole investigation and comparing the results [1].
Question 5 Report
A persistent pesticide, known as compound Z, was sprayed onto farmland near a lake to control insect pests. Compound Z is not broken down easily and is not excreted by animals once it enters their bodies. Some time later, scientists measured the concentration of compound Z in the lake water and in each of the organisms of the food chain shown in the diagram. Their measurements, in parts per million (ppm), are given in the table.
| Component of food chain | Concentration / ppm |
|---|---|
| Lake water | 0.001 |
| Algae | 0.04 |
| Small fish | 0.5 |
| Large fish | 4.0 |
| Fish-eating bird | 32.0 |
(a) State what is meant by the term food chain. [2]
(b) Using the table, describe how the concentration of compound Z changes along this food chain. [2]
(c) Explain why the concentration of compound Z increases at each higher trophic level. [3]
(d) Calculate how many times greater the concentration in the fish-eating bird is than the concentration in the algae. [2]
(e) Suggest two harmful effects that the high concentration of compound Z could have on the fish-eating birds. [2]
(f) Name the top consumer in this food chain. [1]
(a) What is meant by a food chain [2]. A sequence showing the feeding relationships between organisms [1], showing the flow of energy from a producer through a series of consumers [1].
(b) How the concentration of compound Z changes along the chain [2]. The concentration increases along the chain [1], from the algae up to the fish-eating bird, that is from 0.04 ppm to 32.0 ppm [1].
(c) Why the concentration increases at each higher level [3]. Compound Z is not broken down and not excreted [1], so it is stored and builds up in the tissues of each organism [1]; each consumer eats many organisms from the level below, so the compound accumulates and becomes more concentrated at each step (bioaccumulation/biomagnification) [1].
(d) How many times greater the bird's concentration is than the algae's [2].
\[\frac{32.0}{0.04}=800\]So the concentration in the bird is 800 times greater than in the algae: \(\dfrac{32.0}{0.04}\) [1] \(=800\) times [1].
(e) Two harmful effects on the birds [2]. Any two of: death or reduced numbers; reduced breeding success/thin eggshells; general poisoning or illness [2], because the highest, most toxic concentration builds up in the top consumer.
(f) The top consumer [1]. The fish-eating bird [1], as nothing in this chain feeds on it.
Answer Details
(a) What is meant by a food chain [2]. A sequence showing the feeding relationships between organisms [1], showing the flow of energy from a producer through a series of consumers [1].
(b) How the concentration of compound Z changes along the chain [2]. The concentration increases along the chain [1], from the algae up to the fish-eating bird, that is from 0.04 ppm to 32.0 ppm [1].
(c) Why the concentration increases at each higher level [3]. Compound Z is not broken down and not excreted [1], so it is stored and builds up in the tissues of each organism [1]; each consumer eats many organisms from the level below, so the compound accumulates and becomes more concentrated at each step (bioaccumulation/biomagnification) [1].
(d) How many times greater the bird's concentration is than the algae's [2].
\[\frac{32.0}{0.04}=800\]So the concentration in the bird is 800 times greater than in the algae: \(\dfrac{32.0}{0.04}\) [1] \(=800\) times [1].
(e) Two harmful effects on the birds [2]. Any two of: death or reduced numbers; reduced breeding success/thin eggshells; general poisoning or illness [2], because the highest, most toxic concentration builds up in the top consumer.
(f) The top consumer [1]. The fish-eating bird [1], as nothing in this chain feeds on it.
Question 6 Report
A person whose kidneys have stopped working properly can be kept healthy by a treatment called dialysis, in which a machine does some of the work normally done by the kidneys. The diagram shows part of a dialysis machine, in which the patient's blood flows close to a fresh dialysis fluid. The table shows the concentration of several substances in the blood before dialysis and in the fresh dialysis fluid.
