An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 Cambridge International A Level Biology paper. Not affiliated with or reproduced from Cambridge.
Paper 1B
Answer all questions. Calculators and rulers are permitted. Total marks: 110.
36 PastPaper.question · 118 PastPaper.marks
PastPaper.question 1 · multiple-choice
1 PastPaper.marks
During eutrophication in a river, what is the direct cause of the death of fish?
A.Toxic chemicals released by rapid algal growth
B.A decrease in dissolved oxygen due to aerobic respiration by decomposers
C.An increase in carbon dioxide blocking the fish's gills
D.Direct competition with algae for food sources
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PastPaper.workedSolution
Eutrophication leads to an algal bloom which blocks sunlight, causing aquatic plants to die. Decomposers (bacteria) multiply rapidly and break down the dead organic matter. These decomposers respire aerobically, depleting the dissolved oxygen levels in the water, which leads to the suffocation and death of fish.
PastPaper.markingScheme
1 mark for the correct option (B).
PastPaper.question 2 · multiple-choice
1 PastPaper.marks
Which statement correctly describes the structure or function of a blood vessel?
A.Arteries have thin elastic walls to carry blood under low pressure
B.Veins have valves to prevent the backflow of blood
C.Capillaries have walls that are several cells thick to allow rapid diffusion
D.Veins carry only deoxygenated blood away from the heart
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PastPaper.workedSolution
Veins carry blood back to the heart under low pressure and contain valves to prevent the backflow of blood. Arteries have thick, elastic walls to withstand high pressure. Capillaries have walls that are only one cell thick to allow rapid diffusion of substances.
PastPaper.markingScheme
1 mark for the correct option (B).
PastPaper.question 3 · multiple-choice
1 PastPaper.marks
Which of the following correctly pairs an enzyme with its site of production and the products of its reaction?
A.Amylase | Salivary glands | Glucose
B.Protease (pepsin) | Stomach | Amino acids
C.Lipase | Pancreas | Fatty acids and glycerol
D.Maltase | Liver | Maltose
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PastPaper.workedSolution
Lipase is produced by the pancreas and released into the small intestine, where it breaks down lipids into fatty acids and glycerol. Amylase produces maltose, pepsin produces peptides, and maltase is produced in the small intestine (not liver).
PastPaper.markingScheme
1 mark for the correct option (C).
PastPaper.question 4 · multiple-choice
1 PastPaper.marks
In the commercial production of bread, which product of the anaerobic respiration of yeast is responsible for causing the dough to rise?
A.Ethanol
B.Carbon dioxide
C.Lactic acid
D.Water
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PastPaper.workedSolution
Yeast respires anaerobically to produce ethanol and carbon dioxide. The carbon dioxide gas bubbles get trapped in the dough, causing it to expand and rise. The ethanol evaporates during baking.
PastPaper.markingScheme
1 mark for the correct option (B).
PastPaper.question 5 · multiple-choice
1 PastPaper.marks
A heterozygous plant with red flowers (Rr) is crossed with a homozygous plant with white flowers (rr). What is the expected ratio of phenotypes in the offspring?
A.3 red : 1 white
B.1 red : 1 white
C.All red
D.1 red : 3 white
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PastPaper.workedSolution
A cross between Rr (heterozygous) and rr (homozygous recessive) produces offspring with genotypes Rr and rr in a 1:1 ratio. Therefore, the expected phenotypic ratio is 1 red to 1 white.
PastPaper.markingScheme
1 mark for the correct option (B).
PastPaper.question 6 · multiple-choice
1 PastPaper.marks
Which of the following processes requires energy in the form of ATP to move substances across a cell membrane?
A.The absorption of mineral ions into plant root hair cells from a low concentration in the soil
B.The diffusion of oxygen from the alveoli into the blood capillaries
C.The movement of water from a dilute solution to a concentrated solution through a partially permeable membrane
D.The entry of carbon dioxide into palisade mesophyll cells during the day
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PastPaper.workedSolution
Active transport is the movement of substances from a dilute solution to a more concentrated solution (against a concentration gradient) and requires energy from respiration in the form of ATP. Root hair cells use active transport to absorb mineral ions from the low concentration in the soil.
PastPaper.markingScheme
1 mark for the correct option (A).
PastPaper.question 7 · short structured answer
3 PastPaper.marks
An agricultural worker accidentally spills a large volume of nitrate-rich fertiliser into a slow-moving freshwater stream. Explain the sequence of events that leads to the death of fish in this stream.
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PastPaper.workedSolution
1. The high concentration of nitrates causes a rapid growth of algae on the water surface, known as an algal bloom. 2. This algal layer blocks light from reaching submerged aquatic plants, preventing photosynthesis and causing them to die. 3. Decomposer bacteria break down the dead plant matter, multiplying rapidly and respiring aerobically. This uses up the dissolved oxygen in the water, leading to the death of fish due to a lack of oxygen.
PastPaper.markingScheme
MP1: Run-off of nitrates causes rapid algal growth / algal bloom which blocks sunlight (1) MP2: Submerged plants die due to lack of photosynthesis and are broken down by decomposer bacteria (1) MP3: Bacteria respire aerobically, using up dissolved oxygen in the water, causing fish to die (1)
PastPaper.question 8 · short structured answer
3 PastPaper.marks
Describe three structural differences between the pollen grains and the stigmas of wind-pollinated flowers compared to insect-pollinated flowers.
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PastPaper.workedSolution
Wind-pollinated flowers are adapted to disperse and capture pollen using air currents, whereas insect-pollinated flowers rely on insects to transfer pollen. 1. Pollen: Wind-pollinated flowers have light, smooth and small pollen grains to easily float in the wind, while insect-pollinated flowers have sticky or spiky pollen to attach to insect bodies. 2. Stigma structure: Wind-pollinated flowers have large, feathery stigmas to provide a large surface area to trap drifting pollen, while insect-pollinated flowers have small, sticky stigmas to scrape pollen off insects. 3. Stigma position: Wind-pollinated flowers have stigmas that hang outside the flower to catch pollen, while insect-pollinated flowers have stigmas inside the flower.
PastPaper.markingScheme
MP1: Comparison of pollen: wind has light/smooth pollen OR insect has sticky/spiky pollen (1) MP2: Comparison of stigma structure: wind has feathery/large surface area stigma OR insect has sticky/lobed stigma (1) MP3: Comparison of stigma position: wind stigma is exposed/outside the flower OR insect stigma is enclosed/inside the flower (1)
PastPaper.question 9 · short structured answer
3 PastPaper.marks
Explain the roles of restriction enzymes and DNA ligase in the production of genetically modified bacteria.
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PastPaper.workedSolution
In genetic engineering: 1. Restriction enzymes are used to cut open the bacterial plasmid (vector) and also to cut out the desired gene (such as the human insulin gene) from the donor DNA. 2. This cutting occurs at specific base sequences, leaving short, single-stranded sections called complementary sticky ends. 3. DNA ligase is then used to join the sticky ends of the gene and the plasmid together by forming covalent bonds in the sugar-phosphate backbone, producing a recombinant plasmid.
PastPaper.markingScheme
MP1: Restriction enzymes cut the plasmid/DNA and the desired gene at specific base sequences (1) MP2: This creates complementary sticky ends (1) MP3: DNA ligase joins the sticky ends of the gene and plasmid together to form recombinant DNA (1)
PastPaper.question 10 · short structured answer
3 PastPaper.marks
Explain why an increase in air humidity decreases the rate of transpiration from a plant leaf.
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PastPaper.workedSolution
Transpiration is the loss of water vapour from plant leaves by diffusion through the stomata. 1. Humidity represents the concentration of water vapour in the air surrounding the leaf. 2. An increase in external humidity reduces the difference in water vapour concentration between the air spaces inside the leaf and the external atmosphere (reduces the concentration gradient). 3. Since diffusion rate depends on the steepness of the concentration gradient, a shallower gradient results in a slower rate of diffusion of water vapour out of the stomata.
