Biology 0610 | Paper 3 Mock 01 | Theory (Core)
Question 1 Report
Farmers who rear animals for meat try to obtain as much growth as possible from the food the animals are given, because feed is expensive and much of its energy is lost rather than turned into body mass. A cattle rancher in Argentina wanted to produce more beef from the same quantity of feed. To do this he kept his cattle in barns so that they moved about less, and he chose the fastest-growing animals to breed from over several generations.
Fig 3.1 is a chart comparing the feed conversion of two different breeds of cattle. A lower value on the chart means that the breed turns its feed into growth more efficiently. Using the chart and your understanding of energy transfer and selective breeding, answer questions about which breed uses its feed better, comparing values taken from the chart, explaining why keeping the cattle in barns increases growth, describing how the fastest-growing animals can be bred together, and considering the concerns raised by this way of keeping animals.
Fig. 3.1
(a) Which breed uses its feed more efficiently [2]: Breed Y [1]; it needs less feed, only 6 kg of feed per kg of gain (a lower value means more efficient conversion) [1].
(b) Difference between the two breeds [1]: \(8-6=2\text{ kg}\) of feed per kg of gain [1].
(c) Why keeping the cattle in barns increases growth [3]: keeping them in barns reduces their movement / activity [1]; so less energy is used in respiration / movement / keeping warm [1]; therefore more of the energy from the food is available for growth / gain in body mass [1].
(d) How the fastest-growing animals are bred together [4]:
- Select the fastest-growing animals as the parents [1].
- Breed these selected animals together [1].
- Select the fastest-growing offspring from the next generation [1].
- Repeat over many generations so the desired characteristic increases in the herd [1].
(e) Two concerns about this way of keeping animals [2]: any two: keeping animals indoors with restricted movement raises animal-welfare concerns [1]; the build-up of waste / methane, or the faster spread of disease in the crowded conditions [1].
Answer Details
(a) Which breed uses its feed more efficiently [2]: Breed Y [1]; it needs less feed, only 6 kg of feed per kg of gain (a lower value means more efficient conversion) [1].
(b) Difference between the two breeds [1]: \(8-6=2\text{ kg}\) of feed per kg of gain [1].
(c) Why keeping the cattle in barns increases growth [3]: keeping them in barns reduces their movement / activity [1]; so less energy is used in respiration / movement / keeping warm [1]; therefore more of the energy from the food is available for growth / gain in body mass [1].
(d) How the fastest-growing animals are bred together [4]:
- Select the fastest-growing animals as the parents [1].
- Breed these selected animals together [1].
- Select the fastest-growing offspring from the next generation [1].
- Repeat over many generations so the desired characteristic increases in the herd [1].
(e) Two concerns about this way of keeping animals [2]: any two: keeping animals indoors with restricted movement raises animal-welfare concerns [1]; the build-up of waste / methane, or the faster spread of disease in the crowded conditions [1].
Question 2 Report
Potato tissue contains the enzyme catalase, which breaks hydrogen peroxide down into water and oxygen. In this investigation equal-sized discs of potato were placed in hydrogen peroxide buffered to hold it at a chosen pH. The oxygen produced formed a foam, and the height of this foam after two minutes was measured with a ruler. The experiment was repeated at a range of pH values, and the results are given in the table. Use the diagram and the table to answer the questions.
| pH | 3 | 5 | 7 | 9 | 11 |
|---|---|---|---|---|---|
| Height of foam / mm | 5 | 20 | 45 | 25 | 8 |
(a) State the pH at which catalase is most active. [1]
(b) Describe the effect of pH on the activity of catalase shown by these results. [3]
(c) Measure the difference in foam height between pH 5 and pH 7. [1]
(d) Explain why catalase is important in living cells. [2]
(e) Suggest why discs of equal size were used. [2]
(f) Before the experiment some discs were left in a concentrated salt solution and became soft and floppy. Explain this change in terms of osmosis. [3]
(g) State the class of biological molecule that catalase is made from. [1]
This question tests how pH affects enzyme activity, using catalase breaking down hydrogen peroxide. Taller foam means faster oxygen release, so foam height is a measure of enzyme activity.
(a) pH of greatest activity [1] - pH 7, where the foam is tallest (45 mm).
(b) Effect of pH on catalase activity [3]
- As the pH increases from 3 to 7, the activity (foam height) increases (1).