| Substance | Blood before dialysis / g per dm3 | Fresh dialysis fluid / g per dm3 |
|---|---|---|
| Glucose | 1.0 | 1.0 |
| Urea | 1.8 | 0.0 |
| Salts | 3.6 | 3.6 |
| Protein | 7.2 | 0.0 |
(a) State the name of the organ that has failed in this person. [1]
(b) Name the process by which urea passes from the blood into the dialysis fluid and, using the table, explain why it moves in this direction. [3]
(c) Explain why the dialysis fluid contains glucose at the same concentration as normal blood. [2]
(d) Explain why the plasma proteins are not lost into the dialysis fluid. [2]
(e) Explain why the used dialysis fluid is continually replaced with fresh fluid. [2]
(f) State one other treatment for kidney failure and give one advantage of it. [3]
(a) The organ that has failed is the kidney. [1]
(b) Urea passes into the fluid by diffusion [1]. From the table it is at a higher concentration in the blood (1.8 g per dm\(^3\)) than in the fresh fluid (0.0 g per dm\(^3\)) [1], so it diffuses down the concentration gradient across the partially permeable membrane into the fluid [1]. [3]
(c) The fluid has glucose at the same concentration as blood (both 1.0), so there is no concentration gradient for glucose [1]; this means glucose is not lost from the blood by diffusion, keeping the patient's blood sugar normal [1]. [2]
(d) The plasma protein molecules are too large [1] to pass through the tiny pores of the partially permeable membrane, so they stay in the blood [1]. The table confirms protein remains at 7.2 in the blood. [2]
(e) Replacing the used fluid keeps it free of urea, which maintains a steep concentration gradient for urea [1]; so urea keeps diffusing out of the blood until enough has been removed [1]. [2]
(f) One other treatment: a kidney transplant [1]; an advantage is that the person no longer needs regular dialysis sessions [1] and can eat a more normal diet / live more freely [1]. [3]
Answer Details
(a) The organ that has failed is the kidney. [1]
(b) Urea passes into the fluid by diffusion [1]. From the table it is at a higher concentration in the blood (1.8 g per dm\(^3\)) than in the fresh fluid (0.0 g per dm\(^3\)) [1], so it diffuses down the concentration gradient across the partially permeable membrane into the fluid [1]. [3]
(c) The fluid has glucose at the same concentration as blood (both 1.0), so there is no concentration gradient for glucose [1]; this means glucose is not lost from the blood by diffusion, keeping the patient's blood sugar normal [1]. [2]
(d) The plasma protein molecules are too large [1] to pass through the tiny pores of the partially permeable membrane, so they stay in the blood [1]. The table confirms protein remains at 7.2 in the blood. [2]
(e) Replacing the used fluid keeps it free of urea, which maintains a steep concentration gradient for urea [1]; so urea keeps diffusing out of the blood until enough has been removed [1]. [2]
(f) One other treatment: a kidney transplant [1]; an advantage is that the person no longer needs regular dialysis sessions [1] and can eat a more normal diet / live more freely [1]. [3]
Question 7 Report
A species of land snail has two forms of shell, a banded form and a plain form. The snails are eaten by thrushes, which are birds that hunt the snails by sight as the snails move across areas of bare soil. Fig 3.1 is a diagram showing the shells of the snails that were present in one population living on bare soil over the course of three successive generations.
(a) State what is meant by natural selection. [2]
(b) Using the diagram, describe how the proportions of the two shell forms change over the three generations. [3]
(c) Explain, in terms of natural selection, why this change in the proportions of the shell forms occurs. [5]
(d) Name the source of any new shell-colour alleles in the population. [1]
(e) Suggest why the change in the proportions of the two forms happens gradually rather than in a single generation. [2]
(f) Give two ways in which this natural selection differs from the selective breeding of farm animals. [2]
This question tests natural selection using shell-form data over three generations. On bare soil, the form that is harder for thrushes to see survives better, so its proportion rises.
(a) [2] Natural selection is the process by which the organisms best adapted to their environment survive and reproduce [1], passing on their advantageous alleles to the next generation [1].
(b) [3] Over the three generations the banded form decreases in proportion / becomes rarer [1] while the plain form increases in proportion / becomes commoner [1]; for example the banded form falls from 3 out of 5 in generation 1 to 1 out of 5 in generation 3 [1].