PastPaper.markingScheme
MP1: Humidity represents the concentration of water vapour in the air outside the leaf (1) MP2: High humidity decreases/reduces the water vapour concentration gradient between the inside of the leaf and the outside air (1) MP3: This decreases the rate of diffusion of water vapour out of the leaf through the stomata (1)
PastPaper.question 11 · short structured answer
3 PastPaper.marks
In a woodland ecosystem, thousands of caterpillars feed on a single large oak tree. Blue tits then feed on these caterpillars. Describe the shape of a pyramid of numbers for this food chain and explain why a pyramid of biomass for the same food chain would have a different shape.
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PastPaper.workedSolution
1. The pyramid of numbers has an inverted or diamond shape. This is because there is only one producer (the single oak tree) at the base of the pyramid, supporting a very large number of primary consumers (thousands of caterpillars). 2. The pyramid of biomass is upright (a traditional pyramid shape). This is because biomass represents the dry mass of living material at each trophic level. 3. The single oak tree has an extremely large mass, which is far greater than the combined biomass of all the caterpillars feeding on it, and biomass decreases at each subsequent level due to energy losses.
PastPaper.markingScheme
MP1: Pyramid of numbers is inverted / narrow at the base / diamond-shaped because there is only one producer (1) MP2: Pyramid of biomass is upright / typical pyramid shape (1) MP3: Because the single oak tree has a much larger dry mass/biomass than all of the caterpillars combined (1)
PastPaper.question 12 · short structured answer
3 PastPaper.marks
Bile is produced in the liver and stored in the gall bladder. Describe how bile assists in the chemical and physical digestion of lipids in the small intestine.
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PastPaper.workedSolution
1. Chemical role / pH neutralisation: Bile is alkaline. It neutralises the highly acidic chyme (stomach contents mixed with hydrochloric acid) as it enters the duodenum. This creates the optimum alkaline/neutral pH (around pH 7-8) for pancreatic lipase enzymes to function. 2. Physical role / Emulsification: Bile contains bile salts that emulsify large lipid droplets into millions of tiny lipid droplets. This physical breakdown increases the total surface area of lipids exposed to the lipase enzymes, allowing chemical digestion into fatty acids and glycerol to occur much faster.
PastPaper.markingScheme
MP1: Bile is alkaline and neutralises hydrochloric acid/stomach acid (1) MP2: This provides the optimum pH for lipase enzymes to function (1) MP3: Bile emulsifies lipids, breaking large droplets into smaller droplets, increasing the surface area for lipase (1)
PastPaper.question 13 · short structured answer
3 PastPaper.marks
A sheep breeder wants to determine whether a black-faced sheep, which has the dominant phenotype for face colour, is homozygous dominant or heterozygous. Describe how the breeder could use a test cross to find out the genotype of this sheep.
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PastPaper.workedSolution
1. The breeder must perform a test cross by mating the black-faced sheep with a sheep that shows the recessive phenotype (a white-faced sheep, which must have the homozygous recessive genotype, e.g., ff). 2. If the black-faced parent is heterozygous (Ff), there is a 50% chance for each offspring to inherit the recessive allele from both parents and have a white face. Thus, if even a single white-faced offspring is produced, the parent is proven to be heterozygous. 3. If the black-faced parent is homozygous dominant (FF), all offspring will receive a dominant allele (F) and will have black faces. If a large number of offspring are produced and all have black faces, the breeder can conclude the parent is homozygous dominant.
PastPaper.markingScheme
MP1: Cross the black-faced sheep with a homozygous recessive / white-faced sheep (1) MP2: If any of the offspring have white faces, the parent sheep must be heterozygous (1) MP3: If all offspring have black faces (in a large sample), the parent sheep is homozygous dominant (1)
PastPaper.question 14 · short structured answer
3 PastPaper.marks
Explain two methods used by fish farmers to manage sea lice infections in their fish pens, and state one disadvantage of using chemical treatments to control these pests.
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PastPaper.workedSolution
1. Biological control: Fish farmers introduce biological control agents, such as wrasse or other cleaner fish, into the salmon cages. These cleaner fish eat the parasitic sea lice off the skin of the salmon without harming the salmon. 2. Chemical control: Farmers can add chemical pesticides or therapeutic baths to the water to kill the sea lice directly. 3. Disadvantage of chemical treatments: Chemical pesticides can wash out of the cages and pollute the surrounding marine environment, potentially killing non-target organisms (such as wild crabs or lobsters) or leading to the evolution of pesticide resistance in the sea lice.
PastPaper.markingScheme
MP1: Biological control: introduction of cleaner fish / wrasse that eat sea lice (1) MP2: Chemical control: use of pesticides / chemical treatments in the water to kill lice (1) MP3: Disadvantage: risk of bioaccumulation / toxicity to non-target marine organisms / development of resistance in sea lice (1)
PastPaper.question 15 · short answer
3 PastPaper.marks
Explain how the leaching of mineral ions from agricultural fertilisers into a lake leads to a reduction in the concentration of dissolved oxygen.
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PastPaper.workedSolution
1. The leached fertilisers (containing nitrates and phosphates) cause rapid growth of algae on the water surface, known as an algal bloom, which blocks sunlight from reaching deeper water. 2. Submerged aquatic plants and algae die because they cannot photosynthesise without light. 3. Decomposing bacteria feed on the dead organic matter and multiply rapidly. These bacteria respire aerobically, consuming the dissolved oxygen in the water.
PastPaper.markingScheme
1 mark: Reference to algal bloom / rapid growth of algae blocking sunlight. 1 mark: Reference to death of submerged plants due to lack of photosynthesis. 1 mark: Reference to decomposers / bacteria feeding on dead matter and consuming oxygen through aerobic respiration. [Maximum 3 marks]
PastPaper.question 16 · short answer
3 PastPaper.marks
Explain how three structural features of a wind-pollinated flower adapt it for successful pollination.
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PastPaper.workedSolution
1. It has a feathery or branched stigma which provides a large surface area to trap drifting pollen grains from the air. 2. The anthers hang outside the flower on long, flexible filaments, allowing pollen to be easily dislodged and carried away by the wind. 3. The pollen grains are small, light, and smooth, which allows them to be easily carried by light air currents over long distances.
PastPaper.markingScheme
Award 1 mark for each clear feature coupled with its explanation (up to a maximum of 3 marks): - Feathery/branched stigma (1) to increase surface area/trap pollen (1). - Long filaments/anthers hanging outside (1) to easily release pollen into the wind (1). - Light/smooth/large quantities of pollen (1) to be easily carried by wind / increase chances of pollination (1). - Small/inconspicuous/green petals (1) to not block the wind / avoid wasting energy on attracting insects (1).
PastPaper.question 17 · short answer
3 PastPaper.marks
Explain the roles of restriction enzymes and DNA ligase in the production of genetically modified bacteria that contain the human insulin gene.
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PastPaper.workedSolution
1. Restriction enzymes are used to cut the human insulin gene out of human DNA, and also to cut open the bacterial plasmid vector. 2. Cutting with the same restriction enzyme creates complementary sticky ends (single-stranded DNA overhangs) on both the gene and the plasmid. 3. DNA ligase is then used to join the sticky ends of the human insulin gene and the plasmid together by forming covalent bonds in the sugar-phosphate backbone, creating a recombinant plasmid.
PastPaper.markingScheme
1 mark: Restriction enzyme cuts out the desired gene / cuts open the plasmid vector. 1 mark: Reference to creating complementary sticky ends (using the same enzyme). 1 mark: DNA ligase joins/ligates the gene and plasmid together to form a recombinant plasmid. [Maximum 3 marks]
PastPaper.question 18 · short answer
3 PastPaper.marks
A student sets up a potometer to measure the rate of transpiration of a leafy shoot. Describe three precautions the student must take when setting up this apparatus to ensure accurate and valid measurements.