- Activity is highest at the optimum, pH 7 (1).
- Above pH 7 the activity decreases again as the pH rises to 11 (1).
(c) Difference in foam height between pH 5 and pH 7 [1]
\( 45 - 20 = 25 \ \text{mm} \) (1).
(d) Why catalase is important in living cells [2]
Hydrogen peroxide is a toxic waste product of metabolism (1); catalase breaks it down into harmless water and oxygen, protecting the cells from damage (1).
(e) Why discs of equal size were used [2]
Discs of equal size have the same surface area, and so the same amount of enzyme exposed (1); this keeps the test fair so that only pH affects the result (1).
(f) Why discs in concentrated salt solution became soft and floppy [3]
- The salt solution has a lower water potential than the cell contents (1).
- Water moves out of the cells by osmosis through the partially permeable membrane (1).
- The cells lose turgor and become flaccid, so the tissue becomes soft (1).
(g) Class of molecule catalase is made from [1] - protein.
Exam tip: enzyme activity peaks at an optimum pH and falls either side because extreme pH changes the shape of the active site (denaturation). Describe the trend on both sides of the optimum for full marks.
Answer Details
This question tests how pH affects enzyme activity, using catalase breaking down hydrogen peroxide. Taller foam means faster oxygen release, so foam height is a measure of enzyme activity.
(a) pH of greatest activity [1] - pH 7, where the foam is tallest (45 mm).
(b) Effect of pH on catalase activity [3]
- As the pH increases from 3 to 7, the activity (foam height) increases (1).
- Activity is highest at the optimum, pH 7 (1).
- Above pH 7 the activity decreases again as the pH rises to 11 (1).
(c) Difference in foam height between pH 5 and pH 7 [1]
\( 45 - 20 = 25 \ \text{mm} \) (1).
(d) Why catalase is important in living cells [2]
Hydrogen peroxide is a toxic waste product of metabolism (1); catalase breaks it down into harmless water and oxygen, protecting the cells from damage (1).
(e) Why discs of equal size were used [2]
Discs of equal size have the same surface area, and so the same amount of enzyme exposed (1); this keeps the test fair so that only pH affects the result (1).
(f) Why discs in concentrated salt solution became soft and floppy [3]
- The salt solution has a lower water potential than the cell contents (1).
- Water moves out of the cells by osmosis through the partially permeable membrane (1).
- The cells lose turgor and become flaccid, so the tissue becomes soft (1).
(g) Class of molecule catalase is made from [1] - protein.
Exam tip: enzyme activity peaks at an optimum pH and falls either side because extreme pH changes the shape of the active site (denaturation). Describe the trend on both sides of the optimum for full marks.
Question 3 Report
The speed at which blood flows depends on how much room it has to move through. A scientist investigated this by measuring, for each main type of vessel, the total cross-sectional area of all the vessels of that type added together, and the average speed at which blood flowed through them. Total cross-sectional area matters because blood leaving one aorta later spreads out through millions of tiny capillaries. The measurements are set out in the table and the same figures are plotted on the graph, so the pattern linking area and speed can be seen clearly.
