Biology (9-1) 0970 | Paper 3 Mock 01 | Theory (Core)
Question 1 Report
Leaves are the main organs of photosynthesis and their internal tissues are arranged to capture light and exchange gases efficiently. The diagram shows the internal structure of a spinach leaf, seen in cross-section under a high-power microscope. The section reveals the waxy cuticle and epidermis covering the outer surfaces, a densely packed upper layer of elongated cells, a looser middle layer containing large air spaces, and the vein tissues that carry water and sugars. A student was asked to relate each of these features to the leaf's function in trapping light and obtaining the raw materials needed for photosynthesis. Use the diagram together with your own knowledge to answer the questions that follow.
(a) Name the layer of cells in which most photosynthesis takes place. [1]
(b) Explain two ways in which the cells of this layer are adapted for photosynthesis. [3]
(c) Describe how carbon dioxide in the air outside the leaf reaches the mesophyll cells. [3]
(d) Explain the function of the waxy cuticle on the upper surface. [2]
(e) Explain why a broad, thin, flat leaf is efficient at photosynthesis. [2]
(a) Layer where most photosynthesis occurs [1]
The palisade mesophyll (the densely packed, elongated cells just below the upper epidermis).
(b) Two adaptations of these cells for photosynthesis [3, max 3]
The palisade cells are column-shaped and packed close to the upper surface [1]; they contain many chloroplasts to trap light [1]; and their position near the top of the leaf lets them receive the most light [1].
(c) How carbon dioxide reaches the mesophyll cells [3]
Carbon dioxide diffuses in from the outside air through the stomata [1]; it moves into the air spaces of the spongy mesophyll [1]; and then diffuses into the mesophyll cells down a concentration gradient (its concentration inside the cells is kept low by photosynthesis) [1].
(d) Function of the waxy cuticle on the upper surface [2]
The cuticle is waterproof [1]; so it reduces water loss by evaporation from the upper leaf surface, which is most exposed to sunlight [1].
(e) Why a broad, thin, flat leaf is efficient [2]
Being thin gives a short diffusion distance for gases moving in and out [1]; being broad and flat gives a large surface area to absorb sunlight [1].
Answer Details
(a) Layer where most photosynthesis occurs [1]
The palisade mesophyll (the densely packed, elongated cells just below the upper epidermis).
(b) Two adaptations of these cells for photosynthesis [3, max 3]
The palisade cells are column-shaped and packed close to the upper surface [1]; they contain many chloroplasts to trap light [1]; and their position near the top of the leaf lets them receive the most light [1].
(c) How carbon dioxide reaches the mesophyll cells [3]
Carbon dioxide diffuses in from the outside air through the stomata [1]; it moves into the air spaces of the spongy mesophyll [1]; and then diffuses into the mesophyll cells down a concentration gradient (its concentration inside the cells is kept low by photosynthesis) [1].
(d) Function of the waxy cuticle on the upper surface [2]
The cuticle is waterproof [1]; so it reduces water loss by evaporation from the upper leaf surface, which is most exposed to sunlight [1].
(e) Why a broad, thin, flat leaf is efficient [2]
Being thin gives a short diffusion distance for gases moving in and out [1]; being broad and flat gives a large surface area to absorb sunlight [1].
Question 2 Report
An experiment measured how quickly oxygen diffused across artificial membranes of different thickness, using the same concentration gradient each time. The diagram compares a thin and a thick membrane, and Table 1 shows the results.
| Membrane thickness / units | Oxygen diffusing per minute / arbitrary units |
|---|---|
| 1 | 40 |
| 2 | 20 |
| 4 | 10 |
(a) Describe the relationship between membrane thickness and the amount of oxygen that diffuses across it. [2]
(b) Calculate how many times faster oxygen diffused across the 1 unit membrane than across the 4 unit membrane. [2]
(c) Explain why a thinner membrane allows faster diffusion. [2]
(d) Explain how the structure of the wall of an alveolus is related to these results. [3]
The table shows oxygen diffusing per minute against membrane thickness (1 unit gives 40, 2 units give 20, 4 units give 10 arbitrary units), with the concentration gradient kept constant each time.
(a) As membrane thickness increases, the amount of oxygen diffusing across decreases (the thicker the membrane, the slower the rate). This is an inverse relationship. [2]
(b) Compare the 1-unit and 4-unit membranes: \[ \frac{40}{10} = 4 \] So oxygen diffuses 4 times faster across the 1-unit membrane. [2] (working \(40 \div 10\), answer 4)
(c) A thinner membrane allows faster diffusion because there is a shorter distance for the particles to travel [1], so they cross the membrane more quickly [1]. [2]
(d) Linking this to a real alveolus [3]: the wall of the alveolus is very thin / only one cell thick [1], giving a short diffusion distance [1], so gases (oxygen and carbon dioxide) diffuse across it rapidly [1], exactly as these results predict.