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PastPaper.workedSolution
1. Cut the leafy shoot under water to prevent air bubbles from entering the xylem vessel, which would break the continuous water column. 2. Assemble the apparatus under water and seal all connections/joints with waterproof petroleum jelly (Vaseline) to ensure the system is completely airtight. 3. Allow the shoot to equilibrate to the conditions and ensure the leaves are dried completely before beginning measurements, as wet leaves will artificially reduce transpiration.
PastPaper.markingScheme
Award 1 mark for each valid precaution with its reason (up to a maximum of 3 marks): - Cut shoot underwater (1) to prevent air entering xylem / breaking water column (1). - Seal joints with Vaseline/petroleum jelly (1) to make the apparatus airtight / prevent leaks (1). - Dry the leaves before starting (1) to avoid blocking stomata / allowing normal transpiration (1). - Introduce only a single air bubble (1) to track movement accurately (1).
PastPaper.question 19 · short answer
3 PastPaper.marks
Explain why only about 10% of the energy in the biomass of a primary consumer is transferred to the secondary consumer in a food chain.
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PastPaper.workedSolution
1. Not all parts of the primary consumer are eaten or consumed (for example, bones, hair, claws, or woody parts are left behind). 2. Some parts that are consumed are indigestible and are egested as faeces, meaning their energy is not absorbed by the secondary consumer. 3. A large amount of the absorbed energy is used by the primary consumer for its own life processes, such as muscle contraction and active transport, and is lost to the environment as heat during aerobic respiration.
PastPaper.markingScheme
1 mark: Uneaten parts (e.g., bones/teeth/claws) mean not all biomass is consumed. 1 mark: Indigestible parts are egested as faeces / lost in excretion (urine). 1 mark: Energy is lost as heat released during respiration (for movement/warmth/metabolism). [Maximum 3 marks]
PastPaper.question 20 · short answer
3 PastPaper.marks
Explain how three features of a villus in the small intestine adapt it for the efficient absorption of digested food molecules.
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PastPaper.workedSolution
1. The surface of the villus epithelial cells is covered with microscopic folds called microvilli, which significantly increase the total surface area available for diffusion and active transport. 2. The wall of the villus (epithelium) is only one cell thick, which provides an extremely short diffusion pathway for nutrients to pass into the bloodstream. 3. There is a dense network of blood capillaries inside each villus, which rapidly carries away absorbed nutrients (like glucose and amino acids) to maintain a steep concentration gradient between the lumen of the gut and the blood.
PastPaper.markingScheme
Award 1 mark for each structural feature linked to its explanation (up to a maximum of 3 marks): - Microvilli / folded surface (1) to increase surface area for faster absorption (1). - One-cell-thick epithelium (1) to provide a short diffusion distance (1). - Rich capillary network / good blood supply (1) to maintain a steep concentration gradient (1). - Lacteal (1) to absorb and transport fatty acids and glycerol (1).
PastPaper.question 21 · short answer
3 PastPaper.marks
In a plant species, flower colour is controlled by codominant alleles. The allele for red flowers is \(C^R\) and the allele for white flowers is \(C^W\). Heterozygous plants (\(C^R C^W\)) have pink flowers. Predict the genotypes and the phenotypic ratio of the offspring produced by crossing two pink-flowered plants.
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PastPaper.workedSolution
1. The parents are both pink-flowered, which means their genotype is \(C^R C^W\). 2. A genetic cross (\(C^R C^W \times C^R C^W\)) produces offspring with the genotypes \(C^R C^R\), \(C^R C^W\), and \(C^W C^W\) in a 1:2:1 ratio. 3. Since the alleles are codominant, \(C^R C^R\) plants are red, \(C^R C^W\) plants are pink, and \(C^W C^W\) plants are white. This results in a phenotypic ratio of 1 red : 2 pink : 1 white.
PastPaper.markingScheme
1 mark: Identify parental genotypes as \(C^R C^W\). 1 mark: Determine offspring genotypes as \(C^R C^R\), \(C^R C^W\), and \(C^W C^W\) (or show correct Punnett square). 1 mark: State phenotypic ratio of 1 red : 2 pink : 1 white (must match correct phenotypes). [Maximum 3 marks]
PastPaper.question 22 · short answer
3 PastPaper.marks
During bread-making, yeast is mixed with flour, water, and sugar to form dough. Explain the biological processes that cause the dough to rise before baking.
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PastPaper.workedSolution
1. Yeast cells use the sugar (glucose) in the mixture to perform respiration. 2. Initially, yeast respires aerobically. Once oxygen inside the dough is depleted, yeast switches to anaerobic respiration (fermentation), which produces carbon dioxide and ethanol. 3. The carbon dioxide gas is trapped in the elastic dough, forming bubbles. As the gas accumulates, the bubbles expand, causing the dough to rise.
PastPaper.markingScheme
1 mark: Yeast respires (initially aerobically, then anaerobically/fermentation) using the sugars. 1 mark: Respiration produces carbon dioxide gas. 1 mark: Carbon dioxide gas gets trapped in the dough and expands, causing the dough to rise. [Maximum 3 marks]
PastPaper.question 23 · short answer
3 PastPaper.marks
Explain how the run-off of fertilizers into a freshwater pond leads to the death of fish.
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PastPaper.workedSolution
1. Fertilizer run-off causes rapid growth of algae (an algal bloom) on the water surface, blocking sunlight from reaching plants deeper in the pond. 2. Submerged plants die due to a lack of light for photosynthesis. 3. Decomposer bacteria feed on the dead plant matter and multiply rapidly, consuming dissolved oxygen through aerobic respiration. This creates anoxic conditions, causing fish to die from a lack of oxygen.
PastPaper.markingScheme
1 mark for algal bloom blocking light or preventing photosynthesis of deeper plants; 1 mark for decomposition of dead plants by bacteria; 1 mark for bacteria using up oxygen during aerobic respiration leading to fish suffocation.
PastPaper.question 24 · short answer
3 PastPaper.marks
Describe how the structure of the human placenta is adapted for the efficient exchange of substances between mother and fetus.
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PastPaper.workedSolution
1. The placenta features numerous chorionic villi that project into the maternal blood spaces, greatly increasing the surface area available for diffusion. 2. The membrane separating the maternal and fetal blood is extremely thin, keeping the diffusion distance to a minimum. 3. There is a rich blood supply on both sides, which constantly moves blood away from the exchange surface, thereby maintaining a steep concentration gradient for substances like oxygen, glucose, and carbon dioxide.
PastPaper.markingScheme
1 mark for large surface area due to villi or microvilli; 1 mark for thin wall/membrane giving a short diffusion distance; 1 mark for rich blood supply maintaining a steep concentration gradient.
PastPaper.question 25 · short answer
3 PastPaper.marks
Explain the roles of restriction enzymes and DNA ligase in the production of recombinant DNA.
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PastPaper.workedSolution
1. Restriction enzymes recognize specific target sequences of DNA and cut the double-stranded DNA at these points. This is done to isolate the desired gene and to open up the vector plasmid, leaving complementary single-stranded overhanging sequences called sticky ends. 2. DNA ligase is then introduced to join the sugar-phosphate backbones of the desired gene and the plasmid together. 3. This joining forms a single, continuous recombinant DNA molecule.
PastPaper.markingScheme
1 mark for restriction enzymes cutting plasmid or DNA at specific sequences to produce sticky ends; 1 mark for DNA ligase joining or bonding the DNA fragments together; 1 mark for reference to complementary base pairing of sticky ends or forming a recombinant plasmid.
PastPaper.question 26 · short answer
3 PastPaper.marks
Describe how guard cells control the rate of transpiration from a leaf.
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PastPaper.workedSolution
1. When there is plenty of water, guard cells actively transport ions in, causing water to enter them by osmosis. This causes the guard cells to swell and become turgid. 2. Because the inner cell wall of each guard cell is thicker and less flexible than the outer wall, the cells curve outwards as they swell, opening the stomatal pore and increasing transpiration. 3. When water is scarce, the guard cells lose water, become flaccid, and straighten, closing the stomatal pore to limit transpiration.