| Vessel type | Total cross-sectional area / cm2 | Speed of blood flow / cm s-1 |
|---|---|---|
| Aorta | 4 | 40 |
| Arteries | 20 | 10 |
| Capillaries | 4500 | 0.03 |
| Veins | 40 | 5 |
| Vena cava | 18 | 18 |
(a) State the type of vessel in which blood flows most slowly. [1]
(b) Using the table, describe the relationship between total cross-sectional area and the speed of blood flow. [2]
(c) Explain why slow blood flow in the capillaries is an advantage. [3]
(d) Calculate the difference in the speed of blood flow between the aorta and the veins. [1]
(e) Explain why blood flows at high speed in the aorta. [2]
(f) Suggest why the speed of blood flow increases again in the vena cava compared with the capillaries. [1]
(a) Blood flows most slowly in the capillaries \((0.03\ \text{cm s}^{-1})\). [1]
(b) As the total cross-sectional area increases, the speed of blood flow decreases [1]; it is an inverse (opposite) relationship [1]. The capillaries have the largest area (4500 cm\(^2\)) and the slowest flow, while the aorta has a small area (4 cm\(^2\)) and the fastest flow. [2]
(c) Slow flow in the capillaries gives more time for the exchange of substances [1], so oxygen and glucose can diffuse out to the tissues and carbon dioxide and other wastes can diffuse in [1] before the blood leaves the capillary [1]. [3]
(d) Difference between the aorta (40) and the veins (5): \( 40 - 5 = \mathbf{35\ \text{cm s}^{-1}} \). [1]
(e) Blood is pumped straight from the ventricle at high pressure [1]; and the aorta has a small total cross-sectional area, so the blood is forced through a narrow space and moves quickly [1]. [2]
(f) Moving on from the capillaries, the many capillaries join into fewer, larger veins, so the total cross-sectional area decreases and the blood speeds up again in the vena cava. [1]
Answer Details
(a) Blood flows most slowly in the capillaries \((0.03\ \text{cm s}^{-1})\). [1]
(b) As the total cross-sectional area increases, the speed of blood flow decreases [1]; it is an inverse (opposite) relationship [1]. The capillaries have the largest area (4500 cm\(^2\)) and the slowest flow, while the aorta has a small area (4 cm\(^2\)) and the fastest flow. [2]
(c) Slow flow in the capillaries gives more time for the exchange of substances [1], so oxygen and glucose can diffuse out to the tissues and carbon dioxide and other wastes can diffuse in [1] before the blood leaves the capillary [1]. [3]
(d) Difference between the aorta (40) and the veins (5): \( 40 - 5 = \mathbf{35\ \text{cm s}^{-1}} \). [1]
(e) Blood is pumped straight from the ventricle at high pressure [1]; and the aorta has a small total cross-sectional area, so the blood is forced through a narrow space and moves quickly [1]. [2]
(f) Moving on from the capillaries, the many capillaries join into fewer, larger veins, so the total cross-sectional area decreases and the blood speeds up again in the vena cava. [1]
Question 4 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 5 Report
Fig. 4.1 shows a synapse, the junction between two neurones, labelled X and Y, in the spinal cord. Where they meet the two neurones do not quite touch, and a very small gap, labelled G, separates the end of neurone X from the start of neurone Y. When a nerve impulse travelling along neurone X reaches this point, it cannot simply jump across the gap as an electrical signal. Instead the signal is carried across by a chemical released from neurone X. Fig. 4.1 shows the ending of neurone X, the gap G and the membrane of neurone Y beyond it.
(a) Name the small gap labelled G between the two neurones. [1]
(b) Name the chemical that is released to carry the signal across gap G. [1]
(c) Describe how an impulse is passed from neurone X to neurone Y across the synapse. [4]
(d) Explain why an impulse can cross this synapse in one direction only. [2]
(e) Suggest one reason why some drugs change a person's mood by acting at synapses. [1]
(a) The small gap G is the synaptic gap (synaptic cleft). [1]
(b) The chemical released to carry the signal across the gap is a neurotransmitter. [1]
(c) Four linked steps (1 mark each): the impulse reaches the end of neurone X; the vesicles release a neurotransmitter into the gap; the neurotransmitter diffuses across the gap; and it binds to receptors on neurone Y, triggering a new impulse. [4]
(d) The neurotransmitter is released only from neurone X [1], and the receptors are only on neurone Y [1], so the signal can pass from X to Y but never the other way. [2]
(e) Many mood-changing drugs work because they add to, imitate or block the neurotransmitter (or its receptors) at the synapse, altering the impulses that reach the brain. [1]
Answer Details
(a) The small gap G is the synaptic gap (synaptic cleft). [1]
(b) The chemical released to carry the signal across the gap is a neurotransmitter. [1]
(c) Four linked steps (1 mark each): the impulse reaches the end of neurone X; the vesicles release a neurotransmitter into the gap; the neurotransmitter diffuses across the gap; and it binds to receptors on neurone Y, triggering a new impulse. [4]
(d) The neurotransmitter is released only from neurone X [1], and the receptors are only on neurone Y [1], so the signal can pass from X to Y but never the other way. [2]
(e) Many mood-changing drugs work because they add to, imitate or block the neurotransmitter (or its receptors) at the synapse, altering the impulses that reach the brain. [1]
Question 6 Report
A doctor carried out an investigation to find out where glucose is absorbed along the small intestine. One hour after a person had eaten a meal of bread, the doctor took samples of fluid from the gut lumen and from a nearby blood capillary at three points along the small intestine: the duodenum, the middle region and the ileum. The glucose concentration of each sample was then measured. Fig. 5.1 shows the results that were obtained, and Fig. 5.2 shows the three regions of the gut from which the samples were taken.