Answer Details
The table shows oxygen diffusing per minute against membrane thickness (1 unit gives 40, 2 units give 20, 4 units give 10 arbitrary units), with the concentration gradient kept constant each time.
(a) As membrane thickness increases, the amount of oxygen diffusing across decreases (the thicker the membrane, the slower the rate). This is an inverse relationship. [2]
(b) Compare the 1-unit and 4-unit membranes: \[ \frac{40}{10} = 4 \] So oxygen diffuses 4 times faster across the 1-unit membrane. [2] (working \(40 \div 10\), answer 4)
(c) A thinner membrane allows faster diffusion because there is a shorter distance for the particles to travel [1], so they cross the membrane more quickly [1]. [2]
(d) Linking this to a real alveolus [3]: the wall of the alveolus is very thin / only one cell thick [1], giving a short diffusion distance [1], so gases (oxygen and carbon dioxide) diffuse across it rapidly [1], exactly as these results predict.
Question 3 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 4 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 5 Report
A student examined a thin slice cut through the blade of a leaf from a flowering plant and viewed it under a microscope to see how the leaf is built. The diagram shows this cross-section through the leaf. Several tissue layers can be seen, from the protective covering on the outer surfaces to the packed cells in the middle of the leaf and the small openings on one surface through which gases move. Water that is drawn up from the roots reaches the leaf and travels through its cells before some of it is lost to the surrounding air. This loss of water from the plant is called transpiration, and the rate at which it happens depends on the structure of the leaf and on the conditions around the plant. The student used the diagram, together with knowledge of how water travels through a plant, to answer the following questions about the leaf and about the loss of water from it.
(a) Define the term transpiration. [2]
(b) Name the tissue that transports water up the stem to the leaf. [1]
(c) State two environmental conditions that would increase the rate of transpiration. [2]
(a) Transpiration is the loss of water vapour from the leaves (aerial parts) of a plant [1], by evaporation from the mesophyll cells and diffusion out through the stomata [1]. [2]
(b) Water is carried up the stem to the leaf in the xylem. [1]
(c) Any two conditions that increase the rate (1 mark each): higher temperature, increased air movement / wind, lower humidity, or higher light intensity (which opens the stomata wider). [2]
Answer Details
(a) Transpiration is the loss of water vapour from the leaves (aerial parts) of a plant [1], by evaporation from the mesophyll cells and diffusion out through the stomata [1]. [2]
(b) Water is carried up the stem to the leaf in the xylem. [1]
(c) Any two conditions that increase the rate (1 mark each): higher temperature, increased air movement / wind, lower humidity, or higher light intensity (which opens the stomata wider). [2]
Question 6 Report
A biologist collected four small animals from the leaves and soil of a garden and photographed them side by side for comparison. All four animals are arthropods, so each has a hard external skeleton, a body built from repeating segments and legs made of several jointed sections. Even so, the four specimens belong to different arthropod groups. Instead of using features that vary a great deal, such as colour or overall size, the biologist identified each animal from clear external features that are easy to see and count, including the number of pairs of legs and the number of separate regions that make up the body.
(a) State the number of pairs of legs found on an insect. [1]
(b) One of the specimens has four pairs of legs. Name the arthropod group to which it belongs. [1]
(c) State two features, other than the number of legs, that could be used to tell an arachnid from an insect. [2]
(d) Describe three features that are shared by all arthropods. [3]
(e) Explain why an arthropod must shed its exoskeleton in order to grow larger. [3]
Labelled answer diagram:
This question tests the shared features of arthropods and how the subgroups are told apart by countable external features rather than by variable ones like colour or size.
(a) Pairs of legs on an insect [1] - three pairs (six legs).
(b) Group with four pairs of legs [1] - arachnids (for example spiders), which have four pairs (eight legs).
(c) Two features, other than legs, separating an arachnid from an insect [2] - one mark each for any two of:
- An arachnid has two body regions (cephalothorax and abdomen) whereas an insect has three (head, thorax, abdomen).
- An arachnid has no antennae, whereas an insect has one pair.
- An arachnid has no wings, whereas many insects have wings.
(d) Three features shared by all arthropods [3] - one mark each:
- Jointed legs (jointed appendages) (1).
- A hard exoskeleton, made of chitin (1).
- A segmented body (1). (Bilateral symmetry is also accepted.)
(e) Why an arthropod must shed its exoskeleton to grow [3]
- The exoskeleton is hard and cannot stretch or grow with the animal (1).
- So it is shed (moulting, or ecdysis) (1).
- The animal then increases in size before the new, larger exoskeleton hardens (1).
Exam tip: classify arthropods by counting body regions and pairs of legs, because these features are constant within a group, unlike colour or overall size.
Answer Details
Labelled answer diagram:
This question tests the shared features of arthropods and how the subgroups are told apart by countable external features rather than by variable ones like colour or size.