PastPaper.markingScheme
1 mark for osmosis of water into guard cells making them turgid (or out making them flaccid); 1 mark for the uneven thickness of the guard cell walls causing them to curve and open the stoma; 1 mark for stomata closing when guard cells are flaccid to reduce water loss.
PastPaper.question 27 · short answer
3 PastPaper.marks
An oak tree supports thousands of caterpillars, which are then eaten by fifty blue tits. Explain why a pyramid of biomass for this food chain is shaped differently from a pyramid of numbers.
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PastPaper.workedSolution
1. A pyramid of numbers shows the physical quantity of individual organisms at each trophic level. Since one large oak tree has enough tissue to feed thousands of small caterpillars, the base of the numbers pyramid is narrow, representing only one organism, while the second tier is very wide. 2. A pyramid of biomass shows the total dry mass of living material at each level. The biomass of one massive oak tree is vastly greater than the collective biomass of all the caterpillars it supports, making the biomass pyramid wide at the base and progressively narrower at higher levels. 3. This is because biomass is lost at each trophic level due to factors like respiration, movement, and undigested waste.
PastPaper.markingScheme
1 mark for explaining that the pyramid of numbers is inverted or narrow at the base because a single oak tree is very large and supports many smaller organisms; 1 mark for explaining that the pyramid of biomass is upright because the total dry mass of the tree is larger than the total mass of caterpillars; 1 mark for noting that biomass always decreases up the trophic levels due to energy or material losses.
PastPaper.question 28 · short answer
3 PastPaper.marks
Explain how the structure of a root hair cell is adapted to absorb mineral ions against a concentration gradient.
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PastPaper.workedSolution
1. The cell membrane of the root hair cell extends outwards to form a long, narrow hair-like projection. This significantly increases the surface area in contact with the soil solution, allowing for more membrane transport proteins to be present. 2. The cell contains a high density of mitochondria. 3. These mitochondria carry out aerobic respiration to produce ATP, which provides the energy required to transport mineral ions against their concentration gradient (active transport).
PastPaper.markingScheme
1 mark for long projection or hair-like structure increasing surface area; 1 mark for high density of mitochondria to release energy or ATP; 1 mark for linking the energy to active transport against the concentration gradient.
PastPaper.question 29 · short answer
3 PastPaper.marks
Describe the steps involved in using micropropagation (tissue culture) to produce large numbers of identical plants.
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PastPaper.workedSolution
1. Small pieces of plant tissue, called explants, are cut from the growing tips or side shoots of a parent plant and sterilized to kill any microorganisms. 2. The explants are placed onto a sterile nutrient agar medium containing plant hormones (like auxins and cytokinins) to promote cell division and development, alongside glucose and mineral ions. 3. Once the explants grow into small plantlets with roots and shoots, they are transferred to compost or soil in a greenhouse to grow into fully developed, genetically identical cloned plants.
PastPaper.markingScheme
1 mark for cutting explants and sterilizing them; 1 mark for placing them on a sterile nutrient agar medium containing minerals/sugar and plant growth hormones; 1 mark for growing them into plantlets and transferring them to soil/compost.
PastPaper.question 30 · short answer
3 PastPaper.marks
Explain how bile assists in both the chemical and physical digestion of lipids in the small intestine.
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PastPaper.workedSolution
1. Physical digestion: Bile is produced in the liver and stored in the gall bladder. When released into the duodenum, it emulsifies lipids, breaking down large oil/fat droplets into thousands of microscopic droplets. This drastically increases the surface area of lipids available for lipase enzymes to attack. 2. Chemical digestion: Bile is alkaline, containing sodium hydrogencarbonate. It neutralizes the highly acidic gastric juice entering the duodenum from the stomach. 3. This creates the optimum slightly alkaline pH (around pH 7-8) for pancreatic lipase to function efficiently, accelerating the breakdown of lipids into fatty acids and glycerol.
PastPaper.markingScheme
1 mark for physical emulsification or breaking large lipid droplets into smaller droplets to increase surface area; 1 mark for being alkaline and neutralizing stomach acid; 1 mark for providing the optimum pH for lipase activity.
A student investigates the effect of temperature on the rate of anaerobic respiration in yeast. She sets up a boiling tube containing a mixture of yeast and glucose solution. She records the volume of carbon dioxide gas produced by the yeast over a period of 10 minutes at different temperatures. The results are shown in the table:
| Temperature (°C) | Volume of carbon dioxide gas produced in 10 minutes (cm³) | | :---: | :---: | | 10 | 2.4 | | 20 | 6.2 | | 30 | 12.8 | | 40 | 18.0 | | 50 | 7.6 | | 60 | 0.0 |
(a) Plot a line graph to show the effect of temperature on the volume of carbon dioxide gas produced. (4 marks)
(b) Explain why the rate of carbon dioxide production is higher at 40 °C than at 20 °C. (2 marks)
(c) State two variables that the student should control in this investigation. (2 marks)
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PastPaper.workedSolution
(a) A line graph is plotted with Temperature (°C) on the x-axis and Volume of carbon dioxide gas produced in 10 minutes (cm³) on the y-axis. The scale is linear and occupies more than half of the grid. Points are plotted accurately and joined with a neat curve or straight lines.
(b) At 40 °C, the yeast cells and glucose molecules have more kinetic energy than at 20 °C. This results in more frequent successful collisions between the enzymes inside the yeast and the glucose substrates, forming more enzyme-substrate complexes and increasing the rate of respiration.
(c) Control variables include the concentration of the glucose solution, the volume of the yeast suspension, and the pH of the mixture.
PastPaper.markingScheme
(a) Graph plotting [4 marks]: - S (Scale): linear, regular, and uses at least half of the grid on both axes. (1 mark) - A (Axes): x-axis is labelled 'Temperature (°C)' and y-axis is labelled 'Volume of carbon dioxide gas produced in 10 minutes (cm³)' or 'Volume of carbon dioxide (cm³)'. (1 mark) - P (Plotting): all 6 points plotted accurately to within half a small square. (1 mark) - L (Line): points connected with a smooth curve or straight, ruled line-to-line segments. (1 mark)
(b) Explanation [2 marks]: - Yeast/enzymes and glucose molecules have more kinetic energy. (1 mark) - Leads to more frequent successful collisions / more enzyme-substrate complexes formed. (1 mark)
(c) Control variables [2 marks]: - Concentration of glucose solution / volume of glucose solution. (1 mark) - pH of the mixture / concentration of yeast suspension / volume of yeast suspension. (1 mark) [Reject: 'amount' of glucose/yeast]
A student investigates the effect of wind speed on the rate of transpiration in a leafy shoot using a potometer. The wind speed is varied by placing a fan at different distances from the plant shoot. The wind speed is measured in metres per second (m/s) using an anemometer. The table shows the results of this investigation:
(a) Plot a line graph to show the relationship between wind speed and the distance moved by the bubble. (4 marks)
(b) Explain why increasing the wind speed increases the rate of transpiration. (3 marks)
(c) State how the student could reset the bubble to the start of the capillary tube in this investigation. (1 mark)
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(a) A line graph is plotted with Wind speed (m/s) on the x-axis and Distance moved by air bubble in 5 minutes (mm) on the y-axis. The scale is linear and occupies more than half of the grid. Points are plotted accurately and joined with a neat curve or straight lines.
(b) Increasing wind speed blows water vapour away from the surface of the leaf / stomata. This maintains a steep water vapour concentration gradient (or water potential gradient) between the air spaces inside the leaf and the external air. As a result, water vapour diffuses out of the stomata more rapidly.
(c) The student can reset the bubble by opening the tap on the reservoir (or syringe) to force water into the capillary tube, pushing the bubble back to the start.