| Region | Glucose in gut lumen (mg / 100 cm3) | Glucose in capillary blood (mg / 100 cm3) |
|---|---|---|
| Duodenum | 180 | 90 |
| Jejunum | 120 | 95 |
| Ileum | 40 | 98 |
Fig. 5.2
(a) Using Fig. 5.1, describe how the glucose concentration in the gut lumen changes from the duodenum to the ileum. [2]
(b) Name the region of the alimentary canal where most glucose is absorbed. [1]
(c) Suggest why the glucose concentration in the capillary blood rises along the gut. [2]
(d) State two features of the small intestine that make it efficient at absorbing glucose. [2]
(e) State the meaning of the term active transport. [1]
(a) How lumen glucose changes from duodenum to ileum [2]: the glucose concentration in the lumen falls / decreases [1]; from 180 down to 40 mg per 100 cm\(^3\) (any correct data quote) [1]. It drops as the food moves along because glucose is being taken out of the gut.
(b) Where most glucose is absorbed [1]: the small intestine (accept ileum) [1].
(c) Why capillary blood glucose rises along the gut [2]: glucose is absorbed from the lumen into the blood along the length of the gut [1]; so the further along the sample is taken, the more glucose has entered the capillary [1] (98 mg per 100 cm\(^3\) at the ileum compared with 90 at the duodenum).
(d) Two features making the small intestine efficient [2]: any two of: many villi / microvilli giving a large surface area; a thin (one-cell-thick) epithelium for a short diffusion distance; a dense capillary network / good blood supply to carry glucose away and maintain the gradient [1 each, max 2].
(e) Meaning of active transport [1]: the movement of molecules against a concentration gradient using energy released by respiration [1]. This lets the gut absorb glucose even when its concentration in the blood is already higher than in the lumen.
Answer Details
(a) How lumen glucose changes from duodenum to ileum [2]: the glucose concentration in the lumen falls / decreases [1]; from 180 down to 40 mg per 100 cm\(^3\) (any correct data quote) [1]. It drops as the food moves along because glucose is being taken out of the gut.
(b) Where most glucose is absorbed [1]: the small intestine (accept ileum) [1].
(c) Why capillary blood glucose rises along the gut [2]: glucose is absorbed from the lumen into the blood along the length of the gut [1]; so the further along the sample is taken, the more glucose has entered the capillary [1] (98 mg per 100 cm\(^3\) at the ileum compared with 90 at the duodenum).
(d) Two features making the small intestine efficient [2]: any two of: many villi / microvilli giving a large surface area; a thin (one-cell-thick) epithelium for a short diffusion distance; a dense capillary network / good blood supply to carry glucose away and maintain the gradient [1 each, max 2].
(e) Meaning of active transport [1]: the movement of molecules against a concentration gradient using energy released by respiration [1]. This lets the gut absorb glucose even when its concentration in the blood is already higher than in the lumen.
Question 7 Report
In an oak woodland a single oak tree is fed on by very large numbers of caterpillars. The caterpillars are in turn eaten by blue tits, and the blue tits are eaten by a sparrowhawk. This simple food chain is shown in the diagram below. Ecologists working in one study area counted the number of individual organisms present at each stage of the chain, and their results are given in the table.
| Organism | Number counted |
|---|---|
| Oak tree | 1 |
| Caterpillars | 3000 |
| Blue tits | 40 |
| Sparrowhawks | 2 |
(a) State what information a pyramid of numbers is designed to show. [1]
(b) State which organism in this food chain is the producer. [1]
(c) Using the figures in the table, describe the shape of the pyramid of numbers that would be drawn from these data. [2]
(d) Explain why this pyramid of numbers does not form a regular pyramid shape. [3]
(e) Calculate the mean number of caterpillars available to each blue tit, assuming that all of the caterpillars are eaten. [2]
(f) Suggest why a pyramid of biomass would give a more regular shape than the pyramid of numbers for this food chain. [1]
This question is about pyramids of numbers, using the oak tree to caterpillars to blue tits to sparrowhawk food chain.