(a) Pairs of legs on an insect [1] - three pairs (six legs).
(b) Group with four pairs of legs [1] - arachnids (for example spiders), which have four pairs (eight legs).
(c) Two features, other than legs, separating an arachnid from an insect [2] - one mark each for any two of:
- An arachnid has two body regions (cephalothorax and abdomen) whereas an insect has three (head, thorax, abdomen).
- An arachnid has no antennae, whereas an insect has one pair.
- An arachnid has no wings, whereas many insects have wings.
(d) Three features shared by all arthropods [3] - one mark each:
- Jointed legs (jointed appendages) (1).
- A hard exoskeleton, made of chitin (1).
- A segmented body (1). (Bilateral symmetry is also accepted.)
(e) Why an arthropod must shed its exoskeleton to grow [3]
- The exoskeleton is hard and cannot stretch or grow with the animal (1).
- So it is shed (moulting, or ecdysis) (1).
- The animal then increases in size before the new, larger exoskeleton hardens (1).
Exam tip: classify arthropods by counting body regions and pairs of legs, because these features are constant within a group, unlike colour or overall size.
Question 7 Report
Yoghurt is made by adding a culture of Lactobacillus bacteria to warm milk and keeping the mixture at a suitable temperature. As the bacteria feed on the milk, the milk gradually thickens and turns into yoghurt. A student made yoghurt in this way and measured the pH of the mixture every hour over a period of four hours. The results are plotted in Fig. 10.1, which shows the pH of the milk against time in hours. Use the graph to answer the questions about the changes that took place as the yoghurt formed.
(a) State the type of microorganism used to make yoghurt. [1]
(b) Using Fig. 10.1, describe how the pH of the milk changed over the four hours. [2]
(c) Explain why the pH of the milk falls. [3]
(d) Explain why the milk thickens as the pH falls. [2]
(e) State the sugar present in milk that is used by the bacteria. [1]
(f) Explain why the milk is heated before the bacteria are added. [2]
(g) Describe why yoghurt keeps for longer than fresh milk. [1]
(a) Type of microorganism [1]. Yoghurt is made using bacteria (Lactobacillus).
(b) How the pH changed [2]. Reading Fig. 10.1, the pH fell throughout the four hours, becoming more acidic: it dropped from about \(6.6\) at the start to about \(4.4\) after four hours. One mark for the direction (pH falls / becomes more acidic), one for supporting it with values read from the graph.
(c) Why the pH falls [3]. The bacteria feed on and respire (ferment) the sugar in the milk; this produces lactic acid; as the acid builds up in the mixture it lowers the pH. Any three of these linked points score. The cause is the acid the bacteria make, not the bacteria themselves.
(d) Why the milk thickens [2]. The lactic acid makes the dissolved milk protein coagulate (clot); the clotted protein turns the liquid milk into a semi-solid, so it thickens. One mark for coagulation of the protein by the acid, one for the resulting thickening.
(e) Sugar used by the bacteria [1]. Lactose, the sugar naturally present in milk.
(f) Why the milk is heated first [2]. Heating kills any unwanted or harmful microorganisms already in the milk; this stops them spoiling the yoghurt or competing with the added Lactobacillus culture. One mark for killing existing microorganisms, one for the reason.
(g) Why yoghurt keeps longer than fresh milk [1]. The low pH (acidic conditions) inhibits the growth of the decay and spoilage bacteria that would turn milk bad, so yoghurt lasts longer.
Answer Details
(a) Type of microorganism [1]. Yoghurt is made using bacteria (Lactobacillus).
(b) How the pH changed [2]. Reading Fig. 10.1, the pH fell throughout the four hours, becoming more acidic: it dropped from about \(6.6\) at the start to about \(4.4\) after four hours. One mark for the direction (pH falls / becomes more acidic), one for supporting it with values read from the graph.
(c) Why the pH falls [3]. The bacteria feed on and respire (ferment) the sugar in the milk; this produces lactic acid; as the acid builds up in the mixture it lowers the pH. Any three of these linked points score. The cause is the acid the bacteria make, not the bacteria themselves.
(d) Why the milk thickens [2]. The lactic acid makes the dissolved milk protein coagulate (clot); the clotted protein turns the liquid milk into a semi-solid, so it thickens. One mark for coagulation of the protein by the acid, one for the resulting thickening.
(e) Sugar used by the bacteria [1]. Lactose, the sugar naturally present in milk.
(f) Why the milk is heated first [2]. Heating kills any unwanted or harmful microorganisms already in the milk; this stops them spoiling the yoghurt or competing with the added Lactobacillus culture. One mark for killing existing microorganisms, one for the reason.
(g) Why yoghurt keeps longer than fresh milk [1]. The low pH (acidic conditions) inhibits the growth of the decay and spoilage bacteria that would turn milk bad, so yoghurt lasts longer.
Question 8 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.