PastPaper.markingScheme
(a) Graph plotting [4 marks]: - S (Scale): linear, regular, and uses at least half of the grid on both axes. (1 mark) - A (Axes): x-axis is labelled 'Wind speed (m/s)' and y-axis is labelled 'Distance moved by air bubble in 5 minutes (mm)'. (1 mark) - P (Plotting): all 6 points plotted accurately to within half a small square. (1 mark) - L (Line): points connected with a smooth curve or straight, ruled line-to-line segments. (1 mark)
(b) Explanation [3 marks]: - Wind moves / blows water vapour away from the leaf surface / stomata. (1 mark) - Maintains a steep water potential gradient / concentration gradient. (1 mark) - Speeds up diffusion of water vapour out of the stomata. (1 mark)
(c) Resetting the bubble [1 mark]: - Open the tap on the reservoir / squeeze water from the syringe. (1 mark) [Accept: open reservoir tap to let water in]
PastPaper.question 33 · practical
6 PastPaper.marks
An aquatic plant, such as *Cabomba*, can be used to study the factors affecting the rate of photosynthesis.
Design an investigation to find the effect of different concentrations of carbon dioxide on the rate of photosynthesis of *Cabomba*.
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To design a valid investigation, the CORMS framework should be followed:
- **C (Change):** Describe how the independent variable is manipulated. Here, the concentration of carbon dioxide is varied by using different, specific concentrations of sodium hydrogencarbonate solution (e.g., 0.2%, 0.4%, 0.6%, 0.8%, 1.0%). - **O (Organism):** State the control variable relating to the living organism. Use the same species of aquatic plant (*Cabomba*) and control its initial state by using pieces of identical length (e.g., 5 cm) or mass. - **R (Repeat):** State that the experiment must be repeated at least three times at each concentration to ensure reliability, allowing for the calculation of a mean and the identification of any anomalous results. - **M1 (Measure 1):** Detail how the dependent variable is measured. This can be counting the number of gas bubbles released, or more accurately, measuring the volume of oxygen collected in a gas syringe. - **M2 (Measure 2):** Define the time period for the measurement (e.g., bubbles per minute, or volume of gas collected over 5 minutes). - **S (Same):** Identify at least two abiotic variables that must be kept constant to ensure a fair test. These include: 1. Light intensity (controlled by keeping the lamp at a fixed distance from the boiling tube, or using an LED lamp to avoid transferring heat). 2. Temperature (controlled by placing the boiling tube inside a water bath).
PastPaper.markingScheme
Any 6 points from the following for a maximum of 6 marks: - **C**: Use at least five different concentrations of sodium hydrogencarbonate solution / carbon dioxide (1) - **O**: Use the same species of *Cabomba* / same initial length or mass of plant (1) - **R**: Repeat at each concentration at least three times to calculate a mean / identify anomalies (1) - **M1**: Count the number of bubbles / measure the volume of gas collected (1) - **M2**: State a specific time period (e.g. per minute / for 5 minutes) (1) - **S**: Control temperature using a water bath (1) - **S**: Control light intensity by keeping the lamp at a fixed distance / use an LED bulb (1) - **S**: Use the same pH (using a buffer solution) (1)
PastPaper.question 34 · Discussion
6 PastPaper.marks
A scientist investigated the effect of discharging organic sewage into a river. They measured the dissolved oxygen concentration and the number of freshwater invertebrate species at different distances upstream and downstream from the sewage discharge pipe.
Here are their results:
* 50m upstream (control): Dissolved oxygen = 8.5 mg/l; Number of species = 12 * At discharge point (0m): Dissolved oxygen = 2.1 mg/l; Number of species = 2 * 50m downstream: Dissolved oxygen = 1.5 mg/l; Number of species = 1 * 100m downstream: Dissolved oxygen = 3.4 mg/l; Number of species = 3 * 200m downstream: Dissolved oxygen = 6.2 mg/l; Number of species = 8 * 400m downstream: Dissolved oxygen = 8.3 mg/l; Number of species = 11
A student evaluates this data and concludes: 'Discharging organic sewage into the river only has a harmful effect on aquatic organisms within 100 metres of the discharge pipe.'
Discuss this conclusion, using the data and your biological knowledge.
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To evaluate this conclusion, we must look at both sides of the argument using the data.
Arguments supporting the conclusion: - At 50m and 100m downstream, the dissolved oxygen levels (1.5 mg/l and 3.4 mg/l) and species counts (1 and 3) are extremely low compared to the upstream control (8.5 mg/l and 12 species). - Beyond 100m, recovery is clearly visible: at 200m downstream, the oxygen level rises to 6.2 mg/l and species count increases to 8. - By 400m, the river has almost entirely recovered (8.3 mg/l oxygen and 11 species).
Arguments against the conclusion: - At 200m downstream, the number of species (8) is still lower than the upstream control (12), which proves that harmful effects still persist beyond 100m. - The oxygen level at 200m (6.2 mg/l) is still lower than the control (8.5 mg/l). - There is no data collected between 100m and 200m, so we do not know exactly where recovery starts. - Only freshwater invertebrates were sampled; other organisms like fish might be more sensitive to reduced oxygen and could be affected much further downstream. - The study does not mention replication, so the results may not be fully reliable.
PastPaper.markingScheme
Award 1 mark per point up to a maximum of 6 marks (with a maximum of 4 marks for either supporting or opposing arguments).
Supporting arguments (max 4 marks): - (MP1) Oxygen levels increase after 100m (from 3.4 to 6.2 mg/l at 200m). - (MP2) Number of invertebrate species increases after 100m (from 3 to 8 at 200m). - (MP3) Near-complete recovery occurs by 400m downstream (oxygen at 8.3 mg/l / 11 species). - (MP4) Biological explanation: decomposers/bacteria break down organic waste, using up oxygen near the pipe, but as waste is depleted downstream, oxygen levels recover.
Opposing/Limiting arguments (max 4 marks): - (MP5) Harmful effects still persist at 200m because species diversity (8) is still lower than the upstream control (12). - (MP6) Dissolved oxygen at 200m (6.2 mg/l) is still lower than the upstream control (8.5 mg/l). - (MP7) No data is available between 100m and 200m to pin-point the recovery zone. - (MP8) The study only monitored invertebrates, not fish or plants which may be affected differently/further downstream. - (MP9) No evidence of replicates/repeats to show data is reliable.
PastPaper.question 35 · Discussion
6 PastPaper.marks
A student investigated the effect of stocking density (number of fish per \(m^3\)) on the growth of farmed salmon in cages over a six-month period. They set up three cages:
* Low density (5 fish per \(m^3\)): Mean mass increase per fish = 1.8 kg; Mortality rate = 2% * Medium density (15 fish per \(m^3\)): Mean mass increase per fish = 1.5 kg; Mortality rate = 5% * High density (30 fish per \(m^3\)): Mean mass increase per fish = 0.9 kg; Mortality rate = 18%
The student concluded: 'To maximise the profit from fish farming, salmon should always be kept at the lowest stocking density because they grow the fastest and have the lowest death rate.'
Evaluate the student's conclusion using the data provided and your biological knowledge of fish farming.
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To evaluate the conclusion, we must analyze the biological and economic trade-offs:
Arguments supporting the student's conclusion: - Low stocking density results in the highest mean mass increase per fish (1.8 kg compared to 0.9 kg at high density). - Low stocking density has the lowest mortality rate (2% compared to 18% at high density), minimizing waste and preserving animal welfare. - High stocking densities lead to overcrowding, which promotes the rapid spread of parasites (e.g., lice) and pathogens, and increases competition for food and oxygen.
Arguments against the student's conclusion: - Profit depends on total yield (total mass of fish sold) rather than individual growth rates. - Let us calculate the total mass increase of surviving fish per unit volume (\(m^3\)): - Low density: \(5 \text{ fish/m}^3 \times (1 - 0.02) \times 1.8 \text{ kg} = 8.82 \text{ kg/m}^3\) - Medium density: \(15 \text{ fish/m}^3 \times (1 - 0.05) \times 1.5 \text{ kg} = 21.38 \text{ kg/m}^3\) - High density: \(30 \text{ fish/m}^3 \times (1 - 0.18) \times 0.9 \text{ kg} = 22.14 \text{ kg/m}^3\) - Both medium and high densities produce more than double the total mass of salmon per cubic metre compared to low density. - Keeping fish at low densities requires much larger cages/more space, which greatly increases capital and operational costs (reducing profit).