- What a pyramid of numbers shows [1]. The number of organisms at each trophic level of a food chain [1].
- The producer [1]. The oak tree [1], because it makes its own food by photosynthesis.
- Shape of the pyramid [2]. The bar for the oak tree (producer) is very narrow because there is only one tree [1]; the caterpillar bar is by far the widest, and the bars then become smaller for the blue tits and smallest for the sparrowhawks [1].
- Why it is not a regular pyramid [3]. A single very large producer (the oak) supports a very large number of small primary consumers (caterpillars) [1]; a pyramid of numbers counts organisms and ignores their size or mass [1]; so the producer bar is narrower than the level above it and the pyramid is irregular (inverted at its base) [1].
- Mean caterpillars per blue tit [2]. \(3000 \div 40\) [1] \(= 75\) caterpillars per blue tit [1].
- Why a pyramid of biomass would be more regular [1]. The single oak has a very large biomass, so its bar would become the largest at the base, giving the normal pyramid shape [1].
Answer Details
This question is about pyramids of numbers, using the oak tree to caterpillars to blue tits to sparrowhawk food chain.
- What a pyramid of numbers shows [1]. The number of organisms at each trophic level of a food chain [1].
- The producer [1]. The oak tree [1], because it makes its own food by photosynthesis.
- Shape of the pyramid [2]. The bar for the oak tree (producer) is very narrow because there is only one tree [1]; the caterpillar bar is by far the widest, and the bars then become smaller for the blue tits and smallest for the sparrowhawks [1].
- Why it is not a regular pyramid [3]. A single very large producer (the oak) supports a very large number of small primary consumers (caterpillars) [1]; a pyramid of numbers counts organisms and ignores their size or mass [1]; so the producer bar is narrower than the level above it and the pyramid is irregular (inverted at its base) [1].
- Mean caterpillars per blue tit [2]. \(3000 \div 40\) [1] \(= 75\) caterpillars per blue tit [1].
- Why a pyramid of biomass would be more regular [1]. The single oak has a very large biomass, so its bar would become the largest at the base, giving the normal pyramid shape [1].
Question 8 Report
The kinds of tooth an animal has, and their shapes, are closely related to the type of food that the animal eats. Fig. 5.1 compares the teeth of a carnivore, which eats mainly meat, with the teeth of a herbivore, which eats mainly plants. The diagram allows the size and shape of the different teeth in each animal to be compared. In both animals the teeth carry out the physical digestion of the food, but the way in which the teeth are used depends on the animal's diet.
Fig. 5.1
(a) State the function of the large, pointed canine teeth of the carnivore. [1]
(b) Explain how the molar teeth of the herbivore are adapted to its plant diet. [3]
(c) A human eats a mixed diet of both plants and animals. Describe how a human uses its different kinds of tooth to deal with this diet. [3]
(d) Explain how the physical action of the teeth aids the chemical digestion of the food. [3]
This links tooth shape to diet and to physical digestion. Physical (mechanical) digestion by the teeth speeds up chemical digestion by increasing the surface area exposed to enzymes.
- (a) The large pointed canines are used for tearing / gripping flesh (meat) / holding prey [1].
- (b) The herbivore's molars are large, flat and ridged [1], giving a large surface area for grinding [1], which breaks down tough plant material / cellulose cell walls [1].
- (c) A human uses different teeth in turn: the incisors bite / cut [1], the canines tear / grip [1], and the premolars and molars grind / crush the food [1].
- (d) The teeth break the food into smaller pieces [1]; this increases the surface area of the food [1]; so enzymes can act faster and digestion is more complete [1].
Answer Details
This links tooth shape to diet and to physical digestion. Physical (mechanical) digestion by the teeth speeds up chemical digestion by increasing the surface area exposed to enzymes.
- (a) The large pointed canines are used for tearing / gripping flesh (meat) / holding prey [1].
- (b) The herbivore's molars are large, flat and ridged [1], giving a large surface area for grinding [1], which breaks down tough plant material / cellulose cell walls [1].
- (c) A human uses different teeth in turn: the incisors bite / cut [1], the canines tear / grip [1], and the premolars and molars grind / crush the food [1].
- (d) The teeth break the food into smaller pieces [1]; this increases the surface area of the food [1]; so enzymes can act faster and digestion is more complete [1].