(c) [5] There is variation, with banded and plain forms in the population [1]; on bare soil the plain snails are better camouflaged, so fewer are seen and eaten by thrushes [1]; the banded snails stand out and are eaten more [1]; the surviving plain snails reproduce and pass on the alleles for the plain shell [1]; so the proportion of the plain form increases over the generations [1].
(d) [1] New shell-colour alleles arise by mutation [1].
(e) [2] The change is gradual because only a proportion of the less-camouflaged snails is eaten in each generation [1]; so the allele frequency shifts only a little at a time, taking several generations of reproduction to change greatly [1].
(f) [2] Any two differences: natural selection is driven by the environment, whereas selective breeding is driven by humans choosing the parents [1]; natural selection has no chosen/planned outcome, whereas selective breeding aims at a chosen feature [1]; (accept: selective breeding is usually faster) [max 2].
Answer Details
This question tests natural selection using shell-form data over three generations. On bare soil, the form that is harder for thrushes to see survives better, so its proportion rises.
(a) [2] Natural selection is the process by which the organisms best adapted to their environment survive and reproduce [1], passing on their advantageous alleles to the next generation [1].
(b) [3] Over the three generations the banded form decreases in proportion / becomes rarer [1] while the plain form increases in proportion / becomes commoner [1]; for example the banded form falls from 3 out of 5 in generation 1 to 1 out of 5 in generation 3 [1].
(c) [5] There is variation, with banded and plain forms in the population [1]; on bare soil the plain snails are better camouflaged, so fewer are seen and eaten by thrushes [1]; the banded snails stand out and are eaten more [1]; the surviving plain snails reproduce and pass on the alleles for the plain shell [1]; so the proportion of the plain form increases over the generations [1].
(d) [1] New shell-colour alleles arise by mutation [1].
(e) [2] The change is gradual because only a proportion of the less-camouflaged snails is eaten in each generation [1]; so the allele frequency shifts only a little at a time, taking several generations of reproduction to change greatly [1].
(f) [2] Any two differences: natural selection is driven by the environment, whereas selective breeding is driven by humans choosing the parents [1]; natural selection has no chosen/planned outcome, whereas selective breeding aims at a chosen feature [1]; (accept: selective breeding is usually faster) [max 2].
Question 8 Report
A student studied a diagram of a human kidney during a lesson on excretion. The kidneys are a pair of organs that remove certain waste substances from the blood and help control the amount of water in the body. The diagram shows a section through one human kidney, with the blood vessels and the tube joined to it. One blood vessel, labelled A, carries blood into the kidney, and a tube labelled B leads away from it. When the kidney is cut open, two distinct regions, C and D, can be seen, and within them are large numbers of microscopic units that filter the blood.
(a) Name the blood vessel labelled A that carries blood into the kidney. [1]
(b) Name the tube labelled B that carries urine away from the kidney. [1]
(c) Two regions, C and D, can be seen when a kidney is cut open. Name region C and region D. [2]
(d) Name the microscopic units, found in regions C and D, that filter the blood. [1]
(e) State one waste substance that the kidney removes from the blood. [1]
(a) Blood vessel A carrying blood into the kidney [1]
The renal artery.
(b) Tube B carrying urine away [1]
The ureter.
(c) Regions C and D [2]
C is the cortex (the outer region) [1]; D is the medulla (the inner region) [1]. (The funnel-shaped space shown may be accepted as the pelvis.)
(d) Microscopic filtering units [1]
Nephrons (kidney tubules).
(e) One waste substance the kidney removes [1]
Urea (excess water or excess salts are also accepted).
Answer Details
(a) Blood vessel A carrying blood into the kidney [1]
The renal artery.
(b) Tube B carrying urine away [1]
The ureter.
(c) Regions C and D [2]
C is the cortex (the outer region) [1]; D is the medulla (the inner region) [1]. (The funnel-shaped space shown may be accepted as the pelvis.)
(d) Microscopic filtering units [1]
Nephrons (kidney tubules).
(e) One waste substance the kidney removes [1]
Urea (excess water or excess salts are also accepted).