PastPaper.markingScheme
Award 1 mark per point up to a maximum of 6 marks (with a maximum of 4 marks for either supporting or opposing arguments).
Supporting arguments (max 4 marks): - (MP1) Individual growth/mass increase is greatest at low density (1.8 kg). - (MP2) Mortality rate is lowest at low density (2%). - (MP3) High stocking density causes stress/competition for food or oxygen, which reduces individual growth. - (MP4) High stocking density increases disease/parasite transmission (leading to 18% mortality).
Opposing/Limiting arguments (max 4 marks): - (MP5) Profit is determined by total yield/biomass sold, not individual growth rate. - (MP6) Calculates total yield per \(m^3\) (e.g., showing Medium yield (approx 21.4 kg) or High yield (approx 22.1 kg) is much greater than Low yield (approx 8.8 kg)). - (MP7) Low density requires more space/cages, increasing infrastructure costs. - (MP8) No statistical testing/replicates mentioned to determine if the differences are significant.
PastPaper.question 36 · Discussion
6 PastPaper.marks
A student used a potometer to investigate the effect of wind speed on the rate of water uptake by a leafy shoot. They placed a fan at different distances to simulate wind speeds of 0 m/s (fan off), 2 m/s (low), and 5 m/s (high). For each wind speed, they recorded the distance moved by the bubble in 10 minutes, repeating the measurement three times.
* 0 m/s (Fan off): Run 1 = 12 mm, Run 2 = 11 mm, Run 3 = 13 mm (Mean = 12.0 mm) * 2 m/s (Low): Run 1 = 28 mm, Run 2 = 29 mm, Run 3 = 18 mm (Mean = 25.0 mm) * 5 m/s (High): Run 1 = 45 mm, Run 2 = 44 mm, Run 3 = 46 mm (Mean = 45.0 mm)
The student concluded: 'Wind speed always increases the rate of transpiration because moving air removes water vapour from around the leaves, increasing the concentration gradient.'
Evaluate the student's investigation and their conclusion.
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Let us evaluate both the investigation's design/data and the conclusion:
Strengths/Support: - The data shows a clear positive correlation: as wind speed increases from 0 m/s to 5 m/s, the mean distance moved by the bubble increases from 12.0 mm to 45.0 mm. This indicates a higher rate of water uptake. - The student's biological reasoning is correct: wind sweeps away humid air (boundary layer) from the leaf surface, maintaining a steep water vapour concentration gradient between the air spaces inside the leaf and the external air, promoting faster diffusion/transpiration. - The student repeated the experiment three times for each condition, which helps to identify anomalies and calculate an average.
Weaknesses/Limitations: - There is a clear anomaly in the 2 m/s group: Run 3 is 18 mm, which is much lower than Run 1 (28 mm) and Run 2 (29 mm). Including this anomaly makes the mean of 25.0 mm less reliable. - A potometer measures water *uptake*, not direct water loss (transpiration). Some water taken up is used by the plant for photosynthesis or to maintain cell turgidity. - The word 'always' is incorrect: at very high wind speeds, plants close their stomata to prevent excessive water loss, which actually decreases the rate of transpiration. - Environmental factors like temperature, light intensity, and room humidity must be controlled to ensure a fair test, but these are not mentioned.
PastPaper.markingScheme
Award 1 mark per point up to a maximum of 6 marks (with a maximum of 4 marks for either supporting or opposing arguments).
Supporting arguments/strengths (max 4 marks): - (MP1) Trend support: Mean distance increases as wind speed increases (from 12.0 mm to 45.0 mm). - (MP2) Biological mechanism: Wind removes the boundary layer of water vapour, maintaining a steep concentration gradient. - (MP3) Design strength: The student performed three trials (replicates) for each wind speed to allow a mean to be calculated.
Opposing arguments/weaknesses (max 4 marks): - (MP4) Anomaly identification: Run 3 at 2 m/s (18 mm) is an anomaly compared to Run 1 (28 mm) and Run 2 (29 mm). - (MP5) Inclusion of this anomaly makes the mean of 25.0 mm less accurate. - (MP6) Equipment limitation: A potometer measures water uptake, not transpiration directly (some water is used for photosynthesis/turgidity). - (MP7) Physiological limitation: 'Always' is incorrect because extremely high winds will cause stomatal closure, reducing transpiration. - (MP8) Controlled variables: No mention of controlling light intensity, temperature, or humidity.
Paper 2B
Answer all questions. Calculators and rulers are permitted. Total marks: 70.
6 PastPaper.question · 70 PastPaper.marks
PastPaper.question 1 · structured
18 PastPaper.marks
Read the passage below and answer the questions that follow.
**The Role of Engineered Microorganisms in Combatting Plastic Pollution**
Polyethylene terephthalate (PET) is one of the most widely used synthetic polymers, with over 300 million tonnes produced annually for packaging and clothing. Due to its highly stable chemical bonds, PET is resistant to natural biodegradation. Most plastic waste ends up in landfills or aquatic environments, where it can take up to 450 years to decompose, compared to paper which takes about 1.5 years.
In 2016, scientists discovered a bacterium, *Ideonella sakaiensis*, growing at a waste recycling site. This bacterium produces a specialized enzyme called PETase. PETase breaks down PET into its constituent monomers: terephthalic acid and ethylene glycol. The bacterium absorbs these monomers and uses them as a source of carbon to produce ATP through aerobic respiration.
To scale up this process, genetic engineers have successfully transferred the gene for PETase into *Escherichia coli* bacteria. This genetic modification allows for rapid, controlled production of the enzyme in industrial fermenters. Furthermore, scientists have modified the active site of the enzyme to increase its efficiency at \(40\ ^\circ\text{C}\), the typical operating temperature of industrial waste vessels.
Despite the benefits, ecological risks exist. If genetically modified bacteria escape into the environment, they could degrade vital infrastructure, such as synthetic cable insulation. Additionally, the rapid breakdown of plastics releases chemical additives like phthalates. Phthalates are known endocrine disruptors that mimic hormones, potentially interfering with the reproduction of aquatic organisms and disrupting food webs.
(a) Calculate how many times longer it takes for PET to decompose naturally than paper. Show your working. (2)
(b) State the names of the two biological monomers produced when PETase breaks down PET. (2)
(c) Explain how *Ideonella sakaiensis* uses these monomers to produce ATP. (3)
(d) Describe how scientists could genetically modify *Escherichia coli* to produce the PETase enzyme. (5)
(e) Explain why operating the industrial vessel at \(40\ ^\circ\text{C}\) increases the rate of plastic breakdown, but temperatures above \(60\ ^\circ\text{C}\) would prevent the reaction. (3)
(f) Suggest the potential effects of endocrine-disrupting chemicals, such as phthalates, on the populations of fish in aquatic food webs. (3)
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PastPaper.workedSolution
(a) \(450 \div 1.5 = 300\) times longer.
(b) The two monomers are terephthalic acid and ethylene glycol.
(c) 1. The monomers are absorbed/transported into the bacterial cell. 2. They act as a carbon source / respiratory substrate (equivalent to glucose). 3. They are broken down in aerobic respiration, reacting with oxygen to release energy in the form of ATP.
(d) 1. Use a restriction enzyme to cut out/isolate the PETase gene from *Ideonella sakaiensis*. 2. Cut an *E. coli* plasmid using the same restriction enzyme to produce complementary sticky ends. 3. Use DNA ligase enzyme to join the gene and the plasmid together to form a recombinant plasmid. 4. Insert the recombinant plasmid back into *E. coli* cells. 5. Select/identify the transgenic bacteria using marker genes and grow them in fermenters.
(e) 1. Raising the temperature to \(40\ ^\circ\text{C}\) increases the kinetic energy of the enzyme and substrate molecules, leading to more frequent successful collisions and the formation of more enzyme-substrate complexes. 2. At temperatures above \(60\ ^\circ\text{C}\), the thermal energy breaks chemical bonds holding the enzyme's tertiary structure together. 3. This denatures the enzyme, changing the shape of its active site so the substrate (PET) can no longer fit.
(f) 1. Endocrine disruptors mimic natural hormones, causing reproductive failure, reduced fertility, or developmental abnormalities in fish. 2. This leads to a decline in the fish population size. 3. A decrease in fish populations reduces the food source for higher-level predators, causing their populations to decline, and may cause bioaccumulation of toxic chemicals up the food chain.
(c) [3 marks] - Monomers are absorbed/transported into the cytoplasm of the bacterium [1 mark] - Used as a respiratory substrate / source of carbon [1 mark] - Reacted with oxygen / aerobically respired to yield ATP [1 mark]
(d) [5 marks] - Use restriction enzyme to isolate/cut out PETase gene [1 mark] - Use same restriction enzyme to cut plasmid vector to produce complementary sticky ends [1 mark] - Use ligase enzyme to join DNA fragments / insert gene into plasmid [1 mark] - Produce recombinant plasmid [1 mark] - Transform/insert recombinant plasmid into *E. coli* bacteria [1 mark]
(e) [3 marks] - Higher kinetic energy leads to more frequent successful collisions / enzyme-substrate complexes [1 mark] - High temperature (above \(60\ ^\circ\text{C}\)) causes denaturation [1 mark] - Active site changes shape / substrate can no longer bind [1 mark]
(f) [3 marks] - Mimicking hormones reduces reproduction / causes infertility in fish [1 mark] - Decreases fish population / number of individuals [1 mark] - Disrupts food chain (e.g., starvation of predators / bioaccumulation / bioamplification) [1 mark]
PastPaper.question 2 · structured
10 PastPaper.marks
A student uses a bubble potometer to investigate the effect of wind speed on the rate of transpiration of a leafy shoot.
(a) Describe three precautions the student should take when setting up the potometer to ensure the measurements are valid. (3 marks)
(b) Explain how the rate of water uptake can be calculated using the bubble potometer. (3 marks)
(c) Explain why the rate of water uptake measured by the potometer is not exactly equal to the rate of transpiration. (2 marks)
(d) Explain how an increase in wind speed affects the rate of transpiration. (2 marks)
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PastPaper.workedSolution
(a) To ensure validity and prevent air locks, the shoot must be cut underwater to stop air from entering the xylem vessels. Cutting the stem at an angle increases the surface area for water absorption. Sealing all joints with petroleum jelly (Vaseline) ensures the apparatus is completely airtight, preventing any external leaks. Drying the leaves prevents liquid water from blocking stomatal pores, which would artificially lower initial transpiration.
(b) To calculate the rate, measure the distance \(d\) moved by the bubble along the scale in a set time interval using a stopwatch. Calculate the volume of water absorbed using the cylinder formula \(\text{Volume} = \pi r^2 d\) (where \(r\) is the inner radius of the capillary tube). Divide this volume by the time taken to determine the rate of water uptake (e.g., in \(\text{mm}^3\text{/minute}\)).
(c) The potometer measures water uptake, not direct transpiration. Not all water taken up is transpired because some is chemically consumed in photosynthesis, some is used to keep plant cells turgid (cell hydration/support), and a small amount is produced during aerobic respiration.
(d) Increased wind speed blows away the boundary layer of moist air/water vapour surrounding the stomatal openings on the leaf surface. This maintains a steep concentration gradient of water vapour between the air spaces inside the leaf and the external air, resulting in a faster rate of diffusion of water vapour out of the stomata.
PastPaper.markingScheme
Part (a): [Maximum 3 marks] - Cut shoot underwater to prevent air bubbles entering / blocking xylem (1) - Cut shoot at an angle to increase surface area for water uptake (1) - Seal joints with petroleum jelly / Vaseline to make the apparatus airtight (1) - Dry leaves before starting to prevent blockage of stomata (1) - Allow plant to acclimatise before taking readings (1)
Part (b): [Maximum 3 marks] - Measure distance moved by the bubble using a ruler/scale (1) - Measure the time taken using a stopwatch/timer (1) - Use the formula \(\pi r^2 d\) / cross-sectional area multiplied by distance to calculate volume (1) - Divide volume by time taken to find rate (1)
Part (c): [Maximum 2 marks] - Not all water absorbed is lost by transpiration (1) - Some water is used in photosynthesis (1) - Some water is used to maintain cell turgidity / turgor pressure / cell expansion (1)
Part (d): [Maximum 2 marks] - Transpiration increases because wind removes water vapour / humid air from around the leaf (1) - This maintains / steepens the concentration / diffusion gradient (1) - Leading to faster diffusion of water vapour out of stomata (1)
PastPaper.question 3 · Physiological explanation
9 PastPaper.marks
During intense physical activity, the body's demand for energy increases. Explain how the cardiovascular system of a human responds physiologically during exercise to ensure that respiring muscle cells receive a sufficient supply of oxygen and glucose.
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The response involves several coordinated cardiovascular mechanisms: 1) The heart rate increases due to stimulation by the sympathetic nervous system and adrenaline. 2) The stroke volume (volume of blood pumped per beat) increases as the heart contracts more forcefully. 3) This results in an increased cardiac output, supplying blood more rapidly. 4) Arterioles leading to skeletal muscles dilate (vasodilation), decreasing resistance and increasing blood flow to these tissues. 5) Arterioles leading to organs like the gut and kidneys constrict (vasoconstriction) to divert blood away from non-essential functions. 6) Venous return increases due to the squeezing action of skeletal muscles on veins. 7) This increased circulation delivers more glucose and oxygen to active muscles to sustain aerobic respiration and meet the demand for ATP, while also removing waste carbon dioxide.
PastPaper.markingScheme
Award 1 mark for each of the following up to a maximum of 9 marks: 1. Heart rate increases; 2. Stroke volume increases / heart contracts with greater force; 3. Cardiac output increases (reference to CO = HR x SV or equivalent description); 4. Vasodilation of arterioles supplying active skeletal muscles; 5. Vasoconstriction of arterioles supplying non-essential organs (e.g. gut/kidneys); 6. This redistributes/redirects blood flow preferentially to working muscles; 7. Increased venous return (skeletal muscle pump action); 8. Delivers more glucose and oxygen for aerobic respiration / ATP production; 9. Facilitates faster removal of carbon dioxide / lactic acid; 10. (Accept) Reference to the Bohr effect: higher carbon dioxide concentration/lower pH causes hemoglobin to release oxygen more readily.
PastPaper.question 4 · Physiological explanation
9 PastPaper.marks
The menstrual cycle is regulated by a complex system of hormones released from both the brain and the ovaries. Describe and explain the physiological roles and feedback interactions of follicle-stimulating hormone (FSH), luteinising hormone (LH), estrogen, and progesterone during a typical 28-day human menstrual cycle.
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PastPaper.workedSolution
The menstrual cycle is controlled by endocrine interactions: 1) FSH is released from the pituitary gland and causes a follicle in the ovary to mature. 2) The developing follicle secretes estrogen. 3) Estrogen acts to repair and thicken the endometrium (uterus lining). 4) Rising estrogen levels exert positive feedback on the pituitary gland, triggering a sudden release (surge) of LH. 5) The LH surge causes ovulation (release of the egg from the follicle). 6) The empty follicle develops into the corpus luteum. 7) The corpus luteum secretes progesterone. 8) Progesterone maintains the thickness and vascularization of the endometrium. 9) Progesterone exerts negative feedback on the pituitary, inhibiting FSH and LH secretion to prevent further follicle development. 10) If fertilization fails, the corpus luteum degenerates, progesterone levels decline, causing the endometrium to break down (menstruation).
PastPaper.markingScheme
Award 1 mark for each of the following up to a maximum of 9 marks: 1. FSH stimulates follicle development/maturation in the ovary; 2. Maturing follicle secretes estrogen; 3. Estrogen stimulates repair/thickening of the endometrium; 4. Estrogen triggers a surge in LH release from the pituitary gland (positive feedback); 5. LH surge causes ovulation / release of the egg; 6. Ruptured follicle becomes the corpus luteum; 7. Corpus luteum secretes progesterone (and estrogen); 8. Progesterone maintains the lining of the uterus (endometrium) / prepares it for implantation; 9. Progesterone inhibits FSH and LH secretion (negative feedback); 10. Decline in progesterone (and estrogen) causes menstruation / shedding of the lining.
PastPaper.question 5 · Physiological explanation
9 PastPaper.marks
A person walking in a hot desert has not consumed water for several hours and has become dehydrated. Explain the physiological processes that occur in the human body to restore water balance, focusing on the roles of the brain, the pituitary gland, and the nephrons in the kidneys.
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PastPaper.workedSolution
The homeostatic response to dehydration is as follows: 1) Loss of water causes blood water potential to decrease (blood becomes more concentrated/hypertonic). 2) This change is detected by osmoreceptor cells in the hypothalamus of the brain. 3) The hypothalamus stimulates the posterior pituitary gland. 4) The pituitary gland secretes more antidiuretic hormone (ADH) into the bloodstream. 5) ADH is transported via the blood to the kidneys. 6) ADH binds to receptors on the cells lining the collecting ducts (and distal convoluted tubules). 7) This increases the permeability of these duct walls to water by causing aquaporins (water channels) to insert into the membranes. 8) Water is reabsorbed out of the filtrate/urine into the hypertonic medulla. 9) Water moves by osmosis down its concentration gradient into the surrounding capillary network (vasa recta). 10) This results in the production of a small volume of highly concentrated urine, helping to restore normal blood water potential.
PastPaper.markingScheme
Award 1 mark for each of the following up to a maximum of 9 marks: 1. Dehydration decreases blood water potential / increases blood osmolarity; 2. Detected by osmoreceptors in the hypothalamus; 3. Pituitary gland is stimulated to release more ADH (antidiuretic hormone); 4. ADH travels through the blood to the kidneys; 5. ADH targets the collecting ducts (and/or distal convoluted tubules) of nephrons; 6. ADH increases the permeability of the collecting duct walls to water; 7. More water channels / aquaporins insert into the cell membranes; 8. Water moves out of the tubule/filtrate by osmosis; 9. Water moves down a water potential gradient into the blood/capillaries; 10. A smaller volume of highly concentrated (darker) urine is produced.
PastPaper.question 6 · structured
15 PastPaper.marks
Scientists investigated the effect of growing genetically modified (GM) cotton containing a gene for Bt toxin, which is toxic to bollworm insect pests. They recorded the mean crop yield (in kg per hectare) and the mass of chemical pesticides applied (in kg per hectare) over a 5-year period. The results are shown in the table.
| Year | Yield of non-GM cotton (kg/ha) | Yield of GM cotton (kg/ha) | Mass of pesticide used on non-GM cotton (kg/ha) | Mass of pesticide used on GM cotton (kg/ha) | | :--- | :---: | :---: | :---: | :---: | | 1 | 1200 | 1500 | 12.4 | 3.2 | | 2 | 1150 | 1580 | 13.0 | 2.8 | | 3 | 1300 | 1620 | 11.8 | 2.5 | | 4 | 1050 | 1600 | 14.2 | 2.1 | | 5 | 1250 | 1650 | 12.6 | 2.0 |
(a) Describe how a gene for Bt toxin from a bacterium is cut out and transferred to plant cells to produce GM cotton plants. (5 marks)
(b) Calculate the percentage increase in the mean yield of GM cotton compared to the mean yield of non-GM cotton over the 5-year period. Show your working and give your answer to 3 significant figures. (4 marks)
(c) (i) Describe the trend in the mass of pesticide used for GM cotton compared to non-GM cotton over the 5-year period. (2 marks)
(c) (ii) Evaluate the widespread use of GM cotton on the local environment and the economy of farming communities, using the data provided and your own biological knowledge. (4 marks)
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PastPaper.workedSolution
(a) To transfer the Bt gene into cotton plants: 1. Use a restriction enzyme to cut the Bt gene out of the bacterial DNA. 2. Cut an appropriate vector (e.g. a plasmid from Agrobacterium tumefaciens) with the same restriction enzyme to produce complementary sticky ends. 3. Use the enzyme DNA ligase to join the Bt gene and the plasmid together, creating a recombinant plasmid. 4. Insert the recombinant plasmid/vector into cotton plant cells. 5. Grow the transformed plant cells into full plants using tissue culture (micropropagation).
(b) Calculation of percentage increase: 1. Calculate the mean yield of non-GM cotton: \( \text{Mean Non-GM} = \frac{1200 + 1150 + 1300 + 1050 + 1250}{5} = \frac{5950}{5} = 1190 \text{ kg/ha} \) 2. Calculate the mean yield of GM cotton: \( \text{Mean GM} = \frac{1500 + 1580 + 1620 + 1600 + 1650}{5} = \frac{7950}{5} = 1590 \text{ kg/ha} \) 3. Calculate the increase: \( 1590 - 1190 = 400 \text{ kg/ha} \) 4. Calculate the percentage increase relative to non-GM: \( \text{Percentage Increase} = \left( \frac{400}{1190} \right) \times 100 \approx 33.6134\% \) 5. Round to 3 significant figures: \( 33.6\% \)
(c) (i) Trend description: - The pesticide use for GM cotton starts low (3.2 kg/ha) and decreases steadily every year to 2.0 kg/ha. - The pesticide use for non-GM cotton is much higher overall and fluctuates between 11.8 and 14.2 kg/ha with no downward trend.
(c) (ii) Evaluation: - Environmental benefits: Less chemical pesticide means reduced damage to non-target insect species (biodiversity), reduced bioaccumulation in local food chains, and less pollution of waterways. - Economic benefits: Higher crop yield (mean 1590 vs 1190 kg/ha) increases profit margins for farmers, and spending less money on chemical pesticides reduces production costs. - Environmental risks: Pests may develop resistance to the Bt toxin over time; the toxin might harm beneficial insects; gene flow/transfer could occur to wild relative plants, producing 'superweeds'. - Economic risks: GM seeds are typically more expensive to buy, and farmers may be forced to buy new seeds each year, increasing dependency on biotechnology companies.
PastPaper.markingScheme
(a) Max 5 marks: - restriction enzyme used to cut gene (from DNA) (1) - plasmid/vector cut with the same restriction enzyme (1) - complementary sticky ends produced (1) - ligase used to join gene and plasmid/vector (1) - recombinant plasmid formed (1) - vector introduced into plant cell / plasmid put into Agrobacterium (1) - tissue culture/micropropagation used to grow transgenic plants (1)
(b) Max 4 marks: - Correct mean of non-GM cotton = 1190 (1) - Correct mean of GM cotton = 1590 (1) - Correct percentage calculation method: \( \frac{\text{GM} - \text{non-GM}}{\text{non-GM}} \times 100 \) (1) - Correct final answer to 3 s.f. = 33.6% (1) [Award 4 marks for correct final answer with working; award 3 marks for correct final answer without working]
(c)(i) Max 2 marks: - pesticide use for GM cotton decreases (over the 5 years) / is always lower than non-GM (1) - pesticide use for non-GM cotton fluctuates / remains high / does not show a downward trend (1) - credit use of data values to support trend (1)
(c)(ii) Max 4 marks (must include at least one environmental and one economic point for full marks): - Less chemical pesticide reduces risk of bioaccumulation / biomagnification / toxicity to non-target or beneficial organisms (1) - Less chemical pesticide runoff prevents water pollution / protects aquatic ecosystems (1) - Higher yield increases farmer income (1) - Reduced expenditure on chemical pesticides (1) - (Risk) Bollworms may evolve/develop resistance to the Bt toxin (1) - (Risk) GM seeds are expensive / farmers must purchase new seeds annually (1) - (Risk) Risk of cross-pollination/gene transfer to wild relatives (creating resistant weeds) (1)